# Math Help - taylor series question

1. ## taylor series question

Hi everybody, I'm having trouble with this problem, if you could help it would be greatly appreciated...
find the Taylor series summation for f(x) = x/((2x+1)(3x-2)) based at 1. Then give an interval where the Taylor series converges.

2. Originally Posted by juntheking@yahoo.com
Hi everybody, I'm having trouble with this problem, if you could help it would be greatly appreciated...
find the Taylor series summation for f(x) = x/((2x+1)(3x-2)) based at 1. Then give an interval where the Taylor series converges.

by parital fractions we get

$f(x)= \frac{1}{7} \left(\frac{2}{3x-2}+\frac{1}{2x+1} \right)$

$f'(x)=\frac{1}{7} \left(\frac{2\cdot 3(-1)}{(3x-2)^2}+\frac{1 \cdot 2 (-1)}{(2x+1)^2} \right)$

$f''(x)=\frac{1}{7} \left(\frac{2\cdot 3^2(-1)(-2)}{(3x-2)^3}+\frac{1 \cdot 2^2 (-1)(-2)}{(2x+1)^3} \right)$

$f^{n}(x)=\frac{1}{7} \left(\frac{2\cdot 3^n(-1)^nn!)}{(3x-2)^{n+1}}+\frac{1 \cdot 2^n (-1)^nn!}{(2x+1)^{n+1}} \right)$

you can use this to generate the coeffeints for x=1 to get

$\frac{1}{3}-\frac{8}{9}(x-1)+\frac{70}{27}(x-1)^2-\frac{626}{81}(x-1)^3+...$

3. I understand why you did that, but how do I describe this in terms of a summation comparing this to the known summation of 1/1-x based at 0?

thanks again

4. ## Ok

Originally Posted by TheEmptySet
by parital fractions we get

$f(x)= \frac{1}{7} \left(\frac{2}{3x-2}+\frac{1}{2x+1} \right)$

$f'(x)=\frac{1}{7} \left(\frac{2\cdot 3(-1)}{(3x-2)^2}+\frac{1 \cdot 2 (-1)}{(2x+1)^2} \right)$

$f''(x)=\frac{1}{7} \left(\frac{2\cdot 3^2(-1)(-2)}{(3x-2)^3}+\frac{1 \cdot 2^2 (-1)(-2)}{(2x+1)^3} \right)$

$f^{n}(x)=\frac{1}{7} \left(\frac{2\cdot 3^n(-1)^nn!)}{(3x-2)^{n+1}}+\frac{1 \cdot 2^n (-1)^nn!}{(2x+1)^{n+1}} \right)$

you can use this to generate the coeffeints for x=1 to get

$\frac{1}{3}-\frac{8}{9}(x-1)+\frac{70}{27}(x-1)^2-\frac{626}{81}(x-1)^3+...$
Assuming the faux emptyset is correct what you should do is once you have split the partial fractions divide the bottom to get it into a form similar to $\frac{1}{1-x}$ and go from there..

5. Originally Posted by juntheking@yahoo.com
I understand why you did that, but how do I describe this in terms of a summation comparing this to the known summation of 1/1-x based at 0?

thanks again
You do not need to use the geometric series for this problem.

The form of the taylor series centered at x=1 is

$\frac{f(1)}{0!}(x-1)^0+ \frac{f'(1)}{1!}(x-1)^1 + \frac{f''(1)}{2!}(x-1)^2+... =\sum_{n=0}^{\infty}\frac{f^{n}(1)}{n!}(x-1)^n$

evaluating a few gives us

$f(1)=\frac{1}{3}$

$f'(1)=-\frac{8}{9}$

Take note that if you want to do something cool evaluate the nth derivative at x=1 and we get...

$f^{n}(1)=\frac{1}{7}\left[ \frac{2 \cdot 3^n}{1^{n+1}}+\frac{2^n}{3^{n+1}} \right](n!)(-1)^n= \frac{(-1)^n n!}{7}\left[ \frac{2 \cdot 3^{n+1}+2^n}{3^{n+1}}\right]$

Plugging this into the sum formula for the Taylor Series gives

$\sum_{n=0}^{\infty}\frac{\frac{(-1)^n n!}{7}\left[ \frac{2 \cdot 3^{n+1}+2^n}{3^{n+1}}\right]}{n!}(x-1)^n= \sum_{n=0}^{\infty}\frac{(-1)^n}{7}\left[ \frac{2 \cdot 3^{n+1}+2^n}{3^{n+1}}\right](x-1)^n$

I hope this helps

6. I guess I need to work on my algebra, but how did you do the partial fraction and extract a (1/7) term?

thanks

7. Originally Posted by juntheking@yahoo.com
I guess I need to work on my algebra, but how did you do the partial fraction and extract a (1/7) term?

thanks
since they are both linear factors we get

$\frac{x}{(2x+1)(3x-2)}=\frac{A}{(2x+1)}+\frac{B}{(3x-2)}$

Clearing the fractions we get

$x=A(3x-2)+B(2x+1)$
if we evalue this at $x=\frac{2}{3},x=-\frac{1}{2}$ we get

$\frac{2}{3}=A(2-2)+B(\frac{4}{3}+1) \iff \frac{2}{3}=\frac{7}{3}B \iff B=\frac{2}{7}$

Now the otherone

$-\frac{1}{2}=A(\frac{-3}{2}-2)+B(-1+1) \iff -\frac{1}{2}=-\frac{7}{2}A \iff A=\frac{1}{7}$

$\frac{\frac{1}{7}}{(2x+1)}+\frac{\frac{2}{7}}{(3x-2)}=\frac{1}{7} \left[ \frac{1}{2x+1}+\frac{2}{3x-2}\right]$

Good luck.