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--- gibson:teaching:fall-2016:math753:lagrangepoly [2016/11/11 12:29]
gibson
+++ gibson:teaching:fall-2016:math753:lagrangepoly [2016/11/11 12:34]
gibson
@@ Line 6: / Line 6: @@
 \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_1)}
+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*}

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