I took an idea from econ and treated it topologically and it took me to an unsolved problem in mathematics which I will definitely need some homology to solve
We say that $V$ is a vector space over the field $k$ if you can scalar multiply vectors in $V$ by $\alpha \in k$. A module is the same thing except generalized. They've replaced field with ring. A ring is a field without the multiplicative inverse requirement. In other words every vector space is actually an example of a module. So any thing you prove in module theory applies to vector spaces.
When you delete an axiom, in this case the axiom(s) of invertibility of elements of $k$, you end up generalizing a theory. What can you prove without inverses, etc.
If you delete any more axioms from ring axioms, you end up with more exotic structures that are hard to find information about.
e.g. Semiring
@geocalc33 does that make sense?
Sometimes there are math questions that require a ring over a field. E.g. diophantine equation solving. However they sometimes map over to associated fields like the field of fractions of $\Bbb{Z}$ is just $\Bbb{Q}$, and answer all they can there as well.
The first step is to google around for a copy of Weibel or purchase it
Not the shitty print kind
I'll have to inspect your copy to make sure it matches mine
Then the first page talks about vector spaces and measuring the defect that $f(x) = 0$ but there exists no $y$ such that $g(y) = x$. I.e. $X \xrightarrow{g} Y \xrightarrow{f} Z$ forms what's called a sequence or chain complex of length $2$ or $3$ depending upon how you count length
And some times $f(g) = 0$ the 0 $R$-module map, which means by definition that $\text{im g} \subset \ker f$.
Those two things on either side of $\subset$ are vector spaces themselves!
So to measur this defect in exactness we'll call it, you simply take $\ker f / \text{im g}$ a quotient of the $R$-module $\ker f$.
That quotient is called Homology, but usually you deal with SES (short exact sequences of a certain form) or infinitely long sequences.
When they're infinitely long you get what's called chain complex. The finite length ones are sometimes simulateable in an infinite length one by filling in the rest of the module entries with $0$ the $0$ module.
@geocalc33 there was a proof in the book (Weibel). I didn't understand their method of proof. I made an alternative proof that I've never seen anywhere. Not of the whole theorem, but just one small part. It involves pasting commuting diagrams along a common subdiagram (last step). Then everything commutes. It's for showing existence of a lift of a first map between two projective resolutions.
Into a chain map. So everything in Homological Algebra recurses.
We have Chain complexes, but also maps between those, and even chain complexes OF chain complexes!!!!!
A family of maps between two chain complexes such that all the squares formed commute is called a chain map. I will teach you this stuff to get you up to speed.
franklin: (d) can be seen to be false because of what happens as x goes to pi/2 (left hand side approaches cos(1), which is strictly less than 1, while the right hand side approaches 1). if you trust that the problem has a solution, then it must be (a)
@leslietownes but the statement you made "(d) can be seen to be false because of what happens as x goes to pi/2 (left hand side approaches cos(1), which is strictly less than 1, while the right hand side approaches 1)."- this is just a risky guess without a caculator isn't it? We know, those things approaches to them, as you mention, but we really don't know how fast they approach to a value?
I think this a question of the type " do the guess and see how lucky you are( intuitions might not lead you that far you have to take risks) "-what do you think ?
uh, i'm very much using that cos(1) is strictly smaller than 1 up above, but other than that, no, it's not a guess. see if you can prove: "if f(x) and g(x) are continuous on [a,b] and f(x) > g(x) holds for all x in (a,b), then f(b) >= g(b)"
the fact that cos(1) < 1 thus tells us that the inequality (d) cannot be true
@leslietownes Hmm...but "if f(x) and g(x) are continuous on [a,b] and f(x) > g(x) holds for all x in (a,b), then f(b) >= g(b)" is true, then the reasoning you stated is completely valid in all sense!
that sort of analysis would be inconclusive if the values of the LHS and RHS happened to be equal at the endpoint pi/2, but at least in (d), they're not
@leslietownes Again, a nitpick seems to be there in my case: If we use, the definition of continuity and limits as: the Left hand limit at a point must be equal to Right hand limit at that point and those limits should be vary well, equal to the value of f(at that point itself) . But, then, this implies that f can never ever be continuous at the end point of a closed interval unless defined or stated otherwise?
what are you talking about? cos(sin(x)) and sin(x) are continuous everywhere
i could imagine some hypothetical problem where maybe you don't know the LHS and RHS are continuous, or the continuity is hard to analyze, but that's not this problem
you often seem to want to generalize far beyond whatever is in front of you, as if it will make what is in front of you simpler
@leslietownes "if f(x) and g(x) are continuous on [a,b] and f(x) > g(x) holds for all x in (a,b), then f(b) >= g(b)" -is indeed an useful assetion/lemma as one might call it. I will keep this in mind! This conclusion, might be helpful in my case
@leslietownes But one thing, how did you come up with it? Was that a standard fact ? Or was it an intution then you verifies it to be true?
If the later is the case, then that sort of intuition is completely lacking in me at this stage
the intuition is that the graphs of continuous functions don't abruptly swap places with one another. if the graph of f(x) is strictly above the graph of g(x) for all x right up to b, the graph of f(x) isn't going to somehow jump under the graph of g(x) right at x = b. the graph of f(x) couldn't do that without crossing the graph of g(x) at some point before b, which it can't, by the assumption that it doesn't
i dunno
it's great to look at consequences of continuity, such as the intermediate value theorem, as things that need to be proved, but don't get so wrapped up in abstractions that those consequences seem counterintuitive
@leslietownes so that lemma i.e "if f(x) and g(x) are continuous on [a,b] and f(x) > g(x) holds for all x in (a,b), then f(b) >= g(b)" is a standard known fact, right ?
i think you are focusing way too much on formalism and/or the status of a formalized statement as a "standard known fact"
i think that someone familiar with limits and continuity should be able to convince themselves that that's true, whether or not they have a proof in mind
but i wouldn't assign any particular importance to that specific set of hypotheses
this is like something from a while ago (maybe even your question), that if a sequence of nonnegative numbers has a limit, that limit is also nonnegative
i see that statement and this one as manifestations of the same thing
@leslietownes you wont believe this maybe: but after reading this I tried up to picture some continuous graphs of cos and sin intutively and now, it seems, crystal clear that this is indeed true!
@leslietownes this makes more sense to me, now :)
@leslietownes It's true that I really dont have a proof in mind, but I can see that this might be true. Or rather seems kinda obvious about the truth of the assertion you made
I think the only need that requires in me : is to focus more and use mathematical intuitions deeply and carefully while solving problems of these sort.
I feel that's the only possible solution...
Consider the function f(x) = max{x, 1/x}/min{x, 1/x} when x โ 0; f(x) = 1 when x = 0. Then (a) lim f(x) = 0 as x -> 0+ (b) lim f(x) = 0 as x -> 0- (c)f(x) is continuous for all x โ 0 (d) f is differentiable for all x โ 0.
Can anyone please validate this: I found that options B,C,D are correct.
First, let me state the proposition in Hatcher's textbook
(a) If $Y$ is obtained from $X$ by attaching 2-cells as described above, then the inclusion $X\hookrightarrow Y$ induces a surjection $\pi_1(X,x_0)\to\pi_1(Y,x_0)$ whose kernel is $N$. Thus $\pi_1(Y)\simeq\pi_1(X)/N$
Note that $N$ is a n...
Never mind my question. I have another. Consider the function $f(x)=\frac{\sqrt{x}}{\ln{x}}$. Apparently it is increasing for $x>e^2$ according to this comment. How can you conclude that?
@Koro $\varphi_a$ becomes nullhomotopic after the disk is attached, so $\gamma_a\varphi_a\overline{\gamma}_a$ is homotopic to $\gamma_a\overline{\gamma_a$ and this is nullhomotopic (even though $\gamma_a$ is not a loop, this is the exact same argument as showing that $\pi_1$ has inverses)
no, that's not what I mean. I'm saying you prove that $\gamma_a\overline{\gamma}_a$ is nullhomotopic as a loop even when $\gamma_a$ is not a loop in the exact same way that you prove $\eta\overline{\eta}$ is a nullhomotopic loop when $\eta$ is a loop (and that's what you did when proving the existence of inverses)
it's the same argument word for word
a homotopy can be constructed like this: at time $t$, you follow $\gamma$ from $\gamma(0)$ up to $\gamma(t)$ and then you walk the same path backwards. for $t=0$, this is the constant path at $\gamma(0)$, for $t=1$ it is $\gamma\overline{\gamma}$.
Given a set $A$ with an equivalence relation $\sim$, we can consider the quotient set $A/\sim$. Given a $n$-dimensional closed ball $B^n$ and the sphere $S^{n-1}$, what does $B^n/S^{n-1}$ mean? What is the implied equivalence relation?
$\sqrt{\frac{\sum(x-\overline{x})^2}{n-1}}$ becomes $\sqrt{\frac{n[\sum x^2-\sum x{\overline{x}} +\sum \overline{x}^2]}{n(n-1)}}$ I'm having trouble seeing how this reduces to $\sqrt{\frac{n(\sum x^2)-(\sum x)^2}{n(n-1)}}$
A collection of geometric figures is said to satisfy Helly property if the following condition holds : For any choice of three figures A, B, C from the collection satisfying AโฉB โ ะค, BโฉC โ ะค and CโฉA โ ะค, one must have AโฉBโฉC โ ะค. Which of the following collections satisfy Helly property?
(a) A set of circles (b) A set of hexagons (c)A set of squares with sides parallel to the axes (d) A set of horizontal line segments.
My thoughts: drawing figures I eliminate, a and b
Also d
I am only left with c. What do you all think on this?
@Franklin Well, that doesn't prove that you can't come up with something that works. You've only shown that there exists a configuration which doesn't satisfy your hypotheses.
@XanderHenderson I think we really can't come up with something that works for d otherwise. But I feel that's not the case. Any ideas how to approach this sort of weird problems?
@Franklin Suppose that you have three intervals $A = [a_1, a_2]$, $B = [b_1, b_2]$, and $C = [c_1, c_2]$ such that all of the pairwise intersections are nonempty. What can you say about the three way intersection?
(My chain suggestion was intended as a counterexample to the squares with sides parallel to the axes, unless those sets include the interiors of the squares)
Yes but. If I have constructed a set of coordinates that are all at distance 1 in any dimension and at angle 60 degrees to each other, does that mean anything? @XanderHenderson
Given some set $E \subseteq \mathbb{R}^3$, and spheres of fixed diameter $d$, how many such spheres can you pack into $E$ so that any two spheres are disjoint (except maybe they touch at a point on their boundaries), and each sphere is contained entirely in $E$?
@XanderHenderson I think after getting exhausted with attempt to give counterexamples, rather than unsuccessful attempts, I conclude c and d as answers. I think this sort of reasoning (intutive) is used in solving mcqs
As it seems, the answer key suggests them as the answers!
It's also a question about the specific geometry of $E$. Like, suppose $E$ is a prism that is a hexagon with side length 1 on one one end and a triangle with side length 3 on the other. What is the maximum number of oranges of radius 1 that you can fit inside that prism?
@Rithaniel But why start with that way with a surrounding shape, why not just add oranges for each new dimension as you go along? n+1 oranges for n dimensions.
Yeah, that's what makes the puzzle difficult. Like, if you have a collection of oranges, you can define an $E$ that is optimally filled by that arrangement (just take $E$ to be the oranges themselves), but the reverse direction is the interesting one
Like, start with an $E$ that we get from something else, and then try to pack oranges in
@MatsGranvik Then you are missing the entire point.
The point is not to simply find an way of arranging balls in a void, but to pack balls into a prexisting, given set.
For example, consider trying to back spheres into a box with a square base that is big enough to pack in four balls (but no more), but a little too short to pack in another layer using your scheme. But maybe you can spread out the bottom layer a bit to pack in a second layer.
There you go. If your set is the triangle bounded by the x-axis, the y-axis, and the line $x+y=\frac{11}{2}$, then your left arrangement is (I believe) optimal: desmos.com/calculator/ev0p8jyhfy
Another example, in 2D: imagine a rectangle with width $4+\varepsilon_1$ and height $2+\sqrt{3}-\varepsilon_2$, where $\varepsilon_1$ and $\varepsilon_2$ are small. How many balls of radius 1 can you pack in there?
alright i found the formulation i want: i want to prove that we can pick row vectors of $A$ to form a set of at most $n$ linearly independent vectors if and only if its reduced row echelon form has $n$ pivots
rank is defined as number of pivots, i avoided it above because what i'm trying to do is prove row rank understood as number of linearly independent vectors is the same as row rank understood as number of pivots in reduced row echelon form
I don't understand how to relate the space with Sausage link. This is the closest food that comes to mind (assuming the central hole to be a point at which north and south poles of the sphere are being identified.)
But I think I understand the spirit of what you mean: I take the 'centre' as basepoint and consider a loop which is a circle.
yeah that much is clear Ted, but that tells us something about the size of the basis, not the maximal number of linearly independent rows of $A$, since we can make a row space basis that contains no rows of $A$
@user123234 I'm not the only person here, so someone who happens by might be able to help, but if you have a good question including some efforts and context, you should indeed post it on the main site.
@shintuku True enough. Indeed, unlike the column space situation, there's no algorithm for saying which particular rows are guaranteed to give a basis.
I think van Kampen is easy enough on the picture I'm saying. Cut it into two pieces (one containing the pinched point, the other a plain cylinder). The pinched point piece is contractible, the overlap is two circles.
Deformation retraction argument and knowing wedge of spaces (path connected) has fundamental group =free product of their fundamental groups does it too.
yeah so i'm wondering, which way would we proceed if we wanted to prove the equivalence of maximal linearly independent set of rows and number of pivots
Can anyone give me a hint to evaluate this integral?
$$\int_0^\infty \frac{dx}{1+x^4}$$
I know it will involve the gamma function, but how?
First, let me state the proposition in Hatcher's textbook
(a) If $Y$ is obtained from $X$ by attaching 2-cells as described above, then the inclusion $X\hookrightarrow Y$ induces a surjection $\pi_1(X,x_0)\to\pi_1(Y,x_0)$ whose kernel is $N$. Thus $\pi_1(Y)\simeq\pi_1(X)/N$
Note that $N$ is a n...
