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Find the first three nonzero terms of the Maclaurin series for each function.$$f(x)=\cos x-(2 /(1-x))$$
$-1-2 x-\frac{5}{2} x^{2}-\cdots,-1<x<1$
Calculus 2 / BC
Chapter 9
Infinite Sequences and Series
Section 8
Taylor and Maclaurin Series
Sequences
Series
Missouri State University
Oregon State University
Harvey Mudd College
University of Michigan - Ann Arbor
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So we need to figure out the MacLaurin series for both for both coastline of X and one divided by one minus X. So we'll start with consent of X. So once we find out the derivatives of G of X and the value of those derivatives at X equals zero weaken generate the MacLaurin series, so it's gonna look like one plus zero times x minus X squared to barter by two factorial, plus zero times X cubed plus X. The fourth developed for factorial plus dot, dot, dot plus and then a general negative one to the part of end times, actually part to end, divided by two in factorial plus doctor dark. And so this is equal to the Siri's form and equals zero or sigma form to infinity. They could have wants to part in intense X, the part you into but by two in factorial and that is just equal to co sign of X or G of X. Other those to the same for one divided by one minus X. So let's that h of X equal to one's about about one minus X and then figure out the derivatives again at X equals zero and let's generate the MacLaurin series. So that is equal to one plus X plus X squared, plus X cubed plus doctor and then in Sigma form. That's equal to some from zero to infinity in the next part in and so that is equal to again. Want to ride by one minus X or H backs? And so, since our original function is given as co son of Ex minus two, divided by one minus X, well, that means we can just rewrite this, which is the given in terms of our new calculated Siri's So X is equal to remember Coast and an ex. We found that it was Sigma permanent and from n equals zero to infinity, the negative one that wants a part of end times X apart to end bye bye to in factorial Last to remember. It's too, because there is a numerator of to appear so last two times the signal that we just found in equal zero to Infinity Ex apartment. And so this sequel, Teoh Negative one must to x slash five hives X squared. You might have started thought once you expand, so this converges on the intersection of negative infinity to positive affinity and negative 1 to 1. And so the interval convergence is negative one one, whereas the 1st 3 terms are up here. Negative one minus two x minus five halves X squared.
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