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acoth

Symbolic inverse hyperbolic cotangent function

Syntax

Description

acoth(X) returns the inverse hyperbolic cotangent function of X.

example

Examples

Inverse Hyperbolic Cotangent Function for Numeric and Symbolic Arguments

Depending on its arguments, acoth returns floating-point or exact symbolic results.

Compute the inverse hyperbolic cotangent function for these numbers. Because these numbers are not symbolic objects, acoth returns floating-point results.

A = acoth([-pi/2, -1, 0, 1/2, 1, pi/2])
A =
 -0.7525 + 0.0000i     -Inf + 0.0000i   0.0000 + 1.5708i...
   0.5493 + 1.5708i      Inf + 0.0000i   0.7525 + 0.0000i

Compute the inverse hyperbolic cotangent function for the numbers converted to symbolic objects. For many symbolic (exact) numbers, acoth returns unresolved symbolic calls.

symA = acoth(sym([-pi/2, -1, 0, 1/2, 1, pi/2]))
symA =
[ -acoth(pi/2), Inf, -(pi*1i)/2, acoth(1/2), Inf, acoth(pi/2)]

Use vpa to approximate symbolic results with floating-point numbers:

vpa(symA)
ans =
[ -0.75246926714192715916204347800251,...
Inf,...
-1.5707963267948966192313216916398i,...
0.54930614433405484569762261846126...
 - 1.5707963267948966192313216916398i,...
Inf,...
0.75246926714192715916204347800251]

Plot Inverse Hyperbolic Cotangent Function

Plot the inverse hyperbolic cotangent function on the interval from -10 to 10.

syms x
fplot(acoth(x),[-10 10])
grid on

Figure contains an axes object. The axes object contains an object of type functionline.

Handle Expressions Containing Inverse Hyperbolic Cotangent Function

Many functions, such as diff, int, taylor, and rewrite, can handle expressions containing acoth.

Find the first and second derivatives of the inverse hyperbolic cotangent function:

syms x
diff(acoth(x), x)
diff(acoth(x), x, x)
ans =
-1/(x^2 - 1)
 
ans =
(2*x)/(x^2 - 1)^2

Find the indefinite integral of the inverse hyperbolic cotangent function:

int(acoth(x), x)
ans =
log(x^2 - 1)/2 + x*acoth(x)

Find the Taylor series expansion of acoth(x) for x > 0:

assume(x > 0)
taylor(acoth(x), x)
ans =
x^5/5 + x^3/3 + x - (pi*1i)/2

For further computations, clear the assumption on x by recreating it using syms:

syms x

Rewrite the inverse hyperbolic cotangent function in terms of the natural logarithm:

rewrite(acoth(x), 'log')
ans =
log(1/x + 1)/2 - log(1 - 1/x)/2

Input Arguments

collapse all

Input, specified as a symbolic number, variable, expression, or function, or as a vector or matrix of symbolic numbers, variables, expressions, or functions.

Version History

Introduced before R2006a

See Also

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