Using everything on what we said above, is it impossible to pick out any "building blocks" from the cantor set, unlike how in rationals you can always pick a rational singleton from the set itself, and an interval from the real line?
One thing I am interested in is whether it is possible to zoom to the region where the cantor function increases (because it is said the increase took place at the points in the cantor set) so I can see how exactly that increase took place without any jump. I am not even sure if there's a mathematical expression that describe that
Guys, say we have the following differential equation: $$ y^2\frac{dy}{dx}=\frac{1}{x^3}. $$ By separating variables, we can write the following: $$ y^2dy=\frac{1}{x^3}dx. $$ Now I would like to make it explicit that $y$ is a function of $x$. I'm guessing I could write the following: $$ y(x)^2d(y(x))=\frac{1}{x^3}dx. $$ Now I understand that we can integrate the RHS, choosing limits as we like. But what happens to the LHS? This seems quite odd to me: $$ \int_{x_0}^{x_1}y(x)^2d(y(x))=\int_{x_0}^{x_1}x^{-3}dx.
Recently I was enumerating ordinals and I suspect the epsilon mapping grows slower than a pentation by considering how when I expand $\epsilon_2$ in terms of $\omega$ and hyperoperations, I got a bunch of left associative stuff thus suggesting it actually grew slower than pentation
And then I stumbled upon the recent PBS infinity which talked about the cantor function, and that get me ponder about the cantor set
while all that time, I should be doing my chemistry PhD stuff and not procrastinating in other fields
@Secret I think the cantor function increases in a jumping way where the jumps have measure zero , you should look at the series form of the cantor function
Actually it works, but it does not in that link snipplet because \end{align} is snipped out. This means the full message has to be read directly by clicking its link
There's a series representation of the cantor function? I so far only found iterative algorithms for it
Hamilton, the son of a Cincinnati doctor, defied the math profession’s nerdy stereotype. Brash and irreverent, he rode horses, windsurfed, and had a succession of girlfriends. He treated math as merely one of life’s pleasures. At forty-nine, he was considered a brilliant lecturer, but he had published relatively little beyond a series of seminal articles on the Ricci flow, and he had few graduate students.
I think it's useful to recap exactly what Griffiths is saying there. First, you've got $$\langle x \rangle = \int_{-\infty}^\infty \Psi^*(x)x \Psi(x)\,dx=\int_{-\infty}^\infty x\rho(x,t)\,dx$$ where $\rho(x,t)=|\Psi(x,t)|^2$.
That differentiates to $$\dfrac{d\langle x\rangle}{dt}=\int_{-\infty}^\infty x\frac{\partial \rho}{\partial t} \,dx=\int_{-\infty}^\infty x \frac{\partial J}{\partial x}\,dx$$ where $J(x,t)$ is the so-called probability current
With that in mind, it's useful to note that $$J/\rho = \frac{i\hbar}{2m}\left(\frac{1}{\Psi}\frac{\partial \Psi}{\partial x}-\frac{1}{ \Psi^{\star} }\frac{\partial \Psi^\star}{\partial x}\right)=\frac{i\hbar}{2m}\frac{\partial}{\partial x}\ln \frac{\Psi}{\Psi^\star}.$$
(if you compare the expression Griffiths has in his exercise on J and what he writes in the text re: the derivative of $|\Psi|^2$, you'll find that the two are actually different by a sign due to the order in which he writes the derivatives)
So therefore it's actually $\frac{d\langle x\rangle}{dt}=[-xJ]_{-\infty}^\infty +\int_{-\infty}^\infty J(x,t)\,dx$
and $J/\rho = \frac{\hbar}{2mi}\frac{\partial}{\partial x}\ln\frac{\Psi}{\Psi^*}$
This tells us that if we want to get a weird example, we'll want to have $x\rho\to0$ (in order for $\langle x\rangle$ to exist) but $\partial S/\partial x\to \infty$ in such a way that the boundary term doesn't vanish.
Now, we need more than just $x\rho\to 0$ of course. We really need it to go to zero faster than $1/x$ at $x=\infty$, since it won't be integrable if it goes to zero as $1/x$.
A figure shows the footprints of a walking person with a step length of 0.7 m. How can we calculate how far the person comes about when she walks x minutes. How could we do that? Do we have enough information to do that?
If we look at the integral term after integration by parts, we can write this as $\int_{-\infty}^\infty J(x,t)\,dx=\int_{-\infty}^\infty \rho(x,t)\frac{\partial S}{\partial x}\,dx$
aside from a factor of 1/m that i'm ignoring because to hell with units
So the term of interest is therefore $$\langle \partial S/\partial x\rangle = \langle x^3\rangle=\frac{2}{\pi}\int_{-\infty}^\infty \frac{x^3}{(1+x^2)^2}\,dx$$
And this isn't a convergent improper integral owing to the $1/x$ behavior at infinity
Instead of doing that, why don't you just multiply the whole expression by $y$ and the let $y^2=t$.
