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Dido525Best ResponseYou've already chosen the best response.1
I can't figure out the damn matrix lol.
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
Are we missing part (a)?
 one year ago

Dido525Best ResponseYou've already chosen the best response.1
Okay, I guess I should include that.
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
i think it is just \([\frac{8}{3},2,2]\)
 one year ago

Dido525Best ResponseYou've already chosen the best response.1
i think so too. It's 5 marks though. Is it really that easy?
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
because if you multiply that by \([a,b,c]\) written of course as a column, you get just what you want
 one year ago

Dido525Best ResponseYou've already chosen the best response.1
Here is part a. I don't think it's required though.
 one year ago

Dido525Best ResponseYou've already chosen the best response.1
Hmm, Feels too easy lol.
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
it does seem easy, but i am fairly sure it works right?
 one year ago

Dido525Best ResponseYou've already chosen the best response.1
Can you answer another question similar to that?
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
i dunno my linear algebra is sometime weak, but i can try
 one year ago

Dido525Best ResponseYou've already chosen the best response.1
So f(0)=c f(1)=a+b+c f(2)=4a+2b+c
 one year ago

Dido525Best ResponseYou've already chosen the best response.1
so: dw:1359170790953:dw
 one year ago

Dido525Best ResponseYou've already chosen the best response.1
How did you deduce that matrix?
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
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 \(f\) ?
 one year ago

Dido525Best ResponseYou've already chosen the best response.1
But we are given: dw:1359171043668:dw
 one year ago

Dido525Best ResponseYou've already chosen the best response.1
So yeah, How did you deduce that? @tkhunny
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
then i guess it is the same as the other one, because the integral is \(\frac{8}{3}a+2b+2\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
oops i meant \(\frac{8}{3}a+2b+2c\)
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
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).
 one year ago

Dido525Best ResponseYou've already chosen the best response.1
Yeah, I guess because f(x) isn't given.
 one year ago

tkhunnyBest 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.
 one year ago

Dido525Best ResponseYou've already chosen the best response.1
Well I posted the question. It dosen't make sense to me though :( .
 one year ago

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

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

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

tkhunnyBest 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!
 one year ago

Dido525Best ResponseYou've already chosen the best response.1
Sorry I was eating lol.
 one year ago

Dido525Best ResponseYou've already chosen the best response.1
Well I will attempt it.
 one year ago

Dido525Best ResponseYou've already chosen the best response.1
Well We know: dw:1359173248519:dw
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
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 : \[B = \left(\begin{matrix}\dfrac{8}{3} & 2 & 2\end{matrix}\right)\]
 one year ago

Dido525Best ResponseYou've already chosen the best response.1
So it would be: dw:1359174107470:dw
 one year ago

Dido525Best ResponseYou've already chosen the best response.1
Like the inverse of A.
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
No, that's \(B\cdot A\) you need \(B\cdot A^{1}\).
 one year ago

Dido525Best ResponseYou've already chosen the best response.1
yeah. I mean Once I find the inverse I set it up like that right?
 one year ago

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

Dido525Best ResponseYou've already chosen the best response.1
Alright. Lets grind this out...
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
You may find the result surprising. It's a bit, shall we say, regular!
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
Big C, not little c, but yes. Are you surprised?
 one year ago

Dido525Best ResponseYou've already chosen the best response.1
How did you make that substitution?
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
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!
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
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.
 one year ago

Dido525Best ResponseYou've already chosen the best response.1
And shoudn't the second row of matrix a say 1 1 1 ? Instead of 0 1 1.
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
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.
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
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!
 one year ago

Dido525Best ResponseYou've already chosen the best response.1
Except I didn't bother factoring out 1/2 .
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
Perfect. The factoring just makes it simpler to write.
 one year ago

Dido525Best ResponseYou've already chosen the best response.1
Still lost on the substitution... Or else I got you just fine.
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
Just calculate \(B\cdot A^{1}\)
 one year ago

Dido525Best ResponseYou've already chosen the best response.1
I did. I got it correct and everything. I just don't get why that works.
 one year ago

Dido525Best ResponseYou've already chosen the best response.1
OOO!!!! I see it! Nvm.
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
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. :)
 one year ago

Dido525Best ResponseYou've already chosen the best response.1
Yep, you just substituted for that right?
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
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.
 one year ago

Dido525Best ResponseYou've already chosen the best response.1
Thank you so much! You helped me out immensely :) .
 one year ago
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