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I am trying to derive Muller's method. I see that if I have three points, I can come up the parabola to interpolate these three points using the Lagrange formula.

After I get this very scary looking formula, I try to find the roots by simplifying some of the constant terms a little bit, but how do I know which roots to take?

  • Write out what you've done so far instead of just stating you've done it so that we can help you more specifically with your problem. Are you trying to solve for the root of this quadratic? If so, what's preventing you from doing such like any other quadratic? – Simply Beautiful Art Apr 27 '20 at 11:53
  • Next, solve $y(x)=0$. –  Apr 27 '20 at 12:03
  • @SimplyBeautifulArt Yes, I think the I am confused about how to proceed with the quadratic, since it is not so simple. And once I have it, how do I know which root to take? –  Apr 27 '20 at 12:06
  • I find the computation not that scary, did you check the wikipedia page (mathematics is politically still uncontroversial enough to remain usable). Also see my answers in https://math.stackexchange.com/q/3548653/115115 and https://math.stackexchange.com/q/1202918/115115 on the implementation of the method. – Lutz Lehmann Apr 27 '20 at 14:38

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WLOG, $x_0=0$ because you can shift all abscissas.

The interpolation can be written $$\begin{vmatrix}y&x^2&x&1\\y_2&x_2^2&x_2&1\\y_1&x_1^2&x_1&1\\y_0&0&0&1\end{vmatrix}=0,$$

You find the roots by setting $y=0$,

$$\begin{vmatrix}0&x^2&x&1\\y_2&x_2^2&x_2&1\\y_1&x_1^2&x_1&1\\y_0&0&0&1\end{vmatrix}=0.$$

This expands in a quadratic polynomial, the coefficients of which are three cofactors of the matrix. $$- \begin{vmatrix}y_2&x_2&1\\y_1&x_1&1\\y_0&0&1\end{vmatrix}x^2+ \begin{vmatrix}y_2&x_2^2&1\\y_1&x_1^2&1\\y_0&0&1\end{vmatrix}x- \begin{vmatrix}y_2&x_2^2&x_2\\y_1&x_1^2&x_1\\y_0&0&0\end{vmatrix}=0,$$

$$- \begin{vmatrix}y_2-y_0&x_2-x_0\\y_1-y_0&x_1-x_0\end{vmatrix}x^2+ \begin{vmatrix}y_2-y_0&x_2^2-x_0^2\\y_1-y_0&x_1^2-x_0^2\end{vmatrix}x- \begin{vmatrix}x_2^2&x_2\\x_1^2&x_1\end{vmatrix}y_0=0.$$