12 Super Derivative (Non-integer times Derivative)

12.1 Super Derivative and Super Differentiation

Defintion 12.1.1

f( )p ( )x obtained by continuing analytically the index of the differentiation operator of Higher Derivative of

a function f( )x to a complex plane [ ]0 ,p from a natural number interval [ ]1 ,n is called Super Derivative

of f( )x .

Example

( )sin x ( )p = sin x+2

p + cp( )x cp( )x is an arbitrary function.

12.1.2 Super Differentiation

Definition 12.1.2

We call it Super Differentiation to differentiate a function f with respect to an independent variable xnon-integer times continuously. And it is described as follows.

dxp

d p

f( )x = dxd

dxd

f( )x dxd

: p pieces

Example

dxp

d p

cosx = cos x+2

p

12.1.3 Fundamental Theorem of Super Differentiation The following theorem holds from Theorem 7.1.3 in 7.1.

Theorem 12.1.3

Let f ( )r r[ ]0 , p be an continuous function on the closed interval I and be arbitrary the r-th order

derivative function of f . And let a( )r be a continuous function on the closed interval [ ]0 , p .

Then the following expression holds for a( )r , x I .

dxp

d p

f( )x = f( )p ( )x + dxp

d p

Σr=0

p -1

f( )r a( )p -r

a( )p

x

a( )p-r

x

dxr(1.1)

Especially, when a( )r = a for all k[ ]0 , p ,

dxp

d p

f( )x = f( )p ( )x + dxp

d p

Σr=0

p -1

f( )r ( )a ( )1+r( )x-a r

(1.2)

Proof Theorem 7.1.3 in 7.1 can be rewritten as follows.

f < >p ( )x = a( )p

x

a( )0

x

f( )x dxp + Σr=0

p -1

f < >p-r a( )p -r

a( )p

x

a( )p-r+1

x

dxr

- 1 -

Especially, when a( )r = a for all k[ ]0 , p ,

f < >p ( )x = a

x

a

x

f( )x dxn + Σr=0

p -1

f < >p-r ( )a ( )1+r( )x-a r

Differentiating these both sides with respect to x p times,

dxp

d p

f < >p ( )x = f< >0 ( )x + dxp

d p

Σr=0

p -1

f < >p-r a( )p -r

a( )p

x

a( )p-r

x

dxr

dxp

d p

f < >p ( )x = f< >0 ( )x + dxp

d p

Σr=0

p -1

f < >p-r ( )a ( )1+r( )x-a r

Shifting by -p the index in the integration operator <> and replacing <> by differentiation operator () ,

we obtain the desired expression.

Constant-of-differentiation Function

We call dxp

d p

Σr=0

p -1

etc. Constant-of-differentiation Function of f( )x . Since p is a real number,

generally it is difficult to obtain this. However, it becomes easy exceptionally at the time of f( )x = e x.

That is, Constant-of-integration Function in 7.1.3 was as follows.

Σr=0

p -1

ea

( )1+r( )x-a r

= eaΣr=0

( )1+r( )x-a r

- ( )1+r+p( )x-a r+ p

Differentiating both sides with respect to x p times,

dx p

d p

Σr=0

p -1

ea

( )1+r( )x-a r

= eaΣr=0

dx p

d p

( )1+r( )x-a r

- ( )1+r+p( )x-a r+ p

From the 12.3.1 mentioned later, the following expressions hold.

dxp

d p

( )x-a r = ( )1+r-p( )1+r

( )x-a r-p , dxp

d p

( )x-a r+p = ( )1+r( )1+r+p

( )x-a r

Substituting this for the above, we obtain the following expression.

dxp

d p

Σr=0

p-1

ea

( )1+r( )x-a r

= eaΣr=0

( )1+r-p( )x-a r-p

- ( )1+r( )x-a r

12.1.4 Lineal and Collateral

In the case of the higher differentiation, since the constant-of-integration polynomial was degree n -1 ,

the constant-of-differentiation function which differentiated this n times became 0. However, in the case of

the super differentiation, since the constant-of-integration function is expressed by a series in general, the

constant-of-differentiation function which differentiated this p times does not become 0. This shows that

there are lineal and collateral in the super differentiation.

Definition 12.1.4

dxp

d p

f( )x = f( )p ( )x + dxp

d p

Σr=0

p -1

f( )r a( )p -r

a( )p

x

a( )p-r

x

dxr(1.1)

In this expression,

when Constant-of-differentiation Functin is 0,

- 2 -

we call dx p

d p

f( )x Lineal Super Differentiation and

we call the function equal to this Lineal Super Derivaive Function.

when Constant-of-differentiation Functin is not 0,

we call dx p

d p

f( )x Collateral Super Differentiation and

we call the function equal to this Collateral Super Derivaive Function .

These are the same also in (1.2) .

In short , Lineal Super Derivaive Function is what differentiated f( )x with respect to x continuously

without considering the constant-of-differentiation function.

Example: lineal derivative and collateral derivative of ex

In the case of easier fixed lower limit, from (1.2) in the theorem

dxp

d p

ex = ex + dxp

d p

Σr=0

p-1

ea

( )1+r( )x-a r

Here, using the former expression i.e.

dxp

d p

Σr=0

p-1

ea

( )1+r( )x-a r

= eaΣr=0

( )1+r-p( )x-a r-p

- ( )1+r( )x-a r

we obtain

dxp

d p

ex = ex + eaΣr=0

( )1+r-p( )x-a r-p

- ( )1+r( )x-a r

= eaΣr=0

( )1+r-p( )x-a r-p

When a- , since the constant-of-differentiation functin can not be 0, this is a collateral differentiation.

When a =-, since ea =0 , we obtain the following lineal differentiation.

dxp

d p

ex = ex

When p =1/2 , a =0 , if the differential quotients on x =0.3 are compared with the calculation result

by Riemann-Liouville differintegral (later 12.2.1), it is as follows.

- 3 -

And if the lineal super derivative and the collateral super derivative are illustrated side by side, it is as follows.

Remark It is thought that this collateral super derivative is an asymptotic expansion. And this collateral super derivative

is corresponding with the termwise super differentiation. In general, a termwise super differentiation seems to

become a collateral super differentiation.

12.1.5 The basic formulas of the Super Differentiation The following formulas hold like the higher differentiation.

c f( )x ( )p = c f( )p ( )x c0 : constant multiple rule

f( )x + g( )x ( )p = f( )p ( )x + g( )p ( )x : sum rule

- 4 -

12.2 Fractional Derivative

12.2.1 Riemann-Liouville differintegral

Among the super integrals of function f(x), the super integral whose lower limit function a( )k is a constant

a was calculable by Riemann-Liouville integral. The super derivativeof such a function f(x) is calculable by

Riemann-Liouville integral and integer times differentiation. It is as follows.

Let n = p =ceil( )p . First, integrate with f (x) n -p times. Next, differentiate it n times. And,

since the result is n- (n-p), it means that f(x) was differentiated p times.

f( )p ( )x = ( )n -p1

dxn

d n

a

x( )x-t n-p-1f()t dt n = p (2.0)

This expression is called Riemann-Liouville differintegral. "differintegral" is a coined word which combined

"differential" and "integral".

Although the numerical integration and the numerical differentiation are possible for (2.0) with this, the

accuracy of numerical differentiation is bad and the desired result may not be obtained. In this case, the

following formula which replaced the calculation order of integration and differentiation is effective.

f( )p ( )x = ( )n -p1

a

x( )x-t n-p-1 dtn

d n

f()t dt n = p (2.0')

Although this formula has a possibility of cutting off a constant of integration as a result of differentiating

previously, in many cases, it is correctly calculable. This (2.0') is often used in the following chapters.

12.2.2 Riemann-Liouville differintegral expressions of super derivatives of elementary functions Riemann-Liouville differintegral expressions of super derivatives of some elementary functions are as follows.

In the right side, super derivatives obtained by super differentiation are shown in advance. Needless to say,

Riemann-Liouville differintegral holds only if the lower limit function a( )k is a constant a . In addition, p is

a positive non-integer and n = p =ceil( )p in all the expressions.

x ( )p =

( )n -p1

dx n

d n

0x( )x -t n-p-1 t dt =

1+-p( )1+

x -p ( ) 0

= ( )n -p

1

x( )x -t n-p-1 dt n

d n

t dt = ( )-1 -p

( )-( )-+p

x -p ( ) < 0

e x ( )p = ( )n -p

1

dxn

d n

x( )x-t n-p-1 e tdt = ( )1 -p

e x

( )log x ( )p = ( )n -p1

dxn

d n

0

x( )x-t n-p-1 log t dt = ( )1-p

log x -( )1-p - x-p

Note

When n = p , -( )n-p < <0 , the following expression does not hold.

x ( )p = ( )n -p

1

dxn

d n

x( )x-t n-p-1 t dt (2.1)

In this case, n -p times integral of x ( )<0 is carried out, and super primitive function x + n-p

+n -p >0 is obtained. According to this, the zero of the super primitive function changes from to 0. The integral with a fixed lower limit is inapplicable to such an integral.

In this case, the following formula which replaced the calculation order of integration and differentiationis effective.

- 5 -

x ( )p = ( )n-p

1

x( )x-t n-p-1 dtn

d n

t dt n =[ ]p (2.1')

If this formula is used, n -p times integral of x -n ( )-n <0 is carried out, and super primitive function

x -p( )-p <0 is obtained. Since the zero of the super primitive function does not change, the integral

with a fixed lower limit is applicable. (See Example 5 in the following chapter).

Super Derivative & Fractional Derivative In Super Derivative which I developed, first, we obtain the higher derivative, next, extending the index of

the operator to real number, we obtain the super derivative.

On the contrary, in traditional Fractional Derivative , the super derivative is directly drawn from Riemann-

Liouville differintegral. However, the calculation is very difficult.

Three examples are shown below. In each example, the 1st is Super Derivative , and the 2nd is Fractional

Derivative .

Example 1

x1 21

= 1+1-1/2( )1+1

x1- 2

1 = 3/2

( )2x 2

1

=

2x 2

1

x1 21

= ( )1-1/21

dx1

d 1

0

x( )x-t

1- 21

-1 t1 dt

= ( )1/21

dxd

0

x( )x-t

- 21

t1 dt =

1

dxd -

32

x-t ( )2x+t0

x

=

1dxd 3

4x 2

3

=

2x 2

1

Example 2

e-x 21

= ( )-1- 2

1

e-x = i1

e-x

e-x 21

= ( )1-1/21

dx1

d 1

+

x( )x-t

1- 21

-1 e-tdt

=

1dxd

+

x( )x-t

- 21

e-tdt =

1dxd -e x erf i x-t

x

=

1dxd e-x erfi x- =

i1

dxd e-xerf i x-

= i1

dxd e-xerf - =

i1

dxd -e-x =

i1

e-x

Example 3

( )log x 2

1

= 1-1/2log x - 1-1/2 -

x- 2

1

= 1/2log x - 1/2 -

x- 2

1

- 6 -

=

log x - ( )--2 log2 - x

- 21

=

log x + 2 log 2 x

- 21

( )log x 2

1

= ( )1-1/21

dx1

d 1

0

x( )x-t

1- 21

-1 log t dt

=

1dxd

0

x( )x-t

- 21

log t dt

=

1dxd -4 x tanh -1 x

x-t - 2 x-t ( )log t-2

0

x

Here

tanh -1

nx

= logn-xn+x

= 21 log( )n +x - log( )n -x | |x < n

Then

tanh -1 x

x-t = 2

1 log x + x-t - log x - x-t

From this

4 x tanh -1 x

x-t = 2 x log x + x-t - log x - x-t

= 4 x log x + x-t - 2 x log x + x-t + log x - x-t

= 4 x log x + x-t - 2 x log x-( )x-t

= 4 x log x + x-t - 2 x log tTherefore

4 x tanh -1 x

x-t + 2 x-t ( )log t-2

= 4 x log x + x-t - 2 x log t + 2 x-t log t - 4 x-t

= 4 x log x + x-t - 2 x - x-t log t - 4 x-tUsing this

( )log x 2

1

=

1dxd -4 x tanh -1 x

x-t - 2 x-t ( )log t-2

0

x

=

1dxd -4 x log x + x-t +2 x - x-t log t +4 x-t 0

x

=

1dxd -4 x log x + 2 x log x

+

1dxd 4 x log x + x -2 x - x log0-4 x

=

1dxd 4 x log 2 x - 4 x 2 x log x = x log x

- 7 -

Furthermore

4 x log 2 x - 4 x = 4 x log 2 + 4 x log x - 4 x

= 4 x log 2 + 2 x log x - 4 x

= 4 x ( )log 2-1 + 2 x log xThus

( )log x 2

1

=

1dxd 4 x ( )log 2-1 +2 x log x =

log x +2 log2

x- 2

1

Reference

( )x-t- 2

1 t dt = - 3

2x-t ( )2x+t

( )x-t- 2

1 e t dt = -e x erf x-t erf( )z =

2

0

ze-t2

dt

( )x-t- 2

1 e-t dt = -e-x erf i x-t erf i( )z = erf( )iz /i

( )x-t- 2

1 log t dt = -4 x tanh -1 x

x-t - 2 x-t ( )log t-2

12.2.3 The demerit and the strong point of Fractional Derivative As seen in three upper examples, Fractional Derivative based on Riemann-Liouville differintegral is difficult

like this. Although Example 3 was the 1/2 times derivative of a logarithmic function, for this calculation, the

delicate technique was used abundantly, and one day was required. When p is a real number, the p times

derivative of this function is hopelessly difficult. Thus, in Fractional Derivative , it is very difficult to obtain the

derivative from Riemann-Liouville differintegral. Moreover, how to take a lower limit is not clear, and it cannot

treat trigonometric functions etc. These are the same as that of Fractional Integral .

A peculiar problem to Riemann-Liouville differintegral is not to be able to use it when p is an integer. It is

because of becoming ( )0 = of n=p at this time. In this case, it will rely on the Higher Derivative.

However, (2.0) and (2.0') are the powerful tools of numerical computation and can obtain super derivative of

the arbitrary points of arbitrary functions easily. And super derivatrive can be verified numerically with these.

- 8 -

12.3 Super Derivative of Power Function

12.3.1 Formula of Super Derivative of Power Function

Analytically continuing the index of the differentiation operator in Formula 9.2.1 (9.2 ) to [ ]0 ,p from [ ]1,n

we obtain the following formula. In addition, Rieamnn-Liouville differintegrals are also expressed together

Formula 12.3.1

When ( )z is gamma function and n = p is ceiling function, the following expressions hold

(1) Basic form

x ( )p =

( )1+-p( )1+

x -p = ( )n-p

1

dx n

d n

0x( )x-t n-p-1 t dt ( ) 0

= ( )-1 -p

( )-( )-+p

x -p = ( )n-p

1

x( )x-t n-p-1 dt n

d n

t dt ( ) <0

(2) Linear form

( )ax+b ( )p= a

1 -p

( )1+-p( )1+

( )ax+b -p ( ) 0

= ( )n-p1

dxn

d n

-

ab

x( )x-t n-p-1( )at+b dt

( )ax+b ( )p= -

a1 -p

( )-( )-+p

( )ax+b -p ( ) <0

= ( )n-p1

x( )x-t n-p-1 dtn

d n

( )at+b dt

Caution !

Do not describe a1 -p

to be a pin the upper formula. Because, since the law of exponents a p q

= a pq

does not hold for the arbitrary real numbers p,q at the time a<0,

a1 -p

= (a -1)-p a( )-1 ( )-p = a p i.e. a1 -p

a p

If it is described as a1 -p

= a p, the rotation direction on the complex plane becomes reverse and the

mistaken result is caused.

For example, when a=-3 , p=1/2 ,

a1 -p

= -31

- 21

= ( )-1 31

- 21

= ( )-1- 21

3 21

= -i 3

a p = ( )-3 21

= ( )-1 3 21

= ( )-1 21

3 21

= i 3

- 9 -

Example 1 > p

2x2 10

1

= 2 1+2-1/10

( )1+2x2- 10

1 = 171 9/10

100x 1019

=0.547239 x 1019

2x2 10

9

= 2 1+2-9/10

( )1+2x2- 10

9 = 11 1/10

100x 1011

= 0.955579 x 1011

When x 2/2 , x 2/2< >1/10

, x 2/2< >9/10

, x 1 are drawn on a figure side by side, it is as follows.

Example 2 = p

x1 ( )1 = 1+1-1

( )1+1x1-1 = 1

( )2x0 = 1

1x0 = 1

x 21

21

= 1+

21-21

1+21

x 21- 21

= 1

23

x0 = 2

Example 3 0 < p

x0 21

= 1+0-1/2( )1+0

x0- 2

1

= ( )1/2( )1

x- 21

= 1

x- 21

x1( )2

= 1+1-2( )1+1

x1-2 = ( )0( )2

x-1 = 1

x-1 = 0

Example 4 < 0

x-1 21

= ( )-1- 21

-( )-1 -( )-1 +1/2

x-1- 2

1

= -i 2

x- 23

x-1( )1

= ( )-1 -1

-( )-1 -( )-1 +1

x-1-1 = - ( )1( )2

x-2 = -x-2

x- 21 ( )2

= ( )-1 -2

-( )-1/2 -( )-1/2 +2

x- 21-2

= ( )1/2( )5/2

x- 25

= 43

x- 25

- 10 -

Example 5 - p -p < <0

x-31 2

1

= ( )-1-21

- -31

- -31

+21

x-31-21

= -i ( )1/3( )5/6

x-65

= -2.67891.1287

i x-65

= -0.42135 i x-65

x-31

21

= 1-

21

1

x( )x -t

-21

dt 1d 1

t-31

dt = -

3 1

x( )x-t

-21

t-34

dt

This last integral cannot be expressed with an elementary function. Then, if the value on x=1 is numerically

integrated by mathematical software, it is as follows. This result is consistent with the previous value

-0.42135i exactly.

Riemann-Liouville differintegral a := -1/3: p := 1/2: df := diff(t^a,t)

1

3 t43

fl := x-> 1/gamma(ceil(p)-p) *int((x-t)^(ceil(p)-p-1)*df, t=infinity..x)

x 1(p p)

x(x t)p p 1 df d t

float(fl(1))

0.4213560763 i

12.3.2 Half Derivative of a power function Especially, Super Derivative of order 1/2 is called Half Derivative.

Formula 12.3.2

Let n be a non-negative integer, -1!! 1, ( )2n-1 !! 135( )2n-1 ,

0!! 1, 2n!! 2462n ,then following expressions hold.

(1) Basic form

xn 21

= ( )2n-1 !!

( )2n !! x

n- 21

xn+ 2

1 21

= 2( )n+1 !!

( )n+1 ( )2n+1 !! xn

- 11 -

(2) Linear form

( )ax+b n 21

= ( )2n-1 !!( )2n !!

a

( )ax+bn- 2

1

( )ax+bn+ 2

1 21

= 2( )n+1 !!

( )n+1 ( )2n+1 !!a ( )ax+b n

Proof

Upper rows 0 of the linear form of Formula 12.3.1 were as follows.

( )ax+b n 21

= a1

- 21

1+n- 21

( )1+n( )ax+b

n- 21

( )ax+bn+ 2

1 21

= a1

- 21

( )1+n

1+n+ 21

( )ax+b n

Here, when n is a non-negative integer,

n+ 21

= 2n

( )2n-1 !! , ( )2n !! = 2n n!

Then

1+n- 21

( )1+n =

n+ 21

( )1+n =

( )2n-1 !! 2n n!

= ( )2n-1 !!

( )2n !!

( )1+n

1+n+ 21

= ( )1+n

( )1+n +21

= n!1

21+n

2( )1+n -1 !!

= 2n+1 ( )n+1 !

( )n+1 ( )2n+1 !! =

2( )n+1 !!( )n+1 ( )2n+1 !!

Substituting these for the previous formula, we obtain the linear form, andgiving a=1, b=0 to them,

we obtain the basic form.

Example 1

x0 21

= ( )-1 !!

0!! x- 21

= 1

x- 21

( a0x- 21

)

x1 21

= 1!! 2!!

x 21

= 2

x 21

( a1x 21

)

x2 21

= 3!! 4!!

x 23

= 3 8

x 23

( a2x 23

)

x3 21

= 5!! 6!!

x 25

= 5 16

x 25

( a3x 25

)

- 12 -

Example 2

x 21 2

1

= 2!!11!!

x0 = 2

x0 ( = a1

1x0 )

x 23 2

1

= 4!!23!!

x1 = 43

x1 ( = a2

2x1 )

x 25 2

1

= 6!!35!!

x2 = 1615

x2 ( = a3

3x2 )

x 27 2

1

= 8!!47!!

x3 = 3235

x3 ( = a4

4x3 )

12.3.3 Half Derivative of an integer power function (Fractional Derivative) Next, using Riemann-Liouville differintegral, we obtain the Half Derivative of an integer power function.

Formula 12.3.3 When n denots a natural number, the following expression holds.

xn 21

=

2n+1Σk=0

n

2k+1( )-1 k

n

k x

n- 21

Proof

Let n be a natural number, =n , p=1/2 . Then, since >p , from Formula 12.3.1 we obtain

the following expression.

xn 21

= ( )1/21

dx1d 1

0

x( )x-t

- 21

t n dt =

1

dxd

0

x

x-t

tn

dt

Here, according to「岩波数学公式Ⅰ」p96, the following expressions hold

x-t

tn

dt = ( )-1 n+1

2 x-tΣr=0

n

( )-x r n

r 2n-2r+1( )x-t n-r

Then

0

x

x-t

tn

dt = ( )-1 n+1

2 x-tΣr=0

n

( )-x r n

r 2n-2r+1( )x-t n-r

0

x

= ( )-1 n

2x 21

Σr=0

n

( )-1 r n

r 2n-2r+1xn

Differentiate both sides of this with respect to x as follows.

dxd

0

x

x-t

tn

dt = dxd ( )-1 n

2Σr=0

n

( )-1 r n

r 2n-2r+11

xn+ 2

1

= ( )-1 n

2 n+ 2

1Σr=0

n

( )-1 r n

r 2n-2r+11

xn- 2

1

- 13 -

= ( )2n+1 Σr=0

n

2n-2r+1( )-1 r-n

n

r x

n- 21

xn 21

=

2n+1Σr=0

n

2n-2r+1( )-1 r-n

n

r x

n- 21

Here, we devise further,

Σr=0

n

2n-2r+1( )-1 r-n

n

r = Σ

r=0

n

2( )n-r +1( )-1 n-r

n

n-r = Σ

k=0

n

2k+1( )-1 k

n

kUsing this, we obtain

xn 21

=

2n+1Σk=0

n

2k+1( )-1 k

n

k x

n- 21

By-product Comparing Formula 12.3.2 and Formula 12.3.3 , we obtain the following formula.

Σk=0

n

2k+1( )-1 k

n

k = ( )2n+1 !!

( )2n !!

This is the same as the by-product in 7.3.3 .

12.3.4 Fractional Derivative of an integer power function Generalizing Formula 12.3.3 , we calculate a fractional derivative of an integer power function. First, we

prepare the following lemma.

Lemma When m,n are natural numbers, the following expression holds.

( )x-t-

m1

tn dt = ( )-1 n+1

m( )x-t mm-1

Σr=0

n

( )-x r n

r m( )n-r +( )m-1( )x-t n-r

(4.0)

Proof Let

F()t = ( )-1 n+1

m( )x-t mm-1

Σr=0

n

( )-x r n

r m( )n-r +( )m-1( )x-t n-r

Differentiate this with respect to t. Then

dtd

F()t = dtd

( )-1 n+1

m( )x-t mm-1

Σr=0

n

( )-x r n

r m( )n-r +( )m-1( )x-t n-r

+ ( )-1 n+1

m( )x-t mm-1

dtdΣr=0

n

( )-x r n

r m( )n-r +( )m-1( )x-t n-r

= -( )-1 n+1

( )m-1 ( )x-t-

m1

Σr=0

n

( )-x r n

r m( )n-r +( )m-1( )x-t n-r

- 14 -

- ( )-1 n+1

m( )x-t mm-1

Σr=0

n

( )n-r ( )-x r n

r m( )n-r +( )m-1( )x-t n-r-1

= ( )-1 n

( )x-t-

m1

Σr=0

n

( )m-1 +m( )n-r ( )-x r n

r m( )n-r +( )m-1( )x-t n-r

= ( )-1 n

( )x-t-

m1

Σr=0

n

n

r( )-x r( )x-t n-r =

( )-1 n

( )x-t-

m1

( )-t n

= ( )x-t-

m1

t n

Using this Lemma, we obtain the following formula.

Formula 12.3.4

When n, m are natural numbers such that n1, m2 , the following expressions hold.

xn m1

= - ( )-1/mm( )mn+m-1

Σk=0

n

mk+( )m-1( )-1 k

n

k x

n-m1

(4.1)

= ( )1-1/mmn+( )m-1

Σk=0

n

mk+( )m-1( )-1 k

n

k x

n-m1

(4.1')

Σk=0

n

mk+( )m-1( )-1 k

n

k = -

m( )mn+m-1( )1+n , -1/m

() : beta function (4.2)

Proof

x ( )p = 1+-p

( )1+x-p = ( )k-p

1

dxk

d k

0

x( )x-t k-p-1 t dt ( ) p

Give =n , p=1/m to this. Then since k = 1/m =1 ,

xn m1

= ( )1-1/m1

dx1d 1

0

x( )x-t

-m1

t n dt

= - ( )-1/mm

dxd

0

x( )x-t

-m1

t n dt

Using the above Lemma, we calculate as follows.

0

x( )x-t

-m1

t n dt = ( )-1 n+1

m( )x-t mm-1

Σr=0

n

( )-x r n

r m( )n-r +( )m-1( )x-t n-r

0

x

= ( )-1 n

m x mm-1

Σr=0

n

( )-x r n

r m( )n-r +( )m-1xn-r

- 15 -

= mΣr=0

n

m( )n-r +( )m-1( )-1 r-n

n

r x

n-m1+1

Furthermore, since r,n are integers,

Σr=0

n

m( )n-r +( )m-1( )-1 r-n

n

r = Σ

r=0

n

m( )n-r +( )m-1( )-1 n-r

n

n-r

= Σk=0

n

mk+( )m-1( )-1 k

n

kUsing this,

0

x( )x-t

-m1

t n dt = mΣ

k=0

n

mk+( )m-1( )-1 k

n

k x

n-m1+1

Differentiating this with respect to x,

dxd

0

x( )x-t

-m1

t n dt = ( )mn+m-1 Σ

k=0

n

mk+( )m-1( )-1 k

n

k x

n-m1

Thus

xn m1

= - ( )-1/mm( )mn+m-1

Σk=0

n

mk+( )m-1( )-1 k

n

k x

n-m1

(4.1)

Moreover, (4.1') follows immediately from this.

Next,

xn m1

= ( )1+n-1/m( )1+n

xn-

m1

= ( )-1/m( )1+n

( )1+n-1/m( )-1/m

xn-

m1

Since this have to be equal to (4.1) ,

-m( )mn+m-1 Σk=0

n

mk+( )m-1( )-1 k

n

k = 1+n-1/m

( )1+n ( )-1/m

From this

Σk=0

n

mk+( )m-1( )-1 k

n

k = -

m( )mn+m-1( )1+n , -1/m

(4.2)

Remark (4.2) suggests that (4.1) can be expressed with a beta function and n,m can be real numbers. Actuality,

( )1+-p( )1+

= ( )-p( )1+

( )1+-p -p

= ( )-p( )1+ , -p

Then,

x ( )p = ( )1+-p

( )1+ x-p = ( )-p

( )1+ , -p x-p ( ) 0

12.3.5 Super Derivative of an integer power function

Replacing 1/m with p in Formula 12.3.4 , we obtain the following formula.

Formula 12.3.5

When n is a natural number, the following expressions hold for 0< p n .

- 16 -

xn ( )p = ( )1-p

n+1-pΣk=0

n

k+1-p( )-1 k

n

k xn-p

(5.1)

( )n , -p = -p

n-pΣk=0

n-1

k+1-p( )-1 k

n-1

k () denots beta function. (5.2)

Example 1

x2 3 =

1+2- 3

( )1+2 x2- 3 =

3- 3

2 x2- 3 = 2.2151 x2- 3

x2 3

= 1- 3

3- 3 1- 3

1 2

0-2- 3

1 2

1+3- 3

1 2

2 x2- 3

= 2.2151 x2- 3

Example 2

2 , - 3 = -3

2- 3 1- 3

1 1

0-2- 3

1 1

1 = 0.788675

( )3 , -e = -e3-e 1-e

1 2

0-2-e1 2

1+3-e1 2

2 = -0.596137

12.3.6 Super Derivative of a positive power function Since Formula 12.3.5 are binomial forms, the further generalization is possible.

Formula 12.3.6

The following expressions hold for p ,q such that 0< p q .

xq ( )p = ( )1-p

q+1-pΣr=0

r+1-p( )-1 r

q

r xq-p

(6.1)

( )q , -p = -p

q-pΣr=0

r+1-p( )-1 r

q-1

r() denots beta function. (6.2)

12.3.7 Super Derivative of a polynomial

In the case of a polynomial f( )x =Σk=0

m

cm-k xm-k

, as for the zero of the super derivative, it is good to

perform it as follows. It is based on experience of a writer.

(1) When f( )x is factored by primary formula ax+b , let -b/a be the zero.

(2) When f( )x is not factored by primary formula ax+b , let 0 be a the zero.

For example, in the case of the 1/2 times derivative of f( )x = x 2-2 x+ 2, the following calculation is

right in many case.

x2-2x+2 21

= ( )x- 2 21

= 3 8

( )x- 23

If this is calculated termwise as follows, the result is different from the former.

- 17 -

x2-2x+2 21

= x2 21

- 2 x1 21

+ 2 x0 21

= 1

38

x 23

- 4 x 2

1 + 2x

- 21

Needless to say, this cause is the difference between

x

x

x2-2x+2 dx- 21

and 0

x

0

x

x2-2x dx- 21

+

x

x2dx

- 21

That is, it is because the latter regarded it as 0 and although the former regarded the zero of the super

derivative as pi.

The latter is right when there is a special reason why the zero of the super derivative should be 0. However,

such a case is rare, and in almost all cases the former is right.

- 18 -

12.4 Super Derivative of Exponential Function

Analytically continuing the index of the differentiation operator in Formula 9.2.2 (9.2 ) to [ ]0 ,p from [ ]1,nwe obtain the following formula. In addition, Rieamnn-Liouville diffeintegrals are also expressed together

Formula 12.4.1

When n = p is ceiling function, the following expressions hold

(1) Basic form

e x ( )p = ( )1 -p

e x = ( )n -p1

dxn

d n

x( )x-t n-p-1 e tdt

(2) Linear form

eax+b ( )p= a

1 -p

eax+b = ( )n -p1

dxn

d n

x( )x-t n-p-1eat+bdt

a >0:- , a <0:+(3) General form

ax+b ( )p= a log

1 -p

ax+b = ( )n -p1

dxn

d n

x( )x-t n-p-1 at+b

dt

a >0:- , a <0:+

Proof of the general form

Let c=a log , d =b log , then

ecx+d = exalog+b log = eaxlogeblog = elog ax elog b

= ax b = ax+b

Applying this to (2) Linear form, we obtain (3) General form immediately.

Example

e-x 21

= ( )-1- 2

1

e-x = i1

e-x = -ie-x

e3x-4 2 = 3

1 - 2

e3x-4 = 4.728804 e3x-4

2x ( )i = log 2

1 -i

2x = ( )0.933582 - 0.358362 i 2x

( )-3 x 21

= log( )-31

- 21

( )-3 x

= ( )1.487742 + 1.05582 i ( )-3 x

3x 21

= log 31 -1/2

3x = 1.048147 3x

When 3x ( )1 , 3x ( )1/2

, 3x are drawn on a figure side by side, it is as follows.

- 19 -

3^x*ln(3)3^x/(1/ln(3))^(1/2)3^x

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0

2

4

6

8

10

x

y

- 20 -

12.5 Super Derivative of Logarithmic Function Reversing the sign of the index of the integration operator in Formula 7.5.1 (7.5 ), we obtain the following

formula. In addition, Rieamnn-Liouville differintegrals are also expressed together

Formula 12.5.1

When ( )z , ( )z , n = p denote zeta function, psi function, ceiling function respectively, the followingexpressions hold.

(1) Basic form

( )log x ( )p = ( )1-plog x -( )1-p -

x-p = ( )n -p1

dxn

d n

0

x( )x-t n-p-1 log t dt

(2) Linear form

log( )ax+b ( )p = ( )1-plog( )ax+b -( )1-p -

x+ab -p

= ( )n -p1

dxn

d n

-

ab

x( )x-t n-p-1log( )at+b dt

Where, ( )1-p( )1-p

= ( )-1 p( )p -1 ! for p =1,2,3,

Proof Formula 7.5.1 (7.5 ) was as follows.

0

x

0

x

log xdxp = ( )1+plog x -( )1+p -

xp

-

ab

x

-ab

x

log( )ax+b dxp = ( )1+plog( )ax+b -( )1+p -

x+ab p

Since differentiation is the reverse operation of integration, replacing the index p of the integration operator

with -p , we obtain the desired expressions. And Formula 1.3.1 (1.3 ) was

( )-n( )-n

= ( )-1 n+1n! , n =0,1,2,3,

Then, replacing n with p -1 , we obtain

( )1-p( )1-p

= ( )-1 p( )p -1 ! , p =1,2,3,

Example

( )log x ( )1 = 1-1log x - 1-1 -

x-1 = -( )-1 10!x-1 = x-1

log( )3x+4 21

= 1-

21

log( )3x+4 - 1-21

-

x+34

- 21

= 0.5641( )log x + 1.3862 / x +1.3333

- 21 -

( )log x 10

1 =

1-101

log x - 1-101

- x

- 101

= 0.9357x-

101

( )log x +0.1777

( )log x 10

9 =

1-109

log x - 1-109

- x

- 109

= 0.1051 x-

109

( )log x +9.8465

When log x , ( )log x ( )1/10 , ( )log x ( )9/10 , ( )log x ( )1 are drawn on a figure side by side,

it is as follows.

ln(x)-1/x^(1/10)/gamma(9/10)*(EULER - ln(x) + psi(9/10)-1/x^(9/10)/gamma(1/10)*(EULER - ln(x) + psi(1/10)1/x

1 2 3 4 5 6 7

-6

-4

-2

0

2

x

y

- 22 -

12.6 Super Derivative of Trigonometric Function

12.6.1 Super Derivatives of sin x , cos x

Analytically continuing the index of the differentiation operator in Formula 9.2.4 (9.2 ) to [ ]0 ,p from [ ]1,nfwe obtain the following formula.

Formula 12.6.1

(1) Basic form

( )sin x ( )p = sin x+2

p

( )cosx ( )p = cos x+2

p

(2) Linear form

sin( )ax+b ( )p = a1 -p

sin ax+b+2

p

cos( )ax+b ( )p = a1 -p

cos ax+b+2

p

Example

( )sin x 2

1

= sin x+21

2

= sin x+4

((sin x) 21

) 21

= (sin (x+4

)) 21

= 11

- 21

sin x+4

+21

2

= sin x+2

= cos x

sin x-2 ( )1

= 11 -1

sin x-2

+2

1 = sin x

( )cos x 2

1

= cos x+21

2

= cos x+4

( )cos x

2

= cos x+ 2

2

= cos( )x+1

When cos x , ( )cos x ( )1/2 , -sin x are drawn on a figure side by side, it is as follows. Red shows

1/2 order super derivative. It is clear also in the figure that super derivative which is the easiest to understand

is super derivative of trigonometric functions.

- 23 -

12.6.2 Termwise Super Derivative of sin x , cos x Reversing the sign of the index of the operator of the collateral super integrals of sin x , cos x in 7.6.2 (7.6 ) ,

we obtain the following termwise super derivatives. These are collateral super derivatives as understood from

the constant-of-differentiation function in the right side.

( )sin x ( )p =Σk=0

( )2k+2-p( )-1 k

x2k+1- p = sin x+2

p + C( )p,x

C( )p,x =Σk=1

p

( )1-p-kx-p-k

sin 2k

( )cosx ( )p =Σk=0

( )2k+1-p( )-1 k

x2k- p = cos x+2

p - C( )p,x

C( )p,x =Σk=1

p

( )1-p-kx-p-k

cos 2k

When the 1/2th order collateral super derivative of cos x is drawn as cos x and -sin x side by side, it is as

follows. Red shows the 1/2th order collateral super derivative.

- 24 -

Compared with the upper figure, the collateral super derivative is curving unnaturally near the coordinate origin

in this figure. Since this is similar also in sin x, it is thought that the termwie super derivatives of sin x and cos x

are asymptotic expansions of the lineal super derivatives.

- 25 -

12.7 Super Derivative of Hyperbolic Function

12.7.1 Super Derivatives of sinh x , cosh x

Analytically continuing the index of the differentiation operator in Formula 9.2.5 (9.2 ) to [ ]0 ,p from [ ]1,nwe obtain the following formula.

Formula 12.7.1

(1) Basic form

( )sinh x ( )p = i-p sinh x+2

p i = 2

ex - ( )-1 -pe-x

( )cosh x ( )p = i-p cosh x+2

p i = 2

ex + ( )-1 -pe-x

(2) Linear form

sinh( )ax+b ( )p = ai -p

sinh ax+b+2

p i

= 21 a

1 -n

eax+b - ( )-1 -pe-( )ax+b

cosh( )ax+b ( )p = ai -p

cosh ax+b+2

p i

= 21 a

1 -n

eax+b + ( )-1 -pe-( )ax+b

Example 1

( )sinh x 2

1

= i- 2

1

sinh x+21

2 i

= i- 2

1

sinh x+4 i

((sinh x) 21

) 21

= (i- 2

1

sinh(x+4 i

)) 21

= i- 2

1

i- 2

1

sinh x+4 i

+21

2 i

= i-1 sinh x+2 i

= cosh x

cosh x+2 i ( )1

= 1i -1

cosh x+2 i

+2

1 i

= -i cosh( )x+ i = i cosh x

( )cosh x 10

i

= i- 10

i cosh x-

20

= 1.170088 cosh x-20

( )cosh x 10

9 i

= i- 10

9 i cosh x-

209

= 4.111207 cosh x-209

- 26 -

Super derivative of hyperbolic function is the most incomprehensible in super derivatives. The reason is that

the super derivative turns into a complex function except the order p is an integer or a purely imaginary number.

Then, when cosh x , ( )cosh x ( )i/10 , ( )cosh x ( )9 i/10 , sinh x which can be displayed on a real number

domain are drawn on a figure side by side, it is as follows.

cosh(x)1/(I)^(1/10*I)*cosh(x - 1/20*PI)1/(I)^(9/10*I)*cosh(x - 9/20*PI)sinh(x)

-4 -3 -2 -1 1 2 3 4 5 6

100

200

300

400

x

y

All of four curves have overlapped in the positive area. It is natural that ( )cosh x ( )i/10 is near cosh x in

a negative area. But ( )cosh x ( )9 i/10 is far apart from sinh x in why.

Example 2

( )sinh x 2

1

= 2ex - ( )-1

- 21

e-x

= 2ex + i e-x

((sinh x) 21

) 21

= 2ex + i e-x 2

1

= 2ex

+ 2i e-x 2

1

= 2ex

+ 2i -i e-x = 2

ex + e-x

= cosh x

cosh x+2 i ( )1

= 21 1

1 -1

ex+ 2

i + ( )-1 -1e

- x+ 2 i

= 21 e 2

i ex - e

- 2 i

e-x = 2

1 i ex -

i1

e-x

= 2i ex + e-x = i cosh x

( )sinh x

i

= 2ex - ( )-1

- i

e-x

= 2ex - ei

- i

e-x

= 2ex - e1-x

- 27 -

12.7.2 Termwise Super Derivative of sinh x , cosh x Reversing the sign of the index of the operator of the collateral super integrals of sinh x , cosh x in 7.7.2

we obtain the following termwise super derivatives. These are collateral super derivatives as understood from

the constant-of-integration function in the right side.

( )sinh x ( )p =Σk=0

( )2k+2-px2k+1- p

= i-p sinh x+2

p i - C( )p,x

C( )p,x Σk=1

p

( )1-p -kx-p-k

i-k sinh 2k i

( )cosh x ( )p =Σk=0

( )2k+1-px2k- p

= i-p cohs x+2

p i - C( )p,x

C( )p,x Σk=1

p

( )1-p -kx-p-k

i-k cosh 2k i

When the values of the lineal and the collatera l .7th order derivatives of cosh x on x=1, x=6 are calculated

respectively, it is as follows.

Though the difference of both is large where x is small, both are almost corresponding ing where x is large.

Since this is similar also in sinh x, it is thought that the termwise super derivatives of sinh x and cosh x are

asymptotic expansions of the lineal super derivatives.

- 28 -

12.8 Super Derivative of Inverse Trigonometric Function

12.8.1 Super Derivatives of tan-1x , cot -1x Reversing the sign of the index of the integration operator in Formula 7.8.1 (7.8 ), we obtain the following

formula.

Formula 12.8.1

When ( )x ,( )x denote gamma function and digamma function respectively, the following expressions

hold for x 1 .

tan -1x( )p

= ( )1-ptan -1x

Σk=0

( )-1 k

-p

-p-2kx-p-2k

+ 2( )1-p

log 1+x2Σk=1

( )-1 k

-p

-p+1-2kx-p+1-2k

- ( )1-p1

Σr=1

(-1)r

-p

-p+1-2r ( )1-p -( )2r x-p+1-2r

cot-1x( )p

= ( )1-px-p

cot-1x - ( )1-ptan -1x

Σk=1

( )-1 k

-p

-p-2kx-p-2k

- 2( )1-p

log 1+x2Σk=1

( )-1 k

-p

-p+1-2kx-p+1-2k

+ ( )1-p1

Σr=1

(-1)r

-p

-p+1-2r ( )1-p -( )2r x-p+1-2r

Example: 1/2th order derivative of cot -1x

Riemann-Liouville differintegral g := x-> 1/gamma(1-p)*int((x-t)^(1-p-1)*arccot(t), t=0..x)

- 29 -

x 1(1 p)

0

x(x t)1 p 1 arccot(t) d t

2010.07.07

K. Kono

Alien's Mathematics

- 30 -