Post AS2Gqdbu9XP74ftnRQ by j@mathstodon.xyz
 (DIR) More posts by j@mathstodon.xyz
 (DIR) Post #AS2GqZsm0sWLVplDBA by robinhouston@mathstodon.xyz
       2023-01-02T22:41:14Z
       
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       According to Oded Margalit in https://youtu.be/CYKXBkGkpUI, it is an open problem to explain why the value of \[\int_0^\infty\cos(2x)\prod_{n=1}^\infty\cos(\frac xn)\ dx\] is almost, but not exactly, \(\frac\pi8\)
       
 (DIR) Post #AS2GqaJiOi4UrOMjAm by robinhouston@mathstodon.xyz
       2023-01-02T22:54:16Z
       
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       I’m not even sure how to evaluate the product numerically. Does anyone know a good way?*Edit*: Oh, never mind! It’s quite easy to approximate, since the terms tend to 1, just by evaluating a finite product up to some bound.
       
 (DIR) Post #AS2GqasoIE90cEmksK by robinhouston@mathstodon.xyz
       2023-01-02T23:23:18Z
       
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       It‘s equal to π/8 until the 42nd decimal place. It would require some very careful numerical analysis to realise that this is not just numerical error! I’d love to know more about how it was discovered.
       
 (DIR) Post #AS2GqbEmyVj1iP4J8K by johncarlosbaez@mathstodon.xyz
       2023-01-03T00:24:53Z
       
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       @robinhouston - I'd like to know how someone did the integral, since if you replace the product by a finite product you get an integral of a periodic function from 0 to infinity.  Such an integral does not converge!Maybe the infinite product makes the integral converge.  But it's a bit delicate.
       
 (DIR) Post #AS2Gqbbpaq9mrrqi36 by robinhouston@mathstodon.xyz
       2023-01-03T00:26:00Z
       
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       @johncarlosbaez That’s what I want to know! It looks like a fairly challenging exercise in numerical analysis.
       
 (DIR) Post #AS2Gqc2PzzQMCKHwUS by johncarlosbaez@mathstodon.xyz
       2023-01-03T01:45:48Z
       
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       @robinhouston - actually, I'm going to be suspicious of  Margalit's claim that this integral is very close but not quite equal to π/8 until I see some more evidence!   It resembles the Borwein integrals, but differs from them in some crucial ways.
       
 (DIR) Post #AS2GqcTiMVG5Yz3k2K by gregeganSF@mathstodon.xyz
       2023-01-03T05:52:55Z
       
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       @johncarlosbaez @robinhouston Amusingly (and maybe even usefully, though I haven’t found a way to leverage it yet):\[ \prod_{n=1}^\infty (1-(x/n)^2/2) = \frac{\sqrt{2} \sin \left(\frac{\pi  x}{\sqrt{2}}\right)}{\pi  x} \]  where the terms in this product are the second-order Taylor series for the cosines in the true product, and serve as lower bounds on those cosines for sufficiently small x/n.
       
 (DIR) Post #AS2Gqd3sC4BLN7ycOe by j@mathstodon.xyz
       2023-01-03T06:16:10Z
       
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       @gregeganSF @johncarlosbaez @robinhouston That's getting somewhere toward actually evaluating the integral, if you expand it to  4th order terms and integrate,\[ \begin{align}&\int_0^\infty \mathrm{d}{x} \cos (2x) \prod_{n=1}^\infty\left(1 - \frac{(x/n)^2}{2} + \frac{(x/n)^4}{24}\right) \\ &= \frac{\sqrt{3 + \sqrt{6}}}{2} - \frac{\sqrt{6}}{\pi}\end{align}\]\( \approx 0.38751 \approx \pi/8 \)
       
 (DIR) Post #AS2Gqdbu9XP74ftnRQ by j@mathstodon.xyz
       2023-01-03T06:25:38Z
       
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       @gregeganSF @johncarlosbaez @robinhouston starting to wonder if it actually is equal to \(\pi / 8\) and can be proved with a simple exchange of limits ± the arguments for why its valid to do  so.
       
 (DIR) Post #AS2Gqe3YUjWQSQpsXY by gregeganSF@mathstodon.xyz
       2023-01-03T06:46:14Z
       
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       @j @johncarlosbaez @robinhouston It looks as if (just from messing around in Mathematica, I have no rigorous explanation for this) the infinite product of the order-2p Taylor series for cos(x/n) is always a product of p sinc functions, e.g. for order 6 it is:sinc(a x) sinc(b x) sinc(c x)where a,b,c are 3 roots of the polynomial:720𝑧⁶−360𝑧⁴+30𝑧²−1  Because sinc is the FT of a rectangular pulse, the cos(2x) gives us a certain point on the convolution of several rectangular pulses.
       
 (DIR) Post #AS2Gqkuwz7nXjlrbbE by gregeganSF@mathstodon.xyz
       2023-01-03T07:26:14Z
       
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       @j @johncarlosbaez @robinhouston Associated with the infinite product of the order-2p Taylor approximation for cos(x/n) is the polynomial:\[ f(z)=\sum _{i=0}^p \frac{(-1)^{p+i} (2 p)! z^{2 i}}{(2 (p-i))!} \] Empirically, it looks like:\[ \prod_{n=1}^\infty \cos_{2p}(x/n) = \prod_{i=1}^p \frac{\sin(\pi a_i x)}{\pi a_i x} \] where the \( a_i \) are roots of f(z) and \( \cos_{2p} \)  is the order-2p Taylor approximation for cos.
       
 (DIR) Post #AS2GqqtOlo7cGhBS0u by gregeganSF@mathstodon.xyz
       2023-01-03T11:53:32Z
       
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       @j @johncarlosbaez @robinhouston For p=500, and making a somewhat arbitrary choice to cut off the integral at x=400 (where the magnitude of the integrand has fallen to around 10^{-200} before it starts to blow up because the Taylor approximation starts to fail), the integral divided by π/8 computed with 500-digit precision is:0.99999999999999999999999999999999999999999811373468...That’s 41 9s.Coincidence? I’ll see if I can replicate with p=1000.
       
 (DIR) Post #AS2GqxhbS3qVO8id6G by gregeganSF@mathstodon.xyz
       2023-01-03T21:16:38Z
       
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       @j @johncarlosbaez @robinhouston I got the same result for the first 50 digits of the ratio with p=1000 and integrating out to x=500. So the difference from π/8 seems robust.