October 19

If a solid dot on a graph '•' means a point is there, an open dot 'o' means it's not included, what's an 'x'? --24.76.248.193 04:06, 19 October 2007 (UTC)[reply]

These marks do not have fixed meanings. It means whatever the person making the graph wanted it to mean.  --Lambiam 09:48, 19 October 2007 (UTC)[reply]

The dot and the open dot are pretty standard meanings for coordinate grids and number lines, but the x is not as standard and what Lambian said. [Mac Δαvιs] ❖ 10:31, 19 October 2007 (UTC)[reply]

Might represent a point from a separate function or data set.. squares and triangles are also used in this respect.87.102.7.57 13:08, 19 October 2007 (UTC)[reply]

There are also some programs that may extrapolate extra points in order to fit a curve, in which case a dot, o, or x may not represent an actual data point, rather something in between two actual points.--VectorPotential^Talk 13:10, 19 October 2007 (UTC)[reply]

I'm reading a paper and the following is asserted:

The decision to stop selecting new subsets of P can be based upon the expected number of trials ${\displaystyle k}$ required to select a subset of ${\displaystyle n}$ good data points. Let ${\displaystyle w}$ be the probability that any selected data point is [a good data point]. Then we have:

${\displaystyle E(k)=b+2*(1-b)*b+3*(1-b)^{2}*b\ldots +i*(1-b)^{i-1}*b+\ldots }$

where ${\displaystyle E(k)}$ is the expected value of ${\displaystyle k}$ and ${\displaystyle b=w^{n}}$.

I don't see how this follows so directly. I must be missing some general principle. Could somebody point me in the right direction? Sancho 04:31, 19 October 2007 (UTC)[reply]

Seems to be saying you have a probability b of getting a good set of data points on the first trial; a probability (1-b)b of getting the first good set of data points on the second trial (one bad trial followed by one good trial); a probability (1-b)²b of getting the first good set of data points on the third trial (two bad trials followed by one good trial) etc. So the probability that k takes a given value n is (1-b)^n-1b.

As you would expect, this infinite sum simplifies to E(k)=1/b. Gandalf61 09:43, 19 October 2007 (UTC)[reply]

That makes sense! Thanks. Sancho 15:30, 19 October 2007 (UTC)[reply]

Please help me understand degree of freedom by answering these questions Max obtains a set of six numbers by rolling a die six times: 2, 3, 3, 4, 6, 6. This set of numbers (let’s call them ‘Set A’) has how many degrees of freedom? b) Max converts these six numbers to deviations from the mean. The resulting set of six deviation scores (let’s call them ‘Set B’) has many degrees of freedom? c) Max obtains another set of six numbers by rolling a die six times. He tells you that this set of numbers (let’s call them ‘Set C’) have the same sum as the number is Set A. Explain how many degrees of freedom there are in Set C? —Preceding unsigned comment added by Madisalman (talk • contribs) 05:25, 19 October 2007 (UTC)[reply]

Back in high school, we argued some time about whether Pythagorean triples are a two- or three parameter set. On one hand, as two numbers in a triple determine the other, they're quite obviously not a three-parameter set. On the other hand, as (k(m^2 - n^2), 2kmn, k(m^2 + n^2)) gives all the triples and the three parameters are uniquely determined by the triple, which shows that it's a three-parameter triple. (Similarly, even the set of natural numbers can be said to be a two-parameter set, because they can be written uniquely as ((n+k)^2+n+k)/2+k, where n and k are natural numbers.) – b_jonas 09:22, 19 October 2007 (UTC)[reply]

"Degrees of freedom" has a specific meaning in statistics. To understand this concept, a good place to start is to read our article on Degrees of freedom (statistics). To answer part (b) you will need to know that the sum of deviations from the mean is always 0 (assuming they are signed deviations). Gandalf61 09:30, 19 October 2007 (UTC)[reply]

I must prove that ${\displaystyle (P\to Q)\land (Q\to R)\to (P\to R)\equiv T.}$

Here's where I got, starting from the initial proposition:

${\displaystyle (\neg P\lor Q){\color {Red}\land }(Q\land \neg R)\lor (\neg P\lor R)}$

${\displaystyle (\neg P\lor Q)\land (Q\land \neg R)\lor (R\lor \neg P)}$ Commutative Law

${\displaystyle (\neg P\lor Q)\land (Q\lor R)\land (\neg R\lor R)\lor \neg P}$ Distributive Law

${\displaystyle (\neg P\lor Q)\land (Q\lor R)\land T\lor \neg P}$ Domination for (!R or R)

${\displaystyle (\neg P\lor Q)\land (Q\lor R)\land T}$ Domination for (T or !P)

Trouble is, I can't see how to take it from here. I would be interested to see how "Quine's Method" is applied to this problem but I don't have any literature pertaining to it. The article seems to come from a binary logic viewpoint and I would like the method broken down into a traditional logical approach. 91.84.143.82 14:13, 19 October 2007 (UTC)[reply]

Well, for starters, we need some brackets in the proposition to be proved. You have to show that ${\displaystyle ((P\to Q)\land (Q\to R))\to (P\to R)}$ is a tautology. Then we translate this in the rather horrifying way you seem to have been told to use, replacing ${\displaystyle P\to Q}$ with ${\displaystyle (\neg P\lor Q)}$. Applying De Morgan's laws and eliminating double negations, we end up having to prove ${\displaystyle (P\land \neg Q)\lor (Q\land \neg R)\lor (\neg P\lor R)}$. Algebraist 14:48, 19 October 2007 (UTC)[reply]

Ah, I see where I went wrong - I only negated ${\displaystyle (Q\to R)}$ when I should have negated ${\displaystyle ((P\to Q)\land (Q\to R))}$. 91.84.143.82 14:55, 19 October 2007 (UTC)[reply]

What you do from now on will depend on exactly what rules of inference you're allowed, which I obviously don't know. The way I (as a mathematician) learnt logic, you would either semantically show that the given proposition is a tautology by truth tables, or syntactically prove it from (an extremely minimal set of) axioms, depending on exactly what you wanted to do. Of course, in practice you would do the syntactic version by semantically showing it to be a tautology and appealing to the completeness theorem for propositional logic. Algebraist 14:53, 19 October 2007 (UTC)[reply]

(edit conflict)

First, check the red operator. It seems to me it should be \or, not \and. --CiaPan 14:54, 19 October 2007 (UTC)[reply]

Is there an (easy) way to solve d² f(t)/dt² = k sin ( f(t) ) when d f(t=0)/dt = 0 and f(0) = a

ie Is there a page or a link to get a solution (excluding the simplified solution for when oscillations are small - when |f(t)| is small..?87.102.7.57 15:01, 19 October 2007 (UTC)[reply]

As mentioned at pendulum (mathematics), the solution to this equation (for a!=0) is an elliptic integral which cannot be expressed in terms of elementary functions. So you might be out of luck, depending on what you mean by 'solve'. Algebraist 17:36, 19 October 2007 (UTC)[reply]

It seemed that a series eg f(t) = k Sum (n=0 to infinity) a_ncos(2pint/λ) would be an expression for it since the function is period in time, putting that (as a power series) into sin(f(t)) gave another power series .. eventually giving infintite linear equations with infinite unknowns.. and no obvious way forwards ... so solving would mean finding the values a_n analytically..87.102.7.57 17:56, 19 October 2007 (UTC)[reply]

I'd like to know the name of a certain "math magic trick" so I can read more about it. It is typically taught to grade school kids, I think. You have a set of 10 or so cards each with a different matrix on it. You tell a friend to pick a secret number and then find all the cards that contain that number in the matrix. Then you add up the numbers in the upper left hand corner of the cards, and you "magically" determine your friend's secret number.

I have two reasons for asking: 1) I'd like to know how it works, 2) I know an old guy who claims to have invented this and I think he's full of it. ike9898 18:43, 19 October 2007 (UTC)[reply]

Well, he may have invented it, in fact. He probably just wasn't the first person to invent it. But I suppose I can't be sure even of that.

The way this trick works is, each card has those numbers whose binary expansion has a 1 in a particular location. The smallest such number, the one in the corner, has 0 in every other location. The cards chosen determine the binary representation of the number; then when you add them up, looking at it in binary, it's easy to see that you recover the original number. --Trovatore 19:49, 19 October 2007 (UTC)[reply]

Was it anything like this?

   +-----------+-----------+-----------+-----------+-----------+
   | 7 27 21  1|11 13 15  8| 7 31 10  2|22 20 27 16|22 20  7  4|
   |13 31 23 17|10 27 28 31|11  6 30 19|19 28 24 21|28 21 31 12|
   | 9 29  5 19|30 14 24 25|18 15 27  3|31 29 30 17|29 13 23 14|
   |11 25 15  3|12 26  9 29|22 23 26 14|25 26 23 18| 5 15 30  6|
   +-----------+-----------+-----------+-----------+-----------+

Think of any natural number up to 31. – b_jonas 19:53, 19 October 2007 (UTC)[reply]

I don't know a common name but I found descriptions [1] and [2] with a broad Google search. I have a version where the cards have holes in different positions. A card is turned upside down if the number is not there. At the end, look through all the cards at a solution card where only the chosen number is visible. No computation is involved and people can play it on their own without knowing how it works. PrimeHunter 20:13, 19 October 2007 (UTC)[reply]

Ok, let's make it more misterious, the powers of two thing might be too obvious.

Think of any natural number up to 99. Search all of its occurrences in the non-parenthisized numbers. Add the parenthisized numbers from those cards where you've found that number.

   +--------------+--------------+--------------+--------------+--------------+
   |           (4)|           (2)|          (16)|          (28)|          (40)|
   |84 12 88 69 24|28 89 63 45 60|64 97 88 65 83|63 39 68 64 80|82 83 96 85 72|
   |43 60 93 59 47|99 19 98 2  92|98 44 16 86 91|74 38 65 60 91|70 73 47 86 75|
   |83 67 34 37 7 |64 58 55 77 86|66 96 62 58 51|89 42 29 87 79|40 90 52 76 71|
   |39 46 26 94 41|37 82 76 90 52|59 19 89 82 78|54 85 49 78 88|69 61 58 97 77|
   |76 97 73 30 57|87 12 8  6  66|36 43 69 41 50|73 75 98 92 34|80 67 56 92   |
   |80 62 18 50 13|17 81 3  31 95|74 57         |84 55         |              |
   |95 85 25 61 27|80 56 47 30 84|              |              |              |
   |78 86 21 4  44|35 57 34 36 53|              |              |              |
   |75 5  33      |46 72 75 85 24|              |              |              |
   +--------------+--------------+--------------+--------------+--------------+
   |          (11)|          (22)|          (61)|           (3)|          (15)|
   |18 45 91 88 61|60 46 88 53 31|99 81 94 95 93|92 76 84 17 82|90 79 44 23 57|
   |92 89 67 95 76|50 59 68 71 35|              |63 10 62 67 9 |51 98 94 77 82|
   |46 17 72 28 59|93 66 48 84 37|              |89 71 6  95 87|32 74 36 86 25|
   |57 66 51 87 55|98 63 62 43 91|              |78 45 18 36 7 |30 70 41 87 26|
   |84 77 74 65 20|96 33 79 54 87|              |64 53 96 90 11|48 89 97 81 64|
   |53 78 24 98 70|83 27 22 97 90|              |30 98 74 52 54|45 99 76 78 49|
   |99 62         |              |              |99 68 21 65 48|72            |
   |              |              |              |38 72 70 32 85|              |
   |              |              |              |57 60 51 13 81|              |
   |              |              |              |35            |              |
   +--------------+--------------+--------------+--------------+--------------+
   |           (8)|           (6)|           (1)|
   |79 44 53 14 89|53 31 52 14 24|6  47 33 88 70|
   |28 32 55 69 45|88 93 61 99 15|68 83 72 67 29|
   |95 94 35 85 15|30 71 82 66 62|52 27 37 15 5 |
   |86 11 66 90 42|33 21 45 12 57|3  60 1  28 39|
   |50 20 96 37 92|9  84 79 32 28|26 31 69 65 54|
   |84 91 23 77 48|51 89 55 87 56|77 46 19 44 99|
   |68 56 67 63 21|95 94 42 25 80|78 98 10 86 74|
   |              |38 96 13 46 68|38 96 17 24 66|
   |              |41 91 10 8  59|75 53 73 20 43|
   |              |39 26 65 49   |76            |
   +--------------+--------------+--------------+

E.g. if you think of 28, you find it on the cards with (2) (11) (8) (6) (1). You don't count the card with (28), because it only has 28 in parenthesis, not among the other numbers. And, miraculously, 2+11+8+6+1 = 28. – b_jonas 20:21, 19 October 2007 (UTC)[reply]

The trick is obvious once you construct it in the right order.

First, invent the parenthisized numbers. Here, I've chosen 1 2 3 4 6 8 11 15 16 22 28 40 61, but they can be just about anything, except there should be enough small numbers (though you could avoid that too if you really wanted to).

Then, for each number between 1 and 100, find a way to write it as the sum of some of those parenthisized numbers. Make sure to use each only once in the sum. Then, write each number on the cards whose parenthisized number occurs in that sum.

Finally, shuffle and tabulate the numbers on each card, and shuffle the cards. – b_jonas 20:28, 19 October 2007 (UTC)[reply]

October 20

i need halp to do this can some one halp me place.thank you 1) -5p-7=-27 2)4n-12=28 3)2s+6=0

 —Preceding unsigned comment added by 4.236.132.174 (talk) 00:40, 20 October 2007 (UTC)[reply] 

Isolate the variables. By performing the same actions on both sides of the equation, try to get the numbers on just one side and the variable on the other. Strad 01:27, 20 October 2007 (UTC)[reply]

Furthermore, do you remember the order of operations??? Keep in mind that when solving an equation you use the reverse order, Addition/Subtraction first, Multiplication/Division 2nd, and so on. parenthesis are worked on outside to inside vs the usual inside-out method. This is because you are altering both expressions that are on either side of the equals sign. Think about it this way, when trying to solve for a variable you must undo what the expression (on the left in this case) is doing to that variable. So what happens last?? Addition/Subtraction, then next to last... etc. So add 7 to both sides, in problem 1) to eliminate the -7 from the left side, remember to add 7 to both sides or you lose your equality and thus you would not get a correct solution. We add, not subtract because we must use an opposite to eliminate the -7 term. Then you have -5p=-20, next divide both side by -5. So now you have p=4. And it is generally a good idea to check your answer by substituting it in place of p (use parenthesis so as, to insure proper order of operations) and show the the left and right sides are equal. That is not necessary but if you're not compete in your answer being correct that can verify whether or not it is. A math-wiki 05:45, 20 October 2007 (UTC)[reply]

i need halp plase —Preceding unsigned comment added by 4.236.132.174 (talk) 00:50, 20 October 2007 (UTC)[reply]

Somewhere in the region of ${\displaystyle {\sqrt {france}}\,}$, I would expect. 86.141.145.45 01:45, 20 October 2007 (UTC)[reply]

I find it hard to believe someone can properly spell "profession" and "veterinary" but misspell "help" and "please". --KSmrq^T 06:28, 20 October 2007 (UTC)[reply]

They make ${\displaystyle {\sqrt {-i}}}$ dollars a week plus tips. 210.49.155.124 07:29, 20 October 2007 (UTC)[reply]

ok - this person is asking on the science desk similar questions - so why here? Anyway just out of respect please don't take the piss. Please.87.102.17.46 10:07, 20 October 2007 (UTC)[reply]

The maths desk is not really for financial advice - such as estimated wages - the answer to your question depends very much on which country you are thinking about - which you need to tell us.87.102.17.46 10:09, 20 October 2007 (UTC)[reply]

Typically a vetinarian earns above average wages or is well paid.87.102.17.46 10:48, 20 October 2007 (UTC)[reply]

I read the article on the Vysochanskiï-Petunin inequality, and am trying to understand it, and its implications on the interpretation of control charts using 3-sigma limits, i.e. lambda=3. The article states: "The sole restriction the distribution is that it be unimodal and have finite variance. (This implies that it is a continuous probability distribution except at the mode, which may have a non-zero probability.)" I am having problems with the above, because up until now I have always thought that a probability distribution is either continuous or discrete, but the statement in parenthesis seems to imply that it can be both (continuous for some arguments, discrete for others). To me, this raises several questions:

• Is there such a thing as a unimodal discrete probability distribution? The statement cited above seems to indicate that the answer is "no", but oviously many discrete probability distributions have only one mode.
• If the answer to the preceding question is "yes", does the Vysochanskiï-Petunin inequality apply to a unimodal discrete probability distribution?
• What is the probability density of a "continuous" probability distribution at the mode, if the probability is non-zero? Undefined?
• Could someone give an example of a reasonably occurring probability density with mode=0, nonzero probability at 0, and arguments that are positive or zero, which would either satisfy nor not satisfy the requirements of the inequality?

Thank you! --NorwegianBlue ^talk 12:24, 20 October 2007 (UTC)[reply]

Probability distributions can be quite general. Given your sample space, say ${\displaystyle {\mathcal {X}}=\mathbb {R} }$ and any σ-algebra on it, you can define any function from that σ-algebra to [0, 1] which satisfies a few axioms, and you get a probability distribution. Continuous and discrete are just two narrow, yet common, kinds. The distribution kind alluded to is not really one of them, but resembles one or the other in different regions.

• No (unless there is just one point where the probability is positive), and this becomes clear if you take a look at the definition of a Unimodal function. The probability density for a discrete distribution is 0 for most real numbers. Hence, for any number where the probability is positive, you have a local maximum, and thus the distribution is not unimodal.
• At the basic level it is undefined, but the Dirac delta function can be introduced to deal with it.
• That "reasonably occurring" bit is tricky... Can't think of any right now.

-- Meni Rosenfeld (talk) 13:06, 20 October 2007 (UTC)[reply]

Thanks a lot, Meni, I think I understood. To verify, I'll try to give examples of "reasonably occuring" probability distributions with nonzero probability at 0, and which satisfy and don't satisfy the requirement. Am I correct in thinking that the distribution

p(x<0) = 0
p(x=0) = 0.3
p(x>0) = 0.7*exp(-x)

satisfies the requirement, whereas the distribution

p(x<0) = 0
p(x=0) = 0.2
p(x>0) = 0.7*exp(-x)+0.1*g(x)

where g(x) has the properties

g(x) = 0 when x<0
The integral of g from -inf to +inf is 1
g(5) > exp(-5)

does not? --NorwegianBlue ^talk 15:34, 20 October 2007 (UTC)[reply]

The idea for the first looks good, but the notation isn't. The distribution you mean can be described by:

${\displaystyle P(X\leq a)=\left\{{\begin{array}{ll}0&a<0\\1-0.7\exp(-a)&a\geq 0\end{array}}\right.}$

As for the second, same problem with the notation, and I'm not sure we know enough about g to deduce anything. If, on the other hand, we know that ${\displaystyle g'(5)>7\exp(-5)\;\!}$, I think we can conclude the distribution isn't unimodal. -- Meni Rosenfeld (talk) 16:49, 20 October 2007 (UTC)[reply]

Thanks again. I'm sorry about the notation, I knew it was awful and should have apologized beforehand. I also realize that I failed to make a clear distinction between densities and probabilities, and that I failed to take into account the factors 0.7 and 0.1 that I was multiplying exp(-x) and g(x) with. By cutting and pasting from the wiki math notation in your reply, I'll have a second go at rephrasing what I was trying to express:

The first distribution was intended to be:

${\displaystyle P(a\leq X\leq b)=\left\{{\begin{array}{ll}0&a\leq b<0\\0.3&a=b=0\\0.7\int _{a}^{b}\exp(-t)\,dt&0<a\leq b\end{array}}\right.}$

The second distribution was intended to be:

${\displaystyle P(a\leq X\leq b)=\left\{{\begin{array}{ll}0&a\leq b<0\\0.2&a=b=0\\0.7\int _{a}^{b}\exp(-t)\,dt+0.1\int _{a}^{b}g(t)\,dt&0<a\leq b\end{array}}\right.}$

As you point out, the requirements to g should have been:

${\displaystyle \,g(x)=0\;{\mbox{when}}\;x<0}$

${\displaystyle \,\int _{0}^{\infty }g(x)\,dx=1}$

${\displaystyle \,g(5)>{\color {red}\mathbf {7} }exp(-5)}$

Sorry about persevering, I just want to make sure I understand:

• Did I get the notation right?
• If so, is my first distribution just a more awkward way of expressing what you did in two lines?
• Does the first distribution satisfy the requirements of the Vysochanskiï-Petunin inequality?
• Does the second distribution violate the requirements of the Vysochanskiï-Petunin inequality?

-NorwegianBlue ^talk 19:53, 20 October 2007 (UTC)[reply]

• Much better. It seems a bit unconventional, but it gets the necessary information across.
• Pretty much, yes. Basically, by calculating ${\displaystyle P(X\leq b)-P(X\leq a)}$ using my formula, you will get the same results as your formula (there are some technical details involved).
• Yes.
• I think not. The factor of 7 wasn't the only thing I have changed - I have given a condition on the derivative of g rather than g itself. To ensure being non-unimodal, there must be some point where the derivative of the density function (which is also the second derivative of the cumulative density function) is positive. Ensuring that with a condition on g is trickier.

-- Meni Rosenfeld (talk) 22:23, 20 October 2007 (UTC)[reply]

I understand. Your replies have been very helpful. Thank you! --NorwegianBlue ^talk 11:16, 21 October 2007 (UTC)[reply]

could you please give me a comprehensive list of all types of numbers, ( names only ) i mean to say not inly the kinds of numbers like natural, whole.integers,.etc but also the others like prime, perfect, abundant, kapreskar,ghosh number,etc..... 59.93.102.39 23:55, 20 October 2007 (UTC)[reply]

Try Category:Integer sequences. —Keenan Pepper 00:09, 21 October 2007 (UTC)[reply]

Or Category:Numbers and the sub-categories therein.87.102.16.28 09:00, 21 October 2007 (UTC)[reply]

This link starts to answer your question http://mathworld.wolfram.com/search/?query=number there's over 100- pages to go through though..87.102.16.28 09:00, 21 October 2007 (UTC)[reply]

http://www.research.att.com/~njas/sequences/Seis.html is also pretty usefull. 48v 23:15, 21 October 2007 (UTC)[reply]

October 21

Dear all,

I have a question.

How advanced were the multiplication and division skills of people in the 16th century? I found the following paragraph very strange. Did they really not know how to multiply? Did they not have mathematical tables? --Kushal^t 20:39, 21 October 2007 (UTC)[reply]

I quote:

Its immediate applications were obvious: [...]given a number through his projected 'Universal Characteristic ' […] (Ross, 2000). Instead of fruitless arguing, people would say, 'Let us calculate ' – and they could do so by setting the dials and cranking the handle of his machine (one of a number of Leibnizian schemes satirised in Swift's Voyage to Balnibarbi) (Ross, 2000).

Ross, G. M. (2000, July). Leibniz. Retrieved October 20, 2007, from University of Leeds Electronic Text Centre: http://www.etext.leeds.ac.uk/leibniz/leibniz.htm

Thank you very much.

Regards,

Kushal --Kushal^t 20:39, 21 October 2007 (UTC)[reply]

The quote you give is referring to Leibnitz's characteristica universalis, by which he hoped to reduce all logical reasoning to calculation (which could then be carried out mechanically). As for arithmetical skills in the 17th century (note this is when Leibniz lived, not the 16th), I expect that then, as now, most educated people were fairly innumerate ^{[citation needed]}, but certainly it was known how to multiply, mathematical tables of many kinds were in use, and (without calculators) some people had to perform huge numbers of calculations in the course of their work. Algebraist 21:48, 21 October 2007 (UTC)[reply]

Reading the quoted paragraph, it speaks about the mechanical calculation device, not about the characteristica universalis. It says its usefullness is due to the fact that "at the time even educated people rarely understood multiplication", which is probably what prompted this question. -- Meni Rosenfeld (talk) 22:25, 21 October 2007 (UTC)[reply]

Yes, Meni. You are correct. BTW, does the paragraph talk about the ability to multiply with a table or the understanding of the concept of multiplication of two numbers?

I wonder if it could this be a minority view or a widely accepted one. In case of the latter, could my fellow Wikipedians provide me a resource right off the bat (hopefully one that I can cite). --Kushal^t 00:57, 22 October 2007 (UTC)[reply]

One educated person living at the time, who did not know his multiplication tables, and complained in his diary about the difficulty of multiplication, was Samuel Pepys. See [3]. Whether this was the exception, or the rule, I do not know. --NorwegianBlue ^talk 17:30, 24 October 2007 (UTC)[reply]

I'm interest in calculating the probabilities of events. I though I look towards the experts in the field. According to this article Mathematics of bookmaking it says

This is achieved primarily by adjusting what are determined to be the true odds of the various outcomes of an event in a downward fashion (i.e. the bookmaker will pay out using his actual odds, an amount which is less than the true odds would have paid; thus hopefully ensuring a profit).

But it does not say how the hell do the expert bookmakers determine the true odds of events. Isn't that the absolutely crucial bit of bookmaking? For example, if bookmaker A uses method A to determine the true probability of an event to be 0.2 while bookmaker B uses method B to determine the true probability of an event to be 0.5 while bookmaker C uses method C to determine the true probability of an event to be 0.8 then who is right and who is wrong? I am so confused. 202.168.50.40 22:55, 21 October 2007 (UTC)[reply]

If I'm reading that right, it would seem that the bookmaker is (like most successful business people) bottomline minded. That is to say that he will pay out more for a bid which will more likely yield a profit for him, and vice versa. A math-wiki 10:46, 22 October 2007 (UTC)[reply]

See also Odds#Gambling odds versus probabilities. "True odds" are often not "true" in a mathematical sense but the term reflects what the bookmaker really thinks the probability is, and that is different from the odds he offers to gamblers. Note that often there is coordination between the odds offered by different bookmakers, so the underlying "true odds" may not reflect what the individual bookmaker believes. If there was no coordination then situations could arise where gamblers had a guaranteed profit by simultaneously playing on one outcome at a bookmaker giving better odds on that, and the opposite outcome at another bookmaker. PrimeHunter 14:06, 22 October 2007 (UTC)[reply]

Such opportunities do exist. I think arbitrage betting covers this idea. Basically with so many bookies around there will be instances where betting on both home, away and the draw will guarantee a return when you cover the cost of all 3 bets. I understand my friend, who is a professional gambler, often looks out for these. He also looks out for wildly-unusual odds - apparently sometimes the sites accidently input the odds incorrect into the web-update and if you get a bet in before they fix if you will be given it at the odds specified (i.e. the whole sold as seen style setup). It's actually a very interesting area of business I find. ny156uk 18:53, 22 October 2007 (UTC)[reply]

October 22

I was wondering if there was a way to solve the Schrödinger equation numerically with a user-supplied potential function on a certain interval. I know about the parts that match normal solution of differential equations, but I was wondering how to set boundary conditions and how to numerically find a valid value of the energy of a bound state. --Zemyla^t 17:25, 22 October 2007 (UTC)[reply]

From experience except in the simplest cases the differential equation is not analytically solvable.

I was wondering what you meant by 'set boundary conditions' - did you mean ways to create energy boundaries in V(x)? (This can easily be done with equations similar to V(x) = fn(exp((x-a)ⁿ)) where n is large odd integer eg 1, 'a' is the position of the boundary ( x must be less than a ) and f(x) can be exp(x²) or x²ⁿ or just x etc .. for boundaries such as a<x<b use V(x)=fn(exp(a-x)²ⁿ) +fn(exp(x-b)²ⁿ) etc . An alternative energy barrier for x<a is V(x) = fn( (x/a)²ⁿ )

f(x) is there just to make the energy barrier 'more sharp' - without having to resort to discontinuous functions....

The energy of a state comes easily if you can solve the differential - I'm sure it's covered at that page.

The differential might be solvable for a specific case of V(x) - if you have that and are stuck on it someone (probably not me!) might be able to help you.. Otherwise you need to ask about getting approximate solutions to differential equations...87.102.83.3 20:40, 22 October 2007 (UTC)[reply]

Yes, I'm talking about a numerical solution, not an analytic solution. I remember a piece of software a while back called MP Desktop had a numerical solver for Schrödinger's equation, and from what I could tell, it integrated from either end toward the middle, and if the energy value given wasn't a valid bound state, the wave function would be discontinuous. But I don't know exactly how that worked. --Zemyla^t 14:07, 23 October 2007 (UTC)[reply]

So the software checked if a particular function was a valid solution of the equation? (Or did it work the other way round and try to find a wavefunction for a particular energy).. The second way is much more difficult..

Your original question asks how to get the energy - for this you need a wavefunction - if you can give more details - such as any values/function/potential energys you want to include - that would help a lot.87.102.7.135 18:23, 23 October 2007 (UTC)[reply]

It just occurred to me that what you where discribing was the Particle in a box problem see also (http://ww.google.co.uk/search?hl=en&sa=X&oi=spell&resnum=0&ct=result&cd=1&q=particle+box+schrodinger&spell=1 first few links) - this is a special case - in fact this problem is more like the equation of motion of a vibrating string.. In this case if the box edges (boundaries) are at A and B then the wavefunction at A and B is zero. (technically this specific problem isn't really suited to modelling with the schroedinger equation since the function is discontinuous - and there are various problems associated with that..)87.102.10.72 22:07, 23 October 2007 (UTC)[reply]

What in the world do you mean "this specific problem isn't really suited to modelling with the schroedinger equation"? Solving the Schrödinger equation for the particle in an infinite square well is one of the first exercises in any quantum mechanics course. —Keenan Pepper 03:37, 25 October 2007 (UTC)[reply]

Have you looked at the WKB approximation? —Keenan Pepper 03:10, 24 October 2007 (UTC)[reply]

Isn't that almost totally useless after (in the text) the point where it says "Next, the semiclassical approximation is invoked" - since it basically 'bodges' the solution by ignoring all the terms in the series except A0, and ignores the differentials etc etc87.102.94.157 12:19, 24 October 2007 (UTC)[reply]

No, the WKB approximation is not "almost totally useless", and it doesn't "bodge" anything. It's a perfectly valid limiting approximation for a large class of potentials. You think you're smarter than Wentzel, Kramers, and Brillouin? —Keenan Pepper 03:32, 25 October 2007 (UTC)[reply]

fuck off87.102.94.16 12:54, 25 October 2007 (UTC)[reply]

my son has a project and it is about a project of the year which is math and needs to know something about the number 20 and its vocabulary i guess that is what it is. please help if you can. he is in 5th grade —Preceding unsigned comment added by 76.182.221.170 (talk) 22:20, 22 October 2007 (UTC)[reply]

We have the article 20 (number). PrimeHunter 22:33, 22 October 2007 (UTC)[reply]

October 23

what is the mean value theorem, and please explain with reference to integrals and limits. and plz do provide a set of solved examples. and are there any online notes available for reviewing these concepts (of the book cALCULUS BY THOMAS AND FINNEY 9TH EDITION) —Preceding unsigned comment added by 203.128.4.231 (talk) 06:17, 23 October 2007 (UTC)[reply]

Have you seen Mean value theorem? If you're looking for worked examples, you can also look at b:Calculus/Mean Value Theorem for Functions. Confusing Manifestation 06:55, 23 October 2007 (UTC)[reply]

October 24

A flyer from Scientific American invites me to buy books about 0, φ, π, e and i. It says that π was mentioned in testimony in the murder trial of O. J. Simpson. Anyone know how? —Tamfang 02:31, 24 October 2007 (UTC)[reply]

I found it with a little Googling. Search "FBI" in [4]. It quotes The Joy of Pi by David Blatner. An FBI Special Agent is quizzed on pi and the area of a circle. His suggestions for pi were 2.12 and 2.17. Judge Ito got closer with 3.1214. PrimeHunter 22:17, 24 October 2007 (UTC)[reply]

At [5], in the upper left hand corner, there's a picture of a crumpled up surface. What is it? Black Carrot 05:44, 24 October 2007 (UTC)[reply]

I believe I've seen that same picture labeled as a minimal surface, but I can't find a reference. 75.3.82.67 06:30, 24 October 2007 (UTC)[reply]

It is indeed: namely a gyroid, made by Bathsheba Grossman. I have one beside my monitor. —Tamfang 08:42, 24 October 2007 (UTC)[reply]

Is it really metal? does it weigh a bit.. Wow - has anyone got any links to how these are made - it's not immediately apparent how I would print with metal - are chemicals involved - or some sort of ion beam?87.102.94.157 12:58, 24 October 2007 (UTC)[reply]

These things can be made with Rapid prototyping systems. Basically the shape is built up lay by layer using special chemicals which are hardened by lasers (or similar). The end result is a plastic model which can then be used to cast a metal object. An alternative could be CNC (computer numerical control). --Salix alba (talk) 15:38, 24 October 2007 (UTC)[reply]

Actually it might be done with Direct metal printing, Direct metal deposition is the closest WP article. --Salix alba (talk) 15:42, 24 October 2007 (UTC)[reply]

the artist's description of the technology. I have two of her pieces: the mini gyroid weighs perhaps an ounce, and the mini Nexus a bit more. They're quite sturdy. —Tamfang 16:22, 24 October 2007 (UTC)[reply]

Thanks those links answered it for me.. ... from the picture it looked like the thing was six inches across and weighed about a kilo... (not 28g)87.102.94.157 19:16, 24 October 2007 (UTC)[reply]

Please help me in knowing the process of writing a number as sum of four squares ( if there is any ) according the Lagrange's theorem. Will such splitting is unique? Also let me know the way of writing the following numbers as the sum of four squares.

33102; 33215; 103993; 104348 Kasiraoj 13:29, 24 October 2007 (UTC)[reply]

I think the proof of Lagrange's four-square theorem given here is constructive, and can be turned into a method of finding a four squares representation of any positive integer. The representation is not usually unique, even if we disregard differences of order, because identities such as

${\displaystyle 7^{2}+4^{2}=8^{2}+1^{2}}$

often allow you to replace one pair of squares with another pair. If you want a simple pragmatic method of finding a representation, I would try a greedy algorithm approach - start with the largest square less than n, then take the largest square less than the remainder, and see if that leaves you with something that is obviously a sum of two squares. If not, backtrack and try a smaller square on either of the first two steps. For example:

${\displaystyle 33102-181^{2}=341}$

${\displaystyle 341-18^{2}=17}$

${\displaystyle 17=4^{2}+1^{2}}$ Gandalf61 16:34, 24 October 2007 (UTC)[reply]

The applet at [6] does it, explained at [7]. PrimeHunter 22:01, 24 October 2007 (UTC)[reply]

What are MCA's? Are they worth while for most home owners or are they only good in a few situations?

Thanks

Don Dutton —Preceding unsigned comment added by 130.76.32.182 (talk) 14:27, 24 October 2007 (UTC)[reply]

A quick google search brings up lots of relevant results. The short version is that your paycheck deposits directly against your mortgage (instead of into a traditional checking account) and you're allowed to make withdrawals against your mortgage. Money that would otherwise be unspent counts toward your mortgage, paying it off faster (and reducing the interest you're paying). Additionally, there are suggestions that you can fiddle with credit card balances and such to gain an edge in interest accumulation. A significant caveat is that the system demands careful money management, particularly with regard to long-term savings. As for their overall merit, you would be well-advised to consult a professional. — Lomn 15:02, 24 October 2007 (UTC)[reply]

I wouldn't do something like that. Unless you have a very high mortgage rate, the benefits of the liquidity of cash-equivalent investments such as CDs outweigh the savings of paying down a mortgage in most circumstances. In a declining real estate market, it especially makes little sense to focus on paying down a mortgage ahead of schedule as those who are in danger of being underwater on their mortgage are also unlikely to be able to pay off the mortgage at a sufficient rate to fix their situation. I would imagine that in an MCA situation the bank is going to look very unfavorably at someone's mortgage balance exceeding the market value of the home so you could see the house next door being sold in a foreclosure auction quickly eliminating any savings that you might have managed to store in an MCA. Having a stash of ready cash will be helpful for handling such situations as getting into an apartment (between security deposit, first/last month's rent and moving costs, it can easily be $5-6K to move into an apartment in Los Angeles). Basic money management would put paying down the mortgage at the bottom of the priority list. Donald Hosek 16:33, 24 October 2007 (UTC)[reply]

I performed the same Google search as Lomn, and after seeing a lot of words but not a lot of information, I added the word "scam" to my search. It seems that a lot of money is being charged for these accounts, and there are also a lot of "paid referrers". I don't have any statistics at hand, but I'd guess that most mortgage loans these days allow additional payments of principal at any time, without resorting to some "new" scheme that has a bit of on odor to it. --LarryMac | Talk 17:52, 24 October 2007 (UTC)[reply]

First of all, I'm not sure if this is the right place to ask this. But here it is anyway.

Wikipedia can be represented as a graph (where pages are vertices, and links are edges). Has anyone ever done any research about wikipedia in terms of graph properties? For example, it would be interesting to know whether the English wikipedia has more than one nontrivial connected component. Since it's more closely represented by a digraph, maybe "strongly connected" would be a better term. I find graph theory quite interesting, and I wonder if this type of analysis has ever been done with wikipedia. This sparked my interest when I heard about a game in which you pick any two topics, and try to find the shortest path from one to the other by only clicking links in the article. --BennyD 15:15, 24 October 2007 (UTC)[reply]

• Hello BennyD, did you try this one : wikimindmap ? Just take a look and tell me if it helps. -- DLL ^{.. T} 17:11, 24 October 2007 (UTC)[reply]

I have to work out Taylor series of various functions, and there's one that seems a little tricky. It's not that I can't do Taylor series, but the only way I can think of to differentiate ${\displaystyle {\frac {1}{2+3x}}}$ is to use the quotient rule, which works fine for the first differential, but is totally unwieldy for the 3rd (having to square a massively long polynomial is a complete pain). To make things harder, I'm not making the series around 0, so I can't cancel out all the x terms later. Is there a simpler way to differentiate this function (it seems really simple, so I'm sure I'm missing something obvious...) Laïka 16:49, 24 October 2007 (UTC)[reply]

Firstly, it's easier to use the chain rule rather than the quotient rule for this: the derivative of 1/f(x) is -f'(x)/(f(x)^2) (of course, the quotient rule gives the same answer). Secondly, don't expand the polynomial in the denominator. So the first derivative is ${\displaystyle {\frac {-3}{(2+3x)^{2}}}}$, and the 2nd derivative is ${\displaystyle {\frac {-3*-3*2}{(2+3x)^{3}}}}$ - can you take it from there? AndrewWTaylor 17:14, 24 October 2007 (UTC)[reply]

Differentiation is the hard way to find the Taylor series of this function. I'll give a hint for the easy way: The Taylor series of ${\displaystyle {\frac {1}{1-x}}}$ around 0 is ${\displaystyle \sum _{n=0}^{\infty }x^{n}}$. -- Meni Rosenfeld (talk) 17:21, 24 October 2007 (UTC)[reply]

Thanks; I'd tried the chain rule but a stupid mistake in my working screwed up the answer; I knew there was something obvious I'd missed. Laïka 19:46, 24 October 2007 (UTC)[reply]

This is probably a bit of a dunce-ish question but never mind. If I know the cross section of a river, assumed constant, and the speed of the water flowing in the river, again assumed constant, how can I estimate the volume of water flowing in a given time frame. (This is a homework question but I tried to be vague. If it doesn't make sense as it is, I can give more specific info.) Thanks asyndeton 20:08, 24 October 2007 (UTC)[reply]

What do you get if you multiply the area of the section by the speed? Note the units. —Tamfang 20:24, 24 October 2007 (UTC)[reply]

The speed is m/s, the area is m^2, so I would get m^3/s? And then I just mulitply by, in my case, 60 to get the volume in a minute. Thanks asyndeton 20:44, 24 October 2007 (UTC)[reply]

Yep. It's somewhat impressive how many problems reduce to simple multiiplications and divisions once you look at the units. Donald Hosek 23:12, 24 October 2007 (UTC)[reply]

A tad embarassing really. But then again, I suppose we all learn from our mistakes. asyndeton 23:14, 24 October 2007 (UTC)[reply]

Blindly trusting the power of dimensional analysis is dangerous, as there might be other physical quantities or constants effecting the problem, the units of which can change the result. I will always remember a book I've read which uses dimensional analysis to "prove" that drag in a liquid is proportional to the square of the speed, where in practice the relation is much more similar to linear. The discrepancy is due to ignoring a unit-bound property of the liquid.

This particular problem can be solved easily without resorting to such unsound methods. -- Meni Rosenfeld (talk) 12:24, 25 October 2007 (UTC)[reply]

Well I'm looking to learn. How would you go about it? asyndeton 12:26, 25 October 2007 (UTC)[reply]

There are several equivalent ways to think about it - one is to consider the "end" of the river, where the water enters the ocean. The volume of water flowing in a given time frame (${\displaystyle \Delta t}$) is the volume of water exiting the river in that period. The water that will exit the river from now until ${\displaystyle \Delta t}$ from now is exactly the water that can reach the end in time less than ${\displaystyle \Delta t}$, and since it flows at speed v, this is exactly the water at distance up to ${\displaystyle v\Delta t}$ from the end - and it forms a prism of cross section S and height ${\displaystyle v\Delta t}$. The volume of this is ${\displaystyle Sv\Delta t}$. -- Meni Rosenfeld (talk) 12:38, 25 October 2007 (UTC)[reply]

where to find the 64 analyzable prisoner's dilemmas ? 71.113.165.99 11:58, 25 October 2007 (UTC)[reply]

Our Prisoner's dilemma article describes one of them, and perhaps can get you started on the search for the other 63. -- Meni Rosenfeld (talk) 12:14, 25 October 2007 (UTC)[reply]