Mathematics - 2022-05-20

12:39 AM

Evaluating the Taylor's polynomial of third degree with center in 0 of the inverse of f:R->R defined by f(x)=x+sinh(x), I found out that it is P(y)=y/2-y^3/(3!*8)+o(y^3), however my textbook says that it is P(y)=y/2-y^3/(3!*16)+o(y^3). I checked my calculations a lot of times and I can't find that 1/2 factor that is missing in the y^3 term, can someone tell me if the textbook is wrong or am I wrong because I am going insane?)

leslie townes

1:14 AM

i get /(3!*4), so now we have 3 answers.

hooray :)

Sonozaki

1:36 AM

Hahaha that made me laugh, it was unexpected, thanks for this and the help leslie)

robjohn

2:01 AM

@Sonozaki Do you know what the inverse function of $x+ax^3$ is to order $x^3$?

Try $\left(x-ax^3\right)+a\left(x-ax^3\right)^3$

or $\left(x+ax^3\right)-a\left(x+ax^3\right)^3$

Akiva Weinberger

2:29 AM

@Sonozaki Checking it numerically, your textbook is right

leslie townes

booo

you could have voted for me, akiva. i'll remember this.

Akiva Weinberger

I think the key is an extra factor of $1/(1+\cosh(0))=1/2$? @Sonozaki

from a chain rule somewhere

I have $x=y+\sinh y$, $~dx/dy=1/(1+\cosh y)$, $~d^2x/dy^2=-\sinh y/(1+\cosh y)^3$

and that last one near $(x,y)=(0,0)$ is around $-y/8$ or $-x/16$

meaning $d^3x/dy^3$ at $x=0$ is $-1/8\cdot dx/dy=-1/16$

Wait

I meant $dy/dx$ etc for all of those

Sorry I was doing the logic with the letters reversed and I decided to switch them at the last moment

but didn't switch them fully

*I have $x=y+\sinh y$, $~dy/dx=1/(dx/dy)=1/(1+\cosh y)$, $~d^2y/dx^2=d/dx(dy/dx)=d/dy(dy/dx)\cdot dy/dx=-\sinh y/(1+\cosh y)^3$

robjohn

3:29 AM

$f(2x)=x+\sinh(x)=2x+\frac{x^3}6$

$f(x)=x+\frac{x^3}{48}$

$f^{-1}(x)=x-\frac{x^3}{48}$

$f^{-1}(x)/2=\frac{x}2-\frac{x^3}{96}$

bumblebee

10:43 AM

0

Q: What are some nice topics in undergraduate mathematics to write about?

As a high school student, I will soon be going on to do the IB(college equivalent). As part of the IB curriculum we are expected to write an Extended Essay(EE) that counts towards our grade. For the EE we are expected to motive a question on a subject, or particular topic, and write a paper answe...

reference-request

Any suggestions would be great!

Calvin Khor

10:59 AM

@bumblebee "what hill would a ball roll down fastest from"?

Prithu biswas leftmse

@bumblebee Axiom of choice.

bumblebee

11:18 AM

@CalvinKhor Sorry but I can't go with that. Malpractice.

@Prithubiswasleftmse interesting...let me think about.

Prithu biswas leftmse

@bumblebee Maybe that topic is a bit too boring, so I take back my suggestion =(

Calvin Khor

11:36 AM

@bumblebee lol what do you mean malpractice

@bumblebee it was a serious suggestion lol and the answer may surprise you but I don't know good references

linear_combinatori_probabi

Hola chat

I'm trying to re-learn Linear Algebra. A quick question:

Say there is a point denoted as $(3,2)$. It can represent a vector, where its tail is at the plane's origin. (If the standard basis is used, then it can represent "the only" corresponding vector on the plane.) So far so good. Now the video I'm watching said that: So the coordinate $(3, 2)$ can be seen as scalars that multiply the basis, and the resulting vector is the result of adding all the vectors together. E.g. if standard basis is used then $(3, 2)=(1,0)\cdot3 + (0,1)\cdot2$.

So, I would like to know more about the part I bolded, i.e. "adding all the vectors together". What does it imply?

By intuition, all components(?not sure about the correct term, for a component I mean either $3$ or $2$ of $(3,2)$) of a given vector are mutually independent

They're independent, but we can add them. Sounds like a contradiction.

@robjohn: Help me...

Sorry for the ping, if it's annoying...

robjohn

12:10 PM

@linear_combinatori_probabi $\vec{x}$ and $\vec{y}$ are dependent if there are constants $a$ and $b$, not both $0$, so that $a\vec{x}+b\vec{y}=\vec{0}$.

They are independent otherwise.

I see no contradiction by saying that if two vectors are independent, we can add them.

linear_combinatori_probabi

Thanks for your explanation. I need some time to think about this.

I was implementing a machine learning model, and each input of the model can be represented by a bunch of vectors. More specifically, these vectors are column vectors. Thus, each batch is now represented by a matrix.

Let me pick an example, possibly nonsense, but anyway:

Say the dataset is about cats. A cat has many features: height, weight, color of skin, etc.

So we can use a vector $(1,0,2,3,...)$ to represent all the features of a cat, given that each category is mapped to a non-negative number for each feature.

For me, it's no meaning to "adding all the vectors together", when I was considering my statement above.

Which leads to my confusion.

robjohn

Say you want to compute the "average cat". Then you would add all the vectors for all the cats together, then divide that vector by the number of cats.

linear_combinatori_probabi

12:28 PM

That's reasonable. But how about each vector itself when it can be represented as "adding all the (coordinate) vectors (each scaled by a scalar)"?

robjohn

$3\begin{bmatrix}1\\0\end{bmatrix}+5\begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix}3\\5\end{bmatrix}$

linear_combinatori_probabi

Sorry that my cat example might be not easy to generate some examples. This question might be a bad question too.

But this kind of confusion always there in my head when I was learning math...

graffe

could someone cleverer than me explain why all the mice can't just be unrelated ibb.co/Yh8tx88 ?

in other words, how does this puzzle make sense?

linear_combinatori_probabi

So, by adding $\begin{bmatrix}1\\0\end{bmatrix}$ and $\begin{bmatrix}0\\1\end{bmatrix}$, independent things/objects are added. While the result is "one vector", it doesn't actually combine anything. It just "organize" indepent things decently.

robjohn

@linear_combinatori_probabi what do you mean by "combine"? it looks like there are two vectors being combined here.

linear_combinatori_probabi

12:42 PM

a vector of "A cat has two ears", adding one "A cat has four legs", resulting in one vector "A cat has two ears and four legs". But it's strange to use the word "add" on legs and ears

graffe

is it just really obvious?

robjohn

@linear_combinatori_probabi you are confusing the vector with the interpretation of the data represented by the vector. A vector is simply a collection of numbers, interpretation can be given to those numbers, but a vector does not have anything to do with any of those interpretations.

Just because I say that the pressure is 14.7 psi, does not imbue the number 14.7 with any given units. I can add 14.7 to 4.2 as they are just numbers.

graffe

I think I worked it out

linear_combinatori_probabi

Yes, reality has restricted my imagination.

robjohn

@graffe Is there some assumption not given in that passage?

linear_combinatori_probabi

12:47 PM

@graffe Sorry for ignoring your question, let me take a look now

It looks like a question that intended to be solved pigeon-hole theorem

@graffe I got exactly same question as you pointed out

I think the word "colony" should resolve this confusion(? or not)

Wow, this question is actually interesting to me.

$a_1\dots a_{n+1}$ has provided ${n+1\choose 2}-n$-many mice satisfying (a).

graffe

1:04 PM

@robjohn that's the whole question

@linear_combinatori_probabi thanks!

linear_combinatori_probabi

OK, I need to take out the trash first, then I will be back to try solving this question

graffe

@linear_combinatori_probabi cool

bumblebee

1:16 PM

@CalvinKhor I have seen the animations of balls moving down different shaped lines before. Sorry, I didn't mean to say malpractice. I've just been warned against getting research questions from external parties. The person who warned me called it malpractice.

@Prithubiswasleftmse DON'T TAKE IT BACK! I don't think it's boring, I just mean't that I wanted a few more options to consider before choosing! This is the problem with chatting online, it's hard to judge tone.

Akiva Weinberger

1:45 PM

Observation: both 4'20" (aka 5'8") and 69' (aka 5'9") are relatively normal heights for a person to be

@bumblebee Perhaps discrete differential geometry

youtube.com/playlist?list=PL9_jI1bdZmz0hIrNCMQW1YmZysAiIYSSS

^a good source

Discrete Differential Geometry - Welcome Video

►

Questions like: how to define and measure things like mean curvature and Gaussian curvature of polyhedra?

bumblebee

Thanks Akiva. I'll take a look

Akiva Weinberger

I suppose it's helpful if you learn what mean curvature and Gaussian curvature are for smooth surfaces first

I don't remember if it's in the series or not (probably is)

This one probably goes into it

Lecture 15: Curvature of Surfaces (Discrete Differential Geometry)

►

Nico Tripeny

@bumblebee here are some comments/ideas. Hope you don't mind the inundation 1) RSA and public key cryptography. Not super original here but an interesting topic with a lot of resources. 2) square the circle/double the cube/ Galois theory. Some really old and famous questions proved ~100 yeah ago that are interesting.

I have like 8 more but they length capped me. Let me know if you want them

bumblebee

Please do share Nico!

Nico Tripeny

3) Knot coloring and other simple knot/link invariants. This is a fun topic with results you can definitely grasp but that are still meaningful (I.e. taught in medium level college courses). 4) Fibonacci numbers, identities, and different definitions. Again not original but interesting and with a lot of room to grow as a mathematician. There are tons of definitions of the Fibonacci numbers that are different and allow you to see them in a new light+ tons of simple theorems.

5) same as the last but Catalan numbers. Probably more creative and just as interesting with easy resources. 6) Basics of Combinatorial Game Theory. If you don’t know what that is, that’s totally normal. Look it up on Wikipedia and see what you think. There are a lot of ways to take it, Trina deep dive into a game like chess to a general overview of a certain type of game. If I had done one it would have probably been on something like this.

7) Poker/games of chance. Combinatorial game theory is games of no chance. Have a game of chance you like? Look into the strategy. 8) non-Euclidean geometry. Yeah not creative as a topic but again a lot to explore+ history. Could focus on Euclid’s fifth postulate or something like that. 9) Bertrands paradox. Now been on numberphile twice so might not be original. If you have any interest in probability this could be interesting.

10) Russell’s paradox and foundations of math. Could look at different “solutions” to the problem (I. E. Intuitionism, ZFC, other axiomatic systems…). As a note, axiom of choice is definitely a good choice too, but may not be quite as motivated for a high schooler. Just depends on how much experience you have

Akiva Weinberger

1:52 PM

I can also recommend knot theory

Knot Knotes is a good place to learn about it

bumblebee

Thank so much you guys! ☺️

Akiva Weinberger

Suppose we have a diagram of a knot or a link (a loop or multiple loops), colored in three colors (above: blue, orange, and white) such that at every crossing, there are either three colors or only one color. As a knot or link moves in space, its diagram (projection onto a plane) can change. How does that sort of transformation affect its coloring?

If two knots have a different number of ways they can be colored (with the above rule), can they be the same knot?

@bumblebee Hilbert's third problem is also a fun topic. I think Numberphile's done a video on it? Basically: it turns out that of I give you any two polygons with the same area, you can cut one into pieces and rearrange it to get the second polygon. (They are called "scissors-equivalent".) Is the same true in 3D? Can, for example, a cube and a regular tetrahedron of the same volume be cut and rearranged into each other? Turns out, no! The proof is pretty fun.

Another thing (you're gonna be drowning in topics, sorry) is Niven's one-page proof that pi is irrational. It's a very condensed proof, and an expanded explainer could be nice. It does require calculus, though - I don't know if you know it already or not.

Calvin Khor

@bumblebee well I dont see what the difference is between what I did and what prithu did ._.

bumblebee

@CalvinKhor wdym?

Akiva Weinberger

Another idea: the Mercator projection. What is it (formulas, properties, history)?

For example, it is the only projection that preserves angles and such that north is always up.

bumblebee

2:04 PM

@AkivaWeinberger If you don't mind, could you continue to spoil me with more ideas. Thanks.

Akiva Weinberger

(If Google maps used a projection that doesn't preserve angles, if you zoom into your city you might find that rectangular buildings become parallelograms!)

If you know about complex numbers, this sentence might make sense (don't worry if not): the Mercator projection is the complex logarithm of the stereographic projection.

You can also talk about other map projections and their properties.

Calvin Khor

@bumblebee I was suggesting an idea like your Q asked for, but its "warned against getting research questions from external parties" for my suggestion and thanks for the other suggestion. But whatever, its not important

Akiva Weinberger

Fun fact: in 2D, there's a lot of angle-preserving projections, but in higher dimensions (from hyperspheres), there's only a small family of angle-preserving projections (basically just stereographic from different points). There is no hyper-Mercator

bumblebee

@CalvinKhor Forgive me.

Akiva Weinberger

Speaking of spheres, you could do some spherical geometry

Like this thing

Pentagramma mirificum

Pentagramma mirificum (Latin for miraculous pentagram) is a star polygon on a sphere, composed of five great circle arcs, all of whose internal angles are right angles. This shape was described by John Napier in his 1614 book Mirifici Logarithmorum Canonis Descriptio (Description of the Admirable Table of Logarithms) along with rules that link the values of trigonometric functions of five parts of a right spherical triangle (two angles and three sides). The properties of pentagramma mirificum were studied, among others, by Carl Friedrich Gauss. == Geometric properties == On a sphere, both...

Calvin Khor

2:07 PM

@bumblebee no worries m8 keep on keeping on

Akiva Weinberger

and the fact that if a sphere has curvature $k$ (defined to be $1/R^2$ where $R$ is the radius), the area of a triangle is completely determined by its angles: $kA=\alpha+\beta+\gamma-\pi$ where $\alpha,\beta,\gamma$ are the angles in radians

bumblebee

Any ideas @CalvinKhor

Akiva Weinberger

(This technically works on the plane, too, with curvature $k=0$.)

Calvin Khor

you mean besides the calculus of variations problem?

Akiva Weinberger

Graph theory could be fun. On any connected planar graph, #vertices-#edges+#faces=1 ($V-E+F=1$). This is Euler's formula. It also has uses in topology (on a sphere you get $2$, on a torus you get $0$)

bumblebee

2:11 PM

@CalvinKhor yup!

Calvin Khor

oh, do you know any probability? random graphs are cool

Akiva Weinberger

Or number theory. Start with modular arithmetic and Fermat's little theorem (not to be confused with Fermat's last theorem), try to see if you can build up to something like the AKS primality test

bumblebee

@CalvinKhor only GCSE/IB probability. That doesn't matter though. I have a whole year to research and learn.

Akiva Weinberger

Another topic: Penrose tilings and related things like Ammann bars (two 'n's) and the Gummelt decagon

bumblebee

@Prithubiswasleftmse share some ideas too!

Akiva Weinberger

2:15 PM

It's an "aperiodic tiling", meaning it never repeats. Despite that, it has "quasi-10-fold rotational symmetry", meaning if you're dropped on a random part of it and can only see a finite neighborhood around you, you can only determine your orientation up to a tenth of a turn

(and also true 5-fold rotational symmetry from one point. You can't have 5-fold rotational symmetry from multiple points, though)

Calvin Khor

theres a cool proof method where you prove something exists by showing it happens with positive probability, but without any way of finding it

Probabilistic method

The probabilistic method is a nonconstructive method, primarily used in combinatorics and pioneered by Paul Erdős, for proving the existence of a prescribed kind of mathematical object. It works by showing that if one randomly chooses objects from a specified class, the probability that the result is of the prescribed kind is strictly greater than zero. Although the proof uses probability, the final conclusion is determined for certain, without any possible error. This method has now been applied to other areas of mathematics such as number theory, linear algebra, and real analysis, as well as...

Akiva Weinberger

You can talk about things like computability theory, the halting problem, and the Church–Turing thesis

Calvin Khor

@bumblebee or perhaps you could discuss why a drunk man will eventually find his way home, but a drunk bird will be lost forever (almost surely)

Akiva Weinberger

> The concept 'computable' is in a certain definite sense 'absolute', while practically all other familiar metamathematical concepts (e.g. provable, definable, etc.) depend quite essentially on the system to which they are defined. - Gödel

bumblebee

@AkivaWeinberger @CalvinKhor Things are probably gonna get very crowded here. I might also end up losing track. If you do have any more ideas, please share them here: chat.stackexchange.com/rooms/136459/ajay-personal-room

@TedShifrin Do you have any interesting ideas?

Haven't talked to you for a while now.

Calvin Khor

2:33 PM

haha ok :) @bumblebee

Jam

3:25 PM

consider the polynomial with roots $i\sqrt{3}$ and $1+i\sqrt{3}$ $f(x)=x^4 -2x^3+7x^2-6x+12$. does there exist a Q isomorphism that send $i\sqrt{3}$ to $1+i\sqrt{3}$ ?

i know the rest roots are the conjugate numbers

and my polynomial splits at $\mathbb{Q}(i\sqrt{3})$

which is a 2 degree extension so there are 2 Q automorphisms

one is the id

why the other cant it be the one that is beeing asked for?

does there exist Q-automorphism**

leslie townes

but isn't the other complex conjugation? long time since i studied any of this.

the main site has a few questions like this every couple of months. might be worth looking for others of this type. it isn't easily amenable to search terms, but maybe 'splitting' 'automorphism' 'root' or the like.

Jam

well since it will be a group of two elements makes sense the other one to be to the conjugate

so applying the map again i get the id

well it has to be a root of the minimal polynomial i used

which doesnt have root the 1+isqrt3

so ok thats it

is short 1+isqrt(3) and isqrt3 are not the roots of a single minimal polynomial

but my theory says it permutes the roots of my polynomial not the minimal necessary

Ted Shifrin

The polynomial fails to be irreducible, so the Galois group won't be transitive.

leslie townes

it is no surprise that ted remembers this stuff.

i even still have my galois theory book/notes, both undergrad and grad version. survived all USPS mishandling. for all the good it did me.

Calvin Khor

3:43 PM

I did all my galois homework following algorithms with virtually no understanding :') and got a pretty good score. Didn't push through to taking the exams..

Jam

Can i use the argument that the splitting field of the minimal polynomial with roots +-isqrt3 is the same as the splitting field of the given polynomial then a Q-automorphism must take a root of the minimal to root of the minimal again so i cant have such a Q -automorphism

cant find the transitive condition on the galois group on my notes

leslie townes

there were a few months in around the year 2000 when i understood the basics and even some stuff about extensions that are 'bad' for the basic theory. had MSE existed i could have answered those questions.

i don't want to blame it all on 9/11, but it didn't help

Akiva Weinberger

@Jam $i\sqrt3$ satisfies $t^2+3=0$. $~1+i\sqrt3$ does not. Therefore no automorphism of $\Bbb Q$ sends one to the other

Jam

thanks i think i got it

Akiva Weinberger

Sometimes I think writing $\sqrt{-3}$ rather than $i\sqrt3$ is clearer, in this sort of thing

linear_combinatori_probabi

3:56 PM

@graffe Did you solve it? I fell asleep.

Akiva Weinberger

In Galois theory, at least, I don't particularly care what the real and imaginary parts are, but I care about its algebraic properties (it squares to $-3$)

Jam

how to prove $Q(\sqrt{2})$ and$ Q(\sqrt{3}) $are no Q-isoimorphic using the fact they are splitting fields of $x^2-2$and $x^2-3$

leslie townes

akiva: yeah. any 'ambiguity' seen in the notation is reflected in, well, there's a Q-automorphism sending one to the other. i read a smart thread on either MSE or MO about this once.

it's not hard to do that 'bare-handed' by showing that Q(sqrt(2)) doesn't have anything that squares to 3, or vice versa. that may not be the most illuminating way of doing the exercise.

Jam

there must be a theorem saying again a Q isomorphism would take sqrt2 to itself or negative and since it is a linear map should take a basis to a basis so it cant map any element to sqrt3 ?? maybe i dont know

Calvin Khor

i feel like that would have made a certain lightbulb in me flicker if the wires joining it to mains didnt already decay years ago

Akiva Weinberger

4:06 PM

@Jam You need to show $(a+b\sqrt2)^2\ne3$ for all rational $a,b$

Jam

ya thats one way

i wanted to use the machinary with splitting fields and extentions

extensions*

Akiva Weinberger

I think there isn't, because $\Bbb Q(\sqrt2)$ and $\Bbb Q(\sqrt8)$ are isomorphic and so you need something that works for your case but not these

It turns out that $\Bbb Q(\sqrt a)=\Bbb Q(\sqrt b)$ ($a,b\in\Bbb Q$) iff $ab$ is a square in $\Bbb Q$

Jam

i think i can say there are at most 2 Q isomorphisms

Akiva Weinberger

I think

Ted Shifrin

Again: Any $\Bbb Q$-auto sends a root of an irreducible poly to another root.

Jam

4:10 PM

ok nice and sqrt2 doesnt belong

leslie townes

as ted notes, we may be ignoring that the automorphism is a Q-automorphism.

Jam

to my second field

ok thanks Ted

Akiva Weinberger

@Jam You don't know that. You claim $\sqrt2\notin\Bbb Q(\sqrt3)$ but it's not obvious

Jam

Akiva you are right ofc ill prove it

you cant write sqrt2 as a+bsqrt3

Akiva Weinberger

The key is that $2$, $3$, and $6$ are all not squares

leslie townes

4:11 PM

you're not a square.

Jam

the elements of that field are of the form a+bsqrt3 so i can prove there doesnt exist rationals a,b such that sqrt2 is a combination

ye probably ill square the equation

and get to something impossible

Ted Shifrin

That just says the fields are different. Why could they not be isomorphic?

Jam

as you said x^2-2 is irreducible and sqrt2 must be mapped to itself

or negative

wait i know Q automorphism permutes roots does that hold for Q-isomorphisms

Akiva Weinberger

Define the words automorphism and isomorphism for me

Jam

i mean the the target field is the same

on autos

Akiva Weinberger

4:16 PM

Oh, right, sure

Jam

but i think Q isomorphism permute roots also

so we good

it works ^_^

Akiva Weinberger

Yeah as long as you're both subsets of some larger field like $\Bbb C$ you're good

Jam

would like a nice counterexamle on that

i think it works between any Fields containing Q . A Q-isomorphism will permute a root of the minimal of the one field

leslie townes

akiva: this reminds me of a moment in a class on this subject where our professor had a field diagram with two different extensions of a ground field and said: "although we're just drawing this part, remember that they're all contained in [context field]. they're not just flapping in the breeze."

that stuck in my head forever.

far longer than the reason why it mattered that they weren't just doing that

Jam

$f(\sigma((a))=0$ for any $\sigma$ Q isomorphism from E to L fields containing Q and $f$ minimal polynomial of E so $\sigma(a) $ also a root

do i need them to be subsets of some algebraic closure?

Akiva Weinberger

4:25 PM

@Jam If we have $\Bbb Q(\alpha)/\langle\alpha^2=2\rangle$ and $\Bbb Q(\sqrt2)$, in the former the roots of $x^2-2$ are $\alpha,-\alpha$ and in the latter the roots are $\sqrt2,-\sqrt2$

The elements of a field don't have to be (complex) numbers, they can be whatevers, as long as we have a rule for addition and multiplication

Jam

is this a counterexample? trying to keep up

these 2 fields are isomorphic but a root doesnt go to a root?

Akiva Weinberger

A root goes to a root but it's not a permutation

'cause the starting and ending sets of roots are different

Jam

do you mean Q/<x^2-2> ?

so one has classes as roots and the other real nbumbers

but they are the same

i think i got it so i have to keep in mind they live inside a common bigger field

good example i have in mind the one with the ponynomials Q/<x^2-2> works too right?

ponynomials hahaha thats cute

Lucas Henrique

4:54 PM

hey there chat

sanity check: if $K/k$ is a field extension with transcendence basis $S$, then $K/k(S)$ is an algebraic extension

is this correct?

leslie townes

today's field theory day

Ted Shifrin

You going to have a field day?

Lucas Henrique

@leslietownes I wish. I'm studying this transcendence basis stuff superficially because I need it to prove stuff about rational maps of algebraic varieties

(varieties over alg. closed $k$ + dominant rational maps) $\cong$ (category of finitely generated extensions of $k$)${}^\text{op}$. cute

leslie townes

5:20 PM

ahh! varieties

Akiva Weinberger

6:02 PM

I don't really know what a transcendence basis is

leslie townes

akiva: you need them to prove stuff about rational maps of algebraic varieties

Ted Shifrin

Start by adjoining (independent) transcendental elements to your field.

leslie townes

6:35 PM

you can also stop there.

geocalc33

6:57 PM

What does integration really mean in $\Bbb R^2_{\gt 0}?$ From what I understand you're using an additive linear operator in a multiplicative space. Whereas you could just use the geometric integral. What does it mean to mix this additive operator into this multiplicative space? Or is there really nothing to understand?

leslie townes

lots of ways to view integration but if you're talking about real valued functions on a space X, you can do something like that as long as X has a sufficiently nice notion of numerical size of subsets of X. it's fine if X has other structure but integration doesn't need to know that. any more than integrals of functions over subsets of R need to care about intervals with only positive numbers vs. those that might contain negative numbers, although that difference might be significant elsewhere.

geocalc33

okay

geocalc33

11:05 PM

So analytic continuation doesn't work for functions that are in a quasi analytic class

Derivative

For which n is the group $(\mathbb{Z}[i]/n\mathbb{Z}[i])^\times$ cyclic?

Transcript for

$Mathematics$ Mathematics