[math-fun] An Unusual Integral?
3==Integrate[Hypergeometric2F1[1/3,2/3,1,x-(1/4)*x^2],{x,0,2}] ( Cf. https://oeis.org/A006480 and thereafter ) Gosper said that this calculation was "weird". I think it would be rare to get an integer value from an integral of 2F1(a,b,c,z), with non-cancelling (a,b,c), z=P(x) in Q[[x]], and z over 0...1, between D.E. singular points. Does anyone else have an opinion? --Brad
For comparison, let's keep the same parameters and try the more simple P(x)=x^2. Then, Equal[ Integrate[Hypergeometric2F1[1/3, 2/3, 1, x^2], {x, 0, 1}], 2 (3 ArcSinh[1/(2 Sqrt[2])] + Log[8])/(Sqrt[3] Pi)] N[% /. Equal -> List, 20] Maybe other list readers do not find so much fun when "chasing a trail of smoke and reason" (though it does go even higher). So let's look at another geometry: Show[ContourPlot[ -p^2 + q^2 - (4/27)*(-3*q*p^2 - q^3)^2 == 0, {q, -2, 2}, {p, -2, 2}]] https://0x0.st/zbVV.png Should we be surprised to find out that the area interior to sextic lemniscate zbVV, up to harmonic scale factor pi/sqrt(3), equals a rational number 3/2? --Brad On Fri, Apr 26, 2019 at 11:01 AM Brad Klee <bradklee@gmail.com> wrote:
Does anyone else have an opinion? --Brad
Surprises come in different flavors. There are scads of theorems in which complicated integrals evaluate to surprisingly simple numbers, but I’ve seen a lot of those, so in a way I’ve stopped finding such surprises surprising. On the other hand, there aren’t a lot of nontrivial theorems about digit-patterns in integers or real numbers, so when I see something that looks like a digit-pattern, I’m surprised. (But I’ve looked at other Fibonacci numbers, and I don’t see anything going on with their digits, so I’ve become convinced that the “pattern” I saw was just a fluke.) Jim On Sat, Apr 27, 2019 at 10:35 AM Brad Klee <bradklee@gmail.com> wrote:
For comparison, let's keep the same parameters and try the more simple P(x)=x^2. Then,
Equal[ Integrate[Hypergeometric2F1[1/3, 2/3, 1, x^2], {x, 0, 1}], 2 (3 ArcSinh[1/(2 Sqrt[2])] + Log[8])/(Sqrt[3] Pi)] N[% /. Equal -> List, 20]
Maybe other list readers do not find so much fun when "chasing a trail of smoke and reason" (though it does go even higher). So let's look at another geometry:
Show[ContourPlot[ -p^2 + q^2 - (4/27)*(-3*q*p^2 - q^3)^2 == 0, {q, -2, 2}, {p, -2, 2}]]
Should we be surprised to find out that the area interior to sextic lemniscate zbVV, up to harmonic scale factor pi/sqrt(3), equals a rational number 3/2? --Brad
On Fri, Apr 26, 2019 at 11:01 AM Brad Klee <bradklee@gmail.com> wrote:
Does anyone else have an opinion? --Brad
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
There's something like F_{5^k} \equiv 25 (mod 100) or 625 (mod 1000) for all large enough values of k (this was something like 20+ years ago, so don't quote me too hard on the statement). While at first it is surprising, I hear it boils down to the mu/lambda loop algorithm running on F_k (mod 100). On 4/27/19 10:59 , James Propp wrote:
Surprises come in different flavors. There are scads of theorems in which complicated integrals evaluate to surprisingly simple numbers, but I’ve seen a lot of those, so in a way I’ve stopped finding such surprises surprising. On the other hand, there aren’t a lot of nontrivial theorems about digit-patterns in integers or real numbers, so when I see something that looks like a digit-pattern, I’m surprised.
(But I’ve looked at other Fibonacci numbers, and I don’t see anything going on with their digits, so I’ve become convinced that the “pattern” I saw was just a fluke.)
Jim
On Sat, Apr 27, 2019 at 10:35 AM Brad Klee <bradklee@gmail.com> wrote:
For comparison, let's keep the same parameters and try the more simple P(x)=x^2. Then,
Equal[ Integrate[Hypergeometric2F1[1/3, 2/3, 1, x^2], {x, 0, 1}], 2 (3 ArcSinh[1/(2 Sqrt[2])] + Log[8])/(Sqrt[3] Pi)] N[% /. Equal -> List, 20]
Maybe other list readers do not find so much fun when "chasing a trail of smoke and reason" (though it does go even higher). So let's look at another geometry:
Show[ContourPlot[ -p^2 + q^2 - (4/27)*(-3*q*p^2 - q^3)^2 == 0, {q, -2, 2}, {p, -2, 2}]]
Should we be surprised to find out that the area interior to sextic lemniscate zbVV, up to harmonic scale factor pi/sqrt(3), equals a rational number 3/2? --Brad
On Fri, Apr 26, 2019 at 11:01 AM Brad Klee <bradklee@gmail.com> wrote:
Does anyone else have an opinion? --Brad
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Look at this: https://mathoverflow.net/questions/87692/leading-digits-of-fibonacci-sequenc... On Sat, Apr 27, 2019 at 14:39 Andres Valloud < avalloud@smalltalk.comcastbiz.net> wrote:
There's something like F_{5^k} \equiv 25 (mod 100) or 625 (mod 1000) for all large enough values of k (this was something like 20+ years ago, so don't quote me too hard on the statement). While at first it is surprising, I hear it boils down to the mu/lambda loop algorithm running on F_k (mod 100).
On 4/27/19 10:59 , James Propp wrote:
Surprises come in different flavors. There are scads of theorems in which complicated integrals evaluate to surprisingly simple numbers, but I’ve seen a lot of those, so in a way I’ve stopped finding such surprises surprising. On the other hand, there aren’t a lot of nontrivial theorems about digit-patterns in integers or real numbers, so when I see something that looks like a digit-pattern, I’m surprised.
(But I’ve looked at other Fibonacci numbers, and I don’t see anything going on with their digits, so I’ve become convinced that the “pattern” I saw was just a fluke.)
Jim
On Sat, Apr 27, 2019 at 10:35 AM Brad Klee <bradklee@gmail.com> wrote:
For comparison, let's keep the same parameters and try the more simple P(x)=x^2. Then,
Equal[ Integrate[Hypergeometric2F1[1/3, 2/3, 1, x^2], {x, 0, 1}], 2 (3 ArcSinh[1/(2 Sqrt[2])] + Log[8])/(Sqrt[3] Pi)] N[% /. Equal -> List, 20]
Maybe other list readers do not find so much fun when "chasing a trail of smoke and reason" (though it does go even higher). So let's look at another geometry:
Show[ContourPlot[ -p^2 + q^2 - (4/27)*(-3*q*p^2 - q^3)^2 == 0, {q, -2, 2}, {p, -2, 2}]]
Should we be surprised to find out that the area interior to sextic lemniscate zbVV, up to harmonic scale factor pi/sqrt(3), equals a rational number 3/2? --Brad
On Fri, Apr 26, 2019 at 11:01 AM Brad Klee <bradklee@gmail.com> wrote:
Does anyone else have an opinion? --Brad
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
You know, the Fibonacci numbers are really fun to study, especially in relation to the golden mean, Penrose tiling, golden spiral, etc. Analysis via Benford's law, though somewhat interesting, unfortunately makes the Fibonacci idea look terribly one sided. Now if you take the integrals On Sat, Apr 27, 2019 at 6:37 PM Victor Miller <victorsmiller@gmail.com> wrote:
Look at this: https://mathoverflow.net/questions/87692/leading-digits-of-fibonacci-sequenc...
(sorry accidental misfire, my dominant wrist is injured) Continuing... If you take the integrals: Integrate[Hypergeometric2F1[1/2, 1/2, 1, x], {x, 0, 1}] == 4/Pi Integrate[Hypergeometric2F1[1/2, 1/2, 1, 1 - x], {x, 0, 1}] == 4/Pi And compare with Harold Edwards' normal form for elliptic curves, then it is easy to interpret the integrals as the area 4 of a 2x2 square, divided by Pi. Transformation of the elliptic curve explains why the quartic lemniscate along one complex dimension would again have a simple area integral. You could then take the geometry as a diversion to physics or even these days to Elliptic Curve Cryptography. Instead, how about a few more maths curiosities: Integrate[Hypergeometric2F1[1/2, 1/2, 1, x^2], {x, 0, 1}] ==4*Catalan/Pi Integrate[Hypergeometric2F1[1/2, 1/2, 1, 1 - x^2], {x, 0, 1}] == Pi/2 There is no digital pattern, nor any pattern in continued fraction digits, but both of these numbers have relatively simple Dirichlet series. Compare with: Integrate[Hypergeometric2F1[1/6, 5/6, 1, x^2], {x, 0, 1}] ==3*Sqrt[3]*ArcCsch[Sqrt[2]]/Pi Integrate[Hypergeometric2F1[1/6, 5/6, 1, 1 - x^2], {x, 0, 1}] ==3*Sqrt[3] /4 So, hmm... the z=1-x^2 integrals differ--only one algebraic--while the x^2 integrals are similar with a factor 1/Pi. If we follow Bruce Berndt and define signature s, then rewrite parameters a=1/s, b=(s-1)/s, c=1, then we have Elliptic K for s=2, and Ramanujan K for s=3,4, and 6. Elliptic curves are well known for z=x, z'=1-x. When we go to quadratic argument: z=x^2, z'=1-x^2, then the aforementioned integrals determine areas interior to the following few lemniscate curves: Show[ ContourPlot[ p^2 - q^2 + (4/27) (q^6) == 0, {q, -2, 2}, {p, -2, 2}, ContourStyle -> Red], ContourPlot[ -p^2 + q^2 - (4/27) (-3 q p^2 - q^3)^2 == 0, {q, -2, 2}, {p, -2, 2}, ContourStyle -> Blue ] /. {x_, y_} :> {x, y + 4}, ContourPlot[ p^2 - q^2 + (1/4) (q^4 + 6 q^2 p^2 + p^4) == 0, {q, -2, 2}, {p, -2, 2}, ContourStyle -> Green ] /. {x_, y_} :> {x, y + 2}, PlotRange -> {{-2.5, 2.5}, {-1.5, 5}}] https://0x0.st/zcZ0.png At this point we can start to distinguish between Elliptic K and Ramanujan K. Why is it so easy to find an algebraic model for the Ramanujan integrals with quadratic argument? Does an algebraic model exist for Elliptic K with quadratic argument? This last question seems huge, because it would give another (possibly new) definition of Catalan's constant. Or focusing on R's three chosen parameter sets, why should the periods on higher-genus curves work out to satisfy such a simple D.E.? The situation is similar to what we have seen with the Klein Quartic, just one curve. But here we have energy modulus z and three entire families!
From reading the archives, I see that N O A M E L K I E S sometimes writes the list. His article "Klein Quartic in Number Theory" covers similar ground, and seems to show a sense of fun for symmetry analysis and period integrals.
So why shouldn't Noam Elkies respond with a perspective on the decomposition symmetries of Riemannian surfaces? Or is this request a geopolitical no-go? Or a class issue? Because I'm new? Or what? According to the "Man who Knew Infinity", near where Ramanujan meditated for three days to obtain adesh: "the air was full of smoke, and the stone walls black with incense". What a nice detail giving a sense of time, where three days does not seem like too long to wait! The history around these integrals is also quite fun and enjoyable to learn! Cheers, Brad On Mon, Apr 29, 2019 at 1:33 PM Brad Klee <bradklee@gmail.com> wrote:
You know, the Fibonacci numbers are really fun to study, especially in relation to the golden mean, Penrose tiling, golden spiral, etc. Analysis via Benford's law, though somewhat interesting, unfortunately makes the Fibonacci idea look terribly one sided.
Now if you take the integrals
On Sat, Apr 27, 2019 at 6:37 PM Victor Miller <victorsmiller@gmail.com> wrote:
Look at this: https://mathoverflow.net/questions/87692/leading-digits-of-fibonacci-sequenc...
Surprises come in different flavors . . . so when I see something that looks like a digit-pattern, I’m surprised.
If you need to see everything in terms of digits, then why not read the original post as inherently about digit patterns? The number 3 can also be written 3.000000000...etc, with a pattern of repeating zeros that never ends. The other integral value I mentioned, sqrt(3)*log(8)/pi = 1.1464562082685322958... doesn't repeat at all. This is more typical of what I would expect given the constraints. Here is another tendril about changing parameters: Integrate[Hypergeometric2F1[1/4,3/4,1,x^2], {x,0,1}]==(1/2)*(ArcCosh[17]+4*ArcSinh[1])/Pi N[% /. Equal -> List, 20] Integrate[Hypergeometric2F1[1/4,3/4,1,1-x^2], {x,0,1}]==Sqrt[2] N[% /. Equal -> List, 20] Neither of these have discernible decimal digit patterns. However sqrt(2) does have a repeating continued fraction, so if you are looking for digit patterns, there you go, another easy one. It's easy to say that you have "scabs of theorems", but I can't make any reasonable response to a statement with no particular meaning. Does one of your "scats of theorems" have anything to do with Ramanujan's elliptic integrals K1, K2, and K3? --Brad
participants (4)
-
Andres Valloud -
Brad Klee -
James Propp -
Victor Miller