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# Number of integer solutions to $3i^2 + 2j^2 = 77 \cdot 6^{2012}$

here is another problem I did not manage to solve in the contest I mentioned in my previous question.

Determine the number of integer solutions $(i, j)$ of the equation: $3i^2 + 2j^2 = 77 \cdot 6^{2012}$.

Applying logarithms is not useful, since on the left hand side we have a sum; I also tried some algebraic manipulations that led me to nothing useful. Is there a simple solution to the problem?

Thank you,
rubik

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$A)~ i^2=25x^2 ~\text {and}~ j^2=x^2$

$3 \cdot 25x^2+2x^2=77 \cdot 6^{2012} \Rightarrow x^2=6^{2012}$ , hence :

$i= \pm 5x \Rightarrow i = \pm 5 \cdot 6^{1006}$

$j= \pm x \Rightarrow j = \pm 6^{1006}$

$R_A :$

$(i,j) \in \{(-5 \cdot 6^{1006},-6^{1006}),(-5 \cdot 6^{1006},6^{1006}),(5 \cdot 6^{1006},-6^{1006}),(5 \cdot 6^{1006},6^{1006})\}$

$B)~ i^2=9x^2 ~\text {and}~ j^2=25x^2$

$3 \cdot 9x^2+2 \cdot 25x^2=77 \cdot 6^{2012} \Rightarrow x^2=6^{2012}$ , hence :

$i= \pm 3x \Rightarrow i = \pm 3 \cdot 6^{1006}$

$j= \pm 5x \Rightarrow j = \pm 5 \cdot 6^{1006}$

$R_B :$

$(i,j) \in \{(-3 \cdot 6^{1006},-5 \cdot 6^{1006}),(-3 \cdot 6^{1006},5 \cdot 6^{1006}),(3 \cdot 6^{1006},-5 \cdot 6^{1006}),(3 \cdot 6^{1006},5 \cdot 6^{1006})\}$

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Wow, this is brilliant! Simple and neat! I am choosing your answer as best answer because in my opinion it's the simplest and intuitive one. – rubik Mar 11 '12 at 10:51
I don't understand. So what if the ratio $i/j$ is neither 5 nor 5/3? – Zsbán Ambrus Mar 11 '12 at 11:10
I'm with @ZsbánAmbrus I can't follow this at all. It seems completely unintuitive. – Pureferret Mar 11 '12 at 13:56
@ZsbánAmbrus: Can you elaborate a bit more? – rubik Mar 12 '12 at 13:33

Observe that $i$ has to be even, so do $j$ because $3i^2$ and the right side can be divided by 4. Let $i'=2i$, $j'=2j$; the equation becomes $3i'^2+2j'2=77 \cdot 3^{2012} \cdot 2^{2010}$. The same argument applies 1006 times, and we get something like $3i^2+2j^2=77 \cdot 3^{2012}$. Now do the same looking the divisibility by 3. It works. Finaly check $3i^2+2j^2=77$

Edit: It's my first posting here, so hello to all from France !

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Welcome to math.stackexchange ama! – Ragib Zaman Mar 11 '12 at 9:49

Considering the equation modulo $3$, we must have $2j^2 \equiv 0$ which implies that $j=3k$ for some $k\in \mathbb{N}$.

$3i^2 + 18k^2 = 77 \cdot 6^{2012}$ so $$i^2 + 6k^2 = 77 \cdot 2^{2012} \cdot 3^{2011}.$$

which now implies $i=6l$ for some $l\in \mathbb{N}.$ Thus

$$6l^2 + k^2 = 77 \cdot 2^{2011} \cdot 3^{2010}.$$

Similarly, $k=6m$ for some $m\in\mathbb{N}$ and

$$l^2 + 6m^2 = 77 \cdot 2^{2010} \cdot 3^{2009}.$$

A pattern is emerging that we will keep on having this divisibility by 6 until we reach some $a$ and $b$ such that $$a^2 + 6b^2 = 77 \cdot 2 =154$$

which (by checking the small number of possible cases) has solutions $(a,b) = (10,3)$ and $(2,5)$ in the naturals, or by varying all possible sign combinations, there are 8 solutions in the integers. Now to find the solutions of the original equation, simply back substitute the factors of 6 we kept extracting. For this problem however that is not required since only the number of solutions is asked, and the answer is 8.

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+1: I like your method because it's intuitive and very understandable. – rubik Mar 11 '12 at 10:56

The righthand side is even, as is $2j^2$, so $i$ must be even. Say $i=2^nm$, where $m$ is odd and $n\ge 1$, and $j=2^k\ell$, where $\ell$ is odd and $k\ge 0$; then $3m^22^{2n}+\ell^2 2^{2k+1}=77\cdot6^{2012}$. Let $a=\min\{2n,2k+1,2012\}$; then $3m^22^{2n-a}+\ell^2 2^{2k+1-a}=77\cdot3^{2012}2^{2012-a}$, where the exponents on the $2$’s are all non-negative, and at least one is $0$. Clearly it’s not possible for exactly two of these exponents to be positive, and since the first two have opposite parity, it’s not possible for all three to be zero, so exactly one is positive and two are zero. The first and third have the same parity, so we must have $2n-a=2012-a=0$ and $2k+1-a>0$. In particular, $a=2012$, and $n=1006$.

To recapitulate, at this point we know that $i=2^{1006}m$ for some odd $m$, and $j=2^k\ell$ for some odd $\ell$ and $k\ge 1006$, and we can let $k\,'=k-1006$ and divide through by $2^{2012}$ to get $$3m^2+2^{2k\,'+1}\ell^2=77\cdot3^{2012}\;.\tag{1}$$

Now perform a similar analysis considering factors of $3$ instead of $2$. If $3^b$ and $3^c$ are the highest powers of $3$ dividing $m$ and $\ell$, respectively, then $3m^2$ has $2b+1$ factors of $3$, $2^{2k\,'+1}\ell^2$ has $2c$, and the righthand side has $2012$. Reasoning as before, we see that this is possible only if $2c=2012$ and $2b+1>2012$, so that $c=1006$ and $b\ge 1006$. Thus, if we set $b\,'=b-1006$ and divide through by $3^{2012}$, we can further reduce $(1)$ to

$$3^{2b\,'+1}u^2+2^{2k\,'+1}v^2=77\tag{2}\;,$$

where $m=3^{2b+1}u$, $\ell=3^{1006}v$, and $u$ and $v$ are not divisible by $2$ or $3$.

Plainly $0\le b\,'\le 1$ and $0\le k\,'\le 2$, so there are at most six solutions. Moreover, the only possible values for $u$ and $v$ are $1$ and $5$: $7$ is already too large.

• If $u=v=1$, the lefthand side is either too large or at most $3^3+2^5=59$, so we get no solutions.

• If $u=5$, $b\,'$ must be $0$, and we get the solution $3\cdot25+2\cdot1=77$.

• If $v=5$, $k\,'$ must be $0$, and we get the solution $3^3\cdot1+2\cdot25=77$.

The original equation therefore has two solutions in positive integers and therefore eight in integers (since we can change the sign of $i$ or $j$ independently in each solution).

This can probably be done more elegantly; I wrote it up as I was working it out.

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Doesn't 2 solutions in the naturals imply 2*4 = 8 solutions in the integers, since for each (+,+) we also have (-,+), (+,-) and (-,-). – Ragib Zaman Mar 11 '12 at 9:51
@Ragib: Yep; momentary slip of the mind. Thanks. – Brian M. Scott Mar 11 '12 at 9:56
+1: Thank you; your method is a bit more complicated than the other ones but you showed how to reason. Probably your last lines served as basis for pedja's answer (I'm referring to $3\cdot25+2\cdot1=77$ and $3^3\cdot1+2\cdot25=77$). – rubik Mar 11 '12 at 11:11

I find it easiest to proceed as follows. First note that $j$ must be divisible by $3$ and $i$ must be even. But then $3i^2$ is divisible by $9,$ so $i$ is divisible by $3,$ and similarly $j$ is even. Suppose then that $6^k$ divides ${\rm gcd}(i,j)$ but $6^{k+1}$ does not. Then $1 \leq k \leq 1006.$ Write $i = 6^{k}s$ and $j = 6^k t.$ If $k < 1006,$ we have $36$ divides $3s^2 + 2t^2,$ and again $3$ divides both $s$ and $t$, and $s$ and $t$ are both even, contadicting the choice of $k,$ since $6$ then divides ${\rm gcd}(s,t).$ Hence we are reduced to finding the number of integer solutions to $3s^2 + 2t^2 = 77.$ Now $s$ must be odd, so $3s^2 \equiv 3$ (mod $8$) and hence $2t^2 \equiv 2$ (mod $8$), and $t$ is also odd. Thus $s$ an $t$ are both odd. Clearly $|s| \leq 5.$ If $|s| = 5,$ then $|t| = 1,$ while if $|s| = 3$ then $|t| = 5.$ If $|s| = 1,$ then there is no solution for $t.$ Hence the solutions for $s$ and $t$ are $(5,1),(5,-1),-5,1),(-5,-1), (3,5),(3,-5),(-3,5),(-3,-5),$ and each solution for $(s,t)$ yields the unique solution $6^{1006}(s,t)$ for $(i,j).$

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Hint $\ \ \$ Prime $\rm\:p\nmid a,\:\ p^2\ |\ n = a\: x^2 + p\:y^2\:\Rightarrow\: p\ |\ x^2\:\Rightarrow\: p\ |\ x\:\Rightarrow\: p\ |\ y^2\:\Rightarrow\: p\ |\ y$

Therefore, we deduce $\rm\:\ n/p^2 = a\:\bar x^2 + p\:\bar y^2,\$ for $\rm\ \bar x = x/p,\ \bar y = y/p \in\mathbb Z$

So for $\rm\: p = 3,2,\ \ \ 3\:i^2 + 2\:j^2 = 77 \cdot 6^{2012}\:\Rightarrow\: 3\:\bar i^2 + 2\:\bar j^2 = 77\$ for $\rm \bar i = i/6^{1006},\:\bar j = j/6^{1006}\in\mathbb Z$

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### The Kelly Criterion and a Sure Thing

The Kelly Criterion is an alternative to standard utility theory, which seeks to maximize expected utility. Instead, the Kelly Criterion seeks to maximize expected growth. That is, if we start out with an initial bankroll $$B_0$$, we seek to maximize $$\mathrm{E}[g(t)]$$, where $$B_t = B_0\cdot e^{g(t)}$$.
As a simple example, consider the following choice. We can have a sure $3000, or we can take the gamble of a $$\frac{4}{5}$$ chance of$4000 and a $$\frac{1}{5}$$ chance of \$0. What does Kelly say?
Assume we have a current bankroll of $$B_0$$. After the first choice we have $$B_1 = B_0+3000$$, which we can write as $\mathrm{E}[g(1)] = \log\left(\frac{B_0+3000}{B_0}\right);$for the second choice we have $\mathrm{E}[g(1)] = \frac{4}{5} \log\left(\frac{B_0+4000}{B_0}\right).$And so we want to compare $$\log\left(\frac{B_0+3000}{B_0}\right)$$ and $$\frac{4}{5} \log\left(\frac{B_0+4000}{B_0}\right)$$.
Exponentiating, we're looking for the positive root of ${\left({B_0+3000}\right)}^5 - {B_0}\cdot {\left({B_0+4000}\right)}^4=0.$Wolfram Alpha now tells us that we should go with the sure thing if $$B_0 < 4942.92$$, and take the gamble otherwise.

### A Bayes' Solution to Monty Hall

For any problem involving conditional probabilities one of your greatest allies is Bayes' Theorem. Bayes' Theorem says that for two events A and B, the probability of A given B is related to the probability of B given A in a specific way.

Standard notation:

probability of A given B is written $$\Pr(A \mid B)$$
probability of B is written $$\Pr(B)$$

Bayes' Theorem:

Using the notation above, Bayes' Theorem can be written: $\Pr(A \mid B) = \frac{\Pr(B \mid A)\times \Pr(A)}{\Pr(B)}$Let's apply Bayes' Theorem to the Monty Hall problem. If you recall, we're told that behind three doors there are two goats and one car, all randomly placed. We initially choose a door, and then Monty, who knows what's behind the doors, always shows us a goat behind one of the remaining doors. He can always do this as there are two goats; if we chose the car initially, Monty picks one of the two doors with a goat behind it at random.

Assume we pick Door 1 and then Monty sho…

### What's the Value of a Win?

In a previous entry I demonstrated one simple way to estimate an exponent for the Pythagorean win expectation. Another nice consequence of a Pythagorean win expectation formula is that it also makes it simple to estimate the run value of a win in baseball, the point value of a win in basketball, the goal value of a win in hockey etc.

Let our Pythagorean win expectation formula be $w=\frac{P^e}{P^e+1},$ where $$w$$ is the win fraction expectation, $$P$$ is runs/allowed (or similar) and $$e$$ is the Pythagorean exponent. How do we get an estimate for the run value of a win? The expected number of games won in a season with $$g$$ games is $W = g\cdot w = g\cdot \frac{P^e}{P^e+1},$ so for one estimate we only need to compute the value of the partial derivative $$\frac{\partial W}{\partial P}$$ at $$P=1$$. Note that $W = g\left( 1-\frac{1}{P^e+1}\right),$ and so $\frac{\partial W}{\partial P} = g\frac{eP^{e-1}}{(P^e+1)^2}$ and it follows $\frac{\partial W}{\partial P}(P=1) = … ### Solving a Math Puzzle using Physics The following math problem, which appeared on a Scottish maths paper, has been making the internet rounds. The first two parts require students to interpret the meaning of the components of the formula $$T(x) = 5 \sqrt{36+x^2} + 4(20-x)$$, and the final "challenge" component involves finding the minimum of $$T(x)$$ over $$0 \leq x \leq 20$$. Usually this would require a differentiation, but if you know Snell's law you can write down the solution almost immediately. People normally think of Snell's law in the context of light and optics, but it's really a statement about least time across media permitting different velocities. One way to phrase Snell's law is that least travel time is achieved when \[ \frac{\sin{\theta_1}}{\sin{\theta_2}} = \frac{v_1}{v_2},$ where $$\theta_1, \theta_2$$ are the angles to the normal and $$v_1, v_2$$ are the travel velocities in the two media.

In our puzzle the crocodile has an implied travel velocity of 1/5 in the water …

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# zbMATH — the first resource for mathematics

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Innovation diffusion model in patch environment. (English) Zbl 1052.91076

A mathematical model is proposed to describe the dynamics of users of one product in two different patches. Advertisement force, contact rate between users and non-users, population dispersal rate and returning rate from user class to non-user class are chosen as key parameters. For the product with a long life-span, it is assumed that the returning rate is proportional to the number of users. It is shown that this model has a unique positive equilibrium which is a globally stable. For the product with a short life-span, time delays are introduced to represent the duration of the product in two different markets and the stability of a positive equilibrium is analyzed in one reasonable case. Periodic advertisements are also incorporated and the existence and uniqueness of positive periodic solutions are investigated.

The model designs in assuming, that two patches are isolated, and the dynamics of both classes are governed by

$\begin{array}{cc}\hfill \frac{d{N}_{1}}{dt}& =-{\gamma }_{1}{N}_{1}-{\lambda }_{1}{A}_{1}{N}_{1}-{\delta }_{1}{N}_{1}+{\nu }_{1}{A}_{1}+{\beta }_{1},\hfill \\ \hfill \frac{d{A}_{1}}{dt}& ={\gamma }_{1}{N}_{1}+{\lambda }_{1}{A}_{1}{N}_{1}-{\delta }_{1}{A}_{1}-{\nu }_{1}{A}_{1},\hfill \end{array}$

where ${N}_{1}$ is the number of non-users in the first patch, ${A}_{1}$ is the number of users in the first patch, ${\beta }_{1}$ is the birth rate of the population in the first patch, ${\delta }_{1}$ is the death rate of the population in the first patch, ${\gamma }_{1}$ represents the intensity of advertisement for promoting product-users in the first patch, and ${\lambda }_{1}$ is the valid contact rate of users of the product with non-users in the first patch. Similarly, it is supposed that the dynamics of non-user-class and user-class in the second patch are governed by

$\begin{array}{cc}\hfill \frac{d{N}_{2}}{dt}& =-{\gamma }_{2}{N}_{2}-{\lambda }_{2}{A}_{2}{N}_{2}-{\delta }_{2}{N}_{2}+{\nu }_{2}{A}_{2}+{\beta }_{2},\hfill \\ \hfill \frac{d{A}_{2}}{dt}& ={\gamma }_{2}{N}_{2}+{\lambda }_{2}{A}_{2}{N}_{2}-{\delta }_{2}{A}_{2}-{\nu }_{2}{A}_{2},\hfill \end{array}$

where ${N}_{2}$ is the number of non-users in the second patch, ${A}_{2}$ is the number of users in the second patch, ${\beta }_{2}$ is the birth rate of the population in the second patch, ${\delta }_{2}$ is the death rate of the population in the second patch, ${\gamma }_{2}$ represents the intensity of advertisement for promoting product-users in the second patch, and ${\lambda }_{2}$ is the valid contact rate of users of the product with non-users in the second patch.

But in case, when the two patches are connected, ${\theta }_{1}$ is the probability that an individual migrates from the first patch to the second patch, ${\theta }_{2}$ the probability that an individual migrates from the second patch to the first patch. Further, it is assumed that ${k}_{2}$ is the fraction of new immigrants from the second patch who remain in user-class and that ${k}_{1}$ is the fraction of new immigrants from the first patch who remain in user-class. Under these assumptions, the model is:

$\begin{array}{cc}\hfill \frac{d{N}_{1}}{dt}& =-{\gamma }_{1}{N}_{1}-{\lambda }_{1}{A}_{1}{N}_{1}-{\delta }_{1}{N}_{1}+{\nu }_{1}{A}_{1}+{\beta }_{1}+{\theta }_{2}\left({N}_{2}+\left(1-{k}_{2}\right){A}_{2}\right)-{\theta }_{1}{N}_{1},\hfill \\ \hfill \frac{d{A}_{1}}{dt}& ={\gamma }_{1}{N}_{1}+{\lambda }_{1}{A}_{1}{N}_{1}-{\delta }_{1}{A}_{1}-{\nu }_{1}{A}_{1}-{\theta }_{1}{A}_{1}+{k}_{2}{\theta }_{2}{A}_{2},\hfill \\ \hfill \frac{d{N}_{2}}{dt}& =-{\gamma }_{2}{N}_{2}-{\lambda }_{2}{A}_{2}{N}_{2}-{\delta }_{2}{N}_{2}+{\nu }_{2}{A}_{2}+{\beta }_{2}+{\theta }_{1}\left({N}_{1}+\left(1-{k}_{1}\right){A}_{1}\right)-{\theta }_{2}{N}_{2},\hfill \\ \hfill \frac{d{A}_{2}}{dt}& ={\gamma }_{2}{N}_{2}+{\lambda }_{2}{A}_{2}{N}_{2}-{\delta }_{1}{A}_{2}-{\nu }_{2}{A}_{2}-{\theta }_{2}{N}_{2}+{k}_{1}{\theta }_{1}{A}_{1}·\hfill \end{array}$

In order to incorporate duration of the product into the model, described above was modified. The assumption that the returning rate is proportional to the number of users seems reasonable when the length of the life-span of the product is large. Otherwise, the effect of the length will be significant. Because of this, time delays due to the life-span of the product was introduced into the model and it was obtained:

$\begin{array}{c}\frac{d{N}_{1}}{dt}=-{\gamma }_{1}{N}_{1}\left(t\right)-{\lambda }_{1}{A}_{1}\left(t\right){N}_{1}\left(t\right)-{\delta }_{1}{N}_{1}\left(t\right)+{e}^{-{\delta }_{1}{\tau }_{1}}\left({\gamma }_{1}{N}_{1}\left(t-{\tau }_{1}\right)\hfill \\ \hfill +{\lambda }_{1}{A}_{1}\left(t-{\tau }_{1}\right){N}_{1}\left(t-{\tau }_{1}\right)\right)+{\beta }_{1}+{\theta }_{2}\left({N}_{2}+\left(1-{k}_{2}\right){A}_{2}\right)-{\theta }_{1}{N}_{1},\end{array}$ $\begin{array}{c}\frac{d{A}_{1}}{dt}={\gamma }_{1}{N}_{1}\left(t\right)+{\lambda }_{1}{A}_{1}\left(t\right){N}_{1}\left(t\right)-{\delta }_{1}{A}_{1}\left(t\right)-{e}^{-{\delta }_{1}{\tau }_{1}}\left({\gamma }_{1}{N}_{1}\left(t-{\tau }_{1}\right)\hfill \\ \hfill +{\lambda }_{1}{A}_{1}\left(t-{\tau }_{1}\right){N}_{1}\left(t-{\tau }_{1}\right)\right)-{\theta }_{1}{A}_{1}\left(t\right)+{k}_{2}{\theta }_{2}{A}_{2}\left(t\right),\end{array}$ $\begin{array}{c}\frac{d{N}_{2}}{dt}=-{\gamma }_{2}{N}_{2}\left(t\right)-{\lambda }_{2}{A}_{2}\left(t\right){N}_{2}\left(t\right)-{\delta }_{2}{N}_{2}\left(t\right)+{e}^{-{\delta }_{2}{\tau }_{2}}\left({\gamma }_{2}{N}_{2}\left(t-{\tau }_{2}\right)\hfill \\ \hfill +{\lambda }_{2}{A}_{2}\left(t-{\tau }_{2}\right){N}_{2}\left(t-{\tau }_{2}\right)\right)+{\beta }_{2}+{\theta }_{1}\left({N}_{1}+\left(1-{k}_{1}\right){A}_{1}\right)-{\theta }_{2}{N}_{2},\end{array}$ $\begin{array}{c}\frac{d{A}_{2}}{dt}={\gamma }_{2}{N}_{2}\left(t\right)+{\lambda }_{2}{A}_{2}\left(t\right){N}_{2}\left(t\right)-{\delta }_{2}{A}_{2}\left(t\right)-{e}^{-{\delta }_{2}{\tau }_{2}}\left({\gamma }_{2}{N}_{2}\left(t-{\tau }_{2}\right)\hfill \\ \hfill +{\lambda }_{2}{A}_{2}\left(t-{\tau }_{2}\right){N}_{2}\left(t-{\tau }_{2}\right)\right)-{\theta }_{2}{A}_{2}\left(t\right)+{k}_{1}{\theta }_{1}{A}_{1}\left(t\right)·\end{array}$

Here ${\tau }_{1}$ represents the life-span of the product in the first patch and ${\tau }_{2}$ represents the life-span of the product in the second patch.

##### MSC:
 91D25 Spatial models (mathematical sociology) 34K60 Qualitative investigation and simulation of models
##### Keywords:
stability; periodics; delay


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# Manifold Atlas:Wiki syntax

This page collects some simple but useful tips about the Wiki syntax. The MediaWiki Handbook is a good reference for the wiki software: see especially the section for editors.

## Contents

You create hyperlinks inside the Manifold Atlas by writing [[page name|link text]]. For example

[[Surfaces|2-dimensional manifold]]

If you wish to link to a Wikipedia page, use the above format with the page name: Wikipedia:page_name. For example

[[Wikipedia:Differentiable_manifold|smooth manifold]]

### Links to sections and subsections

You can refine your hyperlinks to sections and even subsections of web-pages. For example

[[5-manifolds: 1-connected#Invariants|linking form]]

produces the link linking form, which goes directly to the invariants section of the page on simply-connected 5-manifolds.


text:
## Peter Dankelmann ; Alex Alochukwu - Wiener index in graphs with given minimum degree and maximum degree

dmtcs:6956 - Discrete Mathematics & Theoretical Computer Science, June 3, 2021, vol. 23 no. 1
Wiener index in graphs with given minimum degree and maximum degree

Authors: Peter Dankelmann ; Alex Alochukwu

Let $G$ be a connected graph of order $n$.The Wiener index $W(G)$ of $G$ is the sum of the distances between all unordered pairs of vertices of $G$. In this paper we show that the well-known upper bound $\big( \frac{n}{\delta+1}+2\big) {n \choose 2}$ on the Wiener index of a graph of order $n$ and minimum degree $\delta$ [M. Kouider, P. Winkler, Mean distance and minimum degree. J. Graph Theory 25 no. 1 (1997)] can be improved significantly if the graph contains also a vertex of large degree. Specifically, we give the asymptotically sharp bound $W(G) \leq {n-\Delta+\delta \choose 2} \frac{n+2\Delta}{\delta+1}+ 2n(n-1)$ on the Wiener index of a graph $G$ of order $n$, minimum degree $\delta$ and maximum degree $\Delta$. We prove a similar result for triangle-free graphs, and we determine a bound on the Wiener index of $C_4$-free graphs of given order, minimum and maximum degree and show that it is, in some sense, best possible.

Volume: vol. 23 no. 1
Section: Graph Theory
Published on: June 3, 2021
Submitted on: December 3, 2020
Keywords: Mathematics - Combinatorics,05C12 (primary) 92E10 (secondary)


text:
# Graham number

The Graham number or Benjamin Graham number is a figure used in securities investing that measures a stock's so-called fair value.[1] Named after Benjamin Graham, the founder of value investing, the Graham number can be calculated as follows:

$\sqrt{22.5\times(\text{earnings per share})\times(\text{book value per share})}$

The final number is, theoretically, the maximum price that a defensive investor should pay for the given stock.[2] Put another way, a stock priced below the Graham Number would be considered a good value, if it also meet a number of other criteria.

This number applies only to certain types of stocks in combination with a number of other criteria. The complete Graham selection procedure is much more elaborate. No decision should be made based on this number alone.[3]

## Alternative calculation

Earnings per share is calculated by dividing net income by shares outstanding. Book value is another way of saying shareholders' equity. Therefore, book value per share is calculated by dividing equity by shares outstanding. Consequently, the formula for the Graham number can also be written as follows:

$\sqrt{22.5 \times \left(\frac{\text{net income}}{\text{shares outstanding}}\right) \times \left(\frac{\mathrm{shareholders'\ equity}}{\text{shares outstanding}}\right)}$


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# Expectation of semi-orthogonal random projection

Suppose I want to compute the expectation of $$\|x'O\|_2$$ where $$O$$ is a random matrix having the uniform distribution on the set of $$n \times r$$ $$(r \leq n)$$ semi-orthogonal matrices (i.e., $$O'O = I_r$$.); $$x \in \mathbb{R}^n$$ is fixed.

My intuition tells me that, $${\rm E}(\|x'O\|_2) = \sqrt{\frac{r}{n}}\|x\|_2$$ (where the expectation is taken wrt the distribution for $$O$$) but I'm having a difficult time proving this. How would I go about this? Are there any references I am missing?

• Is $x$ also random, or is it a constant? Dec 11, 2020 at 17:20
• It is fixed -- updated my question to make this clear. Thanks! Dec 11, 2020 at 17:29
• It's known that you can generate matrices $O$ by using $O=Q(Q^{T}Q)^{-1/2}$ where the entries of the $n$ by $r$ matrix $Q$ are iid $N(0,1)$. Dec 11, 2020 at 18:00
• This is true -- but I could not think of an obvious way to use this fact in a proof. Dec 11, 2020 at 18:09
• Also, I have been able to verify my conjecture experimentally using this construction. Dec 11, 2020 at 18:31


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# Should you trust modern medicine?

A general belief in progress has led the majority to trust modern medicine, but is this trust misguided? In his thought-provoking book Medical Nihilism, Dr. Jacob Stegenga rigorously argues that we ought to have little confidence in modern medicine.

Jacob’s stance is summarized by his master argument, which is captured in a clever, mathematical formula leveraging Bayes Rule. If we define $$H$$ as the truth of whether a drug is effective or not, then $$H = \{ 0, 1 \}$$. Similarly we can define $$E$$ as the evidence either supporting or refuting the effectiveness of the drug, then $$E = \{ 0, 1 \}$$. He comes to this conclusion above by critically dissecting the three components on the right of the equal sign below, which are defined as follows:

$P(H|E) = \frac{P(E|H)\cdot P(H)}{P(E)}$

P(H): low due to the dearth of magic bullets; thin theoretical basis for many interventions; covered in Part I of book.
P(E|H): low due to the ubiquity of small effect sizes; based on this there is little empirical reason to believe modern medicine is effective; touched on in Part II.
P(E): high due to the financial inclinations of pharmaceutical companies to exploit the malleability of research methods to bring their drug to market; methodological and societal bent toward favorable evidence; this is the key focus of the book and is covered extensively in Part II and some of Part III.

Given that both of the numerators are small and the denominator is large, Jacob concludes that the probability that a drug is effective given available evidence is small, i.e. $$P(H=1|E) <<<P(H=0|E)$$. Consequently, we ought to have little confidence in the effectiveness of novel medical interventions, even in light of supporting evidence.

As intriguing and clever as his multifaceted argument is, it is ultimately misguided. In this post I’ll pick apart his argument at face-value before I dive into a deeper critique of his real argument. After this I’ll propose a more constructive and realistic framework for evaluating drugs on a case-by-case basis before concluding this post.

## Critiquing the critic

“We should have little confidence in the effectiveness of medical interventions.” - Jacob Stegenga

This is what Jacob said, and his argument is carefully poised to defend this general statement above. __Effectiveness of medical interventions is made relative

## Deeper counterarguments

Most modern pharmaceutical drugs are ineffective.

This is what Jacob wanted to say, and ultimately explains why he titled his book Medical Nihilism, why he blanketed his critique on novel drugs with the ambiguous term medical interventions, and why he is dead-set against pharmaceutical companies for researching, producing, and marketing drugs. What motivates a person to unreasonably compare on an entire industry to its greatest triumphs?

Should we spit on the food industry because not all meals are Michelin-starred masterpieces (effectiveness)? And… another example pointing to purposeful deceit

Unknowingly nor not, his argument is motivated by a misguided anti-pharma sentiment; a sentiment enjoying great popularity in our button-pushing, instant-gratification culture. To show where he errs, let’s look at some of his stronger more central points and show why even these crumble under scrutiny.


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# A group-theoretic perspective on Frankl's union closed problem

Here is a group theoretic phrasing of a special case of the union closed conjecture:

Question: Given a finite group $G$, is there an element of prime power order which is contained in at most half the subgroups of $G$?

Motivation: Frankl's union closed sets conjecture has an equivalent phrasing in terms of lattices. It says that in every finite lattice there is a join irreducible element which is less than or equal to at most half the elements in the lattice.

Finite lattices are always isomorphic to intervals of subgroups $[H,G]$ for groups $H,G$ (i.e. the lattice of subgroups $H\subseteq K \subseteq G$, with the subgroup relation). It is not known whether it suffices to take $H$ and $G$ to be finite. I wonder if anything about Frankl's conjecture is known for the case when $H$ is the trivial group. And that is precisely what is asked above. Notice that the elements of prime power order are in correspondence with the join irreducibles of the lattice of subgroups of $G$.

Is the answer to the question above known? Is this known for special classes of groups?

• Is there any group $H_0$ such that the answer in the case $H=H_0$ is known? – Noah Schweber Jan 9 '14 at 4:12
• @Noah, not that I know of. – Gjergji Zaimi Jan 9 '14 at 7:19
• Is it conceivable that the following reduction might hold: Let $N$ be a normal subgroup of a group $G$. If $N$ contains an element $g$ of prime power order that is contained in at most half the subgroups of $N$, then $g$ is contained in at most half the subgroups of $G$? – Nick Gill Jan 10 '14 at 12:03
• Were such a reduction possible, then one might be able to proceed by proving the statement for quasisimple groups (generalizing @Alireza's comment below), then proving it for central products of isomorphic quasisimples, and then the result would be true for groups with non-nilpotent generalized Fitting subgroup of $G$. In other words you'd be left with the situation $F^*(G)=F(G)$... – Nick Gill Jan 10 '14 at 12:05
• ... at which point the direction that @Russ suggests might work - you'd need to prove the statement for nilpotent groups. Having typed all that, I start to doubt that the reduction is really possible... – Nick Gill Jan 10 '14 at 12:07

In a somewhat different direction from Alireza: the conjecture is true for a large family of groups, including all abelian groups and many supersolvable groups.

Let me start with the abelian case. Pick an element $g$ of highest possible prime-power order in an abelian group $G$. Then $\langle g \rangle$ has a complement: that is, there is a subgroup $K$ such that $K \langle g \rangle = G$ and $K \cap \langle g \rangle = 0$. In particular, $K \cong G / \langle g \rangle$ by the Isomorphism Theorems, and for any subgroup $X$ with $\langle g \rangle \subseteq X$ there is a corresponding subgroup $X \cap K$ which does not contain $\langle g \rangle$.

In fact, the same argument applies to any $G$ and prime-power order element $g$ if 1) $g$ generates a normal subgroup, and 2) we can find a complement $K$ to $\langle g \rangle$ in G. In this situation, $[\langle g \rangle, G] \cong [1,K]$.

(Edit: deleted discussion of supersolvable groups, which is irrelevant in light of update below.)

UPDATE: The conjecture is true for all finite solvable groups.

Proof: Let $G$ be a solvable group. Then $G$ has a normal subgroup $N$ of prime index, and some element of prime-power order $g \notin N$. Since $N$ is maximal in $G$, we have $\langle g,N \rangle = G$, and since $N$ is normal we have $\langle g,N \rangle = \langle g \rangle N$. Then by Dedekind's identity, we get that $\langle g \rangle (H \cap N) = H \cap G = H$ for any $H$ containing $\langle g \rangle$.

The last tells us that the map from the interval $[\langle g \rangle,G] \rightarrow [1,N]$ given by $H \mapsto H \cap N$ is an injection. Since $N$ doesn't contain $g$, we get the conjectured statement. $\square$

Indeed, the above works whenever $G$ has a maximal normal subgroup of prime index. (E.g., for symmetric groups.)

(Thanks to John Shareshian for several useful comments and discussion.)

UPDATE 2: By combining my argument for solvable groups with Alireza's argument for (certain) finite simple groups, we can prove the conjecture for all groups having a quotient satisfying a certain condition. This very likely holds for all finite groups, but as I'll explain, a little bit of work remains on the group theory side.

Theorem 1: If $G$ is a finite group with a normal subgroup $N$ such that $G/N$ is generated by (at most) two elements of prime power order, then $G$ satisfies the Frankl condition.

Proof: A routine argument shows that we can pick $g,h \in G$ of prime power order such that $g$ and $h$ project in the quotient to generators of $G/N$. (If $G/N$ is cyclic, assume $g=h$.) Let $K = \langle g, h \rangle$, and notice that $KN = \langle N, g, h\rangle = G$.
Then wlog there are at least as many subgroups containing $h$ as containing $g$ (else switch $h$ and $g$). This gives an injection $\varphi_1$ from the set of subgroups containing $g$ but not $h$ to the set of subgroups containing $h$ but not $g$.
Moreover, if $H$ is a subgroup containing both $h$ and $g$, then $H \supseteq K$. It then follows from the Dedekind identity that $K (H \cap N) = H \cap KN = H$, hence the map $\varphi_2$ sending $H \mapsto H \cap N$ is an injection from $[K,G] \rightarrow [1,N]$.
Since the intervals $[\langle h \rangle, G]$ and $[1,N]$ are disjoint, combining the two maps gives an injection from $[\langle g \rangle, G]$ to its complement, and we see the Frankl condition to hold. $\square$

Remark 1: The solvable case (above) is the case when $G/N$ is cyclic -- here, $K = \langle g,h \rangle = K$. The case where $N=1$ solved by Alireza has $K = G$, in which case we map $G$ to $1$ and all other subgroups containing $g$ to a subgroup containing $h$.

Remark 2: The proof has a particularly pleasing explanation if we can take $g$ and $h$ to be conjugate. In this case, the intervals $[\langle g \rangle, G]$ and $[\langle h \rangle, G]$ are isomorphic lattices, with the isomorphism given by conjugation of subgroups. As we shall see below, this stronger conjugacy condition often does seem to occur.

Now, if $N$ is a maximal normal subgroup, then $G/N$ is simple, so to verify the conjecture for all groups it suffices to verify that every finite simple group is generated by two elements of prime-power order. I managed to put together the following from results in the literature:

Theorem 2: All but possibly finite many of the finite simple groups are generated by two elements of prime power order. The two elements may be taken to be conjugate. Any exceptions are classical groups.

The conjugacy part will come from the following lemma:

Lemma: If a finite simple group $G$ is generated by $x$ and $g$, where $x$ has order 2, then $G = \langle g, g^x \rangle$.

Proof (of Lemma): Suppose not, and let $H = \langle g, g^x \rangle$. Then $H$ is easily seen to permute with $\langle x \rangle$, hence $H$ has index 2, contradicting simplicity of $G$. $\square$

We say a group is $(2,q)$-generated if it is generated by an element of order 2, together with an element of order $q$. To prove Theorem 2, it suffices to show that (modulo possible exceptions) every finite simple group is (2,$q$)-generated for some prime-power $q$. We go through the list of groups from the Classification Theorem.

1. Many finite simple groups are known to be (2,3)-generated, and this condition has been well-studied. ((2,3)-generated groups can be represented as quotients of $PSL_2(\mathbb{Z})$, explaining the high level of interest in this condition.)

A. Liebeck and Shalev in "Classical groups, probabilistic methods, and the (2,3)-generation problem" showed that, excluding the groups $PSp_4(q)$, all but finitely many of the classical groups are (2,3)-generated.

B. Di Martino and Cazzola in "$(2,3)$-generation of $PSp(4,q),\ q=p^n,\ p\neq 2,3$" showed that $PSp_4(q)$ is (2,3)-generated, except in characteristic 2 or 3.

C. Malle, and Lübeck and Malle in a series of papers culminating in "(2,3)-generation of exceptional groups" show that all exceptional groups of Lie type are (2,3)-generated, except for the Suzuki groups and $G(2)' \cong PSU_2(9)$.
(Nick Gill also referenced this paper in his comment to Alireza's answer.)

D. Woldar in "On Hurwitz generation and genus actions of sporadic groups" showed the sporadic groups to be (2,3)-generated, except for $M_{11}$, $M_{22}$, $M_{23}$, and $McL$

E. GA Miller proved in 1901 that the alternating groups $A_n$ are (2,3)-generated for $n \neq 6,7,8$.

2. $A_6, A_7, A_8$ are easily verified to all be (2,5)-generated.

3. Liebeck and Shalev in Proposition 6.4 of the same paper referenced above show that all but finitely many of the groups $PSp_4(2^n)$ and $PSp_4(3^n)$ are (2,5)-generated.

4. Suzuki, in his original paper on the topic, showed that the Suzuki groups are (2,4)-generated.

5. For the remaining sporadics, $M_{11}$ and $M_{22}$ are (2,4)-generated, while $M_{23}$ and $McL$ are (2,5)-generated. (The best place I found to read about this is in a series of papers on symmetric genus of groups. These papers are by various authors, frequently including Conder and Woldar.)

Theorem 2 follows from the list above, together with the Classification of Finite Simple Groups.

There is very likely a nicer approach to some of the simple group stuff than what I do above -- I'm far from an expert. (Maybe one of the real experts around here can remove the possible exceptions from the statement!!)

I found the slides linked here of Maxim Vsemirnov on (2,3)-generated groups very helpful for understanding the exceptions to (2,3)-generation.
In particular, he has a list of groups that are not (2,3)-generated on page 15 (which he conjectures to be complete). I've verified with GAP that all are (2,$q$)-generated for some prime $q$, except for the PSp's and $\Omega_8^+(2), P\Omega_8^+(3)$.

Corollary: The conjecture holds for any $G$ with a maximal normal subgroup $N$ such that $G/N$ is not one of the exceptional (small) classical groups in Theorem 2.

• Russ, thank you for this great answer. I will need some time to digest your latest update, but it would be wonderful if the answer turned up to be positive for all groups :-) – Gjergji Zaimi Jan 21 '14 at 7:28

If $G$ is a non-trivial finite group which can be generated by two non-trivial elements of prime power orders, then the answer to the question is affirmative. Let $\mathcal{G}$ be the set of all subgroups of $G$ and let $G=\langle x,y\rangle$, where $x$ and $y$ are non-trivial elements of $G$ of prime power oders. Assume $A=\lbrace H\in \mathcal{G} \;|\; x\in H, y\not\in H \rbrace$, $B=\lbrace H\in \mathcal{G} \;|\; x\not\in H, y\in H \rbrace$, $C=\lbrace H\in \mathcal{G} \;|\; x\not\in H, y\not\in H \rbrace$ and $E=\lbrace H\in \mathcal{G} \;|\; x\in H, y\in H \rbrace$. So $E=\{G\}$ and $|C|\geq 1$ as $\{1\}\in C$. It follows that either $|A|\leq |\mathcal{G}|/2$ or $|B|\leq |\mathcal{G}|/2$.

I think (if I am not wrong) that every (if not "most" of!) finite simple groups can be generated by two elements of prime power orders. So the question has affirmative answer for finite simple groups and $2$-generated groups of prime power orders.

UPDATE (17/January/2014): Inspired by Russ, one can show that the question has affirmative answer for finite groups having a maximal subgroup(not necessarily normal) of prime index. Let $M$ be a maximal subgroup of $G$ of prime index $p$. If there exists an element $g\in G\setminus M$ of $p$-power order, then $G=M\langle g\rangle$. The latter holds since $$|M\langle g\rangle|=\frac{|M||g|}{|M\cap \langle g\rangle|}.$$ Now the map $H\mapsto H\cap M$ is injective for all subgroups $H$ containing $g$. So we may assume that all $p$-power order elements are in $M$ which implies that $M$ contains all Sylow $p$-subgroups of $G$, a contradiction. This slightly improves the result of Russ, as we do not assume $M$ is normal in the group.

• To understand the situation mentioned in the final paragraph, refer to Lubeck & Malle, (2,3)-generation of exceptional groups, in which Corollary 1.3 states: Let $G$ be a non-abelian finite simple group not equal to $Sp_4(2^n)$, $Sp_4(3^n)$ or ${^2B_2}(2^{2n+1})$, then, up to a finite number of exceptions, $G$ is generated by an element of order 2 and an element of order $3$... – Nick Gill Jan 10 '14 at 11:59
• ... If you want to know what the list of exceptions is, then you should refer to the references given near the statement of Corollary 1.3. So I'm guessing that @Alireza's statement is true (all FSGs are generated by two elements of prime power order) - one just needs to prove it for the three infinite families just mentioned, plus the finite list. – Nick Gill Jan 10 '14 at 12:01
• ... One can probably deal with the three infinite families directly, or using ideas in On triangle generation of finite groups of Lie type by Claude Marion. – Nick Gill Jan 10 '14 at 12:18


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# How do we know quantum entanglement works no matter the distance?

It is said quantum entanglement works regardless of distance. 2 particles can be entangled and information is shared instantaneously, even if they are lightyears away from each other.

But how do we know this still works with such a vast distance between both particles? I can image experiments in a lab, or even on opposites sides of the planet, but not with light years between them. So how do we know?

• Related: physics.stackexchange.com/q/3158/2451 and links therein. – Qmechanic Nov 7 '18 at 15:19
• "information is shared instantaneously" That is outright rejected by many QM interpretations, and at the very least considered misleading by most. – Acccumulation Nov 7 '18 at 16:45
• I've removed a highly-upvoted comment that should have been an answer, and about six replies to it. Please use comments only to suggest improvements on the post they are attached to. – rob Nov 8 '18 at 17:54
• physics.stackexchange.com/questions/104050/… – alanf Jun 20 at 13:51

Distance is not a relativistic invariant. Let A be the event marked by my beginning to write this answer, and B the event where I finish writing it. In the frame of reference of my desk, the distance between A and B is zero, but in the frame of reference of an observer moving at $$0.9999999c$$ relative to the earth, the distance between A and B is millions of kilometers.

If the separation between two events is spacelike, then all we can say is that in other frames, they are also spacelike-separated. The distance between them can be made arbitrarily large by picking an appropriate frame of reference.

So all we need is one experiment showing that entanglement works over spacelike intervals, and it follows that it works over arbitrarily large distances. One such experiment is Guerreiro et al., http://arxiv.org/abs/1204.1712 .

• In your first example the separation between the events is timelike. And you argue that the distance between them can be made arbitrarily large by observing them from a suitable reference frame. That part sounds right. But how does it follow from that that the same is true for events with a spacelike separation? Isn't length contraction going to cause it to be the other way around such that the spacelike separation can be made an arbitrarily small distance, but not larger? – kasperd Nov 7 '18 at 16:52
• Wait, what? That doesn't appear to answer OP at all, just add more detail to his question. Guerreiro shows a statistically significant distribution at whatever distance he measured at; OP asks essentially "how can we know that distribution is invariant with distance?" – fectin Nov 7 '18 at 19:02
• @kasperd: The timelike and spacelike cases are not the same. I discussed the timelike case just as an illustration of the fact that distances are not invariant. For the spacelike case, the fact that the distance can be made arbitrarily large follows by writing down the Lorentz transformation. When the events are simultaneuos in some frame, then their separation in another frame is greater by a factor $\gamma$, which can be made arbitrarily large. – Ben Crowell Nov 7 '18 at 21:04
• I argue, in the spirit of many of the other answers posted here, that this question is fundamentally about empirical testing of models. Viewed in this way this answer is not sufficient. This answer says that if you believe the predictions of special relativity and quantum mechanics up to very high velocities then current experiments are already consistent with quantum entanglement "working" over arbitrarily large distances. But the spirit of this question doubts the mathematical theories until empirical evidence is provided... – jgerber Nov 7 '18 at 23:40
• @jgerber: There has not yet been empirical evidence that quantum mechanics and special relativity should work as expected at these large velocity and distance scales What velocities and scales do you have in mind? Lorentz invariance has been tested to incredibly high precision at many different velocities and scales. An example at velocities very close to c would be the CERN FTL neutrino debacle, which was later resolved and found to be in agreement with Lorentz invariance. – Ben Crowell Nov 7 '18 at 23:45

The term "quantum entanglement exists" is a subset of the term "there exists a single quantum mechanical wavefunction describing the system".

If you know the wavefunction, i.e. have a mathematical description of it, the theory of quantum mechanics has constraints on quantum numbers and their conservation, so quantum numbers are "entangled" because there exists a single wavefunction describing the system, as ayc also states.

Let us take a simple example, the pi0 decay. It decays into two photons/gammas. The pi0 has spin zero. From conservations of quantum numbers ( in this case angular momentum conservation) in the decay wavefunction, one of the photons must have spin +1 and the other -1 the direction being the direction of their motion.

Once the decay has happened and there are no interactions, the spins are unknown for the experimenter, but they are defined by the wavefunction.

If I measure one in the lab and the other reach Alpha Centauri, I know its spin by having measured the one in the lab and using angular momentum conservation.

and information is shared instantaneously, even if they are lightyears away from each other

The information is not shared by the photons, the information is in the head of the experimenter, who sits in the lab.

How do we know this is true? Because we have validated quantum mechanical theory with innumerable experiments, trust mathematics, and extrapolate, all the information here in the lab.

• Entanglement is about more than just a common wavefunction describing the system; the point is the state cannot be decomposed as a tensor product over Hilbert spaces for the subsystems. If it can, then there's no entanglement. For example, $(|\uparrow\downarrow\rangle+|\downarrow\uparrow\rangle)/\sqrt{2}$ is entangled, but $|\uparrow\downarrow\rangle$ is not. – J.G. Nov 7 '18 at 20:26
• @J.G. I have given an experimental example. quantum number conservation in the observable outcome holds whatever the mathematical formulation of the wavefunction. "Entanglement" describes an extra constraint using conservation laws . How conservation laws are imposed on wavefunctions is a mathematical problem. – anna v Nov 8 '18 at 7:57
• The last paragraph of this answer could have been said about Newtonian mechanics some two hundred years after the Principia. – Martin Argerami Nov 8 '18 at 12:56
• @MartinArgerami . Sure, it is only measurements that lead us by the nose to quantum mechanics. If new, more accurate (or whatever) measurements invalidate quantum mechanics, it will become , instead of the underlying layer of all classical models, one of the layers with its region of validity. It is "turtles all the way down". – anna v Nov 8 '18 at 13:04
• But there is still the "spooky" part when you devise the experiment such that the method of observation determines which direction the spin is observed in. Until that observation is made, what can the wave function say other than the other particle has the opposite of an unknown spin? – Michael Nov 8 '18 at 23:00

The short answer is "No". In fact, the basic fact of life in science is that we never know, so to speak: We have a theory, it makes predictions, which we test, and if the test fails, then we know for certain that the prediction was wrong, roughly speaking. If it doesn't fail, we don't actually know that we were right - but we can feel justified in believing that we are. So, strictly speaking, even if we have experienced $$10^7$$ times that a rock falls down, we still don't know that it will do so next time.

Of course, some - most - would say that this is unnecessary sophistry, and I would tend to agree in most cases, within reasonable limits. We have confirmed in experiments that entanglement appears to work over distances of several hundred km, I believe, but claiming that it will work over lightyears is still very much speculation or wishful thinking, IMO.

===EDIT===

In view of the comments so far, I think I should perhaps clarify my position. The fundamental insight that is formalized in what we call the scientific method is, that we can never achieve absolute certainty about the validity of theory through experimentation - the only absolute truth comes in the form of falsification; a successful theory is one in which we have consistently failed to falsify the predictions. This is something we always must keep in mind when we extrapolate based on our theory - just think of Newton's theory of gravity: it works wonderfully for (nearly) all practical purposes, we use it when we hurl our pebble sized satellites out and around several planets and still manage to hit the target most times - but it failed to fully explain the precession of the perihelion of Mercury.

One day - soon, one hopes - we will find situations in which both GR and QM fail. Then we will take the next step forward.

===EDIT2===

I see that my answer got voted down - that's OK with me, but please, if you feel that way, add a comment, so I can learn if I made a mistake, was rude or anything else. I won't hold it against you.

• I was with you until the last sentence. It is neither. It is a consequence of the theory that it does. We have established that the theory has merit. It is as true as entanglement at exactly 2.050421 cm. Untested, but we assume it to be so untill we find reason to improve the theory. – Stian Yttervik Nov 7 '18 at 17:08
• I would upvote this answer 100 times if I could. I would agree that extrapolation of anything from several hundred kilometers to lightyears (or billions of light years) is a significant extrapolation beyond the conditions under which the theory has been experimentally validated that should significantly reduce our confidence in the result still holding. It doesn't mean the theory is wrong under longer distances, just that we really don't know as we have no evidence for or against at such distances. And intellectual humility requires us to acknowledge this. – bob Nov 7 '18 at 22:00
• I'm glad to inform you there are billions of people (including me) who know that rocks fall down. I am sorry you've got distracted from reality at some point and accepted axiomatic definition of knowledge. It is only useful in mathematics and religion, but not in natural sciences, for they lack axioms. – kubanczyk Nov 7 '18 at 23:17
• The answer would be "we don't" not "no", since the question is "how do we know?" – Kat Nov 8 '18 at 4:00
• "We all know that rocks fall down", until we discover an unaccelerated frame. – amI Nov 8 '18 at 8:20

It is said quantum entanglement works regardless of distance. 2 particles can be entangled and information is shared instantaneously, even if they are lightyears away from each other.

To be perfectly clear: entanglement cannot be used to communicate, no matter how long or how short the spatial separation. This is explained in more depth in this thread, but the basic principle is this:

• Suppose you have two entangled particles, say, spins in the up-down entangled state $$|{\uparrow}{\downarrow}\rangle + |{\downarrow}{\uparrow}\rangle$$.
• Suppose further that you measure the first particle along the up-down basis, and you get the result $$s$$. Then that will also project the second spin onto the state $$-s$$.
• Furthermore, depending on your interpretation of QM and on your overall stance on its foundational issues, it is possible to interpret this as an instantaneous action on the second qubit regardless of the distance between them.
• However: you have no control over whether you will get the result $$s={\uparrow}$$ or $$s={\downarrow}$$, so you have no control over the "message" that gets sent.

This can be further formalized in the No-Communication Theorem, which basically says that if the system obeys the rules of quantum mechanics, then no shenanigans that you could conceivably pull along these lines can be used to communicate faster than light. Quantum mechanics is a fully causal theory, i.e. there are no scenarios where the effects of a given cause can be observed outside of its future light cone.

On the other hand, it is possible to interpret the situation described in the bullet points above as the particles "communicating" with each other in a non-causal faster-than-light way and then "conspiring" to make that FTL communications channel unavailable to any macroscopic experiment ─ but the second half of that combo is crucial and can never be left out. Some people are OK with that, but I find it deeply unsatisfying as a philosophical position.

Also, to be perfectly clear, the protocol I described above does not actually rely on entanglement, and it is susceptible to a "Bertlmann's socks" explanation where you just put an $$\uparrow$$ and a $$\downarrow$$ spins in unlabeled boxes and ship them off. However, this kind of Local Hidden Variable explanation is insufficient to explain the full set of measurement results possible using entangled states: hidden-variable theories are constrained by Bell's theorem to satisfy a set of inequalities on the types and amounts of correlations that they can show, and there are multiple experiments showing that quantum mechanical systems routinely break those inequalities.

However, with all of those precisions in place:

It is said quantum entanglement works regardless of distance.

Yes, this is correct: as far as we know, all of the above works regardless of the spatial separation between the particles.

But how do we know this still works with such a vast distance between both particles? I can image experiments in a lab, or even on opposites sides of the planet, but not with light years between them. So how do we know?

We have no evidence for this beyond the fact that the theory has worked to explain every experiment we've asked it to explain on all the scales that we've managed to construct working tests of the theory, and the fact that all of our astronomical observations of physics from places that are inaccessible to us (from the presence of helium in the Sun to the thermal spectrum of the cosmic microwave background) can be explained using the same laws of physics that we test using earth-bound laboratories.

It is perfectly possible that any given law of physics will break outside of the range where we've tested it, which is why we keep testing them in newer and bigger regimes; indeed if we do find such deviations, they would be much more interesting than the finding that it's all the same everywhere. However, until and unless we do find such a result, there is no evidence to suggest that it's not the case.

Over time experimentors have been succesful in increasing distance over which entanglement could be demonstrated, up to many kilometers.

Let me make a comparison with the history of electromagnetism:

When Maxwell published the theory of electromagnetism we now know as Maxwell's equations one of implications of the theory was that it should be possible to generate a propagating electromagnetic wave, and receive that electromagnetic wave. Heinrich Hertz set out to verify that. The emitter he constructed was kicked to carry a very brief, very strong alternating current by having a spark jumping over. The strong alternating current caused an electromagnetic wave. The receiver was located pretty much next to the emitter. The receiver picked up the electromagnetic wave.

Of course, the distance that this first received wave had travelled was very underwhelming, just a meter or so. But the thing was: there was nothing in the theory to suggest any limit to how far electromagnetic waves can propagate. Indeed, it was soon discovered that there is a range of frequencies that reflects against the ionosphere, and so within a very short time after Hertz' demonstration there were broadcasting antannas broadcasting a signal that could be received all over the world.

Something similar applies in the case of quantum entanglement. The first demonstrations were over relatively short distances, the dimensions of a single bench setup. But to elementary particles even distances of centimeters are vast. So: once quantum entanglement was demonstrated on a single bench setup the picture was pretty clear: nothing in the theory to suggest any cut-off distance, and a demonstration over a distance of centimeters is already a demonstration over a vast distance.

Of course, the longer the distance the smaller the yield. The longer the transit time the larger the probability of losing the entanglement along the way. So to have an acceptable number of confirmations per second you have to generate and send larger populations of entangled particles. But we have every reason to expect that there is no distance beyond which the yield goes to zero.

• Interestingly, Maxwell summarised all the equations into a single quaternion form (Art 623, IIRC), which essentially is the 4d space time contraction. It depends on whether one believes the metric comes first, or the 4d space. – Philip Oakley Nov 8 '18 at 11:36

The answer is "we never know." There is no way to know for certain whether the laws of physics operate the same on the other side of Jupiter. Of course, the observations we have made beyond Jupiter have all supported the conclusion that the laws of physics are the same everywhere, but we'll never know for sure.

Accordingly, all we need to really do is have "empirically sufficient" evidence that the effects work at long distances. For that, we note that distances really don't seem to be the major limiting factor. The more interesting limit is the speed of light. If entanglement does a good job of modeling this, then it's a useful theory.

Speed is easier than distance. If I can time two measurements to be mere nanoseconds apart, my experiment can take place within a room and still exhibit entanglement like effects which cannot be explained via transmissions that travel at the speed of light. Trivially, if the measurement is triggered by a common signal half way between the parts of my experiment, we can see that I can interact with an entangled pair in ways which are much faster than light-speed communication would permit because there's no time for round trips.

So is that proof that the effects work over the distance of light years? No. If I did it in a 10ft room, it proves nothing about whether entanglement works in an 11ft room. But all of our experiments so far have shown that distance has no effect at all.

Unfortunately, I don't recall all the details, but it's the experiments that were performed in response to the EPR Paradox, which forced us to accept the non-locality of an entangled state.

These experiments basically create some pairs of entangled particles, mostly photons, send them some distance apart, and measure both in a random way. It can be shown that the precise statistics of these measurements require that one measurement actually influences the other. This result is, what quantum mechanics predicted, and what Einstein, Podolsky and Rosen found absurd (which is why it's called a paradox). Afaik, the consequence of these results is, that we must either accept the non-locality of entangled states, or we must question our notion of physical reality.

These experiments have been performed with various distances, I don't know what the current record distance is. But it's basically pointless to argue about distances once the two measurements are far enough apart that light cannot reach the location of the other measurement before that measurement actually takes place. That's easy for photons to achieve: If the entangled photons go off in opposite directions, and you place a measuring station just after a distance of a single meter, the measurements will take place at the same instant in the lab's frame of reference, but two meters apart. That's enough to rule out any light-speed communication between the entangled particles.

• This might be of interest: Extending Wheeler's delayed-choice experiment to Space – Dhruv Saxena Nov 8 '18 at 9:41
• For example, look up Ghost Imaging, which uses twin photons to detect objects seen by one photon by observing the effect on the other. en.wikipedia.org/wiki/Ghost_imaging The majority of the problem is deciding if we have a 4d space time (all points unique) or that the 4d measurement is just a projection from a different dimensionality (e.g. all points on the light cone being equal, when within a very small origin distance - the entanglement distance). – Philip Oakley Nov 8 '18 at 11:33

The premise that "quantum entanglement works" at all is mistaken - there is no action involved - as is the one that "information is shared" at all (regardless of whether it's instantaneous or not). Quantum entanglement is not in any way a vector for transmission of information, and it is really no deeper than taking two balls, one red and one blue, putting each in a box without looking at them, separating the boxes, opening one box, seeing a red ball, and concluding that there is a blue ball in the other box.

When I originally posted the above analogy as a comment, an objection was raised that it's misleading and implies that quantum entanglement is explainable with (local) hidden variables. That is of course impossible by Bell's Theorem as well as many experimental results, and the "no deeper than" analogy should not be interpreted as implying there's any hidden variable. If you really want, you could imagine setting it up as a Schrödinger's cat style thought experiment, distributing the balls in an (approximately) isolated part of the system according to some true random process that's not observed outside. But this is aside from the point.

The point is that, like in the ball analogy, quantum entanglement is about drawing conclusions from a relationship where, conditioned on prior observation of the setup, there is a correlation between certain observables on the outcome.

This is all a consequence of a general principle that, in QM, there's really nothing deeper than conditional probabilities on a giant model that would be meaningless to make any predictions without conditioning on observations.

I consider the question of how far quantum entanglement works this way:

For as long as entanglement is considered a mysterious Quantum phenomenon and discussed in terms of wave functions or other such mathematical descriptions it is hard to think of it as something that can survive great distances across space (and time).

However, if you go right back to the root of it - we discovered entanglement when we were experimenting with correlated properties. Properties we know to be correlated because - if they were not - then conserved quantities - like angular momentum - would not be conserved. Now, while GR leaves some loopholes and the universe can make energy - there are no indications in the universe that conservation is strictly local and energy and momentum and whatever other else conserved properties can be trivially broken if you just add a bit more space to the middle of the experiment.

i.e. any distance we can show is too short to cause a symmetry violation is also a distance we can expect Quantum entanglement to work over.

Personally I would like to say the statement is not exact or at least not clear so far. If generally we believe the structure of spacetime is closely related with entanglement, then we should be very careful to talk about 'distance/time/velocity' in a case when entanglement is involved.

My opinion is that when we are playing with two entangled particles, ER=EPR seems to claim that the 'normal' spacetime structure is broken (or at least perturbed) by the entanglement. Under this situation, it's not very strict to say 'the two particles are far away' since the tiny wormhole between them might just provide a shortcut so that their 'distance' is 0. In fact, the observed fact that entangled particles show an instantaneous correlation seems to claim that they ARE the same point.

Finally, I believe before we can complete the quantum gravity to integrate QM and GR, there should be no definite answer to this question.

## protected by Emilio PisantyNov 9 '18 at 17:27

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text:
Online Seminar "Geometric Analysis"

Since the coronavirus pandemic forces us to cancel workshops, conferences and seminars, we need new ways to stay connected. So lets make the best out of it and join forces to build an online seminar able to compete with all our local research seminars.

How it (should) work

• We will use zoom to make this seminar work. You can join through a link we will send you through the mailing list you have to subscribe to before.
• Meeting id: 693-964-597. You will get the password through the mailinglist.
• Please be patient, in case the system breaks down

Mailing list

There is a mailing list which announces the speakers for each week and shares the password to the meeting. To subscribe, send a blank email with the subject "subscribe osga YOUREMAILADRESS YOURNAME" to listserver@sbg.ac.at. You can unsubscribe in a similar way.

List of Upcoming Talks

13.10.2020, 19:00 (CEST)
Renan Assimos (MPI Leipzig, Germany) On a spherical Bernstein theorem by B. Solomon
Abstract:
Joint work with J. Jost: A result of B.Solomon (On the Gauss map of an area-minimizing hypersurface. 1984. Journal of Differential Geometry, 19(1), 221-232.) says that a compact minimal hypersurface $M^k$ of the sphere $S^{k+1}$ with $H^1(M)=0$, whose Gauss map omits a neighborhood of an $S^{k−1}$ equator, is totally geodesic in $S^{k+1}$. In this talk, I will present a new proof strategy for Solomon's theorem which allows us to obtain analogous results for higher codimensions. If time permits, we sketch the proof for codimension 2 compact minimal submanifolds of $S^{k+1}$.
20. 10. 2020, 19:00 (CEST)
Daniel Campbell (University of Hradec Kralove, Czeck Republic) Pathological Sobolev homeomorphisms in GFT and NE
Abstract:
Sobolev homeomorphisms are the natural choice for minimization problems in non-linear elasticity. For the regularity of these problems it would be useful to be able to approximate these maps by smooth homeomorphisms in their corresponding Sobolev space (the so-called Ball-Evans problem). We construct a pair of homeomorphisms for which is impossible simultaneously solving the Hajlasz problem. That is we construct a Sobolev homeomorphism equalling identity on the boundary of a cube but with negative Jacobian almost everywhere.
27. 10. 2020, 19:00 (CEST)
Simon Brendle (Columbia University, USA) The isoperimetric inequality for minimal surfaces
03. 11. 2020, 19:00 (CEST)
Henrik Matthiesen New minimal surfaces from shape optimization
10. 11. 2020, 19:00 (CEST)
Julia Menzel (University Regensburg, Germany) Boundary Value Problems for Evolutions of Willmore Type
01. 12. 2020, 19:00 (CEST)
Marius Müller (University Ulm, Germany)
08. 12. 2020, 19:00 (CEST)
Ursula Ludwig (University Duisburg, Germany) An Extension of a Theorem by Cheeger and Müller to Spaces with Isolated Conical Singularities
Abstract:
An important comparison theorem in global analysis is the comparison of analytic and topological torsion for smooth compact manifolds equipped with a unitary flat vector bundle. It has been conjectured by Ray and Singer and has been independently proved by Cheeger and Mu ̈ller in the 70ies. Bismut and Zhang combined the Witten deformation and local index techniques to generalise the result of Cheeger and Mu ̈ller to arbitrary flat vector bundles with arbitrary Hermitian metrics. The aim of this talk is to present an extension of the Cheeger-Mu ̈ller theorem to spaces with isolated conical singularities by generalising the proof of Bismut and Zhang to the singular setting.
12. 01. 2021, 19:00 (CEST)
Rupert Frank
19. 01. 2021, 19:00 (CEST)
Wilderich Tuschmann

Complete List of Talks

03. 04. 2020, 19:00 (CEST)
Philipp Reiter (University Halle, Germany) A bending‐twist model for elastic rods
07. 04. 2020, 19:00 (CEST)
Remy Rodiac (University Paris-Saclay, France) Inner variations and limiting vorticities for the Ginzburg-Landau equations video
14. 04. 2020, 19:00 (CEST)
Daniel Steenebrügge (RWTH Aachen, Germany) A speed preserving Hilbert gradient flow for generalized integral Menger curvature video
21. 04. 2020, 19:00 (CEST)
Andrew Sageman-Furnas (TU Berlin, Germany) Navigating the space of Chebyshev nets video
21. 04. 2020, 20:00 (CEST)
Siran Li (Rice University, USA) Isometric Immersions of Riemannian Manifolds into Euclidean Spaces, Revisited video
28. 04. 2020, 19:00 (CEST)
Bastian Käfer (RWTH Aachen, Germany) A Möbius invariant energy for sets of arbitrary dimension and codimension video
05. 05. 2020, 19:00 (CEST)
Huy The Nguyen (Queen Mary University London, United Kingdom) High Codimension Mean Curvature Flow and Surgery video
05. 05. 2020, 20:00 (CEST)
Elena Mäder-Baumdicker (TU Darmstadt, Germany) The Morse index of Willmore spheres and its relation to the geometry of minimal surfaces
12. 05. 2020, 19:00 (CEST)
Jesse Ratzkin (University Würzburg) On constant Q-curvature metrics with isolated singularities and a related fourth order conformal invariant video
19. 05. 2020, 19:00 (CEST)
Volker Branding (University Vienna, Austria) Higher order generalizations of harmonic maps video
26. 05. 2020, 19:00 (CEST)
Katharina Brazda (University Vienna, Austria) The Canham-Helfrich model for multiphase biomembranes video
02. 06. 2020, 19:00 (CEST)
Lynn Heller (University Hannover) Area Estimates for High genus Lawson surfaces via DPW video
09. 06. 2020, 19:00 (CEST)
Sven Pistre (RWTH Aachen, Germany) The Radon transform and higher regularity of surfaces minimising a Finsler area video
16. 06. 2020, 19:00 (CEST)
Marc Pegon (University Paris-Diderot, France) Partial regularity for fractional harmonic maps into spheres video
16. 06. 2020, 20:00 (CEST)
Myfanwy Evans (University Potsdam, Germany) Periodic tangling
Abstract:
This talk will introduce the use of geometric ideas in the characterisation and analysis of tangled biophysical systems. It will introduce the construction of idealised tangled structures using ideas of both symmetry and homotopy of tangled lines on surfaces. These structures provide an extensive set of tangling motifs for the exploration of the behaviour of tangled microstructures in liquids, and I will show preliminary results working towards this goal, including an example of the geometry-driven swelling of human skin cells.
23. 06. 2020, 19:00 (CEST)
Miles Simon (University Magdeburg, Germany) On the regularity of Ricci flows coming out of metric spaces. video
Abstract:
Joint work with Alix Deruelle, Felix Schulze We consider solutions to Ricci flow defined on manifolds M for a time interval $(0,T)$ whose Ricci curvature is bounded uniformly in time from below, and for which the norm of the full curvature tensor at time $t$ is bounded by $c/t$ for some fixed constant $c>1$ for all $t in (0,T)$. From previous works, it is known that if the solution is complete for all times $t>0$, then there is a limit metric space $(M,d_0)$, as time t approaches zero. We show : if there is a open region $V$ on which $(V,d_0)$ is *smooth*, then the solution can be extended smoothly to time zero on $V$.
30. 06. 2020, 19:00 (CEST)
Peter Topping (University Warwick, United Kingdom) Uniqueness of limits in geometric flows video
Abstract:
Quite often when considering long-time behaviour of geometric flows, or considering blow-ups of singularities in geometric PDE, we extract limits using soft compactness arguments. For example, a flow might easily be seen to converge to a limit at a *sequence* of times converging to infinity. The more subtle question is then whether the flow converges as time converges to infinity, without having to restrict to a sequence of times. I will outline some of the issues that arise in this subject, focussing on gradient flows for the harmonic map energy, and sketch some recent work with M.Rupflin and J.Kohout.
07. 07. 2020, 19:00 (CEST)
Ruben Jakob (Technion, Israel) Generic full smooth convergence of the elastic energy flow in the 2-sphere video
Abstract:
The speaker is going to present his recent investigation of the Moebius invariant Willmore flow'' (MIWF) in the 3-sphere and of some particular version of the elastic energy flow'' (EEF) in the 2-sphere. We will discuss the interaction between these two geometric flows via the Hopf fibration and the resulting possibility to transfer particular insights about the EEF'' to the MIWF'', and vice versa special insights about the MIWF'' back to the EEF''. A big motivation for this parallel investigation is the announced proof (by the speaker) of the ''generic full smooth convergence'' of the EEF'' in the 2-sphere.
14. 07. 2020, 19:00 (CEST)
Christian Bär (University Potsdam, Germany) Counter-intuitive approximations video
Abstract:
The Nash-Kuiper embedding theorem is a prototypical example of a counter-intuitive approximation result: any short embedding of a Riemannian manifold into Euclidean space can be approximated by *isometric* ones. As a consequence, any surface can be isometrically $C^1$-embedded into an arbitrarily small ball in $R^3$. For $C^2$-embeddings this is impossible due to curvature restrictions. We will present a general result which will allow for approximations by functions satisfying strongly overdetermined equations on open dense subsets. This will be illustrated by three examples: real functions, embeddings of surfaces, and abstract Riemannian metrics on manifolds.
21. 07. 2020, 19:00 (CEST)
Carla Cederbau (University Tübingen, Germany) On CMC-foliations of asymptotically flat manifolds video
Abstract:
In 1996, Huisken and Yau proved existence of foliations by constant mean curvature (CMC) surfaces in the asymptotic end of an asymptotically Euclidean Riemannian manifold. Their work has inspired the study of various other foliations in asymptotic ends, most notably the foliations by Willmore surfaces (Lamm, Metzger, Schulze) and by constant expansion/null mean curvature surfaces in the context of asymptotically Euclidean initial data sets in General Relativity (Metzger). I will present a new foliation by constant spacetime mean curvature surfaces (STCMC), also in the context of asymptotically Euclidean initial data sets in General Relativity (joint work with Sakovich). The STCMC-foliation is well-suited to define a notion of total center of mass in General Relativity.
28. 07. 2020, 19:00 (CEST)
Nadine Große (University Freiburg, Germany) Boundary value problems on singular domains: an approach via bounded geometries video
Abstract:
In this talk, we consider boundary value problems on domains with non smooth boundaries. We approach this problem by transferring it to non-compact manifolds with a suffiently nice geometry -- the bounded geometry. This gives a more general framework that allows to handle Dirichlet (or Dirichlet-Neumann mixed) boundary value problems for domains with a larger class of singularities on the boundary and gives a nice geometric interpretation. This is joint work with Bernd Ammann (Regensburg) and Victor Nistor (Metz).
04. 08. 2020, 19:00 (CEST)
Melanie Rupflin (University Oxford, UK) Łojasiewicz inequalities near simple bubble trees for the $H$ surface equation
Abstract:
In this talk we discuss a gap phenomenon for critical points ofthe $H$-functional on closed non-spherical surfaces when $H$ is constant, and in this setting furthermore prove that sequences of almost critical points satisfy Łojasiewicz inequalities as they approach the first non-trivial bubble tree. To prove these results we derive sufficient conditions for Łojasiewicz inequalities to hold near a finite-dimensional submanifold of almost-critical points for suitable functionals on a Hilbert space. The presented results are joint work with Andrea Malchiodi and Ben Sharp.
11. 08. 2020, 19:00 (CEST)
Julian Scheuer (University Cardiff, UK) Concavity of solutions to elliptic equations on the sphere video
Abstract:
An important question in PDE is when a solution to an elliptic equation is concave. This has been of interest with respect to the spectrum of linear equations as well as in nonlinear problems. An old technique going back to works of Korevaar, Kennington and Kawohl is to study a certain two-point function on a Euclidean domain to prove a so-called concavity maximum principle with the help of a first and second derivative test. To our knowledge, so far this technique has never been transferred to other ambient spaces, as the nonlinearity of a general ambient space introduces geometric terms into the classical calculation, which in general do not carry a sign. In this talk we have a look at this situation on the unit sphere. We prove a concavity maximum principle for a broad class of degenerate elliptic equations via a careful analysis of the spherical Jacobi fields and their derivatives. In turn we obtain concavity of solutions to this class of equations. This is joint work with Mat Langford, University of Tennessee Knoxville.
18. 08. 2020, 19:00 (CEST)
Richard Bamler (University of California Berkeley, USA) Ricci flow in higher dimensions video
Abstract:
I will present new results concerning Ricci flows in higher dimensions
25. 08. 2020, 19:00 (CEST)
Max Engelstein (University of Minnesota Winding for Wave Maps
Abstract:
Wave maps are harmonic maps from a Lorentzian domain to a Riemannian target. Like solutions to many energy critical PDE, wave maps can develop singularities where the energy concentrates on arbitrary small scales but the norm stays bounded. Zooming in on these singularities yields a harmonic map (called a soliton or bubble) in the weak limit. One fundamental question is whether this weak limit is unique, that is to say, whether different bubbles may appear as the limit of different sequences of rescalings. We show by example that uniqueness may not hold if the target manifold is not analytic. Our construction is heavily inspired by Peter Topping's analogous example of a winding bubble in harmonic map heat flow. However, the Hamiltonian nature of the wave maps will occasionally necessitate different arguments. This is joint work with Dana Mendelson (U Chicago).
01. 09. 2020, 19:00 (CEST)
Peter McGrath (North Carolina State University) Quantitative Isoperimetric Inequalities on Riemannian Surfaces video
Abstract:
In this talk, we introduce a scattering asymmetry which measures the asymmetry of a domain by quantifying its incompatibility with an isometric circle action. We prove a quantitative isoperimetric inequality involving the scattering asymmetry and characterize the domains with vanishing scattering asymmetry by their rotational symmetry. We also give a new proof of the sharp Sobolev inequality for Riemannian surfaces which is independent of the isoperimetric inequality. This is joint work with J. Hoisington.
08. 09. 2020, 19:00 (CEST)
Ryan Alvarado (Amherst College) A characterization of the Sobolev embedding theorem in metric-measure spaces.
Abstract:
Historically, the Sobolev embedding theorem on domains has played a key role in establishing many fundamental results in the area of analysis and it is well known that the geometry of the underlying domain is intimately linked to the availability of these embeddings. In fact, certain geometrical characterizations of domains which support Sobolev embeddings have been obtained in the Euclidean setting, albeit in the plane. In this talk, we will revisit these embedding theorems in the more general context of metric-measure spaces and discuss some recent work which identifies a measure theoretic condition that is both necessary and sufficient to ensure their veracity. A measure characterization of Sobolev extension domains in the metric setting as well as applications of our methods to spaces supporting $p$-Poincaré inequalities will also be discussed. This talk is based on joint work with Przemysław Górka (Warsaw University of Technology), Piotr Hajłasz (University of Pittsburgh).
22.09.2020, 19:00 (CEST)
Fritz Hiesmayr (University College London, UK) A rigidity theorem for the Allen-Cahn equation in $S^3$
Abstract:
We present a recent rigidity theorem for the Allen-Cahn equation in the three-sphere: critical points with Morse index are symmetric and vanish on a Clifford torus. One key ingredient is a novel Frankel-type property we establish for the nodal sets of any two distinct solutions: they intersect if they are connected. This in fact holds in all manifolds with positive Ricci curvature. Time permitting we will discuss additional rigidity results in higher-dimensional spheres.
29. 09. 2020, 19:00 (CEST)
Anna Siffert (University Münster, Germany) Constructing explicit p-harmonic functions
Abstract:
The study of $p$-harmonic functions on Riemannian manifolds has invoked the interest of mathematicians and physicists for nearly two centuries. Applications within physics can for example be found in continuum mechanics, elasticity theory, as well as two-dimensional hydrodynamics problems involving Stokes flows of incompressible Newtonian fluids. In my talk I will focus on the construction of explicit $p$-harmonic functions on rank-one Lie groups of Iwasawa type. This joint work with Sigmundur Gudmundsson and Marko Sobak.
06. 10. 2020, 19:00 (CEST)
Jonas Hirsch (University Leipzig, Germany) On the regularity of area minimzing currents mod(p)
Abstract:
Joint work with C. De Lellis, A Marches and S. Stuvard In this talk I would like to give a glimpse on the regularity of area minimzing currents mod(p). Motivation: If one considers real soap films one notice that from time to time one can find configurations where different soap films join on a common piece. One possibility to allow this kind of phenomenon is to consider flat chains with coefficients in $mathbb Z_p$. For instance for $p = 2$ one can deal with unoriented surfaces, for $p = 3$ one allows triple junctions. Considering area minimzing currents within this class the aim is to give a bound on the Hausdorff dimension of the singular set sing(T) in the interior. These are alle points where the precise representative of the minimiser T is not even locally supported on a piece of a $C^{1,lpha}$ regular surface.
After a short introduction into general theory of currents mod(p), I will give you glimpse on the previously known results and on our new bound on the Hausdorff dimension of the set. If time permits I will give a short outlook of what we would be the expected result.
13.10.2020, 19:00 (CEST)
Renan Assimos (MPI Leipzig, Germany) On a spherical Bernstein theorem by B. Solomon
Abstract:
Joint work with J. Jost: A result of B.Solomon (On the Gauss map of an area-minimizing hypersurface. 1984. Journal of Differential Geometry, 19(1), 221-232.) says that a compact minimal hypersurface $M^k$ of the sphere $S^{k+1}$ with $H^1(M)=0$, whose Gauss map omits a neighborhood of an $S^{k−1}$ equator, is totally geodesic in $S^{k+1}$. In this talk, I will present a new proof strategy for Solomon's theorem which allows us to obtain analogous results for higher codimensions. If time permits, we sketch the proof for codimension 2 compact minimal submanifolds of $S^{k+1}$.
20. 10. 2020, 19:00 (CEST)
Daniel Campbell (University of Hradec Kralove, Czeck Republic) Pathological Sobolev homeomorphisms in GFT and NE
Abstract:
Sobolev homeomorphisms are the natural choice for minimization problems in non-linear elasticity. For the regularity of these problems it would be useful to be able to approximate these maps by smooth homeomorphisms in their corresponding Sobolev space (the so-called Ball-Evans problem). We construct a pair of homeomorphisms for which is impossible simultaneously solving the Hajlasz problem. That is we construct a Sobolev homeomorphism equalling identity on the boundary of a cube but with negative Jacobian almost everywhere.
27. 10. 2020, 19:00 (CEST)
Simon Brendle (Columbia University, USA) The isoperimetric inequality for minimal surfaces
03. 11. 2020, 19:00 (CEST)
Henrik Matthiesen New minimal surfaces from shape optimization
10. 11. 2020, 19:00 (CEST)
Julia Menzel (University Regensburg, Germany) Boundary Value Problems for Evolutions of Willmore Type
01. 12. 2020, 19:00 (CEST)
Marius Müller (University Ulm, Germany)
08. 12. 2020, 19:00 (CEST)
Ursula Ludwig (University Duisburg, Germany) An Extension of a Theorem by Cheeger and Müller to Spaces with Isolated Conical Singularities
Abstract:
An important comparison theorem in global analysis is the comparison of analytic and topological torsion for smooth compact manifolds equipped with a unitary flat vector bundle. It has been conjectured by Ray and Singer and has been independently proved by Cheeger and Mu ̈ller in the 70ies. Bismut and Zhang combined the Witten deformation and local index techniques to generalise the result of Cheeger and Mu ̈ller to arbitrary flat vector bundles with arbitrary Hermitian metrics. The aim of this talk is to present an extension of the Cheeger-Mu ̈ller theorem to spaces with isolated conical singularities by generalising the proof of Bismut and Zhang to the singular setting.
12. 01. 2021, 19:00 (CEST)
Rupert Frank
19. 01. 2021, 19:00 (CEST)
Wilderich Tuschmann

Organizers

Simon Blatt (University Salzburg)

Philipp Reiter (University Halle)

Armin Schikorra (University Pittsburgh)

Guofang Wang (University Freiburg)


text:
A and B are two numbers. 6 times of square of B is 540 more than the square of A. If the respective ratio

Question Asked by a Student from EXXAMM.com Team

Q 1635167062.     A and B are two numbers. 6 times of square of B is 540 more
than the square of A. If the respective ratio between A and B
is 3 : 2, what is the value of B?
IBPS-CLERK 2013 Mains 14/12/2013
A

10

B

12

C

16

D

8

E

14

HINT

(Provided By a Student and Checked/Corrected by EXXAMM.com Team)

Access free resources including

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text:
The guessing game in R (with a twist, of course)

May 27, 2010
By

(This article was first published on Probability and statistics blog » r, and kindly contributed to R-bloggers)

Maybe you remember playing this one as a kid. If you are about my age, you may have even created a version of this game as one of your first computer programs. You guess a number, the computer tells you if you if you are too low or high. I’ve limited the number of maximum guesses, and randomized the computer’s choice based on the Poisson distribution (more on that later).

Here’s the code. This was part of my attempt to understand how R reads input from the command line. One of the things I learned: you may need to save this to a file and run it with “source()”, instead of running it directly from the console, line by line.

# Classic guessing game with twist x = 0 gotRight = 0 failed = 0   # Initial lambda for our random var correct = 2000 initial = correct   # How many guesses should we allow per number maxGuesses = 7   while(x != Inf) { # The +1 part makes sure we never get zero, which would trigger 0's forever correct = rpois(1,correct) + 1   # The advantage of using "cat" instead of "print" is that you remove those pesky quotation marks cat("I am thinking of a number between 1 and infinity. What is it? (Type Inf to quit)n")   # Solicit input from the user x = scan(n=1) # Just one item in this vector   # Be nice and let the user quit. if(x == Inf) { cat("The correct answer was", correct, "n") cat("You got", gotRight, "right and failed", failed, "times. Maximum allowed guesses was", maxGuesses, "and initial lambda was", initial, ". Goodbye.n") cat("Post your score to http://www.statisticsblog.com/2010/05/the-guessing-game-in-r-with-a-twist-of-course/#comments n") break }   for(i in 1:maxGuesses) { if(x == correct) { print("You rock!") gotRight = gotRight + 1 break } else { if(i == maxGuesses) { cat("You ran out of guesses. I will pick a new random number based on the last one.n") failed = failed + 1 } else { if(x < correct) { cat("You are too low. Guess again.n") } else { cat("You are too high. Guess again.n") }   x = scan(n=1) } } } }

Note 1: My code makes a couple uses of the aparently controversial “break” function. I can still recall a heated debate I had with a CS professor who believed that calling “break” (in Python) was as bad as crossing the streams of your Proton Pack. That said, I have sucessfully used it on several occasions now without any appearance by Stay Puft Marshmallow Man or changing the natural order between dogs and cats. In R, the biggest problem with using constructs like “break” and “while” is that, for reasons clear only to readers of this blog but not myself, if you ask R for help about either of these tokens using

?term

you get an sent an error or to purgatory, respectively.

Hint: Because the random guesses are Poisson based, using a “half the distance” strategy for guessing may not be the best way to go. The hardcore amongst yourselves might want to calculate the median of the expected value conditional on having guessed too low or high.

Note 2: The Poisson isn’t a very good distribution for for this. Maybe you can find a better one, or at least jack up the dispersion like an overzealous offroader tweaking the suspension of his 4Runner.

R-bloggers.com offers daily e-mail updates about R news and tutorials on topics such as: visualization (ggplot2, Boxplots, maps, animation), programming (RStudio, Sweave, LaTeX, SQL, Eclipse, git, hadoop, Web Scraping) statistics (regression, PCA, time series, trading) and more...


text:
# Extending a plot over a secondary variable

Apologies for the vague title; couldn't really think of a better way of putting it.

I have a function $F$ of two variables, $\rho\in[0,\infty)$ and $f\in[0,1]$, that I wish to create a plot over (e.g. with CountourPlot or DensityPlot). I can numerically generate the entire curve of $F$ over $f$ for a particular value of $\rho$ relatively cheaply, but I can't directly compute $F$ for individual values of $f$.

Is there an efficient way of taking such a curve and plotting it for a range of $\rho$ without Mathematica having to evaluate the function independently at every $(\rho,f)$?

Here's how I compute $F$:

Lorentz[data_] := Module[{xs, ys},
xs = Range[0, Length[data]];
ys = Prepend[Accumulate[Sort[data]], 0];
Transpose[{xs/Last@xs, ys/Last@ys}]];

L = Interpolation[Lorentz[data], InterpolationOrder -> 1];
F = 1 - L[1 - #] &;

data is a list of real numbers generated according to the value of $\rho$.

Essentially, I generate a Lorentz curve numerically as a list of $(x,y)$ tuples, turn it into a function via Interpolation, reflect it a couple of times, and that's my function $F$.

-
For this specific problem it will be nice if you can supply us the code for $F(\rho,f)$. –  PlatoManiac Feb 7 '13 at 12:35
@PlatoManiac I've added the details. –  Will Vousden Feb 7 '13 at 12:47
related: wolfram demonstrations Lorenz Curve and Gini Coefficient –  kguler Feb 7 '13 at 12:58

Lets change the definition of your function a bit. here n is the number of points you take in the direction of variable f.

Lorentz[rho_, n_: 100] := Module[{data, xs, ys},
data = RandomVariate[NormalDistribution[1, 3], n];
xs = Range[0, n];
ys = Prepend[Accumulate[Sort[data]], 0];
xs /= Last@xs;
ys /= Last@ys;
N@Transpose@{Transpose@{xs, ConstantArray[rho, n + 1]}, ys}
];
(* Discretize f direction with 1000 *)
(* Discretize rho direction say {0,5} with 0.1 *)
L = Interpolation[Flatten[Table[Lorentz[i, 1000], {i, 0, 5, .1}], 1],
InterpolationOrder -> 3];
F[f_, rho_] := 1 - L[1 - f, rho];

Now lets plot what you want!

GraphicsGrid[{{Plot3D[F[f, rho], {f, 0, 1}, {rho, 0, 5},
Mesh -> False, PlotPoints -> 100, PlotRange -> All,
Boxed -> False,ColorFunction -> "Rainbow"],
ContourPlot[F[f, rho], {f, 0, 1}, {rho, 0, 5}, Mesh -> None,
PlotPoints -> 100, PlotRange -> All,ColorFunction -> "Rainbow",
ContourLabels -> True]}}, ImageSize -> 800]

The plot takes some time to evaluate. For fast execution you can reduce the PlotPoints.

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This is perfect; thanks! –  Will Vousden Feb 7 '13 at 14:44


text:
## Change in total kinetic energy in a conservation of momentum problem?

Question:

On a frictionless, horizontal air table, puck A (with mass $$m_1$$) is moving toward puck B (with mass $$m_2$$), that is initially at rest. After the collision, puck A has a velocity of $$v_1$$ to the left, and puck B has velocity $$v_2$$ to the right.

1. What was the speed of puck A before the collision?

My answer to this part of the question was correct. It is $$(m_2*v_2)/(m_1) - (v_1)$$.

2. Calculate the change in the total kinetic energy of the system that occurs during the collision.

I think this depends on the first part of the question.

So, I'm thinking:

$$\Delta K= K_f - K_i$$
$$(1/2)(m)(v_f^2 - v_i^2)$$

I should sum up the velocities, right?

$$(1/2)(m_1+m_2)((v_2 - v_1)^2 - ((m_2*v_2)/(m_1) - (v_1))^2)$$

But, that is not correct.
 Recognitions: Homework Help I'm not very clear with what was meant by your last equation. In anycase, you basically do: Change in kinetic energy = Kinetic energy of puckA before collision - Kinetic energy of puckA after collision - kinetic energy of puckB after collision. KE is a function of only mass and velocity. Since all are known, carefully plugging in the numbers should do the trick.
 Recognitions: Homework Help Science Advisor erik-the-red, When calculating kinetic energy, you first square the velocity of each particle, multiply by the particles mass divided by two and then add. You have added the final velocities first and then squared and this is wrong.

## Change in total kinetic energy in a conservation of momentum problem?

I inputed $$(1/2)(m_1)((m_2*v_2)/(m_1) - (v_1))^2 - (1/2)(m_1)(v_1)^2 - (1/2)(m_2)(v_2)^2$$ and it was incorrect.

My understanding of change in total kinetic energy, $$\Delta K$$, is the difference between the final and initial kinetic energies.

Physics Monkey, you told me to add. Why is this so?
 Mentor Blog Entries: 1 $\Delta K = K_f - K_i$, not $K_i - K_f$.

Try applying conservation of momentum and kinetic energy.

 A perfectly elastic collision is defined as one in which there is no loss of kinetic energy in the collision. An inelastic collision is one in which part of the kinetic energy is changed to some other form of energy in the collision. Any macroscopic collision between objects will convert some of the kinetic energy into internal energy and other forms of energy, so no large scale impacts are perfectly elastic. Momentum is conserved in inelastic collisions, but one cannot track the kinetic energy through the collision since some of it is converted to other forms of energy.
http://hyperphysics.phy-astr.gsu.edu.../elacol.html#4

Consider the momentum of the masses to left and right. Remember momentum is proportional to velocity, so it is a vector quantity. Energy is a scalar.
 Recognitions: Homework Help From the looks of it, the problem now isn't with the physics, but rather the math. Try cleaning it up, do things step by step and plug in the numbers. Like what Physics Monkey said, you are doing some incorrect operations by trying to rush the calculation. One by one, $$KE_1 = \frac{1}{2}m_1v_1^2$$
 It looks like I definitely did $$K_i - K_f$$, which obviously is not correct. So, I decided to analyze $$K_f$$ first. I got $$K_f = (1/2)(m_1)(-v_1)^2 + (1/2)(m_2)(v_2)^2$$. Then, I decided to analzye $$K_i$$. I got $$K_i = (1/2)(m_1)((m_2*v_2)/(m_1) - v_1)$$ This simplifies to $$(1/2)(m_2*v_2 - m_1*v_1)$$. I took the difference $$K_f - K_i = (1/2)(m_1)(-v_1)^2 + (1/2)(m_2)(v_2)^2 - (1/2)(m_2*v_2 - m_1*v_1)$$ and it was not correct. I really think I did everything right this time. ?

Mentor
Blog Entries: 1
 Quote by erik-the-red It looks like I definitely did $$K_i - K_f$$, which obviously is not correct.
Other than that (which gives the wrong sign) I don't see anything wrong with what you posted in #4.

 So, I decided to analyze $$K_f$$ first. I got $$K_f = (1/2)(m_1)(-v_1)^2 + (1/2)(m_2)(v_2)^2$$.
OK.

 Then, I decided to analzye $$K_i$$. I got $$K_i = (1/2)(m_1)((m_2*v_2)/(m_1) - v_1)$$ This simplifies to $$(1/2)(m_2*v_2 - m_1*v_1)$$.
You forgot to square the speed.
 Thanks so much, Doc Al. I wanted to simplify my answer, but I forgot that it's $$v^2$$.