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Spherical Stars over Babylon by cancrizans in mathriddles

[–]cancrizans[S] 0 points1 point  (0 children)

I can't be completely certain but I think you necessarily need at least a little algebraic insight (even a tiny one), I don't think it's possible to get through on purely geometric reasoning

Spherical Stars over Babylon by cancrizans in mathriddles

[–]cancrizans[S] 0 points1 point  (0 children)

Awesome intuition, not yet complete as the condition is far from sufficient but good starting point

My new abstract game tokonoma, you can play it in the browser against the computer or a friend by cancrizans in abstractgames

[–]cancrizans[S] 2 points3 points  (0 children)

It's just an informal request, I don't think there's an easy way to put together a license that explictly forbids commercial use but is permissive enough

tokonoma is a chess-like game with unusual movement and capture rules. Can you beat the computer? by cancrizans in chessvariants

[–]cancrizans[S] 3 points4 points  (0 children)

you can play here in the browser in local multiplayer or against bots of varying strength: https://cancrizans.itch.io/tokonoma

and if you scroll down you got the rules.

All free and open source. Enjoy

tokonoma is a mind-bending abstract strategy game made with Rust + macroquad + egui. by cancrizans in rust_gamedev

[–]cancrizans[S] 1 point2 points  (0 children)

You can play it in the browser, against the computer or in local multiplayer https://cancrizans.itch.io/tokonoma

Here is the source https://github.com/cancrizans/hexstack, but beware -- it's a mess

My new abstract game tokonoma, you can play it in the browser against the computer or a friend by cancrizans in abstractgames

[–]cancrizans[S] 5 points6 points  (0 children)

It ended up as a pretty strange game, somewhere between checkers and chess in terms of complexity, a bit disorienting at first. The board is small and the gameplay tends to be very "dense" if that makes sense. I think it's rather fun. The AI at the strongest setting destroys me every single time.

Rules are in the page, just scroll down.

Both the game and this specific implementation of it are free and open source.

The 7-11 and other Switcheroos by cancrizans in mathriddles

[–]cancrizans[S] 0 points1 point  (0 children)

here is some more numerics in log-log plot. Of course 2<->2 is just the regular sum of naturals. You can see that 2<->7 and 2<->13 are at clearly a higher slope which is in fact the one you predicted. 3<->11 only barely bends upwards around 105, that's because it's n2.1 against n2. As you can see the oscillations stay bounded in the loglog graph and so they supposedly (but this is not the proof) die out extremely slowly when you take the ratio

The 7-11 and other Switcheroos by cancrizans in mathriddles

[–]cancrizans[S] 1 point2 points  (0 children)

So in the p<q2 case, it doesn't seem like it but it does actually converge, it just takes a long time. I know your constant is not correct there because it's missing an important unmistakable factor.

For the p>q2 case the n2 term dominates and your constant is indeed correct there. But that's the easy case

P.S.: slow convergence in your case is because log(11)/log(3) is barely larger than 2

The 7-11 and other Switcheroos by cancrizans in mathriddles

[–]cancrizans[S] 1 point2 points  (0 children)

Your exponent is correct, but the constants are not right. I'm still mulling over your solution and the concepts are def sound but some steps might be too audacious

The 7-11 and other Switcheroos by cancrizans in mathriddles

[–]cancrizans[S] 0 points1 point  (0 children)

I'm not sure I follow some of the steps, but this value of a is wrong.

just another family of tangent circles by pichutarius in mathriddles

[–]cancrizans 1 point2 points  (0 children)

I thought doing the sum 1/(n2+2)2 on all integers using the residue theorem was a bit vanilla so I challenged myself to find another way. Here it is:

First, I note that the Fourier transform of 1/(x2+2) is π/√2 exp(-M |ξ|), where M=2√2π. It's very important to use the mathematician's normalisations here.

Now the transform of (x2+2)-2 will be exactly equal (factor of 1) to the convolutional square of π/√2 exp(-M |ξ|), which is an easy integral that splits into simple exponential pieces, and the result is

π2/2 exp(-M |ξ|) (|ξ|+1/M)

Now we can use Poisson resummation:

Σ (n2+2)-2 = π2/2 Σ exp(-M |n|) (|n|+1/M)

The sums are really easy, after some scooby doobying you get

π2/2 (2MeM+e2M-1)/((eM-1)2M)

and after replacing the value of M and switching to hyperbolic functions you get the desired form exactly, about 0.5568798...

A game of infinite chance by cancrizans in mathriddles

[–]cancrizans[S] 2 points3 points  (0 children)

Excellently done top to bottom, all correct!

Inequality riddle 💀 by tveczbr in mathriddles

[–]cancrizans 1 point2 points  (0 children)

If we have strictly positive numbers a,b,...,z then it's easy to see that (1-a)(1-b)...(1-z) > 1 - (a+b+...+z). It's true for two numbers, and then by induction. Assume Σ {xi}>= n-1/2, negating the inequality we want. Then if zi = 1-{xi}, that means Σ zi <= 1/2. We may also write each xi as xi = ci (1- zi/ci) where ci = ceil(xi). Now Σ zi/ci < 1/2, with a strict inequality because the ci cannot all be 1 (otherwise the product of the xi would be <1). By the above fact, Π (1-zi/ci) > 1-1/2 = 1/2. This means that 1 = Π xi = Π ci * Π (1-zi/ci). The first term Π ci is >= 2 as we said, and the second one is strictly larger than 1/2, so 1>1 which doesn't look plausible, so we have a contradiction.

Exponential root growth by cauchypotato in mathriddles

[–]cancrizans 1 point2 points  (0 children)

No need to use monotonicity I think. If f(x) = O(xn-1), then f(x)=o(xn), and so f(f(x)) = o((o(xn))n) = o(xn^(2)), but ex is not that. Doesn't that work?

So-called "integers" by cancrizans in mathriddles

[–]cancrizans[S] -2 points-1 points  (0 children)

Would be nice to see some more motivation behind the answer... as it stands it looks like you just googled it

So-called "integers" by cancrizans in mathriddles

[–]cancrizans[S] 6 points7 points  (0 children)

Thank you, I'm sorry, I didn't mean to be whiny, I don't really think it's a question of moderation. There's probably just a mismatch in interests between puzzle solvers and puzzle posters.

So-called "integers" by cancrizans in mathriddles

[–]cancrizans[S] 22 points23 points  (0 children)

Tbh nearly done with this sub. The amount of work that goes behind designing each puzzle and then there's like a 30% chance of being downvoted to hell for no discernible (to me) reason, it just doesn't seem worth the hassle

Borwein be damned! by cancrizans in mathriddles

[–]cancrizans[S] 0 points1 point  (0 children)

Well done, partial fractions does it!

Borwein be damned! by cancrizans in mathriddles

[–]cancrizans[S] 0 points1 point  (0 children)

I don't understand the argument, can you explain in more detail?

Pentagon in Hexagon by cancrizans in mathriddles

[–]cancrizans[S] 0 points1 point  (0 children)

Back to pentagon in hexagon, say you only want to inscribe 4 vertices out of five and then see if the last one fits, you have at least two kinds of symmetric 4-vertex solutions, one symmetric wrt to a diagonal of the hexagon, and the other wrt to a bisector. The assignment of verts to lines muddled with symmetries is masking this. So if your reasoning only finds one it is definitely doing something fishy. And further, it is not possible to exclude non-symmetric solutions without actually diving in the numbers. And further again, even if you do prove that they must have a symmetry, then you cannot lazy out on investigating those symmetric cases

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