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Axoren
Axoren
1:17 AM
Does anyone know why Munkres avoids starting his definitions with 0? Going out of his way to start with 1 in the set of $\mathbb Z_+$ rather than 0 in the set of $\mathbb N$? Is there a good reason why some mathematicians don't use 0 in their set of counting numbers?
 
user19405892
user19405892
1:29 AM
hi
does anyone know how to prove that $2^x = 3 \cdot 9^m+5$ has no positive integer solutions for $m \geq 2$?
 
Axoren
Axoren
@user19405892 Rewrite it as $2^x = 3^{2m+1} + 5$
Does that seem easier to solve?
 
user19405892
user19405892
Yes it does, but I don't know how to prove the statement
 
Eric Stucky
Eric Stucky
Axo, some people do, some people don't. I don't know much more than that; trying to think of a reason why a topologist would choose not.
I mean, $\Bbb Z_+$ is pretty unambiguously starting at 1. It's $\Bbb N$ that gets you into trouble.
 
user19405892
user19405892
@EricStucky Do you know of a way to prove $2^x = 3 \cdot 9^m+5$ has no positive integer solutions for $m \geq 2$?
 
PVAL
PVAL
@Axoren None of the content in that book depends on this.
 
Axoren
Axoren
1:44 AM
Assume it was true for some other $x = 2k$, you could factor $3^{2m+1} + 5$ into two factors $(2^k)(2^k)$. $2^k$ is an integer, is $\sqrt{3^{2m+1} + 5}$?
Rather, when is $\sqrt{3^{2m+1} + 5}$ an integer?
 
Eric Stucky
Eric Stucky
(u19: no)
 
Axoren
Axoren
@EricStucky using the @ symbol notifies people that you've replied to them.
 
Eric Stucky
Eric Stucky
You know
as mortifying as it was
 
Axoren
Axoren
Just so you don't have to skimp on @user19405892's numbers and at the same time not have to type them all
 
Eric Stucky
Eric Stucky
to have the "stop tagging me" comment starred
at least people stopped pinging me for a while...
 
Axoren
Axoren
1:48 AM
Not a fan of it? Luckily, they've made a mute button in the top right, now.
 
Eric Stucky
Eric Stucky
ooh
 
Axoren
Axoren
Mortification: optional.
 
Eric Stucky
Eric Stucky
u r hero :D
 
Axoren
Axoren
But yeah, it's just good habit to tag everyone else.
 
user19405892
user19405892
@Axoren Can you explain how you you get $(2^k)(2^k) \cdot 2^k$?
 
Eric Stucky
Eric Stucky
1:49 AM
idk, for me it's like
we're talking, right?
 
Axoren
Axoren
@user19405892 That's a period, not a cdot.
EricStucky: Yeah, but if I walk away and come back? Chat scrolls. Your message is lost.
 
user19405892
user19405892
$2^k \cdot 2^k = 3^{2m+1}+5$
why did you take the square root?
 
Axoren
Axoren
$\sqrt{2^k 2^k} = 2^k$, do you agree?
 
user19405892
user19405892
yes
 
Axoren
Axoren
So, then square-root both sides.
 
user19405892
user19405892
1:52 AM
right, so we need $3^{2m+1}+5$ to be a perfect square
but it turns out that $x$ must be odd
not even
 
Axoren
Axoren
Why must $x$ be odd?
Was that part of the original problem?
 
user19405892
user19405892
No, but see that $3 \cdot 9^m + 5 \pmod 9 \equiv 5$, which implies $x \equiv 5 \pmod 6$
thus, $x$ is odd
 
Axoren
Axoren
Right, either way, we've shown that $x$ can't be even.
 
user19405892
user19405892
yes, the tricky part is dealing with $x$ is odd
 
Axoren
Axoren
You said it like you knew it before analyzing it.
 
user19405892
user19405892
1:56 AM
yes, i had known $x$ is odd from the time i saw this question first
 
Axoren
Axoren
Consider now $2 \cdot 2^k 2^k = 3^{2m+1} + 5$
 
user19405892
user19405892
ok
 
Axoren
Axoren
This is simply $x = 2k +1$, now.
The only other case.
 
user19405892
user19405892
how do we deal with it?
 
Axoren
Axoren
Actually, this may be kind of silly, but I think it's easier to just show that $2^x = 3^y + 5$ only has 2 integer solutions.
Than to do this with radicals.
Because if you do that, it's a stronger claim.
 
user19405892
user19405892
2:04 AM
Actually, that's how i got this question in the first place. I was trying to prove that result
Now how to prove that $2^{2k+1} = 3 \cdot 9^m+5 $ has no solutions?
 
Double AA
Double AA
What's the mathjax way to write the alpha-like proportionality relation?
 
Axoren
Axoren
You've said above that $x \equiv 5 \mod 6$, so we have instead $2k \equiv 4 \mod 6$, then $k \equiv 2 \mod 6$.
 
Eric Stucky
Eric Stucky
2:21 AM
@DoubleAA: \propto
Related, detexify is great
 
Axoren
Axoren
Actually, scratch that. I don't like doing it the modulo way. Assume we know it is true for $k=2$. If $k=3$, we have $2^7 = 3 \cdot 9 ^ m + 5$ for some $m$. But we also know the following: $2^5 = 3 \cdot 9^1 + 5 = 32$
 
Double AA
Double AA
@EricStucky That does it! Thanks a bunch
 
Eric Stucky
Eric Stucky
npnp
 
Axoren
Axoren
So, we're really trying to say that $2^2 = (3\cdot 9^m + 5) / 32$
We also know that it held for $2^3 = 3 \cdot 9^0 + 5 = 8$
So we have to also show that $2^4 = (3 \cdot 9^m + 5)/8$
We know that for even $x$, there's no integer solution, so $3 \cdot 9^m + 5 \not \equiv 0 \mod 8$
Since we know that, we know that it can't be divisible by $32$, because it would have to be divisible by $8$.
 
Eric Stucky
Eric Stucky
2:58 AM
Hey Mike :)
just wanted to say, I'm interested in what you have to say, and I'll hit you up soon
But I have a final to do tonight, gotta get to it ;)
see you around
 
Mike Miller
Mike Miller
Good luck!
 
 
1 hour later…
Joshua Lamusga
Joshua Lamusga
4:21 AM
I have to find the normal vector of <7cos(t), 7sin(t), 7ln(cos(t))> at point (7, 0, 0). That point is t = 0. The normal is T/T' and T is r/r' where r is the vector I was given. I get r’: <-7sin(t), 7cos(t), -7tan(t)>
|r’|: 7sqrt(sec^2(t))
T(0) : <0, 1, 0>. This is correct.
T: 1/7sqrt(sec^2(t)) * <-7sin(t), 7cos(t), -7tan(t)>
So only the center coordinate can change since sin(0) and tan(0) are 0, but the derivative of sqrt(sec^2(t))cos(t) is 0. How is the normal not <0, 0, 0>?
 
 
2 hours later…
TheGreatDuck
TheGreatDuck
6:09 AM
does anyone know how to solve this?
4
Q: Explicit formula for floor(x)?

mickIn number theory we have so-called explicit formula's in terms of the Riemann zeta zero's. For instance to count the sum of the logarithms of the primes below some given integer. ( second Chebyshev Function ) Consider the floor function : http://mathworld.wolfram.com/FloorFunction.html Is there...

intuition analytic-number-theory floor-function
 
anon
anon
6:20 AM
the explicit formula for psi(x) in terms of zeta zeros is essentially a fourier series. by asking us not to give a fourier series, OP is essentially telling us not to answer the question. his loss.
and yours too I guess.
 
TheGreatDuck
TheGreatDuck
well then that sentence needs to be ignored
if it can be answered without using the arctan formula that was given in the other question (simply because people said it was flawed) then it should be and not left to sit and rot
 
Tobias Kildetoft
Tobias Kildetoft
6:39 AM
@AlexClark So, is it done or are those last cases taking a long time?
 
Balarka Sen
Balarka Sen
6:59 AM
@anon curious; in what sense is Riemann's formula a fourier series?
 
anon
anon
with $e^{a+bi}=e^ae^{ib}$, the imaginary parts of zeta zeros are frequencies and the real parts contribute to their amplitudes. if there were a real part >1/2, it would mean there are zeta zeros that contribute more than others to the growth of primes.
 
Balarka Sen
Balarka Sen
aha, I see.
 
anon
anon
(of course it's actually $x$ as the base of the exponentials, but that's reaslly $e^{\ln x}$, so things are just a bit logarithmic in $x$)
 
Balarka Sen
Balarka Sen
right
interesting aspect
I have never had a particular intuition for Riemann's formula, myself. I "know" how to prove it, but...
 
Vrouvrou
Vrouvrou
7:51 AM
hello
 
user 1618033
user 1618033
@robjohn hey! Did you have a chance to work on my problem?
(the limit)
 
robjohn
robjohn
8:29 AM
@user1618033 Not yet. I've been out of town for a while, and tomorrow I am proctoring an exam at UCLA. I will look tomorrow when I get a chance.
 
user 1618033
user 1618033
@robjohn No hurry with that. I wanted more to know how it seems to you in terms of difficulty.
 
robjohn
robjohn
8:45 AM
@user1618033 At first glance, it looks pretty hard. I can comment better when I work on it.
 
user 1618033
user 1618033
@robjohn Thanks. OK.
 
 
2 hours later…
user 1618033
user 1618033
10:41 AM
@robjohn Soon, if all is fine I'll start some painting courses for which I have a call too. Hope to be able to put in pictures some of my nicest creations, kind of abstract art.
Pehaps it will take some months to finish a picture, but that is fine.
 
 
2 hours later…
justin77
justin77
12:33 PM
hi every body
 
user243301
user243301
12:57 PM
@TheGreatDuck , I add here a comment about mick question in which you are interested, it is only an opinion that the problem to find such formula is in Generalized Möbious Inversion (it's like a dead end, a street withou exit) $G(x):=\sum_{n\leq x}\mu(n) \left\{ \frac{x}{n} \right\} $ ( since $\mu(n)$ is completly multiplicative) if and only if $ \left\{ x \right\} =\sum_{n\leq x}(\mu(n))^2G \left( \frac{x}{n} \right) $. It isn't neccesary a response of this opinion. . Thanks.
 
 
2 hours later…
Mats Granvik
Mats Granvik
2:41 PM
@user243301 You should learn to define the Möbius function as a recurrence instead of doing Möbius inversions.
 
 
1 hour later…
Aritra Das
Aritra Das
3:52 PM
Hey guys
I have a problem
I posted a meta question
and now it has 3 answers
but I have no clue as to which one to accept
Since this is a meta question, and the accepted answer could potentially be viewed as the verdict on the topic
Hence I want your (the community's) opinion on making this choice
Here's the question I'm talking about
25
Q: Is coloring equations good practice or bad?

Aritra DasI've noticed that though there is a way to color text or equations, in mathjax, though very few users use it. On the otherhand, those who do, use it regularly. To me, coloring seems a nice way to remove the barrier that digital or printed text entails - in a class, a professor might show with t...

discussion formatting best-practices typesetting
Which answer should I accept?
 
Mats Granvik
Mats Granvik
@AritraDas Do as I do, accept the answer with the most votes. Then for a while feel how it is to have a bad conscience, and wait for the feeling to pass. Then you know that that behaviour has become part of your character.
 
user147690
4:33 PM
@TobiasKildetoft Done. All have failed. Congratulations!
 
Aritra Das
Aritra Das
5:02 PM
@MatsGranvik That was some solid advice right there.
Thanks
 
Brody
Brody
5:41 PM
I have thoughts on a certain question, namely
1
Q: Functions and Derivatives

Jawad AnwarGeneraly curious: Let there be a set of functions: Will the sum of the derivatives of the functions be equal to the derivative of the sums?

functions derivatives
I want to say, that for finite sets of (arbitrary) functions (of one variable), it is not necessarily true that the sum of derivatives is the derivative of the sum
take the counterexample: let $f_1(x)$ be the Dirichlet function of $x$ and $f_2(x)$ be the same with arguments reversed (1 for $x$ irrational, 0 elsewhere)
isn't their sum $f_1(x)+f_2(x)=1$ for all real $x$? therefore the derivative of their sum is 0 everywhere, but the sum of their derivatives does not exist?
 
Obliv
Obliv
in D&F it says for n>1 that $\mathbb{Z}/n\mathbb{Z}$ is not a group under multiplication of residue classes and to prove it. I want to argue that for $n$ is prime, $\mathbb{Z}/n\mathbb{Z}$ is a group under multiplication as every element in $\mathbb{Z}/n\mathbb{Z}$ has a multiplicative inverse and an identity. Multiplication is also associative so it should be a group, no?
 
Brody
Brody
@BalarkaSen, @MikeMiller, poking if you're here ^.^
 
Obliv
Obliv
nvm 0 doesn't have an inverse
 
s.harp
s.harp
5:57 PM
@Brody I imagine if somebody is talking about the sum of derivatives he is looking a functions for which the derivatives exist. If you want you can leave a comment though.
 
Brody
Brody
@s.harp thanks. i thought as much, though the counterexample seems of interest. will probably leave a comment
 
Mike Miller
Mike Miller
6:16 PM
I'm here-ish, but working on a question on main, then probably going to do my own work. Sorry.
 
Semiclassical
Semiclassical
@MikeMiller which question are you working on?
 
Mike Miller
Mike Miller
@Semiclassical This one. I had a great time with it, probably my favorite MSE question in a while.
 
Semiclassical
Semiclassical
ahh
 
Mike Miller
Mike Miller
Wow, that was energizing. Now I'm motivated to work.
 
Obliv
Obliv
6:48 PM
anyone know what is meant by $x * y$ is the fractional part of $x + y$ i.e$(x * y = x+y - [x+y]$ where $*$ is a binary operation on the set of reals.
 
Mike Miller
Mike Miller
What is unclear about it?
 
Obliv
Obliv
and where $[a]$ is the greatest integer less than or equal to a.
I don't understand what $x*y$ does. It looks like it just equals 0
 
Mike Miller
Mike Miller
What is it for x=y=1/2?
 
Obliv
Obliv
oh..
integer
wait it is still 0. 1/2 + 1/2 - [1/2 + 1/2] where the greatest integer less than or equal to the evaluated brackets is 1.
 
Mike Miller
Mike Miller
Oh.
Okay, x=y=1/4. (I misread your question.)
 
Obliv
Obliv
6:52 PM
now it's 1/2
I see. The distinction of integer is very important :p
 
user243301
user243301
@MatsGranvik feel free to tell me about definitions of Möbius function as a recurrence, I am following some of your post, still I don't understand but it is apprecite that you always put your codes, very thanks much. Also if some day you or other user want say that perhaps I am boring or with many revolutions it is best an advice early than later. I'm changing my attitude , thank you.
 
Mats Granvik
Mats Granvik
@user243301 The Möbius function as a recurrence is done similarly as for the von Mangoldt function. math.stackexchange.com/a/164829/8530
 
Tobias Kildetoft
Tobias Kildetoft
@AlexClark Awesome. Thanks for the help. I can now finally remove that exception from the paper. I'll also add a line thanking you in the acknowledgements.
 
Mike Miller
Mike Miller
cool collaboration
 
Mats Granvik
Mats Granvik
@user243301 oeis.org/A271047
 
DeNiSkA
DeNiSkA
7:37 PM
A seed for war between mathematician and Physicist!
in The h Bar, 7 mins ago, by ramsay
2
Q: help: in applying calculus in physics

ramsayI am new to calculus and during my mathematics class my sir defined $\dfrac{dx}{dt}$ as $$dx/dt=\lim_{t\to t_1}\dfrac{f(t)-f(t_1)}{t-t_1}$$ and my sir made a clear statement that $\dfrac{dx}{dt}$ is not a fraction it only behaves like a fraction! (it means $\dfrac{dx}{dt}$ is just a notati...

kinematics calculus
 
Obliv
Obliv
what does his math teacher mean with $\frac{dx}{dt}$ is not a fraction, it only behaves like a fraction?
it's an infinitesimally small change over an infinitesimally small interval how is that not a fraction..
 
DeNiSkA
DeNiSkA
@Obliv yes! even my mathematics teacher
 
Mike Miller
Mike Miller
Without doing a lot of work that your course no doubt has not done, there is no such mathematical object as an 'infinitesimally small change'.
One can make rigorous sense of infinitesimals; there are a couple users in this room who are really into that. I'm not, and most mathematicians are not, and no calculus course is.
 
DeNiSkA
DeNiSkA
haha! physicist believe $dx$ to infin.. small change
although newton was a 'physicist'
 
Mike Miller
Mike Miller
Sure, that's the intuition.
 
Obliv
Obliv
7:47 PM
what are the qualifications for treating $dx$ as a mathematical object?
 
Mike Miller
Mike Miller
Some people will tell you differential forms; I don't really like that answer. I prefer to take manipulations of $dx$ as philosophy, from which you can prove actually correct results.
 
Obliv
Obliv
reminds me of xeno's paradox actually
 
DeNiSkA
DeNiSkA
@MikeMiller now, i have not done rigorous integral calculus but then why do we integrate $dy=\int f(x)dx$ what does dx signify there
 
Mike Miller
Mike Miller
Notation.
Like, you shouldn't feel bad about thinking of these things as infinitesimals. That's the idea. But the reason people say not to think of $dy/dx$ as a fraction is twofold. 1) Mathematically, that's not literally what it is. It's a limit etc etc. 2) Sometimes, if you are too uncareful with treating these terms as variables, you can come to a result that's just actually wrong. I don't have an example off the top of my head, but when I was teaching calculus, I saw this happen now and then.
On the other hand, I don't have a problem with people reading this as "infinitesimal change of y over infinitesimal change of x". That's good by me. The moral one should take is to be a bit careful with these manipulations, that's all.
 
Obliv
Obliv
noted.
 
DeNiSkA
DeNiSkA
7:54 PM
so do I!
 
Obliv
Obliv
does $\mathbb{R} / [0,1)$ mean $\mathbb{R}~(\mod 1)$?
 
Lord_Farin
Lord_Farin
8:11 PM
Evening.
 
Mike Miller
Mike Miller
hey
 
Lord_Farin
Lord_Farin
@MikeMiller Your answer to the currently hot post on Diff Geo went completely over my head :/.
 
Mike Miller
Mike Miller
:(
 
Lord_Farin
Lord_Farin
Relax, it's not your fault.
How's life? @Mike
 
Mike Miller
Mike Miller
I can explain the details if you want. It's good. I gotta go for a bit.
 
robjohn
robjohn
8:36 PM
@r9m I don't know if you ever got a response to this question, but the answer is yes, as long as $f$ does not have any singularities in the wedge between $\arg(z)=0$ and $\arg(z)=\arg(\lambda)$. It is pretty easy to show with contour integration.
 
Mike Miller
Mike Miller
ok, I'm back
 
robjohn
robjohn
@MikeMiller let the party resume!
 
Mike Miller
Mike Miller
:)
 
robjohn
robjohn
@MikeMiller On campus for a midterm today. It is nicer here than it is in the Valley. It is supposed to be in the low 90's there.
Now the NWS says it is only supposed to get to 85° in the Valley today (71° here).
but we have a 10% chance of rain by the morning... storm watch!
 
trilolil
trilolil
Hello guys very easy and quick question
$K(x) = \frac{Z(x)+1}{Z(x)-1}$
$Z(x) = \frac{1+K(x)}{1-K(x)} $
 
Mike Miller
Mike Miller
8:46 PM
I almost wore a sweatshirt this morning but decided against it... good decision.
 
trilolil
trilolil
how do you get to the second equation?
 
Mike Miller
Mike Miller
I'd offer to grab a cup of coffee but despite appearances I have to run around today.
 
trilolil
trilolil
any idea guys?
 
Lord_Farin
Lord_Farin
@MikeMiller I realised it's good. But at this point it's probably not going to be anything more than a stimulus for me to learn homology theory.
But thanks :).
 
Mike Miller
Mike Miller
Ah, without a little homology this will be a tough sell, sorry.
 
Lord_Farin
Lord_Farin
8:53 PM
Such is life.
 
trilolil
trilolil
??
 
Mike Miller
Mike Miller
But maybe you can appreciate the result? :)
 
Lord_Farin
Lord_Farin
Definitely. It's pretty nice that one can avoid blatant discontinuities in these situations.
Even nicer that we can actually classify when this happens.
@trilolil Write the first equation's RHS as $2+(Z(x)-1)^{-1}$ and rearrange to find out it's actually $Z(x) = \frac{K(x)-1}{K(x)-2}$.
 
Mike Miller
Mike Miller
I forget, what do you do? Some flavor of accounting? (Forgive me if I got that wrong.)
 
Lord_Farin
Lord_Farin
@MikeMiller Software engineering.
And I forgive you.
:)
 
trilolil
trilolil
9:03 PM
@Lord_Farin not sure that s correct...
 
Mike Miller
Mike Miller
Ah, accounting for programming errors.
 
Lord_Farin
Lord_Farin
@trilolil You're right. I meant $1+\frac2{Z(x)-1}$.
 
trilolil
trilolil
even worse x). It deffinitely should be: Z(x)=1+K(x)/1−K(x)
 
Lord_Farin
Lord_Farin
@trilolil I meant for the rewriting of the RHS.
Then you'll end up with your stated answer.
@MikeMiller I guess you could put it that way :).
How about yourself? Still aiming to stay in academia?
 
trilolil
trilolil
@Lord_Farin I really don t see how you d get that rhs form
 
Mike Miller
Mike Miller
9:08 PM
I don't think it's fair to talk about aims when the job market is so impossible, but yes, that would be desirable.
 
Lord_Farin
Lord_Farin
@trilolil $Z(x)+1 = 2 + (Z(x)-1)$
@MikeMiller Oh? I thought it was still relatively tolerable in the US? Am I mistaken?
 
trilolil
trilolil
@Lord_Farin Where is K? and how did you end up with "2"?
 
Mike Miller
Mike Miller
Some flavor of mistaken, yes.
 
Lord_Farin
Lord_Farin
@trilolil $K(x) = \frac{Z(x)+1}{Z(x)-1} = \frac{(Z(x)-1)+2}{Z(x)-1} = \frac{Z(x)-1}{Z(x)-1} + \frac2{Z(x)-1}$
@trilolil Did you actually try anything?
 
trilolil
trilolil
@Lord_Farin wut... hm ok. ANd is that really necessary? (splitting it in that way)
yes ofc I did...
 
Lord_Farin
Lord_Farin
9:13 PM
@MikeMiller Hm... I'm sad to hear that :/. Are you already considering alternatives or is it still possible to really strive for a job in pure maths?
 
Mike Miller
Mike Miller
I have escape plans. I'm not going to think about it in detail until a year before I go on the market.
I'm not really sure whether I'd enjoy other things. It'd probably have to be something I have to throw myself into, lest I spend my time doing the stuff I failed to get a job in. ;)
 
Lord_Farin
Lord_Farin
@trilolil You could also multiply by $Z(x)-1$ and then do a similar trick writing $1-K(x)$ in brackets and then dividing by it. But I consider that much uglier, even.
You could also do it all in your head.
@MikeMiller Hmm. After graduating for MSc. it took me half a year to accept I wouldn't find a job in maths. As I gather from my friends it helps tremendously if you have a clear picture of the field you want to work in (I'm guessing DiffGeo), the desire to work on it deeply, and the right connections.
For me, it turned out I'm not suited to the deep work. I'm a generalist.
 
Mike Miller
Mike Miller
I know precisely what I like to think about (which some might call a flavor of differential geometry). But I'm avowed to not thinking about this yet.
 
Lord_Farin
Lord_Farin
Why?
 
Mike Miller
Mike Miller
Because there is no tangible benefit to doing so two and a half years before I'm on the job market and it causes stress to do so.
 
Lord_Farin
Lord_Farin
9:26 PM
Ah, you mean whether you'll get a position in this field. I mistook your "this" for thinking about the field you like to think about -- which I found strange :).
 
Mike Miller
Mike Miller
Oh, sorry, that was extremely poorly worded.
 
Lord_Farin
Lord_Farin
't Happens.
 
Steven Stewart-Gallus
Steven Stewart-Gallus
10:12 PM
Is there a way to simplify $\left(\vec{s} \cdot \vec{v}\right) \left(\vec{s} \times \vec{v} \right)$?
 
Sosi
Sosi
Anyone here knows basics in R? I have a stupid question :(
 
Mambo
Mambo
10:35 PM
GO AHEAD
 
Akiva Weinberger
Akiva Weinberger
10:51 PM
Disappointed that the plural of algebra isn't algebrae
…or that the plural of Balarka isn't Balarkae
2
 
ramsay
ramsay
Suppose we have a positive real number $x$ and $y$. Then does Archimedean property of real number mean $xn>y$ where $n \in \mathbb N$
 
Karl Kronenfeld
Karl Kronenfeld
nah
 
ramsay
ramsay
Ok! That edited one?
 
trilolil
trilolil
@AkivaWeinberger I googled your name
it didn't tell me much, you r not the only one with that name...
 
Akiva Weinberger
Akiva Weinberger
I know what you're referencing; I wrote it.
 
Karl Kronenfeld
Karl Kronenfeld
11:02 PM
@ramsay Suppose you were to look it up
 
Mambo
Mambo
@ramsay Is $\Bbb N$ bounded?
 
DeNiSkA
DeNiSkA
@Obliv a mathematician named arturo magidin defeated my views about derivative !
 

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