Bisection Method-- 4 Iterations by Hand (example)

Bisection Method-- 4 Iterations by Hand (example)
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[@carlosgarzon58]

I do not get why professors go back in time... all the way until 500 BC , and they still do not explain like this....Thanks a lot!

[@jashael]

This is a very good explanation. Thanks!

[@michaelhambozi6726]

weoooooooow amazing

[@bog4240]

When you swap a or b with p, is it dependent on the value of f(a) / f(b) being positive/negative or is it just a patent rule that you always use where if < 0 swap a if > 0 swap b?

[@tugi007]

😂😂missed my lecture but its all good

[@mohammedferas4393]

im still looking for the hard example

[@yazanstar5806]

Thank you. What if i want to find the relative error using only these data? PLEASE I NEED AN ANSWER.

[@ScreenedInstructor]

To do that you can first solve x^2-5=0 | x^2=5 | x = +/- sqrt of 5, but the problem says that the value is from [0,3] so we just take the positive value so sqrt(5)= 2.236. Now compare that value with the value we got using the bisection method and 4 iterations. Then use the relative error formula (since we know both the actual and approximate value)

[@yazanstar5806]

@@ScreenedInstructor thank you so much you deserve a subscribe. Just to be clear, the relative error for this problem of the last iteration will be as this => |2.063-2.236| / |2.236| * 100

[@hayubinspiration4069]

What if the range is -1,1 instead of 0,1 how do we do it

[@yasinkaancopluoglu4494]

when u put into equation -1 and 1 the two equation is -4 two values are negative you can not find any root in that interval