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
@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)
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
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
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
I'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...
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 $F(t_0) = F_0$?"
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
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'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.'
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
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@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.
@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 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 $f(a) = 0$
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 $\frac{dy(x)}{dx} = f(x,y)$
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.
@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.
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
@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
@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"
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)
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
Suppose 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...
@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
@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
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
@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
"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
@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 $f(a)=0$
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]... '
@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
@EmilioPisanty Well, determining that an integral always gives a unique particular solution relies on integrability implying uniqueness and existence of particular solutions, no?
@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.
The 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...
In mathematics, an abelian integral, named after the Norwegian mathematician Niels Henrik Abel, is an integral in the complex plane of the form
∫
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{\displaystyle \int _{z_{0}}^{z}R(x,w)\,dx,}
where
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@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?
@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.
@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.
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
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...
I 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?
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