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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 :confused: :confused:](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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 $$c_{i0}=y_{i}-\frac{h^{2}}{6}s''_{i}-\frac{y_{i}-y_{i-1}}{h}x_{i}+\frac{h}{6}(s''_{i}-s''_{i-1})x_{i}$$
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
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 $$c_{i0}=y_{i}-\frac{h^{2}}{6}s''_{i}-\frac{y_{i}-y_{i-1}}{h}x_{i}+\frac{h}{6}(s''_{i}-s''_{i-1})x_{i}$$