Bisection Method

Theory For BISECTION METHOD: In mathematics, the bisection method is a root-finding algorithm which repeatedly divides an  interval in half and then selects the subinterval in which a root exists. It is a very simple and  robust method, but it is also rather slow.Suppose we want to solve the equation f(x)=0,where f is a continuous function. The bisection method starts with two points xl and xu such that f(xl) and f(xu) have opposite  signs. The intermediate value theorem says that f must have at least one root in the interval  [xl,xu]. The method now divides the interval in two by computing xr = (xl+xu) / 2.   There are now two possibilities: either f(xl) and f(xu) have opposite  signs, or f(xr) and f(xl) have opposite signs. The bisection algorithm is then  applied recursively to the sub-interval where the sign change occurs. To determine in which sub-interval the root lies:     If f(xl).f(xr)<0,then we set xu=xr   If f(xl).f(xr)>0, then we set xl=xr   If f(xl).f(xr)=0, then we set xt=xr According to the condition we have to iterate again and again till then whenever |%error| or  |ea|=es. Then we have to find out the %error. The rule for determine %error  is %error or ea = (xrnew-xrold)/  xrnew*100%

MATLAB CODE FOR BI-SECTION METHOD OF GIVEN EUATION: FUNCYION OF RLC: function f_value=RLC(R) t=0.05; q0=5e-4; q=5e-6; L=5; C=1e-4; f_value=(exp(-R*t/(2*L)))*cos(sqrt((1/(L*C))-(R/(2*L))^2)*t)-q/q0; MAIN PROGRAM: clc; clear all tic Rl=200; Ru=400; es=.0001; i=0; imax=25; Rr=(Rl+Ru)/2; fl=RLC(Rl) while(1)   Rrold=Rr   Rr=(Rl+Ru)/2   fr=RLC(Rr)   if Rr~=0   ea=abs((Rr-Rrold)/Rr)*100   end   test=fl*fr   if test<0 ru="Rr">0   Rl=Rr;   fl=fr   else   ea=0;   end   if(ea=imax   break   end      fp=fopen('iea.txt','a+');  [err,msg]=ferror(fp,'clear');  if err~=0   disp('An error occurred while appending data to the file')   disp(msg)  end  fprintf(fp,'%0.0f\t%8.5e\n',i,ea);  err=fclose(fp);  if err~=0   disp('Could not close the file')  end  i=i+1; end disp('The value of resistance is:') disp(Rr) disp('The value of i is:') disp(i) toc

OUTPUT fl = -0.1631 Rrold =  300 Rr =  300 fr = -0.0295 ea =   0 test =   0.0048 fl =   -0.0295 Rrold =   300 Rr =   350 fr =   0.0209 ea =   14.2857 test =  -6.1704e-004 Ru =   350 Rrold =   350 Rr =   325 fr =   -0.0032 ea =   7.6923 test =   9.3070e-005 fl =   -0.0032 Rrold =   325 Rr =   337.5000 fr =   0.0091 ea =   3.7037 test =  -2.8865e-005 Rr =   328.1513 fr =  -1.0717e-007 ea =   1.1625e-004 test =   5.2177e-014 fl =  -1.0717e-007  Ru =   337.5000 Rrold =   337.5000 Rr =   331.2500 fr =   0.0031 ea =   1.8868 test =  -9.6751e-006 Ru =   331.2500 Rrold =   331.2500 Rr =   328.1250 fr =  -2.6307e-005 ea = 0.9524 test =   8.2989e-008 fl =  -2.6307e-005 Rrold =   328.1250 Rr =   329.6875 fr =   0.0015 fl =  -2.0056e-006 Rrold =   328.1494 Rr =   328.1616 fr =   1.0144e-005 ea =   0.0037 test =  -2.0345e-011 Ru =   328.1616 Rrold =   328.1616 Rrold =   328.1513 Rr =   328.1515 fr =   8.2673e-008 ea =   5.8124e-005 test =  -8.8602e-015  Rr =   328.155 fr =   4.0694e-006 ea =   0.0019 test =  -8.1616e-012 Ru =   328.1555 Rrold =   328.1555 Rr =   328.1525 fr =   1.0319e-006 ea =   9.2998e-004 test =  -2.0696e-012 Ru =   328.1525 Rrold =   328.1525 Rr =   328.1509 fr =  -4.8686e-004 ea =   4.6499e-004 test =   9.7646e-013 fl =  -4.8686e-007 Rrold =   328.1509 Rr =   328.1517 fr =   2.7252e-007 ea =   2.3250e-004 test =  -1.3268e-013 Ru =   328.1517 Rrold =   328.151 Ru =   328.1515 The value of resistance is:   328.1515 The value of i is:   19 elapsed_time =   0.1410 >>

Disadvantages:   • It is relatively slow when compared with other root finding methods we will study,   especially when the function f (x) has several continuous derivatives about the root  xr.   • The algorithm has no check to see whether the e is too small for the computer   arithmetic being used. [This is easily fixed by reference to the machine epsilon of the   computer arithmetic.]   • When we do this method by software (matlab) then need more time of execution for this method  because this method takes more number of iteration (i). • It cannot be used to  find roots when the function is tangent is the axis and does not pass through the axis.   For example: f(x) = x2

Discussion: In this experiment we learn the MATLAB code of bi-section method for finding root by numerical  method. We also clarify the basic theory of bi-section method. In MATLAB code we use a user define  function named RLC so we learn the rule to define a ‘user defined function’ in MATLAB. Finally we  can say that we properly follow our teacher instruction that’s why we do not face any problem to  solve our problem.