Dido525
Integrals and linear algebra?
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Dido525
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Dido525
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I can't figure out the damn matrix lol.
Dido525
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Please help.
tkhunny
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Are we missing part (a)?
Dido525
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Okay, I guess I should include that.
anonymous
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i think it is just \([\frac{8}{3},2,2]\)
Dido525
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i think so too. It's 5 marks though. Is it really that easy?
anonymous
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because if you multiply that by \([a,b,c]\) written of course as a column, you get just what you want
Dido525
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Here is part a. I don't think it's required though.
Dido525
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Hmm, Feels too easy lol.
Dido525
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Thanks!
anonymous
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it does seem easy, but i am fairly sure it works right?
Dido525
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Yeah it does.
anonymous
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ok
Dido525
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Can you answer another question similar to that?
anonymous
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i dunno my linear algebra is sometime weak, but i can try
Dido525
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Dido525
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@tkhunny
Dido525
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So f(0)=c
f(1)=a+b+c
f(2)=4a+2b+c
Dido525
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so:
|dw:1359170790953:dw|
Dido525
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right?
Dido525
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How did you deduce that matrix?
anonymous
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i 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
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But we are given:
|dw:1359171043668:dw|
Dido525
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So yeah, How did you deduce that? @tkhunny
anonymous
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then i guess it is the same as the other one, because the integral is \(\frac{8}{3}a+2b+2\)
anonymous
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oops i meant \(\frac{8}{3}a+2b+2c\)
tkhunny
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Never 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
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Yeah, I guess because f(x) isn't given.
Dido525
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Any ideas?
tkhunny
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\(\int\limits_{0}^{2}p(x)\;dx = \int\limits_{0}^{2}ax^{2} + bx + c\;dx = \dfrac{8}{3}a + 2b + 2c \approx \int\limits_{0}^{2}f(x)\;dx\)
I'm not sure what else we are looking at, here.
Dido525
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Well I posted the question. It dosen't make sense to me though :( .
Dido525
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tkhunny
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I see it. Sorry, it took me a minute. It will take a moment to code and it requires Part A!
tkhunny
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Part 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
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Part B
\[i = \left(\begin{matrix}\dfrac{8}{3} & 2 & 2\end{matrix}\right)\cdot \left(\begin{matrix}a \\ b \\ c\end{matrix}\right)\]
tkhunny
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@Dido525 Are you SURE you don't want to do the glorious Part C? It's now staring at you!
Dido525
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Sorry I was eating lol.
Dido525
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Well I will attempt it.
Dido525
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Well We know:
|dw:1359173248519:dw|
Dido525
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right?
Dido525
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@tkhunny
tkhunny
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No, \(A\) 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
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So it would be:
|dw:1359174107470:dw|
Dido525
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Like the inverse of A.
Dido525
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Which I have to find.
tkhunny
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No, that's \(B\cdot A\) you need \(B\cdot A^{-1}\).
Dido525
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yeah. I mean Once I find the inverse I set it up like that right?
tkhunny
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Yes, but for future reference, don't write \(A\) as a place holder for \(A^{-1}\). Very confusing.
Dido525
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Alright. Lets grind this out...
tkhunny
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You may find the result surprising. It's a bit, shall we say, regular!
Dido525
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2/3 2/3 2/3 ?
Dido525
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for c?
tkhunny
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Big C, not little c, but yes. Are you surprised?
Dido525
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I am actually! :P .
Dido525
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Can I ask something?
Dido525
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How did you make that substitution?
tkhunny
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What 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
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You 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
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And shoudn't the second row of matrix a say 1 1 1 ? Instead of 0 1 1.
Dido525
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Matrix A*
tkhunny
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Whoops! 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
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You 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
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Yep Yep, I did XD .
Dido525
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Except I didn't bother factoring out 1/2 .
tkhunny
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Perfect. The factoring just makes it simpler to write.
Dido525
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Still lost on the substitution... Or else I got you just fine.
tkhunny
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Just calculate \(B\cdot A^{-1}\)
Dido525
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I did. I got it correct and everything. I just don't get why that works.
Dido525
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OOO!!!! I see it! Nvm.
Dido525
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Thanks so much!!!
tkhunny
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Hmmm...\(\left(\begin{matrix}a \\ b \\ c\end{matrix}\right)\) is the key. Stare at it until it soaks in. That's all I did while you were having dinner. :-)
Dido525
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Yep, you just substituted for that right?
tkhunny
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That'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
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Thank you so much! You helped me out immensely :) .