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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!

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    $\begingroup$ The Taylor coefficients should be $f^{(n)}(a)/n!$. So you have to divide your derivatives by the corresponding factorial. $\endgroup$ – Robert Z Apr 22 at 10:32
  • $\begingroup$ 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? $\endgroup$ – Shaun Weinberg Apr 22 at 10:37
  • $\begingroup$ All the coefficients are $1$ and the polynomial is $1 + x + x^2 + x^3 + x^4 + x^5$. $\endgroup$ – Robert Z Apr 22 at 10:40
  • $\begingroup$ 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 $\endgroup$ – Shaun Weinberg Apr 22 at 10:48
  • $\begingroup$ $n!$ is the factorial of $n$ see en.wikipedia.org/wiki/Factorial $\endgroup$ – Robert Z Apr 22 at 10:50
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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.

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  • $\begingroup$ I got it thank you $\endgroup$ – Shaun Weinberg Apr 22 at 10:57
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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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  • $\begingroup$ 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? $\endgroup$ – Shaun Weinberg Apr 22 at 10:49
  • $\begingroup$ ! is given by : $$n!=n\times (n-1)\times...\times 1 \quad \forall n \in N$$ $\endgroup$ – Dicordi Apr 22 at 10:49
  • $\begingroup$ Ok got it thanks $\endgroup$ – Shaun Weinberg Apr 22 at 10:57

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