throwaway080383

Joined 8 years ago

Karma
396
Submissions
175
Member since
Jan 4
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Recent submissions

Comment · 8 years ago
Are the upper class not simply the descendants of those who were once the upper echelon?
Comment · 8 years ago
Fair enough. I guess I presumed the preference since he uses an iPhone.
Comment · 8 years ago
I'm surprised that despite being able to concisely explain why Apple is preferable to Google due to the privacy implications of their respective business models, they're perfectly content with a Google Home in their kitchen!

Maybe I'm paranoid, but isn't it obvious how the whole Home Speaker story ends? "We're not spying on you, we're learning your behaviors to offer a better experience!"

Comment · 8 years ago
I disagree. It's a cliche at this point that when Wal-Mart enters a town, all the mom and pop stores close up. The profit margins from distributing these goods now contributes to $WMT rather than staying in the local economy.
Comment · 8 years ago
I'm curious how much it would cost to say double the worker's salaries (or halve their hours at the same pay).

How many Foxconn man-hours does it take to produce an iPhone, and what is the average hourly wage?

Comment · 8 years ago
Ship it and sell it for $15!
Comment · 8 years ago
I never say anything about d being even. Rather, as soon as d > 0, x{d} is even, so "almost all" x{d} are even. As you vary x and d, you cover all integers exactly once, so from this perspective, "almost all" integers are even.

My point is that this is very similar to how elliptic curves are being counted here.

Comment · 8 years ago
The twist x{d} is even as soon as d>0.
Comment · 8 years ago
Even the phrase almost all is a bit misleading here, since there is no canonical choice of measure on the set of elliptic curves in question.
Comment · 8 years ago
To back up my last claim, let me prove that 100% of positive integers are even in the same spirit:

Fix any odd positive integer x, and consider its "twists" x{d}, which for positive d I define to be x*2^d. Every integer is of the form x{d} for a unique choice of x and d. Now, for fixed N, if I consider all "twists" for which d<N, the proportion which are even is (N-1)/N. Thus, as N tends to infinity, the proportion tends to 1.

Comment · 8 years ago
It's not accurate. Looking at the arxiv link (https://arxiv.org/abs/1702.02325), the idea is you fix some elliptic curve E, and you look at its "twists" E^d, where the parameter d is a nonzero integer.

If you only look at twists where |d| < N, you can ask "What proportion of these are rank zero, rank one, rank two, ...?" The Theorem in the paper is that as you let N go off to infinity, these proportions tend to 1/2, 1/2, 0, 0, 0, ... respectively.

Here's the thing: this does not correspond to a measure on the set of rational elliptic curves, and indeed, there is no reasonable way to define a uniform probability measure on a countable set. Consequently, statements like "half of all elliptic curves..." are kind of misleading and meaningless.

Comment · 8 years ago
More severe than death?
Comment · 8 years ago
Pretty sure books are going to be around for a while.
Comment · 8 years ago
That website doesn't call it crazy, it just points out many of the difficulties, and the comments refute many of them.

I do have to admit, it did not occur to me that the work week would still effectively enforce an "ordering" on longitudes which somehow takes the beauty out of it.

Comment · 8 years ago
I'm not sure if you were serious, but I actually think completely getting rid of timezones is not that crazy of an idea. It would take some getting used to, but naively I don't think it'd be any harder than say, switching to metric in the US.

That is, I like the idea in principle, but I know it'll never happen.

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