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(cos(x), x) -sin(x) >>> diff(exp(x**2), x) ⎛ 2⎞ ⎝x ⎠ 2⋅x⋅ℯ
>>> 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
>>> 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 >>> 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 >>> 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 >>> 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(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 To compute a definite integral, pass the argument \[\int_0^\infty e^{-x}\,dx,\] 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 >>> expr = integrate(x**x, x) >>> print(expr) Integral(x**x, x) >>> expr ⌠ ⎮ x ⎮ x dx ⌡ As with >>> expr = Integral(log(x)**2, x) >>> expr ⌠ ⎮ 2 ⎮ log (x) dx ⌡ >>> expr.doit() 2 x⋅log (x) - 2⋅x⋅log(x) + 2⋅x
>>> 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⋅S⎜────⎟⋅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 Limits#SymPy can compute symbolic limits with the is >>> limit(sin(x)/x, x, 0) 1
>>> expr = x**2/exp(x) >>> expr.subs(x, oo) nan >>> limit(expr, x, oo) 0 Like >>> 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 \[\lim_{x\to 0^+}\frac{1}{x},\] 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 >>> 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 \(x\) where \(x \rightarrow \infty\)). 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 >>> 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 >>> expr.series(x, 0, 4).removeO() 2 x ── + x + 1 2 The >>> 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 >>> 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 you already have a >>> 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 If using >>> 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 |