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September 22

Bouncing balls

Hi. Imagine a ball propelled upwards from the origin with unit speed, with gravity of magnitude 2 units acting on it. The origin is a hard surface, and a coefficient of restitution between it and the ball of e. On any bounce, the time taken to reach the ground again satisfies:

$-u=u+at$

$2u=2t$

$t=u$

And after n bounces, the initial speed has been multiplied by e n times:

$u_{n}={e^{n}}u_{1}=e^{n}$ Hence: $t_{n}=e^{n}$

Now, let us consider the total time it spends bouncing. A coefficient of restitution is some number, $0\leq e\leq 1$. Assuming it is less than 1, then our series of times $t_{0},t_{1},t_{2},...$ are a geometric progression, whose sum tends to a limit:

$\sum _{n=0}^{\infty }t_{n}={\frac {1}{1-e}}$

In other words, the total time that the ball spends bouncing is finite, even though there are an infinite number of bounces. But how can it ever stop bouncing? The ball's velocity never reaches zero. After each bounce, it is only multiplied by some number, and so merely tends towards zero. 79.78.99.44 (talk) 16:05, 22 September 2009 (UTC)

See Zeno's paradoxes. — Emil J. 16:10, 22 September 2009 (UTC)

Moreover, see geometric sequence. ~~ Dr Dec (Talk) ~~ 18:45, 22 September 2009 (UTC)

will you please pinpoint what in geometric sequence is of use for the above question, because the article is quite long. Thx! --84.220.118.29 (talk) 19:27, 22 September 2009 (UTC)

Not sure I see what the problem is. The ball only covers a finite distance in its infinite number of bounces. The theory is consistent, although at some point this idealised model obviously ceases to be an accurate representation of reality. Gandalf61 (talk) 16:16, 22 September 2009 (UTC)

You seem to miss several idelizations here. First, the 'bounce' is not an event but rather a process: it involves interaction between a flexible ball's surface and a hard surface below it. It consists of two phases, namely compressing the ball material (say, rubber) and relaxing it. So it takes some time and space (vertical movement from the moment the ball touches the table until they separate). So an answer to your question could be: 'bouncing stops (or becomes not observable) when the jump height becomes less than the rubber grain size'. --CiaPan (talk) 19:53, 22 September 2009 (UTC)

It takes an infinite number of bounces for the speed to reach 0, but as you showed those infinite bounces take only a finite amount of time to elapse in total. So considering the speed over time, it reaches 0 in a finite amount of time. As Emil J pointed out, it's Zeno's paradox. Rckrone (talk) 21:58, 22 September 2009 (UTC)

... and, in reality, all but the first dozen (or maybe score) of "bounces" are just increasingly rapid oscillations within the rubber of the ball. Dbfirs 13:52, 24 September 2009 (UTC)

September 23

Advanced geometry homewrok

Yes I know you dont do homework, but just point me in the right direction. I have been given a pig of a problem i dont have any idea how to start: Prove that in any planar triangle, the centroid, orthocenter, circumcenter and center of the 9 point circle all lie on a straight line??? Any pointers welcome.

Euler line may be a good start.

Consider where the 9 point center must lie between the altitude and perpendicular bisector from each side. See if you can show that it must be exactly halfway between the circumcenter and orthocenter. Use the fact that the centroid falls 1/3 of the way along each median to make a similar argument about where the centroid falls between the circumcenter and orthocenter. Rckrone (talk) 04:43, 23 September 2009 (UTC)

If getting started at all is a problem, start by deriving formula for each type of point mentioned above for a triangle (x1,y1) (x2,y2) etc..

Once you've got 2 points you have a line - the other points (in terms of x1,x2,y1,y2 etc) should satisfy the eqaution of this line (if it's true)83.100.251.196 (talk) 19:32, 23 September 2009 (UTC)

Ignoring the nine point centre for the moment, one nice way of proving that O, G and H are collinear is to consider a dilation of the large triangle by a factor of -1/2 centred at the centroid (i.e. shrink it to half its size and reflect it about the centroid). Then see what happens to the point O of the larger triangle. AMorris (talk)●(contribs) 11:01, 24 September 2009 (UTC)

Recruiting math exam questions...

Someone confirm that either I'm crazy and have forgotten basic math, or the answer key for this recruiting exam has errors:

1. In a 100m race, A can beat B by 25m and B can beat C by 4m. In the same race, A can beat C by:
Me: 29m || Key: 28m

2. A man bought fruit at the rate of 16 for $24 and sold them at the rate of 8 for $18. What is the percent profit?
Me: 50% || Key: 40%

3. Jack has two children, and at least one of them is a girl. What is the probability that both children are girls?
Me: 1/2 || Key: 1/3

4. How many times are the hands of a clock at right angles in a day?
Me: 48 || Key: 44

Help please? 218.25.32.210 (talk) 06:07, 23 September 2009 (UTC)

I think I understand number 3 now -- based on the Punnet square, Jack could have BB | BG | BG | GG, but because at least one is a girl, you discard the BB result and are left with a 1/3rd chance that he has 2 girls.
I'm still clueless about the rest of them though. 218.25.32.210 (talk) 06:32, 23 September 2009 (UTC)

(ec) 1. Key is right. B runs at 0.75x A's speed and C runs at 0.96x B's speed => C runs at 0.72x A's speed so A beats C by 28m.

2. (edited) hmm i'm confused.

3 Key is right. Possibilities are boy-boy, boy-girl, girl-boy, or girl-girl. Boy-boy is eliminated leaving three equal other possibilities. girl-girl is one of these 3.

4. (edited) Key is right. Hour hand turns at 1/12 rph (revolutions per hour). Minute hand turns at 1 rph. The first time the hands are at right angles (i.e. 1/4 revolution apart) is at t hours, where t/12=t-0.25 => t=12t-3 => t=3/11 hours = 16.36 minutes after 12. Your error is in thinking it was at 15 minutes after 12 which would only be right if the hour hand didn't move continuously as the minute hand turns. These right-angles occur every 6/11th hours so in 24h it happens 44 times. 70.90.174.101 (talk) 06:50, 23 September 2009 (UTC)

The key is right for 1, 3 & 4. Let me give you some clues for 4. Think of two times when the hands are obviously perpendicular. How long from one of these times to the next time? It's not going to be just an hour is it, because in one hour when the minute hand is in its original position, the hour hand will have moved forward 30°. In °/hr or revolutions/hr, what is the speed of the hour hand and the minute hand? Subtract the smaller of these from the larger. What is the significance of this number? It might be easier to work it out for the situation when there is no angle between the hands, eg. 12 noon, and then extrapolate from then. RupertMillard (Talk) 07:49, 23 September 2009 (UTC)

The key is wrong for question 2. Depending on which value you choose for base you either arrive at 50% or 33.3%, there seems no way to get 40% as an answer. Taemyr (talk) 11:24, 23 September 2009 (UTC)

To understand the clock problem in an obvious way: what is the angle between the hands at 12:15? The big hand is on the 3 and the small hand is 1/4 of the way between the 12 and the 1, so it's not 90 degrees. That should make the situation clear. 70.90.174.101 (talk) 23:44, 23 September 2009 (UTC)

40% would be if you assume your base was (24+36)/2, i.e. the average, but that seems a dumb thing to use. Dragons flight (talk) 11:40, 23 September 2009 (UTC)

For number 3, the article boy or girl paradox may be of interest. Pallida  Mors 14:16, 23 September 2009 (UTC)

The key is right in #3. The three equally probable possibilities are:

• girl, boy
• boy, girl
• girl, girl

The probability of the third one is 1/3. Michael Hardy (talk) 15:57, 23 September 2009 (UTC)

The key is wrong in #2. The simplest way to see that may be to imagine that 16 is the exact number he bought and sold. He paid $24; he got $36. The other way is to say that in the first case the price per item was $24/16 = $1.5, and in the second case it was $18/8 = $2.25. Michael Hardy (talk) 16:01, 23 September 2009 (UTC)

Problem 2 is not just wrong, it is poorly stated. In the real world, you usually don't sell more than you buy, but you certainly can sell less than you buy. In fact, it's common. The problem doesn't state that you sold all you bought. It is arguably an implicit assumption, but I would argue that implicit assumptions are acceptable if they match likely outcomes. For example, it is an implicit assumption that the value of the dollar hasn't changed between the date of the purchase and the date of the sale. Were I to challenge the problem on the basis that the implicit assumption isn't spelled out, I would be properly upbraided for nit-picking. However, it is not only common, it is usual for fruit vendors to sell fewer than they buy, and spoilage should be included in the calculation. The wording always makes a point of avoiding the issue - I think in this case it was incompetence, but if it was a finance class, not an arithmetic class, I'd suspect that the teacher wanted you to identify the missing assumption. If we accept that purchase count equal sales count the right answer is 33.3...%. 50% is the markup, not the profit.--SPhilbrickT 21:20, 24 September 2009 (UTC)

Introduction to number theory

Does anyone know of any good introductions to number theory (am I misusing the term)? I am trying to understand the Number Field Sieve but I keep realizing I just don't have the background to fully understand it. I would really like it if someone knew of an online resource. My math education stopped at calculus. Thank you. PvsKllKsVp (talk) 21:48, 23 September 2009 (UTC)

Tom Apostol's Analytic Number Theory is a nice introduction.--pma (talk) 21:58, 23 September 2009 (UTC)

Elementary Number Theory by Kenneth Rosen StatisticsMan (talk) 22:14, 23 September 2009 (UTC)

Thank you. PvsKllKsVp (talk) 22:34, 23 September 2009 (UTC)

I suggest "A Course in Number Theory and Cryptography" by Neal Koblitz. It starts at an undergraduate level and gets up through the quadratic sieve. I don't know whether the new edition includes the number field sieve, but it at least gets most of the way there. 70.90.174.101 (talk) 23:39, 23 September 2009 (UTC)

Calculating integrals (for use in probability theory)

I'm stuck on some integrals here. One problem is to find all the moments for

$f(x)={\frac {1}{{\sqrt {2\pi }}x}}e^{-(\log x)^{2}/2},\qquad 0\leq x<\infty ,$

or, more accurately, verify that they are $EX^{r}=e^{r^{2}/2}$ for r = 0, 1, ... .

I have two main options, as far as I can see. One is to just calculate the integral

$\int _{0}^{\infty }x^{r}f(x)\,dx$

but I don't know how to do that. I know how for r = 0. That is an easy u-substitution. But, for r > 0, I am not seeing it. I thought, for r = 1, I could use integration by parts with u = x and dv = f(x) dx. But, I don't really know how to do that integration by parts because it would end up with v as the integral of $e^{-x^{2}/2}\,dx$, which has no closed-form antiderivative. The other way would be to use the moment generating function but that just adds in an $e^{tx}$ which doesn't seem to make things any easier as I think I'd end up with the same v. Any ideas? Thanks. StatisticsMan (talk) 22:23, 23 September 2009 (UTC)

Follow the first option: the integral is not that difficult. Make the substitution $t=\log(x)$, and complete the square in the argument of the exponential that results as integrand; finally make a translation. Is it OK?--pma (talk) 22:36, 23 September 2009 (UTC)

I see what you mean. That's pretty simple. But, I'm off by a factor of 2. Ah, I think I just figured it out. As x goes to 0, log x goes to -infinity but I just left the limits of integration as 0 to infinity. Thanks. As always, you are very helpful. This will help on another problem as well probably. StatisticsMan (talk) 23:01, 23 September 2009 (UTC)

Resolved

September 24

My mathematical ability.

I have a mathematical ability and wonder if there is an actual name for it? Putting a number to every letter of the alphabet e.g. A is 26 to Z is one i can give the sum of a word faster than than people can work it out on paper or on a computer. My average for a ten letter word would be under five seconds and people have been quite shocked when i have demonstrated this ability.So, can you tell me please is there a name i can put to this? Thanks, Harry Remfrey.

I doubt there is a name for it. It is just mental arithmetic. You have most likely memorised (perhaps intentionally, perhaps not) the numbers for each letter (oddly, you have them backwards from the way most people do such associations - A is usually 1 and Z is 26). Being good at mental arithmetic usually involves learning lots of tricks to make calculations quicker (among the most common are things like adding one to a number and then minusing the appropriate amount at the end to compensate, eg. 5x9=5x10-5=45, but there are more complicated tricks). You may also have memorised (again, intentionally or otherwise) the totals for certain combinations - 10 letter words are usually built up from smaller words with prefixes and suffixes, so if you already know what "ing" adds up to, you can speed up a lot of long words. --Tango (talk) 01:14, 24 September 2009 (UTC)

Furthermore being able to associate two different symbols by rote memorization is no indication of mathematical ability; although you seem to be able to calculate long sums quickly, it is not the case that you could match a computer program in terms of raw speed. On top of this, being able to calculate sums and work with integers is obviously beneficial, but being a mathematician is about problem solving, hard work and having an artistic streak too. Do you know how to prove things? That is real math. 94.171.225.236 (talk) 12:16, 24 September 2009 (UTC)

I'm not suggesting that you suffer from this, but you might be interested in our article on Savant syndrome. Many mathematicians (such as Carol Vorderman?) and other "normal" people such as magicians have taught themselves similar tricks. Dbfirs 13:42, 24 September 2009 (UTC)

My average arithmetic speed is 50% that of a normal person, and I always make mistakes. Does this mean I'm suffering from something? And yes I agree with 94.171.225.236 proving things is a totally different matter from arithmetics Breath of the Dying (talk) 16:35, 24 September 2009 (UTC)

You probably just haven't practised enough and learnt all the tricks. You may also be misjudging other people's abilities. --Tango (talk) 17:35, 24 September 2009 (UTC)

Moment Generating Function

Alright, now I need to show the moment generating function for that same function as my last post,

$f(x)={\frac {1}{{\sqrt {2\pi }}x}}e^{-(\log x)^{2}/2},\qquad 0\leq x<\infty ,$

does not exist. I think I have done it but I just want to make sure this makes sense. I don't have any of my undergrad analysis books here.

I calculated the derivative of $e^{tx}f(x)$. It is positive for x such that $tx-1-\log x$ is positive. Depending on t, I know this will eventually be positive for some x and also every x bigger than that x since x grows faster than $\log x$. So, the derivative is eventually positive which means the function is eventually increasing. And, the function itself is always positive. So, it can not tend toward 0 so the integral must be divergent. Does this make sense? Thanks. StatisticsMan (talk) 02:09, 24 September 2009 (UTC)

Yes. (Detail: to be precise, I wouldn't just say it can not tend toward 0, but rather it is bounded away from 0 for large x, that also follows from your argument. Because a function may be integrable over R even if it doesn't tend to 0 at infinity). You may also check that the moments $m_{r}$ grow too fast: $m_{r}/r!$ are not the coefficients of a convergent power series. To this end you may use the Stirling formula for $r!$, but in fact it is sufficient to know $m_{r}=\exp(r^{2}/2)$ and $r!\leq r^{r}$ so $(m_{r}/r!)^{1/r}\geq r^{-1}\exp(r/2)$, that diverges as r goes to infinity, so the Cauchy test gives a null radius of convergence. --pma (talk) 07:18, 24 September 2009 (UTC)

Easy question

Hi I have a easy question that I need someone to confirm I'm right. Let X be a set, and F a collection of subsets of X satisfying the finite intersection property. By Hausdorff's maximality principle, we can show there exists a maximal collection E of subsets of X satisfying the f.i.p and is a superset of F. Now take the intersection of all elements of E, does this intersection contain at most one element? I say yes because if it's not empty and contains x then {x} is in the collection E since E is maximal. Am I right? This stuff is really confusing Breath of the Dying (talk) 12:29, 24 September 2009 (UTC)

You are not asking if an ultrafilter E (containing the filter generated by F, which is not relevant for the question) is necessarily a principal ultrafilter, that is, the collection of all supersets of a singleton {x}. The answer to the question you didn't ask is no (otherwise the theory of filters would be totally trivial): just note that if the intersection of all members of F is empty, the same is true a fortiori for the intersection of all members of E. So take for instance F:=all co-finite subsets of N.--pma (talk) 15:40, 24 September 2009 (UTC)

Actually, the OP is asking whether the intersection of an ultrafilter contains at most one point, not at least. The answer is yes, and the given reasoning is correct. — Emil J. 15:59, 24 September 2009 (UTC)

ops, sorry I misread the question. Thanks Emil --pma (talk) 17:21, 24 September 2009 (UTC)

Funniest fix I've seen in a while :-) -- Meni Rosenfeld (talk) 20:05, 24 September 2009 (UTC)

you should see my Beetle ;-) --pma (talk) 21:32, 24 September 2009 (UTC)

Solving systems of equations in three variables

Resolved

By my teacher

7x+5y+ z=  0
-x+3y+2z= 16
 x-6y- z=-18

I got this correct

    z=4
7x+5y+4=0
 7(-x-3y+2(4))=16

  7x+5y+4=0
 -7x-21y+56=112

 -16y+60=112

   -16y=52

Y needs to be 2 (check by using calculator.) Accdude92 (talk) (sign) 13:38, 24 September 2009 (UTC)

• OK. Do you have a question ? Rkr1991 ^{(Wanna chat?)} 14:04, 24 September 2009 (UTC)

As I recall sometime solutions of three equations with three variables don't exist, this looks like it might be a case of that.83.100.251.196 (talk) 15:53, 24 September 2009 (UTC)

Try solving for z=4 into the 3rd equation - see what happens.83.100.251.196 (talk) 15:55, 24 September 2009 (UTC)

The OP has indeed correctly solved the set of equations... I don't understand what more there is to say... Rkr1991 ^{(Wanna chat?)} 18:10, 24 September 2009 (UTC)

The OP wanted to know what was the mistake in the given calculation. Which is in the line $7(-x-3y+2(4))=7\cdot 16$, where the sign of $3y$ was reversed. -- Meni Rosenfeld (talk) 19:58, 24 September 2009 (UTC)

I get x = − 2, y = 2 and z = 4. ~~ Dr Dec (Talk) ~~ 19:27, 24 September 2009 (UTC)

Complex number

OK so this is a homework question but I'm only asking to have my work checked.

"If a complex number z is defined by $z=z_{1}+z_{2}$ and $|z_{1}|=1,|z_{2}|=2$ then sketch the subset of the Argand diagram represented by z."

So, $z_{1}$ must be a circle of radius one centred at the origin and $z_{2}$ must be a circle of radius 2 centred at the origin. So does this mean that z is a circle of radius 3 centred at the origin? Thanks 92.4.52.65 (talk) 18:54, 24 September 2009 (UTC)

No it does not: for you only have an inequality : $|z|=|z_{1}+z_{2}|\leq |z_{1}|+|z_{2}|=1+2=3$. Actually you can obtain all z in the closed disk of radius 3, not just the circle. --84.221.209.230 (talk) 19:03, 24 September 2009 (UTC)

As this is homework, I'll say only this: 84's answer isn't right either, but the general logic is sound. --Tardis (talk) 19:21, 24 September 2009 (UTC)

Indeed: 84.221.209.230's solution isn't quite right. Here are a few tips:

• We know that |z₁| = 1 and so z₁ = e^iθ for some 0 ≤ θ < 2π.
• We know that |z₂| = 2 and so z₂ = 2e^iφ for some 0 ≤ φ < 2π.
• SInce z₁ + z₂ = e^iθ + 2e^iφ, what can we say about |z₁ + z₂|?

You'll find that |z₁ + z₂| can only get so small. For example, can z₁ + z₂ = 0? ~~ Dr Dec (Talk) ~~ 19:45, 24 September 2009 (UTC)

I think there's an even easier solution. I'm pretty sure that the following inequality always holds in a normed vector space.
Please, correct me if I'm wrong (add any missing assumptions, etc)

$||z_{1}|-|z_{2}||\leq |z_{1}+z_{2}|\leq |z_{1}|+|z_{2}|\ .$

This shows why 84's solution was not quite right: he only had the right hand part of the inequality. ~~ Dr Dec (Talk) ~~ 20:54, 24 September 2009 (UTC)

It does always hold. The right inequality, the triangle inequality of course, is part of the definition of a normed linear space. The left inequality comes from using the triangle inequality twice. Here is once

$|z_{1}|=|z_{1}+z_{2}-z_{2}|\leq |z_{1}+z_{2}|+|z_{2}|\Rightarrow |z_{1}|-|z_{2}|\leq |z_{1}+z_{2}|$

Just do it again with $z_{1},z_{2}$ switched and combine the two to get the left inequality. StatisticsMan (talk) 21:08, 24 September 2009 (UTC)

Good. OP: and don't forget the other inclusion (if z has 1 ≤ |z| ≤ 3, it can be written as z = z₁ + z₂). --pma (talk) 21:23, 24 September 2009 (UTC)

So is the defined area the area between a circle of radius one and a circle of radius 3? 92.4.52.65 (talk) 17:13, 25 September 2009 (UTC)

Yes. However above there's only the proof of one inclusion. --pma (talk) 23:30, 25 September 2009 (UTC)

September 25

Number of possible permutations in the lottery

I've been trying to calculate the odds of winning our national lottery, but there is a rule in the permutations and I want to know if there's a formula that works for it.

In the South African lottery, the ball set consists of numbers 1 through 49.
A single lottery ticket/game has 6 of your chosen numbers.
So how many possible combinations of numbers are there?

The problem is that in your series of 6 numbers, a number cannot be <= ^{less than or equal to} the previous number.

In other words the formula isn't 49 ^ 6 ^{49 to the power of 6} nor is it 49 x 48 x 47 x 46 x 45 x 44.

Thanks Rfwoolf (talk) 11:39, 25 September 2009 (UTC)

It's not a permutation, it's an unordered selection. Ie. you select 6 numbers, but the order of those does not matter. The actual number you are looking for is 49!/(43!*6!). Taemyr (talk) 11:43, 25 September 2009 (UTC)

• It's     $C_{6}^{49}={49 \choose 6}=13,983,816$   — see Combination#Number of k-combinations from a set. --CiaPan (talk) 12:29, 25 September 2009 (UTC)

The above answers are correct, but I feel that a little explanation might be in order. You have 49 balls and you draw 6 of them. You have 49 to choose from for your first ball, 48 for the second, 47 for the third, 46 for the fourth, 45 for the fifth, and 44 for the sixth. That gives 49 × … × 44 ways. But you're not interested in the order that they came out in. So, how many ways can you re-arrange the order that the six balls came out in? Well you have six balls to choose for the first ball to come out, five for the second, four for the third, three for the fourth, two for the fifth, and one for the sixth. There are 6 × 5 × … × 1 orders in which the six balls could have been drawn. So in total there are

${\frac {49\times \cdots \times 44}{6\times 5\times \cdots \times 1}}=13,983,816$

ways of choosing six balls from 49 when the order doesn't matter. ~~ Dr Dec (Talk) ~~ 22:57, 25 September 2009 (UTC)

I would guess that they draw 6 balls out of a pool of 49, then arrange them from highest to lowest, just for convenience, as the actual order doesn't matter. StuRat (talk) 13:22, 26 September 2009 (UTC)

Of course, that's exactly what they do. ~~ Dr Dec (Talk) ~~ 22:13, 26 September 2009 (UTC)

with the size of the Lotto jackpots recently who would want to play anyway? even the mathematically stupid don't want to play for "only" R3 million. Of course the above answers demonstrate WHY playing the lotto is a surefire way to lose money, your chances are slim-to-none. Zunaid 22:41, 26 September 2009 (UTC)

I like the story of a someone (Dogbert ?) who sold "day-old" (that is, expired) lottery tickets for half the original price. He argued that he was doing people a favor, since the original $2 ticket only returned 10%, so, represented a $1.80 loss, while the half price expired tickets cost $1, all of which was lost. So, it could be argued that you saved 80 cents by buying an expired ticket. (Of course, if you buy two expired tickets instead of a single good one, that logic breaks down.) StuRat (talk) 15:19, 27 September 2009 (UTC)

Zunaid: playing the lottery is not a "surefire way to lose money". I can only find a stat from May 2006, but up to then the British lottery had made almost 2,000 people millionaires. If it were "surefire" then surely that number should be zero? That's not to mention the smaller prizes given out and, in Britain at least, the £9.5 billion given to good causes by the end of 2008. Besides that, it's quite fun... it only costs £1 to play and you get 10 or 15 minutes of fun thinking about what you might do with the money if you did happen to win. Compare that to almost £3 for a pint and it seems like a healthy, and possibly profitable, bit of fun. ~~ Dr Dec (Talk) ~~ 16:27, 27 September 2009 (UTC)

OK, talking about dreams, I have a dream too. Lotteries: suppressed. Horoscopes: suppressed. Television: suppressed. Discos: suppressed. Soccer: suppressed, with the exception of recreational local clubs. Functional Analysis: mandatory for everybody in the public administration. Don't worry: just a re-educational program for the after-berlusconi Italy. And I had an idea about the Vatican too... oh yes: Vatican: suppressed.--pma (talk) 19:54, 27 September 2009 (UTC)

• Thanks for the answers folks, but are you sure they take into account that numbers combinations aren't repeated? You will see what i mean if you open a spreadsheet (e.g. excel) and start playing around. In our scenario, when we eventually get our first ball to '10', our second ball cannot be 1 to 10, it can only be 11 to 49 because all combinations of 1 and 10 have already been listed at this point, and only combinations between 10 and 11to49 are incomplete. Capish? Rfwoolf (talk) 22:28, 27 September 2009 (UTC)

You really need to explain what you mean. Each lottery draw is an independent event, so that numbers drawn this week do not depend on the numbers drawn last week and they do now effect the numbers drawn next week. What do you mean by getting a first ball to '10'? I think you need to specify the rules of you lottery because it doesn't seem to be like any lottery I have every seen. ~~ Dr Dec (Talk) ~~ 22:35, 27 September 2009 (UTC)

I am not talking about the actual drawing itself, I was talking about writing out each and every ball combination to calculate the number of possible calculations. You said that with the first ball you have 49 options, with the second only 48 and the third only 47, etc, but that logic starts to fail because, when writing out each and every possible combination, start in numerical order from 1, you cannot draw a ball less than the previous number -- something you will understand properly if you do try to write it all out. Let's do a quick example, here we go: 1, 2, 3, 4, 5, 6; 1, 2, 3, 4, 5, 7; 1, 2, 3, 4, 5, 8... Fast forward to 10, 11, 12, 13, 14, 15. Here we have 39 options for ball one, only 38 for ball two, 37 for ball three etc, and you will note that each ball cannot be lower than the previous ball. In our example, we would never go 10, 11, 12, 13, 14, 1; because the combination of 1, 10, 11, 12, 13, 14 was done already a long time back, same as 10, 9, 12, 13, 14, 15; because 9, 10, 12, 13, 14, 15 has already been done. Rfwoolf (talk) 23:38, 27 September 2009 (UTC)

The logic doesn't fail at all. Just read what I wrote carefully. You start of by assuming that the order in which the balls are drawn matters, i.e. 1, 2, 3, 4, 5, 6 is different to 1, 3, 2, 4, 5, 6, and they are both different to 2, 1, 3, 6, 5, 4. In this case there are 49 × 48 × … × 44 possible combinations: you have 49 balls to chose from for the first ball, then there are only 48 balls left from which you choose the second ball, then there are only 47 balls from which you choose the third ball, etc. So there are 49 × 48 × … × 44 possible draws that can be made, there are 49 × 48 × … × 44 different ways the balls can come out of the machine. But, after the balls have been drawn the numbers are put into ascending order, because it's not actually the order that counts, just the balls that came out: If you chose 1, 2, 3, 4, 5, 6 as you numbers then you would win if the draw came 1, 2, 3, 4, 5, 6 or if it came 1, 3, 2, 4, 5, 6 or if it came 2, 1, 3, 6, 5, 4. Well, how many ways are there to reorder the numbers 1, 2, 3, 4, 5, 6? Well we have 6 balls to choose as the first ball, then there are five left to choose the second ball, then there are four left from which to choose the third ball, etc. There are 6 × 5 × … × 2 × 1 ways or reordering 1, 2, 3, 4, 5, 6. So there are, as we have all said above, (49 × 48 × … × 44) / (6 × 5 × … × 2 × 1) = 13,983,816 ways of chosing 6 balls from 49 when the order doesn't matter. This is also, by definition, the number of ways you can write ascending 6-tuples of numbers from 1 to 49. If you still think that the logic fails, try your spreadsheet method with a simpler example: choose 2 balls from 4. Reapply my logic above and you will get that there are (4 × 3) / (2 × 1) = 6 ways you can do it when the order doesn't matter. And look: {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4} − 6 ways! If you still don't believe me then check out the national lottery article where it says the odds are 13,983,815-to-1, i.e. 13,983,815 losing combinations and 1 winning one, i.e. a total of 13,983,816 combinations. If you still don't get it then, as CiaPan suggested, read the combination article and in particular this section. ~~ Dr Dec (Talk) ~~ 08:59, 28 September 2009 (UTC)

Null hypothesis "trueness"?

If those familiar would comment on this, it would be helpful. Thanx! DRosenbach ^{(Talk | Contribs)} 12:35, 25 September 2009 (UTC)

If I understand you correctly, you claim that the definition of p-value ("the probability of rejecting the null hypothesis when the null hypothesis is true") should be replaced with the definition "the probability of rejecting the null hypothesis when the null hypothesis is unrejectable." I don't believe that your definition is correct (apart from the likely circular argument that arises from the use of the word "unrejectable"). The p-value is predicated on the *assumption* that the null hypothesis is true. Hence, we report the probability of observing what we did observe given the assumption. Wikiant (talk) 15:22, 25 September 2009 (UTC)

Straightforwardly, the equivalent "doing something when it can't be done" is nonsense. So live with the usage that has been commonly accepted for decades. We're talking about a conditional probability, the "given" basis is that the null hypothesis is true. I'll make the further point that the reference desk, as has been said before, is for readers' questions and not for editors to discuss the fine detail of articles.→81.153.220.122 (talk) 08:17, 26 September 2009 (UTC)

Isomorphism between the space of infinite R-valued sequences and the space of polynomials on R

How would one begin showing whether or not there exists an isomorphism between the space of polynomials: $\mathbb {R} \to \mathbb {R}$ and $\mathbb {R} ^{\mathbb {N} }$, the space of infinite sequences in the reals? I can see how the polynomials, which must have a maximum degree for any polynomial, would be isomorphic to the space of sequences with only finitely many nonzero terms, with the obvious bijection between coefficients and terms in the sequence, and this must be contained in $\mathbb {R} ^{\mathbb {N} }$, but I have a feeling the latter may have an uncountable basis whilst the former has the obvious countable basis e_i=(0,0,...,1,0,0,...) with the 1 in the i-th position - but how would I show such a thing if so?

Otherlobby17 (talk) 15:51, 25 September 2009 (UTC)

Hey I don't know much algebra but I think I have a few ideas for you. Like you said, the polynomial space is isomorphic to all eventually 0 sequence space, so if you can show there's no isomorphism between R^N and the space of all eventually 0 space you're done. A good reason for me to believe this is true is: if we replace R, the real numbers, by Q, the rationals, then the cardinality of Q^N is greater than the cardinality of the space of all eventually 0 sequences with rational entries, the first being continuum size and the latter being countable, hence there can be no isomorphism. Breath of the Dying (talk) 16:31, 25 September 2009 (UTC)

You're mistaken. It's easy to show there's no isomorphism between Rⁿ and the space of all eventually zero sequences. But that doesn't answer the question at all. Michael Hardy (talk) 16:33, 25 September 2009 (UTC)

Oh, you meant R^N. Michael Hardy (talk) 16:34, 25 September 2009 (UTC)

An isomorphism is a structure-preserving bijection. But which structure do you want to preserve? Addition of polynomials? Multiplication by scalars? If just those two, then it's vector-space isomorphism. Then there's multiplication of polynomials by each other, making it an algebra isomorphism. The set of sequences with only finitely many nonzero terms is algebra-isomorphic to the algebra of polynomials in an obvious way (see Cauchy product). But if you want to include sequences with infinitely many nonzero terms, then there's the question of how you could multiply them in a way that corresponds to multiplication of polynomials. I don't have an answer to that one. But now let's look at your mention of dimension. The vector space spanned by the countable basis is the set of all finite linear combinations of the basis vectors. If there's no countable basis for the vector space of all sequences of reals, then there's no vector-space isomorphism. And then a fortiori no algebra isomorphism. Right now I don't have that answer either—it seems as if there should be some obvious way to show there's no countable basis. Maybe later.... Michael Hardy (talk) 16:33, 25 September 2009 (UTC)

Let S be an independent system on N of size 2^ω (meaning a set $S=\{X_{\alpha }\mid \alpha <2^{\omega }\}$ of subsets of N such that for every finite sequence $\alpha _{1}<\alpha _{2}<\dots <\alpha _{k}<2^{\omega }$ and $e_{1},\dots ,e_{k}\in \{0,1\}$, $X_{\alpha _{1}}^{e_{1}}\cap \dots \cap X_{\alpha _{k}}^{e_{k}}\neq \varnothing$, where $X_{\alpha }^{0}=X_{\alpha }$, $X_{\alpha }^{1}=\mathbb {N} \setminus X_{\alpha }$.). Then characteristic functions of elements of S form a linearly independent subset of R^N, hence the latter linear space has dimension at least 2^ω. — Emil J. 16:38, 25 September 2009 (UTC)

In fact, one does not even need the system to be fully independent, it suffices that no element of S is covered by a finite union of other elements of S. Any almost disjoint system also has this property, for instance. — Emil J. 16:43, 25 September 2009 (UTC)

ec Indeed, they are not linearly isomorphic, for the space $\mathbb {R} ^{\mathbb {N} }$ has uncountable dimension. While waiting for the algebraic proof here's a topological proof: the space $\mathbb {R} ^{\mathbb {N} }$ is (admits a structure of) a Fréchet space: in particular, is a topological linear space topologized by a complete metric. Hence it can't have countable dimension. Proof: if such a space has countable dimension, it is union of countably many linear finite dimensional subspaces, which are in particular closed subsets with empty interior: but that's impossible by the Baire category argument.(If you prefer to use Banach spaces or Hilbert spaces, just observe that $\mathbb {R} ^{\mathbb {N} }$ has $\ell ^{2}$ as a subspace, already with uncountable dimension for the same reason) --pma (talk) 16:49, 25 September 2009 (UTC)

Emil, does this work? for any real number $\alpha$ in [0,1] let $u_{\alpha }:\mathbb {N} \to 2$ be a binary sequence such that the frequency of 1's tend to $\alpha$. (it can be choosen canonically, it seems to me). Then, it seems that these are a linearly independent set. --pma (talk) 17:06, 25 September 2009 (UTC)

You mean that the asymptotic density of $u_{\alpha }^{-1}(1)$ is $\alpha$? I don't think that this is a sufficient condition to ensure linear independence. For example, assume that the choice is made in such a way that $u_{1/2}^{-1}(1)=\{6n+0,1,2\mid n\in \omega \}$, $u_{1/3}^{-1}(1)=\{6n+3,4\mid n\in \omega \}$, $u_{1/6}^{-1}(1)=\{6n+5\mid n\in \omega \}$, $u_{1}^{-1}(1)=\omega$. Then $u_{1}=u_{1/2}+u_{1/3}+u_{1/6}$, so they are not linearly independent. — Emil J. 17:22, 25 September 2009 (UTC)

Yes, I apologize: I was as late and in a hurry as the March Hare, so I posted without thinking, and left, and when I started thinking, it was too late. Anyway, the correct thing is: define, for all real $\alpha ,$ the sequence $u_{\alpha }$ as a binary sequence with the k-th 1 at the place $\lfloor e^{\alpha k}\rfloor$. Then no finite nontrivial linear combination of these $u_{\alpha }$ may be zero. (their supports $u_{\alpha }^{-1}(1)$ have the property stated by Emil 2h:20 ago, at least in this weak form, still sufficient: in any finite collection of them, at least one -the one with smaller α- is not covered by the union of the others). --pma (talk) 19:24, 25 September 2009 (UTC)

PS: also, we can fix the first construction choosing the family $u_{\alpha }$ to be monotone wrto $\alpha .$ To make the construction, it is sufficient to define the monotone $u_{\alpha }$ for all dyadic $\alpha$, then extend. --pma (talk) 18:51, 26 September 2009 (UTC)

But in fact, an even simpler uncountable linearly independent set is just the family of sequences $v_{\alpha }\in \mathbb {R} ^{\mathbb {N} }$ for all $\alpha >0$, where $v_{\alpha }$ is the exponential sequence with base $\alpha$, that is the sequence $v_{\alpha }(k):=\alpha ^{k}$ for all $k\in \mathbb {N}$. Then it is standard that they are linearly independent, for a proper linear combination:

$\sum _{i=1}^{n}c_{i}v_{\alpha _{i}},$

say with $\alpha _{1}>\alpha _{2}..>\alpha _{n}$ and $c_{1}\neq 0$ can't be zero, just because

$\sum _{i=1}^{n}c_{i}\alpha _{i}^{k}=c_{1}\alpha _{1}^{k}(1+o(1))$ as $k\to \infty$. --pma (talk) 19:24, 25 September 2009 (UTC)

And if you want to see the linear independence of the above $v_{\alpha }$ algebrically, the fact is that n of them are already independent in the projection over the first n coordinates k=0,1,..,n-1 : use the Vandermonde matrix (and note that the wiki-article has it already written with the $\alpha _{i}.$ ). --pma (talk) 23:26, 25 September 2009 (UTC)

Or, for another topological proof: the subspace $\ell ^{\infty }$, has uncountable dimension, for otherwise it was separable with its standard norm. But it is not. (It contains the uncountable family of characteristic functions of all subsets of $\mathbb {N}$, that have uniform distance 1 from each other). --pma (talk) 23:26, 25 September 2009 (UTC)

Also, the subspace $\ell ^{2}$ with its Hilbert norm is isomorphic to $L^{2}([0,1])$, that admits the uncountable chain of subspaces $L^{2}([0,\alpha ])$, $0\leq \alpha \leq 1$--pma (talk) 18:51, 26 September 2009 (UTC)

Summary for the OP. Now you have: a combinatorial proof, following the lines of EmilJ's post. It is the most general, for it also works for $k^{\mathbb {N} }$; it exploits the fact that linear indipendence of a special family of sequences may be established just from the combinatorics of the collection of their supports. To exhibit such a family you also have some hints above. You then have a couple of topological proofs, that also come from more general facts: (I) no complete metric topological linear space has countable dimension, and (II) non-separable topological linear spaces have uncountable dimension. A Hilbert space proof is also available, using the isomorphism of $\ell ^{2}$ with $L^{2}$, where the uncountable dimension is more evident. Finally, a simple linearly independent uncountable family is given by exponential sequences; to show the independence you have an asymptotic order argument and a purely algebraic proof. In fact, it is now clear that you can make a proof out of virtually any result vaguely linked with the topic. --pma (talk) 12:33, 27 September 2009 (UTC)

Convergence of random variables in mean

Is there a simple example of a sequence of random variables that converge in mean to some random variable or constant, but don't converge in mean square? 128.237.235.59 (talk) 19:57, 25 September 2009 (UTC)

Take $\Omega =[0,1]$ with the Lebesgue measure, and $f_{n}:=a_{n}\chi _{[0,1/n]}$, where $\chi _{[0,1/n]}$ is the characteristic function of [0,1/n]. Choose conveniently the coefficients $a_{n}$. --pma (talk) 23:41, 25 September 2009 (UTC)

September 26

Chess outcomes

If draws by agreement or timeout (other than those known to have been book draws) are excluded, what percentage of international-tournament chess games are won by white, and what percentage by black? NeonMerlin[1] 03:06, 26 September 2009 (UTC)

There is an article on First-move_advantage_in_chess. Bo Jacoby (talk) 08:05, 26 September 2009 (UTC).

There are also stalemates, draws due to repeated board position, draws due to insufficient material to mate, and draws due to no piece taken in (50?) moves. Did I miss any ? StuRat (talk) 13:12, 26 September 2009 (UTC)

No, you didn't miss any. It is 50 moves, though some exceptional material balances were tried with more moves for a while, and also a pawn move restarts the count.Julzes (talk) 15:28, 26 September 2009 (UTC)

Interesting. I wonder what the most number of moves is, in a chess game that eventually resulted in a win. And, in the case of running out the clock, doesn't the player who still has time left win ? StuRat (talk) 15:12, 27 September 2009 (UTC)

If the player with time left does not have mating material, it is a draw. These really are not mathematics questions, by the way.Julzes (talk) 16:49, 27 September 2009 (UTC)

September 27

Fictional Solar Eclipse Calculation

This is an inquiry related to the calculations for solar eclipses. I am pretty good with scientific concepts but not so good with the actual math. Any help is appreciated.

I know the basics of how solar eclipses cause lunar shadows (umbra and penumbra) to fall on small portions of the Earth. I also know that the distance between the sun, moon, and earth affect these shadows. As a kid I learned that I could create an quasi-eclipse (ie: "blot out the sun") with a simple penny if I held it the right distance between my eye and the sun. This is because of something known as "apparent size" I think.

Assuming (just for the sake of simplifying the discussion) that it is possible to change the size and position of the moon without affecting gravity here is what I would like to know:

1. If I wanted to create a penumbra big enough to cover the whole Earth (well, the sun facing side of course) how big would the moon need to be? What about for an umbra that big?

2. Like the penny, would moving the moon closer or farther from the Earth help make the umbra/penumbra bigger? If so what is the optimal combination of size (smallest diameter) and distance (between the sun and earth) to put the moon so the whole planet is in shadow?

Of course I would love to see the math involved though I make no promises that I will understand it. Thanks again. 66.102.205.169 (talk) 06:04, 27 September 2009 (UTC

This is simple geometry. Draw a line (proportional to) the diameter of the sun. Draw a line parallel to it the diameter of the earth at the right distance. Draw 4 lines from the both sides of the sun to both sides of the earth. Choose a position for the moon and the outermost two lines determine the minimum diameter of the moon for an umbra that is visible every where on the facing earth simultaneously. (It will be more practical to draw it not to scale and to use equations...) -- SGBailey (talk) 11:09, 27 September 2009 (UTC)

Derivative of a complex function

I'm reading through a book on complex analysis, and have got stuck on one of the first problems of the text ("Introduction to Complex Analysis" by Nehari). It reads as follows:

"If the function F(z) is regular in a domain D, show that the function Re[F(z)] has a derivative only at those points of D at which F'(z)=0."

The term "regular" may be taken to mean "differentiable". According to my following "proof", however, one has G'(z)=Re[F'(z)], where G=Re[F]. Indeed, let F(z)=u(z)+iv(z), where u and v are real functions; as F is regular on its domain D, for any z in D it must be that u and v are differentiable at z and hence F'(z)=u'(z)+iv'(z). Then one has

$\lim _{h\to 0}{\frac {G(z+h)-G(z)}{h}}=\lim _{h\to 0}{\frac {u(z+h)-u(z)}{h}}=u'(z);$

the above seems to hold without any need whatsoever for F'(z)=0 (which would then imply that u'(z)=v'(z)=0). Either the book is wrong or I'm missing some fundamental piece of information here; any help on ascertaining which of the two is so would be appreciated. -- Nm420 (talk) 17:14, 27 September 2009 (UTC)

He (Nehari) means (...) the function Re[F(z)] has a complex derivative only at those points of D at which F'(z)=0. At any point z, say with F'(z)=a , as in your computation you have, for all h:

ReF(z+h) = ReF(z) + Re(ah) + o(h).

This only shows that the map Re F(z) is R-differentiable and the R-differential is the R-linear map taking the complex h into Re(ah): but you want it C-differentiable, that is, C-linear (and not only R-linear), which simply means Re(ah)=Re(a)h even for h=i : clearly, this happens if and only if a=0 . --pma (talk) 18:40, 27 September 2009 (UTC)

September 28

prove ~A /\ ~B from ~(A \/ B)

I know that they are equivalent statements from truth tables and common sense, but I have to come to this conclusion via only six mechanisms: elimination of an and; elimination of an or; introduction of an and; introduction of an or; introduction of a contradiction; elimination of a contradiction. To me, this result is so "duh!" but I can't figure out a valid proof. I can't use anything like de Morgan's laws as ... that would be beyond the mechanisms given me. John Riemann Soong (talk) 00:43, 28 September 2009 (UTC)

Does this help? I don't really understand your allowed operations but it looks to me like you're trying to prove a restricted statement of De Morgan's Laws of which the former link looks to provide a valid treatment. (Technically it shows the other part of De Mogan's Laws but it does suggest a way for you to proceed.) Martlet1215 (talk) 01:15, 28 September 2009 (UTC)

I think he means prove ¬(A ∨ B) ⇒ ¬A ∧ ¬B. ~~ Dr Dec (Talk) ~~ 11:13, 28 September 2009 (UTC)

(1−A)(1−B) = 1−(A+B−AB) is an algebraic identity equivalent to your logical identity, as ~A = 1−A and A/\B = AB and A\/B = A+B−AB. Bo Jacoby (talk) 06:21, 28 September 2009 (UTC).

Could you explain in more detail these "mechanisms" you are permitted to use? It's unclear what proof technique you are using. Is it a form of natural deduction? If so, what are "introduction of a contradiction" and "elimination of a contradiction"? Maelin (Talk | Contribs) 11:41, 28 September 2009 (UTC)

What is this "theory"/scheme called?

I remember reading an article where a person would sell a $1 coin at a market/auction/something and people would "buy" the $1 coin. It shows how a person is forced to outbid the other bidder and thus eventually ends up paying more than the amount of the coin. Know what I'm talking about? :) Deon555 (talk) 06:49, 28 September 2009 (UTC)

Yep - and a lucky guess led me to Dollar auction. AndrewWTaylor (talk) 07:41, 28 September 2009 (UTC)

Haha, thanks Andrew - I kept searching dollar coin (instead of dollar 'note') and thus returning nothing :). Thanks for your help. Deon555 (talk) 10:54, 28 September 2009 (UTC)