Calculus

This section covers how to do basic calculus tasks such as derivatives, integrals, limits, and series expansions in SymPy. If you are not familiar with the math of any part of this section, you may safely skip it.

>>> from sympy import *
>>> x, y, z = symbols('x y z')
>>> init_printing(use_unicode=True)

Derivatives

To take derivatives, use the diff function.

>>> diff(cos(x), x)
-sin(x)
>>> diff(exp(x**2), x)
     ⎛ 2⎞
     ⎝x ⎠
2⋅x⋅ℯ

diff can take multiple derivatives at once. To take multiple derivatives, pass the variable as many times as you wish to differentiate, or pass a number after the variable. For example, both of the following find the third derivative of \(x^4\).

>>> diff(x**4, x, x, x)
24⋅x
>>> diff(x**4, x, 3)
24⋅x

You can also take derivatives with respect to many variables at once. Just pass each derivative in order, using the same syntax as for single variable derivatives. For example, each of the following will compute \(\frac{\partial^7}{\partial x\partial y^2\partial z^4} e^{x y z}\).

>>> expr = exp(x*y*z)
>>> diff(expr, x, y, y, z, z, z, z)
 3  2 ⎛ 3  3  3       2  2  2                ⎞  x⋅y⋅z
x ⋅y ⋅⎝x ⋅y ⋅z  + 14⋅x ⋅y ⋅z  + 52⋅x⋅y⋅z + 48⎠⋅ℯ
>>> diff(expr, x, y, 2, z, 4)
 3  2 ⎛ 3  3  3       2  2  2                ⎞  x⋅y⋅z
x ⋅y ⋅⎝x ⋅y ⋅z  + 14⋅x ⋅y ⋅z  + 52⋅x⋅y⋅z + 48⎠⋅ℯ
>>> diff(expr, x, y, y, z, 4)
 3  2 ⎛ 3  3  3       2  2  2                ⎞  x⋅y⋅z
x ⋅y ⋅⎝x ⋅y ⋅z  + 14⋅x ⋅y ⋅z  + 52⋅x⋅y⋅z + 48⎠⋅ℯ

diff can also be called as a method. The two ways of calling diff are exactly the same, and are provided only for convenience.

>>> expr.diff(x, y, y, z, 4)
 3  2 ⎛ 3  3  3       2  2  2                ⎞  x⋅y⋅z
x ⋅y ⋅⎝x ⋅y ⋅z  + 14⋅x ⋅y ⋅z  + 52⋅x⋅y⋅z + 48⎠⋅ℯ

To create an unevaluated derivative, use the Derivative class. It has the same syntax as diff.

>>> deriv = Derivative(expr, x, y, y, z, 4)
>>> deriv
     7
    ∂     ⎛ x⋅y⋅z⎞
──────────⎝ℯ     ⎠
  4   2
∂z  ∂y  ∂x

To evaluate an unevaluated derivative, use the doit method.

>>> deriv.doit()
 3  2 ⎛ 3  3  3       2  2  2                ⎞  x⋅y⋅z
x ⋅y ⋅⎝x ⋅y ⋅z  + 14⋅x ⋅y ⋅z  + 52⋅x⋅y⋅z + 48⎠⋅ℯ

These unevaluated objects are useful for delaying the evaluation of the derivative, or for printing purposes. They are also used when SymPy does not know how to compute the derivative of an expression (for example, if it contains an undefined function, which are described in the Solving Differential Equations section).

Derivatives of unspecified order can be created using tuple (x, n) where n is the order of the derivative with respect to x.

>>> m, n, a, b = symbols('m n a b')
>>> expr = (a*x + b)**m
>>> expr.diff((x, n))
  n
 ∂ ⎛         m⎞
───⎝(a⋅x + b) ⎠
  n
∂x

Integrals

To compute an integral, use the integrate function. There are two kinds of integrals, definite and indefinite. To compute an indefinite integral, that is, an antiderivative, or primitive, just pass the variable after the expression.

>>> integrate(cos(x), x)
sin(x)

Note that SymPy does not include the constant of integration. If you want it, you can add one yourself, or rephrase your problem as a differential equation and use dsolve to solve it, which does add the constant (see Solving Differential Equations).

Quick Tip

\(\infty\) in SymPy is oo (that’s the lowercase letter “oh” twice). This is because oo looks like \(\infty\), and is easy to type.

To compute a definite integral, pass the argument (integration_variable, lower_limit, upper_limit). For example, to compute

\[\int_0^\infty e^{-x}\,dx,\]

we would do

>>> integrate(exp(-x), (x, 0, oo))
1

As with indefinite integrals, you can pass multiple limit tuples to perform a multiple integral. For example, to compute

\[\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{- x^{2} - y^{2}}\, dx\, dy,\]

do

>>> integrate(exp(-x**2 - y**2), (x, -oo, oo), (y, -oo, oo))
π

If integrate is unable to compute an integral, it returns an unevaluated Integral object.

>>> expr = integrate(x**x, x)
>>> print(expr)
Integral(x**x, x)
>>> expr
⌠
⎮  x
⎮ x  dx
⌡

As with Derivative, you can create an unevaluated integral using Integral. To later evaluate this integral, call doit.

>>> expr = Integral(log(x)**2, x)
>>> expr
⌠
⎮    2
⎮ log (x) dx
⌡
>>> expr.doit()
         2
x⋅log (x) - 2⋅x⋅log(x) + 2⋅x

integrate uses powerful algorithms that are always improving to compute both definite and indefinite integrals, including heuristic pattern matching type algorithms, a partial implementation of the Risch algorithm, and an algorithm using Meijer G-functions that is useful for computing integrals in terms of special functions, especially definite integrals. Here is a sampling of some of the power of integrate.

>>> integ = Integral((x**4 + x**2*exp(x) - x**2 - 2*x*exp(x) - 2*x -
...     exp(x))*exp(x)/((x - 1)**2*(x + 1)**2*(exp(x) + 1)), x)
>>> integ
⌠
⎮ ⎛ 4    2  x    2        x          x⎞  x
⎮ ⎝x  + x ⋅ℯ  - x  - 2⋅x⋅ℯ  - 2⋅x - ℯ ⎠⋅ℯ
⎮ ──────────────────────────────────────── dx
⎮               2        2 ⎛ x    ⎞
⎮        (x - 1) ⋅(x + 1) ⋅⎝ℯ  + 1⎠
⌡
>>> integ.doit()
                 x
   ⎛ x    ⎞     ℯ
log⎝ℯ  + 1⎠ + ──────
               2
              x  - 1

>>> integ = Integral(sin(x**2), x)
>>> integ
⌠
⎮    ⎛ 2⎞
⎮ sin⎝x ⎠ dx
⌡
>>> integ.doit()
                ⎛√2⋅x⎞
3⋅√2⋅√π⋅fresnels⎜────⎟⋅Γ(3/4)
                ⎝ √π ⎠
─────────────────────────────
           8⋅Γ(7/4)

>>> integ = Integral(x**y*exp(-x), (x, 0, oo))
>>> integ
∞
⌠
⎮  y  -x
⎮ x ⋅ℯ   dx
⌡
0
>>> integ.doit()
⎧ Γ(y + 1)    for re(y) > -1
⎪
⎪∞
⎪⌠
⎨⎮  y  -x
⎪⎮ x ⋅ℯ   dx    otherwise
⎪⌡
⎪0
⎩

This last example returned a Piecewise expression because the integral does not converge unless \(\Re(y) > 1.\)

Limits

SymPy can compute symbolic limits with the limit function. The syntax to compute

\[\lim_{x\to x_0} f(x)\]

is limit(f(x), x, x0).

>>> limit(sin(x)/x, x, 0)
1

limit should be used instead of subs whenever the point of evaluation is a singularity. Even though SymPy has objects to represent \(\infty\), using them for evaluation is not reliable because they do not keep track of things like rate of growth. Also, things like \(\infty - \infty\) and \(\frac{\infty}{\infty}\) return \(\mathrm{nan}\) (not-a-number). For example

>>> expr = x**2/exp(x)
>>> expr.subs(x, oo)
nan
>>> limit(expr, x, oo)
0

Like Derivative and Integral, limit has an unevaluated counterpart, Limit. To evaluate it, use doit.

>>> expr = Limit((cos(x) - 1)/x, x, 0)
>>> expr
     ⎛cos(x) - 1⎞
 lim ⎜──────────⎟
x─→0⁺⎝    x     ⎠
>>> expr.doit()
0

To evaluate a limit at one side only, pass '+' or '-' as a fourth argument to limit. For example, to compute

\[\lim_{x\to 0^+}\frac{1}{x},\]

do

>>> limit(1/x, x, 0, '+')
∞

As opposed to

>>> limit(1/x, x, 0, '-')
-∞

Series Expansion

SymPy can compute asymptotic series expansions of functions around a point. To compute the expansion of \(f(x)\) around the point \(x = x_0\) terms of order \(x^n\), use f(x).series(x, x0, n). x0 and n can be omitted, in which case the defaults x0=0 and n=6 will be used.

>>> expr = exp(sin(x))
>>> expr.series(x, 0, 4)
         2
        x     ⎛ 4⎞
1 + x + ── + O⎝x ⎠
        2

The \(O\left(x^4\right)\) term at the end represents the Landau order term at \(x=0\) (not to be confused with big O notation used in computer science, which generally represents the Landau order term at \(x=\infty\)). It means that all x terms with power greater than or equal to \(x^4\) are omitted. Order terms can be created and manipulated outside of series. They automatically absorb higher order terms.

>>> x + x**3 + x**6 + O(x**4)
     3    ⎛ 4⎞
x + x  + O⎝x ⎠
>>> x*O(1)
O(x)

If you do not want the order term, use the removeO method.

>>> expr.series(x, 0, 4).removeO()
 2
x
── + x + 1
2

The O notation supports arbitrary limit points (other than 0):

>>> exp(x - 6).series(x, x0=6)
            2          3          4          5
     (x - 6)    (x - 6)    (x - 6)    (x - 6)         ⎛       6       ⎞
-5 + ──────── + ──────── + ──────── + ──────── + x + O⎝(x - 6) ; x → 6⎠
        2          6          24        120

Finite differences

So far we have looked at expressions with analytic derivatives and primitive functions respectively. But what if we want to have an expression to estimate a derivative of a curve for which we lack a closed form representation, or for which we don’t know the functional values for yet. One approach would be to use a finite difference approach.

The simplest way the differentiate using finite differences is to use the differentiate_finite function:

>>> f, g = symbols('f g', cls=Function)
>>> differentiate_finite(f(x)*g(x))
-f(x - 1/2)⋅g(x - 1/2) + f(x + 1/2)⋅g(x + 1/2)

If we want to expand the intermediate derivative we may pass the flag evaluate=True:

>>> differentiate_finite(f(x)*g(x), evaluate=True)
(-f(x - 1/2) + f(x + 1/2))⋅g(x) + (-g(x - 1/2) + g(x + 1/2))⋅f(x)

This form however does not respect the product rule.

If you already have a Derivative instance, you can use the as_finite_difference method to generate approximations of the derivative to arbitrary order:

>>> f = Function('f')
>>> dfdx = f(x).diff(x)
>>> dfdx.as_finite_difference()
-f(x - 1/2) + f(x + 1/2)

here the first order derivative was approximated around x using a minimum number of points (2 for 1st order derivative) evaluated equidistantly using a step-size of 1. We can use arbitrary steps (possibly containing symbolic expressions):

>>> f = Function('f')
>>> d2fdx2 = f(x).diff(x, 2)
>>> h = Symbol('h')
>>> d2fdx2.as_finite_difference([-3*h,-h,2*h])
f(-3⋅h)   f(-h)   2⋅f(2⋅h)
─────── - ───── + ────────
     2        2        2
  5⋅h      3⋅h     15⋅h

If you are just interested in evaluating the weights, you can do so manually:

>>> finite_diff_weights(2, [-3, -1, 2], 0)[-1][-1]
[1/5, -1/3, 2/15]

note that we only need the last element in the last sublist returned from finite_diff_weights. The reason for this is that the function also generates weights for lower derivatives and using fewer points (see the documentation of finite_diff_weights for more details).

If using finite_diff_weights directly looks complicated, and the as_finite_difference method of Derivative instances is not flexible enough, you can use apply_finite_diff which takes order, x_list, y_list and x0 as parameters:

>>> x_list = [-3, 1, 2]
>>> y_list = symbols('a b c')
>>> apply_finite_diff(1, x_list, y_list, 0)
  3⋅a   b   2⋅c
- ─── - ─ + ───
   20   4    5