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Let $\mathbb{P}_{3}$ have the inner product given by evaluation at $-3,-1$ $1,$ and $3 .$ Let $p_{0}(t)=1, p_{1}(t)=t,$ and $p_{2}(t)=t^{2}$a. Compute the orthogonal projection of $p_{2}$ onto the sub- space spanned by $p_{0}$ and $p_{1}$ .b. Find a polynomial $q$ that is orthogonal to $p_{0}$ and $p_{1},$ such that $\left\{p_{0}, p_{1}, q\right\}$ is an orthogonal basis for $\operatorname{Span}\left\{p_{0}, p_{1}, p_{2}\right\} .$ Scale the polynomial $q$ so that its vec- tor of values at $(-3,-1,1,3)$ is $(1,-1,-1,1)$

A. 5B. $q=\frac{1}{4}\left(t^{2}-5\right)$

Calculus 3

Chapter 6

Orthogonality and Least Square

Section 7

Inner Product Spaces

Vectors

Missouri State University

Harvey Mudd College

Idaho State University

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there's for this exercise. Let's consider this three point on those here zero p one and p two. What we need to do in this exercise is well, first, let's consider into super space of you that is given by the span off the polynomial P zero on P one. Okay, so w is a subspace off p two that is given by the span of the polynomial P zero B one. Okay, so this exercise is asking for the compute the projection off the victory be too, on W which is the span of the vectors busy from the war. So how to do this? Well, we need thio it me. Read this here. So the projection off to on the view. Okay, So this is going to be the inner product off, too, with each off the vectors that expanded the space. So it's p two p zero divided by the inner product off busier with itself. Times busy. Go toe us the inner product off p two with p one divided by the inner product off the vector p one. We did. So times be one. Okay, so this is the formula for computer projection. Thes vector between on this punch off the other two x of the other two. Poor economic ALS. So now we need to compute each off these components off this in their product here to how to do that. Well, we need to remember that the inner product off two point on mules, Two arbitrary points on those Let me break this better so P Q. Are two point on peaches The inner pros off P and Q East on Also, let's consider three riel values. So these are just evaluations off. The point is equals to the summation off the T I que t I, with our equals to zero from 0 to 2. Okay, so this is the way to compute these in my brother. So let the start. So let a story with this one. B two b zero. So the inner product off the factory p two on zero is equals to consider with the product off T two on one. Okay, the evaluations for for the image products on this exercise is given by zero equals two minus three. She wants equals to one. T two equals two. His mind is one one on T three equals two, three. Okay, so in this case, right here, we're considering not B two, but three. So this is the summation from zero the three years three. And here we need a next to a tree. So these are just really place. Okay, so these are going to be the evolution of our Polina mules. So let's compute the inner problem. So this becomes nine plus one plus one plus nine on this is just 20. Okay, then we need to compute the inner product off busier with itself. This is just 11 We got that. This is just one plus one plus one plus one. That is just four. Okay, now we need to compute the inner product off to would be one. This is taking t squared t and this result to be minus three cubed plus minus one cube. Tow us one Q plus three. Cute. And he would happen is that these two elements cancel on these two other also councils. So at the end, this is just zero that implies here on the formula, this is zero. So this whole terms become zero. So we just need to focus on this term here, so at the end. Well, the protection off the vector B two on W where w is the span off the vector. PCO on P one is equals 2 20 divided four times. Easy room, but zero is just one. So this is equal to five to the production off YouTube on the span of the other two. Millennials is just fine. Yes. Now we got another part. So the party, I mean, they ask for, uh, or terminal basis for these for the for the span for the space spanned by these vectors. B zero p one on BT. Okay. So because this space, let's call it the for simplicity. All right, on, we need thio find an or terminal phases. How to do that? Well, we need to use the ground Schmidt procedures. So first we need to note that B zero B one is equals to zero. So that means that actually p zero on the one are are tonal already. So let's let's illustrate this. So this in the product is just one with t can. The values of evaluation is minus three minus 11 three. So this in their products become minus three minus one plus one plus three. This zero. So this implies that we want is there someone out to be zero? So these two are already or tonal. Now PT is not or tonal on. Here's where we're going to use the garnishment procedure. So what the eagerness Schmidt procedure Thus is that Let's suppose that we got two vectors that's supposed be to. And here's some are other point vector. So this just for industry we're not dealing with vectors. Well, yes, but the these vectors are pulling animals. Okay, so let's say that here, we got another vector. Cute. So how to make this pity or tonal took you? Well, we know that the the projection off PTO on cue. It's going to be this victory in green here. So this is going to be the protection off the victor. P two on cues. So what happened if we took B two unsubs tracked the protection. Oh, B two on cube. Well, what we're doing is that we're substructure these components off the vector on. We remain just with the or tone or component to Cuba. So these black vector here corresponds to taking p two a minus. The projection off B two. Thank you. Okay, so this is the idea. So that's what X. That is exactly what we're going to do. So we're going to define point of cute that is defined as taking this point on the O. P. Two minus the projection off Peter on the server space dolphin. Why? Because we know already that p zero on p one R or tonal. And we got another vector here. Be two that are not necessarily or terminal to these two factors. So we took this pun given by these two. Victor's here. That is this w So remember that we define W as the spun off B zero with p one. So here we took the whole space spun by these two vectors on. Now, by doing this procedure here that is similar to Grant Smith. Ah, well, technically is gonna Smith were obtaining a vector cube that is or tonal to this, uh, to this space over here. Okay, so that is the idea in this case that this this is more or less the geometric visualization off the procedure. Onda, we have already compute this value here. So be to let remember the B two is equals to t square. So at the end on this value here is five. So departmental cute is equals toe T square minus five. So with this you obtained on maternal bases, so their total basis is given by zero p one on cute. Where Q. Is this value here? So this is for the first part off the party. Okay, Now we need to find a different scale for Q because we are going to evaluate it. So let's consider our vector V. That is minus three minus one 13 on. We need that you evaluated at V is equals to 21 minus one, minus one. I know what Okay. So Well, we need a different cue. That is re scale. So let's see what happens if we evaluate you at this value at this factory. Okay, so if we took cute at minus three, minus 11 three, we obtain. Okay, four minus four minus 44 So we're here is clear. What is going to be the risk scale? So cute. Tilda. Yeah, Team. It's going to be taking q A T divided four, but this is just taking ts square minus five divided four. So at the end, you think that the eternal bases your terminal bases at which the Vector cube at minus three minus one or three, equals toe one minus one minus 11 is the set zero be one que Where Q is It goes to T square minus five, divided four.

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