TITLE: Is there any meaning to a path integral of path integrals?
QUESTION [1 upvotes]: Is there any useful results, or even any meaning to integration over a space of functionals? For example, consider
$$\int\mathcal{D}S\,e^{-Z[S]},$$
where $Z[S]=\int\mathcal{D}\phi\,e^{-S[\phi]}$ is the partition function of a (Euclidean) quantum field theory with action $S[\phi]$. Such an expression would perhaps carry some fundamental problems like

Would it converge for general $Z[S]$? (I think not)
What does it even means?
Is it at least well-defined?

We could take our wisdom from (ordinary) path integrals and try to remediate problem 1 by taking quotients, like
$$\frac{\int\mathcal{D}S\,e^{-Z[S]}S_{\text{Some action of interest}}[\phi]}{\int\mathcal{D}S\,e^{-Z[S]}},$$
but the expression still looks either useless, meaningless, or hopelessly complicated. So,
is there any meaning to these "higher order" functional integrals?

Edit: One could go more general and define $\int\mathcal{D}S\,W[S]$, for arbitrary $W[S]$, but I think the above less general expression may be a better entry point in thinking about usefulness, e.g. In some very non-rigorous way it's like a quotient of QFTs.

REPLY [3 votes]: Note that even in traditional path integrals over fields, $\mathcal{D}\phi$ is meaningless. There is no Lebesgue measure over infinite-dimensional spaces. So $\mathcal{D}S$ should also be meaningless. One can define (field) path integrals as honest integrals with respect to a probability measure, namely, justify an equation like
$$
\frac{\int \mathcal{D}\phi\ F(\phi)e^{-S(\phi)}}{\int \mathcal{D}\phi\ e^{-S(\phi)}}\ =\ \int_{S'(\mathbb{R}^d)} d\mu(\phi) \ F(\phi)\ .
$$
Here $\mu$ should be a Borel probability measure on the space of temperate Schwartz distributions on $d$-dimensional spacetime where the fields $\phi$ live. In your case, one would first need a suitable space $W$ of test-functionals on say $S'(\mathbb{R}^d)$, and then you could try to construct your path integrals using Borel probability measures on $W'$. For this you need $W$ to have nice properties like being nuclear and Fréchet. The issue tough is the lack of explicit examples. There are examples in the area of probability theory called "white noise calculus" (see keywords like Kondratiev triple, Hida distribution). But I don't know if the kind of (functional) path integrals you are thinking about arise naturally anywhere.


TITLE: Solve $y^2=x^4+x+2$ in $x,y\in \mathbb{Z}$
QUESTION [2 upvotes]: I have to solve $y^2=x^4+x+2$ in $x,y\in \mathbb{Z}$ and right now I am stuck. This is how far I came:
A little manipulation yields $y^2-2=x(x+1)(x^2-x+1)$. $x=1$ and $y=\pm 2$ are solutions. Assume from now on that $y$ is positive and $x>1$. I somehow want to find a prime $p=\pm 3\mod 8$ which  divides this expression so that $y^2=2\mod p$ and this is not possible then. But I really don't know how to do this. I really need some hints. Thanks.

REPLY [7 votes]: hint: $x^4+x+2$ is between $(x^2)^2$ and $(x^2+1)^2$ for $x>2$.


TITLE: A problem about Lebesgue measurable set
QUESTION [1 upvotes]: I'm doing exercise in "REAL ANALYSIS" of Folland and got stuck on this problem. I got no clue on how to find the set $I$. Hope someone can help me solve this. Thanks so much

Suppose $m$ is Lebesgue measure and $L$ is its domain. If $E \in L$ and $m(E) \gt 0$, for any $\alpha < 1$, prove that there is an open interval $I$ such that $m(E \bigcap I) \gt \alpha m(I)$

REPLY [4 votes]: You have
$$
m(E)=\inf\{\sum_j b_j-a_j:\ E\subset\bigcup_j(a_j,b_j)\}.
$$
So, given $\alpha\in(0,1)$, there exist intervals $\{(I_j)\}$ with
$$
m(E)\geq\alpha\sum_jm(I_j).
$$
Then
$$
\sum_j m(E\cap I_j)\geq m(E)\geq\alpha\sum_jm(I_j).
$$
For this last inequality to hold, it needs to hold for at least one $j$: for such $j$,
$$
m(E\cap I_j)\geq\alpha m(I_j).
$$


TITLE: Grade 10 Level - Cube Split into 3 Equal volumes
QUESTION [0 upvotes]: first of all, sorry for the unspecific question, I couldn't think of a specific enough title, if there is any suggestion, I can edit the title.

This question seems simple enough, but I can't seem to solve it? Two of the sections are clearly a right based triangular pyramid missing a chunk of a similar triangular pyramid. So if we continue the line through P, the side profile would look like this, and if we label PV x we can get the other lengths through similar triangle shenanigans.

Then, $\frac{1}{1-x}\cdot 1 \cdot 1 /3 - \frac{x}{1-x}\cdot x \cdot x / 3 = 1/3$ and rearranging would give $x^2-1=0$, which is nonsensical. Could someone please point out my mistake?
*Question is from 2019 Senior Division Australian Mathematics Competition

REPLY [0 votes]: Let $PV = x$, then extending the triangular pyramid chunk below the cube, it will form the full pyramid at $z = -h$, where $z = 0$ is the horizontal base of the cube.
From similar triangles, we have the following ratios
$ \dfrac{h}{x} = \dfrac{1 + h}{1} $
From which
$ h = (1 + h) x $
$ h(1 - x) = x $
and
$ h = \dfrac{x}{1 - x} $
The volume of the pyramidal chunk (inside the cube) is
$ V = (1 - \bigg(\dfrac{h}{h+1}\bigg)^3 ) V_0 $
Where V_0 is the volume of the all the pyramid, which is one third of the base area times the total height
$V_0 = \dfrac{1}{3} \dfrac{1}{2} (1 + h) $
Now $V = \dfrac{1}{3} $, so
$  (1 - \bigg(\dfrac{h}{h+1}\bigg)^3 ) \dfrac{1}{2} (1 + h) = 1 $
Now, $1 + h = \dfrac{1}{1 - x} $ and $ \dfrac{h}{h+1} = x $
So,
$ \dfrac{ 1 - x^3 }{1 - x} = 2 $
And this simplifies to
$ 1 + x + x^2 = 2 $
So that
$ x^2 + x - 1 = 0$
whose postive solution is
$ x = \dfrac{-1 + \sqrt{5}}{2} $
So the correct answer is $(E)$


TITLE: Term for a fully connected balanced graph (Rock paper scissor)
QUESTION [2 upvotes]: Is there a mathematical, graph theory, game theory term for a graph that is fully connected and balanced evenly with each other node.
I'm thinking in situations like Rock paper scissors where each action is equally balanced by each other option?

REPLY [2 votes]: A regular tournament $T=(V,E)$ is a complete digraph on $|V|=2n+1$ vertices where each vertex has the same score as each other vertex. Each vertex is connected to each other vertex by exactly one directed edge and each vertex dominates exactly $\frac{|V|-1}{2}$ other edges and is dominated by the $\frac{|V|-1}{2}$ other edges.