Discussion Week 15

Bonnie Cooper

8.8.29

use the Taylor series given in Key Idea 8.8.1 to create the Taylor series of the function: \[f(x)=e^x \sin x\] for this problem show just the first 4 terms

Definition 8.8.1 Taylor and Maclauren Series

Let \(f(x)\) have derivatives of all orders at x=c.

  1. The Taylor Series of \(f(x)\), centered at c is \[\sum_{n=0}^{\infty} \frac{f^{n}(c)}{n!}(x-c)^n\]

It is given in the table of Important Taylor Series Expansions (pg491) that:

  1. \(e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!} \longrightarrow 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}\)
  2. \(\sin x= \sum_{n=0}^{infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!} \longrightarrow x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!}\)

Applying Theorem 8.8.2, we find that: \[e^x \sin x = \left( 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}\ \right) \left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} \right) \]

Distributing the right hand expression across the left:

\[ = 1 \cdot \left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} \right) + x \cdot \left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} \right)\] \[+ \frac{x^2}{2!} \cdot \left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} \right) + \frac{x^3}{3!} \cdot \left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} \right) + \ldots\] distributing again gives a crazy mess of terms:

\[x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + x^2 - \frac{x^4}{3!} + \frac{x^6}{5!} - \frac{x^8}{7!} +\] \[\frac{x^3}{2!} - \frac{x^5}{12} + \frac{x^7}{240} - \frac{x^9}{10080} + \frac{x^4}{3!} - \frac{x^6}{36} + \frac{x^8}{720} - \frac{x^{10}{30240} \ldots\]

Collect like terms and show just the first 4: \[x + x^2 + \frac{x^3}{3} -\frac{x^5}{30}\]