# Power Series Representation-Quick question

bcjochim07

## Homework Statement

Is there a way to check if a power series representation of a function is correct?

## The Attempt at a Solution

Homework Helper
If f(x) is the function, you might want to check if f(0) is equal to the value of the power series at x=0. You may also want to check if f'(0) is equal to the derivative of the power series at x=0. You can continue for higher derivatives if you are in doubt. If you check all of the infinite number of derivatives you KNOW you are right.

bcjochim07
Ok so if my function is x/(1-x)^2 and my power series is sum from 1 to infinity of nx^n, I can check it using derivatives and values of x, but what do I put in for n?

bcjochim07
Ok f(0) = the value of the power series at x=0

The first derivative of the function is (-x^2 + 1)/(1-x)^4 so f'(0)=1
The first derivative of the power series is sum from n=1 to infinity of n^2*x^(n-1) and if I plug in x, I will get zero. What am I not understanding?

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Homework Helper
Ok f(0) = the value of the power series at x=0

The first derivative of the function is (-x^2 + 1)/(1-x)^4 so f'(0)=1
The first derivative of the power series is sum from n=1 to infinity of n^2*x^(n-1) and if I plug in x, I will get zero. What am I not understanding?

The power series is x+2x^2+3x^3+... The derivative of that at x=0 is most certainly 1. What went wrong with your reasoning above?

bcjochim07
Oh, ooops I wasn't thinking right

bcjochim07
I forgot the first x term

$$f(x)= \begin{array}{c}e^{-\frac{1}{x^2}} x\ne 0 \\ 0 x= 0\end{array}$$
is infinitely differentiable, has all derivatives at 0 equal to 0 and so has Taylor series expansion about 0 $\Sum 0\cdot x^n$ but is not equal to that anywhere except at x= 0.