If yesterday were tomorrow, then today would be Friday. (S) If yesterday were tomorrow, then today would be Friday.
Question: What day is today?

This seems to be an old puzzle, and depending on the interpretations, the answers are Wednesday or Sunday (or perhaps Friday as well?). I would like to understand the logic required to analyze and answer the above question.
The following is my attempt in formalizing the analysis. Please let me know if (and where) I err.
Model
Let the actual "today" be $t$, so that "yesterday" in the antecedent of (S) is $t-1$ and "tomorrow" in the antecedent is $t+1$.
Let $D(\tau)$ denote the day of week of date $\tau$.
The subjunctive "today" in the consequent of (S) can be formalize in two ways: as (i) "the yesterday of tomorrow", or (ii) "the tomorrow of yesterday".
The two interpretations thus lead to two ways to translate (S):
$$(t-1)=(t+1) \quad\Rightarrow\quad D(t+1)-1=\text{Friday}\tag{1}$$
$$(t-1)=(t+1) \quad\Rightarrow\quad D(t-1)+1=\text{Friday}\tag{2}$$
From $(1)$, we have $D(t-1)=\text{Saturday}\Rightarrow D(t)=\text{Sunday}$.
From $(2)$, we have $D(t+1)=\text{Thursday}\Rightarrow D(t)=\text{Wednesday}$.
But both interpretations also seem to suggest Friday as a solution, which is implausible (?) and indicates that the model I've proposed has flaws.
What's wrong, and how can we improve the model?

Yesterday was not tomorrow.  From a false assumption, any conclusion is possible.

But one interpretation of the puzzle goes like this. The only day of the week $x$ for which is would be correct to say "If yesterday was $x$, then today would be Friday"
is Thursday.  So on that interpretation, $x = $Thursday, which actually happens to be tomorrow, so today is Wednesday.

Alternatively, "If $y$ was tomorrow, then today would be Friday" is true if $y$ is Saturday.  On that interpretation, Saturday is actually yesterday instead of tomorrow, and today is Sunday.

As mentioned by others, yesterday is never tomorrow, assuming a linear timeline. Thus, it has no bearing on the question.

In other words, the answer is the same as "Zip-A-Dee-Doo-Dah. What day is today?"

According to my phone, today is February 3rd. In general, the answer is "[insert today's date here]".


Expectation of geometric mean reversion process The second part to a question I asked here in which I had to show that the solution to

$dX_t = \kappa\left(\alpha-\ln X_t \right)X_t dt + \sigma X_t dB_t$

was

$ X_t = \exp \left( \mathbb{e}^{-k t}\ln x+(a-{\sigma^2\over2k})(1-\mathbb{e}^{k t})+\sigma \mathbb{e}^{-k t}\int_0^t \mathbb{e}^{k s}\mathbb{d}B_s\right), \space X_0 = x > 0$

Part two asks to show that:

$\mathbb{E}\left[X_t\right] = \exp\left( \mathbb{e}^{-\kappa t} \ln x + (\alpha - {\sigma^2 \over 2 \kappa})(1-\mathbb{e}^{-\kappa t}) + {\sigma^2(1-\mathbb{e}^{-2 \kappa t})\over 2 \kappa}\right)$

I thought that I could take the expectation of the first 2 terms in the exponential, the final term would be zero since:

$\mathbb{E}\left[ \int_S^T f(t,\omega) dB_t \right] = 0$

If $f(t,\omega)$ is $\mathscr{B}\times\mathscr{F}-$measurable, $\mathscr{F}_t$-adapted and  $\mathbb{E}\left[\int_S^T f^2dt\right]<\infty.$

How do you obtain the last term in the exponent?

Edit from comments:

Need to compute $\mathbb{E}\left[\exp(\sigma\mathbb{e}^{-kt} \int_0^t \mathbb{e}^{ks} dB_s)\right]$

The mean of the integral in the exponent is 0 and the variance:

$\mathbb{Var} \left[\int_0^t \mathbb{e}^{ks} dB_s \right] = \mathbb{E}\left[(\int_0^t \mathscr{e}^{ks} dB_s)^2\right] = \frac{1}{2\kappa}(\mathbb{e}^{2 k t}-1)$

...by Ito-isometry and computing a trivial intergral.

Then since the co-efficient is $\sigma \mathbb{e}^{-kt}$, to get the variance of the non deterministic term in the exponent multiply the variance by the square since it is a normal i.e.:

$\mathbb{Var}\left[ \sigma \mathbb{e}^{-k t}\int_0^t \mathbb{e}^{k s}\mathbb{d}B_s \right]=\frac{\sigma^2}{2\kappa}(1-\mathbb{e}^{-2\kappa t})$

$\mathbb{E}\left[ \exp(Z) \right]=\mathbb{e}^{\mu+{\sigma'}^2/2}, Z \sim \mathscr{N}(0, \frac{\sigma^2}{2\kappa}(1-\mathbb{e}^{-2\kappa t}))$

$\mathbb{E}\left[ \exp\left( \sigma \mathbb{e}^{-kt} \int_0^t \mathbb{e}^{ks} dB_s\right)\right]= \exp(\frac{\sigma^2}{4k}(1-\mathbb{e}^{-2kt}))$

This is out by a factor of 1/2.

Take a function, $f(t, X(t)) = \exp( t)\log X(t)$ then apply Ito's lemma and solve where $f(t) =  \exp{ t} \log X(t) ; f(X(t)) = \exp( t)/ X(t)$
and $f(X^2(t)) = -\exp( t)/(2 * X(t)^2))$


Normal Ring and Prime Ideal whose Square is Principal Let $k$ be a field of characteristic $\neq2$ and consider $R=k[x,y]/(y^2-x^3+x)$. Then

(a) Show that $R$ is normal.

(b) Let $P=(x,y)$ be a prime ideal of $R$. Show that $P^2$ is a principal ideal.

(c) Show that $P$ is not a principal ideal.

I came across this in some commutative algebra reading recently. Here are my thoughts so far (really only for part (c)). Assume that $P$ is generated by $\alpha=f(x)+g(x)y$. Then if $\sigma$ is the nontrivial element in the Galois group, $\sigma(\alpha)$ will generate $\sigma(P)=P$, so the norm of $\alpha$ will generate $P^2$. I thought to compare this supposed generator with the one found in part (b) and then arrive at a contradiction.

Any help with this problem would be appreciated!

This answer continues the OP's thoughts for part (c).

(c) Let $\sigma:k[X,Y]\to k[X,Y]$ be the $k$-automorphism given by $\sigma(X)=X$, $\sigma(Y)=-Y$ (which is probably the automorphism mentioned by the OP in his attempt). Since $\sigma(Y^2-X^3+X)=Y^2-X^3+X$, we can "extend" $\sigma$ to a $k$-automorphism of $R$. Obviously, $\sigma(P)=P$.

Now note that the elements of $R$ can be uniquely written as $u(x)+v(x)y$ and such an element is in $P$ if and only if $ x\mid u(x)$.

If $P=(xa(x)+b(x)y)$, then $P=\sigma(P)=(\sigma(xa(x)+b(x)y))=(xa(x)-b(x)y)$. It follows $(x)=P^2=(x^2a^2(x)-b^2(x)y^2)=(x^2a^2(x)-b^2(x)(x^3-x))$ and therefore there are $u,v\in k[X]$ and $w\in k[X,Y]$ such that $$X=[u(X)+v(X)Y][X^2a^2(X)-b^2(X)(X^3-X)]+w(X,Y)(Y^2-X^3+X).$$ By taking the degree in $Y$ of both sides we get $w=0$. In the end we have $$X^2a^2(X)-b^2(X)(X^3-X)=1 \text{ or } X,$$ and both cases are clearly impossible.

a) Regular implies normal.

Okay, maybe that's too high-tech.  If $f=g+hy$ lies in the integral closure, then it satisfies the quadratic polynomial $(z-g)^2 - h^2(x^3-x)$.  If the coefficients are polynomials (which they must be, by Gauss's Lemma), then $g$ is a polynomial, and $h^2 (x^3-x)$ is a polynomial.  But $x^3-x$ is square-free (here we use that the characteristic is not $2$), so $h$ is a polynomial and $f\in R$.

b) Since $y^2,x^3 \in P^2$, it is straightforward that $x\in P^2$ and $P^2 = (x)$.

c) (Note: it is sufficient to address this problem for $R\otimes_k \overline{k}$, so we assume that $k$ is algebraically closed to avoid some technicalities.)

First of all, $X=\operatorname{Spec}k[x,y]/(y^2-x^3+x)$ is an open subset of an elliptic curve $\overline{X} = Y\cup \{\infty\}$.  This is a rather important observation, because it implies that $R=k[x,y]/(y^2-x^3+x)$ doesn't have any principal prime ideals whatsoever (except, of course, for $(0)$).

We will demonstate this, real mathematician-style, by looking at the global behavior of a hypothetical generator of $P$.  If $P=(f)$ is principal, then $f$ extends to a rational function on $\overline{X}$ with a simple pole at $\infty$.  Therefore $1/f$ is a non-constant rational function on $\overline{X}$ with only a simple pole at $P$.  This implies that $l(P) \geq 2$.

Here $l(P)$ is shorthand for the dimension of the space of rational functions on $\overline{X}$ which have at worst a simple pole at $P$, and are regular elsewhere.  In general, $l(D)$ is defined for $D$ a Weil divisor, i.e. a formal sum of points.  For example, $l(P+2Q-3R)$ counts the number of linearly independent functions which must have a triple zero at $R$, and are allowed to have a pole of order $1$ at $P$, and $2$ at $Q$.  The sum of the coefficients of $D$ is called its degree.

We will calculate $l(P)$ exactly using the Riemann-Roch theorem, which says that, for any Weil divisor $D$, $l(D)-l(K-D) = \operatorname{deg}{D} + 1 - g$, where $g$ is the genus of $\overline{X}$, and $K$ is the divisor associated to any differential form on $\overline{X}$.  For an elliptic curve, we must have $K=0$ and $g=1$, so this equation takes the simple form $l(D)-l(-D) = \operatorname{deg}{D}$.

We set $D=P$.  Since $\operatorname{deg}{P} = 1$, and $l(-P)=0$ (in fact, $l(-D)=0$ for any divisor of positive degree), we conclude that $l(P) = 1$.  Since earlier we found that $l(P)\geq 2$ for any principal maximal ideal on $X$ (in fact, the proof works for any projective curve with a single point removed), we have our contradiction.

(The above does have a nice, concrete interpretation: when you strip away the machinery, the argument is that, if $P=(f)$, then the map $k[x]\to R$ sending $x$ to $f$ is an isomorphism.  We use basic facts about divisors to avoid some difficult calculations, and genus to show that $R$ cannot be a polynomial ring.)


For $p$ prime, every prime divisor of $p!+1$ is greater than $p$ I am not sure how to start this queston. Could you give me some hint?

Show that for any prime number $p$, every prime divisor of $p!+1$ is greater than $p$.

Thanks

Notice that all primes less than or equal to $p$ divide $p!$. Therefore, they cannot also divide $p!+1$. Therefore, $p!+1$ can be represented as a product of primes which are all greater than $p$. This is actually one of the nice proofs showing that there are infinitely many primes.


Prove $x^2-y^2<z^2$ is open I'm trying to prove $x^2-y^2<z^2 $ is open. I believe this is a cone shape, can I simply assert that $∀ x ∈ P$ there exists $r > 0$ such that $B_r(x) ⊆ P$, where $P$ is my subset? Or is there an easier way to explicitly prove why it is open?

Let $(x_0,y_0,z_0)$ be a point of your set. Because the inequality is strict, you have some margin around the first two coordinates: if you add, say, $\epsilon$, to the first two coordinates the point you get is still in the set. This shows a small (2D) circle around the point is still in the set .
Then, for each of the points in the set, if you add the same $\epsilon$ (or even more!) to $z_0$ you get a looser inequality and thus you get another point of your set.
In this way we get a whole (3D) box around our point. Inside this box we can inscribe a ball, thus the openness.

HINT

Try to find an obviously open set that is mapped to your set by a continuous function. Then it will be open by definition of a continuous function, since it must map open sets to open sets.


Existence of two independent random variables on a finite sample space. The sample space $\Omega$ contains five elements with positive probability. Are there two independent random variables on $\Omega$, each of which takes five different values?

My solution is the following. Suppose there are random variables $X$ and $Y$, each of which takes five different values. Let $\omega_0$ be one of the outcomes and $X(\omega_0) = x_0, Y(\omega_0) = y_0 $ and $\forall i X(\omega_i) \neq x_0, Y(\omega_i) \neq y_0 $. Then $P(x = x_0, y = y_0) = p $ and $P(x = x_0) \cdot P(y= y_0) = p^2 $. Therefore the independence criteria is not satisfied and the answer to question is no.

Is that a correct solution?

It suffices to assume that there exists $\omega\in \Omega$ with $0<\mathsf{P}(\{\omega\})<1$, where $\mathsf{P}$ is a probability measure on $(\Omega,2^{\Omega})$ because for any injective functions $X,Y:\Omega\to\mathbb{R}$ (random variables taking distinct values for each $\omega$), $\sigma(X)=\sigma(Y)=2^{\Omega}$.

I think, as I wrote in the comment, that the proposed solution from the OP, after the changes he made, is correct. Here I propose my approach, which is based on conditional probabilities,  that maybe shows a different point of view. I try to be a bit heavy with the notation to make the argument clearer.

We can consider $\Omega=\{1,2,3,4,5\}$, with probabilities $p_1,..p_5$, as sample space.

Let $X$,$Y$ two generic random variables, so that $X(i)=x_i$ and $Y(i)=y_i$. We observe that we can define $Z=(X,Y)$ so that:

$Z(i)=(x_i,y_i)$

It is clear than that:

$p(\{X=x_i\} \cap \{Y=y_k\})=p(Z=(x_i,y_k))=\delta_{i,k}p_i$ [1],

because if $i$ is not equal to $k$, $(x_i,y_k)$ does not belong to the codomain of $Z$. If $i$ is equal $k$ than the event $\{X=x_i\}$ happens only for $\omega \in \Omega=i$, for which $Y$ will be automatically equal to $y_i$. Thus the probability in [1] is simply $p_i$.

Using [1], $p(X=x_i|Y=y_k)=\frac{p(\{X=x_i\} \cap \{Y=y_k\})}{p(Y=y_k)}=\delta_{i,k}$ [2]

, where $\delta$ is the Kronecker symbol. If the two variables were independent [1] could be factored and we would obtain $p(X=x_i|Y=y_k)=p_i$.

Therefore the two variables cannot be independent. In fact given the value of $Y$ one can reconstruct in a unique way the value of X and viceversa (this is the interpretation of [2]: when conditioned on the value of $Y$, $X$ becomes deterministic). Told another way the value of X is highly "correlated" with the value of Y.


Bayes rule for continuous variables Student S has a flu 20% the time. Suppose we can only detect whether he has
flu or not by his body temperature, which is a continuous random variable.
Without flu, the temperature of a person is N ~ (98.6, 0.4)
With flu, the body temperature of a person is N ~ (101,2)
If S’s body temperature today is 100, what is the probability that he has the flu?

I have tried to solve it this way:
$Z_Nflu$ = $ {(X-98.6)/ 0.6}$

$Z_flu$ = $(X-101/ 1.4)$

P($Z_flu$  = 100) = (-0.71) = 1- 0.76 = 0.24

P(flu | temp) = (0.24* 0.2) /(0.2 * 0.24  + 0.8 *0.99)

Is this correct?

Should not it be P(flu | temp) = (0.99* 0.2) /(0.2 * 0.24 + 0.8 *0.99), since we need to find the probability of temperature 100 belonging to with flu temperature data set ?


Expectation inequality Let $X, Y$ be random variables with $0 \leq X \leq Y$ and $\mathbb E[Y]=1$. Let $t>0$. Does the inequality

$$
\mathbb E[e^{tX}] \leq x\, \mathbb E[e^{tY}]+1-x
$$

where $x=\mathbb E[X]$ hold?

This inequality does not hold in general. For example, take $X$ and $Y$ such that:

with probability $\frac{2}{3}$, $X = 0$ and $Y = \dfrac{1}{2}$,
with probability $\frac{1}{3}$, $X = 2$ and $Y = 2$.

In this case, the inequality would write:
$$
\frac{2}{3}+\frac{e^{2t}}{3} \leq \frac{2}{3}\left(\frac{2e^{t/2}}{3} + \frac{e^{2t}}{3}\right) + 1 - \frac{2}{3}
$$
As $t$ tend to $+\infty$, this would lead to $1 \leq \frac{2}{3}$, wich is of course absurd.

However, it can be made true under certain hypotheses: namely, if we assume that $X = ZY$ (hence $0 \leq Z \leq 1$) where $Z$ is independent of $Y$.

The convexity of the function $\exp$ then gives:
$$
e^{tX} = e^{t(Z\times Y + (1-Z)\times 0)} \leq Z e^{tY} + (1-Z)e^0
$$
for every $t > 0$. Taking the expectations, the independence gives
$$
E(e^{tX}) \leq x E(e^{tY}) + 1 - x
$$
since $x = E(YZ)=E(Z)\times 1$.

By series expansion $e^{tX} = 1 + tX + \frac{t^2X^2}{2!} + \frac{t^3X^3}{3!} + \cdots +\frac{t^nX^n}{n!} + \cdots.$

$M_X(t) = E(e^{tX}) = 1 + tm_1 + \frac{t^2m_2}{2!} + \frac{t^3m_3}{3!}+\cdots + \frac{t^nm_n}{n!}+\cdots$

Thus, $E(e^{tX}) > 1+tE[X]$ and hence $E(e^{tX})>1+tx$ since $E[X]=x$.

Similarly, $E(e^{tY})>1+t$

Now,
\begin{eqnarray}
 xE(e^{tY})+1-x &>& x(1+t)+1-x      \nonumber \\
   &>& x+tx+1-x \nonumber \\
   &>&1+tx \nonumber \\
   &>& E(e^{tX})
\end{eqnarray}

Since $X\leq Y$, thus, $E(e^{tX})\leq E(e^{tY})$ and hence the inequality follows.


The definition of Determinant in the spirit of algebra and geometry The concept of determinant is quite unmotivational topic to introduce. Textbooks use such "strung out" introductions like axiomatic definition, Laplace expansion, Leibniz'a permutation formula or something like signed volume.

Question: is the following a possible way to introduce the determinant?

Determinant is all about determing whether a given set of vectors are linearly independent, and a direct way to check this is to add scalar multiplications of column vectors to get the diagonal form:

$$\begin{pmatrix}
a_{11} & a_{12} & a_{13} & a_{14} \\
a_{21} & a_{22}  & a_{23} & a_{24} \\
a_{31} & a_{32}  & a_{33} & a_{34} \\
a_{41} & a_{42}  & a_{43} & a_{44} \\
\end{pmatrix} \thicksim \begin{pmatrix}
d_1 & 0 & 0 & 0 \\
0 & d_2  & 0 & 0 \\
0 & 0  & d_3 & 0 \\
0 & 0  & 0 & d_4 \\
\end{pmatrix}.$$

During the diagonalization process we demand that the information, i.e. the determinant, remains unchanged. Now it's clear that the vectors are linearly independent if every $d_i$ is nonzero, i.e. $\prod_{i=1}^n d_i\neq0$. It may also be the case that two columns are equal and there is no diagonal form, so we must add a condition that annihilates the determinant (this is consistent with $\prod_{i=1}^n d_i=0$), since column vectors can't be linearly independent.

If we want to have a real valued function that provides this information, then we simply introduce an ad hoc function $\det:\mathbb{R}^{n \times n} \rightarrow \mathbb{R}$ with following properties:

$$\det (a_1,\ldots,a_i,\ldots,a_j,\ldots,a_n)=\det (a_1,\ldots,a_i,\ldots,k\cdot a_i+a_j,\ldots,a_n).$$
$$\det(d_1\cdot e_1,\ldots,d_n\cdot e_n)=\prod_{i=1}^n d_i.$$
$$\det (a_1,\ldots,a_i,\ldots,a_j,\ldots,a_n)=0, \space \space \text{if} \space \space a_i=a_j.$$

From the previous definition of determinant we can infer the multilinearity property:

$$[a_1,\ldots,c_1 \cdot u+c_2 \cdot v,\ldots,a_n]\thicksim diag[d_1,\ldots,c_1 \cdot d'_i+c_2 \cdot d''_i ,\ldots,d_n],$$ so $$\det[a_1,\ldots,c_1 \cdot u+c_2 \cdot v,\ldots,a_n]=\prod_{j=1:j\neq i}^n d_j(c_1 \cdot d'_i+c_2 \cdot d''_i)$$ $$=c_1\det(diag[d_1,\ldots, d'_i,\ldots,d_n])+c_2\det(diag[d_1,\ldots, d''_i,\ldots,d_n])$$ $$=c_1\det[a_1,\ldots,u,\ldots,a_n]+c_2\det[a_1,\ldots, v,\ldots,a_n].$$

Note that previous multilinearity together with property $(1)$ gives the property $(2)$, so we know from the literature that the determinant function $\det:\mathbb{R}^{n \times n} \rightarrow \mathbb{R}$ actually exists and it is unique.

Obviously, the determinant offers information how orthogonal a set of vectors is. Thus, with Gram-Schmidt process we can form an orthogonal set of vectors form set $(a_1,\ldots, a_n)$, and by multilinearity and property $(2)$ the absolute value of determinant is the volume of parallelepiped spanned by the set of vectors.

Definition.
Volume of parallelepiped formed by set of vectors $(a_1,\ldots, a_n)$ is $Vol(a_1,\ldots, a_n)=Vol(a_1,\ldots, a_{n-1})\cdot |a_{n}^{\bot}|=|a_{1}^{\bot}|\cdots |a_{n}^{\bot}|$, where $a_{i}^{\bot} \bot span(a_1,\ldots, a_{i-1}).$

This approach to determinant works equally well if we begin with the volume of a parallelepiped (geometric approach) or with the search of invertibility (algebraic approach). I was motivated by the book Linear algebra and its applications by Lax on chapter 5:

Rather than start with a formula for the determinant, we shall deduce it from the properties forced on it by the geometric properties of signed volume. This approach to determinants is due to E. Artin.

$\det (a_1,\ldots,a_n)=0$, if $a_i=a_j$, $i\neq j.$
$\det (a_1,\ldots,a_n)$ is a multilinear function of its arguments, in the sense that if all $a_i, i \neq j$ are fixed, $\det$ is a linear function of the remaining argument $a_j.$
$\det(e_1,\ldots,e_n)=1.$

That seems quite opaque: It's a way of computing a quantity rather than telling what exactly it is or even motivating it. It also leaves completely open the question of why such a function exists and is well-defined. The properties you give are sufficient if you're trying to put a matrix in upper-triangular form, but what about other computations? It also gives no justification for one of the most important properties of the determinant, that $\det(ab) = \det a \det b$.

I think the best way to define the determinant is to introduce the wedge product $\Lambda^* V$ of a finite-dimensional space $V$. Given that, any map $f:V \to V$ induces a map $\bar{f}:\Lambda^n V \to \Lambda^n V$, where $n = \dim V$. But $\Lambda^n V$ is a $1$-dimensional space, so $\bar{f}$ is just multiplication by a scalar (independent of a choice of basis); that scalar is by definition exactly $\det f$. Then, for example, we get the condition that $\det f\not = 0$ iff $f$ is an isomorphism for free: For a basis $v_1, \dots, v_n$ of $V$, we have $\det f\not = 0$ iff $f(v_1\wedge \cdots \wedge v_n) = f(v_1) \wedge \cdots \wedge f(v_n) \not = 0$; that is, iff the $f(v_i)$ are linearly independent. Furthermore, since $h = fg$ has $\bar{h} = \bar{f}\bar{g}$, we have $\det(fg) = \det f \det g$. The other properties follow similarly. It requires a bit more sophistication than is usually assumed in a linear algebra class, but it's the first construction of $\det$ I've seen that's motivated and transparently explains what's otherwise a list of arbitrary properties.

A rationale i could give to the determinant function is to generalise the property of the power, i.e we know that for any scalars $x,a,b,c$ it holds:
$x^{a + b + c \ldots} = x^a x^b x^c\ldots$

But, lets assume that $a,b,c\ldots$ are actually the diagonal entries of a matrix $\mathbf{A}$ then it holds that $x^{trace(\mathbf{A})} = det(x^\mathbf{A})$ , with $x^A$ the matrix exponential (of base $x$);


Homotopy group of the conformal group I would like to know which are the first three homotopy groups of the conformal group SO(4,2):
$$
\pi_n(SO(4,2))=? \quad n=1,2,3
$$

According to ncatlab, the maximal compact subgroup of (the connected component of) $SO(4,2)$ is $SO(4)\times SO(2)$. Any connected Lie group retracts onto its maximal compact subgroup, so your question is about $SO(4)\times SO(2)$.

Since $SO(2)$ is just a circle, we have $\pi_n(SO(2))=\mathbb{Z},0,0,\ldots$.

$SO(4)$ is a semi-simple group, $SO(4)\simeq (SU(2)\times SU(2))/\mathbb{Z_2}$, so it is easy to compute the first homotopy groups.

Indeed because the second homotopy of any simple group is trivial, and the third is $\mathbb Z$, we have $\pi_2(SO(4))=0$ and $\pi_3(SO(4))=\mathbb{Z}^2$ (for $n>4$, $\pi_3(SO(n))=\mathbb{Z}$, since then $SO(n)$ is actually simple).

The fundamental group of $SO(n)$, $n\geq 3$ is $\mathbb{Z}_2$, as can be seen in this specific case from the isomorphism above.

Thus we find finally (for the connected component)
$$
\pi_n(SO(4,2))=\mathbb{Z}_2\times \mathbb{Z},0,\mathbb{Z}^2,\mathbb{Z}_2^2,\mathbb{Z}_2^2,\mathbb{Z}_{12}^2,\ldots,
$$
or more generally for $n>1$
$$
\pi_n(SO(4,2))=\pi_n(SU(2))\times \pi_n(SU(2))=\pi_n(S^3)\times \pi_n(S^3),
$$
where the homotopy groups of spheres can be found on Wikipedia. Note that we do not know all the homotopy groups of $S^3$ in full generality, as noted by Mariano in the comments.

Edit: fixed the 3rd homotopy group and added the general relation to $S^3$.

The group $SO(n) \subset SO(n+1)$ by an $n-1$-connected map. Consequently for $k < n-1$ $\pi_k(O(n+1)) = \pi_k(O(n))$. I am euclideanizing $SO(4,2) \rightarrow SO(6)$, and not considering for the time the hyperbolic aspects. So all we have to consider is the fundamental group $\pi_1(SO(4,2))$ The Serre fibration
$$
SO(n) \rightarrow SO(n+1) \rightarrow SO(n+1)/SO(n) \sim S^n
$$
gives the sequence of homotopies
$$
\pi_k(SO(n)) \rightarrow \pi_k(SO(n+1)) \rightarrow \pi_k(SO(n+1)/SO(n))
$$
has $\pi_k(S^n) = 0$ this demonstrates the equality. I will now state that it is known that the fundamental group of Lie algebras are abelian.

To continue this, sorry I had to post due to interruption, I now appeal to Bott periodicity. I now use the fact from Bott periodicity theorem that $\pi_k(Sp) = \pi_{k+4}(O)$. Now we can focus in on $\pi_1(sp(2))$ and the knowledge that $sp(2) \sim U(1)$. The homotopy is abelian, which means it is equal to its homology group, which for the circle is $\mathbb Z$

As for not going hyperbolic, it is the case with physics problems that one looks at the Euclidean case first.


For some integers a and b, the same integer satisfies both of the equations, what is the greatest possible value? For some integers $a$ and $b$, the same integer satisfies both of the equations $$x^2 - ax + 2014 = 0$$ and $$x^2 - bx + 2015 = 0.$$ What is the greatest possible value of this common solution?

Let $x_1$ and $x_2$ be roots of the first equation and $x_1$ and $x_3$ be roots of the second.

Thus, $x_1x_2=2014$ and $x_1x_3=2015$, which gives $2014$ and $2015$ divided by $x_1$,

which gives $x_1=1$ or $x_1=-1$, which gives the answer: $1$.

Both solutions are integers (this is the big hint), so let factoring them out
For x^2 - ax + 2014,there are only two possible factors
(x±1)(x±2014) or (x±2)(x±1007)
the solutions will be ±1 & ±2014, or ±2 & ±1007

For x^2 - bx + 2015,
(x±1)(x±2015) or (x±5)(x±403) the solutions are ±1 & ±2015 or ±5 & ±403

So, the greatest possible common solution is 1


liminf of sequence of iid random variables if $P(\liminf (X_n>a))=0$, does that mean $\liminf X_n <a$ almost surely? Where $X_n$ are iid random variables.

Important inequalities

Williams - Probability with Martingales

Deduced similarly:

(iii) If $\liminf x_n > z$, then

$ \ \ \ \ \ \ \ (x_n > z)$ eventually (that is, for infinitely many n)

(iv) If $\liminf x_n < z $, then

$ \ \ \ \ \ \ \ (x_n < z)$ infinitely often (that is, for infinitely many n)

  if $P(\liminf (X_n>a))=0$, does that mean $\liminf X_n <a$ almost surely?

Note that $P(\liminf (X_n>a))=0 \iff P(\limsup (X_n \le a))=1$

Now it is not true that

$$P(\limsup (X_n \le a))=1 \to P(\liminf X_n <a) = 1$$

Consider $X_n = a - 1/n$.

However

$$P(\limsup (X_n \le a))=1 \to P(\liminf X_n \le a) = 1$$

The contrapositive is

$$P(\liminf X_n > a) = 1 \to P(\liminf (X_n > a))=1$$

This can proven by proving

$$\liminf X_n > a \subseteq \liminf (X_n > a)$$

Pf: Suppose

$$\sup_{m \ge 1} \inf_{n \ge m} X_n := \beta > a$$

Then $\forall \epsilon > 0, \exists m \ge 1$ s.t.

$$\inf_{n \ge m} X_n := \alpha_m \in (\beta - \epsilon, \beta]$$

Choose $\epsilon > 0$ s.t.

$$\inf_{n \ge m} X_n := \alpha_m > a$$

$$\to \inf (X_m, X_{m+1}, ...) := \alpha_m > a$$

$$\to X_n \ge \alpha_m > a \ \forall n \ge m \ QED$$


Dimension of the space of symmetric traceless matrices A set of symmetric $n \times n$ matrices have $ \frac{1}{2}n^{2} - \frac{1}{2}n$ independent elements.  But how do you get to this result?  I understand that a general $n \times n$ matrix would have $n^{2}$  independent elements.  How exactly does the constraint of symmetry lead to the piece above?  Furthermore, how would I determine the number of elements for a symmetric traceless matrix?

An $n\times n$ matrix, $\mathbf T$, has, as you said, $n^2$ independent components, namely $[\mathbf{T}]_{ij}$, where each index can run from $1$ to $n$.
Symmetry, in this context, means that the matrix is equal to its transpose, $\mathbf{T}=\mathbf{T}^T$. What does this mean about the components? Let's see...

$[\mathbf{T}]_{ij}=[\mathbf{T}^T]_{ij}\\
   \hspace{0.85cm}=[\mathbf{T}]_{ji}$

That was simple enough! Symmetry implies that the $(i,j)$ component of the matrix equals the $(j,i)$ component of the matrix. What does this look like visually?

Neat - when you write out a symmetric matrix you'll notice that almost every component has a twin on the other side of the matrix. A helpful way to visualize a transpose is to flip the matrix across its diagonal and realize that symmetry means doing so does not change the matrix whatsoever. Long story short, when you get to counting independent components, be careful to not count both twins as if they were independent - only count pairs of twins.

Now, which elements DON'T have a twin? If you stare at the matrix, you'll soon figure out that it's the elements on the diagonal. That's 'cause $[\mathbf{T}]_{ii}=[\mathbf{T}]_{ii}$ trivially - symmetry did not impose any additional pairing up for these elements. Going back to our visual representation, if you flip a matrix across its diagonal, the diagonal elements stay put.
Phew, that was a lot, but now we're ready to count the number of independent components of a symmetric matrix. First the diagonal - there are $n$ independent components there. Now the pairs of twins. Removing the diagonal elements, we now have $n²-n$ components. If we twin them up, then we'll have $\frac{n²-n}{2}$ independent pairs. Then we just have to add the independent diagonal components back to get the total number of independent components:

$\frac{n²-n}{2}+n=\frac{n²-n+2n}{2}=\frac{n(n+1)}{2}$

Voila!
If you didn't like how we counted, then we can totally count them another way. Let's look at each row of the matrix and count how many independent components there are. Remember we're only counting the diagonal components and those on one side of the diagonal. The first row has $n$ independent components. The second row has $n-1$ independent components. We continue in this fashion till we get to the $n^{th}$ row which has but one independent component. Now we just have to add all these up:

$1+2+3+\cdots+(n-1)+n = \frac{n(n+1)}{2}$

If you're interested in how this sum was done, check out the really cool triangle numbers.

Now, about those traceless matrices. The trace is equal to the sum of all the diagonal components of the matrix. For a matrix to be traceless means that its trace is equal to zero. But that means that you can rewrite one of the diagonal components in terms of the others, so we have lost an independent component! Hence the number of independent components of a traceless symmetric matrix will be one less than the number of independent components of a symmetric matrix, namely

$\frac{n(n+1)}{2}-1=\frac{n^2+n-2}{2}=\frac{(n+2)(n-1)}{2}$

Hope that helps!

I'll$^1$ take an example from $SO(N)$:

Consider $T^{ij}$ containing in total $N\times N = N^2$ indep. objects$^2$. But one can show that the symmetric, antisymmetric and the trace of $T^{ij}$ transform within themselves i.e. they don't mix with each other:
$$\tag{def. trans. rule for each index}T^{ij}\rightarrow T^{'ij} = O^{il}O^{jm}T^{lm}$$
$$\tag{the symmetric part}S^{ij}\rightarrow S^{'ij} = O^{il}O^{jm}S^{lm}$$
$$\tag{the antisymmetric part}A^{ij}\rightarrow A^{'ij} = O^{il}O^{jm}A^{lm}$$
where $S^{ij} = 1/2(T^{ij}+T^{ji})$ and $A^{ij} = 1/2(T^{ij}-T^{ji})$.

Finally the trace $T:=\delta^{ij}T^{lm}$ transforms as
$$T\rightarrow T' = \delta^{ij}T^{'ij} = \delta^{ij}O^{il}O^{jm}T^{lm} = (O^T)^{li}\delta^{ij}O^{jm}T^{lm} = \delta^{lm}T^{lm} = T$$ where we used $O^TO=\mathbf{1}$. Thus we can define the symmetric traceless tensor from $T^{ij}$
$$Q^{ij} := S^{ij}-\frac{1}{N}\delta^{ij}T$$ which contains $\frac{1}{2}N(N+1)-1$ objects.

In other words, one can split up the $N^2$ objects as

$$N\otimes N = [\frac{1}{2}N(N+1)-1]\oplus 1\oplus \frac{1}{2}N(N-1), $$
where the $1$ comes from the trace.

To count the objects in the symmetric and asymmetric tensors just use (draw an $N$ by $N$ matrix and count the entries including the diagonal for the symmetric one, but not for the asymmetric one):

$$\tag{for the symmetric part}\sum_{j=1}^{N} j = \frac{1}{2}N(N+1). $$
While for the asymmetric part we get
$$\tag{for the antisymmetric part}\sum_{j=1}^{N} j - N= \frac{1}{2}N(N-1). $$
where the $N$ comes from the $N$ zeros on the diagonal.

See also this PSE entry.

$^1$Disclaimer: I'm not a mathematician so the below might contain errors and wrong terminology. Feel free to teach/correct me.

$^2$With the number of objects contained in a tensor we mean the dimension of the representation.

$^3$Much of this is taken from "QFT in a nutshell" by A. Zee.


Prove that equation $x^5-5x^3+4x-1=0$ has exactly 5 roots Prove that equation $x^5-5x^3+4x-1=0$ has exactly 5 root.

By using intermediate value theorem, I can show that $x^5-5x^3+4x-1=0$ has at least one root in each following intervals: $(-2,-1.5),\ (-1.5,-1),\ (-1,0.5),\ (0.5,1),\ (1,3)$. So , it has exactly 5 roots.
But I wonder that there is some logical ways we can find intervals that contain roots without guessing?

Given $f(x)=x^5-5x^3+4x-1$, differentiate first:
$$ f'(x) = 5x^4 - 15x^2 +4 . $$
Setting the derivative equal to $0$, we have
$$ 0 = 5x^4 - 15x^2 +4 . $$
This is quadratic in $x^2$, so we can use the quadratic formula:
$$ x^2 = \frac{15 \pm \sqrt{145}}{10} . $$
Since $\sqrt{145}<15$, $x^2>0$, so our solutions are
$$ x = \pm \sqrt{\frac{15 \pm \sqrt{145}}{10}} . $$
So, we have four distinct local extrema for $f$. If there are five distinct roots, each must be on one of the following intervals:
$$ \left(-\infty, -\sqrt{\frac{15+\sqrt{145}}{10}}\right) , $$
$$ \left(-\sqrt{\frac{15+\sqrt{145}}{10}}, -\sqrt{\frac{15-\sqrt{145}}{10}}\right) , $$
$$ \left(-\sqrt{\frac{15-\sqrt{145}}{10}}, \sqrt{\frac{15-\sqrt{145}}{10}}\right) , $$
$$ \left(\sqrt{\frac{15-\sqrt{145}}{10}}, \sqrt{\frac{15+\sqrt{145}}{10}}\right) , $$
$$ \left(\sqrt{\frac{15+\sqrt{145}}{10}}, \infty\right) . $$

I would start with the rational root theorem, which fails but gets you values at $x=\pm1$.
Because the coefficients are small, the roots are all close to $0$.  Checking values at the integers close to $0$ is a good approach.  Once you go from $-3$ to $+3$ you have found one root in $(2,3)$.  You should be able to convince yourself there are not roots outside $(-3,3)$ because the $x^5$ term dominates.  Now checking at half integers is a good next step and finds the roots.  There was no guessing involved.  The real question comes if you don't find the roots with the half integral evaluations.  Do you trust the problem setter to guarantee there are five roots?  If so, trying quarters is the next logical step.  Or do you worry the question is wrong?  I have no advice here.


If $2(bc^2+ca^2+ab^2) = b^2c+c^2a+a^2b+3abc, $ then prove that triangle $ABC$ is equilateral In a triangle $ABC,$ if $2(bc^2+ca^2+ab^2) = b^2c+c^2a+a^2b+3abc, $ then prove that triangle

$ABC$ is equilateral

$\displaystyle \frac{2(bc^2+ca^2+ab^2)}{abc} = \frac{b^2c+c^2a+a^2b+3abc}{abc}$

$\displaystyle 2\left(\frac{c}{a}+\frac{a}{b}+\frac{b}{c}\right) = \left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)+3$

let $\displaystyle \frac{a}{b} = x,\frac{b}{c}=y,\frac{c}{a}=z$

so $\displaystyle 2(x+y+z) = \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+3$

wan,t be able to proceed after that, could some help me

Since $a,b,c$ are sides of a triangle we may write $(a,b,c) = (y+z,z+x,x+y)$
with $x,y,z>0$.  The difference between the two sides is then
$$
x^3 + y^3 + z^3 \; + \; x^2 y + y^2 z + z^2 x
- 2 (xy^2 + yz^2 + zx^2).
$$
By cyclic symmetry we may assume $x \geq z$ and $y \geq z$, and write
$x = (1 + \alpha) z$ and $y = (1 + \beta) z$ with $\alpha,\beta \geq 0$
to find $z^3$ times
$$
2(\alpha^2 - \alpha\beta + \beta^2) + \beta(\alpha-\beta)^2 + \alpha^3,
$$
in which each term is nonnegative, and all are zero iff
$\alpha = \beta = 0$ $-$ which in turn is equivalent to $x=y=z$,
and thus $a=b=c$, so triangle $ABC$ is equilateral as claimed.
$\Box$

[Added later: The solution above applies a general technique for such
problems; but I see that for the present question
Hari Shankar posted (in a comment) a link to an
AoPS item
that gives an even simpler conclusion that retains the cyclic symmetry:
$$
x^3 + y^3 + z^3 \; + \; x^2 y + y^2 z + z^2 x
- 2 (xy^2 + yz^2 + zx^2) = x(x-z)^2 + y(y-x)^2 + z (z-y)^2,
$$
again with all terms nonnegative; so equality implies that
they all vanish and the rest follows as before.  That link also
gives the example $(a:b:c) = (10:3:24)$ of a rational point on the cubic
$2(bc^2 + ca^2 + ab^2) = b^2c + c^2a + a^2b + 3abc$ with positive variables
that do not satisfy the triangle inequality.]

$x+y+z\geq 3(xyz)^{\frac{1}{3}}$,when $x=y=z$ ,equal sign is established.So:$$2\sum\frac{c}{a}\geq\sum\frac{c}{a}+3$$ i.e.$$\frac{c}{a}+\frac{b}{c}+\frac{a}{b}\leq\frac{a}{c}+\frac{b}{a}+\frac{c}{b}$$Without loss of generality, we assume:$c\geq b\geq a$,so $\frac1{a}\geq\frac1{b}\geq\frac1{c},$then we have:$$\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\leq\frac{c}{a}+\frac{b}{c}+\frac{a}{b},$$so $a=b=c$.I'm using an incomplete sort to derive a contradiction.


Evaluate $\int_0^{\infty}\frac{\sqrt{x}}{x^2+1}dx$ without Complex Analysis I found  this question on a Complex Analysis Qualifying Exam:

How would you evaluate: $$\int_0^{\infty}\frac{\sqrt{x}}{x^2+1}dx$$

I am interested in all methods including Complex Analysis, I'm just less familiar with it.

For instance, since the integral is from $0$ to $\infty$ do we only consider the $x-i$ factor in $x^2+1=(x-i)(x+i)$ since $i$ would be the only root in the upper half plane?

Write $x=\tan t$ so your integral is $$\int_0^{\pi/2}\tan^{1/2}tdt=\frac12\operatorname{B}\left(\frac14,\,\frac34\right)=\frac{\pi}{2}\csc\frac{\pi}{4}=\frac{\pi}{\sqrt{2}}.$$See here and here if any of the theory I used is unfamiliar.

Let $x^2=y$ we obtaine :

$\int_0^{\infty}\frac{x^{-\frac{1}{4}}}{2(1+x)}dx$

Then we use beat function :

$\beta(\frac{3}{4},\frac{1}{4})$


Verify that two line segments do not cross, or projected intersection is not on either line I'm not a math guy, so I'm looking for concrete formulas without a lot of symbols or jargon, if possible.

I have two line segments, and need to determine if they intersect (true or false). The closest I've found is this: given coordinates of beginning and end of two intersecting line segments how do I find coordinates of their intersection?
 where the original poster provides a solution for the projected intersection, and from there I can compare the intersection to the x,y endpoints of one of the lines to see if it is within a line segment.

(a) is there an easier way to do this overall? I suspect I could just create a y=mx+b equation for both and solve for a common x,y value, but I'm working with screen coordinates in powerpoint and I'm not sure what to do about the b parameter since the lines are on different parts of the screen and I don't have a true 'zero' point. Also, coordinates are measured from the upper left corner, which just adds to my confusion when thinking about a traditional line formula.

(b) if I were to use the equation in the referenced post, is there an easier way to check the projected intersect against the lines, other than just comparing to see if the point is within the x and y range of one of the lines?

Thanks!

Here is another approach: somewhat similar to Zen's, but more complicated.

The segments $AB$ and $CD$ intersect if and only if both of the following are true:

$A$ and $B$ lie on different sides of the line through $C$ and $D$
$C$ and $D$ lie on different sides of the line through $A$ and $B$

These two conditions can be tested for using the notion of a scalar cross product (formulas below).

is true if and only if the scalar cross products $\vec{CA}\times \vec{CD}$ and $\vec{CB}\times \vec{CD}$ have opposite signs.
is true if and only if the scalar cross products $\vec{AC}\times \vec{AB}$ and $\vec{AD}\times \vec{AB}$ have opposite signs.

Formulas for scalar cross products (in which you will see a pattern):
$$
\begin{split}
\vec{CA}\times \vec{CD} &= (x_A-x_C)(y_D-y_C)-(y_A-y_C)(x_D-x_C) \\
\vec{CB}\times \vec{CD} &= (x_B-x_C)(y_D-y_C)-(y_B-y_C)(x_D-x_C) \\
\vec{AC}\times \vec{AB} &= (x_C-x_A)(y_B-y_A)-(y_C-y_A)(x_B-x_A) \\
\vec{AD}\times \vec{AB} &= (x_D-x_A)(y_B-y_A)-(y_D-y_A)(x_B-x_A) \\
\end{split}
$$
Conclusion: the line segments intersect if and only if both of the following numbers are negative:
$$\big((x_A-x_C)(y_D-y_C)-(y_A-y_C)(x_D-x_C)\big)\big((x_B-x_C)(y_D-y_C)-(y_B-y_C)(x_D-x_C)\big)$$
$$\big((x_C-x_A)(y_B-y_A)-(y_C-y_A)(x_B-x_A)\big) \big((x_D-x_A)(y_B-y_A)-(y_D-y_A)(x_B-x_A)\big)$$

What do you consider jargon? Have you had at least some basic linear algebra?
If you have two infinitely long lines in 2-d,they always intersect unless they're parallel.
That is to say, some values of s and t exist that.
a+(b-a)t == c+(d-c)s
Where your line points are a & b and c & d as 2d vectors.
This is the same as saying:
a-c+(b-a)t == (d-c)s
Which is a simple set of two linear equations.
If both s and t are between 0 and 1, then that means the two intersect, if not, then the intersection is on the extension of the line.


How to solve a puzzle with numerals Any insights are welcome for this puzzle.

  The following equation is wrong: $103 - 102 = 3$. Move one numeral to make it correct. The numeral moved is: $0,1,2$ or $3$?

Konstantinos was on the right track.

   $$103 - 10^2 = 3$$

Since the question was "how to solve it", for any interested...

I solved it by first seeing that all the obvious a + b = c can be eliminated by seeing the range of the reduced or increased numbers.  Konst suggested exponents, which is the only other operator I can think of, $1$ is too small to be helpful, but 2 or 3 could work.

The solution indeed uses exponents:

$103-10^2=3$


Using the comparison Test to see if a series converges or diverges ∑1/(1+(n^3)):

Im trying to use the limit comparison test, but I'm struggling to find the comparing equation.  I would appreciate if someone could either give me advice to finding the comparing equations and/or the equation of the this problem.


Cox proof of product rule - step explanation I'm going through "Probability Theory - The logic of science" written by E.T. Jaynes and I have a problem with one step on page 27/28 in the proof of the product rule.
The idea here is that we have a function $F$ for which:
$$(AB|C)=F[(B|C), (A|BC)]$$
We want to proof its associativity:
$$F[F(x, y), z] = F[x, F(y,z)]$$
We assume that F is differentiable. We denote:
$$u = F(x,y)$$
$$v = F(y,z)$$
$$F_1(x,y) = \frac{\partial F}{\partial x}$$
$$F_2(x,y) = \frac{\partial F}{\partial y}$$
With this assumptions, we want to prove:
$$F(x, v) = F(u, z)$$
We differentiate w.r.t. x and y:
$$F_1(x,v)=F_1(u,z)F_1(x,y)$$
$$F_2(x,v)F_1(y,z)=F_1(u,z)F_2(x,y)$$
We then use:
$$G(x,y)=\frac{F_2(x,y)}{F_1(x,y)}$$
And get equation E1:
$$G(x,v)F_1(y,z)=G(x,y)$$
Which can be rewritten as equation E2:
$$G(x,v)F_2(y,z)=G(x,y)G(y,z)$$
Now there's the part is hard for me to understand. We differentiate E1 w.r.t. z and E2 w.r.t. y. The proof states that left hand sides of both derivatives are the same. How is that? When I'm using product rule of calculus and chain rule, I get summation in E1 left side:
$$G_2(x,v)F_2(y,z)F_1(y,z)+G(x,v)F_{12}(y,z)$$
and for equation E2, left side equals:
$$G_2(x,v)F_1(y,z)F_2(y,z)+G(x,v)F_{21}(y,z)$$
So there's the quation - where's my mistake? I can see that first product of both equations derivative are the same but how this equations are the same? I can see two options: first, I shouldn't use product rule here (why?), second, for some reason
$$F_{12}(y,z)=F_{21}(y,z)$$
but I can't see reason for any of these.
I've checked original Cox proof in his book "The algebra of probable inference" but it's quite the same as in Jaynes book.

Upon reading this chapter again, I believe the symmetry of the second derivatives was part of the assumptions Cox made. Jaynes suggests looking at Aczél's Lectures on Functional Equations and Their Applications for a derivation of the general solution without assumptions. He also revisited it here. Unfortunately, it's paywalled but his lack of assumptions is mentioned in the abstract:

  Abstract Associativity is regarded as functional equation. Two ways of obtaining its general continuous strictly increasing solutions on real intervals are described. No differentiability or existence of neutral element or commutativity is pre‐assumed. Comments are made, among others on applications, and a generalization is also presented.

Yes, the double derivatives are the same whatever the performance order.

$$\begin{align}F_{12}(s,t) ~=~& \frac{\mathrm d ~}{\mathrm d s}\frac{\mathrm d ~}{\mathrm d t}F(s,t) \\[1ex] =~& \frac{\mathrm d ~}{\mathrm d t}\frac{\mathrm d ~}{\mathrm d s}F(s,t) \\[1ex]=~& F_{21}(s,t)\end{align}$$


In how many possible ways can we write $3240$ as a product of $3$ positive integers $a$, $b$ and $c$? In how many possible ways can we write $3240$ as a product of $3$ positive integers $a$, $b$ and $c$?

This is the question where I've been stuck. The answer is $450$, but I don't know why. I've tried taking out the number of factors, then applying the combination formula in different ways.

3420 can be written as 2^3 . 3^4 . 5^1

We have 3 places _ _ _ or lets denote them as x y z.

We have to distribute 3 '2's, 6 '3's and 1 '5' into 3 places, where '2','3','5' are objects.

Now, How many ways to distribute 3 '2's ? You observe that no matter how you distribute '2's the total number of '2's will always be 3.

Lets say you distribute the '2's such that x gets 1 '2', y gets 1 '2', z gets 1 '2' like so ----- 2|2|2

The bars (|) separate what x, y, z gets like so ----------------------------------------------- x|y|z

Again you can distribute such that x gets no '2', y gets 2 '2's, z gets 1 '2' like ----------- |2.2|2

-------------------------------------------------------------------------------------------------- x| y |z

Again you can distribute such that x gets 3 '2's, y gets no '2', z gets no '2' like --------- 2.2.2||

------------------------------------------------------------------------------------------------------- x|y|z

Therefore you see this is an arrangement problem of 5 items with 3 similar items (3 '2's) and 2 similar items (2 bars(|))

Therefore no. of ways to distribute 3 '2's to 3 places will be 5!/(3! . 2!) = 10 ways

Similarly for distributing 4 '3's to 3 places will be arranging ||3.3.3.3 ; which is 6!/(4! . 2!) = 15 ways.

No. of was to distribute 1 '5' to 3 places is 3 ways

Therefore the total no. of ways is 10.15.3 = 450 ways.


How to tell if a set of vectors spans a space? I want to know if the set $\{(1, 1, 1), (3, 2, 1), (1, 1, 0), (1, 0, 0)\}$ spans $\mathbb{R}^3$. I know that if it spans $\mathbb{R}^3$, then for any $x, y, z, \in \mathbb{R}$, there exist $c_1, c_2, c_3, c_4$ such that $(x, y, z) = c_1(1, 1, 1) + c_2(3, 2, 1) + c_3(1, 1, 0) + c_4(1, 0, 0)$.

I've looked around the internet, but all the answers I found involve setting up a matrix and finding the determinant, and I can't do that here because my matrix isn't square. What am I missing here?

There are several things you can do. Here are four:

You can set up a matrix and use Gaussian elimination to figure out the dimension of the space they span. They span $\mathbb{R}^3$ if and only if the rank of the matrix is $3$. For example, you have
$$\begin{align*}
\left(\begin{array}{ccc}
1 & 1 & 1\\
3 & 2 & 1\\
1 & 1 & 0\\
1 & 0 & 0
\end{array}\right) &\rightarrow \left(\begin{array}{ccc}
1 & 0 & 0\\
3 & 2 & 1\\
1 & 1 & 0\\
1 & 1 & 1
\end{array}\right) &&\rightarrow \left(\begin{array}{ccc}
1 & 0 & 0\\
0 & 2 & 1\\
0 & 1 & 0\\
0 & 1 & 1
\end{array}\right)\\
&\rightarrow \left(\begin{array}{ccc}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 2 & 1\\
0 & 1 & 1
\end{array}\right) &&\rightarrow \left(\begin{array}{ccc}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
0 & 0 & 1
\end{array}\right).
\end{align*}$$
(Sequence of operations: exchanged rows 1 and 4; subtracted first row from other rows to make $0$s in first column; exchanged second and third rows; added multiples of the second row to third and fourth row to make $0$s in the second column).

At this point, it is clear the rank of the matrix is $3$, so the vectors span a subspace of dimension $3$, hence they span $\mathbb{R}^3$.

See if one of your vectors is a linear combination of the others. If so, you can drop it from the set and still get the same span; then you'll have three vectors and you can use the methods you found on the web. For example, you might notice that $(3,2,1) = (1,1,1)+(1,1,0)+(1,0,0)$; that means that
$$\mathrm{span}\Bigl\{(1,1,1),\ (3,2,1),\ (1,1,0),\ (1,0,0)\Bigr\} = \mathrm{span}\Bigl\{(1,1,1),\ (1,1,0),\ (1,0,0)\Bigr\}.$$

Determine if the vectors $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$ lie in the span (or any other set of three vectors that you already know span). In this case this is easy: $(1,0,0)$ is in your set; $(0,1,0) = (1,1,0)-(1,0,0)$, so $(0,1,0)$ is in the span; and $(0,0,1) = (1,1,1)-(1,1,0)$, so $(0,0,1)$ is also in the span. Since the span contains the standard basis for $\mathbb{R}^3$, it contains all of $\mathbb{R}^3$ (and hence is equal to $\mathbb{R}^3$).

Solve the system of equations
$$\alpha\left(\begin{array}{c}1\\1\\1\end{array}\right) + \beta\left(\begin{array}{c}3\\2\\1\end{array}\right) + \gamma\left(\begin{array}{c}1\\1\\0\end{array}\right) + \delta\left(\begin{array}{c}1\\0\\0\end{array}\right) = \left(\begin{array}{c}a\\b\\c\end{array}\right)$$
for arbitrary $a$, $b$, and $c$. If there is always a solution, then the vectors span $\mathbb{R}^3$; if there is a choice of $a,b,c$ for which the system is inconsistent, then the vectors do not span $\mathbb{R}^3$. You can use the same set of elementary row operations I used in 1, with the augmented matrix leaving the last column indicated as expressions of $a$, $b$, and $c$.

If you check throw away $(3,2,1)$, you are left with 3 easily checked vectors. In fact, if $a(1,1,1)+b(1,1,0)+c(1,0,0)=(0,0,0) $ then we must have $a=0$ because only the first vector has a last coordinate. The same argument again gives $b=0$.

Three linearly independent vectors in a 3-dimensional space spans the space.


Aid solving logical puzzle I have a logical puzzle here that I need to solve. I'm not after just the answer of this puzzle, but also the reasoning and thought-process behind solving it. Any assistance is greatly appreciated.

What number between 1 and 7 do the following equal? A=, B=, C=, D=, E=, F=, G=

Given that:

1.  A ≠ 2
2.  A + B = F
3.  C – D = G
4.  D + E = 2F
5.  E + G = F

The rules are:

1. All the variables (A, B, C, D, E, F, G) are equal to integer values between 1 and 7
2. None of the variables (A, B, C, D, E, F, G) are equal to each other i.e. all seven values will be used, no repeat use of integers

Notice that if you have positive numbers, the relation $x+y=z$ implies $x<z$ and $y<z$. So properties 2., 3., 5. give $A<F$, $B<F$, $D<C$, $G<C$, $E<F$, $G<F$.
Moreover since $D+E = 2F$ and $E<F$ you get $F<D$.
So $A,B,E,G < F < D < C$ which means that $F = 5, D = 6, C = 7$.

Then from 4. $E=4$, from 3. $G=1$ from 1. $A=3$, $B=2$.


How do you prove $10^n + 3(4)^{n+2} + 5$ is divisible by 9? I was able to do it making the $10^n$ substitution ($10^{k+1} = 10^k (10)$). However, I cant prove it making the $3(4)^{k+2}$ substitution. It should definitely be possible, but I am stuck at $10^{k+1} + 4(9J) - 4(10^k) - 15$ (where J is an integer). I can continue on this path to factories out a $3$, but this is not a $9$. Any help?

Just reduce modulo $9$ :
$$10^n + 3(4)^{n+2} + 5 \equiv 1^n + 3\times 16 \times 4^n + 5 \equiv 1+(3\times-2)\times4^n + 5 \equiv 6(1-4^n)$$
And $4^n = 1$ modulo $3$, so $1-4^n$ is dividible by $3$, so $6(1-4^n)$ is dividible by $9$.

Because $$10^n+3\cdot4^{n+2}+5=10^n-1+3\left(4^{n+2}-1\right)+9.$$
We see that $10^n-1$ is divisible by $9$ and $4^{n+2}-1$ is divisible by $3$.
The inductive step:
By the assumption of the induction we obtain:
$$10^{n+1}+3\cdot4^{n+3}+5=10\left(10^n+3\cdot4^{n+2}+5\right)-18\cdot4^{n+2}-45$$ is divisible by $9$.


Why is the "finitely many" quantifier not definable in First Order Logic? In First Order Logic with Identity (FOL+I), one can express the proposition that there are exactly 3 items that have the property P.

Why is it not possible to express the proposition that there is a finite number of items that have the property P (in FOL+I)?

We can define formula $P_i$ that says "there are at most $i$ elements satisfying $P$". Now, if the infinite disjunction of the $P_i$ was definable in FO, it would (by compactness) imply a conjunction of some finite subset of the $P_i$, hence it would imply $P_i$ for some $i$. That is not true, if $P$ can have (say) $i+1$ elements satisfying it.

I think you can express "forall but finitely many" in FOL, which means "for all sufficiently large". But I am not sure whether this is equivalent to "finitely many", it seems to be a little bit different. See http://arxiv.org/pdf/math/0602415.pdf.


Is this theorem concerning the change of bases and my proof thereof correct? I seek verification of my proof for this change of bases theorem. Any help would be greatly appreciated!

The theorem is stated below:

  Let $V, W$ be finite-dimensional vector spaces over a field $F$.

  Let $T: V\to W$ be a linear mapping.

  Let $B_{0}, B_{1}$ be distinct ordered bases for V and let $B_{2}, B_{3}$ be distinct ordered bases for $W$.

  Let $M^{2}_{0}(T)$ be the matrix for $T$ with respect to $B_{0}, B_{2}$ and let $M^{3}_{1}(T)$ be the matrix for $T$ with respect to $B_{1}, B_{3}$.

  Let $M^{0}_{1}(I_{V})$ be the matrix of the identity map $I_{V}:V \to V$  with respect to $B_{0}, B_{1}$ and let $M^{2}_{3}(I_{W})$ be the matrix of the identity map $I_{W}:W \to W$ with respect to $B_{2}, B_{3}$.

  Then $ M^{3}_{1}(T) = M^{3}_{2}(I_{W})M^{2}_{0}(T)M^{0}_{1}(I_{V}) $.

My proof:

Take any  $ v \in V$ and let $w = T(v) \in W$. I write $[v]_{0}, [v]_{1}$ for $v$ expressed in bases$B_{0}, B_{1}$, respectively. Similarly for $[w]_{2}, [w]_{3}$.

The following relations hold:

$1)$ $[v]_{0} = M^{0}_{1}(I_{V})[v]_{1} $

$2)$ $[w]_{3}= M^{3}_{2}(I_{W})[w]_{2}$.

It follows that

$ M^{3}_{1}(T)[v]_{1}=[w]_{3}= M^{3}_{2}(I_{W})[w]_{2} = M^{3}_{2}(I_{W})  M^{0}_{2}(T) [v]_0 =  M^{3}_{2}(I_{W})  M^{0}_{2}(T) M^{0}_{1}(I_{V})[v]_{1}$.

Since the choice of $v$ was arbitrary, it must be that

$M^{3}_{1}(T)= M^{3}_{2}(I_{W})  M^{2}_{0}(T) M^{0}_{1}(I_{V})$.   $ \blacksquare $

[Added: Essentially I have tried to rewrite the following theorem in a way that is more intuitive to me. Have I succeeded? Is my proof also correct?]

I would like to point few things about your proof:
$a)$ $v$ should be a non-zero vector for the conclusion of the statement $M^{3}_{1}(T)= M^{3}_{2}(I_{W})  M^{2}_{0}(T) M^{0}_{1}(I_{V})$.
$b)$ This statement $ M^{3}_{1}(T)[v]_{1}=[w]_{3}$  doesn't follow from relation $1)$ and $2)$. It should be made an explicit statement.
$c)$ You should take linear transformation $T$ to be a non-zero linear transformation which is inherently assumed in Theorem $45$..

Rest Your Proof works fine for me.


Intuitive explanation for why $\left(1-\frac1n\right)^n \to \frac1e$ I am aware that $e$, the base of natural logarithms, can be defined as:

$$e = \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$$

Recently, I found out that

$$\lim_{n\to\infty}\left(1-\frac{1}{n}\right)^n = e^{-1}$$

How does that work? Surely the minus sign makes no difference, as when $n$ is large, $\frac{1}{n}$ is very small?

I'm not asking for just any rigorous method of proving this. I've been told one: as $n$ goes to infinity, $\left(1+\frac{1}{n}\right)^n\left(1-\frac{1}{n}\right)^n = 1$, so the latter limit must be the reciprocal of $e$. However, I still don't understand why changing such a tiny component of the limit changes the output so drastically. Does anyone have a remotely intuitive explanation of this concept?

The point is that $1-\frac{1}{n}$ is less than $1$, so raising it to a large power will make it even less-er than $1$. On the other hand, $1+\frac{1}{n}$ is bigger than $1$, so raising it to a large power will make it even bigger than $1$.

There's been some brouhaha in the comments about this answer. I should probably add that $(1-\epsilon(n))^n$ could go to any value less than or equal to $1$, and in particular it could go to $1$, as $n$ increases. It so happens that in this example, it goes to something less than $1$. The reason it goes to something less than $1$ is because we end up raising something sufficiently less than $1$ to a sufficiently high power.

The first definition of $e$ is $$ \lim_{n \to \infty} \left(1 + \frac{1}{n} \right)^{n} = e^1 $$ which is basically just answering the question of what happens if you take the limit of discrete compounded growth by 100% to continuous growth. Note that $e>2$, i.e. this limit of discrete compounded growth asymptotically approaches a value that is greater than one we would have arrived at with the initial rate of growth. This means that even though we're chipping away at the rate we're growing by with every step of compounding, the aggregate effect is more growth. Also, I just want to note that $e$ is the universal constant from continuous growth by a certain rate - meaning that simply raising $e^{rt}$ will give the effect of continuously growing at a rate $r$ for $t$ units of time.

If we decide instead to see what happens when we take the limit of discrete compounded decay instead of growth, $$ \lim_{n \to \infty} \left( 1- \frac{1}{n} \right)^n = e^{-1} $$ we see that the opposite happens. Going from discrete compounded decay to instantaneous decay lessens the amount to which we are decaying by with every step of compounding, and asymptotically approaches a value $1/e> 0$, the value we would have been at if we did a single step of decay at a rate of 100%.

Don't know if this helps at all.


Question on the continuity of a function Here's a question from a real analysis textbook:

  Let $f\colon \mathbb{R} \to \mathbb{R}$ be continuous taking values in
  $\mathbb{Z}$ or in $\mathbb{Q}$. Then show that $f$ is a constant
  function.

What is the author trying to say here? The domain of $f$ is clearly $\mathbb{R}$, so it can accept values from both $\mathbb{Q}$ and $\mathbb{Z}$. Are they trying to say that it is continuous on either $\mathbb{Z}$ or $\mathbb{Q}$ and discontinuous on $\mathbb{R}\setminus \mathbb{Z}$ and $\mathbb{R}\setminus \mathbb{Q}$ respectively? The question is quite vague.

The author is saying that $f$ is a continuous function from $\Bbb R$ to $\Bbb R$ such that $f(x)\in\Bbb Z$ (respectively, "$\in \Bbb Q$") for all $x\in \Bbb R$.

And that you must prove that, with either of these assumptions, $f$ can only be constant.

Let $f$ be non-constant, and assume $f(x_0)\ne f(x_1)$. Continuity allows to apply the intermediate value theorem so that $f(x)=y$, where $y$ is an irrational number in $((f_0), f(x_1))$, must have a solution. A contradiction.


How many integers less than 2015 are multiples of 2 or 3 (or both)? Here is what I did.

To find all the multiples of 2 that is less than 2015 all we need to do is
divide by 2. The same can be done for multiples of 3 that is less than 2015:

2015 / 2 = 1007 (Integers multiples of 2)

2015 / 3 = 671 (Integers multiples of 2)

2015 / 6 = 335

( 1007 + 671 ) - 355 = 1344 Answer

Your method (using the in-and-out formula) is correct, although your work (at least the version you posted) is kind of sloppy. Here is another way:

Out of every six consecutive integers, four are divisible by $2$ or $3$; namely, $6n+2,6n+3,6n+4,6n+6$ are divisible by $2$ or $3$ (or both), while $6n+1,6n+5$ are not. Simce $2014=335\times6+4,$ the number of integers in $[1,2015)$ which are divisible by $2$ or $3$ is $335\times4+3=1343.$

an alternate method would be using AP

for example divisors of 2

2,4,6........2014

using

nth term = first term +(number of terms - 1)common difference

2014=2+(n-1)2

n=1007

similarly you can easily solve for

3,6,9........,2013
and thus for
6,12,18........2010


2 heads or more in 3 coin toss formula what is the formula to calculate the probabilities of getting 2 heads or more in 3 coin toss ?
i've seen a lot of solution but almost all of them were using method of listing all of possible combination like HHT,HTH,etc

what i am trying to ask is the formula and/or method to calculate this using a formula and
no need to list all of the possible combination.

listing 3 coin toss combination is easy(8 possible combination),but suppose i change the coins to dice or say 20-side dice. that would take a long time to list all the possible combination.

The probability of exactly $k$ heads in $n$ tosses is
$$
p_{n,k} = \left(\frac{1}{2}\right)^n \binom{n}{k}
$$
so, the probability of getting at least $k$ heads is
$$
\sum_{i=k}^n p_{n,i} = \frac{1}{2^n}\sum_{i=k}^n \binom{n}{i}
$$
or
$$
1 - \sum_{i=0}^{k-1} p_{n,i} = 1-\frac{1}{2^n}\sum_{i=0}^{k-1} \binom{n}{i}
$$
For $n=3$ and $k=2$ you get $\frac{1}{8}\left(\binom{3}{2}+\binom{3}{3}\right)=\frac 12$.

The question is about the chances of $~2~$ or more success occurring so. $~x\ge 2~$.

Not $~x=2~$  so the $~1$-binomial$(n,p,x-1)~$ function is supposed to give $~.5~$ but it doesn't, because Aliens wear pure hats.


Is it true that any two of the graphs of cubic functions are similar? It is evident that any two lines are congruent. In addition, any two parabolas are similar. I wonder whether every graph defined by a cubic function $y=ax^3+bx^2+cx+d$ can be rewritten as $Y=X^3$ via a similarity transformation.
It can be easily demonstrated that a graph defined by a cubic function can be rewritten as $y=x^3+Ax$ via a similarity transformation. Therefore, the question boils down to whether the graph of $y=x^3+Ax$ is similar to that of $y=x^3$ for every real number $A$.
It seems counterintuitive to think the graph of $y=x^3-3x$ is similar to that of $y=x^3$ since the former has two "prominent" points. Nonetheless, I cannot prove or disprove the proposition that any two of the graphs of cubic functions are similar. Any helpful advice is cordially appreciated.

A short disclaimer/warning: I haven't studied these types of problems formally, so there may be a standard, simpler way to deal with this.  This is just the method that occurred to me.
Also, in the general proof, I abuse the symbol $y'$ a fair bit to mean the derivative on different curves, even in the same equation.  Hopefully the context gives enough clues to what I mean, but if anyone knows a standard notation for this application, I'd appreciate comments to that effect.

For the moment, let's consider $y=x^3$ and $y=x^3-3x$ as you have done.  Note that the range of the derivative of each function is of the form $[c_0,\infty)$ for different values of $c_0$ ($0$ and $-3$ respectively).  This difference is what we want to capture, but we need to define a quantity that both makes sense and is independent of similarity.
To do this, first we make the interval bounded by applying $\arctan$ so the range of $\arctan(y')$ is of the form $[c_1,\pi/2)$ for different values of $c_1=\arctan(c_0)$.  Then take [Lebesgue] measures, giving $$\lambda(\arctan(y'(\mathbb{R}))) = \frac{\pi}{2} - c_1.$$  This gives us the quantity we want.

More generally, we can consider any smooth curve $C \subset \mathbb{R}^2$ and let $[C]$ denote the equivalence class of all curves which are similar to $C$.  Then I claim that the function $$[C] \mapsto \lambda(\arctan(y'(C)))$$ is well-defined.  To see this, it's enough to see how it behaves under translation, uniform scaling, reflection, and rotation.

Translation and uniform scaling: Note that $y'(C)$ is unchanged under both, so obviously the above value is unchanged.
The particular reflection $R_y(x,y) = (x,-y)$ satisfies
$$\begin{align*}\lambda(\arctan(y'(R_yC))) &= \lambda(\arctan(-y'(C))) \\ &= \lambda(-\arctan(y'(C))) \\ &= \lambda(\arctan(y'(C)))\end{align*}$$
and any other reflection can be written as $T\rho R_y \rho^{-1}T^{-1}$ where $T$ is a translation and $\rho$ is a rotation
For $0 \leq \theta < \pi$, any counterclockwise rotation $\rho_{\theta}$ by $\theta$ about some point satisfies, for $(x,y) \in C$, $$\arctan(y'(\rho_\theta(x,y))) = \begin{cases}\arctan(y'(x,y)) + \theta & \text{ if } \arctan(y'(x,y)) + \theta \leq \frac{\pi}{2} \\ \arctan(y'(x,y)) + \theta - \pi & \text{ otherwise}\end{cases}$$ Using this, we may show directly that $\lambda(\arctan(y'(\rho_\theta C))) = \lambda(\arctan(y'(C))).$
For $\pi \leq \theta < 2\pi$, just note $\rho_\theta =
   (\rho_{\theta/2})^2$ and $0 \leq \theta/2 <\pi$.

Finally, applying the above to $C_A : y=x^3 + Ax,$ we find that $$\lambda(\arctan(y'(C_A))) = \frac{\pi}2 - \arctan(A)$$ so none of these curves are similar to one another.

There are four operations we can perform to convert two similar 2D graphs: translation, reflection, rotation, and uniform scaling
It is obvious that the first three of these cannot change the existence of local extreme points.
If we uniformly scale an equation by a factor of $\lambda$, we essentially replace $x$ with $\lambda x$ and $y$ with $\lambda y$.  If we take a cubic with no local extreme points, i.e. one of the form $y-A=C(x-B)^3$ and rescale it, we get $\lambda y-A= C (\lambda x-B)^3$.  If we redefine $A'=\frac{A}{\lambda}, B'=\frac{B}{\lambda}<, C'=C\lambda^2$, this is equivalent to $y-A'=C'(x-B')^3$, which is also a cubic with no local extreme points


How can a stopping time be independent of its stochastic process? I was reading about a special case of Wald's equation, which led me to the following question:

If $X_t$ is a sequence of iid RV's, and $\tau$ is a stopping time for this discrete stochastic process, how is it possible for $\tau$ to be independent of $X_i$'s? Wouldn't this violate the definition of a stopping time which says that we should be able to tell if the stopping time is equal to $k$ by looking only at $X_1$,...,$X_k$? Doesn't a stopping time by definition depend on the process?

I'm asking because I saw an example of an independent stopping time in my text: do a random walk on the integers, and at every step flip a coin, when the coin is heads for the first time, stop. But I cannot see why this is a stopping time because if you were given $X_1,...,X_k$ here, you don't have enough information to conclude whether $k$ is the time to stop.

Indeed $\tau$ need not be independent of the process $\{X_n\}$. In the example where $\mathbb P(X_1=1)=1-\mathbb P(X_1=-1)=p$ and $$\tau=\inf\{n>0:X_n=1\}, $$ we have $$\tau = \sum_{n=1}^\infty p(1-p)^{n-1}\mathsf 1_{\{X_n=1\}\cap\{X_{n-1}=\cdots=X_1=-1\}}, $$ so $\sigma(\tau)\subset\sigma\left(\bigcup_{n=1}^\infty X_n \right)$ as $\tau$ is a measurable function of $\{X_n\}$.

In general, a nonnegative integer-valued random variable $T$ is a stopping time with respect to the filtration $\{\mathcal F_n\}$ if $\{T= n\}\in\mathcal F_n$ for all $n$. This implies that $$\sigma(T)=\sigma\left(\bigcup_{n=0}^\infty \{T=n\}\right)\subset\sigma\left(\bigcup_{n=0}^\infty \mathcal F_n\right),$$ so $T$ is independent of $\{X_n\}$ if and only if $\mathbb P(A)\in\{0,1\}$ for all $A\in\sigma(T)$, i.e. $\mathbb P(T=n)=1$ for some $n$.

The crux is in the measurability condition. Independent T is not adapted to the natural filtration of the process, of course, rather requires additional randomisation. You can think of larger filtration, generated at step n by X_1,...,X_n and another independent variable Y (which can be Y=T itself), then T qualifies as so-called randomised stopping time. Note that for the proof of Wald's identity we only need independence of {T>n} and X_j's with j>n.


answer to iq test with colored squares What is the best (whatever this means) answer and why?

Well, that's funny, 'cause you can get an answer basing only on several conventions (resonable ones). But they are not unique, so guess you can get different answers. Let's try.
 In the left column we have initial states (I've placed an axis, running along the colored blocks from blue to red).
 In the middle colum we have the state that shows us the direction of transformations (rotations). I've placed an axis along the colored in gray blocks in such a way that it always points upwards. Plus I've added an azure line which coincides in direction with the initial state and has two parts: the inital and mirrored one.
 In the right column we have final states which are obtained by the rotation of the initial state along the direction of transformations.
So following this logic I guess the answer is $\mathrm{G}$.

The answer is undoubtedly B. If you consider the entire group of blocks along with their rhythm,  you will clearly see the answer. I've taken some of the most complex and in depth IQ test and have a consistent score of 192
I belong to several elite high IQ societies also. The answer to that question is simply B.


Hilbert And Banach dual spaces Are, and in general, all Hilbert spaces identified to their dual ?, i.e $H'\equiv H$
What about Banach spaces? And can someone please state the weak convergence in each?

A vector space (a fortiori, Hilbert space, Banach space) is isomorphic to its dual if, and only if, it is finite dimensional. For any infinite dimensional vector space, the dual space is always "larger" than the predual, in the precise technical sense that there exists an injection from the predual into the dual, but never a bijection.
Additionally, the isomorphism between a finite dimensional vector space and its dual is dependent on a choice of vector basis. It is not a "natural" isomorphism, in the language of category theory. However, the isomorphism between a vector space and its double dual (the dual of the dual) IS natural, in that it can be defined without a choice of basis.
Studying this phenomenon is anecdotally what pushed Saunders MacLane and Samuel Eilenberg to invent category theory, by the way.
Edit: I was mistaken in the above, due to an ambiguity on the notion of "dimension" as used by an algebraist, and as used by an analyst; and have provided in the comments below a document which clarifies my point, and explains my mistake in your specific context.


How do you calculate the semi-minor axis of an ellipsoid? Given the semi-major axis and a flattening factor, is it possible to calculate the semi-minor axis?

Possibly something like this. Correct me if I'm wrong.

$j$ = semi-major
$n$ = semi-minor
$e$ = eccentricity

$n = \sqrt{(j\sqrt{1 - e^{2}}) \times (j(1 - e^{2}))}$

Where,
$a$ = transverse radius = semi-major axis (for ellipse/oblate spheroid);
$b$ = conjugate radius = semi-minor axis ("     "     ");
$oe$ = angular eccentricity = $\arccos(\frac{b}{a})$;
$f$ = flattening = $\frac{a-b}{a} = 1 - \frac{b}{a} = 1 - \cos(oe) = 2\sin(\frac{oe}{2})^{2}$;

then $b = a\cos(oe) = a(1-f)$.


Show that if $a^{n-1} \equiv 1 \pmod{n}$ for all $a$ such that $\gcd(n,a) = 1$, then $a^{n} \equiv a \pmod{n}$ for all $a$. In Burton's Elementary Number Theory, when introducing absolute pseudoprimes (also called Carmichael numbers), the author states that they are composite numbers $n$ such that $a^{n-1} \equiv 1 \pmod n$ for all integers $a$ with $\gcd(a,n) = 1$.

When showing that 561 is an absolute pseudoprime, he then concludes that $a^{561} \equiv a \pmod{561}$ for all $a$.

I searched a bit on the internet, and it seems that these two are actually equivalent definitions of a Carmichael number. However I have not been able to find a proof; the fact that the implication in the title of the post is stated so plainly in my textbook makes me think it is quite simple to see, but I can't seem to understand the argument.

The equivalent definitions of Carmichael numbers are the special case $\,e = n\,$ below.

Theorem $\ $  The following are equivalent for  integers  $\,n,e>1$.
$(1)_{\phantom{|_{|_.}}}\  n\mid a^e\ -\ a\ \ $ for all $\,a\in\Bbb Z^{\phantom{|^|}}\!\!,\: $ and $\ (e\!-\!1,n)=1$
$(2)_{\phantom{|_{|_.}}}\ n\mid a^{e-1}\!-1\ $ for all $\,a\in\Bbb Z\,$ with $\, \color{#90f}{(a,n)=1}= (e\!-\!1,n)$
$(3)\ \ \ \:\! n\,$ is squarefree, $ $ prime $\,p\mid n\,\Rightarrow\,   \color{#0a0}{p\!-\!1\mid e\!-\!1},\ p\nmid e\!-\!1$

Proof $\ \ (1\Rightarrow 2)\ \ \ (1)\,\Rightarrow\, \color{#90f}{n\mid a}(a^{e-1}-1)\,\Rightarrow\, n\mid a^{e-1}-1\,$ by $\,\color{#90f}{n,a}\,$ coprime & Euclid's Lemma.

$(2\Rightarrow 3)\ \ $ Suppose prime $\,p\mid n,\,$ so  $\,n = j\, p^k$ with $\,k\ge 1,\ p\nmid j.\,$ Let $\,g\,$ be a primitive root $\!\bmod p^k,\,$ i.e. $\,g\,$ has order $\,(p\!-\!1)p^{k-1}.\,$ By CRT there's $\,a\in\Bbb Z\,$ with $\,\ a\equiv 1\pmod{\!j},\,a\equiv g\pmod{\!p^k},\,$ thus $\,a\,$ is coprime to $\,j,p\,$ so also to $\,n = j\,p^k.\,$ So $\,(2)\Rightarrow\,a^{e-1}\equiv 1\,$ holds $\!\bmod n\,$ so also $\!\bmod p^k,\,$ thus Order Theorem $\Rightarrow\,(\color{#0a0}{p\!-\!1})p^{k-1}\!\mid \color{#0a0}{e\!-\!1}\,\Rightarrow\,k=1\,$ (else $\,p\mid e\!-\!1,n\,$ contra $\,(e\!-\!1,n)=1)$.

$(3\Rightarrow 1)\ \ $ Let prime $\,p\mid n.\,$  If $\,p\mid a\,$ then $\,p\mid a^e-a\,$ by $\,e>1.\,$ Else $\,p\nmid a\,$ so by little $\rm\color{#c00}{Fermat}$ $\!\bmod p\!:\ a^{\large\color{#0a0}{e-1}}\equiv \smash[t]{\color{#c00}{(a^{\color{#0a0}{\large p-1}})}}^{\large\color{#0a0} k}\equiv \color{#c00}1^{\large k}\!\equiv 1\,$ so $\,p\mid a^{e-1}-1\mid a^e-a.\,$ So $\, a^e-a\,$ is divisible by all primes $\,p\mid n\,$ so also by their lcm = product = $\,n,\,$ by $\,n\,$ squarefree. $\,(e\!-\!1,n)=1\,$ by $\,p\mid n\,\Rightarrow\,p\nmid e\!-\!1$.

Remark $\,\ (3)\, $ for $\,e=n\,$ is known as Korselt's criterion for Carmichael numbers.

Let a^(n-1) be congruous to 1 (mod n).
You can also see that a is congruous to a (mod n), that means n divides a-a=0(This resulta could be also explained noticing that a congruence is an equivalence relation, so it satisfies the reflective property).
Thus (a^(n-1) times a) is congruous (1 times a) (mod n). This result comes from this property: a congruous b(mod n), c congruous d(mod n)  implies ac congruous bd (mod n).

I’m sorry for the language and the lack of notation.


Simplest proof that $\zeta(s) \to \infty$ as $s \to 1$? For homework I had to prove the divergence of the series $1/(k\log^p k)$ for all real $p$ (it is simple to do so via integration.)  However a more elegant means would be to appeal to the behavior of the zeta function $\zeta(s)=\sum\limits_{k=1}^\infty 1/k^s$ on the real line.  What is the most elementary proof that $\lim\limits_{s\to 1}\zeta(s)=\infty$?

For every $k\geqslant1$ and every $x\geqslant k$, $\dfrac1{k^s}\geqslant\dfrac1{x^s}$, hence $\displaystyle\frac1{k^s}\geqslant\int_k^{k+1}\frac{\mathrm dx}{x^s}$.
Summing this over $k\geqslant1$, one gets $\zeta(s)\geqslant\displaystyle\int_1^{+\infty}\frac{\mathrm dx}{x^s}=\frac1{s-1}$. Since $\dfrac1{s-1}\to+\infty$ when $s\to1^+$, $\zeta(s)\to+\infty$.

It seems that the most elegant proof is Euler's proof via comparison
\begin{align*}
\zeta (1) &= 1 + \left( {\frac{1}{2}} \right) + \left( {\frac{1}{3} + \frac{1}{4}} \right) + \left( {\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}} \right) + \cdots \\
&> 1 + \left( {\frac{1}{2}} \right) + \left( {\frac{1}{4} + \frac{1}{4}} \right) + \left( {\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}} \right) + \cdots \\
&= 1 + \left( {\frac{1}{2}} \right) + \left( {\frac{1}{2}} \right) + \left( {\frac{1}{2}} \right) + \cdots \\
&= \infty.
\end{align*}


How to do rational expressions Never was much of a math student but I am brushing up on my arithmetic and algebra for college.

I am using sample questions from Accuplacer and then using video lectures and practice on Khan Academy. It is going well but sometimes the examples in the lecture are very trivial and I am currently doing very poorly with rational expressions in the form:

$$ \frac{1}{x+3} + \frac{1}{x}  = 2x +\frac{3}{x(x+3)}$$

or

$$ \frac{u}{x} + \frac{5u}{x} - \frac{u}{5x} = \frac{29u}{5x}$$

The answers that I supplied are from a multiple choice list, so I need to be able to be able to get those answers on my own.

I have looked up the answers to these problems and I do not understand how it works. I am looking for any killer examples that would really help me grasp it.

This video lecture is where I am at when it comes to rational expressions. I can do some work when it is a 1 coefficient and when it is > 1. I lose confidence in my ability to teach myself the above two problems.
Thank you.

I'll take your two problems themselves. The key to do what you want to do is multiplying the numerator of fraction 1 by the denominator of fraction 2, and the numerator of fraction 2 by the denominator of fraction 1. Your first problem. $$\dfrac{1}{x+3} + \dfrac{1}{x}$$ Now how do we do this? Well firstly you have to make one common denominator, and no better way to do that then multiplying the two denominators. That means our denominator will be $x(x+3)$. But notice what we are doing here to get this denominator. We are multiplying fraction 1 by $\dfrac{x}{x}$ and fraction two by $\dfrac{x+3}{x+3}$ (because only multiplication by $1$ is allowed, otherwise the problem changes). From here we get: $\dfrac{x}{x(x+3)} + \dfrac{x+3}{x(x+3)}$. And we add these together to get $\dfrac{2x+3}{x(x+3)}$

In an equation you may do anything (except multiply or divide with 0) as long as you do it to both sides: so in this case the denominators can be canceled by multiplying every term with factors that will cancel the denominators: for example in the first expression you multiply EVERY TERM with $x(x+3)$. This will cancel factors in the with terms that have it in the denominator and you will be left with fraction-less expressions:

$$\frac{1}{x}\times x(x+3)+3\times x(x+3)+\frac{1}{x}\times x(x+3)=2x\times x(x+3)+\frac{3}{x(x+3)}\times x(x+3) $$
$$(x+3)+3\times x(x+3)+(x+3)=2x\times x(x+3)+3$$
$$...$$

Since we multiplied with x(x+3) its just important to note that it may never equal 0. So x may not equal 0 or -3 in the end. If it does label that answer as not applicable.

To determine what to multiply with factorize the denominators first and then multiply each term with the lowest common multiples of the denominators.


What is the derivative of $f(x)$ with respect to $g(x,y)$ 2 Suppose that we have a function $f(x)$ of one variable, like $f(x)=x$.

And I want to derive this function with respect to another multivariate function $g(x,y)=x-y$.

My hypothesis is that the derivative is equal to $0$:

$$\frac{{\partial f(x)}}{{\partial g(x,y)}}=0$$

My reason to think that is that $f(x)$ can never be expressed in terms of $g(x,y)$, so $g(x,y)$ does not impact directly $f(x)$ in any way. Also $x$ and $y$ are independent, meaning they are not related, there is no function between them.

Or maybe is undetermined?

But if that is true, I would need to prove it. I would appreciate if you could help me.

Basically it's a normal derivative, but the fact that it does operate on the variable $(x-y)$.

For example:

$$\frac{\partial (x-y)^2}{\partial (x-y)}$$

By calling $(x-y) = z$,  you can find the usual derivative:

$$\frac{\text{d} z^2}{\text{d}z} = 2z \longrightarrow 2(x-y)$$

as it shall be.

In your case your function is simply $x$.

By a change of variable like $x-y = z$ you find that $x = z-y$ hence your derivative is like

$$\frac{\text{d} (z-y)}{\text{d}z} = \frac{\text{d}}{\text{d}z}(z-y) = 1$$

In any case, let me tell you that that is a very ill notation.


Show linear independence of vector addition Let $\{v_1, v_2, v_3\}$ be a linearly independent set of vectors in a vector space $V$. Show that the set $\{v_1 + v_2, v_2 + v_3, v_1 + v_3\}$ is also a linearly independent set.

I put the equations in the form:
$c_1(v_1 + v_2) + c_2(v_2 + v_3) + c_3(v_1 + v_3) = 0$, to see if it implies that $c_1 = c_2 = c_3 = 0$.

Then I distributed to change it to the form:
$v_1(c_1 + c_3) + v_2(c_1 + c_2) + v_3(c_2 + c_3) = 0$

I'm not sure what to do from here. I need to show the linear independence of:
$c1 + c3 = 0$,
$c1 + c2 = 0$, and
$c2 + c3 = 0$

Do I just put this in reduced row echelon form?

The set $\{v_1,v_2,v_3\}$ is linearly independent.

Since you've reduced it to the form $$(c_1+c_3)v_1+(c_1+c_2)v_2+(c_2+c_3) = 0,$$ you know that this is only true if all $c_i = 0$, since we assumed linear independence i.e. let $d_1 = (c_1+c_3)$, $d_2 = (c_1+c_2)$, $d_3=(c_2+c_3)$. Then we have $$d_1v_1 + d_2v_2 + d_3v_3$$ which is only possible if the $d_i =0$. Ergo, your second set is linearly independent, since they may only sum to $0$ if the coefficients are all $0$. Is this clear?

So once you're at the step
$$(c_1+c_3)v_1+(c_1+c_2)v_2+(c_2+c_3)v_3=0$$
because you know the set $\{v_1,v_2,v_3\}$ is linearly independent, we must have all these coefficients equal to zero, i.e.
$$c_1+c_3=c_1+c_2=c_2+c_3=0$$

From the first equality we can say we must have $c_2=c_3$ (by subtracting $c_1$ from both). Then
$$0=c_2+c_3=c_3+c_3=2c_3\implies c_3=0$$
and since $c_2=c_3$, we must have $c_2=0$. Then $0=c_1+c_3=c_1$, so all the $c_i$ are zero and the set is linearly independent.


Frechet Differentiability versus Strict Differentiability We recall two concepts of differentiability of a mapping. Let $X, Y$ be real Banach spaces and $f: X\rightarrow Y$
be a mapping.

$f$ is said to be Frechet differentiable at $\bar{x}$ if there
exists a linear continuous operator $\nabla f(\bar{x}): X\rightarrow Y$ such that
$$
\lim_{x\rightarrow \bar{x}}\frac{f(x)-f(\bar{x})-\nabla f(\bar{x})(x-\bar{x})}{\|x-\bar{x}\|}=0.
$$
$f$ is said to be strictly differentiable at $\bar{x}$ if $f$ is Frechet differentiable
at $\bar{x}$ and
$$
\lim_{\substack{x\rightarrow \bar{x}\\ u\rightarrow \bar{x}}}\frac{f(x)-f(u)-\nabla f(\bar{x})(x-u)}{\|x-u\|}=0.
$$

It is known that if $f$ is continuously Frechet differentiable in a neighborhooh of $\bar{x}$
 then $f$ is strictly differentiable at this point but not vice versa.

I would like to find a mapping $f$ such that $f$ is strictly differentiable at $\bar{x}$
but it is not differentiable at points near $\bar{x}$.

What you call strict differentiability is what Schechter calls strongly differentiable. See chapter 25 of his book.

Section 25.23 proves that

  If $f$ is Frechet differentiable on an open set, then $f$ is continuously differentiable if and only if $f$ is strongly differentiable on the same set.

Now to answer your question:

Let $f:\mathbb{R}\to\mathbb{R}$ be given by (below we have $j\in\mathbb{Z}$):
$$ f(t) = \begin{cases} 2^{2j} + (|t| - 2^{j}) ( 2^{j+2} - 2^j) & |t|\in [2^j, 2^{j+1}) \\ 0 & t = 0 \end{cases} $$
Our function $f(t)$ is piecewise linear and continuous, as can be easily checked. It is not differentiable at any $t = 2^j$ since the slope at $(2^j)_+$ and $(2^j)_-$ are different.

One can, however, easily check that $f$ is strongly differentiable at 0 with derivative 0. We see easily that for $x,y$
$$ \frac{|f(x) - f(y)|}{|x-y|} \leq \frac{|f(|x|) - f(|y|)}{||x| - |y||} $$
so we can assume WLOG $x,y$ has the same sign. Then we have that using the convexity of the construction (the slope increases), that $f(x) - f(y)$ (assuming $x > y$) is at most $3x(x-y)$. And hence as $x,y \to 0$ the limit exists and is 0.

$f: [-1,1] \rightarrow \mathbb R , \bar x=0$, defined by
$f(x):=\begin{cases} x^2 & x=1/k , k\in \mathbb N \\ 0 & x=0\\ \text{linear} & \text{otherwise} \end{cases}$


Non-reflexive module isomorphic to its double dual Could you give me an example of a non-reflexive module isomorphic to its double dual?

I found an example here but I cannot understand it, do you have any simpler examples?

By this question we should only use not finitely generated modules.

What about $l^1(\mathbb{R})$ as an $\mathbb{R}$ vector space?
The natural map $J_1\colon l^\infty \to (l^1)^\vee$, which reads a bounded sequence as the multiplication by said sequence, is an isomorphism, while the dual map $J_2\colon l^1 \to (l^\infty)^\vee$ is not. We then get, composing the natural bidual map with the traspose of $J_1$, that
$$l^1 \to (l^1)^{\vee\vee} \to (l^\infty)^{\vee}$$
is, again, $J_2$! Hence $l^1$ cannot be reflexive. Still, we can find bases of the same cardinality for the two spaces.


English translation of Erdős proof of: $\sum_{k=0}^{n} \frac{1}{a+kd} $ is not an integer. In his paper, Erdős proved that:

  For positive integers $a$ and $d$, with $\gcd(a, d)=1$, and positive integer $n$, the expression
  $$\frac{1}{a}+\frac{1}{a+d}+\dots+\frac{1}{a+nd}$$
  is not an integer.

Do you know a reference for an English language of this proof?

I can't understand Hungarian but I have an idea of what it's talking about.

This is a proof by contradiction. Suppose that for some $n$, the fractional sum posted in the OP is an integer.

Let $p > n$ be a prime such that $p^{\alpha} || a + kd$ (Erdos will prove that such a prime exists later in the paper). It's obvious that $(p, d) = 1$ because if not, then $(a, d) \neq 1$.

He starts the proof making the statement:

Claim:
There cannot exist $k'$ such that $k \neq k'$, $k' < n$ and $p^{\alpha} | a + k'd$

Proof:
Suppose that such a $k'$ exists. Then by simple modular arithmetic,
$$k \equiv k' \pmod{p^\alpha}$$ This proves the claim since $|k - k'| < n < p^\alpha$ is absurd.
$\square$
$$\\$$
Denote
$$\prod_{i = 0}^n (a + id) = !(a + nd)$$

The fractional sum (let's call it $S$) is thus equal to
$$S = \frac{!(a + nd) + \frac{a \cdot !(a + nd)}{a + d} + \cdots + \frac{a \cdot !(a + nd)}{a + nd}}{a \cdot !(a + nd)}$$

It's clear that $v_p\left(\frac{a \cdot !(a + nd)}{a + kd}\right) < v_p\left(\frac{a \cdot !(a + nd)}{a + ld}\right)$ for $l \neq k$ so $v_p(S) = v_p\left(\frac{a \cdot !(a + nd)}{a + kd}\right) < v_p(a \cdot !(a + nd))$ so it's clear that $S$ cannot be an integer. This is a contradiction.

Next is the step that takes up most of the proof: the proof of the prime $p > n$. Note that $p$ does not necessarily have to be greater than $n$; only $p^\alpha$ has to necessarily be greater than $n$.

Suppose that for all $a + id$ that if a prime $p$ divides it, then the maximal power of $p$ that divides it is less than or equal to $n$. Now consider the fraction
$$\frac{!(a + nd)}{n!}$$
If $q$ is a prime such that $\sqrt{n} < q \le n$, then at most $\lfloor \frac{n}{q} \rfloor$ numbers less than or equal to $n$ are divisible by $q$. Since $q > \sqrt{n}$, each of those numbers can be divisible by $q$ at most once. Hence, there are $\lfloor \frac{n}{q} \rfloor$ factors of $q$ in $n!$. Since $q^2 > n$, under the assumption of the statement we are trying to prove false, there are at most $\lfloor \frac{n}{q} + 1 \rfloor$ (Exercise for the reader: why $ + 1$?) factors of $q$ in $!(a + nd)$. Using similar logic, Erdos deduces that

$$\frac{!(a + nd)}{n!} \le \prod_{\sqrt{n} < q \le n} q \prod_{n^{1/3} < q \le \sqrt{n}} q^2 \prod_{n^{1/4} < q \le n^{1/3}} q^3 \cdots = \prod_{q \le n} q \prod_{q \le \sqrt{n}} q \prod_{q \le n^{1/3}} q \cdots$$

In addition, we have that
$$\frac{!(a + nd)}{n!} > d^n \ge 4^n$$
assuming that $d \ge 4$ (Exercise for the reader: prove the original statement by the OP for $d < 4$).

Hence,

$$4^n < \prod_{q \le n} q \prod_{q \le \sqrt{n}} q \prod_{q \le n^{1/3}} q \cdots$$

The rest of the proof is devoted to showing that such an inequality is impossible. Let
$$a_i = \lceil \frac{n}{2^i} \rceil$$
where $a_i \ge 1$. We can easily prove that $a_i \le 2a_{i + 1}$. Hence, the intervals, $(a_i^{1/\alpha}, (2a_i)^{1/\alpha})$ for all (valid) $i$ covers the interval $(1, n^{1/\alpha})$. Consider the binomial: ${2n \choose n}$. Any prime number $n^{1/\alpha} < p \le (2n)^{1/\alpha}$ (\alpha \ge 1) must divide the binomial at least once because $\lfloor \frac{2n}{p} \rfloor > 2 \lfloor \frac{n}{p} \rfloor$ (and also by Legendre's formula).

Thus,
$$\prod_{q \le n} q \prod_{q \le \sqrt{n}} q \prod_{q \le n^{1/3}} q \cdots < \prod {2a_i \choose a_i}$$

We will show that $\prod {2a_i \choose a_i} < 4^n$. This is easy since by simple induction (or by Stirling's formula if you want to drop the sledgehammer), we have that ${2n \choose n} < 4^{n - 1}$ if $n \ge 5$. Summing up all of the powers of $4$ in the RHS of
$$\prod {2a_i \choose a_i} < \prod 4^{a_i - 1}$$
we have that
$$\prod 4^{a_i - 1} < 4^n$$
(Proof is left as an exercise to the reader). Hence,
$$\prod {2a_i \choose a_i} < 4^n$$
This is a contradiction to an earlier inequality. So there must exist a prime power $p^\alpha | a + kd$ such that $p^\alpha > n$.

QED
$$\\$$

The rest of the paper is Erdos dealing with $d = 1, 2, 3$.

Here is what Google translate produces:

Page 1
Budapest, 1932. Sonderabdruck aus «Mathematical and Physical Tabs Band XXXIX. Budapest 1932. AN EXCELLENT CHURCH-EARTH BETWEEN DIALOGUE Generalizations. THEISINGER proved that the harmonic series of tranches m = n . 1 Sums, m , Can not be integers.1 OBLÁΤY general m = 2 mn It has re-established this item, 2 if it has proved that am m = k ni Can not be an integer if the ams are positive integers and (am, m) = 1 Kürschák 3 gave it a completely complete proof of it m = n 1 . The batch of that m Can not be an integer, no matter m = k The integer integer is ak and n. . This item will be generalized if k, k + 1,. . .K + n Instead of a special arithmetic sequence instead of the more general arithmetic lines, Show them a batch Let a, d, be any positive integers; the Expression can not be an integer. The bizonyításnál assume that d and relative prime numbers, Because if they are not the reciprocal of their greatest common divisor (1), the other factor is again (1), but a and d They are now relative primes. Thus, if a line (a, d) = 1, it is true, Otherwise it is valid. 1 T Heisinger Monatshefte fϋr Math. u. Ρhys., 26, k. (1915) 135. 1. 2. Oblath: Mathematical and Physical Journal, 27 k. (1918) 93. 1. 3 KURSCHAT Mathematical and Physical Journal, 27 k. (1918) 299. 1. 1
Page 2
h PALE FOREST. First, d> 4; The other cases at the end of our We'll take care of it separately. Evidence is based on the following auxiliary: the One of the numbers can be divided by a p prime number that is Greater than n. If the auxiliary evidence has been proved, further evidence Nete the next. Let kd + pa is divisible, where p '> n; They can not exist in a variety of k from k ' is not possible because n. Let's introduce it now nominations. You can then write (1) The counter members are integers; As well as The tag can be divided by the lower power of p, As the other members of the counter and the denominator, so (4) can not be integer. Q. ed That is the only proof we need to prove. Contrary to this entry, suppose a + d, a + 2d, ... a + nd
Page 3
GENERATION OF A BACKGROUND OF AN EXTRAORDINARY BALANCE ELEMENTS. Series members. and only p ' n, up Under theorem every element that only up to - first Power, even up to [~] + 1 times nyezőként. After the acronym, no more than p remains. Let 1in  n, under the terms of each element that No more than a second power, even in total Up to [~] +1, up to [~] + 1 times, Such as p2, so after the abbreviation, the maximum p2 remains in the counter. In the same way it will be appreciated that if "~ + n  4
Page 4
ERDŐS PAL. We will prove that this is not possible. To this end, Let's play it Binomial coefficients. (9) The register is divided into every n  2 [k]) p p Let us denote the first integer with> xa {x} Where the arbitrary integer; then furthermore So since k and ak + 1 are integers,
Page 5
GENERATION OF A BACKGROUND OF AN EXTRAORDINARY BALANCE ELEMENTS. Be already m the first number for which it is 2m <1; then m = 1. Clearly, 2a1> n. Furthermore, (10) it is Am  10 s suppose that the theorem is true min- For less than n; at Which is achieved by applying the batch to 2a2-1; 2a2 - from one item to your true since n> 2a2 - 1
Page 6
PALE FOREST. With full induction, we can easily assure that n> 5 1 ~ n) < 4n-1, so (13) by Now, it is obvious that 2a1  k + 2a,
Page 7
Ι Ι Ι Ι Ι GENERATION OF A BACKGROUND OF AN EXTRAORDINARY BALANCE ELEMENTS. Our contradictions are minimal. Further proof is so Occurs as in the general case. Finally, let d = 2; Since (1) is the first member the And the rest Smaller than this, so that (i) integers can be most one of the members must be below + to stand, ie that n> the be. So it is The last member of a series of numbers 2n +> = a + 2a and 3a of the series- There are all odd numbers for which  a and 3a occur among the members (15); but Then another member of the series can not be divided Let be 3a, because 3a + 2.3a = 3a + 1 would also occur (15) - Which is impossible because of the maximality of 3a. The further Conclusion is made as in the general case. Similarly, but with slightly longer calculations, We will also consider the following item for the special cases discussed later: The arithmetic line there is a + kd member, a prime number p It contains more power than the other members. It can also be proven for special cases, ter- Of course, with the exception of d = 1, it is proven to be a general case Our auxiliary. Generally, the OBLÁTH theorem is that t. i. can not be an integer, if (k, a + kd) = 1, the method of his demonstrated. Pál Erdős.
Page 8
q Ρ. ERDŐS. VERALLGEMEINERUNG EINES ELEMENTAR-ZAHLENTHEORETISCHEN SATZES VON KIRCSCHAKK. 1 mn Herr THEISINGER bewies, dass ~ -für keinen Wert von n eine Ganze Zahl darstellt. Herr OBLQTH gab eine Verallgemeinerung dieses m = n Satzes, Inde bewies er, dass ~ am Keine ganze Zahl sein kann, 17 m = n i wenn (am, m) = 1 ist. KÜRSCHÁK bewies elementar, dass ~ -bei m = k Keinem Werte von k und n eine ganze Zahl sein kann. Hier wird dieser Satz folgendermassen verallgemeinert, indem statt Der speziellen arithmetischen Reihe Allgemeinere arithmetische Reihen betrachtet werden. E seien a, d, n beliebige positive ganze Zahlen, dann ist Keine ganze Zahl. Besteht der Grundgedanke des Beweises Darin, dass ein glied a + kd Angegeben wird, welches durch eine höhere Potenz einer Primzahl Teilbar ist, als die übrigen Glieder. Dies ergibt sich aus der Analyzes der Primteiler der Ausdrücke:. Paul Erdős.


Explaining why $\displaystyle {d \over dt}e^{\int_0^tA(s)\,ds} \ne A(t)e^{\int_0^tA(s)\,ds}$ by definition I've been asked to explain why $\displaystyle {d \over dt}e^{\int_0^tA(s)\,ds} \ne A(t)e^{\int_0^tA(s)\,ds}$ using the definition of $e^A$, where $A$ is an $n\times n$ matrix.

I don't fully understand how they aren't equal. Someone please point out where I went wrong:

$$A(t)e^{\int_0^tA(s)\,ds} = A(t) \sum_{n = 0}^\infty{1 \over n!}\left(\int_0^tA(s)\,ds\right)^n$$

We have been given that

  $${d \over dt}\left(\int_0^tA(s)\,ds\right)^n = nA(t)\left(\int_0^tA(s)\,ds\right)^{n-1},$$

which I use to write $$\begin{align}{d \over dt}e^{\int_0^tA(s)\,ds} &= {d\over dt}\left(\sum_{n=0}^\infty{1 \over n!}\left(\int_0^tA(s)\,ds\right)^n\right) \\ & = \sum_{n=0}^\infty{1 \over n!}{d\over dt}\left(\int_0^tA(s)\,ds\right)^n\\&=\sum_{n=1}^\infty{1\over(n-1)!}A(t)\left(\int_0^tA(s)\,ds\right)^{n-1}\end{align}$$

They look equal to me, but likely there's a mistake somewhere. Can anyone see what's wrong?

Notice that, according to Leibniz's rule,

$$\frac {\Bbb d} {\Bbb d t} \left(\int \limits _0 ^t A(s) \ \Bbb d s \right) ^n = \frac {\Bbb d} {\Bbb d t} \underbrace {\int \limits _0 ^t A(s) \ \Bbb d s \dots \int \limits _0 ^t A(s) \ \Bbb d s} _{n \text{ times}} = \color{red} {A(t)} \underbrace{ \int \limits _0 ^t A(s) \ \Bbb d s \dots \int \limits _0 ^t A(s) \ \Bbb d s} _{n-1 \text{ times}} + \int \limits _0 ^t A(s) \ \Bbb d s \color{red} {A(t)} \underbrace{ \int \limits _0 ^t A(s) \ \Bbb d s \dots \int \limits _0 ^t A(s) \ \Bbb d s} _{n-2 \text{ times}} + \dots \underbrace{\int \limits _0 ^t A(s) \ \Bbb d s \dots \int \limits _0 ^t A(s) \ \Bbb d s} _{n-1 \text{ times}} \color{red} {A(t)} ,$$

where $A(t)$ occupies, in turn, all the positions from $1$ to $n$ in that product, and it is not at all obvious why $A(t)$ should commute with $A(s)$ for $0 \le s < t$.

If, on the other hand, these matrices do commute, then indeed, the result is as you have written, namely

$$\color{red} {A(t)} \ n \left( \int \limits _0 ^t A(s) \ \Bbb d s \right) ^{n-1} .$$

Calling the whole integral at the exponent as $A$ you get:

$$\frac{\text{d}}{\text{d}t} e^A = \frac{\text{d}}{\text{d}t}\sum_{k = 0}^{+\infty}\frac{A^k}{k!} = \sum_{k = 0}^{+\infty}\frac{1}{k!}\frac{\text{d}}{\text{d}t}(A^k)$$

Now

$$\frac{\text{d}}{\text{d}t} A^k = k\cdot A^{k-1}\cdot \frac{\text{d}A}{\text{d}t}$$

Hence

$$\frac{\text{d}}{\text{d}t} e^A = \sum_{k = 0}^{+\infty}\frac{1}{k!}\frac{\text{d}}{\text{d}t}(A^k) = \sum_{k =0}^{+\infty}\frac{1}{k!}k\cdot A^{k-1}\frac{\text{d}A}{\text{d}t} = A\cdot\frac{\text{d}A}{\text{d}t} \sum_{k = 0}^{+\infty}\frac{k\cdot A^k}{k!}$$

And you can see that it's proven.


The domain of lsc convex functions How to see that $int(dom(f)) \neq \emptyset$ where $f$ is a proper convex lower semi-continuous function from a Banach space to $\mathbb{R} \cup \{+\infty\}.$

A function $$f(x)=\begin{cases}+\infty&\text{for }x\ne 0\\0&\text{for }x=0\end{cases}$$ is proper convex lsc function while its domain is a singleton $\{0\}.$


If $(x,y,z)$ is a Pythagorean triple such that each of $x,y,z$ can be written as sum of two squares then prove that $180|xyz$ This is a problem based on Pythagorean Triples.

If $(x,y,z)$ is a Pythagorean triple such that each of $x,y,z$ can be written as sum of two squares then prove that $180|xyz$

Any ideas of  how to start solving it?

I tried substituting $x = a^2 + b^2 \  , y = c^2 + d^2 \ , z = e^2 + f^2$ and so on and I am not getting any idea how to show it is divisible by $180$ , or divisible by $4,5,9$

Outline:

By considering $x^2+y^2=z^2$ modulo $8$, deduce that one of $x,y,z$ must be a multiple of $4$.
By considering $x^2+y^2=z^2$ modulo $5$, deduce that one of $x,y,z$ must be a multiple of $5$.
By considering $x^2+y^2=z^2$ modulo $3$, deduce that one of $x,y,z$ must be a multiple of $3$.
Confirm that any multiple of $3$ that is the sum of two squares must be a multiple of $9$. (This is the only step where being a sum of two squares is used.)

These four results combine to give you the divisibility by $180$ you desire.

$$x^2= y^2+z^2\;\;\;(*)$$

Divisibilty with $4$:

If $y$ and  $z$ are even then we are done.
If $y$ and  $z$ are odd then $x $ is even so $x=2a$ and $y=2b+1$ and $z=2c+1$ so $$4a^2 = 4b^2+4b+4c^2+4c+2$$ which is impossible
If $y=2b+1$ and  $z=2c$ then $x=2a+1$ so $$4a^2+4a+1= 4b^2+4b+1+4c^2$$ so $$\underbrace{a(a+1)}_{even}-\underbrace{b(b+1)}_{even}=c^2\implies 2\mid c$$ and we are done since $4\mid z$.

Divisibilty with $5$: For each $n$ we have $$n^2\equiv 0,1,-1 \pmod 5$$. So if $x,y, z\ne 0$ then we can not have $(*)$

Divisibilty with $9$: Hint, observe the equation modulo $3$.


Fourier transform of KPZ equation What is the Fourier transform of $|\partial f(x,t)/\partial x|^{2}$?

To put the question in some context, I was trying to write the KPZ equation in Fourier space. The KPZ equation is given by $\partial h(x,t)/\partial t=\nabla^{2}h(x,t)+|\nabla h|^{2}+\eta(x,t)$. While the Fourier transform of linear parts is done, I was stuck at deriving the Fourier transform of the non-linear term. I referred to the literature on this and the Fourier transform was given to be

$-i\omega h(k,w)=-k^{2}h(k,w)-\int\int \frac{d\Omega \: dq}{4\pi^{2}}\; q(k-q)\;h(q,\Omega)\;h(k-q,\omega-\Omega)\; +\eta(k,\omega)$

As I have said, the linear parts can be obtained by multiplying with $e^{i(kx-\omega t)}$ throughout and then integrating over $x$ and $t$. I am unable to figure out how such an operation would generate the above given non-linear term.

Use the convolution theorem for Fourier transform!