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October 19

Graphing

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]

expected value

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 $k$ required to select a subset of $n$ good data points. Let $w$ be the probability that any selected data point is [a good data point]. Then we have:

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

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

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]

Degree of freedom

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]

Proving a tautology

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

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

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

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

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

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

$(\neg P\lor Q)\land (Q\lor R)\land T$ 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

((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

P\to Q

with

(\neg P\lor Q)

. Applying De Morgan's laws and eliminating double negations, we end up having to prove

(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

(Q\to R)

when I should have negated

((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]

pendulum

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]

Name of a math trick

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]