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Prove that the set {1, cos X, cos 2x, '} is orthogonal on the interval [~n,n].Express the polynomial z2 + x _ 1 in terms of the base {1, €, (3x2 _ 1)}...

Question

Prove that the set {1, cos X, cos 2x, '} is orthogonal on the interval [~n,n].Express the polynomial z2 + x _ 1 in terms of the base {1, €, (3x2 _ 1)}

Prove that the set {1, cos X, cos 2x, '} is orthogonal on the interval [~n,n]. Express the polynomial z2 + x _ 1 in terms of the base {1, €, (3x2 _ 1)}



Answers

Determine an orthogonal basis for the subspace of $C^{0}[a, b]$ spanned by the given vectors, for the given interval $[a, b] .$ Use the inner product given in Equation $(5.1 .5)$. $$\begin{array}{l} f_{1}(x)=1, \quad f_{2}(x)=\sin x, \quad f_{3}(x)=\cos x \\ a=-\pi / 2, \quad b=\pi / 2 \end{array}$$

We have three functions in complex space function. One is one function to is the sign of pie X and function three is the co sigh of pie X. We need to show that they are orthogonal first. So the inner product of F one F two would be the integral over the interval. Um, from negative 1 to 1 of ah f of x, have one of xf two of x, which is just gonna be the sign of pie X. So we take the into grill and we get, um, the sign of pi X that should be co sign opposite of CO sign of pi X over pie from negative 1 to 1. So, um, the co sign of pie is negative one. And the coastline of the opposite of pie is also negative one. So we get negative one over pi minus negative one over a pie, which is zero, uh, one, the inner product of F one. Enough three in a girl from negative 1 to 1 of the co sign of Pi X, which would be the sign of pi X over Hi from negative one toe one. Um, the sign off pie is here with the sign of negative pious zero that's going to give a zero minus zero, which is zero um, F two at three. The inner product that's going to get be the integral from negative 1 to 1 of sign of pi x co sign of pie axe DX should have been writing the X on all these, which is the same as the integral from negative 1 to 1 using, um, double angle identities. Um, to sign x co sign X equals a sign of two x so we could just rate in sign of two pi axe over to DX, which is going to be negative co sign of two pi axe over, um, two times two pi, which would be four hi from negative 1 to 1. So that's gonna be the co sign of two pi to the same as the co sign of zero. And the Kusenov zero is one. And then the 2nd 1 is gonna be the question of negative two pi, which is going to the same thing one. So it's going to give us negative 1/4 pi minus negative. 1/4 pi, which is zero. Now we need to make an Ortho. Oh, normal set. And so first thing we need to dio is we need to dio the norm of F one, which is going to be the square root of the integral from negative 1 to 1 of F one is just one one square is just gonna be one squared D x. Okay, so that's going to give us the integral of one is X from negative one toe one. Okay. Oh, good. And that's going to give us the square root of one minus negative one. I thought we're gonna end up with zero, which would be not good. Um, so that's the square root of two. The norm of f two of axe square root Negative one toe one. This time it's the sign of pi X time. The sign of pie axe, which would be the sine squared of pi X. We take the integral, and that gives us, um, X over to minus the sign. Uh, two pi axe over four, huh? And that's going to go from negative 1 to 1. I just want to verify that I got this. Correct. Um, pie. Okay. Very good. Okay. The sign of two pi times one is gonna be zero. So this second term is gonna be zero for for both one and negative one. So we end up with the square root. Ah, 1/2 minus negative, 1/2 Whoops. So we put in one and we get 1/2 minus put in negative. One negative, 1/2 1 half minus negative. 1/2 is one. So the square root of one is one. So that one's already normal. That will find the distance normal too. So this time it's the co sine squared of Pi X, whose son of pie extends co sign up I X Okay, that's going to give us X over two plus the sign of two pi X over four pi. So this really ends up being the same thing because that second term is zero. And so the beginning is the same square root of 1/2 minus negative 1/2 which is one and so our Ortho normal set is the norm of F one is the square root of two. So one over the square to two is just the square to to over to times off one which is one one over the norm of F two is just one and one over. The norm of F three is just one so we can just write down Sign of Pie Exco Zenapax And now that we know that sign of pi X and co sign pot of Pi X, our normal, um, we can just leave them alone and not have to get the norm of those two functions in the future.

Of the integral from negative pie pie off co sign MX Co sign and X This if we use the formula, is 1/2 times the signed off M plus and X, divided by M plus and plus the sign of M minus and X divided by and minus end and my eyes and and we're evaluating this from negative pie. Hi. So in the case where Alan does not equal and we're going to say that this is equal to zero all right, the 2nd 1 is the same thing. But in terms of cosa, in terms of science later pipe, I sign on X times the sign of an ax. So this is equal to 1/2 times the integral from negative pied hi off co sign M minus and X minus co sign and plus and axe the ex. So this is just equal to 1/2 times It's the sign of M minus and acts divided by M minus and minus minus, the sign of M plus and the spine of M plus and acts divided by M plus end. And this is the same thing as up here is evaluated from negative nakedness. Hi too high. And since it's the same thing is up here, we're going to note that this also equal zero when M does not equal to end. Okay, so for the last one, we have the integral from negative pie High off the sign of an ex tires the co sign off an X This is equal to one over m times the sine squared off on AKs, divided by two. And we're gonna evaluate this from negative pied high playing in numbers. Use me realize that again. This also just equals zero because if you have one over has the sine squared of some multiple off pie minus one over m times the sign square off another multiple off pie a pie and negative fire. The same thing in terms of sign almost was a pious give you zero said zero minus. Here is eagle this year

In this video gonna go through the answer to question number 26 for chapter 10.3. So it has to show that this set our functions is off the normal over the domain from minus 11 Using the wave function to be your backs is equal to one. So what this means is ready to show two things were personally show this off. Agonal, which is t say that, uh, this is integral of any two. The product by two of these functions equal to zero. If they're funny, two distinct functions on, We also need to show they're often normal. So this means that additional me to show that when integrate the product of two of the same function from this list, it's gonna be able to walk. That's the first to show that it's often normal. We need to show this integral for whenever F n and M are both causes. Well, sheriff, when they're both signs, only show one is causing one is aside. So first up, when they both causes, it's gonna look something like this. In this case, we can use a triggered a metric identity which takes the product of causes to be sums of causes. So it takes us too, huh? At times because of and minus m my ex close because I've and plus and plus M minus one by X D x. Okay, So what this is, is in here? We've got the some off these two into grounds on Dhere. Rather. Sorry. And here we've got this interview and mine. Is this interim? Okay, let me sell that again. In this argument, we've got this argument minus this argument and this argument we've got this. I have implicit sighting. Okay, so then we can integrate this. Um, we're gonna have, uh, one of the n minus m. Hi. And this could be a sign of Ah, Well, this and this is gonna pay endless M minus one times high. Is that the derivative of the argument? Maybe that sign of well, this. Okay, so we'll evaluate that between minus 11 But of course, what we have in here and here is just And when we evaluate their X equal minus 11 is we just have institute falls high side of Incheon. Most qualified. You know it. Zero. So we have that. Uh, this is gonna be zero. This is gonna be zero. So therefore, the whole thing zero next up when we you're in tune different signs and I am different again. We could write this as the product that sorry, the some off to causes. And that's gonna be a because en months Mm pie. Thanks. Minus because And plus 10 minus one. Hi. Ex The ex and then following. The same idea is before we can show that the secret to they were because this cause in this cause when it's great to be signed at sign evidence, you number pi, it's gonna be Sarah. So whenever the causes for the signs next, we need to do it when one is Because I wanted to sign. Um so in this case ended them can be the same because, uh, before, which is a different, because it's cause inside. But this is an easy one, because because off to a man is one of its who Hi X is gonna be even function sign of two n minus one over to my act is gonna be our function, function, size, and even function. Um, is not function on the interpreter, not function between minus any number and plus the same number on dhe, the positive part of the vertical negative body into a pencil out s. Oh, that's just gonna be zero. Okay, so now we know that we are dealing with an old bull. Fucking old. Yes, sir. We need to now show that that letting off a normal sir. So we need to show that the ah magnitude or the norm of any of these individual functions is people to walk S O s fishing to show that bird square of the norm is equal to one. Because this just lasted not to deal with any square ese. So we need to show that cause off too, and minus one over to high X. Ah, the norm of that is equal to one. It's the norm of it. Uh, that function is written here. We can use another triggered a metric identity to show that this is gonna be equal to take half of the front. We can write it in terms off causes rather than cost squared. Sze that's gonna be from minus 1 to 1 one. Plus, it's gonna be two times the argument. That's good, because off, too, and minus one pie. Thanks. The ex on then, Just as before, this bitch is gonna integrate To be signed, Signed, evaluates it outside of and inside your love pie is gonna be zero. So that law is all gonna go to zero. So what we're left with is just the integral of one. So got half. It's world minus one, the 11 day X This is clearly just one Sidaway weaken, Right? Uh, the norm squared norm of sign to end months one of the two pi axe in this way. So weakened. Use a no, the trigonometry identity To write this as it's taken. Half of the friend we can write as one minus cause too, and minus one. Hi X, The ex just like before This bit's gonna integrate to be signed. That's gonna be signed off into multiple pie, which could be Xaver s. So all we have is this one being integrated on DDE just before this is gonna be equal to one. Do that for that. This fucking all set is and all the normal set and that completes

Okay. First part of this question wants us to show that these vectors are an orthogonal set and a basis for our three. So if they're in orthogonal set, that means that they're dot products between each other will always be zero. So we'll test u one u two, u one and u three and you too. And you three So you won dotted with you to that would be six minus six plus zero, giving us zero. You one daughter with you three would be three minus three plus zero, giving us zero. And then you two dotted with you three would be two plus two minus four, giving us again. Zero. So since we have no orthogonal set, we know this does span are three. So now we're looking to solve an equation of the Forum. X is equal to a constant times. You won, plus another constant times. You too, plus another constant times. You three. And in our case, filling in for everything. X would be five negative 31 you want would be three negative 30 You too would be 2 to 1, and gamma would be 114 and we can actually rewrite this nicely as a matrix. So weaken right? 321 negative. 321 014 times the unknowns Alfa Beta and Gamma would give us five negative 31 And since we know that these vectors air spanning, set in this matrix we know it has an inverse. So we could just use a calculator to get that. And we know that Alfa Beta Gamma would just be equal Thio this matrix once we take the inverse of it. So you take the inverse of the matrix and multiply that bye five negative 31 And doing this calculation tells us that Alfa Beta and Gamma are equal to four thirds, one third and one third, respectively. So now that we have our alpha, beta and gamma, we can just plug in. So X is four thirds you one plus one third you too. Plus one third you three. And that's our answer.


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