Would it be on-topic here to ask the number of possible combinations of inputs for a function that takes an input length of anywhere from 1 to 2^128 bits? I know the answer for a function that takes exactly 2^128 bits (it's 2^2^128), but I'm wondering if this basic question is acceptable here.
There's surely a better way to calculate it than 2^(2^1 + 2^2 + 2^3 + ... + 2^128).
Can anyone confirm his claim that n leq 23 from a 90 degree bend radius? I cannot confirm his result from any of my calculations https://defproc.co.uk/blog/lattice-hinge-design-minimum-bend-radius/
So $\sum\limits_{i=1}^{128} 2^{2^i}$ is correct? For information, this is because I'm trying to correct an answer I have on another site where I incorrectly claimed that SHA-512, a hash function that takes an input message up to 2^128 bits, had an input message space of 2^2^128.
It's just a (mathematical) function that takes an input that's typically considered to be arbitrarily sized, but is actually limited at 2^128. Why would that be over-counting?
Then I guess there's no simpler way to say "There are X possible distinct inputs" without having $\sum\limits_{i=1}^{128} 2^{2^i}$ in place of X? I should calculate it...
There is one zero-length message, two 1 bit long messages, four 2 bit long messages, eight 3 bit long messages, and so on. Since these messages are distinct, you can simply use the formula:
$$1 + 2 + 4 + 8 + \dots + 2^{n-1} = 2^n-1$$
Note that the maximum message length is the same as the maxim...
Trying to hash this out in some code I'm writing that models the probability of die rolls in a pool of die rolls (like "roll 6d6") - how do I model an conditional effect like "remove a die and reroll any amount of remaining dice"? Do I have to split the entire pool by conditional probability, like P(using the effect) * P(die outcomes minus the die removed by reroll) + P(not using the effect) * P(current outcomes)? Sorry if this makes no sense, my probability skills are way out of date.
The WA solution is a bit misleading: the Lambert-W function has an infinite number of branches, so you need to say which one you’re using. It’s analogous to saying that $y=x^2$ implies that x is the square root of y: Which square root?
A Poincaré map, by definition, is supposed to be a reduction of a phase diagram on a codimensional plane cutting normal to the original trajectory. In this form, one should plot points from the phase space (xdot vs. x), maybe in this case it makes sense to plot a-dot versus a (semi-major axis). B...
@TobiasKildetoft i mean that could you prove that $x ≈ 0.3181315052047640944061139753... \pm 1.337235701430689415856543101 i...$ are not the only two solutions?
Assume u satisfies $e^u=u$ then $\ln (e^u)=\ln u$ which means $u=\ln u$ which we get $e^u=\ln u$
Anyway, even once you pick a branch, you can not just go back again, so you have a correct implication, but no reason to expect to get all solutions (which would then be obtained from other branches)
The wiki page for the W-function (en.m.wikipedia.org/wiki/Lambert_W_function) gives an argument in the section “Special values” that W(x) is transcendental for any nonzero algebraic x
@Tobias I haven't had enough exposure to other areas of mathematics because my undergrad was so inadequate, so maybe I'll find something else interesting :)
$H^n(K(\Bbb Z, 2), \Bbb Z) = [K(\Bbb Z, 2), K(\Bbb Z, n)]$ is in bijection with natural transformations $H^2 \Rightarrow H^n$ by Yoneda lemma. If $n = 2k$ this has to be generated by $\alpha \mapsto \alpha^k$, but if $n$ is odd there's no such thing.
@LeakyNun This is a small part of what's called the Hurewicz mod C theorem. In fact any simply connected CW complex $X$ with finitely many cells in each dimension necessarily has all homotopy groups finitely generated. (Clearly this extends to finite fundamental group.) It is not true in general: $\pi_2(S^1 \vee S^2) = \pi_2(\vee_{\Bbb Z} S^2) = \Bbb Z^\infty$.
@LeakyNun In the approach you're using, you compute them by knowing what the $E^\infty$ page is, and so get through formal arguments that the differentials have to do something or another
Using that to compute a lot more differentials is why Balarka is saying to use cohomology; you have a ring structure and the differentials satisfy a Leibniz rule
Spectral sequences seem like a mathematical equivalent of Cthulhu, in the sense that if you don't know enough algebraic topology then you can go on in life blissfully ignorant that such things exist.
(I count myself in the blissfully-ignorant crowd.)
heck, this line out of Hermite is very nearly Lovecraftian: “I turn away with fear and horror from the lamentable plague of continuous functions which do not have derivatives.”