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November 26

Incorrect theorem

Has there ever been a theorem which was accepted by mathematicians and used in other proofs etc. but was later shown to be false? Thanks! 194.80.32.9 (talk) 12:32, 26 November 2008 (UTC)[reply]

I do not know of relevant cases; it may happen. But important new theorems are so deeply checked by mathematicians that it is very unlikely that a mistake surives for a long time. On the contrary, unimportant theorems may survive forever even with mistakes; reviews are full of such cases: but who cares? After all for an unimportant theorem it is not so a serious defect to be false. The only historic case that I recall, close to your query, of a mistake in maths with a relevant consequence, is the mistake made by William Shanks in the computation of Pi: he computed by hand 707 digits of Pi, making a mistake at around the 500th --thus loosing all the following; the mistake was discovered only in 1944, when computers became powerfull enough to check Shank's computation. In the meanwhile the wrong decimal expansion of Pi was copied and printed in all numeric tables, and also in the (of course, circular) room "Salle Pi" of the Palais de la Découverte, in Paris (they had to re-paint the walls, as a consequence).--PMajer (talk) 13:55, 26 November 2008 (UTC)[reply]

Why would anyone need that many digits of

\pi

in a table? If you have computers to process it with, I can see that it could be relevant, but in that age? Bromskloss (talk) 00:06, 28 November 2008 (UTC)[reply]

I think this is commonplace. Tables very often have omissions or duplications, so most classification style results are usually a little wrong. John Thompson's N-groups paper left out one of the groups and this error is reproduced in new work. The proof of uniqueness for Thompson's sporadic group relied on a smaller uniqueness theorem that was false, so a correction had to be written. Holt and Plesken's Perfect Groups classification is missing two groups, and Sims's Primitive Groups classification switched two groups. Robinson's classification of PT groups includes a typo that drastically alters its impact, and several papers are having corrigenda submitted. The classification of 2-generated p-groups of class 2 omitted some groups, again causing a small flurry of corrigenda. JackSchmidt (talk) 14:28, 26 November 2008 (UTC)[reply]

The Gibbs phenomenon is interesting as even when somebody saw it he didn't believe it. Dmcq (talk) 14:35, 26 November 2008 (UTC)[reply]

See List of published false theorems, but it doesn't have anything juicy. PrimeHunter (talk) 19:45, 26 November 2008 (UTC)[reply]

Malfatti circles are a curious example of solving a mistaken problem. —Tamfang (talk) 01:25, 27 November 2008 (UTC)[reply]

See Henri Poincaré#The three-body problem for where he made a mistake, and was awarded a prize for it. He paid for a new issue of his paper to be issued. He thought the solution was stable but it was an early example for chaos theory. See also Henri Poincaré. A Life in the Service of Science Dmcq (talk) 19:49, 1 December 2008 (UTC)[reply]

How often (if ever) do lotteries have a positive expectation?

I am willing to buy a lottery ticket only when the lottery has a positive expectation. How often, if ever, does this happen in practice please? Some lotteries such as Euromillions have "roll overs" that might lead to this.

There are two aspects to this: estimating a positive expectation in advance of the draw, and estimating some time after the event when some statistics may have been made available (although I think the operators may be deliberately vague about them - I am not a lottery expert). I'd be interested in either. Thanks. 89.243.205.213 (talk) 17:27, 26 November 2008 (UTC)[reply]

(edit conflict) I have seen statements that such roll-over lotteries (such as Powerball in the US) have positive expectations at times. However, those huge jackpots (and associated media coverage) attract large numbers of players, which increases the likelihood that the winner will have to split the prize. I'm not sure if the "positive expectation" claims take this effect into account. -- Coneslayer (talk) 17:35, 26 November 2008 (UTC)[reply]

You don't need to take it into account. The expectation is just the prize fund divided by the number of tickets sold, it doesn't matter whether people are winning the jackpot or smaller wins or splitting wins or whatever. For something like the UK lottery (which gives a set portion of each ticket to the prize fund, I don't know how others work) without a roll over, that's easy, 50p from every ticket goes into the prize fund (the rest is tax, good causes, administration and profit), so the expectation is 50p (ie. losing half your money). With a rollover, you have to take the amount rolled over from last week, divide it by the number of tickets bought this week (you can estimate that pretty well from historical numbers - just remember to compare with other rollover weeks, not standard weeks) and add that to the expectation. Whether that's enough to reach positive expectation ever, I don't know, but it's theoretically possible (it would require multiple rollovers, though, since you need to more than double the expectation and one rollover won't even get close since not all the prize fund is rolled over and more tickets are bought than usual, so the extra expectation will be less than the usual 50p). --Tango (talk) 19:49, 26 November 2008 (UTC)[reply]

Why don't you need to take spitting wins into account?? I mean, imagine a lottery where there are 10 numbers, and you simply have to guess which number comes up, with the prize being $10. If only 1 person is playing, their expected winnings is $1. If two people are playing, then you expect to get the right number 1/10 times, and in this case you expect the other person to have the same number as you 1/10 times (so that you get $5) and a different non-winning number 9/10 times (so you get $10), hence your expected winnings is 1/10*9/10*$10 + 1/10*1/10*$5 = 95c. Also, why would the expected winnings be the prize fund divided by the number of tickets sold, since in the lottery above, with one player $10/1 ticket = $10, yet, if I'm correct, you would only expect $1 in winnings? 202.37.62.209 (talk) 03:35, 27 November 2008 (UTC)[reply]

Indeed, the possibility of rollovers means the expectation is slightly lower than it would seem at first glance. The expectation is the expected total amount of prizes paid out, divided by the number of tickets, but the expected amount paid out would not be the entire prize fund. With my example of the UK lottery below, it's not an issue with a triple rollover (which is pretty much the only time you could conceivably get a positive expectation) because there is no possibility of more rollovers, if no-one wins the jackpot the prize money just gets paid to the next best tickets. For other draws, the expectation is even worse than my calculations would suggest. --Tango (talk) 13:18, 28 November 2008 (UTC)[reply]

Thanks, where can I get historical stats about the number of entrants and total jackpot for lotteries available in the UK please? (Of course a lottery would have a positive expectation if the total of all the prizes divided by the number of tickets sold was greater than the cost of the ticket). I expect more tickets are bought when rollovers occur, making the figures less easy to estimate. Thanks. 78.151.111.2 (talk) 20:00, 26 November 2008 (UTC)[reply]

Rollovers and double rollovers happen frequently enough for there to be plenty of datapoints to estimate from, if it requires more than that to become profitable then you might struggle to find enough stats. I don't know if the stats are particularly easy to come by, but the prize fund is announced before each draw (so should be easy enough to find somewhere) so if you know how much was rolled over you can just minus that and then double it to get the number of tickets sold. --Tango (talk) 21:01, 26 November 2008 (UTC)[reply]

Ok, I've done some research. The UK Lotto will only rollover 3 times, if no-one wins in 4th consecutive week it goes to the next place down (incidentally, the possibility of a rollover reduces your expectation whenever it isn't a triple-rollover, since there is a chance not all the prize fund will be paid out, so it's actually less than 50p usually). The prizes and number of winners for each draw are on the main lottery website ([1]), I've found a double-rollover from August. The second time, the amount rolled over was £8,483,403 and in the 3rd week the total prize fund was £26,478,178. That corresponds to (26,478,178-8,483,403)*2=35,989,550 tickets. The expectation that week was therefore £26,478,178/35,989,550=73.6p, so a net loss. I can't be bothered to search for a triple rollover, so I'll just make some numbers up - if the jackpot in the 3rd week was £13,582,850, so if that rolled over and say 40 million tickets were sold (a conservative estimate), that would give an expectation of 50p+£13,582,850/40,000,000=84p, still a net loss. Those are just numbers from one example (and my head), so they may not be typical, but it doesn't look like the UK Lotto is ever worth buying a ticket on. --Tango (talk) 21:23, 26 November 2008 (UTC)[reply]

The expectation for Powerball is

\$0.20+{Jackpot \over 146107962}

per dollar. It follows therefore that the contest has a positive expectation whenever it rolls above ~$117,000,000 (or really ~$230,000,000 if you want to include taxes). Since Powerball rolls over without limit, this is entirely possible. The four largest single ticket Powerball jackpots were each over $300M. Dragons flight (talk) 01:43, 27 November 2008 (UTC)[reply]

That's assuming there will be at most one winner. When the jackpot is that high, many more people play, meaning that even if you win you might win only a fraction of the jackpot (if someone else wins as well). Since you can't know, at the time you buy a ticket, how many other people are also buying tickets, it's not possible to calculate an exact expectation. You could come up with a probability distribution of the number of other players and probably calculate a reasonably meaningful number, but it's going to have a subjective (Bayesian) component. --Trovatore (talk) 01:50, 27 November 2008 (UTC)[reply]

You also need to consider the difference between the advertised jackpot and the free-and-clear present-day value of the actual award. I don't know about the UK Lotto, but typical US lotteries pay out over 20 to 25 years, to the present value of that payout is only about 50% of the advertised value. On top of that, the payout is considered taxable income, from which the government will take back a slice. Say that amounts to a 25% tax. That leaves you with 50% x 75% = 37.5% of that advertised jackpot. So if the odds are 1 in 100 million, the jackpot would need to be on the order or 100 million / 37.5% = about 266 million dollars/pounds/euros to be worth you 1 dollar/pound/euro ticket price (ignoring the lesser prizes and assuming only one winner). On top of that, you might consider the diminishing value to you of each dollar/pound/euro when you're a millionaire compared to what it's worth to you now. A hot cinnamon roll at the mall is much more of a treat when you're a poor man than when you can have your personal chef make a fresh batch daily. -- Tcncv (talk) 05:12, 27 November 2008 (UTC)[reply]

The UK Lotto is a tax free lump sum. (It's definitely tax free, I'm 90% sure it's a lump sum.) --Tango (talk) 11:26, 27 November 2008 (UTC)[reply]

One other way there could be a positive return, on average, is if the lottery advertises a fixed prize, but then fails to get enough players to cover the cost of the prize. This would be bad news for the lottery, especially in the case of a cash prize, so large state lotteries rarely operate that way. However, a small lottery for charity, say a church, which raffles off prizes donated by the community, might operate that way. Say a restaurant donates a free meal for 2, which they claim has a $100 retail value, to the church. The church then has a lottery and sells 90 tickets at $1 each. The actual cost for the restaurant to produce the food may be far lower, say only $20. In theory, the church could have just sold the donated restaurant meal for more than $90, but they would be unlikely to get any takers, especially if the meal has restrictions to times when the restaurant is never busy. Also, the church may have hoped to sell more than 100 raffle tickets. Finally, the restaurant may have only been willing to donate the meal in exchange for the free advertising they get from it being raffled off, hoping some of the losers will elect to pay for their own meals there. StuRat (talk) 15:34, 27 November 2008 (UTC)[reply]

Sure. We're talking about the kind of lottery where the prize fund comes out of the revenue, if the prize fund can be greater than the revenue (which is what could theoretically happen with a rollover) then the expectation can be positive. --Tango (talk) 16:53, 27 November 2008 (UTC)[reply]

Algebra for college

How do you know when an equation has infinitely many solutions? —Preceding unsigned comment added by 70.197.205.85 (talk) 17:38, 26 November 2008 (UTC)[reply]

That depends on what sort of equation you're talking about. What did you have in mind? Algebraist 17:41, 26 November 2008 (UTC)[reply]

He's probably talking about a system of several linear equations in several variables. Michael Hardy (talk) 23:03, 26 November 2008 (UTC)[reply]

Well, if it's linear equations, you'd get infinitely many solutions when the system has at least one solution and the number of variables is strictly larger than the number of linearly independent equations.155.198.201.14 (talk) 19:38, 2 December 2008 (UTC)[reply]

A hypothesis in probability and statatistics

I have a hypothesis: Given the probability densities of two random variables x and z, there exists a never-negative random variable y such that the cumulative distribution of (x+y) equals the cumulative distribution of z, if and only if the cumulative distribution of x is never lesser than the cumulative distribution of z.

Does this make sense, is it true, does it have a name, does it have generalizations? --Masatran (talk) 18:14, 26 November 2008 (UTC)[reply]

Counterexample: x is distributed uniformly on [0,1] and z is always 2. You at the least need some sort of condition on the entropy of the probability distributions, since adding an independent random y to x can't produce a distribution with less entropy. --Tardis (talk) 18:56, 26 November 2008 (UTC)[reply]

Then let us define a partial order among random variables. xRz, if and only if the cumulative distribution of x is never lesser than the cumulative distribution of z.

Defining such a partial order might be helpful, in say, evaluating actions which lead to different probability densities for the utility (utility as in economics). In that case, this partial order could be used to form a partial order between actions.

Does this make sense, does it have a name? --Masatran (talk) 19:10, 26 November 2008 (UTC)[reply]

It's equivalent (for non-discrete distributions) to saying that

z=f(x)

with

f(x)\geq x

, although only in the sense that z has that distribution, not that it is dependent on x. (We are talking only about independent variables here, right?) So you could say that it is an "improvement" on x in that it replaces each value of x with one that is no smaller. It seems related to the notion of a dominating strategy. Note that it may be very rare in practice: for instance, no two normal distributions with different variances will be ordered. --Tardis (talk) 19:27, 26 November 2008 (UTC)[reply]

There is a Wikipedia article about the partial order described above: Stochastic ordering. Michael Hardy (talk) 23:02, 26 November 2008 (UTC)[reply]

In case you're interested, Masatran, this (first-order) stochastic dominance is a useful concept in choice under uncertainty. A good microeconomic reference for this topic is Mas-Colell, Whinston and Green: Microeconomic Theory (1995), chapter 6. Pallida Mors 14:46, 27 November 2008 (UTC)[reply]

Circles

How do I get the equation of the large circle here [2]? I was able to get it fairly close by zooming in but I would like to know how to get it mathematically for a more accurate description of it. TIA, Ζρς ι'β' ^¡hábleme! 18:58, 26 November 2008 (UTC)[reply]

It's 2+2.sqrt(2), the distance to the centre of a little circle is 2.sqrt(2) because it is the diagonal of a square of side 2. Then you just add the radius of the small circle as it is on the same diagonal. Dmcq (talk) 19:08, 26 November 2008 (UTC)[reply]

[edit conflict: this looks more hint-worthy than solution-worthy to me] By symmetry, the line to each point of tangency passes through the center of the relevant small circle, which means its length is the radius of that small circle plus the distance from its center to the origin. That gives you a radius of the large circle, and I think you know its center. --Tardis (talk) 19:11, 26 November 2008 (UTC)[reply]

Well, I knew that all I needed was the length of the line from the origin to the point of tangency on one circle, but how do I find the point of tangency. That's really all I need because if I can get that, I can just use the distance formula. Ζρς ι'β' ^¡hábleme! 19:16, 26 November 2008 (UTC)[reply]

But as explained above, you don't need to know the point. All you need is the length of the big circle's radius which is the distance from origin to centre of small circle + radius of small circle. Zain Ebrahim (talk) 19:23, 26 November 2008 (UTC)[reply]

Ah, I got it now. Thanks all, Ζρς ι'β' ^¡hábleme! 20:03, 26 November 2008 (UTC)[reply]

Logarithm problem

I'm on my school's Math Team, and on a recent practice round we got the following question:

3^x - 3^x-2 = 24√3

Is there any formulaic/algebraic way to solve this? I last did logarithms two years ago and don't remember much. I solved it by assuming that x is positive and ends in .5 (because of the √3), and testing all possibilities (the answer's 3.5, by the way), but problems like this are frequent and they don't always have nice small numbers that make these strategies possible.

I got another question on another practice round, and again, I solved it by trial and error but am at a loss to find any algebraic way of solving it, and the general form is quite frequent on the Math Team tests.

log₂x + log_x16 = 4

(The answer's 4, by the way.) 69.177.191.60 (talk) 19:29, 26 November 2008 (UTC)[reply]

3^{x}-3^{x-2}=3^{x-2}(3^{2}-1)=8\times 3^{x-2}

. Divide by 8 and take base 3 logarithms.

\log _{2}x+\log _{x}16=\log _{2}x+{\frac {\log _{2}16}{\log _{2}x}}

, so you get a quadratic in

\log _{2}x

. Algebraist 19:35, 26 November 2008 (UTC)[reply]

I think you mean

\log _{3}x

throughout, no? --Tango (talk) 19:41, 26 November 2008 (UTC)[reply]

Actually, I don't follow at all... after taking base 3 logs you just have x-2=3/2, where did the log x and log 16 come from? --Tango (talk) 19:41, 26 November 2008 (UTC)[reply]

I was asking about two separate problems. Sorry if there was any confusion.69.177.191.60 (talk) 21:16, 26 November 2008 (UTC)[reply]

Sorry, I skipped to the other question there. Algebraist 19:42, 26 November 2008 (UTC)[reply]

{\begin{aligned}&3^{x}-3^{x-2}=24{\sqrt {3}}\\\\&9\cdot 3^{x-2}-3^{x-2}=24{\sqrt {3}}\\\\&(9-1)\cdot 3^{x-2}=24{\sqrt {3}}\\\\&8\cdot 3^{x-2}=24{\sqrt {3}}\\\\&3^{x-2}=3^{1}{\sqrt {3}}\\\\&3^{x-2-1}={\sqrt {3}}\\\\&3^{x-3}=3^{1/2}\\\\&x-3=1/2\\\\&x=7/2.\end{aligned}}

Michael Hardy (talk) 22:53, 26 November 2008 (UTC)[reply]

Elliptic curve

I've trying to solve algebraically the equation on the right here, $y^{2}=x^{3}-x+1$ .

To differentiate, I took the square root of each side to get y = ±(x^(3/2) - x^(1/2) + 1) then differentiated.

$0={\frac {3{\sqrt {x}}}{2}}-{\frac {1}{2{\sqrt {x}}}}$
$0={\frac {3x-1}{2{\sqrt {x}}}}$
So a turning point occurs at $x={\frac {1}{3}}$

Therefore, putting x=1/3 back into $y^{2}=x^{3}-x+1$ , you get ± ${\frac {\sqrt {57}}{3}}$ Which is approx ±0.83. Is this right so far?

So co-ordinates of the turning points are $({\frac {1}{3}},{\frac {\sqrt {57}}{3}})$ and $({\frac {1}{3}},-{\frac {\sqrt {57}}{3}})$ .

In order to determine the nature, I done the second derivative, but you have a ± equation as the original was y^2, so both plus and negative. Therefore, the derivative of the above equation, ${\frac {3x-1}{2{\sqrt {x}}}}$ , is (3x+1)/(4x^(3/2)) and (-3x-1)/(4x^(3/2))

So putting x=1/3 into each of the equations gives a positive and negative answer, suggesting that there are maximum and minimum turning points. All the above still seems to fit the diagram (even though it doesn't have a scale). Solved by iteration, to get the x-axis cutting point, $x=-1.3$ .
To get y-axis cutting poing, easy, ± $y=1$ .

So, is all this right? Thanks 86.159.230.56 (talk) 21:41, 26 November 2008 (UTC)[reply]

{\sqrt {a+b}}\neq {\sqrt {a}}+{\sqrt {b}}

. You're probably better off differentiating both sides of the original equation, using the chain rule for the left hand side, rather than square rooting. You could do it that way if you wanted, though, but you need to use the chain rule to differentiate the square root. Oddly, all your numbers seem pretty plausible, except for the fact that the graph has 4 points with horizontal tangent, you've only found two (I almost missed your mistake until I noticed that!). Also, on your last line, it should be y=+/-1. You may also want to observe when dy/dx goes to infinity, since that indicates a vertical tangent. --Tango (talk) 22:00, 26 November 2008 (UTC)[reply]

Ah yes, thank you. I've fixed the y=±1. I'm now having problems with the implicit differentiation on

y^{2}=x^{3}-x+1

.

I get: $2y{\frac {dy}{dx}}=3x^{2}-1$ then ${\frac {dy}{dx}}=0$ , so: $3x^{2}-1=0$
$3x^{2}=1$
$x^{2}={\frac {1}{3}}$
x=±(1/√3) But this gives values that I'm not too sure of... 86.159.230.56 (talk) 22:41, 26 November 2008 (UTC)[reply]

That looks right, 2 values of x, each of which will correspond to two values of y, giving 4 points with horizontal tangents, which is what we see on the graph (they look like the x co-ordinates are symmetric about the y-axis, which is what you've found, as well). What makes you unsure? --Tango (talk) 23:31, 26 November 2008 (UTC)[reply]