00:00 - 19:0019:00 - 23:00

12:06 AM
Would it be on-topic here to ask the number of possible combinations of inputs for a function that takes an input length of anywhere from 1 to 2^128 bits? I know the answer for a function that takes exactly 2^128 bits (it's 2^2^128), but I'm wondering if this basic question is acceptable here.
There's surely a better way to calculate it than 2^(2^1 + 2^2 + 2^3 + ... + 2^128).

12:19 AM
Can anyone confirm his claim that n leq 23 from a 90 degree bend radius? I cannot confirm his result from any of my calculations

I guess what I'm asking is whether there's a better way to say $2^{\sum_{i=1}^{128}2^i}$ .

@forest: Of course there's the formula for the sum of a geometric series.

@TedShifrin Not sure what that is, or if it'd be relevant here (my math needs work).

$\sum\limits_{i=0}^n r^i = \dfrac{r^{n+1}-1}{r-1}$

Yeah that's not what I need.

12:26 AM
So, in your case, with $r=2$, you get $2^{129}-1$.

oh

Then $2^{2^{129}-1} = \frac12 2^{2^{129}}$, which is not what you typed originally.

Would that answer how many possible inputs there are to a function that takes a 1 to 2^128 bit input?

No, I think you need to put the sum sign somewhere else for that.
If you have a $2^k$-bit input, then you get $2^{2^k}$ possible inputs, right?

Isn't $2^{\sum_{i=1}^{128}2^i}$  the correct formula then?
@TedShifrin Yes, but you can also have an input length anywhere from 1 to 2^k.

12:29 AM
No, I think it's $\sum\limits_{i=1}^{128} 2^{2^i}$.
Which is WAY huger. Or is it?

d'oh

We're comparing $2^{a+b}$ with $2^a+2^b$. No, your original sum formula was way huger.
Aha, so now $2^{2^{128}}$ gives the right order of magnitude.

Yeah that's the correct order of magnitude, but it's not exact.

(Not the right answer exactly, of course.)
I don't know a closed-form expression for adding $2^i$th powers.

So $\sum\limits_{i=1}^{128} 2^{2^i}$ is correct? For information, this is because I'm trying to correct an answer I have on another site where I incorrectly claimed that SHA-512, a hash function that takes an input message up to 2^128 bits, had an input message space of 2^2^128.

12:33 AM
I don't know the computer jargon.
It seems to me this formula is wrong, though. You're way over-counting.

It's just a (mathematical) function that takes an input that's typically considered to be arbitrarily sized, but is actually limited at 2^128. Why would that be over-counting?

Isn't having $N$ bits the same as having $N-k$ bits with the last $k$ bits known to be $0$?

Not for this function (it encodes the length internally).

Because I think a $2$-bit function gets counted when you count $4$-bit functions.
Yeah, but encoding just the length is a tiny order of magnitude.
So you're saying that you can't absorb a $2$-bit function into a $4$-bit or $8$-bit function? If so, the summation formula I wrote down is correct.

What do you mean?

12:37 AM
I said overcounting because you can think of a length 2-sequence as a length 4 (or 8 or whatever) sequence where the last digits are all $0$s.
If you say that's wrong, then the summation formula is correct.

The original one I posted or the corrected formula you gave me?

The corrected one.

$\sum\limits_{i=1}^{128} 2^{2^i}$ right?

Yup.

Then I guess there's no simpler way to say "There are X possible distinct inputs" without having $\sum\limits_{i=1}^{128} 2^{2^i}$ in place of X? I should calculate it...

12:39 AM
I don't know a formula for that, as I said.

oh
If there are $\sum\limits_{i=1}^{128} 2^{2^i}$ possible lengths in bits, then that would be the exponent for 2. Unless I misunderstood.

No, you have to count the functions for each length and then add up.

For length 1 it's 2^1, for length 2 it's 2^2, for length 2^128 it's 2^2^128.
Then I just sum those, right?

That was the formula I wrote down.

But then that is the formula that I'm looking for (the number of combinations).

12:42 AM
I think you're wrong, but I am not continuing this.
I am packing for a move.

Ok. Thanks for taking your time and putting up with a novice like me. :P

Good luck.

thanks

1:15 AM
It was a lot simpler than I expected:
1

There is one zero-length message, two 1 bit long messages, four 2 bit long messages, eight 3 bit long messages, and so on. Since these messages are distinct, you can simply use the formula: $$1 + 2 + 4 + 8 + \dots + 2^{n-1} = 2^n-1$$ Note that the maximum message length is the same as the maxim...

2^(2^128 - 1) - 1

1:53 AM
Trying to hash this out in some code I'm writing that models the probability of die rolls in a pool of die rolls (like "roll 6d6") - how do I model an conditional effect like "remove a die and reroll any amount of remaining dice"? Do I have to split the entire pool by conditional probability, like P(using the effect) * P(die outcomes minus the die removed by reroll) + P(not using the effect) * P(current outcomes)? Sorry if this makes no sense, my probability skills are way out of date.

4 hours later…

1 hour later…
7:11 AM
Morning all

7:46 AM
Morning @ÍgjøgnumMeg

Hey @Alessandro, you in Italy yet? :P

Yep, I landed around midnight

Cool, have fun :) How long you there for?

Until the winter semester begins I guess

Fair

2 hours later…
9:56 AM
@BalarkaSen spectral sequence computation of K(Z,2) is dank

1 hour later…
11:00 AM
The solution of $e^x=x$ is expressed as a complex number of the form $a+bi$. Are $a, b$ transcendental?

@Mathphile Are you saying it has a unique solution?

Interesting. Is that obvious?

according to WA it is $-W(-1)$
although i do not know how we get that result
also i wonder if there is a closed form for $a$ and $b$
@Semiclassical any ideas?

11:21 AM
no it has infinitely many solutions
WA says it is $-W_n(-1)$
I'm thinking of using Rouche's theorem to prove it
anyway
@Mathphile if a and b are both algebraic, then $z=a+bi$ is also algebraic, so $e^z$ is transcendental (Lindemann theorem), contradiction
so at least one of them must be transcendental

@LeakyNun okay
@LeakyNun would $-W(-1)$ be one of the infinite solutions?

sure

btw what does $n$ mean in $-W_n(-1)$?

The WA solution is a bit misleading: the Lambert-W function has an infinite number of branches, so you need to say which one you’re using. It’s analogous to saying that $y=x^2$ implies that x is the square root of y: Which square root?

@Semiclassical i see
If we take logarithm on both sides of the first equation:$$\ln e^x = \ln x \implies x=\ln x \implies e^x = \ln x$$
also WA shows only two solutions to $\ln x=e^x$
does this mean that there are only two solutions to $e^x=x$?

11:39 AM
There are two real-valued solutions

@Semiclassical what do you mean by real-valued?

solutions which are real
There’s complex solutions as well

well aren't both of those solutions complex numbers?
$x ≈ 0.3181315052047640944061139753 \pm 1.337235701430689415856543101 i...$
i don't see any real solutions
also i don't think there plots can ever intersect

Yeah, I was being careless

@Semiclassical so are there only two complex-solutions to $x=e^x$?

12:34 PM
@Semiclassical how do you prove that there are infinitely many solutions?
(saying that they are $-W_n(-1)$ is cyclic)

@LeakyNun Well, if you can show that those are solutions, then that gives you a proof

I thought they're just defined as the $n$-th solutions

So if $x$ is a solution and $a$ is some number such that $ax = x^a$ and $e^a = a$ then $ax$ is also a solution. Right?
Probably not a great way to proceed

12:53 PM
5

A Poincaré map, by definition, is supposed to be a reduction of a phase diagram on a codimensional plane cutting normal to the original trajectory. In this form, one should plot points from the phase space (xdot vs. x), maybe in this case it makes sense to plot a-dot versus a (semi-major axis). B...

if there are infinitely many solutions then why does wolframalpha only show two?

I've just added a +200 bounty to the Poincaré map question. It needs a practical (rather than pure mathematical) answer.

@TobiasKildetoft i am really confused. Are there two solutions or infinte?
could anyone explain why would there be infinite solutions to $e^x=\ln x$?

@Mathphile When did you change the equation?

1:04 PM
@Mathphile You started with $e^x = x$

as i showed above there are equivalent
1 hour ago, by Mathphile
If we take logarithm on both sides of the first equation:$$\ln e^x = \ln x \implies x=\ln x \implies e^x = \ln x$$

ahh, right, sure
but why complicate it further?

@TobiasKildetoft because wolframalpha only gives an exact solution to $e^x=\ln x$

Can anyone confirm his claim that n \geq 23 when Theta = 90deg and t=3? defproc.co.uk/blog/lattice-hinge-design-minimum-bend-radius

@TobiasKildetoft i.e. $x ≈ 0.3181315052047640944061139753... \pm 1.337235701430689415856543101 i...$

1:07 PM
@Mathphile How is that an exact solution?
And didn't it give a solution involving the Lambert omega before?

Didn't it say that $-W_n(-1)$ were solutions?

you can check it out above
@TobiasKildetoft that was when i typed in $e^x=x$ not $\ln x=e^x$

Right, so how did this in any way improve things?
You went from a nice complete solution to an approximate one

what i am trying to say is that i think there may be only two solutions and not infinite
and the only two solutions are : $x ≈ 0.3181315052047640944061139753... \pm 1.337235701430689415856543101 i...$

1:11 PM
@Mathphile Why would you conclude that from this?
You have even taken a logarithm, which in itself basically ensures you get infinitely many values

@TobiasKildetoft because according to wolframalpha, there are the only two solutions that exist for $e^x=\ln x$

@Mathphile Does WA claim it to be a complete solution? Because it does only give it as numerical

thus there are only two solutions to $x=e^x$

@Mathphile Getting from the first to the last involved taking a logarithm, which is something you have not accounted for yet

@TobiasKildetoft could you give me an example of another numerical example other than this?

1:16 PM
@Mathphile Not sure what you mean. I am just saying you have not shown the equations to actually be equivalent.

@TobiasKildetoft i mean that could you prove that $x ≈ 0.3181315052047640944061139753... \pm 1.337235701430689415856543101 i...$ are not the only two solutions?
Assume u satisfies $e^u=u$
then $\ln (e^u)=\ln u$
which means $u=\ln u$
which we get $e^u=\ln u$
i don't understand how is this not true?

@Mathphile Probably they are the only solutions to the new equation. But you are taking a logarithm to get from one to the other

@TobiasKildetoft yes but as long as i am taking the logarithm on both sides it should be fine

no, it will not, since these are complex numbers
Anyway, even once you pick a branch, you can not just go back again, so you have a correct implication, but no reason to expect to get all solutions (which would then be obtained from other branches)

$x=\ ln x$ also seems to have the same solutions
@TobiasKildetoft will doing something like $e^(x)=e^{(\ln x})$ also change the solutions due to complex numbers

1:33 PM
Btw, if I type “solve x==e^x” into wolfram alpha, I get “-W_n(-1) for integer n”
Which is to say: infinitely many solutions, indexed by n

@Semiclassical okay but are there only two solutions for $e^x=\ln x$?

Dunno. WA only reports two solutions to “solve x==ln x” tho
Both in terms of the W function

@Semiclassical but if click on approximate forms you would get $x ≈ 0.3181315052047640944061139753... \pm 1.337235701430689415856543101 i...$

@Semiclassical Presumably one solution per branch (it just picks one before solving)

My guess is that it depends on which branch of the log function you apply to both sides

1:40 PM
so we can prove that there are infinite solutions?

The solutions it reports are $e^{-W(-1)}}$ and $e^{-W_{-1}(-1)}$

@Semiclassical yes

For the rest I’d guess one must consider x +2pi i n = ln(x) for integer n
Any solution to one of those will be a solution to x==e^x

@Semiclassical okay

There is an interesting question there: which branches of the log function yield which solutions of x==e^x?

1:47 PM
@Semiclassical are there an infinite number of branches which yield solutions?

Yep
Looking at the numbers for the x==e^x solution, the rule of thumb seems to be “which multiple of 2pi is the imaginary part closest to”
The 0.318 +/- 1.33i solutions are closest to 0, so they get grouped together upon taking logs

@Semiclassical you could ask this on MSE if you want

The next pair of solutions are 2.06 +/- 7.56i, and the imaginary parts are closest to +/- 2pi respectively
So they’ll get grouped differently
So it’s not quite “one per branch of the log” but the only exception are the first two which land in the same branch @TobiasKildetoft

2:05 PM
leakynun proved that if $a+bi$ is the solution for $x=e^x$ then at least one of $a, b$ must be transcendental
Can we prove that both $a,b$ are transcendental?

@Mathein Mein Zulassungsbescheid ist gekommen :D

@ÍgjøgnumMeg My German is a bit rusty. Something about a message related to lassos?
(or does it mean letter of approval?)

The wiki page for the W-function (en.m.wikipedia.org/wiki/Lambert_W_function) gives an argument in the section “Special values” that W(x) is transcendental for any nonzero algebraic x
Via Lindemann-Weierstrass

@Semiclassical But that still only accounts for one coordinate

2:13 PM
@Semiclassical it only implies that at least one of $a$ or $b$ is trancendental, not that they both are

yes

@Tobias letter of acceptance :)

@ÍgjøgnumMeg Ahh, so my guess was not that far off

I had a big problem with some bureaucracy that meant I received a rejection letter late last month
lol

2:16 PM
@ÍgjøgnumMeg Actually, would zulassung not be closer to "admission" than "acceptance"? Or is my German truly rusty?

it does imply that a+bi and a-bi are both transcendental, but yeah—no guarantee that both coordinates are

@Tobias well Zulassung is admission yeah
but letter of acceptance is more standard in Englisch

Eg, e+i and e-i are both transcendental
But 1 certainly isn’t

right, the bescheid means "message", right?

Hello, people

2:17 PM
@Semiclassical it does guarentee that at least one of the coordinates is transcendental

Bescheid geben means "notify"

Sure
And if I had to guess, it’d be that both coordinates are transcendental for the x==e^x solutions. But I have little hope to prove that

@Semiclassical yes
and that's what i want to prove

@ÍgjøgnumMeg Anyway, congrats. Where did you get admitted to?

although i think it may fall in the same category of difficulty as proving $\pi+e$ as transcendental

2:22 PM
@Tobias Thanks :) Heidelberg

cool

Yup! So my scholarship is now officially takes effect
since it was tied to admission to Heidelberg

So this is for a master's, or for PhD?

For a Masters

neat

2:24 PM
2 year masters, hopefully I can get a PhD position somewhere nearby, or in Heidelberg afterwards
Otmar Venjakob is a big name in Iwasawa Theory so that would be neat

@Semiclassical i wonder if an example exists where one of $a,b$ is algebraic
@Semiclassical also if $e^{-W(-1)}=a+bi$, then are there some closed forms for $a, b$?

@ÍgjøgnumMeg I could not seem to find a list of their math faculty anywhere. Who else is there?

Gebhard Boeckle (Mathein's supervisor)
@Tobias mathinf.uni-heidelberg.de/personen.html this is the list of faculty I believe

I see. No names I recognize, so presumably not anyone in my (previous) field.

2:39 PM
I guess so

@ÍgjøgnumMeg Do you know who your advisor will be, or will you not decide that until you start the thesis in a year or so?

@Mathphile one has to be a bit careful with the statement here. There are certainly values of x for which x==W(x) e^W(x) has nice solutions

@Tobias I don't know how it works yet, presumably one picks later in the year

ok, that makes sense

For instance, if x==2e^2 then W(2e^2)=2 is algebraic

2:46 PM
(was the same way it worked when I did my masters in Denmark)

But 2e^2 isn’t

Ah fair, I'm hoping to link my master thesis to my undergrad
so it would be nice to have Venjakob
but obv one can't necessarily guarantee that lol

By contrast, -1 is certainly algebraic and so W(-1) is transcendental for any branch
The question is whether any of the real/imaginary parts happen to be nice

@ÍgjøgnumMeg There is also something to be said for branching out a bit from your undergrad, in order to gain a broader base.

My guess would be no but I can’t back that up

2:48 PM
@Tobias well my undergrad was mostly applied mathematics, I mean from my undergrad dissertation

Still

from which all the material was self-taught lol

3:03 PM
I'll see what I like lol
@Tobias I haven't had enough exposure to other areas of mathematics because my undergrad was so inadequate, so maybe I'll find something else interesting :)

3:14 PM
@LeakyNun Try K(Z, 3)

@BalarkaSen doesnt the same approach work
since ΩK(Z,3) = K(Z,2)?

Looped K(Z, 2) is K(Z, 1) = S^1, which has much simpler homology groups
Compute it, you'll see

oh I can foresee the problem
but let me revise K(Z,2) first
so we take the fibration ΩK(Z,2) -> PK(Z,2) -> K(Z,2)

@RyanUnger Lol @ Mochizuki

somehow PK(Z,2) is contractible
and ΩK(Z,2) is S^1

3:16 PM
You should be able to prove PX is contractible for any X.

just contract through the paths right

Ya
"Slurping the spaghetti"

lol I like that
so there's a spectral sequence E with $E_{p,q}^2 = H_p(K(Z,2); H_q(S^1))$ that converges to $H_\ast(PK(Z,2))$

Mhm

since $PK$ is contractible, we have $E_{p,q}^\infty = 0$ except $E_{0,0}^\infty = \Bbb Z$
and beyond $E^3$, the arrows either point to 0 or start from 0, so $E^3 = E^4 = E^5 = \cdots$, so $E^3 = E^\infty$

3:21 PM
@BalarkaSen did you look at it
IUTT is just the arithmetic version of computing the Gaussian integral

I saw some Gaussian integral stuff

turns out Scholze and others just don't understand it because they never learned calculus properly
smdh

so $E^3_{p,q} = 0$ except $E^3_{0,0}=\Bbb Z$

Yup

this means that $E^2_{p,1} = E^2_{p+2,0}$

3:26 PM
What's the news

so $H_{p+2}(K(Z,2)) = H_p(K(Z,2))$

Leaky is learning spectral sequences.

so it is $\Bbb Z$ when $p$ is even and $0$ when $p$ is odd
and in the case of K(Z,3) we don't have finite support on K(Z,2)... this is not good

but on even pages the differentials all point to 0 / away from 0
so $E^{2r} = E^{2r+1}$?
how do we know its limit even exists

3:28 PM
Ya but that doesn't tell you much.
I don't actually know if K(Z, 2) -> PK(Z, 3) -> K(Z, 3) spectral sequence even degenerates at a finite page. I doubt.
@MikeMiller surely knows
Ya it doesn't.
You're going to need the ring structure for this

but this is homology not cohomology
page 2: $\begin{array}{c} H_0(K(Z,2)) & H_1(K(Z,2)) & H_2(K(Z,2)) \\ 0&0&0 \\ H_0(K(Z,2)) & H_1(K(Z,2)) & H_2(K(Z,2)) \\ 0&0&0 \\ H_0(K(Z,2)) & H_1(K(Z,2)) & H_2(K(Z,2)) \\ \cdots\end{array}$

Right, that's a problem, I guess. For K(Q, 3) with rational coefficients, homology and cohomology are the same. The Ext term doesn't exist.

how do you know the rank is finite...?

K(Q, 3) is a rational space. It's integral homology is a Q-vector space

but the spectral sequence must converge to the homology right, according to the theorem
so it must degenerate...?

3:34 PM
No, why?
Degeneration is like the sequence eventually becomes constant
It can converge without being degenerate

oh ok I see

3:45 PM
$H^n(K(\Bbb Z, 2), \Bbb Z) = [K(\Bbb Z, 2), K(\Bbb Z, n)]$ is in bijection with natural transformations $H^2 \Rightarrow H^n$ by Yoneda lemma. If $n = 2k$ this has to be generated by $\alpha \mapsto \alpha^k$, but if $n$ is odd there's no such thing.

using yoneda lemma unironically?

It's so useful m8

@BalarkaSen if you want to read a quick paper today, check out "the unknottedness of minimal embeddings" by Lawson.
Closed minimal surfaces in $S^3$ are isotopic to the standard genus $g$ surfaces.

page 2: $\begin{array}{c} Z&Z&Z \\ 0&0&0 \\ Z&Z&Z \\ 0&0&0 \\ Z&Z&Z \\ \cdots\end{array}$
page 3 the same
how to compute the differentials? @BalarkaSen

You need to do cohomology spectral sequence.

3:53 PM
@BalarkaSen a related idea is that in any Ricci positive manifold, compact minimal surfaces must intersect

@RyanUnger That sounds very nice. I will look at it soon, but not today.

a cool application is that a minimal surface in $S^3$ has to intersect every equator

@LeakyNun Hold on a second. It's H_*(Base, H_*(Fiber)). What did you write there?

yeah I know lol
I didn't bother to correct it

try to be careful if you want my friendly advice
i usually make tons of errors
and it takes me ages to figure out what went wrong
spectral sequences are the worst tool of mankind

3:59 PM
this is also a quick read
and uses similar ideas
it's all about the morse theory for normal geodesics

i gotta scurry. i'll check these out later for sure
thanks

@BalarkaSen im preparing you to read my work in preparation

ok so there's a spectral sequence converging to Z on (0,0) and 0 otherwise, whose page 2 is: $\begin{array}{c} H_0(K(Z,3)) & H_1(K(Z,3)) & H_2(K(Z,3)) & \cdots \\ 0&0&0 \\ H_0(K(Z,3)) & H_1(K(Z,3)) & H_2(K(Z,3)) \\ 0&0&0 \\ H_0(K(Z,3)) & H_1(K(Z,3)) & H_2(K(Z,3)) \\ \vdots\end{array}$
or cohomology idk
I haven't learnt about the cohomology one
@BalarkaSen I still don't know how you compute the differentials
anyway from this we already can deduce that $H_0(K(Z,3)) = \Bbb Z$, $H_1(K(Z,3)) = 0$, $H_2(K(Z,3)) = \Bbb Z$

5:02 PM

@LeakyNun This is a small part of what's called the Hurewicz mod C theorem. In fact any simply connected CW complex $X$ with finitely many cells in each dimension necessarily has all homotopy groups finitely generated. (Clearly this extends to finite fundamental group.) It is not true in general: $\pi_2(S^1 \vee S^2) = \pi_2(\vee_{\Bbb Z} S^2) = \Bbb Z^\infty$.
@LeakyNun In the approach you're using, you compute them by knowing what the $E^\infty$ page is, and so get through formal arguments that the differentials have to do something or another
Using that to compute a lot more differentials is why Balarka is saying to use cohomology; you have a ring structure and the differentials satisfy a Leibniz rule

5:46 PM
@LeakyNun Are you looking at the Leray-Serre spectral sequence?

Spectral sequences seem like a mathematical equivalent of Cthulhu, in the sense that if you don't know enough algebraic topology then you can go on in life blissfully ignorant that such things exist.
(I count myself in the blissfully-ignorant crowd.)

@Perturbative I think so

@Semiclassical I feel like this applies to most subjects in math

Another example would be near-rings

6:14 PM
@loch yeah
heck, this line out of Hermite is very nearly Lovecraftian: “I turn away with fear and horror from the lamentable plague of continuous functions which do not have derivatives.”

You know, Lovecraft was commented as having a "constitution too weak for math."
He was sent to some secondary school to specialize in the subject and ended up leaving because he just couldn't handle it, if I recall correctly.
Which sheds light on his writing about "impossible, four dimensional geometries that drive people mad to look upon them."

6:51 PM
Hah

imagine lovecraft reading Freedman's Poincare paper

I should read some Lovecraft. I know very little about him. He seems to have a lot of influence on modern horror.
@LeakyNun Do you want me to walk you through cohomology spectral sequences using this example?

I get enough horror trying to read IUTT