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Re: The official ask math questions thread [Re: 4HO-DMT]
    #20031259 -

physicist said:
I have had trouble getting the tex feature working. for example:
Formula: 0
should read ei pi = -1

But it clearly doesn't.

It used to work quite well, so I don't know what happened.



I've noticed the same problem.  Some work, others do not.  I suspect certain packages might be broken.

Formula: 1

Formula: 2

Formula: 3

Formula: 4

:shrug:

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Re: The official ask math questions thread [Re: DieCommie]
    #20031278 -

I'll submit a ticket, because there have been times when I wanted to use it.

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Re: The official ask math questions thread [Re: 4HO-DMT]
    #20031304 -

Actually, I now think it is not rendering any "new" equations.  The ones I posted above that worked are common and have probably been rendered before and the resulting image stored.  The ones that did not work have not been stored.  I alter them a little and they dont work.  Just a guess...

Formula: 0

Formula: 1

Formula: 2

Edited by DieCommie (05/23/14 07:21 PM)

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Re: The official ask math questions thread [Re: 4HO-DMT]
    #20031381 -

i have been unable to get it to work.


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:heart: :todcasil: :heart:

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Re: The official ask math questions thread [Re: Acidic_Sloth]
    #20031492 -

I think that the people that teach math are the problem, alot of them don't care and that rubs off on the students. I've had some horrible math teachers over the past 2 years. It was pretty obvious they didn't want to be there

This is in response to something you had said before if your wondering where the hell this came from :lol:


--------------------
The more you know, the less you need.
May as well be here we are as where we are.


                               
                  <img src="https://files.shroomery.org/smileys/imagine.gif" alt=":imagine:" title=":imagine:"/>                                   

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Re: The official ask math questions thread [Re: Acidic_Sloth]
    #20031513 -

Acidic_Sloth said:
i have been unable to get it to work.



I submitted a ticket. Maybe Sellyoursoul knows how to fix it.

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Re: The official ask math questions thread [Re: 4HO-DMT]
    #20031639 -

Prove Fermat's Last Theorem. Take all the time you need.


Edited by Confucian (05/23/14 08:50 PM)

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Re: The official ask math questions thread [Re: gg92507]
    #20032938 -

Let the compact set be called C, delta_x mean (the delta for point x that satisfies continuity at x for a given epsilon).

and consider set S = {y in C | (y - delta_y, y + delta_y) is an element of the finite subcover on C}. Let delta_min = min of the delta for each element in S.

There may exist a point z not in S such that both the delta_z is less than delta_min, and there does not exist an element y in S such that (delta_y - abs(z - y)) is greater than delta_min. This means that the final delta is <= delta_min.

As for finding the final delta, I think it is the min of inf( max((delta_y - abs(z - y))/2 | y in S) | z in C ) and delta_min.

I was trying to prove Lemma 1.2 on page 2 of this document
http://www.math.lsa.umich.edu/~kesmith/nov5notes.pdf

Edited by Satira (05/24/14 02:38 AM)

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Re: The official ask math questions thread [Re: Mr.GuessWork]
    #20032969 -

I've posted some on math exchange, I was just curious about assessing interest in math from the community here/I'm a huge novice at math (just learning calculus right now)

Edited by Satira (05/24/14 02:36 AM)

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Re: The official ask math questions thread [Re: Satira]
    #20033002 -

Quote:
After a few math classes you learn to stop crying, eh... mostly. You conserve your tears at least. The WTF thoughts are reoccurring though. The notation is a real bitch at first because it's necessarily strict. Same thing with the vocabulary, the methodology, and the assumptions. The strictness confines the ideas, so that they can be developed through deduction without being ambiguous enough to create inconsistencies, but it still allows enough ambiguity to make the ideas useful anywhere that the strict confines are satisfied. In math a contradiction means something is falsely assumed to be true. That's powerful, since the world seems to have lots of contradictions, and it kicks ass when those contradictions can be resolved with a little bit of mathy thinking and modeling.



:thumbup:

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Re: The official ask math questions thread [Re: Satira]
    #20033256 -

Sorry--I must not have been clear.

Here are more details of how I would do it.

Let C be a compact subset of R, the set of real numbers.  Let f be a continuous function from C to R.  Fix epsilon>0.  For each point, x in C, let delta_x>0 be such that (|y-x|<=delta_x implies |f(y)-f(x)|<epsilon).  Let U_x be the subset of C where delta_x works.  Notice U_x is an open set (we will prove this below).  U_x form an open cover.  Get finite subcover.  Then use the minimum of the delta_x of this finite subcover.  It'll work because it works on all of the open sets in the finite subcover.

I don't entirely understand your method (I am just a beginner), but using an inf is suspicious.  How do you know the inf is not zero?

To prove that U_x is open, let V be the set {(x,y) in C*C such that |f(y)-f(x)|>=epsilon and |y-x|<=delta_x}.  V is closed--it is the intersection of two sets defined by closed conditions.  Thus its image under projection to C is closed because C*C is compact.  So the complement of image, which is exactly the set on which delta_x works, is open.  It looks like we used compactness of C to prove U_x is open, but we actually don't need C to be compact because R is locally compact.

This stuff seems more advanced than your average calculus stuff.  However, it is seems pretty much exactly like the stuff I learned in Math 1a, freshman Calculus.

Here is a cute math problem--find two 3 by 3 real orthogonal matrixes with no multiplicative relations.  (This is part of the proof of the cutting an orange and reassembling into two oranges paradox.)  Terence Tao glosses over this with a dimension counting argument that doesn't work.  He says the matrixes that don't work are a countable union of algebraic sets, thus nowhere dense.  But this doesn't work unless, for each possible relation, you can find at least one pair of matrixes that don't satisfy the relation.  I think I solved this while mushroom hunting, but I never wrote it down, so I am not sure.

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Re: The official ask math questions thread [Re: gg92507]
    #20033517 -

I thought about orange cutting problem I mentioned above.  I must have made a mistake while mushroom hunting, because I can't think of a solution anymore.  We can boil the problem down to this--given a non trivial multiplicative relation with two letters, that is, a non trivial element of F_2, the free group on two letters, find two 3 by 3 real orthogonal matrixes that do not satisfy the relation.  It is probably easy--that's why everybody seems to gloss over it.  But I must be dumb--I can't think of an answer.  Hopefully a kind Shroomery reader can help me.

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Re: The official ask math questions thread [Re: gg92507]
    #20036853 -

Hey, thanks for the reply, and yeah, I kind of "guessed" that the inf would not be 0 via the way I defined it in my previous post, but guessing in math rarely works :P

But I see why I was confused--you defined the cover on C to be the union of U_x for each x in C, where U_x = {y in C | f(B_delta_x(y)) is subset of B_epsilon(f(y))}, where B_a(x) is the a-ball about point x and delta_x is the delta that satisfies continuity at x. 

The cover on C I (wrongly) assumed you were defining was the union of the B_delta_x(x) for each x in C.

Ok, so now I see, using the cover you defined, the steps in your proof:
1. Prove U_x is open.
  a. Prove that if C is compact, then C*C is compact. (I still need to prove this to myself)
  b. Prove that V_x = {(y,z) in C*C | |y-z| < delta_x and |f(y) - f(z)| >= epsilon} is closed because it is the intersection of two closed sets.
  c. Prove that the projection of V_x onto C is closed because of compactness of C*C and because a closed subset of a compact set maps to a closed subset under a continuous map (and projection is a continuous map.).  (sorry if I made a mistake in this step, I still need to work it out clearly myself)
  d. Prove projection of V_x onto C is the complement of U_x. (easy).
2. There exists a finite subcover on C of the collection of U_x--just take the min of the deltas of the elements used to define this finite subcover, and thus, the delta_min works for every element y in C by membership of y to some element in the subcover.
---

I'm still unsure on how to prove the Lemma 1.2 in the document referenced in my previous post, but I think I need to address this myself since it is labeled as an easy exercise.

---

I don't know any linear algebra at this point, but I should learn enough to try your question out, thanks for the motivation.

Edited by Satira (05/25/14 01:44 AM)

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Re: The official ask math questions thread [Re: Satira]
    #20041365 -

Yes--I wasn't clear about what open cover I meant.  I should have stated it explicitly.

I guess the details have gotten a bit involved.  I took way too much for granted and forgot that there are all these little details to prove.  It might be fun and educational to prove those details, at least in the case of subsets of R.

Probably books have better ways of proving these things.  I just wrote down the first thing I thought of.

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Re: The official ask math questions thread [Re: Satira]
    #20041381 -

Here is a spoiler for lemma 1.2.

You might look at the finite set of points ({y_i+delta_i} union {y_i-delta_i} union {a,b}).  Is there a delta that is less than the distance between any pairs of these points?

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