Engineering Mathematics: 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)

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

Engineering Mathematics