Research Article A Numerical Test of Padé …downloads.hindawi.com/archive/2014/587430.pdfResearch Article A Numerical Test of Padé Approximation for Some Functions with Singularity

Research ArticleA Numerical Test of Padeacute Approximation forSome Functions with Singularity

Hiroaki S Yamada1 and Kensuke S Ikeda2

1 Yamada Physics Research Laboratory Aoyama 5-7-14-205 Niigata 950-2002 Japan2Department of Physics Ritsumeikan University Noji-higashi 1-1-1 Kusatsu 525 Japan

Correspondence should be addressed to Hiroaki S Yamada hyamadauranusdtinejp

Received 17 July 2014 Accepted 14 October 2014 Published 20 November 2014

Academic Editor Don Hong

Copyright copy 2014 H S Yamada and K S IkedaThis is an open access article distributed under the Creative CommonsAttributionLicense which permits unrestricted use distribution and reproduction in anymedium provided the originalwork is properly cited

The aim of this study is to examine some numerical tests of Pade approximation for some typical functions with singularities suchas simple pole essential singularity brunch cut and natural boundary As pointed out by Baker it was shown that the simple poleand the essential singularity can be characterized by the poles of the Pade approximation However it was not fully clear how thePade approximation works for the functions with the branch cut or the natural boundary In the present paper it is shown that thepoles and zeros of the Pade approximated functions are alternately lined along the branch cut if the test function has branch cut andpoles are also distributed around the natural boundary for some lacunary power series and random power series which rigorouslyhave a natural boundary on the unit circle On the other hand Froissart doublets due to numerical errors andor external noisealso appear around the unit circle in the Pade approximation It is also shown that the residue calculus for the Pade approximatedfunctions can be used to confirm the numerical accuracy of the Pade approximation and quasianalyticity of the random powerseries

1 Introduction

Pade approximation was introduced in mathematics by Her-mite and Pade and it has been used in physics for morethan 40 years ago [1ndash3] In particular there have been severalimportant examples in physics to which the Pade approx-imation was applied such as summation of the divergentRayleigh-Schrodinger perturbation series in scattering the-ory [4] critical phenomena in statistical physics [5ndash9]denoising from noisy data of time-series [10ndash14] and detec-tion of singularity of phase space trajectories of Hamiltoniandynamical systems [15ndash19]

Mathematically the Pade approximation can be used toestimate analyticity of functions Indeed the Pade approx-imation is usually superior to the truncated Taylor expan-sions when the original function contains any singularityLet us consider a simple example The function 119891(119911) =

radic(1 + 2119911)(1 + 119911) has brunch points at 119911 = minus1 and 119911 = minus12The domain of convergence is |119911| lt 12 Nevertheless we canobtain an exact solution radic2 = 14142 for 119911 rarr infin when weapply [3 | 3] diagonal Pade approximation to the function

(See Section 33 for more details on this example) Althoughthe mathematical validity of the Pade approximation hasnot been exactly proved yet the Pade approximation ispractically very useful to continue a singular function beyondthe domain of convergence

Let us consider a critical phenomenon for Ising modelas a simple example in the statistical physics [5 7 8] Weassume that at the critical point 119906 = 119906

119888the exact magnetic

susceptibility 120594 has a singularity as

120594 sim (119906 minus 119906119888)minus120574

(1)

where 119906 is a function of temperature and interactions and soon In this case we sometimes use the logarithmic derivativeof120594whenwe estimate the critical point 119906 = 119906

119888and the critical

exponent 120574 as a pole-type singularity as

119889 log120594119889119906

=1205941015840

120594=

120574

(119906 minus 119906119888) (2)

Hindawi Publishing CorporationInternational Journal of Computational MathematicsVolume 2014 Article ID 587430 17 pageshttpdxdoiorg1011552014587430

2 International Journal of Computational Mathematics

where 1205941015840 denotes the derivative with respect to the variable 119906In the low temperature expansion for the magnetic suscepti-bility 120594with some coefficients 119886

119899 we assume that the approx-

imated susceptibility 120594119873and the logarithmic derivative can

be obtained as follows

120594 sim 120594119873=119873

sum119899=0

119886119899119906119899

1205941015840

120594sim119873

sum119899=0

119887119899119906119899 (3)

Then we can estimate the critical point 119906 = 119906119888and the critical

exponent 120574 by the Pade approximation to the truncatedexpansion 120594

119873 Note that the singularity on |119906| = 119906

119888will be

infinitely differentiable if the coefficients 119886119899fall off sufficiently

rapidlyIn general analytic continuation over the singular point

is possible along any other path in complex plane even ifthe function diverges at the singular point determining theradius of convergence as seen in the above singularity of theIsing model Therefore the Pade approximation is useful toimprove the convergence of the power series and approximatethe exact solution

Furthermore the Pade approximation has been usedto investigate convergence of Fourier series [2 3] and thebreakdown of KAMcurves in complex plane forHamiltonianmap systems which is described as the analytic domains ofLindstedt series for standard map [15ndash19]

In addition we can see an interesting example concerningthe Pade approximation in noisy data analysis [10ndash14] Thepower series with finite random coefficient ldquoalmost alwaysrdquohas a natural boundary on the unit circle in the complexplane [20ndash24] In a finite time-series Froissart has shownthat a natural boundary generated by the random time-seriesis approximated by doublets of poles and zeros (Froissartdoublets) of the Pade approximated function which aresurrounding the vicinity of the unit circle Taking advantageof the characteristic the Pade approximation has been usedin order to remove the noise and extract the true poles asso-ciated with damping modes from the observed noisy time-series

The main purpose of the present paper is to investigatewhether the Pade approximation is numerically useful fordetecting the singularity of some test functions In particularwe examine the usefulness for the functions with a naturalboundary such as a lacunary power series and a randompower series [2 13 18]

The organization of the paper is as follows In Section 2we give a brief explanation of the Pade approximation andsome important reminders in the numerical calculation InSection 3 we present some numerical results of the Padeapproximation for test functions with branch cut essentialsingularityWe also try to apply the Pade approximation to anentire function In Section 4 application of the Pade approx-imation to some lacunary series which is known to have anatural boundary is given In Section 5 numerical resultsof the application to random power series with a naturalboundary and some test functionswith randomnoise are alsoshown In Section 6 we discuss the residue calculus for thePade approximated functions to confirm the numerical errors

and quasianalyticity of the random power series In the lastsection we give summary and discussion

In Appendix A the general result for Fibonacci generatorused in Section 32 is given In Appendix B some theoremsconcerning zeros of polynomials are given In Appendix Csome mathematical theorems for lacunary series whichare useful in reading the main text are summarized InAppendix D exact Pade approximated functions to somelacunary power series with a natural boundary are givenResidue analysis for quasianalytic functions of Carlemanclass is given in Appendix E Furthermore some theoremsconcerning the randompower series are given inAppendix F

2 Padeacute Approximation

In this section we give preliminary instructions to the Padeapproximation and some important reminders on the numer-ical calculation

For a given function 119891(119911) a truncated Taylor expansion119891[119873](119911) of order119873 about zero is given as

119891 (119911) sim 119891[119873] (119911) =119873

sum119899=0

119888119899119911119899 (4)

where 119888119899denotes the coefficients of the Taylor expansion

Pade approximation is more accurate approximation for119891(119911)up to order 119874(119911119873) than the Taylor expansion The Padeapproximation is a rational function namely a ratio of twopolynomials which agrees with the highest possible order119874(119911119873) with a truncated polynomial 119891[119873](119911) as follows

119891[119873] (119911) =1198860+ 1198861119911 + 11988621199112 + sdot sdot sdot 119886

119871119911119871

1 + 1198871119911 + 11988721199112 + sdot sdot sdot 119887

119872119911119872

equiv119875119871(119911)

119876119872(119911)

equiv 119891[119871|119872] (119911)

(5)

where 119875119871(119911) is a polynomial of degree less than or equal to 119871

and119876119872(119911) is a polynomial of degree less than or equal to119872

Note that 1198870= 1 (normalized) here A unique approximation

can be specified for all choices of 119872 and 119871 such that 119873 =119871 + 119872 when it exists The coefficients 119886

119899 119887119899 can be

obtained from the condition that the first (119871 + 119872 + 1) termsvanish in the Taylor series in (4) The difference between thePade approximation and the original function satisfies thefollowing equation

119891 (119911) minus 119891[119871|119872]

(119911) = 119874 (119911119871+119872+1) (6)

In this paper we sometimes use ldquo[119871 | 119872] Pade approx-imationrdquo or ldquo[119871 | 119872] Pade approximated functionrdquo for119891[119871|119872](119911) Solving the problem in (6) is called a linear Teoplitzproblem which is generally ill-conditioned We used full LUdecomposition for the Teoplitz matrix in the problem as wellas iterative improvement in order to eliminate the ill-posedproblem [25 26] In addition hereafter we use the diagonalPade approximation that is 119871 = 119872 of order 119872 le 65

International Journal of Computational Mathematics 3

to estimate the singularity of the test functions because ofthe convergence and limitation due to the round-off errorsand other sources of numerical errors Here the singularityof the function 119891(119911) is approximated by configuration ofthe poles and zeros of the [119872 | 119872] order diagonal Padeapproximated function As mentioned in Introduction ingeneral the Pade approximations are useful for representingunknown functions with possible poles The application ofthe diagonal Pade approximation is insured for the functionswith isolated singular points and rational-type functionsHowever it is not fully clarified how the poles and zeros of thePade approximated function describe brunch cut and naturalboundary [27ndash29]

Generally the magnitude of the residues associated withthe spurious poles is much smaller than that with the truepoles and they are close to machine precision Very recentlyGonnet et al suggested an efficient algorithm for the Padeapproximations [30 31]The algorithmdetects and eliminatesthe spurious pole-zero pairs caused by the rounding errorsby means of singular value decomposition for the Teoplitzmatrix

Before closing this section we list up some importantreminders when we numerically apply the Pade approxima-tion to unknown functions as follows

(1) More accurate calculation becomes possible by scal-ing the expansion variable 119911 if there is a simplepole with large magnitude 120588(≫1) That is we shouldchange the order of radius of convergence into 119874(1)by the scaling the expansion variable as 119911 rarr 119911120588 inorder to keep the numerical accuracyThis procedureis effective when we apply the Pade approximation toexponentially decaying coefficients 119886

119899 with fluctua-

tion(2) Poles (ie roots of 119875

119872(119911) = 0) and zeros (ie roots

of 119876119872(119911) = 0) are sometimes cancelled (zero-pole

ghost pairs) We can remove the effects of the ghostpairs and confirm the singularity of the functions byusing the residue analysis of the Pade approximatedfunction

(3) Poles and zeros of the Pade approximation to thetruncated random power series accumulate aroundthe unit circle as Froissart doublets It is difficult todistinguish whether the poles of the Pade approxi-mation originated from a natural boundary of theoriginal function or the natural boundary generatedby the numerical error andor noise Therefore thenumerical accuracy will be important to determinethe coefficients of the Pade approximation

(4) In general the denominators of the diagonal Padeapproximated functions to the lacunary power seriesand the random power series become lacunary andrandom polynomials respectively Accordingly thedistribution of the poles and zeros of the approxi-mated functions are similar to the distribution of thezeros corresponding to the original lacunary powerpolynomial and random power polynomial In par-ticular it is well known that the zeros of the random

polynomials uniformly distribute about the unit circle(see Appendix B)

3 Examples of Padeacute Approximation forSome Functions

In this section we investigate the configuration of the polesand zeros of the Pade approximation to some test functionswith singularity

31 Comparison of Pade Approximation with Taylor Expan-sion First we try the Pade approximation to the followingtest function 119891

1(119911) with a brunch point at 119911 = minus1

1198911(119911) =

log (1 + 119911)119911

(7)

The truncated Taylor expansion of the order 119873 = 4 around119911 = 0 is

119891[4]1

(119911) = 1 minus1

2119911 +

1

31199112 minus

1

41199113 +

1

51199114 (8)

The [2 | 2] Pade approximation is given as

119891[2|2]1

(119911) =1 + (710) 119911 + (130) 1199112

1 + (65) 119911 + (310) 1199112 (9)

Figure 1 shows the approximated functions the truncatedTaylor series 119891[4]

1(119911) and the original test function 119891

1(119911)

The approximated function 119891[4]1(119911) by the truncated Taylor

expansion converges only within |119911| lt 1 and deviates fromthe exact function 119891

1(119911) for |119911| gt 1 On the other hand it

follows that the Pade approximated function 119891[2|2]1

(119911) wellapproximates the original function 119891

1(119911) with very high pre-

cession even beyond the radius of convergence up to Re 119911 =119909 sim 10 Therefore it is found that the divergent power seriesexpansion (Taylor expansion) does still contain informationabout the original function outside the convergence radiusand rearranging the coefficients of the expansion into thePade approximation recovers the information As a result theconversion from the Taylor form to the Pade form usuallyaccelerates the convergence and often allows good accuracyeven outside the radius of convergence of the power series

32 Pade Approximation for Fibonacci Generating FunctionLet us consider the Fibonacci generating function 119891

119865(119911)

where Fibonacci sequence 119865119899 is encoded in the power series

as the coefficientsThe Fibonacci sequence 119865119899 is given by the

following recursion relation

119865119899= 119865119899minus1

+ 119865119899minus2

(119899 ge 2) (10)

where 1198650= 0 and 119865

1= 1 Then the Fibonacci generating

function 119891119865(119911) becomes


86420

Exact Truncated Taylor exp

30

25

20

15

10

05

00

z

f1(z

)

[2|2] Pad e app

Figure 1 The Pade approximated function 119891[2|2]1

(119911) the truncatedTaylor series 119891[4]


1(119911)

119891119865(119911) =

infin

sum119899=0

119865119899119911119899

=119911

1 minus 119911 minus 1199112

=1

radic5

1

1 minus 120601+119911minus

1

radic5

1

1 minus 120601minus119911

(11)

where 120601+ equiv ((1 + radic5)2)(= 161803 ) and 120601minus equiv ((1 minusradic5)2)(= minus061803 ) The generating function has poles at119911 = 120601+ and 119911 = 120601minus Generating functions by more generalrecursion relation is given in Appendix A

It should be noted that the diagonal Pade approximation119891[1198732|1198732]119865

(119911) for the truncated Fibonacci generation function119891[119873]119865

(119911) = sum119873

119899=0119865119899119911119899 has the following form

119891[1198732|1198732]119865

(119911) =119911

1 minus 119911 minus 1199112+ 119874 (119911119873+1) (12)

for even number 119873 ge 2 This means that the diagonal Padeapproximation can detect the exact poles of the generatingfunction irrespective of the order

33 Examples of Some Test Functions with Pole Brunch Cutand Essential Singularity In this subsection we use some testfunctions in applying of the Pade approximation

1198912(119911) = 119890minus119911

1198913(119911) = radic

1 + 2119911

1 + 119911

1198914(119911) = 119890minus119911(1+119911)

1198915(119911) = tan 1199114

(13)

Here1198912(119911) has no singularity for |119911| lt infin119891

3(119911) has a brunch

cut along a line on [minus1 minus12] and 1198914(119911) has an essential

singularity at 119911 = minus1 1198915(119911) has eight poles at points on the

unit circle 119911 = exp119894120587(1198984) (119898 = 0 1 2 7)First let us apply Pade approximation to119891

2(119911) In this case

the explicit form of the Pade approximated function can beobtained in the following form [33]

119875119872(119911) =

119872

sum119896=0

(2119872 minus 119896)119872

(2119872)119896 (119872 minus 119896)(minus119911)119896

119876119872(119911) =

119872

sum119896=0

(2119872 minus 119896)119872

(2119872)119896 (119872 minus 119896)119911119896

(14)

Note that the coefficients of the numerator 119875119872(119911) have

always alternative sign and the zeros and poles of the Padeapproximated function are symmetrical to the imaginary axiswith each other because 119875

119872(119911) = 119876

119872(minus119911) Figure 2(a) shows

the numerical results in the complex 119911-plane All poles areon the left half-plane Re 119911 gt 0 and all zeros are on theright half-plane Re 119911 lt 0 The poles and zeros of the Padeapproximated functions for the regular function 119891

2(119911) go

infinite and disappear as119872rarrinfin because the function 1198912(119911)

is an entire functionIn Figure 2(b) distribution of the zeros and poles of the

Pade approximated function to 1198913(119911) is shownThe poles and

zeros make a line alternately between the two branch pointsof the 119891

3(119911) 119911 = minus1 and 119911 = minus12

Figure 2(c) shows the distribution of the zeros and polesof the Pade approximation to 119891

4(119911) The Pade approximation

clusters the poles and zeros at the singular point for 1198914(119911)

As119872 rarr infin the poles and zeros approach the singular point119911 = minus1 that reflects the essential singularity of the originalfunction 119891

4(119911)

Figure 3(a) shows the poles and zeros of the Pade approxi-mated function to the test function119891

5(119911)The poles and zeros

are alternatively distributed in eight directions from the ori-gin It seems that the distance between the poles andor zeroson the same line becomes small as they approach the locationof the true poles Some ldquospurious polesrdquo appear around theunit circle with the increase of the order of the Pade approxi-mation as seen in Figure 3(b) which are irrelevant polesdue to insufficient numerical accuracy It is found that thenumerical accuracy of the Pade approximation fails for thehigher order of the Pade approximationWediscuss the spuri-ous poles in Sections 4 and 5 again

4 Natural Boundary of Lacunary Power Series

In this section we examine the applicability of the Pade appr-oximation to investigate the analyticity of some well-knowntest functions with a natural boundary on the unit circle|119911| = 1 This will provide a preliminary information aboutwhat occurs in the Pade approximated functions for some


0

20

40

0 20 40minus40

minus40

minus20

minus20

Re zM = 10 polesM = 10 zeros

M = 20 polesM = 20 zeros

Im z

(a)

Re z

Im z

PolesZeros

000

minus10 minus09 minus08 minus07 minus06 minus05

(b)

Re z

Im z

000

005

010

minus010

minus005

minus110 minus105 minus100 minus095 minus090



(c)

Figure 2 Distributions of poles (e ⃝) and zeros (+ times) of the [119872 | 119872] diagonal Pade approximated functions for some test functions (a)1198912(119911) (b) 119891

3(119911) and (c) 119891

4(119911)

test functions with a natural boundary Indeed we do knowonly a very few numerical examples which have a naturalboundary and allow an exact diagonal Pade approximation

The following are famous lacunary power series with anatural boundary on the unit circle |119911| = 1119891Jac(119911) = sum

infin

119899=01199112119899

119891Wie(119911) = sum

infin

119899=0119911119899 and 119891Kro(119911) = sum

infin

119899=01199111198992

where the 119891Jac(119911)119891Wie(119911) and 119891Kro(119911) are called after Jacobi Weierstrass andKronecker Some theorems for the lacunary series with anatural boundary are given in Appendix C [20ndash22]

Here we use polar-form 119891119903(120579) for the function 119891(119911) by

changing the variable that is 119911 = 119903119890119894120579 in order to simply

display the functions as

119891119903(120579) = 119891 (119911 = 119903119890119894120579) =

infin

sum119899=0

119888119899(119903119890119894120579)

119899

(15)

Then note that the modulus 119903 works as a convergence factorof the series because it well converges for 119903 lt 1 Typically wetake 119903 = 1 on the unit circle or 119903 = 098 inside the circle inthe following numerical calculations

41 Example 1 Jacobi Lacunary Series We try to applyPade approximation to the function 119891Jac(119911) with a natural


0

1

2

0 1 2minus2

minus2

minus1

minus1

Re z

Im z

(a)

0

1

2

0 1 2minus2

minus2

minus1

minus1

Re z

Im z

(b)

Figure 3 Distributions of poles ( ⃝) and zeros (times) of the [119872 | 119872] diagonal Pade approximated functions for the test function 1198915(119911) (a) The

[50 | 50] Pade approximation (b) The [75 | 75] Pade approximation The unit circle is drawn to guide the eye

boundary on the unit circle |119911| = 1 The Pade approximatedfunction exactly has the following form

119891[2119873

]

Jac (119911) sim 119891[2119873minus1

|2119873minus1

]

Jac (119911)

=119860119873Jac (119911)

1 + sum119873minus2

119896=01199112119896

minus 1199112119873minus1

(16)

where the explicit form of the numerator 119860119873Jac(119911) is given inAppendix D Accordingly the poles of the [2119873minus1 | 2119873minus1] Padeapproximated function are given by roots of the polynomial

1 +119873minus2

sum119896=0

1199112119896


= 0 (17)

This is also just a lacunary polynomial In Figure 4 thenumerical result of the Pade approximation for 119891Jac(119911) isshown The poles and zeros are plotted for the [64 | 64]Pade approximation in Figure 4(a) Inside the circle |119911| = 1some cancellations of the ghost pairs appear The poles andzeros accumulate around |119911| = 1 with the increase of orderof the Pade approximation In the case of the 119872 = 64 thepoles accumulate around |119911| = 1 with making the zero-polepairings Figure 4(b) shows the Pade approximated functionsin the polar-formwith 119903 = 1 It well approximates the originalfunction 119891Jac(119911) when the order of the Pade approximationincreases

It is also shown that the complex zeros of the polynomial(17) cluster near unit circle |119911| = 1 and distribute uniformlyon the circle as 119865

119873rarrinfin by Erdos-Turan-type theorem given

in Appendices C and B [34ndash41]

42 Example 2 Fibonacci Lacunary Series As a second exa-mple we would like to apply Pade approximation to the fol-lowing lacunary series

119891Fib (119911) =infin

sum119899=0

119911119865119899 (18)

where 119865119899is 119899th Fibonacci number This function also has

a natural boundary on |119911| = 1 The Pade approximatedfunction exactly has the following form

119891[119865119873]

Fib (119911) sim 119891[1198651198732|1198651198732]

Fib (119911)

=119860119865119873

Fib (119911)

1 + 119911119865119873minus4 minus 119911119865119873minus2

(19)

The explicit form of the numerator 119860119865119873

Fib(119911) is given inAppendix D The poles of the [119865

1198732 | 119865

1198732] Pade approxi-

mated function are given by zeros of the lacunary polynomial

1 + 119911119865119873minus4 minus 119911119865119873minus2 = 0 (20)

In Figure 5 the numerical result of the Pade approxima-tion to 119891Fib(119911) is shown The poles and zeros are plotted forthe [55 | 55] Pade approximation in Figure 5(a) The polesand zeros accumulate around |119911| = 1 with the increase of theorder of the Pade approximation No pole appears inside theunit circleThe original function is also well approximated bythe [56 | 56] Pade approximation (see Figure 5(b))


00

05

10

100500minus10 minus05

Re z

minus10

minus05

Im z

(a)

6

4

2

0

3025201505 1000

Exact120579

minus2

minus4

M = 64

M = 32

fJa

c(120579

)

(b)

Figure 4 (a) Distribution of poles ( ⃝) and zeros (times) of the [64 |64] Pade approximation for the test function 119891Jac(119911) with a naturalboundary on |119911| = 1 The unit circle is drawn to guide the eye (b)ThePade approximated functions119891[32|32]Jac (120579)119891[64|64]Jac (120579) and the exactfunction 119891Jac(120579) in the polar-form with 119903 = 10 (after [32])

5 Natural Boundary of Random Power Seriesand the Noise Effect on Padeacute Approximation

In this section we apply Pade approximation to the randompower series with a natural boundary with probability 1 andinvestigate how the approximation detect the singularity ofthe series In addition we examine the effect of noise on thecoefficients of the power expansion for some test functionsSome related theorems for the natural boundary of thefunction generated by the random power series are given inAppendix F

00

05

10

10minus10

minus10

minus05

minus05

00 05

Re z

Im z

(a)

6

4

2

0

Exact

20151005

00

120579

minus2

M = 21

M = 34

M = 55

fFi

b(120579

)

(b)

Figure 5 (a) Distribution of poles ( ⃝) and zeros (+) of the [55 |55] Pade approximated function for the test function 119891Fib(119911) witha natural boundary on |119911| = 1 The unit circle is drawn to guidethe eye (b) The Pade approximated functions 119891[21|21]Fib (120579) 119891[34|34]Fib (120579)and119891[55|55]Fib (120579) corresponding to 119873 = 119865

9= 55 119873 = 119865

10= 89 and

119873 = 11986511

= 144 respectively and the exact function 119891Fib(120579) in thepolar form with 119903 = 10 (after [32])

51 Random Power Series and Natural Boundary Let us con-sider a random power series

119891noise1 (119911) =infin

sum119899=0

120598119903119899119911119899 (21)

Here the coefficients 1199030 1199031 1199032 are iid random variables

which take a value within 119903119899isin [0 1] and 120598 is the strength

of the randomness It is shown that in general the randompower series has a natural boundary on the unit circle |119911| = 1with probability one Figure 6(a) shows distribution of polesand zeros of the [50 | 50] Pade approximated function for


100500minus05minus10

minus10

minus05

00

05

10

Re z

Im z

(a)

2

1

0

654321

Exact120579

minus1

minus2

fno

ise(120579

)

M = 60

(b)

Figure 6 (a) Distribution of poles ( ⃝) and zeros (times) of thePade approximated function 119891[50|50]noise1 (119911) for a random power series119891noise1(119911) with 120598 = 1 The unit circle is drawn to guide the eye (b)The Pade approximated function 119891[50|50]noise1 (120579) and the exact function119891Fib(120579) in the polar form with 119903 = 10

119891noise1(119911) Some pairs of poles and zeros are perfectly can-celled inside the circle |119911| = 1 On the other hand almostall the poles and zeros of the Pade approximated functionassemble around the circle |119911| = 1 and not cancelledThe pairof poles and zeros around the circle |119911| = 1 is called ldquoFroissartdoubletsrdquo and it well corresponds to the natural boundary of119891noise1(119911) The original function is also well approximated bythe [50 | 50] Pade approximation (see Figure 6(b))

Figure 7 shows an example of the coefficients 119888119899 = 120598119903

119899

of the random power series and the coefficients 119886119899 and 119887

119899

of the [50 | 50] Pade approximated function The fluctua-tion of the coefficient 119887

119899 that determines the poles of the

Pade approximated function is smaller than that 119886119899 of the

numeratorNote that the truncated random series is a random poly-

nomial As for the random polynomial it is well known thatthe distribution of the zeros converges on the uni circle when

020015010005000

50403020100n

c n

(a)

04

50403020100n

minus4

anb

n

an

bn

(b)

Figure 7 (a) The coefficient 119888119899 = 120598119903

119899 of a truncated random

power series 119891[100]noise1(119911) with 120598 = 01 (b) The coefficients 119886119899 and

119887119899 of the Pade approximated function 119891[50|50]noise1 (119911) for 119891

[100]

noise1(119911)

the order of the random polynomial increases (Erdos-Turan-type theorem) [34 35 40] Accordingly we can generallyinterpret that in the Pade approximated function to therandom power series the distribution of poles and zeros alsoaccumulates around the unit circle when the order of thePade approximation increases The dependence of the zerosof the randompolynomial and the zeros and poles of the Padeapproximation has been studied by Gilewicz and Kryakin[42] and Ding and Xiao [43]

52 Effect of Noise on a Function with a Simple Pole In thefollowing subsections we investigate influences of noise onthe Pade approximation for some constructed noisy test func-tions as follows

119891test+noise2 (119911) = 119891test (119911) + 119891noise2 (119911) (22)

where 119891test(119911) = suminfin

119899=0119886119899119911119899 and 119891noise2(119911) = sum

infin

119899=0120598119899119911119899 The

120598119899 is iid random variables within [minus120598 120598] where 120598 is

the noise strength Essentially 119891noise2(119911) is the same as therandom power series 119891noise1(119911) First of all in this subsectionwe consider a truncated functionwith a simple poleNote thatif 119886119899= 119862 (constant) and 120598 = 0 that is in noise-free case

119891pole+noise(119911) = 119862suminfin

119899=0119911119899 = 119862(1 minus 119911) with a simple pole

at 119911 = 1 In [2] by Baker Jr the noise effect is summarizedas follows the [119872 | 119872] Pade approximation has an unstablezero at the distance of order 120598minus1 from the origin and the otherzeros make (119872 minus 1) Froissart doublets (zero-pole pairs)

Next we consider a function

119891pole2+noise2 (119911) = 119891pole2 (119911) + 119891noise2 (119911)

=infin

sum119899=0

(1

2119899+ 120598119899) 119911119899

(23)


Re z

Im z 0

1

2

3

0 1 2 3

Unit circle

minus3

minus3

minus2

minus2

minus1

minus1

Zeros 120576 = 0

Poles 120576 = 0

Poles 120576 = 001Zeros 120576 = 001

Figure 8 Distribution of poles ( ⃝) and zeros (times +) of the [10 |10] Pade approximated function 119891[10|10]pole2+noise2(119911) with a stable pole at119911 = 2 for noise strength 120598 = 0 120598 = 001 The unit circle is drawn toguide the eye

with the noise strength 120598 lt 1 Note that

119891pole2 (119911) =2

(2 minus 119911)(24)

with a simple pole at 119911 = 2 to clearly show the shift of thepoles of the approximated function due to the noisy series

Figure 8 shows distribution of the poles and zeros of the[10 | 10] Pade approximated functions It clearly shows thepole shift by the noise effect In the noise-free case (120598 = 0)a pole of the Pade approximation appears at 119911 = 2 and theother poles are cancelledwith zeros (zero-pole ghost pairs) Ina case when the relatively small noise (120598 = 001) is added thepoles and zeros move toward |119911| = 1 with making Froissartdoublets although a pole at 119911 = 2 is quite stable It becomesimpossible to detect the true pole at 119911 = 2 when the noisestrength is relatively large (120598 = 01) not shown in Figure 8

As a result it is found that the locations of the ghost pairsare unstable for noise and the residues for the poles aremuchsmaller than one corresponding to the true poleWe can guessthat the proximity of the nonmodal poles and zeros of thePade approximated function can be understood in a sensethat the poles due to the noise need zeros to cancel with eachother as 120598 rarr 0

53 Effect of Noise on a Functionwith a BranchCut We inves-tigate the effect of the noise on functions with a branch cutFirst let us consider a function

119891branch1 (119911) = radic3 + 119911

1 + 119911(25)

0

1

2

3

0 1 2 3

minus3

minus3

minus2

minus2

minus1

minus1

Re z

Zeros 120576 = 001

Poles 120576 = 001Im

z Unit circleZeros 120576 = 0

Poles 120576 = 0

Figure 9 Distribution of poles ( ⃝) and zeros (times +) of the [10 |10] Pade approximated function 119891[10|10]branch1+noise2(119911) with a brunch cutfrom 119911 = minusinfin to 119911 = 0 for the noise strength 120598 = 0 120598 = 001 Theunit circle is drawn to guide the eye

with an algebraic branch points at 119911 = minus1 and 119911 = minus3 andwiththe branch cut in [minus3 minus1] Distribution of the poles and zerosof the Pade approximated function 119891[10|10]branch1+noise2(119911) is shownin Figure 9 In a case with relatively small noise (120598 = 001)some poles make a line on the branch cut and some polesand zeros move toward the unit circle |119911| = 1 It is impossibleto detect the branch cut when the noise strength is relativelylarge (120598 = 01)

Next let us consider a function

119891branch2 (119911) = log(65minus 119911) (26)

with a logarithmic branch point at 119911 = 65 and with a brunchcut from 119911 = 65 to 119911 = infin The distribution of the polesand zeros of the Pade approximated function119891[10|10]branch2+noise2(119911)for the 119891branch2(119911) with the noisy perturbation is shown inFigure 10 Some poles and zeros are making a line alterna-tively on the branch cut in the noise-free case (120598 = 0) Itassembles around the unit circle |119911| = 1withmaking Froissartdoublets when the noise with strength 120598 = 001 is added

54 Effect of Noise on a Function with a Natural BoundaryFigure 11 shows distribution of the poles and zeros of the [50 |50] Pade approximated function for

119891Jac+noise (119911) = 119891Jac (119911) + 119891noise2 (119911) (27)

which has a natural boundary on |119911| = 1


10

05

00

6420

Unit circle

minus05

minus10

Re z

Poles eps = 0

Poles eps = 001

Zeros eps = 001

Im z

Zeros eps = 0

Figure 10 Distribution of poles ( ⃝) and zeros (times +) of the [10 |10] Pade approximated function 119891[10|10]branch2+noise2(119911) with a brunch cutfrom 119911 = 65 to 119911 = infin for the noise strength 120598 = 0 120598 = 001 Theunit circle is drawn to guide the eye

In the noise-free case the pairs of poles and zeros of thePade approximated function are perfectly cancelled inside theunit circle |119911| = 1 The other poles and zeros of the Padeapproximated function assemble around the circle |119911| = 1without cancellation In the relatively small noise case (120598 =001) the location of the poles is not significantly changedcompared with the zeros shifted outside the unit circle dueto the noise effect And again the poles and zeros movetoward |119911| = 1withmaking Froissart doublets when the noisestrength is relatively large (120598 = 01) It is closely related to afact that fluctuation of the coefficients of the numerator of thePade approximated function is much larger than those in thedenominator as seen in Pade approximation to the randompower series in Figure 7 As a result the singularity of thePade approximated function for the function with a naturalboundary ismore sensitive to the noisy perturbation than thatin the functions with the other type singularity such as simplepoles and branch points

It is very difficult to effectively distinguish whether thepoles of the Pade approximation originated from the naturalboundary on |119911| = 1 of the original function 119891Jac(119911) or fromthe other natural boundary on |119911| = 1 generated by noisyseries 119891noise2(119911) or numerical errors Actually the round-offerror affects the distribution of the poles and zeros of thePade approximated function Accordingly to determine theexpansion coefficients 119888

119899with adequate accuracy becomes

very important in the numerical calculation This is a draw-back of the Pade approximation when we use it for functionswith unknown singularities

55 Numerical Accuracy and Spurious Poles As we observedin the last subsection the effect of rounding error andaccuracy limit of computers work in the numerical results ofthe Pade approximation As the result of accumulation of theround-off error the ldquospurious polesrdquo appear around the unitcircle |119911| = 1 as the pole-zero pairs when the order of Padeapproximation increases (we used a term ldquoFroissart doubletsrdquofor the poles-zero pairs generated by random power seriesconveniently although we cannot numerically distinguish itfrom the spurious poles due to the round-off errors in thenext section we will discuss the Froissart doublets again)

00

05

10

0500 10

Unit circle

minus10

minus10

minus05

minus05

Re z

Im z

eps = 0

eps = 0

eps = 0001

eps = 0001

eps = 001

eps = 001

Figure 11 Distribution of poles (e ⃝ ) and zeros (times+ lowast) ofthe Pade approximated function 119891[50|50]Jac+noise(119911) for the lacunary series119891[100]Jac+noise(119911) with a natural boundary on |119911| = 1 The noise strengthsare 120598 = 0 120598 = 0001 and 120598 = 001 respectively The unit circle isdrawn to guide the eye

However we can roughly distinguish between true polesand the spurious poles by ldquoresidue analysisrdquo of the Padeapproximated function because the spurious poles-zero pairsare unstable for the change of the order In this subsectionwe try to investigate the residues of the Pade approximationfor some test functions Up to now the residue analysis hasbeen mainly used for performance comparison between thedifferent algorithms of the Pade approximation of the sameorder [30 31] On the other hand it seems that the study byusing the information of the residue analysis is still rare in thePade approximation [10 12]

Generally the rational polynomials of the diagonal Padeapproximation can be uniquely identified by the poles 119911

119896

and the corresponding residues 119860119896as follows

119876119872(119911)

119875119872(119911)

=119872

sum119896

119860119896

119911 minus 119911119896

(28)

where the residues are given by

119860119896=

119876119872(119911119896)

prod119872

119895( =119896)(119911119896minus 119911119895) (29)

Here we investigate the convergence property of themagnitude of residues |119860

119896| arranged in descending order

Figure 12 shows the absolute value of the residues |119860119896|

of some Pade approximated functions for the test function


706050403020100

M = 50

M = 75

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

10minus7

10minus8

10minus9

10minus10

10minus11

10minus12

10minus13

|Ak|

k

Figure 12 Absolute values of the residues |119860119896| of the [50 | 50] and

[75 | 75] Pade approximated functions for the test function 1198915(119911)

without noise The |119860119896| are arranged in descending order

1198915(119911) which are arranged in descending order (note that they

are noise-free cases)Thedistribution of the poles and zeros ofthe Pade approximated functions is given in Figure 3 In a caseof119872 = 50 themagnitude of the all residues |119860

119896| is larger than

119874(10minus3) which correspond to the relevant poles arrangedradially in eight directions from the true poles On the otherhand in a case of 119872 = 75 the spurious poles appear anddistribute around the unit circle |119911| = 1 (see Figure 3(b)) It isfound that the absolute values of the residues correspondingthe spurious poles are several order of magnitude smallerthan the relevant poles

Distribution of poles and zeros of the Pade approximatedfunction 119891[20|20]branch2(119911) for the test function 119891branch2(119911) is shownin Figure 13The stable poles and zeros are lined on [65infin]and the spurious poles appear around |119911| = 1Themagnitudeof the residues of the spurious poles is also enormously smallcompared with that of the stable poles remaining with theincrease of the order of the Pade approximation

Figure 14 is also the result of the residues analysis for thePade approximated function for the test function 119891Jac(119911)witha natural boundary on the unit circle |119911| = 1 In the [50 | 50]Pade approximated function the magnitude of the residues|119860119896| is shown in changing the noise strengths 120598 = 0 001 01

corresponding to poles-zeros distribution in Figure 11In the small noise case (120598 = 001) the results of the residue

analysis for 119891Jac+noise2(119911) is almost the same as the noise-freecase (120598 = 0) and in the case with relatively strong noise (120598 =01) the noise shifts themagnitude of the residues with largervalue In addition the result of the residue analysis of thenoise-free cases for some different orders of the Pade approxi-mation is shown in Figure 14(b)We should have inmind that

00051015

6420

minus15

minus10

minus05

Re z

Im z

(a)

15105k

101

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

10minus7

10minus8

10minus9

10minus10

|Ak|

M = 10

M = 20

(b)

Figure 13 (a) Distribution of poles ( ⃝) and zeros (times) of the [20 |

20] Pade approximated function 119891[20|20]branch2(119911) The unit circle is drawnto guide the eye (b) Absolute values of the residues |119860

119896| of the [10 |

10] and [20 | 20] Pade approximated functions for the test function119891branch2(119911) without noise The residues are arranged in descendingorder

the order is important when we apply Pade approximationto the lacunary power series because we should not take theorder of the approximation in the gap of the series

6 Froissart Doublets

The problem of constructing the 119885-transform 119885(119911) of afinite time-series is a standard problem in mathematics [10ndash14] For example it is shown that for a sum of oscillatingdamped signals the 119885-transform associated with the time-series can be characterized by a sum of the poles of the Padeapproximated function The position of each pole is simplylinked to the damping factor and the frequency of each of theoscillators Also it is important to note that all these poleslie strictly outside the unit circle because it corresponds tothe damping [10ndash13] In addition we will consider quasian-alyticity property of the random power series by the residueanalysis of the Pade approximation


40302010

0 001 01

k

101

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

|Ak|

M = 50

(a)

605040302010k

M = 50

M = 32

M = 64

102

101

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

|Ak|

(b)

Figure 14 (a) Absolute values of the residues |119860119896| of the [50 | 50] Pade approximated functions to the noise added test function 119891Jac+noise(119911)

with the strength 120598 = 0 001 01 (b) Absolute values of the residues |119860119896| of the Pade approximated functions 119891[50|50]Jac (119911) 119891[32|32]Jac (119911) and

119891[64|64]Jac (119911) for the test function 119891Jac(119911) without the noise The residues are arranged in descending order

61 Noise Attractor In signal processing we can use the factthat the poles and zeros of the Pade approximated functionto the noisy series distribute around the unit circle |119911| = 1when we remove the noise from the observed data through119885-transform andor Fourier transform of the data Let asequence 119904

0 1199041 119904

119899 be a sample signal without noise

Then we define the 119885-transform of the sequence as

119885 (119911) =119873

sum119899=0

119904119899119911119899 (30)

The function119885(119911) is analytic interior of |119911| lt 1 if the numberof signals119873 is finite [44]Note that discrete Fourier transformis a special case of the 119885-transform

Next let us consider a signal sequence in 119905 isin [0 119879]consisting of the superimposed damping oscillators as

119904119896= sumℓ

119860ℓ119890119894120596ℓ(119896119873)119879 119896 = 0 1 119873 minus 1 (31)

where 119860ℓis the amplitude of the ℓth oscillator and 120596

ℓ=

2120587119891ℓ+119894120572ℓ Here119891

ℓand 120572ℓare the frequency and the damping

factor of the ℓth oscillator Then the 119885-transform is

119885 (119911) =infin

sum119899=0

119904119899119911119899

=infin

sum119899=0

sumℓ

119860ℓ119890119894120596ℓ(119899119873)119879119911119899

= sumℓ

119860ℓ

1 minus 119911119911ℓ

(32)

where we take a limit 119899 rarr infin keeping119879119873 and 119911ℓequiv 119890119894120596ℓ(119879119873)

Accordingly the singularity of 119885(119911) appears as the poles at119911 = 119911minus1

ℓequiv 119890minus119894120596ℓ(119879119873) outside the unit circle |119911| gt 1 and the

residue is Re 119904(119911minus1ℓ) = 119911minus1ℓ119860ℓ

On the other hand let us consider a noise-added sequ-ence 119878

0 1198781 119878

119899 Then the Froissart pointed out that

there are two types of the poles stable poles and unstablepoles when we apply the diagonal Pade approximation tothe unknown data set In general the 119885-transform 119885(119911) =

sum119873

119899=0119878119899119911119899 of the noisy sequence has a natural boundary on

the unit circle |119911| = 1 with probability 1 In fact the poles andzeros (Froissart doublets) of the Pade approximated functionoften distribute around the unit circle when the numericalerror andor noise are mixed into the Taylor series of theanalytic functions as seen in the last sectionThat is to say wesometimes call the unit circle |119911| = 1 noise attractor in a sensethat the poles and zeros are attracted to the circle as the Frois-sart doublets [45] Accordingly it is found that Pade approx-imated function for the function 119885(119911) has stable poles asso-ciated with the damping modes and unstable spurious polesassociated with the noisy fluctuation After elimination ofthe spurious poles around the noise attractor from the noisysequence we can reconstruct the noise-free sequence consist-ing of the stable poles located in the domain |119911| gt 1 Anotherremarkable feature of the nonmodal poles is that the absolutevalues of the Cauchy residues associated with them areusually much smaller than those associated with true poles

62 Random Power Series and Quasianalytic Function Wei-erstrass defined the analytic function by direct analytic con-tinuation of function Then apparently the analytic continu-ation is impossible beyond the natural boundary even if we


can uniquely define the function and it is analytic outsidethe analytic domain Borel and Gammel extended the narrowcondition for the analyticity and gave a definition of quasian-alytic functions [46 47] Gammel conjectured the followingfor the random power series [10 45]

Gammel Conjecture (1973)The random power series belongsto the Borel class of quasianalytic functions as the followingform

119891Gammel (119911) =infin

sum119896=0

119861119896

1 minus 119908119896119911 (33)

where119908119896= 1198901198942120587119883119896 and 119883

119896 are real numbers in the interval in

119883119896isin [0 1] and 119861

119896decreases rapidly with 119896 Then the natural

boundary in the Weierstrass sense can be crossedThe function (33) is a simple example that poles are

densely distributed on the unit circle Then the convergenceproperty of the sequence |119861

119896| is important for the analyticity

of the function Carleman proved that 119891Gammel(119911) is quasian-alytic if 119861

119896satisfies the following condition

10038161003816100381610038161198611198961003816100381610038161003816 lt 119862119890minus119896

1+119888

119888 gt 0 (34)

This is Carleman class of quasianalytic functions See Gam-melrsquos paper [45] for the details Moreover Gammel and Nut-tall proved that the quasianalytic functions can be exactlyapproximated by the Pade approximation [45]

Gammel-Nuttall Theorem (1973) If 119861119896in (33) satisfies the

condition (34) and |120596119896| = 1 then the sequence of [119873+119869 | 119873]

Pade approximation to the 119891Gammel(119911) converges in measureto the function119891Gammel(119911) as119873 rarrinfin in any closed boundedregion of the complex plane where 119869 is a natural number thatequals119873 or less

Is the Gammel Conjecture True We try to examine thevalidity of the Gammel conjecture by applying residue anal-ysis of the Pade approximated function to the random powerseries 119891noise2(119911) Figure 15 shows the absolute values of theresidues |119860

119896| of the Pade approximated functions 119891[45|45]noise (119911)

for three different samples in descending order |119860119896| roughly

exponentially decreases with respect to 119896 as1003816100381610038161003816119860119896

1003816100381610038161003816 sim exp (minus120573119896) (35)

where120573 is the decay exponent It shows exponential decay (orfaster) and on the surface supports the Gammal conjecture

However it is not nearly so simple We should check thestability of the exponential-like decay of the magnitude of theresidues by changing the order of the Pade approximationFigure 16 shows the result for the three different orders119872 =15 119872 = 45 and 119872 = 55 It expresses an indication thatthe decay exponent 120573 does not converge to a positive certainvalue It seems that the exponent behaves 120573 rarr 0 as a limit119872 rarr infin On the other hand if we directly apply the Padeapproximation to the quasianalytic function 119891Gammel(119911) with119861119896= 119890minus119896 the exponent 120573 is stable for changing the order of

the Pade approximation (seeAppendix E)These facts suggestthat the random power series does not belong to Carleman

0001

001

01

1

40302010

|Ak|

k

Figure 15 Absolute values of the residues |119860119896| of the Pade approx-

imated functions 119891[45|45]noise2 (119911) for three samples of the truncatedrandom power series of order 119873 = 90 and 120598 = 1 The |119860

119896| are

arranged in descending order

0001

001

01

1

40302010

|Ak|

k

M = 15

M = 25

M = 45


imated functions 119891[45|45]noise2 (119911) 119891[25|25]

noise2 (119911) and 119891[15|15]

noise2 (119911) for a sampletruncated random series and 120598 = 1 The |119860

119896| are arranged in

descending order

class of quasianalytic functions although it has a naturalboundary on the unit circle and it has the form (33) Asa result we can say that no optimism is warranted on theGammel conjecture

How does the residue analyses of the Pade approximationfor the analyticity andor quasianalyticity of unknown func-tion work It is an interesting and future problem


7 Summary and Discussion

In the present paper we numerically examined the effective-ness of the Pade approximation for some test functions withbranch point essential singularity and natural boundary bywatching the singularities of the Pade approximated func-tions For the functions with a branch cut the poles and zerosof the Pade approximated function are lined along the truebranch cut The poles and zeros are distributed around thetrue natural boundary if the original test function has a nat-ural boundary In addition we gave the explicit Pade appro-ximated functions for some lacunary power series which areuseful to check the numerical result It was shown that inparticular the distribution of poles and zeros of the Padeapproximated function for lacunary power series and therandom power series accumulated around the unit circlewhen the order of the approximation increases

We often suffer from the difficulty to distinguish whetheror not the poles of the Pade approximation are intrinsicallyoriginated from the natural boundary of the original powerseries because the numerical errors contained in the expan-sion coefficients also yield a false natural boundaryThereforethe expansion coefficients with adequate numerical accuracyare necessary when we apply the Pade approximation tofunctions with unknown singularities

Furthermore the residue calculus of the Pade approxi-mated function is useful when we detect the singularity ofthe original power series from the asymptotic behavior of thetruncated series It is useful also for estimating the accuracy ofthe approximation As a result the residue calculus suggestedthat the random power series does not obey Gammelrsquosconjecture that is it does not belong to Borel class of the qua-sianalytic functions

We finally remark that the most serious problem to beimproved is the numerical accuracy due to the limitationof the order in the Pade approximation when we use it fordetecting unknown singularities of wave functions in quan-tum physics [32]

Appendices

A General Recursion Relation

We can construct a power series that has some pole-type sin-gularities in the following form

1198891199112 + 119890119911 + 119891

1198861199112 + 119887119911 + 119888=infin

sum119899=0

119886119899119911119899 (A1)

where 119886 119887 119888 119889 119890 and 119891 are real and 119888 = 0 for simplicityThen the coefficients 119886

119899 can be obtained by rearranging and

comparing with the coefficients of the both sides in the sameorder as follows

1198891199112 + 119890119911 + 119891 = (1198871198860119911 + 119888119886

0+ 1198881198861119911)

+infin

sum119899=2

(119886119886119899minus2

+ 119887119886119899minus1

+ 119888119886119899) 119911119899

(A2)

As a result the power series with the pole-type singulari-ties can be constructed by the recursion relation

119886119896= minus

119887

119888119886119896minus1

minus119886

119888119886119896minus2

119896 ge 2 (A3)

with 1198881198860= 119891 119887119886

0+ 1198881198861= 119890 and 119886119886

0+ 1198871198861+ 1198881198862= 119889

It becomes Fibonacci sequence when we set 1198860= 0 119886

1=

1 and 119886119896= 119886119896minus1

+ 119886119896minus2

B Random Polynomial

The following theorems concerning the random power seriesare well known

Erdos-Turan-TypeTheorem (1950) Let us define a polynomial

119891 (119911) =119873

sum119899=0

119886119899119911119899 (B1)

where coefficients 119886119899are randomly distributed and 119886

0119886119873

=0 for simplicity Then the zeros of the random polynomialcluster uniformly around the unit circle |119911| = 1 if ldquosize of thetruncated seriesrdquo 119871

119873(119891) is small compared to the order119873 of

the polynomial where

119871119873(119891) = log(

sum119873

119899=0

10038161003816100381610038161198861198991003816100381610038161003816

radic100381610038161003816100381611988601198861198731003816100381610038161003816

) (B2)

Note that this theorem also holds for the polynomialswith deterministic coefficients 119886

119899such asNewman-type poly-

nomial having coefficients in the sets 0 1 or 0 plusmn1

Peres-Virag Theorem (2005) Let 119886119899 be iid Gaussian-type

random variables then the distribution 119870(119911) of the complexzeros 119911

119896 of the power series

119891 (119911) =infin

sum119899=0

119886119899119911119899 (B3)

is

119870(119911119896) =

1

120587

1

(1 minus1003816100381610038161003816119911119896

10038161003816100381610038162

)2 (B4)

C Some Gap Theorems ofLacunary Power Series

Weierstrass considered the analyticity of the power series

119891 (119911) =infin

sum]=0119886]119911119887] 119887 isin 119873 119887 = 1 (C1)

where 119886] is a positive number In the main text we set 119886] =1 119887 = 2 for 119891Wei(119911) Then it is proved that the function(C1) has a natural boundary on the unit circle |119911| = 1 ifthe convergence radius of the function is unity based on thefollowing theorems for the lacunary power series


Hadamard-Barck GapTheorem (1892) Let

119891 (119911) =infin

sum]=0119886]119911120582] (C2)

where 119886] is a positive number and 120582] denote a strictlyincreasing sequence of the natural numbers satisfying aninequality 119902120582] le 120582]+1 for 119902 gt 1 Then the function 119891(119911) has anatural boundary on the unit circle |119911| = 1

Fabryrsquos Gap Theorem (1899) Power series

119891 (119911) =infin

sum]=0119886]119911120582] (C3)

with radius of convergence 119877 = 1 has a natural boundary onthe unit circle |119911| = 1 provided that it is Fabry series that is

lim]rarrinfin

120582]

]= infin (C4)

D Numerators of Diagonal PadeacuteApproximations for 119891Jac(119911) and 119891Fib(119911)

The diagonal Pade approximation for the truncated lacunarypower functions119891Jac(119911) and119891Fib(119911) can be exactly executed asgiven in themain textThe numerators119860119873Jac(119911) and119860

119865119873

Fib(119911) ofthe Pade approximated functions can be given as follows

119860119873Jac (119911) = 119911 + 21199112

+ 2119873minus1

sum119899=2

119911119867119899 (119911 + 1199112 +119899minus2

sum119896=1

119911119867119896+2) (D1)

where119867119899= 2119899minus1

Numerator of the diagonal Pade approximated functionfor 119891Fib(119911) is

119860119865119873

Fib (119911) = 119878119873minus4

(119911)

+ [119878119873minus8

(119911) + 119911] (119891119873minus4

(119911) minus 119891119873minus2

(119911))

+ [2119891119873minus3

(119911) + 2119891119873minus2

(119911) + 119891119873minus3

(119911) 119891119873minus6

(119911)]

(D2)

where 119878119871(119911) = sum

119871

119896=0119891119896(119911) 119891

119896(119911) = 119911119865119896 119865

119873means 119873th

Fibonacci number and we set 119865minus1= 119865minus2= sdot sdot sdot = 0

We have inductively obtained above results by means ofMathematica

E Residue Analysis for Carleman Class ofQuasianalytic Functions

In this appendix we give a direct result of residue analysisfor ldquoCarleman classrdquo of the quasianalytic functions for com-parison with the other residue analyses in the main text We

5040302010

10minus1

10minus2

10minus3

10minus4

|Ak|

k


imated functions for a truncated Carleman function 119891Carleman(119911) oforder119872 = 15 25 45 which is artificially constructed by (E2)Theyare arranged in descending order in each case

apply the Pade approximation to the quasiperiodic function119891Carleman(119911) of the Carleman class which is artificially con-structed by a set of the poles 119911

119896 as follows

119891Carleman (119911) =119870

sum119896=1

(1

1 minus 119911119896119911+

1

1 minus 119911lowast119896119911) 119890minus119896 (E1)

= 2infin

sum119899=0

119870

sum119896=1

119890minus119896 cos (2120587119883119896119899) 119911119899 (E2)

where we set the poles at 119911119896= exp(plusmn2120587119894119883

119896) (119896 = 1 2 119870)

on the unit circle 119883119896 are iid random variables in the

interval119883119896isin [01] and we take119870 = 100 Figure 17 shows the

absolute values of the residues |119860119896| of the Pade approximated

functions of order 119872 = 15 119872 = 25 and 119872 = 45 for119891Carleman(119911) They are arranged in descending order

As a result it seems that |119860119896| exponentially decreaseswith

a stable exponent regardless of the order of the Pade approxi-mation This supports that certainly the Pade approximationis applicable to the quasianalytic functions in the Gammelconjecture as given in Gammel-Nuttall theorem The Padeapproximation for the quasianalytic function converges to thefunction even outside the unit circle It should be also notedthat in all cases the tails of |119860

119896| are rapidly decay because the

ldquotruncatedrdquo series are essentially analytic functions


F Some Results for Natural Boundary inNoisy Series

In this appendix some theorems for the random power seriesare given See for example [21] for the proofs

Steinhausrsquos Theorem (1929) Suppose that the power series

119891 (119911) =infin

sum119899=0

119886119899119911119899 (F1)

has radius of convergence 119877 = 1 Let 1198830 1198831 119883

119899be a

sequence of iid random variables in the interval119883119894isin [0 1]

Then with probability one the random power series

119891Steinhaus (119911) =infin

sum119899=0

119886119899119908119899119911119899 (F2)

has a natural boundary on |119911| = 1 where 119908119896= 1198901198942120587119883119896

Paley-Zygmund Theorem (1932) Suppose that the powerseries (F1) has the radius of convergence 1 Let 119903

0 1199031 119903

119899

be a sequence of binary stochastic variables taking minus1 or 1with equal probability Then with probability one the ran-dom power series

119891119875minus119885

(119911) =infin

sum119899=0

119903119899119911119899 (F3)

has a natural boundary on the unit circle |119911| = 1The similar theorems can hold for random power series

suminfin

119899=0119903119899119911119899 with a sequence of stochastic variables obeying iid

in the interval 119903119894isin [minus1 1] or 119903

119894isin [0 1] [48]

KahanersquosTheorem (1985)The circle of convergence is the nat-ural boundary for randomTaylor series (F1) if the coefficients119886119899 are independent and symmetric random variablesThe more generalized version has been given in the

following form [22]

Breuer-Simon Theorem (2011) Suppose that the power series(F1) has the convergence radius 1 Then for ae 120596 119891(119911) =suminfin

119899=0119886119899(120596)119911119899 has a strong natural boundary on |119911| = 1 if the

119886119899(120596) is a stationary ergodic bounded and nondeterministic

process

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This paper was partially written for ldquoInternational Sympo-siumof ComplexifiedDynamics Tunnelling andChaosrdquo heldon 2005 in KusatsuThis work is partly supported by Japanesepeoplersquos tax viaMEXT and the authorswould like to acknowl-edge them They are also very grateful to Dr T Tsuji andto Koike Memorial House for using the facilities during thisstudy

References

[1] G A Baker and J L GammelThePade Approxima tion inTheo-retical Physics Academic Press New York NY USA 1970

[2] G A Baker Jr Essentials of Pade Approximants AcademicPress New York NY USA 1975

[3] J Baker and P Graves-Morris Pade Approximants CambridgeUniversity Press Cambridge UK 2nd edition 1996

[4] F Sasagawa Scattering Theory Syoukampbou 1991 (Japanese)[5] H Stanly Introduction to Phase Transitions and Critical Phe-

nomena Clarendon Press Oxford UK 1971[6] C N Yang and T D Lee ldquoStatistical theory of equations of

state and phase transitions I Theory of condensationrdquo PhysicalReview vol 87 p 404 1952

[7] B Nickel ldquoOn the singularity structure of the 2D Ising modelsusceptibilityrdquo Journal of Physics A vol 32 no 21 pp 3889ndash3906 1999

[8] R Kubo M Toda and N Hashitsume Statistical Physics IISpringer Berlin Germany 1985

[9] B M McCoy ldquoDo hard spheres have natural boundariesrdquohttparxivorgabscond-mat0103556

[10] D Bessis ldquoPade approximations in noise filteringrdquo Journal ofComputational andAppliedMathematics vol 66 no 1-2 pp 85ndash88 1996

[11] H Stahl ldquoThe convergence of diagonal Pade approximants andthe Pade conjecturerdquo Journal of Computational and AppliedMathematics vol 86 no 1 pp 287ndash296 1997

[12] J Gilewicz and M Pindor ldquoPade approximants and noise acase of geometric seriesrdquo Journal of Computational and AppliedMathematics vol 87 no 2 pp 199ndash214 1997

[13] D Bessis and L Perotti ldquoUniversal analytic properties of noiseintroducing the 119869-matrix formalismrdquo Journal of Physics A vol42 no 36 2009

[14] L A Barbosa Coelho and L A Baccala ldquoPade approximationsas a modal identification techniquerdquo in Proceedings of the 27thIMAC Orlando Fla USA February 2009

[15] A Berretti and L Chierchia ldquoOn the complex analytic structureof the golden invariant curve for the standard maprdquo Nonlinear-ity vol 3 no 1 pp 39ndash44 1990

[16] C Falcolini and R de la Llave ldquoNumerical calculation ofdomains of analyticity for perturbation theories in the presenceof small divisorsrdquo Journal of Statistical Physics vol 67 no 3-4pp 645ndash666 1992

[17] R de la Llave and S Tompaidis ldquoComputation of domainsof analyticity for some perturbative expansions of mechanicsrdquoPhysicaDNonlinear Phenomena vol 71 no 1-2 pp 55ndash81 1994

[18] A Berretti and S Marmi ldquoScaling perturbative renormaliza-tion and analyticity for the standard map and some generaliza-tionsrdquo Chaos Solitons and Fractals vol 5 no 2 pp 257ndash2691995

[19] A Berretti C Falcolini and G Gentile ldquoShape of analyticitydomains of Lindstedt series the standardmaprdquo Physical ReviewE vol 64 no 1 Article ID 015202 2001

[20] T W Korner Exercises for Fourier Analysis Cambridge Univer-sity Press Cambridge UK 1993

[21] R Remmert Classical Topics in Complex Function TheorySpringer New York NY USA 1st edition 1998

[22] J Breuer and B Simon ldquoNatural boundaries and spectral the-oryrdquo Advances in Mathematics vol 226 no 6 pp 4902ndash49202011


[23] O Knill and J Lesieutre ldquoAnalytic continuation of Dirichletseries with almost periodic coefficientsrdquo Complex Analysis andOperator Theory vol 6 no 1 pp 237ndash255 2012

[24] O Costin and M Huang ldquoBehavior of lacunary series at thenatural boundaryrdquoAdvances in Mathematics vol 222 no 4 pp1370ndash1404 2009

[25] WH Press S A TeukolskyW T Vetterling and B P FlanneryNumerical Recipes in C Cambridge University Press 1988

[26] W H Press and S A Teukolsky ldquoPade approximantsrdquo Comput-ers in Physics vol 6 p 82 1982

[27] J Nuttall ldquoThe convergence of Pade approximants of meromor-phic functionsrdquo Journal of Mathematical Analysis and Applica-tions vol 31 no 1 pp 147ndash153 1970

[28] C Pommerenke ldquoPade approximants and convergence in capa-cityrdquo Journal of Mathematical Analysis and Applications vol 41no 3 pp 775ndash780 1973

[29] H Stahl ldquoSpurious poles in Pade approximationrdquo Journal ofComputational and Applied Mathematics vol 99 no 1-2 pp511ndash527 1998

[30] P Gonnet R Pachon and L N Trefethen ldquoRobust rationalinterpolation and least-squaresrdquo Electronic Transactions onNumerical Analysis vol 38 pp 146ndash167 2011

[31] P Gonnet S Guttel and L N Trefethen ldquoRobust Pade Approx-imation via SVDrdquo SIAM Review vol 55 no 1 pp 101ndash117 2013

[32] H S Yamada and K S Ikeda ldquoAnalyticity of quantum states inone-dimensional tight-binding modelrdquo The European PhysicalJournal B 2014

[33] E B Saff and R S Varga ldquoOn the zeros and poles of Padeapproximants toezrdquo Numerische Mathematik vol 25 no 1 pp1ndash14 1975

[34] M Kac ldquoOn the average number of real roots of a randomalgebraic equationrdquo Bulletin of the American MathematicalSociety vol 49 pp 314ndash320 1943

[35] P Erdos and P Turan ldquoOn the distribution of roots of polyno-mialsrdquo Annals of Mathematics vol 51 pp 105ndash119 1950

[36] F Amoroso and M Mignotte ldquoOn the distribution of the rootsof polynomialsrdquo Annales de lrsquoInstitut Fourier vol 46 no 5 pp1275ndash1291 1996

[37] A Odlyzko and B Poonen ldquoZeros of polynomials with 0 1coefficientsrdquo LrsquoEnseignement Mathematique vol 39 pp 317ndash348 1993

[38] B Simon Orthogonal Polynomials on the Unit Circle Part lClassical Theory American Mathematical Society 2004

[39] B Simon Orthogonal Polynomials on the Unit Circle Part 2Spectral Theory American Mathematical Society ProvidenceRI USA 2004

[40] Y Peres and B Virag ldquoZeros of the iid Gaussian powerseries a conformally invariant determinantal processrdquo ActaMathematica vol 194 no 1 pp 1ndash35 2005

[41] B Simon Szegorsquos Theorem and Its Descendants Spectral Theoryfor L2 Perturbations of Orthogonal Polynomials PrincetonUniversity Press 2010

[42] J Gilewicz and Y Kryakin ldquoFroissart doublets in Pade approx-imation in the case of polynomial noiserdquo Journal of Computa-tional and Applied Mathematics vol 153 no 1-2 pp 235ndash2422003

[43] X Ding and Y Xiao ldquoNatural boundary of random DirichletseriesrdquoUkrainian Mathematical Journal vol 58 no 7 pp 1129ndash1138 2006

[44] ldquoWe can also define the 119885-transform by negative power 119911minus119899Then the function 119885(119911) is analytic in outer domain of |119911| = 1the poles corresponding to damping oscilla tions appear in theinside the unit circle |119911| lt 1rdquo

[45] J L Gammel and J Nuttall ldquoConvergence of Pade approximantsto quasianalytic functions beyond natural boundariesrdquo Journalof Mathematical Analysis and Applications vol 43 no 3 pp694ndash696 1973

[46] A Shenitzer and N Luzin ldquoFunction part Irdquo The AmericanMathematical Monthly vol 105 no 1 pp 59ndash67 1998

[47] N Luzin ldquoFunction Part IIrdquo The American MathematicalMonthly vol 105 no 3 pp 263ndash270 1998

[48] J-P Kahane Some Random Series of Functions vol 5 of Cam-bridge Studies in Advanced Mathematics Cambridge UniversityPress Cambridge UK 2nd edition 1985

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of


Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of


Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of


Mathematical PhysicsAdvances in

Complex AnalysisJournal of


OptimizationJournal of


CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of


Operations ResearchAdvances in

Journal of


Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences


The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014


Algebra

Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

Discrete MathematicsJournal of


Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


where 1205941015840 denotes the derivative with respect to the variable 119906In the low temperature expansion for the magnetic suscepti-bility 120594with some coefficients 119886

119899 we assume that the approx-

imated susceptibility 120594119873and the logarithmic derivative can

be obtained as follows

120594 sim 120594119873=119873

sum119899=0

119886119899119906119899

1205941015840

120594sim119873

sum119899=0

119887119899119906119899 (3)

Then we can estimate the critical point 119906 = 119906119888and the critical

exponent 120574 by the Pade approximation to the truncatedexpansion 120594

119873 Note that the singularity on |119906| = 119906

119888will be

infinitely differentiable if the coefficients 119886119899fall off sufficiently

rapidlyIn general analytic continuation over the singular point

is possible along any other path in complex plane even ifthe function diverges at the singular point determining theradius of convergence as seen in the above singularity of theIsing model Therefore the Pade approximation is useful toimprove the convergence of the power series and approximatethe exact solution

Furthermore the Pade approximation has been usedto investigate convergence of Fourier series [2 3] and thebreakdown of KAMcurves in complex plane forHamiltonianmap systems which is described as the analytic domains ofLindstedt series for standard map [15ndash19]

In addition we can see an interesting example concerningthe Pade approximation in noisy data analysis [10ndash14] Thepower series with finite random coefficient ldquoalmost alwaysrdquohas a natural boundary on the unit circle in the complexplane [20ndash24] In a finite time-series Froissart has shownthat a natural boundary generated by the random time-seriesis approximated by doublets of poles and zeros (Froissartdoublets) of the Pade approximated function which aresurrounding the vicinity of the unit circle Taking advantageof the characteristic the Pade approximation has been usedin order to remove the noise and extract the true poles asso-ciated with damping modes from the observed noisy time-series

The main purpose of the present paper is to investigatewhether the Pade approximation is numerically useful fordetecting the singularity of some test functions In particularwe examine the usefulness for the functions with a naturalboundary such as a lacunary power series and a randompower series [2 13 18]

The organization of the paper is as follows In Section 2we give a brief explanation of the Pade approximation andsome important reminders in the numerical calculation InSection 3 we present some numerical results of the Padeapproximation for test functions with branch cut essentialsingularityWe also try to apply the Pade approximation to anentire function In Section 4 application of the Pade approx-imation to some lacunary series which is known to have anatural boundary is given In Section 5 numerical resultsof the application to random power series with a naturalboundary and some test functionswith randomnoise are alsoshown In Section 6 we discuss the residue calculus for thePade approximated functions to confirm the numerical errors

and quasianalyticity of the random power series In the lastsection we give summary and discussion

In Appendix A the general result for Fibonacci generatorused in Section 32 is given In Appendix B some theoremsconcerning zeros of polynomials are given In Appendix Csome mathematical theorems for lacunary series whichare useful in reading the main text are summarized InAppendix D exact Pade approximated functions to somelacunary power series with a natural boundary are givenResidue analysis for quasianalytic functions of Carlemanclass is given in Appendix E Furthermore some theoremsconcerning the randompower series are given inAppendix F

2 Padeacute Approximation

In this section we give preliminary instructions to the Padeapproximation and some important reminders on the numer-ical calculation

For a given function 119891(119911) a truncated Taylor expansion119891[119873](119911) of order119873 about zero is given as

119891 (119911) sim 119891[119873] (119911) =119873

sum119899=0

119888119899119911119899 (4)

where 119888119899denotes the coefficients of the Taylor expansion

Pade approximation is more accurate approximation for119891(119911)up to order 119874(119911119873) than the Taylor expansion The Padeapproximation is a rational function namely a ratio of twopolynomials which agrees with the highest possible order119874(119911119873) with a truncated polynomial 119891[119873](119911) as follows

119891[119873] (119911) =1198860+ 1198861119911 + 11988621199112 + sdot sdot sdot 119886

119871119911119871

1 + 1198871119911 + 11988721199112 + sdot sdot sdot 119887

119872119911119872

equiv119875119871(119911)

119876119872(119911)

equiv 119891[119871|119872] (119911)

(5)

where 119875119871(119911) is a polynomial of degree less than or equal to 119871

and119876119872(119911) is a polynomial of degree less than or equal to119872

Note that 1198870= 1 (normalized) here A unique approximation

can be specified for all choices of 119872 and 119871 such that 119873 =119871 + 119872 when it exists The coefficients 119886

119899 119887119899 can be

obtained from the condition that the first (119871 + 119872 + 1) termsvanish in the Taylor series in (4) The difference between thePade approximation and the original function satisfies thefollowing equation

119891 (119911) minus 119891[119871|119872]

(119911) = 119874 (119911119871+119872+1) (6)

In this paper we sometimes use ldquo[119871 | 119872] Pade approx-imationrdquo or ldquo[119871 | 119872] Pade approximated functionrdquo for119891[119871|119872](119911) Solving the problem in (6) is called a linear Teoplitzproblem which is generally ill-conditioned We used full LUdecomposition for the Teoplitz matrix in the problem as wellas iterative improvement in order to eliminate the ill-posedproblem [25 26] In addition hereafter we use the diagonalPade approximation that is 119871 = 119872 of order 119872 le 65


















1198911(119911) =

log (1 + 119911)119911

(7)


119891[4]1

(119911) = 1 minus1

2119911 +

1

31199112 minus

1

41199113 +

1

51199114 (8)


119891[2|2]1

(119911) =1 + (710) 119911 + (130) 1199112

1 + (65) 119911 + (310) 1199112 (9)



1(119911)









119865(119911)




119865119899= 119865119899minus1

+ 119865119899minus2

(119899 ge 2) (10)

where 1198650= 0 and 119865




86420


30

25

20

15

10

05

00

z

f1(z

)

[2|2] Pad e app




1(119911)

119891119865(119911) =

infin

sum119899=0

119865119899119911119899

=119911

1 minus 119911 minus 1199112

=1

radic5

1

1 minus 120601+119911minus

1

radic5

1


(11)




(119911) = sum119873


119891[1198732|1198732]119865

(119911) =119911

1 minus 119911 minus 1199112+ 119874 (119911119873+1) (12)



1198912(119911) = 119890minus119911

1198913(119911) = radic

1 + 2119911

1 + 119911

1198914(119911) = 119890minus119911(1+119911)

1198915(119911) = tan 1199114

(13)








119875119872(119911) =

119872

sum119896=0

(2119872 minus 119896)119872

(2119872)119896 (119872 minus 119896)(minus119911)119896

119876119872(119911) =

119872

sum119896=0

(2119872 minus 119896)119872

(2119872)119896 (119872 minus 119896)119911119896

(14)



119872(119911) = 119876



2(119911) go





3(119911) 119911 = minus1 and 119911 = minus12





4(119911)







0

20

40

0 20 40minus40

minus40

minus20

minus20



Im z

(a)

Re z

Im z

PolesZeros

000


(b)

Re z

Im z

000

005

010

minus010

minus005




(c)


3(119911) and (c) 119891

4(119911)



infin

119899=01199112119899

119891Wie(119911) = sum

infin

119899=0119911119899 and 119891Kro(119911) = sum

infin

119899=01199111198992





119891119903(120579) = 119891 (119911 = 119903119890119894120579) =

infin

sum119899=0

119888119899(119903119890119894120579)

119899

(15)




0

1

2

0 1 2minus2

minus2

minus1

minus1

Re z

Im z

(a)

0

1

2

0 1 2minus2

minus2

minus1

minus1

Re z

Im z

(b)




119891[2119873

]

Jac (119911) sim 119891[2119873minus1

|2119873minus1

]

Jac (119911)

=119860119873Jac (119911)

1 + sum119873minus2

119896=01199112119896


(16)


1 +119873minus2

sum119896=0

1199112119896


= 0 (17)






119891Fib (119911) =infin

sum119899=0

119911119865119899 (18)



119891[119865119873]

Fib (119911) sim 119891[1198651198732|1198651198732]

Fib (119911)

=119860119865119873

Fib (119911)


(19)



1198732 | 119865






00

05

10


Re z

minus10

minus05

Im z

(a)

6

4

2

0

3025201505 1000

Exact120579

minus2

minus4

M = 64

M = 32

fJa

c(120579

)

(b)




00

05

10

10minus10

minus10

minus05

minus05

00 05

Re z

Im z

(a)

6

4

2

0

Exact

20151005

00

120579

minus2

M = 21

M = 34

M = 55

fFi

b(120579

)

(b)


9= 55 119873 = 119865

10= 89 and

119873 = 11986511



119891noise1 (119911) =infin

sum119899=0

120598119903119899119911119899 (21)






minus10

minus05

00

05

10

Re z

Im z

(a)

2

1

0

654321

Exact120579

minus1

minus2

fno

ise(120579

)

M = 60

(b)




119899


119899






020015010005000

50403020100n

c n

(a)

04

50403020100n

minus4

anb

n

an

bn

(b)





[100]

noise1(119911)





119899=0119886119899119911119899 and 119891noise2(119911) = sum

infin

119899=0120598119899119911119899 The








=infin

sum119899=0

(1

2119899+ 120598119899) 119911119899

(23)


Re z

Im z 0

1

2

3

0 1 2 3

Unit circle

minus3

minus3

minus2

minus2

minus1

minus1

Zeros 120576 = 0

Poles 120576 = 0

Poles 120576 = 001Zeros 120576 = 001



119891pole2 (119911) =2

(2 minus 119911)(24)





119891branch1 (119911) = radic3 + 119911

1 + 119911(25)

0

1

2

3

0 1 2 3

minus3

minus3

minus2

minus2

minus1

minus1

Re z

Zeros 120576 = 001

Poles 120576 = 001Im


Poles 120576 = 0




119891branch2 (119911) = log(65minus 119911) (26)






10

05

00

6420

Unit circle

minus05

minus10

Re z

Poles eps = 0

Poles eps = 001

Zeros eps = 001

Im z

Zeros eps = 0







00

05

10

0500 10

Unit circle

minus10

minus10

minus05

minus05

Re z

Im z

eps = 0

eps = 0

eps = 0001

eps = 0001

eps = 001

eps = 001




119896


119876119872(119911)

119875119872(119911)

=119872

sum119896

119860119896

119911 minus 119911119896

(28)


119860119896=

119876119872(119911119896)

prod119872

119895( =119896)(119911119896minus 119911119895) (29)






706050403020100

M = 50

M = 75

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

10minus7

10minus8

10minus9

10minus10

10minus11

10minus12

10minus13

|Ak|

k












00051015

6420

minus15

minus10

minus05

Re z

Im z

(a)

15105k

101

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

10minus7

10minus8

10minus9

10minus10

|Ak|

M = 10

M = 20

(b)



119896| of the [10 |






40302010

0 001 01

k

101

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

|Ak|

M = 50

(a)

605040302010k

M = 50

M = 32

M = 64

102

101

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

|Ak|

(b)





0 1199041 119904



119885 (119911) =119873

sum119899=0

119904119899119911119899 (30)



119904119896= sumℓ

119860ℓ119890119894120596ℓ(119896119873)119879 119896 = 0 1 119873 minus 1 (31)


ℓ=

2120587119891ℓ+119894120572ℓ Here119891



119885 (119911) =infin

sum119899=0

119904119899119911119899

=infin

sum119899=0

sumℓ

119860ℓ119890119894120596ℓ(119899119873)119879119911119899

= sumℓ

119860ℓ

1 minus 119911119911ℓ

(32)






0 1198781 119878



sum119873







119891Gammel (119911) =infin

sum119896=0

119861119896

1 minus 119908119896119911 (33)

where119908119896= 1198901198942120587119883119896 and 119883


119883119896isin [0 1] and 119861







10038161003816100381610038161198611198961003816100381610038161003816 lt 119862119890minus119896

1+119888

119888 gt 0 (34)









1003816100381610038161003816 sim exp (minus120573119896) (35)




0001

001

01

1

40302010

|Ak|

k



119896| are


0001

001

01

1

40302010

|Ak|

k

M = 15

M = 25

M = 45



noise2 (119911) and 119891[15|15]



descending order









Appendices



1198891199112 + 119890119911 + 119891

1198861199112 + 119887119911 + 119888=infin

sum119899=0

119886119899119911119899 (A1)




1198891199112 + 119890119911 + 119891 = (1198871198860119911 + 119888119886

0+ 1198881198861119911)

+infin

sum119899=2

(119886119886119899minus2

+ 119887119886119899minus1

+ 119888119886119899) 119911119899

(A2)


119886119896= minus

119887

119888119886119896minus1

minus119886

119888119886119896minus2

119896 ge 2 (A3)

with 1198881198860= 119891 119887119886

0+ 1198881198861= 119890 and 119886119886

0+ 1198871198861+ 1198881198862= 119889


1=

1 and 119886119896= 119886119896minus1

+ 119886119896minus2

B Random Polynomial



119891 (119911) =119873

sum119899=0

119886119899119911119899 (B1)


0119886119873




119871119873(119891) = log(

sum119873

119899=0

10038161003816100381610038161198861198991003816100381610038161003816

radic100381610038161003816100381611988601198861198731003816100381610038161003816

) (B2)







119891 (119911) =infin

sum119899=0

119886119899119911119899 (B3)

is

119870(119911119896) =

1

120587

1

(1 minus1003816100381610038161003816119911119896

10038161003816100381610038162

)2 (B4)



119891 (119911) =infin

sum]=0119886]119911119887] 119887 isin 119873 119887 = 1 (C1)




119891 (119911) =infin

sum]=0119886]119911120582] (C2)



119891 (119911) =infin

sum]=0119886]119911120582] (C3)


lim]rarrinfin

120582]

]= infin (C4)



119865119873


119860119873Jac (119911) = 119911 + 21199112

+ 2119873minus1

sum119899=2

119911119867119899 (119911 + 1199112 +119899minus2

sum119896=1

119911119867119896+2) (D1)

where119867119899= 2119899minus1


119860119865119873

Fib (119911) = 119878119873minus4

(119911)

+ [119878119873minus8

(119911) + 119911] (119891119873minus4

(119911) minus 119891119873minus2

(119911))

+ [2119891119873minus3

(119911) + 2119891119873minus2

(119911) + 119891119873minus3

(119911) 119891119873minus6

(119911)]

(D2)

where 119878119871(119911) = sum

119871

119896=0119891119896(119911) 119891

119896(119911) = 119911119865119896 119865

119873means 119873th





5040302010

10minus1

10minus2

10minus3

10minus4

|Ak|

k




119896 as follows

119891Carleman (119911) =119870

sum119896=1

(1

1 minus 119911119896119911+

1


= 2infin

sum119899=0

119870

sum119896=1

119890minus119896 cos (2120587119883119896119899) 119911119899 (E2)


119896) (119896 = 1 2 119870)













119891 (119911) =infin

sum119899=0

119886119899119911119899 (F1)


119899be a




sum119899=0

119886119899119908119899119911119899 (F2)



0 1199031 119903

119899


119891119875minus119885

(119911) =infin

sum119899=0

119903119899119911119899 (F3)


suminfin



119894isin [0 1] [48]


following form [22]




process



Acknowledgments


References

























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra


























1198911(119911) =

log (1 + 119911)119911

(7)


119891[4]1

(119911) = 1 minus1

2119911 +

1

31199112 minus

1

41199113 +

1

51199114 (8)


119891[2|2]1

(119911) =1 + (710) 119911 + (130) 1199112

1 + (65) 119911 + (310) 1199112 (9)



1(119911)









119865(119911)




119865119899= 119865119899minus1

+ 119865119899minus2

(119899 ge 2) (10)

where 1198650= 0 and 119865




86420


30

25

20

15

10

05

00

z

f1(z

)

[2|2] Pad e app




1(119911)

119891119865(119911) =

infin

sum119899=0

119865119899119911119899

=119911

1 minus 119911 minus 1199112

=1

radic5

1

1 minus 120601+119911minus

1

radic5

1


(11)




(119911) = sum119873


119891[1198732|1198732]119865

(119911) =119911

1 minus 119911 minus 1199112+ 119874 (119911119873+1) (12)



1198912(119911) = 119890minus119911

1198913(119911) = radic

1 + 2119911

1 + 119911

1198914(119911) = 119890minus119911(1+119911)

1198915(119911) = tan 1199114

(13)








119875119872(119911) =

119872

sum119896=0

(2119872 minus 119896)119872

(2119872)119896 (119872 minus 119896)(minus119911)119896

119876119872(119911) =

119872

sum119896=0

(2119872 minus 119896)119872

(2119872)119896 (119872 minus 119896)119911119896

(14)



119872(119911) = 119876



2(119911) go





3(119911) 119911 = minus1 and 119911 = minus12





4(119911)







0

20

40

0 20 40minus40

minus40

minus20

minus20



Im z

(a)

Re z

Im z

PolesZeros

000


(b)

Re z

Im z

000

005

010

minus010

minus005




(c)


3(119911) and (c) 119891

4(119911)



infin

119899=01199112119899

119891Wie(119911) = sum

infin

119899=0119911119899 and 119891Kro(119911) = sum

infin

119899=01199111198992





119891119903(120579) = 119891 (119911 = 119903119890119894120579) =

infin

sum119899=0

119888119899(119903119890119894120579)

119899

(15)




0

1

2

0 1 2minus2

minus2

minus1

minus1

Re z

Im z

(a)

0

1

2

0 1 2minus2

minus2

minus1

minus1

Re z

Im z

(b)




119891[2119873

]

Jac (119911) sim 119891[2119873minus1

|2119873minus1

]

Jac (119911)

=119860119873Jac (119911)

1 + sum119873minus2

119896=01199112119896


(16)


1 +119873minus2

sum119896=0

1199112119896


= 0 (17)






119891Fib (119911) =infin

sum119899=0

119911119865119899 (18)



119891[119865119873]

Fib (119911) sim 119891[1198651198732|1198651198732]

Fib (119911)

=119860119865119873

Fib (119911)


(19)



1198732 | 119865






00

05

10


Re z

minus10

minus05

Im z

(a)

6

4

2

0

3025201505 1000

Exact120579

minus2

minus4

M = 64

M = 32

fJa

c(120579

)

(b)




00

05

10

10minus10

minus10

minus05

minus05

00 05

Re z

Im z

(a)

6

4

2

0

Exact

20151005

00

120579

minus2

M = 21

M = 34

M = 55

fFi

b(120579

)

(b)


9= 55 119873 = 119865

10= 89 and

119873 = 11986511



119891noise1 (119911) =infin

sum119899=0

120598119903119899119911119899 (21)






minus10

minus05

00

05

10

Re z

Im z

(a)

2

1

0

654321

Exact120579

minus1

minus2

fno

ise(120579

)

M = 60

(b)




119899


119899






020015010005000

50403020100n

c n

(a)

04

50403020100n

minus4

anb

n

an

bn

(b)





[100]

noise1(119911)





119899=0119886119899119911119899 and 119891noise2(119911) = sum

infin

119899=0120598119899119911119899 The








=infin

sum119899=0

(1

2119899+ 120598119899) 119911119899

(23)


Re z

Im z 0

1

2

3

0 1 2 3

Unit circle

minus3

minus3

minus2

minus2

minus1

minus1

Zeros 120576 = 0

Poles 120576 = 0

Poles 120576 = 001Zeros 120576 = 001



119891pole2 (119911) =2

(2 minus 119911)(24)





119891branch1 (119911) = radic3 + 119911

1 + 119911(25)

0

1

2

3

0 1 2 3

minus3

minus3

minus2

minus2

minus1

minus1

Re z

Zeros 120576 = 001

Poles 120576 = 001Im


Poles 120576 = 0




119891branch2 (119911) = log(65minus 119911) (26)






10

05

00

6420

Unit circle

minus05

minus10

Re z

Poles eps = 0

Poles eps = 001

Zeros eps = 001

Im z

Zeros eps = 0







00

05

10

0500 10

Unit circle

minus10

minus10

minus05

minus05

Re z

Im z

eps = 0

eps = 0

eps = 0001

eps = 0001

eps = 001

eps = 001




119896


119876119872(119911)

119875119872(119911)

=119872

sum119896

119860119896

119911 minus 119911119896

(28)


119860119896=

119876119872(119911119896)

prod119872

119895( =119896)(119911119896minus 119911119895) (29)






706050403020100

M = 50

M = 75

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

10minus7

10minus8

10minus9

10minus10

10minus11

10minus12

10minus13

|Ak|

k












00051015

6420

minus15

minus10

minus05

Re z

Im z

(a)

15105k

101

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

10minus7

10minus8

10minus9

10minus10

|Ak|

M = 10

M = 20

(b)



119896| of the [10 |






40302010

0 001 01

k

101

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

|Ak|

M = 50

(a)

605040302010k

M = 50

M = 32

M = 64

102

101

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

|Ak|

(b)





0 1199041 119904



119885 (119911) =119873

sum119899=0

119904119899119911119899 (30)



119904119896= sumℓ

119860ℓ119890119894120596ℓ(119896119873)119879 119896 = 0 1 119873 minus 1 (31)


ℓ=

2120587119891ℓ+119894120572ℓ Here119891



119885 (119911) =infin

sum119899=0

119904119899119911119899

=infin

sum119899=0

sumℓ

119860ℓ119890119894120596ℓ(119899119873)119879119911119899

= sumℓ

119860ℓ

1 minus 119911119911ℓ

(32)






0 1198781 119878



sum119873







119891Gammel (119911) =infin

sum119896=0

119861119896

1 minus 119908119896119911 (33)

where119908119896= 1198901198942120587119883119896 and 119883


119883119896isin [0 1] and 119861







10038161003816100381610038161198611198961003816100381610038161003816 lt 119862119890minus119896

1+119888

119888 gt 0 (34)









1003816100381610038161003816 sim exp (minus120573119896) (35)




0001

001

01

1

40302010

|Ak|

k



119896| are


0001

001

01

1

40302010

|Ak|

k

M = 15

M = 25

M = 45



noise2 (119911) and 119891[15|15]



descending order









Appendices



1198891199112 + 119890119911 + 119891

1198861199112 + 119887119911 + 119888=infin

sum119899=0

119886119899119911119899 (A1)




1198891199112 + 119890119911 + 119891 = (1198871198860119911 + 119888119886

0+ 1198881198861119911)

+infin

sum119899=2

(119886119886119899minus2

+ 119887119886119899minus1

+ 119888119886119899) 119911119899

(A2)


119886119896= minus

119887

119888119886119896minus1

minus119886

119888119886119896minus2

119896 ge 2 (A3)

with 1198881198860= 119891 119887119886

0+ 1198881198861= 119890 and 119886119886

0+ 1198871198861+ 1198881198862= 119889


1=

1 and 119886119896= 119886119896minus1

+ 119886119896minus2

B Random Polynomial



119891 (119911) =119873

sum119899=0

119886119899119911119899 (B1)


0119886119873




119871119873(119891) = log(

sum119873

119899=0

10038161003816100381610038161198861198991003816100381610038161003816

radic100381610038161003816100381611988601198861198731003816100381610038161003816

) (B2)







119891 (119911) =infin

sum119899=0

119886119899119911119899 (B3)

is

119870(119911119896) =

1

120587

1

(1 minus1003816100381610038161003816119911119896

10038161003816100381610038162

)2 (B4)



119891 (119911) =infin

sum]=0119886]119911119887] 119887 isin 119873 119887 = 1 (C1)




119891 (119911) =infin

sum]=0119886]119911120582] (C2)



119891 (119911) =infin

sum]=0119886]119911120582] (C3)


lim]rarrinfin

120582]

]= infin (C4)



119865119873


119860119873Jac (119911) = 119911 + 21199112

+ 2119873minus1

sum119899=2

119911119867119899 (119911 + 1199112 +119899minus2

sum119896=1

119911119867119896+2) (D1)

where119867119899= 2119899minus1


119860119865119873

Fib (119911) = 119878119873minus4

(119911)

+ [119878119873minus8

(119911) + 119911] (119891119873minus4

(119911) minus 119891119873minus2

(119911))

+ [2119891119873minus3

(119911) + 2119891119873minus2

(119911) + 119891119873minus3

(119911) 119891119873minus6

(119911)]

(D2)

where 119878119871(119911) = sum

119871

119896=0119891119896(119911) 119891

119896(119911) = 119911119865119896 119865

119873means 119873th





5040302010

10minus1

10minus2

10minus3

10minus4

|Ak|

k




119896 as follows

119891Carleman (119911) =119870

sum119896=1

(1

1 minus 119911119896119911+

1


= 2infin

sum119899=0

119870

sum119896=1

119890minus119896 cos (2120587119883119896119899) 119911119899 (E2)


119896) (119896 = 1 2 119870)













119891 (119911) =infin

sum119899=0

119886119899119911119899 (F1)


119899be a




sum119899=0

119886119899119908119899119911119899 (F2)



0 1199031 119903

119899


119891119875minus119885

(119911) =infin

sum119899=0

119903119899119911119899 (F3)


suminfin



119894isin [0 1] [48]


following form [22]




process



Acknowledgments


References

























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










86420


30

25

20

15

10

05

00

z

f1(z

)

[2|2] Pad e app




1(119911)

119891119865(119911) =

infin

sum119899=0

119865119899119911119899

=119911

1 minus 119911 minus 1199112

=1

radic5

1

1 minus 120601+119911minus

1

radic5

1


(11)




(119911) = sum119873


119891[1198732|1198732]119865

(119911) =119911

1 minus 119911 minus 1199112+ 119874 (119911119873+1) (12)



1198912(119911) = 119890minus119911

1198913(119911) = radic

1 + 2119911

1 + 119911

1198914(119911) = 119890minus119911(1+119911)

1198915(119911) = tan 1199114

(13)








119875119872(119911) =

119872

sum119896=0

(2119872 minus 119896)119872

(2119872)119896 (119872 minus 119896)(minus119911)119896

119876119872(119911) =

119872

sum119896=0

(2119872 minus 119896)119872

(2119872)119896 (119872 minus 119896)119911119896

(14)



119872(119911) = 119876



2(119911) go





3(119911) 119911 = minus1 and 119911 = minus12





4(119911)







0

20

40

0 20 40minus40

minus40

minus20

minus20



Im z

(a)

Re z

Im z

PolesZeros

000


(b)

Re z

Im z

000

005

010

minus010

minus005




(c)


3(119911) and (c) 119891

4(119911)



infin

119899=01199112119899

119891Wie(119911) = sum

infin

119899=0119911119899 and 119891Kro(119911) = sum

infin

119899=01199111198992





119891119903(120579) = 119891 (119911 = 119903119890119894120579) =

infin

sum119899=0

119888119899(119903119890119894120579)

119899

(15)




0

1

2

0 1 2minus2

minus2

minus1

minus1

Re z

Im z

(a)

0

1

2

0 1 2minus2

minus2

minus1

minus1

Re z

Im z

(b)




119891[2119873

]

Jac (119911) sim 119891[2119873minus1

|2119873minus1

]

Jac (119911)

=119860119873Jac (119911)

1 + sum119873minus2

119896=01199112119896


(16)


1 +119873minus2

sum119896=0

1199112119896


= 0 (17)






119891Fib (119911) =infin

sum119899=0

119911119865119899 (18)



119891[119865119873]

Fib (119911) sim 119891[1198651198732|1198651198732]

Fib (119911)

=119860119865119873

Fib (119911)


(19)



1198732 | 119865






00

05

10


Re z

minus10

minus05

Im z

(a)

6

4

2

0

3025201505 1000

Exact120579

minus2

minus4

M = 64

M = 32

fJa

c(120579

)

(b)




00

05

10

10minus10

minus10

minus05

minus05

00 05

Re z

Im z

(a)

6

4

2

0

Exact

20151005

00

120579

minus2

M = 21

M = 34

M = 55

fFi

b(120579

)

(b)


9= 55 119873 = 119865

10= 89 and

119873 = 11986511



119891noise1 (119911) =infin

sum119899=0

120598119903119899119911119899 (21)






minus10

minus05

00

05

10

Re z

Im z

(a)

2

1

0

654321

Exact120579

minus1

minus2

fno

ise(120579

)

M = 60

(b)




119899


119899






020015010005000

50403020100n

c n

(a)

04

50403020100n

minus4

anb

n

an

bn

(b)





[100]

noise1(119911)





119899=0119886119899119911119899 and 119891noise2(119911) = sum

infin

119899=0120598119899119911119899 The








=infin

sum119899=0

(1

2119899+ 120598119899) 119911119899

(23)


Re z

Im z 0

1

2

3

0 1 2 3

Unit circle

minus3

minus3

minus2

minus2

minus1

minus1

Zeros 120576 = 0

Poles 120576 = 0

Poles 120576 = 001Zeros 120576 = 001



119891pole2 (119911) =2

(2 minus 119911)(24)





119891branch1 (119911) = radic3 + 119911

1 + 119911(25)

0

1

2

3

0 1 2 3

minus3

minus3

minus2

minus2

minus1

minus1

Re z

Zeros 120576 = 001

Poles 120576 = 001Im


Poles 120576 = 0




119891branch2 (119911) = log(65minus 119911) (26)






10

05

00

6420

Unit circle

minus05

minus10

Re z

Poles eps = 0

Poles eps = 001

Zeros eps = 001

Im z

Zeros eps = 0







00

05

10

0500 10

Unit circle

minus10

minus10

minus05

minus05

Re z

Im z

eps = 0

eps = 0

eps = 0001

eps = 0001

eps = 001

eps = 001




119896


119876119872(119911)

119875119872(119911)

=119872

sum119896

119860119896

119911 minus 119911119896

(28)


119860119896=

119876119872(119911119896)

prod119872

119895( =119896)(119911119896minus 119911119895) (29)






706050403020100

M = 50

M = 75

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

10minus7

10minus8

10minus9

10minus10

10minus11

10minus12

10minus13

|Ak|

k












00051015

6420

minus15

minus10

minus05

Re z

Im z

(a)

15105k

101

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

10minus7

10minus8

10minus9

10minus10

|Ak|

M = 10

M = 20

(b)



119896| of the [10 |






40302010

0 001 01

k

101

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

|Ak|

M = 50

(a)

605040302010k

M = 50

M = 32

M = 64

102

101

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

|Ak|

(b)





0 1199041 119904



119885 (119911) =119873

sum119899=0

119904119899119911119899 (30)



119904119896= sumℓ

119860ℓ119890119894120596ℓ(119896119873)119879 119896 = 0 1 119873 minus 1 (31)


ℓ=

2120587119891ℓ+119894120572ℓ Here119891



119885 (119911) =infin

sum119899=0

119904119899119911119899

=infin

sum119899=0

sumℓ

119860ℓ119890119894120596ℓ(119899119873)119879119911119899

= sumℓ

119860ℓ

1 minus 119911119911ℓ

(32)






0 1198781 119878



sum119873







119891Gammel (119911) =infin

sum119896=0

119861119896

1 minus 119908119896119911 (33)

where119908119896= 1198901198942120587119883119896 and 119883


119883119896isin [0 1] and 119861







10038161003816100381610038161198611198961003816100381610038161003816 lt 119862119890minus119896

1+119888

119888 gt 0 (34)









1003816100381610038161003816 sim exp (minus120573119896) (35)




0001

001

01

1

40302010

|Ak|

k



119896| are


0001

001

01

1

40302010

|Ak|

k

M = 15

M = 25

M = 45



noise2 (119911) and 119891[15|15]



descending order









Appendices



1198891199112 + 119890119911 + 119891

1198861199112 + 119887119911 + 119888=infin

sum119899=0

119886119899119911119899 (A1)




1198891199112 + 119890119911 + 119891 = (1198871198860119911 + 119888119886

0+ 1198881198861119911)

+infin

sum119899=2

(119886119886119899minus2

+ 119887119886119899minus1

+ 119888119886119899) 119911119899

(A2)


119886119896= minus

119887

119888119886119896minus1

minus119886

119888119886119896minus2

119896 ge 2 (A3)

with 1198881198860= 119891 119887119886

0+ 1198881198861= 119890 and 119886119886

0+ 1198871198861+ 1198881198862= 119889


1=

1 and 119886119896= 119886119896minus1

+ 119886119896minus2

B Random Polynomial



119891 (119911) =119873

sum119899=0

119886119899119911119899 (B1)


0119886119873




119871119873(119891) = log(

sum119873

119899=0

10038161003816100381610038161198861198991003816100381610038161003816

radic100381610038161003816100381611988601198861198731003816100381610038161003816

) (B2)







119891 (119911) =infin

sum119899=0

119886119899119911119899 (B3)

is

119870(119911119896) =

1

120587

1

(1 minus1003816100381610038161003816119911119896

10038161003816100381610038162

)2 (B4)



119891 (119911) =infin

sum]=0119886]119911119887] 119887 isin 119873 119887 = 1 (C1)




119891 (119911) =infin

sum]=0119886]119911120582] (C2)



119891 (119911) =infin

sum]=0119886]119911120582] (C3)


lim]rarrinfin

120582]

]= infin (C4)



119865119873


119860119873Jac (119911) = 119911 + 21199112

+ 2119873minus1

sum119899=2

119911119867119899 (119911 + 1199112 +119899minus2

sum119896=1

119911119867119896+2) (D1)

where119867119899= 2119899minus1


119860119865119873

Fib (119911) = 119878119873minus4

(119911)

+ [119878119873minus8

(119911) + 119911] (119891119873minus4

(119911) minus 119891119873minus2

(119911))

+ [2119891119873minus3

(119911) + 2119891119873minus2

(119911) + 119891119873minus3

(119911) 119891119873minus6

(119911)]

(D2)

where 119878119871(119911) = sum

119871

119896=0119891119896(119911) 119891

119896(119911) = 119911119865119896 119865

119873means 119873th





5040302010

10minus1

10minus2

10minus3

10minus4

|Ak|

k




119896 as follows

119891Carleman (119911) =119870

sum119896=1

(1

1 minus 119911119896119911+

1


= 2infin

sum119899=0

119870

sum119896=1

119890minus119896 cos (2120587119883119896119899) 119911119899 (E2)


119896) (119896 = 1 2 119870)













119891 (119911) =infin

sum119899=0

119886119899119911119899 (F1)


119899be a




sum119899=0

119886119899119908119899119911119899 (F2)



0 1199031 119903

119899


119891119875minus119885

(119911) =infin

sum119899=0

119903119899119911119899 (F3)


suminfin



119894isin [0 1] [48]


following form [22]




process



Acknowledgments


References

























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










0

20

40

0 20 40minus40

minus40

minus20

minus20



Im z

(a)

Re z

Im z

PolesZeros

000


(b)

Re z

Im z

000

005

010

minus010

minus005




(c)


3(119911) and (c) 119891

4(119911)



infin

119899=01199112119899

119891Wie(119911) = sum

infin

119899=0119911119899 and 119891Kro(119911) = sum

infin

119899=01199111198992





119891119903(120579) = 119891 (119911 = 119903119890119894120579) =

infin

sum119899=0

119888119899(119903119890119894120579)

119899

(15)




0

1

2

0 1 2minus2

minus2

minus1

minus1

Re z

Im z

(a)

0

1

2

0 1 2minus2

minus2

minus1

minus1

Re z

Im z

(b)




119891[2119873

]

Jac (119911) sim 119891[2119873minus1

|2119873minus1

]

Jac (119911)

=119860119873Jac (119911)

1 + sum119873minus2

119896=01199112119896


(16)


1 +119873minus2

sum119896=0

1199112119896


= 0 (17)






119891Fib (119911) =infin

sum119899=0

119911119865119899 (18)



119891[119865119873]

Fib (119911) sim 119891[1198651198732|1198651198732]

Fib (119911)

=119860119865119873

Fib (119911)


(19)



1198732 | 119865






00

05

10


Re z

minus10

minus05

Im z

(a)

6

4

2

0

3025201505 1000

Exact120579

minus2

minus4

M = 64

M = 32

fJa

c(120579

)

(b)




00

05

10

10minus10

minus10

minus05

minus05

00 05

Re z

Im z

(a)

6

4

2

0

Exact

20151005

00

120579

minus2

M = 21

M = 34

M = 55

fFi

b(120579

)

(b)


9= 55 119873 = 119865

10= 89 and

119873 = 11986511



119891noise1 (119911) =infin

sum119899=0

120598119903119899119911119899 (21)






minus10

minus05

00

05

10

Re z

Im z

(a)

2

1

0

654321

Exact120579

minus1

minus2

fno

ise(120579

)

M = 60

(b)




119899


119899






020015010005000

50403020100n

c n

(a)

04

50403020100n

minus4

anb

n

an

bn

(b)





[100]

noise1(119911)





119899=0119886119899119911119899 and 119891noise2(119911) = sum

infin

119899=0120598119899119911119899 The








=infin

sum119899=0

(1

2119899+ 120598119899) 119911119899

(23)


Re z

Im z 0

1

2

3

0 1 2 3

Unit circle

minus3

minus3

minus2

minus2

minus1

minus1

Zeros 120576 = 0

Poles 120576 = 0

Poles 120576 = 001Zeros 120576 = 001



119891pole2 (119911) =2

(2 minus 119911)(24)





119891branch1 (119911) = radic3 + 119911

1 + 119911(25)

0

1

2

3

0 1 2 3

minus3

minus3

minus2

minus2

minus1

minus1

Re z

Zeros 120576 = 001

Poles 120576 = 001Im


Poles 120576 = 0




119891branch2 (119911) = log(65minus 119911) (26)






10

05

00

6420

Unit circle

minus05

minus10

Re z

Poles eps = 0

Poles eps = 001

Zeros eps = 001

Im z

Zeros eps = 0







00

05

10

0500 10

Unit circle

minus10

minus10

minus05

minus05

Re z

Im z

eps = 0

eps = 0

eps = 0001

eps = 0001

eps = 001

eps = 001




119896


119876119872(119911)

119875119872(119911)

=119872

sum119896

119860119896

119911 minus 119911119896

(28)


119860119896=

119876119872(119911119896)

prod119872

119895( =119896)(119911119896minus 119911119895) (29)






706050403020100

M = 50

M = 75

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

10minus7

10minus8

10minus9

10minus10

10minus11

10minus12

10minus13

|Ak|

k












00051015

6420

minus15

minus10

minus05

Re z

Im z

(a)

15105k

101

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

10minus7

10minus8

10minus9

10minus10

|Ak|

M = 10

M = 20

(b)



119896| of the [10 |






40302010

0 001 01

k

101

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

|Ak|

M = 50

(a)

605040302010k

M = 50

M = 32

M = 64

102

101

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

|Ak|

(b)





0 1199041 119904



119885 (119911) =119873

sum119899=0

119904119899119911119899 (30)



119904119896= sumℓ

119860ℓ119890119894120596ℓ(119896119873)119879 119896 = 0 1 119873 minus 1 (31)


ℓ=

2120587119891ℓ+119894120572ℓ Here119891



119885 (119911) =infin

sum119899=0

119904119899119911119899

=infin

sum119899=0

sumℓ

119860ℓ119890119894120596ℓ(119899119873)119879119911119899

= sumℓ

119860ℓ

1 minus 119911119911ℓ

(32)






0 1198781 119878



sum119873







119891Gammel (119911) =infin

sum119896=0

119861119896

1 minus 119908119896119911 (33)

where119908119896= 1198901198942120587119883119896 and 119883


119883119896isin [0 1] and 119861







10038161003816100381610038161198611198961003816100381610038161003816 lt 119862119890minus119896

1+119888

119888 gt 0 (34)









1003816100381610038161003816 sim exp (minus120573119896) (35)




0001

001

01

1

40302010

|Ak|

k



119896| are


0001

001

01

1

40302010

|Ak|

k

M = 15

M = 25

M = 45



noise2 (119911) and 119891[15|15]



descending order









Appendices



1198891199112 + 119890119911 + 119891

1198861199112 + 119887119911 + 119888=infin

sum119899=0

119886119899119911119899 (A1)




1198891199112 + 119890119911 + 119891 = (1198871198860119911 + 119888119886

0+ 1198881198861119911)

+infin

sum119899=2

(119886119886119899minus2

+ 119887119886119899minus1

+ 119888119886119899) 119911119899

(A2)


119886119896= minus

119887

119888119886119896minus1

minus119886

119888119886119896minus2

119896 ge 2 (A3)

with 1198881198860= 119891 119887119886

0+ 1198881198861= 119890 and 119886119886

0+ 1198871198861+ 1198881198862= 119889


1=

1 and 119886119896= 119886119896minus1

+ 119886119896minus2

B Random Polynomial



119891 (119911) =119873

sum119899=0

119886119899119911119899 (B1)


0119886119873




119871119873(119891) = log(

sum119873

119899=0

10038161003816100381610038161198861198991003816100381610038161003816

radic100381610038161003816100381611988601198861198731003816100381610038161003816

) (B2)







119891 (119911) =infin

sum119899=0

119886119899119911119899 (B3)

is

119870(119911119896) =

1

120587

1

(1 minus1003816100381610038161003816119911119896

10038161003816100381610038162

)2 (B4)



119891 (119911) =infin

sum]=0119886]119911119887] 119887 isin 119873 119887 = 1 (C1)




119891 (119911) =infin

sum]=0119886]119911120582] (C2)



119891 (119911) =infin

sum]=0119886]119911120582] (C3)


lim]rarrinfin

120582]

]= infin (C4)



119865119873


119860119873Jac (119911) = 119911 + 21199112

+ 2119873minus1

sum119899=2

119911119867119899 (119911 + 1199112 +119899minus2

sum119896=1

119911119867119896+2) (D1)

where119867119899= 2119899minus1


119860119865119873

Fib (119911) = 119878119873minus4

(119911)

+ [119878119873minus8

(119911) + 119911] (119891119873minus4

(119911) minus 119891119873minus2

(119911))

+ [2119891119873minus3

(119911) + 2119891119873minus2

(119911) + 119891119873minus3

(119911) 119891119873minus6

(119911)]

(D2)

where 119878119871(119911) = sum

119871

119896=0119891119896(119911) 119891

119896(119911) = 119911119865119896 119865

119873means 119873th





5040302010

10minus1

10minus2

10minus3

10minus4

|Ak|

k




119896 as follows

119891Carleman (119911) =119870

sum119896=1

(1

1 minus 119911119896119911+

1


= 2infin

sum119899=0

119870

sum119896=1

119890minus119896 cos (2120587119883119896119899) 119911119899 (E2)


119896) (119896 = 1 2 119870)













119891 (119911) =infin

sum119899=0

119886119899119911119899 (F1)


119899be a




sum119899=0

119886119899119908119899119911119899 (F2)



0 1199031 119903

119899


119891119875minus119885

(119911) =infin

sum119899=0

119903119899119911119899 (F3)


suminfin



119894isin [0 1] [48]


following form [22]




process



Acknowledgments


References

























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










0

1

2

0 1 2minus2

minus2

minus1

minus1

Re z

Im z

(a)

0

1

2

0 1 2minus2

minus2

minus1

minus1

Re z

Im z

(b)




119891[2119873

]

Jac (119911) sim 119891[2119873minus1

|2119873minus1

]

Jac (119911)

=119860119873Jac (119911)

1 + sum119873minus2

119896=01199112119896


(16)


1 +119873minus2

sum119896=0

1199112119896


= 0 (17)






119891Fib (119911) =infin

sum119899=0

119911119865119899 (18)



119891[119865119873]

Fib (119911) sim 119891[1198651198732|1198651198732]

Fib (119911)

=119860119865119873

Fib (119911)


(19)



1198732 | 119865






00

05

10


Re z

minus10

minus05

Im z

(a)

6

4

2

0

3025201505 1000

Exact120579

minus2

minus4

M = 64

M = 32

fJa

c(120579

)

(b)




00

05

10

10minus10

minus10

minus05

minus05

00 05

Re z

Im z

(a)

6

4

2

0

Exact

20151005

00

120579

minus2

M = 21

M = 34

M = 55

fFi

b(120579

)

(b)


9= 55 119873 = 119865

10= 89 and

119873 = 11986511



119891noise1 (119911) =infin

sum119899=0

120598119903119899119911119899 (21)






minus10

minus05

00

05

10

Re z

Im z

(a)

2

1

0

654321

Exact120579

minus1

minus2

fno

ise(120579

)

M = 60

(b)




119899


119899






020015010005000

50403020100n

c n

(a)

04

50403020100n

minus4

anb

n

an

bn

(b)





[100]

noise1(119911)





119899=0119886119899119911119899 and 119891noise2(119911) = sum

infin

119899=0120598119899119911119899 The








=infin

sum119899=0

(1

2119899+ 120598119899) 119911119899

(23)


Re z

Im z 0

1

2

3

0 1 2 3

Unit circle

minus3

minus3

minus2

minus2

minus1

minus1

Zeros 120576 = 0

Poles 120576 = 0

Poles 120576 = 001Zeros 120576 = 001



119891pole2 (119911) =2

(2 minus 119911)(24)





119891branch1 (119911) = radic3 + 119911

1 + 119911(25)

0

1

2

3

0 1 2 3

minus3

minus3

minus2

minus2

minus1

minus1

Re z

Zeros 120576 = 001

Poles 120576 = 001Im


Poles 120576 = 0




119891branch2 (119911) = log(65minus 119911) (26)






10

05

00

6420

Unit circle

minus05

minus10

Re z

Poles eps = 0

Poles eps = 001

Zeros eps = 001

Im z

Zeros eps = 0







00

05

10

0500 10

Unit circle

minus10

minus10

minus05

minus05

Re z

Im z

eps = 0

eps = 0

eps = 0001

eps = 0001

eps = 001

eps = 001




119896


119876119872(119911)

119875119872(119911)

=119872

sum119896

119860119896

119911 minus 119911119896

(28)


119860119896=

119876119872(119911119896)

prod119872

119895( =119896)(119911119896minus 119911119895) (29)






706050403020100

M = 50

M = 75

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

10minus7

10minus8

10minus9

10minus10

10minus11

10minus12

10minus13

|Ak|

k












00051015

6420

minus15

minus10

minus05

Re z

Im z

(a)

15105k

101

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

10minus7

10minus8

10minus9

10minus10

|Ak|

M = 10

M = 20

(b)



119896| of the [10 |






40302010

0 001 01

k

101

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

|Ak|

M = 50

(a)

605040302010k

M = 50

M = 32

M = 64

102

101

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

|Ak|

(b)





0 1199041 119904



119885 (119911) =119873

sum119899=0

119904119899119911119899 (30)



119904119896= sumℓ

119860ℓ119890119894120596ℓ(119896119873)119879 119896 = 0 1 119873 minus 1 (31)


ℓ=

2120587119891ℓ+119894120572ℓ Here119891



119885 (119911) =infin

sum119899=0

119904119899119911119899

=infin

sum119899=0

sumℓ

119860ℓ119890119894120596ℓ(119899119873)119879119911119899

= sumℓ

119860ℓ

1 minus 119911119911ℓ

(32)






0 1198781 119878



sum119873







119891Gammel (119911) =infin

sum119896=0

119861119896

1 minus 119908119896119911 (33)

where119908119896= 1198901198942120587119883119896 and 119883


119883119896isin [0 1] and 119861







10038161003816100381610038161198611198961003816100381610038161003816 lt 119862119890minus119896

1+119888

119888 gt 0 (34)









1003816100381610038161003816 sim exp (minus120573119896) (35)




0001

001

01

1

40302010

|Ak|

k



119896| are


0001

001

01

1

40302010

|Ak|

k

M = 15

M = 25

M = 45



noise2 (119911) and 119891[15|15]



descending order









Appendices



1198891199112 + 119890119911 + 119891

1198861199112 + 119887119911 + 119888=infin

sum119899=0

119886119899119911119899 (A1)




1198891199112 + 119890119911 + 119891 = (1198871198860119911 + 119888119886

0+ 1198881198861119911)

+infin

sum119899=2

(119886119886119899minus2

+ 119887119886119899minus1

+ 119888119886119899) 119911119899

(A2)


119886119896= minus

119887

119888119886119896minus1

minus119886

119888119886119896minus2

119896 ge 2 (A3)

with 1198881198860= 119891 119887119886

0+ 1198881198861= 119890 and 119886119886

0+ 1198871198861+ 1198881198862= 119889


1=

1 and 119886119896= 119886119896minus1

+ 119886119896minus2

B Random Polynomial



119891 (119911) =119873

sum119899=0

119886119899119911119899 (B1)


0119886119873




119871119873(119891) = log(

sum119873

119899=0

10038161003816100381610038161198861198991003816100381610038161003816

radic100381610038161003816100381611988601198861198731003816100381610038161003816

) (B2)







119891 (119911) =infin

sum119899=0

119886119899119911119899 (B3)

is

119870(119911119896) =

1

120587

1

(1 minus1003816100381610038161003816119911119896

10038161003816100381610038162

)2 (B4)



119891 (119911) =infin

sum]=0119886]119911119887] 119887 isin 119873 119887 = 1 (C1)




119891 (119911) =infin

sum]=0119886]119911120582] (C2)



119891 (119911) =infin

sum]=0119886]119911120582] (C3)


lim]rarrinfin

120582]

]= infin (C4)



119865119873


119860119873Jac (119911) = 119911 + 21199112

+ 2119873minus1

sum119899=2

119911119867119899 (119911 + 1199112 +119899minus2

sum119896=1

119911119867119896+2) (D1)

where119867119899= 2119899minus1


119860119865119873

Fib (119911) = 119878119873minus4

(119911)

+ [119878119873minus8

(119911) + 119911] (119891119873minus4

(119911) minus 119891119873minus2

(119911))

+ [2119891119873minus3

(119911) + 2119891119873minus2

(119911) + 119891119873minus3

(119911) 119891119873minus6

(119911)]

(D2)

where 119878119871(119911) = sum

119871

119896=0119891119896(119911) 119891

119896(119911) = 119911119865119896 119865

119873means 119873th





5040302010

10minus1

10minus2

10minus3

10minus4

|Ak|

k




119896 as follows

119891Carleman (119911) =119870

sum119896=1

(1

1 minus 119911119896119911+

1


= 2infin

sum119899=0

119870

sum119896=1

119890minus119896 cos (2120587119883119896119899) 119911119899 (E2)


119896) (119896 = 1 2 119870)













119891 (119911) =infin

sum119899=0

119886119899119911119899 (F1)


119899be a




sum119899=0

119886119899119908119899119911119899 (F2)



0 1199031 119903

119899


119891119875minus119885

(119911) =infin

sum119899=0

119903119899119911119899 (F3)


suminfin



119894isin [0 1] [48]


following form [22]




process



Acknowledgments


References

























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










00

05

10


Re z

minus10

minus05

Im z

(a)

6

4

2

0

3025201505 1000

Exact120579

minus2

minus4

M = 64

M = 32

fJa

c(120579

)

(b)




00

05

10

10minus10

minus10

minus05

minus05

00 05

Re z

Im z

(a)

6

4

2

0

Exact

20151005

00

120579

minus2

M = 21

M = 34

M = 55

fFi

b(120579

)

(b)


9= 55 119873 = 119865

10= 89 and

119873 = 11986511



119891noise1 (119911) =infin

sum119899=0

120598119903119899119911119899 (21)






minus10

minus05

00

05

10

Re z

Im z

(a)

2

1

0

654321

Exact120579

minus1

minus2

fno

ise(120579

)

M = 60

(b)




119899


119899






020015010005000

50403020100n

c n

(a)

04

50403020100n

minus4

anb

n

an

bn

(b)





[100]

noise1(119911)





119899=0119886119899119911119899 and 119891noise2(119911) = sum

infin

119899=0120598119899119911119899 The








=infin

sum119899=0

(1

2119899+ 120598119899) 119911119899

(23)


Re z

Im z 0

1

2

3

0 1 2 3

Unit circle

minus3

minus3

minus2

minus2

minus1

minus1

Zeros 120576 = 0

Poles 120576 = 0

Poles 120576 = 001Zeros 120576 = 001



119891pole2 (119911) =2

(2 minus 119911)(24)





119891branch1 (119911) = radic3 + 119911

1 + 119911(25)

0

1

2

3

0 1 2 3

minus3

minus3

minus2

minus2

minus1

minus1

Re z

Zeros 120576 = 001

Poles 120576 = 001Im


Poles 120576 = 0




119891branch2 (119911) = log(65minus 119911) (26)






10

05

00

6420

Unit circle

minus05

minus10

Re z

Poles eps = 0

Poles eps = 001

Zeros eps = 001

Im z

Zeros eps = 0







00

05

10

0500 10

Unit circle

minus10

minus10

minus05

minus05

Re z

Im z

eps = 0

eps = 0

eps = 0001

eps = 0001

eps = 001

eps = 001




119896


119876119872(119911)

119875119872(119911)

=119872

sum119896

119860119896

119911 minus 119911119896

(28)


119860119896=

119876119872(119911119896)

prod119872

119895( =119896)(119911119896minus 119911119895) (29)






706050403020100

M = 50

M = 75

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

10minus7

10minus8

10minus9

10minus10

10minus11

10minus12

10minus13

|Ak|

k












00051015

6420

minus15

minus10

minus05

Re z

Im z

(a)

15105k

101

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

10minus7

10minus8

10minus9

10minus10

|Ak|

M = 10

M = 20

(b)



119896| of the [10 |






40302010

0 001 01

k

101

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

|Ak|

M = 50

(a)

605040302010k

M = 50

M = 32

M = 64

102

101

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

|Ak|

(b)





0 1199041 119904



119885 (119911) =119873

sum119899=0

119904119899119911119899 (30)



119904119896= sumℓ

119860ℓ119890119894120596ℓ(119896119873)119879 119896 = 0 1 119873 minus 1 (31)


ℓ=

2120587119891ℓ+119894120572ℓ Here119891



119885 (119911) =infin

sum119899=0

119904119899119911119899

=infin

sum119899=0

sumℓ

119860ℓ119890119894120596ℓ(119899119873)119879119911119899

= sumℓ

119860ℓ

1 minus 119911119911ℓ

(32)






0 1198781 119878



sum119873







119891Gammel (119911) =infin

sum119896=0

119861119896

1 minus 119908119896119911 (33)

where119908119896= 1198901198942120587119883119896 and 119883


119883119896isin [0 1] and 119861







10038161003816100381610038161198611198961003816100381610038161003816 lt 119862119890minus119896

1+119888

119888 gt 0 (34)









1003816100381610038161003816 sim exp (minus120573119896) (35)




0001

001

01

1

40302010

|Ak|

k



119896| are


0001

001

01

1

40302010

|Ak|

k

M = 15

M = 25

M = 45



noise2 (119911) and 119891[15|15]



descending order









Appendices



1198891199112 + 119890119911 + 119891

1198861199112 + 119887119911 + 119888=infin

sum119899=0

119886119899119911119899 (A1)




1198891199112 + 119890119911 + 119891 = (1198871198860119911 + 119888119886

0+ 1198881198861119911)

+infin

sum119899=2

(119886119886119899minus2

+ 119887119886119899minus1

+ 119888119886119899) 119911119899

(A2)


119886119896= minus

119887

119888119886119896minus1

minus119886

119888119886119896minus2

119896 ge 2 (A3)

with 1198881198860= 119891 119887119886

0+ 1198881198861= 119890 and 119886119886

0+ 1198871198861+ 1198881198862= 119889


1=

1 and 119886119896= 119886119896minus1

+ 119886119896minus2

B Random Polynomial



119891 (119911) =119873

sum119899=0

119886119899119911119899 (B1)


0119886119873




119871119873(119891) = log(

sum119873

119899=0

10038161003816100381610038161198861198991003816100381610038161003816

radic100381610038161003816100381611988601198861198731003816100381610038161003816

) (B2)







119891 (119911) =infin

sum119899=0

119886119899119911119899 (B3)

is

119870(119911119896) =

1

120587

1

(1 minus1003816100381610038161003816119911119896

10038161003816100381610038162

)2 (B4)



119891 (119911) =infin

sum]=0119886]119911119887] 119887 isin 119873 119887 = 1 (C1)




119891 (119911) =infin

sum]=0119886]119911120582] (C2)



119891 (119911) =infin

sum]=0119886]119911120582] (C3)


lim]rarrinfin

120582]

]= infin (C4)



119865119873


119860119873Jac (119911) = 119911 + 21199112

+ 2119873minus1

sum119899=2

119911119867119899 (119911 + 1199112 +119899minus2

sum119896=1

119911119867119896+2) (D1)

where119867119899= 2119899minus1


119860119865119873

Fib (119911) = 119878119873minus4

(119911)

+ [119878119873minus8

(119911) + 119911] (119891119873minus4

(119911) minus 119891119873minus2

(119911))

+ [2119891119873minus3

(119911) + 2119891119873minus2

(119911) + 119891119873minus3

(119911) 119891119873minus6

(119911)]

(D2)

where 119878119871(119911) = sum

119871

119896=0119891119896(119911) 119891

119896(119911) = 119911119865119896 119865

119873means 119873th





5040302010

10minus1

10minus2

10minus3

10minus4

|Ak|

k




119896 as follows

119891Carleman (119911) =119870

sum119896=1

(1

1 minus 119911119896119911+

1


= 2infin

sum119899=0

119870

sum119896=1

119890minus119896 cos (2120587119883119896119899) 119911119899 (E2)


119896) (119896 = 1 2 119870)













119891 (119911) =infin

sum119899=0

119886119899119911119899 (F1)


119899be a




sum119899=0

119886119899119908119899119911119899 (F2)



0 1199031 119903

119899


119891119875minus119885

(119911) =infin

sum119899=0

119903119899119911119899 (F3)


suminfin



119894isin [0 1] [48]


following form [22]




process



Acknowledgments


References

























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











minus10

minus05

00

05

10

Re z

Im z

(a)

2

1

0

654321

Exact120579

minus1

minus2

fno

ise(120579

)

M = 60

(b)




119899


119899






020015010005000

50403020100n

c n

(a)

04

50403020100n

minus4

anb

n

an

bn

(b)





[100]

noise1(119911)





119899=0119886119899119911119899 and 119891noise2(119911) = sum

infin

119899=0120598119899119911119899 The








=infin

sum119899=0

(1

2119899+ 120598119899) 119911119899

(23)


Re z

Im z 0

1

2

3

0 1 2 3

Unit circle

minus3

minus3

minus2

minus2

minus1

minus1

Zeros 120576 = 0

Poles 120576 = 0

Poles 120576 = 001Zeros 120576 = 001



119891pole2 (119911) =2

(2 minus 119911)(24)





119891branch1 (119911) = radic3 + 119911

1 + 119911(25)

0

1

2

3

0 1 2 3

minus3

minus3

minus2

minus2

minus1

minus1

Re z

Zeros 120576 = 001

Poles 120576 = 001Im


Poles 120576 = 0




119891branch2 (119911) = log(65minus 119911) (26)






10

05

00

6420

Unit circle

minus05

minus10

Re z

Poles eps = 0

Poles eps = 001

Zeros eps = 001

Im z

Zeros eps = 0







00

05

10

0500 10

Unit circle

minus10

minus10

minus05

minus05

Re z

Im z

eps = 0

eps = 0

eps = 0001

eps = 0001

eps = 001

eps = 001




119896


119876119872(119911)

119875119872(119911)

=119872

sum119896

119860119896

119911 minus 119911119896

(28)


119860119896=

119876119872(119911119896)

prod119872

119895( =119896)(119911119896minus 119911119895) (29)






706050403020100

M = 50

M = 75

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

10minus7

10minus8

10minus9

10minus10

10minus11

10minus12

10minus13

|Ak|

k












00051015

6420

minus15

minus10

minus05

Re z

Im z

(a)

15105k

101

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

10minus7

10minus8

10minus9

10minus10

|Ak|

M = 10

M = 20

(b)



119896| of the [10 |






40302010

0 001 01

k

101

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

|Ak|

M = 50

(a)

605040302010k

M = 50

M = 32

M = 64

102

101

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

|Ak|

(b)





0 1199041 119904



119885 (119911) =119873

sum119899=0

119904119899119911119899 (30)



119904119896= sumℓ

119860ℓ119890119894120596ℓ(119896119873)119879 119896 = 0 1 119873 minus 1 (31)


ℓ=

2120587119891ℓ+119894120572ℓ Here119891



119885 (119911) =infin

sum119899=0

119904119899119911119899

=infin

sum119899=0

sumℓ

119860ℓ119890119894120596ℓ(119899119873)119879119911119899

= sumℓ

119860ℓ

1 minus 119911119911ℓ

(32)






0 1198781 119878



sum119873







119891Gammel (119911) =infin

sum119896=0

119861119896

1 minus 119908119896119911 (33)

where119908119896= 1198901198942120587119883119896 and 119883


119883119896isin [0 1] and 119861







10038161003816100381610038161198611198961003816100381610038161003816 lt 119862119890minus119896

1+119888

119888 gt 0 (34)









1003816100381610038161003816 sim exp (minus120573119896) (35)




0001

001

01

1

40302010

|Ak|

k



119896| are


0001

001

01

1

40302010

|Ak|

k

M = 15

M = 25

M = 45



noise2 (119911) and 119891[15|15]



descending order









Appendices



1198891199112 + 119890119911 + 119891

1198861199112 + 119887119911 + 119888=infin

sum119899=0

119886119899119911119899 (A1)




1198891199112 + 119890119911 + 119891 = (1198871198860119911 + 119888119886

0+ 1198881198861119911)

+infin

sum119899=2

(119886119886119899minus2

+ 119887119886119899minus1

+ 119888119886119899) 119911119899

(A2)


119886119896= minus

119887

119888119886119896minus1

minus119886

119888119886119896minus2

119896 ge 2 (A3)

with 1198881198860= 119891 119887119886

0+ 1198881198861= 119890 and 119886119886

0+ 1198871198861+ 1198881198862= 119889


1=

1 and 119886119896= 119886119896minus1

+ 119886119896minus2

B Random Polynomial



119891 (119911) =119873

sum119899=0

119886119899119911119899 (B1)


0119886119873




119871119873(119891) = log(

sum119873

119899=0

10038161003816100381610038161198861198991003816100381610038161003816

radic100381610038161003816100381611988601198861198731003816100381610038161003816

) (B2)







119891 (119911) =infin

sum119899=0

119886119899119911119899 (B3)

is

119870(119911119896) =

1

120587

1

(1 minus1003816100381610038161003816119911119896

10038161003816100381610038162

)2 (B4)



119891 (119911) =infin

sum]=0119886]119911119887] 119887 isin 119873 119887 = 1 (C1)




119891 (119911) =infin

sum]=0119886]119911120582] (C2)



119891 (119911) =infin

sum]=0119886]119911120582] (C3)


lim]rarrinfin

120582]

]= infin (C4)



119865119873


119860119873Jac (119911) = 119911 + 21199112

+ 2119873minus1

sum119899=2

119911119867119899 (119911 + 1199112 +119899minus2

sum119896=1

119911119867119896+2) (D1)

where119867119899= 2119899minus1


119860119865119873

Fib (119911) = 119878119873minus4

(119911)

+ [119878119873minus8

(119911) + 119911] (119891119873minus4

(119911) minus 119891119873minus2

(119911))

+ [2119891119873minus3

(119911) + 2119891119873minus2

(119911) + 119891119873minus3

(119911) 119891119873minus6

(119911)]

(D2)

where 119878119871(119911) = sum

119871

119896=0119891119896(119911) 119891

119896(119911) = 119911119865119896 119865

119873means 119873th





5040302010

10minus1

10minus2

10minus3

10minus4

|Ak|

k




119896 as follows

119891Carleman (119911) =119870

sum119896=1

(1

1 minus 119911119896119911+

1


= 2infin

sum119899=0

119870

sum119896=1

119890minus119896 cos (2120587119883119896119899) 119911119899 (E2)


119896) (119896 = 1 2 119870)













119891 (119911) =infin

sum119899=0

119886119899119911119899 (F1)


119899be a




sum119899=0

119886119899119908119899119911119899 (F2)



0 1199031 119903

119899


119891119875minus119885

(119911) =infin

sum119899=0

119903119899119911119899 (F3)


suminfin



119894isin [0 1] [48]


following form [22]




process



Acknowledgments


References

























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










Re z

Im z 0

1

2

3

0 1 2 3

Unit circle

minus3

minus3

minus2

minus2

minus1

minus1

Zeros 120576 = 0

Poles 120576 = 0

Poles 120576 = 001Zeros 120576 = 001



119891pole2 (119911) =2

(2 minus 119911)(24)





119891branch1 (119911) = radic3 + 119911

1 + 119911(25)

0

1

2

3

0 1 2 3

minus3

minus3

minus2

minus2

minus1

minus1

Re z

Zeros 120576 = 001

Poles 120576 = 001Im


Poles 120576 = 0




119891branch2 (119911) = log(65minus 119911) (26)






10

05

00

6420

Unit circle

minus05

minus10

Re z

Poles eps = 0

Poles eps = 001

Zeros eps = 001

Im z

Zeros eps = 0







00

05

10

0500 10

Unit circle

minus10

minus10

minus05

minus05

Re z

Im z

eps = 0

eps = 0

eps = 0001

eps = 0001

eps = 001

eps = 001




119896


119876119872(119911)

119875119872(119911)

=119872

sum119896

119860119896

119911 minus 119911119896

(28)


119860119896=

119876119872(119911119896)

prod119872

119895( =119896)(119911119896minus 119911119895) (29)






706050403020100

M = 50

M = 75

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

10minus7

10minus8

10minus9

10minus10

10minus11

10minus12

10minus13

|Ak|

k












00051015

6420

minus15

minus10

minus05

Re z

Im z

(a)

15105k

101

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

10minus7

10minus8

10minus9

10minus10

|Ak|

M = 10

M = 20

(b)



119896| of the [10 |






40302010

0 001 01

k

101

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

|Ak|

M = 50

(a)

605040302010k

M = 50

M = 32

M = 64

102

101

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

|Ak|

(b)





0 1199041 119904



119885 (119911) =119873

sum119899=0

119904119899119911119899 (30)



119904119896= sumℓ

119860ℓ119890119894120596ℓ(119896119873)119879 119896 = 0 1 119873 minus 1 (31)


ℓ=

2120587119891ℓ+119894120572ℓ Here119891



119885 (119911) =infin

sum119899=0

119904119899119911119899

=infin

sum119899=0

sumℓ

119860ℓ119890119894120596ℓ(119899119873)119879119911119899

= sumℓ

119860ℓ

1 minus 119911119911ℓ

(32)






0 1198781 119878



sum119873







119891Gammel (119911) =infin

sum119896=0

119861119896

1 minus 119908119896119911 (33)

where119908119896= 1198901198942120587119883119896 and 119883


119883119896isin [0 1] and 119861







10038161003816100381610038161198611198961003816100381610038161003816 lt 119862119890minus119896

1+119888

119888 gt 0 (34)









1003816100381610038161003816 sim exp (minus120573119896) (35)




0001

001

01

1

40302010

|Ak|

k



119896| are


0001

001

01

1

40302010

|Ak|

k

M = 15

M = 25

M = 45



noise2 (119911) and 119891[15|15]



descending order









Appendices



1198891199112 + 119890119911 + 119891

1198861199112 + 119887119911 + 119888=infin

sum119899=0

119886119899119911119899 (A1)




1198891199112 + 119890119911 + 119891 = (1198871198860119911 + 119888119886

0+ 1198881198861119911)

+infin

sum119899=2

(119886119886119899minus2

+ 119887119886119899minus1

+ 119888119886119899) 119911119899

(A2)


119886119896= minus

119887

119888119886119896minus1

minus119886

119888119886119896minus2

119896 ge 2 (A3)

with 1198881198860= 119891 119887119886

0+ 1198881198861= 119890 and 119886119886

0+ 1198871198861+ 1198881198862= 119889


1=

1 and 119886119896= 119886119896minus1

+ 119886119896minus2

B Random Polynomial



119891 (119911) =119873

sum119899=0

119886119899119911119899 (B1)


0119886119873




119871119873(119891) = log(

sum119873

119899=0

10038161003816100381610038161198861198991003816100381610038161003816

radic100381610038161003816100381611988601198861198731003816100381610038161003816

) (B2)







119891 (119911) =infin

sum119899=0

119886119899119911119899 (B3)

is

119870(119911119896) =

1

120587

1

(1 minus1003816100381610038161003816119911119896

10038161003816100381610038162

)2 (B4)



119891 (119911) =infin

sum]=0119886]119911119887] 119887 isin 119873 119887 = 1 (C1)




119891 (119911) =infin

sum]=0119886]119911120582] (C2)



119891 (119911) =infin

sum]=0119886]119911120582] (C3)


lim]rarrinfin

120582]

]= infin (C4)



119865119873


119860119873Jac (119911) = 119911 + 21199112

+ 2119873minus1

sum119899=2

119911119867119899 (119911 + 1199112 +119899minus2

sum119896=1

119911119867119896+2) (D1)

where119867119899= 2119899minus1


119860119865119873

Fib (119911) = 119878119873minus4

(119911)

+ [119878119873minus8

(119911) + 119911] (119891119873minus4

(119911) minus 119891119873minus2

(119911))

+ [2119891119873minus3

(119911) + 2119891119873minus2

(119911) + 119891119873minus3

(119911) 119891119873minus6

(119911)]

(D2)

where 119878119871(119911) = sum

119871

119896=0119891119896(119911) 119891

119896(119911) = 119911119865119896 119865

119873means 119873th





5040302010

10minus1

10minus2

10minus3

10minus4

|Ak|

k




119896 as follows

119891Carleman (119911) =119870

sum119896=1

(1

1 minus 119911119896119911+

1


= 2infin

sum119899=0

119870

sum119896=1

119890minus119896 cos (2120587119883119896119899) 119911119899 (E2)


119896) (119896 = 1 2 119870)













119891 (119911) =infin

sum119899=0

119886119899119911119899 (F1)


119899be a




sum119899=0

119886119899119908119899119911119899 (F2)



0 1199031 119903

119899


119891119875minus119885

(119911) =infin

sum119899=0

119903119899119911119899 (F3)


suminfin



119894isin [0 1] [48]


following form [22]




process



Acknowledgments


References

























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










10

05

00

6420

Unit circle

minus05

minus10

Re z

Poles eps = 0

Poles eps = 001

Zeros eps = 001

Im z

Zeros eps = 0







00

05

10

0500 10

Unit circle

minus10

minus10

minus05

minus05

Re z

Im z

eps = 0

eps = 0

eps = 0001

eps = 0001

eps = 001

eps = 001




119896


119876119872(119911)

119875119872(119911)

=119872

sum119896

119860119896

119911 minus 119911119896

(28)


119860119896=

119876119872(119911119896)

prod119872

119895( =119896)(119911119896minus 119911119895) (29)






706050403020100

M = 50

M = 75

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

10minus7

10minus8

10minus9

10minus10

10minus11

10minus12

10minus13

|Ak|

k












00051015

6420

minus15

minus10

minus05

Re z

Im z

(a)

15105k

101

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

10minus7

10minus8

10minus9

10minus10

|Ak|

M = 10

M = 20

(b)



119896| of the [10 |






40302010

0 001 01

k

101

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

|Ak|

M = 50

(a)

605040302010k

M = 50

M = 32

M = 64

102

101

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

|Ak|

(b)





0 1199041 119904



119885 (119911) =119873

sum119899=0

119904119899119911119899 (30)



119904119896= sumℓ

119860ℓ119890119894120596ℓ(119896119873)119879 119896 = 0 1 119873 minus 1 (31)


ℓ=

2120587119891ℓ+119894120572ℓ Here119891



119885 (119911) =infin

sum119899=0

119904119899119911119899

=infin

sum119899=0

sumℓ

119860ℓ119890119894120596ℓ(119899119873)119879119911119899

= sumℓ

119860ℓ

1 minus 119911119911ℓ

(32)






0 1198781 119878



sum119873







119891Gammel (119911) =infin

sum119896=0

119861119896

1 minus 119908119896119911 (33)

where119908119896= 1198901198942120587119883119896 and 119883


119883119896isin [0 1] and 119861







10038161003816100381610038161198611198961003816100381610038161003816 lt 119862119890minus119896

1+119888

119888 gt 0 (34)









1003816100381610038161003816 sim exp (minus120573119896) (35)




0001

001

01

1

40302010

|Ak|

k



119896| are


0001

001

01

1

40302010

|Ak|

k

M = 15

M = 25

M = 45



noise2 (119911) and 119891[15|15]



descending order









Appendices



1198891199112 + 119890119911 + 119891

1198861199112 + 119887119911 + 119888=infin

sum119899=0

119886119899119911119899 (A1)




1198891199112 + 119890119911 + 119891 = (1198871198860119911 + 119888119886

0+ 1198881198861119911)

+infin

sum119899=2

(119886119886119899minus2

+ 119887119886119899minus1

+ 119888119886119899) 119911119899

(A2)


119886119896= minus

119887

119888119886119896minus1

minus119886

119888119886119896minus2

119896 ge 2 (A3)

with 1198881198860= 119891 119887119886

0+ 1198881198861= 119890 and 119886119886

0+ 1198871198861+ 1198881198862= 119889


1=

1 and 119886119896= 119886119896minus1

+ 119886119896minus2

B Random Polynomial



119891 (119911) =119873

sum119899=0

119886119899119911119899 (B1)


0119886119873




119871119873(119891) = log(

sum119873

119899=0

10038161003816100381610038161198861198991003816100381610038161003816

radic100381610038161003816100381611988601198861198731003816100381610038161003816

) (B2)







119891 (119911) =infin

sum119899=0

119886119899119911119899 (B3)

is

119870(119911119896) =

1

120587

1

(1 minus1003816100381610038161003816119911119896

10038161003816100381610038162

)2 (B4)



119891 (119911) =infin

sum]=0119886]119911119887] 119887 isin 119873 119887 = 1 (C1)




119891 (119911) =infin

sum]=0119886]119911120582] (C2)



119891 (119911) =infin

sum]=0119886]119911120582] (C3)


lim]rarrinfin

120582]

]= infin (C4)



119865119873


119860119873Jac (119911) = 119911 + 21199112

+ 2119873minus1

sum119899=2

119911119867119899 (119911 + 1199112 +119899minus2

sum119896=1

119911119867119896+2) (D1)

where119867119899= 2119899minus1


119860119865119873

Fib (119911) = 119878119873minus4

(119911)

+ [119878119873minus8

(119911) + 119911] (119891119873minus4

(119911) minus 119891119873minus2

(119911))

+ [2119891119873minus3

(119911) + 2119891119873minus2

(119911) + 119891119873minus3

(119911) 119891119873minus6

(119911)]

(D2)

where 119878119871(119911) = sum

119871

119896=0119891119896(119911) 119891

119896(119911) = 119911119865119896 119865

119873means 119873th





5040302010

10minus1

10minus2

10minus3

10minus4

|Ak|

k




119896 as follows

119891Carleman (119911) =119870

sum119896=1

(1

1 minus 119911119896119911+

1


= 2infin

sum119899=0

119870

sum119896=1

119890minus119896 cos (2120587119883119896119899) 119911119899 (E2)


119896) (119896 = 1 2 119870)













119891 (119911) =infin

sum119899=0

119886119899119911119899 (F1)


119899be a




sum119899=0

119886119899119908119899119911119899 (F2)



0 1199031 119903

119899


119891119875minus119885

(119911) =infin

sum119899=0

119903119899119911119899 (F3)


suminfin



119894isin [0 1] [48]


following form [22]




process



Acknowledgments


References

























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










706050403020100

M = 50

M = 75

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

10minus7

10minus8

10minus9

10minus10

10minus11

10minus12

10minus13

|Ak|

k












00051015

6420

minus15

minus10

minus05

Re z

Im z

(a)

15105k

101

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

10minus7

10minus8

10minus9

10minus10

|Ak|

M = 10

M = 20

(b)



119896| of the [10 |






40302010

0 001 01

k

101

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

|Ak|

M = 50

(a)

605040302010k

M = 50

M = 32

M = 64

102

101

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

|Ak|

(b)





0 1199041 119904



119885 (119911) =119873

sum119899=0

119904119899119911119899 (30)



119904119896= sumℓ

119860ℓ119890119894120596ℓ(119896119873)119879 119896 = 0 1 119873 minus 1 (31)


ℓ=

2120587119891ℓ+119894120572ℓ Here119891



119885 (119911) =infin

sum119899=0

119904119899119911119899

=infin

sum119899=0

sumℓ

119860ℓ119890119894120596ℓ(119899119873)119879119911119899

= sumℓ

119860ℓ

1 minus 119911119911ℓ

(32)






0 1198781 119878



sum119873







119891Gammel (119911) =infin

sum119896=0

119861119896

1 minus 119908119896119911 (33)

where119908119896= 1198901198942120587119883119896 and 119883


119883119896isin [0 1] and 119861







10038161003816100381610038161198611198961003816100381610038161003816 lt 119862119890minus119896

1+119888

119888 gt 0 (34)









1003816100381610038161003816 sim exp (minus120573119896) (35)




0001

001

01

1

40302010

|Ak|

k



119896| are


0001

001

01

1

40302010

|Ak|

k

M = 15

M = 25

M = 45



noise2 (119911) and 119891[15|15]



descending order









Appendices



1198891199112 + 119890119911 + 119891

1198861199112 + 119887119911 + 119888=infin

sum119899=0

119886119899119911119899 (A1)




1198891199112 + 119890119911 + 119891 = (1198871198860119911 + 119888119886

0+ 1198881198861119911)

+infin

sum119899=2

(119886119886119899minus2

+ 119887119886119899minus1

+ 119888119886119899) 119911119899

(A2)


119886119896= minus

119887

119888119886119896minus1

minus119886

119888119886119896minus2

119896 ge 2 (A3)

with 1198881198860= 119891 119887119886

0+ 1198881198861= 119890 and 119886119886

0+ 1198871198861+ 1198881198862= 119889


1=

1 and 119886119896= 119886119896minus1

+ 119886119896minus2

B Random Polynomial



119891 (119911) =119873

sum119899=0

119886119899119911119899 (B1)


0119886119873




119871119873(119891) = log(

sum119873

119899=0

10038161003816100381610038161198861198991003816100381610038161003816

radic100381610038161003816100381611988601198861198731003816100381610038161003816

) (B2)







119891 (119911) =infin

sum119899=0

119886119899119911119899 (B3)

is

119870(119911119896) =

1

120587

1

(1 minus1003816100381610038161003816119911119896

10038161003816100381610038162

)2 (B4)



119891 (119911) =infin

sum]=0119886]119911119887] 119887 isin 119873 119887 = 1 (C1)




119891 (119911) =infin

sum]=0119886]119911120582] (C2)



119891 (119911) =infin

sum]=0119886]119911120582] (C3)


lim]rarrinfin

120582]

]= infin (C4)



119865119873


119860119873Jac (119911) = 119911 + 21199112

+ 2119873minus1

sum119899=2

119911119867119899 (119911 + 1199112 +119899minus2

sum119896=1

119911119867119896+2) (D1)

where119867119899= 2119899minus1


119860119865119873

Fib (119911) = 119878119873minus4

(119911)

+ [119878119873minus8

(119911) + 119911] (119891119873minus4

(119911) minus 119891119873minus2

(119911))

+ [2119891119873minus3

(119911) + 2119891119873minus2

(119911) + 119891119873minus3

(119911) 119891119873minus6

(119911)]

(D2)

where 119878119871(119911) = sum

119871

119896=0119891119896(119911) 119891

119896(119911) = 119911119865119896 119865

119873means 119873th





5040302010

10minus1

10minus2

10minus3

10minus4

|Ak|

k




119896 as follows

119891Carleman (119911) =119870

sum119896=1

(1

1 minus 119911119896119911+

1


= 2infin

sum119899=0

119870

sum119896=1

119890minus119896 cos (2120587119883119896119899) 119911119899 (E2)


119896) (119896 = 1 2 119870)













119891 (119911) =infin

sum119899=0

119886119899119911119899 (F1)


119899be a




sum119899=0

119886119899119908119899119911119899 (F2)



0 1199031 119903

119899


119891119875minus119885

(119911) =infin

sum119899=0

119903119899119911119899 (F3)


suminfin



119894isin [0 1] [48]


following form [22]




process



Acknowledgments


References

























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










40302010

0 001 01

k

101

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

|Ak|

M = 50

(a)

605040302010k

M = 50

M = 32

M = 64

102

101

100

10minus1

10minus2

10minus3

10minus4

10minus5

10minus6

|Ak|

(b)





0 1199041 119904



119885 (119911) =119873

sum119899=0

119904119899119911119899 (30)



119904119896= sumℓ

119860ℓ119890119894120596ℓ(119896119873)119879 119896 = 0 1 119873 minus 1 (31)


ℓ=

2120587119891ℓ+119894120572ℓ Here119891



119885 (119911) =infin

sum119899=0

119904119899119911119899

=infin

sum119899=0

sumℓ

119860ℓ119890119894120596ℓ(119899119873)119879119911119899

= sumℓ

119860ℓ

1 minus 119911119911ℓ

(32)






0 1198781 119878



sum119873







119891Gammel (119911) =infin

sum119896=0

119861119896

1 minus 119908119896119911 (33)

where119908119896= 1198901198942120587119883119896 and 119883


119883119896isin [0 1] and 119861







10038161003816100381610038161198611198961003816100381610038161003816 lt 119862119890minus119896

1+119888

119888 gt 0 (34)









1003816100381610038161003816 sim exp (minus120573119896) (35)




0001

001

01

1

40302010

|Ak|

k



119896| are


0001

001

01

1

40302010

|Ak|

k

M = 15

M = 25

M = 45



noise2 (119911) and 119891[15|15]



descending order









Appendices



1198891199112 + 119890119911 + 119891

1198861199112 + 119887119911 + 119888=infin

sum119899=0

119886119899119911119899 (A1)




1198891199112 + 119890119911 + 119891 = (1198871198860119911 + 119888119886

0+ 1198881198861119911)

+infin

sum119899=2

(119886119886119899minus2

+ 119887119886119899minus1

+ 119888119886119899) 119911119899

(A2)


119886119896= minus

119887

119888119886119896minus1

minus119886

119888119886119896minus2

119896 ge 2 (A3)

with 1198881198860= 119891 119887119886

0+ 1198881198861= 119890 and 119886119886

0+ 1198871198861+ 1198881198862= 119889


1=

1 and 119886119896= 119886119896minus1

+ 119886119896minus2

B Random Polynomial



119891 (119911) =119873

sum119899=0

119886119899119911119899 (B1)


0119886119873




119871119873(119891) = log(

sum119873

119899=0

10038161003816100381610038161198861198991003816100381610038161003816

radic100381610038161003816100381611988601198861198731003816100381610038161003816

) (B2)







119891 (119911) =infin

sum119899=0

119886119899119911119899 (B3)

is

119870(119911119896) =

1

120587

1

(1 minus1003816100381610038161003816119911119896

10038161003816100381610038162

)2 (B4)



119891 (119911) =infin

sum]=0119886]119911119887] 119887 isin 119873 119887 = 1 (C1)




119891 (119911) =infin

sum]=0119886]119911120582] (C2)



119891 (119911) =infin

sum]=0119886]119911120582] (C3)


lim]rarrinfin

120582]

]= infin (C4)



119865119873


119860119873Jac (119911) = 119911 + 21199112

+ 2119873minus1

sum119899=2

119911119867119899 (119911 + 1199112 +119899minus2

sum119896=1

119911119867119896+2) (D1)

where119867119899= 2119899minus1


119860119865119873

Fib (119911) = 119878119873minus4

(119911)

+ [119878119873minus8

(119911) + 119911] (119891119873minus4

(119911) minus 119891119873minus2

(119911))

+ [2119891119873minus3

(119911) + 2119891119873minus2

(119911) + 119891119873minus3

(119911) 119891119873minus6

(119911)]

(D2)

where 119878119871(119911) = sum

119871

119896=0119891119896(119911) 119891

119896(119911) = 119911119865119896 119865

119873means 119873th





5040302010

10minus1

10minus2

10minus3

10minus4

|Ak|

k




119896 as follows

119891Carleman (119911) =119870

sum119896=1

(1

1 minus 119911119896119911+

1


= 2infin

sum119899=0

119870

sum119896=1

119890minus119896 cos (2120587119883119896119899) 119911119899 (E2)


119896) (119896 = 1 2 119870)













119891 (119911) =infin

sum119899=0

119886119899119911119899 (F1)


119899be a




sum119899=0

119886119899119908119899119911119899 (F2)



0 1199031 119903

119899


119891119875minus119885

(119911) =infin

sum119899=0

119903119899119911119899 (F3)


suminfin



119894isin [0 1] [48]


following form [22]




process



Acknowledgments


References

























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra












119891Gammel (119911) =infin

sum119896=0

119861119896

1 minus 119908119896119911 (33)

where119908119896= 1198901198942120587119883119896 and 119883


119883119896isin [0 1] and 119861







10038161003816100381610038161198611198961003816100381610038161003816 lt 119862119890minus119896

1+119888

119888 gt 0 (34)









1003816100381610038161003816 sim exp (minus120573119896) (35)




0001

001

01

1

40302010

|Ak|

k



119896| are


0001

001

01

1

40302010

|Ak|

k

M = 15

M = 25

M = 45



noise2 (119911) and 119891[15|15]



descending order









Appendices



1198891199112 + 119890119911 + 119891

1198861199112 + 119887119911 + 119888=infin

sum119899=0

119886119899119911119899 (A1)




1198891199112 + 119890119911 + 119891 = (1198871198860119911 + 119888119886

0+ 1198881198861119911)

+infin

sum119899=2

(119886119886119899minus2

+ 119887119886119899minus1

+ 119888119886119899) 119911119899

(A2)


119886119896= minus

119887

119888119886119896minus1

minus119886

119888119886119896minus2

119896 ge 2 (A3)

with 1198881198860= 119891 119887119886

0+ 1198881198861= 119890 and 119886119886

0+ 1198871198861+ 1198881198862= 119889


1=

1 and 119886119896= 119886119896minus1

+ 119886119896minus2

B Random Polynomial



119891 (119911) =119873

sum119899=0

119886119899119911119899 (B1)


0119886119873




119871119873(119891) = log(

sum119873

119899=0

10038161003816100381610038161198861198991003816100381610038161003816

radic100381610038161003816100381611988601198861198731003816100381610038161003816

) (B2)







119891 (119911) =infin

sum119899=0

119886119899119911119899 (B3)

is

119870(119911119896) =

1

120587

1

(1 minus1003816100381610038161003816119911119896

10038161003816100381610038162

)2 (B4)



119891 (119911) =infin

sum]=0119886]119911119887] 119887 isin 119873 119887 = 1 (C1)




119891 (119911) =infin

sum]=0119886]119911120582] (C2)



119891 (119911) =infin

sum]=0119886]119911120582] (C3)


lim]rarrinfin

120582]

]= infin (C4)



119865119873


119860119873Jac (119911) = 119911 + 21199112

+ 2119873minus1

sum119899=2

119911119867119899 (119911 + 1199112 +119899minus2

sum119896=1

119911119867119896+2) (D1)

where119867119899= 2119899minus1


119860119865119873

Fib (119911) = 119878119873minus4

(119911)

+ [119878119873minus8

(119911) + 119911] (119891119873minus4

(119911) minus 119891119873minus2

(119911))

+ [2119891119873minus3

(119911) + 2119891119873minus2

(119911) + 119891119873minus3

(119911) 119891119873minus6

(119911)]

(D2)

where 119878119871(119911) = sum

119871

119896=0119891119896(119911) 119891

119896(119911) = 119911119865119896 119865

119873means 119873th





5040302010

10minus1

10minus2

10minus3

10minus4

|Ak|

k




119896 as follows

119891Carleman (119911) =119870

sum119896=1

(1

1 minus 119911119896119911+

1


= 2infin

sum119899=0

119870

sum119896=1

119890minus119896 cos (2120587119883119896119899) 119911119899 (E2)


119896) (119896 = 1 2 119870)













119891 (119911) =infin

sum119899=0

119886119899119911119899 (F1)


119899be a




sum119899=0

119886119899119908119899119911119899 (F2)



0 1199031 119903

119899


119891119875minus119885

(119911) =infin

sum119899=0

119903119899119911119899 (F3)


suminfin



119894isin [0 1] [48]


following form [22]




process



Acknowledgments


References

























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra















Appendices



1198891199112 + 119890119911 + 119891

1198861199112 + 119887119911 + 119888=infin

sum119899=0

119886119899119911119899 (A1)




1198891199112 + 119890119911 + 119891 = (1198871198860119911 + 119888119886

0+ 1198881198861119911)

+infin

sum119899=2

(119886119886119899minus2

+ 119887119886119899minus1

+ 119888119886119899) 119911119899

(A2)


119886119896= minus

119887

119888119886119896minus1

minus119886

119888119886119896minus2

119896 ge 2 (A3)

with 1198881198860= 119891 119887119886

0+ 1198881198861= 119890 and 119886119886

0+ 1198871198861+ 1198881198862= 119889


1=

1 and 119886119896= 119886119896minus1

+ 119886119896minus2

B Random Polynomial



119891 (119911) =119873

sum119899=0

119886119899119911119899 (B1)


0119886119873




119871119873(119891) = log(

sum119873

119899=0

10038161003816100381610038161198861198991003816100381610038161003816

radic100381610038161003816100381611988601198861198731003816100381610038161003816

) (B2)







119891 (119911) =infin

sum119899=0

119886119899119911119899 (B3)

is

119870(119911119896) =

1

120587

1

(1 minus1003816100381610038161003816119911119896

10038161003816100381610038162

)2 (B4)



119891 (119911) =infin

sum]=0119886]119911119887] 119887 isin 119873 119887 = 1 (C1)




119891 (119911) =infin

sum]=0119886]119911120582] (C2)



119891 (119911) =infin

sum]=0119886]119911120582] (C3)


lim]rarrinfin

120582]

]= infin (C4)



119865119873


119860119873Jac (119911) = 119911 + 21199112

+ 2119873minus1

sum119899=2

119911119867119899 (119911 + 1199112 +119899minus2

sum119896=1

119911119867119896+2) (D1)

where119867119899= 2119899minus1


119860119865119873

Fib (119911) = 119878119873minus4

(119911)

+ [119878119873minus8

(119911) + 119911] (119891119873minus4

(119911) minus 119891119873minus2

(119911))

+ [2119891119873minus3

(119911) + 2119891119873minus2

(119911) + 119891119873minus3

(119911) 119891119873minus6

(119911)]

(D2)

where 119878119871(119911) = sum

119871

119896=0119891119896(119911) 119891

119896(119911) = 119911119865119896 119865

119873means 119873th





5040302010

10minus1

10minus2

10minus3

10minus4

|Ak|

k




119896 as follows

119891Carleman (119911) =119870

sum119896=1

(1

1 minus 119911119896119911+

1


= 2infin

sum119899=0

119870

sum119896=1

119890minus119896 cos (2120587119883119896119899) 119911119899 (E2)


119896) (119896 = 1 2 119870)













119891 (119911) =infin

sum119899=0

119886119899119911119899 (F1)


119899be a




sum119899=0

119886119899119908119899119911119899 (F2)



0 1199031 119903

119899


119891119875minus119885

(119911) =infin

sum119899=0

119903119899119911119899 (F3)


suminfin



119894isin [0 1] [48]


following form [22]




process



Acknowledgments


References

























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119891 (119911) =infin

sum]=0119886]119911120582] (C2)



119891 (119911) =infin

sum]=0119886]119911120582] (C3)


lim]rarrinfin

120582]

]= infin (C4)



119865119873


119860119873Jac (119911) = 119911 + 21199112

+ 2119873minus1

sum119899=2

119911119867119899 (119911 + 1199112 +119899minus2

sum119896=1

119911119867119896+2) (D1)

where119867119899= 2119899minus1


119860119865119873

Fib (119911) = 119878119873minus4

(119911)

+ [119878119873minus8

(119911) + 119911] (119891119873minus4

(119911) minus 119891119873minus2

(119911))

+ [2119891119873minus3

(119911) + 2119891119873minus2

(119911) + 119891119873minus3

(119911) 119891119873minus6

(119911)]

(D2)

where 119878119871(119911) = sum

119871

119896=0119891119896(119911) 119891

119896(119911) = 119911119865119896 119865

119873means 119873th





5040302010

10minus1

10minus2

10minus3

10minus4

|Ak|

k




119896 as follows

119891Carleman (119911) =119870

sum119896=1

(1

1 minus 119911119896119911+

1


= 2infin

sum119899=0

119870

sum119896=1

119890minus119896 cos (2120587119883119896119899) 119911119899 (E2)


119896) (119896 = 1 2 119870)













119891 (119911) =infin

sum119899=0

119886119899119911119899 (F1)


119899be a




sum119899=0

119886119899119908119899119911119899 (F2)



0 1199031 119903

119899


119891119875minus119885

(119911) =infin

sum119899=0

119903119899119911119899 (F3)


suminfin



119894isin [0 1] [48]


following form [22]




process



Acknowledgments


References

























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra













119891 (119911) =infin

sum119899=0

119886119899119911119899 (F1)


119899be a




sum119899=0

119886119899119908119899119911119899 (F2)



0 1199031 119903

119899


119891119875minus119885

(119911) =infin

sum119899=0

119903119899119911119899 (F3)


suminfin



119894isin [0 1] [48]


following form [22]




process



Acknowledgments


References

























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











































Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









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