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I have a question on the geometric series being represented by the Maclaurin series.

Wiki defines the Taylor series as: $\sum_{n=0}^\infty \frac {f^{(n)}(a)} {n!}(x-a)^n$

where $a = 0$ is the Maclaurin series.

Wiki then states "The Maclaurin series for $\frac 1 {1-x}$ is the geometric series $1 + x + x^2 + ...$

My first question is, what is $f(a)$ for the Taylor Series that results in this MacLaurin series?

Wiki then goes into the Taylor series for $\frac 1 x$ at $a = 1$ is $1 - (x - 1) + (x - 1)^2 - (x - 1)^3 + ...$

Does anyone know how this is derived?

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If we take the function $f(x) = 1/(1-x)$ for $|x|<1$, then $f^{(n)}(x) = n! /(1-x)^n$ for all $n$ [by induction], so $f^{(n)}(0) = n!$ for all $n$, so the Maclaurin series is $$ \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}(x-0)^n \sum_{n=0}^\infty \frac{n!}{n!}(x-0)^n= \sum_{n=0}^\infty x^n. $$

If we take $f(x) = \frac{1}{x}$ for $x>0$, then $f^{(n)}(x) = \frac{(-1)^n n!}{x^{n+1}}$ and $f^{(n)}(1) = (-1)^n$. The Taylor series at $a=1$ is $$ \sum_{n=0}^\infty \frac{f^{(n)}(1)}{n!}(x-1)^n = \sum_{n=0}^\infty \frac{(-1)^n n!}{n!}(x-1)^n = \sum_{n=0}^\infty (-1)^n(x-1)^n . $$

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To start with, let us consider the geometric sequence whose first term equals one with ratio $|x| < 1$.

The we can express the sum of its first $n+1$ terms by \begin{align*} s_{n}(x) = 1 + x + x^{2} + \ldots + x^{n} & \Rightarrow xs_{n}(x) = x + x^{2} + x^{3} + \ldots + x^{n+1}\\\\ & \Rightarrow (1-x)s_{n}(x) = 1 - x^{n+1}\\\\ & \Rightarrow s_{n}(x) = \frac{1 - x^{n+1}}{1-x} \end{align*}

Since $x^{n}\to 0$ for $|x| < 1$ as $n$ approaches infinity, one concludes that \begin{align*} \lim_{n\to\infty}s_{n}(x) = \lim_{n\to\infty}\frac{1-x^{n+1}}{1-x} = \frac{1}{1 - x} = 1 + x + x^{2} + x^{3} + \ldots \end{align*}

Consequently, the desired function is given by $f(x) = \dfrac{1}{1-x}$.

Based on such results, it is possible to give an answer to your second question as well.

More precisely, if $|x - 1| < 1$, we can conclude that

\begin{align*} \frac{1}{x} = \frac{1}{1 - (1-x)} = 1 + (1 - x) + (1-x)^{2} + (1-x)^{3} + \ldots \end{align*}

Hopefully this helps!

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  • $\begingroup$ Using $f(x) = \frac 1 {1-x}$ and taking $f'(x) = \frac 1 {(x-2)^2}$ and substituting into the Taylor series, I do not arrive at the second term $x$. $\endgroup$
    – Nick
    Dec 16 '20 at 2:36
  • $\begingroup$ The derivative of $f(x)$ is given by $$f'(x) = \frac{1}{(1-x)^{2}}$$ More generally, as pointed out by @GEdgar, one has that $$f^{(n)}(x) = \frac{n!}{(1-x)^{n}}$$ $\endgroup$
    – user0102
    Dec 16 '20 at 2:37

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