0
$\begingroup$
In[1]:= Solve[{\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = 0\), \(10\)]\(
\*FractionBox[\(4\), \(\((2*\ k + 1)\)*\ \[Pi]\)]*
     MittagLefflerE[\[Beta], \((\((\(-d\))\)*\ 
\*SuperscriptBox[\((2\ *k + 1)\), \(2\)]\ *
\*SuperscriptBox[\(\[Pi]\), \(2\)]\ *
\*SuperscriptBox[\(398\), \(\[Beta]\)])\)]\ *
     Sin[\((2\ *k + 1)\)*\ \[Pi]*\ 0.5]\)\) == 0.58, \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = 0\), \(10\)]\(
\*FractionBox[\(4\), \(\((2*\ k + 1)\)*\ \[Pi]\)]*
     MittagLefflerE[\[Beta], \((\((\(-d\))\)*\ 
\*SuperscriptBox[\((2\ *k + 1)\), \(2\)]\ *
\*SuperscriptBox[\(\[Pi]\), \(2\)]\ *
\*SuperscriptBox[\(1019\), \(\[Beta]\)])\)]\ *
     Sin[\((2\ *k + 1)\)*\ \[Pi]*\ 0.5]\)\) == 0.42, \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = 0\), \(10\)]\(
\*FractionBox[\(4\), \(\((2*\ k + 1)\)*\ \[Pi]\)]*
     MittagLefflerE[\[Beta], \((\((\(-d\))\)*\ 
\*SuperscriptBox[\((2\ *k + 1)\), \(2\)]\ *
\*SuperscriptBox[\(\[Pi]\), \(2\)]\ *
\*SuperscriptBox[\(7709\), \(\[Beta]\)])\)]\ *
     Sin[\((2\ *k + 1)\)*\ \[Pi]*\ 0.5]\)\) == 0.15}, {d, \[Beta]}]

During evaluation of In1:= Solve::inex: Solve was unable to solve the system with inexact coefficients or the system obtained by direct rationalization of inexact numbers present in the system. Since many of the methods used by Solve require exact input, providing Solve with an exact version of the system may help.

Out[1]= Solve[{0.0606305 MittagLefflerE[\[Beta], -441 398^\[Beta] d \
\[Pi]^2] - 
    0.0670126 MittagLefflerE[\[Beta], -361 398^\[Beta] d \[Pi]^2] + 
    0.0748964 MittagLefflerE[\[Beta], -289 398^\[Beta] d \[Pi]^2] - 
    0.0848826 MittagLefflerE[\[Beta], -225 398^\[Beta] d \[Pi]^2] + 
    0.0979415 MittagLefflerE[\[Beta], -169 398^\[Beta] d \[Pi]^2] - 
    0.115749 MittagLefflerE[\[Beta], -121 398^\[Beta] d \[Pi]^2] + 
    0.141471 MittagLefflerE[\[Beta], -81 398^\[Beta] d \[Pi]^2] - 
    0.181891 MittagLefflerE[\[Beta], -49 398^\[Beta] d \[Pi]^2] + 
    0.254648 MittagLefflerE[\[Beta], -25 398^\[Beta] d \[Pi]^2] - 
    0.424413 MittagLefflerE[\[Beta], -9 398^\[Beta] d \[Pi]^2] + 
    1.27324 MittagLefflerE[\[Beta], -398^\[Beta] d \[Pi]^2] == 0.58, 
  0.0606305 MittagLefflerE[\[Beta], -441 1019^\[Beta] d \[Pi]^2] - 
    0.0670126 MittagLefflerE[\[Beta], -361 1019^\[Beta] d \[Pi]^2] + 
    0.0748964 MittagLefflerE[\[Beta], -289 1019^\[Beta] d \[Pi]^2] - 
    0.0848826 MittagLefflerE[\[Beta], -225 1019^\[Beta] d \[Pi]^2] + 
    0.0979415 MittagLefflerE[\[Beta], -169 1019^\[Beta] d \[Pi]^2] - 
    0.115749 MittagLefflerE[\[Beta], -121 1019^\[Beta] d \[Pi]^2] + 
    0.141471 MittagLefflerE[\[Beta], -81 1019^\[Beta] d \[Pi]^2] - 
    0.181891 MittagLefflerE[\[Beta], -49 1019^\[Beta] d \[Pi]^2] + 
    0.254648 MittagLefflerE[\[Beta], -25 1019^\[Beta] d \[Pi]^2] - 
    0.424413 MittagLefflerE[\[Beta], -9 1019^\[Beta] d \[Pi]^2] + 
    1.27324 MittagLefflerE[\[Beta], -1019^\[Beta] d \[Pi]^2] == 0.42, 
  0.0979415 MittagLefflerE[\[Beta], -13^(2 + \[Beta])
         593^\[Beta] d \[Pi]^2] + 
    0.0606305 MittagLefflerE[\[Beta], -441 7709^\[Beta] d \[Pi]^2] - 
    0.0670126 MittagLefflerE[\[Beta], -361 7709^\[Beta] d \[Pi]^2] + 
    0.0748964 MittagLefflerE[\[Beta], -289 7709^\[Beta] d \[Pi]^2] - 
    0.0848826 MittagLefflerE[\[Beta], -225 7709^\[Beta] d \[Pi]^2] - 
    0.115749 MittagLefflerE[\[Beta], -121 7709^\[Beta] d \[Pi]^2] + 
    0.141471 MittagLefflerE[\[Beta], -81 7709^\[Beta] d \[Pi]^2] - 
    0.181891 MittagLefflerE[\[Beta], -49 7709^\[Beta] d \[Pi]^2] + 
    0.254648 MittagLefflerE[\[Beta], -25 7709^\[Beta] d \[Pi]^2] - 
    0.424413 MittagLefflerE[\[Beta], -9 7709^\[Beta] d \[Pi]^2] + 
    1.27324 MittagLefflerE[\[Beta], -7709^\[Beta] d \[Pi]^2] == 
   0.15}, {d, \[Beta]}]

enter image description here as you can see in the picture, what does the inexact numbers mean in the warning?

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2
  • 1
    $\begingroup$ Even you use Rationalize to eliminate inexact numbers to convert to exact number, Solve can't solve, because you have a Transcendental equations. See: en.wikipedia.org/wiki/Transcendental_equation. Another problem you have a 3 equations and 2 variables to find.Try: Solve[{x + y == 1, x - 2 y == -3, 3 y - 3 x == -1}, {x, y}]. Solve gives NO solutions. $\endgroup$ Commented May 12, 2022 at 15:26
  • 1
    $\begingroup$ Any one of the three equations can be removed, still, no solution is obtained. Is this caused by 'the Transcendental equation'? $\endgroup$
    – Stan
    Commented May 13, 2022 at 7:01

2 Answers 2

2
$\begingroup$
Clear["Global`*"]

Use exact numbers in equations (e.g., 1/2 rather than the inexact 0.5)

eqns = {Sum[(4/((2*k + 1)*Pi))*
             MittagLefflerE[β, (-d)*(2*k + 1)^2*Pi^2*
                 398^β]*Sin[(2*k + 1)*Pi*1/2], 
      {k, 0, 10}] == 29/50, 
       Sum[(4/((2*k + 1)*Pi))*
             MittagLefflerE[β, (-d)*(2*k + 1)^2*Pi^2*
                 1019^β]*Sin[(2*k + 1)*Pi*1/2], 
      {k, 0, 10}] == 21/50, 
       Sum[(4/((2*k + 1)*Pi))*
             MittagLefflerE[β, (-d)*(2*k + 1)^2*Pi^2*
                 7709^β]*Sin[(2*k + 1)*Pi*1/2], 
      {k, 0, 10}] == 3/20} //
   Simplify;

Length@eqns

(* 3 *)

Variables[Level[eqns, {-1}]]

(* {d, β} *)

There are three equations with only two variables. The system is overdetermined. Approach it as a minimization problem.

The constraints are

Cases[eqns, MittagLefflerE[a_, _] :> a > 0, Infinity] // Union

(* {β > 0} *)

Minimizing,

{min, arg} = NMinimize[
  {Total[(#[[1]] - #[[-1]])^2 & /@ eqns], β > 0}, {d, β}, 
  AccuracyGoal -> 10, PrecisionGoal -> 9, WorkingPrecision -> 20]

(* General::ovfl: Overflow occurred in computation.

General::ovfl: Overflow occurred in computation.

General::ovfl: Overflow occurred in computation.

General::stop: Further output of General::ovfl will be suppressed during this calculation.

NMinimize::cvmit: Failed to converge to the requested accuracy or precision within 100 iterations.

{0.25640235187367699755, {d -> 0.70884195297684036358, β -> 
   0.0029224816962366914900}} *)

(#[[1]] - #[[-1]] & /@ eqns) /. arg

(* {-0.42886059095864565614, -0.26922282559129573706, 0.000124807520} *)

Ideally these would be zero. This is as close as the system gets.

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6
  • $\begingroup$ I have studied the code for some days, still, what the meaning of '#[[1]] ' or '#[[-1]]'? I use the minimized d and β to fit some data, which doesn't work well. ClearAll["*"]; Remove["*"]; data = {{43.68, 1.}, {206.42, 0.76}, {398.11, 0.58}, {1019.47, 0.42}, {1910.44, 0.33}, {2964.39, 0.26}, {4116.84, 0.22}, {5318.17, 0.19}, {6505.9, 0.16}, {7709.48, 0.15}, {8827.55, 0.13}, {9984.18, 0.12}, {11015.08, 0.11}}; model = Sum[ 4/((2*k + 1)*Pi)*1(c0)* MittagLefflerE[[Beta], ((-d)*(2*k + 1)^2*[Pi]^2*t^[Beta])]* Sin[(2*k + 1)*Pi*1/2], {k, 0, 10}]; $\endgroup$
    – Stan
    Commented May 19, 2022 at 15:20
  • $\begingroup$ M = NonlinearModelFit[data, model, {{[Beta], 0.0029}, {d, 0.71}}, t]; Show[{ListPlot[data, PlotStyle -> Black], Plot[M // Normal, {t, data[[1, 1]], data[[-1, 1]]}, PlotStyle -> Red, PlotRange -> All]}] (* NonlinearModelFit::nrlnum: The function value {Overflow[],Overflow[],Overflow[],Overflow[],Overflow[],Overflow[],Overflow[],Overflow[],Overflow[],Overflow[],Overflow[],Overflow[],Overflow[]} is not a list of real numbers with dimensions {13} at {[Beta],d} = {0.0467468,-0.359457}.*) The picture can't be pasted, where no fitting curve appears on the data point. $\endgroup$
    – Stan
    Commented May 19, 2022 at 15:37
  • $\begingroup$ If you highlight # in Mathematica and press F1 for help you get a link to the documentation for Slot ("represents the first argument supplied to a pure function"). (#[[1]] - #[[-1]])^2& /@ eqns maps the pure function for (square of First part - Last part) onto each equation, i.e., (LHS - RHS)^2. To satisfy an equation this should be zero. Minimizing the sum (Total) of all of the squares is the closest to a solution (in a minimum mean square error sense). If the equations were consistent, the result would have been zero. $\endgroup$
    – Bob Hanlon
    Commented May 19, 2022 at 17:05
  • $\begingroup$ Thanks Bob. I used {β=0.0029, d= 0.71} to fit the data which doesn't work well. while {d=0.0029, β= 0.71} fit the data very well. I wonder whether the value of β and d is opposite. $\endgroup$
    – Stan
    Commented May 21, 2022 at 10:06
  • $\begingroup$ Fitting data has nothing to do with the original question (solving three simultaneous equations in two unknowns). If you need help with fitting data, post a new question. Convert your cells to InputForm prior to copy and paste into MSE. $\endgroup$
    – Bob Hanlon
    Commented May 21, 2022 at 13:53
-1
$\begingroup$

I believe that the reason you get inexact coefficients is due to your use of decimals. Replacing all the decimals with their equivalent fractions results in this equation:

Solve[{\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k\.01 = \.010\), \(10\)]\(
\*FractionBox[\(4\), \(\((2*k\.01 + \.011)\)*\[Pi]\)]*
     MittagLefflerE[\[Beta], \((\((\(-d\))\)*\ 
\*SuperscriptBox[\((2*k\.01 + \.011)\), \(2\)]\ *
\*SuperscriptBox[\(\[Pi]\), \(2\)]\ *
\*SuperscriptBox[\(398\), \(\[Beta]\)])\)]\ *
     Sin[\((2\ *k + 1)\)*\ \[Pi]/2]\)\) == 58/100, \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k\.01 = \.010\), \(10\)]\(
\*FractionBox[\(4\), \(\((2*k\.01 + \.011)\)*\[Pi]\)]*
     MittagLefflerE[\[Beta], \((\((\(-d\))\)*\ 
\*SuperscriptBox[\((2*k\.01 + \.011)\), \(2\)]\ *
\*SuperscriptBox[\(\[Pi]\), \(2\)]\ *
\*SuperscriptBox[\(1019\), \(\[Beta]\)])\)]\ *
     Sin[\((2\ *k + 1)\)*\ \[Pi]/2]\)\) == 42/100, \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k\.01 = \.010\), \(10\)]\(
\*FractionBox[\(4\), \(\((2*k\.01 + \.011)\)*\[Pi]\)]*
     MittagLefflerE[\[Beta], \((\((\(-d\))\)*\ 
\*SuperscriptBox[\((2*k\.01 + \.011)\), \(2\)]\ *
\*SuperscriptBox[\(\[Pi]\), \(2\)]\ *
\*SuperscriptBox[\(7709\), \(\[Beta]\)])\)]\ *
     Sin[\((2\ *k + 1)\)*\ \[Pi]/2]\)\) == 15/100}, {d, \[Beta]}]

It returns a warning saying that This Equation

I suspect that this equation is beyond the scope of Solve, but the solving algorithm is not sure if it is or not, so it asks you for symbolic representations instead of decimals. I tried using Reduce as well, but it doesn't return, so I would assume you would have to do some manual solving first before feeding it to Solve.

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2
  • $\begingroup$ That message is not any error, but a warning. $\endgroup$
    – user64494
    Commented May 12, 2022 at 16:22
  • $\begingroup$ Fixed. Thanks for feedback. $\endgroup$
    – Romanp
    Commented May 12, 2022 at 16:40

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