Why can't NSolve find values that FindInstance can?

Question

I am trying to solve for \[Epsilon], given three input paramters n, x and eps. Only FindInstance provides a solution for me whereas NSolve and NSolveValues fail.

Why does this happen?

$PreRead = (# /. 
     s_String /; 
       StringMatchQ[s, NumberString] && 
        Precision@ToExpression@s == MachinePrecision :> s <> "`50." &);
kldFunc[x_, y_] := x Log[x/y] + (1 - x) Log[(1 - x)/(1 - y)];
n = 10.^13.; x = 6. 10.^9.; eps = 10.^-20.;
Clear[[Epsilon], [Mu], p]; 
[Mu] = x + n [Epsilon];
p = [Mu]/n; 
[Epsilon] = [Epsilon] /. 
    Flatten[FindInstance[
      Exp[-kldFunc[p - [Epsilon], p] n] == eps && 
       1. > [Epsilon] > 0., [Epsilon], Reals] ]
Clear[[Epsilon], [Mu], p]; 
[Mu] = x + n [Epsilon];
p = [Mu]/n; 
[Epsilon] = 
   NSolveValues[
    Exp[-kldFunc[p - [Epsilon], p] n] == eps && 
     1. > [Epsilon] > 0., [Epsilon], Reals]
Clear[[Epsilon], [Mu], p]; 
[Mu] = x + n [Epsilon];
p = [Mu]/n; 
[Epsilon] = 
   NSolve[Exp[-kldFunc[p - [Epsilon], p] n] == eps && 
     1. > [Epsilon] > 0., [Epsilon], Reals]

EDIT

I tried plotting the two function to find the intersection

$PreRead = (# /. 
     s_String /; 
       StringMatchQ[s, NumberString] && 
        Precision@ToExpression@s == MachinePrecision :> 
      s <> "`100." &);
kldFunc[x_, y_] := x Log[x/y] + (1 - x) Log[(1 - x)/(1 - y)];
n = 10.^13.; x = 6.  10.^9.; eps = 10.^-20.;
Clear[[Epsilon], [Mu], p]; 
[Mu] = x + n [Epsilon];
p = [Mu]/n; 
Plot[{10^-20, Exp[-kldFunc[p - [Epsilon], p] n]}, {[Epsilon], 
  7 10^-8, 8  10^-8}, WorkingPrecision -> 50]

I tried to extract the intersection point using


plot1 = Plot[
   Exp[-kldFunc[p - \[Epsilon], p] n], {\[Epsilon], 7 10^-8, 
    8  10^-8}, WorkingPrecision -> 50];
plot2 = Plot[10^-20, {\[Epsilon], 7 10^-8, 8  10^-8}, 
   WorkingPrecision -> 50];
intersections = GraphicsMeshFindIntersections[{plot2, plot1}]

but it does'nt give an output.(I'm running on 13.3. See related)

Per documentation for NSolve, Details and Options bullet item: "NSolve deals primarily with linear and polynomial equations.` It has only limited capabilities for handling transcendental equations. — Daniel Lichtblau, Sep 26 '23 at 14:51
So what can I use to solve such equations?. FindInstance works most of the time but sometimes it runs for a couple of minutes without being able to find a solution. — Dotman, Sep 26 '23 at 15:01
Here is is your equation E^(1.0...00*10^13 (-((1 +\[Epsilon] - 1.0...00*10^-13 (6.00...0*10^9 + 1.000...000*10^13\[Epsilon])) Log[( 1 + \[Epsilon] - 1.0000..0000*10^-13 (6.0000...00*10^9 + 1.0000...000*10^13 \[Epsilon]))/( 1 - 1.0000..000*10^-13 (6.00...0*10^9 + 1.000...0000*10^13 \[Epsilon]))]) - (-\[Epsilon] + 1.0000..0000*10^-13 (6.0000...0000*10^9 + 1.00...000*10^13 \[Epsilon])) Log[( 1.000...000*10^13 (-\[Epsilon] + 1.0000...0000*10^-13 (6.00...000*10^9 + 1.00...00000*10^13 \[Epsilon])))/( 6.00000...0000*10^9 + 1.000..000000*10^13 \[Epsilon])])) == 1.0..0000000*10^-20. — user64494, Sep 26 '23 at 16:15
FindRoot would be my go-to for this. `kldFunc[x_, y_] := x Log[x/y] + (1 - x) Log[(1 - x)/(1 - y)]; n = 10^13; x = 6 *10^9; eps = 10^-20; [Mu] = x + n [Epsilon]; p = [Mu]/n;
In[59]:= FindRoot[ N[Exp[-kldFunc[p - [Epsilon], p] n] - eps, 50] == 0, {[Epsilon], 0, 1}, WorkingPrecision -> 30]

Out[59]= {[Epsilon] -> 7.43192053584110156413007451123*10^-8}` — Daniel Lichtblau, Sep 26 '23 at 20:09
See the comment for the answer. I modified my input values to n = 10^15; x = 6*10^9; eps = 10^-20; \[Mu] = x - n \[Epsilon]; p = \[Mu]/n;, the code failed to converge to a solution. — Dotman, Sep 26 '23 at 22:15
Also see my edit. Is it possible to get reliable solutions from the plot intersection? — Dotman, Sep 26 '23 at 22:16

Bob Hanlon · Answer 1 · 2023-09-26T16:39:38.937

Use FindRoot

Clear["Global`*"]
kldFunc[x_, y_] := x Log[x/y] + (1 - x) Log[(1 - x)/(1 - y)];
n = 10^13; x = 6*10^9; eps = 10^-20;
μ = x + n ϵ;
p = μ/n;
expr = N[Exp[-kldFunc[p - ϵ, p] n] - eps, 100] // Simplify;

With FindRoot you need to provide a starting value and only a single result is returned irrespective of the number of roots.

(sol1 = FindRoot[expr == 0, {ϵ, 10^-8}, WorkingPrecision -> 80]) // N
(* {ϵ -> 7.4319210^-8} )
expr /. sol1
(* 0.10^-94 )
(sol2 = FindRoot[expr == 0, {ϵ, 10^-7}, WorkingPrecision -> 80]) // N
(* {ϵ -> 7.4319210^-8} )
expr /. sol2
(* 0.10^-94 )

With two starting values, FindRoot avoids the use of derivatives

(sol3 = FindRoot[expr == 0, {ϵ, 10^-9, 10^-6}, 
    WorkingPrecision -> 80]) // N
(* {ϵ -> 7.4319210^-8} )
expr /. sol3
(* 0.10^-94 )

EDIT: For your revised values given in the comment

n = 10^15; x = 6*10^9; eps = 10^-20;
μ = x - n ϵ;
p = μ/n;

Increase the precision for expr

expr = N[Exp[-kldFunc[p - ϵ, p] n] - eps, 105] // Simplify;

The starting value must be close to the actual value

(sol4 = FindRoot[expr == 0, {ϵ, 7*10^-10}, WorkingPrecision -> 80]) //
  N
(* {ϵ -> 7.4332110^-10} )
expr /. sol4
(* 0.10^-92 )

This is nice and works a lot faster than my code.Thanks. However, as I modified my input values to n = 10^15; x = 6*10^9; eps = 10^-20; \[Mu] = x - n \[Epsilon]; p = \[Mu]/n;, the code failed to converge to a solution — Dotman, Sep 26 '23 at 16:26

cvgmt · Answer 2 · 2023-09-26T23:50:07.057

NSolve work if we use infinite precision number n,x,eps.

Clear["Global`*"];
kldFunc[x_, y_] := x Log[x/y] + (1 - x) Log[(1 - x)/(1 - y)];
n = 10^13; x = 6*10^9; eps = 10^-20;
μ = x + n ϵ;
p = μ/n;
NSolve[{Exp[-kldFunc[p - ϵ, p] n] == eps, 
  0 < ϵ < 10^-1}, ϵ]

{{ϵ -> 7.43192*10^-8}}

test another values.

NSolve[{Exp[-kldFunc[p - ϵ, p] n] == 10^-3, 
  0 < ϵ < 10^-1}, ϵ]

{{ϵ -> 2.8783*10^-8}}

Why can't NSolve find values that FindInstance can?

2 Answers2