Re: minimizing sum{x(i)*log(x(i))}
- From: Walter Roberson <roberson@xxxxxxxxxxxx>
- Date: Sun, 21 Dec 2008 20:54:16 -0600
Ming wrote:
This is because log(0) = -inf and 0*-inf = NaN while mathematically x*log(x) = 0 when x
takes on zero.
No, mathematically x*log(x) is indeterminate when x takes on 0, and if your
code requires 0 * infinity to be any particular real number, the code is
mathematically broken.
One argument sometimes given is:
Let c be any real number. Then limit c/x as x approaches infinity is 0.
Now consider limit c/x * x as x approaches infinity. Do the usual "cancellation"
of the numerator and denominator, to find that limit c/x * x as x approaches infinity
"must be" c. But c was any arbitrary real constant, so 0 * infinity must be simultaneously
equal to all real constants, and therefore the answer to 0 * infinity is indeterminate.
--
..signature note: I am now avoiding replying to unclear or ambiguous postings.
Please review questions before posting them. Be specific. Use examples of what you mean,
of what you don't mean. Specify boundary conditions, and data classes and value
relationships -- what if we scrambled your data or used -Inf, NaN, or complex(rand,rand)?
.
- Follow-Ups:
- Re: minimizing sum{x(i)*log(x(i))}
- From: Roger Stafford
- Re: minimizing sum{x(i)*log(x(i))}
- From: Ming
- Re: minimizing sum{x(i)*log(x(i))}
- References:
- minimizing sum{x(i)*log(x(i))}
- From: Ming
- minimizing sum{x(i)*log(x(i))}
- Prev by Date: Re: minimizing sum{x(i)*log(x(i))}
- Next by Date: Re: How to integration a polynomial on a polygon?
- Previous by thread: Re: minimizing sum{x(i)*log(x(i))}
- Next by thread: Re: minimizing sum{x(i)*log(x(i))}
- Index(es):
Relevant Pages
|
Loading