Posts for OmnipotentEntity

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Posted: 12/15/2022 4:26 PM
(Edited: 12/15/2022 4:27 PM)
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OmnipotentEntity
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Shame about having to play Rostov 1. Good thing it's short though! I love this game. Yes vote for vault. I hope to see a max medals route in the future, which would hopefully show off a bit more gameplay. Was the Port Hedland route tested? Or did you just take my word for it? Also, I recognize that graphic! :O
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Posted: 12/9/2022 7:47 PM
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OmnipotentEntity
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I would be interested to know as well, r57shell. I played around for a very short time with it, and it seemed complicated to me. Challenge. There is a box that magically produces mangos. You may check it at most once a day, and every day it has a 5% chance of spawning a mango, unless there is already a mango in the box. However, if you open the box to check, regardless of whether or not you find a mango inside, you cannot open the box for 3 days, and the box will not spawn a mango for those 3 days. How often should you check the box to maximize your expected rate of mango acquisition?
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Posted: 11/17/2022 2:57 PM
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OmnipotentEntity
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That is an unbelievably cool change in perspective.
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Posted: 11/14/2022 5:04 PM
(Edited: 11/14/2022 5:10 PM)
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OmnipotentEntity
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Challenge There is a wheel that costs 1 coin to spin, and it rewards you according to the following probability table: 5/9 : 0 coins 1/3 : 2 coins 1/9 : 3 coins You have 5 coins and you need at least 9 coins to purchase the prize you want. Using this wheel, what is the (exact) probability that you win those 9 coins (before you run out of coins and cannot play anymore)? Observation: Because the EV is exactly 1 coin gained to every coin lost, this is more or less a martingale. However, I don't know much about martingales, so I'm not sure what to do with this information, other than to justify the approximation of x/y as the probability where x is the number of starting coins and y is the number of goal coins. Solution method spoiler: I've already solved this problem and I have an exact answer using Markov chains, but because it relies on the analysis of a particular n+1 by n+1 matrix, it's not very amenable to extrapolation to large n, so I was hoping someone else had a more clever approach. Solution form spoiler: The form of the solution is the ratio of two 7 digit numbers that roughly approximates 5/9.
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Posted: 11/7/2022 7:46 PM
(Edited: 11/7/2022 10:14 PM)
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OmnipotentEntity
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He defines exponentiation for positive integers (but only really by example) and then once real numbers are established to exist he defines roots and proves they exist. But Rudin doesn't actually mention negative powers, despite proving some other very simple properties of addition and multiplication, like if x + y = x + z, then y = z. But I don't know that I'm not "allowed" to assume they exist. Perhaps the intention was I am allowed, it was just unclear to me, so I operated conservatively.
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Posted: 11/7/2022 5:47 PM
(Edited: 11/7/2022 5:51 PM)
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OmnipotentEntity
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p4wn3r wrote:
Since this looks like a homework problem, I'll only give a hint.
Not actually a homework problem. I've been out of school for a while. I'm just doing this independently for fun. I was just wondering, because I know people here *have* actually gotten math degrees, if 1.6 was somehow known for being involved. Here's what I've done so far. 1. Define exponentiation for negative numbers (and 0). This hasn't been done in the book yet, but is critical to justifying many things. 2. Show that addition and multiplication formulas of exponents holds for all integers. a^b a^c = a^(b+c) and (a^b)^c = a^bc 3. Show that for any base, b > 0 we can choose an n != 0, such that (b^n)^(1/n) 4. Show that roots combine in the manner we'd expect: (b^(1/m))^(1/n) = b^(1/mn) 5. Show that roots commute with powers: (b^m)^(1/n) = (b^(1/n))^m 6. Finally, we can complete part (a) and show that if m/n = p/q then (b^m)^(1/n) = (b^p)^(1/q), defining the rational exponents. 7. Complete part (b): b^r b^s = b^(r+s) for rational r and s 8. Show step 3, but for rational, rather than integer, powers and roots. 9. Show that if x and y are both larger than 1, then xy > x. 10. Show that if b>1 and n is positive, then b^n >= b. 11. Using 9 and 10, show that if x >= y > 1, then x^n >= y^n. (for positive integer n) 12. Prove for b>1, x>1, there's some positive integer n such that b < x^n. 13. Using 12, prove that given b>1, x > y there exists some rational number r such that x > b^r > y. 14. Using 11, Show that b^r is an upper bound of B(r) 15. Using 13, show that b^r is the least upper bound of B(r), defining the real exponents, completing part (c). Then for part (d), I'll need to formalize and extend the argument alluded to in the appendix of chapter 1, where multiplication of reals is defined, but I haven't completed that quite yet (life is getting in the way), so I can't give you the steps, but I think I've got this covered. I'm sure I'll come up with *a* proof. Just wanted to know if 1.6 is notorious, or if I can expect the rest of the problem in Rudin to be like this, and 1-5 were the odd ones out, or if I'm begin entirely too careful, and this isn't expected, or if there's a fundamentally easier way that I'm missing.
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Posted: 11/2/2022 3:12 PM
(Edited: 11/2/2022 3:39 PM)
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OmnipotentEntity
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So I'm putting myself through Baby Rudin, and I was working exercises in the very first chapter. Is it just me, or is problem 1.6 significantly more involved than problems 1-5? More involved than each individually sure, but also significantly more involved than all 5 of them combined. I just finished part (c), so I still have the final part to go, but I currently have 16 numbered statements (Defs, Lemmas, etc), and it spans 8 pages of latex so far. And I think I might be overthinking it. For reference, Problem 1.6 is the problem where they ask you to define rational and then real exponentiation and prove some things about it.\ ETA: I guess since this is the math challenges thread, I might as well post the problem statement itself: 6. Fix b > 1. (a) If m, n, p, q are integers, n > 0, q > 0, and r = m/n = p/q, prove that (bm)1/n = (bp)1/q. Hence it makes sense to define br = (bm)1/n. (b) Prove that br + s = br bs if r and s are rational. (c) If x is real, define B(x) to be the set of all numbers bt, where t is rational and t <= x. Prove that br = sup B(r) when r is rational. Hence it makes sense to define bx = sup B(x) for every real x. (d) Prove that bx+y = bx by for all real x and y. ---- For this problem, you are armed only with the axioms of a field, the definition of an nth root. The fact that any set of reals with an upper bound have a least upper bound, and the Archimedean property.
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Posted: 5/13/2022 6:47 PM
(Edited: 5/14/2022 8:28 PM)
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OmnipotentEntity
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Here's my attempt. Let f(n) be a random function that takes a positive integer, and we let Xn be a random variable representing a discrete uniform distribution from 1 to n-1 inclusive, and f(n) = n - Xn. This represents a single iteration of our process. Because there is no valid random variable X1, we must define the f(1) = 1. By applying this function multiple times we get the various iterations. f2(n) is the second iteration, f3(n) is the third, and so on. Because each step of the process is guaranteed to reduce the value, lim m->inf fm(n) = 1 for any finite value of n. We can find the expected value of the smallest m such that fm(n) = 1 by noting that this process has no memory. So E(m | fm(n) = 1 and fm-1(n) != 1) is always the same given the same value of n. Let g(n) = E(m | fm(n) = 1 and fm-1(n) != 1) (for n > 2) First let's find some base cases: g(1) = 0, because 1 is already equal to 1. g(2) = 1, because f(2) = 1 always g(3) = 1/2 * g(1) + 1/2 * g(2) + 1 = 1.5. This final line is because half of the time, f(3) = 1, which needs g(1) additional steps to reach 1. The other half f(3) = 2, which needs g(2) additional steps to reach 1, and the final + 1 represents the current step. And the inductive case: g(n) = 1 + sum from i=1 to n-1 1/(n-1) g(i) This recursion is unfortunately not a simple kind of recursion. It is linear sure, but it is non-homogenous, non-constant coefficient, and of non-constant dimensionality. I don't have any idea how to tackle this recursion explicitly. As I mentioned in my previous post, I assume this value is strongly linked to the logarithm. I misremembered though, I reviewed my notes and it seems like it should be base e rather than base 2. So I assume the solution is ln(n) + O(1) Empirical evidence seems to suggest that O(1) approaches the Euler-Mascheroni Constant. And in fact, it seems that, again empirically that the value of g(n) is simply the n-1(th) harmonic number. Knowing this, there is almost certainly a really elegant way to restate this problem and arrive at this solution cleanly. But I still cannot see it. EDIT: Well, let's try induction then. If g(n-1)...g(1) are Hn-2...H0 (where H0 is taken to simply be 0), then g(n) = 1 + 1/(n-1) sum from i=1 to n-2 Hi = 1 + 1/(n-1) sum from i=1 to n-1 (n-1-i)/i This last transformation was recasting the sum as "columnwise" instead of "rowwise" To explain, if you have: H3 = 1/1 + 1/2 + 1/3 H2 = 1/1 + 1/2 H1 = 1/1 Then H3 + H2 + H1 = 3/1 + 2/2 + 1/3 Then if we bring the 1/(n-1) into the sum again, and perform partial fractions we get: g(n) = 1 + sum i=1 to n-1 (-1)/n-1 + 1/i We can see that the first term in the sum, after summation is simply -1, which cancels with the +1. And we can see that the second term is simply the n-1(th) harmonic number. So our induction holds! Because we have g(1) = H0, and if that's not satisfying g(2) = H1 and g(3) = H2 just from the worked examples up there, we have our full proof. It's not very satisfying though, because it took a guess and check method. If you have a way of arriving at this from first principles please post it!
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Posted: 5/13/2022 4:26 PM
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OmnipotentEntity
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Dacicus wrote:
Finally, the expected number of moves is m(N) / M(N) = N / 2.
I haven't fully digested the content of your post. But I have worked a problem very similar to this with respect to neutron slowing down in my college courses. So I would expect them to have the same answer, which is log_2(N). And in fact, when simulating this using 1000000 it only take handful of steps rather than 500000.
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Posted: 4/27/2022 9:04 PM
(Edited: 4/27/2022 9:18 PM)
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OmnipotentEntity
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There are two rules that allow the combination of functions (in a way that increases the total terms): the chain rule and the product rule. (The quotient rule is just the product rule in funny clothing, and so doesn't increase the complexity more quickly than the product rule alone.) Both of these have the effect of a quadratic increase in the number of terms. Can we do better than quadratic by composing the two methods of combining functions? If we consider the set of abstract functions f_I (where I is an arbitrary dimensional multi-index) which are collectively differentiable over some common interval, which cannot be simplified through combination of themselves or their first derivatives, then we can consider some recursive definition of composite functions g_I, where the example given in the last post is structured similar to: g_0(x) = f_0(x) g_n(x) = g_{n-1}(f_n(x)) And maybe a more complex example might be where at every level of function nesting we have a product of two functions: g_0(x_0, x_1=x_0) = f_0(x_0) f_1(x_1) g_1(x_00, x_01=x_00, x_10=x_00, x_11=x_00) = f_0(f_00(x_00) f_01(x_01)) f_1(f_10(x_10) f_11(x_11)) g_n({x_{I0}, x_{I1}=x_{I0}}) = g_{n-1}({f_{I0}(x_{I0}) f_{I1}(x_{I1})}) Where the notation {x_{I0}, x_{I1}} denotes a set of 2^{n+1} arguments with I being a valid n-dimensional multi-index running from 0 to 1 in each dimension. The indexing notation here is getting a bit dizzying but the upshot is these arbitrary recursive composite functions are themselves just plain functions. Because these functions are themselves differentiable over the same common interval, then we can just examine the recursive case. And because the recursive case is just a plain function the best we can do is quadratic as compared to the input. I hope that's careful enough.
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Posted: 4/20/2022 8:13 PM
(Edited: 4/20/2022 8:18 PM)
Post subject: Primes generated by Euclid's Algorithm
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OmnipotentEntity
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It is, of course, possible to prove that there are an infinite number of primes by observing that for any list of primes $\{p_i\}$, you can form the product and add one to generate a number that is either prime or has a prime factorization that only contains primes not on your list. If we actually attempt to generate primes using this algorithm we get the following sequence: 2, 3, 7, 43, 1807, 3263443, ... If we factorize these numbers we receive the following sequence: 2, 3, 7, 13, 43, 73, 139, 181, 547, 607, 1033, 1171, ..., which are the only primes that we will generated by this algorithm. (Other primes get trapped into loops mod p, for instance 5 gets trapped in the cycle 2, 3, 2, 3, ... mod 5). If we examine the ratio of the generated primes vs the total primes up to n we seem to have a falling trajectory, and it would make sense if the ratio converges to 0 as n goes to infinity, these primes seem to be quite rare. This makes intuitive sense if we think about this particular algorithm mod p as needing to generate a 0 mod p before generating any other value it's already generated, we can loosely consider the action of the algorithm as a random draw with replacement, and it's easy to see that the probability of drawing a 0 is much less than drawing any two numbers in a row due to the Birthday Paradox. (That being said, the Euclidean algorithm definitely is not a random process mod p, and I'm not even certain it is particularly well-approximated by one.) (x-axis is natural numbers and y-axis is the ratio) (x-axis counts the nth prime and y-axis is the ratio) Can we prove this ratio tends to zero? Also beginning with other seed values will generate different sequences. All odd primes will immediately generate 2, so 2 will appear in all such sequences. Are there other primes that appear more often than chance would dictate? Less often?
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Posted: 2/15/2022 5:48 PM
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OmnipotentEntity
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I marked the forums as read before I left the site yesterday. The "View posts since last visit" button disappeared when I did that. Today I refreshed on the same tab, new posts, yay. The "view posts since last visit" button is also back. I clicked on the button, and I have posts in the list dating back since January 9, still. :(
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Posted: 2/12/2022 7:30 PM
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OmnipotentEntity
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adelikat wrote:
My question is that if the definition of "last visit" were more accurate, does the "post since last visit" feature work? Or are there other issues with that page?
Sorry about the delay in responding, I actually completely lost this thread in the shuffle. As I mentioned above, I would prefer if the ordering were by reverse chronological ordering and if results were displayed as threads rather than posts. As it is, I have to click into the thread anyway to see context, so I don't see the value of dumping a bunch of unrelated posts on the same page. And because the results aren't really clearing themselves (they didn't either on the old site, to be fair) having the sort be in reverse chronological order makes more sense because it shows me the most recent posts and I can "dig" as much as I care to. What might also be useful is a "mark forum as read" button
Follow up question is if the "recent posts" page had the same UI as "post since last visit" would that be better or at least acceptable?
I am not certain I follow the question. AFAICT, the difference between "recent posts" and "posts since last visit" is that "recent posts" have a hard time out (whatever recent means), and they're sorted in reverse chronological order. Anyway, my gripes are surely not terribly important either way. I'm a long term, but not terribly active user. I just thought I'd voice my desires because that's what this thread is for.
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Posted: 2/1/2022 4:11 PM
Post subject: Most recent unread post search
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OmnipotentEntity
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When I use posts since last visit search, it's sorting in chronological order, rather than reverse chronological order as it was in the previous iteration of the site. Furthermore, it's returning in terms of posts rather than topics. Also because my session is apparently infinite, I now have 50 pages of unread posts, going back 23 days, despite having visited the site several times since early January. I'm sure there's something I'm doing wrong, but I have no idea what.
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Posted: 12/7/2021 7:11 PM
(Edited: 1/17/2022 4:51 PM)
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OmnipotentEntity
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Actually modeling this problem will be rather difficult. Especially because shape has a lot to do with how convective flows move around. But a simple model with a fixed convective coefficient is probably good enough for a first order approximation. Let's consider the two extreme cases, a cube and the sphere which inscribes the cube (both of ice). The cube will have a momentary heat transfer which is related directly to the surface area via the coefficient, so the ratio of heat transfer is dependent on the ratio of the surface areas. In this case the ratio of surface area between the sphere and cube is 24/4pi = 6/pi. Whereas the ratio of volume between the two is 8/(4/3 pi) = 6/pi. Wait what? (Of course it makes sense, because the surface area is the differential of the volume, and that's linear.) So assuming the ice melts perfectly evenly, both the cube and the sphere will finish melting at the same time. How unsatisfying. But the ice won't melt perfectly evenly on the cube, because the ice is not perfectly at 0 degrees all of the way through, so corners heat up faster, and because, outside of our idealized single convection coefficient world, we have interesting edge effects in convection that cause edges to be affected more than corners. These complications have the effect of rounding off the corners, reducing the surface area of the cube, slowing it down. So our square will melt slower overall. I agree that the question seems ill-posed.
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Posted: 5/3/2021 3:41 PM
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OmnipotentEntity
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Warp, I think the disconnect you're having is that you're assuming tacitly that both limits will approach 0 at the same rate. But that is not necessarily the case. Take a really stupid example where f(x) = e^x and g(x) = 1/x Now it's clear that limit x->0 f(x)^g(x) is just e. However, the two limit case is different. To make this a bit more clear, I'm going to do a change of variables: limit x->0 f(x)^(limit y->0 g(y)). Now it's clear that if we have a free variable k that relates x and y (y = kx), then we can make the two stage limit equal any complex number (given k complex), because the limit in that case will be e^k. (In this case I also selected the indeterminate form 1^inf)
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Posted: 4/14/2021 3:46 PM
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OmnipotentEntity
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NxCy wrote:
How is the trick performed?
So if we look at this from the perspective of information theory. The passed hand contains the identities of the four cards and their order. There are 5 choose 4 ways of choosing which cards to pass, and 4! ways of choosing how to pass them. The receiver of the cards only has partial access to this information though. (which is 6.9 bits of information). He or she obviously has access to the 4! ordering. But this is only 4.6 bits of information, and you need 5.6 bits to specify 1 of 48 remaining cards. You can add the final bit by handing the cards either face up or face down, which is very magic-trick-like, in that there's a trick. And that gives you the exact number of bits required to specify 1 of 48 unique cards. I don't know if they do this though. But here's how I'd run an encoding: First, define any unambiguous ordering to the cards. (0=ace of clubs, 1=two of clubs, etc up to 51=king of spades) Second, choose any card as the secret card and find the number of this card (after removing the other 4 cards from the ordering.) Call this number n. Third, find the modulo 24 of n, call this m. Four, construct the mth permutation of the remaining four cards from sorted order. Fifth, if n >= 24 then hand the cards face up, otherwise hand them face down.
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Posted: 2/3/2021 5:21 PM
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OmnipotentEntity
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Warp wrote:
You are right. There isn't enough information. The two blue shapes should be semi-circles. (I'll change the picture to be more accurate.)
Given that they are semicircles, the answer is half the area of the circle. You can arrive at this by using the diameter of the circle (2r) is equal to the height of the lower segment (ha) plus the radius of the lower semicircle (a) plus the radius of the upper circle (b) plus the height of the upper segment (hb) ha = r - sqrt(r^2 - a^2) hb = r - sqrt(r^2 - b^2) So: 2r = r - sqrt(r^2 - a^2) + a + b + r - sqrt(r^2 - b^2) a + b = sqrt(r^2 - a^2) + sqrt(r^2 - b^2) Then do a lot of tedious algebraic manipulation to solve for r^2. And you arrive at r^2 = a^2 + b^2 And the area of the two circles is 1/2 pi a^2 + 1/2 pi b^2 = pi/2 (a^2 + b^2) = pi/2 r^2 There is probably a more elegant approach, possibly involving the line connecting the left (or right) edge of both segments, which creates a segment that directly has length a^2 + b^2. But I didn't immediately see an easy way to directly relate this with the radius, and my cat sat down on my paper, so I couldn't continue working.
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Posted: 11/9/2020 6:44 PM
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OmnipotentEntity
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So I help out from time to time on a homework help discord channel. Sometimes we get trolls who pose as high schoolers and pretend that they have very difficult problems. This is one such problem. Find the area bounded by the implicit curve x^6 + y^4 = 36xy + 4
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Posted: 10/27/2020 8:50 PM
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OmnipotentEntity
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Yes, and yes, but the effect is very slight (though it's much more noticeable as a chemical difference, charged batteries are more bouncy). It's also part of the reason why in-spiraling black holes release so much energy (in the form of gravitational waves). When you bring two objects together that form a very strong bond of some kind, the energy of the bond is released somehow. Whether that's electric (chemistry, in the form of photons typically), strong force (nuclear fusion), or gravity (black holes coalescing).
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Posted: 10/19/2020 6:37 PM
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OmnipotentEntity
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Binding energy generally goes the other way. It manifests as a mass defect.
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Posted: 10/12/2020 9:52 PM
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OmnipotentEntity
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The Lambert W function is multivalued, so is the natural logarithm. The two surfaces representing these functions embedded in the 4d output space touch at an infinite number of points. Specifically, at the point specified by each branch of the W function. These are: 
 k | intersection
----------------------
-5 | 3.28777 +26.5805 i
-4 | 3.02024 +20.2725 i
-3 | 2.65319 +13.9492 i
-2 | 2.06228 +7.58863 i
-1 | 0.318132 +1.33724 i
 0 | 0.318132 -1.33724 i
 1 | 2.06228 -7.58863 i
 2 | 2.65319 -13.9492 i
 3 | 3.02024 -20.2725 i
 4 | 3.28777 -26.5805 i
 5 | 3.49852 -32.8807 i
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Posted: 10/8/2020 4:13 AM
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OmnipotentEntity
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Here's an older problem that's been giving me some difficulty: With the clarification that if more than one student fits this criteria then one will be selected at random uniformly. The answer to the problem posed in the question is known, and is a large irreducible fraction for the given prompt using number of students = n = 10. I also know the values for n up to 20 using a brute force algorithm in their irreducible fraction forms. However, I've been completely unable to analyze this problem farther. I have no idea how to even set up a recursive definition for the probability of n, let alone a closed form solution. Best of luck!
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Posted: 10/5/2020 5:27 AM
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OmnipotentEntity
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If e^z = ln(z), because e^z and ln(z) are inverse functions, then it is the case that also e^z = z. Then solving using the Lambert W function is rather trivial: ze-z = 1 -ze-z = -1 -z = W(-1) z = -W(-1)
Build a man a fire, warm him for a day, Set a man on fire, warm him for the rest of his life.
Posted: 9/20/2020 4:20 PM
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OmnipotentEntity
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Holonomic constraints are in the form of: f(p1, p2,...; q1, q2,...) = 0 So I guess if you had a system of differential equations that formed your constraint set, then if they were exact (optionally with an integrating factor), you could recover f. That makes sense. Thanks.
Build a man a fire, warm him for a day, Set a man on fire, warm him for the rest of his life.
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