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# Note AccumulationBounds has an alias: AccumBounds
AccumulationBounds represent an interval `[a, b]`, which is always closed
at the ends. Here `a` and `b` can be any value from extended real numbers.
The intended meaning of AccummulationBounds is to give an approximate
location of the accumulation points of a real function at a limit point.
Let `a` and `b` be reals such that `a \le b`.
`\left\langle a, b\right\rangle = \{x \in \mathbb{R} \mid a \le x \le b\}`
`\left\langle -\infty, b\right\rangle = \{x \in \mathbb{R} \mid x \le b\} \cup \{-\infty, \infty\}`
`\left\langle a, \infty \right\rangle = \{x \in \mathbb{R} \mid a \le x\} \cup \{-\infty, \infty\}`
`\left\langle -\infty, \infty \right\rangle = \mathbb{R} \cup \{-\infty, \infty\}`
``oo`` and ``-oo`` are added to the second and third definition respectively,
since if either ``-oo`` or ``oo`` is an argument, then the other one should
be included (though not as an end point). This is forced, since we have,
for example, ``1/AccumBounds(0, 1) = AccumBounds(1, oo)``, and the limit at
`0` is not one-sided. As `x` tends to `0-`, then `1/x \rightarrow -\infty`, so `-\infty`
should be interpreted as belonging to ``AccumBounds(1, oo)`` though it need
not appear explicitly.
In many cases it suffices to know that the limit set is bounded.
However, in some other cases more exact information could be useful.
For example, all accumulation values of `\cos(x) + 1` are non-negative.
(``AccumBounds(-1, 1) + 1 = AccumBounds(0, 2)``)
A AccumulationBounds object is defined to be real AccumulationBounds,
if its end points are finite reals.
Let `X`, `Y` be real AccumulationBounds, then their sum, difference,
product are defined to be the following sets:
`X + Y = \{ x+y \mid x \in X \cap y \in Y\}`
`X - Y = \{ x-y \mid x \in X \cap y \in Y\}`
`X \times Y = \{ x \times y \mid x \in X \cap y \in Y\}`
When an AccumBounds is raised to a negative power, if 0 is contained
between the bounds then an infinite range is returned, otherwise if an
endpoint is 0 then a semi-infinite range with consistent sign will be returned.
AccumBounds in expressions behave a lot like Intervals but the
semantics are not necessarily the same. Division (or exponentiation
to a negative integer power) could be handled with *intervals* by
returning a union of the results obtained after splitting the
bounds between negatives and positives, but that is not done with
AccumBounds. In addition, bounds are assumed to be independent of
each other; if the same bound is used in more than one place in an
expression, the result may not be the supremum or infimum of the
expression (see below). Finally, when a boundary is ``1``,
exponentiation to the power of ``oo`` yields ``oo``, neither
``1`` nor ``nan``.
Examples
========
>>> from sympy import AccumBounds, sin, exp, log, pi, E, S, oo
>>> from sympy.abc import x
>>> AccumBounds(0, 1) + AccumBounds(1, 2)
AccumBounds(1, 3)
>>> AccumBounds(0, 1) - AccumBounds(0, 2)
AccumBounds(-2, 1)
>>> AccumBounds(-2, 3)*AccumBounds(-1, 1)
AccumBounds(-3, 3)
>>> AccumBounds(1, 2)*AccumBounds(3, 5)
AccumBounds(3, 10)
The exponentiation of AccumulationBounds is defined
as follows:
If 0 does not belong to `X` or `n > 0` then
`X^n = \{ x^n \mid x \in X\}`
>>> AccumBounds(1, 4)**(S(1)/2)
AccumBounds(1, 2)
otherwise, an infinite or semi-infinite result is obtained:
>>> 1/AccumBounds(-1, 1)
AccumBounds(-oo, oo)
>>> 1/AccumBounds(0, 2)
AccumBounds(1/2, oo)
>>> 1/AccumBounds(-oo, 0)
AccumBounds(-oo, 0)
A boundary of 1 will always generate all nonnegatives:
>>> AccumBounds(1, 2)**oo
AccumBounds(0, oo)
>>> AccumBounds(0, 1)**oo
AccumBounds(0, oo)
If the exponent is itself an AccumulationBounds or is not an
integer then unevaluated results will be returned unless the base
values are positive:
>>> AccumBounds(2, 3)**AccumBounds(-1, 2)
AccumBounds(1/3, 9)
>>> AccumBounds(-2, 3)**AccumBounds(-1, 2)
AccumBounds(-2, 3)**AccumBounds(-1, 2)
>>> AccumBounds(-2, -1)**(S(1)/2)
sqrt(AccumBounds(-2, -1))
Note: `\left\langle a, b\right\rangle^2` is not same as `\left\langle a, b\right\rangle \times \left\langle a, b\right\rangle`
>>> AccumBounds(-1, 1)**2
AccumBounds(0, 1)
>>> AccumBounds(1, 3) < 4
True
>>> AccumBounds(1, 3) < -1
False
Some elementary functions can also take AccumulationBounds as input.
A function `f` evaluated for some real AccumulationBounds `\left\langle a, b \right\rangle`
is defined as `f(\left\langle a, b\right\rangle) = \{ f(x) \mid a \le x \le b \}`
>>> sin(AccumBounds(pi/6, pi/3))
AccumBounds(1/2, sqrt(3)/2)
>>> exp(AccumBounds(0, 1))
AccumBounds(1, E)
>>> log(AccumBounds(1, E))
AccumBounds(0, 1)
Some symbol in an expression can be substituted for a AccumulationBounds
object. But it does not necessarily evaluate the AccumulationBounds for
that expression.
The same expression can be evaluated to different values depending upon
the form it is used for substitution since each instance of an
AccumulationBounds is considered independent. For example:
>>> (x**2 + 2*x + 1).subs(x, AccumBounds(-1, 1))
AccumBounds(-1, 4)
>>> ((x + 1)**2).subs(x, AccumBounds(-1, 1))
AccumBounds(0, 4)
References
==========
.. [1] https://en.wikipedia.org/wiki/Interval_arithmetic
.. [2] https://fab.cba.mit.edu/classes/S62.12/docs/Hickey_interval.pdf
Notes
=====
Do not use ``AccumulationBounds`` for floating point interval arithmetic
calculations, use ``mpmath.iv`` instead.
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