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Dido525
 2 years ago
Best ResponseYou've already chosen the best response.1I can't figure out the damn matrix lol.

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0Are we missing part (a)?

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.1Okay, I guess I should include that.

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1i think it is just \([\frac{8}{3},2,2]\)

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.1i think so too. It's 5 marks though. Is it really that easy?

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1because if you multiply that by \([a,b,c]\) written of course as a column, you get just what you want

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.1Here is part a. I don't think it's required though.

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.1Hmm, Feels too easy lol.

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1it does seem easy, but i am fairly sure it works right?

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.1Can you answer another question similar to that?

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1i dunno my linear algebra is sometime weak, but i can try

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.1So f(0)=c f(1)=a+b+c f(2)=4a+2b+c

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.1so: dw:1359170790953:dw

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.1How did you deduce that matrix?

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1i am a bit confused because those are \(f\) values not \(p\) values. \(f(0)=0 f(1)=? , f(2)=?\) are you supposed to use \(p\) instead of \(f\) ?

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.1But we are given: dw:1359171043668:dw

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.1So yeah, How did you deduce that? @tkhunny

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1then i guess it is the same as the other one, because the integral is \(\frac{8}{3}a+2b+2\)

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1oops i meant \(\frac{8}{3}a+2b+2c\)

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0Never mind that silly thing, I still haven't figured out what we're doing, here. I believe the premise is that we CANNOT find an antiderivative of f(x), but we CAN find one for p(x).

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.1Yeah, I guess because f(x) isn't given.

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0\(\int\limits_{0}^{2}p(x)\;dx = \int\limits_{0}^{2}ax^{2} + bx + c\;dx = \dfrac{8}{3}a + 2b + 2c \approx \int\limits_{0}^{2}f(x)\;dx\) I'm not sure what else we are looking at, here.

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.1Well I posted the question. It dosen't make sense to me though :( .

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0I see it. Sorry, it took me a minute. It will take a moment to code and it requires Part A!

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0Part A \[A = \left[\begin{matrix}0 & 0 & 1 \\ 0 & 1 & 1 \\ 4 & 2 & 1\end{matrix}\right]\] \[\left(\begin{matrix}f(0) \\ f(1) \\ f(2)\end{matrix}\right) = A * \left(\begin{matrix}a \\ b \\ c\end{matrix}\right)\] \[A^{1} = \dfrac{1}{4}\left[\begin{matrix}1 & 2 & 1 \\ 4 & 4 & 0 \\ 4 & 0 & 0\end{matrix}\right]\] \[A^{1\cdot}\left(\begin{matrix}f(0) \\ f(1) \\ f(2)\end{matrix}\right) = \left(\begin{matrix}a \\ b \\ c\end{matrix}\right)\]

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0Part B \[i = \left(\begin{matrix}\dfrac{8}{3} & 2 & 2\end{matrix}\right)\cdot \left(\begin{matrix}a \\ b \\ c\end{matrix}\right)\]

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0@Dido525 Are you SURE you don't want to do the glorious Part C? It's now staring at you!

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.1Sorry I was eating lol.

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.1Well I will attempt it.

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.1Well We know: dw:1359173248519:dw

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0No, \(A\) is no longer useful. Look at the very bottom of Part A and compare that carefully to Part B, right below it. Make the substitution and you are done. \(C = B\cdot A^{1}\) I guess I forgot to say : \[B = \left(\begin{matrix}\dfrac{8}{3} & 2 & 2\end{matrix}\right)\]

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.1So it would be: dw:1359174107470:dw

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0No, that's \(B\cdot A\) you need \(B\cdot A^{1}\).

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.1yeah. I mean Once I find the inverse I set it up like that right?

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0Yes, but for future reference, don't write \(A\) as a place holder for \(A^{1}\). Very confusing.

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.1Alright. Lets grind this out...

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0You may find the result surprising. It's a bit, shall we say, regular!

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0Big C, not little c, but yes. Are you surprised?

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.1How did you make that substitution?

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0What a strange result. Sample the function at both ends and in the middle and then use 2/3 of those values and you've created a quadratic approximation. Too simple!

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0You have to come to terms with the very last equation in my Part A. This gives you an expression for (a b c )^T in terms of f(0), f(1), and f(2). We are simply substituting this expression for (a b c)^T from Part B.

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.1And shoudn't the second row of matrix a say 1 1 1 ? Instead of 0 1 1.

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0Whoops! You are right. This changes things and that silly 2/3 things IS as weird as it looked. I like \(\left(\begin{matrix}\dfrac{1}{3} & \dfrac{4}{3} & \dfrac{1}{3}\end{matrix}\right)\) much better.

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0You should get \(A^{1} = \dfrac{1}{2}\left[\begin{matrix}1 & 2 & 1 \\ 3 & 4 & 1 \\ 2 & 0 & 0\end{matrix}\right]\) Important lesson, never mess up Matrix \(A\) when you will need it later!

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.1Except I didn't bother factoring out 1/2 .

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0Perfect. The factoring just makes it simpler to write.

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.1Still lost on the substitution... Or else I got you just fine.

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0Just calculate \(B\cdot A^{1}\)

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.1I did. I got it correct and everything. I just don't get why that works.

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0Hmmm...\(\left(\begin{matrix}a \\ b \\ c\end{matrix}\right)\) is the key. Stare at it until it soaks in. That's all I did while you were having dinner. :)

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.1Yep, you just substituted for that right?

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0That's it, but I used Part A to develop the substitution. Do we remember someone saying Part A was not helpful? Good work! Way to hang in there.

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.1Thank you so much! You helped me out immensely :) .
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