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March 3

Variations on the Leibniz formula for π

This Mathologer video draws a connection between the series $1-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}-\cdots ={\frac {\pi }{4}}$ and the area of a circle using the Sum of two squares theorem. (Actually it uses a stronger result which gives the number of ways that an integer can be written as the sum of two squares. This doesn't seem to be mentioned in our article. It is mentioned, uncited, in the article on the Sum of squares function.) Mathologer cites his source as the book Geometry and the Imagination by Hilbert & Cohn-Vossen. (p. 37–...)

It's not hard (assuming the relevant number theoretic results) to find variations on this theme: Relate partial sums of a series to the number of lattice points satisfying a certain inequality, interpret this number as the approximate area of a geometrical figure, then compute the area of the figure. The geometrical figure is either an ellipse or hyperbolic sector. Some examples are: $1+{\frac {1}{3}}-{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}+{\frac {1}{11}}-{\frac {1}{13}}-{\frac {1}{15}}+\cdots ={\frac {\pi }{2{\sqrt {2}}}}$ $1-{\frac {1}{3}}-{\frac {1}{5}}+{\frac {1}{7}}+{\frac {1}{9}}-{\frac {1}{11}}-{\frac {1}{13}}+{\frac {1}{15}}+\cdots ={\frac {\ln({\sqrt {2}}+1)}{\sqrt {2}}}$ $1-{\frac {1}{2}}+{\frac {1}{4}}-{\frac {1}{5}}+{\frac {1}{7}}-{\frac {1}{8}}+\cdots ={\frac {\pi }{3{\sqrt {3}}}}$ $1-{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{11}}+{\frac {1}{13}}-{\frac {1}{17}}-{\frac {1}{19}}+{\frac {1}{23}}+\cdots ={\frac {\ln({\sqrt {3}}+2)}{\sqrt {3}}}$

I assume all of these are well known results, and there seem to be several connections to number theory (analytic and otherwise). I'm wondering if this is covered in Wikipedia at all other than what is mentioned above. I'm also wondering how far these generalizations can be taken. Closed forms for the sums can probably be found in other ways, but the connection to quadratic forms and geometry seems to get more complex as numerator sequence period gets longer. Perhaps it breaks down entirely at some point. In other words, are these connections part of some more general theory or are there only a few random examples? --RDBury (talk) 09:45, 3 March 2023 (UTC)[reply]

Did you try doing a search of the academic literature? You might also add the Mercator series evaluated at 1, which is log(2). –jacobolus (t) 21:03, 3 March 2023 (UTC)[reply]

As always, you could start with Euler. See here page 167. –jacobolus (t) 22:59, 3 March 2023 (UTC)[reply]

@Jacobolus:The Mercator series doesn't really fit the pattern, at least as far as I could tell. The numerators are multiplicative, meaning n(pq) = n(p)n(q), but that doesn't hold with the Mercator series. I'm not sure what to search for in the literature; all I have is some unnamed identities. In any case, I doubt it would be in anything modern since I'd guess they were already well known 100 years ago. I tried OEIS for one of the values, but the paper it linked to only showed the value with nothing relating it to quadratic forms. The book mentioned above was based on lectures Hilbert gave more than 100 years ago, so it figures that it was already "settled" math at that time.

Like I said above, there are probably ways to find closed forms for the sums in other ways, and the series mentioned by Euler seem to be a starting point. (I don't read Latin so I can't be entirely sure.) But they don't connect to sums to quadratic forms. In modern language they seem to be the series expansions for ln(1+e^it), basically well known trigonometric series.

Anyway, for a given odd prime p there are three non-trivial characters for the multiplicative group mod 4p. That gives three sums using the characters as numerators in the series. Of these, one is easily obtained from the Leibniz formula by crossing out every pth term, That leaves 2 which allows for a possible interpretation in terms of the forms a²±pb². There also 2 forms for p=2 and I guess you could call the Leibniz formula the p=1 case. So p=1, 2, and 3 are given (more or less) by the series above. I thought p=5 might be a deal breaker because z[√-5] is not a u.f.d, but the pattern carries on regardless. I'll start on p=7 next; each sum requires a certain amount of ad hoc computation and I only have enough energy to do one or two per day. Maybe eventually I'll discover on my own where the pattern breaks down or hit on a general formula. --RDBury (talk) 23:47, 3 March 2023 (UTC)[reply]

These might also be of interest

https://www.mathpages.com/home/kmath477/kmath477.htm

http://numbers.computation.free.fr/Constants/Pi/piSeries.html (click the link for the postscript version)

–jacobolus (t) 01:27, 4 March 2023 (UTC)[reply]

There are some relevant looking references at https://mathworld.wolfram.com/GausssCircleProblem.html and https://arxiv.org/pdf/math/0410522.pdf is apparently a survey of recent research in similar topics with almost 200 references. –jacobolus (t) 02:31, 4 March 2023 (UTC)[reply]

This is an English translation of the piece by Euler; page 167 in the Latin version corresponds to §12 on page 12. --Lambiam 03:42, 4 March 2023 (UTC)[reply]

@Jacobolus:I think most of what I'm looking for is in the Mathpages link about half way down -- Dirichlet's Class number formula. It's a special case of he more general Class number formula and it's mentioned in that article. I don't know how Dirichlet proved it, but it seems like this geometric approach would be a reasonable. Basically you'd be summing over each form with the given discriminant, and each area seems to be the same, so you get the class number in the numerator. Each value is repeated a number of times given by the number of units in the corresponding quadratic field, giving the denominator. The rest is just the area of the corresponding ellipse. That's the D<0 case anyway, for D>0 it's similar but instead of ellipses you look at hyperbolic sectors. So the upshot is that there is a general formula there, or really two if you consider D<0 and D>0 as different formulas. If you know the class number you can find the sum of the series, and if you can estimate the sum of the series then you can get the slass number. --RDBury (talk) 08:16, 4 March 2023 (UTC)[reply]

If you can read German, Dirichlet's (lengthy) investigations can be read online here. --Lambiam 14:24, 4 March 2023 (UTC)[reply]

I can read German better than Latin at least :) Formulas 21 (page 52) and 24 (page 57) seem to be equivalent to the class number formula. I didn't get into the weeds much but so far it looks like his methods are more analytical than geometrical. --RDBury (talk) 06:41, 5 March 2023 (UTC)[reply]

The title of the treatise promises us various applications of infinitesimal calculus, so one expects to see analysis in action. I think it is conceivable that various connections with geometric area estimators have gone unnoticed or have not been deemed worthy of publication, in particular when they are isolated results for a series of a less popular constant than

π

. --Lambiam 11:05, 5 March 2023 (UTC)[reply]

circular arc question.

With three identical circular Arcs placed with A & B touching at ends and B&C touching at ends so that the curvature of A & C is opposite of B. What percentage of the circle would these arcs have to be so that A&C share exactly one point. (I'm looking at the non-degenerate case here)

(If the percentage was exactly 50% A,B & C would sort of look like a wave. I'm trying to figure out how much the arc has to increase so that A&C are brought together just enough to touch.Naraht (talk) 18:20, 3 March 2023 (UTC)[reply]

If I'm reading your description correctly, the answer is never. See the diagram I (badly) created: the ends of A and C only touch when the gap becomes exactly 0, so for any arc of less than 360 degrees, the will never touch. The will only touch in the limit of a fill circle.--Jayron 32 18:51, 3 March 2023 (UTC)[reply]

Unless you're asking when the sides (rather than the ends) touch. In that case, it's going to be identical to the "three identical circles touching" problem. In that case, it will be 1/6 of the circle missing (60 degrees), so it will be 5/6 of the circle (300 degree arc). --Jayron 32 19:00, 3 March 2023 (UTC)[reply]

I too interpreted the question as having 5/6 = 300° as the answer, and constructed basically the same visual proof. --Lambiam 19:47, 3 March 2023 (UTC)[reply]

Jayron32, Lambian Original poster thanks!