bio | website | gowers.wordpress.com |
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location | ||
age | ||
visits | member for | 5 years, 5 months |
seen | Mar 16 '14 at 23:40 | |
stats | profile views | 31,728 |
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? |