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fsolve No Solution found (but there is one)

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Federico Maglione
Federico Maglione 2017 年 8 月 31 日
編集済み: Matt J 2017 年 9 月 1 日
Hello everyone,
I have checked in other posts having the same problem but, given the specificity of those, the proposed solutions do not apply to my case. However, I believe my situation is even simpler. I just need a little guidance.
Basically I am using fsolve for solving one non-linear equation F(x)=0. I am sure the solution exists as when I plot the function I can see the point.
However I receive this message
fsolve stopped because the problem appears regular as measured by the gradient, but the vector of function values is not near zero as measured by the default value of the function tolerance.
I suppose it is happening as, despite the look of the graph given the plot range, the solution is in a region where the function is almost flat (between 2.5*10^9 and 3.5*10^9).
I tried to change the starting point, increase the TolFun/TolX but it still produces the same error.
What should I do? Increase TolFun and decrease TolX maybe?
Thank you

  2 件のコメント

José-Luis 2017 年 8 月 31 日
Could you post your code?
dpb 2017 年 8 月 31 日
Specifically at least the function. You might try standardizing the variables to increase dynamic range somewhat but looks like fsolve should have no problem unless there's something funky going on in the neighborhood of the solution that is hidden by the scaling.
Try blowing up the plot in the ROI and see if the smoothness apparent here is really not so smooth...



Matt J
Matt J 2017 年 8 月 31 日
編集済み: Matt J 2017 年 8 月 31 日
For starters, fsolve is overkill for a one-variable monotonic function. You could just as well use fzero.
Second, you could adjust TolX or TolFun, but I would instead rescale the function so that the factors of 10^9 both in input and output space go away.

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Matt J
Matt J 2017 年 8 月 31 日
Well, I had better luck with fzero than with fsolve,
>> [x,fval]=fzero(f4_new,[1,5])
x =
fval =
>> [x,fval]=fsolve(f4_new,2.8)
x =
fval =
The reason fsolve is doing poorly is probably due to the lack of smoothness in the function observed below. So dpb was right, however, it's not at all clear to me what's causing the non-smoothness.
Federico Maglione
Federico Maglione 2017 年 8 月 31 日
Thank you very much!
dpb 2017 年 9 月 1 日
The roughness comes from the iterative nature of the solution in prior steps. I've not tried, but I'd guess a better solution in the previous would lead to smoother results later on although that's not guaranteed--it might be a pathological case that the discontinuities become more and more pronounced instead.
The indication of an error in erfc is evidence of how sensitive the internals are.


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