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Question 1 (a) (5 marks) Find the Laurent series expansion off(z)at z = 1. On which domain does the series converge? (b) 5 marks) Without writing down the expansio...

Question

Question 1 (a) (5 marks) Find the Laurent series expansion off(z)at z = 1. On which domain does the series converge? (b) 5 marks) Without writing down the expansion, find the radius of convergence for the Taylor series of f(z) = Log(z) + Log(iz) at z = -l+i.

Question 1 (a) (5 marks) Find the Laurent series expansion of f(z) at z = 1. On which domain does the series converge? (b) 5 marks) Without writing down the expansion, find the radius of convergence for the Taylor series of f(z) = Log(z) + Log(iz) at z = -l+i.



Answers

a. Use the binomial series and the fact that
$$\frac{d}{d x} \sin ^{-1} x=\left(1-x^{2}\right)^{-1 / 2}$$
to generate the first four nonzero terms of the Taylor series for sin $^{-1} x .$ What is the radius of convergence?
b. Series for $\cos ^{-1} x$ Use your result in part (a) to find the first five nonzero terms of the Taylor series for $\cos ^{-1} x .$

In part A. Were asked to use the binomial series and the fact that the derivative with respect to X. Of the inverse sine of X equals one minus X squared to the negative one half power to generate the first four non zeros terms of the tailored series for inverse sine of X. Were also asked to find the radius of convergence of this series. Hello? Well first of all, one minus X squared to the negative one half will expand using binomial series. The first four terms. Well we get one plus um negative one half times negative X squared which is X squared over two plus negative one half times negative three halves times negative X squared squared. And he was like yeah yes going on this side over keith factorial Which is three times X to the 4th over eight the most Plus and then the last term. But this is going to be negative 1/2 times -3/2 times negative five halves over three factorial. What? This is uh 5:16. Yeah And then we multiply by negative X squared to the third power. So negative X. D. A. This is positive 5/16 xd eight the same day nope. Sorry X to the 6th. My mistake. And you know the second you cool. Mhm bunch. So using these terms we find that the inverse sine. Well this is approximately right. You've never heard his place. The integral of this expression one plus X squared over two plus three X. To the fourth over eight plus 5 16 X. To the sixth D. X. Sorry Which is approximately yeah. Uh X plus X cubed over two times three or six plus. Let me have extra fifth times 3/40 plus number of five times X. To the 7/7 times 16 which is five X to the seven over 112. Don't have videos approx. Another these are the first four terms in the series expansion for english sign. Not quite as simple as inverse tangent, but we do have a way of finding them now to find the root radius of convergence. Consider the limit as an approaches infinity. Um Well, so now it's going 200 were called, definitely how we calculated this will have the ratio of consecutive terms the sex. It's one times three times five times all the way up to two, two and minus one times two N plus one times X to the two n plus three over two times 4 times six times all the way up to to end times two N plus two. I did I wanted to work and then we also divide bye uh to N plus three because we integrated. Yeah, I thought it two. Yeah, he did times and we have the term before this. But we flip it so we have two times four times six. All the way up to two in. And then two N plus 1/1 times three times five. All the way up to two and -1 times extra the two N plus one. It's when Well we can cancel out factors. We get the limit as N approaches infinity of the absolute value of. See we have two n plus one reach for like nick. Are you going to do that? It's like squared. Yeah. Over that's problems that sometimes both parts. Yes. Sometimes yes. Is the people that are like Mhm. Mhm. Usually two n plus two Times two n Plus three. Yeah. Thank you all times X squared. This commercial director. It's always the client. Yeah. It's always like the can bullshit. Like more and we want this to be less than one for it to converge. So while this limit is the same as one, this is the same as when X squared is less than one. Which is only the case When the absolute value of X is less than one. Yeah. Therefore by the ratio test we should find him. I got in trouble. Stick around. It follows that the radius of convergence is one. That's who that food is. Mhm. Not the people that are going to go out anyways and have a 6000 new ones compare this with the different exercise exercise 69 then in part B where asked the same question essentially. But for inverse Cosine effects breast user result in part A to find the first five non zero terms in the taylor series for inverse co sign fx. Yeah. Right. Well, we know that the derivative with respect to X of the inverse cosine of X is equal to the opposite of one minus x squared to the negative one half. And therefore it follows the coast inverse Cosine of X is equal to well integrating. It's uh pi over two minus the inverse sine effects. This is just an identity. Just because And this is approximately using our formula from before high over 2 -2 X plus X cubed over six Plus three x lifted over 40. Yeah Plus five extra 7th Over 112. Which is approximately a constant pi over two minus x minus X cubed over three -3 X to the 5th over 40 -5 X to the 7th over 112. Yeah. So these are the first five terms of the expansion of inverse Cosine effects. Yeah, john.


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