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January 1

Graph Theory

I was solving problems from a Graph theory text book and came across this problem: (1) Prove that removing opposite corner squares from an 8 by 8 checkerboard leaves a subboard that cannot be partitioned into 1 by 2 and 2 by 1 rectangles. (I assume the problem asks for removing only 1 pair of corner squares). (2) Using the same argument, make a general statement about all bipartite graphs. Can someone help me with this please. Thanks -Shahab (talk) 14:29, 1 January 2011 (UTC)

Explicitly making this a question about a checkerboard (rather than an arbitrary 2n by 2n grid) actually makes it easier, since the checkerboard comes pre-equipped with a helpful colouring. Algebraist 14:33, 1 January 2011 (UTC)

I keep thinking of Ike somehow... --Stephan Schulz (talk) 14:42, 1 January 2011 (UTC)

You think he saw everything as black and white? ;-) Dmcq (talk) 15:22, 1 January 2011 (UTC)

To expand on Algebraist's comments:

1) Opposite corners come in the same color on an 8×8 checkerboard, and initially there are the same number of black squares as white (32 each).

2) Therefore, removing one opposite set leaves you with a color imbalance (30 white squares and 32 black or vice-versa).

3) Since removing a 1×2 rectangle will always remove both a black and a white square, that will maintain the color balance. So, after one rectangle is removed, a full checkerboard would have 31 black squares and 31 whites, for example.

4) If you do the math on removing 1×2 squares from a checkerboard with one pair of opposite corners removed, you find you eventualy end up with only two black squares or two whites, which can't posibly form a rectangle, due to the way the board is laid out.

OK, so how do we generalize this ?

A) The general rule appears to be that you can't remove balanced color rectangles (or squares) from a larger imbalanced-color square (or rectangle), and end up with nothing.

B) Note that this doesn't necessarily mean that you always can remove balanced color rectangles (or squares) from a larger balanced-color square (or rectangle), such as removing 6×1 rectangles from a chessboard. At a minimum, the larger product must be a multiple of the smaller product. So, in our example, the 62 squares we had left did divide evenly by the 2 squares we removed at a time, while it would not divide evenly by the 6 squares in a 6×1 removal scheme. Of course, being a multiple, alone, isn't sufficient to ensure that the configuration allows them to be removed exactly. There's the color imbalance issue, but also the problem of physically fitting the removal patches on the board so they don't overlap or fall off an edge.

C) Think of which rectangular or square boards have opposite corners the same color (it helps to have a checkerboard in front of you when you do this, so here's one: chessboard). Now, can you come up with a formula to describe the relationship between the X and Y directions that gives you opposite corners the same color ? Hint: It's more complex than just X=Y.

D) Similarly, think of squares or rectangles that can be removed which have "balanced color". Is there any square or rectangle that has more blacks than white or vice-versa ? (And, if so, is it always imbalanced in the same direction, or not ?).

StuRat (talk) 16:11, 1 January 2011 (UTC)

January 2

Countable union of countable sets cannot have cardinality $\aleph _{2}$

Could anyone direct me to a proof, using Zermelo-Fraenkel axioms, that a countable union of countable sets cannot have cardinality $\aleph _{2}$? I've searched online but can't manage to track one down! Thankyou :) Simba31415 (talk) 03:24, 2 January 2011 (UTC)

Suppose ω₂ (the smallest ordinal of size $\aleph _{2}$) is a countable union of countable sets. If we had choice, we could simultaneously count all the sets to get an injection from ω₂ to ω×ω, which of course is impossible, since ω×ω is countable. Without choice, we can't do that, but what we do have is explicit well-orders for all the sets (since they're subsets of the well-ordered set ω₂). The Mostowski collapse lemma gives explicit bijections from these well-ordered sets to the corresponding ordinals. Since the sets are countable, these are countable ordinals, and so we have a injection from ω₂ to ω×ω₁, which is impossible since the latter set has cardinality $\aleph _{1}$. Algebraist 04:31, 2 January 2011 (UTC)

I really like this proof! 86.205.29.53 (talk) 23:30, 3 January 2011 (UTC)

Method of Characteristics

I am trying, without success, to understand the method of characteristics for solving PDEs. I am working with the following example.

"Consider $e^{x}u_{x}+u_{y}$ with $u(x,0)=cosh(x)$. The initial curve B is the x axis (x=t, y=0) and the initial data along b is h(t)=cosh(t). The characteristics satisfy

${\frac {dx}{ds}}=e^{x}$ and ${\frac {dy}{ds}}=1$, $x(0)=t$ and $y(0)=0$." (There is more to the problem but this should be sufficient for my question.)

What I cannot understand is where the initial conditions for x(0) and y(0) come from. I have looked through my notes and other examples countless times but just cannot see it. I have every confidence I am being incredibly dim but could someone please enlighten me? Thank you. asyndeton talk 20:16, 2 January 2011 (UTC)

The initial curve B is the x axis (x=t, y=0)... :) Hence, your initial conditions. 86.164.58.246 (talk) 12:07, 3 January 2011 (UTC)

I'm such a fool. Thank you. asyndeton talk 17:01, 4 January 2011 (UTC)

January 3

Percentages and Probability

Let's say you hold an election and the sample size is s₁. You achieve an approval rating of A%. Then, you analyse the results from randomly selected subsets of the electorate. Assume that you randomly selected s₂ people and they tell you how they actually voted.

• What is the probability of getting an approval rating of A% in this smaller sub-sample?
• What is the probability of the approval rating being within ±d% of the original approval rating?

— Fly by Night (talk) 00:19, 3 January 2011 (UTC)

If you want an exact result, you need to sum the values of the Hypergeometric distribution. Note that if $s_{1}$ and $s_{2}$ are coprime and $0<A<1$, then the probability of having exactly A is 0.

For an approximation, you can use the normal distribution with $\mu =A$ and $\sigma ={\frac {A(1-A)}{s_{2}}}$. -- Meni Rosenfeld (talk) 09:25, 3 January 2011 (UTC)

Thanks for that Meni. — Fly by Night (talk) 15:13, 3 January 2011 (UTC)

January 4

statistics

I'm having difficulty with my thesis. can "knowledge or awareness of people to statistics" be subjected to study? what statistical tool can be used? — Preceding unsigned comment added by Rionsgeo (talk • contribs) 02:06, 4 January 2011 (UTC)

If I understand you, you want to measure people's knowledge and awareness of statistics, using statistical methods. I suppose you could do a poll/quiz where you ask people how often they use statistics, then ask them to solve some statistics problems, and use that data to calculate standard deviations, confidence intervals, etc.

One suggestion for a refinement of your thesis: "Resolved, that people who are ignorant or distrustful of statistics tend to engage in statistically unhealthy habits and thus shorten their lives." You could design a way to test this assertion and either prove or disprove it. StuRat (talk) 05:43, 4 January 2011 (UTC)

Why does this remind me to Correlation from xkcd? – b_jonas 10:12, 7 January 2011 (UTC)

Permutations

How can I compute the number of ways to choose n elements in sets of size k (with replacement), so that no element occurs in each set more than x times? 70.162.9.144 (talk) 07:30, 4 January 2011 (UTC)

I don't know if this helps, but it should be equal to the coefficient of $y^{k}$ in $(y^{x+1}-1)^{n}/(y-1)^{n}$, if I followed your notation correctly. I'd guess there isn't any nice closed-form solution. Are you looking for a way to efficiently compute it? Eric. 82.139.80.114 (talk) 01:39, 5 January 2011 (UTC)

Sorry, could you explain what you mean by coefficient and how it is derived from $(y^{x+1}-1)^{n}/(y-1)^{n}$? 70.162.9.144 (talk) 04:37, 5 January 2011 (UTC)

${\frac {(y^{x+1}-1)^{n}}{(y-1)^{n}}}=\left(\sum _{k=0}^{x}y^{k}\right)^{n}$ is expanded by the Multinomial theorem. Bo Jacoby (talk) 12:56, 5 January 2011 (UTC).

Identifying a quotient

Let A be free abelian on 3 generators a,b,c and K the subgroup (n+m)a+(n-m)b+(m-n)c for all integers n,m. Is A/K just Z x Z/2Z? This seems like a very trivial task a computer should be able to do, is there any software to identify stuff like this? Money is tight (talk) 08:26, 4 January 2011 (UTC)

K is generated by the elements a + b − c and 2a, it thus consists of elements of the form na + mb + kc where k = −m and $n\equiv m{\pmod {2}}$. From this it follows easily that K = Ker(f), where f: A → Z × (Z/2Z) is defined by f(na + mb + kc) = (m + k, n + m mod 2). Since f is clearly onto, A/K is indeed isomorphic to Z × (Z/2Z).—Emil J. 12:41, 4 January 2011 (UTC)

Thanks, I'm used to doing "show/prove" questions and this question wasn't one of those, just needed some confirmation to check my understanding is correct. Money is tight (talk) 00:56, 5 January 2011 (UTC)

I think the convention is to use the symbol ⊕ to denote the direct sum of abelian groups. (Instead of using the direct product, which is often used with non-abelian groups.) So it would be normal to write Z ⊕ Z/2Z. If A₁, …, A_n are abelian groups then the direct sum A₁ ⊕ … ⊕ A_n is the set of n-tuplues (a₁, …, a_n), where a₁ ∈ A₁, …, a_n ∈ A_n, under the binary operation

$(a_{1},\ldots ,a_{n})+(a_{1}',\ldots ,a_{n}')=(a_{1}+a_{1}',\ldots ,a_{n}+a_{n}').$

This turns A₁ ⊕ … ⊕ A_n into an abelian group. — Fly by Night (talk) 13:43, 5 January 2011 (UTC)

January 5

Special magic squares

I'm reading a book that has a chapter on magic squares, and it gives the following special magic square:

 5 22 18
28 15  2
12  8 25

If you write the names for the numbers out in English and then count the number of letters in each name, you get another magic square:

 4  9  8
11  7  3
 6  5 10

It then says that, for totals of less than 200, English has seven of these squares, while French only has one. What are they? --75.60.13.19 (talk) 03:08, 5 January 2011 (UTC)

non-trivial irreducible character of primitive permutation character is faithful?

Hello,

I am struggling with what should be an easy exercise...

let G be a group acting primitively on a set $\Omega$. Let $\chi$ be the permutation character( let's say over $\mathbb {C}$). Hence $\chi$ maps every$g\in G$ to the number of elements in $\Omega$ it fixes.

Let$\chi _{1}$ be any irreducible constituent of $\chi$, different from the trivial character. Prove that $\chi _{1}$ is faithful (i.e. the corresponding representation of $G$ maps only the trivial element of $G$ to the trivial matrix).

I think it should be easy, but it appears I am missing a crucial observation. I already know why primitivity is important, because otherwise the non-faithful permutation character on the blocks of imprimitivity would be contained in the character.

Groups of order 24

Do all the 15 groups of order 24 have a subgroup of order 12? How can I prove or disprove this.-Shahab (talk) 09:41, 5 January 2011 (UTC)

Take a look at the Sylow theorems article. As the article says, they are "a collection of theorems... that give detailed information about the number of subgroups of fixed order that a given finite group contains.". Since 24 = 2³⋅3, Burnside's theorem tells us that G is solvable. There's lots more information to be had in the Solvable group article. — Fly by Night (talk) 15:48, 5 January 2011 (UTC)

I think that the semidirect product $G=C_{3}\ltimes Q_{8}$ has no subgroup of order 12, where Q₈ is the quaternion group, and C₃ acts on Q₈ by cyclic permutation of i, j, k. Any subgroup of G of index 2 would be normal, hence it would induce a nontrivial homomorphism f: G → C₂. However, f vanishes on C₃ (as it has order 3) and on at least one of i, j, k (as ij = k), and therefore on all of i, j, k (as they are conjugated in G, and C₂ is abelian). Thus f is trivial, a contradiction.—Emil J. 16:00, 5 January 2011 (UTC)

This is a central extension of A₄ by C₂. A₄ has no subgroups of order 6 and this implies that the extension has no subgroups of order 12. This is also the only group of order 24 that has no subgroup of order 12. I got this by checking generators and relations for all 15 groups, these are well known, but I suppose it wouldn't be too hard to prove it from scratch using Sylow etc.--RDBury (talk) 16:54, 5 January 2011 (UTC)

1+1=2

What were the definitions and axioms used in Principia Mathematica that required such a long proof of 1+1=2 and other basic arithmetic facts? A link to a page or website describing the foundations is sufficient; I just don't know where to find such a reference. --24.27.16.22 (talk) 15:54, 5 January 2011 (UTC)

At the very end of that page about Principia Mathematica you can find proposition 54.43 which is that 1+1=2 validated using a modern theorem checker Metamath. Dmcq (talk) 16:06, 5 January 2011 (UTC)

Qualitatively speaking, the issue is that the axioms are very low-level. For instance, 1+1=2 does not require proof if you are using the Peano axioms. But if you don't want to do that, you could start with Zermelo_Fraenkel_set_theory and derive the Peano system from there. This would be a much longer proof than that quoted on the PM page. Both Peano and ZFC schemes pre-date PM, and Whitehead even used much of Peano's notation. I think that PM axioms are lower-level / weaker than ZFC, but hopefully someone else can elaborate on that. I can't easily find an short itemized list of axioms in the PM, but you can download the whole book through google books here [1]. SemanticMantis (talk) 17:03, 5 January 2011 (UTC)

If we start with ZFC, and identify natural numbers with finite cardinals and define addition as a disjoint union of sets (using some silly trick like A+B = (Ax∅)∪(Bx{∅})), would a complete formal proof of 1+1=2 require a similar length? --24.27.16.22 (talk) 17:49, 5 January 2011 (UTC)

No, but that's an awful lot of "if"s. Russell and Whitehead were working at a time when there was no settled notions of things such as a logical theory, and so they had to build up everything from the very ground. Many tricks and tools that would be used in a modern work to simplify and streamline the presentation simply hadn't been invented yet (or had been used once or twice but their general applicability not appreciated). Furthermore, the theory they designed is/was markedly more cumbersome to work with and reason about than ZFC, and as a result didn't really catch on.

Also, nobody's saying that all of the 360 pages that preceded *54.43 were necessary prerequisites to that particular proposition. As a comparison with a newer text, Mendelson Introduction to Mathematical Logic (4th ed., 1997) reaches page 258 before it defines "$X{+_{c}}Y\mathrm {~for~} (X\times \{\emptyset \})\cup (Y\times \{1\})$". However, that includes 70 pages that develop formal number theory from the Peano axioms and are not used for the axiomatic set theory, so "number of pages before addition is defined" is not really a meaningful metric. –Henning Makholm (talk) 22:08, 5 January 2011 (UTC)

Does anyone know any source where real everyday mathematics is developed formally in ZFC? By real math I mean calculus, algebra, analysis etc. Money is tight (talk) 23:22, 5 January 2011 (UTC)

http://us.metamath.org/mpegif/mmset.html 67.122.209.190 (talk) 08:50, 6 January 2011 (UTC)

What constitutes a valid solution?

One of my profs made an interesting remark today. He was doing an example on the board, and the answer came out to $x={\sqrt {2}}$. But he stopped, and said that he didn't feel this was a complete solution, and that the fullest answer possible would be to say that $x=1.4142...$.

As an example, he said, suppose we had to solve the equation $x^{3}=4$. Saying that $x={\sqrt[{3}]{4}}$ is just tautological. The only meaningful solution would be to write out cube root of 4 as a decimal expansion.

At first I agreed, but thinking about it now, I'm not so sure. A decimal expansion of a number is, by definition, the expression of the number as an finite or infinite series of fractions with a denominator of a power of ten. How is this any more valid of an answer? —Preceding unsigned comment added by 74.15.138.87 (talk) 16:56, 5 January 2011 (UTC)

What constitutes a "complete solution" in mathematics is somewhat arbitrary, and most professors would be perfectly happy to accept $x={\sqrt {2}}$ in my experience, since that is the most precise answer one can provide. It's really not clear exactly what your professor is looking for in a complete solution, except possibly that he expects final answers to be written out in decimal notation? --COVIZAPIBETEFOKY (talk) 17:14, 5 January 2011 (UTC)

√2 is exact, whereas the decimal form is approximate. Of course all solutions of equations are in a sense tautological. Suppose the equation had been

$3x-5=0.\,$

Would he object that x = 5/3 was "just tautological? Michael Hardy (talk) 18:30, 5 January 2011 (UTC)

Right on Michael. Much of math can be seen as a search for tautologies. In my experience, "just a tautology" is used as a dismissal much more commonly in fields outside of math. SemanticMantis (talk) 19:02, 5 January 2011 (UTC)

Incidentally: I took two classes taught by Alphonse Vasquez, and remember he used to say that everything that's been proven in math is tautological, and if one doesn't see something as such then he hasn't sufficiently wrapped his head around it.—msh210℠ 19:16, 5 January 2011 (UTC)

When I teach math, I ask students for exact answers. Partially this is to make it easier on the grader, and partially because I think it is important to grasp that root two cannot be expressed exactly in decimal notation. But this is just personal preference. Also, asking for decimal approximations implicitly encourages students to use calculators when they are not required, and I believe this is counterproductive to really learning math. --But on to your prof's comments. I disagree completely that the decimal approximation is the "fullest possible answer", but perhaps this is not a direct quote from the instructor. Rather than talk of 'meaningful' or 'valid', we may consider whether an answer is *informative*. Consider an application where two quantities are to be compared. If solved exactly, it is not easy to see how ${\sqrt[{5}]{2}}$ compares to ${\sqrt[{10}]{5}}$. However, it's quite easy to see that 1.1487... < 1.1746... So really, the best form for an answer depends on why you're quantifying something in the first place. SemanticMantis (talk) 18:55, 5 January 2011 (UTC)

There's an important point that my prof made that I forgot to mention. In the final solution, he doesn't want decimals; he wants radicals, because decimals aren't exact. His point was that the only reason $x={\sqrt {2}}$ is an acceptable answer is because someone could look up the decimal expansion to the desired precision. Presumably, he also means to say that if mathematicians were dumber, and had made the notation $x={\sqrt {2}}$ but couldn't figure out how to expand it, then $x={\sqrt {2}}$ would be meaningless. But I don't see why decimal representation should have validity than other representations. At the same time, if we accept, as Michael Hardy said (and which I agree with) that all solutions are tautological, then that would mean that when we say $x={\sqrt {2}}$, we are really saying "I have found that the solution to such-and-such equation also happens to be the positive solution to the equation $x^{2}-2=0$". I guess this makes sense, but it makes the whole business of solving equations kinda...arbitrary, no?74.15.138.87 (talk) 20:53, 5 January 2011 (UTC)

It depends why you're solving something. Do you need to know how many people it will take to change your lightbulb? Then you need a decimal representation (actually, the ceiling of one, usually). Do you need a number you can substitute into another expression? Then a form like${\sqrt {2}}$ is usually best. Do you need it as an element of ℂ? Then you want (perhaps) ${\sqrt {2}}+0i$. Etc.—msh210℠ 21:00, 5 January 2011 (UTC)

That's very dismissive of tautologies. I try and make everything I say logical and as nearly tautological as possible. Like for instance 'If I don't get some sleep I'll never wake up in the morning' ;-) Dmcq (talk) 21:30, 5 January 2011 (UTC)

I grappled with the same problem when initially introduced to square roots and logarithms at school (and arcsin...in fact, ANY inverse function). It seemed retarded and meaningless to say that the solution of $10^{x}=2$ is $x=\log(2)$ (decimal log obviously). What's the point of just inventing notation and calling these answers "solutions"? They are no more useful than the original equations. In a sense, ALL inverse functions are just tautologies, useless for ACTUALLY solving anything. :That was until I reached university and learned about Taylor series, which allows you to actually compute and evaluate those expressions in a meaningful way. That was when it all made sense. Zunaid 12:20, 6 January 2011 (UTC)

Math and prejudice

How many cases should you consider until you come up to the conclusion that an ethnic group has this or that feature? For example, if you take nations with 300 millions or 100 millions, is my personal experience of 200 interactions each year enough? Quest09 (talk) 17:03, 5 January 2011 (UTC)

There's a question above about elections that is related. What if the 200 people you meet all had a property that no-one else in the population had? I guess you want to answer the following question: If p% of a sample has a certain property, then what's the probability that (p±d)% of the whole population has that property. You need to decide what percentage of the sample/population needs to posses a property before you call it a characteristic. Take a look at my question, and Meni's answer, here. — Fly by Night (talk) 17:19, 5 January 2011 (UTC)

This is a question of statistics. A simple random sample of 1000 or so people is generally enough to establish with reasonable certainty the rough proportion of people in a given population who possess a feature/hold a particular opinion/etc, independently of the size of the population (assuming only that it is significantly larger than 1000). Your personal experiences most likely do not constitute a simple random sample, and therefore cannot be used for this purpose. --COVIZAPIBETEFOKY (talk) 17:24, 5 January 2011 (UTC)

What COVIZAPIBETEFOKY said. We all have social interactions strongly biased by our income, locality, profession, etc. Our own experience is never a good source from which to extrapolate to the general population. See Sampling bias#Historical examples for some famous cases where samples of millions that were not truly random and unbiased, were insufficient. Ray^Talk 18:35, 5 January 2011 (UTC)

The Real Answer

Let me put it this way. Let's say your IQ is 300, and you've just made a handful of major breakthroughs in your chosen field, however you are just an undergraduate at a huge state school. Let's say that you are able to prove yourself to be an extremely valuable researcher to a professor at your University, if he will speak to you for 10 minutes. Then he would be convinced by your ideas, be instantly swayed, and want to publish with you. As a result of this, you would be able to get admitted to the graduate program of your choice, even if you did nothing more than flesh out the ideas you just published as an undergrad, you would be set to get tenure based on that, if not at Harvard, then at least in some respectable state school such as the one you're attending. There's just one little problem: your state school is in Misouri, has low standards of admission, you are of a minority race, and the professor has had a LOT of experience with semi-literate members of that minority!! He might refuse the 10-minute interview on that grounds alone, just from remembering your face among the 300 faces he teaches at any one time!! So, let me ask you the question this way: how many members of your race that he had experiences with, who were semi-literate and had an IQ of more like 60 (one fifth of yours), would make you say: You know what, he shouldn't waste 10 minutes on an interview with me, it's just not a reasonable request. 100 such people? 1000? A million? How about if you're Indian, and there are one BILLION people who are all different from you, and you're the only Indian who can do Italian opera in all the world? Would you agree that the director of La Scala should refuse to even listen to you on that basis? 87.91.6.33 (talk) 19:05, 5 January 2011 (UTC)

I would argue that such an interview should never be refused. However, there are cases where the "judge" is in such demand that he can't spend the time, so then underlings should be enlisted to do a "pretest", to determine if you are anything worth bothering the "judge" about. StuRat (talk) 21:56, 5 January 2011 (UTC)

Exacrtly. the OP should come to the same conclusion, and, therefore, realize that he should not become racist even after a billion examples confirming his suspicions about a group. 87.91.6.33 (talk) 22:02, 5 January 2011 (UTC)

(ec) Please, there is no need for emotional language; just state your assertion clearly and with justification (and without fallacies like appeal to consequences).

The question Quest09 asked ("is my personal experience of 200 interactions each year enough [... to] come up [with] the conclusion that an ethnic group has this or that feature") is a simple question of statistics for which COVIZAPIBETEFOKY gave a direct answer ("Your personal experiences most likely do not constitute a simple random sample, and therefore cannot be used for this purpose."). Eric. 82.139.80.114 (talk) 22:05, 5 January 2011 (UTC)

Yes, this was a question about mathematics in the math RD. I was not asking about moral implications and actually not even thinking about discriminating people. Quest09 (talk) 23:04, 5 January 2011 (UTC)

Of course there also is an implicit assumption about why the interview was rejected. Maybe the professor can only grant so many interviews and cuts it off at an arbitrary number. Or maybe he grants no non-class-related interviews to undergrads at all. Or maybe he has already seen common misconceptions during the initial request for the interview. All of these are arguably beyond the realm of mathematics, of course.Unless you consider "There are only 16 working hours per day, if I grant one interview to every of my 600 students I'll never finish that research" as maths ;-) --Stephan Schulz (talk) 10:41, 6 January 2011 (UTC)

600 student won't ask you for an interview. And even if they do, if you spend 15 minutes with each, that would only make 30'/day each year. Quest09 (talk) 12:10, 6 January 2011 (UTC)

With 600 what I'd do is I wouldn't even bother reading the first line of the application for most. I'd do a random draw of a selection to check through and then winnow down to an even smaller number to interview. Actually I think interviews are very overrated so its mainly to eliminate the unsuitable rather than to pick the best. Dmcq (talk) 13:15, 6 January 2011 (UTC)

let me spell it out for you guys

"the conclusion that an ethnic group has this or that feature" is the definition of racism. Sorry. That thought is the definition of racism. In other words, this is a question about what level of statistical confidence justifies racism. The answer is: none. Even if you have a hundred billion examples of a member of an ethnic group with a certain feature, you still can't come to "the conlcusion that an ethnic group has this or that feature". Is that clear enough for you? How about this way: I grew up in a very poor part of Boston. I met literally thousands of black kids who were way below the required level in their grade. How many should I have met before I concluded that a black kid has features that make them, say, not qualified for a Presidential-track education? The answer is, there is no such number. (Math: Not a Number, NaN). Because even if you have 300,000,000 black kids who can't be president, because they're too stupid and all the ritalin in the world would not make them smart enough: it only takes one. You can never induce the rule. The rule is the definition of racism. Clear enough for you? 87.91.6.33 (talk) 20:38, 6 January 2011 (UTC)

in yet different terms, an ethnic group can't have any features (besides the tautological ones*). Except in the mind of a racist. A racist can list dozens of features for any given ethnic group. Go try a racist sometime, I am not making this up. 87.91.6.33 (talk) 20:43, 6 January 2011 (UTC)

* tautological = obviously if you group people in races based on criteria, the "race" will have member meeting that criteria... but this says nothing about them, and only about you and your grouping choices -- it's "begging the question".

It's not racist to believe or state a fact about demographics if there's truth behind it. Chances are, however, that the true statement will not be a universal; it will merely be that a certain percentage of the population shares a characteristic, and that this percentage happens to be unusually high compared to other demographics. Also, statement of a fact does not necessitate any particular response, positive or negative, to the fact.

For instance, it is definitely true that a higher percentage of black people in the United States are below poverty line than the US population as a whole. This does not necessarily mean that black people are somehow inherently poor, and that we should discriminate against them for being incompetent or, at the other end of the spectrum, that we should send more money to the cause of improving the welfare of black people; it is merely a statement of fact.

We may then speculate as to whether black people are in some way rendered incapable of finding and keeping a self-sustaining job as a consequence of genetic predispositions not directly/obviously related to the color of their skin, or if there are social and environmental pressures that cause them to fall below the poverty line, and they could have done better if they had been brought up differently. I think most people agree that the latter explanation is the more significant factor, in this case. We would also want to speculate what the proper response would be to encourage improvement in the welfare of black people, a problem which remains unsolved.

It is also important to be able to make observations about various characteristics of a population in order to better understand a variety of phenomena. For instance, again in the US, non-whites as a direct percentage are more likely to vote democrat than whites are. But if you correct for socioeconomic status, which is also seen to affect a person's vote, you actually find that non-whites are more likely to vote republican than whites are. Direct percentages are misleading, but you can't see this until you are willing to take in 'racist' data. --COVIZAPIBETEFOKY (talk) 02:40, 7 January 2011 (UTC)

It is important to separate what is true from what we believe. For example, I know it is true that black Americans are dumber -- have a lower IQ -- than white Americans. See the book called "The Bell Curve". However, even though I am aware of this fact, I do not believe it. I do not believe that black Americans are dumber than white Americans. Why? Because I'm not a racist. It's that simple guys. When it comes to groups of people, you simply can't internalize statistically "true" statements about that group, unless you're a racist. Only a racist would agree that black Americans are dumber than white Americans. Your very first sentence "It's not racist to believe or state a fact about demographics if there's truth behind it" is, simply false. 87.91.6.33 (talk) 11:32, 7 January 2011 (UTC)

January 6

Tetrahedral angles

If I take a regular tetrahedron and draw segments from each of the vertices to the center, what's the angle between two of those segments? --75.60.13.19 (talk) 00:54, 6 January 2011 (UTC)

If you look at the Tetrahedron article, i.e. click here, then you'll find out everything you need to know, any many things you don't. — Fly by Night (talk) 01:33, 6 January 2011 (UTC)

Learning to read and write proofs

What books would you recommend for learning the general techniques of mathematical proofs? 74.14.111.188 (talk) 07:17, 6 January 2011 (UTC)

I suppose it really depends on the level of mathematics you're concerned with, but one course I found very helpful in the early years of my undergraduate studies used the textbook "A Transition to Higher Mathematics," by Smith, Eggen, and St. Andre. I found it be very illuminating at the time, particularly in regards to learning methods of proof. Nm420 (talk) —Preceding undated comment added 15:41, 6 January 2011 (UTC).

How to read and do proofs by Daniel Solow is another one to look at. There are probably others in the same vein.--RDBury (talk) 00:54, 7 January 2011 (UTC)

Websites first postings

I need to find out when the sites listed below first established a website/page on the internet. Thanks

Taylor & Francis Group: an informa business - http://www.taylorandfrancisgroup.com/

- www.tandf.co.uk

Association for Childhood Education International - http://acei.org/

- www.acei.org/cehp.htm

National Council of Teachers of Mathematics - http://www.nctm.org/

- http://my.nctm.org

National Association for the Education of Young Children - http://www.naeyc.org/yc/

- www.journal.naeyc.org —Preceding unsigned comment added by 24.210.25.124 (talk) 13:30, 6 January 2011 (UTC)

Question reformatted for legibility. The Wayback machine should be able to help - e.g. it suggests that www.tandf.co.uk first appeared in early 1997. AndrewWTaylor (talk) 20:50, 6 January 2011 (UTC)

Remainder term

If the Taylor series becomes arbitrarily close to the original function for all analytic functions (in other words, all function that we are normally interested in), what is the purpose of the remainder function? 24.92.70.160 (talk) 21:34, 6 January 2011 (UTC)

Analytic functions define a very small subset in the space of all function. Asking a function to have continuous derivatives of all orders and then asking for a series to converge is a very, very big ask. High school functions like 1/x aren't analytic; it fails to be continuous at x = 0. Take a look at the article on flat functions. The exponential $\scriptstyle f(x)\,=\,e^{-1/x^{2}}$ is an example of a flat function. It is well defined at x = 0 because both the positive and negative limits x → 0 give ƒ(x) → 0. In fact, it is a smooth function because each of its derivatives exist and are continuous for all x (particularly at x = 0). But you will find that each and every derivative vanishes at x = 0. So the function is always contained in the remainder function; no matter how far along the Taylor series you go. As for non-smooth functions, only the first few derivatives may be continuous, so we can only define the Taylor series up to a certain, finite order. Then the remainder term takes up the slack. — Fly by Night (talk) 22:03, 6 January 2011 (UTC)

Basically its purpose is to provide you with language for reasoning about whether or not the function you're considering at any given time happens to be one of the (in practice) rarely occurring exceptions to analyticity. –Henning Makholm (talk) 23:15, 6 January 2011 (UTC)

Numerical analysts often use a finite series as an approximation for a function; having an upper bound on the remainder term allows them to draw conclusions about how accurate their final results will be. Knowing that the infinite series converges exactly is not good enough if you can't compute an infinite series. Eric. 82.139.80.114 (talk) 01:54, 7 January 2011 (UTC)

Just to clarify a few points: analytic functions are a small subset theoretically (measure zero in L^2 I think...), but a huge subset of functions commonly used (i.e. in practice outside pure math research). Another thing to consider is the domain. Everyone seems to be assuming the whole real line, but 1/x and exp(-1/x^2) are both analytic on R\{0}. SemanticMantis (talk) 02:06, 7 January 2011 (UTC)

As a tangent here, are there smooth functions that are not analytic almost everywhere? The smooth function article claims that they exist (even nowhere analytic ones), but does not give details. Can they be constructed without the axiom of choice? –Henning Makholm (talk) 03:14, 7 January 2011 (UTC)

Here for instance is an explicit example. Algebraist 03:18, 7 January 2011 (UTC)

And it turns out we have an article on it. Algebraist 03:25, 7 January 2011 (UTC)

Oops, didn't notice you had already added a link to the smooth function article. It wasn't my intention to have two of them. –Henning Makholm (talk) 07:36, 7 January 2011 (UTC)

Another example. Algebraist 03:30, 7 January 2011 (UTC)

Interesting, thanks. Those appear to be impeccably constructible. –Henning Makholm (talk) 03:42, 7 January 2011 (UTC)

LaTeX

Can someone suggest a good LaTeX writing program for Windows. I mean a program where I type the code and it compiles it, turns it into .dvi, .ps or .pdf. I use Kile on Linux but it needs some fiddling with to run on Windows and I don't know how. A nice user friendly interface would be perfect, with some symbol buttons that substitute the LaTeX code when you click them, etc. — Fly by Night (talk) 22:25, 6 January 2011 (UTC)

there aren't any. just install miktex like everyone else. 87.91.6.33 (talk) 22:42, 6 January 2011 (UTC)

There's a commercial program called PCTeX that some people like. Personally I don't like it; if I recall correctly (but this was years ago) it has its own style and/or class files, and it's a bit of a pain to produce TeX source that other people can use without the program. But maybe they've fixed that for all I know. --Trovatore (talk) 22:48, 6 January 2011 (UTC)

I use TeXnicCenter; some of my friends use LEd. But there are plenty of editors (free or not) listed here. The Menu for Inserting Symbols column may be of interest to you. Invrnc (talk) 22:52, 6 January 2011 (UTC)

I use Lyx, quite good and free editor and writer. It can give output in .div, .ps and .pdf formats. Anyway, the list given above indicates many options. I haven't used it yet, but have heard good comments of Scientific WorkPlace. Pallida  Mors 00:53, 7 January 2011 (UTC)

January 7

Differentiation w/ respect to complex conjugate

My prof defined partial differentiation of a (single-variable) complex function with respect to the complex conjugate as follows:

If z = x + iy, and f(z) = u(x,y) + iv(x,y), then ${\frac {\partial f}{\partial {\overline {z}}}}=\left({\frac {\partial u}{\partial x}}-{\frac {\partial v}{\partial y}}\right)+i\left({\frac {\partial u}{\partial y}}+{\frac {\partial v}{\partial x}}\right)$

Is there an intuitive way of seeing the origin of this definition, other than an after-the-fact observation that it behaves as a partial derivative w/ respect to ${\overline {z}}$? 74.15.138.87 (talk) 01:56, 7 January 2011 (UTC)

I suppose the observation you refer to is something like

$f(z)=f(z_{0})+(z-z_{0}){\frac {\partial f}{\partial z}}(z_{0})+{\overline {(z-z_{0})}}{\frac {\partial f}{\partial {\overline {z}}}}(z_{0})+o(|z-z_{0}|)$

given a suitable companion definition of $\partial f/\partial z$ (beware: I haven't checked whether this is in fact true; some factors of -1 or 2 or 1/2 may be needed to make it true). There's nothing wrong with "after-the-fact observations"; they are only "after-the-fact" by virtue of the more or less arbitrary order in which your text presents things. The $\partial f/\partial {\overline {z}}$ symbol could equally well be defined as the complex number that makes the equation above hold, except in that case you would still need to prove that such a number is unique if exists, etc. etc. Most authors seem to feel that, all other things being equal, it is easiest to understand the formal development if definitions are chosen such that there is minimal doubt about the definition actually defining something.

As an alternative to either of these two characterizations, you could interpret $\partial f/\partial {\overline {z}}$ as a way to quantify how far f is from satisfying the Cauchy-Riemann equations. –Henning Makholm (talk) 02:54, 7 January 2011 (UTC)

There might not be anything wrong with an after-the-fact definition, but it's nice to have different perspectives on things, and certainly being able to see the logic behind the definition (before seeing its consequences) has some advantages.

I was looking for a way to go from

${\frac {\partial }{\partial {\overline {z}}}}f(z,{\overline {z}})=\lim _{h\to 0}{\frac {f(z,{\overline {z}}+h)-f(z,{\overline {z}})}{h}}$

to the above equation (h is, obviously, a complex number). Would you know how? 74.15.138.87 (talk) 03:43, 7 January 2011 (UTC)

I'm not sure your limit even makes sense to me; previously you said that f is a single-variable function but here you give it two arguments? In any case, a limit of this kind would probably not exist unless the function f happened to be an ordinary holomorphic function composed with an explicit conjugation, which is not a very interesting situation.

I would suggest that the property of ordinary differentiation that your definition generalizes is not the high-school limit definitions, but more the property of being the coefficient in a linear approximation. Does the characterization I gave above make sense to you? –Henning Makholm (talk) 03:56, 7 January 2011 (UTC)

(Also, I think this is one of the not uncommon cases where "the logic behind the definition" is that it happens to be what gives the desired consequences) –Henning Makholm (talk) 04:00, 7 January 2011 (UTC)

The equation you wrote above is familiar to me from real-valued differentiation, but my problem for ${\frac {\partial f}{\partial {\overline {z}}}}$ is the same as for ${\frac {\partial f}{\partial z}}$ (which, just to make sure I understand, is different than ${\frac {df}{dz}}$, right?). At any rate, I haven't seen that formalism for complex numbers, so I can't say I understand it entirely.

As for my differentiation thing, what I meant is something like this: suppose $f(z)=|z|=z{\overline {z}}$. Evidently, f is a function of z alone, but you could pretend that z and ${\overline {z}}$ are seperate variables. Then, ${\frac {\partial f}{\partial z}}={\overline {z}}$ and ${\frac {\partial f}{\partial {\overline {z}}}}=z$. Does that make sense? Probably not ... I'm a physics major, so there's a good chance I just broke like ten rules of math. But I like to have an intuitive understanding of the math I'm using, and when I see a symbol like ${\frac {\partial f}{\partial {\overline {z}}}}$, this is what I think of. So, for me at least, it's nice to see how this perhaps non-rigorous understanding of the math fits into the overall picture. 74.15.138.87 (talk) 04:40, 7 January 2011 (UTC)

I'm not sure how well your idea of pretending that z and z* are different works. What if f is given by some arbitrary expressions for u(x,y) and v(x,y)? Then we couldn't say which x's and y's came from z and which came from z*. It might work OK in those cases where you can express f as a complex differentiable function of z and z*, if you add a factor of 1/2 to your definition as I suggest below (or at least it seemed to work in the few examples I worked out), but it still seems a bit shifty to me. –Henning Makholm (talk) 07:26, 7 January 2011 (UTC)

Okay, first beware that I've actually never seen the notation $\partial f/\partial {\overline {z}}$ before; I'm making this up as I go along! If you're doing this for the purpose of physics, it may be that it's all meant to be used for some kind of Hermitean-ish form and my suggestions are completely off. But:

Usually, $df/dz$ is only defined when $f$ satisfies the Cauchy-Riemann equations, and in that case your $\partial f/\partial {\overline {z}}$ would be identically zero. So I'm assuming that there is a $\partial f/\partial z$ to go along with it, such that both are somehow meaningful for a non-differentiable $f$.

My idea now is to go back to multivariate real analysis and look at the real functions u and v. Let's keep $z_{0}$ fixed and look at the differential

$df=du+i\,dv={\frac {\partial u}{\partial x}}dx+{\frac {\partial u}{\partial y}}dy+i{\frac {\partial v}{\partial x}}dx+i{\frac {\partial v}{\partial y}}dy$

(from the definition of f, and the chain rule). Now, if we can write the left-hand side of this in the form

$df=A(dx+i\,dy)+B(dx-i\,dy)$

for some appropriate complex numbers A and B (which depend on $z$ but not on $dx$ and $dy$), then it would make some sense to call A and B $\partial f/\partial z$ and $\partial f/\partial {\overline {z}}$, respectively, because then the whole thing would look sort of like the chain rule. Expressing A and B in terms of the partial derivatives of u and v is a matter of simple (real) linear algebra. Calculate, calculate ... it turns out that B becomes half of your definition for $\partial f/\partial {\overline {z}}$. No matter; this just means that the pseudo-chain rule that works for your definitions will be

$df={\frac {1}{2}}\left[{\frac {\partial f}{\partial z}}dz+{\frac {\partial f}{\partial {\overline {z}}}}d{\overline {z}}\right]$

which is not quite an unreasonable convention either, though it does have the strange consequence that $\partial f/\partial z$ is two times $df/dz$ when the latter is defined. Alternatively, perhaps there is an 1/2 in your notes that you forgot to copy?

Clearer now? –Henning Makholm (talk) 06:45, 7 January 2011 (UTC)

limit

how would I prove that $\lim _{x\to \infty }{\frac {x!}{n^{x}}}=\infty$ for any n, therby proving that the factorial function grows faster than any exponential? Is this even true? 24.92.70.160 (talk) 02:43, 7 January 2011 (UTC)

See Factorial#Rate_of_growth. Staecker (talk) 02:45, 7 January 2011 (UTC)

Basic idea: When you increase x by 1, the numerator increases by a factor of about x, whereas the denominator increases by a constant factor of n. Eventually x is greater than n. Work it out from there. --Trovatore (talk) 02:48, 7 January 2011 (UTC)

If part of your quandary is how to deal with the factorial, you might try converting Stirling's approximation into a bound on x!, and use that to derive the limit. Alternatively, you can substitute the Gamma function for the factorial, as the Gamma function is the continuous version of the factorial. -- 174.21.250.227 (talk) 03:06, 7 January 2011 (UTC)

I don't think that is any easier than keeping it as a discrete sequence and working directly from the definition. One easily sees that from a certain $x$ onwards, the sequence is strictly increasing, and it is then also easy for any $N$ to find an $x$ such that $x!/n^{x}>N$ (note that it is not necessary to be able to pinpoint the first such $x$). –Henning Makholm (talk) 03:22, 7 January 2011 (UTC)