MHB Interpolation with cubic splines

AI Thread Summary
The discussion centers on the proof of a theorem regarding cubic splines, specifically the existence and uniqueness of a function that interpolates a given function while maintaining continuity in its first and second derivatives. The user is attempting to derive constants for the spline function using conditions from the linear Lagrange polynomial but encounters difficulties in matching the textbook solution. Clarifications are provided regarding the correct form of the constants and the relationship between the coefficients in the spline equation. The conversation emphasizes the importance of correctly bundling terms to derive the constants accurately. The user is guided toward recognizing the proper structure of the spline function to resolve their confusion.
evinda
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Hello! :) I am looking at the proof of this theorem:
Let $f \in C^{1}([a,b]),P:a=x_{0}<x_{1}<...<x_{n}=b $ uniform partition of $[a,b]$.Then there is exactly one function $s \in S_{3}(P)$ so that $s(x_{i})=f(x_{i}),i=0,...,n$ and $s,s',s''$ continuous at $x_{i}$.Also,$s'(x_{0})=f'(x_{0}),s'(x_{n})=f'(x_{n})$.

but at some point I got stuck :confused:

Let $s'_{i}=s'(x_{i}),s''_{i}=s''(x_{i}),s_{i}=s(x_{i}),y_{i}=f(x_{i})$

From the linear Lagrange polynomial,we find that:
$$s''(x)=\frac{1}{h}[s''_{i}(x-x_{i-1})-s''_{i-1}(x-x_{i})] ,x \in [x_{i-1},x_{i}]$$

Then,integrating two times,we get:
$$s(x)=\frac{1}{6h}[s''_{i}(x-x_{i-1})^{3}-s''_{i-1}(x-x_{i})^{3}]+c_{i1}x+c_{i0}$$
Then using the conditions $s_{i-1}=y_{i-1},s_{i}=y_{i}$, we want to find the constants $c_{i1},c_{i0}$ .In my textbook,there is only the solution,which is $s(x)=\frac{1}{6h}[s''_{i}(x-x_{i-1})^{3}-s''_{i-1}(x-x_{i})^{3}]+(\frac{y_{i}}{h}-s''_{i}\frac{h}{6})(x-x_{i-1})-(\frac{y_{i-1}}{h}-s''_{i-1}\frac{h}{6})(x-x_{i})$

I tried to find the constants..For $c_{i1}$ I found $\frac{y_{i}-y_{i-1}}{h}-\frac{h}{6}(s''_{i}-s''_{i-1})$ ,but replacing the conditions and $c_{i1}$, I don't get the right result! Have I calculated wrong the value of $c_{i1}$ ?
From the condition $s_{i-1}=y_{i-1} \text{ and } c_{i1}$ I get $$c_{i0}=y_{i-1}-\frac{h^{2}}{6}s''_{i-1}-\frac{y_{i}-y_{i-1}}{h}x_{i-1}+\frac{h}{6}(s''_{i}-s''_{i-1})x_{i-1}$$
and from the condition $s_{i}=y_{i} \text{ and } c_{i1}$ I get $$c_{i0}=y_{i}-\frac{h^{2}}{6}s''_{i}-\frac{y_{i}-y_{i-1}}{h}x_{i}+\frac{h}{6}(s''_{i}-s''_{i-1})x_{i}$$
 
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Hey evinda! :D

evinda said:
Hello! :) I am looking at the proof of this theorem:
Let $f \in C^{1}([a,b]),P:a=x_{0}<x_{1}<...<x_{n}=b $ uniform partition of $[a,b]$.Then there is exactly one function $s \in S_{3}(P)$ so that $s(x_{i})=f(x_{i}),i=0,...,n$ and $s,s',s''$ continuous at $x_{i}$.Also,$s'(x_{0})=f'(x_{0}),s'(x_{n})=f'(x_{n})$.

but at some point I got stuck :confused:

Let $s'_{i}=s'(x_{i}),s''_{i}=s''(x_{i}),s_{i}=s(x_{i}),y_{i}=f(x_{i})$

From the linear Lagrange polynomial,we find that:
$$s''(x)=\frac{1}{h}[s''_{i}(x-x_{i-1})-s''_{i-1}(x-x_{i})] ,x \in [x_{i-1},x_{i}]$$

Then,integrating two times,we get:
$$s(x)=\frac{1}{6h}[s''_{i}(x-x_{i-1})^{3}-s''_{i-1}(x-x_{i})^{3}]+c_{i1}x+c_{i0}$$
Then using the conditions $s_{i-1}=y_{i-1},s_{i}=y_{i}$, we want to find the constants $c_{i1},c_{i0}$ .In my textbook,there is only the solution,which is $s(x)=\frac{1}{6h}[s''_{i}(x-x_{i-1})^{3}-s''_{i-1}(x-x_{i})^{3}]+(\frac{y_{i}}{h}-s''_{i}\frac{h}{6})(x-x_{i-1})-(\frac{y_{i-1}}{h}-s''_{i-1}\frac{h}{6})(x-x_{i})$

I tried to find the constants..For $c_{i1}$ I found $\frac{y_{i}-y_{i-1}}{h}-\frac{h}{6}(s''_{i}-s''_{i-1})$ ,but replacing the conditions and $c_{i1}$, I don't get the right result! Have I calculated wrong the value of $c_{i1}$ ?

Nope. Your value of $c_{i1}$ is all good. (Nod)
Note that the given solution has factors $(x-x_i)$ and $(x-x_{i-1})$.
If you bundle the factors of $x$ from the solution, you get your $c_{i1}$.
From the condition $s_{i-1}=y_{i-1} \text{ and } c_{i1}$ I get $$c_{i0}=y_{i-1}-\frac{h^{2}}{6}s''_{i-1}-\frac{y_{i}-y_{i-1}}{h}x_{i-1}+\frac{h}{6}(s''_{i}-s''_{i-1})x_{i-1}$$
and from the condition $s_{i}=y_{i} \text{ and } c_{i1}$ I get $$c_{i0}=y_{i}-\frac{h^{2}}{6}s''_{i}-\frac{y_{i}-y_{i-1}}{h}x_{i}+\frac{h}{6}(s''_{i}-s''_{i-1})x_{i}$$

Well... I guess the solution should be of the form:
$$... + C(x-x_i) + D(x-x_{i-1}) + E$$
instead of
$$... + c_{i1}x+c_{i0}$$
I haven't checked, but presumably your $c_{i0}$ will be equal to $-Cx_i - Dx_{i-1}+E$.
 
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