yes, so two extremes can occur, that the whole population is infected, and that noone is anymore infected, and you are worried that the peak could be at the, for example, 10 percent (or 20) of the whole population?
there will presumably be differences due to the varying state responses, but the interconnected nature of the states means that one state's poor response will affect others
from wikipedia: "Summer high temperatures in Minnesota average in the mid-80s F (30 °C) in the south to the upper-70s F (25 °C) in the north, with temperatures as hot as 114 °F (46 °C) possible."
activity is good for every kind of disease/illness, if we keep being stuck in our houses/apartments by not doing much then we are more likely to be infected
Hi! Can anybody help with a question about Gauss curvature? Q: I have calculated the Gauss curvature, K=-\frac{1}{2}\frac{\frac{\partial^{2}G}{\partial t^{2}}}{tG}+\frac{1}{4}\left(\frac{\frac{\partial G}{\partial t}}{t^{2}G}+\frac{(\frac{\partial G}{\partial t})^{2}}{tG^{2}}\right)=-\frac{2Gt\cdot\frac{\partial^{2}G}{\partial t^{2}}-\frac{\partial G}{\partial t}(\frac{\partial G}{\partial t}\cdot t+G)}{4G^{2}t^{2}}, and apparently it diverges for G^{2}t^{2}=0. I am trying to figure out how G:=G(t,x) (which is defined for all t,x\in\mathbb{R}) has to look like in order to avoid a curvature …
To say something slightly more productive, open, connected subsets of the plane can be classified completely. This probably does not generalize to higher dimensions. Connected, closed 2-manifolds are also classified. I don't know what can be said about closed, connected sets generally and this also gets vastly more complicated higher dimensions. Generally, a classification of open subsets or manifolds in a reasonable sense is not possible in higher dimensions since that has something to do with the word problem via algebraic topology, so I wouldn't be surprised if the same holds for connect…
@Thorgott @LeakyNun Take any finite presentation $\langle S | R \rangle$, and consider $M = (S^1 \times S^3) \# \cdots \# (S^1 \times S^3)$ where connected sum has happened $|S|$ many times. Observe $\pi_1 M$ is the free group $F_S$. For each relator $r$ in $R$, choose a smooth representative loop $\gamma_r$ in $M$ whose homotopy class is $r$. You can make this choice such that $\gamma_r$ are all disjoint by transversality. Take tubular neighborhood of $\gamma_r$ in $M$, homeomorphic to $S^1 \times D^3$ with boundary $S^1 \times S^2$ and replace the interior with $D^2 \times S^2$ (this is p…
You can explicit build a $4$-manifold with a given unimodular symmetric billinear form, by taking connected sums of $\Bbb{CP}^2$, $\overline{\Bbb{CP}^2}$, $S^2 \times S^2$s and $\Bbb{CP}^2 \# \overline{\Bbb{CP}^2}$'s. Alternatively, take $N' = N \# (\Bbb{CP}^2 \# \overline{\Bbb{CP}^2})$ which has intersection form indefinite and of odd type. Then $p \Bbb{CP}^2 \# q \overline{\Bbb{CP}^2}$ for some appropriate choice of $p, q$ has the same intersection form as $N'$
As everything is smooth here, Freedman's theorem applies: Smoothable simply connected $4$-manifolds are homeomorphic iff they have the same intersection forms. This implies $N' \cong p\Bbb{CP}^2 \# q\overline{\Bbb{CP}^2}$ if $P$ is the trivial presentation. If $P$ is not the trivial presentation, these two manifolds are not homeomorphic because one is simply connected the other isn't
So if you had an algorithm to determine if two 4-manifolds are homeomorphic, you would have an algorithm to determine if a given presentation of a fp group is trivial or not.
If you had an algorithm to determine if two connected subsets of $\Bbb R^n$ are homeomorphic, you would have an algorithm to determine if two connected $4$-manifolds are homeomorphic, so there isn't anything like that either
@LeakyNun A simple example is as follows; consider $F_2 \times F_2 \to \Bbb Z$, sending all $4$ generators to $1$. The kernel is finitely generated, not finitely presented.
@LeakyNun They are. When you connect sum with other stuff, usually they're not diffeomorphic. Eg, CP^2 # CP^2 and CP^2 # bar CP^2 are not diffeomorphic
Not even homotopy equivalent; completely different intersection forms
> Some candidates proposed for exotic 4-spheres are the Cappell–Shaneson spheres (Sylvain Cappell and Julius Shaneson (1976)) and those derived by Gluck twists (Gluck 1962). Gluck twist spheres are constructed by cutting out a tubular neighborhood of a 2-sphere S in S4 and gluing it back in using a diffeomorphism of its boundary S2×S1. The result is always homeomorphic to S4. Many cases over the years were ruled out as possible counterexamples to the smooth 4 dimensional Poincaré conjecture. For example, Cameron Gordon (1976), José Montesinos (1983), Steven P. Plotnick (1984), Gompf (1991),…
come on don't cut out a part of me -- S^4, probably
Let $N$ be the kernel of $f : F_2 \times F_2 \to \Bbb Z$. Then we have a function $f : K(F_2 \times F_2, 1) \to S^1$ which lifts to a function $\widetilde{f} : K(N, 1) \to \Bbb R$. $\widetilde{f}$ is what is known as a PL Morse function on $K(N, 1)$, and if you "discrete gradient flow" using $\widetilde{f}$ then topology changes at the vertices with a contribution of the ascending and descending parts of the link of the vertex in $K(N, 1)$.
The ascending and descending links of a vertex in $K(F_2, 1) = 8$ is $S^0$, so the ascending and descending links of a vertex in $K(F_2 \times F_2, 1)$ is $S^0 * S^0 = S^1$, as $\text{Lk}_{(v, w)}(X \times Y) = \text{Lk}_v(X) * \text{Lk}_w(Y)$, where $*$ is join of spaces. $K(N, 1)$ covers $K(F_2 \times F_2, 1)$, so links remain the same.
That means the ascending and descending links of $K(N, 1)$ are both circles, i.e, it has infinitely many $2$-cells, implying $H_2 K(N, 1) = H_2 N$ is infinitely generated aka $N$ is not finitely presented.
But $N$ is clearly finitely generated, if $f : F_2(a, b) \times F_2(x, y) \to \Bbb Z$, $f(a) = f(b) = f(x) = f(y) = 1$, then $ax^{-1}, ay^{-1}, bx^{-1}, by^{-1}$ is a generating set.
> In this approach, the algorithm selects as the pivot element the entry that is largest relative to the entries in its row. This strategy is desirable when entries' large differences in magnitude lead to the propagation of round-off error.
(That depends on the definitions to be fair, so you should check yours, some authors require the involution in Banach *-algebras to be an isometry, while in the $C^\ast$ case that's not needed since it follows from the $C^\ast$ condition)
Question: The maximum height allowed for the throw of a ball is $D$ and the magnitude of initial velocity should be $\sqrt{ 6gD}$. The motion ought to be projectile. Find the maximum horizontal distance the ball can traverse.
My attempt: *Let the angle of throw be $\theta$, so we have $$ u_x = \sqrt{6gD} \cos \theta \\ u_y = \sqrt{6gD} \sin \theta$$ Time of flight is $$ t= \frac{2 ~\sqrt{6D}}{\sqrt g} \sin\theta $$ And maximum height can be found by $$ h_{max} = 3D \sin^2\theta \\ h_{Max} \leq D \\ 3D \sin^2\theta \leq D \\ \theta \leq 0.61549163$$
Range of projectile $$ R = u_x t \\ s = \sqrt{6gD} \cos\theta ~2 ~ \frac{\sqrt{6D}}{\sqrt{g}} \sin\theta \\ R = 6D \sin 2\theta $$
Now, we got to maximise $R$ so just differentiate it and set it to zero $$ \frac{d}{d\theta}\left ( 6D \sin 2\theta \right) =0 \\ \cos 2\theta =0 \\ \theta = \frac{\pi}{4}$$
But $\pi/4$ is greater than $0.615491$
But we can see that $$R = 6D \sin 2\theta$$ is an increasing function so for maximum limit of $\theta$ would give us the maximum range. Therefore, $$R_{max} = 6D \sin(2~ 0.615491) \\ R_{Max} = 5.6568 D$$
But the answer given is $\frac{4}{\sqrt 2} D$ Which is about $2.8288 D$
But the answer given is $\frac{4}{\sqrt 2} D$ Which is about $2.8288 D$
@AbhasKumarSinha Actually Abhas I have a intrinsic dislikeness for competitive exams, and IIT is one of them. I have seen many questions of IIT they all involve some very weird tricks which I think a man doesn’t need.
@Student404Mus Sorry. I dun have any knowledge of Rotation Matrices and it's integration. I want to sincerely express my helplessness to you. The best I can do is to star your messages so that the correct expert would help you with the problem...
@Knight I started programming when I was 11 or 12 because back then I wanted to become a software engineer and I was used to be fascinated by building apps and stuff... So I started with Swift. But then I discovered Stack Overflow and Code Golf and that made me want to learn Python and web languages... So now I know Swift, Python, bits and pieces of HTML & JS and a lot of esolangs (and of course C++ bc we study it in HS)
@AbhasKumarSinha Well, I wouldn't say I "master" either, but thank you :)
The picture here is not quite as bad so far. But part of that is due to a huge boost from the government, paying 75% of the wages of people who need to be sent home, under the condition that the employer still pays the remaining 25%, promises not to fire the person, and that the person uses at least 5 vacation days as part of this
Same. If anything, it has been positive for my situation not having a commute. Really feel for all the people who have suddenly been out of work. I don't know what I would do if that happened to me.
Probably a starting point for the model will need to assume that each person gets into contact with some fixed number $n$ other people, randomly and uniformly chosen, in any given time interval
@ChantryCargill For example, I probably had the virus last week (sick for 4 days with shortness of breath and flu symptoms), but not bad enough to get tested, so never confirmed
@Bob Well, we also need the infection rate among people encountered
It may be simpler to ignore that last and make a model for the total number of people who are or have been infected, in which case you end up with a logistic model
s=Def. A ring R is a ring. [Def. A **ring** is a] s=
Output to my console app
Outputs vocab in bolded or italicized depending on setting.
I'm not storing the original version of the source English, but standardizing the output when I convert it to string
I'm using TextBlob for parsing out parts of speech
I want to do string-based "type theory". For instance if string "a" is known to be an element of a ring, then you can concatenate it with other elements in certain ways, etc.
This treats all of existing type theory implementations as a sort of black-box area. If I can input the same strings or equivalent and output what is necessary, then the goal is met
Hello again! Can anyone help with a question regarding Gauss curvature? My co-advisor doesn't have a phone or a computer (=no home office) ... :-( ... I am isolated without anyone to talk about math.
@TedShifrin I have calculated the Gauss curvature, K=-\frac{1}{2}\frac{\frac{\partial^{2}G}{\partial t^{2}}}{tG}+\frac{1}{4}\left(\frac{\frac{\partial G}{\partial t}}{t^{2}G}+\frac{(\frac{\partial G}{\partial t})^{2}}{tG^{2}}\right)=-\frac{2Gt\cdot\frac{\partial^{2}G}{\partial t^{2}}-\frac{\partial G}{\partial t}(\frac{\partial G}{\partial t}\cdot t+G)}{4G^{2}t^{2}}, and apparently it diverges for G^{2}t^{2}=0. I am trying to figure out how G:=G(t,x) (which is defined for all t,x\in\mathbb{R}) has to look like in order to avoid a curvature singularity at t=0.
This answer to Is there a name for the great circle where latitude and longitude are equal? says (in part):
More generally, a clélie is the name given to any spherical curve where the longitude $\varphi$ and colatitude $\theta$ have the relationship $\varphi=c\theta,\quad c>0$, and the curve ...
