@Chris'ssis The asymptotic behavior of the sequence in this question does not happen until $n\ge3.2197\times10^{70683}$. As this comment notes, that number is much larger than the number of atoms in the universe (by a LOT).
how is orientation defined for n-simplices in general? (i'm only looking for the restriction of the orientation to discrete types of things, no crazy manifold stuff if possible)
@robjohn is there a trick to get the polynomial $\frac{M}{6}(x-x^3)$ in math.stackexchange.com/questions/353947/… .. ? I tried to apply your way to a similar problem and didn't know how to get the poly
@AlexanderGruber $[abc]$ is a $2$-simplex;) We have $$\partial [a_0,a_1,\dotsc,a_n] = \sum_{i=0}^n (-1)^i [a_0,\dotsc, \widehat{a_i},\dotsc, a_n],$$ where $\widehat{a_i}$ means that vertex is omitted. Whether that is functorial, maybe I can answer if I know what functorial means here.
@PedroTamaroff if at some point in the future you were inclined to use your combinatorial knowledge to try to count the number of unlabeled graphs on $n$ vertices with 3-colorable, triangle-free complements, that's something i would be quite excited about. :p
@DrJonesYu The way you proceed in your inductive proof doesn't work. Not every graph (of your form) can be constructed in that manner.
In general, when you're doing an inductive proof in graph theory, you start with a graph on $k+1$ vertices and somehow reduce it to the case of a graph with $k$ vertices - not the other way around.
So lets say I have a $k$-manifold $M$ in $\mathbb{R}^n$, and I cover it up with coordinate patches $\{\alpha_i\}$. I can find a set of partitions of unity $\{\phi_1,...\phi_l\}$ on $M$ which is dominated by the coordinate patches (that is each of their supports has to be a subset of a coordinate ...
@Mike I think I was told that I can find a collection that would constitute as a partition of unity...that is $\phi_1 \dot h^_{-1},...,$\phi_l \dot h^_{-1}$ would be a partition of unity on N
A,B and C are three commodities. A packet contains 5 pieces of A, 3 of B and 7 of C costs $24.50 . A packet containing 2,1 and 3 of A, B and C respectively costs $17.00 The cost of a packet containing 16, 9 and 23 items of A,B and C respectively is......" and answer is $100 But i can't understand how the answer is derived. it is 3 variables in 2 equations! so isn't the info lacking?
only some journals care about publishing quality papers. some make money from the authors who just want to get their papers published, or some will publish anything in general
it takes time to learn which journals are of a good quality and which aren't. the journal "Chaos, Solitons, and Fractals" which you probably got that paper out of isn't a good one.
the fact that Elsevier was glad to allow El Naschie to edit this journal that he frequently published in with no supervision is a problem and speaks poorly to Elsevier
i was reading another paper i found on arxiv by someone name Paul Bracken ( this does not to seen a 'bad' mathematician )... and at that paper he mentioned the word 'compacton'... and i googled 'compacton' and found ji huan-he between many results
A,B and C are three commodities. A packet containing 5 pieces of A,3 of B and 7 of C costs $24.50 . A packet containing 2,1 and 3 of A,B and C respectively costs $17.00 The cost of a packet containing 16,9 and 23 items of A,B and C respectively is $100. @Mike, but how tio find it?
@AlexanderGruber the condition on the matrix decomposition of Aut(A x B x C x ...) is that the summation is sufficiently refined, i.e. the summands share no common direct factors. so given that, a necessary and sufficient condition is that none of the factors have nontrivial images in the other factors' centers
in particular this shows that if the $P_k$s have coprime orders then the Auts work out like that
A student studying the weather for d days observed that 1. it rained on 7 days, morning or afternoon; 2. when it rained in the afternoon, it was clear in the morning; 3. there were 5 clear afternoons; 4. there were 6 clear mornings, then d=9. How did we get d=9?
the only possibilities for morning/afternoon states are C/R or C/C or R/C (where C=clear and R=raining). so the number of occurences of each of these states is x, y and z respectively. now translate all of the information into equations in these variables and solve.
you should get x+y+z = ? x+z = ? y+z = ? x+y = ?
(fill in the blanks)
add the last three up and compare to the first equation
I'm looking for words to google. Fitting a polynomial to a set of points using least squares. The problem is that outliers gets so strong using least squares
@robjohn I am here with an official complaint. I got one downvote on an old answer on each of the last three days. Could you look into it and find the culprit? I take this as a malicious act.
Is the following statement true?
Suppose, $f:\mathbb C\to \mathbb C $ be an entire function. $ |f(z)| $ is bounded in a region where $ \alpha\le \arg(z)\le \beta $ with $|\beta-\alpha|>\pi $. Then $f(z) $ is constant.
@JasperLoy OK, so they are "streetwise," my point is don't give them what they want by letting them bother you pal.
@JasperLoy Obviously they want to annoy you, that is clearly the intent, so by not caring you will show them that they wasted their time and annoy them.
@ParthKohli Yeah, that is what I asking you! "To derive simple induction from these axioms, we must show that if P(n) is some proposition predicated of n, and if: P(0) holds and whenever P(k) is true then P(k+1) is also true then P(n) holds for all n."
This is the q I askd you, and you proved it with induction which is circular!
@Sawarnik I expanded all the cos terms and added them and saw a very interesting symmetry with the given conditions, further manipulation yielded $p(\cos^2(x+y+z)+\sin^2(x+y+z))$ or something like that.
@Hawk Here is one more, if you like it: Prove that its an integer: $$(\frac{9\sqrt{6} - 19}{6\sqrt{6}})^\frac{1}{3} + (\frac{9\sqrt{6} + 19}{6\sqrt{6}})^\frac{1}{3}$$
Hey guys, quick question if you have the time. I need to calculate the probability that 15 or more of a random set of 20 integers are a multiple of 5. How would I go about doing that? Bit lost at how to approach it.
Is the following statement true?
Suppose, $f:\mathbb C\to \mathbb C $ be an entire function. $ |f(z)| $ is bounded in a region where $ \alpha\le \arg(z)\le \beta $ with $|\beta-\alpha|>\pi $. Then $f(z) $ is constant.
