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How I can solve the equation :

$$R -\tan^{-1}[(m1/m3)*(k3/k1)] - \tan^{-1}[(m1/m2)*(k2/k1)]=0$$

I tried on it using findroot as following :

$$
F = 0.56; 
w0 = 4*10^9*3.14;
wp = 10*10^9*3.14;
gama = 0.03*wp;
T = 0.03*w0;
e1 = -3.7;
m1 = -1;
e2 = 1;
m2 = 1;
e3 = -9.5;
m3 = 1;
w = 4.6*10^9*3.14;
c = 3*10^8;
k0 = w/c;
k1 = Sqrt[(k0^2*e1*m1) - b^2];
k2 = Sqrt[b^2 - (k0^2*e2*m2)];
k3 = Sqrt[b^2 - (k0^2*e3*m3)];
R = 3.45;

bbValue = 
 FindRoot[R - ArcTan[(m1/m3)*(k3/k1)] - ArcTan[(m1/m2)*(k2/k1)] , {b,0.5}]$$

But it give me an error :

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.

Regards

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8
  • $\begingroup$ How I can solve the equation Solve it for what? $\endgroup$
    – Ghoster
    Commented Feb 2, 2023 at 8:06
  • $\begingroup$ @Ghoster i want to solve it for b $\endgroup$
    – A.J.H
    Commented Feb 2, 2023 at 8:08
  • $\begingroup$ There’s no $b$ in your equation. What is the point of writing the equation you are solving but not giving all the necessary info to check whether your code is correct? $\endgroup$
    – Ghoster
    Commented Feb 2, 2023 at 8:11
  • $\begingroup$ it is not an error, just a warning on accuracy of root found. b -> 2.34938*10^-13 - 3.75897 I} $\endgroup$
    – Nasser
    Commented Feb 2, 2023 at 8:35
  • $\begingroup$ @Ghoster there are b in k1 , k2 and k3 ... $\endgroup$
    – A.J.H
    Commented Feb 2, 2023 at 8:46

1 Answer 1

3
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Have a look at this plot

Plot[{Re[R - ArcTan[(m1/m3)*(k3/k1)] - ArcTan[(m1/m2)*(k2/k1)]], 
  Im[R - ArcTan[(m1/m3)*(k3/k1)] - ArcTan[(m1/m2)*(k2/k1)]]}, {b, -120, 120}]

Which root do you want to find?

enter image description here

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