i have this paper
http://www.sysmath.com/jweb_xtkxyfzx/CN/article/downloadArticleFile.do?attachType=PDF&id=10691
and i dont understand how to prove in page 3 that $\overline{c}$ is a critical value
please help me
Thank you.
@Vrouvrou this is beyond my scope of knowledge , there are surly people here that could give you good answers, did you ask on the Forum? they usually answer very fast
polysign numbers, I seem to recall those are a crank reinvention of the wheel
based on my skimming of the question, I'd say it would take me 20 times more effort to understand what your question is, translated into standard terms and concepts, than it would be to answer the question itself once understood!
Presently I have a calculus book where I want to come up with new solutions for each of the author's solutions. Maybe I send him my collection after I finish all his questions.
@robjohn that form it's OK for me. Well, I don't know any other simpler form.
@robjohn Which of the 2 questions seemed more difficult to you? This one or that one you solved yesterday? How about a difficulty scale? (I think you missed this row I also wrote above)
@mick also, that is a quotient of a polynomial ring (you meant to write R[x]/(1+x+x^2)); a group ring for example would be R[x]/(x^3-1), which is R[C3]
@robjohn Ah, yesterday you weren't here at the moment I saw his comment on meta. I was so angry ... he didn't have! One that loves calculus would be so so proud to post such a solution!
@robjohn I also love to find proofs, but when one is simply keener than me and finds a nicer way, I recognize that and want to learn. That comment on meta was so ridiculous ...
C is isomorphic to R[x]/(x^2+1). the ring R[x]/(x^3-1) has zero divisors and idempotents (which C does not), and is dimension 3 as an R-vector space (as opposed to C, which has real dimension 2)
I think you are confusing ring with group ring ... Imho - wrong perhaps - : C is isomorphic to the ring R[x]/(x^2 + 1) and isomorphic to the group ring R[x]/(1 + x + x^2).
@mick yes, C is isomorphic to R[x]/(1+x+x^2) (as well as R[x]/(x^2+1)), but the second thing is not a group ring, and is not isomorphic to R[x]/(x^3-1) (which is a group ring, it is R[C_3]).
@mick I thought we were talking about R[x]/(1+x+x^2). if you mean the polysign numbers P3 as in the link, the discussion of those is rather confusing and I can't make heads or tails of it. I am not arguing with you; I am trying to teach you math!
@mick no, I said R[C_3] is iso to R[x]/(x^3-1) (not the brackets, not the parentheses), although I guess it is also isomorphic to R[x]/(x^3+1) (as it happens)
@robjohn absolutely! - I hate driving on ice with a passion. I remember once driving (at 5 mph) towards another car and having absolutely no control ...
@anon ok let me try to explain 1) I think this is a group ring , but maybe its something else 2) Let R(x) be the semiring of polynomials with positive real coef. 3) If we take R[C_3] is that a group ring ? 4) If we take R[x]/(1+x+x^2) is that a group ring ? 5) if 3) and 4) are not group rings , what are they ??
@mick (3) R[C_3] is a group ring, (4) no R[x]/(1+x+x^2) is not a group ring (it is not presented as one, nor is it in any way isomorphic to a group ring over the reals), (5) the ring R[x]/(1+x+x^2) is a quotient ring of a polynomial ring by a principal ideal; it is isomorphic to C by sending the coset x+(1+x+x^2) to either primitive cube root of unity
@anon but from what I have found, it seems to behave more hideously than any holomorphic function I have ever come across
@DanielFischer Yep - I reckon it has no radial limits anywhere, no non-tangential limits anywhere, and probably no (finite) asymptotic values ... really nasty (but a fantastic counter-example)
@mick you do not need to define R[x], it is a polynomial ring. how in the world are you even talking about quotient rings such as R[x]/(1+x+x^2) when you don't know what a coset is!?
@mick no, you are talking about quotient rings (although group rings can be characterized as certain types of quotient rings). what did you think R[x]/(p(x)), where p(x) is a polynomial, meant?
@DanielFischer I find it fascinating - generally holomorphic functions are really nicely behaved, but as you move towards the boundary, they can become incredibly badly behaved - and as you say, it is likely that the last word has not yet been written about the situation
@robjohn Are you sure that you think supporting asciimath in chat is a bad idea ? I agree that in the beginning there may be some confusion, but if it is promoted a little bit that chatjax has an update, then sooner or later most people will have the new version. And it wouldn't be a disaster is some people don't have the new version. The good thing about asciimath, is that it looks readable, even if it isn't rendered.
For example, if I look at the things just written in asciimath, that is the reason asciimath exists. They are automaticly converted to good looking math: \`1^(1/3)\` \`exp(2pi*i/3)\` \`(x,y) * (a,b) = (xa,yb)\`
@mick in my answer I even say that group semiring would be a better term (for polysign numbers, anyway - I don't talk any more about those). I can give you the necessary background information to answer almost all possible versions of questions you could actually have in mind.
@90intuition we already use the ` symbols around latex in order to prevent latex rendering by our chatjax code (in order to teach latex for example); having another usage that actually renders could potentially create issues
@mick I never said you have to do the track according to your age
@mick I meant for example get a basic "modern algebra" or "abstract algebra" text and learn the standard concepts and definitions in the correct order, not be satisfied with the classes you're offered
@90intuition Once you are familiar with LaTeX it becomes a lot easier to read. There is no reason to have another typesetting format, as it would cause confusion (not to mention take the fun and control away from writing formulas). Sometimes \frac12 wins over 1/2
I remember when I was around 14, in addition to learning calculus myself from my dad's college text, I dabbled in stuff beyond my ability and a couple of borderline crankish things (in number theory). I would have been much better served by simply picking up a basic abstract algebra or elementary number theory book and doing it that way from the get-go.
@Alizter Just to be sure I'm clear, it's not that you couldn't use LaTeX anymore. You could also use LaTeX if you prefer. It is just that you could look intuitive notation like a*b or 1/2 look like a \cdot b or \frac12 just by putting it in escaped backticks.
But maybe something else then escaped backticks would be better.
Terry tao is to advanced for me :) but ill read the link :)
@anon who do you think is the best mathematician ? some told me Terry tao ... well people have their expertise field ... but then again some have more than one expertise ...
best mathematician is both subjective and hard to judge, because nobody is versed in enough fields to be sure anymore. there are the big names of history - Gauss, Ramanujan, Euler, Riemann, Archimedes, and so on. perhaps the best one in recent times is Grothendieck.
Next week I am to give a presentation to (oldish) amateur mathematicians on the topic of n-dimensional space - I have a basic talk organised (mostly from things I have used in the past to fill in bits of lessons) - does anyone have any favourite facts/curiosities about n-dimensions?
some things that might be of interest: http://math.stackexchange.com/questions/48301/examples-of-results-failing-in-higher-dimensions http://math.stackexchange.com/questions/8794/intuitive-explanation-of-the-difference-between-waves-in-odd-and-even-dimensions http://math.stackexchange.com/questions/3093/why-is-the-3d-case-so-rich
@DanielFischer Well. First, since every point has a compact nbhd we may work locally in a compact Hausdorff space. But a compact Hausdorff space is normal. And then I can mimic the proof for complete metric spaces. Namely, construct a decreasing sequence of nested closed = compact sets that will have nonempty intersection and this point will be in $\bigcap O_i$ and I win. Left or right?
@PedroTamaroff You don't actually need that. Just take a point in $U_{i+1} = \overset{\circ}{K}_i \cap O_{i+1}$, and let $K_{i+1}$ be a compact neighbourhood contained in $U_{i+1}$.