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August 17

Tuk-tuks

How many Tuk-Tuk's (three wheeled vehicles) are in India? —Preceding unsigned comment added by Lanceboe (talk • contribs) 01:01, 17 August 2009 (UTC)

Since I doubt that there are accurate figures on the number of tuk-tuks in India, this sounds like a Fermi problem. In which case, how would you set up the problem, and which factors do you have trouble estimating? Confusing Manifestation(Say hi!) 05:12, 18 August 2009 (UTC)

QQ vs. YoY

I’m pulling my hair out trying to calculate quarter-to-quarter annualized real economic growth such that my results match those published by statistical authorities. Singapore, for example, contracted 3.5% in the second quarter (year-on-year), or it grew 20.7% on a quarter-to-quarter annualized basis. In Excel, my formula for year-on-year growth is =sum((Q2year2 –Q2year1)/Q2year1)*100, which give me, say, 4.3 or a 4.3% rise.

What formula should I use for quarter-to-quarter calculations? DOR (HK) (talk) 03:49, 17 August 2009 (UTC)

I don't understand the word "sum". The expression ((Q2year2 − Q2year1)/Q2year1)*100 is just one term. What terms are you adding? Michael Hardy (talk) 04:42, 17 August 2009 (UTC)

=sum is commonly used in Excel equations, although the "sum" may be optional. DOR (HK) (talk) 07:39, 17 August 2009 (UTC)

Similar terms are "=count" and "=average," if that helps. DOR (HK) (talk) 08:24, 17 August 2009 (UTC)

I am not certain what you are trying to do here - it might be clearer if you can show us your underlying data. Cheap and cheerful way to annualise quarterly returns is to multiply by 4, so 4.3% quarterly growth would be 17.2% annual growth. More accurate method is to compound quarterly growth using the formula (1 + r)⁴ - 1. So 4.3% quarterly growth annualises to (1.043)⁴ - 1 = 0.183 = 18.3% annual growth. Gandalf61 (talk) 08:45, 17 August 2009 (UTC)

OK, some real-world data:

US GDP, in chained 2005 $ billion

Qtr/Yr	$ billion	YoY Growth	Qtr-Qtr Growth
Q1 2008	13,367	NA	NA
Q2 2008	13,415	NA	+1.5%
Q3 2008	13,325	NA	-2.7%
Q4 2008	13,142	NA	-5.4%
Q1 2009	12,925	-3.3%	-6.4%
Q2 2009	12,892	-3.9%	-1.0%

In the last line, year-on-year is calculated thus:

=((12,892 – 13,415) / 13,415) = -0.03898 (i.e., -3.9%)

The quarter-to-quarter annualized growth rate is reported as -1.0%. Question: what is the formula for arriving at -1.0% (or, any of the right-hand column numbers) using this data? Thanks. DOR (HK) (talk) 03:09, 18 August 2009 (UTC)

I think the Qtr-Qtr growth figures are calculated by finding the percentage growth since the previous quarter, then annualising this with compounding. For example, in Q2 2009 we have:

• % growth since previous quarter = (12,892 - 12,925) / 12,925 = -0.00255 = -0.255 %
• Annualised % growth = (1 - 0.00255)⁴ - 1 = -0.01017 = -1.02 %

Using this method, I get values of +1.44%, -2.66%, -5.38%, -6.44%, -1.02% as compared to the published figures of +1.5%, -2.7%, -5.4%, -6.4%, -1.0%. Gandalf61 (talk) 11:20, 18 August 2009 (UTC)

I agree. Michael Hardy (talk) 16:15, 18 August 2009 (UTC)

Many thanks for that. What I have not been able to do is to construct an Excel formula that duplicates the results. Any thoughts? DOR (HK) (talk) 01:33, 19 August 2009 (UTC)

Could it be that you're rounding too early? A small rounding error in an intermediate step can in some cases result in a large error in the bottom line. Unless you've got a really good handle on how big the effect of rounding at some intermediate stepp will be on the bottom line, you should not round beyond what the machine forces you do to, until the last step. Michael Hardy (talk) 01:57, 19 August 2009 (UTC)

...or it could be that the published figures are based on rounding too early. Michael Hardy (talk) 01:58, 19 August 2009 (UTC)

Got it! =(((Q2-Q1)/Q1)*4)*100 Been driving me crazy for the longest time. Many thanks! DOR (HK) (talk) 03:45, 19 August 2009 (UTC)

quick maths question

if ab > 0 and b > 0, can you take out b from the inequality, leaving a > 0? —Preceding unsigned comment added by 59.189.62.104 (talk) 05:46, 17 August 2009 (UTC)

Yes. If ab > 0 then either {a > 0 and b > 0} or {a < 0 and b < 0}. If you know that b > 0 then you also have a > 0. Gandalf61 (talk) 08:34, 17 August 2009 (UTC)

Alternative method, you can divide both sides by b. You can't do that without knowing it is positive since when you divide by a negative number you have to flip the inequality, but you do know it is positive, so that's ok. --Tango (talk) 13:23, 17 August 2009 (UTC)

How many eighth degree monic polynomials are such that...

I don't know if any programmer out there wants to tackle this, but my question is: How many monic polynomials of eighth degree have remaining coefficients in the set {0, 1, 2, ..., 41} and are composite when the variable is between 1 and 41 (inclusive), but prime when it is 42? I have one example, and I would like to know how many if any others there are. My example is

$P(n)=n^{8}+15n^{7}+13n^{6}+17n^{5}+7n^{4}+40n^{3}+3n^{2}+34n+25$.Julzes (talk) 15:17, 17 August 2009 (UTC)

How accurate do you want this figure? A Monte Carlo estimate shows that about 0.2% of the polynomials matching the coefficients criterion also match the primality criterion. Since there are $42^{8}$ of those, this turns out around $2\cdot 10^{10}$. One pleasant-looking example I've found is $n^{8}+9n^{7}+41n^{5}+3n^{2}+37$. -- Meni Rosenfeld (talk) 20:29, 17 August 2009 (UTC)

I guess I'd like a slightly better estimate with the range expanded to between 0 and 68 with the exception at 42, as 69 is where the next prime comes in in my particular example. Ideally, I'm curious about just how far up the most extreme case goes (what the polynomial which presents itself as a prime the second time at the highest value is), but I don't suppose that is a practical inquiry. I'm surprised that my initial question got any kind of answer and so quickly, and that the answer is on the order of one in every five hundred.Julzes (talk) 21:15, 17 August 2009 (UTC)

The high proportion should be no surprise at all - most numbers are composite, so it's not very demanding to require that many values of the polynomial will be. Neither should getting an answer so quickly - this is WP:RD/math. :)

About 0.022% of polynomials are prime for 42 and composite elsewhere between 0 and 68, which is about $2.1\cdot 10^{9}$.

If I understood correctly that your "ideal inquiry" is to find a polynomial P such that:

• The coefficients criterion is met.
• $P(42)$ is prime.
• $P(n)$ is composite for $0\leq n\leq N,n\neq 42$.
• N is as high as possible.

Then $n^{8}+17n^{7}+8n^{6}+22n^{5}+14n^{4}+5n^{3}+3n^{2}+37n+25$ satisfies this with $N=1145$. Of course, there may be polynomials with even higher N; only an exhaustive search (which would take quite a while) or some intelligent reasoning can find the absolute best. -- Meni Rosenfeld (talk) 16:14, 18 August 2009 (UTC)

PS: "Quite a while" = 5 years with my current program and hardware. So far I've improved N to 1697. -- Meni Rosenfeld (talk) 16:29, 18 August 2009 (UTC)

Ok. Thanks for all that. How about prime at n=-41 and n=42, but composite in between, if you're interested in continuing? Of course, you'll need either the broad definition of primality or to take absolute values. I'm not at all surprised it would take 5 years to answer the bigger question.Julzes (talk) 22:05, 18 August 2009 (UTC)

I just noticed that the constant term has to be 25 (if P(0) is to be composite but P(42) prime). Also, the condition on negatives to but not including -41 should be more stringent than the condition involving some number greater than or equal to 69. Julzes (talk) 03:17, 19 August 2009 (UTC)

question about probability

The time on digital clock is 10.38 .where here 1 is in column p ,0 is in q ,3 is in r and 8 is in column s .after x hours time is noted. find the probability that the number in

1:- column q is "9"

2:- column q is less than 5.

—Preceding unsigned comment added by True path finder (talk • contribs) 16:58, 17 August 2009 (UTC)

As you described it, the situation is perfectly deterministic, there is no probability involved. The answer is thus either 0 or 1 depending on the value of x. — Emil J. 17:15, 17 August 2009 (UTC)

(ec) Unless there is some constraint on x them it doesn't matter what the time now is. Is the clock a 12 hour or a 24 hour clock? 1:- 1/12 (09) or 2/24 (the same value) (09, 19) 2:- 7/12 (01, 02, 03, 04, 10, 11, 12) or 14/24 (01, 02, 03, 04, 10, 11, 12, 13, 14, 20, 21, 22, 23, 24) -- SGBailey (talk) 17:18, 17 August 2009 (UTC)

SGBailey, you seem to be assuming x is a uniformly distributed random variable. Why are you assuming that. This is a very vaguely stated question where we have to guess what was meant, so it doesn't seem safe to just assume things. And you didn't even state that assumption. Michael Hardy (talk) 17:37, 17 August 2009 (UTC)

AFAIK, time is uniformly distributed. I take that as a given unless something else is stated. I did say "unless there is some constraint on x" which was intended to specify the conditions of my answer. With no x constraint then the time after x is equally likely to be anywhere in a 24 hour cycle. The OP has now constrained x to be from T+1 to T+25 - this happens to be a 24 hour period, so I think my answer still applies. -- SGBailey (talk) 20:24, 19 August 2009 (UTC)

As a minor aside (obviated now that OP has clarified) it is not even possible for x to be uniformly distributed. As I said minor, but relevant to the solution of some apparent paradoxes.--SPhilbrickT 20:11, 18 August 2009 (UTC)

sorry, condition on x is 1 <x <25.and answer on book is 0.1 of first and 0.5 of second. —Preceding unsigned comment added by True path finder (talk • contribs) 19:55, 17 August 2009 (UTC)

Okay, so the time at which we next look at the clock is later than 11:38 today and earlier than 11:38 tomorrow. Let's assume, for the sake of argument, that the time is uniformaly distributed between those limits. Then I don't think I can see how the answer you gave can be reached. Are you sure you have told us the right column - column q is the least significant digit of the hours number, yes ? Gandalf61 (talk) 12:13, 19 August 2009 (UTC)

round robin tournament

I'm wondering about ways of proving that "a round robin tournament is always possible to construct for an even number of players"

The algorthym in section Round-robin_tournament#Scheduling_algorithm shows that it is always possible. However it's not clear how I could obtain that algorthym without having a 'flash of inspiration'. ie to me it seems that the algorhtym is non-obvious... Is there a more workmanlike proof of do-ability.?83.100.250.79 (talk) 19:43, 17 August 2009 (UTC)

Number the players p1,p2,...p(2n). Draw a point for each player and draw lines (edges) between every pair of points (complete graph on 2n nodes). There will be (2n)(2n-1)/2 = n(2n-1) such edges where each edge represents a game between the two players it connects. Then just enumerate the edges in any way you like, and take n of them in every round, giving a 2n-1 round tournament. 70.90.174.101 (talk) 03:14, 19 August 2009 (UTC)

Yet taking the edges off in certain orders can result in having unplayable rounds - ie certain combinations don't work.

Is there a way to proof that there will always be a valid combination of games for all rounds for any n?83.100.250.79 (talk) 12:36, 19 August 2009 (UTC)

I've designed such tournaments from time to time, using the following method. Whether it is equivalent to the algorithm you cite I can't say, but it works for any number of players (for an odd number, create an extra dummy player to denote an idle round). To avoid confusing (to me, anyway) generality, I'm taking the case of 8 players from A to H, but I can see that by simple extension it will work for any number. A table will be constructed showing the number of the round in which each pairing will occur, with rows from A to G and columns from B to H - ultimately, only the "northeast" part will apply. From the top, put 1 and 2 in col B, 2, 3 and 4 in col C, ..., 6, 7, 1, 2, 3, 4 and 5 in col G and 7 in col H. Col H is then completed by transferring the "below diagonal" entry in each row, i.e. 2 in row B, 4 in row C, etc, giving col H the successive values 7, 2, 4, 6, 1, 3 and 5 from the top. You'll see that this process guarantees the 28 matches will be played in 7 rounds.

For any number of players, the columns for all but the last one are filled immediately, that for the last one is completed by transferring the numbers below the stepped diagonal. The quick way to fill the last column, once you've seen the pattern, is to put the highest round number (odd) at the top, then to follow it with the even rounds from 2 upwards then the other odd ones from 1 upwards.→217.43.210.186 (talk) 19:19, 19 August 2009 (UTC)

It sounds basically the same in that it gives one solution, but apart from working - doesn't explain how...

I was wondering if anyone had found how to calculate the number of different sets of complete rounds constructable for every n (clearly each round has (n/2)! degenerate permutations, and the sets of rounds assuming order of play is irrelevent is also (n-1) degenerate).

But ignoring the degenerate cases is there ever more than one way to contruct an entire set of rounds of games?83.100.250.79 (talk) 22:30, 19 August 2009 (UTC)

Yes, there are additional ways. For example, with players 1, 2, 3, 4, 5, 6, we could have a set of rounds containing (1, 2)(3, 4)(5, 6) and another set of rounds containing (1, 2)(3, 5)(4, 6), and these two sets are not related by the degeneracies you identified. But that raises the question of whether there are additional ways after accounting for degeneracies from permuting players. Eric. 216.27.191.178 (talk) 02:17, 20 August 2009 (UTC)

Yes thanks, that's obvious though I didn't think of it.83.100.250.79 (talk) 11:36, 20 August 2009 (UTC)

There are distinct ways to create a tournament with 8 players even after identifying tournaments related by permutation of round order, permutation of games within a round, and permuting the identity of the players. Eric. 216.27.191.178 (talk) 02:27, 20 August 2009 (UTC)

Thanks, I was wondering about that, but haven't suceeded in finding a way to calculate the number.

In all honesty I imagine any solution (if known) might be too complex for me to understand.83.100.250.79 (talk) 19:19, 21 August 2009 (UTC)

August 18

Combinatorics question

I have a list of N categories with Nk members each. (i.e. the kth category has Nk members, and these number of members may be unique, but not necessarily distinct.) I'd like to take a random member from only 3 of the categories. How many ways can I do this? The brute force way to do select 3 groups (and there are obviously N choose 3 ways to do this) and then multiply the cardinality of each of those 3 groups to get the number of ways to choose an item from each. Iterate over all distinct groups of 3 categories. Is there an analytical way to do this? or a combinatorial formula? Any relevant literature?

Here's an example if my above explanation was confusing.

|A| = 4, |B| = 3, |C| = 5, |D| = 1.
A, B, C = 4*3*5 ways = 60
A, B, D = 4*3*1 ways = 12
A, C, D = 4*5*1 ways = 20
B, C, D = 3*5*1 ways = 15
                     
Sum                   107

Also, if anyone knows how to do this in R (I'm sure it's really simple, that would be much appreciated.)

Thanks, --Rajah (talk) 03:32, 18 August 2009 (UTC)

Notice that, for instance, (a+b+c+d)*(a+b+c+d) = aa+bb+cc+dd+2(ab+ac+ad+bc+bd+cd). So, if the sum of a through d is s1, and the sum of the squares of a through d is s2, ((s1^2)-s2)/2 is the sum of all pairs of different letters. If S_n is the sum of all the nth powers of the letters, I think what you want is ${\frac {1}{6}}S_{1}^{3}-{\frac {1}{2}}S_{1}S_{2}+{\frac {1}{3}}S_{3}$. That would be, in this case, (1/6)(a+b+c+d)³-(1/2)(a+b+c+d)(a²+b²+c²+d²)+(1/3)(a³+b³+c³+d³) = (1/6)(13)³-(1/2)(13)(51)+(1/3)(217) = 107. It doesn't look too helpful for a small example, but imagine how much easier it would be for a list of 20 numbers. Black Carrot (talk) 05:25, 18 August 2009 (UTC)

Some related ideas: multiplicative functions (such as the divisor function) and symmetric polynomials. Black Carrot (talk) 05:34, 18 August 2009 (UTC)

Yes, you want the 3rd elementary symmetric polynomial, and as BC shows you can express it in other bases of symmetric polynomials. If you need to deal more heavily with the combinatorics of symmetric polynomials you may find useful R.P.Stanley's book "Enumerative Combinatorics", (chapt 7, vol 2). [1]. PS: what do you mean by how to do this in R? --pma (talk) 06:01, 18 August 2009 (UTC)

Yeah, I meant R, the programming language, as Bo Jacoby pointed out. I was just being lazy, sorry about that. Elementary symmetric polynomials are exactly what I wanted, nice work. --Rajah (talk) 15:12, 18 August 2009 (UTC)

R (programming language) probably. Note that f(x)=(x+a)(x+b)(x+c)(x+d)= x⁴+(a+b+c+d)x³+(ab+ac+ad+bc+bd+cd)x²+(abc+abd+acd+bcd)x+abcd, so the number you want is the coefficient of x in the taylor expansion of f(x). A solution coded in J (programming language) looks like this

  (<-4 3 5 1)&p. t. 1
107

Bo Jacoby (talk) 14:11, 18 August 2009 (UTC).

Awesome, thanks. All of these replies are very helpful! --Rajah (talk) 15:00, 18 August 2009 (UTC)

Limit of integral

Show that if

$\lim _{h\to 0}{\frac {1}{h}}\int _{a}^{b}|f(x+h)-f(x)|\,dx=0$,

then there is some constant c such that $f(x)=c$ a.e.

Here's the thing. I have a proof, but I just read it through again and I have a statement in the middle that I am not sure is true. The first thing to do is, for any $a\leq x_{1}<x_{2}\leq b$, given $\epsilon >0$

$|\int _{x_{1}}^{x_{2}}{\frac {1}{h}}(f(x+h)-f(x))\,dx|\leq \int _{x_{1}}^{x_{2}}{\frac {1}{h}}|f(x+h)-f(x)|\,dx\leq \int _{a}^{b}{\frac {1}{h}}|f(x+h)-f(x)|\,dx<\epsilon$

for $h\in (-\delta ,\delta )$ (where $\delta$ comes from $\epsilon$ since the limit is 0). What this proves is

$\lim _{h\to 0}{\frac {1}{h}}\int _{x_{1}}^{x_{2}}f(x+h)-f(x)\,dx=0$

for any such $a\leq x_{1}<x_{2}\leq b$. From here, what I have down is set $F(x)=\int _{a}^{x}f(t)\,dt$ and that since f is integrable on [a, b] for $h\in (-\delta ,\delta )$, then $F'(x)=f(x)$ a.e. But, I never showed f is integrable and I am not sure if it is. Any ideas?

Just so you know, the next step I have, assuming that is right:

$0=\lim _{h\to 0}{\frac {1}{h}}\int _{x_{1}}^{x_{2}}f(x+h)-f(x)\,dx=\lim _{h\to 0}{\frac {F(x_{2}+h)-F(x_{2})}{h}}-{\frac {F(x_{1}+h)-F(x_{1})}{h}}$

Assuming f is integrable, F is absolutely continuous so that we end up getting the limit to be $0=F'(x_{2})-F'(x_{1})=f(x_{2})-f(x_{1})$ for those $x_{1},x_{2}$ where $F'(x)=f(x)$ and thus it is equal to the same constant almost everywhere. StatisticsMan (talk) 21:37, 18 August 2009 (UTC)

Ok, so we only have $f:\mathbb {R} \to \mathbb {R}$ with no assumptions of integrability nor measurability; for all $a<b$ in $\mathbb {R}$ that limit is 0 and the thesis is f(x)=0 a.e. (Alternatively, one can assume the limit to be 0 of a fixed pair a<b, and then the conclusion is that f(x)=0 a.e. in the interval [a,b]. One can easily prove each of the two statements from the other).

Maybe there are simpler ways, but one possibility is doing your program for the function $\textstyle g_{y}(x):=f(x+y)-f(x)$ for any fixed $\textstyle y\in \mathbb {R}$ and draw the consequences. By hypothesis, each $g_{y}\in L_{\mathrm {loc} }^{1}(\mathbb {R} )$, and since we have

${\frac {g_{y}(x+h)-g_{y}(x)}{h}}={\frac {f(x+y+h)-f(x+y)}{h}}-{\frac {f(x+h)-f(x)}{h}}$

you can do your complete argument and conclude that the function $g_{y}(x)$ is a constant (depending on y). Going back to the hypothesis on f(x), this tells us that the integral in $x$ is an integral of a constant function, thus just multiplication by (b-a), that is, f is a function with $f^{\prime }(x)=0$ in the classical sense for all x, and the conclusion follows. --pma (talk) 07:07, 19 August 2009 (UTC)

But in fact in the above argument I assumed $\scriptstyle x\mapsto g_{h}(x):=f(x+h)-f(x)$ to be locally integrable. Assuming also w.l.o.g. that f is compactly supported, one can write e.g. $\scriptstyle f(x)=-g_{1}(x)+f(x+1)=-\sum _{k=1}^{N}g_{1}(x+k)$ for N large enough, so f is $L^{1}$ too. If we assume the minimal requirement only, that is, that $|f(x+h)-f(x)|$ is (loc) integrable for all h, I am not sure about whether f is measurable. For sure the original statement is already of interest in the assumption of f measurable or integrable; a result under a more general hypothesis wouldn't necessarily be more useful.--pma (talk) 12:06, 19 August 2009 (UTC)

Well, I have seen a "solution" from a professor where he provides almost no detail. It took me a long time to fill them in, perhaps wrongly. But, he seems to be assuming f is measurable, though it is not in the problem. I actually did not notice that I was assuming that without being told, so good point. His entire solution is: "If the limit holds at the end points, then for every a < x1 < x2 < b it is surely true that $\lim _{h\to 0}{\frac {1}{h}}\int _{x_{1}}^{x_{2}}|f(x+h)-f(x)|\,dx\to 0$ as h goes to 0 and hence the limit without the absolute value signs inside also tends to zero with h. However

${\frac {1}{h}}\int _{x_{1}}^{x_{2}}(f(x+h)-f(x))\,dx={\frac {1}{h}}\left(\int _{x_{2}}^{x_{2}+h}f(x)\,dx-\int _{x_{1}}^{x_{1}+h}f(x)\,dx\right)\to f(x_{2})-f(x_{1})$

and hence the result." Perhaps he is also assuming that it is integrable. It seems so, right? StatisticsMan (talk) 12:25, 19 August 2009 (UTC)

Nice; I suppose you mean the last expression to be written without the big parentheses. For the last equality, f should be assumed to be integrable, yes. After all it seems that the point of the exercise is not that. NOte that your proof with F(x) is almost identical to your professor's one (in both the key point is using that (F(x+h)-F(x))/h converges to f(x) as h tends to 0; you are not using anything else about absolute continuity, and you may even avoid mentioning it) --pma (talk) 14:23, 19 August 2009 (UTC)

What you said would make it what I meant, but what I really meant to do was the big parentheses and not have the 1/h on the inside. So, I just took that out. Thanks for the info. I guess if he is assuming those things as well, I am fine with assuming them. I just didn't know for sure. My test starts tomorrow morning! So, I must go study some more. StatisticsMan (talk) 15:19, 19 August 2009 (UTC)

very good; as your coach, I recommend you to stop studying in the afternoon and relax till tomorrow. You have all to do it. and everybody in the RD/M roots for you!  ;) --pma (talk) 16:33, 19 August 2009 (UTC)

Hehe, thanks. Unfortunately, I am not going to take the afternoon off because I am trying to read through every solution I have written up to qual problems. This will end up taking about 2 full days and I still have 30 problems to read. Then, I have some other problems with solutions I want to look through as well. Thanks for the support. StatisticsMan (talk) 18:58, 19 August 2009 (UTC)

Well, I did enough to pass if everything I did was correct, or so I think. And, I think it was but I could be wrong. They aren't real clear about what it takes to pass. But, the first problem on the real analysis side was to show if the integral of a function is 0 on ever interval, then the function is 0 a.e. So, thanks for explaining to me how to do that problem, pma. Without it, I definitely did not pass. I will find out in a week or so if I passed. StatisticsMan (talk) 20:34, 20 August 2009 (UTC)

August 19

Is this polynomial most extreme?

The polynomial

$P(x):=x^{32}+x^{31}+x^{30}+x^{29}+x^{28}+x^{26}+x^{24}+x^{22}+x^{21}+x^{20}+x^{19}+x^{16}+x^{14}+x^{13}+x^{12}+x^{11}+x^{10}+x^{4}+1$

takes on prime or almost-prime (product of two primes) values for x=1*,2*,7*,16,29,30,31,32,33,36,37*,..., where it is prime for the numbers marked (*). Is it likely that there is another polynomial with coefficients in {0,1} such that over half of the values of x for which it is an almost-prime or prime up to some point at least as high as x=33 occur in a string at least five long ending at that point? Julzes (talk) 06:57, 19 August 2009 (UTC)

I took the liberty of formatting your formula with LaTeX --pma (talk) 07:09, 19 August 2009 (UTC)

Note that I suspect that "five" could probably be replaced with "two" and the value 33 reduced also. Julzes (talk) 09:33, 19 August 2009 (UTC)

?Julzes (talk) 04:00, 29 August 2009 (UTC)

Vector perpendicular to plane

So given the equation of a plane. Say... 3x+y-2z=10 How would you find the unit vector orthogonal to it? I ask because I'm not very familiar with vector algebra and this would help me finish a few proofs.--Yanwen (talk) 20:44, 19 August 2009 (UTC)

See Surface normal#Calculating a surface normal. —JAO • T • C 21:47, 19 August 2009 (UTC)

(ec) The vector v = (3, 1, -2) is a normal vector for the plane $P=\{(x,y,z)|3x+y-2z=10\}$. This is why: suppose $p_{1}=(a_{1},b_{1},c_{1})$ and $p_{2}=(a_{2},b_{2},c_{2})$ are two points on the plane P; thus we know $3a_{1}+b_{1}-2c_{1}=3a_{2}+b_{2}-2c_{2}=10$. Then the vector connecting $p_{1},p_{2}$ is $p_{2}-p_{1}=(a_{2}-a_{1},b_{2}-b_{1},c_{2}-c_{1})$, and the dot product of $p_{2}-p_{1}$ with v is

$v\cdot (p_{2}-p_{1})=3(a_{2}-a_{1})+1(b_{2}-b_{1})-2(c_{2}-c_{1})$

$=(3a_{2}+b_{2}-2c_{2})-(3a_{1}+b_{1}-2c_{1})=10-10=0\,$,

so v and $p_{2}-p_{1}$ are orthogonal. Therefore v is a normal vector of the plane P. To get a unit normal vector, just divide v by its length. Eric. 216.27.191.178 (talk) 21:52, 19 August 2009 (UTC)

Let's say the plane has equation $ax+by+cz=d$ for some, not all zero, real numbers a, b, c, and d. A normal vector would be (a,b,c), and so the two unit normal vectors are

${\mathbf {N} }^{\pm }:={\frac {\pm (a,b,c)}{\sqrt {a^{2}+b^{2}+c^{2}}}}\ .$

In fact, if you have some surface S given by an equation f = 0, (where f is a smooth function). So

$S=\{(x,y,z)\in \mathbb {R} ^{3}:f(x,y,z)=0\}\ ,$

then a normal vector to S is given by (f_x,f_y,f_z) where f_x, f_y and f_z are the partial derivatives of f with respect to x, y and z respectively. The surface S will be singular at a point $(x_{0},y_{0},z_{0})\in S$ when

$f_{x}(x_{0},y_{0},z_{0})=f_{y}(x_{0},y_{0},z_{0})=f_{z}(x_{0},y_{0},z_{0})=0\ .$

Assuming that not all three partial derivatives are zero we have two unit normals given by

${\mathbf {N} }^{\pm }:={\frac {\pm (f_{x},f_{y},f_{z})}{\sqrt {f_{x}^{2}+f_{y}^{2}+f_{z}^{2}}}}\ .$

In the plane example f(x,y,z) = ax + by + cz - d, and so f_x = a, f_y = b and f_z = c. In your example a = 3, b = 1 and c = -2 so

${\mathbf {N} }^{\pm }=\pm \left({\frac {3}{\sqrt {14}}},{\frac {1}{\sqrt {14}}},{\frac {-2}{\sqrt {14}}}\right).$

~~ Dr Dec (Talk) ~~ 15:56, 20 August 2009 (UTC)

So for a surface $S=\{(x,y,z)\in \mathbb {R} ^{3}:x^{2}-y^{2}+z^{2}=18\}\ ,$

Would the normal vector at (3,4,5) be ${\mathbf {N} }^{\pm }={\frac {\pm (6,-8,10)}{10{\sqrt {2}}}}\ .$--Yanwen (talk) 21:57, 20 August 2009 (UTC)

That looks right to me. Eric. 98.207.86.2 (talk) 06:31, 21 August 2009 (UTC)

Thanks, All!--Yanwen (talk) 12:46, 21 August 2009 (UTC)

No, that's not right! First of all ${\mathbf {N} }^{\pm }$ are two vectors: there's a choice of sign. Both ${\mathbf {N} }^{-}$ and ${\mathbf {N} }^{+}$ are unit normal vectors. They are both unit length and they are both perpendicular to the plane. Notice that ${\mathbf {N} }^{-}=-{\mathbf {N} }^{+}$. The expression

${\mathbf {N} }^{\pm }=\pm \left({\frac {3}{\sqrt {14}}},{\frac {1}{\sqrt {14}}},{\frac {-2}{\sqrt {14}}}\right)$

gives two choices of unit normal vector at each point of the plane. Think about the plane z = 0; both (0,0,1) and (0,0,−1) are unit normal vectors. There is always a choice of two. Notice that ${\mathbf {N} }^{\pm }$ is independent of x, y and z. So if a vector is a unit normal at one point of the plane then it will be normal vector at every point of the plane. So, the two choices of unit normal vector would be

${\mathbf {N} }^{-}=\left({\frac {-3}{\sqrt {14}}},{\frac {-1}{\sqrt {14}}},{\frac {2}{\sqrt {14}}}\right),$

${\mathbf {N} }^{+}=\left({\frac {3}{\sqrt {14}}},{\frac {1}{\sqrt {14}}},{\frac {-2}{\sqrt {14}}}\right).$

I'm not sure why you've re-written the vectors with $10{\sqrt {2}}$ in the denominator. It looks more complicated. The simpler form would, IMHO, be the one already given. I hope this helps... ~~ Dr Dec (Talk) ~~ 22:30, 21 August 2009 (UTC)

I'm not sure exactly what you think is not right here, but you do realize that Yanwen's $\scriptstyle {\frac {\pm (6,-8,10)}{10{\sqrt {2}}}}$ vectors were explicitly supposed to be normals of the surface $\scriptstyle x^{2}-y^{2}+z^{2}=18$, not of the plane in the original question, right? —JAO • T • C 00:02, 22 August 2009 (UTC)

What isn't right is that he said the unit normal vector is $\scriptstyle {\frac {\pm (6,-8,10)}{10{\sqrt {2}}}}$. Once you make a choice of sign you get a unit normal vector. The expression $\scriptstyle {\frac {\pm (6,-8,10)}{10{\sqrt {2}}}}$ gives two unit normal vectors. He seems to have misunderstood my notation and has taken ${\mathbf {N} }^{\pm }$ to be a single vector, which it is not: it is a pair of vectors differing by sign. I thought I made that quite clear in my last post. ~~ Dr Dec (Talk) ~~ 10:32, 22 August 2009 (UTC)

Ah, I see now that Yanwen wrote "vector" in place of "vectors". I think it's a fair bet that he already understood that there are two unit normal vectors, but I see your point. —JAO • T • C 11:41, 22 August 2009 (UTC)

August 21

Quadratic turd

My friend tell me there is a number called "quadratic turd? Why is mathematicians so vulgar to call a number turd? 67.101.25.201 (talk) 01:28, 21 August 2009 (UTC)

Quadratic surd Black Carrot (talk) 04:34, 21 August 2009 (UTC)

"Surd" from Latin surdus meaning deaf or mute - see Jeff Miller's "Earliest Known Uses of Some of the Words of Mathematics". Gandalf61 (talk) 12:12, 21 August 2009 (UTC)

Significant digits

Does the fact that 304.52 has 5 significant figures while 4.52 has 3 mean that the former is any more accurate? DRosenbach ^{(Talk | Contribs)} 03:59, 21 August 2009 (UTC)

If they've been rounded, they're equally precise relative to a given unit of measure, and the first is more precise relative to its size. Significant digits is a loose measurement of precision relative to size. Black Carrot (talk) 04:35, 21 August 2009 (UTC)

Quite right, but precision and accuracy are not the same. Accuracy is closeness of a measurement to the "actual" value. Precision is how specific the measurement is i.e. how many digits after the decimal point are displayed. For instance, my computed value of pi at 9.7564648389271381932 isn't very accurate, but it's pretty precise by most standards!--Leon (talk) 11:23, 21 August 2009 (UTC)

But since they share the same number of post-decimal digits, I figure the former measurement expresses no greater precision. If I am measuring the lengths of two pieces of wire, it's merely a reflection of our 10-base scale that makes a length of 9 one significant figure but 10 two significant figures. I mean, let's say we had an 11 base system with the symbol ¿ serving to signify the number "one greater than 9" and "one less than 10." Is measuring with our new number system make my measurement any less accurate? At first blush, I reflect that the smaller one's scale, the more precise a measurement should be, because it's easier to think about one unit, half a unit, a third of a unit, a sixth of a unit, etc., and each of these successive fractions is a smaller measurement for a smaller scale unit. I'm going off on a tangent here with my philosophical queries (I suppose there are about 4 different topics I've touched on above) but my main issue in asking this question is why adding pre-decimal digits adds precision when there doesn't seem to be any greater specificity in my measurement...It's just that one runs out of equivalent number of digits in our counting system and one must therefor carry it over to the next space. I hope this makes sense. DRosenbach ^{(Talk | Contribs)} 13:28, 21 August 2009 (UTC)

The reasons you mention are part of why significant figures aren't used in serious work. They are only really used in schools. Actual scientists will specify the precision in ways that don't depend on what base you are writing the numbers in. For example, you could say "10 +/- 2". You can measure precision either absolutely ("+/- 5") or relatively ("+/- 5%"), which is more useful depends on what you are doing with the numbers. --Tango (talk) 13:41, 21 August 2009 (UTC)

Are you talking absolute accuracy or relative accuracy? As written, both have the same absolute errors: ±0.005 units. They differ in their relative precision, though. 4.52±0.005 has a 1100 part per million error, while 304.52±0.005 has a 16 ppm error. Another complication is if the two numbers are in different units: 4.52±0.005 light years has much larger errors in both absolute and relative terms than 304.52±0.005 km. -- 128.104.112.102 (talk) 16:00, 21 August 2009 (UTC)

Yes -- that distinction definitely answers my query. Wow...that was spectacular. Thank you! DRosenbach ^{(Talk | Contribs)} 02:12, 23 August 2009 (UTC)

Wait...could you explain the 1100 vs. 16 ppm portion, I didn't get that. DRosenbach ^{(Talk | Contribs)} 02:15, 23 August 2009 (UTC)

"ppm" essentially means ${\frac {1}{1000000}}$. A relative error of 1 ppm would mean that the error is a millionth of the actual value. In the 4.52±0.005 case, the relative error is

${\frac {0.005}{4.52}}\approx 0.0011={\frac {1100}{1000000}}=1100\mathrm {ppm} .$

-- Meni Rosenfeld (talk) 10:57, 23 August 2009 (UTC)

That much I understood -- my followup was more so directed at how the other calculation possesses a denominator of greater magnitude yet results in ppm of much lesser magnitude. Was it a typo? Forgive my temporary idiocy. Got it, thanx! DRosenbach ^{(Talk | Contribs)} 13:06, 23 August 2009 (UTC)

August 23

Pi

In the same vein as my question above on significant figures, how can we say that pi extends indefinitely if it is the answer to a division problem, named 22/7, in which the LCD of significant figures is 1? Is it because we are not measuring anything with the 22 and the 7, and we really therefor mean 22.000000000000000000000000000000000000000000000000000/7.0000000000000000000000000000000000000000000, but of course with the zeros continuing forever? DRosenbach ^{(Talk | Contribs)} 02:18, 23 August 2009 (UTC)

π ≠ 22/7 Staecker (talk) 02:21, 23 August 2009 (UTC)

So how do we know what pi is other than dividing a circumference by a diameter, neither of which we are able to measure to an infinite number of decimal places? DRosenbach ^{(Talk | Contribs)} 02:56, 23 August 2009 (UTC)

Pi also can be derived with a variety of other techniques. The most common ones that come to mind are some easy integrals, which are part of elementary calculus. You can read pi, and Proof that π is irrational. I think you might be stuck on the "repeating decimals" thing - you have to recognize that irrational number is a more subtle mathematical concept than just the number of decimal-places. The fact that the digits never repeat is an artifact of the way we represent numbers with decimal place-values, but in reality, the concept of irrational numbers is much more precisely defined than the non-repeating decimal representation. There is no number system (not binary, not octal, not rational numbers, nothing)... which can express the exact value of an irrational number; it can not be exactly bound by two other numbers greater and less than it (because for any arbitrarily sized distance, there are smaller numbers which make a better bound). There's a lot of subtle advanced mathematics here - the best place to start would be to really read and understand our irrational number and real number articles. That being said, we can both prove that pi is irrational (meaning we can not find an exact value for it); and at the same time, we have lots of techniques and algorithms to approximate its value to any desired level of precision (if we are willing to work out the math to that accuracy). In any case, 22/7 is a terribly inaccurate approximation - they differ by more than 0.001 (which is a lot!) Nimur (talk) 03:02, 23 August 2009 (UTC)

A small nitpick: there are number systems in which pi can be expressed exactly. Base pi is the obvious one. Not that I would ever want to use base pi for anything. Rckrone (talk) 05:53, 23 August 2009 (UTC)

Rckrone, the nitpick wasn't very constructive. I think Nimur gave a very nice reply to the earlier post. I have only ever come across number systems to integer bases. Talking about base π seems a little wishy-washy to me. What's the expansion of 7 in base π? It would be irrational! Infact every number except multiples of π would be irrational. Would it even be well defined? ~~ Dr Dec (Talk) ~~ 17:06, 23 August 2009 (UTC)

Some irrational bases are perfectly sound—Golden ratio base comes to mind. But π is a quite different matter, of course. —JAO • T • C 17:28, 23 August 2009 (UTC)

Looking at the Golden ratio base article, it seems to support what I was saying. The arcticle says that 11_φ = 100_φ. So in this base 11 = 100. So there's no uniqueness without talking a standard form. ~~ Dr Dec (Talk) ~~ 18:47, 23 August 2009 (UTC)

A standard convention for eliminating these duplicates is to insist that there are no consecutive 1's. -- Meni Rosenfeld (talk) 19:29, 23 August 2009 (UTC)

That is no worse than base 10 where, for example, 0.999...=1. This kind of notation is usually non-unique, regardless of the base. --Tango (talk) 19:39, 23 August 2009 (UTC)

But the article also says that golden base ratio has this problem too, i.e. equalities along the lines of 0.999... = 1. It seems that every choice of base gives this same problem with infinite decimal expansions. But what base 10 doesn't say is that 11 = 100. ~~ Dr Dec (Talk) ~~ 12:30, 24 August 2009 (UTC)

Yes, this is topological. Infinite strings from a finite alphabet are homeomorphic to Cantor space and therefore totally disconnected (indeed, zero-dimensional), whereas the reals are connected and one-dimensional.

On another note, there is, of course, an exact decimal representation of pi. It happens to contain infinitely many digits. But that's nothing special. Even the decimal representation of the real number zero has infinitely many digits. It just so happens that they're all 0. --Trovatore (talk) 20:57, 24 August 2009 (UTC)

Dr. Dec, I'm sure you know this but "irrational" means not a ratio of integers, which is a property of the number rather than its representation. 7 is rational whether you express it in base 10, 2 or π - "having no repeating expansion in base π" is the correct term.

Now, whether Rckrone's comment is central to the discussion is irrelevant - Nimur has made a factually incorrect statement, and correcting it is appropriate. -- Meni Rosenfeld (talk) 19:29, 23 August 2009 (UTC)

Meni Rosenfeld, number theory isn't really my thing so you're going to have to help me out. I'm getting a bit confussed. What is π in base-π? Well one representation would be that π_π = 1. Now π₁₀ has a non-repeating decimal expansion where as π_π has a repeating decimal expansion. Now π is irrational base 10, but π is rational base π. ~~ Dr Dec (Talk) ~~ 12:39, 24 August 2009 (UTC)

There is no such thing as "rational in base 10" or "rational in base π". A real number is either rational or not. From Rational number:

"a rational number is any number that can be expressed as the quotient a/b of two integers, with the denominator b not equal to zero".

The number 7 can be expressed as the quotient 7/1, where 7 and 1 are integers. Therefore, 7 is a rational. It doesn't matter at all how you choose to represent the number. You can represent it as "7", "3+4" or "$111_{2}$" - it is the same number, and it is rational.

As it happens, there is a theorem that says that a rational number has a repeating expansion in any fixed integer base numeral system. However, the word "rational" does not mean "has a repeating expansion", and the distinction becomes important when we discuss non-integer bases.

So the number 7 has an expansion in base π which starts with "20.202112..." and happens to be non-repeating. But the number 7 is still rational.

Similarly π has a base-π expansion "10", which is repeating (terminating, even). But that does not make it any more rational, since it is still not a ratio of integers.

Anyway, you have also mixed up your notation - digits_base means "the number whose expansion in the given base consists of the given digits". So 10_π = π rather than π_π = 10. -- Meni Rosenfeld (talk) 13:22, 24 August 2009 (UTC)

Meni Rosenfeld, I've already confessed that number theory isn't my thing, so I would ask you to kindly tone down your rhetoric a couple of notches, and to show patience and good humour. We have that π₁₀ = 10_π. Now 10 is an integer and so a rational number. Now 10 is a rational number, but π₁₀ = 10_π, so why isn't π₁₀ writen to the base π a rational number? ~~ Dr Dec (Talk) ~~ 13:46, 24 August 2009 (UTC)

I guess you need to say that a number is rational when it can be writen as the quotient of two integer, base-10. Like we have seen: the rational number 7₁₀ written in base-π gives a number that would be irrational if it were to base-10. Similarily the irrational number π₁₀ written in base-π gives a number that would be rational if it were to base-10. ~~ Dr Dec (Talk) ~~ 13:53, 24 August 2009 (UTC)

I'm not sure what you mean by "tone down my rhetoric". I'm using whatever rhetoric I think would be effective, while remaining civil.

Please read User:Meni Rosenfeld/Numbers (work in progress), and then the rest of my post.

"π₁₀" doesn't make much sense. The subscript "10" means that we use the decimal system to represent a number with a sequence of decimal digits. "π" is not a sequence of decimal digits, it is a Greek letter. This letter denotes the number π in a way which does not involve any radix system.

Now, 10 is an integer, not "10". 10 is the number of slashes in "//////////", and is an integer with any conventional way of defining integers. "10" is a string consisting of the symbol "1" followed by the symbol "0", which can mean anything at all. Usually this string is interpreted using the decimal system, in which case it means the number 10. But the string itself is not a number and certainly not an integer (unless we choose to embed all finite ASCII strings as integers, but that is beside the point). 10 is of course also a rational number, while "10" is not.

As you have noted, the representation of the number π in base π is the string "10". But the fact that the string "10" denotes a rational number in some other system doesn't say anything about the rationality of the number it denotes in base π. And indeed, we know that the latter is π, which is an irrational number.

A rational number is the ratio of two integers. I can represent these integers in decimal, in Roman numerals, in Gematria, in base π or whatever - they are the same numbers in any case.

"the rational number 7₁₀ written in base-π" does not give a number, it gives a string of symbols (an infinite string). In theory we can take this string and interpret it in base 10, resulting in a number which is indeed irrational. But doing this doesn't make any sense. -- Meni Rosenfeld (talk) 20:50, 24 August 2009 (UTC)

I didn't see anything offensive in Meni's language. —Tamfang (talk) 17:05, 27 August 2009 (UTC)

I apologize; I should not have said "there is no number system..." which can represent pi. What I meant to say is, there is no number system which can represent all irrational numbers exactly. If you switch to a number system based on an irrational number, you will be mapping a different subset of real numbers into rationals and irrationals; but there will still not be an exact representation for all real numbers. These are moot points though - I shouldn't have made such a strong statement about pi. In any case, the OP is clearly unaware of the concept of rational and irrational numbers - so let's try to phrase our responses to help him/her understand that, before diving into such subtleties. Nimur (talk) 21:55, 23 August 2009 (UTC)

We can't measure a diameter and a circumference with infinite precision, true; we also can't know π with infinite precision. To put it another way, we can (in principle) measure a circle to as many digits as we want, and compute π to as many digits as we want. —Tamfang (talk) 17:05, 27 August 2009 (UTC)

DrDec: "Rational" does not mean rational relative to some base. "Rational" means a ratio of two integers. Thus 22/7 is a rational number. What base you express numbers is has absolutely nothing to do with it. Michael Hardy (talk) 23:26, 24 August 2009 (UTC)

Michael you're not being very clear. There's no point being curt and just barking a definition at me without giving an explanation. The integer article does not specify a base, and says that 10 is an integer. When you talk about integers, you seem to mean those numbers with integer values when expressed to base 10. So the integers base 10 seem to hold a special place in our arithmetic. I asked this question earlier, but didn't get a reply. If not, well then isn't 10_π an integer? If so then 10_π/1_π would be a rational number. But hang on: π₁₀ = 10_π/1_π. Any real number x can be expressed as an integer, since 10 is a (baseless) "integer" and x₁₀ = 10_x. Once again, I would appreciate patience and good humour in any following posts. ~~ Dr Dec (Talk) ~~ 11:09, 25 August 2009 (UTC)

The integers are the closure under addition and negation of {1}. Just as with rationals, no mention of bases. --COVIZAPIBETEFOKY (talk) 12:34, 25 August 2009 (UTC)

People keep saying the same thing without ever actually addressing my question. I think we do need a base. Why do we say that 10₁₀ is an integer, but 10_π is not an integer? Base 10 holds a special place. ~~ Dr Dec (Talk) ~~ 12:48, 25 August 2009 (UTC)

I think you have a very good point here: even the introduction of our Integer article seems to say that the concept "integer" can be defined as a number that can be written without a period. As you have discovered, this definition only works in base 10, 2, 16, etc—or in other words, only in bases which are, themselves, integers. A much better definition is "an integer is an element of Z", after Z has been constructed as in Integer#Construction (with the natural numbers given by the Peano axioms). Or more loosely but more intuitively, n is an integer iff at least one of the numbers n and −n is a natural number/a finite cardinal/a finite ordinal. Regardless of whether we write π as 3.14159...₁₀, as 10_π, or just as the symbol π (with no need for a base), we cannot create a set with that many elements, and that's why it's not an integer. (By the way, I apologize if my earlier answers have offended you, and I hope you'll find this one better.) —JAO • T • C 13:07, 25 August 2009 (UTC)

Are you familiar with the constructions of natural numbers, integers, rational numbers and real numbers? Very few of these constructions make any mention of base representations. A base-ten representation of a number may look pretty, but, really, it's a rather clunky representation to go by for the purposes of mathematics (partly because it contains multiple representations for the same number, and partly because the definitions of addition and multiplication aren't exactly trivial). The integers are not defined by their decimal representation.

If you have the complete set of real numbers without having taken any of the intermediate steps in defining natural numbers, integers and rationals, you can still produce the set of all integers, as I said above, as the smallest possible closure of the set {1} (where 1 is the multiplicative identity) under addition and negation. --COVIZAPIBETEFOKY (talk) 13:11, 25 August 2009 (UTC)

Jao, thanks for your reply. The construction of the integers starts of by saying that they are equivalence classes of pairs of natural numbers. And this is okay because I understand what COVIZAPIBETEFOKY is saying about the integers being generated by {1}. Since 1_x = 1_y for all real numbers x and y. And indeed the closure of {1_n} under addition and negation, where n is an integer, gives the set of integers; but there's the problem - I'm going around in circles. We need to know what an integer is, before we can understand what an integer is. I know I sound insane, and I'm getting quite worried. But, for example, what does 10_π actually mean?

$10_{\pi _{10_{\ddots }}}\ ,$

What base is the base in? And what base is the base of the base in? We have to take base 10 as a starting point and move on. Otherwise I find myself going round and around in circles. I guess that's the problem with expressing the integers in any base. It reminds me of relativity, and there being no idea of absolute space or absolute time; only inertial frames. I'm going for a lie down...~~ Dr Dec (Talk) ~~ 13:29, 25 August 2009 (UTC)

As for that specific example, remember that "π" is not the base-10 representation of the number π; the base-10 representation of π is "3.14159...". π is just a symbol whose value does not depend on base—just like the symbols 1, 2 and 5 denote the same numbers in octal, decimal or hexadecimal. Base dependence only comes into play when two or more such symbols are juxtaposed. But with another example, your remark is quite apt. When we write "30A6₁₆", for instance, we assume the subscript to be in decimal, and trying to explicitly state this by subscripting another "10" leads nowhere. You can go around this, of course, by writing the subscript as "ten", "dec" or even "5+5", all of which always mean the number ten, which "10" does not necessarily do. (The convention that numbers are in decimal unless otherwise stated is pretty useful.) —JAO • T • C 17:07, 25 August 2009 (UTC)

You make a good point about π! As for the point about bases, it shows that base 10 does indeed play a special role. We always assume that numbers are base 10, and then within this framework we can talk about other bases (base 10). To define integers and the like without any mention of a base requires more sophisticated machinery. It's just like coordinate geometry and coordinate-free geometry. Jao, thanks for your patience :o) ~~ Dr Dec (Talk) ~~ 18:24, 25 August 2009 (UTC)

No, base 10 is not "special", its just customary. You confuse numbers with their representation in a certain writing system. 10100₂ is the same number as 4₁₀, and is (e.g.) the cardinality of the set S={{{{{}}}}, {{{}}}, {{}}, {}}. We can specify or construct (many) numbers and all integers without resorting to a certain base X notation. --Stephan Schulz (talk) 18:36, 25 August 2009 (UTC)

Of course, you meant "100₂". -- Meni Rosenfeld (talk) 19:11, 25 August 2009 (UTC)

Well, of course I meant "10₄" ;-). --Stephan Schulz (talk) 19:47, 25 August 2009 (UTC)

Dr Dec - I have already addressed all of your concerns. Please read User:Meni Rosenfeld/Numbers like I asked you. If you missed the post in which I asked this, please go over this thread to locate it. I'm sure your confusion will be dispelled if you do, or at least we will be able to intelligently discuss this. -- Meni Rosenfeld (talk) 18:02, 25 August 2009 (UTC)

Meni, your rhetoric was quite strong. I felt like you were speaking down to me and trying to belittle me. Just stop and read what you've written. Phrases like "As it happens, there is a theorem...", "Anyway, you have also mixed up your notation" and "I have already addressed all of your concerns" and quite strong, and could be written, IMHO, in a much more friendly way. (Although I can be quite sensative.) Many mathematicians – and I have sometimes been guilty of this myself – seem to suffer from a superiority complex. There is no need for it. I'm sure that there are areas of mathematics where our levels of knowledge are reversed, and I would take great pleasure in fostering your interest in those areas by being patient, warm, helpful, and respectful. ~~ Dr Dec (Talk) ~~ 18:24, 25 August 2009 (UTC)

Frankly, I think you really are too sensitive. I completely agree that we should be friendly and respectful, but I still don't see where I failed that.

I have no doubt that there are areas of mathematics where our levels of knowledge are reversed - but here, in this discussion, you have made several statements which reflect very fundamental misunderstandings about the subject matter. These misunderstandings don't resolve themselves, and the more deeply they are rooted, the stronger the rhetoric required.

The "I have already addressed all of your concerns" comment was inevitable after I explained that numbers are distinct from their representations, that you do not need a base to represent numbers, that "π₁₀" doesn't make sense, and what the "10" in the subscript of "7₁₀" means -- and yet you continued to repeat the same mistakes. -- Meni Rosenfeld (talk) 19:09, 25 August 2009 (UTC)

But that oversight on my part exemplifies my point: I took π as a given, i.e. 3.14159265... = π. Most people always assume that the numbers are given to base 10. I agree that numbers exist independently of their representations. My problem seems to lie with the representations themselves. I have read you user page section on numbers and it does help a little. I am begining to think that my problem is not (thankfully) with the idea of integers and natural numbers, but with the notation used to give representations of these numbers. I have learned something over the last few days; but it has been quite painful. I'm sure people could have been more gentle. Anyway... ~~ Dr Dec (Talk) ~~ 19:20, 25 August 2009 (UTC)

Perhaps you will be interested in the last post here, where I try to define what a decimal expansion really is. If you're interested in more general radix representations, you can read pages such as Numeral system, Non-standard positional numeral systems and those linked by them. -- Meni Rosenfeld (talk) 19:46, 25 August 2009 (UTC)

DRosenbach: See:

• Computing π
• Bailey–Borwein–Plouffe formula
• Numerical approximations of π
• Chronology of computation of π
• Leibniz formula for pi
• Liu Hui's π algorithm
• Proof that π is irrational
• Proof that π is transcendental
• Gauss–Legendre algorithm
• Software for calculating π
• Viète's formula
• Wallis product
• Borwein's algorithm
• List of topics related to π

Michael Hardy (talk) 23:39, 24 August 2009 (UTC)

Lagrangian is invariant

What does that mean? Let's for simplicity take a system where the Lagrangian L is invariant under U(1) (which I understand contains the complex numbers with absolute value 1). But invariance can not mean -L = L. 93.132.179.71 (talk) 07:58, 23 August 2009 (UTC)

Just to fix the vocaboulary: if $f:X\to Y$ is any function (usually with $Y=\mathbb {R}$) and if G is a group with an action on $X$, one says that $f$ is invariant for the action of G, or G-invariant, iff for all $x\in X$ and $\gamma \in G$ there holds $f(\gamma \cdot x)=f(x)$. One also says that G is a group of symmetries of f. So in any case there's no "-L=L". Maybe you have in mind a situation where there is a G action on Y too, (usually, but not necessarily, the same action, with X=Y), and $f$ satisfies $f(\gamma \cdot x)=\gamma \cdot f(x)$ for all $\textstyle x$ and $\textstyle \gamma$; in this case one says that $f$ is equivariant wrto the two G actions, or G-equivariant. --pma (talk) 10:19, 23 August 2009 (UTC)

Thank you, that excludes some of the possibility for misinterpretation. So f is constant on the equivalence classes of X induced by G. Now with the Lagrangian itself operating on functions I'm still not clear if in this case G acts on these functions (as the domain of L) or on the domain of each of these functions. 93.132.179.71 (talk) 13:26, 23 August 2009 (UTC)

(By the way, the equivalence classes induced by a group action have a special term, "orbits", for those days when you get tired of writing "equivalence class induced by the group action". Eric. 98.207.86.2 (talk) 21:01, 23 August 2009 (UTC))

Proof?

Hi, indeed I mostly reply questions at Arabic Wikipedia, and one question stopped me where I couldn't think where to classify (regardless whether or no being a homework). I'll try to translate it in English symbols as follows: For any natural numbers n, k; Prove that

$3n^{2}+3n+7\neq k^{3}$

always holds.

I guessed it can be done by mathematical induction but failed. This is the original question in Arabic. Thanks in advance, --Email4mobile (talk) 11:11, 23 August 2009 (UTC)

These problems are usually best done with modular arithmetic. Take the equation mod 6 ( $\scriptstyle n(n+1)$ is even), what does this tell you about $\scriptstyle k$ mod 6? If $\scriptstyle k$ is $\scriptstyle m$ mod 6 then you have $\scriptstyle 3n(n+1)+6+1=(m+6r)^{3}=m^{3}+18m^{2}r+108mr^{2}+216r^{3}$ for some $\scriptstyle r$... 129.67.186.200 (talk) 11:40, 23 August 2009 (UTC)

Taking the equation mod 6 and probing for the possible values of x^3 mod 6 shows k = 1 mod 6. But I can't see any use of this fact. 93.132.179.71 (talk) 13:31, 23 August 2009 (UTC)

Me neither. Maybe there is an elementary reason, maybe not so elementary; for instance the classical diophantine equation $\scriptstyle x^{2}+2=y^{3}$ is studied factorizing the LHS in $\scriptstyle \mathbb {Z} [{\sqrt {-2}}]$ (so here one would work in $\scriptstyle \mathbb {Z} [{\sqrt {-3}}]$ instead). --pma (talk) 13:51, 23 August 2009 (UTC)

Once you know that $\scriptstyle k=1+6r$, you get $\scriptstyle 3n(n+1)+6+1=1+18r+108r^{2}+216r^{3}$. This gives $\scriptstyle n(n+1)+2=6r(1+6r+12r^{2})$, but $\scriptstyle n(n+1)+2$ has no roots mod 6. 129.67.186.200 (talk) 15:38, 23 August 2009 (UTC)

I'm not very good with number theory: it's not my thing. I've given this a once over and here are my thoughts. The LHS $3n^{2}+3n+7$ is always congruent to 1 modulo 6. That means that $3n^{2}+3n+7$ can always of the form $6m+1$ for some integer m. This has reduced the problem from a quadratic to a linear problem. Why not try more modular arithmetic on this linear expression. ~~ Dr Dec (Talk) ~~ 17:19, 23 August 2009 (UTC)

p.s. I've just tested it for $0\leq n\leq 1,000,000$ and it holds. ~~ Dr Dec (Talk) ~~ 17:30, 23 August 2009 (UTC)

Yes, seeing that it's 6m + 1 is certainly the first step. Seeing why m can't be 57 (or any other number making 6m + 1 a cube) is the second. (This is what 129.67.186.200 did two hours ago.) —JAO • T • C 17:49, 23 August 2009 (UTC)

Really? If he did then it's very convoluted. I didn't see that he said that. If I didn't see that then I'm sure the person making the original post didn't see it either. Jao, please do me a favour and calm down. From this post, and earlier ones, you seem to be quite confrontational. So what if someone said something earlier? Does it really matter? Is there any need to make a sly comment? No! ~~ Dr Dec (Talk) ~~ 18:09, 23 August 2009 (UTC)

Hi, 129 here, I don't think my solution is convoluted. All I did was to observe 1) the LHS is 1 mod 6 so k must be 1 mod 6 (I left this as a hint, 93 spelled it out) 2) knowing this you are lead to the equation n(n+1)+2 = 0 mod 6 which is easily seen to have no solutions (in my 15:38 post). Jao is only saying (I think!) that you should read the thread carefully before you reply, it gets confusing otherwise. Anyway, the problem is done. 87.194.213.98 (talk) 19:33, 23 August 2009 (UTC)

I did read the thread carefully, it wasn't very long. You gave very little explanation and just wrote a lot of maths. I found that convoluted. Clearly, you wouldn't think that your solution is convoluted: you wrote it! People ask questions because they don't understand the topic. There's no point writing a reply that assumes they have the required knowledge to solve the problem. If they did then they would have solved it, and if they'd have solved it then they wouldn't be asking the question. And whatever Jao was trying to say, he didn't say it in a very nice way. Scroll up and read more of his posts, he's not the most civil or polite. ~~ Dr Dec (Talk) ~~ 20:54, 23 August 2009 (UTC)

Injecting myself into this conversation, I'd like to say that Jao has always been civil in the lengthy period of time I've seen him contributing to the RD. I can sort of see how Jao's comment in the unit normal discussion appears sharp, but that can't possibly be a big deal, can it? Eric. 98.207.86.2 (talk) 21:14, 23 August 2009 (UTC)

I found Jao's remark a gentleman post. Very opportune, too. --pma (talk) 06:57, 24 August 2009 (UTC)

Added emphasis to "appears". Eric. 98.207.86.2 (talk) 07:53, 25 August 2009 (UTC)

PS: I just meant to agree with your remark, it was no objection --pma (talk) 15:37, 26 August 2009 (UTC)

Ah, cool. Eric. 216.27.191.178 (talk) 19:36, 26 August 2009 (UTC)

I'd like to thank you all for you wonderful cooperation and interaction. To tell you the truth, I feel such proofs are totally new to me because I didn't study about "modular arithmetic"; although I think the one who asked this in Arabic Wikipedia must have studied it unless there's another way to treat such a problem.--Email4mobile (talk) 18:27, 23 August 2009 (UTC)

Well, 129's proof is clever and clear. It reduces to an elementary fact (i.e., n(n+1)+2 is never divisible by 6) that one can easily check. Most likely you are able to understand it perfectly, but of course, if any point is still not clear to you, I guess you are welcome to ask for further details. --pma (talk) 06:57, 24 August 2009 (UTC)

I started reading about modular arithmetic in order to learn more, and I hope that you give me some more recommendations to study this subject. --Email4mobile (talk) 10:14, 24 August 2009 (UTC)

Average

In a philosophically mathematical manner, is it ridiculous to ask if "the average person is average"? I suppose there could be some term switching in that both averages would not refer to the same method of obtaining an average (e.g. mean vs. median), but that would sort of be a fallacy of four terms. My sister-in-law asked me this question, and at first I thought it was insightful, but shortly thereafter figured that it's probably not that insightful at all. Any comments? DRosenbach ^{(Talk | Contribs)} 13:38, 23 August 2009 (UTC)

I'd say not; I guess it can be translated in a more mathematical language as asking whether the variance is small. An average gives only a first description of a distribution; then you can ask how close the individuals are to the average value. For instance, in a society of clones, everybody is very close to the average person (and I'm trying to recall a sentence in a novel by Oscar Wilde, asserting that the typical Englishman is not typical...)--pma (talk) 14:16, 23 August 2009 (UTC)

If "the average person" refers to someone chosen at random, there is a 1 in 2 chance that he is not average because only 50 of the population placed on a bell curve would be considered average. So perhaps it's not such a bad statement after all. DRosenbach ^{(Talk | Contribs)} 14:36, 23 August 2009 (UTC)

I've never heard of "average" meaning "random" or "in the middle 50%". --Tango (talk) 19:43, 23 August 2009 (UTC)

Another interpretation would be to relate this to ergodicity. (My understanding of ergodic theory isn't great, so please correct me if this makes no sense). For instance, if you were interested in which parks in a city were well-liked by people, you could can do two things. You can look at a lot of people in a single instant and see which parks are busy and which are empty. You can also follow a single person for a year, and see which parks the person frequents.

These method will likely give very different results, whatever person you choose. This means that even if you choose the "most average" person in the city (by whatever definition) his path (the parks he visits) will still diverge from the average. In that sense human beings have a tendency to be non-average (which, I think is more subtle than just having a high variance, although that may be an effect). risk (talk)

A game with positions expressable as vectors

I'm looking for a reasonably complex game of perfect information, where each position (ie. the state of the game after a player has made her move) can be fully expressed as an n-dimensional vector in a straightforward manner. I'm sure that the board-positions of Chess and Go can be encoded in this way, but any method I can think of is rather convoluted, 'hiding' the relevant information, or leads to a massive, very empty space.

Can anybody think of a game that fits this description? I'm sure such a game would be very useful in Game Theory (relating game theory to geometry), so I figured someone might have invented one just for that purpose. risk (talk) 16:02, 23 August 2009 (UTC)

A very simple example would be naughts and crosses. There are nine positions (represent this by $\{0,\ldots ,8\}=\mathbb {Z} _{9}$), and each place can have a naught, a cross, or be empty (represent this by $\{0,1,2\}=\mathbb {Z} _{3}$). So any game would be a vector in $\mathbb {Z} _{9}\times \mathbb {Z} _{3}.$ Alternatively, a game could be given by a mapping $f:\mathbb {Z} _{9}\to \mathbb {Z} _{3}.$ Of course not all of these vectors and maps would represent a valid game: for example, you couldn't have a cross in all nine positions. As to your problem about big empty spaces. The layout of a game has myriad posibilities, and so the space of all game states would be huge – if not infinite. So you would only ever have a single vector for a given game state sat in a massive space of game states. I've thought about something similar with snooker. The pocket is on a two-dimensional table and the trajectories of the cue ball which lead to a pot are limited. But when you add side, top, bottom and power and other such factors you can think of your pockets being multi-dimensional pockets on a multi-dimensional table. ~~ Dr Dec (Talk) ~~ 18:34, 23 August 2009 (UTC)

I was thinking along similar lines for Chess or Go, but I figure that even if the space of all games is near infinite, the space between 0 and 1 in even one dimension is actually infinite, so it could easily accommodate all those positions. Consider, for instance a representation of a chess position as a binary string. You can interpret each binary string as a number in (0, 1), which gives you the required densely packed representation (informally; it wouldn't actually be a dense set, I guess).

The problem with this is that the transformation removes too much information and the geometric interpretation of this set is fairly meaningless. I was trying to think of a game where there is a more natural transformation from positions to euclidean points, so that things like the euclidean distance are more meaningful. risk (talk) 19:37, 23 August 2009 (UTC)

Wouldn't $\mathbb {Z} _{3}^{9}$ be a more obvious vector space for describing games states in noughts and crosses? Each position is represented by a dimension and what is filling that position is given by the coordinate in that dimension. I'm not sure how your suggestion would work - you would need 9 points in that space to determine a game state and the first coordinate is completely redundant as long as those points are ordered (and the coordinates in a vector are always ordered). --Tango (talk) 20:03, 23 August 2009 (UTC)

Yeah, I'm not sure why I wrote $\mathbb {Z} _{9}\times \mathbb {Z} _{3}.$ I was thinking one thing and writing another. $\mathbb {Z} _{3}^{9}$ would indeed be the vector space to use. An empty board would be prepresented by the zero vector. A board filled with naughts (although not a valid game state) would be (1,…,1) and a board filled with crosses (although not a valid game state) would be (2,…,2). Tango, thanks for the correction. ~~ Dr Dec (Talk) ~~ 20:26, 23 August 2009 (UTC)

Risk, I see what you're saying about the interval. Is there really a difference? Is a point any more or less isolated in the interval [0,1] as a vector is in $\mathbb {Z} _{3}^{9}?$ I would say that a point is more isolated in the interval than a vector in $\mathbb {Z} _{3}^{9}.$ The vector $(v_{1},\ldots ,v_{9})$ where $v_{i}\in \{0,1,2\}$ could be set to correspond to the number $0.v_{1}v_{2}\ldots v_{9}.$ What about irrational numbers like π/4 ≈ 0.7854? They don't correspond to any element of $\mathbb {Z} _{3}^{9}.$ In fact the numbers in [0,1] which don't correspond to elements of $\mathbb {Z} _{3}^{9}$ form an open and dense subset of [0,1]. ~~ Dr Dec (Talk) ~~ 20:42, 23 August 2009 (UTC)

That's a good point. In $\mathbb {Z} _{3}^{9}$ every point is a legal board position. In that sense, you're absolutely right, the space is fully packed with positions (though not all legal). But the structure that those points make when you view the set as a subset of $\mathbb {R} ^{9}$, isn't very interesting (essentially a 3x3x3x... grid), and in that sense it does create more 'empty space'. Of course, a carefully designed game could get around this by making all the real or rational vectors possible/legal moves, which is more or less what I'm looking for.

I should perhaps elaborate a little on what I'm trying to do. One of the things I'm toying with is the idea that in such a representation you can talk about the (fractal) dimension of the total set of board positions, or the dimension of the sets of winning/losing positions, or the dimension of the boundary between these points. (You could do that anyway by just defining a metric on the positions, but that would introduce the possibility that the choice of metric is more important to the dimension than the game). Of course any game with a finite number of positions has dimension zero, but there are ways around that.

Another option would be to treat a move as a map. This would open up the possibility of treating the game like a discrete dynamical system. risk (talk) 21:40, 23 August 2009 (UTC)

Are you restricting yourself to vector spaces or is any manifold acceptable? There is an additional requirement that the subsets you are interested in actually be subspaces/submanifolds, otherwise the dimensions you talk about aren't well defined. I'm sure we can invent a game that works, but finding an existing one (other than the kind of strange games mathematicians invent for various purposes) seems unlikely. Most games have a finite number of possible game states since there is usually a finite number of possible moves at any given point and a finite amount of time to play the game (although I suppose it is possible that time could be unbounded, leading to countably infinitely many game states, but that isn't much help anyway). If we want to have interesting geometry over the real numbers we obviously need uncountably many game states, that means uncountably many decisions at a given point. The only such games I can think of are sports where you have things moving around in space, but they aren't usefully described in the way you want. For football (soccer) you can have the position of the ball represented by a 2D vector, the possible game states are a rectangle and the winning game positions are a line segment at each end of the pitch. Not very interesting, but it fits your requirements technically. You could try snooker with 2 dimensions per ball, the possible game states would be those where the balls are in a rectangle and there is a minimum distance between them (the diameter of the balls). The end game states would be a discrete set where all the balls are in the pockets (the minimum distance rule doesn't apply there), and you need 2 extra discrete dimensions for the scores. So, you can do the kind of things you are talking about, but they aren't useful. I think it is only going to work nicely if you invent a game for that purpose. --Tango (talk) 22:29, 23 August 2009 (UTC)

I guess you're right. Specifically, the difference between a countable and an uncountable number of positions may not be so easy to gloss over as I had thought. I gues I have some more thinking to do. Thank you both for the insights. risk (talk) 11:11, 24 August 2009 (UTC)

If it helps, most variations of Nim, and most take-away games in general, can be viewed as motion in a discrete subset of a vector space in an obvious way. I've also seen an analysis of Peg Solitaire that viewed playing the game as vector addition in a high power of Z₂. And of course, there's the classic probabilistic games in economic Game Theory, which are explicitly viewed as points in Rⁿ. Black Carrot (talk) 06:31, 26 August 2009 (UTC)

Those are very helpful suggestions, thank you. risk (talk) 15:08, 27 August 2009 (UTC)

BN-pairs: conjugate generators weyl group give double cosets of same size

Hello,

I read (B, N) pair, but I still have a question about the correspondence described there between elements of the Weyl group and double cosets.

Suppose $w_{1}$ and $w_{2}$ are both generators of the Weyl group, and suppose that they are conjugate (within the Weyl group). Is it correct, and trivial to see(?), that the double cosets $Bw_{1}B$ and $Bw_{2}B$ have the same cardinality? (I'm rather sure it should be, because in geometric language, this is the number of chambers through one panel in a thick building).

However, I'm afraid it's not like we can prove that by proving that both double cosets are conjugate as well (because I think they aren't). Do you have any ideas? Thanks!

Evilbu (talk) 21:25, 23 August 2009 (UTC)

So by w_1 and w_2 being generators, I'm assuming you mean they are both part of a single (Coxeter) generating set. The Weyl group of type An is a little weird because individually all of its generators are conjugate (they are just two cycles in the symmetric group), but within the particular Coxeter system they are quite different. For instance, take the group to be SL(4,p^f) of type A3. B is a Borel subgroup, the normalizer of a Sylow p-subgroup. B is a semidirect product of T and the Sylow p-subgroup, and N is the normalizer of T in G. N/(BnN) is isomorphic to the symmetric group on 4 points, a Coxeter group of type A3 with Coxeter generating set w1=(1,2), w2=(2,3), and w3=(3,4). Letting n1, n2, n3 be preimages of w1, w2, and w3, one has that the double cosets Bn1B and Bn2B have different sizes. For instance, when p^f=3, Bn1B has size 157464 and Bn2B has size 17496. Now, if w1 and w2 are conjugate under a graph automorphism of the Weyl group, then I think the group itself should have a graph automorphism so that Bn1B and Bn2B are conjugate. The graph automorphism is unlikely to be an inner automorphism, so probably they are not conjugate within the group itself, but they would be in the holomorph. JackSchmidt (talk) 00:34, 24 August 2009 (UTC)

Hello, thanks for your reply. Indeed, I mean that they are both part of a single coxeter generating set (sorry if that was unclear). I must admit that I do not know what a "graph automorphism of a group" is. I do know "automorphisms of a graph" (and there is plenty on that on the internet) but that's something else (it seems). I was looking at the numbers you gave me so I could try it myself. Do you also happen to know the size of B? If $p^{f}=3$, I find that it should be 2^3 3^6=5832. In general, if I let q denote p^f, I find that: $B=(q-1)^{3}q^{6},Bn_{1}B=Bn_{2}B=Bn_{3}B=q^{7}(q-1)^{3}$ (which is 17496 if q=3). So I don't understand what is going on. Note that the cardinality of each double coset must be a multiple of the cardinality of a coset of B. (And to be honest : it is in fact the quotient that I am interested in). Many thanks.ıı

Oops, sorry, my explicit calculation was wrong (it took a B and an N, but not form the same (B,N) pair, so my n1 was actually of length 3 in the true Coxeter generators). The general calculation made it clear:

For SL(n+1,q) of type An, B has order (q-1)^n*q^(n(n+1)/2) and it has a very important subgroup U of order q^(n(n+1)/2). U is made up of subgroups X_i, each of order q, called root subgroups, that are in one to one correspondence with the positive roots. For a given element w of the Weyl group, there is a subgroup U_w of w which is generated by all X_i such that w(i) is a negative root. So U_1 = 1, and U=U_{w0} where w0 is the longest element of the Weyl group. The double coset BwB is also equal to BwU_w and has cardinality |B|*|U_w|. In particular, for w=w1, U_w1 = X_1 has order q, and for w=w2, U_w2 = X_2 has order 3, so the quotient |BwB|/|B| = |U_w| is equal to q whenever w has length 1 (that is, whenever it is one of the generators from the Coxeter generating set). Indeed, if w has length l (it is a product of l generators, but not a product of fewer than l generators), then |U_w|=q^l.

For general (B,N) pairs I don't know a formula for the size, but for so called "split (B,N) pairs of characteristic p satisfying the commutator relations" this same thing occurs, BwB = BwU_w and |BwB|/|B| = |U_w|. However, even for algebraic groups, the size of U_{w1} can depend on which root you take, though this only happens in the twisted groups. Most of this can be found in Carter's Simple Groups of Lie Type. Hopefully I'll have time later today to write out the SU(n,q) case. JackSchmidt (talk) 13:06, 24 August 2009 (UTC)

Hello, I certainly enjoy "debating" with you (and I think we have also crossed paths elsewhere). If it suits you better, this discussion can also be continued on my talk page. I somewhat recognise the stuff you wrote about U_w. I read that in general, there is a q_i associated with each generator w_i of the coxeter generating set. The correct formula for the size of U_w would then be $q_{1}^{a_{1}}\ldots q_{n}^{a_{n}}$ where a_i is the number of positive roots, conjugate to $w_{i}$, turned negative by w. (so we use weighted lengths instead. So for SU(6,q) there would be 6 positive roots associated with $q^{2}$, and 3 positive roots associated with q. So the size of $U=U_{w_{0}}$ would be $(q^{2})^{6}(q)^{3}=q^{1}5$.

It is exactly those q_i that I am after. They should also be the cardinalities |B w_i B|, divided by |B| itself. And they should be equal if the corresponding generators are conjugate. (so that is why there is only one q involved in the projective case, the A_n case). A lot of people have told me that Carter's book is a good reference, but I couldn't find an answer in there. That is in generaly my problem with theory like this, too much stuff I picked up "from the street". I was very curious if there is a really quick way to see that those q_i must be equal for conjugate generators Evilbu (talk)

It's nice to have the question asked this way. I'll try to check the SU case today (I was thinking that this would be an example where the wi could be conjugate but have different $q_{i}$, but perhaps they happen to be). Unfortunately the Weyl group is type B_n in the SU case, and so there does seem to be a good chance the q_i will respect conjugacy. The size of the q_i though is supposed to depend on the size of the orbit of the root under the graph automorphism that defined the twisted group, and I don't see why that should be particularly related to conjugacy in the Weyl group. If it is true (in some generality), then Carter's Finite Groups of Lie Type (the thick brown one, not the thin black/gray one) might have the appropriate setup in its early chapters. I prefer the thin "Simple" book for most things, but it treats twisted and untwisted separately. JackSchmidt (talk) 15:52, 24 August 2009 (UTC)

Perhaps you know all this, but in $A_{n}$ all generators (and hence all roots) are conjugate, in $B_{n}$ they all are except that one on the end. In F_4 the two on the left are conjugate, and so are the two on the left. The $q_{i}$ can be equal- as far as I understand it- even if the generators are not conjugate, but conjugacy does imply that they are equal. I must have a look at that other book by Carter then. But in the mean time, could you tell me what exactly you mean by "graph automorphism"?Evilbu (talk)

Cool, I checked all the twisted groups and it is always true that if w1 and w2 are conjugate in the Weyl group then q1=q2 for the root subgroups. I still don't quite understand why this is true other than as coincidence, but it seems to work. By graph automorphism I mean five related things: an automorphism of the Coxeter diagram, an automorphism of the Weyl group described by permuting the generators according to the diagram, an automorphism of the algebraic group (over the algebraically closed field) inducing that automorphism on the Weyl group, an automorphism of the finite untwisted group that is just the restriction of the one on the algebraic group, and, when talking of a twisted type group, the automorphism of the algebraic group that gets combined with the frobenius endomorphism to define the twisted group. I think the fourth and the fifth don't conflict, since I think a twisted group doesn't have any graph automorphisms in the fourth sense, though 2E6 as F4 worries me a little.

Is there a simple reason why conjugate generators have equal q_i? JackSchmidt (talk) 21:33, 24 August 2009 (UTC)

August 24

Probability

I've got a question on Probability which I found on one of my reference books, which I couldn't figure out. Guys, this is not homework, just an interesting question which I ran into. Say a police officer is investigating a case. He is 60% convinced that X is the culprit. Suddenly, he finds a new piece of evidence, which is a mark on the culprit's face, which 90% of the population don't have. If X has the mark, what is the probability that X is the prisoner ? Actually, though this question looks simple enough, I just can't crack it ! Excuse me if I had asked a very obvious question, but I need some help. Rkr1991 ^{(Wanna chat?)} 10:54, 24 August 2009 (UTC)

Let A be the event "X is the culprit" and B "X has a mark". We have $P(A)=0.6,P(B|A)=1,P(B|A^{C})=0.1\,\!$. By Bayes' theorem the posterior probability of A given that we know B is

$P(A|B)={\frac {P(A)P(B|A)}{P(B)}}={\frac {0.6\cdot 1}{P(A,B)+P(A^{C},B)}}={\frac {0.6}{0.6+0.04}}=0.9375$

-- Meni Rosenfeld (talk) 12:16, 24 August 2009 (UTC)

Either there's a typo in the question or it's a trick question. The probability that X is the prisoner cannot be calculated from the police officer's certainty that X is the culprit and the ratio of people carrying a certain facial mark. We're not told anything about any trial turning a suspected culprit into a prisoner. ~~ Dr Dec (Talk) ~~ 12:24, 24 August 2009 (UTC)

Great, that's the right answer. Thanks a lot !!!! Rkr1991 ^{(Wanna chat?)} 13:25, 24 August 2009 (UTC)

General formula for power sum

Today I am the one who asks about a general formula for the following sum, of integers.

$\sum _{i=1}^{n}i^{k}$

For example, the simple form is given by the arithmetic progression sum, in which k = 1, is given as

$\sum _{i=1}^{n}i={\frac {n+n^{2}}{2}}$

If there exists an article, I'd be glad to be guided to.--Email4mobile (talk) 16:27, 24 August 2009 (UTC)

Have a look at Bernoulli number which answers your question. 129.67.186.200 (talk) 16:39, 24 August 2009 (UTC)

Thank you very much, 129.67.186.200. That is what I need. I did try to find a general solution yesterday, didn't know this formula exists. That's my try, User:Email4mobile/sum of power arithmetic progression.--Email4mobile (talk) 21:46, 24 August 2009 (UTC)

Also: Faulhaber's formula. Michael Hardy (talk) 23:15, 24 August 2009 (UTC)

Conceptually the simplest way to figure out that summation is through the Lagrange interpolation formula. You know that it's obviously a polynomial of degree k+1, so plot some points and then fit the curve through it. 70.90.174.101 (talk) 00:14, 25 August 2009 (UTC)

Greater, lesser or non-comparable

Would it be proper to say that negative numbers are less than positive numbers -- because an angle of -7 is a greater angle than an angle of 5? Or are angles different, because in regard to angles magnitude is all that matters? DRosenbach ^{(Talk | Contribs)} 18:55, 24 August 2009 (UTC)

Yes: the negative numbers are less than the positive numbers, with repect the the usual ordering. What exactly do you mean when you say that "−7 is a greater angle than an angle of 5"? I assume you mean something along the lines of an angle of −7 degrees is larger than an angle of 5 degrees. Well, this is very different. The number −7 is further from 0 than 5 is. This is because |−7| > |5|. It all depends on how you chose to order your set, and once an ordering has been chosen then it doesn't really make any sense to compare orderings. For example, given a natural number k, the metric norm

$||(x,y)||_{k}:=(x^{k}+y^{k})^{1/k}\,$

have very different "unit disks". The unit disk for k = 2 (i.e. in the usual Euclidean metric) is the usual circular Euclidean unit disk. In the case where $k\to \infty$, the unit ball is the Euclidean unit square! ~~ Dr Dec (Talk) ~~ 19:27, 24 August 2009 (UTC)

Wow...I got everything up until your example of "k" -- from then on, I would say it's all Chinese to me. If you insist it's English, then I don't know what I'll say... :) DRosenbach ^{(Talk | Contribs)} 21:59, 24 August 2009 (UTC)

I think it was just an analogy. The real numbers with different orders are completely different things in the same way that the plane with different metrics/norms are completely different things. It doesn't really matter, but basically that formula is a new definition of the distance between two points, so you end up with up different definitions of discs (since a disc is all the points less than or equal to a certain distance from a certain point, so you chance the definition of distance you change the definition of a disc). If it doesn't make sense, just ignore it! The important bit is everything before "For example". --Tango (talk) 22:14, 24 August 2009 (UTC)

Well, there are many ways to measure distance, and to measure the length of something. A metric is something that we use to measure the distance between two points. Let's consider the plane, denoted by $\mathbb {R} ^{2}$, as an example. How do we define the distance between two points, say a and b? Well, if a = (a₁,a₂) and b = (b₁,b₂) then the distance, let us denote it by d(a,b), in the good old fashioned using-a-ruler sense is given by

$d({\mathbf {a}},{\mathbf {b}})={\sqrt {(a_{1}-b_{1})^{2}+(a_{2}-b_{2})^{2}}}\ .$

The idea of distance has some nice properties. Let a, b and c be points in the plane (not nevessarily distinct).

1. The distance from a point to itself is zero, i.e. d(a,a) = 0.
2. The distance between points is always non-negative, i.e. d(a,b) ≥ 0.
3. Informally, the shortest distance between two points is a straight line, i.e. d(a,b) + d(b,c) ≥ d(a,c).

This last one is called the triangle inequality. Well, what we do is to abstract this idea. We say that any function d that satisfies these three conditions is a metric, and gives us some idea of distance. One such metric would be

$d_{4}({\mathbf {a}},{\mathbf {b}})=((a_{1}-b_{1})^{4}+(a_{2}-b_{2})^{4})^{1/4}\ .$

In fact, for any positive integer k we have have a metric given by

$d_{k}({\mathbf {a}},{\mathbf {b}})=(|a_{1}-b_{1}|^{k}+|a_{2}-b_{2}|^{k})^{1/k}\ .$

Good old fashioned distance comes from k = 2. But for all positive integers k we have a metric, i.e. a function that takes any two points in the plane and gives us a number in a way that satisfies our three conditions above. Now, the unit disk, for a metric d, centred at some point a is given by all the points b judged to be at a distance of less than or equal to 1 from a, i.e.

$D_{\mathbf {a}}=\{{\mathbf {b}}\in \mathbb {R} ^{2}:d({\mathbf {a}},{\mathbf {b}})\leq 1\}\ .$

In the case of d₂ the unit disk, centred at the origin is just the set of all points at a distance at most 1 from the origin, i.e. the filled in circle. This matches our idea of equidistance: the points at a fixed distance from you will form a circle - a circle has a constant radius. But the unit disks for the metrics d_k are far from circular disks. In fact, as k gets bigger and bigger, the disks get more and more square shaped. In fact the limiting shape is exactly a square. So, according to the metric d_∞, the points a fixed distance from the origin actually form a square! The point is that there are many ways of talking about size and distance. Some are very natural, like d₂; but some are most counterintuative, like d_∞. You need to use the right metric for the right problem, and be very careful when you're trying to compare answers in one metric to answers in another metric, or else you might start saying that a square is a circle (and it is, topologically speaking... but that's another story). ~~ Dr Dec (Talk) ~~ 22:49, 24 August 2009 (UTC)

With angles it depends on what you are doing. 350 degrees and -10 degrees are often completely equivalent, but have very different magnitudes. Sometimes you want to consider the magnitudes, sometimes you want to consider them modulo 360 degrees, sometimes you want the actually real number itself. --Tango (talk) 20:07, 24 August 2009 (UTC)

To some extent, the original question appears to be a language-usage question rather than a mathematical one. Do we say −1000000 is a bigger or smaller negative number than −1? I'm afraid authors are probably inconsistent on this point. But if you see the phrase large negative number it most likely means "negative number of large absolute value" rather than "negative number far to the right (among negative numbers) in the number line". --Trovatore (talk) 22:54, 24 August 2009 (UTC)

August 25

Topology

Is there a simple English equivalent meaning for the word "topology" (even into a simple sentence)? Because I'd like to make a related meaning in Arabic beside the Arabic use of Latin name. If not it is still Ok. Thanks, --Email4mobile (talk) 08:31, 25 August 2009 (UTC)

To be precise, the word topology is from Greek, not Latin. The Latin expression would be analysis situs (yet with half Greek origin). --79.38.22.37 (talk) 08:37, 25 August 2009 (UTC)

Thanks for the correction, but can you give me an alternative meaning in English?--Email4mobile (talk) 09:02, 25 August 2009 (UTC)

See -ology. "topos" means place, and "logos" means "word". By analogy,

• biology is the study/theory/science of life (more literally, the word or talk about life)
• sociology is the study of society/social structure

so topology is the theory of "place" or "space". As -ology explains, X-ology can also a subject itself, not only its study. See network topology for example.

But please remember that Wikipedia is not a place where words should be invented.

--Aleph4 (talk) 09:37, 25 August 2009 (UTC)

Inner model theorists are topologists, because they study mice.

(It's sort of a pun.) --Trovatore (talk) 09:46, 25 August 2009 (UTC)

In Tuscany there is also a coarser pun, linked to the jargon word "topa" --pma (talk) 10:02, 25 August 2009 (UTC)

(ec) Do you mean, an other meaning of the word "topology"? As far as I know, the word "topology" is used in maths only, with these meanings: a) the branch of maths devoted to the study of topological spaces; b) the family of all open sets of a topological space: "the Euclidean topology of the real line is generated by all open intervals"; c1) sligthly more generally, the complex of all topological properties of a space: "a horned sphere has a weird topology"; c2) sometimes with a point of view of algebraic topology: "a disk has no topology"; "the projective space has a rich topology"; "a non-orientable manifold has no topology at the top level".--pma (talk) 09:52, 25 August 2009 (UTC)

Thank you all for explanation. Though, topology is also used outside Maths fields; for example: network topology. I'm not sure if the term "the study of space" or "the theory of space" would be the nearest meaning). How about "preserved space"?--Email4mobile (talk) 10:43, 25 August 2009 (UTC)

People probably assumed you were asking about the mathematical usage of the term, as this is the Mathematics Reference Desk. For other meanings see Topology (disambiguation). Don't see how you get to "preserved space", but for general translations (rather than mathematical meanings) you could also ask at the Language Reference Desk. Gandalf61 (talk) 11:07, 25 August 2009 (UTC)

Indeed network topology is still in the mathematical sense, even if it is more used in applied maths. "Study of space" is quite ok as informal explanation of "topology". Preserved space apparently has no clear sense. Still it is not clear if you want other meanings of the same word "topology", or other words, sinonyms or phrases with the same meaning. The original question seems to ask for "an equivalent meaning for the word topology" (the answer then would be: "topology" I guess). --pma (talk) 11:12, 25 August 2009 (UTC)

A network topology is more like a graph (in the graph-theory sense) than a topology. I suppose you could code it as a topology, but it takes a bit of work, especially to make sure you can distinguish nodes of degree less than 3. --Trovatore (talk) 19:39, 25 August 2009 (UTC)

For a lot of uses "connectivity" or "study of connectivity" has a very similar meaning. You could clarify(?) by calling it the "abstracted connectivity"...maybe...83.100.250.79 (talk) 11:40, 25 August 2009 (UTC)

I used this online translator and got طوبولوجيا. It seems okay, when I translated it back from Arabic to English it came up with topology. It even translated طوبولوجيا from Arabic into Spanish as topología which is correct. ~~ Dr Dec (Talk) ~~ 13:54, 25 August 2009 (UTC)

Indeed if you switch languages from English to Arabic (العربية) at the en.wikipedia page topology, you immediately get طوبولوجيا at the ar.wikipedia page. But I fear this is the information that the OP needs to know the less... --pma (talk) 15:06, 25 August 2009 (UTC)

Sorry, but what's an OP? ~~ Dr Dec (Talk) ~~ 15:14, 25 August 2009 (UTC)

OP=Original post/Original poster. It was clear that طوبولوجيا is a transliteration (and that of course the OP (=Email4mobile) knows it very well) --pma (talk) 15:34, 25 August 2009 (UTC)

The word "طوبولوجيا" already exists in the article but what you aren't aware of is that it is the literal translation of "Topology". "طوبولوجيا" is not an original Arabic word and not even containing a similar word. We use this kind of translation when we can't get any Arabic alternative. For this reason I said in the beginning it would be Ok if there is no alternative. —Preceding unsigned comment added by Email4mobile (talk • contribs) 15:13, 25 August 2009 (UTC)

I see. Well then I doubt very much that there will be an Arabic word for it. In English and all other Eurpoean languages (using the Roman alphabet) the word is a loan word from Greek: Topology in English, Topología in Spanish, Topologie in French, Topologia in Italian, etc. Even in other languages, like Russian, it's a loan word: Топология. They are all more or less just transcriptions of one another. ~~ Dr Dec (Talk) ~~ 15:20, 25 August 2009 (UTC)

You mean transliteration ;) --pma (talk) 15:38, 25 August 2009 (UTC)

I wrote transliterations in the first place, but then after reading the article on transliteration decided that transcription might be better. Oh well... ~~ Dr Dec (Talk) ~~ 15:42, 25 August 2009 (UTC)

"Topology" describes what elements of a set or system are connected to each other, for whatever meaning of "connected" you want to apply to that system. For example, you could describe a topology on the system of Wikipedia articles, in which two articles are connected if any individual editor has edited both of them. Why not look up the Arabic word for topology (I'm sure there is one) and use it? 67.122.211.205 (talk) 05:30, 26 August 2009 (UTC)

The real numbers and subsets of the natural numbers

The natural numbers are countably infinite and are the smallest infinite set, i.e. they have cardinality ℵ₀. The set of all subsets of the natural numbers has cardinality 2^ℵ₀: assign to each natural number a 0 or a 1 depending on whether the element is in or out of a given subset. Now, we can also show that the cardinality of the real numbers, i.e. the cardinality of the continuum is exactly 2^ℵ₀. The proof I have seen is a proof by contrandiction. Does anyone know of an explicit proof which finds a bijection between the real numbers (or an open subset thereof) and the set of all subsets of the natural numbers? ~~ Dr Dec (Talk) ~~ 11:21, 25 August 2009 (UTC)

I've seen a "proof" that puts a number into binary. Then for each string of 0s and 1s we get a subset of the natural numbers. For example 101 ~ {0,2} and 111101 ~ {0,1,2,3,5}. But this seems to be giving a bijection between the natural numbers and the set of all subsets of the natural numbers... that can't be right! Any ideas?! And besides, what about numbers numbers in the interval [0,1]? Won't we get some overlap? Is it even a bijection? I should stick to my own field and stop playing with fire. My head hurts! ~~ Dr Dec (Talk) ~~ 11:30, 25 August 2009 (UTC)

• When representing natural numbers in binary each natural number is a finite string. So you get a bijection between the natural number and the finite subsets of the natural numbers. Taemyr (talk) 11:37, 25 August 2009 (UTC)
• (ec)You only get finite sets that way. Finite subsets of natural numbers are countable. I suspect what you (miss-)remember is the proof that reals are uncountable (by trying to enumerate binary values from [0,1] and showing via diagonalization that that fails). --Stephan Schulz (talk) 11:45, 25 August 2009 (UTC)

Ah, I think you're right about that one! ~~ Dr Dec (Talk) ~~ 13:36, 25 August 2009 (UTC)

Oh, okay, cool. Do you have any idea of an explicit bijection from the real numbers to the set of all subsets of the natural numbers? ~~ Dr Dec (Talk) ~~ 11:42, 25 August 2009 (UTC)

Well, you can map each real in binary to a set of natural numbers (n is in the set if the nth bit of the binary representation of real number r is 1). Real numbers have infinite representations (the few with finite ones can be extended with zeros to infinity). --Stephan Schulz (talk) 11:59, 25 August 2009 (UTC)

Nice, straightforward idea but sadly not 1-1, as {1} maps to 0.1₂ and {2,3,4...} maps to 0.0111...₂ and these binary expansion represent the same real number, namely 1/2. Gandalf61 (talk) 12:25, 25 August 2009 (UTC)

True, but it is 1-1 if we only consider mapping infinite binary expansions to infinite subsets of natural numbers.

There probably isn't any pretty bijection between an interval of real numbers and the set of all subsets of natural numbers. That doesn't mean it can't be done, though. For instance, let the above bijection be Y=f(X), where X is a real number, and Y is an infinite subset of natural numbers. We define g(X)=f(X) if X cannot be represented as 1/n for some natural number n. For those numbers, letting n range from 1 to infinity, we define g(1/2n)=f(1/n), and g(1/(2n-1)) to be an enumeration of finite subsets of natural numbers. If we define g(0) to be the empty set, then g is a bijection between [0, 1] and the set of all subsets of natural numbers. --COVIZAPIBETEFOKY (talk) 12:50, 25 August 2009 (UTC)

My suggestion above is simple enough. To get around the 0.1*=1 problem, just define a unique representation for each rational number (by forbidding the ones that ends in 1*). Remember, we start with the numbers, not the strings. The problem I found by now, however, is that it does not map R, but only [0;1[ (or equivalent intervals). But it should be possible to get around this somehow by using a bijection from R to [0;1[. I think x->(1/(x+1)+0.5) (if positive or 0), x->(1/(-x+1)) (if negative) should do the job. --Stephan Schulz (talk) 14:14, 25 August 2009 (UTC)

So you represent 1/2 uniquely by 0.1, then map this to {1} - but now you have no real number that maps to {2,3,4,...}. So some subsets of natural numbers do not have a pre-image, and you still don't have a bijection. (Not saying this can't be patched up somehow to make it into a bijection - just saying it's not as simple as you originally said). Gandalf61 (talk) 15:55, 25 August 2009 (UTC)

Point taken. --Stephan Schulz (talk) 15:59, 25 August 2009 (UTC)

Another example: To each infinite subset of the natural numbers, say $0\leq a_{1}<a_{2}<\cdots$ assign the sequence $n_{1}=a_{1},n_{2}=a_{2}-a_{1},n_{3}=a_{3}-a_{2},\ldots$ of positive natural numbers; interpret this sequence as a continued fraction, and you will get a positive irrational number ${n_{1}+{\frac {1}{n_{2}+{\frac {1}{\ddots }}}}}$. Every positive irrational is obtained in this way. This is a (reasonably nice, I think) bijection between the (positive) irrationals and the infinite subsets of the natural numbers.

All that remains is to map the finite subsets of the natural numbers bijectively onto the rational positive numbers. As both these sets are countable, this is easy to do, but it can even be done "nicely", again using continued fractions.

--Aleph4 (talk) 22:08, 25 August 2009 (UTC)

On a different, but related topic. Is there anyway of giving a cananonical example for a set with cardinality ℵ_n for some natural number n? ~~ Dr Dec (Talk) ~~ 16:43, 25 August 2009 (UTC)

More or less by definition, ℵ_n is the cardinality of the set of ordinals of cardinality at most ℵ_n-1. Algebraist 16:46, 25 August 2009 (UTC)

Could you please explain that in a little more detail, if you don't mind? The ordinal article says that an ordinal are is the order type of a well-ordered set. So they are numbers. So the cardinality of an ordinal will be 1. So all ordinals have cardinality less than ℵ₀. ~~ Dr Dec (Talk) ~~ 17:00, 25 August 2009 (UTC)

(if you didn't mean to make a joke) "the cardinality of the ordinal x", refers to x as an ordered set, e.g. the set of all smaller ordinals, according to the von Neumann representation. --79.38.22.37 (talk) 17:18, 25 August 2009 (UTC)

A standard definition for ordinal numbers is very similar to the one for natural numbers, which defines each natural number as the set of all natural numbers less than it (where 0 is the empty set). That is, an ordinal number is often defined as the set of all ordinal numbers less than it. The finite ordinal numbers are natural numbers. The first infinite ordinal number is the set of all natural numbers. The second infinite ordinal number would be the set containing all natural numbers and the set of all natural numbers. And so on. --COVIZAPIBETEFOKY (talk) 17:12, 25 August 2009 (UTC)

Hold on, forgot to make my point (I guess you got it, but I'll put it explicitly anyways). So, the first infinite ordinal number has cardinality equal to ℵ₀, and so do many many many more after that. However, there will eventually be an ordinal number with cardinality ℵ₁, and ℵ₂, and so on. --COVIZAPIBETEFOKY (talk) 17:15, 25 August 2009 (UTC)

I see! ~~ Dr Dec (Talk) ~~ 17:15, 25 August 2009 (UTC)

Two more examples of sets of size aleph1:

• Example 1: Let P the set of all subsets of the rational numbers. (So P has size continuum.)

On the set P define an equivalence relation E by declaring two sets A and B (both subsets of the rationals) to be in relation E if they are order isomorphic (i.e., there is an order preserving bijection from A onto B), or neither of the two is a well-order. (For example, the sets {0,1,2} and {-1, 1/2, 7} are in relation E.)

The set of E-equivalence classes has cardinality aleph1. This is easy to see as each equivalence class (except for the class of non-well-ordered sets) naturally and bijectively corresponds to a countable ordinal. (So this example is just a thinly disguised version of the first uncountable ordinal, mentioned above.)

• Example 2: Let P be the same set as above. Now define a relation L as follows:

(A,B) is in L iff there is an order preserving embedding from A into B, and there is an order-preserving embedding of B into A.

Laver's theorem says that there are exactly aleph1 equivalence classes. ("Fraisse's conjecture"; highly non-trivial.)

--Aleph4 (talk) 21:19, 25 August 2009 (UTC)

Is this correct?

Resolved

 – Accdude92 already has the correct answer

Find F(x)+g(x) F(x)=X/X+2 G(X)=2X=3

I got 2x(squared)+8X+6\X+2 Accdude92 (talk) (sign) 14:12, 25 August 2009 (UTC)

I'm sorry, but I don't understand the problem. Do you mean find F(x) + G(x) when

$F(x)={\frac {x}{x+2}}\ ,$

$G(x)=2x-3?\,$

In that case it's just addition of fractions and some simplification. I get

$F(x)+G(x)={\frac {2(x^{2}+x-3)}{x+2}}\ .$

If x and X are different and g and G are different then I have no idea. ~~ Dr Dec (Talk) ~~ 14:18, 25 August 2009 (UTC)

Oh, I see, you have G(x) = 2x + 3, and in this case you're right:

${\frac {x}{x+2}}+(2x+3)={\frac {2(x^{2}+4x-3)}{x+2}}\ .$ ~~ Dr Dec (Talk) ~~ 14:20, 25 August 2009 (UTC)

Darn keyboard. Its really Find F(x)+g(x) F(x)=X/X+2 G(X)=2X+3 —Preceding unsigned comment added by Accdude92 (talk • contribs) 14:21, 25 August 2009 (UTC)

fibonacci spiral

for what value of $\omega$ does $|z|=re^{\omega arg(z)}$ plot a fibonacci spiral in the argand plane? —Preceding unsigned comment added by 92.1.60.82 (talk) 23:55, 25 August 2009 (UTC)

See Golden spiral. Black Carrot (talk) 06:09, 26 August 2009 (UTC)

August 26

Computer-aided proofs since Appel-Haken

This is less about mathematics than about mathematics history. I'm interested in knowing what notable questions in mathematics there have been whose solution have relied upon an effectively not-by-hand-checkable computer program since Appel and Haken proved the Four-Color Theorem using a computer program in 1976. —Preceding unsigned comment added by Julzes (talk • contribs) 08:22, 26 August 2009 (UTC)

What do you accept as "proved"? The Kepler conjecture has most likely been proved by computer, but nobody is completely sure if that proof is sound. There are several proofs now that have been found by automated reasoning systems - the most famous one is probably the Robbins conjecture proved by EQP. However, that proof was hard to find, but easily checkable. --Stephan Schulz (talk) 09:03, 26 August 2009 (UTC)

I'm mostly interested in the question of exhaustive searches and other sorts of things where the acceptance of effectively uncheckable computer calculations is involved. The Kepler Conjecture is a good example, but the fact that it's controversial sort of goes against what I would have proposed saying about the acceptance of things similar to the proof of the Four-Color Theorem having become somewhat routine.Julzes (talk) 10:08, 26 August 2009 (UTC)

Well, sorry if reality does not support your proposed opinions ;-). If you are interested in searches: The claim that the game-theoretic value of Checkers is 0 comes from exhaustive calculations by the Chinook team, and seems to be widely accepted. --Stephan Schulz (talk) 10:33, 26 August 2009 (UTC)

A lot of work has been done on formalizing proofs that have already been found by humans (the Flyspeck project is an example, as is Gonthier's Coq implementation of the 4-color theorem). There is also lots of exploratory math done with computers--see some of Doron Zeilberger's articles on experimental math for info. But I think not all that much has been done using computers to find proofs of traditional math conjectures. WZ theory might be an example though--a scheme for automatically solving certain hypergeometric sums. Hmm, another example, the Bailey-Borwein-Plouffe formula for computing digits of pi was found with a big computer search involving the PSLQ algorithm (there are related such formulas too, found in similar ways). This page is a reasonable place to look for info about formalized math in general, and there was a special issue of AMS Notices about formalized math, that has several good articles that you might like.[2] —Preceding unsigned comment added by 67.122.211.205 (talk) 23:39, 26 August 2009 (UTC)

I should add that the point of Gonthier's proof of 4CT is that (unlike Appel-Haken's) it can be checked by a program that itself is supposed to be hand checkable (Coq's proof checking core is based on a small, hand checkable kernel, per the de Bruijn criterion). 67.122.211.205 (talk) 00:01, 27 August 2009 (UTC)

Further: OK, I think I understand your question better now. The issue with the Appel-Haken proof is it was done with 1970's computer and software technology, and was a huge complicated set of C assembly language programs that were potentially full of bugs and was too large to be inspected by hand (being small enough for hand verification is called the de Bruijn criterion), so nobody could be sure it was really right. These days, programming itself (at least at this pointy-headed level) is getting more formalized, so in a language like C you can write a program with the intention that it computes X, and make informal arguments that it actually does compute X, but maybe your arguments have a gap and your program has bugs. But in a language like Coq (you can view Coq as a programming language), the programs are stated rigorously enough that you can prove theorems about their properties, and mechanically verify the proofs, using a hand checkable verifier. So you can write a Coq program that computes X, and state what X is, and prove as a theorem (formally verified) that your program really computes X. That is essentially Gonthier's proof of 4CT, I think. (There is an article about it in the AMS issue I linked to). In principle you could repeat Schaffer's calculation of the outcome of checkers, by writing a Coq program that computes the outcome of checkers, proving its correctness, and running it for however long it took (hmm, I guess that involves trusting the compiler, but we are getting closer to having completely verified compilers), and then saying the output is verified. In that way we're getting away from the acceptance gap that the Appel-Haken proof faced. Anything we can compute unverifiably, we can do verifiably with enough extra work. 67.122.211.205 (talk) 00:12, 27 August 2009 (UTC)

Thanks for the thoughtful answers. Aside from helping me help someone else write a short piece on the 4CT, it's a subject of some personal interest, so I will be having a look at what you have offered up.Julzes (talk) 03:56, 27 August 2009 (UTC)

Definition of Kähler manifolds

From the Wikipedia article about Kähler manifolds, a Kähler manifold is a complex manifold with a Hermitian metric $h=g+i\omega$ where $g$ is a Riemannian metric, $i$ an almost complex structure and $\omega$ a symplectic form, which must satisfy an additional condition.

The article gives two of such equivalent additional conditions:

• $d\omega =0$, the symplectic form is closed.
• Parallel transport induced by the Levi-Civita connection corresponding to $g$ actually induces $\mathbb {C}$-linear mappings of tangent spaces, not only $\mathbb {R}$-linear.

I've been thinking about this for a while now but I can't seem to figure out the equivalence. For example, starting with the second condition, I have no idea how to involve $\omega$.

I'd be very grateful if anyone had a few tips or a good reference. Thanks. -Xedi^Talk 13:52, 26 August 2009 (UTC)

can u solve this

1=19-20=60

one straight line --America vega (talk) 21:23, 26 August 2009 (UTC)

What is the question? You have given a false arithmetic statement and a geometric object, you haven't actually given us anything to solve. --Tango (talk) 21:29, 26 August 2009 (UTC)

I know my brother gave it to me to solve but I can't I think it might be a riddle--America vega (talk) 21:39, 26 August 2009 (UTC)

is it (1,20) (20,60) which is y=40x/19+17+17/19 ? 83.100.250.79 (talk) 21:45, 26 August 2009 (UTC)

Commutivity of addition

Has anyone any idea how one might prove the commutivity of addition?

• Let's start with a + b = b + a where a and b are real numbers. ~~ Dr Dec (Talk) ~~ 23:14, 26 August 2009 (UTC)

This will depend on your preferred Construction of the real numbers. Algebraist 23:15, 26 August 2009 (UTC)

Good point! Okay, let's list the possible constructions and their respective proofs... ~~ Dr Dec (Talk) ~~ 23:21, 26 August 2009 (UTC)

The reals are a Field (mathematics). The additive operation of a field is commutative by definition. QED. ;-) --Stephan Schulz (talk) 23:24, 26 August 2009 (UTC)

This is a joke, right?! PROVE that addition is commutative! ~~ Dr Dec (Talk) ~~ 23:27, 26 August 2009 (UTC)

Only slightly. The most common definition of the reals is as the essentially unique ordered field satisfying a completeness property (exactly how the completeness property is phrased varies); with this definition, commutativity of + is automatic. Algebraist 23:30, 26 August 2009 (UTC):

There are a great many constructions of the reals, and I doubt we could find them all, let alone list them. For many constructions, the commutativity of addition is a trivial consequence of the commutativity of rational or integer addition; this is the case for every construction listed in our article, for example. Algebraist 23:30, 26 August 2009 (UTC)

Okay, my first example was that of the reals. What about the addition of connexions? What about group rings? (Can you PROVE the commutivity instead of ASSUME it? Can you PROVE it from the properties that are not derived from the assumption itself?) ~~ Dr Dec (Talk) ~~ 23:40, 26 August 2009 (UTC)

I think "prove addition is commutative" makes the most sense when you're discussing addition of the natural numbers. More complicated structures like the reals are constructed and artificial and commutativity of addition is usually taken as an axiom (Archimedean complete ordered field) rather than something to be proved (unless you're discussing a specific construction of reals, such as Dedekind cuts). The basic properties of the natural numbers are usually given as the Peano axioms. They say that every natural number N has a successor S(N), and they define addition by saying A+0=A and A+S(B)=S(A+B), plus a few other things. From these axioms, the commtativity of addition should be provable as a theorem. 67.122.211.205 (talk) 23:49, 26 August 2009 (UTC)

67.122.211.205, you're half way there. Now prove it... ~~ Dr Dec (Talk) ~~ 23:55, 26 August 2009 (UTC)

It needs an inductive proof. Addition is "only" an inductive theorem of the natural numbers. Via induction, the proof is fairly simple. --Stephan Schulz (talk) 23:59, 26 August 2009 (UTC)

Even with the naturals, there are alternative approaches. For example, if we instead define the naturals to be the finite cardinals, then commutativity of addition is a triviality. If we define the naturals to be the nonnegative (or positive, to taste) part of the essentially unique well-ordered integral domain, then commutativity of addition is again true by definition. Algebraist 00:05, 27 August 2009 (UTC)

But that's my hang-up: you're ASSUMING it as a definition. Something doesn't sit very well with me about this axiomatic aproach to mathematics. Lots and lots of assuming makes many many things matters of definition. For example, as the axiom article says: "...non-logical axioms (e.g., a + b = b + a) are actually defining properties for the domain of a specific mathematical theory (such as arithmetic)." So the natural numbers are defined, arithmetic is defined. They are not provable; simply defined and accepted. ~~ Dr Dec (Talk) ~~ 00:22, 27 August 2009 (UTC)

Yes, so? This is the usual way of defining mathematical systems since Hilbert; you define a system by its key properties, and call those properties axioms for the system. It is of course possible to go the other way, and define your system to be some specific object in some other system (usually a set theory of some sort) and prove that the object you've chosen has the key properties you want, but this is rather artifical. After all, if the object you'd decided was the real numbers turned out not to have commutative addition, then that wouldn't say anything interesting about the real numbers. It would just mean you'd picked the wrong object to be them. Algebraist 00:33, 27 August 2009 (UTC)

It may be artificial, yes, but on the flip side, actually constructing the object of interest (set of real numbers, set of natural numbers) in terms of other, well-accepted theory, provides one clear way of proving the existence of that object. Declan does have a point that simply making a list of properties of real numbers is not very satisfying until you can prove that a set of numbers, together with appropriately defined addition and multiplication, actually exists.

Of course, this has been done time and time again with the real numbers, most notably with Dedekind cuts and Cauchy sequences. --COVIZAPIBETEFOKY (talk) 00:45, 27 August 2009 (UTC)

Another point that should perhaps be made is that those constructions do depend on some framework of set theory, so if you don't accept set theory, then these hardly suffice as constructions. The fact is, however, that you have to start somewhere, and in the predominant view, ZFC is accepted as the basis for nearly all of mathematics. --COVIZAPIBETEFOKY (talk) 00:50, 27 August 2009 (UTC)

I didn't mean to disparage embedding things in set theory in general, but rather the relevance of the practice to the question at hand. Implementing real numbers as Cauchy sequences is a worthwhile pursuit for a number of reasons, but it doesn't do much to elucidate the commutativity of real addition, and shouldn't really be considered a proof of said commutativity. Algebraist 00:55, 27 August 2009 (UTC)

I think you're missing Declan's point here. Sure, you can take the synthetic approach to defining real numbers, and then - duh! - addition is commutative. However, the synthetic approach makes the assumption that the set of real numbers satisfying all of those properties exists in the first place. This has to be proven first, which can easily be done by way of Dedekind cuts or Cauchy sequences.

The biggest hurdle in proving commutativity of addition in the set-theoretic constructions of natural numbers, integers, rationals and reals is proving that it is so for the natural numbers, and Carl has outlined that part below. --COVIZAPIBETEFOKY (talk) 01:10, 27 August 2009 (UTC)

From a certain perspective, "constructions" of number systems in set theory do not increase our confidence in the consistency of those number systems. Set theory is important as a foundational system because many systems of interest can be formalized within it. For example, we may argue in favor of set theory because we can formalize the real numbers within it. But this argument presupposes that we already know what the real numbers are, so that we can test whether they can be formalized within set theory. — Carl (CBM · talk) 01:18, 27 August 2009 (UTC)

(edit conflict) I take the point, but I don't think there's any avoiding it. The same applies to constructive approaches: you make up a definition, carefully making sure your definition ensures all the properties you want, and – duh! – addition is commutative. A better question might be "What is it about what we use the real numbers for that makes us want to force addition to commute?". One possible answer would be that a fundamental use of real numbers is to measure physical magnitudes, such as lengths of line segments, and there's a natural geometric way of adding lengths of line segments which is commutative. I have not thought much about this and have no idea if this is a very good answer. Algebraist 01:20, 27 August 2009 (UTC)

Declan, could you clarify exactly what your hang-up is here? If it's that you're assuming that a set of "real numbers" exists that satisfies all of its supposed properties, this can be proven - under the assumption that we accept the axioms of set theory. The fact that we have to assume something is an inescapable fact of mathematics, due to Godel's theorem.

If it's closer to what Algebraist is saying, in that you want justification for the assumption of commutativity of addition, well, you have Algebraist's answer here. --COVIZAPIBETEFOKY (talk) 01:36, 27 August 2009 (UTC)

Expanding on what Algebraist said: In the axiomatic approach, we take an object and choose a system of axioms that describe it to a greater or lesser extent. If we can prove some property of the object from those axioms, all that this shows is that the property in question is a logical consequence of those axioms, and thus no additional information about the object is needed, apart from the axioms, for us to know it has the property in question. In a certain sense, however, such a proof does not really tell us anything "new" about the system, because we already knew (or assumed) that the object satisfies the axioms, and thus we had already implicitly assumed that the object has every property that is a consequence of the axioms. In this sense, we only learn "new" things about an object when we learn that some property possessed by the object is not a consequence of the axioms we had been studying. — Carl (CBM · talk) 01:12, 27 August 2009 (UTC)

To prove that addition is commutative in Peano arithmetic, one proceeds in the only way possible, which is by induction. One first proves that addition is associative and that, for all a, 1 + a = a + 1. Then one proves by induction on b that b + a = a + b. The key string of equations is:

$a+(c+1)=a+(1+c)=(a+1)+c=(1+a)+c=1+(a+c)=1+(c+a)=(1+c)+a$

— Carl (CBM · talk) 00:51, 27 August 2009 (UTC)

To get at what I think the point of Declan's question is, we arrive at mathematical axioms through abductive reasoning and I don't know of any escape from that. There are various systems of logic like ludics that try to be somehow more primitive than Hilbert-style deduction but AFAIK that doesn't help at the very bottom. 02:28, 27 August 2009 (UTC) —Preceding unsigned comment added by 67.122.211.205 (talk)

Associativity of composition

Has anyone any idea how one might prove the associativity of composition?

• Let's start with real valued functions f, g and h. How do we prove that $(f\circ g)\circ h=f\circ (g\circ h)?$ ~~ Dr Dec (Talk) ~~ 23:14, 26 August 2009 (UTC)

This is true essentially by definition: $((f\circ g)\circ h)(x)=(f\circ g)(h(x))=f(g(h(x)))=f((g\circ h)(x))=(f\circ (g\circ h))(x)$. Algebraist 23:18, 26 August 2009 (UTC)

Yeah, I agree. (I was going to leave a note saying that this one was easier, but I was busy reading your post on the addition problem). Good work Algebraist.

...of course, it misses an "for all x" quantifier, and you need to assume Extensionality . --Stephan Schulz (talk) 23:28, 26 August 2009 (UTC)

It's not extensionality that is used but the definition of equality of functions. The proof does not depend on functions being implemented as sets in any particular way, or indeed at all. Algebraist 00:11, 27 August 2009 (UTC)

You do need to assume extensionality, but not in the set theoretic sense. What is ordinarily called "equality of functions" is more precisely extensional equality of functions. In systems with intensional equality, function composition generally not associative in the sense described above. In systems with extensional equality, the proof is immediate, as shown above. — Carl (CBM · talk) 00:43, 27 August 2009 (UTC)

Good point. Algebraist 00:47, 27 August 2009 (UTC)

August 27

A weird thing

Consider the set of all continuous functions f:[0,1]->R in the sup metric, i.e.

d(f,g)=sup{|f(x)-g(x)|}

Show that this space has a countable dense subset and therefore has a countable basis.

The proof is very easy the book gives a hint (which I won't say for those of you who want to do it yourselves), but immediately I noticed something WEIRD.

Since one point sets in hausdorff spaces are closed, there are at least as many open sets as points in the space (x->({x} complement)). Now the above space has a countable base so there are as many continuous function as real numbers. Is this true??? It's totally outrageous!!! Standard Oil (talk) 05:45, 27 August 2009 (UTC)

Why is it outrageous? It seems perfectly reasonable. I feel like there should be a simple elementary proof but I'm spacing on it right now. 67.122.211.205 (talk) 06:18, 27 August 2009 (UTC)