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

  • Thread starter uzman1243
  • Start date
In summary: From x = (z^2 - 1) / (z^2 +1), you can multiply both sides by (z^2 + 1) to get x(z^2 + 1) = z^2 - 1. Then expand and rearrange to get xz^2 + x = z^2 - 1. Then bring all the terms with z to one side and all the constants to the other: xz^2 - z^2 = -1 - x. Then factor out z^2: z^2(x-1) = -1 - x. Finally, divide both sides by (x-1) and take the log: z = ln(-1-x)/(x-1
  • #1
uzman1243
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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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  • #2
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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  • #3
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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  • #4
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
 
  • #5
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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Likes 1 person

What is the formula for x=tanh(y)?

The formula for x=tanh(y) is x = (e^y - e^-y) / (e^y + e^-y).

What does the tanh function represent?

The tanh function represents the hyperbolic tangent function, which is a mathematical function used to map real numbers to values between -1 and 1.

What is the relationship between x and y in the equation x=tanh(y)?

The equation x=tanh(y) represents a relationship between the independent variable y and the dependent variable x, where x is equal to the hyperbolic tangent of y.

What are the properties of the tanh function?

The tanh function is an odd function, meaning that it is symmetric about the origin. It also has a range of -1 to 1, and its graph is asymptotic to the lines y=1 and y=-1 as x approaches positive and negative infinity, respectively.

How is the tanh function used in science?

The tanh function is commonly used in science to model and analyze data that exhibits a sigmoidal (S-shaped) pattern, such as in biology, economics, and physics. It is also used in artificial neural networks and in the study of differential equations.

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