equation solving - Is FindRoot wrong about its WorkingPrecision?

EDIT: I filed a bug report and after a small back-and-forth the support person agreed this is a bug. He said: "I have filed a report with the developers to issue warning messages when the arithmetic can affect the results." I'm satisfied with that.

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I noticed an odd mismatch between the behaviors of NDSolve and FindRoot.

If you give NDSolve an equation of a certain precision, say 20, and tell it to use a higher WorkingPrecision, say 30, it will complain (and rightly so), since it can't maintain a precision greater than it started with. For example:

NDSolve[{y'[x] == 2.0``20, y[0] == 0}, {y}, {x, 0, 1}, WorkingPrecision -> 30];


NDSolve::precw: The precision of the differential equation ({{(y^\[Prime])[x]==2.0000000000000000000,y[0]==0},{},{},{},{}})
is less than WorkingPrecision (30.`). >>

However, FindRoot seems to be more forgiving than it should be in a similar situation. For instance, if you have a function y[x] = 2x to precision 20...

Clear[y]
sol = NDSolve[{y'[x] == 2.0``21, y[0] == 0}, {y}, {x, 0, 1}, WorkingPrecision -> 21];
y[x_] = Evaluate[y[x] /. sol[[1]]];
Precision[y[1]]


20.699

...and you use it in FindRoot with WorkingPrecision->30...

root = FindRoot[y[x] == 1.0``30, {x, 0.4``20}, WorkingPrecision -> 30]

{x -> 0.500000000000000000000000000000}

...it doesn't complain at all! It even claims that it's able to keep a full precision of 30:

Precision[root]


30.

Which is seemingly does, despite having an equation less precise than that. It does complain if both sides of the given equation don't have high enough precision:

root = FindRoot[y[x] == 1.0``20, {x, 0.4``20}, WorkingPrecision -> 30]

FindRoot::precw: The precision of the argument function (InterpolatingFunction[{{0,1.00000000000000000000}},{5,3,1,{16},{4},0,0,0,0,Automatic,{},{},False},{{0,<<14>>,1.00000000000000000000}},{{0,2.00000000000000000000},{5.02973371873174322338*10^-6,2.00000000000000000000},<<12>>,{1.87545103551469488009,2.00000000000000000000},{2.00000000000000000000,2.00000000000000000000}},{Automatic}][x]==1.0000000000000000000)
is less than WorkingPrecision (30.`). >>

but I can't see why it would have a problem with the latter case but not the former.

Answer

Based on a discussion with the developers, FindRoot does not take into account the precision of heads when issuing the message, because it isn't necessarily related to the actual function values.

For an example,

x[1.][z_?NumberQ] := z^2 - 2

FindRoot[x[1.][z] == 0, {z, 2}, WorkingPrecision -> 20]

(* {z -> 1.4142135623730950488} *)

should not issue any warning, despite x[1.] having only machine precision. Since x[1.][z] only evaluates for numeric z, the objective function is simply a black-box to FindRoot.

Also, one should be careful in the interpretation of the Precision of a result, which is just a property of the number itself.

It is determined by the precision of the input numbers and the arithmetic operations done with them, but does not necessarily say much about how accurate the result might be in terms of convergence to an exact solution.

I would also take this opportunity to recommend reading the answer to the Is manual adjustment of AccuracyGoal and PrecisionGoal useless? question.