Limits of Sequences
Provides methods to compute limit of terms having sequences at infinity.
- sympy.series.limitseq.difference_delta(expr, n=None, step=1)[source]
Difference Operator.
Explanation
Discrete analog of differential operator. Given a sequence x[n], returns the sequence x[n + step] - x[n].
Examples
>>> from sympy import difference_delta as dd >>> from sympy.abc import n >>> dd(n*(n + 1), n) 2*n + 2 >>> dd(n*(n + 1), n, 2) 4*n + 6
References
- sympy.series.limitseq.dominant(expr, n)[source]
Finds the dominant term in a sum, that is a term that dominates every other term.
Explanation
If limit(a/b, n, oo) is oo then a dominates b. If limit(a/b, n, oo) is 0 then b dominates a. Otherwise, a and b are comparable.
If there is no unique dominant term, then returns
None
.Examples
>>> from sympy import Sum >>> from sympy.series.limitseq import dominant >>> from sympy.abc import n, k >>> dominant(5*n**3 + 4*n**2 + n + 1, n) 5*n**3 >>> dominant(2**n + Sum(k, (k, 0, n)), n) 2**n
See also
- sympy.series.limitseq.limit_seq(expr, n=None, trials=5)[source]
Finds the limit of a sequence as index
n
tends to infinity.- Parameters
expr : Expr
SymPy expression for the
n-th
term of the sequencen : Symbol, optional
The index of the sequence, an integer that tends to positive infinity. If None, inferred from the expression unless it has multiple symbols.
trials: int, optional
The algorithm is highly recursive.
trials
is a safeguard from infinite recursion in case the limit is not easily computed by the algorithm. Try increasingtrials
if the algorithm returnsNone
.
Admissible Terms
The algorithm is designed for sequences built from rational functions, indefinite sums, and indefinite products over an indeterminate n. Terms of alternating sign are also allowed, but more complex oscillatory behavior is not supported.
Examples
>>> from sympy import limit_seq, Sum, binomial >>> from sympy.abc import n, k, m >>> limit_seq((5*n**3 + 3*n**2 + 4) / (3*n**3 + 4*n - 5), n) 5/3 >>> limit_seq(binomial(2*n, n) / Sum(binomial(2*k, k), (k, 1, n)), n) 3/4 >>> limit_seq(Sum(k**2 * Sum(2**m/m, (m, 1, k)), (k, 1, n)) / (2**n*n), n) 4
See also
References
- R736
Computing Limits of Sequences - Manuel Kauers