Using FindRoot in a Table of Iterations

Question

So I have a tanh curve and a straight line that cross at the point:

msol=FindRoot[Tanh[x] == (T/Tc03) x, {x, 1}]

This gives me the roots that I need at a certain temperature, T. I now need these roots to iteratively get a value for chi at different temperatures between 0K and 4K. I am currently using a Table to do this like so:

(*constants*)

mu = 6.0281*10^-23;
Tc03 = 0.00459;
k = 1.3806488*10^-23;

(*calculation*) 

chi = Table[(mu^2 Sech^2[(Tc03/(mu T)) FindRoot[
     Tanh[x] == (T/Tc03) x, {x, 1}]])/(k T (1 - (Tc03/
      T) Sech^2[(Tc03/(mu T)) FindRoot[
        Tanh[x] == (T/Tc03) x, {x, 1}]])), {T, 1*10^-6, 4, 1}]

This gives me the values but where the FindRoot is, it says x->y where y is the root at that temperature, i.e. not numerical values that I can plot. The aim is to plot T vs chi iteratively and then join up the dots and get the form of the curve.

The table also does not tell me what value of T the iteration has taken place at, which I need for the plot.

Any help would be much appreciated.

FindRoot[] returns replacement rules. You need to understand how to use them to program in Mathematica — Dr. belisarius, Mar 13 '15 at 19:29
See for example here: http://mathematica.stackexchange.com/a/18706/193 and please read all the answers therein — Dr. belisarius, Mar 13 '15 at 19:30
Welcome to Mathematica.SE! I suggest the following: 0) Browse the common pitfalls question 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Read the [faq]! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! — Dr. belisarius, Mar 13 '15 at 19:34

score 0 · Accepted Answer · answered Mar 15 '15 at 15:15

Your expression for chi should be rewritten as

chi = 
  Table[
    With[{
      val = Sech[Tc03/(mu T) ((T/Tc03) x /. 
              FindRoot[Tanh[x] == (T/Tc03) x, {x, 1}])]},
      (mu^2 val^2)/(k T (1 - (Tc03/T) val^2))], 
    {T, 1.*10^-6, 4., 1.}]

{-5.73535*10^-20, 2.64408*10^-22, 1.319*10^-22, 8.78661*10^-23}

However, FindRoot will always find the root 0 at x = 0, val will always evaluate to 1., so I question your formulation of your problem.

Using FindRoot in a Table of Iterations

1 Answers1