@leslietownes Apologies in advance if this isn't related but, how about if one restricts attention to only a 1-parameter family of real functions on G=(0,1)^2 whose Mellin transform on compact support exists on (0,1). Then mapping each of the functions in the family via Mellin transform gives rise to new functions on the new function space which has the new domain R=(0,\infty)X(0,1). I'm looking for an isometry between G and R, where G has the standard euclidean norm.
And the reason is because I have some problems in (0,1)^2 that I'd like to view from a different vantage point.
One problem: a vector field X on G with orbits I. transport the space of all orbits to the region R via the (potential) isometry.
please have no follow up questions to this, but i think for hatcher, an n-cell is either an open n-disc (i.e. a space homeomorphic to {x in R^n: |x| < 1} where |x| is the usual euclidean length, or its closure. and he often 'attaches' them to other things via maps on the boundary of the disc. and a CW complex is generally a union of those things under a bunch of identifications.
putting aside whether or not hatcher says that, if you pretend that he did say that, would you still have questions?
i.e., i think a cell is a set and not a map, but that you could associate at least the interior of a "cell" with a subset of a CW complex that it is attached to.
and maybe you can carry out this association for the whole cell, if you are willing for the boundary to be part of the cell and for it to be crunched into something that maybe doesn't look like the boundary of a disc anymore under the attaching map.
this might be getting into the definitionology of, is a concrete realization of the result of a bunch of identifications the same, or different from, a presentation of the same topological space as an actual disjoint union with actual identifications.
i didn't think of it as too productive for me to personally discuss what hatcher does or doesn't define since i don't have the book in front of me, but that does come as a relief
with a book like hatcher that is maybe slightly less formal than other texts, its more about, does the vibe make sense or not, than, is this packaged in a box labeled definition, but i do recall the vibe making sense in hatcher
CW complexes are kind of weird because as most people define them, the attaching maps can be pretty "bad"
but that also makes them easier to construct and work with
e.g. you need some silly number of triangles just to triangulate a 2-torus
I want to find a minimal triangulation of torus in $\mathbb{R}^3$
To do this we need $14$
triangles How can we construct the triangulation ?
(cf. 35p. in the book From Euclid to Alexandrov a guided tour - Petrunin and Yashinski)
I do not want a proof but
triangulation
I find that minimal triang...
In the image, the reason is mentioned in the last part as " It is the presence of this multiplier that makes the L'Hospital rule inapplicable in this case, since it simultaneously nullifies the derivatives of the functions being compared." -But this doesn't make any sense to me!
it might be more helpful to reframe this as, why would it apply? what version of l'hopital's rule do you have (there are many out there) and what are its hypotheses?
@leslietownes The hypothesis are that the ratio of the functions should be as infinity/infinity or 0/0 and the ratio of the limit of the derivative should exist.
Also, the derivatives must be defined in a particular neighborhood of the limit point considered.
@leslietownes This is precisely the definition of L Hospital I use.
the function g'(x) here is 4 cos^2(x) + (2x + sin(2x)) cos(x), which is zero whenever cos(x) is 0, and in particular, it is not the case that g'(x) is nonzero in a deleted neighborhood of a.
you miss this if you just form f'(x)/g'(x) and cancel, because the thing that cancels is, in this cooked up example, the thing that makes g'(x) fail to be nonzero in a neighborhood of a.
so here g'(x) has zeros at a sequence of points that approaches a. in this example, f'(x) also has zeros at those points, and it turns out that when you form the quotient, and cancel out this cos(x) term, what's left in the denominator does not have zeros anymore.
this is what they are saying with the "simultaneously nullifies the derivatives" business. i wonder when this book was written.
ok, i googled that book. the translator is using weird words there. they were probably trying to do something word for word, in a way where it sounds way more natural in russian than it does in english.
i'd say something like, when you compute f'(x) and g'(x) and form the quotient f'(x)/g'(x), some cancellation can take place that may cause you to fail to notice that g'(x) is not of a form that the rule applies to.
@leslietownes Ok, so that means if we have, say f'(x)/g'(x) and we do some cancellations to obtain \psi(x)/\phi(x) then after the cancellation, we must check that whether L Hospital applies to psi(x)/\phi(x)
or put another way, the hypothesis on g'(x) not vanishing is a genuinely separate condition, that cannot be skipped or blended into the limit computation. unless you are the sort of person who would somehow notice, after the cancellation but before computing the limit, that the domain on which you have your simplified expression for the quotient is not one that contains a neigborhood of infinity, and maybe that rings an alarm bell.
franklin: good question. i would say that the hypothesis on g'(x) is not something you can circumvent or avoid by simply working with the quotient f'(x)/g'(x). it is something that needs separate verification.
i'm surprised that we don't have one of those complaints about l'hopital's rule having hypotheses in the top starred comments. there's usually one of those near the top, and here we had a question exactly about this.
@leslietownes This might be the complete case , so now, the takeaway must be: To apply L Hospital, we must check, whether (1) the given ratio of function f(x)/g(x) has the indeterminate form infty/infty or 0/0, (2) Next we have to ascetain that the limit of f'(x)/g'(x) exists
@leslietownes (1) the given ratio of function f(x)/g(x) has the indeterminate form infty/infty or 0/0, (2) Next we have to ascetain that the limit of f'(x)/g'(x) exists(3) that g'(x) is nonzero in a neighborhood of where you are taking the limit. -These are the 3 cases to check, right?
If I get a green singmal in all of them then only I will go ahead and apply L Hospital...
@leslietownes More confidence of course. I was so confused! You won't believe it! How much taxation I did on my brain to understand this weird thing!!!
@leslietownes This indeed gives me confidence 😂😂😂
@user4539917 Yes! This is a heck of an example.
I think as I will know more about limits as I study more over many years I will be able to see these clearly
(LIKE A CRYSTAL!!!
But for now, I am forgetting whether this at all happened!😂😂😂
it might be illuminating to see a proof of the stated version of l'hopital's theorem, so that you can step through it and highlight the place where the assumption that g'(x) is nonzero is actually used for something.
yeah, a lot of calculus books will only sketch a picture of why you might be able to believe l'hopital's rule. or give a proof under some easier set of hypotheses, and say "just trust us on this that it also works in the generality we have stated"
franklin: maybe you did and didn't notice? it's a similar thing actually. the usual motivation of the proof is that the difference quotient [f(g(t+h)) - f(g(t))]/h can be written as a product [f(g(t+h)) - f(g(t))]/[g(t+h) - g(t)] * [g(t+h) - g(t)]/h, and the annoyance is that, well, OK, only when h is such that g(t+h) - g(t) is nonzero, and there's no guarantee that that can't happen for a sequence of h approaching 0 and what then.
i say it's similar in that it's this "what if that thing in that denominator is zero infinitely often in every neighborhood of where we want to take the limit"
at the level of that vibe, it's the same concern presented by your example above.
@leslietownes In short, I just have to forget it until the day I become more knowledgeable and grow 100 years in experience? (THIS IS NOT SARCASM, I WROTE IT LITERALLY!,WARNING: LET THIS NOT BE MISUNDERSTOOD OTHERWISE)
1) A n-cell is a map from $D^n$ to the topological space $X$ you're working with.
2) Attaching a space $X$ to a space $Y$ along a map $\phi : E \to Y$, assuming $E \subseteq X$, means to create the quotient space over the disjoint union of $X$ and $Y$ by identifying a point $e \in E$ with its ima...
Leslie: I understand your definition: n-cell is homeomorphic to $D^n$
We call an interval [0,1] a 1-cell sometimes, a square a 2-cell etc. so it makes sense.
Dude I have this assignment I don't know what to do. I know how to start the proof but I don't understand this function. To me it is not one to one but only onto. For 0.1, 0.11,0.111,... It maps to empty space.
lee mosher's answer, which seems to follow the book more closely than the accepted answer, makes it sound like maybe sometimes hatcher uses the "set" definition of cell, and sometimes identifies the set with what mosher's post says is identified in an appendix as a "characteristic map"
not having the book in front of me, i can't say, but it's interesting that one of the answers calls out the accepted answer as not being quite what hatcher says
not tfue: i do agree that the function f they define sends the element of E that has only 1s in its decimal representation to the emptyset element of P(N). i'm not sure what this has to do with the rest of your question/difficulty
not tfue: there's definitely a theorem that says if you have an injection from A to B and an injection from B to A, then there is a bijection between A and B. i don't see that theorem being suggested by the hint, however, which indicates that a specific map is a bijection between A and B
not: first, formality check, elements of E are real numbers. so the element of E whose decimal expansion contains only 1s is maybe not a sequence of truncations 0.1, 0.11, 0.111. it's just the real number with 0.111..... as its decimal expansion. but putting that formality aside, why would this thing being mapped to the emptyset under f prevent f from being one-to-one?
not: yeah. the real number 0.1 is not an element of E. the restrictive definition of E is pretty important to f being well-defined, and to f being one-to-one
in particular, E doesn't contain any elements with finite decimal expansion, so the issue of different digit strings potentially representing the same real number (like 0.1 and 0.099999.... being the same thing but having different expansions) doesn't come up.
you can ask 'analysis' around like here. Almost everyone here has studied/are still studying analysis so you can get an answer. But algebraic topology, not everyone may have studied it so that makes it more difficult for me.
@user7269591 Truly, that was unknown in my vocabulary!
Can anyone please help me with the problem posed at the end of the image in a similar flavor used in proving the former inequality ?
I did manage to prove the Right inequality(i.e sinx <x) by the same way analogous to the previous problem, but I dont get how to prove the left inequality .
let g(x) = sin(x) - (x - x^3/6), note that g(0) = 0, convince yourself that g'(x) is positive for positive x, and convince yourself that all of this implies that g(x) is positive for positive x
if you get stuck, one way of continuing might be to use the same idea as the example to prove g'(x) > 0 for x > 0, but before you go there let's see what it is
does it? let h(x) = cos(x) - 1 + (x^2)/2. note that h(0) = 0 and h'(x) is positive for all positive x...
i might add that generally, inequalities can be very difficult to prove, and the technique being used here maybe isn't as general as a few examples might make it seem
often all it will immediately get you is something useful in only some interval, perhaps a small interval, even if outside of that small interval the inequality might still be true for some other, more difficult to see reason
but cool here that these things hold for all positive x
Let $A=(0,1)\subset \mathbb R$ and $f$ be a function $A\to \mathbb R$. I heard by a mathematician that one can't write $\lim_{x\to 1}f(x)$ but instead one should write $\lim_{x\to 1-}f(x)$. Is that his personal opinion or is it generally used way to use one sided limit in that kind of situation?
@copper.hat No that's not at all fun, I know what rheumatoid arthritis is! No, I don't have it, but I seen people suffering from it. And I never ever want to have osteoarthritis. Wish you best of luck !😂😂😂😂
I just heard that one of my college's math major undergraduate students published a paper in combinatorics (specifically, graph theory). He already uploaded three papers in combinatorics in arxiv. Stunning.
if x is a non-zero element in any field, x^{-1} denotes its multiplicative inverse. x^a denotes x * ... * x multiplied a times, as you say. (x^{-1})^a and (x^a)^{-1} are always the same thing and denoted x^{-a}
again, the exponent a here has to be understood as an integer
That blows ones mind a little bit, because for a,b in Z_11 (g^a)^b = g^{ab}. In fact, we say that -1 is in Z_11, because Z_11 is a ring of equivalence classes of integer numbers. So, setting b = -1 is actually allowed, because -1 is the same as 10 in Z_11 and (g^a)^{-1} = g^{-a} look normally.
I got your point, however. I think the problem is to start looking at -1 as at some notation, not as if this was an integer, or a number.
If I carefully prove each theorem with original thought it will take me decades to invent my own version of real analysis. Most post tells to prove those theorem yourself I don't agree because it is like trying to reinvent the wheel.
@SineoftheTime I mean the way we are taught to study analysis is like to reinvent analysis itself. My professor goes each proof of book and within hour it is like only 2 to 3 theorems.
then she skips and says let's do more important one other are assignment.
I am having problem with solution of part a. It considers 0 as a critical point. But why? Since the definition of a critical points are those points where the derivative is either 0 or do not exist.
But at 0 the derivative is infinity
Another important result, used in this solution is
@Arthur Ok. So note that this is just my personal opinion- I did not like the teaching here. The teaching here did not meet my expectations. The way they teach here is not even 1% of how Gilbert Strang would teach linear algebra. But again, this is just my personal opinion.
in any case i agree with koro, usually approaches infinity suggests positive infinity, otherwise they say approaches negative infinity or something of the sort
At 0, clearly the derivative does not exist ! Hence, 0 is considered as the critical point. This is the true reason. Not because that :at 0 the derivative is infinity(, which is completely false as seen evidently) and hence it's a critical point(This is what I was thinking prev and getting confused)
@SineoftheTime I would say wrong!
Ok, now the next thing.
They essentially used the third image, to calculate the sign of f'(x) in the interval (-1,0) and (0,1)
But 0 is not a stationary point at all !
So how did they apply the result in the 3rd image for a non-stationary point ?
Is the result in the 3rd image valid for non-differentiable points also ? Just in case...
[This is the final ambiguity I am facing miserably !]
I want to find a minimal triangulation of torus in $\mathbb{R}^3$
To do this we need $14$
triangles How can we construct the triangulation ?
(cf. 35p. in the book From Euclid to Alexandrov a guided tour - Petrunin and Yashinski)
I do not want a proof but
triangulation
I find that minimal triang...
