asech
Symbolic inverse hyperbolic secant function
Syntax
Description
Examples
Inverse Hyperbolic Secant Function for Numeric and Symbolic Arguments
Depending on its arguments, asech
returns
floating-point or exact symbolic results.
Compute the inverse hyperbolic secant function for these numbers. Because these
numbers are not symbolic objects, asech
returns floating-point
results.
A = asech([-2, 0, 2/sqrt(3), 1/2, 1, 3])
A = 0.0000 + 2.0944i Inf + 0.0000i 0.0000 + 0.5236i... 1.3170 + 0.0000i 0.0000 + 0.0000i 0.0000 + 1.2310i
Compute the inverse hyperbolic secant function for the numbers converted to symbolic
objects. For many symbolic (exact) numbers, asech
returns
unresolved symbolic calls.
symA = asech(sym([-2, 0, 2/sqrt(3), 1/2, 1, 3]))
symA = [ (pi*2i)/3, Inf, (pi*1i)/6, acosh(2), 0, acosh(1/3)]
Use vpa
to approximate symbolic results with floating-point
numbers:
vpa(symA)
ans = [ 2.0943951023931954923084289221863i,... Inf,... 0.52359877559829887307710723054658i,... 1.316957896924816708625046347308,... 0,... 1.230959417340774682134929178248i]
Plot Inverse Hyperbolic Secant Function
Plot the inverse hyperbolic secant function on the interval from 0 to 1.
syms x fplot(asech(x),[0 1]) grid on
Handle Expressions Containing Inverse Hyperbolic Secant Function
Many functions, such as diff
,
int
, taylor
, and
rewrite
, can handle expressions containing
asech
.
Find the first and second derivatives of the inverse hyperbolic secant function.
Simplify the second derivative by using simplify
.
syms x diff(asech(x), x) simplify(diff(asech(x), x, x))
ans = -1/(x^2*(1/x - 1)^(1/2)*(1/x + 1)^(1/2)) ans = -(2*x^2 - 1)/(x^5*(1/x - 1)^(3/2)*(1/x + 1)^(3/2))
Find the indefinite integral of the inverse hyperbolic secant function:
int(asech(x), x)
ans = atan(1/((1/x - 1)^(1/2)*(1/x + 1)^(1/2))) + x*acosh(1/x)
Find the Taylor series expansion of asech(x)
around x =
Inf
:
taylor(asech(x), x, Inf)
ans = (pi*1i)/2 - 1i/x - 1i/(6*x^3) - 3i/(40*x^5)
Rewrite the inverse hyperbolic secant function in terms of the natural logarithm:
rewrite(asech(x), 'log')
ans = log((1/x - 1)^(1/2)*(1/x + 1)^(1/2) + 1/x)
Input Arguments
Version History
Introduced before R2006a