Number of roots of tanh(ax) = x

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SUMMARY

The discussion centers on the equation x = tanh(βx) and how the number of solutions changes as β varies from 1/2 to 3/2. It is established that when β < 1, the equation has only one root at x = 0. Conversely, when β > 1, the equation has three roots: x = 0, x = r, and x = -r, where 0 < r < 1. The behavior of the function tanh(βx) is characterized by its slope at the origin being equal to β, influencing the intersection points with the line y = x.

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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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