<s>[INST] Find Taylor Polynomial order 5 of $f(x) = \frac{1}{(1-x)}$, at $a = 0$ Find Taylor Polynomial order 5 of $f(x) = \frac{1}{(1-x)}$ , at $a = 0$
So I start of with:
$f(0) = \frac{1}{(1-0)}=1$
$f'(0) = \frac{1}{(1-0)^2 }= 1$
$f''(0) = \frac{2}{(1-0)^3 }=  2$
$f'''(0) = \frac{6}{(1-0)^4 }= 6$
$f^{(4)}(0) = \frac{24}{(1-0)^5 }= 24$
$f^{(5)}(0) = \frac{120}{(1-0)^6 }=  120$
and I get
$1 + x + \frac{2x^2}{2} + \frac{6x^3}{3} + \frac{24x^4}{4}  + \frac{120x^5}{5} $ 
Answer = $1 + x + x^2 + 2x^3 + 5x^4 + 24x^5 $
The problem is that a calculator i used to check it up said the answer was
answer = $1 + x + x^2 + x^3 + x^4 + x^5 $
So im confused, im I right or is the calculator wrong? Thanks!
 [/INST] i see a problem. The taylor series in this case is given by : 
$$f(x)=f(a)(x-a)+f'(a)(x-a)+f''(a)\frac{(x-a)^2}{2!}+f'''(a)\frac{(x-a)^3}{3!}+...$$
I think you forgot the $!$...
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