ATANH
Returns the inverse hyperbolic tangent of a number. The number must be between -1 and 1 (exclusive).
ATANH(number)Arguments
numberA real number strictly between -1 and 1.
=ATANH(0.5)Inverse hyperbolic tangent of 0.5
=ATANH(0)ATANH of 0 is 0
- •Returns #NUM! if number ≤ -1 or number ≥ 1
- •ATANH(x) = 0.5 × LN((1+x)/(1-x))
- •ATANH is the inverse of TANH
- •Passing a number equal to or outside the range -1 to 1, which returns a #NUM! error since the inverse hyperbolic tangent is only defined strictly between -1 and 1, excluding the endpoints themselves
- •Confusing ATANH with ATAN - ATANH is restricted to inputs strictly between -1 and 1, while ATAN accepts any real number, and the two are unrelated despite the similar-looking name
- •Expecting values very close to 1 or -1 to return a reasonable result - as the input approaches either boundary, ATANH's output grows toward positive or negative infinity, which can produce extremely large numbers
TANHPerforms the inverse operation - calculates the hyperbolic tangent of a number, undoing what ATANH calculates.ASINHCalculates the inverse hyperbolic sine, a related function but defined for all real numbers, unlike ATANH's restricted -1 to 1 domain.ATANCalculates the regular (non-hyperbolic) inverse tangent, a different function despite the similar name, accepting any real number as input.Why does ATANH return a #NUM! error?
The input is at or beyond the boundaries of -1 and 1 - the inverse hyperbolic tangent is only defined strictly between those two values, not including them.
Why does ATANH return an enormous number for an input close to 1?
That's expected behavior - as the input approaches 1 (or -1), ATANH's result grows without bound, approaching positive (or negative) infinity.
How can I calculate ATANH manually without the function?
ATANH(x) equals 0.5 times LN((1+x)/(1-x)) - useful if you need to replicate the calculation elsewhere.
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