Basics

Here we discuss some of the most basic aspects of expression manipulation in Diofant.

Assumptions

The assumptions system allows users to declare certain mathematical properties on symbols, such as being positive, imaginary or integer.

By default, all symbols are complex valued. This assumption makes it easier to treat mathematical problems in full generality.

>>> sqrt(x**2)
   ____
  ╱  2
╲╱  x

Yet obviously we can simplify above expression if some additional mathematical properties on x are assumed. This is where assumptions system come into play.

Assumptions are set on Symbol objects when they are created. For instance, we can create a symbol that is assumed to be positive.

>>> p = symbols('p', positive=True)

And then, certain simplifications will be possible:

>>> sqrt(p**2)
p

The assumptions system additionally has deductive capabilities. You might check assumptions on any expression with is_assumption attributes, like is_positive.

>>> p.is_positive
True
>>> (1 + p).is_positive
True
>>> (-p).is_positive
False

Note

False is returned also if certain assumption doesn’t make sense for given object.

In a three-valued logic, used by system, None represents the “unknown” case.

>>> (p - 1).is_positive is None
True

Substitution

One of the most common things you might want to do with a mathematical expression is substitution with subs() method. It replaces all instances of something in an expression with something else.

>>> expr = cos(x) + 1
>>> expr.subs({x: y})
cos(y) + 1
>>> expr
cos(x) + 1

We see that performing substitution leaves original expression expr unchanged.

Note

Almost all Diofant expressions are immutable. No function (or method) will change them in-place.

Use several method calls to perform a sequence of substitutions in same variable:

>>> x**y
 y
x
>>> _.subs({y: x**y}).subs({y: x**x})
 ⎛ ⎛ x⎞⎞
 ⎜ ⎝x ⎠⎟
 ⎝x    ⎠
x

Use flag simultaneous to do all substitutions at once.

>>> (x - y).subs({x: y, y: x})
0
>>> (x - y).subs({x: y, y: x}, simultaneous=True)
-x + y

Numerics

To evaluate a numerical expression into a floating-point number with arbitrary precision, use evalf(). By default, 15 digits of precision are used.

>>> expr = sqrt(8)
>>> expr.evalf()
2.82842712474619

But you can change that. Let’s compute the first 70 digits of \(\pi\).

>>> pi.evalf(70)
3.141592653589793238462643383279502884197169399375105820974944592307816

Complex numbers are supported:

>>> (1/(pi + I)).evalf()
0.289025482222236 - 0.0919996683503752⋅ⅈ

If the expression contains symbols or for some other reason cannot be evaluated numerically, calling evalf() returns the original expression or a partially evaluated expression.

>>> (pi*x**2 + x/3).evalf()
                  2
3.14159265358979⋅x  + 0.333333333333333⋅x

You can also use the standard Python functions float and complex to convert symbolic expressions to regular Python numbers:

>>> float(pi)
3.141592653589793
>>> complex(pi + E*I)
(3.141592653589793+2.718281828459045j)

Sometimes there are roundoff errors smaller than the desired precision that remain after an expression is evaluated. Such numbers can be removed by setting the chop flag.

>>> one = cos(1)**2 + sin(1)**2
>>> (one - 1).evalf(strict=False)
-0.e-146
>>> (one - 1).evalf(chop=True)
0

Discussed above method is not effective enough if you intend to evaluate an expression at many points, there are better ways, especially if you only care about machine precision.

Substitution may be used to evaluate an expression for some floating-point number

>>> expr = sin(x)/x
>>> expr.subs({x: 0.1})
0.998334166468282

but this method is slow.

The easiest way to convert an expression to the form that can be numerically evaluated with libraries like numpy or the standard library math module — use the lambdify() function.

>>> f = lambdify(x, expr, 'math')
>>> f(0.1)
0.9983341664682815

Using the numpy library gives the generated function access to powerful vectorized ufuncs that are backed by compiled C code.

>>> f = lambdify(x, expr, 'numpy')
>>> f(range(1, 5))
[ 0.84147098  0.45464871  0.04704    -0.18920062]