Re: minimizing sum{x(i)*log(x(i))}

Hi Walter,

I don't know what you meant by "indeterminate". If we define f(x) = x*log(x) and f(0) = lim{x->0} x*log(x), then f(0) = 0. This way, f(x) is continuous from right at 0.

In your argument, you made a mistake. limit{f(x)*g(x)} does not equal to limit{f(x)}*limit{g(x)}. But the point, 0*inf is indeterminate, is correct. We don't know what value it should take unless we know at what rate 0 and inf are approached.

thanks,
Ming



Walter Roberson <roberson@xxxxxxxxxxxx> wrote in message <2jD3l.8328$cL7.952@xxxxxxxxxxxx>...
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)?
.

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