why it is a projective space? The unite sphere bundle of $TS^2$ , the tangent bundle of the 2-sphere, is the real projective space, $RP^3$. I can not understand the reason of it. And why the complement of the unit disc bundle of $TS^2$ is diffeomorphic to $(1,∞)×RP^3$

You should first convince yourself that the unit tangent bundle of $S^2$ is diffeomorphic to $SO(3)$. (Hint: There is a natural action of $SO(3)$ on the manifold of unit tangent vectors to $S^2$. What's the stabilizer subgroup of a point?)

Then it is a standard argument that $SO(3)\cong \Bbb RP^3$. Every nontrivial rotation has an axis and rotating $\pi$ around $\xi$ is the same as rotating $-\pi$ around $-\xi$. So $SO(3)$ can be identified with the unit ball $D^3$ with points on the boundary sphere identified to their antipodes. (Think "polar coordinates" with radius $\pi$.)

One places a plane separate to the sphere, and straight lines are formed by a projection of a feature when viewed from a point U.  Where U is interior to the sphere, the projection onto the map for a line running theough U gives a point on the map, and a pair of points or a through line in the sphere.

Where U is on the surface of the sphere, the projection gives the line at infinity for the plane tangent to the sphere at U, and a straight line equates to a circle through U.  The sphere is showing the inversion model, and the plane a general projective model.  When the tangent plame is parallel to the map, the model realy is euclidean geometry.

When U is outside the sphere, the results are the poincare model on the sphere, and the Beltrami Klein model on the map.  The cone with the apex at U and tangential to the sphere gives the hyperbolic Horocycle at infinity.

You can see from this model, that it is possible to build a geometry, where it is not known the relation of U to the surface of the sphere, leaving it undetermined if real space is one of these three geometries


Help with a prime number spiral which turns 90 degrees at each prime I awoke with the following puzzle that I would like to investigate, but the answer may require some programming (it may not either).  I have asked on the meta site and believe the question to be suitable and hopefully interesting for the community.

I will try to explain the puzzle as best I can then detail the questions I am interested in after.

Imagine squared paper.  In one square write the number $1.$  Continue to write numbers from left to right (as normal) until you reach a prime.  The next number after a prime should be written in the square located $90$ degrees clockwise to the last.  You then continue writing numbers in that direction. This procedure should be continued indefinitely.

Here is a sample  of the grid:

$$\begin{array}{} 7&8&9&10&11&40&41 \\6&1&2&&12&&42\\5&4&3&14&13&44&43\\&&34&&26\\&&33&&27\\&&32&&28\\&&31&30&29\end{array}$$

Note that the square containing 3 also contains 15 (I couldn't put it in without confusing the diagram.  In fact some squares contain multiple entries.
I would have liked to see an expanded version of the diagram.  I originally thought of shading squares that contain at least one number.

Questions
Does the square surrounded by $2,3,9,10,11,12,13,14$ ever get shaded?
If so, will the whole grid ever be shaded?
Is there a maximum number of times a square can be visited?  I have got to 4 times but it is easy to make mistakes by hand.
Are there any repeated patterns in the gaps?
I have other ideas but this is enough for now as I have genuinely no idea how easy or difficult this problem is.

Please forgive me for not taking it any further as it is so easy to make mistakes.
I hope this is interesting for the community and look forwards to any results.
Thanks.

Any questions I'll do my best to clarify.

Side note: I observed that initially at least the pattern likes to cling to itself but I suspect it doesn't later on.

Just for visual amusement, here are more pictures. In all cases, initial point is a large red dot.

Primes up to $10^5$:

Primes up to $10^6$:

Primes up to $10^6$ starting gaps of length $>6$:

Primes up to $10^7$ starting gaps of length $>10$:

Primes up to $10^8$ starting gaps of length $>60$:

For anyone interested, all the images were generated using Sage and variations of the following code:

d = 1
p = 0
M = []
prim = prime_range(10^8)
diff = []
for i in range(len(prim)-1):
    diff.append(prim[i+1]-prim[i])
for k in diff:
    if k>60:
        M.append(p)
    d = -d*I
    p = p+k*d
save(list_plot(M,aspect_ratio = 1,axes = false,pointsize = 1,figsize = 20)+point((0,0),color = 'red'),'8s.png')

As @ajotatxe explained, the point between $2$ and $12$ can never be crossed, but as regards your grid, the first time the point between $33$ and $27$ is reached at $6\ 716\ 606$:

Position[AnglePath[-Pi/2 Boole[PrimeQ[Range[10^7]]]],{2,-3}][[1, 1]]

With[{a = AnglePath[-Pi/2 Boole@PrimeQ@Range@6716606]}, Graphics[{Black, PointSize@0.002, Point@Last@a, Blue, Line@a, Thick, Red, Line@Take[a, 45]}, ImageSize -> 5000]]


A question on Taylor Series and polynomial Suppose $ f(x)$ that is infinitely differentiable in $[a,b]$.

For every $c\in[a,b] $ the series $\sum\limits_{n=0}^\infty \cfrac{f^{(n)}(c)}{n!}(x-c)^n $ is a polynomial.

Is true that $f(x)$ is a polynomial?

I can show it is true if for every $c\in [a,b]$, there exists a neighborhood $U_c$ of $c$, such that
$$f(x)=\sum\limits_{n=0}^\infty \cfrac{f^{(n)}(c)}{n!}(x-c)^n\quad\text{for every }x\in U_c,$$
but, this equality is not always true.

What can I do when $f(x)\not=\sum\limits_{n=0}^\infty \cfrac{f^{(n)}(c)}{n!}(x-c)^n$?

As I confirmed here, if for every $c\in[a,b] $, the series $\sum\limits_{n=0}^\infty \cfrac{f^{(n)}(c)}{n!}(x-c)^n $ is a polynomial, then for every $c\in[a,b]$ there exists a $k_c$ such that $f^{(n)}(c)=0$ for $n>k_c$.

If $\max(k_c)$ is finite, we're done: $f(x)$ is a polynomial of degree $\le\max(k_c)$.

If $\max(k_c)=\infty$ it means there is an infinite number of unbounded $k_c$'s, but $f$ is infinitely differentiable, so (hand waving) the $c$'s can't have a limit point, i.e. although $\max(k_c)=\infty$ it can't be $\lim_{c\to c_\infty}k_c=\infty$ for some $c_\infty\in[a,b]$ because that would mean $k_{c_\infty}=\infty$, i.e. not a polynomial.

So the infinite number of unbounded $k_c$'s need to be spread apart, e.g. like a Cantor set.

Does this suggest a counterexample or can a Cantor-like distribution of $k_c$'s never be infinitely differentiable?

All polynomials are Power Series but not all Power Series are not polynomials.
For a certain Power Series $\displaystyle f(x) = \sum_{k=0}^\infty a_k \left( x-c \right)^k = a_0 + a_1 (x-c)^1 + a_2 (x-c)^2 + a_3 (x-c)^3 + \cdots$ to be a Polynomial of degree $n$, then for all $k>n$, $a_k = 0$.
If $ f(x)$ is infinitely differentiable in the interval $[a,b]$, then for every $k \in \mathbb{N}$, $f^{(k)}(x) \in \mathbb{R}$ i.e. exists as a finite number. The Taylor Series of $f(x)$ in the neighbourhood of $c$ is $\sum\limits_{k=0}^\infty \cfrac{f^{(k)}(c)}{k!}(x-c)^k $ and
If the remainder, $R_N(x) = f(x) - \sum\limits_{k=0}^N \cfrac{f^{(k)}(c)}{k!}(x-c)^k $ for a certain $N \in \mathbb N$, converges to $0$ then $f(x) = \sum\limits_{k=0}^\infty \cfrac{f^{(k)}(c)}{k!}(x-c)^k $
Taylor's inequality: If $|f^{(N+1)}(x)|≤ B$ for all $x$ in the interval $[a, b]$, then the remainder $R_N(x)$ (for the Taylor polynomial to $f(x)$ at $x = c$) satisfies the inequality $$|R_N(x)|≤ \cfrac {B}{(N+ 1)!}|x − c|^{N+1}$$ for all $x$ in $[c − d, c + d]$ and if the right hand side of this inequality converges to $0$ then $R_N(x)$ also converges to $0$.

According to your question, supposing that $\sum\limits_{k=0}^\infty \cfrac{f^{(k)}(c)}{k!}(x-c)^k $, $\forall c \in [a,b]$ is a polynomial which translates to  $$\text{given } c\in[a,b],\ \ \exists n_c\in \mathbb N \ (\text{ $n_c$ depends on c}) \quad|\quad\sum\limits_{k=0}^\infty \cfrac{f^{(k)}(c)}{k!}(x-c)^k=P_{n_c}(x)$$ $$\quad \quad \quad\quad \quad \quad \quad\quad \quad \quad\quad \quad \quad\quad \quad \quad\quad \quad \text { and} \  \forall k>n_c, \ k\in \mathbb N, \ {f^{(k)}(c)}=0$$

This is true because if one looks at the finite sum $N\ge n_c$,   $$\displaystyle\sum^N_{k=0} a_k(x-c)^k=\sum^N_{k=0}\sum^k_{i=0}a_k\binom ki(-1)^{k-i} c^{k-i}x^{i}=\sum^N_{i=0}x^{i}\sum^N_{k=i}a_k\binom ki(-1)^{k-i} c^{k-i}$$ if this is a polynomial $P_{n_c}(x)$ of degree $n_c$, then $$\forall i>n_c,\ \ \displaystyle \sum^N_{k=i}a_k\binom ki(-1)^{k-i} c^{k-i}=0$$ Solving this system of equations gives that $\forall n_c<k\le N, \ \ a_k=0$ and

$$a_k=\cfrac{f^{(k)}(c)}{k!}=0\implies f^{(k)}(c)=0, \ \  \forall k>n_c$$
This holds when $N\rightarrow \infty$

Since $n_c$ depends on each $c\in[a,b]$, it is sufficient to take $\displaystyle n=\max_{c\in[a,b]} (n_c)$ such that for any $c\in [a,b]$ and for any $k>n,\ \ k\in \mathbb N$, we have $f^{(k)}(c)=0$.

Thus, the Taylor series is of $f$ is a polynomial of degree $\displaystyle n=\max_{c\in[a,b]} (n_c)$ because $\displaystyle f(x) = \sum_{k=0}^\infty a_k \left( x-c \right)^k=P_n(x)$.

At this point it is sufficient to prove that $\displaystyle f(x) = \sum_{k=0}^\infty a_k \left( x-c \right)^k=P_n(x)$ using the Taylor Remainder Theorem (#4).

We've already found out that $f^{(k)}(c) = 0,\space  \forall k>n$, thus $ f^{(n+1)}(x) = 0$ or simply $ f^{(n+1)}(x) \le 0$ (to work with inequalities) which implies that $B = 0$. At this point it is clear that $|R_N(x)|≤ \cfrac {B}{(N+ 1)!}|x − c|^{N+1} = 0$ and we can conclude that $R_N(x)$ converges to $0$ and that $f(x) = \sum\limits_{k=0}^\infty \cfrac{f^{(k)}(c)}{k!}(x-c)^k = P_n(x)$.

$f$ is a polynomial.


How to solve this congruence of triangles geometrically? Find x.

Solution with trigonometry:

  First there are the missing angles ...

  Superior = $180 - 13 - 32 = 135$

  Bottom = $180 - 13 - 24 = 143$

  By the breast theorem we have to
  $$\frac{9 \sqrt{2}}{sen(13)}=\frac{Diagonal}{sen(135)} \to Diagonal=\frac{9 \sqrt{2} \cdot sen(135)}{sen(13)}$$
  Also in other triangle...
  $$\frac{x}{sen(13)}=\frac{Diagonal}{sen(143)} \to x = \frac{Diagonal \cdot sen(13)}{sen(143)}$$
  Using what we know diagonally
  $$x = \frac{9 \sqrt{2} \cdot sen(135) \cdot sen(13)}{sen(13) \cdot sen(143)}=\frac{9 \sqrt{2} \cdot sen(135)}{sen(143)}$$

Maybe there is a shorter way:

Here, you can use what you called breast theorem on $\Delta ABC$, you will find the same answer.

In the figure $ x = 3 $. However, the lenght that you seek is $ 5x = 15 $

In your case $$x = \frac{9 \sqrt{2} \cdot sen(135)}{sen(143)} = \frac{9 \sqrt{2} \cdot \frac{1}{\sqrt 2}}{\frac{3}{5}} = 15 $$


Statistics - Binomial and Poisson Distribution Problem I am given:

I get 11 text messages per hour according to a Poisson process.
The probability that a given text message is from my mother is $0.62$.

I then have to find the probability that I receive exactly two text messages from my mother in a 40 minute period.
My thought process is:

Find the expected number, $E(X)$, of text messages in a 40 minute period.
Use this value as the number of trials, $n$, in a binomial distribution to calculate $P(Y = 2)$ where $Y =$ "the number of text messages I receive from my mother in a 40 minute period".

However, $E(X) = 11 \cdot \frac{40}{60} = \frac{22}{3}$ meaning my idea won't work without rounding the number of trials - giving two wildly different answers (as expected). Am I meant to round up or down, or am I doing this completely wrong? I did try averaging the probabilities of the two answers I obtained, but I'm not sure if this is correct either.

If the probability that a given text message is from my mother is independent from other text messages, you can split the poisson process: you can say that the stream of messages from my mother is a poisson process too with rate $0.62 \cdot 11 = 6.82$. This is known as 'splitting'.
The number of text messages in a 40 minute interval follows a poisson distribution with parameter $6.82 \cdot 40/60 = 341/75$. The probability that I receive exactly two text messages from my mother in a 40 minute period is:
$$\frac{e^{-341/75} (341/72)^2}{2!} \approx 0.12.$$


Prove that if $\sum u_n$ is convergent, then $\sum\sqrt[n]n\,u_n$ is also convergent If $\sum u_n$ is a convergent series then can it be concluded that
  $\sum\sqrt[n]n\,u_n$ is also convergent

I think that by abels theorem since  $\sqrt[n]n$ is monotone bounded hence the above theorem holds true

Please help from there

If $u_n \geq 0$ for all sufficiently large $n$, yes. And this is easy.

Proof: without loss of generality, assume $u_n > 0$ for all $n$.
$
\lim_{n\to \infty} \sqrt[n]{n} = 1
$, so $\sqrt[n]{n}u_n \sim_{n\to\infty}u_n$ and you can conclude by the limit comparison test.
If $(u_n)_n$ is allowed arbitrary signs, yes. But this is less obvious (though not horrendous either), and as you suggested follows from Abel/Dirichlet's test.
(What follows is not the shortest proof, I reckon, but I find it instructive.)
Note that since $$\sqrt[n]{n} = e^{\frac{\ln n}{n}} = 1+ \frac{\ln n}{n}+O\left(\frac{\ln^2 n}{n^2}\right)\tag{1}$$
when $n\to \infty$, and $\lim_{n\to\infty}{u_n} = 0$ we have
$$
\sqrt[n]{n}u_n = u_n + \frac{\ln n}{n}u_n + \varepsilon_n \tag{2}.
$$
with $\varepsilon_n = o\left(\frac{\ln^2 n}{n^2}\right)$.
By assumption, $\sum_n u_n$ is a convergent series. By comparison with a $p$-series, the series $\sum_n \varepsilon_n$ is absolutely convergent. So overall, we have that
$$
\sum_n \sqrt[n]{n}u_n
$$
converges if, and only if,
$$
\sum_n \frac{\ln n}{n}u_n
$$
does. So it suffices to prove the latter: this is where Dirichlet's test comes in handy: we apply it with $a_n = \frac{\ln}{n}$, $b_n=u_n$, so that (i) $(a_n)_n$ is non-increasing with limit $0$; (ii) the partial sums of $\sum_n b_n$ are uniformly bounded (since $\sum_n u_n$ is convergent. $\square$


If $\gcd( a, b ) = 1$, then is it true to say $\gcd( ac, bc ) = c$? I tried to prove this theorem, but I'm really confused about its correctness.
Let's say

$$\gcd( a, b ) = 1 \Rightarrow am + bn = 1 \text{ where } m, n\in\mathbb Z$$

Then by multiplying both side by $c$, we have: $amc + bnc = c$.
From here, is it sufficient to say $c = \gcd( ac, bc )$?
I feel like this proof needs some more explanation, but I can't think of one. Any idea?

Thanks,

What you have shown is that there is a linear combination of $ac$ and $bc$ that equals $c$. Now, for any $x$, $y$, the gcd of $x$ and $y$ divides any linear combination of $x$ and $y$. In this case, this means that the gcd of $ac$ and $bc$ divides $c$.

You are almost done, but you also need to argue that $c$ divides the gcd of $ac$ and $bc$, to conclude the equality (if $z$ and $w$ are positive, and $z$ divides $w$, and $w$ divides $z$, then they are equal).

But, any linear combination of $ac$ and $bc$ is (obviously) a multiple of $c$ since $c$ divides both $ac$ and $bc$. So we are in fact done.

Since $\gcd(a,b)=1$, there exist two integer $x$ and $y$ such that $ax+by=1$.

Now, $acx+bcy=c$.

It follows that $\gcd(ac,bc)=c.$


A question regarding the proof of the second existence theorem for weak solutions of elliptic equations A question arised when I was self-studying the PDE book of Evans.

Let $Lu = - \sum _{i,j=1}^n (a^{ij} u_{x_i} ) _{x_j} + \sum_{i=1}^n b^{i} u_{x_i} + cu$ be a uniformly elliptic operator on a bounded domain $U$ with $a^{ij}, c \in L^{\infty} (U)$ and $b^i \in C^1 ( \bar U)$. Here, $a^{ij} = a^{ji}$.

The formal adjoint of $L$, denoted $L^*$, is defined as follows.

$L^* v = - \sum_{i,j=1}^n (a^{ij} v_{x_j})_{x_i} - \sum_{i=1}^n b^i v_{x_i} +(c-\sum_{i=1}^n b^i_{x_i} )v$.

It is easy to check that the associated bilinear form to $L^*$ satisfies $B^* [ u,v] = B[v,u]$,  where $B$ is associated to $L$. ($u, v \in H^1_0 (U)$)

Now, here is (a part of) the proof stated in the book. Here, $\gamma>0$ is chosen so that
$\beta ||u||^2 _ {H^1_0} \leq B[u,u] + \gamma ||u||^2_{L^2}$ for all $u \in H^1_0 (U)$.

Now, the author claims that: $v \in L^2(U) $ satisfies $v=K^* v$ if and only if $v \in H^1_0 (U) $ is a weak solution of the system [$L^*v = 0 $ on $U$, $v =0$ on $\partial U$]. (Here, $K^* : L^2 (U) \rightarrow L^2 (U)$ is the adjoint operator of the compact operator $K$.)

How can I prove this claim?  (Note: The analogous claim for $L$ and $K$ is indeed obvious.)

In my proof I will let ( , ) to denote the $L^2(U)$ inner product

Let  $v∈L^2(U)$

$v = K^*v $ iff

($v$,$u$) = ($K^*v$,$u$) = ($v$,$Ku$) for any $u∈L^2(U)$

$=\gamma$($v$,$L_{\gamma}^{-1}$$u$) by Definition of $K$ in (18)

iff (this holds because we can replace $u$ with $L_\gamma$$u$)

($v$,$L_{\gamma}u$) = $\gamma(v,u)$ for any $u∈L^2(U)$

iff

$0$ =  ($v$,$Lu$) = ($L^*v$,$u$) = $B^*$[$v$,$u$] for any $u∈L^2(U)$

iff $[L^∗v=0$ on $U$, $v=0$ on $∂U]$

Hope that can help.

I just guess the solution from the uppest and the lowest logcally equivalence and luckily got the second and third


Integral equation/ODE I have to find all the functions $f(x)$ such that
$$f(x)=xe^{(1-x^{2})/2}-xe^{-x^{2}/2}\int_{1}^{x}t^{-2}e^{t^{2}/2}dt$$
which satisfies
$$f(x)=1-x\int_{1}^{x}f(t)dt$$

I tried to equal both, but when I derivate, the integral keeps there.

What should I do?

PS: the answer is: only the function given first satisfies the equation.

The question is bizarre, one should probably solve the following:

  Consider the function $f$ defined by
  $$f(x)=xe^{(1-x^{2})/2}-xe^{-x^{2}/2}\int_{1}^{x}t^{-2}e^{t^{2}/2}dt.\tag{1}$$
  Show that $f$ solves
  $$f(x)=1-x\int_{1}^{x}f(t)dt.\tag{2}$$

To show this, consider the function $g$ defined by
$$
g(x)=xe^{(1-x^{2})/2},$$ then $(1)$ is equivalent to
$$f(x)=g(x)-g(x)\int_{1}^{x}\frac{dt}{tg(t)}.\tag{3}$$
Hence, if $f$ solves $(1)$, then, differentiating $(3)$, one gets
$$
f'(x)=g'(x)-g'(x)\int_{1}^{x}\frac{dt}{tg(t)}-g(x)\frac1{xg(x)}.
$$
Using $(3)$ to identify the integral in the RHS, one gets
$$
f'(x)=g'(x)+g'(x)\frac{f(x)-g(x)}{g(x)}-\frac1{x}=\frac{g'(x)}{g(x)}f(x)-\frac1x.
$$
Identifying the function $g'/g$, this shows that every solution of $(1)$ solves
$$
f'(x)=\frac{1-x^2}{x}f(x)-\frac1x.
$$
On the other hand, $f$ solves $(2)$ if and only if $f(1)=1$ and
$$
f'(x)=-xf(x)-\int_1^xf(t)dt=-xf(x)+\frac{f(x)-1}x=\frac{1-x^2}{x}f(x)-\frac1x.
$$
Thus, $(1)\implies(2)$.

Here is how you advance. First equate the two equations

  $$ 1-x\int_{1}^{x}f(t)dt=xe^{(1-x^{2})/2}-xe^{-x^{2}/2}\int_{1}^{x}t^{-2}e^{t^{2}/2}dt \longrightarrow (*).$$

then differentiate both sides w.r.t. $x$ and you will need the product rule for differentiation and Leibnize rule

  $$ {\mathrm{d}\over \mathrm{d}x} \left( \int_{f_1(x)}^{f_2(x)} g(t) \,\mathrm{d}t \right )= g[f_2(x)] {f_2'(x)} - g[f_1(x)] {f_1'(x)} . $$

Note: To get rid of the integral which appears after differentiation on the left hand side of the equation $(*)$ you need to use the second equation which you have been given as

  $$ \int_{1}^{x}f(t)dt = \frac{1-f(x)}{x} .$$


The p-adic numbers as an ordered group So I understand that there is no order on the field of $p$-adic numbers $\mathbb{Q}_p$ that makes it into an ordered field (i.e.) compatible with both addition and multiplication.

Now, from the responses to a couple of my previous questions,

$\mathbb{Q}_p$ is a divisible abelian group under addition (being a field of characteristic $0$).
$\mathbb{Q}_p$ is torsion-free.
It admits an order compatible with the group operation (addition), since every torsion-free abelian group is orderable.

My question is, can I write an explicit ordering of $\mathbb{Q}_p$ compatible with the group operation? By "explicit", I mean an ordering in which, given two $p$-adic numbers, I can decide which is greater.

P.S.: I was not sure how to classify this problem, so please feel free to change the tags.

EDIT:  I thought it would be appropriate, given the perhaps unexpected descriptive set-theoretic nature of this answer, to give a foreword explaining the focus on the Baire property (BP for short).  A subset of a topological space has the BP if it differs from an open set by a meager set (a set contained in the union of countably many nowhere dense sets).  In the setting of a Polish space (such as $\mathbb{Z}_p$ or the real numbers), the BP sets contain the Borel sets (and continuous images of Borel sets) and satisfy some nice regularity properties.  In this context BP sets should be considered analogous to measurable sets, with meager sets serving as analogs of the null sets.  For example, the Polish space itself is not meager, and the Kuratowski-Ulam theorem asserts that a subset of the plane is meager if and only if only a meager set of vertical sections are nonmeager (this is basically a Fubini theorem).

One particularly noteworthy fact is that it's consistent with the axioms $\mathtt{ZF+DC}$ that every subset of a Polish space has the BP.  Recall that $\mathtt{ZF}$ is the standard axiomatization of set theory without choice, and $\mathtt{DC}$ is the axiom of dependent choice.  Intuitively, $\mathtt{DC}$ gives you the freedom to make a countable sequence of choices, where each choice is allowed to refer to properties of your previous choices (so they aren't "independent").  Dependent choice is enough to do almost all of common mathematics: you can perform most of analysis, carry out typical inductive constructions, Borel sets behave reasonably, the first uncountable cardinal is not a countable union of countable sets, etc.  Some contexts in which $\mathtt{DC}$ doesn't bestow the full power of $\mathtt{AC}$ include performing wildly nonconstructive acts like building a Vitali set or choosing bases from huge vector spaces.  So, I think that $\mathtt{ZF+DC}$ is a reasonable framework in which to carry out your request for an "explicit" linear order of $(\mathbb{Z}_p,+)$.  Once we rule out the existence of such a linear order with the property of Baire, we're therefore forced to concede that no argument producing this order may be carried out in $\mathtt{ZF+DC}$, dashing our hopes of an explicit construction.

By the way, I focus on $(\mathbb{Z}_p, +)$ rather than $(\mathbb{Q}_p,+)$ simply for convenience.  It should be clear that any order of the latter induces an order of the former, so if anything the problem is harder for $\mathbb{Z}_p$.

There is no linear order of the additive group $(\mathbb{Z}_2,+)$ of $2$-adic integers which has the property of Baire (with respect to the usual Polish topology).  In particular, it is consistent with $\mathtt{ZF+DC}$ that no such order exists at all, so a large fragment of the axiom of choice is indeed required to build such an order.  Analogous arguments will work for all $\mathbb{Z}_p$, with slightly more obnoxious notation.

From here on we will identify elements of $\mathbb{Z}_2$ with infinite binary strings, that is, elements of $2^\omega$ with the product topology.  We define an equivalence relation $E_0$ on $2^\omega$ by setting two strings equivalent iff they differ in finitely many coordinates.  This $E_0$ has a nice interpretation in $\mathbb{Z}_2$: $x$ and $y$ are $E_0$ related iff their difference is a (standard) integer.  More precisely, $x E_0 y$ iff for some $n$, $x + 1 + 1 + \cdots + 1 (n \mbox{ times}) = y$ or vice-versa, where $1$ denotes the standard integer $1$ (i.e., the sequence $10000\ldots$).

(That last part isn't literally true, since the constant $1$ sequence plus $1$ equals the constant $0$ sequence.  But it is true off of the eventually constant sequences, which is enough to make the below argument go through (since there are only countably many eventually constant sequences).)

We use without proof two standard facts about $E_0$:

If $A \subseteq 2^\omega$ has the Baire property (from now on abbreviated BP) and meets each $E_0$-class in at most one point, then $A$ is meager (this is essentially the Vitali argument);
If $A \subseteq 2^\omega$ has the BP and is $E_0$-invariant (i.e., $x \in A$ and $x E_0 y$ implies $y \in A$), then $A$ is either meager or comeager (this is a form of ergodicity) (*).

Now, given a putative order $<$ with the BP, we can partition $2^\omega$ into three $E_0$-invariant BP pieces:

$X_- = \{x \in 2^\omega : \forall y (y E_0 x \implies y < 0)\}$;
$X_+ = \{x \in 2^\omega : \forall y (y E_0 x \implies y > 0)\}$;
$X_0 = 2^\omega \backslash (X_- \cup X_+)$

(here $0$ is the identity element of the group: the constant $0$ sequence).  So $X_-$ is the union of the $E_0$-classes which are entirely negative, $X_+$ the union of those entirely positive, and $X_0$ the union of those which are sometimes positive and sometimes negative.

(Technically, these pieces might not have the BP, but by Kuratowski-Ulam there's some element in $2^\omega$ we can use in place of $0$ to make the pieces have the BP.  For ease of notation, let's assume $0$ works.)

We first observe that $X_0$ is meager.  Note that the set $\{x : 0 \leq x < 1\}$ (which is meager by Fact 1) hits each $X_0$ class in exactly one point, so $X_0$ is the union of countably many homeomorphic translations (namely, the standard integer shifts) of a meager set, thus is meager.

Now let $f: 2^\omega \to 2^\omega$ denote the bitflipping homeomorphism, so $f(01001110\ldots) = 10110001\ldots$.  We note that $x \in A_-$ iff $f(x) \in A_+$, since $x + f(x) + 1 = 0$ for all $x$.  This means that $A_-$ cannot be comeager, else $A_+ = f[A_-]$ would be a disjoint comeager set.  But then by Fact 2, $A_-$ is meager, thus so is $A_+ = f[A_-]$, and consequently we've written $2^\omega$ as the union of three meager sets.  So we've hit a contradiction.

(*) By request, here is a reference for Fact 2: Theorem 3.2 of G. Hjorth: Classification and Orbit Equivalence Relations, Mathematical Surveys and Monographs, 75, American Mathematical Society, Providence, RI, 2000.  Although actually this theorem is overkill for this special case -- here's a sketch of a more elementary argument that works here.

Suppose that $B \subseteq 2^\omega$ is nonmeager; we want to show that $[B]_{E_0} = \{x: \exists y \in B\ (x E_0 y)\}$ is comeager.  By localization, there's a basic open set $U$ such that $B \cap U$ is comeager in $U$.  We can find a finite binary string $s$ let's say of length $n$ such that $U$ contains all elements of $2^\omega$ beginning with $s$.  Now look at the $2^n$ homeomorphisms of $2^\omega$ which flip some subset of the first $n$ bits of a string and leave the rest unchanged.  These maps send each $x$ to something $E_0$-related to $x$, so it follows that $[B]_{E_0}$ is comeager in the union of $U$'s images under these maps.  But the union of these images is all of $2^\omega$!

$1$ is either positive or negative.  Say it is positive (otherwise, reverse the order).  Then $0<1<2<3<...$.  But there is a subsequence of these that converges to $0$.  Impossible.


Candies withdrawal probability for a particular subsequence You are taking out candies one by one from a jar that has 10 red candies, 20 blue candies, and 30 green candies in it. What is the probability that there are at least 1 blue candy and 1 green candy left in the jar when you have taken out all the red candies?

Indeed.   We consider what would happen were we not to stop after drawing the last red candy.

The probability that the last of sixty candies drawn is one of the thirty green candies is $\tfrac{30}{60}$.   Ignoring the green candies, the probability that the last of the thirty other candies drawn is one of the twenty blue is $\tfrac{20}{30}$.

Likewise the probability that the last draw would be blue and the last not blue draw would be green is $\tfrac{20}{60}\tfrac{30}{40}$ .

Thus the probability that you seek is clearly: $\tfrac{30}{60}\tfrac{20}{30}+\tfrac{20}{60}\tfrac{30}{40} =\tfrac 7 {12}$

$P(T_r<T_b)$ is not that simple to calculate.

So I was counting paths. Basically for given number of $r,b,g$, we have $\frac{(r+b+g)!}{r!\cdot b!\cdot g!}$ paths. Denote number of all path where $T_r<T_b$ as $F(r,b,g)$, we have:
$$
F(r,b,g)=P_g(T_r<T_b)\frac{(r+b+g)!}{r!\cdot b!\cdot g!}
$$
and
$$
F(r,b,g)=F(r-1,b,g)+F(r,b-1,g)+F(r,b,g-1)
$$
We can easily verify that $P_g(T_r<T_b)\neq \frac{b}{r+b}$


Is the concatenation operation on a list associative? Is the concatenation operation on a list associative?

[] = empty list
l and l' = lists

Let's assume that we have to prove that:

l concat (l' concat []) = (l concat l') concat []

How can I prove it using a prof by induction?

I was thinking that:
base case is: [] concat l' = l'
Step case  is = (s:l) concat l' = l'

I am not sure if what I wrote is correct and if the base and step cases are correct.

Thank you

The easiest way to prove that concatenation is associative is using indices: prove that if $a,b,c$ are lists and $i$ is an index, then
$$((a\ \mathrm{concat}\ b)\ \mathrm {concat}\ c)[i]=(a\ \mathrm{concat}\ (b\ \mathrm {concat}\ c))[i]$$
Proving it this way does not require induction.


Evaluating $ \lim_{x \to 0^+} (\ln (x)+e^{\frac{-1}{x}})/( \frac{1}{x^2} + x\sin \left(\frac{-1}{x^2} \right))$ Any ideas?
$$ \lim_{x \to 0^+} \frac{\ln (x)+e^{\frac{-1}{x}}}{ \frac{1}{x^2} + x\sin \left(\frac{-1}{x^2} \right)}$$
L'Hopital didn't really help, maybe taylor?

Thanks

An idea

$$0\xleftarrow[0^+\leftarrow x]{}\frac{\log x}{\frac1{x^2}+1}\le\frac{\log x+e^{-1/x}}{\frac1{x^2}+x\sin\left(-\frac1{x^2}\right)}\le\frac{\log x+1}{\frac1{x^2}-1}\xrightarrow[x\to0^+]{}0$$

The right-hand and left-hand limits follow from applying l'Hospital, say:

$$(*)\stackrel{l'H}=\lim_{x\to0^+}\frac{\frac1x}{-\frac2{x^3}}=-\lim_{x\to0^+}\frac{x^2}2=0$$

and the inequalities follow since for $\;x\;$ pretty close to $\;0^+\;$ we have that

$$\begin{cases}0<e^{-1/x}<1\\{}\\-1\le-x\le x\sin\left(-\frac1{x^2}\right)\le x\le 1\end{cases}$$

Using KfSsOc's excellent hint, we can see that $\lim_{x \to 0^{+}} e^{-1/x} = 0$ and $\lim_{x\to 0^{+}} x \sin \left(-1/x^2\right) = 0$ Hence for small $x$ we have $$\frac{\ln x+e^{\frac{-1}{x}}}{ \frac{1}{x^2} + x\sin \left(\frac{-1}{x^2} \right)} \approx \frac{\ln x + 0}{1/x^2 + 0}$$

Hence we need only determine $$\lim_{x \to 0^{+}} \frac{\ln x}{1/x^2} = \lim_{x\to 0^{+}} -\frac{1/x}{2/x^3} = \lim_{x\to 0^{+}} \frac{x^2}{2} = 0$$

Where we used L'Hopital once above.


Least Squares understanding I am trying to understand the least squares method, I have found a simple enough example, but it doesn't fully explain the last part of it.

http://www.emathzone.com/tutorials/basic-statistics/example-method-of-least-squares.html

The part I am struggling with is this part

"Eliminate ‘a’ from equation (1) and (2), multiply equation (2) by 3 and subtract form equation (2), we get the values of ‘a’ and ‘b’."

First of all, why do you multiple by 3? Is this it just some arbitrary number that works in this least squares method. Secondly how do I go about this calculation. I have tried it, and I don't understand how I am meant to get 2 values. It does not show the working out, and I fail to see how you can get 2 values from a single subtraction. If I multiple equation 2 by 3, then subtract from equation 2, I would only get one value. How is 1.1 and 1.3 retrieved?

Math is not exactly my strongest point, but I need to understand it for task I am working on. Would someone be able to show me exactly how this solution was worked out

Thanks in advance

OK Adam, if you go through what the Least Squares actually does you end up needing to find two numbers $a$ and $b$ that satisfy the simultaneous equations

$$\begin{align}5a+15b&=25\\ 15a+55b&=88.\end{align}$$
Now there are many different ways of solving such a pair of equations (a linear system if you will). Here I will show you the way you might have solved them in school all those years ago.

First we need to understand what these equations are saying. When I write down

$$2x+3=11,$$

what we do first is we assume that $x$ solves the equation.

The second thing you need to realise is that if you have an equation, for example,

$$2x+3=11,$$
then you are allowed to add something to both sides... because when two numbers are equal, for example five and five, then adding three to both of them leaves you with two numbers that are still equal. So for example you could add $-3$ to both sides of $2x+3=11$ and you get

$$2x=8,$$

and these two numbers are still equal.

We are also able to multiply both sides by the same thing and we will end up with numbers that are still equal... in fact, if we are careful, we can do a lot of different things to the two numbers that are equal and they will remain equal. If we multiply both sides by a half here we end up with

$$x=4.$$

So what we do is we assume that $x$ solves the original equation and we are lead then to the conclusion that $x$ must equal four and indeed $2x+3=2(4)+3=8+3=11$.

The next thing we need to realise is that if you have two equations, then you can add these equations together. For example suppose I have
$$\begin{align}
2x&=1
\\y&=0.
\end{align}$$
What I can do is add $y$ to $2x$ and add zero to one. Why? Well because $y$ is equal to zero I am actually only adding the same thing to both sides.

Now if we look at our equations

$$\begin{align}
5a+15b&=25
\\ 15a+55b&=88,
\end{align}$$
the problem is we have two numbers to find... i.e. we assume that $a$ and $b$ solve both equations and we want to go off and conclude that they must equal this or that...

The next problem is that, in general, one equation in two unknowns can have an infinite number of solutions. For example, if $x$ and $y$ are whole numbers, then

$$x+y=0,$$
has infinite number of solutions. You can have $x=1$, $y=-1$, $x=2$, $y=-2$, etc. Now the way you want to think is... if only I could find an equation in only one unknown. You are able to do that as follows.

Try and match up the number of $a$s (or $b$s), by multiplying both sides of the first equation by three:

$$3(5a+15b)=3(25).$$

Remember. $5a+15b$ is equal to 25 so if you multiply it by three you get 75. Now you need the little fact that $x(y+z)=xy+xz$... I assume and hope this is OK for you:

$$15a+45b=75$$.

Now we have that these are equal and allied to the second equation we know have

$$\begin{align}
15a+45b&=75
\\ 15a+55b&=88.
\end{align}$$
Similarly to equation one, just to make things nice for us later, we can multiply both sides of the second equation by minus one to get the two pairs of equal numbers:

$$\begin{align}
15a+45b&=75
\\ -15a-55b&=-88.
\end{align}$$

Now if we add the equations together --- i.e. add -88 to 75 and an equal thing ($-15a-55b$) to $15a+45b$ and then we get another pair of equal numbers (the $\Rightarrow$ stands for 'means that' or 'implies that'):

$$\begin{align}
15a+45b+(-15a-55b)&=75+(-88)
\\ \Rightarrow -10b&=-13
\end{align}$$
Now multiply both sides by minus one:

$$10b=13,$$
and multiply both sides by $\frac{1}{10}$:

$$b=\frac{13}{10}=1.3.$$

Great! We know $b$. Now we know that the numbers $5a+15b$ and $25$ are equal from the very start... but $b$ is equal to 1.3 so that:

$$5a+15(1.3)=25.$$

I will leave the rest up to you.

Make sure to end by presenting your answer nicely with

$$Y(X)=a+bX.$$

In general, you end up with two equations, of the form

$$
C = na + rb \\
D = ra + sb
$$

Here $n$ is the number of points, $r$ is the sum of the $x$-values, and $s$ is the sum of the squared $x$-values. In the example, $n = 5$, $r = 15$, and $s = 55$. I'm using letters rather than numbers so that you can see the general pattern.

I'm going to show how to perform algebraic operations on these two equations until we have formulas that express $a$ and $b$ in terms of $n$, $r$, and $s$; these formulas are the only ones you then need.

What I'm going to do is to multiply each equation by a constant, like changing $1 = 3 - 2$ into $4 = 12 - 8$ by multiplying everything by 4. Doing so doesn't change the solutions of the equations, as long as the constant I'm using isn't zero, but it can help make them easier to work with.

First, I'll multiply the top one by $s$:

$$
sC = sna + rs b \\
D = ra + sb
$$

That's allowed, because the number $s$, which is the sum of the squares of a bunch of different numbers, will always be positive, so it's not zero.

Now I'm going to multiply the bottom equation by $r$, which might be zero. If it is, the new equation that I'll get will be just $0 = 0$, in disguise. So I'll keep both the new equation and the old one, to make sure I haven't lost any information:

\begin{align}
sC &= sna + rs b \\
D &= ra + sb\\
rD &= r^2a + rsb
\end{align}

Now I can subtract the bottom equation from the top one, to get this
\begin{align}
sC - rD &= (sna - r^2a)  + rs b - rsb\\
D &= ra + sb.
\end{align}

Notice how the two $rsa$ terms in the first equation cancel -- that's why I chose $s$ and $r$ as multipliers in the earlier steps. So now we have

\begin{align}
sC - rD &= (sn - r^2)a\\
D &= ra + sb.
\end{align}

From the first equation, we can solve for $a$ by dividing both sides by $r^2 - ns$. That's only valid if $r^2 - ns$ is nonzero. It's a slightly tricky fact about the least-squares set-up that this number will always be nonzero, as long as you have at least two distinct $x$-values. (If you had only one, it'd be tough to choose a single line to "best" fit through it, wouldn't it?) Anyhow, here's the result:

\begin{align}
\frac{sC - rD}{sn - r^2} &= a\\
D &= ra + sb.
\end{align}

Now that we have a formula for $a$, in the second equation, the values $D, r, a,$ and $s$ are all known, so we can find $b$. To do that in algebra, I'll bring the $ra$ to the other side:

\begin{align}
\frac{sC - rD}{sn - r^2} &= a\\
D -ra &= sb.
\end{align}
and then divide through by $s$ (which is nonzero, so this is OK):
\begin{align}
\frac{sC - rD}{sn - r^2} &= a\\
\frac{D -ra}{s} &= b.
\end{align}

Now we can substitute the answer we got for $a$ into the second equation to get a final result:

\begin{align}
\frac{sC - rD}{sn - r^2} &= a\\
\frac{D -r\frac{sC - rD}{sn - r^2}}{s} &= b
\end{align}
which can be simplified a little bit (with some algebraic steps I'm not going to go through) to get

\begin{align}
\frac{sC - rD}{sn - r^2} &= a\\
\frac{-rC +nD}{sn - r^2} &= b
\end{align}

And that is the general solution for any least squares problem. Written out in terms of the things in the original article, you get

\begin{align}
Q = n(\sum x_i^2) - (\sum x_i)^2 \\
\frac{(\sum x_i^2)(\sum y_i) - (\sum x_i)(\sum y_i^2)}{Q} &= a\\
\frac{-(\sum x_i)(\sum y_i) + n(\sum x_i y_i)}{Q} &= b
\end{align}
where I've defined the number $Q$ to make the second and third lines a little simpler.

Just to finish this off, let's look at the original problem, where
$$
r = 15 \\
s = 55 \\
n = 5 \\
C = 25 \\
D = 88
$$

Then
\begin{align}
\frac{sC - rD}{sn - r^2} &= a\\
\frac{-rC +nD}{sn - r^2} &= b
\end{align}

becomes

\begin{align}
\frac{55 \cdot 25 - 15 \cdot 88}{55 \cdot 5 - 15^2} &= a\\
\frac{-15 \cdot 25 +5 \cdot 88}{55 \cdot 5 - 15^2} &= b
\end{align}
which simplifies to
\begin{align}
\frac{55}{50} &= a\\
\frac{65}{50} &= b
\end{align}
\begin{align}
1.1 &= a\\
1.3 &= b
\end{align}


Prove that this function is bounded. I am trying to prove that this function is bounded for all $x$ $$\frac{e^{-ix}-1}{x}$$

I found that :

$|\frac{e^{-ix}-1}{x}| \le \frac{|cos(x)|}{|x|}+\frac{|sin(x)|}{|x|}+\frac{1}{|x|}$

So when $x \rightarrow (-)\infty $, the function goes to $0$

and in $x=0$, we have that : $\lim_{x \to 0} |\frac{e^{-ix}-1}{x}| = \lim_{x \to 0} |\frac{e^{-ix}-e^(i0)}{x-0}| = (e^{-ix})'(0) = 0$

and by continuity of the function in $]0,\infty[$ it cannot diverge between $0$ and $(-) \infty$, so it must be bounded

Is it correct ?

I'll write $f(x)= \frac{e^{-ix}-1}{x}$, $x \in \mathbb{R} \backslash \{0\}$. As you've noted, a bound like
$$|\frac{e^{-ix}-1}{x}| \leq \frac{|e^{ix}|+1}{|x|} = \frac{2}{|x|}$$
implies that $f$ is bounded at infinity. In particular, say, $|f(x)| \leq 2$ for $|x| \geq 1$.
In addition, you've effectively noted that
$$\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{e^{-ix}-e^{-i0}}{x-0} = -ie^{-ix}|_{x=0} = -i,$$
and this implies that $\exists \epsilon > 0$ such that $|x| < \epsilon \implies |f(x)+i| < 1$, which by the triangle inequality implies $|f(x)| \leq 2$.

So, $|f(x)| \leq 2$ for $|x|<\epsilon$ and $|x|>1$. Meanwhile, the set $A=\{x \in \mathbb{R} \backslash \{0\} : \epsilon \leq |x| \leq 1\}$ is compact, so continuity of $x \mapsto |f(x)|$ implies that it achieves its maximum value $M \geq 0$ on $A$. Hence, $\forall x \in \mathbb{R} \backslash \{0\}$ we have the bound $|f(x)| \leq M+2$.

A similar argument yields that any continuous function on $\mathbb{R}^n \backslash \{p\}$ for which the limits as $x \to \infty$ and $x \to p$ exist is bounded. More generally, the existence of the $x \to \infty$ limit may be relaxed to $f$ being bounded outside of a sufficiently large sphere.

Hint: $$ \left|e^{ix} - 1\right|\le \min(2, |x|) $$


How many resulting regions if we partition $\mathbb{R}^m$ with $n$ hyperplanes? This is a generalization of this question. So in $\mathbb{R}^2$, the problem is illustrated like so:

Here, $n = 3$ lines divides $\mathbb{R}^2$ into $N_2=7$ regions. For general $n$ in the case of $\mathbb{R}^2$, the number of regions $N_2$ is $\binom{n+1}{2}+1$. But what about if we consider the case of $\mathbb{R}^m$, partitioned using $n$ hyperplanes?

Is the answer $N_m$ still $\binom{n+1}{2}+1$, or will it be a function of $m$?

Hint: Let's consider how to prove that in $\mathbb{R}^2$, the maximum number of distinct regions is $\frac { n^2 + n + 2 } { 2} $.

First, since there are finitely many lines, we can tilt any configuration such that no line is parallel to the horizontal. Now, let's consider the 'lowest' point of any region.

If the region is bound below, than such a point must exist, is unique, and is a point of intersection of 2 lines. In fact, there is a one-to-one relation between regions that are bounded below, and these points of intersections, which gives us $ { n \choose 2 } = \frac{n^2-n} {2} $ regions.

What happens if the regions are not bounded below? How many are there? Well, let's simply insert a horizontal line way down below, and count the number of unbounded regions, by associating them to these new bounded regions. Now, let's consider the 'lowest, leftmost' point of any region. If the region is bound to the left, then such a point must exist, is unique, and is appoint of intersection of the original $n$ lines with the new horizontal line. In fact, there is a one-to-one relation between regions that are unbounded below, bounded to the left, and these points of intersections, which gives us ${n \choose 1} = n$ regions.

Finally, how many regions are not bounded below, and not bounded to the left? You should convince yourself that there is ${n \choose 0} = 1 $ regions.

Hence, the total number of ways is ${n \choose 2} + {n\choose 1} + {n \choose 0}$.

Now, you can easily see how this shows that the number of regions for the hyperplane version of your problem is $$\sum_{i=0}^m {n \choose i}. $$

I will try to solve the problem in terms of "how many regions add if we insert an extra hyperplane L (Let) in a space of     k-1 hyperplanes in n-dimensions".

Let T (n,k) be the maximum number of regions formed in R^n by k hyper-planes.

Now we need to find T (n,k) - T(n,k-1)=S (Let).

Infact the answer will come out to be  T (n-1,k-1).

Let us see how.

I have tried to give a purely Geometrical argument.

Some steps:

The number S is actually the maximum number of regions that  L can pass through.[Because every region it passes through divides into 2 and hence the count for that region is 1]
Now , After  drawing the new hyperplane, in your mind's eye ,consider only the intersections of L with the remaining (k-1) hyperplanes on L.
So, L which is a (n-1) dimensional Space now have a set of (k-1) intersections [ so (k-2) dimensional lines drawn on it.]

The next step is the crucial one.
4. The number of regions formed by the (k-1) lines in on  L (n-1 dimensional space) is bijective to the number of regions L passes through.

So only we need to verify why 4 happens.Others are easy to infer.

Now every region formed by the (k-1) hyperplanes in R^n has a boundary of atleast 2 hyperplanes.

Now consider L ,as L passes through such a region.Observe what happens on L, at least 2  (k-2) dimesional lines occur on L.

Now to maximize the number of regions L passes through it will  intersect with all the (k-1) hyperplanes.

So consider what happens when L enters one region to another.

New lines appear.

Call 2 regions adjacent in R^n if they have atleast one common hyperplane.

Now,See that L always passes through one region to its adjacent region.
As a result the intersections all give rise to adjacent regions.

And as the  passes from one region to other the one- one correspondence with regions formed by the lines on L and the regions L passes through happens (due to this adjacent behaviour of the regions,L passes through)

So the maximum number of regions that the (k-1) lines on L form is actually the number T (n-1,k-1),because L is n-1 dimensional and the lines on L are its k-1 hyperplanes.

Now we need to define T (0,k) and T (k,0).

Define T (0,k)=T (k,0)=1.(quite natural)

This seems like a  kind of a hand waving argument because we cant actaully visualize n>4.

But this can be easily checked in 3 dimensions and 2 dimensions.

Note: I have mentioned lines and planes to let you visulaize in 3 dimesions.

NOTE:

Now in accordance with the previous answer we know that there exists a similar recursion in the combinatorial domain.

If we solve the recursion,it comes out to be the same as above.

Moreover you can substitute and check that it holds.

Will like to hear how to verify 4 in other different ways in the comments!

Will like to hear any good suggestions to this answer!

I am new to this ,If you want to downvote my answer (please mention the reason,otherwise I will  repeat the same mistakes again).


Evaluate $\lim_{n\to\infty} \frac{b_n}{n}$ define a sequence $a_n = (1-a_{n-1})(1-a_{[n/2]})$ where $[k]$ represents the largest integer $s\leq k$ and $a_1 = 1$. Furthermore, define $b_n = \sum_{k = 1}^{n} a_k$. Evaluate $$\lim_{n\to\infty} \frac{b_n}{n}.$$
For now I have been able to prove by contradiction that $a_k = 1 \implies k \equiv 0$ (mod 2) and by writing a simple python program I realised the expression seems to tend to $1/3$. How can this result be proven mathematically?

This is not an answer, just notes too long for comments. You have already proved that the odd terms (except from 1) are zero. So, using the recurrence, you can compute the even terms as
$$
a_{2s} = (1-\underbrace{a_{2s-1}}_{=0})(1- a_s) = 1 - a_s
$$
Every even number can be written, in a unique way, in the form $n = 2^k m$, where $m$ is an odd number. Let's look at a few examples:

$10 = 2^1 \times 5$:
$$
a_{10} = a_{2\times 5} = 1-a_5 = 1-0 = 1
$$

$20 = 2^2\times 5$:
$$
a_{20} = a_{2\times 10} = 1 - a_{10} = 1 -a_{2\times 5} = 1 - (1-a_5) = 0
$$

$8 = 2^3 \times 1$:
$$
a_{2^3} = 1-a_4 = 1-(1-a_2) = 1-(1-(1-a_1))) = 0
$$

$16 = 2^4 \times 1$:
$$
a_{16} = 1 - a_8 = 1-0 = 1
$$

So, the power $k$ tells you how many times you flip between 1 and 0.
In general, if $m \ne 1$, $a_{2^k m} = \frac{1 - (-1)^k}{2}$. When $m=1$, $a_{2^k} = \frac{1+(-1)^k}{2}$.
The nonzero terms are exactly the ones for which $ n = 2^{2k-1} m$, with odd $m$.
With these tools we are able to compute $b_n$ for any $n$. If $b_n$ converges, its limit will coincide with the limit of any subsequence. for instance, for $k \ge 2$ you can establish that
$$
b_{2^{2k}} = \frac{1}{2^{2k}}\sum_{n=1}^{2^{2k}} a_n =
\frac{1}{2^{2k}}(\frac 13 4^{k}+\frac 23)  \to \frac 13  $$
It remains to show that the sequence converges.

Observe $a_n$ series peacefully that series look like as shown below:
$10010100010001010100010101000100010001010100010001000101010001010...$
So when we will be solving for limit $n→∞$ it should be equal to $W/n$ where $W$ is the total number of '1' that occurred in the series so our aim is to generate a formula that can help us to find $W$.
We will try different values of n to understand how limit approaches to 1/3 as shown below:
TRIAL 1 $n=10$ then W=4 hence $lim_{n→∞}{{b_n/n}} $ approximate to 4/10
TRIAL 2 $n=100$ then W=34 hence $lim_{n→∞}{{b_n/n}} $ approximate to 34/100
TRIAL 3 $n=1000$ then W=334 hence $lim_{n→∞}{{b_n/n}} $ approximate to 334/1000
TRIAL 4 $n=10000$ then W=3333 hence $lim_{n→∞}{{b_n/n}} $ approximate to 3333/10000
TRIAL 5 $n=100000$ then W=33335 hence $lim_{n→∞}{{b_n/n}} $ approximate to 33335/100000
Ok so now we understood why answer is $0.3333333333333333333333.....$ for n approaching to infinity.
Now assume you have calculated out $f(n)$ which help us to find $W$ so divide it by n to get your answer :).
Try for $f(n)$ if you are not getting I am still here to help in the comment section.


All possible combinations of pixels in an image. Forgive me if this is a rudimentary question but It's been about 4 years since I took stats, so it's pretty much all gone.

Say I have an image which is nxm pixels, and each pixel has 3 layers (RGB) that have 256 levels each, how do I calculate all possible combinations of images representable in this format?

Is it simply (I doubt it):

$$ 256^3 \cdot n \cdot m $$

Or something like:

$$ \frac{(nm)!}{256!256!256!}$$

Thanks in advance.

Each layer has $256$ possible values. There are $3$ layers per pixel, and there are $nm$ pixels. This gives you

$$256^{3mn}.$$

$256^{3+m+n}$ since there are $256^3$ possible colors in each pixel


"Modus moron" rule of inference? This is an exercise I got from the book "First Order Mathematical Logic" by Angelo Margaris (1967). I have never heard of this rule before, the question is whether what Margaris calls the modus moron rule of inference is correct or not and to explain why I think so.

$$\frac{P\Rightarrow Q, Q}{\therefore P}\qquad \text{(modus moron)}$$

It seems correct to me, my reasoning is that if $P\Rightarrow Q$ and $Q$ it does not matter whether $P$ or $\neg P$ since a false antecedent makes a true conditional, which I would show by the rows of the truth table of $(P\Rightarrow Q)$ where $Q$ is true.

Is this a valid argument?

It's true that if $P$ is false then $P\Rightarrow Q$ is true. But the question is not asking if $P\Rightarrow Q$ is true, it's asking you if you can infer $P$ from $P\Rightarrow Q$ and $Q$.

Let's be concrete. Suppose "if Mark is drunk, then Mark is happy" is a true statement, because Mark is a happy drunk. Given that Mark is presently happy, may we infer that Mark is drunk? No; there may be other circumstances in which Mark is happy besides being drunk.

(This example is drawn from a real life discussion between friends about putative alcoholism which devolved into a debate about logical implication.)

This will, in all likelihood, be drowned out by the other answers. But, a mathematical version of this, is as follows:

$$p>3,p\in \mathbb P\implies p\in\{x:\text{Mod}(x,6)\in\{1,5\}\},25\in\{x:\text{Mod}(x,6)\in\{1,5\}\}\over\therefore 25\in\mathbb P$$


What's the difference between $E(K)$ and $E/K$ in elliptic curves? For a field $K$, sometimes I see elliptic curves denoted as $$E(K) : y^2 = x^3 + ax+b$$ and other times as $$E/K :y^2 = x^3 + ax+b$$

What is the difference? I believe that $E/K$ denotes that the coefficients $a,b$ are pulled from the field $K$ only, whereas in $E(K)$ they may be pulled from the algebraic closure of $K$. Is this correct? I.e. if $\bar{K}$ is the algebraic closure, then $E(K) = E/\bar{K}$ ? Also, is the $E/K$ notation at all related to the notation used for extension fields or quotient groups?

The notation $E/K$ means that $E$ is an elliptic curve defined over the
field $K$.

The notation $E(K)$ refers to the group formed by the points on $E$
with coordinates in the field $K$.

The set of points of $E $ is a group,  with addition defined geometrically. ..$E (K)=\{(x,y)\in E|x,y \in K \}$ , for a field K, is a subgroup of  $E $,by a theorem of poincaré...(my interpretation is that  $E $ is originally taken over a field $L $, where $L $ is an extension of  $K $.)


Problem in Divisibility and Functions Suppose that $ n $ is an integer, $ A $ is the set of $ n $'s divisors and $ f :A\rightarrow A $ be a function with this property: if $ a $ and $ b $ belong to $ A $ and $ a\mid b $, then $ f (a)\mid f (b) $. Prove that there is an integer $m$ in $ A $ that $f (m)=m $.

If $n$ is a prime number than it is clear, because either $f$ is constant or the identity.

If $n$ is not a prime number:Assume that f does not have a fixed point and  let $f(1) = m$. Then we can define $f'$ for $n/m$ by $f'(k) = f(k)/m$ for all divisors $k$ of $n/m$. Since $f$ has no fixed point, also $f'$ has no fixed point. Inductively we will arrive at a prime and get a contradiction.

I'm assuming that you intended to say that "... there is an integer $m$ ...", as @BrianTung pointed out.

However, the statement then isn't true.  Take $n=1,$ so that $A = \lbrace 1,-1 \rbrace.$ Define $f(x)=-x$ for a counterexample.

I suspect that the problem was intended to be about positive integers.


Help with proof that $ \sum_{i=0}^k{k \choose i}{N-k \choose n-i} = {N \choose n} $ I would appreciate if somebody could give me a hint for the following problem. I'm seeking a combinatorial proof for:
$ \sum_{i=0}^k{k \choose i}{N-k \choose n-i} = {N \choose n} $

Suppose  $k \leq N-k $. Then if you want to choose $k$ people, you can choose $i$ men in $\displaystyle{k\choose i}$ ways and $(k-i)$ women in $\displaystyle{N-k\choose k-i}$ ways , for $i\in\{0,1,\dots n\}.$
Then we have $$\sum_{i=0}^{k}{k\choose i}{N-k\choose k-i}={N\choose k}.$$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{i = 0}^{k}{k \choose i}{N - k \choose n - i} & =
\sum_{i = 0}^{k}{k \choose i}\bracks{z^{n - i}}\pars{1 + z}^{N - k} =
\bracks{z^{n}}\pars{1 + z}^{N - k}\sum_{i = 0}^{k}{k \choose i}z^{i}
\\[5mm] & =
\bracks{z^{n}}\pars{1 + z}^{N - k}\pars{1 + z}^{k} =
\bracks{z^{n}}\pars{1 + z}^{N} = \bbx{N \choose n}
\end{align}


Factoring a cubic equation but I want a way to figure out the factor of the function $(5-x)(x-2)(x-2)-4=0$
I want to expand this, subtract the four from the constant then factor this again. How do I find the linear factor? Are there easier ways to do this?
If it helps, the original question is that we have two equations $y=4/(x-2)^2$ and $y=5-x.$ One intersection is at point $(1,4)$, so I have to find the coordinates of the other points of intersection.

Because we are searching were $y=4/(x−2)^2$ and $y=5−x$ intersects, we have the equation: $5-x- \dfrac{4}{(x−2)^2} =0$, which is equal to $\dfrac{(5-x)(x-2)^2 -4}{(x-2)^2} =0$ expanding this we get that this is also equal to $\dfrac{-(x^3-9x^2+24x-16)}{(x-2)^2}$ as we can see, x cannot be equal to 2, and also, this fraction only will be 0 if the numerator is zero, so, we are searching where $x^3-9x^2+24x-16=0$, as we know, 1 is a solution, so we can factor this as $(x-1)(ax^2+bx+c)$. As we can obtain, $a = 1$, $b = -8$ and $c= 16$ so, finally, we have the equation $(x-1)(x^2-8x+16)$. Now, looking at the roots of the quadratic part, we see that 4 is the only solution, so $y=4/(x−2)^2$ and $y=5−x$ intersects at $(1,4)$ and $(4,1)$.

Expand $(5-x)(x-2)(x-2)-4=0$ and use the fact that $x=1$ is solution.
$$\begin{aligned}(5-x)(x-2)(x-2)-4=&-x^3+9x^2-24x+16\\=&-(x-1)(x^2-8x+16)\\=&-(x-1)(x-4)^2\end{aligned}$$
The second point has coordinates $(4,1)$ and it is a point of tangency due to the square (it is a double point of contact).


Limit of cosines product Let $\displaystyle P_n:=\prod_{k=1}^n \cos\frac{\pi}{k+2}$. Evaluate $\displaystyle\lim_{n\to \infty} P_n$.

I've only shown that the limit is positive.

Let $\vartheta_k:=\pi/(k+2)$. We have $\log P_n=\sum_{k=1}^n \log\cos\vartheta_k$. Now,
$$\log\cos\vartheta_k=-\frac{1}{2}\log(1+\tan^2\vartheta_k)>-\frac{1}{2}\tan^2\vartheta_k=-\frac{1}{2}\frac{\sin^2\vartheta_k}{\cos^2\vartheta_k}>-2\vartheta_k^2.$$
Therefore we can write that
$$\log P_n>-2\pi^2\sum_{k=1}^n \frac{1}{(k+2)^2}>-2\pi^2\int_2^{\infty}\frac{dx}{x^2}=-\pi^2$$
i.e. $P_n>e^{-\pi^2}$.

We may use the product for $\sin x$ and $\cos x$ to conclude that
\begin{align}
\prod_{n = 1}^\infty \cos\frac{x}{n} &= \prod_{n = 1}^\infty \prod_{k = 0}^\infty \left( 1 - \frac{4x^2}{(2k + 1)^2 n^2 \pi^2} \right)\\
&= \prod_{k = 0}^\infty \prod_{n = 1}^\infty \left( 1-\frac{4x^2}{(2k + 1)^2 n^2 \pi^2} \right)\\
&= \prod_{k = 0}^\infty \frac{\sin\frac{2x}{2k + 1}}{\frac{2x}{2k + 1}}
\end{align}
which could be used to approximate your product.


Convergent Sequence from Introduction to Analysis Consider the sequence of real numbers
  $$\frac 12, \cfrac 1{2+\cfrac 1 2}, \cfrac 1{2+\cfrac 1{2+\cfrac 12}}, \ldots.$$

  Show that this sequence is convergent and find its limit by first showing that the two sequences of alternate terms are monotonic and finding their limits.

(From Introduction to Analysis by Maxwell Rosenlicht)

I am not sure if I have burned out or what, but I cannot do this. Any help would be much appreciated.

So do we have

$$x_{n+1} = {1\over 2 + x_n}?$$

If so, I can edit this answer to produce a solution.
It appears so.

So, in the steady state, we have $x_n = x_{n+1}$. We solve this equation

$$t= {1\over (2 + t)}.$$

Multiplying,
$$ t^2 + 2t = 1.$$
so
$$ t^2 + 2t -1 = 0.$$
Availing ourselves of the quadratic formula we get
$$ t = {-2\pm \sqrt{4 + 4}\over 2} = -1\pm \sqrt{2}.$$
In this case, you will be attracted to the positive root $\sqrt{2} - 1$.


Hilbert's Hotel, cardinality, and equality One of the absurdities illustrated by Hilbert's Hotel, some say, is infinity + infinity = infinity. This is absurd in the sense that "After the infinitely many new guests check in, the number of guests will remain the same as before." I think what is meant by "the number of guests is the same as before" is that the infinities in question have the same cardinality - the same number of guests before and after.

Question: Is it true that if two infinite sets have the same cardinality, they are equal? Or is it possible that two sets have the same cardinality, yet have different numbers of members?

If it is true that infinity + infinity = some greater infinity, in what sense is the claim true that "After the infinitely many new guests check in, the number of guests will remain the same as before."

Note: The context of this question is a discussion on an objection to Hilbert's Hotel on Philosofize

The term "infinity" means nothing in modern mathematics. It is just a term which represents "not finite".

In the context of the Hilbert hotel, and it seems that you discuss this in the context of infinite cardinals, rather than ordinals, or some other notion of infinity, infinities are cardinals. So when you say that "two infinities have the same cardinality they are equal" you really just say "two infinites are equal then they are equal".

On the other hand, if you mean infinities in the sense of ordinal numbers, which are another form of infinitary numbers, then addition is not idempotent. $\alpha+\alpha\neq\alpha$. But ordinals carry more structure than cardinals, and the addition of ordinals is different than the addition of cardinals because of that extra structure.

Glancing over the comments in the "objection" that you have linked, it seems to me that a particularly important point is amiss:

The notion of "number" simply means some "measurable quota" in some aspect. Cardinality of sets is a way to measure size of sets in a particular way. So cardinal numbers are numbers. It is true that those are not real numbers, but this is why we don't use the term "infinities" so wildly, and we tend to be very clear in what sort of context it is applied.

If one thinks of numbers as real numbers, and infinity as the infinity we know from calculus (a length which is longer than any finite length), then of course there is merits to this apparent objections. But this is not what Hilbert was trying to show when he opened his grand hotel to the unsuspecting public. No, he was talking about cardinal numbers which are a whole other system of numbers.

According to the conventional definition, the number of elements in a set is defined to be its cardinality. If you want to insist that the number of elements in a set is always larger than the number of elements in a proper subset of it, we have to break the assumption that the number of elements in any set is its cardinality. I think there is a consistent system that does that.


Can you find the resultant force between these two vectors? Determine the magnitude of the resultant force on an object if force $A$ is pulling the object with $150$ lbs of force and force $B$ is pulling with $300$ lbs, and the angle between the two forces is $110^\circ$.

1) Draw a diagram with coordinate axes:

2) Determine the components of the two forces:

$${F_A}_x = F_A = 150 lbs \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space {F_A}_y = 0$$
$$\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space{F_B}_x = -F_B\cos70° = -103 lbs  \space\space\space\space\space\space\space\space\space\space{F_B}_y = F_B\sin70° = 282 lbs$$
3) Find the components of the resultant force
$$R_x = {F_A}_x + {F_B}_x = 47 lbs$$
$$R_y = {F_A}_y + {F_B}_y = 282lbs$$
4) Apply Pythagoras' theorem to get the magnitude of the resultant:
$$R = \sqrt {R_x^2 + R_y^2} = 286lbs$$

If you want to know more, here's a good article which explains each step in more detail: How to find the resultant force acting on an object

If we assume $f_A$ points in the positive $x$ direction and $f_B$ is located in quadrant II, we have,

$$f_A =
\left[ \begin{array}{c}
150\\
0 \end{array} \right]$$

and

$$f_B =
\left[ \begin{array}{c}
300\cos(110) \\
300\sin(110) \end{array} \right]$$

The resultant is

$$f_R = f_A + f_B =
\left[ \begin{array}{c}
150 + 300\cos(110) \\
300\sin(110) \end{array} \right]$$

then

$$\|f_R\| = \sqrt{(150 + 300\cos(110))^2 + (300\sin(110))^2}=285.86$$

Note:  Notice that $\cos(110)$ is negative.


Prove that $23$ does not divide $2^n + 3^m$, for any $m, n \in \mathbb{N}$ Prove that $23$ does not divide $2^n + 3^m$, for any $m, n \in \mathbb{N}$.

I've tried to prove that the equation $2^n + 3^m = 0$ has no solutions in $\mathbb{Z}_{23}$, but didn't succed.

Thank you!

Hint:  Render $2\equiv 5^2$, $3\equiv 7^2$ mod $23$.  Can a sum of nonzero squares have zero residue mod $23$ when $23$ is a $4k+3$ prime?
Another approach, if the above is "slightly overlevel" as in the comments:
If $2^n+3^m$ were divisible by $23$ then so is $8^n+27^m$ as $a+b$ divides $a^3+b^3$.  Then $8^n+4^m=2^{3n}+2^{2m}$ would also be a multiple of $23$.  Through division by the smaller power of $2$ in $\{8^n,4^m\}$ we find that $2^k+1$ should be a multiple of $23$ for some whole number $k(=|3n-2m|)$.  Now try to find a value of $k$ that makes $2^k\equiv-1\bmod 23$.

There are only 11 modului that $2^n$ and $3^n$ can be, and the other remaining 11 modulii are the 11 numbers required to add be added to give 23.

The fact that 23 is 3, mod 4, but is -1, mod 24, means that if $2^m+3^n$ gives a value, mod 23, then the negative is not in that this form,

The same applies to any prime that gives 23 mod 24, such as 47, 71, 167, ...


Proving that if $ab=e$ then $ba=e$ Suppose that instead of the property $ab=ba=e$ a group G has the condition that for every element $a$ there exists an element $b$, such that $ab=e$. Prove that $ba=e$. Is the following a valid proof?

Since $ab=e$ then under the condition of the group there exists an element $k$ such that $bk=e$ for some $k$ in the group.

Now $bk=e$ so $abk=ae$ therefore $(ab)k=a$ and finally $ek=a$ and $k=a$.

Is this a valid proof?

I assume that you mean if for SOME $a,b$ (not every) $ab = e$ then $ba = e$.
Your proof  is valid, but you could write it much easier without playing with $k$. $$ab = e \Rightarrow  bab = b.$$ If you already know the cancellation law, then we are done.  Otherwise you may continue by writing $baba = ba$, so that $(ba)^2 =ba$. Just note that the only idempotent in a group is $e$.

Theorem $(A)$: For every $\{a, b\} \subset \mathbb{R}, \ ab = c \ $ and $\ a \neq 0$
$$\implies b = \frac{c}{a}$$
Proof: $$\begin{align} ab &= c \\ \iff \bigg(\frac{1}{a}\bigg)(ab) &= \bigg(\frac{1}{a}\bigg)c \\ \iff \bigg(\frac{1}{a} \times a\bigg)b &= \bigg(\frac{1}{a}\bigg)c \\ \iff 1\times b &= \bigg(\frac{1}{a}\bigg)c \\ \iff b &= \bigg(\frac{1}{a}\bigg)c \\ \iff b &= \frac{c}{a} \end{align}$$

$(A)$ is derived from the multiplicative cancellation law, which proves that for every $\{a, b, c\} \subset \mathbb{R}, \ ac = bc \ $ and $\ c \neq 0 \Rightarrow a = b$. So now all you have to do is substitute $c = e$ and $b = a$ in the proof of $(A)$ and you can prove that $\ ab = e \Rightarrow ba = e \Rightarrow ab = ba$.


how to calculate integral of square of a function When doing differentiation, I know that if $f$ is a function on $x$, then

$$ { d \over dx } f^2 = 2 f {df \over dx} $$

so the opposite in integration is also clear:

$$ \int 2 f { df \over dx } dx = f^2 $$

I also know that

$$ \int x^2 dx = { x^3 \over 3} $$

But I'm not sure as to how I can evaluate:

$$ \int f^2 dx $$

I mean is there any identity for this? That the above is equal to another function of $f$ (such as $f^3 \over 3$ times something)? Is there any method to find this? I googled some but perhaps I wasn't using proper search terms so I didn't get any clear results so I'm asking here. [I hope my question is clear enough :-(]

Can't be done, in general. For example, it is easy to do $$\int xe^{x^2}\,dx$$ but there is no expression for $$\int x^2e^{2x^2}\,dx$$ in terms of the familiar functions of undergraduate mathematics.

i suppose if you substitute the function as any arbitrary variable X u can solve this integration. for example while integrating (1+x)^2 you substitute 1+x=t then differentiate both with respect to t. this gives you dx=dt. here, no matter what your function, make dx the subject of the equation so that you can replace dx in your original integral. after that you integrate the function as integration of t^2 which is t^3/3 and then resubstitute t=1+x in the answer to get your answer.This is of course a very simple example but the process is the same.


Prove: for $x$ large enough, there are more primes in the interval $(1,x]$ than in $(x,2x]$ I want to show that there are more primes in the interval $(1,x]$ than in $(x,2x]$, when $x$ is large enough. Let $\pi$ be the prime counting function, ie. $\pi(x)=\# \{ p \leq x | p \text{ is a prime} \}$. Then the above statement is equivalent to showing:
$\pi (2x)-2\pi(x)=-\log(2)\frac{x}{\log ^2x}+O\left(\frac{x}{\log ^3 x}\right)$
I have a hint saying to use the prime number theorem. I'm trying to use that $\pi(x)= \frac{x}{\log x} + \frac{x}{\log ^2 x} + O \left( \frac{x}{\log ^3 x} \right)$, but can't seem to rewrite it fitting. Maybe I need to use other aspects of the prime number theorem.
Added: It's useful to see alternative ways of solving the problem, but I'm still specifically seeking to show the equation above using PNT.

From Dusart we know that $\frac{x}{\ln x}\left(1+\frac{1}{\ln x} \right) < \pi(x )<\frac{x}{\ln x}\left(1+\frac{1.3}{\ln x} \right) $ for all $x \geq 600$ and so you need to prove that $\pi(2x) -2 \pi(x) \leq \frac{2x}{\ln 2x} \left(1+\frac{1.3}{\ln 2x} \right)-2\frac{x}{\ln x}\left(1+\frac{1}{\ln x} \right) << 0  $
Or Equivalently $\frac{1}{\ln 2x} \left(1+\frac{1.3}{\ln 2x} \right) << \frac{1}{\ln x}\left(1+\frac{1}{\ln x} \right) => \ln x (1+\frac{1.3}{\ln 2x}) < (\ln x +\ln 2)(1+\frac{1}{\ln x}) => \ln x +\frac{1.3 \ln x}{\ln 2x} < ln x +1.3< \ln x+ 1 + \ln 2 < \ln x+1+\ln 2 + \frac{\ln 2}{\ln x} $
Checking for smaller cases the inequality is true for all $x\geq 11$

There is no need for error estimates.  Use the PNT in the form $\pi(x) \sim x/\log x$. That implies $\pi(2x)/\pi(x) \sim (2x/\log(2x))/(x/\log x) = 2(\log x)/\log(2x)$.  Now convince yourself that this tends to $2$ as $x\to \infty$, so the difference
$$
\pi(2x) - \pi(x) = \pi(x)\left(\frac{\pi(2x)}{\pi(x)} - 1\right)
$$
is large for large $x$.  Similar reasoning shows for each $c > 1$ that $\pi(cx) - \pi(x)$ is large for large $x$.


Derivation of Joint distribution Property I have the following note in my Statistics Script used for a derivation of a uniform minimum risk estimator.

Given  a $Y \equiv X - Z$, with (X,Z) being f(x,z) distributed the density of y is:

$f_Y(y) = \int f(y + z, z) dz$

Can anyone help me to derive the expression above. It is not trivial to me.

Thanks

$P(Y \leq y)=P(X\leq Z+y)=\int_{-\infty}^{\infty}\int_{-\infty}^{v+y}  f(u,v)dudv$. Putting $w=u-v$ we can write this as  $$\int_{-\infty}^{\infty}\int_{-\infty}^{y}  f(w+v,v)dwdv.$$ Interchanging the order of integration this becomes $$\int_{-\infty}^{y} \int_{-\infty}^{\infty} f(w+v,v)dvdw.$$ Differentiating this we see that the density of $Y$ is $$ \int_{-\infty}^{\infty} f(y+v,v)dv.$$ which is same as $$ \int_{-\infty}^{\infty} f(y+z,z)dz.$$


Find the derivative of $\int_a^{g(x)} f(t)dt $ wrt $x$ Question :
Let $$f:[a,b]\rightarrow \mathbb R$$ be continuous  and $$g:[c,d]\rightarrow \mathbb R$$ be  differentiable .  Define  $$\psi(x) := \int_a^{g(x)} f(t)dt $$ . Prove  that $\psi$ is  differentiable and  compute  the  derivative .
My Attempt :
Define $$F(x)=\int_a^x f(t)dt$$
Then by Second Fundamental Theorem   of  Calculus , we  have $F$  is  differentiable  and  $F'=f$ at  the points  of  continuity  of  $f$ .
Then $$\psi(x)=(g\circ F)(x)$$
Both  being  differentiable ,  their  composition  is  so   and  hence  $\psi$  is  differentiable. And  by  Chain Rule  we  have , $$\psi'(x)=g'(F(x))F'(x) \\ =(g'(F(x)))\cdot f(x)$$
Upto  this  is  I  think  ok. But  now  how to  simplify $g'(F(x))$ $?$

Your function is  $\psi(x)=F(g(x))-F(a)$ Where $F(x)$ is a primitive of $f(x)$ so the derivative is:
$$
\psi'(x)=F'(g(x))g'(x)=f(g(x))g'(x)
$$

You did the composition wrong! The integral is $F\circ g(x)$ and not $g\circ F(x)$.


Existence of the bijective map from set of points to set of lines Let $P$ be a set of all points in a plane and $L$ set of all lines in that plane.
Is there a bijective map $f:P\to L$ such that for all colinear points $A,B,C$ the lines$f(A),f(B),f(C)$ are parallel or concurrent?

If a plane is projective then a polar map (with respect to any circle) does the job. But what if the plane is euclidean? I have no idea how to approach.

That's not possible.

A way to prove this is to form a 2-bijection from $P$ to the antipodally punctured unit sphere (APUS) with axis perpendicular to $P$ that has contradictory properties.

We form this by selecting a point $a$ outside $P$. We take a point in $P$ maps it to a line, then we take the plane through $a$ containing that line and finally the two unit normals of that plane (being two values, but they are each others negation or opposites).

This mapping has some properties that aren't really compatible:

It's bijective (meaning that it never maps to the same pair of opposite vectors twice)
It maps lines to great circles (allowing any intersection of the APUS and a plane through the center of the APUS including those that actually aren't circles - that is they're punctured)

These are contradictory since two lines that doesn't intersect must map to great circles that doesn't intersect which is only possible if they are punctured. On the other side two lines that do intersect must map to great circles that do intersect.

This is the contradiction since any line must map to a great circle that is both punctured  and not punctured.

As for your example with the projective plane I guess that the mapping constructed above would work out since it would map to the non-punctured unit sphere instead, lines in $P$ would always intersect and great circles would always intersect.

Most of the proof is quite straight forward and there's nothing really difficult, but there are some points that's probably not that obvious.

The fact that lines would map to great circles follows from three bits put together:

If three lines are parallel in $P$ they won't intersect and therefore the intersections of the planes through $a$ containing the lines would be a lines and that lines has to be parallel to $P$. Or put in another way the cross product of the normals of the planes must be parallel to $P$. Or put in another way two of the normals of those planes and the normal of $P$ are linearly dependent which makes the normals of those planes linearly dependent which in turn puts them all in the same plane.
If three lines are concurrent in $P$ the planes through $a$ containing these line will both contain $a$ and the intersection point of the lines. This means that there is a line common to all these planes and their normal must be perpendicular to this line. This puts the normals all in the same plane.
The image of a line is therefore contained in some great circle. If the image was not the entire great circle there would be a point in the great circle whose preimage wouldn't be on the line. Now pick any point in $P$ then that point, the preimage mentioned and a point on the line would be colinear and must map to the great circle with the exception that the mentioned preimage and the picked point is on the parallel line or that the point on the line is antipodal (to the image of the mentioned preimage). This exception is not enough to make the mapping surjective.

Having a bijective map implies that it is invertible. If you were given parallel lines $f(A),f(B),f(C)$, then $f^{-1} : L \rightarrow P$ should give you colinear points $A,B,C$. However there is not one obvious choice for which colinear points should be chosen (there are uncountably many). This should tell you that a map from the set of points to the set of lines and vice versa would not be bijective.


Size of the closure of a set Why in a Hausdorff sequentially compact space the size of the closure of a countable subset is less or equal than $c$ ? I can see why this is true when the space if first countable but we are not assuming so.

The statement is consistently false.

In vaughn's article "Countably compact and sequentially compact spaces" in the Handbook of Set-Theoretic Topology, he gives the following equivalences for an infinite cardinal $\kappa$:

$\{0,1\}^\kappa$ is sequentially compact
$\kappa < \mathfrak{s}$ (where $\mathfrak{s}$ is the "splitting number")
every compact space of weight $\leq\kappa$ is sequentially compact

Thus if we are in a universe where $\omega<\omega_1=\kappa<\mathfrak{s}\leq\mathfrak{c}<2^\kappa$ then the space
$\{0,1\}^{\omega_1}$ gives a separable space which is sequentially compact but has cardinality greater than $\mathfrak{c}$.

Added: including the forcing construction to get such a model (from van Douwen's article):

Start with ground model
$M\models(\mathfrak{c}=\omega_2 \wedge 2^{\omega_1}=\omega_3)$
then obtain an iterated ccc extension $\langle M_\eta : \eta\in\omega_2\rangle$ by adding $X_\eta\in[\omega]^\omega$ at stage $\eta\in\omega_2$ s.t. $\forall Y\in M\cap[\omega]^\omega$ either $X_\eta\subseteq^* Y$ or $X_\eta\subseteq^* (\omega\setminus Y)$.  Then in $M_{\omega_2}$ we have $\mathfrak{c}=\omega_2<2^{\omega_1}$, $\mathfrak{t}=\omega_1$, $\mathfrak{s}=\omega_2$, exactly as desired.  (Details are left as an exercise).

It seems like you don't need all the conditions, so maybe I am missing something.  There are only continuum many sequences of points from a countable set.  So the limit points are at most continuum many.


Deriving formula from Fourier series: $\frac{\pi^2}{12} = \sum\limits_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}$ The equation/formula $$ \frac{\pi^2}{12} = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}$$ is to be derived.

I know that the Fourier expansion of $f(x)=x$ for $x \in (-\pi,\pi)$ is $$f(x)=x=\sum_{n=1}^{\infty} \frac{2\,(-1)^{n+1}}{n}\,\sin{nx}$$ and there was an example of how for $x=\frac{\pi}{2}$ you'd get $\frac{\pi}{4} = 1-\frac{1}{3}+\frac{1}{5}- \cdots = \sum_{l=0}^\infty \frac{(-1)^{l}}{2l+1}$, but here it's not the square of $n$. I got a Fourier series for $$f_1(x)=x^2=\sum_{n=1}^{\infty} \frac{4\,(-1)^n}{n^2}\,\cos{nx}$$ but then the exponent of the $(-1)$ isn't the same.

I'm obviously missing something, so any hints are appreciated.

We have that:
$$ \sum_{n\geq 1}\frac{2(-1)^{n+1}}{n}\,\sin(nx) $$
is the Fourier series of the $2\pi$-periodic extension of $f(x)=x$, so, by termwise integration,
$$\begin{eqnarray*} \forall x\in(-\pi,\pi),\qquad x^2&=&\sum_{n\geq 1}\frac{4(-1)^{n+1}}{n^2}\left(1-\cos(nx)\right)\\&=&\color{blue}{\left(\sum_{n\geq 1}\frac{4(-1)^{n+1}}{n^2}\right)}-\sum_{n\geq 1}\frac{4(-1)^{n+1}}{n^2}\cos(nx)\end{eqnarray*}$$
and it follows that the blue term is the mean value of the function $g(x)=x^2$ over the interval $(-\pi,\pi)$, so:
$$\color{blue}{\sum_{n\geq 1}\frac{4(-1)^{n+1}}{n^2}}=\frac{1}{2\pi}\int_{-\pi}^{\pi}x^2\,dx = \color{blue}{\frac{\pi^2}{3}}$$
and

  $$\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^2}=\color{red}{\frac{\pi^2}{12}}$$

readily follows.

Consider the Riemann zeta function: $$\zeta(s)=1+\frac1{2^s}+\frac1{3^s}+\frac1{4^s}+\ldots +\frac1{n^s}+...$$
$$\frac2{2^s}\zeta(s)=\frac2{2^s}+\frac2{4^s}+\frac2{6^s}+\ldots+\frac2{(2n)^s}+...$$
subtracting $2$nd line from the first we get
$$\zeta(s)\left(1-\frac2{2^s}\right)=1-\frac1{2^s}+\frac1{3^s}-\frac1{4^s}+\ldots +(-1)^{n+1}\cdot \frac1{n^s}+...$$
Substitute $s=2$ in the above equation
we get
$$\frac12 \cdot \zeta(2)=1-\frac1{2^s}+\frac1{3^s}-\frac1{4^s}+\ldots+(-1)^{n+1}\cdot \frac1{n^s}+\ldots$$
$\zeta(2)$ is well known as $\frac{\pi^2}6$ due to Euler
so the RHS$=\frac{\pi^2}{12}$


Why does Wolfram|Alpha make a mistake here? We want to evaluate $$\lim_{x \to -8}\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}}.$$The solving process can be written as follows:\begin{align*}\lim_{x \to -8}\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}}&=\lim_{x \to -8}\left[\frac{(\sqrt{1-x}-3)(\sqrt{1-x}+3)}{(2+\sqrt[3]{x})(4-2\sqrt[3]{x}+\sqrt[3]{x^2})}\cdot \frac{4-2\sqrt[3]{x}+\sqrt[3]{x^2}}{\sqrt{1-x}+3}\right]\\&=\lim_{x \to -8}\left[\frac{-(x+8)}{x+8}\cdot \frac{4-2\sqrt[3]{x}+\sqrt[3]{x^2}}{\sqrt{1-x}+3}\right]\\&=-\lim_{x \to -8} \frac{4-2\sqrt[3]{x}+\sqrt[3]{x^2}}{\sqrt{1-x}+3}\\&=-2.\end{align*}

But when I input this

lim\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}} as x to -8

into Wolfram|Alpha, it gives the limit $0$.

Why is Wolfram|Alpha making a mistake here?

WolframAlpha understands the expression $\sqrt[3]{x}$ for negative x in a different way than you expect.

Try this: lim\frac{\sqrt{1-x}-3}{2+surd(x,3)} as x to -8

I think Wolfram Alpha does the wrong thing here, no matter what everybody else says. The cube root of a negative real number should be real by default. But anyway you can work around it by asking Wolfram Alpha for

$$\lim_{x \to -8}\frac{\sqrt{1-x}-3}{2-\sqrt[3]{-x}}$$

Inputting

lim\frac{\sqrt{1-x}-3}{2-\sqrt[3]{-x}} as x to -8

gives $-2$, as expected.


Young's inequality or something similar I was trying to prove the following inequality:

Take an arbitrary $p\in\mathbb{N}$, then

$\forall\varepsilon>0\,\forall{}a,b\in\mathbb{R}_{+}\exists{}C_{\varepsilon}>0\,:\,(a+b)^p\leq{}(1+\varepsilon)a^p\,+\,C_{\varepsilon}b^p$

This looks an awful lot like young's inequality to me, but I couldn't quite figure out how to find an analogy between the two that could help.

I've also considered proving this via induction, but for $b\leq{}a$ it doesn't seem to work. I'd greatly appreciate any tips or suggestions.

Thanks a lot.

Dividing over $a^p$, we get an equivalent inequality $(1+x)^p \le 1+\varepsilon + C_\varepsilon x^p$. When $x\le x_\varepsilon:=(1+\varepsilon)^{1/p}$, the inequality is evident. For $x>x_\varepsilon$, it is evident as well with $C_\varepsilon = (1+x_\varepsilon^{-1})^p$.

$\varepsilon$ and a, b positifs numbers
you can take

$C_\varepsilon \geq \frac{(a+b)^p-(1+\varepsilon)a^p}{b^p}$


Discrete Mathematics (Prove or Find a Counterexample of a Proposition) Let f : X → Y be a function. For each of the following propositions, prove or find a counterexample. (In constructing a counterexample, please specify what your domain and codomain are as well as the function itself.)

If S,T ⊆ X then f(S ∩ T) = f(S) ∩ f(T).

Can somebody please show me how to solve this type of problem? I have other similar problems and need to learn the process for solving them. Thank you!

Usually what I do, if I'm not sure whether a statement is true or not is I start trying to prove it and if I hit a spot where I feel like I can't finish my proof because some condition doesn't seem to hold then I try and come up with an example where that happens.  My hope is that the example will either be a counterexample or it will indicate to me how to get around my difficulty.

For your problem you want to prove two sets are equal so you prove that each is contained in the other.  We'll just start proving and see if we get stuck...

Step 1) Assume $x \in f(S \cap T)$ and prove that $x \in f(S) \cap f(T)$.

If $x \in f(S \cap T)$ then there is a $y \in S \cap T$ such that $f(y) = x$.  Now $y \in S \cap T$ means $y \in S$ and $y \in T$.  That $y \in S$ and $f(y) = x$ means $x \in f(S)$.  Similarly $y \in T$ gives $x \in f(T)$.  Now we have $x \in f(S)$ and $x \in f(T)$ so $x \in f(S) \cap f(T)$.  Done.

Step 2) Assume $x \in f(S) \cap f(T)$ and prove that $x \in f(S \cap T)$.

Assume $x \in f(S) \cap f(T)$.  Then $x \in f(S)$ and $x \in f(T)$.  That $x \in f(S)$ means there is a $y \in S$ such that $f(y) = x$.  That $x \in f(T)$ means there is a $z \in T$ such that $f(z) = x$... hmmm.  I need the element that hits $x$ to be in both $S$ and $T$ at the same time, but right now I have two different elements.  If they were the same, $y = z$, then I'm done, but why does this need to be the case?  If the function were injective we would get that, but there are functions that are not injective.  In fact, now that I think about it, why would $S$ and $T$ need to intersect at all?  Maybe this means it's not true.

Counterexample:

Ok, we got stuck.  If the function isn't injective then I can't seem to get any further.  So for my counterexample I want a $z$ and a $y$ that both map to $x$, but $z \neq y$.  So what about the function $f\colon\{1, 2\} \to \{1\}$ defined by $f(1) = f(2) = 1$.  Then I choose $S = \{1\}$ and $T = \{2\}$.  Now with this explicit choice for $f, S, T$ figure out what $f(S \cap T)$ and $f(S) \cap f(T)$ are.  Are they the same?  If not you have a counterexample.  If yes, then maybe the example you chose wasn't the right one to be a counterexample, or maybe the theorem is true and there is no counter example.  There's no sure fire way to know which case your in, this is where experience and intuition are helpful.

When I have a problem that asks me to prove something, I first try to convince myself if it is true. If I don't believe it, then I start out by trying to find a counterexample. When I look at your problem statement, I notice that we do have $f(S \cap T) \subseteq f(S) \cap f(T)$. So the only question is whether or not $f(S) \cap f(T) \subseteq f(S \cap T)$. At this point, I would notice that we did not assume that the function $f$ is one-to-one. Hence, I would build a counterexample: Pick an $s \in S$ for which $s \not\in T$ and a $t \in T$ for which $t \not\in S$. What happens if $f(s) = f(t)$? We would get a counterexample.

However, if I have trouble finding a counterexample, I may then try to prove the statement instead. (This is extremely similar to what Jim wrote at the beginning of his answer.)


Function Spaces: Characterizations of Positivity Context

The problem here is about the characterization of positivity for real or complex valued functions:
$$\sigma(f)\geq 0\iff\sigma(f(x))\geq 0\text{ for all }x\in X\iff  f(x)\geq 0\text{ for all }x\in X$$

The point is that the first characterization reflects its intrinsic properties allowing to apply general results in the study of positive operators as integrals...

Problem 0

Let $A$ be an algebra with identity and $\Omega$ a plain space.
Consider the algebra of functions:
$$\mathcal{F}:=\{F:\Omega\to A\}$$
together with pointwise operations.

Is it true that the following characterizations of positivity are equivalent:
$$\sigma(F)\geq 0\iff\sigma(F(\omega))\geq 0\text{ for all }\omega\in\Omega$$
where $\sigma(F)$ is the spectrum of $F$ in $\mathcal{F}$ whereas $\sigma(F(\omega))$ the spectrum of $F(\omega)$ in $A$?

Problem 1a

Now, let $B$ be a Banach algebra with identity.
Consider the (Banach) algebra of bounded functions:
$$\mathcal{B}:=\{F:\Omega\to B:F\text{ bounded}\}$$
again with pointwise operations.
(Note that $F$ is not necessarily continuous.)

Is it still true that the characterizations of positivity are equivalent:
$$\sigma(F)\geq 0\iff\sigma(F(\omega))\geq 0\text{ for all }\omega\in\Omega$$
where $\sigma(F)$ again is the spectrum of $F$ in $\mathcal{B}$ whereas $\sigma(F(\omega))$ the spectrum of $F(\omega)$ in $B$?
(Here, a function is invertible iff it has a bounded inverse!)

Problem 1b

Next, let $X$ be also topological space.
Consider the algebra of continuous functions:
$$\mathcal{C}:=\{F:X\to B:F\text{ continuous}\}$$
again with pointwise operations.
(Note that $F$ is possibly unbounded.)

Is it still true that the characterizations of positivity are equivalent:
$$\sigma(F)\geq 0\iff\sigma(F(x))\geq 0\text{ for all }x\in X$$
where $\sigma(F)$ again is the spectrum of $F$ in $\mathcal{C}$ whereas $\sigma(F(x))$ the spectrum of $F(x)$ in $B$?
(Here, a function is invertible iff it has a continuous inverse!)

Problem 2

Finally, consider the (Banach) algebra of bounded continuous functions:
$$\mathcal{BC}:=\{F:X\to B:F\text{ bounded and continuous}\}$$
again with pointwise operations.

Is it still true that the characterizations of positivity are equivalent:
$$\sigma(F)\geq 0\iff\sigma(F(x))\geq 0\text{ for all }x\in X$$
where $\sigma(F)$ again is the spectrum of $F$ in $\mathcal{BC}$ whereas $\sigma(F(x))$ the spectrum of $F(x)$ in $B$?
(Here, a function is invertible iff it has a bounded and continuous inverse!)

I'm no expert on this, so I might say stupid things, but lets have a go. Let me assume that $A$ is unital (with unit $\mathbb{1}$). The answer to your first question is positive, since $$\sigma(F)=\bigcup_{\omega\in\Omega} \sigma(F(\omega)).$$ Indeed, the inverse $h$ (if it exists) of the function $\lambda I-F$ satisfies $h(\lambda I-F)=(\lambda I-F)h\equiv 1$ pointwise and is therefore necessarily given by \begin{equation}
h(\omega):=(\lambda 1-F(\omega))^{-1}.
\end{equation} From this one sees that if $\lambda \in \sigma(F(\omega))$ for some $\omega$, then $\lambda \in \sigma(F)$ and conversely, if $\lambda \in \sigma(F)$ then for some $\omega$ the existence of $(\lambda 1-F(\omega))^{-1}$ must fail.

As for the continuous case, the collection of all continuous functions do not form a Banach algebra because you cannot define a norm there (unless your topological space is compact). You need to impose a boundedness condition: $$\|F\|:=\sup_{x\in X}\|F(x)\|_B<\infty.$$ This quantity defines a norm which makes your space a Banach algebra.

Now while an inverse $h$ of some $(\lambda I-F)$ must still satisfy the identity written above, the identity itself does not necessarily define a continuous and bounded function. An example illustrating this can be seen by taking $X=\mathbb R=B$ and considering the Banach algebra $C_b(\mathbb R)$ of bounded continuous functions.

Still, the case where $X$ is compact forms an exception for the following reason. The inclusion $$\bigcup_{x\in X} \sigma(F(x))\subset \sigma(F)$$ always holds. Suppose then that $\lambda\notin \bigcup_{x\in X} \sigma(F(x))$. Since $Im(\lambda I-F)$ is a compact subset of the open set $G$ of all invertible elements of $B$, and the inversion $x\mapsto x^{-1}$ is continuous on $B$ it follows that the inverses of all the elements in $Im(\lambda I-F)$ also form a compact set (see this for all these properties), thus lying in some bounded ball of radius $a<\infty$. In other words, $$\|(\lambda I-F)^{-1}\|=\sup_{x\in X}\|(\lambda 1-F(x))^{-1}\|_B \le a,$$ thus giving a bounded, continuous inverse of $\lambda I-F$. This shows the opposite inclusion.

EDIT

Problem 0: As discussed above, the equivalence holds true in this case.

Problem 1a: In this case, for the same reasons as in the continuous and bounded case (see above) the equivalence is not true (even with compactness, since we do not have continuity).

Problem 1b: Here the equivalence is again true, since in the continuous & bounded case the only obstruction was boundedness.

Problem 2: In general no, in case $X$ is compact yes.

Due to the pointwise structure it formally holds:
$$\sigma(F)=\bigcup_{x\in X}\sigma(F(x))$$

The difficulties arise as soon as additional requirements are tied upon the functions:

0) The formal inverse exists.
1a) The formal inverse is not necessarily bounded. (See below!)
1b) The formal inverse is continuous due to the Neumann series. (See Contuity of inversion!)
2) The formal inverse is again continuous but not necessarily bounded.

Especially for complex valued functions it holds: $\sigma(f(x))=f(x)$

Consider the bounded function:
$$f:(0,1]\to\mathbb{R}:x\mapsto x$$
Then zero does not belong to its range but its formal inverse is unbounded:
$$f^{-1}:(0,1)\to\mathbb{R}:x\mapsto \frac{1}{x}$$


Does the Cauchy-Schwartz inequality apply to integrals? The integral $\int f(t) g(t) \, \mathrm{d} t$ is like the dot product $\vec{u} \cdot \vec{v}$.

The Cauchy-Schwartz inequality is

$$ \left\lvert \vec{u} \cdot \vec{v} \right \rvert \le \left\lvert u \right\rvert \left\lvert v \right\rvert$$

Is the following true?

$$ \left\lvert \int^b_a f(t) g(t) \, \mathrm{d} t \right \rvert \le \left\lvert \int^b_a f(t) \, \mathrm{d} t\right\rvert \left\lvert \int^b_a g(t) \, \mathrm{d} t \right\rvert$$

I'd imagine that if this works it only applies for very carefully defined functions such as smooth functions and only on finite intervals.

This could be useful for integral transforms:

$$ \left\lvert \int^b_a K(s, t) f(t) \, \mathrm{d} t \right \rvert \le \left\lvert \int^b_a K(s, t) \, \mathrm{d} t\right\rvert \left\lvert \int^b_a f(t) \, \mathrm{d} t \right\rvert$$

But I don't think this applies to most common generalized functions used for such transforms.

Yes it does (although not in the way you stated):
$$\left|\int_{\mathbb{R}^n} f(x) \overline{g(x)}\,dx\right|^2 \leq \int_{\mathbb{R}^n} |f(x)|^2\,dx \cdot \int_{\mathbb{R}^n} |g(x)|^2 \,dx.$$
 https://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality#L2

Let $(\Omega,\Sigma,\mu)$ be a measure space and $f^2,g^2$ be summable ($ \int_{\Omega} f(x)^2 \, d\mu(x) < \infty $). Then
$$ 2\left( \int_{\Omega} f(x)^2 \, d\mu(x) \right) \left( \int_{\Omega} g(x)^2 \, d\mu(x) \right) - 2\left( \int_{\Omega} f(x)g(x) \, d\mu(x) \right)^2 \\
= \int_{\Omega} \int_{\Omega} (f(x)^2g(y)^2+f(y)^2g(x)^2-2f(x)g(x)f(y)g(y)) \, d\mu(x) \, d\mu(y) \\
= \int_{\Omega} \int_{\Omega} (f(x)g(y)-f(y)g(x))^2 \, d\mu(x) \, d\mu(y) \geq 0, $$
since a square is nonnegative and the integral of a nonnegative number is nonnegative. Rearranging and taking the square root gives Cauchy–Schwarz for integrals. Equality holds when $ f(x)g(y)-f(y)g(x)=0 $ $\mu$-almost everywhere, which needs $f(x)=cg(x)$. This is easily the most straightforward method of proof. This applies in particular to any interval $(a,b)$ with ordinary Lebesgue measure $dx$, whether $a,b$ are finite or not.

The inequality you give does not work even for the simplest case of a function with two values: $a_1b_1+a_2b_2$ has no particular relationship with $(a_1+a_2)(b_1+b_2)$, because the difference is
$$ (a_1+a_2)(b_1+b_2)-(a_1b_1+a_2b_2) = a_1b_2+a_2b_1, $$
which can have whatever value we like: $(1-1)(1-1)-|(1)(1)+(-1)(-1)| = -2<0 $, for example.


Determinant of a $4\times4$ matrix without using triangular method if possible Find the unique value of $x$ for which
$$\det\begin{pmatrix}-6 & -1 & -3 & 6\\
5 & -5 & x & 6\\
-8 & -4 & -8 & 2\\
-5 & 1 & -6 & -4
\end{pmatrix}=0.$$


What is the difference between $\mathbb E[Z|\mathcal G]=Y$ and $\mathbb E[Z|\mathcal G]\stackrel{\text{a.s.}}{=}Y$? I'm somewhat confused by the definition of martingale:

Let $(\Omega, \mathcal F, \mathcal F_n, \mathbb P)$ be a filtered probability space. We call $(X_n)_{n\in\mathbb N}$ martingale if for all $n\in\mathbb N$ holds:

$X_n \in \mathcal L^1(\mathbb P)$

$X_n$ is $\mathcal F_n$-measurable

$\mathbb E[X_{n+1}|\mathcal F_n]\stackrel{\text{a.s.}}{=}X_n$

Why do we need "a.s." in the 3rd part? Isn't $\mathbb E[Z|\mathcal G]$ always unique only up to "a.s." anyway? How would we check the 3rd part without knowing that the 2nd part holds? (I.e. doesn't the 3rd imply the 2nd?) Why do we separate the points 2 and 3? Wouldn't it be enough to just ask for $\mathbb E[X_{n+1}|\mathcal F_n]=X_n$ instead?

In general, 3 does not imply 2

Suppose $g$ is $\mathcal{F}$-measruable and $h=g$ almost surely, it does not imply $h$ is also  $\mathcal{F}$-measurable.

Take a set $A$ which is contained in $B \in \mathcal{F}$ with $P(B) = 1$, but $A$ is not itself in $\mathcal{F}$. By definition, $h = 1_{A}$ is alomst surely equal to the constant variable $g \equiv 0$. But $h$ is not $\mathcal{F}$-measurable, since $h^{-1}(\{1\}) = A \not\in \mathcal{F}$.

If $\mathcal{F}$ has been completed and the filtration is augmented by $\mathbb{P}$, then (2) is indeed unnecessary. That's one of the reason of completion and augmentation: to make sure modification on a null set doesn't cause measurability problem.

Let's generalise how 3 doesn't imply 2:

  All versions of a conditional expectation are almost surely equal to each other, but any version is free to be equal to another random variable that is not a version of said conditional expectation.

Precisely:

  Let $(\Omega,\mathcal F, \mathbb P)$ be a probability space with random variable $X \in \mathscr L^1$ and a sub-$\sigma$-algebra $\mathcal G$. Versions of $\mathbb E[X|\mathcal G]$ are almost surely equal to each other, but any version is free to be equal to another random variable, $Y$, that is not $\mathcal G$-measurable.

Consider $X=c$, $Y=c1_A$, $P(A)=1$ and $A \notin \mathcal G$. Then

$$\mathbb E[X|\mathcal G] := \mathbb E[c|\mathcal G] = c \stackrel{\text{a.s.}}{=}c1_A=:Y,$$ but $Y$ is not $\mathcal G$-measurable. You could say that $Y$ is a non-$\mathcal G$-measurable version of $\mathbb E[X|\mathcal G]$, i.e. it satisfies $E[Y1_G] = E[X1_G] \ \forall G \ \in \ \mathcal G$ but not $\sigma(Y) \subseteq \mathcal G$, if that's even a thing in probability.


Jacobian matrix of a composition of functions with chain rule Consider the differentiable functions $f:\mathbb{R}^{\ell}\to\mathbb{R}^{m}$, $g:\mathbb{R}^{m}\to\mathbb{R}^{n}$, and $h:\mathbb{R}^{n}\to\mathbb{R}^{o}$. Let $F:\mathbb{R}^{\ell}\to\mathbb{R}^{o}$ be the composition of these functions, i.e., $F=h\circ g\circ f$. We define the Jacobian matrix of $F$ evaluated at $\boldsymbol{x}\in\mathbb{R}^{\ell}$ as $J_{F}(\boldsymbol{x})\in\mathbb{R}^{o\times \ell}$, and similarly, we define the Jacobian matrices $J_{f}$, $J_{g}$, and $J_{h}$ for $f$, $g$, and $h$, respectively.
Using the chain rule, is it safe to say that $ J_F(x)=J_h(g\circ f(x))\cdot J_g(f(x))\cdot J_f(x) $ ?
If so, how can that be converted in matrix form ?

$$
J_F(x)=J_h(g\circ f(x))\cdot J_g(f(x))\cdot J_f(x)
$$


Probability of a path of a given length between two vertices of a random graph Suppose that in  random graph $G$ on $n$ vertices any $2$ vertices can be connected by an edge with probability $\dfrac{1}{2}$, independently of all other edges.  What is the probability  $P_n(k)$ that two arbitrary vertices are connected by a simple path of length $k$, $0<k \leq n-1$?

My attempt. Fix $2$ vertices. To build a path of length $k$ we have to  choose $k-1$ vertices from the remaining $n-2$ vertices. Since order is important we can do it in $(k-1)! {n-2 \choose k-1}$  ways. There are $2^k$ configurations of the edges for a path of length $k$. Thus the probability seems to be
$$
P_n(k)=\frac{(k-1)! {n-2 \choose k-1}}{2^k}.
$$
The sum over all path lengths, $\displaystyle \sum_{k=1}^{n-1}P_n(k)$, must equal $1$, but my calculations for small $n$ show that it is not $1$.

Where is my mistake?

Your mistake is that you assume that the probability of a union of events is the sum of their probabilities. This is only true if the events are disjoint, which isn't the case here.

Your formula for the probability that there is a path of length $k$ is wrong.  Note for example that for $k=2$ your expression for $P_n(k)$ is $\frac{n-2}{2}$, which is larger than 1. Probabilities are always between zero and one.

In fact, your formula is the expected number of paths of length $k$.

When you add the probabilities together, once again you are computing an expectation. What your formula actually computes is the expected number of paths between two vertices.

Lets try the case k = 2. Pick any two vertices, without loss of generality you can call them 1 and 2. To construct a 2-path between them we first need to choose a third vertex, call it $i$. Clearly there are n-2 choices for $i$. Thus

$$\text{ total number of possible paths from 1 to 2 } = n-2 $$

Now for any $k$, there is an edge from 1 to $k$ with probability $p$. There is an edge from $k$ to $2$ with probability $p$ also. These are independent events, so the probability of both occuring is $p^{2}$. Thus the expected number of 2-paths from $1$ to $2$ is $p^{2}(n-2)$. and so the probability of observing one is:

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

I suspect that in general the probability of observing a $k$ path between any two fixed vertices in such an random graph (i.e. an Erdos-Renyi random graph) is p^{k}