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12:51 AM
Q: open set in $\mathbb{R}^n$ property of diameter

MathematicallyInterestedProblem: 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...

i found an incredible sequence
1:19 AM
Q: Prove the sequences are even aside from 1st term

geocalc33Consider 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...

it's not actually that cool.. but kinda neat
2 hours later…
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!
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
'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.
Hm, if I want to do all rational numbers in $[0, 1]$, then could this be ordered to be monotonically increasing?
To be more concise: If I have a particular rational number, can you tell me “the next” rational number?
You’re not listening!
4:25 AM
so then I cannot reference the order really at all of such sequences
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.
You can choose increasing sequences of rationals, but they will omit plenty!
"sequence" is extra structure. a countable set is not a "sequence." sometimes this distinction is skipped over when it is not the point.
the extra structure of the sequence on a countable set is the order endowed by the mapping of the sequence?
you can make a countable subset of R into all sorts of different sequences.
4:29 AM
Bleh. Okay well then maybe I will try to take a different approach to this problem :P.
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.
Luckily, we do not know the problem.
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.
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
You can do this totally explicitly.
Increasing sequences are far, far from the right intuition.
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.
Think about what it takes to have a subsequence converging to any given number.
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.
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 $x \in [0, 1]$
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.
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.
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.
mm, maybe stick with ted's request, and work with explicitly defined sequences before branching out to slightly less explicit descriptions.
Well, the sequence containing all rationals in $[0, 1]$ works I think
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.
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
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
5:13 AM
Let’s try not to reuse any numbers in our sequence.
@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.
5:31 AM
well I don't know :P guess I got to try some other sequence exercises
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.
@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].
6:11 AM
Can anyone please validate this: " A differential equation can have infinitely many GENERAL SOLUTIONS " ?
Hmm... We can order the rationals ℚ = {r0,r1,r2,r3,r4,r5,r6,...}.
I think that assertion is absurd
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?
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.
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.
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,...}."
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.
@leslietownes Yeah, 'enumerate' seems to be better =)
1 hour later…
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....
3 hours later…
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 $2x+2$
11:29 AM
@Franklin been looking at it a bit more, maybe you meant something like this: math.stackexchange.com/questions/3697476/…
@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.
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).
12:37 PM
@Franklin see here for a concrete example: math.stackexchange.com/questions/2139416/…
@TedShifrin So zeta values?
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.
@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
@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.
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
1:46 PM
Yes for that case I can prove it using the fact that it is monotonic
nice, can you find an interval where the statement would be false in a quadratic function?
using size 2 vs size 3 partitions
I can't find such an interval, should I have been able to?
no there aren't any, can you prove it?
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
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
@shintuku That's a good one! I think that infinite general solutions might exist for ODE (Ordinary Differential Equations)'s as well :)...
@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.
2:18 PM
@ephe casework is how I do it too, have you shown it for both a convex and a concave quadratic?
@shintuku I forgot about the concave case but I have done that now.
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.
@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
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.
Let $G$ be a countable discrete abelian group. Is the pontryagin dual of $G$ polish? If so, are you able to provide a reference?
Then I guess complex manifold should do it
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
@Ajay We have the compkex-geometry tag, which sadly gets misused most of the time.
its proper use is, of course, to talk about complicated shapes
2:46 PM
@TedShifrin what? is it not for Euclidean geometry in the complex plane?
@shintuku False.
@shintuku No, that is how you misuse the fractal geometry tag.
@Thorgott Ah, very nice! Separable and metrizable, so it is polish. Thank you!
Polish ≠ polish. :P
@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.
3:01 PM
ok, on that same function can you do the sumfor n-1 is smaller than the sum for n?
@XanderHenderson I remember a really funny big bang theory episode about this
otherwise, go back an do it for a linear function, then for a quadratic concave, then convex, and then get to this one @ephe
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
@Ajay I don't believe you. You are positing a world where Big Bang Theory is funny.
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 ...
@PrithuBiswas Just saw that last night.
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
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.
3:29 PM
@A016090 This is UC break week. Do you have classes this week?
(University of California)
@Ajay (1) Humor is not a democracy, (2) an affirmative vote is an "aye", not an "I".
@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!
@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.
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.
@XanderHenderson eye, eye, sir!
Just counting facial features.
3:36 PM
@ephe once you got the sine version down you should be able to generalize
or admiralize
@XanderHenderson Do parliamentary motions then follow Newton's first law?
@ephe if your function is continuous and nonconstant, there's no more general case than a nonrepeating wavy function
@A016090 One combined office vs an office per department?
@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.
3:39 PM
yes do the most concrete case first to see if you can prove it, then work your way up
@A016090 Sure.
@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.
@A016090 efficiency at the cost of usefulness.
Unfortunately so.
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.
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.
bad singularity near $x=1$
It is its own inverse, though
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 $x=0$?
@A016090 1. $\log(1) = 0$, hence $1/\log(1)$ is problematic.
@A016090 There is a singularity at $x=1$, but the edge of the domain of definition is $x=0$
Though regarded as a complex function, there is an essential singularity at 0.
4:01 PM
@XanderHenderson I think the singularity at $x=1$ is an essential singularity. $x=0$ is the end of a branch cut
@XanderHenderson I have blundered in not realizing this, thank you.
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?
I kept looking at just $\log(x)$
@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.
If one considers too many branch cuts of log, does it become sawdust?
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.
@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
@XanderHenderson What have you been doing, if not math?
math - actual math = imaginary math?
Finally, we can find the argument of math.
And the absolute value of math
4:18 PM
Would that be the distance of math from the origin?
Math has absolutely zero value.
@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.
@TedShifrin Does that mean that math is at the origin of everything?
@robjohn It must.
@TedShifrin or that it is as cold as it gets.
4:22 PM
Keep Kelvin out of this :)
@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.
It only takes a permutation to turn a cheater into a teacher.
@robjohn ... :(
And inversely.
indeed, as permutations are injective and surjective
4:26 PM
That was clever, though
I feel like invective comes next.
I get that a lot
4:39 PM
Hmm, I miss Kelvin and Hobbs.
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 $b$
Well, since $1\equiv 0 \pmod b$ is nonsense ... you know that's a typo
good grief, 20 min wasted. I knew it didn't make sense.
Yeah, I know what you mean.
What they may have had was $a^{\phi(b)}-1\equiv 0$.
Ooooh, you're probably right.
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?
well, he doesn't exclude that in the definition
but it's nonetheless impossible by uniqueness of power series
@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.
Skepticism through the roof bruv, but it just might be correct
@A016090 I'll start doing that!
yep I missed the factor of y oof
5:10 PM
@TedShifrin thank u very much sir for such an easy and comprehensive answer
Q: Image of curves in the complex plane

Snickett I'm not really sure what I'm being asked in this question. If $x=C,y=C$ doesn't that mean $z=C+iC$?

5:26 PM
If I have two integers $x$ and $y$, what's the easiest way to get -1 from them, or perhaps even just one of them, via extrinsic operations?
why doesn't $x-x-1$ work
@shintuku because what I actually have is some $\frac x y$, and I need to compute $\frac{x -1}{y}$
See the image, second equation from the bottom
So what I need is necessarily some transform $T$ such that $T(\frac x y) = \frac{x-1}{y}$
$1 - \frac x y = \frac{y-x}{y}$, etc., but that's all I got for now.
why not $f(k, y) = k - \frac{1}{y}$
We don't have $\frac 1 y$
It's easy enough if $y$ is a Mersenne number as that guarantees $x - (y - x) = 1$, but I'm interested in the general solution here.
Should say really what I have is $\frac {2^{\lfloor\log_2(y)\rfloor}} y$
So you have that, $2^a y = z$, and every mathematical transform in existence. How do I get $\frac 1 y$?
5:59 PM
heyo @Thorgott, can I ask you a topology question?
@AMDG Then $\frac{2-1}3=T\left(\frac23\right)=T\left(\frac46\right)=\frac{4-1}6$. Since $\frac13\ne\frac12$, this cannot happen.
@robjohn error in your reasoning :(
@AMDG where is the error?
$2-1\neq 4-1$
You said that $T\left(\frac xy\right)=\frac{x-1}y$
and $\frac23=\frac46$
6:09 PM
And what you have in the middle is an equation $T(\frac x y) = T(\frac{2x}{2y})$
No, if there is a function $T$ with the given property you say, then $T\left(\frac23\right)=T\left(\frac46\right)$.
since $\frac23=\frac46$
They're the exact same number in essence but with different factors
No, they are the same number
There is no such $T$
i.e., two unique representations of the same number
It's complex-valued if you wish :P
what does that matter?
If $x=y$ then $T(x)=T(y)$
for any function $T$
6:14 PM
Well I said transform, not function.
Here obviously not bijective
I never said bijective
nor injective
nor surjective
just a function
Don't functions have to be bijective by definition tho anyways?
If all you have is $\frac xy$, then there is no way to get $\frac{x-1}y$
unless you know more
that would enable you to compute both $x$ and $y$
@robjohn Oh ok. Thanks for the correction; but in any case...
never mind, I figured it out!
6:22 PM
@robjohn I'm probably not explaining well, and I think I just got confused by what you were doing. I'm saying $T(\frac x y) = \frac{x-1}{y}$ is the transform itself, so using T as you have here, getting a contradiction $\frac 1 3 \neq \frac 1 2$ is expected.
Can you extract useful information about the primes from a function that converges to a function that counts primes?
Indeed for $\frac 4 6$, we want $\frac 3 6$, not $\frac 2 3$
To evaluate that "transform", you need to know both $x$ and $y$
And I mean that the differences go to zero as $n\to \infty$
@robjohn Well what we have is $\frac x y$ itself indirectly as a whole, but seemingly none of its parts.
6:25 PM
Then there is no way to compute $\frac{x-1}y$
If you have both $\frac xy$ and $\frac{x-1}y$, you can compute both $x$ and $y$
There is is my point, but I'm literally trying to solve a diophantine equation, so... :D
We know what these integers are indirectly, but not their values. It's finding the side lengths of a rectangle of area $z$ and the side lengths are $2^a$ and $y$ respectively with $2^a y = z$
If you know that $\frac xy\in\mathbb{Q}$, then you can write $\frac xy$ as $\frac pq$ where $p,q\in\mathbb{Z}$ and $(p,q)=1$ and then you can write $T\left(\frac xy\right)=\frac{p-1}q$
What does the notation $(p,q)=1$ mean?
$p,q\in\mathbb{Z}$ then $(p,q)=1$ means that $p$ and $q$ are relatively prime.
We do indeed know p/q is rational
I guess that's a gcf shorthand?
6:31 PM
@robjohn So yeah, then this is what I meant to say.
This function will not have a nice looking graph
Meh, as long as I can avoid using gcd directly to compute these values by way of a closed form , it's all good.
@AMDG you'll need to use the gcd, I believe
If you look at the second equation from the bottom in the image I posted, you can see what I'm solving for.
Specifically $y$ in that equation
If we had either $a$, $2b + 1$, or $y$ in that equation, then we'd have this solved, and a closed form (without GCD) for computing the factors of $b$ in any $z$ for all integers $b$ and $z$, but all we have is $z$, $2^{\lfloor\log_2(z)\rfloor}$, and every mathematical operation under the sun to be applied to $z$ which is also in closed form.
And if we could compute $\frac {p - 1} q$, then we'd already have the answer.
6:45 PM
See now if only there were an easy way to construct a pair of integers instantly for any arbitrary rational number given...
The differences between the green and blue go to zero. Blue counts primes, so I'm not sure what I'm missing here...how could a function converge to some prime counting function
Because yes
says every undergraduate math teacher ever, probably
6:59 PM
@robjohn Phrased differently, $2^{n + \log_2(m) - \lfloor\log_2(m)\rfloor} = m$
is what we are equivalently solving.
The "fractional" part of the logarithm is identical with the value which produced it through the logarithm, so it's ultimately saying you can recover $m$ just from the "fractional" part somehow.
At least for integers
but $2^{\log_2(m) - \lfloor\log_2(m)\rfloor} = 2^{-\lfloor\log_2(m)\rfloor} m$
In the case of the $\frac p q$ mentioned earlier, well we have some $2^a m$, and we can compute $2^{-\lfloor\log_2(m)\rfloor} m$ via $2^{-\lfloor\log_2(2^a m)\rfloor} (2^a m)$, but we don't know $a$ nor $m$, only $2^a m$ as a whole integer.
With $m$ a Mersenne number, computing $2^a$ is trivial as we can negate $2^a m$ and take its logarithm to compute the factors of two directly since it corresponds with the number of bits in this case.
Negate as in $2^{\lfloor\log_2(2^a m)\rfloor + 1} - 2^a m$.
Therefore $\lfloor\log_2(2^{\lfloor\log_2(2^a m)\rfloor + 1} - 2^a m)\rfloor = a$
How to generalize that, I don't know, but its generalization would give the true value for all $m$.
@ShaVuklia glad you figured it out then
7:25 PM
FWIW, on Physics.SE, we recently had a HNQ on negative probability:
Q: Do we actually need negative probabilities in quantum mechanics?

Mikayla Eckel CifreseI was reading this thread and I'm a bit confused. The answer says negative probabilities can account for destructive wave interference and the events cancelling out. But if events just cancel out, shouldn't that make the probability zero? Why would it be negative? Additionally, my (possibly in...

Im a bit surprised. Is this not an interesting question/phenomenon ?
Q: Special numerical method for $\sqrt 2$ with rational functions

mickNumerical methods for approximating Pythagoras' constant $t =\sqrt 2$ by fractions. (This is an idea from my mentor while he was barely $13$ yo, as a response to a challenge). We all know Newton's method for finding $t$. It converges quadratically meaning like $o( C x^{2^n} )$ where $C$ is a cons...

about as many upvotes as downvotes
opinions differ !
@PM2Ring interesting !
Well, it certainly attracted a lot of attention. OTOH, many physicists are notoriously sloppy with mathematics. Or, at least, are sloppy with their mathematical notation / verbiage. ;)
7:52 PM
I wanted to ask a HNQ but only got 2 upvotes and 35 views
Is there any way that maybe digamma could be used to compute p and q? mathworld.wolfram.com/GausssDigammaTheorem.html
These p and q hold, so this closed form can be used.
8:05 PM
How did Gauss find this anyways?
@geocalc33 HNQ ?
@mick Hot Network Question
@geocalc33 LINK
What makes the question hawt?
@AMDG i assume recursion and mod arithmetic
8:12 PM
I mean your gauss is as good as mine (bah dum tiss).
What I'd rather have is their techniques and intuition
Gauss, Euler, etc.
Save for finding it by accident, they had to have intuition prior due to the linear nature of reasoning. I mean that's how I realized $xy = 2^p (2^q - 1)$: I began by understanding rationals themselves, then the algebra naturally followed to model them algebraically.
Rational philosophy gets hijacked by rational analysis, nice.
every mathematician had good and bad parts.
Mertens theorems were for instance much more formal than cauchy and co who already gave " nonmodern " proofs
So I would think that Gauss et al. had moreso a grasp on the integers and the relations they worked with as essences more than just that they had a fine set of tools.
I still don't get what Zeta is, but I think Gauss nearly had its comprehension, and had he found it before his death, he probably would have solved his own conjecture.
gauss had many conjectures , Riemann made the zeta conjecture so euhm pardon me , which conjecture ?
8:37 PM
I came up with a conjecture. Does anyone want to hear my conjecture?
9:03 PM
@geocalc33 Yes
@A016090 Geocalc's (vanishing holes) conjecture: There exists a decomposition of $X$ by once punctured planes with no singularities.
@mick lol I can't believe I made that mixup :joy:
And now for something completely different
Suppose I have an NxN matrix of zeros and ones. All but the diagonal are ones. I fill in one of the zeros at random. Can that change be detected including which row/column was changed?
I do not think anyone can solve RH
9:20 PM
Now let's take an equivalent formulation: I have the sum of all possible integer configurations $$\sum_{n=1}^k \frac 1 2 k (k - 1) - n = \frac 1 2 k (k^2 - 1)$$. I add some $a + \log_2(x)$. Can I detect what the value of $a$ was?
Can you point us to a research paper where this is explored?
Well you're reading it I guess :P
In any case, it isn't hard to see that the above sum permits arranging as a $k\cdot k$ square of terms of the integers with zeros along the diagonal. Adding some $a + \log_2(x)$ for which it is given that $k \geq a + \log_2(x)$ makes restores one of the zeroed terms along the diagonal while adding $\log_2(x)$ onto the whole.
Each row is the integers in the range $[1, k]$
Now given these constraints, we know that $a$ is in fact going to make one of the zeroed terms non-zero exactly, and it doesn't matter which row it is: if we then subtract $\frac 1 2 k(k - 1)$ from this new sum, we zero out that row (or column if you tilt your head).
And that's about where I'm at right now
neat. I hope you continue to pursue these ideas, find new perspectives and advance your mathematical knowledge over time
9:35 PM
Thenks, me too
9:48 PM
@AMDG Are the indices on the sum correct? That formula does not look correct. Perhaps it is just that the ${}-n$ might need parentheses, otherwise the $n$ is not in the scope of the sum.
Parens it is then
$$\sum_{n=1}^k \left(\frac 1 2 k (k - 1) - n\right) = \frac 1 2 k (k^2 - 1)$$
Is that what you meant?
9:56 PM
It is equivalent to $$\frac12k\left(k^2-2k-1\right)=\frac12k(k^2-1)$$
which is only true when $k=0$
I'm confused. What is the LHS relating this sum on the RHS to?
I added the parentheses and evaluated
I just used wolfram alpha. How do you evaluate sums for closed forms?
Oh I see, I guess I used the wrong sign for the constant term in the sum
lol so then what I really meant was $$\sum_{n=1}^k \left(\frac 1 2 k(k + 1) - n\right) = \frac 1 2 k(k^2 - 1)$$
10:31 PM
I think I figured it out.
10:45 PM
Yes, that is correct
It's a square. The diagonal parts are in fact the integers in $[1, 5]$. $\frac 1 2 k(k^2 - 1) + a + \log_2(x) = c$ fills in one of the zeroed terms in the diagonal. If we fill in the rest of the spaces of the square and subtract the terms of the square, then we get $\log_2(x)$.
It should be possible to construct this square without $x$ as a number $b$, and then $b - c = -\log_2(x)$, right?
Err wait sorry.
That should give the unfilled parts of the square of terms and $-\log_2(x)$ as the difference $b - c$.
11:35 PM
But is there a way to compute a finite sum back from the closed form?
I'd like to know about $\frac 1 2 k (k - 1)$ and $\frac 1 2 k(k^2 + 1)$'s sums

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