AngouriMath
StirlingRewritten(AngouriMath.Entity,AngouriMath.Entity.Variable,AngouriMath.Entity)
Method (no overloads)
Summary
The expression with the logarithm of every diverging factorial in it replaced by
Stirling's expansion, ornull where there is none or the error the
expansion drops would not vanish out of the answer.
Stirling's expansion, or
expansion drops would not vanish out of the answer.
Remarks
StirlingExponent(AngouriMath.Entity,AngouriMath.Entity,AngouriMath.Entity.Variable,AngouriMath.Entity) can state its guard as power / f -> 0 because
it knows the shape it is working in: the answer there is
e^(power * ln(base)) , so power *is* the coefficient the dropped
1/(12f) gets multiplied by. Anywhere else that coefficient has to be found
rather than read off, and it is found by putting a variable where the logarithm is
and differentiating with respect to it.
x * (ln(x!) - (x*ln(x) - x + ln(2*pi*x)/2)) is 1/12 -- an expression
built out of the dropped term itself -- and a rewrite that did not ask would answer
it0 . There the coefficient is x and x / x does not vanish, so
it is refused. Forln(x!) / x the coefficient is 1/x and the ratio is
1/x^2 , so it is allowed.
expression need not be linear in it: forln(x!)^2 the derivative is
2*ln(x!) , which grows like 2*x*ln(x) and is refused -- correctly, as
the difference of the squares isln(x)/6 and diverges.
#765
it knows the shape it is working in: the answer there is
rather than read off, and it is found by putting a variable where the logarithm is
and differentiating with respect to it.
built out of the dropped term itself -- and a rewrite that did not ask would answer
it
it is refused. For
expression need not be linear in it: for
the difference of the squares is
#765
Angouri © 2019-2023 · Project's repo · Site's repo · Octicons · Transparency · 1953 pages online