i think, actually, that there's kind of a morse theory story going on in the example I quoted
namely, if you pick sufficiently small energy in that case, the point is that the lines of constant $H=E$ in the $(x,p)$ plane are three distinct circles (whose crossings with the $x$-axis define the above roots)
anyways. as you increase the energy, eventually the adjacent circles will 'touch' to form a caustic. and above that you've instead got a single large circle which encloses the initial three
which is equivalent, i should think, to taking a particular genus-2 torus, intersecting with an appropriate plane, and watching how the intersection changes as you drag the plane out. @mike
Do you think a "What are some exceptional numbers?" question would be a good and fun question to ask? There's $\pi, e$ and a bunch of others, but there's probably a bunch of stealth ones, like $\sum\limits_{i=0}^\infty \frac 1 {10^{n!}}$.
Wait, that's the wrong formula.
It's the one who's got digits of 1's that get wider and wider gaps between them
kind've a sloppy response, but: if there's no reason to prefer one point over the other, then there's not a single direction in the sense that a ray does. instead you have two possible directions. @johan
now, you could choose to describe those two different directions as 'equivalent' since they describe the same line. and that's a valid perspective to take, but it's one which comes with surprises.
to get a sense of the strangeness that can result: suppose i draw a sphere and then pick two points, one on the sphere's surface and one at its center
i can use that to define both a line between these points and a ray emanating from the center, and by changing the point on the sphere i change what ray + line i get
@Jasper I don't really know. She did not say anything to me before she left. She came here in mid February, but I was either not here, or did not recognize the name she was using.
the weirdness is this: if i move my point on the sphere from, say, the north pole to the south pole, the ray reverses direction. but the line doesn't change.
so according to the former, the north and south pole are different. but not according to the latter
I am convinced of a result but I do not know where to start in proving it: $$\int_{1}^{x}{{{\left \lceil z \right \rceil}\over{z}}\;dz}=\log \left({{x^{\left \lfloor x \right \rfloor+1}}\over{\left \lfloor x \right \rfloor\,\Gamma\left(\left \lfloor x \right \rfloor\right)}} \right)$$
I guess I know where to start... I used a finite difference of a derivative of the Hurwitz zeta function combined with some identities for sums-of-logarithms to derive this, but I believe at some point I exchanged the order of a sum and an integration without proving that it was correct to do so.
@Evinda Do you remember I said that $\int_{\mathbb{R}^n}E(\epsilon,x)\,\mathrm{d}x$ is independent of $\epsilon$? Read up on approximations of the identity.
I want to find how many different necklaces we can create with $m$ symmetric beads if we have $k$ colours.
Burnside's Formula is the following: $$\text{ # orbits } =\frac{1}{|G|}\sum_{g\in G} X (g)$$
In this case $G$ is the group of the permutations of the symmetric beads, or not? And $X(g)$ is the number of elements that $g$ leaves unchanged, right? But how can we find this number in this case? Could you give me a hint?
Something more specific, can't remember the name of it.
But yeah. The $m$ case is relatively easy. Now, there's the other cases $m - i$, where $i$ elements are changed. There are $\binom m i$ ways to pick the elements which get changed and there are $i! - 1$ ways to change them.
@robjohn Ah, I see... :) But how do we use the fact that $\int_{\mathbb{R}^n} E(\epsilon, x) dx$ is independent on x, at the calculation of $\lim_{\epsilon \to 0} \int_{\mathbb{R}^n} E(\epsilon, x) \phi(\epsilon, x) dx$ ?
@Evinda Do you remember I said that $\int_{\mathbb{R}^n}E(\epsilon,x)\,\mathrm{d}x$ is independent of $\epsilon$? Read up on approximations of the identity.
@MaryStar After selecting the $i$ elements to change, one of the $i!$ orderings is the order they're already in.
For example, in $1, 2, 3, 4, 5$, if we select $1, 2, 3$ for $i = 3$, one of the orderings is $1, 2, 3$. This permutation no longer changes $i$ elements, it changes $0$.
@robjohn You mean that even if we write it like that : $\int_{\mathbb{R}^n} e^{-\frac{x_1^2+ \dots+ x_n^2}{4}} dx_1 \cdots dx_{n}$ the integral is equal to $\int_{\mathbb{R}^{n-1}} \left( \int_{-\infty}^{+\infty} e^{-\frac{x_1^2+ \dots+ x_n^2}{4}} dx_1 \right) dx_2 \cdots dx_{n}$ and at the inner interval $x_2, \dots, x_n$ are considered to be constants, right?
@MaryStar Do you mean how many elements these permutations leave unchanged?
If there are $m$ beads, and we change $i$ of them, $m - i$ beads are left unchanged.
So, if we enumerate all the permutations that change $i$ beads. Let's call this set $C(i)$, the set of permutations that change $i$ elements. If we know there are that many of them, and we know they leave $m - i$ elements unchanged, then the sum $\sum C(i) (m - i) = \sum X(g)$
@robjohn Ah I see... Could you maybe also explain further to me what change of variables we could do in order to show that the integral $\int_{\mathbb{R}^n} \frac{1}{(2 \sqrt{\pi \epsilon})^n} e^{-\frac{|x|^2}{4 \epsilon}} \phi(\epsilon,x) dx$ is also 1?
@Evinda Do you remember I said that $\int_{\mathbb{R}^n}E(\epsilon,x)\,\mathrm{d}x$ is independent of $\epsilon$? Read up on approximations of the identity.
@Evinda Look, the integral of $E$ is one for all $\epsilon$, as $\epsilon\to0$, the function tends to a point mass at $0$. Use this idea to show the convolution limit above.
Find all continuous functions $f$ over reals satisfying $f(x)^2+x^2=1$.
I thought to solve we can just do $f(x) = \pm \sqrt{1-x^2}$. Doesn't that solve it? I think it doesn't since they aren't defined over the reals.
@robjohn So do we just say that since $E(\epsilon, x)$ is not defined for $\epsilon=0$ and $\int_{\mathbb{R}^n} E(\epsilon, x) dx=1$ that $E(\epsilon, x) \to \delta(x)$ ?
I'm told the mods here are "msh210", "MonicaCellio" and "DoubleAA".
How do I calculate the equivalency classes defined by the congruence relation a (mod p) of a ∈ ℤ for p ∈ {"msh210", "MonicaCellio", "DoubleAA"}?
@MaryStar The original sum is $\sum\limits_{g \in G} X(g)$. For every $g \in G$, add its $X(g)$, which is the number of elements it leaves unchanged.
2
Instead of going over one $g$ at a time, we instead separate them into groups because on how many elements they change.
i.e., we partition them under $X$
By doing so, we have slices of the partition $P_i = \{g \mid X(g) = i\}$
If we take the union over all $P_i$, we get $G$, so we're not leaving out any group elements, we're not including ones that weren't there before, and we're only counting each group element once. Exactly the same as the original summation.
However, calculating the partial sums in each of these slices of the partition is much easier.
$\sum\limits_{g\in P_i} X(g) = i|P_i|$
Do you see that all members of $P_i$ have the same $X$ value and that that value is $i$? Now, simply sum over all $i$ to get the total for the full set $G$. $$\sum\limits_{g \in G} X(g) = \sum\limits_i \sum\limits_{g \in P_i} X(g) = \sum\limits_i i|P_i|$$
Find all continuous functions $f$ over reals satisfying $f(x)^2+x^2=1$.
I thought to solve we can just do $f(x) = \pm \sqrt{1-x^2}$. Doesn't that solve it? I think it doesn't since they aren't defined over the reals.
I'm told the mods here are "msh210", "MonicaCellio" and "DoubleAA".
How do I calculate the equivalency classes defined by the congruence relation a (mod p) of a ∈ ℤ for p ∈ {"msh210", "MonicaCellio", "DoubleAA"}?
