Mathematics - 2019-10-09

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user193319

01:22

Let $f : H \to G$ be a Lie group homomorphism, and $\alpha_{v} : \Bbb{R} \to H$ that unique one parameter subgroup such that $\dot{\alpha}_{v}(0)=v$. How do I compute $\frac{d}{dt} \big|_{t=0} f(\alpha_{v}(t))$?

We have this functor $L$ which I'm not sure how to define. I want to show that $\frac{d}{dt} \big|_{t=0} f(\alpha_{v}(t)) = Lfv$, but I don't see how to compute the derivative...I need some sort of chain rule.

All I know is that $L$ is a map from the Lie algebra $LH$ associated to $H$ to the Lie algebra $LG$ associated to $G$.

2 hours later…

Mike Miller

03:11

the chain rule

it's $df(\alpha'_v(0)) = df_e(v)$. your functor "$L$" sends a Lie group to $T_e G$ and a homomorphism to $df_e$

Hawk

03:59

If a geometer is present, I'd like to know on Riemannian Manifold, we have connection and hence covariant differentiation. Then what is the use of Lie Derivative? As covariant differentiation is more precise.

Sayan Chattopadhyay

04:48

@Hawk Vector fields are very crucial in manifolds. Given one vector field you want to know how another vector field "flows" with respect to the other. Here you need the Lie derivative

Note that you need to define a connection to make sense of the covariant derivative. A lie derivative doesn't need that.

So in some sense it's natural on a manifold

Hawk

05:13

@SayanChattopadhyay and covariant derivative does not tell you how vector field flow wrt another? I am just asking because on a Riemannian manifold the covariant derivative seems to completely replace Lie derivative.

Sayan Chattopadhyay

05:37

@Hawk they encode different sets of information(Though there's a formulae relating the two). For instance consider the fact that any connection compatible with the metric should satisfy $\Del_{X}g = 0$. This means that tangent vectors along X a flowing in a manner such that the metric on them remains the same. Whereas there are tons and tons of vector fields such that $L_{X}g \neq 0$. If it were it would mean that it's a killing vector field and generates an isometry.

4 hours later…

Alessandro Codenotti

09:25

Suppose that $f$ is a measurable function on $\Bbb R$, must there be a Borel function $g$ that differs from $f$ on a measure $0$ set?

MatheinBoulomenos

10:05

@AlessandroCodenotti write $f$ as a pointwise limit of simple functions and use that every Lebesgue measurable set is a union of a Borel set and a null set. I think this should work out

Alessandro Codenotti

10:19

I think that works, I'll write down the details

Sayan Chattopadhyay

11:16

Howdy@BalarkaSen

Balarka Sen

Hi @Sayan

Sayan Chattopadhyay

Came back home?

Balarka Sen

For the weekend, yeah. I'm back in ISI now

Sayan Chattopadhyay

11:34

Cool. Is this your second year at ISI?

Balarka Sen

Yeah

Sayan Chattopadhyay

Great. Do you have any plan for masters?

Mike Miller

you think too far ahead

Rithaniel

Having a plan for the future is always good

Sayan Chattopadhyay

@MikeMiller lol yeah, I have to plan for stuff given that I do not have much economic stability

Rithaniel

11:47

(Even if it doesn't come to fruition)

Mike Miller

@SayanChattopadhyay fair enough, though to a degree it's not clear to me what planning on your soph year impacts

Balarka Sen

@SayanChattopadhyay No clue man

Sayan Chattopadhyay

Yeah, I guess, just some convo. Though I am not in soph year so I thought of asking that question

Cool@BalarkaSen

skullpetrol

12:06

those that do not plan ahead get pushed aside by those that do have a plan

that is the nature of competition

aggressive as it sounds

Rithaniel

A fair dire outlook: kill or be killed, eh?

skullpetrol

Darwinism at its finest :P

Mike Miller

12:40

perhaps not a consistent outlook with spending your time on a math chatroom

Rithaniel

The chatroom is part of the plan

Clearly, if you master the chatroom then you master the world

5

skullpetrol

also there are worse inconsistencies in life...

Astyx

Hi chat

skullpetrol

welcome

skullpetrol

13:02

@Rithaniel you don't need to be "killed," just being pushed around all your life

or at least until you realize how to push back

Balarka Sen

@Rithaniel Clearly!

You get pushed around all your life until you realize how to pullback

2

please clap

skullpetrol

::clap::

Astyx

Ok, but just once

skullpetrol

the sound of a one hand clap?

Balarka Sen

Drawing fiber diagrams obviously takes you higher up the status hierarchy

Secret

13:16

push forward in life, pull back when the commute breaks down

Ted is the master of this chatroom

Astyx

13:58

@BalarkaSen You get pushed forward rather hmmm ?

skullpetrol

are you taking any classes for advanced students @BalarkaSen

Astyx

What's an advanced student ?

skullpetrol

honours classes

Secret

14:34

Why do prime numbers make these spirals?

This actually raise a much deeper question about exploring mathematics and the connection between formal mathematics and recreational mathematics:

> Given a mode of presentation $P$ of a problem $q$ with deep theorems $d$, which has artefacts $A$ such that there is some map $f$ where $f(A)=d$, how to find such $P, A$

In this particular case, it just happens the geometric artifact of the topology of a circle allows the residual classes, which is part of dirichlet's theorem to be singled out because they force the mathematical concidence $\frac{a}{b}\approx n\pi$

Likewise, we saw very similar things happening in $\Bbb{R}^2$ such as how certain curves can compute transcendentals while others cannot, and how certain family of lines can be used to prove a cantor diagonal theorem of the cardinality of $\Bbb{R}^2$

SilenceOnTheWire

Hi

Secret

The very act of visualising mathematical problems, actually imposes a constraint $P$ onto the problem $q$ such that $P(q)$ contains artefacts which contains parts of the underlying deep theorem

SilenceOnTheWire

Can I get some help with homework?

It's linear algebra

Astyx

Ask away

I don't guarantee an answer though

SilenceOnTheWire

Okay

Secret

14:40

But other than rely solely on neural networks, is there a efficient way to find those $P,A$ given problems $q$?

SilenceOnTheWire

Consider the sets A:={1,2} and B:={a,b,c}. Do AxB.

I think it's {a+b+c,2a+2b+2c} is this correct?

Astyx

Is that the cartesian product of the sets A and B ?

SilenceOnTheWire

Yes

Astyx

Then that's not it

SilenceOnTheWire

How do I do it?

Astyx

14:44

AxB is the set of the (x,y) where x is an element of A and y an element of B

for instance (1,c)

SilenceOnTheWire

Actually I am not sure if it's the cartesian product. That's all it says

Astyx

Does what I said help ?

SilenceOnTheWire

Not really

Rithaniel

Is that the full statement of the problem? "Do AxB" isn't even a full sentence

SilenceOnTheWire

"Determine A x B."

It's all it says.

Rithaniel

14:57

Alright, that's more clear, at least. You've probably been given a description of what x is supposed to represent, but, without that, the best we can give you is what we would suspect x to represent

SilenceOnTheWire

x is the multiplication sign. Sorry.

*

Astyx

How is it defined in your course ?

SilenceOnTheWire

A and B?

Rithaniel

Which I would say, in agreement with Astyx, is the cartesian product. After all, both A and B are sets and, if they were vectors, they are of mismatched dimensions and so multiplication doesn't make sense.

Unless of course A is meant to be a column vector, in which case AxB would be a 2x1 matrix multiplied by a 1x3 matrix

But if they are expressed as sets, then I'd still favor the cartesian product

Astyx

hmm, considering it's meant to be linear algebra that's probably it

Mats Granvik

15:07

The answer to all your math problems is chia seeds. Just remember to soak them in plenty of water over night.

Rithaniel

Tangentially, if that's what is wanted, the entries of the resulting matrix correspond quite clearly to the ordered pairs of the cartesian product

Mats Granvik

Birds of a feather flock together.

Secret

My habit:

Give me a maths video

And I will create a problem so meta it blow up people's mind

41

Q: Are there mathematical objects that have been proved to exist but cannot be described in words?

This might be a very stupid, and possibly philosophical question, but attempt to apply mathematics to everything plus inspired by this question caused me to ask this question Is there any mathematical object that has been proved to exist but cannot be described in words? If the answer is...

And yes I have done that before

3 hours later…

Ultradark

17:46

The logarithm function is a multiple-valued function when it takes complex arguments, so you'll have to be careful and define what you mean by logx in this case. I'm not even sure that well-defined exponential and logarithm functions exist when you go to the quaternions and octonion

0

Q: Spatial network whose vertices are actually graphs when viewed at the proper distance?

Let's say you have a spatial network, and when you zoom in on one of the vertices, you actually find another spatial network, and when you zoom in on one of the vertices of that spatial network you find another spatial network, and so on. I'm not sure how to capture the notion of "zooming in." I...

graph-theory reference-request soft-question fractals network

Thoughts?

Semiclassical

18:17

@Ultradark Do you know what the adjacency matrix of a graph is?

for an undirected graph, it’s a matrix whose i,j entry counts the edges between vertices i and j (if there are any)

If you apply that idea to your networks, then what you’d get is a block matrix—that is, a matrix whose elements are themselves matrices

You can do multiple levels of this nesting, and I know I’ve seen that discussed in regards to block matrices

So it seems plausible that that sort of hierarchical view of networks has been explored

Mary Star

18:31

Hello!! Is someone of you familiar with Gödel number of Turing machines?

1 hour later…

Shine On You Crazy Diamond

19:46

Brain teaser for you all:

0

Q: Computing magmas for the smallest grammar problem.

Background: Wikipedia: the smallest grammar problem Now imagine there were a space representing all ways you could subdivide an arbitrary strings $S = abcdefg\dots z$ (not necc. 26 letters), using parentheses, and for the sake of computing smallest grammars. So for instance splitting $(ab)$ fur...

formal-languages quotient-spaces context-free-grammar discrete-optimization magma

If you don't understand smallest grammar it's easy

$\text{SG}(a^6) = \{S \to AA, A \to aaa\}$ And there's another one with instead $2$ a's.

But you're only required to compute one of them to be an exact smallest grammar algorithm, by definition.

I think though that most algorithms that are making use of symmetries will have a way built-in to enumerate all of them.

The math is more elegant when you include all of them. The cartesion product to "any one of these" has no natural standardization.

dondeman

20:03

Is it possible to link images here?

I have a quick math question :) Thought it would be a waste if I created a post

dondeman

20:20

imgur.com/a/JxBCplr is it correct that this matrix system is consistent for all values of a, b and c?

ÍgjøgnumMeg

20:40

@Mathein @Alessandro alright nobody told me about the knocking on the tables thing

Semiclassical

Yes, seems right. I think you only have to go as far as your third matrix to check consistency tho

At that point, it should be obvious that there’s no way the rows are linearly dependent

Alessandro Codenotti

@ÍgjøgnumMeg Lmao it caught me by surprise too last year

Ted Shifrin

@dondeman: As long as you have no rows of zeroes in the echelon form, it will be consistent for every right-hand side vector.

hi @ÍgjøgnumMeg, @Semiclassic, demonic @Alessandro

Balarka Sen

Hi @Ted

Alessandro Codenotti

Hi @Ted

dondeman

20:42

@TedShifrin Thank you very much

ÍgjøgnumMeg

@Alessandro we had the first Iwasawa theory talk today, at the end I'm getting my hands ready to clap and everyone just starts hitting the table, felt so cultish hahaha

Ted Shifrin

oh, hi a @Balarka

ÍgjøgnumMeg

Hey @Ted @Balarka @Alessandro @@@@@@@@@@@@@

Balarka Sen

Hi @ÍgjøgnumMeg

Lol they knock their hands on the table as applause?

ÍgjøgnumMeg

@Balarka yeah! Well with their knuckles, not just brutally smacking the table with their palms

Balarka Sen

20:43

Loool

so weird

ÍgjøgnumMeg

Very funny, I literally opened my hands to clap and that happened, so I awkwardly just went along with it

Ted Shifrin

I wonder if @Rithaniel ever thought about his group theory question.

Alessandro Codenotti

For some reasons that's the German version of clapping in an academic context

ÍgjøgnumMeg

@Alessandro apparently it's done after business meetings etc.

schn

Given a transformation $F$ that projects onto $\text{im} \ F$ along $\text{ker} \ F$, to find the equation of the plane that is projected onto (i.e. $\text{im} \ F$), one can find the kernel of $\text{ker} \ F-I$, since this is equal to $\text{ker} \ I-F = \text{im} \ F$. However, if given the matrix of $F$, would this be equivalent to determines the number of column vectors of the matrix that are linearly independent and write them into a parametric equation for the plane?

Rithaniel

20:47

Not yet, unfortunately. Spent a total of nine hours on campus already and I've gotta leave in 30 minutes to go up and take a test.

Alessandro Codenotti

@ÍgjøgnumMeg Ah I see

Ted Shifrin

Poor @Rithaniel :(

Rithaniel

In my one hour of free time, I've been watching dumb YT videos to refresh my brain

Ted Shifrin

@schn: Do you know this is literally a projection linear map? Or are you using that word?

Rithaniel

Nah, I'm sure you guys have had rougher semesters.

Ted Shifrin

20:48

Not counting cancer, I only made life rough on my students :)

ÍgjøgnumMeg

heavy

Alessandro Codenotti

They do that at the end of every lecture, it's kinda weird but you get used to it @ÍgjøgnumMeg

ÍgjøgnumMeg

@Alessandro fair enough hahaha

at least the talk was super interesting

schn

@TedShifrin What do you mean by 'projection linear map'?

Ted Shifrin

You used the word projects. Tell me what that means.

Alessandro Codenotti

20:49

Yesterday I discovered that my seminar talk won't be at the beginning of December as I thought but in 20 days instead lol

Rithaniel

Well, I know that grading assignments can be just as difficult as filling out assignment

ÍgjøgnumMeg

@Alessandro oops

Ted Shifrin

I spent 40+ years grading a lot, @Rithaniel, but I considered it an important part of my job, even though most professors don't grade much (especially in Europe).

I graded homeworks and exams for all my upper-level courses.

Exams for everything, of course.

@Alessandro: You're experienced. You'll do fine.

Rithaniel

It seems the norm at my university that first and second semester grad students do the grading for a lot of the professors.

Balarka Sen

There's a guy from my university who joined Bonn this year I believe

Alessandro Codenotti

20:51

I talked with the professor supervising the seminar, should be doable to prepare it in a couple of weeks, I just need to go through this short set theory paper

Ted Shifrin

Just what you've always wanted! (Yuck.) :D

Alessandro Codenotti

@BalarkaSen I haven't met the new students yet

schn

@TedShifrin Probably meant projection.

Alessandro Codenotti

@TedShifrin Pfff it's actually very nice, it's about the independence from ZFC of a strong version of Fubini's theorem

Ted Shifrin

Strong how?

I'm asking you to explain, @schn.

dondeman

20:54

imgur.com/a/At8S7aN, How do I find all values for P so that the matrix is invertible

Thought that I could calculate the Det(B)

Which gives me $p^3 + 5p^2 - 2p - 8$

Then I can solve it for $p^3+5p^2−2p−8 = 0$, however don't I have to use the Rhapson method for solving this?

Ted Shifrin

You need to learn how to put actual questions here without graphics, @dondeman. Anyhow, you need to find roots.

Can you find an integer root by inspection?

anakhro

Hi all.

Ted Shifrin

Hi @anakhro.

Alessandro Codenotti

Suppose $F\colon \Bbb R\times\Bbb R\to[0,\infty)$ is such that for almost all $x$, $F_x$ and $F^x$ are measurable and such that the functions $x\mapsto\int F_x dx$ and $y\mapsto \int F^ydy$ are measurable. Must the two iterated integrals be equal?

The main thing is that it doesn't ask for $F$ to be measurable

Leaky Nun

@AlessandroCodenotti hey that sounds like something I asked you a few months ago :P

Ted Shifrin

20:57

I don't know what $F_x$ is, as opposed to $F^x$.

Alessandro Codenotti

$F_x$ is $F$ with the first coordinate fixed to $x$, $F^x$ is fixing the second one

dondeman

@TedShifrin, by using the Newton Method?

Leaky Nun

4

Q: A counterexample in measure theory on $\sigma$-infinite spaces

Usually measure theory books include the following theorem (citing Proposition 5.1.3 in Cohn's measure theory book) Let $(X, \mathcal A , \mu )$ and $(Y, \mathcal B, \nu )$ be $\sigma$-finite measure spaces. If $E$ belongs to the $\sigma$-algebra $\mathcal{A}\times\mathcal{B}$, then the funct...

analysis measure-theory examples-counterexamples

anakhro

A painful notational convention.

Ted Shifrin

Blah @noation, Alessandro. OK.

@dondeman: No, do you know what integers can possibly be roots of a polynomial with integer coefficients?

Alessandro Codenotti

20:59

@LeakyNun It's kinda similar

schn

@TedShifrin If given a vector space $V$ that is the direct sum of $U'$ and $U''$, and given $\textbf{u} \in V$, then the unique vectors $\textbf{u'}$ and $\textbf{u''}$ such that $\textbf{u}=\textbf{u'}+\textbf{u''}$ are called the projections of $\textbf{u}$ on $U'$ and $U''$ respectively.

dondeman

@TedShifrin I am sorry, English is not my primary language. I do not seem to understand your question

Ted Shifrin

So the iterated integrals in different orders can be equal without having the double integral exist? @Alessandro

Alessandro Codenotti

Anyway it's easy to construct models of ZFC in which that theorem fails (any model of CH will do), my talk will be about this result of Friedman in which he constructed a model where the theorem holds

anakhro

Why do non-identity elements of a cyclic group generate the same cyclic subgroup?

Am I being an idiot and this is obvious for some reason?

Ted Shifrin

21:01

OK, @schn, so nothing about orthogonality.

@dondeman: Look up Rational Root Theorem.

@anakhro: That's very false unless the order is prime.

anakhro

Yeah the order here is prime minus 1.

schn

@TedShifrin If say $U''$ is the orthogonal complement to $U'$, then it's an orthogonal projection.

Ted Shifrin

@schn: So you're looking for equations for the image, i.e., the subspace.

anakhro

So what are they meaning........hmmmm

schn

@TedShifrin Yes.

Ted Shifrin

21:03

OK, @schn. With orthogonality, it would be easier. So you're asking how to find equations for a subspace if you know a basis for the subspace?

ÍgjøgnumMeg

@Alessandro for basically the whole talk there was a restriction to all primes $p < 163 \times 10^6$

which was somewhat humorous

schn

No, the basis isn't given. Only the matrix of the projection linear map.

Alessandro Codenotti

@ÍgjøgnumMeg Is that as far as computers got or?

Ted Shifrin

Yes, but you can always find a basis for the column space of the matrix.

dondeman

@TedShifrin Would that be 1,2,4,8 for 8?

ÍgjøgnumMeg

21:05

@Alessandro yeah, some guys did some calculations and that's as far as they went I guess

Ted Shifrin

$\pm$ @dondeman

dondeman

Ah, and for p^3 It is only 1?

Ted Shifrin

@schn: And you can find the kernel of the matrix, too.

Alessandro Codenotti

@ÍgjøgnumMeg Makes sense

Ted Shifrin

@dondeman: No, it's only $0$ !!

ÍgjøgnumMeg

21:06

it was just kinda funny "let $p$ be an odd prime less than 163 million"

Alessandro Codenotti

I suppose it's conjectured to hold for all $p$, whatever it was?

Ted Shifrin

So @schn you had the right idea, I think. A vector is in the image of $F$ if and only if that vector is in the kernel of $I-F$.

schn

@TedShifrin So if one finds the number of linear independent columns of the matrix, one could deduce the equation of the image of the transformation?

dondeman

@TedShifrin wouldn´t that make p/q impossible?

schn

Right

dondeman

21:08

(1,2,4,8)/0?

Ted Shifrin

No, just knowing dimensions doesn't tell you the image, just its dimension.

Balarka Sen

Linear optimization is A+ stuff

Ted Shifrin

No, @dondeman. I don't know what you're reading. Here $p$ is a factor of the constant term and $q$ is a factor of $1$, the coefficient of $p^3$. So you're dividing $\pm (1,2,4,8)$ by $1$.

@schn: You have the answer from what you wrote (and what I wrote just up there).

ÍgjøgnumMeg

@Alessandro I think the relevant conjecture was something about the $p$ primary part of the class group of some number field was always elementary abelian

and for every prime less than that bound it's true

dondeman

So all the possible p's so that the equation equals 0 is $1, 2, 4, 8, -1, -2, -4 , -8$?

Ted Shifrin

21:11

Possible rational roots only.

But if you find one obvious rational (integer) root $r$, then you can divide by $p-r$ and solve a quadratic.

schn

@TedShifrin Okay, so finding the kernel of $I-F$ is equivalent to finding the independent column vectors of the matrix (since they span the image)?

dondeman

Oh so! I have to try each value to check if it equals zero

Got it

Ted Shifrin

Yes, but you have more information than that. What does it mean to give equations for a subspace, @schn?

@dondeman: I was able to see one obvious one doing arithmetic in my head :)

schn

@TedShifrin In which case do I have more information?

Ted Shifrin

In your problem. Answer my question. What does it mean "to give equations for" a subspace $V\subset \Bbb R^n$?

As opposed to giving a basis for it ...

schn

21:17

Okay. One could determine if a point belongs to the plane?

dondeman

@TedShifrin None of these values inserted into the equation equal zero? Or am I doing something wrong?

Ted Shifrin

How is that "giving equations"?

Hmm, @dondeman. Maybe I screwed up. What about $-1$?

Oh, you're right. My apologies.

dondeman

Gives -1

-2*

Ted Shifrin

Sorry ... my fault.

OK, so have you double-checked your computation of the determinant?

dondeman

No problem, just curious as to how I solve it

I will double check once again

Ted Shifrin

21:20

I just did it. You should see a factorization immediately from computing that.

Your polynomial is wrong.

schn

@TedShifrin If $U'$ is given the equation $x_1+x_2+x_3=0$, that equation determines if a point belongs to the plane and the normal vector to the plane.

dondeman

Hmm is it?

Will try again

Ted Shifrin

$(p+2)(p^3+3p-4) = (p+2)(p+4)(p-1)$.

@schn: OK. So how do you find such an equation (or several equations) for your question?

dondeman

Yeah, I messed up the polynomial

Sorry

Ted Shifrin

Best thing to do is to try to keep things factored in case you can combine terms in a smart way.

dondeman

21:25

Thank you very much for your help

Ted Shifrin

@schn: I'll give you a big hint. Giving equations for a subspace is expressing the subspace as the set of solutions to $Ax=0$ for some matrix $A$.

You're welcome, @dondeman. And that stuff about integer roots may come in handy later on :)

dondeman

Yeah, that was really cool stuff

Have never been taught that at school which is quite weird

Ted Shifrin

I always taught my linear algebra students this when I got to eigenvalues/eigenvectors. But it should be in high school math.

dondeman

We are at eigenvalues right now

Ted Shifrin

So you will probably find it useful with homework on $3\times 3$ integer matrices.

schn

21:31

@TedShifrin Right, if one knows the linear independent vectors that span the subspace, one also has a parametric system of equations, which one could solve for a single equation.

Ted Shifrin

No, not a single equation, in general. It depends on dimensions.

dondeman

@TedShifrin Yeah, thanks for mentioning it

Ted Shifrin

But think carefully about what I just typed and what you and I both typed a while ago. @schn

You can tell your friends in the class, @dondeman :)

dondeman

Will do!

schn

@TedShifrin Every plane can be represented by an equation of the form $ax+by+cz+d=0$, where $x,y,z$ are the components of the coordinate vectors in 3-space. Isn't this equation derived from a system of equations, where each coordinate $x,y,z$ is expressed in terms of parameters $r,t \in \textbf{R}$?

dondeman

21:37

Last question, will write this here and not as a graphic question. Given a set of linearly independent vectors $ S = (v1, v2, v3, v4, v5) $ and u1 to u5 is given as $u1 = v1$, $u2 = -v1 + v2 $ , $u3 = v1 - v2 + v3$, $u4 = -v1 + v2 -v3 + v4$ , $u5 = v1 - v2 + v3 - v4 + v5$. How can I show that $ T = (u1, u2, u3, u4 , u5) $ is a set of linearly independent vectors as well?

My first thought was the zero vector, but I guess that is not relevant here

schn

You're right, it depends on dimensions. A subspace in $\textbf{R}^4$ spanned by two vectors may not be a plane...

Orb

22:03

Hello, fellow mathematicians!

I am here to ask a question about infinite-dimensional polar coordinates.

hi loch

helo\

Corellian

22:52

If $f$ is odd on some real interval centered at $0$ and we're given the Fourier sine series for $f$, then integrating the series term by term will yield the cosine series for an antiderivative of $f$ on that interval (assuming the function is polite enough for all this)

But how does the initial term $a_0$ come about :thonk: Indeed, ignoring it and just integrating the sines will leave the series function off by that much

Ted Shifrin

@dondeman: You have to start with the obligatory sentence: "Suppose $c_1u_1+\dots+c_5u_5=0$. I must show that $c_1=\dots=c_5=0$." You might find some of my lectures on YouTube helpful. For this, see lecture 35.

Corellian

Hi @ted

Ted Shifrin

Any antiderivative has a $+C$ appended to it, @Corellian.

So you'll get the average value of your function (antiderivative) appearing as the $+C$.

Corellian

Good point

Ted Shifrin

Does that set your mind at ease? :)

Corellian

23:04

hmm

I'm given that $a_0$ is the average value by definition

Ted Shifrin

Yes, so it's going to depend on your particular even function ...

Corellian

although in this case with the antidifferentiation I'm wondering why it's determined that way

Ted Shifrin

It's not determined by antidifferentiation.

Corellian

oh

Ted Shifrin

You're getting the general odd function as a sine series and the general even function as a cosine series. The particular coefficients depend, of course, on the particular function.

Corellian

23:09

That makes more sense

Ted Shifrin

OK.

Corellian

Simple example: if $x \;=\; \sum b_n\sin u(x)$ then we can get the cosine series for $x^2$ by integration.

So $x^2 \;=\; a_0 + 2\,b_n\sum\int\sin u(x)\,dx$

Ted Shifrin

Huh? So, you need to pay attention to details. What is $u(x)$? This isn't right. And what interval are you using?

Corellian

a real interval centered at $0$ with length (say) $2L$

so then $u(x) = \dfrac{2\pi x}{L}$

Ted Shifrin

No, no, no. You need $n$'s.

Corellian

23:15

oops right brain fart, $\dfrac{n\pi x}{L}$

Ted Shifrin

Sit down and do everything carefully.

I've got other things to do right now.

Corellian

sure np, thanks

DarkRunner

Hi, I need help with a problem

I've been fiddling with it for the past three weeks, but I just can't figure it out

It's basically a system of equations:

$\frac { 1 }{ 4 } =UI(sec\theta -DE)$

$(\frac { 1 }{ 2 } -{ tan }^{ 2 }\theta )=DE(sec\theta -UI)$

I seek to express $DE$ and $UI$ in terms of $\theta$

It would be great if someone could help out; I've tried various methods, but to no avail.

William Sun

23:48

Quick question for the map from a ring $A$ to its prime localization $A_p$ defined by $a\mapsto a/1$ why does the maximal ideal of $A_p$ have preimage p?

I can’t see that immediately although I know I can prove it using the fact that the image of $A-p$ are all units in $A_p$.

Sorry about the notation above: $p$ denotes a prime ideal in $A$.

Konformist Liberal

Spaces $Spec (S^{-1}R)$ and $\{p \in Spec R | p \cap S = \emptyset \}$ are homeomorphic.

@WilliamSun (Here S is a multiplicative subset of R)

I am trying to find the derivative of the function $f(x) = x / {\|x\|_2}$ defined from non-zero real numbers.

William Sun

Thank you for the reply. Yes that makes sense to me. I came up with this question after reading a proof in Atiyah’s Intro to commutative algebra. It goes as follows:

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