12:37 AM
@TedShifrin standard metric, in $\mathbb{R}$.

3 hours later…
4:03 AM
@TedShifrin All right! Means my main doubt arises from using Fundamental Theorem of Calculus when it should not be used blindly, okay!

4:43 AM
@adeshmishra since you reference Feynman: the most obvious instance of a line integral in physics is the work to move a particle along a path through a force field
so $W=\int_a^b \vec{F}\cdot d\vec{r}$
In the particular case of $\vec{F}$ being a conservative force, one can find a (scalar) potential energy such that $\vec{F}=-\nabla U$
and therefore the work done is just $W=-\Delta U$
combine that with the work-energy theorem $W=\Delta K$, and you get energy conservation: $\Delta E = \Delta K+\Delta U = 0$
(also, in that case the work to move a particle along a closed path is zero: if the force is conservative, then when the particle returns to its initial state then the initial potential energy is the same as the final, i.e., no change in potential energy)

5:15 AM
Hello all. I have a quick yes or no question I was hoping someone could answer for me

@Alex lets hear it

I was asked the following: Let W be the set of all real valued polynomials of any degree and U be the set of all exponential functions. I'm asked whether U+W is a direct sum. I said yes because their intersection is empty. Is this correct reasoning?
It was an exam question but I thought it was really simple, so I'm having second thoughts maybe there's more to it
It's from linear algebra

@Alex im way out of my depth here ,I'm a high school student :p

5:37 AM
Alex aren't constants both exponential $ce^{0x}$ and polynomial $cx^0$

U is not a vector space

Is there a term for, like, linear fractional transformations except quadratic. Like z->2z/(1+z^2)

The question is: What is the set of all exponential functions?
In order to have a vector space, you need all constant multiples, for starters, and that includes the zero function. You also need the zero function in $W$ (and it's not a polynomial of any (finite) degree). So the question is very sloppy. Did your exam question state it more carefully?

Hi chat
Im trying to solve this question
how many roots does x^y = 1 (mod p)
we know that there are p-1 elements in F_p so that is an upperbound
also we have that order of x has to divide y
the answer is supposed to be gcd (p-1, y-1)
But I do not see how
the original question is this x^y -x = 0 modulo n
but by the CRT I can split the solutions mod primes

5:57 AM
Well yes, sorry. The question is: Consider the vector space $V$ the set of all real-valued continuous functions, and subspaces $U$, the set of all exponential functions, their sums and scalar multiples, and $W$, the set of all real-valued polynomials of any degree. Prove whether $U+W$ is a direct sum.

6:08 AM
Jack the multiplicative group mod p has p-1 elements, and any solution to that equation generates a subgroup of order y-1

@KrullDimension can you explain further?
the way i thought about it is that , those elements that have order y-1 or order dividing y-1 would be a solution
we have p-1 elements in total
I do not understand the gcd part

By Lagrange's theorem, the order of any subgroup must divide the order of the group, so if y-1 does not have a common divisor with p-1, we only get one solution
If there is a common divisor, then there must be as many solutions as that common divisor, because the multiplicative group mod p is cyclic

I see
@KrullDimension can you tell me how any solution generates a subgroup?

So for example, if you have p=13, y=21, you could just take all the powers of 2 (mod 13), pick every third one, and get four solutions to your equation

it can be the case that x^(y-1) = 1
but order of x is just a divisor of y-1
why would that be a subgroup of order y-1 ?

6:20 AM
Yeah, that's true
I misspoke, you generate a subgroup of order dividing y-1

Still not seeing the full picture of the proof

6:50 AM
The group of solutions to x^(y-1)=1 (mod p) is isomorphic to the group of solutions to x*(y-1)=0 (mod p-1). These solutions are all multiples of (p-1)/gcd(p-1,y-1), of which there are precisely gcd(p-1,y-1) less than (p-1)
@JackOhara Sorry, took me a moment to collect my thoughts

@KrullDimension thanks and no worries
let me see if that makes sense to me

7:15 AM
@KrullDimension works thanks!

3 hours later…
9:46 AM
Does the calculator that comes with Microsoft Windows have a bug? It evaluates 2^3^4 to 4,096 but WolframAlpha evaluates it to something much greater wolframalpha.com/input/?i=2%5E3%5E4
Isn't it normally accepted that $^$ is right associative?

In fact, a^b^c is usually considered to be a^(b^c)

@Peter so put simply MS calculator is wrong
@Peter so put simply MS calculator is wrong

10:08 AM
@Semiclassical Thank you

2 hours later…
12:02 PM
Hello,
I encountered a nice thing about matrices but I can't prove it.
For block matrices of the form [I, A; A' I] (I use MATLAB's notation) there is some special property to the $V$ matrix. So if C = [I, A; A', I] then the $V$ matrix of the SVD of $C$ has the property that $V(1:n / 2, :)' V(1:n / 2, :) = 0.5 I$.
Anyone have seen this?

12:13 PM
What's Matlab notation and what is the V matrix ?

OK. $C = U S {V}^{T}$.
So ${V}_{1: \frac{n}{2}, :}^{T} {V}_{1: \frac{n}{2}, :}^{T}$ has $0.5$ in its diagonal.

What is U ? and S ?

I use 1:r, 1:q to mean a sub set of the 1 to r rows and 1 to q columns.
$U$ ans $S$ are from the SVD decomposition of C.

So U and V are unitary and S is diagonal ?

It is an interesting property I haven't seen anywhere.
Yep. We are on real matrices so $U$ and $V$ are orthonormal.

12:23 PM
Is I the identity matrix ?

Now, let's say $\boldsymbol{v}_{i}$ is the vector composed by the $i$ column of $V$. What I see is that the norm of its first $n$ elements is $0.5$ (Where the vector length is $2 n$). Always.
Actually the squared norm.
It comes form the special structure of $C$ but I can't prove why.

C is symmetrical, which means it U = V

No necessarily.

What do you mean ?

I think $U$ and $V$ are equal for matrices which are not only symmetric but are result of some quadratic form.
Have a look at - pastebin.com/FbcRfg23.

12:43 PM
In mathematics, particularly linear algebra and functional analysis, a spectral theorem is a result about when a linear operator or matrix can be diagonalized (that is, represented as a diagonal matrix in some basis). This is extremely useful because computations involving a diagonalizable matrix can often be reduced to much simpler computations involving the corresponding diagonal matrix. The concept of diagonalization is relatively straightforward for operators on finite-dimensional vector spaces but requires some modification for operators on infinite-dimensional spaces. In general, the spectral...

I know the spectral theorem. do you want to pin point me to what you meant?

The spectral theorem tells you that if your matrix is symmetrical it decomposes as $USU^T$ with S diagonal and U unitary

That's not the SVD

henlo people can i ask a question here to recieve a tip to solve a problem

The SVD has a lot to do with the Eigen Decomposition. They collide for the case the matrix is in the form ${A}^{T} A$ or $A {A}^{T}$.

12:58 PM
Oh right

in an Eigendecomposition $USU^T$, the diagonal of $S$ are the eigenvalues whereas in an SVD $USV^T$, the diagonal of $S$ are the singular values
the singular values are the square roots of the Eigenvalues of $A^TA$ and the SVD is obtained from the Eigendecomposition of $A^TA$ (which is always symmetric), explaining the relationship

@Thorgott, do you have any idea about what I saw?

1:15 PM
Was the structure of $C$ clear?
I will try again. $C \in \mathbb{R}^{2 n \times 2 n}$ where $C = \begin{bmatrix} I & A \\ {A}^{T} & I \end{bmatrix}$
The matrix $I$ is the identity matrix (Size $n \times n$) and $A$ is any $n \times n$ matrix.
The SVD of $C$ is given by $C = U S V$.
So far so good?

yeah, I just don't follow the ${V}_{1: \frac{n}{2}, :}^{T} {V}_{1: \frac{n}{2}, :}^{T}$, it looks like these don't even have compatible dimensions to me

Now, let's have $\boldsymbol{v}_{i}$ as the $i$ -th column of V.
Clearly ${v}_{i} \in \mathbb{R}^{2n}$.
What's amazing in that case is: $\sum_{j = 1}^{n} {\left( \boldsymbol{v}_{i}[j] \right)}^{2} = 0.5$.

Theorem: Let x0, y0 and t0 be real and F(x,y,t) and G(x,y,t) be continuous with continuous partial derivatives. Then, there are unique functions x(t), y(t) defined on an interval around t0 satisfying:

x'(t) = F(x,y,t)
y'(t) = G(x,y,t)
x(t0) = x0
y(t0) = y0

Does this theorem have a name? Is there a name for the general, n-dimensional version of this (this would be the 2-dimensional version)? I can only recall Picard-Lindelöf from analysis, but I don't need it for Lipschitz-continuous, just continuously differentiable is sufficient.

Where $\boldsymbol{v}_{i}[j]$ is the $j$ -th element of the vector.
Needless to say $\sum_{j = n + 1}^{2 n} {\left( \boldsymbol{v}_{i}[j] \right)}^{2} = 0.5$ as well.
Is it clear now? The indices?

1:56 PM
But what if $A=0$? Then you can take $C=U=S=V=1$ and then the property isn't satisfied.
@Huy continuously differentiable implies locally Lipschitz

I never said anything else

You're right.
Let's say it is not the case.
As in my actual case $C$ has no zero entries off the block diagonal.

@Huy well, it's a corollary of Picard-Lindelöf then and that has generalizations to $n$ dimensions
@Royi a column of $V$ will be an eigenvector of $C^TC$ to some Eigenvalue $\lambda$, if you write the column as $\begin{pmatrix}v\\w\end{pmatrix}$, then $\lVert v\rVert^2+\lVert w\rVert^2=1$ by normality and then your observation is equivalent to $\lVert v\rVert^2=\lVert w\rVert^2$.
What I am able to deduce from the properties is that we always have $\lambda\lVert v\rVert^2=\lVert v\rVert^2+\lVert A^Tv\rVert^2+2\langle Aw,c\rangle$ and $\lambda\lVert w\rVert^2=\lVert w\rVert^2+\lVert A^Tw\rVert^2+2\langle Aw,v\rangle$. Not sure how to proceed.

I'm not sure how you got the property you mention.

2:13 PM
I split the Eigenvalue equation into parts for $v,w$, then paired them up with $v,w$ and rearranged

2:23 PM

2:50 PM
Hello
I have a doubt about it:
If $\alpha<\betha$,
then the lipchitz function set $C(a,b)^\alpha$ = $C(a,b)^\betha$, I´m not sure if is equal or only contention

3:05 PM
@LukasHeger Hi Lukas , how to find the number of solutions to x^y = x (mod p) I have a solution but I wonder if there is a neater way that you have.
the original problem is about to find x^y =x mod (N) where N is product of two primes
but since Z/N is isomorphic to Z/p x Z/q we can solve that modulo each prime

if x is not zero that's equivalent to x^{y-1} = 1 mod p. Now the multiplicative group of Z/p is cyclic of order p-1. Suppose for a moment that y-1 divides p-1, then there's an element of order y-1 in the multiplicative group of Z/p, as that is cyclic, so we have y-1 solutions and y in total if we add 0 back in. If y-1 does not divide p-1, then the subgroup of elements Z/p^x satisfying x^{y-1}=1 is the same as the subgroup of elements satisfying x^{gcd(y-1,p-1)}=1,
so as in the previous case, there are gcd(y-1,p-1) of those, so if we add 0 back in, we have gcd(y-1,p-1)+1 total solutions

@LukasHeger Thank you!
@LukasHeger in the third line. the subgroup of elements Z/ p^x
where did that come from?
oh wait , let me try to figure it out

the solutions to x^{y-1}=1 mod p is a subgroup of the multiplicative group of Z/p

3:25 PM
I want to know about the history of Dirac Delta Function
I read Wikipedia but it was not that easy to comprehend. I want to know how Heaviside found the divergence of $\mathbf E$

@Royi $C^TC=C^2=\begin{pmatrix}I+AA^T&2A\\2A^T&I+AA^T\end{pmatrix}$, so the Eigenvalue equation is equivalent to $(I+AA^T)v+2Aw=\lambda v$ and $(I+AA^T)w+2A^Tv=\lambda w$. Apply $\langle\cdot,v\rangle$ to the first and $\langle\cdot,w\rangle$ to the second equation.

3:52 PM
@LukasHeger if we have y-1 dividing p-1 , why does that gurantee existence of elelemnt of order y-1 ?

because the multiplicative group is cyclic

so it is a result about cyclic groups

yes

the way I think about it is like this

if $\alpha$ is a generator (hence of order $p-1$), then $\alpha^{\frac{p-1}{y-1}}$ will have order $y-1$

3:54 PM
the solutions are those elements that has order dividing y-1
is that the wrong approch ?

no
that's fine

the GCD of p-1 and y-1 is the part that am not understanding well
if we try it using that idea
or wait no

by Lagrange, the order of any element divides p-1
now if the order divides both p-1 and y-1, it divides the gcd

yes this part is clear
I know that the order of elements has to divide gcd ( p-1, y-1)
but why are they that many elements

because the group is cyclic
same result as befoe
gcd(p-1,y-1) divides p-1

3:58 PM
ohhh
I see thanks !

I'm looking for a theorem that says something like "if there is a continuous map $X\to\Delta^n$ (or $S^n$) which is essential (can't be homotoped to a map on the boundary relative to the boundary), then the Lebesgue covering dimension of $X$ is at least $n$" or something similar characterizing the dimension in terms of maps to spheres. Does anyone know something similar?

4:22 PM
@Lukas Alright so I'm gonna be talking about the four squares theorem on Tuesday, tell me if this strategy is the correct one: define a function $f(z) := \sum_{n=0}^\infty r_nq^n$ where $r_n$ counts the number of representations of $n$ as a sum of 4 squares, show this is an element of $M_2(\Gamma_0(4))$, so it's expressible as a linear combination of Eisenstein series of the relevant type, equate coefficients

@EdwardEvans yes, this is the correct approach

Cool :) I'll be waving my hands so hard that I might take off
will probably have to weigh myself down

note that the space $M_2(\Gamma_0(4))$ is two-dimensional, so you only need to compute two Fourier coefficients
@EdwardEvans you can philosophize about how remarkable the fact that all spaces of modular forms of fixed weight and level are finite-dimensional and that we can efficiently compute (or bound) its dimension.
This means that if we have a counting problem (such as counting number of reps by sums of four squares) and it turns out that the generating function for the counting problem is a modular form, we can rejoice, as we only need to compute finitely many terms of the generating series and by finite-dimensionality, we will get some linear independence with other modular forms

I could but I only have a short window of time in which to put across an interesting concrete fact about modular forms so I'm just gonna prove something that they'll understand

<s>just talk about how every cuspidal Hecke eigenform gives you a Galois representation</s>

4:38 PM
rofl it would probably be good for me to talk about Hecke theory since it'd force me to learn it properly (which I still need to do for exams rofl)
Alas, I only have 1 hour to get stuff across

nah, Hecke theory is not the right thing for a one hour talk
just do one hour of unmotivated double coset computations

4:52 PM
hahahaha
I actually wanted to do two proofs of quadratic reciprocity; one elementary proof and one by cyclotomic fields
but they don't know what a field is so
loool

DSolve[{x'[t] == -y[t]^3, y'[t] == x[t]^3, x[0] == 1, y[0] == 0}, {x,
y}, t] is this not the right syntax for Mathematica to solve a system of DEs?
Mathematica evaluates the first two equations and returns "True" for both of them, returns no solution

@EdwardEvans derive QR from Artin reciprocity

Well my actual plan was to build up, starting from elementary proofs, then a proof by cyclotomic fields, and then derivation from Artin reciprocity, and was expecting only to give a talk to my lecturers rofl
but unfortunately they want me to talk to students too, so I went for something they'll understand
hence only focussing on some elementary results in the theory of modular forms

now Artin reciprocity is just a special case of Langlands reciprocity (which is conjectural even for GL_2)

special case where the extensions are abelian right?

4:58 PM
why do they not know what a field is?

I mean yeah kinda. You have (conjectured) Langlands for every reductive group over a number field, if you plug in GL_1, you get CFT

Nice :)
also @Thorgott my previous university is just.. focussed on churning out faceless office droids
so the "mathematics" degrees are designed to produce statistical analysts for industry

very roughly, some kinds of representations of GL_n will correspond to some n-dimensional Galois representations
so for n=1, you get 1-dimensional Galois representations, but a 1-d rep of a group will always factor throug the abelianization

coooooool

that's why CFT, or Langlands for GL_1 only says something about abelian extensions

5:01 PM
that's nice

@EdwardEvans I plan to sketch some of this stuff in my thesis defense, as a motivation
I'll invite you ofc

Nice, thanks

oof

modular forms can be regarded as certain "automorphic repesentations of GL_2" (that's a more complicated way of looking at things, a priori) and you can associate a 2-d Galois rep to them (construction due to Eichler-Shimura-Deligne-Serre), so that's a special case of global Langlands for GL_2 that we actually kind of understand
now modularity results (e.g. Wiles/Taylor proof of FLT or the Serre conjecture) prove that certain kinds of Galois reps (e.g. the one coming from the Tate module of an elliptic curve) actually arise via this construction
so the point of the Eichler-Shimura-Deligne-Serre construction is just setting up the map from some kind of automorphic representations of GL_2 to 2-d Galois reps, which is hard enough. And modularity theorems now basically just tell you that some stuff is contained in the image of that map
which is really hard somehow
sorry for the rambling

that's cool, I jsut went for a coffee and a cigarette lol
It's something that I'm interested in learning in the course of my studies lol
although I'm still really interested in going down the Iwasawa theory road

5:25 PM
Is there an "intuitive" way to see why the integral representation of the arithmetic-geometric mean works?
I can follow and reproduce the proof, but it involves a weird change of variables which makes it hard to see what's "really" going on, if that makes sense

5:55 PM
@Huy I suspect you're going to need NDSolve. But don't you need {x[t],y[t]},t at the end?

2 hours later…
8:11 PM
I have some issue with the following claim (Something I noticed playing with matrices in MATLAB):
Given a Symmetric Matrix $C \in \mathbb{R}^{2 n \times 2 n}, \; C = \begin{bmatrix} I & A \\ {A}^{T} & I \end{bmatrix}$ where $I$ is the identity matrix and $A \in \mathbb{R}^{n \times n}$. Its Eigen Decomposition is given by $C = Q \Lambda {Q}^{T}$. Let $Q = \begin{bmatrix} U \\ V \end{bmatrix}$ then for any $i$ the following equality holds ${\left\| \boldsymbol{u}_{i} \right\|}_{2}^{2} = {\left\| \boldsymbol{v}_{i} \right\|}_{2}^{2} = \frac{1}{2}$ where $\boldsymbol{u}_{i}$ and $\boldsymbol{v}_{i}$ are the $i$ -th column of $U$ and $V$ respectively.
Now, I created the following proof:
By the Eigen Decomposition we have $C Q = Q \Lambda$ hence for a given $i$ we have $C \boldsymbol{q}_{i} = {\lambda}_{i} \boldsymbol{q}_{i}$ which means $C \begin{bmatrix} \boldsymbol{u}_{i} \\ \boldsymbol{v}_{i} \end{bmatrix} = {\lambda}_{i} \begin{bmatrix} \boldsymbol{u}_{i} \\ \boldsymbol{v}_{i} \end{bmatrix}$.
This yields the equations, $\boldsymbol{u}_{i} + A \boldsymbol{v}_{i} = {\lambda}_{i} \boldsymbol{u}_{i}$ and ${A}^{T} \boldsymbol{u}_{i} + \boldsymbol{v}_{i} = {\lambda}_{i} \boldsymbol{v}_{i}$.
Multiplying the first on left by $\boldsymbol{u}_{i}^{T}$ and the second by $\boldsymbol{v}_{i}^{T}$ will yield $\boldsymbol{u}_{i}^{T} A \boldsymbol{v}_{i} = 2 {\lambda}_{i} {\left\| \boldsymbol{u}_{i} \right\|}_{2}^{2}$ and $\boldsymbol{v}_{i}^{T} {A}^{T} \boldsymbol{u}_{i} = 2 {\lambda}_{i} {\left\| \boldsymbol{v}_{i} \right\|}_{2}^{2}$.
Since $\boldsymbol{u}_{i}^{T} A \boldsymbol{v}_{i}$ is a scalar it means $\boldsymbol{u}_{i}^{T} A \boldsymbol{v}_{i} = \boldsymbol{v}_{i}^{T} {A}^{T} \boldsymbol{u}_{i}$ which means, if ${\lambda}_{i} \neq 0$ that ${\left\| \boldsymbol{u}_{i} \right\|}_{2}^{2} = {\left\| \boldsymbol{v}_{i} \right\|}_{2}^{2}$. Now since their sum is 1 it means they equal to $\frac{1}{2}$.
While I don't find any mistake in it, it seems the requirement I derived isn't enough. any idea?
@Thorgott, Any idea?

8:44 PM
multiplying the first by $u_i^T$ on the left, I get $\lVert u_i\rVert2^+u_i^TAv_i=\lambda_i\lVert u_i\rVert^2$ instead of what you claim and similarly for the other equation. I also don't understand the conclusion after the "since"

I think you found it. Always those small errors.
So we have $\boldsymbol{u}_{i}^{T} A \boldsymbol{v}_{i} = \left( {\lambda}_{i} - 1 \right) {\left\| \boldsymbol{u}_{i} \right\|}_{2}^{2}$ and $\boldsymbol{v}_{i}^{T} {A}^{T} \boldsymbol{u}_{i} = \left( {\lambda}_{i} - 1 \right) {\left\| \boldsymbol{v}_{i} \right\|}_{2}^{2}$.

yeah, I agree

how do you call subset of R3 that is basically a pyramid that extends from apex to infinity?

that's at least more elegant than the ad hoc expression I computed earlier

So the requirement is the matrix $C$ not to have any unit Eigen Vector and then the proof will hold. Am I missing something?

9:24 PM
i think you should exclude $\pm1$, but other than that I'm not seeing anything wrong right now

Why do you say $\pm 1$?

cause $\lambda_i$ is a singular value, not an Eigenvalue

9:43 PM
Actually in the case above I used them as Eigen Values.
Before that we talked about ${C}^{T} C$ then they were Singular Values.

ohh
I overlooked that
yeah, you're right then

Can anyone recommend any lecture notes on spectral sequences they'd say is better than the usual textbook treatments? (PS: if possible more from an algebraic geometry motivation, rather than the super abstract category theory expositions.)

9:59 PM
@JamalS have you tried Hatcher?

10:15 PM
Let's move. let's move through this relationship like i cruise on the safety ship, and i don't know if i can get to outer space with you. but imma say i could and ill play it cool, cause i know that you are the real deal too, and you look so pretty, and i wanna make it through wit you. i wanna make it through wit you. girl.
Ur a 10 outta 10 in my book and nobody else book really matters, to me, cause they don't have the right mindset, and they don't have the right penmanship, and I don't see the right way to phrase it, but you're amazing, girl, let's play wit it
0

What is going on in this image mathematically, and are there some neat underlying mathematical themes relating to this image? I did a little experiment by holding a translucent plastic container to some closed blinds on a bright sunny day. I think the even discretization of the light ...

10:31 PM
how large primes can we find if we let mathematica run for 1 minut? approx
and how large a prime can we deterministically check using the command ProvablePrimeQ during one minut?
it does depend on what computer we are using but if we just want an approximation to both , what would be the answer ?

@JamalS Weibel's homological algebra book does them early so that you actually see applications of them early on.
And Aluffi's Chapter 0 has an okay supplementary section on them.