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Find the first three nonzero terms of the Maclaurin expansion of the given functions.$$f(x)=\sqrt[3]{1+x}$$

$1+\frac{\frac{1}{3}}{1 !} x+\frac{-\frac{2}{9}}{2 !} x^{2}+\frac{\frac{10}{27}}{3 !} x^{3}$

Calculus 2 / BC

Chapter 30

Expansion of Functions in Series

Section 2

Mclaurin Series

Series

Harvey Mudd College

University of Nottingham

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so the Mekele in Syria's generated by side of exes Sigma from n equals zero to infinitely negative went to the party end, but by two and plus one factorial times x the part of two in post one. This converges are negative. Infinity to infinity and the Siri's generated by, um, the natural log of one plus X is sigma for men equals one to infinity Negative once a part of n minus one. But my Aunt Times X apartment, which converges on native 1 to 1. So again this Siri's we're here is equal son X and this Siri's right here is natural log of one plus X. And so the Siri's generated by F of X, which is just these two functions multiplied together, is going to equal Sigma an equal zero to infinity. Negative wants the part and divided by two in plus one factorial times X to the power to and plus one and then this multiplied by Sigma n equals zero n equals one Teoh infinity the negative wants the part of and minus 12 by two by end times X apartment. And so once reasonable fire we should get This is equal to x squared minus 1/2 x cubed +16 x Afford this convergence on the intersection of hold on X squared. So this is X squared. This Siri's right here converges on the intersection of negative infinity to infinity and negative one toe one. So the actual interval convergence is negative ones toe one. Remember, this is going to be minus Doctor dog. And so the first terms are going to be represented right here. So X squared minus what happened? X cubed +16 x to the fourth and then the interval. Convergence is negative 1 to 1.

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