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The function $ A $ defined by $ A (x) = 1 + \fr…

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Problem 35 Hard Difficulty

The function $ J_1 $ defined by
$ J_1(x) = \sum_{n = 0}^{\infty} \frac {(-1)^n x^{2n + 1}}{n! (n + 1)! 2^{2n + 1}} $
is called the Bessel function of order 1.
(a) Find its domain.
(b) Graph the first several partial sums on a common screen.
(c) If your CAS has built-in Bessel Functions, graph $ J_1 $ on the same screen as the partial sums in part (b) and observe how the partial sums approximate $ J_1. $

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Infinite Sequences and Series

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Video Transcript

we use the ratio test here, so limit as n goes to infinity of absolute value of a in plus one over a N for this whole thing here is our A M value and we want for this to be less than one. This is limit as n goes to infinity of absolute value of X to the two times in plus one plus one over in plus one factorial times 10 plus two factorial times two to the two n plus one plus one and then notice that we didn't do anything with this minus one to the end. And that's because of the absolute value signs absolute value. It doesn't matter whether or not and it is positive or negative hopes. And this should be it should be a plus one there. Okay, so this is a n plus one, except for the minus one to the end, isn't there? And now we're dividing by a n So we're multiplying by the reciprocal of a M. So we'll have an infect. Oh, real top. We'll have an n plus one factorial up top as well, and we'll have a two times two to the end plus one up top. Okay, so whoops. And then we also have this next to the two n plus one down here. All right, so if we expand this, this is too in plus two plus one. So two n plus one will cancel out with this two n plus one. So these powers of X are just going to simplify the X squared in factorial divided by N plus one factorial is going to simplify the one over n plus one in plus one factorial divided by in plus two factorial is one over n plus two, two n plus one plus one. This is the same thing as to one plus two plus one, 21 plus one is going to cancel it with this two n plus one. So then we're just going to have to to the to and the denominator. So it doesn't matter what X is. As n goes to infinity, this is going to go to zero, which is less than one. So we get convergence regardless of what X is. So the domain is everything minus infinity to infinity. So if we look at the somewhere we have, let's see s one of X So this is this guy when we plug in an equal zero plus this guy, when we plug in n equals one and if you graph it, you know, you get something kind of squiggly. This should be passing through. The origin is something kind of squiggly like this Here, this is somewhere around three. This is somewhere around 0.6. Okay, so that's, you know, that'd be the second partial sum you took. You could also have the partial somewhere. You're just looking at the value you get when it is equal to zero and, uh, the whole function from in equal 02 Infinity is gonna look kind of like this, except it's going to be more squiggly. Okay, so it's still going to have this type of thing happening. Except now we have more squiggles. So in both directions and again, this value is going to be some were pretty close to three and appear this is somewhere pretty close to 0.6. So you'll see that the partial sums do start to look like the function that we're evaluating. And the more partial sums that you do, the more squiggly it's going to get, and then eventually it's going to turn out to be this really squiggly thing here

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Problem

The function $ A $ defined by $ A (x) = 1 + \fr…

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Video Transcript

James Stewart

Calculus: Early Transcendentals

Grace He

Anna Marie Vagnozzi

Michael Jacobsen

Joseph Lentino

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Grace He

Anna Marie Vagnozzi

Michael Jacobsen

Joseph Lentino

James StewartCalculus: Early Transcendentals

Grace He

Anna Marie Vagnozzi

Michael Jacobsen

Joseph Lentino

James Stewart

Calculus: Early Transcendentals