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12:00 AM
i was previously aware that it exists. does that count as 'familiar with'?
 
LOL. Just wondering if someone had read it and if it was worthwhile in terms of understanding number systems
 
i mildly discourage people from reading books that are that old, if only because they may not use terminology that would make clearer connections to terms that people use today.
 
that's well taken
 
there's maybe also a worthwhile distinction to make between constructing number systems in terms of various formal systems, and "understanding" number systems. i think you can understand real analysis pretty thoroughly without knowing about any particular set theoretic construction.
 
do you have a favorite reference for number systems nevertheless?
very fair re: real analysis
 
12:04 AM
there's a more modern book that is kind of interesting, a series of essays on various topics, called "numbers" and edited by a guy named ebbinghaus, whose hilarious name came up in chat recently. you might check that out.
it uses the modern language for everything but is also really historical in terms of citing the original papers for stuff, etc.
 
That was the other one I saw referenced! I think it's at too high a level for me though :(
that was why I went looking elsewhere
 
sometimes history of math books will have OK treatments of at least some aspects of number systems. maybe not the real numbers as dedekind cuts though. that's just boring as hell and hard to find outside of a book specifically about that.
i took an undergrad class out of 'history of math' by victor katz, it is a very readable book, high school level in a good way.
 
I will take a peek, thanks for the tip :)
 
12:38 AM
arithmetic with the natural numbers is such a damn pain
i never thought i'd say i'm dying to learn about the integers here
 
> $|x_j-x_k|\to 0$ when $\min \left(j{,}k\right)\to \infty $
When the minimum of $j$ and $k$ tend to infinity...suppose this is $j$, then must $k$ tend to infinity as well? I'm confused.
 
You don’t necessarily know that $j$ is always the smaller, but yes, sure, both go to $\infty$.
this is the Cauchy criterion, eh?
 
yes
 
just say $j,k\to\infty$.
 
I agree, much clearer
 
1:29 AM
A parabola is a curve, but it's also a group
:>
In a natural way - I don't mean imposing some random group structure
 
 
1 hour later…
2:37 AM
Does the statement that any linear transformation is unqiquely specified by its action on a basis need choice in infinite dimensions?
The proofs I see on the main site all just state that the proof goes through the same
But I am curious as to why my book would state only the finite dim case for the domain if so
 
Because the notion of basis in infinite dimensions is a hornet’s nest.
 
2:56 AM
For sure, but still good for me to have a sense no?
4
Q: Existence of a linear transformation in an infinite dimension vector space.

OmarIf $V$ and $W$ are vector spaces, $\beta=\{v_1, \ldots , v_n\}$ is a finite a basis for $V$ and $\{w_1, \ldots , w_n\}\subset W$, we know there is an unique linear transformation $T:V\rightarrow W$ such that $T(v_i)=w_i$ for $i=1, 2, \ldots , n$ Is this valid when $V$ is not finite-dimensional?

In something like this it’s not clear to me if choice slips in unwittingly
 
you need something like choice to know that an arbitrary vector space has a basis, but that's the only hurdle, really.
often in the infinite dimensional setting you see people with something that is less spacey than 'linear algebra basis' in that context (i.e. something kinda like a linear algebra basis, but where you do not need AC to ensure one exists). maybe at the cost of not always having finite linear combinations anymore but allowing "infinite linear combinations" (e.g. a sequence of linear combinations that converges in some topology)
 
A schauder basis?
 
@leslietownes Hamel vs Schauder
Ninja'd.
 
I see Leslie, so you’re saying that the theorem i quoted above sort of slips in choice by the “if there is a basis of V” premise being true only because of choice
 
3:12 AM
in general, yeah. "is a vector space with basis" and "is a vector space" are generally distinguishable concepts without AC.
some infinite dimensional vector spaces do "come with" obvious choices of basis of course, without AC, but not all of them.
 
Gotcha, that’s well taken. Thanks very much
 
X4J
4:07 AM
Suppose (a_n) is a sequence of reals such that the infinite sum a_1 + a_2 + a_3 + .... converges. If for each n, we denote L(n) as the finite sum: a_n + a_{n+1} + a_{n+2} + ..., then is the sum L(1) + L(2) + L(3) + L(4) .... is always finite?
 
So, does $\sum na_n$ converge?
 
yeah. maybe some clean examples with telescoping series. e.g. a_k = 1/k - 1/(k+1)
 
X4J
Yes, more clearly: Let $(a_n)_{n=1}^{\infty}$ be an infinite sequence of reals. Suppose that $\sum{a_n}$ converges and denote $\forall n \in \mathbb{N} : L(n) = \sum_{k=n}^{\infty} a_k$. Is the sum of $\sum{L_n}$ always finite?
 
1
Q: Can this subgroup of the positive rational numbers be quotiented to a non-trivial finite group?

MathCrackExchangeDefine $G = \Bbb{Q}_{\gt 0}$ to be the (multiplicative) group of positive rational numbers. Suppose that there is a group surjection: $$ f : G \twoheadrightarrow \Bbb{Z} $$ and we took its kernel $N = \ker f$. Then of course $G/N \simeq \Bbb{Z}$ by first isomorphism theorem. By the third isom...

@Shaun group theory and twin primes holding hands
 
X4J
@leslietownes thanks that's interesting, what would be an intuitive way to think about it so it could provide me some idea if I want to prove it?
 
4:19 AM
Potato Schauder bases
 
prove what? that the example is a counterexample? (the sum of the L(k)'s isn't always finite under those hypotheses.) in my example, you can evaluate each of the L(k) explicitly.
 
someone say potato?
 
1
A: Sandbox for Proposed Challenges

Dannyu NDosAdd two rational numbers... esoterically Objective Build a binary operator \$\star : \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}\$ such that, there exists a bijection \$i : \mathbb{Z} \to \mathbb{Q}\$ such that, for every integer \$M\$ and \$N\$, \$M \star N = i^{-1}(i(M) + i(N))\$ holds. Or in m...

Would this challenge be easier than it seems?
In layman's terms, "Find the one-to-one correspondence between the integers and the rationals that is the easiest to compute."
 
X4J
4:47 AM
@leslietownes Oh I see it clearly now. In fact the case I initially observed was more specific: $(a_n)$ is a sequence of reals and the sum $\sum \frac{|a_{k+1} - a_{k}|}{a_{k+1}}$ is finite. We denote $L(n) = \sum_{k=n}^{\infty}\frac{|a_{k+1} - a_{k}|}{a_{k+1}}$, is the sum $\sum L_n$ always finite?
 
if f(1)=1, f'(1)=1, f''(1)=1 then f(1/2) is ....
I was thinking to use FTC but couldn't then I used Taylor Series expansion of f upto its 2nd derivate and got f(1/2)=5/8. Is this method correct?
 
the given info is not enough to determine the value of f(1/2). it is enough to determine the value of the second degree taylor polynomial (at 1) for f at 1/2.
 
Now I see, Thank you
 
5:19 AM
0
Q: Is this proof of Darboux's Theorem valid?

Thomas FinleyLet $f:\Bbb [a,b]\to \Bbb R$ 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...

Need some help with this issue...
 
5:30 AM
@X4J Did you see my question? That’s the answer to your question.
 
X4J
@TedShifrin In what case? I understood the counter-example given at the first case, currently trying to figure the second case I mentioned
 
I did not look at the second version since you never responded to my response.
It’s easier to analyze the series $\sum na_n$ than to look at your formulation.
 
X4J
Yes, I see now it's actually nan
 
@ThomasFinley Your proof is not complete. How do you know the minimum is at an interior point. Notice you haven’t used any hypotheses, so you should be completely suspicious.
You should be asking yourself that question.
 
Just to check before posting on CS SE: The set of 2-adic integers is an additive group, right?
 
5:42 AM
wiki said it's an integral domain
 
So a ring. Nice.
 
@TedShifrin Thanks! I understood the problem. That's the point where my proof fails in general.
 
I remember that p-adic number was introduced in commutative algebra class which I forgot almost everything.
I remember Artin-Rees lemma only the name.
 
Thanks. I wondered whether each arithmetic hierarchy is a subgroup. Or rather, a subring.
That's what I'm gonna post on CS SE.
 
X4J
@TedShifrin Unfortunately I struggle to figure out how the convergence of \sum na_n effects the convergence of \sumL_(n)
 
 
2 hours later…
7:19 AM
0
Q: Is NP a subring of the 2-adic integers?

Dannyu NDosLet me take the set $A = \{1, 2\}$ as the alphabet. By the bijective binary numeral system, $A^*$ has one-to-one correspondence to the set of nonnegative integers $\mathbb{N}$. As such, each language $L \subset A^*$ corresponds to a subset of $\mathbb{N}$. Now consider the power set of $\mathbb{N...

*cough*
Too sadge if this proof is correct, because that would imply that I didn't solve the P-versus-NP.
 
7:36 AM
0
Q: Is this proof of Rolle's Theorem a valid one?

Thomas FinleyLet $f:\Bbb [a,b]\to \Bbb R$ satisfy the following : (i) $f(x)$ is continuous in $[a,b]$ (ii) $f(x)$ is derivable in $(a,b)$ (iii) $f(a)=f(b)$ then, $\exists c\in (a,b)$ such that $f'(c)= 0.$ I tried to prove this theorem as follows: Since, $f(x)$ is continuous in $[a,b]$ so by Maximum-Minimum Th...

Need some help with this.
 
 
1 hour later…
X4J
8:40 AM
Oh well a_1 + 2a_2 + … + Na_N +… = (a_1+a_2+…) + (a_2 + a_3 + …) + … + (a_N + a_{N+1} + …) +… = L(1) + L(2) +…+L(N)+…
I guess @TedShifrin?
 
I consume way too much sugar with my tea
 
9:09 AM
How many teaspoons per cup?
 
a lot of em.
trying to imagine where you'd need to scale down from cups to teaspoons. trying to recreate a recipe you learned in a prison kitchen at home, maybe.
 
☕🥄🥄🥄
^that's my intake for 4 cups of tea, each
 
oh ahha i hadn't seen hades' question, i thought you were performing an unusual unit conversion.
 
I didn't learn much in the prison kitchen.
 
i spent most of my time sharpening the teaspoons
 
9:19 AM
making weapons out of nothing at all
btw google says: 1 cup = 48.7 tsp which doesn't seem right to me
 
9:36 AM
Let $X$ be a metric space then $f:X\to\Bbb R$ is lower semicontinuous iff it is the limit of an increasing sequence of continuous functions.
Is it true? because in some places they assume bounded belowness of $f$ and in some places they don't assume that.
e.g. (with the boundedness) here math.stackexchange.com/questions/165764/… and here arxiv.org/pdf/2002.07205.pdf [see the section 2]
e.g. (without boundedness) en.wikipedia.org/wiki/Semi-continuity#Properties see theorem of Baire. and here mathoverflow.net/questions/280537/…
 
10:04 AM
Strangely or not strangely, they dont specify anything about $\lim_{x\to a+}f(x)$
I think, $\lim_{x\to a+}f(x)=\pm \infty$. Am I correct?
I think they wanted to consider the rule $\infty/\infty$ and $\-\infty/-\infty$
But not mentioning it, makes things complicated and mysterious!
@leslietownes Can you please help me with this?
 
@ThomasFinley If the denominator goes to infty then you don't need any lim of numerator.
But if you have g(x)\to0 then you need f(x)\to0.
 
10:24 AM
@PNDas Oh, thanks a ton! So, basically they wanted to say, that if, $\lim_{x\to a+}g(x)=\pm \infty$ and $f,g$ are differentiable then, we can directly apply this theorem. Is it?
 
Yes (you also need: g' doesn't vanish)
 
@PNDas Just one more thing: Is the above theorem valid if $a=\pm\infty$ ?
@PNDas K
 
@ThomasFinley Yes
 
@PNDas Just a second:
@PNDas This is the full image
They mention that this theorem is valid if $a=-\infty$
But they never mention whether it's valid for $a=\infty$ or not?
So, should I consider that the rule in the above image is valid for $a=\pm\infty$ as well, @PNDas?
 
What is $\infty+$?
 
10:31 AM
@PNDas No idea
@PNDas Similar to $a+$ when a is real ?
 
World moves too fast for me
 
@PNDas I think I never came accross that notation tho.
But my question whether the theorem in the image above is valid if $a=\infty$ ?
@PNDas I take this statement correct and proceed with my study :)
 
@Jakobian What is the weather like in Bernstein set? Too windy?
 
Will you advocate this approach below ?
33 secs ago, by Thomas Finley
@PNDas I take this statement correct and proceed with my study :)
 
Since they are talking about RHLs so they discarded infty.
 
10:36 AM
@PNDas You know, that makes sense!
@PNDas So, the theorem in the image above is valid even if $a=\infty$ , right?
Sorry, if I sound repititive but I really want to get things clarified :?)
 
@ThomasFinley If you modify the hypothesis and ignore + then yes it's true.
 
@PNDas sorry, but which hypothesis should I modify and which + should I ignore?
 
Since they are talking about RHls at $a$ so they only considered $(a,b)$ with $a<b$ and bla bla.
So for a general statement you have to change the statement appropriately.
You can see the wiki page.
 
@onepotatotwopotato non-measurable
2
 
@PNDas Ok, now I understand it. So, the thing is I need to delete the expression , "$-\infty\leq a\lt b\leq \infty$" and replace it with "$a=\infty$ " and replace all the symbols "x\to a+$ with $x\to\infty$ throughout the theorem. If this is done then the theorem is true for $a=\infty.$ Did I get it correctly?
 
10:52 AM
You missed a dollar sign somewhere and now it's unreadable.
 
@PNDas ok, wait.
Repost, @PNDas : O k, now I understand it. So, the thing is I need to delete the expression , "$-\infty\leq a\lt b\leq \infty$" and replace it with "$a=\infty$ " and replace all the symbols "$x\to a+$ with $x\to\infty$ throughout the theorem. If this is done then the theorem is true for $a=\infty.$ Did I get it correctly?
Is it readable now, @PNDas ?
 
What would you do for the line "f,g are differentiable on (a,b)"?
 
@PNDas "f,g are differentiable on some interval $(p,\infty)$, where, $p\in\Bbb R$ " ?
 
Yeah.
 
@PNDas So, then the theorem in the pic will be completely valid for $a=\infty$ , right?
 
11:05 AM
@Jakobian lol
I like your response
 
 
3 hours later…
1:51 PM
> The series $\sum_{k=1}^\infty \pmb x_k$ of elements in $\mathbf R^n$ is convergent if and only if $$\sum_{k=m}^n \pmb x_k\to 0 \quad\text{when }\min(m,n)\to\infty.$$
This is how it's stated in my textbook. I'm curious, we could also require that the norm of the series converges to $0$, right? As it stands, I'm interpreting the $0$ here to mean the $0$ vector.
 
@sunny $y_n\in\mathbb{R}^n$, then $y_n\to 0$ iff $\|y_n\|\to 0$
 
ah, ok.
👍
 
yeah so this is really a statement about real numbers
 
@MathCrackExchange hi, the exercise is tricky.
It depends upon parity of n.
For n=3, the statement is false.
how to do the chess one??
suppose players have to play two moves instead of one. Then, there exists a non losing strategy for white.
 
2:24 PM
Struggling to understand the argument in their last paragraph of Prop 6.3 proof here. How are they using the induction hypothesis in "In view of the induction hypothesis..."?
The usual argument I've seen separates out some size $n$ subset of this $n+1$ element set, but I don't see how that's done here
 
@EE18 there are fixing where a_j goes first.
 
Ahhhhh is it a typo?
Do they mean $a_j$ to $a_j$?
 
no!
@EE18 no, I mean there are looking at the set {f is a permutation of Y: f(a_j)=a_1}
for a fixed j.
so one just has to permute the remaining a_i's in Y. That is, one has to permute Y-{a_j}, a set of n elements, hence you get fact(n) from here.
 
I'm thinking they are arguing as follows: there are $n+1$ choices for which $a_m$ to delete from $Y$. For each fixed deletion, we have an $n$ element subset and so the IH applies and there are $n!$ such permutations. Each deletion (I will need to show this) corresponds precisely to a permutation of $Y$ in which the deleted element is mapped to itself.
Then, by Exercise 5, since these subsets of permutations form a partition of all permutations of $Y$, and there are $n+1$ such disjoint subsets, we arrive at $(n+1)n!$ by that formula (since equality obtains when we have pairwise disjointness)
But that seems different than what you're saying their argument is
the specification of $a_1$ is what's confusing me
Never mind, my argument above must be wrong since it misses e.g. cyclic permutations
 
you could take a_2 instead of a_1.
 
2:34 PM
@Koro I think this message may be what makes it click for me
If I fix where $a_j$ goes, then I can consider the given permutation $f$ of $Y$ (the one which sends $a_j \mapsto a_1$) as the union of that function with just the $\{(a_j,a_1)\}$ pair and the restriction of $f$ to the subset $Y \setminus \{a_j\}$. There are $n!$ permutations $f$ of $Y$ with this $j$ which can be built in this way
Now I just need to convince myself that I am not repeating or missing any :)
 
Define $A_j=\{f\in S_Y: f(a_j)=a_1\}$. Then, do you see that $S_Y=\cup_j A_j$?
then, the critical observation is: $A_j$'s are pair-wise disjoint.
@EE18 something like that
 
yup, disjointness I see. I also see $S_Y \subseteq \cup_jA_j$ because any $f \in S_Y$ is onto and so has some $a_j$ with $f(a_j) = a_1$, so $f \in A_j$. Does my argument above sound right for getting the size of each $A_j$?
 
@EE18 yes for the equality of S_Y and \cup_j A_j.
then, you just use |A\cup B|= |A|+|B|, A, B disjoint.
 
Yup, agreed :) the bit I just want to confirm is how one rigorously argues for the size of each $A_j$. It's "obvious", but to confirm does my argument above sounds reasonable? That there's a bijection between $A_j$ and the set of restrictions to $Y \setminus \{a_j\}$, and it's to that latter set of restrictions to which we can apply the IH?
PS, perhaps it should have been but it wasn't obvious to me that defining $A_j=\{f\in S_Y: f(a_j)=a_j\}$ (as I initially did) would have been wrong. Is there any way someone can point out where the flaw in such a definition (and subsequent argument that $S_Y=\cup_j A_j$) is
 
2:51 PM
@EE18 it looks okay to me.
 
Thank you very much Koro! This was very helpful for me :) gonna write it all up now
 
@EE18 disjointness will create problem.
 
The $A_j$ I
 
A_1 and A_2 will have identity map common.
 
Ahhhh I see.
I guess they will also miss some permutations of $Y$ right? The ones where no element is invariant?
So the argument I gave above for $S_Y \subseteq \cup_j A_j$ would not work for $f$ such that $f(a_j) \neq a_j$ for all $j$
Got it. Thanks again Koro!
 
2:55 PM
@EE18 yes, there is no guessing here.
yw!
 
If $U$ and $V$ are unit vectors and $U\cdot V=1$, does that mean $U=V$? I know $U\cdot V=U\cdot U=V\cdot V=1$. But I also know $U\cdot V=U\cdot U$ doesn't necessarily mean $V=U$
 
3:16 PM
@Koro I am sadly realizing my argument is for the size of the $A_j$ is incorrect
I wrote "
If I fix where $a_j$ goes, then I can consider the given permutation $f$ of $Y$ (the one which sends $a_j \mapsto a_1$) as the union of that function with just the $\{(a_j,a_1)\}$ pair and the restriction of $f$ to the subset $Y \setminus \{a_j\}$. There are $n!$ permutations $f$ of $Y$ with this $j$ which can be built in this way"
But "the restriction of $f$ to the subset $Y \setminus \{a_j\}$" is not itself a permutation (the set of which I can apply the IH too). Well, maybe it is if I define the restriction also with a corestriction to delete $\{a_1\}$, but now then my domain and codomain are different (contradicting the definition of permutation)?
 
3:40 PM
Can anyone explain what's the meaning of this question? Is this implying $\theta=f(a+h)$ ?
 
By definition, a manifold has no boundary (a manifold with boundary is not a manifold). — Robert Israel Sep 29, 2016 at 22:53
Is this a standard convention?
 
@SoumikMukherjee I think so yeah
 
@EE18 Their proof is not right. The induction hypothesis is being applied not to a permutation of a set with $n$ elements, but to a bijection friom one set with $n$ elements to another. So perhaps they had a lemma or exercise to make this translation ?
@SoumikMukherjee Yes
Unless stated explicitly otherwise.
 
@EE18 yes, but you said something about bijections. Probably better would be to strengthen up your IH to "no. of bijections between two n element sets is n!".
 
3:51 PM
@ThomasFinley No. Read carefully and you’ll find $\theta$ appearing. This looks like another old-fashioned text.
@EE18 That’s an error I would expect from a student, not from a book. What book are you reading?
 
:fire:
 
That's well taken @Koro and @TedShifrin Do you see any way to stitch up the current proof (using the current induction hypothesis)? I suppose somehow I'd need to show a bijection between the $A_j$ as defined above and the set of permutations on some $n$ element set $X$?
It's Amann Escher Analysis I, Prof. Shifrin. I thought it was highly regarded so hopefully this is just one slip up...
Maybe I can email them and ask what they had in mind, or is this truly an error do you think?
 
Bad. I don’t know the book, but this is not inspiring me with confidence. Yes, prove the lemma to that the number of bijections blah blah equals the same number.
 
Ugh...!
 
They also should have defined binomial coeffs when $k<0$ to be $0$. Sloppy authors.
 
3:57 PM
I don't wanna work today!
 
@XanderHenderson Huh, didn’t you say a job where you look forward to not doing it is bad?
 
I went away from Baby Rudin because I didn't feel like it gave enough motivation, so I wasn't getting the "big picture". I suppose an advantage of Baby Rudin is that one can be assured of there being zero errors :)
 
@冥王Hades This is not an ongoing, existential thing. Only today.
I don't want to work today.
 
@TedShifrin Of this I am not sure though -- at this point they have only defined the natural numbers, so $k<0$ isn't even in the space of things we know about. I assume they do that later once integers are introduced?
 
No pictures at all in Rudin. One of my complaints.
Oh good grief.
 
3:59 PM
@TedShifrin No pictures, the proof of the MVT is completely unmotivated, and the wheels fall of the book starting around chapter 9. ;)
 
Most proofs are not motivated, and, yes, the multivariable treatment is as ungeometric as it can be.
 
@TedShifrin Was this in regard to the natural numbers comment?
 
But for gifted students with a good teacher, it’s still a good book!
 
@TedShifrin Yeah, but there are very motivated proofs of the MVT, starting from Rolle, which I think is a theorem which can be well-motivated.
 
EE, yes.
 
4:02 PM
My complaint is not about unmotivated proofs in general, only the quality of that particular proof, in light of other texts with much better exposition on the topic.
@TedShifrin Agreed. I do think that it is a "must read" for anyone serious about analysis.
 
unfortunately, not every student is a gifted student (nor every teacher a good teacher)
which places Rudin firmly in the "probably not a good book for teaching analysis" camp
 
Death, taxes, and debates about the merit of Baby Rudin :)
 
Anyhow, EE, state and prove the lemma I suggested.
 
I have no idea about the man, but I always picture Rudin and Jackson (John D) as of the same tough, stern, but fair ilk
Yup, working on that right now Prof. Shifrin. Just writing out why their proof idea fails first in my notes
 
No clue who that is.
 
4:07 PM
Jackson was a physics prof who wrote the most famous EM text
Famously hard too
 
Oh, OK.
 
Ha! Never seen this. At the end "A recording of this song was played at jackson's 60th birthday party" is too funny also
The adaptation of the classic scene in the movie Downfall to be about Jackson is hilarious too
 
@TedShifrin yes, theta appears there, but is theta really a function?
 
That's an excellent question.
Certainly it's not if $f^{(n)}$ is a constant function.
Definitely old-fashioned and sloppy.
 
4:23 PM
@TedShifrin why $\int_{|z|=1} \frac{dz}{z} = 2 \pi i \, ?$ I know we can do it via substitution but see function explodes at the origin so how??
 
Why does that matter, Lucky?
You're not integrating on a path that goes through the origin.
 
But function is taking large values at smaller inputs near zero so
that's why I asked :'(
 
But your path stays far away.
 
Oh so we're just integrating on boundary of circle which is centered at the origin and has radius one? Don't have anything to do with interior part
 
It is a line integral, i.e., an integral over a curve.
If you wanted to integrate over the entire unit disk you would need to integrate $dx\,dy$ or $dz\,d\bar z$.
 
4:30 PM
Oh now I got. So if I have more than one singularities in my counter integral i.e $int_{|z|=2} \frac{dz}{z(z-1)}$ then Can I use partial fraction and do via substitutions?
I have recently started complex and I have lots of questions and doubts. I need to learn more about Complex function. Thank you @TedShifrin
 
Yes, use partial fractions. You might need to know that any two simple closed paths (with the same orientation) that go around $z=a$ result in the same integral $\int \frac{dz}{z-a}$. It doesn't just have to be a circle centered at $z=a$. I don't know how your book has discussed this so far.
 
X4J
@TedShifrin could you notice the observation I stated
 
What observation?
 
X4J
You hinted me to consider \sum na_n
 
Indeed, it equals your awkward sum.
Whoever gave you the question intended you to figure this out for yourself.
 
X4J
4:38 PM
I asked this
 
@TedShifrin Yeah, actuallly book hasn't discussed anything like what I am asking. In "Complex Analysis by Joseph Bak and Donald Newman" they just done integral of 1/z-a by parametrization. I was trying to see what is going on there that's how.
@X4J what was your question?
 
I don't know the book, so I can't say for sure what they have or have not done.
@X4J You don't see why it's true?
 
@TedShifrin Ok, but if $f$ is not a constant function, then can we really say, that $\theta$ will be a function?
 
No need to look at it Prof. Shifrin, but just to show that I "did the work", so to speak :) thanks again for the help!
 
4:40 PM
@Thomas No, I doubt it.
 
X4J
@TedShifrin you mean, why the convergence of \sumna_n will imply convergence of \sum L(n)?
 
@EE18 I would be a bit more pedantic and say the bijection goes from $X$ to $Y$.
They are equal, X4j.
 
Where would you write that Prof. Shifrin? In the Prop statement?
 
Yes.
I haven't looked at the rest.
 
Oh "from" rather than "between"
 
4:42 PM
A bijection has to be a function.
 
No worries on the rest, it was just a bunch of annoying legwork in fact 3 in particular. So all good after the discussion with you and Koro
 
@TedShifrin yes, and this is the reason why I think I should skip this question. Will this be a justified action on my part?
 
Is the English phrasing "between" frowned upon generally?
 
As I said, Thomas, this looks like an old-fashioned book (again), so expect non-modern sloppiness.
 
How do you keep track of 3 different conversations at once, Ted? 4 if you count this.
 
4:43 PM
No, @EE18. But it has its place.
 
LOL!
 
I don't, @Hades.
 
Touche. Thanks again!
 
@TedShifrin Ok, then I think the best thing to do is, solve this problem assuming $\theta$ is really a function depending upon $h$ i.e $\theta=\phi(h).$
 
X4J
@TedShifrin Indeed the way I wrote the sum simplifies it but how can one reason this (namely, reason that \sum L(n) converges
 
4:45 PM
I think you might advocate this.
 
It doesn't necessarily converge. That's the whole point, X4J.
@Thomas I've never seen anything remotely like this before. But if you're going to work on it, you should make such an assumption, yes.
 
X4J
Ah I’m confused - isnt the convergence of \sum na_n equivalent?
 
Your hypothesis was that $\sum a_n$ converged, not that $\sum na_n$ converged.
 
X4J
Yes, I already understood that it isnt sufficient and thanks to you I could ask a more precise question
 
is Princeton lectures in analysis series good?
 
4:50 PM
So, yes, @X4J, your series converges if and only if $\sum na_n$ converges.
You're talking about Stein/Stakarchi, Lucky?
 
Yeah, sir Stein/Shakarchi
 
They are good books. There are some mistakes. I taught a graduate complex analysis course out of the complex analysis book and found that I did not like some of the things they did. Some important things to me were not emphasized until way too late in the book.
 
oh, so which book should I take? I want to get intuitive insight.
 
X4J
I see Ted. What would be one approach to prove it?
 
We're going around and around in circles. They are equal. You said you understood that.
Needham's Visual Complex Analysis is very popular, Lucky. I personally have not looked at it;
 
4:57 PM
I miss Texas sometimes. Only sometimes.
 
Ugh.
 
X4J
@TedShifrin I see it now, thanks
 
I should be more specific. I miss Brisket
 
Ah. That might be worth missing.
 
And my paren— no I don’t
 
5:12 PM
@冥王Hades Hi
 
@PrithuBiswas howdy
 
@冥王Hades Can I ask you something regarding computers?
Specifically, computer ram.
 
@PrithuBiswas sure ask away
 
@冥王Hades I want to upgrade my ram from 2GB to 8GB. I am currently using a 2GB 800MHz ram stick. Should I stick with that same speed while upgrading or change it to something faster?
Also, do I have to worry about motherboard compatibility?
 
@PrithuBiswas depends entirely on what platform you are on yes. Compatibility is a concern, with 800Mhz it sounds like you’re on a DDR2 system, and finding the exact same DDR2 RAM sticks as you’re using currently is very difficult since they’re quite old at this point
also I hope you’re running dual channel at least, if not, you definitely should if or when you go 8GB
 
5:22 PM
How is Ahlfors as a text @TedShifrin ? Thinking of eventually using that as my complex text after Amann Escher
 
@TedShifrin I think it's excellent but one should not use this as a textbook.
 
@冥王Hades Maybe in my country I can easily find ddr2 memory XD
 
Ugh... in the last 90 minutes, I have only managed to complete one item off of my to-do list. :(
I really thought that I could get both probation reports done before class. I was even hoping to get some actual grading done. And I have quizzes to write. I think it's going to be a late one. :(
 
@冥王Hades Yeah I will run it dual channel. Not sure why it is good, but still I will do it.
 
5:56 PM
@XanderHenderson did someone undergo disciplinary measures or is probation something else here
 
does someone know about interest rate in finance?
 
@user123234 what about it
 
@shintuku If I have an interest rate $r$ for one period, and I start with a value $K$ on my bank account then it is clear that at the end of the period I will have $K+K\cdot r$ on my bank account. Now if the interest rate $r$ is compounded at $t=1/2$ why do I then have $K+K\cdot r/2+(K+K\cdot r/2)\cdot r/2$? so I don't get where $(K+K\cdot r/2)\cdot r/2$ comes from
 
That's a question of how "interest rate" is defined here. Your $r$ there is a so-called APR. $r/2$ is then the effective rate for the $t/2$ compounding period.
Again, this is just a matter of definitions (in the realm of finance).
 
@EE18 but even if $r$ is the annual percentage rate, why do I have this additional term for $1/2=t$? why isn't it only $K+K\cdot r/2$
 
6:07 PM
@user123234 you need to distinguish the nominal rate period from the conversion period. usually the nominal rate period is annual (APR)
 
In the second case about which you are confused, you have $K+K\cdot r/2$ at time $t = 1/2$, and at time $t = 1$ you have that same money plus interested earned thereon: $[K+K\cdot r/2]+(K+K\cdot r/2)\cdot r/2$.
 
to get the rate per interest period you divide the nominal rate (can be APR) by the number of conversions per nominal rate period
 
@EE18 ah so I think I missunderstood what t=1/2 means. This does not mean that I want to know what I have after half a period?
 
no because conversion doesn't happen continuously, it happens at regular intervals
t=1/2 means conversion will happen at half of the nominal rate period
 
sorry english is not my mother tongue, could you explain what conversion mean?
because I have not found a good translation
 
6:12 PM
uh i meant compounding
 
does this mean that the interest is payed twice in the period?
 
so if the interest period has length 1/2 of the nominal rate period, at t=0 and t=1/4 you have the same amount of money
@user123234 distinguish these terms: nominal rate period, nominal rate, interest period (also called conversion period), interest rate
 
ah I see!
Perfect thanks!
 
np
 
I'm looking at some probability reasoning like the following: for some threshold $\alpha > 0$, $P(X > Y) \leq P(X \geq \alpha) + P(Y \leq \alpha)$. Maybe I'm missing something obvious, but how does this happen?
Maybe you shoehorn $\alpha$ between $X$ and $Y$, turn that into a union somehow and then take the union bound?
(Well, that would obviously require that $X > \alpha > Y$)
 
6:21 PM
My book gives the following proof for this standard theorem, but it's not clear to me how Proposition 6.7 means we can immediately reduce the problem to the case of all $X_n$ countably infinite and pairwise disjoint -- any hints?
NB that this book uses "countable" for "at most countable" (finite or countably infinite)
 
6:33 PM
I think ProofWiki (proofwiki.org/wiki/…) has my answer, so all good!
Surprised my book wouldn't mention choice is needed here tho...
 
@XanderHenderson one whole item in 90 minutes?! That’s one more than me.
 
Consider a vector space V over $\Bbb C$, and suppose $A$ is a basis of $V$ over $\Bbb C$. How can I prove that $A \cup iA$ is a basis of $V$ over $\Bbb R$? My problem is when the dimension of V is not finite
 
You know I’ve started to realize Japan isn’t for me. When someone trips and falls in public, most people would help that person immediately. My first reaction however is always a laughter
 
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