a repository of mathematical know-how

Recent comments

  • Anonymous (not verified) 10 years 28 weeks ago Mathematicians need to be metamathematicians

    "Think about the converse" should be added as a link in this article.

  • alex 10 years 30 weeks ago Bounding probabilities by expectations

    Here is a point which I feel needs to be emphasized more in Example 3.

    If we try to make the argument work without leaving \lambda to be optimized later, but by, say, initially setting \lambda=1, the argument will fail: we will get a useless bound. The reason: we need to pick a function F which will grow rapidly while E F(X) will grow slowly. In this case, E e^{X_1 + \ldots X_n} grows pretty fast with n.

    What kind of functions F work for this? Well, F(X)=X^2 works: out of the n^2 terms in (X_1 + \cdots + X_n)^2, almost all have expectation 0. Similarly, F(X)=X^{2k} works fine as well: in the sum (X_1 + \cdots + X_n)^{2k}, we will be able to easily cancel many terms because they have expectation 0.

    The central insight behind the proof in Example 3 is that picking F(x)=(1+t)^x for really, really small t works. The reason: the average of 1+t and 1/(1+t) is really close to 1. That is: near one, inversion is about the same as subtraction. This insight can be pushed a little further to argue that if X has mean 0, the quantity E (1+t)^X should be pretty close to 1. Thus one might hope that E[(1+t)^{X_1 + \cdots X_n}] = (E[(1+t)^{X_1}])^n grows slowly - after all, it is "approximately" 1^n = 1.

  • Anonymous (not verified) 10 years 37 weeks ago How to compute the (co)homology of a space

    Maybe it'd be nice to mention that if your space is being decomposed into U and V and these have non-empty intersection, then the end terms (H_0 and such) can be taken as reduced homology instead of just normal homology, simplifying some calculations.

  • Anonymous (not verified) 10 years 37 weeks ago How to compute the (co)homology of a space

    If I recall correctly, the Mayer-Vietoris sequence tells you something about a space decomposed as two sets _whose interiors cover the space_. So I think U and V must intersect in two intervals (which are homotopically trivial, so the rest of the proof runs the same).

  • Anonymous (not verified) 10 years 38 weeks ago How to solve cubic and quartic equations

    what is uv?

  • Nicolai H (not verified) 10 years 42 weeks ago Dimension arguments in combinatorics

    I would suggest using the less eponymous term "Shatter function lemma" (that Matousek uses in his Lectures on Discrete Geometry) in addition to "Sauer-Shelah lemma". I personally have known this lemma for a long time, but always under the more descriptive name "Shatter function lemma".

  • Marius Overholt (not verified) 10 years 45 weeks ago Smoothing sums

    It should be Acta Arithmetica, of course!

  • Marius Overholt (not verified) 10 years 45 weeks ago Smoothing sums

    It was actually Waclaw Sierpinski who obtained the exponent 1/3 in the Gauss circle problem, as part of his doctoral dissertation, and who published the result in 1906 (in Polish). His adviser Georgy Voronoi had obtained the same exponent in the Dirichlet divisor problem in 1903. There is an article by Andrzej Schinzel about Sierpinski's papers in number theory in volume XXI of Acta Mathematica (1972). Also there is supposed to exist (but which I have never seen) a description of Sierpinski's proof in German, possibly in some Jahrbuch der Deutsche Mathematikerverein from those years. Schinzel states that the proof is by Voronoi's geometric method. That would mean that the circle is approximated by a polygon, and Euler-Maclaurin summation applied on each piece.

  • maexe (not verified) 10 years 50 weeks ago How to use tensor products and evaluation maps in representation theory

    In Proof of Theorem 2:
    ... use Theorem 1 to identify \mathrm{Hom}_{k(G)}(U,U \otimes_k V) with V (not U),
    we see that we may regard S as a subspace of V (not U).
    Beyond this, a very nice article.

  • Dia Sharma (not verified) 10 years 51 weeks ago How to solve linear equations in one variable

    Very good but it would be nice if variable seperation method would be included.

  • olof 11 years 3 days ago Complete the square

    I don't find the article unintelligible at all; perhaps you could clarify what you find problematic.

    Also, editing a page to correct typos (like the missing coefficient you spotted) is very simple: just click on edit at the top of the article. I've corrected it here.

  • boumbi 11 years 6 days ago Complete the square

    This article is unintelligible.

    How can it be possible to transform the method of completing the square into such a "charabia".

    By the way, a(x+x_0)^2 = ax^2 + 2ax x_0 + x_0^2 is false.

  • neroden 11 years 1 week ago I want to find a non-trivial solution to a system of equations

    This should probably be restructured to match the form of the articles under "What kind of problem am I trying to solve?". I might even do so myself if I have the time and energy...

  • neroden 11 years 1 week ago Transform equation to reach a solved pattern

    This describes one of the most fundamental and general mathematical techniques there is, one so basic that probably most math majors find it second nature already. As such, the description of it as a general technique may be most useful for lower-level pedagogy; failing to grasp this general technique can make math very hard for primary and secondary school students. However, I would love to have examples of its use at all levels and fields, since it is so general-purpose.

  • neroden 11 years 1 week ago Transform equation to reach a solved pattern

    Blech, yes, I got stuff reversed. Fixing.

  • neroden 11 years 1 week ago How to solve cubic and quartic equations

    It's a Wiki. Usual advice is to rewrite the style yourself. :-)

  • Anonymous (not verified) 11 years 2 weeks ago How to solve cubic and quartic equations

    My personal favorite way of understanding cubics is the following. In analogy to quadratic equations, try a solution of the form u + v, figuring once we get a single root we can get the other two in short order. Then get rid of the bx^2 term like you did, obtaining x^3 + cx + d = 0. The idea now is to cancel the cx term too. Namely, x^3 = u^3 + 3uv(u + v) + v^3, and cx = c(u + v), so to cancel the cx
    term and the 3uv(u + v) term at the same time one stipulates that 3uv = -c.

    As a result, one has two equations now: u^3 + v^3 + d = 0, and 3uv = -c. Plugging
    v = -c/3u into the first equation gives a quadratic equation in u^3, which has
    six solutions u. To each of these three is a single v = -c/3u that works and then one obtains six solutions u + v. But of course u and v are symmetric here so we really just have 3 solutions.

  • tao 11 years 3 weeks ago The tensor power trick

    For consistency I've now decided to use \chi throughout that section.

  • Anonymous (not verified) 11 years 4 weeks ago The tensor power trick

    Before (1), the equation for the Fourier transform has \chi when it should have \xi.

  • emerton 11 years 5 weeks ago A good way of proving that a set is countable

    I don't think one really needs to apply the axiom of choice in this context: Since the multifunction is taking values in positive integers, we could just define the corresponding function by taking its value to be the least member of the set of values of the multifunciton. (This process will then, I guess, secretly take us back to the original function construction, where we write each rational in lowest terms — but we don't need to pay attention to this if we don't want to.)

  • Anonymous (not verified) 11 years 5 weeks ago How to solve linear equations in one variable

    This article is should not be part of the Tricki.

  • Michael Burge (not verified) 11 years 5 weeks ago A good way of proving that a set is countable

    I like how you snuck in the Axiom of Choice here, and hid it under the rug since its use is "clear".

  • Anonymous (not verified) 11 years 7 weeks ago Transform equation to reach a solved pattern

    Should "difficult" be changed to "easy"?

  • Didier Piau (not verified) 11 years 9 weeks ago How to use martingales

    There is a typo in the second paragraph of section Example 1, continued: one should cancel the symbol \mathbb{E} in the two occurrences of

    the expected value of \mathbb{E}\phi(\sigma) given that

    And, reading the last paragraph of this section, I wondered why you picked 100\sqrt{n} as an example of extremely unlikely deviation. First, 10\sqrt{n} would already yield a rather respectable (i.e., small) bound, namely, something like 10^{-11}. Second, there is nothing specific about 100\sqrt{n} here, although the formulation seems to indicate otherwise: you might want to add a for example somewhere.

    Thank you for this article.

  • mckeown_j.c 11 years 9 weeks ago If you are getting stuck, then try to prove rigorously that your approach cannot work

    Does it mean the re-write needed to see "variance" is small? Or is it because "partition" naturally suggests "statistics"?

    There are reasons to prefer habitually testing for Cauchy-Schwartz, or even Jensen's inequality: for instance they are both more flexible than "variance".