Re: Roots of AX^2 + BX + C
- From: Gerry Myerson <gerry@xxxxxxxxxxxxxxxxxxxxxxxxx>
- Date: Mon, 27 Oct 2008 03:34:55 GMT
In article <87vdvff16y.fsf@xxxxxxxxxxxxxxxxxxxx>,
Phil Carmody <thefatphil_demunged@xxxxxxxxxxx> wrote:
Gerry Myerson <gerry@xxxxxxxxxxxxxxxxxxxxxxxxx> writes:
In article <87abcrizth.fsf@xxxxxxxxxxxxxxxxxxxx>,
Phil Carmody <thefatphil_demunged@xxxxxxxxxxx> wrote:
nils_von_nostrand@xxxxxxxxx writes:
Recently a student and I talked about the probability of the roots a
quadratic equation (AX^2 + B*X + C) to be complex. As we all know, the
roots will be complex if (B^2-4*A*C) is less than zero.
Can someone please explain why three (uniformly distributed) randomly
chosen numbers, A, B, C
Uniformly distributed, eh? On an infinite space?
How does that work?
There is no problem with a random variable uniformly distributed
on, say, the real numbers.
What would the (closed form for the) PDF be?
The probability of hitting any finite interval would have to be 0,
and of hitting any semi-infinite interval would have to be a
constant (or two, one for -> -oo, one for -> +oo).
Surely you need finite measure?
Yes, I think I was thinking of the finite measure case.
--
Gerry Myerson (gerry@xxxxxxxxxxxxxxx) (i -> u for email)
.
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- Roots of AX^2 + BX + C
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- Re: Roots of AX^2 + BX + C
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- From: Gerry Myerson
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