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November 15

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.

1. Who invented the
   Life International
   in 1962, maybe the first one of its kind?
2. 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]

Find a surjective group homomorphism ${\displaystyle S_{5}\rightarrow S_{4}}$ 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 ${\displaystyle S_{5}}$ 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]

Yup. There's a stronger fact that's worth mentioning: apart from a few (two/three?) exceptions, each symmetric group has only one proper, non-trivial normal subgroup: its alternating group. (And we know what size it has.) Icthyos (talk) 09:54, 16 November 2012 (UTC)[reply]

(Just to be precise: ${\displaystyle S_{5}}$ does have subgroups of order five, but no normal subgroups of order 5.) Icthyos (talk) 12:30, 16 November 2012 (UTC)[reply]

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]

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:

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

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

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.)

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

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 ${\displaystyle f(x)=x^{2}+ax+b}$ for both ${\displaystyle x+\Delta }$ and ${\displaystyle x-\Delta }$, and see that the result ${\displaystyle m=2x+a}$ is independent of ${\displaystyle \Delta }$ (for ${\displaystyle \Delta }$ 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]

Does there exist a transitive action of ${\displaystyle S_{5}}$ 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]

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

I can derive that ${\displaystyle IJ\subset I\cup J}$ (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 ${\displaystyle R=\mathbb {Z} }$ , ${\displaystyle I=5\mathbb {Z} }$ and ${\displaystyle J=7\mathbb {Z} }$, say. A typo in the statement, probably because someone was standing on their head.John Z (talk) 21:50, 15 November 2012 (UTC)[reply]

Yes, that is a typo. The statement is actually ${\displaystyle R=I+J\implies I\cap J=IJ}$. That's what you get for standing on your head. Anyway ${\displaystyle IJ\subset I\cap J}$ as well.--AnalysisAlgebra (talk) 03:58, 16 November 2012 (UTC)[reply]

Well, the other way is pretty easy too. Consider what R = I+J means for explicit elements of R, I & J, by the definition of the sum of two ideals. And then use that to get the other inclusion.John Z (talk) 05:53, 16 November 2012 (UTC)[reply]

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 ${\displaystyle \mathrm {Pr} (X=1|X<4)={\tfrac {1}{3}}}$. 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]

And is there a third or further position that is not one of these two (Bayesian or frequentist)? Comploose (talk) 00:51, 17 November 2012 (UTC)[reply]

None that I'm aware of. -- Meni Rosenfeld (talk) 17:16, 17 November 2012 (UTC)[reply]