anonymous 3 years ago Integrals and linear algebra?

1. anonymous

2. anonymous

I can't figure out the damn matrix lol.

3. anonymous

4. tkhunny

Are we missing part (a)?

5. anonymous

Okay, I guess I should include that.

6. anonymous

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

7. anonymous

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

8. anonymous

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

9. anonymous

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

10. anonymous

Hmm, Feels too easy lol.

11. anonymous

Thanks!

12. anonymous

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

13. anonymous

Yeah it does.

14. anonymous

ok

15. anonymous

Can you answer another question similar to that?

16. anonymous

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

17. anonymous

18. anonymous

@tkhunny

19. anonymous

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

20. anonymous

so: |dw:1359170790953:dw|

21. anonymous

right?

22. anonymous

How did you deduce that matrix?

23. 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 $$f$$ ?

24. anonymous

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

25. anonymous

So yeah, How did you deduce that? @tkhunny

26. anonymous

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

27. anonymous

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

28. 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).

29. anonymous

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

30. anonymous

Any ideas?

31. 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.

32. anonymous

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

33. anonymous

34. tkhunny

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

35. tkhunny

Part A $A = \left[\begin{matrix}0 & 0 & 1 \\ 0 & 1 & 1 \\ 4 & 2 & 1\end{matrix}\right]$ $\left(\begin{matrix}f(0) \\ f(1) \\ f(2)\end{matrix}\right) = A * \left(\begin{matrix}a \\ b \\ c\end{matrix}\right)$ $A^{-1} = \dfrac{1}{4}\left[\begin{matrix}1 & -2 & 1 \\ -4 & 4 & 0 \\ 4 & 0 & 0\end{matrix}\right]$ $A^{-1\cdot}\left(\begin{matrix}f(0) \\ f(1) \\ f(2)\end{matrix}\right) = \left(\begin{matrix}a \\ b \\ c\end{matrix}\right)$

36. tkhunny

Part B $i = \left(\begin{matrix}\dfrac{8}{3} & 2 & 2\end{matrix}\right)\cdot \left(\begin{matrix}a \\ b \\ c\end{matrix}\right)$

37. tkhunny

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

38. anonymous

Sorry I was eating lol.

39. anonymous

Well I will attempt it.

40. anonymous

Well We know: |dw:1359173248519:dw|

41. anonymous

right?

42. anonymous

@tkhunny

43. 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 : $B = \left(\begin{matrix}\dfrac{8}{3} & 2 & 2\end{matrix}\right)$

44. anonymous

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

45. anonymous

Like the inverse of A.

46. anonymous

Which I have to find.

47. tkhunny

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

48. anonymous

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

49. tkhunny

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

50. anonymous

Alright. Lets grind this out...

51. tkhunny

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

52. anonymous

2/3 2/3 2/3 ?

53. anonymous

for c?

54. tkhunny

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

55. anonymous

I am actually! :P .

56. anonymous

57. anonymous

How did you make that substitution?

58. 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!

59. 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.

60. anonymous

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

61. anonymous

Matrix A*

62. 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.

63. 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!

64. anonymous

Yep Yep, I did XD .

65. anonymous

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

66. tkhunny

Perfect. The factoring just makes it simpler to write.

67. anonymous

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

68. tkhunny

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

69. anonymous

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

70. anonymous

OOO!!!! I see it! Nvm.

71. anonymous

Thanks so much!!!

72. 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. :-)

73. anonymous

Yep, you just substituted for that right?

74. 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.

75. anonymous

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