How do you find the Maclaurin series for e^x sin x?

To find the Maclaurin series for the function \( e^x \sin x \), we first need to recall the Maclaurin series expansions for \( e^x \) and \( \sin x \) separately and then combine them.

### Step 1: Maclaurin Series for \( e^x \)
The Maclaurin series for \( e^x \) is given by:
\[
e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots
\]
Or more generally:
\[
e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}
\]

### Step 2: Maclaurin Series for \( \sin x \)
The Maclaurin series for \( \sin x \) is:
\[
\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots
\]
Or more generally:
\[
\sin x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}
\]

### Step 3: Combine the Series
To find the Maclaurin series for \( e^x \sin x \), we multiply the series for \( e^x \) and \( \sin x \) term by term.

\[
e^x \sin x = \left( \sum_{n=0}^{\infty} \frac{x^n}{n!} \right) \cdot \left( \sum_{m=0}^{\infty} (-1)^m \frac{x^{2m+1}}{(2m+1)!} \right)
\]

Now we perform the multiplication of these two series. The general term of the product is obtained by multiplying the terms from both series. The power of \( x \) in the general term will be the sum of the powers from both series.

The general term for \( e^x \sin x \) is:
\[
\sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \frac{(-1)^m}{n! (2m+1)!} x^{n + 2m + 1}
\]

### Step 4: Simplify the Expression
We express the result as a single sum:
\[
e^x \sin x = \sum_{k=0}^{\infty} \left( \sum_{n=0}^{k} \frac{(-1)^{(k-n-1)/2}}{n! \left( \frac{k-n+1}{2} + 1 \right)!} x^k \right)
\]
Thus, this represents the Maclaurin series expansion for \( e^x \sin x \).