Just ask Cleo
My real name is Cleo, I'm female. I have a medical condition that makes it very difficult for me to engage in conversations, or post long answers, sorry for that. I like math and do my best to be useful at this site, although I realize my answers might be not useful for everyone.
There's a website called Math StackExchange where people ask and answer questions. When hard integrals come up, Cleo often does them - with no explanation! She has a lot of fans now.
The integral here is a good example. When you replace ln³(1+x) by ln²(1+x) or just ln(1+x), the answers were already known. The answers involve the third Riemann zeta value:
ζ(3) = 1/1³ + 1/2³ + 1/3³ + 1/4³ + ...
They also involve the fourth polylogarithm function:
Li₄(x) = x + x²/2⁴ + x³/3⁴ + ...
Cleo found that the integral including ln³(1+x) can be done in a similar way - but it's much more complicated. She didn't explain her answer... but someone checked it with a computer and showed it was right to 1000 decimal places. Then someone gave a proof.
The number
ζ(3) = 1.202056903159594285399738161511449990764986292...
is famous because it was proved to be irrational only after a lot of struggle. Apéry found a proof in 1979. Even now, nobody is sure that the similar numbers ζ(5), ζ(7), ζ(9)... are irrational, though most of us believe it. The numbers ζ(2), ζ(4), ζ(6)... are much easier to handle. Euler figured out formulas for them involving powers of pi, and they're all irrational.
But here's a wonderful bit of progress: in 2001, Wadim Zudilin proved that at least one of the numbers ζ(5), ζ(7), ζ(9), and ζ(11) must be irrational. Sometimes we can only snatch tiny crumbs of knowledge from the math gods, but they're still precious.
For Cleo's posts, go here:
https://math.stackexchange.com/users/97378/cleo
For more on ζ(3), go here:
https://en.wikipedia.org/wiki/Apery's_constant
This number shows up in some physics problems, like computing the magnetic field produced by an electron! And that's just the tip of an iceberg: there are deep connections between Feynman diagrams, the numbers ζ(n), and mysterious mathematical entities glimpsed by Grothendieck, called 'motives'. Very roughly, a motive is what's left of a space if all you care about are the results of integrals over surfaces in this space.
The world record for computing digits of ζ(3) is currently held by Dipanjan Nag: in 2015 he computed 400,000,000,000 digits. But here's something cooler: David Broadhurst, who works on Feynman diagrams and numbers like ζ(n), has shown that there's a linear-time algorithm to compute the nth binary digit of ζ(3):
• David Broadhurst, Polylogarithmic ladders, hypergeometric series and the ten millionth digits of ζ(3) and ζ(5), available at https://arxiv.org/abs/math/9803067
He exploits how Riemann zeta values ζ(n) are connected to polylogarithms... it's easy to see that
Liₙ(1) = ζ(n)
but at a deeper level this connection involves motives. For more on polylogarithms, go here:
https://en.wikipedia.org/wiki/Polylogarithm
Thanks to +David Roberts for pointing out Cleo's posts on Math StackExchange!
#bigness
My real name is Cleo, I'm female. I have a medical condition that makes it very difficult for me to engage in conversations, or post long answers, sorry for that. I like math and do my best to be useful at this site, although I realize my answers might be not useful for everyone.
There's a website called Math StackExchange where people ask and answer questions. When hard integrals come up, Cleo often does them - with no explanation! She has a lot of fans now.
The integral here is a good example. When you replace ln³(1+x) by ln²(1+x) or just ln(1+x), the answers were already known. The answers involve the third Riemann zeta value:
ζ(3) = 1/1³ + 1/2³ + 1/3³ + 1/4³ + ...
They also involve the fourth polylogarithm function:
Li₄(x) = x + x²/2⁴ + x³/3⁴ + ...
Cleo found that the integral including ln³(1+x) can be done in a similar way - but it's much more complicated. She didn't explain her answer... but someone checked it with a computer and showed it was right to 1000 decimal places. Then someone gave a proof.
The number
ζ(3) = 1.202056903159594285399738161511449990764986292...
is famous because it was proved to be irrational only after a lot of struggle. Apéry found a proof in 1979. Even now, nobody is sure that the similar numbers ζ(5), ζ(7), ζ(9)... are irrational, though most of us believe it. The numbers ζ(2), ζ(4), ζ(6)... are much easier to handle. Euler figured out formulas for them involving powers of pi, and they're all irrational.
But here's a wonderful bit of progress: in 2001, Wadim Zudilin proved that at least one of the numbers ζ(5), ζ(7), ζ(9), and ζ(11) must be irrational. Sometimes we can only snatch tiny crumbs of knowledge from the math gods, but they're still precious.
For Cleo's posts, go here:
https://math.stackexchange.com/users/97378/cleo
For more on ζ(3), go here:
https://en.wikipedia.org/wiki/Apery's_constant
This number shows up in some physics problems, like computing the magnetic field produced by an electron! And that's just the tip of an iceberg: there are deep connections between Feynman diagrams, the numbers ζ(n), and mysterious mathematical entities glimpsed by Grothendieck, called 'motives'. Very roughly, a motive is what's left of a space if all you care about are the results of integrals over surfaces in this space.
The world record for computing digits of ζ(3) is currently held by Dipanjan Nag: in 2015 he computed 400,000,000,000 digits. But here's something cooler: David Broadhurst, who works on Feynman diagrams and numbers like ζ(n), has shown that there's a linear-time algorithm to compute the nth binary digit of ζ(3):
• David Broadhurst, Polylogarithmic ladders, hypergeometric series and the ten millionth digits of ζ(3) and ζ(5), available at https://arxiv.org/abs/math/9803067
He exploits how Riemann zeta values ζ(n) are connected to polylogarithms... it's easy to see that
Liₙ(1) = ζ(n)
but at a deeper level this connection involves motives. For more on polylogarithms, go here:
https://en.wikipedia.org/wiki/Polylogarithm
Thanks to +David Roberts for pointing out Cleo's posts on Math StackExchange!
#bigness

View 37 previous comments
+Linas Vepstas - you asked about the decidability and computational complexity of problems associated to definite integrals. There seems to be more known about indefinite integrals, yet still some fundamental questions remain known. Let me quote a bit from Wikipedia:
The Risch algorithm is used to integrate elementary functions. These are functions obtained by composing exponentials, logarithms, radicals, trigonometric functions, and the four arithmetic operations (+ − × ÷).
They explain it's an algorithm that finds an indefinite integral of an elementary function if the indefinite integral is elementary. They then admit:
The Risch algorithm applied to general elementary functions is not an algorithm but a semi-algorithm because it needs to check, as a part of its operation, if certain expressions are equivalent to zero (constant problem), in particular in the constant field. For expressions that involve only functions commonly taken to be elementary it is not known whether an algorithm performing such a check exists or not (current computer algebra systems use heuristics); moreover, if one adds the absolute value function to the list of elementary functions, it is known that no such algorithm exists; see Richardson's theorem.Jun 30, 2017
Computerized (poly)Logarithmic Equation Optimizer. And also an amazing troll. Some people there get astoundingly upset.Jun 30, 2017
The equivalence of group presentations is (famously) not decidable, so it should not be a surprise that other algebraic problems would fall into this class. None-the-less, I cannot help avoiding thinking of these problems as involving some sort of landscape or lattice where some points "touch" i.e. are solvable and most don't.
I also can't avoid thinking of the substrate to this landscape as being extremely hyperbolic. For example, a sequence of algebraic manipulations branch like crazy, (with a choice of which manipulation to apply next) but still take the form of trees, and trees are embeddable into hyperbolic spaces. (the plane if you wish) So if I perform such an embedding, there will be certain "spots" where the algebraic manipulation resulted in a closed solution. What is the geometric structure of this thing? What is it's "shape"?Jun 30, 2017
Looking at the polylog function, and Li_n (1) = ζ(n), I thought I would try to calculate some values, but there are singularities at Li_n (1).
For instance, Li_0 (x) = x/1-x
But (for example) if I want Li_0(1) , then this is undefined. What is the most straightforward way to handle that? It doesn't seem to be a residue. I don't see a way to use L'Hopital.
This seems to be a problem at most x=1Jul 1, 2017
+David Foster Computing the polylog is not so easy. Here's a reference, and source code for computing it: arxiv.org - [math/0702243] An efficient algorithm for accelerating the convergence of oscillatory series, useful for computing the polylogarithm and Hurwitz zeta functions
arxiv.org - [math/0702243] An efficient algorithm for accelerating the convergence of oscillatory series, useful for computing the polylogarithm and Hurwitz zeta functions The source code: https://github.com/linas/anant It works fairly well across some reasonable chunk of the complex plane, but gets real tough to ascend the imaginary axis.
Well, Li_n(1/2) is real easy to compute, because the series converges rapidly. It gets harder to extend outside of the unit circle. There are two cuts in the complex plane and a cool-looking monodromy between them. For LI_2, the monodromy is the Heisenberg group. Which is kind-of cool. It all kind-of shows up in affine Lie groups, and the Moyal product on universal enveloping algebras, but that's something else again.Jul 1, 2017
Thanks, +Linas Vepstas I will take a look at that.Jul 2, 2017
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