Hi everyone. I'm going to answer some questions on the homework problems from 1,110.11 These were asked by some of the students in class. Are the problems to be done are question 18 from 11, 11 and the additional problems from which are 2.3 and then two questions from 1110, which are 89.90 I will start with the question 18 from 11, 11, okay? And it asks for the following. It says that you're given the function f of x equals natural log of one plus two x, and you're going to do a Taylor expansion, A equals one up to three terms, okay? And then there are other questions which I will come to in a second, all right? Okay. So parts find, compute, compute the third degree Taylor poor normal. Okay? Here, notice that A equals one, because the expansion is around A equals one. You cannot use this for the geometric series because when we start with 1/1 plus two x, that expansion gives you in powers of x, not around powers of x minus one. Here, everything is in powers of x minus one. Okay? We actually have to use the Taylor coefficients. X, natural log of one plus two x. So this is important here, okay, is one. Okay? So, you know, you'll be working with powers of x minus one, not x to the park. That's why we cannot start this with a geometric series. All right, though, log one plus x, we can do a round x equal to zero x is that A is one of which is f of one is log one plus two, which is natural log of three. Then I want one x, it's 1/1 plus two x times two, right? Which I will write as two times one plus two x to the far minus 11 of A, which is one in this case is 2/3 Then two of x is two times minus one, and then there's another two. And then one plus two x to the power minus two, which is, this time you can multiply because there's not going to be an obvious pattern. Well, there is, but we won't need it. Okay? And then two at a, which is one, that's -4/9 Okay? And then we need three, that's eight times two. -3.3, at one, Okay? I'm sorry, I'm repeating this, but A is one, okay? Three at one is 16/27 We will actually also need four x for part B. I'm just computing this. That's 16, that's -48 times 21 plus two x to the power minus four. I won't need the value. Okay, So that's four of X. Okay? So what is three of x? I'm doing part A here. Yeah, three of x is the value of t, then it's one minus a to the power 101 factorial. Let's put the factorial under. X minus a square plus three, a three factorial x minus a. Cute. Okay, hopefully there are no mistakes in these derivative calculate. Let me just check through xk. Let's plug in the values. Three x is equal to f is natural log of three plus 12/3 x minus 12 is -4/9 x minus one squared. And F 316/27 three x. I'll just simplify this. 2/3 x minus one, -2/9 x minus one squared three factorial. That's three times two, that's plus 8/81 x minus one. Okay. This completes part A of the problem which is just find the Taylor polynomial associated with this function. This function around A equals one and up to degree three. All right. Now part says, asks estimate how much is the error if we approximate log off, which was one plus two x minus three x. Okay? For x n. Now remember the interval a is one, so it'll be some interval around one. And the question says for x 0.5 to 1.5 Okay? That's how big can this error? We know Taylor's Theorem, right? N three. Taylor theorem says x minus three x is four. P four factorial times x minus e to the power four or some between a and x. Okay? So now we've already found four. Take the absolute value is equal to absolute four factorial, absolute x minus one to the power four, right? Now, four was already found here. This is -90 61 plus two x to the power minus four. This is absolute value is 96, the absolute value of one plus two to the par -4/4 factorial, absolute x minus 14. Let's simplify this four factorial is 24, this is 4/1 plus two the four into absolute x minus four, okay? Now, x is between one point, sorry, 0.5 to 1.5 and is between A, which is one and x, okay? So is also in 0.51 0.5 Okay. So therefore, absolute text minus one, you know the four is the largest. Next is 0.5 or 1.5 right at the ends. That's the furthest of you are. And we two p is positive, so I'm not even going to bother with the absolute value. This is largest when gas is in the denominator in pallet that is at equal to 0.5 Okay? Absolute x minus three x x log one plus two x is less than equal to when x is 0.51 0.5 for that, and p is 0.5 for this 4/1 plus two times 0.5 to the power four. X minus one is 0.51 0.5 is the same. Let's simplify that. It's four and then three to the power four. This is 0.5 to the power 44/81 and this is a half, it's 1/16 so that's 1/324 Okay, So just to remind you, this is Taylor's Theorem. Okay? Now, four was already calculated. Four x was calculated here. Okay? So I just, instead of x, I put a there, and of course there is absolute value there is four and that's four factorial. Four factorial is 24, so that's four. Okay? Firstly, for x is 0.1 0.5 okay? We know what this is, right? Maximum. It can be x minus one is maximum. Absolute value is 0.5 is between a and x. P is also in that interval. This is largest here. This is then is one plus two smallest, let's write that, okay? The smallest, that's at 0.5 We plug it in, okay? So is 0.5 and x is 1.5 0.5 it gives you the same value. We do that, that's three, and that's 0.5 which is a half. That's 324. Okay. Part is just, you can say plot. You can do that in mathematica, Natural log of one plus two x minus three x on the interval 0.515 Okay? You can use Mathematica. It's basically telling you how is it that's what you're checking, right? And we know that it's at least as close as this. Okay? So that's question 18. We are done, okay? I'm going to do additional problems for 11,112.32 B is just a verification of Smit. So let's just focus on three a question three says, in this case your function is f of x, which is cosinxr given a is zero. And it says, part says, compute the Taylor series of cocinex for A equal to zero. Okay? Now, we cannot get this from a geometric series. It's not like obtained by integrating 1/1 plus x or anything like that. We really have to go and do the Taylor series thing carefully, okay? But this is not hard. I don't know if I did this in class. X is zero, is cosine zero, which is one. That's minus sinx 10 is zero, minus cosine x. Three x is Cx is equal to zero, and then four x cosine x. And that repeats, this will also repeat that. You can say that several things. X is plus or minus sinex or cocinex. That's number 1.0, is one. Okay? If n is 04, it's minus one. If n is 2610, and it's zero otherwise, okay? We will need both of these things later on. Well, let me call this A. Now we are a position to write the Taylor series. Okay? Remember the Taylor series, The sigma n going from zero to infinity, NA over n factorial x minus A to the power n or you know, which is, let's write it here. X minus eight, power 11, the one factorial, x sinus data, the power one factorial, and so on. Right now we've found all these values. So it starts with one, then that is zero, then that is minus one, then that will be zero, and then that'll be one. All these as a zero, sorry. Then they'll repeat itself. 10 minus 10. So it'll be zero. 10 minus 1010. Then it'll be minus one. This is five, this is six factorial then. Okay. Just let's do this. Plus one, it's one minus x square over two factorial x to the power 4/4 minus x to the power 6/6 factorial. And so that's a Taylor series which we can write in summation notation. Notice only even powers, so it will only be x to the part two n even numbers. Okay? And then you have to put minus one. Okay? Now, before we do that, series will start from n equal to zero because to get the one you need zero. Okay? Now we have to fix the signed, it's either n or n plus one. Right? N plus one will give you the wrong sine. N is the correct one. Here, you can check it. This is the series, notice I did not put any quality. Okay. Now, so what's the radius of convergence? So, you know and is minus one to the power end factorial minus n plus one. Wherever there is an you put a parentheses n plus one. Wherever there is an put a parentheses n plus one, N plus one over n, okay? And then you put the corresponding thing under the next two, n plus two. And then this goes on top and this comes on, okay? So this is minus one x square, and this has two extra terms. Two n plus 2.2 n plus one. If you take absolute value, you get absolute x square over, then this goes to zero as en goes to infinity. Okay? So therefore, zero always for all x O. Therefore, the series is convergent for all x. Okay? So we now have cosin ex and its Taylor series is this for all legs. Okay? Now, it did not ask us to show that the sum of the Taylor series is equal to cosx. It doesn't ask that part B says, part B says, okay, I got mixed up here. That part B says actually show that cosine effect is that sum of the Taylor series. So you have to show, now there is equality. So what we have to show is that the sequence of partial sums of this is cosine effect. We have to limit n going to infinity in x minus nth degree Taylor polynomial. This goes to zero the error for all x. Okay, How do we do that? We do that with Taylor's theorem. This is different from the project. Okay. Remember that in the project we already had the equality. Okay, for the log one. Because the series came from a geometric series where there was already an equality. We did part was relatively easy in the project. Okay, because part already came with an equality for the project. Because we started with a series whose sum we already knew. In this case, we did not start with any series whose sum we knew. All right, there was no series associate, no series which was equal to cosine x. To begin with, we had to compute the Taylor series, this, but we didn't know whether it was cosine x or not, unlike the project case which started with the geometric series. Okay, here, now we have to actually show that the cosine x is the sum of the series that comes from Taylor's theorem. It says cosine x minus t, and x is f of n plus one n plus one factorial x minus A to the par n plus one for some between a and x. Remember this is, I'm fixing x here. Okay? So how big is the error? Absolute value of N plus one? Absolute text to the power n plus one. Okay? Now, we showed that that of any x is just plus or minus sine x or x. Okay? N of N is less than equal to one, or an n and x, this quantity is utmost one, that is one. This is true for all x and for all n. If you look at go to infinity, how much is the error, right? Is less than equal to the limit as n goes to infinity. Okay? Now, this is zero. Why? We had argued this before. Remember that this series, this series is convergent, right? It's in fact it's convergent because for all legs, we showed this for exponential. It is the exponential x. Therefore, term test it's limit. I had argued this in class. I'd shown you this once before. Okay. This is the series for the power x, and we've shown this conversion, that's why it is zero. This goes to zero. Therefore yax is actually equal to the smut, it's Taylor series. Okay, In some cases, the fact that a function is equal to the Sailor series comes directly if you started with the geometric series and did some things with it. But in other cases, you have to work hard. You have to estimate the error. And for general end, show that the error goes to zero as n goes to infinity. Okay. That is the case here. What helped our case was that the nth derivative was just plus or minus sinex cocinex. It was easy to estimate this N plus one because just one. We got that. And then you may have forgotten this, but I showed in class that this limit is always zero. Okay? All right. Now, I want you to mark this because this is not going to be needed in Part C. Okay? Part C says estimate, not estimate what end should be used if he want x minus t and x to be less than. Ten to the minus eight for all x minus pi over two pi to how many terms of the Taylor series should we take if you want cosxNx to be an approximation for cosin x for every x in minus pi over two to that, right? Well, we've already got this estimate. Cosin x minus x is less than absolute x n plus one over n plus one factorial for all x. Okay? Let me call this star from star we know is less than the absolute text to the p n plus one over n plus one factoria. This is for all x n, okay? Now, x minus pi over two pi over two absolute x is less than pi over two pi over two is 3.14 something. This is like 1.51 0.57 I'll just take it less than equal to two. Okay? I'll just take it to absolute minus x. The error is less than two to the power n plus one over n plus one factorial For all x minus pi over two to pi over two, all you want to make this quantity less than ten to the power minus eight. All right? We've shown that the error in cosine t and x is less than two n plus one over n plus one factorial if you keep your x in minus p over two. Right? We had this error from Taylor's theorem. Okay. We showed that in the part B or part. Yeah, red. We showed that in part. Now x is between minus pi. You can restrict absolute text. That's what I did here. Absolute this is the error. X is in -90 to 90. So absolute ex is less than equal to two. I get two. I want that the power n plus one over n plus one factorial is ten to the power minus eight. So you can make a table. We need to make this table, and I just copied, I've done this before. So when n is ten, you can compute using Mathematica and chest multiplication division, that it's, this, that's not good enough, right? It's, it's not good enough. It's ten to the power minus eight. This is really ten to the power minus four type thing. A minus five, maybe. Then I said, okay, let's choose something really just to see where to start from. I said 100 and then you get ten to the power minus 130. I only want minus eight. I cut it in half, 50. This quantity turns out to be two to the power endals. One over is about 1.5 ten to the power -51 It works, but why do you take 50 times? Maybe you need fewer. I tried roughly half, I did 20, I got 4.1 ten to the power -14 and that's only ten to the power minus eight. Then I did 15 and you got ten to the power nine, but there is three. It's really ten to the power minus eight, it's correct. And this is probably good. N equals 15 or higher basically, or higher. Will work. Okay? So therefore, for sine x -15 of x is less than ten to the power minus eight for all x minus pi over two pi. Okay? So that's the problem. Three in the extra handout, extra stuffs. Okay? All right, so that takes care of three. Okay? That was long, but this is closer to your project, but slightly different, okay? There's a hard part here which is proving the convergence of the Taylor series to sine effect. That was much easier in the project, okay? But in the project, the error estimate was harder. Okay, Now let's look at question two and the additional problems. That's fairly straightforward. It says, verify that six to the upper 1/3 is greater than equal to by hand. It's like what I was doing here with two effect is in between that how big is absolute Texas Pi over two? It's hard to work with. I just said I will just do is an upper bound. We could work with P when I'm doing this table here. I could use pi over two. Right? But I just said, let me just do it with 3.14 something. It's 1.57 which is less than equal to two. Okay? This is the same kind of estimate that I'm trying to do here which we needed for question two. How do I verify that? Well, I want to show this is less than 613, right? Well, this is going to be true if and only if I don't want to take cube roots. So you take the cube, this will be true, this is true, right? Which is true if and only if that's 27/8 is less than equal to six, which is true if and only if 27 is less than six times eight, which is 48, and this is true. All right. I wanted to verify this, which I've written it the other way. Well, I don't want to deal with cube roots. I take the cube on both sides. This is true if anomally, if they're equivalent. Okay? And then cubes are easy is true, this is true. So the original must be true. All right. We verified this is true. And this was used actually somewhere in two when we're trying to estimate cube roots, okay? Okay, so now we go to 1110, and that's question 89. And the question is, what is the sum of the series? They give it to you as 39/2 factorial, 27/3 factorial, 81/4 factorial, and so on. Okay? The goal is basically related, to, related to the series whose sum we know. All right, so what are the series whose sons we know we actually have written on some of the series geometric series then We know the series for this, we know the series for this. Well, let's put it like this. We know the series for one on one minus R, you've done natural log of one plus x. I think I did another one. Yeah, I did arc tan of x. Okay. All these are actually given on. The table. Actually, yeah, this is one table. One on page four. When you're doing 89.90 just have this table in front of you so you can see it. Let's go back now. Firstly, we see there's a pattern here, right? This is nine is three squared, 2073 cubed. This series is really 313 square over two factorial, three cubed over three factorial. And so this is really 11 factorial n, right? This looks just like the series for exponential effects n factorial, right there. Except that this one is not there, but doesn't matter, Write it. All right, But I don't have the one, so well subtracted the square over two factorial appears. So that is the sum sets. Okay? So that was question 89. Firstly to recognize what pattern it is. There are powers, so you get that it looks like the exponential, right? The exponential series. Then see what's the connection cubed, but the one is not there, subtracted. All right, let's look at number 90. In 11.10 the series is 1/1 times two, -1/3 times two squared plus 1/5 times, sorry, this is too cute. 1/5 times two to the power five, -1/7 to two to the power seven, and so on. All right, so now you can see that it is okay. So you can see the connection here is this and this go hand in hand. All right, actually here, there is a one. Let's try it. Let's write it in submission notation. It's one over about the plus minus in a second. Let's leave room for it now. It's only going through odd, right? The way you generate odd is you put two n plus one. It's two part two n plus one. I have to fix the well, before I fix that, where does it begin? N equal to zero or one, right? We can see that when equals zero, you get one years. It goes from, okay. Now we need to put the sign there. Does this work or do I need n one? Because it's alternating. All right? N equal to zero would make it negative. It's really actually, okay. Okay. So, you know, this is looking like a series. Sorry, uh, I made an error. This is like the x is in the numerator. X to the part two N plus one with x equals half, all right? Okay. I pause the recording to think for a moment. Okay, This series looks like this. For x equals a half. Now you notice that this two n plus one is there and you can get rid of this two n plus one. This is creating a problem without it. It's like a geometric series, I'll show you. All right, without it, but it's a problem with this two n plus one. But you notice that this two plus 1.2, plus one are the same here on the top. You can get rid of them by differentiating it. Here is what I suggest. This is what I want to know, right? I get a function of x, which is the sum of this series minus one to the par n over two N plus one X to the part two N plus one. Okay? We want what is set a half. That's what we want and then that's the answer. Okay? But I'm going to do the following. Differentiate X P, x is zero to infinity, minus one to the P2n, plus 12n, plus one x to the part two n, these two cancel out. Minus one to the x to the part, x squared to the part to. Okay. Now, this is the geometric series. Well, you have to think about it. What is the R there? All right. If you reorganize this, you will see it. You have to write it in the power r to the power n, right? You take the minus, and you have to take the x squared, that's what you have to do. It becomes N going from zero to infinity, minus x squared to the Po N. Okay? Geometric series. R equals minus x squared convergent absolute of minus x squared less than one. Okay? And this is primex 1/1 minus R, which is 1/1 plus six squared. Actually, I didn't look at it. Okay, let me finish this, then I will tell you that there was a shortcut. So therefore you integrate x is tan of x plus a constant. Now, how do you get the constant? X is the series f. X equal to 00 is 100 plus c is zero, right? Which is the series. I made it a little longer. There's a short a way, which I'll tell you in a second. Okay, so C zero, therefore some of the series tenex we wanted this, which is ten on a half, that's what it is. Actually, there is a shortcut. Let's just go over this again and then I will show you the shortcut. Okay, so this is the original series which I write submission notation. The short cut actually was that if you go and look at one of the series, you look at the table, you will see that they are time series is like this. Really, you can ignore all this. What I did here, I should have put through this quicker. Let's write down this thing again. This is really the real solution. My question was, I'm going to copy this again. Okay? To find the sum. And this is the part two n plus 1/2 n plus one. Okay? This is the minus one to the P N, X to the part two N plus 1/2 N plus one, and X is a two. This is the series for ten inverse sex table on page 84. This is tie in six text equals a half, it's ten for half. All right, sorry about that, I didn't remember this series. I basically re derive this result essentially the R series for art ten. Okay? So this is the simple solution there. Okay? So these problems you just have to look at the series and see where it matches. Okay, I think we are done now.
Some HW problems from 11.10, 11.11 Math 242, Fall 2023
From Rakesh Rakesh November 28, 2023
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