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Show that the second-order Taylor series approximation of quadratic form q(x) = xTAx is in fact exact. ( Hint: do the expansion about 0)_...

Question

Show that the second-order Taylor series approximation of quadratic form q(x) = xTAx is in fact exact. ( Hint: do the expansion about 0)_

Show that the second-order Taylor series approximation of quadratic form q(x) = xTAx is in fact exact. ( Hint: do the expansion about 0)_



Answers

Using known Taylor series, find the first four nonzero terms of the Taylor series about 0 for the function. $$\sinh t$$

For this problem we are asked to show that a polynomial p of x equals a knot plus a one x plus dot dot dot plus a n times x power of n is its own taylor series. So what we want to do first is remember our taylor series definition? Where it's going to be the sum from K equals zero up to um if we're looking at the end of taylor series, so it will be from K equals zero up to end of the cave derivative of our function evaluated at zero divided by K factorial times X to the power of K. Now for P of X, if we consider its first derivative we know that the constant term a not will drop off and we'll be left with a one plus two A two x plus. It would be three A 3 x squared plus dot dot dot. Which means that if we evaluate that at zero we'll just have a one. Now if we take the second derivative we'll have to a two plus three times to a three x plus dot dot dot. So if we evaluate that at zero We'll have to a two. And now the point where the pattern becomes very clear if we take the third derivative We'll have three times to also known as three factorial. That's three factorial. A three plus The next term would be four factorial, A four times x plus dot dot dot. So when we evaluate it at the 0.0 we're just left with three factorial A three. Which shows us that if we take the 10th derivative of our function and evaluated at zero, we would expect to end up with n. Factorial times a N. Okay, so that means then that are some is going to be the sum from K equals zero to n of k. Factorial times a sub K divided by K factorial, divided by K. Factorial times X to the power of K. Which we can see that the K factorial cancel each other out. And we're just left with the some from K equals zero up to end of a k time to expire of K, which we'll get a zero plus a one X plus a two X squared plus dot dot dot, which gives us back our original polynomial.

This problem we are asked to find the first four non zero terms of the taylor series. About zero for the function coach of tea. So to begin we know or should know that coach of tea is the same thing as E. To the power of tea over to plus eat our negative T. Over to. So what we want to do then recognize that we just have to add up the taylor series or taylor expansions of E. To the power of tea and eat the power of negative T. Then divide everything by two. So I'll put the one half out front taylor series of eat power of tea. I'll just write down the first for non zero terms obviously. So we would have one plus T. Plus he squared over to plus t cubed over six plus dot dot dot. Then we start the one for negative T. So then we'll have one minus T. Plus t squared over two. Then it would be minus T cubed over three factorial or over six. Yeah so that would give us one half of, well we'd have one plus one is we have to we have plus t minus T. So the T. S. Will go to zero. Then we have plus t squared over two plus t squared over two. So we have two plus t squared. Then the tea cubes will add to zero which means that we actually need to grab a few more terms here but we should be able to see what the sort of forming pattern is going to be. Um All of the odd terms will be disappearing. All the evens will be getting multiplied by two. So the next one that we would expect would be two times T to the power of four or four factorial. And then we should have plus two times T to the power of 6/6 factorial. And then since we're dividing everything by two, we should get one plus T squared over two plus T to the power of 4/4 factorial, plus teeth are of six over six factorial plus dot dot dot.

For this problem, we want to find the first four non zero terms of the taylor series. About zero for the function arc tanne, also known as tan inverse, are 10 of R squared using known taylor series. So over to the side here, I have some known taylor series and we can see you that's cheating. Can't use that one. Um mm But I believe at least for the purpose of this problem, intention because I say we can't use this one because of the textbook doesn't actually give this to you. I found this on the internet. Okay, So the intention for this I believe is that is to see, okay, this isn't a known taylor series according to the textbook, but if we look at a derivative table, we should be able to see that the derivative of arc tanne X. It's going to be 1/1 plus X squared, which we can make the necessary adjustments. We find that the derivative of arc tanne of R squared should be d by D R, R R squared. It's going to give us to our over one plus R to the power of four. And now this doesn't immediately look like it but it's actually comparable to this form up here. The 1/1 minus X where the necessary changes would be. We can write this as two times are over our two times are times 1/1 minus negative R. To the power of four. Where we clearly have that X equals negative R to the power of four. So having that we can do, yeah, essentially what we'll do is do the taylor expansion of the derivative, then integrate. So the tailored expansion of the derivative. I'm just going to write it down here with the equal sign. That's the taylor expansion of the derivative. Arkan, R squared 20 to our times. That will be one minus R. To the power of four. It would be plus R to the power of eight then it would be minus R. To the power of 16 not 16. Excuse me? Art of our of 12. Yeah. One second here. Okay. Just wanted to double check. So we're playing in that to our we'd have to ar minus two R. To the power of five plus two R. To the power of nine minus two R. To the power of 13. So now to get our taylor series of arc tanne of R squared, we integrate this. So we want to integrate two ar minus dot dot dot. I'm just going to write that down there. Expedience. So integral of two. R is going to be our two R squared over two. So that's just going to go to R squared and we have minus two R. To the power of 6/6 plus two R. To the power of 10/10 minus two R. To the power of 14 over 14 plus a constant plus dot dot dot. So the constant since we're expanding about R equals zero, we need arc 10 of zero. Let's take care. The concept of integration would be our 10 of zero, but that's just zero. So the constant zero. So we can simplify daisy can simplify this down. The final result that we should get is just going to be R squared minus R to the power of 6/3 plus R to the power of 10 over five minus r to the power of 14. I was 14/7 plus dot dot dot.

All right. So we want to calculate the taylor series of this function, the root of one plus X. Um by taking three or four derivatives. So we're using this formula and we're centering at zero. So the formula looks pretty nice. Um So we need to take those derivatives, put them into the coefficients and we'll be able to get a taylor series and there may not be a pattern. So first let's take the derivatives. Um This is a power of one half. So we're going to take down a one half and then the new power will be negative one half. So there'll actually be a route on the bottom and then there is a chain role but the derivative of one plus X is one. So we're just multiplied by one. So that's good Second derivative, this is now a power of negative 1/2. So we're taking that down so negative one quarter And then the power will now be a -3/2. So it will be a 3/2 on the bottom and then the third derivative. Um we're taking down a negative three have so it's now positive 3/8 and then the power will be negative five have so five halves on the bottom. Okay? Um and then we need all the values at zero, so F zero is the root of one which is one F prime of zero is 1/2. F double prime of zero is uh negative 1/4. Okay, so these are actually pretty nice. F triple prime of zero is positive 3/8 great. And so then we can write the taylor series using the formula up there by putting in all these coefficients. So we get one plus one half x minus a quarter. And then we have this two on the bottom, so minus 1/8 x squared plus 3/8. And then we're dividing by six. So actually it's plus 1 16 X. Cubed. So I think I can see a pattern. Um The next one is probably going to be minus, probably gonna be minus 1/32 X the fourth and so on. So this is actually half decent way of calculating the taylor series if pattern starts to emerge like this.


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