|WikiProject Mathematics||(Rated B-class, Mid-importance)|
- 1 Headline text
- 2 Overlooking the obvious?
- 3 Picture as proof
- 4 You can prove it easily by using Fourier transform
- 5 Basel problem extended
- 6 Irrelevant topic
- 7 Fundamental Theorem of Algebra unnecessary
- 8 What you need to know
- 9 Coprime?
- 10 Long proof
- 11 A slicker proof from Fourier series
- 12 Problem in proof.
- 13 Flaw in Euler's method
- 14 Proof through Complex Analysis
- 15 wouldt it be...
- 16 broken link
- 17 the record holder's rectangles area will always be greater than the ideal rectangle - citation needed
I'm thinking of renaming this article "The Basel Problem" and adding a bunch of historical remarks and putting in broader context. I still think the proof can stay here, it would just be a part (about half) of the whole article. I don't think it would make it too long, and people not interested in the proof could just skip that section. I thought of having 2 articles, but it seemed redundant and just giving the proof without any remarks about the problem seemed strange. Revolver 15:31, 26 Feb 2004 (UTC)
- Possibly...I find it hard to imagine someone linking to it, but it can't hurt. Revolver 02:41, 29 Feb 2004 (UTC)
That's a good idea...I forget simple things like, "why does this sum converge in the first place", assuming everyone has seen p-series in calculus, or something. There may be some other simple ways to approximate this that people used before Euler that might be interesting. Revolver 22:18, 3 Mar 2004 (UTC)
Overlooking the obvious?
It isn't clear to me from the article why this problem is called the Basel problem. Dataphile 19:49, Aug 22, 2004 (UTC)
- Basel is the name of a town, somehow the town is related to the origin of the problem. I don't know exactly -- in math, so many things have so many names, math people often don't know why something's called what it is. Revolver 17:51, 24 Aug 2004 (UTC)
The Bernoulli family, which worked on the problem for a long time, were located in Basel. Johann Bernoulli taught Euler at the University of Basel.
Picture as proof
"First note that 0 < sin x < x < tan x. This can be seen by considering the following picture:"
It's definitely a cool picture. Can its owner change the 'theta' to an x? Do we need to point out that the angle is equal to the length of the circular arc AD?
I can 'see' that sin x < x, but I can only convince myself that x < tan x using calculus (e.g., the derivative of (tan x - x) is tan2x, which is > 0 for x in (0, pi/2), ...). Is that a failure of imagination on my part? Perhaps it's obvious that tan x grows faster than x.
- The area of the triangle OAE is tan(θ)/2. The area of the circle sector OAD is θ/2. The circle sector is contained within the triangle, so θ < tan(θ). Fredrik Johansson - talk - contribs 18:18, 24 January 2006 (UTC)
Thanks. I've added this to the article. Buster79 23:08, 2 September 2006 (UTC)
You can prove it easily by using Fourier transform
I will not go into details, but the proof starts with the function absolute value of x on the interval [-pi,+pi] You calculate the Fourier transform of the function. Later you can seperate the sum into odd and even numbers and find out that the sum equals pi^2/6.
The Riemann zeta formula can be approached by Parseval's theorem.
The point of the proof in this article is that it needs only elementary methods. There are several proofs of this problem (there is an article called "Six ways to sum a series" which gives 6 of them), but most of them require more advanced machinery.
Basel problem extended
The Basel problem can be extended to find the closed forms for every N. An approximate sequence can be found in the OEIS, A111510. Included is an expression of Pi where the odd and even terms of Triangular(n)define the differences. Would the contributers to the Basel problem pages care to comment and to suggest the best way to include this? Marc M. 20-6-06
The last topic in the article on the Basel problem reads as follows:
"Use in the calculation of π: In 1881, Ernesto Cesaro showed that the probability of two integers being relatively prime equals 6 / π2, which is the reciprocal of ζ(2). By the above proof, Cesaro's theorem thus allows a value for π to be calculated from a large collection of random integers, by determining the proportion of them which are relatively prime."
I don't see how this uses the result of the Basel problem, ζ(2) = π2 / 6, at all; it just mentions it in passing. Likewise, the phrase "By the above proof" is unnecessary, since the application of Cesaro's theorem goes along just fine without using the above proof at all. Moreover, it's very misleading to say that π can be calculated by this method. If you pick 100 pairs of random numbers from 1 to 10, and every possible pair shows up once, you will "calculate" π ≈ 3.086. (There are 63 pairs that are relatively prime.) No matter how many pairs are selected, the result will be an algebraic number, and π is transcendental. I suggest the entire paragraph be removed from the article. Did I miss anything? Gwil 20:03, 27 September 2006 (UTC)
Fundamental Theorem of Algebra unnecessary
I object to the following:
If p(t) is a polynomial of degree m, then p has no more than m distinct roots. Proof: This is a consequence of the fundamental theorem of algebra.
The statement "If p(t) has degree m, then p has no more than m distinct roots" can be proven by elementary methods, whereas the FTA is much more difficult (it's usually proven with complex analysis). Specifically, the "no more than m distinct roots" theorem follows from the factor theorem ( (x - r)|p(x) if and only if p(r) = 0, a consequence of polynomial division), together with an induction argument to show that the product of (x - rj) over all the roots rj of p(x) divides p(x), and an application of the fact that degree(f * g) = degree(f) + degree(g) for nonzero polynomials f and g.
Moreover, the "no more than m distinct roots" theorem holds in the polynomial ring of any integral domain, whereas the FTA is a very special fact about the complex numbers.
I would recommend that the line "Proof: This is a consequence of the fundamental theorem of algebra." be replaced by an outline of the actual proof from the factor theorem. --Gene496 08:35, 27 July 2007 (UTC)
While Euler is credited with the solution, i.e., an exact answer, is pi^2/6 an exact number? pi itself is the sum of an infinite series as is the original Basel series! SEIBasaurus 22.214.171.124 17:41, 7 August 2007 (UTC)
What you need to know
Just a comment on the "What you need to know" sub-section... love this, extremely helpful, should be present in more articles! Error792 03:20, 6 August 2007 (UTC)
Would it be worthwhile to add a wikilink to coprime, regarding the application of ζ(2) in evaluating the probability that two randomly chosen numbers are coprime? I just added a wikilink in the other direction, from that article to here. —David Eppstein (talk) 22:11, 15 December 2007 (UTC)
This article has bothered me for a long time. I have looked for elementary proofs that were also simple, but found none. Perhaps it is not surprising that it took Euler quite a while to figure it out.
I like the elementary proof precisely because it is elementary. However, I think it is also very complicated and in most cases will not be interesting to the reader. On the one hand, I think it is too lengthy to be readable by most laypersons. On the other hand, I would think that almost anyone who is familiar with Fourier series would prefer that proof, because one can understand it at a glance, and remember it forever more (whereas I could not reproduce this proof without some effort, and I do math research...)
Because this proof is so well written and also elementary, I think it should be moved to its own article. I like Euler's "proof", and it should stay, and for completeness, we could leave the Fourier Series argument which I just put in, because that would give a very short airtight argument. However, right now, I think that most of our readers must stop reading somewhere in the "What you need to know" section.
- I think that several proofs on the same page is a good idea. The Fourier series argument is a very welcome addition, it's the proof usually taught in calculus courses. There are more proofs given in the external links. I prefer to have good outlines or section titles for the proofs instead of distributing them over several articles. So the reader can make a choice according to his/her level of mathematical education or specific interest, compare with Proof that the sum of the reciprocals of the primes diverges and Inequality of arithmetic and geometric means. Schmock (talk) 00:37, 24 July 2008 (UTC)
Nono, what I'm proposing is to segregate most of the proofs to an article, maybe, Proof that zeta(2)=π^2/6, much like Proof that the sum of the reciprocals of the primes diverges. This article could be more about the history of the Basel problem and maybe one proof as short as possible.
- I prefer to have the elementary proof on the same page as the history of the problem, because this proof is potentially readable for people without a university education in mathematics. When we have several (at least 4) advanced proofs, a separate article might be a good idea. Personally, I prefer to have some proofs in the article zeta constant, in particular for ζ(2n). As far as I remember, Fourier methods also work in this more general case, but other proofs might be shorter. Schmock (talk) 08:06, 24 July 2008 (UTC)
My understanding of zeta(2n) is that you find an nth antiderivative of f(x)=x. So for instance, assume that . So g is periodic and n times differentiable. The constant terms are prescribed by the homology or something. Clearly, g(x) is a polynomial and for a fixed n, you can compute it. Hence, you can compute \int g^2. Furthermore, and that gives you the necessary formula.
But I don't know if there's an explicit formula for g. Do you know?
A slicker proof from Fourier series
What is meant with "worked out in that article"? If it's the last link than the link is dead. Furthermore does the Parseval's identity on wikipedia not give the desired result for this prove. Mostly because here the Parseval's identity is stated wrong. 126.96.36.199 (talk) 08:59, 8 November 2008 (UTC)
Problem in proof.
It seems to me that the proof shown in the section "Euler attacks the problem" has a fallacy. Before I explain it, it should be said that the proof naturally can be used after it has been modified a bit.
It is stated that:
Actually this is only true for where since both sides equals 0 (which has been mentioned earlier in the proof). Generally we have:
But as noted before the proof is still valid. It should just be noted that the following calculations in the proof only are valid for where .
I haven't changed the article since I want to be sure I'm not making a mistake, but if you agree then please correct it.
You are making a mistake. The formula is correct with a=1. The only problem is that Euler had no proof of this fact; it was compelling because it worked to solve the problem, but Euler died before the tools to prove it were developed. Nowadays this product expansion can be derived from Weierstrass's factorization theorem in complex analysis. 188.8.131.52 (talk) 06:36, 16 October 2009 (UTC)
Flaw in Euler's method
The fact that sin(x)/x has zeroes at +/-(pi, 2pi, 3pi, ...) doesn't mean sin(x)/x can be expressed as the product of terms in the form (1 - x/pi).
Maybe I'm missing something here but when you express a finite polynomial in terms of its linear factors (which is the assumption Euler is making here and the basis of his proof), it is of the form: (x - root1)(x-root2)...
- For a polynomial of degree n with roots z1, z2, ..., zn; p(z) = a (z-z1) (z-z2) ... (z-zn) -- that is, the polynomial is determined by its roots only up to a constant multiple. When 0 is not a root we can equally write p(z) = b (1-z/z1) (1-z/z2) ... (1-z/zn) where b = a (-z1) (-z2) ... (-zn). This form has its advantages, one being that b = p(0). -- ToET 16:45, 4 July 2010 (UTC)
Proof through Complex Analysis
- There's a proof using complex analysis already in the article, as the first proof: Basel_problem#A_rigorous_proof_using_Fourier_series. For many other proofs, see Robin Chapman's collection, also linked in the article. Shreevatsa (talk) 09:01, 12 August 2010 (UTC)
wouldt it be...
the record holder's rectangles area will always be greater than the ideal rectangle - citation needed
"Because π is a transcendental number, the record holder's rectangles area will always be greater than the ideal rectangle."
What's the source for this? I don't see how π being transcendental, i.e. not a solution to an algebraic equation - implies that a well defined packing algorithm has to be greater than the ideal rectangle. For instance is easy to write out a program that can print out all the digits of π one after another. That a number is transcendental doesn't normally have much by way of implications for computability. Am going to add a citation needed tag to this. If it is "obvious" for some reason do explain and add some hint to the text to help readers who don't get it. Thanks! Robert Walker (talk) 18:26, 7 September 2014 (UTC)
- A transcendental number has the implication of not being able to be computable exactly only approximately. Because of this, "the record holder's rectangles area will always be greater than the ideal rectangle". Even if you find pi to 10^1000 digits it is only approximately correct (the 10^1000 + n digit number would yet to be found). 184.108.40.206 (talk) 02:59, 11 September 2014 (UTC)
- This is as confused as the sentence in the article. Either we have a proof that the rectangle with the exact area can be packed or we don't. If we do have such a proof, then there is a packing that is NOT greater than the ideal area. If we don't have such a proof, then any packing we do have uses greater area. This would be true regardless of whether the number in question were rational, algebraic, or transcendental. And whether or not such a packing exists is also different than whether the packing is computable. Additionally, transcendental numbers such as π are no more computable or uncomputable than rational numbers like 1/3 — neither is exactly representable in binary but both can be computed easily to arbitrarily high precision. In short, I agree with Robertinventor that being transcendental doesn't have much to do with the inequality between known and ideal packings. I think we should just remove this sentence. —David Eppstein (talk) 03:19, 11 September 2014 (UTC)