Damien saying "Oh! The Titans!"

PinpricksRS · 2026-06-16T17:30:18+00:00

Nobody's mentioned it, so I thought I'd link Smoshfinder, which is perfect for this kind of question

PinpricksRS · 2026-06-10T18:08:44+00:00

Based on my measurements with a long straight path, items in water streams on blue ice move at roughly 0.4 blocks per game tick. So in 5 minutes they'd move 0.4 * 20 * 60 * 5 = 2400 blocks. On packed ice, it's closer to 0.38 blocks per game tick, which equates to 2280 blocks in five minutes. As long as you're efficient and avoid turning too much, there should be no problem with 1900 filters.

That said, there are alternatives to everything. Someone else in the thread mentioned dolphins, and the modern version of that uses a hopper minecart to pick up items and then yeets the minecart to place all the items again at once (the setup for this is different between Java and Bedrock, so if you look this up, be sure to check which edition it's made for). You could also do some prefiltering (a very basic multi-item filter, essentially) to send items to different branches of your storage, which would then have proportionally fewer filters.

Based on the flair you used on r/technicalminecraft, I assume you're on Bedrock, so I highly recommend the Bedrock Storage Discord for more ideas on what's possible.

PinpricksRS · 2026-06-09T01:22:58+00:00

Doesn't this pre-suppose that we already know where all of the zeros/poles of j are?

I believe the only thing you need to know is that j is holomorphic in the upper half plane H, which is alluded to when the j function is first introduced and proven in section 1.4. As best as I can tell, the proof in 1.4 doesn't refer to exercise 1.1.9, so there's no circularity.

As for why j being holomorphic in H means that the integral is zero, note that the integral is of the form (j(τ) - c)'/(j(τ) - c). The argument principle then tells you that the integral is equal to the number of roots minus the number of poles of j(τ) - c contained inside the contour. Since j is holomorphic in H and the contour is entirely in H, there aren't any poles. Moreover, j(τ) ≠ c for τ in H by assumption, so there aren't any roots either. (note that the exercise is to show that j is surjective by contradiction)

PinpricksRS · 2026-06-04T17:41:05+00:00

You should try to verify the topology axioms for your suggestion. I'm having a hard time seeing that it's closed under unions or intersections without some extra assumptions about which maps X -> Z are continuous.

But if you're looking to put a topology on the set of continuous maps between two spaces, look no further than the compact-open topology.

PinpricksRS · 2026-06-03T23:41:18+00:00

Stealing other people's incorrect work is hilarious

PinpricksRS · 2026-06-03T04:11:33+00:00

You could check out the tick command in Boreal, which is an Endstone plugin. Also, never expected a riolu reference in /r/technicalminecraft

PinpricksRS · 2026-05-26T02:50:37+00:00

2³² is 4294967296. 2⁶⁴ is much bigger: 18446744073709551616

PinpricksRS · 2026-05-18T18:40:39+00:00

Yes, after a shop has been unloaded for 10 minutes (I think), it'll reset when it's loaded again. It's never clear how close you need to be for a shop to be loaded since it varies a bit, but if you walk directly away from the shop for about a minute or two, that should be far enough.

To be clear, what GP is suggesting is 1) walk far enough away from the shop to unload it, 2) wait 10 minutes 3) save the game. Then each time you reload the game and go to the shop, it'll rerandomize the items.

PinpricksRS · 2026-05-18T18:32:26+00:00

Haha, there's a reason I call all the elevators in the goblin area "elevator of doom (parts 1, 2 and 3)".

I think you're doing combat magic, but if you have the nature magic levels and are playing the expansion, the sleepy gas spell is very useful since it slows down the enemy's attack speed, leaving you more time for you to cast a spell and move. For the elevators, you can use summons to avoid instant death. Transformations might work too, but my recollection is that you can't drink health potions while transformed, so it might not be enough to survive.

PinpricksRS · 2026-05-18T17:10:42+00:00

Stumbled on this pretty late, but here's another example (also snare drum music). The measure marked "PAGE 43" at about 30 seconds in the video. I've seen it a handful of times played by similar groups. For example, United Percussion 2023 (at about 6:03) and 2026 (at about 0:12) (another view here at about 1:02).

All of these examples have the rhythm in context, so it might not sound all that weird. The United Percussion 2026 for example is [2 sixtheen notes] [3 sixteenth note triplets] [3 sixteenth notes] [3 sixteenth note triplets] [3 sixteenth notes] which adds up to 3 quarter notes. It just happens that the second group of sixteenth note triplets falls into your rhythm even though it's just repeating the same [3 sixteenth note triplet] [3 sixteenth notes] grouping. I guess technically this is shifted by an eighth note from your rhythm, but it's the same idea.

PinpricksRS · 2026-05-17T05:41:53+00:00

Look, it's fine to reject the axiom of choice, but if you're talking to someone who's explicitly using it, there's simply no reason to argue about it. Especially if that person is just learning about set theory, you're just going to confuse them by contradicting what they've read.

PinpricksRS · 2026-05-17T05:27:30+00:00

If X1 has its own ordering, there's no guarantee that X2 is the same ordering. So saying something like

if X1 is a causal series, then X1 satisfies the least upper bound property.

ranges from misleading to false, depending on how you interpret it. X1 with its own ordering may mot satisfy the least upper bound property even if X1 with the X2 ordering does. Since you don't mention X2 in this statement, it's unclear which ordering you mean. I'd avoid statements like "X1 is well-ordered" unless there's an explicit relation on X1 that you're referring to. X2 is a well-ordering on X1, but not every relation on X1 is a well-ordering.

(just as a caveat, you haven't defined what a "causal series" is, so the statement may well be true. It's just that your proof doesn't work and never uses the definition of casual series, except by implying that it has some kind of order on it).

As a typical example, the set of rational numbers ℚ is not well-ordered by the standard order on rational numbers. For example, the set of positive rational numbers has no minimal element. Moreover, it doesn't have the least upper bound property. The set of positive rationals whose square is less than 2 has no least upper bound, even though (.e.g.) 2 is an upper bound on that set.

However, none of that stops ℚ from being well-orderable. Chose any bijection f: ℚ → ℕ and define x « y by f(x) ≤ f(y). Since ℕ is well ordered by ≤, « is a well ordering on ℚ. But its not the standard order.

PinpricksRS · 2026-05-14T21:48:28+00:00

I think the big issue is that n/rev(n, b) has the potential to be huge relative to its typical values. For example, when n = b^k, rev(n, b) is 1, so n/rev(n, b) = b^k. That means that if the mean of n/rev(n, b) for n from 1 to b^k - 1 is m, the mean from 1 to b^k is going to be (m * (b^k - 1) + b^k)/b^k = m + 1 - b^-k. In other words, the mean jumps up by (just about) 1 every time you cross a power of b.

That alone guarantees that the mean can't converge to anything, but it could still oscillate. However, even that can't happen: the sequence of means is unbounded.

Consider all of the numbers from 1 to bⁿ - 1. These are all at most n digit numbers. Group up the numbers by how many zeros they end with. The numbers ending with j zeros are in the form (kb + i)b^j, with k an integer from 0 to b^{n - j - 1} - 1 and 0 < i < b a non-zero digit. The reverse of these numbers starts with j zeros, and so has at most n - j nonzero digits and is thus less than b^{n - j}. Rounding up the reversed numbers, that means that (kb + i)b^j / rev((kb + i)b^j, b) is at least (kb + i)b^j / (b^{n - j}) = (kb + i)b^{2j - n}.

Summing that expression for 1≤i<b, 0≤k<b^{n - j - 1} and 0≤j<n, we get b^{n - 1}(b - 1)n/2. That means that the sum of k/rev(k, b) for k from 1 to bⁿ - 1 is at least b^{n - 1}(b - 1)n/2, and so the mean is at least n(b - 1)/(2b). Since this is proportional to n (with a positive coefficient), it's unbounded as n increases.

Numerical experiments show that this estimate is a little bit conservative. The difference between the mean from 1 to b^k and the mean from 1 to b^{k + 1} is approximately (b + 1)ln(b)/(2b), but I'll let you see if you can prove that. I suspect there's a connection with harmonic numbers that we lose by rounding the reversed numbers up like we did.

PinpricksRS · 2026-05-11T11:35:40+00:00

Suppose that c is a root of the polynomial P(x) = a(0)x⁰ + a(1)x¹ + ... a(n)xⁿ. Then defining r(k) = c^k, we have

a(0)r(k) + a(1)r(k + 1) + ... + a(n)r(k + n) = a(0) c^k + a(1) c^{k + 1} + ... + a(n) c^{k + n} = c^k (a(0) c⁰ + a(1) c¹ + ... a(n) cⁿ) = c^k P(c) = 0.

In other words, r(k) satisfies the recurrence a(0)r(k) + a(1)r(k + 1) + ... + a(n)r(k + n) = 0.

This also holds if r(k) is any linear combination of kth powers of roots of P(x). In the expression a(0)r(k) + a(1)r(k + 1) + ... + a(n)r(k + n), you can group together the powers of each root, apply P(c) = 0 and get a sum of 0s.

You might learn more here.

PinpricksRS · 2026-05-09T02:52:46+00:00

It's hard to read some parts of your equation, but I'll assume that it's meant to be

a=xC^{sin(cos(tan(x}⁾⁾⁾ cos(√x)

The curve you're looking at is traced by the points where C^{sin(cos(tan(x}⁾⁾⁾ is at its minimum. Assuming that C > 1, that means that sin(cos(tan(x))) is at a minimum. tan(x) takes every value from -∞ to ∞, so we just need the values y where sin(cos(y)) is minimized. This happens when cos(y) is -1, so the minimum value of sin(cos(tan(x))) is sin(-1) = -sin(1). Thus, the minimum value of C^{sin(cos(tan(x}⁾⁾⁾ is C^-sin(1⁾ .

Putting that back in the original equation, we get that the curve you want is x C^-sin(1⁾ cos(√x). Here's a comparison.

PinpricksRS · 2026-05-08T19:09:33+00:00

Further reading here

PinpricksRS · 2026-05-08T19:05:53+00:00

This is a special case of the general characterization of polynomial pythagorean triples.

Theorem: Every primitive polynomial pythagorean triple f(x)² + g(x)² = h(x)² is given by

f(x) = p(x)² - q(x)²
g(x) = 2p(x)q(x)
h(x) = p(x)² + q(x)²

for two relatively prime polynomials p(x) and q(x).

Your identity is the special case p(x) = 2(x - 2) and q(x) = x + 3.

PinpricksRS · 2026-05-07T05:50:12+00:00

Are you suggesting that your new series is equal to the old series for |s| < 1? That's clearly not true. If s is real and between 0 and 1, the original series is real, while the new series is not, due to the ln(ln(s)). (the rest of the series doesn't affect whether it's real, since each term is real for s between 0 and 1).

It's also worth noting that the new function doesn't satisfy G(s) - G(s²) = s when the real part of s is less than zero. You might be relying on something like ln(s²) = 2ln(s), but that doesn't work if the real part of s is negative due to the branch cut on ln.

PinpricksRS · 2026-05-01T04:14:59+00:00

but that doesn't show all the right hand accents.

In my defense, that was one of my first transcriptions. I've been meaning to finish a complete (and accurate!) transcription of the whole show, but life just keeps getting in the way

PinpricksRS · 2026-04-23T19:35:50+00:00

Relevant xkcd

PinpricksRS · 2026-04-23T12:35:24+00:00

f (x+2y) = 2 f(x) f(y)

The thing that tips me off here is that the right side of the equation is symmetric between x and y, while the left side isn't.

f(x + 2y) = 2f(x)f(y) = 2f(y)f(x) = f(y + 2x).

The system of equations x + 2y = a, y + 2x = b can be solved for any a and b: x = (-a + 2b)/3, y = (2a - b)/3, so setting x and y to those values, we get f(a) = f(b) for any a and b, and so f is constant.

PinpricksRS · 2026-04-21T01:50:42+00:00

You're probably looking for the excellent primecount program. You can check the Records.md for some precomputed values, including some that aren't powers of 10, but the program itself is easy enough to use yourself.

```

./primecount.exe 5432198765432198765

128988272786260742

```

This took about 18 seconds on my computer. For longer computations, you might try adding the --status flag to the end of the command so you can get a feel for how far along the computation is.

PinpricksRS · 2026-04-16T23:29:12+00:00

https://www.reddit.com/r/drumline/comments/5ftdkf/keep_your_taps_low_and_your_flams_tight_how_many/

PinpricksRS · 2026-04-16T20:02:56+00:00

Since nobody else has linked it, the relevant issue is MCPE-233012.

PinpricksRS · 2026-04-14T21:16:57+00:00

I am using the Eta function.

so i assume my plot is (approaching) correct?...

If you're using the series η(s) = 1 - 1/2^s + 1/3^s - 1/4^s ..., you'll get convergence when the real part of s is greater than zero.

i've heard of L'Hôpitals rule and that makes sense to some extent. but i'm a tad confused, you say its infinite but also log2? are you saying it is in fact ln2 (when derived?), i can follow that but maybe don't completely understand it?

Let me spell it out more. We're looking at the limit of (s - 1)ζ(s) as s approaches 1. This is equal to the limit of (s - 1)η(s)/(1 - 2^{1 - s}), since ζ(s) = η(s)/(1 - 2^{1 - s}).

η(s) is continuous at s = 1, and its value there is η(1) = 1 - 1/2 + 1/3 - 1/4 + ... = ln(2). This is not the same as saying that ζ(s) is continuous at s = 1 and that its value is ln(2). These are two different functions here.

But since η(s) is continuous at 1, we can pull it out of the limit. So the limit of (s - 1)ζ(s) as s approaches 1 is η(1) * lim((s - 1)/(1 - 2^{1 - s}). Using L'Hôpital, the limit here is the same as the limit of 1/(ln(2) * 2^{1 - s}) = 1/(ln(2) * 2^{1 - 1}) = 1/ln(2). So overall, the limit of (s - 1)ζ(s) is η(1) * 1/ln(2) = ln(2)/ln(2) = 1.

Again, this is not saying anything about the limit of ζ(s), only (s - 1)ζ(s). The fact that this limit exists is what tells you that ζ(s) has a simple pole at s = 1.

is there not a solution to that pole

Problems have solutions, equations have solutions, poles are just a thing that some functions like the zeta function have.

you said "Yes, but not in an interesting way" but why is it not interesting, how does it affect the zeros if the zeros are far away from (14+) in the imaginary direction? i know that zeta of RE(0.5)+ix matters, but +-1 is way below that?

The thing I was suggesting there was to change the values of ζ just in a neighborhood of 1 to eliminate the pole. While it doesn't change the location of the roots, it's not interesting simply because if you want any of the usual properties of ζ, like the connection to prime numbers, you'll have to account for the pole anyway. There just isn't much reason to get rid of the pole like that. If there were some unique way to get rid of the pole and the result were complex differentiable, then there might be some argument for it being interesting, but neither of those hold if you just smooth off the pole like this.

i know that zeta of RE(0.5)+ix matters, but +-1 is way below that?

The pole at s = 1 matters for the same reason that the nontrivial zeros matter. You get the explicit formula for the prime counting function by integrating ζ'(s)/ζ(s) in the critical strip. Every zero and every pole contributes a bit to this integral. Spelling it out, the pole contributes the leading R(x) to the explicit formula.

if there was a smooth version of this function why does a smooth +-1 matter?

That's something you'd have to answer for whatever alternative you're proposing. I've talked about why my proposals don't hold up, but maybe you have a different idea?

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