Question

use power series operations to find the taylor series at ( x = 0 ) for the following function.
( 12xe^{x} )
the taylor series for ( e^{x} ) is a commonly known series. what is the taylor series at ( x = 0 ) for ( e^{x} )?
( sum_{n = 0}^{infty}\frac{x^{n}}{n!} ) (type an exact answer.)
use power series operations and the taylor series at ( x = 0 ) for ( e^{x} ) to find the taylor series at ( x = 0 ) for the given function.
( sum_{n = 0}^{infty}square ) (type an exact answer.)

Explanation:

Step1: Recall the Taylor series of \(e^{x}\)

The Taylor series of \(e^{x}\) at \(x = 0\) is \(\sum_{n=0}^{\infty}\frac{x^{n}}{n!}=1 + x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\cdots\)

Step2: Multiply by \(12x\)

We have the function \(y = 12xe^{x}\). Multiply the Taylor series of \(e^{x}\) by \(12x\):

$$
LATEXBLOCK0
$$

Let \(m=n + 1\), when \(n = 0\), \(m = 1\). Then \(n=m - 1\) and the series can be rewritten as \(\sum_{m=1}^{\infty}\frac{12x^{m}}{(m - 1)!}\)

Answer:

\(\sum_{n = 0}^{\infty}\frac{12x^{n+1}}{n!}\) (or \(\sum_{m = 1}^{\infty}\frac{12x^{m}}{(m - 1)!}\))