Number of roots of tanh(ax) = x

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Homework Help Overview

The discussion centers around the equation x = tanh(βx) and how the number of solutions changes as the parameter β varies from 1/2 to 3/2. Participants explore the behavior of the functions involved and their intersections.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the shapes of the graphs of y = tanh(βx) and y = x, questioning how these shapes affect the number of intersections, or roots, of the equation.

Discussion Status

Some participants have offered insights into the slopes of the functions at the origin and how these slopes influence the number of roots. There is an ongoing exploration of the implications of varying β, with multiple interpretations being considered.

Contextual Notes

Participants note constraints such as the lack of graphing tools and the nature of the examination question, which may limit their approaches to finding solutions.

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Homework Statement



what happens to the number of solutions of the equation
x = \tanh(\beta x)

When
\beta is varied from \frac{1}{2} to \frac{3}{2}
[/tex]



a) unchanged
b) increase by 1
c) increase by 2
d) increase by 3

Homework Equations



\tanh(ax) =.... -\frac{17}{315} \, a^{7} x^{7} + \frac{2}{15} \, a^{5} x^{5} -<br /> \frac{1}{3} \, a^{3} x^{3} + a x


tanh(ax)= \frac{e^{ax}-e^{-ax}}{e^{ax}+e^{-ax}}<br />

The Attempt at a Solution



Tried to Apply the two above formulas without any success. This is an examination question and too bad, Graphing and programming calculators are not allowed :(
 
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hi shakgoku! :smile:

what's the shape of y = tanhßx?

what's the shape of y = x? :wink:
 
hi tiny-tim,

The shape of tanh(βx) is nearly straight line near origin and has horizontal asymptotes at y = 1 and y = -1
The shape of x is straight line through origin , slope = 45 degrees.

so x = 0 is definitely a root. how to find it there are more roots?
 
hi shakgoku! :smile:

(just got up :zzz: …)
shakgoku said:
The shape of tanh(βx) is nearly straight line near origin and has horizontal asymptotes at y = 1 and y = -1

yes, but which ways does it bend? :wink:
 
What is the slope of the line tangent to tanh(ßx) at the origin?
 
What is the slope of the line tangent to tanh(ßx) at the origin?

yes, but which ways does it bend?

Thanks , its so obvious now. The slope is at x = 0 is beta. So, if beta<1 the slope keeps decreasing and never gets a chance to intersect y = x again.

But beta > 1, tanh(beta*x) stays > (y = x) but due to decreasing slope, intersects y = x again at two places x = + and - r
(if r is a root).


I've also found out that 0 < r < 1

by taking f = tanh(beta*x) - x and checking its value for 0 and 1. for one value its +ve and for other it was -ve.
 
shakgoku said:
Thanks , its so obvious now. The slope is at x = 0 is beta. So, if beta<1 the slope keeps decreasing and never gets a chance to intersect y = x again.

But beta > 1, tanh(beta*x) stays > (y = x) but due to decreasing slope, intersects y = x again at two places x = + and - r

yup!

just sometimes we can prove something that looks complicated with hardly any maths! :biggrin:
 
hi!
I didnt get that beta>1 part. You mean to say, the equation has 3 roots-0,1,-1? Please let me know, am too stuck with the same problem.
 
draw the graph!

what does it look like?​
 

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