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Numerade Educator



Problem 5 Easy Difficulty

Use the definition of a Taylor series to find the first four nonzero terms of the series for $ f(x) $ centered at the given value of $ a. $

$ f(x) = xe^x, $ $ a = 0 % $


$$x+x^{2}+\frac{1}{2} x^{3}+\frac{1}{6} x^{4}$$


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Video Transcript

given a function F of X. Which is equal to X times erase X. We are to find the first four non zero terms of the taylor series centered at equals zero. Know that in taylor series you can write the function F of X. S. A. Some of the length derivative of the function times x minus a race to the end power over. And factorial. Now since it's just zero then we have summation from 02 infinity. You have anti derivative at zero times x rays the end over N. Factorial. And so to find the first foreign and zero terms we simply find the and the derivative of the function F evaluated at zero. And then we plug it in the um taylor series formula Only the non zero values. So for better understanding we will do a table showing the and the derivatives of F. And then the one which is evaluated at Which is equal to zero. So when any zero we have F of X which is X times erased. X & F of zero will be zero times erased. zero, that's just zero. And then for n equals one. We have F prime of X. That will leave a product rule X times erase two. X plus erase X times one and F. Prime of zero. That will be zero plus erases zero. That is just one. Not that we can factor this out the F prime of X into erase the X times X plus one. So we will use this to find the derivative of um F prime. And so when it is too we have F double prime. That's equal to erased X times one. Plus we have express one times erase the X. This is just erase the X times X plus two. And so F double prime of zero. This will be erased zero times one plus two. That's Rather zero Plus 2. This is equal to two. So in in S three we have f triple prime of X. That's just erase the X times the derivative of experts to which is one plus X plus two times the derivative of E which is dressed itself. And so factoring we get erased X times X plus three. So f triple prime at zero will be erased zero times zero plus three. That's just three. And because we need four terms we need at least 4 non-0 values of The ends derivative evaluated at zero. So When it is four we have F To the 4th power X. That's equal to The race to x times derivative of exports three which is one plus You have X-plus three times derivative of E. Which is itself. So we get erase the X times X plus four. So Evaluating at zero we get E raised to zero times zero plus four or that's just for And so by taylor series definition we have F of X. This is just summation from And he called 0 to Infinity of as derivative evaluated zero times X rays. And over in factorial or in expanded form. This is just when an zero we have F of zero plus you have F prime of zero over one factorial times X rays to one or X. And then plus we have F double prime, zero over two factorial times X squared plus F triple prime at zero Over three factorial times x rays to the 3rd power. And then lastly we have plus F forced derivative at zero Over four factorial times x rays to the 4th power. And so from here we have zero plus, you know, F prime of zero, that's one Over one factorial which is one times X plus F double prime of zero. That's just to over two factorial times X squared plus we have 3/3 factorial times X cubed plus For over four factorial Times X to the 4th power. So simplifying this, we have X plus to over two, that's one times X squared will be X squared plus we have three over three factorial is just six and then we have X to the third power plus we have 4/4 factorial is four times three times two times one, which is 24 times X to the 4th power. And simplifying further, we get X plus X squared plus 1/2 x cube plus 1/6 X to the 4th power. And so these are the first four non zero terms of F of X in taylor series format