How to Show x=tanh(y) is Equivalent to y=tanh^-1(x)?

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Homework Help Overview

The discussion revolves around demonstrating the equivalence between the equations x = tanh(y) and y = tanh-1(x). Participants are exploring the algebraic manipulation of hyperbolic functions and their inverses.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to manipulate the equation x = tanh(y) into the form y = tanh-1(x) using various algebraic steps. There are questions about the validity of certain transformations, particularly concerning logarithmic properties and algebraic substitutions.

Discussion Status

Some participants have provided hints and alternative methods for approaching the problem, such as suggesting substitutions and algebraic manipulations. There is an ongoing exploration of different approaches without a clear consensus on the best method to proceed.

Contextual Notes

Participants express uncertainty about specific algebraic steps and the application of logarithmic laws. There is mention of using different methods to arrive at the solution, indicating a variety of interpretations and approaches being considered.

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Homework Statement


attachment.php?attachmentid=68506&stc=1&d=1397116840.png


Homework Equations


(above)


The Attempt at a Solution


I know that x=tanh(y) can be shown as y=tanh^-1(x). The problem is how do i get from there to the next part. I'm kinda stuck here.

I can show that
x = sinh(y)/cosh(y)
x = (e^y - e^-y) / (e^y + e^-y)

x = lny (1+y)/(1-y)
I know something is terribly wrong in the last step. and I don't think this is the way to proceed with the question. Can you help me out?
 

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uzman1243 said:

Homework Statement


attachment.php?attachmentid=68506&stc=1&d=1397116840.png


Homework Equations


(above)

The Attempt at a Solution


I know that x=tanh(y) can be shown as y=tanh^-1(x). The problem is how do i get from there to the next part. I'm kinda stuck here.

I can show that
x = sinh(y)/cosh(y)
x = (e^y - e^-y) / (e^y + e^-y)

Up to here, it's correct. I assumed you canceled the 1/2 from top and bottom.

x = lny (1+y)/(1-y)
I know something is terribly wrong in the last step. and I don't think this is the way to proceed with the question. Can you help me out?

No idea how you made the jump to the last step. No law of logs allows that.

Hint: let z = ey and solve algebraically for z in terms of x. Then take the log of both sides to get y in terms of x.

Remember that e-y is the reciprocal of ey.
 
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x = (ey - e-y)/(ey + e-y)
Another easy way to finish from there is:
xey + xe-y = ey - e-y
and then multiply each of the 4 terms by ey,
which will lead (after a couple of steps) to
(x-1) e2y = (-1-x)
and the rest is quite easy.
 
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Curious3141 said:
Up to here, it's correct. I assumed you canceled the 1/2 from top and bottom.



No idea how you made the jump to the last step. No law of logs allows that.

Hint: let z = ey and solve algebraically for z in terms of x. Then take the log of both sides to get y in terms of x.

Remember that e-y is the reciprocal of ey.

So far I have got x = (z^2 - 1) / (z^2 +1)
Im stuck from there though.

I was able to get the answer using az_lender method but I want to know how to do it using yours.
Can you guide me from where I am stuck? thank you
 
uzman1243 said:
So far I have got x = (z^2 - 1) / (z^2 +1)
Im stuck from there though.

I was able to get the answer using az_lender method but I want to know how to do it using yours.
Can you guide me from where I am stuck? thank you

It's pretty much the same sequence of steps.
 
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