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gibson:teaching:fall-2016:math753:lagrangepoly

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

gibson/teaching/fall-2016/math753/lagrangepoly.1478896490.txt.gz · Last modified: 2016/11/11 12:34 by gibson