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The derivation of the cubic spline interpolation (from this MIT OCW lecture, page 13) starts off with second derivatives, which are piecewise linear functions. It's written in the form

$$ s''(x) = \frac{x_j - x}{h_j}\sigma_{j-1} + \frac{x - x_{j-1}}{h_j}\sigma_{j}, h = x_j - x_{j-1}, \sigma_{j} = s''(x_j) $$

which is confusing to me, because I think of it as

\begin{align*} s''(x) = f(x) &= \frac{f_{j} - f_{j-1}}{x_j - x_{j-1}}(x - x_j) \\ &= \frac{f_{j}}{h_j}(x - x_j) - \frac{f_{j-1}}{h_j}(x - x_j) \\ &= \frac{f_{j}}{h_j}(x - x_j) + \frac{f_{j-1}}{h_j}(x_j - x) = \frac{\sigma_{j}}{h_j}(x - x_j) + \frac{\sigma_{j-1}}{h_j}(x_j - x) \\ &= \frac{x - x_j}{h_j}\sigma_{j} + \frac{x_j - x}{h_j}\sigma_{j-1} \end{align*}

which is not quite the same, and I'm confused by how they obtained that form, mainly where $x_{j-1}$ came from.

After that, they take the integral of the function twice, obtaining a function with integration constants (?)

$$\alpha_j (x - x_{j-1}) + \beta_j(x_j - x)$$

and I was under the impression one integral would give integration constant $\alpha_j$, then a second would give

$$\alpha_j (x) + \beta_j$$

The pdf's explanation doesn't really do it much justice, and so I'm confused how such an integration is possible (why is $\beta_j$ a coefficient when it should be a constant).

Could someone clarify these steps for me? Thanks!

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$$ \frac{(x_{j-1}-x)f_{j-1}+(x-x_{j-1})f_{j}}{x_j-x_{j-1}} = \frac{f_j-f_{j-1}}{x_j-x_{j-1}}(x-x_{j-1}) $$

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(I change the names on purpose, with the hope it's simpler to understand the problem, as it's really related to linear interpolation, not splines)

You want a linear function on $[a,b]$ that equals $p$ at $x=a$ and $q$ at $x=b$.

Consider the function $g(x)=\dfrac{x-a}{b-a}q$. It's linear and $g(a)=0$, $g(b)=q$. With a carefully chosen linear combination of two similar functions, you get the linear function you want.

Consider now

$$f(x)=\dfrac{x-a}{b-a}q+\dfrac{b-x}{b-a}p$$

The function $f$ is linear, and $f(a)=p$, $f(b)=q$.


Now, the $f$ you considered is

$$f(x)=\frac{q-p}{b-a}(x-b)$$

This can't work, as $f(b)=0$, not $q$, and $f(a)=p-q$, not $p$. You forgot one term in the linear interpolation, the function should be:

$$f(x)=\frac{q-p}{b-a}(x-b)+q$$

Of course, it's equivalent to the linear interpolation above:

$$f(x)=\frac{q-p}{b-a}(x-b)+q=\frac{q(x-b)-p(x-b)+q(b-a)}{b-a}\\=\frac{q(x-a)-p(x-b)}{b-a}=\frac{x-a}{b-a}q+\frac{b-x}{b-a}p$$

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