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# 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).

## 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).

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 third 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

Currently, `O` only supports orders at 0, so series expansions at points other than 0 are computed by first shifting to 0 and then shifting back.

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

This means that if you compute the series expansion at a point other than 0, the result will be shifted to 0. You can easily shift it back with `subs`.

    >>> exp(x - 6).series(x, 6).removeO().subs(x, x - 6)
               5          4          3          2
        (x - 6)    (x - 6)    (x - 6)    (x - 6)
    x + ──────── + ──────── + ──────── + ──────── - 5
          120         24         6          2

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