/math/: Math general

We suppose that [math]\operatorname ker f = \operatorname ker f^2[/math] and [math]f^2[/math] is diagonalizable. The assumption on kernels allows us to assume WLOG that 0 is not an eigenvalue of [math]f[/math] through a standard calculation with Jordan form. Suppose we have a [math]v \in V, k>0, \lambda \in \mathbb C^*[/math] for which [math](f - \lambda I)^k v = 0[/math]. Observe that [math](f^2 - \lambda^2 I)^k v = \left( (f + \lambda I) (f - \lambda I) \right)^k v = 0[/math] using that these two operators commute. Thus [math](f + \lambda I) (f - \lambda I) v = 0[/math] because [math]f^2[/math] is diagonalizable, so all eigenvalues are in fact proper. We conclude that [math](f - \lambda I) v = 0[/math] (I don't have time to expand on this), so [math]f[/math] is indeed diagonalizable.

if i have a cube that is 100m^3
how many smaller cubes could fit inside it if they werent?

Newfag fucker here.

I want to learn mathematics out of sheer curiosity. How deep does the rabbit hole go? Is it worth the effort?

How do I approach my math major optimized for graduate school towards getting a job and a career outside of academia? How do I start graduate school and leave with an industry job?

yes

Scrub here. Mind explaining that briefly?

Let [math]f:\mathbb{C}\left[ {x,y} \right] \to \mathbb{C}\left[ {x,y} \right][/math] be the map such that [math]f\left( {{x_i}} \right) = {f_i}[/math] and [math]\det \left( {\partial {f_j}/\partial {x_i}} \right) \in \mathbb{C} - \left\{ 0 \right\}[/math].

Show [math]f[/math] is an isomorphism.

Nice, that'll work.

As far as you are willing go. Obviously, no one has seen the end

Some analysis:
Characterize the functions [math]f: \mathbb R \to \mathbb R[/math] such that, for each [math]x[/math] and each sequence [math](u_n)[/math] such that [math]\left(\frac{u_1 + \dots + u_n}{n}\right)[/math] converges to [math]x[/math], [math]\left(\frac{f(u_1) + \dots + f(u_n)}{n}\right)[/math] converges to [math]f(x)[/math]

Publish papers which have applications to industry; theoretical comp sci oriented and the like, and you should recieve job offers towards the latter end of your phd. If youre getting a masters, just apply for interviews; math degrees are highly employable.

An isomorphism of vector spaces ? of rings ?

No, because if the cubes want water

Trying to understand cyclical linear logic (CyLL). If it's modeled by a (nonsymmetric) monoidal category, why does the cyclic rule not imply exchange? You have [math]A \otimes B \cong B \otimes A[/math] so if you have associativity then you have [eqn](A \otimes B) \otimes C \cong (B \otimes A) \otimes C \cong B \otimes (A \otimes C) \cong B \otimes (C \otimes A) \cong (B \otimes C) \otimes A \cong (C \otimes B) \otimes A[/eqn] which isn't a cyclic permutation. Or is it modeled differently?

under the assumption that multiplication is commutative

Huh?

For random variables I am trying to understand what the below equation means:

Pr[X=v] := Pr[X^-1(v)]

Is it basically saying the probability of v in set V is the same as the probability of falling into the pre-image of v in set U?

Here's a problem i've been thinking about:
Let's consider strings of digits, which all have the same length n. There are 10^n such strings.
From those 10^n strings we pick a set of strings S, such that any two strings in S:
(a) differ at least on two places
(b) differ at least on three places
(c) differ at least on k places
What is the maximum possible number of elements in S? I can prove that for (a) it's 10^{n-1}, but i don't know about the others.

Basically my question is how do you model the cyclic permutation of two things without making the product commutative? Even if you do something like [math]Hom(A, B \otimes C) \cong Hom(A, C \otimes B)[/math] that should imply commutativity by Yoneda.

Is it possible to use only chords to divide a circle into equal area pieces with no two pieces congruent?

Why make the distinction with linear maps over real vectorspaces?
Or does diagonalisable mean real eigenvalues there?

Of course. Do you mean "with ruler and compass only"?

Yes, it means diagonalizable with real eigenvalues

I see a solution was posted already but I just want someone to check my reasoning:
[math]f[/math] is diagonalizable iff its jordan matrix is diagonal.
If f is diagonalizable, then the implication holds easily.
If f is not diagonalizable, then it has a 2x2 (or bigger) jordan block. If this block has nonzero eigenvalue, then the jordan matrix [math]f^2[/math] contains exactly this block squared, so [math]f^2[/math] is not diagonalizable. On the other hand, if the block had eigenvalue = 0, then kernel of [math]f^2[/math] is bigger than ker f.

These are exactly the linear functions. It's easy to check that these work.
Now assume that f is not linear; then we can find three points [math]a

You're a smart guy

I wish we had an irc or skype group or something so we could do higher level textbooks together as a group-study/group-mentoring community thing. I like learning with other people. I wanted to do some stuff like sheaves and schemes or something. Of course open to all topics at the 4th year undergrad/grad level.

I'm down, do you wanna create it?

Sure, skype or irc?

Up to you. Maybe both if you want, then link to each other in the info.

I'd help but I'm at work right now and I don't get off for another 8 hours.

I can create a irc channel and post the info here later. I will be able to do it after I finish studying. It won't be for another few hours.

Scratch that, I am making one now. Will post info soon.

Sounds good, I will see if I can get irc working on my phone.

If you are truly busy I can make one.

Join the channel ##sci on freenode.

webchat.freenode.net/

/join ##sci

I'm part of a group that's already been doing this (as of around the middle of March), would you like an invite? Just a slight issue though, we're not using IRC.

Oh, whatcha usin?

let's use irc for now, it's simple to use and anyone can join.

We're using a slack group consisting of multiple channels dedicated to projects/subjects of study.

Unfortunately we're using Slack. It's sort of like a glorified IRC but it helps us have nice features like private channels and etc.

Does that cost money?

That seems right. The probability of outcome v for random variable X is the same as the sum of probabilities over all arrows leading to v.

The function X is not injective and not necessarily even surjective (all points could map to 1, i.e. X is always true), while X^(-1) should be both surjective (every point in U has an outcome) and injective (each point in U has only one assigned outcome).

Please check this yourself, don't take my word for it.

No it's free, there are paid options but the one we're using is completely free.

No, it's a free service. If you're interested in joining, send us an e-mail. (I recommend you create a throwaway email because I wouldn't post my personal email on here)

is it? show please

Thanks!

Mathematicians join

##sci

webchat.freenode.net

Draw one chord followed by a perpendicular bisector of that chord using arcs and a straight-line or something followed by yet another bisector of that bisector. This is my best guess.

you guys should come to the slack the others are commenting
it's a nice interface and simple enough to use
we've had people going through math for a while

I don't think that's going to work

Oh shit I didn't notice that user said "no conrguent pieces".

I heard it was only basic algebra and logic with only a few people doing functional.

If [math]A[/math] is involutory, show [math]\frac{1}{2}(I+A)[/math] and [math]\frac{1}{2}(I-A)[/math] are idempotent, and [math]\frac{1}{2}(I+A) \frac{1}{2}(I-A) = 0[/math].

You should be able to solve this.

The square of a matrix in Jordan form need not be in Jordan form, I believe. Consider

[math]\displaystyle \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} [/math]

it's hard to say what "a few people" means, because it's a small group.

it's divided in channels though, so if you stick to the math channel you only talk to math people, simple.

i say it's fine to have a separate slack and irc channel.

so far everyone in the irc channel is doing higher level math and no plebs

with congruent pieces in the sense of circular sectors it won't work anyways
for example you can't construct a regular heptagon with only circle and straightedge, which is equivalent to finding 7 equal circular sectors

That doesn't matter much anyway, since the reason why the discussions are rather low-level is because most people are exactly at that level. The slack group is just a way for people to gather and study together. If grad students come together, then they'll talk about grad mathematics. We can always create new projects if people are interested.

So how many people are working on functional then? Exactly?

two I think, maybe someone else is

Yep around that out of the five-six people that are in galois (the higher-level math channel). The last project was, well, galois theory.

[eqn]P = \frac{v^{\top} v}{vv^{\top}}[/eqn] Prove [math]P^2 = P[/math].

You might try this sort of thing. I'm not sure is there any solution but why not))

v * vtranspose is symmetric

so is vtranspose * v

symmetric means equal to transpose

(vtranspose*v)transpose = v*vtranspose

because v*vtranspose and vtranspose*v are symmetric and tranpositions of each other they are the same.

P = I, I^2 = I

[math]v[/math] is a vector, not a matrix. The result is a scalar.

I guess you mixed up the transposes.
There is a shorter way to prove it, but I like this one, since it shows, what kind of map [math] P [/math] actually is:

Take any orthogonal base of vectors including [math] v [/math].
For any vector we have
[math]Pw= \frac{vv^T}{v^Tv}w = \frac{\left \langle v,w \right \rangle}{v^Tv} v [/math]
and with that we have [math] Pv = v [/math] and [math] Pw = 0 [/math] for every other base vector.
This also gives us [math] P^2v = P(Pv)=P(v) [/math] and [math] P^2w = P(Pw)=P(0) = 0 = Pw [/math]

Hm, sorry, that one doesn't work.

Veeky Forums join ##sci on rizon.

rizon.net/chat

/join #sci

We need more Veeky Forums folks

That is the Jacobian Conjecture...

This might work

for you

okay, you might have done it for 7.
now show a general straightedge and compass construction for n pieces

##sci on webchat.freenode.net

Real channel:

#sci on rizon.net/chat

freenode's is a aimed for math sci's

Is anyone into HoTT? Your thoughts? Is it another fancy idea or does it have immediate applications allowing one to easily prove easy things?

It would be extremely painful (like when Wiles pulled off Fermat's last theorem).

there is no alpha

hes a smart guy

[email protected] for the slack group please

bumb

Why is {(x, y) | x is an odd integer, and y is an even integer} not a function?

a function has one output for every input
where would 1 go under that rule? that set has (1,2) (1,4) for example, so multiple outputs

I see...I hadn't thought of a function that would output two different values I was thinking in terms of one output...
Thnx

That's the thing. The entire notion of a function captures the idea of assigning a unique output to every input. So given some guy in the domain, I can speak unambiguously about its value in the image.

Just write down the definitions bro.

[math]\frac{1}{2}(I + A)\frac{1}{2}(I + A) = \frac{1}{4}(I + A + I + A) = \frac{2}(I + A)[/math] etc.

oops, that last term should be [math]\frac{1}{2}(I + A)[/math]

Combination formula:

The calcul for find k-combination of a set S who has n elements is: n! / k!(n-k)!

Whether a set with 3 elements (a, b, c): there are 6 combinations of 2 elements: ab, ac, ba, bc, ca, cb.

But with the formula: 3!/ 2!*1! = 6/2 = 3, so 3 elements.

What is wrong ?

>Whether a set with 3 elements (a, b, c): there are 6 combinations of 2 elements: ab, ac, ba, bc, ca, cb.
Those are permutations. Combinations are where order doesn't matter, so ac = ca.

Consider the xy plane and color every integer coordinates with 1 of 3 colors. Prove that there exists a rectangle such that all 4 corners are the same color.

No one cares about true rigorous computer math? :(

Right now it mainly has practical applications for some mathematicians, like people who want to prove things about infinity groupoids and homotopy. They've found new proofs of old results like [math]\pi(S^1) = \mathbb{Z}[/math] which is pretty cool.

Then again, if Mochi had proved abc in Coq, no one would have been able to dispute its correctness. It decouples proof verification from understanding (which obviously has its pros and cons). But of course you don't need HoTT for that.