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Problem 41 Medium Difficulty

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

$ f(x) = \frac {x}{\sqrt {4 + x^2}} $

Answer

$\sum_{n=0}^{\infty}\left(\frac{\frac{1}{2}}{n}\right) \frac{x^{2 n+1}}{2^{2 n+1}}$

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Video Transcript

So using table one from the McLaurin Siri's, we have that one plus X to the K is equal to the some from an equal zero to infinity. This is K no fraction bar That's equal toe one plus k x plus K times K minus one over two factorial plus okay, times K minus one to over three factorial x cubed. And that goes on and on. And this also is X squared right here. So with that, we know the radius of convergence is R equals one. But in this case, we have to manipulate the given expression we have, because what we're given is f of X is equal to X over the square root of four plus x squared. So to manipulate it, how we're gonna end up writing this will transform it to be X over to times one plus X squared over four to the negative one half. And you could check this out for yourselves to ensure that it is correct. Then it is those check out to be the same thing. So now we actually have our values. K equals negative one half and we'll replace every instance of X with X squared over four. Lastly, we want to multiply the whole thing by X over to, um So what we're gonna end up getting as a result is going to be When we multiply everything out, we'll get X over to Since we multiply that by the one, then we'll get ah minus X cubed over 16 because we multiplied it by four and we had two factorial. Um, so that's what we'll end up getting for that. And then we'll have plus another value. It'll be one times three. So one times three xff This was just one times, um over to to the seventh. Two factorial minus one times three times five x to the seventh over to to the 10th, three factorial. So that's gonna keep going on. And we can recognize this pattern and we see that it's gonna be X over to plus the summation. It's an ugly summation. Here we go. And it's gonna be from n equals zero, actually, and equals one because we're starting after this term to infinity. We wanted to be negative one to the end because it's alternating the positive and negative, and then it will have one times three times five all the way until two and minus one over to to the three and plus one and factorial, and that's times X to the to end plus one. So with that in mind, we can use the ratio test. When we do the ratio test, we'll end up getting as our final result is the limit as an approaches infinity of two plus one over n over eight times, one plus one over end. So we do that. We see that there's also Times X squared so we'll end up getting is to over eight times X squared. So it's just gonna give X squared over four. So as long as the absolute value of X is less than two, because we can just set that less than one Aziz, long as it's less than two. So we see that X is between negative, too, and to so the radius of convergence would certainly be, too. As a result of that