Finding Entire Functions Satisfying Specific Conditions

[DanniHuang]

Homework Statement

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

Homework Equations

How many functions can be found?

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?

 

[STEMucator]

Homework Statement

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

Homework Equations

How many functions can be found?

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?

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

;)

 

[Dick]

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

 

[DanniHuang]

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

;)

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

 

[STEMucator]

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

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

Those are all even functions as well.