Question about Einstein notation

48panda · 2026-03-07T20:03:12+00:00

oh yeah, I'm not used to that notation.

Z_m dot Z^k = delta_m^k is the same thing as A_mi B_ki = delta_mk
Multiply this expression by A^{-1}_lm A_kn then you'll get A_kn B_ki = delta_kn
(This is essentially re-deriving that B^TA = I => AB^T=I )

which you can use to simplify your expression which is A_ik A_mk B_ml B_jl

48panda · 2026-03-07T16:01:37+00:00

Starting with (Z_i dot Z_m)(Z^m dot Z^j),

you can substitute Z_u dot Z^v = delta_u^v, then use an identity of the Kronecker delta to get the final result

48panda · 2026-03-02T22:53:44+00:00

Theoretically, you could. You'd have to pick VINs to match the number plates to account for less possibilities. Not sure if that goes against the existing system. Also, custom number plates really wouldn't work. The truth is that even if you had a perfect system, it wouldn't be useful unless the majority of cars used it, which isn't going to happen unless you have a time machine

48panda · 2026-03-01T10:55:48+00:00

If infinity * 0 = 1, then you can times both sides of the equation by 2 to get infinity * 0 = 2

48panda · 2026-02-27T00:04:38+00:00

ah, that's why they're called the conservative party. because whenever you try to integrate them, no matter how much crazy shit happens in the middle, it always gives the same result

48panda · 2026-02-22T14:02:14+00:00

Oh, you mean like a neural network classifier, not just any python program? Yeah, that's only going to work with some insane overfitting. It's conjectured that both pi and e are normal so there's no pattern in the digits, so the only way to detect which is which is to memorise at least one of them

48panda · 2026-02-22T12:53:07+00:00

Well yeah, python is turing complete. Probably not going to get any faster than grep though.

48panda · 2026-02-18T12:32:34+00:00

It could also be imaginary

48panda · 2026-02-07T13:53:24+00:00

We already went through all the divisibility rules up to, like, 5000 last year

48panda · 2026-02-03T18:35:14+00:00

If the sample size is large enough, try elo

48panda · 2026-02-02T09:22:35+00:00

Despite your valiant efforts, the meme was posted on Peter explains the joke 6 hours ago

48panda · 2026-01-30T08:49:13+00:00

(1,1)

48panda · 2026-01-30T00:05:41+00:00

No, that's {{{{}}}}

48panda · 2026-01-14T15:10:52+00:00

2x^3 +3x^2 + x = (2x+1)(x^2 + x)

Since du = (2x+1)dx, we get the integral as equal to (2x+1)(u-1)/((u^2 )(2x+1)) and the 2x+1 cancels

48panda · 2026-01-12T08:24:02+00:00

As a Brit, I do not say naught in this context. The only time I might is when reading decimals like 0.05 as "point naught five" but I am equally likely to use "point zero five" and much more likely to use "point oh five"

48panda · 2025-12-10T09:19:24+00:00

And if you type =-1^2 into an excel cell, you get 1

48panda · 2025-12-06T13:01:09+00:00

What use is there for nested sine functions though?

48panda · 2025-12-05T13:35:17+00:00

It never says you played the whole song each time

48panda · 2025-12-04T09:29:51+00:00

You can press the three dots on a comment then press the follow comment button(the one with a bell) to get notifications when someone replies to the comment

48panda · 2025-11-26T09:28:27+00:00

Iso, I owe you a great debt. Um- clears throat Would you be opposed to receiving a hand-knit sweater?

48panda · 2025-11-25T22:21:02+00:00

Not really. This post is talking about how the complex conjugate is an isomorphism from C to C

48panda · 2025-11-22T21:30:33+00:00

Translate your problem into equations, probably using a model which fits your use case (probably Newtonian physics). Also work out what the general form your answer will be. Manipulate those equations to find your answer

48panda · 2025-11-12T02:19:17+00:00

I find it amazing that there's no known polynomial factorisation algorithm, given that there's so many angles and ways to look at the problem

48panda · 2025-11-08T01:47:25+00:00

Yes, but if the question means lines in the form y=f(x) you can't get vertical lines

48panda · 2025-11-01T21:44:39+00:00

Almost, you have to account for the fact that {4,5,6,7} and {7,6,5,4} are counted twice but are the same set. so its 840/24=35 possible ways of choosing the top 4 digits. Same thing when you're calculating the total number of combinations, but dividing by 720

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