问:
$P(x) = x^3 – ax^2 + bx + c$. has integral roots such that $P(6) = 3$ then sum of all possible values of $a$ is

Consider the cubic polynomial $P(x) = x^3 – ax^2 + bx + c$. If the equation $P(x) = 0$ has integral roots such that $P(6) = 3$ then sum of all possible values of $a$ is

My try  Let $\eta$ be the root of $P(x)=0$ then by rational root theorem $\eta|c$ also $$P(6)=3\Rightarrow 216-36a+6b+c=3$$
$$\Rightarrow 3|c$$
What do i do next?

Please provide hints rather than complete solutions (it helps me)

答:
A very much thanks to @kelenner to put me in the right direction:

we have $$(6-u)(6-v)(6-w)=3$$
we see that for $xyz=3$ we can heve the following unordered  pairs $$(6-u,6-v,6-w)=(1,1,3),(1,-1,-3),(-1,-1,3)$$
which means the unorederd triplet can take $$(u,v,w)=(5,5,3),(5,7,9),(7,7,3)$$
Thus by vieta $$a=u+v+w=13,17,21$$


问:
How does $axy+bx+cy+d=0$ represent a rectangular hyperbola?
I'm told a curve of this form $$axy+bx+cy+d=0$$ is a rectangular hyperbola.

But I don't seem to understand where are it's $x^2$ and $y^2$ terms since $x^2-y^2=a$ is also a rectangular hyperbola.

Sorry for the boring question, but did I learn something wrong here?

答:
Yes, $x^2-y^2=a$ with $a\neq 0$ is equation of a rectangular hyperbola.

Its asymptotes are $y=\pm x$ and axes of symmetry $x=0$ and $y=0.$

I assume you know that $y={1 \over x}$ also defines a rectangular hyperbola obtained by a rotation of the first one, where $a=\pm 1.$ The asymptotes are now $x=0$ and $y=0,$ the axes are $y=\pm x.$

The equation $axy+bx+cy+d=0$ (here $a$ is not that from the above part) defines a hyperbola of the second type: its axes are parallel to $y=\pm x,$ thus they are orthogonal.

They pass through the center of the hyperbola $\left(-\frac{c}a,-\frac{b}a\right).$

For more information see this wikipedia article.

ContentLicense: CC BY-SA 4.0


答:
You need to rotate the coordinates by $45°$ to see the correspondence, which is achieved by the transformation

$$x=u+v,\\y=u-v.$$

If we plug this in $x^2-y^2$, we obtain

$$(u+v)^2-(u-v)^2=4uv,$$ which gets closer to the desired form. (Conversely, $xy=u^2-v^2$.)

Title: How does $axy+bx+cy+d=0$ represent a rectangular hyperbola?

Body:
I'm told a curve of this form $$axy+bx+cy+d=0$$ is a rectangular hyperbola.

But I don't seem to understand where are it's $x^2$ and $y^2$ terms since $x^2-y^2=a$ is also a rectangular hyperbola.

Sorry for the boring question, but did I learn something wrong here?


问:
Are there any increasing or decreasing intervals of a quadratic equation with no real values of $x$?

Using first derivative test, find the increasing/decreasing intervals and local maximum and local minimum values of the function $f(x) = x^3 -3x^2 - 6x$

$$f'(x)= 3x^2 -6x -6=0 $$
what do I write for this type of question ?

答:
$$f(x)=x^3-3x^2-6x$$

$$f'(x)=3x^2-6x-6=0$$

$$f'(x)=x^2-2x-2=0$$

$$D=(-2)^2-4\times 1\times (-2)=12$$

$x_1=1-\sqrt{3}$, $x_2=1+\sqrt{3}$

The function is increasing (because $f'(x)>0$) in $(-\infty,1-\sqrt{3})$ and increasing (because $f'(x)>0$) in $(1+\sqrt{3},\infty)$ , decreasing (because $f'(x)<0$) in $(1-\sqrt{3},1+\sqrt{3})$

And there is a local minimum for $x=1+\sqrt{3}$, a local maximum for $x=1-\sqrt{3}$.


问:
How to prove by induction that given a set $S$ of $n$ elements, the number even subsets of $S$ equals to the number of odd subsets?
How do I prove by induction that given a positive integer $n\ge1$ and a set $S$ of $n$ elements, the number even subsets of $S$ equals to the number of odd subsets? For example, for $S=\{g,c\}$ there are two subsets with even cardinality: the empty set $\varnothing$ and $\{g,c\}$, as well as two subsets with odd cardinality: $\{g\}$ and $\{c\}$.

答:
Let ${\cal P}_n$ be the power set of $[n]$. Then $|{\cal P}_0|=1$.

For any $A\in{\cal P}_n$ we obtain two elements of ${\cal P}_{n+1}$, namely $A$ and $A\cup\{n+1\}$. One of them is even, the other odd.


问:
Matrix of a transformation relative to a basis using eigenvalues and eigenvectors
Suppose $A$ is a 3x3 matrix and $X=\{x_1,x_2,x_3\}$ is a basis of $R^3$ where:

$x_1$ is an eigenvector of $A$ with eigenvalue 6

$x_2$ is an eigenvector of $A$ with eigenvalue 2

$x_3$ satisfies $(A-2I)x_3=x_2$

Let $T: R^3 \to R^3$ be the linear transformation given by $T(y)=Ay$ and write the matrix of $T$ relative to the basis $X$. Show that $x_2$ and $x_3$ are both elements of the null space of $(A-2I)^2$.

I'm thinking that since $x_3$ satisfies $(A-2I)x_3=x_2$, then the eigenvalue 2 must be of multiplicity 2, telling me that the diagonalization of $A$ would be

$$
\begin{bmatrix}
6 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 2 \\
\end{bmatrix}
$$

which also tells me that the characteristic polynomial would be

$$ (x-6)(x-2)^2 = x^3-10x^2+28x-24$$

What I don't know how to do is find the matrix of $T$ relative to the basis $X$ without any further information, or how to find the null space of $(A-2I)^2$ without knowing what $A$ is. I would appreciate any guidance.

答:
Your reasoning is flawed. In fact, the existence of such a vector $x_3$ tells us that $A$ cannot be diagonailzed.

I would suggest that trying to "find the null space" is the wrong way to approach the problem. Instead, consider this: if I have a matrix $M$ and a vector $v$, how can I test whether $v$ is in the null space of $M$? It would be helpful to think about the definition of "the null space" of a matrix.


Title: How $\sum_{n=1}^{\infty} \left( \frac{1}{(10n-9)^2} + \frac{1}{(10n-1)^2} \right)$ equals to $\frac{{\pi}^2}{50} \frac 1 { 1-\cos{\frac {\pi} 5} }$

Body:
Why is
$$
\sum_{n=1}^{\infty} \left( \frac{1}{(10n-9)^2} + \frac{1}{(10n-1)^2} \right) = \frac{\pi^2}{50} \frac{1}{1- \cos \frac{\pi}{5}}
$$
?

Each form of
$$
\sum_{n=1}^{\infty} \frac{1}{(10n-9)^2}
$$
and
$$
\sum_{n=1}^{\infty} \frac{1}{(10n-1)^2}
$$

couldn't be calculated but the sum of them has a closed form

Body:
Using
$$\sum_{n\in\mathbb{Z}}\frac{1}{(n-a)^2+b^2}=\frac{\pi\sinh2\pi b}{b\left(\cosh2\pi b-\cos2\pi a\right)}$$
from this, one has
\begin{eqnarray}
&&\sum_{n=1}^{\infty} \left( \frac{1}{(10n-9)^2} + \frac{1}{(10n-1)^2} \right)\\
&=&\sum_{n\in\mathbb{Z}}\frac{1}{(10n-1)^2}=\frac{1}{10^2}\sum_{n\in\mathbb{Z}}\frac{1}{(n-\frac1{10})^2}\\
&=&\frac1{10}\lim_{b\to0}\frac{\pi\sinh(2\pi b)}{b\left(\cosh(2\pi b)-\cos(\frac{2\pi}{10})\right)}\\
&=&\frac{\pi^2}{50} \frac{1}{1- \cos \frac{\pi}{5}}.
\end{eqnarray}


问:
Find an expression of the generalised kronecker delta in terms of the determinant of a square matrix.
Definition

The determinant $\det A$ of a square matrix $A$ with coefficients in a field $\Bbb K$ is the scalar
$$
\det A:=\sum_{\sigma\in\mathfrak S_n}\text{sgn}(\sigma)a_{1,\sigma(1)}\cdots a_{n,\sigma(n)}
$$
where the summation is extended to all permutation of $n$ elements.

So if $\delta^{i_1,...,i_k}_{j_1,...,j_k}$ is the generalised kroneker delta (see here for details about its definition) I ask to prove that
$$\delta^{i_1,...,i_k}_{j_1,...,j_k}=\text{det}\begin{pmatrix}\delta^{i_1}_{j_1}&\cdots&\delta^{i_1}_{j_k}\\
\vdots&\ddots&\vdots\\
\delta^{i_k}_{j_k}&\cdots&\delta^{i_k}_{j_k}\end{pmatrix}
$$
for each choice of indices. So could someone help me, please?

答:
First of all we put
$$
\Delta^{i_1,...,i_k}_{j_1,...,j_k}:=\text{det}\begin{pmatrix}\delta^{i_1}_{j_1}&\cdots&\delta^{i_1}_{j_k}\\
\vdots&\ddots&\vdots\\
\delta^{i_k}_{j_k}&\cdots&\delta^{i_k}_{j_k}\end{pmatrix}
$$
for each choice of indices so that we have to prove that
$$
\delta^{i_1,...,i_k}_{j_1,...,j_k}=\Delta^{i_1,...,i_k}_{j_1,...,j_k}
$$
for each choice of indices.

So if $\delta^{i_1,...,i_k}_{j_1,...,j_k}$ is zero then either one between the $k$-tuples $(i_1,...,i_k)$ and $(j_1,...,j_k)$ has duplicate indices or the indices of such $k$-tuples are not equal so that in the first case the matrix $\Delta^{i_1,...,i_k}_{j_1,...,j_k}$ has two rows duplicate and in the second case the same matrix has at least one column zero and thus the equality tritivally holds by notable properties of the determinant. Conversely if  $\delta^{i_1,...,i_k}_{j_1,...,j_k}$ is not zero then the indices of the $k$-tuples are distinct and the indices of the first $k$-tuple are the same of the second so that the first $k$-tuple is a permutation of the second and vice versa. So we know that
$$
\Delta^{i_1,...,i_k}_{j_1,...,j_k}=\sum_{\sigma\in\mathfrak S_k}\text{sgn}(\sigma)\delta^{i_1}_{j_{\sigma(1)}}\delta^{i_2}_{j_{\sigma(2)}}...\delta^{i_k}_{j_{\sigma(k)}}
$$
and so we analyse which and how many terms of the previous sum are not zero. So by the zero-product property we know that $\text{sgn}(\tau)\delta^{i_1}_{j_{\tau(1)}}\delta^{i_2}_{j_{\tau(2)}}...\delta^{i_k}_{j_{\tau(k)}}\neq0$ for some $\tau\in\mathfrak S_k$ if and only if the quantity $\delta^{i_h}_{j_{\tau(h)}}$ is not zero and this happens if and only if $i_h=j_{\tau(h)}$ for each $h=1,...,k$ so that the only term not zero in the previous summ is that related to the perutation $\tau$ and thus
$$
\Delta^{i_1,...,i_k}_{j_1,...,j_k}=\sum_{\sigma\in\mathfrak S_k}\text{sgn}(\sigma)\delta^{i_1}_{j_{\sigma(1)}}\delta^{i_2}_{j_{\sigma(2)}}...\delta^{i_k}_{j_{\sigma(k)}}=\\\text{sgn}(\tau)\delta^{i_1}_{j_{\tau(1)}}\delta^{i_2}_{j_{\tau(2)}}...\delta^{i_k}_{j_{\tau(k)}}=\text{sgn}{\tau}=\delta^{i_1,...,i_k}_{j_1,...,j_k}
$$
as we desired.


答:
By the binomial theorem,
$$0=(1-1)^n = \sum_{k=0}^{n}{(-1)^k \binom{n}{k}},$$ which leads to $$\binom{n}{0}+\binom{n}{2}+\binom{n}{4} +\cdots=\binom{n}{1}+\binom{n}{3}+\binom{n}{5} +\cdots.$$
The left side is the number of ways of choosing an even subset and the right side is the number of ways of choosing an odd subset. They are equal.

Title: How to prove by induction that given a set $S$ of $n$ elements, the number even subsets of $S$ equals to the number of odd subsets?

Body: How do I prove by induction that given a positive integer $n\ge1$ and a set $S$ of $n$ elements, the number even subsets of $S$ equals to the number of odd subsets? For example, for $S=\{g,c\}$ there are two subsets with even cardinality: the empty set $\varnothing$ and $\{g,c\}$, as well as two subsets with odd cardinality: $\{g\}$ and $\{c\}$.


问:
The span of two parallel line segments
I'm trying to understand how to think about the span of two parallel line segments.

Let $\Pi_0$ be the vector space of all line segments through the origin. Then, the span of two line segments through the origin $\vec{o}$ is, $${span}(\vec{x}, \vec{y})= a\vec{x}+b\vec{y} = \Pi_0.$$
However, what happens when both vectors do not intersect at $\vec{o}$ or are parallel. I understand that this is not a subspace of $\Pi_0$ but, is it even defined, or can it be a vector space and/or a subspace of another vector space. How does a linear combination of these two vectors look like geometrically?

Your feedback will be much appreciated.

答:
Perhaps to answer your question, you need to remember a very useful theorem in linear algebra. This theorem states that:

Theorem: Let $(V,\oplus, \odot, (F,+,\cdot))$ a vector space over a field $F$ with the operations $\oplus$ and $\odot$ sums, and scalar-vector product, respectively. Let $H\subseteq V$. We will say that $H$ is a vector subspace if, and only if,

$H\not=\emptyset \iff \vec{0}_{V}\in H$.

$\forall h_{1},h_{2} \in H: h_{1}\oplus h_{2} \in H$.

$\forall \alpha \in F, \forall h \in H: \alpha \odot \in H$.

Now, I have placed emphasis on identifying the sum and product operation, because here is the key to the answer to one of your questions. I will try to answer your questions one by one.

For example, I will explain this using the vector space $\mathbb{R}^{2}$.

Indeed, let the vector space of $\mathbb{R}^{2}$ and let $\Pi_{0}\subseteq \mathbb{R}^{2}$ with de conventional operations so $$\Pi_{0}=\left\{\begin{pmatrix} x \\ y \end{pmatrix} \in \mathbb{R}^{2}: ax+by=0, a,b \in \mathbb{R} \right\}=\text{span} \left\{ \begin{pmatrix} -b \\ a \end{pmatrix}\right\}$$
is a vector subspace of $\mathbb{R}^{2}$

If the vectors do not intersect in $(0,0)$ that means that the zero vector of $\mathbb{R}^{2}$ isn't in $\Pi_{0}$ for example and that contradicts our first property of the theorem, therefore it would not be a vector subspace.

The reason I already mentioned and emphasized the notation associated with operations in vector space is the key to the answer to your question.
You can define a vector space with unconventional operations where you can change the geometry of vector spaces where some "weird" things happen.

Take for example $\mathbb{R}^{2}$ is a vector space rich in geometry and surely you will be able to observe how the linear combination of the vectors that are subject to the condition that I indicated in $\Pi_{0}$.

Title: The span of two parallel line segments


问:
Value of a line integral
What is the value of the line integral  $\oint\limits_{C}(sin(y)dx+xcos(y)dy)$ , where C consists of the circle $x^2+y^2=2 $ from $(-1,1)$ to $(1,1)$ and the line segment from $(1,1)$ to $(-1,1)$?

What I have done so far:

Using Green's Theorem I have $$\oint\limits_{C}(sin(y)dx+xcos(y)dy)= \iint\limits_{\Omega}2cos(y)dxdy$$

However I don't know what the limits should be for the integrals.

答: The vector field $(\sin y, x \cos y)$ is conservative. You can check this by noting that $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$. So, the value of the integral will be independent from the path and, in particular, if you end up returning to the starting point, the integral is zero.

Title: Value of a line integral


问:
What are the various respects under which a logic can deviate from classical logic, thus being " non-classical"?
Is it possible to get a synoptic view of the ways in which a logic can, so to say, deviate from classical logic?

I think one can find rather easily a list ( though maybe incomplete) of non-classical logics.

But it seems more difficult to find a presentation of the field that exhibits in a systematic fashion under which respects a logic can be non classical.

The respects I can think of are the following :

(1) type of objects over which quantifiers range --> first order/ second order logic

(2) validity of "ex falso" or not  --> paraconsistent logics

(3) use of modal operators, or not --> modal logics

(4) finite or infinite number of premises --> compactness ( not sure of this)

There is an attempt at such a presentation in Theodore Sider's book Logic For Philosophy, but I'd be much interested in other references.

Note : I'm not asking for an absolutely complete list of points of departure from clasical logic; I suppose it would be too long. Rather, what interests me is the systematicity of the presentation.

答:
One attempt at classifying ways to alternate logic could be:

extending or restricting the language: This covers the transition to logics of higher order (introducing new types of variables and predicates) or, in the other direction, monadic logic (restricting the language to 1-place predicates), as well as e.g. modal logic (introducing additional modal operators). Of course, an extension of the language always comes with an extension of the semantics.

altering the range of semantic values: e.g. many-valued logics, fuzzy logics.

altering the semantics of the logical constants: e.g. intuitionistic logic and minimal logic, where $\neg$ does not adhere to the classical truth table semantics, as well as paraconsistent logic, where $\bot$ does not "mean" a proposition from which anything may be inferred.


问:
Do we really need polynomials (In contrast to polynomial functions)?
In the following I'm going to call

a polynomial expression an element of a suitable algebraic structure (for example a ring, since it has an addition and a multiplication)
that has the form $a_{n}x^{n}+\cdots+a_{1}x+a_{0}$, where $x$ is
some fixed element of said structure,

a polynomial function a function of the form $x\mapsto a_{n}x^{n}+\cdots+a_{1}x+a_{0}$ (where the same algebraic considerations as above apply)

a polynomial an element of the form $a_{n}X^{n}+\cdots+a_{1}X+a_{0}$ with indeterminates $X$ (these
can be formalized, if we are in a ring, with strings of $0$s and
a $1$).

Note that when we are very rigorous/formal, polynomial expressions
and polynomials are something different (although in daily life we
often use them synonymously). Polynomial functions and expressions
are also different from each other although in this case the relationship
is a closer one, since every polynomials expression can be interpreted
as a polynomial function evaluated at a certain point (thus "polynomial
functions" are something more general than "polynomial expressions").

My question is: Why do we use polynomials ? It seems to me that every
piece of mathematics I have encountered so far, one could replace
every occurrence of polynomials with polynomial expressions/functions
without any major change in the rest of the proof/theorem/definition
etc.

My thoughts:

The only reasons that I can see to use polynomials are the following
two:

After one makes the idea precise that one can plug "suitable"
elements into polynomials (which may lie a ring containing the ring
in which the coefficients live in), one can save time in certain setting,
by handling the "plugging of suitable elements into polynomials"
more elegantly: For example in case of the theorem of Cayley-Hamilton,
which in its "polynomial function" version would look like:

Let $A$ be an $n\times n$ matrix over $K$, whose characteristic
polynomial (function) is $x\mapsto a_{n}x^{n}+\cdots+a_{1}x+a_{0}$.
Then
$$
a_{n}A^{n}+\cdots+a_{1}A+a_{0}I=0.
$$

whereas the "polynomial" version looks more elegant:

Let $A$ be an $n\times n$ matrix over $K$, whose characteristic
polynomial is $p_{A}\in K\left[X\right]$. Then
$$
p_{A}\left(A\right)=0.
$$

The only thing that polynomials can "do", but algebraic expressions/functions can't, is to be different, when the algebraic expressions/functions are the same (i.e. there's a theorem that tells us that the mapping of polynomials to polynomials expressions/functions
isn't injective, if the field is finite). Maybe this small difference makes a big enough difference
to consider polynomials after all, but as I said, I haven't encountered
any situation in which this difference could manifest itself.

(I'm guessing that maybe cryptography or higher number theory really needs
polynomials and not just polynomial expressions/functions. Since I don't know
anything about these subjects, I would be very happy with an example
of a theorem (whose content isn't merely technical as it is the case with the theorem above) involving polynomials, where these absolutely cannot
be replaced by polynomial expressions/functions. Conversely I would also be happy with an authoritative statement from someone knowledgeable that indeed we could dispense of polynomials, if we wanted to.)

答:
Constructing field extensions is a very important operation. This is most conveniently done as follows: Let $F$ be a field and $p(X)\in F[X]$. Suppose we wish to adjoin to $F$ an element $\alpha$ that will satisfy a certain algebraic equation. That is, there is a polynomial $P(X)\in F[X]$ such that we wish $P(\alpha)=0$ (notice here we switched between the polynomial $P(X)$ and its associated function evaluated at $\alpha$). If $P(X)$ is irreducible, then such a field extension is constructed as $F[X]/(P(X))$. It is crucial for this construction to take $F[X]$ to be the ring of polynomials, not polynomial functions.

Another reason is the convenient fact that a ring homomorphism $R[X]\to S$, from the ring of polynomials with coefficients in $R$ to a ring $S$, is the same thing as a homomorphism $R\to S$ together with a choice of an arbitrary element in $S$. Thus, polynomial rings allow for a systematic study of homomorphism+an element. It turns out that many properties of the chosen element are reflected by properties of the homomorphism, which in turn are related to properties of the domain, that is of polynomials (not polynomial functions).

One more reason is that polynomials are combinatorial beasts. They can be manipulated purely combinatorially. But since they give rise to functions, it means that we can use combinatorics to study functions. For instance, given functions $f,g$, can you find functions $a,b$ such that $af+bg=1$ is quite easily solved using the Euclidean algorithm if all these functions are required to be polynomials (the solution is actually constructive).

ContentLicense: CC BY-SA 3.0


答:
You have incorrectly done Greens.

$P= \sin(y)$
$ Q = x \cos(y)$

$$ \frac{ \partial Q}{\partial x} = \cos(y)$$

$$ \frac{ \partial P}{\partial y} =  \cos(y)$$

Now greens say:

$$ \oint Pdx + Qdy = \int \frac{ \partial Q}{\partial x} - \frac{\partial P}{\partial y} dA= \int 0 dA$$

Comment:

The line integral being zero for any loop is equivalent to field being conservative, for any loop we could have done the green's conversion is zero, hence the criteria for if vector field is conservative or not is that the:

$$ \frac{ \partial Q}{\partial x} - \frac{ \partial P}{\partial y} = 0$$

As mentioned by @PierreCarre

Title: Value of a line integral

Body:
What is the value of the line integral  $\oint\limits_{C}(sin(y)dx+xcos(y)dy)$ , where C consists of the circle $x^2+y^2=2 $ from $(-1,1)$ to $(1,1)$ and the line segment from $(1,1)$ to $(-1,1)$?

What I have done so far:

Using Green's Theorem I have $$\oint\limits_{C}(sin(y)dx+xcos(y)dy)= \iint\limits_{\Omega}2cos(y)dxdy$$

However I don't know what the limits should be for the integrals.


问:
On the nearest-square function - Part 2 and the quantity $m^2 - p^k$ where $p^k m^2$ is an odd perfect number
(Note:  This question has been cross-posted to MO.)

This question is an offshoot of this earlier one and this other question.

Let $n = p^k m^2$ be an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.

It was conjectured in Dris (2008) and Dris (2012) that the inequality $p^k < m$ holds.

Brown (2016) showed that the Dris Conjecture (that $p^k < m$) holds in many cases.

It is trivial to show that $m^2 - p^k \equiv 0 \pmod 4$.  This means that $m^2 - p^k = 4z$, where it is known that $4z \geq {10}^{375}$.  (See this MSE question and answer, where the case $m < p^k$ is considered.)  Note that if $p^k < m$, then $$m^2 - p^k > m^2 - m = m(m - 1),$$
and that
$${10}^{1500} < n = p^k m^2 < m^3$$
where the lower bound for the magnitude of the odd perfect number $n$ is due to Ochem and Rao (2012).  This results in a larger lower bound for $m^2 - p^k$.  Therefore, unconditionally, we have
$$m^2 - p^k \geq {10}^{375}.$$
We now endeavor to disprove the Dris Conjecture.

Consider the following sample proof argument:

Theorem If $n = p^k m^2$ is an odd perfect number satisfying $m^2 - p^k = 8$, then $m < p^k$.

Proof

Let $p^k m^2$ be an odd perfect number satisfying $m^2 - p^k = 8$.

Then $$(m + 3)(m - 3) = m^2 - 9 = p^k - 1.$$

This implies that $(m + 3) \mid (p^k - 1)$, from which it follows that
$$m < m + 3 \leq p^k - 1 < p^k.$$
We therefore conclude that $m < p^k$.

QED

So now consider the equation $m^2 - p^k = 4z$.  Following our proof strategy, and the formula in the accepted answer to the first hyperlinked question, we have:

$$m^2 - \bigg(\lfloor{\sqrt{m^2 - p^k} + \frac{1}{2}}\rfloor\bigg)^2 = p^k + \Bigg(4z - \bigg(\lfloor{\sqrt{m^2 - p^k} + \frac{1}{2}}\rfloor\bigg)^2\Bigg).$$

So the only remaining question now is whether it could be proved that
$$\Bigg(4z - \bigg(\lfloor{\sqrt{m^2 - p^k} + \frac{1}{2}}\rfloor\bigg)^2\Bigg) = -y < 0$$
for some positive integer $y$?

In other words, is it possible to prove that it is always the case that
$$\Bigg((m^2 - p^k) - \bigg(\lfloor{\sqrt{m^2 - p^k} + \frac{1}{2}}\rfloor\bigg)^2\Bigg) < 0,$$
if $n = p^k m^2$ is an odd perfect number with special prime $p$?

(Additionally, note that it is known that $m^2 - p^k$ is not a square, if $p^k m^2$ is an OPN with special prime $p$.  See this MSE question and the answer contained therein.)

If so, it would follow that
$$\Bigg(m + \lfloor{\sqrt{m^2 - p^k} + \frac{1}{2}}\rfloor\Bigg)\Bigg(m - \lfloor{\sqrt{m^2 - p^k} + \frac{1}{2}}\rfloor\Bigg) = p^k - y$$
which would imply that
$$\Bigg(m + \lfloor{\sqrt{m^2 - p^k} + \frac{1}{2}}\rfloor\Bigg) \mid (p^k - y)$$
from which it follows that
$$m < \Bigg(m + \lfloor{\sqrt{m^2 - p^k} + \frac{1}{2}}\rfloor\Bigg) \leq p^k - y < p^k.$$

Update (November 11, 2020 - 10:21 PM Manila time) Please check out the recently posted answer for a minor adjustment to the logic that should make the general proof argument work.

答:
If you don't have a proof that the smallest square larger than $m^2-p^k$ is not $m^2$, then your method does not work.

Otherwise, your method works.

Using your idea, one can prove that if $\lfloor\sqrt{4z}+1\rfloor\lt m$, then $m\lt p^k$.

Proof :

Subtracting $\lfloor\sqrt{4z}+1\rfloor^2$ which is the smallest square larger than $4z$ from the both sides of
$$m^2=p^k+4z$$
gives
$$m^2-\lfloor\sqrt{4z}+1\rfloor^2=p^k-\lfloor\sqrt{4z}+1\rfloor^2+4z$$
which can be written as
$$(m-\lfloor\sqrt{4z}+1\rfloor)(m+\lfloor\sqrt{4z}+1\rfloor)=p^k-\lfloor\sqrt{4z}+1\rfloor^2+4z\tag1$$

So, we can say that
$$m+\lfloor\sqrt{4z}+1\rfloor\mid p^k-\lfloor\sqrt{4z}+1\rfloor^2+4z\tag2$$

If $\lfloor\sqrt{4z}+1\rfloor\lt m$, then LHS of $(1)$ is positive, so RHS of $(1)$ is positive. So, we can say that$$(2)\implies m+\lfloor\sqrt{4z}+1\rfloor\le p^k-\lfloor\sqrt{4z}+1\rfloor^2+4z$$from which we can have$$m\lt m+\lfloor\sqrt{4z}+1\rfloor\le p^k-\lfloor\sqrt{4z}+1\rfloor^2+4z\lt p^k.\quad\blacksquare$$

If $m=\lfloor\sqrt{4z}+1\rfloor$, then letting $\sqrt{4z}=N+a$ where $N\in\mathbb Z$ and $0\le a\lt 1$, we have
$$p^k-m=(N+1)^2-(N+a)^2-N-1=(1-2a)N-a^2$$
whose sign depends on $a$ and $N$.


问:
Proving a square set is convex
Assume there is a set $S$ in $\mathbb{R}^2 $ that is a square with $x \in [-1 ,1]$ and $y\in [-1,1]$. I need to prove that this set is convex. Hence, I thought of the following:

Suppose $a, b \in S$ such that $a = (x_1, y_1)$, $b = (x_2, y_2)$ and $\forall \theta \in [0,1]$ the following must hold:

$$(x,y) = \theta a + (1-\theta) b \in S$$

So I need to prove that:

$$||(x,y)|| = ||(\theta x_1 + (1-\theta) x_2, \theta y_1 + (1-\theta ) y_2 )|| \leq \sqrt{2}$$

Is this correct and how can I continue?

答:
It is not correct because the condition $||(x,y)|| \leq \sqrt{2}$ decribes a circular disk with center $0$ and radius $\sqrt 2$.

You know that $x_1,x_2,y_1,y_2 \in [-1,1]$. Hence $\theta x_1 + (1-\theta)x_2 \in [-1,1]$, $\theta y_1 + (1-\theta)y_2 \in [-1,1]$. Thus $\theta a + (1-\theta)b \in S$.

Title: Proving a square set is convex


问:
Banach spaces $X^*=Y^*$ implies $X=Y$?
Let $X$ and $Y$ be two Banach space. If $X^*$ is isometry to $Y^*$, then $X$ is isometry to $Y$? Here $^*$ means the dual space.

If one of $X$ and $Y$ is reflective,the answer is affirmative. Here is my proof. (“=” means isometry)Suppose $X$ is reflective, then $Y^*=X^*$ which is reflective, so $Y^{**}$ is reflective. The classical embedding $Y \mapsto Y^{**}$ shows $Y$ is (isometry to) a closed subspace of $Y^{**}$. Hence $Y$ is reflective by the result that any closed subspace of reflexive space is reflective. So $Y=X^{**}=Y^{**}=Y$.

For non-reflective case, I try to prove but failed. And I guess this claim is false. But It seems to be difficult to give counter examples.
Can someone give me some hint or references? Thanks!

答:
A simple counterexample is
$$
X = c = \{ \text{convergent sequences}\}
$$
and
$$
Y = c_0 = \{ \text{null sequences}\}.
$$
Then, one can check that
$$
X^* = \ell^1 = Y^*
$$
(isometric isomorphisms).

The spaces $X$ and $Y$ are isomorphic but not isometrically isomorphic (look at the extremal points of the unit ball).

Title: Banach spaces $X^*=Y^*$ implies $X=Y$?


问:
Proof that two sets have the same cardinality
Look in the complex plane at the set $\mathbb{R}\cdot1\cup\mathbb{Z}\cdot i\subset\mathbb{C}$, the union of the real line and all whole multiples of $i$. Prove that this set has the same cardinality as $\mathbb{R}$. I'm having trouble finding a bijection. I would appreciate any help!

答:
Consider $f :\left(\mathbb{R} \times \lbrace 0 \rbrace\right) \cup \left(\mathbb{Z} . i \right) \rightarrow \mathbb{R}$ defined for all $x \in \mathbb{R}$, $n \in \mathbb{Z}$ by
$$
f(x,0) = \left\{
    \begin{array}{ll}
      \frac{1}{2k-1}   & \mbox{if }  x=\frac{1}{k} \text{ for an integer }k\in \mathbb{Z}_{>0}\\
 \frac{-1}{2k-1}   & \mbox{if }  x=\frac{1}{k} \text{ for an integer }k\in \mathbb{Z}_{<0}\\
       x  & \mbox{otherwise.}
    \end{array}
\right.
$$
$$
f(n.i) = \left\{
    \begin{array}{ll}
      0   & \mbox{if }  n=0\\
      \frac{1}{2n}  & \mbox{otherwise.}
    \end{array}
\right.
$$

Show that $f$ is bijective.


问:
Derivation of geodesic on $m$-sphere $\gamma : \mathbb{R} \rightarrow S^m$ is given by $\gamma(t) = \cos(t |v|)p \ + \frac{\sin(t |v|)}{|v|} v$
Given $p \in S^m$ and $0 \neq v \in T_pS^m = p^{\perp}$, the geodesic $\gamma : \mathbb{R} \rightarrow S^m$ seems to take the form by $\gamma(t) = \cos(t |v|)p \ + \frac{\sin(t |v|)}{|v|} v$, where $t \in \mathbb{R}$.

This is pretty much a given as fact in all differential geometry books I've seen so far. I assume this is some kind of supposedly trivial parameterization obtained by intersecting $S^m$ with the plane spanned by $\big\{p, \frac{v}{|v|} \big\}$, but I haven't been able to connect that with the actual closed form of the geodesic $\gamma$.

I'd tremendously appreciate a derivation of this result, as I haven't been able to find this anywhere else.

答:
Consider $\mathbb{S}^m \subset \mathbb{R}^{m+1}$ with the induced metric. What is to be known is that for a riemannian submanifold $(N,g,\nabla) \subset (M,\bar g,\bar \nabla)$, the Levi-Civita connection $\nabla$ is the orthogonal projection of the ambiant Levi-Civita connexion $\bar \nabla$ on the tangent space. Thus, if $c$ is a parametrized curve in $\mathbb{S}^m$, $\nabla_{c'} c' = \left(\bar \nabla_{c'} c'\right)^{\perp}$. Using this, one can show that the curve
\begin{align}
\gamma (t) = \cos(t\|v\|)p +  \sin(t\|v\|)\frac{v}{\|v\|}
\end{align}
is a geodesic of $\mathbb{S}^m$. Indeed, in the ambiant space $\mathbb{R}^{m+1}$, the connexion is trivial, so the computation shows that $\bar\nabla_{\gamma'(t)}\gamma'(t) = \gamma''(t)=-\|v\|\sin(t\|v\|)p + \cos(t\|v\|)v$ is orthogonal to $T_p\mathbb{S}^m$, and $\nabla_{\gamma'}\gamma' = 0$. Thus, $\gamma$ is a geodesic.

To conclude, by uniqueness of geodesics (because $\nabla_{\gamma'}\gamma'$ is a linear second order differential equation in charts), if $c$ is a geodesic with $c(0)=p$ and $c'(0) = v$, $c = \gamma$.

Edit Here is another proof, more constructive. The idea is to show that a geodesic has to lie inside a linear plane of $\mathbb{R}^{m+1}$. For that, let $p \in \mathbb{S}^m$ and $v \in T_p\mathbb{S}^m$ be a non-zero tangent vector. Let $P = \mathrm{span}(p,v)$. Consider the linear isometry $u$ of $\mathbb{R}^{m+1}$ that is the reflexion with respect to $P$. One can show it is an isometry of $\mathbb{S}^m$ because it stabilizes the sphere and preserves the scalar product on tangent spaces.

Now that we know $u$ is an isometry of $\mathbb{S}^m$, let $\gamma$ be the geodesic of $\mathbb{S}^m$ with $\gamma(0) = p$ and $\gamma'(0)=v$. Then $u\circ \gamma$ is also a geodesic (easy computation) with the same initial data. Thus, $u\circ \gamma = \gamma$ and $\gamma$ has to lie in the set of points in $\mathbb{S}^m$ unvariant by $u$, that is, $P \cap \mathbb{S}^m$. So $\gamma$ is a curve lying insinde the great circle $\mathbb{S}^m\cap P$. As it is a circle and as a geodesic has to be parametrized with constant velocity, is has to be of the form given above.


问:
Probability Theory - Independent Uniform Variables
Suppose $X, Y$ are independent uniform(0,1) random variables. For some arbitrary $t$, I want to find $P(X/Y \le t)$. I am trying to draw a picture to figure this out, but I don't think I am accomplishing too much. Could I have some direction on how to do this?

For clarity, $t$ is arbitrary. I have no information on what it is.

答:
First of all note that $T=\frac{X}{Y} \in(0;+\infty)$

The result is the following

$$\mathbb{P}[T\leq t]=\begin{cases}
0,  & \text{if $t<0$ } \\
\frac{t}{2}, & \text{if $0\leq t<1$}\\
1-\frac{1}{2t}, & \text{if $t\geq 1$}\\
\end{cases}$$

To understand this result take a look at the following drawing

$$\mathbb{P}\Bigg[\frac{X}{Y} \leq t\Bigg]=\mathbb{P}\Bigg[Y > \frac{X}{t}\Bigg]$$

thus the probability you are looking for is the purple area. When $t<1$ it is easy to calculate 1 minus white triangle's area


问:
Injective $C^1$ curve $[0,1)\rightarrow \mathbb{R}^2$ such that $c([0,1))=S^1$ and $c'$ has exactly 7 zeros
This is an exercise for the reader from my differential geometry course:

Find an injective $C^1$ curve $c:[0,1)\rightarrow \mathbb{R}^2$ such that its image is the unit circle and its derivative has exactly 7 zeros. Does this work with $C^\infty$ curves?

For the first part of the question, I think it's enough to find a monotonously increasing function $\phi:[0,1)\rightarrow [0,2\pi)$ and then plug that in to the usual $(cos(\phi(t)),sin(\phi(t)))$. I don't know how to get $\phi'$ to have exactly $7$ zeros though. The obvious choice would be a degree $7$ polynomial, but at the same time $\phi'\ge 0$ must hold.

Because this is just a "little" exercise, I feel I'm missing something much simpler. Also, what I've got so far doesn't really help with the second part of the question.

答:
I found an answer to both parts of the question.
$$\phi(t) = \frac {\sin(14\pi t)}{14\pi}+t$$
Then $(\cos(2\pi \phi(t)), \sin(2\pi \phi(t)))$ satisfies the requirement and is obviously smooth, i.e., $C^\infty$


问:
Find linear transformation given a kernel and image
Find a linear transformation $T: R^2 \rightarrow R^3$ where

$Im(T) = sp \{(1, 0, 0), (0, 1, 0)\} .$

$Ker(T) = sp \{(0, 1)\}$

So, the first thing I did was to expand $Ker(T)$ to form a basis of $R^2$:

$B = sp\{(0, 1), (1, 0)\}$

Next, I mapped the basis elements of $B$ to $Im(T)$:

$T(B_1) = 0$ (since $B_1 \in Ker(T)$)

$T(B_2) = (1, 0, 0)$

But since $B$ only has 2 elements, I can't map another element to $(0, 1, 0)$ (the last element in $Im(T)$).

Does that mean that this linear transformation doesn't exist?

答:
Hint: For a linear transformation $T:V\rightarrow W$, the rank-nullity theorem gives $\dim V= \dim kernel(T) + \dim image(T)$.
Now you can do book keeping: $2\ne 1+2$.


答:
Since that $$\dim(Ker(T))+\dim(Im(T))=\mathbb{R}^{2}$$
but, here $\dim(Im(T))=2$ and $\dim(Ker(T))=1$, so the rank-null theorem is not satisfied, so such a linear transformation does not exist.

Title: Find linear transformation given a kernel and image

Body:
Find a linear transformation $T: R^2 \rightarrow R^3$ where

$Im(T) = sp \{(1, 0, 0), (0, 1, 0)\} .$

$Ker(T) = sp \{(0, 1)\}$

So, the first thing I did was to expand $Ker(T)$ to form a basis of $R^2$:

$B = sp\{(0, 1), (1, 0)\}$

Next, I mapped the basis elements of $B$ to $Im(T)$:

$T(B_1) = 0$ (since $B_1 \in Ker(T)$)

$T(B_2) = (1, 0, 0)$

But since $B$ only has 2 elements, I can't map another element to $(0, 1, 0)$ (the last element in $Im(T)$).

Does that mean that this linear transformation doesn't exist?


问:
number of solutions to an equation?
Given $x$ and $y$ are multiples of $2$ satisfying
$$x^2 - y^2 = 27234702932$$
Find the number of solutions to $x$ and $y$.</p>

答:
HINT $(X-Y)(X+Y)=X^2-Y^2$

(then factor)

Title: number of solutions to an equation?

Body:
HINT $(X-Y)(X+Y)=X^2-Y^2$

(then factor)


问:
When is 2 a quadratic residue mod p
I'm trying to prove that 2 is a QR when $p = \pm \ 1 \text{ mod } 8.$ I know how to prove via a "factorial" proof, butI'm given the hint to use the fact that if $ \zeta $ is a primitive 8-root of unity, then $(\zeta + \bar{\zeta})^2 = 2$, and would like to solve the exercise using the hint. This is where I got so far: $$ \text{ I can use the fact that } 2 \text{ is a quadratic residue if and only if } 2^{\frac{p-1}{2}}=1 \text{ mod p}$$
$$ 2^{\frac{p-1}{2}} = (\zeta + \bar{\zeta})^{p-1} = \sum_{k = 0} ^{p-1} {p-1 \choose k} \bar\zeta^k \zeta^{p-1-k} = \sum_{k = 0} ^{p-1} {p-1 \choose k}\zeta^{p-1+6k}$$ Now I guess I should use the fact that $ \zeta^4 = -1$, and thus the symmetry of the sum cancels many factors, but I'm stuck at this point, and even computing the sum for small p didn't help much. Can someone help me out? Thanks!

答:
Another method is based on this answer. It considers $\sqrt2$ times an eighth root of unity.

When $p$ is prime and $k\not\in\{0,p\}$, $\binom{p}{k}\equiv0\pmod{p}$. Thus, the Binomial Theorem says that
$$\newcommand{\Re}{\operatorname{Re}}
(1+i)^p\equiv1+i^p\pmod{p}\tag1
$$
We also have that
$$
\begin{align}
(1+i)^p
&=\left(\sqrt2\,e^{\frac{\pi i}4}\right)^p\\
&=2^{\frac{p-1}2}\sqrt2\,e^{\frac{\pi i}4p}\tag2
\end{align}
$$
Comparing $(1)$ and $(2)$, we get that
$$
2^{\frac{p-1}2}\Re\left(\sqrt2\,e^{\frac{\pi i}4p}\right)\equiv1\pmod{p}\tag3
$$

Then, we only need note that $\Re\left(\sqrt2\,e^{\frac{\pi i}4p}\right)=1$ when $p\equiv\pm1\pmod8$ and that $\Re\left(\sqrt2\,e^{\frac{\pi i}4p}\right)=-1$ when $p\equiv\pm3\pmod8$.

Title: When is 2 a quadratic residue mod p


问:
Prove alternate form of Littlewood-Offord
I'm self-studying Bollobás' Combinatorics textbook and am stuck on a question regarding a reformulation of Littlewood-Offord. We are given a vector $x\in {\bf R}^d$ and $n$ other vectors $x_1,\ldots,x_n$. All of these vectors have length at least $1$. Now we consider all $2^n$ sums of the form
$$\sum_{i=1}^n \epsilon_ix_i$$
where $\epsilon_i\in\{-1,1\}$ and the goal is to show that at most ${n\choose \lfloor n/2\rfloor}$ of these sums can be at a distance $\leq 1$ from $x$.

For $d=1$, we can assume that all the $x_i$ are positive by multiplying by $-1$ if necessary. For $A\subseteq [n]$, we let
$$x_A = \sum_{i\in A} x_i - \sum_{i\notin A} x_i.$$
Let ${\cal F}$ be the set of all $A\subseteq [n]$ such that $|x_A - x| < 1$. Let $A$ be a proper subset of $B\subseteq [n]$ and consider $|x_A - x| + |x_B-x|$. By the triangle inequality, we have
$$\eqalign{
|x_A - x| + |x_B - x|&\geq |x_B - x_A| \cr
&= \Big| \sum_{i\in B} x_i -\sum_{i\in A}x_i + \sum_{i\notin A}x_i - \sum_{i\notin B} x_i\Big| \cr
&= 2\Big|\sum_{i\in B\setminus A} x_i\Big|\cr
&\geq 2 \Big(\sum_{i\in B\setminus A}x_i - \sum_{i\in B\setminus A} x_i \Big)\cr
&= 2x_{B\setminus A}\cr
&\geq 2.
}$$
So one of $x_A$ and $x_B$ is not in ${\cal F}$, meaning we can apply Sperner's theorem. But in arbitrary dimension, the last bit does not work, because introducing the part of the sum that is subtracted, we could actually make the vector longer. There doesn't seem to be an analogue of assuming all the vectors are "positive", like we did at the beginning.

Instead, I'm guessing that the author intended us to use the Littlewood-Offord theorem statement found in the chapter. I'll rephrase it here:

Theorem 2. Let $B$ be a normed space and let $x_1, \ldots, x_n\in B$ be vectors of norm $\geq 1$. Consider all $2^n$ possible sums (where the null sum has value $0$). If we pick a subset $A$ of these sums such that every pair $x,y\in A$ is such that $|\!|x - y|\!| < 1$, then the subset must have size $\leq{n\choose \lfloor n/2\rfloor}$.

In fact, the author says that these two statements are equivalent, and indeed they seem like they should be, but I'm stuck on the details.

答:
After a rumble and a tumble, I think I got it figured out. (I followed a construction I found on Wikipedia, but the details still took a bit of work and I'm spelling them out here for completeness.) Fix some vectors $x_i$ as well as $x$, all of norm $\geq 1$, let $x_A$ be as above and let ${\cal F}$ be the family of $A\subseteq S = [n]$ such that $\lVert x_A -x\rVert < 1$. Let $f : {\bf R}^d\to {\bf R}^d$ be the affine transformation given by
$$f(x) = {1\over 2}\Big(x+\sum_{i\in S} x_i\Big).$$
Note that for any $A\subseteq S$,
$$\eqalign{
f(x_A) &= {1\over 2} \Big( \sum_{i\in A} x_i - \sum_{i\notin A} x_i + \sum_{i\in S} x_i\Big)\cr
&= \sum_{i\in A} x_i;\cr
}$$
these vectors are exactly of the form prescribed by Theorem 2. Let $x_A$ and $x_B$ be vectors such that $A,B\in \cal F$. By the triangle inequality applied with $x$ as the intermediate point, we have
$\lVert x_A - x_B\rVert <2$. Now we can compute
$$\eqalign{
\big\lVert f(x_A) - f(x_B)\big\rVert &= \bigg\lVert {1\over 2} \Big( \sum_{i\in A} x_i - \sum_{i\notin A} x_i + \sum_{i\in S} x_i\Big) - {1\over 2} \Big( \sum_{i\in B} x_i - \sum_{i\notin B} x_i + \sum_{i\in S} x_i\Big) \bigg\lVert \cr
&= {1\over 2}\lVert x_A - x_B\rVert \cr
&<1,
}$$
so the set $\{ f(x_A) : A\in {\cal F}\}$ consists of vectors that are at distance $<1$ of each other. But this is a set of sums of $x_i$, and since $f$ is injective, it has exactly the same size as ${\cal F}$, so by Theorem 2, ${\cal F}$ must have had size $\leq {n \choose \lfloor n/2\rfloor}$. (In fact, $f$ is a bijection, so the argument can be reversed to show that the proposition is equivalent to Theorem 2.)


问:
If $\varphi$ is continuous, find a continuous branch of $t\mapsto\ln(\varphi(t))$
Let $E$ be a $\mathbb R$-Banach space and $\varphi\in C^0(E,\mathbb C\setminus\{0\})$.

Let $x\in E$. How can we show that "there is an unique branch $h(t)$, $0\le t\le 1$, of $\ln\varphi(tx)$ such that $h(0)=0$ and $h$ is continuous" and what's the precise meaning of this claim?

It' clear to me that if $z\in\mathbb C\setminus\{0\}$, then $\phi\in\mathbb R$ is called argument of $z$ if $$z=|z|(\cos\phi+{\rm i}\sin\phi)\tag1.$$ If $\operatorname{arg}z$ denotes the set of all arguments of $z$, then the unique element $\operatorname{Arg}z\in\operatorname{arg}z\cap(-\pi,\pi]$ is called the *principal value of the argument of $z$ and we define $$\ln z:=\ln|z|+{\rm i}\operatorname{Arg}z\tag2.$$

I guess the main issue we deal with in the question is the lack of continuity of $[0,1]\ni t\mapsto\operatorname{Arg}\varphi(tx)$.

答:
This is a classic result and one proof could go like this: first note that $t \to \phi(tx)$ is a continuous function from $[0,1] \to \mathbb C^*$ so the space $E$ is irrelevant for our purposes and we can just consider the case $t \to z(t)$ a continuous function from $[0,1] \to \mathbb C^*$.

Since the image is compact and avoids zero, there is $\delta >0, |z(t)| \ge \delta, t \in [0,1]$ and since the function is uniformly continuous, there is $\lambda>0, |z(t)-z(s)| < \delta, |t-s| < \lambda$ and we can choose a partition $0=t_0<t_1<...t_m=1, t_{k+1}-t_{k} < \lambda$

Consider $z(t)=|z(t)|e^{\psi(t)}$ where $\psi(t)$ is the principal argument in $-\pi <\psi(t) \le \pi$ and we construct by induction $\phi(t)$ continuous argument.

We define $\phi(a)=\psi(a)$ and if $\pi \notin \psi [a,t_1]$ (equivalently $z(t) \notin (-\infty, 0)$ there) then $\psi$ is continuous there (the only discontinuity is at $\pi$ by our choice) so we don't change anything and let $\phi=\psi$ there.

If $\pi \in \psi [a,t_1], \psi(s)=\pi$ we claim that there is no $s_1 \in [a,t_1], \psi(s_1)=0$ as otherwise $\delta > |z(s_1)-z(s)|=|z(s_1)|+|z(s)| > |z(s_1)| > \delta$ by our choices which is impossible. This means that we can take a continuous argument $\psi_1(t) \in (0, 2\pi)$ now there and if $\psi(a)-\psi_1(a)=2m_0\pi$ we define $\phi(t)=\psi_1(t)+2m_0\pi$, $\phi(a)=\psi(a)$ as required.

But now it is clear that we can repeat this on $[t_1,t_2]$ with $t_1=a$ now and find $\phi$ continuous argument, where the only change now is that in either case ($\psi$ continuous on $[t_1,t_2]$ or $\psi_1$ with values in $(0, 2\pi)$ continuous argument there), we shift by the appropriate integer multiple of $2\pi$ to preserve the value $\phi(t_1)$ and keep it continuous from the left there too, so by induction we can continue until we reach $t_k=1$ and we are done and can find $\phi$ continuous argument!


问:
Polar decomposition of real matrices
Can we decompose real matrices in a way that is analogous to polar decomposition in the complex case?

What I mean is: given an invertible real matrix $M$, can we always write:
$$
M = OP,
$$

maybe uniquely, where $O$ is orthogonal, and $P$ is symmetric positive-definite?

Thanks.

答:
Theory of Polar Decomposition is described in Wikipedia
For the sake of simplicity we shall restrict our attention to real-valued non-singular square (invertible) matrices.

Theorem. Every such a matrix B can be decomposed as follows and the decomposition is unique:
$$
B = U Q
$$
Where $U =$ orthogonal matrix and $Q=$ symmetric positive definite matrix.

Instead of providing still another proof of the above theorem, I have decided to be practical and describe how to numerically
obtain the decomposition.
First define the transpose of the matrix multiplied with the original:
$$
P = B^T B = \left(B^TB\right)^T = P^T \quad \mbox{: symmetric}
$$
It follows that $P$ is symmetric and positive definite.
Then what we need is the square root of the matrix $P$. In order to understand how to obtain it, let's take a look at
Newton's method for obtaining the square root
of a real positive number $p$:
$$
f(x) = x^2 - p\quad \Longrightarrow \\ x_{n+1} = x_n - \frac{x_n^2-p}{2x_n} = \left(x_n + p\,x_n^{-1}\right)/2
$$
Let's do the same for our matrix $P$, iterations starting with the unit matrix:
$$
X_0 = I \quad ; \quad X_{n+1} = \left(X_n + P X_n^{-1}\right)/2 \quad ; \quad \sqrt{P} = \lim_{n\to\infty} X_n
$$
According to a
MSE reference
$X_n^{-1}$ is positive definite and symmetric, provided that $X_n$ is positive definite and symmetric. Products and sums
of positive definite and symmetric matrices are positive definite and symmetric too. Therefore,
by induction to $n$, it can be proved that $\sqrt{P}$ is positive definite and symmetric as well. Now define $Q=\sqrt{P}=Q^T$.
At last define $U = B Q^{-1}$ and prove that $U$ is orthogonal:
$$
U^{T}U = \left(B Q^{-1}\right)^T\left(B Q^{-1}\right) = \left(Q^{-1}\right)^T\left(B^TB\right)Q^{-1} = \\
\left(Q^T\right)^{-1} Q^2 Q^{-1} = \left(Q^{-1}Q\right)\left(QQ^{-1}\right) = II = I
$$
The conclusion looks like trivial:
$$
B = \left[B\left(B^TB\right)^{-1/2}\right]\left[\left(B^TB\right)^{1/2}\right]
$$
So far so good about theory. I want to talk about practice now, which in modern times is computer programming:

Accompanying software

My favorite programming language is (Delphi) Pascal.
First we need some standard matrix manipulation routines and a Newton-Raphson routine for calculating the matrix
square root. These are implemented in a Unit called 'Wiskunde'. The software is completed by testing if theory
works in practice as expected. The number of iterations in procedure 'Newton' has been determined as follows. A suitable stopping criterion is:
$$
\left|\left| X^2 - B^TB\,\right|\right| < \epsilon
$$
with $\epsilon = 10^{-10}$ and the norm of a $n \times n$ matrix $A$ defined as:
$$
\left|\left|A\right|\right| = \sqrt{\sum_{i=1}^n \sum_{j=1}^n A_{ij}^2}
$$
Convergence is very fast. When cast in MathJax format the output looks like this:
$$ \small
\begin{bmatrix} 0.361048 & -0.297419 & -0.227079 \\ 0.171654 & -0.181309 & -0.338205 \\ -0.127762 & -0.074326 & -0.417988 \end{bmatrix} = \\ \small
\begin{bmatrix} 0.491571 & -0.868969 & -0.057015 \\ 0.599413 &  0.385125 & -0.701701 \\ -0.631714 & -0.310760 & -0.710187 \end{bmatrix}
\begin{bmatrix} 0.361082 & -0.207928 & -0.050301 \\ -0.207928 & 0.211719 & 0.196967 \\ -0.050301 & 0.196967 & 0.547115 \end{bmatrix}
$$

Title: Polar decomposition of real matrices


问:
"Standard" ways of telling if an irreducible quartic polynomial has Galois group C_4?
The following facts are standard: an irreducible quartic polynomial $p(x)$ can only have Galois groups $S_4, A_4, D_4, V_4, C_4$. Over a field of characteristic not equal to $2$, depending on whether or not the discriminant $\Delta$ is a square and whether or not the resolvent cubic $q(x)$ is irreducible, we can distinguish four cases:

If $\Delta$ is not a square and the resolvent cubic is irreducible, then the Galois group is $S_4$.

If $\Delta$ is a square and the resolvent cubic is irreducible, then the Galois group is $A_4$.

If $\Delta$ is a square and the resolvent cubic is reducible, then the Galois group is $V_4$.

If $\Delta$ is not a square and the resolvent cubic is reducible, then the Galois group is either $D_4$ or $C_4$.

How can we resolve the ambiguity in the last case? For $p$ a monic polynomial in $\mathbb{Z}[x]$, I know the following approaches:

In the simplest cases one can work directly with the splitting field. But this is rare, although it can work if $p = x^4 + ax^2 + b$ for some $a, b$.

If $p$ has two complex roots (equivalently, if the discriminant is negative), there is a transposition in the Galois group, so the Galois group is $D_4$.

If $p$ factors as the product of two linear factors and an irreducible quadratic factor modulo a prime, there is a transposition in the Galois group, so the Galois group is $D_4$.

In practice, the last two often work to identify a Galois group of $D_4$ (and in principle it must eventually work by the Frobenius density theorem). But I don't know a corresponding practical way to identify a Galois group of $C_4$. There is a criterion due to Kappe and Warren which I learned about from one of Keith Conrad's expository notes here. However, in an upcoming exam I'm taking, I'll only be able to cite results proven in the course, and this criterion isn't one of them.

So what are my other options in general?

答:
For what it's worth, here's a special case that can be done with standard tools. Again we restrict to a base field $k$ of characteristic not $2$. Suppose that $f(x) = (x^2 - a)^2 - b$ is irreducible. (This is Corollary 4.5 in the linked notes by Keith Conrad, but there it is proven as a corollary of Kappe-Warren.) We compute that $f'(x) = 4x(x^2 - a)$; letting the roots of $f$ be $\alpha_1, ... \alpha_4$, the discriminant is then

$$(-1)^{ {4 \choose 2} } \prod f'(\alpha_i) = 64 (a^2 - b) b^2$$

which is square if and only if $a^2 - b$ is. Since the splitting field of $f$ is given by a tower of quadratic extensions, the Galois group is one $V_4, D_4, C_4$. Casework:

If $a^2 - b$ is a square, then the Galois group is $V_4$.

If $a^2 - b$ is not a square and neither is $b(a^2 - b)$, then the splitting field contains distinct quadratic subfields $k(\sqrt{b})$ and $k(\sqrt{a^2 - b})$, hence the Galois group is $D_4$.

If $a^2 - b$ is not a square but $b(a^2 - b)$ is, write $u = \sqrt{a + \sqrt{b}}$ in a splitting field for $f$. Then $u \sqrt{a - \sqrt{b}} = \sqrt{a^2 - b} = c \sqrt{b} \in k(u)$ for some $c \in k$, hence $u$ generates the splitting field, which must have degree $4$. Hence the Galois group is $C_4$.


问:
Sum of kth roots ($\sum\sqrt[k]{m}$)
I'm trying to find an asymptotic to $$S(n) = \sum_{k=1}^n\sqrt[k]{m}$$ From computational tests, it seems to grow nearly as slowly as $n$. However even $$\sum_{k=1}^\infty\sqrt[k]{m}-1$$ diverges (for $m\neq1$) by the comparison test.

I'm thinking it might be something like $n\log{\log n}$, but I don't know how to show it.

Update: So it turns out to be closer to $n\sqrt[n]{m}$. Does anybody know if there is a nice formula?

答:
For small enough $x>0$ we have $e^x-1\in(x,2x)$. Furthermore, $\sqrt[k]{m}-1=m^{1/k}-1=e^{(\ln m)/k}-1$. As $k$ increases $x$ will approach $0$ and hence for large enough $k$ we have
$$ \frac{\ln m}{k}<\sqrt[k]{m}-1<\frac{2\ln m}{k}\,.$$
Since
$$ \sum_{k=1}^n\frac1k \sim \ln n \longrightarrow \infty $$
for $n\to\infty$, by the sandwich lemma, your sum will also go to infinity for $n\to\infty$ with logarithmic speed.</p>

Title: Sum of kth roots ($\sum\sqrt[k]{m}$)


问:
Show that $T_1=T_2S$ if $R(T_1)=R(T_2)$
Suppose $W$ is finite-dimensional and $T_1,T_2 \in L(V,W)$. Prove that
if $\operatorname{range}(T_1) = \operatorname{range}(T_2)$, then there exists an invertible operator $S \in L(V,V)$ such that $T_1 = T_2S$.

Attempt: If $V$ is finite-dimensional the problem can be solved by taking basis for $V$, $N(T_1)$ and $N(T_2)$. But since the dimension of $V$ is not mentioned, how can the problem be solved when $\dim(V)$ is infinite? That's where I'm struggling.

Any help on the latter case would be appreciated!

答:
One approach is as follows: with the assumption of Zorn's lemma (or equivalently the axiom of choice), we can assume that $\ker(T_j)$ has a basis (for $j=1,2$), and that this basis may be extended to produce a basis of $V$. Let $\mathcal K_j = (u^j_k)_{k \in \alpha_j}$ denote a basis for the kernel of $T_j$ with $\alpha_j$ denoting an associated index set. Similarly, let $\mathcal B_j = (v^j_k)_{k \in \beta_j}$ denote a set of vectors that can be added to $\mathcal K_j$ to produce a basis $\mathcal B_j \cup \mathcal K_j$ of $V$.

Argue that the vectors $(T_j v^j_k)_{k \in \beta _j}$ must form a basis of the range of $T_j$, and so $\beta_j$ must be finite. In particular, we can take $\beta_j = \{1,\dots,r\}$ without loss of generality.

Define $Q_j = \operatorname{span}\{v_1^j,\dots,v_d^j\}$ and $R = \operatorname{range}(T_1) = \operatorname{range}(T_2)$. Argue that the maps $T_j|_{Q_j} : Q_j \to R$ are invertible (for $j=1,2$) ($T|_Q$ denotes the restriction of $T$ to $Q$).

Define $\bar S:Q_1 \to Q_2$ by $\bar S = [T_2|_{Q_2}]^{-1}T_1|_{Q_2}$. Verify that $T_2|_{Q_2}\bar S = T_1|_{Q_1}$.

To define $S$, extend the definition of $\bar S$ to a map on $V$ that is zero for all elements in $\ker(T_1)$. Equivalently, define $S$ on the basis elements with
$$
S(v^1_k) = \bar S(v^1_k) \quad k= 1,\dots,d, \qquad
S(u^j_k) = 0 \quad k \in I_1.
$$
Because we have $T_2S(v) = T_1(v)$ for all elements $v$ from our basis $\mathcal B_1 \cup \mathcal K_1$, we have $T_2 S = T_1$ as desired.

Title: Show that $T_1=T_2S$ if $R(T_1)=R(T_2)$


问:
Torsion in the first Homology group
I was trying to solve an exercise which says the following:

Let $U\subset\mathbb{R}^3$ be an open subset. Then $H_1(U;\mathbb{Z})$ has no torsion.

I think that the universal coefficient theorem in homology has to be used, but I don't know how.

Moreover, this result seems very strange to me since we do not have any other hypothesis on the open subset $U$. For example in dimension $4$ we have the smooth embedding of $ \ \mathbb{RP}^2$ in $ \ \mathbb{R}^4$ and so it's enough to take a tubular neighborhood of $ \ \mathbb{RP}^2$ (regarded as a closed submanifold of $\mathbb{R}^4$) to obtain an open subset $U\subset\mathbb{R}^4$ which retracts by deformation on $ \ \mathbb{RP}^2$ and so $H^1(U;\mathbb{Z})=\mathbb{Z}_2$.

If the proposition at the beginning is true then it is saying that no "strange" phenomena can happen in $\mathbb{R}^3$, as for example a smooth embedding of some projective space. It seems to be a non trivial result, but it was given to me as an exercise during my Algebraic Topology course.

答:
Hint: If $H_1(U)$ has torsion, find a nice compact subset $K\subseteq U$ such that $H_1(K)$ also has torsion, and then use Alexander duality.

More details are hidden below

 Triangulate $U$ and consider a nonzero torsion class in $H_1(U)$.  There is then a finite subcomplex $K$ of $U$ which contains a 1-cycle $\alpha$ representing that torsion class, and also a 2-cycle whose boundary is a multiple of $\alpha$ to witness that $[\alpha]$ is torsion.  So, $[\alpha]$ is also a torsion class in $H_1(K)$.  But by Alexander duality, $H_1(K)\cong H^1(S^3\setminus K)$, and $H^1$ can never have torsion (this follows from the universal coefficient theorem since $H_0$ is always free).

Title: Torsion in the first Homology group


问:
Interchanging integral with real and imaginary operators?
$$ \int \sin(3x) \cos(nx) dx \to \Re \int \sin(3x) e^{inx} dx$$

$$ \sin(3x) \to \Im e^{i3x}$$

Hence,

$$ \Re \left( \Im \int e^{i(3+n)x} dx \right) $$

Or,

$$ \Re \left(\Im \frac{e^{i(3+n)}}{i(3+n)} \right) \to \Re \left(\Im \frac{-ie^{i(3+n)}}{(3+n)} \right)$$

Considering,

$$ \Im \frac{-ie^{i(3+n)}}{(3+n)} \to -\frac{\cos(3+n)}{3+n}$$

Hence,

$$ \int \sin(3x) \cos(nx) dx = - \frac{ \cos(3+n)}{3+n}$$

Now this is wrong.. why?
Btw I am using result from here

答:
You have indeed

$$\sin 3x = \Im e^{i3x}$$ And therefore:

$$(\sin 3x) e^{inx} = (\Im e^{i3x}) e^{inx}$$ and
$$\Re[(\sin 3x) e^{inx}] = \Re[(\Im e^{i3x}) e^{inx}]$$
which is not equal to
$$\Re[\Im (e^{i(3+n)x})]$$

In general for two complex numbers

$$\Im(zz^\prime) \neq \Im(z) z^\prime$$

Example

$$\Im( i \cdot i) = 0 \neq  i = \Im(i) \cdot i$$


问:
On the sup-norm and the $L^\infty$-norm on the space of bounded sequences
Let $\mathbb{N}$ be the set of natural numbers. We denote by $\#$ the counting measure on $(\mathbb{N},2^{\mathbb{N}})$.

It is well known that $L^\infty(\mathbb{N},\#)$ (this space is often written as $l^\infty$) becomes a Banach space with the norm $\|u\|_\infty=\sup\{a \in \mathbb{R} \mid \#(\{n \in \mathbb{N} \mid |u_n|>a\})>0\}$, $u \in L^\infty(\mathbb{N},\#)$. It holds for any $u \in L^\infty(\mathbb{N},\#)$
\begin{align*}
(1)\quad \|u\|_\infty=\sup_{n \in \mathbb{N}}|u_n|.
\end{align*}
However the proof for (1) I know is roundabout. I use the relation
\begin{align*}
\|u\|_{\infty}=\lim_{p \to \infty}\|u\|_p=\sup_{n \in \mathbb{N}}|u_n|,\quad u \in L^\infty(\mathbb{N},\#) \cap L^q(\mathbb{N},\#).
\end{align*}
Here $q\ge 1$ and $\|u\|_p$ denotes the $L^p$-norm of $u$.

Do you know a simpler proof? If so, please let me know.

答:
Write
$$
\alpha := \sup_{n\in\mathbb{N}} |u_n|
$$
Let $\epsilon > 0$. By your definition of $\|u\|_\infty$ there is some $a\in\mathbb{R}$ such that
$$
\|u\|_\infty - \epsilon < a \leq \|u\|_\infty
$$
and
$$
\#(\{n\in \mathbb{N} \mid |u_n|>a\}) >0.
$$
This implies that for some $n\in\mathbb{N}$ you have that $|u_n|>a$ and thence $\alpha > a$. Thus
$$
\alpha > a > \|u\|_\infty - \epsilon
$$
and making $\epsilon \to 0$ we obtain that
\begin{equation}\tag{1}
\alpha \geq \|u\|_\infty.
\end{equation}

For the reverse inequality, take again $\epsilon > 0$, by the definition of $\alpha$ there is some $n$ such that
$$
\alpha - \epsilon < |u_n| \leq \alpha.
$$
This implies that
$$
\alpha - \epsilon \in \{a \in\mathbb{R} \mid \#(\{n\in\mathbb{N} \mid |u_n|>a\}) >0\}
$$
and as consequence, by definition of $\|u\|_\infty$, we have that
$$
\|u\|_\infty \geq \alpha - \epsilon.
$$
Making $\epsilon \to 0$ we have that
\begin{equation}\tag{2}
\|u\|_\infty \geq \alpha.
\end{equation}

Equations (1) and (2) prove the result.


问:
Understanding derivative as slope to the curve
I am a novice in calculus and have been learning differentiation lately. I read many books and saw you tube videos. There I noticed that to find the slope of a point in a curve,people take a point $P=(x+h,f(x+h))$ and determine the slope $\frac{f(x+h)-f(x)}{h}$. Now to find the slope more accurately, they take the point $P$ close to $(x,h)$ and say that it boils down to a tangent line. But the tangent by definition will intersect the curve at only one point,in that case shouldn't $h=0$? I know that gives indeterminate form and the people use limits by taking $h$ close to $0$. My question is how is that line a tangent? Even if it is close to $0$,it is still not a $0$,right? So that line still intersects. In that case isn't the derivative an approximation instead of an exact answer. I may be wrong,so please forgive me if I made any mistake.

答:
Relevant answer I wrote on PSE

The simple answer is that you are making the 'h' exactly to zero, you are going to down a chain of decreasing the size of the quantity till you can approximate the slope of tangent line.

Here is a simple exercise to understand this:

Give me any number, I will give you a number less than it. For example, you give me $.1$ then I will say $.01$, if you give me $.001$ then I can give you $.00001$ .. so it an endless limiting process. And we can keep having this exchange till we reach the $h$ value where the rate of change of the function with the input is more or less the slope of tangent line.


答: As @MarkViola noted, the problem is nothing to do with calculus, only with how you manipulate complex numbers. To wit, with $w:=\exp3ix,\,z:=\exp inx$ you seem to argue $\Im w\Re z=\Re\Im(wz)$ (which would be $\Im(wz)$, by the way). But$$w=a+ib,\,z=c+id,\,a,\,b,\,c,\,d\in\Bbb R\implies\Im w\Re z=bc,\,\Im(wz)=ad+bc.$$The original problem can be solved without complex numbers using$$2\sin3x\cos nx=\sin[(n+3)x]-\sin[(n-3)x].$$

Title: Interchanging integral with real and imaginary operators?

Body:
$$ \int \sin(3x) \cos(nx) dx \to \Re \int \sin(3x) e^{inx} dx$$

$$ \sin(3x) \to \Im e^{i3x}$$

Hence,

$$ \Re \left( \Im \int e^{i(3+n)x} dx \right) $$

Or,

$$ \Re \left(\Im \frac{e^{i(3+n)}}{i(3+n)} \right) \to \Re \left(\Im \frac{-ie^{i(3+n)}}{(3+n)} \right)$$

Considering,

$$ \Im \frac{-ie^{i(3+n)}}{(3+n)} \to -\frac{\cos(3+n)}{3+n}$$

Hence,

$$ \int \sin(3x) \cos(nx) dx = - \frac{ \cos(3+n)}{3+n}$$

Now this is wrong.. why?
Btw I am using result from here


问:
Mean value theorem for $\displaystyle f(x)=\frac{3-x^{2}}{x-2}$
Given the following function
$$f(x)=\frac{3-x^{2}}{x-2}$$
does the Mean value theorem applies to if in the interval $[1,3]$?

Since $f$ is not defined for $x=2$ then $f$ is continuous and differentiable in $[1,3]-\{2\}$. So, the theorem is valid. But I can't find a number $c\in(1,3)$ such that
$$f'(c)=-2$$

答: Mean Value Theorem doesn't apply. $f$ is not continuous on $[1,3]$ nor differentiable on $(1,3)$ as it is not defined at $2$!


问:
Show that $\sum\limits_{n=1}^{\infty}\frac{\left(1+\frac{1}{n}\right)^n}{n}(-1)^n$ converges
How would you show that  $\sum\limits_{n=1}^{\infty}\frac{\left(1+\frac{1}{n}\right)^n}{n}(-1)^n$ converges?

My approach

At first glance I would go for the alternate series criterion and show that $\frac{\left(1+\frac{1}{n}\right)^n}{n}$ is a monotone sequence with $\lim\limits_{n\to\infty}\frac{\left(1+\frac{1}{n}\right)^n}{n}=0$. So after a few algebraic manipulations I get:

$$
\frac{\frac{\left(1+\frac{1}{n}\right)^n}{n}}{\frac{\left(1+\frac{1}{n+1}\right)^{n+1}}{n+1}}=
\frac{\left(1+\frac{1}{n}\right)^n(n+1)}{\left(1+\frac{1}{n+1}\right)^{n}\left(1+\frac{1}{n+1}\right)n}=
\left(\frac{1+\frac{1}{n}}{1+\frac{1}{n+1}}\right)^{n}\cdot \frac{n+1}{\frac{n^2+2n}{n+1}}=
\cdots=\frac{n^2+2n+1}{n^2+2n}\left(\frac{n^2+2n+1}{n^2+2n}\right)^n\geq1.
$$
So the expression is monotonically decreasing. Further, we know that $\lim\limits_{n\to\infty}\left(1+\frac{1}{n}\right)^n=e^1$. Hence, $\left(1+\frac{1}{n}\right)^n$ is bounded, say $S\geq\Big| \left(1+\frac{1}{n}\right)^n\Big|$. This allows us to conclude:
$$
\Big| \frac{\left(1+\frac{1}{n}\right)^n}{n}-0\Big|\leq \frac{S}{n}<\epsilon,\text{ where n is sufficiently large enough}.
$$
Hence, $\lim\limits_{n\to\infty}\frac{\left(1+\frac{1}{n}\right)^n}{n}=0$ and by the alternate series crterion the series $\sum\limits_{n=1}^{\infty}\frac{\left(1+\frac{1}{n}\right)^n}{n}(-1)^n$ converges.

Is this correct? Is there a faster way?

答:
HINT

We have that

$$\left(1+\frac{1}{n}\right)^n= e^{n\log\left(1+\frac{1}{n}\right)}=e^{1-\frac1{2n}+O\left(\frac1{n^2}\right)}=e-\frac e{2n}+O\left(\frac1{n^2}\right)$$

therefore

$$\frac{\left(1+\frac{1}{n}\right)^n}{n}=\frac e n-\frac e{2n^2}+O\left(\frac1{n^3}\right)$$


问:
Stability of complex vector bundles
In the definition of the stability for a holomorphic vector bundle, where exactly do we use the fact that the bundle is holomorphic and not only complex? Why couldn't we define the slope of any complex vector bundle?

The only place I can see we actualy use holomorphicity is that we require that the slope should be smaller for any holomorphic subbundles. Is there anything that I am missing?

答:
I will try to give you an explanation only considering complex and holomorphic vector bundles over Riemann Surfaces. In general it is true the following:

Theorem

Let $X$ be a compact connected Riemann Surface of genus $g\ge 2$ and let $E,F$ be two smooth complex vector bundles over $X$. Denote with $r=$rk$(E)$, $ \ r'=$rk$(F)$, $ \ d=$deg$(E):=\int_Xc_1(E)$, $ \ d'=$deg$(F):=\int_Xc_1(F)$. Then $E$ and $F$ are isomorphic if and only if $r=r'$ and $d=d'$. Moreover for any pair $(r,d)\in\mathbb{Z}_+\times\mathbb{Z}$ there exist a smooth complex vector bundle over $X$ with rank $r$ and degree $d$

This classification can be found in:

[M. Thaddeus : An introduction to the topology of the moduli space
of stable bundles on a Riemann Surface, 1997]

In particular it says that from a smooth (topological) viewpoint, isomorphism classes of complex vector bundles over $X$ are parametrized by discret invariants. It must be noted that for this classification we didn't need to invoke the slope-stability.

Everything change if we now consider smooth complex vector bundle $E$ with an holomorphic structure, which can be given by an operator (this is not the usual definition but an equivalent one):

\begin{equation*}\bar{\partial}_E:\Omega^0(X;E)\longrightarrow\Omega^0(X;E\otimes (T^{0,1}X)^*)\end{equation*} such that:

$\bar{\partial}_E\circ\bar{\partial}_E=0$

$\bar{\partial}_E(f\cdot s)=\bar{\partial}f\cdot s + f\cdot\bar{\partial}_E(s), \qquad$ for $f\in C^\infty(X)$ and $s\in\Omega(X;E)$

If you try to classify all the possible holomorphic structure on $E$ up to isomorphism (two holomorphic structures are isomorphic if they are conjugated by an automorphism of $E$) then you can realize that strange things can happen. In fact there are examples in which the moduli space of holomorphic structures is not Hausdorff and hence it can not carry a manifold structure, which is exactly what one wants when defining a moduli space.

In order to solve this problem, the notion of slope-stability was introduced. In fact, thanks to the works of Mumford, Donaldson, Atiyah, Uhlenbeck and many other very good mathematicians fantastic results have been obtained regarding the moduli space of stable holomorphic bundles over projective manifolds.

Title: Stability of complex vector bundles


答:
The mean value theorem will apply individually in $[1,2),(2,3]$ but not over $[1,3]-\{2\}$ since it is not an interval.

But the conditions of the theorem are sufficient conditions, that is to say, the conclusion of the theorem may hold even for discontinuous functions/domains. In short, the conclusion may hold in cases where MVT does not apply. For example, it holds for $f(x)=1,x\in[-1,1]-\{0\}$ but MVT doesn't apply.

You can see that the MVT does not hold in our case either since for $x\in(1,3)-\{2\}$,$$f'(x)=\frac{(3-x)(x-1)}{(x-2)^2}>0$$so there is no $c\in(1,3)-\{2\}$ for which $f'(c)=-2=\frac{f(3)-f(1)}{3-1}.$

Title: Mean value theorem for $\displaystyle f(x)=\frac{3-x^{2}}{x-2}$

Body:
Given the following function
$$f(x)=\frac{3-x^{2}}{x-2}$$
does the Mean value theorem applies to if in the interval $[1,3]$?

Since $f$ is not defined for $x=2$ then $f$ is continuous and differentiable in $[1,3]-\{2\}$. So, the theorem is valid. But I can't find a number $c\in(1,3)$ such that
$$f'(c)=-2$$


问:
If an 8 digit number is randomly generated, what are the odds that all 8 digits are different/distinct?
If an 8 digit number is randomly generated, what are the odds that all 8 digits are different/distinct?

Assume that we can use the 10 numbers 0-9.
What I got myself is this:

(*)    $\frac{10!}{(10-8)!}$=$\frac{10!}{2!}$=1814400

The total number of 8 digit numbers is $10^8$=100,000,000.

So dividing those numbers gives:

$\frac{1814400}{100,000,000}$=0,018144

What I'm not sure of is whether all possibilities in the step with the (*) are distinct from each other, so there is no repetition of numbers within an 8-digit number. Could someone explain this to me whether it is or not and why? Then I know that my final answer is correct or incorrect.

Maybe good to know, I am just starting with probability theory...

Thanks in advance for your help!

答:
Think of your numerator as $10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3$; that is, you have $10$ choices for the first digit, but then only $9$ for the second, $8$ for the third, and so on down to $3$ for the eighth digit. The decreasing number of choices reflects the fact that you are not repeating digits that have already been used. Alternatively, $10!/2!$ means you permute the ten digits in any of $10!$ ways, drop the final two digits, and group them in pairs ($2!$) according to the two digits that get dropped. In either interpretation there is no repetition of digits.

In comments below the OP, it's pointed out that you are allowing your eight-digit numbers to begin with a $0$, which may or may not be what the problem poser intended. As it turns out, however, it doesn't matter for the final result. If starting $0$'s are allowed, the probability is

$$10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\over10\cdot10\cdot10\cdot10\cdot10\cdot10\cdot10\cdot10$$

whereas if starting $0$'s are not allowed, the probability is

$$9\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\over9\cdot10\cdot10\cdot10\cdot10\cdot10\cdot10\cdot10$$

In either case, cancellation leaves

$$9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\over10\cdot10\cdot10\cdot10\cdot10\cdot10\cdot10$$


答:
Since
$$
2\le\left(1+\frac1n\right)^n\le3
$$
and
$$
\sum_{n\ge1}\frac{(-1)^n}n=\log2
$$
Calling $S$ the desired limit (that is, your sum), you can immediately get that
$$
2\log 2\le S\le3\log 2\;.
$$

Title: Show that $\sum\limits_{n=1}^{\infty}\frac{\left(1+\frac{1}{n}\right)^n}{n}(-1)^n$ converges

Body:
How would you show that  $\sum\limits_{n=1}^{\infty}\frac{\left(1+\frac{1}{n}\right)^n}{n}(-1)^n$ converges?

My approach

At first glance I would go for the alternate series criterion and show that $\frac{\left(1+\frac{1}{n}\right)^n}{n}$ is a monotone sequence with $\lim\limits_{n\to\infty}\frac{\left(1+\frac{1}{n}\right)^n}{n}=0$. So after a few algebraic manipulations I get:

$$
\frac{\frac{\left(1+\frac{1}{n}\right)^n}{n}}{\frac{\left(1+\frac{1}{n+1}\right)^{n+1}}{n+1}}=
\frac{\left(1+\frac{1}{n}\right)^n(n+1)}{\left(1+\frac{1}{n+1}\right)^{n}\left(1+\frac{1}{n+1}\right)n}=
\left(\frac{1+\frac{1}{n}}{1+\frac{1}{n+1}}\right)^{n}\cdot \frac{n+1}{\frac{n^2+2n}{n+1}}=
\cdots=\frac{n^2+2n+1}{n^2+2n}\left(\frac{n^2+2n+1}{n^2+2n}\right)^n\geq1.
$$
So the expression is monotonically decreasing. Further, we know that $\lim\limits_{n\to\infty}\left(1+\frac{1}{n}\right)^n=e^1$. Hence, $\left(1+\frac{1}{n}\right)^n$ is bounded, say $S\geq\Big| \left(1+\frac{1}{n}\right)^n\Big|$. This allows us to conclude:
$$
\Big| \frac{\left(1+\frac{1}{n}\right)^n}{n}-0\Big|\leq \frac{S}{n}<\epsilon,\text{ where n is sufficiently large enough}.
$$
Hence, $\lim\limits_{n\to\infty}\frac{\left(1+\frac{1}{n}\right)^n}{n}=0$ and by the alternate series crterion the series $\sum\limits_{n=1}^{\infty}\frac{\left(1+\frac{1}{n}\right)^n}{n}(-1)^n$ converges.

Is this correct? Is there a faster way?


问:
Sum the series : $\frac{1}{9\sqrt11 + 11\sqrt9} +\frac{1}{11\sqrt13 + 13\sqrt11} +\ldots$

$$\frac{1}{9\sqrt11 + 11\sqrt9} + \frac{1}{11\sqrt13 + 13\sqrt11} + \frac{1}{13\sqrt15 + 15\sqrt13} + \ldots + \frac{1}{n\sqrt{n+2} + (n+2)\sqrt{n}} = \frac{1}{9}$$

Find the value of $n$.

I got the summation as $$\sum_{r=1}^n{\frac{1}{(2r+7)\sqrt{2r+9} + (2r+9)\sqrt{2r+7}}} = \sum_{r=1}^n{\frac{1}{\sqrt{2r+7}\sqrt{2r+9}(\sqrt{2r+7} + \sqrt{2r+9})}}$$ but I'm not able to simplify it further. I think I should've expressed it as difference of two summations but I'm not able to figure out which summations. Please help.

Body:
$$S_m=\sum_{r=1}^m{\frac{1}{(2r+7)\sqrt{2r+9} + (2r+9)\sqrt{2r+7}}} =\sum_{r=1}^m{\frac{(2r+7)\sqrt{2r+9} - (2r+9)\sqrt{2r+7}}{(2r+7)^2(2r+9)-(2r+9)^2(2r+7)}}$$
$$= \sum_{r=1}^m{\frac{(2r+7)\sqrt{2r+9} - (2r+9)\sqrt{2r+7}}{-2(2r+7)(2r+9)}} = -\frac{1}{2}\sum_{r=1}^m{\frac{\sqrt{2r+9}}{2r+9}} - {\frac{\sqrt{2r+7}}{2r+7}}$$ $$=\frac{1}{2}\sum_{r=1}^m{\frac{1}{\sqrt{2r+7}}} - {\frac{1}{\sqrt{2r+9}}} = \frac{1}{2}\left({\frac{1}{3}} - {\frac{1}{\sqrt{2m+9}}}\right)  $$

So $$S_m = \frac{1}{9} \Longleftrightarrow  {\frac{1}{\sqrt{2m+9}}} = \frac{1}{9} \Longleftrightarrow \sqrt{2m+9} = 9 \Longleftrightarrow m=36$$


问:
Collinearity of three points
In my book its given , three points A,B,C with position vectors a,b,c are collinear if and only if there exists scalars x,y,z not all zero simultaneously such that xa + yb + zc = 0, where x + y + z = 0.

Surprisingly, this is also the condition for coplanarity of three vectors. But All COPLANAR vectors are NOT collinear . So where is the the flaw in my argument ?

答:
As you write, vectors $a,b,c$ are coplanar iff there are scalars $x,y,z$, not all $0$, such that $xa+yb+zc=0$, i.e. iff they are linearly dependent.

The endpoints of $a,b,c$ are collinear iff $c-b$ is parallel to $c-a$, that is, $c-a=t(c-b)$ for some scalar $t$, assuming $b\ne c$.

But then $1a+(-t)b+(t-1)c=0$ and these coefficients sum up to $0$.

Conversely, if $xa+yb+zc=0$ with $x+y+z=0$, then either $x=0$ whence $y=-z$ and $b=c$, or we can divide by $x$ and set $t=-y/x$ to conclude $c-a=t(c-b)$.

Title: Collinearity of three points


问:
Can we find a lower bound for $P(V>1)$ when we know $EV=1$ and $P(V \le 1- \eta) \ge \delta? V \ge 0$ is non-constant.
Let $V \ge 0$ be a non-negative, non-constant, random variable and assume that we know $EV=1$ and $P(V \le 1- \eta) \ge \delta.$ I'm trying to see if I can find a lower bound for $P(V>1),$ in terms of $\eta, \delta?$

Intuitively, speaking, since $EV=1,$ it stands to reason that $P(V>1)$ and $P(V<1)$ both must be strictly positive as $V$ is non-constant. Now in this problem: we've a lower bound on $P(V \le 1- \eta),$ but does this guarantee a lower bound for the probability of the "other side", namely $P(V>1),$ which intuitively seems correct as this probability has to be big enough to balance the expectation of $V$ to be equal to $1.$ Intuitively, if this probability is too small, then $P(V>1)$ is close to $0,$ so $P(V<1)$ is close to $1,$ and as a result, it seems the $EV$ will be strictly less than $1.$

EDIT: Is it possible to have this kind of uniform lower bound if we instead work with a uniformly bounded or uniformly integrable family of random variables $\{V_{\epsilon}\}?$ If yes, what'd be the bound?

答: Unless $V$ is bounded, $P(V > 1)$ can still be arbitrarily small.  For fixed $0 < \eta, \delta < 1$ let $0 < \varepsilon < 1-\delta$.  Then define $V = \frac{1}{\varepsilon}$ with probability $\varepsilon$ and $V = 0$ with probability $1-\varepsilon$.  Then $\mathbb{E}[V] = 1$ and $P(V \le 1-\eta) = 1-\varepsilon > \delta$, but $P(V > 1) = \varepsilon$.


问:
number of solutions to an equation?
Given $x$ and $y$ are multiples of $2$ satisfying
$$x^2 - y^2 = 27234702932$$
Find the number of solutions to $x$ and $y$.</p>

答:
As $27234702932=2^2\cdot181\cdot37616993$ where the last two factors are primes

If $x=2X,y=2Y$

$$X^2-Y^2=181\cdot37616993$$

If $X^2-Y^2=p\cdot q$ where $p,q$ are primes,

the possible cases for $X+Y,X-Y$ are $\pm pq, \pm 1$  and $\pm p,\pm q $

For example,  if $X+Y=1,X-Y=pq$

The number of positive factors of $p\cdot q$ is $(1+1)\cdot(1+1)=4$

So, there should be $4\cdot2 =8$ solutions in integers.

Title: number of solutions to an equation?


问:
Find expected number of draws from first principles
This is a variation of a common probability problem: Given a bag of N blue marbles and M red marbles what's the number of expected draws until you get a blue marble?

With $p_R = \frac{M}{M+N}$ and $p_B = \frac{N}{M+N}$, the expected number of trials before the first success is a discrete RV distributed according to a Geometric distribution. This distribution's expected value is known and in this case is $\mathbb{E}[X]= \frac{1}{p_{B}}=\frac{M+N}{N}$.

Given that this is correct (which I think it is), is there a way to derive this result from first principles, e.g. if we didn't know about the Geometric distribution?

答:
Let $X$ be the number of trials until the first success, with each trial having success probability $p$. Then
$$E(X)=\sum_{r=1}^{\infty}P(X\geq r)=\sum_{r=1}^{\infty}(1-p)^{r-1}=\frac{1}{1-(1-p)}.$$


问:
Simple question on differentiation of cdf
Suppose we have $n$ $i.i.d$ random variables $X_{1},\ldots,X_{n}$  all distributed uniformly, $X_{i} \sim \mathrm{Uniform}\left(0,1\right)$ .

We want to find the expected value of $\mathbb{E}[Y_{n}]$  where $Y_{n} =
\max\left\{X_{1},\ldots,X_{n}\right\}$.

Here I got $F_Y\left(y\right) =
\int\mathrm{d}x_{1}\ldots\int\mathrm{d}x_{n} = \left[F_{X}\left(y\right)\right]^n $ but to get $f_{Y}\left(y\right)$ i need to differentiate $F_{Y}\left(y\right)$ with respect to $y$ but how do i differentiate $\left[F_{X}\left(y\right)\right]^{n}$ with respect to $y\,?$.

答:
Define $Y = \max_{1 \leq i \leq n} X_i$ so that for any $0 \leq y \leq 1$, we have

$$F_{Y}(y) = P(Y \leq y ) = \prod_{i = 1}^{n} P(X_i \leq y) = [P(X_1 \leq y)]^{n} = y^{n}.$$

Differentiating, one can find

$$f_{Y}(y) = \frac{d}{dy} F_{Y}(y) = ny^{n - 1},$$

which yields

$$f_{Y}(y) = \begin{cases}
ny^{n - 1} & \text{ if } 0 \leq y < 1 \\
0 & \text{ otherwise.}
\end{cases} $$

The expected value is computed as follows:

$$\mathbb{E}[Y] = \int_{0}^{1} y \cdot f_{Y}(y) \mathop{dy} =  \int_{0}^{1} ny^{n} \mathop{dy} = \boxed{\frac{n}{ n + 1}}$$


问:
{$ \rightarrow, \wedge $} and Quantifiers

Implication goes with $ \forall $

Conjunction goes with $ \exists $

Does the first statement mean that a conjunction should not appear in the scope of a $ \forall $, or does it mean that a conjunction can appear as long as it is not the main connective?

e.g. $ \forall x (P(x) \wedge Q(x) \rightarrow R(x))$

Same question for the second statement

答:
No, there are no rules that forbide the use of $\forall$ with conjunction.

$\forall x (A(x) \land B(x))$ is a correct formula.

Maybe you are alluding to the so-called restricted (or: bounded) quantifiers:

$\forall x \ \in A : P(x)$ and $\exists x \ \in A :P(x)$

that are abbreviations respectively for the two formulae :

$\forall x (x \in A \to P(x))$ and $\exists x (x \in A \land P(x))$.


答:
First factor $27234702932=2^2\times181\times37616993$. Since both are even, so is $x+y$ and $x-y$. Each solution is uniquely determined by $x+y$ and $x-y$.<br>
Hence the number of solutions is equal to the number of solutions to $ab=27234702932$ where both $a,b$ are even. </p>

Title: number of solutions to an equation?

Body:
Given $x$ and $y$ are multiples of $2$ satisfying
$$x^2 - y^2 = 27234702932$$
Find the number of solutions to $x$ and $y$.</p>

Body:
First factor $27234702932=2^2\times181\times37616993$. Since both are even, so is $x+y$ and $x-y$. Each solution is uniquely determined by $x+y$ and $x-y$.

Hence the number of solutions is equal to the number of solutions to $ab=27234702932$ where both $a,b$ are even.


答:
Differentiation of $F(y)$ is not necessary.

In fact,

$$F_Y(y)=y^n$$

But as per the fact that Y is non negative, its expectation can be derived in the following way

$$\mathbb{E}[Y]=\int_0^1 [1-y^n]dy=\frac{n}{n+1}$$

Title: Simple question on differentiation of cdf

Body:
Suppose we have $n$ $i.i.d$ random variables $X_{1},\ldots,X_{n}$  all distributed uniformly, $X_{i} \sim \mathrm{Uniform}\left(0,1\right)$ .

We want to find the expected value of $\mathbb{E}[Y_{n}]$  where $Y_{n} =
\max\left\{X_{1},\ldots,X_{n}\right\}$.

Here I got $F_Y\left(y\right) =
\int\mathrm{d}x_{1}\ldots\int\mathrm{d}x_{n} = \left[F_{X}\left(y\right)\right]^n $ but to get $f_{Y}\left(y\right)$ i need to differentiate $F_{Y}\left(y\right)$ with respect to $y$ but how do i differentiate $\left[F_{X}\left(y\right)\right]^{n}$ with respect to $y\,?$.


问:
Trouble evaluating a limit
I have this following limit:
$$
\operatorname{f}\left(x\right) =
\lim_{x \to 2k}\,\,
\frac{x\left(x - 2k\right)}
{\cos\left(\pi x\right) - 1},\quad
k \in \mathbb{Z}
$$
The obvious solution is to apply L'Hospital's rule twice, which seems to result in a constant, and hence the limit can be evaluated. However, I am told that this limit is indeterminate, and I am unsure as to how this may be possible.

答:
Applying L'Hospital once, we get

$$\lim_{x\to 2k}\frac{2x-2k}{-\pi\sin(\pi x)}$$

which is not an $\dfrac00$ indeterminate form and you can't apply L'Hospital twice.

Title: Trouble evaluating a limit


答:
Observe $$\dfrac{1}{9\sqrt{11} + 11\sqrt9}$$

$$ =\dfrac{1}{\sqrt 9 \sqrt {11} }\cdot\dfrac{1}{\sqrt9 + \sqrt{11}}$$

$$ =\dfrac{1}{\sqrt 9 \sqrt {11} }\cdot \dfrac{\sqrt{11} - \sqrt{9}}{2}$$

$$ =\dfrac{1}{2}\cdot \dfrac{\sqrt{11} - \sqrt{9}}{\sqrt 9 \sqrt {11} }$$

$$ =\dfrac{1}{2}\cdot \Big(\dfrac{1}{\sqrt{9}} - \dfrac{1}{\sqrt{11}}\Big)$$

Thus LHS

$$ \dfrac{1}{2}\cdot \Big(\dfrac{1}{\sqrt{9}} - \dfrac{1}{\sqrt{11}}\Big) + \dfrac{1}{2}\cdot \Big(\dfrac{1}{\sqrt{13}} - \dfrac{1}{\sqrt{11}}\Big) + \ldots + \dfrac{1}{2}\cdot \Big(\dfrac{1}{\sqrt{n}} - \dfrac{1}{\sqrt{n+2}}\Big)$$

telescopes to

$$ \dfrac{1}{2}\cdot \Big(\dfrac{1}{\sqrt{9}}  - \dfrac{1}{\sqrt{n+2}}\Big) = \dfrac{1}{9}$$

$$ \Rightarrow \dfrac{1}{3}  - \dfrac{1}{\sqrt{n+2}} = \dfrac{2}{9}$$

$$ \Rightarrow n+2 = 81 $$

$$ \Rightarrow \boxed{n=79} $$

Note  The $n$ in your question and the $n$ in your attempt are different. I have found $n$ in your question. Indeed final term is $\frac{1}{\sqrt{81}}$ as in the other answer. The $n$ in your attempt is $36$. Indeed $2\cdot36+9=81$.

Title: Sum the series : $\frac{1}{9\sqrt11 + 11\sqrt9} +\frac{1}{11\sqrt13 + 13\sqrt11} +\ldots$

Body:
$$\frac{1}{9\sqrt11 + 11\sqrt9} + \frac{1}{11\sqrt13 + 13\sqrt11} + \frac{1}{13\sqrt15 + 15\sqrt13} + \ldots + \frac{1}{n\sqrt{n+2} + (n+2)\sqrt{n}} = \frac{1}{9}$$

Find the value of $n$.

I got the summation as $$\sum_{r=1}^n{\frac{1}{(2r+7)\sqrt{2r+9} + (2r+9)\sqrt{2r+7}}} = \sum_{r=1}^n{\frac{1}{\sqrt{2r+7}\sqrt{2r+9}(\sqrt{2r+7} + \sqrt{2r+9})}}$$ but I'm not able to simplify it further. I think I should've expressed it as difference of two summations but I'm not able to figure out which summations. Please help.


问:
Colouring $2n$ Points in $\mathbb{R}^2 $
Definition: A strip is the area in the plane enclosed between some two parallel lines. An axis-
parallel strip is a strip whose bounding lines are parallel to one of the axes.

Let $P$
be a set of $2n$ points in the plane (for some integer $n$). Prove that the points of $P$
can be colored with 'red' or `blue' such that in any axis-parallel strip the difference
between the number of red points and the number of blue points is at most 2.

The direction I was thinking about is definig monotone non-decreasing seq of the points, relatively to the $x$-axis, and another non-decreasing monotone sequence relatively to y-axis, from the given points. Then I would like taking a graph, such that the vertices are the points and the edges connect consecutive terms of both sequences. Im pretty stuck right now.

答:
The next answer is not mine' but of my proffesor.

As been offered in the answer above me , we counstruct a graph which it's vertices are our points, and the edges constructed by pairs of points- by a non dec seq in x- axis and y-axis.

Now, since our edges are realy a $ matching$ , every cycle of it has even length.

Our graph is , by the bipartite graph characterization, bipartite, i.e $V= X \cup Y $ where X and Y are disjoint sets. Color X with blue and Y with red.

For given parellel strip, we can see that between every pair of one color points, there is a point colored by the opposite color. Hence, exept the "end pints" of the strip, we have split our strip into pairs of different colors. There are at most 2 end pints to a given strip- so we are done :)

My proffesor mentioned that it's just an example to much deeper idea in math, called Discrepancy.


问:
What means this mathematical notation?
Which members of the set this set contains?

I cannot understand these symbols like ≡ and mod.

答:
$B$ is the set of integers $x$ satisfying $x\equiv0\bmod4$, which means that $4|x$.

$B$ is thus the set of integers divisible by $4$ ($\{0,\pm4,\pm8,\pm12,...\}$).

Title: What means this mathematical notation?


答: Factorize the number and use $x^2-y^2=(x-y)(x+y)$? </p>

Title: number of solutions to an equation?


答:
As noticed we can apply l'Hospital only for one step to obtain

$$\lim_{x\to 2k}\frac{2x-2k}{-\pi\sin(\pi x)}$$

which doesn't exist since RH and LH limits are different.

Note also that for this kind of simple limits, it is preferable to proceed without l'Hospital's rule, notably by $x-2k=y \to 0$ we have that

$$\frac{x(x-2k)}{\cos(\pi x) -1}=\frac{(y+2k)y}{\cos(2\pi k+\pi y) -1}= (y+2k) \frac{y}{\cos(\pi y) -1}$$

with $y+2k \to 2k$ and since

$$\frac{y}{\cos(\pi y) -1}=\frac1{\pi^2}\frac{(\pi y)^2}{\cos(\pi y) -1}\frac1y$$

with $\frac{(\pi y)^2}{\cos(\pi y) -1} \to -2$ by standard limits, we conclude that the given limit doesn't exist since

$$\lim_{x\to 2k^+}\frac{x(x-2k)}{\cos(\pi x) -1} =-\infty \neq \lim_{x\to 2k^+}\frac{x(x-2k)}{\cos(\pi x) -1} =\infty$$

Title: Trouble evaluating a limit

Body:
I have this following limit:
$$
\operatorname{f}\left(x\right) =
\lim_{x \to 2k}\,\,
\frac{x\left(x - 2k\right)}
{\cos\left(\pi x\right) - 1},\quad
k \in \mathbb{Z}
$$
The obvious solution is to apply L'Hospital's rule twice, which seems to result in a constant, and hence the limit can be evaluated. However, I am told that this limit is indeterminate, and I am unsure as to how this may be possible.


问:
To find kernel of given transformation
Question says "Let $T:R^3\to R^3$ be a projection onto P, where P is a plane through origin with normal vector $\vec{n}\neq \vec{0}$. Let $S:R^3 \to R^3$ be a reflection across L, where L is the line given by span$(\vec{v})$ with $\vec{v} \neq \vec{0}$, a vector in P." I needed to find kernel of $T\circ S$ and $S\circ T$.

My thought for $T\circ S$ is that $T$ will send the vectors along $\vec{n}$ to origin. But I am not sure how to find those vectors which will get mapped to scalar multiple of $\vec{n}$ by $S$. How to find them?

For $S\circ T$, I think kernel is precisely $\alpha \vec{n}$. Am I correct?

答:
Kernel of $T \circ S$

$x \in \ker T \circ S$ means $T(S(x))=0$, i.e. $S(x)$ collinear with $n$. It exists $\lambda \in \mathbb R$ such that $S(x) = \lambda n$. Now remember that a reflexion is an involution. So

$$x=(S \circ S)(x) = \lambda S(n)$$

Conclusion: $\ker T \circ S$ is the image under $S$ of the line orthogonal to $P$ passing through the origin.

Kernel of $S \circ T$

$S(T(x)) = 0$ implies that $T(x) = 0$ as $S$ is bijective. Hence the kernel is equal to the line normal to $P$ passing through the origin.

Title: To find kernel of given transformation


问:
how to prove a matrix is symmetric
Suppose square matrix B has nxn dimensions, has $Null(B) = Col(B)^\perp$ and rank(B)=1, how can we prove the matrix B is equal to its transpose?

I'm not sure how to get started I think I have to start with the eigen vectors and then somehow prove its symmetry but I'm not sure

答:
Hint: If the rank of $B$ is $1$, then there are non-zero column-vectors $u,v$ for which
$$
B = uv^T.
$$
What is the relationship between the vectors $u,v$ and the column/null spaces of $B$? Note that $B$ will be a symmetric matrix and only if $u$ is a multiple of $v$.

Title: how to prove a matrix is symmetric


问:
$\mathbf{u}$ and $\mathbf{v}$ are unit vectors with angle $\theta$ between them. Show that $\sin(\theta /2 ) = \| \mathbf{u} + \mathbf{v} \| /2 $
I was trying this problem but I can't solve with the $\sin(\theta /2)$.

I try this:

$$<\mathbf{u}+\mathbf{v},\mathbf{u}>\ = \|\mathbf{u}+\mathbf{v} \| \|\mathbf{u} \| \cos\alpha$$
where $\alpha$ is the angle between $\mathbf{u}+\mathbf{v}$ and $\mathbf{u}$.

$$ <\mathbf{u}+\mathbf{v},\mathbf{v}>\ = \| \mathbf{u}+\mathbf{v} \| \|\mathbf{v}\| \cos\alpha$$

Then I add the equations:

$$<\mathbf{u}+\mathbf{v},\mathbf{u}+\mathbf{v}>\ =  2 \| \mathbf{u}+\mathbf{v} \| \cos\frac{\theta}{2}$$

Because $\| \mathbf{u}\| = 1$ and $\| \mathbf{v}\| = 1$. So,

$$\frac{\| \mathbf{u}+\mathbf{v} \|}{2} = \cos \frac{\theta}{2}$$
Maybe I forgot some explanations, but that is my idea however, I can't see how to go from $\cos$ to $\sin$.

Maybe there is another form; thanks for your help in advance!

答:
There might be an error in the problem. Consider this simple counter example:

Let $\mathbf{u}=\mathbf{v}$, be arbitrary unitary vectors. Then, $\sin\theta=\sin 0=0\Rightarrow \sin(\theta/2)=0$.

However,
$$\frac{||2\mathbf{u}||}{2}=\frac{2||\mathbf{u}||}{2}= 1$$

For the cosine, it works. As far as I can tell, your answer looks correct.


答:
Without L'Hospital:

$$\lim\limits_{x\to 2k}\frac{x(x-2k)}{\cos(\pi x) -1} = \lim\limits_{x\to 2k}\frac{x(x-2k)}{\cos(\pi x-2 \pi k) -1}=\\
=-\lim\limits_{x\to 2k}\frac{x(x-2k)}{2\sin^2(\frac{\pi}{2}(x-2k))}$$
last one have $\sin$ in second power in denominator.

Title: Trouble evaluating a limit


问:
Find the required gcd

Question: Let $\gcd(a,n)=d_1\gt 1$ and $\gcd(b,n)=d_2\gt 1$, given that $\gcd(d_1,d_2)=1$, find $\gcd(a+b,n)$.

My try
Since $\gcd(a,n)=d_1$ so $a=pd_1$ and since $\gcd(b,n)=d_2$ so $b=qd_2$.

My guess is since the greatest common divisors of $a,b$ have gcd =1, so $\gcd(a+b,n)=1$.
But I cant prove it in a compact and well defined fashion

Can someone please help me out?

The question boils down to this:

How is $\gcd(a+b,n)$ related to $\gcd(a,n)$ and $\gcd(b,n)$?

答:
I don't think this is answerable.

Deep breath. Let's break everything down to "components".

$\gcd(n,a) = d_1$ so let $\alpha = \frac a{d_1}$ so $a = d_1\alpha$ and let $\nu = \frac n{d_1}$ so $n = d_1\nu$.  It's easy to verify that $\alpha$ and $\nu$ are relatively prime.  But $\alpha$ and $d_1$ and $\nu$ and $d_1$ may have  factors in common, but the factors that $d_1$ his in common with $\alpha$ must be relatively prime to the factors in common with $\nu$.

Let $k_1 =\gcd(d_1, \alpha)$ and $m_1\gcd(d_1, \nu)$ so that we may express $a = d_1k_1a'$ and $n=d_1m_1\overline n$ where: $d_1,d_2$ are relatively prime;  $j_1,m_1$ are relatively prime but $k_1,m_1|d_1$; and $a',\overline n$ are relatively prime.

We can do the same for $b$ so that $b=d_2k_2b'$ and $n = d_2m_2\overline{\underline n}$.

Now $d_1$ is relatively prime to $d_2$ and $m_1|d_1$ and $m_2|m_2$ so $d_1m_1$ is relatively prime to $d_2m_2$ so $d_1m_1|\overline{\underline n}$.

And so we can write $n = d_1m_1d_2m_2n'$.

So $\gcd(a+b,n)=$

$\gcd(d_1k_1a' + d_2k_2b', d_1m_1d_2m_2n')=$.

$\gcd(d_1k_1a' + d_2k_2b', n')$

(The last step is acceptable as $d_1,m_1,d_2,m_2$ each divide one of the summands but is relatively prime to the other summand, therefore $d_1m_1d_2m_2$ is relatively prime to $d_1k_1a' + d_2k_2b'$)

And this can be pretty much any thing we want provided we chose the terms so that they are mutually relatively prime.

Suppose, for example, I want $\gcd(a+b, n) = 39$.

I can let $n'=39$ and $a+b=39$ and $a=d_1k_1a'= 20$ with $d_1=4$ and $k_1=1;a'=5$ and $b=d_2k_2b'=19$ with $d_1=19;k_1=b'=1$.

Then $n=4\cdot 19\cdot 39 = 2964$

And thus we have $\gcd(20,2964)= 4$ and $\gcd(19,2964)=19$. And $\gcd(20,19)=1$.  And $\gcd(20+19,2964)=\gcd(39,2964)= 39$.

Title: Find the required gcd


问:
Isomorphism of directed graphs
Consider the following directed graphs:

One is obtained from the other by reversing the direction of all edges.

Are they isomorphic as directed graphs ?

On the one hand, I would answer: no because there is no pair of maps between the vertices and the edges respectively that preserves the adjacency relation.

On the other hand, if one forgets about the ``labels'' on say, the edges, then the graphs are the same (just exchange the label 'e' with label 'f').

答:
Your comment helps clarify the source of your confusion:

A graph morphism is a pair of maps between the respective set of vertices $p:V \to V$ and and between the respective set of edges $q:E \to E$. If I set $q(e) = f$, $q(f) = e$ and $q(l) = l$ then because of the adjacency relation, I have to set: $w = \text{initial vertex of } f = \text{initial vertex of } q(e) = p(\text{initial vertex of e})= p(v)$. So $p$ has to exchange $v$ and $w$. Now $q(l) = l$ so again by the adjacency relation one must have: $v = \text{initial of vertex of } l = \text{initial vertex of } q(l) = p(\text{initial vertex of } l) = p(v)$. Hence $p$ has to fix the vertex $v$. This is a contradiction.

You implicitly (and wrongly) assume that edge $f$ in the left graph is the same as edge $f$ in the right graph. The confusion is only due to name clash. Renaming should save you from this mistake.

If we named edges in the right graph $e'$ and $f'$, then instead of writing this statement (original):

$q(e) = f$, ... $w$ = initial vertex of $f$

we would have written this statement:

$q(e) = f'$, ... $w$ = initial vertex of $f$

Title: Isomorphism of directed graphs


问:
Uniqueness continuity equation optimal transport (using transport equation)
In the following
$\mu: [0,1] \to P_2(\mathbb{R}^d)$ is a curve in space of probability measures having finite second moment in $\mathbb{R}^d$ and $v:[0,1] \times \mathbb{R}^d \to \mathbb{R}^d$ is a vector field.
The continuity equation is
$$
\partial_t \mu_t + \nabla\cdot (\mu_t v_t) =0 \\
\mu(0)=\mu_0
$$
in the sense of distribution.

Suppose that the for every $T>0$ and $\psi: \mathbb{R}^+ \times\mathbb{R}^d \to \mathbb{R}$ the transport equation TE admits a solution $f$, where TE is
$$
\frac{d}{dt} f+ v\cdot \nabla f = \psi \\
f(T,\cdot)=0
$$

Then prove that the continuity equation admits at most one absolutely continuous solution in the sense of distribution.

I do not know where to start, I wrote the continuity equation as the condition $ \forall \phi \in C_c^\infty (\mathbb{R}^d) $
$$
\frac{d}{dt} \int \phi \, d\mu_t = \int \nabla \phi\, \cdot v_t d\mu_t
$$
and try to think of a smart choice of $\psi$ to get something interesting, but had no ideas. Any help?

答:
I'm not really a PDE guy, so I'm deliberately neglecting regularity issues. Yet, I guess a rigorous proof will rely on an argument as follows:

Assume there is another solution to the CE, say $\mu'$. Then by linearity the difference $\nu=\mu-\mu'$ also satisfies CE, but with $\nu(0)=0$. Now let $f$ be a solution to TE. Then
\begin{equation}
\frac{d}{dt}\int f_{t}d\nu_{t}=\int(\dot{f_{t}}+\nabla f_{t}\cdot v_{t})d\nu_{t}=\int\psi d\nu_{t}.
\end{equation}
Integrate from $0$ to $T$ you observe that
\begin{equation}
0=\int_{0}^{T}\int\psi d\nu_{t}.
\end{equation}
This being true for all $T>0$ and $\psi$ should imply $\nu\equiv 0$ which yields the claim.


答: The $\forall$ usually goes with $\to$, and the $\exists$ usually goes with $\land$, but this is not a hard rule: it's perfectly ok to have any kind of operator inside any quantifier, even as the main operator ... it's just that many sentences we use in real life, when translated to logic, end up with the pairings you indicated.

Title: {$ \rightarrow, \wedge $} and Quantifiers

Body:
Implication goes with $ \forall $

Conjunction goes with $ \exists $

Does the first statement mean that a conjunction should not appear in the scope of a $ \forall $, or does it mean that a conjunction can appear as long as it is not the main connective?

e.g. $ \forall x (P(x) \wedge Q(x) \rightarrow R(x))$

Same question for the second statement


答:
With these questions I feel it is best to avoid L'Hospital's rule.  I often set exam questions that are impractical to solve  that way, but relatively trivial if you use common sense.  I know other lecturers do the same - they want students to understand what is going on, rather than using a black box.

Anyway:
\begin{align*}
 \lim_{x\to 2k}\frac{x(x-2k)}{\cos(\pi x) -1}=\lim_{x\to 2k}\frac{x(x-2k)}{\cos(\pi (x-2k)) -1}=\lim_{y\to 0}\frac{(y+2k)y}{\cos(\pi y) -1}
\end{align*}

setting $y=x-2k$.  However $\cos(\pi y)=1-\frac {\pi^2y^2}2+o(y^3)$

So our limit is: \begin{align*}
 \lim_{y\to 0}\frac{y+2k}{\frac {-\pi^2y}2+o(y^2)}=\lim_{y\to 0}\frac1y \frac{y+2k}{\frac {-\pi^2}2+o(y)}=\frac{-4k}{{\pi^2}}\lim_{y\to 0}\frac1y
\end{align*}

which diverges.

Title: Trouble evaluating a limit


问:
Proof by Induction that the n-th triangular number is $\frac{(n(n+1))}{2}$
Prove by induction that the $n$-th triangular number is $\dfrac{(n(n+1))}{2}$.
Triangular numbers are numbers that make a triangular dot pattern.

答:
Let $P(n)$ denote the following statement.

The $n^{th}$ triangular number is $\frac{n(n+1)}{2}$.

Now the proof for $P(1)$:

$1^{st}$ trianular number is $\frac{1(1+1)}{2} =1$. So, $P(1)$ is true.

Let $P(n)$ be true. We'll prove $P(n+1)$ is true.

Firstly, note that $(n + 1)^{th}$ triangular number is $(n + 1) + n^{th}$ triangular number.

By this fact, and because $n^{th}$ triangular number is $\frac{n(n+1)}{2}$ (assumed), $(n + 1)^{th}$ triangular number is $n + 1 + \frac{n(n + 1)}{2} = \frac{(n + 1)(n + 2)}{2}$.

Hence, $P(n + 1)$ is true. And therefore the statement is true for all natural numbers.


问:
A binomial identity from Mathematical Reflections
Here is the problem:

Let $m,n$ be positive integers with $n>m$. Prove that

$\displaystyle\sum_{k=0}^{\lfloor\frac{n+m}2\rfloor} (-1)^{k}\binom{n}{k}\binom{m+n-2k}{n-1}=\binom{n}{m+1}$

This problem is O243 of Mathematical Reflections. A solution had been published using complex integration (https://www.awesomemath.org/assets/PDFs/MR5sol(1).pdf). However, I would like to see a solution using the difference operator, if any, since the form of the summand brings this to mind.

答:
Sledgehammer is Gosper summation... see Petkovsek, Wilf, Zeilberger "A = B". That works, guaranteed. Most computer algebra packages implement this.

Otherwise I'd grab Graham, Knuth, Patashnik "Concrete mathematics", techniques to handle such sums by clever index manipulation are clearly explained there.

Another venue would be Wilf's "snake oil method" (see his "generatingfunctionology").


答:
Neither. What is meant is that the "all" sentences you typically want to express translate as $\to$, and typical "there is" sentences translate as $\land$.

It is perfectly legal to write

$$\forall x (Swan(x) \land White(x))$$

-- if what you want to say is that all that your model contains is white swans.

Likewise,

$$\exists x(Swan(x) \to White(x))$$

is a perfectly grammatical sentence -- which becomes true as soon as there is anything which is not a swan, or anything which is white. It's just that this is not something you will very often find yourself in the need to say.

What is meant by these rules of thumbs that when expressing an "all" statement, what you mean is usually $\forall x (\phi(x) \to \psi(x))$, and sentences with "there is" usually mean what is formalized as $\exists x (\phi(x) \land \psi(x))$. If you are sure it's something else you want to express, then as far as the language of predicate logic is concerned, you are allowed to do so: $\forall$ and $\exists$ can range over arbitrary formulas with arbitrary connectives.

Also note that the two statements left and right of $\to$ resp. $\land$ can be "compound" expressions:

For instance, adjective + noun or noun + relative clause translate as $\land$, and predications with "or" may also occur, so e.g.

$$\text{Every old swan is white}\\
= \forall x ((Swan(x) \land Old(x)) \to White(x))$$

$$\text{Every swan is an animal which is white}\\
= \forall x (Swan(x) \to ((Animal(x) \land White(x)))$$

$$\text{Every swan is white or gray}\\
= \forall x (Swan(x) \to ((White(x) \lor Gray(x)))$$

and the same for existential expressions.