# 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!

• The Taylor coefficients should be $f^{(n)}(a)/n!$. So you have to divide your derivatives by the corresponding factorial. – Robert Z Apr 22 at 10:32
• I did and it gave me answer 1. The calculator gave me answer 2. So I am not sure if I am wrong or the calculator wrong? – Shaun Weinberg Apr 22 at 10:37
• All the coefficients are $1$ and the polynomial is $1 + x + x^2 + x^3 + x^4 + x^5$. – Robert Z Apr 22 at 10:40
• Apologies i'm a part-time student so I study from a workbook provided by uni. The workbook explains that the polynomials are the number you are evaluating so hence why I divided by 1, 2, 3, 4, 5 in each term – Shaun Weinberg Apr 22 at 10:48
• $n!$ is the factorial of $n$ see en.wikipedia.org/wiki/Factorial – Robert Z Apr 22 at 10:50

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.

• I got it thank you – Shaun Weinberg Apr 22 at 10:57

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 $$!$$...

• Yes, but what is the ! meaning please. In my book its just a value hence why i divided by 1, 2, 3, 4, 5. But apparently this is incorrect? – Shaun Weinberg Apr 22 at 10:49
• ! is given by : $$n!=n\times (n-1)\times...\times 1 \quad \forall n \in N$$ – Dicordi Apr 22 at 10:49
• Ok got it thanks – Shaun Weinberg Apr 22 at 10:57