Wikipedia:Reference desk/Archives/Mathematics/2011 March 16

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March 16

2nd order difference equations vs 2nd order differential Equations

I need to clarify the distinction between difference and differential equations.

Am I correct in saying that given a characteristic equation for a second order difference or differential equation

$\lambda ^{2}+a\lambda +b=0$

The differential equation is stable if both roots are negative, but the difference equation is stable if the absolute value of both roots is less than one?

A differential equation will have a saddle point equilibrium if there is one positive and one negative root, the difference equation will have a saddle point if there is one root less than one in absolute value and the other root greater than one in absolute value?

Both the differential and difference equations will have complex roots (spirals) if the discriminant is negative?

Both the differential and difference equations will have pure imaginary roots (centres) if a=0 and b is negative. 118.208.133.15 (talk) 01:19, 16 March 2011 (UTC)

In the difference equation y_n+2+ay_n+1+by_n = 0 substitute y_n = λⁿ to get λ²+aλ+b = 0.

In the differential equation y' '(t)+ay'(t)+by(t) = 0 substitute y(t) = e^λt to get λ²+aλ+b = 0.

The λs have different interpretations in the two cases. Bo Jacoby (talk) 12:43, 18 March 2011 (UTC).

Probability

Prize: 5

Participant: 14 people (include my friend and i)

What is the probability of either of us win the prize? (and both of us win as well?)

Thanks!

218.188.16.142 (talk) 01:43, 16 March 2011 (UTC)

Assuming that each of the participants is equally likely to win a prize (i.e. this is something like a random draw, not a game of skill) and no participant can win two prizes, then the probability that exactly one out of you and your friend wins a prize is

${\frac {2\times 5\times 9}{14\times 13}}={\frac {90}{182}}\approx 49.5\%$

and the probability that you both win a prize is

${\frac {5\times 4}{14\times 13}}={\frac {20}{182}}\approx 11\%$

so the probability that one or both of you win a prize is just over 60%. Gandalf61 (talk) 09:12, 16 March 2011 (UTC)

[ec] There are ${\binom {14}{5}}=2002$ ways to distribute 5 prizes among 14 people. Of these, $2\cdot {\binom {12}{4}}=990$ give exactly one of you two a prize, and ${\binom {12}{3}}=220$ give both of you a prize. Thus the probability that at least one gets a prize is ${\frac {990+220}{2002}}={\frac {55}{91}}\approx 0.604$, and the probability that both get one is ${\frac {220}{2002}}={\frac {10}{91}}\approx 0.110$. See also Combination. -- Meni Rosenfeld (talk) 09:16, 16 March 2011 (UTC)

Wow Thanks a lot guys! I can never understand probability :S 218.188.16.142 (talk) 01:54, 17 March 2011 (UTC)

Don't tell yourself (nor anybody else) that there is something you can never understand. Tell yourself (and anybody else) that of course you can understand it when having it properly explained. Keep asking questions. Bo Jacoby (talk) 11:54, 17 March 2011 (UTC).