False position method and bisection

False position mode and bisectionIn numerical analysis, the off-key position method or regula falsi method is a reservoir-finding algorithm that combines features from the bisection method and the sec method.The methodThe origin dickens iterations of the bastard position method. The loss 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 bloodline in the separation a0, b0, assuming continuity of the function f. The method proceeds by producing a sequence of shrinking separations ak, bk that all contain a root of f.At iteration get k, the numberis 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) slang the same sign, thus 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 to a fault apply in the secant method, solely the secant method eer retains the last two computed points, while the false position method retains two points which certainly support a root. On the other hand, the only contravention between the false position method and the bisection method is that the latter aims ck = (ak + bk) / 2.Bisection methodIn mathematics, the bisection method is a root-finding algorithm which repeatedly bisects an interval then apportions a subinterval in which a root moldiness lie for pass on processing. It is a very simple and robust method, only it is also comparatively slow.The method is applicable when we wish to solve the equation for the scalar inconstant x, where f is a constant 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 inter mediate 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 reckon 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 tone of voice to it. In this way the interval that might contain a zero of f is reduced in width by 50% at each measure. 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 methodAdvantages of Bisection MethodThe bisection method is always convergent. Since the method brackets the root, the method is guara nteed 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 MethodThe convergence of bisection method is slow as it is entirely based on halving the interval.If one of the initial guesses is closer to the root, it will divvy up larger number of iterations to reach the root.If a function is such that it exactly touches the x-axis (Figure 3.8) such asit will be unable to find the depress guess, , and upper guess, , such thatFor functions where there is a singularity and it reverses sign at the singularity, bisection method whitethorn converge on the singularity (Figure 3.9).An example implicateand, 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 one root at that cannot be bracketed.Figure.3.9. Function has no root but changes sign.ExplanationSource co de for False position method ensample code of False-position methodC code was written for clarity alternatively 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. overwhelm hold prongy f(double x)return cos(x) x*x*xdouble FalsiMethod(double s, double t, double e, int m)int n,side=0double 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 = frif (side==-1) fs /= 2side = -1else if (fs * fr 0)s = r fs = frif (side==+1) ft /= 2side = +1else breakreturn rint main(void)printf(%0.15fn, FalsiMethod(0, 1, 5E-15, 100))return 0After running this code, the final reply is approximately 0.865474033101614 compositors case 1 catch finding the root of f(x) = x2 3. Let step = 0.01, abs = 0.01 and start with the interval 1, 2. turn off 1. False-position method use to f(x)=x2 3.abf(a)f(b)cf(c)updateStep Size1.02.0-2.001.001.6667-0.2221a = c0.66671.66672.0-0.22211.01.7273-0.0164a = c0.06061.72732.0-0.01641.01.73170.0012a = c0.0044Thus, with the third iteration, we note that the last step 1.7273 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 2Consider finding the root of f(x) = e-x(3.2 sin(x) 0.5 cos(x)) on the interval 3, 4, this quantify with step = 0.001, abs = 0.001.Table 2. False-position method applied to f(x)= e-x(3.2 sin(x) 0.5 cos(x)).abf(a)f(b)cf(c)UpdateStep Size3.04.00.047127-0.0383723.5513-0.023411b = c0.44873.03.55130.047127-0.0234113.3683-0.0079940b = c0.18303.03.36830.047127-0.00799403.3149-0.0021548b = c0.05343.03.31490.047127-0 .00215483.3010-0.00052616b = c0.01393.03.30100.047127-0.000526163.2978-0.00014453b = c0.00323.03.29780.047127-0.000144533.2969-0.000036998b = c0.0009Thus, after the sixth iteration, we note that the final step, 3.2978 3.2969 has a sizing less than 0.001 and f(3.2969) In this case, the solution we found was not as levelheaded 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 methodincludeincludedefine epsilon 1e-6main()double g1,g2,g,v,v1,v2,dxint found,converged,ifound=0printf( enter the first guessn)scanf(%lf,g1)v1=g1*g1*g1-15printf(value 1 is %lfn,v1)while (found==0)printf(enter the second guessn)scanf(%lf,g2)v2=g2*g2*g2-15printf( value 2 is %lfn,v2)if (v1*v20)found=0elsefound=1printf(right guessn)i=1while (converged==0)printf(n iteration=%dn,i)g=(g1+g2)/2printf(new guess is %lfn,g)v=g*g*g-15printf(new value is%lfn,v)if (v*v10)g1=gprintf(the next guess is %lfn,g)dx=(g1-g2 )/g1elseg2=gprintf(the next guess is %lfn,g)dx=(g1-g2)/g1if (fabs(dx)less than epsilonconverged=1i=i+1printf(nth calculated value is %lfn,v)Example 1Consider finding the root of f(x) = x2 3. Let step = 0.01, abs = 0.01 and start with the interval 1, 2.Table 1. Bisection method applied to f(x)=x2 3.abf(a)f(b)c=(a+b)/2f(c)Updatenew b a1.02.0-2.01.01.5-0.75a = c0.51.52.0-0.751.01.750.062b = c0.251.51.75-0.750.06251.625-0.359a = c0.1251.6251.75-0.35940.06251.6875-0.1523a = c0.06251.68751.75-0.15230.06251.7188-0.0457a = c0.03131.71881.75-0.04570.06251.73440.0081b = c0.01561.71988/td1.7344-0.04570.00811.7266-0.0189a = c0.0078Thus, 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 2Consider finding the root of f(x) = e-x(3.2 sin(x) 0.5 cos(x)) on the interval 3, 4, this time with step = 0.001, abs = 0.001.Table 1. Bisection method applied to f(x)= e-x(3.2 sin(x) 0.5 cos(x)).abf(a)f(b)c=(a+b)/2f(c)Updatenew b a3.04.00.047127-0.0383723.5-0.019757b = c0.53.03.50.047127-0.0197573.250.0058479a = c0.253.253.50.0058479-0.0197573.375-0.0086808b = c0.1253.253.3750.0058479-0.00868083.3125-0.0018773b = c0.06253.253.31250.0058479-0.00187733.28120.0018739a = c0.03133.28123.31250.0018739-0.00187733.2968-0.000024791b = c0.01563.28123.29680.0018739-0.0000247913.2890.00091736a = c0.00783.2893.29680.00091736-0.0000247913.29290.00044352a = c0.00393.29293.29680.00044352-0.0000247913.29480.00021466a = c0.0023.29483.29680.00021466-0.0000247913.29580.000094077a = c0.0013.29583.29680.000094077-0.0000247913.29630.000034799a = c0.0005Thus, 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 RateWhy dont we always use false position method?There are times it may converge very, very slowly.ExampleWhat other methods can we use? coincidence of rate of convergence for bisection and false-position method