That is up to you based on the accuracy that you need for your application. If you have a desired accuracy, you can calculate the tolerance value you need from whatever equation you are using to evaluate the "error" each iteration. In general, I try to get at least one more digit of precision from my calculation than I think I need. Hope this helps!
I believe I've written a simple function in MATLAB to capture this method, as it outputs what I expect for simple f(x) although I haven't fed it something crazy difficult yet. I did notice however that I don't need to calculate fU or fL, though, since I never actually use it after its definition. So this means I can save the "two days" according to your abstract example you covered in the video, correct?
fU and fL are not recalculated every time, only one new point is calculated and the other values are adjusted and set equal to old values. That is the typical method. Hopefully I conveyed that in my video.
Please how do we get the tolerance value?
That is up to you based on the accuracy that you need for your application. If you have a desired accuracy, you can calculate the tolerance value you need from whatever equation you are using to evaluate the "error" each iteration. In general, I try to get at least one more digit of precision from my calculation than I think I need. Hope this helps!
I believe I've written a simple function in MATLAB to capture this method, as it outputs what I expect for simple f(x) although I haven't fed it something crazy difficult yet. I did notice however that I don't need to calculate fU or fL, though, since I never actually use it after its definition. So this means I can save the "two days" according to your abstract example you covered in the video, correct?
fU and fL are not recalculated every time, only one new point is calculated and the other values are adjusted and set equal to old values. That is the typical method. Hopefully I conveyed that in my video.