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Q. 21

Found in: Page 692


Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

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Short Answer

Use an appropriate Maclaurin series to find the values of the series in Exercises 17–22.


The required answer is k=0(-1)kπ2k+1(2k+1)!=0

See the step by step solution

Step by Step Solution

Step 1. Given Information    

The given series is k=0(-1)kπ2k+1(2k+1)!

Step 2. Explanation    

The maclaurin series for the function f(x)=sinx is sinx=k=0(-1)kx2k+1(2k+1)!

So, the given series is the maclaurin series for sin x at x=π

Since, k=0(-1)kx2k+1(2k+1)!=sinx



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