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Consider the basis S = {1,1,22 in the vector space Pz, the set of all polynomials of degree at most 2, where Pz is a subspace of L?[0,1]. (a) Show that S = {1,2,22}...

Question

Consider the basis S = {1,1,22 in the vector space Pz, the set of all polynomials of degree at most 2, where Pz is a subspace of L?[0,1]. (a) Show that S = {1,2,22} is not an orthogonal set in L?[0,1]. (b) Use the Gram-Schmidt algorithm to generate an orthonormal basis E = {e1,e2, e3_ from S Use E to find the polynomial p = p(z) of degree at most 2 that best approximates 1/2 < x < 1 f() = 0 0 < r < 1/2 (i.e , find the (orthogonal) projection of f onto the space Pz spanned by E)_ (d S

Consider the basis S = {1,1,22 in the vector space Pz, the set of all polynomials of degree at most 2, where Pz is a subspace of L?[0,1]. (a) Show that S = {1,2,22} is not an orthogonal set in L?[0,1]. (b) Use the Gram-Schmidt algorithm to generate an orthonormal basis E = {e1,e2, e3_ from S Use E to find the polynomial p = p(z) of degree at most 2 that best approximates 1/2 < x < 1 f() = 0 0 < r < 1/2 (i.e , find the (orthogonal) projection of f onto the space Pz spanned by E)_ (d Sketch p and f on the same set of axes.



Answers

Let $V$ be the vector space of polynomials over $\mathbf{R}$ of degree $\leq 2$ with inner product defined by $\langle f, g\rangle=\int_{0}^{1} f(t) g(t) d t .$ Find a basis of the subspace $W$ orthogonal to $h(t)=2 t+1$.

We have your focus now. Basis zero p one Batou, given by the polynomial is won T and T Square Minister. We know these from example Five. So I mean, let me right zero bar for the vector obtained by computing P zero in minus two. Imara swan in zero in one and into So this is just the vector off ones and then be one bar is minus two, minus 10 and two and finally P to bar. He's two minus one minus two minus one into. Now we write these vectors because we're called the dinner plucked between fully normals. So in this case, for instance, PCR against zero he's given by p zero bar don't p zero bar. Sonny's get is five, and similarly, we compute the one against the one to be 10 and p two against me, too, to be 14. We complete these numbers now because they're going to use them now. We'll soon when we're doing the Grand Street process. So let's go see there The Perimeter Cube, which is given by D Cubed. Let's right, Q bar, of course, is minus eight minus 101 and eight and we want to apply the garnish meat process to Q to obtain well obtain up vector riches or for colonel to both be zero p one MP too. So now the computation is simplified by the fact that the vectors p zero be wanting me to our world vectors. The polynomial sze p zero we won't be too are already or for Colonel. So all we need to do is to compute the projection off cue unwto p zero p one MP to Palestine would be zero, which is Q against zero divided by P 00 And of course, we recall that these is Q bar dot p zero bar divided by five the fire we computed previously and then P zeros of the constant function one and now you can compute a cube are 0.0 bars zero So this whole projection is the no vector. Then we compute the projection on Topi one off you seeming our lady this cure against the one divided by the one against the one you compute his number. It is the result to be 34 divided by 10 and then be wanted as the constant Well, the function t That's what these 3.4 t and finally the projection onto P to off you He's again Q against me, too divided by P two against me. To that multiplies the polynomial P too. And again you against see two turns out to be just zero. So this whole number is really bothered by 14. That multiplies the vector with very normal D squared minus two and he's just zero. And therefore we can put the three to make you minus all these projections. And the fourth is too cute miners, while the only thing they were subtracting 3.40. Finally, we're asked to scale this polynomial so that p three bar is minus 1202 and one. And of course, we see that in our particular case, be three bodies given my minus 1.220 months 2.4 and 1.2. So all we need to do is to scale this vector by one over 1.2, which is exactly 5/6. So if we divide by 1.2, we get the scale. Vector is carefully. Normal is 56 off a cube minus 17 6th off T. And in fact, this bully normal where we take the vector, Sit through it with the bar. He's exactly minus 1 to 0 minus two and one has wanted.

Hello there. Okay. For this exercise, we are going to consider the space defined, um, over the continuous functions on the interval, minus 11 with the inner product equipped when an inner product defined as the interval between minus one and one off the multiplication of the functions Mhm Andi were in particular. In this exercise, we need to find some orthogonal basis. So we need to find another tonal vases the span by the following polynomial one t n t square. Okay, so how to do this? Well, we're we need Thio. Think of it not as functions, but as a vectors. Okay, so we're going to use the Grinch meet procedure to obtain or through no basis. So let's start to see what happened with these two. Paulino knows here one anti So there are tonal. Well, let's see, by computing the inner product. So let us start by considering the inner product within one anti. This is just the integral within minus one and one off t. Sorry. Here is steep. T v t. Okay, So he we got on this interval minus 11 on T is an off function. So this is equal to zero, or you can just integrate us the usual way and you will find that is equal to do. Okay, this is due to the symmetry. You can think off. This also geometrically t is just a straight line. So the integral in this symmetric interval minus 11 you can see that the area here in the area here are the same. So they're going to cancel between them. There's going to be zero. Yeah. Okay. So this happened with odd functions in general. Or you can just integrate this and you will see that you will. Let's make it explicitly. Just this case. This will be t squared, divided to evaluate. On the interval one minus one, this will be one half minus one half and this is equal to this year. So this inner product is zero. So this means that so we got that won t is equal to zero. So that implies. Actually, that means that there is an equivalence that say that one and t r or phono. Okay, so we have the first part off or thrown of basis. So so far we got that foreign of places. Let's say Earth, this is until now will be a span by one t. And there is some other function here, but we don't know yet. Okay. Off course is need to have going to be a polynomial, the guy that have a t square on it. Okay, so now let's do that. Okay. So we got to also know elements in our basis so far, so we can use DeGrange mitt procedure here. Okay, So you can think this geometrically as like the usual vectors. As we got some vector here, it's going to be one. And here there is another vector t. So these two vectors here spun the space, don't you? Okay, so we're going to find W as the span off one anti. Okay, so we got this space here. Now we're going to consider DTs Square, okay? We don't know if they if this T square is orthogonal to one anti, Okay, But we without knowing that we can convert it Thio or Thorne. Well, you will see the actually Tsk. Square is not tonal toe. These two. We need need to add something. Okay. So what the Grant Smith procedures say is that we take the projection here on the space of you. I'm going to create this ASUs. It's purple. So this that you're here will be the projection off the square on the span off one. Until that is, these super space W Okay, so this is the the the projection on this soft space w on we can subtract. So what happened if we take T Square and we subtract the projection? Well, we're going to obtain. We're going to eliminate this part off this vector here on. We're going toe end with a North Onal function polynomial. Uh, that is our tonal to this super space dog. You Okay, so that is the Grand Schmidt procedure in equations. That means that this GT that is going to be or next function is defined a steel square minus the projection off T square on the subspace off you. Where did this Super space W is defined as the spine off one empty. Okay, So what we need to do now is compute that projection. So the projection off T square on the subspace w is defined as inner product off the square with one. And here the in the product off one with one times one plus the product off the square tee. Times t divided the inner product off Tiv ity with itself. Okay, so we need to compute this. So let's consider each in their product separately. So let's start with this one. So the inner product off to square one is the integral within minus one and one off square DT again. Here we got on, um symmetric interval on. In this case, the square is an even function, so we can just consider the half off the interval on multiplied by two. So we're going to consider the interval between zero on one times two times the interval between zero and one off square ditty on this is just two thirds. Okay, so the inner product between two square and one is to felt. Then we need thio Compute the inner product off one with 10 this is trivial. His minus 11 off won t t This is just too Okay now let's see what happened with the other inner product that is between t square here indeed with itself. Okay, so the inner product between t square t is equals to the interval between minus 11 off the Cube deity. So we have seen before that if we got a symmetric interval on Dhere, we got on what function then? This integral is opposed to zero. And again, you can think off this geometrically if you like, do the plot off. This this is more or less in this way. Here is defined minus one and here is defined one. So the integral here will cancel with this part off integral. So at the end is just zero or you can just integrate in the usual way and you will obtain that is equals to zero. But it is always good to know that if you got a symmetric interval like this one on what you have insight is a not function immediately. There's all will be easier. Okay, so we can just eliminate this part here because these inner product is equal to zero. So we all this term is going to Syria. So the projection. So the projection off T square on the subspace W is equals to is equal to two thirds divided too. But this is just one third. Okay, so this is the projection. Okay, So with this information, we can get the rial value off GT, that is or for vernal function to the subspace W is to find a sti square minus the projection off the square over the subspace W where w is the span. Great. This below where this w is the span. Just remember of one T, Okay? And we got here. The value of the projection is just one third. So at the end, GT is just T square minus one third. Okay, so from this or author onal basis is a spine. Bye. The functions one d and D square, minus one third. This is or phone all basis.

Were given a subspace W. Which is the set of all polynomial of degree at most. two P. Two with an inner product inner product of F. And G is the integral from 0 to 1 of fft times. G M T E T. So oh yeah, interested in the projection of the function F f T equals T cute onto the space. W. As a hint, we're told you the orthogonal pollen or meals one To T -1 -1 and 60 squared minus 60 plus one. And then you can So we found that these were orthogonal in exercise 722. Yeah. Now because these are our dog kennel, we can just calculate the fourier coefficients to find a projection of that's under the space. Yeah, these are an orthogonal basis. Mm mm for W. Okay. Yeah. So before a coefficient C one is in a product of t cubed with one Over the inner product of one with itself and Yes. Right. Yeah. So this is the integral from 0 to 1 of t cubed D. T Over the integral from 0 to 1 of one gt. This is uh 1/4 Over one which is 1/4. Yeah the second fourier coefficients. C two is the inner product of t cubed with two t minus one Over the inner product of two T -1 with itself. This is the integral from 0 to 1 of well, t cubed times to t minus one DT over. The integral from 0-1 of to t minus one squared E T. This is equal to the integral from 0 to 1 of two T to the fourth minus t cube pt over the integral from 0 to 1 of four, T squared plus minus 40 plus one B two. Right. And this is equal to Take me up to derivatives and evaluating to 5th drugs. 14 over four thirds minus two plus one. This is to Fix. -14 is 8 20th -5, 20 assists 3 20th We were 4/3 -2 4/3 6/3 is negative. 2/3 plus one plus three thirds is positive. One third which is mm 9/20 right. You know, I like like in the field and finally for a coefficient C3 this is the inner product of she accused, I will be with uh then, I mean New Orleans, these Chicago, the 21st and 22nd, 60 squared -60 plus one. Yeah, pleased by the over the inner product of 60 squared minus 60 plus one with itself. What he said This is the integral from 0 to 1 of t cubed times 60 squared minus 60 plus one. Bt over the integral from 0 to 1 of 60 squared -60 plus one squared E. T. Yeah back I mean mhm. The this is the integral from 0 to 1 of uh 60 to the fifth minus 62. The fourth plus T cubed E. T. over. The integral from 0 to 1 of this is a little tougher. Doctor said 36 T. to the 4th. Mhm. 36 T. to the 4th -72 T Cubed. I do act plus 48 T squared minus 12 T. Mhm. No it's somebody that cares. Plus one. Well he's not taking anti derivative and evaluating we get one minus 6/5 plus 1/4. Yeah. Yeah. Mhm. 36 5th minus. Yeah. Yes. Thinking of buddy 18. Mhm. Yeah. Plus uh 16 minus six plus one. This is uh 40th minus 24/20 is negative. 4 20th Plus 5 20th says positive. 1 20 over. Uh six in Melbourne, Melbourne that 36/5 95th. Yeah, This is 1 5th he looked Which is 5 20th so or 1/4. Mhm. And therefore, oh okay, projection of our function F onto our space W This is going to be C1 times are first function one Plus C, two times their second function to T -1 Plus C. three times our third function 60 squared minus 60 plus one. So, plugging in this is uh 1/4 plus 9/20 Times to T -1 plus 1/4 times 60 squared minus 60 plus one. Mhm. This sympathize too. Three halves, t squared minus 3/5 T plus 1/20.

That was. Mhm. Well, given a vector w. and R. four second W has components 1 -2 -1, 3. Just it's no reason to try and in part a were as defined in the cardinal basis for W. That's welcome back to Jersey in our four. We'll say that some parents X, Y, Z. T. If this is in W perp this implies the product of V. Or W. Perp society you Kirk? Which is W. The equal 30 and authorized. Mm. So we have X minus two, Y minus Z. Plus through T equals zero. We have a lot of freedom here. When Children basis vectors 3° of three. New Precise, let's take X. And Y. To be equal to zero and keep it equal to one. Then it follows that Z is equal to three. And so we get detective. Yes, you won. Yes. With coordinates. 00 31 Yeah. Against. Mhm. Just mm. Now we want to find an Ortho dogma basis for W. For we want to find another vector. Mhm. That's that. Um He is not only their parker. V. With W. Zero but also Being a product of v. With you won at the 4-0. So now we get the equation um When we had the four X -2 line, mine is easy. Uh Plus three two equals zero. And we also have the equation three Z plus two equals zero. You can simplify this system. I will Well not only have 2° of freedom, so I'll take uh michigan J issue jewish wow. We get these new systems three x Uh -6. Y mm. Yeah. His muscles Plus 10 t. equals zero from the Seinfeld Kramer. Yeah. Or if you want to keep this a little bit differently, X -2 Y. And they subtract nine, Z -10. Z. equals zero and three Z plus T equals zero. Like sixties. Mhm. So now ah it's like wow let's take TP equals 2, 3. Well then it follows an x equal to zero and it follows that Z. Is equal to negative one. Yeah. And therefore why is you put too Positive five. So we get Mhm. The other vector U. two with coordinates zero five negative 13 Mhm. This is orthogonal who you want. So texting we still have one more degree of freedom. Now a last system is that our vector V. Has to be orthogonal www. But also you won. Um You too. And so we have the above system X minus two, Y -10, z equals zero And three z plus T. But now we are all equal zero. And now we also have that five Y minus Z plus three T equals zero. It's a good take a class seriously, she said especially as a matter of fact once again I'll take two to be three. That's the only degree of freedom they have though. This tells us that Z. Is negative one. And therefore from our 3rd equation, why have you booked through? Mhm mm negative two. Yeah I'm and therefore from our first equations X. Is equal to chords toys. R. Us negative 14. Yes. And so we get a third bacteria. Use three. With the opponents negative 14 negative two. Yeah negative one and three. Well they don't matter. All right. It's right there in that song. That's so and therefore the vectors U. One U. Two and V. Three to form an orthogonal basis for soviet. Perk in the. Yeah. Well I yes in parties were asked to sign an Ortho normal basis for W. Proof. This is similar to part A. But now we want to normalize each of you one through three. Now the normal view one squared is the sum of the squares of the components. Mm So this is a nine plus one which is 10. The norm of U two squared is 25 plus one plus 9 Which is 35. And the normal view three squares, this is 14 squared plus four plus one plus nine, Which is equal to 210. Therefore, a set of Ortho normal basis is given by you once over 10, You two over route 35 And you three over route to 10. This forms an Ortho normal basis for the subspace. W perk. All right.


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