问:
how many abelian groups of order $8$ and $21$ exist?
I have to find how many abelian groups of order $8$ and order $21$ exist. There this theorem that states that every abelian group is a direct sum of cyclic groups. So what I did is this: $8=2\cdot2\cdot2$, now I look to all possible divisors of $8$, that is $2$, $4$ and $8$. Then $G$ is isomorphic to $\mathbb{Z}/8\mathbb{Z}$ or $\mathbb{Z}/4\mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}$ or $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$. For $G$ of order 21=7*3 we have $G$ is isomorf with $\mathbb{Z}/21\mathbb{Z}$ or $\mathbb{Z}/7\mathbb{Z}\oplus\mathbb{Z}/3\mathbb{Z}$. Thus there are $3$ abelian groups of order $8$ and $2$ abelian groups of order $21$.

My question is: is this the right way to attack this problem (if it's correct)? Thank you.

答:
The second answer needs to be modified, since the two groups you mention are isomorphic. The direct sum is cyclic of order $21$, the element $(1,1)$ is a generator.

For completeness, one should prove that the three groups of order $8$ that you list are non-isomorphic. One quick way is to observe that the first group has an element of order $8$. The other two groups do not. And the second group has an element of order $4$, while the third does not.


问:
Describing the plane given by a line and a vector
I'm studying lines and planes and am stumped by the following problem:

If $L$ is the line through the point $A = (3,2,1)$ and parallel to the
vector $v = (-2,1,3)$, what's the equation of the plane that contains
$L$ and the point $B = (-2,3,1)$?

I'm trying to describe the plane in terms of $A$, $v$, and $B$ and then find a normal vector. The line $L$ is the span of $v$ shifted by $A$, so it should be $L = \{A + a v : a \in \mathbb R\}$. Then the plane itself is formed by adding in $B$, so the plane is $H = \{A + av + bB : a,b \in \mathbb R\}$. This looks like the plane through the origin $\mathrm{span}\{v, B\}$ shifted by $A$, so I can describe the plane by getting a normal vector $\mathbf n$ to $\mathrm{span}\{v, B\}$ and then a vector $x$ is in the plane exactly when $\mathbf n^T(x - A) = 0$. I found $\mathbf n$ by picking something parallel to $v \times B$ (I got $\mathbf n = (2,1,1)$). But when I do this I get $2x + y + z = 9$ as the equation which my book says is wrong. But I can't see a flaw in my argument. Am I wrong about $H$ or $L$ or something else?

答:
If you have another point along the line, like $\boldsymbol{C} = \boldsymbol{A} + \boldsymbol{v} = \pmatrix{1 & 3 & 4}$ then you can define the plane from three points.

Vector normal to plane $$\boldsymbol{n} = (\boldsymbol{A}\times\boldsymbol{B}) + (\boldsymbol{B}\times\boldsymbol{C})+(\boldsymbol{C}\times\boldsymbol{A}) = \pmatrix{3 \\ 15 \\ -3} $$

Equation of plane based on an arbitrary point on the plane $\boldsymbol{P} = \pmatrix{x & y & z}$ and a known point $A$ (also on the plane). $$ \boldsymbol{n}\cdot \boldsymbol{P} = \boldsymbol{n} \cdot \boldsymbol{A}$$ $$3x+15y-3z=36$$

You can also define the normal by

$$ \boldsymbol{n} = ( \boldsymbol{B}-\boldsymbol{A} ) \times \boldsymbol{v} = \pmatrix{3 \\ 15 \\-3} $$ and proceed with the equation as above.

Note that $(B-A)\times(C-A) = (B-A)\times C - (B-A) \times A = B \times C - A \times C - B \times A = A\times B+B\times C + C \times A$

ContentLicense: CC BY-SA 4.0


问:
Find $\lim_{t \rightarrow 0} \int_{0}^{t} \frac{\sqrt{1+\sin(x^2)}}{\sin t} dx$
Find $$\lim_{t \rightarrow 0} \int_{0}^{t} \frac{\sqrt{1+\sin(x^2)}}{\sin t} dx$$

One easy step is replacing $\sin t = t$, since $\lim_{t \rightarrow 0} \sin t / t = 1$. How do I continue with the rest?

答:
The mean of a function over a domain $x\in[a,b]$ is given by:
$$\bar{f}=\frac{1}{b-a}\int_a^bf(x)dx$$
so if we have a domain of $x\in[0,t]$ then we can say:
$$\bar{f}=\frac1t\int_0^tf(x)dx$$
and as $t\to0$ the interval we are finding the average over is getting smaller until it is down to the point of $x=0$ and so we can say:
$$\lim_{t\to0}\int_0^t\sqrt{1+\sin(x^2)}dx=\lim_{t\to0}t\sqrt{1+\sin(t^2)}$$
and so for your integral you have:
$$\lim_{t\to0}\frac{t\sqrt{1+\sin(t^2)}}{\sin(t)}$$
and we can break this up and it is known that:
$$\lim_{t\to0}\frac{\sin(t)}{t}=1$$
so we have:
$$\lim_{t\to0}\sqrt{1+\sin(t^2)}=1$$


问:
Function in quotient space
Let $V$ and $W$ be both K-linear vector spaces. Now $Y$ is a linear subspace of $V$ and $Z$ is a linear subspace of $W$. Let $\alpha: V \rightarrow W$ be a $K$-linear function with $\alpha(Y)\subseteq Z$.

I want to prove that $\alpha$ induces linear functions $\alpha|_Y: Y \rightarrow Z$ with $\alpha|_Y(y)=\alpha(y)$ and $\bar{\alpha}: V/Y \rightarrow W/Z$ with $\bar{\alpha}(v+Y)=\alpha(v)+Z$

My idea:
As $\alpha(Y)\subseteq Z$, $\exists z\in Z: \forall y\in Y:z=\alpha(y)$. So $\alpha$ induces $\alpha|_Y(y)=\alpha(y)$. $\square$

Now as $\alpha$ is linear, we can write $\bar{\alpha}(v+Y)=\bar{\alpha}(v)+\bar{\alpha}(Y)$ but now I do not know how to proceed.

Is this the right way to proof this? Thanks in advance for any kind of help.

答:
Your  idea is erroneously formulated: it should be
$$\forall y\in Y, \exists z\in Z: z=\alpha(y),$$
which is more or less a tautology.

As to the quotient map, you only have to prove that it is well-defined, i.e. that if $y'\equiv y\mod Y$, then $\alpha(y')\equiv \alpha(y)\mod Z$, which shouldn't be very hard with the hypotheses that $\alpha$ is linear and maps $Y$ into $Z$.

Title: Function in quotient space


问:
Lambda Calculus Beta reductions
I have two questions about $\beta$-reduction in the $\lambda$-calculus. Please find them below. I have also included some background information about how lists (i.e. finite sequences) can be encoded in the $\lambda$-calculus.

Background information

Lists can be encoded in the $\lambda$-calculus in the following way: $$[N_1,N_2,...,N_k] ≜ λc.λn.c N_1 (c N_2 (...(c N_k n)...))$$

Intuitively, the variable $c$ represents the cons operation (which adds an item to the head of the list) and the variable $n$ represents the empty list, nil: note that the term for the empty list is $[\,] = λc.λn.n.$, according to the definition above.

Questions

Q1.
Show that the following term $β$-reduces to $6$: $[3,2,1] \, \mathsf{times} \, 1$.

Here the natural numbers $1 , 2 , 3 , \dots$ denote the corresponding Church numerals, that is, the $\lambda$-terms encoding the corresponding natural numbers in the $\lambda$-calculus, and $\mathsf{times}$ is the $\lambda$-term for multiplication. You may use the fact that the $\lambda$-term $\mathsf{times} \, n \, m$ (i.e. $\mathsf{times}$ applied to the Church numerals $n$ and $m$) $\beta$-reduces to the numeral $n×m$.

Q2.
The term $\mathsf{cons}$ appends an element to the front of a list.

$$\mathsf{cons} ≜ \lambda x. \lambda l. \lambda c. \lambda n. c x (l c n)$$

Show that it works by reducing $\mathsf{cons} \, 3 \, [2, 1]$ to $[3, 2, 1]$. Make sure to include every $\beta$-step, and try to use abbreviations for readability.

答:
Question 1. Just apply the definition of list and the fact that $\mathsf{times} \, \underline{n} \, \underline{m} \to_\beta^* \underline{n \times m}$, where $\underline{n}$ is the Church numeral of $n$, for every natural number $n$.

\begin{align}
 [\,\underline{3}, \underline{2}, \underline{1}\,] \, \mathsf{times} \, \underline{1} &= \big( \lambda c. \lambda n. c\,\underline{3} ( c \, \underline{2} (c \, \underline{1} \, n) )  \big) \mathsf{times} \, \underline{1}
\\
 &\to_\beta \big( \lambda n. \mathsf{times} \,\underline{3} ( \mathsf{times} \, \underline{2} (\mathsf{times} \, \underline{1} \, n) )  \big) \underline{1}
\\
 &\to_\beta \mathsf{times} \,\underline{3} ( \mathsf{times} \, \underline{2} (\mathsf{times} \, \underline{1} \, \underline{1}) )
\\
 &\to_\beta^* \mathsf{times} \,\underline{3} ( \mathsf{times} \, \underline{2} (\underline{1 \times 1}) )
\\
&= \mathsf{times} \,\underline{3} ( \mathsf{times} \, \underline{2} \, \underline{1} )
\\
&\to_\beta^* \mathsf{times} \,\underline{3} ( \underline{2 \times 1} )
\\
&= \mathsf{times} \,\underline{3} \, \underline{2}
\\
&\to_\beta^* \underline{3 \times 2}
\\
&= \underline{6}
\end{align}

Question 2. Just apply the definitions of $\mathsf{cons}$ and of list.
\begin{align}
 \mathsf{cons} \, 3 \, [2, 1] &= (\lambda x. \lambda l. \lambda c. \lambda n. cx (lcn)) \, 3 \, [2,1]
\\
 &\to_\beta (\lambda l. \lambda c. \lambda n. c\,3 (lcn))   [2,1]
\\
 &\to_\beta \lambda c. \lambda n. c\,3 ([2,1] cn)
\\
&= \lambda c. \lambda n. c\,3 \big((\lambda c'. \lambda n'. c' 2 (c'  1 \, n')) c\,n \big)
\\
 &\to_\beta \lambda c. \lambda n. c\,3 \big((\lambda n'. c \, 2 (c \, 1 \, n')) n \big)
\\
 &\to_\beta \lambda c. \lambda n. c\,3 ( c \, 2 (c \, 1 \, n) )
\\
&= [3,2,1]
\end{align}

Note that in Question 2, the fact that $1, 2, 3$ are Church numerals do not play any role.
Actually, $1, 2, 3$ here could be replaced by any $\lambda$-terms.

Title: Lambda Calculus Beta reductions


问:
How to integrate the following extreme value type function?
I am trying to integrate the following function.
$$\int_{x}^{\infty}yf(y) =\int_{x}^{\infty}y\bigg(1 + \xi\big(\frac{y-\mu}{\sigma}\big)\bigg)^{-(1+1/\xi)}\exp\bigg(-\bigg(1 + \xi\big(\frac{y-\mu}{\sigma}\big)\bigg)^{-1/\xi}\bigg)dy$$
where $f(y)$ is Generalized Extreme Value disstribution. I did a substituition that $u = 1 + \xi\big(\frac{y-\mu}{\sigma}\big)$ . Then I got the following.

$$\int_{s}^{\infty} \bigg(\frac{\sigma(u-1)}{\xi} + \mu\bigg)u^{-(1+1/\xi)}\exp(-u^{-1/\xi}) \frac{\sigma}{\xi} du $$

where $s = 1+ \xi \big(\frac{x-\mu}{\sigma}\big)$. I tried different substitutions, but I couldn't solve this integral. Could you please give me a hint? Thanks.

答:
Complete solution

For $\xi\neq0$ and $\xi<1$, and by introducing
$$\tau_x=\left ( 1+\xi \frac{x-\mu}{\sigma} \right )^{-1/\xi}$$
then,
$$\int_{x}^{\infty}yf(y)dy=\left ( \mu - \frac{\sigma}{\xi} \right )\left ( 1-e^{-\tau_x} \right ) + \frac{\sigma}{\xi}\cdot\gamma\left ( 1-\xi,\tau_x \right )$$
Where $\gamma()$ is the lower incomplete gamma function. It can be also written according to Gamma distribution CDF, if you environment can easily compute it
with a shape equal to $(1-\xi)$ and a rate equal to 1
$$\left ( \mu - \frac{\sigma}{\xi} \right )\left ( 1-e^{-\tau_x} \right ) + \frac{\sigma}{\xi}\cdot CDF_{gam}\left ( 1-\xi,1,\tau_x \right ) \cdot \Gamma\left ( 1-\xi \right )$$
Where $\Gamma(1-\xi)$ is the classic gamma function.

PROOF

By supposing you want to compute this integrale for a statistical issue. According to this,

if $\xi<0$, then $-\infty < x \leq \mu- \frac{\sigma}{\xi}$, and $\int_{x}^{\infty}yf(y)dy=\int_{x}^{\mu-\sigma/\xi}yf(y)dy$

else if $\xi>0$, then $\mu- \frac{\sigma}{\xi} \leq x < \infty$, no change.

Let introduce,
$$t(y)=\left ( 1+\xi \frac{y-\mu}{\sigma} \right )^{-1/\xi}$$
so,
$$dt=-\frac{1}{\sigma} \left ( 1+\xi \frac{y-\mu}{\sigma} \right )^{-1/\xi-1} dy$$
and finally (note that in your question the factor $1/\sigma$ is missing to define GEV PDF),

$$\int yf(y)dy=-\int \left ( \left ( t^{-\xi}-1 \right )\frac{\sigma}{\xi}+\mu \right )e^{-t}dt$$

Integration boundaries have to be considered differently according to $\xi$ value.

For $\xi<0$, $t(\mu-\sigma/\xi)=0$, and $\tau_x=t(x)$. Furthermore, x can tend to limit
$$\lim_{x \rightarrow -\infty} \tau_x = +\infty$$

For $\xi>0$, we can keep $\tau_x=t(x)$ definition, but limit to the infiny has to be considered for the 2nd boundary
$$\lim_{y \rightarrow +\infty} t(y) = 0$$
And in case that x tend to the GEV lower boundary
$$\lim_{x \rightarrow \mu-\sigma/\xi} \tau_x = +\infty$$

Finally, for any $\xi\neq0$,
$$\int yf(y)dy=-\int_{\tau_x}^{0} \left ( \left ( t^{-\xi}-1 \right )\frac{\sigma}{\xi}+\mu \right )e^{-t}dt$$

We can reverse boundaries thanks to "-" sign and separate the two parts,
$$\int yf(y)dy=\left (\mu -\frac{\sigma}{\xi}\right )\int_{0}^{\tau_x} e^{-t}dt + \frac{\sigma}{\xi} \int_{0}^{\tau_x} t^{-\xi}e^{-t}dt$$

And the really end,
$$\int yf(y)dy=\left (\mu -\frac{\sigma}{\xi}\right )\left ( 1-e^{-\tau_x} \right ) + \frac{\sigma}{\xi} \int_{0}^{\tau_x} t^{(1-\xi)-1}e^{-t}dt$$

Where the remaining integrale is by definition the lower incomplete gamma function $\gamma(1-\xi,\tau_x)$, and finite for $\xi<1$
It can be compute by Gamma distribution CDF with a shape equal to $(1-\xi)$ and a rate equal to 1 if needed
$$Gam_{CDF}(1-\xi,1,\tau_x)=\frac{\gamma(1-\xi,\tau_x)}{\Gamma(1-\xi)}$$

What about $\xi=0$

By a similar approach, with $\theta_x=e^{-\frac{x-\mu}{\sigma}}$
$$\int yf(y)dy=\mu-\sigma \int_{0}^{\theta_x}\ln(t) e^{-t}dt$$

To do so, it seems possible to have some consideration on derivative incomplete gamma function, based on Meijer G-function... out of my skills!


问:
In proving something, is it better to use $\iff$ even if you only ‘need’ $\implies,$ or should you only $\implies$ what you need?
Suppose we want to prove a result $R.$ In the process of proving $R,$ we have a biconditional $P \iff Q,$ which is true by observation and doesn't require an explicit proof taking up space.

However, to prove $R,$ we only really needed $P \implies Q,$ which is also obviously true by inspection, and therefore a proof was not given for it.

In such a situation, is it better to write (stylistically) that $P \iff Q$ or $P \implies Q ~ $?

答: An arrow that points both ways means an "if and only if" statement - which means it has more weight to it than an arrow that points one way. I'd prefer to use an arrow that is directed to my final destination (avoid meddling with both-ways arrows that require more explanation, unless trivial or are part of the proof). To conclude, it depends on the level of triviality/ sort of question:), but if it's enough to use an arrow that points one way in order to complete your assignment, why bother with the two sided one.</p>


问:
Prove that, if $\frac{\tau_i}{\tau_j}\in\Bbb Q\space j\ne i,\space i,j\in\{1,\ldots,n\},$ the function $f(x)=\sum_{k=1}^nf_k(x)$

Let $f_1,f_2,\ldots,f_n:\Bbb R\to\Bbb R$ be (non-constant) periodic functions with periods $\tau_1,\tau_2,\ldots,\tau_n\in\Bbb R^+$ respectively. Prove that, if $\dfrac{\tau_i}{\tau_j}\in\Bbb Q\space j\ne i,\space i,j\in\{1,\ldots,n\},$ the function $f(x)=\sum\limits_{k=1}^nf_k(x)$ is periodic.

My attempt:

I gave it a try by induction.

Let's look at the base case $n=2$ and let's take $g(x)=(f_1+f_2)(x),$ where $f_1(x)=f_1(x+\tau_1),\space\forall x\in\Bbb R$ and $f_2(x)=f_2(x+\tau_2)\forall x\in\Bbb R$. If $\dfrac{\tau_1}{\tau_2}\in\Bbb Q,$ then we either have $$\tau_1\in\Bbb Q^+\space\land\space\tau_2\in\Bbb Q^+$$ or $$\tau_1=q_1r\in\Bbb R\setminus\Bbb Q\space\land\space\tau_2=q_2r\in\Bbb R\setminus\Bbb Q,\quad q_1,q_2\in\Bbb Q^+,\space r\in\Bbb R^+\setminus\Bbb Q$$

In the first case, let $\tau_1=\dfrac{a}b,\space\tau_2=\dfrac{c}d$ and, WLOG, let's assume $\gcd(a,b)=1,\space\gcd(c,d)=1,\space a,b,c,d\in\Bbb N$ (I don't include $0$). Then:
$$\begin{aligned}f_1(x)&=f_1\left(x+\frac{a}b\right)=f_1(x+a),\forall x\in\Bbb R\\f_2(x)&=f_2\left(x+\frac{c}d\right)=f_2(x+c),\forall x\in\Bbb R\end{aligned},$$

but then, we can find $m,n\in\Bbb N$ s. t. $ma=nc=\gcd(a,c)=k,$ so $f_1(x)=f_1(x+k),\space\forall x\in\Bbb R$ and $f_2(x)=f_2(x+k),\space\forall x\in\Bbb R$ and therefore, $g(x)=(f_1+f_2)(x)=(f_1+f_2)(x+k)=g(x+k)\space\forall x\in\Bbb R$.

In the other case, analogously, let $q_1=\dfrac{a}b,q_2=\dfrac{c}d,\space\gcd(a,b)=1,\gcd(c,d)=1,a,b,c,d\in\Bbb N$ and let $r\in\Bbb R^+$. Then:
$$\begin{aligned}f_1(x)&=f_1\left(x+\dfrac{a}br\right)=f_1(x+ar),\forall x\in\Bbb R\\f_2(x)&=f_2\left(x+\dfrac{c}dr\right)=f_2(x+cr),\forall x\in\Bbb R\end{aligned}$$

and we can again find $\gcd(a,c)=ma=nc=k$ s. t. $g(x)=(f_1+f_2)(x)=(f_1+f_2)(x+kr)=g(x+kr)\forall x\in\Bbb R$.

Then, if we assume the statement holds for any $2\le m\le n$, i. e., that $g(x)=\sum\limits_{k=1}^m f_k(x)$, where the period of $f_k$ is $\tau_g=\dfrac{a_k}{b_k}\in\Bbb Q^+$ or $\tau_g=\dfrac{a_k}{b_k}r\in\Bbb R^+$ and therefore $g(x)=g(x+\gcd(a_1,a_2,\ldots,a_m))$ or $g(x)=g(x+\gcd(a_1,a_2,\ldots,a_m)r),2\le m\le n$, then:

if $$\tau_{n+1}=\dfrac{a_{n+1}}{b_{n+1}}\in\Bbb Q\space\land\space\tau_g\in\Bbb Q$$ or $$\tau_{n+1}=\dfrac{a_{n+1}}{b_{n+1}}\in\Bbb R^+\setminus\Bbb Q\space\land\space\tau_g\in\Bbb R^+\setminus\Bbb Q$$  $f(x)=g(x)+f_n(x)$ will be periodic with a period $\tau=\gcd(a_1,a_2,\ldots,a_n,a_{n+1})$ or $\tau=\gcd(a_1,a_2,\ldots,a_n,a_{n+1})r$ respectively.

May I ask for verification of my work so far and advice on what I have to correct/improve?

Thank you in advance!

答:
Some remarks concerning your proof:

This part is wrong or misleading:
$$
ma=nc=\gcd(a,c)=k \, .
$$
$\gcd(a,c)$ is the greatest common divisor of $a$ and $c$. What you need is the least common multiple $\operatorname{lcm}(a, c)$ (or any common multiple) in order to get a common period of $f_1$ and $f_2$.

The distinction whether $\tau_1$ and $\tau_2$ are both rational or both irrational is not needed.

The assumption that the fractions $\tau_1 = a/b$ and $\tau_2 = c/d$ are in reduced form is not needed.

I would simplify the proof as follows: With $\tau = \tau_1$ and $r_k = \tau_k/\tau_1 \in \Bbb Q^+$ the periods of $f_1, f_2, \ldots, f_n$ are
$$
 r_1 \tau, \, r_2 \tau, \, \ldots, \, r_n \tau \,,
$$
respectively. The positive rational numbers $r_k$ can be written as $p_k/q$ with $p_k \in \Bbb N$ and a common denominator $q \in \Bbb N$.

Let $p = \operatorname{lcm}(p_1, p_2, \ldots, p_n)$ be the least common multiple of the numerators (or any common multiple, $p=p_1p_2 \cdots p_n$ would work as well). Then
$$
 \forall x: f_k(x) = f_k(x + p_k \frac{\tau}{q}) \\
 \implies \forall x: f_k(x) = f_k(x + p \frac{\tau}{q})
$$
so that $p \tau/q$ is a common period of all functions $f_1, \ldots, f_n$. It follows that $\sum_{k=1}^n f_k$ is periodic.


问:
A question about the relation between linear transformations and their representing matrices
I would like to have some clarification regarding the notion that all linear transformations can be represented by "representing matrices". There's this lemma that the space of all linear transformations from spaces V to U are isomophic to m×n matrices (m being dimV and n being dimU). Which leads to my question: given a matrix m×n, can I deduce that there exists some transformation from V to U (dimV = m and dimU =n) that is represented by this very matrix (since isomophic relationships go both ways)?</p>

答:
Yes that's right... though it seems your convention for matrix representations is different from what I'm used to. Lets say $V,U$ are vector spaces over a field $F$ of dimension $n,m$ respectively. Then, yes we can construct an isomorphism $L(V,U)\cong M_{m\times n}(F)$, but this isomorphism depends heavily on the choice of basis.

More explicitly, if $\beta$ is a basis of $V$, and $\gamma$ is a basis of $U$, then we can construct a linear isomorphism $\Phi_{\beta,\gamma}:L(V,U)\to M_{m\times n}(F)$. But, like I've indicated in the notation, the isomorphism $\Phi_{\beta,\gamma}$ very much depends on the basis $\beta,\gamma$. If you change the basis to a different one, say $\beta', \gamma'$, you will get a different isomorphism $\Phi_{\beta',\gamma'}$. So you always have to be careful when dealing with matrices and always make sure to explicitly mention what the bases being used are.


问:
Show that the method of moments estimator of $\theta$ is biased
Let ${X_1},...,{X_n}$  a random sample from the probability distribution with pdf $f(x) = \theta {x^{\theta  - 1}}$, $0 < x < 1$ where $\theta  > 0$

I've found the method of moments estimator for $\theta$ to be : ${\hat \theta _{mom}} = \frac{{\overline X }}{{1 - \overline X }}$

Next, I'm asked to show that ${\hat \theta _{mom}}$ is biased if $n=1$ and $\theta=1$. I'm struggling to understand the solution provided, which follows:
If $n=1$ and $\theta=1$ then $f(x)=1$ and $X_1=\overline X$

$$E[{\hat \theta _{mom}}] = E\left[ {\frac{{{X_1}}}{{{X_1} + 1}}} \right] = 1 - E\left[ {\frac{1}{{{X_1} + 1}}} \right] = 1 - \int\limits_0^1 {\frac{1}{{1 - x}}dx =  - \infty  \ne \theta }$$

I understand why a result of $- \infty $ not being equal to theta proves it is biased, however I don't understand how the expectations and the integral have been formulated.

If ${\hat \theta _{mom}} = \frac{{\overline X }}{{1 - \overline X }}$ then why is $E[{\hat \theta _{mom}}]$ not equivalent with $E\left[ {\frac{{{X_1}}}{{1 - {X_1}}}} \right]$? And how is the expectation equivalent with the integral?

I hope somebody can shed some light on what it is that I'm missing here.

答:
There are 2 typos in the solution from your question.

$\theta_{mom} \ne {\frac{{{X_1}}}{{{X_1} + 1}}} $ because $\theta_{mom} = \frac{X_1}{1-X_1}$.

And if we will work with $\theta^* =  {\frac{{{X_1}}}{{{X_1} + 1}}} $, then instead of

$$E \theta^* = E\left[ {\frac{{{X_1}}}{{{X_1} + 1}}} \right] = 1 - E\left[ {\frac{1}{{{X_1} + 1}}} \right] = 1 - \int\limits_0^1 {\frac{1}{{1 - x}}dx }$$
we should have
$$1 - \int\limits_0^1 {\frac{1}{{1 + x}}dx}$$


答:
For small $t$, we know that $\sin (t)=t+\mathcal{O}(t^3)$, meaning that $\frac{1}{\sin(t)}=\frac{1}{t}\left[1+\mathcal{O}(t^2)\right]$ . Your limit is
$$\lim_{t\to\ 0}\frac{\int_0^t \sqrt{1+\sin(x^2)}\,dx - \int_0^0 \sqrt{1+\sin(x^2)}\,dx}{t}\left[1+\mathcal{O}(t^2)\right].$$
Hence, the limit is also
$$\frac{d}{dt}\Bigg|_{t=0}\int_0^t\sqrt{1+\sin(x^2)}\,dx=1.$$

Title: Find $\lim_{t \rightarrow 0} \int_{0}^{t} \frac{\sqrt{1+\sin(x^2)}}{\sin t} dx$

Body:
Find $$\lim_{t \rightarrow 0} \int_{0}^{t} \frac{\sqrt{1+\sin(x^2)}}{\sin t} dx$$

One easy step is replacing $\sin t = t$, since $\lim_{t \rightarrow 0} \sin t / t = 1$. How do I continue with the rest?


问:
Graph Theory: what are the two vertices and how to draw them?

Define a graph $G$ such that $V(G) = \{2,3,4,5,11,12,13,14\}$ and two
  vertices $s$ and $t$ are adjacent if and only if $\gcd\{s,t\} = 1$. Draw a
  diagram of $G$ and find its size $e(G)$.

I can understand V(G) = {2,3,4,5,11,12,13,14} but what are "two vertices $s$ and $t$ are adjacent if and only if $\gcd\{s,t\} = 1$"?

答:
Here's a drawing of the graph in question.

The $7$ dashed lines represent non-edges, and the $\binom{7}{2}-7=14$ solid lines are the present edges.

ContentLicense: CC BY-SA 3.0


问:
Indefinite Integral of $\dfrac {x^{p-1}}{x^{2m} - a^{2m}}$
I am looking at a bunch of related integrals found in Spiegel's "Mathematical Handbook of Formulas and Tables", (Schaum, 1968), item $14.336$.

It's a complicated and messy old beast:

For $m \in \mathbb Z$ such that $m \ge 1$:

$\displaystyle \int \dfrac {x^{p-1} \, \mathrm d x}{x^{2m} - a^{2m}} = \frac 1 {2 m a^{2 m - p}} \sum_{k=1}^m \cos \frac {k p \pi} m \ln \left({x^2 - 2 a x \cos \frac {k \pi} m + a^2}\right)$

$\displaystyle - \frac 1 {m a^{2 m - p}} \sum_{k=1}^m \sin \frac {k p \pi} m \arctan \left({\frac {x - a \cos \frac {k \pi} m}{a \sin \frac {k \pi} m}}\right)$

$\displaystyle + \frac 1 {2 m a^{2 m - p}} (\ln (x - a) + (-1)^p \ln (x + a)) + C$

for all $0 < p \le 2 m$.

(Note that he glosses over the sign of $x$, presenting the expression for positive $x$ only -- I will explore the negative $x$ case later.)

I thought of $2$ different approaches:

Factorise the denominator using the standard result:

$\displaystyle x^{2 n} - y^{2 n} = (x - y) (x + y) \prod_{k \mathop = 1}^{n - 1} \left({x^2 - 2 x y \cos \dfrac {k \pi} n + y^2}\right)$

and then explore the possibility of doing a partial fraction expansion. However, this did not work out so well, as I was then unable to see how to simplify the expression on the top of the resulting terms in all but the $x - a$ and $x + a$ denominators, so I didn't proceed further with that.

Prove it by induction on $m$, which first concerns establishing the base case:

$\displaystyle \int \frac {\mathrm d x} {x^2 - a^2} = \frac 1 {2 a} \ln \left({\dfrac {x - a} {x + a}}\right) + C$ (that is: $m = 1$, $p = 1$)

$\displaystyle \int \frac {x \, \mathrm d x} {x^2 - a^2} = \ln \left({x^2 - a^2}\right) + C$ (that is: $m = 1$, $p = 2$)

But when I put $m = 1$ into the above expression, I get:

$\dfrac 1 {2 a^{2 - p} } ( (-1)^p \ln (x^2 + 2 a x + a^2) ) + \dfrac 1 {2 a^{2 - p} } (\ln (x - a) + (-1)^p \ln (x + a) )$

But that pesky term on the left, in $(-1)^p \ln (x^2 + 2 a x + a^2)$ is clearly incorrect.

This arises from the $\cos \dfrac {k p \pi} m \ln \left({x^2 - 2 a x \cos \dfrac {k \pi} m + a^2}\right)$ expression in the given solution.

But for integer $p$ and $m = 1$, that $\cos \dfrac {k p \pi} m$ term, which I would hope to vanish, defiantly remains in place.

It doesn't help much changing it to $\cos \dfrac {k p \pi} {2 m}$, because this time it does not vanish for $p = 2$.

So either Spiegel has reported this result wrong, or I'm misunderstanding something.

Should that cosine in fact be a sine? Then it would vanish away like I want it to.

For $m = 2$ and $p = 1$ you get this:

$\dfrac 1 {4 a^3} \ln \left( {\dfrac {x - a} {x + a} }\right) - \dfrac 1 {2 a^3} \arctan \dfrac x a$

which is consistent with the result quoted. I have gone a little further, trying $p = 2, 3, 4$, but now it is looking as though the cosine term is completely erroneous and perhaps should not be there at all, because any contribution it makes is consistently absent.

I have gone no further with higher $m$ values.

答:
The sane way to do this is using complex functions.  The integrand is a rational function with simple poles at $x = \omega_k a$ where $\omega_k =e^{\pi i k /m} $, $k = 0,1,\ldots, 2m-1$ are the  $2m$'th roots of unity:

$$ f(x) = \frac{x^{p-1}}{x^{2m} - a^{2m}} = \frac{x^{p-1}}{\prod_{k=0}^{2m-1} (x - \omega_k a)}$$

Its residue at $x=\omega_k$, is then $$\text{res}(f, \omega_k a) = \frac{(\omega_k a)^{p - 2m}}{2m}$$
so that
$$f(x) = \sum_{k=0}^{2m-1} \frac{(\omega_k a)^{p-2m}}{2m (x - \omega_k a)} $$
which has an indefinite integral
$$ F(x) = \sum_{k=0}^{2m-1} \frac{(\omega_k a)^{p-2m}}{2m} \log (x - \omega_k a) $$
For positive real $x$ and $a$ we can write
$$\log(x - \omega_k a) = \frac{1}{2} \log(x^2 - 2a \cos(2\pi k /m) x + a^2) - i \arctan \left(\frac{a \sin(2 \pi k/m)}{x - a \cos(2\pi k /m)}\right)$$
$$ (\omega_k a)^{p-2m} = a^{p-2m} \left(\cos(p k \pi/m) + i \sin(p k \pi/m)\right)$$
and the formula follows.


问:
How many final placements are possible if the leader changed 0...5 times?
Six competitors A-F take part in cycling time trial. The race is about covering several dozen kilometers as quickly as possible. Following the rules of this sporting event, competitors start one by one, at regular intervals. Let's call the cyclist, who has ridden the entire route and at the moment he is in the lead (so he had the best time so far), the leader. The final placement is the list obtained by ranking the cyclists according to the final places they took in the race. How many final placements are there in which:

A) the leader didn't change from the start;

B) the leader changed 2
times;

C) the leader changed 3 time;s

D) the leader changed 4
times;

E)the leader changed 5 times;

I believe the key is understanding the mechanic of leadership. Also they don't wait for the previous one to finish so for example B can finish before A. Then A can no longer take the leadership. Subproblems A and E are trivial, A is just 5! and E is 1, if I am not mistaken.

EDIt: I got more info, answers for 3 changes and 4 changes are 85 and 15 respectively. Looking for a way that isn't bruteforcing it.

答:
You are correct for A.  The first rider has to stay in front, but the others can be in any order.

For B, any of the riders except A can be the eventual winner.  If B is the eventual winner, the others can finish in any order, so there are $5!$.  If C is the eventual winner, B cannot beat A, so there are $\frac 12 \cdot 5!$.  If D is the eventual winner, B and C cannot beat A, so there are $\frac 13 \cdot 5!$.  If E is the eventual winner, $\frac 14\cdot 5!$ and if F is the eventual winner, $\frac 15 \cdot 5!$ for a total of $\frac {137}{60}\cdot 120=274$.  $\frac {137}{60}=H_5$, the fifth harmonic number.

For C you need to look at the two riders that take the lead.  As in B, riders before the first to take the lead cannot beat A.  Now riders between the first to take the lead and the eventual winner can beat A but not the interim leader.


问:
Is there a $\triangle ABC$, with $D$ on $BC$, such that $AB$, $BC$, $CA$, $AD$, $BD$, $CD$ are all integers, but only one is even?
This is just a question I came up with while working through another problem.

If we have a triangle $ABC$ and we draw a cevian from $A$ to $D$ on side $BC$, is it possible to assign integer values to lengths $AB$, $BC$, $AC$, $AD$, $BD$ or $CD$ such that only one of the listed lengths has an even valued length? If it is possible, what are these lengths?

I have done some parts of this problem on my own. I figured out that this even length could only be $BC$, $CD$, or $BD$, with $BD$ and $CD$ being the same case.

With $BC$ being the only even length I computed for $BC$ using Stewart's Theorem and got
$$BC = \frac{(AC^2)(BD)+(AB^2)(CD)}{(BD)(CD)+(AD)^2}$$ where every length but $BC$ is odd. The equation seems to work but I don't know how to obtain its integer solutions.

Similarly with $BD$/$CD$ case, I applied Stewart's theorem and got $$BD=\frac{(AB^2)(CD)-(BC)(AD^2)}{(CD)(BC)-(AC)^2}$$
Once again this seems to work, but I don't know how to obtain its integer solutions.

Pretty much all I need help is verifying whether or not my work is right, and if it  is how to find the integer solutions of my provided equations. If there is a better way or my work has flaws please let me know.

答:
Too long for a comment but here is food for thought.

It would appear, incorrectly, that we can find what you need if we join two dissimilar Pythagorean triples on a matching side-A. This side and be any odd number greater than one. We begin with Euclid' formula here shown as.

$$A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2$$
If we solve the A-function for k, we can find triples that match a length "A" by testing a finite range of m-values to see which ones yield integers.

\begin{equation}
A=m^2-k^2\implies k=\sqrt{m^2-A}\qquad\text{for}\qquad  \lfloor\sqrt{A+1}\rfloor \le m \le \frac{A+1}{2}
\end{equation}
The lower limit ensures $k\in\mathbb{N}$ and the upper limit ensures $m> k$.
$$A=15\implies \lfloor\sqrt{15+1}\rfloor=4\le m \le \frac{15+1}{2} =8\quad\land\quad m\in\{4,8\}\implies k \in\{1,7\} $$
$$F(4,1)=(15,8,17)\qquad \qquad F(8,7)=(15,112,113) $$

If these two are joined at side-A, we have an acute triangle with sides $$(17,113,120)$$ with a cevian $(15)$ going from the $(17,113)$ vertex to the. 120-side, splitting it into lengths of $8$ and $112$. Unfortunately, this yields $3$ even lengths. If we were to join the even sides of two triples such as

$$(5,12,13)\qquad (35,12,37)$$ we would have $(13,37,40)$ with $40$ split into $(5,35)$. That would still leave two even lengths: the base $(40)$ and the cevian $(12)$. This probably eliminates Pythagorean triples from the solution. There are methods of finding integer cevians in a right triangle from a non-right vertex to an adjacent side but the question remains about the split-length of the target side being integers.


问:
Martingale Representation Theorem Applied to Integral of Squared Brownian Motion wrt time
I am trying to find the martingale representation of $F = \int_0^TW_t^2dt $.

I wrote that $ \int_0^TW_t^2dt = TW_T^2 - \frac{T^2}{2} - \int_0^T 2tW_t dW_t$ and I'm stuck.

Any ideas on how to proceed?

答:
As $\frac12 W_{T}^2 - \frac{T}2= \int_{0}^T W_t dW_t$ we may proceed:

$$TW_T^2 - \frac{T^2}{2} - \int_0^T tW_t dW_t
=2T(\frac12 W_{T}^2 - \frac{T}2) + \frac{T^2}2  - \int_0^T tW_t dW_t =$$
$$ = \frac{T^2}2 + 2T \int_{0}^T W_t dW_t  - \int_0^T tW_t dW_t=$$
$$=\frac{T^2}2  +\int_0^T (2TW_t - tW_t) dW_t .$$

But there's a typo in your formula.

Indeed, by Ito formula $dW_t^2 = 2W_t d W_t + dt$ and thus

$$\int_0^TW_t^2dt = TW_T^2 - 0W_0^2 - \int_0^T t dW_t^2 = TW_T^2 - 0W_0^2 - \int_0^T t  (2W_t d W_t + dt) =$$
$$= TW_T^2 -   \frac{T^2}{2} - 2\int_0^T tW_t dW_t  $$

Thus $$\int_0^TW_t^2dt    = TW_T^2 -   \frac{T^2}{2} - 2\int_0^T tW_t dW_t  = \frac{T^2}2  + 2\int_0^T (T-t)W_tdW_t .$$


问:
Probability question - Bayes' Theorem biased coins
A bag contains 120 coins: 30 coins come up heads $40\%$ of the time (group a), 30 coins come up heads $20\%$ of the time (group b), and 60 coins have tails on both sides (group c).

I've calculated:

I pick a coin at random, toss it, and it comes up tails.  What is the probability that it is from group b?  0.2353

I toss that same coin for the second time and it shows tails again. Now hat is the probability that it is from group b? 0.2215

The question I'm stuck on is:
Flipping the same coin for the third time, now it comes up heads. What is the probability that is from group b?  I calculated the answer to be 0.2953

However, my professor has marked this wrong with the following working and answer:

Flipping the same coin for the third time. Now it comes up heads. What is the probability that is from group b?

P(Head | B) = 0.2

P(Tail) = 0.6 * 0.5  + 0.8 * 0.5 = 0.7

P(B | Tail) = 0.8 * 0.25 / 0.7 = 0.2887

P(Head) = 0.2*0.5 + 0.4 * 0.5 = 0.3

P(B | Tail, Tail) = P(Tail | B) / P(Tail) * P(Tail | B) = 0.8 / 0.7 * 0.2887 = 0.3299

P(B | Head, Tail, Tail) = P(Head | B) / P(Head) * P(B | Tail, Tail) = 0.2 / 0.3 * 0.3299 = 0.2199

I really don't understand how the answer is 0.2199? can someone explain this please? Thank you

答:
$$P(B|TTH) = \frac{P(TTH|B)P(B)}{P(TTH)}$$

$$ = \frac{P(TTH|B)P(B)}{P(TTH|A)P(A) + P(TTH|B)P(B) + P(TTH|C)P(C)}$$

Here $P(A)= P(B) = 1/4$, and $P(C) = 1/2$.

$P(TTH|A) = .8 \cdot .8 \cdot .2$

$P(TTH|B) = .6 \cdot .6 \cdot .4$

$P(TTH|C) = 0$

Substituting I get $P(B|TTH) = 0.470588$

And thus, of course $P(A|TTH) = 0.529412$ and $P(C|TTH) = 0$.

Your error was your denominator, i.e., your calculation of $P(TTH)$.  Check that (conceptually), realizing the process:  FIRST choose a coin, THEN flip it three times.  You either chose an $A$ or a $B$ or a $C$ coin, and then were "stuck" with it for all the flippings.  See?

The professor is simply confused.  You never have to go through all those intermediate states (e.g., $TT$).


问:
If $f$ is continuous then $f$ is uniformly continuous iff $|f|$ is uniformly continuous
If $f:\Bbb R^n \to \Bbb R$ is continuous then $f$ is uniformly continuous iff $|f|$  is uniformly continuous.

A map $f$ from a metric space $M=(M,d)$ to a metric space $N=(N,\rho)$ is said to be uniformly continuous if for every $\epsilon>0$, there exists a $\delta>0$ such that $\rho(f(x),f(y))<\epsilon$ whenever $x,y \in M$ satisfy $d(x,y)<\delta$.

Clearly if $f:\Bbb R^n \to \Bbb R$ is uniformly continuous then $|f|$  is uniformly continuous as $|f|(x)-|f|(y)|\leq |f(x)-f(y)|$ but I am having a real trouble showing the converse part. In the region where $f$ is always positive or negative, we will not have any problem but how to deal with the points where $f$ is changing sign. If the zeros of $f$ are finite then also we can take a minimum of all $\delta$s and conclude the result. What will happen if zeros of $f$ are infinite?

答:
As mentioned in the comments, the proof given here may easily be modified to work for the whole of $\mathbb{R}^n$.

Since $\lvert f \rvert$ is uniformly continuous, there exists a $\delta > 0$ such that
\begin{align*}
d(x,y) \leq \delta \Rightarrow \lvert \lvert f \rvert (x) - \lvert f \rvert (y) \rvert \leq \frac{\epsilon}{2}.
\end{align*}
Note that if $f(x)f(y) > 0$, then
\begin{align*}
\lvert f(x)-f(y)\rvert
\leq \lvert \lvert f \rvert(x) - \lvert f \rvert (y) \rvert,
\end{align*}
which is less than $\epsilon/2$ whenever $d(x,y) \leq \delta$. Unsurprisingly, this case was quite trivial. We now turn our attention to the case where $f(x)f(y) \overset{\star}{\leq} 0$. Since it always holds that
\begin{align*}
\lvert f(x)-f(y)\rvert
\leq \lvert f \rvert(x) + \lvert f \rvert (y).
\end{align*}
it suffices to show that $\star$ implies the existence of a $z$ such that $d(x,z) \vee d(y,z) \leq d(x,y)$ and $f(z) = 0$. Because then
\begin{align*}
\lvert f(x) - f(y) \rvert
&\leq
\lvert \lvert f \rvert(x) - \lvert f \rvert(z) \rvert
+
\lvert \lvert f \rvert(y) - \lvert f \rvert(z) \rvert \\
&\leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon
\end{align*}
whenever $d(x,y) \leq \delta$. Since $f$ is continuous, the existence of a suitable $z$ follows from the continuity of $f$ and $\star$ (as a consequence of the Intermediate Value Theorem, see e.g. here).


问:
Proving A Combination Statement By Forming committees
$ \binom{n}{m}\binom{m}{m} + \binom{n}{m+1}\binom{m+1}{m} + \binom{n}{m+2}\binom{m+2}{m} + \cdots + \binom{n}{n}\binom{n}{m} = \binom{n}{m} 2^{n-m}$

I am completely lost on where to go. Any hints? Thank you very much!

答:
There are $n$ people. From these, some will go to a party. From people that go to the party, $m$ will be singers.

Now you can either choose who go to party first (at least $m$), then from the chosen, you choose who will sing.

Or you can choose who will go to party to sing first, and then the remaining $n-m$ people may or may not go to party


问:
Evenly distributing $n^2$ ball between $n$ pockets by chosing $2$ of them at a time.
We have $n^2$ balls distributed among $n$ pockets (some of them may be empty). We can choose any two pockets of which sum of balls is even and then distribute the balls equally among these two pockets. We can repeat it as many times as we wish. For every natural, even number $n$ such that $2 ≤ n ≤ 10$ decide if by using this method we can always distribute the balls so that at the end every bag contains n balls, no matter how many balls there are in each pocket at the beginning.

I think and I am informed it is true that we can do it for every n number of pockets which is a power of two. I suppose, we split them in two until we are left with $n/2$ pairs of pockets. Then we make a move on each pocket and make our way out merging the subsets of pairs. For $n=4$ we have two pairs which we first want to even out and then make the move on the first elements from the each pair an then on the second elements from each pair. However the number of ball combined in the selected pockets has to be even. That feels to be the case as if $n$ is a power of two, $n^2$ is even so we have an even number of odd numbers of balls. How would a proper proof of that look? How to prove we cannot do it if $n$ is not a power of two?

答:
To rule out cases where a fair distribution is impossible it suffices to show there is one initial allocation that can't be distributed.

$n$ is odd

Let all the balls be in the same pocket. Then no "move" is permitted. Thus the allocation is impossible.

$n = 2^d\cdot m,\ m$ is odd $>1$

Split the $2^{d+1}\cdot m^2$ balls evenly into $2^{d+1}$ pockets. Note that since $m\ge3$ there are enough pockets. Now, we can "move" forever but no move can change the allocation, apart from change which pockets contain the balls.

$n = 2^d$

Proof by induction:

Base case: Check $n = 2$ by hand.

Inductive hypothesis: Assume that it holds for some $d$.

Inductive step: Pick any starting allocation. Then, if there are any pockets with odd number of balls paired them with each other until only pockets with even are left. Now, if you treat any two balls as one, the series of "moves" that solves the inductive case allocates 2 balls in $2^d$ pockets. To finish, you simply match all two ball pockets with all the empty ones.

EDIT:
Case 2 is wrong. My idea is to split it into $2^i m^2$ in such way that you will always be stuck with numbers of the same form. Working on it.


问:
Need Help Solving Logarithmic Equation $2(5)^x=3^{x+1}$
Need Help Solve Logarithmic Equation. Thanks Gary

$$2(5)^x = 3^{x+1}$$

答:
Assuming you mean $2(5)^x = 3^{x + 1}$, you can take the natural log of both sides.

$$\ln(2(5)^x) = \ln(3^{x + 1})$$

Then the product/sum rule:
$$\ln(2)+\ln(5^x) = \ln(3^{x + 1})$$
Then the exponents:
$$\ln(2)+x \ln(5) = (x+1)\ln(3)$$
Distribute:
$$\ln(2)+x \ln(5) = x\ln(3)+\ln(3)$$
Collect like terms:
$$x\ln(5) - x\ln(3) = \ln(3)-\ln(2)$$
Factor:
$$x(\ln(5) - \ln(3)) = \ln(3)-\ln(2)$$
Difference/quotient:
$$x \ln(\frac{5}{3}) = \ln(\frac{3}{2})$$
Divide:
$$x  = \dfrac{\ln(\frac{3}{2})}{\ln(\frac{5}{3})}$$

Not pretty, but that's one way to write the solution.


问:
Classical References on Stability of Autonomous and Non-autonomous Systems and Lyapunov Stability
I ask for classical, but not much advanced books or articles in Stability of Autonomous and Non-autonomous Systems and Lyapunov Stability.

I mean: a book where the definitions and examples are in easy access, but the kind of book that we could reference/cite without fear..!

Thank you!

答:
There are two standard textbooks:

Khalil, Hassan K., and Jessy W. Grizzle. Nonlinear systems. Vol. 3. Upper Saddle River, NJ: Prentice hall, 2002.

Slotine, Jean-Jacques E., and Weiping Li. Applied nonlinear control. Vol. 199. No. 1. Englewood Cliffs, NJ: Prentice hall, 1991.

Note that the first book is somewhat detailed, and a simplified "student" edition was recently published:

Khalil, Hassan K. Nonlinear control. Pearson Higher Ed, 2014.


问:
Is my thought proces right? Equation of tangent plane at level surface at given point.
So the question is: Give the equation of the tangent plane at the surface level of the function $$f(x, y, z)=\cos (x+2 y+3 z)$$ at the point $(x, y, z)=(\pi / 2, \pi, \pi)$

I don't actually want you to calculate it (but if you like, the bettter :) ), I really want to know if my thought process is correct (If I didn't make a mistake e.g. my equation for a tangent plane).

So first I calculated the partial derivatives (at the given point)

$f_x=1$
$f_x=2$
$f_x=3$

and then I used the equation for a tangent plane (don't really  know if this is correct)
$$z=f(a,b,c)+f_x(a,b)(x-a)+f_y(a,b)(y-a)+f_z(a,b)(z-a)$$
and got $x+2y+3z-\frac{11}{2}\pi$

If you can help me, it would be very much appreciated.

Thanks in advance

答:
This is just alternative solution without using Calculus

Notice that $f(x,y,z) = k_{1} \implies x+2y+3z=k_{2}$ i.e. the tangent plane equation is simply $x+2y+3z=\frac{\pi}{2}+2\pi+3\pi$


问:
What is the probability that precisely $z$ cars are yellow?
There are 100 cars in a field. $x$ cars are yellow.

We are choosing $y$ cars randomly to a test drive.

What is the probability that precisely $z$ cars are yellow?

I thought about something like $\frac{\binom{z}{y}}{100!}$, but I probably have to use the fact that $x$ cars are yellow.

答:
Hints

First see combinations.

Label the cars $1,2,\cdots, 100$.

WLOG the yellow cars are the ones labeled $1,2,\cdots, x.$

Probability = $$\frac{N\text{(umerator)}}{D\text{(enominator)}}.$$

$N = A \times B$ where $A = $ the total number of ways of selecting exactly $z$ numbers from the set $\{1,2,\cdots, x\}$

and $B = $ the total number of ways of selecting exactly $(y - z)$ numbers from the set $\{(x+1), (x+2), \cdots, 100\}.$

$D = $ the total number of ways of selecting $y$ numbers from the set $\{1,2,\cdots, 100\}.$


答:
What you want is the hypergeometric distribution.

The answer will be $$\frac{\binom{x}{z}\binom{100-x}{y-z}}{\binom{100}{y}}.$$

Body:
There are 100 cars in a field. $x$ cars are yellow.

We are choosing $y$ cars randomly to a test drive.

What is the probability that precisely $z$ cars are yellow?

I thought about something like $\frac{\binom{z}{y}}{100!}$, but I probably have to use the fact that $x$ cars are yellow.


问:
Function in quotient space
Let $V$ and $W$ be both K-linear vector spaces. Now $Y$ is a linear subspace of $V$ and $Z$ is a linear subspace of $W$. Let $\alpha: V \rightarrow W$ be a $K$-linear function with $\alpha(Y)\subseteq Z$.

I want to prove that $\alpha$ induces linear functions $\alpha|_Y: Y \rightarrow Z$ with $\alpha|_Y(y)=\alpha(y)$ and $\bar{\alpha}: V/Y \rightarrow W/Z$ with $\bar{\alpha}(v+Y)=\alpha(v)+Z$

My idea:
As $\alpha(Y)\subseteq Z$, $\exists z\in Z: \forall y\in Y:z=\alpha(y)$. So $\alpha$ induces $\alpha|_Y(y)=\alpha(y)$. $\square$

Now as $\alpha$ is linear, we can write $\bar{\alpha}(v+Y)=\bar{\alpha}(v)+\bar{\alpha}(Y)$ but now I do not know how to proceed.

Is this the right way to proof this? Thanks in advance for any kind of help.

答:
$\alpha(Y) \subseteq Z$ means that $\alpha(y) \in Z$ for each $y \in Y$, so, defining $\alpha|_Y : Y \to Z$ by $\alpha|_Y(y) := \alpha(y)$ for all $y \in Y$ makes sense simply because the image of an element of the domain is an element of the codomain. Note that, writting $(\exists z\in Z)(\forall y\in Y) \ z=\alpha(y)$ you are saying that every element of $Y$ goes to a same element of $Z$ under $\alpha$ (in this case, $z$) and this is not necessarily true.

An element in the quotient $V/Y$ is an expression of the form $v+Y$, where $v \in V$. So, $v+Y$ is not the sum of $v$ with $Y$, it is the representant of $v$ in $V/Y$, the class of $v$. Two classes $v_1+Y$ and $v_2+Y$ (that is, two elements of $V/Y$) are equal whenever $v_1-v_2 \in Y$. Now, you need to prove that the rule $\overline{\alpha}(v+Y) := \alpha(v)+Z$ for every $v \in V$ is independent of the representants, in other words, no matter what class of $v$ you pick, under $\overline{\alpha}$ goes to $\alpha(v)+Z$, always. Thus, you need to prove that $$v_1+Y = v_1+Y \implies \alpha(v_1)+Z=\alpha(v_2)+Z$$ i.e. $v_1-v_2 \in Y \implies \alpha(v_1)-\alpha(v_2) \in Z$. Can you see this is true? Recall that $\alpha(Y) \subseteq Z$.

Title: Function in quotient space


问:
Find $A$ such that $f(0) = A$ and there exists $f'(0)$
Let
$$f(x)=
\begin{cases}
(e^{3x}-3e^x+2)/x^2 & :x \neq 0\\
A & :x=0\\
\end{cases}$$
I'm supposed to find $A$ such that there exists $f'(0)$. My attempt:

First, I found $A$ such that $f$ is continuous at $x=0$, using de l'Hospital's rule to calculate $\lim_{x \to 0^{+}}f(x)$ and $\lim_{x \to 0^{-}}f(x)$.
$$f(x)=
\begin{cases}
(e^{3x}-3e^x+2)/x^2 & :x \neq 0\\
3 & :x=0\\
\end{cases}$$
Then I calculated one-sided limits of the difference quotient at $x=0$ using de l'Hospital's rule (but I'm not sure if I'm allowed to do so - I've read that applying this rule to the definition of derivative may lead to some kind of circular reasoning).
$$\lim_{h \to 0^{+}}\frac{f(0+h)-f(0)}{h}=\lim_{h \to 0^{-}}\frac{f(0+h)-f(0)}{h}=\lim_{h \to 0}\frac{f(0+h)-f(0)}{h}=f'(0)=4$$
Is this solution correct? Wolfram Alpha and Desmos tell me that $f'(0)=0$ for $A=3$. However, when I set $A$ to a different value (meaning the function isn't continuous at $x=0$ and hence not differentiable) I still get $f'(0)=0$, which is quite confusing.

答:
What do you mean by "same result" for$A\ne 3$?  In that case $f'(0)$ does not exist.

To get continuity, use power series for the numerator: $1+3x+\frac{9x^2}{2}-3(1+x+\frac{x^2}{2})+2=3x^2$  Therefore $\lim_{x\to 0}3$.


问:
Rotate the axis of rotation of a quaternion by another quaternion
I have two quaternions, $q_1$ and $q_2$. I want to rotate the axis of rotation of $q_1$ by $q_2$. Is there any way of doing this directly, without extracting the axis of rotation from $q_1$, rotating it and re-inserting it again?

Please note that I am not trying to perform a combined rotation by quaternion multiplication ($q' = q_2q_1$). Instead I want to rotate the axis of rotation of $q_1$.

答:
The whole thing can be done with quaternion arithmetic.

The axis of $q_1$ is $a=\frac{q_1-\overline{q_1}}{2}$, and by computing $q_2aq_2^{-1}$ you will be rotating the vector $a$ into  a new direction.


问:
Inverse transformation sampling: Logistic CDF
We have a generator $X$ that selects numbers from a uniform distribution on $(0,1)$ denoted $\text{Unif}(0,1)$.

We have to show how it can be used with the function $x \mapsto \log\left(\frac{x}{1-x}\right)$ to generate a random number from a distribution with cumulative distribution function $F(x)=\frac{e^x}{1+e^x}$.

The calculation required is to verify that $F^{-1}\left(X\right)$ has cumulative distribution function $F$.

I try it this way:

Let random variable $X \sim \text{Unif}(0,1)$.

$P\left(\log\left(\frac{X}{1-X}\right)\le x\right)=P\left(\frac{X}{1-X}\le e^x\right)=P\left(X\le \frac{e^x}{1+e^x}\right)$.

I think that this is the way but I don't know how to continue and how to justify that:

$P\left(X\le \frac{e^x}{1+e^x}\right)=\frac{e^x}{1+e^x}$.

答:
We have $X \sim \text{Unif}(0,1)$, $Y \sim F$, where $F(x) = \frac{e^x}{1+e^x}$.

We want to show that $F^{-1}(X)$ has the same distribution as $Y$. In other words, we must verify that
\begin{align*}
\mathbb{P}(F^{-1}(X) \leq y) = \mathbb{P}(Y \leq y) = F(y).
\end{align*}
By straightforward calculations we find that $F^{-1}(x) = \log\big(\frac{x}{1-x}\big)$. Thus
\begin{align*}
\mathbb{P}(F^{-1}(X) \leq y)
&=
\mathbb{P}\!\left(\log\left(\frac{X}{X-1}\right) \leq y \right) \\
&=
\mathbb{P}\left(X \leq \frac{e^y}{1+e^y}\right) \\
&= \frac{e^y}{1+e^y} = F(y),
\end{align*}
as desired. We have used that $\mathbb{P}(X \leq x) = x$ for $x\in(0,1)$. This holds since $X \sim \text{Unif}(0,1)$.


问:
Do we call this a flow? and why so.
Let $\Phi:\mathbb{R}^d\to \mathbb{R}$ be a differentiable.

Suppose we have a curve $\gamma_t$ in $\mathbb{R}^d$, i.e $\gamma : [0,T]\to \mathbb{R}^d$, which evolves according to

\begin{equation}
    \partial_t \gamma_t = -\nabla \Phi(\gamma_t).
\end{equation}

Do we call $\gamma_t$ the flow of $\Phi$? If so why?

答: This type of problem, when we have a "potential" function $\Phi : \mathbb{R}^d \to \mathbb{R}$, is sometimes called the gradient flow problem. This is because the solution $\gamma$ to the problem flows along the gradient of the potential. It is in contrast to the Newtonian problem when the second derivative is equated to the gradient, i.e. $\partial_t^2 \gamma_t = \pm \nabla \Phi(\gamma_t)$.

Title: Do we call this a flow? and why so.


问:
"Set of all formal products" - what does this mean?
List the set of all formal products of $(1+x^2+x^4)^2(1+x+x^2)^2$ with exponents summing to $4$.

What is this question asking exactly? What is a "formal product"? Does it have anything to do with formal power series? How can be different products of the same expression? What counts as a valid "formal" product of an expression? Obviously if I just simplify the expression I get one thing, not a set. The book seems to not define "formal product", only "product of formal series" which yields a simplified form that I don't think is a set of all possible products.

答:
This is not the usual meaning of the term formal product; vadim123 has given that in his answer, but it’s not what appears to be intended here.

If you multiply out $(1+x^2+x^4)^2$ without collecting terms or simplifying products, you get

$$1\cdot1+1\cdot x^2+\underline{1\cdot x^4}+x^2\cdot1+\underline{x^2\cdot x^2}+x^2\cdot x^4+\underline{x^4\cdot1}+x^4\cdot x^2+x^4\cdot x^4\;;$$

the individual terms of this expression are the formal products in question. Three of them, which I’ve underlined, have exponents summing to $4$: $1\cdot x^4$, $x^2\cdot x^2$, and $x^4\cdot 1$. Of course this is only part of the product in your problem, which has many more terms. However, you don’t have to write them all out to pick out the ones with exponents summing to $4$. Just observe that each formal product in

$$(a_1+a_2+\ldots)(b_1+b_2+\ldots)(c_1+c_2+\ldots)\ldots$$

is a product of one term from each of the factors.


答:
$f(x)= 4x+3$ if $x = 0$

(filled needed 7  more characters)

Title: Find $A$ such that $f(0) = A$ and there exists $f'(0)$

Body:
Let
$$f(x)=
\begin{cases}
(e^{3x}-3e^x+2)/x^2 & :x \neq 0\\
A & :x=0\\
\end{cases}$$
I'm supposed to find $A$ such that there exists $f'(0)$. My attempt:

First, I found $A$ such that $f$ is continuous at $x=0$, using de l'Hospital's rule to calculate $\lim_{x \to 0^{+}}f(x)$ and $\lim_{x \to 0^{-}}f(x)$.
$$f(x)=
\begin{cases}
(e^{3x}-3e^x+2)/x^2 & :x \neq 0\\
3 & :x=0\\
\end{cases}$$
Then I calculated one-sided limits of the difference quotient at $x=0$ using de l'Hospital's rule (but I'm not sure if I'm allowed to do so - I've read that applying this rule to the definition of derivative may lead to some kind of circular reasoning).
$$\lim_{h \to 0^{+}}\frac{f(0+h)-f(0)}{h}=\lim_{h \to 0^{-}}\frac{f(0+h)-f(0)}{h}=\lim_{h \to 0}\frac{f(0+h)-f(0)}{h}=f'(0)=4$$
Is this solution correct? Wolfram Alpha and Desmos tell me that $f'(0)=0$ for $A=3$. However, when I set $A$ to a different value (meaning the function isn't continuous at $x=0$ and hence not differentiable) I still get $f'(0)=0$, which is quite confusing.


问:
How to find the inverse of a permutation
I'm struggling a lot with understanding the steps to find the inverse of a permutation. I'm hoping I can get an extremely simple explanation outlining the process. Let's use an easy example,

Let $P_3 = (5, 1, 4, 2, 3)$

Find $P_3^{-1}$

I know that if we can define each element of $P$ based on its $n$-th position. That is, $P(1)=5, P(2)=1,$ and so on. So, obviously, we can define each individual inverse as $P^{-1}(5)=1,P^{-1}(1)=2,$ and so on. What would be the next step to finding the inverse set? I think my confusions occurs with what exactly the inverse of a permutation set is supposed to be.

答:
Well if $P=(5,1,4,2,3)$ then $P^{-1}=(2,4,5,3,1)$ so that $P(P^{-1}(x))=P^{-1}(P(x))=x$ for each $x\in 1,2,3,4, 5$.

Inverses are always defined in the sense that if you apply a procedure or function to a given element and then the inverse function or procedure to that result, it equates to applying the identity operation and you get the same element you started with.

Method: Begin with
$$\bigg(\begin{matrix} 1 & 2 & 3 & 4 & 5 \\ 5 & 1 & 4 & 2 & 3\end{matrix}\bigg)$$

Flip it over: $$\bigg(\begin{matrix}  5 & 1 & 4 & 2 & 3 \\1 & 2 & 3 & 4 & 5\end{matrix}\bigg)$$
and rearrange so that the top row is ordered again: $$\bigg(\begin{matrix}  1&2&3&4&5 \\2 & 4 & 5 & 3 & 1\end{matrix}\bigg)$$


答: You are right, wolfram says f'(0)=4

Body: You are right, wolfram says f'(0)=4


答:
That is a cycle permutation, to find the inverse of a cycle, just run the cycle backwards.

i.e. $P=(5,1,4,2,3)$ $\rightarrow$ $P^{-1}=(3,2,4,1,5)$.

Title: How to find the inverse of a permutation

Body:
I'm struggling a lot with understanding the steps to find the inverse of a permutation. I'm hoping I can get an extremely simple explanation outlining the process. Let's use an easy example,

Let $P_3 = (5, 1, 4, 2, 3)$

Find $P_3^{-1}$

I know that if we can define each element of $P$ based on its $n$-th position. That is, $P(1)=5, P(2)=1,$ and so on. So, obviously, we can define each individual inverse as $P^{-1}(5)=1,P^{-1}(1)=2,$ and so on. What would be the next step to finding the inverse set? I think my confusions occurs with what exactly the inverse of a permutation set is supposed to be.


问:
Sum with central trinomial coefficients
I am trying to evaluate the following sum:
\begin{equation}
 \sum_{n=0}^\infty \frac{(-ix)^n}{(n+2)!} T_n
\end{equation}

where $T_n$ represent the central trinomial coefficients , with the following convetion: $T_0 = 1, T_1 =1, T_2 = 3, T_3 = 7 ...$. I know that this sum converges, however I do not know if there is a way to evaluate it. If it is not possible (or too hard) it would be nice if I could obtain at least the asymptotic behavior when $x \to \infty$ .

答:
We can use the exponential generating function (see OEIS A002426) $$\sum_{n=0}^\infty T_n \frac{z^n}{n!}=e^z I_0(2z)$$ to obtain an integral representation for the sum. Here $I_0(x)$ is the zeroth modified Bessel function, but the substitution $z=-i x$ converts it to

$$\sum_{n=0}^\infty T_n \frac{(-i x)^n}{n!}=e^{-i x} I_0(-2i x)=e^{-i x}J_0(2x)$$
where $J_0(x)$ is the regular zeroth Bessel function. Integrating both sides twice (with base point $0$) and dividing by $x^2$ produces

\begin{align}
 \sum_{n=0}^\infty T_n \frac{(-i x)^{n}}{(n+2)!}
&=\frac{1}{x^2}\int_0^x\int_0^{x''} e^{-i x''}J_0(2x'')\,dx''\,dx'\\
&=\frac{1}{x^2}\int_0^x (x-t)e^{-i t}J_0(2t)\,dt.
\end{align}
where in the second line I've made use of the Cauchy formula for repeated integration. This is as far as I know how to go, unfortunately: Mathematica does not yield a closed form. But this may be sufficient for asymptotics or numerical exploration.

Title: Sum with central trinomial coefficients


问:
How to do numerical integration of a function with values known at a given point set (finite and discrete) over an area bounded by discrete points?
Let $D$ be the area bounded by a series of points $(x_i,y_i)_{i=1}^{N}$.(The area need not to be convex and the points are supposed to go along the boundary curve.)

Let $f$ be a function defined on $D$ but we only know its values on a given point set (finite and discrete), say $(x'_i,y'_i,f(x'_i,y'_i))_{i=1}^{N'}$.(The given data set need not to be "dense" in $D$.)

How can I do numeric integration of $f$ over $D$?

Here is what I think:

1) First we should approximate the boundary of $D$ by segments between those series of points.

2) Then we should do some interpolation on the given data set. However, interpolation in two-dimension is not always possible. Then I get stuck.

Can you please help? Thank you.

EDIT: The function value are only known on ($(x'_i,y'_i)$ in the interior of $D$.
Its values on the boundary of $D$ (where the points are $(x_i,y_i)$, without prime) are not known.

答: For point 2), you should create a triangle mesh using the interior and the boundary points. Then you can perform linear or higher order interpolation in each triangle. Mesh generation is a huge area and there are many algorithms. See for instance Chapter 14 of Computational Geometry: Algorithms and Applications. For software, see for instance Triangle.</p>


问:
Question on definition of a continuous function using example
I have a question on the definition of a continuous function.  I provide some definitions, an example and questions.

Wiki says sufficiently small changes in input of a continuous function result in small changes in output and that Cauchy defined continuity of $y = f(x)$ per above as $f(x + \alpha) - f(x)$ over all values.  Then Weierstrass defined it using $(\epsilon, \delta)$ ranges.  Finally, the function is continuous at some point $c$ of its domain if $\lim_{x \rightarrow c} f(x) = f(c)$

My question is application of these definitions to the following example.  Let $f(c)$ be the first point on the right closed interval.

It's my understanding this function is discontinuous at $c$.  Applying the Cauchy definition to the image: $c \in \mathbb{R}$ approaches from the left there is no change in the output, therefore the function is discontinuous.  Is this correct analysis?  I get confused because $f(c)$ is a change in output from the left side of the function.

Similarly applying $\lim_{x \rightarrow c} f(x) = f(c)$ one finds $f(x) \neq f(c)$ also showing discontinuity at $f(c)$.

Finally applying $(\epsilon, \delta)$ to above, $\lim_{x \rightarrow c} f(x) = f(c) \iff (0 \lt |x - c| \lt \delta) \implies |f(x) - f(c)| < \epsilon)$.  In this case I'm assuming the inequalities are false?

答:
For reference, the discontinuity is at $c \in \mathbb{R}$. Let's say that earlier $f(x)$ takes value $1$ and after $c$ it takes value $2$.

Taking them one by one:

Cauchy's definition: This is the definition that a small change in $x$ gives a small change in $f(x)$. At $c$ above, no matter how small you change $x$ around $c$, $f(x)$ is gonna jump by $2-1$ units. Meaning you can't ever achieve a smaller change than $1$ unit. No matter how small you "wiggle" $x$ around $c$, $f(x)$ will always wiggle at least $1$ unit around $f(c)$. The idea of continuity is that you should always be able to make the wiggling of $f$ as small as you'd like (by making the wiggling of $x$ small enough).

Limit definition: You seem to have gotten this one.

$\epsilon-\delta$ definition: If for instance, I take $\epsilon = 0.5$, then no matter how small I take $\delta$, for every point $x \in (c - \delta, c)$, $|f(c) - f(x)| = |2-1| = 1 > \epsilon = 0.5$.

Of course all three definitions ultimately prove discontinuity by basically the same means, which is to be expected as they are equivalent.


问:
Clarification of a logical deduction example
I would like to ask for some clarification on an assignment of logical deduction. Specifically, the question is Which of the following are logically correct deductions? and below is an example:

If an International Weapons Limitation Agreement is signed, or
the United Nations approve a disarmament plan, then shares in
the arms industry will slump. But armament shares will not slump,
so an International Weapons Limitation Agreement will not be
signed.

Presumably, the above statements can be converted accordingly:

P - International Weapons Limitation Agreement is signed

Q - the United Nations approve a disarmament plan

R - shares in the arms industry will slump

The first statement (If an International Weapons ... will slump) then becomes $(P \lor Q) \Rightarrow R$ and the second one (But armament shares ... so not be signed) $\neg R \Rightarrow \neg P$.

Assuming that the first statement is true, it can only be true if $P, Q, R$ are respectively:

$t, t, t$

$t, f, t$

$f, t, t$

$f, f, t$

$f, f, f$

In all of the above cases of values of $P, Q, R$ also the second statement $\neg R \Rightarrow \neg P$ is true.

Does this then mean that the second statement is indeed a correct logical deduction from the first one? If yes, is it because a deduction has to be true whenever an original statement is true?

答:
Your analysis is not quite right. Instead of the conclusion being $\neg R \Rightarrow \neg P$, the $\neg R$ is actually stated as fact, i.e. $\neg R$ is a second premise. And, the conclusion is simply $\neg P$

But yes, this is a deductively valid argument. To quickly see this, note that if the IWLA does get signed, then given the first premise it is clear that arnament shares will slump ... but given that the second premise tells us that the shares don't slump, it cannot be true either that the IWLA gets signed.


问:
Define a recursive function
How to Define a recursive function is-bst : EBT(N) → Bool in such a
way that is-bst(t) = true if, and only if, t is a binary search tree.</p>

Title: Define a recursive function

Body:
Here's a possible solution
$$\operatorname{isBST}(x)=\left\{\begin{array}{ll}
\mbox{true},&x\lt0\\
\mbox{false},&\operatorname{prev}(x)\ge0\land\operatorname{value}(x)\le\operatorname{value}(\operatorname{prev}(x))\\
\mbox{false},&\operatorname{next}(x)\ge0\land\operatorname{value}(x)\ge\operatorname{value}(\operatorname{next}(x))\\
\operatorname{isBST}(\operatorname{prev}(x))\land\operatorname{isBST}(\operatorname{next}(x)),&\mbox{otherwise}
\end{array}\right.$$


问:
Show that the method of moments estimator of $\theta$ is biased
Let ${X_1},...,{X_n}$  a random sample from the probability distribution with pdf $f(x) = \theta {x^{\theta  - 1}}$, $0 < x < 1$ where $\theta  > 0$

I've found the method of moments estimator for $\theta$ to be : ${\hat \theta _{mom}} = \frac{{\overline X }}{{1 - \overline X }}$

Next, I'm asked to show that ${\hat \theta _{mom}}$ is biased if $n=1$ and $\theta=1$. I'm struggling to understand the solution provided, which follows:
If $n=1$ and $\theta=1$ then $f(x)=1$ and $X_1=\overline X$

$$E[{\hat \theta _{mom}}] = E\left[ {\frac{{{X_1}}}{{{X_1} + 1}}} \right] = 1 - E\left[ {\frac{1}{{{X_1} + 1}}} \right] = 1 - \int\limits_0^1 {\frac{1}{{1 - x}}dx =  - \infty  \ne \theta }$$

I understand why a result of $- \infty $ not being equal to theta proves it is biased, however I don't understand how the expectations and the integral have been formulated.

If ${\hat \theta _{mom}} = \frac{{\overline X }}{{1 - \overline X }}$ then why is $E[{\hat \theta _{mom}}]$ not equivalent with $E\left[ {\frac{{{X_1}}}{{1 - {X_1}}}} \right]$? And how is the expectation equivalent with the integral?

I hope somebody can shed some light on what it is that I'm missing here.

答:
With a different approach it is easy to show that the result is valid in general,  $\forall n, \theta$

Observe that you are not asked to explicitly calculate $E(\hat{\theta}_{MM})$ but only to show that it is not $\theta$

Your estimator is

$$\hat{\theta}_{MM}= \frac{\overline{X}_n}{1-  \overline{X}_n  }$$

Thus, using Jensen's inequality you get

$$E(\hat{\theta}_{MM})\ne  \frac{E(\overline{X}_n)}{1-E(\overline{X}_n)  }=\frac{E(X)}{1-E(X)}=\frac{\frac{\theta}{\theta+1}}{1- \frac{\theta}{\theta+1}}  =\theta$$

and you are all set!

On the other hand, as sample mean is consistent for $E(X)=\frac{\theta}{1+\theta}$ and $g(z)=\frac{z}{1-z}$ is continuous in $z$, then $\hat{\theta}_{MM}$ is consistent for $\theta$


问:
Compute Casimir operator for $su(n)$ representations via the Weyl formula
While learning general representation theory of Lie algebra, I am having hard times to compute myself the value of the quadratic Casimir operator for some basic examples.

To fix the notation, according to Ref. [1], the quadratic Casimir operator is defined by

$$ C = g^{\mu\nu} T_\mu T_\nu
\tag{1}\label{def}
$$

where $T_\mu$ are the generators of the algebra, satisfying
$$ [T_\mu, T_\nu] = C^\rho_{\mu\nu}T_\rho $$

and the metric $g_{\mu\nu}$ is defined as

$$ g_{\mu\nu}\equiv C^\alpha_{\mu\beta}C^\beta_{\nu\alpha}$$.

$g^{\mu\nu}$ is the inverse of $g_{\mu\nu}$, i.e., $g^{\mu\nu}g_{\nu\rho}=\delta^{\mu}_\rho$.

Ref. [1] reports the following Weyl's formula for the eigenvalues $C(\Lambda)$ of the representation of $C$

$$C(\Lambda) = \langle \Lambda |\Lambda + \delta\rangle
\tag{2}\label{weyl}
$$

where $\Lambda$ is the maximum weight of the representation and $\delta$

$$\delta\equiv \frac{1}{2} \sum_{\alpha>0} \alpha$$

is half of the sum of the positive roots.

(In other words, $C$ is represented by $C(\Lambda)I$, $I$ the identity matrix.)

Now let me take as a basic example the defining representation of $su(4)$. With the "physicists" convention, the algebra is the one of hermitian traceless matrix, and is spanned by 3 commuting elements generating the Cartan subalgebra

$$(T_\alpha)_{ij}=\delta_{ij}(\delta_{\alpha,i}-\delta_{\alpha+1,i})\qquad \alpha=1,2,3
\tag{3}\label{cartan}
$$

and the remaining $4^2-1-3=12$ generators can be chosen to be matrices like these

$$\begin{pmatrix}0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{pmatrix},\qquad
\begin{pmatrix}0 & 0 & i & 0\\ 0 & 0 & 0 & 0\\ -i & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{pmatrix},\quad \text{etc.}
$$

Computing by brute force the sum of Eq. (\ref{def}) I get $C=15/32$ for the defining representation.

On the other hand, the maximum weight of the defining representation is $\Lambda=(1,0,0)$ according to Ref. [1] and also to a direct evaluation from Eq. (\ref{cartan}). Also, the positive roots are $(1,-1,0)$, $(1,0,-1)$, $(0,1,-1)$ from which I get $\delta=(1,0,-1)$. This also agrees with a formula reported in Ref. [1]

$$
\delta_\rho = n/2-\rho,\qquad su(n)
\tag{4}\label{delta}
$$

With Eq. (\ref{delta}), and using Eq. (\ref{weyl}) one gets $C(\Lambda)=2$, different than the explicit computation.

More generally, for the defining representation of $su(n)$, which has $\Lambda=(1,0,\ldots,0)$ [1], I would get $C(\Lambda)=\frac{n^2-1}{2n^2}$ by summing explicitly Eq. (\ref{def}), whereas apparently with Eq. (\ref{weyl}) I would get $C(\Lambda)=n/2$.

I also noticed that Ref.[2] cites a slightly different Weyl formula, perhaps a different normalization:

$$
C(\Lambda) = \langle \Lambda + \delta|\Lambda + \delta\rangle-\langle \delta| \delta\rangle = \langle \Lambda |\Lambda + 2\delta\rangle
\tag{5}\label{weyl2}
$$

What is the source of the discrepancy between $C$ as obtained with an explicit summation, and the (apparent) result of Eq. (\ref{weyl}) or Eq. (\ref{weyl2})?

What is the correct way of using Eq. (\ref{weyl}) (or perhaps Eq. (\ref{weyl2})) to compute $C(\Lambda)$, for a generic representation with maximum weight $\Lambda$?

[1]
J.D. Vergados, "Group and Representation Theory", World Scientific

[2]
B. C. Hall, "Lie Groups, Lie Algebras and Representations", Springer

答:
$\DeclareMathOperator{\Tr}{Tr}\DeclareMathOperator{\ad}{ad}$
Before doing the calculations for $\mathfrak{su}(4)$, here are the corresponding ones for $\mathfrak{su}(2)$ first.

Let $x = -\frac{i}{2} \sigma_1$, $y = -\frac{i}{2} \sigma_2$ and $z = -\frac{i}{2} \sigma_3$, where the $\sigma_i$ are the Pauli matrices. Besides using Pauli matrices, I am using mathematicians' notations, so that elements in $\mathfrak{su}(2)$ are skew-hermitian, rather than hermitian as in physics texts. We then have

$$ [x,y] = z \text{ and cyclic}.$$

Let us denote the Killing form by $B(-,-)$, so that

$$ B(x_1,x_2) = \Tr(\ad_{x_1} \circ \ad_{x_2}), $$

where $x_1, x_2 \in \mathfrak{su}(2)$. We then have $B(x,x) = B(y,y) = B(z,z) = -2$ and $x$, $y$ and $z$ are $B$-orthogonal.

Let $\Lambda$ be the highest weight for the fundamental representation of $\mathfrak{su}(2)$. In that representation, we then have

$$ x^2 = y^2 = z^2 = -\frac{1}{4} I.$$

So the Casimir element is therefore

$$ C = -\frac{1}{2}\left( - \frac{1}{4}I - \frac{1}{4}I - \frac{1}{4}I\right) = \frac{3}{8}I.$$

Let $e_1$ and $e_2$ denote the following diagonal matrices

$$ e_1 = \operatorname{diag}(i,0) \text{  and  } e_2 = \operatorname{diag}(0,i).$$

Let $h = e_1 - e_2$, which spans a Cartan subalgebra of $\mathfrak{su}(2)$. Note that $B(h,h) = -8$, by a straight-forward calculation, so that $B = -4 g_{Euc}$, where $g_{Euc}$ is the Euclidean inner product induced on the real span of $h$ from the Euclidean inner product on the real span of $e_1$ and $e_2$, so that for instance $g_{Euc}(e_1-e_2,e_1-e_2) = 2$.

If $D$ is the real span of $e_1$ and $e_2$, so that $D$ consists of pure imaginary diagonal $2$ by $2$ matrices, let $D^*$ be its dual. Moreover, let $e^1, e^2$ be the dual basis of $e_1$, $e_2$, so that we have

$$ (e_i, e^j) = \delta_i^j$$

where $(-,-)$ here denotes the natural pairing between $D$ and its dual space $D^*$, and $\delta_i^j$ is the Kronecker delta.

We then have $\delta = \Lambda = \frac{1}{2}(e^1-e^2)$. We now have to be careful and use $B^{-1}$ on the real span of $e^1-e^2$, rather than $B$, as we are now looking at elements in the dual of the span of $h$, i.e. in the dual of the chosen Cartan subalgebra. Most authors just write $B$, rather than $B^{-1}$, as it is implicitly understood that one should use the inverse inner product, but I will write it as $B^{-1}$ to stress this point. Moreover, it is customary to introduce a minus sign and consider minus $B^{-1}$ rather than $B^{-1}$, in order to get a positive-definite inner product, rather than a negative-definite one.

Then

$$-B^{-1}(\Lambda+2\delta, \Lambda) = \frac{1}{4}g^{-1}_{Euc}(\frac{3}{2}(e^1-e^2),\frac{1}{2}(e^1-e^2)) = \frac{1}{4} \times \frac{3}{4} \times 2 = \frac{3}{8}.$$

Note that if we had used instead $\Lambda + \delta$ rather than $\Lambda + 2 \delta$, we would have gotten the incorrect value $1/4$, so the correct formula is the one which has $\Lambda + 2 \delta$.

(Note that I did not go over the references themselves, so there is a chance that their notation may be a little different).

Edit 1: I then did the computation for $\mathfrak{su}(4)$ and I obtained that, for the fundamental representation,

$$C = \frac{15}{32}I$$

from the definition of $C$. On the other hand,
$$\delta = \frac{3}{2}e^1 +\frac{1}{2}e^2 -\frac{1}{2}e^3 -\frac{3}{2}e^4$$
and
$$\Lambda = \frac{1}{4}\left( 3e^1-e^2-e^3-e^4 \right)$$
so that
$$\Lambda + 2 \delta = \frac{1}{4}\left( 15e^1 + 3e^2 - 5e^3 - 13e^4 \right).$$

Moreover, the Killing form $B$ is
$$B = -8 g_{Euc},$$
so that
$$-B^{-1}(\Lambda, \Lambda+2\delta) = \frac{15}{32}$$
which is the correct answer indeed.

Remark: the calculation of the Killing form and its inverse were kind of long, but luckily, there were many zeros. It would take a long time to type everything, but let me know if something is not clear.

Edit 1: the positive roots of $\mathfrak{su}(4)$ with respect to our choice of Cartan subalgebra and simple roots, is $e^i - e^j$ for $1 \leq i<j \leq 4$.

If we let $T_i = e_i-e_{i+1}$, for $i = 1, \ldots, 3$, then the $T_i$ form a basis of the chosen Cartan subalgebra of $\mathfrak{su}(4)$. If we adjoin to them $T_4 = (1/4)(e_1+e_2+e_3+e_4)$, then the $T_i$, for $1 \leq i \leq 4$, form a basis for the diagonal matrices in $\mathfrak{u}(4)$, and their dual basis, is given by

$$T^1 = \frac{1}{4}(3e^1-e^2-e^3-e^4)$$
$$T^2 = \frac{1}{2}(e^1+e^2-e^3-e^4)$$
$$T^3 = \frac{1}{4}(e^1+e^2+e^3-3e^4)$$ and
$$T^4 = \frac{1}{2}(e^1+e^2+e^3+e^4).$$

Note that $T^i$ for $i = 1, \ldots, 3$ form a basis dual to the $T_i$ for $i = 1, \ldots, 3$. One can then check, that with respect to that basis, one has

$$e^1 - e^2 = (2,-1,0)$$
$$e^2 - e^3 = (-1,2,-1)$$
$$e^3 - e^4 = (0,-1,2)$$ and so on.

One can thus check that $\delta = (1,1,1)$ and $\Lambda = (1,0,0)$. Therefore $\Lambda + 2\delta = (3,2,2)$.

Moreover, with respect to the $T_i$ basis ($1 \leq i \leq 3$), one has
$$ g_{Euc} = \left( \begin{array}{ccc} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2 \end{array} \right).$$
Inverting it gives
$$ g_{Euc}^{-1} = \frac{1}{4} \left( \begin{array}{ccc} 3 & 2 & 1 \\ 2 & 4 & 2 \\ 1 & 2 & 3 \end{array} \right).$$

We then deduce that
$$\frac{1}{8} g_{Euc}^{-1}(\Lambda+2\delta, \Lambda) = \frac{15}{32},$$
as required.

Edit 2: a reference I personally find very helpful, though it is perhaps best as a second book on the topic, is Knapp's "Lie groups beyond an introduction". In the appendices at the end of the book, if I remember well, you have various tables for various complex semisimple Lie algebras, their root systems, their $\delta$ etc. I am personally mostly self-taught on this beautiful topic, and I find that Knapp's book is great for looking up various notions and theorems regarding the structure and representation theory of Lie groups/Lie algebras (especially the complex semisimple ones).


问:
does a matrix like this exist?
Question:

Does a matrix $A \in M_{3 \times 3}(F)$ exist s.t. $A^4=
\begin{bmatrix} 0&0&1\\0&0&0\\0&0&0\end{bmatrix}$

What I thought:

I think it doesn't. How do you start a proof of such a thing (prefer hints at first).
Thx

答:
Hints:

Suppose such a matrix exists, then

(1) $\;A^8=0\;$ ;

(2) $\,A\,$ is nilpotent and of order $\,3\times 3\implies A^3=0\;$ ,  contradiction.

Title: does a matrix like this exist?


问:
Solve $ T'' + 2aT' - c^2 \lambda T = 0, T'(0) = 0 $
I need to solve the following PDE:
$$
u_{tt} + 2au_t - c^2u_{xx} = 0
$$
$$
u(0,t) = u(\pi,t) = 0
$$
$$
u(x,0) = x(\pi-x)
$$
$$
u_{t}(x,0) = 0
$$

Now, i define:
$$
u(x,t) = X(x)T(t)
$$
getting:
$$
\frac{T''+2aT'}{T} = c^2 \frac{X''}{X}
$$

Therefore:

$$
\frac{X''}{X} = \lambda
$$

$$
\frac{T''+2aT'}{T} = c^2 \lambda
$$

So now i want to solve:

$$
T'' + 2aT' - c^2 \lambda T = 0
$$
$$
T'(0) = 0
$$

So:

$$
r^2 + 2ar - c^2 \lambda  = 0
$$

Therefore i get:
$$
r_{1,2} = -a \pm i\sqrt{c^2n^2 - a^2}
$$

Therefore we get:
$$
T_n = e^{-at}(A \cos \sqrt{c^2n^2-a^2}t + B \sin \sqrt{c^2n^2 - a^2}t)
$$

The problem now is that i only have 1 condition: $T'(0) = 0$ but 2 unknowns $A,B$, how can i solve this?

答:
Since the problem has gone this long unsolved there must be issues. The first is that we have not been guaranteed that $a<c$, although I think it's safe to assume that $a>0$ because that would correspond to a guitar string clamped at both ends, plucked and released at $t=0$ with viscous damping. So for the modes with $a>cn$ we have
$$T_n(t)=e^{-at}\left(A\cosh\sqrt{a^2-c^2n^2}t+B\sinh\sqrt{a^2-c^2n^2}t\right)$$
For such modes,
$$T_n^{\prime}(0)=-aA+\sqrt{a^2-c^2n^2}B=0$$
So we can say that
$$T_n(t)=e^{-at}\left(\cosh\sqrt{a^2-c^2n^2}t+\frac a{\sqrt{a^2-c^2n^2}}\sinh\sqrt{a^2-c^2n^2}t\right)$$
If there is a critically damped mode with $a=cn$ it is
$$T_n(t)=e^{-at}(A+Bt)$$
Where
$$T_n^{\prime}(0)=-aA+B=0$$
So that
$$T_n(t)=e^{-at}(1+at)$$
For the underdamped modes $a<cn$
$$T_n(t)=e^{-at}\left(A\cos\sqrt{c^2n^2-a^2}t+B\sin\sqrt{c^2n^2-a^2}t\right)$$
And in this case
$$T_n^{\prime}(0)=-aA+\sqrt{c^2n^2-a^2}B=0$$
With solution
$$T_n(t)=e^{-at}\left(\cos\sqrt{c^2n^2-a^2}t+\frac a{\sqrt{c^2n^2-a^2}}\sin\sqrt{c^2n^2-a^2}t\right)$$
So our solution thus far is
$$\begin{align}u(x,t)&=\sum_{0<na/c}b_ne^{-at}\left(\cos\sqrt{c^2n^2-a^2}t+\frac a{\sqrt{c^2n^2-a^2}}\sin\sqrt{c^2n^2-a^2}t\right)\sin nx\end{align}$$
And now comes the second issue: @EditPiAf seems to have been naughty and changed the initial condition from $u(x,0)=x(\pi-x)$ to $u(x,0)=u(\pi-x)$, which doesn't even make any sense. Thus we claim that the initial condition reads
$$u(x,0)=x(\pi-x)=\sum_{n=1}^{\infty}b_n\sin nx$$
And if we integrate
$$\begin{align}\int_0^{\pi}x(\pi-x)\sin mx\,dx&=\left[-\frac{x(\pi-x)}n\cos nx+\frac{\pi-2x}{n^2}\sin nx-\frac2{n^3}\cos nx\right]_0^{\pi}\\
&=\frac{2\left(1-(-1)^n\right)}{n^3}=\sum_{n=1}^{\infty}b_n\int_0^{\pi}\sin nx\sin mx\,dx\\
&=\sum_{n=1}^{\infty}b_n\frac{\pi}2\delta_{nm}=\frac{\pi}2b_m\end{align}$$
We have our final solution
$$\begin{align}u(x,t)&=\frac4{\pi}e^{-at}\left\{\sum_{0<na/c}\frac{1-(-1)^n}{n^3}\left(\cos\sqrt{c^2n^2-a^2}t+\frac a{\sqrt{c^2n^2-a^2}}\sin\sqrt{c^2n^2-a^2}t\right)\sin nx\right\}\end{align}$$

Body:
Since the problem has gone this long unsolved there must be issues. The first is that we have not been guaranteed that $a<c$, although I think it's safe to assume that $a>0$ because that would correspond to a guitar string clamped at both ends, plucked and released at $t=0$ with viscous damping. So for the modes with $a>cn$ we have
$$T_n(t)=e^{-at}\left(A\cosh\sqrt{a^2-c^2n^2}t+B\sinh\sqrt{a^2-c^2n^2}t\right)$$
For such modes,
$$T_n^{\prime}(0)=-aA+\sqrt{a^2-c^2n^2}B=0$$
So we can say that
$$T_n(t)=e^{-at}\left(\cosh\sqrt{a^2-c^2n^2}t+\frac a{\sqrt{a^2-c^2n^2}}\sinh\sqrt{a^2-c^2n^2}t\right)$$
If there is a critically damped mode with $a=cn$ it is
$$T_n(t)=e^{-at}(A+Bt)$$
Where
$$T_n^{\prime}(0)=-aA+B=0$$
So that
$$T_n(t)=e^{-at}(1+at)$$
For the underdamped modes $a<cn$
$$T_n(t)=e^{-at}\left(A\cos\sqrt{c^2n^2-a^2}t+B\sin\sqrt{c^2n^2-a^2}t\right)$$
And in this case
$$T_n^{\prime}(0)=-aA+\sqrt{c^2n^2-a^2}B=0$$
With solution
$$T_n(t)=e^{-at}\left(\cos\sqrt{c^2n^2-a^2}t+\frac a{\sqrt{c^2n^2-a^2}}\sin\sqrt{c^2n^2-a^2}t\right)$$
So our solution thus far is
$$\begin{align}u(x,t)&=\sum_{0<n<a/c}b_ne^{-at}\left(\cosh\sqrt{a^2-c^2n^2}t+\frac a{\sqrt{a^2-c^2n^2}}\sinh\sqrt{a^2-c^2n^2}t\right)\sin nx\\
&\quad+\sum_{n=a/c}b_ne^{-at}(1+at)\sin nx\\
&\quad+\sum_{n>a/c}b_ne^{-at}\left(\cos\sqrt{c^2n^2-a^2}t+\frac a{\sqrt{c^2n^2-a^2}}\sin\sqrt{c^2n^2-a^2}t\right)\sin nx\end{align}$$
And now comes the second issue: @EditPiAf seems to have been naughty and changed the initial condition from $u(x,0)=x(\pi-x)$ to $u(x,0)=u(\pi-x)$, which doesn't even make any sense. Thus we claim that the initial condition reads
$$u(x,0)=x(\pi-x)=\sum_{n=1}^{\infty}b_n\sin nx$$
And if we integrate
$$\begin{align}\int_0^{\pi}x(\pi-x)\sin mx\,dx&=\left[-\frac{x(\pi-x)}n\cos nx+\frac{\pi-2x}{n^2}\sin nx-\frac2{n^3}\cos nx\right]_0^{\pi}\\
&=\frac{2\left(1-(-1)^n\right)}{n^3}=\sum_{n=1}^{\infty}b_n\int_0^{\pi}\sin nx\sin mx\,dx\\
&=\sum_{n=1}^{\infty}b_n\frac{\pi}2\delta_{nm}=\frac{\pi}2b_m\end{align}$$
We have our final solution
$$\begin{align}u(x,t)&=\frac4{\pi}e^{-at}\left\{\sum_{0<n<a/c}\frac{1-(-1)^n}{n^3}\left(\cosh\sqrt{a^2-c^2n^2}t+\frac a{\sqrt{a^2-c^2n^2}}\sinh\sqrt{a^2-c^2n^2}t\right)\sin nx\right.\\
&\quad+\sum_{n=a/c}\frac{1-(-1)^n}{n^3}(1+at)\sin nx\\
&\quad+\left.\sum_{n>a/c}\frac{1-(-1)^n}{n^3}\left(\cos\sqrt{c^2n^2-a^2}t+\frac a{\sqrt{c^2n^2-a^2}}\sin\sqrt{c^2n^2-a^2}t\right)\sin nx\right\}\end{align}$$


问:
Reference request regarding Mackey topology
I would like to ask for some request on the relationship between Mackey topology and the weak topology. Recently, I have read quite some statements where a property in weak topology (such as compactness) implies that in Mackey topology. The basic connection I have now is that the topology is identitical if we are talking about an equicontinuous subset. However, I have a feeling that there must be something more. The reference I got is schaefer, Topological Vector Space and Bourbaki, Topological Vector Space. However, both of the books are quite brief on this matter.

Thanks in advance.

答:
The basic reference is Topological Vector Space, the second is Bourbaki's classic. Other books on Topological Spaces also can be mentioned. However, I want to mention another direction which might interest you.

The Banach Lattice Theory (or more general form), makes it very easy to find Mackey topology. The most important space $L$-space and $M$-space can be identify with $L_1$ and $C(X)$, therefore, it is extremely easy get from weak to Mackey thanks to Banach–Alaoglu theorem
, the basic reference for lattice theory is enter link description here. However, it doesn't reveal the full power. For the full power, see this notorious hard-to-read book, from chatper 1 to chapter 10, as well as the appendix.


问:
Rule for indices of expected values
Is it forever and always the case that for a given distribution, to use the E[X] formula as well as LOTUS, that the formula is only valid if you set the indices to the domain of the original distribution? I realize I might be using the wrong language, but as an example to illustrate:

Let's say I'm trying to figure out something like $\sum_{n=1}^{\infty}5\cdot\frac{e^{-2}2^n}{n!}$

Clearly the right hand side of the equation represents $Pois(2)$

In order for me to solve this using E[N], would it be necessary to set the beginning index to n=0?

Likewise is this true for E[X] for all probability distributions?

答:
In R, the first 101 terms of the expression for $E(X)$ are as follows:

```<code>lam = 2; i = 0:100;  pdf=exp(-lam)*lam^i/factorial(i)
</code>```

The vector pdf [of probabilities $f(i)\,]$ can also be computed using the R function dpoisfor a Poisson PDF.

```<code>PDF = dpois(i, lam)
</code>```

Here is a table of the first six values:

```<code>lam = 2; i = 0:100;  pdf=exp(-lam)*lam^i/factorial(i)
PDF = dpois(i, lam)
head(cbind(i, pdf, PDF))
     i        pdf        PDF
[1,] 0 0.13533528 0.13533528
[2,] 1 0.27067057 0.27067057
[3,] 2 0.27067057 0.27067057
[4,] 3 0.18044704 0.18044704
[5,] 4 0.09022352 0.09022352
[6,] 5 0.03608941 0.03608941
</code>```

The exact computation of $E(X) = \sum_{i=0}^\infty if(i) = \lambda = 2,$
where $f(i)$ is the PDF. But we can get very close by summing the first 101 terms:

```<code>sum(i*pdf)
[1] 2
</code>```

Even the first 10 terms would be good enough for most practical purposes.

```<code>sum((0:9)*pdf[1:10])
[1] 1.999525
</code>```

You can leave off the first term because it is multiplied by $0.$  That is, $\sum_{i=0}^\infty if(i) = 0\cdot f(0)+ \sum_{i=1}^\infty if(i) = \sum_{i=1}^\infty if(i).$

However, $$P(X \le 5) = P(X=0)+P(X-1)+\cdots+P(X=5)\\ = \sum_{i=0}^{5} f(i) = 0.9834364$$ and you must add all six terms--including the first--to get the right answer.

In R, you can uxe:

```<code>sum(dpois(0:5, 2))
[1] 0.9834364
ppois(5, 2)    $ 'ppois' is a Poisson CDF
[1] 0.9834364
</code>```

Title: Rule for indices of expected values

Body:
In R, the first 101 terms of the expression for $E(X)$ are as follows:

lam = 2; i = 0:100;  pdf=exp(-lam)*lam^i/factorial(i)

The vector pdf [of probabilities $f(i)\,]$ can also be computed using the R function dpoisfor a Poisson PDF.

PDF = dpois(i, lam)

Here is a table of the first six values:

lam = 2; i = 0:100;  pdf=exp(-lam)*lam^i/factorial(i)
PDF = dpois(i, lam)
head(cbind(i, pdf, PDF))
     i        pdf        PDF
[1,] 0 0.13533528 0.13533528
[2,] 1 0.27067057 0.27067057
[3,] 2 0.27067057 0.27067057
[4,] 3 0.18044704 0.18044704
[5,] 4 0.09022352 0.09022352
[6,] 5 0.03608941 0.03608941

The exact computation of $E(X) = \sum_{i=0}^\infty if(i) = \lambda = 2,$
where $f(i)$ is the PDF. But we can get very close by summing the first 101 terms:

sum(i*pdf)
[1] 2

Even the first 10 terms would be good enough for most practical purposes.

sum((0:9)*pdf[1:10])
[1] 1.999525

You can leave off the first term because it is multiplied by $0.$  That is, $\sum_{i=0}^\infty if(i) = 0\cdot f(0)+ \sum_{i=1}^\infty if(i) = \sum_{i=1}^\infty if(i).$

However, $$P(X \le 5) = P(X=0)+P(X-1)+\cdots+P(X=5)\\ = \sum_{i=0}^{5} f(i) = 0.9834364$$ and you must add all six terms--including the first--to get the right answer.

In R, you can uxe:

sum(dpois(0:5, 2))
[1] 0.9834364
ppois(5, 2)    $ 'ppois' is a Poisson CDF
[1] 0.9834364


问:
real mechanism behind addition, subtraction, multiplication, division
we all know the basic rules for operations of addition, subtraction, multiplication and division.

but what i don't know is why these rules (of addition, subtraction, multiplication and division) works.

as if we have been given algorithm to do these operations but not explained how they were derived.

if anyone can explain please expain .

答:
$$
\begin{array}{cccccc}
& & 4 & 3 \\
& \times & 7 & 9 \\
\hline & 3 & 8 & 7 \\
3 & 0  & 1 \\
\hline 3 & 3 & 9 & 7
\end{array}
$$

Why is this algorithm, taught in elementary school, justified?

It's because multiplication distributes over addition:
$$
9\cdot(43) = 9\cdot(4\cdot10 \quad+\quad 3) = (9\cdot4)(10)\quad+\quad(9\cdot3)
$$
Hence
$$
\begin{array}{ccc}
& 2 & 7 \\
3 & 6 \\
\hline
3 & 8 & 7
\end{array}
$$

Similarly one multiplies $43$ by $7$, getting $301$, but the $301$ is moved one place to the left because it is $70$ rather than $7$ by which $43$ was to be multiplies; thus one multiplies by $10$ by moving one place to the left.

A think and explanation of long division might be more involved.  Or might not$\ldots$

Later edit: $1352/43$.  There's a quotient and a remainder: $1352 = 31\cdot 43 + 19$. The quotient is $31$ and the remainder is $19$.  So $\dfrac{1352}{43} = 31 + \dfrac{19}{43}$.  The remainder must be less than $43$ so the fraction must be less than $1$.  So if we move the decimal point over to get a one-digit quotient, we have
$$
\frac{135.2}{43} = 3.1 +\text{less than $0.1$} = 3 + \text{less than $1$}
$$
Since $43$ goes $3+\text{fraction}$ times into $135.2$, and $43\cdot3$ must be an integer, we have $43$ goes $3+\text{some fraction}$ times into $135$.  In particular,
$$
135 = 43\cdot3 + 6.\quad \text{The quotient is $3$ and the remainder is $6$.}
$$
Now look at the usual algorithm:

$$
\begin{array}{cccccccccc}
& & & & 3 \\
\hline
43 & ) & 1 & 3 & 5 & 2 \\
   &   & 1 & 2 & 9 \\
\hline & & & & 6
\end{array}
$$
Why should we now bring down the $2$ and then divide the resulting $62$ by $43$?

$$
\frac{135.2}{43} = \frac{135}{43} + \frac{0.2}{43} = 3 + \underbrace{\frac{6}{43} + \frac{0.2}{43}} = 3 + \frac{6.2}{43}.
$$

So
$$
\frac{1352}{43} = 30 + \frac{62}{43}.
$$
So that's why.


问:
On open subsets of the subspace topology of a set and of the subspace topology of its closure
Let $(X, ||.||)$ be a an infinite dimensional normed space (with the norm $||.||$), and let $A$ be a bounded subset of $X$. Let $\tau$ be a locally convex topology on $X$ such that $||.||\geq \tau$, that is, the norm topology is stronger than the topology $\tau$.

Now, consider the topological spaces $(\overline{A}^{||.||}, \tau)$ and $(A, \tau)$. Do these spaces have the same open sets ? Any answer/comment will be appreciated.

EDIT: After the comments I reformulated the question as follows:

Let $(X, ||.||)$ and $A$ be as above. Let $\tau$ be a locally convex topology on $X$ such that $||.||> \tau$, that is, the norm topology is strictly stronger than the topology $\tau$. Consider the topological spaces $(\overline{A}^{||.||}, \tau)$ and $(A, \tau)$. Is it true that each open subset of $(A, \tau)$ is also open in $(\overline{A}^{||.||}, \tau)$? If not, what kind of conditions should be imposed on the set $A$? For example, if the set $A$ is, additionally, convex, is the answer positive?

答:
I cannot yet give you the full answer as it is too long. However, I would like to give you a more broad aspect since you question is linked to comparing topologies.  Two references will help you. First, you want to check the last chapter of enter link description here on $\mathfrak{G}$-topology. Second, you might want to check the this whole book on topological vector space.

I am sorry I could not type the whole answer or anything, but I think these will have things that are not only related to your question, but also other things you are working on.


问:
Minimal polynomials and Galois extensions
I'm back again with a question, but this time I am only curious about one thing. Here's how it goes:

Let $K$ be a Galois extension of a field $F$. By the theorem of the primitive element, we know that $K = F(\alpha_1)$ for some $\alpha_1 \in K.$ Suppose that $f(X)$ is the minimal polynomial of $\alpha_1$ over $F$. Now $K$ is the splitting field for $f(X)$ as $K$ is separable and normal. We also know that $F(\alpha_1, \alpha_2, \ldots, \alpha_n)$ for the distinct roots $\alpha_i$ of $f(X)$  is a splitting field for $f(X)$. This means that $F(\alpha_1, \alpha_2, \ldots, \alpha_n) = F(\alpha_1)$. Since $f(X)$ has $n = deg f(X)$ distinct roots it must be the case that all roots of $f(X)$ are linear combinations of $\alpha_1$. This also means that $F(\alpha_k) = F(\alpha_d)$ for some $k, d \leq n$.

However, I am not sure if my arguments are correct. I am also quite afraid of drawing my own conclusions as I am currently self-studying Galois theory with basically no prior knowledge of algebra. I hope that someone can correct me if I am wrong!

答:
This is all good, except that $F(\alpha_1)$ will be the set of all linear combinations of powers $\{\alpha_1^j\}_{j=0}^{n-1}$ where $n = \deg(f)$, not just linear combinations of $\alpha_1$ itself.

You can see that this set of powers is linearly independent over $F$ because if there did exist a nontrivial linear combination of these equaling zero$^\dagger$—i.e. if we had $\displaystyle \sum_{k=0}^{n-1} c_k\alpha_1^k = 0$ for some collection of not-all-zero coefficients $c_k$ from $F$—then  the polynomial $g(x) = \displaystyle \sum_{k=0}^{n-1} c_kx^k$ would be of smaller degree than $f$ despite having $\alpha_1$ as a root.  But this can't be since $f$ was the minimal polynomial of $\alpha_1$.

Moreover, this set of powers must be an exhaustive list of the generating elements for $F(\alpha_1)$ because the degree of the extension $[F(\alpha_1):F]$ is equal to the degree of the minimal polynomial of $\alpha_1$, and the degree of the extension is literally how many "basis vectors" $F(\alpha_1)$ has when viewed as a vector space over $F$.

Finally, we'd actually have $F(\alpha_1) \cong F(\alpha_j)$ for any $1 \leq j \leq n$. To see this, first recall that $F(\alpha_1) \cong F[\alpha_1]$ because $\alpha_1$ is algebraic over $F$ (see my post here). Thinking of $F(\alpha_1)$ as $F[\alpha_1]$, we can show that $F[\alpha_1] \cong F[\alpha_j]$ via different homomorphisms out of $F[x]$:

For each $j$, define a homomorphism $\phi_j:F[x] \to F[\alpha_j]$ to be the map $g(x) \mapsto g(\alpha_j)$; that is, we evaluate each polynomial at $\alpha_j$.  You can prove for yourself (for instance, using the fact that univariate polynomial rings over fields are principal ideal domains) that the kernel of each $\phi_j$ must be the principal ideal generated by the minimal polynomial of $\alpha_j$, namely $f$.  Applying the first isomorphism theorem for rings to each homomorphism, we see that $F[\alpha_j] \cong F[x] / \langle f \rangle$ for every $j$ between $1$ and $n$.  By transitivity of congruence, then, we must have $F[\alpha_1] \cong F[\alpha_j]$ for every $j$.

$^\dagger$ Just applying the fact that a set of vectors $\{ \mathbf{v}_k \}_{k=1}^n$ in a vector space $V$ over a field $F$ is linearly dependent $\iff$ there exists elements $c_k \in F$, not all zero, such that $\displaystyle \sum_{k=1}^n c_k \mathbf{v}_k = \mathbf{0}$.


问:
Proving an inequality regarding the surfaces of triangles
Let $S_{\mathcal T}$ be the surface of triangle $\mathcal T$.
Let $3$ isometrical circles meet at a certain point, and name $A$, $B$, $C$ their $3$ pairwise intersections. Let $XYZ$ be the triangle containing the $3$ circles such that each side is tangent to $2$ of them. Show that $S_{XYZ}\ge 9S_{ABC}$

After thinking about this problem for a bit, and looking at a drawing of it, the circumradius of $XYZ$ seems to be $r_1+r_2$, where $r_1$ is the common radius of the $3$ isometrical circles and $r_2$ is the inradius of the triangle formed by the centers of the circles (which I believe is also the inradius of $ABC$, since both triangles seem isometrical). If this is proven, then we can easily conclude with Euler's relation...

答:
Let us assume that the radius of each original circle is $1$, and let us denote with $D,E,F$ the centers of such circles and with $P$ their common point.

A homotethy with center $P$ and factor $\frac{1}{2}$ maps $ABC$ into the medial triangle of $DEF$, hence $[ABC]=[DEF]$. $XYZ$ and $DEF$ have parallel sides and the inradius of $XYZ$ is just one plus the inradius of $DEF$, so

$$ \frac{[XYZ]}{[ABC]}=\frac{[XYZ]}{[DEF]}=\left(\frac{r_{DEF}+1}{r_{DEF}}\right)^2 $$
and it is enough to show that $r_{DEF}\leq \frac{1}{2}$.
On the other hand the circumradius of $DEF$ is $1=PD=PE=PF$ and by Euler's inequality $r_{DEF}\leq \frac{1}{2}R_{DEF}$: we are done.

Since $DEF$ and $ABC$ are isometrical, we have also proved Johnson's theorem $R_{ABC}=1$.


问:
Conditional expectation of a smaller filtration
I am trying to solve the following problem.

Let $\mathcal{F}_t$ be the filtration generated by a Brownian motion $W$, and $\theta$ be a random variable on the same sample space $\Omega$, independent of $W$. Define an initially enlarged filtration $\mathcal{G}_t = \mathcal{F}_t \vee \sigma(\theta)$.

Now let $X$ be a $\mathcal{G}_s$-meaurable random variable. My question is, do I have
$\mathbb{E}[X\vert \mathcal{F}_t] = \mathbb{E}[X\vert \mathcal{F}_s]$? I tried to use $\pi$-$\lambda$ theorem but failed.

Any hint is greatly appreciated!

答:
I presume we have $0\leq s\leq t < \infty$.

Note that trivially,
$$
\mathcal{F}_s \, \perp \!\!\! \perp \, \mathcal{F}_t \, \mid \, \mathcal{F}_s.
$$
Since $\theta \perp \!\!\! \perp W$ and $\mathcal{F}_s \subset \mathcal{F}_t$, it also holds that
$$
\sigma(\theta) \, \perp \!\!\! \perp \, \mathcal{F}_t \, \mid \, \mathcal{F}_s.
$$
Collecting results,
$$
\mathcal{G}_s \, \perp \!\!\! \perp \, \mathcal{F}_t \, \mid \, \mathcal{F}_s.
$$
In other words, for any $A\in\mathcal{G}_s$ it holds that
$$
\mathbb{P}(A \, | \, \mathcal{F}_t) = \mathbb{P}(A \, | \, \mathcal{F}_s).
$$
By classic extension (monotone class argument),
$$
\mathbb{E}[X \, | \, \mathcal{F}_t] = \mathbb{E}[X \, | \, \mathcal{F}_s]
$$
for any $\mathcal{G}_s$-measurable random variable $X$ with finite expectation.

If you're not too familiar with conditional independence, please let me know.


问:
Integration with respect to the empirical process?
Let $X_1, X_2, \dots, X_n$ be i.i.d from distribution $F(x)$. Let $v_n (x)$ be the corresponding empirical process

$$v_n (x) = \sqrt{n} \left[ F_n (x) - F(x) \right]$$

where $F_n (x)$ is the empirical distribution function.

Then I would like to understand the integral of a deterministic function $g(x)$ with respect to $v_n (x)$. That is, the integral

$$ \int_\mathbb{R} g(y) dv_n (y)$$

Intuitively, when I first saw this integral I interpreted it as

$$ \int_\mathbb{R} g(y) dv_n (y) = \sqrt{n} \left( \int_\mathbb{R} g(y) dF_n (y) - \int_\mathbb{R} g(y) dF(y) \right)$$

Which in turn gives

$$ \frac{1}{\sqrt{n}} \sum_{i=1}^n g(X_i) - \sqrt{n} \int_\mathbb{R} g(y) dF(y)$$

Is this the correct interpretation?

答:
First, the empirical process, in general, is not what you mean. The empirical process is a process index by a function space. Check this classical book chapter 1 for an understanding.

Return to your question though, you are integrating w.r.t to a cdf function. Therefore, if you are lucky it can be well defined, it is just a Lesbesgue integration. Therefore, everything you know about Lesbesque measure applies.


问:
Show Diagonal Matrix is not possible
Consider the real matrix $\alpha = \begin{pmatrix}
7 &3  &-4 \\
 -2&-1  &2 \\
 6&2  &-3
\end{pmatrix}$. Show/explain that there does not exist a invertible real matrix $\beta$, so that $\beta ^{-1} \alpha \beta$ is a diagonal matrix

My effort so far

I know that to diagonalize a matirx $A$

Find the eigenvalues of $A$ using the characteristic polynomial.

For each eigenvalue $λ$ of $A$ , compute a basis $B λ$ for the $λ$-eigenspace.

If there are fewer than $n$ total vectors in all of the eigenspace bases $B λ$ , then the matrix is not diagonalizable. I am losing it when it comes to the specific case "Show/explain that there does not exist a invertible real matrix $\beta$, so that $\beta ^{-1} \alpha \beta$ is a diagonal matrix". How would I go about this one in question?

答: The characteristic polynomial of $\alpha$ is $(X^2+1)(X-3)$ therefore $i$ and $-i$ are eigenvalues of $\alpha$ and thus $\alpha$ is not diagonalizable in $\mathbb{R}$ (but is in $\mathbb{C}$).

Title: Show Diagonal Matrix is not possible


问:
Euler lagrange equation is a constant
I'm working through exercises which require me to find the Euler-Lagrange equation for different functionals.
I've just come across a case where the Euler Lagrange equation simplifies to
$$1=0.$$
Please could someone explain what can be concluded about the set of extremals to problems where this is the case.</p>

答:
This is breeze can no man do calculus of variations no more? Kmt

$$
L[u]=\int(f(u)u_{x}+u)dx
$$

$$
\frac{d}{dx}\left( \frac{\partial F}{\partial u_{x}}\right)-\frac{\partial F}{\partial u}=0
$$

$$
f'(u)u_x +1 -f'(u)u_x=0
$$

$$
1=0
$$

so for any $u$, the E-L equations is 1=0, thus the set of extremals is empty - hence the functional has no stationary value - consider the graph of $e^x$. Infimum, not minimum.
(Do you even math?)

Title: Euler lagrange equation is a constant