Question

use substitution to find the taylor series at x = 0 of the function ( e^{-9x} ).
what is the general expression for the nth term in the taylor series at x = 0 for ( e^{-9x} )?
( sum_{n = 0}^{infty}square ) (type an exact answer.)

Explanation:

Step1: Recall the Taylor series of \(e^u\)

The Taylor series of \(e^u\) at \(u = 0\) is \(\sum_{n = 0}^{\infty}\frac{u^n}{n!}\).

Step2: Substitute \(u=-9x\)

Substitute \(u=-9x\) into the Taylor series of \(e^u\). We get \(\sum_{n = 0}^{\infty}\frac{(-9x)^n}{n!}\).
Using the property \((ab)^n=a^n b^n\), where \(a=-9\) and \(b = x\), we can rewrite \((-9x)^n\) as \((-1)^n9^n x^n\). So the series becomes \(\sum_{n = 0}^{\infty}\frac{(-1)^n9^n x^n}{n!}\).

Answer:

\(\sum_{n = 0}^{\infty}\frac{(-1)^n9^n x^n}{n!}\)