AngouriMath
StirlingByItsOwnLogarithm(AngouriMath.Entity,AngouriMath.Entity.Variable,AngouriMath.Entity)
Method (no overloads)
Summary
The logarithm of an expression holding a diverging factorial, with Stirling's
expansion written into it -- so that the limit ise to the limit of this --
ornull where there is no such factorial or the expansion would
not be sound.
expansion written into it -- so that the limit is
or
not be sound.
Remarks
StirlingRewritten(AngouriMath.Entity,AngouriMath.Entity.Variable,AngouriMath.Entity) reaches a factorial only where a logarithm is
already written down, andx! / x^x has none, so that shape had no limit at
all. It is read here by supplying the logarithm: for a positive expression
lim H is e^(lim ln H) , and ln H is where the expansion applies.
**Not by substituting e^(Stirling(f)) for the factorial**, which is the
obvious move and was measured to be much worse. It puts ane to a large
exponent into the expression, the machinery evaluates that constant to a
hundred-digit decimal, and everything downstream carries it:
lim x->+oo (x!/e^x)^(1/x) went from half a second to over a minute, on an
expression StirlingExponent(AngouriMath.Entity,AngouriMath.Entity,AngouriMath.Entity.Variable,AngouriMath.Entity) already answers. Going through the
logarithm keeps the onlye in the final answer.
The guard is PowerItAppearsTo(AngouriMath.Entity,AngouriMath.Entity,AngouriMath.Entity.Variable): it is the power of the factorial the
expression depends on, which is exactly the coefficientln(f!) carries in
ln H , so the dropped 1/(12f) has to vanish against it as everywhere
else.(x!)^x gives x , and x/(12x) does not vanish.
already written down, and
all. It is read here by supplying the logarithm: for a positive expression
obvious move and was measured to be much worse. It puts an
exponent into the expression, the machinery evaluates that constant to a
hundred-digit decimal, and everything downstream carries it:
expression StirlingExponent(AngouriMath.Entity,AngouriMath.Entity,AngouriMath.Entity.Variable,AngouriMath.Entity) already answers. Going through the
logarithm keeps the only
expression depends on, which is exactly the coefficient
else.
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