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November 9

Definitions of logarithmic and trigonometric functions

What's the definition of sine function? If I wanted to know to what evaluates Sin(50) how could I do it? (Same with logarithm function) 190.60.93.218 (talk) 16:34, 9 November 2012 (UTC)[reply]

I can't do any better than refer you to the article sine. Duoduoduo (talk) 17:03, 9 November 2012 (UTC)[reply]

Which yields the infinite series

sin(x)=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)!}}x^{2n+1}

. The input must be in radians. --Melab±1 ☎ 03:10, 11 November 2012 (UTC)[reply]

Also logarithm. And if you just want to find the value of say sin (50), you can just use a technical calculator. Make sure you put it in "deg" mode if you mean 50 degrees, or "rad" mode if you mean 50 radians. Also, the technical calculator will have two different logarithm keys: one called "log" for the log to the base 10, and one called "ln" for the natural log. Duoduoduo (talk) 17:15, 9 November 2012 (UTC)[reply]

I read the article, I think it doesn't show it's function definition, It shows how it uses Taylor series to get an approximation of the value, but I don't think that's the definition, just a way-around. So what's the function definition of Sine? (How would you manually evaluate sine?) (Same for logarithm)190.60.93.218 (talk) 17:21, 9 November 2012 (UTC)[reply]

Page 1 and further of "Real And Complex Analysis" by Walter Rudin. You first define

$\exp(z)\equiv \sum _{n=0}^{\infty }{\frac {z^{n}}{n!}}$

The series coverges absolutely for every z and converges uniformly on every bounded subset of the complex plane. This then proves that exp is a continuous function. And because it converges absolutely, you can multiply exp(z1)*exp(z2) by substituting the series in each factor, do a change in variables in one summation to obtain exp(z1+z2).

Then the following theorem follows:

a) for every z, $\exp(z)\neq 0$

b) exp is its own derivative

c) The restriction of exp to the real axis is a monotonically increasing function, exp(x) tends to infinity for x to infinity and to zero for x to minus infinity.

d) There exists a positive number pi such that exp(i pi/2) = i and such that exp(z) = 1 if and only if z/(2 pi i) is an integer.

e) exp is a periodic function with period 2 pi i.

f) The mapping t to exp(i t) maps the real axis to the unit circle.

g) If w is a complex number and $w\neq 0$ , then w = exp(z) for some z.

The proofs are very simply, I can leave that as a exercise to the OP. For real t, one defines cos(t) and sin(t) to be the real and imagnary parts of the function exp(i t). This fixes all the properties of these functions. Count Iblis (talk) 17:48, 9 November 2012 (UTC)[reply]

(ec)The article sine says In a right triangle, sine gives the ratio of the length of the side opposite to an angle to the length of the hypotenuse. That's the definition -- the sine of e.g. 50° is found by forming a triangle with one 90° angle and one 50° angle. The sine of 50° is the ratio of the length of the side opposite the 50° angle to the length of the hypotenuse (the side opposite the 90° angle). Usually the exact value of the sine is irrational, so you're not going to be able to express it exactly in a finite number of decimal places. But you can get as close as possible by using a Taylor series expansion.

Likewise, the definition of logarithm is given by the article as if $x=b^{y}$ , then y is the logarithm of x to base b, and is written $y=log_{b}(x)$ , so $log_{10}(1000)=3.$ That's the definition: the logarithm is the value of y that solves that equation. To compute it you can use a Taylor expansion to get an arbitrarily close approximation, but again, since it's usually irrational you're never going to be able to express it exactly in a finite number of decimal places. Duoduoduo (talk) 17:58, 9 November 2012 (UTC)[reply]

Here is a possible definition of sine. There is exactly one mapping e from R to C which: 1) is a morphism, i.e., e(x + y) = e(x)e(y); and 2) satisfies e'(0) = i. (In fact, this mapping is e(t) =e^it.) Then sin t is defined as the imaginary part of e(t). The logarithm function can be defined as the inverse of the exponential function exp(t) = e^t from R to (0,+∞). The function exp can be defined like e above, except that its derivative at 0 is to be 1 instead of i.

To evaluate sin 50 (in radians), you might use the Taylor series above. To evaluate log 50, you might find an approximate value for the integral of 1/t from 1 to 50, perhaps by using Simpson's formula.

If you meant sine of 50 degrees, you might let x = 50 degrees, and note that sin 3x = 1/2. Since sin 3x can be expressed easily in terms of sin x via the addition formulas, you can write down a third-degree equation one of whose solutions is sin x. Then use Newton's method, for example, to find the appropriate root. 96.46.202.24 (talk) 10:55, 11 November 2012 (UTC)[reply]

Log(50) = Log(2) + 2 Log(5)

For for $-1<x\leq 1$ , you can use the series expansion:

\log(1+x)=\sum _{n=1}^{\infty }(-1)^{n+1}{\frac {x^{n}}{n}}

This converges very slowly for x = 1, so it can't be used directly. However, for large x, we can write Log(1+x) = Log(x) + Log(1+1/x), which yields:

\log(1+x)=\log(x)+\sum _{n=1}^{\infty }(-1)^{n+1}{\frac {1}{nx^{n}}}

For x = 2 this is still not an improvement, but for x = 5 you obtain a reasonably fast converging series in terms of Log(4)= 2 Log(2). To improve things more, we put $1+{\frac {1}{x}}={\frac {1+y}{1-y}}$ . Then $y={\frac {1}{2x+1}}$ and we have

\log(1+x)=\log(x)+\log(1+y)-\log(1-y)=\log(x)+2\sum _{n=0}^{\infty }{\frac {y^{2n+1}}{2n+1}}

So, we have:

\log(1+x)=\log(x)+2\sum _{n=0}^{\infty }{\frac {1}{(2n+1)(2x+1)^{2n+1}}}

This converges fast for large x, even for x = 1 the convergence is quite fast. The logarithms of the integers can be computed using only fast converging series by exploiting relations such as 3^4 = 81 = 1 + 80. Putting x = 80 in the above expression gives

4\log(3)=4\log(2)+\log(5)+2\sum _{n=0}^{\infty }{\frac {1}{(2n+1)(161)^{2n+1}}}

The relation 7^4 = 1 + 2400 gives:

4\log(7)=5\log(2)+\log(3)+2\log(5)+2\sum _{n=0}^{\infty }{\frac {1}{(2n+1)(4801)^{2n+1}}}

With a few such relations you can solve for the logarithms of the prime numbers. Count Iblis (talk) 18:08, 11 November 2012 (UTC)[reply]

$\log(50)=\sum _{n=0}^{\infty }{\frac {1}{2n+1}}\left(350\left({\frac {1}{99}}\right)^{2n+1}-44\left({\frac {1}{161}}\right)^{2n+1}+156\left({\frac {1}{251}}\right)^{2n+1}+136\left({\frac {1}{4801}}\right)^{2n+1}\right)$

Count Iblis (talk) 22:53, 11 November 2012 (UTC)[reply]

Hello! I would define logarithm as

\ln(x):=\int _{1}^{x}{\frac {1}{t}}dt

.--AnalysisAlgebra (talk) 05:31, 13 November 2012 (UTC)[reply]

November 10

Conway puzzle

In conway puzzle, the diagram is wrong. This was pointed out 2 years ago, but nothing has happened. Does anyone here have the ability (and inclination) to redraw it? -- SGBailey (talk) 07:36, 10 November 2012 (UTC)[reply]

I think the diagram is actually correct, and that it is being misperceived. Looie496 (talk) 15:50, 10 November 2012 (UTC)[reply]

A slightly different angle would make it more obvious. the middle block touches the other two blocks at the corners. If you can describe it better then try editing the article to give a short description under the diagram. Dmcq (talk) 15:59, 10 November 2012 (UTC)[reply]

I suggest adding a normal projection onto the two vertical planes, to make it's location obvious. StuRat (talk) 20:52, 12 November 2012 (UTC)[reply]

The nearest block occupies positions (1,1,1), (2,1,1) and (3,1,1). The vertical block occupies (4,2,2), (4,2,3) and (4,2,4). The farthest block occupies (5,3,5), (5,4,5) and (5,5,5). The vertical block looks like it starts at (5,3,1) but it's an illusion - it starts at (4,2,2), because it is higher it looks from this perspective as if it's farther. -- Meni Rosenfeld (talk) 19:14, 13 November 2012 (UTC)[reply]

Euler axis/angle to Tait–Bryan angles (yaw, pitch, roll) conversion

I have a rotation vector for a camera where the length represents the angle and the direction the axis about which I rotate. While this is an elegant way to store rotation it is not very user friendly. One typically wants to know the yaw (pan), pitch (tilt) and roll of the camera. The article on rotation formalisms in three dimensions gives a bunch of conversion formulae for some cases, but not this one. I preferably do not want to go via expressing the rotation as a matrix (combining the first two conversions given), that seems cumbersome if both representations use just 3 values. 41.164.7.242 (talk) 13:20, 10 November 2012 (UTC) Eon[reply]

Axis-angle representation and Rodrigues' rotation formula may well help you here as they describe your way of representing a rotation, and conversion to a rotation matrix. I think using matrices will help you, essentially these are a way of storing three linear combinations of three values.--Salix (talk): 08:03, 11 November 2012 (UTC)[reply]

Thanks, so I'll go (Euler axis/angle) -> (rotation matrix) -> (yaw, pitch, roll). I don't care for the matrix though. 105.236.57.198 (talk) 14:06, 11 November 2012 (UTC) Eon[reply]

November 11

Calculus notation

What type of function is a "d" with a forward-slanted strikethrough, I think it is a type of differencial.

An example: d(U - TS) less than or equal to (d̸)w_expansion + (d̸)w_nonexpansion. Plasmic Physics (talk) 00:36, 11 November 2012 (UTC)[reply]

In thermodynamics the state functions have perfect differentials (d), and then there are differentials d̸ which are not perfect, meaning that they are not differentials of state functions. For example, in a fluid the increase of energy dE = TdS−PdV where T is temperature, dS is increase in entropy, P is pressure, and dV is increase in volume. Here dE and dS and dV are perfect differentials, because E and S and V are state functions. For reversible processes TdS=d̸ Q is the heat transferred to the fluid, and PdV=d̸ W is the work done by the fluid. But 'heat content' Q and 'work content' W are not state functions. Bo Jacoby (talk) 01:28, 11 November 2012 (UTC).[reply]

Explicit solution that gives the n^th Harmonic number

Does one exist without integrals or other such operations that require the further determination of a solution? --Melab±1 ☎ 03:07, 11 November 2012 (UTC)[reply]

I'm fairly certain the harmonic number function is not elementary. As such there is no "explicit solution" for it in the form you seem to want. If you're satisfied with an approximation, taking a few terms from the asymptotic expansion works well. -- Meni Rosenfeld (talk) 06:08, 11 November 2012 (UTC)[reply]

What does f ⊑ g mean?

I recently read a paper that used the notation "f ⊑ g" without defining the symbol "⊑". It looks like a kind of square "subset" symbol. f and g were functions. Does this notation mean anything to anyone? --Doradus (talk) 22:54, 11 November 2012 (UTC)[reply]

Could they have been partial functions? It often means "extends".--80.109.106.49 (talk) 00:42, 12 November 2012 (UTC)[reply]

Unicode Character 'SQUARE IMAGE OF OR EQUAL TO' (U+2291) – though this probably does not help much. — Quondum 01:06, 12 November 2012 (UTC)[reply]

Thanks for the tip on partial functions. When I get a chance, I'll re-read that passage and see if that explains it. --Doradus (talk) 18:46, 12 November 2012 (UTC)[reply]

Of course it would have been boring to tell us the title of the paper or even its topic. That would have made the question much too easy. Looie496 (talk) 03:33, 12 November 2012 (UTC)[reply]

Of course it would be silly just to ask for the information you want. Much better to post a sarcastic remark insinuating that every newcomer asking a question ought to know the right amount of context to provide. --Doradus (talk) 18:46, 12 November 2012 (UTC)[reply]

To me it just means "f is a subset of g", but I could be wrong. Perhaps they are used in Cauchy spaces, but then again, perhaps not. Is there a way to search for such a symbol encyclopedia-wide in Wikipedia math-notation Special:Search function? ~AH1 ^(discuss!) 04:01, 13 November 2012 (UTC)[reply]

Actually Wikipedia search turns up zero hits for that symbol. Perhaps the Search function has some trouble with unusual unicode characters? --Doradus (talk) 13:38, 13 November 2012 (UTC)[reply]

For anyone who is interested, the paper is this one: "Strictness and Binding-Time Analyses: Two for the Price of One". It doesn't seem to be available for free online. The sentence is:

A domain projection γ on a domain D is a continuous function γ: D → D such that (i) γ ⊑ ID, and (ii) γ ∘ γ = γ (idempotence).

I take it that "ID" is the identity function over D. (perhaps it should have been I_D?) --Doradus (talk) 13:49, 13 November 2012 (UTC)[reply]

(edit conflict) This is the unicode character U+2291 (Square Image Of Or Equal To) (see for example this website). The German Wikipedia lists it in the article Unicodeblock Mathematische Operatoren. According to page 5 of this pdf file it denotes some kind of relation. ~~but I don't know to what mathematical objects this is applied~~ It seems Doradus found something. -- Toshio Yamaguchi (tlk−ctb) 13:52, 13 November 2012 (UTC)[reply]

Seeing the word “domain”: ⊑ is commonly used to denote partial orders (domains) in domain theory. If D, E are domains, the order on the function space D → E is defined pointwise, i.e., f ⊑ g iff f(x) ⊑ g(x) for every x ∈ D. So, if ID is indeed the identity function as you suggest, then (i) translates to γ(x) ⊑_D x for every x, where ⊑_D is the order supplied by the definition of D being a domain.—Emil J. 14:30, 13 November 2012 (UTC)[reply]

I had a look at the paper, and indeed it seems to be talking about domain theory (with some category theory mixed in). However, the author is very reader-unfriendly when it comes to providing context, or definitions of the concepts he’s using (let alone notation).—Emil J. 14:53, 13 November 2012 (UTC)[reply]

That sounds like the definition of a retract (in the sense used in topology, not in the sense used in category theory, see retract (category theory) for that). — Tobias Bergemann (talk) 14:32, 13 November 2012 (UTC)[reply]

Terrific! Thanks again, everyone, for all the leads. --Doradus (talk) 20:59, 17 November 2012 (UTC)[reply]

Fibonacci numbers

Hello, could anyone explain me Fibonacci sequence in short? And what is the meaning? Thanks. Bennielove (talk) 23:43, 11 November 2012 (UTC)[reply]

The article Fibonacci numbers starts at a fairly basic level. Please let us know what information there you would like more information on.Naraht (talk) 00:12, 12 November 2012 (UTC)[reply]

1 + 0 is 1, 1 + 1 is 2, 2 + 1 is 3, 3 + 2 is 5, 5 + 3 is 8, 8 + 5 is 13, 13 + 8 is 21, 21 + 13 is 34, and so on. Placing these numerical values into the side lengths of adjacent squares gives you the Fibonacci spiral, which is found everywhere from Ammonites to pinecones to the fractal-like interactions of tropical cyclones and just everything else found in nature. See Golden ratio and natural computing for some direct applications. ~AH1 ^(discuss!) 04:04, 13 November 2012 (UTC)[reply]

Ahem no. Please read Fibonacci Flim-Flam to find most of these statements to be debunked myths. "Nature has many spiral forms. None of them are golden spirals. Many don't even come close. None of them are 'explained' by Fibonacci mathematics." --KnightMove (talk) 12:32, 14 November 2012 (UTC)[reply]

November 12

Lagrangian density

File:Generalized coordinates 1df.svg

What happens to a lagrangian in vector space

Can someone explain to me Lagrangian density used to derived field equations like how ordinary lagrangian derives the trajectory of a particle? For example I don't understand what L is a function defined on for the density version, I know for ordinary Lagrangian L is defined on R^2n for n dimensional space, but I don't understand the field version. And also what's the motivation behind it? Ordinary Lagrangian wants the stationary point of the action due to a trajectory, so what about Lagrangian density? Money is tight (talk) 08:44, 12 November 2012 (UTC)[reply]

Perhaps the attached image will help. ~AH1 ^(discuss!) 04:06, 13 November 2012 (UTC)[reply]

This is a bit beyond me, but try looking at [1] - see Section 3 (p22) 'Lagrangian Mechanics' which leads into 3.2 'Principles of Lagrangian Field Theory' Christopherlumb (talk) 17:19, 14 November 2012 (UTC)[reply]

November 13

General algorithm for finding quotient rings.

Let $R$ be a ring and $(f_{1},...,f_{n})$ be the ideal generated by the elements $f_{1},...,f_{n}\in R[x]$ . Is there a general algorithm for finding $R[x]/(f_{1},...,f_{n})$ up to isomorphism?--AnalysisAlgebra (talk) 05:27, 13 November 2012 (UTC)[reply]

November 14

2012x2015

Hello all. This is a puzzle that was in the newspaper (no prize or anything like that, don't worry); usually they are pretty easy but this one had me stumped. It went: Consider a 2012x2015 rectangle. Divide it into unit squares (1x1). How many of these unit squares does a diagonal of this rectangle pass through? I did some back of the envelope work with simpler squares (2x3, 3x5) and estimated that it had to be in the ballpark of 4030, but I couldn't find an exact answer (as 2012 and 2015 are coprime). I think I'm missing something really simple but my brain isn't working today. Help? 24.92.74.238 (talk) 03:49, 14 November 2012 (UTC)[reply]

Well, if it was exactly square, the answer would be the length of a side. Since it's slightly off square, I'd say it should be about double the longest length, or 4030. I assume that's how you got that answer. Not sure how to find the exact answer, though. StuRat (talk) 06:32, 14 November 2012 (UTC)[reply]

The answer is 4026. If you look at the line

y={\frac {2012}{2015}}x

, a segment of

x\in [m,m+1]

will usually pass through 2 squares, but only one if for some integer n,

n\leq {\frac {2012}{2015}}x<{\frac {2012}{2015}}(x+1)\leq n+1

. This happens if

2012m\in \{0,1,2,3\}\ \mathrm {mod} \ 2015

or equivalently

3m\in \{0,2012,2013,2014\}\ \mathrm {mod} \ 2015

. The solutions are

m\in \{0,671,1343,2014\}

. -- Meni Rosenfeld (talk) 06:35, 14 November 2012 (UTC)[reply]

I just ran a simulation and got 4025. You didn't count both (0,0) and (2012,2015) as squares, did you ? StuRat (talk) 07:07, 14 November 2012 (UTC)[reply]

No, I didn't. My result is also confirmed by 1Ouch0's solution, which is much more elegant (and general) than mine and easily verifiable for some simple special cases such as nXn and nX1.

Can you share your code? -- Meni Rosenfeld (talk) 07:45, 14 November 2012 (UTC)[reply]

Included below. StuRat (talk) 08:26, 14 November 2012 (UTC)[reply]

Simulation code (FORTRAN).

     program DIAGONAL
     implicit none

     logical*1     TILE_OCCUPIED(0:2016,0:2016)
     integer*2     I,J,K,COUNT
     real*8        X,Y

Initialization:

     do I = 0,2016
       do J = 0,2016
         TILE_OCCUPIED(I,J) = .false.
       enddo
     enddo
     COUNT = 0

Body:

     do K = 1,2015

       X = 1.0*K - 0.00000001
       Y = X*2012/2015
       I = X ! Truncates.
       J = Y ! Truncates.
       if (.not. TILE_OCCUPIED(I,J) .and. (I .gt. 0) .and. (J .gt. 0)) then
         TILE_OCCUPIED(I,J) = .true.
         COUNT = COUNT + 1
         print *,"Tile",COUNT,"= (",I,",",J,")"
       endif

       X = 1.0*K + 0.00000001
       Y = X*2012/2015
       I = X ! Truncates.
       J = Y ! Truncates.
       if (.not. TILE_OCCUPIED(I,J) .and. (I .gt. 0) .and. (J .gt. 0)) then
         TILE_OCCUPIED(I,J) = .true.
         COUNT = COUNT + 1
         print *,"Tile",COUNT,"= (",I,",",J,")"
       endif

     enddo

Termination:

     print *
     print *,"Count = ",COUNT

end

Simulation solution set.

Tile      1 = (      1 ,      1 )
Tile      2 = (      2 ,      1 )
Tile      3 = (      2 ,      2 )
Tile      4 = (      3 ,      2 )
Tile      5 = (      3 ,      3 )
Tile      6 = (      4 ,      3 )
Tile      7 = (      4 ,      4 )
Tile      8 = (      5 ,      4 )
Tile      9 = (      5 ,      5 )
Tile     10 = (      6 ,      5 )
Tile     11 = (      6 ,      6 )
  .
  .
  .

(See the rest here: [2].)

  .
  .
  .
Tile   4016 = (   2010 ,   2007 )
Tile   4017 = (   2010 ,   2008 )
Tile   4018 = (   2011 ,   2008 )
Tile   4019 = (   2011 ,   2009 )
Tile   4020 = (   2012 ,   2009 )
Tile   4021 = (   2012 ,   2010 )
Tile   4022 = (   2013 ,   2010 )
Tile   4023 = (   2013 ,   2011 )
Tile   4024 = (   2014 ,   2011 )
Tile   4025 = (   2015 ,   2012 )

Count =    4025

Your solution set is missing (2014,2012). Which is a result of using an incorrect offset for the diagonal. In your calculation x=2 corresponds to the right edge of the leftmost square, which means the corresponding y-value is 1 + 2012/2015, and not 2*2012/2015 as in your code. By the time it gets to x=2015-ε, it is thinking it is exiting (2014,2011) through a corner.

It's better to think of leftmost square as [0,1]X[0,1], and since you round down the real values, it should be denoted (0,0). So you should let K run from 0 to 2015, and instead of

I>0,\ J>0

require that

0\leq I\leq 2014,\ 0\leq J\leq 2011

. -- Meni Rosenfeld (talk) 12:51, 14 November 2012 (UTC)[reply]

The part which confused me is that my values seemed to match your statement that "The solutions are

m\in \{0,671,1343,2014\}

". That is, I only get one tile on each of those columns, including the 2014 column. If I shift to allow tiles (0,0) and (0,1), and exclude tile (2015,2012), then I do get 4026 tiles, but, oddly, there are now two tiles in column 2014, at (2014,2011) and (2014,2012). StuRat (talk) 19:47, 14 November 2012 (UTC)[reply]

If (0,0) is the bottom-left square, then (0,1) shouldn't be in the solution set, so something is still off. I think you shifted by decreasing both X and Y by 1 (which gives you the same wrong offset), where what you should have done is decrease X by 1 and calculate Y based on the new X (meaning it decreases by 2012/2015). -- Meni Rosenfeld (talk) 22:03, 14 November 2012 (UTC)[reply]

(ec) Sturat and Meni beat me to it.

You were almost there...

X and Y are coprime, that means that the diagonal will only enter one square at a vertex, it will enter (Y - 1) squares via the bottom line, and (X - 1) squares via the left - so the answer is X + Y - 1, whether you are looking at 2x3, 3x5, or 2012x2015.

If X and Y are not coprime, subdivide the problem into n rectangles of size p times q, where p and q are coprime. So it'll be n (p + q - 1) which reduces to X + Y - n, n being the gdc of X and Y.

Cheers. - ¡Ouch! (^{hurt me} / _{more pain}) 06:59, 14 November 2012 (UTC)[reply]

November 15

Logic puzzle questions

These are questions rather for the "humanities" or "miscellaneous" desk, but I have a feeling that my chances to get an answer are best here.

Who invented the logic grid puzzle, and when? Was the Zebra puzzle, brought to a wider public by Life International in 1962, maybe the first one of its kind?
The text published in there does not mention any involvement of Einstein, which implies that this "information" was added later. However, it could be possible that it was mentioned in the removed images. Has anybody seen the original and can tell for sure?

--KnightMove (talk) 07:53, 15 November 2012 (UTC)[reply]

Algebra questions of a certain type.

Find a surjective group homomorphism $S_{5}\rightarrow S_{4}$ or prove that one does not exist. What is the strategy in general for finding surjective homomorphisms?--AnalysisAlgebra (talk) 09:43, 15 November 2012 (UTC)[reply]

Say we wish to map a group G onto a group H. If we know a generating set for H, clearly we want to make sure that it is contained in the image of our map - then we're done. This can be easier said than done, however! In the case of finite groups, you can search for normal subgroups of size |G|/|H|, since the kernel of your proposed map will have to be such a thing. For the symmetric group, this is not *such* a Herculean task, because we know that conjugation by the symmetric group acts transitively on the set of elements of a fixed cycle-type. (So, if the normal subgroup contains a 3-cycle, it actually contains *all* 3-cycles, etc.) Our article symmetric group discusses just what normal subgroups a symmetric group can have. Sorry if this answer was too specific to your one example - it can be quite ad hoc at times, finding epimorphisms. Icthyos (talk) 10:45, 15 November 2012 (UTC)[reply]

So the kernel has to be a normal subgroup of size 5. So the answer is that it's not possible, since no subgroup of

S_{5}

has 5 elements. This is because there is no way to sum the orders of selected conjugacy classes to 5.--AnalysisAlgebra (talk) 18:59, 15 November 2012 (UTC)[reply]

Statistical analysis of exam passes

Hi, I have some data about some students who have to take a series of 13 exams and are expected to pass a certain number after a set number of 6 month periods. So I have the following table of "expected" passes each 6 month period:

Exam session                   1 2 3 4 5 6 7 8  9 10 11 12 13 14 15 16 17 18 
                               =============================================
Number of expected exam passes 0 1 3 5 7 8 8 9 10 11 12 13

Then I have the dates of the month that each student passed each exam (which can easily be converted to total number of exams passed at each 6 month period for each student). Eg:

Exam Period 1 2 3 4 5 6 7  8  9 10 11 12 13 14 15 16 17 
            ===========================================
Passes      0 0 0 1 3 5 7 10 10 10 11 11 12 12 13 13 13

(This student clearly not passing fast enough compared to the above table)

What statistical technique would be suitible to analyse whether the above table is reasonable compared to the actual performance of the student, that is, is the expected numbers too harsh, or too lenient given actual students performance?

I am just looking for the name of a relevant technique, wikipedia should suffice for the rest as I am a fairly competent mathematician (just not statistician).

Thanks for any help in advance,

Matt 80.254.147.164 (talk) 11:08, 15 November 2012 (UTC)[reply]

Maybe I'm misunderstanding your set-up. It seems to me that as of the end of each of the 12 six-month periods for which data are given, the student has equaled or exceeded (actually exceeded, until right near the end) expectations. Why do you say This student clearly not passing fast enough compared to the above table? Duoduoduo (talk) 15:50, 15 November 2012 (UTC)[reply]

Sorry, I copied it from another source and didnt have zeroes in the right places, edited and fixed.80.254.147.164 (talk) 16:22, 15 November 2012 (UTC)[reply]

I edited your post to line up the columns, although I had to make a guess on the first part, since the number of columns don't match. Are you saying all 13 exams should be satisfactorily completed by the 12th period ? If so, I suggest you just list the number of periods behind or ahead for each student, and then find the difference in this number from period to period. In this case:

Exam Period  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 
             ====================================================
Passes       0  0  0  1  3  5  7 10 10 10 11 11 12 12 13 13 13 13
Expectation  0  1  3  5  7  8  8  9 10 11 12 13 13 13 13 13 13 13
             ====================================================
Comparison   0 -1 -3 -4 -4 -3 -1 +1  0 -1 -1 -2 -1 -1  0  0  0  0
Difference    -1 -2 -1  0 +1 +2 +2 -1 -1  0 -1 +1  0 +1  0  0  0

You could then total all the differences in the comparisons for all the students. This suggests that you need to give more time where the differences are negative, and less where they are positive. However, this doesn't directly account for knowledge that some exams might be harder than others. To do this, you might want to break it down by exam:

Exam                             1    2    3    4    5    6    7    8    9   10   11   12   13
                               ===============================================================
Periods allocated to pass exam   2  1/2  1/2  1/2  1/2  1/2  1/2    1    0    1    1    1    1
Periods taken to pass exam       4  1/2  1/2  1/2  1/2  1/2  1/2  1/3  1/3  1/3    3    2    2
                               ===============================================================
Difference                      -2    0    0    0    0    0    0 +2/3 -1/3 +2/3   -2   -1   -1

This suggests you need to allocate more time for the first exam and last 3 (of course, this assumes the student you listed is typical, you need to total your results to see the real patterns). StuRat (talk) 16:44, 15 November 2012 (UTC)[reply]

Property of the parabola

I'm looking for a textbook (or website) that mentions a property of the parabola. If it exists, it is likely to be a textbook on analytic geometry or an introductory textbook on differential calculus. I can describe this property using a theorem. It might say something like the following.

If (x_A, y_A) and (x_B, y_B) are the coordinates of two points, A and B, on a parabola, the line joining these two points is parallel to the tangent at point C where the x coordinate of C is:

x_{C}={\frac {x_{A}+x_{B}}{2}}

The value of this theorem would be that it enables students who are about to be introduced to differential calculus to calculate the slope of a tangent to a parabola (ie the derivative of a quadratic function) without requiring an understanding of the limit or the derivative. If asked to find the slope of the tangent to a given parabola at the point where x=x₁ students can select any two values of x equidistant from x₁, find the values of the function for those two values of x, and then find the slope of the line joining the two points.

slope:

m={\frac {f(x_{1}+\Delta )-f(x_{1}-\Delta )}{2\Delta }}

This theorem is reminiscent of the mean value theorem but of course the MV theorem is unnecessarily advanced for students about to be introduced to differential calculus.

The only proof I can offer is one based on differential calculus - the following equation is true for any second degree polynomial but not for any higher-degree polynomial. (It is also trivially true for linear functions and constants.)

{\frac {f(x_{B})-f(x_{A})}{x_{B}-x_{A}}}=f^{\prime }\left({\frac {x_{A}+x_{B}}{2}}\right)

Any suggestions about a textbook? Dolphin (t) 14:50, 15 November 2012 (UTC)[reply]

A non-calculus proof goes like this: use your expression for the arc slope m , plug in

f(x)=x^{2}+ax+b

for both

x+\Delta

and

x-\Delta

, and see that the result

m=2x+a

is independent of

\Delta

(for

\Delta

not equal to zero, but arbitrarily small). That provides the intuition. Is that what you wanted? Duoduoduo (talk) 16:24, 15 November 2012 (UTC)[reply]

Maybe this passage from Polynomial long division will help:

Polynomial long division can be used to find the equation of the line that is tangent to a polynomial at a particular point.[Strickland-Constable, Charles, "A simple method for finding tangents to polynomial graphs", Mathematical Gazette 89, November 2005: 466-467.] If R(x) is the remainder when P(x) is divided by (x – r )² — that is, by x² – 2rx + r ² — then the equation of the tangent line to P(x) at x = r is y = R(x) (regardless of whether or not r is a root of the polynomial).

This gives a non-calculus way to get the tangent's slope, although I don't know if that method can be made intuitive for a student who doesn't know calculus. Duoduoduo (talk) 16:34, 15 November 2012 (UTC)[reply]

Another way, using this Theorem: Lets have a family of parallel lines which intersect the parabola. On each line we mark the middle point between the two points of intersection of the line with the parabola. Then all the points we marked are on a single vertical line. --84.229.146.110 (talk) 18:35, 15 November 2012 (UTC)[reply]

Transitive actions

Does there exist a transitive action of $S_{5}$ on a set with 7 elements?--AnalysisAlgebra (talk) 15:21, 15 November 2012 (UTC)[reply]

Try using the orbit-stabiliser theorem. It's very helpful for this sort of problem, especially since you require the action to be transitive, so there is only one orbit. Icthyos (talk) 15:30, 15 November 2012 (UTC)[reply]

So again the answer is no, since 7 does not divide 120.--AnalysisAlgebra (talk) 19:00, 15 November 2012 (UTC)[reply]

Rings.

Let $R$ be a ring and $I,J$ ideals. Show that $R=I+J\implies I\cup J=IJ$

I can derive that $IJ\subset I\cup J$ (it is true in general without the given condition, in fact) but that's all.--AnalysisAlgebra (talk) 19:03, 15 November 2012 (UTC)[reply]

That's because it isn't true. Consider

R=\mathbb {Z}

,

I=5\mathbb {Z}

and

J=7\mathbb {Z}

, say. A typo in the statement, probably because someone was standing on their head.John Z (talk) 21:50, 15 November 2012 (UTC)[reply]

Is any conditional probability a Bayesian probability?

Are both synonym? Comploose (talk) 19:06, 15 November 2012 (UTC)[reply]

No, the terms are orthogonal. A conditional probability can be interpreted in either a frequentist or Bayesian way. If X is the result of throwing a fair die, then

\mathrm {Pr} (X=1|X<4)={\tfrac {1}{3}}

. The frequentist interpretation of this is that if you toss a die many times, and consider only those time which resulted in less than 4, about a third of those will result in 1. The Bayesian interpretation is that if I toss one die, and observe that the result is less than 4, my subjective probability for it being 1 is updated to 1/3.

Conversely, if you are a Bayesian agent who has not yet observed any evidence and is working by your prior, your Bayesian subjective probability will not be a conditional probability. -- Meni Rosenfeld (talk) 19:34, 15 November 2012 (UTC)[reply]

I have question about prices and gains

Hi, well, my questions is this: I sold 75 boxes for a price of $58 a unit and my gain is 15%, what would be my total gain and gain per unit? Thanks