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12:02 AM
Is it conventional to define $\mathrm{gcd}(0,0) = 0$?
 
yes
 
@DannyuNDos in general $\text{gcd}$ is about the non-zero elements under equivalence relation $x\sim y$ iff $x, y$ differ by a unit, and the induced partial order of division
 
And $\mathrm{lcm}(m,0) = \mathrm{lcm}(0,n) = 0$, right?
 
if we allowed $0$, then it will be a top element in that partial order, so indeed $\text{gcd}(0, 0) = 0$
@DannyuNDos yes
this will be the "join" of $0$ and $n$ (or $m$) in this partial order, so naturally will be equal to the greater element, which is $0$
$\text{gcd}(x, y) = x \wedge y$ and $\text{lcm}(x, y) = x\vee y$
under the order $x\leq y \iff x|y$ of the equivalence classes
(I've assumed this is an integral domain so $x|y$ and $y|x$ iff $x \sim y$)
 
Thanks!
 
12:57 AM
I'm reading about the inversion theorem for the Fourier transform. The theorem reads:
> Theorem. Suppose that $f\in L^1(\mathbb R)$, that $f$ is continuous except for a finite number of finite jumps in any finite interval, and that $f(t)=\frac12 (f(t+)+f(t-))$ for all $t$. Then $$ f(t_0)=\lim_{A\to\infty}\frac1{2\pi}\int_{-A}^A\hat{f}(\omega)e^{i\omega t_0} \ d\omega \tag1$$ for every $t_0$ where $f$ has (generalized) left and right derivatives.
Following this proof, there is a remark which I don't quite understand.
> Remark. If $\int_{\mathbb R} |\hat{f}(\omega) |\ d\omega$ is convergent, i.e. $\hat{f}\in L^1(\mathbb R)$, then $(1)$ can be written as the absolutely convergent integral $$f(t_0)=\frac1{2\pi}\int_{-\infty}^\infty \hat{f}(\omega)e^{i\omega t_0} \ d\omega,\tag2$$ but in general one has to make do with the symmetric limit in $(1)$.
I don't see what's the difference between (1) and (2)? if someone could explain, I'd be grateful.
 
i dont think that theorem is true as stated
 
@psie If $\int_{-\infty}^\infty$ were an improper Riemann integral, then would you see a difference?
@Thorgott there is one version of this on wikipedia but they assume piecewise smooth
 
1:13 AM
yeah, that's how it's often stated
you can weaken the condition, but not drop it entirely afaik
best I can prove is that you can weaken it to assume that the one-sided difference quotients at each point are integrable near that point
 
In the theorem you assume that for $t_0$, $f$ has generalized left and right derivatives
 
I think I understand now. I was thinking all the time that $\lim_{A\to\infty}\int_{-A}^A =\int_{-\infty}^\infty $, but this is not the case, as e.g. $\sin x$ demonstrates.
 
oh, i overlooked that part, my bad
 
@psie for improper Riemann integrals this would be $\lim_{A, B\to\infty} \int_{-A}^B$
here we are dealing with Lebesgue integrals which are a little different but the idea is similar to that of improper Riemann integration here
 
1:27 AM
@Thorgott so, $\mathbb{A}^n$ can be identified with $\text{MaxSpec } k[t_1, ..., t_n]$ using Zariski's lemma, right?
for $k$ algebraically closed
 
it depends on what you mean by $\mathbb{A}^n$
 
$k^n$, the affine $n$-space
 
then yes
 
and this is called weak Hilbert's Nullstellensatz?
while strong Hilbert's Nullstellensatz is about the whole Galois connection between ideals of $k[t_1, ..., t_n]$ and subsets of $\mathbb{A}^n$
 
hmm no
the weak Hilbert Nullstellensatz is the statement that polynomials generating a proper ideal in $k[t_1,\dotsc,t_n]$ have a common zero
which, of course, isn't far off from the statement you gave
 
1:40 AM
I took my terminology from the Polish wikipedia
in red, weak Nullstellensatz, and in green, strong Nullstellensatz
 
i guess it's fair to call either statement the weak Nullstellensatz, I've just only ever seen it used for the former
but they're easily equivalent
 
2:16 AM
G-d, I wish Arizona had a "none of the above" option like Nevada. :/
 
2:26 AM
more like Dullstellensatz am i right
 
3:10 AM
Maybe Grabemsnatch ?
Always raising the tone.
 
3:44 AM
haha
 
 
2 hours later…
5:28 AM
@XanderHenderson do you have to choose one, or can you just not vote in that race?
 
 
1 hour later…
6:56 AM
Suppose $X,X',Y,Y'$ are objects in a category $C$ and $f:X\to Y$ and $g:X'\to Y'$. Suppose $F:C\to D$ is a functor such that $F(X)=F(X')$ and $F(Y)=F(Y')$. It may happen that $F(f)=F(g)$ but the functor is still faithful right?
 
 
3 hours later…
10:04 AM
0
Q: Sufficient condition for being a topological line

Luca T. CastrillónLet $S\subset \mathbb{R}^n$ a (path-)connected subspace in the usual topology with the property that, for all $x\in S$, $S-\{x\}$ has two connected components. Is $S$ homeomorphic to $R$ with the usual topology? My attempts at (dis)proving this have been unfruitful, with the tools that I now have...

calling all weirdos
 
 
1 hour later…
11:09 AM
@SoumikMukherjee yeah
 
11:32 AM
Thanks for confirming
 
np
 
@robjohn I could choose not to vote. But I REALLY want to vote against Trump. I ended up voting Haley, as she seems to be the only one still running with a chance of beating Trump.
The whole point of changing my party affiliation was to vote against Trump in a vain effort to keep him off the ballot in the general election.
 
12:14 PM
Voting for someone else would add to someone else's vote total, but I am missing how voting for "none of the above" would be different than simply not voting in that race.
 
12:28 PM
Though I do like the idea of voting against Trump in the primaries.
 
 
1 hour later…
1:28 PM
@leslietownes connected, locally connected separable metric space, such that every point is a strong cut point
overdone topic
 
1:48 PM
this part is about Cauchy's theorem and cross cuts, not residues or any further topics
I wonder how should I approach $b$, I thought about possible standard cross-cuts to find an equivalent contour, but I could only come up with something which is not convenient in the slightest.
if someone has any hint, I'd be glad to hear
 
@leslietownes turns out this one was non-trivial
3
A: Homeomorphic characterization of the real line?

D.S. LiphamTheorem. Suppose $X$ is a connected separable metric space such that $X\setminus \{x\}$ has exactly two (non-empty) open path-connected components for every $x\in X$. Then $X\simeq \mathbb R$. Proof. We claim that each point $x\in X$ is contained in an arc $A$ with endpoints $a$ and $b$ such tha...

we don't actually need $X$ to be locally connected
 
2:04 PM
@robjohn The Nevada system is a little weird. In general, if "none of the above" wins, they have to run a new election.
I used that option a lot when I lived there.
Though at higher levels, a "none of the above" vote is perhaps the same as not voting at all.
(In local elections, it can force a new election---in statewide races, I think that the next candidate down is given the office, though with the understanding that they are hated.)
 
@XanderHenderson Did that ever happen actually?
That nota won
 
@SoumikMukherjee Not while I lived there.
But I always voted "none of the above" in judicial races---I don't think that judges should be elected.
 
always go too far for that is where you will find the truth
 
@Jakobian I'm really not. In hindsight, it doesn't take a genius to realize that a message saying "Here's all the answers to the test" with a link attached is probably a prank/bait.
 
2:21 PM
@冥王Hades I managed to beat pvz zombotany with only sunflower and chompers.
And also without loosing a single lawnmower.
 
 
2 hours later…
4:40 PM
@XanderHenderson your guys' judges are elected by public voting!? What in the world..
 
i wonder if that thing rolls
probably not
Sean carrol you dirty dog!
 
5:46 PM
@Sahaj it varies from state to state.
But yes, it is nonsense.
 
6:12 PM
Xander, as a professor have you ever dealt with a prodigy student? Young, extremely talented for their age, etc? Do these things actually happen or are they for the movies?
 
@Sahaj It's not about extreme talent, but about extreme interest/motivation towards the subject.
 
if we decompose what talent is, won't it be the same as talent
but "prodigy" is a bit you know
no offence but that word is a bit stupid to describe anyone
 
@Jakobian I agree.
 
I don't know why it really exists. Its just your own expectations that make someone a prodigy
I suppose that might be the point? To make someone famous. For people to expect something from them
say, "boxing prodigy"
sports might have been a huge influence here
 
6:27 PM
@Jakobian I think most people considers talented people to possess some supernatural powers, totally ignoring their individual efforts.
 
at least in media
 
This reminds me the case of Soborno Isaac Bari, a media hyped 'prodigy".
 
6:43 PM
@SoumikMukherjee he inspired my question. I just saw him on a post saying "XYZ celebrates Dr. Bari's achievement in US Olympiad". No details on what Olympiad.
 
@Sahaj (1) I am not a professor. While I hold a PhD, I have a teaching position at a community college. My job title is officially "Faculty of Mathematics".
 
As a faculty of mathematics, have you come across any prodigy?
 
It's quite unlikely..
 
(2) No, I do not believe that I have ever "dealt with" a prodigy. By the time they get to me, they are likely too old to be considered prodigies any longer, and any "true" prodigy is going to end up in a much more esteemed institution than any where I have worked.
I have had some very bright students, but "prodigy" suggests something far beyond that. It is perhaps possible that one of the guys I went to high school with was in that category (he graduated even faster than I did, ended up with some named fellowship in astro-physics at Berkeley, and, last I heard, was at University of Arizona---I keep meaning to try to get together with him when I am in Tucson, but my timing is always off).
 
6:54 PM
I feel bad for him, it's not his fault that his parents are using him for publicity.
 
When I speak, most don't listen. When they listen, they don't understand. Who am I?
 
I also think that it is very hard to judge the nature of a "prodigy" in the moment. A lot of very bright kids burn out in their teens, or never manage to make anything out of their perceived potential as children, or just turn out to be very average people who were somewhat precocious as children.
@Jakobian A woman in America? hey-yo!
 
Because of politics?
 
@Jakobian Jakobian
I only answered the last question.
You are Jakobian
 
I definitely never understand him, so I can vouch.
I'm far too low on the altitudes of mathematical knowledge to understand whatever he talks about
 
7:05 PM
@SoumikMukherjee I expected this answer and I'm happy someone said it :)
I'll wait a little before revealing the intended answer
 
@Jakobian :)
 
@TedShifrin hi Ted. Did you ever see notation $\text{RSpec}(K)$ for the set of orders of a field $K$?
 
I've never seen RSpec, no. And nor do I know what "orders" of a field are.
 
I'm actually wondering that too. The only hint I have is that $\text{RSpec}(K)$ should be closed in $\mathcal{P}(K)\cong \{0, 1\}^K$
oh, they must be talking about total orders
 
This doesn't sound like it has anything to do with commutative algebra/algebraic geometry. Spec is usually the "spectrum" of prime ideals in a ring.
 
7:16 PM
The embedding I wrote is also wrong
$P\in\text{RSpec}(K)$ is a function $P:K^\times\to \{-1, 1\}^{K^\times}$ where $P(x) = 1$ for $x > 0$ and $P(x) = -1$ for $x < 0$
@TedShifrin I am reading notes by an arithmetic geometer
 
I wonder what this has to do with the conventional notion of Spec($R$).
 
we can try to ask Pete Clark
 
1
Q: Sum of two subharmonic functions is subharmonic.

Anacardium Definition $:$ Let $G \subseteq \mathbb C$ be a domain. A function $s : G \longrightarrow \mathbb R \cup \{-\infty\}$ is said to be subharmonic if $(1)$ $s$ is upper semicontinuous. $(2)$ given $D \subset \subset G$ and a continuous function $h : \overline D \longrightarrow \mathbb R$ with $h \r...

Any idea on how should I proceed?
 
@Jakobian what is it?
 
Pete is my former colleague and a friend. But I'm sure he must explain it if he's the one writing it.
 
7:26 PM
There's no explanation
 
I have just finished Chapter 9 of Naive Set Theory wherein he defines the Cartesian product of some family of sets (a family being a function from an index set $I$ to some collection of sets $\{A_i\}$) as the set of all families (this time, functions from $I$ to $\cup_{i \in I} A_i$) with the restriction (axiom of specification) that each such family (call them $x$) have their values in a given set, $x_i \in X_i$
i am vaguely aware that this construction leads to AC, but want to be sure I understand. We are allowed by the axioms I've seen so far (everything except choice, regularity, and infinity) to define this set. It's just that AC guarantees it's not empty?
 
@EE18 Set of functions $f:I\to \bigcup_{i\in I} A_i$ with $f(i)\in A_i$
 
oh yes
sorry, i changed up my notation halfway through
I should have written $x_i \in A_i$
Yup, I agree with that
 
@EE18 this doesn't lead to AC
yes its just that for arbitrary sets, for products to be non-empty you need AC
but you can still talk about those without AC
 
"lead to" may be too strong, but "is related to" perhaps
Is this correct then: We are allowed by the axioms I've seen so far (everything except choice, regularity, and infinity) to define this set (the Cartesian product of an arbitrary family). It's just that AC guarantees it's not empty?
 
7:37 PM
AC is equivalent to "arbitrary products are non-empty"
with a proper interpretation for the latter
people that study ZF and so on, for them its not much of a restriction anyway
the idea is to study concrete objects like $\mathbb{R}$
 
Totally fair, but I just want to confirm if my understanding/capitulation of the state of things above is correct. i.e. i'm not wondering about the statement of AC but rather about what the "current state of affairs is" in the book at this point, where this definition has been made but AC has as yet not been given
 
well, were they correct?
I'm not a dictator for thought but merely a guide
 
that's very fair, but what was not clear about the question (so that i can ask a clearer one)?
 
did I say that
or otherwise for that matter
 
i am very confused about what you're asking then
 
7:53 PM
if you were correct
isn't it?
 
you're asking if i am correct?
i'm asking the same question :)
 
I'm asking if you were correct
 
yes but that is my question to you. i am asking you if i am correct
 
@EE18 then why won't you answer your question?
 
Around and around the mulberry bush we go ...
 
7:55 PM
Pop goes the weasel!
 
well i think i am, but that's why i come to the experts to ask if indeed i am. surely you've seen many a student be certain they're correct when, in fact, they're not. so in order to be sure that i'm not falling prey to this, i am asking here?
 
@EE18 I am not going to give you my judgement
I'd rather you form your own judgement based on what I've already said
 
British spelling :P
 
after all, you are the one asking the question, and you know yourself the best
so you'll know if your question was answered and not me
 
i'm not sure that's how it works to be honest though Jakobian...like i said, many a student thinks they "know" the question is answered when they are in fact wrong, so surely this method isn't foolproof?
 
7:59 PM
@EE18 There's no visible to the eye misconceptions in what you are saying to me
 
That is encouraging :)
 
but I'm not saying you don't have them either
 
Oh I doubtless do
 
@EE18 I'll get back to you after I acquire telepathy
or "mind reading"
 
I already have it. Cosmic telepathy.
 
8:02 PM
i can't tell if i am being pranked but i will leave it to others to let me know if my original question above was ill-formed or required mind reading :)
 
Do you think there is a foolproof method of education?
 
certainly not
 
well then I'm not sure what you're saying, you seem to be contradicting yourself
 
put it this way (and then i am going to get back to reading): you have very graciously answered many of my questions before and never before mentioned telepathy was required, so i am unsure about what is different about the question i asked this time
 
You've said that "surely this isn't foolproof"
so I've said, it will be once I acquire telepathy or a similar ability
 
8:08 PM
@Jakobian the "this" referred to this
 
yes, it isn't foolproof but there is nothing that can be done
 
i.e. to the method of knowing
 
I'm not asking you to verify all the details yourself
I've already gave you answers
you have opportunity to answer if you were correct
since you're asking me this question, to verify if you were correct, maybe you ignored them
 
Goo' day. I'm working the following exercise, which I'm stuck on.
> Assume that $f$ is differentiable and has the Fourier transform $$\hat{f}\left(\omega \right)=\frac{1+i\omega }{1+\omega ^6}.$$ Compute $f'(0)$.
There is a hint that says I do not even have to find a formula for $f(t)$. I've been looking at the time derivative property, but this determines a formula for the FT of $f'(t)$, which is not what I want I assume. Any ideas on how to proceed?
 
@psie do you know the formula for $\widehat{f'}$?
 
8:19 PM
indeed, $\widehat{f'}(\omega)=i\omega\hat{f}(\omega)$, right?
 
something like that.
Now, Fourier inversion formula
 
good idea
 
8:34 PM
@Jakobian I get $f'(0)=\lim_{A\to\infty}\frac1{2\pi}\int_{-A}^A i\omega\hat{f}(\omega) \ d\omega$. I guess integration by parts then...
 
@psie no, substitution
treat real and imaginary part separately
 
ok, will try
 
Hell would freeze over before I get MathJax to function properly on my computer
 
@TedShifrin I found people often say that the set of orders on a group is called a spectrum
13
A: The set of orders of elements in a group

Stefan KohlObviously not for every set $A \subset \mathbb{N}$ there is a group $G$ with $A$ as set of orders of its elements (usually called 'spectrum') -- for example if $G$ has an element of order $n$, then $G$ also has an element of order $d$ for every divisor $d$ of $n$. For a survey of what is known o...

 
8:49 PM
@冥王Hades is it MathJax or ChatJax that is not behaving on your computer?
 
oh wait, its not those kind of orders
 
@冥王Hades Done.
Current temperature is 30°F, which is freezing. If only barely.
 
@Jakobian Group element orders, silly.
 
I get that $\int_{-A}^A i\omega\hat{f}(\omega) \ d\omega$ is $$\int_{-A}^A \left(\frac{i\omega}{1+\omega^6}-\frac{\omega^2}{1+\omega^6}\right) \ d\omega.$$ In the first integral we can make the substitution $u=\omega^2$ and in the second $u=\omega^3$. Then we get $$i\int_{-A}^A \frac{du}{2(1+u^3)}-\int_{-A}^A \frac{du}{3(1+u^2)}.$$ The second is OK, the first one is...painful.
 
Your substitution for the first one is wrong
Either way, you don't need any substitutions for the first integral
 
9:00 PM
@XanderHenderson looks to be warming up... 32° now
 
your substitution for second one is wrong too actually
 
Ah, Google and Apple have a difference of 2°
 
@robjohn It's those cheap processors. Apples run hot. :P
 
@Jakobian yeah, I forgot about the integration limits. How would one solve the first one without substitution?
 
I'm wrong with my comment, you can technically use substitution for the first integral
But you will obtain integral from $A^2$ to $A^2$
and if you look closely, you are integrating an odd function
 
9:08 PM
aha! neat.
if I change the integration limits to $-A^3$ to $A^3$, then the substitution in the second integral is correct, right?
 
maybe it is we who integrate the function who are odd
 
@leslietownes So true. *snaps*
 
@psie yep
that's just some arctan
 
9:25 PM
@leslietownes Odd ducks need apply.
 
ted, the recent rains have brought the duck pond to us. there are some big puddles that haven't gone away and the ducks come to them. munchkin is in heaven
 
How literally to you?
 
well, a few blocks away. normally we have to go elsewhere to see ducks
 
So no driving now required?
I’m surprised the ducks break old habits …
 
yeah they were right on it too. like the day the puddle showed up
 
9:40 PM
quarry lake in tilden is full again. however, the place where one walked to from the farm is more or less completely grown over now.
one of my surgery bills came to an unexpected \$4.5k. i went to the business office to discuss in downtown oakland and after 10 mins or so the bill was reduced to \$1.8k.
makes you wonder how many folks are getting ripped off.
 
we're still fending off a four-figure bill for the birth of munchkin 2
 
wow. i would have expected insurance to cover most
they did not charge for my daughter because i think they knew it would trigger me into action (they screwed up royally in many, many ways).
you do not want to hear that story, even with the new case of claret i acquired.
 
for some reason, their current position is that his coverage only started a few months after he was born
 
fight it. if you have the energy & time.
sounds like total bs to me.
 
oh of course. they just got a date wrong somewhere in their system
i'm not sure they even know it's a baby, they are treating it like we just started submitting claims for some random person whose SSN we didn't have for a while
 
9:46 PM
unbelievable.
 
it is enough to drive one to strong language
 
you have relevant skills. is your bar still active in calif?
write to them, sign esq.
make a fake llp and go with that.
not fake, just not one you will actually use.
 
i have thought about this
 
do it. you will regret not doing it later.
my energy levels are returning now that i can walk properly, etc. just making a list of things to discuss with insurance & alta bates on monday.
if i had personal injury credentials i would start a business helping folks with medical billing :-)
 
i just need my bookmaking business to pick up a little
 
9:55 PM
decades ago, my wife's elderly aunt was having difficulty with her citizenship because the fbi would not clear her fingerprints. i wrote to fbi director vedder and magically the fingerprints that she had already provided were suddenly acceptable.
can you imagine if turned i maga.
 
the copper hat is already a step in the, uh, "right" direction :)
 
exercising my right to write letters
 
@copper.hat The Bar in Arizona tried to suspend him for nonpayment.
A year after he died.
 
:-). the irony is that my life is an open book now
 
And 9 months after they were informed that he had died. My mother was pissed.
 
9:57 PM
unbelievable.
i can see why folks get pissed with bureacracy
large companies, same thing.
had some eejit come by last night trying to sell at&t newest super fast fibre
 
Indeed.
 
i was very restrained
 
@copper.hat Heh.
 
i think the years of my hip bothering me took some of my "friendly" energy away.
but i'm back :-)
 
I wish that we had better options for internet here. There is only one service provider. And they sell exactly one package in the area. For the past several months (after installing a couple of streaming apps on my brother's television), we have hit the cap for data, and had to pay an extra fee for more data.
 
10:00 PM
unbelieveble really.
 
There is a message which pops up and says "You should get a better internet package."
 
would starlink suffice?
 
Which... like... okay, fine... but you don't offer one.
 
not that i am encouraging musk in any way
 
@copper.hat Likely not, and it would likely be much more expensive, were it available.
Though they do keep talking about bringing Starlink to the Rez for super cheap in the next couple of years.
 
10:02 PM
every read Michael Lewis flash boys?
 
It is kind of a kick in the guts to be told "You're out of internets! You should get a better package. Oh, but we don't offer one! Ha ha! F*ck you!"
@copper.hat Don't think so.
 
sad that we can plough through whereever to save a few pico seconds, but basic service is wanting
my god, i'm turning into a communist
wait, then i can't use "my god"
 
my typing violates some law of probability, i mess up the single most likely character to change meaning on a frequent basis
$800+ after insurance for 4 physical therapy visits. wtf
 
Probably just a series of Freudian slips.
Ouch...
Ugh... I need to make dinner. But I don't want to.
And to make dinner, I have to do a bunch of dishes. But I don't want to do that, either.
 
10:05 PM
we sold my moms house and my sibs just cleared the house today. they found my copy of desmond morris' man watching
speaking of freudian
 
And before I can do the dishes, I need to put away the clean dishes on the rack. And I don't want to do that, either.
Ugh...
 
no trader joes nearby?
 
@copper.hat Let me check...
 
they have some frozen stuff that is ok, some microwaveable Indian meals if you like that sort of thing (i do)
 
So, the good news is that I could get to the nearest Trader Joes before it closes.
But it would take about three hours. And I don't really feel like the drive.
 
10:07 PM
there's your plan :-)
 
Oh, and then another three hours to get home.
 
its warm here today (change from the last few weeks), 70+
forget that! 3 hrs
im going to try taking my bike up a hill to see what happens to my leg.
 
Good luck.
I'm going to start working on dinner. Because I need to. :/
 
thx. not really a concern, i can always stop & walk
my fast dinner is sausages & hot dog buns :-)
our 6 mos car insurance just jumped from \$1.8k (for 4 drivers, 3 cars) to \$2.7k.
 
speaking of outdoor activities, does anyone by any chance know of good sandals that you can take long walks with, even on concrete? :) For days now, I've been looking for some comfortable sandals, since I found weather wise this to be the best shoe (so far), but they are not so kind to my feet
weather wise...except for when it is snowing and freezing
 
10:16 PM
What if it is freezing but not snowing? or snowing but not freezing?
 
well, winter in general I avoid them :) I have yet to find a waterproof shoe that is actually waterproof, but when it comes to sandals, it doesn't matter
 
10:29 PM
my dad came in to home all dirty, with bloody face and slightly drunk, he must have fell
I just hope he doesn't have any internal injuries
 
no fun.
hope he is ok.
 
10:42 PM
me too
I need to wait till tomorrow to see
 
11:25 PM
I have a basic question. If a function has a Fourier transform, does this mean it is in $L^1(\mathbb R)$? I know that if a function is in $L^1(\mathbb R)$, then the integral that defines its Fourier transform is absolutely convergent, but I kind of wonder about the converse. The reason; because of my problem earlier, i.e. I was given a Fourier transform and I applied the inversion theorem, but this requires the function to be in $L^1(\mathbb R)$.
 
@psie sounds context dependent
you can take Fourier transform of any function $f\in L^1(\mathbb{R})+L^2(\mathbb{R})$
this is way more than just $L^1$
I'm sure this is more general too, but I digress. You don't need to be in $L^1$ to take Fourier transform of something
Note that if you take $f\in L^2$ then the Fourier transform $\hat f$ won't in general be the same as $\int f(x)e^{-itx}dx$
 
hmm ok, the problem I was working on was:
3 hours ago, by psie
> Assume that $f$ is differentiable and has the Fourier transform $$\hat{f}\left(\omega \right)=\frac{1+i\omega }{1+\omega ^6}.$$ Compute $f'(0)$.
and we applied the inversion formula
 
the Fourier transforms I'm talking about aren't really defined using integrals
not explicitly anyway
 
I still do not understand why we were justified in using the inversion formula when we didn't know the function was in $L^1(\mathbb R)$
the version of the theorem I'm referring to is this one
 
let me ask you this
did you learn any other Fourier transforms than the one on $L^1$?
 
11:39 PM
no :)
 
then here you have your answer
the sole fact that they're writing $\hat f(\omega)$ means they must be assuming $f$ is in $L^1$
 
ah, ok, I grok it I think, thank you for clarifying
 
i thought you were boycotting wikipedia
 
yeah sorry :) I only refer to it casually...don't really read through the details
 
Signs of a true addict.
 
11:49 PM
i only take this cocaine because i am curious to see if it still smells that way
 
11:59 PM
Once bitten, twice shy.
 

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