Vrftsjtryk

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?

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
$$frac{1}{1+frac{x-a}{b}}=sum_{k=0}^{infty}(-1)^kleft(frac{x-a}{b}right)^k,$$
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)^kleft(frac{x-9}{5}right)^k+frac{1}{104}(-1)^kleft(frac{x-9}{13}right)^k\
&=sum_{k=0}^{infty}(-1)^kleft(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*}

Decompose:
$$frac{1}{16-x^2}=frac{1}{(4-x)(4+x)}=frac18left[frac1{4-x}+frac1{4+x}right]$$
Expand each term separately:
$$begin{align}f(9)&=frac1{4-9}=-frac15; \
f'(9)&=frac1{(4-9)^2}=frac1{5^2};\
f''(9)&=-frac2{(4-9)^3}=-frac2{5^3};\
&vdots\
frac1{4-x}&=-frac15+frac{1}{1!cdot 5^2}(x-9)-frac2{2!cdot 5^3}(x-9)^2+cdots+(-1)^{n+1}frac{(n-1)!}{(n-1)!cdot 5^{n+1}}(x-9)^n+cdots\
=====&========================================== \
f(9)&=frac1{4+9}=frac1{13};\
f'(9)&=-frac1{(4+9)^2}=-frac1{13^2};\
f''(9)&=frac2{(4+9)^3}=frac2{13^3};\
&vdots\
frac1{4+x}&=frac1{13}-frac{1}{1!cdot 13^2}(x-9)+frac2{2!cdot 13^3}(x-9)^3+cdots+(-1)^nfrac{(n-1)!}{(n-1)!cdot 13^{n+1}}(x-9)^n+cdots\
end{align}$$
Hence:
$$begin{align}frac1{16-x^2}&=frac18left[sum_{n=0}^{infty} frac{(-1)^{n+1}}{5^{n+1}}(x-9)^n+sum_{n=0}^{infty} frac{(-1)^{n}}{13^{n+1}}(x-9)^nright]=\
&=frac18sum_{n=0}^{infty} frac{1}{65^{n+1}}left[(-1)^{n+1}13^{n+1}+(-1)^n5^{n+1}right](x-9)^n=\
&=frac18sum_{n=0}^{infty} 65^{-n-1}left[(-13)^{n+1}+(-1)^n5^{n+1}right](x-9)^n.
end{align}$$
See WolframAlpha's answer.