if we have some drinks to sell. If it costs $1$ to buy coke (C) per pint that we can sell for $2$. And it costs $1.5$ dollar per pint for sprite (S). We also have twice as much of C pints for each 1 pint of S, meaning that $C\ge 2S$. Also, the company does only cell 3000 of pints from both C and S per week.
I am trying to build up constraints here please: why the $Profit = C + 1.5S$ and not $profit = 2C + 3S$ based on how much we can sell them?
your statement is a bit ambiguous to me. how much does Sprite sell for? also, i don't understand "only Sell 3000 of pints from both C and S per week". does that mean $C+S \le 3000$ or $C \le 3000, S \le 3000$?
@TedShifrin It's some kind of geometric description of cup product. It's written in Hatcher. I didn't understand it fully so doubt my argument. I thought you already know that.
Now I have one question about graph though. X-axis represents amounts (pints) of C and Y-axis represents amoung (pints) of S. So, how is it possible that we draw profit to be $3500 on x-axis. Not sure if my question is clear pls?
math.stackexchange.com/a/187417/668308 I have some question in this answer. It says $H_n(M\# N,S^{n-1})\to H_{n-1}(S^{n-1})$ is surjective because the boundary of $M$ minus a disk has boundary exactly a fundamental class of $S^{n-1}$. But why this is true? I mean the topological meaning of boundary same as image of cycle under boundary map?
Company will only sell Moe one pint of Sprite (S) for every two pints or MORE of Coke (C) bought. Company does not sell more than 3,000 pints per week.
@love_sodam I know the material very well, but your sentence was just plain wrong. Yes, you can choose disjoint $\Bbb CP^1$s in the two summands. When I inquire, you should reflect carefully on what you’ve written.
@TedShifrin $X=CP^2\# CP^2$. As in Example 3.7 in Hatcher, I chose dual bases in $Hom(H_1(X),\Bbb Z)$ correspond to two $2$-cells in $X$. Then since it's a connected sum, two duals don't intersect. So trivial cup product. Does it make sense?
Have you reread the sentence I complained about? You need to be precise with language. What you just wrote now is wrong, but it’s restating what I saud.
An industrial factory produces pieces of which 1/5 of them are defective. Two buyers A and B, rate these acquired pieces in two categories: I and II, paying $1.20 and $0.80 respectively as the following style:
Buyer A: Gets five pieces sample. If he finds more than a defective piece, he rate the batch as II. Buyer B: Get ten pieces sample. If he finds more than two defective pieces, he rate the batch as II.
What is the probability of the buyers A and B found good pieces?
@Avra I'm not sure what your question was. There is a line indicating the points that generate equal profit, one of them is for \$3,000 profit. The line intersects the $S=0$ line at $C=3,000$ which shows that (as expected) there is a profit of 1 for each $C$ sold. The same line intersects the $C=0$ line at $S=2,000$ which shows that (contrary to the statement above) that the profit per unit $S$ is 1.5.
@Semiclassical suppose he gets a batch with pieces (of anything). If he finds more than one defective piece, he says this batch is bad batch quality. So, He will pay only $0.80 per piece.
@Avra Each point on the plot corresponds to a particular $(C,S)$ pair, where each value is a number of pints. The profit in dollars for a given $(C,S)$ pair is $C+ {3 \over 2} S$. The lines shows points of equal profit. But the dimensions on the axes are pints.
@TedShifrin I think I should've said that not because they intersect transversely, but : Let $\alpha\in Hom(H_2(X),Z)$ be the dual of $a$ assigning the value $1$ to $a$ and $0$ to $b$ where $a,b$ are two $2$-cells in $X$ and similarly to $\beta$. Representing $\alpha,\beta$ by counting the number of intersections of each $2$-cells, $\alpha\cup\beta = 0$ since $2$-cells are disjoint.
Then there is a formula for the profit $P(C,S) = C + 1.5S$. This is just a formula. If you pick a particular profit, say \$3,000, then $P(C,S) = 3,000$ will corresponds to some points on the graph.
@Semiclassical I didn't think this because I thought when he picks up the pieces, doesn't matter in what order he gets because the act of get bad or good pieces is independent.
this must have been part of a general cheapening of the culture. i heard the exploratorium got dumbed down a bit when they moved it to the waterfront.
although some of the smart stuff is dumb too. those pendulums that knock over dominoes once an hour (or whatever) as they rotate are moronic and every science museum has one.
it's the perfect answer to, what makes the world's most visually boring exhibit.
or is it that we're rotating? whatever the point is, i missed it.
the fun of the hyatt is throwing paper airplanes from the upper floors of the indoor atrium until people from the hotel catch on and try to find who is doing it and eject them.
i was in the lobby of the hyatt the morning of december 31 1999 and saw more balloons being filled and loaded into nets than i have ever seen elsehwere.
lots of goofy memories of that time. the dot com bubble hadn't burst yet. some of those startups threw parties that would have been great, had they not been at the office.
another good one would be a gimmicked pendulum that physics teachers use. you know, where they take this really heavy weight, hold it right up to their nose, then let it go. gimmick one of those so it can hit back every once in a while.
there is some gootube video with a physics prof (i think Dutch) who shows some conservation principle with a large & potentially dangerous pendulum. he is an antidote to flat earthers.
yes, i want a pendulum with some chaos built in. in fact, i think it would be a good addition to congress.
@Ted: we have to let our pup go tomorrow. She has two forms of cancer and a decubitous ulcer (like a bedsore on a joint) on her left front elbow from leaning on her left side (due to the edema in her right rear leg). The ulcer can't heal while chemo is being given and the cancer is growing too rapidly to wait for the ulcer to heal. We are bringing her home from the hospital tomorrow and sending her off from here.
I think I finally cracked the case. I'll finalize things later today if not tomorrow assuming there aren't still any glaring mistakes, but I revisited the identity of $a\bmod bc$ and the results do appear more promising compared to before. Just take a look at this and it should give a hint of what I'm doing, but I'm too tired to make any meaningful progress right now. desmos.com/calculator/fs2sqj5bj0
Pardon my interruption. If $c$ is the modulus we want to compute, and $b$ is a power of two, then after computing $a \bmod bc$, we can compute a smaller modulus as $\left(a\bmod bc\right) \bmod b\prime c$. Simplest idea I have is to halve $a\bmod bc$ and find $b\prime$ such that $2^{\lfloor\log_2(b)\rfloor - \lfloor\log_2(b\prime)\rfloor} b\prime c = bc$. Is there any error in my reasoning?
I've been exposed to plenty of identities concerning a constant modulus, but not really any identities of the modulus itself and what holds for elementary arithmetic applied to it.
err the idea is to halve $a\bmod bc$ and find the largest multiple of $c$ less than $bc$ and for which $\lfloor\frac{a\bmod bc}{b\prime c}\rfloor = 1$ I believe is more descriptive. Something around that description is what I want.
I'm already aware that $\lfloor\frac{2^x}{2^{x-\lfloor\log_2(y)\rfloor - [\lfloor\log_2(y)\rfloor \neq \log_2(y)]}y}\rfloor = 1$. (Brackets are Iverson bracket notation)
im likely missing something obvious, im trying to show that for some fixed compact set $E \subset \mathbb{C}$, there is some monic polynomial of degree $n$ with a minimum maximum modulus on $E$
oh, maybe I just identify the space of monic polynomials of degree $n$ with $\mathbb{C}^n$, and endow it with the norm that is the maximum modulus of any polynomial with associated coefficients, this norm is necessarily equivalent to the usual one, so bolzano-weierstrass still holds with respect to this norm, so we take a sequence of polynomials (i.e. their coefficients) each with norms approaching the minimum, so this is bounded in norm and has a convergent subsequence
the polynomial whose coefficients are the limit of this sequence is the minimizing one
or at least a minimizing one
oh wait, its not quite a norm though, because there can be nonzero polynomials with zero maximum modulus on $E$ as long as $E$ is finite
but if $E$ is finite the minimum maximum modulus is just zero so i don't think this is too much of an issue
uh, its only zero if the size of $E$ is less than or equal to $n$ I think..
@leslietownes Thanks for the reply. The domain of $y(n)$ should be the natural numbers. The reason for the question was that, if $f_n(x)$ converges uniformly to $f(x)$, then one should be able to find an $N(\epsilon)$ given any $\epsilon$ such that $n>N(\epsilon)$ implies $|f_n(x) - f(x)|<\epsilon$ for all $x$ in the domain of $f_n(x)$ and $f(x)$.
By analogy with showing some function is continuous, I would try to obtain $n>N(\epsilon)$ from $|f_n(x) - f(x)|<\epsilon$; this only seems possible if $|f_n(x) - f(x)|$ is bounded above by some function only depending on $n$, so one can in turn bound it from above by $\epsilon$ and solve for $n$.
Is someone around that can help me understand modular congruence better?
I'm trying to find a value that is modularly congruent to $38 \cdot (70 + 133)$ with modulus $n = 40$. As far as I understand, that means find a value $a$ such that $a - (38 \cdot (70 + 133)$ divides $40$, but is in also in the set {1, 2, ..., 40}, correct?
hi everyone, I am wondering if there's someone who's familiar with the min-cost max-flow problem, particularly in the context of integer programming and heuristics
basically I'd like to know the name of a similar problem: where instead of optimizing flow, one is given a flow and finds a minimum-cost capacity / topology that satisfies it
I am currently not sure whether the problems are identical, or whether what I'm saying even makes sense as a problem, but I'd like to know whether it's been considered and what is its name
vrouvrou, if it's theorem 1.2.4 of something, there's at least one data point suggesting that it's true. :) in english you might try searching for sobolev embedding theorems. they usually come out of inequalities involving integrals.
context: i've got a particular set $S$ of permutations $\in S_7$. i'm told (and can readily believe) that this set generates the entire group. given a particular permutation, how do we find the simplest representation as a product of the generators? @BalarkaSen
you can solve it by a computer by first drawing the Cayley graph of $S_7$ wrt $S$ then taking the geodesic between $e$ and your permutation in the graph
interesting question: randomly choose $S \subset S_n$ and $x \in S_n$. in expectation, how much of $Cayley(S_n, S)$ do you need to sample to compute a geodesic joining $e$ and $x$? most of the graph should be typically irrelevant given size of $S$
@BalarkaSen i think the point is (mostly) this. suppose $a,b\in S$. then you can draw a van Kampen diagram to visualize some portion of the full Cayley graph
something like: vK's are to Cayley(S_n,S) as Cayley(S_n,S) is to Cayley(S_n)
i'm dubious that any of this can be made easy to do by hand, though
@BalarkaSen for specifics, my case is $S=\{(67), (57),(37),(46)(57),(23)(67),(45)(67),(26)(37),(13)(57),(15)(37)\}$
i think the difficulty is that most discussions of the word problem are in terms of Dehn's algorithm, which seems simple enough if you have a small presentation
For contradiction compound statement that is always false. Suppose we have contradiction statement $c$ and another statement $p$, then if $c=F \land p=F$, then why $c \land p = F$ and not $T$?
Got you. So how about $F$ statement that stands for $F$? Does it mean only FALSE, so we have for $P \lor F$ the following: P -- F ------- T -- F F -- F F -- F
Is this how the truth table looks like for $P -- F$ before taking $P\lor F$?
