## Dido525 Group Title Integrals and linear algebra? one year ago one year ago

1. Dido525 Group Title

2. Dido525 Group Title

I can't figure out the damn matrix lol.

3. Dido525 Group Title

4. tkhunny Group Title

Are we missing part (a)?

5. Dido525 Group Title

Okay, I guess I should include that.

6. satellite73 Group Title

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

7. Dido525 Group Title

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

8. satellite73 Group Title

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

9. Dido525 Group Title

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

10. Dido525 Group Title

Hmm, Feels too easy lol.

11. Dido525 Group Title

Thanks!

12. satellite73 Group Title

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

13. Dido525 Group Title

Yeah it does.

14. satellite73 Group Title

ok

15. Dido525 Group Title

Can you answer another question similar to that?

16. satellite73 Group Title

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

17. Dido525 Group Title

18. Dido525 Group Title

@tkhunny

19. Dido525 Group Title

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

20. Dido525 Group Title

so: |dw:1359170790953:dw|

21. Dido525 Group Title

right?

22. Dido525 Group Title

How did you deduce that matrix?

23. satellite73 Group Title

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 Group Title

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

25. Dido525 Group Title

So yeah, How did you deduce that? @tkhunny

26. satellite73 Group Title

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

27. satellite73 Group Title

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

28. tkhunny Group Title

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 Group Title

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

30. Dido525 Group Title

Any ideas?

31. tkhunny Group Title

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

32. Dido525 Group Title

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

33. Dido525 Group Title

34. tkhunny Group Title

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

35. tkhunny Group Title

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 Group Title

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 Group Title

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

38. Dido525 Group Title

Sorry I was eating lol.

39. Dido525 Group Title

Well I will attempt it.

40. Dido525 Group Title

Well We know: |dw:1359173248519:dw|

41. Dido525 Group Title

right?

42. Dido525 Group Title

@tkhunny

43. tkhunny Group Title

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 Group Title

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

45. Dido525 Group Title

Like the inverse of A.

46. Dido525 Group Title

Which I have to find.

47. tkhunny Group Title

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

48. Dido525 Group Title

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

49. tkhunny Group Title

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

50. Dido525 Group Title

Alright. Lets grind this out...

51. tkhunny Group Title

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

52. Dido525 Group Title

2/3 2/3 2/3 ?

53. Dido525 Group Title

for c?

54. tkhunny Group Title

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

55. Dido525 Group Title

I am actually! :P .

56. Dido525 Group Title

57. Dido525 Group Title

How did you make that substitution?

58. tkhunny Group Title

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 Group Title

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 Group Title

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

61. Dido525 Group Title

Matrix A*

62. tkhunny Group Title

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 Group Title

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 Group Title

Yep Yep, I did XD .

65. Dido525 Group Title

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

66. tkhunny Group Title

Perfect. The factoring just makes it simpler to write.

67. Dido525 Group Title

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

68. tkhunny Group Title

Just calculate $$B\cdot A^{-1}$$

69. Dido525 Group Title

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

70. Dido525 Group Title

OOO!!!! I see it! Nvm.

71. Dido525 Group Title

Thanks so much!!!

72. tkhunny Group Title

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 Group Title

Yep, you just substituted for that right?

74. tkhunny Group Title

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 Group Title

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