问:
How to calculate this PV complex integral?
$$P.V. \int_{-\infty}^{\infty}\frac{xe^{ix}}{x^2-\pi^2}.dx$$
I find this a challenging integral to solve. I tried solving by extending to the complex plane and integrating over a semicircle contour (real axis, but infinitesimally small hoops (radii $\rho$) around points $x=-\pi$ and $x=\pi$,  plus the arc $Re^{i\phi}$, $\phi \in [0,\pi]$), but this leads me nowhere as i find that the integral on all of the three arcs vanishes, in the limit $R->\infty$, $\rho ->0$ .

答:
Problem resolved. As @Svyatoslav mentioned (thank you kindly), the integrals over the two hoops do not vanish.
These integrals can be evaluated by rewriting $f(z)$ (after coordinate transformation of $z'= z \pm \pi$)  as $\frac{1}{z} g(z)$, where $g(z)$ is analytic in the relevant domain and can therefore be written as a power series:
$$\int f(z) dz = \int \frac{1}{z}(a_0 + a_1z + a_2z^2 ... ) dz = \int \frac{a_0}{z}dz +\int (a_1 + a_2z + a_3z^2 ... ) dz  $$
In the limit of radius $\rho$ approaching $0$, the latter integral vanishes (as the begining point and end point of the path of integration overlap). The integral that remains can be easily evaluated:
$$\int \frac{a_0}{z} dz = i\int_{\pi}^{0} a_0 dt.$$


问:
How to show lim sup is $0$ for $f \in L^1(R)$
For any $f \in L^1(R)$, I am trying to show that

$$\lim_{t\to0}\sup\left\{\frac{1}{|I|} \int
_I |f-f_I| :I \text{ is an interval that contains b with length t}\right\}=0$$

for a.e.  $b \in R$ and $$f_I = \frac{1}{|I|} \int_If$$ where $0 < |I| <\infty$.

I am almost sure that I have to use Lebesgue's Differentiation Theorem which states that
$$ \lim_{t\to0}\frac{1}{2t} \int_{b-t}^{b+t}|f-f(b)|=0 \text{ for a.e. } b \in \mathbb{R} \text{ and  }f\in L^1(R). $$

So I attempted to convert it to this form and  let I = ($b+t/2,b-t/2$) and let $a= t/2.$
By Lebesgue's Differentiation Theorem, I have  $$ \lim_{a\to0}\frac{1}{2a} \int_{b-a}^{b+a}|f-f(b)|=0.$$ If f is continuous at b, I know how to proceed because I can bound it with a difference that goes to 0. But I don't know how to proceed without assuming continuity. Any help is appreciated. Thanks!

答:
To prove this, you can follow the scheme presented in "Real and Complex Analysis" by W. Rudin page 138-139 (the same scheme has already been suggested in the comments) but you need to know some important facts about the Hardy-Littlewood maximal function.

Define, for any $b \in \mathbb{R}$,
$$(T_tf)(b) = \frac{1}{|I_t|} \int_{I_t} |f-f(b)|dx$$
here $I_t$ stands for an interval of lenght $t$ whose center is $b$. Set
$$ (Tf)(b) = \limsup_{t \to 0} \hspace{0.1 cm} (T_tf)(b).$$
To prove that $Tf=0$ a.e., we will prove that for any $y>0$, the set
$$\{x \hspace{0.1cm}| \hspace{0.1 cm} T_f(x)> 2y \}$$
has measure 0.

Your idea about using continuity is very useful:  Given any $n \in \mathbb{N}$, you may find a continuous function $g$ such that
$$\|f-g \|_{1}<\frac{1}{n}.$$
Set $h=f-g.$
Since $g$ is continuous, $Tg=0$ and it is also possible to show that $T_tf \leq T_tg + T_th$ and that
$$T_th \leq Mh+|h|,$$
where $Mh$ is the maximal function of $h$:
$$Mh(b) = \sup_{t} \frac{1}{I_t} \int_{I_t} |h| dx.$$

By the above estimate, you can show that
$$ \{ x \hspace{0.1 cm}|\hspace{0.1 cm} Tf(x)> 2y\} \subset \{ x \hspace{0.1 cm}|\hspace{0.1 cm} Mh(x)> y\} \cup \{ x \hspace{0.1 cm}|\hspace{0.1 cm}|h(x)|> y\}$$
Now you have to estimate the measure of the members of the right hand side of the previous expression. Here is where you need some results regarding the Maximal function of $h$. More precisely, it is possible to show that
$$\left | \{ x \hspace{0.1 cm}|\hspace{0.1 cm} Mh(x)> y\} \right | \leq \frac{3}{y}\|h\|_{1}$$
and
$$ \left|\{ x \hspace{0.1 cm}|\hspace{0.1 cm}|h(x)|> y\} \right| \leq \frac{1}{y}\|h\|_{1}$$
see Rudin's book page 138.

Using the previous estimates and bearing in mind that $\|h\|_1 < \frac{1}{n}$, you find out that
$$\left|  \{ x \hspace{0.1 cm}|\hspace{0.1 cm} Mh(x)> y\} \cup \{ x \hspace{0.1 cm}|\hspace{0.1 cm}|h(x)|> y\}\right| \leq \frac{4}{yn}$$

Finally, you use the above fact to show that the set $ \{ x \hspace{0.1 cm}|\hspace{0.1 cm} Tf(x)> 2y\}$ has measure $0$. You consider the sets
$$E(y,n) = \{ x \hspace{0.1 cm}|\hspace{0.1 cm} Mh(x)> y\} \cup \{ x \hspace{0.1 cm}|\hspace{0.1 cm}|h(x)|> y\}$$
that arise when you take different continuous functions $g$ that approximate $f$. You can show that
$$ \{ x \hspace{0.1 cm}|\hspace{0.1 cm} Tf(x)> 2y\} \subset \bigcap_{n=1}^{\infty} E(y,n)$$
and by all the previous estimates, you can show that the set $\bigcap_{n=1}^{\infty} E(y,n)$ has measure zero and therefore,
$$ \left|\{ x \hspace{0.1 cm}|\hspace{0.1 cm} Tf(x)> 2y\}\right| =0 $$
for any $y>0$.

Rudin, Walter, Real and complex analysis., New York, NY: McGraw-Hill. xiv, 416 p. (1987). ZBL0925.00005.

ContentLicense: CC BY-SA 4.0


问:
Prime ideals in quotients of polynomial ring over finite field
Reading a book, i found this argument ($\mathbb{F}_2$ is the field with 2 elements):

consider the quotient ring:
$$\mathbb{F}_2[x]/(x+1)^2$$

Then it has only one prime ideal, namely the following:
$$(x+1)\mathbb{F}_2[x]/(x+1)^2$$

I've some questions:

what does the notation $(x+1)\mathbb{F}_2[x]/(x+1)^2$ means, i.e. what are the elements of this ideal?

how can i see that this ideal is prime?

how can i exclude that other primes are there, different from that?

Thanks for any help

答:
This is one of those cases where knowing less makes everything a lot simpler. Let $R$ be a commutative ring, and $I$ an ideal in $R$.

If $J$ is another ideal in $R$, the ideal $JR/I$ means the image of the ideal $J$ in $R/I$, given by the natural projection $p:R \to R/I$.

As rschwieb has pointed out below, $JR/I$ is actually just $(J+I)/I$.

In this case, $J=(x+1), R = \mathbb{F}_2[x], I = (x+1)^2$.

There is a one-to-one correspondence between ideals in $R$ containing $I$ and ideals of $R/I$, sending $A \to p(A)$ for $A$ an ideal in $R$, and $B \to p^{-1}(B)$, for $B$ an ideal in $R/I$. This is also a correspondence between prime ideals. Have you seen this result?

Thus what remains to be checked is that the ideal $(x+1)$ is the only prime ideal in $\mathbb{F}_2[x]$ containing $(x+1)^2$. Luckily, $\mathbb{F}_2[x]$ doesn't contain that many elements of small degree, so this can be checked manually.

ContentLicense: CC BY-SA 3.0


问:
Existence condition for rational polynomials
First of all sorry for the stupid question but my knowledge in mathematics is not big. I am studying by myself how determine the existence condition for a polynomial.
In particular I am not sure to perform correctly the following exercise.

My book says: determine the the existence condition of the following expression before simplifying this expression
$$\left(\frac{a^2}{4a^2+4ab+b^2}-\frac{a-b}{6a+3b}\right):\frac{a^3-b^3}{12a+6b}$$
First of all I would rewrite the expression in this way:
$$\frac{\left(\frac{a^2}{\color{red}{(2a+b)^2}}-\frac{a-b}{\color{green}{3(2a+b)}}\right)}{\frac{a^3-b^3}{\color{blue}{6(2a+b)}}}$$
so since I should have all denominators in the expression different from $0$, I have imposed the following:
$$\begin{cases} \color{red}{(2a+b)^2}\neq 0\\
\color{green}{3(2a+b)}\neq 0\\
\color{blue}{6(2a+b)}\neq 0\\
a^3-b^3\neq 0\\
\end{cases}\iff  b\neq -2a; a\neq b $$

$\textbf{My doubt:}$ I am not sure on what I have done. In particular I have imposed, in addition to the first three condition (red, blue and green) referred to the denominator of the fractions in the expression, also that the term $\frac{a^3-b^3}{\color{blue}{6(2a+b)}}$ has to be different from $0$.

I have thought that since it is a factor that the exercise tell us to divide thanks to the symbol "$:$", in addition to the existence condition on this fraction I should be sure that this fraction is not identically $0$ since it represents a denominator for the term $\left(\frac{a^2}{4a^2+4ab+b^2}-\frac{a-b}{6a+3b}\right)$: so I have to impose also the condition in black, right?

Can you help me please?

答:
Everything that you have done is correct, though I'm going to add some extra notes here as you seem a little uncertain and say that you're teaching yourself.

You're looking at what are called rational polynomials as they are the ratio of polynomials (or "fractions") and so the existence condition is basically that these things are not allowed to be undefined.  That happens if you divide by $0$, and so that is why you're checking the conditions that you're checking.

The exercise is first to determine if each of the polynomial fractions exists, and then to determine if their ratio (indicated by the ":" as you noted) exists, which again requires division as you have understood.

Everything else is fine; you can simplify your calculations a little by noticing that the term $(a+2b)$ is present and dominant in the red, green and blue terms, so all three are controlled by requiring $a\not=2b$.

Finally, yes, you need to be sure that $a^3 - b^3 \not= 0$ as well for the final ratio/division check.  This is the first time when it might matter what number field you're working in: $a^3-b^3 = (a-b)(a^2+2ab+b^2)$ and so if you're working over $\mathbb C$ then there will be two additional complex roots that $a$ and $b$ are not allowed to take on.


问:
$X=\{1,2,3\}$, which kind of topology on $X$ can be induced by a metric?
The question starts from a common quiz:

Given $(X,\tau)$, where $X$ is $\{1, 2, 3\},\tau$ is the indiscrete topology (Only $X$ and $\emptyset$), and the question is the existence of a metric which can induce this topology.
I don't think this kind of metric exist, for $\{1\}$ will be open. (If not wrong)

So I start to consider, which kind of topology of $X$ can be induced by a metric?
Just as $\tau=\{X,\emptyset,\{1\}\}$，or other patterns?

答: You're right that the trivial topology is not metrizable on this set.  A metric space is always Hausdorff, for instance, so if the underlying set is finite we get a discrete space.</p>


问:
Find the normalized eigenvectors of a matrix
Let $T=\begin{pmatrix}5 & 0 & 0 \\ 0 & 2 & i\\ 0 & -i & 2 \end{pmatrix}$.

I found the eigenvalues and eigenvectors already and they are $1,3,5$ and $\begin{pmatrix}0\\-i\\1\end{pmatrix}$,$\begin{pmatrix}0\\i\\1\end{pmatrix}$, and $\begin{pmatrix}1\\0\\0\end{pmatrix}$.

According to mathematica, their orthoganalized form will be $\left(
\begin{array}{ccc}
 1 & 0 & 0 \\
 0 & \frac{i}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
 0 & -\frac{i}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
\end{array}
\right)$

But I am getting wildly different answers when I try to find the normalized eigenvectors using the Gram-Schmidt process.

答: You may be right. There is no unique orthonormal basis. Just check that whatever you are getting are eigenvectors, are orthogonal and have norm $1$ each.</p>


答:
Given any  set $X$, containing at least two (distinct) elements, the function $d:X\times X\to\mathbb{R}$ given by
$$d(x,y)=1\ \ \hbox{if $x\neq y$}$$
$$d(x,x)=0 \ \ \forall x\in X$$
is a metric inducing the discrete topology.

Title: $X=\{1,2,3\}$, which kind of topology on $X$ can be induced by a metric?

Body:
The question starts from a common quiz:

Given $(X,\tau)$, where $X$ is $\{1, 2, 3\},\tau$ is the indiscrete topology (Only $X$ and $\emptyset$), and the question is the existence of a metric which can induce this topology.
I don't think this kind of metric exist, for $\{1\}$ will be open. (If not wrong)

So I start to consider, which kind of topology of $X$ can be induced by a metric?
Just as $\tau=\{X,\emptyset,\{1\}\}$，or other patterns?


问:
Show that $A=\{(x,y) \in \mathbb{R}^2 : x^2+y^2< 2y\}$ is open in $\mathbb{R}^2$.

Show that $A=\{(x,y) \in \mathbb{R}^2 : x^2+y^2< 2y\}$ is open in $\mathbb{R}^2$.

The set can be expressed as $$A = \{(x,y) \in \mathbb{R}^2 : x^2+y^2< 2y\} \\
=\{(x,y) \in \mathbb{R}^2 : x^2+y^2 -2y< 0\} = \{(x,y) \in \mathbb{R}^2 : x^2+(y-1)^2 < 1\} = B((0,1),1)$$

and since open balls are open sets $A$ would be open. However I'm trying to use the metric to show this. If $A=\{(x,y) \in \mathbb{R}^2 : x^2+y^2 -2y< 0\}$  and I let $(x,y)\in A$ and $B((x,y), r)$, where $r=\min \{|x|,|y|\}$. Now picking $(x',y') \in B((x,y),r)$ I would want to show that $B((x,y), r) \subset A$, but I'm not entirely sure how to do this?

Body: Let $P=(x,y)\in B_1 (0,1)$. If $r=d(P,\partial B_1(0,1))$, you can prove that $B_{r/2}(x,y)\subset B_1(0,1)$.


问:
Counting principles - why does this approach yield the wrong answer?
For the following question:

The barcode strings of a new product are created from four letters A,
B, C, D and ten digits 0, 1, 2, ..., 9. No three letters may be
written consecutively in a barcode string. There is no restriction on
the order in which the numbers are written. Find the number of barcode strings that can be created.

The correct answer is $7.66 \cdot 10^{10}$, which I can achieve by subtracting the number of barcodes with three or four consecutive letters from the total number of barcodes.

However, I found another approach which yields a slightly different answer (which I suspect is wrong, but I can't explain why):

By calculating:

$10! \cdot 11 \cdot 10 \cdot 9 \cdot 8$ (number of barcodes with four separated letters),

$+ 10! \cdot \binom{4}{2} \cdot 2 \cdot 11 \cdot 10 \cdot 9$ (number of barcodes with two letters grouped together and the other two separated)

$+ 10! \cdot \binom{4}{2} \cdot 2 \cdot 2 \cdot 11 \cdot 10$ (number of barcodes with two pairs of letters)

I get $8.14 \cdot 10^{10}$ which suggests that I've counted some possibilities several times. Where did I make a mistake?

答:
The answer should be -

$10! \cdot {11 \choose 4} \cdot4! + 10! \cdot 2 \cdot {4 \choose 2} \cdot {11 \choose 3} \cdot 3!+ 10! \cdot {4 \choose 2} \cdot {11 \choose 2} \cdot 2! \cdot 2!$

First term - all four alphabets are separate.

Second term - three groups of alphabets separated from each other (one group has two alphabets together)

Third term - two groups of alphabets separated from each other (each group has two alphabets together)

It is the last term where you have a mistake. we first make two ordered groups of alphabets by ${4 \choose 2}$. We then choose $2$ places for two groups out of $11$ places. We then permute both pairs within ($2! \cdot 2!$). Like other terms, we permute digits by $10! \ $.


问:
How to check if topologies are the same?
Let $(X,\mathcal{T})$ be topological space. Let $A$ be a set such that $A \subseteq X$. Show that family $\mathcal{T}_A := \{U \cup (V \cap A) \mid U, V \in \mathcal{T} \}$ is a topology on X. I have show shown that it is a topology, now I have to check if topologies are the same.

答: $A = \emptyset \cup (X \cap A) \in \mathcal{T}_A$. So the topologies are the same iff $A \in \mathcal{T}$ i.e. iff $A$ is open. Otherwise $A \in \mathcal{T}_A\setminus \mathcal{T}$ and the topology is strictly finer (taking $V=\emptyset$ in the definition shows that $\mathcal{T}\subseteq \mathcal{T}_A$ always).


问:
Sum of Random Variables...
Imagine we repeat the following loop some thousands of times:

$$
\begin{align}
& \text{array} = []\\
& \text{for n} = 1: 10 000 \\
& k = 0 \\
& \text{while unifrnd}(0,1) < 0.3 \\
& k = k + 1 \\
& \text{end} \\
& \text{if k} \neq 0 \\
& \text{array} = [\text{array,k}] \\
& \text{end} \\
\end{align}
$$

whereas "unifrnd(0,1)" means a random number drawn from the uniform distribution on the unit interval.

My question is then: What is then the value of k, which is the most often observed - except for k = 0? And is that the expectation of k?

Thanks very much

答: It appears you exit the loop the first time the random is greater than $0.3$.  In that case, the most probable value for $k$ is $0$.  It occurs with probability $0.7$.  The next most probable is $1$, which occurs with probability $0.3 \cdot 0.7$, because you need the first random to be less than $0.3$ and the second to be $0.7$.  In general, the probability of a value $k$ is $0.3^k\cdot 0.7$</p>

Title: Sum of Random Variables...


答: Either use the basic fact that all open balls are open in the metric topology or use that $A = f^{-1}[(-\infty,0)]$ where $f(x,y) = x^2 + y^2 -2y$ is continuous.

Body:
Show that $A=\{(x,y) \in \mathbb{R}^2 : x^2+y^2< 2y\}$ is open in $\mathbb{R}^2$.

The set can be expressed as $$A = \{(x,y) \in \mathbb{R}^2 : x^2+y^2< 2y\} \\
=\{(x,y) \in \mathbb{R}^2 : x^2+y^2 -2y< 0\} = \{(x,y) \in \mathbb{R}^2 : x^2+(y-1)^2 < 1\} = B((0,1),1)$$

and since open balls are open sets $A$ would be open. However I'm trying to use the metric to show this. If $A=\{(x,y) \in \mathbb{R}^2 : x^2+y^2 -2y< 0\}$  and I let $(x,y)\in A$ and $B((x,y), r)$, where $r=\min \{|x|,|y|\}$. Now picking $(x',y') \in B((x,y),r)$ I would want to show that $B((x,y), r) \subset A$, but I'm not entirely sure how to do this?


问:
About the existence of homotopy operator
What shown below is a reference from the text Analysis on Manifolds by James Munkres. I point out that the sybols $\Omega$ and $^*$ are used by Munkres to denote a set of differential forms and the pull back map respectively.

Definition

Let be $A$ and $B$ open sets in $\Bbb R^n$ and $\Bbb R^m$ respectively and let be $g$ and $h$ two $C^r$ map defined from $A$ into $B$. So we say taht $g$ and $h$ are differentiably homotpic of class $C^r$ if there exists a $C^r$ mapd $H$ from $A\times[0,1]$ into $B$ such that
$$
H(x,0):=g(x)\,\,\,\text{and}\,\,\,H(x,1)=h(x)
$$
for any $x\in A$. The map $H$ is called a differentiable homotopy between $g$ and $h$.

Theorem

Let be $A$ and $B$ open sets in $\Bbb R^n$ and $\Bbb R^m$, respectively. Let $g,h:A\rightarrow B$ be $C^\infty$ maps that are differentiably homotopic. Then there is a linear transformation
$$
\mathcal P:\Omega^{k+1}(B)\rightarrow\Omega^k(A)
$$
defined for $k>0$ such that for any form $\eta$ of order $k>0$
$$
d(\mathcal P\eta)+\mathcal P(d\eta)=h^*\eta-g^*\eta
$$
while for a form of $f$ of order $0$,
$$
\mathcal P(df)=h^*f-g^*f
$$

So to follow I summarise the proof of the previous theorem but if you like you can read it
here.

Well first of all Munkres consider a very special case. So given a neighborhood $U$ of $A\times [0,1]$ in $\Bbb R^{n+1}$ he defines the map $\alpha,\beta:A\rightarrow U$ given by the equations
$$
\alpha(x):=(x,0)\,\,\,\text{and}\,\,\,\beta(x):=(x,1)
$$
and thus observing that these two maps are differentiably homotopic he proves the existence of a linear operator $P$ from $\Omega^{k+1}(U)$ to $\Omega^k(A)$ havig the desidred properties. Then the general case follows by this. Inded if $g,h:A\rightarrow B$ are differetiably homotopic with respect $H:A\times [0,1]\rightarrow B$ then he defines the operator $\mathcal P$ through the equation
$$
\mathcal P\eta:=P(H^*\eta)
$$
for any $\eta\in\Omega^{k+1}(B)$ because (Munkres himself states what to follow) $H^*\eta$ is a $(k+1)$-form defined in a neighborhood of $A\times[0,1]$ and thus $P(H^*\eta)$ is a $k$-form defined in $A$.

So I do not really understand why the pull back of $H$ with respect $H$ is a $(k+1)$-form defined in $A\times[0,1]$. Indeed if $H$ goes from $A\times[0,1]$ to $B$ then the pull back $H^*$ goes from $\Omega^{k+1}(B)$ to $\Omega^{k+1}(A\times[0,1])$ and thus this is a very big problem: in particular it seems to me that if we consider the restriction $\tilde H$ of $H$ to $A\times[0,1)$ then the pull back $\tilde H^*$ define a $(k+1)$-form on $A\times[0,1)$ of class $C^r$ that can be extended to a $(k+1)$-form of class $C^r$ defined in an open neighborhood of $A\times[0,1)$ but unfortunately this open neighborhood is not generally a open neighborhood of $A\times[0,1]$. So why $H^*\eta$ is a form defined in an OPEN neighborhood $A\times[0,1]$? Could someone explain this to me, please?

答: One point is that if $S$ is an arbitrary (in particular, not necessarily open) subset of $\mathbb R^m$ and $f: S\to \mathbb R^n$, then to say $f$ is differentiable on $S$ means that $f$ is the restriction of a differentiable function defined on an open neighborhood of $S$: Munkres spells this out in the definition on p199, just before Lemma 23.1. This neighborhood and extended function are often not mentioned explicitly, but in your situation the map $H$ is implicitly taken to be defined on an open neighborhood of $A\times [0,1]$. Hence $H^*\eta$ is also defined and smooth on this neighborhood, which is what it means to say $H^*\eta$ is smooth on the non-open set $A\times [0,1]$.


问:
Solve transcendental equation / maximum likelihood estimation
Consider $(X_1,\dots X_n)$ with $$\mathbb P[X=k]=\begin{cases}p_0,&k=0\\\frac{(1-p_0)\lambda^k}{k!(e^\lambda)-1},&k\geq1\end{cases}$$

I want to determine the maximum likelihood estimator of $\lambda$

I arrive at $$\frac{\partial}{\partial \lambda}\log L((p_0,\lambda))=\sum_{i=1}^{n}1_{\{X_i\geq 1\}}(X_i)\big(\frac{X_i}{\lambda}-\frac{e^\lambda}{e^\lambda-1}\big)\stackrel{!}{=}0$$

Which is equivalent to
$$\frac{1}{\lambda}\frac{e^\lambda-1}{e^\lambda}=\frac{\sum_{i=1}^{n}1_{\{X_i\geq 1\}}(X_i)}{\sum_{i=1}^{n}1_{\{X_i\geq 1\}}(X_i)X_i}$$

How do I solve this equation? When can it be solved? (If I didn't make any mistakes this is a transcendental equation)

答:
Starting from @heropup's answer, there is only one explicit solution of the equation
$$\frac{\lambda}{1-e^{-\lambda}}=k$$
It is given by
$$\lambda=k+W\left(-e^{-k} k\right)$$ where $W(.)$ is the principal branch of Lambert function.

To evaluate it, since the argument is small, use the expansion for small $x$ as given in the linked page.

Otherwise, you can have good estimates using high order iterative methods. I give you a quite good one
$$\lambda\sim\frac{2 \left(2 k^6-26 k^5+72 k^4+562 k^3-313 k^2-285 k-12\right)}{4 k^5-62   k^4+264 k^3+590 k^2+146 k+3}$$
For $k=3$, it gives $\lambda\sim 2.82297$


问:
What is the domain of definition of $f(x,y)=\frac{y-x}{1+\ln(x)}$?
I have some confusion over the domain of definition of a function.

Clearly, in the case above, it must be true that $x>0$ in order for the function to be defined. Also there is a singularity when $x=e^{-1}$.

So am I correct in saying that the domain of definition is: $x>0$, $x\neq e^{-1}$ and $y\in \mathbb{R}$?

However, if we rearrange the function, we get: $$(1+\ln(x))f(x,y)=y-x$$

So $x=e^{-1}$ implies that $$0=y-e^{-1}$$

So this would imply that $f(x,y)$ takes infinitely many values at the point $(e^{-1},e^{-1})$

This is where my confusion lies - so does the point $(e^{-1},e^{-1})$ form part of the domain of definition?

答:
Strictly speaking, the expression for $f(x,y)$ is meaningless unless the precise domain of definition is provided. This is because a function, by definition, is a triple $(f,X,Y)$ where $X$ is the domain of definition, $Y$ is the range, and $f$ is a function.
However, mathematics is often not very strict, so that in "practice" we get an expression such as $f(x,y)$ for which we have to figure out the precise domain of definition. Of course, we must assume something about $(x,y)$. For example, here it is tacitly assumed that $(x,y)\in\mathbb{R}^2$. (but that's only because of convention; in truth, there is nothing to prevent an alien from interpreting $(x,y)$ as an element of the Cartesian product of any two sets). Therefore, we must find the largest possible subset of $\mathbb{R}^2$ for which the expression $f(x,y)$ is well defined. That will be our domain of definition. By inspection, we see that we must avoid nonpositive $x$ and such $x$ for which $\ln x=-1$. The first condition gives the set $A=\{(x,y)\in\mathbb{R}^2| x>0\}$ and the second condition gives $B=\{(x,y)\in\mathbb{R}^2| x\neq e^{-1}\}$. Thus, the domain of definition is $A\cap B$ - the set where both conditions are met.


问:
Evaluate $(2 + \frac{1}{\sqrt 2})^5 + (2 - \frac{1}{\sqrt 2})^5$
My question says:

Without using tables, find the value of
$$\left(2 + \frac{1}{\sqrt 2}\right)^5 + \left(2 - \frac{1}{\sqrt 2}\right)^5$$

I have said:

$$\left(2 + \frac{1}{\sqrt 2}\right) = 2\left(1 + \frac{1}{2\sqrt2}\right)$$

$$\implies\left(2 + \frac{1}{\sqrt 2}\right)^5 = 2^5\left(1 + \frac{1}{2\sqrt2}\right)^5$$

$$\left(2 + \frac{1}{\sqrt 2}\right)^5 = 2^5\Bigg(1 + \frac{5}{2\sqrt2} + \frac{5\cdot4}{2!}\left(\frac{1}{2\sqrt2}\right)^2 + \frac{5\cdot4\cdot3}{3!}\left(\frac{1}{2\sqrt2}\right)^3 + \frac{5\cdot4\cdot3\cdot2}{4!}\left(\frac{1}{2\sqrt2}\right)^4 + \left(\frac{1}{2\sqrt2}\right)^5\Bigg)$$

Similarly,

$$\left(2 - \frac{1}{\sqrt 2}\right)^5 = 2^5\Bigg(1 - \frac{5}{2\sqrt2} + \frac{5\cdot4}{2!}\left(\frac{1}{2\sqrt2}\right)^2 - \frac{5\cdot4\cdot3}{3!}\left(\frac{1}{2\sqrt2}\right)^3 + \frac{5\cdot4\cdot3\cdot2}{4!}\left(\frac{1}{2\sqrt2}\right)^4 -\left(\frac{1}{2\sqrt2}\right)^5\Bigg)$$

Adding gives,

$$\left(2 + \frac{1}{\sqrt 2}\right)^5 + \left(2 - \frac{1}{\sqrt 2}\right)^5 = 32\left(2 + \frac{5}{2} + \frac{5}{64}\right) = 146.5$$

But my text book says the answer is $149$.

Have I adopted the wrong approach or have I merely slipped up somewhere? I have checked my working several times.

Body:
Hint:

Use algebra:

If $a=2+\dfrac1{\sqrt2},b=2-\dfrac1{\sqrt2}$

$a+b=4, ab=\dfrac72$

$$a^2+b^2=(a+b)^2-2ab=?$$

$$a^3+b^3=(a+b)^3-3ab(a+b)=?$$

$$(a^3+b^3)(a^2+b^2)=?$$


问:
Algebra question with sequences.
I dont know how to solve it. Can somebody help but also tell me how to approach these kind of questions.

Terms of the sequence $a_1, a_2, \ldots, a_{2020}$ are obtained by the rule
$$
a_2 = \frac{1+a_1}{1-a_1}, \qquad
a_3 = \frac{1+a_2}{1-a_2}, \qquad \ldots, \qquad
a_{2020} = \frac{1+a_{2019}}{1-a_{2019}}
$$
If $a_{2020} = \frac{1}{5}$, find $a_1$.

答:
In these type of problems there are hidden patterns. So try to find a pattern for this problem. That is the easy way to solve this problem.
So just don't give up and try to find vales for $\mathbf{a_{2019},a_{2018},a_{2017},a_{2016},a_{2015}}$.
So you will get,

$$\mathbf{a_{2020}=\frac{1}{5}}$$
$$\mathbf{a_{2019}=\frac{-2}{3}}$$
$$\mathbf{a_{2018}=-5}$$
$$\mathbf{a_{2017}=\frac{3}{2}}$$
$$\mathbf{a_{2016}=\frac{1}{5}}$$
$$\mathbf{a_{2015}=\frac{-2}{3}}$$
So after that you will realised that this has a pattern and calculation goes as same.Now you can see that $\mathbf{a_{2017} \: to \: a_{2020}}$ there is a pattern $\mathbf{\frac{1}{5},\frac{-2}{3},-5,\frac{3}{2}}$. So then you realise that if we divide the term number by $\mathbf{4}$ then,

if we got mod as $\mathbf{0}$ then its value is $\mathbf{\frac{1}{5}}$,

if we get mod as $\mathbf{1}$ then its value is $\mathbf{\frac{3}{2}}$

if we get mod as $\mathbf{2}$ then its value is $\mathbf{-5}$

if we get mod as $\mathbf{3}$ then its value is $\mathbf{\frac{-2}{3}}$

So we have to calculate $a_{1}$. Lets divide its term number 1 by 4. So we get it's mod as 1. So $$\mathbf{a_{1}=\frac{3}{2}}$$

Title: Algebra question with sequences.


Title: Evaluate $(2 + \frac{1}{\sqrt 2})^5 + (2 - \frac{1}{\sqrt 2})^5$

Body:
My question says:

Without using tables, find the value of
$$\left(2 + \frac{1}{\sqrt 2}\right)^5 + \left(2 - \frac{1}{\sqrt 2}\right)^5$$

I have said:

$$\left(2 + \frac{1}{\sqrt 2}\right) = 2\left(1 + \frac{1}{2\sqrt2}\right)$$

$$\implies\left(2 + \frac{1}{\sqrt 2}\right)^5 = 2^5\left(1 + \frac{1}{2\sqrt2}\right)^5$$

$$\left(2 + \frac{1}{\sqrt 2}\right)^5 = 2^5\Bigg(1 + \frac{5}{2\sqrt2} + \frac{5\cdot4}{2!}\left(\frac{1}{2\sqrt2}\right)^2 + \frac{5\cdot4\cdot3}{3!}\left(\frac{1}{2\sqrt2}\right)^3 + \frac{5\cdot4\cdot3\cdot2}{4!}\left(\frac{1}{2\sqrt2}\right)^4 + \left(\frac{1}{2\sqrt2}\right)^5\Bigg)$$

Similarly,

$$\left(2 - \frac{1}{\sqrt 2}\right)^5 = 2^5\Bigg(1 - \frac{5}{2\sqrt2} + \frac{5\cdot4}{2!}\left(\frac{1}{2\sqrt2}\right)^2 - \frac{5\cdot4\cdot3}{3!}\left(\frac{1}{2\sqrt2}\right)^3 + \frac{5\cdot4\cdot3\cdot2}{4!}\left(\frac{1}{2\sqrt2}\right)^4 -\left(\frac{1}{2\sqrt2}\right)^5\Bigg)$$

Adding gives,

$$\left(2 + \frac{1}{\sqrt 2}\right)^5 + \left(2 - \frac{1}{\sqrt 2}\right)^5 = 32\left(2 + \frac{5}{2} + \frac{5}{64}\right) = 146.5$$

But my text book says the answer is $149$.

Have I adopted the wrong approach or have I merely slipped up somewhere? I have checked my working several times.

Body: Since$$\sum_\pm(a\pm b)^5=2\left(a^5+\binom{5}{2}a^3b^2+\binom{5}{4}ab^4\right)$$you want$$2a_5=2\left(2^5+10\cdot2^3/\sqrt{2}^2+5\cdot2/\sqrt{2}^4\right)=149.$$


问:
the norm of the trace as a linear functional on $\mathbb{C}^{n \times n}$
Let $T:\mathbb{C}^{n \times n} \rightarrow \mathbb{C}$ be a linear functional on the space of all $n \times n$ matrices whose entries are complex numbers. I need to show that the norm of the trace $||trace||$ " as a linear functional" is $\sqrt n$.
I have proved one direction as follows:

$|trace(A)| \leq ||trace||||A||$

this holds for all $A \in \mathbb{C}^{n \times n}$. So, pick $A=I_n$ identity matrix, we get:

$n \leq ||trace|| \sqrt n,   \quad \quad \quad \quad \quad      (1)$

(Here, $ ||I_n||^2=<I_n,I_n>=trace(I_n^{*}I_n)=n$, where $I_n^*$ represents the conjugate transpose of $I_n$.)

For the second direction, I know that

$||trace|| = \sup_{A\in \mathbb{C}^{n \times n}}=\{ |trace(A)| : ||A|| \leq 1$, but I don't know how can I get the upper bound $\sqrt n$ of the definition of the norm.

答:
Hint:

Let $e_1,\dots e_n$ be the standard basis of $\mathbb{C}^n$. Then for any $A\in\mathbb{C}^{n\times n}$, we have
$$\hbox{trace}A=\sum_{i=1}^n\langle Ae_i,e_i\rangle$$
where $\langle\cdot \rangle$ is the standard inner product. The result follows
from the CS inequality.


问:
If $a+b+c=3,$ show $a+b+c\geq ab+bc+ac$ by using AM-GM
A am having issues with a problem that states: let $a, b, c$ be positive real numbers such that $a + b + c = 3$. Show that $$a^bb^cc^a\le1$$

As this was an example problem, a solution path was provided but I am confused as to why the solution works. It writes:
$$1=\frac{a+b+c}{3}$$
$$\ge\frac{ab+bc+ca}{a+b+c}$$
$$\ge (a^bb^cc^a)^{\frac{1}{a+b+c}}$$

I understand how weighted AM-GM could show the second inequality but I can't seem to understand how one can show $a+b+c \ge ab+bc+ca$ which is what the first inequality implies.

Btw: at this point in the text, they have only proved AM-GM and weighted AM-GM. I can think of a way to show this using Chebyshev's Inequality but I was hoping to try to understand what the text meant by this specific solution path.

答:
Hint

$$(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)$$

but, by $MA\ge MG$:  $a^2+b^2\ge 2ab, a^2+c^2\ge 2ac, b^2+c^2\ge 2bc$, what gives you

$$a^2+b^2+c^2\ge ab+ac+bc.$$

Can you finish?


答: I am pretty sure your solution was close to perfect but I think you put that $(\frac{5*4*3*2}{4!}(\frac{1}{2*\sqrt{2}})^4) *2$  is $\frac{5}{64}$ but in reality it is $\frac{5}{32}$. Fix that and your solution gives you 149.


问:
Prove that $n^3+5n$ is always divisible by 6 using induction with mod
So I had to prove this in my calculus exam today

Prove that $n^3+5n$ is always divisible by 6.

I know that there have been other people around here having asked this already, but I chose a different approach and would really appreciate to hear whether my train of thought here was right or wrong. So I am especially not looking for the "right way" to prove it (i.e., I don't need a proper solution) I would just appreciate for somebody to assess my solution, if it's completely wrong or possibly not complete etc.

so, what I did:

Initial case prove that $n^3+5n \bmod 6 = 0$ for $n=1$.
$$1^3+5\cdot 1 \bmod 6 = 0 \Leftrightarrow 6 \bmod 6 = 0$$
which is trivially true.

Induction step Prove that for every $n$, if the statement holds for $n$, then it holds for $n + 1$.
$$((n+1)^3+5(n+1)) \bmod 6 =0 \\ \Leftrightarrow ((n^3+3n^2+3n+1)+5n+5) \bmod 6 =0 \\ \Leftrightarrow (n^3+5n \bmod 6) + (3n^2+3n+6 \bmod 6)=0 \\ \Leftrightarrow 0+(3n^2+3n+6 \bmod 6)=0 \\ \Leftrightarrow 3n^2+3n+6 \bmod 6 =0.$$

And this is true for all $n \in \mathbb{N}$, where from the third to the fourth line, we replaced with the induction hypothesis that $n^3+5n \bmod 6 =0$ . Take for instance $1, 2,3,...$ it is always divisible by 6 without leaving any rest!!

However, my other friends that took the exam said that this is in need of some extension to show that the last line actually holds true for whatever n you insert. But I cannot grasp what they mean; by the definition of the modulus, it should already trivially be given that the last line is right...

I'm really looking forward to your thoughts, do you think I would at least get partial points for this? :( I really don't see how this can be wrong...

Thanks so much,
Lin

答:
In view of the final step, $3n^2+3n+6\equiv 0\mod 6$ is equivalent to $3n^2+3n\equiv 0\mod 6$, i.e.,

$3(n^2+n)\equiv 0\mod 6$.

So it is sufficient to show that

$2\mid n^2+n$.

But $n^2+n = n(n+1)$ is a product of consecutive integers and one of which will be even. Thus $n^2+n$ is a multiple of $2$ and hence $3(n^2+n)$ is a multiple of $6$, as required.


答:
This is correct, but as your friends said, not proving the last line may cause some loss of points. To prove this, it's not very "long" or something. Just show that
$$3n^2+3n+6\equiv 0\ (\textrm{mod}\ 2)$$
Since it is trivial that it is $0\ \textrm{mod}\ 3$, the statement follows.

Hope this helps. Ask anything if not clear :)

Title: Prove that $n^3+5n$ is always divisible by 6 using induction with mod

Body:
So I had to prove this in my calculus exam today

Prove that $n^3+5n$ is always divisible by 6.

I know that there have been other people around here having asked this already, but I chose a different approach and would really appreciate to hear whether my train of thought here was right or wrong. So I am especially not looking for the "right way" to prove it (i.e., I don't need a proper solution) I would just appreciate for somebody to assess my solution, if it's completely wrong or possibly not complete etc.

so, what I did:

Initial case prove that $n^3+5n \bmod 6 = 0$ for $n=1$.
$$1^3+5\cdot 1 \bmod 6 = 0 \Leftrightarrow 6 \bmod 6 = 0$$
which is trivially true.

Induction step Prove that for every $n$, if the statement holds for $n$, then it holds for $n + 1$.
$$((n+1)^3+5(n+1)) \bmod 6 =0 \\ \Leftrightarrow ((n^3+3n^2+3n+1)+5n+5) \bmod 6 =0 \\ \Leftrightarrow (n^3+5n \bmod 6) + (3n^2+3n+6 \bmod 6)=0 \\ \Leftrightarrow 0+(3n^2+3n+6 \bmod 6)=0 \\ \Leftrightarrow 3n^2+3n+6 \bmod 6 =0.$$

And this is true for all $n \in \mathbb{N}$, where from the third to the fourth line, we replaced with the induction hypothesis that $n^3+5n \bmod 6 =0$ . Take for instance $1, 2,3,...$ it is always divisible by 6 without leaving any rest!!

However, my other friends that took the exam said that this is in need of some extension to show that the last line actually holds true for whatever n you insert. But I cannot grasp what they mean; by the definition of the modulus, it should already trivially be given that the last line is right...

I'm really looking forward to your thoughts, do you think I would at least get partial points for this? :( I really don't see how this can be wrong...

Thanks so much,
Lin


问:
How can it be derived? (Law of the unconscious statistician)
On pg. 7 of Financial Calculus: An Introduction to Derivative Pricing by Martin Baxter & Andrew Rennie, it states:

My Understanding

I did try to derive this picture. But I don't understand how this outcome that is $S_0\exp(\mu -0.5\sigma^2)$
is derived. Could I get any hint of it?

答: The stated law is the all-too-familiar definition of $\Bbb E(h(X))$. Since$$X=\ln S_T-\ln S_0=\ln\tfrac{S_T}{S_0},$$we have $S_T=S_0e^X$. This function of $X$ has mean $\int_{-\infty}^\infty\frac{S_0}{\sigma\sqrt{2\pi}}e^{S_0x-(x-\mu)^2/(2\sigma)^2}$, which evaluates to $\Bbb E(S_T)=S_0e^{\mu+\tfrac12\sigma^2}$.


答:
The quick answer is that $Y=e^X\sim \text{Lognormal}$ thus its mean is well known

If you want to do all the calculation with the gaussian distribution, it is not difficult; try, it is a good exercise

Title: How can it be derived? (Law of the unconscious statistician)

Body:
On pg. 7 of Financial Calculus: An Introduction to Derivative Pricing by Martin Baxter & Andrew Rennie, it states:

My Understanding

I did try to derive this picture. But I don't understand how this outcome that is $S_0\exp(\mu -0.5\sigma^2)$
is derived. Could I get any hint of it?


问:
Understating the acceleration equation in terms of velocity
Sorry if this is considered a ‘basic’ question, but I’ve been stuck on this for a while now and can’t find much help. I know $\frac{d}{dx} \left(\frac{1}{2} v^2\right) =$ acceleration. But I do not understand what is wrong with my working out.

$$\left(\frac{d}{dt}\right)\left(\frac{dx}{dt}\right) = \left(\frac{d}{dx}\right)\left(\frac{dx}{dt}\right)\left(\frac{dx}{dt}\right) = \frac{d}{dx} (v^2)$$ I don’t know why I am missing the half that is in the original equation. I have seen the proof for it, but I do not exactly know what wrong assumption I am making. Thank you.

答:
Your working should go like this:

$a = \frac{dv}{dt}= \frac{dv}{dx} \cdot \frac {dx} {dt} = v\frac{dv} {dx}$

and at this point observe that $d(v^2) = 2vdv$, which allows you to rewrite the final expression as:

$ \frac{d(\frac 12 v^2)}{dx}$ as required.

Your error was in thinking $\frac {d}{dx} (\frac{dx} {dt})$ is a multiplication. It is not. It is differentiation of the velocity with respect to the displacement.


问:
Expected value of the sum of two dice.
Came across this question:

We roll two dice. Let $X$ be the sum of the two numbers appearing on the dice.

Find the expected value of $X$.

Find the variance of $X$.

I'm not sure how to do either, but this was my thinking for part 1:

$$E(X) = 2((1/6)^2) + \\
3(2(1/6)^2) + \\
4(2(1/6)^2 + (1/6)^2) + \\
5(2(1/6)^2 + 2(1/6)^2) + \\
6(2(1/6)^2 + 2(1/6)^2 + (1/6)^2) + \\
7(2(1/6)^2 + 2(1/6)^2 + 2(1/6)^2) + \\
8(2(1/6)^2 + 2(1/6)^2 + (1/6)^2) + \\
9(2(1/6)^2 + 2(1/6)^2) + \\
10(2(1/6)^2 + (1/6)^2) + \\
11(2(1/6)^2) + \\
12((1/6)^2)$$

The reason I multiplied some by 2 is because it could possibly switch up or permute. So, for example, for 4, the two sums that could give us 4 are (3,1) and (2,2), so I multiplied one of the probabilities by 2 because (3,1) could come as either (3,1) or (1,3) whereas (2,2) can only come in one form.

Title: Expected value of the sum of two dice.

Body:
$$E(X)=E(X_1+X_2)=E(X_1)+E(X_2)=3.5+3.5=7$$

can you calculate $V(X)$?

$$V(X_1)=E(X_1^2)-E^2(X_1)\approx 2.917$$

thus

$$V(X)=2\times 2.917=5.8\overline{3}$$


答:
What you have is correct so far.

However, a simpler approach is to write $X=X_1+X_2$, where $X_1,X_2$ are the values on the individual dice, and use the facts that:

$\mathbb{E}(X_1+X_2)=\mathbb{E}(X_1)+\mathbb{E}(X_2)$ for any two variables $X_1,X_2$;

$\mathrm{Var}(X_1+X_2)=\mathrm{Var}(X_1)+\mathrm{Var}(X_2)$ provided $X_1,X_2$ are independent.

Title: Expected value of the sum of two dice.

Body:
Came across this question:

We roll two dice. Let $X$ be the sum of the two numbers appearing on the dice.

Find the expected value of $X$.

Find the variance of $X$.

I'm not sure how to do either, but this was my thinking for part 1:

$$E(X) = 2((1/6)^2) + \\
3(2(1/6)^2) + \\
4(2(1/6)^2 + (1/6)^2) + \\
5(2(1/6)^2 + 2(1/6)^2) + \\
6(2(1/6)^2 + 2(1/6)^2 + (1/6)^2) + \\
7(2(1/6)^2 + 2(1/6)^2 + 2(1/6)^2) + \\
8(2(1/6)^2 + 2(1/6)^2 + (1/6)^2) + \\
9(2(1/6)^2 + 2(1/6)^2) + \\
10(2(1/6)^2 + (1/6)^2) + \\
11(2(1/6)^2) + \\
12((1/6)^2)$$

The reason I multiplied some by 2 is because it could possibly switch up or permute. So, for example, for 4, the two sums that could give us 4 are (3,1) and (2,2), so I multiplied one of the probabilities by 2 because (3,1) could come as either (3,1) or (1,3) whereas (2,2) can only come in one form.


问:
Is My understanding of complete relations correct?
I am using "Foundations of Mathematical Economics" by Michael Carter. The problem 1.16 of the book: Consider the relations $<$, $\le$ and $=$ on $\mathbb R$. Which of the above properties (reflexive,complete,transitivity,symmetric,asymmetric,antisymmetric) do they satisfy?

About the complete properties, $<$ is NOT complete because either $A  B$ BUT not both. Also, $=$ is complete because $A=B$ or $B=A$ or both, but the solution manual shows that $<$ IS complete and $=$ is NOT complete.

Is this an error? Truthfully, this book is good but full of convoluted details without mentioning in the book.

I thank you very much.

答:
If is not true that for EVERY value of $A$ and $B$ either $A=B$ or $B=A$ or both.   For example, if $A=1$ and $B=2$, then this is false.  If a statement says "For every . . . '", then the way to show it is false is to show that there is a counterexample.

The relation $\le$ is complete because for every $A$ and every $B$, either $A\le B$ or $B\le A$ or both.  The relation $<$ is not complete because instances in which $A=B$ are counterexamples.

Possibly the authors used a definition of "complete" that said that for every $A$ and every $B$, if $A$ is not the same as $B$, then either $A$ is related to $B$ or $B$ is related to $A$  or both.


答:
Your friends are right.
You need to explain why the last equivalent equation is an identity, meaning it solves for any integer n greater than zero.

To do that you show that $3n^2+3n+6$ is divisible both by 2 and 3.
Divisibility by 3 is obvious since the expression is a multiple of 3. You show that by rewriting the expression into $3(n^2+n+2)$
Divisibility by 2 results from observing that the product $n(n+1)$ of two consecutive numbers is always an even number: $n^2+n+2=n(n+1)+2=2k\cdot(2k\pm1) +2=2[k(2k\pm1)]$


答:
You forgot to multiply the last term by $2$. It should be
$$32\left(2+\frac52+\frac{5}{32}\right)=\boxed{149}$$
as expected.

Hope this helps. Ask anything if not clear :)


答:
Just to add to Deepak's answer, the result he quotes in the first line actually holds for total derivatives as well. Perhaps this proof is useful to the OP

$$\frac{d v(x, t)}{dt}
= \frac{\partial v}{\partial t} + \frac{\partial v}{\partial x} \frac{dx}{dt}
= \frac{\partial v}{\partial t} + \frac{\partial v}{\partial x} v $$

$$\frac{d v(x, t)}{dx}
= \frac{\partial v}{\partial x} + \frac{\partial v}{\partial t} \frac{dt}{dx}
= \frac{\partial v}{\partial x} + \frac{\partial v}{\partial t} \frac{1}{v} $$

Multiplying lower equation by $v$ and subtracting the two yields

$$\frac{dv}{dt} = v \frac{dv}{dx}$$

Title: Understating the acceleration equation in terms of velocity

Body:
Sorry if this is considered a ‘basic’ question, but I’ve been stuck on this for a while now and can’t find much help. I know $\frac{d}{dx} \left(\frac{1}{2} v^2\right) =$ acceleration. But I do not understand what is wrong with my working out.

$$\left(\frac{d}{dt}\right)\left(\frac{dx}{dt}\right) = \left(\frac{d}{dx}\right)\left(\frac{dx}{dt}\right)\left(\frac{dx}{dt}\right) = \frac{d}{dx} (v^2)$$ I don’t know why I am missing the half that is in the original equation. I have seen the proof for it, but I do not exactly know what wrong assumption I am making. Thank you.


问:
Finding $\sum_{r=2}^{\infty}\ln\left(1-\frac{1}{r^3}\right)$

Find $$\sum_{r=2}^{\infty}\ln\left(1-\frac{1}{r^3}\right)$$

I solved the simpler versions of these:

$$\sum_{r=2}^{\infty}\ln\left(1-\frac{1}{r}\right)\;\;,\;\;\sum_{r=2}^{\infty}\ln\left(1-\frac{1}
{r^2}\right)$$

by factorizing which yields a telescoping sum. However the cube version certainly breaks this trend.

What else can be done here? Does a closed form exist?

Body:
Hint:

$$1-\dfrac1{r^3}=\dfrac{(r-1)(r^2+r+1)}{r^3}=\dfrac{f(r-1)g(r-1)}{f^3(r)}$$

where $g(r-1)=r^2+r+1\implies g(r)=(r+1)^2+(r+1)+1=r^2+3r+3$

and $f(r)=r$

Similarly,  $$1-\dfrac1{(r+1)^3}=\dfrac{r(r^2+3r+3)}{(r+1)^3}=\dfrac{f(r)g(r)}{f^3(r+1)}$$

$$\implies\prod_{r=2}^n\left(1-\dfrac1{r^3}\right)=\prod_{r=2}^ng(r)\cdot\dfrac{f(1)}{f^2(2)f^2(3)\cdots f^2(n)f^3(n+1)}=\dfrac{f(1)}{f(n+1)}\prod_{r=2}^n\dfrac{g(r)}{f^2(r)}$$

$\prod_{r=2}^nf(r)=n!$


问:
$X=\{1,2,3\}$, which kind of topology on $X$ can be induced by a metric?
The question starts from a common quiz:

Given $(X,\tau)$, where $X$ is $\{1, 2, 3\},\tau$ is the indiscrete topology (Only $X$ and $\emptyset$), and the question is the existence of a metric which can induce this topology.
I don't think this kind of metric exist, for $\{1\}$ will be open. (If not wrong)

So I start to consider, which kind of topology of $X$ can be induced by a metric?
Just as $\tau=\{X,\emptyset,\{1\}\}$，or other patterns?

答: In a metric topology any set of the form $\{x\}$ is closed. So if $\mathcal{T}$ is a metric topology on $X$, a finite set, any subset of $X$ is finite, and so closed ( as a finite union of singletons), and so any subset of $X$ is open too. It follows that $\mathcal{T}=\mathscr{P}(X)$, the discrete topology. This topology is indeed induced by the so-called discrete metric $d(x,y)=1$ if $x \neq y$, so a metric topology on $\{1,2,3\}$ can only be the discrete topology.


问:
How do you solve the following integral?
How do you solve the following integral
$$\int_0^{2\pi}\sqrt{\frac{5}{4} + \cos(t)} \, dt.$$

It has been a while since I have done integration and I think somehow I could multpily by the conjugate but I am unsure.  Any assistance would be appreciated.

This problem is part of a larger problem to find the length of the cardiod.

$$r(\theta)=\frac{1}{2}+\cos(\theta).$$
This integral is the length integral and I originally just found a numerical approximation but after reviewing my problem my teacher said there is a nice way to antidifferentiate but I was unsure how to go about that.

答:
Your are facing an interesting problem. Not repeating what has been already answered, using Wolfram convention
$$\int_0^{2\pi}\sqrt{k + \cos(t)} \, dt=4 \sqrt{k+1}\, E\left(\frac{2}{k+1}\right)$$ In your case, $k$ is close to $1$. Using expansions
$$E\left(\frac{2}{k+1}\right)=1-\frac{1}{8} (k-1) (\log (k-1)+1-5 \log (2))+$$ $$\frac{1}{256} (k-1)^2 (10 \log
   (k-1)+19-50 \log (2))-$$ $$\frac{1}{2048} (k-1)^3 (31 \log (k-1)+68-155 \log (2))+O\left((k-1)^4\right)$$

Using $k=\frac 54$, this would give
$$\int_0^{2\pi}\sqrt{\frac{5}{4} + \cos(t)} \, dt\sim \frac{3 (127516+26649 \log (2))}{65536} =6.68279$$ which is not bad for a shortcut calculation.


问:
How is a Riemann surface different from an ordinary surface?
How is a Riemann surface different from an ordinary surface? I'm studying Riemann surfaces and as I understand they're just differential manifolds of complex dimension one. How is that different from just a surface, so a differential manifold of real dimension two? Is the difference (if any) only algebraic? Or are they topologically different?

Just to be sure: the sphere and the torus are both Riemann surfaces, but they're also normal surfaces right?

答:
A Riemann surface is just a complex manifold of complex dimension 1.  In other words, a Riemann surface is a complex curve.

Now here's the point.  A complex curve is of course a real smooth orientable manifold of dimension 2, but the converse is also true.

Any orientable 2-dimensional real smooth manifold $M$ is always going to be a complex manifold. Hence, $M$ always admits the structure of a complex curve.  Here's how.

Since $M$ is orientable, it has a volume form $\omega$ which in this case is a nowhere vanishing 2-form.  In particular, $\omega$ is closed since $\dim M=2$ and all $k$-forms on $M$ are zero for $k>2$.  Hence, $(M,\omega)$ is a symplectic manifold.  Its a standard result from symplectic geometry that every symplectic manifold admits an almost complex structure $J$. Now $(M,J)$ is a complex manifold iff $J$ is integrable, that is, its Nijenhuis tensor vanishes.  The Nijenhuis tensor associated to $J$ takes two vector fields and outputs another vector field:
$$
N(X,Y)=J[X,JY]+J[JX,Y]+[X,Y]-[JX,JY]
$$
Let $p\in M$ and let $X$ be any vector field such that $X_p\neq 0$.  Since $\dim M=2$, $X_p$ and $JX_p$ both span $T_pM$.  Since $N$ is a tensor, $N_p=0$ iff $N(X,JX)=0$. By direct calculation,
$$
\begin{align}
N(X,JX)&=J[X,J^2X]+J[JX,JX]+[X,JX]-[JX,J^2X]\\
&=-J[X,X]+J[JX,JX]+[X,JX]+[JX,X]\\
&=0+0+[X,JX]-[X,JX]\\
&=0
\end{align}
$$
Therefore $N_p=0$.  Since $p\in M$ is arbitrary, we have $N\equiv 0$.  By the Nijenhuis theorem, $(M,J)$ is a complex manifold.

So the takeaway from all this is that real smooth orientable manifolds of dimension 2 and Riemann surfaces are one and the same.  (When I say real manifold I also mean without boundary.)

OwnerDisplayName: Mauro Giliberti


问:
Find the polynomial $f(x)$ which have the following property
Find the polynomial $p(x)=x^2+px+q$ for which $\max\{\:|p(x)|\::\:x\in[-1,1]\:\}$ is minimal.

This is the 2nd exercise from a test I gave, and I didn't know how to resolve it. Any good explanations will be appreciated. Thanks!

答:
An answer that uses more advanced mathematics (which is probably not what you want) ...

The monic polynomial of degree $n$ that has minimal sup-norm on $[-1,1]$ is $T_n(x)/2^{n-1}$, where $T_n(x)$ is the Chebyshev polynomial of degree $n$. In fact, this minimum norm property can serve as the defining property of Chebyshev polynomials.

In our case, $n=2$, so the desired polynomial is $T_2(x)/2$ = $(2x^2 - 1)/2 = x^2 - \tfrac12$.

More about Chebyshev polynomials here.


问:
Show that $A=\{(x,y) \in \mathbb{R}^2 : x^2+y^2< 2y\}$ is open in $\mathbb{R}^2$.

Show that $A=\{(x,y) \in \mathbb{R}^2 : x^2+y^2< 2y\}$ is open in $\mathbb{R}^2$.

The set can be expressed as $$A = \{(x,y) \in \mathbb{R}^2 : x^2+y^2< 2y\} \\
=\{(x,y) \in \mathbb{R}^2 : x^2+y^2 -2y< 0\} = \{(x,y) \in \mathbb{R}^2 : x^2+(y-1)^2 < 1\} = B((0,1),1)$$

and since open balls are open sets $A$ would be open. However I'm trying to use the metric to show this. If $A=\{(x,y) \in \mathbb{R}^2 : x^2+y^2 -2y< 0\}$  and I let $(x,y)\in A$ and $B((x,y), r)$, where $r=\min \{|x|,|y|\}$. Now picking $(x',y') \in B((x,y),r)$ I would want to show that $B((x,y), r) \subset A$, but I'm not entirely sure how to do this?

答:
There might some problems with what you are trying... for example $(x,y)=(0,1)\in B((0,1),1)$
then $r=\text{min}\{|0|,|1|\}=0$ and the ball $B((0,1),0)=\emptyset\subset B((0,1),1)$. The radius should be positive.

Even for other points that $x\neq0$ will also have some issues, you can see this by drawing a point near enough to the line of circumference.

A solution could be achieved by setting $r=1-d((0,1),(x,y))$ where $d$ is the distance function.


问:
Why the étale cohomologies of small and big étale sites are coincide
Let $(Sch/S)_{et}$ be the big étale site and $S_{et}$ be the small site, the theorem in StacksProject claims their cohomologies are the same, but the proof there uses many concepts like cocontinues. Are there brief ways to prove it or are there any other references?

答:
you can look at the Milne book about etale cohomology. the point is that we have a natural map $f$ from sheaves on big etale site to sheaves on small etale site(restriction), which is exact because covers are the same and send Injectives
objects to acyclic object: the chech cohomology of injective object in big etale site is zero almost by definition and again because covers are the same the chech cohomology is equal in the small and the big site, hence restriction of an injective object has zero chech cohomology in the small site and it is a famous theorem that such objects are acyclic(you can see the proof in Milne book) hence $f$ of injective resolution for $F$ gives an acyclyc resolution for $f(F)$. we also by definition have $\Gamma(F)=\Gamma(f(F))$ so the cohomologies are the same.

this is the usual technique for proving equality of cohomology on two different site you have to prove that $f_{\star}$ is exact then the rest of argument work.


问:
Given a finite solvable group $G$, prove that a minimal normal subgroup $H$ is a $p$-group
Given a finite solvable group $G$, and a minimal normal subgroup $H$, prove that $H$ is a $p$-subgroup for some prime $p$.

My Attempt:

I am trying to write this proof without using the term "characteristic subgroup". I'm aware to the fact that by proving that $p$-subgroup of $M$ is a characteristic subgroup will finish the proof.

Let $p$ be a prime number such that $p$ is a divisor of the order of $M$. By Sylow theorem, there exists a $p$-subgroup of $M$, let it be called $S$.

I'd like to prove that every element of $M$ is of order $p^n$, for some $n \in \mathbb{N}$.
let $s\in S, \  g\in G: g^{-1}sg \in M$, as $S\le M$ and $M$ is normal in $G$. but why and how I can prove that $=g^{-1}sg$ is of order $p^m$, for some $m\in \mathbb{N}: m\le n$.

答: Since $G$ is solvable, $H$ is solvable, so its derived series must eventually end at $1$ and since $H$ is non-trivial we have $H' \lt H$. But $H' \text { char } H \unlhd G$, so $H' \unlhd G$. But $H$ is minimal normal, whence $H'=1$, that is $H$ is abelian. Since $H$ is finite we can find an element in $H$ of order $p$, a prime (Cauchy!). Since $H$ is abelian the set $S=\{h \in H : h^p=1\}$ is in fact a (non-trivial) characteristic subgroup of $H$ and hence it is normal in $G$. So we must have $H=S$ and thus $H$ is an elementary abelian $p$-group.


问:
Convergence of $\;\sum^\infty_{n=1}\frac{(n+1)^{n-1}}{(-n)^n}\;$
Does the following series converge?
$$\sum^\infty_{n=1}\frac{(n+1)^{n-1}}{(-n)^n}$$
I've tried both the root- and the ratiotest, where the resulting sequence seems to always converge to $1$. I've also tried the Leibniz test where:
$$\left|\frac{(n+1)^{n-1}}{(-n)^n}\right| \;=\; \left(\frac{n+1}{n}\right)^n \;\cdot\; \frac{1}{n+1} \; \longrightarrow\; e\cdot0\;=\; 0$$
but am unable to show that the absolute value of the sequence is also decreasing. Any hints?

Body: Let $a_n=(n+1)^{n-1}n^{-n}$ be the absolute value of the term; then, for $n>1$, $$a_n/a_{n-1}=(n+1)^{n-1}n^{-n}n^{-n+2}(n-1)^{n-1}=(1-n^{-2})^{n-1}<1.$$


问:
Trigonometric identities using $\sin x$ and $\cos x$ definition as infinite series
Can someone show the way to proof that $$\cos(x+y) = \cos x\cdot\cos y - \sin x\cdot\sin y$$ and $$\cos^2x+\sin^2 x = 1$$ using the definition of $\sin x$ and $\cos x$ with infinite series.
thanks...

答:
Let me do a different one.  Begin with
$$
\sin x = \sum_{k=0}^\infty \frac{(-1)^k x^{2k+1}}{(2k+1)!},
\qquad
\cos x = \sum_{k=0}^\infty \frac{(-1)^k x^{2k}}{(2k)!}
$$
So compute
$$\begin{align}
\sin(x+y) &= \sum_{k=0}^\infty \frac{(-1)^k(x+y)^{2k+1}}{(2k+1)!}
\\ &=
\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}\sum_{j=0}^{2k+1} \binom{2k+1}{j} x^j y^{2k+1-j}
\\ &=
\sum_{k=0}^\infty (-1)^k\sum_{j=0}^{2k+1} \frac{1}{j!(2k+1-j)!} x^j y^{2k+1-j}
\\ &=
\sum_{j=0}^\infty \frac{x^j}{j!}\sum_k \frac{(-1)^k}{(2k+1-j)!} y^{2k+1-j}
\end{align}$$
where the inner sum is over all $k$ such that $2k+1 \ge j$.  Consider two cases for the inner sum, $j$ even and $j$ odd.

If $j=2n$, then $2k+1 \ge j$ iff $2k+1 \ge 2n$ iff $k \ge n$.  So the $k$-sum is:
$$
\sum_{k=n}^\infty \frac{(-1)^k}{(2k+1-2n)!} y^{2k+1-2n}
$$
Use change of variables $i=k-n$ to get
$$
\sum_{i=0}^\infty \frac{(-1)^{i+n}}{(2i+1)!} y^{2i+1} = (-1)^n \sin y .
$$

If $j=2n+1$, then $2k+1 \ge j$ iff $2k+1 \ge 2n+1$ iff $k \ge n$.  So the $k$-sum is
$$
\sum_{k=n}^\infty \frac{(-1)^k}{(2k+1-2n-1)!}y^{2k+1-2n-1}
$$
Again use change of variables $i=k-n$ to get
$$
\sum_{i=0}^\infty \frac{(-1)^{i+n}}{(2i)!} y^{2i} = (-1)^n \cos y.
$$
So finally we have
$$
\sin(x+y) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!} \sin  y
+\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}\cos y
= \cos x \sin y + \sin x \cos y.
$$


Title: Convergence of $\;\sum^\infty_{n=1}\frac{(n+1)^{n-1}}{(-n)^n}\;$

Body:
Does the following series converge?
$$\sum^\infty_{n=1}\frac{(n+1)^{n-1}}{(-n)^n}$$
I've tried both the root- and the ratiotest, where the resulting sequence seems to always converge to $1$. I've also tried the Leibniz test where:
$$\left|\frac{(n+1)^{n-1}}{(-n)^n}\right| \;=\; \left(\frac{n+1}{n}\right)^n \;\cdot\; \frac{1}{n+1} \; \longrightarrow\; e\cdot0\;=\; 0$$
but am unable to show that the absolute value of the sequence is also decreasing. Any hints?

Body:
We have
$$ \frac{|a_{n+1}|}{|a_n|} = \frac{(n+2)^n}{(n+1)^{n+1}}\frac{n^n}{(n+1)^{n-1}} = \frac{(n+2)^n n^n}{(n+1)^{2n}} = \left(\frac{n^2+2n}{n^2+2n+1}\right)^n < 1$$
so $|a_{n+1}|<|a_n|$ and you can use the Leibniz test.


问:
Algebra question with sequences.
I dont know how to solve it. Can somebody help but also tell me how to approach these kind of questions.

Terms of the sequence $a_1, a_2, \ldots, a_{2020}$ are obtained by the rule
$$
a_2 = \frac{1+a_1}{1-a_1}, \qquad
a_3 = \frac{1+a_2}{1-a_2}, \qquad \ldots, \qquad
a_{2020} = \frac{1+a_{2019}}{1-a_{2019}}
$$
If $a_{2020} = \frac{1}{5}$, find $a_1$.

Title: Algebra question with sequences.

Body:
Given equatiobn is:
$$A_{n+1}=\frac{1+A_n}{1-A_n} \implies A_{n+1} (1-A_n)=1+A_n~~~(1)$$
Let $$1-A_n=\frac{B_n}{B_{n-1}} \implies A_{n+1}=1-\frac{B_{n+1}}{B_n}$$Then (1) becomes
$$\frac{B_n-B_{n+1}}{B_n}\frac{B_n}{B_{n-1}}=2-\frac{B_n}{B_{n-1}}$$
$$\implies B_n-B_{n+1}=2B_{n-1}-B_n \implies B_{n+1}+2B_{n-1}-2B_n=0$$
Let $B_n=x^n$, then $x^2-2x+2=0 \implies x=(1\pm i)=a,b$ $$\implies B_n=P a^n +Q b^n\implies A_n=\frac{B_{n-1}-B_n}{B_n}=\frac{P a^{n-1}(1-a)+Qb^{n-1}(1-b)}{Pa^{n-1}+Qb^{n-1}}$$
So $$\implies A_n=-i\frac{a^{n-1}-Rb^{n-1}}{a^{n-1}+Rb^{1-n}}=-i\frac{1-Re^{i\pi(1-n)/2}}{1+Re^{i\pi(1-n)/2}}$$
$$A_{2020}=-i\frac{1-iR}{1+iR}=\frac{1}{5}\implies R=\frac{-1-5i}{i+5} \implies A_1=-i\frac{1-R}{1+R}=\frac{3}{2}$$


问:
Resultant of two polynomials in $K[X,Y]$ is a polynomial in $K[X]$ with degree less than the product of the original ones
I'm trying to prove the following:

Let $f, g$ be two polynomials in $K[X,Y]$ with $K$ a field and consider them as polynomials in one indeterminate with coefficientes in$ K[X]$, ie, as elements of $K[X][Y]$, then the resultant of these two polynomials would be another polynomial $d \in K[X]$ and I want to show that $\operatorname{deg}(d) \leq \operatorname{deg}(f)\operatorname{deg}(g)$.

I found it in a textbook on algebraic geometry as part of some proof but there is no hint or reference on how to prove it and I have no idea on how to get the result.

Could someone give a proof about this result?

Thanks a lot.

Body:
Here is an example. Let $f=x^2y^3+7xy^5+17$ and $g=x^5+6x+5y^2x-1$. Then the resultant is given by
$$
{\rm Res}(f,g,y)=49x^{27} - 70x^{24} + 1470x^{23} - 245x^{22} + 25x^{21} - 1680x^{20} + 17920x^{19} - 5880x^{18}
 + 940x^{17} - 15195x^{16} + 110880x^{15} - 53340x^{14} + 11520x^{13} - 61870x^{12} + 347835x^{11} - 216720x^{10} + 58600x^9 - 99300x^8 + 442199x^7 - 332665x^6 + 1010645x^5 -
17710x^4 + 1470x^3 - 49x^2.
$$
So we see that the degree is less than $\deg(f)\deg(g)=6\cdot 5=30$.
Since the resultant is a determinant polynomial coming from the Sylvester matrix, the estimate follows from it, see here.


问:
Prove that $n^3+5n$ is always divisible by 6 using induction with mod
So I had to prove this in my calculus exam today

Prove that $n^3+5n$ is always divisible by 6.

I know that there have been other people around here having asked this already, but I chose a different approach and would really appreciate to hear whether my train of thought here was right or wrong. So I am especially not looking for the "right way" to prove it (i.e., I don't need a proper solution) I would just appreciate for somebody to assess my solution, if it's completely wrong or possibly not complete etc.

so, what I did:

Initial case prove that $n^3+5n \bmod 6 = 0$ for $n=1$.
$$1^3+5\cdot 1 \bmod 6 = 0 \Leftrightarrow 6 \bmod 6 = 0$$
which is trivially true.

Induction step Prove that for every $n$, if the statement holds for $n$, then it holds for $n + 1$.
$$((n+1)^3+5(n+1)) \bmod 6 =0 \\ \Leftrightarrow ((n^3+3n^2+3n+1)+5n+5) \bmod 6 =0 \\ \Leftrightarrow (n^3+5n \bmod 6) + (3n^2+3n+6 \bmod 6)=0 \\ \Leftrightarrow 0+(3n^2+3n+6 \bmod 6)=0 \\ \Leftrightarrow 3n^2+3n+6 \bmod 6 =0.$$

And this is true for all $n \in \mathbb{N}$, where from the third to the fourth line, we replaced with the induction hypothesis that $n^3+5n \bmod 6 =0$ . Take for instance $1, 2,3,...$ it is always divisible by 6 without leaving any rest!!

However, my other friends that took the exam said that this is in need of some extension to show that the last line actually holds true for whatever n you insert. But I cannot grasp what they mean; by the definition of the modulus, it should already trivially be given that the last line is right...

I'm really looking forward to your thoughts, do you think I would at least get partial points for this? :( I really don't see how this can be wrong...

Thanks so much,
Lin

答: Since your final step implies  $3\mid 3n^2-3n$, it suffices to show $2\mid n^2-n$ for any $n$, which can be seen as trivial. That was all you missed in my opinion.


问:
Equivalence of compact and uniform convergence for sequences of holomorphic functions
Let $D\subset \mathbb{C}$ be a domain and let $(f_n)$ be a sequence of holomorphic functions on $D$. I would like to know if the following equivalence holds: $(f_n)$ converges compactly on $D$ if and only if $(f_n)$ converges uniformly on $D$. I am especially interested into the case $D=B_R(0)$ for some $R>0$, but would also like to know what topological properties of $D$ (maybe simply connectedness, convexity) are needed such that the above equivalence holds. References to literature are also appreciated.

答:
On locally compact spaces, uniform convergence on compacts is equivalent to local uniform convergence. It is not equivalent to uniform convergence: $z^n$ for example converges to $0$ on the unit disc but the convergence is not uniform.

One can prove that, for every domain $D\subseteq \mathbb{C}$, there's a sequence of holomorphic functions on $D$ such that $f_n$ converges to $0$ uniformly on compact subsets of $D$ and yet the convergence is not uniform on $D$:
for $\mathbb{C}$, an example is $\frac{z}{n}$, as marlasca23 noted in the comments to the question. If $D\subsetneq \mathbb{C}$, take $z_0\in\partial D$ and let $f_n:=\frac{1}{n(z-z_0)}$. So the equivalence you mention never holds.


问:
Let $\phi:C^1([a,b])\to \Bbb{R}$ defined by $\phi(f)=\int\limits_a^b L(t,f(t),f'(t))\ dt$ where $L\in C^1(\Bbb{R}^3)$. Find $D\phi(f)(h)$.

Prove that $E=C^1([a,b];\Bbb{R})$ is a Banach space with respect to the norm $\lVert f\rVert=\text{sup}|f|+\text{sup}|f'|$. Let $L\in C^1(\Bbb{R}^3;\Bbb{R})$ be fixed, define $\phi:E\to \Bbb{R}$ by $$\phi(f)=\int\limits_a^b L(t,f(t),f'(t))\ dt$$ Prove that, $$D\phi(
f)(h)=\int\limits_a^b (D_2L(t, f(t),f'(t))h(t)+D_3L(t, f(t), f'(t))h'(t))\ dt$$

I have proved the first part of the problem.

Let $\{f_n\}$ be a cauchy sequence in $E$. Let, $\epsilon>0$ then $\exists N\in\Bbb{N}$ such that $\lVert f_n-f_m\rVert <\epsilon/2\ \forall n,m\ge N$. This implies $\text{sup}|f_n-f_m|<\epsilon/2$ and $\text{sup}|f_n'-f_m'|<\epsilon/2\ \forall n,m\ge N$. This shows that $\{f_n\}$ converges uniformly to a function $f$ on $[a,b]$. Again, $\{f_n'\}$ also converges uniformly to a function $g$ on $[a,b]$ and it is continuous as $f_n'$ is continuous for all $n$. From the property of derivative and uniform convergence we have $f$ is differentiable and $f'=g$. Hence, $f\in E$.

Now we have $|f_n(x)-f_m(x)|<\epsilon/2$ and $|f_n'(x)-f_m'(x)|<\epsilon/2\ \forall x\in [a,b]\ \forall n,m\ge N$. Taking $m\to \infty$ we get $|f_n(x)-f(x)|\le\epsilon/2$ and $|f_n'(x)-f'(x)|\le\epsilon/2\ \forall x\in [a,b]\ \forall n\ge N\implies \text{sup}|f_n-f|+\text{sup}|f_n'-f'|\le\epsilon/2\ \forall n\ge N\implies \lVert f_n-f\rVert\le \epsilon\ \forall n\ge N$.

So, $f_n\to f$ in the NLS $(E,\lVert\cdot\rVert)$. Therefore, $E$ is a banach space.

But I have no idea how to differentiate this $\phi$. Can anyone help me in this regard? Thanks for your help in advance.

答:
Hint

By definition of the (Fréchet) derivative, it is sufficient to prove that

$$\lim\limits_{h \to 0} \frac{\vert \phi(f+h) - \phi(h) - D \phi(f)(h) \vert}{\Vert h \Vert} = 0.$$

Then, you have to use the following elements:

As $L$ is supposed to be $\mathcal C^1(\Bbb{R}^3;\Bbb{R})$, $D_2L, D_3L$ are bounded on $[0,1]$.

The derivative of the linear map $f \mapsto f^\prime$ is itself as it is linear and we're using the appropriate norm.

The chain rule.


Title: Show that $A=\{(x,y) \in \mathbb{R}^2 : x^2+y^2< 2y\}$ is open in $\mathbb{R}^2$.

Body:
Choose $(x_0,y_0)\in A$ and then choose $r\gt0:r\lt \min\{\sqrt{x_0^2+(y_0-1)^2},1-\sqrt{x_0^2+(y_0-1)^2}\}$.

For any $(x',y')\in B((x_0,y_0),r), (x'-x_0)^2+(y'-y_0)^2\lt r^2 $

By triangle inequality,

$\begin{align}
\sqrt {x'^2+(y'-1)^2}&\le \sqrt{(x'-x_0)^2+(y'-y_0)^2} + \sqrt{x_0^2+(y_0-1)^2}\\
                     &\lt r+\sqrt{x_0^2+(y_0-1)^2}\\
                     &\lt (1-\sqrt{x_0^2+(y_0-1)^2})+\sqrt{x_0^2+(y_0-1)^2}=1\\
                   &\implies (x',y')\in B((0,1),1)\implies  B((0,1),1)\subset B((x_0,y_0),r)\end{align}$.


问:
Show that a p-group has a faithful irreducible representation over $\mathbb{C}$ if it has a cyclic center
A p-group is a group of order $p^d$ where p is a prime.

If the center has order $p^m$ (since its order must divide the order of the group) then we have a one dimensional faithful irreducible representation of the center which would map a generator of the center to $e^{2\pi i/m}$. Could we then induce a representation on the whole group? If so, how do we know this is faithful and irreducible? If not, how else could we prove this?

答: The induced representation is faithful, but not necessarily irreducible. But, since the centre of the $p$-group $P$ is cyclic, the unique subgroup $K$ of $Z(P)$ with $|K|=p$ is contained in every nontrivial normal subgroup of $P$. Since the induced representation is faithful, at least one of its irreducible constituents does not have $K$ in its kernel, and then that consituent must be faithful.</p>


问:
Is submodel of atomic model atomic
I think the submodel (elementary or arbitrary) of an atomic model may not be atomic. But I do not know an example.</p>

答:
Noah had shown that (1) every elementary substructure of an atomic model is atomic, (2) not every substructure of an atomic model is atomic.

One might ask: What about a substructure of an atomic model which is elementarily equivalent to it? That is, for a complete theory $T$, can a non-atomic model of $T$ embed (non-elementarily) in an atomic model of $T$?

The answer is yes. Consider the theory $T$ of forests (simple graphs with no cycles) augmented by axioms asserting that every vertex has infinitely many neighbors. This theory is complete.

A countable model of $T$ consists of a disjoint union of copies of the complete countably branching tree (at least one copy and at most countably many copies). The atomic model $M$ is the connected one (just one copy of the tree). Now pick an arbitrary vertex $v\in M$ and remove it (and all edges out of it). The resulting graph $N$ is disconnected - each of the neighbors of $v$ is now in its own connected component, and each connected component is a copy of the complete countably branching tree.

So $N$ is a non-atomic model of $T$ which is a substructure of the atomic model $M$.

OwnerDisplayName: anon

LastEditorDisplayName: anon

Title: Is submodel of atomic model atomic


问:
Two variables limit with weird requirements
I currently have a piecewise-defined function:

$$f(x,y)=\begin{cases}\frac{\sin(xy)}{xy}&\text{ if }xy\ne0\\1&\text{ if }xy=0.\end{cases}$$

and it requires me to discuss the continuity of $f(x,y)$. I found out the way to do this is to find the limit of function when $xy$ approaches $0$. However, I wonder how should I compute the limit: should I just directly replace $xy$ with some variable $m$ and say $m$ tends to $0$, so the limit is $1$ and it is continuous? Thank you.

答: I think that the natural approach is to express your function as $g\circ h$, with$$\begin{array}{rccc}h\colon&\Bbb R^2&\longrightarrow&\Bbb R\\&(x,y)&\mapsto&xy\end{array}$$and$$\begin{array}{rccc}g\colon&\Bbb R&\longrightarrow&\Bbb R\\&x&\mapsto&\begin{cases}\frac{\sin(x)}x&\text{ if }x\ne0\\1&\text{ if }x=0.\end{cases}\end{array}$$And now, since $\lim_{(x,y)\to(0,0)}h(x,y)=0$ and $\lim_{x\to0}g(x)=1$, you have $\lim_{(x,y)\to(0,0)}f(x,y)=1$.


问:
general question about infinity induction
I'm a computer science student and during the lecture, the lecturer prove that a set
$\cup_{n=0}^\infty  A_n \subseteq B$ by induction on $n$. i.e $\forall n$  $A_n \subseteq B$

I remember from my mathematic for computer science that kind of induction doesn't hold.

Who wrong?

答:
Induction allows you to prove (in a certain formal way that often isn't needed in ordinary mathematical proofs) that the infinitely many statements $A_0 \subseteq B,$ $A_1 \subseteq B,$ $A_2 \subseteq B, \, \ldots$ all hold, and hence that $(\forall n)(A_n \subseteq B)$ holds.

However, once you have this, it follows easily from the definition of union that $\cup_{n=0}^\infty  A_n \subseteq B$ holds:

$x \in \cup_{n=0}^\infty  A_n$ implies there exists $N$ such that $x \in A_N$ (by definition of union), and hence from the already established $A_N \subseteq B$ we have $x \in B.$ Thus, we have $x \in \cup_{n=0}^\infty  A_n$ implies $x \in B,$ and hence (by definition of subset) $\cup_{n=0}^\infty  A_n \subseteq B.$

That said, it sounds like the lecturer is confused about induction, because it doesn't establish an "infinite case" from all finite cases. Instead, it simply allows you to "simultaneously in a finite amount of time, so to speak" prove all infinitely many of the finite cases.


问:
the norm of the trace as a linear functional on $\mathbb{C}^{n \times n}$
Let $T:\mathbb{C}^{n \times n} \rightarrow \mathbb{C}$ be a linear functional on the space of all $n \times n$ matrices whose entries are complex numbers. I need to show that the norm of the trace $||trace||$ " as a linear functional" is $\sqrt n$.
I have proved one direction as follows:

$|trace(A)| \leq ||trace||||A||$

this holds for all $A \in \mathbb{C}^{n \times n}$. So, pick $A=I_n$ identity matrix, we get:

$n \leq ||trace|| \sqrt n,   \quad \quad \quad \quad \quad      (1)$

(Here, $ ||I_n||^2=<I_n,I_n>=trace(I_n^{*}I_n)=n$, where $I_n^*$ represents the conjugate transpose of $I_n$.)

For the second direction, I know that

$||trace|| = \sup_{A\in \mathbb{C}^{n \times n}}=\{ |trace(A)| : ||A|| \leq 1$, but I don't know how can I get the upper bound $\sqrt n$ of the definition of the norm.

答:
Your norm seems to be induced by the inner product $\langle A,B\rangle = \operatorname{Tr}(B^*A)$
so we can use Cauchy-Schwarz inequality

$$\left|\operatorname{Tr}(A)\right|^2 = \left|\operatorname{Tr}(I^*A)\right|^2 =|\langle A,I\rangle|^2 \le \|A\|^2\|I\|^2 = n\|A\|^2$$
and hence $\left|\operatorname{Tr}(A)\right| \le \sqrt{n}\|A\|$.

Furthermore, CS states that equality holds if and only if $A$ and $I$ are proportional, so the bound $\sqrt{n}$ is attained.


问:
Extremely confused about Riemann Integrability definitions
EXTREMELY CONFUSED RIGHT NOW. I explain everything down below.

Let $f(x)$ be a function on the closed interval $[a,b]$.  We define the limit of Riemann Sums $S_{P}=\sum_{k=1}^{n} f(c_{k})\Delta x_{k}$ on $[a,b]$ as the mesh $\mu{(P)}\rightarrow 0$ to be the number $I$ if

Given $\epsilon>0$, $\exists\delta$ such that for every partition $P$ of the interval $[a,b]$

$$\mu{(P)}<\delta \Longrightarrow |S_{P}-I|<\epsilon$$

for any $c_{k}$ we choose in the subintervals $[x_{k-1},x_{k}]$.}

\noindent If the definition holds, then the limit of a Riemann Sum would exist and we can write

$$\lim_{\mu{(P)}\to 0} \sum_{k=1}^{n} f(c_{k})\Delta x_{k}=I$$

THEN WE HAVE...

A bounded function is integrable on $[a,b]$ if and only for every $\epsilon>0$ there exists a partition $P_{\epsilon}$  of $[a,b]$ such that
$$U(f,P_{\epsilon})-L(f,P_{\epsilon})<\epsilon$$

There are so many criteria for integrability, and when I read different textbooks I find different arguments -- at least that's what I have experienced. From two "definitions" above, which one is correct? Are both correct and Riemann Integrability can be defined in many ways? Is one definition better than the other even though all definitions are correct? How should I interpret this concept?

Thank you!

答:
The two definitions are equivalent but the second one involving the upper and lower limits $(U(f,P_{\epsilon}),L(f,P_{\epsilon}))$ was given by Darboux.

We can show that a function is Riemann integrable iff it is Darboux integrable, thus the two are equivalent.

For a proof of this you can check this post-How to prove that a function is Riemann integrable if and only if it is Darboux integrable? .

Because of Riemann's contribution to the development of the theory of defined integrals, in bibliography the dominant term is Riemann Integrability.


问:
The Riemann Surface $G/H$
Let $G$ be a subgroup of $SL(2,Z)$ that is of finite index and $H$ be the upper half-plane.

How is the quotient topological space $G/H$ defined (understood)?

答:
As a topological space $G\backslash\Bbb{H}$ is just the space whose points are subsets of $\Bbb{H}$ of the form $\{ gz,g\in G\}$ and whose open sets are the image of $G$-invariant open subsets of $\Bbb{H}$.

It becomes a Riemann surface with its (pre)sheaf of holomorphic functions: the functions on some open of $G\backslash\Bbb{H}$ that come from an holomorphic function on the corresponding open subset of $\Bbb{H}$.

$gs = s$ for some $g\ne \pm I\in SL_2(\Bbb{Z})$ implies that $s=\gamma(i)$ or $s=\gamma(e^{2\pi /3})$ for some $\gamma\in SL_2(\Bbb{Z})$. Thus for a small enough disk $D\ni s\not \in SL_2(\Bbb{Z})i,SL_2(\Bbb{Z})e^{2i\pi /3}$ any holomorphic function $f$ on $D\subset \Bbb{H}$ gives an holomorphic function $F(Gz)=f(z)$ on $GD\subset G\backslash\Bbb{H}$.

When $f$ has a simple zero at $s$ then it is a chart for the Riemann surface $G\backslash\Bbb{H}$ around $Gs$.

The case $s=i$:
When $S=\pmatrix{0&1\\-1&0}\in G$ then $f$ must be of the form $h(z)h(-1/z)$ with $h$ holomorphic for $F(Gz)$ to be well-defined around $Gi$, and it is a chart when $h$ has a simple zero at $i$.

The case $s=e^{2i\pi/3}$: When $ST\in G$ then $f$ must be of the form $h(z)h(STz)h((ST)^2 z)$ to be well-defined at $e^{2i\pi/3}$ and again it is a chart when $h$ has a simple zero at $e^{2i\pi /3}$.

Title: The Riemann Surface $G/H$


问:
I can't find the equation of common tangents to these two circles in the 2-d plane .
What are the equations of the tangents common to the circles x² + y²=1 , and (x-1)² + (y-3)² = 4?

I first used the T²=SS1 form which gives the joint equation of two tangents to a curve dream from an external point , but the equation of becoming too complex to solve ,please help me out .
Thank you

答: A quick sketch will show you that lines $x=-1$ and $y=1$ are two common tangents. The other two tangents are the reflections of these about the line through the centres.


问:
Number of solutions of $x(x^x-x)=0$
How many solutions the equation $x(x^x-x)=0$ has in real numbers set?
$$1)\text{zero}\quad\quad\quad\quad\quad2)\text{one}\quad\quad\quad\quad\quad3)\text{two}\quad\quad\quad\quad\quad4)\text{three}\quad\quad\quad\quad\quad5)\text{infinity}$$

Here is my work:

To solve this problem I considered two cases: $x=0$ or $x^x-x=0$. for $x=0$ I think $0(0^0-0)$ is undefined in real numbers( because we have $0^0$). so we should solve $x^x=x$ however I'm not sure how we should solve this but it is easy to see $x=\pm1$ are the roots. so I think the equation has two roots in real numbers.

Is my answer right?

答:
The answer is either "two roots" or "three roots" depending on if you consider $0^0$ to be $1$ or undefined, and both are common, so it's a poorly written question.

Ignoring the $x=0$ case, you are right that $x^x-x=0$ is only possible when $x=\pm1$. We can divide through by $x$ to get $x^{x-1} = 1$. Some casework, now:

One way to have $x^y = 1$ is if $y=0$; here, that means $x-1=0$, so $x=1$.

If the power is nonzero, then only powers of $\pm 1$ can be $1$.

Title: Number of solutions of $x(x^x-x)=0$


答:
First, in the general case, there may be several distinct criteria for an object to have an interesting property.  The different criteria may each correctly identify some of the objects having that property, but be mute on the other objects having that property.  This leads to summary theorems of the form "If any of the following conditions hold: (1) ... (2) ... ... (13) ...  Then (object) has (property)."  So it is entirely reasonable to have several if-then criteria for an object to have a property.

This is not what is happening here.

Here, you have if-and-only-if criteria.  (All definitions are if-and-only-if, even though the mathematical style is to write them as if-thens.  For instance, from https://en.wikipedia.org/wiki/Continuous_function , "a function is continuous if arbitrarily small changes in its output can be assured by restricting to sufficiently small changes in its input".  Strictly interpreting this sentence, this is "if arbitrarily small changes in a function's, $f$'s, output can be assured by restricting to sufficiently small changes in its input, then $f$ is continuous" and only gives an inference "... $\implies$ $f$ is continuous".  In contraposition, it also gives an inference "$f$ not continuous $\implies$ ...".  Notice that, strictly, there is no inference given of the form "$f$ is continuous $\implies$ ...".  This obvious absurdity is removed when we establish that definitions written as implications are actually biimplications.)

In general, a text takes one definition as "the definition", then proves equivalence between that definition and other characterizations of the property under study.

This could be what is happening here, but it isn't.  So what is happening here?

Here, you have the same criterion written in apparently different ways.  Can we show this equivalence?

Let $\varepsilon > 0$ and let $\delta$ be chosen so that any partition, $P$, of $[a,b]$ satisfies
$$  \mu(P) < \delta \implies |S_P - I| < \varepsilon/2  \text{.}  $$
One detail that has been omitted is the choice of $c_k$ in each interval of $P$.  In the definition we have been reciting through so far, the choice of the $c_k$ is universally quantified.  For all choices of the $c_k$, $S_P$ and $I$ are close in value.  This means the inequality holds if we make specific choices.
Suppose $c_k$ is chosen to give the minimum (or, if $f$ is so wild it doesn't have a minimum, infimal) value of $f$ in each interval of $P$.  Then $S_P = L(f,P)$.  Choosing instead the maximal (or supremal) points, $S_P = U(f,P)$.  Our inequality promises
$$  |L(f,P) - I| < \varepsilon/2  $$
and
$$  |U(f,P) - I| < \varepsilon/2  $$
giving
$$  |U(f,P) - L(f,P)| = U(f,P) - L(f,P) < \varepsilon  \text{.}  $$

So the first definition is saying pretty much the same thing as the second: "any choice of the $c_k$" gives small difference between the Riemann sum and $I$" implies "the most extreme difference between choices of $c_k$, the upper sum choice of the $c_k$ and the lower sum choice of the $c_k$, yields small differences in the Riemann sum (because both sums are close to $I$)".

Going in the reverse direction: we have that $L(f,P_\varepsilon) \leq S_{P_\varepsilon} \leq U(f, P_\varepsilon)$ for any choice of the $c_k$,  and we apply the squeeze theorem (as $\varepsilon \rightarrow 0$) to show that all three of these (or actually $L(...)$, $U(...)$, and all infinitely many $S_P$ as we range over the possible choices of the $c_k$) converge to the same value, $I$.

One might ask why any particular criterion is "natural" (i.e., is one we might choose as the definition), since once we have all these equivalences, any of them could be the definition.  In the cases you cite, the definitions are modeled on two different ways to approach convergence.  The first definition is modeled on the definition of convergence of a sequence in the limit, so can be taught very early during rigorous analysis.  The second definition is modeled on an alternative definition of convergence of a sequence: that the limit infimum and limit supremum agree.  (One nice property of these two limits is that they always exist, so we don't have to say "... both exist and agree".)  Why do we have both ideas?

Convergence of a sequence requires a sequence, a collection indexed by a countable set.  So, in some sense, the first definition is only useful if you are interested in convergence of a small (ordered) set of values.  Limits infimum and supremum can be defined for uncountably large sets, so are useful in more settings that the first definition.

Title: Extremely confused about Riemann Integrability definitions

Body:
EXTREMELY CONFUSED RIGHT NOW. I explain everything down below.

Let $f(x)$ be a function on the closed interval $[a,b]$.  We define the limit of Riemann Sums $S_{P}=\sum_{k=1}^{n} f(c_{k})\Delta x_{k}$ on $[a,b]$ as the mesh $\mu{(P)}\rightarrow 0$ to be the number $I$ if

Given $\epsilon>0$, $\exists\delta$ such that for every partition $P$ of the interval $[a,b]$

$$\mu{(P)}<\delta \Longrightarrow |S_{P}-I|<\epsilon$$

for any $c_{k}$ we choose in the subintervals $[x_{k-1},x_{k}]$.}

\noindent If the definition holds, then the limit of a Riemann Sum would exist and we can write

$$\lim_{\mu{(P)}\to 0} \sum_{k=1}^{n} f(c_{k})\Delta x_{k}=I$$

THEN WE HAVE...

A bounded function is integrable on $[a,b]$ if and only for every $\epsilon>0$ there exists a partition $P_{\epsilon}$  of $[a,b]$ such that
$$U(f,P_{\epsilon})-L(f,P_{\epsilon})<\epsilon$$

There are so many criteria for integrability, and when I read different textbooks I find different arguments -- at least that's what I have experienced. From two "definitions" above, which one is correct? Are both correct and Riemann Integrability can be defined in many ways? Is one definition better than the other even though all definitions are correct? How should I interpret this concept?

Thank you!


问:
Are $(1+p)^{p^{n-1}} - 1, (1+p)^{p^{n-2}} - 1$ divisible by $p^n$?
Is it true that $p^n$ divides $(1+p)^{p^{n-1}}-1$ if $p$ is an odd prime ? Furthermore, does $p^n$ divide $(1+p)^{p^{n-2}}-1$?

I tried to show the following, but made no progress:

$$ p^n \text{ divides } \binom{p^{n-1}}{k}p^k \text{ for each } k=1,\,2,\,\cdots,\,p^{n-1}$$

How to show this?

答:
You can use the logarithm map (at least for $p > 2$ and for $p=2$, it works if you replace $(1+2)^{2^n}$ with $(1+4)^{2^n}$). The idea is the following:

We can define a group homomorphism $\log_p: (1 + p\mathbb Z_p)^\times \to \mathbb pZ_p$ by the usual infinite series:
$$\log_p(1+x) = -x - \frac{x^2}{2} + \dots$$
It's not hard to check that this always converges and takes multiplication to addition. Moreover, it has an inverse $\exp_p$ but this only converges for $p \geq 3$ or on $4\mathbb Z_2$ for $p = 2$.

Anyway, once you have the idea to consider these maps, it's a routine calculation. $\log_p(1+p) = p\times \text{unit}$ so $\log_p(1+p)^{p^n} = p^{n+1}\times\text{unit} $ so taking exponentials, we see that $(1+p)^{p^n} - 1$ is exactly divisible by $p^{n+1}$.


问:
What is the autocovariance of $x_t = U_1 \sin(2 \pi \omega_0 t) + U_2 \cos(2 \pi \omega_0 t)$
A time series with a periodic component can be constructed from
$$x_t=U_1\sin(2\pi \omega_0t)+U_2\cos(2\pi \omega_0t),$$
where $U_1$ and $U_2$ are independent random variables with zero means and
$$E(U_1^2)=E(U_2^2)=\sigma^2$$
Show this series is weakly stationary with autocovariance
$\gamma(h) = \sigma^2 \cos(2 \pi \omega_0 h)$.

So I set $p = 2\pi\omega_0$

I have that the mean function is

$E(x_t) = E(U_1 \sin(pt)) + E(U_2 \cos(pt)) = 0$

which is not dependent on $t$

What I have for covariance is based on the following property:

If the random variables

$U=\sum^m_{j=1}a_jX_j$ and $V=\sum^r_{k=1}b_kY_k$

are linear combinations of (finite variance) random variables ${X_j}$ and ${Y_k}$, respectively, then

$cov(U,V)=\sum^m_{j=1}\sum^r_{k=1}a_jb_k\operatorname{cov}((X_j,Y_k)$

Furthermore, $\operatorname{var}((U)=\operatorname{cov}((U,U)$

Then setting

$U = U_1 \sin(p(t+h)) + U_2 \cos(p(t+h))$

and

$V = U_1 \sin(p(t)) + U_2 \cos(p(t))$

with

$a_1 = \sin(p(t+h))$ and $a_2 = \cos(p(t+h))$

and

$b_1 = \sin(pt)$ and $b_2 = \cos(pt)$

and

$X_1 = Y_1 = U_1$ and $X_2 = Y_2 = U_2$

And noting that $\operatorname{cov}((U_1, U_2) = 0$ and $\operatorname{var}(U_1) = \operatorname{var}(U_2) = \sigma^2$

we have

$\gamma(h) = \operatorname{cov}(U, V) $

$= \sigma^2 [\sin(p(t+h))\sin(pt) + \cos(p(t+h)) \cos(pt)]$

see rest of solution below

答:
By trig identities

$\sin(pt + ph)\sin(pt) + \cos(pt +ph)\cos(pt) = \cos(pt + ph -pt) = \cos(ph)$

and the result is that
$\gamma_x(h) = \sigma^2\cos(2 \pi \omega_0 h)$

so neither the mean nor the autocovariance is dep on t and the series is stationary.


问:
Sum of Random Variables...
Imagine we repeat the following loop some thousands of times:

$$
\begin{align}
& \text{array} = []\\
& \text{for n} = 1: 10 000 \\
& k = 0 \\
& \text{while unifrnd}(0,1) < 0.3 \\
& k = k + 1 \\
& \text{end} \\
& \text{if k} \neq 0 \\
& \text{array} = [\text{array,k}] \\
& \text{end} \\
\end{align}
$$

whereas "unifrnd(0,1)" means a random number drawn from the uniform distribution on the unit interval.

My question is then: What is then the value of k, which is the most often observed - except for k = 0? And is that the expectation of k?

Thanks very much

答:
if you want the number that is most likely to be made out of a sum like this. you need combinations to get that sum * probability of each combination. So,

$${n\choose m}*.3^m*.7^{n-m}$$

should be maximized. according to wolfram alpha this is not exactly .3*n http://www.wolframalpha.com/input/?i=maximise+%28choose%28200%2Cn%29.3^n.7^%28200-n%29%29

that's the best I can do.

Edit: You changed your question, and I anticipate one more change at least. I think you are looking for the expected value of k, rather than the most often observed value. So here is how you calculate that:

$$\sum_{n\to \infty }{n*Pr(n)}$$ where $$Pr(n) = .3^n$$

let $r=.3$ and we can psuedoexpand our series (to make it clearer what we are doing) to: $$ r+2r^2+3r^3+...$$ or $$r+r^2+r^3+...$$ $$r^2+r^3+...$$ $$r^3+...$$
then we have ${\frac{r}{1-r}}+r{\frac{r}{1-r}}+r^2{\frac{r}{1-r}}+... = {\frac{r}{(1-r)^2}}=30/49$ which is probably the answer you want...

Title: Sum of Random Variables...


问:
Finding a confidence interval for different of proportions
Let two independent random variables, $Y_1$ and $Y_2$ that have binomial distribution have parameters $n_1 = n_2 = 100$, $p_1$ and $p_2$, respectively, be observed to be equal to $y_1 = 50$ and $y_2 = 40$. Determine an approximate $90\%$ confidence interval for $p_1 - p_2$.

I'm pretty new to confidence intervals and wanted help on this problem from my book. I have tried to apply the following definition:

Let $X_1, X_2, \ldots, X_n$ be a sample on a random variable $X$, where $X$ has p.d.f. $f(x;\theta)$, where $\theta\in \Omega$. Let $0 < \alpha < 1$ be specified. Let $L = L(X_1, \ldots, X_n)$ and $U = U(X_1, \ldots, X_n)$ be two statistics. We say that $(L, U)$ is a $(1 - \alpha)100\%$ confidence interval for $\theta$ if $1 - \alpha = P_{\theta}(\theta \in (L, U))$.

However, I have been having trouble applying this definition directly. I can identify $\alpha = 0.1$ here, which means that I think we want $P(L < p_1 - p_2 < U) = 0.9$. Now I'm really not sure how to use this information to get the answer; I have no clue where the observed values would come into play.

Any help is appreciated.

答:
The idea is that the sample averages $Y_1/n_1$ and $Y_2/n_2$ should be good approximations of the proportions $p_1$ and $p_2$ respectively.  We can use the observed value $y_1/n_1 - y_2/n_2$ as a point estimate of $p_1 - p_2$, and get a confidence interval by figuring out the approximate distribution of $Y_1/n_1 - Y_2/n_2$.

Because of the Central Limit Theorem, we know that the distribution $\operatorname{Bin}(n, p)$ is approximately normal with mean $np$ and variance $np(1-p)$.  Therefore since $Y_1$ and $Y_2$ are independent, the random variable $Y_1/n_1 - Y_2/n_2$ is approximately distributed like a normal random variable with mean $p_1 - p_2$ and variance $p_1(1-p_1)/n_1 + p_2(1-p_2)/n_2$.  After standardizing this tells us that the test statistic
$$
\frac{(Y_1/n_1 - Y_2/n_2) - (p_1 - p_2)}{\sqrt{p_1(1-p_1)/n_1 + p_2(1-p_2)/n_2}}
$$
approximately follows a standard normal distribution.  Using this, a little bit of algebraic manipulation will show that
$$
\mathbb{P} \left( | (p_1 - p_2) - (Y_1/n_1 - Y_2/n_2)| \leq 1.645 \sqrt{p_1(1-p_1)/n_1 + p_2(1-p_2)/n_2} \right) = 90\%
$$
(the number $1.645$ comes from the $95$th percentile of a standard normal distribution)  So an approximate $90$% confidence interval would be
$$
(Y_1/n_1 - Y_2/n_2) \pm 1.645 \sqrt{p_1(1-p_1)/n_1 + p_2(1-p_2)/n_2}.
$$
The only problem is that $p_1$ and $p_2$ still appear in this expression, but they are unknown.  One standard way to fix this is just to replace $p_1$ and $p_2$ by their point estimates $Y_1/n_1$ and $Y_2/n_2$ in the above expression, and you are still left with a pretty good approximation.


问:
What about $\mathbb{Z}[\sqrt[n]{2}]$?
Most textbooks talk about $\mathbb{Z}[\sqrt{d}]$ but I haven't succeed in my search of direct references about $\mathbb{Z}[\sqrt[n]{d}]$.

I would like to ask about $\mathbb{Z}[\sqrt[n]{2}]$ for $n>2$. I know it inherits the norm from $\mathbb{Q}(\sqrt[n]{2})$, but what is it known about $\mathbb{Z}[\sqrt[n]{2}]$? Is it a euclidean domain, or a unique factorization domain, or at least a GCD domain? Can their prime or irreducible elements be clasiffied as nicely as in $\mathbb{Z}[\sqrt{2}]$ and $\mathbb{Z}[\sqrt{-1}]$? I will be very grateful with answers or references for this questions, even for just $\mathbb{Z}[\sqrt[3]{2}]$.

Thank you.

答:
According to LMFDB - The L-functions and Modular Forms Database:

$\Bbb Z\!\left[\sqrt[3]2\right]$is a Principal Ideal Domain.

$\Bbb Z\!\left[\sqrt[4]2\right]$ is a Principal Ideal Domain.

$\Bbb Z\!\left[\sqrt[5]2\right]$ is a Principal Ideal Domain.

$\Bbb Z\!\left[\sqrt[6]2\right]$ is a Principal Ideal Domain.

$\Bbb Z\!\left[\sqrt[7]2\right]$ is a Principal Ideal Domain.

$\Bbb Z\!\left[\sqrt[8]2\right]$ is a Principal Ideal Domain.

$\Bbb Z\!\left[\sqrt[9]2\right]$ is a Principal Ideal Domain.

$\Bbb Z\!\left[\sqrt[10]2\right]$ is a Principal Ideal Domain.

$\Bbb Z\!\left[\sqrt[11]2\right]$ is a Principal Ideal Domain.

$\Bbb Z\!\left[\sqrt[12]2\right]$ is a Principal Ideal Domain.

$\Bbb Z\!\left[\sqrt[20]2\right]$ is a Principal Ideal Domain assuming GRH.

$\Bbb Z\!\left[\sqrt[40]2\right]$ is a Principal Ideal Domain assuming GRH.

There must be something fishy going on...

Methodology: the link says that an integral basis is $1, a, a^2, \cdots$ which means that $\Bbb Z\!\left[\sqrt[n]2\right]$ is the ring of integers of $\Bbb Q\!\left(\sqrt[n]2\right)$, and it also says that the class number is $1$, which means that the ring is a Principal Ideal Domain.

Title: What about $\mathbb{Z}[\sqrt[n]{2}]$?


问:
The "best" simple and accurate approximation for $x - \cos(x) = w$

the given equation is $x-\cos(x) = w$

given some $w$ values $-100, -1.9827359827356, 50, 75.2982735, 100\ldots$ maybe many more between, say $-10000\ldots 10000$

Please find some

a) simple !!
b) enough accurate

approximation to find $x$

I tried

$a_1  \sin{(z_1  w + z_2)} + a_2  \sin{(z_3  w + z_4)} + a_3 \sin{(z_5  w + z_6)} ... + w + constant$

and found some good but it was something like at least $9 \sin$ things to get a good approximation.

Can you find more simple still somewhat accurate approx equation for this ?

答:
A rather good approximation is
$$x=w-\frac{6 (-1766 \sin (2 w)+\sin (4 w)-3676 \cos (w)+156 \cos (3 w))}{45380 \sin
   (w)-380 \sin (3 w)-8916 \cos (2 w)+\cos (4 w)+37483}$$ which is the result of one single iteration of a method of order $8$.

Trying for $w=\pi^k$
$$\left(
\begin{array}{ccc}
k & \text{estimate} & \text{solution} \\
 1 & 2.402303939 &  2.402507520 \\
 2 & 8.967638069 &  8.970865828 \\
 3 & 31.89675206 &  31.89410931 \\
 4 & 96.66186737 &  96.66210022 \\
 5 & 305.0248861 &  305.0781580 \\
 6 & 962.1032272 &  962.1030947 \\
 7 & 3019.254913 &  3019.325573 \\
 8 & 9488.852896 &  9488.852896 \\
 9 & 29809.05062 &  29809.05062 \\
 10 & 93647.19019 &  93647.19169
\end{array}
\right)$$

Edit

In a comment, the questioner wonders if this is a Padé approximant. The funny answer is that it is and it is not at the same time.

As done, it was the first iterate of an high order iterative method.

But, if we consider the $[1,n]$ Padé approximant (built around $x=w$) of the function
$$f(x)=x-\cos(x)-w$$ it would write

$$P_n(x)=\frac{-\cos(w)+ a^{(n)} (x-w) } {1+\sum _{i=1}^n b_i^{(n)} (x-w)^i }$$ and the approximate solution is
$$x_{(n)}=w +\frac{\cos(w)}{  a^{(n)} }$$ but $a^{(n)}$ depends on the function and derivative values at $x=w$, we find again Newton $(n=0)$, original Halley $(n=1)$, original Householder $(n=2)$ formulae. For example
$$P_0=(x-w) (\sin (w)+1)-\cos (w)$$
$$P_1=\frac{\frac{(x-w) \left(2 \sin ^2(w)+4 \sin (w)+\cos ^2(w)+2\right)}{2 (\sin
   (w)+1)}-\cos (w)}{1-\frac{(x-w) \cos (w)}{2 (\sin (w)+1)}}$$

The given result would correspond to $[1,6]$ which is order $8$.

What is amuzing is that for some $n$ appears also a cotangent terms. For example $[1,3]$ gives
$$x=w+\frac{2}{9} \left(1-\frac{2}{\cot \left(\frac{w}{2}\right)+1}+\frac{17 \sin (2
   w)+302 \cos (w)}{60 \sin (w)-\cos (2 w)+131}\right)$$


问:
Unbounded second derivative of continuous bounded function
Does a continuous function $f$ exist where:

$f$ is continuous,

$f$ is known to be bounded with a codomain of $\left[0,1\right]$

$f''$ (second derivative) is unbounded with a codomain of $(-\infty,+\infty)$

答:
$$f(x)=\begin{cases}
\frac{1}{2}+\frac{1}{2}\sqrt{1-x^2} & \text{ if } 0\le x\le 1 \\
\frac{1}{2}-\frac{1}{2}\sqrt{1-(2-x)^2} & \text{ if } 1< x\le2
\end{cases}$$
This is bounded—$0\le f(x)\le 1$—and its second derivative is unbounded in both directions (though I think $|f''(x)|\ge 1$, which may not be quite what you wanted).</p>


问:
A simple $n\times n$ system of linear equations
Can anyone provide the analytical solution of the following $n\times n$ system of linear equations

$\underbrace{\begin{pmatrix}
 a & -b & -b & ... & -b\\
 -b & a & -b & ... & -b\\
 -b & -b & a & ... & -b\\
  : &  : & : & ... & : \\
  : & : & : & ... &: & \\
 -b & -b & -b & ... & a \\
 \end{pmatrix}}_{A}\underbrace{\begin{pmatrix}x_1\\
x_2\\
x_3\\
:\\
:\\
x_n
\end{pmatrix}}_{X}=\underbrace{\begin{pmatrix}y_1\\
y_2\\
y_3\\
:\\
:\\
y_n
\end{pmatrix}}_{Y}
$
where $A_{(n\times n)}$ is a symmetric matrix and $a$, $b$, $y_i$ are positive real numbers.

Note: I would appreciate if someone could provide more details about the theory of eigenvalues-eigenvectors related to the solutuion of a linear system of eaquations.

答:
You can write your equations as
$$
\left((a+b)I-b\mathbb{1}\mathbb{1}^T\right)x=y\ ,$$
where $\ \mathbb{1}\ $ is the column vector whose entries are all $1$.  Multiplying this equation on the left by the row vector $\ \mathbb{1}^T\ $ gives
\begin{align}
(a+b)\mathbb{1}^Tx-b\mathbb{1}^T\mathbb{1}\mathbb{1}^Tx&=\mathbb{1}^Ty\\
&=\big(a-(n-1)b\big)\mathbb{1}^Tx\ ,\\
\hspace{-18em}\text{or}\\
\mathbb{1}^Tx&=\big(a-(n-1)b\big)^{-1}\mathbb{1}^Ty\ ,
\end{align}
provided $\ a-(n-1)b\ne0\ $. Now rewriting the first equation above as
$$
(a+b)x=y+b\mathbb{1}\mathbb{1}^Tx\ ,
$$
and substituting $\ \big(a-(n-1)b\big)^{-1}\mathbb{1}^Ty\ $ for $\ \mathbb{1}^Tx\ $ gives
\begin{align}
x&=\left(\frac{1}{a+b}\right)y+\left(\frac{b}{(a+b)\big(a-(n-1)b\big)}\right)\mathbb{1}\mathbb{1}^Ty\\
&=\left(\left(\frac{1}{a+b}\right)I+\left(\frac{b}{(a+b)\big(a-(n-1)b\big)}\right)\mathbb{1}\mathbb{1}^T\right)y\ ,
\end{align}
provided $\ (a+b)\ne0\ $, which gives you the solution to your problem.


问:
Prove that if $\beta$ is an orthonormal basis for $V$ then $T(\beta)$ is an orthonormal basis for $V$
Let $T$ be a linear operator on a finite diemnsional inner product space $V$.

Prove that if $\beta$ is an orthonormal basis for $V$ then $T(\beta)$ is an orthonormal basis for $V$.

Proof

 Let $\beta=\{v_1,\dots,v_n\}$ be an orthonormal basis for $V$ so $T(\beta)$.

$\langle T(v_i),T(v_j)\rangle=\langle v_i,v_j\rangle =\delta_{ij}$ , therefore $T(\beta)$ is an orthonormal basis for $V$.

The first equality is from the equivalent theorem $\langle T(x),T(y)\rangle=\langle x,y\rangle$.

I wonder why the last sentence means that $T(\beta)$ is an orthonormal basis for $V$.

They are linearly independent, of course, but they do span $V$? How can show that?

Sorry to make you be confused. I omitted this information : $TT^*=T^*T=I$.

答: The statement ⟨T(x),T(y)⟩=⟨x,y⟩ means that T is an isomorphism, the result isn't true for an arbitrary linear operator T. </p>


问:
Convex hulls in an infinite dimensional simplex
I have the following problem:

Let

$$S=\{x\in R_{+}^\infty \mid \sum_{n=1}^\infty x_n=1\}$$

be an infinite simplex.

Suppose $\hat{S}$ is a countable subset of $S$ such that

for each $x\in \hat{S}$ only at most finitely many coordinates are non-zero, denote the set of those coordinates $N(x)$;

for any $x\neq x’$, $x, x’ \in \hat{S}$, $N(x)\cap N(x')=\emptyset$.

Claim: No $x\in \hat{S}$ can be represented as a convex combination of other elements of $\hat{S}$.

I find the result rather obvious, but I do not know how to write the proof formally.

答:
The set $\hat{S}$ is linearly independent. To see this assume:
$$
v = a_1 x_1 + \ldots + a_n x_n = 0
$$
where $x_1, \ldots, x_n \in \hat{S}$. Let $\pi_i : S \to \Bbb{R}^\infty$ be the projection onto the span of the $e_i$ with $i \in N(x_i)$, where the $e_i$ are the standard basis for $\Bbb{R}^\infty$. Then, by condition 2:
$$
\pi_i(x_j) = \left\{
   \begin{array}{l@\quadl}
      0 & \mbox{if $i \neq j$}\\
      x_j & \mbox{if $ i = j$}
   \end{array}\right.
$$
whence, for $i = 1, \ldots, n$, $\pi_i(v) = a_i x_i = 0$ so that $a_i = 0$ (since, by definition of $S$, $x_i \neq 0$).
As $\hat{S}$ is linearly independent, no $x\in \hat{S}$ can be written as a convex combination of other elements.


问:
Can the graph of a continuous real function be a totally disconnected subset of $\mathbb{R}^2$?
Can the graph of a continuous real function $f:\mathbb{R}\to\mathbb{R}$ be a totally disconnected subset of $\mathbb{R}^2$?

Certainly there can be many disconnected points (in $\mathbb{R}^2$), like the point $(0,0)$ in the graph of:

$
f(x)=
\begin{cases}
 x\sin\frac{1}{x}&\text{if}\, x\neq 0\\
 0&\text{if}\, x=0\\
\end{cases}
$

And perhaps it isn't necessary for the graph of $f$ to be connected either side of these "disconnected points".

Whether or not there is a real function $f$ and dense subset $X$ of $\mathbb{R}$ so that $\{\ (x,f(x)): x\in X\}$ is a totally disconnected subset of $\mathbb{R}^2,\ $ I am not sure. Perhaps this is not possible?

答:
The graph of any continuous function $f:X\to Y$ is homeomorphic to $X$ via $x\mapsto \big(x,f(x)\big)$ (see this). In particular any continuous function $\mathbb{R}\to\mathbb{R}$ has a (path)connected graph.

It is unclear why you think that your particular $f$ has a disconnected graph. It certainly is even path connected.