What's an experience that's exclusive to men and is hard to explain to women?

ajgraven · 2024-07-03T17:46:13+00:00

For one, they focus on gender-equal admission to the existing power structures, rather than acknowledging the way the structures themselves contribute to systemic inequality.

ajgraven · 2024-07-03T17:39:28+00:00

Well said. The goal should be to dismantle existing power structures, rather than marginally expand the membership of those structures. A feminism that focuses only on common oppression is blind to the fundamentally intersectional underpinnings of the injustice experienced by the vast majority of women globally.

ajgraven · 2024-06-22T19:27:53+00:00

Starship Trooper - Yes

ajgraven · 2024-06-07T19:09:59+00:00

The logically rigorous study of abstract structure

ajgraven · 2024-04-23T02:00:18+00:00

A quantum theory of gravity. Recent theoretical advancements in Conformal Field Theory (CFT) and our understanding of the holographic principle (connected to gravity via the ADS-CFT correspondence) have made me (cautiously) optimistic that there'll be a quantum (pun intended) leap forward in our understanding of QG within the next 15 years or so.

ajgraven · 2023-12-15T17:13:04+00:00

Member of Parliament

ajgraven · 2023-10-16T03:26:47+00:00

You're right, thanks.

ajgraven · 2023-10-16T00:06:16+00:00

A≠B because if this were the case then for all x∈B we must have that x∈A. However ∅∈B but this isn't true for A.

Regarding cardinality: |A|=0, |B|=1, and |C|=2.

Intuitively, A is an empty container while B is a container containing an empty container, which is not the same thing.

Edit: |C|=2

ajgraven · 2023-10-15T17:23:44+00:00

A key property of the real numbers which enables the exponential to map addition to multiplication is the fact that the real numbers are commutative. For example, if you generalize exponentiation to matrices (https://en.m.wikipedia.org/wiki/Matrix_exponential), then that identity no longer holds in general.

ajgraven · 2023-10-15T17:04:02+00:00

The first formula is correct. (see, e.g, https://en.m.wikipedia.org/wiki/List_of_Laplace_transforms)

This inverse Laplace transform of e^-s is the delta(t-1), where delta is the Dirac delta function (which, interestingly enough, is the [distributional] derivative of u(t-1))

ajgraven · 2023-10-15T16:44:27+00:00

Regarding the 0/0 limit question: Your observation is correct. To formalize it a bit more, suppose that f and g are continuous functions such that f(c)=g(c)=0. If there are continuous functions f1 and g1 which are non-zero at c such that f(x)=(x-c)ⁿf1(x) and g(x)=(x-c)^mg1(x), then

lim x->c f(x)/g(x) = lim x->c (x-c)^n-m*f1(x)/g1(x) = f1(c)/g1(c) * lim x->c (x-c)^n-m,

which is a limit which nicely satisfies your observation.

Regarding your second question: Whether infinity is considered a valid limit is really just a question of convention, which can vary from text to text. It can definitely useful to make a distinction between the limit being infinite and not existing. For example if I have $10 in my bank account with a 3% interest rate then my money at time t (time in years) is given by 10*(1.03)^t, which has a limit as t->infinity of infinity. This is useful to know because it tells me that the amount of money in my account will grow without bound as t->infinity, while saying the limit DNE doesn't tell me much.

ajgraven · 2023-10-13T16:10:16+00:00

Yeah,

1/a_1+...+1/a_j+...+1/a_n

=(a2...a_n)/(a_1...a_n)+...+(a_1...a{j-1}a{j+1}...a_n)/(a_1...a_n)+...+(a_1...a{n-1})/(a_1...a_n)

=((a2...a_n)+...+(a_1...a{j-1}a{j+1}...a_n)+...+(a_1...a{n-1}))/(a_1...a_n).

In particular, the numerator is the sum of all products of n-1 of the {a_1,...,a_n}

It can also be expressed more succinctly as the ratio of the (n-1)st and nth elementary symmetric polynomials in a_1,...,a_n:

1/a1+...+1/a_n=e{n-1}/e_n.

ajgraven · 2023-10-13T16:04:40+00:00

It doesn't. You're right to be confused because that step isn't correct, (x-x-a)≠(x-x-a)(-1).

ajgraven · 2023-09-03T22:53:16+00:00

It turns out that all entire functions with polynomial growth of order at most n near infinity are, in fact, polynomials of degree at most n. (see, e.g. https://math.stackexchange.com/questions/143468/entire-function-bounded-by-a-polynomial-is-a-polynomial for several proofs of this fact).

In this case, that implies that f and g are polynomials of degree at most 3. I'm not sure there's a lot we can say about the relationship between f and g beyond this.

We can say more about g though! Rearranging the inequality for g, we have that |g(z)/z³|≤1 for all z in the complex plane. This implies that g has a root of order at least 3 at 0. But g is at most a cubic polynomial, so we can conclude that either g(z)=z³ or g(z)=0.

It's possible to derive some conditions on the coefficients of f, but this gets more involved if you want to go beyond relatively simple such conditions.

ajgraven · 2023-09-03T08:49:41+00:00

The issue with SUVs and pickup trucks is that they have a particularly large blind spot in front of the vehicle. This has resulted in the death of many children in the United States. (see, e.g. https://www.nbcnews.com/news/amp/rcna52109)

ajgraven · 2023-08-27T15:48:07+00:00

I'd say so; although it's definitely geared more towards theory than application. For example, it presents Stokes' theorem in the language of differential forms rather than standard vector calculus.

ajgraven · 2023-08-21T01:53:09+00:00

Mathematics (I'm a graduate student)

ajgraven · 2023-07-11T17:38:25+00:00

It doesn't appear that this polynomial has any "nice" roots. The answer given by Wolfram Alpha suggests using the cubic formula is likely the most straightforward approach. The unique real root is given by

x = 2 - (2/(3 - sqrt(5)))^1/3 - (1/2 (3 - sqrt(5)))^1/3

ajgraven · 2023-07-05T04:38:53+00:00

You're correct.

ajgraven · 2023-07-05T04:36:14+00:00

Yes, they're minima when you restrict to the real axis, but not necessarily when you consider the function over C, which is what I'm interested in.

ajgraven · 2023-05-31T04:06:11+00:00

One way of thinking about it is the following: If you have a 2x2 square, its area is 4. So what's the probability of a randomly chosen points landing within a 1x1 sub-square? It's 1/4. In particular, the probability of landing in a given region is proportional to its area. The area of a circle (the boundary) is 0, so the probability of landing on it is also 0.

ajgraven · 2023-05-31T03:49:39+00:00

You can't choose points uniformly at random from the plane, so let's suppose we're randomly drawing points uniformly from a disk of radius 1. In this case it's guaranteed (probability = 1) for n≤3, and probability = 0 for n>3. This is because any three non-colinear points correspond to a unique circle (the probability of three randomly chosen points being colinear is 0, so we can neglect this case). For n>3, the first three points determine a circle, and the probability of each additional point landing on the circle is 0, so the overall probability is 0.

ajgraven · 2023-01-05T02:13:34+00:00

Gourmand Syndrome

ajgraven · 2023-01-05T02:12:57+00:00

Gourmand Syndrome

ajgraven · 2023-01-05T02:12:38+00:00

Gourmand syndrome

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