oxeimon

I've spent over 3 hours on this

and that's today

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

and that's today

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.

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

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

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

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

:\

oh

vinay, it's sort of a confusing question

oh

vinay, it's sort of a confusing question

vinay

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

Kasadkad: do you have any ideas?

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

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

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

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

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

then there are at most p identity elements

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

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

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.

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

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

I think I saw that in Herstein though

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

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

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?

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

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

hm

ah

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

so it's a normal subgroup

maybe it doesn't work

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

so it would be cyclic by induction

hm

ah

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

so it's a normal subgroup

maybe it doesn't work

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

so it would be cyclic by induction

BrainDeadGebril

proof is simplier

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

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

but it assumes it's abelian

but it assumes it's abelian

kingfishr

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

I'm so tired of thinking about it

i hate not knowing though

I'm so tired of thinking about it

i hate not knowing though

BrainDeadGebril

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

very much of*

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

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

http://mathbin.net/41759

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

http://mathbin.net/41759

asn

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

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

(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

asn: also Calculus, by Michael Spivak

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.

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

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 calcĂșlus

asn

found it! I'll read from there then

CESSMASTER: Robba: does it also have convergence examples?

CESSMASTER: Robba: does it also have convergence examples?

Robba

Yes.

asn

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

nice, thanks.

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

right?

actually this seems ot suggest it does

well hold on hmm

okay, nevermind, I generalized too much

but it holds if the function is a norm

:-D

right?

actually this seems ot suggest it does

well hold on hmm

okay, nevermind, I generalized too much

but it holds if the function is a norm

:-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!

(by counting over all divisors of |G|)

(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

good night

good night

oxeimon

:-(

night

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

enjoy

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

enjoy

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

vinay

hehe, ok

here's the problem:

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

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

n each of the following parts we define

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

under that operation.

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

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?

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

n each of the following parts we define

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

under that operation.

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

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?

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

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

then we're done, right?

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

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

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

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

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}

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

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...

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

that does it :)

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.

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

that does it :)

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

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

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

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

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

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

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?

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

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

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

hm

maybe i need to think about this in NFA land

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

but i don't think that works...

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

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

hm

maybe i need to think about this in NFA land

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

but i don't think that works...

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

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

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

kingfishr

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

vinay

yeah...

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

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

er

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

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

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

er

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

kingfishr

vinay, yes

vinay

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

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

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

kingfishr

vinay, DROPOUT(a) = epsilon

vinay

err.. i did it again

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

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

kingfishr

vinay, yep yep

vinay

DROP-OUT(a U b) = epsilon

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

i wonder if we can generalize these to regex?

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

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

i wonder if we can generalize these to regex?

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 ?