21 Super Calculus of the product of many functions

21.1 Super Integrals of the product of many functions

(1) Generalized binomial theorem and Super integral of the product of 2 functions According to the generalized binomial theorem in 3.2 , the following expression holds for the real numbers

x1 , x2 such that x1 > x2 and a positive number p .

x1+x2-p = Σ

r=0

-p

rx1

-p-rx2 r

On the other hand, according to Formula 17.1.2 in 17.1 , the following expression holds for the functions f1 , f2of x and a positive number p .

a

x

a

x

f1 f2< >p dx2= Σ

r=0

m -1

-p

rf1< >p+r f ( )r

2 + Rmp

Rmp = ( )p ,m

( )-1 m

Σk=0

m+ k1

p -1

k a

x

a

x

f< >m+ k1 f( )m+ k

2 dxp

Reversing the sign of the index of the differentiation operator ( )p in the Leibniz Rule about the Super

Differentiation in 19.1 ,

f1 f2( )-p = Σ

r=0

m -1

-p

rf1( )-p-r f ( )r

2 + Rm-p

Rm-p = ( )p ,m

( )-1 m

Σk=0

m+ k1

p -1

k f< >m+ k1 f( )m+ k

2( )-p

And, replacing ( )-p with the intagration operator < >p , we obtain the above formula.

(2) Generalized multinomial theorem and Super Integral of the product of many functions According to the generalized multinomial theorem in 3.4 , the following expression holds for real numbers

x1 , x2 , , x such that x1 > x2 + x3 ++ x and a positive number p .

x1+x2++x-p

=Σr1=0

Σr2=0

r1

Σr-1=0

r-2

-p

r1 r1

r2

r-2

r-1

x1-p-r1 x2

r1-r2 xr-1

Therefore, the following expression must hold for functions f1, f2 , , f of x and a positive number p .

a

x

a

x

f1 f2 f dxp =Σr1=0

m-1

Σr2=0

r1

Σr-1=0

r-2

-p

r1 r1

r2

r-2

r-1

f p+ r11 f r1-r2

2 f r-1

+ Rmp

21.1.1 Super Integral of the product of many functions

Theorem 21.1.1

Let p , r are positive numbers, m be a natural number, f( )rk be the r th order derivative function of

fk( )x k =1,2,, , f < >rk be arbitrary r th order primitive function of fk( )x and ( )p ,m be

the beta function. At this time, if there is a number a such that

f< >r1 ( )a = 0 r[ ]0 ,m+p or f( )s

k ( )a = 0 s[ ]0 ,m+p -1 for at least one k >1 ,

the following expression holds

- 1 -

a

x

a

x

f1 f2 f dxp =Σr1=0

m-1

Σr2=0

r1

Σr-1=0

r-2

-p

r1 r1

r2

r-2

r-1

f p+ r11 f r1-r2

2 f r-1

+ Rmp

Rmp = ( )p ,m

( )-1 m

Σk1=0

Σk2=0

m +k1

Σk3=0

k2

Σk-1=0

k-2

m+ k1

1

p-1

k1 m+k1

k2 k2

k3

k-2

k-1

a

x

a

x

f m+ k11 f m+k1-k2

2 f k2-k33 f k-1

dxp

Proof

Analytically continuing the index of the integration operator in Formula 20.2.1 in 20.2 to [ ]0 ,p from[ ]1 ,nwe obtain the desired expression.

Example computation of the product of three functions

Example1 Super Integral of x ex sinx

Since the zero of the higher integral of x e x sin x is x =- , the zero of the super integral is also

the same. Then let f1 = x , f2 = e x , f3 = sin x

x < >p+ r = 1++p +r

( )1+x+p+ r , x < >m+ r

= 1++m+r( )1+

x+m+ r

ex ( )r-s = ex ( )m+ r-s

= ex , ( )sin x ( )s = sin x+2s

Substituting these for Theorem 21.1.1 , we obtain

-

x

-

x

x ex sin xdxp

= Σr=0

m-1

Σs=0

r

-p

r r

s 1++p +r( )1+

x+p+ r ex sin x+2s

+ Rmp

(1.1)

Rmp = ( )p ,m

( )-1 m

Σr=0

Σs=0

m + r

m+ r1

p -1

r m+ r

s

-

x

-

x

1++m+r( )1+

x+m+ rexsin x+2s

dxp(1.1r)

As seen in Theorem 20.2.1 in 20.2 , this Rmp

is not converged on 0 at the time of m . That is, this

polynomial is an asymptotic expansion. When =3 , p =5/2 , the values of the both sides on arbitrary

point x =20 are as follows.

- 2 -

Example2 Super Integral of x ex log x

Since the zero of the higher integral of x e x log x is x =- , the zero of the super integral is also

the same. Then let f1 = x , f2 = e x , f3 = log x

x < >p+ r=

1++p +r( )1+

x +p+ r , x < >m+ r=

1++m+r( )1+

x +m+ r

e x ( )r-s = e x , e x ( )m+ r-s

= e x

( )log x ( )0 = log x ( )s =0 , ( )log x ( )s = ( )-1 s-1( )s -1 !x -s ( )s0

Separating the terms containing f( )03 from Theorem 21.1.1 and substituting these fot it,

-

x

-

xx e x log x dx p = e x log xΣ

r=0

m -1

-p

r 1++p +r( )1+

x +p+ r

- Σr=1

m -1

Σs=1

r( )-1 s

-p

r r

s 1++p +r( )1+ ( )s

x +p+ r-se x + Rmp

(1.2)

Rmp =

( )p ,m( )-1 m

Σr=0

m+ r1

p -1

r m+r

0 -

x

-

x

1++m+r( )1+

x +m+ r e x log x dx p

+( )p ,m( )-1 m

Σr=0

Σs=1

m + r

m+ r( )-1 s-1

p -1

r m+r

s -

x

-

x

1++m+r( )1+ ( )s

x +m+ r-s e x dx p

(1.2r)

When =3/2 , p =2.7 , the values of the both sides on arbitrary point x =17 are as follows. This also

seems to be asymptotic expansion.

21.1.2 Super Integral of the power of a function

Especially, if f1 = f2 = = f in Theorem 21.1.1 , the following theorem holds immediately.

Theorem 21.1.2

Let p , r are positive numbers, m be a natural number, f( )rbe the r th order derivative function of f( )x ,

f < >rbe arbitrary r th order primitive function of f( )x and ( )p ,m be the beta function. At this time,

if there is a number a such that

f< >r ( )a = 0 r[ ]0 ,m+p or f( )s ( )a = 0 s[ ]0 ,m+p -1

- 3 -

the following expression holds for = 2,3,4, .

a

x

a

x

f dxp =Σr1=0

m-1

Σr2=0

r1

Σr-1=0

r-2

-p

r1 r1

r2

r-2

r-1

f p+ r1 f r1-r2 f r-1 + Rmp

Rmp = ( )p ,m

( )-1 m

Σk1=0

Σk2=0

m +k1

Σk3=0

k2

Σk-1=0

k-2

m+ k1

1

p-1

k1 m+k1

k2 k2

k3

k-2

k-1

a

x

a

x

f m+ k1 f m+k1-k2 f k1-k2 f k-1 dxp

Example Super Integral of log 3x Since the zero of the higher integral of log 3x is x =0 , the zero of the super integral is also the same.

Then let f = log x

( )log x < >p+ r = 1+p +r

log x - 1+p +r -x p+ r , ( )log x < >m+ r =

1+m+r

log x - 1+m+r -x m+ r

( )log x ( )r-s = log x ( )r =s , ( )log x ( )r-s = ( )-1 r-s-1( )r -s -1 !x -r+s ( )rs

( )log x ( )0 = log x ( )s =0 , ( )log x ( )s = ( )-1 s-1( )s -1 !x -s ( )r0

Separating the terms containing f( )0from Theorem 21.1.2 and substituting these fot it,

0x0

x( )log x 3dx p =

1+plog x - 1+p -

x p( )log x 2

- 2x p log xΣr=1

m -1( )-1 r

-p

r 1+p +r

log x - 1+p +r - r

+ x p Σr=2

m -1

Σs=1

r-1( )-1 r

-p

r r

s 1+p +r

log x - 1+p +r - r -s ( )s

+ Rmp

(2.1)

Rmp = -

( )p ,m1

Σr=0

Σs=0

m + r

m+ r( )-1 r-s

p -1

r m+ r

s

0x0

x

1+m+r

log x - 1+m+r -( )m+r -s x sdx p

(2.1r)

And limm

Rmn = 0 holds although the proof is difficult. Then

0x0

x( )log x 3dx p =

1+plog x - 1+p -

x p( )log x 2

- 2x p log xΣr=1

( )-1 r

-p

r 1+p +r

log x - 1+p +r - r

+ x pΣr=2

Σs=1

r-1( )-1 r

-p

r r

s 1+p +r

log x - 1+p +r - r -s ( )s

(2.1')

However, the convergence speed is very slow.

When p =3/2 , m=400 , the values of the both sides on arbitrary point x =0.2 are as follows. Even if

calculated so far, both sides are corresponding only to 1 digit below the decimal point. Although (2.1') is not

suitable for the calculation, it is meaningful that the integral can be expressed by the double series.

- 4 -

21.1.3 Super Integrls of cosm x , sinm x

Formula 21.1.3

Let m be a natural number, p be a positive number and a( )s s[ ]0 ,p be the zero of the lineal super

primitive function of cos 2m+1x or sin 2m+1x . Then the following expressions hold.

a( )p

x

a( )0

x

cos 2m+1x dx p = 22m

1Σr=0

m

( )2m-2r +1 p

Ｃ2m +1 r cos ( )2m-2r +1 x -

2p

(3.c)

a( )p

x

a( )0

x

sin 2m+1x dx p = 22m

1Σr=0

m

( )2m-2r +1 p

( )-1 m-r Ｃ2m +1 r sin ( )2m-2r +1 x -

2p

(3.s)

Proof Analytically continuing the index of the integration operator in Formula 20.2.3c (1) and Formula 20.2.3s (1)

in 20.2 to [ ]0 ,p from[ ]1 ,n , we obtain the desired expressions.

Note

For example, in the case of cos 2m+1x , a( )s is a solution of the following transcendental equation.

Σr=0

m

( )2m-2r+1 s

Ｃ2m +1 rcos ( )2m-2r+1 x-

2s

= 0 s[ ]0 , p

Although it is difficult to calculate this solution, the necessity does not exist in this formula.

Example The 1.5th order intagral of cos 3x

- 5 -

- 6 -

21.2 Super Derivatives of the product of many functions

(1) Generalized binomial theorem and Leibniz Rule about Super Differentiation According to the generalized binomial theorem in 3.2 , the following expression holds for the real numbers

x1 , x2 such that x1 > x2 and a positive number p .

x1+x2p = Σ

r=0

p

rx1

p-rx2 r

On the other hand, according to Formula 19.1.1 in 19.1 , the following expression holds for the functions f1 , f2of x and a positive number p .

f1 f2( )p = Σ

r=0

m -1

p

rf1( )p-r f ( )r

2 + Rmp

Rmp = ( )-p ,m

( )-1 m

Σk=0

m+ k1

-p -1

k f< >m+ k1 f( )m+ k

2( )p

(2) Generalized multinomial theorem and Super Derivative of the product of many functions According to the generalized multinomial theorem in 3.4 , the following expression holds for real numbers

x1 , x2 , , x such that x1 > x2 + x3 ++ x and a positive number p .

x1+x2++x p

=Σr1=0

Σ r2=0

r1

Σr-1=0

r-2

p

r1 r1

r2

r-2

r-1 x1

p-r1 x2r1-r2 x

r-1

Therefore, the following expression must hold for functions f1 , f2 , , f of x and a positive number p .

f1 f2 f( )p = Σ

r1=0

m-1

Σr2=0

r1

Σr-1=0

r-2

p

r1 r1

r2

r-2

r-1

f p- r11 f r1-r2

2 f r-1 + Rm

p

21.2.1 Super Derivative of the product of many functions

Theorem 21.2.1

Let p , r are positive numbers, m be a natural number, f( )rk be the r th order derivative function of fk( )x

k =1,2,, , f < >rk be arbitrary r th order primitive function of fk( )x and ( )p ,m be the beta function.

At this time, if there is a number a such that

f< >r1 ( )a = 0 r[ ]0 ,m-p or f( )s

k ( )a = 0 s[ ]0 ,m-p -1 for at least one k >1 ,

the following expression holds

f1 f2 f( )p =Σ

r1=0

m-1

Σr2=0

r1

Σr-1=0

r-2

p

r1 r1

r2

r-2

r-1

f p- r11 f r1-r2

2 f r-1 + Rm

p

Rmp = ( )-p ,m

( )-1 m

Σk1=0

Σk2=0

m +k1

Σk3=0

k2

Σk-1=0

k-2

m+ k1

1

-p -1

k1 m+k1

k2 k2

k3

k-2

k-1

f m+ k11 f m+k1-k2

2 f k2-k33 f k-1

( )p

Proof Reversing the sign of the index of the integration operator < >p in Formula 21.1.1 ,

- 7 -

a

x

a

x

f1 f2 f dx-p =Σr1=0

m-1

Σr2=0

r1

Σr-1=0

r-2

p

r1 r1

r2

r-2

r-1

f -p+ r11 f r1-r2

2 f r-1

+ Rm-p

Rm-p = ( )-p ,m

( )-1 m

Σk1=0

Σk2=0

m +k1

Σk3=0

k2

Σk-1=0

k-2

m+ k1

1

-p -1

k1 m+k1

k2 k2

k3

k-2

k-1

a

x

a

x

f m+ k11 f m+k1-k2

2 f k2-k33 f k-1

dx-p

Then, replacing the integration operator < >-p with the differentiation operator ( )p , we obtain the desired

expression.

Example computation of the product of three functions

Example1 Super Derivative of x ex sinx Let f1 = x , f2 = e x , f3 = sin x , then

x ( )p- r =

1+-p +r( )1+

x -p+ r , x < >m+ r =

1++m+r( )1+

x +m+ r

e x ( )r-s = e x ( )m+ r-s

= e x , ( )sin x ( )s = sin x +2s

Substituting these for Theorem 21.2.1 , we obtain

x e xsin x( )p

= Σr=0

m-1

Σs=0

r

p

r r

s 1+-p +r( )1+

x -p+ r e x sin x +2s

+ Rmp

(1.1)

Rmp =

( )-p ,m( )-1 m

Σr=0

Σs=0

m + r

m+ r1

-p -1

r m+ r

s

1++m+r( )1+

x +m+ re xsin x +2s ( )p

(1.1r)

When =3/2 , p =1/2 , the values of the both sides on arbitrary point x =1.4 are asfollows.

Example2 Super Derivative of x ex log x Let f1 = x , f2 = e x , f3 = log x , then

x ( )p- r =

1+-p +r( )1+

x -p+ r , x < >m+ r =

1++m+r( )1+

x +m+ r

e x ( )r-s = e x , e x ( )m+ r-s

= e x

- 8 -

( )log x ( )0 = log x ( )s =0 , ( )log x ( )s = ( )-1 s-1( )s -1 !x -s ( )s0

Separating the terms containing f( )03 from Theorem21.2.1 and substituting these for it,

x e x log x( )p

= exlog xΣr=0

m-1

p

r ( )1+-p +r

( )1+x -p+ r

- exΣr=1

m-1

Σs=1

r( )-1 s

p

r r

s ( )1+-p +r

( )1+ ( )sx -p+ r-s + Rm

p(1.2)

Rmp =

( )-p ,m( )-1 m

Σr=0

m+ r1

-p -1

r ( )1++m+r

( )1+x +m+ rexlog x

( )p

- ( )-p ,m

( )-1 m

Σr=0

Σs=1

m + r

m+ r( )-1 s

-p -1

r m+r

s ( )1++m+r

( )1+ ( )sx +m+ r-sex

( )p

(1.2r)

Polynomials in (1.2) seem to be asymptotic expansion. when =3/2 , p =1/2 , the values of the both sides

on arbitrary point x =16 are as followas.

21.2.2 Super Derivative of the power of a function

Especially, if f1 = f2 = = f in Theorem 21.2.1 , the following theorem holds immediately.

Theorem 21.2.2

Let p , r are positive numbers, m be a natural number, f( )rbe the r th order derivative function of f( )x ,

f < >r be arbitrary r th order primitive function of f( )x and ( )p ,m be the beta function. At this time,

if there is a number a such that

f< >r ( )a = 0 r[ ]0 ,m-p or f( )s ( )a = 0 s[ ]0 ,m-p -1the following expression holds for = 2,3,4, .

f ( )p = Σ

r1=0

m-1

Σr2=0

r1

Σr-1=0

r-2

p

r1 r1

r2

r-2

r-1

f p+ r1 f r1-r2 f r-1 + Rmp

Rmp = ( )-p ,m

( )-1 m

Σk1=0

Σk2=0

m +k1

Σk3=0

k2

Σk-1=0

k-2

m+ k1

1

-p -1

k1 m+k1

k2 k2

k3

k-2

k-1

f m+ k1 f m+k1-k2 f k2-k3 f k-1( )p

Example Super Derivative of log 3x Let f = log x , then

- 9 -

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

log x - 1-p +r -x -p+ r , ( )log x < >m+ r =

1+m+r

log x - 1+m+r -x m+ r

( )log x ( )r-s = log x ( )r =s , ( )log x ( )r-s = ( )-1 r-s-1( )r -s -1 !x -r+s ( )rs

( )log x ( )0 = log x ( )s =0 , ( )log x ( )s = ( )-1 s-1( )s -1 !x -s ( )r0

Separating the terms containing f( )03 from Theorem21.2.2 and substituting these for it , we obtain

( )log x 3 ( )p =

1-plog x - 1-p -

x -p( )log x 2

- 2x -p log xΣr=1

m -1( )-1 r

p

r ( )1-p +r

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

+ x -p Σr=2

m -1

Σs=1

r-1( )-1 r

p

r r

s ( )1-p +r

log x - ( )1-p +r -( )r -s ( )s

+ Rmp

(2.1)

Rmp = -

( )-p ,m1

Σr=0

( )m+ r 2

( )-1 r 2

-p -1

r log x - 1+m+r - log x

( )p

+ ( )-p ,m

1Σr=0

Σs=1

m + r-1

m+ r( )-1 r

-p -1

r m+ r

s

1+m+r

log x - 1+m+r -( )m+r -s ( )s

( )p

(2.1r)

And limm

Rmn = 0 holds although the proof is difficult. Then

( )log x 3 ( )p =

( )1-p

log x - ( )1-p -x -p( )log x 2

- 2x -p log xΣr=1

( )-1 r

p

r ( )1-p +r

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

+ x -p Σr=2

Σs=1

r-1( )-1 r

p

r r

s ( )1-p +r

log x - ( )1-p +r -( )r -s ( )s

(2.1')

However, the convergence speed is very slow.

When p =1/2 , m=120 , the values of the both sides on arbitrary point x =4 are as follows. Even if

calculated so far, both sides are corresponding only to 1 digit below the decimal point.

- 10 -

21.2.3 Super Derivatives of cosm x , sinm x

Formula 21.2.3

When m is a natural number, p is a positive number and is the floor function,

cosmx( )p

= 2m-1

1Σr=0

m /2Ｃm r ( )m-2r p cos ( )m-2r x+

2p

(3.c)

sin mx( )p

= 2m-1

1Σr=0

m /2Ｃm r ( )m-2r p cos ( )m-2r x-

2

+2

p(3.s)

Proof

Analytically continuing the index of the differentiation operator in Formula 20.1.3 in 20.1 to [ ]0 ,p from

[ ]1 ,n , we obtain the desired expressions.

Example The 2.5th order derivative of cos 4x

Collateral Super Derivatives of cosm x , sinm x Unlike the higher derivative, in the super derivative, the lineal super derivative and the collateral one exist.

Although this was described in 12.1.4 , I describe it once now.

- 11 -

For example, if we differenciate cos 3x with respect to x p times according to Theorem 21.2.2 , it is as

follows.

cos3x( )p

=Σr=0

m -1

Σs=0

r

p

r r

scos x +

2( )p -r

cos x +2

( )r -s cos x +

2s

+ Rmp

(4.c')

Rmp =

( )-p ,m( )-1 m

Σr=0

Σs=0

m + r

m+ r1

-p -1

r m+r

s

cos x -2

( )m+r cos x +

2( )m+r -s

cos x +2s p

(4.cr')

Then, this is a collateral super derivative. If it is why, this theorem was drawn out as a reverse-operation of the

super integral of the product of many functions as seen during the proof of Theorem 21.2.1 . And the base was

the super integral with a fixed lower limit.

However, the lineal super integral of cos 3x is one with a variable lower limit. Therefore, (4.c') and (4.cr') derived

based on the fixed lower limit cannot be the lineal super derivative. Althogh this holds as a equation, the

polynomial of (4.c') is not well behaved.

By reference, let us compare (4.c') with the following lineal super derivative (4.c) derived from Formula 21.2.3 .

cos3x( )p

= 43

cos x+2

p + 4

3p

cos 3x+2

p(4.c)

Althogh (4.c') fits (4.c) most at the time of m=3 , a big difference is still seen by both.

However, since 1/( )-p ,m =0 at p= m -1 , m =1,2,3, , Rmp = 0 . Therefore

cos 3x( )m -1

=Σr=0

m -1

Σs=0

r

m -1

r r

scos x +

2( )m -1-r

cos x +2

( )r-s cos x +

2s

Furthermore, replacing m-1 with n ,

cos3x( )n

=Σr=0

n

Σs=0

r

n

r r

scos x +

2 n -r

cos x +2

( )r -s cos x +

2s

And this results in the following lineal super derivative. ( The proof is long, then omitted. )

cos3x( )n

= 43

cos x+2

n + 4

3n

cos 3x+2

n

2010.12.25

K. Kono

Alien's Mathematics

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