Regula Falsi

The regular falsi, or method of false position, is a method to solve an equation in one unknown, from an initial “guess”. Guesses are progressively refined, by a rather simple method as we will see, until the exact answer is reached or until convergence is reached. The method is useful when you don’t really know how to divide by arbitrary values (as it was the case in Ancient Times) or when the equation is cumbersome.

Let’s start with a simple example:

8a=13, find a.

Armed with today’s schoolboy algebra, we readily find \displaystyle a=\frac{13}{8}. But what if we don’t have algebra?

Well, we could start with a guess, say, a=2. Clearly, it’s too large because now 8a=8\times{2}=16. But that overshoot tells us something important: our guess is \displaystyle \frac{16}{13} times too large. Our second guess will be:

\displaystyle a'=\frac{a}{\frac{16}{13}}=a\frac{13}{16}=\frac{13}{8},

which is now the correct answer, since, clearly, \displaystyle 8\times\frac{13}{8}=13.

* *

According to Michel [1], the Ancient Egyptians sometimes used the method of false position to solve simple linear equations in one unknown. This version works best when there’s a linear relationship between the unknown and the value given by the equation. Above, in 8a=13, we arrived at the correct solution in only one iteration, and if we guess correctly, it must be so. If we guess incorrectly, say, we’d guess an adjustment by \frac{16}{12}=\frac{4}{3} rather than \frac{16}{13}, we would have had a couple more iterations.

However, the method still works when the relationship isn’t linear, but, just as an incorrect guess lengthens the process, the non-linearity introduces errors (as the guesses are linear), and the procedure will need many more iterations to converge.

Let’s take for example x^2+x=4, to solve for x. The first thing is to choose a good initial guess for x. Here, using commonsense, we see that x_0=2 can’t be that far off, because the dominating term, x^2, will be x_0^2=2^2=4. So let the first guess be x_0=2. We find:


giving us the second guess \displaystyle x_1=x_0\frac{4}{6}=\frac{4}{3}. Now:

\displaystyle x_1^2+x_1=\frac{28}{9},

and \displaystyle x_2=x_1\frac{4}{\frac{28}{9}}=\frac{12}{7}, thus

\displaystyle x_2^2+x_2=\frac{228}{29},

giving \displaystyle x_3=x_2\frac{4}{\frac{228}{29}}, so that

\displaystyle x_3^2+x_3=\frac{1316}{361},

and so on… If we’re quite patient, we find:

\displaystyle x_5=\frac{188}{123}=1.52846\ldots,

\displaystyle x_{10}=\frac{21036}{13447}=1.56436\ldots,

\displaystyle x_{20}=\frac{255572812}{163663727}=1.56157\ldots,

while the exact answer is \displaystyle x=\frac{1}{2}\left(\pm\sqrt{17}-1\right), which is either -2.56155\ldots or 1.56155\ldots.

* *

The rule to adjust the guess is fundamentally linear, because the scaling is a simple ratio of the desired result to the guess. Of course, to ensure faster convergence, we’d have to use at least the first order derivative, but that would bring us into methods that are quite like the Newton-Raphson methods for finding roots (i.e., values such that f(x)=0, for a given function f(x)). These methods are however still sometimes called regula falsi!

* *

This simple method isn’t quite sufficient as a general numerical method to solve equations in one unknown. Some equations are especially problematic as they will crash the (simple) method. Take for example x^2=2, with x_0=\frac{3}{2}, which seems like a good starting point. In this case, the method described above cycles through two values: x_t=\frac{3}{2} and x_{t+1}=\frac{4}{3}, and, therefore, never comes close to the correct answer.

[1] Marianne Michel — Les mathématiques de l’Égypte ancienne. Numération, métrologie, arithmétique, géométrie et autres problèmes &msdash; Éditions Safran (2014).

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