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Compute the following inner products o Uo [-1,1] where f,9 >= fl1 f(c)g(w) dr:f(c) =r? _ %, g() =r8 3r + 2 f() =e", g(r) = 22 f(z) = sin(z) , 9(x) = 1...

Question

Compute the following inner products o Uo [-1,1] where f,9 >= fl1 f(c)g(w) dr:f(c) =r? _ %, g() =r8 3r + 2 f() =e", g(r) = 22 f(z) = sin(z) , 9(x) = 1

Compute the following inner products o Uo [-1,1] where f,9 >= fl1 f(c)g(w) dr: f(c) =r? _ %, g() =r8 3r + 2 f() =e", g(r) = 22 f(z) = sin(z) , 9(x) = 1



Answers

(Calculus required) use the inner product $$\langle\mathbf{f}, \mathbf{g}\rangle=\int_{0}^{1} f(x) g(x) d x$$ on $C[0,1]$ to compute $\langle\mathbf{f}, \mathbf{g}\rangle$ $$\mathbf{f}=x, \mathbf{g}=e^{x}$$

It's question when you've been function, have every year. Next, Why zing you go, Jew Go! Signed up a new minus Do ecstasy off. Why breast See? So the 1st 1 to computer half under 01234 Get any coaching course NYSE Oh, my Enough! Uh, June times one times three on. But you're plus three And then we get called you one Mine us. This will be the six out of five sewing and ico June minus one out of five. The next one to find the informed of Hae Ju zero minus 1/2 organic Because I'm the I minus, they will have a Jew times Teoh, Disappointment Your writing. By the zero last one, you get equal to minus one and then, uh minus the far and then we get included a minus five

Hello. So, in this exercise, we're going to consider, um, space of extra space V that will be defined by the continuous function on the intervals. Here one Andi, uh, also we're going to define and in their problem in new space. So let's suppose that we took F and G functions that on this season in the intervals here one. So in this case, the inner product off F and G is defined as the integral between the between the zero between 01 off f off T times de off T DJ. So this is how is the finding the problem between functions? This is one way to define animal problem. So having in mind this way to compute the inner product within functions, now we're going to consider to specific functions, So f will be one mines three T square and G is defined us. T sorry. Look like zita t minus t Q. Okay, so these our our function. So let's compute being their product between them on the intervals. Your one that's important. So thes two functions are elements off C zero. Okay, Okay. So the inner product between F and G will be the integral between 01 off the function F one minus three T square on the function G T minus D cube on all these, Did he? Okay, so we'll do this. Very straightforward. We just need Thio Compute the integral. We're going to spend the expression here, so we obtain three t mhm to the five power minus 40 q plus t beauty on. Then we just need to integrate this. So this becomes one half t to the sixth minus t four. Pull us on Health T Square evaluated between zero on one. Okay, so after we do the evaluations on the interval with teen that this is one half minus one blast one half so clearly this is equals to zero. So that means that the inner product between F and G is a close to zero. These implies, actually, just azan extra part off the analysis off dysfunction. This means that f and D r or phono or showing off because the inner product between them is equals to zero. So that is just a neck stra part of analysis of the function off the solution

Yeah, the inner product is to finance this integral. And here are vehicles coastline, two packs and G equals signed two pi X. So we need to compute this integral notes that this formula holds for the triangular function, and hence the single girl because this one and then we need to find the anti derivative outside four pi X. So it has just 1/4 pi times miners coastline, £4 ends. Then they vary the function value at 10 Yes, he calls 1/8 pi and the function value at one, which is miners causing four pi is my nurse one and a function vary at zero, which is miners course, and zero is also minus one. And that is miners one miners, miners one which equals zero.

All right, so in this problem, we need to determine a basis to the solution space, that is, or Fogg, it'll so we first need to find any basis for the solution set for decisions base of this differential equation. So to do so first, we're going to find the auxiliary equation. That's for me. P r p R is equal to R squared plus R minus two. Okay. And we said that equal to zero. Now, we need to, uh, factor this. This could be factored into our plus two and R minus one is equal to zero. So that gives us two routes. Are is equal to native to and one that means basis for our two solutions. Our for our solution space is going to be e to the X and e to the negative two X like So now we need to or fog analyze this vector space. So to do that, we're going to use a Gram Schmidt ORF organization. We're gonna let um, you the X BV one and then even negative two x BV, too. Some are V one perpendicular is just going to be V one. And then v two perpendicular is going to be able to v Tu minus And then in a product V one perpendicular V two divided by V one perpendicular comma V one perpendicular V one perpendicular Like so, Remember, this is the projection of V two on to be one perpendicular. So now we're going to, um, calculate V one, the inner product V one perpendicular view, too. So that, um, inner product V one perpendicular V two. That's just going to be that in integral from 0 to 1 of so V one perpendicular is just gonna be e to the X. So that's either X on then V two is either the negative two x the X they gives us. Um so this becomes e to the negative X here, even the negative acts like that. So the integral of that is just negative e to the negative X and then this is from 0 to 1. So you have negative e to the negative one minus, uh, one like so. So either zero or sorry plus one. Either zero is just one, remember? Okay, so, uh, that can simplify to one minus e to the negative one one minus e to the negative one like that, Okay. And the next we need to calculate the one particular V two. This is gonna be the integral from 0 to 1 of all sorry V one particular V one particular. So we eat of the x times e to the X dx, and then this is just gonna become either the two X. So the integral of either two X is gonna be 1/2 each of the two X and then again from 0 to 1. So this becomes 1/2 east squared minus 1/2. Either the zero is just once or just want minus 1/2. So then now here we can even simplify this to be 1/2 e squared, minus one. So fucking all of those in here V two get is need to the to Ah, all right, negative two X and then minus. And then we have one minus e to the negative, one divided by 1/2. And then the 1/2 can actually go up to the top and become too. So that me, right. That too one minus ease of the negative one, guarded by E squared, minus one like that. And then times e to the X. Okay, so we can just leave this, um, like this. So then our are a functional basis. Gonna B e to the X comma in the negative two X minus 21 minus e to the negative one. Uh, it's vital by e squared minus one e to the X, and this is our or FAQ. It'll basis.


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