Question

use power series operations to find the taylor series at x = 0 for the following function.
17xe^{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}\square (type an exact answer.)

Explanation:

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

The Taylor series for \(e^{x}\) at \(x = 0\) (Maclaurin series) is given by the formula \(e^{x}=\sum_{n = 0}^{\infty}\frac{x^{n}}{n!}\), where \(n!=n\times(n - 1)\times\cdots\times1\) and \(0!=1\).

Step2: Multiply the Taylor series of \(e^{x}\) by \(17x\)

We have \(17xe^{x}=17x\sum_{n = 0}^{\infty}\frac{x^{n}}{n!}\).
Using the property of series \(a\sum_{n = 0}^{\infty}u_{n}=\sum_{n = 0}^{\infty}(au_{n})\) and \(x\times x^{n}=x^{n + 1}\), we get \(17xe^{x}=\sum_{n = 0}^{\infty}\frac{17x^{n+1}}{n!}\).

Answer:

The Taylor series for \(e^{x}\) at \(x = 0\) is \(\sum_{n=0}^{\infty}\frac{x^{n}}{n!}\).
The Taylor series for \(17xe^{x}\) at \(x = 0\) is \(\sum_{n = 0}^{\infty}\frac{17x^{n + 1}}{n!}\)