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Regula–Falsi Method  is a numerical method for estimating the  roots of a polynomial   f(x) . EXAMPLE:   Consider  f(x) = x 3  + 3x – 5 , where  [ a = 1, b = 2 ]  and  DOA = 0.001 . i a x b f(a) f(x) f(b) 1 1 1.1 2 – 1 – 0.369 9 2 1.1 1.13544668587896 2 – 0.369 – 0.129797592130931 9 3 1.13544668587896 1.14773797024856 2 – 0.129797592130931 – 0.0448680509813286 9 4 1.14773797024856 1.15196570867269 2 – 0.0448680509813286 – 0.0154155863909917 9 5 1.15196570867269 1.15341577448 2 – 0.0154155863909917 – 0.0052852985292482 9 6 1.15341577448 1.15391264384212 2 – 0.0052852985292482 – 0.00181077883487646 9 7 1.15391264384212 1.15408284038531 2 – 0.00181077883487646 – 0.000620231485743084 9 Source

  Find the root between (2,3) of x3+ - 2x - 5 = 0, by using regular falsi method. The Regula falsi method is an oldest method for computing the real roots of an algebraic equation. Examples: Find the root between (2,3) of x 3 + - 2x - 5 = 0, by using regular falsi method. Given f(x) = x 3  - 2 x - 5 f(2) = 2 3  - 2 (2) - 5 = -1 (negative) f(3) = 3 3  - 2 (3) - 5 = 16 (positive) Let us take a= 2 and b= 3. The first approximation to root is x 1  and is given by x 1  = (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) = 2.058 3  - 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 x 2  = (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) = 2.081 2  - 2 x 2.081 - ...

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 pr...

  Explain BISECTION METHOD (BOLZANO METHOD).  Engineering maths The method is applicable for numerically solving the equation  f ( x ) = 0 for the  real  variable  x , where  f  is a  continuous function  defined on an interval [ a ,  b ] and where  f ( a ) and  f ( b ) have opposite signs. In this case  a  and  b  are said to bracket a root since, by the  intermediate value theorem , the continuous function  f  must have at least one root in the interval ( a ,  b ). At each step the method divides the interval in two by computing the midpoint  c  = ( a + b ) / 2 of the interval and the value of the function  f ( c ) at that point. Unless  c  is itself a root (which is very unlikely, but possible) there are now only two possibilities: either  f ( a ) and  f ( c ) have opposite signs and bracket a root, or  f ( c ) and  f ( b...