Question

use power series operations to find the taylor series at ( x = 0 ) for the following function.

( 5 x e ^ { 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 series for \(e^{x}\) by \(5x\)

We have \(5xe^{x}=5x\sum_{n = 0}^{\infty}\frac{x^{n}}{n!}\). Using the property of series \(a\sum_{n = 0}^{\infty}u_{n}=\sum_{n = 0}^{\infty}(a\times u_{n})\) (where \(a = 5x\) and \(u_{n}=\frac{x^{n}}{n!}\)), and \(x\times x^{n}=x^{n + 1}\), we get \(5xe^{x}=\sum_{n = 0}^{\infty}\frac{5x^{n+1}}{n!}\).

Answer:

The Taylor series for \(e^{x}\) at \(x = 0\) is \(\sum_{n=0}^{\infty}\frac{x^{n}}{n!}\), and the Taylor series for \(5xe^{x}\) at \(x = 0\) is \(\sum_{n = 0}^{\infty}\frac{5x^{n + 1}}{n!}\)