Mathematics

12:02 AM

people do not seem to have liked this one:
http://math.stackexchange.com/questions/422322/how-much-wood-could-a-woodchuck-chuck

Peter Tamaroff

@WillJagy Hello. I am bad at counting.

Peter Tamaroff

12:14 AM

@WillJagy How are you?

@anon Will this do?

pourjour

12:29 AM

does anybody know a good book for functional analysis with a lot of problems

Peter Tamaroff

1:07 AM

@pourjour You're taking a functional analysis course?

AlanH

1:33 AM

$H \leq G$ s.t. $H$ is not normal. Show there exists cosets $aH$ and $bH$ such that $(aH)(bH)$ is not a coset:

Let $b = a^{-1}$. Since $H$ is not normal, not all products of $aHa^{-1}$ are in $H$. By definition of coset, $xH$ is a coset for all $x \in G$. So $(aH)(bH)$ is not necessarily a coset.

Is this correct?

Peter Tamaroff

1:56 AM

@AlanH If $H$ is not normal, there exists $g\in G$ and $h\in H$ such that $ghg^{-1} \notin H$. You ought to work with that.

In what you write, what is $b$? What is $a^{-1}$?

AlanH

@PeterTamaroff I meant to write $aHa^{-1}H$ somewhere in there. $b$ is the inverse of $a$

uh okay let me think about it again

robjohn

@PeterTamaroff It's neat, but on the border of obvious. It takes a minute to figure out what is going on.

Peter Tamaroff

@robjohn Heh, true.

@robjohn Now I have to prove every natural $n$ can be written in $2^{n-1}-1$ ways as a sum of positive naturals less than $n$.

robjohn

2:12 AM

The basic idea is $(1+x)(1+x^2)(1+x^4)(1+x^8)(1+x^{16})\dots=\dfrac1{1-x}$

@PeterTamaroff Case it on the number of $0$s on the right in binary

no $0$s on the right, there are just as many ways as $n-1$

Peter Tamaroff

@robjohn Come again?

robjohn

Sorry, I was thinking of something else with a similar solution. Case on the smallest number in the sum...

Peter Tamaroff

@robjohn "Case on..."?

robjohn

Break up the sums based on the smallest number in the sum and equate that with the number of ways to make up a previous number.

AlanH

@PeterTamaroff So if $H$ is not normal, there exists some$ g $and$ h$ such that $ghg^{-1}$ is not in $H$. Then since $(ghg^{-1})H \in (gH)(g^{-1}H)$, so $(ghg^{-1})H$ cannot be a left coset?

Peter Tamaroff

2:24 AM

@AlanH Could you $\TeX$ that up?

AlanH

@PeterTamaroff Nevermind. Don't bother with it

Peter Tamaroff

@AlanH Wait. What you want to show is that the equivalence relation fails to be a congruence.

Ah, you call $mH$ a right coset, yes?

OK.

AlanH

@PeterTamaroff I think I got it. Thanks.

Peter Tamaroff

@AlanH Good. Byes.

AlanH

@PeterTamaroff See ya. Thanks for your help again today

Vincent Tjeng

3:01 AM

In 2-dimensional geometry, is it correct to say that the perimeter of the convex hull of a set of points is the curve with the minimum perimeter bounding all those points?

What's the name of a curve of minimum area bounding all points in 2-dimensions?

MRS1367

4:49 AM

Hi all

Bageer

Hello

MRS1367

Hell @Bageer

I need some help

I've need to a formula for achieve to this:

Bageer

Well you can ask and someone might help.

MRS1367

My question is for calculating place of an object in an computer app

Alexander Gruber

wtf lol

Bageer

4:55 AM

Looks like someone is swimming in rep

MRS1367

1. Maximum mode = when one of the objects is placed in 4792.7 px

my other objects must place in 2266.7 px

2. Middle mode = when one of the objects is placed in 1826.496 px
my other objects must place in 685.85 px

Alexander Gruber

@Bageer i leave MSE alone for half a day and look what happens.

this was not exactly what i imagined my highest voted answer would look like.

Bageer

Anyone know if there is some sort of and operation for tags, for example suppose I don't have a problem with looking at questions that have homework or calculus but not both, so I would have in my ignore list something like calculusANDhomework.

@AlexanderGruber Yah, in a way it is sort of sad.

MRS1367

3. Minimum mode = when one of the objects is placed in 219.15 px
my other objects must place in -167.6 px

Alexander Gruber

@Bageer i don't think this exists, but you should ask on meta (maybe a feature request? i'd use that.)

Bageer

5:00 AM

@AlexanderGruber I guess I will do that.

MRS1367

totally, I must say that when the first object moves, the second object must change its position

I check many formula for find a formula for do that in the coorect method

but I couldn't find a good method

I can't find a similar formula for doing that for changing the second object position

I check for finding a same multiple for doing that

but I can't

I check for doing that with some other formulas

but I can't find anything

Dominic Michaelis

5:50 AM

Does someone here have Mathematica?

anon

....

Vincent Tjeng

@DominicMichaelis actually, me ... want me to run some code?

Dominic Michaelis

@VincentTjeng something like that. This `FindDistributionParameters[
Table[PDF[BinomialDistribution[10000, 0.2], x], {x, 0, 10000, 1}],
NormalDistribution[\[Mu], \[Sigma]]]` gives an unexpected result and I don't know why :/

FindDistributionParameters[
 Table[PDF[BinomialDistribution[10000, 0.2], x], {x, 0, 10000, 1}],
 NormalDistribution[\[Mu], \[Sigma]]]

Vincent Tjeng

okay, I run mma 9.0 on windows

Dominic Michaelis

Same here

Vincent Tjeng

5:53 AM

running...

{[Mu] -> 0.00009999, [Sigma] -> 0.000833768}

Dominic Michaelis

The result should be mu -> 2000, and sigma -> 40 in my opinion

but I get the same result as you

Vincent Tjeng

hmm

Dominic Michaelis

I mean the mean of a binomial distribution with n=10000 and p=0.2 is clearly 2000 isn't it ?

Vincent Tjeng

yep. reading through the documentation right now

@DominicMichaelis - now I'm confused ... what does this mean?

FindDistributionParameters[
Table[PDF[NormalDistribution[100, 0.2], x], {x, 0, 200, 1}],
NormalDistribution[\[Mu], \[Sigma]]]

I get {[Mu] -> 0.00992401, [Sigma] -> 0.140346}

Dominic Michaelis

Ah I see my input makes no sense. I try to make a distribution over the given propabilities not over the the data

Vincent Tjeng

5:59 AM

I see.

Dominic Michaelis

I thought I just give the propabilities of the old distribution to estimate another one ...

FindDistributionParameters[
 RandomVariate[BinomialDistribution[10000, 0.2], 10^5],
 NormalDistribution[\[Mu], \[Sigma]]]

this would be a correct input

Orangutango

Hello, I have an off topic problem I would like to introduce. It is a question I posed yesterday which has not gotten a response.

I do not wish to violate chat etiquette, so stop me if I am.

Here is the link to my problem.

Dominic Michaelis

6:19 AM

your problem is not well defined

you sum up to $f(n)$ but $n$ isn't in the Domain of $f$

Orangutango

ah thats a typo, thank you.

Alexander Gruber

@Orangutango this does violate chat ettiquette (see point 3 here), but at least you're being polite about it, so i'm ok with it this time

Orangutango

but is the question otherwise clear? i am searching for an algorithm to maximize the sum.

Bageer

@Orangutango Does it not maximize if your order smallest^0 +largest + second largest^2+...+smallest^n-1?

Dominic Michaelis

@Bageer it does

Bageer

6:24 AM

Maybe I am just being naive

or not...

Orangutango

The optimal way to sum $\{ \frac{1}{10},\frac{1}{2},\frac{4}{7},\frac{3}{5},\frac{2}{3}\}$ does not follow that algorithm.

user1

I almost had an answer to this question, using a bit of abstract algebra + analysis.

Orangutango

I verified this in Mathematica.

In my post, the last sum is the optimal one.

I can post the Mathematica code to verify if anyone doubts this.

Dominic Michaelis

I am honestly surprised

Orangutango

thats why this problem caught my attention! my intuition was totally wrong, and I still don't know exactly why

Bageer

6:30 AM

That is cool

Dominic Michaelis

But post the Mathematica Code :)

Bageer

was the optimal one the second one you posted

?

Vincent Tjeng

@DominicMichaelis someList = {1/10, 1/2, 4/7, 3/5, 2/3};
Last@SortBy[{N[Total[#^(Range[Length@someList] - 1)]], #} & /@
Permutations[someList], First]

:)

Orangutango

beat me to it

Dominic Michaelis

@VincentTjeng thats a Neat code :)

Orangutango

6:34 AM

and very slickly done

cleaner than my code XD

@Bageer yes, it is

Vincent Tjeng

thanks :)

anyway @Orangutango - I think part of the reason why your question didn't receive so much attention is that some of us didn't understand it! :)

Orangutango

well at least I'm not alone then! it's been troubling me for almost a week now.

Maisam Hedyelloo

@MRS1367: hi

Orangutango

well, i'll let you all toy with it as you will. it's very late my time and i've been thinking about this problem for far too long today. thanks for hearing me out, i hope one of you gains some insight on the problem.

Vincent Tjeng

7:09 AM

@Orangutango - doing a study for sets of 5 random reals between 0 to 1...

data = Table[someList = RandomReal[{0, 1}, 5];
Ordering@
Ordering@(Last@
SortBy[{N[Total[#^(Range[Length@someList] - 1)]], #} & /@
Permutations[someList], First])[[2]], {i, 1, 10000}];

Tally[data]

I got {{{1, 3, 4, 5, 2}, 3605}, {{1, 4, 5, 3, 2}, 2664}, {{1, 4, 3, 5, 2},
452}, {{1, 5, 4, 3, 2}, 659}, {{1, 2, 3, 4, 5},
1834}, {{1, 3, 4, 2, 5}, 450}, {{1, 3, 2, 4, 5},
112}, {{1, 4, 3, 2, 5}, 59}, {{1, 2, 4, 3, 5},
48}, {{1, 3, 5, 4, 2}, 85}, {{1, 2, 3, 5, 4}, 11}, {{1, 2, 4, 5, 3},
21}} for 10000 runs

Dominic Michaelis

That doesn't look good :/

Vincent Tjeng

strangely, only 12 possible orderings pop out ... not 1, 4, 2, x, x for instance

not sure whether i coded it wrongly or what

Dominic Michaelis

Show[
 Plot[
  Evaluate@
   Table[
    x^j - x^k,
    {k, 1, Length@someList},
    {j, 1, k}],
  {x, 0, 1}]]

Ethan

how can you select 'random reals' between [0,1]

Dominic Michaelis

RandomReal[{0,1}]

Ethan

7:14 AM

lol.

Whats your definition of random?

Dominic Michaelis

Pseudorandom ;)

Bageer

Is there an easy way to figure out the parity of the permutation?

Ethan

why do you ask

Dominic Michaelis

@VincentTjeng well we can say for sure that the first one is fixed

the smallest number is always the first

Vincent Tjeng

oh well... i run 100k times and look what pops out

{{{1, 3, 4, 5, 2}, 35695}, {{1, 4, 5, 3, 2}, 26819}, {{1, 3, 4, 2, 5},
4708}, {{1, 2, 3, 4, 5}, 18862}, {{1, 5, 4, 3, 2},
6257}, {{1, 3, 5, 4, 2}, 959}, {{1, 4, 3, 5, 2},
4388}, {{1, 3, 2, 4, 5}, 1011}, {{1, 4, 3, 2, 5},
623}, {{1, 2, 4, 5, 3}, 213}, {{1, 2, 3, 5, 4},
58}, {{1, 2, 4, 3, 5}, 403}, {{1, 2, 5, 4, 3}, 4}}

@bageer what do you mean by parity? I think it can be done

Bageer

7:22 AM

The permutation sign (whether the permuations is odd or even)

I have test, almost all the ones you put up and so far they have all been even.

Dominic Michaelis

{1,3,4,5,2}-> {1,2,4,5,3}-> {1,2,3,5,4}-> {1,2,3,4,5} is odd isn't it ?

Vincent Tjeng

@DominicMichaelis do help me check through my code... I don't want us to be trying to find reasons for wrong results :)

Bageer

Damn, I have been putting in the cycles into wolfram alpha... its late, okay!

Vincent Tjeng

{Signature[#[[1]]], #[[2]]} & /@ Tally[data]

Bageer

(cycles as in the list)

Vincent Tjeng

7:27 AM

{{-1, 35695}, {-1, 26819}, {1, 4708}, {1, 18862}, {1, 6257}, {1,
959}, {1, 4388}, {-1, 1011}, {-1, 623}, {1, 213}, {-1, 58}, {-1,
403}, {-1, 4}}

does anyone find it curious that 99% of the lists have either the 2nd or the 5th smallest as the last term?

Bageer

Oh no particularly good pattern. I have to say that his problem is surprising me.

Dominic Michaelis

no that is expectable

Look at the plot

Show[
 Plot[
  Evaluate@
   Table[
    x^j - x^k,
    {k, 1, Length@someList},
    {j, 1, k}],
  {x, 0, 1}, PlotLegends -> Automatic]]

if your number is very small or very big raising it's power don't hurt the sum

MRS1367

Hi @MaisamHedyelloo

Vincent Tjeng

oh, i see.

Dominic Michaelis

and by the plot the second smallest is most likely to become the last

which is with our data at over 70 % of the cases true

@VincentTjeng Why do you use two orderungs?

Vincent Tjeng

7:42 AM

@DominicMichaelis Ordering@{1, 5, 2, 3, 4} gives you {1, 3, 4, 5, 2}

applying Ordering to that again gives you the original order. a bit difficult to explain but i think its correct from the documentation

Dominic Michaelis

I am running for 300 k

in your code you make at first a list of 5 randomreals

Then you make a list of its permutations and on the permutationslist you apply a list of functions namely the sum we are trying to minimize and the identiy

this list you sortby the sum,

but this should give the smallest as the first ?

Vincent Tjeng

yes

Dominic Michaelis

ah right so the largest is in the end and you take the last one

Vincent Tjeng

mm thats why i take the last

quick and dirty code

Dominic Michaelis

afterward you take the second part which gives you the permuted list of randomreals

your code seems to work

i mean that it looks correct oto me

Vincent Tjeng

8:04 AM

hmm ok

Dominic Michaelis

@VincentTjeng I gonna write an answer with our intermediate results

Vincent Tjeng

thanks! interested to see the details from a big run :)

Dominic Michaelis

10:07 AM

hi

I posted some things

Vincent Tjeng

10:27 AM

@DominicMichaelis nice work. hope it inspires some further thinking from other people

Dominic Michaelis

i will try with longer permutations

for longer permutations we need another code

Vincent Tjeng

i think we can force the first one to be smallest

not sure wha other simplifications we can make

if you can you should extract some examples for the rare cases like {1,2,5,4,3}

Dominic Michaelis

yeah we need to do more analysis before improving the code

if we take your command for a list of length 9 it takes 100 seconds at me

Vincent Tjeng

same.

Dominic Michaelis

length 10 *

Vincent Tjeng

10:39 AM

around there

Dominic Michaelis

we could reduce it by faktor 10 with forcing the smallest to be first

in the best case

this sounds totally like a project euler problem to me :D

maybe there is something like that on project euler

Vincent Tjeng

perhaps :)

Chris's wise sister

10:58 AM

Greetings all.

@robjohn are you around?

I'm looking for some nice example of limits where $\lim_{n\to\infty} n(b_{n+1}-b_{n})=l>0$, $\lim_{n\to\infty} b_n=b>0$

Ethan

$\ln(n)=b_{n}$ works for the first one..

@Chris'swisesister I don't think any exist

Chris's wise sister

11:14 AM

@Ethan are you sure?

Ethan

@Chris'swisesister If $n(b_{n+1}-b_{n})$ is bounded

Then $b_{n+1}-b_{n}$ over $\frac{1}{n}$ is bounded

so that $\frac{c_1}{n}<b_{n+1}-b_{n}<\frac{c_2}{n}$

Sum both sides from $n=1$ to $n=m-1$

$c_1\ln(n)<b_{n}-b_{1}<c_2\ln(n)$

For sufficiently large $n$

Sense $\sum_{n=1}^m\frac{1}{n}=\ln(n)+O(1)$

Chris's wise sister

That seems right.

Ethan

Assuming $b_1$ is defined and bounded, if its not we can always start the sum at a different number and use the same argument

in which case regaurdless we get $c_1\ln(n)<b_{n}<c_2\ln(n)$

for constants fixed constants $c_1$, $c_2$, and for large enough $n$

and it follows $\lim_{n\to\infty}b_n$ is not bounded

Though if we lift the second constraint

The only $b_{n}'s$ that work for the first one, are non zero multiples of the natural logarithm I think

Chris's wise sister

I felt it's something wrong with this question ...

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$Mathematics$ Mathematics