atanh
Symbolic inverse hyperbolic tangent function
Syntax
Description
Examples
Inverse Hyperbolic Tangent Function for Numeric and Symbolic Arguments
Depending on its arguments, atanh returns
floating-point or exact symbolic results.
Compute the inverse hyperbolic tangent function for these numbers. Because these
numbers are not symbolic objects, atanh returns floating-point
results.
A = atanh([-i, 0, 1/6, i/2, i, 2])
A = 0.0000 - 0.7854i 0.0000 + 0.0000i 0.1682 + 0.0000i... 0.0000 + 0.4636i 0.0000 + 0.7854i 0.5493 + 1.5708i
Compute the inverse hyperbolic tangent function for the numbers converted to symbolic
objects. For many symbolic (exact) numbers, atanh returns
unresolved symbolic calls.
symA = atanh(sym([-i, 0, 1/6, i/2, i, 2]))
symA = [ -(pi*1i)/4, 0, atanh(1/6), atanh(1i/2), (pi*1i)/4, atanh(2)]
Use vpa to approximate symbolic results with floating-point
numbers:
vpa(symA)
ans = [ -0.78539816339744830961566084581988i,... 0,... 0.1682361183106064652522967051085,... 0.46364760900080611621425623146121i,... 0.78539816339744830961566084581988i,... 0.54930614433405484569762261846126 - 1.5707963267948966192313216916398i]
Plot Inverse Hyperbolic Tangent Function
Plot the inverse hyperbolic tangent function on the interval from -1 to 1.
syms x fplot(atanh(x),[-1 1]) grid on

Handle Expressions Containing Inverse Hyperbolic Tangent Function
Many functions, such as diff,
int, taylor, and
rewrite, can handle expressions containing
atanh.
Find the first and second derivatives of the inverse hyperbolic tangent function:
syms x diff(atanh(x), x) diff(atanh(x), x, x)
ans = -1/(x^2 - 1) ans = (2*x)/(x^2 - 1)^2
Find the indefinite integral of the inverse hyperbolic tangent function:
int(atanh(x), x)
ans = log(x^2 - 1)/2 + x*atanh(x)
Find the Taylor series expansion of atanh(x):
taylor(atanh(x), x)
ans = x^5/5 + x^3/3 + x
Rewrite the inverse hyperbolic tangent function in terms of the natural logarithm:
rewrite(atanh(x), 'log')
ans = log(x + 1)/2 - log(1 - x)/2
Input Arguments
Version History
Introduced before R2006a