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Inverse tanh proof

I'm trying to show that tanh−1z=12^{-1}z = \frac{1}{2}ln1+z1−z\frac{1+z}{1-z}

I started off by writing zz = tanh xx

Then z=sinhxcoshx=ex−e−xex+e−xz = \frac{sinhx}{coshx} = \frac{e^x-e^{-x}}{e^x+e^{-x}}

I'm not sure where to go from here?
Original post by Monaa123
I'm trying to show that tanh−1z=12^{-1}z = \frac{1}{2}ln1+z1−z\frac{1+z}{1-z}

I started off by writing zz = tanh xx

Then z=sinhxcoshx=ex−e−xex+e−xz = \frac{sinhx}{coshx} = \frac{e^x-e^{-x}}{e^x+e^{-x}}

I'm not sure where to go from here?


Multiply by exex\frac{e^x}{e^x} then multiply by the denominator.

Collect e2xe^2x terms and factorise ...
Original post by Monaa123
I'm trying to show that tanh−1z=12^{-1}z = \frac{1}{2}ln1+z1−z\frac{1+z}{1-z}

I started off by writing zz = tanh xx

Then z=sinhxcoshx=ex−e−xex+e−xz = \frac{sinhx}{coshx} = \frac{e^x-e^{-x}}{e^x+e^{-x}}

I'm not sure where to go from here?


Multiply right hand side by exex\frac{e^x}{e^x} then multiply both sides by the denominator.

Collect e2xe^{2x} terms and factorise ...
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Monaa123
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Original post by Mr M
Multiply right hand side by exex\frac{e^x}{e^x} then multiply both sides by the denominator.

Collect e2xe^{2x} terms and factorise ...


Thanks, I got it!

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