@OldJohn yeah it's the same

then $\int_1^2 z dz$

@PeterTamaroff not mean

then you see that it doesn't matter whether we have $t$ or $\tau$ in our example

@OldJohn you're totally right I get it thanks for the clarification :D

@pourjour OK - glad to be of help

@Charlie He cannot answer.

He's banned.

@PeterTamaroff But I can say hi

@Charlie He says "Hi."

@OldJohn thanks :D

@Charlie And to stop being an annoyance.

@PeterTamaroff :D

@PeterTamaroff ok

@Charlie Just kidding.

I wrote that.

@PeterTamaroff >8(

just another thing how to change this $\int_{-x}^{-2x}$ to $\int_x^{2x}$

we have

$x\le t \le 2x$

when $t=-x$, we have $\tau = x$ and similarly for $2x$

@Hayaku still there?

@OldJohn ok great so it's just about substitution

:D

@Hayaku Let me check my answer, I get $\frac{4-e}{2}\doteq0.640859085770477$

@robjohn yes

@Hayaku is that the answer we're supposed to get?

yes

Cool :-)

So the thing about the probability that the sum of $n$ is less than $x$ is $\frac{x^n}{n!}$ is key...

@pourjour Yep - just an exercise in substitution

@OldJohn thanks you very much you solved a big problem in understanding integration :)

@pourjour laa shukr 3ala waajib :)

The density of the sum of $n-1$ trials is then $\frac{x^{n-2}}{(n-2)!}$

@OldJohn :D

interesting

The probability of trial $n$ then being the end is $x$

i still have no idea how to use that equatioh

so the probability that we end on $n$ is $\int_0^1\frac{x^{n-2}}{(n-2)!}x\,\mathrm{d}x$

and if we compute that we get $\frac1{(n-1)!}-\frac1{n!}$

so we are good there

now the average value that will put us over is then $1-\frac{x}{2}$

So we compute $\int_0^1\frac{x^{n-2}}{(n-2)!}x\left(1-\frac{x}{2}\right)\,\mathrm{d}x$

and compute $\sum_{n=2}^\infty$

@robjohn ROb.

and that gives $\frac{4-e}{2}$

@PeterTamaroff yes?

@robjohn Does this look fine to you?

@PeterTamaroff why would that family be equicontinuous?

@robjohn It is not equicontinuous.

@robjohn Oh.

@PeterTamaroff Oh, I thought you were concluding it were and then deriving a contradiction

Because it is a sequence of continuous functions over a compact $K$ that allegedly converges uniformly.

i think my problem is that I don't understand what you mean by x and n because sometimes you use it as a turn number and other times a sum

like when you say sum of n <= x

@robjohn So, we're good?

@PeterTamaroff okay

@PeterTamaroff i have this professor who doesn't lecture, instead he assigns days and makes students take turns lecturing. meanwhile he sits in the back and "asks questions" (heckles them). he asks them to prove things and students will get flustered, and he'll try to guide them to the answer, but when they just don't know he just ends up shouting "JUST THINK. JUST THINK. JUST THINK." over and over.

@Hayaku no, x is a real number denoting the sum of the $n-1$ trials. $n$ is the number of trials

@Hayaku I am sorry, but I have to leave. I will be back later.

thank you for the explanation, i'll have to look it over to understand

@Hayaku Look over the things I've said.

good

@Hayaku I will page you when I get back

@AlexanderGruber hehehehe =)

@Lord_Farin @PeterTamaroff See cs.elte.hu/blobs/diplomamunkak/mat/2009/hubai_tamas.pdf, in particular section 1.3

@anon how cool!

:)

why do people do what derek holt does here, posting an answer as a comment?

i see this all the time. i don't get it. it leaves the question unanswered.

some people, perhaps depending on their mood, have high enough standards on their own answers, according to their perception, that they feel they only should contribute a comment given what they know or have time to divulge on a particular topic or question

@anon so where does that leave the question? it is unanswered forever. should someone copy the content (if not the exact words) of his comment into a CW answer?

@AlexanderGruber I was thinking of doing that.

the hypothetical user leaves those sorts of questions and concerns to others.

@anon 5hs for that? =)

I said 5 hours because that's how long I'd be busy for :)

@anon the commenter, you mean?

mmhmm

well, what do you think? is that the solution?

@anon Anyhow, I solved the problem.

someone else CWing a comment as an answer? well, asking the expert commenter for an answer first would be nice, and if that fails then sure.

hi

hi userboos

@anon who's userboos?

you

@anon Do you think that deconvolution simply division in frequency domain?

should there be "is" or some other verb in that question? :)

but sure, since fourier transform turns convolution into multiplication, it'd be fine to think of it that way

*is simply

how would you call the functions that you divide?

I don't know a term for it.

@anon Did you upvote me here?

f*g=h and g=h/f

didn't read your answer

@anon Oh, OK...

do you want me to?

@anon I cannot make you read it. Do as it pleases you.

@anon Should I just say function f and g or function f and g in frequency domain or something?

@PeterTamaroff what do you think?

@user8005 About what?

just be clear and describe things in straightforward ways, don't worry about jargon or cleverness

@PeterTamaroff this: math.stackexchange.com/questions/380720/…

@anon Who are you addressing?

userboos

@user8005 I cannot make heads or tails of what you're saying.

yes, it's a good answer.

head or tails?

What does that mean? You don't understand? Which part???

@PeterTamaroff head or tails? What does that mean? You don't understand? Which part???

@user8005 $\hat f$ is $f$'s Fourier transform, yes?

yes

@user8005 Talking about $g$ and $\hat{g}$ as if they are the same function just from different perspectives (time domain vs frequency domain) is useful but not entirely conventional ("notate a function in frequency domain with a hat above the letter"). Indeed, it makes sense to say that $h$ is a function in the frequency domain in which case its fourier transform $\hat{h}$ is a function in the time domain.

Simply describe them as fourier pairs, and say $\hat{g}$ is the fourier transform of $g$ (read out loud as"g hat" if you wish).

Ok. And $$(f* g)(t) =\int_0^t f(t-x)g(x)dx \text{ ? }$$

integrated over all of R in this context, I believe

@PeterTamaroff Try a notation for deconvolution, please?

eh?

Peter is describing convolution, and there is no reason to use notation for deconvolution (whatever that is) for convolution.

The thing is that $\mathcal F(f*g)=\mathcal F(f)\cdot \mathcal F(g)$

I think that is what you're saying.

yes

so that means solving $f*g=h$ for $f$ (i.e. deconvolution) is the same thing as dividing $\hat{f}\cdot\hat{g}=\hat{h}$ by $\hat{g}$.

@anon OK, yes.

But then what you want is to take $\mathcal F^{-1}$.

@anon " $\hat{g}$ is the fourier transform of $g$" <- I can't say function $g$ in frequency domain?

you can say things that are true when they are true, no more and no less

@anon then, is my statement true?

you can't say every function there is is in the frequency domain, no, because not every function is in the frequency domain.

@anon I was talking about: function $\hat{g}$ is function $g$ in frequency domain??

Anyone having problem with this statement?

How about the statement: A spatial domain is a 2D time domain?

If $g$ is in the frequency domain, then $\hat{g}$ is not in the frequency domain. Taking fourier transforms switches time/frequency domain, but a hat over top a letter doesn't tell you which domain you're talking about automatically.

@anon do you mean that there's no rule that says when to use the hat and when not, when talking about Fourier transforms?

the only obstacle to taking a fourier transform is insufficient decay, not which domain you are in.

you can take the fourier transform of a function that exists in the time domain, you can also take the fourier transform of a function that exists in the frequency domain. simply seeing the symbols "$\hat{f}$" in the wild does not automatically tell you which domain $f$ or $\hat{f}$ is considered to exist in.

I can define $\hat{g}$ as a function of $g$ in frequency domain, if I want to do so?

if you mean "I can define the fourier transform $\hat{g}$ of the function $g$, given that $g$ is in the frequency domain," then yes

and if $g$ is in the time domain?

you can in fact take the fourier transform more than once, and you get $\widehat{\widehat{f}}(x)=f(-x)$

@user8005 and what?

given that $g$ is NOT in the frequency domain, but in the time domain

I've already answered the question of what information is relevant to whether one can take a fourier transform of a function. which domain $g$ is defined in is irrelevant to whether one can define $\hat{g}$

in fact, much more generally, you can define fourier transforms of functions whose domains are not even R, but other locally compact hausdorff topological groups. the domain will switch from G to the dual group of G and back.

please note that we multiply $\hat{f}$*$\hat{g}$

for the convolution

by * do you mean pointwise multiplication of functions?

yes

poor choice of notation for standard multiplication in the context of convolution

I would use $\cdot$ (\cdot)

So, you say that $\hat{f}$ is not in frequency domain in this case?

in what case?

the convolution case just above

what is $f$?

multiplication in frequency domain = convolution in time domain or something?

that equation is correct

$\hat{f}$ $\cdot$ $\hat{g}$ | You say that $\hat{f}$ and $\hat{g}$, that we are multiplying are not in frequency domain?

we are multiplying $\hat{f}$ and $\hat{g}$ in whatever domain they are defined on

they could be in frequency domain?

sure. I feel like I've been repeating myself quite a bit now.

I'm sorry, I saw a lot of 'whatever! ' in there...