问:
Can there be different particular solutions for a differential equation since the constant MAY change?
Say, we had the following differential equation:

$dy/dx = 2(y^2)$

Now, when we solve it, we may get one of the following equations, depending on IF we decide to multiply something with the constant OR leave it as it is:

$y = 1/-2x-2c$ (if we decide to multiply -2 with the constant)

$y = 1/-2x+c$ (if we decide to leave the constant alone)

If we're asked to find the particular solution for when x=2 and y=1, obviously, the value of the constant and the particular solution in each case will be different. Now, I'm a bit confused here. Some of the teachers like to leave the constant alone, explaining that it has a particular value which can not change and hence, we leave it as it is. Other teachers like to multiply stuff with the constant as well.

What's the right way to approach this, and why?

EDIT: The value of C maybe different, but the constant we get at the end, after solving for the particular solution and plugging the value of the constant into the general solution, will exactly be the same! Thanks, Joe!

答:
\begin{align}
\frac{dy}{dx} &= 2y^2 \\
\int 1 \, dx &= \int \frac{1}{2}y^{-2} \, dy \\
x + C &= -\frac{1}{2}y^{-1} \\
y(x+C) &= -\frac{1}{2} \\
y &= -\frac{1}{2(x+C)}
\end{align}
What the solution to this differential equation shows is that any function $y$ of the form
$$
y(x)=-\frac{1}{2(x+\text{some arbitrary constant})}
$$
has the derivative $2y^2$. It doesn't matter how you label that constant. For example, you could use $C$, you could use $K$, you could even use $17C-398$; it's all the same. It is very common to not even bother multiplying the constant as you are solving the equation, because we just need some letter that represents a member of the real numbers. For example, you might see
\begin{align}
2y &= 5x + C \\
y &= 5x + C
\end{align}
More formally, this could be justified in the following way:
\begin{align}
2y &= 5x + C \\
y &= \frac{5}{2}x + C/2
\end{align}
Let $K=C/2$. Then we have $y=\frac{5}{2}x+K$. However, since the value of $K$ is arbitrary, we may relabel it as $C$. (This is despite the fact that we are using the letter $C$ to represent something that earlier we would denote $C/2$.) Again, we just need to use a letter to emphasise that it is a family of solutions to the differential equation that we have found. The constant $C$ is just an arbitrary member of the real numbers.


问:
Definition of a subcomplex of a $\Delta$-complex
I am taking the following as the definition of a $\Delta$-complex.

(i) one starts with an indexing set $I_n$ for each $n \in \mathbb{Z}_{\ge 0}$.

(ii) for each $\alpha \in I_n$, one takes a copy $\sigma_\alpha^n$ of the standard $n$-simplex.

(iii) one forms the disjoint union of all of these simplices, for all $n\geq 0$.

(iv) now require that for each $(n-1)$-dimensional face of $\sigma_\alpha^n$, there is an associated $\sigma_\beta^{n-1}$ for some $\beta \in I_{n-1}$.

(v) now form the quotient space by identifying each $(n-1)$-dimensional face of each $\sigma_\alpha^n$ with $\sigma_\beta^{n-1}$ using the canonical homeomorphism. In particular, these homeomorphisms preserve the ordering of vertices.

The sources I've consulted don't make it clear what a subcomplex is. Can someone give me a rigorous definition?

答:
With that (slightly strange way of phrasing the) definition, a subcomplex should be a sequence $(J_n)_{n\geq0}$ such that

for each $n\geq0$ we have $J_n\subseteq I_n$, and

if $\alpha\in J_n$ and $\beta\in I_{n-1}$ are such that $\sigma_b^{n-1}$ is associated to one of the faces of $\sigma_\alpha^n$, then $\beta\in J_{n-1}$.

N.B.: you did not tell us where you got that definition from, nor what other sources you consulted. There are lots of nice expositions of this subject... With time, I've learned to appreciate C. R. F. Maunder's Algebraic Topology, for example.

ContentLicense: CC BY-SA 3.0


问:
How to solve coupled differential equations with additional constant vectors?
I know how to solve coupled differential equations in the form:
$$\frac{d\vec{x}}{dt}=A\vec{x}$$ Where $A$ is a constant matrix. But how would I go about solving a differntial equation like:
$$\frac{d\vec{x}}{dt}=A\vec{x}+\vec{a}$$ Where $\vec{a}$ is a constant vector.

答:
Another trick is to reformulate your problem from the affine form into a linear form.
You can do that by expanding your linear system of n dimensions by one dimension.

$$ A' =\left[
\begin{array}{c|c}
A & \vec{a}\\
\hline
  \vec{0}^\intercal & 0
\end{array}
\right]$$

The new initial value is set $x_{n+1}(t_0) = 1$. By doing that we introduce a constant 1 (because $\frac{dx_{n+1}}{dt} = 0$) which gets multiplied by the vector.

The resulting system $\frac{d\vec{x}}{dt} = A' \vec{x}$ you already know how to solve. This transformation is useful if you are dealing with numerical integration code which only takes the linear form.

ContentLicense: CC BY-SA 4.0


问:
Orthogonal Projection onto a Variation of the Unit Simplex

Given the set $T_{\alpha}=\{x\in\mathbb{R}^n:\sum x_i=1,0\leq x_i\leq \alpha\}$

For which $\alpha$ the set is non-empty?

Find a dual problem with one dual decision variable to the problem of finding
the orthogonal projection of a given vector $y ∈\mathbb{R}^n$ onto $T_α$.

I found that $\alpha\geq1/n$ and when i tried to calculate the dual function  I did the following.

$\min$ $||x-y||^2$

s.t

$\sum x_i=1$

$0\leq x_i\leq\alpha\to X=\{x:0\leq x_i\leq\alpha\}$

Setting the lagrangian to be $L(x,\lambda)=||x-y||^2+2\lambda(\sum x_i-1)$,we would like to minimize the lagrangian w.r.t x then :

$$\underset{x\in X}{\min}L(x,\lambda)=\sum x_i^2+2(\lambda-y_i)x_i+||y||^2-2\lambda$$

$$\frac{\partial L}{\partial x_i}=2x_i+2(\lambda-y_i)=0\to x_i^*=\left\{\begin{array}{rcl} y_i-\lambda&0\leq y_i-\lambda\leq\alpha\\
\alpha&y_i-\lambda>\alpha\\
0&y_i-\lambda<0\end{array}\right.$$

and the dual function is $q(\lambda)=-2\lambda+||y||^2+\left\{\begin{array}{rcl}n\alpha^2+2n\alpha\lambda-2\alpha\sum y_i&y_i-\lambda>\alpha\\
-\sum [y_i-\lambda]_+^2&\mbox{else}\end{array}\right.$
and the derivative of $q$ is:

$$q'(\lambda)=-2+\left\{\begin{array}{rcl}2n\alpha &y_i-\lambda>\alpha\\
2\sum[y_i-\lambda]_+^2&\mbox{else}\end{array}\right.$$

I'm not sure if I've found the right dual function and if I've found the right derivative and write matlab code that solves the dual and finds the projection of $y$

答:
This is basically Projection onto the Simplex with some modifications.

The problem is given by:

$$
\begin{alignat*}{3}
\arg \min_{x} & \quad & \frac{1}{2} \left\| x - y \right\|_{2}^{2} \\
\text{subject to} & \quad & 0 \leq {x}_{i} \leq \alpha \\
& \quad & \boldsymbol{1}^{T} x = 1
\end{alignat*}
$$

The problem is valid for $ \alpha \geq \frac{1}{n} $ otherwise the constraint $ \boldsymbol{1}^{T} x = 1 $ isn't feasible.

For $ \alpha \geq 1 $ the problem matches the Projection onto the Simplex as the upper boundary can not be an active constraint (Well it is for 1, but then it is equivalent for the equality constraint and the non negativity).

The Lagrangian in that case is given by:

$$ \begin{align}
L \left( x, \mu \right) & = \frac{1}{2} {\left\| x - y \right\|}^{2} + \mu \left( \boldsymbol{1}^{T} x - 1 \right) && \text{} \\
\end{align} $$

The trick is to leave non negativity constrain implicit.

Hence the Dual Function is given by:

$$ \begin{align}
g \left( \mu \right) & = \inf_{0 \leq {x}_{i} \leq \alpha} L \left( x, \mu \right) && \text{} \\
& = \inf_{0 \leq {x}_{i} \leq \alpha} \sum_{i = 1}^{n} \left( \frac{1}{2} { \left( {x}_{i} - {y}_{i} \right) }^{2} + \mu {x}_{i} \right) - \mu && \text{Component wise form}
\end{align} $$

Taking advantage of the Component Wise form the solution is given:

$$ \begin{align}
{x}_{i}^{\ast} = { \left( {y}_{i} - \mu \right) }_{0 \leq \cdot \leq \alpha}
\end{align} $$

Where the solution includes the inequality constrains by Projecting onto the box $ \mathcal{B} = \left\{ x \mid 0 \leq {x}_{i} \leq \alpha \right\} $.

The solution is given by finding the $ \mu $ which holds the constrain (Pay attention, since the above was equality constrain, $ \mu $ can have any value and it is not limited to non negativity as $ \lambda $).

The objective function (From the KKT) is given by:

$$ \begin{align}
0 = h \left( \mu \right) = \sum_{i = 1}^{n} {x}_{i}^{\ast} - 1 & = \sum_{i = 1}^{n} { \left( {y}_{i} - \mu \right) }_{0 \leq \cdot \leq \alpha} - 1
\end{align} $$

The above is a Piece Wise linear function of $ \mu $.

Since the function is continuous yet it is not differentiable due to its piece wise property theory says we must use derivative free methods for root finding. One could use the Bisection Method for instance.

MATLAB Code

I wrote a MATLAB code which implements the method with Bisection Root Finding. I verified my implementation vs. CVX. The MATLAB Code which is accessible in my StackExchange Mathematics Q3972913 GitHub Repository.


问:
Find all natural numbers $n$ where $n<1000$ such that cube of sum of the digits of $n$ be equal to $n^2$

Find all natural numbers $n$ where $n<1000$ such that cube of sum of
the digits of $n$ be equal to $n^2$

For one digit numbers $n=1$ is the only answer. For two digits numbers $\overline{ab}$:

$$(10a+b)^2=(a+b)^3$$
$$100a^2+20ab+b^2=a^3+3a^2b+3ab^2+b^3$$

But it seems this method doesn't work because I can't simplify above equation.

答:
Use @lulu's notation for $S(n)$ to denote the sum of digits of $n$.  Using @NeatMath's observation that $(S(n))^3 = n^2$, we see that $n$ is a perfect cube, so it suffices to check the set $\{1^3,2^3,\dots,9^3\}$, due to the constraint $n < 1000 = 10^3$.  You might include $0$ if you have an affirmative answer to Is $0$ a natural number?.  $0$ and $1$ are trivial solutions.

@lulu's criterion $n^3 \equiv n^2 \pmod9$ restricts the searching to $1^3$, $3^3$, $6^3$ and $9^3$ because if $\gcd(n,9) = 1$, then the cancellation law gives $n\equiv1\pmod9$.

$n$
$S(n)$
$(S(n))^3$
$n^2$

$1^3 = 1$
$1$
$1$
$1$

$3^3 = 27$
$2+7 = 9$
$9^3 = 729$
$729$

$6^3 = 216$
$2+1+6 = 9$
$9^3 = 729$
$46656$

$9^3 = 729$
$7+2+9 = 18$
$18^3 = 5832$
$531441$

Hence, there's only two solutions $1$ and $27$.


问:
Find ${\partial z\over \partial x}$ and ${\partial z \over \partial y}$ for $xy + yz = xz$
I'm currently learning how to use partial differentiation. Can someone give me a hint?

find  ${\partial z\over \partial x}$ and ${\partial z \over \partial y}$ for $xy + yz = xz$

答:
Assuming that $x$ and $y$ are independent variables, differentiating both sides with respect to $x$ using product rule, we get,
\begin{equation*}
y+y\dfrac{\partial z}{\partial x} = z+x\dfrac{\partial z}{\partial x}
\end{equation*}
Solving this we get the desired result for $\dfrac{\partial z}{\partial x}$. Similarly for $y$.

You can also check that answer obtained this way is correct answer by explicit differentiation:
\begin{align}
&z=\dfrac{xy}{x-y}\\
\implies &\dfrac{\partial z}{\partial x} = -\dfrac{y^2}{(x-y)^2}
\end{align}

And answer obtained above is
\begin{equation}
\dfrac{\partial z}{\partial x} = \dfrac{y-z}{x-y}
\end{equation}

Substituting $z=\dfrac{xy}{x-y}$ in above equation, we get,
\begin{equation}
\dfrac{\partial z}{\partial x} = -\dfrac{y^2}{(x-y)^2}
\end{equation}


问:
The value of $\lim_{x\to +\infty} \dfrac{e^{x^2}}{10^{|x|}}$
I want to find the value of
$$\lim_{x\to +\infty} \dfrac{e^{x^2}}{10^{|x|}}.$$

Since $x \rightarrow + \infty$, I only consider the value of the function for $x \ge 0$, i.e.
$$\lim_{x\to +\infty} \dfrac{e^{x^2}}{10^{x}}.$$

Noticing that the numerator and denominator both approaches $+\infty$ as $x$ approaches $+\infty$, I tried using L'Hôpital's Rule. However, it makes things worse. I did check at Wolfram Alpha, and found out that the value of the limit is $\infty$.

I then noticed that both the numerator and the denominator has powers of some function of $x$. Taking natural logarithm of the numerator and denominator, we have $$\dfrac{\ln e^{x^2}}{\ln 10^{x}}=\dfrac{x^2 \ln e}{x \ln 10}=\dfrac{x}{\ln 10}.$$

This seems really silly, but would I be justified if I say that $$\lim_{x\to +\infty} \dfrac{e^{x^2}}{10^{x}} = \lim_{x\to +\infty}\dfrac{x}{\ln 10}= + \infty?$$

If this is incorrect, kindly give me a hint to tackle this problem.

Thank you for your time.

答: The original problem could be handled more directly, but to pick up after what you have done so far: If you have shown that $\lim_{x\to\infty}\frac{\ln(f(x))}{\ln(g(x))}=\infty$, then you have shown that for any $M$, there is an $x_M$ such that for any $x>x_M$, that $$\ln(f(x))>M\cdot\ln(g(x))$$ This implies $$f(x)>g(x)^M$$ and that $$\frac{f(x)}{g(x)}>g(x)^{M-1}$$ Now if $g(x)$ itself approaches $\infty$, then it suffices to settle for any $M>1$ and you have your desired result.</p>


问:
Let $I_1$, $I_2$ are fractional ideals, if $a$, $b$ are non-zero elements with $aI_1 \subset R$ and $b \in I_2$ then $ab \in R$?
First, $R$ is a integral domain, and $K$ is a field of fractions of $R$

Let $I_1$, $I_2$ are fractional ideals, and

$J = \{ x \in K \ | \ x I_2 \subset I_1 \}$

if $a$, $b$ are non-zero elements with $aI_1 \subset R$ and $b \in I_2$ then $abJ \subset R$. So $J$ is a fractional ideal.

It means $ab$ is a element of $R$, and I know $a$ is in $R$ by difinition. Despite $b$ is a element of $I_2$, not $I_2 \cap R$, why $ab$ is a element of $R$?

答:
This is not true. Let $a=1$, choose $b\in K-R$, and take $I_1=R$ and $I_2=bR$. Then $aI_1\subseteq R$ (in fact $aI_1=R$) but $ab=b\notin R$.

It is true in general by the definition of the fractional ideal $J$ that, under the stated conditions, $abJ\subseteq R$. But $ab$ may not be an element of $J$ (note that $1\in J$ if and only if $I_2\subseteq I_1$).


问:
function is continuous and y→f(x,y) is integrable
$f:\mathbb{R^{2}} \to \mathbb{R}:\mathcal{f(x,y)} = \left\{ \begin{matrix} \mathcal{x^{2}\sin{\frac{1}{x}}\exp(-x^{2}|y|)} & x\neq0 \\ \mbox{0} & x=0 \end{matrix}\right.$

I want to prove that this function is continuous and $y\to f(x,y)$ is integrable.

To prove that $f$ is continuous , I think it's enough to prove that $\lim_{x\to 0} f=0$.
So $\lim_{x\to 0} f =$$ \lim_{x \to 0}x^{2}\sin{\frac{1}{x}}\exp(-x^{2}|y|)$. Because $\lim_{x \to 0}x^{2}\exp(-x^{2}|y|)=0$ and $-1 \leq \sin(\frac{1}{x}) \leq 1$ follows that $\lim_{x\to 0} f=0$.

Now to prove that $y \to f(x,y)$ is integrable for every $x \in \mathbb{R}$ I think it's enough to prove that the integrand is smaller than$\infty$. Can someone confirm this?

EDIT: new idea. I can say that $|f(x)| \leq x^{2}\exp(-x^{2}|y|)$ because of the same reason I explained with the sinus above this EDIT.

EDIT2: I don't know. Can someone explain how I can prove that the function $y\to f(x,y)$ is integrable.

OFFICIAL EDIT: Is it possible that I can try to solve the integral , because $f$ is always a positiv , so measurable, function. And that I can switch the limit and integral because $f$ is continuous? So the integral would be $0$ which is $<\infty$?

答:
As you need to prove $y\to f(x,y)$ is integrable, here $x$ is effectively a constant for you. Thus, we effectively need to prove the integrability of
$$ g(y) =
\begin{cases}
Ae^{-B|y|} & x \neq 0 \\
0 & x = 0 \\
\end{cases}
$$
where $A=x^2\sin\frac{1}{x}$ and $B=x^2$ are constants.

Now, as $g(y)$ is continuous for each $x$ (easy to see), it is integrable.


问:
Suppose that $\underset{x\rightarrow0}{\lim}\frac{f\left(x\right)}{\left|x\right|}=1$ , may $f$ be differentiable at $0$?
Prove or disprove:

Let $f$ be a function that is defined on a neighbourhood of $0$.
Suppose that $\underset{x\rightarrow0}{\lim}\frac{f\left(x\right)}{\left|x\right|}=1$ , then $f$ is not differentiable at $0$

I somehow think its false but couldn't find a counter example. So I tried proving it and this is my prove of it being true:

let us assume by contradiction, that $f$ is differentiable at $0$. thus:

$\underset{x\rightarrow0}{\lim}\frac{f\left(x\right)-f\left(0\right)}{x}=f'\left(0\right)$

Which implies:

$f'\left(0\right)=\underset{x\rightarrow0^+}{\lim}\frac{f\left(x\right)-f\left(0\right)}{x}=\underset{x\rightarrow0^+}{\lim}\underbrace{\left[\frac{f\left(x\right)}{\left|x\right|}-\frac{f\left(0\right)}{\left|x\right|}\right]}$

and:

$f'\left(0\right)=\underset{x\rightarrow0^-}{\lim}\frac{f\left(x\right)-f\left(0\right)}{x}=\underset{x\rightarrow0^-}{\lim}{\left[\frac{f\left(x\right)}{-\left|x\right|}-\frac{f\left(0\right)}{-\left|x\right|}\right]}=\underset{x\rightarrow0^-}{\lim}-\left[\frac{f\left(x\right)}{\left|x\right|}-\frac{f\left(0\right)}{\left|x\right|}\right]=\underbrace{-\underset{x\rightarrow0^+}{\lim}{\left[\frac{f\left(x\right)}{\left|x\right|}-\frac{f\left(0\right)}{\left|x\right|}\right]}}$

They must be equal, thus:

$\underset{x\rightarrow0}{\lim}{\left[\frac{f\left(x\right)}{\left|x\right|}-\frac{f\left(0\right)}{\left|x\right|}\right]}=0$

As we know $\underset{x\rightarrow0}{\lim}\frac{f\left(x\right)}{\left|x\right|}=1$,
then:

$\underset{x\rightarrow0}{\lim}\frac{f\left(0\right)}{\left|x\right|}=1$

which cant be, because if $f(0)=0$, then the limit is $0$, and
if $f(0)\ne0$, the the lim is $\pm\infty$

Which in both cases, doesn't equal $1$,

Contradiction.

.

.

Was there a problem separating to $0\pm$ and change $x$ to $|x|$ without modifying $f(x)$? Or was there any other problem ?

First post, be gentle.

Thank you!

答: No. Take $f(x)=|x|$. Then $f$ is not differentiable at 0.


问:
How to graph $f(x)= \frac{1}{2} (e^x + e^{-x})$ without graphing app?

The function $f$ is defined by $f(x)= \frac{1}{2} (e^x + e^{-x})$, $x \in \Re$.
Show that $f$ has no inverse.

If I am to show that function $f$ has no inverse, I would graph function $f$ and show that it does not pass the horizontal-line test, as a horizontal line passing through the graph would have more than one intersection point.

The problem is, I am not sure how to graph $f(x)= \frac{1}{2} (e^x + e^{-x})$ without using a graphing application. I know how a $y=e^x$ graph looks like, but how do I combine $e^x + e^{-x}$?

I thought of using Piecewise function, but I do not know how to proceed.

答:
We have that

$$\lim_{x\to0^+}\frac{e^x}2=\frac12\;,\;\;\lim_{x\to0^+}\frac{e^{-x}}2=\frac12$$

and then

$$\lim_{x\to\infty}\frac{e^x}2=\infty\;,\;\;\lim_{x\to0^+}\frac{e^{-x}}2=0$$

And all this basically tells you what happens to the graph of the function for $\;x\ge0\;$.

If we add to this that the function is monotonically ascending since

$$f'(x)=\frac12\left(e^x-e^{-x}\right)=\frac{2^{2x}-1}{2e^x}>0\,,$$

Since the exponential is ascending and $\;e^{2x}>e^x\;$, you can see how the positive (and thus also the negative) branch of this function ascend pretty quickly to $\;\infty\;$


问:
Is it common notion to call limits that can be evaluated with algebra "intermediate limits"?
Recently, I just found myself calling limits that can be solved using algebra as "intermediate limits". Example:

$$ \lim_{x \to 1} \frac{x^2-1}{x-1}$$

The expression evaluates to an indeterminate form.

But, if we factor out:

$$ \lim_{x \to 1} \frac{(x-1)(x+1)}{x-1} = \lim_{x \to 1} {(x+1)} = 2$$

Intermediate limit: limit that evaluate to an indeterminate form but can be simplified with algebra, to a limit which evaluates to a number.

The reason why I call them "intermediate limits", is that they're not so easy as so you can just evaluate, but they're not so hard they can't be solvable with just plain algebra. Hence, the "intermediate".

Question: Is it common notion? And is it misleading or correct?

答:
I've never heard the term 'intermediate limit' before, but the limit you used an example is known as indeterminate form. An indeterminate form is where you don't have enough information to compute a limit just by looking at the individual terms. For example, $0/0$ is an indeterminate form, because
$$
\lim_{x \to c}\frac{f(x)}{g(x)}
$$
cannot be evaluated simply with the knowledge that $f(x) \to 0$ and $g(x) \to 0$. Compare this with
$$
\lim_{x \to c}\frac{u(x)}{v(x)}
$$
where $u(x) \to 5$ and $v(x) \to 3$ (as $x$ tends towards $c$). This limit is easy to solve because we can use the quotient law:
$$
\lim_{x \to c}\frac{u(x)}{v(x)} = \frac{\lim_{x \to c}u(x)}{\lim_{x \to c}v(x)}=\frac{5}{3} \, .
$$
Regarding the limit you gave as an example,
$$
\lim_{x \to 1}\frac{x^2-1}{x-1} \, ,
$$
you might find this thread to be of interest: 'Why does factoring eliminate a hole in limit?'.


问:
Proving $(A\times B) \cap (C\times D) = (A\cap C) \times (B\cap D)$
So there is a similiar question in the archives which I looked at after I attempted my proof: Proving that for any sets $A,B,C$, and $D$, if $(A\times B)\cap (C\times D)=\emptyset $, then $A \cap C = \emptyset $ or $B \cap D = \emptyset $

But it is not exactly the same, so I wanted to write my proof out from start to finish to see if my thought process was correct.

1) First I just experimented with some sets to see if anything came about: let $A = \{1\}$, $B = \{2\}$, $C = \{3\}$, $D = \{4\}$:

$(A\times B) \cap (C\times D)$ side:

$$(1,2)\cap (3,4) = (\emptyset,\emptyset)$$ [not sure if i could state this, but it is what I said in my solution]

$(A\cap C)\ \times (B\cap D)$ side:
$$\emptyset\ \times \emptyset = (\emptyset, \emptyset)$$

Ok so I established what appears to be equality, so now I have to prove it.

let $(x,y) \in (A\ X\ B) \cap (C\ X\ D)$
--> $(x\in A \cap y\in B) \cap (x\in C \cap y\in D)$
--> $(x\in A \cap x\in C) \cap (y\in B \cap y\in D)$
--> $x\in (A\cap C)\ X\ y\in (B\cap D)$
--> $ (x,y)\in (A\cap C)\ X\ (B\cap D)$
 Done.

Then I would have to do the other way as well but it would amount to a similar argument.

P.S: How to get lines of my proof to line up with arrows?

答:
$$(A\times B)\cap(C\times D)=(A\cap C)\times (B\cap D) ?$$
Def.:
$$ X\cap Y= \left\{x|x\in X\wedge y\in Y\right\}.$$
$$ X\times Y=\left\{<x,y>|x\in A\wedge y\in B\right\}. $$
$$ <x, y> = \{ \{x\}, \{x, y\} \}$$
Proof:
\begin{align*}
       (A\times B) \cap(C\times D)&= \left\{<x,y>|x\in A\wedge y\in B\right\}\cap  \left\{<x,y>|x\in C\wedge y\in D\right\} \\
       &=   \left\{<x,y>|(x\in A\wedge y\in B)\wedge (x\in C\wedge y\in D)\right\} \\
       &= \left\{<x,y>|(x\in A\wedge x\in C)\wedge (y\in B\wedge y\in D)\right\}\\
       &=(A\cap C)\times (B\cap D).
       \end{align*}
$ \dashv $


问:
All Covering spaces of the Annulus
Consider the annulus $$A:=\{(x,y)\in\mathbb{R}^{2}\mid 1\leq x^{2}+y^{2}\leq 4\}$$
I have to find all (connected) covering up to isomorphism.

My proof so far:

First of all, we have a homeomorphism $A\cong S^{1}\times [1,2]$. Therefore, we find $\pi_{1}(A)\cong \pi_{1}(S^{1}\times [1,2])\cong \pi_{1}(S^{1})=\mathbb{Z}$. All the subgroups of $\mathbb{Z}$ are of the form $\mathbb{Z}n$ for $n\in\mathbb{N}$ and since $\mathbb{Z}$ is abelian, all of them define distinct conjugacy classes of subgroups.

Now there is a theorem, which says that,if $A$ admits an universl cover, the then there are for each $n$ a covering $(\widetilde{A},p)$ such that $p_{\ast}(\pi_{1}(\widetilde{A}))\cong n\mathbb{Z}$ and this are all coverings up to isomorphism.

Now, I know that for $S^{1}$, we can define the maps $f_{n}:z\mapsto z^{n}$, which define covering maps of $S^{1}$, such that $(f_{n})_{\ast}(\pi_{1}(S^{1}))\cong n\mathbb{Z}$. So therefore, my idea was to define the maps $g_{n}:S^{1}\times [1,2]\to A\cong S^{1}\times [1,2]$ exactly by $g_{n}(z,\lambda):=(f_{n}(z),\lambda)$. This should then be a covering with the claimed property $(g_{n})_{\ast}(\pi_{1}(S^{1}\times [1,2]))\cong n\mathbb{Z}$.

Now to my question: Does this look right so far? and furthermore, the annulus $A$ is not simply connected, therefore all of my constructed coverings above are not universal. So I still have to find a universal cover, in order to show that I can apply the theorem above.

答:
Thanks to the comments, I think I can answer it by myself.

All the coverings of the annulus are (up to isomorphism):

(1) The universal cover
$$p:\mathbb{R}\times [1,2] \to  A, \\ (t,\lambda)\mapsto (\lambda \cos(2\pi t),\lambda\sin(2\pi t))$$

(2) For each $n\in\mathbb{N}$ the cover
$$g_{n}:S^{1}\times [1,2] \to A, \\ (e^{i\theta},\lambda)\mapsto (\lambda \cos(\theta n ),\lambda\sin(\theta n ))$$

Title: All Covering spaces of the Annulus


问:
Find the limit $ \lim_{n \to \infty}\left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)^n$
For $a,b,c>0$, Find $$ \lim_{n \to \infty}\left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)^n$$

how can I find the limit of sequence above?
Provide me a hint or full solution.
thanks ^^

答:
$$\lim_{n \to \infty}\ln \left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)^n=\lim_{n \to \infty}\frac{\ln \left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)}{\frac{1}{n}}=\lim_{x\to 0^+}\frac{\ln \left(\frac{a^{x}+b^{x}+c^{x}}{3}\right)}{x}$$

L'Hospital or the definition of the derivative solves it. Actually since the limit is just the definition of the derivative of $\ln \left(\frac{a^{x}+b^{x}+c^{x}}{3}\right)$ at $x=0$, it would be wrong to use L'H :)


问:
Using the cosine law: Given a $\triangle ABC$ with $AB=3$, $AC=8$, and median $AD=5$, find $BC$.
I'm trying to solve the following question but somehow got stuck with no clue how to continue:

Given a $\triangle ABC$ with
$$AB=3, AD=5, AC=8$$
where $AD$ is the median of $BC$ (meaning $BD=DC$). Find $BC$.

I've used cosine law and reached that: $$ BD=\sqrt{34-30\cdot \cos y}$$
$$CD= \sqrt{89-80\cdot \cos a}$$
Not sure that I can assume that angle $a=y$ ($\angle DAB=\angle DAC$).

Any help would be appreciated.

Thanks!

答:
By Applonius Theorem suggested by @AlbusDumbledore,
\begin{align}
AB^2 + AC^2 &= 2(AD^2+BD^2) \\
3^2 + 8^2 &= 2(5^2 + BD^2) \\
BD &= \frac12\sqrt{3^2 + 8^2 - 2\cdot 5^2} = \frac{\sqrt{23}}{2} \\
BC &=  2BD = \sqrt{23}
\end{align}

Not sure that I can assume that angle $a=y$ ($\angle DAB=\angle DAC$).

No, $AD$ is median, not angle bisector.


问:
$B(x,y)\Gamma(x+y)=\Gamma(x) \Gamma(y)$ for all $x,y \in \mathbb{C}_{Re>0}$
Define $\Gamma: \mathbb{C}_{Re>0} \to \mathbb{C}$ with $\Gamma(x):=\int_0^\infty\gamma_x(t)dt=\int_0^\infty t^{t-1} \exp(-t)dt$ and $B: \mathbb{C}^2_{Re>0} \to \mathbb{C}$ with $B(x,y):=\int_0^1 \beta_{x,y}(t)dt=\int_0^1t^{x-1}(1-t)^{y-1}dt$.

How do I show that $B(x,y)\Gamma(x+y)=\Gamma(x) \Gamma(y)$ for all $x,y \in \mathbb{C}_{Re>0}$?

答: In $\operatorname{B}$, substitute $t=\cos^2\theta$; in the three $\Gamma$s, substitute $t=r^2$, $t=X^2$, $t=Y^2$. The claim becomes$$\begin{align}&2\int_0^{\pi/2}\cos^{2x-1}\theta\sin^{2y-1}\theta d\theta\cdot2\int_0^\infty r^{2x+2y-1}e^{-r^2}dr\\&=2\int_0^\infty X^{2x-1}e^{-X^2}dX\cdot2\int_0^\infty Y^{2y-1}e^{-Y^2}dY.\end{align}$$Rewrite both sides of a double integral, then use polar coordinates $r=X\cos\theta,\,Y=X\sin\theta$ over the first quadrant so $rdrd\theta=dXdY$.


问:
Find the multiplicative order of $10$ modulo $2003$

Find the multiplicative order of $10$ modulo $2003$, i.e the least positive integer $k$ such that $10^k \equiv 1 \pmod {2003}$.

I haven't learned about this yet, does this mean I have to find $k = \text{ord}_{2003}10$?
Any hints to solve this? Thanks for your help!

答:
Find the multiplicative order of $10$ modulo $2003$, i.e the least positive integer $k$ such that $10^k \equiv 1 \pmod {2003}$.

...does this mean I have to find $k = \text{ord}_{2003}10$?

Yes, that's the definition of $\text{ord}_n$.

First to note/check is that $2003$ is prime. This means under Fermat's little theorem that $10^{2002}\equiv 1 \bmod 2003$, and thus we know that $k$ divides $2002 = 2\cdot 7 \cdot 11 \cdot 13$, giving $16$ options, $k \in \{1,2,7,11,13,14,22,26,77,91,143,154,182,286,1001,2002\}$,  two of which are obviously wrong ($1$ and $2$).

Calculating, we get $10^7 \equiv 1024 \bmod 2003$ which is interesting since $1024 =2^{10}$ is a square, so - since we were looking at an odd power of $10$ - there must be some value $r$ such that $r^2\equiv 10 \bmod 2003$ (we don't need to find it), meaning $r^{2002} \equiv 10^{1001}\equiv 1 \bmod 2003$, so $k\mid 1001$ and all the even possibilities for $k$ are eliminated.

So I leave you to calculate which of $\{11,13,77,91,143,1001\}$ is the answer sought.


问:
How should I interpret these parametric equations of orbit?

In my math book there are these parametric equations of the orbit/trajectory of a point $P$:
$$\begin{cases}
x(t)=\sin t\\
y(t)=\sin 4t\\
\end{cases}
$$
along with a graph/picture of it on the $y$ and $x$ axis.
Could anyone explain why it follows the path it does?

I unfortunately cant plot it on desmos or show a picture, so if anyone could do this, that would be great.

I thought about it, and since the x values are given by $sin(t)$, does this mean that the trajectory of $P$ first goes to the right, but after $sin(\pi)$ the $x$ values suddenly get negative, and the trajectory switches to the left of the $y$ axis?

So my question is in which direction does the orbit exactly go?

Thanks in advance

答:
The figure you're looking at is determined by looking at an $x$-$y$ plane and by drawing out a path $(x(t),y(t))$ that depends on a parameter $t\in \mathbb{R}$, which you can intuitively understand to be time.

The coordinates $x(t)$ and $y(t)$ are functions of the parameter $t$ given in your case by $x(t) =\sin(t)$ and $y(t)=\sin(4t)$. The figure $$\{(\sin(t),\sin(4t)\mid t\in\mathbb{R}\}$$

is actually called a Lissajous curve. The direction and speed at which the figure is "drawn" depends on how you go trough the parameter $t$. If you start at $t=0$, you start at the origin $(0,0)$ and as $t$ increases, the point $(x(t),y(t))$ starts moving to the Northeast. However, you could just as well look at $$\{(\sin(-t),\sin(-4t)\mid t\in\mathbb{R}\}.$$

The curve in the $x$-$y$ plane will still be the same Lissajous curve, but if you start at $t=0$ and let $t$ increase and follow $(x(t),y(t))$, the point will move to the Southwest instead. In this case you're following the curve in the other direction.

So it all depends on $t$. In your original parametric equations, you are following the points by "walking" along the curve starting at the origin $(0,0)$ and moving to the Northeast. This video might help you as well to visualize this.


问:
Question about little-o
I'm studying little-o properties and I'm stuck on this case in particular.

Is it correct to say $\lim\limits_{x \rightarrow 0} \dfrac{ x^3+o(x^3) }{ x^4+o(x^4) }$ $=$ $\lim\limits_{x \rightarrow 0} \dfrac{ x^3 }{ x^4 } = +\infty$ ?

答:
It may be more readable if you rewrite it as$$\lim\limits_{x \rightarrow 0} \dfrac{ x^3+o(x^3) }{ x^4+o(x^4) } = \lim_{x\to 0} \frac{x^3 + \alpha(x)x^3}{x^4 + \beta(x)x^4}$$
where both $\alpha$ and $\beta$ go to $0$ as $x\to 0$.
$$\lim_{x\to 0} \frac{x^3 + \alpha(x)x^3}{x^4 + \beta(x)x^4} = \lim_{x\to 0} \frac{1 + \alpha(x)}{x + \beta(x)x} = \infty \tag{divided by $x^3$}$$
Note that $\alpha(x) = o(1)$ (as $x\to 0$) is another notation for infinite-small expressions around $0$; that is,
$$\lim\limits_{x \rightarrow 0} \dfrac{ x^3+o(x^3) }{ x^4+o(x^4) } = \lim_{x\to0} \frac{1 + o(1)}{x + o(x)} \tag{divided by $x^3$}$$
 [More info about little-o]

Title: Question about little-o


答: You can note that the function $10^x$ is increasing less rapidly than $e^{x^2}$ so that the limit is absolutely $\infty$</p>

Title: The value of $\lim_{x\to +\infty} \dfrac{e^{x^2}}{10^{|x|}}$

Body:
I want to find the value of
$$\lim_{x\to +\infty} \dfrac{e^{x^2}}{10^{|x|}}.$$

Since $x \rightarrow + \infty$, I only consider the value of the function for $x \ge 0$, i.e.
$$\lim_{x\to +\infty} \dfrac{e^{x^2}}{10^{x}}.$$

Noticing that the numerator and denominator both approaches $+\infty$ as $x$ approaches $+\infty$, I tried using L'Hôpital's Rule. However, it makes things worse. I did check at Wolfram Alpha, and found out that the value of the limit is $\infty$.

I then noticed that both the numerator and the denominator has powers of some function of $x$. Taking natural logarithm of the numerator and denominator, we have $$\dfrac{\ln e^{x^2}}{\ln 10^{x}}=\dfrac{x^2 \ln e}{x \ln 10}=\dfrac{x}{\ln 10}.$$

This seems really silly, but would I be justified if I say that $$\lim_{x\to +\infty} \dfrac{e^{x^2}}{10^{x}} = \lim_{x\to +\infty}\dfrac{x}{\ln 10}= + \infty?$$

If this is incorrect, kindly give me a hint to tackle this problem.

Thank you for your time.


问:
When Haar measure is written as $U^{-1}dU$, what does $dU$ mean?
On the real line, I understand that $x^{-1}dx$ is a measure invariant by scaling, where $dx$ is the usual measure.

However, for the unitary group, for example, I have seen Haar measure written analogously as $U^{-1}dU$. But what does the symbol $dU$ stand for, in this expression? Is $dU$ unitary? Is $U+dU$ unitary? Does it mean that I can use any parametrization whatsoever and implement small changes in the parameters?

答: You (or/and your sources) are confusing the Haar measure and the Maurer–Cartan form on Lie groups, which for matrix groups indeed can be written as $g^{-1}dg$. Here $g$ is understood as a smooth map (local parameterization) to the Lie group and $dg$ is the differential of this map.


问:
maximum distance graph
I have big trouble with graph theory. I just dont really know how to prove this theorem correctly. My idea was to do it by induction but I am not so sure.

Prove that in every simple, directed, connected graph G with n vertices there is a vertex whose distance from all other vertices is at most n / 2.

1)First, prove the following implication: If the above statement holds for every tree on n vertices, then it holds for every connected graph on n vertices.
2)
Complete the proof by showing the statement for the case that G is a tree.

I

答:
Induction is less useful in graph theory than you might initially think.  Using a maximality or minimality argument with contradiction is much more powerful and intuitive.

In this problem (ignoring the strategy suggested in the hint), you can start by considering a path $P$ in your graph of maximal length.  Then, yadda yadda yadda you nominate a vertex in the graph that has the desired feature.  To see that, assume we could choose a vertex in the graph that was further from the candidate then that.  Then, yadda yadda yadda there is a path longer than $P$ in the graph.  Since that violates the choice of $P$, the candidate must meet the desired criteria.

Title: maximum distance graph


问:
Finding the possible Jordan canonical forms of a matrix, given its characteristic polynomial and minimal polynomial
Say we have a complex square matrix $M$ with:

Characteristic polynomial: $(\lambda-3)^4(\lambda+2)^5(\lambda-5)^2$

Minimal polynomial: $(\lambda-3)^2(\lambda+2)^3(\lambda-5)$

$\dim(E_3)=2$ and $\dim(E_{-2})=3$ where $E_\lambda$ is the eigenspace for $\lambda \in \mathbb{R}$

Would $\dim(E_5)=1$? Is it really so simple as that the dimension of the eigen space is the multiplicity of that eigenvalue in the minimal polynomial?

However, my main reason for making this post is I would like someone to help me gain insight into the possible Jordan canonical forms of $M$. Thank you.

Would

答:
We know that $\forall \lambda_i$, $m_g(\lambda_i)$ is equal to the number of Jordan blocks relative to the eigenvalue $\lambda_i$, while the powers that appear in the minimal polynomial represent the biggest Jordan block relative to a specific eigenvalue $\lambda_k$.

The algebraic multiplicity $m_a(\lambda_i)$ is equal to the sum of the orders of the blocks relative to the eigenvalue $\lambda_i$.

We have $$\chi(t)=(t-3)^4(t+2)^5(t-5)^2, \mu(t)=(t-3)^2(t+2)^3(t-5)$$
$$\text{and }m_g(3)=2,m_g(-2)=3.$$
The informations that we know are that the sum of the orders of the Jordan blocks relative to the eigenvalue $3$ is $4$ and the biggest block has dimension $2$.

The sum of the orders of the Jordan blocks relative to $-2$ is $5$ and the biggest of them has dimension $3$.

Finally, we have that the sum of Jordan blocks relative to $5$ is $2$ and the biggest of them has order $1$.

The possibilities that we have are:
$$J_3\to\begin{cases}(\textbf{J}_\textbf{3,2},J_{3,2})\\(\textbf{J}_\textbf{3,2}J_{3,1},J_{3,1})\text{ not acceptable} \end{cases}$$
$$ J_{-2}\to\begin{cases}(\textbf{J}_\textbf{-2,3},J_{-2,2})\text{ not acceptable}\\(\textbf{J}_\textbf{-2,3},J_{-2,1},J_{-2,1}) \end{cases}$$
$$J_5\to (J_{5,1},J_{5,1})$$
In the notation "$J_{i,k}$", $i$ is the eigenvalue and $k$ is the dimension of the block.

The idea is that, after we fix the biggest block, we study the possible partitions of the number of the remaining blocks.


答:
Apply the cosine rule as follows instead
$$AB^2= AD^2+ BD^2 -2AD\cdot BD \cos \theta $$
$$AC^2= AD^2+ CD^2 - 2AD\cdot CD \cos (\pi-\theta )$$
Substitute $BD=CD=\frac12BC$ and eliminate the variable $\theta$ to obtain $BC$.

Title: Using the cosine law: Given a $\triangle ABC$ with $AB=3$, $AC=8$, and median $AD=5$, find $BC$.

Body:
I'm trying to solve the following question but somehow got stuck with no clue how to continue:

Given a $\triangle ABC$ with
$$AB=3, AD=5, AC=8$$
where $AD$ is the median of $BC$ (meaning $BD=DC$). Find $BC$.

I've used cosine law and reached that: $$ BD=\sqrt{34-30\cdot \cos y}$$
$$CD= \sqrt{89-80\cdot \cos a}$$
Not sure that I can assume that angle $a=y$ ($\angle DAB=\angle DAC$).

Any help would be appreciated.

Thanks!


问:
Direct product of periodic elements is periodic.
Definition: An element $p$ of a ring $R$ is called periodic if, there exists distinct positive integer $k$ and $l$ such that $p^k=p^l$.

My Question: Let $x,y$ be two periodic elements of a ring $R$. Then $(x,y)$ is a periodic element of the ring $S:=R×R$.

My attempt: Let $x,y$ be two periodic elements of a ring $R$. Then, there exist $k\neq l,m\neq n\in \mathbb{N}$ such that $x^k=x^l$ and $y^m=y^n$. Hence $(x,y)^{km}=(x^{lm},y^{kn})$. Now, what should I do?

答:
I found an answer of my question which is as follows:

Suppose that $x^k=x^l (k>l)$ and  $y^m=y^n (m>n)$. Then $x^l=x^k=x^{k-l+l}=x^{2(k-l)+l}=\cdots=x^{r(k-l)+l}$ and $y^n=y^m=y^{m-n+n}=y^{2(m-n)+n}=\cdots=y^{r(m-n)+n}$ for any positive integer $r$. Let $s=(k-l)(m-n)$ and $t=max\{l,n\}$. Then $(x,y)^{s+t}=(x,y)^{t}.$


问:
If e is idempotent, prove that 1-e is as well.
Let $(R,+,·)$ be a ring with unity. If $e$ is idempotent, prove that $1-e$ is also idempotent.

Here's my attempt:
If $e$ is idempotent then $e^n=e$. Then
$$e=e^n$$
$$e-e^n=0$$
$$e(1-e^{n-1})=0$$
$$1-e^{n-1}=0$$
But $e^{n-1}=e$, so:
$$1-e=0$$
which means $1-e$ is idempotent. Is this correct?

答:
I will assume that throughout your question, $n = 2$.

No, that is not correct, because given that $e (1 - e) = 0$, we cannot conclude that either $e = 0$ or $1 - e = 0$. It's possible for both $e$ and $1 - e$ to be nonzero, and yet for $e (1 - e)$ to be $0$ nevertheless.

For example, in the ring of $2 \times 2$ matrices (where $0$ denotes the zero matrix and $1$ denotes the identity matrix),

$$1 - \begin{bmatrix} 1 & 0 \\ 0 & 0\end {bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 1\end {bmatrix} \ne 0,$$

and yet

$$\begin{bmatrix} 1 & 0 \\ 0 & 0\end {bmatrix} \left (1 - \begin{bmatrix} 1 & 0 \\ 0 & 0\end {bmatrix} \right) = \begin{bmatrix} 1 & 0 \\ 0 & 0\end {bmatrix} \begin{bmatrix} 0 & 0 \\ 0 & 1\end {bmatrix} = 0.$$

However, in any ring, $(1 - e)^2 = (1 - e)(1 - e) = 1 - 2 e + e^2$. Since $e^2 = e$, what can you conclude from this?


Title: Find the limit $ \lim_{n \to \infty}\left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)^n$

Body:
For $a,b,c>0$, Find $$ \lim_{n \to \infty}\left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)^n$$

how can I find the limit of sequence above?
Provide me a hint or full solution.
thanks ^^

Body:
$$\begin{align}\lim_{n \to \infty}\left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)^{n} &=\lim_{r \to 0^+}\left(\frac{a^{r}+b^{r}+c^{r}}{3}\right)^{1/r}\\ &=\lim_{r \to 0^+}\frac{1}{r}\cdot \exp\left( [\log \left(a^{r}+b^{r}+c^{r}\right)]\right)
\\
&=\exp\left( [\lim_{r \to 0}\frac{d}{dr}\log(a^{r}+b^{r}+c^{r})]\right)
\\
&=\exp\left( \frac{[\log(a)+\log(b)+\log(c)]}{3}\right)
\\
&= (abc)^{1/3}
\end{align}$$


问:
If $G$ is finitely generated and $G/N$ is finitely presented , then $N$ is normal closure of finite set.

Suppose $G$ is a finitely generated group and $N$ is a normal subgroup of $G$ such that $G/N$ is a finitely presented .Then show that $N$ is a normal closure of some finite set of $G$.

[My attempt]

Let $\pi : G \to G/N$ be a natural quotient map.

Since $G/N$ is a finitely presented, there exist a exact sequence of group
$$ 1  \; \to \; \langle R\rangle \; \to \; F(S) \; \stackrel{f}{\to} \; G/N \; \to \; 1 $$
where, $F(S)$ is free group on $S$ and $R$ is a relation set. ($F$, $S$ are finite)
($\langle \cdot\rangle$: normal closure notation)

Since $F(S)$ is free, we construct map $\phi: F(S) \to G $ such that $f=\pi \; \circ \; \phi$.

Therefore, we obtain $\phi(R) \subset \ker\pi = N$ so $\langle \phi(R)\rangle \;\subseteq N$.

I'm stuck here.

How can I prove that $ N \subseteq \; \langle \phi(R)\rangle $ ?

Thank you for your attention.

答:
I believe there's no escaping the result, which I learned from the paper

B. H. Neumann, Some remarks on infinite groups, J. London Math. Soc. 12 (1937),
120–127,

but which I understand is a basic fact in universal algebra, that

if a group is finitely presented with respect to a particular finite set of generators, then it is so with respect to any finite set of generators.

Suppose $G$ is generated by the finite set $S$. Let $F$ be the free group on $S$. Then we have natural surjective morphisms
$$
F \overset{f}{\to} G \overset{\pi}{\to} G/N.
$$
Let $K$ the kernel of the composite map. By the above result, there is a finite subset $T$ of $F$ such that $K$ is the normal closure in $F$ of $T$. Now the image of $K$ under $f$ is $N$, so $N$ is the normal closure in $G$ of the finite set $f(T)$ .

Addendum Contrary to my expectations, you can indeed avoid the result I quoted, see Derek Holt's comment below.


答:
Let $f(x) = \frac12(e^x + e^{-x}).$

$f''(x) = \frac12(e^x + e^{-x}) > \frac12(0+0)= 0$ for all $x$, i.e. $f(x)$ is strictly concave.

Also, as pointed out before, $f(x)$ is even: $f(x) = f(-x)$, and this means that the graph can be reflected in the $y-$axis. Concavity along with evenness of $f$ suggests that $f(0)$ is a minimum.

Let's confirm the stationary points of the graph for good measure. $f'(x) = 0 \implies e^x - e^{-x} = 0 \implies e^x = \frac{1}{e^x} \implies e^{2x} = 1 \implies 2x = 0 \implies x=0,$ so $(0,f(0)) = (0,1)$ is the only stationary point and it is a minimum because $f$ is concave.

Also, $\lim_{x \to \infty}f(x) = \lim_{x \to -\infty}f(x) = \infty $, but we are still not done, because we haven't shown that our graph doesn't have asymptotes. But $\lim_{x \to \infty}f'(x) = \lim_{x \to \infty}{ \left(e^x - e^{-x}\right)} = \infty,$ and $\lim_{x \to -\infty}f'(x) = \lim_{x \to -\infty}{ \left(e^x - e^{-x}\right)} = -\infty.$

And now we know what the graph looks like.

Title: How to graph $f(x)= \frac{1}{2} (e^x + e^{-x})$ without graphing app?

Body:
The function $f$ is defined by $f(x)= \frac{1}{2} (e^x + e^{-x})$, $x \in \Re$.
Show that $f$ has no inverse.

If I am to show that function $f$ has no inverse, I would graph function $f$ and show that it does not pass the horizontal-line test, as a horizontal line passing through the graph would have more than one intersection point.

The problem is, I am not sure how to graph $f(x)= \frac{1}{2} (e^x + e^{-x})$ without using a graphing application. I know how a $y=e^x$ graph looks like, but how do I combine $e^x + e^{-x}$?

I thought of using Piecewise function, but I do not know how to proceed.


问:
Show that $\beta_{x,y}:(0,1) \to \mathbb{C}, \beta_{x,y}(t)=(1-t)^{x-1}t^{y-1}$ is integrable
Let $\mathbb{C}_{Re>0}=\{x \in \mathbb{C} \space | \space Re(x)>0 \}$.

For all $x,y \in \mathbb{C}_{Re>0}$ the function $\beta_{x,y}:(0,1) \to \mathbb{C}, \beta_{x,y}(t)=(1-t)^{x-1}t^{y-1}$ is integrable.

How do I show this?

答:
First, note that for all $x,a,b\in\mathbb{R}:x^{a+ib} = x^ae^{ib\ln(x)}$.

Now we can use this because $$\left|\int_0^1(1-t)^{x-1}t^{y-1}dt \right|\leq\int_0^1|(1-t)^{x-1}t^{y-1}|dt,$$

and for all $t\in\mathbb{R}:|t^x|=|t^a|$, where $x=a+ib$. Therefore it is sufficient to show that the function $$f:t\mapsto (1-t)^{\alpha-1}t^{\beta-1}$$
is integrable on $(0,1)$ when $\alpha,\beta>0$.

To do this, first note that $f$ is bounded on any closed interval $[a,b]\subseteq (0,1)$ because $f$ is continuous. Furthermore, note that $$\lim_{t\to 0}\dfrac{f(t)}{t^{\beta-1}}=1 \hspace{0.5cm} \text{and} \hspace{0.5cm} \lim_{t\to 1}\dfrac{f(t)}{(1-t)^{\alpha-1}}=1$$
so $$f(t) = \Theta\left(\frac1{t^{1-\beta}}\right)\hspace{0.5cm}\text{when}\hspace{0.5cm} x\to 0$$
and $$f(t) = \Theta\left(\frac1{(1-t)^{1-\alpha}}\right)\hspace{0.5cm}\text{when}\hspace{0.5cm} x\to 1.$$
So, since $\alpha,\beta>0$, $f$ is integrable on $(0,1)$.


问:
Number of pairs $(i,j)$ less than or equal to $m$
Given a number $m\in\mathbb{N}$, how many pairs $(i,j)\in \mathbb{Z}^2$are there such that $$|i|+|j|\leq m?$$

For example, for $m=1$ one has $5$ pairs: $(0,0),(1,0),(0,1),(-1,0),(0,-1)$. What about in the general case?

答: If $i>0$, $j\ge0$ and $i+j\le m$, then $(j,-i),(-i,-j),(-j,i)$ are distinct solutions; these four sets of solutions cover the entire solution space save for $(0,0)$. Thus the number of solutions is four times the number of solutions in a quadrant plus one or $4\left(\frac{m(m+1)}2\right)+1=2m^2+2m+1$.


答:
Purely Enumerative Approach: Do this in two cases, either you have a $0$ or not. If you have a $0,$ there are $1+2\cdot 2\cdot m$ ways to do this (why?). If you do not have a $0,$ you have $4$ ways to choose the signs and in how many ways can you have $i+j\leq m$ for $i,j>0$? Use stars and bars or simply fix $i$ and iterate $j.$

You should get $1+4m+4\binom{m}{2}=1+4\left (\binom{m}{1}+\binom{m}{2}\right )=1+4\binom{m+1}{2}.$

For the fun (Enumerative Geometry Approach): Use Pick's theorem. Notice that $2m^2$ is the area of the square and Pick's theorem says that $2m^2=i+b/2-1,$ where $i$ are the number of reticular internal points, and $b$ the reticular boundary points. Notice further that the number of boundary points are $4m$ points. So $4m^2=2i+b-2.$ Notice that you want $i+b$ to be your answer(why?), call it $x$ so $4m^2-i+2=4m^2-x+4m+2=x$ implying the result.

Title: Number of pairs $(i,j)$ less than or equal to $m$

Body:
Given a number $m\in\mathbb{N}$, how many pairs $(i,j)\in \mathbb{Z}^2$are there such that $$|i|+|j|\leq m?$$

For example, for $m=1$ one has $5$ pairs: $(0,0),(1,0),(0,1),(-1,0),(0,-1)$. What about in the general case?


问:
Intuition behind partial fractions
I was trying to solve the following integral:

$$\int \frac{dx}{(x^2 + 2x + 1)(x^2 + 1)}$$

But I'm having some trouble doing partial fractions decomposition with this one. What I did was the following: Firstly I simplified the denominator of the fraction:

$$\int \frac{dx}{(x^2 + 1)(x + 1)^2}$$

So, I know that, to do partial fractions with this we are suposed to write it as:

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

The part I'm having touble with is figuring out what should go in the $*$. I checked in WolframAlpha and it's supossed to be written as:

$$\frac{1}{(x^2 + 1)(x + 1)^2} = \frac{Ax + B}{x^2 + 1} + \frac{C}{x + 1} + \frac{D}{(x + 1)^2}$$

My question is: How can I conclude this? Is there a general rule for doing this? What is the thought process behind choosing what to put in the numerator of the fractions? My teacher only showed us some examples and I also find myself having some trouble doing this when it's a type of fraction I've never seen before, can you give me some general tips when doing this?

答:
If I were you I would get an online reference for partial fractions. It isn't hit or miss.

The theory works out pretty much goes as follows.

Assuming $p$ and $q$ polynomial, and $\deg{p}<\deg{q}$, factorise $q$ fully. Over the reals you can factor into a product of linear and quadratic terms. Ensure that you group terms in the sense that $(x+1)(2x+2)=2(x+1)^2$ or $(x^2+1)(x^2+1)(4x^2+4)=4(x^2+1)^3$, etc. (basically group terms with same roots).

Now, each factor in the factorisation begets a term in the partial fraction expansion according to four rules.

I: a single linear factor $ax+b$ begets a $A/(ax+b)$.
II: a repeated linear factor $(ax+b)^n$ begets:

$$\frac{A_1}{ax+b}+\frac{A_2}{(ax+b)^2}+\cdots+\frac{A_n}{(ax+b)^n}.$$

Note $(ax+b)^n$ is $n$ factors multiplied together and so there are $n$ terms.

This reduces to rule I if $n=1$.

III: a quadratic $ax^2+bx+c$ begets a term $(Ax+B)/(ax^2+bx+c)$

and rule IV follows a similar pattern to Rule II.

WHY this works I would suggest you find a reference for that. The one thing that might help you understand is the following:

Say your $1/((x^2+1)(x+1)^2)$... so you have a $x^2+1$ and a $x+1$ and another $x+1$. What won't work is:

$$\frac{1}{(x^2+1)(x+1)^2}=\frac{Ax+B}{x^2+1}+\frac{C}{x+1}+\frac{D}{x+1}$$

because those last two terms give $(C+D)/(x+1)$... and there are examples where you cannot get the rational function on the left here without a $1/(x+1)^2$ term... it just doesn't work. So you change the $D/(x+1)$ to $D/(x+1)^2$ and the theory (to be found in a reference) says that this will always work.

Hopefully this is some kind of help. It isn't a scatter gun approach, there is theory behind the rules.

Title: Intuition behind partial fractions


问:
Beta Distribution Sufficient Statistic
So I have this homework problem that I am struggling a little bit with coming to a solid answer on.
The problem goes like this:

Suppose X~Beta($\theta,\theta), (\theta>0)$, and let $\{X_1, X_2 , \ldots , X_n \}$ be a sample.
Show that T=$\Pi_i(X_i*(1-X_i)$ is a sufficient statistic for $\theta$.

I started out with my Beta distribution as:

$f(x_i,\theta)=\frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha)\Gamma(\beta)}x^{(\alpha-1)}(1-x)^{(\beta-1)}$

$=\frac{\Gamma(\theta + \theta)}{\Gamma(\theta)\Gamma(\theta)}x_1^{(\theta-1)}(1-x_1)^{(\theta-1)} ***\frac{\Gamma(\theta + \theta)}{\Gamma(\theta)\Gamma(\theta)}x_n^{(\theta-1)}(1-x_n)^{(\theta-1)} $

$=\frac{\Gamma(2\theta)}{\Gamma(\theta)^2}x_1^{(\theta-1)}(1-x_1)^{(\theta-1)} ***\frac{\Gamma(2\theta)}{\Gamma(\theta)^2}x_n^{(\theta-1)}(1-x_n)^{(\theta-1)}$

$={(\frac{\Gamma(2\theta)}{\Gamma(\theta)^2})}^n \Pi_i (x_i)(1-x_i)^{(\theta-1)}$

I know that in order for my statistic to be sufficient by factorization, I need to have a $g(T,\theta)$ and a $h(x_1,x_2,...,x_n)$.

What I have above is my $g(T,\theta)$, but I am not so sure about my $h(x_1,x_2,...,x_n)$.
I have seen other places where the suggestion is to use 1 for my $h(x_1,x_2,...,x_n)$.
Could I do this here with this problem?
It just seems a little too easy to do that, but I will be happy if it is that easy.

If anyone could let me know, that would be greatly appreciated.

答:
(The above answer did not answer minimal sufficiency. So I show it here.)

Note that beta distribution with p.d.f $$\delta_{\alpha,
\beta}(x) = cx^{\alpha-1}(1-x)^{\beta-1}dx$$ is a exponential distribution, because it can be rewrite as

$$\delta_{\alpha,
\beta}(x) = \exp{\left[\alpha\log(x)+\beta\log(1-x)\right]}\mu(dx)$$

We know that, for example from equation 2.5 from this reference,

a minimal sufficient statistic for a distribution belongs to the
full-rank exponential family is $T(X)$, where the p.d.f of exponential
family is $\exp{[T(X)'\eta(\theta)]c(\eta(\theta))}\mu(dx)$.

Thus we conclude that $$\left(\sum_{i=1}^n\log(x_i),\sum_{i=1}^n\log(1-x_i)\right)$$
is a minimal sufficient statistic for $(\alpha, \beta)$.

When $\alpha=\beta$, the distribution with p.d.f is still from a full-rank exponential family because the p.d.f can be written as
$$\delta_{\alpha}(x) = \exp{\left\{\alpha[\log(x)+\log(1-x)]\right\}}\mu(dx)$$
This means $$\sum_{i=1}^n\left[\log (x_i), \log(1-x_i)\right]$$ is a minimal sufficient statistic when $\alpha=\beta$.

Title: Beta Distribution Sufficient Statistic


答:
Let
$a_n=\{(i,j)\in \mathbb Z^2: |i|+|j|\leq n\}$ the number integer-coordinate points inside the $\ell_1$ ball with radius $n$. We know that $a_0=1$.
Based on this geometric intuition, it is not hard to see that
$$a_{n+1}=a_n+4n.$$
By induction, we have
$$a_n=1+4.1+4.2+...+4.n=1+4\frac{n(n+1)}{2}=1+2n(n+1).$$


问:
Showing a functional is coercive
This question is from Calculus of Variations by Rindler (problem 2.2):

Let $\Omega \subset \mathbb{R}^d$ be a bounded Lipschitz domain. Define

$V := \{ u \in W^{1,2}(\Omega) : \int_\Omega u(x)dx = 0 \}$.

Assume that $f : \Omega \times \mathbb{R}^d \rightarrow \mathbb{R}$ is continuously differentiable with
\begin{align}
\mu \vert A \vert^2 &\leq f(x,A) \quad \quad \quad \, \,  \text{for some }\mu > 0 \text{ and all } (x, A) \in \Omega \times \mathbb{R}^d, \quad \ (1) \\
\vert D_A f(x,A)\vert &\leq M(1 + \vert A \vert^2) \quad \text{ for some  }M > 0 \text{ and all }(x, A) \in \Omega \times \mathbb{R}^d, \quad (2)
\end{align}
and that $A \mapsto f(x,A)$ is convex for all $x \in \Omega$.

Let $g \in L^2(\Omega)$.

Show that the functional

$\mathcal{F}[u] := \int_\Omega f(x,\nabla u(x)) - g(x)u(x) $

is coercive on $V$; that is, show there exists $\mu > 0$ such that for all $u \in V$ we have

$\mathcal{F}[u] \geq \mu \vert\vert u \vert\vert_{W^{1,2}}^2 - \mu^{-1}$.

I can deal with the first part of the integrand: we have
\begin{align}
\mathcal{F}[u] &\geq \mu\int_\Omega  \vert\nabla u(x) \vert^2 - \int_\Omega gu \quad\,\,\,\,\text{ by }(1) \\
&\geq \frac{\mu}{1+C_p} \vert\vert u \vert\vert_{W^{1,2}}^2 - \int_\Omega gu \quad\text{ by the Poincare-Wirtinger inequality}
\end{align}
using the condition that $\int_\Omega u = 0$. I haven't been able to deal with the $-\int_\Omega gu$ term on the end though. I suspect this is where the $g \in L^2$ condition is meant to come into play; I've tried to use Holder's inequality and Young's inequality, but neither of them have got me anywhere and I can't see how I'm meant to tackle this question.

Any suggestions on how to proceed?

Thanks

答:
So (I think) I've managed to solve this using Holder's inequality-- details below in case anyone's interested. I had tried this approach before but mistakenly "worked out'' that from the fourth line it wasn't possible to obtain $a_2$ and $c$.

\begin{align*}
\mathcal{F}[u] &\geq \mu\int_\Omega  \vert\nabla u(x) \vert^2 - \int_\Omega gu \, \text{ by } (1)\\
&\geq \frac{\mu}{1+C}\int_\Omega  \vert\nabla u(x) \vert^2 + \frac{\mu}{1+C} \int_\Omega u^2 - \int_\Omega gu \,\text{ by the Poincare-Wirtinger inequality} \\
&\geq \frac{\mu}{1+C}\int_\Omega  \vert\nabla u(x) \vert^2 + \frac{\mu}{1+C} \int_\Omega u^2 - \left(\int_\Omega g^2\right)^{1/2} \left(\int_\Omega u^2\right)^{1/2} \text{ by Holder's inequality} \\
&= \frac{\mu}{1+C}\int_\Omega  \vert\nabla u(x) \vert^2 + a_1x^2 - bx \,\text{ setting }x=\left(\int_\Omega u^2\right)^{1/2} \\
&\geq \frac{\mu}{1+C}\int_\Omega  \vert\nabla u(x) \vert^2 + a_2x^2 - c \, \text{ for some }a_2, c > 0\quad (*) \\
&\geq m \vert\vert u \vert\vert_{W^{1,2}}^2 - c \, \text{ setting }m=\min \left( \frac{\mu}{1+C}, a_2 \right) > 0.
\end{align*}
A quick computation shows that this is greater than or equal to either $m \vert\vert u \vert\vert_{W^{1,2}}^2 - \frac{1}{m}$ or $\frac{1}{c} \vert\vert u \vert\vert_{W^{1,2}}^2 - c$, which gives us what we wanted.

$(*)$ This step is valid since the inequality $a_1 x^2 - bx \geq a_2 x^2 - c$ is equivalent to $(a_1 - a_2)x^2 - bx + c \geq 0$. If $a_1 > a_2$, this holds so long as $b^2 -  4(a_1 - a_2)c < 0$, which we can guarantee for positive $a_2$, $c$ by taking an arbitrarily small $\varepsilon > 0$ then choosing $a_1 - a_2 = \frac{b^2}{4c} + \varepsilon$ and $c$ large enough so that $a_2$ is positive ($a_1$ is positive since $\mu$ and $C$ are; then when $a_2$ is close enough to $a_1$ it is also forced to be positive).

Title: Showing a functional is coercive


问:
show f is differentiable at c if and only if there exists a function g on I that is continuous at c and satisfies f(x) − f(c) = g(x)(x − c) ∀x ∈ I.
Let f be defined on an interval I containing the point c. Then f is differentiable at c if and only if there exists a function g on I that is continuous at c and satisfies
f(x) − f(c) = g(x)(x − c) ∀x ∈ I.
In this case1, we have g(c) = f'(c).

1This result is also known as Caratheodory’s Theorem

答:
Assume $f$ to be differentiable at $c$. Define $g:I\to\mathbb{R}$ by $g(x)=\frac{f(x)-f(c)}{x-c}$ for $x\ne c$ and $g(c)=f'(c)$. By the definition of the derivative $g$ is continuous at $c$ and it is clear that we have $f(x)-f(c)=g(x)(x-c)$.

Now assume there is such a function $g$. Then for $x\ne c$ we have:$$\frac{f(x)-f(c)}{x-c}=g(x)$$
Since $g$ is continuous at $c$ the limit $x\to c$ of the left side exists. But this precisely means that $f$ is differentiable at $c$.


问:
How do you prove that the mean of the co-primes of a number is half the number?
Say $n = 6$, The set of co-primes is $\{1, 5\}$, $\text{mean} = 3$

For $n = 9$, the set of co-primes is $\{1, 2, 4, 5, 7, 8 \}, \text{mean} = 4.5$

Question: Prove that the mean of co-primes of $n$, which are less than $n$ is half the number itself.

I computed all values until $10000$ and it seems to hold good.

答:
Let $\{a_1, \dots a_{ \phi(n)} \}$ be set of all co-primes of $n$, you notice that $a_{\phi(n)}=n-1( Why?)$ and $a_1=1$.

Let $n= p_1^{\alpha_1} \cdots p_k^{\alpha_j}$ ,  $p_i \nmid a_k$ where $1 \le i \le k$ and $1 \le j \le \phi(n)$

As rightly noted by Battacharjee in his comment, $\gcd({n,a_i})=1=\gcd(a_i,n-a_i)$

And the proof for that:

$(n,a_i)=d \implies n=df, a=dt \implies \gcd(dt,d(f-t))=d=\gcd(a_i,n)$,since $\gcd(t,f)=1$.

Here $\gcd(a_i, n)=1 \implies \gcd(a_i,n-a_i)=1$

Note that: $a_i, n-a_i \in \{a_1, \dots a_{ \phi(n)} \}$

$$\sum_1^{\phi(n)}a_i= \dfrac{\phi(n) n}{2}$$

Now the result is quite direct.

$$\text{Mean}= \dfrac{\sum_1^{\phi(n)}a_i}{\phi(n)}=\dfrac{n}{2}$$

Some things that need to be explained:

$\phi(n)$ denotes the Totient function. Number of co-primes of $n$ , less than $n$. You should read more about it.

$a_i, n-a_i \in \{a_1, \dots a_{ \phi(n)} \}$(Why?)

Because: $\gcd(a_i, n-a_i)=1$, and all numbers which are co prime to $n$ are in the set, which means they are in the set.

The answer looks too complicated only because of notations. Look at Goos' answer.


答:
Rule of thumb: "The partial fraction $\frac{P(x)}{Q(x)}$ has the following property: the degree of $P(x)$ must be one less than the degree of $Q(x)$. So we have a constant over a linear, a linear over a quadratic, a quadratic over a cubic, etc." and this always works (Taken from Joe's comment)

Actually, you take linear in the numerator even for $(x+1)^2$ but it gets reduced

$$\frac{1}{(x^2 + 1)(x + 1)^2} = \frac{Ax+B}{x^2 + 1} + \frac{C}{x + 1} + \frac{Dx+E}{(x + 1)^2}$$

$$\frac{1}{(x^2 + 1)(x + 1)^2} = \frac{Ax+B}{x^2 + 1} + \frac{C}{x + 1} + \frac{D(x+1)+(E-D)}{(x + 1)^2}$$
$$\frac{1}{(x^2 + 1)(x + 1)^2} = \frac{Ax+B}{x^2 + 1} + \frac{C+D}{x + 1} + \frac{(E-D)}{(x + 1)^2}$$

And you get the form,
$$\frac{1}{(x^2 + 1)(x + 1)^2} = \frac{*}{x^2 + 1} + \frac{*}{x + 1} + \frac{*}{(x + 1)^2}$$

So the rule of thumb always works, WolframAlpha gives you a simplified version.

Hope this helped.:)

Title: Intuition behind partial fractions

Body:
I was trying to solve the following integral:

$$\int \frac{dx}{(x^2 + 2x + 1)(x^2 + 1)}$$

But I'm having some trouble doing partial fractions decomposition with this one. What I did was the following: Firstly I simplified the denominator of the fraction:

$$\int \frac{dx}{(x^2 + 1)(x + 1)^2}$$

So, I know that, to do partial fractions with this we are suposed to write it as:

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

The part I'm having touble with is figuring out what should go in the $*$. I checked in WolframAlpha and it's supossed to be written as:

$$\frac{1}{(x^2 + 1)(x + 1)^2} = \frac{Ax + B}{x^2 + 1} + \frac{C}{x + 1} + \frac{D}{(x + 1)^2}$$

My question is: How can I conclude this? Is there a general rule for doing this? What is the thought process behind choosing what to put in the numerator of the fractions? My teacher only showed us some examples and I also find myself having some trouble doing this when it's a type of fraction I've never seen before, can you give me some general tips when doing this?


问:
Roots of a matrix equation
Let $X \in \mathbb{C}^{n \times m} $ be a rectangular matrix of full rank, and $X^*$ its hermitian conjugate, let $A \in \mathbb{C}^{m \times m}$ a square matrix, and let $f: \mathbb{C} \to \mathbb{C}$ be defined by
$$
f(z) = \det \left( I_n + X \frac{1}{A - z I_m} X^* \right)
$$
where $I_n$ is the $n \times n$ identity matrix.

Show that for $n \leq m$ solutions of $f(z)= 0$ are given by $z$ equal to an eigenvalue of $B : = A + X^* X$.

My attempt: for $n = m$, $f(z)$ can be straightforwardly rearranged to
$$
f(z) = \det \left( X \frac{1}{B-A} \big( B - I_m z \big) \frac{1}{A - z I_m} X^* \right) = \frac{\det(X) \det(B - I_m z) \det(X^*)}{\det(A - z I_m) \det(B - A)}
$$
where the desired result follows from the factor $\det(B - I_m z)$.

However I cannot factorise the determinant in this way for $n < m$, and I am unsure how to proceed.

答:
By the Schur complement
$$\det(z I_m-A)\,\det\left(I_n - X \frac{1}{z I_m-A} X^*\right) = \det(I)\det \left(z I_m-A-X^*X\right).$$
So $z$ should be the the eigenvalues of $A+X^*X$ but not of those of $A$.

Title: Roots of a matrix equation


问:
Strange algebra question: is this matrix invertible?
Let $m>0$ be an integer. Given non-zero real numbers $b_1, \ldots, b_m$, can we always find real numbers $a_{i,j}, i=1,\ldots,m-1$, $j=1,\ldots, m-1$ such that

$$\mathbb \det
\begin{pmatrix}
a_{1,1} & \ldots & a_{1,m}\\
\vdots &  & \vdots \\
a_{m-1,1} & \ldots & a_{m-1,m}\\
\sum\limits_{i=1}^{m-1}  a_{i,1} b_i & \ldots & \sum\limits_{i=1}^{m-1}  a_{i,m} b_i\\
\end{pmatrix} \neq 0$$

If so, how can we prove that ? And could we relax the restrictions on $b_i$ (all non-zero) ?

答:
The matrix will never be invertible as the last row is a linear combination of the others. This also applies to the case $m=1$ in which case we simply have the zero matrix (because the sum is then empty).

And if the matrix is not invertible its determinant will be $0$


问:
What can you conclude about $A \setminus B = B \setminus A$?
I have the following identity involving sets $A$ and $B$:

$$A \setminus B = B \setminus A$$

Now, a solution says that we can conclude that $A=B$, but what is wrong with a solution that says that this means that $A\subseteq B$? I mean if we remove all elements of $B$ from $A$, we would have what is left from $B$ and same for if we remove all elements of $A$ from $B$ as it will yield also elements of $B$. Please correct me if I am wrong.

Thanks.

答:
Notice that $A\cap B = (A\cup B) \setminus [(A\setminus B) \cup ( B\setminus A) ]=  \begin{cases} (A\cup B) \setminus (A\setminus B) = B \\
(A\cup B) \setminus (B\setminus A) = A\end{cases} $

But this can be true if and only if $A=B$


问:
CAS with a standard language
I hope this question is suitable for the site.

I recently had to work with Mathematica, and the experience was, to put it kindly, unpleasing. I do not have much experience with similar programs, but I remember not liking much Matlab or Maple either. The result is that I am a mathematician who likes programming, but I never managed to learn how to work with a computer algebra system.

Does there exist a CAS which can be programmed using a standard language? I guess the best thing would be just an enormous library of mathematical algorithms implemented for C or Python or whatever.

I know SAGE is based on Python, but as far as I understand (which is not much) it just collects preexisting open source software, so (I assume) one has to learn how to use a new tool for every different problem.

答:
If you're a big fan of C numerous libraries exist for doing certain tasks. MPIR/GMP are two libraries for doing arbitrary precision arithmetic. On top of these you'll find more specialised libraries like MPC which adds support for the complex field.

Flint is a library that sits on top of MPIR, for example, providing an array of number theory functions. PARI/GP I believe provides a similar foundation but via the GMP library.

I'm sure if you googled "library for linear algebra" or "library for x" you would find it. However, a CAS unites all of this functionality.

Now we come to sage. From what I understand of the sage project, it is not so much a collection of open source libraries but an interface uniting them. For example, from the tutorial:

```<code>sage: a = 5   # a is an integer
sage: type(a)
<type 'sage.rings.integer.Integer'>
sage: a = 5/3  # now a is a rational number
sage: type(a)
<type 'sage.rings.rational.Rational'>
sage: a = 'hello'  # now a is a string
sage: type(a)
<type 'str'>
</code>```

Is a sage session. As you're probably aware sage.rings.integer.Integer is not a built-in python type.

I would recommend giving sage a go. I think it is the closest thing out there to what you're after. I personally don't find Mathematica too bad.

ContentLicense: CC BY-SA 2.5

Title: CAS with a standard language

Body:
If you're a big fan of C numerous libraries exist for doing certain tasks. MPIR/GMP are two libraries for doing arbitrary precision arithmetic. On top of these you'll find more specialised libraries like MPC which adds support for the complex field.

Flint is a library that sits on top of MPIR, for example, providing an array of number theory functions. PARI/GP I believe provides a similar foundation but via the GMP library.

I'm sure if you googled "library for linear algebra" or "library for x" you would find it. However, a CAS unites all of this functionality.

Now we come to sage. From what I understand of the sage project, it is not so much a collection of open source libraries but an interface uniting them. For example, from the tutorial:

sage: a = 5   # a is an integer
sage: type(a)
<type 'sage.rings.integer.Integer'>
sage: a = 5/3  # now a is a rational number
sage: type(a)
<type 'sage.rings.rational.Rational'>
sage: a = 'hello'  # now a is a string
sage: type(a)
<type 'str'>

Is a sage session. As you're probably aware sage.rings.integer.Integer is not a built-in python type.

I would recommend giving sage a go. I think it is the closest thing out there to what you're after. I personally don't find Mathematica too bad.


问:
What is the logical consequence of a tautology
I know that the logical consequence of a contradiction is the whole Language $\mathcal{L}$ : $Cn(\bot) = \mathcal{L}$, but what is the logical consequence of a tautology is it $Cn(\top)$ ?

答:
The logical consequence of a tautology must be a tautology.

If $A$ is a tautology, it means that $A$ is true in every interpretation, i.e. every interpretation is a model of $A$.

But $A \vDash B$, we have that $B$ must be true in every model of $A$, and thus also $B$ is true in every interpretation.


问:
What's the limit of the function $f(x)=\ln x-ax$ $(a>0)$, when $x$ approach $0$ and infinity?
I'm a beginner in calculus.

I know when $x \to 0$, $\ln x \to -\infty$ and $-ax \to 0,$ then $\ln x-ax \to -\infty$.

But when $x$ approaches infinity, $\ln x \to \infty$ and $-ax \to -\infty$.

Which one changes faster?

Thanks a lot.

答:
Hint

Calculate and think about $f'(x)$ and $f''(x)$.


问:
Calculating the time of a Brachistochrone
I derived the general equation of a Brachistochrone, which is a cycloid.

$y=A(1-\cos\theta)$

$x=A(\theta-\sin\theta)$

I'm now trying to calculate the time needed to go from the origin to a point $(x,y)$.

From previous analysis I found that the time is equal to

$T=\sqrt{\frac{1}{2g}}\int_0^x\sqrt{\frac{1+(y')^2}{y}}dx$

I'm struggling to solve this. This is what I found so far

$\frac{dx}{d\theta}=A(1-\cos\theta)$

$\frac{dy}{d\theta}=A\sin\theta$

Applying the chain rule: $\frac{dy}{dx}=\frac{dy}{d\theta}\frac{d\theta}{dx}=\frac{A\sin\theta}{A(1-\cos\theta)}=\frac{\sin\theta}{1-\cos\theta}$

Therefore, $(\frac{dy}{dx})^2=(\frac{\sin\theta}{1-\cos\theta})^2$

Therefore,

$T=\sqrt{\frac{1}{g}}\int_0^x\sqrt{\frac{1-\cos^2\theta}{(1-\cos\theta)^2}}d\theta$

Is this right? How do I go on from here? I have the coordinates of two points (and therefore I could derive the equation of the brachistochrone curve between them) and I would like to find the time taken to fall from the initial to the final point along the brachistochrone under acceleration g. Thank you!

答:
From $ x = A(\theta - \sin\theta) $ and $ y = A(1 - \cos\theta) $ we have:
$$ v = \frac{ds}{d\theta} = \sqrt{\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2}  = A\sqrt{(1 - \cos\theta)^2 + (\sin\theta)^2} = A\sqrt{(2(1 - \cos\theta)} = \sqrt{2Ay}$$

By the law of conservation of energy we get:  $$ \frac{ds}{dt} = v = \sqrt{2gy}  . $$
Therefore: $$ \frac{d\theta}{dt} = \frac{\frac{ds}{dt}}{\frac{ds}{d\theta}} = \frac{\sqrt{2gy}} {\sqrt{2Ay}} = \sqrt{\frac{g}{A}} \quad \text{and} \quad T = \int_0^\theta\sqrt{\frac{A}{g}} \mathrm{d}t = \sqrt{\frac{A}{g}}·\theta $$


问:
Linear dependency proof with LES
Problem:

Let $V = \mathbb{R}^n$ be a vector space and let $v_i$ be vectors for indices $i \in \{1,\dots,n+1\}$.

Show that for $i \leq i \leq n+1$ there exist scalars $\lambda_i \neq 0$ so that for at least one $i$

$$ \lambda_1v_1 + \lambda_2v_2 + \dots + \lambda_{n+1}v_{n+1} = 0 $$

holds.

Question:

I would like to know, how to split this equation in a linear equation system, as well as in more generally how to find the proof in this. Essentially what I understand from this, you put this into a linear equation system and try to shorten it down with the help of the Gauß-Algorithm. So that i can read from the rank of the matrix that there has to be at least one scalar so that this equation holds.

答:
A linear equation system is:

$$\mathbf V \mathbf \lambda = \mathbf 0$$
or
$$ \left( \array{v_{1,1} & v_{2,1} &\cdots &v_{n+1,1} \\
v_{1,2} & v_{2,2} &\cdots &v_{n+1,2} \\
\vdots & \vdots &\ddots &\vdots \\
v_{1,n} & v_{2,n} &\cdots &v_{n+1,n}} \right)\left(\array{\lambda_1 \\
\lambda_2 \\
\vdots \\
\lambda_{n+1}}\right) = \mathbf 0
$$

where $\mathbf V$ is the matrix with $n+1$ columns and $n$ rows made up of $v_1, v_2, \cdots , v_{n+1}$ and $\mathbf \lambda$ is the vector with unknown elements $\lambda_1, \lambda_2, \cdots,\lambda_{n+1}$.  Thus you have $n$ equations and $n+1$ unknowns so there will always be a non-trivial solution.  You can apply the algorithm to find one.

Another way to think of it is that if there were $n+1$ linearly independent vectors in a vector space then the space must have dimension $n+1$ or higher.  We know $\mathbf R^{n}$ is $n$-dimensional, so it cannot have $n+1$ linearly independent vectors.

Title: Linear dependency proof with LES


问:
ex falso quodlibet in natural deduction (ND)?
My understanding of inference rules is that they should be intuitively acceptable (like axioms). But my only intuition for ex falso quodlibet in natural deduction is not immediate but comes from its proof, which rests on disjunctive syllogism (which apparently holds in classical propositional logic and intuitionistic logic).
I am probably missing something obvious. Is the principle of explosion really basic/intuitive and why?

The closest justification I see, based on the current answers, is that Ex Falso encodes a (metalogical) desire that the proof system be consistent. But also see the references to intuitionism in the answers.

Edit: as pointed out below, it may have been a distraction for me to raise "intuitionism" in the original question, although Ex Falso arises in classical and intuitionistic logic. This does not exclude that I would like axioms and rules to be intuitively acceptable (to me).

" There  is  surprisingly  little  agreement  about  the  exact  status  of EFQ in  the literature.  Gentzen groups the ex  falso rule alongside the other operational rules(1969,  p.  77).   Prawitz  follows  this  practice.   He  treats ⊥ as  a  ‘0-place  sentential operation’ (Prawitz 1978, p. 38), one ‘for which there is no canonical proof’ (Prawitz1977, p. 26). EFQ is then understood as the elimination rule for ⊥. Others, including Dummett, present ⊥ as an elimination rule for ¬ (e.g. Dummett 1991, p. 291).  I contend that both of these approaches are mistaken...." ref.

Ex Falso has an ‘anomalous position [. . . ]  inside the scheme of introduction/elimination rules’, tarnishing the neat symmetry of the intuitionistic system" (Weir 1986, p. 461).

According to here: "Kolmogorov’s criterion whether to keep an axiom was whether a proposition has an “intuitive foundation” or “possesses intuitive obviousness” (van Heijenoort 1967: 421, 422) ... Kolmogorov said that, just like PEM, Ex Falso “has no intuitive foundation” (van Heijenoort 1967: 419). In particular, he says that Ex Falso is unacceptable for the reason that it asserts something about the consequences of something impossible (van Heijenoort 1967: 421)."

... "I submit that these difficulties can be overcome simply by treating⊥as a punc-tuation marker, as Tennant suggests.  Of course this means thatEFQis left withouta  logical  constant  which  it  serves  to  eliminate.   But  on  the  view  I  am  proposingno such constant is needed becauseEFQis not an operational rule at all.  Its roleis  best  accounted  for  by  according  it  the  status  of  astructural  rule.   It  simply  isthe structural rule that tells us that any sentence whatsoever follows from a con-tradiction.  Such a license isnotspecific to any logical constants, but amounts to ablanket policy.  Our reclassification ofEFQthus fits neatly with our characterizationof structural rules as global rules that assign properties to our deducibility relation." ref.

Edit: I came across this reformulation of my initial uneasiness.

"... adopting disjunctive syllogism as an axiom or as a primitive rule of inference is not an option.  Hence, it is ex  falso that is naturally treated as primitive, not the principle of disjunctive syllogism. Therefore, rather than thinking of ex  falso as the product of the mean-ings of disjunction and negation, we must treat disjunctive syllogism as a product of the ex falso rule.  This suggests that the law of disjunctive syllogism is not a con-sequence only of the meanings of the constants involved, but also hinges on one’s stance concerning the general principle expressed by EFQ." ref.

答:
Ex Falso, in the form:

$$\dfrac {\bot}{\varphi} \ (\bot \text E)$$

is a rule of Natural Deduction, both classical and intuitionsitic.

In the Hilbert-style version it amounts to: $\lnot A \to (A \to B)$.

The formula is consistent with the operational interpretation of Intuitionistic Logic: there is no proof of $\bot$  (the absurdity).

See Arendt Heyting, Intuitionism. An Introduction (1st ed, 1956), page 106:

"Axiom X [the propositional form of EFQ]  may not seem intuitively clear."

And see Andrei N. Kolmogorov, On the principle of excluded middle (1925):

"Axiom 5 [the propositional form of EFQ] is used only in symbolic presentation of the logic of judgments; therefore it is not affected by Brouwer's critique, especially since it has no intuitive foundation either. [...]

Hilbert's first axiom of negation, "Anything follows from the false",  made its appearance only with the rise of symbolic logic. [...] the axiom does not have and cannot have any intuitive foundation since it asserts something about the consequence of something impossible: we have to accept $B$ if the true judgment $A$ is regarded as false."

For a system that rejects EFQ, see Minimal Logic.


答:
This equation reads: "If $x$ is in $A$ but not $B$, then $x$ is also in $B$ but not $A$

The set of elements which can be seen as inside and not inside an arbitrary set $A$ simultaneously is $\emptyset$.

Hence $$A\setminus B=\emptyset \implies A=B$$

Title: What can you conclude about $A \setminus B = B \setminus A$?

Body:
I have the following identity involving sets $A$ and $B$:

$$A \setminus B = B \setminus A$$

Now, a solution says that we can conclude that $A=B$, but what is wrong with a solution that says that this means that $A\subseteq B$? I mean if we remove all elements of $B$ from $A$, we would have what is left from $B$ and same for if we remove all elements of $A$ from $B$ as it will yield also elements of $B$. Please correct me if I am wrong.

Thanks.


问:
Euclidean algoritm to get two polynomials
Determine two polynomials in $\mathbb{Z}_5$ so that $$p(x)(x^3+2x^2+3x)+q(x)(x^2+3x+2)=1$$

I know the answer, but not where it comes from.
$$p(x)=3x\\
q(x)=2x^2+3x+3$$

With Euclidean algoritm I've got that
$$(x^3+2x^2+3x)=(x^2+3x+2)(x+4)+(4x+2)\\
(x^2+3x+2)=(4x+2)(4x)+(2)\\
(4x+2)=(2)(2x+1)+(0)$$

What am I missing?

答:
Substituting the equations you got from bottom to top, we find an expression of the gcd in terms of the two original polynomials:
\begin{align*}
2 &= (x^2+3x+2) - (4x+2)(4x) \\
&= (x^2+3x+2) - \left[ (x^3+2x^2+3x) - (x^2+3x+2)(x+4)\right](4x) \\
&= (x^2+3x+2) - (x^3+2x^2+3x)(4x) + (x^2+3x+2)(x+4)(4x) \\
&= (x^2+3x+2)\cdot(1+(x+4)(4x)) + (x^3+2x^2+3x)\cdot (-4x) \\
&= (x^2+3x+2)\cdot\underbrace{(4x^2+x+1)}_{2\,q(x)} + (x^3+2x^2+3x)\cdot \underbrace{x}_{2\,p(x)}.
\end{align*}
This can also be achieved by more efficient book keeping using the Extended Euclidean Algorithm.


问:
Do strict inequalities hold when taking limits?
I'm a bit confused on the following statement.

Suppose $f(x) < g(x)$ for all $x$ in the open interval $(a,b)$ and that $\lim_{x\to a}f(x)$ and $\lim_{x\to a}g(x)$ both exist.

In this case is it  true that $\lim_{x\to a}f(x) \leq \lim_{x\to a}g(x)$.

Or is it true that $\lim_{x\to a}f(x) < \lim_{x\to a}g(x)$.

I am not quite sure which one is true i think its the first one but I don't quite know why.
Thanks in advance.

答:
In general, strict inequalities are not preserved when taking limits. For instance, when $x>0$,
$$
\dfrac{x}{x+1} > \dfrac{x}{x+2}
$$

but

$$
\lim_{x\to +\infty} \dfrac{x}{x+1} = \lim_{x\to +\infty} \dfrac{x}{x+2}.
$$

Or, to keep within the exact scope of the question, if you consider $f(x)=x$ and $g(x)=x^2$ defined on $(0,1)$, you'll see that even though $g(x) < f(x)$ you have that $\lim_{x\to 0} f(x)= \lim_{x \to 0} g(x)=0$.


问:
Using a poset or directed graph as input for a neural network.
As the title states, I'm trying to train a neural network using some unconventional input. I'm wondering if anyone has any experience or has read any papers that involve using a partially ordered set or alternatively a directed graph as an input for a neural network? Is there a way something like a directed graph can be effectively embedded as a vector? If so, I would greatly appreciate some guidance. Thanks</p>

答:
A bit late, but have you looked at Graph Neural Networks? The representation of each node is updated from its neighbours representations. The neighbours could be the parent of a node only for example. There is a lot to it, so it's hard to summarize. The library PyTorchGeometric is well written with good documentation.

After N layers of GNN you can aggregate all the nodes together to get a fixed size vector representation. Alternatively you can add a "god" node that connects to everything and then use the representation of that node subsequently.

Not sure if it answers your question, but I stumbled here when looking more into these specific topics.


问:
Convex open set
Let $X$ be a normed linear space and let $C$ be a compact convex subset of $X$. Is it true that for every open set $V$ (in $X$) containing $C$, there exists a convex open set $U$ contained in $V$ and contains $C$?

Thanks in advance

答:
Suppose $C+{1 \over n} B(0,1) \cap V^c$ is non empty for all $n$. Then there are $c_n\in C, u_n \in B(0,1)$ such that $c_n+{1 \over n}u_n \in V^c$. By
passing to a subsequence as necessary, we can assume that $c_n \to c \in C$
and since $V^c$ is closed we see that $c \in V^c$ which is a contradiction.

Hence there is some $n$ such that $C+{1 \over n} B(0,1)$ (which is convex and open) is contained in $V$.

Title: Convex open set


问:
Integration by Parts. Get original expression
One of the engineering proofs I am trying to understand has the following line. It says using integration by parts the below can be shown:

$$\int(\sigma_{ij},_j * u_i)dV = \int((\sigma_{ij} * u_i),_j)dV - \int(\sigma_{ij} * u_i,_j) dV$$

N.b. The above is in einstein notation in which the comma represents partial diff.

I pulled out my notes but could not perform integration by parts for whatever reason. I arrived at the original expression. Any help appreciated!

答:
This looks like the standard version of partial integration. Partial integration in one variable says the following:

$$\int_a^bf(x)g'(x)dx = \int_a^b[f(x)g(x)]'dx - \int_a^bf'(x)g(x)dx.$$

Applied to your question, with some nicer notation (sometimes these shorthands like Einstein notation can cloud our vision):

$$\int \partial_j\sigma_{ij}\cdot u_idV = \int \partial_j(\sigma_{ij}u_i)dV-\int\sigma_{ij}\cdot\partial_ju_idV.$$


问:
$ e^{At}$ for $A = B^{-1} \lvert \cdots \rvert B $
For a homework problem, I have to compute $ e^{At}$ for

$$ A = B^{-1}  \begin{pmatrix}
-1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 3 \end{pmatrix} B$$

I know how to compute the result for $2 \times 2$ matrices where I can calculate the eigenvalues, but this is $3 \times 3$, and I cannot compute eigenvalues, so is there any identity or something which allows computing such exponentials?
Thanks!

答:
Note that $A$ has been diagonalized, and that: $$e^{At}=I+At+\frac{1}{2!}(At)^2+\frac{1}{3!}(At)^3+\cdots  $$ $$ \Longrightarrow e^{BDB^{-1}t}=I+BDB^{-1}t+\frac{1}{2!}(BDB^{-1}t)^2+\frac{1}{3!}(BDB^{-1}t)^3+\cdots $$ $$ =BB^{-1}+B(Dt)B^{-1}+\frac{1}{2!}B(Dt)^2B^{-1}+\frac{1}{3!}B(Dt)^3B^{-1}+\cdots $$ $$ =B\left(I+Dt+\frac{1}{2!}(Dt)^2+\frac{1}{3!}(Dt)^3+\cdots\right)B^{-1}$$ What is the expression in the middle?

Edit: Forgot the $t$.


Title: Do strict inequalities hold when taking limits?

Body:
I'm a bit confused on the following statement.

Suppose $f(x) < g(x)$ for all $x$ in the open interval $(a,b)$ and that $\lim_{x\to a}f(x)$ and $\lim_{x\to a}g(x)$ both exist.

In this case is it  true that $\lim_{x\to a}f(x) \leq \lim_{x\to a}g(x)$.

Or is it true that $\lim_{x\to a}f(x) < \lim_{x\to a}g(x)$.

I am not quite sure which one is true i think its the first one but I don't quite know why.
Thanks in advance.

Body: Take $(a,b)=(0,1)$, $f(x)=0$ and $g(x)=x$, then $f(x)<g(x)$ but $\lim_{x\to 0} f(x) = 0 = \lim_{x\to 0} g(x)$.


问:
Usage of intermediate value theorem in proof
Edit: As explained in the comments and in the answer, the IVT was not used to show the derivative but for the statement right before it ($g(t)<g(v)$)...

In another question I came across a reasoning which used the intermediate value theorem and somehow makes conclusions about the derivate of the function. I have summed up the part of the answer which I think is crucial to possibly follow the reasoning:

($g$ is differentiable.) Since $g(t)>g(v)$ for $t\in[c,v)$ by the intermediate value theorem, it follows that $g′(v)\le 0$.

Please refer to the full answer if more context is needed. Unfortunately, I don't see how the IVT is linked to the derivative.

In addition I tried to clarify the answer by suggesting the following:

Since $g(t) > g(v)$ for $t\in [c,v)$ we know $\frac{g(v)-g(t)}{v-t}<0$, thus with $t\to v^-$ and because $g$ is differentiable, it follows that $g'(v) \le 0$.

However, my edit was rejected and I also can't comment on the answer (reputation).

Can you help me understand why the IVT proves the point in the first place?

Is my alternative reasoning correct?

答:
In that part of the answer we have

$a < c < b$ with $g(a) \le g(b) < g(c)$,

$M = \{t\in (c,b] : g(t) = g(a)\}$,

$v = \min M > c$,

and the claim is that $g(t) > g(a)$ for $c \le t < v$.

It is clear from the definition of $v$ as the minimum of $M$ that $g(t) \ne g(a)$ for $c \le t < v$, so we must exclude the case that $g(t_1) < g(a)$ for some $t_1 \in [c, v)$. This is the part where the IVT is used: Since $g(c) > g(a)$, the IVT would imply that $g(t_0) = g(a)$ for some $t_0$ between $c$ and $t_1$, contradicting the definition of $M$.

The next step is to conclude that $g'(v) \le 0$, and that is done exactly as you argued.


问:
Can every isomorphism between a vector space and its dual be written as a non-degenerate bilinear form?
If I have understood correctly, non-degeneracy of a bilinear form:

$$\omega:V\times V\rightarrow \Bbb F \tag{1},$$

in which $\Bbb F$ the underlying field of $V$, is a strong enough condition to conclude that $\omega$ provides an isomorphism between $V$ and $V^*$: non-degeneracy $\Rightarrow$ trivial kernel $\Rightarrow$ injectivity $\Rightarrow$ surjectivity if dimesions are the same.

My question is does the converse hold, can all isomorphisms between a vector space and its dual be induced by some non-degenerate bilinear form on $V$?

答:
If $f:V\to V^*$ is an isomorphism, and we define $\omega:V\times V\to \Bbb{F}$ as $\omega(v_1,v_2):= [f(v_1)][v_2]$, then $\omega$ is bilinear, and note that from the definition, we have that $\omega(v_1,\cdot)= f(v_1)$. In other words, the mapping $v_1\mapsto \omega(v_1,\cdot)$ is equal to $f$, which by assumption is an isomorphism of $V$ onto $V^*$.

If $V$ is infinite-dimensional, there are no isomorphisms $V\to V^*$, so the statement above becomes vacuous.

Edit:
The previous version of my answer made it seem like there was a distinction to be made (because it wasn't obvious to me in the case of non-symmetric/skew-symmetric bilinear forms). But, thanks to @Marc van Leeuven for pointing out that the distinction between left/right non-degeneracy is unnecessary.

For the sake of completeness, here's the proof of the equivalence. Let $\omega_l:V\to V^*$ be the left map, $x\mapsto \omega(x,\cdot)$, and let $\omega:V\to V^*$ be the right map $y\mapsto \omega(\cdot, y)$. Also, let $\iota:V\to V^{**}$ be the canonical map into the double dual. One can easily verify by unwinding definitions that
\begin{align}
\omega_r^*\circ \iota=\omega_l\quad\text{and} \quad\omega_l^*\circ\iota=\omega_r
\end{align}
where the stars on the maps indicate the dual/transposed mapping.
Now, make the assumption that $V$ is finite-dimensional. Then, $\iota$ is an isomorphism.
So, if $\omega_r$ is assumed an isomorphism, then so is $\omega_r^*$, and thus (by first equation) $\omega_l$ being the composition of two isomorphisms is as well. Similarly, $\omega_l$ being an isomorphism implies by the second equation that $\omega_r$ is.

One can also rephrase the argument using matrices since we're in finite dimensions, but I'd rather not :)


问:
Any nice/good way to allow an "or" option in a linear program?
I am looking for a way to express an "or" option in a system of linear inequalities for a linear program I am working on.

I will explain what I mean precisely: Lets say I have a set of inequalities $eq_1$ to $eq_n$, which must hold.  But, additionally, I have several more inequalities $eqa_1,....,eqa_k$ and $eqb_1,...,eqb_k$, that can be partitioned to couples (of the form $\left< eqa_i ,\, eqb_i\right>$ for all $1\leq i \leq k$), such that either $eqa_i$ holds or if it doesn't then $eqb_i$. Nonetheless, it is also possible for both of them to hold. What I am looking for is an option to encode such an "or" condition on a set of inequalities.

Ofcourse, if we say that $eqa_i$ is lets say $f\left(...\right) \leq K$ and $eqb_i$ is $g\left(...\right) \leq M$ I can write $f\left(...\right)+g\left(...\right) \leq K+M$, but this demand is too strong, because in my original problem I may allow $f\left(...\right) > K+M$ (for example) as long as $g\left(...\right) \leq M$, but then $f\left(...\right)+g\left(...\right) > K+M$. So requiring $f\left(...\right)+g\left(...\right) \leq K+M$ is too strong for me.

My question can be divided to 2 questions:

Is there any idea for a nice trick here or for some sort of a reduction to Linear Programming? (A reduction from an instance in which you have couples of inequalities with an or, to a regular instance of LP)

Perhaps there exists a software or an online site which allow this kind of "or"? I am using this site: https://online-optimizer.appspot.com/. However, if there are other sites or software which have such a built in option for an "or", it'd be great as well.

If my inequalities are too generic, I know that most of them  seem from the form $x_{i_1}+x_{i_2}+x_{i_3} \leq K$, where $K$ is the same $K$ for all the inequalities, and the rest are either $x_{i_1}+x_{i_2} \leq K$ or $x_{i_1}+x_{i_2}+x_{i_3}+x_{i_4} \leq K$.

答:
You can enforce $f \le K \lor g \le M$ by introducing a binary decision variable $x$ and linear "big-M" constraints:
\begin{align}
f - K &\le M_1 x \tag1 \\
g - M &\le M_2(1-x) \tag2
\end{align}
Here $M_1$ is a constant upper bound on $f-K$ when $x=1$, and $M_2$ is a constant upper bound on $g-M$ when $x=0$.

If $x=0$, then constraint $(1)$ forces $f \le K$ and constraint $(2)$ is redundant.
If $x=1$, then constraint $(2)$ forces $g \le M$ and constraint $(1)$ is redundant.


问:
The convergence of $\int_0^{\infty} \sin(x)(x-[x])/x^{\alpha}$
I need to evaluate the following integral

$$\int_0^{\infty} \sin(x)(x-[x])/x^{\alpha}dx$$
where $\alpha\in (0,1)$ and $[x]$ is the floor function.
Without $x-[x]$, we can evaluate the integral easily using analytical continuation.
$$\int_0^{\infty} \sin(x)/x^{\alpha}dx=\Im \int_0^{\infty} e^{ix}/x^{\alpha}dx=\Im [\Gamma(1+\alpha)/i^{\alpha-1}]$$

In the presence of $x-[x]$, my idea is that one could use inequality to upper bound the integral if it converges or lower bound the integral if it diverges. But I'm not sure what kind of inequality I should use.

I tried Hölder's inequality, but it will square the sin function in the upper bound, which makes it diverge.

答:
Let's settle (as in the title of the OP) just the convergence of the integral. The obvious choice is Dirichlet's test: we only have to show that $\int^N_0\sin(x)(x-[x])\,dx$ is uniformly bounded. Since $\sin(x)(x-[x])$ is bounded, it's sufficient to show that for integer $N$. But then,
$$\int^N_0\sin(x)(x-[x])\,dx=\sum^{N-1}_{n=0}\int^{n+1}_n\sin(x)(x-[x])\,dx=\int^1_0y\,\sum^{N-1}_{n=0}\sin(n+y)\,dy,$$ and
$$\sum^{N-1}_{n=0}\sin(n+y)=\cos y\,\sum^{N-1}_{n=0}\sin n+\sin y\,\sum^{N-1}_{n=0}\cos n.$$
Since $$\left|\sum^{N-1}_{n=0}\sin n\right|=\left|\frac{\cos(1/2)-\cos(N-1/2)}{2\,\sin(1/2)}\right|\le\frac1{\sin(1/2)}$$ and $$\left|\sum^{N-1}_{n=0}\cos n\right|=\left|\frac{\sin(N-1/2)+\sin(1/2)}{2\,\sin(1/2)}\right|\le\frac1{\sin(1/2)},$$ this is uniformly bounded. $x^{-\alpha}\to0$ as $x\to\infty$ guarantees the convergence of the integral.

BONUS: If $\alpha>0$, $e^{-bx}\,x^{-\alpha}\to0$ as $x\to\infty$ uniformly in $b\ge0$, so $\displaystyle\int_0^{\infty} e^{-bx}\,\sin(x)(x-[x])\,x^{-\alpha}\,dx$ converges uniformly in $b\ge0$.


问:
Showing that a real number is an algebraic integer
For what values of $x,y,z\in\mathbb{Z}$, such that $0\leq x,y,z\leq 2, $ the real number $$\alpha:=\frac{1}{3}\left(x+\sqrt[3]{175} \cdot y+\sqrt[3]{245}\cdot z\right)$$ is an algebraic integer i.e. root of a monic polynomial in $\mathbb{Z}[x]$ ?

I tried to compute the norm but the computation is getting ugly. Also if we can find those $x,y,z $  for which the $\mathbb{Z}$-module generated by $1,\alpha ,\alpha^2,\ldots$ is finitely generated, then those are the required algebraic integers. Can this be done ? Also I would like to know if there are any other simpler approach.

答:
Let $\vartheta=\sqrt[3]{175}$, and let $K=\mathbb{Q}(\vartheta)$. Then it can be shown that $1, \vartheta, \frac{\vartheta^2}{5}$ is an integral basis for $K$. This should prompt you towards the discovery that $\sqrt[3]{245}=\frac{\vartheta^2}{5}$, and hence $\alpha=\frac{1}{3}(x+y\vartheta+\frac{z}{5}\vartheta^2)$.

So you would require $3\mid x,y,z$ if $\alpha$ was to be an algebraic integer. Given that $0\le x,y,z\le 2$, this can only happen if $x=y=z=0$.

Edit. Again let $K=\mathbb{Q}(\vartheta)\cong\mathbb{Q}[x]/(x^3-175)=\mathbb{Q}[x]/(f)$. Recall that $\Delta(\vartheta)=[\mathcal{O}_K:\mathbb{Z}[\vartheta]]^2\Delta_K$, and $\Delta(\vartheta)=-27(-175)^2=-826875=-3^3\cdot 5^4\cdot 7^2$. So the only rational primes that could divide the index are $3$, $5$, and $7$, and we can factor $(3)$, $(5)$, and $(7)$ to see whether or not they do..

($p=3$) Then $f(x)=(x-2)^3$ modulo $3$. Since $3^2\nmid f(2)$, $(3)=\mathfrak{p}_3^3$, where $\mathfrak{p}_3=(3,\vartheta-2)$ is invertible: so $3$ does not divide the index.

($p=7$) Then $f(x)=x^3$ modulo $7$. Since $7^2\nmid f(0)$, $(7)=\mathfrak{p}_7^3$, where $\mathfrak{p}_7=(7,\vartheta)$ is invertible: so $7$ does not divide the index.

($p=5$) Then $f(x)=x^3$ again modulo $5$. But $5^2\mid f(0)$, so we know that we need to enlarge our ring, and we can take an element in $\mathcal{O}_K\setminus\mathbb{Z}[\vartheta]$. When dividing $f(x)$ by $x$ (it's irreducible factor modulo $5$), we get $f(x)=\color{red}{x^2}\cdot x-175$, hence $\frac{\vartheta^2}{5}\in\mathcal{O}_K\setminus\mathbb{Z}[\alpha]$.

Now the minimal polynomial of $\frac{\vartheta^2}{5}$ is $g(x)=x^3-245$, and $g(x)=x^3$ modulo $5$. Since $5^2\nmid g(0)$, we know that this is the only element we need to add to $\mathbb{Z}[\vartheta]$, and $\mathcal{O}_K=\mathbb{Z}[\vartheta, \frac{\vartheta^2}{5}]$.


问:
Volume of body bounded with surfaces
I need to find volume bounded with $x=2y-y^2-z^2$ and $2y-x-1=0$. So first one is paraboloid that is located on x-axis and its "maximum" is in $(0,2,0)$ and the second is plane paralele to z-axis.
I just want to check do we get the volume when substract volume of $2y-x-1=0$ from paraboloid.(Also using polar coordinates it is easy to get $0\le r\le 1, 0\le \phi \le 2\pi$). I'm just not sure whether I sketched this correctly, precisely is given plane above paraboloid?

答: Sometimes viewing the volume helps:

Title: Volume of body bounded with surfaces

Body: Sometimes viewing the volume helps:


问:
How do I check if a state is reachable when I have a nonlinear system?
I have a nonlinear system and I am doing some simulations in order to reach a situation in which all the states are equal to zero.

I have done a lot of simulations and tried many things, but I couldn't reach the desired state. At this point, I am starting to think that the state I want to reach is not reachable.

So, my question is:

How do I check if a state is reachable when I have a nonlinear system?

For example, consider the equations of the unicycle robot which are the simplest that comes to my mind:

$\dot{x}=vcos\theta$

$\dot{y}=vsin\theta$

$\dot{\theta}=\omega$

In this case I know that the state $(0,0,0)$ is reachable, but how do I show it mathematically?

答:
Just like stable linear systems usually converge to some stable equilibrium, non-linear systems do to. However, non-linear systems can have any amount of equilibria. Undamped systems often get stuck in cycle (for instance the undamped inverted pendulum). This means that every reachable state of the system must be inside this cycle. the cycles shape (doesnt have to be a circle) is undetermined and can be influenced by the initial conditions and the external input. To continue to your system, suppose the velocity $v$ and the angular velocity $\omega$ are constants. Additionally, since you wrote down the difference equations, I suppose your state vector represents $\left[x, y, \theta \right]^T$. From what it appears, I can integrate each of these equations over time directly:
$$\theta = \int \omega~~ dt = \omega t + \theta_0$$
$$x = \int vcos\theta~~ dt = \frac{v}{\omega}sin\theta+x_0$$
$$y = \int vsin\theta~~ dt = -\frac{v}{\omega}cos\theta+y_0$$
Now as you might already imagine, these equations describe a circle with radius $\frac{v}{\omega}$ at center $\left[ x_0, y_0 \right]$. This means that only if the constants $v$, $\omega$ and / or the initial positions are chosen correctly, the state $(0,0,0)$ is reachable:
$x_0 = 0, y_0 = \frac{v}{\omega} $
Which represents a circle that crosses the origin at $\theta = 0$.
This entire story does indeed change if $v$ or $\omega$ are dissipative functions over time, at which the position $x$ and $y$ will always return to zero regardless of $\theta$.

So that concludes your specific example. In an abstract sense, determining reachability of a non-linear system is a very interesting topic that usually comes down to determining the stable equilbria and finding the one it will converge to from the given initial state. I do note that even for dissipative systems, some systems converge to a limit cycle. To make it more interesting, some systems converge to a limit cycle if it is in a certain range or initial states, and to a stable equilibrium point for others.


答: Given the convex subsets $C$ and $B_\epsilon(0)$, then $C+B_\epsilon(0)$ is also convex since $\forall a,b\in C, \forall x,y\in B_\epsilon(0)$, $$(1-t)(a+x)+t(b+y)=(1-t)a+tb+(1-t)x+ty\in C+B_\epsilon(0)$$ Since $C\subset V$, each of its points $x\in C$ is covered by some $B_r(x)\subset V$, so by a standard compactness argument there is a finite subcover and an $\epsilon>0$ such that $C+B_\epsilon(0)\subset V$.

Title: Convex open set

Body:
Let $X$ be a normed linear space and let $C$ be a compact convex subset of $X$. Is it true that for every open set $V$ (in $X$) containing $C$, there exists a convex open set $U$ contained in $V$ and contains $C$?

Thanks in advance


问:
Prove that $a^ab^b+a^bb^a\le 1$.

Let $a$ and $b$ be positive real numbers such that $a+b=1$. Prove that $a^ab^b+a^bb^a\le 1$.

This is RMO 2012.

I got a solution, but I think it's flawed.

Consider $a+\dots +a (a$ $ \text{times})$ and $b+\dots +b(b$ $ \text{times})$. We get $a\cdot a$+ $b\cdot b \\ $.

Applying AM-GM , we get  $$\sqrt[a+b]{a^a\cdot b^b}=a^a\cdot b^b\le a\cdot a+b\cdot b.$$

Considering $a+\dots +a (b$ $\text{times})$ and $b+\dots+b(a$ $\text{times})$, we get
$$\sqrt[a+b]  {a^bb^a}=a^bb^a \le 2\cdot ab \Rightarrow a^a\cdot b^b +a^b\cdot b^a\le (a-b)^2=1$$

I have a lot of flaws. Like $a$ is real so we can't actually consider this $a+\dots +a (b$ $\text{times})$ . Can this be corrected somehow?

答:
Your idea is correct however its representation has to be fixed as you gussed the representation is only valid when $a,b$ are natural

Actually we  have as a subcase the weighted AM-GM inequality as follows $$xa+yb\ge (x+y)\sqrt[x+y]{a^xb^y}$$ where $a,b,x,y$ are positive reals.
For more reference see here.

So instead of writing $$a+a+a...(a \space \text{times})+b+b+.....(b \space \text{times})=a^2+b^2\ge \sqrt[a+b]{a^ab^b}$$ i would just write it as by weighted AM-GM $$a\cdot a+b\cdot b\ge \sqrt[a+b]{a^ab^b}$$

Similarly for the other...:)

Title: Prove that $a^ab^b+a^bb^a\le 1$.