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(a) Use the power series for find power series representation of f (z) = In([ ~T) What is the radlius of convergence? (Note: YOu don need to use the ratio test here...

Question

(a) Use the power series for find power series representation of f (z) = In([ ~T) What is the radlius of convergence? (Note: YOu don need to use the ratio test here because we know the radlius of convergence o the series Xae0- Use part (a) to fiuc] power" series for f () =cln( - #) By putting ^ = in vour result from part (a), express In 2 as the Sum of an infinite sericsUsepOwer' sericsapproximate the dlefinite integraldr 6o six dlecimal 0+r'pacos_

(a) Use the power series for find power series representation of f (z) = In([ ~T) What is the radlius of convergence? (Note: YOu don need to use the ratio test here because we know the radlius of convergence o the series Xae0- Use part (a) to fiuc] power" series for f () =cln( - #) By putting ^ = in vour result from part (a), express In 2 as the Sum of an infinite serics Use pOwer' serics approximate the dlefinite integral dr 6o six dlecimal 0+r' pacos_



Answers

Find the interval of convergence of the power series. (Be sure to include a check for convergence at the endpoints of the interval.) $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^{n}}{6^{n}}$$

These represent this as a power series and then find it Bridget It's internal. And so First we need to get this into the form of 1- are here. And so we're going to manipulate this little bit by doing X over one minus negative two X squared. We'll have like this that means our is negative. So this is our our here and then that's for a laptop here. Okay. So we have the summation from an equal zero to infinity of X times negative two X squared to the end of power. So that would translate into The submission from n equals 0 to infinity of -2 to the end times X two to it plus one. Yeah. So let's use the ratio test to be about the interval of convergence. We'll have negative two to the end plus one next to the times X to the two N plus three Over a -2 to the end Times X to the two n Plus one. Okay, so take the absolute value their out of here. left with two x squared. You must end one here. Okay. Means that X squared is less than one half studying taxes between One over radical two. Single too radical too of her too. So that means our interval of convergence is going to be a firm radical two or 2 Negative radical to over to medical to over two and the police won't work for that. So because some corporations test. So this is our interval of convergence

You notice that we have one of the one plus X, We could get a submission off the X O and from zero to infinity minus one power in here we explain this one which you get equal to the one minus X plus X square minus X poetry does expelled for and so on. Now, similarly, we were one other one minus X that we're getting equal to the submission on the export and And this was going to go to the one plus x on then plus X square plus expire agree plus expel fall and so on. Now, if we put the line now and then we do the subjection and then we should get the one other +11 plus X minus one other one minus x on the left hand side. On the right hand side, we have the submission of the most one about an expel AM, then minus submission of the expelled and and it will see that it was to check them. Doesn't we cancel out? Doesn't minus two X. Doesn't we cancel out then we have a minus two x poetry. Similarly, God managed to expel five and so on. We can register into the minus two that we have this way can register into the submission off the X power to a minus one, starting from one up to infinity. And here we will have have we should rewrite everything here. One plus X minus 1/1, minus x. You got to manage to submission of the expert Thio and minus one on one up to infinity. And if we do the integral on both sides D x t X t x, This one here exactly will be the function F X ego to the and then, uh, one minus X square. It will equal to here. We have managed to outside so much no one to infinity on tiny river doing this when you go to the expert, um, you had invited to and and it will be valid for on absolutely thanks more than one. And it means that actually between the minus one and one notice that every, uh we both either the minus under endpoint is to be divergent. Therefore, we have the power series of the function F X on the interval, The convergence here

It is to find a power series representation for this function. And in order to find that we're gonna need to do is well, the fact of the matter is that here, this is similar to 1/1 minus are here. And we're going to use the geometric series to be able to reconstruct this here. So that means we put this in that format, we have 1/1 minus negative X here. Which case that would go into submission format of from an equals infinity to n equals mhm 02 Infinity here. The reason we say zero is because a one that they went up here. So Hey, right. All right. So there'll be one times and they will be negative X to the nth power here. And so then our objective from here is to this is equal to Submission from an equal 0 to community of negative want its power some sex to the end here. Okay, so from here we'll we'll have to do is use the ratio. Test be able to get the interval convergence and I'm going to skip the negative one to the end cause I'm taking up some value. So this is just equal to the value of X is less than one. So basically the interval coverages from -1, 21 just like this. And the end points are not uh conversion here. So then this is our in trouble here is why the endpoints are not convergence because it would just give us Uh either positive or negative one or an oscillating between oscillation between -1 and one. So that's the interval that this is our serious power series representation.

We want to find a convergence set for the given power series. That is the series one plus X was 1/2 X one squared over two square was excellent, expose one cube to cube and so on. This Question of child your knowledge of convergence of power series. We did the first by the end of the series. The pattern the episode was clearly excellence. One of the endeavor to do yet we can plug this and turn into the absolute ratio test to find the convergence set step one of the absolute ratio test is to find P equals limit and approaching infinity absolute value N plus one over A. M. That's plugging in our term. This is limit as N approaches infinity expose one of the N plus 1/2 to the N plus one divided by X plus one of the N divided by two PM or limit as N approaches. And for the expos one times one half or absolute value X plus 1/1 half. We have convergence on this test for P less than one. So we know that our series must converge for X between negative three and one, we check our endpoints X equals negative three and one at both end points. The series diverges so our convergence that is negative 321 un inclusive of the endpoints.


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