>first semester grad school Topology (Munkres)
>haven't done a single assignment before the day it was due
I am simply built different>>16503733
They started with Munkres? I like Munkres but my school started with Lee's topological manifolds book and used Munkres as an "undergrad reference.">>16503868
I wanna kill myself so bad
I live in such a shitholeI don't care if you're a PhD developing the newest computational automorphisms on stacks, if you're struggling with depression you ought to stay out of /mg/, this isn't a thread for diary posting>>16503984
Please don't. I'm sorry you don't like where you live/are from. If you feel like you aren't getting enough because of where you're from, just take a deep breath and try to snap out of it. People can follow all sorts of life paths after all sorts of universities, and if you do good work people will barely even think about where your degree is from.

If it's just grad school depression, I understand that in spades. I've been getting my ass kicked by a problem that I probably could have solved a while ago if I just faced the music and spent time learning random matrix theory.Approximately four years ago I saw a youtube comment where a carpenter turned programmer was saying that they need to use algebraic geometry and differential geometry to program a CNC machine to make certain wooden parts of expensive staircase handrails, because the act of making a banister profile smoothly follow a curve in three dimensions to the specifications of some clients was a complicated mathematical problem.
They went on to say that they needed the math to make the CNC software but that subsequent carpenters would be able to just input the requirements into the machine and never have to learn anything beyond precalculus.

The argument vaguely made sense but I have never heard of anything to corroborate it since. Was the whole thing made up or are there really staircases where the shapes of some of the components can't be made without university math?>>16505126
Probably just typical programmer moment where they use big words to sound more fancy. I needed to solve a system of polynomial equations to do a thing? I did a le hecking algebraic geometry.Are there any good proofs in linear algebra that rely on the IVT? I've only ever seen the IVT used in analysis and topology.>>16505293
Maybe that some real matrix has eigenvalue if its characteristic polynomial is a polynomial of odd degree>>16483163
Usually the opposite. At my uni, Math & Physics guys hung out with the Phil and Lit guys
>>16485720
CS/CE/EEs are one track minds uncultured nerds
>>16486329
>Neither are high-art literature but they aren't slop either.
Those are all slop>>16505126
If the closed figure is embeddable in R^2, made up of finitely many algebraic curve pieces (eg lines, quartics, circle sectors) joined at angles which are algebraic multiples of pi and it does not intersect, the boundary of the solid produced by smoothly transporting it along a smooth curve in R^n is numerically computable by combining linear algebra and the newton method. You don't need anything more advanced.>>16505504
Not true. You can construct turing machines with only the limited set of concepts you use.>>16505410
Before the Dawn is not slop. Get out of here.Are there any almost continuous functions that are Lebesgue integrable but not absolute (improper) Riemann integrable or vice versa?>>16505983
for the latter, all riemann integrable -> lebesque integrable.>>16477194
you mean "number system"?
mod 2?>>16477217
lol wut?Yo what are hypergeometric series? They came up on a thing I put into Wolfram but wikipedia does a bad job of explaining them.>>16505983
Every Riemann integrable function is Lebesgue integrable. That is not true in the other direction. There are many Lebesgue integrable functions are not Riemann integrable.Putnam Competition is over!
anyone ITT take it?
im thinking about B2, which asks about quadralaterals:
if you have quadralateral ABCD and take the diagonal AC then reflect D across the perpendicular bisector of AC (call it E) then ABCD and ABCE are called partners.
is there an infinite sequence of quadralaterals a_k s.t. a_n and a_{n+1} are partners and there does not exist any i≠j where a_i is congruent to a_j

this problem seemed too easy for me, and i fear i got it wrong. heres the sketch of my solution:
call |AB|=a |BC|=b |CD|=c and |DA|=d
note that CD and CE are congruent. and AD and AE are congruent, so each quadralateral that is a partner shares all the same side lengths a,b,c,d, albeit in a different order. the number of orderings of a,b,c,d is 4!, which is finite, so any infinite sequence must have multiples of some ordering.>>16489313
This doesn't have any exercises.>>16506542
Lang has exercises. Read Artin + Lang (first edition)How do you prove that any endomorphism of a f.g. module over an arbitrary ring satisfies a monic polynomial, without using determinants and adjugate matrices?>>16475762
in a flat circular coordinate system, can I use pi radians to do trigonometry? if so, how?
are pi radians and plain radians (already multiplied by pi) different in this regard?>>16506609
Why first edition?>>16506672
https://link.springer.com/book/10.1007/978-1-4613-0077-9Hey math enthusiasts. Let I={(0,n) in R^2: -1<=n<=1} and A={(x,sinx) in R^2: x>0}. I need to prove that A union I is connected but not path connected. Any ideas? Or a counterexample would be greatly appreciated.>>16506807
It's path connected. You made an error writing down the problem.

The correct version of the problem is solved here.
https://proofwiki.org/wiki/Closed_Topologist%27s_Sine_Curve_is_not_Path-Connected>>16506807
As stated your set is both connected and path connected. In the definition of A, you probably meant to write (x, sin1/x) which is what I'm going to assume.
First note that both I and A are connected and path connected.
>connected
Take a continuous function X -> {0,1}. Suppose w.l.o.g. I maps to zero (I is connected so maps to the same set). Take any point x in I. By continuity, there is an open ball around x that maps to 0. But inside that ball there are points in A, thus A also maps to zero. Thus there is no surjective map X-> {0,1}.
>not path connected.
Cannot connect (0,0) with (1, sin(1)). This is because any path that starts at (0,0) stays in I, because to leave it it would have to jump too much to be continuous.>>16506827
Why is it path connected? I mean, my prof wrote it like that, using (x,sinx). Maybe he made a mistake.>>16506900
And why is it connected?>>16506900
>>16506909
>Continuous functions map path-connected sets to path-connected sets
>[math]\mathbb{R}[/math] is path-connected
>[math]f(x) = (\max(0,x) ,\sin(x))[/math] is continuous and [math]f(\mathbb{R}) = A \cup I[/math]
>Path-connected sets are connectedI'm retarded. Will an Nspire CAS calculator help my limp through a calc 2 class. I roughly remember some stuff but was always kind of shit at math>>16507063
Plotting and numerically calculating integrals can help, but it won't help you learn unless you are actually committed. Having a calculator that you can use to check definite integrals is quite nice though.Doesn't the uncountability of Dedekind cuts on the rationals contradict the partition principle?>>16507487
Dedekind cuts are countable. Any subset of a countable set is countable.

/mg/ - mathematics general

Thomae's function edition[eqn] f(x) = \begin{cases}\frac{1}{q} & \text{if } x = \frac{p}{q}, \text{where } p,q \in \mathbb{Z} \text{ are co-prime} \\0 & \text{if }x \text{ is irrational}\end{cases}[/eqn]previous thread>>16432396

< Go back

Thomae's function edition
[eqn]
f(x) =
\begin{cases}
\frac{1}{q} & \text{if } x = \frac{p}{q}, \text{where } p,q \in \mathbb{Z} \text{ are co-prime} \\
0 & \text{if }x \text{ is irrational}
\end{cases}
[/eqn]
previous thread>>16432396