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Thread: Quadratic spline interpolation: S'(x_i) = m_i + [(m_{i+1}-m_i)/h][x-x_i], x in [x_i,

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    Quadratic spline interpolation: S'(x_i) = m_i + [(m_{i+1}-m_i)/h][x-x_i], x in [x_i,

    Hi, I have been working on this 3 hours, and I just can't figure it out. I have this article https://books.google.ee/books?id=5fxu15HQRJ0C&pg=PA60&lpg=PA60&dq=ruutspla in+pedas&source=bl&ots=8h47blCY7y&sig=ce8u4FB5U18A kvFDEyke4ChKnOw&hl=et&sa=X&ved=0ahUKEwiYjuL80dLZAh WFDCwKHf9JCPkQ6AEIJjAA#v=onepage&q=ruutsplain%20pe das&f=false and I need to derive the formula (10) from page 50. Can anyone show me how it's done?

    I know that

    . . . . .[tex]S'(x_i)\,=\,m_i,\quad i\,=\,0,\,1,\,...,\,n\,-\, 1[/tex]

    and

    . . . . .[tex]S'(x)\,=\,m_i\,+\, \dfrac{m_{i+1}\, -\, m_i}{h}\, (x\, -\, x_i),\quad x\, \in\, [x_i,\, x_{i+1}][/tex]

    so integrate from [tex]x_i[/tex] to [tex]x[/tex]

    . . . . .[tex]S(x)\,-\,S(x_i)\, =\, m_i(x-x_i)\, +\, \dfrac{m_{i+1}\, -\, m_i}{h} \, \dfrac{(x\, -\,x_i)(x\, -\, x_i)}{2}[/tex]
    Last edited by stapel; 03-07-2018 at 05:01 PM. Reason: Tweaking LaTeX.

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    Still haven't figured it out, can someone help me or give me a hint?

  3. #3
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    Sorry, my Estonian isn't too good. My Finnish is passable.

    Write out the integral: [tex]\int\limits_{x_{i}}^{x}S'(t)\;dt[/tex]

    Substitute your other expression for [tex]S'(t)[/tex] and see where it leads.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

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    It's actually in English. Just a little section in Estonian. Well, I tried but it didn't get me anywhere.

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    Quote Originally Posted by theMR View Post
    It's actually in English. Just a little section in Estonian. Well, I tried but it didn't get me anywhere.
    Nonresponsive. Show what you tried, please.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

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    First of all, for me [tex]\eta=\frac{1}{2}[/tex] and [tex]h_i=h[/tex] for every [tex]i[/tex], so

    . . . . .[tex](P_ny)(t)=y_{i+1}+ \left[\dfrac{h}{8}-\dfrac{(t_{i+1}-t)^2}{2h}\right]m_i+\left[\dfrac{(t-t_i)^2}{2h}-\dfrac{h}{8}\right]m_{i+1}[/tex]

    I tried like this

    . . . . .[tex]S'(t)=m_i -\dfrac{m_{i+1}-m_i}{h}(t-t_i)[/tex]

    now if I integrate it from [tex]x[/tex] to [tex]x_i[/tex], I get

    . . . . .[tex]S(x)=S(x_i)+m_i(x-x_i)+\dfrac{m_{i+1}-m_i}{2h}(x-x_i)(x-x_i)[/tex]

    now if [tex]x=x_{i+1}[/tex]

    . . . . .[tex]S(x_{i+1})=S(x_i)+hm_i+\dfrac{(m_{i+1}-m_i)h}{2}[/tex]

    and from that

    . . . . .[tex]S(x_i)=y_{i+1}-hm_i-\dfrac{m_{i+1}-m_i}{2}h[/tex]

    . . . . .[tex]S(x)=y_{i+1}-hm_i-\dfrac{m_{i+1}-m_i}{2}h+m_i(x-x_i)+\dfrac{m_{i+1}-m_i}{2h}(x-x_i)(x-x_i)[/tex]
    Last edited by stapel; 03-07-2018 at 05:04 PM. Reason: Tweaking LaTeX.

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