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1:12 AM
@satan29: look at the actual solution: $\frac\pi4e^{-2|t|}$. It's first derivative at $0$ is discontinuous. Is it surprising that the second derivative has problems?
 
EM4
1:30 AM
hello
 
2:11 AM
@EM4. Hello
 
2:29 AM
Hello, I am trying to understand why if we are trying to prove a topology is contained in another, i.e., $\tau_1\subseteq\tau_2$ why is it sufficient to show that any basic open neighborhood of any point in $\tau_1$ contains an open neighborhood of this point in $\tau_2$?
 
@robjohn I thought it should give $\cdot \ldots \cdot$ but I didn't know about amsmath :-)
 
Yeah, \dots decides where it should put the \ldots and \cdots
if you want one that is not chosen automatically, you need to explicitly use \ldots or \cdots
 
2:45 AM
Hi, could anyone please give me some hints for my question?
 
What does it mean to say one topology is contained in another?
 
That every open set in one topology is open in the other
 
OK, so what does this have to do with your question?
How do I check?
 
So I can take a basic open set in the first topology, and show that it is open in the other topology - then I will be done, since any open sets are unions of basic open sets
 
Yes.
So for any point in that open set, you must ….
 
2:50 AM
For any point in the basic open set, there must contain an open set from the other topology contained or equal to that basic open set
 
Where did the point go?
 
I am not sure what you mean
 
You want a basis element of the second topology containing the point and contained in the original open set.
How does this show the desired result?
 
That is equivalent to what the claim is right?
 
Yes, you need a last sentence to finish .
 
3:00 AM
So I take any basic open set, and any point in it. Then I need to show that that basic open set is open in the other topology. It is open in the other topology if it is a union of basic sets from that topology. If there is an open set containing that point in the other topology...then I am stuck
Do I first take an arbitrary point, or an arbitrary basic open set?
 
3:22 AM
You fixed the first basic open set. Can you write it as a union of basic open sets for the other topology?
 
I think I got it. If you take any basic open neighborhood in the first topology, and for each point in it, there exists an open neighborhood of that point contained in that topology, then that basic open neighborhood must be open (as a union perhaps) of those contained neighborhoods
Is my reasoning correct?
 
Take the union over all $p$ … of the basis elements for the second topology containing $p$.
 
What are $p$?
 
3:39 AM
Points.
 
The union of all basis elements is the set itself?
 
3:56 AM
Yes, because you have one for each point in the set.
 
4:41 AM
are there any interesting avenues for analysis of extrema when the second multivariable derivative test is inconclusive?
or is it totally and completely a case-by-case analysis
 
5:23 AM
@robjohn i'm working my way through "the salmon of doubt" at the moment. his "last chance to see" inspired me to circumnavigate the globe.
 
6:13 AM
Prove $\lim \limits_{x \to 1} x^2 + 3 = 4$, i.e., $|x-1| < \delta \rightarrow |x-1||x+1| < \epsilon$.
If $\delta < 1$, then $x$ is not $0$ and there exists an $M$ with $|x+1| < M|x-1| < \delta$.
Suppose $\delta = \frac{\sqrt{\epsilon} -1}{M} < 1$. Then, if $|x-1| < \delta$, we have $|x+1| < M|x-1| < \sqrt{\epsilon} - 1 < \sqrt{\epsilon}$.
So, $|x+1||x-1| < \sqrt{\epsilon}\sqrt{\epsilon} = \epsilon \ \blacksquare$
I'm scared $M$ might not always exist
does it always exist?
hm actually I just realized $\delta < 1$ is not sufficient to prevent $x$ from being $0$
oh wait the definition of the limit requires $0 < |x-c| < \delta$ i'm safe
 
6:33 AM
@shintuku you are scaring me a little :-). If $\delta\le 1$ then $|x^2+3-4| \le 2|x-1|$ and so $\delta = \min(1, { 1\over 2} \epsilon)$ will suffice. I do not understand where the square roots are coming from.
 
ohhhh right, wait I might not need $\delta < 1$
 
yes you do, or some limitation on $|x-1|$.
 
how about $\delta = \frac{\sqrt{\epsilon} - 1}{M}$, with $M$ such that $|x+1|<M|x-1| < \delta$?
then, for a given $\epsilon$, suppose $|x-1| < \delta$
 
i don't want to even think about that. where is the square root coming from or why is it necessary?
you are making things much more complicated than needed
 
it just makes (I think?) the proof work, you get $|x-1| < \sqrt{\epsilon}$ and $|x+1| < \sqrt{\epsilon}$
 
6:37 AM
i think you need to step back a little
you want to show $|x^2+3-4| < \epsilon$.
$=|x-1||x+1|$.
 
right, so we want to see how a bound on $|x-1|$ binds $|x+1|$
 
if you choose $\delta \le M$ then $|x-1| \le M$
then $|x+1| \le 1+M$.
and $|x^2+3-4| \le (1+M) |x-1|$.
then you choose $\delta \le {1 \over 1+M} \epsilon$ to get the desired result.
so pick $\delta = \min(M, {1 \over 1+M}\epsilon)$.
no reason not to pick $M=1$.
i really do not see how you got a square root in there
 
i noticed people do that, i understand how it's happening and the reasoning process, and the proof does end up working
 
it just seems strange to make it more complicated
 
give me a chance i'll explain it might make very intuitive sense, but I think I could also be totally wrong I hope you have a minute
since $x$ is never $0$ and is near one, we get $|x-1| < |x+1|$
 
6:43 AM
why is x never 0?
 
$0 < |x-1| < \delta$ is required for the condition to apply
if $x = 0$, then the condition does not apply
 
and $\delta$ is some arbitray number?
 
$\delta$ is defined $\delta > 0$
 
if you want me to follow, you need to be precise
if i don't follow i give up
one last chance
 
alright, I'm terribly sorry I'm just not too good at this yet
thanks for helping by the way
 
6:44 AM
if $\delta = 2$ then $0 < |x-1| < \delta$ does not imply $x \neq 0$
i don't even know why you care.
if you give me your proof i will try to follow, but the above is much simpler
 
suppose $\delta = 2$, then $0 < |x-1| < 2$, and therefore, if $x=1$, $0 < |1-1|$ is a contradiction
 
i don't know what you mean
that would be the interval $(-2,3)$ whcih certainly contains zero
you said $x \neq 0$. which is it?
i am setting a timer for 5 mins.
 
suppose $\delta = 2$. Then it is the interval $(-2, 3)$ without $x=0$, I think, since we require $0 < |x-1| < \delta$.
 
that is wrong
clearly $0 < |0-1| < 2$.
 
oh! my god, you are totally right
 
6:49 AM
i don't know what you are trying to achieve
 
I completely missed the point
there necessarily has to be $\delta < 1$ otherwise $x$ could be $0$
 
if you want $x>0$ (i don't know why) you need $\delta \le 1$.
($\le$, not $<$ but that's fine)
 
you're completely right, I had missed that thanks
then, first $\delta < 1$. Therefore, there is an $M$ such that $|x+1| < M|x-1|$
 
ok, suppose $\delta <1 $, go ahrad
1:45 left
what??????
 
since $x$ is not $0$
so if $|x-1| < \delta$, we have $|x+1| < M|x-1| < M\delta$
 
6:52 AM
are you trying to claim that ${x+1 \over x-1}$ is bounded for $x $ near $1$????
 
which means $|x+1| < M\delta$ and $|x-1| < M\delta$, which means that if $(M\delta)^2 < \epsilon$, we have $|x+1||x-1| < \epsilon$
 
i just really have no clue what you are trying to do
please
how can such an $M$ possibly exist???
i think you need to review limits.
what is $\lim_{x \to 1} {x+1 \over x-1}$?
 
argh, there's no such $M$? but since $|x-1|$ is nonzero, isn't there necessarily such an $M$?
2 sec properly calculating the one you mentioned
 
its $\infty$
 
right, ok I catched that
 
6:57 AM
well, i should have $| \cdot| $ around it
ok, so no such (finite) $M$ exists. what now?
 
oh, I see
you proved the $M$ doesn't exist
 
well, that is not the point.
how are you going to prove the original limit?
 
we'll have to do it the $\delta \leq M$ way
 
there are lots of ways.
 
I understood where I went wrong, I thought there existed an $M$ s.t. $|x+1| < M|x-1|$
 
7:02 AM
if you know $|x-1| < \delta$ then $|x+1| = |x-1+2| \le \delta+2$.
then $|x^2+3-4| \le \delta (\delta+2)$.
So you need to choose $\delta$ so that $\delta(\delta+2) < \epsilon$.
 
right, it makes perfect sense
thank you very much for the help, it really is super appreciated
 
glad to help. i am a bit cranky before i got to sleep :-) sorry
 
no prob I was not being very clear
 
Note that if $|x-1|\le1$, then triangle inequality says that $|x+1|\le3$, so $\left|x^2-1\right|=|(x+1)(x-1)|\le3|x-1|$
 
that is why we are here to help (and not be cranky :-))
 
7:07 AM
so $\delta=\min\left(1,\frac\epsilon3\right)$ is sufficient
 
i think i made a minor arithmetic error above, if $|x-1| \le M$ then $|x+1| \le M+2$ not $M+1$ as i had written.
 
@robjohn oh that's a very quick way to solve it
 
so the above formula shoudl have been $\delta = \min(M, {1 \over 2+M}\epsilon)$
@shintuku we covered that above already
except for my arithmetic boo boo
 
i'm a bit slow hehe, thanks
 
you should be checking what i write as well, if you don't follow then stop me.
good night folks!
 
7:16 AM
g'night!
 
good night, thanks a lot!
 
7:30 AM
@copper.hat sorry, I didn't see this until just now. I have not read the Salmon of Doubt, but my wife has, and I intend to. Globetrotting sounds fun! Glad you were inspired.
I read the Hitchhiker trilogy when it was still only 3 books ;-)
 
 
2 hours later…
9:20 AM
If a professor told us how to do an answer of an physics assignment(which is a part of an active exam) and if I believe that there is a flaw in the logic of the professors answer and if everyone is writing professors answer for their assignment (except me) then:
(i)Should I write down professor answer in my own assignment even though I believe that there is a flaw.
(ii)Can I ask question about the flaw on this forum?
 
@Prithubiswas if you have room, you should (1) write the professor's answer (2) explain why you think it is wrong and (3) write your own
i do this all the time with my homework
for some reason my profs always give me homeworks full of damn typos
 
@shintuku this assignment is a part of an active public exam (countrywide) , and the school (the professor and a second examineer) will check it and submit it to the education board. I think that format of (1) and (2) might not be exceptable by school.
 
mathse has people from all countries, so you could ask if you're allowed to do this in your particular exam
you might be risking plagiarism or related accusations if you post a question about a specific answer in the exam however
 
9:35 AM
It is okay if I dont get marks for my answer if I am convinced that my answer is right.But it is kind of weird to me if I get full marks for an answer(provided by the professor during an active exam) which I "believe" is wrong.
 
you have no room to do all three above suggestions (at the same time, not only one or two of them)? also, are you not allowed to call/email the school to ask if you're allowed to correct the question?
 
@shintuku Well , because it is a public exam in out country, probably it is not allowed to ask questions.
 
you should try asking, the worst that can happen is they say "we're not allowed to answer"
better than to risk plagiarism accusations
 
What about if I ask the question , but change all of the numbers to variables? Since , my question is conceptual , not numerical.
 
you can still risk plagiarism, you really should consult your school authorities if this is something like a ministerial exam
 
9:42 AM
Oh ok,I wont ask then :). By the way , I am not allowed to correct the question . I did ask about this dought of mine , but the professor still showed the same reasoning which I believe is false.
Now that I think of it , technically the professor himself is not allowed to show the students how to answer the assignment because it is a part of an active public exam. But the professor is still showing the answer (right or wrong).Isn't this also plagiarism?
 
 
2 hours later…
11:42 AM
how come the author uses $|f - L| < \epsilon x$ instead of just $< \epsilon$? is it an equivalent formulation?
noooo I had not seen the $x$ dividing $f$, ignore the above
 
12:29 PM
if I get a $<2\epsilon$ on a limit, what do I do to fix my degenerate $<\epsilon$
this is probably a case by case answer
sigh
If I needed to show $\lim \limits_{x \to 0} \frac{f(x)}{x} = 0$ and I got $|\frac{f(x)}{x}| < 2\epsilon$, did I succeed?
after assuming $|x| < \delta$
 
1:17 PM
I think I just had an aha moment with computing quotients in base two.
 
1:44 PM
I am stuck at this question: It is given that $a_1=a, b_1=b, a_{n+1}=\frac{a_n+b_n}2, b_{n+1}=\frac{a_{n+1}+b_n}2$. It is to be shown that $\lim a_n=\lim b_n$
I showed that: $a_{n+1}-a_n=\frac{b_n-a_n}2$ and $b_{n+1}-a_{n+1}=\frac{a_{n+1}-a_n}4=\frac{b_n-a_n}4$
whence it follows that $\lim (a_{n+1}-a_n)=0=\lim(b_{n+1}-a_{n+1})$
Considering that $a_{n+1}-a_n=\frac{b_n-a_n}2=\frac{b-a}{2.4^{n-1}}$ and assuming $b\le a$, it is clear to me that $(a_n)$ is decreasing but then I have problem proving it bounded from below to confirm its convergence
Once convergence of either $a_n$ or $b_n$ is confirmed then the convergence of the other one follows.
Any suggestions on this? Thanks.
Hi Leslie, how are you?
@shintuku @shin:if your $\epsilon\gt 0$ arbitrary and for each choice of $\epsilon$, you have a $\delta$ corresponding to it (to emphasize this point, let's say $\delta_{\epsilon}\gt 0$) then yes.
 
@Koro yeah, my $\epsilon$ was arbitrary and there exists a $\delta$ for each epsilon
I'm working on the formal proof that $<g(\epsilon)$, with $\lim \limits_{y\to 0+}g(y) = 0$ implies $<\epsilon$
 
further to my earlier comment :why so? because due to arbitrary nature of epsilon, you can say let $\epsilon =\frac{\epsilon'}2$, where $\epsilon'\gt 0$. :)
 
2:00 PM
Sorry to interrupt, but as I just realized, you can represent every larger power of $2^{-n}$ in terms of the smallest power of two that you desire. Then $\sum_{n=1}^\infty 2^{-n} = \sum_{n=0}^\infty 2^{n-M}\text{ for some constant } M\in\mathbb{Z}$.
 
@Koro prove by induction that both $(a_n)$ and $(b_n)$ are bounded above by $\max(a,b)$ and below by $\min(a,b)$
 
you have said it @Thorgott, it will be proven now. how did you see that?
I really want to know that.
is it obvious from the recursive definitions stated above?
 
$a_{n+1}$ is the average of $a_n$ and $b_n$. The average of two numbers is smaller than the larger of the two and larger than the smaller of the two. $b_{n+1}$ is a weighted average of $a_n$ and $b_n$ and the same observation applies. So, inductively, everything stays in the interval between $a$ and $b$.
 
I would think that $a_{n+1}=$ A.M. ($a_n,b_n)$ and since $b_n-a_n\to 0$
then I had difficulty further
 
(contd.) We can exploit this for $a$-bit dividend and $a$-bit divisor to compute quotients using the inequality $\sum_{n=0}^{\leq a} 2^{yn - M} \leq 1$ to compute the reciprocal $\frac{1}{y}$, then multiply by $x$ to approximate $\frac{x}{y}$.
This makes the divisors homogeneous, simplifying calculations and making it possible to parallelize the operation, and I haven't thought far enough into this yet so I'm uncertain about the requirements of the remainders for each individual term of the sum and how relevant they are.
 
2:18 PM
@Koro have you had linear algebra?
 
Hi professor Rob, how are you?
yes, I have had LA during my first year at college.
I have figured out that question now professor Rob.
Instead of trying to analytically prove the whole thing, I should have plotted the given info. to see what it meant. I think that was the key to solve the exercise.
Thanks @Thorgott
 
2:53 PM
@Koro I was working on another problem, but I think that there is a way to apply LA to this.
 
I didn't think of LA on this problem. How can we apply LA to this?
we can try finding nth term using recursive sequence discussion we had few days back
 
3:26 PM
i think the idea is to set up recursion problems as application of a matrix to a state vector so that applying the matrix to the $n$th state vector gives you the $(n+1)$th state vector. Then the long term behavior of the sequence can be read off from the eigenvalues and eigenvectors of said matrix.
 
I tried to find nth term of $a_n$ and found it as $a_n=k\frac14 (\frac 1{4^{n-1}}-1)+a$, where $k$ is a constant. It follows that $a_n$ converges and therefore from relation deduced between $a_n$ and $b_n$ above, that $\lim b_n=\lim a_n$
@Quin i see, thanks. i have never seen that before :-(
 
3:46 PM
@Koro sorry it took so long:
The equations
$$
\begin{align}
a_{n+1}&=\frac{a_n+b_n}2\\
b_{n+1}&=\frac{a_{n+1}+b_n}2
\end{align}
$$
are simply
$$
\begin{align}
a_{n+1}&=\frac{a_n+b_n}2\\
b_{n+1}&=\frac{a_n+3b_n}4
\end{align}
$$
which gives the recursive relation
$$
\begin{bmatrix}\frac12&\frac12\\\frac14&\frac34\end{bmatrix}\begin{bmatrix}a_n\\b_n\end{bmatrix}=\begin{bmatrix}a_{n+1}\\b_{n+1}\end{bmatrix}
$$
The matrix has eigenvector $\begin{bmatrix}1\\1\end{bmatrix}$ with eigenvalue $1$ and eigenvector $\begin{bmatrix}-2\\1\end{bmatrix}$ with eigenvalue $\frac14$.
From this it is easy to compute the limit
 
4:25 PM
@robjohn I don't understand what is "eigenvector components from $[a,b]^T$" but I got the idea which I think is $Ax_n=x_{n+1}$ whence $x_{n+1}=A^n x_1$ and then distinct eigenvalues of A ensure that $A$ can be diagonalized hence there exists an invertible matrix $M$ such that $A=M \Lambda M^{-1}$, where $\Lambda$ is a diagonal matrix with eigenvalues and it is also to be noted that $A^n=M\Lambda^n M^{-1}$
And I also realize now that difference equations which you suggested in past also follows from this method.
 
@Koro If you can write a vector as a linear combination of eigenvectors $u=\sum\limits_ka_kv_k$, you can write $M^nu=\sum\limits_ka_kv_k\lambda^n$
 
What are eigenvalues and eigenvectors and what are their purposes?
 
I have now also determined $k$ here as $\frac 83(a-b)$ and therefore the expression obtained for $a_n$ is same as the $a_n$ expression obtained in method suggested by you above. And this is as expected as that difference method as I realize now is a a special case of this matrix approach.
 
yeah, this is the classy approach to solving recurrence relations
 
@AMDG one of the nice applications is to compute higher powers of a matrix (diagnoalizable)
 
4:32 PM
And what about in relation to recurrence relations?
 
for that please refer above nice answer by professor Rob and then my subsequent comment.
 
Oh wait I didn't see the rest of the text
 
@AMDG If you know some physics, it's all about normal modes.
 
Well long division can be defined as a recursive relation, though I'm having trouble with it. My brain is half-fried.
 
@Koro Okay, so do you understand why I wanted the eigenvector components?
the coefficients $a_k$ to get the decomposition $u=\sum\limits_ka_kv_k$
 
4:37 PM
@robjohn i don't :-( Sadly, I don't understand the term "eigenvector components" :-(.
 
sorry for the $a_k$ overlap
 
@robjohn I'm thinking about this one right now.
 
we want to write $(a,b)$ as a linear combination of the eigenvectors $(1,1)$ and $(-2,1)$.
 
ahh. alright then!
now the terminology is clear to me.
 
okay
 
4:40 PM
@robjohn did you mean $A$ here instead of M?
 
What is $A$?
$M$ is the matrix $\begin{bmatrix}\frac12&\frac12\\\frac14&\frac34\end{bmatrix}$
 
16 mins ago, by Koro
@robjohn I don't understand what is "eigenvector components from $[a,b]^T$" but I got the idea which I think is $Ax_n=x_{n+1}$ whence $x_{n+1}=A^n x_1$ and then distinct eigenvalues of A ensure that $A$ can be diagonalized hence there exists an invertible matrix $M$ such that $A=M \Lambda M^{-1}$, where $\Lambda$ is a diagonal matrix with eigenvalues and it is also to be noted that $A^n=M\Lambda^n M^{-1}$
@robjohn ahh okay.
that was my $A$. sorry about that.
 
I didn't read far enough into your statement. I stopped at the confusion and tried to deal with that, sorry
 
You can see all this discussed in my videos, of course, @Koro.
 
@robjohn wow wow wow wow. it's fantastic. I understand it now. So we had $x_{n+1}=Mx_n$ whence we got $ x_{n+1}=M^nx_1=M^n [a,b]^T=M^n(x [1,1]^T+y[-2,1]^T)$, where x and y are eigenvector components of $[a,b]^T$. And then we know that $M^nv=\lambda^n v$. And thus the last expression follows. Thanks a lot professor Rob. :)
 
4:59 PM
I'm pretty sure I'm converting this recursive relation to an explicit relation incorrectly because I don't know how to get rid of $f\left(n-1\right)$. desmos.com/calculator/nwzbxc3kgt
 
@TedShifrin Hi professor Ted, how are you? Your videos are surely in my to watch list :).
I'll watch those soon. Thanks a lot for making the videos available to everyone professor Ted.. :)
 
5:18 PM
does anyone know if there is some sort of computational logic system that has real analysis proofs?
I'm trying to figure out explicitly how nested limits work
dependence and availability of which variable at each point in the proof
 
5:30 PM
@shintuku So are you saying you don't understand what $\lim\limits_{x\to 0}\lim\limits_{y\to 0} f(x,y)$ actually means?
 
well, intuitively yes
and I can do limit algebra
but what happens when you start naming the epsilons involved
for instance, is there two epsilons involved in the above nested limits, or only one?
 
@Koro Yay! Each problem you understand is a stepping stone for the next!
 
ah, this is the stuff you learn proving limit algebra! the proofs look related to this
 
I don't think the epsilonics are the interesting thing here. It's a question of understanding why the iterated limit is different from the joint limit. Can you do actual examples?
I would break the problem into pieces: Let $g(x)=\lim\limits_{y\to 0} f(x,y)$, and consider $\lim\limits_{x\to 0} g(x)$.
Now write that epsilonically if you insist.
 
ah that makes things simpler
 
5:45 PM
That's the way you should think about it, too :)
 
hm does the nested limit have it's own epsilon and delta in $\forall \epsilon \exists \delta: 0<|x|<\delta \implies |\lim \limits_{y\to0}f(x,y)-L| < \epsilon$?
 
Of course you are missing the $0<|x|$ ...
 
$\lim\limits_{x\to 0}\lim\limits_{y\to 0} f(x,y)=\lim\limits_{x\to 0}\left(\lim\limits_{y\to 0} f(x,y)\right)$
 
whoops my bad
 
I don't understand your question.
 
5:49 PM
evaluate the inner parens first
 
Yes, so that's what I wrote as $g(x)$ above.
 
@robjohn true indeed. Thanks a lot professor Rob for all the help :)
 
There is no relation of the epsilons and deltas for the limits
 
right, from which we get $\forall \epsilon \exists \delta: 0<|x|<\delta \implies |g(x) - L| < \epsilon$
oh, then we might as well use different symbols for them!
 
Yes, now you can separately write a definition of $g(x)$ with (different) $\epsilon$ and $\delta$ (controlling $y$).
 
5:51 PM
@shin: you should also see what happens if you do $\lim x\to 0$ first and then $\lim y\to 0$ that is $\lim_{y \to 0} \lim _{x\to 0} f(x,y)$
 
As I said a while ago, it's best to understand some basic examples where, for example, the iterated limits are different.
 
once you have established a particular limit then forget about the epsilons & deltas which are scaffolding used as part of the proof.
 
Hi copper
 
hi @Koro!
 
@copper.hat easier living through abstraction
 
5:53 PM
the issue is that I'm trying to prove a specific theorem about precisely the epsilon, that $|f(x) - L| < g(\epsilon)$, where $\lim \limits_{y\to 0} g(y) = 0$ and $g(y) > 0$ implies $|f(x) - L| < \epsilon$
 
@robjohn that remark is in a category by itself
 
so I can't, I think, ignore the epsilons here
 
As that's stated, it's not correct. What is $g(\epsilon)=\sqrt\epsilon$?
Of course, you're going to have to incorporate monkeying with $\delta$s in the statement.
 
oh, the limit has to be from the positive side, so $y \to 0^+$
 
OK, not my point.
 
5:57 PM
hm
 
@shintuku: Perhaps a specific example of why you want to have such a special form of existence of a limit is in place.
 
the idea is to prove that you can make epsilon proofs without worrying too much about what epsilon you find
i'm working out the statement for this still
i was stuck doing epsilon proofs with $<\epsilon$ until someone told me it wasn't actually important
@TedShifrin sorry, I'm not ignoring but I'm not finding it
 
Your sentence needs $\delta$s in it, @shin.
 
oh, right
 
Certainly saying $|z|<\sqrt\epsilon$ does not give you $|z|<\epsilon$.
 
6:01 PM
the complete statement is:
$$\forall \epsilon\ \exists \delta\ \forall x \ \ |x-x_0|<\delta\Rightarrow |f(x)-L|<g(\epsilon)\tag{1}$$
where $\displaystyle \lim_{y\to 0+}g(y)=0$ and $g(y) > 0$, implies that
$$\forall \epsilon\ \exists \delta\ \forall x \ \ |x-x_0|<\delta\Rightarrow |f(x)-L|<\epsilon\tag{2}$$
 
And the $\delta$s are most definitely not the same.
 
that's precisely what I wanted to ask
 
You should understand that if you end up with $<5\epsilon$, it's just as good as ending up with $<\epsilon$.
 
so I guess the epsilon and deltas in (1) and (2) are different
 
It makes no sense to say the epsilons are different. Look at quantifiers.
 
6:05 PM
hm, maybe I should read up on this, or in any case, that's why I was looking for a rigorous formalization. these things aren't at all obvious to me
 
@shin: please note that in limit definition $\epsilon\gt 0$ is arbitrary so when you put $g(\epsilon)$, you are restricting the arbitrary nature of $\epsilon$ and if restricted nature of $\epsilon$ implies something, there is no reason to believe that the implication will hold for all arbitrary values of $\epsilon$.
 
All that matters is small $\epsilon$, @Koro, and you can always throw in a min, so I disagree with you.
 
i think you are fixating a little on the symbols $\epsilon,\delta$ rather than what they represent in the context of the limit.
 
@shin You get too obsessed with abstract generalities. You have to understand concrete examples (e.g., the one I gave with $5\epsilon$).
 
well, in the $5\epsilon$ case, there's a definite relationship with any other arbitrary constraint you might want to set on the image of the function
 
6:07 PM
@Ted: why the disagreement? Did I say something that is wrong? :(
 
if you reduce the $5\epsilon$ fivefold, you get a constraint on the image five times smaller
 
@Koro Well, maybe I didn't understand your complicated sentence. I think it holds for arbitrary $\epsilon>0$.
@shin And so what?
I suggest you write out the concrete proof in this case.
 
ahh, I just wanted to write in language avoiding symbols to help shin understand and I guess that didn't work @Ted :(
 
@TedShifrin there's some other constraint that works with the given $\delta$ which is $\frac{5\epsilon}{5}$
i'll work on the proof, thanks for the tip
 
mmh, might not fit directly here. But is Turing equivalence even possible?
 
6:16 PM
Hello, do you know a better algorithm than robust mahalanobis distance ?
that performs well on large dimensions $n$ please?
 
@shintuku elaborating what Ted wrote (and before dealing with the $g(.)$ lemma), try showing that if $\lim_{x\to x_0} f(x) = L$ then for all $\epsilon>0$ there exists some $\delta>0$ such that if $|x-x_0| < \delta$ then $|f(x)-L| < {1 \over 5} \epsilon$.
 
noted! thanks a lot
 
@SAJW surely a Turing machine is equivalent to itself, so trivially the answer must be yes.
 
touché
 
@shintuku did you resolve your ${f(x) \over x}$ issue? it is related to the immediately preceding discussion.
 
6:25 PM
?
is there a way with new notations to always be able to write (readable) a bigger number on the same area like a DIN A3 piece of paper or whatever size your favourite is.
the piece of paper must be finite (...)
but you of course always get a new paper if you want to write down a bigger number
 
you would need a more precise definition of notation, symbol, etc for this to have an answer. otherwise i can just introduce a new symbol each time i want to represent a bigger number.
 
that's true
 
Suppose $\lim \limits_{x\to x_0} f(x) = L$, i.e., suppose:
$\forall \epsilon>0, \exists \delta>0, \forall x$ such that $0<|x-x_0| < \delta \implies |f(x) - L| < \epsilon$.
Let $\epsilon$ be given. Then, $\exists \delta>0$ such that $0<|x-x_0|<\epsilon \implies |f(x)-L| < \epsilon$.
Let $\alpha_1 = \frac{\epsilon}{5}$. Then, $\exists \alpha_2$ s.t. $0<|x-x_0|<\delta \implies |f(x)-L|<a_1=\frac{\epsilon}{5}$.
right?
@copper.hat i think i just did a few seconds ago
 
what is $\alpha_2$ doing?
 
it was destined to be that delta, but the typo got there first
the delta after $\alpha_2$ is a typo
 
6:39 PM
that is the idea. the same general approach applies with the $g(.)$.
 
i feel the power of analysis flowing through my veins
thanks for the help
 
6:53 PM
 
No, it’s quite stupid.
 
if that's a meme, the meaning of the word has changed. It seems now to mean joke.
 
the power of precedence
what happened to dawkins meme?
 
it literally became a meme
 
@copper.hat THAT is the meme I know
 
6:55 PM
order of operations?
 
@robjohn reminds me of a response i have received in a few meetings "that's just semantics"
 
@SAJW If you take the standard order of operations the answer is $5!$, if you take the first operation first order, the answer is 5 (!)
 
i suppose it is cute in some sort of way !
 
since it has two meanings, it is a bit
 
nevermind
 
6:59 PM
@SAJW it's all about order of operations
and whether $!$ is a function or punctuation
 
so there is a point missing in one case
 
?
 
You won't believe it, but the answer is $5!$.
or another exclamation mark
 
in one reading, yes, there is a period missing.
so it's a semi-periodic joke
 
:D
 
7:03 PM
the use of good english is went
 
Even though a period is the lower half of a colon, I would not call a periodic function a lower-colonic function
 
alimentary my friend
quick which bond movie...
 
good night sazzle
The new film of Sherlock Holmes?
with Pierce Brosnan as a Cameo, wonder when that comes in the cinema
 
@copper.hat Diamonds are Forever, but I cheated
 
:-). i'm a total bond addict
 
7:07 PM
I went ogle
 
i still remember watching that particular scene
a looong time ago
sorry, brosnan as bond was a total disaster. even if he is irish
he redeemed himself in november man
 
I liked Brosnan, it was George Lazenby I didn't like
 
the married movie bond
fleming was a total bigot, but typical of the times
 
How can James Bond be married?!
 
that's why tracy had to go
 
7:15 PM
Thanks for the suggestion, @robjohn.
You may be right about looking for a generalization -- it feels like your answer and other people's answer before that have used contextual arguments to motivate $o\!\left((hu)^m\right)=u^mo\!\left(h^m\right)$. For example, you write that "...the article is looking at letting $h\to0$ in the estimate, while $u$ is integrated over $\mathbb{R}$...". This makes sense, however, with this argumentation it seems like one has determined $\textit{from context}$ what the definition of little-o is (i.e. which variable the limit should be taken with respect to). Thi
 
If you read this, you survived 365*(your age) days plus somemore, congratz!
 
I just got notified that I received a Yearling badge (another year). Which reminds me: I've been here 10 years.
 
congrats :-)
 
@schn The way it is used in Taylor's theorem is right. Give me a variable, Vasily. One variable only.
@copper.hat: what movie? ^^^
 
enemy at the gates :-)
 
7:22 PM
@robjohn What is Vasily? :)
 
I was thinking of Hunt for Red October
Give me a ping, Vasily. One ping only.
Another Bond actor
 
sorry, showing my dark nature :-)
 
Got it :)
 
8:16 PM
i have a mild fever today. knock on wood.
 
If $X$ is an $n$-mfd, then $H_q(X,X-x_0) \neq 0$ iff $q = n$ where $x_0\in X$. Is there similar fact for $H_q(X)$?
 
when the pandemic it's easy to get nervous and forget that sometimes people get normal-sick too.
 
hope you recover quickly.
 
@leslietownes. wish you become healthier and stronger
 
i'm drinking ginger 'tea.' usually does the trick.
 
8:24 PM
Hello, are two topologies the same iff they have the same basis; also if they have the same subbasis?
 
hot toddy.
 
@love_sodam Have you tried any examples? Before asking?
@leslietownes I’ve just been through this. Scary.
Must be something sinus/cough/fever going around.
 
8:54 PM
yeah, a guy at my office had it a week ago. i wasn't there at the time. i probably got it from my daughter, who never shows symptoms.
 
that's one possible advantage of having kids around. most of my relatives who were frequently surrounded by kids lasted a lot longer (and in better mental health) than those who were more isolated.
 
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