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