Mathematics

12:51 AM

0

Q: open set in $\mathbb{R}^n$ property of diameter

Problem: Assume $W$ is an open set in $\mathbb{R}^n$. Assume $X$ is a metric space $f:X\rightarrow \mathbb{R}^n$ is continuous. Suppose $g:X\rightarrow \mathbb{R}^{>0}$ is a continuou function such that $g(x)<d(f(x),\mathbb{R}^n\backslash W)$ for all $x\in X$. Show that if $F$ is closed in $X$, t...

geocalc33

i found an incredible sequence

geocalc33

1:19 AM

0

Q: Prove the sequences are even aside from 1st term

Consider the following two alternating sequences: $$ A=\bigg\lbrace f^{(1)1}(x),\cdot\cdot\cdot \bigg \rbrace \bigg|_{x=1/e}=\bigg \lbrace-1,2,-6,32,-320,4452,-70798, \cdot\cdot\cdot \bigg \rbrace$$ $$B=\bigg\lbrace f^{(1)1}(x),\cdot\cdot\cdot \bigg \rbrace \bigg|_{x=e}=\bigg \lbrace-1,10,-150,30...

calculus sequences-and-series elementary-number-theory functions derivatives

it's not actually that cool.. but kinda neat

A016090

3:14 AM

@geocalc33 it's way late here but I'll take a crack at it in the morning if no one beats me to it!

Silly Goose

4:09 AM

Let's say I have a monotonically increasing sequence $\{s_n\}$. If I cannot explicitly write out the starting point, can I still work with the sequence?

What I have in mind is the sequence of all rational numbers in $(0, 1]$ ordered so that it is monotonically increasing

leslie townes

'work with' is going to need some elaboration. at the moment, you do not have a sequence.

there's no map from the positive integers onto the rationals in (0,1] that does what you want.

you can turn the rationals in (0,1] into a sequence by choosing some bijection between the positive integers and those rationals. you can maybe make that choice fairly explicit, but it's not going to be monotonically increasing.

this isn't specific to your countable set being maybe difficult to explicitly enumerate via a bijection with the positive integers n, either. as perhaps a simpler example, there's no monotonically increasing map from the positive integers onto {1/n: n is a positive integer}.

so "in bijection with the positive integers" , i.e. countability, is not the same property as "in an order preserving bijection with the positive integers," even if you're considering subsets of R which everybody knows is a linearly ordered set.

Silly Goose

Hm, if I want to do all rational numbers in $[0, 1]$, then could this be ordered to be monotonically increasing?

Ted Shifrin

To be more concise: If I have a particular rational number, can you tell me “the next” rational number?

You’re not listening!

Silly Goose

4:25 AM

so then I cannot reference the order really at all of such sequences

leslie townes

well, you have the order that "rationals in [0,1]" gets as a subset of R. but it isn't isomorphic, as an order, the order on the set of positive integers.

Ted Shifrin

You can choose increasing sequences of rationals, but they will omit plenty!

leslie townes

"sequence" is extra structure. a countable set is not a "sequence." sometimes this distinction is skipped over when it is not the point.

Silly Goose

the extra structure of the sequence on a countable set is the order endowed by the mapping of the sequence?

leslie townes

you can make a countable subset of R into all sorts of different sequences.

Silly Goose

4:29 AM

Bleh. Okay well then maybe I will try to take a different approach to this problem :P.

leslie townes

well, the indexing of the set by, this one is labeled 1, this one is labeled 2, etc. if you think of the indexing as carrying an order, then that too.

Ted Shifrin

Luckily, we do not know the problem.

leslie townes

it's very common to describe a countable set as the range of some function on the set of positive integers, but worth keeping in mind that the function and the set are different things.

Silly Goose

Well, so the problem concerns finding a sequence with values in [0, 1] such that the sequence has subsequences that converge to every real number in [0, 1]. My intuition is that we construct the reals from dedekind cuts of the rationals whose limit points are the real numbers. I think the better approach is just to deal with things in terms of limit points rather than show convergence explicitly. But, I would expect that you can translate the result written in terms of limit points into

the language of explicit convergence

Ted Shifrin

You can do this totally explicitly.

Increasing sequences are far, far from the right intuition.

leslie townes

4:33 AM

silly: you can definitely do that with enumerations of rationals in [0,1], but why you'd need anything to be monotonic is beyond me.

Ted Shifrin

Think about what it takes to have a subsequence converging to any given number.

leslie townes

also note that once you are willing to pass to subsequences you're no longer obligated to have your subsequence cover all rationals in [0,1]. that could be helpful.

Silly Goose

well the part that i used (and subsequently found out is wrong) monotonicity for is the $n \geq K$ part of the definition of convergence. Because certainly a rational in whatever subsequence exists such that $d(x, q) = x - q < \epsilon$ for all $\epsilon$ where $x \in [0, 1]$

Ted Shifrin

You need to work on writing sentences that make sense.

Let’s do this. Give me a sequence so that both $1/2$ and $5/6$ are limits of subsequences.

leslie townes

consider for example the sequence of all one-digit decimals 0.0 - 0.9, in order, followed by all two digit decimals 0.00 to 0.99, in order, followed by all three digit decimals 0.000 to 0.999, in order, followed by ... . you could find a lot of interesting monotonic subsequences of this mess, although this mess itself is not monotonic.

Silly Goose

4:52 AM

Hm, so you can do like {1/n: n = 2, 3, ...}, which converges to 0. Then add $1/2$ to each element and get something that converges to 1/2.

leslie townes

mm, maybe stick with ted's request, and work with explicitly defined sequences before branching out to slightly less explicit descriptions.

Silly Goose

Well, the sequence containing all rationals in $[0, 1]$ works I think

leslie townes

there's more than one sequence that does that. in the absence of uniqueness, to refer to "the" sequence [having some property] is confusing. do you mean, "if (a_n) is any sequence with {a_n: n in N} = Q, then (a_n) will have the property i want" ?

you could also be less ambitious and start with ted's simpler example.

Silly Goose

since an at most countable union of countable sets is still countable, maybe take the union of a countable number of 1/2s and a countable number of 5/6s and make some function which hits all of the 1/2s before hitting the 5/6s

leslie townes

you're jumbling set operations, on the one hand, with defining a sequence, on the other. i sort of get what you're driving at, but you need to think and write more concretely.

very close to "the decimal that looks like 0 point (a sequence of infinitely many zeros), then followed by a 1, is a positive number smaller than any rational" here

Ted Shifrin

5:13 AM

Let’s try not to reuse any numbers in our sequence.

ryang

@TedShifrin that Question came up in the RHS 'Related' column of a newly posted question. (anyway, i think that Answer is grasping at something like, " 'Px doesn't imply Qx' doesn't preclude Qx from being true in particular cases" (i.e., trying to explain about the implicit universal quantification), but sounds incoherent because this explanation is inappropriate in this case. but the

explanation is vague enough that i wasn't entirely sure if i was reading it correctly.) thanks for your input! P.S. that answer has now been deleted.

Silly Goose

5:31 AM

well I don't know :P guess I got to try some other sequence exercises

leslie townes

5:54 AM

ted: munchkin got to swim with ducks today. a duck was in the pool when she got in, and two more landed in it while she was swimming.

Prithu Biswas

@SillyGoose So we have to find a sequence a : ℕ→[0,1] such that Given any real number c, there is some sub-sequence g of a such that limit[n→∞] (g(n)) = c ?

*Given any real number c ∈ [0,1].

Franklin

6:11 AM

Can anyone please validate this: " A differential equation can have infinitely many GENERAL SOLUTIONS " ?

Prithu Biswas

Hmm... We can order the rationals ℚ = {r0,r1,r2,r3,r4,r5,r6,...}.

Franklin

I think that assertion is absurd

Prithu Biswas

And then construct the sequence a like this:

a = (r0,r0,r1,r0,r1,r2,r0,r1,r2,r3,r0,r1,r2,r3,r4,r0,r1,r2,r3,r4,r5,....)

Then if we are given a real number c ∈ [0,1],

Let s be the sequence on [0,1]∩ℚ such that s converges to c.

Then I think s is also a sub-sequence of a.

@leslietownes Maybe this could work?

Ted Shifrin

6:34 AM

@leslietownes so she knows how to be quiet and calm when she wants to be!

@Franklin Very non-absurd. They don’t say independent.

leslie townes

it seems potentially gibberish-y to me. at least, for some books, the 'general solution' to a DE is its full set of solutions (ignoring initial values or boundary whatever), and there's only one of those. although it is usually an infinite set, it's only one set.

ted: anything to recreate with ducks.

Prithu Biswas

6:52 AM

A more general problem could be, find a sequence X : ℕ→ℚ such that Given any sequence Y : ℕ→ℚ , Y is a sub-sequence of X.

Oh wait.... My solution is completely wrong.

I should have started with "We can order the rationals ℚ∩[0,1] = {r0,r1,r2,r3,r4,r5,r6,...}."

leslie townes

i might use the term 'enumerate' rather than 'order.' the rationals already have a very familiar order (coming from R), and the difference between that order and what you might get out of an enumeration is what was confusing whoever was asking the question. but yes, lots of variations on that idea do work.

Prithu Biswas

@leslietownes Yeah, 'enumerate' seems to be better =)

Franklin

8:07 AM

@TedShifrin So you mean, if an ordinary differential equation of order n , with independent variable as x, have a general solution say a(x), (thus, a(x) has n arbitary constant )then it might have another general solution b(x) ( thus b(x) has n arbitary constants as b(x) is a general solution) such that $b(x)\neq a(x)$ ?

To be frank: This is what I want to know....

shintuku

11:15 AM

@Franklin it's simply the fact that there are infinitely many curves that can have the same derivative, if for example they're only shifted by a constant

@Franklin the general solution of $\frac{dy}{dx} = 2$ is $2x+C$, with examples of particular solutions being $2x+1$ and $2x+2$

shintuku

11:29 AM

@Franklin been looking at it a bit more, maybe you meant something like this: math.stackexchange.com/questions/3697476/…

Franklin

@shintuku that is true, but they are the particular solutions of a single general solution. But I specifically, want to know, whether multiple general solutions exists or not like, say for example in a 2nd order differential equation, we get two general solutions $ a(x)=c_1e^x+c_2a^{x^2}$ and $b(x)=c_1e^x+c_2a^{x^2-x}$, where $c_1,c_2$ are arbitary constants: Is this, at all possible? Please dont take this example literally, I just used those equations to just picture the situation clearly.

one potato two potato

12:11 PM

There was a seminar at my college about some applications of homogenous dynamics in number theory. Although I understood almost nothing after the first fifteen minutes, I took notes and draw pictures (well I did my best).

shintuku

12:37 PM

@Franklin see here for a concrete example: math.stackexchange.com/questions/2139416/…

Astyx

@TedShifrin So zeta values?

ephe

1:28 PM

If P_n denotes a partition of [a,b] into n equal lengths, is it necessarily true for a function integrable on [a,b] that value of lower Riemann sums increase as n increases? I know that we have bigger lower sums for refinement of partitions but I can't figure out a way to prove this one.

shintuku

@ephe not true for constant functions

maybe a good condition is: there exists at least one sequence of two intervals of the partition $n-1$ where the lower sum differs

but that might not be general enough

@ephe see first if you can prove: for any partition of size 2 of a nonconstant function, the partition of size 3 will result in a greater lower sum

ephe

@shintuku Sorry, I notice that I have misphrased the question. I should perhaps more accurately ask whether we have L(f,P_a) \leq L(f,P_b) whenever a<b. I've been trying to work with smaller numbers n like in your suggestion but I can't figure out how to compare the lower sums because all the intervals in the partitions are different from each other.

shintuku

can you prove this for a linear function, in the case of partition of size 2 vs. partition of isze 3?

say, from 0 to 10, for a strictly positive linear function

ephe

1:46 PM

Yes for that case I can prove it using the fact that it is monotonic

shintuku

nice, can you find an interval where the statement would be false in a quadratic function?

using size 2 vs size 3 partitions

ephe

I can't find such an interval, should I have been able to?

shintuku

no there aren't any, can you prove it?

ephe

2:02 PM

I can't formulate it perfectly but I think I can show this using the fact that the maximum/minimum will be only in the middle interval in the size 3 case but in both intevals in the size 2 case

shintuku

you'll have to formulate it perfectly, you'll need at least that level of abstraction for the other proof you want. show it for an arbitrary interval chosen on a strictly positive section of a quadratic function

Franklin

@shintuku That's a good one! I think that infinite general solutions might exist for ODE (Ordinary Differential Equations)'s as well :)...

ephe

@shintuku Okay so using the fact that there is a minimum in each interval I've shown this but I had to consider many many cases (about where this minimum might be) so I'm guessing there must be a better way to show this.

shintuku

2:18 PM

@ephe casework is how I do it too, have you shown it for both a convex and a concave quadratic?

ephe

@shintuku I forgot about the concave case but I have done that now.

Ajay

Do you guys think there should be a tag for Calabi–Yau Manifolds?

I mean, they're quite important in string theory and mirror symmetry.

shintuku

@ephe cool, now do the same we did but with an interval of $\sin(x) + 10$ that contains both a minimum and a maximum

@ephe so, an interval of a strictly positive sine function that has both a concave and a convex part

Ted Shifrin

2:34 PM

@ephe You can easily give a counterexample with $n=2$ and $n=3$. Think about step functions.

@Ajay once you start down that path, you’ll need zillions more tags.

user193319

Let $G$ be a countable discrete abelian group. Is the pontryagin dual of $G$ polish? If so, are you able to provide a reference?

Ajay

Then I guess complex manifold should do it

Thorgott

the pontryagin dual of a discrete is compact and since G is hemicompact, the pontryagin dual is completely metrizable

not sure about separability

oh wait, C(G,T) should be second-countable, hence so should be the pontryagin dual of G

Ted Shifrin

@Ajay We have the compkex-geometry tag, which sadly gets misused most of the time.

shintuku

its proper use is, of course, to talk about complicated shapes

Thorgott

2:46 PM

@TedShifrin what? is it not for Euclidean geometry in the complex plane?

Ted Shifrin

@shintuku False.

Xander Henderson

@shintuku No, that is how you misuse the fractal geometry tag.

user193319

@Thorgott Ah, very nice! Separable and metrizable, so it is polish. Thank you!

Xander Henderson

Polish ≠ polish. :P

ephe

@shintuku Okay I've done this too, separating the cases based on where the max/min occur and then using the fact that there are intervals where the function increases/decreases.

shintuku

3:01 PM

ok, on that same function can you do the sumfor n-1 is smaller than the sum for n?

Ajay

@XanderHenderson I remember a really funny big bang theory episode about this

shintuku

otherwise, go back an do it for a linear function, then for a quadratic concave, then convex, and then get to this one @ephe

Ajay

Sheldon talked about Polish sausages and how Madame Curie was Polish

But it was actually Penny

Who pointed out that it was nail polish

Due to the small p

Xander Henderson

@Ajay I don't believe you. You are positing a world where Big Bang Theory is funny.

Prithu Biswas

The Big Bang Theory - Nail Polish - Pictionary Game - Learn English

►

Franklin

3:13 PM

Can anyone please help me with the usage of D as a symbol when D actually means differentiation operator. I am having a hard time with it in differential equations ...

robjohn

@PrithuBiswas Just saw that last night.

Ajay

We have a motion: "I am positing a world where Big Bang Theory is funny". All in favor of the motion say "I"... As there are no votes to back up @XanderHenderson's motion, the motion is denied. muahahaha

A016090

Can you elaborate? $\frac{d}{dx}$ for example can be expressed as $D$ for simplicity.

ie) $\frac{d}{dx}\left(x^2-5x\right)=D\left(x^2-5x\right)$

Also greetings from the empty tutoring office, sanity is in short supply today.

robjohn

3:29 PM

@A016090 This is UC break week. Do you have classes this week?

(University of California)

Xander Henderson

@Ajay (1) Humor is not a democracy, (2) an affirmative vote is an "aye", not an "I".

ephe

@shintuku I'll be working on this for some time. Could you tell me the things I should prove after this one to step by step get closer to proving the statement? By the way, thank you so much for all your help!

A016090

@robjohn The break for the college office I tutor at doesn't start until next Wednesday. However, we've been drying up as the college has chipped down our hours and removed our campus advertising in favor of a general tutoring office. Now we only see consistent souls during common exam times.

Xander Henderson

Also, a motion only succeeds if there are sufficient affirmative votes. If there are no votes, then the motion fails. In this case, you declared voting over before any votes could be registered, hence the motion fails. So if you believe that humor is a democracy, you have failed to declare that BBT is funny via a proper democratic process.

robertsrules.com

robjohn

@XanderHenderson eye, eye, sir!

Just counting facial features.

shintuku

3:36 PM

@ephe once you got the sine version down you should be able to generalize

robjohn

or admiralize

A016090

@XanderHenderson Do parliamentary motions then follow Newton's first law?

shintuku

@ephe if your function is continuous and nonconstant, there's no more general case than a nonrepeating wavy function

robjohn

@A016090 One combined office vs an office per department?

ephe

@shintuku Oh I see, thank you! This seems like a very cool method to go about proving things I can't wrap my head around.

shintuku

3:39 PM

yes do the most concrete case first to see if you can prove it, then work your way up

Xander Henderson

@A016090 Sure.

A016090

@robjohn Precisely, the trouble is that our general tutoring office doesn't have anyone who can assist in most courses above the 200/2000's levels, which normally is where we'd take them in, but they've stopped referring them to us.

robjohn

@A016090 efficiency at the cost of usefulness.

A016090

Unfortunately so.

robjohn

It's like getting rid of the different tracts of classes in earlier schooling because it makes those in the slower tracts feel bad, but the removal impairs the education of those in the faster tracts.

A016090

3:47 PM

@robjohn Precisely, also unfortunately one of the bags of bricks that led to me resigning from teaching at the local school.

Complaints aside, I find it really interesting that $e^{\frac{1}{\ln{x}}}$ feels almost teasing in the fact that it can't be further reduced or simplified.

robjohn

bad singularity near $x=1$

It is its own inverse, though

A016090

I know and it's absolutely vile, I love it.

@robjohn Wait, perhaps I'm not as well-educated as I should be, is the singularity at $x=1$ or $x=0$?

Xander Henderson

@A016090 1. $\log(1) = 0$, hence $1/\log(1)$ is problematic.

robjohn

@A016090 There is a singularity at $x=1$, but the edge of the domain of definition is $x=0$

Xander Henderson

Though regarded as a complex function, there is an essential singularity at 0.

robjohn

4:01 PM

@XanderHenderson I think the singularity at $x=1$ is an essential singularity. $x=0$ is the end of a branch cut

A016090

@XanderHenderson I have blundered in not realizing this, thank you.

Gwyn

Can you suggest an analysis textbook that defines the elementary functions with integrals and proceed to prove all the basic limits like $\lim_{x \to 0} \frac{\log(1+x)}{x}=1$ only using integral definition of elementary functions and derivatives like $(\sin x)'=\cos x$ using the fundamental theorem of calculus?

A016090

I kept looking at just $\log(x)$

Xander Henderson

@robjohn They are both essential singularities, are they not? I was only focusing on the one at zero, as you didn't seem to be regarding the function as a complex function prior to my comment.

robjohn

If one considers too many branch cuts of log, does it become sawdust?

Xander Henderson

4:04 PM

@robjohn Ouch.

Oh... wait... there is no way to make things holomorphic near zero. Nevermind.

Gawd... it's been too long since of done actual math. :(

None of the definitions are just immediate any more.

robjohn

@XanderHenderson Yeah, the definition of an essential singularity at $a$ I was using requires an open $U$ where $f$ is defined on $U-\{a\}$. I don't know if there are others that do not require an open neighborhood of $a$.

singularities are hard

A016090

@XanderHenderson What have you been doing, if not math?

robjohn

math - actual math = imaginary math?

A016090

Finally, we can find the argument of math.

robjohn

And the absolute value of math

user4539917

4:18 PM

Would that be the distance of math from the origin?

Ted Shifrin

Math has absolutely zero value.

Xander Henderson

@A016090 Teaching.

The highest level course offered at my current institution is multivariable calculus, and we haven't had enough students to teach even that for a long time. So I teach mostly precalc and calc I/II.

So my complex anal muscles don't get flexed very often.

3

robjohn

@TedShifrin Does that mean that math is at the origin of everything?

Ted Shifrin

TMI.

@robjohn It must.

robjohn

@TedShifrin or that it is as cold as it gets.

Ted Shifrin

4:22 PM

Keep Kelvin out of this :)

A016090

@XanderHenderson The fact that you're in education but still miss higher math, even recreationally is a breath of fresh air. Where I used to work, every soul didn't want to talk about anything math, or even try a fun problem of the week or so.

robjohn

It only takes a permutation to turn a cheater into a teacher.

Xander Henderson

@robjohn ... :(

Ted Shifrin

And inversely.

robjohn

indeed, as permutations are injective and surjective

A016090

4:26 PM

That was clever, though

Ted Shifrin

I feel like invective comes next.

robjohn

I get that a lot

user4539917

🙈🙉🙊

Ted Shifrin

4:39 PM

Hmm, I miss Kelvin and Hobbs.

UnderMathUate

I'm learning about Euler phi function. There's one theorem that states if $gcd(a, b) = 1$, then $a^{\phi(b)} \equiv 1 (mod \text{ } b)$.

But then, I also saw another claim that $a^{\phi(b)} \equiv 1 \equiv 0 (mod \text{ } b)$. I don't understand where the zero is coming from.

For context $\phi(b)$ is the number of integers that are relatively prime with $b$

Ted Shifrin

Well, since $1\equiv 0 \pmod b$ is nonsense ... you know that's a typo

UnderMathUate

good grief, 20 min wasted. I knew it didn't make sense.

Ted Shifrin

Yeah, I know what you mean.

What they may have had was $a^{\phi(b)}-1\equiv 0$.

UnderMathUate

Ooooh, you're probably right.

PNDas

4:55 PM

Rudin says "$f$ is representable by power series in $\Omega$ if to every disc $D(a;r)\subset\Omega$ there corresponds a series $\sum c_n(z-a)^n$ which converges to $f$ for all $z\in D(a;r)$." If I'm understanding it correctly, it means that if $D(a;r_1)\subset D(a;r_2)$, then we may have two different power series on those discs. right?

Thorgott

well, he doesn't exclude that in the definition

but it's nonetheless impossible by uniqueness of power series

A016090

@UnderMathUate When those moments come up, look online and see if the company posted an errata document for your version. It has saved me hours of grief.

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$Mathematics$ Mathematics