TITLE: Einstein's elevator : factor 2 missing between SR and GR
QUESTION [2 upvotes]: Reading Semon et al. (2009) (https://doi.org/10.1119/1.3088883), I have found this:
"The reason the historical approach
is not generally used in advanced texts is that although
the examples of a uniformly accelerating elevator and
a rotating material disk were instrumental in the development
of Einstein’s thought as he was creating the theory of
general relativity, the examples are problematic when analyzed
with the mathematical formalism of the completed
theory. For example, using the accelerating elevator to predict
the bending of light in a gravitational field leads to a
result that differs by a factor of 2 from that predicted by the
field equations of general relativity (and verified by experiment
)."
Do you understand what the author is saying ? I would really appreciate to compare the calculations and see the missing factor 2. And understand why apparently it is not a problem.
Thank you !
Francois

REPLY [0 votes]: "The reason the historical approach is not generally used in advanced texts is that although the examples of a uniformly accelerating elevator and a rotating material disk were instrumental in the development of Einstein’s thought as he was creating the theory of general relativity, the examples are problematic when analyzed with the mathematical formalism of the completed theory. For example, using the accelerating elevator to predict the bending of light in a gravitational field leads to a result that differs by a factor of 2 from that predicted by the field equations of general relativity (and verified by experiment )."
That above paragraph is silly. We have developed advanced equations and now we use them, that's all.
Let's say a small laboratory is hovering near the sun. A ray of light comes in through a small hole on the wall. A physicist in the lab compares the path of the ray to a straight rod. He says the ray curves x degrees.
Another physicist on the earth measures how much that same ray curves, she gets two times larger number. She explains the difference by saying that the rod in the lab is not straight.
The curvature of space-time is the reason that the rod is not straight. All rods in the lab are non-straight, that is why the physicist in the lab can't find out the 'real' curving rate of the light ray.


TITLE: What does this problem mean?
QUESTION [0 upvotes]: I have this probability exercise from Grimmett and Welch, which states that
If $A_{1}, A_{2}, \dots , A_{m} \in \mathcal{F}$ and $k$ is positive integer, show that the set of points in $\Omega$ which belong to exactly $k$ of the $A_{i}$ belongs to $\mathcal{F}$.
where $\Omega$ and $\mathcal{F}$ are sample and event spaces respectively. I am not asking anyone to solve this problem for me, I am simply asking what this problem means precisely.

REPLY [0 votes]: One way of reformualting this is: Show that $\{\omega: \sum\limits_{i=1}^{n} I_{A_i}(\omega)=k\}$ belongs to $\mathcal F$. (Note that the sum here is $k$ precisely when $\omega$ belongs to $k$ of the sets $A_1,A_2,..,A_n$).


TITLE: Complement of a tree $T$
QUESTION [0 upvotes]: Let $T$ be a tree having $n$ nodes with $n\geq 2$. Show that if the complement of $T$ is also a tree, then $n=4$. Note that if $T = (V,E)$ is a tree, then the complement of $T$ is $\overline{T} = (V,K\setminus E)$, where $K$ consists of all 2-element subsets of $V$.

REPLY [1 votes]: Hint: (i) If $T$ is a tree with $n$ vertices, then the number of edges is $e=n-1$.
(ii) Since the union of a graph $G$ with $n$ vertices and its complement $\bar{G}$ is the complete graph $K_n$, the number of edges of $\bar{G}$
is equal to $\frac{n(n-1)}{2}-e$, where $e$ is the number of edges in $G$.


TITLE: Find the $x$ values for which my function has a horizontal tangent.
QUESTION [0 upvotes]: I want to find the $x$ values for which the following function has a horizontal tangent line:
$$\sqrt{x^2-2x+2}+\sqrt{x^2-6x+12}$$
My attempt so far has been to compute the derivative, which I found to be $$\frac{x-1}{\sqrt{x^2-2x+2}}+\frac{x-3}{\sqrt{x^2-6x+12}}$$However, I wasn't able to find the $x$ values for which this expression is $0$.

REPLY [1 votes]: Set your expression of $f'(x)$ equal to $0$  to express that you have an extremum.
Write this condition under the form:
$$\frac{x-1}{\sqrt{x^2-2x+2}}=-\frac{x-3}{\sqrt{x^2-6x+12}}\tag{1}$$
Square the two sides of (1) giving
$$\frac{N_1(x)}{D_1(x)}=\frac{N_2(x)}{D_2(x)}$$
giving you a polynomial equation:
$$N_1(x)D_2(x)-N_2(x)D_1(x)=0$$
otherwise said:
$$(x-1)^2(x^2-6x+12)-(x-3)^2(x^2-2x+2)=0$$
$$(x-1)^2((x-3)^2+3)-(x-3)^2((x-1)^2+1)=0\tag{2}$$
Let us expand (2) under the form:
$$(x-1)^2 . (x-3)^2+(x-1)^2 . 3 \ - \ (x-3)^2(x-1)^2-(x-3)^2 . 1=0$$
By cancellation of terms $(x-1)^2 . (x-3)^2$:
$$3(x-1)^2-(x-3)^2=0$$
a quadratic equation with two real roots.
Take care: check the solutions against the initial equation (1); indeed a spurious solution has been introduced by the squaring step.


TITLE: Subset of Permutations of distinct symbols, with positional count constraint/
QUESTION [1 upvotes]: I am working a personal project, and I need to derive subsets of permutations of unique elements.
For example:  I have a set of 5 unique elements {A, B, C, D, E}.   There are 60 permutations tuples of size 3 of this: {A,B,C}, {A,B,D}...{E,C,D}, {E,D,C}.  What I want is to select tuples from this list such that the count of each symbol in the same position across all tuples is constrained as 0 < A# < B# < C# < D# < E# the count of each symbol is the same in each position.
E.g.
{A, B, C}
{A, B, D}
⋮
{E, C, D}
{E, D, C}

For the full set of tuples, there are 12 of each symbol in each position within the tuple, or specifically for what is explicitly written above there are quantity A2 and E2 in position 0, B2 C1 D1 in position 1, and C2 and D2 in position 2, so this does not meet the criteria.
With the example of 5 permute 3, as there are only 60 permutation tuples I was able to create such a desired subset by brute force, finding a collection of 20 permutation tuples resulting in counts of A2, B3, C4, D5, E6 in each position.
My question is:   Is there a known algorithmic way of selecting subsets with this type of criteria, and is there a predictive formula to calculate how many valid solutions can be derived?  I will need this to be for quantities of symbols up to 15 and up to tuple size 5 so brute force is sub-optimal.
For completeness, the solution subset I found for 5 symbols permut 3 is the following set:
AEC
AED
BCE
BDE
BED
CBE
CDB
CDE
CED
DBC
DBE
DCE
DEB
DEC
EAC
EAD
ECA
ECD
EDA
EDB

Edit:  Using logic inspired by Ross Millikan, I verified that 4 permut 3 has no solution, as the permut domain is too small.  I did find via brute force a solution for 6 permut 3: A3,B4,C5,D6,E7,F8 for a total of 33 tuples:
ABC
ABE
ACD
BCF
BDF
BEF
BFE
CAD
CED
CEF
CFD
CFE
DCE
DCF
DEC
DEF
DFC
DFE
ECF
EDB
EDC
EDF
EFB
EFC
EFD
FAD
FAE
FBA
FBE
FDA
FDB
FEA
FEB

REPLY [1 votes]: Not a complete answer, however some interesting research results
Using the setup of 5 distinct symbols {A,B,C,D,E} permut 3, 60 permutations.  I have run exhaustive checks on all permutations for the positional symbol count criteria form of A#1, B#2, C#3, D#4, E#5.
For ease of computation I arbitrarily grouped the set based upon the symbol in position 0, thus having 5 subgroups of 12 tuples each, e.g. ABC ABD ABE ACB ACD ACE ADB ADC ADE AEB AEC AED ...
With this, the maximum number of checks can be calculated as
$$
\binom{12}{1}\binom{12}{2}\binom{12}{3}\binom{12}{4}\binom{12}{5}=68,309,049,600
$$
For optimization, I recursively worked through the initial symbol sub groups starting with symbol A, and could abandon a recursion before reaching symbol E subgroup as soon as the remaining symbol count violated the constraint.   With this optimization, when I ran this the code only had to perform 12,788,832 checks to find there were in total only 130 solutions that met the criteria.  Since it is such a small set, below is the exhaustive list by row of the valid tuple subsets.
ACE BDE BEC CBE CDE CED DCE DEA DEB DEC EAD EBD ECD EDB EDC
ACE BDE BEC CDE CEA CED DBE DCE DEB DEC EAD EBD ECD EDB EDC
ACE BDE BEC CDE CEB CED DBE DCE DEA DEC EAD EBD ECD EDB EDC
ACE BDE BEC CDE CEB CED DBE DCE DEB DEC EAD EBD ECD EDA EDC
ACE BDE BED CAE CDE CED DCE DEA DEB DEC EBC EBD ECD EDB EDC
ACE BDE BED CBE CDE CED DCE DEA DEB DEC EAC EBD ECD EDB EDC
ACE BDE BED CBE CDE CED DCE DEA DEB DEC EAD EBC ECD EDB EDC
ACE BDE BED CDE CEA CED DAE DCE DEB DEC EBC EBD ECD EDB EDC
ACE BDE BED CDE CEA CED DBE DCE DEB DEC EAC EBD ECD EDB EDC
ACE BDE BED CDE CEA CED DBE DCE DEB DEC EAD EBC ECD EDB EDC
ACE BDE BED CDE CEB CED DAE DCE DEA DEC EBC EBD ECD EDB EDC
ACE BDE BED CDE CEB CED DAE DCE DEB DEC EBC EBD ECD EDA EDC
ACE BDE BED CDE CEB CED DBE DCE DEA DEC EAC EBD ECD EDB EDC
ACE BDE BED CDE CEB CED DBE DCE DEA DEC EAD EBC ECD EDB EDC
ACE BDE BED CDE CEB CED DBE DCE DEB DEC EAC EBD ECD EDA EDC
ACE BDE BED CDE CEB CED DBE DCE DEB DEC EAD EBC ECD EDA EDC
ACE BEC BED CAE CDE CED DBE DCE DEB DEC EBD ECD EDA EDB EDC
ACE BEC BED CBE CDE CED DAE DCE DEB DEC EBD ECD EDA EDB EDC
ACE BEC BED CBE CDE CED DBE DCE DEB DEC EAD ECD EDA EDB EDC
ACE BEC BED CDE CEB CED DAE DBE DCE DEC EBD ECD EDA EDB EDC
ADE BCE BDE CDE CEB CED DCE DEA DEB DEC EAD EBC EBD ECD EDC
ADE BCE BEC CBE CDE CED DCE DEA DEB DEC EAD EBD ECD EDB EDC
ADE BCE BEC CDE CEA CED DBE DCE DEB DEC EAD EBD ECD EDB EDC
ADE BCE BEC CDE CEB CED DBE DCE DEA DEC EAD EBD ECD EDB EDC
ADE BCE BEC CDE CEB CED DBE DCE DEB DEC EAD EBD ECD EDA EDC
ADE BCE BED CAE CDE CED DCE DEA DEB DEC EBC EBD ECD EDB EDC
ADE BCE BED CBE CDE CED DCE DEA DEB DEC EAC EBD ECD EDB EDC
ADE BCE BED CBE CDE CED DCE DEA DEB DEC EAD EBC ECD EDB EDC
ADE BCE BED CDE CEA CED DAE DCE DEB DEC EBC EBD ECD EDB EDC
ADE BCE BED CDE CEA CED DBE DCE DEB DEC EAC EBD ECD EDB EDC
ADE BCE BED CDE CEA CED DBE DCE DEB DEC EAD EBC ECD EDB EDC
ADE BCE BED CDE CEB CED DAE DCE DEA DEC EBC EBD ECD EDB EDC
ADE BCE BED CDE CEB CED DAE DCE DEB DEC EBC EBD ECD EDA EDC
ADE BCE BED CDE CEB CED DBE DCE DEA DEC EAC EBD ECD EDB EDC
ADE BCE BED CDE CEB CED DBE DCE DEA DEC EAD EBC ECD EDB EDC
ADE BCE BED CDE CEB CED DBE DCE DEB DEC EAC EBD ECD EDA EDC
ADE BCE BED CDE CEB CED DBE DCE DEB DEC EAD EBC ECD EDA EDC
ADE BDE BEC CBE CDE CED DCE DEA DEB DEC EAD EBD ECB ECD EDC
ADE BDE BEC CDE CEA CED DBE DCE DEB DEC EAD EBD ECB ECD EDC
ADE BDE BEC CDE CEB CED DBE DCE DEA DEC EAD EBD ECB ECD EDC
ADE BDE BEC CDE CEB CED DBE DCE DEB DEC EAD EBD ECA ECD EDC
ADE BDE BED CAE CDE CED DCE DEA DEB DEC EBC EBD ECB ECD EDC
ADE BDE BED CBE CDE CED DCE DEA DEB DEC EAC EBD ECB ECD EDC
ADE BDE BED CBE CDE CED DCE DEA DEB DEC EAD EBC ECB ECD EDC
ADE BDE BED CDE CEA CED DAE DCE DEB DEC EBC EBD ECB ECD EDC
ADE BDE BED CDE CEA CED DBE DCE DEB DEC EAC EBD ECB ECD EDC
ADE BDE BED CDE CEA CED DBE DCE DEB DEC EAD EBC ECB ECD EDC
ADE BDE BED CDE CEB CED DAE DCE DEA DEC EBC EBD ECB ECD EDC
ADE BDE BED CDE CEB CED DAE DCE DEB DEC EBC EBD ECA ECD EDC
ADE BDE BED CDE CEB CED DBE DCE DEA DEC EAC EBD ECB ECD EDC
ADE BDE BED CDE CEB CED DBE DCE DEA DEC EAD EBC ECB ECD EDC
ADE BDE BED CDE CEB CED DBE DCE DEB DEC EAC EBD ECA ECD EDC
ADE BDE BED CDE CEB CED DBE DCE DEB DEC EAD EBC ECA ECD EDC
ADE BEC BED CAE CDE CED DBE DCE DEA DEC EBD ECB ECD EDB EDC
ADE BEC BED CAE CDE CED DBE DCE DEB DEC EBD ECA ECD EDB EDC
ADE BEC BED CAE CDE CED DBE DCE DEB DEC EBD ECB ECD EDA EDC
ADE BEC BED CBE CDE CED DAE DCE DEA DEC EBD ECB ECD EDB EDC
ADE BEC BED CBE CDE CED DAE DCE DEB DEC EBD ECA ECD EDB EDC
ADE BEC BED CBE CDE CED DAE DCE DEB DEC EBD ECB ECD EDA EDC
ADE BEC BED CBE CDE CED DBE DCE DEA DEC EAD ECB ECD EDB EDC
ADE BEC BED CBE CDE CED DBE DCE DEB DEC EAD ECA ECD EDB EDC
ADE BEC BED CBE CDE CED DBE DCE DEB DEC EAD ECB ECD EDA EDC
ADE BEC BED CDE CEA CED DAE DBE DCE DEC EBD ECB ECD EDB EDC
ADE BEC BED CDE CEB CED DAE DBE DCE DEC EBD ECA ECD EDB EDC
ADE BEC BED CDE CEB CED DAE DBE DCE DEC EBD ECB ECD EDA EDC
AEC BCE BDE CBE CDE CED DCE DEA DEB DEC EAD EBD ECD EDB EDC
AEC BCE BDE CDE CEA CED DBE DCE DEB DEC EAD EBD ECD EDB EDC
AEC BCE BDE CDE CEB CED DBE DCE DEA DEC EAD EBD ECD EDB EDC
AEC BCE BDE CDE CEB CED DBE DCE DEB DEC EAD EBD ECD EDA EDC
AEC BCE BED CAE CDE CED DBE DCE DEB DEC EBD ECD EDA EDB EDC
AEC BCE BED CBE CDE CED DAE DCE DEB DEC EBD ECD EDA EDB EDC
AEC BCE BED CBE CDE CED DBE DCE DEB DEC EAD ECD EDA EDB EDC
AEC BCE BED CDE CEB CED DAE DBE DCE DEC EBD ECD EDA EDB EDC
AEC BDE BED CAE CDE CED DBE DCE DEA DEC EBD ECB ECD EDB EDC
AEC BDE BED CAE CDE CED DBE DCE DEB DEC EBD ECA ECD EDB EDC
AEC BDE BED CAE CDE CED DBE DCE DEB DEC EBD ECB ECD EDA EDC
AEC BDE BED CBE CDE CED DAE DCE DEA DEC EBD ECB ECD EDB EDC
AEC BDE BED CBE CDE CED DAE DCE DEB DEC EBD ECA ECD EDB EDC
AEC BDE BED CBE CDE CED DAE DCE DEB DEC EBD ECB ECD EDA EDC
AEC BDE BED CBE CDE CED DBE DCE DEA DEC EAD ECB ECD EDB EDC
AEC BDE BED CBE CDE CED DBE DCE DEB DEC EAD ECA ECD EDB EDC
AEC BDE BED CBE CDE CED DBE DCE DEB DEC EAD ECB ECD EDA EDC
AEC BDE BED CDE CEA CED DAE DBE DCE DEC EBD ECB ECD EDB EDC
AEC BDE BED CDE CEB CED DAE DBE DCE DEC EBD ECA ECD EDB EDC
AEC BDE BED CDE CEB CED DAE DBE DCE DEC EBD ECB ECD EDA EDC
AED BCE BDE CAE CDE CED DCE DEA DEB DEC EBC EBD ECD EDB EDC
AED BCE BDE CBE CDE CED DCE DEA DEB DEC EAC EBD ECD EDB EDC
AED BCE BDE CBE CDE CED DCE DEA DEB DEC EAD EBC ECD EDB EDC
AED BCE BDE CDE CEA CED DAE DCE DEB DEC EBC EBD ECD EDB EDC
AED BCE BDE CDE CEA CED DBE DCE DEB DEC EAC EBD ECD EDB EDC
AED BCE BDE CDE CEA CED DBE DCE DEB DEC EAD EBC ECD EDB EDC
AED BCE BDE CDE CEB CED DAE DCE DEA DEC EBC EBD ECD EDB EDC
AED BCE BDE CDE CEB CED DAE DCE DEB DEC EBC EBD ECD EDA EDC
AED BCE BDE CDE CEB CED DBE DCE DEA DEC EAC EBD ECD EDB EDC
AED BCE BDE CDE CEB CED DBE DCE DEA DEC EAD EBC ECD EDB EDC
AED BCE BDE CDE CEB CED DBE DCE DEB DEC EAC EBD ECD EDA EDC
AED BCE BDE CDE CEB CED DBE DCE DEB DEC EAD EBC ECD EDA EDC
AED BCE BEC CAE CDE CED DBE DCE DEB DEC EBD ECD EDA EDB EDC
AED BCE BEC CBE CDE CED DAE DCE DEB DEC EBD ECD EDA EDB EDC
AED BCE BEC CBE CDE CED DBE DCE DEB DEC EAD ECD EDA EDB EDC
AED BCE BEC CDE CEB CED DAE DBE DCE DEC EBD ECD EDA EDB EDC
AED BCE BED CAE CDE CED DBE DCE DEB DEC EBC ECD EDA EDB EDC
AED BCE BED CBE CDE CED DAE DCE DEB DEC EBC ECD EDA EDB EDC
AED BCE BED CBE CDE CED DBE DCE DEB DEC EAC ECD EDA EDB EDC
AED BCE BED CDE CEB CED DAE DBE DCE DEC EBC ECD EDA EDB EDC
AED BDE BEC CAE CDE CED DBE DCE DEA DEC EBD ECB ECD EDB EDC
AED BDE BEC CAE CDE CED DBE DCE DEB DEC EBD ECA ECD EDB EDC
AED BDE BEC CAE CDE CED DBE DCE DEB DEC EBD ECB ECD EDA EDC
AED BDE BEC CBE CDE CED DAE DCE DEA DEC EBD ECB ECD EDB EDC
AED BDE BEC CBE CDE CED DAE DCE DEB DEC EBD ECA ECD EDB EDC
AED BDE BEC CBE CDE CED DAE DCE DEB DEC EBD ECB ECD EDA EDC
AED BDE BEC CBE CDE CED DBE DCE DEA DEC EAD ECB ECD EDB EDC
AED BDE BEC CBE CDE CED DBE DCE DEB DEC EAD ECA ECD EDB EDC
AED BDE BEC CBE CDE CED DBE DCE DEB DEC EAD ECB ECD EDA EDC
AED BDE BEC CDE CEA CED DAE DBE DCE DEC EBD ECB ECD EDB EDC
AED BDE BEC CDE CEB CED DAE DBE DCE DEC EBD ECA ECD EDB EDC
AED BDE BEC CDE CEB CED DAE DBE DCE DEC EBD ECB ECD EDA EDC
AED BDE BED CAE CDE CED DBE DCE DEA DEC EBC ECB ECD EDB EDC
AED BDE BED CAE CDE CED DBE DCE DEB DEC EBC ECA ECD EDB EDC
AED BDE BED CAE CDE CED DBE DCE DEB DEC EBC ECB ECD EDA EDC
AED BDE BED CBE CDE CED DAE DCE DEA DEC EBC ECB ECD EDB EDC
AED BDE BED CBE CDE CED DAE DCE DEB DEC EBC ECA ECD EDB EDC
AED BDE BED CBE CDE CED DAE DCE DEB DEC EBC ECB ECD EDA EDC
AED BDE BED CBE CDE CED DBE DCE DEA DEC EAC ECB ECD EDB EDC
AED BDE BED CBE CDE CED DBE DCE DEB DEC EAC ECA ECD EDB EDC
AED BDE BED CBE CDE CED DBE DCE DEB DEC EAC ECB ECD EDA EDC
AED BDE BED CDE CEA CED DAE DBE DCE DEC EBC ECB ECD EDB EDC
AED BDE BED CDE CEB CED DAE DBE DCE DEC EBC ECA ECD EDB EDC
AED BDE BED CDE CEB CED DAE DBE DCE DEC EBC ECB ECD EDA EDC
AED BEC BED CBE CDE CED DAE DBE DCE DEC ECB ECD EDA EDB EDC

Counting the total tuples in this collection of the 130 sets of 15 tuples that met the criteria revealed some interesting insights.  Below is this count, sorted by first position symbol and then by quantity found in the total collection of tuples:
ACE:          20
AEC:          20
ADE:          45
AED:          45
--------------------
BCE:          45
BEC:          45
BDE:          85
BED:          85
--------------------
CAE:          20
CEA:          20
CBE:          45
CEB:          45
CDE:         130
CED:         130
--------------------
DAE:          45
DEA:          45
DBE:          85
DEB:          85
DCE:         130
DEC:         130
--------------------
EAC:          20
ECA:          20
EAD:          45
EBC:          45
ECB:          45
EDA:          45
EBD:          85
EDB:          85
ECD:         130
EDC:         130

The first insight that jumped out to me is clearly there is a reflexive pair pattern.  This makes sense, since effectively one could swap all the symbols in position 1 and 2 to yield the mirror result.
The next insight that surprised me is that subgroups C, D, and E shows tuples pairs that must appear in a valid solution, e.g. CDE/CED, DCE/DEC, ECD/EDC, and some tuple pairs in every sub group that never appear.
To see how the pattern extends, I repeated this process with the same initial collection of 5 permute 3, changing the criteria to A#2 B#3 C#4 D#5 E#6.  This had a possibility of requiring 5.2E12 comparisons, though the optimizations reduced this to only ~1E9 tests required to find 50,202 valid solutions.  The tuple count in the valid solutions, below, shows the same reflective pattern, as well as perhaps some other insights.
ABC:         561
ACB:         561
ABD:       1,144
ADB:       1,144
ABE:       3,131
AEB:       3,131
ACD:       5,140
ADC:       5,140
ACE:      13,792
AEC:      13,792
ADE:      26,434
AED:      26,434
--------------------
BAC:         561
BCA:         561
BAD:       1,144
BDA:       1,144
BAE:       3,131
BEA:       3,131
BCD:       9,889
BDC:       9,889
BCE:      24,351
BEC:      24,351
BDE:      36,227
BED:      36,227
--------------------
CAB:         561
CBA:         561
CAD:       5,140
CDA:       5,140
CBD:       9,889
CDB:       9,889
CAE:      13,792
CEA:      13,792
CBE:      24,351
CEB:      24,351
CDE:      46,671
CED:      46,671
--------------------
DAB:       1,144
DBA:       1,144
DAC:       5,140
DCA:       5,140
DBC:       9,889
DCB:       9,889
DAE:      26,434
DEA:      26,434
DBE:      36,227
DEB:      36,227
DCE:      46,671
DEC:      46,671
--------------------
EAB:       3,131
EBA:       3,131
EAC:      13,792
ECA:      13,792
EBC:      24,351
ECB:      24,351
EAD:      26,434
EDA:      26,434
EBD:      36,227
EDB:      36,227
ECD:      46,671
EDC:      46,671

Research is ongoing to see if I can derive a predictive formula, as well as ongoing optimizations to the exhaustive checks (taking advantage of the mirror reflection at least).


TITLE: When is $G^n$ a subgroup of $G$?
QUESTION [3 upvotes]: Let $G$ be a group and $n > 1$ be a positive integer. I want to know when $G^n = \{ g^n | g \in G \}$ is a subgroup of $G$? A general answer would be most helpful, though in the context I am using this, the most useful things we can say is that $G$ is nilpotent and $n$ is prime, I am not sure if that is useful.

REPLY [1 votes]: Every divisible group has this property: $G^n=G$ in this case. Examples of divisible groups exist among abelian groups, such as ${\mathbb Q}$, and even finitely generated infinite nonabelian groups...


TITLE: When to use $\mathbf{P}$ , and when to use $\mathbb{P}$ as the symbol for probability?
QUESTION [3 upvotes]: I am trying to make myself some notes in preparatino for my FRM examination. Because I did a little mathematics and mathematical statistics at uni, I am getting very impatient with the inconsistent notations usen in FRM prescribed learning material. Having glanced through some many a probability theory books, and financial mathematics books, i have noticed that the mathematicians often use the $\mathbb{P}$ and $\mathbf{P}$ symbols, while most statisticians just stick to using $\mathbf{P}$ , or $\mathrm{P}$ . At first I thought that its just a matter of choice, but then I noticed that in some financial mathematics texts, that both $\mathbb{P}$ and $\mathrm{P}$  are sometimes used in the same expression? Typo, choice, or are the two symbols used for different purposes?

REPLY [0 votes]: Bland $P$ can also clash with event names. $M, N, P$ are out as event names, otherwise you get $P(P)$. Same goes for e-xpectation operator. $E$ can also be an e-vent (or the full space). Consider $E[1_E]$ and $E[1_P]=P(P)$. But, yes: "De gustibus non est disputandum".


TITLE: How to Prove Triangle Centers in Tetrahedra
QUESTION [2 upvotes]: How would you prove the existence of triangle centers in tetrahedra, for example, the incenter, circumcenter, or centroid?

REPLY [0 votes]: By construction. For example the incenter, if it exists, is the point equidistant from all 4 faces. Construct it as the intersection of 3 of the 6 bisector planes, to be chosen so that the 3 corresponding edges do not meet in one vertex, and conclude that this intersection must lay at the same distance to all 6 faces (so that it automatically lies on the other 3 bisector planes as well).


TITLE: How do I find the composition of two functions?
QUESTION [1 upvotes]: In a question sheet for my university course I am given the sets:

$A = \{1, 2, 3, 4\}, B = \{x, y, z, w\}$ and $C = \{a, b, c\}$.

I am then told to work out the functions:

$f: A → B$ which is defined by $f(1) = y, f(2) = w, f(3) = z, f(4) = x$.
$g: C → A$ which is defined by $g(a) = 2, g(b) = 3, g(c) = 4$.
$h: A → C$ which is defined by $h(1) = a, h(2) = c, h(3) = a, h(4) = b$.

The ordered pairs of these functions is:

$f: A → B = \{(1, y), (2, w), (3, z), (4, x)\}$
$g: C → A =  \{(a, 2), (b, 3), (c, 4)\}$
$h: A → C = \{(1, a), (2, c), (3, a), (4, b)\}$

The above is easy enough but I am then asked to find the composition of these functions. The actual process of doing the composition seems easy enough but deciding which domains/codomains are mapped to each other confuses me. The answers are provided by the sheet I'm working from as:

$f ∘ g: C → B$
$g ∘ h: A → A$
$h ∘ g: C → C$

What is considered when choosing what the domain/codomain maps to? Also, take $f ∘ g$ for example. Why is g's domain used first when it makes sense that $f$ is first so shouldn't $f$'s domain be used?

REPLY [1 votes]: The definition of (say) $f \circ g$ is to "do $g$ first, then $f$" even though we're used to reading left to right.
This is to agree with our notation that $f(x)$ applies the function $f$ to input $x$.
So $(f \circ g)(x)$ means that $g$ gets to "go first" since it's closer; we have $(f \circ g)(x) = f(g(x))$; "apply $f$ to what we get applying $g$ to $x$."
Since $g$ is going first, the domain of $f \circ g$ will be the domain of $g$. Likewise, since $f$ is the final function applied (to the outputs of $g$), our overall codomain will be wherever $g$ sends things.


TITLE: If the series $\sum_0^\infty a_n$ converges, then so does $\sum_1^\infty \frac{\sqrt{a_n}}{n} $
QUESTION [16 upvotes]: Problem:

Suppose that for every $n\in\mathbb{N}$, $a_n\in\mathbb{R}$ and $a_n\ge 0$. Given that
  $$\sum_0^\infty a_n$$
  converges, show that
  $$\sum_1^\infty \frac{\sqrt{a_n}}{n} $$
  converges.

Source: Rudin, Principles of Mathematical Analysis, Chapter 3, Exercise 7.

REPLY [23 votes]: We have for all real numbers $2ab\leq a^2+b^2$ hence
$$0\leq \frac{\sqrt{|a_n|}}n\leq \frac{|a_n|+\frac 1{n^2}}2.$$
Since the series $\sum_n|a_n|$ and $\sum_n\frac 1{n^2}$ are convergent, we get the convergence of $\sum_n\frac{\sqrt{|a_n|}}n$.


text:
\section{Bounds on the Smallest Eigenvalue of $L_g$ }
 \label{sec:spectralprop}

The following theorem provides our core bounds on the smallest eigenvalue of the grounded Laplacian; in subsequent sections, we will characterize graphs where these bounds become tight.

\begin{theorem}
Consider a graph $\mathcal{G}= \{\mathcal{V},\mathcal{E}\}$ with a set of grounded nodes $S \subset \mathcal{V}$.  Let $\lambda$ denote the smallest eigenvalue of the grounded Laplacian $L_g$ and let $\mathbf{x}$ be a corresponding nonnegative eigenvector, normalized so that the largest component is $x_{max} = 1$.
Then
\begin{equation}
\frac{|\partial S|}{n-|S|}x_{min} \leq \lambda \leq \min_{X\subseteq \mathcal{V}\setminus S} \frac{|\partial X|}{|X|}\leq \frac{|\partial S |}{n-|S|},
\label{eqn:maineq}
\end{equation}
where $x_{min}$ is the smallest eigenvector component in $\mathbf{x}$.
\label{thm:main}
\end{theorem}

\begin{IEEEproof}
From the Rayleigh quotient inequality \cite{Horn}, we have
$$
\lambda \leq z^T L_g z ,
$$
for all $z\in \mathbb{R}^{n-|S|}$ with $z^Tz=1$. Let $X \subseteq \mathcal{V}\setminus{S}$ be the subset of vertices for which $\frac{|\partial X|}{|X|}$ is minimum, and assume without loss of generality that the vertices are arranged so that those in set $X$ come first in the ordering.  The upper bound  $ \min_{X\subseteq \mathcal{V}\setminus S} \frac{|\partial X|}{|X|}$ is then obtained by choosing $z=\frac{1}{\sqrt{|X|}}[\mathbf{1}_{1\times |X|} \quad \mathbf{0}_{1\times |\mathcal{V}\setminus \{X \cup S\}|}]^T$, and noting that the sum of all elements in the top $|X| \times |X|$ block of $L_g$ is equal to the sum of the number of neighbors each vertex in $X$ has outside $X$ (i.e., $|\partial X|$).   The upper bound $\frac{|\partial S |}{n-|S|}$ readily follows by choosing the subset $X = \mathcal{V}\setminus{S}$.

For the lower bound, we left-multiply the eigenvector equation $L_g \mathbf{x} = \lambda \mathbf{x}$ by the vector consisting of all $1$'s to obtain
\begin{equation*}
\sum_{i=1}^{n-|S|}\alpha_ix_i =\lambda\sum_{i=1}^{n-|S|} x_i,

\end{equation*}
where $\alpha_i$ is the number of grounded nodes in node $v_i$'s neighborhood.  Using the fact that the eigenvector is nonnegative, this gives
\begin{equation*}
x_{min}\sum_{i=1}^{n-|S|}\alpha_i\leq \sum_{i=1}^{n-|S|}\alpha_ix_i=\lambda\sum_{i=1}^{n-|S|} x_i \leq \lambda(n-|S|)  x_{max}=\lambda(n-|S|).
\end{equation*}
Since $\sum_{i=1}^{n-|S|}\alpha_i=|\partial S|$, the lower bound is obtained.
\end{IEEEproof}

\begin{remark}
For the case that $|S|=1$ we have
\begin{equation*}
\frac{d_s x_{min}}{n-1}\leq \lambda\leq \frac{d_s}{n-1},

\end{equation*}
where $d_s$ is the degree of the grounded node.  Note that the smallest eigenvalue of the grounded Laplacian for a set $S$ of grounded nodes is always upper bounded by $|S|$ (since $|\partial S| \le |S|(n-|S|)$), with equality if and only if all grounded nodes connect to all other nodes (it is easy to see that the smallest eigenvector component $x_{min} = 1$ in this case).
\end{remark}

\begin{example}
Consider the graph shown in Figure~\ref{fig:dumbbell} consisting of two complete graphs on $\frac{n}{2}$ nodes, joined by a single edge.  Suppose the black node in the figure is chosen as the grounded node.  In this case, we have $|\partial S| = \frac{n}{2} - 1$, and the extreme upper bound in \eqref{eqn:maineq} indicates that $\lambda \le \frac{|\partial S|}{n-1} \approx \frac{1}{2}$ for large $n$.  Now, if we take $X$ to be the set of all nodes in the left clique, we have $|\partial X| = 1$ and $|X| = \frac{n}{2}$, leading to $\lambda \le \frac{2}{n}$ by the intermediate upper bound in \eqref{eqn:maineq}.
\label{ex:dumbbell}
\end{example}

\input{dumbbell}

In the next section, we will characterize graphs under which $x_{min}$ (the smallest eigenvector component) converges to $1$, in which case the lower and upper bounds in \eqref{eqn:maineq} coincide and yield a tight characterization of $\lambda$.  As seen in the above example, the presence of bottlenecks among the non-grounded nodes will cause $x_{min}$ to go to zero;  in certain graphs with good expansion properties, however, we will see that this will not occur.

config: arxiv


TITLE: Kolmogorov distance properties
QUESTION [1 upvotes]: Suppose that $X,Y,Z$ are random variables. Does the following holds without assuming any further conditions:
$$
|\mathsf{P}(X+Z\le x)-\mathsf{P}(Y+Z\le x)|\le \sup_{x\in\mathbb{R}}|\mathsf{P}(X\le x)-\mathsf{P}(Y\le x)| \quad?
$$

It is easy to show that the inequality holds when $Z$ is independent of $X$ and $Y$ because
$$
\mathsf{P}(X+Z\le x)-\mathsf{P}(Y+Z\le x)=\mathsf{E}[\mathsf{P}(X\le x-Z\mid Z)-\mathsf{P}(Y\le x-Z\mid Z)]
$$
and
$$
|\mathsf{P}(X\le x-z)-\mathsf{P}(Y\le x-z)|\le \sup_{x\in\mathbb{R}}|\mathsf{P}(X\le x)-\mathsf{P}(Y\le x)|.
$$

REPLY [1 votes]: In general no: take $X$ such that $-X$ and $X$ have the same distribution, $Y=-X$ and $Z=X$. For a negative $x$, $|\mathbb{P}(X+Z\le x)-\mathbb{P}(Y+Z\le x)|=\mathbb P(2X\leq x)$ while the right hand side of the wanted inequality is zero.


TITLE: Existence of the Limit of a Sequence of Characteristic Functions is not sufficient for Convergence in Distribution of a Sequence of R.V
QUESTION [0 upvotes]: Let $X_1, X_2,$ ... denote some sequence of random variables uniformly distributed on the interval $(-n,n)$ and let $\varphi_1, \varphi_2$, ... be their characteristic functions. Show that the existence of the limit $\varphi(t):= \lim_{n\to\infty} \varphi(t)$ is not sufficient in order for the sequence $(X_n)_{n\ge1}$ to converge in distribution.
This is what I have so far:
Let ${X_n}$ be uniformly distributed on the interval $(-n,n)$ then the characteristic function will be $$ \varphi_{X_n}(t) = \frac{\sin(tn)}{tn}$$ and so:
$$\varphi_{X_n}(t)=\begin{cases}\frac{\sin(tn)}{tn} &t \ne0\\ 1&t = 0\end{cases}.$$
and then:
$$\lim_{n\to\infty} \varphi(t)=\begin{cases}0 &t \ne0\\ 1&t = 0\end{cases}.$$
Now, I can see that when $t = 0$ for  $t \in \mathbb{R}  $, the limiting characteristic function $\varphi(t)$ is discontinuous. So the limit exists except for when $t=0$.
Now,  $X_n$ converges in distribution to a random variable $X$ if  $\lim_{n\to\infty}F_{X_n}(x) \rightarrow F(x)$ so for every $x \in \mathbb{R}  $ the cumulative distribution function $F_X$ is continuous.
I'm not very sure if I'm on the right track, and I'm also not very sure how to prove that the sequence of random variables $X_n$ fails to converge in distribution to X, also, I think that the continuity Theorem could help me but I still don't see how to make the link. I would appreciate any hint! Thanks.

REPLY [1 votes]: You have made couple of mistakes but they must just be typos.
$\phi_n(t)=\frac {\sin (nt)} {nt}$ if $t \neq 0$ and $1$ if $t=0$. [All characteristic functions have the value $1$ at $0$].
The limit of this is $0$ for $ t \neq 0$ and $1$ for $t=1$.
If random variables converge in distribution then (by the continuity theorem) their characteristic functions necessarily converge to the characteristic function of the limit which is necessarily a continuous function. Since the limit is not continuous in this case it follows that your sequence $(X_n)$ does not converge in distribution.