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12:40 AM
@Koro Thanks
3 hours later…
3:40 AM
picked up my walker today. 19 days before i go bionic
@copper.hat 6 million dollar mathematician?
Wow, you’ll be unwaveringly vicious now!
1 hour later…
5:11 AM
When we have two sets say A={1,2} and B={Ann, Jack, Janet}, the set operation - union, when applied on these two sets give A union B = {1,2,Ann, Jack, Janet} right?
I mean the set A union B can have elements of different types?
I have seen in some places they say a set can have elements of a same type only. But I feel doubtful about this
And if we can take union like above, we ca have elements of different types too right?
hasini: the usual pure math presentations of set theory do not have a notion of 'type,' let alone forbid sets from containing elements of different types. so yes, A union B would be expected to be what you wrote up above.
Okay. So it will only occur if we specially define it? :)
Thank you very very much @leslietownes :) :)
5:27 AM
sometimes domains of functions or other objects can impose restrictions that are analogous to restrictions on "type," but distinction based on "type" is not baked into the set concept in pure math, the way it is baked into many computer programming languages.
if you had some reason for forbidding someone from considering {1,2,Ann,Jack,Janet} in a math class, you would do it with something other than the concept of 'set'
Okay. Now I get it :)
Thanks a million again @leslietownes
Have a nice day!!!
you too :)
@Hasini define Ann
i don't think the definition of Ann would matter under any formalism in which the A and B above are themselves sets
maybe it would matter if you wanted to decide whether Ann = 1 or Ann = Jack or something
5:42 AM
Well in this case B={x| x is a student in class whose height is more than 6 feet} or something like that
And hence it includes names like this
Then it wouldn't matter like @leslietownes also has explained
I think :)
Thanks a lot for replying to my question too @Jakobian
yeah. my basic starting point is if it makes sense to write A = {x,y} and B = {c, d, e, f, g} in your world of "sets" then A union B is {x, y, c, d, e, f, g}, and this would be true no matter what x, y, c, d, e, f, g are. in particular, no matter what "types" they had or did not have.
Did we understand your reply correctly @Jakobian?
somewhat tangential to this, hasini, but kind of in the same ball park, is that in the usual pure math world of sets, {1,1} and {1} are the same set, and this is also unlike what you might expect from at least some data types in some computer programming languages.
Hmmm okay @leslietownes then in computer programming languages, they usually take them to have the same data type
Even without defining specially?
{1,2} and {2,1} are also the same set. a lot of concepts relating to order and repetition that derive from our intuitive understanding of "lists" are not baked into the set concept.
5:47 AM
Because to run a program, they should be having either string or integer? Aren't there programs which can be adjusted to have any type?
languages impose all kinds of rules, or not, depending on how they are set up. i am trying to keep the focus mainly on the math concept of set, which is somewhat more stable than what language X might or might not allow.
Hmm yeah okay
it's pretty common for programmers to use "list-like" data types to model math sets, but for those data types to impose restrictions that the math set concept does not have.
and also for those data types to have structure that the set concept does not retain.
I like to understand the math concept too
Are you familiar with B programming too?
Using Atelier B and Pro B @leslietownes?
like it's pretty inherent in listing something, whether or not you are programming, just as a human being making a list, like a shopping list, to specify an order, even if you understand and the reader understands that the order doesn't matter.
this is also kind of inherent in what you do when you write something like "let A = {1,2,3,4,5}," like, those numbers are going to come in some order as you write them, and of course numbers are naturally placed in a particular order.
5:51 AM
@leslietownes Okay :) cool
but that isn't baked into the set concept either.
which is actually nice, it makes it somewhat easier to define something like A union B in such a way that you can be sure that A union B and B union A are the same thing, for example.
i don't know B programming.
@leslietownes Yeah I see :)
@leslietownes Okay
Many many thanks again @leslietownes :) :) :)
@robjohn more like two buck engineer from trader joes
6:08 AM
So your name is Chuck now?
2 hours later…
7:42 AM
how can I get from the upper red line to the lower red line?
2 hours later…
9:30 AM
Hi. My question was deleted by a bot (again). Please help:
A: Requests for Reopen & Undeletion Votes (volume 01/2022 - today)

ShaunUndeleted by users then deleted by bot. Please undelete and reopen this: https://math.stackexchange.com/q/4618260/104041 The question is mine. I put an awful lot of effort into it. The downvotes confuse me and I have no clue why it was deleted (beyond the automatic deletion when there's enough ...

I asked in CURED too.
1 hour later…
10:31 AM
Is my answer correct?
11:13 AM
@Hasini $1$ has definition in set theory, but Ann does not
@SineoftheTime correct
You're basically saying that the limit is $0$ because its bounded by $2\sqrt{h^2+k^2}$, no need for polar coordinates
@leslietownes this confused me for awhile because I took the analogy of a set as an actual "box containing members" too literally.
11:37 AM
12:29 PM
exam went terribly wrong today. I may have to write it again. I was clueless/blank for the most part during the exam.
1:19 PM
@Koro you probably did better than most of your peers
1:41 PM
that's not the point.
Q: Proof that $\cos(t^2)$ is not periodic

peter.petrovI am trying to prove that $f(t) = \cos(t^2)$ where $t \in \mathbb{R}$ is not periodic. Suppose this function has a period $B > 0$, i.e. suppose that $\cos (t^2) = \cos((t+B)^2)$ for every $t$. Set $t=0$, then it follows that $B^2 = 2k\pi$ for some $k \in \mathbb{N}$ But I am unable to continue fu...

there is a very lengthy answer too there. I don't see why it had to be so lengthy.
@Koro I don't know if my reasonig is correct, but if $f\neq 0$ and $f$ is periodic, then $\int_0^{+\infty} f(x) dx$ diverges. But the integral of $\cos (x^2)$ converges so it's not periodic
If $U\otimes U$ is a unitary matrix, where $U$ is a complex square matrix, does this imply that $U$ is also unitary?
2:06 PM
$\cos(t^2)$ has only a discrete set of zeros. if it were periodic with period $P$, then every interval of length $P$ would only contain $n$ zeros for some fixed number $n$. but the number of zeros of $\cos(t^2)$ in an interval of length $P$ grows to infinity as the interval travels towards either infinity.
2:32 PM
@Koro My point is that even though you did bad, your peers are probably still worse, so the grades will be better than you expect
@SineoftheTime Not quite. The integrals $\int_0^t f$ for $0\leq t \leq T$ where $T$ is the period has to be zero, which impllies that $f = 0$ a.e. but this equality doesn't have to be everywhere
If $f$ is not $0$ a.e. then you're right that this implies the integral diverges
Of course this doesn't matter if $f$ is continuous
What I'd write is that, if I was trying to make this kind of argument, if $f(t) = \cos(t^2)$ was periodic, then since $f$ is continuous and non-zero, $\int_0^\infty f$ is divergent, but this is well-known to be convergent
thanks for the clarification
@Jakobian do you like Fourier Transform?
2:54 PM
Let $f$ be Riemann-integrable function at the interval $[0,1]$. Prove/disprove: For every point $c \in [0,1)$, $\lim_{h \to 0^{+}} \int_{0}^{c}(f(x+h)-f(x)) dx = 0$
Intuitively this seems to be true, but I struggle to figure how to prove it which again makes me wonder if it is false
Simplest case is when $f$ is continious it is true
Also when $f$ is integrable and has an anti-derivative I think it is satisfied
3:17 PM
@X4J one way to prove it continuity of indefinite integral
Another way to prove it is using Lebesgue dominated convergence + f has set of discontinuities of measure zero
The second approach doesn't work with e.g. Lebesgue integrable functions
@Koro These days, MSE is run by only down-votes and closed-questions. MSE is becoming toxic day by day. sigh...
3:43 PM
@LuckyChouhan One could just as easily say that Math SE is being overrun by low-quality homework questions and answerers who are enabling that style of question. Gosh, Math SE sure is becoming toxic...
(Perhaps it would be good to lay off the hyperbole, and try to actually understand the dynamics that exist on the site...)
complaining is uninteresting
It snowed so much, I love snow but its so cold
@XanderHenderson yeah, would you like to tell us what is dynamics?
It was an obvious question. Now Jack please don't say that "you're becoming a rude." He finds everything rude :((
@LuckyChouhan I feel like I just did explain some of the dynamics, by pointing out a differing point of view.
Downvoting is part of the process of sorting low quality content out.
@XanderHenderson According to you what is the purpose of this site?
@LuckyChouhan The goal of the SE network is to create a repository of high quality questions and answers. That repository is meant to be searchable, and to provide value to a broad range of people who might have a question (rather than just the person who posts a question).
3:51 PM
@XanderHenderson Yeah this sounds plausible. Sometimes MSE and AoPS help me a lot when I use ApproachZero :)
Sometimes I use math.se to search for arguments I'm too lazy to come up with on my own
> Passively searching and reading highly ranked Stack Overflow answers as they appear in web search results is arguably the primary goal of Stack Overflow. If Stack Overflow is working like it's supposed to, 98% of programmers should get all the answers they need from reading search result pages and wouldn't need to ask or answer a single question in their entire careers. This is a good thing! Great, even!
Is the transpose of an elementary row operation matrix also an elementary row operation matrix? This answer here says yes, but I don't think so/ would like to see a more involved proof math.stackexchange.com/questions/2520599/…
I don't think so
But that isn't what the question is asking.
It is asking if the transpose of an elementary matrix is an elementary matrix.
@XanderHenderson It was such a good article. Thank you for sharing :-)
3:58 PM
An elementary matrix is not a matrix which corresponds to an elementary row operation.
@XanderHenderson well technically that is true in the broader sense, if you include both elementary column and row operations on the identity
@XanderHenderson doesn't it still scale a row?
@nickbros123 That is the definition with which I am familiar.
The elementary matrix for scaling a row is a diagonal matrix, and the only difference is which side we multiple the other matrix with
And elementary matrix differs from the identity by a single elementary row or column operation.
4:01 PM
@XanderHenderson well we do the extensive and exhaustive proofs regarding invertibility and it's links with being product of elementary matrices in the context of rows, and I suppose this can be extended to the column domain without much fuss. What I set out to prove was that the transpose of an elementary row operation matrix was an invertible matrix
@nickbros123 Oh, well, that's true.
The transpose of an invertible matrix is invertible; elementary row operation matrices are invertible. N'est-ce pas?
Wait I guess we can show this rather easily.....
@XanderHenderson yeahhh....
@Jakobian Oh, derp. I was thinking of scaling a row and then adding it to a different row.
I haven't come to the rigorous treatment of transposes to be fair, (I'm doing Hoffman kunze), I wanted to think about a few theorems pertaining to both row and column operations, as I was trying my hand at proving the Hilbert matrix is invertible myself
Although, now that I think of that, that might also be left-multiplication vs right-multiplication.
It has been such a long time since I've thought about these.
(Like... uh... 20 years?)
@nickbros123 Transposing a matrix swaps the column space and the row space, no? Think about rank-nullity...
4:07 PM
@XanderHenderson I haven't come that far 😅
Well, what is the notion of invertibility that you have?
Row reductible (or column reductible) to identity, product of elementary row and or column matrices, Ax=y has solution for all y, Ax=0 has only trivial solution...
@nickbros123 Well, if a matrix is row-reducible to the identity, then it's transpose is column-reducible to the identity, no?
The notion of transpose I have is simply a mental picture of flipping the matrix
@XanderHenderson yes, but I don't really wanna talk about transposing the matrix A itself, but rather the transpose of elementary row operation matrices. For context, I noticed transposing elementary row operation matrices gave elementary column operation matrices which when post multiplied constituted the same operation on the column. Then I set out to show that elementary column op matrices are invertible.. that's it
In fact now I have a feeling I can exactly "construct" a column operation matrix with a bunch of row operations on identity...
Good book suggestions to learn about functions and the function series?
4:16 PM
@nickbros123 That seems like overkill. If $E$ is an elementary row operation matrix, then $AE$ is the row-operation applied to $A$, yes? What is $EA$?
@XanderHenderson $EA$ would be the row operation on A, and $AE$ would be the corresponding column operation on A
@Jakobian How do we prove it using the continiousity of the definite integral?
We only know $f$ is integrable
@nickbros123 No, $AE$ is the row operation applied to $A$.
You can express the elementary matrices in the following form $1+be_{ij}$, $1+(u-1)e_{ii}$, $1-e_{ii}-e_{jj}+e_{ij}+e_{ji}$, since all of them are invertible and their transposes are also elementary matrices which means invertible. Isn't it?
@oscarmetalbreak Sure, but the transposes don't have anything to do with the transpose of $A$.
4:21 PM
And, unless I am misunderstanding what @nickbros123 is trying to do, they are attempted to prove something about the transpose of $A$ in terms of the transposes of the elementary row operations.
I thought you were talking about elementary matrix
But maybe I am completely misunderstanding what they are asking about.
oh I see
@XanderHenderson wait, the $E$ you're talking about here is a row operation done on the identity right?
4:23 PM
@nickbros123 $E$ is the matrix such that $AE$ is the row operation applied to $A$.
@XanderHenderson no, to be honest my end goal has already been proved... It was to show that an elementary column operation matrix was invertible..
So it is the same as if the elementary matrix is invertible or not?
@nickbros123 Okay. But an elementary column operation matrix is an elementary row operation matrix. They are the same matrices.
It is just a matter of left-multiplication vs right-multiplication, no?
The column operation or row operation just depends on the left or right multiplication
@XanderHenderson well I am confused now because this is not how I learnt elementary matrices. Hoffman and kunze define an elementary matrix as a matrix obtained by performing exactly one row operation on identity. And a lemma that follows is that, pre multiplication of an elementary row matrix with $A$ constitutes a row operation on A..
4:30 PM
I'll admit that I might have it backward (again, it's been 20 years since I've thought about this), but does it matter?
The difference is between left- and right-multiplication.
I need some time to think about this...
Ugh... it's cold in my office.
I think they turn the heat off over the weekend.
@X4J write the first terms as $\int_h^{h+c} f(x) dx$ by substitution.
@XanderHenderson what temperature is too cold for you?
for me temperature below 22 deg C is a problem.
I prefer to keep my home at around 62°F in the winter.
But I like my office to be a little warmer---66°F is good.
that's about 16 C-too cold for me.
4:36 PM
@XanderHenderson here are some of what i understand: row operation on $I$ is an elementary row matrix, and column operation on $I$ is an elementary column matrix. Row operation on A is just an elementary row matrix left multiplied on A, column operation on A is an elementary column matrix right multiplied with A. The relation between an elementary row and its corresponding column operation matrices are that they're transposes of each other
@nickbros123 Swapping two rows is symmetric. Scaling a row is diagonal. The transpose does nothing to these.
Adding a row to another row has one entry off of the diagonal. The transpose of this switches which row is added to another.
The transpose of a row-operation is a row-operation.
@XanderHenderson yeah, i noticed this, but the situation can technically be salvaged I think cos it can also be written as row operation still
It needn't be the same row-operation, but it is still a row operation.
Yeah 👍
@X4J indefinite integral
4:40 PM
So I guess invertibility can be shown as such too
That it's also a row operation at the end of the day
Is dating hard nowadays? I just read an article.
basically what Koro said, $\int_h^{h+c} f - \int_0^c f$, now rewrite this as $\int_c^{c+h} f - \int_0^h f$
use continuity of the indefinite integral $F(x) = \int_0^x f$ at the points $x = 0$ and $x = c$
@nickbros123 If you are not sure you can always check the exact form of the elementary matrix though.
this approach works for gauge integrals as well
4:43 PM
@oscarmetalbreak Yeah my issues been sorted more or less, thanks
@LuckyChouhan What kind of dating? If you have access to a spectrometer, radiocarbon dating isn't so bad. And dendrochron is pretty straightforward (though tedious).
@Koro Thanks
@Jakobian don't you find gauge integrals as boring. I mean when we can see things obviously then why we're doing all that stuff. For example, we can intuitively see that f(x)=x is an integrable function, but books do uninteresting stuff with this
@XanderHenderson 🤣🤣, I wasn't talking about Carbon dating I was talking about dating which boy and girl do :))
I only mentioned carbon dating as one form of dating. Dendro is completely different.
What about potassium-argon?
@XanderHenderson yeah. Have you ever dated someone? Actually I got to know that "Even girls having hard time in understanding that why so many body chase them?"
4:48 PM
Or something on the uranium decay chain?
@XanderHenderson We need to think about this. I want to do some experiments with radium and polonium in Poland.
@LuckyChouhan (1) You are into a super hetero-normative framework (why can't a boy date a boy? or a girl and two other boys?).
(2) I am not really sure that it is any of your business as to whether I have dated or not. That being said, I was married for 15 years.
@XanderHenderson that's too complicated of a concept to grasp for Lucky
@Jakobian Seems so.
@XanderHenderson I'm straight so...........
@XanderHenderson nice, so you got married in your 20's.
4:53 PM
@LuckyChouhan Insofar as I care about dating or sex, I am a Kinsey 0. That doesn't mean that I don't acknowledge other kinds of relationships.
@LuckyChouhan Why "nice"? Not that I'm bitter, but that woman has been profoundly bad for my mental health over the last several years, not to mention my economic circumstances. Not that I'm bitter, or anything.
@Jakobian 😅, Actually It makes me think when we talk about 'boy date a boy' that how both males are not attracted toward opposite gender? It is very strange phenomenon, and we can't see just after seeing a person that whether they are hetro,gay, or so and so
@XanderHenderson Oh so sorry, I am acknowledging them from now.
@LuckyChouhan I think that you maybe need to read up on human sexuality a bit more.
@XanderHenderson Oh my god, what she has done to you. Was it love marriage?
@LuckyChouhan It really isn't any of your business.
4:58 PM
@XanderHenderson I agree,
@XanderHenderson what you wish you should know in your 20s?
@LuckyChouhan Never marry a shiksa?
@XanderHenderson I am sorry to say but "Shiksa" means "Education" in Hindi. So here what does your Shiksa mean?
Is your wife Indian?😥
@LuckyChouhan A non-Jewish woman. it is mildly disparaging. You probably should not use the term.
@XanderHenderson Now I see, okay I won't marry. Any other advice?
don't be so inquisitive about other people's affairs is my advice
consider that what you're trying to say might be rude is my second advice
5:08 PM
@Jakobian Okay fine. But my questions aren't inquisitive they are just natural. For example, as Xander sir said "His wife is not good for his mental health." then isn't it obvious to ask 'what she has done to him'??
@Jakobian I will try my best to not ask any personal questions. Thanks Jack😊
@LuckyChouhan Are you familiar with the idea of "reading between the lines"? How about "reading the room"?
@XanderHenderson Yeah, I know about 'reading between the lines' means understanding something without being told. Reading the room I don't know.
You've been told explicitly not to bother people with personal questions on different occasions, no need to read between the lines.
Now I see.
I am an literal idiot. I won't be asking any personal questions.
@XanderHenderson I just realized. does it mean that "Marry a Jewish."??😭😢
5:20 PM
@LuckyChouhan You should be careful about using adjectives to describe people in English. The correct phrase would be "Marry a Jewish woman."
Which is really more a comment about marrying a person who is not too culturally dissimilar from yourself. It can work, but it adds stress to a relationship.
what book on complex analysis do you recommend?
@SineoftheTime I'll be honest, none of the ones that I have learned out of were anything that I would consider to be all that amazing. (In grad school we used Conway; I TA'd out of... uh... Churchill? and undergraduate complex was taught out of Sarason).
I don't think that any of those books distinguish themselves all that much.
I like other books in Stein and Shakarchi's set, and am tempted to recommend their complex analysis, but I haven't actually worked out of that one, so I can't recommend it out of experience.
@SineoftheTime I recommend (no particular order) 1) Ponnusamy's 2) Freitag and Busam
I have also heard that Lang's book is quite good---again, I don't know the book from personal experience, but I like other stuff that he's done.
3) Stein and Shakarachi's: here I found that they don't focus much on multivalued functions/branches etc..
5:29 PM
@XanderHenderson they say that about his abstract algebra book too, but its terse as heck
and Conway's of course.
@Koro I feel like that is okay for a first course. I'd rather hold off on getting into branches until you are ready to talk about complex manifolds / Riemann surfaces. These don't feel like introductory topics to me. Though others might differ.
The solution is to not learn complex analysis
Q: What is a good complex analysis textbook, barring Ahlfors's?

MBPI'm out of college, and trying to self-learn complex analysis. I'm finding Ahlfors' text difficult. Any recommendations? I'm probably at an intermediate sophistication level for an undergrad. (Bonus points if the text has a section on the Riemann Zeta function and the Prime Number Theorem.)

@AlessandroCodenotti non mi sembra una buona idea
5:30 PM
I've been learning a bit from complex analysis made simple
thank you Xander and Koro
I like this book. But I forgot the contents very fast
I agree with Alessandro on this one, the solution is to not learn complex analysis
@Jakobian That book looks interesting, but it doesn't seem to be very elementary (e.g. it looks like it assumes that the reader already has some background in analysis, topology, differential equations, and (perhaps) complex analysis).
@SineoftheTime Do try Ponnusamy's or Freitag and Busam's.
@XanderHenderson maybe a bit of real analysis, but there's the appendix
I don't remember it requiring much else... but I also come with a different mindset to those kind of things
5:34 PM
Ahlfor's book is a stone-cold classic, by the way. I really love that book, but I never learned out of it. I have only used it as a reference.
@SineoftheTime so when you speak it out loud, do you stress the end? Do you speak it as "non mi sembra una buooona"?
I have no idea if it is a good book for a novice learner.
> This book is suitable for a first-year course in complex analysis. The exposition is aimed directly at the students, with plenty of details included. The prerequisite is a good course in advanced calculus or undergraduate analysis.
^ about Complex made simple
Yes, but that means "first year of graduate school.
He is presuming familiarity with undergraduate real analysis.
@Koro no, you don't stress the end
5:35 PM
(and, again, from the table of contents, knowing some topology and diffy-q would help)
I know basic topology and ode
@SineoftheTime ok. I saw videos of a politician and their interviewers speak like that.
politicians are clowns
@XanderHenderson I think the book is only about 130 pages or so...
5:38 PM
@SineoftheTime I'd try to pick one and see if I can stick to it
I think it's like saying 'May's algebraic topology' is for a beginner.
If you want a physical copy, then personally I'd download the book first anyway and then buy it if I like it
from the index, the books you suggestest seems interesting @jak
or borrow it from university library
@Jakobian the author also happens to be an mse user.
5:40 PM
wait... yeah, its the same person thats on m.se
@Koro Yeah, that's part of what makes it great. :D
that's why the name looks familiar
It looks more familiar if you have watched the show 'Vikings'. 😅
5:41 PM
@Koro the "uo" in "buono" is a single /w/ sounds in the phonetic alphabet
seems like he stopped frequenting m.se since last year
hope he's good
krm also hasn't been seen for about 2 years now.
But he's active on his own website.
I suspect (just a speculation) that a new user is him with a different account and name.
Their writing styles seem similar -short and insightful.
the new user appeared when krm stopped visiting mse.
but who knows...
who is krm?
Kavi Rhama Murty - might have butchered his name a little
Kavi Rama Murthy - I was pretty close
5:46 PM
yes :)
he doesn't have LaTeX on his site
I never saw an answer of him that's more than 4 or 5 lines.
there is krm in their nickname
sounds plausible
yeah, and also their answers seem similar style.
5:53 PM
and profile picture
it's like how you can tell if an answer is by Robjohn without even looking at the author's name. I don't know. I think at least I can tell as I learn so much from them.
dude is so powerful in analysis
@Jakobian their tags (in which they have answered) are also almost same.
Ugh... I came into the office today specifically to meet with a student. That student never showed (though she did email me 10 minutes before out meeting to let me know that "something came up").
It is probably good that I grade work anonymously, because this has seriously soured me on that student. X(
I'm going home. It is cold here.
That sucks. Go and warm yourself up at home
6:15 PM
@LuckyChouhan Now, coming back to your question. No I don't find gauge integrals boring. If you were to pay attention to my posts, I find them really interesting. Your comment makes little to no sense to me, and sounds like a comment of someone that doesn't understand what gauge integration is about.
Either that, or you were trying to be rude
@robjohn what do you think of this answer? I came up with the idea but I don't know if it's rigorous
1 hour later…
7:30 PM
I don't know why people insist on slamming complex analysis. It is beautiful mathematics and ties in with so many different fields and topics.
it's one of the most elegant theories
@TedShifrin Yeah, I don't get it either. I really like complex analysis.
7:48 PM
Lots of people deleting of late.
I dislike complex analysis for simple reason of arguments being unintuitive to me
I prefer real analysis
8:14 PM
I dunno why this site is messing up my link just above.
@TedShifrin You closed the link with a square brace, not a round brace. I fixed it for you.
moreover I feel like I'm at a stage where learning complex analysis would take a lot out of me
8:45 PM
I have no intuition for complex analysis, it all seems mysterious to me
Let's see if this works...
\mathcal{L}(f * g)(s)&=\int_{t=0}^{\infty} e^{-s t} \left(\int_{\theta=0}^t f(t-\theta) g(\theta) d \theta\right) d t \tag1 \\
&=\iint_{0 \leq \theta \leq t<\infty} e^{-s(t-\theta)} f(t-\theta) e^{-s \theta} g(\theta) d \theta d t \tag2 \\
& =\iint_{\substack{0 \leq \theta<\infty \\
0 \leq \varphi<\infty}} e^{-s \varphi} f(\varphi) e^{-s \theta} g(\theta) d \theta d \varphi \tag3 \\
&=\mathcal{L}\{f\}(s) \mathcal{L}\{g\}(s) \tag4
if you told me that there's a subject in math that, right now, could be upended by a flaw that hasn't been discovered yet that ruins most of it to the point that all textbooks require a rewriting, i'd bet that it was complex analysis
@psie Ok, so this is how the convolution theorem for the Laplace transform is proved in a handout of mine. I have a question about the equality going from $(2)$ to $(3)$.
It looks like the change of variables is simply $t-\theta=\varphi$ and so $d\varphi=dt$, but what's confusing me is that this change of variables seems to occur on the outer integral. Can you do that? Or are they secretly switching the order of integration back and forth?
@leslie complex function have a lot of nice properties and they seem so perfect
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