oxeimon | I've spent over 3 hours on this |
---|---|

oxeimon | and that's today |

oxeimon | I feel like I just don't have good intuition for functional analysis so far |

vinay | oxeimon: beyond me, sorry. perhaps someone else can help you. |

vinay | oxeimon: or perhaps you can help me. i don't know. |

vinay | oxeimon: do you have a moment to take a look at my problem? |

kingfishr | Sorry to spam, but I'm going to bump...Can someone give me slight prodding ? I'd like to show that if G is a finite group such that for n \in (naturals) there are at most n elements x such that x^n=e then G is cyclic. |

Kasadkad | kingfishr, do you mean for all n? |

kingfishr | yep sorry left that word out |

Kasadkad | :\ |

Kasadkad | oh |

Kasadkad | vinay, it's sort of a confusing question |

vinay | Kasadkad: yeah.. i don't really understand it |

vinay | Kasadkad: do you have any ideas? |

BrainDeadGebril | kingfishr: where does this question come from? |

Kasadkad | vinay, well i guess they're saying to show that if you have a rational function or an infinite series in z, z*, then differentiating formally with respect to z or z* is the same as using those partials they defined |

Kasadkad | so check dz/dz = 1, dz*/dz* = 1, and that the product rule and quotient rule and such still work |

kingfishr | BrainDeadGebril, friend's hw...I guess he figured it out, but I can't stop thinking about it :\ |

vinay | Kasadkad: but how do "a" and "b" come into play? |

Kasadkad | eh i think they're just saying what i said |

OxE6 | I like "c" and "d" better |

Kasadkad | "differentiate formally with respect to z or z*" |

vinay | i think that the point is to prove that z and z* behave as independent real variables |

Kasadkad | i don't know what that means |

BrainDeadGebril | hmm |

BrainDeadGebril | kingfishr: ok, let me try to come up with something: if |G| is prime we are done, if not, then say |G|=pq |

BrainDeadGebril | consider H={g^p | g belongs to G} as a subset of G |

BrainDeadGebril | then there are at most p identity elements |

sparr | Is there any database of polyhedra more suited to searching/indexing than the polyhedra family (archimedean, johnson, catalan, etc) pages on wikipedia and mathworld? |

BrainDeadGebril | then consider remaining at least G-p elements, put them to the power q, there can be at most q additional identities, so that we have |G|-p-q non-identity elements though all elements are to the power |G| which forces them to be identities. |

BrainDeadGebril | so we must have |G|=p+q as well as p,q | |G| |

BrainDeadGebril | oh right, I see, sorry missunderstood statment in the problem arrived at contradiction that it is a group:) |

BrainDeadGebril | I think I saw that in Herstein though |

kingfishr | BrainDeadGebril, that might very well be the text they're using...they changed in the two years since I took it |

tmorton | Anyone around familiar with basis polynomials for lagrange interpolation? |

tmorton | i'm trying to figure out why if you add all the basis polynomials, you always get 1 |

Kasadkad | kingfishr: take g in G with maximal order n |

Kasadkad | hm |

Kasadkad | ah |

Kasadkad | then <g> is the only cyclic subgroup of G with order n |

Kasadkad | so it's a normal subgroup |

Kasadkad | maybe it doesn't work |

Kasadkad | i wanted to say G/<g> will still have your property |

Kasadkad | so it would be cyclic by induction |

BrainDeadGebril | proof is simplier |

BrainDeadGebril | as i remember how I done it, but I still can be wrong:) |

Kasadkad | but i don't like that idea anymore, i didn't even use that g had maximal order |

BrainDeadGebril | aha, I found it |

BrainDeadGebril | but it assumes it's abelian |

kingfishr | screw it...I'll ask my friend later |

kingfishr | I'm so tired of thinking about it |

kingfishr | i hate not knowing though |

BrainDeadGebril | I am awake for 24 hours so I am not of very help either |

BrainDeadGebril | very much of* |

asn | I have trouble understanding basic sequence convergence (|a_n -a| < x , etc.). Anyone knows some good reading on it (probably with graphic representation)? |

freeboot | by x, you mean some epsilon? |

oxeimon | sorry to spam, but I'm still stuck on this |

oxeimon | if X is a normed vector space, and X* is separable, then X is separable...why? |

oxeimon | http://mathbin.net/41759 |

asn | freeboot: (exactly, I just didn't have a greek epsilon here :P) |

asn | (and I thought that the absolute value would give what I meant away) |

b4ry0n | asn: try this: http://en.wikibooks.org/wiki/Real_Analysis/Sequences it's some important stuff put in a nutshell... |

asn | b4ry0n: okie, will read it! thanks |

b4ry0n | but a real analysis book is of course more helpful |

CESSMASTER | asn: Real Mathematical Analysis, by Charles Pugh, has a lot of illustrations |

CESSMASTER | asn: also Calculus, by Michael Spivak |

freeboot | just imagine a box that starts at some N and has a side from a+e to a-e and then stretches off to infinity. Convergence just says for any e you want, you can find an N so that all the points after N are in that box |

asn | b4ry0n: I'm reading from a 'real analysis book' (my university one) it just that the formal definition of sequence convergence confuses me (shallow as it may sound, it probably is because it introduces many variables in inequalities) |

Robba | How do I show that 1-x^2 <= e^(-x^2) <= 1/(1+x^2) ? |

asn | freeboot: yes, that's what http://explainingmaths.wordpress.com/2008/12/12/quantifier-packaging-when-teaching-convergence-of-sequences/ explains, by introducing "absorption". |

freeboot | Robba: did you try calculus? |

__penguin__ | what's calculus |

Robba | I dont think I'm allowed to use the series expansion. |

freeboot | no. first and second derivative |

asn | but still when I see convergence in 'real' examples, I really don't understand where the n_o and epsilon's come from. |

asn | CESSMASTER: will try to find a copy of them, they seem interesting |

freeboot | e^(-x^2) +x^2 -1 ... find the minimum of that. if it's > 0 then the equality holds |

b4ry0n | asn: actually there is an awesome algebra and calculus book by R. Wuest. with lots of proofs and examples Unfortunately it is in german, and i'm not sure, if there is a translation... |

CESSMASTER | asn: spivak is widely used, a copy should be easy to find |

asn | "Calculus On Manifolds: A Modern Approach To Classical Theorems Of Advanced Calculus " ? |

Robba | Spivak has another book called "Calculus" |

freeboot | 1+x <= e^x is true so you could just try that and save yourself some extra algebra |

asn | yep found it. thanks. |

CESSMASTER | asn: no, just Calculus |

Robba | thanks, freeboot |

__penguin__ | what's calcúlus |

asn | found it! I'll read from there then |

asn | CESSMASTER: Robba: does it also have convergence examples? |

Robba | Yes. |

asn | like, find if a_n = (3*n + 5)/2n converges? |

asn | nice, thanks. |

Robba | There's a whole chapter on infinite sequences and series |

oxeimon | is it true that any continuous functional on a closed subspace of a normed vector space achieves its minimum and maximum on that space? |

tmorton | Anyone around familiar with basis polynomials for lagrange interpolation? |

asn | thanks Robba, CESSMASTER and freeboot :) |

tmorton | i'm trying to figure out why if you add all the basis polynomials together, you always get 1 |

CESSMASTER | oxeimon: it need not achieve a maximum or minimum at all |

oxeimon | well if the functional is continuous, then it's bounded |

oxeimon | right? |

oxeimon | actually this seems ot suggest it does |

oxeimon | well hold on hmm |

oxeimon | okay, nevermind, I generalized too much |

oxeimon | but it holds if the function is a norm |

oxeimon | :-D |

BrainDeadGebril | kingfishr: but doesn't that condition give just that number of solutions to x^n=I for n<|G| is |G|-Euler Totient(|G|) by just simply counting them? That provides us with Euler Totient(|G|) of generating elements, which is kinda enough! |

BrainDeadGebril | (by counting over all divisors of |G|) |

oxeimon | CESSMASTER: any chance you could take a look at my problem? o.o |

CESSMASTER | i will probably look at it and immediately fall asleep |

oxeimon | lol |

CESSMASTER | but link it |

oxeimon | well, in case you don't....http://mathbin.net/41759 |

freeboot | I completely forget how any of that goes. |

CESSMASTER | surprise it is too late to be doing functional analysis |

CESSMASTER | good night |

oxeimon | :-( |

oxeimon | night |

freeboot | night |

oxeimon | thanks anyway |

kingfishr | BrainDeadGebril, I don't follow the beginning...how does the condition lead to the number of solutions being |G| - phi(|G|)? |

BrainDeadGebril | inclusion-exclusion over divisors |

vinay | would this be the place for formal language theory questions? or is there a better channel for that? |

BrainDeadGebril | I am sorry, I gtg to a lecture, coincidentally on groups rings and modules |

kingfishr | BrainDeadGebril, thanks, I'll think about it |

kingfishr | enjoy |

kingfishr | vinay, ask...I'm so much better at that than algebra :) |

vinay | hehe, ok |

vinay | here's the problem: |

Teknomancer | morning |

vinay | Say that string x is a prefix of string y if a string z exists where xz = y and that x |

vinay | is a proper prefix of y if in addition x != y. |

vinay | n each of the following parts we define |

vinay | an operation on a language A. Show that the class of regular languages is closed |

vinay | under that operation. |

vinay | NOPREFIX(A) = { w in A|no proper prefix of w is a member of A} |

vinay | it seems that it would be wise to start with a DFA that recognizes A for this problem rather than tatking the regex route, right? |

kingfishr | vinay, pretty sure, yeah |

vinay | alright... so we start with the DFA. now we want to construct a DFA such that the goal states represent all the ways to generate A without generating a prefix of A first, right? |

vinay | oh, this is really easy with an NFA isn't it? |

vinay | we start with an NFA with one start state and one goal state. we then remove any outward arrows from the goal state |

vinay | then we're done, right? |

kingfishr | vinay, yep :) |

vinay | awesome :) |

kingfishr | vinay, it's no fun if you solve it by yourself |

vinay | that was far easier than i was making it |

vinay | hehe, well i've got 2 more if you wanna have some fun :) |

vinay | don't feel obligated to work on this one - i haven't thought about it much myself yet. but if you're interested, the next one is NOEXTEND(A) = { w in A|w is not the porper prefix of any string in A} |

kingfishr | vinay, I see the solution...you shouldn't have any problem with that one. |

vinay | the gears are turning... |

vinay | ah.. start w/ NFA w/ 1 goal state, remove any arrows that start in the goal state and point back into the goal state |

vinay | that does it :) |

vinay | kingfishr: here's the last one, which is one i've actually thought about a bit and haven't yet come up with a solution. Let A be any language. Define DROP-OUT(A) to be the language containing all strings that can be obtained by removing one symbol from a string in A. |

asn | wee, finally I understood convergence! Having a nice book makes a total difference. |

vinay | Thus, DROP-OUT(A) = {xz | xyz in A where x, z in Epsilon*, y in Epsilon}. Show that the class of regular languages is closed under the DROP-OUT operation |

kingfishr | vinay, sigma* right... |

vinay | whoops, i'm sorry |

vinay | tired :). yes. x, z in Sigma*, y in Sigma |

iSchool | hey |

vinay | in epsilon* would be quite different huh :P |

iSchool | Fiend two numbers who's sum is 7 and who's product is a maximum |

kingfishr | vinay, you would have defined the language with just the empty string, so it would make the problem pretty easy :) |

vinay | kingfishr: hehe yes |

kingfishr | vinay, again easy |

kingfishr | vinay, wait...no my solution doesn't work. Thinking... |

kingfishr | vinay, ok got it. Let me know if you have trouble. |

vinay | kingfishr: could you give me a hint, perhaps? i've thought about this one and nothing is coming yet |

kingfishr | vinay, regular languages are closed under union |

vinay | so... do we union together all the possible ways to create A, dropping out one character? |

vinay | meaning we drop out one character each time we have a concatenation, keep things the same every time we have a star |

kingfishr | vinay, I think that works. Remember the first and last ones are special cases...I was thinking about it in terms of NFAs |

vinay | yeah.. this was kinda the line of thought that i went down, but it gets messy |

vinay | i.e. if we have (a U b)* |

vinay | hm |

vinay | maybe i need to think about this in NFA land |

vinay | take our NFA, remove an arrow. take the NFA, remove a different arrow. union all these together |

vinay | but i don't think that works... |

vinay | insert "do this for all arrows in the NFA" before "union all these together" |

kingfishr | vinay, you have to do more than remove arrows |

vinay | oh.. insert epsilon move |

vinay | hm.. i'm not sure if that works, either... |

kingfishr | vinay, oh hmm you're right...loops complicate things |

vinay | yeah... |

vinay | when we have a loop, we want a no-op, right? |

vinay | DROP-OUT(a*) = DROP-OUT(a*) |

vinay | er |

vinay | DROP-OUT(a*) = a*, i think |

kingfishr | vinay, yes |

vinay | DROP-OUT(a) = DROP-OUT(epsilon) |

vinay | DROP-OUT(ab) = DROP-OUT( (a U b )) |

kingfishr | vinay, DROPOUT(a) = epsilon |

vinay | err.. i did it again |

vinay | both times, remove the drop-out on the right hand side |

kingfishr | vinay, yep yep |

vinay | DROP-OUT(a U b) = epsilon |

vinay | all of these things only hold true for a, b primitives, though. not a, b regex |

vinay | i wonder if we can generalize these to regex? |

vinay | DROP-OUT(R*) = R*, but we can't do the same thing for concatenation |

kingfishr | vinay, that's not actually true |

vinay | kingfishr: which part? |

kingfishr | vinay, DROPOUT(w*) = w*DROPOUT(w)w*...i think |

Nece228 | hi, how to how much is -2sin135 ? |

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