Ridders' method: Difference between revisions

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==Method==

Given two values of the independent variable, ''x''₀ and ''x''₁, which are on two different sides of the root being sought, i.e.,<math>f(x_0)f(x_2) < 0</math>.The method begins by evaluating the function at the midpoint ''x'' <submath>1x_1 = (x_0 +x_2)/2 < 0</submath> between the two points. One then finds the unique exponential function of the form ''e''<supmath>''e^{ax''}</supmath> which,such whenthat multiplied byfunction ''<math>h(x)=f'',(x)e^{ax}</math> transformssatisfies the<math>h(x_1)=(h(x_0) function at+h(x_2))/2 the three points into a straight line</math>. The false position method is then applied to the transformed values, leading to a new value ''x''₃, between ''x''₀ and ''x''₂, which can be used as one of the two bracketing values in the next step of the iteration.

The other bracketing value is taken to be ''x''₃ if f(''x''₃) has the opposite sign to f(''x''₄), or otherwise whichever of ''x''₁ and ''x''₂ has f(x) of opposite sign to f(''x''₄).