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yes but the mega rich would still find a way to benefit from this, so nothing would fundamentally change. they're not going to let go of their power that easily.

I will leave the base case for you to verify. Inductive hypothesis: a_n = 3a_{n-1}-a_{n-2} Known information: a_n = \frac{1+a_{n-1}^2}{a_{n-2}}\text{ for every integer above 2.} So we write out the definition of a_(n+1): \begin{align*}a_{n+1} &= \frac{1+a_n^2}{a_{n-1}}\\ &=...

i mean i hate the federal gov and i still got my vaxx, it's a matter of what the options are (Δ is simply too transmissible for it to ever go to 0 without a legitimate hard lockdown which is never happening)

that is to say, not at all

it's still as relevant to the conversation as your remark

surviving isn't living

Models suggest Herd Immunity is impossible for the UK, as the theoretical number is higher than the eligible population. I think the same applies to Australia.

i like how this is just a reskin of the BOS Trials' cancerous arctangent integral

oh i'm not here to participate; i'm here to cast judgement on dan as he is casting on the entirety of the non-straight population.

you don't?

it is clear that dan has a naïve or possibly non-existent understanding of the psychology of sexuality (and even if it exists, it is heavily clouded by religious puritanicalism), so i would take anything he says about it with a molecule of salt.

obviously they have access to a hyperbolic time chamber

⁻¹/₁₂

Personally I don't see a problem with this, but you would have to prove that a for f a strictly increasing function, f applied to a strictly increasing sequence is a strictly increasing sequence. However, I feel like this lemma has been stated without proof in past HSC questions, so I don't know.

Random observation I want to make. Let a = \tan{A}, b = \tan{B}, c = \tan{C} Then the problem reduces to maximising \cos(2A) + \cos(2B) + 2\cos C subject to the constraint \cos(A+B+C) = 0 and A,B,C are acute angles. I don't know if this has a nice geometric interpretation, but if anyone...

actually there is a systematic way of doing this and it's basically the discrete version of the derivative Define the forward difference operator Δf(n) = f(n+1)-f(n) the binomial coefficients xCk are the monomial analogues of the continuous monomials x^k /k! in the sense that Δ(xCk) =...

use the greedy algorithm to successively eliminate the highest available power by subtracting the appropriately weighted binomial coefficient this is a lot of expanding and algebra simplification so it's not that much nicer than splitting into cases and arguing carefully, but it doesn't require...

yeah and we didn't sugar coat it by splitting into odd and even cases in the main steps of the proof :biggrin:

Black Magic Proof: Observe the following identity: n^5+n^3+2n \equiv 120\binom{n}{5} + 240\binom{n}{4} + 156\binom{n}{3} + 36\binom{n}{2} + 4\binom{n}{1} All coefficients of the binomial sum are multiples of 4, and all binomial coefficients are integers, so the RHS is divisible by 4, hence...