# Taylor series and Geometric series

#### lucarioimpersius

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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#### Archie

If the Taylor Series converges for some interval about a given point, it is the unique power series representation of the function about the given point.

From the geometric series we get a power series representation of the function, so it has to be the Taylor Series.