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(a) Find the term up to x'in the Maclaurin expansion of f (x)= In(cosx) (6) Use this series t0 find an approximation in terms of T for In2 [7 marks]...

Question

(a) Find the term up to x'in the Maclaurin expansion of f (x)= In(cosx) (6) Use this series t0 find an approximation in terms of T for In2 [7 marks]

(a) Find the term up to x'in the Maclaurin expansion of f (x)= In(cosx) (6) Use this series t0 find an approximation in terms of T for In2 [7 marks]



Answers

Use (7) to find the Maclaurin series for the given function. $$ f(x)=e^{x} $$

For this problem we are asked to find the MacLaurin series for the function F of X equals E. To the power of negative acts. So our first step here is to evaluate our function at X equals zero. E. To the power of zero is just going to be one. Then we would start taking it for some terrific. Our first derivative would be negative E. To the power of negative acts. Our second derivative, I'm going to write in a little bit of a funky way but it will make sense. Our second derivative will be negative one squared times eat power of negative X. Our third derivative will be negative one cubed times Eat the power of negative X dot dot dot. So we can see that our 10th derivative Is going to equal -1. The power of N times E to the power of negative X. So we can then write our function in the MacLaurin series form of F of X is going to equal the sum N equals zero up to infinity of negative one, I N, not E to the negative X negative one to the power of N. Um Then when we plug in X equals zero, then it's part of negative acts just becomes one. So we just have that negative one out front times X. Power of N over N factorial

For this problem we are asked to find the MacLaurin series of the function F of X equals cinch of X. So our first step is to evaluate our function at zero, which cinch of zero is just zero. Then we'll start taking some derivatives. The first derivative is going to be a coach of X. The second is going to be a cinch what's kosher X and that's a cinch of X. The third derivative we'll be cinch of X again or excuse me coach of X again and then it will keep on flopping back and forth like that Evaluating those derivatives at x equals zero. Kohsh affects just one. Sinche FX as I said before is just zero and coach of X is one again. So if we were to look at the first few terms we can see that our series should look something like well we have non zero term at the N equals one, then we have a non zero term at N equals three and then we can expect that we'll have non zero. I should be careful here because we are still dividing by n factorial. So that would be executed over three factorial and then exits are 5/5 factorial plus dot dot dot so we can see the pattern would then be Some from any equals 0 to infinity of now we can write this or one way that we can write this actually we should yeah. Sorry, what we can do is say that well that first term let that be an equal zero and let that be an equals one and let that be N equals two and so on. In which case we can get the correct pattern by putting X power of two N plus one. In dividing by two N plus one factorial. Mhm.

So we know the go sign on the banks. Yes, the Macron stories under form and goes from such infinity and minus one about and experiment you and environments, Jewels and the trial. And we want to have Meg mysteries of the close off. Expelled three. So I will meet you There would be based on the X by the expert tree. So equally executive submission ingots from church infinity minus one about am Then you'll be expelled off six and over to and Tonio so and want to find the half six. I'm zero right? And listen, we correspond to the off. Ah doesn't correspond to the ex powers. Six. So, in the only case to get experts text from this expression will be when and go to one toe. Win anything to one. We have this one be, uh we have a term. Will be I off one. We go Jew minus X power six. Divine by Du diable could get you So means that doesn't can be written down is the, uh, half of 60 times Expos six divine by six foot Toyo. So it from here it conclude under F 60 echo Jew revived by six foot, turning much record. You're minus one off a Jew. So means that doesn't implies if thanks. I have six off. Zero was a Jew. Six for Troy on the vomit you.

Mhm For this problem we are asked to find the MacLaurin series for the function F of X equals lawn of one plus X. So to find the MacLaurin series, we'll have to take some derivatives and then see if we can see a pattern when we evaluate them at zero. So the first derivative of Lawn of one plus X is going to be 1/1 plus X. And the second derivative is going to be negative 1/1 plus X squared. Third derivative will be negative two times 1/1 plus X. The power of three and we'll continue on downwards. So if we evaluate these at zero F from zero will be one F double prime of zero. There will be negative one. Yeah, F triple prime of zero will be hopes that actually should be positive. There will be two times one. And what we find is that the 4th derivative Evaluated at zero would be negative three factorial. So there are a few different ways that we can try to represent this. But first we can see that we have this alternating one or a positive negative positive negative pattern, which we can get by saying that whatever the inthe derivative is when it's evaluated at zero you'll have a coefficient out front of negative one times N plus one. So The first derivative would give us -1 squared which will become positive. Then we can see, well when we have N equals one, we have one. When we have n equals two, we have negative one and then when n equals three we have two times one. So we could attempt writing something like n minus one factorial. And there is the slight problem that's a that will break down when we put in the zeroth derivatives. Um so what we can do then is right our some or our MacLaurin series separating out the initial term. So we would have lawn of one which would actually just be zero. So we don't even need to worry about it. So separating out the initial term would be eliminating that N equals zero term. So instead we just have a sum from N equals one up to infinity of now. The last thing we want to do is actually separate off that. Oh no, never mind. That does work because we have zero factorial one and then um one factorial giving us one again. So it all checks out. We just need to note that when we plug this into our MacLaurin series form You have negative one to the power of n plus one Times N -1 Factorial Divided by N Factorial. Which we must note that N factorial is the same thing as N times n minus one factorial. So those can divide out. So we can simplify this down just negative one to the power of N plus one over N times X to the power event


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