regula falsi, method / false position method / explain

regula falsi, method of false position, or false position method

 In mathematics, the regula falsi, method of false position, or false position method is a very old method for solving an equation with one unknown; this method, in modified form, is still in use. In simple terms, the method is the trial and error technique of using test ("false") values for the variable and then adjusting the test value according to the outcome. This is sometimes also referred to as "guess and check". Versions of the method predate the advent of algebra and the use of equations.

As an example, consider problem 26 in the Rhind papyrus, which asks for a solution of (written in modern notation) the equation x + x/4 = 15. This is solved by false position.^[1] First, guess that x = 4 to obtain, on the left, 4 + 4/4 = 5. This guess is a good choice since it produces an integer value. However, 4 is not the solution of the original equation, as it gives a value which is three times too small. To compensate, multiply x (currently set to 4) by 3 and substitute again to get 12 + 12/4 = 15, verifying that the solution is x = 12.

Modern versions of the technique employ systematic ways of choosing new test values and are concerned with the questions of whether or not an approximation to a solution can be obtained, and if it can, how fast can the approximation be found.

 

if a and b are known. The method begins by using a test input value x′, and finding the corresponding output value b′ by multiplication: ax′ = b′. The correct answer is then found by proportional adjustment, x = b/b′ x′.

Double false position is aimed at solving more difficult problems that can be written algebraically in the form: determine x such that

${\displaystyle f(x)=ax+c=0,}$

if it is known that

${\displaystyle f(x_{1})=b_{1},\qquad f(x_{2})=b_{2}.}$

Double false position is mathematically equivalent to linear interpolation. By using a pair of test inputs and the corresponding pair of outputs, the result of this algorithm given by,^[2]

${\displaystyle x={\frac {b_{1}x_{2}-b_{2}x_{1}}{b_{1}-b_{2}}},}$

would be memorized and carried out by rote. Indeed, the rule as given by Robert Recorde in his Ground of Artes (c. 1542) is:^[2]

The Regula falsi method is an oldest method for computing the real roots of an algebraic equation. This below worksheet help you to understand how to compute the roots of an algebraic equation using Regula falsi method. The practice problems along with this worksheet improve your problem solving capabilities when you try on your own

Examples:

Find the root between (2,3) of x3+ - 2x - 5 = 0, by using regular falsi method.

Given

f(x) = x3 - 2 x - 5

f(2) = 23 - 2 (2) - 5 = -1 (negative)

f(3) = 33 - 2 (3) - 5 = 16 (positive)

Let us take a= 2 and b= 3.

The first approximation to root is x1 and is given by

x1 = (a f(a) - b f(b))/(f(b)-f(a))

=(2 f(3)- 3 f(2))/(f(3) - f(2))

=(2 x 16 - 3 (-1))/ (16- (-1))

= (32 + 3)/(16+1) =35/17

= 2.058

Now f(2.058) = 2.0583 - 2 x 2.058 - 5

= 8.716 - 4.116 - 5

= - 0.4

The root lies between 2.058 and 3

Taking a = 2.058 and b = 3. we have the second approximation to the root given by

x2 = (a f(a) - b f(b))/(f(b)-f(a))

= (2.058 x f(3) - 3 x f(2.058)) /(f(3) - f(2.058))

= (2.058 x 16 -3 x -0.4) / (16 - (-0.4))

= 2.081

Now f(2.081) = 2.0812 - 2 x 2.081 - 5

= -0.15

The root lies between 2.081 and 3

Take a = 2.081 and b = 3

The third approximation to the root is given by

x3 = (a f(a) - b f(b))/(f(b)-f(a))

= (2.089 X 16 - 3 x (-0.062))/ (16 - (-0.062))

= 2.093

The root is 2.09

For an affine linear function,

${\displaystyle f(x)=ax+c,}$

double false position provides the exact solution, while for a nonlinear function f it provides an approximation that can be successively improved by iteration.