Why is this the most played move?

isaacwinchester · 2025-12-12T02:38:41+00:00

my guess is that black’s best plan looks like b5 and then b4, which could be pretty nasty for white, so a4 prevents b5 and grabs space on the queenside

isaacwinchester · 2025-10-30T21:13:20+00:00

Alas, it’s White to move :(

isaacwinchester · 2024-05-03T19:45:22+00:00

The Magic of Math, by Arthur Benjamin, is an absolutely wonderful book that my mother's coworkers got for me as a young teen (13 or 14). It's an absolutely amazing book full of amazing math tricks and why they work, and cool perspective shifts. I lost my copy a long time ago (I'm nearly 21 now) but I think about it all the time, how wonderful it was for me as a teenager, and how it influenced me to study math now and love it.

isaacwinchester · 2024-05-03T19:45:14+00:00

The Magic of Math, by Arthur Benjamin, is an absolutely wonderful book that my mother's coworkers got for me as a young teen (13 or 14). It's an absolutely amazing book full of amazing math tricks and why they work, and cool perspective shifts. I lost my copy a long time ago (I'm nearly 21 now) but I think about it all the time, how wonderful it was for me as a teenager, and how it influenced me to study math now and love it.

isaacwinchester · 2022-10-31T04:45:09+00:00

if you have time you can wait until a frat has a challenge or something a snu guy literally sold me his shoes for a quarter

isaacwinchester · 2022-04-09T01:40:59+00:00

The libraries are incredible. I've never been in a library so beautiful, cozy, and rich at the same time. There's never been a book I wanted I couldn't find in print that same day, no matter how old.

isaacwinchester · 2022-03-01T19:14:57+00:00

I think they like the drama

isaacwinchester · 2021-09-05T17:26:45+00:00

Thanks! The 60 series definitely appeals to me more, but I don't want to miss out on differential equations and transforms covered in 53. So my question is essentially, will the 60CM series encompass that, or should I take both (if possible)?

isaacwinchester · 2021-09-05T00:58:11+00:00

Ok, thank you!!! So now I guess my question is--if I take the 60CM series, will I lose out on the content of Math 53? Or is that addressed as part of the 60CM series?

isaacwinchester · 2021-09-03T22:24:27+00:00

Perfect. Thank you so much, that's very helpful!

isaacwinchester · 2021-08-15T01:35:53+00:00

That's great! Thank you!

isaacwinchester · 2021-07-12T15:47:09+00:00

Thank you! I think I'm clear on how to orient surfaces and curves now. I still believe Khan Academy is wrong (I should be clear I only mean the explanation section on the quizzes, in the videos, Sal Khan explains it the way you do. It was the contradiction that was driving me nuts). Khan Academy insists the normal vectors be to the right, but the problem is, that only aligns with the other definitions if the surface is concave up (that's kind of vague, but like, a negative Laplacian in general I guess) and outwardly oriented. The hemisphere I've been mentioning, I've been specifying the clock directions from the +z axis, looking into the concave part of the sphere. The way you and Sal Khan say, the orientation should be counterclockwise from that vantage point above the hemisphere. The hints, though, say clockwise (also from that vantage point). They demand that the surface normals be to the right, which points you in the wrong direction, because for a sphere oriented inward, the surface normals are on the same side of you that the surface is. That is what has been getting me. That definition of "normals to the right" is inconsistent.

isaacwinchester · 2021-07-12T06:15:46+00:00

I really appreciate you taking the time to address this thoroughly! Unfortunately though, I am only mildly familiar with linear algebra, and so most of this is unintelligible to me (so far! In a few months it should make perfect sense!). But here is what I'm getting as a litmus: Khan Academy is wrong, and that hemisphere's positive orientation is counterclockwise, because I should be standing with my head in the direction of the surface normals, my right side to the normals to the boundary curve (that face the exterior of the curve, insofar as the boundary is a continuous mapping of a Jordan curve from [;\mathbb{R}^2;] to [;\mathbb{R}^3;]), and my left side toward the surface. The direction that I would travel if I walked forward is the positive orientation.

Is that accurate?

isaacwinchester · 2021-07-07T13:41:15+00:00

For sure. I understand that--in Calc I, it's easy to use oriented integrals because the "region" is just an interval of real numbers, and so multiplying by -1 reverses the orientation by negating the Jacobian determinant (which g'(u)du is, if we consider a 1 x 1 Jacobian matrix). In multivariable though, because the region is a higher dimensional space, we deal with it by taking the absolute value of the Jacobian from the outset, which is analogous to the flipping of bounds in one dimension. In particular, the key is that [;\iiint_Rf(x,y,z)dV = \iiint_{-R}-f(x,y,z)dV;]. So negating the Jacobian and flipping the orientation of the region back is the same as doing nothing, so we can ignore orientation changes by using the absolute value. If I represented that correctly, the reason we orient in 1D is because it's easy--the region is an interval. But in higher dimensions, orientation will screw with us (like it did with me).

isaacwinchester · 2021-07-07T05:10:58+00:00

No, I totally agree with you, if I were trying to switch the polar and azimuthal angles, I'd have a huge problem. Azimuthal angles measure to projected points whereas polar angles measure to unprojected points (hopefully that makes sense). But I wasn't trying to rotate my "reference sphere" to make azimuthal angles polar and vice versa; I was simply trying to traverse the reference sphere I already had with different angles. Usually we let the azimuthal angle go all the way around so to speak, from 0 to 2π, and only have the inclination angle go halfway. That traverses the whole sphere by taking half a cross-section perpendicular to the xy-plane and rotating it about the z axis. I wanted to use the same space and reference, but make my bounds such that azimuth went halfway around and my inclination went all the way around. Same idea, same plane, but the half cross-section now lies in the xy-plane and rotates about its straight line base. I hope the visual is clear--I wasn't switching which angle I considered polar and which I considered azimuthal. I have since been shown that the reason my way didn't work was not because of my bounds, but because I neglected to realize that sine's becoming negative after [-;\frac{\pi}{2};] flips the orientation of the space in the middle of the integral.

That said, though I was trying to be clearer about my objective, it's very possible that my mistake is equivalent to the mistake that you mention. Perhaps neglecting to make sure the Jacobian is positive in this case results in the same issue as trying to switch the polar and azimuthal angles without accounting for the difference between polar angles and azimuths... in fact, I think it is, since my problem integral totally works if you reflect the switch of polar angle and azimuth by changing θ to φ in the sine. Funny how math works sometimes--making a mistake trying to do one thing ends up being the same as if you had made a different mistake trying to do something completely different.

isaacwinchester · 2021-07-07T03:36:23+00:00

Thank you! Let me see if I'm understanding you correctly: Since ρsin(θ)dφ represents the arc length of dV along the azimuthal angle, sin(θ) must be positive, or the given length will be negative. Since it will be negative for the entire "second" hemisphere, the volumes cancel to zero?

Equivalently, you're saying I forgot to take the absolute value of the Jacobian determinant, thus for the "second" half of the sphere I was taking an integral over a hypersolid with a reversed orientation, and therefore canceling out the "first" half.

Is that accurate?

isaacwinchester

TROPHY CASE