{"id":4660,"date":"2021-08-03T10:12:20","date_gmt":"2021-08-03T04:42:20","guid":{"rendered":"https:\/\/www.goseeko.com\/blog\/?p=4660"},"modified":"2025-06-02T20:58:29","modified_gmt":"2025-06-02T15:28:29","slug":"what-is-regula-falsi-method","status":"publish","type":"post","link":"https:\/\/www.goseeko.com\/blog\/what-is-regula-falsi-method\/","title":{"rendered":"What is regula-falsi method?"},"content":{"rendered":"\n

Overview<\/h2>\n\n\n\n

The regula-falsi method<\/a> is the oldest method of finding the approximate numerical value of a real root of an equation f(x) = 0.<\/p>\n\n\n\n

This method is also known as method of false position<\/a>. The method used to estimate the roots of a polynomial f(x).<\/p>\n\n\n\n

In this method we suppose that x1<\/sub> and x2<\/sub> are two points where f(x1<\/sub>) and f(x2<\/sub>) are of opposite sign .Let f(x1<\/sub>) < 0 and f(x2<\/sub>) > 0. <\/p>\n\n\n\n

Hence the root of the equation f(x) = 0 lies between x1<\/sub> and x1<\/sub> and so, f(x1<\/sub>)f(x2<\/sub>) < 0.<\/p>\n\n\n\n

Formula of regula-falsi method<\/h2>\n\n\n\n

The Regula Falsi formula  <\/p>\n\n\n

\n
\"\"<\/figure><\/div>\n\n\n

Find f(x3<\/sub>) is positive or negative. If f(x3<\/sub>) > 0 then root lies between x3<\/sub> and x1<\/sub> or if f(x3<\/sub>) < 0 then root lies between x3<\/sub> and x1<\/sub>  similarly we calculate x4<\/sub>.<\/p>\n\n\n\n

Proceed in this manner until the desired accurate root is found.<\/p>\n\n\n\n

Solved example<\/h2>\n\n\n\n

Example:<\/u> Apply Regula Falsi Method to solve the equation<\/strong>– 3x – cosx – 1 = 0.<\/p>\n\n\n\n

Solution:
let f(x) = 3x – cosx – 1 = 0<\/p>\n\n\n\n

by hit and trial<\/p>\n\n\n\n

f(0.60) = 0.010351 > 0<\/p>\n\n\n\n

So the root of the equation lies between x1<\/sub>=0.60 and x2<\/sub> = 0.61, also<\/p>\n\n\n\n

f(0.60) f(0.61) < 0<\/p>\n\n\n\n

By Regula  Falsi Mehtod, the first approximation,<\/p>\n\n\n

\n
\"\"<\/figure><\/div>\n\n\n

now<\/p>\n\n\n

\n
\"\"<\/figure><\/div>\n\n\n

So, root of the equation f(x) = 0  lies between 0.60709 and 0.61 and also f(0.60709) f(0.61) < 0.<\/p>\n\n\n\n

Second approximation<\/p>\n\n\n

\n
\"\"<\/figure><\/div>\n\n\n

now<\/p>\n\n\n

\n
\"\"<\/figure><\/div>\n\n\n

So, root of the equation f(x) = 0 lies between 0.60710 and 0.61 and also f(0.60710) f(0.61) < 0.<\/p>\n\n\n\n

The third approximation,<\/p>\n\n\n

\n
\"\"<\/figure><\/div>\n\n\n

Hence the root of the given equation correct to five decimal place is 0.60710.<\/p>\n\n\n\n

Interested in learning about similar topics? Here are a few hand-picked blogs for you!<\/p>\n\n\n\n