# Power Series Representation-Quick question

1. Apr 24, 2008

### bcjochim07

1. The problem statement, all variables and given/known data
Is there a way to check if a power series representation of a function is correct?

2. Relevant equations

3. The attempt at a solution

2. Apr 24, 2008

### Dick

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.

3. Apr 24, 2008

### 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?

4. Apr 24, 2008

### 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?

Last edited: Apr 24, 2008
5. Apr 24, 2008

### Dick

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?

6. Apr 24, 2008

### bcjochim07

Oh, ooops I wasn't thinking right

7. Apr 24, 2008

### bcjochim07

I forgot the first x term

8. Apr 24, 2008

### HallsofIvy

Staff Emeritus
However, in general checking that derivatives at a specific point are the same does NOT prove that a power series is equal to a function. That is only true if the function in question is "analytic" at the point which means, by definition, that it is equal to its Taylor Series expansion about the point.

For example, it is easy to show that the function
$$f(x)= \begin{array}{c}e^{-\frac{1}{x^2}} x\ne 0 \\ 0 x= 0\end{array}$$
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.

Last edited: Apr 24, 2008
9. Apr 24, 2008

### bcjochim07

So then is there a way to prove that a power series is equal to a function?

10. Apr 24, 2008

### bcjochim07

Anyone have thoughts on this?