• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon

Suggested languages for you:

Americas

Europe

Q-18E

Expert-verified
Found in: Page 434

### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# Question 18: In Problems, find a power series expansion for $f\left(X\right)$ , given the expansion for f(x). role="math" localid="1664283848012" $f\left(x\right)=\mathrm{sin}x=\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^{k}}{\left(2k+1\right)!}{x}^{2k+1}$

See the step by step solution

## Step 1: differentiation of given power series

The differentiation for the series

$f\left(x\right)=\sum _{n=0}^{\infty }{a}_{n}{x}^{n}f\left(x\right)=\sum _{n=0}^{\infty }\left({a}_{n}{x}^{n=1}\right)$

The given series is

$f\left(x\right)=\mathrm{sin}x=\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^{k}}{\left(2k+1\right)!}{x}^{2k+1}$

The differentiation of the above will be

$f\left(x\right)=\sum _{k=0}^{\infty }\left(2k+1\right)\left(\frac{{\left(-1\right)}^{k}}{\left(2k+1\right)!}{x}^{2k}\right)$

## Step 2: Final proof

$f\left(x\right)=\sum _{k=0}^{\infty }\left(2k+1\right)\left(\frac{{\left(-1\right)}^{k}}{\left(2k+1\right)!}{x}^{2k}\right)$