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Muller's method
Muller's method is a root-finding algorithm, a numerical method for solving equations of the form f(x) = 0. It was first presented by David E. Muller in 1956.
Muller's method proceeds according to a third-order recurrence relation similar to the second-order recurrence relation of the secant method. Whereas the secant method proceeds by constructing a line through two points on the graph of f corresponding to the last two iterative approximations and then uses the line's root as the next approximation at every iteration, by contrast, Muller's method uses three points corresponding to the last three iterative approximations, constructs a parabola through these three points, and then uses a root of the parabola as the next approximation at every iteration.
Muller's method uses three initial approximations of the root, and , and determines the next approximation by considering the intersection of the x-axis with the parabola through , and .
Consider the quadratic polynomial
that passes through , and . Define the differences
and
Substituting each of the three points , and into equation (1) and solving simultaneously for and gives
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Muller's method
Muller's method is a root-finding algorithm, a numerical method for solving equations of the form f(x) = 0. It was first presented by David E. Muller in 1956.
Muller's method proceeds according to a third-order recurrence relation similar to the second-order recurrence relation of the secant method. Whereas the secant method proceeds by constructing a line through two points on the graph of f corresponding to the last two iterative approximations and then uses the line's root as the next approximation at every iteration, by contrast, Muller's method uses three points corresponding to the last three iterative approximations, constructs a parabola through these three points, and then uses a root of the parabola as the next approximation at every iteration.
Muller's method uses three initial approximations of the root, and , and determines the next approximation by considering the intersection of the x-axis with the parabola through , and .
Consider the quadratic polynomial
that passes through , and . Define the differences
and
Substituting each of the three points , and into equation (1) and solving simultaneously for and gives