问:
Compute $\frac{1^2 t}{1!}+\frac{2^2 t^2}{3!}+\frac{3^2 t^3}{5!}+\frac{4^2 t^4}{7!}+\ldots+\frac{n^2 t^n}{(2n-1)!}+\ldots$
I have to compute $$\frac{1^2 t}{1!}+\frac{2^2 t^2}{3!}+\frac{3^2 t^3}{5!}+\frac{4^2 t^4}{7!}+\ldots+\frac{n^2 t^n}{(2n-1)!}+\ldots$$ I know that $\sinh t$ can be represented as a series1, but for that I require only odd powers $t^{2n-1}$, but I have no idea how to get them or what to do with the even part. Can anybody give me a hint how to start?

1 The expansion is $$\sinh x = x + \frac {x^3} {3!} + \frac {x^5} {5!} + \frac {x^7} {7!} +\cdots = \sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)!}.$$

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\sum_{n = 1}^{\infty}{n^{2} \over
\pars{2n - 1}!}\,t^{n}} =
{\root{t} \over 4}\sum_{n = 1}^{\infty}{\bracks{\pars{2n - 1} + 1}^{\,2} \over \pars{2n - 1}!}\,\pars{\root{t}}^{2n - 1}
\\[5mm] = &\
{\root{t} \over 4}\sum_{n = 1}^{\infty}{\pars{n + 1}^{\,2} \over n!}
\pars{\root{t}}^{n}\,\,{1 - \pars{-1}^{n} \over 2}
\\[5mm] &\
{\root{t} \over 8}\bracks{%
\sum_{n = 1}^{\infty}{\pars{n + 1}^{\,2} \over n!}
\pars{\root{t}}^{n} -
\sum_{n = 1}^{\infty}{\pars{n + 1}^{\,2} \over n!}
\pars{-\root{t}}^{n}}\label{1}\tag{1}
\end{align}

Note that $\ds{\sum_{n = 1}^{\infty}{x^{n + 1} \over n!} =
x\pars{-1 + \sum_{n = 0}^{\infty}{x^{n} \over n!}} =
-x + x\expo{x}}$ and
\begin{align}
&\left\{\begin{array}{rcl}
\ds{\sum_{n = 1}^{\infty}{n + 1 \over n!}\,x^{n + 1}}
& \ds{=} &
\ds{x\,\totald{\pars{-x + x\expo{x}}}{x}}
\\ & \ds{=} &
\ds{-x + x\expo{x} + x^{2}\expo{x}}
\\[5mm]
\ds{\color{red}{\sum_{n = 1}^{\infty}{\pars{n + 1}^{2} \over n!}\,x^{n}}} & \ds{=} &
\ds{\totald{\pars{-x + x\expo{x} + x^{2}\expo{x}}}{x}}
\\ & \ds{=} &
\ds{\color{red}{\pars{x^{2} + 3x + 1}\expo{x} - 1}}
\end{array}\right.
\end{align}

Then ( see (\ref{1}) ),
\begin{align}
&\bbox[5px,#ffd]{\sum_{n = 1}^{\infty}{n^{2} \over
\pars{2n - 1}!}\,t^{n}}
\\[5mm] = &\
{\root{t} \over 8}\bracks{%
\pars{t + 3\root{t} + 1}\expo{\root{t}} - 1}
\\[2mm] - &\
{\root{t} \over 8}\bracks{%
\pars{t - 3\root{t} + 1}\expo{-\root{t}} - 1}
\\[5mm] = &\
\bbx{{1 \over 4}\root{t}\pars{t + 1}\sinh\pars{\root{t}} +
{3 \over 4}t\cosh\pars{\root{t}}} \\ &
\end{align}


问:
Why ($e^x$)'s derivative i.e. $e^x$ , not equal to $(a^x )$'s derivative while $a$ is any number and $e$ is a number as well?
Why ($e^x$)'s derivative i.e. $e^x$ , not equal to ($a^x$ )'s derivative while $a$ is any number and $e$ is any number as well?

答: Let $y=e^x$. Then $x=\ln y$. Taking the derivative of both sides with respect to $x$ yields $$\frac{dx}{dx}=\frac{d(\ln y)}{dx}$$ so $$1=\frac{1}{y}\frac{dy}{dx}$$ and finally we obtain $$\frac{dy}{dx}=y=e^x.$$ Therefore, $$\frac{d}{dx}e^x=e^x.$$ This is the only number that satisfies this property, since $y=e^x$ gives us $x=\ln y$. If $y=2^x$, for example, $x$ would be a nasty $\log_2 y$, and when we take this derivative, it will not be as nice.

ContentLicense: CC BY-SA 4.0


问:
$H_{I}^{n}(M)\cong H_{I}^{n}(R)\otimes_R M.$
Let $R$ be a Noetherian ring and $I$ an ideal of $R$. If $n$ is the cohomological dimension of $I$, then why is the following isomorphism true:
$$H_{I}^{n}(M)\cong H_{I}^{n}(R)\otimes_R M.$$

The cohomological dimension of $I$ is defined to be the supremum of the set of integers $i$ such that $H_{I}^{i}(M)\neq 0$ for some $R$-module $M$.

答:
I will work with total derived functors, I hope it is OK.

So, let us look at the complex $R\Gamma_I (R)$. Note that because we can compute it using the Čech complex, one always has (over noetherian rings):

$R\Gamma_I(M) \cong R\Gamma_I(R) \otimes_R M$ for any $M$.

Now, let us calculate the $n$th cohomology of both sides:

LHS: $H^n(R\Gamma_I(M)) = H^n_I(M)$ - simply by definition.

RHS: $H^n(R\Gamma_I(R) \otimes_R M) = H^{(n+0)}(R\Gamma_I R \otimes_R M) \cong H^n(R\Gamma_I R) \otimes_R H^0(M)$ $=$ $H^n_I(R) \otimes_R M$. This follows from the Künneth spectral sequence since for $i>n$ we have $H^i_I(R) = 0$, and for $j>0$ we have $H^j(M)=0$, because we consider $M$ as a complex concentrated in degree $0$.

Comparing both, we get the required isomorphism:

$H^n_I(M) \cong H^n_I(R) \otimes_R M$.

ContentLicense: CC BY-SA 3.0


问:
complex numbers structure of Chern classes
I have started to read on Chern classes. I cannot quite yet see where the complex numbers and their properties come in as we define Chern classes and other properties. I do see it in later theorems and computation. It seems we could have defined projectivization for any vector bundle for example or even Chern classes. Am I missing something? Where and when do complex field properties really come in?

I am reading Bott and Tu book differential forms in Algebraic Topology. There the construction is based on line bundles and then projectivization and at least in initial steps I do not see any complex field requirement in definitions or proofs. See pp 267 and 270. https://www.maths.ed.ac.uk/~v1ranick/papers/botttu.pdf

答:
For complex line bundles $L$ the first Chern class $c_1(L)$ takes values in $H^2(-, \mathbb{Z})$ and this is quite specific to working over $\mathbb{C}$. One way of thinking about this class is as coming from the exponential sequence

$$1 \to \mathbb{Z} \xrightarrow{2\pi i} \mathbb{C} \xrightarrow{\exp} \mathbb{C}^{\times} \to 1$$

which induces a long exact sequence with connecting homomorphism $H^1(-, \mathbb{C}^{\times}) \to H^2(-, \mathbb{Z})$; this is exactly the first Chern class. Another way of thinking about this class is as a generator of the cohomology of $\mathbb{CP}^{\infty}$, the classifying space of complex line bundles.

Over $\mathbb{R}$ we instead have the first Stiefel-Whitney class $w_1(L)$ of a real line bundle which takes values in $H^1(-, \mathbb{Z}_2)$ and which generates the mod-$2$ cohomology of $\mathbb{RP}^{\infty}$, the classifying space of real line bundles. This comes directly from $\mathbb{R}^{\times}$ being homotopy equivalent to $\mathbb{Z}_2$, with no need for a long exact sequence.

Above, $\mathbb{C}^{\times}$, which is homotopy equivalent to $S^1$, satisfies $\pi_0 \cong 1, \pi_1 \cong \mathbb{Z}$, whereas $\mathbb{R}^{\times}$, which is homotopy equivalent to $S^0$, instead satisfies $\pi_0 \cong \mathbb{Z}_2, \pi_1 \cong 1$. So their homotopy is quite different and this leads to the difference in their characteristic classes.


问:
Show a function such that $f'(c)\neq \dfrac{f(b)-f(a)}{b-a}$ for any $a<b$
Could you help me with the following please:

Give an example of a differentiable function $f:\mathbb{R} \rightarrow \mathbb{R}$ that has a point $c$ such that $f'(c)$ is not equal to the difference quotient $\dfrac{f(b)-f(a)}{b-a}$ for any $a<b$. Why does this not contradict the mean value theorem?

In a graphical way what can be seen is that this occurs when the second derivative of $ f $ is 0, but trying to find this function, I have considered a polynomial with this property, but I cannot see that it fulfills this characteristic. One attempt has been to consider the $ f(x)=x^3+3x+2 $, but it failed to conclude that this is an example of what the exercise asks of us.

And well, I think this does not contradict the MVT for the domain of said function.

答:
Your attempt is close, but instead try $f(x) = x^3$, so $f'(x) = 3x^2$. Using $c = 0$ then gives $f'(0) = 0$, so for it to equal $\frac{f(b) - f(a)}{b - a}$ for some $a \lt b$ requires $f(b) = f(a)$. However, apart from at $x = 0$, $f(x)$ is a strictly increasing function, so $f(b) = f(a)$ only if $b = a$.

Regarding why this doesn't contradict the mean value theorem, note the theorem only states there exists a $c$ which meets its condition (i.e., $f'(c) = \frac{f(b) - f(a)}{b - a}$) for any specific start & end points of $a$ and $b$, not that the derivative of every point within the range must satisfy the condition for some specific start and end points.


问:
Let $\alpha$ be a root of $(x^2-a)$ and $\beta$ be a root of $(x^2-b)$. Provide conditions over $a$ and $b$ to have $F=K(\alpha+\beta)$.
QUESTION: Let $K$ be a field of characteristic different of $2$. Let $F$ be a splitting field for $(x^2-a)(x^2-b)\in K[x]$. Let $\alpha$ be a root of $x^2-a$ and $\beta$ be a root of $x^2-b$. Provide conditions over $a$ and $b$ to have $F=K(\alpha+\beta)$.

MY ATTEMPT:

Let $\alpha=\sqrt{a}$, $\beta=\sqrt{b}$ and $\gamma=\alpha+\beta$.
First of all, we have $F=K(\alpha, \beta)$ due to the definition of splitting field.
Defining $K(\alpha+\beta)=K(\gamma)$.

Let's show that $K(\alpha, \beta)\subset K(\gamma)$:

From $\gamma=\alpha+\beta$ follows that
\begin{align*}
\gamma^2&=(\alpha+\beta)^2\\
&=\alpha^2+2\alpha\beta+\beta^2\\
&=(\sqrt{a})^2+2\sqrt{a}\sqrt{b} +(\sqrt{b})^2\\
&=(a+b)+2\sqrt{a}\sqrt{b}\qquad (*)\\
\end{align*}

Now we are going to show that $\sqrt{b}\in K(\gamma)$

Indeed, multiplying both sides in $(*)$ by $\sqrt{b}$ we have:

$\gamma^2\sqrt{b}=(a+b)\sqrt{b}+2\sqrt{a}(\sqrt{b})^2$. Then $$\sqrt{b}=\frac{2b\sqrt{a}+(a+b)\sqrt{b}}{\gamma^2}\in K(\gamma)$$

Similarly, $\sqrt{a}\in K(\gamma)$, this is

$\gamma^2\sqrt{a}=(a+b)\sqrt{a}+2(\sqrt{a})^2\sqrt{b}$, then

$$\sqrt{a}=\frac{(a+b)\sqrt{a}+2a\sqrt{b}}{\gamma^2}\in K(\gamma)$$

MY DOUBT:
I guess there is no conditions over $a$ and $b$ such that $\alpha=\sqrt{a}$ and $\beta=\sqrt{b}$, however I'm not sure. And I don't know how to connect this with the hypothesis that $K$ has characteristic different of two. Would you help me please?

答:
We assume that $x^2-a,x^2-b$ are irreducible over $K$ and $b\ne a$, since otherwise the problem is trivial.

If $\sqrt{b}\not \in K(\sqrt{a})$ then show that $\sqrt{a}+\sqrt{b}$ has 4 distinct conjugates (that's where we use $char(K)\ne 2$) which implies that $[K(\sqrt{a}+\sqrt{b}):K] = 4$.

If $\sqrt{b}=u+v\sqrt{a} \in K(\sqrt{a})$ then $v\ne 0$ so $(u+v\sqrt{a})^2\in K$ implies $u=0$. Since $b\ne a$ then $v\ne \pm 1$ and hence $K(\sqrt{a}+\sqrt{b})= K((v+1)\sqrt{a})=K(\sqrt{a})$.


Title: Show a function such that $f'(c)\neq \dfrac{f(b)-f(a)}{b-a}$ for any $a<b$

Body:
Could you help me with the following please:

Give an example of a differentiable function $f:\mathbb{R} \rightarrow \mathbb{R}$ that has a point $c$ such that $f'(c)$ is not equal to the difference quotient $\dfrac{f(b)-f(a)}{b-a}$ for any $a<b$. Why does this not contradict the mean value theorem?

In a graphical way what can be seen is that this occurs when the second derivative of $ f $ is 0, but trying to find this function, I have considered a polynomial with this property, but I cannot see that it fulfills this characteristic. One attempt has been to consider the $ f(x)=x^3+3x+2 $, but it failed to conclude that this is an example of what the exercise asks of us.

And well, I think this does not contradict the MVT for the domain of said function.

Body:
Let $f(x)=|x-2|$. then for $x \in [1,3]$ we can write
$$\frac{f(3)-f(1)}{3-1}=0\ne f'(c)$$
note that $$f(x)=(x-2), x>2, f(x)=-(x-2), x \le 0
\implies f(x)=1, x>2; f'(x)=-1, x\le 2$$


问:
Showing or refuting that $f:X\to X$ injective implies epijective.
If $f:X\to X$ is an a injective function and $X$ set, does this implies that $f$ epijective?

I have that:

$f$ injective implies there exists $g:X\to X$ left inverse to $f$, $gf(x)=x$ for all $x$. Now, let $y\in X$, $f(y)=x$ some $x\in X$, then $gf(y)=g(x)$, therefore $y=g(x)$ but $g$ is not $f$...
The sentence is true?

答: What you have shown is that $g(x)$ is surjective, which isn't much of a surprise, as $f(x)$ is a right inverse to it. You have not shown that $f(x)$ is surjective, which you won't be able to do, as $f : \mathbb{Z} \rightarrow \mathbb{Z}$ given by $f(x) = 2x$ is an injection from a set to itself that is not surjective.


问:
Proving that $a_kA^k+a_{k-1}A^{k-1}+\cdots +a_1A+a_0I = 0_n$
I've been studying linear algebra and came across this question that I have no general idea how to solve. The question is as follows.

Let $A$ be an $n×n$ diagonalizable matrix with distinct eigenvalues $λ_1,λ_2,λ_3$. Suppose $p(λ)=a_kλ^k+a_{k−1}λ^{k−1}+\cdots+a_1λ+a_0$  is a polynomial such that $p(λ_j)=0$ for each $j=1,2,3$. Prove that:
$$a_kA^k+a_{k-1}A^{k-1}+\cdots +a_1A+a_0I = 0_n$$
where $0_n$ denotes the $n×n$ zero matrix.

Any sort of guidance on how to solve this problem would be greatly appreciated

答:
Since $p(x)$ has at least three roots $\lambda_1,\lambda_2,\lambda_3$, it may be written as $$p(x) = (x-\lambda_1)(x-\lambda_2)(x-\lambda_3)q(x)$$ where $q(x)$ is a polynomial with degree $3$ less than that of $p(x)$.

$A$ is diagonalizable, so its minimal polynomial splits into distinct linear factors, the roots of which are the eigenvalues of $A$. So the minimal polynomial $m_A(x) = (x-\lambda_1)(x-\lambda_2)(x-\lambda_3)$

Clearly, $m_A(x)$ divides $p(x)$. Moreover, $$p(x) = m_A(x)q(x)$$

We know that $m_A(A) = 0_n$, i.e. the zero matrix. This helps us conclude $p(A) = 0_n$ from the above relation!


问:
Brouwer degree is independent choice of regular value
I was reading Milnor's differential topology book in page 27,there is a statement :

Let $f:X\to Y$ be smooth map with $X$ and $Y$ be same dimension,such
that both of them are oriented , we further assume $X$ is compact and
$Y$ is connected.

denote the Brouwe degree by $$\text{deg}(f)_y = \sum_{x \in f^{-1}(y)}
 \text{sgn}(\det(df_x))$$ prove it does not depend on the regular value
$y$ we choose.in other words deg as function of y is constant.

The idea is :to prove that deg is locally constant,then by connectess
it's contant everywhere

since we know for the setting above,any regular value $y$.in fact exist a pile of neiborhood $V_1,V_2,...V_n \subset M$ such that diffeomorphic to neiborhood of $y$， that is exist local diffeomorphism $V_i\to U_y$,as y varies in $U_y$ will not change the number of term in the sum

the question is why $ \text{sgn}(\det(df_x))$ will also not change as y varies in $U_y$,the intuition is easy that is as $y$ change a little bit,$x$ will change a liile bit,hence det will not change too much to filp the sign How to prove it ?

答: It's far simpler. If $f$ maps each $V_i$ diffeomorphically to $U$, then $df_x$ is an isomorphism from $T_xV_i$ to $T_{f(x)}U$ for every $x\in V_i$. In particular, $df_x$ is either orientation-preserving for all $x\in V_i$ or orientation-reversing for all $x\in V_i$. (If you choose a parametrizations/charts, then you can look at the square matrices, and their determinants vary continuously; one cannot change from positive determinant to negative determinant without passing through $0$, but this would violate the fact that the matrices are all non-singular.)


问:
Deduce that $X$ has normal distribution with mean $0$ and variance $1$
I've asked before about a Problem from Grimmet$Welsh (and I thank a lot to @angryavian and @Graham Kemp):

"If $X + Y$ and $X - Y$ are independent, show that
\begin{align}
M\left(2t\right) = M\left(t\right)^{3}M\left(-t\right),
\end{align}
where $X,Y$ are independet r.v. with mean $0$, variance $1$ and $M(t)$ finite."

This is it's link: Moment generating function applied in $2t$.

But now there's the "second" part of the problem: To show that $X$ (and $Y$) are a r.v. with normal distribution with mean $0$ and variance $1$.

The book itself sugest to define a function $\psi(t) = \frac{M(t)}{M(-t)}$ and show that $\psi(t)=\psi(2^{-n}t)^{2n}$. Then, show that $\psi(t) = 1 + o(t^{2})$ as $t \to 0$ and $\psi(1)=1$ when $n \to 0$. This will allow us to conclude that $M(t)=M(-t)$ and, when we apply this to the main equation (the one in the link and above) we get $M(t)=M(\frac{1}{2}t)^{4}$. The book then says to repeat the precedure to obtain the desired result. So, I have some questions:

How to show that  $\psi(t)=\psi(2^{-n}t)^{2n}$ ?

What does that "o" in $\psi(t) = 1 + o(t^{2})$ means? (I can't remember of seeing this through the Chapter)

What is the procedure to repeat in order to obtain the desired result? The entire one? The last part?

Of course, if somebody knows some another way to prove this statement, I'll be very aprecciated! Thanks in advance for your help!

答:
Hints:

Using the first part of the question, $\psi(t) = \frac{M(t)}{M(-t)} = \frac{M(t/2)^3 M(-t/2)}{M(-t/2)^3 M(t/2)}$. Do some more work to show $\psi(t) = \psi(t/2)^2$.

$\psi(t) = \psi(t/2)^2$ implies the more general equality $\psi(t) = \psi(t/2^n)^{2n}$.

By a Taylor expansion of $\psi$, we have $\psi(t) = \psi(0) + \psi'(0) t + \frac{1}{2} \psi''(0) t^2 + \frac{1}{6} \psi'''(\xi)t^3$ for some $\xi$ between $0$ and $t$. We know $\psi(0)=1$. We have
$$\psi'(t) = \frac{M'(t)M(-t) + M(t)M'(-t)}{M(-t)^2}$$
so $\psi'(0)=0$ (because $M'(t)=E[X]=0$). We also have
$$\psi''(t) = \frac{d}{dt}\frac{M'(t)M(-t) + M(t)M'(-t)}{M(-t)^2}
= \frac{[M''(t)M(-t) - M(t) M''(-t)]M(-t)^2
+ [M'(t)M(-t)+M(t)M'(-t)] 2 M(-t) M'(-t)}{M(-t)^4}$$
so $\psi''(0)=0$ (since $M''(0)=E[X^2]=1$). So the Taylor expansion becomes
$$\psi(t) = 1 + \frac{1}{6} \psi'''(\xi) t^3.$$
If you show $\psi'''$ is bounded by some constant $C$ for $t$ near zero (I can't think of a simple way to show this, and I am assuming third moments exist... maybe someone else can clean up my mess here), then we have
$$\lim_{t \to 0} \frac{|\psi(t) - 1|}{t^2} = \lim_{t \to 0} \frac{C}{6} |t| \to 0$$ which is the definition of $\psi(t)=1+o(t^2)$.

$\psi(1) = \lim_{n \to \infty} \psi(2^{-n})^{2n} = \lim_{n \to \infty} (1 + o(2^{-2n}))^{2n} = 1$ (I am skipping steps here)

$\psi(1)=1$ implies $M(t)=M(-t)$

$M(2t) = M(t)^3 M(-t) = M(t)^4$

$M(t) = M(t/2)^4$

$M(t) = M(t/2^n)^{4n}$

some more work needs to be done to deduce that $M(t)=e^{-t^2/2}$ is the only possible candidate MGF that satisfies the above recurrence.


问:
In acute $\triangle ABC$, $D$ is midpoint of $BC$, $DE \perp AB$, $DF \perp AC$, and $CE = BF$. Prove "geometrically" that $AB = AC$.
Really seemingly simple geometry problem:

In acute $\triangle ABC$, $D$ is midpoint of $BC$. $DE \perp AB$, $DF \perp AC$. $CE = BF$. Prove that $AB = AC$

I am looking for a more "geometrical" solution. We can apply cosine theorem on $\triangle BCF$ and $\triangle BCE$ to prove that $\cos B = \cos C$. But the computation is annoying. Is there a more pretty solution?

答:
Let $B'$ be the projection of $B$ on $AC$. Then $F$ is the mid point of $B'C$, and we have:
$$
BF^2 =B'B^2 +B'F^2=4FD^2+FC^2 =3FD^2 + DC^2\ .$$
There is a similar relation for $CE^2$. From $DB=DC$ we obtain $FD=ED$. Then $\Delta FDC=\Delta EDB$, giving $\hat B=\hat C$ in $\Delta ABC$.


问:
Proving existence of a square-free sequence
I found this problem and a solution sketch in a MathOverflow answer, and I thought it was nice enough to deserve more attention and a properly written solution.

Problem:

Prove that for each natural number $n$, there is some natural number $r$ for which the $n$ integers $r+1^2,r+2^2,…r+n^2$ are all square-free.

答:
Let $p_1, p_2, \cdots , p_k$ be all the prime numbers less than $n^2$ (for some $k$).  For any $K > k$, let $p_{k+1}, p_{k+2}, \cdots , p_K$ be the next prime numbers after that (larger than $n^2$).

For $1 \le i \le k$, there is at least one value of $r$ $\pmod {p_i^2}$ that satisfies $r + 1^2, r+2^2, \cdots , r+n^2 \not \equiv 0$ (i.e. there is at least one viable possible remainder for $r$.)  To see this, note that the squares modulo $p_i^2$ do not include $p_i$ itself, thus setting $r \equiv -p_i$ is sufficient.

For $k+1 \le i \le K$, the numbers $1^2, 2^2, \cdots , n^2$ are all distinct $\pmod {p_i}$, because $p_i > p_{k+1} > n^2$.  Therefore, there are exactly $(p_i^2 - n)$ viable possibilities for $r \pmod {p_i}$.

Now let $N = p_1^2p_2^2p_3^2\cdots p_K^2$.
By the Chinese remainder theorem, there are at least
$$(1)(1) \cdots (1)(p_{k+1}^2 - n)(p_{k+2}^2 - n)\cdots (p_{K}^2 - n)$$

viable options for $r$ modulo $N$ (by which we mean values of $r$ modulo $N$ that cause $r + 1^2, r + 2^2, \cdots r+n^2$ to be free from squares of all the primes $p_1, \cdots ,p_K$).
For any $K$, we define
$$
x_K := \frac{\text{Number of viable options for } r \text{ mod } N}{N}
$$
which evaluates to
$$
\frac{(p_{k+1}^2 - n)(p_{k+2}^2 - n)\cdots (p_{K}^2 - n)}{p_1^2p_2^2p_3^2\cdots p_K^2}
=
\frac{1}{p_1}\frac{1}{p_2} \cdots \frac{1}{p_k} \left(1 - \frac{n}{p_{k+1}^2}\right)\cdots \left(1 - \frac{n}{p_K^2}\right)
$$

The infinite product $\lim_{K \to \infty} \left(x_K \right)$ converges to a positive number, and the viable options for $r$ modulo $N$ for the first $K+1$ primes are a subset of the viable options for $r$ modulo $N$ for the first $K$ primes for any $K$.
Therefore, there are infinitely many values of $r$ which work for every $K$.
In particular, there is at least one positive integer $r$ such that $r + 1^2, r+2^2, \cdots r+n^2$ are squarefree.


问:
The value of $\lim_{n\rightarrow\infty}[(n+1)\int_{0}^{1}x^{n}$ $\ln(1+x)$ $dx]$
I evaluated it as
$$\lim_{n \to \infty}\left[\left.x^{n + 1}\,
\ln\left(1 + x\right)\,\right\vert_{0}^{1} -
\int_{0}^{1}x^{n + 1}\,\,\left(1 + x\right)^{-1}\,\mathrm{d}x\right],
$$
which comes as
$$
\ln\left(2\right) -
\lim_{n \to \infty}\,\int_{0}^{1}x^{n + 1}
\left(1 - x + x^{2} - x^{3}\cdots\right)\mathrm{d}x = \ln\left(2\right) - 0\ ?
$$
How to evaluate
$\displaystyle\lim_{n \to \infty}\,\,
\int_{0}^{1}{x^{n + 1} \over 1 + x}\,\mathrm{d}x\ ?$.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\lim_{n \to \infty}\bracks{%
\pars{n + 1}\int_{0}^{1}x^{n}\,\ln\pars{1 + x}\,\dd x}}
\\[5mm] \stackrel{x\ \mapsto\ 1 - x}{=}\,\,\,&\
\lim_{n \to \infty}\bracks{%
\pars{n + 1}\int_{0}^{1}\pars{1 - x}^{n}\,\ln\pars{2 - x}\,\dd x}
\\[5mm] = &\
\lim_{n \to \infty}\bracks{%
\pars{n + 1}\int_{0}^{1}\expo{n\ln\pars{1 - x}}\,\,
\ln\pars{2 - x}\,\dd x}
\\[5mm] = &\
\lim_{n \to \infty}\bracks{%
\pars{n + 1}\int_{0}^{\infty}\expo{-nx}\,\,
\ln\pars{2 - 0}\,\dd x}
\\[5mm] = &\
\ln\pars{2}\lim_{n \to \infty}\pars{n + 1 \over n} =
\bbx{\ln\pars{2}} \approx 0.6931\\ &
\end{align}


问:
Use two different ways to calculate OR exlusive permutations
I am asked to solve the following question:

A license plate that is composed of exactly 3 letters and 3 identical. Order matters. So how many arrangements are there that satisfies: 3 letters are identical OR 3 digits are identical

Once I get this question, I thought of the inclusion-exclusion formula. So I calculate first cases where 3 letters are identical plus cases where 3 digits are identical minus cases where both 3 letters and 3 digits are identical:

$$\text{Case where 3 letters are identical}: 10^3*26*C(6,3) = 520000$$
$$\text{Case where 3 digits are identical}: 10*26^3*C(6,3) = 3515200 $$

$$\text{Case where both digits and letters are identical}: 10*26*C(6,3) = 5200 $$

$$\textbf{In total we have}: 520000+3515200-5200 = 4030000 $$

In order to verify my answer, I decide to use another way to calculate this by using a universal set minus the set where all letters and digits are not identical:

$$\text{Without restrictions}: 10^3*26^3*C(6,3) = 351520000$$
$$\text{Case where none of the digits or letters are identical}: 10*9*8*26*25*24*C(6,3) = 224640000 $$

$$\textbf{In total we have}: 351520000-224640000 = 12688000 $$

So the results are not equal. However, I have been using these two methods to verify my answers many times already. It has always worked. So I am not sure where I did wrong.

I even tried to try a smaller problem (where I use only two letters and two numbers so I can actually count one by one), and in that situation, both ways of solving this problem gave me the same result, as it should have.

I am totally lost!

答: In your subtraction, you missed the cases where two of the digits are identical or where two of the letters are identical.</p>


问:
Why doesn't $f$ being Lebesgue integrable imply that the Hardy-Littlewood function is integrable?
Let $f$ be an integrable function on the reals. Define the Hardy-Littlewood function as $$f^{\ast}(x) = \sup\limits_{x \in B} \frac{1}{m(B)}\int_{B}|f(y)|dy,$$
where the supremum is taken over all balls $B$ containing $x.$ Let $$g(x) = \lim_{m(B) \rightarrow 0} \frac{1}{m(B)}\int _{B} f(y)dy,$$ where $B$ contains $x$.

Now here is the question. Suppose $f$ is a nonnegative function. Is $f^{\ast} = g$ almost everywhere? If true then we must have the maximal function is integrable because the differentiation theorem implies $g(x) = f(x)$ almost everywhere. So if $f$ is Lebesgue integrable, $g$ is integrable hence the maximal function is integrable.

答:
No. By Lebesgue differentiation theorem, $g=f$ almost everywhere. Now let $f(x)=x^2 \mathbf 1_{[-10,10]}\in L^1$. $f\ge 0$. Then, for all $x\in[-9,9]$, a lower bound for $f^*$ is given by $\frac12\int_{x-1}^{x+1} y^2 dy = x^2 + \frac13$. In particular, $f^*>f=g$, at least near $x=0$. Here's a Desmos graph to verify. In fact $f^*(0)=100/3$, which is much larger than $f(0)=0$. You can similarly show that $f^*(x)>f(x)$ for all  $x\in(-10,10)$.

In addition, for any $x$ with $|x|>10$, there will always be some (possibly large) ball around $x$ that intersects $[-10,10]$. This implies $f^*(x)>0=f(x)$. In conclusion, $f^*>f$ almost everywhere.

One can make this last part for $|x|>10$ more precise (and this is close to what you might read in a book)： the ball of radius $|x|+10$ around $x$ completely contains $[-10,10]$. Consequently,
$$f^*(x)>\frac1{2|x|+20}\int_{-10}^{10} y^2 dy \ge \frac{C}{1+|x|}.$$ This proves that $f^*$ is not integrable at infinity.


问:
Convergence of $\int_{x=0}^\infty x^8 e^{-\sqrt x} dx$

Does $\int_{x=0}^\infty x^8 e^{-\sqrt x} dx$ converge?

I've tried splitting it into $\int_{x=0}^1 x^8 e^{-\sqrt x} dx$ + $\int_{x=1}^\infty x^8 e^{-\sqrt x} dx$. The first one is a continuous function over a bounded interval so that converges however I'm struggling with the second one. I have tried using limit comparison with $e^{- \frac{\sqrt x}{2}}$ as a test function but ended up going in circles. I have also tried direct comparison and integration by parts.

Thanks

Morgzy

答:
For the second one you can use direct comparison as follows:

Note that for $x>0$ you have

$$e^{-\sqrt x} = \frac 1{e^{\sqrt x}} = \frac 1{\sum_{n=0}^{\infty}\frac{x^{\frac n2}}{n!}} \stackrel{n=20}{<} \frac 1{\frac{x^{10}}{20!}}$$

Hence,

$$\int_{x=1}^\infty x^8 e^{-\sqrt x} dx< 20! \int_{x=1}^\infty \frac 1{x^2} dx < \infty$$


问:
About a theorem that appears in Mrówka's article.
I was reading and article of S. Mrówka and I found a theorem that sounds good but, unfortunately, I can't figure out the proof. The theorem is 1.3 in the article:

1.3 Theorem (The formal theorem) If $h$ is a homeomorphism of $X$ into $Y$, then there exist a superspace $X^{*}$ (i.e., an space that contains $X$ as a subspace) of $X$ which is homeomorphic to $Y$ by a homeomorphism $h^{*}$ wich is an extension of $h$.

I don't know how to prove it and, moreover, I can't find the proof in another articles. In fact, in the article doesn't exist a reference about such theorem.

I though in take $h$ as the identity function because, roughly speaking, we can consider $X$ as a subspace of $Y$ and trivially $h$ extends to $h^{*}=\text{id}:Y\to Y$ and $h^{*}$ is an homeomorphism. I know that this isn't the general case, but, in the practice, an extension $T$ of a space $X$ can always viewed such that $X\subseteq T$. Am I wrong? How can I take the general case? Thanks for your attention.

答:
Suppose that $h$ is a homeomorphism of $X$ into $Y$. Let

$$X^*=X\cup\big(\{X\}\times(Y\setminus h[X])\big)\,,$$

and define

$$h^*:X^*\to Y:z\mapsto\begin{cases}
h(z),&\text{if }z\in X\\
z,&\text{otherwise;}
\end{cases}$$

clearly $h^*$ is a bijection, and $h^*\upharpoonright X=h$. We define the topology $\tau$ on $X^*$ so as to make $h^*$ a homeomorphism:

$$\tau=\left\{(h^*)^{-1}[U]:U\text{ is open in }Y\right\}\,.$$

It only remains to check that the topology that $X$ inherits from $\tau$ is its original topology, which is straightforward.


问:
Can you solve for x for the equation $25+2^{\log_{10} x}=x$?

Solve for $x$ : $$25+2^{\log_{10} x}=x$$

My work

Well, I could not figure out an algebraic solution to this problem.
$$25+2^{\log_{10} x}=x \implies 5^2+x^{\log_{10} 2}=x$$
$$\implies x^{\log_{10}2}-x-25=0$$
which does not seem to be solved further.

I have solved this by using the graphical method by plotting both sides of this equation. And the answer comes near to  $27.7$. I have also verified it by using the desmos graph calculator according to which the answer is $27.718$.

How can I solve this question by the algebraic method?

答:
Consider that you look for the zero of function
$$f(x)=25+2^{\log_{10}( x)}-x$$and use inspection. You have $f(10)=17$ and $f(100)=-71$. Compute the equation of the straight line going through the two points. It is
$$y=\frac{241}{9}-\frac{44 x}{45} \implies x_0=\frac{1205}{44}\approx 27.3864$$ and
$$f\left(\frac{1205}{44}\right)\approx 0.322212$$ We are so close to the solution that any iterative method would converge very fast. Below are some numbers with a ridiculous number of figures starting with $x_0=\frac{1205}{44}$ and using Newton method

$$\left(
\begin{array}{cc}
n & x_n \\
 0 & \color{red}{27.}386363636363636364 \\
 1 & \color{red}{27.7184}63076353301887 \\
 2 & \color{red}{27.71842019257}5559688 \\
 3 & \color{red}{27.718420192574854316}
\end{array}
\right)$$


问:
suppose K is an extension field of finite degree, and L,H are middle fields such that L(H)＝K.Prove that [K:L]≤[H:F]
Suppose $K$ is an extension field of finite degree, and $L,H$ are middle fields such that $L(H)＝K$.Prove that $[K:L]≤[H:F]$

Well, I just have no idea about how to use the condition that $L(H)＝K$. Please give me some help, thanks!

答:
Let $F$ be the base field. Since $[K:F]<\infty$, then $K$ is algebraic over $F$. Similarly by the tower theorem, $H$ is algebraic over $F$ and $L$ is algebraic over $F$. This means that $H=F(\beta_{1},\dots,\beta_{m})$ for some $\beta_{1},\dots,\beta_{m}$ algebraic over $F$, so that $[H:F]=m$, and $L=F(\alpha_{1},\dots,\alpha_{n})$ for some $\alpha_{1},\dots,\alpha_{n}$ algebraic over $F$, so that $[L:F]=n$. Since $L(H)=K$, then
\begin{equation*}
K=L(\beta_{1},\dots,\beta_{m})=F\left(\alpha_{1},\dots,\alpha_{n},\beta_{1},\dots,\beta_{m}\right)
\end{equation*}
so that $[K:F]\leq m+n$, and $[K:L]\leq m=[H:F]$ (draw a diagram to help you visualise this).


问:
Probability of determinants being coprime
I have a question that is not of particular significance, but I would love to understand the underlying principles.

Suppose we have two square $3 \times 3$ matrices, $M_1$ and $M_2$ with

$$M_1 = \begin{pmatrix} a_1 & a_2 & a_3 \\ a_4 & a_5 & a_6 \\ a_7 & a_8 & a_9 \end{pmatrix} \qquad\text{and}\qquad
M_2 = \begin{pmatrix} b_1 & b_2 & b_3 \\ b_4 & b_5 & b_6 \\ b_7 & b_8 & b_9 \end{pmatrix}$$

with the coefficients $a_n,b_n \in \mathbb{Z}$ and $1 \leq a_n,b_n \leq 9$

What is the probability that the matrices' determinants are coprime, when uniformly random coefficients satisfying the conditions are chosen.

I am familiar with the Riemann's $\zeta$ function way to find out the probability of two random integers being coprime, but I have no clue how to apply that here with additional conditions on the numbers given.

I did test it mechanically, using Mathematica and the result is around 30%, but I would like to see a proper way to do it.

I would love to at least get a few pointers as what to research to tackle this problem.

Thank you very much!

答:
This answer was obtained numerical, but it is nevertheless exact

There are $N=9^9$ possibilities to choose the matrix $m$ (each with equal probability). There are 1913 possible outcomes for $\det m$. The most likely is $\det m =0$. The probability for this event is $$P[\det m = 0] = \frac{218605}{14348907} \approx 1.5\%.$$ The rest is distributed over different integer number. The largest determinant is a matrix with $\det m = 1216$. There are 3 possibilities for this matrix (therefore the probability is $P[\det m = 1216] = 3/N$). There are also 3 matrices for which $\det m=-1216$. (in fact this is always true. If there are $n$ matrices with $\det m= x$ then there are exactly $n$ matrices with $\det m=-x$). To find out if the determinants are coprime the sign does not matter, therefore we only need to know the probabilities $$P[|\det m| = x].$$ Generically, the probability decreases for increasing $\det m$.

The absolute value of the determinant may assume 957 values. Therefore, there are $957*(957+1)/2 =458403$ pairs of matrices $m_1$ and $m_2$. Out of them there are 274487 pairs coprime. (The fraction of matrices which are coprime is $274487/458403 \approx 60\%$)

Counting each of the pair of matrices $m_1, m_2$ with the proper probability, we obtain the probability that two matrices are coprime
$$P[ m_1 \text{ and } m_2 \text{ are coprime}]= \frac{7253958722902984}{16677181699666569} \approx 43\%.$$


问:
Questions about projective modules.
Let $P$ be a projective module and $M$ a submodule of $P$. We know that $M$ is also a projective module. Can we conclude that $P=M\oplus N$ for some module $N$? Thank you very much.</p>

答:
As pointed out in the comments that submodules of projective modules are projective does not always hold. If this holds the algebra is called hereditary. An equivalent description of that is that $\operatorname{gldim} A \leq 1$.

But even if it holds, this does not guarantee that submodules of projectives are direct summands. For example take the hereditary path algebra of the quiver $1\to 2$. The projective module to the vertex $1$ has the composition series $0\subset S_2\subset P_1$. The simple module $S_2$ is also projective but the module $P_1$ is indecomposable, hence there is no complement.

Title: Questions about projective modules.


问:
Ceil function proof
I'm sorry if this is a repeated question but I can't find it anywhere. Any help is highly appreciated
How to prove that for any +ve real number

```<code>ceil (a*b) <= ceil(a)*ceil(b)
</code>```

Thankyou very much

答:
By definition of ceil and since $a,b$ are positive, $ab \le \lceil a \rceil \lceil b \rceil$.

Now $\lceil a \rceil \lceil b \rceil$ is an integer greater than or equal to $ab$, while $\lceil ab \rceil$ is the smallest such integer.

Hence $\lceil ab \rceil \le \lceil a \rceil \lceil b \rceil$.

Title: Ceil function proof


问:
Smallest topology on $\mathbb{C}$ such that all singletons are closed
This question was part of an exam for which I am preparing.

Question was related to finding smallest topology T on $\mathbb{C}$ in which all singletons are closed.
and then findind which options are true amog these:

A ($\mathbb{C}$,T) is haursdorff.

B ($\mathbb{C}$,T) is compact.

C ($\mathbb{C}$,T) is connected.

D  $\mathbb{Z}$ is dense in ($\mathbb{C}$,T) .

But  I am just a beginner in topology and self studying from Wayne patty . I was unable to find what topology should be . So, can you please tell what topology should be . rest I would prefer to work by mysely.

Thank you.

答:
I will answer the questions A and D, as you asked in the comments.

A is false because if we have two points and any the neighborhoods of them, say $U$ and $V$, they must satisfy $|\mathbb C\setminus U|,|\mathbb C\setminus V|<\infty$ and their intersection, $U \cap V$, is an open subset. If $U\cap V$ is empty, then its complement is union of finite sets, that is $\mathbb C= \mathbb C\setminus (U\cap V)=(\mathbb C \setminus U)\cap (\mathbb C\setminus V)$ is finite, an absurd!

For proving D, we take any non empty, open subset $U \subseteq \mathbb C$, we have $|\mathbb C\setminus U|<\infty$. If $\mathbb Z \cap U =\emptyset$, then $\mathbb Z \subseteq \mathbb C\setminus U$. That is, $\mathbb Z$ is finite, another absurd! So, $\mathbb Z \cap U \neq \emptyset$. Therefore, the integers are dense in $\mathbb C$ with that topology (in fact, any infinite set).


问:
Multiplication operator with fixed continuous function
Let $u\in C[0,1]$ fixed. The multiplication operator  $M_{u}:L^{2}[0,1]\to L^{2}[0,1]$ defined by $u$ is given by
$$
  M_{u}(f)=u(t)f(t)
$$
I want to show that if $M_{u}$ is compact then $u=0$, the zero function. I know I have to proceed by contradiction, nonetheless, I can't see which particular bounded sequence applied to $M_{u}$ contradicts $u\neq 0$. I've seen many related questions but I ask this question because this one is different from the others since the fixed function is on $C[0,1]$ and the arguments are taken on $L^{2}[0,1]$, so I'm not sure if I can proceed the same way.

答:
A very elementary approach exploits the fact that $L^2([0,1])$ has an orthonormal basis $\{e^{int}\}_{n\in \mathbb Z}$.

So, if we write $u(t)=\sum a_ne^{int},$ then, defining \begin{align} g_k(t):=\sum a_ne^{i(n+k)t}\quad \text{gives}\quad g_k(t)-g_{k+m}(t)=\sum a_ne^{i(n+k)t}-\sum a_ne^{i(n+k+m)t}=\sum a_ne^{i(n+k)t}(1-e^{imt})=(1-e^{imt})e^{ikt}\sum a_ne^{int}=(1-e^{imt})e^{ikt}u(t).\end{align}

If $u\neq 0$ then without loss of generality, $u>\delta>0$ on some $[a,b]\subseteq [0,1].$ Then,

\begin{align} |g_k-g_{k+m}|^2\ge \delta^2\int_a^b|(1-e^{imt})e^{ikt}|^2dt=\delta^2\int_a^b|(1-e^{imt})|^2dt=\delta^2\int_a^b(2-2\cos mt)dt=2\delta^2(b-a-\frac{1}{m}(\sin mb-\sin ma))\end{align}

from which we conclude that no subsequence of $(g_k)$ is convergent.


问:
Probability that $3$ hands of a clock in the same semi-circle?
I am wondering how to calculate the probability that $3$ hands of a clock are in the same semi-circle ? I know a similar question is that if we randomly choose n points in a circle, then the probability that all of them in the same semi-circle will be $n/2^{(n-1)}$ with $n = 3$.

But when it comes to the dial hands, where I guess there are correlations between them, I am not sure whether my question is equivalent to the previous one.
If they are not equivalent, and this question will be hard to calculate, then whether we can determinate this probability will be greater or smaller than $3/4$?

Anyone can help? Thanks.

答:
Take two of the hands. On average they will be a quarter of a circle apart. Distance varies uniformly from zero to half a circle apart.
Turn the dial so that one hand is at 12 and the other hand at 3.
Now when the third hand is in the interval from 9 to 6 all three hands are on the same semicircle.
When the third hand is between 6 and 9 they are not on the same semicircle.
So on average the three hands will be on the same semicircle 3/4 of the time.</p>


问:
Why is the Cantor set uncountable
I am having trouble seeing why Cantor set has uncountably many elements.

A cantor set $C$ is closed. So $[0,1] - C = \bigcup\limits_{n=1}^{\infty} I_n$ is open and is countable union of disjoint open intervals. I can further assume that I can order the $\{I_n\}$ by their left endpoints since there are only countably many of them. So between $I_n=(a_n,b_n)$ and $I_{n+1} = (a_{n+1},b_{n+1})$, we must have $a_n < b_n \leq a_{n+1} < b_{n+1}$. If $b_n < a_{n+1}$, then the Cantor set $C$ consists of an interval, which is a contradiction, so $b_n = a_{n+1}$ for all $n$, and thus Cantor set can have at most countably many points.

答:
I can further assume that I can order the $\{I_n\}$ by their left endpoints since there are only countably many of them.

By this logic, it should also be possible to enumerate the rational numbers in order. But that's absurd.

Title: Why is the Cantor set uncountable


问:
Sequence of Functions Converging to 0
I encountered this question in a textbook. While I understand the intuition behind it I am not sure how to formally prove it.

Define the sequence of functions $(g_n)$ on $[0,1]$ to be $$g_{k,n}(x) = \begin{cases}1 & x \in \left[\dfrac{k}n, \dfrac{k+1}n\right]\\ 0 & \text{ else}\end{cases}$$
where $k \in \{0,1,2,\ldots,n-1\}$.

Prove the following statements:

1) $g_n \to 0$ with respect to the $L^2$ norm.

2) $g_n(x)$ doesn't converge to $0$ at any point in $[0,1]$.

3) There is a subsequence of $(g_n)$ that converges pointwise to $0$.

Thank you for your replies.

答:
1) What is $\lVert g_n\rVert_2$?

2) Show that for any $x$, $g_n(x)=0$ infinitely often and  $g_n(x)=1$ infinitely often.

3) Note that $\frac1n\to 0 $ and $\frac2n\to 0$

Title: Sequence of Functions Converging to 0


答:
The error in your reasoning is the assumption that a countable set of numbers can be ordered. For example, consider the set of rational numbers, countable, but can't be ordered ('ordering' here means enumerating in a sequence such that $\alpha_1<\alpha_2<\dots$).

A simple way to see that the cantor set is uncountable is to observe that all numbers between $0$ and $1$ with ternary expansion consisting of only $0$ and $2$ are part of cantor set. Since there are uncountably many such sequences, so cantor set is uncountable.

Title: Why is the Cantor set uncountable

Body:
I am having trouble seeing why Cantor set has uncountably many elements.

A cantor set $C$ is closed. So $[0,1] - C = \bigcup\limits_{n=1}^{\infty} I_n$ is open and is countable union of disjoint open intervals. I can further assume that I can order the $\{I_n\}$ by their left endpoints since there are only countably many of them. So between $I_n=(a_n,b_n)$ and $I_{n+1} = (a_{n+1},b_{n+1})$, we must have $a_n < b_n \leq a_{n+1} < b_{n+1}$. If $b_n < a_{n+1}$, then the Cantor set $C$ consists of an interval, which is a contradiction, so $b_n = a_{n+1}$ for all $n$, and thus Cantor set can have at most countably many points.


答:
I'm not following your argument well enough to see exactly where it goes wrong... One question you might ask yourself is "does this show that every closed set is countable?" What is special about the cantor set here? I'm not seeing it.

As for why the cantor set is uncountable, consider this:

At each finite level of the cantor set construction, we "throw out" the middle third of each piece. So we have a decision to make at each stage: Do we go left? or do we go right?

E.g., We start in $[0,1]$. Then We have to decide to go into $[0,\frac{1}{3}]$ or into $[\frac{2}{3},1]$. Let's say we go left. Now we have the choice of going into $[0,\frac{1}{9}]$ or $[\frac{2}{9},\frac{1}{3}]$.

You can see that every countable sequence of choices (left or right) gives a unique point of the cantor set. Moreover, every point of the cantor set corresponds to such a sequence of choices. So if we write $0$ for "left" and $1$ for "right, the points of cantor set are in bijection with the infinite strings of $0$s and $1$s.

As a fun aside, the topological structure actually agrees as well! That's why you'll often see people call the cantor set $2^\omega$. In set theoretic language, that basically translates to "infinite sequences of $0$s and $1$s".

Ok, but now there must be uncountably many infinite sequences of $0$s and $1$s by a diagonalization argument. So the cantor set is uncountable too.

I hope this helps ^_^

Title: Why is the Cantor set uncountable


答:
So first of all from the definition like ceil is the lower bound of set of integers such that ceil(x) is greater than or equal to x

so its valid for all real numbers x

so from the definition ceil(a) ≥ a this also implies ceil(b)  ≥ b

so here you said that it is a,b are positive integers so

obviously a*b is an integer this mean ceil(ab)=ab=ceil(a)*ceil(b)

so now lets suppose a,b are two positive reals then let a=ceil(a)-e1
and b=ceil(b)-e2 where e1,e2 are non negative reals less than 1

a*b is obviously less than or equal to ceil(a)*ceil(b) because a <= ceil(a) and b<=ceil(b)

a*b<=ceil(a) * ceil(b)

applying ceil on both sides we get

ceil(a*b) <= ceil( ceil(a)*ceil(b) )

since ceil(a)*ceil(b) is integer ceil( ceil(a)*ceil(b) ) = ceil(a)*ceil(b)

this implies  ceil(a*b) <= ceil(a) * ceil(b)

Title: Ceil function proof

Body:
I'm sorry if this is a repeated question but I can't find it anywhere. Any help is highly appreciated
How to prove that for any +ve real number

```<code>ceil (a*b) <= ceil(a)*ceil(b)
</code>```

Thankyou very much


答:
I can further assume that I can order the $\{I_n\}$ by their left endpoints since there are only countably many of them.

No. Why would you think you can? Consider for instance the countably many numbers
$$
\bigl\{\tfrac1n:\ n\in\mathbb N\bigr\}\cup\bigl\{\tfrac12-\tfrac1n:\ n\in\mathbb N\bigr\}.
$$
As long as there is more than one accumulation point, you cannot expect to order them indexed by integers.

Title: Why is the Cantor set uncountable


问:
A function of a matrix is a projection operator

Let $P_1, P_2, \cdots, P_k$ be projection operators satisfying $\sum_{i=1}^{k} P_i = I$.
The function $\Phi(X)$ is defined as follows:
$$ \Phi(X) = \sum_{i=1}^{k} P_i X P_i$$
Show that $\Phi^2 = \Phi$.

My try: It can be shown that $P_iPj=0, i\neq j$, which gives:
$$\Phi^2(X) = \sum_{i,j}P_i X P_iP_j X P_j = \sum_{i}P_i X P_i X P_i$$
I have no idea on how to proceed further.

答:
Your computation of $\Phi^2(X)$ seems to be incorrect; informally, it shouldn't involve more than one $X$.

We should have

$$\Phi^2(X) = \Phi\left(\sum_{i = 1}^k P_i X P_i\right) = \sum_{j = 1}^k P_j \left(\sum_{i = 1}^k P_i X P_i\right) P_j .$$
Rearranging the sums and applying the definition of projection operator, together with the fact that you mentioned (that $P_i P_j = 0$ for $i \neq j$), yields the identity.


问:
On the consistency of first-order theories with an infinite sequence of proper axioms

Let $T$ be a first-order theory whose proper axioms are an infinite sequence $\alpha_1,\alpha_2,...$ of closed wffs such that $\vdash \alpha_n\to\alpha_m$ iff $n\geq m$. Prove that $T$ is consistent.

So I'm an absolute beginner when it comes to first-order theories, and I've only been introduced briefly to the same. I understand that if $T$ is consistent, then $T$ is negation consistent and absolutely consistent. I'm trying to cook up a proof by contradiction, taking the help of one of these two definitions of consistency.

Let's suppose that $T$ is not consistent, in particular, it is not negation consistent. So there exists a wff $\beta$ such that $T\vdash\beta$ and $T\vdash\lnot\beta$. Here, $T$ refers to the set of proper axioms of the first-order theory $T$, combined with the first-order axioms (seven in number as in Angelo Margaris including Ax-Gen).

What do I do next? I am trying to show that such a $\beta$ cannot exist, while miserably failing at it. How do I find the contradiction? How do I go ahead from here? Thank you!

答: If $T$ is inconsistent, then there is a finite subset that is inconsistent, which are amongst $\{\alpha_1,\alpha_2,\ldots,\alpha_n\}$ for some $n$. But these are all logically implied by $\alpha_n$, so just the single statement $\alpha_n$ is inconsistent. Since it is inconsistent, we can prove anything from it, including $\alpha_{n+1},$ thus we have $\alpha_{n}\vdash \alpha_{n+1}$ and hence $\vdash\alpha_n\to \alpha_{n+1},$ contrary to assumption.


答:
The sequence is given by
$$f_{k,n}(x) = \begin{cases}1 & x \in \left[\dfrac{k}n, \dfrac{k+1}n\right]\\ 0 & \text{ else}\end{cases}$$
where $k \in \{0,1,2,\ldots,n-1\}$.

$1$). Note that $$\int_0^1 \vert f_{k,n} \vert^2 d \mu = \mu\left( \left[\dfrac{k}n, \dfrac{k+1}n\right]\right) = \dfrac1n \to 0 \text{ as }n \to \infty$$

$2$). Consider any $x \in [0,1]$. For any $n \in \mathbb{N}$, we have $x \in \left[\dfrac{k}n, \dfrac{k+1}n \right]$ for some $k = k_x(n)$. (Why?)

$3$). Consider the subsequence $f_{m,n}(x)$, where $m$ is a fixed number in $\{1,2,\ldots,n-2\}$.

EDIT

Adding more details to $2$ and $3$.

$2$). Choose $k_x(n) = \left \lfloor nx\right \rfloor$. By definition, we have $\left \lfloor nx\right \rfloor \leq nx \leq \left \lfloor nx\right \rfloor+1$. Hence, we get that$$\dfrac{\left \lfloor nx\right \rfloor}n \leq x \leq \dfrac{\left \lfloor nx\right \rfloor+1}n$$Hence, for a fixed $x$, if we look at the subsequence, $h_n(x) = f_{k_n(x),n}(x)$, we have $h_n(x) = 1$ for all $n$. Hence, we have subsequence of $f_{k,n}(x)$ converging to $1$.

$3$). Consider the sequence $$h_n(x) = f_{1,n}(x) = \begin{cases}1 & x \in \left[\dfrac1n, \dfrac2n \right]\\ 0 & \text{else}\end{cases}$$ Note that clearly $h_n(0) = 0$, since $0 < \dfrac1n$ for all $n$. Now for any other $x$, note that there exists a $N$ such that $\dfrac2N < x$. Beyond this $N$, for the given $x$, we have $h_n(x) = 0$ for all $n \geq N$. Hence, for any fixed point $x$, we have $$\lim_{n \to \infty} h_n(x) = 0$$

Title: Sequence of Functions Converging to 0

Body:
I encountered this question in a textbook. While I understand the intuition behind it I am not sure how to formally prove it.

Define the sequence of functions $(g_n)$ on $[0,1]$ to be $$g_{k,n}(x) = \begin{cases}1 & x \in \left[\dfrac{k}n, \dfrac{k+1}n\right]\\ 0 & \text{ else}\end{cases}$$
where $k \in \{0,1,2,\ldots,n-1\}$.

Prove the following statements:

1) $g_n \to 0$ with respect to the $L^2$ norm.

2) $g_n(x)$ doesn't converge to $0$ at any point in $[0,1]$.

3) There is a subsequence of $(g_n)$ that converges pointwise to $0$.

Thank you for your replies.


问:
Helly's theorem.
Let F be a finite family of segments in R such that among any n of them there
are two intersecting. Prove that it is possible to divide F into n−1 families such that any two segments in one family are intersecting.</p>

答:
(Fill in the gaps as needed. If you're stuck, write out your working and thought process to demonstrate where you're at.)

Rephrase the statement to

Let F be a finite family of segments in R such that among any $n$ of them there are two intersecting. Prove that there are $n-1$ values, such that each segment contains at least 1 of these values.

You can do a proof by induction on $n$.

The case of $n=2$ is Helly's theorem (or you can prove it directly by considering the left most right endpoint).

Suppose the statement is true for some $k$. Consider $k+1$.

Given any $m \geq k+1$ segments $R_i = [a_i, b_i]$ that satisfy the condition that any $k+1$ segments have 2 segments what intersect.

WLOG, let $R_1$ be the segment with the left-most right endpoint (IE $b_1 = \min(b_i)$)

Any set which intersects $R_1$ must contain $b_1$.

Consider any $k$ sets which do not intersect $R_1$. By the assumption applied on these $k$ sets along with $R_1$, there exists 2 which intersect. $R_1$ cannot be involved, so it's 2 of these $k$ sets. Apply the induction hypothesis, and there are $k-1$ points such that these sets contain at least 1 of these values.
Take the union of these $k-1$ points along with $b_1$, and we are done.

Title: Helly's theorem.


问:
How to show that explicit generator of the fundamental group is the one I found?
I am working on an exercise:

Let $X$ be $\mathbb S^2$ with a straight line segment having the north pole and the south pole as its end points. Compute the fundamental group of $X$ (with the north pole as a base point) of $X$ and find generators.

For the first part, I was able to find that the fundamental group of $X$ is isomorphic to $\mathbb Z$ using Van Campen theorem, and intuitively, I think the generator of the group should be a path that starts from the north pole, goes down the straight line segment, and comes back to the north pole via the path in the sphere, as any path that does not go through the straight line segment would be null homotopic.

However, I am not sure how to rigorously justify that the path I mentioned above is the generator for the fundamental group.

P.S. Also, during my proof for the first part, there was a part where I needed to show that $X \setminus \{ (1, 0, 0) \} \simeq \mathbb S^1$, and I feel like I did not show it rigorously. Is there a way to show this rigorously as well using deformation retraction?

答:
This is too long to post as a comment, so I'm leaving it here instead.

I'm not sure how rigorous an argument you're looking for. Truly rigorous arguments in Algebraic Topology are really hard to write, and are also quite hard to find in the literature. (I have some opinions on this, which I'm putting at the end).

I'll say this, though: If I were grading an algebraic topology class, I would give full credit without hesitation if you said that

$X$ is homotopy equivalent to $S^2 \vee S^1$ (by sliding the south pole to the north pole, say)

Since $S^2$ is simply connected, we have (by Van Kampen) that $\pi_1X =  \pi_1 S^1 = \mathbb{Z}$. (Maybe with some extra justification here, depending on how recently the theorem was introduced)

The generator is the loop that we homotoped from the path connecting the poles in $X$. This is easy to see, because we know what the generator is $\pi_1 S^1$ is.

So we find the generator in $X$ is any old loop that goes through the segment you named (exactly once, of course).

If you're looking for a really rigorous justification, I'm not the right person to give it. Unfortunately, much of Algebraic Topology is hard to really make formal, because you have to make a bunch of choices that don't matter (like a particular path through the sphere connecting the poles to complete your loop). Once you start doing this, the field quickly starts losing its flavor, and a lot of what you're doing is an exercise in analysis instead... HoTT was introduced (in part) to solve these problems. In that setting you can be truly formal, while also reasoning in the way that we do as humans.

All this to say, don't worry too much about being extremely precise. You'll tear your hair out (speaking from experience...) If you really do care about the intricate formalities of what you're doing, then HoTT might be something to look into.

I hope this helps ^_^


问:
How to find the z component of the parameterization of an ellipse that is the intersection of a vertical cylinder and a plane
Given the vertical cylinder with radius 1 ($x^2 + y^2 = 1$) oriented with an outward pointing normal and its intersection with the plane $z = xtan(\phi)$

Consider the coordinates $\tilde{x} = \sqrt{x^2 + z^2}$ and $\tilde{y} = y$

Show that this intersection is an ellipse

Find the curvature, geodesic curvature, and normal curvature of this ellipse (I do not actually need help solving these, but need help to find the parameterization so I can start solving them)

So, I did the following

$\tilde{x} = \sqrt{x^2 + (x tan(\phi))^2} = \sqrt{x^2 + x^2*tan^2(\phi)} = \sqrt{x^2 (1 + tan(\phi))} = xsec(\phi)$

and of course $\tilde{y} = y$

and thus the intersction is $\tilde{x}^2 + \tilde{y}^2 = 1$

and to make it look like an ellipse you can do

$\frac{x^2}{cos^2(\phi)} + \frac{y^2}{1^2} = 1$

where $ a = cos(\phi)$ and $b = 1$

Now that I have found the ellipse, I need to work on finding the curvature, geodesic curvature, and the normal curvature. To do that the hint was given to parameterize the ellipse in terms of the roational angle $\theta$ around the cylinder. So, I looked online and I saw that a general paraneterization for an ellipse is

$x = acos(t)$ and $y = bsin(t)$

However, upon running this information by my instructor I was told that this is only the ellipse, if the ellipse was planar. Therefore my paramterization needs a z-component. I was told that the parameterization is usually something along the lines of

$(x, y, z) = (cos(\theta), sin(\theta), 0)$

But for this I need to rotate through phi in the xz plane to get my parameterization and I'm not exactly sure what to do/what that looks like. Any advice? Thanks!

答:
The surface of the cylinder can be parametrized as ..

$$(x, y, z) = (\cos\theta, \sin\theta, t)$$
intersect with the plane ...

$$ z = x \tan\phi \implies t=\cos\theta \tan \phi $$

so the intersection can be parametrized as ...

$$(x, y, z) = (\cos\theta, \sin\theta, \cos\theta \tan \phi)$$


问:
Constant derivative of a function implies is an affine map.
I am trying to prove the next:

Let $f :\Omega\subset\mathbb{R}^n\rightarrow\mathbb{R}^m$ be differentiable, where $\Omega$ is an open
connected subset of $\mathbb{R}^n$. Suppose $D_{f}(x)$ is constant on $\Omega$, that is,$D_{f}(x) = T$
for all $x\in\Omega.$ Show that $f$ is the restriction to $\Omega$ of an affine transformation.

I was trying to prove that the set $A=\{x\in\Omega: f(x) = f(a) + T(x-a)\},$ where $a\in\Omega$ is fixed, is an open and closed subset of $\mathbb{R}^n,$ then for connected of $\Omega$ implies $\Omega = A$ and the proposition is true, but it seems quite hard to prove that; I was thinking about the definition of differentaiblity of $f$ in $a,$ however the best that we get of such definition is an inequality.

I saw a proof of this using gradient theorem but I do not know yet integration; this proposition appears in the section of differential calculus only, hence it must be possible prove it without integration.

Any kind of help is thanked in advanced.

答:
By considering $f-T$ we can reduce the proof to the case when $D_f(x)=0$ for all $x$. (In this case we prove that $f$ is a constant $c$. In general we get $f(x)=c+Tx$).

Let $x \in \Omega$ and consider a disk $B(x,r) $ contained in $\Omega$. Define $g$ on $[0,1]$ by $g(t)=f(tx+(1-t)y)$. Using Chain Rule show that $g'(t)=0$ for all $t$. It follows that $g$ is constant. Hence $f(x)=g(1)=g(0)=f(y)$. We have proved that each point of $\Omega$ has neighborhood in which $f$ is a constant. Now prove that for any fixed $x_0 \in \Omega$ $\{x \in \Omega: f(x)=f(x_0)\}$ is open and closed in $\Omega$ to finish the proof (using connectedness).


问:
Bloody Poisson question. $X \sim Po(\lambda). P(X>2) = 0.3$. Find $P(X<2)$.
Let $X$ be a random variable with a Poisson distribution, such that $P(X>2) = 0.3$. Find $P(X<2)$.

This is bunched in with some easy Poisson questions in the textbook. I don't see how to answer this easily though.

$ e^{-\lambda} + e^{-\lambda}\lambda + \frac{e^{-\lambda}\lambda^2}{2} = 0.7,\ $ but I don't see where to go from here without going into numerical methods, which is definitely not what they want...

答:
I suppose that graphing you noticed that the solution is close to $\lambda=2$.

On the other hand, you can rewrite the equation as
$$e^\lambda=\frac{5}{7} \left(\lambda ^2+2 \lambda +2\right)$$ Make $\lambda=x+2$ to face
$$e^2\, e^x=\frac{5}{7} \left(x^2+6 x+10\right)$$ Since $x$ is small rewrite it as
$$e^2 \left(1+x+\frac{x^2}{2}+O\left(x^3\right)\right)=\frac{5}{7} \left(x^2+6 x+10\right)$$ Ignoring the higher order terms, you face a quadratic equation in $x$
$$\left(14 e^2-100\right)+\left(14 e^2-60\right) x+\left(7 e^2-10\right)
   x^2=0$$ and the approximate solution is then
$$x=\frac{60-14 e^2+\sqrt{-400+1680 e^2-196 e^4}}{14 e^2-20}\approx -0.0865227$$ which implies $\lambda \approx 1.913473$


答:
Hint

Proving that $A$ is open and closed is a good idea.

The fact that it is closed is trivial as $A$ is the inverse image of $\{0\}$ under the continuous map $g(x)=f(x)-f(a)-T(x-a)$. All linear maps are continuous in a finite dimensional space.

To prove that is is open, take $b \in A$ and prove that   $A$ contains a small open ball centered on $b$. For this use Mean Value Theorem on line segments.

Title: Constant derivative of a function implies is an affine map.

Body:
I am trying to prove the next:

Let $f :\Omega\subset\mathbb{R}^n\rightarrow\mathbb{R}^m$ be differentiable, where $\Omega$ is an open
connected subset of $\mathbb{R}^n$. Suppose $D_{f}(x)$ is constant on $\Omega$, that is,$D_{f}(x) = T$
for all $x\in\Omega.$ Show that $f$ is the restriction to $\Omega$ of an affine transformation.

I was trying to prove that the set $A=\{x\in\Omega: f(x) = f(a) + T(x-a)\},$ where $a\in\Omega$ is fixed, is an open and closed subset of $\mathbb{R}^n,$ then for connected of $\Omega$ implies $\Omega = A$ and the proposition is true, but it seems quite hard to prove that; I was thinking about the definition of differentaiblity of $f$ in $a,$ however the best that we get of such definition is an inequality.

I saw a proof of this using gradient theorem but I do not know yet integration; this proposition appears in the section of differential calculus only, hence it must be possible prove it without integration.

Any kind of help is thanked in advanced.


问:
why $\sup$ come before $(a_n + b_n) ?$
Taken from Rudin real and complex analysis  book     chapter $1$ excercise

let $\{a_n\}$ and $\{b_n\}$ be  sequence  in $[-\infty,\infty]$.Prove the following assertion

$b).\lim_{n \to \infty} \sup(a_n +b_n) \le  \lim_{\to \infty} \sup a_n + \lim_{n\to \infty} \sup b_n$

My attempt : $\sup a_n$ is upper bound  of  $a_n$  and $\sup b_n$ is a upper bound of  $b_n$.

Now $ a_n \le \sup  a_n \tag1$

and  $b_n\le \sup b_n\tag2 $

From $(1)$ and $(2)$  we can say that

$a_n + b_n \le \sup a_n  + \sup b_n$

Put  limit both side

$\lim_{n\to \infty} a_n + b_n \le \lim_{n \to \infty}  \sup a_n  + \lim_{n \to \infty} \sup b_n$

My doubt : why  $\sup$  come  before  $(a_n + b_n) ?$

答:
At this state::
$$a_{n}+b_{n}\le sup\left(a_{n}\right)+sup\left(b_{n}\right)$$
You could say that
$$c_n = (a_n+b_n)$$
is bounded from above by $sup\left(a_{n}\right)+sup\left(b_{n}\right)$
So its sup must be less or equal than this upper bound:
$$sup\left(a_{n}+b_{n}\right)\le sup\left(a_{n}\right)+sup\left(b_{n}\right)$$

Title: why $\sup$ come before $(a_n + b_n) ?$


问:
Solving $|2x-3|+7 \le 3x-3|x-7|$

I was trying to solve this inequality with two absolute values:
$$|2x-3|+7 \le 3x-3|x-7|$$
I've got an empty set of solutions, but it's not correct.

How I've tried to solve it:

I've put in a number line the signs (+ or -) to see what happens in three different cases:

first case:
if $x \le 3/2$, then both of them are negative, so I've rewritten the inequality (by changing the signs) as "$-(2x-3)+7 \le 3x+3x-21$".

if  $3/2<x<7$, then the first absolute value is positive and the other one is negative "$2x-3+7 \le 3x+3x-21$".

if $x \ge 7$, then both of them are positive, so I simply canceled out the absolute values.

Now, I've found the solutions for each system of inequality, by putting them in a number line.

first system of inequalities:
$x<3/2, x<31/8$.

the set of solutions is "$x<3/2$".

second system of inequalities:
$3/2<x<7, x<25/4$.

the set of solutions is "$3/2<x<25/4$".

third system of inequalities:
$x>7, x<17/2$.

by putting on another number line these sets, the solution is an empty set of solution.

--

edit: Okay, thanks to the comment discussion, I've solved it. It was an error (because of distraction), the set of solution is "$25/4<x<17/2$"(not strict) the first system of inequalities has an empty set of solution. I think it is correct. let me know if it's an error.

--

If you know a fastest method to solve inequalities like this, let me know.

答:
$$|2x-3|+7\le 3x-3|x-7|.~~~~(1)$$
$x=3/2,7$ are two nodes here, so three intervals will be created these are $x\in (-\infty, 3/2], (3/2, 7], (7,\infty)$

Re-writing the inequation (1)in these interval we get

Case A:$x\in (-\infty, 3/2]$, we write
$$2(3/2-x)+7-3(7-x) \implies x\ge 31/8.$$
This contradicts with $x\in (-\infty,3/2)$
so no solution in this case.

Case B: $x\in(3/2.7]$, we re-write (1) as
$$(2x-3)+7+3(7-x)-3x\le 0 \implies x\ge 25/4$$
Its overlap with $(3.3,7]$ gives $x \in [25/4,7]$

Case C: $x\in (7,\infty)$, we write
$$(2x-3)+7+3(x-7) \le 0 \implies x\le 17/2.$$
Its overlap with $7, \infty)$ gives $x \in (7, 17/2]$

So all real solutions of (1) are: $x\in [25/4.17/2]$

Title: Solving $|2x-3|+7 \le 3x-3|x-7|$


问:
How do you solve this limit? $ \lim_{x\to \infty} x(e^{2/x} - 1 ) $
I was going through studying with L'Hospital's rule, and went through this question. I've tried hard to solve it, but always couldn't go through all the way to the answer...

$$\lim_{x\to \infty} x(e^{2/x} - 1 ) $$

答:
welcome to MSE. another idea is a substitution like below
$$x\to \infty \\ \frac1x \to 0\\ y=\frac1x \to 0$$now rewrite $$\begin{aligned}
\lim_{x \to \infty}x\left(e^{\frac{2}{x}}-1\right) &= \lim_{ y \to 0}\frac{e^{2y}-1}{y} \\
&= \lim_{ y \to 0}\frac{(1+\frac{2y}{1!}+\frac{(2y)^2}{2!}+...)-1}{2y}\\
&= \lim_{ y \to 0}\frac{y(\frac{2}{1!}+\frac{4y}{2!}+...)}{2y}\\
&= 2
\end{aligned}$$


问:
If $f(x+1/n)$ converges to a continuous function $f(x)$ uniformly on $\mathbb{R}$, is $f(x)$ necessarily uniformly continuous?
Suppose $f:\mathbb{R}\mapsto\mathbb{R}$ is a continuous function. Define $f_n(x)=f(x+1/n)$ for all $n\in\mathbb{N}^+$. If $f_n(x)\to f(x)$ uniformly on $\mathbb{R}$, can we conclude that $f$ is actually uniformly continuous? If not, can you give a counterexample of $f$ to be not uniformly continuous but satisfies all above conditions?

I know that if $f_n(x)$'s are all uniformly continuous and $f_n(x)\to f(x)$ uniformly, then $f(x)$ must be uniformly continuous. However, here we do not assume $f(x)$ to be uniformly continuous, and $f_n(x)$ has specific structure related to $f(x)$, so at least you cannot directly conclude they are uniformly continuous. I guess the statement above is wrong but it's hard for me to find a counter-example. Can anyone help me?

答:
Yes, such an $f$ must be uniformly continuous

Suppose that the sequence $$f_n(x) := f(x + 1/n) \to f(x)$$ uniformly on $\Bbb{R}$. This means for any $\epsilon > 0$, there exists $N_\epsilon$ so that $|f(x + 1/n) - f(x)| < \epsilon$ for all $n \geq N_\epsilon$ and all $x \in \Bbb{R}$.

So fix $\epsilon > 0$, and let $N_k := N_{\epsilon/2^k}$. For any $x \in \Bbb{R}$, define sets $S_k(x)$ via

\begin{align*}
S_0(x) &:= \{ x \} \\
S_1(x) &:= S_0(x) \cup \{ x + 1/k_1: k \geq N_1 \} \\
S_2(x) &:= S_1(x) \cup \{ y_1 + 1/k_2: y_1 \in S_1, k_2 \geq N_2 \} \\
S_3(x) &:= S_2(x) \cup \{ y_2 + 1/k_3: y_2 \in S_2, k_3 \geq N_3 \} \\
\vdots \\
S_j(x) &:= S_{j-1}(x) \cup \{ y_{j-1} + 1/k_j: y_{j-1} \in S_{j-1}, k_j \geq N_j \} \\
\vdots \
\end{align*}

Clearly $S_j(x)$ is closed for all $j \geq 0$, every point in $S_{j-1}(x)$ is a limit point of $S_j(x)$, and if $y_j \in S_j(x)$ for $j \geq 1$, $$|f(y_j) - f(x)| < \epsilon(1 - 2^{-j}).$$ We want to show $$S_\infty(x) := \overline{\cup_{j = 0}^\infty S_j(x)}$$ contains some interval $[x, x + \delta)$ for $\delta > 0$, as this will establish uniform continuity of $f$. If $\sum_{k \geq 1} 1/N_k = \infty$, this is simple, so suppose $\sum_{k \geq 1} 1/N_k$ is convergent. WLOG we can assume $2 \leq N_1 \leq N_2 \leq ...$ Then a simple greedy algorithm establishes that if we set $$\delta = \frac{1}{N_1 N_2} + \frac{1}{N_1 N_2 N_3} + ...$$ then every number $y$ with $x < y \leq x + \delta$ can be written as a sum of the form $\sum_{j = 1}^\infty \frac{a_j}{m_j}$, where $a_j \in \{ 0, 1 \}$ and $m_j \geq N_j$ for each $j$. It follows that for such $y$, $|f(y) - f(x)| < \sum_{j = 1}^\infty \frac{\epsilon}{2^j} = \epsilon$. This establishes uniform continuity of $f$.

I haven't worked out the details, but I'm pretty sure you can use an analogous argument to establish that if $\{ a_n \}$ is any sequence converging to $0$ and $f_n(x) := f(x + a_n)$, then uniform convergence of the $f_n$ to $f$ is equivalent to uniform continuity of $f$.


问:
How to compute infinite limits of summations (Riemann sums) when there are i's in the denominator?
I'm really lost here. I'm trying to use a right Riemann sum to compute:
$$\int_0^1{\frac{x}{x^4+2x^2+1}dx}$$
Eventually I get here:
$$\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n\frac{n^3i}{n^4+2n^2i^2+i^4}$$
Using $\Delta{x}=1/n$ and $x_i=a+i\Delta{x}=0+i/n=i/n$

But now I'm stuck since the only formulas I know to manipulate the summation and get rid of the i's are like $\sum_{i=1}^ni^2$. But since I have i's in the denominator, how can I proceed? Thanks!

答:
I hope you are OK with the substitution $x^2+1=u$ to get the integral as $$\int_{1}^{2}\frac{du}{2u^2}$$ A standard technique now is to use a Riemann sum where the partition points are in geometric progression (in contrast to the more common arithmetic progression).

Let $x_i=2^{i/n}$ be the points of partition and we form the Riemann sum $$\sum_{i=1}^{n}f(x_i)(x_i-x_{i-1})$$ where $f(x) =1/(2x^2)$. The computation of the limit is not difficult and should give you the answer $1/4$.

If on the other hand you wish to use the partition points in arithmetic progression as $x_i=1+(i/n)$ then you need some approximation technique for the Riemann sum. Thus we have $$f(x_i) \Delta x_i=\frac{1}{2n(1+i/n)^2}$$ and $$\frac{1}{2}\left(\frac {1}{1+i/n}-\frac{1}{1+(i+1)/n}\right) \leq f(x_i)\Delta x_i\leq \frac{1}{2}\left(\frac{1}{1+(i-1)/n}-\frac{1}{1+i/n}\right)$$ Adding these inequalities for $i=1,2,\dots, n$ we get $$\frac{1}{2}\left(\frac{1}{1+1/n}-\frac{1}{1+(n+1)/n}\right)\leq \sum_{i=1}^{n}f(x_i)\Delta x_i\leq \frac {1}{2}\left(1-\frac{1}{2}\right)$$ By squeeze theorem the limit of Riemann sum as $n\to \infty $ is $1/4$.

The approximation used above can be applied to the original integral also (without the substitution $u=1+x^2$) but it is slightly more complicated to deal with.


问:
If $H$ is the orthocenter and $D,E,F$ are foot of altitudes prove that $\frac{AD}{HD}+\frac{BC}{HE}+\frac{CF}{HF}≥9$

Let ABC be a triangle with $H$ as its the orthocenter and $AD,BE,CF$ as the three altitudes. Prove that $$\frac{AD}{HD}+\frac{BC}{HE}+\frac{CF}{HF}≥9$$

I substituted $AD$ for $AH+HD$, changing the inequality to $\frac{AH}{HD}+\frac{BH}{HE}+\frac{CH}{HF}≥6$
With a little trigonometry I converted this to $\frac{cosAcosB}{cosC}+\frac{cosBcosC}{cosA}+\frac{cosCcosA}{cosB}≥6$

I tried a few things here, but couldn't make much progress. For example, I took the common denominator and then applied AM GM on the numerator as all the terms were squares. This gave me $$\frac{(cosAcosB)^2+(cosBcosC)^2+(cosCcosA)^2}{cosAcosBcosC}≥3(cosAcosBcosC)^{1/3}$$
I think this is $>0$ but that can't be since we have to prove it to be ≥6.

答:
Hint: Express each individual ratio as the ratio as a ratio of areas.

 There's a natural choice given the setup.

 If you're stuck, state what you've tried, and why it doesn't work.

Then use $ ( a + b + c) ( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} ) \geq 9$.


问:
Bound on difference of eigen projections of positive definite matrices
Suppose I have two positive semidefinite matrices (and their eigendecompositions) $A = U \Lambda_A U'$ and $B = V \Lambda_B V'$. I was wondering if
$$
||U_jU_j' - V_jV_j' ||  \leq C || A - B ||
$$
for some $C$. Here, $U_j$ is the $j$th column of $U$. Hence, I'm asking if the eigenprojections are Lipschitz continuous with respect to the original matrices. I've went through Kato's praised book on perturbation theory, which is great by the way, but the theory there is mostly about when the matrices depend on a single parameter. Any help would be greatly appreciated.

答:
This is in fact related to the perturbation theory for sub-eigenspace.

Intuition: Suppose $\lambda_j(A)$ is the $j$-th largest eigenvalues of $A$. If $\lambda_j(A)$ is in the range $(\lambda_{j+1}(B), \lambda_{j-1}(B))$, then there is a hope to obtain such an upper bound. Otherwise, the eigenvector associated with $\lambda_j(A)$ in $A$ can have no relation with the eigenvector associated with $\lambda_{j}(B)$ in $B$.

Results: Suppose $\lambda_{j+1}(B)\leq \lambda_j(A) \leq \lambda_{j-1}(B)$, we have
$$
\|U_jU_j' - V_jV_j'\|_2 \leq \frac{\sqrt{2}\|B-A\|_2}{\min(\lambda_{j}(A)-\lambda_{j+1}(B), \lambda_{j-1}(B) - \lambda_j(A))}.
$$

Before showing the proof, let us introduce the following famous Davis-Kahan theorem:

Theorem: (Davis-Kahan sin(Θ) theorem) Let $A = E_0A_0E_0'+E_1A_1E_1'$ be the two complementary parts of the eigen-decomposition of $A$. In particular, $E_0 \in R^{n\times r}, E_1 \in R^{n\times (n-r)}, A_0 \in R^{r\times r}, A_1 \in R^{(n-r)\times (n-r)}$ where $E_0'E_0 = I_r, E_1'E_1 = I_{n-r}, E_0'E_1 = 0$ and $A_0, A_1$ are diagonal. Similarly, let $B = F_0B_0F_0' + F_1B_1F_1'$ be the two complementary parts of eigen-decomposition of $B$. If the eigenvalues of $A_0$ are contained in an interval $(a, b)$, and the eigenvalues of $B_1$ are excluded from the interval $(a−\delta, b+\delta)$ for some $\delta > 0$, then
$$
\|F_1'E_0\| \leq \frac{\|F_1'(B-A)E_0\|}{\delta}
$$
for any unitarily invariant norm $\|\cdot\|$.

Corollary: Apply the above theorem to Frobenious norm, we have
$$
\|F_1'E_0\|_{F} \leq \frac{\|F_1'(B-A)E_0\|_{F}}{\delta} \leq \frac{\|F_1'\|_{2} \|B-A\|_{2} \|E_0\|_{F}}{\delta} \leq \frac{\|B-A\|_2 \|E_0\|_2 \sqrt{r}}{\delta} \leq \frac{\sqrt{r} \|B-A\|_2}{\delta}
$$
where $|A|_2$ is the spectral norm and $|A|_{F}$ is the Frobenious norm, and we use the fact that $|E_0|_2 \leq 1, |F_1|_2 \leq 1$ and $|AB|_{F} \leq |A|_2 |B|_F, |A|_{F} \leq \sqrt{min(n,m)}|A|_{2}$ if $A \in n\times m.$ On the other hand, suppose $F_0 \in R^{n\times r}$. Notice that
\begin{align}
\|E_0E_0' - F_0 F_0'\|_{F}^2
&= tr(E_0E_0'E_0E_0' + F_0F_0'F_0F_0') - 2 tr(E_0E_0'F_0F_0')\\
&= tr(E_0E_0') + tr(F_0 F_0') - 2tr(E_0E_0'F_0F_0')\\
&= 2r - 2tr(E_0E_0'F_0F_0')\\
&= 2tr(E_0E_0') - 2tr(E_0E_0'F_0F_0')\\
&= 2tr(E_0E_0'(I - F_0F_0'))\\
&= 2tr(E_0E_0'F_1F_1')\\
&= 2\|F_1'E_0\|_{F}^2.
\end{align}
Therefore,
$$
\|E_0E_0' - F_0F_0'\|_{F} \leq \frac{\sqrt{2r}\|B-A\|_2}{\delta}.
$$
This provides an upper bound for the subspace distance between $E_0E_0'$ and $F_0F_0'$ based on the perturbation $\|B-A\|_2.$

Application:
Now we apply this theorem (corollary) to your problem. Here $E_0 = U_j, E_1 = U_{-j}$ where $U_{-j}$ removes the $j$-th column from $U$ and $A_0 = \lambda_{j}(A)$, $A_1$ is the diagonal matrix consists of eigenvalues $\lambda_1(A), \dotsc, \lambda_{j-1}(A), \lambda_{j+1}(A), \dotsc, \lambda_{n}(A)$ where we assume $\lambda_1(A) \geq \lambda_2(A) \geq \dotsc \geq \lambda_n(A).$ Similarly, $F_0 = V_j, F_1 = V_{-j}$ and $B_1$ consists of eigenvalues $\lambda_1(B), \dotsc, \lambda_{j-1}(B), \lambda_{j+1}(B), \dotsc, \lambda_n(B).$ Therefore, $r = 1$ in your problem.

In a typical case, we shall assume $\lambda_{j+1}(B) \leq \lambda_j(A) \leq \lambda_{j-1}(B)$ otherwise there is no hope to have some continuity from the eigenvector $U_j$ to $V_j$.

Then this corollary implies that
$$
\|U_jU_j' - V_jV_j'\|_{F} \leq \frac{\sqrt{2}\|B-A\|_2}{\delta}
$$
where $\delta = \min(\lambda_{j}(A)-\lambda_{j+1}(B), \lambda_{j-1}(B) - \lambda_j(A)).$ Note that
$$
\frac{1}{\sqrt{2}}\|U_jU_j' - V_jV_j'\|_{F} \leq \|U_jU_j' - V_jV_j'\|_2 \leq \|U_jU_j' - V_jV_j'\|_{F}
$$
since $U_jU_j'-V_jV_j'$ is at most rank 2. This provides (almost tight) bound for your question:
$$
\|U_jU_j' - V_jV_j'\|_2 \leq \frac{\sqrt{2}\|B-A\|_2}{\min(\lambda_{j}(A)-\lambda_{j+1}(B), \lambda_{j-1}(B) - \lambda_j(A))}.
$$

See further references on the proof of Davis-Kahan theorem (the proof is in fact very neat. Highly recommend this!) and the user-friendly variant of Davis-Kahan theorem.


问:
What can be said about the closedness and compactness of the subset $S = \{\sin (2^nx)\ |\ n \in \Bbb N \}$ in $L^2 ([-\pi, \pi] )\ $?

Let $L^2 \left ([-\pi,\pi ] \right )$ be the metric space of Lebesgue square integrable functions on $[-\pi, \pi]$ with a metric $d$ given by $$d(f,g) = \left [\int_{-\pi}^{\pi} \left (f(x) - g(x) \right )^2\ dx \right ]^{\frac {1} {2}},\ \text {for}\ f, g \in L^2 \left ([-\pi, \pi] \right ).$$ Consider the subset $$S = \left \{\sin \left (2^nx \right )\ |\ n \in \Bbb N \right \}\ \text {of}\ L^2 \left ([-\pi, \pi ] \right ).$$ Which of the following statements are true?

$(1)$ $S$ is bounded

$(2)$ $S$ is closed

$(3)$ $S$ is compact

$(4)$ $S$ is non-compact

It is easy to show that $S$ is a compact subset of $\left (L^2 \left ([-\pi, \pi] \right ), d \right ).$ But I can't able to prove or disprove whether it is closed or not. Let $\{x_n \}_{n \geq 1}$ be a convergent sequence in $S.$ Then there exists a strictly increasing sequence of natural numbers $\{k_n\}_{n \geq 1}$ such that $x_n = \sin \left (2^{k_n} x \right ),\ \text {for all}\ n \geq 1.$ Let it converge to some $f \in L^2 \left ([-\pi, \pi] \right ).$ So we have $$\int_{-\pi}^{\pi} \left (\sin \left (2^{k_n} x \right ) - f(x) \right )^2\ dx \to 0\ \text {as}\ n \to \infty.$$ From here how do I conclude that $f \in L^2 \left ([-\pi, \pi ] \right )\ $? Also how to conclude about the compactness or non-compactness of $S\ $? Any help in this regard will be appreciated.

Thanks for your time.

答:
As you mention, the set $S$ itself has no limit point : for every $m \neq n$ , we have $d(\sin(2^n x), \sin(2^m x)) = \sqrt{2 \pi}$. Therefore, the sequence $\sin(2^n x), n=1,2,3,...$ will not have a convergent subsequence, as no subsequence is even Cauchy.

In this regard, certainly $S$ is not compact. (It is not sequentially compact, to be precise, but these mean the same in our space $L^2[-\pi,\pi]$ which is second countable).

However, $S$ is closed. This is because if you take a Cauchy sequence in $S$, then by what we have said it has to be an eventually constant sequence, so of course it converges with the limit in $S$.

Furthermore, $S$ being bounded is obvious, since $|\sin (2^nx)| \leq 1$ for all $n$ and $x$ so the norm of any element in $S$ is just bounded by $\sqrt {2 \pi}$.

Finally, $S$ is a closed, bounded and non-compact subset of $L^2[-\pi,\pi]$. Contrast with the Heine-Borel theorem.

EDIT

We have for any $k \neq l$ positive integers that $$
(\sin kx - \sin lx)^2 = \sin^2 kx + \sin^2 lx - 2 \sin kx \sin lx   \\
= \sin^2 kx + \sin^2 lx - 2(\cos(k-l)x - \cos(k+l)x) \\
= \frac{1 - \cos 2kx}{2} + \frac{1 - \cos 2 lx}{2} - 2(\cos(k-l)x - \cos(k+l)x)
$$

we have for any $t \neq 0$ integer that $\int_{- \pi}^{\pi} \cos(tx) = \left[\frac{\sin(tx)}{t}\right]_{-\pi}^{\pi} = 0 - 0 = 0$. Therefore, provided that $k \neq l$ above, when we integrate $(\sin kx - \sin lx)^2$ from $- \pi$ to $\pi$ only the constants remain as all the cosines integrate to $0$. The constant is just $\frac 12 + \frac 12 = 1$, so the integral is just $\int_{-\pi}^{\pi} 1dx = 2 \pi$.

Set $k = 2^m$ and $l= 2^n $ for $m \neq n$ to see our case. Also we must take the square root since $d$ has a square root, so the final answer is $\sqrt{2 \pi}$.


答:
The fastest way for me is to write down LHS-RHS as a piecewise linear function.

$$f(x)=|2x-3|+7-3x+3|x-7|\\
=\begin{cases}
3-2x+7-3x+3(7-x)=31-8x, & x \leqslant \frac 32\\
2x-3+7-3x+3(7-x)=25-4x, & \frac 32 \leqslant x \leqslant 7\\
2x-3+7-3x+3(x-7)=2x-17, & x\geqslant 7
\end{cases}$$

Note that $f(\frac 32)=19, f(7)=-3$.

When $x \leqslant \frac 32, f$ is decreasing so $f(x)>0$;

When $\frac 32 \leqslant x \leqslant 7, 25-4x \leqslant 0 \implies x \geqslant \frac{25}{4};$

When $x \geqslant 7, 2x-17\leqslant 0 \implies x \leqslant \frac{17}{2}.$

Therefore $\frac{25}{4} \leqslant x \leqslant \frac{17}{2}. \blacksquare$

Title: Solving $|2x-3|+7 \le 3x-3|x-7|$

Body:
I was trying to solve this inequality with two absolute values:
$$|2x-3|+7 \le 3x-3|x-7|$$
I've got an empty set of solutions, but it's not correct.

How I've tried to solve it:

I've put in a number line the signs (+ or -) to see what happens in three different cases:

first case:
if $x \le 3/2$, then both of them are negative, so I've rewritten the inequality (by changing the signs) as "$-(2x-3)+7 \le 3x+3x-21$".

if  $3/2<x<7$, then the first absolute value is positive and the other one is negative "$2x-3+7 \le 3x+3x-21$".

if $x \ge 7$, then both of them are positive, so I simply canceled out the absolute values.

Now, I've found the solutions for each system of inequality, by putting them in a number line.

first system of inequalities:
$x<3/2, x<31/8$.

the set of solutions is "$x<3/2$".

second system of inequalities:
$3/2<x<7, x<25/4$.

the set of solutions is "$3/2<x<25/4$".

third system of inequalities:
$x>7, x<17/2$.

by putting on another number line these sets, the solution is an empty set of solution.

--

edit: Okay, thanks to the comment discussion, I've solved it. It was an error (because of distraction), the set of solution is "$25/4<x<17/2$"(not strict) the first system of inequalities has an empty set of solution. I think it is correct. let me know if it's an error.

--

If you know a fastest method to solve inequalities like this, let me know.


问:
Diffusion equation issue
I need help, I’m stuck on this problem. Below is my progress.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\on{u}_{x}\pars{x,t} =
-\sum_{n = 1}^{\infty}\on{a}_{n}\pars{t}\,n\sin\pars{nx}}$ already satisfies
$\ds{\on{u}_{x}\pars{0,t} = \on{u}_{x}\pars{\pi,t} = 0}$. Note that
\begin{equation}
\int_{0}^{\pi}\cos\pars{mx}\cos\pars{nx}\,\dd x =
{\pi \over 2}\,\delta_{mn}\label{1}\tag{1}
\end{equation}

$\ds{\on{u}\pars{x,t} =
\sum_{n = 1}^{\infty}\on{a}_{n}\pars{t}\cos\pars{nx} + \on{f}\pars{t}}$ which must satisfies the above partial differential equation. Namely,
\begin{align}
&\sum_{n = 1}^{\infty}\dot{\on{a}}_{n}\pars{t}\cos\pars{nx} + \dot{\on{f}}\pars{t} +
\sum_{n = 1}^{\infty}\on{a}_{n}\,n^{2}\pars{t}\cos\pars{nx} =
\expo{-x}\label{2}\tag{2}
\end{align}

Integrate both members of (\ref{2}} over $\ds{x \in \pars{0,\pi}}$:
$$
\dot{\on{f}}\pars{t}\pi = -\expo{-\pi} + 1 \implies
\on{f}\pars{t} = {1 - \expo{-\pi} \over \pi}\,t + c
$$
$\ds{c}$ is a constant.

Use (\ref{1}) and (\ref{2}):
\begin{align}
&\dot{\on{a}}\pars{t} + n^{2}\on{a}\pars{t} =
{2 \over \pi}\int_{0}^{\pi}\expo{-x}\cos\pars{nx}\dd x =
{2 \over \pi}\,{1 - \expo{-\pi}\pars{-1}^{n} \over 1 + n^{2}}
\equiv \varphi_{n}
\\[5mm] &\
\totald{\bracks{\expo{n^{2}\,t}\on{a}_{n}\pars{t}}}{t} =
\varphi_{n}\expo{n^{2}\,t} \implies
\expo{n^{2}\,t}\on{a}_{n}\pars{t} -
\on{a}_{n}\pars{0} =
{\varphi_{n} \over n^{2}}\pars{\expo{n^{2}\,t} - 1}
\\[5mm] &\
\on{a}_{n}\pars{t} = \on{a}_{n}\pars{0}\expo{-n^{2}\,t} +
{\varphi_{n} \over n^{2}}\pars{1 - \expo{-n^{2}\,t}}
\\[5mm] &\
\on{u}\pars{x,t} =
{1 - \expo{-\pi} \over \pi}\,t + c +
\sum_{n = 1}^{\infty}\on{a}_{n}\pars{0}
\expo{-n^{2}\,t}\cos\pars{nx}
\\[2mm] + &\
\sum_{n = 1}^{\infty}
{\varphi_{n} \over n^{2}}\pars{1 - \expo{-n^{2}\,t}}
\cos\pars{nx}
\end{align}

Since $\ds{\on{u}\pars{x,0} = 0}$:
$$
0 = c +
\sum_{n = 0}^{\infty}\on{a}_{n}\pars{0}\cos\pars{nx}
\implies c = 0\ \mbox{and}\ \on{a}_{n}\pars{0} = 0\ \forall n
$$

Finally,
$$
\on{u}\pars{x,t} =
{1 - \expo{-\pi} \over \pi}\,t +
{2 \over \pi}\sum_{n = 1}^{\infty}
{1 - \expo{-\pi}\pars{-1}^{n} \over n^{2}\pars{n^{2} + 1}}\pars{1 - \expo{-n^{2}\,t}}
\cos\pars{nx}
$$

Title: Diffusion equation issue


问:
Equivalency of Modulo function
Is it valid to take modulo of both sides of a complex eqn? I can't seem to pinpoint why, but say we want to find the solutions of this expression

$$2z^2=-4 \\ |2z^2|=|-4| \\ |z^2| = 2 \\ |z| = \sqrt{2}$$

However, it is fairly obvious there are many points on $|z|=\sqrt{2}$ that don't satisfy the first equation? $2(\sqrt{2})^2 \neq -4$

答: Taking modulus is legitimate but it does not yield the final answer. A better way is to write $z^{2}=-2$ which gives $z =\pm \sqrt 2 i$.

Title: Equivalency of Modulo function


问:
Finite Projective Dimension implies non vanishing Ext
Suppose the projective dimension of a module $M$ is $n < \infty$. Does there exist a free $R$-module $F$ such that $\operatorname{Ext}^n(M, F) \not = 0$?

Can't we write the free module as a direct sum of $R$'s and then we just need to show $\operatorname{Ext}^n(M, R) \not = 0$? So does this work actually for any free module?

答:
If you ask the question

if $M$ is an $R$-module of projective dimension $n$, is $\operatorname{Ext}^n(M,R)\neq0$?

then the answer is surpring: it depends. When $R=\mathbb Z$, this is essentially known as Whitehead's problem and Shelah proved that its answer depends on the specific set theory that you choose. Indeed, he showed that depending on the background set theory, there exist abelian groups $M$ which are not free (so that their projective dimension is $1$) such that $\operatorname{Ext}^1(M,\mathbb Z)=0$, or not.

$$\bullet\bullet\bullet$$

For the original question in which you look just for a free module:

Suppose the projective resolution of $M$ ends with $$\cdots\leftarrow P_{n-1}\xleftarrow{\hskip2ex d\hskip2ex }P_n\leftarrow 0$$ and suppose $P_n$ is a direct summand of the free module $F$. Applying the functor $\hom(\mathord-,F)$ we get a complex ending with $$\cdots\to\hom(P_{n-1},F)\xrightarrow{\hskip2ex d^*\hskip2ex }\hom(P_n,F)\to0$$ $\operatorname{Ext}^1(M,F)$ is the cokernel of the map $d$. To show it is not zero, it is enough to show that $d$ is not surjective.

Now let $i:P_n\to F$ be the inclusion and let $j:F\to P_n$ be a retraction of $i$, so that $ji=1_{P_n}$. The element $i\in\hom(P_n,F)$ is not in the image of $d$. Otherwise we'd have a map $r:P_{n-1}\to F$ such that $rd=i$, so composing with $j$ we'd have $jrd=ji=1_{P_n}$. It follows that $jr:P_{n-1}\to P_n$ is a retraction for the map $d$. This is impossible, because the projective dimension of $M$ is exactly $n$.


问:
Ceil function proof
I'm sorry if this is a repeated question but I can't find it anywhere. Any help is highly appreciated
How to prove that for any +ve real number

```<code>ceil (a*b) <= ceil(a)*ceil(b)
</code>```

Thankyou very much

答:
If $a> 0; b> 0$ then let $M = \{n\in \mathbb N|n \ge ab\}$.

Note $0< a \le \lceil a\rceil$ and $0 < b \le \lceil b\rceil$ so $ab \le \lceil a\rceil\lceil b\rceil$ so $\lceil a\rceil\lceil b\rceil \in M$.

And $ab \le \lceil ab\rceil$ so $\lceil ab\rceil\in M$, but not only is $\lceil ab\rceil\in M$, but $\lceil ab\rceil-1 < ab < \lceil ab\rceil$ so $\lceil ab\rceil = \min M$.

And as $\min M \le k$ for all $k \in M$ we must have $\lceil ab\rceil\le \lceil a\rceil\lceil b\rceil$.

The End.

.... to show that $\lceil a\rceil\lceil b\rceil$ need not be equal to $\lceil ab\rceil$

we can not that $(\lceil a\rceil-1) (\lceil b\rceil-1)=\lceil a\rceil\lceil b\rceil- (\lceil a\rceil+\lceil b\rceil-1) < ab \le \lceil a\rceil\lceil b\rceil$ and that is a range that could (doesn't have to be but could) be a range of many integers.

A good exmample of this would be $a=5.1$ and $b=7.1$ so $ab = 36.21$

We have $5 < a < 6$ and $7< b < 8$ so $35 < ab = 36.21 < 48=\lceil 5.1\rceil \lceil 7.1\rceil$.  That is a loose range of $13$ possible integers.

But $36 < ab=36.21 < 37=\lceil 5.1\cdot 7.1\rceil$. That is a tight range of only $1$ possible integer.

Title: Ceil function proof


问:
How to apply definite integral on $\sqrt{1+\sin x}$
I want to solve the following integral:

$$\int_0^\pi \sqrt{1+\sin x} \, dx$$

I want to do this with the substitution:

$$u=\sin x$$

so that:

$$du=\cos x\,dx=\sqrt{1-\sin^2x}\,dx = \sqrt{1-u^2} \, dx$$

$$dx = \frac{du}{1-u^2}$$

$$u(0) = 0, \quad u(\pi/2)=1, \quad u(\pi)=0$$

$$\int\sqrt{1+\sin x} \, dx = \int\frac{1}{\sqrt{1-u}} \, du = -2\sqrt{1-u}$$

So since $u$ isn't defined for 1 (corresponding $x$ is $\pi/2$), I split the integration to ($0, \pi/2$) and ($\pi/2, \pi$)
However when I assign the substitution I get:

$$\int_0^1 \frac{1}{\sqrt{1-u}} \, du + \int_1^0 \frac{1}{\sqrt{1-u}} \, du = 2-2 = 0$$

But the result should be $4.$ It seems like I should flip the sign here for some reason but I can't figure it out, help would be appreciated.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\pi}\root{1 + \sin\pars{x}}\dd x} =
\int_{-\pi/2}^{\pi/2}\root{1 + \cos\pars{x}}\dd x
\\[5mm] = &\
2\int_{0}^{\pi/2}\root{1 + \cos\pars{x}}\dd x =
2\int_{0}^{\pi/2}
\root{1 + \bracks{2\cos^{2}\pars{x \over 2} - 1}}\dd x
\\[5mm] = &\
2\root{2}\int_{0}^{\pi/2}\cos\pars{x \over 2}\,\dd x =
2\root{2}\,{\sin\pars{\pi/4} \over 1/2} = \bbx{4} \\ &
\end{align}


问:
How to prove the projection theorem?
For any point $y$, the projection onto a nonempty and closed convex set $X$ is defined as

$$\Pi_X(y) := \arg\min_{x \in X} \frac{1}{2} \| x - y \|_2^2.$$

I am trying to prove for any point $x\in X$,

$$\langle y - \Pi_X(y) , x - \Pi_X(y) \rangle \le 0$$

First, let $f(x)=\frac{1}{2}\|x-y\|_2^2$, $\nabla f(x)=x-y$. $f(x)$ is a convex function because $\nabla^2f(x)=1\ge0$. So,

$$f(x)\ge f(\Pi_X(y))+\nabla f(\Pi_X(y))^T(x-\Pi_X(y))=f(\Pi_X(y))+(\Pi_X(y)-y)^T(x-\Pi_X(y))\\
\Rightarrow (\Pi_X(y)-y)^T(x-\Pi_X(y)) = \langle \Pi_X(y)-y, x-\Pi_X(y) \rangle \le f(x) - f(\Pi_X(y))\\
\Rightarrow \langle y - \Pi_X(y) , x - \Pi_X(y) \rangle \ge f(\Pi_X(y)) - f(x)$$

And because $\Pi_X(y)$ is the projection of $y$ onto $X$, $f(\Pi_X(y)) \le f(x)$, $f(\Pi_X(y)) - f(x)\le0$. At this point, I can't continue my proof. Is there something wrong with my process?

答:
Let $z:=\Pi_X(y)$, let $x\in X$, $t>0$. Then
$$
0 \le f(z+t(x-z)) - f(z) = t(z-y, x-z) + \frac {t^2}2\|x-z\|^2.
$$
This is the Taylor expansion of the quadratic function $f$. Now divide by $t$ and pass to the limit $t\searrow0$.

Title: How to prove the projection theorem?


答:
A crude but surprisingly effective way to solve for the rate parameter $\lambda$ is to set up a recursion.  Given $$\Pr[X \le 2] = e^{-\lambda}(1 + \lambda + \lambda^2/2) = 0.7,$$ we may write this as $$\lambda = \log \frac{1 + \lambda + \lambda^2/2}{0.7}$$ which induces a recursion relation $$\lambda_{n+1} = \log \frac{1 + \lambda_n + \lambda^2_n/2}{0.7}.$$  Then with an initial guess, say $\lambda_0 = 1$, it is relatively simple to implement this recursion in a typical pocket calculator; e.g. first entering 1, then entering ln((1 + Ans + Ans^2/2)/0.7), then repeating this entry by pressing =.  By the thirtieth iterate, we get at least six digits of precision, sufficient to compute $\Pr[X < 2]$.

We can improve upon the initial guess by noting that for $\lambda = 1$, $\Pr[X \le 2] \approx 0.91$, which is too large; but $\lambda = 2$ gives $\Pr[X \le 2] \approx 0.676$, much closer but on the other side of $0.7$; so we know that the precise value must be slightly less than $2$, say $\lambda_0 = 1.9$.  This saves us about $15$ iterations.

Title: Bloody Poisson question. $X \sim Po(\lambda). P(X>2) = 0.3$. Find $P(X<2)$.

Body:
Let $X$ be a random variable with a Poisson distribution, such that $P(X>2) = 0.3$. Find $P(X<2)$.

This is bunched in with some easy Poisson questions in the textbook. I don't see how to answer this easily though.

$ e^{-\lambda} + e^{-\lambda}\lambda + \frac{e^{-\lambda}\lambda^2}{2} = 0.7,\ $ but I don't see where to go from here without going into numerical methods, which is definitely not what they want...


问:
Proving that $a_kA^k+a_{k-1}A^{k-1}+\cdots +a_1A+a_0I = 0_n$
I've been studying linear algebra and came across this question that I have no general idea how to solve. The question is as follows.

Let $A$ be an $n×n$ diagonalizable matrix with distinct eigenvalues $λ_1,λ_2,λ_3$. Suppose $p(λ)=a_kλ^k+a_{k−1}λ^{k−1}+\cdots+a_1λ+a_0$  is a polynomial such that $p(λ_j)=0$ for each $j=1,2,3$. Prove that:
$$a_kA^k+a_{k-1}A^{k-1}+\cdots +a_1A+a_0I = 0_n$$
where $0_n$ denotes the $n×n$ zero matrix.

Any sort of guidance on how to solve this problem would be greatly appreciated

答: We have $A = T D T^{-1}$, where $D$ is diagonal with $\lambda_i$ on the diagonal. Now $p(D)$ is diagonal, with $p(\lambda_i)$ on the diagonal. So $p(D)=0$. Now notice that $p(A) = T p(D) T^{-1}$.


答:
I think you have made a mistake notice that $$\sum \frac{AH}{HD}=\sum \frac{|\cos A|}{|\cos B\cos C|}\ge 3\sqrt[3]{\frac {1}{|\cos A\cos B\cos C|}}$$

Now $${|\cos A\cos B\cos C|}\le 1/8\tag1$$ the last inequality is proved here

using $(1)$ can you finish it?

Title: If $H$ is the orthocenter and $D,E,F$ are foot of altitudes prove that $\frac{AD}{HD}+\frac{BC}{HE}+\frac{CF}{HF}≥9$

Body:
Let ABC be a triangle with $H$ as its the orthocenter and $AD,BE,CF$ as the three altitudes. Prove that $$\frac{AD}{HD}+\frac{BC}{HE}+\frac{CF}{HF}≥9$$

I substituted $AD$ for $AH+HD$, changing the inequality to $\frac{AH}{HD}+\frac{BH}{HE}+\frac{CH}{HF}≥6$
With a little trigonometry I converted this to $\frac{cosAcosB}{cosC}+\frac{cosBcosC}{cosA}+\frac{cosCcosA}{cosB}≥6$

I tried a few things here, but couldn't make much progress. For example, I took the common denominator and then applied AM GM on the numerator as all the terms were squares. This gave me $$\frac{(cosAcosB)^2+(cosBcosC)^2+(cosCcosA)^2}{cosAcosBcosC}≥3(cosAcosBcosC)^{1/3}$$
I think this is $>0$ but that can't be since we have to prove it to be ≥6.


问:
for every K>1 and natural there is an arithmetic progression with coprime numbers, can't prove this
I have the following question:

Prove that for every $ K  \in \mathbb{N} , K\gt 1$ there exists an arithmetic progression with length K of co prime numbers.
the numbers themselves aren't necessarily prime. but they are co prime.

define how to choose the common difference and why it works ?

I am having a struggle proving this, I had a feeling its related to induction and or euclid

答:
Actually there are infinitely many AP's exist $\forall K$.

Let all elements of the sequence be odd, therefore the increment is even.

Let $r ≥ max(K-1,2)$

and $d = r\#$, where $\#$ is the primorial function, factorial could be used instead.

Let $A = \{1, 2, ..., K - 1\}$.

Let the starting term be $a_1 = d + 1$ (it could be parameterized by multiplying $d$ with an integer, so that even $a_1 = 1$),

and let the increment be $d$ (again, it could be parameterized by multiplying with an integer),

so any other term is $a_i = d + 1 + \alpha d$, $\alpha \in A$.

If the elements have a common prime factor, then it divides their difference, and vice versa, if it divides their difference, then it must divide the elements.

Any two elements' difference is $\beta d$, $\beta \in A$, and no prime factor of this divides any $a_i$ or $a_1$.


问:
Integrals involving sum of differentials.
What would be the solution to such an integral:
$$\int xdx + xdxdx+dxdxdxdx $$
in case of are of a ring we do:
$$\int \pi(r+dr)^2-\pi r^2 $$
$$=> \int 2\pi rdr +drdr $$
Here we neglect drdr but why ?

答:
Welcome to MSE.
I think you are working on a ring from a circle with a radius of $r$ and increasing by $\Delta r$. $$f(r)=\pi r^2\\
\Delta f= f(r+\Delta r)-f(r)
\\=\pi(r+\Delta r)^2-\pi r^2\\=2\pi r \Delta r +(\Delta r)^2$$ usually $\Delta r \to 0 $ or very small, so $(\Delta r)^2$ tends to $0$.

so $ \Delta f\sim 2\pi r \Delta r +\underbrace{(\Delta r)^2}_{\to 0}$  .

Remark:IF we have double integration then $dxdy$ means. for example $$\int\int f(x_1,x_2)dx_1dx_2$$ this is meaningful.


问:
Calderon-Zygmund decomposition
Recently I'm studying something about Calderon-Zygmund decomposition, as in the following link
https://en.wikipedia.org/wiki/Singular_integral_operators_of_convolution_type#Calder%C3%B3n-Zygmund_decomposition.

But I don't know how to  perform the decomposition for a specific function.

Let $f$ be a non-negative integrable on $[-π,π]$, perform the Calderon-Zygmund decomposition for

$f(x)= \delta_0 (x)$ at the level $\alpha =1$.

$f(x)= \delta_0 (x)+\delta_1 (x)$ at the level $\alpha =1$.

where $\delta_a$ equals $1$ at $a$ and $0$ otherwise, and on any interval containing $a$ the integral of $\delta_a$ is $1$.

Any help to solve the problem will be appreciated!

Thanks in advance.

答:
Here's one interpretation of how to carry out the decomposition for $f(x) = \delta_0(x)$, with $\alpha = 1$. I'll leave the other part as an exercise.

The starting average value of $f$, over the full interval $[-\pi, \pi]$, is $(2\pi)^{-1} < \alpha$, so we proceed to subdivide the interval. On the child intervals, $[-\pi, 0]$ and $[0, \pi]$, $f$ has the average value $\pi^{-1} < \alpha$. So we subdivide again, obtaining $[-\pi, -\pi/2]$, $[-\pi/2, 0]$, $[0, \pi/2]$, and $[\pi/2, \pi]$. This time the average values are $0$, $2/\pi$, $2/\pi$, and $0$, still all less than $\alpha$. Subdividing a third time (I'm going to stop listing the intervals) yields averages of $0$, $0$, $0$, $4/\pi$, $4/\pi$, $0$, $0$, and $0$. The middle two intervals, $[-\pi/4, 0]$ and $[0, \pi/4]$, now have averages greater than $\alpha$, so we set those intervals aside. But the other intervals continue to be subdivided forever, with average values never exceeding $\alpha$ (in fact the averages are all $0$).

The definition of $g$ is now potentially a bit hazy. Going back to Stein's exposition in Singular Integrals and Differentiability Properties of Functions (p. 31), we have that $g$ is defined to be equal to $f$ off of the intervals $[-\pi/4, 0]$ and $[0, \pi/4]$. So $g = 0$ on $[-\pi, -\pi/4)$ and $(\pi/4, \pi]$. On the interiors of the set-aside intervals, $g$ is defined to be the average value of $f$ on those intervals. So $g = 4/\pi$ on $(-\pi/4, 0)$ and $(0, \pi/4)$. Stein doesn't specify what should happen at $-\pi/4$, $0$, or $\pi/4$, because for ordinary functions that doesn't matter: that remainder has Lebesgue measure $0$, and for ordinary functions there's no mass on such a null set. But now, because we're dealing with a delta function, there is mass at one of these points. So at the moment it's a bit hazy.

Moving on temporarily, we can safely say that $b$ is $0$ on $[-\pi, -\pi/4)$ and $(\pi/4, \pi]$. Further, though (again per Stein's exposition), the average value of $b$ on each interval $[-\pi/4, 0]$ and $[0, \pi/4]$ is supposed to be $0$. Now, we need to have $b = -4/\pi$ on $(-\pi/4, 0)$ and $(0, \pi/4)$ so that $f = g + b$ there. But then to make the average values work out to be $0$, we also need to include $\delta_0$ as a part of $b$ (and so not as a part of $g$).

With that, we've essentially fully defined $g$ and $b$. There are still a few individual points where we haven't provided specific function values, but these should be immaterial for practical purposes as all of $f$'s mass is accounted for.

Title: Calderon-Zygmund decomposition