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Lagrange Interpolating Polynomial

The Lagrange Interpolating Polynomial is a simple-to-understand but mathematically clunky way to construct an $N-1$-order polynomial interpolant to a set of $N$ data points $(x_1, y_1), (x_2, y_2), \ldots, (x_N, y_N)$.

The best way to see it is by example for a quadratic fit to three data points $(x_1, y_1), (x_2, y_2), (x_3, y_3)$. Let $P(x)$ be

P(x) = y_1 \frac{(x-x_2)(x-x_3)}{(x_1-x_2)(x_1-x_3)} + y_1 \frac{(x-x_1)(x-x_3)}{(x_2-x_1)(x_2-x_3)} + y_3 \frac{(x-x_1)(x-x_2)}{(x_3-x_1)(x_3-x_2)} 

It's easy to see the polynomial goes through each data point. If you plug in $x=x_1$, the second and third terms vanish, and the fraction in the first term is $1$, so that $P(x_1) = y_1$. Similar simplifications occur when plugging in $x=x_2$, to get $P(x_2) = y_2$, and similar for $P(x_3) = y_3$. Also, since everything on the right-hand-side except $x$ is a constant, it's clear that the $P(x)$ is a polynomial in $x$ of order 2. The generalization to higher-order polynomials is straightforward.

For example, let's fit a 2nd order polynomial to $(0,4), (1,3), (2,6)$. Plugging into the formula gives

P(x) = 4 \frac{(x-1)(x-2)}{(-1)(-2)} + 3 \frac{(x)(x-2)}{(1)(-1)} + 6 \frac{(x)(x-1)}{(2)(1)} 

This is mathematically clunky because it expresses a quadratic polynomial $P(x)$ as the sum of three quadratics, and we have a lot of tedious algebra to do to simplify. If we do that, we get

P(x) = 4 - 3x + 2x^2

which can be easily verified as passing through the given data points.

Further reading

gibson/teaching/fall-2016/math753/lagrangepoly.txt · Last modified: 2016/11/11 12:39 by gibson