问:
Does the limit definition of $e$ work if the quantity that approaches 0 (or $\infty$) is complex?
Does the definition

$$e\equiv\displaystyle\lim_{n\to0}\left(1+n\right)^{1/n}$$

or the equivalent

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

work if $n$ is complex? (I'm partial to the first one, as the second one would probably require something like $|n|\to\infty$ and I'm always wary around $\infty$ anyway, but either works.)

I've thought about writing $n$ as $|n|e^{i\arg(n)}$ (I'm not trying to define $e$, just trying to use one of the above definitions to equate something to $e$ in my calculations). For brevity, I'll write $a=|n|,b=\arg(n)$. The first definition of $e$ then becomes

\begin{eqnarray}
\displaystyle\lim_{n\to0}\left(1+n\right)^{1/n} & = & \displaystyle\lim_{a\to0}\left(1+ae^{ib}\right)^{1/ae^{ib}}\\
& = & \displaystyle\lim_{a\to0}e^{ib/ae^{ib}}\left(\left(1+a\right)^{1/a}\right)^{1/e^{ib}}\\
& = & \displaystyle\lim_{a\to0}e^{ib/ae^{ib}}\displaystyle\lim_{a\to0}\left(\left(1+a\right)^{1/a}\right)^{1/e^{ib}}\\
& = & \infty e^{e^{-ib}}\\
& = & \infty,
\end{eqnarray}

but I find it hard to believe that $n$ being complex will turn the limit from $e$ to $\infty$. (I've changed $n\to0$ to $a\to0$ because the only way for $n$ to approach $0$ is for its magnitude to approach $0$; we don't really care about what its phase is doing in this case, right?)

Thanks for any help.

Edit: My maths were wrong because I'm an idiot.

答:
You need the whole complex number to go to the origin, for the definition to hold. One way of doing it is writing $n=re^{i\theta}$ with $r\to 0$.

$$\lim_{r\to 0} (1+re^{i\theta} )^{1/re^{i\theta}} \\ = \bigg(\lim_{r\to 0} (1+re^{i\theta})^{1 /r}  \bigg)^{e^{-i\theta}} \\ =\big( e^{e^{i\theta}} \big)^{e^{-i\theta}} \\ = e$$ where I have used the standard result $\lim_{n\to 0} (1+an)^{1/n} = e^a$ for real $n$.


问:
Trying to understand vector subspaces
I am studying a chapter on vector subspaces and projections. I came across this definition:

We can split the set of all vectors $\mathbb{R}^3$ into two disjoint
sets: vectors entirely contained in $S$ and vectors perpendicular to
$S$. We say $\mathbb{R}^3$ decomposes into the direct sum of the
subspaces $S$ and $S^\bot$:

$$\mathbb{R}^3 = S \oplus S^\bot$$

And then about projection onto a line:

Let $l$ be a line passing through the origin and is in $\mathbb{R}^3$
with direction vector $\vec{v}$:

$$ l: \{(x,y,z) \in \mathbb{R}^3 \mid (x,y,z) = t\vec{v}, t \in \mathbb{R}^3 \} $$

The orthogonal space to the line $l$ consists of all vectors
perpendicular to the direction vector $\vec{v}$:

$$l^\bot : \{(x,y,z) \in \mathbb{R}^3 \mid (x,y,z).\vec{v} = 0 \}$$
...

The orthogonal space for a line $l$ with direction vector $\vec{v}$ is a plane with normal vector $\vec{v}$.

All makes sense except the last sentence which says the orthogonal space is a plane with normal vector $\vec{v}$. What my intuition says is that the orthogonal space should be the set of all planes (not a single plane) in $\mathbb{R}^3$ with normal vector $\vec{v}$ because:

$$\mathbb{R}^3 = S \oplus S^\bot$$

I hope someone can tell me what is it that I am getting wrong here.

答:
Actually you're confusing with vectorial spaces and affines spaces.
With vectorial spaces there exists only one subspace that has a given normal vector, because a plane is define by two non colinear vectors.

But then if you work with affines spaces, you also need a point : that's what we are used to.

ContentLicense: CC BY-SA 4.0

Title: Trying to understand vector subspaces


问:
Identity with the floor function
I'm struggling to complete proof of the following identity:

$$
\Bigl\lfloor \frac{m+ n}{2} \Bigr\rfloor +  \Bigl\lfloor \frac{m - n +1}{2} \Bigr\rfloor =m,
$$
where $m$ and $n$ are both integer.
By definition of floor function, $x-1 < \lfloor x \rfloor \leq x$.

Then
\begin{align}
\frac{m+n}{2} -1 <  & \Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor  \leq  & \frac{m+n}{2} \\
\frac{m-n + 1}{2} -1 < & \Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor  \leq & \frac{m-n +1}{2}.
\end{align}
By adding member to member, we obtain the following result:
$$
 m +\frac{1}{2} -2 <   \Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor + \Bigl\lfloor \frac{m- n +1}{2}\Bigr\rfloor  \leq  m +\frac{1}{2}.
$$
Which implies

$$
 -1.5 < \Bigl\lfloor \frac{m+ n}{2} \Bigr\rfloor + \Bigl\lfloor \frac{m- n +1}{2}  \Bigr\rfloor -m   \leq  .5
$$
This results in

$$
\Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor + \Bigl\lfloor \frac{m- n +1}{2}\Bigr\rfloor \in \left\{0,1\right\}.
$$

How to decide that the result is $0$?

Any help is welcome.

答:
hint

Observe that
$$\frac{m-n+1}{2}=\frac{m+n+1}{2}-n$$

So, you just need to prove that
$$\lfloor a\rfloor +\lfloor a+\frac 12\rfloor=2a$$

where $ a=\frac{m+n}{2}$.

There are two cases

$$a\in \Bbb N \implies \lfloor a\rfloor +\lfloor a+\frac 12\rfloor=a+a$$

$$a\notin \Bbb N \implies a+\frac 12\in \Bbb N$$
$$\implies \lfloor a\rfloor +\lfloor a+\frac 12\rfloor=$$
$$a+\frac 12-1+a+\frac 12=2a$$

Done.

Title: Identity with the floor function


问:
How to prove collinearity of circumcenters
Let $A_{1}A_{2}A_{3}$ be a non-isosceles triangle with incenter I. Let $C_{i}$
, $i = 1, 2, 3$, be the
smaller circle through $I$ tangent to $A_{i}A_{i+1}$ and $A_{i}A_{i+2}$ (the addition of indices being mod 3). Let
$B_{i}$
, $i = 1, 2, 3$, be the second point of intersection of $C_{i+1}$ and $C_{i+2}$. Prove that the circumcenters
of the triangles $A_{1}B_{1}I, A_{2}B_{2}I, A_{3}B_{3}I$ are collinear.

This is an IMO shortlisted question , I have the solution which uses inversion to proof this. However I am not able to visulaize it properly as the diagram, which I have drawn as per my understanding clearly shows that the points are not collinear. Please use a diagram and explain the answer in more elegant manner.
Here is the solution

答:
This answer will be a clarification and simplification of the given solution.  As such it will not be a complete proof in itself, on the assumption that a student of the IMO can fill in some of the steps.

Notation: $(X)$ and $(XYZ)$ will denote the circle with center $X$ and circumcircle of $X,Y,Z$ respectively.  $X'$ will denote the inversion of point $X$ with respect to the incircle $(I)$ (the OP proof uses the notation $X^*$).

The sketch above is the requested diagram.  The aim of the proof is to show that the three circles $(A_iIB_i)$ (red) are coaxal, in which case their centers are collinear.  The circles are coincident at $I$, so they are coaxal if they concur at a second point.  The circles invert in the incircle to become three lines and, if we can prove these lines are concurrent, we are done.

Since the constructions involved can result in a cluttered diagram, the above sketch shows the constructions associated with vertex $A_1$.  It is based on a common construction that shows the relationship of incircle, excircle, angle bisector at $A_1$, and the circle $(A_2IA_3)$ (blue). (see e.g. this note by Even Chen )  Also shown is $C_1$ (red) and the inversions of $C_1$ and $(A_2IA_3)$.  Convince yourself that these lines $KL$ and $A'_2A'_3$ are parallel (hint: they are both perpendicular to $A_1I$), and that the inversion of $C_i$ is the line $B'_{i+1}B'_{i+2}$

Now consider the triangles $\triangle A'_1A'_2A'_3$ and $\triangle B'_1B'_2B'_3$.  As shown in the previous step, these triangles have pairwise parallel sides.  These sides pairwise intersect on the projective line at infinity, so by Desargues theorem there is a point $P'$ such that the triangles are perspective from $P'$.  But this means that the lines $A_iB_i$ are concurrent at $P'$, which means that the circles $C_i$ are concurrent at $P=P''$.  They are concurrent at the two points $I,P$, and thus are coaxal, which is what we wanted to show.


问:
Polynomial $f(X) g(Y)- g(X) f(Y)$ cannot be divided by a non-constant $h(X) \in F[X]$

Let $F$ be a field and $f(X), g(X) \in F[X]$ two coprime
non constant polynomials.
Now the question is how to show that
the symmetric polynomial $P(X,Y):= f(X)  g(Y)- g(X)  f(Y)$
can not be divided by a non constant polynomial
$h(X) \in F[X] \setminus F$ in one variable $X$?

By  this question $P(X,Y)$ need not to be irreducible, so we can probably find divisors in $F[X,Y]$. Now I think that it's not possible to find a divisor in $F[X]$ but I'm not convinced I can show it.

Is there standard approach to show it? I used following
method about which I quite non convinced if this could lead me to
a correct way to prove it: We know that for an arbitrary ring $A$ and ideal $I$ of $A$
there exist a reduction modulo $I$ isomorphism
$A[Y]/I \cdot A[Y] \cong (A/I)[Y]$. For any $f = \sum^n a_i Y^i$
and $a \in A$ this implies that following statements are
equivalent:

$I$ "divides" $f$ in the sense $f \in I \cdot A[Y]$ iff
$a_i \in I$ for every $i $.

Now assume that $A$ is a $F$-algebra (in uor case $A=F[X]$) and non constant
$f(Y),g(Y) \in F[Y]$ are $F$-linear independent.
Let $a, a_f, a_g \in A$.

Is then $a | (a_f f(Y) + a_g g(Y))$ equivalent to
$a | a_f$ & $ a| a_g$? I try think about it as an application
of reduction modulo isomorphism to different basis instead
the standard basis $1, Y, Y^2,...$. Does it make sense?

Otherwise, is there a more conceptional way to prove that $f(X)  g(Y)- g(X)  f(Y)$
can not be divided by a non constant polynomial
$h(X) \in F[X] \backslash F$.

答:
For the sake of completeness I would like to add two alternative proofs which recently came in my mind:

Assume $h(X) \in F[X] \setminus F$ divides $P(X,Y)$. Let us apply the division
algorithm twice and obtain $f(X)= a_f(X)h(X)+r_f(X)$ and
$g(X)= a_g(X)h(X)+r_g(X)$. Then we get $P(X,Y) =
h(X)(a_f(X)g(Y)) - h(X) (a_g(X)f(Y)) + (r_f(X)g(Y)-r_g(X)f(Y))$.

Thus $h(X)$ divides $r_f(X)g(Y)-r_g(X)f(Y)$.
Comparing the degrees in $X$ on both sides we conclude $r_f(X)g(Y)-r_g(X)f(Y)=0$.
But $F[X,Y]$ is a UFD, so $r_f, r_g=0$ and we are done, and we
obtain a contradiction to coprime assumption.

Since $f(X)= \sum^n f_i X^i$ and $g(X)= \sum^m g_i X^i$ are coprime they
are of course $F$-linear independent and we can find two indices $k \neq l$
with the property that the two vectors $(f_k, g_k), (f_l, g_l) \in F^2$
are linearly independent.
Now if $h(X)$ divides $P(X,Y)$, this implies by reduction modulo theorem
that for every $i$ the $h(X)$ divides the linear combinations
$g_if(X) + f_ig(X)$ as we as their own linear combinations.
As by construction $g_kf(X)+ f_k g(X)$ and $g_lf(X)+ f_l g(X)$ are linear
independent and divisible by $h(X)$, the procedure solving linear systems
by building to diagonal shows that $h(X)$ divides also $f(X)$ and $g(X)$,
and this again violates coprimeness of $f$ and $g$.


问:
Is $±x$ the same as $ |x|$?
When solving a quadratic equation, the ± before the square root denotes "plus or minus that value." When solving absolute value equations, you try to solve using the positive and negative versions of the value inside. Does this mean that $y=|x|$ could be rewritten as $y=±x$?

答:
You're confusing implication with bi-implication. It is the case that
$$y=|x| \implies y = \pm x,$$
but they are not equivalent statements.

Title: Is $±x$ the same as $ |x|$?


问:
Show uniform convergence of $\sum\limits_n(1-x^2)^2 x^n$ on $[0,1]$
I'm trying to show that if $u_n(x) = (1-x^2)^2 x^n$ then $\sum u_n$ is uniformly convergent on $[0,1]$.

The only thing I can think to use is Weierstraß M-test to show that $u_n < M_n$ where $\sum  M_n$ converges, but I can't find a way to make the $M_n$ independent of $x$.

答:
By studying the derivative of $u_n(x)$, we can find the maximum value on $[0,1]$. We have:
$$
u'_n(x) = x^{n-1}(x^2 - 1)\left((n + 4)x^2 - n\right)
$$

Thus, $u_n(x)$ reaches its maximum on $[0, 1]$ at:
$$
m_n = \sqrt{\frac{n}{n+4}}
$$

And we have:
$$
u_n(m_n) = \left(\frac{n}{n+4}\right)^{n/2}\frac{16}{(n+4)^2} \le \frac{16}{n^2} = M_n
$$

Since $\sum 1/n^2$ converges, it now follows that the convergence is uniform.

ContentLicense: CC BY-SA 3.0


问:
Minimum volume for cuboid with specific properties in relation to other cuboids
I have lately started to practice some math for fun.
I have come across the following exercise:

"There are three cuboids with the volumes $V_1=10x_1y_1$, $V_2=10x_2y_2$ and $V_3=10x_3y_3$. These must meet the following criterias $V_1\geq V_2$, $V_1\geq V_3$ and $V_1+V_2+V_3=1000$ with $x_n$ and $y_n$ being natural numbers $0\leq x_n$, $y_n\leq 10$.
What is the smallest value for $V_1$?"

Firstly I tried an inelegant exhaustive search through a small script I wrote. With that I came up with the solution of $V_1=350$, which seems plausible but i would like something more elegant than that.
I have also thought of representing the volumes like this:
$V_1=1000-(V_2+V_3)$, $V_2=1000-(V_1+V_3)$, $V_3=1000-(V_1+V_2)$. I hoped to make progress with that but it seems like a dead end to me because all the expressions are dependent on each other so you can't really graph how they affect each other until you already know two volumes.

So I am asking what might be a more elegant approach to the exercise above?

答:
You want $V_1$ to be as small as possible, but you also want it to be greater than $V_2$ and $V_3$. Therefore, it must be at least a third of the combined volume ($\ge 333.\bar{3}$). Since it must be a multiple of $10$, you can check $V_1 = 340, 350, ...$ whether it would work. For $V_1 = 10z$, it must be true that $z$ can equal $xy$ with $x, y \in [1, 10]$. Therefore, only $V_1 = 350, 360, 400, 420, 450, ...$ would work.

For $V_1 = 350$, you have that $V_2 + V_3 = 650$, but since $V_1 \ge V_2, V_3$, it must be true that $(V_2, V_3)$ is in $\{(300, 350), (310, 340), (320, 330)\}$. $V_2, V_3 = 300, 350$ would work because $300 = 10\cdot 5 \cdot 6$ and $350 = 10\cdot 5\cdot 7$. Therefore the minimum $V_1$ is $350$.


问:
Connective symbol similar to implies, but acts like And
This feels like a basic question or a misunderstanding, but I'm struggling to identify a simple concept having to do with the provability of an argument or proof.

Imagine you have a proof that is well-reasoned and believed true, but then you discover that one of its underlying premises is actually false. Still valid, but unsound.

That false premise is conjuncted with other true premises to create lemmas, which no longer hold. These faulty lemmas are then conjuncted with other premises and lemmas to create further lemmas that no longer hold, and so on, until the conclusion is reached. Clearly, the conclusion no longer holds either. The conclusion itself might be true, it might be false, but it's definitely false that it holds.

My desire is to track those truth values of whether the various steps hold. The point is that these truth values (of whether the lemmas and conclusion hold) travel through the proof identically as they would through AND gates. And yet you would not say that the lemmas or conclusion themselves are false, it's just false that they hold. This is not implication, because the truth table behaves differently - if a proof is based off of the conjunction of two lemmas, and the first lemma does not hold, you would not argue that the proof is true.

In other words, look at the truth table for our favorite Socrates syllogism:

A: Socrates is a man

B: All men are mortal

C: Therefore, Socrates is mortal

A
B
A ^ B

T
T
T

T
F
F

F
T
F

F
F
F

I'm not interested in implication (→) here. I am more curious what A ^ B communicates about C. Not the value of C, but the provability of C. Obviously, if B is false, we have not proven Socrates immortal. But A ^ B does mean something about C that is consistent with the above truth table; something that is not implication. It means that the argument for C does not hold if one or more of its premises are false; that it is false that it holds.

What is this logical concept and what symbol should I choose? Are "provable" and "not provable" ( ⊢ and ⊬ ) the best options? i.e.

A
B
A ^ B
A ^ B ⊢ C
i.e.

T
T
T
T
A ^ B ⊢ C

T
F
F
F
A ^ B  ⊬ C

F
T
F
F
A ^ B  ⊬ C

F
F
F
F
A ^ B  ⊬ C

?

It certainly doesn't seem I should use does/doesn't imply/entail (→ ↛ ⇒ ⇏ ⊨ ⊭), because their truth tables are incompatible. Like if A^B is false, I can't say that "it is false that A^B implies C", because "False implies C" is literally true according to the implication truth table.

It's like I want a more colloquial sense of "implies" that means "Since A and B are not both true, A^B (False) does not (imply) that C holds." Is there a symbol/term for that?

答:
Proofs are not "true" or "false". They are simply syntactic objects over a given formal system. There are no invalid proofs. A purported proof that is not valid is simply not a proof. Proofs have nothing to do with the truth-values of the statements occurring in them. A proof in classical logic only guarantees that if all the axioms used by the proof are true (under some interpretation) then the conclusion of the proof is also true (under that same interpretation).

If you wish to reason about proofs of false statements over the system in which you are performing your reasoning, you may want to use provability logic, so you can say things like $¬A ∧ ⬜( A ⇒ B )$ meaning "$A$ is false and provably $A$ implies $B$".
$
\def\dem{\text{Dem}}
$

What you are looking for seems to be "$\dem(P,Q) ≡ P ∧ ⬜(P⇒Q)$" meaning "$P$ is true and provably $P$ implies $Q$" (informally "$P$ demonstrates $Q$"). Its negation is "$¬P ∨ ¬⬜(P⇒Q)$" meaning "either $P$ is false or it is not provable that $P$ implies $Q$".

Thus to say "Since A and B are not both true, A^B (False) does not (imply) that C holds.", we can say "$¬(A∧B)$ and hence $¬\dem(A∧B,C)$. But it does not seem to be of much value to do this...


答:
"We can split the set of all vector in $\mathbb{R}^{3}$ into two
disjoint sets: vectors entirely contained in $S$ and vectors
perpendicular to $S$."

I don't like this line for several reasons, the two main reasons being that

These two sets $S$ and $S^{\perp}$ are not disjoint, they share the $0$ vector;

it seems to suggest that every vector in $\mathbb{R}^{3}$ is either in $S$ or $S^{\perp}$, which is not the case. There are lots of vectors that are neither contained in $S$, nor are orthogonal to $S$. What is true is that every vector in $\mathbb{R}^{3}$ is a sum of a vector in $S$ and a vector in $S^{\perp}$.

The orthogonal space for a line $\ell$ with direction $v$ is a plane with normal vector $v$.

This is absolutely correct. The orthogonal space is a collection of vectors, not a collection of planes. And all vectors are rooted at the origin. So the orthogonal space consists of all vectors in the plane containing the origin which has normal vector $v$ (this is essentially the definition of the normal vector of a plane).

Title: Trying to understand vector subspaces

Body:
I am studying a chapter on vector subspaces and projections. I came across this definition:

We can split the set of all vectors $\mathbb{R}^3$ into two disjoint
sets: vectors entirely contained in $S$ and vectors perpendicular to
$S$. We say $\mathbb{R}^3$ decomposes into the direct sum of the
subspaces $S$ and $S^\bot$:

$$\mathbb{R}^3 = S \oplus S^\bot$$

And then about projection onto a line:

Let $l$ be a line passing through the origin and is in $\mathbb{R}^3$
with direction vector $\vec{v}$:

$$ l: \{(x,y,z) \in \mathbb{R}^3 \mid (x,y,z) = t\vec{v}, t \in \mathbb{R}^3 \} $$

The orthogonal space to the line $l$ consists of all vectors
perpendicular to the direction vector $\vec{v}$:

$$l^\bot : \{(x,y,z) \in \mathbb{R}^3 \mid (x,y,z).\vec{v} = 0 \}$$
...

The orthogonal space for a line $l$ with direction vector $\vec{v}$ is a plane with normal vector $\vec{v}$.

All makes sense except the last sentence which says the orthogonal space is a plane with normal vector $\vec{v}$. What my intuition says is that the orthogonal space should be the set of all planes (not a single plane) in $\mathbb{R}^3$ with normal vector $\vec{v}$ because:

$$\mathbb{R}^3 = S \oplus S^\bot$$

I hope someone can tell me what is it that I am getting wrong here.


问:
$\Phi: X \to \mathscr F (J,Y)$ is continuous $\iff$ $\forall j\in J$ we have that $X\to Y$ given by $x \mapsto (\Phi(x))(j)$ is continuous.
Let $(X, \mathscr T_X)$ and $(Y,\mathscr T_Y)$ be topological spaces.

GOAL: I wish to prove,

Let $J$ be a set and denote by $\mathscr F (J,Y)$ the set of all functions from $J$ to $Y$.

We equip it with the smallest topology containing $\{f \in \mathscr F (J,Y): f(j) \in V \}$ for every $V \in \mathscr T_Y$ and $j \in J$.

Show that a function $\Phi: X \to \mathscr F (J,Y)$ is continuous $\iff $ for each index $j\in J$ we have that $X\to Y$ given by $x \mapsto (\Phi(x))(j)$ is continuous.

Thoughts.

This indexing reminds a bit of the final topology, but it seems to be phrased a little bit differently (is this intuition correct?).  Usually one direction of a proof like this is easy, and the other direction should be a bit more challenging. I am however struggling with understanding the opens of the space $\mathscr F (J, Y)$. Usually the smallest topology containing some subset is referred to as being generated as a subbase. In an earlier question we proved that it is sufficient to check the continuity requirement for elements of the subbase. I have proved the following lemma in the previous question:

Let $Y$ be a set and $S \subset \mathscr P (Y)$ a collection of subsets of $Y$. Equip $Y$ with the smallest topology $\mathscr T_Y$ on $Y$, containing $S$. Show that a function $f: X \to Y$ is continuous if and only if $f^{-1}(U)$ is open in $X$ for every $U \in S$.

It seems that I will be able to use this lemma, but I am a bit overwhelmed as where to start and in which direction to move. Can somebody help me with a rough outline (what is the easy direction, what do I wish to show, how do I use the properties?)

What I have so far (not a lot):

"$\implies$"
Suppose that $\Phi: X \to \mathscr F (J,Y)$ is continuous. Let $j\in J$. We look at $g_j: X \to Y$ defined by $g_j(x)=(\Phi(x))(j)$. Let $V$ be an open set in $Y$, so $V\in \mathscr T_Y$.

"$\impliedby$"
Assume that for each index $j \in J$ we have that the map $g_j:X \to  Y$  defined by $g_j (x) =(\Phi(x)) (j)$ is continuous. Let $V$ be open in $Y$. Then we have $V \in \mathscr T_Y$. Now also let $j\in J$ be fixed. We then have that $g_j: X \to Y$ is continuous so $g_j^{-1}(V) \subset \mathscr F (J,Y)$ is open. ...

答:
HINT: For each $j\in J$ let $Y_j$ be a copy of the space $Y$; then $\mathscr{F}(J,Y)$ with this topology is just the product $\prod_{j\in J}Y_j$ with the product topology. The set

$$\{f\in\mathscr{F}(J,Y):f(j)\in V\}\tag{1}$$

is the same as $\prod_{i\in J}U_i$, where $U_j=V$, and $U_j=Y_j=Y$ for $j\in J\setminus\{i\}$, and those sets are the usual subbase for the product topology. Alternatively, if $\pi_j$ is the usual projection map from this product to the $j$-th factor, the set in $(1)$ is simply $\pi_j^{-1}[V]$.


答:
If $m+n$ is even, then $\Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor = \frac{m+n}{2}$ and as $m-n=m+n -2m$ is also even, $\Bigl\lfloor \frac{m- n+1}{2}\Bigr\rfloor = \frac{m-n-1}{2}$.

So $\Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor + \Bigl\lfloor \frac{m- n +1}{2}\Bigr\rfloor = m$

If $m+n$ is odd, then $\Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor = \frac{m+n-1}{2}$ and as $m-n=m+n -2m$ is also odd, $\Bigl\lfloor \frac{m- n+1}{2}\Bigr\rfloor = \frac{m-n}{2}$.

So $\Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor + \Bigl\lfloor \frac{m- n +1}{2}\Bigr\rfloor = m$

Title: Identity with the floor function

Body:
I'm struggling to complete proof of the following identity:

$$
\Bigl\lfloor \frac{m+ n}{2} \Bigr\rfloor +  \Bigl\lfloor \frac{m - n +1}{2} \Bigr\rfloor =m,
$$
where $m$ and $n$ are both integer.
By definition of floor function, $x-1 < \lfloor x \rfloor \leq x$.

Then
\begin{align}
\frac{m+n}{2} -1 <  & \Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor  \leq  & \frac{m+n}{2} \\
\frac{m-n + 1}{2} -1 < & \Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor  \leq & \frac{m-n +1}{2}.
\end{align}
By adding member to member, we obtain the following result:
$$
 m +\frac{1}{2} -2 <   \Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor + \Bigl\lfloor \frac{m- n +1}{2}\Bigr\rfloor  \leq  m +\frac{1}{2}.
$$
Which implies

$$
 -1.5 < \Bigl\lfloor \frac{m+ n}{2} \Bigr\rfloor + \Bigl\lfloor \frac{m- n +1}{2}  \Bigr\rfloor -m   \leq  .5
$$
This results in

$$
\Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor + \Bigl\lfloor \frac{m- n +1}{2}\Bigr\rfloor \in \left\{0,1\right\}.
$$

How to decide that the result is $0$?

Any help is welcome.


问:
Roots of $P_n(z)=1 + z + z^2 + ... + z^n$
So I know that the factorization of $z^n-1$ is:
$$z^{n}-1=\prod_{k=0}^{n-1}(z-e^{i(\frac{2\pi}{n}k)})$$
But I don't see how to factorize the monic polynomial $1+z+...+z^n$, where none of the coefficients is null.

Body:
Hint:

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


答:
$|x|$ is defined to be $x$ if $x\ge 0$ or $-x$ if $x<0$, so for every number $x$ there is a well-defined number $|x|$. $|\cdot|$ is, in other words, a function.

On the other hand, $\pm$ is not a function, it is a shorthand to avoid writing a long expression twice if the only modification you are making is converting the sign $+$ to $-$. For example, if you say: $y=\sqrt{x\pm\sqrt{x^2-1}}$, what you are really saying is:

$$y=\sqrt{x+\sqrt{x^2-1}}\text{ or }y=\sqrt{x-\sqrt{x^2-1}}$$

So, for example, "$y=\pm x$" means "$y=x\text{ or }y=-x$" - no more and no less. This is equivalent to $|y|=|x|$. In general, this is not equivalent with $y=|x|$ (because in general $y=|y|$ does not hold), but in some cases it may be equivalent if, from the context, you can somehow independently conclude that $y\ge 0$.

Title: Is $±x$ the same as $ |x|$?

Body: When solving a quadratic equation, the ± before the square root denotes "plus or minus that value." When solving absolute value equations, you try to solve using the positive and negative versions of the value inside. Does this mean that $y=|x|$ could be rewritten as $y=±x$?


Title: Roots of $P_n(z)=1 + z + z^2 + ... + z^n$

Body:
So I know that the factorization of $z^n-1$ is:
$$z^{n}-1=\prod_{k=0}^{n-1}(z-e^{i(\frac{2\pi}{n}k)})$$
But I don't see how to factorize the monic polynomial $1+z+...+z^n$, where none of the coefficients is null.

Body: Notice that : $$ P_{n}\left(z\right)=\sum_{k=0}^{n}{z^{k}}=\frac{z^{n+1}-1}{z-1}=\prod_{k=1}^{n}{\left(z-\mathrm{e}^{\mathrm{i}\frac{2k\pi}{n+1}}\right)} $$


答:
As $$\displaystyle{z^{n+1}-1=\prod\limits_{k=0}^{n}(z-e^{i(\frac{2\pi}{n+1}k)})}$$ and $$\displaystyle{z^{n+1} -1 = (z-1)(1+\cdots+z^n)}$$ we get that :
$$\displaystyle{(1+\cdots+z^n) = \prod\limits_{k=1}^{n}\left(z-e^{i(\frac{2\pi}{n+1}k)}\right)}$$


问:
Does this sum involving Champernowne's constants ever converge?
$\sum\limits_{n=2}^{\infty} C_n$, where $C_n$ is the base-$n$ Champernowne constant.
I tried computing it using Desmos, and it appears to grow very slowly. I'm not sure if it actually does converge though, as it's hard to apply any of the tests to this monster of a generating function:

Any ideas/insights?

答: I can't add a comment but $C_n$  > 1/n.

Body: I can't add a comment but $C_n$  > 1/n.


问:
How to prove or statements
How do I prove statements of the following types:

$A \text{ or } B \implies$ C

$A \implies B \text{ or } C$

I don't know how to go about proving statements like this when they have "or". Can someone tell me different ways to prove such statements? And maybe give me some basic examples to help?

Thank-you.

答:
I'll handle a different kind of goal statement first, namely: $P\implies Q$.

To prove this kind of statement you can opt between:

Assume $P$ is true, $(\cdots)$, conclude $Q$.

Prove the logically equivalent statement $\neg Q\implies \neg P$, by using the same technique as above: assume $\neg Q$ is true, $(\cdots)$, conclude $\neg P$.

Assume $P$ is true. Now assume, hoping to find a contradiction, that $Q$ is false. Find a contradiction and conclude that the last assumption you've made ($\neg Q$) is false and conclude $Q$.

Consider the tautology $\neg P\lor P$ and do a proof by cases. If $\neg P$ is true, that $P\implies Q$ is logically true. If $P$ is true, you're back to 1.

To prove a goal statement that looks like $P\lor Q$ you can try:

Assume $\neg P$ to be true, $(\cdots)$, conclude $Q$ or assume $\neg Q$, $(\cdots)$, conclude $P$.

From whatever assumptions you have try to get $P$, from where $P\lor Q$ follows or try to get $Q$, from where $P\lor Q$ follows.

Now there's something you should note. First and foremost the statements $\color{grey}(A\lor B\color{grey})\implies C$ and $A\implies \color{grey}(B\lor C\color{grey})$ are statements of the kind $P\implies Q$. Only at a later stage does the disjunction come into play. Notice the ghost parentheses.

Now note that they are very, very distinct. One of them will eventually have a goal which is a disjunction while the other has the disjuction as an hypothesis. Above I handled the case where the goal is a disjunction.

When you're hypothesis is a disjuction $P\lor Q$ and you want to prove $R$. You can try proof by cases: sinse you're hypothesis is true then either $P$ is true or $Q$ is true:

Suppose $P$ is true, $(\cdots)$, conclude $R$.

Suppose $Q$ is true, $(\cdots)$, conclude $R$.

Finally conclude that $\color{grey}(P\lor Q\color{grey})\implies R$.

You can find lots of examples and a more detailed explanation in this amazing book: How to Prove It: A Structured Approach, by D.J. Velleman.

Title: How to prove or statements


问:
Solve Linear Least Squares with Squared $ {L}_{2} $ Norm Regularization (Tikhonov / Ridge Regression) with Non Negativity Constraint Using FASTA
The open source optimization library FASTA (Fast Adaptive Shrinkage/Thresholding Algorithm) is an efficient, easy-to-use implementation of the proximal gradient method for regularized optimization problems. Embeded solvers could solve LASSO, Ridge Regression and Matrix Complementry problems.

I want to solve the following problem using FASTA:
$$\arg \min_x \{||Ax-b||^2+\lambda||x||^2 \} \\ s.t.\quad  x\ge 0$$

Acordding to the follwing LASSO solver(in MATLAB):

```<code>%%  Define ingredients for FASTA
%  Note: LASSO solver solves min f(Ax)+g(x).
%  A and b are given.
%  f(z) = .5 ||z - b||^2
f    = @(z) .5*norm(z-b,'fro')^2;
grad = @(z) z-b;
% g(z) = 0 for all feasible z, and infinity otherwise.  However, iterations
% always end with a feasbile z, so we need not consider the infinity case.
g = @(x) 0;
% proxg(z,t) = argmin t*mu*|x|+.5||x-z||^2
prox = @(z,t) project_1ball(z,mu);

%% Call solver
[solution, outs] = fasta(A,At,f,grad,g,prox,x0,opts);
</code>```

I write the following nonnegative constraned Ridge Regression Solver:

```<code>%%  Define ingredients for FASTA
%  Note: fasta solves min f(Ax)+g(x).
%  f(z) = .5 ||z - b||^2
f    = @(z) .5*norm(z-b,'fro')^2;
grad = @(z) z-b;

g = @(x) 0.5*mu*norm(x,'fro')^2;
prox = @(x,t) mapping(x,t*mu);

%% Call solver
[solution, outs] = fasta(A,At,f,grad,g,prox,x0,opts);

%% Embeded function
function x = mapping(x,tau)
    x = x+tau;
    x(x<0) = 0;  % for non-negative constraint
end
</code>```

But it did not converge.
How to solve the NON-NEGATIVE constrained ridge regression problem using FASTA?

答:
The FASTA package basically uses some heuristic to better utilize acceleration, momentum methods and line search methods.

Their paper, A Field Guide to Forward Backward Splitting with a FASTA Implementation, describes the method. For background I'd also recommend reading:

Understanding the Role of Momentum in Stochastic Gradient Methods.

Why Momentum Really Works.

I implemented 4 methods for your problem to benchmark different approaches:

Vanilla Projected Gradient Descent.

Projected Gradient Descent with Momentum.

Projected Gradient Descent Accelerated (Nesterov).

Projected Gradient Descent Accelerated (FISTA).

Those are the results I got with comparison to a reference given by CVX:

The MATLAB code is available at my StackExchange Mathematics Q3872982 GitHub Repository.


问:
Show a partition of $\Bbb{R}$ with the following properties (or just show that one exists)
I need to show a partition of $\Bbb{R}$, call it $P$, s.t.:

$|P| = \aleph$

For all $X\in P$ we have $|X| = \aleph $

My aproch was to define an equivilance relation $S$ of $\Bbb{R}$ s.t. for $x,y\in \Bbb{R}$ whe have $xSy$ if and only if $x-y\notin \Bbb{Q}$ or $x=y$. Hence, for $X\in \Bbb{R}/\Bbb{Q}$ we can take $a\in X$ and so $X=\{a+r|r\in \Bbb{R}-\Bbb{Q}\}$ and $|X| = |\Bbb{R}-\Bbb{Q}| = \aleph$. Now the challenging part is showing that $|\Bbb{R}/\Bbb{Q}|=\aleph$, clearly $|\Bbb{R}/\Bbb{Q}|\geq\aleph_{0}$ since for two rational numbers $a,b\in \Bbb{Q}$ we have $a-b\in \Bbb{Q}$ (since $\Bbb{Q}$ is a field) and hence each rational number has its uniqe equivalence class. Also, $\bigcup_{q\in \Bbb{Q}}[q]_{S}\neq\Bbb{R}$ since, for example, $1\notin \bigcup_{q\in \Bbb{Q}}[q]_{S}$.

And here I'm pretty much stuck so any help would be appreciated, I need to show such a partition exists but I decided to try and find one, So if you can help me with that, I would be very thankful.

答: We can give an explicit construction.  It is easier to work on $[0,1)$.  Express each real by its binary representation, using the terminating version where there is ambiguity.  Our sets in $P$ will be identified with the infinite bit strings by taking the odd position bits of each real to decide on the class it belongs in.  The even position bits give continuum many elements in each set, so we have partitioned $[0,1)$ into continuum many sets of size continuum.


答:
$V_1 = 10 x_1 y_1, V_2 = 10 x_1 y_1, V_3 = 10 x_1 y_1$

where $0 \leq x_n, y_n \leq 10\in \mathbb{Z+}$

$V_1 + V_2 + V_3 = 1000$

So reduce the problem to finding $min(x_1y_1)$

and we write for ease of working, $a = x_1 y_1, b = x_2 y_2, c = x_3 y_3$.

So we have $a + b + c = 100$ with $a \geq b, a \geq c$

So $a \geq 34$

Now you should look for two factors of $a$, both $\leq 10$.

So yes the first one is $35 = 5 \times 7$ and we are left with $b + c = 65$ and $b, c \geq 30$ (If any of them is $\lt 30$, the other one becomes $\gt a \, ( = 35)$ which is not allowed. So the possible values with both factors $\leq 10$ are $30 = 3 \times 10$ and $35 = 5 \times 7$ and they meet the criteria of $a \geq b, a \geq c$.

Hence $a = 35$ works with $min (V_1) = 350$.

Title: Minimum volume for cuboid with specific properties in relation to other cuboids

Body:
I have lately started to practice some math for fun.
I have come across the following exercise:

"There are three cuboids with the volumes $V_1=10x_1y_1$, $V_2=10x_2y_2$ and $V_3=10x_3y_3$. These must meet the following criterias $V_1\geq V_2$, $V_1\geq V_3$ and $V_1+V_2+V_3=1000$ with $x_n$ and $y_n$ being natural numbers $0\leq x_n$, $y_n\leq 10$.
What is the smallest value for $V_1$?"

Firstly I tried an inelegant exhaustive search through a small script I wrote. With that I came up with the solution of $V_1=350$, which seems plausible but i would like something more elegant than that.
I have also thought of representing the volumes like this:
$V_1=1000-(V_2+V_3)$, $V_2=1000-(V_1+V_3)$, $V_3=1000-(V_1+V_2)$. I hoped to make progress with that but it seems like a dead end to me because all the expressions are dependent on each other so you can't really graph how they affect each other until you already know two volumes.

So I am asking what might be a more elegant approach to the exercise above?


问:
Can we multiply between two Derivative operator?
I don't know what should be the tile of my question sorry for that ..

I will explain here clearly ...

Suppose this is my integral :

$\int \frac{d^2x}{dt^2}dx$

I have to show that the integral is equal to $(\frac{dx}{dt})^2$

My steps :

$=\int \frac{d}{dt}(\frac{dx}{dt})dx$

$=\int d\frac{d^2x^2}{d^2t^2}$  Note:  Here I am multiplying  between the two operator like $dx \cdot dx=d^2x^2 $ and $dt \cdot dt=d^2t^2 $

$=(\frac{dx}{dt})^2$

So now my question is can we multiply like the way I did according to my note . If my procedure is wrong how can i show $(\frac{dx}{dt})^2$ from the integral

答:
You are aware that
$$\frac{d}{dt}\left(\frac{dx}{dt}\right) = \frac{d^2x}{dt^2},$$
because you used that identity in your first step.
But why isn't it
$$\frac{d}{dt}\left(\frac{dx}{dt}\right) \stackrel?= \frac{d^2x}{d^2t^2},$$
as implied by your note,
"Here I am multiplying  between the two operator like ... $dt \cdot dt=d^2t^2$"?

Also, consider the U-substitution $u = x^2$:
$$
\int du = \int d(x^2) = \int 2x\,dx.
$$
A factor of $2$ shows up that is not accounted for simply by taking factors on the left of the $d$ and multiplying them on the right instead.
Indeed, if we followed your operators-as-factors approach, we would have gotten
$\int x\,dx$ as the last step.

There are a lot of legitimate uses of the $\frac{dx}{dt}$ that look like simple multiplication and/or division of parts of the formulas, but the fact that we have a name for a fact like the Chain Rule,
$\frac{dy}{dx} \frac{dx}{dt} = \frac{dy}{dt},$
should tip you off to the fact that none of these facts about derivatives are simple applications of multiplication and division of the pieces of the notation.
Whatever manipulation you do with derivatives, you need to be able to point back to a formula in a textbook that says you could do exactly what you're doing in exactly the way you're doing it, or you're liable to get zero credit on your homework or exam (and rightly so, because what you did here isn't mathematics, it's just wild guesswork).

The reason you're allowed to use the formula from the book is that someone has proved that the formula actually works that way.
And these proofs never involve just "multiplying  between the two operator" like anything.


问:
Product of random numbers.
Find the constant $p$ such that the product of any (positive) number $N_0$ multiplied by successive random numbers between $0$ and $p$ will, on average, neither diverge to infinity nor converge to zero.

(This is not for homework or anything. This is a problem I "invented" and solved, and now want to see how real mathematicians would think about this. Take your time.)

(Thanks md2perpe for fixing up my tag.)

(Thanks Forester for making the variables render nicely.)

答:
Thanks everyone. It turns out $p=e$ is exactly right, which all hinges on using the logarithm, and then transforming an infinite sum into an integral. My steps were pretty much the same:

Let $\tilde{N}$ denote the product after $j$ iterations. Then, we have $$\frac{\tilde{N}}{N_0} = \left(p \cdot r_1\right) \cdot \left(p \cdot r_2\right) \cdot \left(p \cdot r_3\right) \cdot \cdots \:,$$ where $p$ is constant, and each $r_j$ is a random number between $0$ and $1$. Next, take the natural log of both sides, turning the product on the right into sum:$$\ln\left(\frac{\tilde{N}}{N_0}\right) = \ln\left(p \cdot r_1\right) + \ln\left(p \cdot r_2\right) + \ln\left(p \cdot r_3\right) + \cdots$$

To satisfy the question as stated, we have the left side (on average) resolving to zero, with the number of terms on the right approaching infinity. This effectively makes $r$ a continuous variable: after many many iterations, we have sampled all of the real numbers between zero and one. Point is, we may replace the sum by an integral: $$0 \approx \ln\left(\frac{\tilde{N}}{N_0}\right) = \int_0^1 \left(\ln p + \ln r\right) dr \:,$$ and the answer falls right out. Only $p=e$ satisfies the above.

Title: Product of random numbers.


问:
In a unit circle $x^2 + y^2 = 1$, why can you substitute the x and y with cos and sin?
In this video he talks about a point $(x,y)$ on the unit circle can be rewritten as $(\cos(\theta), \sin(\theta))$.

However, to my understanding, $\cos$ is the adjacent side over the hypotenuse ($\frac{x}{r}$) and sin is the opposite side over the hypotenuse ($\frac{y}{r}$).

Why is it okay to substitute the values of $(x,y)$ with the values of ($\frac{x}{r}$, $\frac{y}{r}$)?

答:
Consider a point $P=(x,y)$ on the unit circle but in the first quadrant, i.e. such that both its coordinates are positive.

Call $Q$ the projection of $P$ on the $x$-axis. The coordinates of $Q$ are $(x,0)$.

Now observe that the triangle $OPQ$ ($O$ being the origin) is right in $Q$.

So if you call $\alpha$ the angle in $O$ and recall that the length of the hypothenuse $OP$ is $1$, since it is the radius, you can recognize that
the length of $OQ$ is $\cos(\alpha)$ and the length of $PQ$ is $\sin(\alpha)$.
But this lengths are also $x$ and $y$ respectively, so that you eventually get
$$
P=(\cos(\alpha),\sin(\alpha)).
$$
So, you can now extend the $\cos$ and $\sin$ functions to angles outside the
limits $(0,\pi/2)$ (where we measure angles in radians instead of degrees) precisely as the coordinates of the point on the unit circle determined by the angle.

This extension of $\cos$ and $\sin$ is consistent with their usage in trigonometry which is explained in (high?) school.


问:
is there a tool for figuring out set combinations with limited numerical entries?
i'm sitting here with a pin-locked pad, trying to figure out which combination clicks, i have been given the hint that the code can only contain the numbers: "0" "9" "8" and that it's a 4-digit Pin.

Now i'm stressing here to figure out how exactly what code to try, someone told me that there's only a 160 possible combinations with those three numbers in a 4 digit required pattern, my question exactly is: How do i figure out all the 160 possible combinations, or is there a tool online which you can input the required numbers, with a set pattern of 4 digits?

Thanks in advance!

答:
If that is what you are after, the following Python program will dump all $3^4=81$ codes for you:

```<code>DIGITS=['0','8','9']

for one in DIGITS:
    for two in DIGITS:
        for three in DIGITS:
            for four in DIGITS:
                print(f"{one}{two}{three}{four}")
</code>```

I am sure you either have Python already on your computer, or you can obtain it, or at worst you can get this code to run on some online Python interpreter.

Body:
If that is what you are after, the following Python program will dump all $3^4=81$ codes for you:

DIGITS=['0','8','9']

for one in DIGITS:
    for two in DIGITS:
        for three in DIGITS:
            for four in DIGITS:
                print(f"{one}{two}{three}{four}")

I am sure you either have Python already on your computer, or you can obtain it, or at worst you can get this code to run on some online Python interpreter.


问:
Identity with the floor function
I'm struggling to complete proof of the following identity:

$$
\Bigl\lfloor \frac{m+ n}{2} \Bigr\rfloor +  \Bigl\lfloor \frac{m - n +1}{2} \Bigr\rfloor =m,
$$
where $m$ and $n$ are both integer.
By definition of floor function, $x-1 < \lfloor x \rfloor \leq x$.

Then
\begin{align}
\frac{m+n}{2} -1 <  & \Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor  \leq  & \frac{m+n}{2} \\
\frac{m-n + 1}{2} -1 < & \Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor  \leq & \frac{m-n +1}{2}.
\end{align}
By adding member to member, we obtain the following result:
$$
 m +\frac{1}{2} -2 <   \Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor + \Bigl\lfloor \frac{m- n +1}{2}\Bigr\rfloor  \leq  m +\frac{1}{2}.
$$
Which implies

$$
 -1.5 < \Bigl\lfloor \frac{m+ n}{2} \Bigr\rfloor + \Bigl\lfloor \frac{m- n +1}{2}  \Bigr\rfloor -m   \leq  .5
$$
This results in

$$
\Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor + \Bigl\lfloor \frac{m- n +1}{2}\Bigr\rfloor \in \left\{0,1\right\}.
$$

How to decide that the result is $0$?

Any help is welcome.

答:
Notice that $\forall x \in \mathbb Z, \left\lfloor \frac x2   \right\rfloor = \frac x2 - \left\lfloor \frac{x \mod 2}{2}\right\rfloor$, therefore

$$\left\lfloor \frac{m+n}{2}\right\rfloor + \left\lfloor \frac{m-n+1}{2}\right\rfloor\\
=\frac{m+n}{2}  - \left\lfloor \frac{m+n \mod 2}{2}\right\rfloor + \frac{m-n+1}{2}- \left\lfloor \frac{m-n+1 \mod 2}{2} \right\rfloor$$
$$= m + \frac 12 - \left\lfloor \frac{m+n \mod 2}{2}\right\rfloor - \left\lfloor \frac{m-n+1 \mod 2}{2} \right\rfloor\tag1$$

Since $m+n$ and $m-n$ have the same parity, the last two terms in $(1)$ sum to $-\frac 12$, and we are done.

Title: Identity with the floor function


问:
Is checking really needed?
I recently bought Pinter's A book of Abstract Algebra and I came to Examples (Chapter 2). Now, I do not understand why are "Checks" in (iii), (iv) needed because we deductively concluded that neutral element $e$ must be $- 1$ and simillary we showed that each inverse $x'$ of $x$ is $x'= - x - 2$ (of course, we concluded all this after assuming that symbol $+$ means regular adittion with all its properties).

Therefore, Checks are just surplus because we’ve already set up to get what we’re “checking” now. (Probably I am just wrong so please correct me.)

Here is the 1st example in part B:

答:
The first line of (iii) establishes the implication "if $e$ has the given property, then $e=-1$." But maybe no number has the given property—this if-then statement hasn't established that $-1$ has the given property, but only that it's the lone candidate. The "Check" actually establishes the implication "$-1$ has the given property".

This is actually a common nuance in high-school algebra in general, not just in this context. Personally I might label the two steps "Explore" and "Proof" instead of "Solve" and "Check" to clarify the difference.

Similar remarks apply to (iv): the first sentence is giving necessary conditions on the possible answer, but the "Check" is what establishes that the proposed answer is actually correct.

Title: Is checking really needed?


问:
prove that $\operatorname{Hom}_\Bbb{Z}(\Bbb{Z}/n\Bbb{Z},A)\cong A_n.$
Let $A$ be any $\Bbb{Z}$-module, let $a$ be any element of $A$ and let $n\in \Bbb{Z}^+$. Prove that the map $\psi_a\colon\Bbb{Z/nZ}\to A$ given by $\psi_a(\overline{k})=ka$ is a well-defined $\Bbb{Z}$-module homomorphism if and only if $na=0$.

Then, prove that $\text{Hom}_\Bbb{Z}(\Bbb{Z}/n\Bbb{Z},A)\cong A_n$, where $A_n=\{a\in A: na=0\}$.

I am struggling to find a map so that i can prove the isomorphism between $\text{Hom}_\Bbb{Z}(\Bbb{Z}/n\Bbb{Z},A)$ and $A_n$.

答:
Another way:
$\newcommand{\Z}{\mathbb Z} \newcommand{\Hom}{\operatorname{Hom}}$

In the previous exercise you show that every $a \in A$ such that $na=0$ (that is, $a \in A_n$) determines a $\Z$-module homomorphism $\psi_a : \Z/n\Z \to A$ (that is, $\psi_a \in \Hom_\Z(\Z/n\Z,A)$) given by $\psi_a(m+n\Z) = ma$. Thus, the map $\psi : A_n \to \Hom_\Z(\Z/n\Z,A)$ given by $\psi(a) = \psi_a$ is well-defined and in fact is a $\Z$-module homomorphism (note that $A_n$ and $\Hom_\Z(\Z/n\Z,A)$ are both $\Z$-modules) because if $a,b \in A_n$, then
$$(\forall k \in \Z) (\forall m \in \Z) \ m(a+kb) = ma+kmb$$ tells us that
$$(\forall k \in \Z) (\forall m \in \Z) \ \psi_{a+kb}(m+n\Z) = (\psi_a+k\psi_b)(m+n\Z)$$ and therefore
$(\forall k \in \Z) \ \psi_{a+kb} = \psi_a+k\psi_b$, that is $$(\forall k \in \Z) \ \psi(a+kb) = \psi(a)+k\psi(b).$$ Also, note that $\psi$ is an isomorphism since $\phi : \Hom_\Z(\Z/n\Z,A) \to A_n$ given by $\phi(f) = f(1+n\Z)$ is a well-defined set-map (just check that $f(1+n\Z) \in A_n$ for every $f \in \Hom_\Z(\Z/n\Z,A)$) such that $$\psi \phi = \operatorname{id}_{\Hom_\Z(\Z/n\Z,A)} \quad \textrm{and} \quad \phi \psi = \operatorname{id}_{A_n}$$ (how is that $\phi$ is a $\Z$-module homomorphism follows from the fact that $\psi$ is a $\Z$-module homomorphism?).


问:
Partitions of $n$: proving $p(n+2)+ p(n) \geq 2p(n+1)$

For $n \geq 2$ give an alternative description of $p(n) - p(n-1)$ as the number of partitions of $n$ which have a certain property.

I have done that part, it is fine. I have not included it here as I am not accustomed with LaTeX but can put it up if wanted. But then it says

Hence or otherwise prove $p(n+2)+ p(n) \geq 2p(n+1)$.

This is the part I do not know how to do. Any help appreciated thank you

答:
The difference $p(n)-p(n-1)$ is the number of partitions of $n$ that do not contain 1/with smallest part (wsp) 2 -- also denoted $p(2,n)$, where $p(\cdot,\cdot)$ is the intermediate function. Using Sharkos's hint, we understand that we aim to prove that $p(2,n)$ is increasing. One way to prove this is to find an injection that, to any partition of $n$ wsp 2, assigns a partition of $n+1$ wsp 2.

Clearly, by adding 1 to any such partition of $n$, we obtain a partition of $n+1$. But doing so by simply adding "+1" to the sum would not make it a valid partition of $n+1$ wsp 2 (the smallest part would be 1).

Instead, consider adding 1 to any of the existing addends of any given partition $P$ of $n$ wsp 2. Our goal is to ensure that we would not obtain the same partition of $n+1$ twice (since it wouldn't be an injection then), and this can be achieved by chosing to systematically add 1 to the greatest element of $P$.

Doing so, we obtain the desired injection, and therefore we prove that there are at least as many partitions of $n+1$ than of $n$ wsp 2.


问:
Primes represented by $x^3-21xy^2+35y^3$.
What do we know about the primes represented by this special binary cubic form $x^3-21xy^2+35y^3$?

I know that my question is very short, but I don't have any idea about it, and I don't know where can I find the answer in the literature.

I searched through the net to find a program to check whether a binary cubic equation $f(x, y)=n$ has a solution or not, but I didn't find anything. In the case of the absence of an answer, or a reference to my question, introducing any program/ engine would be welcome.

Let $\alpha$ be a root of the polynomial $x^3-21x-35=0$, and let $K:=\mathbb{Q}(\alpha)$. Then it is easy to show that $$Norm(x+y\alpha+z\alpha^2)=x^3+35y^3+1225z^3-105xyz-21xy^2+441xz^2+42x^2z-735yz^2.$$ That Binary cubic form is just $Norm(x+y\alpha)$.

Note that the discriminant of $P(x)=x^3-21x-35$ is
$-(4\times(-21)^3+27\times(-35)^2)=3969=3^4\times7^2$, so discriminant of $K$ is a square, so it is a cyclic cubic Galois extension, so we can conclude that $r_1=3$ and $r_2=0$. By Dirichlet's unit theorem, we can conclude that $\mathcal{O}_K^{\times}=\{\pm1\}\times\mathbb{Z}^2$. Also, note that $P(x)=x^3-21x-35$ is $7$-Eisenstein, and $P(x-1)=x^3-3x^2+3x-1-21x+21-35=x^3-3x^2-18x-15$ is $3$-Eisenstein; so we can conclude that $\mathcal{O}_K=\mathbb{Z}[\alpha]=\mathbb{Z}\oplus\mathbb{Z}\alpha\oplus\mathbb{Z}\alpha^2$.

Is the answer to the following question positive?

Assume that $Norm(a+b\alpha+c\alpha^2)=p$. Does there exist a unit $u \in \mathcal{O}_K^{\times}$ such that $(a+b\alpha+c\alpha^2)\times u = A+B\alpha$ for some integers $A, B$? Assume that $a+b\alpha+c\alpha^2$ is given. Can we find a suitable unit, such that after the multiplication, we can write the product as linear combination of $1$, and $\alpha$? without need of $\alpha^2$?

答:
The discriminant of $x^3 - 21 x + 35$ is a square, many things drop out. The primes represented by the full norm form you give will be primes that are
$$ 1, 5, 8, 11, 23, 25, \pmod{63} $$
$$ 62, 58, 55, 52, 40, 38, \pmod{63} $$

There is more restriction, not obvious initially, it is a subgroup of the residues
$$ \color{red}{ 1,  8, 55, 62 \pmod{63} } $$
$$x^3+35y^3+1225z^3-105xyz-21xy^2+441xz^2+42x^2z-735yz^2.$$

What restrictions we get with $z=0$ are anyone's guess.

Note that $x^3 - 21 x + 35$ and $x^3 - 21 x + 28$ give different fields


问:
Are these two series really equal to each other? If so, why?
The conjecture we are trying to prove is the following:

$$ x \in \mathbb{R^+} \implies -\sum_{n=1}^{\infty} \frac{\ln(x)^n}{(2^n-1)n!} = \sum_{n=1}^{\infty} \left( 1 - x^{\frac{1}{2^n}}\right)$$

Note the left hand side is absolutely convergent while the right hand side is conditionally convergent. I have verified these for about 12 decimal digits to be equal. Moreover, they both obey the same functional equation $f(x^2) = f(x) -x + 1$.

The proof of the functional equation for the right hand side is trivial so we show the left hand side

$$  -\sum_{n=1}^{\infty} \frac{\ln(x^2)^n}{(2^n-1)n!}  = - \sum_{n=1}^{\infty} \frac{2^n\ln(x)^n}{(2^n-1)n!} = - \sum_{n=1}^{\infty} \left[\frac{\ln(x)^n}{(2^n-1)n!} + \frac{\ln(x)^n}{n!} \right] =  $$
$$-\sum_{n=1}^{\infty}\frac{\ln(x)^n}{(2^n-1)n!} - \sum_{n=1}^{\infty}  \frac{\ln(x)^n}{n!}= \sum_{n=1}^{\infty}\frac{\ln(x)^n}{(2^n-1)n!}-x+1$$

I have absolutely no reason to believe that just because these functions obey the same equation and are equal for at least 12 decimal digits (which is the limit of the calculator I'm using) that they are totally equal but this is a very odd coincidence to survive for so many digits.

Body:
They are in fact equal.

$$\sum_{n=1}^{\infty} \frac{\ln(x)^n}{(2^n-1)n!} =  \sum_{n=1}^{\infty} \frac{\left(\frac{\ln(x)}{2} \right)^n}{n!} \cdot \frac{1}{1-2^{-n}}= \sum_{n=1}^{\infty} \frac{\left(\frac{\ln(x)}{2} \right)^n}{n!} \cdot \sum_{k=1}^\infty(2^{-n})^{k-1}
$$
$$=\sum_{k=1}^\infty \sum_{n=1}^{\infty} \frac{\left(\frac{\ln(x)}{2^{k}} \right)^n}{n!} = \sum_{k=1}^\infty (e^{\frac{\ln(x)}{2^{k}}}-1)= \sum_{k=1}^\infty (x^{\frac{1}{2^k}}-1).$$


问:
Proof of the derivative of $\sin{x}$ is $\cos{x}$
I recently learned the proof that the derivative of $\sin x$ is $\cos x$ in Stewarts calculus book. However, in his proof he uses preconceived limit laws such as the sum and product law to evaluate the limit. My confusion is that these limit laws can only be used when the limit exists however we do not know the limit in this case we are trying to evaluate it so how can we use these limit laws must we resort to a geometric proof?

答:
There should not be any problem with limits...
$$\begin{align}\lim_{h\to 0} \frac{\sin(x + h) - \sin x}{h} &= \lim_{h\to 0} \frac{2\cos(x + \frac h2)\sin\frac h2}{h} \\&= \lim_{h\to 0} \left[\cos\left(x + \frac h2\right)\cdot\frac{\sin (h/2)}{h/2}\right] \\&\overset*= \lim_{h\to 0}\cos\left(x + \frac h2\right)\cdot\lim_{h\to 0}\frac{\sin (h/2)}{h/2} \\&= \cos x \cdot 1 \\&= \cos x\end{align}$$
Explanation for (*): Limit of product is equal to product of limits in this case since both expressions have limits


问:
Dual space of Bochner space
Let $B$ be a refexive Banach space. I want to show that
$$(L^2(0,T;B))^* = L^2(0,T;B^*)$$ and that
the dual pairing is
$$\langle F,f \rangle_{L^2(0,T;B^*), L^2(0,T;B)} = \int_0^T \langle F(t), f(t) \rangle_{B^*,B}.$$
Can anyone help me with either part? Thanks.</p>

答:
No proof in the generality that is desired is going to be very brief. But all the essential ideas come from the usual, scalar case. First we need the following:

Definition. A Banach space $V$ is said to have the Radon-Nikodym property with respect to a measure space $(X, \Sigma, \mu)$ if, for every $V$-valued vector measure of bounded variation $\nu$ which is absolutely continuous with respect to $\mu$, there exists a Bochner integrable $f \in L^1(X,V)$ such that $\nu(E) = \int_E f \, d\mu$ for every $E \in \Sigma$.

In other words, spaces with the Radon-Nikodym property are precisely those in which the generalization of the Radon-Nikodym theorem for scalar functions holds. Then the key result that one needs to prove is due to Phillips:

Reflexive Banach spaces have the Radon-Nikodym property.

A proof of this requires a decent amount of machinery and a proof of this fact and other related facts is the subject of Chapter 2 of Vector Measures by Diestel and Uhl. Now, with this fact under our belt, we have the following result:

Theorem. Let $(X, \Sigma, \mu)$ be a finite measure space, and let $V$ be a Banach space such that $V'$ has the Radon-Nikodym property. Then, for $p \in [1, \infty)$, $L^p(X,V)' \cong L^q(X,V')$ isometrically, where $1/p + 1/q = 1$.

Proof. Suppose $g \in L^q(X,V')$. Then, for $f \in L^p(X,V)$, the map $\Lambda_g(f) = \int_X \langle f, g \rangle \, d\mu$ is clearly a continuous linear functional, so $L^q(X,V') \subseteq L^p(X,V)'$. To prove the reverse inclusion, let $\Lambda \in L^p(X,V)'$. For $h \in L^{\infty}(X,V)$, $|\Lambda(h)| \le C \|h\|_p \le  C \|h\|_{\infty}$, and hence $\Lambda$ is also a continuous functional on $L^{\infty}(X,V)$. Now, $\Lambda$ induces a $V'$-valued vector measure absolutely continuous with respect with $\mu$ given by
$$ \langle \nu(E), x \rangle = \Lambda(x 1_E)$$
for each $x \in V$. Then, as $V'$ has the Radon-Nikodym property, then there exists $g \in L^1(X,V')$ such that $\nu(E) = \int_E g \, d\mu$. Then, by approximating each $f \in L^p(X,V)$ by simple functions, we have that $\Lambda(f) = \int_X \langle f, g\rangle \, d\mu$. Finally, it is straightforward to show that $g$ is in fact also in $L^q$.

Title: Dual space of Bochner space


Title: Proof of the derivative of $\sin{x}$ is $\cos{x}$

Body: I recently learned the proof that the derivative of $\sin x$ is $\cos x$ in Stewarts calculus book. However, in his proof he uses preconceived limit laws such as the sum and product law to evaluate the limit. My confusion is that these limit laws can only be used when the limit exists however we do not know the limit in this case we are trying to evaluate it so how can we use these limit laws must we resort to a geometric proof?

Body:
Let $$f(x)=\sin(x)$$
and $ a,h \in \Bbb R$.

$$A(h)=\frac{f(a+h)-f(a)}{h}=$$
$$\frac{\sin(a+h)-\sin(a)}{h}=$$

$$\frac{\sin(a)\cos(h)+\cos(a)\sin(h)-\sin(a)}{h}=$$
$$\frac{\cos(h)-1}{h}\sin(a)+\frac{\sin(h)}{h}\cos(a)$$
and,
$$\lim_{h\to 0}A(h)=0+1.\cos(a)=f'(a)$$


问:
In a group of $n>20$ people in which some pairs know each other, 20 people are removed. The number of pairs that know each other is $\frac{n-k}{n}$.
Here's the problem:

In a group of $n>20$ people, there are some (at least one, and possibly all) pairs of people that know each other. (If A knows B, then B knows A). For some $n$ and $k$, this group of people have a property: If any 20 people are removed from the group, the number of pairs of people that know each other is at most $\frac{n-k}{n}$ times that of the original group of people.
For $k=41$ and $k=39$, for what positive integers $n$ could a group exist?

I'm not quite sure where to start. I tried experimenting with different values of $n$ and various setups, but I'm not making any progress. I also tried representing this using a graph, so that might help anyone who's finding a solution. Thanks for the help.

答:
A partial solution:

If you examine the case where all $n$ people know each other, then there are $n\choose 2$ acquaintance pairs. After removing any $20$, there will be $n-20 \choose 2$ acquaintance pairs. So
$$\dfrac {n-20 \choose 2}{n \choose 2}\le \dfrac {n-k}n$$
$$\dfrac{(n-20)(n-21)}{n(n-1)} \le \dfrac {n-k}n$$
$$n^2 - 41n + 420 \le n^2 - (k+1)n + k$$
$$ 420 - k \le (40 - k)n$$
If $40 \le k < 420$, the equation requires a positive value to be $\le$ a non-positive value, which cannot be. But when $k < 40$, we get that if
$$\dfrac{420 - k}{40 - k} \le n$$
then there will be a solution where everyone knows everyone else.

But this does not say anything about solutions where some pairs do not know each other.


问:
Let $(a,b), (c,d)$ be two vectors in $\mathbb R^2$. If $ad-bc=0$ prove that they are linearly dependent
Let $(a,b), (c,d)$ be two vectors in $\mathbb R^2$. If $ad-bc=0$ prove that they are linearly dependent

My attempt: I tried to do it by contradiction: Suppose that they are linearly independent that is:

$\alpha_1(a,b)+\alpha_2(c,d)=\vec0$ then $\alpha_1=\alpha_2=0$; solving the system:
$\begin{cases}
\alpha_1a+\alpha_2c=0 \\[2ex]
\alpha_1b+\alpha_2d=0
\end{cases}$

$$\alpha_1ba-\alpha_1ba+\alpha_2bc-\alpha_2da=0 \Rightarrow \alpha_2(bc-ad)=0$$ hence either $bc-ad=0$ or $bc-ad\neq0$ if it is the latter one then we have a contradiction

I would like you to tell me if this is correct :)

答:
We need to prove that $\alpha(a,b)+\beta(c,d)=(0,0)$ have solutions for scalars $\alpha, \beta$ different of zero.

Then, solving for $\alpha, \beta $ you will have $\begin{bmatrix} a & b \\ c & d \end{bmatrix}
\cdot \begin{bmatrix} \alpha \\ \beta \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

Since a homogeneus system of equations have unique or infinite solutions, your unique option is to have infinite number of solutions and this occurs when $\text{rank}\left(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\right) < 2$ or equivalent $\det\left(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\right) = ad - bc = 0$.

So the solutions for $\alpha, \beta$ are infinite when $ad-bc=0$, that is, is linearly dependent.

You can generalize this for any eucliden vector of $\mathbb{R}^n$.


问:
Explanation of "inclusion reversing".
So I am currently self-studying Galois theory with no one to ask questions to and
that is why I am here. Anyways,
I was reading about the fundamental theorem of Galois theory on this website here https://jhavaldar.github.io/notes/2017/10/19/galoistheory1.html and I am stuck on Proposition 4:

"The above association is inclusion reversing:

If $F_1 \subseteq F_2 \subseteq K$ then $Aut(K/F_2) \leq Aut(K/F_1)$

$\vdots$

"

The proposition continuous and states other things as well. The part that I do not understand is "inclusion reversing", what is that? And what does it mean for $Aut(K/F_2) \leq Aut(K/F_1)$? Is this stating that the cardinality of $Aut(K/F_2)$ is less than $Aut(K/F_1)$?

答:
Recall that a partially ordered set (poset) is a tuple $(X, \leq)$ of a set together with a reflexive, transitive and antisymmetric relation on $X\times X$.

We can now define the following:

DEFINITION If $(X, \leq)$ and $(Y, \preceq)$ are posets, then an anti-monotone map (or antitone map) is a map $f\colon X\to Y$ such that $x\leq x'$ implies $f(x')\preceq f(x)$.

In the case where the partial order on both sets stems from an inclusion, we call an antitone map inclusion-reversing.
In particular, the statement now becomes:

PROPOSITION
Let $K/F$ be a normal and separable[1] field extension.
The map $f\colon F_1 \mapsto \operatorname{Aut}(K|F_1)$ is a well-defined antitone map between the posets

$(\{\text{Intermediate fields of}\ K/F\}, \subseteq)$

$(\{\text{Subgroups of}\ \operatorname{Aut}(K)\ \text{fixing}\ F\}, \leq)$.

Here, $\leq$ is the relation “is a subgroup of”. However, since the subgroups we are concerned with inherit their group structure from the big group $\operatorname{Aut}(F)$ they are all contained in, they are a subgroup if and only if they are included, i.e. $\leq$ is the same as $\subseteq$.

[1] I'm Not sure how much that can be weakened here, the precise demands are however irrelevant for the point I'm making.

Note that we can formulate antitone maps a bit differently using the notion of a reverse order:

Observation If $(X, \leq)$ is a partial order, then the relation $$
\geq\colon x\geq x' \Leftrightarrow x'\leq x
$$ turns $(X, \geq)$ into a partial order as well. We will call that relation the reverse order or opposite order of $(X, \leq)$.

Proposition A map $f\colon (X, \leq) \to (Y, \preceq)$ is antitone if and only if $f\colon (X, \leq) \to (Y, \succeq)$ is monotone.

Title: Explanation of "inclusion reversing".


问:
What does the "$\top\,$" in "$\,y = [x,u,v,y]^{\top}\,$" mean?
There is a diffrential equation which takes the form:

$$\frac{dy}{dt} = f(y,t) $$

$$y = [x,u,v,y]^{\top}$$

What does the superscript of $\top$ in this case mean?

答:
As has been pointed out in the comments, it stands for "transpose":
$$\begin{bmatrix} x & u & v & y \end{bmatrix}^T = \begin{bmatrix} x \\ u \\ v \\ y \end{bmatrix}$$


问:
Prove That $x=y=z$
If $x, y,z \in \mathbb{R}$,

and if

$$ \left ( \frac{x}{y} \right )^2+\left ( \frac{y}{z} \right )^2+\left ( \frac{z}{x} \right )^2=\left ( \frac{x}{y} \right )+\left ( \frac{y}{z} \right )+\left ( \frac{z}{x} \right ) $$

Prove that $$x=y=z$$

答:
Let $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$, and let
$$
G=\sqrt[3]{abc}, A = \frac{a+b+c}{3}, Q=\sqrt{\frac{a^2+b^2+c^2}{3}}
$$
Then the given condition is $Q^2=A$, but by the power mean inequality
$$
Q^2\ge Q \ge A \ge G = 1
$$
with equality in each case only if $a=b=c=1$, i.e. if $x=y=z$.</p>

Title: Prove That $x=y=z$


问:
Matrix with basis and dimensions
I have recently come upon this question and I have no idea how to go on from where I stopped :

$$\text{GivenMatrix}
 :\: A=\begin{pmatrix}
-1&a&3 \\ a & -a^2+a+2 & -4a-2 \\ a-1 & -a^2+2a+2 & -4a+1
\end{pmatrix}
$$

Find for every value of $a$ the basis and the dimension for Nul(A)

Is there a value for a that for it there are $2$ vectors $(\vec{u_1},\vec{u_2})$ not linearly independent that create $A×\vec{u_1}=0$ and $A×\vec{u_2}=0 \,$? If yes, find the value for $a$; if not prove it .

While $a=-2$, prove that for every $2$ vectors $\vec{u},\vec{v} ∈\Bbb R^3$ that create $\vec{u} \in$ Nul$(A)$, $\vec{v} \in$ Nul$(A)$, $\vec{u} \neq \vec{0}$.

I played with the matrix a little bit and reached this point :
$$
A=\begin{pmatrix}
a+2 & 0 & 4a+8 \\ 0 &a+2 & -a-2  \\ 0 & 0 & 0
\end{pmatrix}
$$
I have no clue what to do after this can anyone throw in some useful tips or solutions to help me get going? Thanks in advance.

答:
For the first quesion I get the matrix
\begin{pmatrix}
-1 & a & 3\\
0 & a+2 & -(a+2)\\
0 & 0 & 0
\end{pmatrix}
by $-R_1 + -R_2 + R_3 \Rightarrow R_3$ and $-aR_1 +R_2 \Rightarrow R_2$.

If $a=-2$ then

$$ \left[
\begin{array}{ccc|c}
-1&-2&3&0\\
0&0&0&0\\
0&0&0&0
\end{array}
\right] $$

and $x=-2y+3z$ gives the vector $\begin{pmatrix}
-2y+3z\\
y \\
z
\end{pmatrix}=y\begin{pmatrix}
-2\\
1 \\
0
\end{pmatrix}+z\begin{pmatrix}
3\\
0 \\
1
\end{pmatrix}$ and

$dim($Nul$(A))=2.$

If $a\neq-2$ then you can divide $R_2$ by $a+2$

$$\begin{pmatrix}
-1 & a & 3\\
0 & 1 & -1\\
0 & 0 & 0
\end{pmatrix}\xrightarrow{-aR_2 + R_1 \Rightarrow R_1}\begin{pmatrix}
-1 & 0 & 3+a\\
0 & 1 & -1\\
0 & 0 & 0
\end{pmatrix}$$

$x=(3+a)z$ and $y=z$ gives the vector $\begin{pmatrix}
(3+a)z\\
z \\
z
\end{pmatrix}=z\begin{pmatrix}
3+a\\
1 \\
1
\end{pmatrix}$ and $dim($Nul$(A))=1.$

You cannot divide rows by zero so you need to split it into two cases, $a+2=0$ and $a+2 \neq 0$ and then see what happens.

Title: Matrix with basis and dimensions


问:
Show that two functions have the same minimizer
Define the following two functions:
\begin{align}
g(x)&=1_{\{  \langle v_1,x \rangle  \ge 0 \} }+1_{\{  \langle v_2,x \rangle \ge 0  \} },\\
f(x)&=1_{\{  \langle v_1,x \rangle  \ge 0 \} } \langle v_1,x \rangle +1_{\{  \langle v_2,x \rangle \ge 0  \} } \langle v_2,x \rangle,
\end{align}
where the vectors $v_1$ and $v_2$ are given.  Here, as usual,  $1_{ \{\cdot \}}$, denotes the indicator function. We further restric $x$ to be $\|x\|=1$.

I would like to show that if $x^\star$ is a minimizer of $f(x)$, then it is also a minimizer of $g(x)$.

Intuitively a minimizer of $f$ forces the inner products to be either negative or very close to zero and this, in turn, should minimize $g$.  However, I am not very sure what arguments to use to show this.

Edit: Based on comments this appears to be true up to a set of measure zero for $(v_1,v_2)$.  However, not all details are clear to me yet.

答:
The set of minimisers is almost the same, basically they can differ by a set of measure $0$ when $v_1$ and $v_2$ are not colinear with different sign.

Assume first that $v_1$ and $v_2$ are not colinear with different sign. In essence, we assume that $v_1 \ne -c v_2$ for some $c > 0$. We observe that $g(x)$ and $f(x)$ have image in $[0,\infty)$ so that if there exists $x$ such that $f(x) = 0$, then the minimising set for $f$ is its zero set, and similarly for $g$. Set

$$\hat x = \left(\frac{v_1}{\|v_1\|} +  \frac{v_2}{\|v_2\|}\right)$$
and
$$x^\star = \frac{\hat x}{\|\hat x\|}$$

It is not too hard to see that $f(x^\star) = g(x^\star) = 0$ and that $\|x^\star\| = 1$, the latter condition holding because $v_1 \ne -c v_2$.  Therefore, since both $f$ and $g$ are nonnegative functions, the set which Now, it is clear that $g(x) = 0$ implies $f(x) = 0$. However, there are always vectors such that $f(x) = 0$ while $g(x) \ne 0$. To have $g(x) = 0$, we need two conditions: $\langle x,v_1 \rangle < 0$ and $\langle x,v_2\rangle < 0$, whereas to have $f(x) = 0$ we need $\langle x,v_1\rangle \le 0$ and $\langle x, v_2 \rangle \le 0$. We therefore have that
$$
\{x : f(x) = 0 \text{ and } g(x) \ne 0\} = (\operatorname{span}(v_1)^\perp \cap \{x : \langle x,v_2\rangle \le 0\}) \cup (\operatorname{span}(v_2)^\perp \cap \{x : \langle x,v_1\rangle \le 0\}).
$$
This is the union of two intersections of a half-space with a codimension 1 subspace, so it has codimension 1 in the unit sphere.

If $v_1 = -c v_2$ for some $c > 0$, the picture is completely different. Then,
$$
f(x) = \begin{cases}
0 & \text{if } x \in \operatorname{span}(v_1)^\perp = \operatorname{span}(v_2)^\perp \\
> 0 & \text{otherwise },
\end{cases}
$$
whereas
$$
g(x) = \begin{cases}
2 & \text{if } x \in \operatorname{span}(v_1)^\perp = \operatorname{span}(v_2)^\perp \\
1 & \text{otherwise }.
\end{cases}
$$
This means that in this situation their minimising sets are completely disjoint.


答:
Checking is needed only due to (needless) use of $\rm\color{#c00}{unidirectional}$ (vs. $\rm\color{#0a0}{bidirectional})$ arrows, i.e.

$$\begin{align} &\overbrace{x+e+1}^{\textstyle x*e} = x\ \color{#c00}\Longrightarrow\, e+1 = 0\ \color{#c00}\Longrightarrow\, e = -1\\[.2em]
{\rm Better}\!\!:\ \ &x+e+1 = x\!\color{#0a0}\iff\! e+1 = 0\!\color{#0a0}\iff\! e = -1\end{align}\qquad\qquad$$

The $\rm\color{#c00}{unidirectional}$ inferences yield only that $\,e=-1\,$ is necessary for the identity to hold, not that it is sufficient. For sufficiency (reverse direction) we can either use the $\rm\color{#0a0}{bidirectional}$ arrows, or else explicitly check that the identity holds for  $\,e = -1$.

As in the above inferences (which cancel $\,x\,$ then add $\,-1$) many equational inferences are naturally bidirectional, i.e. they yield equivalent equations, so there is no need to  restrict their power by using them only unidirectionally.

Remark $ $ These painstaking verifications can be eliminated if we simply note that the essence of the matter boils down to the fact that the operation arises simply via  transport of structure.

Title: Is checking really needed?

Body:
I recently bought Pinter's A book of Abstract Algebra and I came to Examples (Chapter 2). Now, I do not understand why are "Checks" in (iii), (iv) needed because we deductively concluded that neutral element $e$ must be $- 1$ and simillary we showed that each inverse $x'$ of $x$ is $x'= - x - 2$ (of course, we concluded all this after assuming that symbol $+$ means regular adittion with all its properties).

Therefore, Checks are just surplus because we’ve already set up to get what we’re “checking” now. (Probably I am just wrong so please correct me.)

Here is the 1st example in part B:


问:
What is the difference between "expectation", "variance" for statistics versus probability textbooks?
It seems that there are two ideas of expectation, variance, etc. going on in our world.

In any probability textbook:

I have a random variable $X$, which is a function from the sample space to the real line. Ok, now I define the expectation operator, which is a function that maps this random variable to a real number, and this function looks like,
$$\mathbb{E}[X] = \sum\limits_{i = 1}^n x_i p(x_i)$$ where $p$ is the probability mass function, $p: x_i \mapsto [0,1], \sum_{i = 1}^n p(x_i) = 1$ and $x_i \in \text{range}(X)$. The variance is,
$$\mathbb{E}[(X - \mathbb{E}[X])^2]$$

The definition is similar for a continuous RV.

However, in statistics, data science, finance, bioinformatics (and I guess everyday language when talking to your mother)

I have a multi-set of data $D = \{x_i\}_{i = 1}^n$ (weight of onions, height of school children). The mean of this dataset is

$$\dfrac{1}{n}\sum\limits_{i= 1}^n x_i$$

The variance of this dataset (according to "science buddy" and "mathisfun dot com" and government of Canada) is,

$$\dfrac{1}{n}\sum\limits_{i= 1}^n(x_i - \sum\limits_{j= 1}^n \dfrac{1}{n} x_j)^2$$

I mean, I can already see what's going on here (one is assuming uniform distribution), however, I want an authoritative explanation on the following:

Is the distinction real? Meaning, is there a universe where expectation/mean/variance... are defined for functions/random variables and another universe where expectation/mean/variance... are defined for raw data? Or are they essentially the same thing (with hidden/implicit assumption)

Why is it no probabilistic assumption is made when talking about mean or variance when it comes to dealing with data in statistics or data science (or other areas of real life)?

Is there some consistent language for distinguishing these two seemingly different mean and variance terminologies? For example, if my cashier asks me about the "mean weight" of two items, do I ask him/her for the probabilistic distribution of the random variable whose realization are the weights of these two items (def 1), or do I just add up the value and divide  (def 2)? How do I know which mean the person is talking about?/

答:
I will add some illustrations to previous answers, especially to the Peter LeFanu Lumsdaine's answer.

Statement: the "statistic" universe from your question is a partial case of "probabilistic" universe.

We need some notation. Suppose that we have a random variable $\xi$ in sense of probability theory and $P(\xi = x_k) = p_k$, $0 \le k \le n$.

For example, consider $\xi$ which is equal to the number of children in "abstract" random family. Put $x_k = k$. Then $E \xi = \sum_{k=0}^n x_k p_k$ where $p_k = P(\xi = x_k)$, and $D\xi = E (\xi - E\xi)^2 = \sum_{k=0}^n p_k (x_k - E\xi )^2$. Moreover, $P(\xi = 69) > 0$ (according to Guinness World Records, I am not sure, but let us  suppose that it's a record nowadays) and as there was people with $69$ children then we may think that for example $P(80) > 0$ - it's natural, because $80$ children is possible, even although Guinness World Records says that still nobody had $80$ children.

In reality we don't have abstract random family, random mothers, random fathers and random children. We have finite number $N$ ( $ 10^6 < N < 10^{100}$) of families, let us number them, and there are $y_1$ children in the first family, there are $y_2$ children in  family number $2$, ..., there are $y_N$ children in  family $N$.

Analogy: $\xi$ corresponds to a fair dice itself and numbers $y_1$, $y_2$, ... corresponds to dice rolls and have form: $5, 1, 6, 6, 3, ...$, these are fixed numbers.

Now consider a random value $\tau$, which has uniform distribution on $\{ 1, 2, \ldots, N\}$. It means that $P(\tau = k) = \frac{1}{N}$ for all $1 \le k \le N$.

Numbers $y_1, \ldots, y_N$ are fixed - suppose they are written in some sociological table. Let us consider a random value $y_{\tau}$. It's not a number of children in abstract random family. It is a number of children in some real family, if a number $\tau$ of such family we chose randomly.

Let us find $E y_{\tau}$ and $D y_{\tau}$. With probability $\frac{1}N$ we have $\tau = k$ and hence $y_{\tau} = y_k$. Thus $$E y_{\tau}=\frac{1}N \sum_{k=1}^N y_k$$ and $$D y_{\tau} = E (y_{\tau} - Ey_{\tau})^2 =  E (y_{\tau} - \frac{1}N \sum_{k=1}^N y_k )^2 =
\frac{1}N  \sum_{k=0}^N (y_k -\frac{1}N \sum_{k=1}^N y_k  )^2.$$

Now the correspondence of "statistical universe" and "probabilistic universe"  is obvious.

Notice that all $y_i \le 69$ and $y_i = 69$ is the maximum value, but the maximum value of $\xi$ is bigger (and not less than $80$).

So, we have shown that the "statistical universe" is a partial case of "probabilistic" universe.

Moreover, there are two moments, when there is randomness:

when we go from abstract random family, corresponding to $\xi$, to numbers $y_1, \ldots, y_N$ from sociological handbook.

As it was already mentioned, there is analogy: $\xi$ corresponds to a fair dice itself and numbers $y_1$, $y_2$, ... corresponds to dice rolls and have form: $5, 1, 6, 6, 3, ...$, these are fixed numbers.

[here we suppose that $y_1, \ldots, y_N$ are fixed numbers]
when we go from the whole sociological handbook to a family with random number $\tau$ and see, how much children is there in this family. The number is $y_{\tau}$.

When from $y_{\tau}$ we go to $E y_{\tau} = \frac{1}N \sum_{k=1}^N y_k$, we get rid of the second randomness, but we still do not get rid of the first randomness. It's connected with the fact that numbers $y_k$ could be another: it could happen, that in some families could be more children, and in some families could be less children, if the circumstances were different. In this sence numbers $y_i$ are not fixed: they are r.v. sampled from distribution, corresponding to $\xi$. And in this sence  $\frac{1}N \sum_{k=1} y_k$ is a random value, and we may write limit theorems such as L.L.N.: $$\frac{1}N \sum_{k=1}^N y_k \to E\xi, \text{ }N \to \infty$$
or CLT or LIL.

Addition: it was shown that if $y_1, \ldots, y_N$ are fixed and $\tau$ is random then "probabilistic" mean (expectation) of $y_{\tau}$ and "probabilistic" variance of $ y_{\tau}$ are well known sample mean and sample variance.

Hope it will be useful. If you have any questions, you are welcome.


问:
Proof of set being $\sigma$-finite
I'm asked to prove:

"$f \in L^1(d\mu) \Rightarrow D= \{x:f(x) \neq0\}$ is $\sigma$-finite"

My approach is:

$f \in L^1(d\mu) \Rightarrow \displaystyle \int_X \mid{f}\mid\mu < \infty \Rightarrow \int_X {s}d\mu < \infty$ for every $s$ step function with $0\leq s(x)\leq \mid f(x)\mid$ $\forall x \in X$.

Let $A_n=\{x \in X: n <\mid f(x)\mid\leq n+1\}$. It is easy to see that $D=\cup_{n=0}^{\infty} A_n$.

Then, $\forall n \geq 1$, we have $\displaystyle \int_X n\chi_{A_n}d\mu=\int_{A_n}nd\mu=n\mu(A_n)$. As $n\chi_{A_n}$ is a step function with the properties listed above, it follows that $\mu(A_n)<\infty$.

However, I do not know how to proceed with $A_0$, because $0\mu(A_0) = 0$ even if $\mu(A_0)=\infty$.

答:
Note that
$$\{x~|~|f(x)|> 0\} = \bigcup_{n = 1}^\infty \{x~|~|f(x)|> 1/n\},$$
and each $\{x~|~|f(x)|> 1/n\}$ has a finite measure by Bienaymé-Tchebychev inequality
$$\mu(\{x~|~|f(x)|> a\}) = \frac{1}{a} \int_X |f(x)|~d\mu(x).$$

Title: Proof of set being $\sigma$-finite


问:
two examples on Graph theory
Could anyone help me construct two examples:
1. a non-regular graph which has the property that the open neighborhoods of any two vertices do not contain each other?
2. minimal asymmetric graph which has the property that the open neighborhoods of any two vertices do not contain each other?

Thanks!

答:
The smallest (non-trivial) asymmetric graphs are listed in Wikipedia; they have 6 vertices, but each can be checked to not satisfy your extra condition:

Property P1: If $N_v$ and $N_u$ are the open neighbourhoods of two distinct vertices $v$ and $u$, then $N_v \not\subseteq N_u$.

(In some of these 6-vertex cases, this can be easily seen:  if there is a leaf node, then the condition is not satisfied.)  So the smallest possible (non-trivial) asymmetric graph that satisfies Property P1 must have 7 vertices; one example is given below.

Example 1: An asymmetric 7-vertex graph satisfying Property P1.

<img src="https://i.stack.imgur.com/uNgXD.png" alt="Asymmetric 7-vertex graph satisfying P1">

This is the smallest (in the sense of number of vertices) asymmetric graph for which Property P1 holds (except for trivial cases).  It's also not regular.  This example is far from unique.

However, we can define a slightly stronger property, which I think is a bit more natural.

Property P2: If $N_v$ and $N_u$ are the open neighbourhoods of two distinct vertices $v$ and $u$, then $N_v \not\subseteq N_u \cup \{u\}$.

Example 1 does not satisfy Property P2, e.g. when $v=1$ and $u=7$.  In fact, an exhaustive computer check reveals that no asymmetric 7-vertex graph satisfies Property P2.  So the smallest possible (non-trivial) asymmetric graph that satisfies Property P2 must have 8 vertices; one example is given below.

Example 2: An asymmetric 8-vertex graph satisfying Property P2.

<img src="https://i.stack.imgur.com/D6vpe.png" alt="Asymmetric 8-vertex graph satisfying P2">

Again, this is the smallest asymmetric graph for which Property P2 holds (except for trivial cases).  It's also not regular.  This is also unlikely to be unique (although I didn't check).

Title: two examples on Graph theory

Body:
The smallest (non-trivial) asymmetric graphs are listed in Wikipedia; they have 6 vertices, but each can be checked to not satisfy your extra condition:

Property P1: If $N_v$ and $N_u$ are the open neighbourhoods of two distinct vertices $v$ and $u$, then $N_v \not\subseteq N_u$.

(In some of these 6-vertex cases, this can be easily seen:  if there is a leaf node, then the condition is not satisfied.)  So the smallest possible (non-trivial) asymmetric graph that satisfies Property P1 must have 7 vertices; one example is given below.

Example 1: An asymmetric 7-vertex graph satisfying Property P1.

This is the smallest (in the sense of number of vertices) asymmetric graph for which Property P1 holds (except for trivial cases).  It's also not regular.  This example is far from unique.

However, we can define a slightly stronger property, which I think is a bit more natural.

Property P2: If $N_v$ and $N_u$ are the open neighbourhoods of two distinct vertices $v$ and $u$, then $N_v \not\subseteq N_u \cup \{u\}$.

Example 1 does not satisfy Property P2, e.g. when $v=1$ and $u=7$.  In fact, an exhaustive computer check reveals that no asymmetric 7-vertex graph satisfies Property P2.  So the smallest possible (non-trivial) asymmetric graph that satisfies Property P2 must have 8 vertices; one example is given below.

Example 2: An asymmetric 8-vertex graph satisfying Property P2.

Again, this is the smallest asymmetric graph for which Property P2 holds (except for trivial cases).  It's also not regular.  This is also unlikely to be unique (although I didn't check).


问:
Reference for Algebraic topology books
Im beginner in algebraic topology.I have read the general topology book by Munkres  and topology without tears by Morris

I have seen the Allen Hatcher book .It is very tough book ,so i need some easier book

I need references for  very basic books  in Algebraic topology and that book  must contain given topics below

$1.$Homotopy Theory.  Covering spaces, homotopy maps, homotopy equivalence,Contractible spaces, deformation retraction.

$2.$Fundamental Groups: Universal cover and lifting problem for covering maps, Fundamental groups of S1 and Sn.

$3$ .Introduction to Homology Theory.

答:
Spanier, Greenburg and Harper, and Munkres are three that come to mind.

There was also one by Fomenko and somebody else (maybe Fuchs) on homotopy theory that had great illustrations.

Title: Reference for Algebraic topology books


问:
Prokhorov theorem in locally compact Hausdorff space?
Prokhorov theorem gives a compactness condition in the space of probability measures on a Polish space. I am wondering whether we have similar conditions for probability measures on more general spaces, say, locally compact Hausdorff spaces, which seems to me to be a more natural setting of measure theory.

However, since the proofs that I have seen for Prokhorov theorem depend heavily on the completeness and separability of the underlying space, they do not help much when one tries to extend the result to more general spaces. And to my best guess such an extension would rely on techniques from functional analysis.

So, do we actually have such a condition for more general spaces?

Thanks!

答:
There is a version of Prokhorov's theorem on locally compact Hausdorff spaces, but it seems surprisingly difficult to locate it in the literature.

One reference (that simultaneously gives both the LCH and Polish space versions) is Bourbaki's Intégration. Chapitre IX, Section 5.5, Théorème 1 + Théorème 2 (each theorem gives one implication). I'll summarize.

Definition (Prokhorov condition) Let $H$ be a collection of complex measures on a topological space $X$. $H$ is Prokhorov if

the total variations $|\mu|(X)$ of $\mu\in H$ are uniformly bounded, and

for every $\varepsilon >0$ there is a compact $K=K_{\varepsilon}\subseteq X$ such that $|\mu|(X-K)<\varepsilon$ for all $\mu\in H$.

Then:

Theorem 1 (Prokhorov $\Rightarrow$ compact) A Prokhorov set of complex measures on a completely regular space $X$ (in particular an LCH space or a metric space) is relatively compact in the weak$^*$ topology on complex measures relative to the bounded continuous functions on $X$.

Theorem 2 (compact $\Rightarrow$ Prokhorov) If $X$ is LCH or complete metric separable then a weak$^*$-compact set of positive finite measures on $X$ is Prokhorov.


问:
Diagonals of quadrilateral $ABCD$ intersect at $E$. Given $\frac{AE}{AC} = λ,\>\frac{BE}{BD} = μ$. Find ratios for sides
In the figure $ABCD$ is a convex quadrilateral. its diagonals $AC$ and $BD$ intersect at $E$, and their midpoints are $P$ and $ Q$, respectively. Given that $\frac{AE}{AC} = λ,\>\frac{BE}{BD} = μ$, (i) Find the ratios $\frac{AR}{RD}$  and $\frac{BS }{SC }$ in terms of λ and μ. (ii) Suppose the area of $ABCD$ is 1, what is the area of $ABSR$?

The attached below contains all my work. I know it is not very neat but I don't think it would matter anyway as I couldn't make any meaningful process either :(

Also, in my book, I am struggling a lot in the topic 'Area lemma' (this question is from that topic) , so is there any suggestion / algorithmic way to approach these problems.

Thanks!

PS; please don't use any fancy stuff (like complex / bary). A solution with Menelaus, Ceva's, Area lemma would be ideal.

EDIT ; I figured out the first part, menelaus on ADE with RS transversal

答:
$\color{blue}{(i)}$. Given the midpoints $P$ and $Q$, establish

$$ \frac{EP}{AP} = \frac{AE-AP}{AP} =2\lambda-1,\>\>\>\>\>\frac{EQ}{DQ} = \frac{BE-BQ}{DQ} =2\mu-1
$$

With [.] denoting areas, derive the ratio $\frac{AR}{RD}$  below and, likewise, $\frac{BS }{SC}$

$$x=\frac{AR}{RD} = \frac{[ARQ]}{[DRQ]}
= \frac{[ARQ]}{\frac{DQ}{EQ} [ERQ]}
= \frac{EQ}{DQ} \frac{AP}{EP}=\frac{2\mu-1}{2\lambda-1}
$$
$$y=\frac{BS }{SC}=\frac{2\lambda-1}{2\mu-1}$$

$\color{blue}{(ii)}$. Note $[ABC]=\mu[ABCD]=\mu$ and $[ADC]=(1-\mu)[ABCD]=1-\mu$
\begin{align}
&[ARP]= \frac x{1+x}[ADP] =\frac x{1+x} \frac12[ADC] = \frac {\frac12x(1-\mu)}{1+x}\\
&[ABSP] = [ABC] - \frac12[ASC]
= [ABC] - \frac{\frac12}{1+y}[ABC]
=\frac {(y+\frac12)\mu}{1+y}
\end{align}
$$\implies [ABSR] = [ARP]+[ABSP]
= \frac{\lambda\mu-\frac14}{\lambda+\mu-1}
$$


问:
When can we interchange the order of limits of the expected value of products of random variables?
Suppose that the sequence of random variables $X_n$ converges to a limiting random variable $X$, and let $Y$ be an arbitrary random variable.

Is there any specific mode of convergence of the sequence $X_n$ that guarantees that
$$\lim_{n \to \infty} E[X_n Y] = E[X Y]?
$$
I'm pretty sure that convergence in distribution is not good enough, but I'm not sure about (e.g.) convergence in probability or convergence a.s.

What about if there are multiple sequences (i.e. $E[X_n Y_n]\stackrel{?}{\to} E[X Y]$), or if $X$ or $Y$ are constant or otherwise constrained? (I know that if $Y$ is constant, then it suffices for $X_n$ to converge in mean.)

答:
Let $||\cdot||_p$ denote the $L^p$-norm. We want $X_nY\to XY$ in $L^1$. Note that
$$||X_nY - XY||_1 = ||(X_n-X)Y||_1 \leq ||X_n-X||_2 ||Y||_2,$$
by the Cauchy-Schwarz inequality. So $X_n\to X$ in $L^2$ suffices, if $||Y||_2<\infty$.

Using Hölder's inequality, you can also get different conditions (stricter for $X_n$ but weaker for $Y$ or vice versa).


问:
Counting The Number Of K-Bounded Good Lists Of Length N
Problem Statement:

A k-bounded list of length n is a sequence [x1, x2, . . . , xn] where each xj
is an
integer between 0 and k, inclusive. A good list is one that does not contain three
consecutive values 0, 1, 0. For example, for n = 4 and k = 2, [2, 0, 0, 1] and [1, 0, 0, 1]
are good lists while [0, 1, 0, 2] and [0, 0, 1, 0] are not.
Let good(n, k) be the number of good k-bounded lists of length n. For instance,
the good 1-bounded lists of length 2 are {[0, 0], [0, 1], [1, 0], [1, 1]}, so good(2, 1) = 4.
Similarly, you can check that good(4, 1) = 12 and good(3, 2) = 26.
Your task is to compute good(n, k) for the following values of n and k.

(a) good(7, 1)

(b) good(7, 3)

(c) good(20, 1)

Note: This is a question from a past contest and the official answers for the parts a, b and c are as follows:

(a) good(7, 1) = 65

(b) good(7, 3) = 15163

(c) good(20, 1) = 97229

My Approach: I came up with the following recurrence relations,

$$p(i,0) = p(i-1,0) + p(i-1,1) - p(i-2,0)$$ $$p(i,1) = [p(i-1,0) + p(i-1,1)]\times k$$
where $p(i,0)$ is the number of good lists of length $i$ ending with 0 and $p(i,1)$ is the number of good lists of length i not ending with 0. The idea behind this is that every good list of length $i$ is an extension of a good list of length $i-1$ but there can also be some bad extensions in the process, hence it is better to handle them separately. Therefore, in case a good list ends with 0,1 then we need to take care of the fact that while extending this list we don't count the lists where we place a 0 at the end, otherwise we will have 0,1,0 in the list. So what we do is that we just count all the lists of length $i$ ending with 0 and then subtract the lists of length $i-2$ that end with a 0. Whereas the case where we do not use a 0 at the end is a bit simpler.

I just filled the table for the above recurrence manually with the initial values as $p(0,0) = 0$, $p(0,1) = 0$, and $p(1,0) = 1$,  $p(1,1) = k$ (since it was a pen and paper exam) and could arrive at the answers which is the sum of $p(n,0)$ and $p(n,1)$ [C++ code for reference: https://ideone.com/wSVhVL]. But then I have been thinking whether this problem has a closed form solution, some function in n and k. I couldn't arrive at one until now. Can anyone provide some combinatorial insight? Is a closed form solution possible?

答:
The following answer is based upon the Goulden-Jackson Cluster Method which provides a convenient technique to solve problems of this kind. We consider the set of words of length $n\geq 0$ built from an alphabet $$\mathcal{V}=\{A_0,A_1,\ldots, A_{k}\}$$ of size $k+1$ and the set $B=\{A_0A_1A_0\}$ containing a bad word of length $3$, which is not allowed to be part of the words we are looking for.

Generating function $f_k(t)$: We derive a generating function $f_k(t)$ with the coefficient of $t^n$ being  the number of wanted words of length $n$.

According to the paper (p.7) the generating function $f_k(t)$  is
\begin{align*}
f_k(t)=\frac{1}{1-(k+1)t-\text{w}(\mathcal{C})}\tag{1}
\end{align*}
with $k+1=|\mathcal{V}|$ the size of the alphabet and $\text{w}(\mathcal{C})$ the weight-numerator of bad words with
\begin{align*}
\text{w}(\mathcal{C})=\text{w}(\mathcal{C}[A_0A_1A_0])\tag{2}
\end{align*}

We calculate according to the paper:
\begin{align*}
\text{w}(\mathcal{C}[A_0A_1A_0])&=-t^{3}-t^2\text{w}(\mathcal{C}[A_0A_1A_0])\tag{3}\\
\end{align*}
where the expression $t^2\text{w}(\mathcal{C}[A_0A_1A_0])$ accounts for overlappings of $A_{0}A_1A_0$ with itself. We obtain from (3):
\begin{align*}
\text{w}(\mathcal{C}[A_0A_1A_0])&=-\frac{t^3}{1+t^2}
\end{align*}
It    follows from (1) - (3):
\begin{align*}
\color{blue}{f_k(t)}&=\frac{1}{1-(k+1)t-\text{weight}(\mathcal{C})}\\
&=\frac{1}{1-(k+1)t+\frac{t^3}{1+t^2}}\\
&\,\,\color{blue}{=\frac{1+t^2}{1-(k+1)t-t^2-kt^3}}\tag{4}
\end{align*}
and conclude the number of valid words of length $n\geq 0$ is the coefficient of $t^n$ in (4).

Coefficient extraction: We use the coefficient of operator $[t^n]$ to denote the coefficient of $t^n$ of a series. Applying standard techniques to extract the coefficient $[t^n]$ from $f_k(t)$ we obtain for $n\geq 2$:
\begin{align*}
\color{blue}{[t^n]f_k(t)}&\color{blue}{=\sum_{j=0}^{n}(-1)^j\sum_{l=0}^j\binom{n-j}{l}\binom{l}{j-l}k^{j-l}(k+1)^{n-j-l}}\\
&\qquad\color{blue}{+\sum_{j=0}^{n-2}(-1)^j\sum_{l=0}^j\binom{n-2-j}{l}\binom{l}{j-l}k^{j-l}(k+1)^{n-2-j-l}}\tag{5}
\end{align*}

Calculation: We derive from (5)

\begin{align*}
\color{blue}{[t^7]f_1(t)}&=\sum_{j=0}^{7}(-1)^j\sum_{l=0}^j\binom{7-j}{l}\binom{l}{j-l}2^{7-j-l}\\
&\qquad+\sum_{j=0}^{5}(-1)^j\sum_{l=0}^j\binom{5-j}{l}\binom{l}{j-l}2^{5-j-l}
\color{blue}{=65}\\
\color{blue}{[t^7]f_3(t)}&=\sum_{j=0}^{7}(-1)^j\sum_{l=0}^j\binom{7-j}{l}\binom{l}{j-l}3^{j-l}4^{7-j-l}\\
&\qquad+\sum_{j=0}^{5}(-1)^j\sum_{l=0}^j\binom{5-j}{l}\binom{l}{j-l}3^{j-l}4^{5-j-l}
\color{blue}{=15\,163}\\
\color{blue}{[t^{20}]f_1(t)}&=\sum_{j=0}^{20}(-1)^j\sum_{l=0}^j\binom{20-j}{l}\binom{l}{j-l}2^{20-j-l}\\
&\qquad+\sum_{j=0}^{18}(-1)^j\sum_{l=0}^j\binom{18-j}{l}\binom{l}{j-l}2^{18-j-l}\color{blue}{=97\,229}\\
\end{align*}

which has been programmatically verified.


问:
Prove using Cauchy criterion that the limit of the function $\cos(\frac{1}{x})+x^2$ doesn't exist
Prove using Cauchy criterion that the limit of the function $\cos(\frac{1}{x})+x^2$ doesn't exist. (when $x_0 \rightarrow 0$)
so we need to prove that:

Exist $\epsilon \ge 0$ such that for all $\delta \gt 0$ exist $x,y$ such that $0 \lt |x-x_0| \lt \delta \wedge 0 \lt |y-y_0| \lt \delta$ And $|f(x)-f(y)| \ge \epsilon$

Well I tried using it but failed in the middle , I don't know how to proceed.

My try :

$|f(x)-f(y)|=|\cos(\frac {1}{x})+x^2 - \cos(\frac {1}{y})-y^2| \ge ?$

Thank you in advance.

答: Take $x_0,y_0 = 0$. Let $\epsilon = 2$, and $\delta>0$. By the Archimedian property of $\mathbb R$, there exist $n,m\in\mathbb N$ such that $$x = \frac{1}{2n\pi}<\delta,\qquad\text{and}\qquad y = \frac{1}{-2m\pi + \pi}<\delta.$$ Without loss of generality, we can assume further that $x<y$. Since $\cos(x^{-1}) = 1$, and $\cos(y^{-1}) = -1$, it follows that $$\begin{align*}\vert f(x) - f(y)\vert &= \vert\cos(x^{-1}) - \cos(y^{-1}) + x^2 - y^2\vert \\ &= \vert 2 + x^2 - y^2\vert \\&> 2 = \epsilon. \end{align*}$$


问:
intuition on the isomorphism between $(c_0)^*$ and $(c)^*$
There are a lot of questions on this topic see for example Show that $(c_{0})'$ and $(c)'$ are isometrically isomorphic. . So I´m not interested in a proof rather I wonder how to think about the result. If I understand it correctly both $c_0^*$ and $c^*$ are isomorphic to $l^1$. I find this very counter intuitive as $c_0$ and $c$ are not isomorphic but rather $c$ is isomorphic to $c_0 + \mathbb{R}$ . This would mean that $(c_0 + \mathbb{R})^*\simeq c_0^*$

In light of this my question is how to think about dual spaces. How is it possible that $c_0^*$ and $c^*$ are isomorphic or am I misunderstanding the answer in the link above?

答:
Because the way you phrase the question is a bit confusing, let me clarify that there is no isometric isomorphism between $c_0$ and $c$; but there is a Banach space isomorphism.

Regardless, it is very intuitive that the dual of $c_0$ and $c_0+\mathbb R$ is the same. After all $c_0+\mathbb R$ is simply the unitization of $c_0$ when seen as a Banach algebra.

Given any $\psi\in (c_0+\mathbb R)^*$, we have $\psi(a+\lambda)=\psi(a)+\lambda\,\psi(1)$. It is not hard to check that
$$
\|\psi\|=\|\psi|_{c_0}\|+|\psi(1)|.
$$
This shows that we can see $(c_0+\mathbb R)'$ as the norm-1 direct sum $c_0^*\oplus_1\mathbb R$. So what we are saying is that $\ell^1\oplus_1\mathbb R\simeq\ell^1$ isometrically. Which is trivial to check with the isometric isomorphism $(a,\lambda)\longmapsto (\lambda,a_1,a_2,\ldots)$.


问:
Propositional Logic "Riddle/Puzzle"
I have this kind of 'riddle' as a question that i need to complete, however I'm not sure what to do of it.

This is the question:

Determine who out of the following is guilty of doping. The suspects are: Sam, Michael, Bill, Richard, Matt.

1) Sam said: Michael or Bill took drugs, but not both.

2) Michael said: Richard or Sam took drugs, but not both.

3) Bill said: Matt or Michael took drugs, but not both.

4) Richard said: Bill or Matt took drugs, but not both.

5) Matt said: Bill or Richard took drugs, but not both.

^ Of these 5 statements, 4 are true, one is false.

6) Tom said: If Richard took drugs, then Bill took drugs.

^  This statement is guaranteed to be true.

From this information, I deduced:

p : Michael took drugs

q : Bill took drugs

r : Richard took drugs

s : Sam took drugs

t : Matt took drugs

So given this I came up with this:

1) (p ^ ~q) v (~p ^ q)

2) (r ^ ~s) v (~r ^ s)

3) (t ^ ~p) v (~t ^ p)

4) (q ^ ~t) v (~q ^ t)

5) (q ^ ~r) v (~q ^ r)

6) (~r v q)

However, I'm not sure where to go from here. I suppose, I could connect each statement with an ^ as that seems like the next step to do. Then that entire equation would essentially tell me who was guilty? The next step, would obviously be to simplify, and come up with a name. However, I'm not sure how to do this.

Could anyone please shed some light and give me some tips on how to do this?

Thanks.

答:
There are two things you need to notice.

First, statements one three and four are mutually contradictory.  So the false statement must be one of those three.

So both statements 5 and 6 are true.  If Richard took drugs then Bill both did and didn't take drugs.  Therefore Richard did not and Bill did.

So by statement 2 (which we know to be true)  Sam also took drugs.

There's no way of solving it from here, we can declare any one of statements 1,3 and 4 to be false and work out the rest.

The only thing you can say for sure is that at least one of Matt or Michael took drugs.

Title: Propositional Logic "Riddle/Puzzle"


问:
How do I prove that a function preserving Cauchy sequences is uniformly continuous?
Let $A \subset \mathbb{R}$ be a bounded interval. Assume a function $f:A \longrightarrow \mathbb{R}$ satisfies that for every cauchy sequence ${x_n}$ where $x_n \in  A$ the sequence ${f(x_n)}$ is also a cauchy sequence. How can I prove that $f$ is uniformly continuous in $A$?

I have tried extending $A$ to be a closed interval $A'=[a,b]$ where $a=\inf A  ,b=\sup A$, and proving that for every convergent series $x_n \in A'$ the series $f(x_n)$ also converges. This would allow me to use the series definition of continuity and thus because this is a closed interval it would also be uniformly continuous. But because the function is not necessarily defined at $f(a)$ I haven't been able to do that.

答:
$A=(a,b)$. Let $(x_n)$ be any sequence in $A$ converging to $b$, for example $x_n := b-(b-a)/n$. Then $(x_n)$ is Cauchy. By our hypothesis $f(x_n)$ is Cauchy, hence converging to say $\kappa$ as $\mathbb{R}$ is complete. Similarly, we can find that for any sequence in $A$ that converges to $a$ converges to say $\lambda$ when $f$ is applied. Extend $f$ to $\tilde{f}$ as $\tilde{f}(x)=f(x)$ when $x \in A$, $\tilde{f}(b)=\kappa$ and $\tilde{f}(a)=\lambda$. Continuity of $\tilde{f}$ follows from our construction and the rest follows from your argument.

I hope this helps.