Take the 2-minute tour ×
Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

I am using Ted Ersek's RootSearch function in Mathematica 9.0 (http://library.wolfram.com/infocenter/Demos/4482/) to create a function that I am using for graphing. Now, I'd like to use this function inside of a FindRoots call to find specific solutions.

To begin, we have a function f[t_, Rb_, N2_]:

 f[t_, Rb_, N2_] =(4.22153*10^6 (45.5182 + 5.226*10^-6 N2^2 + Rb) (0.106 - 
                     0.106 E^(-9.43396 t) - Sin[120 \[Pi] t]/(120 \[Pi])))/N2^2

Then we define the tsat2[Rb_, N2_] function using Ersek's RootSearch function to find the t values where f[t,Rb,N2] = 0.94 and return the minimum value.

tsat2[Rb_, N2_] := N[60*Min[t /. RootSearch[f[t, Rb, N2] - 0.94` == 0, {t, 0, 1}]]]

The above RootSearch function works great, it's this next step that's the trouble. We play a little trick to turn tsat2 into a function of only Rb to satisfy some other conditions to obtain tsat2[Rb,97.65*Rb] Here's the problem: I need to find find f[Rb,97.65*Rb]==1.5 using numerical methods. Before I go and write my own numerical solver, I'd like to see if I can get FindRoots to work:

FindRoots[tsat2[Rb, 97.65*Rb] - 1.5 == 0, {Rb, 10}]

However the problem is Mathematica doesn't even seem to evaluate this ambitious function call and only returns:

FindRoots[-1.5 + 60. (t /. $Failed) == 0, {Rb, 10}]

I suspect that somehow the tsat2 function is not actually being evaluated inside the FindRoot statement, or more specifically, the RootSearch function isn't being passed the values from the higher FindRoots function call. I have tried this also using the RootSearch function (so we essentially have a nested RootSearch) and received the same error:

In[203]:= RootSearch[tsat2[Rb, Rb 97.65] - 1.5 == 0, {Rb, 10}]

Out[203]= RootSearch[-1.5 + 60. (t /. $Failed) == 0, {Rb, 10}]

My question is: How do I get the tsat2 function to act like any other function evaluation inside the FindRoots call?

Thanks in advance!

share|improve this question

1 Answer 1

It turns out that this problem is solved using the solution here.

The answer is that FindRoots needs to have the option Evaluated->False to ensure that the nested RootSearch is evaluated before the FindRoots[expr] takes place.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.