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(a) Use the binomial series to expand $ 1/\sqrt {1 - x^2}. $

(b) Use part (a) to find the Maclaurin series for $ \sin^{-1} x. $

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

b. $\sum_{n=0}^{\infty}\left(\begin{array}{c}-\frac{\frac{1}{2}}{n}\end{array}\right)(-1)^{n} \frac{x^{2 n-1}}{2 n+1}$

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Baylor University

University of Michigan - Ann Arbor

University of Nottingham

Idaho State University

So for this problem, we're gonna be using the binomial Siris in order to expand one over the square root of one minus X squared. So we know that we can simplify this into a term that will make it better for the binomial Siri's. And that's one plus a negative x squared to the negative, one half ongoing. The binomial Siri's All that we can do now is defined Katie equal a negative one half. Andi will replace X with a negative x squared. So when we do that, we'll end up. Getting as a result is going to be one plus one half x squared plus three over two squared times two factorial x to the fourth plus three times five over two cubed times three factorial x to the sixth and then plus three times five times seven over to to the fourth times for factorial X to the eighth plus and so on. So with that in mind, um, what we're gonna end up getting is that this is the same thing as one. Plus these some from an equals one, because we're skipping the first term to infinity of one times three times five times seven times thought that that chew on minus one and then that's all going to be over to to the end times and factorial all times x to the to end. With this in mind, we have the binomial Siri's expanded and now we have it simplified to mean this. So for part B, we want to use this to find McLaurin, Siris for the inverse sign of X. So since the derivative of the inverse sign of X, is that so? Keep in mind that the derivative of the inverse sign of X is in fact one over the square root of one minus X squared. So it follows that if we took the anti derivative of this right here, we would end up getting our answer. And sure enough, we dio setting everything how it is. We end up getting that this is going to be equal to cease, um, constant plus X plus the integral from an equals one to infinity, and we're gonna get something very similar. One times three times started dot two and minus one over two and plus one times to the end times and factorial times X to the to M plus one so very similar. What we end up getting, and then when we set X two equals zero in order to find C, um, we'll end up getting is just zero because this will become a big zero that will get rid of all of this. Um, x will be zero right here, so we'll end up getting zero, which means that C equals zero. Um, eso going back to our original formula, we can just erase the sea right here. And that is going to be our final answer. Andi, remember that? That is our answer for the inverse sign of X.

California Baptist University