Find Taylor Polynomial order 5 of $f(x) = \frac{1}{(1-x)}$, at $a = 0$

Question

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!

Answer 1

Its easy to do, take the inverse of (1-x) and then expand

$(1-x)^{-1}$ = 1 + $x + x^{2} + x^{3}$ + .....

So coefficient of each term is 1.

From your method, you are skipping the factorial term in the denominator. Calculator is right.

Answer 2

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 $!$...