Integrals and linear algebra?

- anonymous

Integrals and linear algebra?

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- schrodinger

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- anonymous

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- anonymous

I can't figure out the damn matrix lol.

- anonymous

Please help.

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## More answers

- tkhunny

Are we missing part (a)?

- anonymous

Okay, I guess I should include that.

- anonymous

i think it is just \([\frac{8}{3},2,2]\)

- anonymous

i think so too. It's 5 marks though. Is it really that easy?

- anonymous

because if you multiply that by \([a,b,c]\) written of course as a column, you get just what you want

- anonymous

Here is part a. I don't think it's required though.

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- anonymous

Hmm, Feels too easy lol.

- anonymous

Thanks!

- anonymous

it does seem easy, but i am fairly sure it works right?

- anonymous

Yeah it does.

- anonymous

ok

- anonymous

Can you answer another question similar to that?

- anonymous

i dunno my linear algebra is sometime weak, but i can try

- anonymous

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- anonymous

@tkhunny

- anonymous

So f(0)=c
f(1)=a+b+c
f(2)=4a+2b+c

- anonymous

so:
|dw:1359170790953:dw|

- anonymous

right?

- anonymous

How did you deduce that matrix?

- anonymous

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 \(f\) ?

- anonymous

But we are given:
|dw:1359171043668:dw|

- anonymous

So yeah, How did you deduce that? @tkhunny

- anonymous

then i guess it is the same as the other one, because the integral is \(\frac{8}{3}a+2b+2\)

- anonymous

oops i meant \(\frac{8}{3}a+2b+2c\)

- tkhunny

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).

- anonymous

Yeah, I guess because f(x) isn't given.

- anonymous

Any ideas?

- tkhunny

\(\int\limits_{0}^{2}p(x)\;dx = \int\limits_{0}^{2}ax^{2} + bx + c\;dx = \dfrac{8}{3}a + 2b + 2c \approx \int\limits_{0}^{2}f(x)\;dx\)
I'm not sure what else we are looking at, here.

- anonymous

Well I posted the question. It dosen't make sense to me though :( .

- anonymous

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- tkhunny

I see it. Sorry, it took me a minute. It will take a moment to code and it requires Part A!

- tkhunny

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

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

@Dido525 Are you SURE you don't want to do the glorious Part C? It's now staring at you!

- anonymous

Sorry I was eating lol.

- anonymous

Well I will attempt it.

- anonymous

Well We know:
|dw:1359173248519:dw|

- anonymous

right?

- anonymous

@tkhunny

- tkhunny

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 : \[B = \left(\begin{matrix}\dfrac{8}{3} & 2 & 2\end{matrix}\right)\]

- anonymous

So it would be:
|dw:1359174107470:dw|

- anonymous

Like the inverse of A.

- anonymous

Which I have to find.

- tkhunny

No, that's \(B\cdot A\) you need \(B\cdot A^{-1}\).

- anonymous

yeah. I mean Once I find the inverse I set it up like that right?

- tkhunny

Yes, but for future reference, don't write \(A\) as a place holder for \(A^{-1}\). Very confusing.

- anonymous

Alright. Lets grind this out...

- tkhunny

You may find the result surprising. It's a bit, shall we say, regular!

- anonymous

2/3 2/3 2/3 ?

- anonymous

for c?

- tkhunny

Big C, not little c, but yes. Are you surprised?

- anonymous

I am actually! :P .

- anonymous

Can I ask something?

- anonymous

How did you make that substitution?

- tkhunny

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

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.

- anonymous

And shoudn't the second row of matrix a say 1 1 1 ? Instead of 0 1 1.

- anonymous

Matrix A*

- tkhunny

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

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!

- anonymous

Yep Yep, I did XD .

- anonymous

Except I didn't bother factoring out 1/2 .

- tkhunny

Perfect. The factoring just makes it simpler to write.

- anonymous

Still lost on the substitution... Or else I got you just fine.

- tkhunny

Just calculate \(B\cdot A^{-1}\)

- anonymous

I did. I got it correct and everything. I just don't get why that works.

- anonymous

OOO!!!! I see it! Nvm.

- anonymous

Thanks so much!!!

- tkhunny

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. :-)

- anonymous

Yep, you just substituted for that right?

- tkhunny

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.

- anonymous

Thank you so much! You helped me out immensely :) .

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