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Minimize cosh(x)+tanh(x)

  1. Dec 2, 2004 #1
    Hi, I need to find the Y-coordinate of the extrema of cosh(x)+tanh(x),

  2. jcsd
  3. Dec 2, 2004 #2


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    You should be able to reduce everything to a fourth order algebraic equation with integer coefficents.
    The condition for extrema (giving the critical points of the graphic) is
    [tex] (\cosh x+\tanh x)'=0 [/tex] which gives an transcedental equation in the hyperbolic functions [tex] \cosh x [/tex] and [tex] \sinh x [/tex].Substitute the definiton of these functions and then make the substitution [tex] \exp x = \lambda [/tex].The equation for lambda is as i said and look for its positive roots.
  4. Dec 2, 2004 #3
    Yes, I found x^3+x+1=0 where x=sinh(x), but there is no simple root, and I need a quite simple solution...
  5. Dec 3, 2004 #4

    Tom Mattson

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    I'd hunt the root down in 2 steps:

    1. The rational root test

    and if no rational roots then...

    2. http://www.math.sc.edu/cgi-bin/sumcgi/Newton.pl [Broken]
    Last edited by a moderator: May 1, 2017
  6. Dec 3, 2004 #5


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    Here is my attempt at the problem:

    [tex]y = \cosh x + \tanh x[/tex]

    [tex]y \cosh x = \cosh^2 x + \sinh x[/tex]

    [tex]y' \cosh x + y \sinh x = 2 \cosh x \sinh x + \cosh \x[/tex]


    [tex]\cosh^2 x + \sinh x = 2 \cosh x \sinh \x + \cosh \x[/tex]

    [tex]\cosh^2 x + (\sinh x - \cosh x) - 2 \cosh x \sinh x = 0[/tex]

    [tex]\frac{e^{2x} + 2e^x - 2e^{-2x}}{4} + e^{-x} - \frac{e^{2x} - e^{-2x}}{2} = 0[/tex]

    [tex]e^{2x} + 2e^x - e^{-2x} + 4e^{-x} - 2e^{2x} + 2e^{-2x}=0[/tex]

    [tex]-e^{2x} + 2e^x + 4e^{-x} + e^{-2x}=0[/tex]

    [tex]-e^{4x} + 2e^{3x} + 4e^x + 1=0[/tex]

    Have I made a mistake or something? Looks pretty hard to me (although solvable exactly).
  7. Dec 3, 2004 #6


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    You're right.This is the quations u should be getting.There is at least one solution in R(which should be negative and in the interval (-1,0)),but to find all of them u have to apply Cardano's formulas.My guess is the other two are complex.And sinh(x) can take only real values,so the path to finding the extremas is hopefully clear from now.
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