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

[theMR]

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=ruutsplain+pedas&source=bl&ots=8h47blCY7y&sig=ce8u4FB5U18AkvFDEyke4ChKnOw&hl=et&sa=X&ved=0ahUKEwiYjuL80dLZAhWFDCwKHf9JCPkQ6AEIJjAA#v=onepage&q=ruutsplain%20pedas&f=false and I need to derive the formula (10) from page 50. Can anyone show me how it's done?

I know that

. . . . .\(\displaystyle S'(x_i)\,=\,m_i,\quad i\,=\,0,\,1,\,...,\,n\,-\, 1\)

and

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

so integrate from \(\displaystyle x_i\) to \(\displaystyle x\)

. . . . .\(\displaystyle S(x)\,-\,S(x_i)\, =\, m_i(x-x_i)\, +\, \dfrac{m_{i+1}\, -\, m_i}{h} \, \dfrac{(x\, -\,x_i)(x\, -\, x_i)}{2}\)

 

[theMR]

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

 

[tkhunny]

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

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

Substitute your other expression for \(\displaystyle S'(t)\) and see where it leads.

 

[theMR]

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

 

[tkhunny]

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.

 

[theMR]

First of all, for me \(\displaystyle \eta=\frac{1}{2}\) and \(\displaystyle h_i=h\) for every \(\displaystyle i\), so

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

I tried like this

. . . . .\(\displaystyle S'(t)=m_i -\dfrac{m_{i+1}-m_i}{h}(t-t_i)\)

now if I integrate it from \(\displaystyle x\) to \(\displaystyle x_i\), I get

. . . . .\(\displaystyle S(x)=S(x_i)+m_i(x-x_i)+\dfrac{m_{i+1}-m_i}{2h}(x-x_i)(x-x_i)\)

now if \(\displaystyle x=x_{i+1}\)

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

and from that

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

. . . . .\(\displaystyle S(x)=y_{i+1}-hm_i-\dfrac{m_{i+1}-m_i}{2}h+m_i(x-x_i)+\dfrac{m_{i+1}-m_i}{2h}(x-x_i)(x-x_i)\)