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Dido525

  • 3 years ago

Integrals and linear algebra?

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  1. Dido525
    • 3 years ago
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  2. Dido525
    • 3 years ago
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    I can't figure out the damn matrix lol.

  3. Dido525
    • 3 years ago
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    Please help.

  4. tkhunny
    • 3 years ago
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    Are we missing part (a)?

  5. Dido525
    • 3 years ago
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    Okay, I guess I should include that.

  6. anonymous
    • 3 years ago
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    i think it is just \([\frac{8}{3},2,2]\)

  7. Dido525
    • 3 years ago
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    i think so too. It's 5 marks though. Is it really that easy?

  8. anonymous
    • 3 years ago
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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

  9. Dido525
    • 3 years ago
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    Here is part a. I don't think it's required though.

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  10. Dido525
    • 3 years ago
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    Hmm, Feels too easy lol.

  11. Dido525
    • 3 years ago
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    Thanks!

  12. anonymous
    • 3 years ago
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    it does seem easy, but i am fairly sure it works right?

  13. Dido525
    • 3 years ago
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    Yeah it does.

  14. anonymous
    • 3 years ago
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    ok

  15. Dido525
    • 3 years ago
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    Can you answer another question similar to that?

  16. anonymous
    • 3 years ago
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    i dunno my linear algebra is sometime weak, but i can try

  17. Dido525
    • 3 years ago
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  18. Dido525
    • 3 years ago
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    @tkhunny

  19. Dido525
    • 3 years ago
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    So f(0)=c f(1)=a+b+c f(2)=4a+2b+c

  20. Dido525
    • 3 years ago
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    so: |dw:1359170790953:dw|

  21. Dido525
    • 3 years ago
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    right?

  22. Dido525
    • 3 years ago
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    How did you deduce that matrix?

  23. anonymous
    • 3 years ago
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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 \(f\) ?

  24. Dido525
    • 3 years ago
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    But we are given: |dw:1359171043668:dw|

  25. Dido525
    • 3 years ago
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    So yeah, How did you deduce that? @tkhunny

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

  27. anonymous
    • 3 years ago
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    oops i meant \(\frac{8}{3}a+2b+2c\)

  28. tkhunny
    • 3 years ago
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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).

  29. Dido525
    • 3 years ago
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    Yeah, I guess because f(x) isn't given.

  30. Dido525
    • 3 years ago
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    Any ideas?

  31. tkhunny
    • 3 years ago
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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.

  32. Dido525
    • 3 years ago
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    Well I posted the question. It dosen't make sense to me though :( .

  33. Dido525
    • 3 years ago
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  34. tkhunny
    • 3 years ago
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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!

  35. tkhunny
    • 3 years ago
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    Part A \[A = \left[\begin{matrix}0 & 0 & 1 \\ 0 & 1 & 1 \\ 4 & 2 & 1\end{matrix}\right]\] \[\left(\begin{matrix}f(0) \\ f(1) \\ f(2)\end{matrix}\right) = A * \left(\begin{matrix}a \\ b \\ c\end{matrix}\right)\] \[A^{-1} = \dfrac{1}{4}\left[\begin{matrix}1 & -2 & 1 \\ -4 & 4 & 0 \\ 4 & 0 & 0\end{matrix}\right]\] \[A^{-1\cdot}\left(\begin{matrix}f(0) \\ f(1) \\ f(2)\end{matrix}\right) = \left(\begin{matrix}a \\ b \\ c\end{matrix}\right)\]

  36. tkhunny
    • 3 years ago
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    Part B \[i = \left(\begin{matrix}\dfrac{8}{3} & 2 & 2\end{matrix}\right)\cdot \left(\begin{matrix}a \\ b \\ c\end{matrix}\right)\]

  37. tkhunny
    • 3 years ago
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    @Dido525 Are you SURE you don't want to do the glorious Part C? It's now staring at you!

  38. Dido525
    • 3 years ago
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    Sorry I was eating lol.

  39. Dido525
    • 3 years ago
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    Well I will attempt it.

  40. Dido525
    • 3 years ago
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    Well We know: |dw:1359173248519:dw|

  41. Dido525
    • 3 years ago
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    right?

  42. Dido525
    • 3 years ago
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    @tkhunny

  43. tkhunny
    • 3 years ago
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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 : \[B = \left(\begin{matrix}\dfrac{8}{3} & 2 & 2\end{matrix}\right)\]

  44. Dido525
    • 3 years ago
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    So it would be: |dw:1359174107470:dw|

  45. Dido525
    • 3 years ago
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    Like the inverse of A.

  46. Dido525
    • 3 years ago
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    Which I have to find.

  47. tkhunny
    • 3 years ago
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    No, that's \(B\cdot A\) you need \(B\cdot A^{-1}\).

  48. Dido525
    • 3 years ago
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    yeah. I mean Once I find the inverse I set it up like that right?

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

  50. Dido525
    • 3 years ago
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    Alright. Lets grind this out...

  51. tkhunny
    • 3 years ago
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    You may find the result surprising. It's a bit, shall we say, regular!

  52. Dido525
    • 3 years ago
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    2/3 2/3 2/3 ?

  53. Dido525
    • 3 years ago
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    for c?

  54. tkhunny
    • 3 years ago
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    Big C, not little c, but yes. Are you surprised?

  55. Dido525
    • 3 years ago
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    I am actually! :P .

  56. Dido525
    • 3 years ago
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    Can I ask something?

  57. Dido525
    • 3 years ago
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    How did you make that substitution?

  58. tkhunny
    • 3 years ago
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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!

  59. tkhunny
    • 3 years ago
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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.

  60. Dido525
    • 3 years ago
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    And shoudn't the second row of matrix a say 1 1 1 ? Instead of 0 1 1.

  61. Dido525
    • 3 years ago
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    Matrix A*

  62. tkhunny
    • 3 years ago
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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.

  63. tkhunny
    • 3 years ago
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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!

  64. Dido525
    • 3 years ago
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    Yep Yep, I did XD .

  65. Dido525
    • 3 years ago
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    Except I didn't bother factoring out 1/2 .

  66. tkhunny
    • 3 years ago
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    Perfect. The factoring just makes it simpler to write.

  67. Dido525
    • 3 years ago
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    Still lost on the substitution... Or else I got you just fine.

  68. tkhunny
    • 3 years ago
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    Just calculate \(B\cdot A^{-1}\)

  69. Dido525
    • 3 years ago
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    I did. I got it correct and everything. I just don't get why that works.

  70. Dido525
    • 3 years ago
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    OOO!!!! I see it! Nvm.

  71. Dido525
    • 3 years ago
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    Thanks so much!!!

  72. tkhunny
    • 3 years ago
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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. :-)

  73. Dido525
    • 3 years ago
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    Yep, you just substituted for that right?

  74. tkhunny
    • 3 years ago
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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.

  75. Dido525
    • 3 years ago
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    Thank you so much! You helped me out immensely :) .

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