Comparing Convergence Of False Position And Bisection Methods Engineering Essay

Comparing Convergence Of False Position And Bisection orders Engineering experimentExplain with congresswoman that rate of convergence of false grade order is faster than that of the bisection manner acting.IntroductionFalse position order actingIn numeral analysis, the false position method or regula falsi method is a etymon- palpateing algorithm that combines features from the bisection method and the secant method.The methodThe first two loop topologys of the false position method. The red curve shows the function f and the blue lines are the secants. same(p) the bisection method, the false position method starts with two points a0 and b0 such(prenominal) that f(a0) and f(b0) are of reversal polaritys, which imp finesses by the intermediate apprise theorem that the function f has a locate in the separation a0, b0, assuming continuity of the function f. The method reward by producing a sequence of shrinking separations ak, bk that all contain a go down of f.At it eration number 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) have the same sign, then we destiny 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 normal is in addition determinationd in the secant method, but the secant method al substances retains the pass away two computed points, while the false position method retains two points which certainly bracket a root. On the other hand, the simply difference between the false position method and the bisection method is that the last mentioned uses ck = (ak + bk) / 2.Bisection methodIn mathematics, the bisection method is a root- finding algorithm which repeatedly bisects an separation then selects a submusical interval in which a root must lie for further processing. It is a genuinely simple and robust method, but it is also relatively sl ow.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 enemy 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 rock-bottom in width by 50% at each step. We stick until we have a bracket sufficiently small for our purposes. This is similar to the estimator science Binary Sea rch, 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 everlastingly 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 fracture in the solution of the equation.Drawbacks of Bisection MethodThe 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 forget 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 asit will be unable to find the lower guess, , and upper guess, , such thatFor 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 includeand, are valid init ial guesses which satisfy.However, the function is non 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.ExplanationSource code for False position methodExample code of False-position methodC code was written for limpidity instead of efficiency. It was designed to solve the same problem as single-minded by the Newtons method and secant method code to find the convinced(p) number x where cos(x) = x3. This problem is trans chassised into a root-finding problem of the form f(x) = cos(x) x3 = 0.include include double f(double x) open 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 utmost decide is approximately 0.865474033101614Example 1Consider finding the root of f(x) = x2 3. permit step = 0.01, abs = 0.01 and start with the interval 1, 2.Table 1. False-position method applied 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 satisfying coif (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 time with step = 0.001, abs = 0.001.Table 2. False-pos ition 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 exam step, 3.2978 3.2969 has a size less than 0.001 and f(3.2969) In this case, the solution we rear was not as good as the solution we found using the bisection method (f(3.2963) = 0.000034799) however, we only(prenominal) 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( picture 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( refreshful 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, ha s 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 time it may converge very, very slowly.ExampleWhat other methods can we use?Comparison of rate of convergence for bisection and false-position method