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I am trying to find a numerical solution to the following transcendental equation

BallooningFile = {0.`, 0.000136`, 0.000572`, 0.001152`, 0.001907`, 
   0.003004`, 0.004199`, 0.005479`, 0.006834`, 0.008256`, 0.008985`, 
   0.009738`, 0.011271`, 0.01285`, 0.013651`, 0.014468`, 0.016119`, 
   0.017797`, 0.019496`, 0.021211`, 0.022069`, 0.022934`, 0.024661`, 
   0.025522`, 0.026386`, 0.028103`, 0.029806`, 0.031492`, 0.033153`, 
   0.034785`, 0.036383`, 0.037942`, 0.039457`, 0.040924`, 0.042338`, 
   0.043695`, 0.04499`, 0.046221`, 0.047382`, 0.048472`, 0.049486`, 
   0.050421`, 0.051276`, 0.052047`, 0.052732`, 0.053329`, 0.053837`, 
   0.054255`, 0.05458`, 0.054813`, 0.054953`, 0.055`, 0.055`, 0.055`, 
   0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 
   0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 
   0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 
   0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 
   0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 
   0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 
   0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 
   0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 
   0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 
   0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 
   0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 
   0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 
   0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 
   0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.054313`, 0.0523`, 
   0.049099`, 0.044931`, 0.040082`, 0.034883`, 0.029689`, 0.024854`, 
   0.020707`, 0.017527`, 0.01553`, 0.015021`, 0.01485`, 0.014888`, 
   0.014998`, 0.015078`, 0.015172`, 0.015281`, 0.0154`, 0.015529`, 
   0.015665`, 0.015807`, 0.01595`};
    BorderValue = 0.0001;    
    AnglesBtm = 
         Table[
          FindRoot[
            Sin[x]/x == 
             1/(If[BallooningFile[[i, 4]] <= BorderValue, 
                 BorderValue, 
                 BallooningFile[[i, 4]]] + 1), {x, .001}][[1, 2]], 
          {i, 1, Length[BallooningFile[[All, 2]]], 1}]

The problem is that I get an error saying:

FindRoot::lstol: The line search decreased the step size to within tolerance specified by AccuracyGoal and PrecisionGoal but was unable to find a sufficient decrease in the merit function. You may need more than MachinePrecision digits of working precision to meet these tolerances.

After some searching on help pages I found out that at least one solution is complex. But I have no idea why on earth would that really be the case? Can somebody explain me this error?

EDIT: Ok, if I change BorderValue from 0.0001 to 0.001 everything works fine. I don't understand this. EDIT2: Corrected code.

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6
  • 2
    $\begingroup$ Please post self-contained, working code. $\endgroup$
    – Yves Klett
    Commented Sep 2, 2015 at 12:28
  • 1
    $\begingroup$ Downloads are awkward and potentially dangerous. Could you hardcode a working example? $\endgroup$
    – Yves Klett
    Commented Sep 2, 2015 at 12:43
  • 1
    $\begingroup$ @YvesKlett There it is, no downloading needed. $\endgroup$
    – skrat
    Commented Sep 2, 2015 at 12:49
  • 1
    $\begingroup$ It is the 3rd search, FindRoot[Sin[x]/x == 0.9994283269969577`, {x, 0.001}]. Increase WorkingPrecision to 20 or decrease PrecisionGoal to 6, following the hints in the warning message. Or simply check the results to see if errors are too great. It's just a warning that the error might be more than you want. $\endgroup$
    – Michael E2
    Commented Sep 2, 2015 at 13:03
  • 3
    $\begingroup$ I feel like this site would benefit from a generic Q&A on how to interpret standard warnings (that are not errors) from numeric solvers. $\endgroup$
    – Michael E2
    Commented Sep 2, 2015 at 13:06

1 Answer 1

Reset to default
8
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Use Sinc[x] rather than Sin[x]/x

BallooningFile = {0.`, 0.000136`, 0.000572`, 0.001152`, 0.001907`, 0.003004`, 
   0.004199`, 0.005479`, 0.006834`, 0.008256`, 0.008985`, 0.009738`, 
   0.011271`, 0.01285`, 0.013651`, 0.014468`, 0.016119`, 0.017797`, 0.019496`,
    0.021211`, 0.022069`, 0.022934`, 0.024661`, 0.025522`, 0.026386`, 
   0.028103`, 0.029806`, 0.031492`, 0.033153`, 0.034785`, 0.036383`, 
   0.037942`, 0.039457`, 0.040924`, 0.042338`, 0.043695`, 0.04499`, 0.046221`,
    0.047382`, 0.048472`, 0.049486`, 0.050421`, 0.051276`, 0.052047`, 
   0.052732`, 0.053329`, 0.053837`, 0.054255`, 0.05458`, 0.054813`, 0.054953`,
    0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 
   0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 
   0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 
   0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 
   0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 
   0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 
   0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 
   0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 
   0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 
   0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 
   0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 
   0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 0.055`, 
   0.055`, 0.055`, 0.055`, 0.055`, 0.054313`, 0.0523`, 0.049099`, 0.044931`, 
   0.040082`, 0.034883`, 0.029689`, 0.024854`, 0.020707`, 0.017527`, 0.01553`,
    0.015021`, 0.01485`, 0.014888`, 0.014998`, 0.015078`, 0.015172`, 
   0.015281`, 0.0154`, 0.015529`, 0.015665`, 0.015807`, 0.01595`};
BorderValue = 0.0001;
AnglesBtm = (x /. 
      FindRoot[Sinc[x] == 
        1/(If[# <= BorderValue, BorderValue, #] + 1), {x, .001}]) & /@ 
   BallooningFile;

ListLinePlot[AnglesBtm]

enter image description here

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1
  • 1
    $\begingroup$ Sinc is a good choice for this local example and avoids the precision loss of Sin[x]/x. (+1) $\endgroup$
    – Michael E2
    Commented Sep 2, 2015 at 13:10

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