# To find an entire function

1. Sep 2, 2012

### DanniHuang

1. The problem statement, all variables and given/known data

To find entire functions which satisfy g($\frac{1}{n}$) = g(-$\frac{1}{n}$) = $\frac{1}{n^{2}}$

2. Relevant equations

How many functions can be found?

3. The attempt at a solution

Because the function is entire, it can be expanded in the Taylor series. But how can I work out the question?

2. Sep 2, 2012

### Zondrina

Hint :

This condition here : g($\frac{1}{n}$) = g(-$\frac{1}{n}$)

Should look oddly familiar to : $g(x) = g(-x)$ which is the definition of an EVEN function.

For example, consider these functions :

f(x) = x^2
f(x) = x^4
...
f(x) = x^2n

;)

3. Sep 2, 2012

### Dick

You should be able to easily find one entire function that satisfies that. A hint might be, 'don't think too hard'.

4. Sep 2, 2012

### DanniHuang

So n can only be even numbers with the Ʃa$_{n}$z$^{n}$=$\frac{1}{n^{2}}$. And then?

5. Sep 2, 2012

### Zondrina

Not necessarily, consider : cos(x), cosh(x), |x|

Those are all even functions as well.

Last edited: Sep 2, 2012