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inducer
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Mar 28, 2026
| #+LATEX: \begin{hidden} | ||
| Easier to build than Lagrange, but: coefficient finding costs \(O (n^2)\). | ||
| \[\varphi _j (x) = \prod _{k = 1}^{j - 1} (x - x_k) . \] | ||
| \[\varphi _j (x) = \prod _{k = 1}^{j} (x - x_k) . \] |
Owner
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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.
Owner
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My counterproposal to this is in #26. Could you take a look?
Owner
|
Closing in favor of #26. |
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