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If we know $f(0),f(1),f(2),\cdots f(n)$, is there a specialized version of the Lagrangian interpolation formula and a shortcut to compute the coefficients ?

(Stability is not a concern.)

  • See section 6.3.1 of E. Isaacson & H. B. Keller, Analysis of Numerical Methods (Wiley 1966, reprinted by Dover 1994), particularly formula (11) on page 265, which agrees with the accepted answer below. – Calum Gilhooley Jul 01 '19 at 18:38
  • @CalumGilhooley: thanks for this reference, also interesting in other respects. –  Jul 02 '19 at 06:38

1 Answers1

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For the $n+1$ points $\{\left(x_0,f(x_0)\right),...,\left(x_n,f(x_n)\right)\}$ the Lagrange interpolating polynomial is defined as $$P(x)=\sum_{j=0}^nP_j(x)=\sum_{j=0}^n f(x_j)\prod_{k=0\\{k\ne j}}^{n}\frac{x-x_k}{x_j-x_k}$$ Now since the $x_j=j,\; (j=0,1,...,n)$ the product can be simplified.

Let $p_n(x):=x(x-1)\cdots(x-n)$ then $$\begin{align}\prod_{k=0\\{k\ne j}}^{n}\frac{x-x_k}{x_j-x_k}&=\prod_{k=0\\{k\ne j}}^{n}\frac{x-k}{j-k} \\&=\frac{p_n(x)}{x-j}\prod_{k=0\\{k\ne j}}^{n}\frac 1{j-k}\\ &=\frac{p_n(x)}{x-j}\cdot\frac 1{j!(-1)^{n-j}(n-j)!}\\ &=\frac{p_n(x)}{n!}{{n}\choose{j}}\frac{(-1)^{n-j}}{x-j} \end{align}$$ Therefore $$P(x)=\frac{p_n(x)}{n!}\sum_{j=0}^n(-1)^{n-j}{{n}\choose{j}}\frac{f(j)}{x-j}$$

polfosol
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    The quantity you denote $p_n(x)$, a "falling factorial", is often denoted $(x)_{n + 1}$. Alternatively, you can regard it as a "rising factor" and denote it using the Pochhammer symbol $(x - n)^{(n + 1)}$. – Travis Willse Jun 25 '19 at 16:03
  • (See https://en.wikipedia.org/wiki/Falling_and_rising_factorials for more.) – Travis Willse Jun 25 '19 at 16:04
  • So the "weights" of the $P_j$ polynomials are essentially the binomial coefficients. –  Jun 25 '19 at 16:21