Concrete

Hypergeometric terms

The center stage, in recurrence solving and summations, play hypergeometric terms. Formally these are sequences annihilated by first order linear recurrence operators. In simple words if we are given term $a(n)$ then it is hypergeometric if its consecutive term ratio is a rational function in $n$.

To check if a sequence is of this type you can use the is_hypergeometric method which is available in Basic class. Here is simple example involving a polynomial:

>>> from sympy import *
>>> n, k = symbols('n,k')
>>> (n**2 + 1).is_hypergeometric(n)
True

Of course polynomials are hypergeometric but are there any more complicated sequences of this type? Here are some trivial examples:

>>> factorial(n).is_hypergeometric(n)
True
>>> binomial(n, k).is_hypergeometric(n)
True
>>> rf(n, k).is_hypergeometric(n)
True
>>> ff(n, k).is_hypergeometric(n)
True
>>> gamma(n).is_hypergeometric(n)
True
>>> (2**n).is_hypergeometric(n)
True

We see that all species used in summations and other parts of concrete mathematics are hypergeometric. Note also that binomial coefficients and both rising and falling factorials are hypergeometric in both their arguments:

>>> binomial(n, k).is_hypergeometric(k)
True
>>> rf(n, k).is_hypergeometric(k)
True
>>> ff(n, k).is_hypergeometric(k)
True

To say more, all previously shown examples are valid for integer linear arguments:

>>> factorial(2*n).is_hypergeometric(n)
True
>>> binomial(3*n+1, k).is_hypergeometric(n)
True
>>> rf(n+1, k-1).is_hypergeometric(n)
True
>>> ff(n-1, k+1).is_hypergeometric(n)
True
>>> gamma(5*n).is_hypergeometric(n)
True
>>> (2**(n-7)).is_hypergeometric(n)
True

However nonlinear arguments make those sequences fail to be hypergeometric:

>>> factorial(n**2).is_hypergeometric(n)
False
>>> (2**(n**3 + 1)).is_hypergeometric(n)
False

If not only the knowledge of being hypergeometric or not is needed, you can use hypersimp() function. It will try to simplify combinatorial expression and if the term given is hypergeometric it will return a quotient of polynomials of minimal degree. Otherwise is will return $None$ to say that sequence is not hypergeometric:

>>> hypersimp(factorial(2*n), n)
2*(n + 1)*(2*n + 1)
>>> hypersimp(factorial(n**2), n)

Concrete Class Reference

class sympy.concrete.summations.Sum(function, *symbols, **assumptions)

Represents unevaluated summation.

Explanation

Sum represents a finite or infinite series, with the first argument being the general form of terms in the series, and the second argument being (dummy_variable, start, end), with dummy_variable taking all integer values from start through end. In accordance with long-standing mathematical convention, the end term is included in the summation.

Finite Sums

For finite sums (and sums with symbolic limits assumed to be finite) we follow the summation convention described by Karr [1], especially definition 3 of section 1.4. The sum:

$$\sum _{m\le i<n}f(i)$$

has the obvious meaning for $m<n$, namely:

$$\sum _{m\le i<n}f(i)=f(m)+f(m+1)+\dots +f(n-2)+f(n-1)$$

with the upper limit value $f(n)$ excluded. The sum over an empty set is zero if and only if $m=n$:

$$\sum _{m\le i<n}f(i)=0\mathrm{for}m=n$$

Finally, for all other sums over empty sets we assume the following definition:

$$\sum _{m\le i<n}f(i)=-\sum _{n\le i<m}f(i)\mathrm{for}m>n$$

It is important to note that Karr defines all sums with the upper limit being exclusive. This is in contrast to the usual mathematical notation, but does not affect the summation convention. Indeed we have:

$$\sum _{m\le i<n}f(i)=\sum _{i=m}^{n-1}f(i)$$

where the difference in notation is intentional to emphasize the meaning, with limits typeset on the top being inclusive.

Examples

>>> from sympy.abc import i, k, m, n, x
>>> from sympy import Sum, factorial, oo, IndexedBase, Function
>>> Sum(k, (k, 1, m))
Sum(k, (k, 1, m))
>>> Sum(k, (k, 1, m)).doit()
m**2/2 + m/2
>>> Sum(k**2, (k, 1, m))
Sum(k**2, (k, 1, m))
>>> Sum(k**2, (k, 1, m)).doit()
m**3/3 + m**2/2 + m/6
>>> Sum(x**k, (k, 0, oo))
Sum(x**k, (k, 0, oo))
>>> Sum(x**k, (k, 0, oo)).doit()
Piecewise((1/(1 - x), Abs(x) < 1), (Sum(x**k, (k, 0, oo)), True))
>>> Sum(x**k/factorial(k), (k, 0, oo)).doit()
exp(x)

Here are examples to do summation with symbolic indices. You can use either Function of IndexedBase classes:

>>> f = Function('f')
>>> Sum(f(n), (n, 0, 3)).doit()
f(0) + f(1) + f(2) + f(3)
>>> Sum(f(n), (n, 0, oo)).doit()
Sum(f(n), (n, 0, oo))
>>> f = IndexedBase('f')
>>> Sum(f[n]**2, (n, 0, 3)).doit()
f[0]**2 + f[1]**2 + f[2]**2 + f[3]**2

An example showing that the symbolic result of a summation is still valid for seemingly nonsensical values of the limits. Then the Karr convention allows us to give a perfectly valid interpretation to those sums by interchanging the limits according to the above rules:

>>> S = Sum(i, (i, 1, n)).doit()
>>> S
n**2/2 + n/2
>>> S.subs(n, -4)
6
>>> Sum(i, (i, 1, -4)).doit()
6
>>> Sum(-i, (i, -3, 0)).doit()
6

An explicit example of the Karr summation convention:

>>> S1 = Sum(i**2, (i, m, m+n-1)).doit()
>>> S1
m**2*n + m*n**2 - m*n + n**3/3 - n**2/2 + n/6
>>> S2 = Sum(i**2, (i, m+n, m-1)).doit()
>>> S2
-m**2*n - m*n**2 + m*n - n**3/3 + n**2/2 - n/6
>>> S1 + S2
0
>>> S3 = Sum(i, (i, m, m-1)).doit()
>>> S3
0

See also

summation, Product, sympy.concrete.products.product

References

[R99]

Michael Karr, “Summation in Finite Terms”, Journal of the ACM, Volume 28 Issue 2, April 1981, Pages 305-350 https://dl.acm.org/doi/10.1145/322248.322255

[R100]

https://en.wikipedia.org/wiki/Summation#Capital-sigma_notation

[R101]

https://en.wikipedia.org/wiki/Empty_sum

euler_maclaurin(

m=0,

n=0,

eps=0,

eval_integral=True,

)

Return an Euler-Maclaurin approximation of self, where m is the number of leading terms to sum directly and n is the number of terms in the tail.

With m = n = 0, this is simply the corresponding integral plus a first-order endpoint correction.

Returns (s, e) where s is the Euler-Maclaurin approximation and e is the estimated error (taken to be the magnitude of the first omitted term in the tail):

>>> from sympy.abc import k, a, b
>>> from sympy import Sum
>>> Sum(1/k, (k, 2, 5)).doit().evalf()
1.28333333333333
>>> s, e = Sum(1/k, (k, 2, 5)).euler_maclaurin()
>>> s
-log(2) + 7/20 + log(5)
>>> from sympy import sstr
>>> print(sstr((s.evalf(), e.evalf()), full_prec=True))
(1.26629073187415, 0.0175000000000000)

The endpoints may be symbolic:

>>> s, e = Sum(1/k, (k, a, b)).euler_maclaurin()
>>> s
-log(a) + log(b) + 1/(2*b) + 1/(2*a)
>>> e
Abs(1/(12*b**2) - 1/(12*a**2))

If the function is a polynomial of degree at most 2n+1, the Euler-Maclaurin formula becomes exact (and e = 0 is returned):

>>> Sum(k, (k, 2, b)).euler_maclaurin()
(b**2/2 + b/2 - 1, 0)
>>> Sum(k, (k, 2, b)).doit()
b**2/2 + b/2 - 1

With a nonzero eps specified, the summation is ended as soon as the remainder term is less than the epsilon.

eval_zeta_function(f, limits)

Check whether the function matches with the zeta function.

If it matches, then return a $Piecewise$ expression because zeta function does not converge unless $s>1$ and $q>0$

is_absolutely_convergent()

Checks for the absolute convergence of an infinite series.

Same as checking convergence of absolute value of sequence_term of an infinite series.

Examples

>>> from sympy import Sum, Symbol, oo
>>> n = Symbol('n', integer=True)
>>> Sum((-1)**n, (n, 1, oo)).is_absolutely_convergent()
False
>>> Sum((-1)**n/n**2, (n, 1, oo)).is_absolutely_convergent()
True

See also

Sum.is_convergent

References

[R102]

https://en.wikipedia.org/wiki/Absolute_convergence

is_convergent()

Checks for the convergence of a Sum.

Explanation

We divide the study of convergence of infinite sums and products in two parts.

First Part: One part is the question whether all the terms are well defined, i.e., they are finite in a sum and also non-zero in a product. Zero is the analogy of (minus) infinity in products as $e^{-\mathrm{\infty }}=0$.

Second Part: The second part is the question of convergence after infinities, and zeros in products, have been omitted assuming that their number is finite. This means that we only consider the tail of the sum or product, starting from some point after which all terms are well defined.

For example, in a sum of the form:

$$\sum _{1\le i<\mathrm{\infty }}\frac{1}{n^2+an+b}$$

where a and b are numbers. The routine will return true, even if there are infinities in the term sequence (at most two). An analogous product would be:

$$\prod _{1\le i<\mathrm{\infty }}{e}^{\frac{1}{n^2+an+b}}$$

This is how convergence is interpreted. It is concerned with what happens at the limit. Finding the bad terms is another independent matter.

Note: It is responsibility of user to see that the sum or product is well defined.

There are various tests employed to check the convergence like divergence test, root test, integral test, alternating series test, comparison tests, Dirichlet tests. It returns true if Sum is convergent and false if divergent and NotImplementedError if it cannot be checked.

Examples

>>> from sympy import factorial, S, Sum, Symbol, oo
>>> n = Symbol('n', integer=True)
>>> Sum(n/(n - 1), (n, 4, 7)).is_convergent()
True
>>> Sum(n/(2*n + 1), (n, 1, oo)).is_convergent()
False
>>> Sum(factorial(n)/5**n, (n, 1, oo)).is_convergent()
False
>>> Sum(1/n**(S(6)/5), (n, 1, oo)).is_convergent()
True

See also

Sum.is_absolutely_convergent, sympy.concrete.products.Product.is_convergent

References

[R103]

https://en.wikipedia.org/wiki/Convergence_tests

reverse_order(*indices)

Reverse the order of a limit in a Sum.

Explanation

reverse_order(self, *indices) reverses some limits in the expression self which can be either a Sum or a Product. The selectors in the argument indices specify some indices whose limits get reversed. These selectors are either variable names or numerical indices counted starting from the inner-most limit tuple.

Examples

>>> from sympy import Sum
>>> from sympy.abc import x, y, a, b, c, d

>>> Sum(x, (x, 0, 3)).reverse_order(x)
Sum(-x, (x, 4, -1))
>>> Sum(x*y, (x, 1, 5), (y, 0, 6)).reverse_order(x, y)
Sum(x*y, (x, 6, 0), (y, 7, -1))
>>> Sum(x, (x, a, b)).reverse_order(x)
Sum(-x, (x, b + 1, a - 1))
>>> Sum(x, (x, a, b)).reverse_order(0)
Sum(-x, (x, b + 1, a - 1))

While one should prefer variable names when specifying which limits to reverse, the index counting notation comes in handy in case there are several symbols with the same name.

>>> S = Sum(x**2, (x, a, b), (x, c, d))
>>> S
Sum(x**2, (x, a, b), (x, c, d))
>>> S0 = S.reverse_order(0)
>>> S0
Sum(-x**2, (x, b + 1, a - 1), (x, c, d))
>>> S1 = S0.reverse_order(1)
>>> S1
Sum(x**2, (x, b + 1, a - 1), (x, d + 1, c - 1))

Of course we can mix both notations:

>>> Sum(x*y, (x, a, b), (y, 2, 5)).reverse_order(x, 1)
Sum(x*y, (x, b + 1, a - 1), (y, 6, 1))
>>> Sum(x*y, (x, a, b), (y, 2, 5)).reverse_order(y, x)
Sum(x*y, (x, b + 1, a - 1), (y, 6, 1))

See also

sympy.concrete.expr_with_intlimits.ExprWithIntLimits.index, reorder_limit, sympy.concrete.expr_with_intlimits.ExprWithIntLimits.reorder

References

[R104]

Michael Karr, “Summation in Finite Terms”, Journal of the ACM, Volume 28 Issue 2, April 1981, Pages 305-350 https://dl.acm.org/doi/10.1145/322248.322255

class sympy.concrete.products.Product(function, *symbols, **assumptions)

Represents unevaluated products.

Explanation

Product represents a finite or infinite product, with the first argument being the general form of terms in the series, and the second argument being (dummy_variable, start, end), with dummy_variable taking all integer values from start through end. In accordance with long-standing mathematical convention, the end term is included in the product.

Finite Products

For finite products (and products with symbolic limits assumed to be finite) we follow the analogue of the summation convention described by Karr [1], especially definition 3 of section 1.4. The product:

$$\prod _{m\le i<n}f(i)$$

has the obvious meaning for $m<n$, namely:

$$\prod _{m\le i<n}f(i)=f(m)f(m+1)\cdot \dots \cdot f(n-2)f(n-1)$$

with the upper limit value $f(n)$ excluded. The product over an empty set is one if and only if $m=n$:

$$\prod _{m\le i<n}f(i)=1\mathrm{for}m=n$$

Finally, for all other products over empty sets we assume the following definition:

$$\prod _{m\le i<n}f(i)=\frac{1}{\prod _{n\le i<m}f(i)}\mathrm{for}m>n$$

It is important to note that above we define all products with the upper limit being exclusive. This is in contrast to the usual mathematical notation, but does not affect the product convention. Indeed we have:

$$\prod _{m\le i<n}f(i)=\prod _{i=m}^{n-1}f(i)$$

where the difference in notation is intentional to emphasize the meaning, with limits typeset on the top being inclusive.

Examples

>>> from sympy.abc import a, b, i, k, m, n, x
>>> from sympy import Product, oo
>>> Product(k, (k, 1, m))
Product(k, (k, 1, m))
>>> Product(k, (k, 1, m)).doit()
factorial(m)
>>> Product(k**2,(k, 1, m))
Product(k**2, (k, 1, m))
>>> Product(k**2,(k, 1, m)).doit()
factorial(m)**2

Wallis’ product for pi:

>>> W = Product(2*i/(2*i-1) * 2*i/(2*i+1), (i, 1, oo))
>>> W
Product(4*i**2/((2*i - 1)*(2*i + 1)), (i, 1, oo))

Direct computation currently fails:

>>> W.doit()
Product(4*i**2/((2*i - 1)*(2*i + 1)), (i, 1, oo))

But we can approach the infinite product by a limit of finite products:

>>> from sympy import limit
>>> W2 = Product(2*i/(2*i-1)*2*i/(2*i+1), (i, 1, n))
>>> W2
Product(4*i**2/((2*i - 1)*(2*i + 1)), (i, 1, n))
>>> W2e = W2.doit()
>>> W2e
4**n*factorial(n)**2/(2**(2*n)*RisingFactorial(1/2, n)*RisingFactorial(3/2, n))
>>> limit(W2e, n, oo)
pi/2

By the same formula we can compute sin(pi/2):

>>> from sympy import combsimp, pi, gamma, simplify
>>> P = pi * x * Product(1 - x**2/k**2, (k, 1, n))
>>> P = P.subs(x, pi/2)
>>> P
pi**2*Product(1 - pi**2/(4*k**2), (k, 1, n))/2
>>> Pe = P.doit()
>>> Pe
pi**2*RisingFactorial(1 - pi/2, n)*RisingFactorial(1 + pi/2, n)/(2*factorial(n)**2)
>>> limit(Pe, n, oo).gammasimp()
sin(pi**2/2)
>>> Pe.rewrite(gamma)
(-1)**n*pi**2*gamma(pi/2)*gamma(n + 1 + pi/2)/(2*gamma(1 + pi/2)*gamma(-n + pi/2)*gamma(n + 1)**2)

Products with the lower limit being larger than the upper one:

>>> Product(1/i, (i, 6, 1)).doit()
120
>>> Product(i, (i, 2, 5)).doit()
120

The empty product:

>>> Product(i, (i, n, n-1)).doit()
1

An example showing that the symbolic result of a product is still valid for seemingly nonsensical values of the limits. Then the Karr convention allows us to give a perfectly valid interpretation to those products by interchanging the limits according to the above rules:

>>> P = Product(2, (i, 10, n)).doit()
>>> P
2**(n - 9)
>>> P.subs(n, 5)
1/16
>>> Product(2, (i, 10, 5)).doit()
1/16
>>> 1/Product(2, (i, 6, 9)).doit()
1/16

An explicit example of the Karr summation convention applied to products:

>>> P1 = Product(x, (i, a, b)).doit()
>>> P1
x**(-a + b + 1)
>>> P2 = Product(x, (i, b+1, a-1)).doit()
>>> P2
x**(a - b - 1)
>>> simplify(P1 * P2)
1

And another one:

>>> P1 = Product(i, (i, b, a)).doit()
>>> P1
RisingFactorial(b, a - b + 1)
>>> P2 = Product(i, (i, a+1, b-1)).doit()
>>> P2
RisingFactorial(a + 1, -a + b - 1)
>>> P1 * P2
RisingFactorial(b, a - b + 1)*RisingFactorial(a + 1, -a + b - 1)
>>> combsimp(P1 * P2)
1

See also

Sum, summation, product

References

[R105]

Michael Karr, “Summation in Finite Terms”, Journal of the ACM, Volume 28 Issue 2, April 1981, Pages 305-350 https://dl.acm.org/doi/10.1145/322248.322255

[R106]

https://en.wikipedia.org/wiki/Multiplication#Capital_Pi_notation

[R107]

https://en.wikipedia.org/wiki/Empty_product

is_convergent()

See docs of Sum.is_convergent() for explanation of convergence in SymPy.

Explanation

The infinite product:

$$\prod _{1\le i<\mathrm{\infty }}f(i)$$

is defined by the sequence of partial products:

$$\prod _{i=1}^{n}f(i)=f(1)f(2)\cdots f(n)$$

as n increases without bound. The product converges to a non-zero value if and only if the sum:

$$\sum _{1\le i<\mathrm{\infty }}\log f(n)$$

converges.

Examples

>>> from sympy import Product, Symbol, cos, pi, exp, oo
>>> n = Symbol('n', integer=True)
>>> Product(n/(n + 1), (n, 1, oo)).is_convergent()
False
>>> Product(1/n**2, (n, 1, oo)).is_convergent()
False
>>> Product(cos(pi/n), (n, 1, oo)).is_convergent()
True
>>> Product(exp(-n**2), (n, 1, oo)).is_convergent()
False

References

[R108]

https://en.wikipedia.org/wiki/Infinite_product

reverse_order(*indices)

Reverse the order of a limit in a Product.

Explanation

reverse_order(expr, *indices) reverses some limits in the expression expr which can be either a Sum or a Product. The selectors in the argument indices specify some indices whose limits get reversed. These selectors are either variable names or numerical indices counted starting from the inner-most limit tuple.

Examples

>>> from sympy import gamma, Product, simplify, Sum
>>> from sympy.abc import x, y, a, b, c, d
>>> P = Product(x, (x, a, b))
>>> Pr = P.reverse_order(x)
>>> Pr
Product(1/x, (x, b + 1, a - 1))
>>> Pr = Pr.doit()
>>> Pr
1/RisingFactorial(b + 1, a - b - 1)
>>> simplify(Pr.rewrite(gamma))
Piecewise((gamma(b + 1)/gamma(a), b > -1), ((-1)**(-a + b + 1)*gamma(1 - a)/gamma(-b), True))
>>> P = P.doit()
>>> P
RisingFactorial(a, -a + b + 1)
>>> simplify(P.rewrite(gamma))
Piecewise((gamma(b + 1)/gamma(a), a > 0), ((-1)**(-a + b + 1)*gamma(1 - a)/gamma(-b), True))

While one should prefer variable names when specifying which limits to reverse, the index counting notation comes in handy in case there are several symbols with the same name.

>>> S = Sum(x*y, (x, a, b), (y, c, d))
>>> S
Sum(x*y, (x, a, b), (y, c, d))
>>> S0 = S.reverse_order(0)
>>> S0
Sum(-x*y, (x, b + 1, a - 1), (y, c, d))
>>> S1 = S0.reverse_order(1)
>>> S1
Sum(x*y, (x, b + 1, a - 1), (y, d + 1, c - 1))

Of course we can mix both notations:

>>> Sum(x*y, (x, a, b), (y, 2, 5)).reverse_order(x, 1)
Sum(x*y, (x, b + 1, a - 1), (y, 6, 1))
>>> Sum(x*y, (x, a, b), (y, 2, 5)).reverse_order(y, x)
Sum(x*y, (x, b + 1, a - 1), (y, 6, 1))

See also

sympy.concrete.expr_with_intlimits.ExprWithIntLimits.index, reorder_limit, sympy.concrete.expr_with_intlimits.ExprWithIntLimits.reorder

References

[R109]

Michael Karr, “Summation in Finite Terms”, Journal of the ACM, Volume 28 Issue 2, April 1981, Pages 305-350 https://dl.acm.org/doi/10.1145/322248.322255

class sympy.concrete.expr_with_intlimits.ExprWithIntLimits(

function,

*symbols,

**assumptions,

)

Superclass for Product and Sum.

See also

sympy.concrete.expr_with_limits.ExprWithLimits, sympy.concrete.products.Product, sympy.concrete.summations.Sum

change_index(

var,

trafo,

newvar=None,

)

Change index of a Sum or Product.

Perform a linear transformation $x\mapsto ax+b$ on the index variable $x$. For $a$ the only values allowed are $\pm 1$. A new variable to be used after the change of index can also be specified.

Explanation

change_index(expr, var, trafo, newvar=None) where var specifies the index variable $x$ to transform. The transformation trafo must be linear and given in terms of var. If the optional argument newvar is provided then var gets replaced by newvar in the final expression.

Examples

>>> from sympy import Sum, Product, simplify
>>> from sympy.abc import x, y, a, b, c, d, u, v, i, j, k, l

>>> S = Sum(x, (x, a, b))
>>> S.doit()
-a**2/2 + a/2 + b**2/2 + b/2

>>> Sn = S.change_index(x, x + 1, y)
>>> Sn
Sum(y - 1, (y, a + 1, b + 1))
>>> Sn.doit()
-a**2/2 + a/2 + b**2/2 + b/2

>>> Sn = S.change_index(x, -x, y)
>>> Sn
Sum(-y, (y, -b, -a))
>>> Sn.doit()
-a**2/2 + a/2 + b**2/2 + b/2

>>> Sn = S.change_index(x, x+u)
>>> Sn
Sum(-u + x, (x, a + u, b + u))
>>> Sn.doit()
-a**2/2 - a*u + a/2 + b**2/2 + b*u + b/2 - u*(-a + b + 1) + u
>>> simplify(Sn.doit())
-a**2/2 + a/2 + b**2/2 + b/2

>>> Sn = S.change_index(x, -x - u, y)
>>> Sn
Sum(-u - y, (y, -b - u, -a - u))
>>> Sn.doit()
-a**2/2 - a*u + a/2 + b**2/2 + b*u + b/2 - u*(-a + b + 1) + u
>>> simplify(Sn.doit())
-a**2/2 + a/2 + b**2/2 + b/2

>>> P = Product(i*j**2, (i, a, b), (j, c, d))
>>> P
Product(i*j**2, (i, a, b), (j, c, d))
>>> P2 = P.change_index(i, i+3, k)
>>> P2
Product(j**2*(k - 3), (k, a + 3, b + 3), (j, c, d))
>>> P3 = P2.change_index(j, -j, l)
>>> P3
Product(l**2*(k - 3), (k, a + 3, b + 3), (l, -d, -c))

When dealing with symbols only, we can make a general linear transformation:

>>> Sn = S.change_index(x, u*x+v, y)
>>> Sn
Sum((-v + y)/u, (y, b*u + v, a*u + v))
>>> Sn.doit()
-v*(a*u - b*u + 1)/u + (a**2*u**2/2 + a*u*v + a*u/2 - b**2*u**2/2 - b*u*v + b*u/2 + v)/u
>>> simplify(Sn.doit())
a**2*u/2 + a/2 - b**2*u/2 + b/2

However, the last result can be inconsistent with usual summation where the index increment is always 1. This is obvious as we get back the original value only for u equal +1 or -1.

See also

sympy.concrete.expr_with_intlimits.ExprWithIntLimits.index, reorder_limit, sympy.concrete.expr_with_intlimits.ExprWithIntLimits.reorder, sympy.concrete.summations.Sum.reverse_order, sympy.concrete.products.Product.reverse_order

property has_empty_sequence

Returns True if the Sum or Product is computed for an empty sequence.

Examples

>>> from sympy import Sum, Product, Symbol
>>> m = Symbol('m')
>>> Sum(m, (m, 1, 0)).has_empty_sequence
True

>>> Sum(m, (m, 1, 1)).has_empty_sequence
False

>>> M = Symbol('M', integer=True, positive=True)
>>> Product(m, (m, 1, M)).has_empty_sequence
False

>>> Product(m, (m, 2, M)).has_empty_sequence

>>> Product(m, (m, M + 1, M)).has_empty_sequence
True

>>> N = Symbol('N', integer=True, positive=True)
>>> Sum(m, (m, N, M)).has_empty_sequence

>>> N = Symbol('N', integer=True, negative=True)
>>> Sum(m, (m, N, M)).has_empty_sequence
False

See also

has_reversed_limits, has_finite_limits

index(x)

Return the index of a dummy variable in the list of limits.

Explanation

index(expr, x) returns the index of the dummy variable x in the limits of expr. Note that we start counting with 0 at the inner-most limits tuple.

Examples

>>> from sympy.abc import x, y, a, b, c, d
>>> from sympy import Sum, Product
>>> Sum(x*y, (x, a, b), (y, c, d)).index(x)
0
>>> Sum(x*y, (x, a, b), (y, c, d)).index(y)
1
>>> Product(x*y, (x, a, b), (y, c, d)).index(x)
0
>>> Product(x*y, (x, a, b), (y, c, d)).index(y)
1

See also

reorder_limit, reorder, sympy.concrete.summations.Sum.reverse_order, sympy.concrete.products.Product.reverse_order

reorder(*arg)

Reorder limits in a expression containing a Sum or a Product.

Explanation

expr.reorder(*arg) reorders the limits in the expression expr according to the list of tuples given by arg. These tuples can contain numerical indices or index variable names or involve both.

Examples

>>> from sympy import Sum, Product
>>> from sympy.abc import x, y, z, a, b, c, d, e, f

>>> Sum(x*y, (x, a, b), (y, c, d)).reorder((x, y))
Sum(x*y, (y, c, d), (x, a, b))

>>> Sum(x*y*z, (x, a, b), (y, c, d), (z, e, f)).reorder((x, y), (x, z), (y, z))
Sum(x*y*z, (z, e, f), (y, c, d), (x, a, b))

>>> P = Product(x*y*z, (x, a, b), (y, c, d), (z, e, f))
>>> P.reorder((x, y), (x, z), (y, z))
Product(x*y*z, (z, e, f), (y, c, d), (x, a, b))

We can also select the index variables by counting them, starting with the inner-most one:

>>> Sum(x**2, (x, a, b), (x, c, d)).reorder((0, 1))
Sum(x**2, (x, c, d), (x, a, b))

And of course we can mix both schemes:

>>> Sum(x*y, (x, a, b), (y, c, d)).reorder((y, x))
Sum(x*y, (y, c, d), (x, a, b))
>>> Sum(x*y, (x, a, b), (y, c, d)).reorder((y, 0))
Sum(x*y, (y, c, d), (x, a, b))

See also

reorder_limit, index, sympy.concrete.summations.Sum.reverse_order, sympy.concrete.products.Product.reverse_order

reorder_limit(x, y)

Interchange two limit tuples of a Sum or Product expression.

Explanation

expr.reorder_limit(x, y) interchanges two limit tuples. The arguments x and y are integers corresponding to the index variables of the two limits which are to be interchanged. The expression expr has to be either a Sum or a Product.

Examples

>>> from sympy.abc import x, y, z, a, b, c, d, e, f
>>> from sympy import Sum, Product

>>> Sum(x*y*z, (x, a, b), (y, c, d), (z, e, f)).reorder_limit(0, 2)
Sum(x*y*z, (z, e, f), (y, c, d), (x, a, b))
>>> Sum(x**2, (x, a, b), (x, c, d)).reorder_limit(1, 0)
Sum(x**2, (x, c, d), (x, a, b))

>>> Product(x*y*z, (x, a, b), (y, c, d), (z, e, f)).reorder_limit(0, 2)
Product(x*y*z, (z, e, f), (y, c, d), (x, a, b))

See also

index, reorder, sympy.concrete.summations.Sum.reverse_order, sympy.concrete.products.Product.reverse_order

Concrete Functions Reference

sympy.concrete.summations.summation(f, *symbols, **kwargs)

Compute the summation of f with respect to symbols.

Explanation

The notation for symbols is similar to the notation used in Integral. summation(f, (i, a, b)) computes the sum of f with respect to i from a to b, i.e.,

                            b
                          ____
                          \   `
summation(f, (i, a, b)) =  )    f
                          /___,
                          i = a

If it cannot compute the sum, it returns an unevaluated Sum object. Repeated sums can be computed by introducing additional symbols tuples:

.. rubric:: Examples

>>> from sympy import summation, oo, symbols, log
>>> i, n, m = symbols('i n m', integer=True)

>>> summation(2*i - 1, (i, 1, n))
n**2
>>> summation(1/2**i, (i, 0, oo))
2
>>> summation(1/log(n)**n, (n, 2, oo))
Sum(log(n)**(-n), (n, 2, oo))
>>> summation(i, (i, 0, n), (n, 0, m))
m**3/6 + m**2/2 + m/3

>>> from sympy.abc import x
>>> from sympy import factorial
>>> summation(x**n/factorial(n), (n, 0, oo))
exp(x)

See also

Sum, Product, sympy.concrete.products.product

sympy.concrete.products.product(*args, **kwargs)

Compute the product.

Explanation

The notation for symbols is similar to the notation used in Sum or Integral. product(f, (i, a, b)) computes the product of f with respect to i from a to b, i.e.,

                             b
                           _____
product(f(n), (i, a, b)) = |   | f(n)
                           |   |
                           i = a

If it cannot compute the product, it returns an unevaluated Product object. Repeated products can be computed by introducing additional symbols tuples:

.. rubric:: Examples

>>> from sympy import product, symbols
>>> i, n, m, k = symbols('i n m k', integer=True)

>>> product(i, (i, 1, k))
factorial(k)
>>> product(m, (i, 1, k))
m**k
>>> product(i, (i, 1, k), (k, 1, n))
Product(factorial(k), (k, 1, n))

sympy.concrete.gosper.gosper_normal(f, g, n, polys=True)

Compute the Gosper’s normal form of f and g.

Explanation

Given relatively prime univariate polynomials f and g, rewrite their quotient to a normal form defined as follows:

$$\frac{f(n)}{g(n)}=Z\cdot \frac{A(n)C(n+1)}{B(n)C(n)}$$

where Z is an arbitrary constant and A, B, C are monic polynomials in n with the following properties:

1. $gcd(A(n),B(n+h))=1\mathrm{\forall }h\in \mathbb{N}$

2. $gcd(B(n),C(n+1))=1$

3. $gcd(A(n),C(n))=1$

This normal form, or rational factorization in other words, is a crucial step in Gosper’s algorithm and in solving of difference equations. It can be also used to decide if two hypergeometric terms are similar or not.

This procedure will return a tuple containing elements of this factorization in the form (Z*A, B, C).

Examples

>>> from sympy.concrete.gosper import gosper_normal
>>> from sympy.abc import n

>>> gosper_normal(4*n+5, 2*(4*n+1)*(2*n+3), n, polys=False)
(1/4, n + 3/2, n + 1/4)

sympy.concrete.gosper.gosper_term(f, n)

Compute Gosper’s hypergeometric term for f.

Explanation

Suppose f is a hypergeometric term such that:

$$s_n=\sum _{k=0}^{n-1}{f}_k$$

and $f_k$ does not depend on $n$. Returns a hypergeometric term $g_n$ such that $g_{n+1}-g_n=f_n$.

Examples

>>> from sympy.concrete.gosper import gosper_term
>>> from sympy import factorial
>>> from sympy.abc import n

>>> gosper_term((4*n + 1)*factorial(n)/factorial(2*n + 1), n)
(-n - 1/2)/(n + 1/4)

sympy.concrete.gosper.gosper_sum(f, k)

Gosper’s hypergeometric summation algorithm.

Explanation

Given a hypergeometric term f such that:

$$s_n=\sum _{k=0}^{n-1}{f}_k$$

and $f(n)$ does not depend on $n$, returns $g_n-g(0)$ where $g_{n+1}-g_n=f_n$, or None if $s_n$ cannot be expressed in closed form as a sum of hypergeometric terms.

Examples

>>> from sympy.concrete.gosper import gosper_sum
>>> from sympy import factorial
>>> from sympy.abc import n, k

>>> f = (4*k + 1)*factorial(k)/factorial(2*k + 1)
>>> gosper_sum(f, (k, 0, n))
(-factorial(n) + 2*factorial(2*n + 1))/factorial(2*n + 1)
>>> _.subs(n, 2) == sum(f.subs(k, i) for i in [0, 1, 2])
True
>>> gosper_sum(f, (k, 3, n))
(-60*factorial(n) + factorial(2*n + 1))/(60*factorial(2*n + 1))
>>> _.subs(n, 5) == sum(f.subs(k, i) for i in [3, 4, 5])
True

References

[R110]

Marko Petkovsek, Herbert S. Wilf, Doron Zeilberger, A = B, AK Peters, Ltd., Wellesley, MA, USA, 1997, pp. 73–100

Concrete Guess Functions Reference

sympy.concrete.guess.guess(

l,

all=False,

evaluate=True,

niter=2,

variables=None,

)

This function is adapted from the Rate.m package for Mathematica written by Christian Krattenthaler. It tries to guess a formula from a given sequence of rational numbers.

Explanation

In order to speed up the process, the ‘all’ variable is set to False by default, stopping the computation as some results are returned during an iteration; the variable can be set to True if more iterations are needed (other formulas may be found; however they may be equivalent to the first ones).

Another option is the ‘evaluate’ variable (default is True); setting it to False will leave the involved products unevaluated.

By default, the number of iterations is set to 2 but a greater value (up to len(l)-1) can be specified with the optional ‘niter’ variable. More and more convoluted results are found when the order of the iteration gets higher:

• first iteration returns polynomial or rational functions;

• second iteration returns products of rising factorials and their inverses;

• third iteration returns products of products of rising factorials and their inverses;

• etc.

The returned formulas contain symbols i0, i1, i2, … where the main variables is i0 (and auxiliary variables are i1, i2, …). A list of other symbols can be provided in the ‘variables’ option; the length of the least should be the value of ‘niter’ (more is acceptable but only the first symbols will be used); in this case, the main variable will be the first symbol in the list.

Examples

>>> from sympy.concrete.guess import guess
>>> guess([1,2,6,24,120], evaluate=False)
[Product(i1 + 1, (i1, 1, i0 - 1))]

>>> from sympy import symbols
>>> r = guess([1,2,7,42,429,7436,218348,10850216], niter=4)
>>> i0 = symbols("i0")
>>> [r[0].subs(i0,n).doit() for n in range(1,10)]
[1, 2, 7, 42, 429, 7436, 218348, 10850216, 911835460]

sympy.concrete.guess.guess_generating_function(

v,

X=x,

types=['all'],

maxsqrtn=2,

)

Tries to “guess” a generating function for a sequence of rational numbers v. Only a few patterns are implemented yet.

Explanation

The function returns a dictionary where keys are the name of a given type of generating function. Six types are currently implemented:

type	formal definition
ogf	f(x) = Sum(            a_k * x^k       ,  k: 0..infinity )
egf	f(x) = Sum(            a_k * x^k / k!  ,  k: 0..infinity )
lgf	f(x) = Sum( (-1)^(k+1) a_k * x^k / k   ,  k: 1..infinity ) (with initial index being hold as 1 rather than 0)
hlgf	f(x) = Sum(            a_k * x^k / k   ,  k: 1..infinity ) (with initial index being hold as 1 rather than 0)
lgdogf	f(x) = derivate( log(Sum( a_k * x^k, k: 0..infinity )), x)
lgdegf	f(x) = derivate( log(Sum( a_k * x^k / k!, k: 0..infinity )), x)

In order to spare time, the user can select only some types of generating functions (default being [‘all’]). While forgetting to use a list in the case of a single type may seem to work most of the time as in: types=’ogf’ this (convenient) syntax may lead to unexpected extra results in some cases.

Discarding a type when calling the function does not mean that the type will not be present in the returned dictionary; it only means that no extra computation will be performed for that type, but the function may still add it in the result when it can be easily converted from another type.

Two generating functions (lgdogf and lgdegf) are not even computed if the initial term of the sequence is 0; it may be useful in that case to try again after having removed the leading zeros.

Examples

>>> from sympy.concrete.guess import guess_generating_function as ggf
>>> ggf([k+1 for k in range(12)], types=['ogf', 'lgf', 'hlgf'])
{'hlgf': 1/(1 - x), 'lgf': 1/(x + 1), 'ogf': 1/(x**2 - 2*x + 1)}

>>> from sympy import sympify
>>> l = sympify("[3/2, 11/2, 0, -121/2, -363/2, 121]")
>>> ggf(l)
{'ogf': (x + 3/2)/(11*x**2 - 3*x + 1)}

>>> from sympy import fibonacci
>>> ggf([fibonacci(k) for k in range(5, 15)], types=['ogf'])
{'ogf': (3*x + 5)/(-x**2 - x + 1)}

>>> from sympy import factorial
>>> ggf([factorial(k) for k in range(12)], types=['ogf', 'egf', 'lgf'])
{'egf': 1/(1 - x)}

>>> ggf([k+1 for k in range(12)], types=['egf'])
{'egf': (x + 1)*exp(x), 'lgdegf': (x + 2)/(x + 1)}

N-th root of a rational function can also be detected (below is an example coming from the sequence A108626 from https://oeis.org). The greatest n-th root to be tested is specified as maxsqrtn (default 2).

>>> ggf([1, 2, 5, 14, 41, 124, 383, 1200, 3799, 12122, 38919])['ogf']
sqrt(1/(x**4 + 2*x**2 - 4*x + 1))

References

[R111]

“Concrete Mathematics”, R.L. Graham, D.E. Knuth, O. Patashnik

[R112]

https://oeis.org/wiki/Generating_functions

sympy.concrete.guess.guess_generating_function_rational(v, X=x)

Tries to “guess” a rational generating function for a sequence of rational numbers v.

Examples

>>> from sympy.concrete.guess import guess_generating_function_rational
>>> from sympy import fibonacci
>>> l = [fibonacci(k) for k in range(5,15)]
>>> guess_generating_function_rational(l)
(3*x + 5)/(-x**2 - x + 1)

See also

sympy.series.approximants.approximants

sympy.concrete.guess.rationalize(x, maxcoeff=10000)

Helps identifying a rational number from a float (or mpmath.mpf) value by using a continued fraction. The algorithm stops as soon as a large partial quotient is detected (greater than 10000 by default).

Examples

>>> from sympy.concrete.guess import rationalize
>>> from mpmath import cos, pi
>>> rationalize(cos(pi/3))
1/2

>>> from mpmath import mpf
>>> rationalize(mpf("0.333333333333333"))
1/3

While the function is rather intended to help ‘identifying’ rational values, it may be used in some cases for approximating real numbers. (Though other functions may be more relevant in that case.)

>>> rationalize(pi, maxcoeff = 250)
355/113

Several other methods can approximate a real number as a rational, like:

• fractions.Fraction.from_decimal

• fractions.Fraction.from_float

• mpmath.identify

• mpmath.pslq by using the following syntax: mpmath.pslq([x, 1])

• mpmath.findpoly by using the following syntax: mpmath.findpoly(x, 1)

• sympy.simplify.nsimplify (which is a more general function)

The main difference between the current function and all these variants is that control focuses on magnitude of partial quotients here rather than on global precision of the approximation. If the real is “known to be” a rational number, the current function should be able to detect it correctly with the default settings even when denominator is great (unless its expansion contains unusually big partial quotients) which may occur when studying sequences of increasing numbers. If the user cares more on getting simple fractions, other methods may be more convenient.

sympy.concrete.guess.find_simple_recurrence(v, A=a, N=n)

Detects and returns a recurrence relation from a sequence of several integer (or rational) terms. The name of the function in the returned expression is ‘a’ by default; the main variable is ‘n’ by default. The smallest index in the returned expression is always n (and never n-1, n-2, etc.).

Examples

>>> from sympy.concrete.guess import find_simple_recurrence
>>> from sympy import fibonacci
>>> find_simple_recurrence([fibonacci(k) for k in range(12)])
-a(n) - a(n + 1) + a(n + 2)

>>> from sympy import Function, Symbol
>>> a = [1, 1, 1]
>>> for k in range(15): a.append(5*a[-1]-3*a[-2]+8*a[-3])
>>> find_simple_recurrence(a, A=Function('f'), N=Symbol('i'))
-8*f(i) + 3*f(i + 1) - 5*f(i + 2) + f(i + 3)

sympy.concrete.guess.find_simple_recurrence_vector(l)

This function is used internally by other functions from the sympy.concrete.guess module. While most users may want to rather use the function find_simple_recurrence when looking for recurrence relations among rational numbers, the current function may still be useful when some post-processing has to be done.

Explanation

The function returns a vector of length n when a recurrence relation of order n is detected in the sequence of rational numbers v.

If the returned vector has a length 1, then the returned value is always the list [0], which means that no relation has been found.

While the functions is intended to be used with rational numbers, it should work for other kinds of real numbers except for some cases involving quadratic numbers; for that reason it should be used with some caution when the argument is not a list of rational numbers.

Examples

>>> from sympy.concrete.guess import find_simple_recurrence_vector
>>> from sympy import fibonacci
>>> find_simple_recurrence_vector([fibonacci(k) for k in range(12)])
[1, -1, -1]

See also

sympy.concrete.guess.find_simple_recurrence

a more user-friendly version.