# Thread: Explanation for Maclaurin Series of X^pi

1. ## Explanation for Maclaurin Series of X^pi

I am confused as how to explain this. Take the function $\displaystyle f(x) = x^{\pi}$. I am fairly certain this does not have a series expansion but is it just because it is already a polynomial? In that case is it a Maclaurin series itself?

2. It's not a polynomial because the power of $\displaystyle x$ is not a nonnegative integer.

I would think that a MacLaurin series is inappropriate here because if you centre the polynomial around $\displaystyle x = 0$, you'll find that your constants all end up being 0, until you get into negative powers of $\displaystyle x$ in which case you end up with them having zero denominators.

I would try a Taylor series centred around $\displaystyle x = 1$.

3. I am not trying to solve this for any practical purpose. The question specifically asks "Why does this function not posses a Taylor series expansion at x=0?"

4. Originally Posted by Planeman
I am not trying to solve this for any practical purpose. The question specifically asks "Why does this function not posses a Taylor series expansion at x=0?"
In that case, it is because when you try to evaluate the coefficients, they all end up being $\displaystyle 0$ or $\displaystyle \frac{\textrm{something}}{0}$.