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How is f(x) evaluated for each iteration using fsolve?

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TS
TS am 14 Jan. 2019
Kommentiert: Torsten am 15 Jan. 2019
I'm using fsolve for an unconstrained non-linear function. And I notice that fval is much different than f(x) for the last iteration. How exactly is f(x) calculated?
  13 Kommentare
TS
TS am 14 Jan. 2019
So it's the squared error? I got confused because it says f(x) even though it's the sum of squared function values.
Thanks so much!
Stephen23
Stephen23 am 14 Jan. 2019
Bearbeitet: Stephen23 am 14 Jan. 2019
"I got confused because it says f(x) even though it's the sum of squared function values."
I agree that the inconsistent labeling is confusing at best, and at worst incorrect. You should make a bug report (and include a link to this thread).

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Martijn
Martijn am 15 Jan. 2019
The values shown under f(x) in the iteration table are indeed not literally the function values at that given iteration, you are seeing the square of the norm of the function value vector here; as documented, see:
The reason for this is that "fsolve" can actually solve whole systems of equations where f(x) would then be a whole vector of values. Displaying this whole vector in the table is not really feasible, there is not enough space for that. Hence, instead, a measure is shown which basically summarizes the whole f(x) vector in a single measure which also actually is representative for how good the current solution is.
We could look into whether we can perhaps change the header to ||f(x)||^2 in future releases to better illustrate this.

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RANJEET YADAV
RANJEET YADAV am 14 Jan. 2019
f(x)=fsolve(@(x) funname(x),x=1);

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