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swissgirl

  • 2 years ago

Cubic Splines

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  1. swissgirl
    • 2 years ago
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    @DanielxAK Can you help me through this problem. It seems quite simple

  2. swissgirl
    • 2 years ago
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    \(S_0(x)=1+2x-x^3 \to 0 \leq x \leq 1\) \(S_1(x)=2+b(x-1)+c(x-1)^2+d(x-3)^3 \to 1 \leq x \leq 2\)

  3. DanielxAK
    • 2 years ago
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    Sorry for the late reply. I was at class and then attended a seminar. Anyway: You know your three points. 0, 1 and 2. You have been given some of your coefficients ( specifically five of them) and need to find the other 3. So, you need 3 equations. Can you tell me which 3 equations you need to use? You're almost given one of them to start.

  4. swissgirl
    • 2 years ago
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    Yes I figured it out \(S_0'(1)=S_1'(1)\) \(S_0''(1)=S_1''(1)\) \(S_1''(2)=0\) THHAANKKSSSS :)

  5. DanielxAK
    • 2 years ago
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    Um. Be careful there. S1''(2) = 0 implies that you have a natural spline. You don't know that. So, that doesn't work. But, you're almost there. Looking at the first two equations you gave me (which are correct), can you find a very similar equation that follows the same pattern as those two?

  6. swissgirl
    • 2 years ago
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    Oh i forgot to mention that it is a natural spline lol

  7. DanielxAK
    • 2 years ago
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    Ah, okay. Well, in case you didn't know, you should still be able to solve it by using the fact that: S0(1) = S1(1), as they have the same y value.

  8. swissgirl
    • 2 years ago
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    Gotcha Can i dirve you crazy with one more question?

  9. DanielxAK
    • 2 years ago
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    Go for it. I'm trying to avoid my own homework as it is.

  10. swissgirl
    • 2 years ago
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    hahaha I do the same. I sit on here answering questions and im like thinking y the hell am I on here when I gotta do my own

  11. swissgirl
    • 2 years ago
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    Construct a natural cubic spline to approximate f(x)=Cos(pi*x) by using the values given by f(x) at x=0, .25,.5, .75 and1 Integrate the spline over [0,1]

  12. swissgirl
    • 2 years ago
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    I think I am allowed to use a computer program but \i am wondering How can this be done by hand

  13. swissgirl
    • 2 years ago
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    Or is it wayyy too time consuming?

  14. DanielxAK
    • 2 years ago
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    Well, you have 5 points. So, you'll end up having 16 coefficients to solve. You're better off solving the equations using a computer program. So, your points will be the x defined above and your five y values will be f(x) for each x. Then, just go back to making cubics on those intervals with unknown coefficients and do all the steps for defining all 16 equations.

  15. swissgirl
    • 2 years ago
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    A(I) B(I) C(I) D(I) 1.00000000 -0.7573593129 0. -6.627416998 0.7071067812 -2. -4.970562748 6.627416998 6.123233996*10^-17 -3.242640687 4.440892099*10^-16 6.627416998 -0.7071067812 -2. 4.970562748 -6.627416998

  16. swissgirl
    • 2 years ago
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    ohhhh it didnt work out uughhhhh

  17. swissgirl
    • 2 years ago
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    whtvrrrr ok once i get the 16 diff values then what?

  18. swissgirl
    • 2 years ago
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    It seems like I am suppossed to integrate but not sure y we are integrating

  19. DanielxAK
    • 2 years ago
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    Don't worry about integrating right now. You want to find your coefficients first so you can find all 4 cubics. Then, after you know your cubics, you integrate them on their intervals. But, not right now. Worry about finding your coefficients using your equations.

  20. swissgirl
    • 2 years ago
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    ok I will write out all the equations then

  21. swissgirl
    • 2 years ago
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    \(S_0(x)=1-.7573593129x-6.627416998x^3\) \(S_1(x)=.707106812-2(x-.25)-4.970562748(x-.25)^2+6.627416998(x-.25)^3\) \(S_2(x)=-3.242640687(x-.5)+6.627416998(x-.5)^3\) \(S_3(x)=-.7071067812-2(x-.75)+4.970562748(x-.75)^2-6.627416998(x-.75)^3\)

  22. swissgirl
    • 2 years ago
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    So there are my four equations

  23. DanielxAK
    • 2 years ago
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    Have you tried plotting them? If those are correct, they should match f(x) = cos(pi*x) pretty well.

  24. swissgirl
    • 2 years ago
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    hmm let me try

  25. swissgirl
    • 2 years ago
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    ummm pretty much sooo

  26. swissgirl
    • 2 years ago
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    Its weird that one of the equations has nothing to do with the graph at all

  27. DanielxAK
    • 2 years ago
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    Yeah, only looks like one or two of them are correct. You may want to double check your equations.

  28. swissgirl
    • 2 years ago
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    ohhh I found my mistake mixed up signs Its only suppossed to match from [0,1] and it does

  29. swissgirl
    • 2 years ago
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    Thats cooollllllllllllllllllll

  30. swissgirl
    • 2 years ago
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    its annoying to work with sooooo mannnyy decimals

  31. swissgirl
    • 2 years ago
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    So now i integrate?

  32. DanielxAK
    • 2 years ago
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    Yes. Now you can integrate using those cubics (so, from 0.25 use S1, 0.25 to 0.5 use S2, etc.).

  33. swissgirl
    • 2 years ago
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    I am just wondering y we are integrating lol I am just not seeing the logic behind that?

  34. DanielxAK
    • 2 years ago
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    The idea is you're able to approximate the integral of any function in the real numbers by building a cubic spline and integrating the pieces of the spline.

  35. swissgirl
    • 2 years ago
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    ohhhhh I get it Ya that makes sense Like I am taking this course online and its def getting annoying wish I had a teacher to explain everything clearly THANKS YOU ARE GREATTTTTTTT

  36. DanielxAK
    • 2 years ago
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    You're welcome.

  37. swissgirl
    • 2 years ago
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    @DanielxAK hahah I have one more question It asks me to find f'(.5) How would I go abt it?

  38. swissgirl
    • 2 years ago
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    Do I add all the equations together and that will be my spline function?

  39. swissgirl
    • 2 years ago
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    Heyyyy :)

  40. DanielxAK
    • 2 years ago
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    It sounds like it's just asking you to find the derivative of f(x) = cos(pi*x) at 0.5.

  41. swissgirl
    • 2 years ago
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    No it asks me to compare both so it cant mean that

  42. DanielxAK
    • 2 years ago
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    Compare both? Ah okay. So, yes. Find the derivative at 0.5 of the original function. Then, find the derivative of 0.5 using the cubic defined at 0.5 (there should be two of them, since it was one of your points you used when building your piecewise function). The values of both should be pretty close.

  43. swissgirl
    • 2 years ago
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    Ohhh what i did for a previous question gottcchhhaaaaaaa

  44. swissgirl
    • 2 years ago
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    I am seriously driving u crazy here. Thanks I really appreciate your help

  45. DanielxAK
    • 2 years ago
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    Ha. That's okay.

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