# Numerical evaluation¶

## Basics¶

Exact SymPy expressions can be converted to floating-point approximations (decimal numbers) using either the .evalf() method or the N() function. N(expr, <args>) is equivalent to sympify(expr).evalf(<args>).

>>> from sympy import *
>>> N(sqrt(2)*pi)
4.44288293815837
>>> (sqrt(2)*pi).evalf()
4.44288293815837


By default, numerical evaluation is performed to an accuracy of 15 decimal digits. You can optionally pass a desired accuracy (which should be a positive integer) as an argument to evalf or N:

>>> N(sqrt(2)*pi, 5)
4.4429
>>> N(sqrt(2)*pi, 50)
4.4428829381583662470158809900606936986146216893757


Complex numbers are supported:

>>> N(1/(pi + I), 20)
0.28902548222223624241 - 0.091999668350375232456*I


If the expression contains symbols or for some other reason cannot be evaluated numerically, calling .evalf() or N() returns the original expression, or in some cases a partially evaluated expression. For example, when the expression is a polynomial in expanded form, the coefficients are evaluated:

>>> x = Symbol('x')
>>> (pi*x**2 + x/3).evalf()
3.14159265358979*x**2 + 0.333333333333333*x


You can also use the standard Python functions float(), complex() to convert SymPy expressions to regular Python numbers:

>>> float(pi)
3.1415926535...
>>> complex(pi+E*I)
(3.1415926535...+2.7182818284...j)


If these functions are used, failure to evaluate the expression to an explicit number (for example if the expression contains symbols) will raise an exception.

There is essentially no upper precision limit. The following command, for example, computes the first 100,000 digits of π/e:

>>> N(pi/E, 100000)
...


This shows digits 999,951 through 1,000,000 of pi:

>>> str(N(pi, 10**6))[-50:]
'95678796130331164628399634646042209010610577945815'


High-precision calculations can be slow. It is recommended (but entirely optional) to install gmpy (http://code.google.com/p/gmpy/), which will significantly speed up computations such as the one above.

## Floating-point numbers¶

Floating-point numbers in SymPy are instances of the class Float. A Float can be created with a custom precision as second argument:

>>> Float(0.1)
0.100000000000000
>>> Float(0.1, 10)
0.1000000000
>>> Float(0.125, 30)
0.125000000000000000000000000000
>>> Float(0.1, 30)
0.100000000000000005551115123126


As the last example shows, some Python floats are only accurate to about 15 digits as inputs, while others (those that have a denominator that is a power of 2, like .125 = 1/4) are exact. To create a Float from a high-precision decimal number, it is better to pass a string, Rational, or evalf a Rational:

>>> Float('0.1', 30)
0.100000000000000000000000000000
>>> Float(Rational(1, 10), 30)
0.100000000000000000000000000000
>>> Rational(1, 10).evalf(30)
0.100000000000000000000000000000


The precision of a number determines 1) the precision to use when performing arithmetic with the number, and 2) the number of digits to display when printing the number. When two numbers with different precision are used together in an arithmetic operation, the higher of the precisions is used for the result. The product of 0.1 +/- 0.001 and 3.1415 +/- 0.0001 has an uncertainty of about 0.003 and yet 5 digits of precision are shown.

>>> Float(0.1, 3)*Float(3.1415, 5)
0.31417


So the displayed precision should not be used as a model of error propagation or significance arithmetic; rather, this scheme is employed to ensure stability of numerical algorithms.

N and evalf can be used to change the precision of existing floating-point numbers:

>>> N(3.5)
3.50000000000000
>>> N(3.5, 5)
3.5000
>>> N(3.5, 30)
3.50000000000000000000000000000


## Accuracy and error handling¶

When the input to N or evalf is a complicated expression, numerical error propagation becomes a concern. As an example, consider the 100’th Fibonacci number and the excellent (but not exact) approximation $$\varphi^{100} / \sqrt{5}$$ where $$\varphi$$ is the golden ratio. With ordinary floating-point arithmetic, subtracting these numbers from each other erroneously results in a complete cancellation:

>>> a, b = GoldenRatio**1000/sqrt(5), fibonacci(1000)
>>> float(a)
4.34665576869e+208
>>> float(b)
4.34665576869e+208
>>> float(a) - float(b)
0.0


N and evalf keep track of errors and automatically increase the precision used internally in order to obtain a correct result:

>>> N(fibonacci(100) - GoldenRatio**100/sqrt(5))
-5.64613129282185e-22


Unfortunately, numerical evaluation cannot tell an expression that is exactly zero apart from one that is merely very small. The working precision is therefore capped, by default to around 100 digits. If we try with the 1000’th Fibonacci number, the following happens:

>>> N(fibonacci(1000) - (GoldenRatio)**1000/sqrt(5))
0.e+85


The lack of digits in the returned number indicates that N failed to achieve full accuracy. The result indicates that the magnitude of the expression is something less than 10^84, but that is not a particularly good answer. To force a higher working precision, the maxn keyword argument can be used:

>>> N(fibonacci(1000) - (GoldenRatio)**1000/sqrt(5), maxn=500)
-4.60123853010113e-210


Normally, maxn can be set very high (thousands of digits), but be aware that this may cause significant slowdown in extreme cases. Alternatively, the strict=True option can be set to force an exception instead of silently returning a value with less than the requested accuracy:

>>> N(fibonacci(1000) - (GoldenRatio)**1000/sqrt(5), strict=True)
Traceback (most recent call last):
...
PrecisionExhausted: Failed to distinguish the expression:

-sqrt(5)*GoldenRatio**1000/5 + 43466557686937456435688527675040625802564660517371780402481729089536555417949051890403879840079255169295922593080322634775209689623239873322471161642996440906533187938298969649928516003704476137795166849228875

from zero. Try simplifying the input, using chop=True, or providing a higher maxn for evalf


If we add a term so that the Fibonacci approximation becomes exact (the full form of Binet’s formula), we get an expression that is exactly zero, but N does not know this:

>>> f = fibonacci(100) - (GoldenRatio**100 - (GoldenRatio-1)**100)/sqrt(5)
>>> N(f)
0.e-104
>>> N(f, maxn=1000)
0.e-1336


In situations where such cancellations are known to occur, the chop options is useful. This basically replaces very small numbers in the real or imaginary portions of a number with exact zeros:

>>> N(f, chop=True)
0
>>> N(3 + I*f, chop=True)
3.00000000000000


In situations where you wish to remove meaningless digits, re-evaluation or the use of the round method are useful:

>>> Float('.1', '')*Float('.12345', '')
0.012297
>>> ans = _
>>> N(ans, 1)
0.01
>>> ans.round(2)
0.01


If you are dealing with a numeric expression that contains no floats, it can be evaluated to arbitrary precision. To round the result relative to a given decimal, the round method is useful:

>>> v = 10*pi + cos(1)
>>> N(v)
31.9562288417661
>>> v.round(3)
31.956


## Sums and integrals¶

Sums (in particular, infinite series) and integrals can be used like regular closed-form expressions, and support arbitrary-precision evaluation:

>>> var('n x')
(n, x)
>>> Sum(1/n**n, (n, 1, oo)).evalf()
1.29128599706266
>>> Integral(x**(-x), (x, 0, 1)).evalf()
1.29128599706266
>>> Sum(1/n**n, (n, 1, oo)).evalf(50)
1.2912859970626635404072825905956005414986193682745
>>> Integral(x**(-x), (x, 0, 1)).evalf(50)
1.2912859970626635404072825905956005414986193682745
>>> (Integral(exp(-x**2), (x, -oo, oo)) ** 2).evalf(30)
3.14159265358979323846264338328


By default, the tanh-sinh quadrature algorithm is used to evaluate integrals. This algorithm is very efficient and robust for smooth integrands (and even integrals with endpoint singularities), but may struggle with integrals that are highly oscillatory or have mid-interval discontinuities. In many cases, evalf/N will correctly estimate the error. With the following integral, the result is accurate but only good to four digits:

>>> f = abs(sin(x))
>>> Integral(abs(sin(x)), (x, 0, 4)).evalf()
2.346


It is better to split this integral into two pieces:

>>> (Integral(f, (x, 0, pi)) + Integral(f, (x, pi, 4))).evalf()
2.34635637913639


A similar example is the following oscillatory integral:

>>> Integral(sin(x)/x**2, (x, 1, oo)).evalf(maxn=20)
0.5


It can be dealt with much more efficiently by telling evalf or N to use an oscillatory quadrature algorithm:

>>> Integral(sin(x)/x**2, (x, 1, oo)).evalf(quad='osc')
0.504067061906928
>>> Integral(sin(x)/x**2, (x, 1, oo)).evalf(20, quad='osc')
0.50406706190692837199


Oscillatory quadrature requires an integrand containing a factor cos(ax+b) or sin(ax+b). Note that many other oscillatory integrals can be transformed to this form with a change of variables:

>>> init_printing(use_unicode=False, wrap_line=False, no_global=True)
>>> intgrl = Integral(sin(1/x), (x, 0, 1)).transform(x, 1/x)
>>> intgrl
oo
/
|
|  sin(x)
|  ------ dx
|     2
|    x
|
/
1
0.504067061906928


Infinite series use direct summation if the series converges quickly enough. Otherwise, extrapolation methods (generally the Euler-Maclaurin formula but also Richardson extrapolation) are used to speed up convergence. This allows high-precision evaluation of slowly convergent series:

>>> var('k')
k
>>> Sum(1/k**2, (k, 1, oo)).evalf()
1.64493406684823
>>> zeta(2).evalf()
1.64493406684823
>>> Sum(1/k-log(1+1/k), (k, 1, oo)).evalf()
0.577215664901533
>>> Sum(1/k-log(1+1/k), (k, 1, oo)).evalf(50)
0.57721566490153286060651209008240243104215933593992
>>> EulerGamma.evalf(50)
0.57721566490153286060651209008240243104215933593992


The Euler-Maclaurin formula is also used for finite series, allowing them to be approximated quickly without evaluating all terms:

>>> Sum(1/k, (k, 10000000, 20000000)).evalf()
0.693147255559946


Note that evalf makes some assumptions that are not always optimal. For fine-tuned control over numerical summation, it might be worthwhile to manually use the method Sum.euler_maclaurin.

Special optimizations are used for rational hypergeometric series (where the term is a product of polynomials, powers, factorials, binomial coefficients and the like). N/evalf sum series of this type very rapidly to high precision. For example, this Ramanujan formula for pi can be summed to 10,000 digits in a fraction of a second with a simple command:

>>> f = factorial
>>> n = Symbol('n', integer=True)
>>> R = 9801/sqrt(8)/Sum(f(4*n)*(1103+26390*n)/f(n)**4/396**(4*n),
...                         (n, 0, oo))
>>> N(R, 10000)
3.141592653589793238462643383279502884197169399375105820974944592307816406286208
99862803482534211706798214808651328230664709384460955058223172535940812848111745
02841027019385211055596446229489549303819644288109756659334461284756482337867831
...


## Numerical simplification¶

The function nsimplify attempts to find a formula that is numerically equal to the given input. This feature can be used to guess an exact formula for an approximate floating-point input, or to guess a simpler formula for a complicated symbolic input. The algorithm used by nsimplify is capable of identifying simple fractions, simple algebraic expressions, linear combinations of given constants, and certain elementary functional transformations of any of the preceding.

Optionally, nsimplify can be passed a list of constants to include (e.g. pi) and a minimum numerical tolerance. Here are some elementary examples:

>>> nsimplify(0.1)
1/10
>>> nsimplify(6.28, [pi], tolerance=0.01)
2*pi
>>> nsimplify(pi, tolerance=0.01)
22/7
>>> nsimplify(pi, tolerance=0.001)
355
---
113
>>> nsimplify(0.33333, tolerance=1e-4)
1/3
>>> nsimplify(2.0**(1/3.), tolerance=0.001)
635
---
504
>>> nsimplify(2.0**(1/3.), tolerance=0.001, full=True)
3 ___
\/ 2


Here are several more advanced examples:

>>> nsimplify(Float('0.130198866629986772369127970337',30), [pi, E])
1
----------
5*pi
---- + 2*E
7
>>> nsimplify(cos(atan('1/3')))
____
3*\/ 10
--------
10
>>> nsimplify(4/(1+sqrt(5)), [GoldenRatio])
-2 + 2*GoldenRatio
>>> nsimplify(2 + exp(2*atan('1/4')*I))
49   8*I
-- + ---
17    17
>>> nsimplify((1/(exp(3*pi*I/5)+1)))
___________
/   ___
1        /  \/ 5    1
- - I*  /   ----- + -
2     \/      10    4
>>> nsimplify(I**I, [pi])
-pi
----
2
e
>>> n = Symbol('n')
>>> nsimplify(Sum(1/n**2, (n, 1, oo)), [pi])
2
pi
---
6
>>> nsimplify(gamma('1/4')*gamma('3/4'), [pi])
___
\/ 2 *pi