How to increase the precision to get the correct roots at the boundaries?

I want to solve an equation and Plot a graph of δ(τ)\delta(\tau), where 0<τ,δ<10<\tau,\delta<1. In principal δ(0)=1,δ(1)=0\delta(0)=1,\delta(1)=0. However, when I solve the equation, the points near τ=1\tau=1 can't get exactly. Below is my sample code: Clear["Global`*"]; eps = 10^-13; stepSize = 0.001; omegaD = 0.5; df = 0.008; c1 = omegaD/df; c0 = c0 /. FindRoot[1/c0 Log[Cosh[c1 c0]] == Sqrt[c1^2 + 1] - 1, {c0, Log[2]}]; list = Table[{Ï„, δ /. FindRoot[Ï„ Log[Cosh[c0 Sqrt[c1^2 + \delta^2]/Ï„]/Cosh[c0 δ/Ï„]] == c0 (Sqrt[c1^2 + 1] - 1), {δ, 1}]}, {Ï„, eps, 1.0,stepSize}]; p1 = ListLinePlot[list] Program complains that: 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.@ Also clearly shown in the graph, the points near τ=1\tau=1 is missing. So I tried to add an option WorkingPrecision->30 at the end of the FindRoot function, the program complain the following this time:

the precision of the argument function … is less than WorkingPrecision (30.)

Even I decrease 30 to other number, it still complaining.

Question is how to add the missing points correctly at the boundary?

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The precision of your arguments must be at least as high as the WorkingPrecision you are requesting. You should change your parameters to exact values, e.g. stepSize = 1/1000; omegaD = 1/2; df = 8/1000; and retry.
– MarcoB
Oct 26 ’15 at 15:41

  

 

As a sidenote to what @MarcoB correctly said, you can also just write 0.01//Rationalize for instance. That might be more convenient for you
– Lukas
Oct 26 ’15 at 16:30

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1 Answer
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First, the Table does not run up to t == 1.:

Table[Ï„, {Ï„, 1/20, 1, stepSize}]
(* {1.*10^-13, 0.001,…, 0.998, 0.999} *)

Update —
Now with the corrected formula (a typo fixed in the OP), let’s try

list = Table[{τ, δ /.
FindRoot[Ï„ Log[Cosh[c0 Sqrt[c1^2 + δ^2]/Ï„]/Cosh[c0 δ/Ï„]] == c0 (Sqrt[c1^2 + 1] – 1),
{δ, 1}]},
{Ï„, Union[Range[eps, 1.0, stepSize], {1.}]}]
p1 = ListLinePlot[list]

We get the desired plot.

  

 

IF you do it analytically, you may find it is indeed 0. That is why I think I need more precision. I will show the analytical derivation when I am at computer.
– buzhidao
Oct 27 ’15 at 4:20

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It’s my mistake, I wrongly typed a character in my code.
– buzhidao
Oct 27 ’15 at 6:53

  

 

@buzhidao Thanks. I’ve updated the answer to your new input.
– Michael E2
Oct 27 ’15 at 11:34

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@downvoter, Why? It would be helpful to the site to indicate a reason you think there is an issue with the answer!
– Michael E2
Oct 27 ’15 at 11:36

  

 

Thanks, I’ve also got that figure after I spot that misprint.
– buzhidao
Oct 27 ’15 at 12:18