And, on one occassion, I wrote an answer like five hours before and he wrote the same answer after a long time. I asked him if he found some details missing and he just downvoted without a comment.
@anon Yes. I told hime he should rise questions about the site in Meta and not in comments. He left a comment saying he did not understand that.
I told him, if he never understood that, it is unlikely he will ever understand it and therefore, should move on...
The reason this annoys me is that if he doesn't delete it in a couple days or so, my accept rate will go down. I will be forced to downvote if it comes to that point.
I doubt the big formula in the middle is going to get me anywhere, it was just an interesting thing I figured out while doing it and felt like including. (It also has a double purpose. I find that symbolically intimidating questions garner an initial two or three upvotes for no reason other than pure awe - and this helps it gain attention.)
Good day! Help me please, if X - is locally compact, is it true that the standard topology at C(X) it is the topology of uniform convergence at each compact in X?
Hi @Didier. May I remind you that I had asked you for a reference to the result you mentioned in the post about getting 6 different sets from closure and interior operations alone. I'd love a reference. Regards,
Here are my preliminary thoughts on the problem. This is not a full solution but perhaps it may lead to one.
I think you can suppose without loss of generality that $A \subsetneqq R \subsetneqq \operatorname{Frac}(A)$. Now we have the inclusion map $\iota\:A \rightarrow R$. Therefore I think it ...
can anyone suggest a beginner's book for propositional calculus,came across 2 of them on amazon and they seem pretty philosophical.Is there one which is purely logic oriented?
@ZhenLin By 'logic',i meant truth based functional logic(true or false).In actuality,i never read a book on propositional calculus,just learnt some techniques during real analysis,but would like to see how a whole book feels sans philosophy.
@MattN you want to assume a few more conditions at least: local convexity is certainly necessary and you should maybe clarify what kinds of characterizations would be acceptable... Most of them will be rather tautological, I'd say, for example: the map $X \to \mathbb{R}^{X'}$ is a topological embedding. Or it's the weakest topology compatible with the dual pairing $(X,X')$, or: it is the $\mathfrak{S}$-topology induced by the bounded subsets contained in finite-dimensional subspaces of $X'$.
@tb Two weeks ago a former girlfriend suddenly became a widow at the age of 34 under exceptionally unpleasant circumstances, and I went out to stay with her for a while. Limited time, and less convenient Internet access than usual.
Is it really so nonsensical to talk about two problems being homomorphic? I mean, my basis for believing that the thing was nonsense was simply the implausibility of connecting algebraic geometry to complexity theory.
@BrianMScott I thought you might like the lemma I use in this answer. It's a little gem that isn't that well-known (there was some advertising of it about ten years ago). It gives the standard consequences of Baire's theorem in basic functional analysis essentially for free. I know that I don't exactly address the question asked, but still, I thought it doesn't miss it entirely either.
I have a set of increasing integers (such as the Fibonacci sequence) and I am trying to figure out a way to derive the number of subsets where the sum is greater than arbitrary k
i've found a lot of information about things summing to *exactly k, but not greater than
Ah, I saw that question. The problem is that it’s far too open-ended: unless the structure of the sequence is known pretty exactly, you simply won’t get a formula. The best you can hope for in general is a reasonably efficient computer algorithm to count them.
I’d have to think about that; it’s a non-trivial problem. If it helps, it is possible to write down a closed form for $a_n$, though it involves solving a cubic equation.
G.f.: 1/(1-x-x^3) and a(n) = floor( d*c^n + 1/2) where c is the real root of x^3-x^2-1 and d is the real root of 31*x^3-31*x^2+9*x-1 ( c=1.465571231876768... and d= 0.611491991950812...)
@Jordan I’ll take a look as soon as your diagram downloads. (I’m on a slow connection.)
Oh, okay, I see now. When you draw a horizontal slice through the shaded region at height $y$, its left end is at $x=y^{1/3}$, and its right end is at $x=1$. When you revolve that slice around the line $x=2$, you get a washer, as shown in the lower picture. The outer edge of the washer is traced out by the left end of the slice; which is $2-y^{1/3}$ units away from the line $x=2$.
(That’s because the distance from $y^{1/3}$ to $2$ is large minus small, which in this case is $2-y^{1/3}$.)
The washer has a hole in it. The inner edge of the washer is traced out by the right end of the slice. That point is $2-1=1$ unit from the line $x=2$. That’s the so-called inner radius of the washer.
If we were to fill in the hole, the washer would be a disk of radius $2-y^{1/3}$, so its area would be $\pi(2-y^{1/3})^2$.
If you look down on a washer, what do you see? A disk with a circular hole cut out of the exact centre, right?
If you look down at the shaded ring in the lower picture from your text, that’s what you see, though in that case the hole is a very large fraction of the washer.
Taking up where I left off, the hole in our washer is also disk-shaped, and it has a nice constant radius of $1$, so its area is $\pi\cdot 1^2=\pi$.
To get the actual area of the washer, not including the hole, I simply subtract: the area is $\pi(2-y^{1/3})^2-\pi$.
@JohnSmith That looks like a fairly long-term project. With the additional information it’s clearly attackable, but I don’t expect it to be easy.
@Jordan: Once you’re okay with the area of the washer, the next step is to take into account its thickness. That’s a tiny bit in the $y$ direction, so it’s $dy$, and the volume of the washer, which I’ll call $dV$, is $$dV=\Big(\pi(2-y^{1/3})-\pi\Big)dy\;,$$ which of course can be simplified quite a bit.
To get the total volume, you ‘add up’ all of these $dV$’s, from a low $y$ value of $0$ to a high of $1$: $$V=\int_0^1\Big(\pi(2-y^{1/3})-\pi\Big)dy\;.$$
That’s exactly what André has in his second answer, except that he’s already factored out the $\pi$.
When one says that a manifold $M$ has "two distinct orientations", does that mean that it may be equipped with two distinct orientations? For instance, I thought that a connected manifold could only have one consistent orientation, since otherwise we could express the $M$ as the union of the open sets in the obvious way. Can anyone clear up my confusion?
@Jordan Subtracting the area of the hole, since it doesn’t contribute to the final volume. Remember, the radius of the hole is $2-1=1$, so its area is $\pi\cdot 1^2=\pi$.
@Jordan Well, here you really need only two. You need the area of a circle, and you need to know that when you fatten up a shape of area $A$ by giving it a thickness $t$, the resulting solid has volume $At$.
You don’t need the volume formula, but it explains why you’re integrating $\text{area}\cdot dy$: $\text{area}\cdot dy$ is the volume of a really skinny fattened-up slice, and the integral effectively adds up those volumes.
If you change every $x$ to $y$ and every $y$ to $x$ throughout the problem. The second boundary curve then becomes $y=1-x^2$.
But experience suggests that in the long run you’ll make fewer careless errors if you practice not making this change and learn to work with functions running either way. (Besides, later in the course you’ll run into problems in which you can’t easily do this sort of interchange.)
You should keep the exact limits until the end. At that point you can replace the final exact answer with a decimal approximation if the problem or the instructor calls for one. As for the integrand, you’re going to get washer’s again.
The outer radius of the washer at height $y$ is the distance from $x=y^2$ to the axis $x=3$, which is $3-y^2$. The inner radius (i.e., the radius of the hole) is the distance from $x=1-y^2$ to the axis, which is $3-(1-y^2)=2+y^2$.
How did I know which was which? The $x=y^2$ curve is further away from the axis $x=3$ than the $x=1-y^2$ curve, so it marks the outer edge of the disk.
You could, but you’d be doing exactly the same work. Let me show you.
If you do it all at once, the area of the washer at height $y$ is $$dA=\pi(3-y^2)^2-\pi(2+y^2)^2\;.$$ Its thickness is $dy$, so its volume is $$dV=\pi\Big((3-y^2)^2-(2+y^2)^2\Big)dy\;.$$ Here I’ve factored out the $\pi$.
Now suppose that you do it in two pieces. The volume including the hole is $$\pi\int_{-1/\sqrt2}^{1/\sqrt2}(3-y^2)^2\dy\;,$$ and the volume of the hole is $$\pi\int_{-1/\sqrt2}^{1/\sqrt2}(2+y^2)^2\dy\;.$$ When you compute the first of these and then subtract the second, you’re doing exactly what you would have done with the single integral above.
@Jordan I explained the $2+y^2$ here. It’s subtracted because it’s the area of the hole in the washer, the part of the disk that isn’t there and therefore generates no volume.
@EricGregor I see. PGF comes with a math engine that can do many things. In particular comes equipped with some math function like sin, cos, exp, logarithms, ceill, floor, and so on. Also it is allowed to multiply, divide, and compose all this functions, so that, for example, we can plot 2d functions easily. What I need is to define a new function to avoid put a ugly-very long expression everywhere when I need it
I force myself to do both. I do all my hand-ins in LaTeX. If my work requires images I am most likely to make them in pgfplots or using tkz/tkz euclide.
@leo Sorry for the spamming, the last one might be a tad hard to see. It is best to open it on a white background. My question is Leo, which of these have been created in Geogebra, and which have been made using tikz?
