View Full Version : Minimize cosh(x)+tanh(x)
Hi, I need to find the Y-coordinate of the extrema of cosh(x)+tanh(x),
Thanks!
dextercioby
Dec2-04, 11:49 AM
Hi, I need to find the Y-coordinate of the extrema of cosh(x)+tanh(x),
Thanks!
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
(\cosh x+\tanh x)'=0 which gives an transcedental equation in the hyperbolic functions \cosh x and \sinh x .Substitute the definiton of these functions and then make the substitution \exp x = \lambda .The equation for lambda is as i said and look for its positive roots.
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...
Tom Mattson
Dec3-04, 03:50 PM
I'd hunt the root down in 2 steps:
1. The rational root test (http://www.purplemath.com/modules/rtnlroot.htm)
and if no rational roots then...
2. Newton's Method (http://www.math.sc.edu/cgi-bin/sumcgi/Newton.pl)
Here is my attempt at the problem:
y = \cosh x + \tanh x
y \cosh x = \cosh^2 x + \sinh x
y' \cosh x + y \sinh x = 2 \cosh x \sinh x + \cosh \x
y'=0
\cosh^2 x + \sinh x = 2 \cosh x \sinh \x + \cosh \x
\cosh^2 x + (\sinh x - \cosh x) - 2 \cosh x \sinh x = 0
\frac{e^{2x} + 2e^x - 2e^{-2x}}{4} + e^{-x} - \frac{e^{2x} - e^{-2x}}{2} = 0
e^{2x} + 2e^x - e^{-2x} + 4e^{-x} - 2e^{2x} + 2e^{-2x}=0
-e^{2x} + 2e^x + 4e^{-x} + e^{-2x}=0
-e^{4x} + 2e^{3x} + 4e^x + 1=0
Have I made a mistake or something? Looks pretty hard to me (although solvable exactly).
dextercioby
Dec3-04, 05:08 PM
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...
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.
vBulletin® v3.8.7, Copyright ©2000-2012, vBulletin Solutions, Inc.