What is the connection between Arctanh x and the logarithmic function?

[paulfr]

I can not seem to figure out why Arctanh x [ Hyperbolic Arctan ]
can be expressed as

Arctanh x = (1/2) [ Log (1+x) - Log (1-x) ]

Note; I know the 1/2 means the expression is a square root
of the ratio of the 2 binomials and has been taken out with the Power Rule

Can anyone show me the connection ?
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Note the above has been corrected for errors from the original post

 

[phyzguy]

You didn't write it correctly. What you wrote reduces to ArcTan(x) = (1/2) Log(2), which is clearly not correct. I think the correct expression is:
\tan ^{-1}(x)=i \log \left(\sqrt{\frac{1-i x}{1+i x}}\right) = \frac{i}{2}\log \left(\frac{1-i x}{1+i x}\right) = \frac{i}{2} (\log (1-i x)-\log (1+i x))

As to why, the best way I know to show the equality is to expand both sides in a Taylor series. Both sides can be expressed as:
\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)}

 

[jackmell]

I think you mean:

\arctan(z)=\frac{i}{2}\log\frac{i+z}{i-z}

Then just let:

w=\frac{\sin(z)}{\cos(z)}

then expand sin and cos in their complex form then solve for z using just basic algebra.

 

[paulfr]

Darn, I was hoping I could correct this before anyone read it.

Sorry
I meant to write arc HYPERBOLIC tangent
And I did write it incorrectly

Arctanh = 1/2 [ Log (1+x) - Log (1-x) ] = Log [Sqrt { (1+x) / (1 - x) } ]

Again please forgive my carelessness.

And thanks for any insight you can offer.

 

[arildno]

Well, we have:
y=tanh(x)=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}=\frac{(e^{x})^{2}-1}{(e^{x})^{2}+1}\to{y}=\frac{u^{2}-1}{u^{2}+1}, u=e^{x}
Now, solve for u in terms of y (almost trivial), then solve for x in terms of y (also trivial).

 

[paulfr]

Arildno
Thanks. That does lead directly to the Log expression.

The similar sequence to find the Log expresion for Arcsinh x and Arccosh x
is not so trivial though.

 

[paulfr]

Sad to say I still can not show or prove this equation to be true.

How can one show that arcsinh x is expressible as this logarithm ?

Thanks for your help

 

[arildno]

Well, let:
y=\frac{e^{x}-e^{-x}}{2}
Rearranging, we get:
e^{2x}-2e^{x}y-1=0

Thus, we get:
e^{x}=\frac{2y\pm\sqrt{4y^{2}+4}}{2}=y+\sqrt{y^{2}+1}
since the other solution is negative (impossible solution for exponential with real exponent).

Take the logarithm of on both sides to find x=arsinh(y)

 

[paulfr]

So it is a direct consequence of Euler's Formula