asinh

Inverse Hyperbolic Sine Function for Numeric and Symbolic Arguments

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

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

A = asinh([-i, 0, 1/6, i/2, i, 2])

A =
   0.0000 - 1.5708i   0.0000 + 0.0000i   0.1659 + 0.0000i...
   0.0000 + 0.5236i   0.0000 + 1.5708i   1.4436 + 0.0000i

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

symA = asinh(sym([-i, 0, 1/6, i/2, i, 2]))

symA =
[ -(pi*1i)/2, 0, asinh(1/6), (pi*1i)/6, (pi*1i)/2, asinh(2)]

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

vpa(symA)

ans =
[ -1.5707963267948966192313216916398i,...
0,...
0.16590455026930117643502171631553,...
0.52359877559829887307710723054658i,...
1.5707963267948966192313216916398i,...
1.4436354751788103012444253181457]

Plot Inverse Hyperbolic Sine Function

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

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

Handle Expressions Containing Inverse Hyperbolic Sine Function

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

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

syms x
diff(asinh(x), x)
diff(asinh(x), x, x)

ans =
1/(x^2 + 1)^(1/2)
 
ans =
-x/(x^2 + 1)^(3/2)

Find the indefinite integral of the inverse hyperbolic sine function:

int(asinh(x), x)

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

Find the Taylor series expansion of asinh(x):

taylor(asinh(x), x)

ans =
(3*x^5)/40 - x^3/6 + x

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

rewrite(asinh(x), 'log')

ans =
log(x + (x^2 + 1)^(1/2))