Problem

Use a Maclaurin series in Table 1 to obtain the M…

01:11

Problem 39 Medium Difficulty

Use a Maclaurin series in Table 1 to obtain the Maclaurin series for the given function.

$ f(x) = x \cos \left( \frac {1}{2} x^2 \right) $

Answer

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

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Video Transcript

So with the McLaren Siris in Table One, we know that the coastline of acts is equal to the sum from an equal zero to infinity of negative one to the end, Times X to the two n over two and factorial. And we could write that out if we want. But it's gonna be one minus x squared over two factorial plus X to the fourth over four factorial minus X to the sixth over six sectorial. So that's gonna keep going on. And we have a radius of convergence of infinity, Then to get X co sign X squared over two. What we're gonna do is just replace every instance of X with X squared over two and then we'll multiply the whole thing by X. So what we end up getting as a result is going to be the summation from an equal zero to infinity of negative one to the end of exports by the thing by X Times X squared to the to end over to to the two n times two and factorial. Then we could simplify this further. If we want. That would ultimately just end up giving us and equal zero to infinity negative one to the end, X to the four and plus one over to to the two n two n factorial. Then, with that in mind, we want to use ratio test. So we're gonna look at the absolute value of a M plus one over a n. When we do that, we end up getting X to the fourth over four times to n plus 2/2 n plus one. Then we want to take the limit of both of those as n goes to infinity. So when we take the limit of that, we end up getting zero. So because of that, the Siri's we know will converge for all X, which means that the radius of convergence is infinity.

Carson M.

California Baptist University

Topics

Sequences

Series

Calculus: Early Transcendentals

Chapter 11

Infinite Sequences and Series

Section 10

Taylor and Maclaurin Series

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