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indexing in the formula for phi_j #25

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indexing in the formula for phi_j #25

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indexing in the formula for phi_j
ShawnL00 Mar 28, 2026
7080930
Use k=1..j indexing for phi_j in Newton interpolation
ShawnL00 Mar 28, 2026
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2 changes: 1 addition & 1 deletion notes/notes.org
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Expand Up @@ -4990,7 +4990,7 @@ Write down the Lagrange interpolant for nodes $(x_i)_{i=1}^m$ and values $(y_i)_
Find a basis so that \(V\) is triangular.
#+LATEX: \begin{hidden}
Easier to build than Lagrange, but: coefficient finding costs \(O (n^2)\).
\[\varphi _j (x) = \prod _{k = 1}^{j - 1} (x - x_k) . \]
\[\varphi _j (x) = \prod _{k = 1}^{j} (x - x_k) . \]
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@inducer inducer Mar 28, 2026

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Technically, that depends on the numbering. I would agree with your change if the numbering started at zero:

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\[\varphi _j (x) = \prod _{k = 1}^{j} (x - x_k) . \]
\[\varphi _j (x) = \prod _{k = 1}^{j} (x - x_k) . \qquad (j\in\{0,\dots,n-1\})\]

but then that's inconsistent with Lagrange.

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@inducer inducer Mar 28, 2026

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My counterproposal to this is in #26. Could you take a look?

(At least) two possibilities for coefficient finding:

- Set up \(V\), run forward substitution.
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