# Homework Help: Taylor Series coefficient of x^n

1. May 7, 2014

### berbek16

1. The problem statement, all variables and given/known data
Find the Taylor series for 0.5x^2[e^x-e^(-x)] around x=0. What is the coefficient of x^n?

2. Relevant equations
e^x=∑x^n/n!

3. The attempt at a solution
I understand how to find the Taylor series for this equation (it being ∑[x^(2n+3)/n!]; x^3+x^5+x^7/2!+...) through manipulation of the Taylor series for e^x; however I'm not quite sure what it means by the "coefficient of x^n?" The answer is apparently 0 when n is 1 or even, and otherwise it is given by 1/[(n-3)/2]!, but how is this found? Thank you so much for any help!

2. May 7, 2014

### jbunniii

The coefficient of $x^n$ is simply the value by which $x^n$ is multiplied when you write out the series. So in your case, assuming your series is correct (which I haven't checked),
$$x^3 + x^5 + \frac{x^7}{2!} + \cdots$$
There is no constant term ($x^0$), so the coefficient of $x^0$ is zero. Similarly, the coefficient for $x^1$ is zero and likewise for all $x^n$ where $n$ is even, since these terms do not appear in the series.

So look at what remains: the coefficient for $x^3$ is $1$, the coefficient for $x^5$ is also $1$, and the coefficient for $x^7$ is $1/(2!) = 1/2$. If you plug $n=3,5,7$ into the formula given in the answer, you can see that it matches these values.

To find the general formula, the question comes down to: what is the coefficient of $x^n$ in your series $\sum x^{2n+3}/(n!)$?

Hint: try using a different letter instead of $n$ to avoid confusion: $\sum x^{2m+3}/(m!)$. What is the $x^n$ term of this series?

3. May 7, 2014

### xiavatar

The coefficient of $x^n$ is precisely what it means. If your given a value $n$ what is the coefficient for that specific term $x^n$.