Question

use power series operations to find the taylor series at ( x = 0 ) for the following function.
( e^{x}+\frac{1}{1 + 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.)
the taylor series for ( \frac{1}{1 + x} ) is a commonly known series. what is the taylor series at ( x = 0 ) for ( \frac{1}{1 + x} )?
( sum_{n = 0}^{infty}square ) (type an exact answer.)

Explanation:

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

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

Step2: Recall the Taylor series for \(\frac{1}{1 + x}\)

We know that the geometric series formula \(\frac{1}{1 - t}=\sum_{n=0}^{\infty}t^{n}\), for \(|t|\lt1\). Replace \(t=-x\), then \(\frac{1}{1 + x}=\sum_{n = 0}^{\infty}(-x)^{n}=\sum_{n=0}^{\infty}(- 1)^{n}x^{n}=1 - x+x^{2}-x^{3}+\cdots\), for \(|x|\lt1\)

Answer:

The Taylor series for \(\frac{1}{1 + x}\) at \(x = 0\) is \(\sum_{n=0}^{\infty}(-1)^{n}x^{n}\)