Calculus

Calculus is hard. So why not let the computer do it for us?

Limits

Let us start by evaluating the limit $$\lim_{x\to 0} \frac{\sin(x)}{x}$$ We first define our variable \( x \) and the expression inside on the right hand side. Then we can use the method limit to evaluate the limit for \( x \to 0 \).

x = sp.symbols('x')
f = sp.sin(x) / x
lim = sp.limit(f, x, 0)
assert lim == sp.sympify(1)

Even one-sided limits like $$ \lim_{x\to 0^+} \frac{1}{x} \mathrm{\ \ vs.\ } \lim_{x\to 0^-} \frac{1}{x} $$ can be evaluated.

f = sp.sympify(1) / x
rlim = sp.limit(f, x, 0, '+')
assert rlim == sp.oo

llim = sp.limit(f, x, 0, '-')
assert llim == -sp.oo

Derivatives

Suppose we want find the first derivative of the following function $$f(x, y) = x^3 + y^3 + \cos(x \cdot y) + \exp(x^2 + y^2)$$ We first start by defining the expression on the right side. In the next step we can call the diff function, which expects an expression as the first argument. The second to last argument (you can use one or more) are the symbolic variables by which the expression will be derivated.

x, y = sp.symbols('x y')
f = x**3 + y**3 + sp.cos(x * y) + sp.exp(x**2 + y**2)
f1 = sp.diff(f, x)

In this we create the first derivative of \( f(x, y) \) w.r.t \(x\). You should get: $$ x^{3}+y^{3}+e^{x^{2}+y^{2}}+\cos (x y) $$

We can also build the second and third derivative w.r.t. x:

f2 = sp.diff(f, x, x)
f3 = sp.diff(f, (x, 3))

Note that two ways of computing higher derivatives are presented here. Of course, it is also possible to create the first derivative w.r.t \( x\) and \(y\):

f4 = sp.diff(f, x, y)

Evaluating Derivatives (or any other expression)

Sometimes, we want to evaluate derivatives, or just any expression at specified points. For this purpose, we could again use Lambda to convert the expression into a function and inserting values by function calls. But SymPy expressions have the subs method to avoid the detour:

f = x**3 + y**3 + sp.cos(x*y) + sp.exp(x**2 + y**2)
f3 = sp.diff(f, (x, 3))
g = f3.subs(y, 0)
h = f3.subs(y, x)
i = f3.subs(y, sp.exp(x))
j = f3.subs([(x, 1), (y, 0)])

This method takes a pair of arguments. The first one is the variable to be substituted and the second one the value. Note that the value can also be a variable or an expression. A list of such pairs can also be supplied to substitute multiple variables.

Integrals

Even the most challenging part in calculus (integrals) can be evaluated $$ \int \mathrm{e}^{-x} \mathrm{d}x $$

f = sp.exp(-x)
f_int = sp.integrate(f, x)

This evaluates to $$ -e^{-x} $$ Note SymPy does not include the constant of integration. But this can be added easily by oneself if needed.

A definite integral could be computed by substituting the integration variable with upper and lower bounds using subs. But it is more straightforward to just pass a tuple as the second argument to the integrate function: Suppose we want to compute the definite integral $$\int_{0}^{\infty} \mathrm{e}^{-x}\ \mathrm{d}x$$ one can write

f = sp.exp(-x)
f_dint = sp.integrate(f, (x, 0, sp.oo))
assert f_dint == sp.sympify(1)

which evaluates to 1.

Also multivariable integrals like $$ \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \mathrm{e}^{-x^2-y^2}\ \mathrm{d}x \mathrm{d}y $$ can be easily solved. In this case the integral

f = sp.exp(-x**2 - y**2)
f_dint = sp.integrate(f, (x, -sp.oo, sp.oo), (y, -sp.oo, sp.oo))
assert f_dint == sp.pi

evaluates to \( \pi \).