4/22/2021 0 Comments Bisection Method Examples
It is quite similar to bisection method algorithm and is one of the oldest approaches.It was developed because the bisection method converges at a fairly slow speed.
![]() To find the value c, we write down two versions of the slope m of the line L. ![]() There the bisection method algorithm required 23 iterations to reach the terminating condition. Here we see that in only 12 iterations we reach the terminating condition and get the root approximation. So in this situation Regula Falsi method conveges faster than Bisection method. As both regula falsi and bisection method are similar there are some common limitaions both the algorithms have. Use the bisection method to approximate the value of frac 1 sqrt5 3. Find the 4th approximation of the positive root of the function f(x) x4 - 7 using the bisection method. The approximations are in blue, the new intervals are in red. The positive root of f(x) x4 - 7 is at approximately x 1.6875. This approximation is off by at most pm 0.0625 units. Find the third approximation of the root of the function f(x) frac 1 2 x-sqrt3x1 using the bisection method. This table indicates the root is between x3 and x 4, so a good starting interval is 3,4. The function has a root at approximately x blue3.125 with a maximum possible error of pm0.125 units. Approximate the negative root of the function f(x) x2-7 to within 0.1 of its actual value. At x -2 the function value is f(-2) -3, and at x -3 the function value is f(-3) 2. The negative root of the function is at approximately x -2.6875 with a maximum error of only pm0.0625 units. Approximate the value of the root of f(x) -3x35x214x-16 near x 3 to within 0.05 of its actual value. Then, since were told that the root is near x 3 we can check that f(3) -10. The root of the function is approximately x 2.65625 and has an associated maximum error of only pm0.03125 units. Find the 4th approximation to the solution of the equation below using the bisection method. The solution to the equation is approximately x 2.8125 with a maximum error of 0.1875 units. Find the 5th approximation to the solution to the equation below, using the bisection method. The solution to the equation is approximately x 1.4375. This approximation has an maximum error of at most 0.0625 units. The equation below should have a solution that is larger than 5. Use the bisection method to approximate this solution to within 0.1 of its actual value. We know the solution is larger than 5, but we dont know how much larger. From this table we can select the first interval and determine the first approximation. The solution to the equation is approximately x 6.0625 with a maximum error of 0.0625 units. Approximate the value of this solution to within 0.05 units of its actual value. Lets set up a table of values to get an idea of where our first interval should be. The equation has a solution at approximately x -3.34275 with a maximum error in the approximation of at most 0.03125 units.
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