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enumaris
enumaris
8:00 PM
@danielunderwood I use my GPUs for my deep learning spell-corrector :D
 
danielunderwood
danielunderwood
Ahh the only NLP I've done was attempting to do sentiment analysis with n-grams and Bayesian something. I don't remember using a gpu on it, but it's been a while
 
enumaris
enumaris
they got deep learning pieces doing named entity recognition and stuff
 
danielunderwood
danielunderwood
I did start doing some MNIST stuff last night because I never tried that when I got started for whatever reason
 
enumaris
enumaris
so there is potential GPU speed ups...but atm the backend I'm using doesn't allow GPU for inference
only for training
MNIST more image recognition tho
 
danielunderwood
danielunderwood
Yeah but what I've heard is that it's kind of like the hello world of ML. I guess the hello world for NLP may be sentiment like I did before?
I remember NLP being kind of disconnected from other ML when I was messing with it, but don't know if that's changed
 
Semiclassical
Semiclassical
8:09 PM
@EmilioPisanty hmm. i was doing some reading on numerical methods for Fourier integrals and was curious if you had some perspective
(basically: most of the split-step Fourier stuff I've seen takes the perspective that one should 'obviously' approximate the continuous Fourier transform by writing it as a Riemann sum and doing the FFT on the sampled points. but the 'Numerical Recipes' bible (this section) specifically warns against this approach. so i'm a bit puzzled)
 
enumaris
enumaris
NLP has had a lot of improvement in terms of end-to-end solutions
so it's definitely become much more entwined with ML
like speech to text used to be all hand engineered features and stuff
nowadays you just put a deep RNN to do it
end to end
directly from an audio .wav file to a text transcription
 
danielunderwood
danielunderwood
I used NLTK when I did it, which I suppose is the old-school way. Looking at the libraries available, I also used spacy but I think it was in the beginning stages then
 
enumaris
enumaris
right
I'm using both
more spacy than NLTK though
 
danielunderwood
danielunderwood
8:25 PM
I should return to doing some of that. I miss programming where I actually have to think lol
 
enumaris
enumaris
lol
 
Cows
Cows
2 hours to demo/presentation lol
5 pc's testing things lolz
muahahahahahaha
still have to drive there too
damn lol
preparing 2 machines (real machines) and putting some vms together too
hopefully i don't crash and sleep during presentation/ or worse collapse lolz
 
danielunderwood
danielunderwood
Speaking of thinking and programming, I recently saw a question something like "Suppose you have two eggs that will break when dropped from a building. What's the worst case performance for finding the floor that they break when dropped from?" I know the question was meant for people who had memorized an algorithms class, but thinking it sounded like quite an optimization problem
I don't know if it's actually boring, but somehow I found it fascinating thinking about how to derive the solution
 
enumaris
enumaris
two eggs...
you only have 2 eggs? lol
 
danielunderwood
danielunderwood
Yeah. Basically the thought is you have to figure it out by dropping them and you're done if they both break
So I guess worse case is actually you decided to go top down and immediately broke both haha
 
enumaris
enumaris
8:32 PM
hmm
something really funky is happening in my code loool
 
danielunderwood
danielunderwood
But really I think the way to do it is start from the bottom. Like you could go up one by one which would be O(N), but you could also skip up 2 at a time and test the one below once one breaks. Or 3 up at a time and so on. I have a hunch that it's log(N) at the end, but not sure. I think you could have a "separation function" that determines how much you skip and optimize that
Yeah code tends to do that
And debugging ML stuff is the worst
 
Cows
Cows
yaasss I can smell the money. . . and access to panda express
hehe
let me just see if this runs :D
 
enumaris
enumaris
woah I'm baffled lol
 
danielunderwood
danielunderwood
If you take a break, there's roughly a 50% chance that the solution magically finds its way into your brain
A sane programmer would use that as a coffee break
 
enumaris
enumaris
oh
it's not a problem
I just couldn't identify the correct points
by eye lol
 
danielunderwood
danielunderwood
8:39 PM
Machine learning, not human learning!
 
enumaris
enumaris
hmmm there appears to be a mismatch of 1 going on somewhere, that's odd
oh no
I can't freaking read...nvm -.-
no bugs in the code
only bugs in my brain lol
time to beam search
 
danielunderwood
danielunderwood
Hey I do that like every day. Especially on SE titles for some reason
 
enumaris
enumaris
upgrading from peasant greedy search to beam greedy search
requires some thinking mmmhm
rubs beard
(unfortunately I have no beard)
 
danielunderwood
danielunderwood
They greatly increase your programming productivity...guess that explains why I'm so lazy during the summer
 
Cows
Cows
@enumaris peasant greedy search to beam greedy search. . . why? lol
i have no idea what does are lol :P
looking at vid on beam search . . . interesting . . . what are you using it for?
 
enumaris
enumaris
8:54 PM
spelling corrector :D
I feel like the deep learning model is doing pretty good...now i have to write some stuff on top of it to get the behavior I want
rather than black box behavior
 
Cows
Cows
lol "i have no idea what those are lol :P"
yeah, lol my brain is dead . . . been begging my pc to build some stuff all day, and it is not cooperating
 
enumaris
enumaris
need music
 
Emilio Pisanty
Emilio Pisanty
@Semiclassical this is news to me
 
Semiclassical
Semiclassical
9:15 PM
see also this question on MSE:
6
Q: How to compute Fourier coefficients using a cubic spline-corrected FFT?

DumpsterDoofusI'm not particularly experienced in numerical analysis, and so I recently had quite a massive shock when I discovered that sampling a smooth function and computing the FFT of the result does not return a correct$^1$ list of it's Fourier coefficients, despite theory saying that it ought to! As an...

numerical-methods fourier-series
I'm working on an answer there b/c (like the OP) I don't entirely follow the details of what they're doing in that section
@EmilioPisanty just to clarify: "don't use FFT naively" is what's new to you? (it was new to me, hence why I was surprised by it)
 
enumaris
enumaris
9:42 PM
progress!
music helps a lot lool
 
Emilio Pisanty
Emilio Pisanty
@Semiclassical correct
@Semiclassical hmmm.
that's one lonely question
 
rob
rob
10:00 PM
@Semiclassical You know the expression "good, fast, cheap: pick any two"? Note that FFT has "fast" in its name.
 
Semiclassical
Semiclassical
@EmilioPisanty Yeah
 
Emilio Pisanty
Emilio Pisanty
@Semiclassical I don't have enough rep on mse to pull the kind of shit I do on pse, but have a sweetener
 
Semiclassical
Semiclassical
Lol
 
Emilio Pisanty
Emilio Pisanty
either you get it, or someone else puts up something even more amazing
 
Semiclassical
Semiclassical
Yeah
 
Emilio Pisanty
Emilio Pisanty
10:06 PM
@rob well, the issue here isn't FFT as such but the discretization
frankly, FFT is a pretty notable exception the the "pick two" rule
it's a goddamn miracle is what it is
 
Sir Cumference
Sir Cumference
Sigh this is the physics chat
 
enumaris
enumaris
didn't read it
 
Semiclassical
Semiclassical
What makes me wonder a bit is whether that poster and I have the same application in mind
 
Emilio Pisanty
Emilio Pisanty
lolz
 
Sir Cumference
Sir Cumference
@enumaris Meant to post that in Math overflow
 
enumaris
enumaris
10:08 PM
I see
 
Sir Cumference
Sir Cumference
Well actually I guess someone here might be able to answer
"Consider the ODE $\frac{df}{dt} = F(t)$ and suppose $F(t)$ is integrable over an interval $I$. Then for all $t_0 \in I$, does there necessarily exist a unique particular solution satisfying the initial condition $F(t_0) = F_0$?"
 
Emilio Pisanty
Emilio Pisanty
in MathOverflow, 13 secs ago, by Emilio Pisanty
::tumbleweeds::
 
Semiclassical
Semiclassical
His seems to more about Fourier coefficients, where the range of integration is finite but the signal is supported on the entire range
 
Emilio Pisanty
Emilio Pisanty
the MO chat isn't the busiest, y'know
 
rob
rob
@EmilioPisanty But ... the "fast" Fourier transform only works on discrete data?
 
Emilio Pisanty
Emilio Pisanty
10:09 PM
@rob correct
 
Semiclassical
Semiclassical
Whereas mine is Fourier transforms, where the range is in principal infinite but the support is finite
The big difference being the behavior near the endpoints
 
bolbteppa
bolbteppa
@SirCumference Picard Existence Theorem?
 
Emilio Pisanty
Emilio Pisanty
@rob as mentioned, there appear to be discretization issues, but absent that, if your data is already discrete, the FFT ticks all three boxes
as we say in my home town, Bueno, Bonito y Barato ;-)
 
bolbteppa
bolbteppa
Picard–Lindelöf theorem
In mathematics, in the study of differential equations, the Picard–Lindelöf theorem, Picard's existence theorem or Cauchy–Lipschitz theorem is an important theorem on existence and uniqueness of solutions to first-order equations with given initial conditions. The theorem is named after Émile Picard, Ernst Lindelöf, Rudolf Lipschitz and Augustin-Louis Cauchy. Consider the initial value problem y ′ ( t ) = f ( t , y ( t ) ) , ...
 
Sir Cumference
Sir Cumference
@bolbteppa Yeah but I first would need Peano's existence theorem to show existence
 
rob
rob
10:10 PM
@EmilioPisanty Yes, I think I agree with you there.
 
Emilio Pisanty
Emilio Pisanty
just substitute Fast for Pretty
 
danielunderwood
danielunderwood
@EmilioPisanty Wouldn't chat.stackexchange.com/rooms/36/mathematics be better? MO is for math researchers, isn't it?
 
Semiclassical
Semiclassical
And it might be the case that the FFT issue being worried about isn’t pertinent when the signal vanishes at the endpoints
 
Sir Cumference
Sir Cumference
@bolbteppa Note I said "integrable", not "continuous"
 
bolbteppa
bolbteppa
'The Picard–Lindelöf theorem shows that the solution exists and that it is unique. The Peano existence theorem shows only existence, not uniqueness, but it assumes only that  f  is continuous in y, instead of Lipschitz continuous.'
 
Emilio Pisanty
Emilio Pisanty
10:11 PM
@Semiclassical that was my gut reaction tbh
 
Semiclassical
Semiclassical
And hence why it’s not emphasized for wave-packet scattering
 
Sir Cumference
Sir Cumference
@bolbteppa Eh, I always viewed Picard's theorem as requiring Peano's anyway
 
bolbteppa
bolbteppa
Carathéodory's existence theorem
In mathematics, Carathéodory's existence theorem says that an ordinary differential equation has a solution under relatively mild conditions. It is a generalization of Peano's existence theorem. Peano's theorem requires that the right-hand side of the differential equation be continuous, while Carathéodory's theorem shows existence of solutions (in a more general sense) for some discontinuous equations. The theorem is named after Constantin Carathéodory. == Introduction == Consider the differential equation y ′ ( t )...
 
rob
rob
@Semiclassical The endpoints are a big part of the problem: iirc, the FFT algorithm assumes that the signal wraps around from the end to the beginning.
 
Semiclassical
Semiclassical
It’s definitely an issue with some subtlety, at least
 
bolbteppa
bolbteppa
10:12 PM
Maybe Caratheodory
 
Emilio Pisanty
Emilio Pisanty
@Semiclassical "edge effects" < "compact support"
 
Sir Cumference
Sir Cumference
@bolbteppa I don't have the background in analysis for that, but does it hold for all integrable right hand sides?
 
rob
rob
So for the math.se question you linked, where the asker took an FFT of $x^2$, there's a bunch of high-frequency garbage associated with the endpoints.
 
bolbteppa
bolbteppa
Maybe in a distribution book there is a theorem
idk I only really know/appreciate Picard's version
 
Semiclassical
Semiclassical
Good point
Whereas for a narrow Gaussian one would probably do better
 
Sir Cumference
Sir Cumference
10:14 PM
@bolbteppa My goal is to determine whether $\int_{a}^{t} F(s) ds$ is always a particular solution to such an ODE for a given constant $a$, such that it satisfies $f(a)=0$
 
bolbteppa
bolbteppa
That looks like the fundamental theorem of calculus?
 
Semiclassical
Semiclassical
Still an interesting issue, but probably not the source of any fundamental issues for wave-packet scattering
 
enumaris
enumaris
ayyyy bro it's not really a diff-eq if you have two different $f$'s....it's just says $F$ is the derivative of $f$
a diff-eq is more like $\dot{f}=f$
 
Sir Cumference
Sir Cumference
@bolbteppa Well sort of. I can certainly integrate, but I think I can only show it satisfies $f(a) = 0$ if integrability implies there exists a unique solution to $f(a) = 0$
 
enumaris
enumaris
There you have to solve for $f$. If you only say $\dot{f}=F$ then that's just defining a new function $F$ which you denote as the derivative of $f$
 
bolbteppa
bolbteppa
10:17 PM
You wrote $\frac{dy}{dx} = F(x)$, surely this means $\frac{dy(x)}{dx} = F(x)$, this is just calculus, differential equations are of the form $\frac{dy(x)}{dx} = f(x,y)$
 
Sir Cumference
Sir Cumference
@enumaris I mean it is an equation involving a derivative. And in that case the antiderivative is just the general solution
 
bolbteppa
bolbteppa
Maybe you meant $\frac{dy}{dt} = f(x(t),y(t))$?
 
Sir Cumference
Sir Cumference
@bolbteppa Well, no, autonomous ODEs for example are of the form $dy/dx = f(y)$
 
enumaris
enumaris
one does not generally think of statements like $v=\dot{x}$ as differential equations...only definitions
 
Sir Cumference
Sir Cumference
I don't think there's any specific criteria, so long as it's an equation involving a derivative
 
Emilio Pisanty
Emilio Pisanty
10:18 PM
@rob endpoints are always a nightmare in fourier analysis
 
Semiclassical
Semiclassical
The only exception is when your signal has the good taste of being analytic on the real line
 
bolbteppa
bolbteppa
@SirCumference you wrote $\frac{dy}{dt} = F(t)$ not $\frac{dy}{dt} = F(y)$
 
enumaris
enumaris
In that case, if given $x$ you can "solve" for $v$ by simply taking the derivative. Given $v$ and an initial condition $x_0$ you can "solve" for $x$ by taking the integral.
That's just the fundamental theorem of calculus
 
Sir Cumference
Sir Cumference
@bolbteppa Yes, but I'm showing that the RHS doesn't need to be of the form $f(x,y)$
 
Emilio Pisanty
Emilio Pisanty
@rob @Semiclassical fun fact, when I was writing this code for calculating high-order harmonics, I was unable to get the code to work with a monochromatic driver for something like two years.
 
Sir Cumference
Sir Cumference
10:20 PM
@enumaris That does not solve. You need an initial condition to get the original function $x$
 
Emilio Pisanty
Emilio Pisanty
turns out that it was almost certainly calculating the response correctly, but basically it sampled from $[0,T]$ instead of $[0,T)$, i.e. the endpoints were effectively repeated, which then ruined the FFT
 
enumaris
enumaris
sorry I typo'd
the initial condition is $x(t_0)=x_0$
 
bolbteppa
bolbteppa
@SirCumference the difference between a calculus problem $\frac{df}{dt} = F(t)$ and a differential equations problem $\frac{df}{dt} = F(t,f) = F(f)$ is the r.h.s. depends on $f$ as well as $t$, you wrote $\frac{df}{dt} = F(t)$ which is calculus, fundamental theorem of calculus realm, not ode realm
 
Sir Cumference
Sir Cumference
@bolbteppa I mean that's a matter of philosophy. IMO the ODE realm deals with "what functions satisfy this statement about their derivatives", which includes the "calculus problem"
 
enumaris
enumaris
well nobody else uses that terminology
so you might run into a lot of confusion
lol
 
Sir Cumference
Sir Cumference
10:23 PM
@enumaris I mean, that just requires an antiderivative, not an integral
 
enumaris
enumaris
they the same bro
 
Emilio Pisanty
Emilio Pisanty
@enumaris no, I think it's fine
 
Sir Cumference
Sir Cumference
@enumaris Ah come on, I don't think there's any specific definition
 
enumaris
enumaris
that's the fundamental theorem of calculus - relating anti-derivatives with integrals
 
Sir Cumference
Sir Cumference
@enumaris Ugggh....
 
bolbteppa
bolbteppa
10:23 PM
It's not philosophy, the behavior of functions $F(x,y,y') = 0$ (ode's) are insanely different to the behavior of functions $F(x,y') = 0$ (calculus problems)
 
Emilio Pisanty
Emilio Pisanty
it's a 'trivial' ODE in some senses, but it's an ODE nonetheless
 
Sir Cumference
Sir Cumference
Antiderivatives and integrals are not the same thing whatsoever
 
Emilio Pisanty
Emilio Pisanty
@enumaris and that's just some pretty hard-headed stubbornness over mere semantic tbh
 
enumaris
enumaris
o.O
 
Sir Cumference
Sir Cumference
The fundamental theorem does not state they are the same. It states that the sum of all the changes in a function equals the net change, i.e. the difference in antiderivatives
 
enumaris
enumaris
10:25 PM
well I could be wrong, but I've never met anyone who would refer to a simple equation like $v=\cdot{x}$ as a differential equation.
 
bolbteppa
bolbteppa
@SirCumference do you not appreciate the radical difference between functions $F(x,y') = 0$ and functions $F(x,y,y') = 0$?
 
enumaris
enumaris
If you insist on calling it an ODE, that's fine
I just think you're going to confuse a lot of people
sigh....
 
Emilio Pisanty
Emilio Pisanty
see e.g. the question below for that viewpoint in action
4
Q: How can I numerically pre-compute an indefinite integral with a parameter?

Emilio PisantySuppose I have a function $f(t)$, and I want to compute its indefinite integral $$F(t)=\int_0^tf(\tau)\mathrm d\tau.$$ Moreover, suppose that, for any of a number of reasons, I require this integral to be done numerically, I need it to be done in a single swoop for all $t$ in a range $[0,T]$ of i...

differential-equations numerical-integration
 
enumaris
enumaris
I mean I'm not here to argue semantics or technicalities.
 
Sir Cumference
Sir Cumference
Sigh, bad connection here
 
Emilio Pisanty
Emilio Pisanty
10:27 PM
@SirCumference if it's not true, and some counter-example exists, then there's a reasonably sizeable chance that it's listed in Counterexamples in Analysis
 
Sir Cumference
Sir Cumference
@bolbteppa I mean, whether I appreciate it or not doesn't really matter. They both try to analyze the derivative, and determine the functions that satisfy that
 
Emilio Pisanty
Emilio Pisanty
it's worth getting anyways
 
Sir Cumference
Sir Cumference
Obtaining the antiderivative (not an integral) is merely getting the general solution of such an ODE
 
bolbteppa
bolbteppa
The general form of a single-variable first order ordinary differential equation with a single dependent variable is that it is an equation of the form $F(x,y,z) = 0$ with $z = \frac{dy}{dx}$, if $F(x,y,y') = 0$ is not of the form $F(x,y,y') = y' - f(x,y) = 0$ then already things become insane, you can't just apply things like Picard's existence theorem, it's a whole separate topic of 'first order, higher degree' equations, with equations like Clairaut as an example etc
You definitely do need to distinguish which case you are discussing, and the case $F(x,y') = 0$ is again insane, if it takes the form $F(x,y') = y' - f(x) = 0$ then you have calculus problems
 
Emilio Pisanty
Emilio Pisanty
@bolbteppa how would a simpler function make the existing formalism not apply?
 
Semiclassical
Semiclassical
10:30 PM
It’s a differential equation in that it relates functuons to derivatives. It’s just a boring one
 
bolbteppa
bolbteppa
So asking for general differential equations theorems for the calculus case is bizarre
 
Sir Cumference
Sir Cumference
@bolbteppa I think it's pointless. The "calculus" problem is merely an ODE that is of the form obtained from differentiation; as such, you can solve it by trying to "reverse" the process of differentiation, i.e. "antidifferentiation"
That's as valid a technique for solving ODEs as any other, and it has a particular use
 
Emilio Pisanty
Emilio Pisanty
wow, OK. This discussion of semantics serves nobody at all
 
bolbteppa
bolbteppa
@EmilioPisanty What do you mean?
 
Emilio Pisanty
Emilio Pisanty
you can call the problem an ODE if you want, or you can refuse to call the problem an ODE if you want. It's irrelevant to the problem itself.
I propose that this discussion re-focus on the actual problem itself instead of what labels do or do not apply to it.
22 mins ago, by Sir Cumference
"Consider the ODE $\frac{df}{dt} = F(t)$ and suppose $F(t)$ is integrable over an interval $I$. Then for all $t_0 \in I$, does there necessarily exist a unique particular solution satisfying the initial condition $F(t_0) = F_0$?"
If $F(t)$ were continuous then $\exists!$ is almost certainly guaranteed
 
Sir Cumference
Sir Cumference
10:32 PM
@EmilioPisanty Yeah, but here I'm only saying it's integrable
 
Cows
Cows
show time. . . . off to client
 
Emilio Pisanty
Emilio Pisanty
if it's merely integrable, then it's not immediately clear to me that it does
 
bolbteppa
bolbteppa
If you ask for a differential equations existence theorem for a calculus problem it will just confuse people
 
Sir Cumference
Sir Cumference
@EmilioPisanty Well, my goal is this
19 mins ago, by Sir Cumference
@bolbteppa My goal is to determine whether $\int_{a}^{t} F(s) ds$ is always a particular solution to such an ODE for a given constant $a$, such that it satisfies $f(a)=0$
 
bolbteppa
bolbteppa
Fundamental theorem of calculus
The fundamental theorem of calculus is a theorem that links the concept of differentiating a function with the concept of integrating a function. The first part of the theorem, sometimes called the first fundamental theorem of calculus, states that one of the antiderivatives (also called indefinite integral), say F, of some function f may be obtained as the integral of f with a variable bound of integration. This implies the existence of antiderivatives for continuous functions. Conversely, the second part of the theorem, sometimes called the second fundamental theorem of calculus, states that...
'We don't need to assume continuity of f on the whole interval. Part I of the theorem then says: if f is any Lebesgue integrable function on [a, b]... '
 
Semiclassical
Semiclassical
10:34 PM
I mean, $\nabla^2 f=0$ is a DE. So is $df/dt=0$
 
Sir Cumference
Sir Cumference
@bolbteppa The fundamental theorem is an easily misinterpreted theorem. It does not solve an ODE in the same way antidifferentiating and using an initial condition would
In fact, I'm not even sure it always gets a particular solution. That's what I'm trying to determine by my question
 
bolbteppa
bolbteppa
@SirCumference did you read the 'generalizations' section of that link and compare it to your question
 
Semiclassical
Semiclassical
And then $df/dt=F(t)$ is just the ingomogeneous form of the last ODE
 
enumaris
enumaris
forget it, I'm not getting into semantics again lol
 
Emilio Pisanty
Emilio Pisanty
@SirCumference that is different, and strictly weaker, than your initial statement
 
Sir Cumference
Sir Cumference
10:36 PM
@Semiclassical Well I mean, I'd say that's a system of PDEs
 
Semiclassical
Semiclassical
What?
 
Sir Cumference
Sir Cumference
$\nabla f = 0$
 
bolbteppa
bolbteppa
Yes technically calculus problems are differential equations if that's the point you're trying to make
 
Sir Cumference
Sir Cumference
Wait, nvm, misread the Laplacian as a gradient
@bolbteppa OK, let's move on to the actual question tho
 
bolbteppa
bolbteppa
But by this logic differential equations problems are simply constrained non-linear equations $f(x,y,z) = 0$ subject to the constraint $z = dy/dx$
 
enumaris
enumaris
10:38 PM
keep going more broad until the language has no meaning anymore :D
 
Sir Cumference
Sir Cumference
@EmilioPisanty Well, determining that an integral always gives a unique particular solution relies on integrability implying uniqueness and existence of particular solutions, no?
 
Emilio Pisanty
Emilio Pisanty
@SirCumference That assertion is false. Let $$F(s) = \begin{cases} 1 & s=0,1,2,\ldots, N \\ 0 & \text{otherwise}.\end{cases}$$ Then $F(s)$ is integrable but $f(t)=\int_0^t F(s) \mathrm d s$ is a differentiable function that does not satisfy $f'(t) = F(t)$.
In other words, Riemann integrability still allows the integrand to have pointwise departures from an otherwise-continuous output, and those do not get caught by the integral.
That counter-example also sinks your initial claim.
 
Sir Cumference
Sir Cumference
@EmilioPisanty Huh, awesome. Thanks
 
Emilio Pisanty
Emilio Pisanty
no worries.
0
A: How to compute Fourier coefficients using a cubic spline-corrected FFT?

robert bristow-johnsonThe FFT is no worse at computing the "Fourier Integral" (what we call in the DSP world the "Fourier Transform") than the Riemann summation is at computing an integral of a "non-pathological" function. Make the FFT larger and larger, sampling the continuous signal with finer and finer sampling pr...

 
bolbteppa
bolbteppa
Interestingly, this kind of thinking motivated Abelian integrals apparently
Abelian integral
In mathematics, an abelian integral, named after the Norwegian mathematician Niels Henrik Abel, is an integral in the complex plane of the form ∫ z 0 z R ( x , w ) d x , {\displaystyle \int _{z_{0}}^{z}R(x,w)\,dx,} where R ( x , w ) {...
 
Semiclassical
Semiclassical
10:45 PM
In other news, I’m typing this from a car and this is what the local weather radar looks like: weather.com/weather/radar/interactive/l/USMN0926:1:US
 
Sir Cumference
Sir Cumference
@EmilioPisanty So just to make sure I'm interpreting this correctly, integrating an integrable derivative does not necessarily give you a function with that derivative, right?
For lack of elegant speaking
 
Emilio Pisanty
Emilio Pisanty
@Semiclassical we're hovering just under 30°C and sunny
I'm struggling with feeling sorry for your weather predicament
@SirCumference No. That's a very different assertion.
If you know that $F(t)$ is a derivative, then you know that there exists (at least one) $f(t)$ such that $F(t)=f'(t)$, which precludes the kind of pointwise jumps in my example.
Neither of your initial claims set any restrictions on $F(t)$ beyond it being integrable.
You need to be extremely careful with how you phrase your claims.
 
Sir Cumference
Sir Cumference
@EmilioPisanty Sigh, right. Didn't think of that
 
Semiclassical
Semiclassical
About 25 C here, but it’s hard to appreciate the difference when it’s raining so much
 
Emilio Pisanty
Emilio Pisanty
but it's close to right. The derivative of the integral of an integrable function need not coincide with the integrand.
 
Sir Cumference
Sir Cumference
10:52 PM
@EmilioPisanty Ah, awesome. That's pretty surprising (assuming you mean the derivative still exists and is unequal with that)
 
Semiclassical
Semiclassical
Main concern is whether the roads are flooded
 
Emilio Pisanty
Emilio Pisanty
well, not particularly, or at least not when you understand the basic mechanism
 
Sir Cumference
Sir Cumference
@EmilioPisanty Basic mechanism?
 
Emilio Pisanty
Emilio Pisanty
@SirCumference that pointwise changes don't affect the integral and are therefore not imprinted onto $f(t)$.
@SirCumference this is where it gets surprising
> A differentiable function whose derivative is positive at a point but which is not monotonic in any neighbourhood of that point
that's the kind of stuff that should knock you, at least at first
 
Sir Cumference
Sir Cumference
@EmilioPisanty Ah, I actually didn't even think of pointwise changes.
 
enumaris
enumaris
10:58 PM
it's a hard knock life
for us
 
Emilio Pisanty
Emilio Pisanty
20 mins ago, by Emilio Pisanty
@SirCumference That assertion is false. Let $$F(s) = \begin{cases} 1 & s=0,1,2,\ldots, N \\ 0 & \text{otherwise}.\end{cases}$$ Then $F(s)$ is integrable but $f(t)=\int_0^t F(s) \mathrm d s$ is a differentiable function that does not satisfy $f'(t) = F(t)$.
 
Sir Cumference
Sir Cumference
@EmilioPisanty Yeah, I looked more closely and realized what you meant
 
Emilio Pisanty
Emilio Pisanty
18 mins ago, by Emilio Pisanty
In other words, Riemann integrability still allows the integrand to have pointwise departures from an otherwise-continuous output, and those do not get caught by the integral.
 
Sir Cumference
Sir Cumference
@EmilioPisanty And I guess I read that too quickly
Welp I deserve a dunce cap at this point
 
bolbteppa
bolbteppa
Comparing to the wiki generalization it seems like if $F(s)$ were continuous at some $s_0$ and $F$ integrable it would force $f'(s_0) = F(s_0)$, but not necessarily otherwise
 
Sir Cumference
Sir Cumference
11:00 PM
@EmilioPisanty Not sure how to interpret "dense"
 
Semiclassical
Semiclassical
That it’s largely a matter of pointwise changes does tell you why it’s not relevant to a lot of typical applications
 
Sir Cumference
Sir Cumference
@EmilioPisanty What
how
 
Emilio Pisanty
Emilio Pisanty
can $f(x)$ be continuous at a point $x_0$ but discontinuous at every point of a neighbourhood of $x_0$?
 
Semiclassical
Semiclassical
For instance, you’d never worry about it for the notion of a particle in classical mech
 
Emilio Pisanty
Emilio Pisanty
@bolbteppa that kind of behaviour might scupper this.
Dense set
In topology and related areas of mathematics, a subset A of a topological space X is called dense (in X) if every point x in X either belongs to A or is a limit point of A, that is the closure of A is constituting the whole set X. Informally, for every point in X, the point is either in A or arbitrarily "close" to a member of A — for instance, every real number either is a rational number or has a rational number arbitrarily close to it (see Diophantine approximation). Formally, a subset A of a topological space X is dense in X if for any point x in X, any neighborhood of x contains at least one...
@SirCumference get the book.
 
Semiclassical
Semiclassical
11:02 PM
(Brownian motion, on the other hand...)
 
Emilio Pisanty
Emilio Pisanty
@SirCumference that one probably isn't very hard to achieve.
$$f(x) = x + x^2 \sin(1/x)$$ should do it
 
Sir Cumference
Sir Cumference
@Semiclassical Heh "applications"
Engineers are obsessed with "applications"
 
Emilio Pisanty
Emilio Pisanty
@SirCumference he's talking about applications to other math
 
Semiclassical
Semiclassical
Yes, and they make bridges stay up :P
 
Emilio Pisanty
Emilio Pisanty
user image
2
 
Sir Cumference
Sir Cumference
11:05 PM
@EmilioPisanty The derivative is positive at zero?
 
Semiclassical
Semiclassical
Lol
 
Emilio Pisanty
Emilio Pisanty
@SirCumference yes
 
Sir Cumference
Sir Cumference
Wait, yes, I'm an idiot
 
Emilio Pisanty
Emilio Pisanty
probably
 
enumaris
enumaris
harsh
 
Emilio Pisanty
Emilio Pisanty
11:06 PM
if not, then up the exponent to $x^4\sin(1/x)$ or something
 
Sir Cumference
Sir Cumference
@EmilioPisanty So basically my claim about the ODE is true for discontinuous $F$ only if it is pointwise discontinuous?
Sigh
 
Emilio Pisanty
Emilio Pisanty
@SirCumference no, that is not what I claimed
 
Sir Cumference
Sir Cumference
Jesus that wasn't at all what you said
I'm out to lunch
 
Emilio Pisanty
Emilio Pisanty
good
@SirCumference Duh, I missed the obvious counter-example, no. 6 on the image above
> a function $f$ such that $g(x) = \int_0^x f(t)dt$ is everywhere differentiable with a derivative different from $f(x)$ on a dense set.
the fact that this is as far as they go makes me suspect that $g'(x)$, thus defined, must agree with $f(x)$ at least some of the time.
 
Sir Cumference
Sir Cumference
@EmilioPisanty Right. But does this apply only when $f(t)$ is pointwise discontinuous?
 
enumaris
enumaris
11:14 PM
not fun beans
 
Emilio Pisanty
Emilio Pisanty
@SirCumference no, why are you mis-reading that in the same way again?
 
Sir Cumference
Sir Cumference
@EmilioPisanty My god...
I need to sleep
 
Emilio Pisanty
Emilio Pisanty
pointwise discontinuities are one way to get this type of behaviour, nobody has even hinted at the possibility that they'd be the only way
get some sleep/food/whatever, and re-approach this with a topped-up energy.
 
Sir Cumference
Sir Cumference
@EmilioPisanty Yep... oye
 
Emilio Pisanty
Emilio Pisanty
and take some time to think carefully about the examples and claims above.
 
Sir Cumference
Sir Cumference
11:16 PM
Evidently I'm not reading them carefully enough
Well I'll get back to them later. Thanks a lot for your time tho
 
Emilio Pisanty
Emilio Pisanty
no worries.
 
Sir Cumference
Sir Cumference
@EmilioPisanty Is that from the book you linked?
 
Emilio Pisanty
Emilio Pisanty
@SirCumference yes. Get it. Read it.
it's printed by Dover for the usual low fare.
on lighter matters
35
Q: Have there been any "Is this a scam?" questions that the answer was "No"?

BlackThornI think most here would agree that whenever someone asks Is this a scam, the answer is almost certainly, Yes, this is a scam. Have there been any instances of questions like these where the answer was actually, No?

discussion
surprisingly, the answer is yes, there have indeed been such questions on the Money SE
 
Sir Cumference
Sir Cumference
I should probably take a real analysis course first tho, to make sure I don't confuse myself with these exceptions before I have an intuition on what usually happens
 
Emilio Pisanty
Emilio Pisanty
@SirCumference there's no harm in getting the book
it may be a good idea to take a solid course first
but please don't think of the entries in that book as "exceptions"
there's no notion of "usually" in analysis
there's either "always" or "not always" and that's it
the counter-examples in that book are, generally, incredibly useful at highlighting the structural patterns in the theory
 
Sir Cumference
Sir Cumference
11:27 PM
@EmilioPisanty I mean yeah, but an intuition generally lessens the chance of misinterpreting them, imo
That and careful reading, which I seem to lack
 
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