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I am trying to solve the following system. Firstly I have a table of numerical values I will call numbers

    numbers={{8.03077, 7.8435, 0.155633}, {8.02983, 7.90858, 0.155633}, {8.02489, 
  7.94988, 0.155633}, {8.0169, 7.97541, 0.155633}, {8.00632, 7.99063, 
  0.155633}, {7.99326, 7.99916, 0.155633}, {7.9777, 8.00329, 
  0.155633}, {7.95953, 8.00449, 0.155633}, {7.93859, 8.00365, 
  0.155633}}

Now I have to solve a system $$\frac{\cos x}{x}=a$$ and $$\frac{\cos y}{y}=b$$ and $$R(y-x)=c$$ where $a,b,c$ are the numerical values in the table above. A is first, b is second and c is third. Basically this is my formula:

Table[FindRoot[{Cos[x]/x == numbers[[j, 1]], 
     Cos[y]/y == numbers[[j, 2]], 
     R*(-x + y) == 
      numbers[[j, 3]]}, {{x, .01}, {y, .05}, {R, .1}}][[1, 2]], {j, 1,
    Length[numbers], 1}];

Now if you run this code, you will find an error saying:

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

I assume I have to change the starting points of x,y or R but I tried many different combinations yet nothing seems to work. So my question here is: What can I do? Is there a nice mathematical way to determine the optimum starting points? Is there a way to see which parameter is good and which one is absolutely horrible?

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    $\begingroup$ Wouldn't it be easier to (i) solve the first equation for $x$, (ii) solve the second equation for $y$, and then (iii) set $R = c/(y-x)$? $\endgroup$
    – user484
    Dec 18, 2015 at 20:25
  • $\begingroup$ i dont if this help...x0 = Min[Table[ Table[Abs[((Cos[x]/x) - numbers[[j, 1]])], {j, 1, Length[numbers], 1}], {x, 0.0001, 2 Pi}]] y0 = Min[Table[ Table[Abs[(Cos[y]/y - numbers[[j, 2]])], {j, 1, Length[numbers], 1}], {y, 0.0001, 2 Pi}]] R0 = Min[Table[ Table[Abs[(R*(-y0 + x0) - numbers[[j, 3]])], {j, 1, Length[numbers], 1}], {R, 0.0001, 2 Pi}]] Table[FindRoot[{Cos[x]/x == numbers[[j, 1]], Cos[y]/y == numbers[[j, 2]], R*(-x + y) == numbers[[j, 3]]}, {{x, x0}, {y, y0}, {R, R0}}][[1, 2]], {j, 1, Length[numbers], 1}] $\endgroup$
    – Diogo
    Dec 18, 2015 at 20:32
  • $\begingroup$ @Rahul Would you believe me that I am trying to find a good solution to this problem for over a month now and that I am mindblown of how simple your solution is? I can't believe I didn't see that! I fell horrible. Thank you! $\endgroup$
    – skrat
    Dec 18, 2015 at 20:50

4 Answers 4

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You can use FindInstance

Table[FindInstance[
Cos[x]/x == numbers[[i, 1]] && Cos[y]/y == numbers[[i, 2]] && 
R == numbers[[i, 3]]/(-x + y), {x, y, R}, Reals], {i, 
Length[numbers]}]

(*  {{{x -> 0.123572, y -> 0.126476, R -> 53.5889}}, {{x -> 0.123586, 
y -> 0.125451, R -> 83.429}}, {{x -> 0.123661, y -> 0.12481, 
R -> 135.465}}, {{x -> 0.123782, y -> 0.124416, 
R -> 245.438}}, {{x -> 0.123943, y -> 0.124183, 
R -> 649.398}}, {{x -> 0.124143, y -> 0.124052, 
R -> -1726.}}, {{x -> 0.124381, y -> 0.123989, 
R -> -397.389}}, {{x -> 0.124661, y -> 0.123971, 
R -> -225.713}}, {{x -> 0.124984, y -> 0.123984, R -> -155.563}}}*)
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  • $\begingroup$ If I substitute R == numbers[[i, 3]]/(-x + y) with R(-x + y) == numbers[[i, 3]] I was surprised to discover that it would not work. Can you comment on why this is and how you came to the realization of using the first form? $\endgroup$
    – Jack LaVigne
    Dec 18, 2015 at 22:21
  • $\begingroup$ I don't know why the original version doesn't work. Actually I was trying Solve, Reduce and FindInstance nothing worked. Then I saw the first comment by @Rahul which led me to try out this form and it worked. So credit goes to him. $\endgroup$
    – Hubble07
    Dec 19, 2015 at 5:00
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This is just a bit faster if you note the equations are decoupled and you separately run FindInstance on the first two and just algebraically solve the third:

f[v_] := {#[[1]], #[[2]] , 
    v[[3]]/(#[[2]] - #[[1]])} &@ (( 
      x /. First@FindInstance[ Cos[x] == # x , x , Reals]   ) & /@ 
    v[[1 ;; 2]])
f /@ numbers // AbsoluteTiming

{0.501644, {{0.123572, 0.126476, 53.5889}, {0.123586, 0.125451, 83.429}, {0.123661, 0.12481, 135.465}, {0.123782, 0.124416, 245.438}, {0.123943, 0.124183, 649.398}, {0.124143, 0.124052, -1726.}, {0.124381, 0.123989, -397.389}, {0.124661, 0.123971, -225.713}, {0.124984, 0.123984, -155.563}}}

Faster still, now you can use FindRoot :

 f[v_] := {#[[1]], #[[2]] , 
    v[[3]]/(#[[2]] - #[[1]])} &@ (( 
          x /. FindRoot[ Cos[x] == # x , {x, 0}]   ) & /@ v[[1 ;; 2]])
 f /@ numbers //AbsoluteTiming

{0.00659699, {{0.123572, 0.126476, 53.5889}, {0.123586, 0.125451, 83.429}, {0.123661, 0.12481, 135.465}, {0.123782, 0.124416, 245.438}, {0.123943, 0.124183, 649.398}, {0.124143, 0.124052, -1726.}, {0.124381, 0.123989, -397.389}, {0.124661, 0.123971, -225.713}, {0.124984, 0.123984, -155.563}}}

For completeness, here is a FindRoot expression that works on the system in one shot:

(FindRoot[{ Cos[x], Cos[y], y - x} == #  { x , y , 1/r} , {{x, 0}, {y,
       0}, {r, 1}} ]) & /@ numbers

This finds the r>0 solutions if you start with {r,1} and the remaining solutions if you start with {r,-1}

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  • $\begingroup$ Very nice, The middle FindRoot approach is blindingly fast compared to the FindInstance approach. $\endgroup$
    – Jack LaVigne
    Dec 19, 2015 at 9:04
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If you only want to work with FindRoot:

numbers = 
  SetPrecision[{{8.03077, 7.8435, 0.155633}, {8.02983, 7.90858, 
     0.155633}, {8.02489, 7.94988, 0.155633}, {8.0169, 7.97541, 
     0.155633}, {8.00632, 7.99063, 0.155633}, {7.99326, 7.99916, 
     0.155633}, {7.9777, 8.00329, 0.155633}, {7.95953, 8.00449, 
     0.155633}, {7.93859, 8.00365, 0.155633}}, 30];

    sol = FindRoot[{Cos[x]/x == #1, Cos[y]/y == #2, R (y - x) == #3}, {x, 0.1}, {y, 0.2}, {R, 100}, 
WorkingPrecision -> 30, MaxIterations -> 1000] & @@@ numbers;

N[sol, 6]

enter image description here

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$\begingroup$ 
PaddedForm[TableForm[
  MapThread[
   Append, {#, numbers[[All, 3]]/-Subtract @@@ #} &[
    Map[FindInstance[Cos[x]/x == #, x, Reals][[1, 1, 2]] &, 
     numbers[[All, ;; 2]], {2}]]],
  TableAlignments -> Right,
  TableHeadings -> {None, {"x", "y", "R"}}], {6, 5}]

enter image description here

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