TITLE: Countable decomposition of a closed subset of real numbers
QUESTION [0 upvotes]: $\newcommand{\R}{\mathbb{R}}$
I am going to demonstrate how I decomposed a closed set into pairwise disjoint closed intervals. (Here, we also take singletons as intervals).
Let $A$ be a closed subset of $\R$. Then, for all $x \in A$, there is the maximal closed interval $I \subseteq A$ containing $x$. We can form it by taking the union of all closed intervals in $A$ containing $x$. The union is closed: 1) it is trivially closed if it is a singleton; 2) Otherwise if one of its endpoints is assumed to be open, the point is a limit point of the interval, and we can add it to the interval, which contradicts to the claim that it is maximal.
Now, let $C$ be the collection of all such intervals. Then, $C$ is disjoint and $\bigcup C = A$. The countability of $C$ directly comes from the proposition, by which we can pair each element of $C$ with an adjacent open interval in the complement of $A$.
Is this correct?

REPLY [2 votes]: The Cantor set is a counterexample. All its maximal subintervals are singletons, and there are uncountably many of them.
Also note  that your subintervals are certainly closed subsets of $\mathbb R$, but they may have the form $\mathbb R$, $(-\infty,a]$ or $[a,\infty)$ with $a \in \mathbb R$. I would not regard as them as closed intervals, but perhaps that is a matter of taste.
All you say about your decomposition is that it can have only countably many non-degenerate maximal subintervals (i.e. non-singletons).
In fact, it impossible that $\mathbb R$ contains uncountably many disjoint intervals.
Let $\{J_\alpha\}$ be an uncountable family of disjoint non-degenerate intervals in $\mathbb R$. Not all of the sets $A_n = \{\alpha \mid J_\alpha \text{ has length } \ge 1/n \}$ can be countable. So let $A_N$ be uncountable. One of the sets $B_n = \{ \alpha \in A_N \mid [-n,n] \cap J_\alpha \ne \emptyset \}$ must be uncountable, thus they are uncountable for $n \ge M$. Take $n = \max(N,M)$. There are at most two $J_\alpha$ with $\alpha \in B_n$ not contained in $[-n,n]$ (one can contain $-n$, another can contain $n$). Thus $[-n,n]$ contains uncountable many disjoint non-degenerate intervals of length $\ge 1/N$. Let their union be $C$. The set $C$ must have infinite Lebesgue measure, which contradicts the fact that it is contained in a set with finite Lebesgue measure.


TITLE: using pumping lemma to prove that a set is not regular
QUESTION [0 upvotes]: A={s11s|s $\epsilon$ {0}^*} so the strings 00011000 and 000001100000 are accepted of A but not 00100 or 001100000.
Demon chooses k.
I choose x = s^k, y=11, z = s^k

however i am not confident i am picking the right variables for y. am I heading down the right path?
the demon chooses u,v,w s.t. y = uvw and v does not equal $\varepsilon$
demon picks u,v,w of length j,m,n with k = j+m+n and m>0
I pick i = 2.
so xuv^2wz = s^k (and here I am stuck, do i say u is just 1 ?)

so the way I am currently learning the pumping lemma:
there exists k>=0 such that for any strings x,y,z with xyz $\epsilon$ A and |y| >= k, there exists strings u,v,w such that y = uvw, v != $\varepsilon$, and for all i>= 0, the string xuv^iwz $\epsilon$ A

REPLY [1 votes]: The Myhill-Nerode theorem shows this immediately.
The infinitely many prefixes $0^i11$ are all pairwise distinguishable. (If $i\ne k$, then $0^i$ is a distinguishing extension: $0^i110^i \in A$ but $0^i110^k\notin A$). Therefore the language cannot be regular.

As for your attempt to use the pumping lemma, I'm not entirely sure how the formulation of the lemma you're working with goes, but you certainly can't expect any demon to split an $y$ of constant length into something that can be pumped. You must have missed some premise in the statement of the lemma.


TITLE: Interpretation of inequality between 2 stopping times.
QUESTION [0 upvotes]: Let $\mathcal{F}_{n}$ be a filtration of $\sigma$-fields, i.e., $\mathcal{F}_{n} \subset \mathcal{F}_{n+1}$ for every $n \ge$ 1. Let $\tau_{1}$ be a stopping time, i.e, a random variable such that the event $\lbrace \tau_{1} \le n\rbrace$ $\in$ $\mathcal{F}_{n}$ for every $n \ge$ 1.
I am given: if $\tau_{1} \le \tau_{2}$ then $\lbrace \tau_{2} \le n \rbrace \subset \lbrace \tau_{1} \le n \rbrace$.
I am having trouble understanding why $\lbrace \tau_{2} \le n \rbrace \subset \lbrace \tau_{1} \le n \rbrace$ and what it means for one stopping time to be smaller than another.
Thanks in advance.

REPLY [1 votes]: $\tau_1\leq \tau_2$ means $\tau_1(\omega)\leq \tau_2(\omega)$ for all $\omega$. From this inequality, you can get $\Omega=\{\tau_1\leq \tau_2\}=\cup_{k\geq 1}\{\tau_1\leq k\leq \tau_2\}$. Then, it follows that $\{\tau_2\leq n\}=\left[\cup_{k\geq 1}\{\tau_1\leq k\leq \tau_2\}\right]\cap \{\tau_2\leq n\}=\cup_{1\leq k\leq n}\{\tau_1\leq k\leq \tau_2\leq n\}\subset \cup_{k\leq n}\{\tau_1\leq k\}=\{\tau_1\leq n\}$.


TITLE: Spectrum of eigenvalues and eigenfunctions
QUESTION [2 upvotes]: Our O.D.Es professor had the "amazing" idea of heavily introducing advanced linear algebra material (which is not an official prerequisite for the course) along with boundary value problems. Not being trained in these types of exercises I am facing quite a few difficulties. If anybody would be willing to help, it would be most appreciated.
For the following sets of boundary conditions, consider the equation:
$$u''(x) + \lambda u(x) = 0, \ \ \ 0 < x < 1$$
and find the spectrum of eigenvalues, in essence the set of values $\lambda$ in the complex plane for which a nontrivial solution exists, and give the
eigenfunction (or eigenfunctions) for each such eigenvalue.
1) $u(0) + u(1) = 0$ and $u'(0) + u'(1) = 0$
2) $u(0) + u(1) = 0$ and $u'(0) - u'(1) = 0$

REPLY [1 votes]: See my solution at my website, http://people.cs.uchicago.edu/~lebovitz/Winter13/Slvdprblms

u(0) + u(1) = 0; u'(0) + u'(1) = 0

SOLUTION: With  $\lambda$  = $\mu^2 \neq$ 0, any solution has the form
u = A cos(x) + B sin(x)
Applying the boundary conditions gives
(1 + cos$\mu$ )A + sin $\mu$B = 0
-$\mu$sin A + $\mu$(1 + cos $\mu$)B = 0
implying that 2$\mu$(1 + cos $\mu$) = 0.
It is easy to see that  $\mu$= 0 leads only
to the solution u(x) = 0 so 0 is not an eigenvalue. The other values,
$\mu_k$ = (2k + 1)$\pi$ or $\lambda_k$ = $(2k + 1)^2\pi^2$
k = 0, 1, 2
are genuine eigenvalues. To each there corresponds a pair of eigenfunctions
$u_k(x)$ = cos($\mu_k$x) and $v_k(x)$ = sin($\mu_k$x)

u(0) + u(1) = 0; u'(0)-u'(1) = 0

SOLUTION: A solution satisfying the first boundary condition is (with $\lambda = \mu^2 \neq$ 0)
$u_1$ =  -sin($\mu$x) + sin ($\mu$(1-x)).
But this also easily seen to satisfy the second boundary condition, without any condition on $\mu$. If  = 0 the function u0(x) = 1  2x satisfies the equation and the boundary conditions. The spectrum therefore consists of the entire complex plane and, for given $\lambda \neq 0$, $u_1$ is the
eigenfunction (or $u_0$ if $\lambda$ = 0).
I don't like your tone of voice, young man. Please see me before the exam.


TITLE: How does opponent's flush affect odds of your full house?
QUESTION [4 upvotes]: A cute probability intuition test:
Let $f$ be the probability of being dealt a full house in a five-card poker hand, from a randomly shuffled standard deck.  ($f \approx 0.001468$).
Now look at the case of two players being dealt hands, and player one shows that she has a flush.  Now what is the probability $f_2$ that the five cards dealt to player 2 is a full house?
It seems clear that we will have $f_2 < f$ because if one player has a full house it must be a tiny bit more likely that the other has pairs and other clumps of the same rank, and a flush has none of those.  But to test your intuition, how small is that effect?
Is $f_2$ more than 99% of $f$?  More than 95%?  More than 90%? or less than 90%?
EDIT I had written $h$ for "house" in the first sentence.  Then I used $f$ for "full" later.

REPLY [4 votes]: Nice question. I'll give my spontaneous intuitive answer, a quick estimate and then a spoiler-proof exact calculation.
My spontaneous intuitive answer was, like you wrote: "a tiny bit" – something like $99\%$ or $98\%$.
Then I did a quick estimate: The hard part of the full house is the three of a kind. Instead of $13$ ranks of $4$, there are now only $8$ ranks of $4$ and $5$ ranks of $3$, and it's only $\frac14$ as likely to get three of a kind in the ranks of $3$, so the factor should be something like $(8+\frac54)\div13\approx71\%$.
Here's the exact calculation:

There are $5\cdot4\cdot\binom33\cdot\binom32=60$ ways to form both the triple and the pair from a depleted rank.
There are $5\cdot8\cdot\binom33\cdot\binom42=240$ ways to form only the triple from a depleted rank.
There are $5\cdot8\cdot\binom43\cdot\binom32=480$ ways to form only the pair from a depleted rank.
There are $8\cdot7\cdot\binom43\cdot\binom42=1344$ ways to form neither the triple nor the pair from a depleted rank.
Thus the probability is
$$\frac{60+240+480+1344}{\binom{47}5}=\frac{708}{511313}\approx0.001385\;,$$
so the factor is about $94\%$. That raises the question why my quick estimate was so wrong. :-)


TITLE: Fermi level in equilibrium and non-equilibrium situations
QUESTION [5 upvotes]: Why do the Fermi level for electrons and holes coincide in equilibrium condition and why do they separate as quasi-Fermi levels in non equilibrium situations?

REPLY [1 votes]: A steady-state equilibrium, different from the thermodynamic equilibrium, can be triggered by external stimulus such as light-shining, which provoke the photoionization and the generation of electron-hole pairs, or current flowing, which injects electrons (or holes) to the system. Under these conditions, the concentrations of electron and holes are no longer governed harmoniously by the mass-action law and the Fermi-Boltzmann thermodynamic equilibrium, but are forced by the external conditions, and pulled-off from their reciprocal equilibrium. By this, the need to separate the two distinct quasi-fermi levels, one for electrons, one for holes, accounting for their out-of-thermodynamic-equilibrium concentrations.


TITLE: Is there a term for this kind of figure? Can it be produced for any n?
QUESTION [0 upvotes]: The Nature Genetics article The draft genome of tropical fruit durian (Durio zibethinus) (open access pdf!) includes an interesting diagram in Figure 2. Figure 2a shows the sharing of genes between five species using a diagram with five-fold rotational symmetry where each wedge includes all possible overlap combinations with the other four. The number of shared genes for each of the 31 possible combinations is shown.
I'd like to ask if there is a term for this kind of figure in Mathematics, if it can be produced in two dimensions for any number $n$, and if there are patterns which are not topologically n-fold symmetric - in other words patterns that can not be reshaped topologically so that they are just an n-fold rotational repeat of a single identical wedges shape as this one is?

Figure 2(a): Sharing of gene families by durian and three other Malvales plants, with A. thaliana as an outgroup. The number in parentheses indicates durian-specific gene families among all 11 plants considered.

Who would have thought that Durian is closely related to Cacao!? See also the BBCs Singapore scientists reveal origins of durian's pungent aroma

REPLY [1 votes]: I'd argue that this is an Euler diagram, as it shows the real world relations rather than all possible (theoretical) relations, which a strict Venn must show. To be a Venn, would it not need to be completely boxed with the space outside the interlocking shapes but inside the box representing plants that fall into none of the categories?


TITLE: Aristotle's wheel
QUESTION [0 upvotes]: two wheels of different diameter attached to each other at the center, roll along a straight line. since they have different diameters, the smaller wheel must slip. (see other stackexchange answers)
Why do I not see the slippage in these videos: https://www.youtube.com/watch?v=mW-0bZwoGwQ
https://www.youtube.com/watch?v=mmBx8HW6qkQ ?

REPLY [1 votes]: The fact that you don't see any slippage does not mean that it's not there.

In the video you see two fixed lines, $a$ (the big circle is rolling along it) and line $b$ which is there just for orientation. If the velocity of circle center $A$ is $v_A$, the velocity $v_B$ of point $B$ currently in touch with line $b$ is much smaller but it's still different to zero (see picture). There is no slippage only if the touching point has velocity equal to zero, which is true for $P_v$ but not for $B$. Therefore, the smaller circle does not roll along the line $b$ without slippage. It's there, it's just hard to spot it because of much smaller magnitude of velocity $v_B$.


TITLE: Showing that the sum of these two closed subsets is a countable dense subset.
QUESTION [3 upvotes]: I am having a really difficult time proving part two of this problem in Baby Rudin.

If$\,\,A\subset\mathbb{R^k}$ and $B\subset\mathbb{R^k}$, define $A+B$ to be the set of all sums $x+y$ with $x\in A$ and $y\in B$.
(b) Let $\alpha$ be an irrational real number. Let $C_1$ be the set of all integers, let $C_2$ be the set of all $n\alpha$ with $n\in C_1$. Show that $C_1$ and $C_2$ are closed subsets of $\mathbb{R^1}$ whose sum $C_1+C_2$ is not closed, by showing that $C_1+C_2$ is a countable dense subset.

I understand that both $C_1$ and $C_2$ are closed, and I understand why $C_1+C_2$ might be dense, but I am just having no luck figuring out how to tackle the problem.
Here are two possible ideas I think might work, but don't know how to make work:

Prove that for every element $z\notin C_1+C_2$ and every $\epsilon$, there exist an element $c\in C_1+C_2$ such that $|z-c|<\epsilon$.

Assuming $z\notin C_1+C_2$, prove that if  $ d_z:\mathbb{R^1}\to \mathbb{R^1}$ where $d_z(x)=|z-x|,$ then $\inf (f(C_1+C_2))=0$

Any ideas as to what might be the best way to tackle this problem?

REPLY [0 votes]: Hint regarding your first idea: you only need one such $z, c$ pair for each $\varepsilon$.


TITLE: How to obtain the sum of an infinite series? is it divergent?
QUESTION [2 upvotes]: $$\lim\limits_{m\rightarrow \infty}\frac{1}{m}\sum\limits_{n=1}^{m} \sum\limits_{i=1}^{20}e^{(i-\tfrac{1}{n})/20}$$
The problem is from a continuous compound interest related calculation (but I made a mistake in deduction, the $\frac{1}{n}$ here should be $\frac{n}{m}$, then the series is expected to be convergent and the guessed value is expected to be $$20(e-1).$$
Though the deduction is wrong, how to obtain the sum of the wrong infinite series seems still interesting.
Update
The following conversion solves my original problem if $\frac{1}{n}$ is $\frac{n}{m}$:
$$\lim\limits_{m\rightarrow \infty}\frac{1}{m}\sum\limits_{n=1}^{m} \sum\limits_{i=1}^{20}e^{(i-\tfrac{n}{m})/20}$$
$$=\int_{0}^1 {\sum\limits_{i=1}^{20}e^{(i-t)/20}}dt$$
So it is quite simple...
thanks to @Did

REPLY [0 votes]: $$\frac{1}{m}\sum_{n=1}^{m}\sum_{i=1}^{20}\exp\left(\frac{i-\frac{1}{n}}{20}\right)=\frac{1}{m}\sum_{n=1}^{m}\exp\left(-\frac{1}{20n}\right)\color{purple}{\sum_{i=1}^{20}\exp\left(\frac{i}{20}\right)}$$
but the last sum is just a constant, so let we evaluate the interesting part:
$$ \frac{1}{m}\sum_{n=1}^{m}\exp\left(-\frac{1}{20n}\right)=\frac{1}{m}\sum_{n=1}^{m}\sum_{k\geq 0}\frac{(-1)^k}{k! 20^k n^k}=1-\frac{H_m}{20 m}+O\left(\frac{1}{m^2}\right).$$
That gives that the wanted limit is just the purple constant.


TITLE: Find all rational solutions of $x^2(x+1)(x^2+1)(x-1)=2(y+1)(y-1)$
QUESTION [4 upvotes]: Find all rational solutions of
$$x^2(x+1)(x^2+1)(x-1)=2(y+1)(y-1).$$
Clearly the following six solutions hold:
$$(x,y)=(1,1),(-1,-1),(-1,1),(1,-1),(0,1),(0,-1)$$
But how to find all rational solutions?

REPLY [17 votes]: This is no complete answer yet, but may get expanded to one in due course.
First we search for points on the curve, which is isomorphic to
$$C \colon y^2 = 2(x^6 - x^2 + 2);$$
this produces points with $x \in \{-3, -1, -4/7, 0, 4/7, 1, 3\}$ as already
mentioned in the comments. It was also mentioned that the rank of the
Mordell-Weil group of the Jacobian variety of $C$ is 3, so that we cannot
apply Chabauty's method directly. What we can try to do is to use a
covering collection plus Elliptic Curve Chabauty.
Let $K = \mathbb Q(a)$ with $a^3 - a + 2 = 0$; then
$$2(x^6 - x^2 + 2) = 2 (x^2 - a) (x^4 + a x^2 + a^2 - 1).$$
We can compute the 2-Selmer set of $C$ and find that it has three
elements that are all images of known rational points on $C$.
If my quick computation is correct, then we obtain two coverings of $C$
of the form
$$D_j \colon u^2 = \delta_j (x^2 - a), \quad
  v^2 = 2 \delta_j (x^4 + a x^2 + a^2 - 1)$$
with $\delta_1 = 1 - a$ and $\delta_2 = -a$, such that each rational
point on $C$ lifts to a $K$-point on $D_1$ or $D_2$ (with the same
$x$-coordinate). Forgetting the first
equation, we get morphisms $D_j \to E_j$, where $E_j$ is an elliptic
curve given by the second equation. We are interested in points in $E_j(K)$
whose $x$-coordinate is in $\mathbb Q$; this is the setting for
Elliptic Curve Chabauty. According to Magma, the Mordell-Weil ranks
of $E_1$ and $E_2$ over $K$ are 2 and 1, so the relevant Chabauty
condition (rank is smaller than degree of $K$) is satisfied.
It remains to actually do the computation, which right now I have no
time to do. Almost certainly the result will be that $C$ only has the
known rational points.
EDIT: I have now done the computation (with Magma), and the result
is as expected: the rational points on $C$ lifting to a $K$-rational
point on $D_1$ have $x$-coordinate in $\{-3, -1, 1, 3\}$ and the
rational points on $C$ lifting to a $K$-rational point on $D_2$ have
$x$-coordinate in $\{-4/7, 0, 4/7\}$. So there are no unknown rational
points on $C$.
Here is Magma code for the computations (it can be run in the
Magma online calculator):
P<x> := PolynomialRing(Rationals());
f := x^2*(x+1)*(x^2+1)*(x-1) + 2;
C := HyperellipticCurve(2*f);
K<a> := NumberField(x^3-x+2);
f1 := ExactQuotient(PolynomialRing(K)!f, Polynomial([-a, 0, 1]));
// compute 2-Selmer set of C
deltas, m := TwoCoverDescent(C);
A<th> := Domain(m);
// find the possible square classes of x(P)^2 - a in K, for P in C(Q)
invol := hom<A -> A | -th>;
ndeltas := [d*invol(d) where d := dd @@ m : dd in deltas];
assert Set(ndeltas) eq {1-th^2, -th^2}; // note that a = th^2, so this is {1-a, -a}
// set up the two elliptic quotients as hyperelliptic curves
D1 := HyperellipticCurve(2*(1-a)*f1);
D2 := HyperellipticCurve(2*(-a)*f1);
// find isomorphic elliptic curve, given by integral Weierstrass model
E1, D1toE1 := EllipticCurve(D1, Points(D1, 1)[1]);
E1i, toE1i := IntegralModel(E1);
// determine E1(K)
MWE1, mMWE1 := MordellWeilGroup(E1i);
P1 := ProjectiveSpace(Rationals(), 1);
// set up x-coordinate map in terms of E1i
E1itoP1 := Expand(Inverse(toE1i)*Inverse(D1toE1)*map<D1 -> P1 | [D1.1, D1.3]>);
// run Elliptic Curve Chabauty
ptsMWE1, N1 := Chabauty(map<MWE1 -> E1i | a :-> &+[Eltseq(a)[i]*mMWE1(MWE1.i) : i in [1..Ngens(MWE1)]]>, E1itoP1);
// make sure known subgroup of E1(K) is saturated at the primes dividing N1
newgens := Saturation([mMWE1(g) : g in OrderedGenerators(MWE1)], Max(PrimeDivisors(N1)));
assert newgens eq [mMWE1(g) : g in OrderedGenerators(MWE1)];
// find the x-coordinates of the points
{E1itoP1(mMWE1(pt)) : pt in ptsMWE1};
// ==> { (3 : 1), (1 : 1), (-1 : 1), (-3 : 1) }
// do the same for the other covering curve
E2 := EllipticCurve(D2, Points(D2, 0)[1]);
E2, D2toE2 := EllipticCurve(D2, Points(D2, 0)[1]);
E2i, toE2i := IntegralModel(E2);
E2itoP1 := Expand(Inverse(toE2i)*Inverse(D2toE2)*map<D2 -> P1 | [D2.1, D2.3]>);
MWE2, mMWE2 := MordellWeilGroup(E2i);
ptsMWE2, N2 := Chabauty(map<MWE2 -> E2i | a :-> &+[Eltseq(a)[i]*mMWE2(MWE2.i) : i in [1..Ngens(MWE2)]]>, E2itoP1);
newgens := Saturation([mMWE2(g) : g in OrderedGenerators(MWE2)], Max(PrimeDivisors(N2)));
assert newgens eq [mMWE2(g) : g in OrderedGenerators(MWE2)];
{E2itoP1(mMWE2(pt)) : pt in ptsMWE2};
// ==> { (0 : 1), (4/7 : 1), (-4/7 : 1) }


TITLE: Two Simultaneous Cumulative Hypergeometric Distributions
QUESTION [0 upvotes]: Suppose we have a standard 52 card deck from which we draw five cards. What are the chances of drawing one or more Aces and one or more Kings? How do I calculate this?
I know that we can calculate the chance of drawing one or more Ace by using cumulative hypergeometric distribution (Population 52, Subpopulation 4, Sample Size 5). This gives us a result of 34.116%.
How do I then find the odds of drawing one or more kings as well?
I mistakenly thought I could "set aside" the Ace, reducing the Population and Sample Size by 1, then calculate the chance for drawing one or more kings out of the rest (28%). Multiplying the two would find the total odds. However, the results off this - 9.765% - is well under the 10% that I continuously get out of a game simulator (multiple tries of 10 million hands).
So then, how do I calculate the above question, and - more directly - what is the correct way to calculate two different cumulative hypergeometric distributions that occur simultaneously?

REPLY [1 votes]: Let $A$ represent the event that you have at least one ace in your hand.
Let $B$ represent the event that you have at least one king in your hand.
You ask how to calculate $Pr(A\cap B)$.
To do this, let us try to calculate the complementary event and apply some common identities.
$Pr(A\cap B)=1-Pr((A\cap B)^c)=1-Pr(A^c\cup B^c)=1-Pr(A^c)-Pr(B^c)+Pr(A^c\cap B^c)$
Each of these terms at the end are relatively easy to calculate once you recognize what they represent.  $Pr(A^c)$ represents the probability that there are no aces in your hand.  $Pr(A^c\cap B^c)$ represents the probability that there are no aces and there are no kings in your hand, etc...
Running the calculations, we arrive at a final answer of:
$$Pr(A\cap B)=1-\dfrac{\binom{48}{5}}{\binom{52}{5}}-\dfrac{\binom{48}{5}}{\binom{52}{5}}+\dfrac{\binom{44}{5}}{\binom{52}{5}}\approx 0.1001785$$


TITLE: What is the main requirement for two arbitrary quantum states to be able to serve as qubit states?
QUESTION [0 upvotes]: Does it need to be orthogonal ?
If now the state is not arbitrary but a coherent state $\lvert\alpha\rangle$, if we use $\lvert0\rangle=\lvert\alpha\rangle$ and $\lvert1\rangle=\lvert-\alpha\rangle$, what condition on the value of $\alpha$ does this requirement set for our qubit coding?

REPLY [2 votes]: Quantum computing with continuous variables is a full field in itself, with many different paradigms for how to actually do the information processing. Therefore, there is not necessarily a single straightforward answer to your question as this depends on what paradigm you're using, how you plan to physically implement this paradigm (i.e. what sort of errors are you expecting), and how you plan to do error-correction. For a reasonable review of part of this field, see Ref [1]. Generally, however, you aren't limited by what values of $\alpha$ you choose to use so much as other factors such as how well you can entangle and how you do your error correction (which includes factors like sources of noise etc).
However, there is a proposal to use coherent states in exactly the way you mention as the computational qubits (Ref [2]). They restrict their values of $\alpha \ge 2,$ utilizing the fact that this makes the overlap of the two states roughly orthogonal as $\left|\langle\alpha|-\alpha\rangle\right|^2\le 1.1\times 10^{-7}$ for $|\alpha|\ge 2$. However, this is not a strict requirement (as pointed out in the paper) as one can modify their scheme to use $|0\rangle = |\alpha\rangle + |-\alpha\rangle$ and $|1\rangle = |\alpha\rangle - |-\alpha\rangle$ as an exactly orthogonal bases (with no limit on the values $\alpha$ can take).
References

C Weedbrook, S Pirandola, R García-Patrón, NJ Cerf, TC Ralph, JH Shapiro, and S Lloyd, "Gaussian quantum information," Reviews of Modern Physics 84, 621–669 (2012).
doi:10.1103/RevModPhys.84.621
arXiv:1110.3234
TC Ralph, A Gilchrist, GJ Milburn, WJ Munro, and S Glancy, "Quantum computation with optical coherent states," Physical Review A 68, (2003).
doi:10.1103/PhysRevA.68.042319
arXiv:quant-ph/0306004


TITLE: Which stable homotopy groups are represented by parallelizable manifolds?
QUESTION [29 upvotes]: The Pontryagin-Thom construction allows one to identify the stable homotopy groups of spheres with bordism classes of stably normally framed manifolds. A stable framing of the stable normal bundle induces a stable framing of the stable tangent bundle.
This means that a framed manifold (one whose tangent bundle is trivial, e.g. a Lie group) represents an element of the stable homotopy groups of spheres.
So some elements are represented by honestly framed manifolds (as opposed to stably framed).
What is known about such elements? Is every element of the stable homotopy groups of spheres represented by an honestly framed manifold (i.e. with a trivial tangent bundle)?

REPLY [13 votes]: Repeating the first part of Oscar's answer and elaborating on comments by Chris and Panagiotis, here is a down-to-earth argument in all cases:
The cases $n=1,3,7$ are fine, since the stable stems are in these degrees generated by $S^1$, $S^3$, $S^7$ with the unstable framing induced by the multiplication in the unit complex numbers, quaternions, or octonions.
In the other cases, we use that the obstruction to destabilising a given stable framing $F$ of an oriented closed manifold $M^n$ lies in $H^n(M,\pi_n(SO/SO(d))$, which is isomorphic (in a preferred way) to $\mathbb{Z}$ if $n$ is even and to $\mathbb{Z}/2$ if $n$ is odd. It is not too hard to see that, with respect to this isomorphism, the obstruction is given by the semi-characterstic: half the Euler characteristic for $n=2d$ and $\sum_{i=0}^d\mathrm{dim}(H_i(M,\mathbb{Z}/2))\text{ mod  }(2)$ for $n=2d+1$ and $n\neq1,3,7$. In particular, the obstruction to destabilising is independent of $F$ which is somewhat surprising.
Originally this was proved by to Bredon and Kosinksi [1] who used a more geometric description of this obstruction: it is the degree (mod $2$ if $n$ is odd) of the Gauss map $M\rightarrow{S^n}$ induced by the stable framing $TM\oplus \varepsilon\cong \varepsilon^{n+1}$ (take the image of the canonical vector field in the trivial line bundle and normalize).
Now observe that, as Oscar explained, by doing a couple of trivial surgeries in a ball corresponding to taking connected sums with $S^1\times S^{n-1}$ or $S^2\times S^{n-2}$ and extending the stable framing, any stably framed bordism class in even dimensions contains a representative with trivial Euler-characteristic. The same works with the semi-characteristic in odd dimensions (here at most one surgery is necessary), so by the discussion above every stably framed bordism class has a representative whose stable framing can be destabilised.
[1] G.E. Bredon and A. Kosinski, Vector fields on $\pi$-manifolds. Annals of Math. 84, 85– 90 (1960).

REPLY [7 votes]: $k \cdot[\mathrm{point}]\in \pi_0^s$ is represented by an honestly framed 0-manifold if and only if $k \geq 0$.


text:
\section{Extending the Spectral Evolution Model}
\label{sec:extmodel}

Consider a network $G$ with $n$ vertices and a sequence of
($n$-by-$n$) adjacency matrices $\MAT{A}_1, \MAT{A}_2, \ldots,
\MAT{A}_t$ representing the evolution of $G$ (i.e., the addition of
new edges among existing nodes) over $t$ time units. One or more edges
can be created from one time unit to the next. The problem of
predicting the formation in $G$ of new edges at time $t+1$ consists in
extending the given sequence of adjacency matrices with a new
adjacency matrix $\MAT{A}_{t+1}$.  The spectral evolution model
introduced in Section~\ref{sec:model} can be used for such a task in
two ways, under the assumption that the eigenvectors of the adjacency
matrices remain constant. On the one hand, a graph kernel can be
chosen as a specific spectral transformation function for obtaining
$\MAT{A}_{t+1}$ from $\MAT{A}_t$. On the other hand, a time $t' < t$
can be picked so that $\MAT{A}_{t+1}$ is extrapolated from
$\MAT{A}_{t'}$ and $\MAT{A}_t$.  In either case, the evolution history
of the network $G$ is not fully considered and mostly neglected. The
working hypothesis is that, by considering the \textit{entire}
sequence $\MAT{A}_1, \MAT{A}_2, \ldots, \MAT{A}_t$, better eigenvalue
predictions could be computed to build $\MAT{A}_{t+1}$. That is, the
more information about how eigenvalues change over the $t$ time units,
the better link prediction for the network would be at time $t+1$. Of
course, the key question to ask is how to use all this information and
yet have an efficient construction method for $\MAT{A}_{t+1}$: this
section introduces an extrapolation-based approach that answers this
question by using approximations computed with the Rayleigh quotient.

\subsection{The Rayleigh Quotient}
\label{sec:rayleigh}

The eigenvalues and eigenvectors of symmetric matrices have many
characterizations. The Rayleigh quotient is a tool from linear algebra
that offers one of such characterizations (see, e.g.,
~\citep{chatelin-eigenvalues-2012}).

\begin{definition}\label{def:rayleigh}
  Let $n \in \nats$ be positive, $\MAT{A}$ an (adjacency) $n$-by-$n$
  matrix, and $\MAT{x}$ a real-valued non-zero $n$-vector. The
  \textit{Rayleigh quotient} of $\MAT{x}$ with respect to $\MAT{A}$ is
  the ratio $R(\MAT{A}, \MAT{x})$ defined by

  \[R(\MAT{A}, \MAT{x}) = \frac{\MAT{x}^T\MAT{A}\MAT{x}}{\MAT{x}^T\MAT{x}}.\]

\end{definition}

Note that the Rayleigh quotient $R(\MAT{A}, \MAT{x})$ is a real number
(i.e., a scalar value). If $\MAT{x}$ is an eigenvector of $\MAT{A}$,
then the ratio $R(\MAT{A}, \MAT{x})$ is the corresponding
eigenvalue.

Also, if a vector $\MAT{y}$ approximates an eigenvector $\MAT{x}$ of
$\MAT{A}$, then $R(\MAT{A}, \MAT{y})$ is an approximation of
$R(\MAT{A}, \MAT{x})$. That is, eigenvalues can be approximated based
on approximations of eigenvectors by using the Rayleigh quotient.

\subsection{Spectral Extrapolation}
\label{sec:spectralextrapolation}

Recall the problem of predicting the formation in $G$ of new edges at
time $t+1$ from the adjacency matrices $\MAT{A}_1, \MAT{A}_2, \ldots,
\MAT{A}_t$ representing the evolution of $G$ over $t$ time units. The
approach is to extrapolate the trajectory of the eigenvalues of the
adjacency matrices in this sequence.

Let $(\MAT{X}_t,\MAT{\Lambda}_t)$ be the spectral decomposition of
$\MAT{A}_t$, and let $\MAT{x}_j$ and $\lambda_j$ be the $j$-th
eigenvector and eigenvalue in $\MAT{X}_t$ and $\MAT{\Lambda}_t$,
respectively (i.e., the latent dimension $j$). Under the assumption of
the spectral evolution model, the trajectory of $\lambda_j$ in
$\MAT{A}_1, \MAT{A}_2, \ldots, \MAT{A}_t$ can be approximated by
$\hat{\lambda}_{j}^{1:t}$ defined as:

\[\hat{\lambda}_{j}^{1:t} = R(\MAT{A}_1,\MAT{x}_j), R(\MAT{A}_2,\MAT{x}_j), \ldots, R(\MAT{A}_t,\MAT{x}_j).\]

Note that $\hat{\lambda}_{j}^{1:t}(t) = R(\MAT{A}_t,\MAT{x}_j) =
\lambda_j$.  The problem of computing the $j$-th eigenvalue of the
matrix $\MAT{A}_{t+1}$ (i.e., of predicting the evolution of the
spectrum for the latent dimension $j$) can then be cast as fitting the
trajectory of $\hat{\lambda}_{j}^{1:t}$. For solving this problem,
e.g., linear and quadratic regressions are available.

Computationally, the following are the calculations: obtaining the
spectral decomposition of $\MAT{A}_t$ takes $O(n^3)$ time, computing
the sequence $\hat{\lambda}_{j}^{1:t}$ takes $O(tn^2)$ (each
application of $R$ can take $O(n^2)$ time), and fitting the sequence,
e.g., $O(tn^2)$ time with least squares regression. Therefore, the
computational cost of computing $\MAT{A}_{t+1}$ can take $O(n^3 +
tn^3) = O(tn^3)$, because the spectral decomposition of $\MAT{A}_t$
can be computed just once, and there are $n$ computations taking
$O(tn^2)$ time, one per latent dimension. Note that $O(tn^3)$ is also
the asymptotic time bound for checking the spectral evolution model
hypothesis for $\MAT{A}_1, \MAT{A}_2, \ldots, \MAT{A}_t$ (namely, that
all eigenvectors of the network remain largely constant over the $t$
time units).

With further experimental evaluation, the constant accompanying the
asymptotic bound $O(tn^3)$ for the construction of $\MAT{A}_{t+1}$ can
be reduced by a fraction. In particular, the approach above can be
carried out for only a selection of the eigenvectors of
$\MAT{A}_t$. For the case study presented in
Section~\ref{sec:results}, it was possible to consider about 8\% of
the eigenvectors and yet reconstruct $\MAT{A}_{t+1}$ with an accuracy
of $99.97\%$. Figure~\ref{fig:adjrq} presents the comparison between
the exact approach (i.e., extrapolating the trajectory of the $n$
eigenvalues) and the above-mentioned approach (i.e., extrapolating the
trajectory of only $8\%$ of the $n$ eigenvalues).

\begin{figure}[htbp!]
	\centering
	\begin{subfigure}[b]{0.45\textwidth}
		\centering
		\includegraphics[width=\textwidth]{fig/VocesDeLaReconciliacion_adjorg-eps-converted-to.pdf}
		\caption{$\mathbf{A}_{t+1}$}
		\label{fig:adjorg}
	\end{subfigure}

	\begin{subfigure}[b]{0.45\textwidth}
		\centering
		\includegraphics[width=\textwidth]{fig/VocesDeLaReconciliacion_adjrec-eps-converted-to.pdf}
		\caption{Reconstructed $\mathbf{A}_{t+1}$}
		\label{fig:adjrec}
	\end{subfigure}
	\caption{Adjacency matrix reconstruction of the mention network
    \#vocesdelareconciliacion. The Rayleigh quotient is used to
    approximate the eigenvalues of a matrix from a set of orthogonal
    vectors (eigenvector approximations). The accuracy of these
    approximations is showed by the comparison between the original
    matrix and its reconstruction (based on the approximations of the
    eigenvalues).}
	\label{fig:adjrq}
\end{figure}

config: arxiv


text:
\begin{document}

\title{Fractional Fokker-Planck equation from non-singular kernel operators}

\author{M. A. F. dos Santos$^{1}$}
\address{$^1$Instituto de F\'isica, Universidade Federal do Rio Grande do Sul, Caixa Postal 15051, CEP 91501-970, Porto Alegre, RS, Brazil}
\ead{santos.maike@ufrgs.br}

\author{Ignacio S. Gomez$^{2}$}
\address{$^2$Instituto de F\'isica, Universidade Federal da Bahia,
			 R.\ Bar\~ao de Jeremoabo s/n, 40170-115 Salvador, Bahia, Brazil}
\ead{nachosky@fisica.unlp.edu.ar}
\vspace{10pt}
\begin{indented}
\item[] \today
\end{indented}

\begin{abstract}
Fractional diffusion equations imply non-Gaussian distributions that generalise the standard diffusive process. Recent advances in fractional calculus lead to a class of new fractional operators defined by non-singular memory kernels, differently from the fractional operator defined in the literature. In this work we propose a generalisation of the Fokker-Planck equation in terms of a non-singular fractional temporal
operator and considering a non-constant diffusion coefficient.
We obtain analytical
solutions for the Caputo-Fabrizio and the Atangana-Baleanu fractional kernel operators, from which non-Gaussian distributions emerge having a long and short tails. In addition, we show that these non-Gaussian distributions are unimodal or bimodal
according if the diffusion index $\nu$ is positive or negative respectively, where a diffusion coefficient of the power law type $\mathcal{D}(x)=\mathcal{D}_0|x|^{\nu}$ is considered.
Thereby, a class of anomalous diffusion phenomena connected with fractional derivatives and with a diffusion coefficient of the power law type

is presented. The techniques employed in this work open new possibilities for studying memory effects in diffusive contexts.
\end{abstract}

\vspace{2pc}
\noindent{\it Keywords}:
Fractional FPE, anomalous diffusion, Caputo-Fabrizio operator, Atangana-Baleanu operator

\section{\label{sec:level1}Introduction}

Gaussian forms are the most studied probability distributions in statistical physics.
Based on a thermal framework, Einstein proposed
the standard diffusion equation and thus he determined the Gaussian forms
that characterise the movement of Brownian particles immersed in an equilibrium liquid \cite{einstein1905molekularkinetischen}. These investigations were also continued
during the period 1908-1914 by Perrin and Nordlund
\cite{perrin1908agitation,perrin1909mouvement,nordlund1914new}, while alternative formulations were proposed by Sutherland, Smoluchowski, and Langevin \cite{sutherland1905lxxv,von1906kinetischen,langevin1908theorie}.
The Brownian motion is characterised by the linear evolution of the mean square displacement (MSD), that is the so-called linear diffusion i.e. $\langle (\Delta x)^2 \rangle \sim t$, and the break of this linearity
defines the
anomalous diffusion (or non-Fickian diffusion) \cite{bouchaud1990anomalous}.
Anomalous diffusion process is characterised by a
power law dependence of the MSD, i.e. $\langle (\Delta x)^2 \rangle \sim t^{\mu}$, for which $0<\mu<1$ and $1<\mu<2$ determine the subdiffusive and superdiffusive regimes respectively.
In some particular cases, ballistic and hyper diffusive processes can occur for
the values $\mu=2$ and
$2<\mu$. Several works show that the anomalous diffusion is strongly connected with non-Gaussian distributions \cite{w1,w2,w3,w4,w5}
that can be also obtained
as solutions of generalisations of the diffusion equation
\cite{bologna2000anomalous,chechkin2017brownian}.

Non-Gaussian distributions are associated intimately with generalisations of the Fokker-Planck equation (FPE) and they
allow to
model multiple and complex phenomena as non-ergodicity, non-locality, memory effects and long range interactions.
Complementary, nonlinear, heterogeneous and fractional approaches also provide generalisations of the FPE.
Nonlinear diffusion equations, whose
solutions exhibit
a non-Gaussian behaviour, has been associated to nonlinear FPE in several scenarios: quantum walks \cite{nonlinear3}, non-extensive statistical mechanics \cite{nonlinear1,nonlinear4,dos2017entropic,nonlinear2,sicuro2016nonlinear},  generalisations of the Central Limit Theorem which imply
non-Gaussian forms (called as $q$-Gaussian distributions) \cite{umarov2007multivariate,umarov2008q,jauregui2015convergence,jauregui2010new}, among others.

In addition,
the study of the FPE with a non-constant diffusion coefficient has allowed to introduce a natural generalisation of the Brownian motion in an heterogeneous medium \cite{risken1996fokker,cherstvy2013anomalous}. This approach was reported in: comb model structures \cite{sandev2017heterogeneous}, reaction--diffusion model \cite{hormuth2017mechanically}, non-ergodicity, criticality of diffusive systems \cite{metzler2014anomalousd} and  intracellular transport \cite{lanoiselee2018model}, etc.

Other way of investigating the Brownian motion in a heterogeneous medium can be achieved by means of the Ito-Langevin equation that links
the type of displacement of the trajectories
with a heterogeneous FPE \cite{risken1996fokker,borland1998microscopic}.
This formalism characterises the diffusion equation through fractional operators, heading to a fractional dynamic. The use of fractional derivatives for describing diffusion equations has been turned out a useful method to analyze non-local phenomena and memory effects
in several systems \cite{li2018time,ervin2018regularity,fractional1,fractional2,fractional3,dos2017anomalous,klages2008anomalous,SUN20094586,yang2017new}.

In this context, Caputo and Fabrizio have recently proposed a non-singular fractional derivative expressed by an exponential kernel \cite{caputo2015new}. Subsequently, Atangana \textit{et. al} also have defined a non-singular kernel in terms by the Mitag-Leffler function \cite{atangana2016new}. These non-singular operators present properties that allow to describe transport phenomena, anomalous diffusion process and dynamical models \cite{gomez2017space,atangana2016chaos,xiao2017general,dokuyucu2018cancer,hristov2017derivation,alkahtani2017generalized,gomez2018analytical} simultaneously.

The main goal of this paper
to propose a generalisation of the FPE,
in terms of a non-singular temporal kernel operator
and a diffusion coefficient of the power law type, that contains all known anomalous (sub-super-ballistic-hyper) diffusive
regimes as special cases.
For accomplish this, we use the Caputo-Fabrizio and the Atangana-Baleanu fractional kernel operators and we obtain non-Gaussian analytical
solutions with unimodal and bimodal characteristics.
Thus, our contribution consist of providing
a unified framework for studying diffusion processes generated by non-singular fractional operators.
The work is structured as follows. In Section \ref{sec2} we review some concepts of the fractional calculus related to the Caputo-Fabrizio and Atangana-Baleanu fractional derivatives along with a generalisation of the FPE within the fractional calculus framework.
Section \ref{sec3} is devoted to obtain the solutions of the generalised FPE from the non-singular temporal kernels of the Caputo-Fabrizio and Atangana-Baleanu fractional derivatives,
and considering a arbitrary variable diffusion coefficient, i.e. $\mathcal{D}(x)$.
We obtain analytical solutions for the Caputo-Fabrizio and Atangana-Baleanu fractional operators.
In Section \ref{sec4} we characterise the type of the diffusion obtained from the calculus of the MSD.
As a consequence, subdiffusion, superdiffusion and hyperdiffusion are obtained as special cases. In Section \ref{sec5} some conclusions and perspectives are outlined.

\section{
Preliminaries}
\label{sec2}
We review some notions and concepts
used
throughout the paper.

\subsection{Fractional calculus}\label{subsec2.1}

Fractional calculus considers the different possibilities of extending powers of the usual
differentiation to the field of the real numbers or complex ones.
The Riemann-Liouville integral constitutes one of the most traditional forms of fractional integration, expressed by
\begin{eqnarray}\label{fractional-derivative-RL}
{}_{a} \mathcal{D}_{t}^{-\alpha} f(t)=
\frac{1}{\Gamma(\alpha)}\int_{a}^{t}(t-s)^{\alpha-1}f(s)ds
\end{eqnarray}
where $f(t)$ is an arbitrary function and $\Gamma(\ldots)$ denotes the Gamma function. By the Laurent series,
from
(\ref{fractional-derivative-RL}) we can see that for $\alpha=0$ and $\alpha=1$ the original function $f(t)$ and its primitive $\int_a^t f(s)ds$ are recovered respectively. In this sense, the fractional derivative of Riemann--Liouville is defined by $\frac{d}{dt} {}_{a} \mathcal{D}_{t}^{-(1-\alpha)} f(t)$, in which $0<\alpha<1$. Other fundamentals  forms to fractional derivative as Gr\"unwald--Letnikov, Caputo, Riesz and other are detailed in Ref. \cite{podlubny1998fractional}.
Recent generalisations of fractional derivatives have been developed \cite{caputo2015new,atangana2016new}. In this work we focus on two types: the Caputo-Fabrizio and the Atangana-Baleanu fractional derivatives. The former does not need
to define fractional order initial conditions as in the
Riemann-Liouville case, while the later uses
a general Mittag-Leffler function as a kernel.
These type of fractional derivatives can be expressed in the compact form
\begin{eqnarray}\label{fractional-derivative-kernel}
\mathcal{O}_t^{\mathcal{K}} f(t)=
\int_{0}^{t}dt^{\prime} \mathcal{K}(t-t^{\prime})\frac{d\ }{dt^{\prime}} f(t^{\prime})
\end{eqnarray}
where $\mathcal{K}$ stands for the kernel employed. For instance, when $\mathcal{K}(t)=\delta(t)$ the usual derivative is recovered. For the cases of the Caputo-Fabrizio and Atangana-Baleanu the kernel is given by
\begin{eqnarray}
\mathcal{K}_{CF}(t)=\frac{f(\alpha)}{1-\alpha} \exp \left[-\frac{\alpha t}{1-\alpha} \right],
\label{Kcaputo}
\end{eqnarray}
for $0<\alpha<1$ \cite{caputo2015new}, and
\begin{eqnarray}
\mathcal{K}_{AB}(t)=\frac{b(\alpha)}{1-\alpha}E_{\alpha}\left[-\frac{\alpha t^{\alpha}}{1-\alpha} \right],
\label{KAtangana}
\end{eqnarray}
for $0<\alpha<1$ with $E_{\alpha}(\cdots)$ the Mittag-Leffler function \cite{haubold2011mittag} which is the kernel associated to the fractional operator of Atangana and Baleanu \cite{atangana2016new}. The AB (Atangana--Baleanu) fractional derivative has two characteristic: a non--singular form (i.e. $\lim_{t \rightarrow 0}\mathcal{K}_{AB}(t) \neq +\infty$) and a power law behaviour for long times, i.e. $\mathcal{K}_{AB}(t) \propto t^{-1-\alpha}$.

\subsection{Fractional Fokker-Planck equation}\label{subsec2.2}

The progress in the fractional calculation was accompanying the developments obtained in the field of diffusion differential equations. Fractional derivatives offer multiples ways of generalising diffusion differential equations as well as for characterising more classes of phenomena than the standard diffusion equation. The unidimensional Fokker-Planck equation with constant diffusion coefficient $D_0$
\begin{eqnarray}\label{FPE}
\frac{\partial \rho(x,t)}{\partial t}=D_0 \frac{\partial^2\rho(x,t)}{\partial x^2}
\end{eqnarray}
describes the evolution of the probability density function (PDF) $\rho(x,t)$ of finding a particle subjected a drag and random forces. The Eq. (\ref{FPE})
describes normal diffusion and further generalisations are needed for taking into account other types of diffusion that can be occur. Even more, the temporal derivative
$\frac{\partial \rho(x,t)}{\partial t}$ is not appropriate to consider memory effects that are present
in a wide class of phenomena. Here is when the fractional generalisations of the FPE enter in order to characterise memory effects of the evolution of the PDF.

For accomplish this, we consider the following generalisation of the FPE (\ref{FPE}) (that we called fractional FPE)
\begin{eqnarray}
\mathcal{O}^{\mathcal{K}}_t\rho(x,t)= \frac{\partial}{\partial x}
\left(\sqrt{\mathcal{D}(x)} \frac{\partial }{\partial x} \left(\sqrt{\mathcal{D}(x)}\rho(x,t)\right)\right),
\label{diffusion}
\end{eqnarray}
where $\mathcal{O}^{\mathcal{K}}_t$  is the
fractional derivative with kernel $\mathcal{K}$ given by the Eq. (\ref{fractional-derivative-kernel})
and $\mathcal{D}(x)$ is a non-constant diffusion coefficient. It is clear that for $\mathcal{K}(t)=\delta(t)$
and $\mathcal{D}(x)=D_0$ the traditional FPE (\ref{FPE}) is recovered.

\section{Fractional FPE with
non-singular kernel operators
}\label{sec3}

In this Section we obtain the solutions of the fractional FPE (\ref{diffusion}) in the context of non-singular operators
with the initial condition $\rho(x,0)=\delta(x)$. The calculations are performed for an arbitrary diffusion coefficient. Then, we consider a power-law dependence of the diffusion coefficient for a detailed analysis. Applying the Laplace transform of Eq. (\ref{diffusion})
we obtain
\begin{eqnarray}
s\mathcal{K}(s)\rho(x,s)-
\mathcal{K}(s) = \frac{\partial}{\partial x}\sqrt{\mathcal{D}(x)}   \frac{\partial}{\partial x} \sqrt{\mathcal{D}(x)} \rho(x,s).
\label{diffusionlanplace}
\end{eqnarray}
where $\mathcal{L}\lbrace \mathcal{O}_t^{\mathcal{K}} f(t) \rbrace = s\mathcal{K}(s) f(s)-
\mathcal{K}(s)f(0)$ is the
Laplace transform of the fractional derivative with kernel $\mathcal{K}$ (Eq. \ref{fractional-derivative-kernel}).
The Laplace transform of a function $f(t)$ is defined by $\int_0^{\infty}dt e^{-st} f(t)=\mathcal{L}\lbrace f(t) \rbrace= F(s)$ and its inversion is given by $\frac{1}{2i \pi}\lim_{T\rightarrow \infty}\int_{\sigma-i T}^{\sigma + i T}ds e^{st} F(s)=\mathcal{L}^{-1}\lbrace F(s)\rbrace =f(t)$, in which $s$ is a complex variable and $\sigma$ is a real number so that the contour path of integration in the region of convergence of $F(s)$.
We search for solutions of the Eq. (\ref{diffusionlanplace}) in the form of the \textit{ansatz}
\begin{eqnarray}
\rho(x,s)= \frac{\Upsilon(s)}{\sqrt{\mathcal{D}(x)}} \exp \left[-\int_{0}^{|x|}dx' \frac{\sqrt{s \mathcal{K}(s)}}{\sqrt{\mathcal{D}(x')}} \right]
\label{ansatz}\end{eqnarray}
where $\rho(x,s)=\rho(-x,s)$ points out the symmetry in the variable $x$. Establishing the change of variable in $x$
\begin{eqnarray}
y(x)=\int_{0}^{|x|}dx' \frac{1}{\sqrt{\mathcal{D}(x)}}, \qquad \textnormal{in which } \qquad \lim_{x \rightarrow \infty}y(x)=\infty,
\end{eqnarray}
and using the \textit{ansatz} (\ref{ansatz}) in Eq. (\ref{diffusionlanplace}) (with the normalization condition $\int_{-\infty}^{\infty} dx \rho(x,t)=1$) we obtain
\begin{eqnarray}
\Upsilon(s)=\frac{1}{2}\sqrt{\frac{\mathcal{K}(s)}{s}},
\label{upslon}
\end{eqnarray}
which implies a general solution in the Laplace coordinate space $s$
\begin{eqnarray}
\rho(x,s)=\frac{\sqrt{\mathcal{K}(s)}}{2 \sqrt{s \mathcal{D}(x)}}  \exp \left[-\int_{0}^{|x|}dx' \frac{\sqrt{s \mathcal{K}(s)}}{\sqrt{\mathcal{D}(x')}}  \right].
\label{ansatzlaplace}
\end{eqnarray}
For the fractional Caputo operator, defined by (\ref{fractional-derivative-kernel}) with the kernel $\mathcal{K}_C(t)=\frac{t^{-\alpha}}{\Gamma[1-\alpha]}$, the Laplace inversion of (\ref{ansatzlaplace})
gives the solution
\begin{eqnarray}
\rho(x,t)=\frac{1}{2 t^{\frac{\alpha}{2}}\sqrt{\mathcal{D}(x)}} {\mbox{\Large{H}}}^{1,0}_{1,1}\left[ \int_{0}^{|x|}dx' \frac{1}{\sqrt{\mathcal{D}(x')}} \frac{1}{t^{\frac{\alpha}{2}}} \Big| ^{\left(1-\frac{\alpha}{2},\frac{\alpha}{2}\right)}_{(0,1)} \right],
\label{caputo}
\end{eqnarray}
where the $\textnormal{\large{H}} (\cdots)$ is the Fox function \cite{mainardi2005fox} that consists of a type of Merlin integration. The solution (\ref{caputo}) is a non-Gaussian distribution. For a constant diffusion coefficient it is a typical solution of the diffusion equation proposed in Montroll works for describing continuous random walks in time \cite{montroll1965random,scher1975anomalous}. On the other hand, the solution (\ref{caputo}) with $\alpha=1$ contains a variety of PDF describing diffusion processes in a heterogeneous medium, which was investigated by Metzler \textit{et al} in \cite{metzler2014anomalousd}. In this framework, our intention is extend this first result to non-singular diffusion equations. This is not a non-trivial issue because, as we have seen, it is necessary to employ the Laplace-inverse transform on the $s$-complex plane and this typically can be a complicated task.

\subsection{General solution for the Caputo-Fabrizio operator}
\label{subsec3.1}

The kernel of the Caputo-Fabrizio fractional derivative can be written in the Laplace coordinate space $s$ as
\begin{eqnarray}
\mathcal{K}(s)=\frac{f(\alpha)}{1-\alpha}\frac{1}{s+\frac{\alpha}{1-\alpha}}.
\label{kernel1s}
\end{eqnarray}
which if replaced
in Eq. (\ref{ansatzlaplace}) allows us to obtain $\rho(x,s)$ in the compact form
\begin{eqnarray}
\rho(x,s)= -\frac{1}{2}\frac{\partial \Phi}{\partial |x|},
\label{pdfs}
\end{eqnarray}
where
\begin{eqnarray}
\Phi (x,s) = \frac{1}{s}\exp \left[ - \int_{0}^{|x|}dx' \frac{1}{\sqrt{\mathcal{D}(x')}} \sqrt{\frac{f(\alpha)}{1-\alpha}} \sqrt{\frac{s}{s+\frac{\alpha}{1-\alpha}}}\right].
\end{eqnarray}
Now, by calculating the inversion Laplace transform of this expression and taking the branch cuts associated with $\sqrt{s}$ and $\sqrt{s+\frac{\alpha}{1-\alpha}}$ along the negative real axis of $s$--plane for $\frac{\alpha}{1-\alpha} >0 $ (see Ref. \cite{widder2015laplace} for more details) we have
\begin{eqnarray}\label{phi-caputo}
\Phi (x,t) &=& 1- \frac{2}{\pi} \int_0^{\infty}d\eta \exp \left[-\frac{\alpha}{1-\alpha} \frac{t\eta ^2}{\eta ^2+1} \right] \frac{\mathcal{S}(t,\eta)}{\eta(1+\eta^2)},
\end{eqnarray}
with
\begin{eqnarray}\label{sin-caputo}
\mathcal{S}(t,\eta) &= & \sin \left[ \int_{0}^{|x|} dx' \frac{1}{\sqrt{\mathcal{D}(x')}} \sqrt{\frac{f(\alpha)}{1-\alpha}}  \eta \right].
\end{eqnarray}
Then, using Eqs. (\ref{phi-caputo}) and (\ref{sin-caputo}) in (\ref{pdfs}), we obtain the general solution of the fractional FPE (\ref{diffusion}) corresponding to the Caputo-Fabrizio kernel
\begin{eqnarray}
\rho(x,t) &=& \frac{1}{\pi}\sqrt{\frac{f(\alpha)}{(1-\alpha)\mathcal{D}(x)}} \int_0^{\infty}d\eta \exp \left[-\frac{\alpha}{1-\alpha} \frac{t\eta ^2}{\eta ^2+1} \right]\frac{\mathcal{C}(t,\eta) }{1+\eta^2},
\label{pdfcaputo}
\end{eqnarray}
with
\begin{eqnarray}
\mathcal{C}(t,\eta) &=& \cos \left[\eta \int_{0}^{|x|}dx' \frac{1}{\sqrt{\mathcal{D}(x')}} \sqrt{\frac{f(\alpha)}{1-\alpha}} \right].
\end{eqnarray}
This PDF implies a generalised non-Gaussian form for any choice of the fractional index $\alpha$, as we shall see.
Henceforward we consider the power-law form for the diffusion coefficient
\begin{eqnarray}
\mathcal{D}(x)=\mathcal{D}_0|x|^{\nu}, \qquad \nu<2.
\label{diffusioncoeficiente1}
\end{eqnarray}
that
was used by Andrey G. Cherstvy \textit{et. al} in a breaking ergodicity context \cite{cherstvy2013anomalous} and in other scenarios in \cite{sandev2017heterogeneous, hormuth2017mechanically,metzler2014anomalousd,lanoiselee2018model}.
Figs \ref{fig1} shows the logarithm of the PDF (\ref{pdfcaputo}) for the values
$\alpha=0.25,0.5,0.7,0.9,0.95,0.99$ and for $\nu=1$ and $\nu=-1$ at the left and right pannels respectively, along with the Gaussian PDF in green dots. We can see that both families of PDF ($\nu=1$ and $\nu=-1$) exhibit a non-Gaussian behavior with a long tail.
For $\nu=1$ every PDF presents an unimodal
shape while for $\nu=-1$ they present a bimodal one.
\begin{figure}[h!]

\includegraphics[scale=0.8]{fabriziodistribution2.pdf}
\includegraphics[scale=0.8]{fabriziodistribution.pdf}
\caption{Log of the solutions of the fractional FPE
for the Caputo-Fabrizio kernel (Eq. \ref{pdfcaputo}) at time $t=10^{-1}$ in function of the coordinate space $x$ for the values of the fractional index $\alpha=0.25,0.5,0.75,0.9,0.95,0.99$ and considering the parameter values $\frac{\mathcal{D}_0}{f(\alpha)} = 10$. For comparing, in both plots the Gaussian PDF (quadratic curve in green dots) with $D_0=10$ is illustrated. Left pannel: family of solutions for $\nu=1$ exhibiting a unimodal behavior. Right pannel: family of solutions for $\nu=-1$ exhibiting a bimodal behavior.}
\label{fig1}
\end{figure}

\subsection{General solution for the Atangana-Baleanu operator}
\label{subsec3.2}

The kernel of the Atangana-Baleanu fractional derivative in the Laplace coordinate space $s$ is given by
\begin{eqnarray}\label{atangana-laplace-kernel}
\mathcal{K}(s)=\frac{b(\alpha)}{\alpha-1}\frac{s^{\alpha-1}}{s^{\alpha}+\frac{\alpha}{1-\alpha}},
\end{eqnarray}
which if replaced
in Eq. (\ref{ansatzlaplace}) allows us to obtain
$\rho(x,s)$, and then the solution $\rho(x,t)$ follows using the inverse Laplace transform.
Analogously to the Caputo-Fabrizio case,
$\rho(x,s)$ admits the representation expressed by the
Eq. (\ref{pdfs}) with $\Phi(x,s)$ given by
\begin{eqnarray}
\Phi (x,s) = \frac{1}{s}\exp \left[ - \int_0^{|x|}dx' \frac{1}{\sqrt{\mathcal{D}(x')}}  \sqrt{\frac{b(\alpha)}{1-\alpha}} \sqrt{\frac{s^{\alpha}}{s^{\alpha}+\frac{\alpha}{1-\alpha}}}\right],
\end{eqnarray}
which can be solved by integrating the branch cut associated with $s^{\alpha}$. Then, from the inverse Laplace transform we have
\begin{eqnarray}
\Phi(x,t) &=& 1-\frac{1}{\pi}\int_0^{\infty} e^{-t \eta}\exp \left \lbrace - \int_0^{|x|}dx' \frac{1}{\sqrt{\mathcal{D}(x')}} \sqrt{\frac{b(\alpha)\eta^{\alpha}}{\gamma_1}} \cos \left[ \frac{\alpha \pi - \gamma_2}{2} \right] \right \rbrace \nonumber \\
& \times &  \sin \left \lbrace \int_0^{|x|}dx' \frac{1}{\sqrt{\mathcal{D}(x')}}  \sqrt{\frac{b(\alpha)\eta^{\alpha}}{\gamma_1}} \sin \left[ \frac{\alpha \pi - \gamma_2}{2} \right] \right \rbrace \frac{d\eta}{\eta}
\label{up2}
\end{eqnarray}
with
\begin{eqnarray}
\gamma_1=\sqrt{\eta^{2\alpha}+\left(\frac{\alpha}{1-\alpha}\right)^2 + 2 \left(\frac{\alpha}{1-\alpha}\right) \eta^{\alpha} \cos[\alpha \pi]}
\end{eqnarray}
 and
\begin{eqnarray}
 \gamma_2= \arctan \left[\eta^{\alpha} \sin[\alpha \pi] \left(\eta^{\alpha}\cos[\alpha \pi]+\frac{\alpha}{1-\alpha}\right) \right].
 \end{eqnarray}
Using Eqs. (\ref{pdfs}) and (\ref{up2})
the general solution of the fractional FPE (\ref{diffusion}) corresponding to the Atangana-Baleanu kernel results
\begin{eqnarray}
\rho(x,t)&=&\frac{1}{2\pi}\int_0^{\infty} e^{-t \eta}\exp \left \lbrace -\int_0^{|x|}dx'\sqrt{\frac{b(\alpha)\eta^{\alpha}}{\mathcal{D}(x')\gamma_1}} \cos \left[ \frac{\alpha \pi - \gamma_2}{2} \right] \right \rbrace  \sqrt{\frac{b(\alpha)\eta^{\alpha}}{\mathcal{D}(x)\gamma_1}}  \nonumber \\
& \times & ( \cos \left[ \frac{\alpha \pi - \gamma_2}{2} \right]  \sin \left \lbrace \int_0^{|x|}dx' \sqrt{\frac{b(\alpha)\eta^{\alpha}}{\mathcal{D}(x')\gamma_1}} \sin \left[ \frac{\alpha \pi - \gamma_2}{2} \right] \right \rbrace \nonumber \\
& + &  \sin \left[ \frac{\alpha \pi - \gamma_2}{2} \right]   \cos \left \lbrace \int_0^{|x|}dx' \sqrt{\frac{b(\alpha)\eta^{\alpha}}{\mathcal{D}(x')\gamma_1}} \sin \left[ \frac{\alpha \pi - \gamma_2}{2} \right] \right \rbrace )   \frac{d\eta}{\eta},
\label{pdfatagana}
\end{eqnarray}
which is valid for all diffusion coefficient $\mathcal{D}(x)$. Assuming
$\mathcal{D}(x)=\mathcal{D}_0|x|^{\nu}$ with $\nu<2$ (as in Eq. (\ref{diffusioncoeficiente1})) and letting $\alpha \rightarrow 1$
the solution becomes in the exponential form
\begin{eqnarray}
\rho(x,t) = \frac{\exp\left[-\frac{\left| x\right| ^{2-\nu}}{ (2-\nu)^2 \mathcal{D}_0 t}\right]}{\sqrt{2 \pi\mathcal{D}_0 |x|^{\nu} t}},
\end{eqnarray}
which reproduces the standard Gaussian solution of the FPE for $\nu=0$.
Other representative case is obtained when
$\mathcal{D}(x)=constant$, which implies an analytical solution with an arbitrary fractional index $\alpha$
\cite{alkahtani2017generalized,hristov2017derivatives}.
Figs \ref{fig2} shows the logarithm of the PDF (\ref{pdfatagana}) for the values
$\alpha=0.25,0.6,0.65$ and for $\nu=1$ and $\nu=-1$ at the left and right pannels respectively along with the Gaussian PDF with green dots. Analogously as in the Caputo-Fabrizio case, we can see that both families of PDF ($\nu=1$ and $\nu=-1$) exhibit a non-Gaussian behavior with a long tail.
For $\nu=1$ every PDF presents an unimodal
shape while for $\nu=-1$ they present a bimodal one.

\begin{figure}[h!]
\centering
\includegraphics[scale=0.8]{Ataganadistribution1.pdf}
\includegraphics[scale=0.8]{Ataganadistribution2.pdf}
\caption{Log of the solutions of the fractional FPE
for the Atangana-Baleanu kernel (Eq. \ref{pdfatagana}) at time $t=10^{-1}$ in function of the coordinate space $x$ for the values of the fractional index $\alpha=0.25,0.6,0.65$ and considering the parameter values $\frac{\mathcal{D}_0}{f(\alpha)} = 10$. For comparing, in both plots the Gaussian PDF (quadratic curve in green dots) with $D_0=10$ is illustrated. Left pannel: family of solutions for $\nu=1$ exhibiting a unimodal behavior. Right pannel: family of solutions for $\nu=-1$ exhibiting a bimodal behavior.}
\label{fig2}
\end{figure}

\section{Types of diffusion from fractional FPE: Caputo-Fabrizio and Atangana-Baleanu kernels}
\label{sec4}

Anomalous diffusion phenomena can be characterised by means of the MSD
of the particles, as was reported by Metzler and Klafter in \cite{metzler2000random}.
In our case, the general solution (\ref{ansatzlaplace}) implies
a null mean displacement $\langle x \rangle =0$, so MSD is equal to $\langle (x-\langle x \rangle)^2 \rangle =\langle x^2 \rangle $.
Using that the MSD is $\langle x^2 \rangle =\int_{-\infty}^{+\infty}x^2\rho(x,t)dx$
with the diffusion coefficient given by (\ref{diffusioncoeficiente1}), we obtain
for the general solution (\ref{ansatzlaplace}) that the MSD in the Laplace
space coordinate $s$ is
\begin{eqnarray}\label{anomalous-laplace}
\langle x^2 \rangle (s) = \frac{\sqrt{\mathcal{K}(s)} 2^{\frac{2 \nu}{\nu -2}} \Gamma \left[\frac{4}{2- \nu}\right] \left \lbrace \frac{\sqrt{\mathcal{K}(s) s}}{\sqrt{\mathcal{D}_0} (2- \nu)}\right \rbrace^{\frac{6-\nu }{\nu -2}}}{(2-\nu)^2 \sqrt{\mathcal{D}_0 s}}.
\end{eqnarray}

Considering the Caputo-Fabrizio kernel in the Laplace space coordinate $s$ (see Eq.(\ref{kernel1s})) and from the inversion Laplace transform of (\ref{anomalous-laplace}) we have
\begin{eqnarray}
\langle x^2 \rangle(t) =\int_0^{t}dt' \frac{e^{-\frac{\alpha  \left(t-t'\right)}{1-\alpha }} \left(\frac{t-t'}{1-\alpha }\right)^{-\frac{2}{2-\nu }-1}}{(1-\alpha ) \Gamma \left(\frac{2}{2-\nu }\right)} \frac{ 2^{\frac{2 \nu}{\nu -2}} \Gamma \left[\frac{4}{2- \nu}\right] t'^{\frac{2  }{2- \nu }} }{\Gamma\left[ \frac{4-\nu}{2-\nu} \right](2-\nu )^{\frac{\nu +2}{2-\nu }} \mathcal{D}_0^{\frac{2}{\nu-2}}},
\end{eqnarray}
that in the asymptotic limit of long times $t\rightarrow\infty$ becomes $\langle x^2 \rangle \sim t^{\frac{2 }{2- \nu }}$. This means that the fractional index of the Caputo-Fabrizio operator does not influence on the type of diffusion in the asymptotic limit, which is a similar behaviour present in clusters of percolation \cite{arkhincheev1991anomalous}.

Regarding the Atangana-Baleanu operator, whose kernel in the Laplace space coordinate $s$ is given by (\ref{atangana-laplace-kernel}), by introducing this in (\ref{anomalous-laplace}) and from its inversion Laplace transform we have
\begin{eqnarray}
\langle x^2 \rangle(t) \sim \left(\frac{1}{\alpha }\right)^{\frac{2}{\nu -2}} \frac{ 2^{\frac{2 \nu}{\nu -2}} \Gamma \left[\frac{4}{2- \nu}\right] t^{\frac{2 \alpha }{2- \nu }} }{\Gamma\left[ 1+\frac{2 \alpha }{2-\nu}  \right](2-\nu )^{\frac{\nu +2}{2-\nu }} \mathcal{D}_0^{\frac{2}{\nu-2}}},
\label{MSDAb}
\end{eqnarray}
where the normal diffusion is recovered for the combination $\alpha + \frac{\nu}{2}=1$.

Hence, from (\ref{MSDAb}) we obtain several types of diffusion

in function of the pair of parameters $(\alpha,\nu)$
\begin{eqnarray}\label{extensionMSD}
\lim_{t\rightarrow +\infty} \langle x^2 \rangle (t) \propto \left \{ \begin{array}{llll}  \mbox{subdiffusion}, & 2\alpha < 2-\nu \\ \mbox{superdiffusion}, & 4-2\nu > 2\alpha > 2-\nu
\\ \mbox{ballistic diffusion} , & \alpha = 2-\nu
\\ \mbox{hyperdiffusion} , & \alpha > 2-\nu.
\end{array} \right.
\end{eqnarray}
Interestingly, for a constant diffusion coefficient ($\nu=0$) we still have a range of
fractional indexes $\alpha$ where the  subdiffusion, superdiffusion, ballistic and hyper diffusion regimes can occur. This marks the degree of generality of the Arangana-Baleanu fractional operator.
Eq. (\ref{extensionMSD}) represents the class of diffusive processes associated to the non-singular Atangana--Baleanu fractional operator, typically applied on the irregular movement of Brownian particles in heterogeneous media.

\section{Conclusion}
\label{sec5}

In this work we have investigated the properties of the solutions of the fractional Fokker-Planck equation with non-singular fractional operators. Our study has been based on two non-singular types of memory kernels: the Caputo-Fabrizio operator and the Atangana-Baleanu one.
In both cases, with the help of the Laplace transform we have obtained the general and analytical solution
of the fractional FPE for an arbitrary diffusion coefficient $\mathcal{D}(x)$.
To illustrate the types of diffusion that can be characterised, we have considered a diffusion coefficient
of the power law type $\mathcal{D}(x)=\mathcal{D}_0|x|^{\nu}$.

The solutions of the Caputo-Fabrizio operator resulted non-Gaussian distributions
that are strongly connected with the index $\alpha$ of the memory kernel,
having a unimodal or bimodal behavior if the diffusion index $\nu$ is positive or negative
respectively (see Fig. \ref{fig1}). However, in the asymptotic limit of long times
$t\rightarrow\infty$ the fractional index $\alpha$ does not affect
the type of diffusion involved.

For the case of the Atangana-Baleanu operator, their solutions
 also resulted non-Gaussian distributions
presenting a unimodal or bimodal behavior according if
the index $\nu$ is positive or negative
respectively (see Fig. \ref{fig2}).
Nevertheless, in the asymptotic limit $t\rightarrow\infty$ the type of diffusion has
a dependence on the pair of indexes $(\alpha,\nu)$ that allows a
characterisation of the normal diffusion on the curve $\alpha+\frac{\nu}{2}=1$ as
 well as the existence of sub-super diffusion and ballistic-hyper diffusion even
 for a constant diffusion coefficient ($\nu=0$).

We consider the results and techniques employed in this work constitute important tools for studying non-Markovian diffusive process with memory effects, thus opening new possibilities in futures researches concerning: diffusion to intermittent movement, reaction-diffusion systems and comb (or grid) structure under influence of new fractional derivative forms.

\section*{Acknowledgements}
M. A. F. dos Santos acknowledges the support of the Brazilian agency CNPq.
I. S. Gomez acknowledges fellowships received from CAPES / INCT-SC (at Universidade
Federal da Bahia).

\section*{References}


TITLE: Is the closure of the span in a Banach space complete?
QUESTION [2 upvotes]: As indicated by the title os the post, I am interested in the following question

Is the closure of the linear span in a Banach space again a Banach space?

If it is true, where I could find the reference, and if it is false, is there a counter example? Thanks!

REPLY [3 votes]: As Theo Bendit nicely stated in the comments, any closed set (and the closure of any set is indeed a closed set) in a complete metric space is complete. Then Theo showed you a nice hint to prove the proposition that I totally endorse. (Note that it is also true that a complete subset of a metric space is closed).
I just wanted to add a remark about the fact that you can use the sequential characterization of closed sets (the definition Theo used) since you are in a Banach space (hence in a metric space). Having your space to be metrizable is a sufficient (though not necessary (!) ) condition to ensure that a closed set (defined as “the complement is open”) is sequentially closed and viceversa.


TITLE: Calculating probabilities of independent events
QUESTION [0 upvotes]: Let $H_1, H_2$ be two independent coin tosses with $P(H_1) = P(H_2) = \frac12.$
Suppose I want to calculate $P(HT).$
Would what I did below make sense?
$P(H_1) = P(H_1)P(H_2|H_1) + P(T)P(H_1|T)  = P(H_1)P(H_2) + P(T)P(H_1) =  \frac12$ and so $P(T)P(H_1)  = \frac14$ since $P(H_1)P(H_2) = \frac14$ because $H_1, H_2$ are independent.
I am just trying to get a feel for doing problems like this.

REPLY [0 votes]: Do you mean to find the probability of obtaining heads in the first trial and tails in the second trial? If so, I would write that as $ P(H_1,T_2) $. Since $ H_1 $ and $ T_2 $ are independent, then:
$ P(H_1,T_2) = P(H_1) \times P(T_2) =  \frac{1}{2}\big(1-\frac{1}{2}\big) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.


TITLE: Are groups with every proper, non-trivial subgroup infinite cyclic simple?
QUESTION [16 upvotes]: In the 1970s Ol'shanskii constructed a non-cyclic finitely generated group $G$ with the following properties:

Every proper, non-trivial subgroup of $G$ is infinite cyclic.
If $X^m=Y^n$ for $X, Y\in G$ with $m,n\neq0$, then $\langle X, Y\rangle$ is cyclic i.e., any two maximal subgroups of $G$ have trivial intersection.

Ol'shanskii gave an easy proof that such a group is simple, which roughly goes: Suppose $N$ is a proper, non-trivial normal subgroup of $G$. If $N$ is maximal then $G/N$ is cyclic of prime order, so $G$ is virtually-$N$, so $G$ is a torsion-free virtually-$\mathbb{Z}$ group, so must itself be cyclic. If $N$ is not maximal then $N$ is contained in a maximal subgroup $M$ such that $M^g\cap M\neq1$ for all $g\in G$, so as $M^g$ is also maximal and as maximal subgroups intersect trivially (by (2)) we have that $M^g=M$ for all $g\in G$, i.e. $M$ is normal in $G$, which is impossible by the previous case.
Property (2) was used here. I was wondering if this can be dropped. So:
Question. Suppose $G$ is a non-cyclic finitely generated group with every proper, non-trivial subgroup of $G$ infinite cyclic. Is $G$ simple?

If $G$ instead satisfies that it is infinite and every proper, non-trivial subgroup has order $p$ for a fixed prime $p$ then $G$ is a "Tarski monster" group and is indeed simple: If $N$ is a proper, non-trivial normal subgroup of $G$ and $g\not\in N$ then $N\cap\langle g\rangle=1$, as both subgroups have prime order, so $N\langle g\rangle=N\rtimes\langle g\rangle$ has order $p^2$, a contradiction. However, this proof uses primality so does not extend to the setting here.

REPLY [8 votes]: The answer is no, there exist non-simple torsion-free Tarski monsters. Theorem 31.4 of Ol’shanskii's book "Geometry of defining relations in groups" ZBL0676.20014 MR1191619 is:
Theorem. There is a non-abelian group all of whose proper subgroups are infinite cyclic and the intersection of any two of them is non-trivial.
Such a group necessarily has infinite cyclic centre: take any non-commuting $x$ and $y$, then the centralizer of $\langle x \rangle \cap \langle y \rangle \cong \mathbb{Z}$ contains $\langle x, y \rangle$, which is the whole group. Thanks to Ashot Minasyan for pointing out Ol’shanskii's theorem.


TITLE: Proving that a matrix is positive semidefinite
QUESTION [3 upvotes]: Let matrices $A, B$ be positive semidefinite. Can we prove that $A(I+BA)^{-1}$ is positive semidefinite?

REPLY [9 votes]: By density, it is enough to prove the property when $A$ is positive definite. Then
$$A(I+BA)^{-1}=A^{1/2}(I+A^{1/2}BA^{1/2})^{-1}A^{1/2}$$
is congruent to $(I+A^{1/2}BA^{1/2})^{-1}$, which itself is positive definite because $I+A^{1/2}BA^{1/2}\succeq I$.