Question

use power series operations to find the taylor series at ( x = 0 ) for the following function.
( x^{3}sinpi x )
the taylor series for ( sin x ) is a commonly known series. what is the taylor series at ( x = 0 ) for ( sin x )?
( sum_{n = 0}^{infty}\frac{(-1)^{n}}{(2n + 1)!}x^{2n+1} ) (type an exact answer.)
use power series operations and the taylor series at ( x = 0 ) for ( sin x ) to find the taylor series at ( x = 0 ) for the given function.
( sum_{n = 0}^{infty}square ) (type an exact answer.)

Explanation:

Step1: Substitute $\pi x$ into the Taylor series of $\sin x$

The Taylor series of $\sin x=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n+1)!}x^{2n + 1}$. Replace $x$ with $\pi x$, we get $\sin(\pi x)=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n+1)!}(\pi x)^{2n + 1}=\sum_{n = 0}^{\infty}\frac{(-1)^{n}\pi^{2n+1}}{(2n+1)!}x^{2n + 1}$

Step2: Multiply by $x^{3}$

Multiply the series of $\sin(\pi x)$ by $x^{3}$. Using the rule of multiplying power - series $a_{n}x^{n}\cdot b_{m}x^{m}=a_{n}b_{m}x^{n + m}$, we have:

$$
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$$

Answer:

$\sum_{n = 0}^{\infty}\frac{(-1)^{n}\pi^{2n + 1}}{(2n+1)!}x^{2n+4}$