How to find a power series for a given function?

Question

For homework I was asked to derive a power series for the function $e^{2x}-\frac{1}{x}$ centered at x=1 and I'm honestly clueless as to how to do it. I tried writing out the derivatives and I couldn't figure out a pattern. I'm not really sure how to do power series, so anything helps! My teacher doesn't go very into depth. Do I use Taylor/Maclaurin series to help me find it?

Answer 1

As far as I know, this function has Laurent Series at point $x = 0$, namely, $$e^{2x}-\frac{1}{x}=\sum_{n = 0}^{\infty}\frac{2^{n}x^{n}}{n!}-\frac{1}{x},$$ its convergent region should be $(-\infty,0)\cup (0,\infty)$, and is very easy to prove.

However, if you want to derive power series for it, let us try to expand this function at point $x=1$.

Since $(e^{2x})^{(n)}=2^{n}e^{2x}$, $$e^{2x}=\sum_{n=0}^{\infty}\frac{2^{n}e^2(x-1)^n}{n!},$$ its convergent region should be $(-\infty,\infty)$.

Regarding $\frac{1}{x}$, since $\frac{1}{x}=\frac{1}{1-(1-x)}$, $$\frac{1}{x}=\sum_{n=0}^{\infty}(1-x)^{n}=\sum_{n=0}^{\infty}(-1)^n (x-1)^n,$$ its convergent region should be $(0,2)$.

Thus, $$e^{2x}-\frac{1}{x}=\sum_{n=0}^{\infty}\frac{2^{n}e^2(x-1)^n}{n!}-\sum_{n=0}^{\infty}(-1)^n (x-1)^n,$$ its convergent region should only be $(0,2)$.

I am not quite sure it is what you mean. Hope so.

Answer 2

A Laurent series, as opposed to a Taylor series, allows positive and negative exponents. If that is the type of series you are after, expand $e^{2x}$ in a Taylor series, which you presumably know how to do, and subtract $\frac 1x$ and you are done. If you want a series in positive powers of $x$ you can't do that around $0$, but you can do that around any other point in the complex plane. The resulting series will converge in a circle with a radius of the distance from your point to the origin.

Answer 3

To get a power series at $x=1$ I would write $x=1+y$ and expand in powers of $y$. In your example, $$e^{2x}-\frac{1}x=e^2e^{2y}-\frac1{1+y}=e^2\sum_{n=0}^\infty\frac{2^n}{n!}y^n -\sum_{n=0}^\infty(-1)^ny^n$$ whenever $|y|<1$. If you like, you can substitute $x-1$ for $y$.