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

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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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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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