acot
Symbolic inverse cotangent function
Syntax
Description
Examples
Inverse Cotangent Function for Numeric and Symbolic Arguments
Depending on its arguments, acot
returns
floating-point or exact symbolic results.
Compute the inverse cotangent function for these numbers. Because these numbers are
not symbolic objects, acot
returns floating-point results.
A = acot([-1, -1/3, -1/sqrt(3), 1/2, 1, sqrt(3)])
A = -0.7854 -1.2490 -1.0472 1.1071 0.7854 0.5236
Compute the inverse cotangent function for the numbers converted to symbolic objects.
For many symbolic (exact) numbers, acot
returns unresolved symbolic
calls.
symA = acot(sym([-1, -1/3, -1/sqrt(3), 1/2, 1, sqrt(3)]))
symA = [ -pi/4, -acot(1/3), -pi/3, acot(1/2), pi/4, pi/6]
Use vpa
to approximate symbolic results with floating-point
numbers:
vpa(symA)
ans = [ -0.78539816339744830961566084581988,... -1.2490457723982544258299170772811,... -1.0471975511965977461542144610932,... 1.1071487177940905030170654601785,... 0.78539816339744830961566084581988,... 0.52359877559829887307710723054658]
Plot Inverse Cotangent Function
Plot the inverse cotangent function on the interval from -10 to 10.
syms x fplot(acot(x),[-10 10]) grid on
Handle Expressions Containing Inverse Cotangent Function
Many functions, such as diff
,
int
, taylor
, and
rewrite
, can handle expressions containing
acot
.
Find the first and second derivatives of the inverse cotangent function:
syms x diff(acot(x), x) diff(acot(x), x, x)
ans = -1/(x^2 + 1) ans = (2*x)/(x^2 + 1)^2
Find the indefinite integral of the inverse cotangent function:
int(acot(x), x)
ans = log(x^2 + 1)/2 + x*acot(x)
Find the Taylor series expansion of acot(x)
for x >
0
:
assume(x > 0) taylor(acot(x), x)
ans = - x^5/5 + x^3/3 - x + pi/2
For further computations, clear the assumption on x
by recreating
it using syms
:
syms x
Rewrite the inverse cotangent function in terms of the natural logarithm:
rewrite(acot(x), 'log')
ans = (log(1 - 1i/x)*1i)/2 - (log(1i/x + 1)*1i)/2
Input Arguments
Version History
Introduced before R2006a