Finding analatical function

Question

As far as i know that in differential equations an analytical function can be represented in terms of the power series and also using power series we can always determine such an analytical function. How to find such an analytical function using this power series.

$$f(x) = \sum\limits_{n=0}^{\infty}{x^n}$$

I mean to say that this is infinite series and have finite sum. If we do not use the power series technique to find such an analytical function is it possible to find out the required function by observing the behaviour and type of the given power series . And discussion on the domain of the function and also the domain of the derivative of the function will be appreciated too :)

Answer 1

I honestly don't understand quite well what you're asking, but I will try to help. You can always represent an analytical function as power series of that type in open discs in which the function is analytic. In your example that power series represent the function $\frac{1}{1-z}$, (we are in $\mathbb{C}$) which is analytic in all $\mathbb{C}$ unless $|z|=1$. That means that we can find power series for that function either in the open disc D(0,1), or in the set C(0;1,$\infty$) (idk how those sets are called in english, sorry). In the latter case we could try to get a different power series by modifyin the function this way: $$\frac{1}{1-z}=\frac{-1}{z-1}=\frac{-1}{z}\frac{1}{1-\frac{1}{z}}$$ The last fraction is again the expression for a geometric sum $\sum1/z^k$ only acceptable when $|1/z|<1\Rightarrow |z|>1$, so we are in the second set, and so our conclusion is that your sum is the expression of $1/(1-z)$ if and only if $|z|<1$, if $|z|>1$, then the same function has a different expansion: $$\sum_{k=1}^\infty \frac{-1}{z^k}$$ Basically, when you have one of those sums, it's only the expression of an analytical function in the set in which it converges. Is that what you're asking?

Answer 2

Assuming the series converges, you can say $$f(x)=\sum_{n=0}^\infty x^n \\ xf(n)=\sum_{n=1}^\infty x^n \\ f(x)=1+xf(x) \\ f(x)=\frac 1{1-x}$$

This converges for $-1 \lt x \lt 1$ or, in the complex plane, for $|x| \lt 1$

This is a classic geometric series