How to find the taylor series for $f(x) = \frac{1}{16-x^2}$ centered at 9 in summation notation

Question

Having trouble finding the taylor series for the following function:

$$f(x) = \frac{1}{16-x^2} \textrm{ centered at c=9}$$

I was trying to look at it in a way such that I could modify $\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$. I wasn't able to come up with anything though.

I found some of the first few terms, but I can't see to find all the patterns.

$$\frac{1}{16-x^2} = \frac{-1}{65} + \frac{18}{4225}(x-9) - \frac{518}{274625}(x-9)^2$$

I noticed the denominators multiply by 65 each time. Can't seem to find a pattern for the numerator though.

Seems to be something like:

$$2x = 2(9) = 18$$

$$6x^2 + 32 = 6(9^2) +32 = 518$$

$$24x^3 + 384x = \textrm{next numerator}$$

Pattern doesn't seem to stay much or gets more complex though. Any idea how to solve this problem?

Answer 1

Write \begin{align*} f(x) & = \frac{1}{16-x^2}\\ & = \frac{1}{8}\left[\frac{1}{4-x}+\frac{1}{4+x}\right]\\ & = \frac{1}{8}\left[\frac{1}{-5-(x-9)}+\frac{1}{13+(x-9)}\right]\\ & = \frac{1}{8}\left[\frac{1}{-5}\left(\frac{1}{1+\frac{x-9}{5}}\right)+\frac{1}{13}\left(\frac{1}{1+\frac{x-9}{13}}\right)\right]\\ & = \frac{1}{-40}\left(\frac{1}{1+\frac{x-9}{5}}\right)+\frac{1}{104}\left(\frac{1}{1+\frac{x-9}{13}}\right) \end{align*} Now use the geometric series expansion $$\frac{1}{1+\frac{x-a}{b}}=\sum_{k=0}^{\infty}(-1)^k\left(\frac{x-a}{b}\right)^k,$$ to get, \begin{align*} f(x) & = \frac{1}{-40}\left(\frac{1}{1+\frac{x-9}{5}}\right)+\frac{1}{104}\left(\frac{1}{1+\frac{x-9}{13}}\right)\\ &=\frac{1}{-40}\sum_{k=0}^{\infty}(-1)^k\left(\frac{x-9}{5}\right)^k+\frac{1}{104}(-1)^k\left(\frac{x-9}{13}\right)^k\\ &=\sum_{k=0}^{\infty}(-1)^k\left(\frac{1}{(-40)5^k}+\frac{1}{(104)13^k}\right)(x-9)^k\\ &=\sum_{k=0}^{\infty}\color{red}{\frac{(-1)^k}{8}\left(\frac{5^{k+1}-13^{k+1}}{65^{k+1}}\right)}(x-9)^k\\ \end{align*}

Answer 2

Decompose: $$\frac{1}{16-x^2}=\frac{1}{(4-x)(4+x)}=\frac18\left[\frac1{4-x}+\frac1{4+x}\right]$$ Expand each term separately: $$\begin{align}f(9)&=\frac1{4-9}=-\frac15; \\ f'(9)&=\frac1{(4-9)^2}=\frac1{5^2};\\ f''(9)&=-\frac2{(4-9)^3}=-\frac2{5^3};\\ &\vdots\\ \frac1{4-x}&=-\frac15+\frac{1}{1!\cdot 5^2}(x-9)-\frac2{2!\cdot 5^3}(x-9)^2+\cdots+(-1)^{n+1}\frac{(n-1)!}{(n-1)!\cdot 5^{n+1}}(x-9)^n+\cdots\\ =====&========================================== \\ f(9)&=\frac1{4+9}=\frac1{13};\\ f'(9)&=-\frac1{(4+9)^2}=-\frac1{13^2};\\ f''(9)&=\frac2{(4+9)^3}=\frac2{13^3};\\ &\vdots\\ \frac1{4+x}&=\frac1{13}-\frac{1}{1!\cdot 13^2}(x-9)+\frac2{2!\cdot 13^3}(x-9)^3+\cdots+(-1)^n\frac{(n-1)!}{(n-1)!\cdot 13^{n+1}}(x-9)^n+\cdots\\ \end{align}$$ Hence: $$\begin{align}\frac1{16-x^2}&=\frac18\left[\sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{5^{n+1}}(x-9)^n+\sum_{n=0}^{\infty} \frac{(-1)^{n}}{13^{n+1}}(x-9)^n\right]=\\ &=\frac18\sum_{n=0}^{\infty} \frac{1}{65^{n+1}}\left[(-1)^{n+1}13^{n+1}+(-1)^n5^{n+1}\right](x-9)^n=\\ &=\frac18\sum_{n=0}^{\infty} 65^{-n-1}\left[(-13)^{n+1}+(-1)^n5^{n+1}\right](x-9)^n. \end{align}$$ See WolframAlpha's answer.