Post on 16-Apr-2018
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
- 12 -