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?
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?
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 $\displaystyle x = 1$.