Mathematics - 2019-10-25

Akiva Weinberger

1:02 AM

@Rithaniel @JohnnyApplesauce The set of integers is not closed under division

but the rationals are (er - they're closed under division by nonzero elements)

Rithaniel

This is a good example

Akiva Weinberger

The rationals are not closed under the operation of taking square roots of positive elements

Rithaniel

Also, division is a good example of a non-associative operation

Akiva Weinberger

The reals are not closed under the operation of taking square roots

Rithaniel

1/(2/3)=3/2 but (1/2)/3=1/6

Akiva Weinberger

1:04 AM

The set of positive integers is closed under multiplication

The set of negative integers is not

The set of irrationals are not closed under the operations of addition, subtraction, multiplication, division, or exponentiation (prove this!)

The reals and the complex numbers have the same cardinality (the same infinity), yes

This is not obvious - the bijection is not continuous

Cantor, when he first discovered it and provided a (messy) proof, is said to have remarked, "I see it but I don't believe it"

It makes a good exercise

@JohnnyApplesauce Consider the operations $(+,-,\times,\div,\sqrt\cdot)$

The closure of the set $\{0,1\}$ under those operations is called the constructible numbers

They represent the lengths you could draw with a straightedge and compass (given you're provided with a line segment of length 1)

It includes $\frac53$ and $\sqrt2$ and $1+\sqrt{1+\sqrt2+\sqrt{\frac53}}$, for example

but it does not include $\sqrt[\Large 3]2$

That's equivalent to saying that the cube root of two is not equal to any expression involving square roots and the four basic operations

It also means you cannot create a line segment with length the cube root of two using a compass and straightedge

(Proof sketch: It can be shown that all constructible numbers have minimal polynomials with degree a power of 2. The minimal polynomial of $\sqrt[\Large3]2$ is $x^3-2$, which has degree 3.)

(The proof uses the fact that we can view $\Bbb Q(\alpha)$ as a vector space over $\Bbb Q$, where $\Bbb Q(\alpha)$ is the field generated by $\alpha$ - that is, the closure of $\{\alpha\}$ under the operations $(+,-,\times,\div)$. We can thus use the tools of linear algebra)

U.S. Military Could Collapse Within 20 Years Due to Climate Change, Report Commissioned By Pentagon Says - Vice

topologicalmagician

1:24 AM

how do I prove that

if a function is integrable on [a,b]

then it is integrable on every compact subset?

Akiva Weinberger

Lebesgue or Riemann

topologicalmagician

Riemann

schn

Why are the functions $x$ and $z$ symmetric over any sphere in the first octant? Where is the symmetry?

This regards triple integrals, and apparently their triple integrals are equal.

Rudi_Birnbaum

@skullpatrol naive set "theory" seems to me to be no "theory" at all, but rather a heuristical vehicle

Akiva Weinberger

It's a reflection across the plane $x=z$

which maps the point $(x_0,y_0,z_0)$ to $(z_0,y_0,x_0)$

schn

1:38 AM

Okay.

topologicalmagician

I

i'm not sure which dissection to use

schn

@AkivaWeinberger Thanks.

Would $y$ and $x$ also be symmetric over that domain?

Akiva Weinberger

Yes

Secret

Year 2039, just about right...

topologicalmagician

Let $\epsilon>0$. As f is integrable, there exists a partition $D$ of $[a,b]$, given by
$D=$ $\{$ $a,x_1,x_2,x_3....,x_n=b$ $\}$
Consider $D'= D \cup$ $\{$ $c,d$ $\}$. Then $D'$ is a refinement of $D$. Let $D''=D'\cap [c,d]$. Then $D''$ is a partition for $[c,d]$.

What should I do next?

oh

I got it

Secret

1:50 AM

Also lol cheeseburger algebra

topologicalmagician

Problem: Suppose $f\in R[a,b]$ then for any $[c,d]\subseteq [a,b]$ we have that $f\in R[c,d]$.

Let $\epsilon>0$. As f is integrable, there exists a partition $D$ of $[a,b]$, given by
$D=$ $\{$ $a,x_1,x_2,x_3....,x_n=b$ $\}$
Consider $D'= D \cup$ $\{$ $c,d$ $\}$. Then $D'$ is a refinement of $D$. Let $D''=D'\cap [c,d]$. Then $D''$ is a partition for $[c,d]$ and is a subset of $D$, therefore $U(f,D'')-L(f,D'')\leq U(f,D')-L(f,D')\leq U(f,D)-L(f,D) < \epsilon$. The first inequality holds because $D''\subseteq D$.

Is the proof correct?

Semiclassical

2:20 AM

@user10478 yep. thank goodness for linearity

Ultradark

2:32 AM

Any interpretations of an undefined expected value for a continuous random variable?

The Cauchy distribution comes to mind

user10478

@Semiclassical Can you also do anything with multiplication, i.e., $(D^2 - 1)[f(x)] * (D + 1)[f(x)] = (D^3 + D^2 - D - 1)[f(x)]$?

Semiclassical

nope

Akiva Weinberger

I want the past tense of vend to be vent

user10478

Okay, thank you

Ultradark

2:53 AM

@AkivaWeinberger I want the past tense of vend to be vent also

Given a probability distribution, does there always exist an application for such?

Leaky Nun

4:03 AM

@AkivaWeinberger "that'll teach you to" means "that will teach you not to"

Akiva Weinberger

5:03 AM

@LeakyNun Hm. Why don't people complain about that as much as they complain about "could care less"?

Ted Shifrin

5:20 AM

@Akiva: DogAteMy, I vent home when I was finished.

@topologicalmagician Except that you forgot to explain what the partition $D$ had to do with $\epsilon$ at the very beginning!

Akiva Weinberger

6:12 AM

If you understand something, but then forget it all, did you derstand it

Akiva Weinberger

7:00 AM

הפרויקט של עידן רייכל - בואי - The Idan Raichel Project - Bo'ee

►

1010011010

11:34 AM

Hello everybody

Ultradark

Hello everybody

1010011010

I hope somebody can shed light on the following

I'm looking at this: math.stackexchange.com/questions/8385/…

Now I'm considering a situation where I don't have a simple case of "$n$ roots of unity"

Rather, I'm picking the roots asymmetrically

So I guess, instead of having the roots $0$ through $n-1$, I'm considering say, $0$ through $m$, then $n-p$ through $n-1$

To name an example

Where obviously, $n-p\neq m$.

How does this affect the result? Can I still use the argument of $z^{n'}-1$? (where $n'$ is the amount of roots I have)?

Certainly the zeroes of my function are given by the product of factors $X-\xi_k$ with $k$ the root, that's certainly clear

Rithaniel

12:13 PM

How would you find a (the?) Sylow 2-subgroup of $\text{GL}_4(\mathbb{Z}_2)$?

Ultradark

12:26 PM

As long as an integral of a function converges can you define a pdf on it

because you can just choose the correct normalizing constant to make the area equal to one

Semiclassical

12:44 PM

@Ultradark A pdf should also be nonnegative everywhere

This means that, for instance, the function sin(x) can be normalized to give a pdf on the interval (0,pi) but not on (0,2pi).

Rithaniel

Meanwhile $\frac{\text{sin}^2(x)}{\pi x^2}$

Semiclassical

Good old sinc squared

Iirc the Fourier transform of that is a triangular pulse

Joakim

anyone know what an "explosive solution" or "entire large solution" is? Google yields plenty of usages but no definitions...

Ultradark

1:01 PM

Can I start a minimizing the arc length contest

or even better, minimize the arc length AND area

simultaneously

Semiclassical

Subject to what conditions?

Ultradark

I'll think about them

but what if you can keep finding a better minimum

Rithaniel

Of note is that $\text{Var}(\frac{\text{sin}^2(x)}{\pi x^2})$ is undefined

At least, using the only method of calculating it which I know off hand

Semiclassical

1:29 PM

@Ultradark If there’s no conditions , the answer is trivial: pick a point and stay there

That’s a perfectly legit curve, with length 0 and area 0

So until you add on some conditions, the answer is boring

Ultradark

I need to transport goods to a friend from (0,1) to (1,0)

but route $\sqrt{2}$ is largely closed for construction

jeea

Hello, is discussion on project euler problems on topic here

I am not asking for answer, only that if my approach is correct

skullpetrol

sure, why not?

jeea

projecteuler.net/problem=29

In this problem, it is given that we need to count all unique $a^b$ for $a\in [2, 100]$, $b \in [2, 100]$

My approach was that since there are 99 possible value for a and 99 for b, we have 99*99 total answers. Now we need to subtract duplicates like $a^b$, $(a^2)^{b/2}$, ... $(a^j)^{b/j}$ if $a^j <= 100$ and $j | b$

Leaky Nun

@Rithaniel its order is $15 \times 14 \times 12 \times 8 = 2^6 3^2 5^1 7^1$

in short, I would not find a/the Sylow 2-subgroup of GL(4,2)

Semiclassical

1:44 PM

@jeea according to mathematica, the answer is 9183

the only issue with your approach I could see is if there's any cases where the same number shows up more than twice

jeea

@Semiclassical yes thats correct, I also got it with brute force approach

@Semiclassical here I try to avoid subtraction again by using a flag (vector "done") : ideone.com/KtMaix

Leaky Nun

@Rithaniel the upper triangular matrices form a subgroup of order $2^6$, there you go

Semiclassical

largest multiplicity I'm seeing is 6 for 1152921504606846976

Leaky Nun

@Rithaniel and this is not a coincidence

Rithaniel

Oh, shoot, I wasn't sure if that was going to be easily answered

Leaky Nun

1:50 PM

well sometimes you just need to try!

Rithaniel

Color me impressed (Also, thank you a lot with the help with the groups of order 36 problem yesterday, reading back on the musings gave me some insight. My proof is probably still a little garbled, but I think I've successfully shown that every group of order 36 must have either a normal Sylow 2 or a normal Sylow 3 subgroup.)

Leaky Nun

wait, the whole point was to show that it has a normal Sylow 2-subgroup or a normal Sylow 2-subgroup?

instead of classifying them?

Rithaniel

I don't have enough experience with the general linear group, but I do know that this gives us that the Sylow 2 subgroup of $\text{GL}_4(\mathbb{Z}_2)$ isn't normal, right?

Leaky Nun

right

Rithaniel

Well, yes, but in order to do that, you have to go through the work of nearly classifying them, so that was the stuff I figured I should be looking for

Leaky Nun

1:54 PM

no you don't...!

Rithaniel

(Also, I prefer broader knowledge over more specific, I ultimately only need to look at a couple of cases, but the other cases are worth looking at, so that you can understand how things work in general)

Semiclassical

2^60 = 4^30 = 8^20 = 16^15 = 32^12 = 64^10 = 1152921504606846976

(whew)

Rithaniel

Well, I mis-spoke a little bit: you have to go through the work of nearly classifying two or three particular cases.

Also, whew indeed, Semi

Ultradark

what METRIC can I put on this 6-d space? $NY:=\Bbb R^4\times\Bbb C^2?$

Alessandro Codenotti

The usual product metric?

Semiclassical

2:03 PM

@Rithaniel thank goodness for mathematica

Leaky Nun

also it's an 8-d space

Ultradark

@AlessandroCodenotti What about if $C^2$ is equipped with the group of complex isometries? Still product metric?

Alessandro Codenotti

I have no idea what that means

Ultradark

so it's 8 dimensional?

Leaky Nun

@Rithaniel $\Bbb R \times \Bbb C$ is not a field

Rithaniel

2:07 PM

Ah, right, zero divisors.

Alessandro Codenotti

It is an 8-dim $\Bbb R$ vector space

It is also 8 dimensional as a topological space regardless of which definition of dimension you prefer (small/large inductive dimension, Lebesgue covering dimension, asymptotic dimension etc.)

Ultradark

so there's no way I can argue that it's six dimensional?

Leaky Nun

no way at all

Rithaniel

Actually, what are some conditions for different types of dimensions of a space to not coincide with each other?

Leaky Nun

having different names

Alessandro Codenotti

2:11 PM

The space needs to be ugly

leaky, what is 4+2

8

lol

That made me actually laugh

Alessandro Codenotti

(Except for asymptotic dimension I guess, which is the weird one out of them, you can very easily construct metrizable spaces whose asymptotic dimension is lower than their topological dimension)

Ultradark

2:13 PM

okay so where is everyone getting 8 from

Rithaniel

(I suppose you could also define new types of dimension, but then you're just cheating the game)

Alessandro Codenotti

From the fact that $\Bbb C$ is two dimensional

Ultradark

oh okay 4+4=8

What is the essence of $\Bbb R \times \Bbb C$?

I really don't know how to conceptualize it

Rithaniel

It depends on in what context you are looking at it. Like, as Leaky pointed out when I made a mistake, $\mathbb{R}\times\mathbb{C}$ isn't a field. It is a ring and a vector space over $\mathbb{R}$, though.

Ultradark

is it the real line in one direction and the complex plane in another direction?

Rithaniel

2:21 PM

It's also potentially a topological space

Leaky Nun

the only time you'll get $6$ from $\Bbb R^4 \times \Bbb C^2$ is when you count the number of fields in the product

Rithaniel

At the very least, it's a set

Wait, Leaky, isn't $\mathbb{Q}$ a field in the product?

Leaky Nun

as in, the number of "summands"

multiplicands?

if a ring can be decomposed into a finite product of fields

then the number of multiplicands is the number of maximal ideals

which is unique to a ring

Ultradark

I want embed some surfaces in $S=\Bbb R \times \Bbb C$

Rithaniel

Ah, well fair enough.

and each maximal ideal corresponds to one of the respective fields being replaced with a 0?

Leaky Nun

2:26 PM

precisely

Alessandro Codenotti

@LeakyNun or in general the ideals correspond to ultrafilters on the index set (of which there are exactly $n$ is the index set is finite with $n$ elements)

Leaky Nun

nice

you mean the maximal ideals

Alessandro Codenotti

Yes

I was so happy when an exercise in AG asked whether an affine scheme has finitely many connected components lol

Ultradark

what structures can one embed in $S$, or is that the wrong question

Alessandro Codenotti

Topologically that's the same as embedding in $\Bbb R^3$

Leaky Nun

2:31 PM

@AlessandroCodenotti any absolutely flat (i.e. von Neumann regular) rings have totally disconnected Spec :P

in particular $\Bbb F_2^\Bbb N$ I presume

did you use ultrafilters :P

Alessandro Codenotti

yep

Of course I did

Leaky Nun

nice

Alessandro Codenotti

Which motivated this question math.stackexchange.com/questions/3014962/…

Leaky Nun

nice

skullpetrol

3:29 PM

;-)

Leaky Nun

because it's Asaf Karagila

genescuba

Hi I am trying to find the number of different ways to partition a number in string format where all splits are prime

For example for '23' : partitions are [2],[3]; [23] so return 2 ways

skullpetrol

@LeakyNun indeed

Leaky Nun

@genescuba that's a finite problem, so you can write a program to do it

genescuba

@LeakyNun, my approach was just generating all the partitions of the list and then just checking wether each split is prime.

Leaky Nun

3:39 PM

that works

genescuba

I had some test cases on which my approach timed out on so I am guessing there is a more efficient solution?

Leaky Nun

well this is math.SE

where programs have infinite memory and infinite time

genescuba

your right, this would be a better question for stack overflow

topologicalmagician

If $A_1,A_2,....$ is a countable family of finite sets, is the union countable or at most countable?

Leaky Nun

but you might want to check how you are generating the partitions

@topologicalmagician "countable" = "at most countable"

genescuba

3:42 PM

Sounds good, thanks @LeakyNun

topologicalmagician

@LeakyNun my definition of countable excludes finiteness

Leaky Nun

@topologicalmagician but if $A_1 = A_2 = \cdots$ then the union is just $A_1$

topologicalmagician

@LeakyNun which is finite, so at most countable

Leaky Nun

ok

topologicalmagician

so then how would I then prove that if $A_1,A_2,A_3.....$ is a countable family of finite sets then the union is at most countable?

if we assume $A_i\neq A_j$ for different $i,j$ then the union is infinite.

Leaky Nun

3:50 PM

you can enumerate the union

topologicalmagician

and then because $A_i\neq A_j$ we can delete of the duplicates, and so we will have a countable set

right?

@LeakyNun

Leaky Nun

at most countable

topologicalmagician

Im confused, when would the union be finite if we assumed $A_i\neq A_j$?

@LeakyNun

Leaky Nun

oh I missed that part

then it's infinite

topologicalmagician

@LeakyNun thanks

so much

Anush

3:58 PM

how can you solve the recurrence T(n) = \sum_{i=1}^{n-1} T(i)T(i-1) ?

T(1) = T(0) = 1

Semiclassical

start by writing the first few instances, I guess

so T(2) = T(1)T(0) = 1, T(3) = T(2)T(1)+T(1) T(0)=2

note that this immediately tells you that T(n+1) = T(n)+T(n-1)T(n-2)

so that's neat

Anush

that is nice!

Semiclassical

T(4) = T(3)+T(2)T(1) = 2+1 = 3
T(5) = T(4)+T(3)T(2) = 3+2 = 5

T(6)=T(5)+T(4)T(3) = 5+6 = 11

hrm

Anush

a few more terms and we can look up the answer :)

Semiclassical

yep, lol

T(7) = T(6)+T(5)T(4) = 11+15 = 26

at which point OEIS gives this: oeis.org/A006888

which, you'll note, doesn't have a closed form

so solving this explicitly is probably out of the question

Anush

4:05 PM

hmm... is there a nice simple exponential lower bound?

topologicalmagician

@LeakyNun considering the case that there exists $\i,j$ such that $A_i=A_j$ is pointless because $A_1,A_2,.....$ is a countable family, right?

Semiclassical

no idea

Anush

T(8) = 26 + 5*11

42?

that must be wrong

Semiclassical

5*11=55

so that's 81

Anush

ah yes :)

81

Semiclassical

4:07 PM

I guess one lower bound for this is as such: T(i)>=1, so T(n)=T(n-1)+T(n-2)T(n-3) >=T(n-1)+T(n-2)

which is the Fibonacci recurrence

so T(n) is at least as big as the nth Fibonacci number

and that grows like (Golden Ratio)^n/sqrt(5)

Anush

ah yes.. good point

Semiclassical

that's not a very good lower bound, of course

Anush

we just need to find an upper bound

Semiclassical

that seems less pleasant

Semiclassical

4:26 PM

@Anush plotting it in mathematica, T(n) seems to grow like log(log(n))

topologicalmagician

Guys

if $A_1,A_2.....$ is a countably infinite family of finite sets

i'm trying to show that their union is likewise countably infinite

how do I enumerate the union?

Anush

@Semiclassical you mean e^e^n?

Semiclassical

derp, yeah

log(log(T(n)) grows linearly

Anush

very odd but interesting

Semiclassical

there's probably a good explanation for it but I don't know it off the top of my head

Anush

4:37 PM

me neither!

@Semiclassical actually.. I am not sure why this is right

@Semiclassical are you sure T(n+1) = T(n)+T(n-1)T(n-2) ?

Semiclassical

Note that T(n+1) = \sum_{i=1}^{n} T(i)T(i-1) , whereas T(n) = \sum_{i=1}^{n-1} T(i)T(i-1)

the only difference between the two sums is that T(n+1) includes the i=n term and T(n-1) doesn't

and the i=n term is just T(n)T(n-1).

so that'd be T(n+1) = T(n)+T(n)T(n-1)...hrm

looks like I may have been off by one, drat

Anush

ah ok

Semiclassical

that recurrence shows up on OEIS as well: oeis.org/A005831

Anush

@Semiclassical T(n+1) = T(n) * (T(n-1) + 1)

Semiclassical

right

Anush

4:51 PM

which still grows doubly exponentially apparently!

Semiclassical

quite

with simpler numbers, apparently?

e^Phi^n

Anush

right

hmm..

Semiclassical

which is consistent with the numbers I get in mathematica

e.g. log(T(21))/log(T(20)) is approximately 1.618... i.e. the golden ratio

Anush

aha

skullpetrol

did you skim over the MO question on proofs I linked in the hbar, sir? @Semiclassical

the suggested textbook states:

> No background beyond standard high school mathematics is assumed.

Semiclassical

5:00 PM

lol

skullpetrol

which has a 2019 edition

40

Q: Teaching undergraduate students to write proofs

In my experience, there are roughly two approaches to teaching (North American) undergraduates to write proofs: Students see proofs in lecture and in the textbooks, and proofs are explained when necessary, for example, the first time the instructor shows a proof by induction to a group of fresh...

How To Prove It: A Structured Approach

last answer^

Semiclassical

the pairing of that book with a discussion-style course seems key

skullpetrol

I agree, it doesn't seem suited for independent study.

I think the two-column proof style is an attempt to add "structure" to proofs.

Semiclassical

5:56 PM

@skullpetrol in two senses, yes. structure for the students writing those proofs, but also for the teacher themself

skullpetrol

yup

win-win

@Semiclassical i think the problem arises with "AHA!" poofs which are mostly found in geometry

skullpetrol

6:18 PM

also, proofs without words

Alessandro Codenotti

@topologicalmagician Do you know how to show that $\Bbb N^2$ is countable?

ÍgjøgnumMeg

6:33 PM

@skullpetrol my university took approach 2 and being at masters level now I can say that it was definitely not beneficial in my case rofl

topologicalmagician

6:44 PM

@AlessandroCodenotti yes

anakhro

Hey hot cats

Ultradark

7:13 PM

Hi^

Ultradark

7:28 PM

hello

anakhro

What are you up to, @Ultradark

Ultradark

drinking an iced mocha latte and getting ready to do some math

anakhro

Less drinks, more math

Ultradark

okay

a commutative diagram leads to one destination whereas a ________ diagram leads to two possible destinations

maybe it's just non-commutative

anakhro

7:47 PM

What a vague fill in the blank

Alessandro Codenotti

8:00 PM

@topologicalmagician For each $A_i$ fix a bijection $f_i\colon A_i\to\Bbb N$. Now you get injections $A_i\to\Bbb N^2$ by sending $a\in A_i$ to $(i,f_i(a))$. Now compose with a bijection $\Bbb N^2\to\Bbb N$

s.harp

@Ultradark what do you mean with destination

Ultradark

no matter which path you take you end up in the same place

anakhro

That's not what a commutative diagram is.

A commutative diagram is about the actual morphism after composition.

By composition, no matter what path you take, you will literally end up in the "same place" if you end at the same object. Commutative or not.

What a commutative diagram tells you is that the morphisms (i.e. the composition of paths) you take are actually equal.

Ultradark

okay I understand

Ultradark

8:17 PM

Anyone know how many probability distributions there are for the support one to infinity off the top of their head?

I perused the wikipedia listing but didn't see anything

anakhro

What do you mean support 1 to infty?

Ultradark

when you integrate the pmf over the support you should get 1

example: Gaussian distribution has a support over the entire real line

beta distribution has support over the unit interval

s.harp

So you want the support to be $[1,\infty)$?

anakhro

The named probability distributions are just named because they have been of particular use and interest.

I think the answer to your question is literally "infinitely many".

Ultradark

like the stars?

I didn't realize there were that many

s.harp

8:26 PM

no, because we dont have any reason to expect there are infintiely many stars, and if there were we would expect there to be countably many whereas you will have a lot more probability measures

Ultradark

@s.harp I see your point.

They never taught us how to find the entropy of a pmf. Any idea how to do that?

anakhro

@Ultradark what is the definition of entropy for a pmf?

Ultradark

@anakhro doesn't look like there's a unified definition

I saw the entropy of a gaussian distribution listed and was wondering how they got that

anakhro

@Ultradark you do know that if you don't have a definition, then you can't exactly "find" it??

Ultradark

probably they used differential entropy

but yeah I'm over this question

I would like to solve this though

0

Q: Looking to solve the dynamical system via differential equation or lack thereof

Given the boundary of the following phase portrait (which the interior can easily be filled in), find the differential equation solution or prove that the solution is independent of a differential equation. I've gone through different scenarios of phase portraits, such as the hyperbolic phase po...

ordinary-differential-equations dynamical-systems

an important constraint I accidentally left out is that the phase portrait should be symmetrical about y=x

and also y=1-x

assuming we're working in $[0,1]^2$

I know I cannot use a $2 \times 2$ matrix as I could for the hyperbolic case (on the unbounded plane)

anakhro

9:06 PM

As with most of your questions, this one suffers from lack of clarity.

Ultradark

@anakhro I agree, but that's because I don't know how to phrase to question properly

Ultradark

9:20 PM

Remember, Grothendieck gave the decisive definition of a scheme, bringing to conclusion a generation of experimental suggestions and partial developments. My question is likewise a partial development

anakhro

I don't think comparing your vague questions to those of research mathematicians is a good analogy.

On your side it seems more of an unwillingness to put in the effort to make it clear.

Ultradark

I work hard and put in as much effort as I can

but most of my questions are just silly little things

genescuba

10:49 PM

Say $x + y + z = 11$ $x,y,z \in \mathbb{Z}^+$. How many solutions exist for $x,y,z$

The simplest way to approach this problem is just to try the different positive ints for which this is true. Which are as follows:

9 + 1 + 1 = 11
8 + 2 + 1 = 11
7 + 3 + 1 = 11
7 + 2 + 2 = 11
6 + 4 + 1 = 11
6 + 3 + 2 = 11
5 + 5 + 1 = 11
5 + 4 + 2 = 11
5 + 3 + 3 = 11
4 + 4 + 3 = 11

We can see one possible solution are the positive integers 9,1,1

However since 1 is used twice here we have only $\frac{3!}{2!}$ solutions for this, if we have distinct numbers we will have 3!.

I am wondering if there is a better way of doing this problem?

anakhro

@genescuba math.stackexchange.com/questions/54635/…

If I understand correctly you want to discuss "integer partitions" of 11.

genescuba

Correct

Thank you for the link, seems to be what I needed

anakhro

I am not sure if it is exactly what you needed.

I didn't see anything about excluding repetition.

genescuba

"Or, in other words, I am interested in the number of solutions with distinct numbers. For 𝑛=2 and 𝑀=5, I would consider solutions (1,4) and (4,1) equivalent"

Seems like author is asking the same question as I

anakhro

Oh great.

Well I am glad it is more spot on than I thought it was. :P

I had done something with Young's tableaux before. Those were neat.

Their usage for the representation theory of symmetric groups is super fun, as well.

I think it's probably one of the best ways to learn about representation theory.

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