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Problem 14 Easy Difficulty

Find the Maclaurin series for $ f(x) $ using the definition of a Maclaurin series. [ Assume that $ f $ has a power series expansion. Do not show that $ R_n (x) \to 0. $] Also find the associated radius of convergence.

$ f(x) = e^{-2x} $

Answer

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

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Video Transcript

So we're gonna be looking into the McLaurin. Siri's using the definition of MacLaurin series. So what we have is f of X being equal to e to the negative two x So first thing we're gonna wanna dio is find some of the first derivatives and calculate their values at X equals zero. So obviously f of as X equals each of the negative two X and f of zero equals one. Then F prime of X is equal to a negative to e to the negative two x and f prime of zero would then give us a negative, too. F double Prime of X is going to be for E to the negative two x f double prime of zero is gonna be four and then we'll do one more. We have f triple Prime of X is going to equal a negative eight e to the negative two x So have triple prime of zero is going to be a negative eight. So we see that we can plug everything into the General Taylor form with the McLaurin being equal zero. So we see that ffx is equal to f of a, but the is equal to zero. So one minus f prime of a over one factorial. So one minus two X and then we're just gonna follow the General Taylor Taylor form using a being equal to zero. So what we're gonna get is, um, this is gonna be then plus for over two factorial X squared minus 8/3 factorial x cubed plus you know, 16/4 factorial X to the fourth and so on. Then with that, we have that This is the summation from an equal zero to infinity of negative one to the end because we wanna alternate negative and positive times two to the end to get the 248 16 going and then times X to the end to get the degree of the X value. And that's obviously going to be over in factorial. Then by the alternating Siri's test, we see that the terms become smaller as an increases. Um, and in fact, Toyo grows faster than the numerator. So it's going to converge for all real numbers. Converges for all real numbers. And we see that the radius of convergence is r equals infinity. Then we could also do a ratio test to test it, but regardless we'll see that it does converge and the race of convergence is r equals infinity.