bio | website | math.berkeley.edu/~sramesh |
---|---|---|
location | Berkeley, CA | |
age | 30 | |
visits | member for | 5 years, 7 months |
seen | yesterday | |
stats | profile views | 3,918 |
I was a graduate student in the Logic program at Berkeley, broadly interested in categorical logic and foundations of mathematics, as well as in applications of category theory to the semantics of programming languages. I work for Google now.
Aug
4 |
comment |
Derivative in terms of finite differences
Thanks for the link! I'll also note the full-text link plouffe.fr/simon/math/…, though perhaps this is of dubious legal status... |
Aug
4 |
comment |
Derivative in terms of finite differences
I suppose my example question may boil down to: Is there simple reason by which one could know that the Newton series for $x \mapsto x^p$ based at any point converges to the correct value within a neighborhood of that point? |
Jul
29 |
comment |
Derivative in terms of finite differences
I see the identity mentioned in Jordan's "Calculus of Finite Differences", but there does not seem to me to be much discussion of the conditions under which the identity holds, which is really the question I am interested in. (For example, is there simple reason by which one could know a priori that this identity is indeed valid at all points for all functions of the form $f(x) = x^p$, even for non-natural $p$?) |
Nov
21 |
comment |
Constructivity of zeros demanded by topological degree
If $f$ is of degree 1, it is homotopic to the identity; thus, we can "augment" $g$ so that it is defined on a slightly larger ball than originally, on whose surface it now acts as the identity. And then, considering this augmented $g$, case (A) cannot occur, since surface points are all mapped to different directions. |
Nov
20 |
comment |
Constructivity of zeros demanded by topological degree
Ah, this is nice. Still, I wonder if there might be some way to push further! (It irks me that there should be such a nice zero-finding algorithm for the degree 1 case and not others). Does there actually exist, for example, so "pathological" a case as that only finitely many surface starting points work out? |
Aug
28 |
comment |
Brouwer fixed points via flow
For what it's worth, the Hirsch/Kellog-style proof of the Brouwer fixed point theorem turns out to be along the lines I was ultimately hoping for (but, of course, different enough from what I asked about to actually work). |
Aug
26 |
comment |
Rationale behind an requirement on Turing machines
Heh, good point. Though we might also consider more sophisticated prefix-free encodings, the simplest is to reserve an "Ok, we're done" character, and that's basically the role the blank symbol plays. |
Aug
26 |
comment |
Rationale behind an requirement on Turing machines
One might note that the proposed machine is Turing-complete using any computable prefix-free encoding of the natural numbers. |
Aug
26 |
comment |
Brouwer fixed points via flow
Alas, this needn't work even in two dimensions, I realize: Take $B$ to be the complex numbers of at most unit magnitude, and consider $f(x) = (1 + \frac{1}{2}e^{i \pi |x|^2}) x$, which smoothly maps $B$ into itself. The unique fixed point of this is at the origin, but I believe the limiting behavior of the flow starting anywhere on the boundary will be to cycle around the circle of squared radius $1/2$. Alas. Well, I'll leave this question here and perhaps someone still will have a good story to tell about conditions under which we should expect this approach to work. |
Aug
26 |
comment |
Brouwer fixed points via flow
@Vidit: Yes, and also the uniqueness. Thus, any solutions defined on an open interval around 0 are compatible, and there is at least one, so we can join them all together and obtain a (unique) solution defined on a maximum open interval around 0 (not necessarily the entire real line), just as claimed, no? |
Aug
26 |
comment |
Brouwer fixed points via flow
@Christian: With the correction, the differential equation becomes $x'(t) = f(x(t)) - x(t) = -2x(t)$ with $x(0) = p$, which has unique solution $x(t) = e^{-2t}p$. The limiting value of this for large $t$ will be $0$, which is a fixed point of $f$. |
Aug
26 |
comment |
Brouwer fixed points via flow
@Christian: $f(x) = 2x$ will not be a map from $B$ to $B$, unless the ball $B$ consists solely of the zero point. |
Jul
23 |
comment |
Reconstructing the argument that yields Graham's number
@TimothyChow: The upper bound given in the paper is $f^7(12)$ where $f(n) = 2 \mathbin{\uparrow}^{n} 3$. That seems to me about exactly as easy to specify as Graham's number $f^{64}(4)$ where $f(n) = 3 \mathbin{\uparrow}^{n} 3$, so I don't know why the latter ever came up. Perhaps Graham actually tried to outline to Gardner the argument for the upper bound, and found it simpler with the weaker bound. |
Jun
22 |
comment |
Additivity of asymptotic density of periodic sets
Ah, nice. Here's a followup question I'm also interested in, which you will perhaps resolve with just as much ease: Suppose now we consider an increasing series of subsets $A_0 \subseteq A_1 \subseteq A_2 \subseteq ...$ of the integers, where each of these is not merely periodic but in fact of the form "Any multiple of a member of F", for some finite set F. Can the density of their union still fail to match the supremum of their individual densities? |
May
14 |
comment |
Commuting limits in relating the harmonic series to coprimality densities
Oh, very nice, thank you. This does settle the question about the existence of the general limit, but, alas, this argument requires that we already know $\prod_{p}(1 - 1/p) = 0$, while my motivating hope remains that there is some way to noncircularly derive this fact from $\prod_{p}(1 - 1/p) = \lim_{q \to \infty} \lim_{n \to \infty} f(n, q)^{-1}$ and $\lim_{n \to \infty}\lim_{q \to \infty} f(n, q)^{-1} = 0$. |
May
14 |
comment |
Commuting limits in relating the harmonic series to coprimality densities
If I'm reading you correctly, you are saying that for sufficiently large $n$ and $q$, we have that $f(n, q)^{-1}$ is at least a quantity which tends to zero. But surely we need to show $f(n, q)^{-1}$ to be at most a quantity which tends to zero, in order to answer the last question? |
May
13 |
comment |
Can the Riemann hypothesis be undecidable?
@Daniel: You're right! I haven't ruled out this possibility! All these years, sitting there unnoticed... That having been said, various sources give Robin's criterion instead as "for all $n > 5040$, $\sigma_1(n) \leq e^{\gamma} n \log \log n$.. To this, exact equality would not serve as a counter-example, and thus falsehood would entail provable falsehood. That having been said, I am not, in fact, familiar enough with the relevant material to verify whether the "$\leq$" form of Robin's criterion is genuine, or, I paranoidly worry, merely the result of careless transcription of the "$<$" form. |
May
24 |
comment |
Avoiding reflexive paradox in set theory
Richard, you are coming now quite close to a principle like "Sets are given by well-founded comprehension"; this is the spirit behind ZF, and once you have that, the rest of its axioms are not far behind [you'll find that the existence of infinite sets, powersets, etc., are not automatically assured, and you may want to assure them]. |
Dec
1 |
comment |
Application of polynomials with non-negative coefficients
Surely, the thing to do is to use q = p(1) + 1 rather than q = p(1), so as not to have to treat the case p(q) = q^n on an ad hoc basis. |
Nov
18 |
comment |
There are two points on the Earth's surface that … ?
What does "separated by the same geodesic distance" mean for two points? |