TITLE: How many triplets of consecutive primes $p$, $q$, $r$ are such that $p^2+q^2+r^2$ is also prime?
QUESTION [0 upvotes]: So we have $p$, $q$, $r$ which are successive prime numbers, we wonder how many such triplets are such that $p^2+q^2+r^2$ is also a prime.

My attempt: I tried to $p^2+q^2+r^2$ refactor it somehow, but to no avail.

REPLY [0 votes]: This OCaml program verifies it!
type 'a stream= Eos| StrCons of 'a * (unit-> 'a stream)

let rec nums_from n = StrCons(n,fun ()-> nums_from (n+1))

let hdStr (s: 'a stream) : 'a =
  match s with
  | Eos -> failwith "headless stream"
  | StrCons (x,_) -> x;;

let tlStr (s : 'a stream) : 'a stream =
  match s with
  | Eos -> failwith "empty stream"
  | StrCons (x, t) -> t ();;

let rec listify (s : 'a stream) (n: int) : 'a list =
  if n <= 0 then []
  else
    match s with
    | Eos -> []
    | _ -> (hdStr s) :: listify (tlStr s) (n - 1);;

let rec filterStream (test : 'a -> bool) (astream :'a stream)=
  match astream with
  |Eos -> Eos
  |StrCons (q,w) -> if test q then StrCons(q, fun () -> filterStream test (w () ))
      else filterStream test (w ())

let removenumbermod n =
  filterStream (fun x -> x mod n <>0)

let rec sieves natural =
  match natural with
  |StrCons(q,w) -> StrCons(q, fun () -> sieves (removenumbermod q (w ())))
  |Eos ->Eos

let isprime n =
  let primes = sieves (nums_from 2) in
  let rec helper nn pp=
    match pp with
    |StrCons(q,w)->if q>nn then false else
        if q=nn then true else helper nn (w ())
    |Eos ->false
  in helper n primes

let primes = sieves (nums_from 2)

let search_threeconsecutiveprimes (primess : int stream) =
  let rec helper current next rest =
    match rest with
    |Eos->Eos
    |StrCons(q,w)-> let nextofnext = q in
        if isprime (current*current+next*next+nextofnext*nextofnext ) then
          StrCons((current, next, nextofnext), fun ()-> helper next nextofnext (w()))
        else helper next nextofnext (w ())
  in let fst = hdStr primes
  in let snd = hdStr (tlStr primes)
  in helper fst snd (tlStr (tlStr primes))

let result = search_threeconsecutiveprimes primes

result:
search_threeconsecutiveprimes primes ;;

- : (int * int * int) stream = StrCons ((3, 5, 7), <fun>)


TITLE: Prove that the biscetors of adjacent suplementary angles are perpendicular?
QUESTION [0 upvotes]: I did the following: Take the vectors $A=(a,0), B=(x,y), -A=(-a,0)$.
Then multiply each vector $X$ by $\cfrac{1}{|X|}$ to obtain the unit vector. (I don't know if this step is really necessary.)
Then we will have:
$$A'=(\frac{a}{\sqrt{a^2}},0)=(1,0) \quad B'=(\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}}  )$$
Then to bissect the first and second angle, I'd use:
$$C=\cfrac{1}{2}(B'-A')+A'\quad \quad \quad D=\cfrac{1}{2}(B'+A')-A' $$
And hence show that it's inner product equals zero. But for some unknown reason, It's not working. I'm computing $\langle C,D \rangle$ it with Mathematica and obtaining:
$$\frac{x^2}{4 \left(a^2+b^2\right)}-\frac{a x}{4 \sqrt{a^2+b^2}}+\frac{x}{4 \sqrt{a^2+b^2}}+\frac{y^2}{4 \left(a^2+b^2\right)}+\frac{a}{4}-\frac{1}{2}$$
I don't know what is wrong.

REPLY [1 votes]: Let see what happened. $$C=\frac{1}{2}(B'+A') = \frac{1}{2}\left(1+\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}}\right).$$
$$D=\frac{1}{2}(B'-A') = \frac{1}{2}\left(-1+\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}}\right).$$
Then, $$CD = \frac{1}{4}\left( \frac{x^2}{x^2+y^2}-1 + \frac{y^2}{x^2+y^2}\right) = 0.$$


TITLE: Show that the transformation T defined by $T(x_1, x_2)\; = \;... $ is NOT linear.
QUESTION [1 upvotes]: I'm studying for a test, and I need help with this problem. I am not sure how to prove that this is not linear due to the notation. The comma is throwing me off.
Show that the transformation $T$ defined by $T(x_1,x_2)$ = $(x_1^2-2x_2, x_1+5x_2)$ is not linear.
I know that the definition of a linear transformation involves:

$T(u+v)=T(u)+T(v)$ for all $u, v$ in the domain of $T$.
$T(c*u) = c*T(u)$ for all scalars $c$ and all $u$ in the domain of $T$.
$T(cu + dv) = cT(u) + dT(v)$ for all vectors $u, v$ in the domain of $T$ and all scalars $c, d$
$T(0) = 0$ if $T$ is linear

However, I'm not sure how to use this definition with the specific function given.
Would $T(x_1 + x_2)$ = $T(x_1) + T(x_2)$ work?

REPLY [0 votes]: The transformation $T$ in fact may be linear.
We are not told what the domain of $T$ and the field of scalars is!
If $F$ is a field of characteristic 2, then $(a+b)^2=a^2+b^2$ holds for all $a,b\in F$. However, $(c\cdot x)^2=c\cdot x^2$ with $x\ne0$ implies $c=0\lor c=1$ in any field. Hence we see that $T\colon F\times F\to F\times F$ is not $F$-linear, but atleast it is $\mathbb F_2$-linear.


TITLE: What is the definition of a generalized eigenvector?
QUESTION [6 upvotes]: I am studying Generalized Eigenvectors. It seems that we can define them as $\mathbf{p}_i$ in this equation:
$$ (\mathbf{A}-\lambda\mathbf{I})^{k}\mathbf{p}_i = \mathbf{0} $$
in which $k$ is the algebraic multiplicity of $\lambda$ in $ |\mathbf{A}-\lambda\mathbf{I}|=0$. Also it can be defined as:
$$ (\mathbf{A}-\lambda\mathbf{I})\mathbf{p}_i = \mathbf{p}_{i-1},~~i=1\ldots,~~\mathbf{p}_{0}=\mathbf{0} $$
or at least this is what I have learned (if sth is wrong, please let me know).
Are these definitions equivalent? I mean
$ (\mathbf{A}-\lambda\mathbf{I})^{k}\mathbf{p}_i = \mathbf{0} $ if and only if $ (\mathbf{A}-\lambda\mathbf{I})\mathbf{p}_i = \mathbf{p}_{i-1} $?
I can prove the first one, if the second one is true (by multiplying both sides in $ (\mathbf{A}-\lambda\mathbf{I})$ k times), but how can I prove the second one, given the first one?
Thanks

REPLY [0 votes]: The definitions are not equivalent. Let $\mathbf{A}$ be the $n$ by n identity matrix. Then $\mathbf {A - I = 0}$.   The problem is that there are several things involved here, the characteristic polynomial, the minimal polynomial, and the elementary divisors. The generalized eigenspace for eigenvalue $\lambda$ has, as you say, dimension $k$ where $k$ is the largest power of $(X-\lambda)$ that divides the characteristic polynomial. The elementary divisors are a special set of powers of  $(X-\lambda)$ whose exponents add up to $k$.
Your second definition works for each elementary divisor $(X-\lambda )^d$: you can find a vector $\mathbf {p}$ which is a generalized eigenvector not appearing in any other space corresponding  to an elementary divisor, such that a basis for the corresponding generalized eigenspace is $$\mathbf p,\  (A-\lambda I)(\mathbf p),  \ \cdots ,\  (A-\lambda I)^{d-1}(\mathbf p).$$
When you get to generalized eigenvectors you will begin working with the second "definition" but until your instructor introduces the topic of elementary divisors and Jordan blocks, concentrate on when all the eigenvectors span the vector space.


TITLE: Definition and (non)equivalent definitions of limit of f
QUESTION [2 upvotes]: so I am preparing for my final and I needed some help verifying my definitions and their justifications. So any help/feedback would be appreciated...
Let $f:(0,1) \to \mathbb{R}$ be a given function, and the limit $\lim\limits_{x \to x_0} f(x) = L$. Then out of the following defnition pick the right definition, and/or equuivalent definition. If the definition is not equivalent provide explanation:
a.) for any $\epsilon \gt 0$, there exists $\delta \gt 0$ such that for all $x \in (0,1)$ and $0 \lt |x-x_0| \lt \delta$, one has $|f(x) - L|\lt \epsilon$.
this definition is true.
b.) for any $\epsilon \gt 0$, there exists $\delta \gt 0$ such that for all $x \in (0,1)$ and $|x-x_0| \leq \delta$, one has $|f(x) - L|\leq \epsilon$.
this definition is false. (I don't quite understand how tho)
c.) for any $\epsilon \gt 0$, for any $\delta \gt 0$ such that for all $x \in (0,1)$ and $0 \lt |x-x_0| \lt \delta$, one has $|f(x) - L|\lt \epsilon$.
this definition is false because for any ϵ>0 there exists some δ>0 that is small enough. It can't be any delta.
d.) for any $\epsilon \gt 0$, there exists $\delta \gt 0$ such that for all $x \in (0,1)$ and $0 \lt |x-x_0| \lt \delta$, one has $0 \lt |f(x) - L|\leq \epsilon$.
this definition is false. suppose $f(x) = 0$ then with $L = 0$ the limit does not exist anywhere  for every$x$ as $|f(x) - L| = 0$ $0$ is not less then $0$.
these are my attempts on the identification. Could you please help me confirm these answers?
Thank you.

REPLY [1 votes]: Note that for French b) is the right definition
$\forall \varepsilon>0,\;\exists \delta>0,\; \forall x\in (0,1),\quad \big[|x-x_0|<\delta \implies |f(x)-L|<\varepsilon\big]$
For other country (UK perhaps) a) is the right definition
$\forall \varepsilon>0,\;\exists \delta>0,\; \forall x\in (0,1),\quad \big[0<|x-x_0|<\delta \implies |f(x)-L|<\varepsilon\big]$
and $b\implies a$


TITLE: Multiple derivative notation
QUESTION [3 upvotes]: Official Leibniz notation for double derivative is:
$$\frac{\mathrm d^2s}{\mathrm dt^2}$$
This term seems inconsistent. Two considerations:

We have infinitesimal change in distance $\mathrm ds$ per infinitesimal change in time $\mathrm dt$: $\mathrm ds/\mathrm dt$. Both terms are a tiny value/interval. Because the $\mathrm d$ symbolizes difference, I would as a change of the change of the distance to time intuitively write:
$$\frac{\mathrm d(\mathrm ds/\mathrm dt)}{\mathrm dt}=\frac{(\mathrm ds^2/\mathrm dt^2)}{\mathrm dt}=\frac{\mathrm ds^2}{\mathrm dt^3}$$
where the extra $\mathrm d$ says that both terms are now "double" infinitesimal differences.
Maybe more properly following mathematical logic and not my intuition, the $\mathrm d$ could be considered a "free" variable in itself that can be multiplied onto this $\mathrm ds/\mathrm dt$ fraction numerator:
$$\frac{\mathrm d(\mathrm ds/\mathrm dt)}{\mathrm dt}=\frac{(\mathrm d^2s/\mathrm dt)}{\mathrm dt}=\frac{\mathrm d^2 s}{\mathrm dt^2}$$
That agrees with the actual notation but doesn't really make physical sense now. $\mathrm d$ means (infinitesimal) difference, so that $\mathrm ds=s_{final}-s_{start}$, and therefore it makes no physical sense to consider the $\mathrm d$ and the $s$ separate. The $\mathrm ds$ is physically just a "name"/"symbol" for one term, which could just as well have been called $x$ or $a$ or anything else.

Now, while searching for an explanation, the answers always tend to consider $\frac{\mathrm d}{\mathrm dt}$ as one symbol in itself, so that a double derivative is $\frac{\mathrm d}{\mathrm dt}\frac{\mathrm d}{\mathrm dt}s=\frac{\mathrm d^2}{\mathrm dt^2}s=\frac{\mathrm d^2s}{\mathrm dt^2}$ - which makes even less physical sense, since the $\mathrm dt$ term has to be a separable term before we can treat $\frac{\mathrm ds}{\mathrm dt}$ as a normal fraction (as done in integration e.g.). $\frac{\mathrm d}{\mathrm dt}$ can't possibly be just "a symbol".
Why is $\frac{\mathrm d^2s}{\mathrm dt^2}$ the correct one in a physical context, where $\mathrm ds$ actually means the infinitesimal difference in $s$? Are my considerations in point 2 correct, and I just can't figure out that splitting $\mathrm d$ and $s$ is allowed?

Update
The answers already given at this time both indicate the use of $\mathrm d/\mathrm dt$ as merely a symbol. So, neither of my two suggestions mentioned above are the case. Sure, I can accept that. But the question still remains of why as well of how come we still treat them as variables then, e.g. in integration?
Let me clarify those two points:
Firstly, if it indeed is the case that $\mathrm d/\mathrm dt$ is merely a symbol and should be thought of as just a symbol, then I do not understand the motivation for this symbol.

Why did Leibniz choose $\mathrm d/\mathrm dt$ as a symbol, which causes the confusion and inconsistency described in the question above?
Why not, say, $\mathrm d/\mathrm d$, in which case we would get a writing-style that at least looks a bit more "consistent": $$\frac{\mathrm d}{\mathrm d}\frac st=\frac{\mathrm ds}{\mathrm dt}\qquad \frac{\mathrm d}{\mathrm d}\frac{\mathrm d}{\mathrm d}\frac st=\frac{\mathrm d^2s}{\mathrm d^2t}\qquad \frac{\mathrm d}{\mathrm d}\frac{\mathrm d}{\mathrm d}\frac{\mathrm d}{\mathrm d}\frac st=\frac{\mathrm d^3s}{\mathrm d^3t}\qquad \cdots$$
Or even better yet, if the two $\mathrm d$ involved in this $\mathrm d/\mathrm dt$ symbol have no meaning as neither a variable nor an indicator of a change in the parameter, then why use this letter at all? Why not stick to, say, the prime-notation throughout and never jump into the Leibniz notation: $$s_t'\qquad s_t''\qquad s_t'''\qquad \cdots$$

And secondly, if the $\mathrm d/\mathrm dt$ really just is a symbol, and that's it, then how come we suddenly can treat it as a fraction again containing a set of variables $\mathrm ds$ and $\mathrm dt$ that we can split apart during for instance integration? Such as here:
$$\frac{\mathrm ds}{\mathrm dt}=v\quad\Leftrightarrow\quad \mathrm ds=v\,\mathrm dt\quad\Leftrightarrow\quad \int  1 \,\mathrm ds=\int v\,\mathrm dt \quad\Leftrightarrow\quad s=\int v\,\mathrm dt$$
I hope to get this notion cleared out and appreciate all comments and answers that can help.

REPLY [0 votes]: The notation came from considering first the finite differences before going to infinitesimal differences.
Thus, the first difference in $s=s(t)$ is $\Delta s =s(t+\Delta t)-s(t),$ which gives $$\frac{\Delta s}{\Delta t}=\frac{s(t+\Delta t)-s(t)}{\Delta t},$$ and the second difference is $$\Delta \Delta s =\Delta (s(t+\Delta t)-s(t))=\Delta s(t+\Delta t)-\Delta s(t)=s(t+2\Delta t)-2s(t+\Delta t)+s(t).$$ Thus the second difference quotient is $$\frac{\Delta \Delta s}{\Delta t\Delta t}\frac{s(t+2\Delta t)-2s(t+\Delta t)+s(t)}{\Delta t\Delta t}.$$
Note that we usually write $\Delta \Delta s=\Delta^2 s$ and $\Delta t \Delta t={\Delta t}^2.$
This shows why that choice of notation was made by Leibniz.
And now a word on differentials. A differential is an infinitesimal change in a function, and an infinitesimal is an infinitely small quantity, as it were. One may semi-formally think of it as a quantity approaching zero. Not the quantity alone, or the zero limit it approaches, but the quantity and its approach to zero. Thus this semiformal definition looks like that of the vector quantities of physics, which capture a notion of directed quantities. That's what the notion of infinitesimal captures -- our intuition of instants of time, points on a line, etc.
Having said that, then it's clear that we may handle differentials separately, and perform operations on them -- and that's how they were used from the beginning (hence, the differential calculus, differential equations, etc.). In particular, if we're thinking of quantities depending on a single independent variable, then we may add, subtract, multiply the differentials of such quantities to get similar differentials. However, with division, even when the differential in the denominator is not that of a constant function, we have to be careful, because the result is no longer always a differential. In many cases it is a function, and this is why the differential calculus is about calculating this type of ratio of differentials.
When we come to integrating, what we're doing is gathering together a continuous stream of differentials, to give a quantity.
The upshot of this all is that you're right. Even those who would banish talk of differentials (even after their justification in the so-called nonstandard analysis) cannot avoid using them when calculating integrals. And I can't imagine them still talking like that if they want to integrate over arbitrary manifolds, for example. So, with care and understanding, one can calculate with differentials, once one understands their provenance and rules of behavior -- they're just infinitesimal differences, that  is we may think of $\rm d$ as the pair $(\Delta, \Delta\to 0),$ and they obey many of the usual rules of a ring (once you're dealing with only one independent variable). With more than one independent variable, things become complicated a little and one may need to run to multilinear algebra for some help. But that's another story.


TITLE: How to find possible closed-form-formula for a given decimal expansion online?
QUESTION [8 upvotes]: What mathematical program or library can guess formula for decimal fraction? For example, for number 2.414213562373095 right guess is 1 + $\sqrt 2$. Are there such tools online? I cannot find one.

REPLY [4 votes]: Using keywords from Peter's comment ("closed form by decimal expansion") I found Inverse Symbolic Calculator
Also see Sympy number identification


TITLE: Prove $\sum_{k=0}^{n}{n \choose k}(-1)^k \frac{1}{k+1} = \frac{1}{n+1} $
QUESTION [1 upvotes]: I am going through a proof of the combinatorial result $$\sum_{k=0}^{n}{n \choose k}(-1)^k \frac{1}{k+1} = \frac{1}{n+1}, $$ and have found something I don't understand.
The proof is as follows:
$$\sum_{k=0}^{n}{n \choose k}(-1)^k \frac{1}{k+1} = \sum_{k=0}^{n}{n \choose k}(-1)^k \int_0^1 t^k\ dt = \int_0^1 (1-t)^n\ dt = \frac{1}{n+1}.$$
I understand this, except from the second to last equality. Why does this equality hold?
Any help would be appreciated!

REPLY [1 votes]: It’s also possible to give a combinatorial proof.
Let $A=\{0\}\cup[n]=\{0,\ldots,n\}$; we’ll count in two ways the permutations of $A$ that begin with $0$.
On the one hand there are of course $n!$ of them, since the remaining $n$ elements can be arranged in any order.
Alternatively, for $k\in[n]$ let $P_k$ be the set of permutations of $A$ in which $k$ precedes $0$; clearly exactly half of the permutations of $A$ are in $P_k$, so $|P_k|=\frac12(n+1)!$. More generally, if $\varnothing\ne I\subseteq[n]$, then $\bigcap_{k\in I}P_k$ is the set of permutations in which $0$ is preceded by every element of $I$, and
$$\left|\bigcap_{k\in I}P_k\right|=\frac{(n+1)!}{k+1}\;.$$
A straightforward inclusion-exclusion argument now yields the equation
$$\left|\bigcup_{k\in[n]}P_k\right|=\sum_{\varnothing\ne I\subseteq[n]}(-1)^{|I|+1}\frac{(n+1)!}{k+1}=\sum_{k=1}^n(-1)^{k+1}\binom{n}k\cdot\frac{(n+1)!}{k+1}\;.$$
But $\bigcup_{k\in[n]}P_k$ is just the set of permutations of $A$ in which $0$ does not come first, so
$$n!=(n+1)!-\sum_{k=1}^n(-1)^{k+1}\binom{n}k\cdot\frac{(n+1)!}{k+1}=(n+1)!\sum_{k=0}^n(-1)^k\binom{n}k\frac1{k+1}\;,$$
and dividing through by $(n+1)!$ yields the desired identity.


TITLE: Which is correct? Group Theory
QUESTION [7 upvotes]: Unfortunately I noticed that all are wrong.
(A) Counterexample: $G=(1)$
(B), (C), (D) Counterexample: $G=\{1,-1\}$
Please help! Where did I go wrong?

REPLY [7 votes]: You are right, and the question is wrong. (And you did not need to ask here to be sure; your counterexamples are convincing).
If it were your task to correct the question (and unfortunately readers of mathematical texts often find themselves in the uncomfortable position of having to correct apparent errors in what they are reading, and guessing the most probable oversight of the author) then you can choose from the additional hypotheses "$|G|>1$" (or $G$ is nontrivial) which makes (A) true, or "$2\nmid|G|$" (or $G$ is of odd order) which makes (D) true.
The first correction seems the most likely one. However it requires (very elementary) ring theory, not just group theory. If $G$ is a finite subgroup of the multiplicative group of a field (or domain) and $a\in G\setminus\{1\}$ then $a\sum_{g\in G} g=\sum_{g\in G}ag=\sum_{g\in G}g$ so $(a-1)\sum_{g\in G}g=0$ and since $a-1\neq0$ one has $\sum_{g\in G}g=0$.
If instead one adds the hypothesis $|G|$ is odd, then $\prod_{g\in G}g=1$. This follows from simple group theory and the fact that $G$ is Abelian (without which the product would make no sense). One can pair up mutually inverse elements in the product, which pairs contribute nothing, leaving only the involutions ($g^2=1$) single, so $\prod_{g\in G}g$ gives the same group restricted to the involutions only. In a group of odd order the identity is the only involution. (When instead $|G|$ is taken even then $\prod_{g\in G}g=-1$, since $-1$ is the unique element of order$~2$ in $G$, and even in $\Bbb C^*$.)
Added. The argument given for (A) with the assumption $|G|>1$ does not use commutativity of multiplication, so it holds for instance for finite subgroups of a skew field (a.k.a. division ring), which do not have to be cyclic. However, it does use the absence of zero divisors, and one might wonder if this can be avoided. The answer is it cannot, as the property sometimes fails in rings with zero divisors: the sum of both invertible elements in $\Bbb F_2[X]/(X^2)$ is $X$ rather than $0$.

REPLY [2 votes]: Yes you are right. So the problem should be modified that $G$ be a nontrivial finite subgroup, in which case any element $g\in G$ has the form
$$g=e^{i\theta}$$
where $\theta$ is rational. Thus the answer is, all add up to 0.


TITLE: Equinumerosity of strings and natural numbers
QUESTION [2 upvotes]: so I'm trying to show that there exists a bijection from the infinite set of finite strings composed of elements from $\{a, b\}$ to $\mathbb{N}$. So I was thinking about showing that $\mathbb{N}$ and the infinite set of finite binary sequences are equinumerous (each have a unique representation), and then constructing a bijection between $\{a, b\}$  and $\{0, 1\}$. Then by the transitive property of equinumerosity, I would be done with my goal.
However, when you're writing something in binary, like 00001 (which would be equivalent to aaaab), and so I think this proof would exclude some cases? How else can I go about this?

REPLY [0 votes]: If your alphabet is $\{0, 1\}$, a simple way to describe the bijection with $\mathbb{N} = \{1, 2, \dotsc\}$ is to stick an 1 in front of the string, and read the result as a binary number. To go the other way around, write the number in binary and chop off the first digit.
The general case for an arbitrary alphabet is called bijective numeration


TITLE: If $M=v_1v_i^TD_i$, compute $v_i$ and $D_i$
QUESTION [1 upvotes]: A known real-valued $n\times n$ matrix $M$ can be written as
$$
M = v_1v_i^TD_i,
$$
(no summation on repeated indices) where the $v_i$ are orthonormal column vectors, $D_i$ are diagonal matrices and
$1\leq i\leq n$.

Knowing only $M$ with the properties stated above, how can the $v_i$ and $D_i$ be computed?

Computing $MM^T$ we have
$$
MM^T=v_1v_i^TD_iD_jv_jv_1^T=(v_i^TD_iD_jv_j)v_1v_1^T=cv_1v_1^T,
$$
where $c$ is a real numberm. As this result must be true whatever $i$ or $j$, then the number $c$ must be the same whatever $i$ or $j$.  Now the vector $v_1$ can be computed through $MM^T=cv_1v_1^T$ and then be used to obtain
$$
v_1^TM=v_i^TD_i.
$$
And here is how far I could go. Obviously the diagonal matrix $D_1$ can also be computed, but I have no clue about how to get the other ones.

Numerical Example
Below there is an example of a matrix $M$with the properties stated above and a solution given by the orthonormal vectors $v_i$. The diagonal matrices can be easily computed.
$$
M=\begin{pmatrix}
   0.14329 &  0.14545 &  0.14999 &  0.14607 &  0.18787  & 0.18648 &  0.14401\\
   0.15788 &  0.16026 &  0.16525 &  0.16094 &  0.20698  & 0.20547 &  0.15866\\
   0.10375 &  0.10532 &  0.10860 &  0.10576 &  0.13603  & 0.13503 &  0.10427\\
   0.15007 &  0.15233 &  0.15708 &  0.15298 &  0.19675  & 0.19531 &  0.15082\\
   0.11313 &  0.11484 &  0.11842 &  0.11533 &  0.14833  & 0.14724 &  0.11370\\
   0.11289 &  0.11459 &  0.11817 &  0.11508 &  0.14801  & 0.14692 &  0.11346\\
   0.11027 &  0.11193 &  0.11540 &  0.11240 &  0.14455  & 0.14350 &  0.11081
\end{pmatrix},
$$
and
$$
[v_1,v_2,\ldots,v_7]=
\begin{pmatrix}
0.4199104 &0.2134484 &-0.3097492& -0.3709365&-0.7358314&-0.0549746&0.0103221\\
0.4626576& 0.4298270& 0.2838626&-0.2988347& 0.4186424& 0.1090201& 0.4941353\\
0.3040402&-0.4154454&-0.1812354&-0.4642904& 0.3939669&-0.4718971&-0.3296302\\
   0.4397767 & -0.6990359 &  0.0094533 &  0.1918160 & -0.0817424 &  0.4438271  & 0.2782053\\
   0.3315402  & 0.1327565 & -0.3266486  & 0.6639579 &  0.0725541&  -0.5220768&   0.2170005\\
   0.3308324 &  0.3013229 & -0.2818176  & 0.1981848  & 0.2493111 &  0.4958764  &-0.6107427\\
   0.3231259&   0.0047608&   0.7774454&   0.2015631&  -0.2323336&  -0.2167151&  -0.3867475
\end{pmatrix}
$$

REPLY [1 votes]: The statement that a matrix can be written as the product of two vectors
$$M = ab^T$$
means that it is a rank-one matrix, i.e. ${\rm rank}(M)=1$.
NB: Your numerical example has ${\rm rank}(M)=7$, so it cannot be the product of two vectors.
The particular form you have chosen
$$M=av^TD$$
reduces to the above canonical form by setting $\,b^T=v^TD$.
Let $m_k$ denote the $k^{th}$ column of $M$.
Then the $a$-vector is equal to any such column, after normalization.
$$a=\frac{m_1}{\|m_1\|}=\frac{m_2}{\|m_2\|}=\ldots$$
The components of the $b$-vector are simply the normalization factors.
$$b^T=\pmatrix{\|m_1\|&\|m_2\|&\|m_3\|&\ldots&}$$
The $b$-vector itself can be factored into a normalized vector and a scale factor.
$$\beta=\|b\|=\|M\|_F,\quad n=\frac{b}{\beta},\quad b=\beta\,n$$
This yields unique factorization of a rank-one matrix in terms of two unit vectors and a scalar factor.
$$M=\beta\,ab^T$$
This is, in fact, the Singular Value Decomposition of the matrix.


TITLE: Simple two variable am-gm inequality
QUESTION [2 upvotes]: Given $x,y \in \Bbb{R}$, show that:$$x^2+y^2+1\ge xy+y+x $$
I tried using the fact that $x^2+y^2 \ge 2xy$ But then I'm not sure how to go on, Also tried factoring but didn't help much, also tried substituting $\frac{x^2+y^2}{2}$ instead of $xy$ but that gave me the same result of the first substitution, i.e. $xy+1\ge x+y$
This inequality seems very easy, I'm feeling dumb for not having solved it yet

REPLY [6 votes]: Since $$a^2-2ab+b^2 = (a-b)^2\geq 0\implies a^2+b^2\geq 2ab$$ we have $$ x^2+y^2\geq 2xy$$
$$x^2+1\geq 2x$$
$$y^2+1\geq 2y$$
Now add all these...

REPLY [2 votes]: We need to prove that
$$y^2-(x+1)y+x^2-x+1\geq0,$$ for which it's enough to prove that
$$(x+1)^2-4(x^2-x+1)\leq0$$ or $$(x-1)^2\geq0.$$


text:
\section{Introduction}

This paper is the second part of a trilogy dedicated to the following problem:
given spherically symmetric characteristic
initial data for the Einstein-Maxwell-scalar
field system with a cosmological constant $\Lambda$,
with the data on the outgoing initial null
hypersurface given by a subextremal Reissner-Nordstr\"{o}m black
hole event horizon,
and the remaining data otherwise free,
study the future extendibility of the corresponding
maximal globally hyperbolic development as a  ``suitably regular'' Lorentzian manifold.
We are motivated by the strong cosmic censorship conjecture and the question of determinism in general relativity.
As explained in detail in the Introduction of Part~1, strong cosmic censorship is one of the most fundamental open problems in
general relativity (see the classic monographs \cite{ChruscielSCC, Earman} and the
discussions in \cite{ChristodoulouGlobalnew, Dafermos1, DafermosBlack} for the general context of this problem).
Although significant developments have been achieved in the last five
decades (from the initial heuristic works \cite{SimpsonInternal, IsraelPoisson} to rigorous mathematical results \cite{Dafermos1, Dafermos2, DafermosProof}), including some recent encouraging progress (see \cite{Franzen, LukInst, DafermosBlack} and references therein), a complete resolution of the conjecture at hand still seems out
of reach. Nonetheless, the spherically symmetric self-gravitating scalar field model has provided considerable insight into the harder problem of vacuum collapse without symmetries \cite{Christodoulou:2008}; this was explored in \cite{LukWeak} to obtain the first promising steps towards understanding the stability of Cauchy horizons without symmetry assumptions.

In Part~1, we established the equivalence (under appropriate regularity conditions for the initial data) between the Einstein equations~\eqref{wave_r}$-$\eqref{wave_Omega}
and the system of first order PDE~\eqref{r_u}$-$\eqref{kappa_at_u}. We proved existence, uniqueness and identified a breakdown criterion for solutions of this system (see Section~\ref{main-one}).

In the current paper we are concerned with the structure of the solutions of the characteristic problem,
and wish to
address the question of existence and stability of the Cauchy horizon
when the initial data is as above. This is intimately related to the issue of global uniqueness for the Einstein equations: it is the possibility of extension of
solutions across this horizon that leads to the breakdown of global uniqueness and, in case the phenomenon persists for generic initial data, to the failure of the strong cosmic
censorship conjecture.

As in~\cite{Dafermos1}, we introduce a certain generic element in the formulation
 of our
problem by
perturbing a subextremal Reissner-Nordstr\"{o}m black
hole (whose Cauchy horizon formation is archetypal) by arbitrary characteristic data along the ingoing null direction. The study of the conditions under
which the solutions can be extended across the Cauchy horizon is left to Part~3.

We take many ideas from~\cite{Dafermos1} and~\cite{Dafermos2} and build on these works.
In particular, we borrow the following three very important techniques.
(i)~The partition of the spacetime domain of the solution into four regions and the construction of a carefully chosen spacelike curve to separate the last two.
(ii)~The use of the Raychaudhuri equation in~$v$ to estimate $\frac{\nu}{1-\mu}$ at a larger~$v$ from its value at a smaller~$v$.
(iii)~The use of BV estimates for the field.

Nonetheless, the introduction of a cosmological constant $\Lambda$ causes a significant
difference that requires deviation
from the original strategies developed in~\cite{Dafermos1} and~\cite{Dafermos2}.
Moreover, we introduce some technical simplifications and obtain sharper and more detailed estimates.
These improvements will be crucial for our arguments in Part~3.

Our approach therefore has three main departures from
the one of Dafermos:
\begin{enumerate}[i)]
	\item First, due to the presence of the cosmological constant $\Lambda$, the curves of constant shift, which are used in~\cite{Dafermos1} and~\cite{Dafermos2},
        are no longer necessarily spacelike for $\Lambda>0$ large. This forces us to find an alternative approach;
	we have chosen to work with curves of constant $r$ coordinate instead of working with curves of constant shift, which turns out to be a simpler approach.
	Furthermore, it allows us to treat the cases $\Lambda<0$, $\Lambda=0$ and $\Lambda>0$ in a unified framework.
	\item Second, we show that the Bondi coordinates $(r,v)$ are the ones most adapted to estimating the growth of the fields
	as we progress away from the event horizon.
	Our approach starts by controlling the field $\frac\zeta\nu$ using~\eqref{triple}. Although this is similar to~\eqref{double}, there is one distinction
	which makes all the difference. It consists of the fact that in the double integral in~\eqref{double} the field $\frac\zeta\nu$ is
	multiplied by the function $\nu$. When we pass to Bondi coordinates this function disappears, making a simple application of Gronwall's inequality,
	such as the one we present, possible. This would not work in the double null coordinate system~$(u,v)$.
	\item Third, our estimates are not subordinate to the division of the solution spacetime into red shift, no shift and blue shift regions. Instead, we consider the regions $\{r \geq \ckrp\}$, $\{\ckrm \leq r \leq \ckrp\}$ and $\{r \leq \ckrm\}$, where $\ckrp$ is smaller than but sufficiently close to the radius $r_+$ of the Reissner-Nordstr\"om event horizon, and $\ckrm$ is bigger than but sufficiently close to the radius $r_-$ of the Reissner-Nordstr\"om Cauchy horizon. These may be loosely thought of as red shift, no shift and blue shift regions of the background Reissner-Nordstr\"om solution, even though the shift factor is not small and indeed changes significantly from red to blue in the intermediate region.
\end{enumerate}

Our first objective is to obtain good upper bounds for $-\lambda$ in the different regions of spacetime. These will enable us to show that
the radius function $r$ is bounded below by a positive constant. However, good estimates for $-\nu$ and the fields $\theta$ and $\zeta$ will also be essential in Part~3.

The main result of this paper is therefore
\begin{Thm}\label{r-stability}
Consider the characteristic initial value problem for the first order system of PDE~\eqref{r_u}$-$\eqref{kappa_at_u}
 with initial data\/~\eqref{iu}$-$\eqref{iv} (so that $\{0\}\times[0,\infty[$ is the event horizon of a subextremal Reissner-Nordstr\"{o}m solution with mass $M>0$).
 Assume that $\zeta_0$ is continuous and $\zeta_0(0)=0$. Then there exists $U>0$ such that
the domain $\mysigma$ of the (future) maximal development contains $[0,\myumax]\times[0,\infty[$. Moreover,
$$
\inf_{[0,\myumax]\times[0,\infty[}r>0,
$$
the limit
$$
r(u,\infty):=\lim_{v \to \infty}r(u,v)
$$
exists for all $u \in \left] 0,U \right]$ and
$$
\lim_{u\searrow 0}r(u,\infty)=r_-.
$$
\end{Thm}
So, under the hypotheses of Theorem~\ref{r-stability},
the argument in~\cite[Section~11]{Dafermos2}, shows that, as in the case when $\Lambda=0$, the spacetime is extendible across the Cauchy horizon with a $C^0$ metric.

We also prove that only in the case of the Reissner-Nordstr\"{o}m solution does the curve $\{ r= r_- \}$ coincide with the Cauchy horizon. As soon as the initial data field is not identically zero, the curve
$\{ r= r_- \}$ is contained in~${\cal P}$ (Theorem~\ref{rmenos}). This is an interesting geometrical condition and it is conceptually relevant given the importance
that we confer to the curves of constant $r$. We also prove that, in contrast with what happens with the Reissner-Nordstr\"{o}m solution,
the presence of any nonzero field immediately causes the integral $\int_0^\infty \kappa(u,v)\,dv$ to be finite for any $u>0$ (Lemma~\ref{l-kappa}).
As a consequence, the affine parameter of any outgoing null geodesic inside the event horizon is finite at the Cauchy horizon (Corollary~\ref{affine}).

config: arxiv


TITLE: Boyd & Vandenberghe, example 2.13 — what is the "convex set of joint probabilities" in this example?
QUESTION [4 upvotes]: I am reading Boyd & Vandenberghe's Convex Optimization. I can't understand the example below:

what is the "convex set of joint probabilities" in this example?
what is the convex set $C$ here?

REPLY [0 votes]: Explanations so far do not do the book justice.
The joint probabilities pij can be seen as n m x n matrix but equivalently as vector of N = m x n nonnegative numbers that sum to one. It helps to picture what is meant if we think of pij as a vector.
As explained in an earlier example 2.5 in the book the set of all possible such probability vectors lie in the probability simplex of affine dimension N-1 embedded in N dimensional space. This simplex is evidently convex, but that is not the point here.
The convex set C in example 2.13 in the book is meant to be any convex subset of this probability simplex.
Each possible vector pij in C then transforms to a vector of conditional probabilities fij of same dimension N by the linear-fractional function given. fij will typically no longer lie in the probability simplex since conditional probabilities need not sum to one.
The claim in the book is that when you sweep out C by varying pij, then fij will sweep out a region f(C) in N-space that is still convex.