Mathematica Mathematica does not like hyperbolic functions

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The discussion revolves around solving the equation cosh(x) = n*x in Mathematica, which presents challenges due to errors indicating non-algebraic dependencies. The critical value of n is identified as approximately 1.51, determining the number of solutions for x. The user initially struggles with Solve and NSolve functions, which fail to provide the expected results. A successful approach is found by using the derivatives of the functions at their intersection points, leading to the equation tanh(x) = 1/x. This method ultimately resolves the issue, demonstrating an effective way to utilize Mathematica for hyperbolic functions.
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[SOLVED] Mathematica does not like hyperbolic functions

So, consider the equation cosh(x)=n*x

For a given n, the equation has 0, 1, or 2 possible values of x. If n is below the critical value, the equation has no solutions. If n is above the critical value, the equation has two solutions. And if n is exactly the critical value, the equation has one solution. My goal is to use Mathematica to show that the critical value is approximately 1.51.

Theoretically, the line Length[Solve[Cosh[x] == n*x, x]] should give the number of solutions for a given n. Then I can make a table of n's and find the point where the number of solutions goes from 0 to 2.

Unfortunately, I keep getting the error:
Solve::tdep: The equations appear to involve the variables to be solved for in an essentially non-algebraic way.

NSolve has the exact same problem. FindRoot always gives exactly one solution, whether there are zero or two solutions to the equation. Is there a way to make Mathematica more cooperative, or another way to go about this problem? Since the TI-89 can handle this problem (but is too slow to be useful), it seems like Mathematica should be able to as well.

Thanks for your help!
 
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Since \cosh x and nx have one intersection point, their derivatives are equal. Along with the original equation, this gives two equations for the two unknowns n and x (at intersection).

\sinh x = n

\cosh x = nx.

Divide these equations and get

tanh x = \frac{1}{x}.

Solve this numerically for x (use only the positive solution), and, to find n, plug this solution into the top equation.
 
Last edited:
Sweet, it worked. Thanks!
 

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