asec

Inverse Secant Function for Numeric and Symbolic Arguments

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

Compute the inverse secant function for these numbers. Because these numbers are not symbolic objects, asec returns floating-point results.

A = asec([-2, 0, 2/sqrt(3), 1/2, 1, 5])

A =
   2.0944 + 0.0000i   0.0000 +    Infi   0.5236 + 0.0000i...
   0.0000 + 1.3170i   0.0000 + 0.0000i   1.3694 + 0.0000i

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

symA = asec(sym([-2, 0, 2/sqrt(3), 1/2, 1, 5]))

symA =
[ (2*pi)/3, Inf, pi/6, acos(2), 0, acos(1/5)]

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

vpa(symA)

ans =
[ 2.0943951023931954923084289221863,...
Inf,...
0.52359877559829887307710723054658,...
1.3169578969248165734029498707969i,...
0,...
1.3694384060045659001758622252964]

Plot Inverse Secant Function

Plot the inverse secant function on the interval from -10 to 10.

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

Handle Expressions Containing Inverse Secant Function

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

Find the first and second derivatives of the inverse secant function:

syms x
diff(asec(x), x)
diff(asec(x), x, x)

ans =
1/(x^2*(1 - 1/x^2)^(1/2))
 
ans =
- 2/(x^3*(1 - 1/x^2)^(1/2)) - 1/(x^5*(1 - 1/x^2)^(3/2))

Find the indefinite integral of the inverse secant function:

int(asec(x), x)

ans =
x*acos(1/x) - log(x + (x^2 - 1)^(1/2))*sign(x)

Find the Taylor series expansion of asec(x) around x = Inf:

taylor(asec(x), x, Inf)

ans =
pi/2 - 1/x - 1/(6*x^3) - 3/(40*x^5)

Rewrite the inverse secant function in terms of the natural logarithm:

rewrite(asec(x), 'log')

ans =
-log(1/x + (1 - 1/x^2)^(1/2)*1i)*1i