6 ms·

Mathematicians revive abandoned approach to the Riemann Hypothesis

91 points by new_guy 7y ago

oneepic 7y ago
Neat. I mean, I don't understand the mathspeak since I stopped taking math after calculus in college, but it's neat to see fun approaches turn into viable tactics for finding a solution. By coincidence I toyed with the problem a bit this past week. Pretty difficult to tackle with a "basic" math toolbox (calc, trig, algebra, and some discrete math/number theory that's taught in many CS undergrad curricula). Much like Fermat's Last Theorem, it seems like this might require much higher-level math to solve what seems like a simple problem.
- auggierose 7y ago
  Oh really? :-D
  - Kurtz79 7y ago
    Well, parent tried to understand the problem, used the tools at his hand to solve it without preconceptions, and realized it was more complex that it seemed. The world would be a better place if more people had the same approach to problem-solving, IMHO.
    - auggierose 7y ago
      Not saying that there is anything wrong with it. It is still funny though :-) It is like somebody saying they tried to go to the moon, but after walking for an hour they discovered it was harder to get there than they initially thought.
  - oneepic 7y ago
    Yes! I had a lot of ideas but didn't get anywhere, including: - reframe the real part of a complex input s as 1/2 + k for -1/2 < k < 1/2, since we're only looking at the critical strip 0 < Re(s) < 1. Then show k has to be zero for the math to work out, so that the real AND complex parts of the value of the function at s are both zero. Eventually failed because I can only show that Re(s) = 1/2 would work out relatively "nicely" so you could more easily pick Imag(s) to find zeroes, but can't prove that a different real value would still let the math work out so you could find a zero in the first place. - come up with a formula to sum the series. I don't have much math know-how in this regard, except for some tricks I learned in discrete math/algos class so I quickly ran out of ideas. Pairing sum terms together (like you can do to prove the sum of telescoping series, or of natural no's from 0 to N) doesn't seem to work. - The original sum formula looks like sum(n=1, infinity, ((-1)^(n+1))/(n^s)) where s is the complex input. I tried to turn it into a continuous function by using -cos(pi*n) for the numerator and then use a definite integral, but the result is nasty before you even plug in the integral bounds. There's an imaginary term as the second argument for the gamma function, which I don't understand how to handle. - show two different real values, r1 and r2, of zeroes at r1 + ci and r2 + di for some imaginary parts c and d. Then show the ratio r1/r2 = 1, and since we know r1 = 1/2 works, we know 1/2 is the only possible real part that works. Much like the summing stuff above, I couldn't figure out how to simplify things enough to even get an equation in terms of either r1 or r2. - show that two zeroes on either side of the line s = 1/2 + iy cannot exist, or at least something like if one exists then the other must not exist, which contradicts the notion that we'd have two zeroes on either side of the line, where one has a real part between 0 and 1/2 (exclusive) and the other is in (1/2, 1). I don't have a good way of actually attacking this idea though. There are some other ideas I tried, but the main problem is I can't simplify the sum. The sum itself plus the fact there's exponents in every denominator makes it especially hard to simplify algebraically, or at least I don't know how to do it.
- signa11 7y ago
  no shit sherlock !
- Ragib_Zaman 7y ago
  Man HR is harsh as hell. While yes, if a proof of RH is found it will probably use advanced tools, ever finding a proof with elementary tools is not totally out of the question. Regarding the Prime Number Theorem, which can be seen as a baby version of RH, experts in the field such as Hardy have been quoted as saying elementary proofs would not exist. Then 50 years after Hadamard and Poussin found the original proofs, Selberg and Erdos published proofs based on elementary methods. No one can say that it's worthless for someone to try the same for RH.
  - pfortuny 7y ago
    Notice also the decidability of primality in polynomial time, which was proved by some Computer Science graduates (Neeraj Kayal, and Nitin Saxena) in 2011, and led Bombieri (one of the leading experts on Number Theory) to say that Number Theory may have gone astray in theoretical complexity in some fields.
    - throwawaymath 7y ago
      That's very interesting, do you know offhand where Bombieri said that?
  - math-fan 7y ago
    I've taken math up to calculus and linear algebra and enjoy watching 3B1B proofs. Is that enough to follow the elementary Prime Number Theorem proof? I've been interested in understanding that for a while. Any good links?
    - dilippkumar 7y ago
      Check out the book “Prime Obsession” It was easy for me to follow along.
    - pmiller2 7y ago
      You can probably slog through it with that level of knowledge, if you’v got a good grasp of summation notation. Here’s one version of the proof: https://www.math.lsu.edu/~mahlburg/teaching/handouts/2014-7230/Selberg-ElemPNT1949.pdf https://www.math.lsu.edu/~mahlburg/teaching/handouts/2014-72... See also: https://www.math.columbia.edu/~goldfeld/ErdosSelbergDispute.pdf https://www.math.columbia.edu/~goldfeld/ErdosSelbergDispute.... https://www.math.columbia.edu/~goldfeld/ErdosSelbergDispute.pdf https://www.math.columbia.edu/~goldfeld/ErdosSelbergDispute.... And, a related, very simple and elementary proof of Chebyschev’s theorem that there is always a prime number between n and 2n: https://en.wikipedia.org/wiki/Proof_of_Bertrand%27s_postulate https://en.wikipedia.org/wiki/Proof_of_Bertrand%27s_postulat...
nutanc 7y ago
Nice detailed lectures on RH along with a suggested proof is here https://www.youtube.com/playlist?list=PLRsxymPrOKUAk3eXhK9FZdGAPmgGGrhfb&disable_polymer=true https://www.youtube.com/playlist?list=PLRsxymPrOKUAk3eXhK9FZ...
- gjm11 7y ago
  I have looked at Eswaran's paper that purports to prove RH and I'm pretty sure it's all wrong. (Currently-latest version of paper is here: https://arxiv.org/pdf/1609.06971v10.pdf https://arxiv.org/pdf/1609.06971v10.pdf) The basic idea is to prove that a certain function L(N) doesn't grow too quickly. It's true that appropriate bounds on L(N) imply RH. But Eswaran's arguments for this point are full of handwaving, and in this field handwaving is just not good enough. L(N) is a sum of terms +1 and -1, and he claims that it behaves like a series of coin tosses, but the specific ways he argues it's like a series of coin tosses are not enough for the conclusion he needs. E.g., in section 5, on page 13, he states a theorem which says: you can pair up the +1 and -1 terms in that series, so that for each +1 there's a -1 and for each -1 there's a +1. This is certainly true, but it doesn't imply that the series is random-like or that the partial sums L(N) don't grow much faster than sqrt(N) (which is the condition he needs). E.g., consider the series that goes ++-++-++-++-++-++- etc., whose partial sums obviously grow like N/3. This satisfies that "pairing" condition just fine. (Pair the first + with the first -, the second + with the second -, etc. There are infinitely many of both, which is all you need to make that work.) In appendix V, he tries a different approach. (The fact that he gives two different "proofs" of the same key result, and both are handwavy, is a Very Bad Sign.) There's lots here that looks wrong, but here's one thing: he claims to prove that if |L(N)| ~ N^a for some a then we must have a=1/2, and having given his proof he claims to have proved that in fact |L(N)| ~ N^1/2. But these are very different propositions. Another possibility is that we don't have |L(N)| ~ N^a for any a at all, which in fact is true because it's known that infinitely often |L(N)|<=1 and infinitely often |L(N)|>=sqrt(N). The lectures appear to be about his alleged proof rather than about RH in general. In any case, I would not trust someone capable of making the errors in his RH paper to teach any research-level material in pure mathematics.
  - nutanc 7y ago
ncmncm 7y ago
I've seen a paper[1] that claims RH, as stated, is undecideable, and then proposes to prove that an analytic continuation of it has the property required. The idea seems to be that you can choose your continuation to have or not have the property. It's not clear to me what could be undecideable about the original proposition; either the continuation matches at the points in question, or it doesn't, and if it doesn't, how is it a continuation? Enlightenment welcome. [1] http://phys.lsu.edu/~fmoxley/r1.pdf http://phys.lsu.edu/~fmoxley/r1.pdf
- pmiller2 7y ago
  Not having read the paper, I’m skeptical, both because of the author’s relative lack of credentials (he’s a physicist with all of 5 publications, from what I can see), and because the core idea seems too simple. However, the paper is clearly a serious attempt by someone inside academia who works in quantum computing, which is a related and heavily mathematical field. You may have just given me a few hours of weekend reading material. :)
- soVeryTired 7y ago
  I've been following the development of the Bender-Brody-Muller approach to the Riemann hypothesis because I knew one of the authors in a previous life. I don't think it's taken that seriously by number theorists [0][1]. The paper was written by physicists, who sometimes play a little fast and loose with mathematical rigour. Maybe this line of argument fixes the flaws in the original BBM paper in some way - I'm not qualified to comment - but I sort of doubt it. [0] https://arxiv.org/abs/1704.02644 https://arxiv.org/abs/1704.02644 [1] https://math.stackexchange.com/questions/2211278/riemann-hypothesis-is-bender-brody-m%C3%BCller-hamiltonian-a-new-line-of-attack https://math.stackexchange.com/questions/2211278/riemann-hyp...
- throwawaymath 7y ago
  I'm not familiar with that particular paper, but I'm aware of a similar line of attack from a decade or so ago. Off the top of my head I can tell you that we know Riemann is provably false if it is false[1]. I'm not aware of any comparable result for the affirmative, but personally I'm a bit skeptical of the undecidability argument. That being said, it would be kind of narratively satisfying if Riemann is undecidable. There are many novel theorems from the past two decades which are conditionally true assuming Riemann is true. Riemann itself has become a useful technique for conditionally proving new theorem. If Riemann is undecidable, it would imply a great deal of mathematics has been developed which compartmentalizes the undecidability of other theorems. ______________________ 1. https://mathoverflow.net/questions/79685/can-the-riemann-hypothesis-be-undecidable https://mathoverflow.net/questions/79685/can-the-riemann-hyp...
twic 7y ago
The idea for the paper was sparked two years ago by a "toy problem" that Ono presented as a "gift" to entertain Zagier during the lead-up to a math conference celebrating his 65th birthday. A toy problem is a scaled-down version of a bigger, more complicated problem that mathematicians are trying to solve. Zagier described the one that Ono gave him as "a cute problem about the asymptotic behavior of certain polynomials involving Euler's partition function, which is an old love of mine and of Ken's—and of about pretty much any classical number theorist." "I found the problem intractable and I didn't really expect Don to get anywhere with it," Ono recalls. "But he thought the challenge was super fun and soon he had crafted a solution." That sounds like a potentially useful problem-solving trick - pull a small but key problem out of your bigger problem, and then nerd-snipe your friends into solving it.
- pmiller2 7y ago
  Definitely. Two fairly standard techniques in mathematics are solving a cut down version of a problem and lifting that to a full solution, and solving a more general version and specializing to a particular case. Sometimes, amazingly, the more general problem is easier to solve, perhaps because fewer irrelevant details stand in the way. Both methods have yielded a lot of progress in mathematics. For example, solving a cut down version of Fermat’s Last Theorem got us the theory of ideals in algebra. Relatedly, Wiles work on the modularity theorem ultimately led to the actual proof of FLT.
  - throwawaymath 7y ago
    > Sometimes, amazingly, the more general problem is easier to solve, perhaps because fewer irrelevant details stand in the way. True, and that's why we like to generalize things. It's usually easier to prove a mathematical object belongs to another class of structure which has certain properties than to prove the object itself has all those properties. Of course as you've noted it's not always like that at all. It's significantly easier to prove many things about finite-dimensional vector spaces over fields versus infinite-dimensional modules over rings. Solving mathematical problems is a lot like working a sponge. It's often useful to alternate between expanding and contracting the scope of your problem.
archgoon 7y ago
A mathematician has spent years trying to prove the Riemann hypothesis – without success. Finally, he decides to sell his soul to the devil in exchange for a proof. The devil promises to deliver a proof within four weeks. Four weeks pass, but nothing happens. Half a year later, the devil shows up again – in a rather gloomy mood. "I’m sorry", he says. "I couldn’t prove the Riemann hypothesis either. But" – and his face lightens up – "I think I found a really interesting lemma…"
butterthebuddha 7y ago
"This method doesn't work and Ken knows that, not sure wtf this article is. At most, this idea can show that the zeroes off the critical line are density zero, which would be a huge result, but could never prove RH. Again, Ken knows this (he says as much when he talks about it) so wtf is this article?" - Someone on Reddit (https://old.reddit.com/r/math/comments/brgp5z/mathematicians_revive_abandoned_approach_to_the/ https://old.reddit.com/r/math/comments/brgp5z/mathematicians...)

91 points by new_guy 7y ago

Neat. I mean, I don't understand the mathspeak since I stopped taking math after calculus in college, but it's neat to see fun approaches turn into viable tactics for finding a solution. By coincidence I toyed with the problem a bit this past week. Pretty difficult to tackle with a "basic" math toolbox (calc, trig, algebra, and some discrete math/number theory that's taught in many CS undergrad curricula). Much like Fermat's Last Theorem, it seems like this might require much higher-level math to solve what seems like a simple problem.

auggierose 7y ago
Oh really? :-D
- Kurtz79 7y ago
  Well, parent tried to understand the problem, used the tools at his hand to solve it without preconceptions, and realized it was more complex that it seemed. The world would be a better place if more people had the same approach to problem-solving, IMHO.
  - auggierose 7y ago
    Not saying that there is anything wrong with it. It is still funny though :-) It is like somebody saying they tried to go to the moon, but after walking for an hour they discovered it was harder to get there than they initially thought.
- oneepic 7y ago
  Yes! I had a lot of ideas but didn't get anywhere, including: - reframe the real part of a complex input s as 1/2 + k for -1/2 < k < 1/2, since we're only looking at the critical strip 0 < Re(s) < 1. Then show k has to be zero for the math to work out, so that the real AND complex parts of the value of the function at s are both zero. Eventually failed because I can only show that Re(s) = 1/2 would work out relatively "nicely" so you could more easily pick Imag(s) to find zeroes, but can't prove that a different real value would still let the math work out so you could find a zero in the first place. - come up with a formula to sum the series. I don't have much math know-how in this regard, except for some tricks I learned in discrete math/algos class so I quickly ran out of ideas. Pairing sum terms together (like you can do to prove the sum of telescoping series, or of natural no's from 0 to N) doesn't seem to work. - The original sum formula looks like sum(n=1, infinity, ((-1)^(n+1))/(n^s)) where s is the complex input. I tried to turn it into a continuous function by using -cos(pi*n) for the numerator and then use a definite integral, but the result is nasty before you even plug in the integral bounds. There's an imaginary term as the second argument for the gamma function, which I don't understand how to handle. - show two different real values, r1 and r2, of zeroes at r1 + ci and r2 + di for some imaginary parts c and d. Then show the ratio r1/r2 = 1, and since we know r1 = 1/2 works, we know 1/2 is the only possible real part that works. Much like the summing stuff above, I couldn't figure out how to simplify things enough to even get an equation in terms of either r1 or r2. - show that two zeroes on either side of the line s = 1/2 + iy cannot exist, or at least something like if one exists then the other must not exist, which contradicts the notion that we'd have two zeroes on either side of the line, where one has a real part between 0 and 1/2 (exclusive) and the other is in (1/2, 1). I don't have a good way of actually attacking this idea though. There are some other ideas I tried, but the main problem is I can't simplify the sum. The sum itself plus the fact there's exponents in every denominator makes it especially hard to simplify algebraically, or at least I don't know how to do it.
signa11 7y ago
no shit sherlock !
Ragib_Zaman 7y ago
Man HR is harsh as hell. While yes, if a proof of RH is found it will probably use advanced tools, ever finding a proof with elementary tools is not totally out of the question. Regarding the Prime Number Theorem, which can be seen as a baby version of RH, experts in the field such as Hardy have been quoted as saying elementary proofs would not exist. Then 50 years after Hadamard and Poussin found the original proofs, Selberg and Erdos published proofs based on elementary methods. No one can say that it's worthless for someone to try the same for RH.
- pfortuny 7y ago
  Notice also the decidability of primality in polynomial time, which was proved by some Computer Science graduates (Neeraj Kayal, and Nitin Saxena) in 2011, and led Bombieri (one of the leading experts on Number Theory) to say that Number Theory may have gone astray in theoretical complexity in some fields.
  - throwawaymath 7y ago
    That's very interesting, do you know offhand where Bombieri said that?
- math-fan 7y ago
  I've taken math up to calculus and linear algebra and enjoy watching 3B1B proofs. Is that enough to follow the elementary Prime Number Theorem proof? I've been interested in understanding that for a while. Any good links?
  - dilippkumar 7y ago
    Check out the book “Prime Obsession” It was easy for me to follow along.
  - pmiller2 7y ago
    You can probably slog through it with that level of knowledge, if you’v got a good grasp of summation notation. Here’s one version of the proof: https://www.math.lsu.edu/~mahlburg/teaching/handouts/2014-7230/Selberg-ElemPNT1949.pdf https://www.math.lsu.edu/~mahlburg/teaching/handouts/2014-72... See also: https://www.math.columbia.edu/~goldfeld/ErdosSelbergDispute.pdf https://www.math.columbia.edu/~goldfeld/ErdosSelbergDispute.... https://www.math.columbia.edu/~goldfeld/ErdosSelbergDispute.pdf https://www.math.columbia.edu/~goldfeld/ErdosSelbergDispute.... And, a related, very simple and elementary proof of Chebyschev’s theorem that there is always a prime number between n and 2n: https://en.wikipedia.org/wiki/Proof_of_Bertrand%27s_postulate https://en.wikipedia.org/wiki/Proof_of_Bertrand%27s_postulat...

nutanc 7y ago

Nice detailed lectures on RH along with a suggested proof is here https://www.youtube.com/playlist?list=PLRsxymPrOKUAk3eXhK9FZdGAPmgGGrhfb&disable_polymer=true https://www.youtube.com/playlist?list=PLRsxymPrOKUAk3eXhK9FZ...

gjm11 7y ago
I have looked at Eswaran's paper that purports to prove RH and I'm pretty sure it's all wrong. (Currently-latest version of paper is here: https://arxiv.org/pdf/1609.06971v10.pdf https://arxiv.org/pdf/1609.06971v10.pdf) The basic idea is to prove that a certain function L(N) doesn't grow too quickly. It's true that appropriate bounds on L(N) imply RH. But Eswaran's arguments for this point are full of handwaving, and in this field handwaving is just not good enough. L(N) is a sum of terms +1 and -1, and he claims that it behaves like a series of coin tosses, but the specific ways he argues it's like a series of coin tosses are not enough for the conclusion he needs. E.g., in section 5, on page 13, he states a theorem which says: you can pair up the +1 and -1 terms in that series, so that for each +1 there's a -1 and for each -1 there's a +1. This is certainly true, but it doesn't imply that the series is random-like or that the partial sums L(N) don't grow much faster than sqrt(N) (which is the condition he needs). E.g., consider the series that goes ++-++-++-++-++-++- etc., whose partial sums obviously grow like N/3. This satisfies that "pairing" condition just fine. (Pair the first + with the first -, the second + with the second -, etc. There are infinitely many of both, which is all you need to make that work.) In appendix V, he tries a different approach. (The fact that he gives two different "proofs" of the same key result, and both are handwavy, is a Very Bad Sign.) There's lots here that looks wrong, but here's one thing: he claims to prove that if |L(N)| ~ N^a for some a then we must have a=1/2, and having given his proof he claims to have proved that in fact |L(N)| ~ N^1/2. But these are very different propositions. Another possibility is that we don't have |L(N)| ~ N^a for any a at all, which in fact is true because it's known that infinitely often |L(N)|<=1 and infinitely often |L(N)|>=sqrt(N). The lectures appear to be about his alleged proof rather than about RH in general. In any case, I would not trust someone capable of making the errors in his RH paper to teach any research-level material in pure mathematics.
- nutanc 7y ago

ncmncm 7y ago

I've seen a paper[1] that claims RH, as stated, is undecideable, and then proposes to prove that an analytic continuation of it has the property required. The idea seems to be that you can choose your continuation to have or not have the property. It's not clear to me what could be undecideable about the original proposition; either the continuation matches at the points in question, or it doesn't, and if it doesn't, how is it a continuation? Enlightenment welcome. [1] http://phys.lsu.edu/~fmoxley/r1.pdf http://phys.lsu.edu/~fmoxley/r1.pdf

pmiller2 7y ago
Not having read the paper, I’m skeptical, both because of the author’s relative lack of credentials (he’s a physicist with all of 5 publications, from what I can see), and because the core idea seems too simple. However, the paper is clearly a serious attempt by someone inside academia who works in quantum computing, which is a related and heavily mathematical field. You may have just given me a few hours of weekend reading material. :)
soVeryTired 7y ago
I've been following the development of the Bender-Brody-Muller approach to the Riemann hypothesis because I knew one of the authors in a previous life. I don't think it's taken that seriously by number theorists [0][1]. The paper was written by physicists, who sometimes play a little fast and loose with mathematical rigour. Maybe this line of argument fixes the flaws in the original BBM paper in some way - I'm not qualified to comment - but I sort of doubt it. [0] https://arxiv.org/abs/1704.02644 https://arxiv.org/abs/1704.02644 [1] https://math.stackexchange.com/questions/2211278/riemann-hypothesis-is-bender-brody-m%C3%BCller-hamiltonian-a-new-line-of-attack https://math.stackexchange.com/questions/2211278/riemann-hyp...
throwawaymath 7y ago
I'm not familiar with that particular paper, but I'm aware of a similar line of attack from a decade or so ago. Off the top of my head I can tell you that we know Riemann is provably false if it is false[1]. I'm not aware of any comparable result for the affirmative, but personally I'm a bit skeptical of the undecidability argument. That being said, it would be kind of narratively satisfying if Riemann is undecidable. There are many novel theorems from the past two decades which are conditionally true assuming Riemann is true. Riemann itself has become a useful technique for conditionally proving new theorem. If Riemann is undecidable, it would imply a great deal of mathematics has been developed which compartmentalizes the undecidability of other theorems. ______________________ 1. https://mathoverflow.net/questions/79685/can-the-riemann-hypothesis-be-undecidable https://mathoverflow.net/questions/79685/can-the-riemann-hyp...

twic 7y ago

The idea for the paper was sparked two years ago by a "toy problem" that Ono presented as a "gift" to entertain Zagier during the lead-up to a math conference celebrating his 65th birthday. A toy problem is a scaled-down version of a bigger, more complicated problem that mathematicians are trying to solve. Zagier described the one that Ono gave him as "a cute problem about the asymptotic behavior of certain polynomials involving Euler's partition function, which is an old love of mine and of Ken's—and of about pretty much any classical number theorist." "I found the problem intractable and I didn't really expect Don to get anywhere with it," Ono recalls. "But he thought the challenge was super fun and soon he had crafted a solution." That sounds like a potentially useful problem-solving trick - pull a small but key problem out of your bigger problem, and then nerd-snipe your friends into solving it.

pmiller2 7y ago
Definitely. Two fairly standard techniques in mathematics are solving a cut down version of a problem and lifting that to a full solution, and solving a more general version and specializing to a particular case. Sometimes, amazingly, the more general problem is easier to solve, perhaps because fewer irrelevant details stand in the way. Both methods have yielded a lot of progress in mathematics. For example, solving a cut down version of Fermat’s Last Theorem got us the theory of ideals in algebra. Relatedly, Wiles work on the modularity theorem ultimately led to the actual proof of FLT.
- throwawaymath 7y ago
  > Sometimes, amazingly, the more general problem is easier to solve, perhaps because fewer irrelevant details stand in the way. True, and that's why we like to generalize things. It's usually easier to prove a mathematical object belongs to another class of structure which has certain properties than to prove the object itself has all those properties. Of course as you've noted it's not always like that at all. It's significantly easier to prove many things about finite-dimensional vector spaces over fields versus infinite-dimensional modules over rings. Solving mathematical problems is a lot like working a sponge. It's often useful to alternate between expanding and contracting the scope of your problem.

archgoon 7y ago

A mathematician has spent years trying to prove the Riemann hypothesis – without success. Finally, he decides to sell his soul to the devil in exchange for a proof. The devil promises to deliver a proof within four weeks. Four weeks pass, but nothing happens. Half a year later, the devil shows up again – in a rather gloomy mood. "I’m sorry", he says. "I couldn’t prove the Riemann hypothesis either. But" – and his face lightens up – "I think I found a really interesting lemma…"

butterthebuddha 7y ago

"This method doesn't work and Ken knows that, not sure wtf this article is. At most, this idea can show that the zeroes off the critical line are density zero, which would be a huge result, but could never prove RH. Again, Ken knows this (he says as much when he talks about it) so wtf is this article?" - Someone on Reddit (https://old.reddit.com/r/math/comments/brgp5z/mathematicians_revive_abandoned_approach_to_the/ https://old.reddit.com/r/math/comments/brgp5z/mathematicians...)