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To find the taylor series of a function you would usually use the formula $\sum_{n=0}^{\infty}\frac{f^{n}(c)}{n!}(z-c)^n$.

However when computing the taylor series for $f(z)=\frac{1}{z+3}$ about $z=1$, I discovered that not only can you compute it using the above formula but you can also compute it using the geometric series formula!

Method 1:

$f(z)=\frac{1}{z+3}, f'(z)=\frac{-1}{(z+3)^2}, f''(z)=\frac{(-1)(-2)}{(z+3)^3}, ..., f^{n}(z)=\frac{(-1)^n n!}{(z+3)^{n+1}}$

$\Rightarrow f^{n}(1)=\frac{(-1)^n n!}{(1+3)^{n+1}} = \frac{(-1)^n n!}{4^{n+1}}$

$\therefore f(z) = \sum_{n=0}^{\infty}\frac{f^{n}(1)}{n!}(z-1)^n = \sum_{n=0}^{\infty}\frac{(-1)^n n!}{4^{n+1}n!} (z-1)^n $

$= \frac{1}{4}\sum_{n=0}^{\infty}\frac{(-1)^n}{4^n} (z-1)^n$ (for $|z-1|<4$)

Method 2:

$f(z)=\frac{1}{z+3}=\frac{1}{(z-1)+1+3}=\frac{1}{(z-1)+4}$

$=\frac{1}{4}\frac{1}{1+\frac{z-1}{4}}$

$=\frac{1}{4}\frac{1}{1-(\frac{-(z-1)}{4})}$

$=\frac{1}{4}\sum_{n=0}^{\infty}(\frac{-(z-1)}{4})^n$ using the formula for geometric series $\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$, $|x|<1$

$=\frac{1}{4}\sum_{n=0}^{\infty}\frac{(-1)^n}{4^n}(z-1)^n$ (for $|z-1|<4$)

So now my question is, why do they give the same result? Is there some relationship between taylor series and geometric series? Also for what type of functions do they give the same result? (i.e. polynomials, trig functions etc.) Because I found computing the taylor series using the geometric series approach a lot quicker.

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The general Taylor series is of the form $$f(x)=\sum_{k=0}^{\infty}{\frac{(x-a)^k}{k!}f^{(k)}(a)}$$

Set $a=0$

$$f(x)=\sum_{k=0}^{\infty}{\frac{x^k}{k!}f^{(k)}(0)}$$

Find some function which satisfies $f^{(k)}(0)=k!$ and you get a geometric series

$$f(x)=\sum_{k=0}^{\infty}{\frac{x^k}{k!}f^{(k)}(0)}=\sum_{k=0}^{\infty}x^k$$

Thus the Taylor series of a function which has infinitely many derivatives at $0$ such that $f^{(k)}(0)=k!$ is a geometric series.

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