SymPy - An Appetizer

SymPyis a module that support symbolic mathematics i.e. things like

• symbolic differentiation,
• symbolic integration,
• solving equations in closed form,
• compute limits,
• solve recurrence equations.

The listing given above is far from complete.

To get started, we need to define some symbolic variables.

We can also define a bunch of variables all at the same time:

Let us verify the first Binomial formula $$ (x + y)^2 = x^2 + 2 \cdot x \cdot y + y^2 $$ using the variables x and y.

Let us generalize this:

The opposite of the command expand is the command factor. Suppose we want to know the solution of the equation $$ x^2 - 8 \cdot x + 15 = 0 $$ and we suspect that the solutions are, in fact, integers. The following command does the trick:

We can also compute the partial fraction decomposition of an expression. The function apart comes in handy for this purpose:

This shows that $$ \frac{2 \cdot x}{x^2 - 8 \cdot x + 15} = \frac{5}{x - 5} - \frac{3}{x - 3}. $$

In order to compute the derivative of $$x \cdot \sin(x)$$ with respect to $x$ we can use the following command:

Symbolic integration is possible, too: To compute the indefinite integral $$\displaystyle\int x \cdot \sin(x)\; dx$$ we use the following command:

SymPy can compute definite integrals. To compute the integral $$\displaystyle\int_{-\infty}^{+\infty} e^{-x^2} \; dx $$ we use the following command:

This shows that $$\displaystyle \int_{-\infty}^{+\infty} e^{-x^2} \; \mathrm{d}x = \sqrt{\pi}. $$

Let us compute the limit $$\lim\limits_{n \rightarrow \infty} \sqrt{n+1} - \sqrt{n}.$$

SymPy can compute the Taylor series of a given function up to a given value of $n$. For example, to compute the Taylor series of the exponential function use the command

In order to compute an analytical expression for the sum $$\displaystyle \sum\limits_{i=1}^n i^3 $$ we can use the following command:

Therefore, we have shown that $$\displaystyle \sum\limits_{i=1}^n i^3 = \frac{n^{4}}{4} + \frac{n^{3}}{2} + \frac{n^{2}}{4} $$ holds. Let's try to sum the first $n+1$ terms of the geometric series:

Note that sympy realizes that we need a case distinction for this formula, as the case $q=1$ is special.

The formula that $$\sum\limits_{i=0}^n q^i = \begin{cases} n + 1 & \text{if}\: q = 1\text{;} \\[0.2cm] \frac{\displaystyle\;1 - q^{n+1}\;}{\displaystyle 1 - q} & \text{otherwise.} \end{cases} $$

Lets try to compute $$ \sum\limits_{i=1}^\infty \frac{1}{i^2}: $$

Let us compute the solution of the following quadratic equation: $$ x^2 - x - 1 = 0 $$ Note that the function solve only takes the term $x^2 - x - 1$ as its argument. It implicitly assumes that we search for that value of the variable $x$, that makes this expressions $0$.

This shows that the two solutions to the quadratic equation $x^2 - x - 1 = 0$ are $$ x_1 = \frac{1}{2} \cdot \bigl(1 - \sqrt{5}\bigr) \quad \mbox{and} \quad x_2 = \frac{1}{2} \cdot \bigl(1 + \sqrt{5}\bigr) $$

In order to solve a system of equations, we need to use sets of equations: $$ \begin{array}[t]{lcr} x + y & = & a \\ x - y & = & b \end{array} $$

Suppose we want to solve the recurrence equation $$ a(n+2) = a(n+1) + a(n) $$
with the initial values $a(0) = 0$ and $a(1) = 1$. First we have to declare that a is a function:

The solution that is computed is $$\displaystyle a(n) = \frac{\sqrt{5}}{5} \cdot\left(\frac{1}{2} + \frac{\sqrt{5}}{2}\right)^{n} - \frac{\sqrt{5}}{5} \cdot\left(- \frac{\sqrt{5}}{2} + \frac{1}{2}\right)^{n}. $$

In order to solve the differential equation $$\displaystyle x \cdot \frac{d\hspace{0.1pt}f}{dx} - f(x) = x^2$$ we first need to declare $f$ as a function.

This shows that for any $C_1 \in \mathbb{R}$ the function $$f(x) = x \cdot \bigl(C_{1} + x\bigr)$$ is a solution of the given differential equation.

Sometimes the output of SymPy is a little hard to read for the untrained eye:

The latex command can help out in cases like these:

If we typeset this in $\LaTeX$, we get $$ \left [ - \frac{- \frac{3 c}{a} + \frac{b^{2}}{a^{2}}}{3 \sqrt[3]{\frac{1}{2} \sqrt{- 4 \left(- \frac{3 c}{a} + \frac{b^{2}}{a^{2}}\right)^{3} + \left(\frac{27 d}{a} - \frac{9 b}{a^{2}} c + \frac{2 b^{3}}{a^{3}}\right)^{2}} + \frac{27 d}{2 a} - \frac{9 b c}{2 a^{2}} + \frac{b^{3}}{a^{3}}}} - \frac{1}{3} \sqrt[3]{\frac{1}{2} \sqrt{- 4 \left(- \frac{3 c}{a} + \frac{b^{2}}{a^{2}}\right)^{3} + \left(\frac{27 d}{a} - \frac{9 b}{a^{2}} c + \frac{2 b^{3}}{a^{3}}\right)^{2}} + \frac{27 d}{2 a} - \frac{9 b c}{2 a^{2}} + \frac{b^{3}}{a^{3}}} - \frac{b}{3 a}, \quad - \frac{- \frac{3 c}{a} + \frac{b^{2}}{a^{2}}}{3 \left(- \frac{1}{2} - \frac{\sqrt{3} i}{2}\right) \sqrt[3]{\frac{1}{2} \sqrt{- 4 \left(- \frac{3 c}{a} + \frac{b^{2}}{a^{2}}\right)^{3} + \left(\frac{27 d}{a} - \frac{9 b}{a^{2}} c + \frac{2 b^{3}}{a^{3}}\right)^{2}} + \frac{27 d}{2 a} - \frac{9 b c}{2 a^{2}} + \frac{b^{3}}{a^{3}}}} - \frac{1}{3} \left(- \frac{1}{2} - \frac{\sqrt{3} i}{2}\right) \sqrt[3]{\frac{1}{2} \sqrt{- 4 \left(- \frac{3 c}{a} + \frac{b^{2}}{a^{2}}\right)^{3} + \left(\frac{27 d}{a} - \frac{9 b}{a^{2}} c + \frac{2 b^{3}}{a^{3}}\right)^{2}} + \frac{27 d}{2 a} - \frac{9 b c}{2 a^{2}} + \frac{b^{3}}{a^{3}}} - \frac{b}{3 a}, \quad - \frac{- \frac{3 c}{a} + \frac{b^{2}}{a^{2}}}{3 \left(- \frac{1}{2} + \frac{\sqrt{3} i}{2}\right) \sqrt[3]{\frac{1}{2} \sqrt{- 4 \left(- \frac{3 c}{a} + \frac{b^{2}}{a^{2}}\right)^{3} + \left(\frac{27 d}{a} - \frac{9 b}{a^{2}} c + \frac{2 b^{3}}{a^{3}}\right)^{2}} + \frac{27 d}{2 a} - \frac{9 b c}{2 a^{2}} + \frac{b^{3}}{a^{3}}}} - \frac{1}{3} \left(- \frac{1}{2} + \frac{\sqrt{3} i}{2}\right) \sqrt[3]{\frac{1}{2} \sqrt{- 4 \left(- \frac{3 c}{a} + \frac{b^{2}}{a^{2}}\right)^{3} + \left(\frac{27 d}{a} - \frac{9 b}{a^{2}} c + \frac{2 b^{3}}{a^{3}}\right)^{2}} + \frac{27 d}{2 a} - \frac{9 b c}{2 a^{2}} + \frac{b^{3}}{a^{3}}} - \frac{b}{3 a}\right ] $$

This looks better.