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Mathematics professor at Cambridge

Feb
16
answered Problem equivalent to “largest square in a cube”
Feb
8
awarded  Nice Question
Feb
7
comment alternative construction of the quotient group
I don't claim it's a good way of doing things, but one could (in desperation) argue that here one is defining quotients in just one situation (what you need to define a group in terms of generators and relations) and getting all quotients out of it. But I wouldn't want to go to the wall on this one ...
Feb
7
answered alternative construction of the quotient group
Feb
5
answered Yet more on distortion
Feb
3
asked Is this a well-known probabilistic model?
Feb
3
comment Why does the Riemann zeta function have non-trivial zeros?
That is a very useful comment -- thanks!
Feb
2
comment Why does the Riemann zeta function have non-trivial zeros?
That was another of the thoughts that lay behind my question. Somehow the fact that the distribution of primes can't be "better than random" feels like a fact that ought to have an elementary proof using some Parseval-like identity. I suspect the zeta function is sort of doing that (with a Mellin transform rather than a Fourier transform), but it doesn't appear to be saying something simple like, "That function has the same L_2 norm and trivially has L_2 norm at least the square root of n."
Feb
1
awarded  Nice Question
Feb
1
comment Why does the Riemann zeta function have non-trivial zeros?
That is a very nice argument, but it also has a magic flavour to it, since you somehow manage to bootstrap a very small error (arising from the fact that $\psi_0(x)$ is discontinuous) into a much bigger one (that the error term in PNT must be more like a square root). But perhaps the bootstrapping is done by the functional equation rather than your argument.
Feb
1
comment Why does the Riemann zeta function have non-trivial zeros?
It's precisely this issue -- why the error term in PNT isn't absolutely tiny -- that I want to understand. E.g. to prove that π(x) does not approximate $Li(x)$ to within $latex x^{1/3}$, the obvious method is to point to the zeros on the critical line. So I'm going round in circles. With the help of the functional equation one can say that if there are no zeros on or to the right of the critical line then there are none at all, but I don't count that as an intuitive argument.
Feb
1
revised Why does the Riemann zeta function have non-trivial zeros?
Added tag
Feb
1
asked Why does the Riemann zeta function have non-trivial zeros?
Jan
26
awarded  Nice Answer
Jan
23
awarded  Nice Answer
Jan
22
comment Proving “almost all matrices over C are diagonalizable”.
Or you could simply upper-triangularize your matrix and do the same.
Jan
22
answered Switching Research Fields
Jan
22
awarded  Nice Answer
Jan
21
revised A random variable: is it a function or an equivalence class of functions?
added 994 characters in body
Jan
21
answered A random variable: is it a function or an equivalence class of functions?