# Thread: Taylor Series 1/x^2 at x=1

1. ## Taylor Series 1/x^2 at x=1

Can i make 1/(x^2) into -1/(((1- ((x-1)^2)+2x)))
and use f(x) = 1/(1-u) where u = ((x-1)^2)+2x))
and get -sigma : ((x-1)^2 + 2x))^n
??
Or is that illegal
How about 1/(x^2) = d/dx [-1/x]
but what is the taylor series for 1/x

2. Originally Posted by khuezy
Can i make 1/(x^2) into -1/(((1- ((x-1)^2)+2x)))
and use f(x) = 1/(1-u) where u = ((x-1)^2)+2x))
and get -sigma : ((x-1)^2 + 2x))^n
??
Or is that illegal
How about 1/(x^2) = d/dx [-1/x]
but what is the taylor series for 1/x
Does $\displaystyle -\Sigma_{n=0}^{\infty} (x^2 + 1)^n$ (which is what your exprssion reduces to) have the form of a Taylor Series expanded around x = 1 (your post title suggests that you want it to .....)?

3. ## yes?

?

4. Originally Posted by khuezy
?
Do you know what a Taylor series looks like? If you do, how can you possibly think that what you posted looks like one.

5. ## zing!

just kidding!

6. Originally Posted by khuezy
Can i make 1/(x^2) into -1/(((1- ((x-1)^2)+2x)))
and use f(x) = 1/(1-u) where u = ((x-1)^2)+2x))
and get -sigma : ((x-1)^2 + 2x))^n
??
Or is that illegal
How about 1/(x^2) = d/dx [-1/x]
but what is the taylor series for 1/x
You can just apply the definition of Taylor series.

$\displaystyle f(x)=\frac{1}{x^2}\implies f(1)=1$

$\displaystyle f'(x)=-\frac{2}{x^3}\implies f'(1)=-2$

$\displaystyle f''(x)=\frac{6}{x^4}\implies f''(1)=6$

$\displaystyle f'''(x)=-\frac{24}{x^5}\implies f'''(1)=-24$

$\displaystyle f^{(4)}(x)=\frac{120}{x^6}\implies f^{(4)}(1)=120$

$\displaystyle \therefore \frac{1}{x^2}=1-2(x-1)+\frac{6(x-1)^2}{2!}-\frac{24(x-1)^3}{3!}+\frac{120(x-1)^4}{4!}+\dots$ $\displaystyle =1-2(x-1)+3(x-1)^2-4(x-1)^3+5(x-1)^4+\dots+(-1)^n(n+1)(x-1)^n$

$\displaystyle \therefore \sum_{k=0}^{\infty}(-1)^k(k+1)(x-1)^{k}=\frac{1}{x^2}$ at the point $\displaystyle x=1$.

Does this make sense?