2
$\begingroup$

I have a very complicated function but only of 1 variable. I want to find the first value for which that function is zero. Mathematica can easily plot it:

func = Det[coeffMatrix]; Plot[func, {\[Beta]1, 0, 3}]

enter image description here

From that plot, one can easily see that the first value would be ~2.556.

To show that $\beta_1$ = 2.556 is actually the approximate solution:

func /. \[Beta]1 -> 2.556

-0.00139597

However, when I try to find it numerically:

NSolve[func == 0 && 0 < \[Beta]1 < 10, \[Beta]1]

...it just runs and runs and runs and never gives an answer. Why ? and how can I fix it ?

The complete code

constants = {b1 -> (-(Cosh[ 0.68*\[Beta]1]*(0.6553600000000004*Cos[0.15*\[Beta]1]* Cos[0.68*\[Beta]1] - 0.6553600000000004*Cos[0.68*\[Beta]1]* Cosh[0.15*\[Beta]1] + 1.2621440000000002*Sin[0.15*\[Beta]1]*Sin[0.68*\[Beta]1] + 0.7378559999999998*Sin[0.68*\[Beta]1]* Sinh[0.15*\[Beta]1])) + Cos[0.68*\[Beta]1]*(-0.7378559999999998*Sin[0.15*\[Beta]1] - 1.2621440000000002*Sinh[0.15*\[Beta]1])* Sinh[0.68*\[Beta]1])/(Cosh[ 0.68*\[Beta]1]*(-0.6553600000000004*Cos[0.68*\[Beta]1]* Sin[0.15*\[Beta]1] + 1.2621440000000002*Cos[0.15*\[Beta]1]*Sin[0.68*\[Beta]1] + 0.7378559999999998*Cosh[0.15*\[Beta]1]*Sin[0.68*\[Beta]1] - 0.6553600000000004*Cos[0.68*\[Beta]1]*Sinh[0.15*\[Beta]1]) + Cos[0.68*\[Beta]1]*(0.7378559999999998*Cos[0.15*\[Beta]1] + 1.2621440000000002*Cosh[0.15*\[Beta]1])*Sinh[0.68*\[Beta]1]), b2 -> (2.5* Sec[0.68*\[Beta]1]*(Cosh[ 0.68*\[Beta]1]*(-0.26214400000000015 + 0.26214400000000015*Cos[0.15*\[Beta]1]* Cosh[0.15*\[Beta]1] - Sin[0.15*\[Beta]1]*Sinh[0.15*\[Beta]1]) + 0.8*(Cosh[0.15*\[Beta]1]*Sin[0.15*\[Beta]1] - Cos[0.15*\[Beta]1]*Sinh[0.15*\[Beta]1])* Sinh[0.68*\[Beta]1]))/((0.7378559999999998* Cos[0.15*\[Beta]1] + 1.2621440000000002*Cosh[0.15*\[Beta]1])*Sinh[0.68*\[Beta]1] + Cosh[0.68*\[Beta]1]*(-0.6553600000000004*Sin[0.15*\[Beta]1] - 0.6553600000000004*Sinh[0.15*\[Beta]1] + 1.2621440000000002*Cos[0.15*\[Beta]1]*Tan[0.68*\[Beta]1] + 0.7378559999999998*Cosh[0.15*\[Beta]1]*Tan[0.68*\[Beta]1])), d2 -> (2.5* Sech[0.68*\[Beta]1]*(-0.26214400000000015*Cos[0.68*\[Beta]1] + 0.8*Cosh[ 0.15*\[Beta]1]*(0.3276800000000002*Cos[0.15*\[Beta]1]* Cos[0.68*\[Beta]1] + Sin[0.15*\[Beta]1]* Sin[0.68*\[Beta]1]) + (Cos[0.68*\[Beta]1]* Sin[0.15*\[Beta]1] - 0.8*Cos[0.15*\[Beta]1]*Sin[0.68*\[Beta]1])* Sinh[0.15*\[Beta]1]))/(-0.6553600000000004* Cos[0.68*\[Beta]1]*Sin[0.15*\[Beta]1] + 1.2621440000000002*Cos[0.15*\[Beta]1]*Sin[0.68*\[Beta]1] - 0.6553600000000004*Cos[0.68*\[Beta]1]*Sinh[0.15*\[Beta]1] + 0.7378559999999998*Cos[0.15*\[Beta]1]*Cos[0.68*\[Beta]1]* Tanh[0.68*\[Beta]1] + Cosh[0.15*\[Beta]1]*(0.7378559999999998*Sin[0.68*\[Beta]1] + 1.2621440000000002*Cos[0.68*\[Beta]1]*Tanh[0.68*\[Beta]1]))} matrix = {{a1 (Sin[u \[Beta]1] - Sinh[u \[Beta]1]), b1 (Cos[u \[Beta]1] - Cosh[u \[Beta]1]), -b2* Cos[y*\[Theta]*\[Beta]1], -d2* Cosh[y*\[Theta]*\[Beta]1]}, {a1 (Cos[u \[Beta]1] - Cosh[u \[Beta]1]), b1 (-Sin[u \[Beta]1] - Sinh[u \[Beta]1]), b2*\[Theta]*Sin[y*\[Theta]*\[Beta]1], -d2*\[Theta]* Sinh[y*\[Theta]*\[Beta]1]}, {a1 (-Sin[u \[Beta]1] - Sinh[u \[Beta]1]), b1 (-Cos[u \[Beta]1] - Cosh[u \[Beta]1]), b2*\[Alpha]^4*\[Theta]^2* Cos[y*\[Theta]*\[Beta]1], -d2*\[Alpha]^4*\[Theta]^2* Cosh[y*\[Theta]*\[Beta]1]}, {a1 (-Cos[u \[Beta]1] - Cosh[u \[Beta]1]), b1 (Sin[u \[Beta]1] - Sinh[u \[Beta]1]), -b2*\[Alpha]^4*\[Theta]^3* Sin[y*\[Theta]*\[Beta]1], -d2*\[Alpha]^4*\[Theta]^3* Sinh[y*\[Theta]*\[Beta]1]}}; testingParam = { \[Theta] -> 0.8, \[Alpha] -> 0.8, u -> 0.15, y -> 1 - 0.15} ; coeffMatrix = (matrix /. a1 -> 1) /. constants /. testingParam ; func = Det[coeffMatrix]; NSolve[func == 0 && 0 < \[Beta]1 < 10, \[Beta]1]
Improve this question
$\endgroup$
2
  • $\begingroup$ Since you can use Plot[] on your function, use the MeshFunctions option, like what was done here. $\endgroup$ Commented Oct 6, 2018 at 9:08
  • $\begingroup$ @J.M.issomewhatokay. Thank you for your comment and suggestion. However, I am not such an advanced user. Would you mind to post an answer of how to use the MeshFunctions ? I did not quite understand it from the example following your link. Thanks ! :) $\endgroup$ Commented Oct 6, 2018 at 9:39

2 Answers 2

Reset to default
4
$\begingroup$

From: About multi-root search in Mathematica for transcendental equations

f[\[Beta]1_] = Det[coeffMatrix]; zeros = Reap[ NDSolve[{y'[x] == D[f[x], x], WhenEvent[y[x] == 0, Sow[{x, y[x]}]], y[1] == f[1]}, {}, {x, 3, 0.01}]][[-1, 1]] Plot[f[x], {x, 0, 3}, Epilog -> {PointSize[Medium], Red, Point[zeros]}]

{{2.61534, -5.0246*10^-18}}

enter image description here

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thanks a lot. This seems to work. I still need to understand it though... $\endgroup$ Commented Oct 6, 2018 at 9:40
3
$\begingroup$ 
FindRoot[func == 0, {\[Beta]1, 2.566}] {\[Beta]1 -> 2.61535}
Improve this answer
$\endgroup$
6
  • $\begingroup$ Thanks ! Do you always need an initial guess ? $\endgroup$ Commented Oct 6, 2018 at 11:25
  • $\begingroup$ When there are many roots, it is advisable to specify the one that you want to get. Otherwise there may be other roots. $\endgroup$ Commented Oct 6, 2018 at 11:45
  • $\begingroup$ I see, but most of the times I just know that I want to get the first root, but I don't know where it is... $\endgroup$ Commented Oct 6, 2018 at 11:49
  • $\begingroup$ In this case, this is not the first root, but the second. $\endgroup$ Commented Oct 6, 2018 at 11:52
  • $\begingroup$ Sorry, yes, I meant the first non-zero root. $\endgroup$ Commented Oct 6, 2018 at 11:52

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.