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October 24

Groups of order 112

How can I show that a group of order 112 cannot be simple? It may have 7 Sylow-2 subgroups which may be non cyclic and also may have 8 Sylow-7 subgroups. The Sylow-7 subgroups will have trivial intersections but the same cannot be said about the Sylow 2-subgroups? Thanks.-Shahab (talk) 05:55, 24 October 2009 (UTC)

If the normalizer mod the centralizer of every 2-subgroup is a 2-group, then the group is 2-nilpotent, and the Sylow 7-subgroup is normal. Assuming the group is not 2-nilpotent, it must have a 2-subgroup whose automorphism group has an element of order 7. However, the only 2-groups of order dividing 16 with an automorphism of order 7 are the elementary abelian groups of order 8 and 16. However, a subgroup of order 8 or 16 is normal in a group of order 16, and since it is also normalized by the element of order 7, it is normal in the whole group. Since the automorphism group of an elementary abelian group of order 8 does not contain a subgroup isomorphic to the dihedral group of order 14, the Sylow 2-subgroup itself is normal. In other words, every group of order 112 is either 2-nilpotent or 7-nilpotent (or both). JackSchmidt (talk) 07:17, 24 October 2009 (UTC)

Another approach: If G is not simple it has 7 Sylow 2-subgroups. 7 is not congruent to 1 mod 4 so two of these subgroups, say P and Q have an intersection of order 8. The normalizer of the intersection contains both P and Q which implies it must be G and so the intersection is normal. This is all assuming you're not allowed to use Burnside p^aq^b.--RDBury (talk) 18:54, 25 October 2009 (UTC)

Subset of an intersection

How to prove $A\subseteq B\cap C\Leftrightarrow A\subseteq B\land A\subseteq C$? --88.78.236.7 (talk) 12:01, 24 October 2009 (UTC)

This question has the appearance of a homework problem. You would want to prove it directly from the definitions of "$A\subseteq B\cap C$", "$A\subseteq B$", and "$A\subseteq C$". — Carl (CBM · talk) 12:07, 24 October 2009 (UTC)

HInt: To show $\Leftrightarrow$, individually show $\Leftarrow$ and $\Rightarrow$. I think this problem is one of the very early examples in Hopcroft/Motwani/Ullman. --Stephan Schulz (talk) 09:00, 25 October 2009 (UTC)

Grades formula

It sounds like a homework assignment, but I'm the teacher, not the student. I've told the class that 'weekly assignments' will count towards 10% of their grade, that 'contribution to seminar discussion, and participation' counts for 30%, and that the 'final paper' counts 60%. Grades will be mainly between 6 (low) and 10 (high), though mainly between 7-9. What formula can I use to calculate an overall grade between 6-10, if a student earns, for example, an '7' on weekly assignments at 10%, a '9' on classroom participation at 30%, and an '8' for the paper and 60%. Thanks if you can make it simple for me. —Preceding unsigned comment added by 80.101.134.43 (talk) 15:26, 24 October 2009 (UTC)

The usual way to do this is to add the grades, each multiplied by its weight (how much it counts towards). Note that $1\%={\frac {1}{100}}$ so $10\%=0.1$, etc. So the final grade will be $0.1\cdot 7+0.3\cdot 9+0.6\cdot 8=8.2$, and you can round it to 8 if necessary.

See also weighted mean. -- Meni Rosenfeld (talk) 17:58, 24 October 2009 (UTC)

A simple spreadsheet will do the calculations for you. Dbfirs 18:08, 24 October 2009 (UTC)

I've always wondered - how do you standardise the scoring? How do you decide what is worth an 8 rather than a 7? How do you ensure that this marking is consistant with the marking another teacher would give of the same work? 89.242.151.212 (talk) 22:24, 24 October 2009 (UTC)

You get other teachers to mark a selection of it and check they agree with you (you can also make sure the distribution of marks doesn't change too much from year to year - if you find you've given out twice as many 10's this year than last then you probably need to go back and change some of them). That may not happen for everything but work that contributes to a university degree, for example, needs to be moderated to ensure consistent standards between universities. At least, that's the way it works in the UK, other countries may have other systems. It doesn't work perfectly, though - UK universities tend to moderate the work of and get their work moderated by universities of a similar standard, which means a 1st from a university between 5th and 15th in the country, say, will be about the same as a 1st from another university in that range, but it can be completely different to a 1st from one of the top unis or one of the "New Universities". --Tango (talk) 22:40, 24 October 2009 (UTC)

Usually, I think that it is undesirable to score work on a large scale (1-10 is reasonable, but 1-30 is not). The level of difficulty in standardizing scores depends on the subject, as well. For instance, in the SAT writing section, two different people will mark one's essay (on a scale of 1-6) and if the two marks differ by more than a point, a third marker will resolve the conflict. This ensures that scores will be more standard in nature. With homework assignments (at university, of course), it is sometimes preferred to give a homework assignment either "no credit", "half credit" or "full credit" rather than on a scale of 1-10. This ensures that there will not be one point differences between good students in terms of homework assignments; rather, exams will discriminate the hard-working students. --PST 00:08, 25 October 2009 (UTC)

On the other hand, I should add that some unprofessional markers give marks such as 7.2/10 or 7.3/10, and then discriminate between students, which I think is quite silly (such marks are, of course, randomly decided). --PST 00:13, 25 October 2009 (UTC)

In my experience, I have found that my instructors give an amount of credit that is determined by a rubric. For example, a rubric for a mathematics assignment might dictate that each question is worth one point. If there were ten questions on the assignment and eight were answered, the instructor would calculate the grade as follows: $8\cdot 1=8$ The total credit for the assignment would be 8/10 or 80%. For an assignment that involves writing an essay, instructors might have a more complex rubric, assigning specific point values for formatting, organization, spelling and mechanics, and so forth. —Dromioofephesus (talk) 02:01, 25 October 2009 (UTC)

Tango - academics submit suggestions as to who might be a suitable external examiner for a programme. In fields where there are many possibilities, this is likely to be someone they know personally and believe is likely to accept the role. They may or may not be from a university perceived as being of a similar standard. For example, I work at a red brick university, and our department has externals from other redbricks, 60s universities, new universities and the Open University. Perhaps that isn't the case everywhere, but external's reports are becoming increasingly visible, and it will start to reflect badly on departments if they only source externals from very similar departments. Warofdreams talk 12:49, 26 October 2009 (UTC)

Maybe things are improving, but fair comparisons between widely different universities can't be taking place at the moment - if they were then there would be a strong correlation between average A-level results of entrants and percentage of firsts handed out (since there is a strong correlation between A-levels and degree classifications for individuals within one uni), and there isn't. --Tango (talk) 19:22, 26 October 2009 (UTC)

I can't believe such an elaborate rigmoral is gone through to mark student essays - you just get someone subjectively choosing what they think the right mark is. Similarly for school marking. 89.240.47.104 (talk) 23:37, 26 October 2009 (UTC)

If the essay doesn't count towards the degree, then sure, but you need to moderate the marking for work that counts otherwise the degree is completely meaningless. --Tango (talk) 23:42, 26 October 2009 (UTC)

Unfortunately, people tend to have extremely poor judgements these days; especially when it comes to marking essays, and especially in low-profile instituitions. Furthermore, marking essays requires experience, and several points of view. If more than one marker marks an essay, it is more likely that the marks will be based on strong writing, rather than on the point of view the essay develops. --PST 07:01, 28 October 2009 (UTC)

Worse judgement than in the past? It would be hard thing to demonstrate, but I've heard some horror stories from a few decades ago. Warofdreams talk 20:44, 28 October 2009 (UTC)

I have frequently had to assess work myself, and all the time its been complete guesswork. I was (now doing something else) supposed to put them in various grades, but there was never any guidance as to what was required for each grade. People overestimate their abilities to judge things objectively - really we are at the mercy of various biases (see List of cognitive biases ) and halo effects. Read books about Human judgement to find out about this. 92.29.91.83 (talk) 20:13, 28 October 2009 (UTC)

It's a worry to hear that you were never given any guidance. But a feeling for which pieces of work fulfil a brief or answer a question better does tend to develop over time. While, no doubt, different markers look for slightly different things, where I work, dissertations and postgraduate work is routinely double marked and the marks then compared; a difference of 5% is common, a difference of more than 10% very unusual. Warofdreams talk 20:44, 28 October 2009 (UTC)

Equalities of means

As far as I recall H^2 = AG where H is the harmonic mean, A is the artithmetic mean, G is the geometric mean and "^2" means squared. Does this only apply to the averages of two numbers or to the averages of any number of numbers please? (If it is not "HAG" then it is G^2=AH ). 89.242.151.212 (talk) 22:18, 24 October 2009 (UTC)

it's G² = AH, but it only works for two numbers. You can see that it fails in the more general case since G² is not guaranteed to be rational, but AH always is. Rckrone (talk) 02:18, 25 October 2009 (UTC)

Presumably you meant to say that G² is guaranteed to be rational for two terms when the terms are rational? --COVIZAPIBETEFOKY (talk) 13:40, 25 October 2009 (UTC)

Yeah sorry, you're right. Rckrone (talk) 15:56, 25 October 2009 (UTC)

Would using the formula for averages of more than 2 numbers be at approximately correct? 84.13.180.244 (talk) 00:12, 28 October 2009 (UTC)

Yes, if the numbers are close enough to one another. The error is cubic in the deviations. -- Meni Rosenfeld (talk) 21:04, 28 October 2009 (UTC)

October 25

Hierarchical probability model

I had a homework question and I already did it and I think I got it right since it sort of tells you what the answer is going to be. But, I'm not sure I understand one step in there which is the crucial one.

I have a three-stage model with distributions $Y|N\sim bin(N,p)\qquad N|\Lambda \sim Poisson(\Lambda )\qquad \Lambda \sim gamma(\alpha ,\beta )$ and I am supposed to find the marginal distribution of Y. Well, my first thought was to find the marginal of N using the conditional on N and the marginal of $\Lambda$. However, that led to a very messy sum and I wasn't sure what to do with it. So, instead I tried finding the conditional of $Y|\Lambda$. What I did seemed to work out but I'm not exactly sure why. I did $f(y|\lambda )=\sum f(y|n)f(n|\lambda )$.

So, my question is pretty simple. Is that equation true? Assuming it is, the rest of the problem seemed to work out. Thanks for any help. StatisticsMan (talk) 02:31, 25 October 2009 (UTC)

I'm not an authority on probability, but I don't think that's true in general since it assumes that Y depends only on N and it's not otherwise affected by Λ. You would need to replace f(y|n) with f(y|n,λ). It works out here since you sum over all the λ's, but doesn't it get you the same result as doing it by finding the marginal of N?

$f(n)=\sum _{\lambda }f(n|\lambda )f(\lambda )$

$f(y)=\sum _{n}f(y|n)f(n)=\sum _{n}f(y|n)(\sum _{\lambda }f(n|\lambda )f(\lambda ))=\sum _{n}\sum _{\lambda }f(y|n)f(n|\lambda )f(\lambda )$

Rckrone (talk) 17:03, 26 October 2009 (UTC)

You've said:

$Y|N\sim {\text{bin}}(N,p)\qquad N|\Lambda \sim {\text{Poisson}}(\Lambda )\,$

and that is enough to entail that

$Y|\Lambda \sim {\text{Poisson}}(p\Lambda )\,$

by doing a bit of algebra (you need to remember the power series for the natural exponential function).

To be continued if I'm not too busy..... Michael Hardy (talk) 19:46, 26 October 2009 (UTC)

First use the law of total probability, then some algebra:

${\begin{aligned}&\Pr(Y=y\mid \Lambda )\\[10pt]&=E(\Pr(Y=y\mid N)\mid \Lambda )=E\left({\binom {N}{y}}p^{y}(1-p)^{N-y}\mid \Lambda \right)\\[10pt]&=\left({\frac {p}{1-p}}\right)^{y}E\left({\binom {N}{y}}(1-p)^{N}\mid \Lambda \right)\\[10pt]&=\left({\frac {p}{1-p}}\right)^{y}\sum _{n=0}^{\infty }{\binom {n}{y}}(1-p)^{n}{\frac {\Lambda ^{n}e^{-\Lambda }}{n!}}\\[10pt]&=\left({\frac {p}{1-p}}\right)^{y}\sum _{n=y}^{\infty }{\binom {n}{y}}(1-p)^{n}{\frac {\Lambda ^{n}e^{-\Lambda }}{n!}}\quad \left({\text{since }}{\binom {n}{y}}=0{\text{ if }}n<y\right)\\[10pt]&=\left({\frac {p}{1-p}}\right)^{y}\sum _{m=0}^{\infty }{\binom {m+y}{y}}(1-p)^{m+y}{\frac {\Lambda ^{m+y}e^{-\Lambda }}{(m+y)!}}\quad \left({\text{by the substitution }}m=n-y\right)\\[10pt]&=p^{y}\Lambda ^{y}e^{-\Lambda }\sum _{m=0}^{\infty }{\binom {m+y}{y}}(1-p)^{m}{\frac {\Lambda ^{m}}{(m+y)!}}\\[10pt]&=p^{y}\Lambda ^{y}e^{-\Lambda }\sum _{m=0}^{\infty }{\frac {\left((1-p)\Lambda \right)^{m}}{y!m!}}\\[10pt]&={\frac {p^{y}\Lambda ^{y}e^{-\Lambda }}{y!}}\sum _{m=0}^{\infty }{\frac {\left((1-p)\Lambda \right)^{m}}{m!}}\\[10pt]&={\frac {p^{y}\Lambda ^{y}e^{-\Lambda }}{y!}}e^{(1-p)\Lambda }\\[10pt]&={\frac {(p\Lambda )^{y}e^{-p\Lambda }}{y!}}.\end{aligned}}$

...and there you have the Poisson distribution with expected value pΛ. Michael Hardy (talk) 20:16, 26 October 2009 (UTC)

... next step. Now that we've established that

$Y\mid \Lambda \sim {\text{Poisson}}(p\Lambda ),\,$

we need to use the distribution of Λ to find the marginal distribution of Y. So:

${\begin{aligned}\Pr(Y=y)&=E(\Pr(Y=y\mid \Lambda ))=E\left({\frac {(p\Lambda )^{y}e^{-p\Lambda }}{y!}}\right)\\&=\int _{0}^{\infty }{\frac {(p\lambda )^{y}e^{-p\lambda }}{y!}}\,f_{\Lambda }(\lambda )\,d\lambda ,\end{aligned}}$

and then you need to put in the gamma density and evaluate the integral. If you know how to find the normalizing constant in the gamma density, plus a bit of algebra, then this one's pretty straightforward.

But you haven't yet told us which of the two commonplace parametrizations of the gamma distribution you're using. Michael Hardy (talk) 20:24, 26 October 2009 (UTC)

I'm still fairly sure that you can't find Y|Λ from only N|Λ and Y|N except in some exceptional cases. For example suppose a bunch of fair coins are flipped and Y, N, and Λ each measure the flip of one of the coins (not necessarily distinct). Suppose f(n|λ) is always 1/2 or in other words N and Λ are measuring different coins, f(y|n) is always 1/2 so Y and N are different coins. You can't say from this information whether Y and Λ measure the same coin or not. You would need Y|N,Λ instead of Y|N to say, but you don't have that. Rckrone (talk) 20:45, 26 October 2009 (UTC)

If you know the distribution of B given A, and that of C given A and B, then you can find the marginal distribution of C. The problem was expressed as (B given A), and (C given B). But the word "hierarchical" was used. I would take that word to imply that "C given B" was intended to mean (C given B and A). Michael Hardy (talk) 02:28, 27 October 2009 (UTC)

Actually here's an example I like better: Suppose a poll finds half of Democrats and half of Republicans approve of the war in Afghanistan, and that half the people who approve of the war and half the people who don't are satisfied with President Obama. That doesn't imply that only half of Democrats are satisfied with the president. Rckrone (talk) 23:36, 26 October 2009 (UTC)

As I mentioned before, the math above will get you to the correct result for Pr(Y=y) even though Y|Λ is not justified. You can get to the same place by using the formula for the marginal distribution twice and exchanging the order of summation:

${\begin{aligned}&\Pr(Y=y)=\sum _{n}\Pr(Y=y|N=n)\Pr(N=n)\\&=\sum _{n}\Pr(Y=y|N=n)\int _{0}^{\infty }f_{N|\Lambda }(n|\lambda )f_{\Lambda }(\lambda )d\lambda \\&=\int _{0}^{\infty }\left(\sum _{n}\Pr(Y=y|N=n)f_{N|\Lambda }(n|\lambda )\right)f_{\Lambda }(\lambda )d\lambda \end{aligned}}$

and then the dirty work for the sum is the same as above. Rckrone (talk) 21:57, 26 October 2009 (UTC)

Thanks for the help! Rckrone, you have two different formulas, one above with a double sum, and one below with a sum and an integral. Is the first one incorrect? Shouldn't the two be the same? Your work shows why that formula "works" even though it's not correct.

Michael Hardy, I am afraid I do not understand some of what you are doing. I do understand the basic idea, as if you use my original, but incorrect, formula, you can "show" that f(y|lambda) is Poisson(Lambda p). I believe you are doing it a correct way, but it may be a bit advanced for me :) For example, I do not get $\Pr(Y=y\mid \Lambda )=E(\Pr(Y=y\mid N)\mid \Lambda )$. I have seen a formula for expectations of conditional expectations, $EX=E(E(X|Y))$. But, I am not sure I have ever seen an expectation of a probability, and I do not understand why such a statement is true. I do understand that the probability you wrote is a random variable since it has N and not n. So, the expectation of it makes sense. But, why can you insert that N in there and put the expectation on the outside? It seems very different to me! Thanks! StatisticsMan (talk) 01:15, 27 October 2009 (UTC)

Oh, also, I did not tell you which Gamma I was using because I did not know there were two :) My book uses the one with the pdf

 $f(x|\alpha ,\beta )={\frac {1}{\Gamma (\alpha )\beta ^{\alpha }}}x^{\alpha -1}e^{-x/\beta },\qquad 0<x<\infty ,\qquad \alpha ,\beta >0$

Thanks again. StatisticsMan (talk) 01:19, 27 October 2009 (UTC)

OK,

$\Pr(Y=y\mid N=n)\,$

depends on n. Thus

${\begin{aligned}\Pr(Y=y\mid N=0),&\\\Pr(Y=y\mid N=1),&\\\Pr(Y=y\mid N=2),&\\\Pr(Y=y\mid N=3),&\\\Pr(Y=y\mid N=4),&\dots \end{aligned}}$

are different numbers. So

$\Pr(Y=y\mid N)$

is a random variable that is equal to

${\begin{aligned}\Pr(Y=y\mid N=0)&{\text{ if }}N=0\\\Pr(Y=y\mid N=1)&{\text{ if }}N=1\\\Pr(Y=y\mid N=2)&{\text{ if }}N=2\\{\text{etc.}}\end{aligned}}$

And as I said, see law of total probability. The expected value of this last-mentioned random variable is equal to the marginal probability that Y = y, i.e. to Pr(Y = y). Michael Hardy (talk) 02:03, 27 October 2009 (UTC) ...or to put it another way, let's say N is either 0, 1, or 2. Then

${\begin{aligned}\Pr(A)&=\Pr([A{\text{ and }}N=0]{\text{ or }}[A{\text{ and }}N=1]{\text{ or }}[A{\text{ and }}N=2])\\[10pt]&=\Pr(A{\text{ and }}N=0)+\Pr(A{\text{ and }}N=1)+\Pr(A{\text{ and }}N=2)\\[10pt]&=\Pr(A\mid N=0)\Pr(N=0)+\Pr(A\mid N=1)\Pr(N=1)+\Pr(A\mid N=2)\Pr(N=2)\\[10pt]&=E(\Pr(A\mid N).\end{aligned}}$

Michael Hardy (talk) 02:22, 27 October 2009 (UTC)

StatisticsMan: The sum version of the formula is for discrete variables and the integral version is for continuous variables. Other than that they're the same formula (see Marginal distribution). In the first post I really wasn't sure what "Poisson" and "bin" and "gamma" were so I didn't know which variables were discrete and which were continuous. I just used the sum version to show the generic case. Then Michael Hardy kindly demonstrated what they look like, so I was more specific the second time with Λ being continuous and N discrete. Rckrone (talk) 02:23, 27 October 2009 (UTC)

OK, here's what you'll need to know about integrals:

$\int _{0}^{\infty }x^{\alpha -1}e^{-x/\beta }\,dx=\Gamma (\alpha )\beta ^{\alpha }$

...to be continued..... Michael Hardy (talk) 02:38, 27 October 2009 (UTC) Continuing: "what you'll need to know about integrals" means

$\int _{0}^{\infty }x^{{\text{something}}-1}e^{-x/{\text{something else}}}\,dx=\Gamma ({\text{something}})({\text{something else}})^{\text{something}}$

and we will changed the "something" and "something else" as we proceed.

We had

$\Pr(Y=y\mid \Lambda )={\frac {(p\Lambda )^{y}e^{-p\Lambda }}{y!}}.$

So now

${\begin{aligned}\Pr(Y=y)&=E(\Pr(Y=y\mid \Lambda ))\\[10pt]&=E\left({\frac {(p\Lambda )^{y}e^{-p\Lambda }}{y!}}\right)\\[10pt]&=\int _{0}^{\infty }{\frac {(p\lambda )^{y}e^{-p\lambda }}{y!}}f_{\Lambda }(\lambda )\,d\lambda \\[10pt]&=\int _{0}^{\infty }{\frac {(p\lambda )^{y}e^{-p\lambda }}{y!}}{\frac {1}{\Gamma (\alpha )\beta ^{\alpha }}}\lambda ^{\alpha -1}e^{-\lambda /\beta }\,d\lambda \\[10pt]&={\frac {p^{y}}{y!\Gamma (\alpha )\beta ^{\alpha }}}\int _{0}^{\infty }\lambda ^{y+\alpha -1}e^{-(p+1/\beta )\lambda }\,d\lambda \\[10pt]&={\frac {p^{y}}{y!\Gamma (\alpha )\beta ^{\alpha }}}\int _{0}^{\infty }\lambda ^{y+\alpha -1}e^{-\lambda /(\beta /(p\beta +1))}\,d\lambda \\[10pt]&={\frac {p^{y}}{y!\Gamma (\alpha )\beta ^{\alpha }}}\cdot \Gamma ({\text{something}})({\text{something else}})^{\text{something}}\\[10pt]&={\frac {p^{y}}{y!\Gamma (\alpha )\beta ^{\alpha }}}\cdot \Gamma (y+\alpha )\left({\frac {\beta }{p\beta +1}}\right)^{y+\alpha }=\left({\frac {p\beta }{p\beta +1}}\right)^{y}{\frac {\Gamma (y+\alpha )}{y!\Gamma (\alpha )}}.\end{aligned}}$

...to be continued.... Michael Hardy (talk) 03:43, 27 October 2009 (UTC)

OK, that last expression should say:

$\left({\frac {p\beta }{p\beta +1}}\right)^{y}\left({\frac {1}{p\beta +1}}\right)^{\alpha }{\frac {\Gamma (y+\alpha )}{y!\Gamma (\alpha )}}.$

Canceling the gammas yields:

$\left({\frac {p\beta }{p\beta +1}}\right)^{y}\left({\frac {1}{p\beta +1}}\right)^{\alpha }{\frac {(y+\alpha -1)(y+\alpha -1)(y+\alpha -3)\cdots (\alpha +1)\alpha }{y!}}.$

This is:

${\begin{aligned}&{}\quad \left({\frac {p\beta }{p\beta +1}}\right)^{y}\left({\frac {1}{p\beta +1}}\right)^{\alpha }{\binom {y+\alpha -1}{\alpha }}\\[10pt]&=(1-s)^{y}s^{\alpha }{\binom {y+\alpha -1}{\alpha }}.\end{aligned}}$

...and there you have a negative binomial distribution. Michael Hardy (talk) 14:36, 27 October 2009 (UTC)

I think this all makes sense now. Sorry I didn't look at the link to total probability. The reason is, as is sort of eluded to in the article, I know the total probability law as that special case of the real total probability law. So, I thought I knew what it was so I didn't look. StatisticsMan (talk) 01:41, 28 October 2009 (UTC)

Pyramids

How many sides does a pyramid have? —Preceding unsigned comment added by 71.50.14.205 (talk) 05:37, 25 October 2009 (UTC)

Depends entirely on how many sides the base area has; see Pyramid (geometry). If you're thinking about the buildings, then pretty much all of them have a rectangular base. Also note that the question is somewhat ambiguous as it's not obvious whether the base itself should count as a "side". —JAO • T • C 06:30, 25 October 2009 (UTC)

"Side" isn't a good word to use about a 3D figure, as it can be interpreted as "face" or as "edge". The former is the likeliest meaning in the OP, which is unfortunate as in 2D the unambiguous meaning is "edge", as in "six-sided polygon".→86.160.55.126 (talk) 11:34, 25 October 2009 (UTC)

It might not be the most exact term, but I don't think there's really any ambiguity. "Side" is often used informally to refer to faces of a polyhedron (as in "a 6-sided die") and not to the edges. It might seem inconsistent to call the edges of a polygon and the faces of a polyhedron the same thing since they have different dimension, but they also both have codimension 1 relative to the whole polytope, which makes them analogous in a lot of ways. Rckrone (talk) 17:31, 25 October 2009 (UTC)

For what it's worth, I initially thought the OP meant edges. So there may not be ambiguity, but there might be a possibility of confusion. -- Meni Rosenfeld (talk) 19:58, 25 October 2009 (UTC)

Well I guess if it's unclear to people then I would be wrong about that. Rckrone (talk) 05:38, 26 October 2009 (UTC)

Surely, sides of a square-base pyramid refers to the four horizontal sides of the square base (which, I agree, are often associated (or even confused) with the four sloping triangular faces). Mathematically, of course, the square-base pyramid has five faces and eight edges, but colloquially one speaks of four sides which one can walk round. Dbfirs 10:50, 26 October 2009 (UTC)

Invariant Reliabilty Index (ϐ)

I am looking for an example of calculating the invariant reliability index (ϐ) for a given function. I can't find anything related in Wikipedia search. —Preceding unsigned comment added by 216.36.86.205 (talk) 19:02, 25 October 2009 (UTC)

Er, after some further searching, it looks like what I'm after is called the first-order reliability method (FORM). Does anyone know where I can find an example of applying the method to a given function? Thanks. 216.36.86.205 (talk) 19:41, 25 October 2009 (UTC)

October 26

Logs

I suck at anything having to do with mathematic logs, so you might have to dumb down your answers to my question... :-)

I want to convert some decibels to w/m². So, how do I do that? I have 110db = 10log(I/10e-12) So, how do I get that log out of there and solve for I? Dismas|^(talk) 04:53, 26 October 2009 (UTC)

Set both sides as an exponent of the log base (which is 10 in this case), so 10¹¹ = 10^{log(I/10e-12)}. Rckrone (talk) 05:32, 26 October 2009 (UTC)

And of course, exponentiation and logs cancel so you end up with $10^{11}={\frac {l}{10e-12}}$. More generally, sound intensity of x decibels is equivalent to $10^{x/10}I_{\textrm {ref}}\,\!$. -- Meni Rosenfeld (talk) 05:42, 26 October 2009 (UTC)

Did you really mean $10e-12=10^{-11}$, or rather $1e-12=10^{-12}$?

PS. I assume you are a logrolling master? -- Meni Rosenfeld (talk) 05:42, 26 October 2009 (UTC)

Now that I see how you do the math markup...To clarify, I meant this: $110=10\log \left({\frac {I}{10^{-12}}}\right)$ I know that you guys think that you're explaining this well but I just don't understand what you're saying.

And yes, I'm pretty good at logrolling, actually.  :-) Dismas|^(talk) 05:55, 26 October 2009 (UTC)

Let's use the letter t to refer to ${\frac {I}{I_{0}}}={\frac {I}{10^{-12}W/m^{2}}}$. You have $110=10\log t$, which means $11=\log t$. Now, "log" here means the common logarithm which is defined as: "$\log x$ is that number y such that $10^{y}=x$". Since $11=\log t$, it means that 11 is the number y such that $10^{y}=t$. In other words, $10^{11}=t$. Since $t={\frac {I}{I_{0}}}$, we have $I=tI_{0}=10^{11}I_{0}=10^{11}10^{-12}W/m^{2}=10^{-1}W/m^{2}=0.1W/m^{2}\,\!$. -- Meni Rosenfeld (talk) 06:10, 26 October 2009 (UTC)

The substitution of t helped a lot! Thanks! Dismas|^(talk) 06:16, 26 October 2009 (UTC)

Diagonalise these 2 QFs at once - must be going wrong somewhere

Hi there all:

I'm looking at the following question -

Find a linear transformation which reduces the pair of real quadratic forms $2x^{2}+3y^{2}+3z^{2}-2yz$ and $x^{2}+3y^{2}+3z^{2}+6xy+2yz-6zx$ to the forms $X^{2}+Y^{2}+Z^{2}$, and $\alpha X^{2}+\beta Y^{2}+\gamma Z^{2}$, where $\alpha$,$\beta$ and $\gamma$ ($\in \mathbb {R}$) should turn out to be integers.

Now that essentially means find a basis for which they're both diagonal, right?

Assuming so, I've found the eigenvalues/vectors of the first form okay, but for the second case I get something like $4,\,{\frac {3\pm {\sqrt {73}}}{2}}$ - is that right? I'm having trouble then calculating the eigenvectors in a non-incredibly-messy way assuming all's okay so far, but I think I might be approaching it wrongly to start with.

Thanks a lot for any help, the more I can get the better!

Spalton232 (talk) 19:08, 26 October 2009 (UTC)

The eigenvalues you specify are correct. The eigenvectors are $(0,1,1),\left({\frac {2}{3}}-{\frac {1}{6}}\left(3\pm {\sqrt {73}}\right),-1,1\right)$, but I have no input on how to find them effectively or other parts of your solution. -- Meni Rosenfeld (talk) 20:54, 26 October 2009 (UTC)

0.9...

There's been some discussion lately at the talk page of 0.999... disputing the accuracy of this article. I am not involved nor do I really plan to be, but I think it's probably better to discuss this here. Is there any way we can make people actually believe the article and its associating proofs? Is it worth investing time into this?

A transcript of the argument:

• First post
• A reply
• Another response

I think it's time we resolved this, personally. ceranthor 19:13, 26 October 2009 (UTC)

We created a page, Talk:0.999.../Arguments, so such arguments don't clog up the article talk page. --Tango (talk) 19:24, 26 October 2009 (UTC)

I don't think that particular discussion is really relevant until we get a response from the OP, explaining his/her stance on the issue, in light of the responses 76.103.47.66 and I have provided.

In general however, there is a clear trend of disbelief in the equality, and I'm honestly not sure how it can be resolved. The trouble is that any proof of the equality has one of two problems: either it is oversimplified and sweeps some important issues under the rug, and the reader may spot these issues, and come to believe that they are inherent flaws in accepting the equality, or it is too complicated for the non-mathematician to follow, and therefore useless for convincing them of the equality.

I will say that I like 76.103.47.66's argument; it's a nice visual proof, not too complicated, and it makes it very clear exactly how we go about deciding that the equality is true (we have the same amount of pie). Does anybody else have any interest in working on getting that argument added to the list of proofs? Obviously, we would have to make it a bit more concise, and maybe add a picture illustrating it.

By the way, shouldn't this discussion be on the 0.999... talk page? --COVIZAPIBETEFOKY (talk) 21:24, 26 October 2009 (UTC)

I don't know where this should or should not be discussed. That much said, I think that proofs are going to be beside the point for many people on this issue simply because they want to accept their wrong intuition that every decimal number representation is for a number distinct from all others. To reach such people, it is important to make it clear that all mathematical objects, even the simple ones with which they themselves are familiar, are abstractions. Then one can say that two points on a line--a kind of abstraction--that have no distance between them (a distance of zero) are really not distinct points and that this number issue is the same exact thing by the correspondence we all (?) learn in grade school. If one wants to go on and pique the curiosity of the person, then (after he or she is convinced and has been shown how a real mathematical proof works in addition) one can talk about how there is a way to make something containing our ordinary numbers but in which infinitely small numbers not equalling zero exist (in a subject called non-standard analysis), etc.Julzes (talk) 07:54, 31 October 2009 (UTC)

Domains wot change

Hi.

$A=\lbrace {y\in \mathbb {R} |y\neq -1}\rbrace$

$f:A\to \mathbb {R}$, with $f(x)={\frac {2x+2}{x+1}}$.

Firstly, can I do this?

$f(x)={\frac {2(x+1)}{x+1}}$,

$f(x)=2\,$

And if so, what's the deal regarding the domain? I assume the original domain is "preserved" so even though it looks like a harmless constant function, we are still denied $x=-1$. In which case, what happens when kinda do the reverse, and turn $f:\mathbb {R} \to \mathbb {R} ,f(x)=2$ into $f(x)={\frac {2x+2}{x+1}}$ by reasonable algebraic manipulations? Does it "force" a domain change, or is it an illegal move altogether? What about if $f:\mathbb {R} \to \mathbb {Z}$? —Preceding unsigned comment added by 94.171.225.236 (talk) 19:47, 26 October 2009 (UTC)

Those are good questions. I'll begin by saying that defining a function by a formula is not enough - you must specify what its domain is. However, usually we assume that if a domain was not specified explicitly, then an implicit specification is taken from the formula. Thus, the formula $f(x)={\frac {2x+2}{x+1}}$ is valid exactly when $x\neq -1$, so we use it to mean that function whose domain is A (at least, if it's clear we're talking about reals and not complexes or something).

By this convention, the formula $f(x)=2$ refers to a function defined on all of $\mathbb {R}$, which is of course different from the previous function which was defined only on A. By itself, the formula is not a valid representation of the original function. However, you still have the option to specify the domain explicitly - so the function defined as $f:A\to \mathbb {R} ,\ f(x)=2$ is the same function you started with.

For the reverse, the manipulation is illegal. From $f(x)=2$ you cannot deduce that $f(x)={\frac {2x+2}{x+1}}$ since the latter is not defined for $x=-1$ which is in f's domain. Of course, you can choose to move on from your original function $f(x)=2$ and discuss the function $f(x)={\frac {2x+2}{x+1}}$ instead - the functions are very similar, having a domain that differs by one point and being equal in their common domain.

Regarding $\mathbb {Z}$ - depends on whom you ask. Some define functions to include a specification of their codomain, while others don't. By the latter convention, it's the same function whether you say $f:A\to \mathbb {Z}$ or $f:A\to \mathbb {R}$ - the domain is the same and the values in this domain are the same. If you use the former convention, they are different as they have different codomains. -- Meni Rosenfeld (talk) 20:23, 26 October 2009 (UTC)

October 27

Is there an Erdős–Ko–Rado theorem for binary codes?

Hello,

I read Erdős–Ko–Rado theorem and I was thinking about alternative forms of this theorem. For binary codes of length $n$, I was thinking of a theorem like this: let $S$ be a set of binary codewords of length n, all being equal in at least one position. Then $|S|\leq 2^{n-1}$ and the bound can only be reached by taking all $2^{n-1}$ codewords with some fixed value in some position.

However, that turns out to be wrong, because for $n=3$, I can take $(0,0,0),(0,0,1),(0,1,0),(0,1,1)$ but also $(0,0,0),(1,0,0),(0,1,0),(0,0,1)$.

In fact, if the length $n$ is odd, I can always do that: I take one codeword, and switch in less than half of the positions, and I will get exactly $2^{n-1}$ codewords, all agreeing in at least one position.

Perhaps this is a very hard and open problem. I was just wondering if this problem has already been considered. Many thanks! — Preceding unsigned comment added by User:Evilbu (talk • contribs)

Okay, so you are asking if S is a set of binary codewords of length n, and each pair of codewords in S are equal in at least one position, then can we prove that $|S|\leq 2^{n-1}$, and when is equality reached ? Well, if s is a binary codeword then the only binary codeword of the same length that does not equal s in at least one position is the binary complement of s. So if we partition the binary codewords of length n into 2^n-1 pairs, each pair containing a codeword and its binary complement, then we can choose one codeword from each pair to give a qualifying set S with 2^n-1 members, and indeed there are 2^{2^{n-1}} ways to do this. But if we have a set with more than 2^n-1 members then, by the pigeonhole principle, it must contain at least one pair of complementary codewords, and this pair is not equal in any position.

For example, if n=3, we can choose either (0,0,0) or (1,1,1); either (1,0,0) or (0,1,1); either (0,1,0) or (1,0,1); and either (0,0,1) or (1,1,0), giving 16 ways of constructing a set S with 4 members. Gandalf61 (talk) 10:50, 27 October 2009 (UTC)

Or in other words : the entire set of codewords just falls apart into pairs, and I just have to take one from each pair. Okay, that's perfectly obvious now, I should have seen that. Thanks a lot! — Preceding unsigned comment added by User:Evilbu (talk • contribs)

Calculating a chord of a sphere

So for amusement, because it's too "early" to go to bed (long story), I'm trying to calculate how far it is from Portland, OR (USA) to Hyderabad (India), and I'm a little stumped as to how to do it. The chord article gives how to calculate the distance between two points in a circle, though calculating theta to plug into that equation is also not trivial (well, for me, I'm not much of a math person). SDY (talk) 12:11, 27 October 2009 (UTC)

You may be interested in this filed thread. The article great circle distance may be the most direct article to start reading. Well, you may well be slept by now. Or awake. Whatever... Pallida  Mors 14:44, 27 October 2009 (UTC)

Once you've realized how to do it, it is quite simple even starting from the fundamental principles, if you assume a spherical Earth with radius r. I assume you know the latitude φ and longitude λ of the places. Triples (r, φ, λ) are vectors in spherical coordinates from the center of Earth to the places, and their scalar product, which can be most easily calculated by first converting the vectors to cartesian coordinates (see this), is, on the other hand, r² cos α, where α is the angle between the two vectors. Finally, αr is the great circle distance sought for. — Pt _(T) 18:17, 27 October 2009 (UTC)

If you are actually looking for the chord length (sorry about my absent-mindedness), then the cos θ that goes into the formulae for chords is just my cos α. — Pt _(T) 18:23, 27 October 2009 (UTC)

I thought a chord meant a straight line (through the earth) between the two points, not an arc (great circle) on the earth's surface. Anyway, same idea. Calculate the rectangular coordinates of both points, then use the Pythagorean distance formula. 69.228.171.150 (talk) 06:41, 28 October 2009 (UTC)

Irrational exponents

A question about exponentiation:

We all know what a rational exponent is. But what about irrational exponents? I can't seem to find a definition anywhere? For example, in 2^π, do we take rational numbers approaching pi, put them into the function 2^x, and find the limit of 2^x as x approaches pi? ²³¹₉₁Pa (talk) 12:45, 27 October 2009 (UTC)

An irrational power of a positive real number can be defined by a limit process, as you suggest. Alternatively, once you have defined the exponential and logarithm functions, then you can use $b^{x}=\exp(x\ln(b))$ as a definition. See Exponentiation#Real powers. Gandalf61 (talk) 12:53, 27 October 2009 (UTC)

It's dealt with in Walter Rudin's Principles of Mathematical Analysis. Michael Hardy (talk) 14:41, 27 October 2009 (UTC)

Intuitively you can think of exponentiation with an irrational exponent as a limit process like that. In calculus, arbitrary powers are usually instead defined in terms of exponentiation, so $2^{\pi }=\exp(\pi \cdot \log 2)$, and exponentiation is defined as the inverse of the logarithm, where the log is defined as an integral (a different sort of limit process): $\log x=\int _{1}^{x}{1 \over x}\,dx$ —Preceding unsigned comment added by 69.228.171.150 (talk) 17:34, 27 October 2009 (UTC)

Power set of a Cartesian product

Is ${\mathcal {P}}(X\times Y)={\mathcal {P}}(X)\times {\mathcal {P}}(Y)$? --88.78.239.155 (talk) 14:07, 27 October 2009 (UTC)

No. There are lots of subsets of $X\times Y$ not of the form $A\times B$ for $A\subset X,B\subset Y$. Try an example where X and Y have size 2. Tinfoilcat (talk) 14:16, 27 October 2009 (UTC)

(after edit conflict) No. The members of ${\mathcal {P}}(X\times Y)$ are sets of pairs, whereas the members of ${\mathcal {P}}(X)\times {\mathcal {P}}(Y)$ are pairs of sets. For example, if X={a} and Y={b} then X x Y = {(a,b)} and the members of P(XxY) are the empty set and {(a,b)} whereas the members of P(X)xP(Y) are (empty set, empty set), ({a}, empty set), (empty set, {b}) and ({a}, {b}). Gandalf61 (talk) 14:19, 27 October 2009 (UTC)

...and just look at the closed unit disk in the plane. It's doesn't look anything like the Cartesian product of two subsets of the line: it's not a rectangle. Michael Hardy (talk) 14:22, 27 October 2009 (UTC)

Is ${\mathcal {P}}(X\times Y)\subseteq {\mathcal {P}}(X)\times {\mathcal {P}}(Y)$? --88.78.239.155 (talk) 14:23, 27 October 2009 (UTC)

Is ${\mathcal {P}}(X\times Y)\supseteq {\mathcal {P}}(X)\times {\mathcal {P}}(Y)$? --88.78.239.155 (talk) 14:24, 27 October 2009 (UTC)

You can work that out by considering the definitions of the sets on each side of the $\subseteq$ sign, as indicated by the comments above. Short, factual questions such as these, with no context, have the appearance of homework problems. — Carl (CBM · talk) 14:34, 27 October 2009 (UTC)

(ec) If you read the above answers carefully, they should settle these questions too. Also, take into account the cardinality of the power set of X. However, notice that you may make the identification ${\mathcal {P}}(X\sqcup Y)={\mathcal {P}}(X)\times {\mathcal {P}}(Y),$ where $X\sqcup Y$ denotes the disjoint union of X and Y, in the obvious sense that each subset of the disjoint union of X and Y is uniquely the disjoint union of a subset of X and a subset of Y. --pma (talk) 14:42, 27 October 2009 (UTC)

What is ${\mathcal {P}}(\{a,b\})\times {\mathcal {P}}(\{c,d\})$? --88.77.252.209 (talk) 14:54, 27 October 2009 (UTC)

P({a,b}) has four members; P({c,d}) has four members; so P({a,b}) x P({c,d}) has sixteen members, each of which is an ordered pair consisting of a subset of {a,b} and a subset of {c,d}. I suggest you list them, then compare them with the sixteen members of P({a,b} x {c,d}). Gandalf61 (talk) 15:02, 27 October 2009 (UTC)

If you know what power sets and Cartesian products are then you can work that our for yourself. If you don't, then what are you doing asking so many questions about that? --Tango (talk) 15:31, 27 October 2009 (UTC)

I agree. We should not be required to answer further questions you pose unless you show us that you understand what you are asking. Mathematics is not simply a list of factual information tied together; one must make connections between different concepts. In this case, it would be of use to establish connections between these various notions. Reading the article power set may be helpful. --PST 07:07, 28 October 2009 (UTC)

Is ${\mathcal {P}}(X\times Y)\cap ({\mathcal {P}}(X)\times {\mathcal {P}}(Y))=\varnothing$? --88.78.15.156 (talk) 15:42, 28 October 2009 (UTC)

Yes. As you were told above, the LHS contains sets of pairs, the RHS contains pairs of sets. They are different things. --Tango (talk) 16:30, 28 October 2009 (UTC)

If you encode pairs as sets, the two concepts might collide. If $(x,y)=\{x,\{x,y\}\}$, consider $a=(\varnothing ,\varnothing ),b=\{a,\varnothing \},X=\{\varnothing ,a,\{\varnothing \}\},Y=\{\varnothing ,a\}$. Then I think the intersection includes $(a,b)$. -- Meni Rosenfeld (talk) 20:42, 28 October 2009 (UTC)

BMO-style question (ish) on sums & products

Hi all -

my friend asked me to help him with this question in his work earlier today and I confess it is bugging me to death, I'm not sure how to approach it at all.

Given real numbers $a_{1},..,a_{n}$, s.t. $\sum _{i}a_{i}=0$ & $\sum _{i}a_{i}^{2}=1$, what is the maximum possible value of $a_{1}a_{2}+a_{2}a_{3}+...a_{n}a_{1}$? By considering the dot product of $(a_{1},..,a_{n})$ and $(a_{2},..,a_{n},a_{1})$ I can see such a sum must $\leq 1$, but I can't see any such $a_{i}$ which would actually make it equal to 1. Can anyone suggest a method to solve this problem?

Thanks a lot Mathmos6 (talk) 16:32, 27 October 2009 (UTC)

That does sound like a homework problem. Consider that the numbers don't all have to be positive. 69.228.171.150 (talk) 17:41, 27 October 2009 (UTC)

I doubt the OP imagined that the numbers are all positive and yet add up to 0. -- Meni Rosenfeld (talk) 19:46, 27 October 2009 (UTC)

Hints. You have a quadratic form $\textstyle (Ax\cdot x)$ in $\textstyle \mathbb {R} ^{n}$, represented by the symmetric matrix $A$ of order n, whose entries $\textstyle a_{i,j}$ are $1/2$ iff $\textstyle i-j=\pm 1\mod n,$ and $0$ otherwise.

Note that the hyperplane $\textstyle \{x\in \mathbb {R} ^{n}\,:\sum _{i=1}^{n}x_{i}=0\}$ is the orthogonal of the vector $u:=(1,1,..,1)$, which is the eigenvector of A corresponding to the maximum eigenvalue 1.

Therefore, your maximum is the second eigenvalue of A (in decreasing order).

Does this make sense to you?--pma (talk) 18:09, 27 October 2009 (UTC)

Is there a nice way to express the solution in the general case?

If anyone's interested, for $n=2,3,4,5,6,7$ the solutions are $-1,{\frac {-1}{2}},0,{\frac {-1+{\sqrt {5}}}{4}},{\frac {1}{2}},0.62348...$. -- Meni Rosenfeld (talk) 19:46, 27 October 2009 (UTC)

I think you skipped n=6 and I took the liberty of inserting it. For n=8 should be $\scriptstyle {\sqrt {2}}/2$... In fact if we work a bit on P_n(x):=2ⁿdet(x+A_n) we find a recurrence saying that P_n(x)/2=T_n(x)-(-1)ⁿ, whence we get all the eigenvalues of A_n using the T_n(cosθ)=cos(nθ).... but the OP and his friend already left... and these computations will be lost in time, like tears in the rain --pma (talk) 22:52, 27 October 2009 (UTC)

I follow all that, yes, so you end up with a very-nearly-tridiagonal matrix for A and you want the second largest eigenvalue - is there a nice way to evaluate det(A-tI) for a general n? Also, does this mean I was wrong about 1 being the maximum value for the sum, since you have eigenvalues greater than 1? Mathmos6 (talk) 23:28, 27 October 2009 (UTC)

Oh, apologies, I see you already said something about that (and I didn't leave, the wireless in my accommodation just broke for the day, just got it back on! :)) - I'm probably being stupid but what does T_n refer to, sorry? And where does my argument about the sum <= 1 fail? Thanks all for the help! Mathmos6 (talk) 23:30, 27 October 2009 (UTC)

well, if A_n were really tridiagonal, that is, if you remove the two 1's in the corners (1,n) and (n,1) (call the resulting tridiagonal matrix B_n), and you expand in the first column, you find a recurrence for det(B_n-tI) giving rise essentially to the Chebyshev polynomials of the second kind, U_n. The A_n have these two more 1's, but you can write det(A_n-tI) in terms of det(B_n-tI), using the linearity on the rows. A relation that you may find useful is U_n -U_n-2=2T_n, and you end up with the relation I wrote (I fixed it). Ask for details in case! cheers --pma (talk) 00:02, 28 October 2009 (UTC)PS: no, you were right: 1 is the maximum eigenvalue of A (the initial matrix with 1/2 and 0) but I think Meni used 2A to get rid of the factor 1/2. I think I used the initial A, but you better check...--pma (talk) 00:13, 28 October 2009 (UTC)

Exactly, I forgot to do the translation halfway through. Sorry for the confusion this has caused. I've fixed it now. -- Meni Rosenfeld (talk) 05:40, 28 October 2009 (UTC)

I think the eigenvalues are cos(2πi/n) with 0 ≤ i < n, so the largest less than 1 is cos(2π/n). If C is the transformation that shifts all the coordinates down one, then C has eigenvalues e^2πi/n. A = (C + C^-1)/2 so it has the same eigenvectors, and the eigenvalues would be (e^2πi/n + e^-2πi/n)/2 = cos(2πi/n). Does that seem correct? Is that what pma already said? I can't really tell. Rckrone (talk) 00:16, 28 October 2009 (UTC)

It seems so... what I found is that the eigenvalues of A_n are the roots of T_n(-x)=(-1)ⁿ --pma (talk) 00:36, 28 October 2009 (UTC)...that is of course the roots of T_n(x)=1 (that are exactly the cos(2πk/n), from T_n(cosθ)=cos(nθ) ). Rckrone's idea of using the (finite dimensional) spectral mapping theorem is definitely better to compute the spectrum of A--pma (talk) 00:46, 28 October 2009 (UTC)

This has been a great help, thankyou very much everyone! Think I've got my head around things now :) Cheers! Mathmos6 (talk) 00:45, 29 October 2009 (UTC)

Any method for solving the following equation...

Given the non-zero parameters: $a_{i},b_{i}$ (for every 1≤ i ≤6), is there any general method for solving the following equation:

$\sum _{i=1}^{3}{\sqrt {a_{i}x+b_{i}}}=\sum _{i=4}^{6}{\sqrt {a_{i}x+b_{i}}}$

HOOTmag (talk) 19:43, 27 October 2009 (UTC)

I think there isn't. Of course, you can find a solution numerically. -- Meni Rosenfeld (talk) 19:51, 27 October 2009 (UTC)

Is there any mathematical discipline dealing with related questions? HOOTmag (talk) 20:32, 27 October 2009 (UTC)

The general method would be to keep isolating one square-root term on the left-hand side and squaring out. You will eventually end up with a (large degree) polynomial, which can be tackled using the usual techniques (e.g. with Newton's method). You will have to check any solutions against the original, since the squaring operation can introduce roots. Zunaid 20:47, 27 October 2009 (UTC)

How can you be sure, this process "to keep isolating one square-root term on the left-hand side and squaring out...will eventually end up with a...polynomial"? HOOTmag (talk) 21:13, 27 October 2009 (UTC)

It's east to find the derivative of $\sum _{i=4}^{6}{\sqrt {a_{i}x+b_{i}}}-\sum _{i=1}^{3}{\sqrt {a_{i}x+b_{i}}}$ do just use Newtons method directly. Taemyr (talk) 21:36, 27 October 2009 (UTC)

I'm looking for an algebraic method. Sorry if this wasn't clear enough.HOOTmag (talk) 09:02, 28 October 2009 (UTC)

This doesn't work when you have more than 4 terms. In particular, if you put 5 square roots in one side and square them, you end up with 10 square roots.

Even if this was possible - if you want to use Newton's method anyway, why do the squaring in the first place? -- Meni Rosenfeld (talk) 21:21, 27 October 2009 (UTC)

Ah, I overlooked that. So it's not as trivial as simply squaring out to end up with a polynomial. To HOOTmag, there is (most likely) no algebraic solution to this problem, (probably) in the same way that there are no general algebraic solutions to polynomials of degree 5 or higher. Zunaid 09:27, 28 October 2009 (UTC)

I'm not sure it's "in the same way" (as you've put it), because solving quintic equations may have some method, in some (rare) cases, e.g. in quintic equations of the form: $(ax+b)^{5}=c$ (when given the non-zero parameters: a,b,c), etc. However I can't think of any equation of the form: $\sum _{i=1}^{3}{\sqrt {a_{i}x+b_{i}}}=\sum _{i=4}^{6}{\sqrt {a_{i}x+b_{i}}}$ (when for every 1≤ i ≤6 one is given the non-zero parameters: $a_{i},b_{i}$, such that every 1≤ j ≤6, not equal to i, satisfies: ${\frac {b_{j}}{a_{j}}}\neq {\frac {b_{i}}{a_{i}}}$), which may have - in some (even rare) cases - any method for solving that equation. HOOTmag (talk) 14:13, 28 October 2009 (UTC)

By my calculations, when you expand out the equation into a polynomial you get a degree 16 equation. A degree 16 equation isn't solvable in general but it may be that it's a special form that is solvable. I don't have a solution at the moment but I haven't seen anything to show it's impossible.--RDBury (talk) 16:02, 28 October 2009 (UTC)

How exactly do you expand out the equation into a polynomial? -- Meni Rosenfeld (talk) 20:12, 28 October 2009 (UTC)

I really don't know how you got a degree 16 equation. By my calculation, you'll never be able to get to any algebraic eqaution. HOOTmag (talk) 20:21, 28 October 2009 (UTC)

Yes, you can get a polynomial equation of which the original equation is a factor, by using Galois theory. We start with $\left(\sum _{i=1}^{3}{\sqrt {a_{i}x+b_{i}}}-\sum _{i=4}^{6}{\sqrt {a_{i}x+b_{i}}}\right)=0$

Each square root term is in a degree two extension of the polynomial ring in $x$. That means, $Y={\sqrt {a_{i}x+b_{i}}}$ is a root of $Y^{2}-(a_{i}x+b_{i})=0$. Since we have six square root terms in the original equation, it is in a degree $2^{6}=64$ extension of the polynomial ring in $x$.

To get a polynomial in $x$, we should multiply your expression by each of its conjugates-- each term ${\sqrt {a_{i}x+b_{i}}}$ is algebraic over the polynomial ring in $x$ and has the conjugate $-{\sqrt {a_{i}x+b_{i}}}$, which is the other solution to $Y^{2}-(a_{i}x+b_{i})=0$. (This is essentially the same trick as multiplying by the conjugate to rationalize the denominator in an expression like ${\frac {3}{2+{\sqrt {7}}}}$. In this case, we have 63 conjugates, since we are in a degree 64 extension.)

So, we need to multiply together all 64 terms of the form $\left(\pm {\sqrt {a_{1}x+b_{1}}}\pm {\sqrt {a_{2}x+b_{2}}}\pm \ldots \pm {\sqrt {a_{6}x+b_{6}}}\right)$, and the result will be a polynomial in x, with coefficients determined by the $a_{i}$ and $b_{i}$. I don't think it will be an easy equation to work with-- it will be a degree 32 polynomial. (In this case, we can obtain RDBury's degree 16 polynomial by omitting the first $\pm$, that is, only taking the 32 expressions with ${\sqrt {a_{1}x+b_{1}}}$ the positive square root. In general, though, we must multiply an expression by all of its conjugates to get a result in the base ring.)

Any solution to your equation will be a root of this polynomial (roots will be complex numbers in general), but most of the roots will be solutions to some conjugate of your equation; and I think it is possible that your equation has no solution. 140.114.81.57 (talk) 01:36, 29 October 2009 (UTC

That's pretty much the argument I had in mind but you don't really need the plus/minus in front of the first term. You can pair off factors in the form ${\sqrt {a_{1}x+b_{1}}}+Y_{1}$ with ${\sqrt {a_{1}x+b_{1}}}-Y_{1}$. So the product only needs 32 factors and the degree of the equation is 16. Not that it makes a difference, once you get a degree over 4 it's not solvable in general. But without knowing the Galois group, it might be a 2-group in which case you'd be possible to solve it with square roots.--RDBury (talk) 13:02, 29 October 2009 (UTC)

(after edit conflict). Substitute $y_{i}^{2}=a_{i}x+b_{i}$ to get 7 algebraic equations in the 7 unknown variables $x,~y_{i}$:

$y_{1}+y_{2}+y_{3}-y_{4}-y_{5}-y_{6}=0\,$

$a_{i}x+b_{i}-y_{i}^{2}=0\quad (i=1\cdots 6)$

Eliminate the 6 y's, using Buchberger's algorithm if everything else fails. The result is a polynomial equation in x alone. This concludes the algebra. The rest is numerical analysis. Bo Jacoby (talk) 02:00, 29 October 2009 (UTC).

October 28

Differential operators of order 4 - help me before I die of boredom

Hi there,

I want to show that the 4th order differential operator ${\mathcal {L}}=\sum _{r=0}^{4}\,p_{r}(x){\frac {d^{r}}{dx^{r}}}$, with $p_{r}$ real functions, is self adjoint iff $p_{3}=2p_{4}',\,p_{1}=p_{2}'-p_{4}'''$ - the only way I've seen of doing this on my course was for the Sturm-Liouville (2nd order) operator, and it involved nothing more than churning through 20 integrations by parts. Not only does this seem like pretty much the ugliest possible approach to the question, it is also long and tedious for order 2 and undoubtedly far more tedious in 4th order operators - can someone please tell me any more concise or cleaner way to solve this problem or at the very least link me to some sort of proof so I can spend my time learning something a little more beneficial rather than sitting up plugging things into "$\int y_{2}{\mathcal {L}}y_{1}=\int y_{1}{\mathcal {L}}y_{2}$"?

I don't mean to sound lazy and I certainly don't just want other people to do all my work for me, I'd just rather actually be learning something new instead of what seems to be little more than an exercise in checking how consistent my algebra is and I'm sure someone out there knows how to save me some time! :)

Thankyou very much for the help, Otherlobby17 (talk) 00:56, 29 October 2009 (UTC)

I understand your point, but if it's just a matter of characterizing when ${\mathcal {L}}$ is symmetric (say on ${\mathcal {D}}(\mathbb {R} )$ wrto the $L^{2}$ inner product), after all the plain computation is not that bad. With a bit of concentration you can just write down immediately what should be the result of: integrating by parts, using the Leibnitz' rule, and equating the integrands of the two sides of $\scriptstyle \int y_{2}{\mathcal {L}}y_{1}=\int y_{1}{\mathcal {L}}y_{2}$ :

$p_{0}=+p_{0}-p_{1}^{(1)}+p_{2}^{(2)}-p_{3}^{(3)}+p_{4}^{(4)}$

$p_{1}=-p_{1}+2p_{2}^{(1)}-3p_{3}^{(2)}+4p_{4}^{(3)}$

$p_{2}=+p_{2}-3p_{3}^{(1)}+6p_{4}^{(2)}$

$p_{3}=-p_{3}+4p_{4}^{(1)}$

$\textstyle p_{4}=+p_{4}$

that gives you the condition you wrote. As a consequence, the bilinear form associated with ${\mathcal {L}}$ via the $L^{2}$ inner product, is symmetric iff it is of the form

$\textstyle \int _{\mathbb {R} }u{\mathcal {L}}v\,dx=\int _{\mathbb {R} }\left(a_{0}uv+a_{1}u'v'+a_{2}u''v''\right)dx.$

If you already know this fact for other reasons (which?), you may also go back to the corresponding operator ${\mathcal {L}}$ and find the condition for being symmetric in another way (and quicker). The interesting question is, how all that generalize to any order. The answer should be: the order is even and you can freely choose the coefficients $p_{r}$ with even indices r, while the ones with odd r are determined. (I guess). --pma (talk) 15:13, 29 October 2009 (UTC)

Logarithmic differentiation

I'll make a comment rather than ask a question, but maybe after that I'll ask a question too, if I feel like it two minutes from now.

Today I found myself before a class of about 25 students, and one of them asked how to use logarithmic differentiation to find something like

${\frac {d}{dx}}\left((2x+1)^{5}(x^{4}-3)^{6}\right),$

and while I'm going through this someone asked a more intelligent question than what I expect of students in this particular course: doesn't this involve taking a logarithm of a negative number? Of course, one can say that

${\frac {d}{dx}}\log |y|={\frac {dy/dx}{y}},$

and then one is taking the logarithm of the absolute value of a negative number.

But there's still the question of dy/dx at points where y = 0. I expect to think about that when I separate variables in a differential equation and find dy/y, and I integrate that to get log |y|, but in such cases I deal with that by asking separately whether "y = 0 everywhere" is a solution.

I could tell the student that the derivative in a case like this must be a polynomial, and a polynomial is determined by its values at all points except the finitely many where it is 0. That still might leave a question of what happens in more general situations than polynomials.

OK, do I feel like ending with a question? Maybe I'll leave it as an exercise to say what questions should be asked at this point...... Michael Hardy (talk) 01:23, 29 October 2009 (UTC)

..... so now I notice that the book does say "If ƒ(x) < 0 for some values of x, then.....". I hardly ever pay close attention to what the book says..... Michael Hardy (talk) 01:39, 29 October 2009 (UTC)

A first remark that comes to my mind is that negative (or even complex) values are not a problem, for we don't need logarithms to define the logarithmic derivative of u as u'/u; nor we need them to prove properties like (uv)'/(uv) = u'/u + v'/v . The zero set of u is clearly a problem in general (e.g. if u is a constant I do not see what use could be u'/u ), but as in your example, a computation with logarithmic derivatives may lead to a perfectly correct result; for instance, one finds an equality that certainly holds at any point where all the functions involved in the computation do not vanish, but then the validity may be extended by continuity everywhere. Of course this works very well with polynomials and more in general with analytic functions. --pma (talk) 09:00, 29 October 2009 (UTC)

That's all very well but you try to explain all that to a class full of freshmen your instructor evals will tank. Pull the one student who asked aside and give them pma's explanation. (You might also see if you can get them become a math major.) For the other students just refer them to the book.--RDBury (talk) 13:39, 29 October 2009 (UTC)

To the 24 freshmen I would just say: the "logarithmic derivative" of u at x is defined as u'(x)/u(x), when u(x)≠0 and u'(x) exists. The name is suggested by the identity (log u)'=u'/u when u is also a positive function. As to the problem with the 25th, it is easily solved telling the class that he is going to convince you to explain everything in terms of operator calculus ;-) --pma (talk) 15:37, 29 October 2009 (UTC)

help this formulae identify itself

For a range data if we apply the formulae (X-min(X))/(max(X)-min(X))=?.What is this referred to as and what is the significance of doing such operation on data sets.(X represents any one data value.) 220.225.98.251 (talk) sci-hunter —Preceding unsigned comment added by 220.225.98.251 (talk) 18:53, 29 October 2009 (UTC)

it's 0 if X is the min; it's 1 if X is the max; it's 1/2 if X is half-way between the min and the max; it's 1/4 if X is 1/4 on the way from the min and the max. So it just tells you the relative position of X in the range between the min and the max in a scale from 0 (X=min) to 1 (X=max). Just a change of scale for jour data, that puts everything in [0,1]. --78.13.140.32 (talk) 19:10, 29 October 2009 (UTC)

Another term for this kind of operation is normalization. -- Meni Rosenfeld (talk) 20:26, 29 October 2009 (UTC)

Algebraic long division

At least, I think that's what I need. I have to divide $b^{3}-a^{3}$ by $b-a$ as part of deriving the variance of the uniform distribution from first principles. Can anyone help? It Is Me Here ^{t / c} 19:00, 29 October 2009 (UTC)

b²+ab+a² --78.13.140.32 (talk) 19:03, 29 October 2009 (UTC)

Thanks, but how did you get there? It Is Me Here ^{t / c} 19:05, 29 October 2009 (UTC)

You're welcome. How I did: because I knew. Check:

(b-a)(bb+ab+aa)=

b(bb+ab+aa)-a(bb+ab+aa)=

bbb+abb+aab-abb-aab-aaa=

bbb-aaa.

sorry for not writing the exponents; just too lazy to clik on the "Supersctipt" thing. Note: you can do as well (b⁷-a⁷)/(a-b) or any other exponent, in the same way (otherwise check polynomial long division and Synthetic division if you want more theory).--78.13.140.32 (talk) 19:22, 29 October 2009 (UTC)

Also, with some algebraic manipulation, ${\frac {b^{3}-a^{3}}{b-a}}$ matches the form of the sum of a geometric series, so you only need to work in reverse to find the corresponding terms of the series. Telescoping series is also related. -- Meni Rosenfeld (talk) 20:24, 29 October 2009 (UTC)

Factorizing small polynomials like that is a standard part of highschool algebra. If you're taking a serious probability course and you're not already reasonably skilled with that sort of thing, you may be in for a difficult time. Anyway, in general,

$a^{n}-b^{n}=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^{2}+\cdots +b^{n-1}).$

If you multiply it all out, you'll see how all the cross-terms in the middle cancel each other. 69.228.171.150 (talk) 21:07, 29 October 2009 (UTC)

First of all, we observe that if a=b then both those expressions are zero, this tells us that a-b is a factor of a³-b³, so we expect to get a polynomial as the answer (as opposed to a Laurent polynomial - one with negative exponents). a-b has both terms having degree 1 and a³-b³ has both terms having degree 3. That tells us we're looking for a polynomial with all the terms having degree 2. There are 3 possible such terms - a², b² and ab (don't forget that last one!). So, the answer will be of the form ma²+nab+pb². The easiest way to find m, n and p is just to multiply it out and equate coefficents - (a-b)(ma²+nab+pb²)=a³-b³. --Tango (talk) 22:22, 29 October 2009 (UTC)

The real refdesk answer, what the poster actually asked for, is the article polynomial long division. Meni mentioned it in his response, but this was the answer that should have been given right off. --Trovatore (talk) 22:24, 29 October 2009 (UTC)

The OP asked how to solve the problem and guessed that polynomial long division might be the best way. I disagree, I think the the method I gave is better in this case. Long division is good if you aren't expecting a perfect polynomial for the answer, but if you are then there are better ways. --Tango (talk) 23:35, 29 October 2009 (UTC)

I tend to think that what the poster asked for was exactly the plain result of $(b^{3}-a^{3})/(b-a)$, not general theories nor tricks... --pma (talk) 00:05, 30 October 2009 (UTC)

This is indeed the first thing the OP was given. The title and his request for clarification indicate he has at least some interest in the general technique. -- Meni Rosenfeld (talk) 06:06, 30 October 2009 (UTC)

Actually, the mention of polynomial long division was part of anon's answer. I was going to mention it before I noticed that anon beat me to it. -- Meni Rosenfeld (talk) 06:06, 30 October 2009 (UTC)

October 29

Piping a pump/filter for minimum number of valves

I am trying to figure what is the best way to pipe a pump and filter so that various operating modes can be achieved with the minimum number of valves. A valve is either open (allowing flow in either direction) or closed (blocking flow).

Using the valves I need to be able to direct flow so these 5 modes of operation that my pump and filter can be used in: 1. Filter (water from pool to be pumped FORWARD through filter and back to pool). 2. Circulate (water from pool to be pumped back into pool). 3. Back-wash (water from pool to be pumped BACKWARD through filter and out to seperate waste-pool). 4. Flush (Water from pool to be pumped FORWARD through filter and out to waste-pool). 5. Waste (water from pool to be pumped directly to waste-pool).

The pump can only pump in one direction. The water must pass the pump before the filter, this is because you cannot pull water thourgh the filter, only push it, I only have 1 filter and 1 pump.

The best I have been able to come up with is 5 valves. But I am hoping there is a way to do it with 4 or less?

       +---Pool--+
       V         |
      Pump       |
       V         |
       +-----+   |
       |     |   |
       X     |   |
       |     |   |
W-+-X--+     X   X
  |    |     |   |
  |  Filter  |   |
  |    |     |   |
  +-X--+-----+---+

X = valve
W = waste
V = pump direction

--Dacium (talk) 06:34, 30 October 2009 (UTC)