Finding Real Solutions for x-tanh ax = 0 with a>1

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Discussion Overview

The discussion revolves around the equation x - tanh(ax) = 0, specifically investigating how many real solutions exist for a > 1. Participants explore various mathematical approaches and reasoning related to the behavior of the function involved.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant asks how many real solutions the equation has for a > 1.
  • Another suggests drawing the graph for 0 ≤ x ≤ 1 to analyze the behavior of the function.
  • There is a discussion about the slope of the function tanh(ax), with multiple participants questioning and refining their understanding of how to find it.
  • Some participants mention that x = 0 is a root of the equation, leading to a discussion about the nature of the solutions.
  • One participant proposes that the first derivative of the equation gives the slope, prompting further exploration of its implications.
  • There are conflicting statements about the correct derivative of tanh(ax), with some participants suggesting it is sech²(ax) while others challenge this assertion.
  • A participant expresses difficulty in understanding the mathematics involved and requests clarification in light of an upcoming entrance exam.
  • Another participant describes the behavior of the function at specific points, noting that when x = 0, y = 1, and as x increases, y decreases.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct derivative of tanh(ax) or the implications for the number of real solutions to the equation. Multiple competing views and uncertainties remain throughout the discussion.

Contextual Notes

There are unresolved mathematical steps regarding the derivative of tanh(ax) and its implications for the solutions of the equation. The discussion also reflects varying levels of mathematical understanding among participants.

physicsblr
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Can anybody help me with this equation?

I need answer for the question, how many real solutions does the equation x-tanh ax = 0 have for a>1?
 
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welcome to pf!

hi physicsblr! welcome to pf! :smile:

draw the graph, for 0 ≤ x ≤ 1 (since tanh can't be ≥ 1) …

what is the slope of tanhax ? :wink:​
 
Is it a?
 
I have a doubt. x=0 is a root of this eq and since its a linear eq, it has only 1 root?
 
tiny-tim said:
what is the slope of tanhax ? :wink:​
physicsblr said:
Is it a?

no, of course not :redface:

how do we find the slope of a graph?​
 
oh! slope is zero.
 
how do we find the slope of a graph?
 
first derivative of the equation = 0 gives the slope.
 
physicsblr said:
first derivative of the equation = 0 gives the slope.

well, first derivative of the equation gives the slope

so the slope of tanhax is … ? :smile:​
 
  • #10
1-tanh2 ax; so for that eq. tanh2 ax = 0 is final result.
 
  • #11
(try using the X2 button just above the Reply box :wink:)
physicsblr said:
1-tanh2 ax

(= sech2ax)

no

use the chain rule​
 
  • #12
am getting 2 cases. either a=1 or sech2 ax=1.
 
  • #13
physicsblr said:
Can anybody help me with this equation?

… how many real solutions does the equation x-tanh ax = 0 have for a>1?
physicsblr said:
am getting 2 cases. either a=1 or sech2 ax=1.

uhh? :redface:

what question are you answering? :confused:​

(and you still need to find the correct derivative of tanh(ax))
 
  • #14
Am unable to get it properly, am no good in mathematics. If u can pls explain it and give an answer since I have an entrance exam next week, I will be very thankful.
 
  • #15
i] draw the graph of y = x for 0≤x≤1 …

that'll be a square with a diagonal​

ii] find the derivative of y = tanh(ax)

(that is not sech2(ax) … do it again, using the chain rule)

iii] the derivative equals the slope of the graph of y = tanh(ax), so plot the values at x = 0 and 1, and use the slope to find how it starts and finishes …

what do you get? :smile:​
 
  • #16
When x=0, y=1 and there on as x is increasing, value of y is decreasing..
 
  • #17
physicsblr said:
When x=0, y=1 and there on as x is increasing, value of y is decreasing..

uhh? :confused:

what was unclear about? …​
tiny-tim said:
i] draw the graph of y = x for 0≤x≤1 …

that'll be a square with a diagonal​

ii] find the derivative of y = tanh(ax)

(that is not sech2(ax) … do it again, using the chain rule)

iii] the derivative equals the slope of the graph of y = tanh(ax), so plot the values at x = 0 and 1, and use the slope to find how it starts and finishes …

what do you get? :smile:​
 

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