TITLE: Linear algebra axler 8.c.16
QUESTION [0 upvotes]: Suppose $V$ is an inner product space and $T$ $\in \mathcal L(V)$. Suppose $$a_0+a_1z+a_2z^2+\cdots+a_{m-1}z^{m-1}+z^m$$ is the minimal polynomial of $T$. Prove that$$\overline{a_0}+\overline{a_1}z+\overline{a_2}z^2+\cdots+\overline{a_{m-1}}z^{m-1}+z^m$$is the minimal polynomial of $T^*$.

Since the first polynomial is a minimal polynomial, we have$$a_0+a_1T+a_2T^2+\cdots+a_{m-1}T^{m-1}+T^m=0$$ I try to use the fact that $M(T)=(\overline{M(T^*)})^t$.
Can anyone give me a hint or tell me what should I consider next? I am stuck here..

REPLY [1 votes]: Recall that the minimal polynomial of $X\in\mathcal L(V)$ is the monic
polynomial $\mu_X(t)$ of least degree that annihilates $X$.
Also recall that the computation
$$
\overline{p}(X)
= \overline{a_0}I+\overline{a_1} X+\dotsb+\overline{a_n}X^n
= (a_0I+a_1X^*+\dotsb+a_n(X^*)^n)^*
= p(X^*)^*
$$
proves that $\overline{p}(X)=p(X^*)^*$ for any $X\in\mathcal L(V)$ and any
polynomial $p(t)$.
Now, you wish to show that $\mu_{T^*}(t)=\overline{\mu_T}(t)$. To do so, suppose
$p(t)$ annihilates $T^*$. Then
$$
\overline{p}(T)=p(T^*)^*=0^*=0
$$
so $\mu_T(t)$ divides $\overline{p}(t)$. It follows that $\overline{\mu_T}(t)$
divides $p(t)$.
That is, $\overline{\mu_T}(t)$ is a monic polynomial that annihilates $T^*$ and
divides every other polynomial that annihilates $T^*$. Hence
$\mu_{T^*}(t)=\overline{\mu_T}(t)$ as desired.


TITLE: A conjecture about the entropy of matrix vector products
QUESTION [5 upvotes]: Consider a random $m$ by $n$ partial circulant matrix $M$ whose entries are chosen independently and uniformly from $\{0,1\}$ and let $m < n$.  Now consider a random $n$ dimensional vector $v$ whose entries are also chosen independently and uniformly from $\{0,1\}$.  Let $N = Mv$ where multiplication is performed over the reals.
The matrices $M$ and $N$ are discrete random variables. Recall that the Shannon entropy for a discrete random variable $Z$ is $H(Z) = -\sum_z P(Z=z)\log_2{P(Z=z)}$. In the case where $P(Z=z)=0$ for some values $z$, the corresponding term in the sum is taken to be $0$.
We therefore know that the (base $2$) Shannon entropy $H(M) = H(v) = n$.  The fact that $H(M) = n$ is a direct result of the fact that the entire matrix is defined by its first row.
If $m = \lfloor 10n/\ln{n} \rfloor$ then I would like to make the following conjecture.

Conjecture: For all sufficiently large $n$, $H(N) \geq n/10$.

The value $10$ is chosen somewhat arbitrarily to be a sufficiently large constant.
Is this a known problem and/or can anyone see a way to approach it? Is it in fact true?

REPLY [2 votes]: This is not an answer, just a long comment.
As pointed out by John Mangual, this game is similar to Mastermind.
In fact, it is even more similar to the game defined by Erdos and Renyi here:
http://www.renyi.hu/~p_erdos/1963-12.pdf.
In your language their game is the following:
We are given the first row of the matrix $M$, denoted by $M'$, (so $m=1$) but what we multiply it with, $X$, is not a vector, but an $n\times a$ matrix.
Our goal is to reconstruct $M$ from $N=M'X$. They show that this can be done whp if $X$ is a random matrix and $a\ge 10n/\log n$.
Of course if $M$ can be reconstructed whp, that also implies that $H(N)\ge (1-\epsilon)n$.
So one possible way to answer your question would be to show that their theorem holds even in the case when $X$ is a circulant matrix, as this is equivalent to $X$ being a vector and $M$ being circulant.
new part
So suppose we want to compute the probability that for two different random vectors, denoted by $v$ and $u$, multiplying them with the rotations of a random vector $r$ we get the same values, i.e., if the rotations of $r$ are denote by $r_1,\ldots,r_k$, then what is the chance that for all $i$ we have $vr_i=ur_i$.
For any $i$, this is like a random walk, as $v$ and $u$ are also random, so $Pr[vr_i=ur_i]\approx 1/\sqrt n$.
Now I claim that this statement is true even in the following conditional form if $k$ is small enough: $Pr[vr_i=ur_i\mid \forall j\ne i: vr_j=ur_j]\approx 1/\sqrt n.$
I don't think this is that hard to prove, but I cannot see a solution now.


TITLE: Clifford PBW theorem for quadratic form
QUESTION [14 upvotes]: $\DeclareMathOperator\Cl{Cl}$Update Feb 3 '12: now with a question 2 which is much more elementary (and should be well-known!).
Let $k$ be a commutative ring with $1$. Let $L$ be a $k$-module, and $g:L\to k$ be a quadratic form, i. e., a map for which the map $L\times L\to k,\ \left(x,y\right)\mapsto g\left(x+y\right)-g\left(x\right)-g\left(y\right)$ is $k$-bilinear and which satisfies
$$g\left(\lambda x\right)=\lambda^2 g\left(x\right)\quad\text{for all $x\in L$ and $\lambda\in k$.}$$
We define the Clifford algebra $\Cl\left(L,g\right)$ as the tensor algebra $\otimes L$ (all tensors are over $k$) divided by the two-sided ideal generated by $x\otimes x-g\left(x\right)$ for all $x\in L$.
Question 1: Is $\Cl\left(L,g\right)$ isomorphic to ${\bigwedge}L$ as a $k$-module? Is the associated graded object of $\Cl\left(L,g\right)$ isomorphic to ${\bigwedge}L$ as a $k$-algebra?
Remark: The answer is yes if the quadratic form $g$ comes from a bilinear (not necessarily symmetric!) form. As a consequence, the answer is yes if $L$ is a free $k$-module or, more generally, a direct sum of quotient $k$-modules of $k$ (in fact, it is easy to see that in this case, every quadratic form on $L$ comes from a bilinear form). (I think that in the case when $L$ is a finite-free $k$-module, the answer "yes" can also be proven by the diamond lemma, though I have not checked.) I am interested in the "perverse" cases when none of these holds, but I suffer from a lack of perversion: I can't name any such case. So here is the question which obviously needs to be addressed first:
Question 2: Find a commutative ring $k$ with $1$ and a $k$-module $L$ with a quadratic form (this is defined as above) which doesn't come from any bilinear (symmetric or not) form on $L$. (A quadratic form $g$ is said to come from a bilinear form $h$ if we have $g\left(v\right)=h\left(v,v\right)$ for every $v\in L$.)
Note: The Clifford algebra of a quadratic form is, in some sense, a "little sister" of the universal enveloping algebra of a Lie or pseudo-Lie algebra. ("Little sister" not in the historical sense, but in the sense of partly having the same properties, but them being easier to prove in the Clifford case than in the universal enveloping algebra case.) The above question asks for a kind of Poincaré–Birkhoff–Witt (PBW) theorem for the Clifford algebra. (Note that the PBW theorem itself requires some niceness conditions such as $L$ being finite-free or $k$ being a $\mathbb Q$-algebra, so it wouldn't surprise me if the Clifford case also doesn't work in full generality. But the opposite case wouldn't surprise me either, because PBW is substantially harder than PBW for Clifford algebras, even in characteristic $0$.)

REPLY [5 votes]: If someone could check the below I'd be very indebted.
Found the counterexample (to Question 1 and thus also to Question 2). It is inspired by the counterexample to ring-theoretical PBW in P. M. Cohn, A remark on the Birkhoff-Witt theorem. J. London Math. Soc. 38 1963 pp. 197-203, MR0148717 (though I still don't know whether Cohn's counterexample is true for all primes $p$ - but here I only need $p=2$).
Let $k$ be the commutative ring $\mathbb F_2 \left[\alpha,\beta,\gamma\right] / \left(\alpha^2,\beta^2,\gamma^2\right)$.
Let $L$ be the $k$-module $\left\langle x,y,z\right\rangle / \left\langle \alpha x - \beta y - \gamma z\right\rangle$.
Define a map $q:L\to k$ by $q\left(\overline{ax+by+cz}\right) = a^2+b^2+c^2$ for all $a,b,c\in k$. This map $q$ is easily seen to be well-defined and a quadratic form; it also satisfies $q\left(\overline x\right)=q\left(\overline y\right)=q\left(\overline z\right)=1$.
In the Clifford algebra $\mathrm{Cl}\left(L,q\right)$, we have $\left(\overline{\alpha x-\beta y-\gamma z}\right)\cdot\overline x = \alpha \underbrace{\overline{x}\cdot\overline{x}}_{=q\left(\overline x\right)=1} - \beta \overline y \overline x - \gamma \overline z \overline x = \alpha - \beta \overline y \overline x - \gamma \overline z \overline x$. Since the left hand side of this equality is $0$ (because $\overline{\alpha x-\beta y-\gamma z}=0$), we thus have $ \alpha - \beta \overline y \overline x - \gamma \overline z \overline x = 0$. Multiplied with $\beta\gamma$, this becomes $\alpha\beta\gamma = 0$ (because $\beta^2=0$ and $\gamma^2=0$ make the other two terms disappear). This must hence hold in $\mathrm{Cl}\left(L,q\right)$. But this does not hold in $k$, and thus not in $\wedge L$ either, so we cannot have an isomorphism.


TITLE: Using Intermediate Value Theorem and Rolle's Theorem to solve for roots.
QUESTION [2 upvotes]: I was having trouble with this question. You are supposed to use the Intermediate Value Theorem and Rolle's Theorem to prove that $x^{4}+4x-8$ has exactly two real roots. How would I do this? Thank you.

REPLY [3 votes]: Let $f(x) = x^{4}+4x-8$. Note $f(1) = -3 < 0$, but $f(2) = 16 > 0$. Hence, $f(x)$ has a root in $(1,2)$. Further note that $f(-3) = 81 - 12 - 8 = 61 > 0$, so $f(x)$ has a root in $(-3, 1)$. (One can observe that $f(-2)=0$, alternatively.) Hence, $f(x)$ has at least two real roots.
Now differentiate $f(x)$ to find $f'(x) = 4x^{3}+4$. Factoring and setting $f'(x) = 0$, we see $f'(x) = 4(x^{3}+1) = 4(x+1)(x^{2}-x+1)$, which has exactly one real root, $x = -1$. Hence, by Rolle's theorem, we conclude...


TITLE: Can I blindly trust WolframAlpha?
QUESTION [1 upvotes]: Recently, my friend gave me the following indefinite integral to evaluate : $$\int \frac{\sin^2 x }{\sin x +\cos x} \mathrm dx$$
I searched it on WolframAlpha, just to be sure, that whether it has an elementary primitive or not. It turn out to be not a elementary one.

Then I asked him, that why did he gave it to me, if it wasn't supposed to be a question for High-schoolers?
His reply surprised me! He had solved it. It goes as follows -
\begin{align}\int \frac{\sin^2 x }{\sin x +\cos x}\mathrm dx &=\int \frac 12 \left(  \frac{\sin^2 x +\color{blue}{\cos^2 x}}{\sin x +\cos x} + \frac{\sin^2 x -\color{blue}{\cos^2 x}}{\sin x +\cos x} \right)\mathrm dx \\
&=\frac 12 \int \frac{1}{\sin x +\cos x}\mathrm dx +\frac 12 \int (\sin x -\cos x)\mathrm dx\\
\end{align}
Now the first part can be solved using $\sin x +\cos x = \sqrt 2 \cos \left( x -\dfrac {\pi}4 \right)$ and $\int \sec x \mathrm dx = \ln (\sec x+ \tan x )$. Second part is trivial.
Did I get something wrong, or Wolfy is the wrong one here?

REPLY [1 votes]: Wolfram Alpha did a perfect job, as usual.
Looking just below the answer, there is a plot clearly showing that the complex version is considered.
And looking a little more below, alternate forms are given, the first of which doesn't have the complex constants.
If you worry about the hyperbolic function, you have to know that WA's answer and yours are equivalent, and the reference to a well-known function is probably a good idea.


TITLE: Expressing recurrence relations in matrix form
QUESTION [0 upvotes]: Consider the following recurrence relations for the sequences $a_n$ and $b_n$:
$ (\gamma_1+ \alpha_1 + n \gamma) a_n = \lambda_1 a_{n-1} + \alpha_2 b_n $
$ (\gamma_2+ \alpha_2 + n \gamma) b_n = \lambda_2 b_{n-1} + \alpha_1 a_n $
subject to the initial condition, say, $a_0 = k_1 $ and $b_0=k_2$.
I want to reexpress the above system in the matrix form, such as
$ A_{n+1} = B A_n $,
where $A_n$ is the vector $(a_n, b_n)^T$ and $B$ is the coefficient matrix. Note that the coefficient matrix commutes with itself for all $n$. Also, if possible, I want to go further and have both a diagonal matrix and a coefficient matrix on the right-hand side of the equality.
Any help would be appreciated!
Edit:
I am looking for a matrix-based generalisation of the solution to the above system. That is, What if I had one more sequence, say $c_n$, and therefore the following system:
$ (\gamma_1+ \alpha_1 + \beta_1 + n \gamma) a_n = \lambda_1 a_{n-1} + \alpha_2 b_n + \alpha_3 c_n$
$ (\gamma_2+ \alpha_2 + \beta_2 + n \gamma) b_n = \lambda_2 b_{n-1} + \alpha_1 a_n + \beta_3 c_n$
$ (\gamma_3+ \alpha_3 + \beta_3 + n \gamma) c_n = \lambda_3 c_{n-1} + \beta_1 a_n + \beta_3 b_n$

REPLY [1 votes]: It's an ugly mess, but I got through it:
$$\begin{align*}&(\gamma_1+\alpha_1+n\gamma)a_n=\lambda_1a_{n-1}+\alpha_2b_n
\\&(\gamma_2+\alpha_2+n \gamma)b_n=\lambda_2b_{n-1}+\alpha_1a_n
\\&\implies a_n=\frac{\lambda_1a_{n-1}+\alpha_2b_n}{\gamma_1+\alpha_1+n\gamma}
\\&\implies b_n=\frac{\lambda_2b_{n-1}+\alpha_1a_n}{\gamma_2+\alpha_2+n\gamma}
\\&\implies a_n(\gamma_1+\alpha_1+n\gamma)=\lambda_1a_{n-1}+\frac{\lambda_2\alpha_2b_{n-1}+\alpha_1\alpha_2a_n}{\gamma_2+\alpha_2+n\gamma}
\\&\implies b_n(\gamma_2+\alpha_2+n \gamma)=\lambda_2b_{n-1}+\frac{\lambda_1\alpha_1a_{n-1}+\alpha_1\alpha_2b_n}{\gamma_1+\alpha_1+n\gamma}\end{align*}$$
For convenience, let:
$$\begin{align}\sigma&=\gamma_1+\alpha_1+n\gamma\\\tau&=\gamma_2+\alpha_2+n\gamma\\\omega&=\sigma\tau-\alpha_1\alpha_2\end{align}$$
Now if we skip over an ugly in-between step,
$$a_n\cdot\omega=a_{n-1}(\lambda_1\cdot\tau)+b_{n-1}(\lambda_2\cdot\alpha_2)
\\b_n\cdot\omega=a_{n-1}(\lambda_1\cdot\alpha_1)+b_{n-1}(\lambda_2\cdot\sigma)$$
And with a pretty sort of symmetry, we get:
$$\begin{bmatrix}a_n\\b_n\end{bmatrix}=\frac{1}{\omega}\begin{bmatrix}\lambda_1\cdot\tau&\lambda_2\cdot\alpha_2\\\lambda_1\cdot\alpha_1&\lambda_2\cdot\sigma\end{bmatrix}\begin{bmatrix}a_{n-1}\\b_{n-1}\end{bmatrix}$$
For factorisation, this is the best I can do:
Let $\sigma'=\gamma_1+\alpha_1$ and $\tau'=\gamma_2+\alpha_2$:
$$\begin{bmatrix}a_n\\b_n\end{bmatrix}=\frac{1}{\omega}\left(\begin{bmatrix}\lambda_1\cdot\tau'&\lambda_2\cdot\alpha_2\\\lambda_1\cdot\alpha_1&\lambda_2\cdot\sigma'\end{bmatrix}+n\gamma\begin{bmatrix}\lambda_1&0\\0&\lambda_2\end{bmatrix}\right)\begin{bmatrix}a_{n-1}\\b_{n-1}\end{bmatrix}$$
And by the distributivity of matrix multiplication:
$$\begin{bmatrix}a_n\\b_n\end{bmatrix}=\frac{1}{\omega}\begin{bmatrix}\lambda_1\cdot\tau'&\lambda_2\cdot\alpha_2\\\lambda_1\cdot\alpha_1&\lambda_2\cdot\sigma'\end{bmatrix}\begin{bmatrix}a_{n-1}\\b_{n-1}\end{bmatrix}+\frac{n\gamma}{\omega}\begin{bmatrix}\lambda_1&0\\0&\lambda_2\end{bmatrix}\begin{bmatrix}a_{n-1}\\b_{n-1}\end{bmatrix}
$$
And if you're running a computer analysis or something that requires you to extrapolate future $a_n$ maybe this helps, since the $1/\omega$ terms can be easily computed and collected as functions of $n$, and the left matrix is constant, and the right matrix is only very trivially a function of $n$.


TITLE: Max. and Min. value of $f(\phi) = \frac{1+\cos \phi}{2+\sin \phi}.$
QUESTION [5 upvotes]: Calculation of Max. and Min. value of $\displaystyle f(\phi) = \frac{1+\cos \phi}{2+\sin \phi}.$

$\bf{My\; Solution::}$ Let $\displaystyle y = \frac{1+\cos \phi}{2+\sin \phi}\Rightarrow 2y+y\sin \phi = 1+\cos \phi$
So $y\cdot \sin \phi - 1\cdot \cos \phi=1-2y\;,$ Now Using cauchy-Schwatrz Inequality, We get
$\displaystyle \left((y)^2+(-1)^2\right)\cdot \left(\sin^2 \phi+\cos^2 \phi\right)\geq \left(2y\sin \phi-\cos \phi\right)^2\Rightarrow \left(y^2+1\right)\geq \left(1-2y\right)^2$
So $\displaystyle y^2+1\geq 1+4y^2-4y\Rightarrow 3y^2-4y\leq 0\Rightarrow 3y\left(y-\frac{4}{3}\right)\leq 0 $
So We get $\displaystyle 0\leq y \leq \frac{4}{3}\Rightarrow y\in \left[0,\frac{4}{3}\right]$
My Question is can we solve it Using Geometrically or using Derivative Test,
If Yes Then plz explain me, Thanks

REPLY [1 votes]: $$\displaystyle f(\phi) = \frac{1+\cos \phi}{2+\sin \phi}\implies f'(\phi)=-\frac{2\sin\phi +\cos \phi +1}{(2+\sin \phi)^2}.$$
Thus,
$$f'(\phi)=0\Leftrightarrow 2\sin\phi +\cos \phi +1=0.$$ Since $\sin^2\phi+\cos^2\phi=1,$ we get that $\sin\phi=-4/5,\cos\phi=3/5$ or $\sin \phi=0,\cos \phi=-1.$
Now,
$$\sin \phi=0,\cos \phi=-1\implies \phi=\pi+2n\pi=(2n+1)\pi, n\in\mathbb{Z}.$$ and
$$\sin \phi=-\frac45,\cos \phi=\frac 35 \implies \phi=-\arcsin(4/5)+2n\pi, n\in\mathbb{Z}.$$
Using the second derivative test we conclude that the function achieves its minimum at $\phi=(2n+1)\pi, n\in\mathbb{Z}$ and its maximum at $\phi=-\arcsin(4/5)+2n\pi, n\in\mathbb{Z}.$ The minimum value is $f(\pi)=0$ and the maximum value is $f(-\arcsin(4/5))=4/3.$


TITLE: How to interpret single photon interference when the two possible paths are different in length?
QUESTION [5 upvotes]: Here is my question.
I struggle with the definition of single photon interference. Let’s assume we have a Michelson interferometer and the interference pattern we observe is a single photon result, then if we have and we almost always have different lengths of the two possible photon paths after the beam splitter then how it could be possible to have the same time arrival of the two photon paths when they are traveling different distances? Maybe my assumption that the beam splitter generates the two paths in the moment the photon is interacting with the beam splitter is false, but I can’t explain the obtained interference unless I accept that there is a time delay in the two possible paths. This is not a paradox only for this type of experiment but for all the experiments that involve some kind of interference. It seems for me that we cannot use the usual assumption of simultaneousness of events.

REPLY [3 votes]: Have a look at my answer to Slit screen and wave-particle duality because this covers a lot of topics relevant to your question.
You're correct that if we imagine the photon as a little ball then if the arms of the interferometer are different lengths the two "halves" of the little ball cannot arrive at the detector at the same time. But this is not how the interference works. It does not work by the photon splitting and its two halves then interfering with each other.
To imagine the light as a little ball (the photon) ricocheting around your interferometer is highly misleading. The behaviour of the light is best explained by quantum field theory, but sticking to regular quantum mechanics we'd have to say that until we interact with the light (e.g. by it hitting a CCD or photographic plate) the light is delocalised over your whole interferometer.
This doesn't mean the photon has a position but we don't know it, it means the light simply has no position in the classical sense. The wavefunction that describes it covers your whole experimental equipment. The probability of detecting the photon at some point in your kit is given by the magnitude squared of its wavefunction at that point. If you change the geometry of your interferometer then you will change the wavefunction and therefore change the probability of detecting the photon at any particular point.


TITLE: Thompson's Conjecture
QUESTION [5 upvotes]: I have heard that the following is a conjecture due to Thompson:
The number of maximal subgroups of a (finite) group $G$ does not exceed the order $|G|$ of the group.
My question is: did Thompson really conjecture this? If so, is there any literature on the subject?

REPLY [8 votes]: I've seen this conjecture attributed to Wall (1961), for example: On a conjecture of G.E. Wall. This is a recent article (journal version appeared in 2007), and it gives a bunch of references. The conjecture remains open. Here is a very recent article which does not attack the conjecture itself, but uses it as an inspiration for a different conjecture.


TITLE: $f$ does not have extrema at the Lagrange multiplier value
QUESTION [0 upvotes]: I have the function $2x^2+y$ and one constraint $x-y^2=1$ and want to find maximum value by lagrange multiplier. Intuitively, I see the point $(2,1)$ satisfies $c$ and have value of  $f(2,1)=9$.
Using Lagrange function $f(z)-\lambda c(z)$, I get $\lambda^3-4\lambda^2-1$ thus $\lambda=4.06$. However, when plug this into the derivatives of $x$ and $y$ I got different values in particular $x=\frac{4.06}{4}=1.015$ which is much lower than $2$. Am I missing something here?

REPLY [1 votes]: Note that your function is unbounded. Your constraint gives $x = 1+y^2$, and plugging this into the objective, you get $2\,(1+y^2)^2 + y$. Now, it is easy to see, that this can attain arbitrary large values. This means that your problem has no global maximum, but it may have a local one.
Further, since the gradient of your constraint is $(1, -2\,y)$, the LICQ is satisfied at every feasible point. This means that every local extremum (may it be a maximum or a minimum) can be found by the Lagrange multiplier method.
As you already found, there is only one solution to Lagrange's method. Moreover, it is easy to see that this is a local minimum (consider, e.g., the second derivative of the reduced objective $2\,(1+y^2)^2+y$ which is positive for $y = -0.123$.
Putting everything together, we find that your problem has no local maximum (and no global one).


TITLE: sum of products of consecutive divisors of a number is smaller than the number squared
QUESTION [3 upvotes]: Let $n>1$ be natural. Let $$1=d_1<d_2<\ldots<d_k=n$$ be its all divisors. Prove that $$s:=d_1d_2+d_2d_3+\cdots+d_{k-1}d_k<n^2$$ and determine when $s$ divides $n^2$.
For example:
$n=12$, divisors: $1,2,3,4,6,12$. Then $$s=1\cdot 2+2\cdot 3+3\cdot 4+4\cdot 6+6\cdot 12 = 2+6+12+24+72=116 < 12^2$$ and $s$ doesn't divide $n^2$.
My attempts
The second part is easy. Namely, $s\geq d_{k-1}d_k = \frac{n^2}{d_2}$, which is the largest proper divisor of $n^2$ (since $d_2$ is the smallest divisor of $n$ greater of $1$ (hence also of $n^2$). Therefore, if $k>2$, that is, if $n$ isn't prime, then $s>\frac{n^2}{d_2}$ and doesn't divide $n^2$. If $n$ is prime than $s|n^2$.
My idea of the first part:

It's easy to show the inequality $s<n^2$ provided $2d_i\leq d_{i+2}$, since then, considering the summands from right to left, one can see that $s\leq n^2/2+n^2/4+\ldots<n^2$.
One can think of $n^2$ as the area of a square and $s$ as the sum of the areas of the rectangles. I don't know how to use it.
Using the inequality of arithmetic and geometric means we obtain
$$s\leq \frac 12+d_2^2+d_3^2+\ldots+d_{k-1}^2+\frac 12d_k^2.$$ This shows that it remains to show that
$$d_1^2+d_2^2+\ldots+d_{k}^2< \frac 12\left(3n^2+1\right).$$ There's a formula for such a sum.
Maybe some induction could be used. I tried to prove that it suffices to show that the conclusion is true for $n/d_2$, without any success.

REPLY [1 votes]: Assume the worst-case scenario: $n$ is enormous, and is divisible by all integers up to some large $k$. If we look at the largest divisors of $n$, they must be $n, n/2, n/3, \cdots n/k$.
Then the largest term in the sum is $n \cdot n/2 = \frac12 n^2$. The next largest term is $n/2 \cdot n/3 = \frac16 n^2$. The next terms will have $12, 20, 30, \cdots$ as their denominators: these are the pronic numbers (of the form $n^2+n$). The sum of the reciprocals of the pronic numbers converges to $1$:
$$\sum^{\infty}_{i=1} \frac{1}{i^2+i} = 1 \implies n^2 = \sum^{\infty}_{i=1} \frac{n^2}{i^2+i} > \sum_{d \mid n} \frac{n^2}{d^2+d} = s$$
The greater than sign appears due to the Fundamental Theorem of Arithmetic; we know $n$ has a non-infinite number of divisors. Therefore $n^2 > s$.


TITLE: Does $\sum^{\infty}_{n=1} \frac{1}{a_n+1}$ with $a_{n+1}=a_n(a_n+1)$ and $a_1=\frac12$ have a closed form?
QUESTION [3 upvotes]: Mr Beast here. I've been working on a complicated math problem, and I obtained a recursive sequence $a_{n+1}=a_n(a_n+1)$ and that $a_1=\frac 12$. I've been thinking, is it possible to convert this to closed form, or otherwise evaluate $\sum^{\infty}_{n=1} \frac{1}{a_n+1}$?
The original question said that $a_{1}=\frac{1}{2}$ and that $a_{n+1}=a_{n}\left(a_{n}+1\right),$ asked me to prove that $\frac{1}{a_{1}+1}+\frac{1}{a_{2}+1}+\frac{1}{a_{3}+1}+\cdots+\frac{1}{a_{n}+1}<2$ for all $n$ larger than $1 .$ I wondered if it was possible to convert into closed form to evaluate the sum definitely. I originally thought to solve it through a telescoping series, but the denominator leaves me unsure of what to do.

REPLY [1 votes]: Note that
$$ \frac1{a_{n+1} } = \frac1{a_n(a_n+1)}=\frac{(a_n+1)-a_n}{a_n(a_n+1)} = \frac1{a_n} - \frac1{a_n+1}$$
Therefore
$$ \frac1{a_n+1}=\frac1{a_n}-\frac1{a_{n+1}}$$
Summing in $n$ gives a telescoping series,
$$ \sum_{k=1}^n\frac1{a_k+1} = \sum_{k=1}^n\frac1{a_k}-\frac1{a_{k+1}} = \frac1{a_1} - \frac1{a_{n+1}}.$$
Plugging in $a_1=1/2$, we see that
$$ \sum_{k=1}^n\frac1{a_k+1} = 2 - \frac1{a_{n+1}}.$$
Note that $a_1=1/2,\ a_2=3/4,\ a_3=21/16>1$, and from that point on the equation gives
$$ a_{n+1} = a_n(a_n+1) \ge a_n+1$$
so that $a_n\to\infty$. Hence $\frac1{a_{n+1}}\to 0$ as $n\to\infty$, which means that
$$ \fbox{$\sum_{k=1}^\infty\frac1{a_k+1} = 2. $}$$
In particular, since $\frac1{a_n+1}>0$ for all $n$, this proves (as is the 'original question' in the comments) that the finite sum of only the first $n$ terms is strictly smaller than $2$ for all $n$.


TITLE: Why should anyone care about computing the Hausdorff Dimension?
QUESTION [0 upvotes]: I am trying to justify an thesis/investigation that will (hopefully) lead to a formula for computing a particular class of Cantor-like sets. The question that I have not satisfactorily answered, according to my department chair, is 'Why should anyone care about computing the Hausdorff Dimension?'

REPLY [0 votes]: I may be shunned by real mathematicians for this but I think it can provide some detail about "how long a coastline is" since they exhibit fractal behaviour. The length keeps getting longer when you measure more closely.
Numberphile did a good video on it https://www.youtube.com/watch?v=7dcDuVyzb8Y

REPLY [0 votes]: For instance, there are various results about Kleinian groups and the hausdorf dimension of their limit sets
http://en.wikipedia.org/wiki/Kleinian_group