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00:02
In Mathematics, a dozen or what they call in chemistry, a mole, would be a unit of measurement. So, a unit.. But what about a thing, like an atom.. what would you call that? e.g. if you write 1 atom = 0.00001g The g(grams) is a unit.. But what is atom?
00:49
@Thorgott thanks. You are right, it is needed that the equality holds for all distinct $a,b\in\mathbb{R}$. I was trying to understand why, because of this, my proposed $g$ doesn't make sense: I concluded that, since the equality must holds for all distinct $a,b\in\mathbb{R}$, the numbers $a,b$ can vary and in my reasoning with $g$ I consider them costant. So I am using a contradictory reasoning, considering simultaneously $a,b$ constants and non constants. You meant this or am I off road?
yeah, that's one way to put it
 
3 hours later…
03:42
hello @robjohn
how do we calculate this
AGM(sqrt2,1)?
04:33
i intuit that a finite subset of the reals cannot be open...but i am having trouble thinking up the argument :P
use the definition of open set in Rudin's
@SillyGoose the way you have stated it, the statement is wrong.
There is a finite subset of the reals which is indeed open.
are you talking about the empty set?
precisely
ah i start the counting numbers at 1 :P
well for this analysis course at least
the complement of a single point is open and so would the intersection of a finite number of such sets.
04:46
Hi @copper.hat!!
Hi @Koro!
i'm about to go to sleep, i think, i had a nice ride on some hill trainls today
nice. I also like to do that but can't as there are no hills nearby :).
04:58
you could let some air out of the front tire to simulate the experience.
or ride with your brakes applied :-)
05:20
riding downhill is really fun. Some would say it's dangerous but I liked it when I did it while returning from our labs. That was so fun.
while the bicycle is going downhill, it's fun to control/balance it :)
I never rode uphill. Some do that as well. But I found it to be way too dangerous.
copper, there used to be a game called pubg. In that there is vehicle called buggy or something. I rode that once in mud!!
used to be, and even still is
sadly, no bicycles in the game
ohh it is banned here.
@Koro can you crack this formula?
0
Q: Will this function $G(n) = \dfrac{p_n\#}{\sum_{d \mid p_n\#}(-1)^{\Omega(d) - (2 \mid d)}\frac{p_n\#}{d}}$ eventually outgrow $p_{n+1}^2 - p_n -3$?

D Left Adjoint to U$$ |A_i| := \vert\Bbb{Z}/(p_1) \times \dots \times \Bbb{Z}/(p_{i-1})\times \{\pm 1\} \times \Bbb{Z}/(p_{i+1}) \dots \times \Bbb{Z}/(p_n)\vert = 2\prod_{i \neq j} p_j $$ is the size of the set of all $z \in \Bbb{Z}$, modulo $p_n\# = p_1 \cdots p_n$ such that $z^2 = 1 \pmod {p_i}$ but $z^2 \neq 1 \...

If always sufficiently bounded, then twin primes is true
as a consequence of this
That is in the case that we can show that it doesn't outgrow that interval length bounded by $p_n, p_{n+1}^2$ roughly.
Can you apply number theory techniques to it?
Specifically those that involve Mertens estimates
05:37
@Obliv Not to sound harsh, but if you're asking these questions you may need to consider doing an introductory linear algebra course.......... 🤷🏿‍♂️
05:56
I know that $(1+ap)^{p^{l-1}}\equiv 1+ap \pmod {p^{l+1}}$
How does it follow from here that: $(1+ap)^{p^{l-1}}\equiv 1 \pmod {p^l}$?
p is an odd prime here, $l\ge 1$.
@DLeftAdjointtoU nope, not right now :(.
:(
Maybe hensel's lifting lemma
to your prob
But I understood 'Sieve of Arestothenes' in your post.
as I studied about it few days back.
I haven't yet studied Hensel's lifting lemma. I'm trying to prove that 1+ap has order $p^{l-1}$ mod ($p^l$).
I am supposed to use the following result: $(1+ap)^{p^{l-1}}\equiv 1+ap \pmod {p^{l+1}}$
This proof is basically corollary 2 in Ireland and Rosen's book's chapter 4. But I don't understand the above step there.
Is $a$ coprime to $p$?
Ahh, I got it now.
Maybe product of rings
06:09
yes, a is coprime to p.
Show me your proof
$p^{l+1}| ((1+ap)^{p^{l-1}}-( 1+ap^l))\implies p^l|p^{l+1}| ((1+ap)^{p^{l-1}}-( 1+ap^l))\implies p^l| ((1+ap)^{p^{l-1}}-( 1+ap^l))$
$(1+ap)^{p^{l-1}}-( 1+ap^l)\equiv 0 \pmod{p^l}\implies (1+ap)^{p^{l-1}}\equiv 1 \pmod{p^l} $
Crap I can't read your proof :)
It's okay though
going to bed
@DLeftAdjointtoU ooh, why not?
I'll word it: We are given that $(1+ap)^{p^{l-1}}\equiv 1+ap^l \pmod{p^{l+1}}$.
(this is known). Assumptions are p does not divide a. $p$ is an odd prime.
So I have simply used the fact that $p^l$ divides $p^{l+1}$. $\ddot\smile$.
I thought that the mathjax didn't render properly, that's why you could not see it, hence my message 'why not?' 😅
@DLeftAdjointtoU Good night :-).
06:32
Night mon
07:12
How do I show that $U(Z/p^a Z)$ is cyclic for odd prime?
probably more than one way to do it. here is one: math.stackexchange.com/questions/1387378/…
I saw that one already. I didn't understand his answer.
Let me write one from the book I'm studying.
8
Q: Other ways to deduce Cyclicity of the Units of certain groups?

quantaThe group of units of the rings $\text{GF}(p^r)$ and $\mathbb{Z}/p^r\mathbb{Z}$ are both cyclic (except for the exception of prime powers are not cyclic when $p=2$ and $r\ge 3$). This is a strong result which I have been using a lot but I don't understand it properly. I would like to know more a...

some more detail on the base case, but basically same argument.
But there are few steps in that which I don't understand.
yeah, maybe a more focused question on a particular argument would attract interest. the general fact has been done to death on MSE.
read arturo's comment to that second linked answer.
it seems like some of what you were doing above may have been setting you up to use that approach.
07:21
hmm, he does mention HL in his comments.
I suppose HL=Hensel's lemma
I haven't studied that yet.
pete's answer above it doesn't use that.
although, i recommend reading up on some version of it. it sure comes up enough.
Hensel's lemma << The main lemma I decided not to study number theory a few years ago.
07:41
@onepotatotwopotato This is something important— not only in number theory, but in algebra.
Good point. Correction: Hensel's lemma << The main lemma I decided not to study algebra a few years ago.
it was really all downhill after hensel's lemma, wasn't it
you can't blame hensel for that
@onepotatotwopotato I noted that onepotatotwopotato also had a question once similar to the one that I was struggling with yesterday.
I wasn't (mathematically) mature enough to digest Hensel's lemma when I first encountered it in number theory class. Anyway, I don't care about Hensel's lemma anymore after commutative algebra class.
07:59
As professor Gilbert Strang says-Newton discovered gravity because he was born early.
implying that if we were born before him, then we would also have discovered gravity. 😃
similarly, if we were born before Hensel, we would have discovered Hensel's lemma too.😬
 
2 hours later…
09:59
@PrateekMourya That AGM value is of historical significance. See en.wikipedia.org/wiki/Lemniscate_constant
@Koro Incidentally, Hensel lifting is essentially Newton's method, but using modular arithmetic.
@Koro I assumd you mean Eratosthenes...
Eratosthenes of Cyrene (; Greek: Ἐρατοσθένης [eratostʰénɛːs]; c. 276 BC – c. 195/194 BC) was a Greek polymath: a mathematician, geographer, poet, astronomer, and music theorist. He was a man of learning, becoming the chief librarian at the Library of Alexandria. His work is comparable to what is now known as the study of geography, and he introduced some of the terminology still used today.He is best known for being the first person known to calculate the circumference of the Earth, which he did by using the extensive survey results he could access in his role at the Library; his calculation was...
9
Q: What is the difference between Hensel lifting and the Newton-Raphson method?

BharatRamSo in the Newton-Raphson method to iteratively approximate a root of a real polynomial, we start with a crude approximation $x_0 \in \mathbb{R}$ for $f(x)=0$ where $f(x) \in \mathbb{R}[x]$. For the next iterate $x_1$, we put $x_1 = x_0 + \epsilon$, and we want to determine $\epsilon$ to get a bet...

I suddenly remember that my teacher mentioned that.
> In the folklore of number theory it has been known for a long time that Hensel's and Newton's method are formally the same (this remark appears in printed form in an article by D. J. Lewis [...]
Pity we can't use the plain Newton's method to find modular square roots. Unfortunately, it generally doesn't converge. I suspect that Hensel discovered his lemma while trying to fix that
10:43
0
Q: Check my teacher's proof for sequential criterion of limits of functions

nickbros123theorem: let $ f:\mathbb{A} \to \mathbb{R} $ where $ \mathbb{A} \subseteq \mathbb{R} $ and let $c$ be a cluster point of $\mathbb{A}$ ,then: if for every sequence $(x_{n})$ in $\mathbb{A}$ that converges to c such that $x_{n} \not=c $ for all $ n \in \mathbb{N}$, the sequence $(f(x_{n}))$ conver...

11:09
Can anyone please take a look at this recently edited question math.stackexchange.com/questions/4642260/… ?
 
1 hour later…
12:10
@leslietownes Thank you for all your travail for clarifying my concerns! I have edited my question a bit :math.stackexchange.com/questions/4642260/… . Mind taking a look at it?
@TedShifrin You might as well like to take a look at it as well, after all, I am so much grateful for your benignant
clarifications ! Thank you 😊
 
2 hours later…
14:36
@PM2Ring yes, indeed.
I am curious why don't instructor explain derivation of inequality even in the most Prestigious university when teaching real analysis
they just give proof
at my college, they sure don't explain anything. If it is written in the book -'it's easy to see that this follows, they also write the same thing on the board- it is easy to follow'. But I'm not sure about other colleges.
like those inequality in rudin I don't know how can something come out of thin air I always need to derive it by reading proof in reverse way
easy to follow and trivial is what they write yes
or this inequality is for assignment
they don't give the big picture or main idea but instead present bunch of proof
may be they already assume you have understand process of making proof in intro to math thinking
 
1 hour later…
16:10
@TedShifrin "My whole issue with "self-study" is that you have no one evaluating your learning and whether you can write up mathematics decently."
What about MathStackExchange (the proof-verification/solution-verification tag)?
For computational problems, the solutions manual should tell you if you're right.
16:33
Can anyone help me with this:math.stackexchange.com/questions/4642951/… ? A fire is raging in the comments! But unfortunately, my issue is not getting resolved...
17:01
I know not really about mathematics , nevetheless , can someone tell me how this statement I found in the net is meant (it is in German , but I can translate it , if someone is interested) : "Frauen warten im Schnitt über sechs Minuten vor Toiletten, Männer elf Sekunden - was auch daran liegt, dass Frauen auf dem Klo länger brauchen als Männer. Bei Unisex-Toiletten wären es für jedes Geschlecht im Schnitt etwa zwei Minuten."
This appears extremely unrealistic. Six minutes waiting time in the average before the ladies loo ? I just do not believe it.
17:28
Gonna change my name again
@Franklin your solution is not correct.
@Franklin move the conversation to chat
@ペガサスSeiya never let them know your next move
@SineoftheTime be unpredictable
17:44
@Koro why?
Where is the solution going wrong?
the answer(s) is/are already there in comments.
Unfortunately, I am unable to decipher it in the huge ocean of comments. What do you think about why it goes wrong?
you are calculating $\Delta x\to 0$ for something, so your $x+\Delta x$ should be close to $x$.
@Koro Yes! But what does that imply?
Do you think the case $1+\Delta x<0$ is wrong?
yes, that is wrong. Think of it like this: It is possible that 1+$\Delta x<0$ for some $\Delta x$. For example: 1-4 is <0 or 1-5<0 etc. But since we want to calculate the limit as $\Delta x\to 0$, we should stay 'close' to 0, i.e., consider 1-0.01 for example which is positive.
that is to say that the quantity $1+\Delta x$ should be treated 'eventually positive' as $\Delta x\to 0$.
and your case 1+$\Delta x>0$ doesn't make sense as your function F is not defined outside the closed interval [-1,1].
17:56
Is this because $1+\Delta x<0\implies \Delta x<-1,$ and thus $\Delta x $ is not nearer to $0$?
Is it the true reason for this to be invalid?
If/when you learn the epsilon-delta definition of limits, you'll understand precisely what 'close to' or 'eventually positive' in my comment mean.
@Koro I have just learnt it. Can you please help me ?
I have done it from IA maron
Problems of Calculus in One Variable
OK. So can you state the definition here now?
Yes!
you may use $h$ instead of $\Delta x$ for brevity.
18:01
For all epsilon >0 there exists a delta>0 such that if |x-c|<delta , then $|f(x)-L|\le \epsilon. This is the definition of lim_{x\to c}f(x)=L
This is what I know about epsilon-delta definition.
To be more precise, we take $0<|x-c|<\delta$ there instead of $|x-c|<\delta$. But anyways, the point is that here if the limit exists, then given any $\epsilon>0$, we can choose a $\delta'< \delta /2$ such that if $0<|x-c|\lt \delta'$, then $|f(x)-L|<\epsilon$.
Here, $\delta$ is the one that you gave me in your comment.
15 mins ago, by Koro
and your case 1+$\Delta x>0$ doesn't make sense as your function F is not defined outside the closed interval [-1,1].
I think this is the real reason! What do you think?since we can never calculate the differential at $\Delta x \to 0_+$ because of tge domain . Thus, RHL dont exist and hence the limit doesn't exist!
As mentioned in the picture I posted
We can define derivative at an end point as a side limit.
in the case of 1, you can actually calculate left hand derivative.
isn't that what you wanted?
18:14
Yes we can calculate LHL but not RHL and thus f(x) won't be differentiable at 1. Since we need RHL=LHL for limit to exist and here, we dont get any RHL. Due to this, limit dont exist.
This is the true reason , right?
that depends upon what conventions you are using. For some people, limit of f: [a,b]-->R, a<b, at b is defined as $\lim_{x\to b-} f(x)$.
Although f is not differentiable at 1, but my proof for this wasn't correct. The concrete proof would be: since if we consider the $\Delta x\to 0_+$, it goes outside domain and hence no limit on the right hand side. Due to this, the limit dont exist. Thus, the function is not differentiable. This is a more rigorous and valid proof, right?
Franklin, the point is: what is your definition of differentiable at an end point?
unless you clarify that in your post, you won't get a unique answer because people use different definitions. For example: Is f(x)=1/x, x in R-{0} discontinuous at 0 or not? The answer depends upon the definition of discontinuity that one uses.
18:29
Hi can someone kindly help with a question in probability cause I've an exam tomorrow and I need to understand this before I got to sleep.
0
Q: Calculating CDF from PDF for jointly random variables?

zoroLet A be the set in $R^2$ bounded between the lines $y = 0$, $x = y$ and $x = 2$. Let $(X, Y)$ be jointly continuous random variables with pdf $f(x, y) = cxy^2$ for $(x, y)$ in $A$. I want to calculate the CDF, I know that: If x is between 0-2 then y is between 0-x $F_{X, Y} (x, y)=\int_{-\infty...

there are upvotes on my post but not even one comment. What's up with that?
out of my 3 questions, I answered one myself but still two remain unanswered.
Can $\mathbb{S}^1$ be regarded as the orbit space of $\mathbb{R}$ by a free action?
How about Z acting on R by translation?
19:04
I cannot see why the colimit of sets is related to this diagram. Isn't it just simply i->i'?
 
1 hour later…
20:08
@WilliamSun that's not necessarily transitive
20:26
Right...not reflexive nor transitive. We need to generate one from that..
20:49
Problem: Let X be a known vector field on R^2 and let I be its family of integral curves. Given the pair (X,I) determine Y in the pair (Y,J), given a known bijective linear map M:I-->J. My question is: Is Y guaranteed to exist and are J guaranteed to be integral curves?
21:05
My attempt is to choose X wisely so that with M being some particular integral transform, possibly on compact support, you can map integral curves on a restricted domain of X to J - since integral transforms preserve properties of a function and it's transformed version I am inclined to believe J will be integral curves satisfying some Y - that must be determined.
but the pre-image of M will not be the complete set of integral curves of X making it unclear whether or not the image of M will capture the entire solution set of the integral curves in the new space
 
1 hour later…
22:17
Do I look nice?
22:32
not bad, didn't recognize you :)
I'm the God of Death and Underworld
 
1 hour later…
23:49
in attempting to get a particular solution for a 2nd order non homogeneous linear diff eq $x'' + 4x = \sin(2t)$ why do we multiply by t if the complementary solution and the particular solution have common terms and not divide by t
some1 said its no longer linear but i dont see it

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