Engineering Mathematics

Friday, April 4, 2014

Reciprocal series and digamma function relationship

DI.ID.3: Reciprocal series and digamma function relationship

$\sum _{n=1}^{\infty }\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{b-a}$

Given

$\Psi (z)= -\gamma -\int _{0}^{1}\frac{t^{z}-1}{1-t}$

(1)

$\frac{1}{(n+a)(n+b)} = \frac{A}{(n+a)} - \frac{B}{(n+b)}$

(2)

$\frac{A}{(n+a)}=\int _{0}^{\infty }e^{-\frac{(n+a)x}{A}}dx$

$\frac{B}{(n+b)}=\int _{0}^{\infty }e^{-\frac{(n+b)x}{B}}dx$

(3)

$\newline A(n+a)-B(n+b)=1 \newline A*n=B*n \newline A*b-B*b=1 \newline A=B \newline A(b-a)=1 \newline A= \frac{1}{b-a}\newline$

(4)

Proof

$\frac{A}{(n+a)}-\frac{B}{(n+b)}=\int _{0}^{\infty }e^{-\frac{(n+a)x}{A}}dx-\int _{0}^{\infty }e^{-\frac{(n+b)x}{B}}dx$

$\frac{A}{(n+a)}-\frac{B}{(n+b)}=\int _{0}^{\infty }e^{-\frac{nx}{A}}e^{-\frac{ax}{A}}dx-\int _{0}^{\infty }e^{-\frac{nx}{B}}e^{-\frac{bx}{B}}dx$

(5)

$\sum _{n=1}^{\infty }\left ( \frac{A}{(n+a)}-\frac{B}{(n+b)} \right )=\sum _{n=1}^{\infty }\left ( \int _{0}^{\infty }e^{-\frac{nx}{A}}e^{-\frac{ax}{A}}dx-\int _{0}^{\infty }e^{-\frac{nx}{B}}e^{-\frac{bx}{B}}dx \right )$

(6)

$\sum _{n=1}^{\infty }\left ( \frac{A}{(n+a)}-\frac{B}{(n+b)} \right )=\left ( \int _{0}^{\infty }\left [ \frac{e^{-\frac{x}{A}}}{1-e^{-\frac{x}{A}}} \right ]e^{-\frac{ax}{A}}dx-\int _{0}^{\infty }\left [ \frac{e^{-\frac{x}{B}}}{1-e^{-\frac{x}{B}}} \right ]e^{-\frac{bx}{B}}dx \right )$

(7)

$\newline e^{-\frac{x}{A}} =t\newline e^{-\frac{x}{A}}=t^{a}\newline \frac{-1}{A}e^{-\frac{x}{A}}dx=dt \rightarrow dx= -\frac{A}{t}\: dt$

(8)

$\sum _{n=1}^{\infty }\left ( \frac{A}{(n+a)}-\frac{B}{(n+b)} \right )=\left ( \int _{0}^{1 }\left [ \frac{t}{1-t} \right ]t^{a}\frac{A}{t}\: dt-\int _{0}^{1 }\left [ \frac{t}{1-t} \right ]t^{b}\frac{B}{t}\: dt \right )$

(9)

$\sum _{n=1}^{\infty }\left ( \frac{A}{(n+a)}-\frac{B}{(n+b)} \right )=\int _{0}^{1 }\frac{(At^{a}-Bt^{b})}{1-t}\: dt$

(10)

from (4) , A=B ,

$\sum _{n=1}^{\infty }\left ( \frac{A}{(n+a)}-\frac{B}{(n+b)} \right )=A\int _{0}^{1 }\frac{((t^{a}-1)-(t^{b}-1)}{1-t}\: dt$

(11)

$\sum _{n=1}^{\infty }\left ( \frac{A}{(n+a)}-\frac{B}{(n+b)} \right )=\frac{1}{b-a}\int _{0}^{1 }\frac{((t^{a}-1)-(t^{b}-1)}{1-t}\: dt$

(12)

from (1)

$\sum _{n=1}^{\infty }\left ( \frac{A}{(n+a)}-\frac{B}{(n+b)} \right )=\frac{\psi(a)-\psi(b)}{b-a}$

(13)

$\sum _{n=1}^{\infty }\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{b-a}$

(14)

Wednesday, July 24, 2013

Bell series

Definition of Bell series

$B(x)= \sum _{n=0}^{\infty }B_{n}\frac{x^{n}}{n!}= =e^{e^{x}-1}$

BS.ID.1: Exponential representation

Given

$\sum _{n=0}^{\infty }\begin{Bmatrix}n\\ k \end{Bmatrix}\frac{x^{n}}{n!} = \sum _{n=0}^{\infty }\frac{1}{k!}\sum _{j=0}^{k}(-1)^{k-j}\binom{k}{j}j^{n}\frac{x^{n}}{n!}$

Proof

$\sum _{n=0}^{\infty }\begin{Bmatrix}n\\ k \end{Bmatrix}\frac{x^{n}}{n!} = \sum _{n=0}^{\infty }\frac{1}{k!n!}\left [\sum _{j=0}^{k}(-1)^{k-j}\binom{k}{j}j^{n}x^{n} \right ]$

(1)

$\sum _{n=0}^{\infty }\begin{Bmatrix}n\\ k \end{Bmatrix}\frac{x^{n}}{n!} = \sum _{n=0}^{\infty }\frac{(jx)^{n}}{n!}\left [\sum _{j=0}^{k}(-1)^{k-j}\binom{k}{j} \right ]\frac{1}{k!}$

(2)

$\sum _{n=0}^{\infty }\begin{Bmatrix}n\\ k \end{Bmatrix}\frac{x^{n}}{n!} = e^{jx}\left [\sum _{j=0}^{k}(-1)^{k-j}\binom{k}{j} \right ]\frac{1}{k!}$

(3)

$\sum _{n=0}^{\infty }\begin{Bmatrix}n\\ k \end{Bmatrix}\frac{x^{n}}{n!} = \left [\sum _{j=0}^{k}(-1)^{k-j}\binom{k}{j}e^{jx} \right ]\frac{1}{k!}$

(4)

Using binomial expansion

$(x-y)^{n} = \sum _{k=0}^{n}x^{k}(-y)^{n-k}=\sum _{k=0}^{n}x^{k}(-1)^{n-k}(y)^{n-k}$

(5)

Combining (4) and (5),

$\sum _{n=0}^{\infty }\begin{Bmatrix}n\\ k \end{Bmatrix}\frac{x^{n}}{n!} = \left [\sum _{j=0}^{k}(-1)^{k-j}\binom{k}{j}e^{jx} \right ]\frac{1}{k!} = \frac{(e^{x}-1)^{k}}{k!}$

(6)

$B_{n}=\sum _{k=0}^{n }\begin{Bmatrix}n\\ k \end{Bmatrix}$

(7)

$B(x)= \sum _{n=0}^{\infty }B_{n}\frac{x^{n}}{n!}= \sum _{n=0}^{\infty } \sum _{k=0}^{n }\begin{Bmatrix}n\\ k \end{Bmatrix} \frac{x^{n}}{n!}$

(8)

$B(x)= \sum _{n=0}^{\infty }B_{n}\frac{x^{n}}{n!}= \sum _{n=0}^{\infty } \sum _{k=0}^{n }\begin{Bmatrix}n\\ k \end{Bmatrix} \frac{x^{n}}{n!}= \sum _{k=0}^{n }\frac{(e^{x}-1)^{k}}{k!} =e^{e^{x}-1}$

(9)

$B(x)= \sum _{n=0}^{\infty }B_{n}\frac{x^{n}}{n!}= =e^{e^{x}-1}$

Monday, February 11, 2013

Polylogarithm

Definition of Polylogarithm

$Li_{s}(z) = \sum _{k=1}^{\infty }\frac{z^{k}}{k^{s}}$

PL.ID.1: Functional Equation of Polylogarithm

$Li_{s}(z) + Li_{s}(-z) = 2^{1-s}Li_{s}(z^{2})$

Proof

$Li_{s}(z) = \sum _{k=1}^{\infty }\frac{z^{k}}{k^{s}}= z+ \frac{z^{2}}{2^{s}}+ \frac{z^{3}}{3^{s}} +\cdot \cdot \cdot$

(1)

$Li_{s}(-z) = \sum _{k=1}^{\infty }\frac{(-z)^{k}}{k^{s}}= -z+ \frac{z^{2}}{2^{s}}- \frac{z^{3}}{3^{s}} \pm \cdot \cdot \cdot$

(2)

summing (1) and (2)

$Li_{s}(z) + Li_{s}(-z) = (z-z)+ (\frac{z^{2}}{2^{s}}+\frac{z^{2}}{2^{s}})+= (\frac{z^{3}}{3^{s}} -\frac{z^{3}}{3^{s}} )+ \cdot \cdot \cdot = 2(\frac{z^{2}}{2^{s}} +\frac{z^{4}}{4^{s}} +\cdot \cdot \cdot)$

$Li_{s}(z) + Li_{s}(-z) = 2\sum _{k=2,4,6\cdot \cdot \cdot }^{\infty }\frac{z^{k}}{k^{s}}$

(3)

$Li_{s}(z) + Li_{s}(-z) = 2\sum _{k=1 }^{\infty }\frac{z^{2k}}{(2k)^{s}}$

(4)

$Li_{s}(z) + Li_{s}(-z) = \frac{2}{2^{s}}\sum _{k=1 }^{\infty }\frac{z^{2k}}{k^{s}}$

(5)

using the definition of polylogarithm , the function equation is:

$Li_{s}(z) + Li_{s}(-z) = 2^{1-s}Li_{s}(z^{2})$

Tuesday, January 1, 2013

Useful Exponential Identity

UEI.ID.1: Series

$(1+x)\,e^{x} = 1 + \sum _{n=0}^{\infty }(\frac{n+1}{n!})\,x^{n}$

Proof

$(1+x)\,e^{x} = e^{x}+ x\,e^{x}$

(1)

$e^{x}+ x\,e^{x} = \sum _{n=0}^{\infty }\frac{x^{n}}{n!} + \sum _{n=0}^{\infty }\frac{x^{n+1}}{n!}$

(2)

$e^{x}+ x\,e^{x} = 1+ \sum _{n=1}^{\infty }\frac{x^{n}}{n!} + \sum _{n=0}^{\infty }\frac{x^{n}}{(n-1)!}$

(3)

$e^{x}+ x\,e^{x} = 1 + \sum _{n=0}^{\infty }(\frac{1}{n!}+\frac{1}{(n-1)!})\,x^{n}$

(4)

$e^{x}+ x\,e^{x} = 1 + \sum _{n=0}^{\infty }(\frac{n+1}{n!})\,x^{n}$

(5)

$(1+x)\,e^{x} = 1 + \sum _{n=0}^{\infty }(\frac{n+1}{n!})\,x^{n}$

Saturday, December 1, 2012

Generating Function

GF.ID.1: Chebyshev polynomial

$\sum _{n=0}^{\infty }T_{n}(x)t^{n}= \frac{1-2xt}{1-2xt+t^{2}}$

Given:

$T_{0}(x) =1$

(1.a)

$T_{1}(x) =x$

(1.b)

$T_{n+1}(x) = 2xT_{n}(x)-T_{n-1}(x)$

(1.c)

Expanding of Chebyshev polynomial as following:

$\sum _{n=0}^{\infty }T_{n}(x)t^{n}= 1+xt + \sum _{n=2}^{\infty }T_{n}(x)t^{n}$

(2.a)

$\sum _{n=0}^{\infty }T_{n}(x)t^{n}= 1+xt + \sum _{n=0}^{\infty }T_{n+2}(x)t^{n+2}$

(2.b)

rewriting (1.c) as :

$T_{n+2}(x) = 2xT_{n+1}(x)-T_{n}(x)$

(3)

from (2.b) and (3)

$\sum _{n=0}^{\infty }T_{n}(x)t^{n}= 1+xt + \sum _{n=0}^{\infty }(2xT_{n+1}(x)-T_{n}(x))t^{n+2}$

(4.a)

$\sum _{n=0}^{\infty }T_{n}(x)t^{n}= 1+xt + \sum _{n=0}^{\infty }2xT_{n+1}(x)t^{n+2} - \sum _{n=0}^{\infty }T_{n}(x)t^{n+2}$

(4.b)

$\sum _{n=0}^{\infty }T_{n}(x)t^{n}= 1+xt + 2xt\sum _{n=0}^{\infty }T_{n+1}(x)t^{n+1} - t^{2}\sum _{n=0}^{\infty }T_{n}(x)t^{n}$

(4.c)

$\sum _{n=0}^{\infty }T_{n}(x)t^{n}= 1+xt + 2xt\sum _{n=1}^{\infty }T_{n}(x)t^{n} - t^{2}\sum _{n=0}^{\infty }T_{n}(x)t^{n}$

(4.d)

similar to (3) , Expanding of Chebyshev polynomial as following:

$\sum _{n=0}^{\infty }T_{n}(x)t^{n} = 1+ \sum _{n=1}^{\infty }T_{n}(x)t^{n}$

(5.a)

$\sum _{n=0}^{\infty }T_{n}(x)t^{n} -1= \sum _{n=1}^{\infty }T_{n}(x)t^{n}$

(5.b)

From (4.d) and (5.b) ,

$\sum _{n=0}^{\infty }T_{n}(x)t^{n}= 1+xt + 2xt\left ( \sum _{n=0}^{\infty }T_{n}(x)t^{n} -1 \right ) - t^{2}\sum _{n=0}^{\infty }T_{n}(x)t^{n}$

(6.a)

$\left ( 1-2xt +t^{2} \right )\sum _{n=0}^{\infty }T_{n}(x)t^{n}= 1-2xt$

(6.b)

rearranging (6.b)

$\sum _{n=0}^{\infty }T_{n}(x)t^{n}= \frac{1-2xt}{ 1-2xt +t^{2} }$

Saturday, November 24, 2012

Dirichlet Series of Divisor Function

DF.ID.1: Dirichlet series of first and second order divisor function:

$\sum _{n=1}^{\infty } \frac{d(n)}{n^{s}} = \zeta ^{2}(s)$

$\sum _{n=1}^{\infty } \frac{d^{2}(n)}{n^{s}} = \frac{\zeta ^{4}(s)}{\zeta (2s)}$

proof
Euler product representation of Dirichlet series of divisor function:

$\sum _{n=1}^{\infty } \frac{d(n)}{n^{s}} = \prod _{p}(1+\frac{d(p)}{p^{s}} + \frac{d(p^{2})}{p^{2s}}+\cdot \cdot \cdot +\frac{d(p^{k})}{p^{ks}}+\cdot \cdot \cdot )$

(1.a)

$\sum _{n=1}^{\infty } \frac{d^{2}(n)}{n^{s}} = \prod _{p}(1+\frac{d^{2}(p)}{p^{s}} + \frac{d^{2}(p^{2})}{p^{2s}}+\cdot \cdot \cdot +\frac{d^{2}(p^{k})}{p^{ks}}+\cdot \cdot \cdot )$

(1.b)

$d(p^{k}) = (k+1)$

(2.a)

$d^{2}(p^{k})= (k+1)^{2}$

(2.b)

according to

$n=p_{1}^{k_{1}}p_{2}^{k_{2}}p_{3}^{k_{3}}\cdot \cdot \cdot p_{r}^{k_{r}}$

from (1.a) and (2.a):

$\sum _{n=1}^{\infty } \frac{d(n)}{n^{s}} = \prod _{p}(1+\frac{2}{p^{s}} + \frac{3}{p^{2s}}+\cdot \cdot \cdot +\frac{k+1}{p^{ks}}+\cdot \cdot \cdot )$

(3.a)

from (1.b) and (2.b):

$\sum _{n=1}^{\infty } \frac{d^{2}(n)}{n^{s}} = \prod _{p}(1+\frac{4}{p^{s}} + \frac{9}{p^{2s}}+\cdot \cdot \cdot +\frac{(k+1)^{2}}{p^{ks}}+\cdot \cdot \cdot )$

(3.b)

The derivative of geometric series (GS.ID.1):

$1+2x+3x^{2}+4x^{3}+\cdot \cdot \cdot = \frac{1}{(1-x)^{2}}$

(4)

$1+4x+9x^{2}+16x^{3}+\cdot \cdot \cdot = \frac{(1-x)^{2}+ 2x(1-x))}{(1-x)^{4}} = \frac{1+x}{(1-x)^{3}}$

(5.a)

$\frac{1+x}{(1-x)^{3}} = \frac{1-x^{2}}{(1-x)^{4}}$

$1+4x+9x^{2}+16x^{3}+\cdot \cdot \cdot = \frac{1-x^{2}}{(1-x)^{4}}$

(5.b)

from ( 3.a) and (4) , the Dirichlet series of first order divisor function:

$\sum _{n=1}^{\infty } \frac{d(n)}{n^{s}} = \prod _{p}\frac{1}{(1-p^{-s})^{2}} =\left ( \prod _{p}\frac{1}{(1-p^{-s})} \right ) ^{2} =\zeta (s)^{2}$

$\sum _{n=1}^{\infty } \frac{d(n)}{n^{s}} = \zeta ^{2}(s)$

(6)

from (3.b) and (5.b) , the Dirichlet series of second order divisor function:

$\sum _{n=1}^{\infty } \frac{d^{2}(n)}{n^{s}} = \prod _{p}\frac{(1-p^{-2s})}{(1-p^{-s})^{4}} =\frac{\zeta (s)^{4}}{\zeta (2s)}$
$\sum _{n=1}^{\infty } \frac{d^{2}(n)}{n^{s}} = \frac{\zeta ^{4}(s)}{\zeta (2s)}$

(7)

DF.ID.2: Dirichlet series of first order divisor function: of squared number

$\sum _{n=1}^{\infty } \frac{d(n^{2})}{n^{s}} = \frac{\zeta ^{3}(s)}{\zeta (2s)}$

proof

$\sum _{n=1}^{\infty } \frac{d(n^{2})}{n^{s}} = \prod _{p}(1+\frac{d(p^{2})}{p^{s}} + \frac{d(p^{4})}{p^{2s}}+\cdot \cdot \cdot +\frac{d(p^{2k})}{p^{ks}}+\cdot \cdot \cdot )$

(8)

$d(p^{2k})= 2k+1$

(8.b)

$\sum _{n=1}^{\infty } \frac{d(n^{2})}{n^{s}} = \prod _{p}(1+\frac{3}{p^{s}} + \frac{5}{p^{2s}}+\cdot \cdot \cdot +\frac{2k+1}{p^{ks}}+\cdot \cdot \cdot )$

(8.c)

given that :

$x+x^{3}+x^{5}+x^{7}+\cdot \cdot \cdot+ x^{2k+1} = \frac{x}{1-x^{2}}$

(9)

The derivative of previous series:

$x+3x^{2}+5x^{4}+7x^{6}+\cdot \cdot \cdot+ 2k+1x^{2k} = \frac{1-x^{4}}{(1-x^{2})^{3}}$

(10)

$x^{2} = p^{-s}$

(11)

from (8.c)(10) and (11),

$\sum _{n=1}^{\infty } \frac{d(n^{2})}{n^{s}} = \prod _{p}\frac{1-p^{-2s}}{(1-p^{-s})^{3}}$

(12)

$\sum _{n=1}^{\infty } \frac{d(n^{2})}{n^{s}} = \frac{\zeta ^{3}(s)}{\zeta (2s)}$

(13)

DF.ID.3: Dirichlet series of sum of higher order divisor function:

$\frac{\zeta (s)\zeta (s-a)\zeta (s-b)\zeta (s-a-b)}{\zeta (2s-a-b)}= \sum _{n=1}^{\infty } \frac{\sigma _{a}(s)\sigma _{b}(s)}{n^{s}}$

Proof

$\zeta (s)= \prod _{p \: \: \: prime} \frac{1}{1-p^{-s}}= \prod _{p \: \: \: prime}\left ( 1+\frac{1}{p^{s}} +\frac{1}{p^{2s}}+... \right )$

(1)

$\zeta (s-a)= \prod _{p \: \: \: prime} \frac{1}{1-p^{a-s}}= \prod _{p \: \: \: prime}\left ( 1+\frac{p^{a}}{p^{s}} +\frac{p^{2a}}{p^{2s}}+... \right )$

(2)

$\zeta (s)\zeta (s-a)= \prod _{p \: \: \: prime}\left ( 1+\frac{1}{p^{s}} +\frac{1}{p^{2s}}+... \right )\left ( 1+\frac{p^{a}}{p^{s}} +\frac{p^{2a}}{p^{2s}}+... \right )$

(3)

$\zeta (s)\zeta (s-a)= \prod _{p \: \: \: prime}\left ( 1+\frac{1+p^{a}}{p^{s}} +\frac{1+p^{a}+p^{2a}}{p^{2s}}+... \right )$

(4)

$\zeta (s)\zeta (s-a)= \prod _{p \: \: \: prime}\left ( 1+\frac{1-p^{2a}}{1-p^{a}}\frac{1}{p^{s}} +... \right )$

(5)

$\frac{\zeta (s)\zeta (s-a)\zeta (s-b)\zeta (s-a-b)}{\zeta (2s-a-b)}= \prod _{p \: \: \: prime} \frac{1-p^{-2s+a+b}}{(1-p^{-s})(1-p^{-s+a})(1-p^{-s+b})(1-p^{-s+a+b})}$

(6)

$p^{-s}=z \: \: ;\: \: p^{-2s}=z^{2}$

(7)

replacing (7) in (6) gives:

$\frac{1-p^{-2s+a+b}}{(1-p^{-s})(1-p^{-s+a})(1-p^{-s+b})(1-p^{-s+a+b})} = \frac{1-p^{a+b}z^{2} }{(1-z)(1-p^{a}z)(1-p^{b}z)(1-p^{a+b}z)}$

(8)

$\frac{1-p^{a+b}z^{2}}{(1-z)(1-p^{a}z)(1-p^{b}z)(1-p^{a+b}z)}= \frac{1}{(1-p^{a})(1-p^{b})}\left \{ \frac{1}{1-z}-\frac{p^{a}}{1-p^{a}z} -\frac{p^{b}}{1-p^{b}z} +\frac{p^{a+b}}{1-p^{a+b}z}\right \}$

(9)

$\frac{1}{(1-p^{a})(1-p^{b})}\left \{ \frac{1}{1-z}-\frac{p^{a}}{1-p^{a}z} -\frac{p^{b}}{1-p^{b}z} +\frac{p^{a+b}}{1-p^{a+b}z}\right \} = \frac{1}{(1-p^{a})(1-p^{b})} \sum _{m=0}^{\infty }\left \{ 1- p^{(m+1)a} - p^{(m+1)b}+ p^{(m+1)(a+b)}\right \} z^{m}$

(10)

$\frac{1}{(1-p^{a})(1-p^{b})} \sum _{m=0}^{\infty }\left \{ 1- p^{(m+1)a} - p^{(m+1)b}+ p^{(m+1)(a+b)}\right \} z^{m} = \frac{1}{(1-p^{a})(1-p^{b})} \sum _{m=0}^{\infty }\left \{ 1- p^{(m+1)a} \right \}\left \{ 1- p^{(m+1)b} \right \} z^{m}$

(11)

$\frac{\zeta (s)\zeta (s-a)\zeta (s-b)\zeta (s-a-b)}{\zeta (2s-a-b)}=\prod _{p \: \: \: prime} \sum _{m=0}^{\infty }\left \{ \frac{1- p^{(m+1)a}}{(1-p^{a})}\cdot \frac{1- p^{(m+1)b}}{(1-p^{b})}. \frac{1}{p^{ms}} \right \}$

(12)

$\frac{\zeta (s)\zeta (s-a)\zeta (s-b)\zeta (s-a-b)}{\zeta (2s-a-b)}= \sum _{n=1}^{\infty } \frac{\sigma _{a}(s)\sigma _{b}(s)}{n^{s}}$

(13)