/mg/ - Math General: Zariski Topology Edition

How do I approach this question? I am really struggling with holomorphic functions expressed as integrals..

No, plug in f(3); that doesnt work

plug in mod4 then

Im sorry
f(x)=(x+2)mod3 +1
This one is right, i guess it could be better

Tips/tricks for solving this problem?

That doesn't fit his graph

Maybe I'm just a brainlet, but I'm not sure this can be defined.

A sawtooth wave under an affine transformation
t. engayneer

I have 4m+2 points on the unit circle with equal angle between them, such that one point is on x-axis.
I'm guessing that the points not on the x-axis all have some similar roots appearing in their x-value, and some distinct roots in y-value.
Distinct in this case means that you can't get any of the x-roots by multiplying some y-root by a rational number.
How would I show this for all positive integers m?

Example: m=1 -> all points have x-value [math]\pm \frac{1}{2}[/math] and y-value [math]\pm \frac{\sqrt 3}{2}[/math].
Example: m=2, the positive y-points are attached, and negative are symmetric.

f(x) = x - 3 ceiling(x/3) + 3

Who /Fourier Analysis/ here?

Ideals of R are direct products of ideals of [math]\mathbb{Z}[/math] and ideals of [math]\mathbb{Z}_{80}[/math]. To see this, multiply each element of an ideal J of R by (1,0) and (0,1). Or note that, since the canonical projections are ring epimorphisms, the images of J under them are ideals of the images of R.

For maximal ideals, note that [math]J = J_1 \times J_2[/math] is maximal in R iff one of [math]J_1[/math] and [math]J_2[/math] is maximal and the other equal to its entire ring, for otherwise you can pick a larger proper ideal in R. So for example [math]2 \mathbb{Z} \times \mathbb{Z}_{80}[/math] is a maximal ideal, because if [math]J_1 \times \mathbb{Z}_{80}[/math] is a larger ideal, [math]J_1[/math] must be an ideal larger than [math]2 \mathbb{Z}[/math], which can only be [math]\mathbb{Z}[/math] itself.

The same holds for prime ideals. [math]J = J_1 \times J_2[/math] is prime in R iff one of [math]J_1[/math] and [math]J_2[/math] is prime and the other is equal to the whole ring. To show J is prime, note that J is the preimage of the prime ideal, either [math]J_1[/math] or [math]J_2[/math], under its canonical projection, and the preimage of a prime ideal under a homomorphism is another prime ideal. You may see by picking elements from J and R why the converse holds.

Since [math]\mathbb{Z}[/math] is a principal ideal domain, every nonzero prime ideal is maximal, but {0} is a prime ideal which is not maximal. Note that since [math]\mathbb{Z}_{80}[/math] is not a domain, {0} is not a prime ideal for [math]\mathbb{Z}_{80}[/math]. So [math]\{ 0 \} \times \mathbb{Z}_{80}[/math] is prime in R by the above, but not maximal.

Similarly you can find nontrivial ideals of R that are neither prime nor maximal by taking any product of two nontrivial proper ideals of [math]\mathbb{Z}[/math] and ideals of [math]\mathbb{Z}_{80}[/math].

Thanks a lot for that mate! Really well explained.

i don't think this makes sense for bigger m
you can't even write sin(pi/7) as an expression involving roots and rational numbers

Let R be a ring and r ∈ R. Then n · r = 0R for every integer n ≥ 0.
Proof: We use a proof by induction on n. The base case n = 0 is obviously true since 0 · r = 0R by definition. Now suppose the claim holds for n − 1 and consider n. Write n = i + j with integers
0 ≤ i, j ≤ n − 1; by the induction hypothesis, we have i · r = j · r = 0R, and hence
n · r = (i + j) · r = i · r + j · r = 0R + 0R = 0R,
as claimed. This completes the proof.
Prove me wrong.
Pro-tip: You can't.

The induction step isn't applicable for n=1, because there are no 0 ≤ i,j ≤ 0 such that i+j = 1.

You are correct, after thinking a bit I managed to reword and reduce the problem into the following:

Let [math]c = \cos(\frac{\pi}{2m+1}), m\in\mathbb{Z}^+[/math]. Do there exist some rational coefficients [math]a_k \in \mathbb{Q}[/math], such that [math]\sqrt{1-c^2}=\sum_{k=0}^N a_k c^k[/math] for some finite [math]N[/math]?
If [math]m=1[/math], left side is irrational and right side is rational, so no coefficients exist.
If [math]m=2[/math], left side has that square root inside square root mess, and right side is always rational plus rational times [math]\sqrt 5[/math], so no coefficients exist.

How did you do this?

Base cases actually have to relate to the induction step, moron. You should be able to move from n=0 to n=1 to n=2 and so on. You've proven n=0, but it isn't a base on which your n=i+j premise holds. Your base needs to be n=0 and n=1, but it clearly fails for n=1.

Read the hints.

1 + (x+2)mod(3)

If I have 2n+1 points inscribed on a circle's circumference (with equal distance between each point), how many points would we expect to be "past" the diameter perpendicular to any one point?

>Test 1, that doesn't work

Using the geometric series I get the integral to be \sum_{n=-1}^{\infty}{z^k}. I can't see how that is the same as the function given..

I'm struggling to figure out b and c of this question. I'm guessing that I should consider the quotient ring R/I, but I'm having difficulty seeing if it's isomorphic to a field.

F[[x]] is local so it only has one maximal ideal. Since there is only one maximal ideal, it must be equal to the Jacobson radical. The Jacobson radical consists of the non-units.

Obviously maximal implies prime.

>I'm guessing that I should consider the quotient ring R/I, but I'm having difficulty seeing if it's isomorphic to a field.
Let [math]a + I \in R/I[/math]. If a is a unit in R, a + I is a unit in R/I. If a is a non-unit in R, [math]a \in I[/math] implies a + I = 0 + I. Therefore any nonzero element in R/I is a unit, and R/I is a field. Of course this rests on the truth of a), which is due to the locality of F[[x]] specifically.

What was the last important math paper to be published?

P ≠ NP

That's compsci, not math

>compsci
>not math

Compsci is a superset of math, so no, P=NP is not math just because it's compsci.

Nice try, but math comprises literally everything.

>Nice try, but math comprises literally everything.
sophomore detected

:^)

The real question is, if x equals 0, does y equal;
A: 0
B: 3
C: N/A (undetermined)

Is OxfordAnon here? I have a question for him.

(I'm an American applicant, so you may be unfamiliar with some of the things I talk about.)

I'm applying to Oxford this year, which is also my gap year (after having finished my 12th year of school). My high school was absolutely terrible; the majority of its graduates end up attending a community college after finishing school. (Enrolling in a community college is kind of like doing foundation courses in the UK). In short, its education standards (and many of its students' prospects) are mediocre. The most rigorous class I've taken is AP Calculus BC. Would you happen to know how this might affect my application? How do I inform Oxford about my circumstances?

It seems that Oxford likes it when A-Level pupils take Further Maths, which I imagine is more difficult/rigorous than AP Calculus BC, right? The most advanced concept we got to in BC Calc was Taylor series. Would that make me seem unqualified and/or hurt my chances?

Unrelated question I was curious about: do the students that apply at age 16 have an edge or a disadvantage in comparison to those who are 17/18 at time of application?

Thanks a lot senpai.

Tips/tricks/hints for this problem?

0

I have a question for OxAnon too: how would you feel about Malala accepting Oxford's offer? Some say her offer exists pretty much solely because of her compelling story, and not because she fits into the Oxonian meritocracy. What do you think?

it was a backhanded complement, user

he did this shit in the last thread too
just ignore him, i'm convinced it's a shitty and unfunny troll

is that aluffi?

Aluffi?

>but I'm not sure this can be defined
you are a brainlet
the graph of the function is right there, it is clearly definable
the poster of that question is just an autistic faggot who doesn't know how to do any of his own work himself

A sawtooth wave with an open gate
t. synthwave gayboard player

I'm trolling because I have a problem you can't solve?

>inb4 there's no answer
gave the solution, you brainlet

Admit it, you're just mad that you weren't smart enough to develop a solution.

you're trolling because between the posts in this thread and last thread you should have been able to do this garbage problem on your own
kys

>asking questions you don't know the answer to IS NOT ALLOWED!!1!
You're embarrassing yourself. Don't worry, I couldn't solve it either (I'm not him).

shut the fuck up Satan

Alright, step aside, brainlets. First off, what you're looking at is what's known as a discontinuous function. Unfortunately, most of you immediately started thinking of the log function rather than the floor and ceiling functions for god knows what fun reason. This guy here knows what he's doing, though it may confuse anyone who failed to learn PEMDAS in school. That's not his fault, though, that's on them.
>b-buh what about x=0
>If you plug that in, you get y = 3
>that doesn't match the graph, reeeee
Shut your pagan whore mouth. That's because you're too dumb to realize that logically, if the function extends into the negatives(which it does), then there's no reason for (0,3) to not be part of the graph. After all, it's pretty clear that for the rest of the drawing, if x divides into 3 with no remainder, then y = 3. Since 0 divides into 3 with no remainder, then if x = 0, y = 3. The OP of the graph drawing made a small mistake; nothing more, nothing less.

>complement

Y = 3 - (((4X/3) mod 4)-1)
Alternatively,
Y = X - (3 × ceiling (X/3)) + 3

What's a p-value?

the chance that your means are actually the means of the population