$f(x)= \frac{1}{x^{2}-3x-4} $The power series expansion at x = 1.

Question

$f(x)= \frac{1}{x^{2}-3x-4} $ The power series expansion at x = 1. I made an answer, but I am not sure if it is right.

$$1/5\left[\sum_{n=0}^∞ (-1)^{(n+1)} -\sum_{n=0}^∞ 1/4^{(n+1)}\right]$$ This is the calculation process, I put it on my blog, this side of the restrictions on sending photos, you strongly demand to, I also edit very slowly.

another part

This is my answer, I do not ask questions often, I hope some of you kind to me, thank you

Because this is homework, I don't want to make mistakes. Can you help me see this picture? Is it centered on 1? wolfram app ???

Answer 1

It looks like you have made an expansion around $x=0$ instead of $x=1$.

You need to add "at x=1" in wolfram Alpha to tell the tool where the expansion occurs:

https://www.wolframalpha.com/input/?i=series+1%2F%28x%5E2-3x-4%29+at+x%3D1

$f(x)=\dfrac 1{x^2-3x-4}$, since we want an expansion around $x=1$, let set $x=1+u$ with $u\to 0$.

$f(1+u)=\dfrac{1}{u^2-u-6}=\dfrac 1{(u+2)(u-3)}=\dfrac {-\frac 1{10}}{1+\frac u2}+\dfrac {-\frac 1{15}}{1-\frac u3}$

Now we can develop each part in power series:

$f(1+u)=-\frac 1{10}\sum\limits_{n=0}^\infty (-\frac u2)^n-\frac 1{15}\sum\limits_{n=0}^\infty (\frac u3)^n=\sum\limits_{n=0}^\infty a_nu^n\quad$ with $a_n=-\frac 15\left(\frac{(-1)^n}{2^{n+1}}+\frac{1}{3^{n+1}}\right)$

Answer 2

Here is what I think the question is asking based on the title of the post:

You start out doing partial-fractions decomposition like you did:

$$ f(x) = \frac{1}{5} \left( \frac{1}{x-4} - \frac{1}{x+1} \right) $$

Usually "power series expansion at $x=1$" means write in the form $\sum_{n=0}^\infty c_n (x-1)^n$. So at this point, re-write the partial fractions expression in terms of $(x-1)$:

$$ f(x) = \frac{1}{5} \left( \frac{1}{(x-1)-3} - \frac{1}{(x-1)+2} \right) $$

Now use your geometric series formula:

$$ \begin{align*} f(x) &= \frac{1}{5} \left( \frac{1}{(x-1)-3} - \frac{1}{(x-1)+2} \right) \\ &= \frac{1}{5} \left( -\frac{1}{3} \, \frac{1}{1-\frac{1}{3}(x-1)} - \frac{1}{2} \, \frac{1}{1 + \frac{1}{2}(x-1)} \right) \\ &= \frac{-1}{5} \left( \sum_{n=0}^\infty \frac{1}{3^{n+1}} (x-1)^n + \sum_{n=0}^\infty \frac{(-1)^n}{2^{n+1}} (x-1)^n \right) \\ &= \frac{-1}{5} \sum_{n=0}^\infty \left( \frac{1}{3^{n+1}} + \frac{(-1)^n}{2^{n+1}} \right)(x-1)^n \end {align*} $$

Answer 3

your expansions are not centered at x = 1

$f(x) = \frac 15(\frac {1}{x-4} - \frac {1}{x+1})$ starts off fine

$\frac 15(\frac {1}{(x-1)-3} - \frac {1}{(x-1)+2})\\ -(\frac 1{15})(\frac {1}{1 - \frac{x-1}{3}} - (\frac {1}{10}) \frac {1}{1+\frac{x-1}{2}})\\ \sum_\limits{k=0}^{\infty} (-\frac {1}{15}\frac {(x-1)^k}{3^k} - \frac 1{10} \frac {(-1)^k(x-1)^k}{2^k})\\ -\frac 15\sum_\limits{k=0}^{\infty} \frac {((-3)^{k+1}+2^{k+1})(x-1)^k}{6^{k+1}}\\-\frac 15\sum_\limits{k=1}^{\infty} \frac {((-3)^k+2^k)(x-1)^{k-1}}{6^{k}}$