# arsinh function

Categories: hyperbolic functions

The arsinh function is a hyperbolic function. It is the inverse of the sinh, and is also known as the *inverse hyperbolic sine* function.

Why is it called the *arsinh* rather than the *arcsinh*? See here.

## Equation and graph

The arsinh function is defined as the inverse of sinh, ie if:

$$ x = \sinh{y} $$

then:

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

There is also a formula for finding arsinh as a logarithm, see below.

Here is a graph of the function:

## arsinh as inverse of sinh

This animation illustrates the relationship between the sinh function and the arsinh function:

The first, blue, curve is the sinh function.

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

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

## Logarithm formula for arsinh

arsinh can be calculated directly, using a logarithm function, like this:

$$ \operatorname{arsinh}{x}= \ln({x+{\sqrt {x^{2}+1}}}) $$

Here is a proof of the logarithm formula for arsinh.

We will use:

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

The sinh of *u* will be *x*, because sinh is the inverse of arsinh:

$$ x = \sinh {u} $$

One form of the formula for sinh is:

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

This gives us:

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

Multiplying both sides by $2e^{u}$ gives:

$$ 2 x e^{u} = e^{2u} - 1 $$

This is a quadratic in $e^{u}$ (using the fact that $e^{2u}=(e^{u})^2$):

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

We use the quadratic formula with $a=1$, $b=-2x$, $c=-1$:

$$ e^{u} = \frac{-b\pm\sqrt{b^2-4ac}}{2a} = \frac{2x\pm\sqrt{4{x^2}+4}}{2} $$

Simplifying and taking the positive solution (since $e^{u}$ must be positive) gives:

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

Taking the logarithm of both sides gives:

$$ ln(e^{u}) = u = ln(x+\sqrt{x^2+1}) $$

And since *u* is $\operatorname{arsinh}{x}$ this gives:

$$ \operatorname{arsinh}{x}= \ln({x+{\sqrt {x^{2}+1}}}) $$

## See also

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