2:20 AM
Hi @robjohn

3 hours later…
4:53 AM
Hi @copper.hat
5:18 AM
why can't a derived function have a jump discontinuity in it's domain ?
Hey @copper.hat How have you been doing? I hope all's good in your side.
5:32 AM
@ThomasFinley what would the slope at that point be?
y = f(x)
y' = f'(x)

1 hour later…
6:56 AM
@user85795 yes and?
7:26 AM
@ThomasFinley See Darboux's theorem

2 hours later…
9:09 AM
ask the guy who posted the following which 'books' he is quoting from :)
1

Let $f:\Bbb [a,b]\to \Bbb R$ satisfy the following : (i) $f(x)$ is derivable in $[a,b]$ (ii) $f'(a)=\alpha\neq f'(b)=\beta$ and (iii) $\exists \gamma\in (\alpha,\beta)$ then, there exists at least one value of $x$ say $\xi$ between $a$ and $b$ such that $f'(\xi)= \gamma.$ I tried to prove the abo...

9:21 AM
oh no :|

2 hours later…
11:30 AM
@ThomasFinley what is a "derived function"? Do you mean a derivative?
11:41 AM
@Jakobian that is probably very late, but I am not sure I understand. would you elaborate on the point when you said from continuity.
11:57 AM
@Serilena all the functions involved are continuous functions of the point (x, y, z) chosen on the plane, the points for which the strict inequality holds being an open set
If it holds on the border of the triangle, then it must if we go a little bit outside of it
12:17 PM
interesting. Thanks!
Tho, I am interested how did you determine the points form an open set?
is this true for any convex shape? or was it something related to triangles specifically.
12:33 PM
If $f, g$ are continuous functions then $\{x : f(x)<g(x)\}$ is an open set
1:06 PM
@XanderHenderson yes
1:23 PM
@ThomasFinley You should endeavor to use the correct terminology---"derived", "derive", and "derivation" all have meanings in mathematics which are distinct from "derivative", "differentiation", and "differentiate" (for example).
1:37 PM
The more I study mathematics, the more interested I am in proving things in different ways (and I don't mean it as "prove this without using x" as undergraduate would understand) rather than just proving them
2:31 PM
Hi @user85795 :-) the room has gone quiet
Hi @ThomasFinley life is good thx :-)
2:52 PM
@Jakobian relatable for me. More proofs the merrier. Recently found 2 proofs for the thrm: compact sets are closed in metric spaces
I was wondering, don't we require $g$ to be continuously differentiable for the inverse function theorem to hold? In other words, is the derivative of $g^{-1}$ well-defined without requiring continuous differentiability?
@XanderHenderson I sometimes ignore answers for upto months cuz sometimes it takes that long to sink in
@SoumikMukherjee turns out I was wrong. I was thinking about number theory theorems stated and proved using real numbers, but turns out we can write them out " arithmetically" atleast in principle

1 hour later…
4:02 PM
@psie I believe $g(x) = x^3$ is a counter-example
for purposes of probability though, it should be enough to be able to differentiate $g^{-1}$ almost surely
all we need to do is apply substitution theorem and for that you don't need differentiable everywhere
there are various "advanced" integration by substitution theorems that you don't really learn about, but this is the idea
4:23 PM
Jakobian is it possible to get another hint for that series problem we were discussing yesterday?
$\sum 1/k(k+1) \leq \sum 1/k^2$
But you are saying I have immediately some upper bound too so that I can use the squeeze theorem?
I still can't see that I'm afraid
4:35 PM
@EE18 no
that would be spoon feeding
@EE18 What are you trying to do?
4:51 PM
Am trying to prove the convergence of the sequence of partial sums $\sum_1^n 1/k^2$. Comparison test has been suggested. I have obviously that $\sum_1^n 1/k(k+1) \leq \sum_1^n 1/k^2$ for each $n$ and i can show the first of these converges to 1 because it telescopes. But I can't see what upper bound should be used.
@EE18 So you want to show that $\lim_{n\to\infty}\sum_{k=1}^{n} \frac{1}{k^2}$ converges? That's it?
Also, don't say "obviously". Nothing is obvious. And if it is obvious, it doesn't add anything to the conversation to add that word in.
Finally, if you want to show that something converges, it doesn't really help to show that it is bigger than something which converges.
You need an inequality which goes in the other direction.
$\sum_2^n 1/k^2 < \sum_1^n 1/k(k+1)$ so that I can argue via some quick epsilon action that $(\sum_2^n 1/k^2)$ is bounded above. Further it's monotone, whence I get that it converges? Then $(\sum_1^n 1/k^2)$ is just an additive constant away so I get that too
Is that what you had in mind Jakobian?
There's no use of comparison test here so maybe not?
@XanderHenderson Thank you, this was helpful
@EE18 Maybe it is not so useful to ask what someone else had in mind. Is your solution correct?
It is, yes. But I would argue knowing what someone else had in mind can be useful, especially if they had in mind a slicker way or a different way :)
@EE18 What do you mean there is no use of the comparison test?
5:00 PM
I guess as with all these named things it's a bit fraught to throw the names out since people mean different things by that
What, precisely, is your statement of the comparison test?
Oh, that version. Yeah, you are definitely using that.
@EE18 yes
How so? If i was using that then I would be able to find the value of my sequence?
But as Jakobian mentioned yesterday that's a much harder problem
5:02 PM
$x_n = \sum_{k=2}^{n} 1/k^2$ and $y_n \le \sum_{k=1}^{n} \frac{1}{k(k+1)}$.
The comparison test only gives you a value if you know the values of $\lim x_n$ and $\lim z_n$, and those values are the same.
not sure what you mean by quick epsilon action but $x_n \leq \sum_{k=1}^\infty \frac{1}{k(k+1)} < \infty$ and the rest is as you said, monotone sequence bounded from above
And then Jakobian types faster than I do. :D
@EE18 this is the comparison test
@XanderHenderson I don't understand, $y_n = \sum_{k=1}^{n} \frac{1}{k^2}$ right?
@EE18 You are getting too caught up on the names of things, and missing their meaning.
That theorem says several things.
One of those things is that if $x_n \le y_n \le z_n$, and $x_n, z_n \to L$, then $y_n \to L$.
But is also says that if $x_n \le y_n$ and $y_n \to L$, then $\lim x_n \le L$ (if the limit exists).
If $x_n$ is monotonic, this further implies that $x_n \le L$ for all $n$.
5:06 PM
@Jakobian By quick epsilon action I mean proving that for $x_n = \sum_{k=1}^{n} \frac{1}{k^2}$ we have $x_n \leq \sum_{k=1}^\infty \frac{1}{k(k+1)}$ which I believe requires fixing (say) $\epsilon = 1$ and then using that $x_n \leq \sum_{k=1}^n \frac{1}{k(k+1)} < \sum_{k=1}^\infty \frac{1}{k(k+1)} + 1$ for $n$ large enough
So you have an upper bound on all of $x_n$. And what do we know about monotonic, bounded sequences?
@EE18 this is called squeeze theorem, usually. Comparison theorem says among others that if $a_n, b_n$ are sequences of non-negative numbers, $a_n\leq b_n$, then $\sum_n b_n < \infty$ implies $\sum a_n < \infty$
2
@XanderHenderson Does the proposition I've given say that?
@Jakobian This is the theorem I had in mind when @EE18 said "comparison test".
(Genuine question as I don't see that)
5:07 PM
@EE18 Yes.
@EE18 no need to, $\sum_{k=1}^n \frac{1}{k(k+1)}$ is an increasing sequence, so its limit is $\leq$ than it i.e. $\sum_{k=1}^n \frac{1}{k(k+1)}\leq \sum_{k=1}^\infty \frac{1}{k(k+1)}$
Sure, ya I guess I have that theorem too. I was just going back to limit definition which is sufficient here, but understood
Xander I still don't see how my Prop 2.9 gives that second statement you mentioned
Possible to explain further?
@EE18 in this case arguing via limits in the way you did is unnatural
@EE18 Look at the proof. If $x_n$ is eventually smaller than $y_n$, then $\lim x_n \le \lim y_n$, assuming that the limits exist.
But, like, this seems like a really awkward way to attack this problem. Isn't it easier to just use the comparison test for series, rather than some knock-off squeeze theorem?
Or, if you want to be fancy, use an integral comparison test to show that $\sum k^s$ is convergent for any $s$ with $\Re(s) < -1$ (or however that works out).
Oh sure, OK you're saying from the proof I can deduce it
I thought you meant from the statement and couldn't see it. But understood now
5:13 PM
@EE18 Yes. Though it is probably stated as a theorem earlier in the text.
Generally, one of the first things that is proved after limits are introduced is that limits play nice with sums, products, and inequalities.
@XanderHenderson I haven't seen series yet is why. So this question came up as an aside in an exercise where what we've just done here (thank you again Jakobian and xander) is enough for what I need
Plus some kind of monotone convergence result.
Perhaps there was a slicker way which didn't go this circuitous route, but I shan't waste your times on that
5:47 PM
@Jakobian If take a group G, subgroup H and define an operation on the set of cosets G/H by picking a representative for each coset and defining the product of two cosets to be the coset containing the product of the representatives then do we get a group if and only if H is normal in G?
That's my understanding Vivaan
@VivaanDaga you can define it only when $H$ is normal
so when $H$ is not normal, this definition of an operation on $G/H$ doesn't even make sense
i.e. if you define $[x]\cdot [y] = [x\cdot y]$ where $x, y\in G$
and $[x] = xH$ is the coset corresponding to element $x$
6:02 PM
@Jakobian why doesn’t h try e operation make sense. We are picking a representative for each coset(say using the axiom of choice) and then defining the product to be the coset in which the product of the representative lies. It seems like a perfectly well defined binary operation?
Sure, but you were unclear about this. If you define it like in my comment then what I said is true
when we define the operation on $G/H$ we usually do it by $[x]\cdot [y] = [x\cdot y]$ where this formula has to hold for all $x, y$, and not by specific representatives
@Jakobian so does this definition lead to a group if and only if H is normal. Assuming H is normal then it’s easy to see this def leads to a group but does the converse hold?
I realise that this is not a great def but am curious about the above Q
@VivaanDaga yes
Its not a bad question, its reasonable
@SoumikMukherjee proof of the hard direction?
6:10 PM
suppose G/H forms a group, f:G->G/H defined by f(g)=gH would be a homomorphism, then H=ker f, which is normal
@SoumikMukherjee why would $f$ be a homomorphism?
We give the set of representatives $S$ a structure of a group $*$ so that $(x*y)H= xyH$
the question is if this must imply that $H$ is normal
6:27 PM
@Jakobian f(a)f(b)=aHbH=[a].[b]=[a.b]=abH=f(ab).
@SoumikMukherjee again, you misunderstood the question. We pick a representative from each of the equivalence classes
In other words there is a map $r:G\to G$ such that $xH = r(x)H$, $r(x)H = r(y)H$ implies $r(x) = r(y)$, and $r(x)r(y)H = r(xy)H$
with operation $x*y = r(xy)$ for $x, y\in S = r[G]$ being a group operation
sorry not $r(x)r(y)H = r(xy)H$ but $r(x)r(y)H = r(r(x)r(y))H$
4 mins ago, by Jakobian
In other words there is a map $r:G\to G$ such that $xH = r(x)H$, $r(x)H = r(y)H$ implies $r(x) = r(y)$, and $r(x)r(y)H = r(xy)H$
i.e. the last condition here is not supposed to be there
@Jakobian We want the operation to be well defined, so if $[a_1],[a_2]\in aH$ and $[b_1],[b_2]\in bH$ then we must have $a_1b_1H=a_2b_2H$ So it doesn't matter which representative we choose
We are picking a representative for each coset(say using the axiom of choice) and then defining the product to be the coset in which the product of the representative lies. It seems like a perfectly well defined binary operation.
It’s slightly confusing but I think the above captures what I am saying perfectly @SoumikMukherjee
6:43 PM
@SoumikMukherjee "We want the operation to be well defined" - no, its always well defined
@VivaanDaga The question is if this is well defined then the subgroup must be normal
No. The binary operation gives a well defined function always. My question is if this operation forms a group then must H be normal?
if we were assering that we are defining the operation on cosets so that $[x]\cdot [y] = [xy]$ for all $x, y\in G$ then what you said would be true and I already said that
but this is not what we are doing
Instead, we are picking representatives $r(x)\in [x]$ using axiom of choice and defining an operation on $r[G] = S$ so that $x*y = r(xy)$ for $x, y\in S$
this is always well-defined. Now asking if the fact that $(S, *)$ is a group implies that $H$ is normal
@Jakobian Maybe I am missing something obvious, but I don't think that the operation is always well defined. Take for example $S_3$ and $H=\{e,(1,2)\}$. Now $\{(1,2,3),(1,3)\}$ is in a coset. But $(1,2,3)H(1,2,3)H=(1,3,2)H$ and $(1,2)H(1,2)H=H$ so the operation is not well defined here.
I don't understand what you mean by not well defined
you haven't even chosen representatives
6:55 PM
@SoumikMukherjee The operation certainly gives us a binary function
Perhaps you misread my question or have a different def of well defined?
a_1=b_1= [(1,2,3)] and a_2=b_2=[(1,2)]
No. You pick one representative from each coset
@VivaanDaga My understanding is that by well defined you mean it doesn't matter which representative from a coset we choose
No…
6:58 PM
That’s not in the question
okay
Okay so the map $r$ assigns each coset the unique representative it contains
@Jakobian yeah I misunderstood the question, sorry
7:27 PM
Working on this one. The idea of course is to show this sequence is Cauchy
I've shown (I think correctly) that for $n > m$ we have
$$d(x_{n},x_m) \leq d(x_m,x_{m-1})\sum_{k=1}^{n-m} \alpha^k \leq \alpha^{n-1}d(x_{1},x_{0})\sum_{k=1}^{m-m} \alpha^k = \alpha^{m-1}d(x_{1},x_{0})\sum_{k=1}^{n-m} \alpha^k = d(x_{1},x_{0})\sum_{k=1}^{n-m} \alpha^{k+ m -1}$$
$$\leq d(x_{1},x_{0})\sum_{k=1}^{n-m} \alpha^{m} = \alpha^m d(x_{1},x_{0}) (n-m-1)$$
But as you can see this isn't a very helpful bound because $n-m$ is unbounded in that last step
Am I on the right track or way off?
7:40 PM
Should be $n-m$ on that last step, not $n-m-1$
@EE18 you need to make more use of the decrease in $d(x_{n-1},x_n)$
I guess I'll keep thinking about it then
8:02 PM
@EE18 geometric sum
Bounds aren't tight enough
8:25 PM
@Jakobian OK. Will try to think about which step I gave up a lot on. I thought I just applied the hypothesis over and over, so not sure where I'm missing though
9:03 PM
Also no question about how to do this one but more about context: "Show that every sequence in R has a monotone subsequence."
Why would they ask me this in the chapter on Cauchy sequences? Is there a proof they had in mind such that some of that machinery is relevant?
As far as I can tell the argument uses Bolzano-Weierstrass and then some casework

1 hour later…
10:20 PM
@EE18 because compactness and completeness relate to each other
so there is some proof involving cauchy sequences?