Mathematics - 2020-02-20

topologicalmagician

00:37

@TedShifrin standard metric, in $\mathbb{R}$.

adesh mishra

04:03

@TedShifrin All right! Means my main doubt arises from using Fundamental Theorem of Calculus when it should not be used blindly, okay!

Semiclassical

04:43

@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)

Alex D

05:15

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

Keyenke

@Alex lets hear it

Alex D

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

Keyenke

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

Krull Dimension

05:37

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

Leaky Nun

U is not a vector space

Krull Dimension

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

Ted Shifrin

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?

Jack Ohara

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

Alex D

05:57

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.

Krull Dimension

06:08

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

Jack Ohara

@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

Krull Dimension

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

Jack Ohara

I see

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

Krull Dimension

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

Jack Ohara

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 ?

Krull Dimension

06:20

Yeah, that's true

I misspoke, you generate a subgroup of order dividing y-1

Jack Ohara

Still not seeing the full picture of the proof

Krull Dimension

06:50

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

Jack Ohara

@KrullDimension thanks and no worries

let me see if that makes sense to me

Jack Ohara

07:15

@KrullDimension works thanks!

Krull Dimension

Glad I could help!

Jack Ohara

:)

northerner

09:46

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?

Peter

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

northerner

@Peter so put simply MS calculator is wrong

adesh mishra

10:08

@Semiclassical Thank you

Royi

12:02

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?

Astyx

12:13

What's Matlab notation and what is the V matrix ?

Royi

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.

Astyx

What is U ? and S ?

Royi

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.

Astyx

So U and V are unitary and S is diagonal ?

Royi

It is an interesting property I haven't seen anywhere.

Yep. We are on real matrices so $ U $ and $ V $ are orthonormal.

Astyx

12:23

Is I the identity matrix ?

Royi

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.

Astyx

C is symmetrical, which means it U = V

Royi

No necessarily.

Astyx

What do you mean ?

Royi

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.

Astyx

12:43

Spectral theorem

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...

Royi

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

Astyx

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

ie U=V in your decomposition

Thorgott

That's not the SVD

Hamzalihi

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

Royi

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} $.

Astyx

12:58

Oh right

I misread what the SVD was my bad

Thorgott

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

Royi

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

Thorgott

I don't really follow your notation

Royi

13:15

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?

Thorgott

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

Royi

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 $.

Huy

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.

Royi

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?

Thorgott

13:56

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

Huy

I never said anything else

Royi

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.

Thorgott

@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.

Royi

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

Thorgott

14:13

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

Royi

14:23

Could you please elaborate?

Andrew

14:50

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

Jack Ohara

15:05

@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

Lukas Heger

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

Jack Ohara

@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

Lukas Heger

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

adesh mishra

15:25

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 $\mathbf E$

Thorgott

@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.

Jack Ohara

15:52

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

Lukas Heger

because the multiplicative group is cyclic

Jack Ohara

so it is a result about cyclic groups

Lukas Heger

yes

Jack Ohara

the way I think about it is like this

Lukas Heger

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

Jack Ohara

15:54

the solutions are those elements that has order dividing y-1

is that the wrong approch ?

Lukas Heger

no

that's fine

Jack Ohara

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

Lukas Heger

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

Jack Ohara

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

Lukas Heger

because the group is cyclic

same result as befoe

gcd(p-1,y-1) divides p-1

Jack Ohara

15:58

ohhh

I see thanks !

Alessandro Codenotti

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?

Edward Evans

16:22

@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

Lukas Heger

@EdwardEvans yes, this is the correct approach

Edward Evans

Cool :) I'll be waving my hands so hard that I might take off

will probably have to weigh myself down

Lukas Heger

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

Edward Evans

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

Lukas Heger

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

Edward Evans

16:38

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

Lukas Heger

nah, Hecke theory is not the right thing for a one hour talk

just do one hour of unmotivated double coset computations

Edward Evans

16:52

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

Huy

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

Lukas Heger

@EdwardEvans derive QR from Artin reciprocity

Edward Evans

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

Lukas Heger

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

Edward Evans

special case where the extensions are abelian right?

Thorgott

16:58

why do they not know what a field is?

Lukas Heger

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

Edward Evans

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

Lukas Heger

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

Edward Evans

coooooool

Lukas Heger

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

Edward Evans

17:01

that's nice

Lukas Heger

@EdwardEvans I plan to sketch some of this stuff in my thesis defense, as a motivation

I'll invite you ofc

Edward Evans

Nice, thanks

I was about to ask if the defense is open lol

Thorgott

oof

Lukas Heger

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

Edward Evans

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

Krull Dimension

17:25

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

Ted Shifrin

17:55

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

Royi

20:11

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?

Thorgott

20:44

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"

Royi

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} $.

Thorgott

yeah, I agree

Stipe Galić

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

Thorgott

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

Royi

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

Thorgott

21:24

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

Royi

Why do you say $ \pm 1 $?

Thorgott

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

Royi

21:43

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

Thorgott

ohh

I overlooked that

yeah, you're right then

JamalS

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.)

Leaky Nun

21:59

@JamalS have you tried Hatcher?

geocalc33

22:15

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

Q: curves on a plastic container using sunlight and blinds

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 ...

Jack Ohara

22:31

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 ?

anakhro

@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.

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