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Description: This video introduces non-linear equations and how to manipulate the equation to solve for its roots. It also discusses cases with asymptotes and multiple roots.
Hello this video is NLE|VL|04 and focuses on the theory of bisection method to solve non-linear equations.
Let's first discuss the principle behind bisection method the root should be between xL, the lower bound and xU, the upper bound. xL represents the smallest number between xL and xU and the estimate of the root known as xR is found by taking the average between xL and xU. This can be done by adding xL and xU together and dividing by two. With xR the region is now divided into two parts first region between xL and xR and the second region between xR and xU. We are now tasked to determine where the root crosses the x-axis. Is it in the first region between xL and xR or in the second region between xR and xU? This can be done by testing the condition if f(xL)f(xR) <0. If it is then the root is between xL and xR and we change it such that xU is now the xR. if it is not within those bounds then it is in within the other bounds where f(xR)f(xU) < 0 and we will change it so that xL is now the new xR. Now that either xL or xU has been changed. We can repeat the process to find the next xR.
Let's now look at this with graphical steps. Let's start with a function at f(x) = 0 that crosses the x-axis. This will be our first iteration i where we have the bounds xL and xU that crosses the x-axis. We know this because f(xL) f(xU) should be less than zero next we find the first estimate of xR by taking the average of xL and xU in this case xL is equal to 0 and xU is equal to 2 then xR, which is the average would be equal to 1. We have now split the bounds into two regions and we can test which region the function crosses the x-axis. So in this case either f(xL)f(xR) < 0 or f(xR)f(xU) <0. For this example, it crosses in the second region. We now adjust the bounds and finish our first iteration. We now have new xL and xU bounds and the process can be repeated. In the second iteration we do the same thing and find the middle between xL and xU and determine where we will move our bounds. You continue doing multiple iterations until the stopping criteria is met depending on how accurate and precise you want to be. The number of iterations could vary between you know small numbers such as five up to thousands. For this case we'll stop after 10 iterations the estimate of the root would be xR in this case 1.65723. What you'll notice is on the y-axis the value is quite close to 0. This could be one of the stopping criteria.
Let's review the algorithm.
This concludes the video. This project could not have been completed without the support of eCampusOntario and the University of Ottawa.
Description: This video discusses a simple trial-and-error method for finding find a root to an NLE equation.
Description: This video implements the theory of incremental search in a practice problem in a step-by-step approach.
Description: This video discusses how to systematically divide a range where the root is present
Description: This video implements the bisection method to solve a problem in a step-by-step procedure.