Upload
manoj302009
View
60
Download
1
Embed Size (px)
Citation preview
Functions of a Complex Variable
Dr. M K SinghAssociate Professor
Jahangirabad Institute of Technology, Barabanki
Functions of A Complex Variables I
Functions of a complex variable provide us some powerful and widely useful tools in theoretical physics.
• Some important physical quantities are complex variables (the wave-function ) • Evaluating definite integrals.
• Obtaining asymptotic solutions of differentials equations.
• Integral transforms• Many Physical quantities that were originally real become complex
as simple theory is made more general. The energy ( the finite life time).
iEE nn0
/1
We here go through the complex algebra briefly.A complex number z = (x,y) = x + iy, Where. We will see that the ordering of two real numbers (x,y) is significant, i.e. in general x + iy y + ix
X: the real part, labeled by Re(z); y: the imaginary part, labeled by Im(z)
Three frequently used representations:
(1) Cartesian representation: x+iy
(2) polar representation, we may write z=r(cos + i sin) or
r – the modulus or magnitude of z - the argument or phase of z
1i
ierz
r – the modulus or magnitude of z - the argument or phase of z
The relation between Cartesian and polar representation:
The choice of polar representation or Cartesian representation is a matter of convenience. Addition and subtraction of complex variables are easier in the Cartesian representation. Multiplication, division, powers, roots are easier to handle in polar form,
1/ 22 2
1tan /
r z x y
y x
212121
ierrzz
212121 // ierrzz innn erz
From z, complex functions f(z) may be constructed. They can be written
f(z) = u(x,y) + iv(x,y)in which v and u are real functions. For example if , we have
The relationship between z and f(z) is best pictured as a mapping operation, we address it in detail later.
)arg()arg()arg( 2121 zzzz
2121 zzzz
xyiyxzf 222
Using the polar form,
2)( zzf
Function: Mapping operation
x
y Z-plane
u
v
The function w(x,y)=u(x,y)+iv(x,y) maps points in the xy plane into pointsin the uv plane.
nin
i
ie
ie
)sin(cos
sincos
We get a not so obvious formula
Since
ninin )sin(cossincos
Complex Conjugation: replacing i by –i, which is denoted by (*),
We then have
Hence
Note:
ln z is a multi-valued function. To avoid ambiguity, we usually set n=0and limit the phase to an interval of length of 2. The value of lnz withn=0 is called the principal value of lnz.
iyxz *
222* ryxzz
21*zzz Special features: single-valued function of a real variable ---- multi-valued function
irez
nire 2
irlnzln
nirz 2lnln
Another possibility
even and 1|cos||,sin|possibly however, x;real afor 1|cos||,sin|
zzxx
Question:
yx
yx
yxiyxiyxyxiyxiyx
iee iziz
222
222
iziz
sinhcos|cosz |
sinhsin|sinz | (b)
sinhsincoshcos)cos( sinhcoscoshsin)sin( (a) show to
2
esinz ;2
ecosz
:identities theUsing
Analytic functions
If f(z) is differentiable at and in some small region around ,we say that f(z) is analytic at
Differentiable: Cauthy-Riemann conditions are satisfied the partial derivatives of u and v are continuous
Analytic function:
Property 1:
Property 2: established a relation between u and v
022 vu
Example:
0zz
0zz 0z
Cauchy-Riemann Equations
0 0 0
0 00 0
0
1 0 0
2 0 0
Let , , be diff. at
then lim exists
with In particular, can be computed along
: , i.e.
: , i.e.
z
f z u x y iv x y z x iy
f z z f zf z
zz x i y
f z
C y y x x z xC x x y y
z i y
Cauchy-Riemann Equations
0 0 0 0
0
0 0 0 0
( , ) ( , )
( , ) ( , )
u vx y i x yx xf z
u vi x y x yy y
Cauchy-Riemann Equations
• We have proved the following theorem.
u vx yu vy x
Theorem
A necessary condition for a fun. f(z)=u(x,y)+iv(x,y)to be diff. at a point z0 is that the C-R eq. hold at z0.Consequently, if f is analytic in an open set G, then the C-R eq. must hold at every point of G.
Theorem
A necessary condition for a fun. f(z)=u(x,y)+iv(x,y)to be diff. at a point z0 is that the C-R eq. hold at z0.Consequently, if f is analytic in an open set G, then the C-R eq. must hold at every point of G.
Application of Theorem
To show that a function is NOT analytic, it suffices to show that the C-R eq. are not satisfied
Cauchy – Riemann conditions
Having established complex functions, we now proceed to differentiate them. The derivative of f(z), like that of a real function, is defined by
provided that the limit is independent of the particular approach to the point z. For real variable, we require that Now, with z (or zo) some point in a plane, our requirement that thelimit be independent of the direction of approach is very restrictive.
Consider
zfdzdf
zzf
zzfzzf
zz
00limlim
oxxxx
xfxfxfoo
limlim
yixz viuf
,
yixviu
zf
Let us take limit by the two different approaches as in the figure. First, with y = 0, we let x0,
Assuming the partial derivatives exist. For a second approach, we set x = 0 and then let y 0. This leads to
If we have a derivative, the above two results must be identical. So,
xvi
xu
zf
xz
00limlim
xvi
xu
yv
yui
zf
z
0lim
yv
xu
,
xv
yu
These are the famous Cauchy-Riemann conditions. These Cauchy-Riemann conditions are necessary for the existence of a derivative, that is, if exists, the C-R conditions must hold.
Conversely, if the C-R conditions are satisfied and the partial derivatives of u(x,y) and v(x,y) are continuous, exists.
xf
zf
Cauchy’s integral Theorem
We now turn to integration. in close analogy to the integral of a real function The contour is divided into n intervals .Let with for j. Then
'00 zz 01 jjj zzz
0
01
limz
z
n
jjj
ndzzfzf
n
The right-hand side of the above equation is called the contour (path) integral of f(z)
. and
bewteen curve on thepoint a is where
, and points thechoosing of details theoft independen
is and existslimit that theprovided
1
j
j
jj
j
zz
z
As an alternative, the contour may be defined by
with the path C specified. This reduces the complex integral to the complex sum of real integrals. It’s somewhat analogous to the case of the vector integral.
An important example
22
11
2
1
,,yx
yxc
z
zc
idydxyxivyxudzzf
22
11
22
11
yx
yx
yx
yxcc
udyvdxivdyudx
c
ndzz
where C is a circle of radius r>0 around the origin z=0 in the direction of counterclockwise.
In polar coordinates, we parameterize and , and have
which is independent of r. Cauchy’s integral theorem– If a function f(z) is analytical (therefore single-valued) [and its partial
derivatives are continuous] through some simply connected region R, for every closed path C in R,
irez diredz i
2
0
11exp
221 dnirdzz
i
n
c
n
1- n for 1-1n for 0
{
0 dzzfc
•Multiply connected regionsThe original statement of our theorem demanded a simply connectedregion. This restriction may easily be relaxed by the creation of a barrier, a contour line. Consider the multiply connected region of Fig.1.6 In which f(z) is not defined for the interior R
Cauchy’s integral theorem is not valid for the contour C, but we canconstruct a C for which the theorem holds. If line segments DE and GA arbitrarily close together, then
E
D
A
G
dzzfdzzf
'2
'1 CEFGCABD
dzzfdzzfEFGGADEABD
ABDEFGAC
0dzzfEFGABD
21 CC
dzzfdzzf
Cauchy’s Integral Formula
Cauchy’s integral formula: If f(z) is analytic on and within a closed contour C then
in which z0 is some point in the interior region bounded by C. Note that here z-z0 0 and the integral is well defined.
Although f(z) is assumed analytic, the integrand (f(z)/z-z0) is notanalytic at z=z0 unless f(z0)=0. If the contour is deformed as in Fig.1.8 Cauchy’s integral theorem applies.So we have
00
2 zifzzdzzf
C
C C
dzzzzf
zzdzzf
2
000
Let , here r is small and will eventually be made to approach zero
(r0)
Here is a remarkable result. The value of an analytic function is given atan interior point at z=z0 once the values on the boundary C are specified.
What happens if z0 is exterior to C?In this case the entire integral is analytic on and within C, so the integral vanishes.
i0 rezz
drie
re
rezfdz
zzdzzf i
C Ci
i
2 2
0
0
00 22
zifdzifC
DerivativesCauchy’s integral formula may be used to obtain an expression for the derivation of f(z)
Moreover, for the n-th order of derivative
0
0 0
12
f z dzdf zdz i z z
C
zfzzdzzf
i exterior z ,0interiorz ,
21
0
00
0
2
000 211
21
zzdzzf
izzdzddzzf
i
1
00 2
!n
n
zz
dzzfi
nzf
. find origin, about the circle a
within andon analytic is a)( If 1.
Examples
0n
n
n
n
a
zzf
jn
jnnj
j zaajzf
1
!
jj ajf !0
12
1!
0n
n
nz
dzzfin
fa
In the above case, on a circle of radius r about the origin,
then (Cauchy’s inequality) Proof:
where
Lowville's theorem: If f(z) is analytic and bounded in the complex plane, it is a constant. Proof: For any z0, construct a circle of radius R around z0,
Mzf
Mra nn
nn
rznn
rM
rrrM
zdzzfa
11 22
21
rfMaxrM rz
22
00
222
1R
RMzzdzzf
izf
R
RM
Since R is arbitrary, let , we have
Conversely, the slightest deviation of an analytic function from a constant value implies that there must be at least one singularity somewhere in the infinite complex plane. Apart from the trivial constant functions, then, singularities are a fact of life, and we must learn to live with them, and to use them further.
R
.const)z(f,e.i,0zf
Laurent Expansion
Taylor ExpansionSuppose we are trying to expand f(z) about z=z0, i.e.,and we have z=z1 as the nearest point for which f(z) is not analytic. Weconstruct a circle C centered at z=z0 with radius
From the Cauchy integral formula,
0n
n0n zzazf
010 zzzz
C 00C zzzz
zdzfi2
1zzzdzf
i21zf
C 000 zzzz1zz
zdzfi2
1
Here z is a point on C and z is any point interior to C. For |t| <1, we note the identity
So we may write
which is our desired Taylor expansion, just as for real variable powerseries, this expansion is unique for a given z0.
0
211
1
n
ntttt
C nn
n
zz
zdzfzzi
zf0
10
0
21
01
002
1
n Cn
n
zz
zdzfzzi
0
00 !n
nn
nzf
zz
Schwarz reflection principleFrom the binomial expansion of for integer n (as anassignment), it is easy to see, for real x0
Schwarz reflection principle:If a function f(z) is (1) analytic over some region including the real axisand (2) real when z is real, then
We expand f(z) about some point (nonsingular) point x0 on the real axisbecause f(z) is analytic at z=x0.
Since f(z) is real when z is real, the n-th derivate must be real.
n0xzzg
*n0
**n0
* zgxzxzzg
** zfzf
0
00 !n
nn
nxf
xzzf
*
0
00
**
!zf
nxf
xzzfn
nn
Laurent SeriesWe frequently encounter functions that are analytic in annular
region
Drawing an imaginary contour line to convert our region into a simply connected region, we apply Cauchy’s integral formula for C2 and C1, with radii r2 and r1, and obtain
We let r2 r and r1 R, so for C1, while for C2, . We expand two denominators as we did before
(Laurent Series)
zzzdzf
izf
CC
2121
00 zzzz 00 zzzz
21000000 112
1
CCzzzzzz
zdzfzzzzzz
zdzfi
zf
zdzfzzzzizz
zdzfzzi
n
n Cn
n Cn
n
001
001
00
21
121
21
n
nn zzazf 0
where
Here C may be any contour with the annular region r < |z-z0| < R encircling z0 once in a counterclockwise sense. Laurent Series need not to come from evaluation of contour integrals. Other techniques such as ordinary series
expansion may provide the coefficients.Numerous examples of Laurent series appear in the next chapter.
Cnn
zzzdzf
ia
102
1
0
2221
mmnimn
i
ner
driei
a
021 2
11
121
mn
mnn
zzdz
izzzd
zia
11zzzf
01,22
21
mmni
i
1- n for 0
-1nfor 1a n
1
32111
1
n
nzzzzzzz
The Laurent expansion becomes
Example: (1) Find Taylor expansion ln(1+z) at point z
(2) find Laurent series of the function
If we employ the polar form
1
1)1()1ln(n
nn
nzz
• Theorem Suppose that a function f is analytic throughout an annular
domain R1< |z − z0| < R2, centered at z0 , and let C denote any positively oriented simple closed contour around z0 and lying in that domain. Then, at each point in the domain, f (z) has the series representation
Laurent Series
0 1 0 20 1 0
( ) ( ) , ( | | )( )
n nn n
n n
bf z a z z R z z R
z z
10
1 ( ) , ( 0,1, 2,...)2 ( )n n
C
f z dza ni z z
10
1 ( ) , ( 1,2,...)2 ( )n n
C
f z dzb ni z z
• Theorem (Cont’)
Laurent Series
0 1 0 2( ) ( ) , ( | | )nn
n
f z c z z R z z R
0 1 0 20 1 0
( ) ( ) , ( | | )( )
n nn n
n n
bf z a z z R z z R
z z
10
1 ( ) , ( 0,1, 2,...)2 ( )n n
C
f z dza ni z z
10
1 ( ) , ( 1, 2,...)2 ( )n n
C
f z dzb ni z z
10
1 ( ) , ( 0, 1, 2,...)2 ( )n n
C
f z dzc ni z z
1 1
00
( )( )
nnnn
n n
bb z z
z z
, 1, 0n
nn
b nc
a n
• Laurent’s Theorem If f is analytic throughout the disk |z-z0|<R2,
Laurent Series
00
( ) ( )nn
n
f z a z z
101
0
1 ( ) 1 ( ) ( ) , ( 1, 2,...)2 ( ) 2
nn n
C C
f z dzb z z f z dz ni z z i
Analytic in the region |z-z0|<R2
0, ( 1,2,...)nb n
( )0
10
( )1 ( ) , ( 0,1, 2,...)2 ( ) !
n
n nC
f zf z dza ni z z n
reduces to Taylor Series about z0
0 1 0 20 1 0
( ) ( ) , ( | | )( )
n nn n
n n
bf z a z z R z z Rz z
• Example 1 Replacing z by 1/z in the Maclaurin series expansion
We have the Laurent series representation
Examples
2 3
0
1 ...(| | )! 1! 2! 3!
nz
n
z z z ze zn
1/2 3
0
1 1 1 11 ...(0 | | )! 1! 2! 3!
zn
n
e zn z z z z
There is no positive powers of z, and all coefficients of the positive powers are zeros.
1
1 ( ) , ( 1, 2,...)2 ( 0)n n
C
f z dzb ni z
1/
1/1 1 1
1 112 ( 0) 2
zz
C C
e dzb e dzi z i
1/ 2z
C
e dz i
where c is any positively oriented simple closedcontours around the origin
• Example 2 The function f(z)=1/(z-i)2 is already in the form of a
Laurent series, where z0=i,. That is
where c-2=1 and all of the other coefficients are zero.
Examples
2
1 ( ) , (0 | | )( )
nn
n
c z i z iz i
30
1 , ( 0, 1, 2,...)2 ( )n n
C
dzc ni z z
3
0, 22 , 2( )n
C
ndzi nz i
where c is any positively oriented simple contouraround the point z0=i
Examples
Consider the following function1 1 1( )
( 1)( 2) 1 2f z
z z z z
which has the two singular points z=1 and z=2, is analytic in the domains
1 :| | 1D z
3 : 2 | |D z
2 :1 | | 2D z
• Example 3 The representation in D1 is Maclaurin series.
Examples
1 1 1 1 1( )1 2 1 2 1 ( / 2)
f zz z z z
11
0 0 0
( ) (2 1) , (| | 1)2
nn n n
nn n n
zf z z z z
where |z|<1 and |z/2|<1
• Example 4 Because 1<|z|<2 when z is a point in D2, we know
Examples
1 1 1 1 1 1( )1 2 1 (1/ ) 2 1 ( / 2)
f zz z z z z
where |1/z|<1 and |z/2|<1
1 1 10 0 1 0
1 1( ) , (1 | | 2)2 2
n n
n n n nn n n n
z zf z zz z
• Theorem 1 If a power series
converges when z = z1 (z1 ≠ z0), then it is absolutely
convergent at each point z in the open disk |z − z0| < R1 where R1 = |z1 − z0|
Some Useful Theorems
00
( )nn
n
a z z
• Theorem Suppose that a function f is analytic throughout a disk
|z − z0| < R0, centered at z0 and with radius R0. Then f (z) has the power series representation
Taylor Series
0 0 00
( ) ( ) , (| | )nn
n
f z a z z z z R
( )
0( ), ( 0,1,2,...)
!
n
nf z
a nn
That is, series converges to f (z) when z lies in the stated open disk.
10
1 ( )2 ( )n n
C
f z dzai z z
Refer to pp.167
Proof the Taylor’s Theorem
( )
0 00
(0)( ) ,(| | )!
nn
n
ff z z z z Rn
Proof:
Let C0 denote and positively oriented circle |z|=r0, where r<r0<R0
Since f is analytic inside and on the circle C0 and since the point z is interior to C0, the Cauchy integral formula holds
0
01 ( )( ) , ,| |
2 C
f s dsf z z z Ri s z
1 1 1 1 1 , ( / ),| | 11 ( / ) 1
w z s ws z s z s s w
Proof the Taylor’s Theorem1
10
1 1 1( )
Nn N
n Nn
z zs z s s z s
0
1 ( )( )2 C
f s dsf zi s z
0 0
1
10
1 ( ) 1 ( )( )2 2 ( )
Nn N
n Nn C C
f s ds f s dsf z z zi s i s z s
( ) (0)!
nfn
Refer to pp.167
0
( )1
0
(0) ( )( )! 2 ( )
n NNn
Nn C
f z f s dsf z zn i s z s
ρN
Proof the Taylor’s Theorem
0
( )lim lim 02 ( )
N
N NN NC
z f s dsi s z s
( ) ( ) ( )1
0 0 0
(0) (0) (0)( ) lim ( ) 0! ! !
n n nNn n n
NN n n n
f f ff z z z zn n n
When
0
00 0
( ) | || | | | 22 ( ) 2 ( )
N N
N N NC
z f s ds r M ri s z s r r r
Where M denotes the maximum value of |f(s)| on C0
0
0 0
| | ( )NN
Mr rr r r
lim 0NN
0
( ) 1rr
Example
expand f(z) into a series involving powers of z. We can not find a Maclaurin series for f(z) since it is not analytic at z=0.
But we do know that expansion
Hence, when 0<|z|<1
Examples
2 2
3 5 3 2 3 2
1 2 1 2(1 ) 1 1 1( ) (2 )1 1
z zf zz z z z z z
2 4 6 82
1 1 ...(| | 1)1
z z z z zz
2 4 6 8 3 53 3
1 1 1( ) (2 1 ...) ...f z z z z z z z zz z z
Negative powers
Residue theoremCalculus of residues
Functions of a Complex Variable
Suppose an analytic function f (z) has an isolated singularity at z0. Consider a contour integral enclosing z0 .
z0
)( sRe22)(
1 ,2)ln(
1 ,01)(
)(
)()()(
01
1'
'01
'
'
10
0
00
zfiiadzzf
niazza
nn
zzadzzza
dzzzadzzzadzzf
C
z
z
z
z
n
n
C
nn
nC
nnC
n
nnC
The coefficient a-1=Res f (z0) in the Laurent expansion is called the residue of f (z) at z = z0.
If the contour encloses multiple isolated singularities, we have the residue theorem:
n
nCzfidzzf )( sRe2)(
z0 z1
Contour integral =2i ×Sum of the residuesat the enclosed singular points
Residue formula:To find a residue, we need to do the Laurent expansion and pick up the coefficient a-1. However, in many cases we have a useful residue formula
)()(lim)!1(
1)(sRe
)!1())(2()1)((lim)(lim
)(lim)!1(
1)()(lim)!1(
1
:Proof
.)()(lim)(sRe
,pole simple afor ly,Particular
)()(lim)!1(
1)(sRe
,order of pole aFor
01
1
01
11
1001
1
01
1
01
1
00
01
1
0
0
00
00
0
0
zfzzdzd
mzfa
mazznmnmnazzadzd
zzadzd
mzfzz
dzd
m
zfzzzf
zfzzdzd
mzf
m
mm
m
zz
n
nnzz
mn
mnnm
m
zz
mn
mnnm
m
zz
mm
m
zz
zz
mm
m
zz
.0 ,)()(lim!
1
:tscoefficien theall find way toa is e that therprovedactually We
.)()(lim)!1(
1 us gives 1 upPick . Also
.)()(lim!
1 ,)()()(
expansionTaylor by analytic, is )()( Because
)()()(
)()(
:#2 Method Proof
0
01
1
1
000
0
0
00
0
0
0
0
kzfzzdzd
ka
a
zfzzdzd
mamkab
zfzzdzd
kbzzbzfzz
zfzz
zzazfzz
zzazf
mk
k
zzmk
mm
m
zzmkk
mk
k
zzkk
kk
m
m
mn
mnn
m
n
mnn
Cauchy’s integral theorem and Cauchy’s integral formula revisited:(in the view of the residue theorem):
.!
)()()!11(
1lim
is at residue its formula, residue the toAccording
1.order of pole a isIt .)(')()( )'3
!)(22)()()( )3
)(2)()(')()( )2
0)(Res2)( )1
))((')()()( :function Analytic
0)(
10
101)1(
1)1(
0
0
01
0
01
0
0)(
100
101
0
00
00
0
0
0
0000
0
0 nzf
zzzfzz
dzd
n
zz
nzzzf
zzzf
zzzf
nzfiiadz
zzzfzza
zzzf
zifdzzzzfzf
zzzf
zzzf
zfidzzf
zzzfzfzzazf
n
nn
n
n
zz
nnn
n
nC nm
nmmn
C
C
m
mm
Evaluation of definite integrals -1Calculus of residues
2222
2
22
2
22
2
22
2
0
220
2
2
2
2
2
0
111
2
11121 ,
111
1211
)1(1)(
1111)( Res )0( Res
.)(
1))((
1lim)( Res
.1))((
1lim)0( Res
circle. theofout is circle, in the is |||| ,1||
.1101/2 ,0 poles, simple 3 have We
)1/2(11
2111
2//112//1
1.|| and real is ,sin1
sin Example
aa
a
aa
aiia
Iaa
a
aa
iaaizzz
zzzz
zzz
zff
zzzz
zzzzzzzzzf
zzzzzzzzzf
zzzzzz
aaizaizzz
dzaizzz
zia
dzaizazz
ziiz
dzizza
izzI
aaa
dI
zz
z
CCC
C
r=1
z+
z-
z0
Evaluation of definite integrals -2Calculus of residues
II. Integrals along the whole real axis:
dxxf )(
Assumption 1: f (z) is analytic in the upper or lower half of the complex plane, except isolated finite number of poles.
∩
R
Condition for closure on a semicircular path:
dzzfdzzfdzzfdzzfdzzfdxxf
RCR
R
RR)(lim)(lim)()()(lim)(
.0 ,1~)(lim0lim) (lim
) (lim ) (lim)(lim
1max0
00
zzfRfRdeRf
deiReRfdeiReRfdzzf
RR
i
R
ii
R
ii
RR
Assumption 2: when |z|, | f (z)| goes to zero faster than 1/|z|.
Then, plane. halfupper on the )( of esiduesR2)(lim)( zfidzzfdxxfCR
.arctan1
Or
.))((
1lim2)( Res2
plane halfupper on the 1
1 of esiduesR2
1
:1 Example
2
2
2
xx
dxiziz
iziifi
ziI
xdxI
iz
.
2'
)()(1lim2)( Res2
plane halfupper on the 1 of esiduesR2
.0 ,
:2 Example
3222
222
222
aaizaiziaziiafi
aziI
aax
dxI
aiz
UNIT - III
MOMENTS, SKEWNESS, AND KURTOSIS
Moment Ratios
• 23 4
1 23 22 2
,
23 4
1 23 22 2
,m mb bm m
NON-CENTRAL MOMENTS
CENTRAL MOMENTS
THEOREMS
SKEWNESS
SkewnessA distribution in which the values equidistant from the mean have equal frequencies and is called Symmetric Distribution.Any departure from symmetry is called skewness.
In a perfectly symmetric distribution, Mean=Median=Mode and the two tails of the distribution are equal in length from the mean. These values are pulled apart when the distribution departs from symmetry and consequently one tail become longer than the other.
If right tail is longer than the left tail then the distribution is said to have positive skewness. In this case, Mean>Median>Mode
If left tail is longer than the right tail then the distribution is said to have negative skewness. In this case, Mean<Median<Mode
KURTOSIS
Kurtosis
•
For a normal distribution, kurtosis is equal to 3.
When is greater than 3, the curve is more sharply peaked and has narrower tails than the normal curve and is said to be leptokurtic.
When it is less than 3, the curve has a flatter top and relatively wider tails than the normal curve and is said to be platykurtic.
44
21 1
4
42
1 1
1 1 ,
1 1 ,
n ni
i i
n ni
i i
xkurt z for population data
n n
x xkurt b z for sample datan n s
CURVE FITTING
Curve Fitting and Correlation
This will be concerned primarily with two separate but closely interrelated processes:(1) the fitting of experimental data to
mathematical forms that describe their behavior and
(2) the correlation between different experimental data to assess how closely different variables are interdependent.
•The fitting of experimental data to a mathematical equation is called regression. Regression may be characterized by different adjectives according to the mathematical form being used for the fit and the number of variables. For example, linear regression involves using a straight-line or linear equation for the fit. As another example, Multiple regression involves a function of more than one independent variable.
Linear Regression
•Assume n points, with each point having values of both an independent variable x and a dependent variable y.
1 2 3The values of are , , ,...., .nx x x x x
1 2 3The values of are , , ,...., .ny y y y yA best-fitting straight line equation will have the form
1 0y a x a
Preliminary Computations
0
1sample mean of the valuesn
kk
x x xn
0
1sample mean of the valuesn
kk
y y yn
2 2
1
1sample mean-square of the valuesn
kk
x x xn
1
1sample mean of the product n
k kk
xy xy x yn
Best-Fitting Straight Line
1 22
xy x ya
x x
2
0 22
x y x xya
x x
0 1Alternately, a y a x
1 0y a x a
Example-1. Find best fitting straight line equation for the data shown below.
x0123456789y4.006.108.309.9012.4014.3015.7017.4019.8022.30
10
1
1 0 1 2 3 4 5 6 7 8 9 45 4.5010 10 10k
k
x x
10
1
1 4 6.1 8.3 9.9 12.4 14.3 15.7 17.4 19.8 22.310 10130.2 13.02
10
kk
y y
Multiple Linear Regression
0 1 1 2 2 ..... m my a a x a x a x
Assume independent variablesm
1 2, ,..... mx x xAssume a dependent variable that is to be considered as a linear functionof the independent variables.
y
m
Multiple Regression (Continuation)
1
Assume that there are values of each of the variables. For , we have
km x
11 12 13 1, , ,....., kx x x xSimilar terms apply for all other variables.For the th variable, we havem
1 2 3, , ,.....,m m m mkx x x x
Correlation
corr( , ) ( )x y E xy xy Cross-Correlation
cov( , ) ( )( )
corr( , ) ( )( )( )( )
x y E x x y y
x y x yxy x y
Covariance
Correlation Coefficient
( )( )( , )
cov( , )cov( , ) cov( , )
x y
E x x y yC x y
x yx x y y
Implications of Correlation Coefficient
• 1. If C(x, y) = 1, the two variables are totally correlated in a positive sense.
• 2. If C(x, y) = -1 , the two variables are totally correlated in a negative sense.
• 3. If C(x, y) = 0, the two variables are said to be uncorrelated.
Binomial Distribution and Applications
Binomial Probability DistributionA binomial random variable X is defined to the number
of “successes” in n independent trials where the P(“success”) = p is constant.
Notation: X ~ BIN(n,p)
In the definition above notice the following conditions need to be satisfied for a binomial experiment:
1. There is a fixed number of n trials carried out.2. The outcome of a given trial is either a “success”
or “failure”.3. The probability of success (p) remains constant
from trial to trial. 4. The trials are independent, the outcome of a trial is
not affected by the outcome of any other trial.
Binomial Distribution• If X ~ BIN(n, p), then
• where
.,...,1,0 )1()!(!
! )1()( nxppxnx
nppxn
xXP xnxxnx
psuccessPnx
nnnn
)"(" trials.in successes""
obtain to waysofnumber the x"choosen " xn
1 1! and 1 0! also ,1...)2()1(!
Binomial Distribution• If X ~ BIN(n, p), then
• E.g. when n = 3 and p = .50 there are 8 possible equally likely outcomes (e.g. flipping a coin)
SSS SSF SFS FSS SFF FSF FFS FFF X=3 X=2 X=2 X=2 X=1 X=1 X=1 X=0 P(X=3)=1/8, P(X=2)=3/8, P(X=1)=3/8, P(X=0)=1/8• Now let’s use binomial probability formula instead…
.,...,1,0 )1()!(!
! )1()( nxppxnx
nppxn
xXP xnxxnx
Binomial Distribution• If X ~ BIN(n, p), then
• E.g. when n = 3, p = .50 find P(X = 2)
.,...,1,0 )1()!(!
! )1()( nxppxnx
nppxn
xXP xnxxnx
83or 375.)5)(.5(.3)5(.5.
23
)2(
ways31)12(
123!1 !2
!3)!23(!2
!323
12232
XP
SSFSFSFSS
The Poisson DistributionThe Poisson distribution is defined by:
!)(
xexf
x
Where f(x) is the probability of x occurrences in an interval is the expected value or mean value of occurrences within an interval
e is the natural logarithm. e = 2.71828
Properties of the Poisson Distribution
1. The probability of occurrences is the same for any two intervals of equal length.
2. The occurrence or nonoccurrence of an event in one interval is independent of an occurrence on nonoccurrence of an event in any other interval
Problem
a. Write the appropriate Poisson distribution
b. What is the average number of occurrences in three time periods?
c. Write the appropriate Poisson function to determine the probability of x occurrences in three time periods.
d. Compute the probability of two occurrences in one time period.
e. Compute the probability of six occurrences in three time periods.
f. Compute the probability of five occurrences in two time periods.
Consider a Poisson probability distribution with an average number of occurrences of two per period.
Problem
!2)(
2
Xexf
x
6
!6)(
6
Xexf
x
27067.2
5413.!2
2)2(22
ef
(a)
(b)
(c)
(d)
Hypergeometric Distribution
rx
nN
xnrN
xr
xf
0 allfor )(
Where
n = the number of trials.
N = number of elements in the population
r = number of elements in the population labeled a success
The Chi-Square Test
Parametric and Nonparametric Tests
It introduces two non-parametric hypothesis tests using the chi-square statistic: the chi-square test for goodness of fit and the chi-square test for independence.
Parametric and Nonparametric Tests (cont.)
• The term "non-parametric" refers to the fact that the chi square tests do not require assumptions about ‑population parameters nor do they test hypotheses about population parameters.
• Previous examples of hypothesis tests, such as the t tests and analysis of variance, are parametric tests and they do include assumptions about parameters and hypotheses about parameters.
Parametric and Nonparametric Tests (cont.)
• The most obvious difference between the chi square tests and the other hypothesis ‑tests we have considered (t and ANOVA) is the nature of the data.
• For chi square, the data are frequencies rather ‑than numerical scores.
The Chi-Square Test for Goodness-of-Fit
• The chi-square test for goodness-of-fit uses frequency data from a sample to test hypotheses about the shape or proportions of a population.
• Each individual in the sample is classified into one category on the scale of measurement.
• The data, called observed frequencies, simply count how many individuals from the sample are in each category.
The Chi-Square Test for Goodness-of-Fit (cont.)
• The null hypothesis specifies the proportion of the population that should be in each category.
• The proportions from the null hypothesis are used to compute expected frequencies that describe how the sample would appear if it were in perfect agreement with the null hypothesis.
The Chi-Square Test for Independence
• The second chi-square test, the chi-square test for independence, can be used and interpreted in two different ways:
1. Testing hypotheses about the relationship between two variables in a population, or
2. Testing hypotheses about differences between proportions for two or more populations.
The Chi-Square Test for Independence (cont.)
• Although the two versions of the test for independence appear to be different, they are equivalent and they are interchangeable.
• The first version of the test emphasizes the relationship between chi-square and a correlation, because both procedures examine the relationship between two variables.
The Chi-Square Test for Independence (cont.)
• The second version of the test emphasizes the relationship between chi-square and an independent-measures t test (or ANOVA) because both tests use data from two (or more) samples to test hypotheses about the difference between two (or more) populations.
The Chi-Square Test for Independence (cont.)
• The first version of the chi-square test for independence views the data as one sample in which each individual is classified on two different variables.
• The data are usually presented in a matrix with the categories for one variable defining the rows and the categories of the second variable defining the columns.
The Chi-Square Test for Independence (cont.)
• The data, called observed frequencies, simply show how many individuals from the sample are in each cell of the matrix.
• The null hypothesis for this test states that there is no relationship between the two variables; that is, the two variables are independent.
The Chi-Square Test for Independence (cont.)
• The second version of the test for independence views the data as two (or more) separate samples representing the different populations being compared.
• The same variable is measured for each sample by classifying individual subjects into categories of the variable.
• The data are presented in a matrix with the different samples defining the rows and the categories of the variable defining the columns..
The Chi-Square Test for Independence (cont.)
• The data, again called observed frequencies, show how many individuals are in each cell of the matrix.
• The null hypothesis for this test states that the proportions (the distribution across categories) are the same for all of the populations
The Chi-Square Test for Independence (cont.)
• Both chi-square tests use the same statistic. The calculation of the chi-square statistic requires two steps:
1. The null hypothesis is used to construct an idealized sample distribution of expected frequencies that describes how the sample would look if the data were in perfect agreement with the null hypothesis.
The Chi-Square Test for Independence (cont.)
For the goodness of fit test, the expected frequency for each category is obtained by
expected frequency = fe = pn(p is the proportion from the null hypothesis and n is the size of the sample)
For the test for independence, the expected frequency for each cell in the matrix is obtained by
(row total)(column total)expected frequency = fe = ─────────────────
n
The Chi-Square Test for Independence (cont.)
2. A chi-square statistic is computed to measure the amount of discrepancy between the ideal sample (expected frequencies from H0) and the actual sample data (the observed frequencies = fo).
A large discrepancy results in a large value for chi-square and indicates that the data do not fit the null hypothesis and the hypothesis should be rejected.
The Chi-Square Test for Independence (cont.)
The calculation of chi-square is the same for all chi-square tests:
(fo – fe)2
chi-square = χ2 = Σ ───── fe
The fact that chi square tests do not require scores ‑from an interval or ratio scale makes these tests a valuable alternative to the t tests, ANOVA, or correlation, because they can be used with data measured on a nominal or an ordinal scale.
Measuring Effect Size for the Chi-Square Test for Independence
• When both variables in the chi-square test for independence consist of exactly two categories (the data form a 2x2 matrix), it is possible to re-code the categories as 0 and 1 for each variable and then compute a correlation known as a phi-coefficient that measures the strength of the relationship.
Measuring Effect Size for the Chi-Square Test for Independence (cont.)
• The value of the phi-coefficient, or the squared value which is equivalent to an r2, is used to measure the effect size.
• When there are more than two categories for one (or both) of the variables, then you can measure effect size using a modified version of the phi-coefficient known as Cramér=s V.
• The value of V is evaluated much the same as a correlation.
The t-test
Inferences about Population Means
Questions
• What is the main use of the t-test?• How is the distribution of t related to the unit
normal?• When would we use a t-test instead of a z-test? Why
might we prefer one to the other?• What are the chief varieties or forms of the t-test? • What is the standard error of the difference between
means? What are the factors that influence its size?
Background
• The t-test is used to test hypotheses about means when the population variance is unknown (the usual case). Closely related to z, the unit normal.
• Developed by Gossett for the quality control of beer.
• Comes in 3 varieties:• Single sample, independent samples, and
dependent samples.
What kind of t is it?
• Single sample t – we have only 1 group; want to test against a hypothetical mean.
• Independent samples t – we have 2 means, 2 groups; no relation between groups, e.g., people randomly assigned to a single group.
• Dependent t – we have two means. Either same people in both groups, or people are related, e.g., husband-wife, left hand-right hand, hospital patient and visitor.
Single-sample z test
• For large samples (N>100) can use z to test hypotheses about means.
• Suppose
• Then
• If
MM est
Xz
.)(
N
NXX
Nsest X
M1
)(
.
2
200;5;10:;10: 10 NsHH X
35.14.14
52005.
Nsest X
M
05.96.183.2;83.235.
)1011(11
pzX
The t DistributionWe use t when the population variance is unknown (the usual case) and sample size is small (N<100, the usual case). If you use a stat package for testing hypotheses about means, you will use t.
The t distribution is a short, fat relative of the normal. The shape of t depends on its df. As N becomes infinitely large, t becomes normal.
Degrees of FreedomFor the t distribution, degrees of freedom are always a simple function of the sample size, e.g., (N-1).
One way of explaining df is that if we know the total or mean, and all but one score, the last (N-1) score is not free to vary. It is fixed by the other scores. 4+3+2+X = 10. X=1.
Single-sample t-testWith a small sample size, we compute the same numbers as we did for z, but we compare them to the t distribution instead of the z distribution.
25;5;10:;10: 10 NsHH X
1255.
Nsest X
M 11
)1011(11
tX
064.2)24,05(. t 1<2.064, n.s.
Interval = ]064.13,936.8[)1(064.211
ˆ
MtX
Interval is about 9 to 13 and contains 10, so n.s.
(c.f. z=1.96)
Difference Between Means (1)
• Most studies have at least 2 groups (e.g., M vs. F, Exp vs. Control)
• If we want to know diff in population means, best guess is diff in sample means.
• Unbiased:• Variance of the Difference:• Standard Error:
22
2121 )var( MMyy
212121 )()()( yEyEyyE
22
21 MMdiff
Difference Between Means (2)
• We can estimate the standard error of the difference between means.
• For large samples, can use z
22
21 ... MMdiff estestest
diffestXX
diffz )()( 2121
3;100;12
2;100;10
0:;0:
222
111
211210
SDNX
SDNX
HH
36.10013
1009
1004. diffest
05.;56.536.2
36.0)1210( pzdiff
Independent Samples t (1)
• Looks just like z:• df=N1-1+N2-1=N1+N2-2• If SDs are equal, estimate is:
diffestyy
difft )()( 2121
21
2
2
2
1
2 11NNNNdiff
Pooled variance estimate is weighted average:
)]2/(1/[])1()1[( 21222
211
2 NNsNsNPooled Standard Error of the Difference (computed):
21
21
21
222
211
2)1()1(.
NNNN
NNsNsNest diff
Independent Samples t (2)
21
21
21
222
211
2)1()1(.
NNNN
NNsNsNest diff
diffestyy
difft )()( 2121
7;83.5;20
5;7;18
0:;0:
2222
1211
211210
Nsy
Nsy
HH
47.13512
275)83.5(6)7(4.
diffest
..;36.147.12
47.10)2018( sntdiff
tcrit = t(.05,10)=2.23
Dependent t (1)Observations come in pairs. Brother, sister, repeated measure.
),cov(2 212
22
12 yyMMdiff
Problem solved by finding diffs between pairs Di=yi1-yi2.
1)( 2
2
N
DDs i
D Nsest D
MD .ND
D i)(
MDestDEDt
.)(
df=N(pairs)-1
Dependent t (2)Brother Sister5 77 83 3
5y 6y
Diff2 11 00 1
1D
58.3/1. MDest
72.158.1
.)(
MDestDEDt
11
)( 2
N
DDsD
2)( DD
Assumptions
• The t-test is based on assumptions of normality and homogeneity of variance.
• You can test for both these (make sure you learn the SAS methods).
• As long as the samples in each group are large and nearly equal, the t-test is robust, that is, still good, even tho assumptions are not met.
UNIT IVThe Bisection Method
Introduction
• Root of a function:
• Root of a function f(x) = a value a such that:
• f(a) = 0
Introduction (cont.)
• Example:
Function: f(x) = x2 - 4
Roots: x = -2, x = 2
Because: f(-2) = (-2)2 - 4 = 4 - 4 = 0 f(2) = (2)2 - 4 = 4 - 4 = 0
A Mathematical Property
• Well-known Mathematical Property:
• If a function f(x) is continuous on the interval [a..b] and sign of f(a) ≠ sign of f(b), then
• There is a value c [∈ a..b] such that: f(c) = 0 I.e., there is a root c in the interval [a..b]
A Mathematical Property (cont.)
• Example:
The Bisection Method
• The Bisection Method is a successive approximation method that narrows down an interval that contains a root of the function f(x)
• The Bisection Method is given an initial interval [a..b] that contains a root (We can use the property sign of f(a) ≠ sign of f(b) to find such an initial interval)
• The Bisection Method will cut the interval into 2 halves and check which half interval contains a root of the function
• The Bisection Method will keep cut the interval in halves until the resulting interval is extremely small The root is then approximately equal to any value in the final (very small) interval.
The Bisection Method (cont.)
• Example:
• Suppose the interval [a..b] is as follows:
The Bisection Method (cont.)
• We cut the interval [a..b] in the middle: m = (a+b)/2
The Bisection Method (cont.)
• Because sign of f(m) ≠ sign of f(a) , we proceed with the search in the new interval [a..b]:
The Bisection Method (cont.)
We can use this statement to change to the new interval:
b = m;
The Bisection Method
• In the above example, we have changed the end point b to obtain a smaller interval that still contains a root
In other cases, we may need to changed the end point b to obtain a smaller interval that still contains a root
The Bisection Method (cont.)
• Here is an example where you have to change the end point a:
• Initial interval [a..b]:
The Bisection Method (cont.)
• After cutting the interval in half, the root is contained in the right-half, so we have to change the end point a:
The Bisection Method
• Rough description (pseudo code) of the Bisection Method:
Given: interval [a..b] such that: sign of f(a) ≠ sign of f(b)
repeat (until the interval [a..b] is "very small") { a+b m = -----; // m = midpoint of interval [a..b] 2
if ( sign of f(m) ≠ sign of f(b) ) { use interval [m..b] in the next iteration
The Bisection Method
(i.e.: replace a with m) } else { use interval [a..m] in the next iteration (i.e.: replace b with m) } }
Approximate root = (a+b)/2; (any point between [a..b] will do because the interval [a..b] is very small)
The Bisection Method
• Structure Diagram of the Bisection Algorithm:
The Bisection Method
• Example execution:
• We will use a simple function to illustrate the execution of the Bisection Method • Function used:
Roots: √3 = 1.7320508... and −√3 = −1.7320508...
f(x) = x2 - 3
The Bisection Method (cont.)
• We will use the starting interval [0..4] since:
The interval [0..4] contains a root because: sign of f(0) ≠ sign of f(4)
• f(0) = 02 − 3 = −3 • f(4) = 42 − 3 = 13
Regula-Falsi Method
Regula-Falsi Method
Type of Algorithm (Equation Solver)
The Regula-Falsi Method (sometimes called the False Position Method) is a method used to find a numerical estimate of an equation.
This method attempts to solve an equation of the form f(x)=0. (This is very common in most numerical analysis applications.) Any equation can be written in this form.
Algorithm Requirements
This algorithm requires a function f(x) and two points a and b for which f(x) is positive for one of the values and negative for the other. We can write this condition as f(a)f(b)<0.
If the function f(x) is continuous on the interval [a,b] with f(a)f(b)<0, the algorithm will eventually converge to a solution.
This algorithm can not be implemented to find a tangential root. That is a root that is tangent to the x-axis and either positive or negative on both side of the root. For example f(x)=(x-3)2, has a tangential root at x=3.
Regula-Falsi Algorithm
The idea for the Regula-Falsi method is to connect the points (a,f(a)) and (b,f(b)) with a straight line.
Since linear equations are the simplest equations to solve for find the regula-falsi point (xrfp) which is the solution to the linear equation connecting the endpoints.
Look at the sign of f(xrfp):
If sign(f(xrfp)) = 0 then end algorithm
else If sign(f(xrfp)) = sign(f(a)) then set a = xrfp
else set b = xrfp
x-axisa b
f(b)
f(a) actual root
f(x)xrfp
equation of line:
axab
afbfafy
)()()(
solving for xrfp
)()(
)()()(
)(
)()()(0
afbfabafax
axafbfabaf
axab
afbfaf
rfp
rfp
rfp
Example
Lets look for a solution to the equation x3-2x-3=0.
We consider the function f(x)=x3-2x-3
On the interval [0,2] the function is negative at 0 and positive at 2. This means that a=0 and b=2 (i.e. f(0)f(2)=(-3)(1)=-3<0, this means we can apply the algorithm).
23
46
31)2(3
)0()2(02)0(0
ff
fxrfp
821
23)(
fxf rfp
This is negative and we will make the a =3/2 and b is the same and apply the same thing to the interval [3/2,2].
2954
5821
23
123
)2(2
23
82121
821
23
23
23
fff
xrfp
267785.02954)(
fxf rfp
This is negative and we will make the a =54/29 and b is the same and apply the same thing to the interval [54/29,2].
Stopping Conditions
Aside from lucking out and actually hitting the root, the stopping condition is usually fixed to be a certain number of iterations or for the Standard Cauchy Error in computing the Regula-Falsi Point (xrfp) to not change more than a prescribed amount (usually denoted ).
Unit - IVInterpolation
• Estimation of intermediate values between precise data points. The most common method is:
• Although there is one and only one nth-order polynomial that fits n+1 points, there are a variety of mathematical formats in which this polynomial can be expressed:– The Newton polynomial– The Lagrange polynomial
nn xaxaxaaxf 2
210)(
Newton’s Divided-Difference Interpolating Polynomials
Linear Interpolation/• Is the simplest form of interpolation, connecting two data
points with a straight line.
• f1(x) designates that this is a first-order interpolating polynomial.
)()()()()(
)()()()(
00
0101
0
01
0
01
xxxx
xfxfxfxf
xxxfxf
xxxfxf
Linear-interpolation formula
Slope and a finite divided difference approximation to 1st derivative
Quadratic Interpolation/• If three data points are available, the estimate is
improved by introducing some curvature into the line connecting the points.
• A simple procedure can be used to determine the values of the coefficients.
))(()()( 1020102 xxxxbxxbbxf
02
01
01
12
12
22
0
0111
000
)()()()(
)()()(
xxxx
xfxfxx
xfxf
bxx
xxxfxfbxx
xfbxx
General Form of Newton’s Interpolating Polynomials/
0
02111011
011
0122
011
00
01110
012100100
],,,[],,,[],,,,[
],[],[],,[
)()(],[
],,,,[
],,[],[
)(],,,[)())((
],,[))((],[)()()(
xxxxxfxxxfxxxxf
xxxxfxxf
xxxf
xxxfxf
xxf
xxxxfb
xxxfbxxfb
xfbxxxfxxxxxx
xxxfxxxxxxfxxxfxf
n
nnnnnn
ki
kjjikji
ji
jiji
nnn
nnn
n
Bracketed function evaluations are finite divided differences
Errors of Newton’s Interpolating Polynomials/• Structure of interpolating polynomials is similar to the Taylor
series expansion in the sense that finite divided differences are added sequentially to capture the higher order derivatives.
• For an nth-order interpolating polynomial, an analogous relationship for the error is:
• For non differentiable functions, if an additional point f(xn+1) is available, an alternative formula can be used that does not require prior knowledge of the function:
)())(()!1(
)(10
)1(
n
n
n xxxxxxn
fR
)())(](,,,,[ 10011 nnnnn xxxxxxxxxxfR
Is somewhere containing the unknown and he data
Lagrange Interpolating Polynomials
• The Lagrange interpolating polynomial is simply a reformulation of the Newton’s polynomial that avoids the computation of divided differences:
n
ijj ji
ji
n
iiin
xxxx
xL
xfxLxf
0
0
)(
)()()(
)(
)()()(
)()()(
21202
10
12101
200
2010
212
101
00
10
11
xfxxxx
xxxx
xfxxxxxxxxxf
xxxxxxxxxf
xfxxxxxf
xxxxxf
•As with Newton’s method, the Lagrange version has an estimated error of:
n
iinnn xxxxxxfR
001 )(],,,,[
Coefficients of an Interpolating Polynomial
• Although both the Newton and Lagrange polynomials are well suited for determining intermediate values between points, they do not provide a polynomial in conventional form:
• Since n+1 data points are required to determine n+1 coefficients, simultaneous linear systems of equations can be used to calculate “a”s.
nx xaxaxaaxf 2
210)(
nnnnnn
nn
nn
xaxaxaaxf
xaxaxaaxf
xaxaxaaxf
2210
12121101
02020100
)(
)(
)(
Where “x”s are the knowns and “a”s are the unknowns.
Spline Interpolation
• There are cases where polynomials can lead to erroneous results because of round off error and overshoot.
• Alternative approach is to apply lower-order polynomials to subsets of data points. Such connecting polynomials are called spline functions.
NEWTON FORWARD INTERPOLATION ON EQUISPACED POINTS• Lagrange Interpolation has a number of disadvantages• The amount of computation required is large• Interpolation for additional values of requires the same amount of effort as the first value (i.e. no part of the previous calculation can be used)• When the number of interpolation points are changed (increased/decreased), the results of the previous computations can not be used• Error estimation is difficult (at least may not be convenient)• Use Newton Interpolation which is based on developing difference tables for a given setof data points
Newton’s Divided Difference Polynomial MethodTo illustrate this method, linear and quadratic interpolation is presented first. Then, thegeneral form of Newton’s divided difference polynomial method is presented. To illustratethe general form, cubic interpolation is shown in Figure
UNIT - VMatrix Decomposition
Introduction
Some of most frequently used decompositions are the LU, QR, Cholesky, Jordan, Spectral decomposition and Singular value decompositions.
This Lecture covers relevant matrix decompositions, basic numerical methods, its computation and some of its applications. Decompositions provide a numerically stable way to solve a system of linear equations, as shown already in [Wampler, 1970], and to invert a matrix. Additionally, they provide an important tool for analyzing the numerical stability of a system.
Easy to solve systemSome linear system that can be easily solved
The solution:
nnn ab
abab
/
//
222
111
Easy to solve system (Cont.)Lower triangular matrix:
Solution: This system is solved using forward substitution
Easy to solve system (Cont.)Upper Triangular Matrix:
Solution: This system is solved using Backward substitution
LU Decomposition
and
Where,
mm
m
m
u
uuuuu
U
00
0 222
11211
mmmm lll
lll
L
21
2221
11
000
LUA
LU decomposition was originally derived as a decomposition of quadratic and bilinear forms. Lagrange, in the very first paper in his collected works( 1759) derives the algorithm we call Gaussian elimination. Later Turing introduced the LU decomposition of a matrix in 1948 that is used to solve the system of linear equation.
Let A be a m × m with nonsingular square matrix. Then there exists two matrices L and U such that, where L is a lower triangular matrix and U is an upper triangular matrix.
A … U (upper triangular) U = Ek E1 A A = (E1)1 (Ek)1 U
If each such elementary matrix Ei is a lower triangular matrices,it can be proved that (E1)1, , (Ek)1 are lower triangular, and(E1)1 (Ek)1 is a lower triangular matrix.Let L=(E1)1 (Ek)1 then A=LU.
How to decompose A=LU?
21336812226
102/1012001
130010001
500240226
21336812226
102/1012001
1120240226
Now, 21336812226
A
U E2 E1 A
Calculation of L and U (cont.)
Now reducing the first column we have
21336812226
A
21336812226
100010001
21336812226
102/1012001
130010001
500240226
21336812226
102/1012001
1120240226
=
If A is a Non singular matrix then for each L (lower triangular matrix) the upper triangular matrix is unique but an LU decomposition is not unique. There can be more than one such LU decomposition for a matrix. Such as
Calculation of L and U
132/1012001
130010001
102/1012001
130010001
102/1012001 11
21336812226
A
132/1012001
500240226
21336812226
A
1330112006
500240
6/26/21
Now
Therefore,
=
=LU=
=LU
Calculation of L and U (cont.) Thus LU decomposition is not unique. Since we compute LU
decomposition by elementary transformation so if we change L then U will be changed such that A=LU
To find out the unique LU decomposition, it is necessary to put some restriction on L and U matrices. For example, we can require the lower triangular matrix L to be a unit one (i.e. set all the entries of its main diagonal to ones).
LU Decomposition in R:• library(Matrix)• x<-matrix(c(3,2,1, 9,3,4,4,2,5 ),ncol=3,nrow=3)• expand(lu(x))
Calculation of L and U
• Note: there are also generalizations of LU to non-square and singular matrices, such as rank revealing LU factorization.
• [Pan, C.T. (2000). On the existence and computation of rank revealing LU factorizations. Linear Algebra and its Applications, 316: 199-222.
• Miranian, L. and Gu, M. (2003). Strong rank revealing LU factorizations. Linear Algebra and its Applications, 367: 1-16.]
• Uses: The LU decomposition is most commonly used in the solution of systems of simultaneous linear equations. We can also find determinant easily by using LU decomposition (Product of the diagonal element of upper and lower triangular matrix).
Calculation of L and U
Solving system of linear equation using LU decomposition
Suppose we would like to solve a m×m system AX = b. Then we can find a LU-decomposition for A, then to solve AX =b, it is enough to solve the systems
Thus the system LY = b can be solved by the method of forward substitution and the system UX = Y can be solved by the method of
backward substitution. To illustrate, we give some examples Consider the given system AX = b, where
and
21336812226
A
17148
b
We have seen A = LU, where
Thus, to solve AX = b, we first solve LY = b by forward substitution
Then
Solving system of linear equation using LU decomposition
132/1012001
L
500
240226
U
17148
132/1012001
3
2
1
yyy
152
8
3
2
1
yyy
Y
Now, we solve UX =Y by backward substitution
then
Solving system of linear equation using LU decomposition
152
8
500240226
3
2
1
xxx
321
3
2
1
xxx
QR Decomposition
If A is a m×n matrix with linearly independent columns, then A can be decomposed as , where Q is a m×n matrix whose columns form an orthonormal basis for the column space of A and R is an nonsingular upper triangular matrix.
QRA
QR-Decomposition Theorem : If A is a m×n matrix with linearly independent columns, then
A can be decomposed as , where Q is a m×n matrix whose columns form an orthonormal basis for the column space of A and R is an nonsingular upper triangular matrix.
Proof: Suppose A=[u1 | u2| . . . | un] and rank (A) = n. Apply the Gram-Schmidt process to {u1, u2 , . . . ,un} and the orthogonal vectors v1, v2 , . . . ,vn are
Let for i=1,2,. . ., n. Thus q1, q2 , . . . ,qn form a orthonormal
basis for the column space of A.
QRA
121
122
2
212
1
1 ,,,
i
i
iiiiii v
v
vuv
v
vuv
v
vuuv
i
ii v
vq
QR-Decomposition
Now,
i.e.,
Thus ui is orthogonal to qj for j>i;
121
122
2
212
1
1 ,,,
i
i
iiiiii v
v
vuv
v
vuv
v
vuvu
112211 ,,, iiiiiiii qquqquqquqvu
},,{ },,,{ 221 iiii qqqspanvvvspanu
112211
223113333
112222
111
,,,
,,
,
nnnnnnnn qquqquqquqvu
qquqquqvu
qquqvu
qvu
Let Q= [q1 q2 . . . qn] , so Q is a m×n matrix whose columns form an orthonormal basis for the column space of A .
Now,
i.e., A=QR. Where,
Thus A can be decomposed as A=QR , where R is an upper triangular and nonsingular matrix.
QR-Decomposition
n
n
n
n
nn
v
quvququvquququv
qqquuuA
0000
,00,,0,,,
33
2232
113121
2121
n
n
n
n
v
quvququvquququv
R
0000
,00,,0,,,
33
2232
113121
QR Decomposition
Example: Find the QR decomposition of
100011001111
A
Applying Gram-Schmidt process of computing QR decomposition 1st Step:
2nd Step:
3rd Step:
Calculation of QR Decomposition
0313131
1
3
11
1
111
aa
q
ar
322112 aqr T
06/1
326/1
ˆˆ1
32ˆ
03/1
3/23/1
0313131
)3/2(
01
01
ˆ
22
2
222
121221122
q
qr
rqaaqqaq T
4th Step:
5th Step:
6th Step:
Calculation of QR Decomposition
313113 aqr T
613223 aqr T
6/26/1
06/1
ˆˆ1
2/6ˆ
12/1
02/1
ˆ
33
3
333
223113332231133
q
qr
qrqraaqqaqqaq TT
Therefore, A=QR
R code for QR Decomposition:x<-matrix(c(1,2,3, 2,5,4, 3,4,9),ncol=3,nrow=3)qrstr <- qr(x)Q<-qr.Q(qrstr)R<-qr.R(qrstr)
Uses: QR decomposition is widely used in computer codes to find the eigenvalues of a matrix, to solve linear systems, and to find least squares approximations.
Calculation of QR Decomposition
2/6006/16/203/13/23
6/2006/16/13/1
06/23/16/16/13/1
100011001111
Least square solution using QR Decomposition
The least square solution of b is
Let X=QR. Then
Therefore,
YXbXX tt
ZYQRbYQRRRbRRYQRRbR ttttttttt
11
YQRYX
RbRQRbQRbQRQRbXXttt
ttttt
Procedure To find out the cholesky decomposition Suppose
We need to solve the equation
nnnn
n
n
aaa
aaaaaa
A
21
22221
11211
TL
nn
n
n
L
nnnnnnnn
n
n
l
lllll
lll
lll
aaa
aaaaaa
A
00
0000
222
12111
21
2221
11
21
22221
11211
Example of Cholesky Decomposition
Suppose
Then Cholesky Decomposition
Now,
2/11
1
2
k
skskkkk lal
5222102224
A
311031002
L
For k from 1 to n
For j from k+1 to nkk
k
sksjsjkjk lllal
1
1
R code for Cholesky Decomposition
• x<-matrix(c(4,2,-2, 2,10,2, -2,2,5),ncol=3,nrow=3)• cl<-chol(x)
• If we Decompose A as LDLT then
and
13/12/1012/1001
L
300090004
D
Application of Cholesky Decomposition
Cholesky Decomposition is used to solve the system of linear equation Ax=b, where A is real symmetric and positive definite.
In regression analysis it could be used to estimate the parameter if XTX is positive definite.
In Kernel principal component analysis, Cholesky decomposition is also used (Weiya Shi; Yue-Fei Guo; 2010)
Jordan Decomposition• Let A be any n×n matrix then there exists a nonsingular matrix P and JK(λ)
a k×k matrix form
Such that
000
010001
)(kJ
)(000
0)(000)(
2
1
1 2
1
rk
k
k
rJ
JJ
APP
where k1+k2+ … + kr =n. Also λi , i=1,2,. . ., r are the characteristic roots And ki are the algebraic multiplicity of λi ,
Jordan Decomposition is used in Differential equation and time series analysis.
Spectral Decomposition
Let A be a m × m real symmetric matrix. Then
there exists an orthogonal matrix P such that or , where Λ is a diagonal
matrix.APPT TPPA
Basic Idea on Jacobi methodConvert the system: into the equivalent system:
• Generate a sequence of approximation
BAx
dCxx
dCxx kk )1()(,..., )2()1( xx
3333132131
2323122121
1313212111
bxaxaxabxaxaxabxaxaxa
33
32
33
321
33
313
22
23
22
231
22
212
11
13
11
132
11
121
abx
aax
aax
abx
aax
aax
abx
aax
aax
Jacobi iteration method
nnnnnn
nn
nn
bxaxaxa
bxaxaxabxaxaxa
2211
22222121
11212111
0
02
01
0
nx
xx
x
)(1 01
02121
11
11 nn xaxab
ax
)(1 011
022
011
1 nnnnnn
nnn xaxaxab
ax
)(1 02
0323
01212
22
12 nn xaxaxab
ax
1
1 1
1 1 i
j
n
ij
kjij
kjiji
ii
ki xaxab
ax
xk+1=Exk+f iteration for Jacobi method
A can be written as A=L+D+U (not decomposition)
00000
0
000000
000000
23
1312
33
22
11
3231
21
333231
232221
131211
aaa
aa
a
aaa
aaaaaaaaa
n
ij
kjij
i
j
kjiji
ii
ki xaxab
ax
1
1
1
1 1 xk+1=-D-1(L+U)xk+D-1bE=-D-1(L+U)f=D-1b
Ax=b (L+D+U)x=b
Dxk+1 =-(L+U)xk+b
kk UxLxDxk+1
Gauss-Seidel (GS) iteration
nnnnnn
nn
nn
bxaxaxa
bxaxaxabxaxaxa
2211
22222121
11212111
0
02
01
0
nx
xx
x
1
1 1
11 1 i
j
n
ij
kjij
kjiji
ii
ki xaxab
ax)(1 0
102121
11
11 nn xaxab
ax
)(1 111
122
111
1 nnnnnn
nnn xaxaxab
ax
)(1 02
0323
11212
22
12 nn xaxaxab
ax
Use the latestupdate
Gauss-Seidel MethodAn iterative method.
Basic Procedure:
Algebraically solve each linear equation for xi
Assume an initial guess solution array
Solve for each xi and repeat
Use absolute relative approximate error after each iteration to check if error is within a pre-specified tolerance.
Gauss-Seidel MethodAlgorithm
A set of n equations and n unknowns:
11313212111 ... bxaxaxaxa nn
2323222121 ... bxaxaxaxa n2n
nnnnnnn bxaxaxaxa ...332211
. . . . . .
If: the diagonal elements are non-zero
Rewrite each equation solving for the corresponding unknown
ex: First equation, solve for x1
Second equation, solve for x2
Gauss-Seidel MethodAlgorithm
Rewriting each equation
11
131321211 a
xaxaxacx nn
nn
nnnnnnn
nn
nnnnnnnnnn
nn
axaxaxac
x
axaxaxaxac
x
axaxaxacx
11,2211
1,1
,122,122,111,111
22
232312122
From Equation 1
From equation 2
From equation n-1
From equation n
Gauss-Seidel MethodAlgorithm
General Form of each equation
11
11
11
1 a
xac
x
n
jj
jj
22
21
22
2 a
xac
x
j
n
jj
j
1,1
11
,11
1
nn
n
njj
jjnn
n a
xac
x
nn
n
njj
jnjn
n a
xac
x
1
Derivation of the Trapezoidal Rule
Method Derived From Geometry
The area under the curve is a trapezoid. The integral
trapezoidofAreadxxfb
a
)(
)height)(sidesparallelofSum(21
)ab()a(f)b(f 21
2)b(f)a(f)ab(
Figure 2: Geometric Representation
f(x)
a b
b
a
dx)x(f1
y
x
f1(x)
Multiple Segment Trapezoidal Rule
f(x)
a b
y
x
4aba
42 aba
4
3 aba
Figure 4: Multiple (n=4) Segment Trapezoidal Rule
Divide into equal segments as shown in Figure 4. Then the width of each segment is:
nabh
The integral I is:
b
adx)x(fI
What is Integration?
Integration
b
adx)x(fI
The process of measuring the area under a curve.
Where:
f(x) is the integrand
a= lower limit of integration
b= upper limit of integration
f(x)
a b
y
x
b
a
dx)x(f
Basis of Simpson’s 1/3rd RuleTrapezoidal rule was based on approximating the integrand by a firstorder polynomial, and then integrating the polynomial in the interval ofintegration. Simpson’s 1/3rd rule is an extension of Trapezoidal rulewhere the integrand is approximated by a second order polynomial.
Hence
b
a
b
adx)x(fdx)x(fI 2
Where is a second order polynomial. )x(f2
22102 xaxaa)x(f
Basis of Simpson’s 1/3rd Rule
Choose
)),a(f,a( ,baf,ba
22
))b(f,b(
and
as the three points of the function to evaluate a0, a1 and a2.
22102 aaaaa)a(f)a(f
2
2102 2222
baabaaabafbaf
22102 babaa)b(f)b(f
Basis of Simpson’s 1/3rd Rule
Solving the previous equations for a0, a1 and a2 give
22
22
0 22
4
baba
)a(fb)a(abfbaabf)b(abf)b(faa
221 22
4332
4
baba
)b(bfbabf)a(bf)b(afbaaf)a(afa
222 22
22
baba
)b(fbaf)a(fa
Basis of Simpson’s 1/3rd Rule
Then
b
adx)x(fI 2
b
adxxaxaa 2
210
b
a
xaxaxa
32
3
2
2
10
32
33
2
22
10abaaba)ab(a
Basis of Simpson’s 1/3rd Rule
Substituting values of a0, a1, a 2 give
)b(fbaf)a(fabdx)x(fb
a 24
62
Since for Simpson’s 1/3rd Rule, the interval [a, b] is broken
into 2 segments, the segment width
2abh
Basis of Simpson’s 1/3rd Rule
0104.203.0 623 xxxxf
)b(fbaf)a(fhdx)x(fb
a 24
32
Hence
Because the above form has 1/3 in its formula, it is called Simpson’s 1/3rd Rule.
Multiple Segment Simpson’s 1/3rd Rule
Just like in multiple segment Trapezoidal Rule, one can subdivide the interval
[a, b] into n segments and apply Simpson’s 1/3rd Rule repeatedly overevery two segments. Note that n needs to be even. Divide interval[a, b] into equal segments, hence the segment width
nabh
nx
x
b
adx)x(fdx)x(f
0
where
ax 0 bxn
Multiple Segment Simpson’s 1/3rd Rule
.
.
Apply Simpson’s 1/3rd Rule over each interval,
...)x(f)x(f)x(f)xx(dx)x(fb
a
64 210
02
...)x(f)x(f)x(f)xx(
64 432
24
f(x)
. . .
x0 x2 xn-2 xn
x
.....dx)x(fdx)x(fdx)x(fx
x
x
x
b
a
4
2
2
0
n
n
n
n
x
x
x
xdx)x(fdx)x(f....
2
2
4
Multiple Segment Simpson’s 1/3rd Rule
...)x(f)x(f)x(f)xx(... nnnnn
64 234
42
64 12
2)x(f)x(f)x(f)xx( nnn
nn
Since
hxx ii 22 n...,,,i 42
Multiple Segment Simpson’s 1/3rd Rule
Then
...)x(f)x(f)x(fhdx)x(fb
a
6
42 210
...)x(f)x(f)x(fh
642 432
...)x(f)x(f)x(fh nnn
642 234
642 12 )x(f)x(f)x(fh nnn
Multiple Segment Simpson’s 1/3rd Rule
b
adx)x(f ...)x(f...)x(f)x(f)x(fh
n 1310 43
)}]()(...)()(2... 242 nn xfxfxfxf
)()(2)(4)(3
2
2
1
10 n
n
evenii
i
n
oddii
i xfxfxfxfh
)()(2)(4)(3
2
2
1
10 n
n
evenii
i
n
oddii
i xfxfxfxfnab
Simpson 3/8 Rule for Integration
The main objective of this chapter is to develop appropriate formulas for approximating the integral of the form
Euler’s Method
We have previously seen Euler’s Method for estimating the solution of a differential equation. That is to say given the derivative as a function of x and y (i.e. f(x,y)) and an initial value y(x0)=y0 and a terminal value xn we can generate an estimate for the corresponding yn. They are related in the following way:
xyxfyy
xxxyx
kkkk
kkkk ),(
),(1
111
The value x = (xn-x0)/n and the accuracy increases with n.
Taylor Method of Order 1
Euler’s Method is one of a family of methods for solving differential equations developed by Taylor. We would call this a Taylor Method of order 1. The 1 refers to the fact that this method used the first derivative to generate the next estimate. In terms of geometry it says you are moving along a line (i.e. the tangent line) to get from one estimate to the next.
Find the second derivative if the first derivative is given to the right.
Set f(x,y) = x2y and plug it into the formula below.
yxdxdy 2
dxdy
yf
xf
dxyd
2
2
yxxxy 222
Here we notice that:
22 xyfandxy
xf
yxxy 42
Higher Derivatives
Third, fourth, fifth, … etc derivatives can be computed with the same method. This has a recursive definition given to the right.
dxdy
dxyd
ydxyd
xdxyd
n
n
n
n
n
n
1
1
Picard IterationThe Picard method is a way of approximating solutions of ordinary differential equations. Originally it was a way of proving the existence of solutions. It is only through the use of advanced symbolic computing that it has become a practical way of approximating solutions.In this chapter we outline some of the numerical methods used to approximate solutions of ordinary differential equations. Here is a reminder of the form of a differential equation.
The first step is to transform the differential equation and its initial condition into an integral
Runge-Kutta 4th Order Method
where
hkkkkyy ii 43211 2261
ii yxfk ,1
hkyhxfk ii 12 2
1,21
hkyhxfk ii 23 2
1,21
hkyhxfk ii 34 ,
For0)0(),,( yyyxf
dxdy
Runge Kutta 4th order method is given by
How to write Ordinary Differential Equation
50,3.12 yeydxdy x
is rewritten as
50,23.1 yyedxdy x
In this case
yeyxf x 23.1,
How does one write a first order differential equation in the form of
yxfdxdy ,
UNIT - II
Fourier Cosine & Sine Integrals
Integral SineFourier :)sin()()(,0)(
)sin()(1odd is f(x) function theIf
Integral CosineFourier :)cos()()(
0)(
)cos()(2)(1)(1
)cos()(1A(w)even is f(x) function theIf
0
0
00
0
dwwxwBxfwA
dvwvvfB(w)
dwwxwAxf
wB
dvwvvfdvdv
dvwvvf
Example
dwwxw
wdwwxwAf(x)
dvwvdvwvvfwB
wwdvwvdvwvvfwA
f(x)
)cos()sin(2)cos()(
is f of integralFourier The
0)sin(1)sin()(1)(
)sin(2)cos(1)cos()(1)(
1xfor 01x1-for 1
Let
00
1
1
1
1
1
1
2 1 0 1 2
0
1
1.5
0.5
f 10 x( )
f 100 x( )
g x( )
22 x
f10 integrate from 0 to 10f100 integrate from 0 to 100g(x) the real function
Similar to Fourier series approximation, the Fourier integral approximation improves as the integration limit increases. It is expected that the integral will converges to the real function when the integration limit is increased to infinity.
Physical interpretation: The higher the integration limit means more higher frequency sinusoidal components have been included in the approximation. (similar effect has been observed when larger n is used in Fourier series approximation) This suggests that w can be interpreted as the frequency of each of the sinusoidal wave used to approximate the real function.Suggestion: A(w) can be interpreted as the amplitude function of the specific sinusoidal wave. (similar to the Fourier coefficient in Fourier series expansion)
Fourier Cosine Transform
)(ˆ of transformcosineFourier inverse theis )(
)cos()(ˆ2)cos()()(
f(x) of transformcosineFourier thecalled is )(ˆ
by x replaced been has ,)cos()(2)(2
)(ˆ
)(ˆ2 Define
.)cos()(2)( where,)cos()()(
:f(x) function even anFor
00
0
00
wfxf
dwwxwfdwwxwAxf
wf
vdxwxxfwAwf
wfA(w)
dvwvvfwAdwwxwAxf
c
c
c
c
c
Fourier Sine Transform
)(ˆ of transformsineFourier inverse theis )(
)sin()(ˆ2)sin()()(
f(x) of transformsineFourier thecalled is )(ˆ
by x replaced been has ,)sin()(2)(2
)(ˆ
)(ˆ2 Define
.)sin()(2)( where,)sin()()(
:f(x) function odd anfor Similarly,
00
0
00
wfxf
dwwxwfdwwxwBxf
wf
vdxwxxfwBwf
wfB(w)
dvwvvfwBdwwxwBxf
S
S
S
S
S
Improper Integral of Type 1a) If exists for every number t ≥ a, then
provided this limit exists (as a finite number).b) If exists for every number t ≤ b, then
provided this limit exists (as a finite number).The improper integrals and are called
convergent if the corresponding limit exists and divergent if the limit does not exist.
c) If both and are convergent, then we define
t
adxxf )(
b
tdxxf )(
t
aa tdxxfdxxf )()( lim
b
t
b
tdxxfdxxf )()( lim
adxxf )(
adxxf )(
adxxf )(
bdxxf )(
a
adxxfdxxfdxxf )()()(
Examples1
111111.1 limlimlim
11 21 2
txdx
xdx
x t
t
t
t
t
1.2 0000
limlimlim
t
tt
x
tt
x
t
x eeedxedxe
22tantan
tantan
11
11
11.3
11
0101
0
0 222
limlim
limlim
tt
xx
dxx
dxx
dxx
tt
t
tt
t
All three integrals are convergent.
1lnlnln11 limlimlim 111txdx
xdx
x t
t
t
t
t
An example of a divergent integral:
The general rule is the following:
1p ifdivergent and 1p if convergent is 11
dxx p
1 2 convergent is 1 that slide previous thefrom Recall dxx
Definition of an Improper Integral of Type 2a) If f is continuous on [a, b) and is discontinuous at b, then
if this limit exists (as a finite number).b) If f is continuous on (a, b] and is discontinuous at a, then
if this limit exists (as a finite number).The improper integral is called convergent if the
corresponding limit exists and divergent if the limit does not exist.
c) If f has a discontinuity at c, where a < c < b, and both and are convergent, then we define
t
a
b
abt
dxxfdxxf )()( lim
b
t
b
aat
dxxfdxxf )()( lim
b
cdxxf )(
c
adxxf )(
b
adxxf )(
b
c
c
a
b
adxxfdxxfdxxf )()()(