False position method and bisection

False position method and bisection In numerical analysis, the false position method or regula falsi method is a root-finding algorithm that combines features from the bisection method and the secant method. The method: The first two iterations of the false position method. The red curve shows the function f and the blue lines are the secants. Like the bisection method, the false position method starts with two points a0 and b0 such that f(a0) and f(b0) are of opposite signs, which implies by the intermediate value theorem that the function f has a root in the interval [a0, b0], assuming continuity of the function f. The method proceeds by producing a sequence of shrinking intervals [ak, bk] that all contain a root of f. At iteration number k, the number is computed. As explained below, ck is the root of the secant line through (ak, f(ak)) and (bk, f(bk)). If f(ak) and f(ck) have the same sign, then we set ak+1 = ck and bk+1 = bk, otherwise we set ak+1 = ak and bk+1 = ck. This process is repeated until the root is approximated sufficiently well. The above formula is also used in the secant method, but the secant method always retains the last two computed points, while the false position method retains two points which certainly bracket a root. On the other hand, the only difference between the false position method and the bisection method is that the latter uses ck = (ak + bk) / 2. Bisection method In mathematics, the bisection method is a root-finding algorithm which repeatedly bisects an interval then selects a subinterval in which a root must lie for further processing. It is a very simple and robust method, but it is also relatively slow. The method is applicable when we wish to solve the equation for the scalar variable x, where f is a continuous function. The bisection method requires two initial points a and b such that f(a) and f(b) have opposite signs. This is called a bracket of a root, for by the intermediate value theorem the continuous function f must have at least one root in the interval (a, b). The method now divides the interval in two by computing the midpoint c = (a+b) / 2 of the interval. Unless c is itself a rootwhich is very unlikely, but possiblethere are now two possibilities: either f(a) and f(c) have opposite signs and bracket a root, or f(c) and f(b) have opposite signs and bracket a root. We select the subinterval that is a bracket, and apply the same bisection step to it. In this way the interval that might contain a zero of f is reduced in width by 50% at each step. We continue until we have a bracket sufficiently small for our purposes. This is similar to the computer science Binary Search, where the range of possible solutions is halved each iteration. Explicitly, if f(a) f(c) Advantages and drawbacks of the bisection method Advantages of Bisection Method The bisection method is always convergent. Since the method brackets the root, the method is guaranteed to converge. As iterations are conducted, the interval gets halved. So one can guarantee the decrease in the error in the solution of the equation. Drawbacks of Bisection Method The convergence of bisection method is slow as it is simply based on halving the interval. If one of the initial guesses is closer to the root, it will take larger number of iterations to reach the root. If a function is such that it just touches the x-axis (Figure 3.8) such as it will be unable to find the lower guess, , and upper guess, , such that For functions where there is a singularity and it reverses sign at the singularity, bisection method may converge on the singularity (Figure 3.9). An example include and, are valid initial guesses which satisfy . However, the function is not continuous and the theorem that a root exists is also not applicable. Figure.3.8. Function has a single root at that cannot be bracketed. Figure.3.9. Function has no root but changes sign. Explanation Source code for False position method: Example code of False-position method C code was written for clarity instead of efficiency. It was designed to solve the same problem as solved by the Newtons method and secant method code: to find the positive number x where cos(x) = x3. This problem is transformed into a root-finding problem of the form f(x) = cos(x) x3 = 0. #include #include double f(double x) { return cos(x) x*x*x; } double FalsiMethod(double s, double t, double e, int m) { int n,side=0; double r,fr,fs = f(s),ft = f(t); for (n = 1; n { r = (fs*t ft*s) / (fs ft); if (fabs(t-s) fr = f(r); if (fr * ft > 0) { t = r; ft = fr; if (side==-1) fs /= 2; side = -1; } else if (fs * fr > 0) { s = r; fs = fr; if (side==+1) ft /= 2; side = +1; } else break; } return r; } int main(void) { printf(%0.15fn, FalsiMethod(0, 1, 5E-15, 100)); return 0; } After running this code, the final answer is approximately 0.865474033101614 Example 1 Consider finding the root of f(x) = x2 3. Let ÃŽÂ µstep = 0.01, ÃŽÂ µabs = 0.01 and start with the interval [1, 2]. Table 1. False-position method applied to f(x)  =  x2 3. a b f(a) f(b) c f(c) Update Step Size 1.0 2.0 -2.00 1.00 1.6667 -0.2221 a = c 0.6667 1.6667 2.0 -0.2221 1.0 1.7273 -0.0164 a = c 0.0606 1.7273 2.0 -0.0164 1.0 1.7317 0.0012 a = c 0.0044 Thus, with the third iteration, we note that the last step 1.7273 à ¢Ã¢â‚¬  Ã¢â‚¬â„¢ 1.7317 is less than 0.01 and |f(1.7317)| Note that after three iterations of the false-position method, we have an acceptable answer (1.7317 where f(1.7317) = -0.0044) whereas with the bisection method, it took seven iterations to find a (notable less accurate) acceptable answer (1.71344 where f(1.73144) = 0.0082) Example 2 Consider finding the root of f(x) = e-x(3.2 sin(x) 0.5 cos(x)) on the interval [3, 4], this time with ÃŽÂ µstep = 0.001, ÃŽÂ µabs = 0.001. Table 2. False-position method applied to f(x)  = e-x(3.2 sin(x) 0.5 cos(x)). a b f(a) f(b) c f(c) Update Step Size 3.0 4.0 0.047127 -0.038372 3.5513 -0.023411 b = c 0.4487 3.0 3.5513 0.047127 -0.023411 3.3683 -0.0079940 b = c 0.1830 3.0 3.3683 0.047127 -0.0079940 3.3149 -0.0021548 b = c 0.0534 3.0 3.3149 0.047127 -0.0021548 3.3010 -0.00052616 b = c 0.0139 3.0 3.3010 0.047127 -0.00052616 3.2978 -0.00014453 b = c 0.0032 3.0 3.2978 0.047127 -0.00014453 3.2969 -0.000036998 b = c 0.0009 Thus, after the sixth iteration, we note that the final step, 3.2978 à ¢Ã¢â‚¬  Ã¢â‚¬â„¢ 3.2969 has a size less than 0.001 and |f(3.2969)| In this case, the solution we found was not as good as the solution we found using the bisection method (f(3.2963) = 0.000034799) however, we only used six instead of eleven iterations. Source code for Bisection method #include #include #define epsilon 1e-6 main() { double g1,g2,g,v,v1,v2,dx; int found,converged,i; found=0; printf( enter the first guessn); scanf(%lf,g1); v1=g1*g1*g1-15; printf(value 1 is %lfn,v1); while (found==0) { printf(enter the second guessn); scanf(%lf,g2); v2=g2*g2*g2-15; printf( value 2 is %lfn,v2); if (v1*v2>0) {found=0;} else found=1; } printf(right guessn); i=1; while (converged==0) { printf(n iteration=%dn,i); g=(g1+g2)/2; printf(new guess is %lfn,g); v=g*g*g-15; printf(new value is%lfn,v); if (v*v1>0) { g1=g; printf(the next guess is %lfn,g); dx=(g1-g2)/g1; } else { g2=g; printf(the next guess is %lfn,g); dx=(g1-g2)/g1; } if (fabs(dx)less than epsilon {converged=1;} i=i+1; } printf(nth calculated value is %lfn,v); } Example 1 Consider finding the root of f(x) = x2 3. Let ÃŽÂ µstep = 0.01, ÃŽÂ µabs = 0.01 and start with the interval [1, 2]. Table 1. Bisection method applied to f(x)  =  x2 3. a b f(a) f(b) c  =  (a  +  b)/2 f(c) Update new b à ¢Ã‹â€ Ã¢â‚¬â„¢ a 1.0 2.0 -2.0 1.0 1.5 -0.75 a = c 0.5 1.5 2.0 -0.75 1.0 1.75 0.062 b = c 0.25 1.5 1.75 -0.75 0.0625 1.625 -0.359 a = c 0.125 1.625 1.75 -0.3594 0.0625 1.6875 -0.1523 a = c 0.0625 1.6875 1.75 -0.1523 0.0625 1.7188 -0.0457 a = c 0.0313 1.7188 1.75 -0.0457 0.0625 1.7344 0.0081 b = c 0.0156 1.71988/td> 1.7344 -0.0457 0.0081 1.7266 -0.0189 a = c 0.0078 Thus, with the seventh iteration, we note that the final interval, [1.7266, 1.7344], has a width less than 0.01 and |f(1.7344)| Example 2 Consider finding the root of f(x) = e-x(3.2 sin(x) 0.5 cos(x)) on the interval [3, 4], this time with ÃŽÂ µstep = 0.001, ÃŽÂ µabs = 0.001. Table 1. Bisection method applied to f(x)  = e-x(3.2 sin(x) 0.5 cos(x)). a b f(a) f(b) c  =  (a  +  b)/2 f(c) Update new b à ¢Ã‹â€ Ã¢â‚¬â„¢ a 3.0 4.0 0.047127 -0.038372 3.5 -0.019757 b = c 0.5 3.0 3.5 0.047127 -0.019757 3.25 0.0058479 a = c 0.25 3.25 3.5 0.0058479 -0.019757 3.375 -0.0086808 b = c 0.125 3.25 3.375 0.0058479 -0.0086808 3.3125 -0.0018773 b = c 0.0625 3.25 3.3125 0.0058479 -0.0018773 3.2812 0.0018739 a = c 0.0313 3.2812 3.3125 0.0018739 -0.0018773 3.2968 -0.000024791 b = c 0.0156 3.2812 3.2968 0.0018739 -0.000024791 3.289 0.00091736 a = c 0.0078 3.289 3.2968 0.00091736 -0.000024791 3.2929 0.00044352 a = c 0.0039 3.2929 3.2968 0.00044352 -0.000024791 3.2948 0.00021466 a = c 0.002 3.2948 3.2968 0.00021466 -0.000024791 3.2958 0.000094077 a = c 0.001 3.2958 3.2968 0.000094077 -0.000024791 3.2963 0.000034799 a = c 0.0005 Thus, after the 11th iteration, we note that the final interval, [3.2958, 3.2968] has a width less than 0.001 and |f(3.2968)| Convergence Rate Why dont we always use false position method? There are times it may converge very, very slowly. Example: What other methods can we use? Comparison of rate of convergence for bisection and false-position method