# artanh function

Categories: hyperbolic functions

The artanh function is a hyperbolic function. It is the inverse of the tanh, and is also known as the *inverse hyperbolic tangent* function.

Why is it called the *artanh* rather than the *arctanh*? See here.

## Equation and graph

The artanh function is defined as the inverse of tanh, ie if:

$$ x = \tanh y $$

then:

$$ y = \operatorname{artanh} x $$

Here is a graph of the function:

The function is only valid for -1 < x < 1.

## artanh as inverse of tanh

This animation illustrates the relationship between the tanh function and the artanh function:

The first, blue, curve is the tanh function.

The grey dashed line is the line $y=x$.

The second, red, curve is the artanh function. As with any inverse function, it is identical to the original function reflected in the line $y=x$.

## Logarithm formula for artanh

There is a also a formula for finding artanh directly:

$$ \operatorname{artanh} x =\frac12\ln\left(\frac{1+x}{1-x}\right) $$

Here is a proof of the logarithm formula for artanh. This follows similar lines to the proof for arsinh.

We will use:

$$ u = \operatorname{artanh}{x} $$

The tanh of *u* will be *x*, because tanh is the inverse of artanh:

$$ x = \tanh {u} $$

One form of the formula for tanh is:

$$ \tanh{u} = \frac{e^{2u}-1}{e^{2u}+1} $$

This gives us:

$$ x = \tanh{u} = \frac{e^{2u}-1}{e^{2u}+1} $$

Multiplying both sides by $(e^{2u}+1)$ gives:

$$ ({e^{2u}+1})x = e^{2u} - 1 $$

Rearranging:

$$ ({e^{2u}+1})x - e^{2u} + 1 = 0 $$

So:

$$ e^{2u}(x - 1) + 1 + x = 0 $$

$$ e^{2u}(x - 1) = -(1 + x) $$

$$ e^{2u} = -\frac{1 + x}{x - 1} = \frac{1 + x}{1-x} $$

Taking the log of both sides:

$$ 2u = \ln\left(\frac{1+x}{1-x}\right) $$

This gives the required result:

$$ u = \operatorname{artanh} x =\frac12\ln\left(\frac{1+x}{1-x}\right) $$

## See also

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