问:
Is there an efficient algorithm to determine if a linear matrix inequality has a solution?
Are there any fast algorithms to determine if a linear matrix inequality (LMI) problem $Ax \leq b$ has a solution?

I am aware that linear programming and the simplex algorithm in particular may be used to solve linear optimization problems subject to LMI constraints ($\min(c^Tx) \text{ subject to: } Ax \leq b$) but I am curious if there are fast methods of proving that an LMI is solvable without necessarily finding a solution. For example, some criterion analogous to $\det(A) \neq 0$ for linear equality equations $Ax=b$.

答:
Not meaningfully. Up until a logarithmic factor the algorithms must be the same speed.

To see why, we can append $c^T$ as a row to $A$, and $q$ to $b$. Then by solving whether $Ax \leq b$ you find whether $q$ is an achievable value for the minimization problem. Then you can try various $q$ in a binary search manner to approach the minimal value quickly.

This holds quite generally, usually. In a lot of cases in computer science the 'decision problem' (does a solution exist) is roughly as hard as the 'optimization problem' (find the optimal solution) because very often you can simply binary search for the optimal solution given a decider. This is also neat for computational complexity results however, since decision problems are easier to study but allow you to put bounds on the optimization problems we are often interested in.


问:
Limiting Poisson behaviour in an urn-ball problem
This is problem from K.R. Parthasarathy's book  "Introduction to probability and measure."

Consider the placement of r indistinguishable balls in n urns. The sample space here consists of the $\displaystyle {n+r-1 \choose r}$ n-tuples of non-negative integers that add up to r, with the ith entry of the n-tuple denoting the number of balls in urn i. Each point in the sample space is assigned an equal probability of $\displaystyle \dfrac{1}{{n+r-1 \choose r}}.$

Let $p_{n,r}(k)$ denote the probability that exactly $k$ urns are empty.

It can be shown that for each $n$ and $r$ we have
$$
p_{n,r}(k) = \sum_{j = k}^n (-1)^{j-k} { j \choose k} {n \choose j} \frac{ {n-j+r-1 \choose r}  }{{n+r-1 \choose r}}.
$$

For each positive integer $n$, we are given a positive integer $r_n$ such that $\displaystyle \frac{n^2}{r_n} \to \lambda > 0$ as $n \to \infty$.

We are asked to show $\displaystyle p_{n,r_n}(k) \to \exp(-\lambda) \frac{\lambda^k}{k!}$, and I need some help in proving this limiting behavior.

Some Observations

One can see that
$$
p_{n,r}(k) = \frac{1}{k!} \sum_{j = k}^\infty \frac{(-1)^{j-k}}{(j-k)!} q_{n,r}(j),
$$
where $$q_{n,r}(j) =
\begin{cases} \dfrac{n!}{(n-j)!}\frac{{\genfrac {(} {)} {0pt}{0} {n-j+r-1} {r} }} {{\genfrac {(} {)} {0pt} {0} {n+r-1}{r} }}, \mbox{} \text{ for } 0 \leq j \leq n-1 \\
0, \mbox{} \text{ otherwise}
\end{cases}
$$
So it is sufficient to show that $q_{n,r_n}(j) \leq M^j$ for some $M > 0$ independent of $n$ and $j$ and $q_{n,r_n}(j) \to  \lambda^{j}$ as $n \to \infty$, for then we can take limit inside the summation sign and see that $\lim_{n\to\infty} p_{n,r_n}(j) = \frac{1}{k!} \sum_{j=k}^{\infty} \frac{(-1)^{j-k}}{(j-k)!} \lim_{n \to \infty} q_{n,r_n}(j) = \frac{1}{k!} \sum_{j=k}^{\infty} \frac{(-1)^{j-k}}{(j-k)!} \lambda^j = \exp(-\lambda) \frac{\lambda^k}{k!}.$

答:
We have, for a fixed $j$,
$$
\begin{align}
q_{n,r}(j) &= \frac{n!}{(n-j)!} \frac{ n-j+r-1 \choose r}{n+r-1 \choose r}\\
&=  \left\{ n (n-1) \dots (n-j+1) \right\}\left\{\frac{ (n-j) (n-j+1) \dots (n-j+r-1) }{ (n)(n+1)\dots (n+r-1)} \right\}\\
& = \prod_{t = 0}^{j-1} (1 - \frac{t}{n}) \times n^j \times \prod_{t=0}^{r-1}(1-\frac{j}{n+t})\tag{1}
\label{e:1}
\end{align}
$$

We will show the third term in the above product satisfies $\displaystyle \prod_{t=0}^{r-1}(1-\frac{j}{n+t}) \leq  \left(\frac{n}{n+r}\right)^j.$

Proof:

From the identity $1-x \leq \exp(-x)$ we have $\displaystyle 0 \leq \prod_{t=0}^{r-1}(1-\frac{j}{n+t}) \leq \exp( -j \sum_{t=0}^{r-1}\frac{1}{n+t})$

For any $p > 0$ we have $\displaystyle \frac{1}{p} \geq \int_{p}^{p+1}\frac{1}{x}dx \geq \log \frac{p+1}{p}$.

This implies $\displaystyle \sum_{t=0}^{r-1}\frac{1}{n+t} \geq \sum_{t=0}^{r-1} \log \frac{n+t+1}{n+t} = \log \frac{ n+ r}{n}$ and hence $\displaystyle\exp(-j \sum_{t=0}^{r-1}\frac{1}{n+t}) \leq \exp(-j \log\frac{n+r}{n} ) \leq \left(\frac{n}{n+r}\right)^j$.

The above inequality implies, on subsitution in $\eqref{e:1}$ $\displaystyle 0 \leq q_{n,r}(j) \leq n^j \left(\frac{n}{n+r}\right)^j = \left( \frac{n^2}{n+r} \right)^j = \left(\frac{ \frac{n^2}{r} } {\frac{n}{r} + 1}\right)^j$.
So, $\displaystyle 0 \leq q_{n,r_n}(j) \leq \lambda_n^j $ where $\displaystyle \lambda_n = \frac{n^2/r_n}{n/r_n+1} \to \lambda$, in particular $\displaystyle q_{n,r_n}(j) \leq M^j$ for some suitable $M > 0$ and $\limsup_{n\to\infty} q_{n,r_n}(j) \leq \lambda^j \tag{2a} \label{e:2a}.$

So it is sufficient to prove $q_{n,r_j}(j) \to \lambda^j.$

We will now show that for all sufficiently large $n$, $$ \displaystyle \prod_{t=0}^{r-1}(1-\frac{j}{n+t}) \geq \left(\frac{n-1}{n+r-1}\right)^j \exp(-j^2\sum_{t=n}^{\infty}\frac{1}{t^2}) \tag{2} \label{e:2}.$$
We may assume $n$ is so large that $\frac{j}{n} \leq \frac{1}{2}$.
For any $0 \leq x \leq \frac{1}{2}$ we have $(1-x) \geq \exp(-x-x^2)$. This follows from exponentiating both sides of the following inequality: $-\log(1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \dots \leq x+\frac{x^2}{2}(1+x+x^2 + \dots + ) \leq x + \frac{x^2}{2}(1+\frac{1}{2}+\frac{1}{2^2}+\dots) = x+x^2.$
Hence, $$
\begin{align}
\displaystyle  \prod_{t=0}^{r-1}(1-\frac{j}{n+t}) &\geq \exp\left( -j \sum_{t=0}^{r-1}\frac{1}{n+t} - j^2\sum_{t=0}^{r-1}\frac{1}{(n+t)^2}\right)\\
& \geq \exp\left( -j \sum_{t=0}^{r-1}\frac{1}{n+t} - j^2\sum_{t=n}^{\infty}\frac{1}{t^2}\right). \tag{3}
\label{e:3}
\end{align}
$$.

For any $p > 1$ we have $\displaystyle \frac{1}{p} \leq \int_{p-1}^{p}\frac{1}{x}dx \leq \log \frac{p}{p-1}$. Hence $\displaystyle \sum_{t=0}^{r-1}\frac{1}{n+t} \leq \sum_{t=0}^{r-1} \log \frac{n+t}{n+t-1} \leq \log\frac{n+r-1}{n-1}$, and $\eqref{e:2} $ follows from substituting this inequality in $\eqref{e:3}$.

So from $\eqref{e:1}$ and $\eqref{e:2}$ we have, $$
\begin{align}
q_{n,r}(j) &\geq \left\{\prod_{t=0}^{j-1}(1-\frac{t}{n})\right\} n^j \left( \frac{n-1}{n+r-1}\right)^j \exp(-j^2\sum_{t=n}^{\infty} \frac{1}{t^2}) \\
&\geq \left\{\prod_{t=0}^{j-1}(1-\frac{t}{n})\right\} \left( \frac{\frac{n^2}{r} -\frac{n}{r}}{\frac{n}{r}+1-\frac{1}{r}}\right)^j \exp(-j^2\sum_{t=n}^{\infty} \frac{1}{t^2}).
\end{align}
$$

So, putting $r = r_n$ above and letting $n \to \infty$ we have,
$\liminf_{n \to \infty} q_{n,r_n}(j) \geq \lambda^j \tag{4} \label{e:4}.$

From $\eqref{e:2a}$ and $\eqref{e:4}$ we have $\lim_{n \to \infty} q_{n,r_n}(j) = \lambda^j$ and the result follows.

ContentLicense: CC BY-SA 4.0


问:
Polynomials - The sum of two roots
If the sum of two roots of
$$x^4  + 2x^3 - 8x^2 - 18x - 9 = 0$$ is
$0$, find
the roots of the equation</p>

答:
Hints:

The constant coefficient (constant term) is $-9$.

$$\pm 1, \pm 3\pm 9 \;\text{divide}\; -9 $$

If there are rational roots to the polynomial, they will be among these values. $\large (\star)$

Two of the roots  $x_i, x_j$ be must be factors of $-9$: and they must be such that $x_i = -x_j$. Lets call this pair $a, \pm a$. That means $(x−a)$ and $x-(-a) = (x + a)$ will be factors of the polynomial.

Now, what do we get when we multiply $(x+a)(x-a)$?

$(\large \star)$ Why do we know this? See the Rational Root Theorem, where you'll see why, and where you'll find additional tremendously useful things to know when you need to factor a polynomial (and you learn how to tell whether or not any rational roots exist!)

ContentLicense: CC BY-SA 3.0

Title: Polynomials - The sum of two roots


问:
Is $f(A_1\bigcup A_2) =f(A_1)\bigcup f(A_2)$ true for countably many $A_i$
I was unable to think about The proof or counterexample of this identity when I was attempting a question.

So, I am asking for help here.

Suppose $A_i$ $i \in \mathbb{N}$ are all non-empty sets , I have proved that $f(A_1 \bigcup A_2 ) = f(A_1) \bigcup f(A_2) $. using mathematical induction the result can be proved for  finite number of $A_i$ but is the result true when $i \in \mathbb{N}$?

Why or why not?

答:
Elaborating on Kavi Rama Murthy's comment, here is a full proof of the formula
$$
f\Bigl(\bigcup_{i \in I}A_i\Bigr) = \bigcup_{i \in I} f(A_i) \qquad (1)
$$
Let $L$ and $R$ be the left and right hand side of $(1)$, respectively.
If $x \in L$, then $x = f(s)$ for some $s \in \bigcup_{i \in I} A_i$. Then there exists $i \in I$ such that $s \in A_i$. It follows that $f(s) \in f(A_i)$ and hence $x = f(s) \in R$. Thus $L \subseteq R$. To prove the opposite inclusion, consider some $x \in R$. Then there exists $i \in I$ such that $x \in f(A_i)$. Consequently, $x = f(s)$ for some $s \in A_i$. It follows that $s \in \bigcup_{i \in I} A_i$ and thus $x = f(s) \in L$. Thus $R = L$.

Observe that no induction is needed for this proof.


问:
What is this curved surface constructed inside a rectangular prism?
Given is a rectangular prism with one side $\square ABCD$ and its opposite side $\square EFGH$. Point $J$ lies on $\square ABCD$ some distance $y$ between diagonal $AC$ and edge $AB$, and point $K$ lies on $\square EFGH$ $y$ between diagonal $EG$ and edge $GH$. The midpoint of line segment $JK$ is the exact center of the prism (i.e., it is also the midpoint of diagonal $AG$), but $JK$ need not be perpendicular to any side of the prism.

Draw curve $AJC$ on side $\square ABCD$ and curve $EKG$ on side $\square EFGH$, so that the curves are geometrically similar. Now you have a curved surface, $AEGC$, with opposite straight sides $AE$ and $GC$, and opposite curved sides $AJC$ and $EKG$. Stipulate also that all points on $JK$ lie on this surface, as do all points on the line segment between the midpoint of $AE$ and the midpoint of $GC$.

What is this three-dimensional, curved surface? Does it have a name and general equation? Possibly an application or analog in real life? If not, how can one find the equation?

Edit: These are images of the surface.

答:
This is my interpretation of the construction, confirmed by OP in a comment to the question:

As point $J$ traces a curve from $A$ to $C$ in plane $ABCD$, the point $K$ (defined as the reflection of $J$ in the center of the prism) traces an identical, but reoriented, curve in plane $EFGH$. The target surface appears by joining corresponding points on these curves in reverse tracing order; ie, parameterizing $J=J(t)$ for $0\leq t\leq 1$, with $K=K(t)$ the reflection of $J(t)$, the surface is the union of segments $\overline{J(t)K(1−t)}$.

Without reverse-tracing, the surface is, of course, a cone; I don't know a name for the reversed version ... "anti-cone"? :)

Now, for a prism with its center at $M$ (so that $\frac12(J(t)+K(t)) = M$), we can parameterize a point $P=P(s,t)$ on the surface by

$$P(s,t) = (1-s) J(t) + s K(1-t) = (1-s) J(t) + s(2M-J(1-t)) \tag{1}$$

So $P(0,t)$ traces the defining curve of $J$, while $P(1,t)$ traces the reflected curve. Note that there's no particular reason here to restrict the defining curve to a plane.

With $M$ conveniently at the origin, $(1)$ reduces to

$$P(s,t) = (1-s) J(t) - s J(1-t) \tag{1'}$$

Some examples ...

If the "curve" is merely a segment joining, say, $J_0$ to $J_1$ (OP's $A$ and $C$), then
$$J(t) = (1-t) J_0 + t J_1 \tag2$$ and $(1')$ becomes
$$\begin{align}
P(s,t) &= (1-s) ((1-t) J_0 + t J_1) - s (t J_0 + (1-t) J_1) \\[4pt]
&= J_0 - s(J_0+J_1)-t(J_0-J_1)
\end{align} \tag{3}$$ giving the plane through $J_0$, $J_1$, and the origin.

If $J_0 = A=(p,p,r)$ and $J_1 = C=(-p,-p,r)$ (giving our prism a square face) and the curve is a semicircle in the plane $z=r$, then
$$J(t) = (0,0,r) + (p,p,0) \cos \pi t + (p,-p,0)\sin\pi t \tag4$$
and we have the cylinder
$$P(s,t) = (1-2s)\;\left(\;p (\cos\pi t + \sin \pi t), p (\cos\pi t - \sin\pi t), r \;\right) \quad\to\quad x^2 + y^2 = 2 p^2 \tag{5}$$

More generally, we can replace $\cos\pi t$ in $(4)$ with any $f(t)$ such that $f(0) = -f(1) = 1$, and $\sin\pi t$ with any $g(t)$ such that $g(0)=g(1)=0$. How far we can get in eliminating the parameters depends upon the exact nature of $f$ and $g$.

 For, say, $$f(t) = 1-2t \qquad g(t) = 2t(1-t) \tag{6}$$ we obtain this cubic surface
$$z(x+y)^2 + 4 p ( r x - r y - p z ) = 0\tag{7}$$ Here's a rendering with $p=r=1$:

In response to a comment: One can solve $(7)$ for, say, $x$ using the Quadratic Formula to get
$$x = \frac{-2 p r - y z \pm 2 \sqrt{p^2 r^2 + 2 p r y z + p^2 z^2}}{z} \tag{7'}$$
For points restricted to the prism, we take the "$\pm$" to be "$+$".


问:
Solve geometrically: Arg(z(z+1)) = 0 mod(pi)
Solve geometrically the following equation :
$arg(z) +arg(z+1)=0 \pmod{\pi}$

My solution:
Let: $z=x+iy$

Then: $\arg(z)+\arg(z+1)=0 \pmod{\pi} \Longleftrightarrow
z(z+1) \in \mathbb{R}$

 Then: $\operatorname{Im}((x+iy) (x+1+iy))=0$

 After simplification we get $x=-1/2 $ or $ y=0$

However, that's an algebraic method and I'm required to solve it geometrically.
Any ideas?

答:
Geometrically speaking, if $z$ is in the first or fourth quadrant, then $z(z+1)$ can never lie on the $x-$axis (i.e., $z(z+1)$ can never be real). E.g. if $z$ is in the first quadrant, then $\arg(z)\in(0,\pi/2)$ and $\arg(z+1)\in(0,\pi/2)$ with $\arg(z+1)<\arg(z)$, so $\arg(z)+\arg(z+1)\in(0,\pi)$.

Notice that we can show the desired $z$ does not lie on $y-$axis (except for the origin) either with a similar way. And if $z$ lies on the $x-$axis, then $z(z+1)$ is obviously real.

Now, suppose $z$ is in the second or the third quadrant. Say $z$ lies in the second quadrant (the third quadrant case is very similar).

Case 1: Suppose $z+1$ lies also in the second quadrant. Then $\arg(z)+\arg(z+1)\in(\pi,2\pi)$, so $z(z+1)$ is not real.

Case 2: Suppose $z+1$ lies in the first quadrant. Then $\arg(z)+\arg(z+1)\in(\pi/2,3\pi/2)$. Thus, if $z(z+1)$ is real, then we must have $\arg(z)+\arg(z+1)=\pi$. This means $z$ and $z+1$ must be symmetric w.r.t. $y-$axis. Hence $Re(z)=-1/2$.


问:
Simplifying $[\{(\lnot p\lor q)\land(\lnot q\lor p)\}\to\{(q\lor\lnot p)\land p\}]\to[(p \leftrightarrow q)\lor(q \bigtriangleup p)]$

Simplify the following expression:

$$\left[\{(\lnot p \lor q)\land(\lnot q \lor p)\}\rightarrow\{(q \lor \lnot p)\land p\}\right]\rightarrow\left[(p \leftrightarrow q)\lor(q \bigtriangleup p)\right]$$

Assume:
$$p = \text{Jenny goes to the movies} \qquad q = \text{Jenny goes to the park}$$
It is known the expression given is a complex statement.

Using the information given simplify the statement and match the right answer. The answers given in my book are as follows:

Jenny either goes to the park or goes to the movies

Jenny goes to the movies if and only if she goes to the park

Jenny goes to the movies or she doesn't go to the movies

Jenny doesn't go to the park but she goes to the movies

My book defines the operator indicated as $\bigtriangleup$ as strong disjunction. This is shown in the truth table from below:

$\begin{array}{|c|c|c|}
\hline
p& q & p \bigtriangleup q \\ \hline
T & T& F\\ \hline
T& F& T\\ \hline
F& T & T\\ \hline
F& F & F\\ \hline
\end{array}$

Then it indicates these identities related with the strong disjunction:

$p \bigtriangleup q \equiv \lnot (p \leftrightarrow q)$

$p \bigtriangleup q \equiv (\lnot p \leftrightarrow q)$

$p \bigtriangleup q \equiv (p \lor q) \land \lnot (p\land q)$

$p \bigtriangleup T \equiv \lnot p$

Considering these how can this expression or statement be simplified?.

Can someone help me with this problem please?. The thing is I'm stuck at the very beginning. I'm assuming that in order to make a simplification it might be needed to use logic algebra but this complex statement it is too long.

Does it exist a way to make it short without a fuss?.

I'm not sure if the identities which should be used here might be absorption or de Morgan's or any of those.

Can someone help me with a detailed step by step on how to use these and solve this riddle? The thing here is that what it confuses me the most is how to approach the biconditional therefore can someone help me?.

答:
$$\left[\{(\lnot p \lor q)\land(\lnot q \lor p)\}\rightarrow\{(q \lor \lnot p)\land p\}\right]\rightarrow\left[(p \leftrightarrow q)\lor(q \bigtriangleup p)\right]$$

First,

$$\begin{align*}
(\neg p\lor q)\land(\neg q\lor p)&\equiv\big((\neg p\lor q)\land\neg q\big)\lor\big((\neg p\lor q)\land p\big)\\
&\equiv(\neg p\land\neg q)\lor(q\land\neg q)\lor(\neg p\land p)\lor(q\land p)\\
&\equiv(\neg p\land\neg q)\lor(q\land p)\\
&\equiv p\leftrightarrow q\,,
\end{align*}$$

and

$$(q\lor\neg p)\land p\equiv(q\land p)\lor(\neg p\land p)\equiv q\land p\,,$$

so

$$\big((\lnot p \lor q)\land(\lnot q \lor p)\big)\to\big((q \lor \lnot p)\land p\big)$$

is equivalent to $(p\leftrightarrow q)\to(p\land q)$. Now

$$\begin{align*}
(p\leftrightarrow q)\to(p\land q)&\equiv\neg(p\leftrightarrow q)\lor(p\land q)\\
&\equiv(p\land\neg q)\lor(\neg p\land q)\lor(p\land q)\\
&\equiv(p\land\neg q)\lor\big((\neg p\lor p)\land q\big)\\
&\equiv(p\land\neg q)\lor q\\
&\equiv(p\lor q)\land(\neg q\lor q)\\
&\equiv p\lor q\,,
\end{align*}$$

so we’ve now simplified the original expression to

$$(p\lor q)\to\big((p\leftrightarrow q)\lor(q\mathop{\triangle}p)\big)\,.$$

Finally,

$$(p\leftrightarrow q)\lor(q\mathop{\triangle}p)\equiv(p\leftrightarrow q)\lor\neg(p\leftrightarrow q)\equiv\top\,,$$

so the original expression simplifies to $(p\lor q)\to\top$, which further simplifies to $\top$.

In other words, it’s true no matter where Jenny goes. The only one of the four statements that is also a tautology (i.e., true no matter what Jenny does), is the third: Jenny goes to the movies or she doesn’t go to the movies.

Note that had I simplified $(p\leftrightarrow q)\lor(q\mathop{\triangle}p)$ to $\top$ first, I could, as I explained in the comments, immediately have concluded that the whole expression is a tautology: every implication of the form $\varphi\to\top$ is true no matter what the truth value of $\varphi$ is. This would have rendered unnecessary any simplification of the lefthand side of the main implication. I included that simplification as an illustration of the kinds of manipulation that might be needed in a different problem.


问:
Problems to understand the definition of the level of significance $\alpha$
I'm having problems to understand the definition of the level of significance $\alpha$. I thought I knew what $\alpha$ is but I realized I don't.

When I stated to study statistics by myself I read this introductory book and everything was fine, the definition is very clear. He says on page 290:

You’re probably wondering, how small does a p-value have to be for us
to achieve statistical significance? If we agree that a p-value of
$0.0001$ is clearly statistically significant and a p-value of $0.50$ is not, there must be some point between $0.0001$ and $0.50$ where we cross
the threshold between statistical significance and random chance. That
point, measuring when something becomes rare enough to be called
“unusual,” might vary a lot from person to person. We should agree in
advance on a reasonable cutoff point. Statisticians call this cutoff
point the significance level of a test and usually denote it with the
Greek letter $\alpha$ (alpha). For example, if $\alpha = 0.05$ we say we are doing
a $5\%$ test and will call the results statistically significant if the
p-value for the sample is smaller than $0.05$. Often, short hand
notation such as $P < 0.05$ is used to indicate that the p-value is less
than $0.05$, which means the results are significant at a $5\%$ level.

Now, I'm studying about statistical inference, a more advanced subject, and I realized there are some concepts that don't exactly have the same definition as I studied before. The level of significance is an example.

I'm reading this book and on page 352 he introduces the Neyman-Pearson lemma as a method to find the UMP test.

Example:

On the basis of a random sample of size $1$ from the p.d.f. $f(x;
 \theta)=\theta x^{\theta-1},\ 0 < x < 1\ (\theta > 1)$:

For $\theta_1>\theta_0$ , the cutoff point is calculated by:

... $C=(1−\alpha)^{\frac{1}{\theta_0}}$

For $\theta_1<\theta_0$ , we have:

... $C = \alpha^{\frac{1}{\theta_0}}$

So in this second book, the cutoff point is not necessarily $\alpha$, I'm confused.

MY ATTEMPT TO UNDERSTAND WITH THE HELP OF THE ANSWERS

The alpha is predetermined, but it doesn't mean I can't have a smaller rejection region. Then I end up having a smaller rejection region using NP lemma with the same level of significance alpha. Some introductory books let the cutoff point to be $\alpha$ by standard (why?), that's the reason of my confusion, I can shrink the rejection region keeping the value of $\alpha$. Can someone say if I'm right?

答:
First remember that you are dealing with simple hypothesis (both simple).

Applying NP Lemma you get that the critical region is

$$\frac{ \theta_0x^{\theta_0-1} }{ \theta_1x^{\theta_1-1}   }\leq k$$

$$x^{\theta_0-\theta_1}\leq k^*$$

Now it is easy to observe that

if $\theta_0>\theta_1$ the critical region can be viewed as $x\leq  c$ and thus, by definition of $\alpha$ you get

$$\int_0^c \theta_0x^{\theta_0-1}dx=\alpha\rightarrow c=\alpha^{1/\theta_0}$$

if $\theta_0<\theta_1$ with similar reasoning (that I left to you as an exercise) the critical region can be viewed as $x\geq c$ that is $c=(1-\alpha)^{1/\theta_0}$

In these formulas $\alpha$ is the given significance level. Usually $5\%$ for a significant test or $1\%$ for a high significant test

Definition of $\alpha$

$$\alpha=\mathbb{P}[\mathbf{x}\in C|H_0]$$

Where $C$ is the critical region calculated with NP Lemma

I suggest you to read Casella Berger or Mood Graybill Boes that are two reference textbook for these arguments


答:
You are misunderstanding what "cutoff" means in the context of the Neyman-Pearson lemma. It is referring to a cutoff of the likelihood ratio in the test, not as a cutoff of what $p$-values are small enough to be significant.

A $p$-value (of some results) represents the probability under the null hypothesis of getting results at least as extreme/unusual as your current results. The idea is that seeing results that are unusual under the null hypothesis should be evidence to reject the null hypothesis. To decide how small a $p$-value is small enough to be significant, you must set a significance level, say $\alpha = 0.05$; if your $p$-value is smaller than $\alpha$, this indicates that your data is unusual under the null, so you reject the null.

Comparing $p$-values against $\alpha$ guarantees that your Type I error is $\le \alpha$: if the null is true, the probability that your test is wrong is $\le \alpha$. More generally, this is how you should be thinking about significance levels: as Type I error (probability that a test rejects the null if null is true). This is important for likelihood ratio tests, where there aren't $p$-values being computed anywhere.

"Cutoff" in Neyman-Pearson

In this context, you are forming a test based on a statistic called the likelihood ratio $\Lambda(x) = \frac{L_1(\theta \mid x)}{L_0(\theta \mid x)}$. (I am following your textbook's convention of putting the null in the denominator.)

The test is of the form

reject $H_0$ if $\Lambda(x) >C$

do not reject $H_0$ if $\Lambda(x) < C$

for some cutoff $C$. (I'm ignoring the case where the likelihood ratio equals $C$ for simplicity.) The intuition is that if the data $x$ seem to support the alternative hypothesis, the numerator of the likelihood ratio would be large so we would lean toward rejecting; likewise if the data $x$ seem to support the null, the denominator would be large and we would lean toward not rejecting. This is $C$ is what the "cutoff" is in the Neyman-Pearson example. It is not a cutoff for what $p$-values are small enough to be significant.

How do we choose the cutoff?
If $C$ is large, then you reject less often (leading to small Type I error); if $C$ is small you reject more often (leading to large Type I error). If you have set a significance level, then you must set $C$ large enough to avoid a large Type I error. The example you are reading is solving for the smallest $C$ that keeps the Type I error below $\alpha$.

Title: Problems to understand the definition of the level of significance $\alpha$

Body:
I'm having problems to understand the definition of the level of significance $\alpha$. I thought I knew what $\alpha$ is but I realized I don't.

When I stated to study statistics by myself I read this introductory book and everything was fine, the definition is very clear. He says on page 290:

You’re probably wondering, how small does a p-value have to be for us
to achieve statistical significance? If we agree that a p-value of
$0.0001$ is clearly statistically significant and a p-value of $0.50$ is not, there must be some point between $0.0001$ and $0.50$ where we cross
the threshold between statistical significance and random chance. That
point, measuring when something becomes rare enough to be called
“unusual,” might vary a lot from person to person. We should agree in
advance on a reasonable cutoff point. Statisticians call this cutoff
point the significance level of a test and usually denote it with the
Greek letter $\alpha$ (alpha). For example, if $\alpha = 0.05$ we say we are doing
a $5\%$ test and will call the results statistically significant if the
p-value for the sample is smaller than $0.05$. Often, short hand
notation such as $P < 0.05$ is used to indicate that the p-value is less
than $0.05$, which means the results are significant at a $5\%$ level.

Now, I'm studying about statistical inference, a more advanced subject, and I realized there are some concepts that don't exactly have the same definition as I studied before. The level of significance is an example.

I'm reading this book and on page 352 he introduces the Neyman-Pearson lemma as a method to find the UMP test.

Example:

On the basis of a random sample of size $1$ from the p.d.f. $f(x;
 \theta)=\theta x^{\theta-1},\ 0 < x < 1\ (\theta > 1)$:

For $\theta_1>\theta_0$ , the cutoff point is calculated by:

... $C=(1−\alpha)^{\frac{1}{\theta_0}}$

For $\theta_1<\theta_0$ , we have:

... $C = \alpha^{\frac{1}{\theta_0}}$

So in this second book, the cutoff point is not necessarily $\alpha$, I'm confused.

MY ATTEMPT TO UNDERSTAND WITH THE HELP OF THE ANSWERS

The alpha is predetermined, but it doesn't mean I can't have a smaller rejection region. Then I end up having a smaller rejection region using NP lemma with the same level of significance alpha. Some introductory books let the cutoff point to be $\alpha$ by standard (why?), that's the reason of my confusion, I can shrink the rejection region keeping the value of $\alpha$. Can someone say if I'm right?


Title: Polynomials - The sum of two roots

Body:
If the sum of two roots of
$$x^4  + 2x^3 - 8x^2 - 18x - 9 = 0$$ is
$0$, find
the roots of the equation</p>

Body: Hint: $x^4+2x^3+x^2-9(x^2+2x+1)=x^2(x+1)^2-9(x+1)^2=(x^2-9)(x+1)^2$.


问:
What is the simplest way to do this combinatorics problem?
Question: Bob is about to hang his eight shirts in the wardrobe. He has 4 different styles of shirt, two identical ones of each particular style. How many different arrangements are possible if no two identical shirts are next to each other?

Here are the solutions:

In this case, we take the first case of all possible combinations: $\frac{8!}{2^4}$
But we see that we counted the possibilities with at least one pair together
so we have to subtract $\frac{7!}{2^3}\cdot \binom{4}{1}$ if we tie the two offending shirts together.

^This I don't get - why and what is being double-counted?

However, we see now that we double subtracted ones that have two pairs thus we must add the chance of two pairs so we must add $\frac{6!}{2^2}\cdot \binom{4}{2}$ and so on

Thus the final answer is equal to $864$

Is there a better way that doesn't involve BOTH adding AND subtracting?

答:
Comment: I think inclusion-exclusion may be the
"best" solution. That is a frequently used combinatorial method.

By whatever method, I'm pretty sure that the number
of allowable arrangements is $864$ out of $8!/16,$ as you say.

Simulating in R, one can use
shirts = c(1,1,2,2,3.3.4,4) and randomly permute
them with the sample procedure:

```<code>shirts = c(1,1,2,2,3,3,4,4)
sample(shirts)
[1] 2 2 4 1 3 1 4 3
sample(shirts)
[1] 2 2 1 4 3 1 3 4
sample(shirts)
[1] 3 2 3 1 4 2 4 1
sample(shirts)
[1] 2 3 1 4 4 1 2 3
</code>```

Some are legal arrangements and some are not.
If you take successive differences, then 0s will
show illegal ones.

```<code>s = sample(shirts);  s;  diff(s)
[1] 4 1 1 3 2 3 4 2
[1] -3  0  2 -1  1  1 -2        # illegal
s = sample(shirts);  s;  diff(s)
[1] 2 3 1 4 3 2 4 1
[1]  1 -2  3 -1 -1  2 -3        # legal
s = sample(shirts);  s;  diff(s)
[1] 1 2 1 4 3 2 4 3             # legal
</code>```

We can count the 0s in each iteration,
and see what proportion of iterations is 'legal' (the ones with no 0s.)

```<code>shirts = c(1,1,2,2,3,3,4,4)
set.seed(2020)
nr.adj = replicate(10^7,
                   sum(diff(sample(shirts))==0))
mean(nr.adj==0)
[1] 0.3428441
</code>```

So about 34.28% of the iterations are legal, suggesting
that there must be about 864 legal arrangements.

```<code> mean(nr.adj==0)*(factorial(8)/16)
[1] 863.9671
</code>```

Body:
Comment: I think inclusion-exclusion may be the
"best" solution. That is a frequently used combinatorial method.

By whatever method, I'm pretty sure that the number
of allowable arrangements is $864$ out of $8!/16,$ as you say.

Simulating in R, one can use
shirts = c(1,1,2,2,3.3.4,4) and randomly permute
them with the sample procedure:

shirts = c(1,1,2,2,3,3,4,4)
sample(shirts)
[1] 2 2 4 1 3 1 4 3
sample(shirts)
[1] 2 2 1 4 3 1 3 4
sample(shirts)
[1] 3 2 3 1 4 2 4 1
sample(shirts)
[1] 2 3 1 4 4 1 2 3

Some are legal arrangements and some are not.
If you take successive differences, then 0s will
show illegal ones.

s = sample(shirts);  s;  diff(s)
[1] 4 1 1 3 2 3 4 2
[1] -3  0  2 -1  1  1 -2        # illegal
s = sample(shirts);  s;  diff(s)
[1] 2 3 1 4 3 2 4 1
[1]  1 -2  3 -1 -1  2 -3        # legal
s = sample(shirts);  s;  diff(s)
[1] 1 2 1 4 3 2 4 3             # legal

We can count the 0s in each iteration,
and see what proportion of iterations is 'legal' (the ones with no 0s.)

shirts = c(1,1,2,2,3,3,4,4)
set.seed(2020)
nr.adj = replicate(10^7,
                   sum(diff(sample(shirts))==0))
mean(nr.adj==0)
[1] 0.3428441

So about 34.28% of the iterations are legal, suggesting
that there must be about 864 legal arrangements.

 mean(nr.adj==0)*(factorial(8)/16)
[1] 863.9671


问:
How can I solve C(n,2)+2×C(n,3)+3×C(n,4)+...+(n-1)×C(n,n)=?
'm able to find an answer but I don't think it's the right one. I tested it for n=2 and it worked but not for n=3.Here's my answer:

$$C(n,2)+2×C(n,3)+3×C(n,4)+...+(n-1)×C(n,n)=[C(n,2)+C(n,3)+...+C(n,n)] + [0×C(n,2)+1×C(n,3)+2×C(n,4)+...+
(n-2)×C(n,n)] =
2^{(n-2)} + 2^{(n-2)}×(n-2)$$

I'm either forgetting something or it's just fully wrong. Sorry for the bad formatting and thank you in advance

答:
You could take a combinatorial approach. As usual, let $[n]=\{1,2,\ldots,n\}$.

$(k-1)\binom{n}k$ is the number of ways to choose a $k$-element subset of $[n]$ and then mark any one of its elements except the largest. Summing from $k=2$ to $k=n$ gives you the number of ways to choose any subset of $[n]$ with at least $2$ elements and then mark any one of its elements except the largest; that’s the sum that you’re trying to evaluate.

Alternatively, you could carry out the same task by the following procedure.

Mark any $k\in[n]$; you can do this in $n$ ways.

Pick any subset $A$ of the remaining $[n]\setminus\{k\}$ to get a set $S=A\cup\{k\}$ with one marked element; you can do this in $2^{n-1}$ ways, so there are $n2^{n-1}$ ways to form a subset $S$ of $[n]$ with one marked element.

If $k=\max S$, discard $S$, since its marked element is its largest element. By construction $S$ could be any non-empty subset of $[n]$; there are $2^n-1$ such subsets, so we’re discarding $2^n-1$ improperly marked sets $S$.

In the end we have $n2^{n-1}-\left(2^n-1\right)=(n-2)2^{n-1}+1$ suitably marked subsets of $[n]$, so

$$\sum_{k=2}^n(k-1)\binom{n}k=(n-2)2^{n-1}+1\,.$$

For $n=4$, for instance, we have

$$\binom42+2\binom43+3\binom44=6+8+3=17=2\cdot2^3+1\,.$$


问:
Calculating expected bonus- what's wrong with my solution?
An auto insurance company is implementing a new bonus system. In each month, if a policyholder does not have an accident, he or she will receive a $5 cashback bonus from the insurer. Among the 1000 policyholders, 400 are classified as low-risk drivers and 600 are classified as high-risk drivers. In each month, the probability of zero accidents for high-risk drivers is 0.8 and that for low-risk drivers is 0.9.

Calculate the expected bonus payment from the insurer to the 1000 policyholders in one year.

The question is what's wrong with my (following) solution?

Probability of zero accidents $= (0.8\times 0.6) + (0.90 \times 0.4) = 0.84$

Then, the expected payment is given by $$\sum_{x=1}^{12} (0.84^x) \times 1000 \times 5x$$

答:
OK here's the full solution:

Each risky customer is Binomial($0.8$), so the mean number of accident-free months is $\mathbf{E}X=np = 9.6$. Bonus is a function of this rv, $^Y=cX$, so $\mathbf{E}Y=9.6 \times 5 = 48$. Expectation is linear, so even if $Y$ are not independent, expected bonus for all risky customers, $S_1=\sum_{k=1}^{600}Y_k$ is:
$$
\mathbf{E}S_1 = \sum_{k=1}^{400}\mathbf{E}Y_k = 600 \times 48=28800
$$
Now, repeat this for non-risky customers,
$$
\mathbf{E}S_2 = 400 \times 0.9 \times 5 \times 12 = 21600
$$
In total,
$$
\mathbf{E}S = 28800+21600 = 50400
$$


问:
Identity map with change of basis
Let a map with the field $K$, from $f:K^3 \rightarrow K^2$ be defined as $f\left (\begin{bmatrix}
x_1\\
x_2\\
x_3
\end{bmatrix}  \right )=\begin{pmatrix}
x_1+2x_2+3x_3\\
3x_1+2x_2+x_3
\end{pmatrix}$

and the basis $\begin{pmatrix}
-1\\
0\\
2
\end{pmatrix}, \begin{pmatrix}
3\\
1\\
-1
\end{pmatrix}, \begin{pmatrix}
1\\
-1\\
1
\end{pmatrix}$ for $K^3$ and the standard basis $(e_1,e_2,e_3)$ for the codomain

Question

Find the matrix that represents $Id:K^3 \rightarrow K^3$ with respect to the given basis for the domain and the standard basis $(e_1,e_2,e_3)$ codomain.

I have found the matrix that represents $f:K^3 \rightarrow K^2$ with respect to the given standardbasis for the domain and the codomain but this identity map is really difficult for me to get.

答:
A general recipe for this:

Suppose that $V$ and $W$ are finite-dimensional vector spaces over a field $k$, and write $n = \dim V$ and $m = \dim W$. Also, consider ordered basis $(v_1,\dots,v_n)$ for $V$ and $(w_1,\dots,w_m)$ for $W$. Now, if $T : V \to W$ is a linear map, the matrix that represents $T$ with respect to $(v_1,\dots,v_n)$ and $(w_1,\dots,w_m)$ is the $m \times n$ matrix with entries in $k$ obtained as follows

First compute the $n$ vectors $Tv_1,\dots,Tv_n$.

For each $j \in \{1,\dots,n\}$ write $Tv_j$ as a linear combination of $w_1,\dots,w_m$, that is, find $a_{1j},\dots,a_{mj} \in k$ such that $Tv_j = a_{1j}w_1+\cdots+a_{mj}w_m$.

Now, for each $j \in \{1,\dots,n\}$ we have that the $j$-th column of the desired matrix is given by $$\begin{pmatrix} a_{1j} \\ \vdots \\ a_{mj} \end{pmatrix}.$$

Title: Identity map with change of basis


答: Your first line computes that the average number of accident free drivers in a month is $840$.  Those people receive $840\cdot 5=4200$ for the month.  Each month is the same, so the annual total is $12\cdot 4200=50400$

Title: Calculating expected bonus- what's wrong with my solution?

Body:
An auto insurance company is implementing a new bonus system. In each month, if a policyholder does not have an accident, he or she will receive a $5 cashback bonus from the insurer. Among the 1000 policyholders, 400 are classified as low-risk drivers and 600 are classified as high-risk drivers. In each month, the probability of zero accidents for high-risk drivers is 0.8 and that for low-risk drivers is 0.9.

Calculate the expected bonus payment from the insurer to the 1000 policyholders in one year.

The question is what's wrong with my (following) solution?

Probability of zero accidents $= (0.8\times 0.6) + (0.90 \times 0.4) = 0.84$

Then, the expected payment is given by $$\sum_{x=1}^{12} (0.84^x) \times 1000 \times 5x$$


问:
If $f(5 + x) = f(5 - x) \forall x \in \mathbb R$ and $f(x) = 0$ has four distinct real roots, then what's the sum of roots?
I have been able to find the 2 roots by the following way :-

```<code>Let a be one root of f(x). Then
f(a) = 0
=>f(5-5+a) = 0
=>f[5-(5-a)] = 0
=>f[5+(5-a)] = 0
=>f(10-a) = 0 i.e. (10-a) and (a) are 2 roots of f(x).
</code>```

How can I find the other 2 roots? Please help...

答: You can do the same thing starting with $b$ as the third root and find $10-b$ for the fourth.  Then the problem asks for the sum of the roots.


答:
Generally, if there's two roots whose sum is zero, then it means that two factors are $x-a$ and $x+a$, which means that $x^2-a^2$ must be a factor. So clearly
$$
(x^2-a^2)(x^2+bx+c)=x^4+2x^3-8x^2-18x-9=0
$$
Find the values of $a^2$, $b$, and $c$ that satisfy the left equality, and you'll have found factors that you can then solve for all roots. (This works even if there's no rational root)</p>

Title: Polynomials - The sum of two roots


问:
How to evaluate the integral: $\int_0^\infty\frac{(-1)^{\lfloor x\rfloor}\sin(2\beta x)}{x}\mathrm dx$?
I'm trying to calculate the following integral:
$$ \int_0^\infty\frac{(-1)^{\lfloor x\rfloor}\sin(2\beta x)}{x}\mathrm dx, $$
where $\lfloor x\rfloor$ is the floor function. My first step was:
$$\sum_{n=1}^\infty (-1)^n\int_n^{n+1}\frac{\sin(2\beta x)}{x}\mathrm dx $$
Since $\displaystyle \int_n^{n+1}\frac{\sin(2\beta x)}{x}\mathrm dx =\text{Si}(2\beta(n+1))-\text{Si}(2\beta n)$, where $\text{Si}(z)$ is the Sine Integral. Then we have the sum:
$$\sum_{n=1}^\infty (-1)^n\left[\text{Si}(2\beta(n+1))-\text{Si}(2\beta n) \right] $$
Here is my problem, I have no idea how to calculate this sum. Perhaps the sum does not fit in terms of Sine Integral, I tried to take the sum into the integral before obtaining Sine Integral, but I was not successful. I will be grateful for help.

Body:
$$I(\beta)=\sum_{n=1}^\infty (-1)^n\Big[\text{Si}(2\beta(n+1))-\text{Si}(2\beta n) \Big]$$
$$I'(\beta)=\sum_{n=1}^\infty (-1)^n\Big[2 (n+1) \text{sinc}(2 \beta  (n+1))-2 n \text{sinc}(2 \beta  n)\Big]$$
Given by a CAS
$$I'(\beta)=\frac{\tan (\beta )-\sin (3 \beta ) \sec (\beta )}{2 \beta }$$
$$I(\beta)=\int\frac{\tan (\beta )-\sin (3 \beta ) \sec (\beta )}{2 \beta }\, d\beta$$ As far as I can see, there is no closed form for the antiderivative.

If $\beta < \frac \pi 2$, we could make a series expansion around $\beta=0$
$$\frac{\tan (\beta )-\sin (3 \beta ) \sec (\beta )}{2 \beta }=\sum_{n=0}^\infty (-1)^n 2^{2n+1}\frac{E_{2 n+1}(-1)+1}{(2n+1)!}\, \beta^{2n}$$
$$I(\beta)=\sum_{n=0}^\infty (-1)^n 2^{2n+1}\frac{E_{2 n+1}(-1)+1}{(2n+1)^2\,(2n)!}\, \beta^{2n+1}$$

Limiting the summation to $1000$ terms, some results
$$\left(
\begin{array}{ccc}
\beta & \text{summation} & \text{integration} \\
 0.1 & -0.0994447100 & -0.0994447100 \\
 0.2 & -0.1955639426 & -0.1955639426 \\
 0.3 & -0.2850622742 & -0.2850622742 \\
 0.4 & -0.3646983033 & -0.3646983033 \\
 0.5 & -0.4312954393 & -0.4312954393 \\
 0.6 & -0.4817299506 & -0.4817299506 \\
 0.7 & -0.5128815984 & -0.5128815984 \\
 0.8 & -0.5215216327 & -0.5215216325 \\
 0.9 & -0.5040901387 & -0.5040901367 \\
 1.0 & -0.4562617463 & -0.4562617462 \\
 1.1 & -0.3720624328 & -0.3720624322 \\
 1.2 & -0.2418977415 & -0.2418977414 \\
 1.3 & -0.0474073540 & -0.0474073533
\end{array}
\right)$$

Edit

I suppose that you will not need too much terms since it is an alternating series. Checking for $\beta=1$ and writing
$$I(1)=\sum_{n=0}^p (-1)^n 2^{2n+1}\frac{E_{2 n+1}(-1)+1}{(2n+1)^2\,(2n)!}+\sum_{n=p+1}^\infty (-1)^n 2^{2n+1}\frac{E_{2 n+1}(-1)+1}{(2n+1)^2\,(2n)!}$$
$$R_p=\Bigg|\frac{2^{2 p+3} (E_{2 p+3}(-1)+1)}{(2 p+3)^2 \Gamma (2 p+3)}\Bigg|$$

Computed for $10 \leq p \leq 100$, a quick and dirty regression
$$\log(R_p)=a+b\, p$$ gives (with $R^2=0.999994$)
$$\begin{array}{clclclclc}
 \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\
 a & -4.54337 & 0.03703 & \{-4.61696,-4.46978\} \\
 b & -0.92421 & 0.00061 & \{-0.92541,-0.92300\} \\
\end{array}$$


问:
How to code a model by a real number (an infinite sequence of bits)?
The proof of Lemma $2.10$ in the paper “Feedback computability on Cantor space” mentions “any fixed standard way of coding a model by a real.”

What does it mean? What can be an example of how to code a model by a real?

答:
For simplicity, consider the case of countably infinite directed graphs, where "directed graph" just means "$\{E\}$-structure" for a binary relation symbol $E$.

To each $X\subseteq\mathbb{N}^2$ we can associate a countably infinite directed graph $\mathcal{A}_X$ as follows: the vertices of $\mathcal{A}_X$ are the natural numbers, and $\mathcal{A}_X$ has an edge between $m$ and $n$ iff $(m,n)\in X$. Every countably infinite directed graph is isomorphic to some $\mathcal{A}_X$. Now if we fix a bijection $b: \mathbb{N}\rightarrow\mathbb{N}^2$ we get a map from sets of naturals to countably infinite directed graphs which is surjective up to isomorphism. It's in this sense that we can code countably infinite directed graphs by reals: each real determines a countably infinite directed graph, and no countably infinite directed graph is unrepresented.

More generally, this same idea works for representing countable(-or-finite) structures in arbitrary countable languages: to each countable language $\Sigma$ we can whip up a construction assigning to each real $r$ a countable $\Sigma$-structure $Struc(r)$ such that every countable $\Sigma$-structure is isomorphic to some $Struc(r)$.


答:
It can help if you think more geometrically about it. The condition that $\color{blue}{f(5+x)=f(5-x)}$ for all real $x$ tells you that the graph of the function is symmetric about the vertical line $x=5$. [Compare this with the definition of even functions: $f(x)=f(-x)$, or equivalently, $f(0+x)=f(0-x)$, whose graphs are symmetric about the $y$-axis, i.e. about the vertical line $x=0$.] This means that, unless $x=5$ is a root, all other roots come in pairs, lying symmetrically around $x=5$.

We're given that there are four roots, so $x=5$ indeed is not a root — otherwise, we would have an odd number of roots, since this one doesn't have a pair. Now, think about any one pair of roots: what is their sum? Since they are symmetric around $x=5$, they lie at the same distance from it, say, $r$ units away. So they are the numbers like $x_1=5+r$ and $x_2=5-r$, and their sum is …

Title: If $f(5 + x) = f(5 - x) \forall x \in \mathbb R$ and $f(x) = 0$ has four distinct real roots, then what's the sum of roots?

Body:
I have been able to find the 2 roots by the following way :-

```<code>Let a be one root of f(x). Then
f(a) = 0
=>f(5-5+a) = 0
=>f[5-(5-a)] = 0
=>f[5+(5-a)] = 0
=>f(10-a) = 0 i.e. (10-a) and (a) are 2 roots of f(x).
</code>```

How can I find the other 2 roots? Please help...


问:
NLA Qualifying Exam Sample Question
I'm curious if anybody has some advice on a problem.

Suppose $A$ is a real symmetric matrix with eigenvalues more or less uniformly distributed over $\left[2,18\right]$ together with an outlier at $\lambda=50$. How many steps of the conjugate gradient iteration must be taken to be sure of reducing the initial error $||{e_{0}}||_{A}$ by a factor of $20^{20}$?

By theorem $38.5$, if all the eigenvalues are in $\left[2,18\right]$, then the A-norms of the error satisfy,$$\frac{||{e_{n}}||_A}{||{e_{0}}||_A}\leq2\left(\frac{\sqrt{\kappa}-1}{\sqrt{\kappa}+1}\right)^{n}.$$
Here $\kappa=\frac{18}{2}=9$ and so, $$\frac{||{e_{n}}||_A}{||{e_{0}}||_A}\leq2\left(\frac{\sqrt{\kappa}-1}{\sqrt{\kappa}+1}\right)^{n}=2\left(\frac{\sqrt{9}-1}{\sqrt{9}+1}\right)^{n}=2\left(\frac{1}{2}\right)^{n}=\left(\frac{1}{2}\right)^{n-1}.$$

Setting $2^{-20}=2^{2-n}$ we find $n=22$ $~~~\blacksquare$

I asked my prof for input and he responded:

You need to take into account the outlier, as you did for Problem 3.

My solution for #3:

$A$ is a diagonalizable matrix with one eigenvalue being $-1$, and others residing in the unit disk centered at $2$ in the complex plane. Prove that the solution to $Ax=b$ through GMRES algorithm has error $$||{e_{n}}||\leq K 2^{-n}$$ for some constant K.

Theorem $35.2$ states,

at step $n$ of the GMRES iteration, the residual $r_{n}$ satisfies, $$\frac{||{r_{n}}||}{||{b}||}\leq\inf_{p_{n}\in P_{n}}||{p_{n}(A)}|| \leq\kappa(V) \inf_{p_{n}\in P_{n}}||{p_{n}}||_{\Lambda(A)}\leq\kappa(v)\inf_{p\in P_{n}} \max_{\lambda\in\Lambda(A)}|{p(\lambda)}|.$$ where $\Lambda(A)$ is the set of eigenvalues of $A$, $V$ is a nonsingular matrix of eigenvectors, and $||{p_{n}}||_{\Lambda(A)}$ is defined by $||{p}||_{S}=\sup_{x\in S}|{p(z)}|$.

We can choose a polynomial $$\hat{p}(z)=\left(1+z\right)q(z)=\left( 1+z \right) \left( 1-\frac{z}{2} \right)^{n-1}$$ so that $$\max_{1\leq\lambda\leq 3} |{q(\lambda)}|\leq\left( 1-\frac{1}{2} \right)^{n-1}$$ or equivalently $$\max_{|{\lambda-2}|\leq 1}|{q(\lambda)}|\leq 2^{1-n}.$$

Thus we have $$||{r_n}|| \leq \kappa(V)||{b}||\max_{\lambda\in\Lambda(A)}|{p(\lambda)}| = \kappa(V)||{b}||\max_{|{\lambda-2}|\leq 1}|{\lambda +1}| |{q(\lambda)}| \leq \kappa(V)||{b}||\max_{|{\lambda-2}|\leq 1} |{\lambda +1}| |{q(\lambda)}|$$
$$\leq \kappa(V)||{b}|| |{3 + 1}|2^{1-n}=\kappa(V)||{b}||2^{3-n}$$

Finally, $$||{e_{n}}||=||{A^{-1}r_{n}}||\leq\Big(8~\kappa(V)||{b}|| ||{A^{-1}}||\Big)2^{n},$$ and as $\Big(8~\kappa(V)||{b}|| ||{A^{-1}}||\Big)$ is constant we have our proof. $~~~\blacksquare$

I'm not sure I see what he's trying to say. Based on his response and the two problems, any thoughts on how he's expecting the first problem to be solved?

答:
The error after $n$ iterations of CGD satisfies $$\dfrac{\|e_n\|_A}{\|e_0\|_A} \le \inf_{\substack{\deg p = n \\ p(0) = 1}}\max_{\lambda \in \Lambda(A)}|p(\lambda)| \le \inf_{\substack{\deg p = n \\ p(0) = 1}}\max_{\lambda \in [2,18] \cup \{50\}}|p(\lambda)|,$$ where I've used the fact that $\Lambda(A) \subseteq [2,18] \cup \{50\}$. Now, we just need to find a polynomial $p(\lambda)$ of small degree such that $p(0) = 1$ and $|p(\lambda)| \le 2^{-20}$ over $\lambda \in [2,18] \cup \{50\}$.

Note that the Chebyshev polynomial of degree $n$ $$T_n(x) = \dfrac{1}{2}\left((x+\sqrt{x^2-1})^n+(x-\sqrt{x^2-1})^n\right)$$ satisfies $|T_n(x)| \le 1$ for $x \in [-1,1]$. Hence, the polynomial $$p_n(\lambda) = \dfrac{T_{n-1}\left(\tfrac{20-2\lambda}{16}\right)}{T_{n-1}\left(\tfrac{20}{16}\right)}\left(1-\dfrac{\lambda}{50}\right)$$ has degree $n$, satisfies $p_n(0) = 1$, $p_n(50) = 0$, and $$|p_n(\lambda)| = \dfrac{\left|T_{n-1}\left(\tfrac{20-2\lambda}{16}\right)\right|}{\left|T_{n-1}\left(\tfrac{20}{16}\right)\right|}\left|1-\dfrac{\lambda}{50}\right| \le \dfrac{\tfrac{48}{50}}{T_{n-1}\left(\tfrac{20}{16}\right)} = \dfrac{\tfrac{48}{50}}{\tfrac{1}{2}\left(2^{n-1}+2^{-(n-1)}\right)} \le \dfrac{96}{25}\cdot 2^{-n}$$ for all $\lambda \in [2,18]$. Hence, $$\dfrac{\|e_n\|_A}{\|e_0\|_A} \le \inf_{\substack{\deg p = n \\ p(0) = 1}}\max_{\lambda \in [2,18] \cup \{50\}}|p(\lambda)| \le \dfrac{96}{25}\cdot 2^{-n}$$ for all integers $n \ge 1$. It's not hard to check that $n = 22$ yields $\dfrac{\|e_n\|_A}{\|e_0\|_A} \le \dfrac{96}{25}\cdot 2^{-22} < 2^{-20}$.

Note that this error bound is roughly a factor of $2$ worse than $2 \cdot 2^{-n}$, which is what we would have had if $\Lambda(A) \subseteq [2,18]$ instead of $\Lambda(A) \subseteq [2,18] \cup \{50\}$. As such, $A$ having one outlier eigenvalue of $50$ increases the number of CGD iterations by one.

Title: NLA Qualifying Exam Sample Question


问:
If $\mathcal B_1\subseteq\tau_2$ is an analytic basis for $\tau_1$, does it follow $\tau_1\subseteq\tau_2$?
Let $E$ be a set, $\tau_i$ be a topology on $E$ and $\mathcal B_1\subseteq\tau_2$ be an analytic basis for $\tau_1$.

Are we able to conclude $\tau_1\subseteq\tau_2$?

Intuitively this should be trivially true, but how can we prove it? Please bear with me if this is rather simple, but I'm still trying to get familiar with general topology.

答: Yes.  This is trivial (he said after leaving the room for half an hour).   $\tau_2$ is closed under unions (because it's a topology), so since everything in $\tau_1$ is a union of elements of $\tau_2$, we have $\tau_1\subseteq\tau_2$.


问:
Joint normal distributions
Let
$ (X_1, X_2)^T $ be MVN $ (\mu, \sum )$, where $\mu = \begin{pmatrix} -1 \\ 1 \\\end{pmatrix}$ and $ \sum = \begin{pmatrix} 2 & -1 \\ -1 & 3 \\\end{pmatrix}$.

Let $ Y_1 = 2X_1 + X_2  $ and $ Y_2 = -X_1 + 2X_2. $

I then need to find the distribution of $ Y=(Y_1, Y_2)^T.$

So, I found that $ X_1 = \sqrt2 Z_1 -1 $ and $ X_2 = \sqrt3 Z_2 +1. $
Where $Z_1$ and $Z_2$ are standard normal.
I rewrote $ Y=(Y_1, Y_2)^T$ in terms of $Z_1$ and $Z_2$ via what I have about Y in terms of $X=(X_1,X_2)^T$.
I obtained:
$ \begin{pmatrix} Y_1 \\ Y_2 \\\end{pmatrix} = \begin{pmatrix} 2\sqrt2 &\sqrt3 \\ -\sqrt2 & 2\sqrt3 \\\end{pmatrix} \begin{pmatrix} Z_1 \\ Z_2 \\\end{pmatrix} + \begin{pmatrix} -1 \\ 3 \\\end{pmatrix}$.

Since $Y_1$ and $Y_2$ is expressed in terms of $Z_1$ and $Z_2$, I directly calculated the joint density of $Y_1$ and $Y_2$.

I obtained that $ \begin{pmatrix} Y_1 \\ Y_2 \\\end{pmatrix} $ ~ $(\begin{pmatrix} -1 \\ 3 \\\end{pmatrix}, \begin{pmatrix} 11 &2 \\ 2 & 14 \\\end{pmatrix})$,

where $\begin{pmatrix} 11 &2 \\ 2 & 14 \\\end{pmatrix}$ = $\begin{pmatrix} 2\sqrt2 &\sqrt3 \\ -\sqrt2 & 2\sqrt3 \\\end{pmatrix} \begin{pmatrix} 2\sqrt2 &\sqrt3 \\ -\sqrt2 & 2\sqrt3 \\\end{pmatrix}^T $.

However, my variance-covariance matrix is different to that of the answer. The answer was

$ \begin{pmatrix} Y_1 \\ Y_2 \\\end{pmatrix} $ ~ $(\begin{pmatrix} -1 \\ 3 \\\end{pmatrix}, \begin{pmatrix} 7 & -1 \\ -1 & 18 \\\end{pmatrix})$.

I don't see where I went wrong either. Any feedback would be appreciated.

答:
Your error is in asserting that $X_1 = \sqrt2 Z_1 -1$ and $X_2 = \sqrt3 Z_2 +1$. If this were true then $X_1$ and $X_2$ would be independent (assuming $Z_1$ and $Z_2$ are independent), but we know they are correlated.

Rather than express $X:=(X_1,X_2)^T$ in terms of $Z_1$ and $Z_2$, express $Y$ directly in terms of $X$. Use the following fact:

If $B$ is a $2\times 2$ matrix of constants, and $X:=(X_1,X_2)^T$ is a random vector, then the expectation of $BX$ is $BE(X)$ and the covariance matrix of $BX$ is
$$\operatorname{Cov}(BX)=B\operatorname{Cov}(X) B^T.$$

Here $B=\begin{pmatrix}2&1\\-1&2\end{pmatrix}$, so plug in $\operatorname{Cov}(X)=\begin{pmatrix}2&-1\\-1&3\end{pmatrix}$ to obtain the mean vector and covariance matrix for $Y:=BX$.

Title: Joint normal distributions


答:
Sorry, but the question as you asked it doesn't make any sense. I'm pretty sure you mixed together two unrelated questions. I suggest you check the source (the textbook or the instructor's original assignment), and I bet these are two separate parts being two separate questions, not a single question.

For starters, if the given function is $f:K^3\to K^2$, then we can't have "the standard basis $(e_1,e_2,e_3)$ for the codomain", because the dimension of the codomain is $2$, not $3$.

My guess is that there's a separate question about finding the matrix representing the given transformation $f$ with respect to some bases in the domain $K^3$ and in the codomain $K^2$, and there's a separate question about finding the matrix representing the identity map $\operatorname{Id}:K^3\to K^3$ with respect to the given basis in the domain $K^3$ and the standard basis in the codomain $K^3$ — but the map $f$ has nothing to do with this question.

To address this latter question, the answer is already in the question. :-) Technically speaking, you need to do what other responses have already said. To find the matrix of a linear transformation $T$, such as e.g. $T=\operatorname{Id}$, you do the following: for each given basis vector $\mathbf{b}_i$ (of the "domain" basis), calculate $T(\mathbf{b}_i)=\operatorname{Id}(\mathbf{b}_i)$, and the components of the resulting vectors with respect to the codomain basis will be the corresponding columns of the desired matrix. And since in the codomain we're using the standard basis, what are we going to see as these columns?

Title: Identity map with change of basis

Body:
Let a map with the field $K$, from $f:K^3 \rightarrow K^2$ be defined as $f\left (\begin{bmatrix}
x_1\\
x_2\\
x_3
\end{bmatrix}  \right )=\begin{pmatrix}
x_1+2x_2+3x_3\\
3x_1+2x_2+x_3
\end{pmatrix}$

and the basis $\begin{pmatrix}
-1\\
0\\
2
\end{pmatrix}, \begin{pmatrix}
3\\
1\\
-1
\end{pmatrix}, \begin{pmatrix}
1\\
-1\\
1
\end{pmatrix}$ for $K^3$ and the standard basis $(e_1,e_2,e_3)$ for the codomain

Question

Find the matrix that represents $Id:K^3 \rightarrow K^3$ with respect to the given basis for the domain and the standard basis $(e_1,e_2,e_3)$ codomain.

I have found the matrix that represents $f:K^3 \rightarrow K^2$ with respect to the given standardbasis for the domain and the codomain but this identity map is really difficult for me to get.


问:
Proof that set is a submanifold
$m$ and $k$ are natural numbers satisfying $1\leq k\leq m$,$M_{k,m}$ is the set of the entire $k\times m$ matrix, and its components are real numbers,and the subset of $M_{k,m}$ is defined as follows
$$V_{k,m}= \{A\in M_{k,m} | AA^T = I_k\}$$
$I_k$ is the k-dimensional identity matrix, $A^T$ is the transposed matrix, take $M_{k,m}$ as $km$-dimesional Euclidean space $\mathbb{R}^{k\times m}$, $V_{k, m}$ is a subset of $\mathbb{R}^{k\times m}$

Problem: Prove that $V_{2, m}$ is a $(2m-3)$-dimensional $C^\infty$ submanifold of $\mathbb{R}^{2m}$

My knowledge of manifolds is not very good. I think need find out the Jacobian and prove that it is a regular value, but I don’t have any ideas and I don’t know how to start,can anyone show me how to do this?

答:
In a lot of problems using matrices like this, it's helpful to use velocity vectors of curves. For example, if $F: M_{k,m} \to M_{k, k}$ is defined by $F(A) = A A^T$, then you can realize an element $v \in T_AM_{k,m}$ of some tangent space of $M_{k,m}$ as an element of $M_{k,m}$ which is realized by a curve $\gamma(t) = A + vt$ for $t \in (-\epsilon, \epsilon)$. Then, $\gamma'(0) = v$ and by the chain rule \begin{align*} d_AF(v) = (F \circ \gamma)'(0)& = \frac{d}{dt}|_{0}(A + vt)(A^T + v^T t) \\&= \frac{d}{dt}|_{0}AA^T + (vA^T + Av^T)t + vv^Tt^2\\&= vA^T + Av^T.\end{align*}

This formula should give you a place to start with regards to proving that $F^{-1}(I_k)$ is a comprised of regular values.

Title: Proof that set is a submanifold


问:
Understanding the solution of a probability question

Automobile policies are separated into two groups: low-risk and high-risk. Actuary Rahul examines low-risk policies, continuing until a policy with a claim is found and then stopping. Actuary Toby follows the same procedure with high-risk policies. Each low-risk policy has a 10% probability of having a claim. Each high-risk policy has a 20% probability of having a claim, The claim statuses of policies are mutually independent. Calculate the probability that Actuary Rahul examines fewer policies than Actuary Toby.

I found a solution online which begins as follows:

The probability that Actuary Rahul examines exactly $n$ policies $=0.9^{n-1}\times 0.1$.

The probability that Actuary Toby examines more than $n$ policies $=0.8^{n}$.

Then, the probability of both these events occurring is $=0.9^{n-1}\times 0.1 \times 0.8^{n}$ and the desired probability is given by $\sum_{n=1}^\infty 0.9^{n-1}\times 0.1 \times 0.8^{n}$.

My question is: in the second line of the solution, why is $0.8^{n}$ not the probability that Actuary Toby examines exactly $n+1$ policies? Is this online solution that I found even correct?

答:
If Toby stops after examining exactly $n+1$ policies, this means that first $n$ policies would be non-claims, and the last (the $n+1$st policy) would be a claim. This has probability $0.8^n\times 0.2$. So $0.8^n$ is not the probability that Toby examines exactly $n+1$ policies.

The event that Toby examines more than $n$ policies is the same as the event that the first $n$ policies examined by Toby are all non-claims. This has probability $0.8^n$, as advertised.


问:
Gauss divergence theorem
I'm trying to solve the following problem using the Gauss divergence theorem. I have to calculate the Flux through a sphere. The sphere is given as $$ x^2 +y^2+z^2==4 $$ where as the z is resticted from $$ \sqrt{3} \; to \; 2$$ I determined the divergence to $$ 4z $$ I first tried to use spherical coordinates which resolve to:

$$
\int _{\sqrt{3}}^2\int _0^{\pi /6}\int _0^{2\pi }4 *r*\text{Cos}[o]*r^2*\text{Sin}[o]dpdodr = \frac{7 \pi }{4}
$$

To confirm this solution I tried to solve this problem using a regular volume integral. Which resolve to: $$ \int _{\sqrt{3}}^2\int _{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}\int _{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}}(4 *z) dxdydz = \pi $$ I guess the error lies in my changed z variable. I replaced it with $$ r*\text{Cos}[o] $$ Could you guys please take a look into it? Thanks

答:
I am taking up this neglected problem, as it left an apparent conflict unresolved, and it seems to present an opportunity to discuss volume integration in the various commonly-used coordinate systems (and the importance of setting integration limits correctly).

The problem calls for evaluating the volume integral $\iiint \ \nabla \cdot \overrightarrow{F} \  dV \  , $ with $ \ \nabla \cdot  \overrightarrow{F} = 4z \ $ , over the interior of the spherical cap $ \ \sqrt{3} \ \le \ z \ \le 2 \ $ of the sphere $ \ x^2 + y^2 + z^2 = 4 \ $ .  We'll be able to show that the set-up by the OP of the integral in spherical coordinates has used an incorrect limit.

For the sake of checking technique, we'll first just calculate the volume of the spherical cap.  One of the lessons here is that, although the region may be based on a sphere, spherical coordinates are not automatically the best choice for working out a problem.

Because the lower bound of the problem is a flat surface parallel to a coordinate plane, and the divergence function is simply proportional to a coordinate value which is oriented perpendicularly to that plane, the symmetry of the region is reduced from "radial" to "axial".   We'll see that cylindrical coordinates are the simplest choice for setting up the computation:

The volume in this case can be treated as a "stack of slices" along the z-axis having circular cross-sections, their radii given by

$$(x^2 + y^2) + z^2 = r^2 + z^2 = 4 \ \Rightarrow \ r = \sqrt{4 - z^2} \ . $$

Because there is axial symmetry of these slices, the "azimuthal angle" integration is separable.  The volume integral becomes

$$\int_{\sqrt{3}}^2 \ \int_0^{2 \pi}  \int_0^{\sqrt{4 - z^2}} \ r \ dr \ \ d\theta \ dz \ = \ \int_0^{2 \pi} d\theta \ \int_{\sqrt{3}}^2 \   \ (\frac{1}{2} r^2) \ |_0^{\sqrt{4 - z^2}}   \ \ dz   $$

$$= \ 2 \pi \ \int_{\sqrt{3}}^2 \   \ \frac{1}{2} (4 - z^2)   \ \  dz \ = \   \pi    \  (4z - \frac{1}{3}z^3) \ |_{\sqrt{3}}^2 \ =  \   \pi  \ (8  -  \frac{8}{3}  -  4\sqrt{3}  + \sqrt{3})   $$

$$=  \   \pi    \ (\frac{16}{3} \ - 3\sqrt{3} ) \ \ \text{or} \ \  \frac{16 \pi}{3} \cdot \ (1 - \frac{9\sqrt{3}}{16} ) \ . $$

The geometric formula for a spherical cap of height $ \ h \ $ from a sphere of radius $ \ R \ $ is  $ \ V = \frac{\pi}{3} h^2 \ (3R - h) \ . $  Here, we have $ \ R = 2 \ $ and $ \ h = 2 - \sqrt{3} \ ; $ hence,

$$V = \frac{\pi}{3} (2 - \sqrt{3})^2 \cdot (3 \cdot 2 - [2 - \sqrt{3}]) \ = \ \frac{\pi}{3} (7 - 4\sqrt{3}) \cdot (4 +  \sqrt{3}) \ =  \ (\frac{16}{3}  - 3\sqrt{3} ) \ \pi \ , $$

which jibes with our integration result (well, after all, the integral we just did is how we might derive the formula).

We will carry out the integration in rectangular coordinates next, which makes the determination of limits as straightforward as possible, at the cost of complicating the integrands produced along the way.  For each circular "slice" along the z-direction, we need to make "strips" along the x-direction, which we will in turn integrate along the y-direction.  The "height" of each "strip" at a particular value of $ \ x \ $ runs from $ \ -\sqrt{(4 - z^2) - y^2 } \ $ to $ \ +\sqrt{(4 - z^2) - y^2 } \ $ , and at each level of $ \ z \ $ , the range for $ \ y \ $ passes from  $ \ -\sqrt{4 - z^2 } \ $ to $ \ +\sqrt{4 - z^2 } \ $ .
So we find the limits that kornnflake shows, leading to the integration

$$ \int_{\sqrt{3}}^2 \ \int_{-\sqrt{4 - z^2}}^{+\sqrt{4 - z^2}} \ \int_{-\sqrt{(4 - z^2) - y^2 }}^{+\sqrt{(4 - z^2) - y^2 }} \  \ dx  \ dy \ dz \ = \ \int_{\sqrt{3}}^2 \ \int_{-\sqrt{4 - z^2}}^{+\sqrt{4 - z^2}} \ 2 \sqrt{(4 - z^2) - y^2  } \ \ dy \ dz  $$

$$= \ 2 \int_{\sqrt{3}}^2 \ \frac{1}{2} \cdot [ \ y \cdot \sqrt{(4 - z^2) - y^2 } \ + \ (4 - z^2) \cdot \arctan( \frac{y}{\sqrt{(4 - z^2) - y^2  }}) \ ] |_{-\sqrt{4 - z^2 }}^{+\sqrt{4 - z^2 }}  \ \ dz  $$

$$= \ \int_{\sqrt{3}}^2 \ [ \ \sqrt{4 - z^2 } \cdot \sqrt{(4 - z^2) - (4 - z^2) } \ + \ (4 - z^2) \cdot \arctan(\frac{\sqrt{4 - z^2 }}{\sqrt{(4 - z^2) - (4 - z^2) }}  ) $$

$$
- \ [ \ -\sqrt{4 - z^2 } \ ] \cdot \sqrt{(4 - z^2) - (4 - z^2) } \ - \ (4 - z^2) \cdot \arctan(\frac{-\sqrt{4 - z^2 }}{\sqrt{(4 - z^2) - (4 - z^2) }}  ) \ ]  \ \ dz  $$

$$= \  \int_{\sqrt{3}}^2 \ [ \ 0 \ + \ (4 - z^2) \cdot (\frac{\pi}{2}) \ - \ 0 \ - \  (4 - z^2) \cdot (-\frac{\pi}{2}) \  ]   \ dz  \ =  \  \int_{\sqrt{3}}^2 \   \pi \ \cdot \ (4 - z^2)   \ \  dz \ , $$

which has led us to the integral which arises in cylindrical coordinates, with the same result.

To carry out the integral in spherical coordinates, the OP has correctly found that since $ \ z = 2 \cos \theta \ $ , with $\theta$ being the "polar angle" here, then the integration must run from $ \ \theta = 0 \ $ to $ \ \theta = \frac{\pi}{6} \ $ .

However, because the spherical cap has a "flat bottom", a ray extending from the origin to the surface of the sphere has imposed upon it a polar angle-dependent length.  The integration in the radial direction extending from $ \ z = \sqrt{3} \ $ to $ \ z = 2 \ $ must be now run from $ \  r \cos \theta = \sqrt{3} \ $ to $ \ r = 2 \ , $ or $ \  r  = \frac{\sqrt{3}}{\cos \theta} \ $ to $ \ r = 2 \ . $  The volume integral for the "cap" becomes

$$\int_0^{2 \pi} \ \int_0^{\pi / 6} \ \int_{\sqrt{3} \sec \theta}^2 \ r^2   \sin \theta \ \ dr \ \ d\theta \ d\phi \ = \ \int_0^{2 \pi}  d\phi \ \int_0^{\pi / 6} \ (\frac{1}{3} r^3) |_{\sqrt{3} \sec \theta}^2 \  \sin \theta  \ \ d\theta  $$

$$= 2 \pi \ \cdot \ \frac{1}{3} \ \int_0^{\pi / 6} \ (8 \ - \ 3 \sqrt{3} \ \sec^3 \theta) \ \sin \theta  \ \ d\theta   $$

$$= \ \frac{2 \pi}{3} \ \int_0^{\pi / 6} \  (8 \sin \theta \ - \ 3 \sqrt{3} \frac{\sin \theta}{\cos^3 \theta})   \ \ d\theta \ = \ \frac{2 \pi}{3}  \cdot  \  (-8 \cos \theta \ - \ 3 \sqrt{3} \cdot \frac{1}{2 \cos^2 \theta}) |_0^{\pi / 6} $$

$$= \ \frac{2 \pi}{3} \ \cdot  \  [ \ (-8 \cos \frac{\pi}{6} \ - \ \frac{3 \sqrt{3}}{2} \sec^2 \frac{\pi}{6} ) \ - \ (-8 \cos 0 \ - \  \frac{3 \sqrt{3}}{2} \sec^2 0 ) \ ]  $$

$$= \ \frac{2 \pi}{3}  \cdot  \   (-4 \sqrt{3} \ - \ \frac{3 \sqrt{3}}{2} \cdot (\frac{2}{\sqrt{3}})^2  \ + \ 8  \ + \  \frac{3 \sqrt{3}}{2} ) $$

$$= \   \frac{2 \pi}{3}  \cdot  \   ( 8  \ - \  \frac{9 \sqrt{3}}{2} ) \ = \   \frac{16 \pi}{3}  \cdot  \  ( 1  \ - \  \frac{9 \sqrt{3}}{16} ) \ ,  $$

as before.

$$ \\ $$

So we have found that the integrations in the three coordinate systems do agree for the volume of the spherical cap, which is $ \ V \approx 0.1372 \pi \ $ .  This would be a good point to make a "sanity check" for the divergence integral:   since  $ \sqrt{3} \ \le \ z \ \le 2 \ , $ we should expect that

$$ 4 \ \cdot \ \sqrt{3} \ \cdot V \ \approx \ 0.951 \pi  \ < \ \iiint  \nabla \cdot \overrightarrow{F} \ dV \ < \ 4 \ \cdot \ 2 \ \cdot V \ \approx \ 1.097 \pi , $$

which indicates that something has gone wrong in OP's first integral (but likely not the second one).  We will now make the divergence integral calculations in each coordinate system, reprising as  briefly as possible the details.

cylindrical coordinates --

$$\int_{\sqrt{3}}^2 \ \int_0^{2 \pi}  \int_0^{\sqrt{4 - z^2}} \ 4z \ \ r \ dr \ \ d\theta \ dz \ = \ 4 \cdot 2 \pi \ \int_{\sqrt{3}}^2 \ \ z \ \cdot \    \frac{1}{2} (4 - z^2)   \ \  dz \ $$

$$= \ \ 4 \pi \ \int_{\sqrt{3}}^2 \   \   (4z - z^3) \ \ dz \ = \   4\pi    \  (2z^2 - \frac{1}{4}z^4) \ |_{\sqrt{3}}^2 \ =  \   4 \pi   \cdot  \ (8  -  \frac{16}{4}  -  2 \cdot 3  +  \frac{9}{4})   $$

$$= \  4 \pi  \ \cdot  \  \frac{1}{4} \ = \ \pi \ . $$

rectangular coordinates --

$$ \int_{\sqrt{3}}^2 \ \int_{-\sqrt{4 - z^2}}^{+\sqrt{4 - z^2}} \ \int_{-\sqrt{(4 - z^2) - y^2 }}^{+\sqrt{(4 - z^2) - y^2 }} \  \ 4z \ \ dx  \ dy \ dz \ = \ 8 \int_{\sqrt{3}}^2 \ \int_{-\sqrt{4 - z^2}}^{+\sqrt{4 - z^2}} \ z \ \sqrt{(4 - z^2) - y^2  } \ \ dy \ dz  $$

$$= \ 4 \int_{\sqrt{3}}^2 \ z \cdot [ \ y \cdot \sqrt{(4 - z^2) - y^2 } \ + \ (4 - z^2) \cdot \arctan( \frac{y}{\sqrt{(4 - z^2) - y^2  }}) \ ] |_{-\sqrt{4 - z^2 }}^{+\sqrt{4 - z^2 }}  \ \ dz  $$
$$ =  \  4 \int_{\sqrt{3}}^2 \   z \ \cdot \ \pi \ \cdot \ (4 - z^2)   \ \  dz \ = \ 4\pi \int_{\sqrt{3}}^2  \ (4z - z^3)   \ \  dz \ , $$

and continuing as for the cylindrical-coordinate integration.

spherical coordinates (using $ \ z \ = \ r \cos \theta \ $ ) --

$$\int_0^{2 \pi} \ \int_0^{\pi / 6} \ \int_{\sqrt{3} \sec \theta}^2 \ 4r \cos \theta \ \cdot \ r^2   \sin \theta \ \ dr \ \ d\theta \ d\phi  $$

$$= \ 4 \int_0^{2 \pi}  d\phi \ \int_0^{\pi / 6}  \ r^3 \  \sin \theta \cos \theta  \ \ dr \ d\theta \ = \ 4 \ \cdot 2\pi \ \int_0^{\pi / 6}  \ (\frac{1}{4} r^4) |_{\sqrt{3} \sec \theta}^2 \  \sin  \theta \cos \theta  \ \ d\theta $$

$$= \ 2 \pi \ \int_0^{\pi / 6}   (16 \ - \ 9 \sec^4 \theta) \ \cdot \  \sin  \theta \cos \theta  \ \ d\theta \ = \ 2 \pi \ \int_0^{\pi / 6}  \ (8 \sin 2\theta \ - \ 9 \frac{\sin \theta}{\cos^3 \theta})   \ \ d\theta$$

$$= \ 2 \pi \ \cdot \ (-4 \cos 2\theta \ - \ \frac{9}{2} \sec^2 \theta)  \ |_0^{\pi / 6} $$

$$= \ 2 \pi \ \cdot \ [ \ (-4 \cos \frac{\pi}{3} \ - \ \frac{9}{2} \sec^2  \frac{\pi}{3}) \ - \ (-4 \cos 0 \ - \ \frac{9}{2} \sec^2 0) \  ]   $$

$$= \ 2 \pi \ \cdot  \   (-4 \cdot \frac{1}{2} \ - \ \frac{9}{2} \cdot \frac{4}{3}  \ + \ 4 \ +  \  \frac{9}{2} \cdot 1 ) \ = \ 2 \pi  \cdot     \frac{1}{2} \ = \ \pi \ .  $$

So we are able to confirm OP's second integration result.  (It is somewhat amusing that though the expression for the volume of the spherical cap is a bit complicated, the divergence integral proves to give a very simple value.

Title: Gauss divergence theorem


问:
Does the quotient of the dyadic and ternary rationals greater than $1$ by their powers of $3$ have the trivial topology?
Consider two topological spaces:

Begin with $X_{>0}$ the positive dyadic and ternary rationals (including products e.g. $\frac n6$), with their topology as a subspace of $\Bbb R$.  Then consider the quotient space $X_{>0}/\langle3\rangle$ given by completing the equivalence relation $x\sim 3x$ or in other words:

$x\sim y\iff\exists i\in\Bbb Z:3^ix=y$

As above, but now begin with $X_{>1}$ the dyadic and ternary rationals (including products) greater than $1$ and take the quotient $X_{>1}/\langle3\rangle$ using the same equivalence relation.

Question

Do these quotient spaces have a nontrivial topology?

Attempt

$\Bbb R$ is a metric space.  Therefore the quotient topology is the same as the toplogy given by the quotient pseudometric, right?  Or are there other Hausdorff quotient topologies?

Then for case 1: its topology IS trivial because I can pick arbitrarily large $i$ such that $d(3^{-i}x,3^{-i}y)$ is arbitrarily small in the vicinity of $0$.

For case 2: There's a prima facie argument that this quotient is Hausdorff because for any pair $x,y$ there's some maximal $i_1,i_2$ such that $\lvert 3^{i_1}x-3^{i_2}y\rvert$ is a difference between two representatives greater than $1$ - guaranteeing disjoint vicinities around them.  But this isn't conclusive because correctly using the quotient metric involves finding the shortest distance between two elements in $X_{>1}/\langle3\rangle$ using any sequence of stepping stones in-between.  I'm pretty much baffled how to do that.  How do I determine a non-recursive function for the pseudometric?

Update

For the reasons given in this answer: https://mathoverflow.net/a/380398/91341 which is largely the same as this problem, I am satisfied that the quotient pseudometric is the trivial $\forall x,y:d(x,y)=0$.  I am of the view however that the quotient topology itself is NOT trivial, although I cannot show it.

答:
Let $[x],[y]\in X_{>1}/\langle3\rangle$ with $x,y\in X_{>1}$ the least
element of their respective equivalence classes. I claim that
\begin{equation}
  \label{eq:pseudometric}
d([x],[y])=|3^{-i}x-3^{-j}y|  \tag{1}
\end{equation}
where $i$ and $j$ are the unique nonnegative intergers for which
$1\leq 3^{-i}x,3^{-j}y<3$. Note that $3^{-i}x$ and $3^{-j}y$ are only
contained in $[x]$ and $[y]$ if they are ternary. However, since the
ternary rationals are dense in $\mathbb{R}$ it suffices to prove (1)
for both $x$ and $y$ ternary. Accordingly, suppose both $x$ and $y$
are both ternary and the least element of their equivalence
class. Thus, $1\leq x, y<3$ so that (1) asserts that
$d([x],[y])=|x-y|$. If $a\sim x$ and $b\sim y$ with $a\neq x$ then
$|a-b|>|x-y|$ so that $|x-y|<|x-a|+|a-b|+|b-y|$. Thus, the infimum of all
the lengths of chains from $x$ to $y$ is $|x-y|$ so that (1) holds.


问:
Why do so many half-circle perimeters contribute $0$ to the answer in contour integration?
Oftentimes, when one is calculating the integral over the real line for an irritating integrand, it is convenient to use contour integration in the complex plane. Then, we very often find ourselves integrating over the half-circle in the upper (or, less canonically, lower) half-plane. These almost always (colloquial, not Lebesgue language) go to zero. What is the intuitive reason behind this? I see no reason to expect it, but of course I understand that we usually end up with

$$\dfrac{R^n}{R^m+\mathcal{O}(R^l)},$$

where $l<m$ and $n<m$, and sending $R\to\infty$ is why it goes to $0$, but what intuitive way of thinking can lead us there? I'm looking for something along the lines of "of course the upper half-circle contributes zero because (short quip)", similar to how I tell my students that the dependence of the implicit derivative on both $x$ and $y$ for a circle makes sense, since circles are not functions and we need both values to determine which point we are talking about.

I am aware of this question:

why does the circle at infinity not contribute the integral?

but it dealt more with the specific problem than the general intuition, I feel.

This question also exists, and is helpful:

Intuitive reason for why many complex integrals vanish when the path is "blown-up"?

But I still think such intuition must exist somewhere, and found the answers there helpful, but not satisfying.

答:
The reason the answer by Ullrich says the phenomenon you are asking about is an illusion is because you only see that method applied in cases where the method works.  It's like students learning calculus for the first time thinking derivatives are "easy" and integrals are "hard" because the actual functions they meet in a calculus course are the kind that they know from school, which all have simple derivative formulas. Analysts know that, in contrast to the impression of calculus students, integration is a better operation than differentiation because the integral of a small function is small (intuitively).

In any case, you apply the method of half-circle contours in cases where the function being integrated gets small along large semicircles as the radius gets large and where that method actually succeeds with the Residue theorem. Imagine trying to prove the Gaussian integral $\int_{-\infty}^{\infty} e^{-x^2/2}\,dx$ is $\sqrt{2\pi}$ by setting up a complex contour integral $\int_{\gamma_R} e^{-z^2/2}\,dz$ along some semicircular contour $\gamma_R$ that includes the real interval $[-R,R]$ and some other piece connecting $-R$ and $R$ through the upper or lower half-plane. You're not going to be able to make progress with this by the Residue theorem, intuitively, because the function is entire: there is no pole at which to compute a nonzero residue!  And also the function $e^{-z^2/2}$ does not get small everywhere when $|z|$ is large: if $z = iy$ is purely imaginary, then $e^{-z^2/2} = e^{y^2/2}$ is huge as $|z| = |y| \to \infty$.

It is possible to compute the Gaussian integral using the Residue Theorem, as shown in the tenth proof here, but it involves a contour integral of a function that is not simply $e^{-z^2/2}$.


问:
Identifying the quotient space.

Let $X$ be a compact Hausdorff space and let $K$ be a compact subset of $X.$ Let $$V = \left \{f \in C(X)\ \big |\ f(x) = 0,\ \text {for all}\ x \in K \right \}.$$ Then identify the quotient normed linear space $C(X)/V.$

For each $x \in K$ I define linear maps $T_x : C(X) \longrightarrow \Bbb C$ defined by $T_x (f) = f(x), f \in C(X).$ Clearly each $T_x$ is a bounded linear map with $\|T_x\|_{\text {op}} = 1,$ for all $x \in K.$ Hence $\text {Ker} (T_x)$ is a closed subspace of $C(X),$ for all $x \in K.$  So $V = \bigcap\limits_{x \in K} \text {Ker} (T_x)$ is also a closed subspace of $C(X).$ So $C(X)/V$ is a valid normed linear space endowed with the quotient norm. But I can't able think of a known normed linear space to which $C(X)/V$ is isomorphic. I am guessing that this quotient space has something to do with $C_c (X).$ But I am unable to figure out how? Any suggestion regarding this will be highly appreciated.

Thanks for your time.

答: $C(X)/V$ can be identified with $C(K)$. If $g \in C(K)$ there is an extension of $g$ to an element $G$ of $C(X)$ preserving the sup norm by Tietze Extension Theorem. If $H$ is another extension then $G-H$ vanishes on $K$ so $G+V=H+V$. Now look at the map $g \to G$. It is easy to check that this is injective. Given any element $f+V$ of $C(X)/V$ check that $f+V$ is the image of $g$ where $g$ is the restriction of $f$ to $K$. You can also verify that $\|G+V\|=\|g\|$ for all $g \in C(K)$.

Title: Identifying the quotient space.


问:
Related Rate problem: Speed of a beam of light on the shoreline
The problem comes from MIT's calculus exercise book (2E-2), and goes as follows:

A beacon light 4 miles offshore (measured perpendicularly from a straight shoreline) is rotating at 3 revolutions per minute. How fast is the spot of light on the shoreline moving when the beam makes an angle of 60* with the shoreline?

The answer from the answer book is:

$tan\theta = x/4$ and $ d\theta/dt=3(2\pi)=6\pi$ with $t$ is measured in minutes and $\theta$ measured in radians. The light makes an angle of $\pi/3$ with the shore when $\theta$ is $pi/6$. Differentiate with respect to $t$ to get $(sec^2 \theta)(d\theta/dt) = (1/4)(dx/dt)$. Since $sec^2 (\theta/6) = 4/3$, we get $dx/dt = 32 \pi$ miles per minute.

My solution is different than theirs:

When the beam rotates by an angle $\theta$, then it travels a distance equal to $l = x\theta$ (x is the radius of the circle whose center is the lighthouse). Differentiating with respect to $t$, we get $dl/dt = x (d\theta/dt)$. With $x = 4 tan(\pi/6)$.

Of course the answers are different. My understanding of the problem is the beam rotates in a circular fashion so the radius of that circle is a constant (which the variable $x$ I used). With that understanding it's only a matter of converting from angular velocity to distance velocity.

However, from the the answer book the distance $x$ is a variable ($x = 4tan\theta$). My struggle is visualising how does this beam of light operate.

答: The rotating beam moves by say $\theta$ but it isn't a constant radius circle that is relevant. Rather, the beam hits a spot on the shore which at different times will be at different distances from the source of the beam. For example when the rotating beam is perpendicular to the shore the distance will be shortest, and as it rotates away from there the distance increases. Draw several pictures and see the relation to $\tan \theta.$


问:
Methods for determining the convergence of $\sum\frac{\cos n}{n}$ or $\sum\frac{\sin n}{n}$
As far as I know, the textbook approach to determining the convergence of series like $$\sum_{n=1}^\infty\frac{\cos n}{n}$$ and $$\sum_{n=1}^\infty\frac{\sin n}{n}$$ uses Dirichlet's test, which involves bounding the partial sums of the cosine or sine terms. I have two questions:

Are there any other approaches to seeing that these series are convergent? I'm mostly just interested to see what other kinds of arguments might be made.

What's the best way to show that these two series are only conditionally convergent? I don't even know the textbook approach to that question.

答:
Hint for 2)

$$\sum_{n=1}^{\infty} \frac{|\cos n|}{n} \geq \sum_{n=1}^{\infty} \frac{\cos^2 n}{n}=\sum_{n=1}^{\infty} \frac{1+\cos {2n}}{2n}$$

Convergence of $\sum_{n=1}^{\infty}\frac{\cos{2n}}{2n}$, and divergence of $\sum_{n=1}^{\infty}\frac{1}{2n}$ gives the divergence.

The same method applies to $\sum_{n=1}^{\infty}\frac{|\sin n|}{n}$.


问:
Chain homotopy as left homotopy and right homotopy
According to Hovey's Model categories (around Theorem 2.3.11), a chain homotopy can equivalently be described as a right homotopy wrt to the standard model structure of the category of chain complexes.

On the other hand, according to this question, it says that chain homotopy can equivalently be described as a left homotopy.

My understanding is that left and right homotopies, in general, do not coincide (when the domain object is cofibrant and the codomain object is fibrant, they do), and I'm confusing about the above two equivalent descriptions of chain homotopies.

答:
A chain homotopy between chain maps $f,g\colon X→Y$
can be defined as a chain map $h\colon I⊗X→Y$,
where $I$ denotes the chain complex of simplicial chains on a 1-simplex,
i.e., ${\bf Z}⊕{\bf Z}←{\bf Z}$ in chain degrees 0 and 1, with the differential $1⊕-1$.
This is a special case of a left homotopy with a cylinder object $I⊗X$.

By the hom-tensor adjunction, $h$ is adjoint to a map $H\colon X→{\rm Hom}(I,Y)$.
Here ${\rm Hom}(I,Y)$ is a path object for $Y$,
so the map $H$ is a right homotopy.


答: Roughly speaking, the textbook interpretation is that the beam from the beacon lights indefinitely far, or rather until it hits an obstacle — such as the shorline. Thus this beam does NOT have a fixed length to be interpreted as its "radius". When it shines towards the shorline, it always reaches the shorline; but then, depending on its direction defined by the angle $\theta$, its length can be different. Thus we have a right triangle here, whose legs are $4$ (the straight distance from the beacon to the shore) and $x$ (the distance on the shore between the base point and the current spot of light) and whose hypotenuse is the length of the beam of light.

Title: Related Rate problem: Speed of a beam of light on the shoreline

Body:
The problem comes from MIT's calculus exercise book (2E-2), and goes as follows:

A beacon light 4 miles offshore (measured perpendicularly from a straight shoreline) is rotating at 3 revolutions per minute. How fast is the spot of light on the shoreline moving when the beam makes an angle of 60* with the shoreline?

The answer from the answer book is:

$tan\theta = x/4$ and $ d\theta/dt=3(2\pi)=6\pi$ with $t$ is measured in minutes and $\theta$ measured in radians. The light makes an angle of $\pi/3$ with the shore when $\theta$ is $pi/6$. Differentiate with respect to $t$ to get $(sec^2 \theta)(d\theta/dt) = (1/4)(dx/dt)$. Since $sec^2 (\theta/6) = 4/3$, we get $dx/dt = 32 \pi$ miles per minute.

My solution is different than theirs:

When the beam rotates by an angle $\theta$, then it travels a distance equal to $l = x\theta$ (x is the radius of the circle whose center is the lighthouse). Differentiating with respect to $t$, we get $dl/dt = x (d\theta/dt)$. With $x = 4 tan(\pi/6)$.

Of course the answers are different. My understanding of the problem is the beam rotates in a circular fashion so the radius of that circle is a constant (which the variable $x$ I used). With that understanding it's only a matter of converting from angular velocity to distance velocity.

However, from the the answer book the distance $x$ is a variable ($x = 4tan\theta$). My struggle is visualising how does this beam of light operate.


问:
Moebius Transformation in the upper half plane
I am studying hyperbolic geometry. But I got stuck a problem. My problem is  as follows:

Let $\mathbb{H}$ be the upper half plane model in hyperbolic geometry. Let $a$ and $b$ be two points on the imaginary axis (one point say, $a$  below $i$ and the other point $b$ a bit above) such that each points have same hyperbolic distance from $i$. Now assume that hyperbolic distance from $a$ to $i$ is $ l_{m}$. Now my question is:

How to find an explicit Moebius transformation  that preserves the imaginary axis and takes $a$ to $b$. (Hint: you have its hyperbolic translation distance $(2l_{m})$).

Please help me. I am looking for a solution using the hint.

Thanking in advanced.

答:
From your hint I am not sure if the distance from $a$ to $i$ is supposed to be $l_m$ or $2l_m$. Let's assume it is $l_m$ (different from the statement of your problem).

Let's assume $a=(0, a)$ (by abusing notation) is below $i=(0, 1)$, then we compute the distance from $a$ to $i$:
$$
\int_a^1\frac{dy}{y}=l_m.
$$
That implies $-\log a=l_m$, that is $a=e^{-l_m}$. Similarly we see $b=(0, e^{l_m})$.

Recall that on ${\mathbb H}$, the "dilation" $(x, y)\to (rx, ry)$ is an isometry. So
$$
\Phi(x, y)=(e^{2l_m}x, e^{2l_m}y)
$$
is the transformation you need.


问:
Stein and Shakarchi, Complex analysis, Proposition $2.3$
page 12(Stein and Shakarchi, Complex Analysis

page 13(Stein and Shakarchi, Complex Analysis

My question has to do with the proof for proposition $2.3$. I am wondering how he arrived at the first equation on Page $13$. Can someone explain the complete proof?

答:
I am not sure whether my understanding is the true...

I think in the first question on Page 13, the left side $J_F(x_0,y_0)(H)$ is a matrix multiplication (it may be better if $H$ is written into column vector)...And the middle side is in 'complex number form'...

So I think the first '$=$' should be a kind of isomorphism...


问:
$ \int_{-\infty}^{+\infty}\sin(x^2)\cos2\alpha x \, \mathrm{d}x $

How to compute
$$
\int_{-\infty}^{+\infty}\sin(x^2)\cos2\alpha x\,\mathrm{d}x
$$

The integrating by parts does not work. Write
$$
I(\alpha)=\int_{-\infty}^{+\infty}\sin(x^2)\cos2\alpha x\,\mathrm{d}x,
$$
then
$$
I'(\alpha)=-\int_{-\infty}^{+\infty}2x\sin(x^2)\sin2\alpha x\,\mathrm{d}x.
$$

I do not kwow how to go on. Appreciate any help!

答:
First, let's notice that
$$
\int_{-\infty}^{\infty}{\sin}\left( x^2 \right) \cos 2ax\,dx=\frac{1}{2}\int_{-\infty}^{\infty}{\sin}\left( x^2+2ax \right) +\sin \left( x^2-2ax \right) \,dx
$$
For the first integral, substitute $u=x+a$, we have
$$
\int_{-\infty}^{\infty}{\sin}\left( x^2+2ax \right) \,dx=\int_{-\infty}^{\infty}{\sin}\left( u^2-a^2 \right) \,du = \int_{-\infty}^{\infty}{\sin}\left( u^2 \right) \cos a^2-\cos \left( u^2 \right) \sin a^2\,du
$$
And a similar process works for the second integral. Finally,
$$\begin{aligned}
\int_{-\infty}^{\infty}{\sin}\left( x^2 \right) \cos 2ax&=\cos a^2\int_{-\infty}^{\infty}{\sin}\left( u^2 \right) \,du-\sin a^2\int_{-\infty}^{\infty}{\cos}\left( u^2 \right) \,du\\&\overset{(1)}=\sqrt{\frac{\pi}{2}}\left( \cos a^2-\sin a^2 \right)\end{aligned}
$$
Proof of (1)


问:
How to integrate $\int_0^1\frac{dx}{1+x+x^2+\cdots+x^n}$
I am interested in finding a solution to the integral
$$I_n=\int_0^1\frac{dx}{\sum_{k=0}^nx^k}$$
Since the denominator is a geometric series with $a=1$ and $r=x$ and it is within the radius of convergence, we should be able to say
$$\sum_{k=0}^nx^k=\frac{1-x^{n+1}}{1-x}=\frac{x^{n+1}-1}{x-1}$$
and
$$I_n=\int_0^1\frac{x-1}{x^{n+1}-1}dx$$
It makes sense to me that, for all values of $n$, $I_n$ is convergent since the bottom of the function is always above zero and the integral exists for $n\to\infty$ however I cannot seem to find a nice closed form for this.

One thought I did have was using:
$$\sum\ln(x_i)=\ln\left(\prod x_i\right)$$
but I cannot seem to make it work. Does anyone have any hints for this type of problem as I would like to try and complete it myself. Thanks :)

答:
Decompose the integrand as

$$\frac{1}{1+x+x^2+...+x^n}=\sum_{k=1}^{n}\frac{a_k}{x-x_k},\>\>\>\>\> a_k=\frac{x_k(x_k-1)}{n+1}
$$
where $x_k= e^{i \frac{2\pi k}{n+1}},\> k=1,2...n$. Then

\begin{align}
I_n&=\int_0^1\frac{dx}{1+x+x^2+...+x^n}\\
&=\int_0^1 \sum_{k=1}^{n}\frac{a_k}{x-x_k}dx
=\frac1{n+1}\sum_{k=1}^{n} x_k(x_k-1)\ln\left(1-\frac1{x_k}\right)
\end{align}
Substitute $x_k= e^{i \frac{2\pi k}{n+1}}$ into above summation to derive the close-form
$$\color{blue}{I_n = \frac\pi{2(n+1)} \csc\frac{2\pi}{n+1}-\frac4{n+1}\sum_{k=1}^{[\frac n2]}
\sin\frac{\pi k}{n+1}\sin\frac{3\pi k}{n+1}\ln\left(\sin\frac{\pi k}{n+1}\right)}
$$
Listed below are some sample results
\begin{align}
I_2 &= \frac{\pi}{3\sqrt3}\\
I_3 &= \frac\pi8 +\frac14\ln2\\
I_4 &= \frac\pi{10}\csc\frac{2\pi}{5}+\frac1{\sqrt5}\ln\left(2\cos\frac\pi5\right)\\
 I_5 &= \frac{\pi}{6\sqrt3}+\frac13\ln2\\
 I_7 &= \frac{\pi}{8\sqrt2}+\frac18\ln2 + \frac{1}{8\sqrt2}\ln \frac{\sqrt2+1}{\sqrt2-1}\\
 I_9 &= \frac\pi{20}\csc\frac{\pi}{5}+\frac15\ln2
+\frac1{2\sqrt5}\ln\left(2\cos\frac\pi5\right)\\
\end{align}
\begin{align}
I_6
&=\frac\pi{14}\csc\frac{2\pi}{7}-\frac1{2\sqrt7}\bigg( \frac{\ln\sin\frac\pi7}{\sin\frac{2\pi}7}
 +\frac{\ln\sin\frac{2\pi}7}{\sin\frac{3\pi}7}-\frac{\ln\sin\frac{3\pi}7}{\sin\frac{\pi}7}\bigg)
\\
\end{align}

\begin{align}
I_8 &= \frac\pi{18}\csc\frac{2\pi}{9}+\frac{2}{3\sqrt3} \bigg(\frac{\ln\csc\frac\pi9}{\csc\frac{\pi}9}
 +\frac{\ln\csc\frac{2\pi}9}{\csc\frac{2\pi}9}-\frac{\ln\csc\frac{4\pi}9}{\csc\frac{4\pi}9}\bigg)
\\
\end{align}


问:
Finding a general form of a recurrence
We have that $T_n^2-T_n+1=T_{n+1}$ for n greater than $1$ ($n$ is an integer)
And we have to find the general form of the series.Also $T_2=2.$

Im new to this topic so i tried substitution but failed,any help would be appreciated.Thanks in advance.

答:
I suggest $T_1=2$ instead of $T_2=2$. Then this is exactly the Sylvester's Sequence. It doesn't have a nice closed form, but one can claim that
$$
T_n=\left [c^{2^{n+1}} + 1/2\right]
$$
More information here.

Title: Finding a general form of a recurrence


问:
Application of derivative - how to calculate change in error
Problem: If the error committed in measuring the radius of the circle is $0.05\%$ then find the corresponding error in calculating the area.

Solution: Let the error be denoted by $\delta r = 0.05\%$, therefore the corresponding error in calculating the area is:

$A = \pi r^2 $

$\frac{\delta A}{\delta r} = 2\pi r$

$\Rightarrow \delta A = 2\pi r \delta r$

Please suggest if this is the right approach. If not, please help me solve this problem along with the concepts.

答: You are on the right track.  When you see $\delta r = 0.05\%$ it really means that $r$ is measured within $0.05\%$, so the proper reading is $\frac {\delta r}r=0.05\%=5\cdot 10^{-4}$.  I don't know if this is in your text, but when errors are quoted as percents, they are relative, not absolute, errors.  Similarly, you are being asked for $\frac {\delta A}A$.  So divide your equation by $A=\pi r^2$ to get $\frac {\delta A}A=2\frac {\delta r}r$</p>


答:
Ahh well that is a good question.

You just have to find the values of some of the first terms

As $T_2$ = 2

You can easily find the values of

$T_3$ = 3

$T_4$ = 7

$T_5$ = 43

As this here at first you would not see any pattern but I have been working on something like this so i recognise this pattern.

It is as follows that you have to multiply the previous term amd add 1 to it to get the next term.

For example

$T_4$ =($T_2$) ($T_3$) + 1

Or I guess if you notice it is a way of finding new primes as suggested by Euclid about the infinite primes his proof include this reccurence of multiplying the primes and adding one to it gives you a new prime.

At last to state there is no general formula for this reccurence as it is dealing with primes I don't think it should but if you got some please do let me know

Title: Finding a general form of a recurrence

Body:
We have that $T_n^2-T_n+1=T_{n+1}$ for n greater than $1$ ($n$ is an integer)
And we have to find the general form of the series.Also $T_2=2.$

Im new to this topic so i tried substitution but failed,any help would be appreciated.Thanks in advance.


问:
Proof by induction on coin problem
Suppose I have an unlimited supply of $k$th and $(k+1)$th cents where $k \in \mathbb{N}$, then I can always pay integer bills starting from $k(k-1)$ cents using only the two types of coins I have. Let this statement be $P(k)$. So,

$P(1):$ I have coins of $1$ and $2$ cents, then I can pay any bill starting from $1(1-1)=0$ cents. (This is true because it does not require any coins).

$P(2):$ I have coins of $2$ and $3$ cents, then I can pay any bill starting from $2(2-1)=2$ cents.

$P(5):$ I have coins of $5$ and $6$ cents, then I can pay any bill starting from $5(5-1)=20$ cents.

Now $P(k)$ seems to hold fine for some arbitrarily chosen positive integer, but I would like to validate this statement using the principle of induction:

Base case:

$P(1)$ holds. (See from above)

Induction on $k$:

Assume $P(i)$ holds for $1\leq i \leq k$. Using this assumption I have to prove $P(k+1)$ holds. I am stuck here, how should I approach it? Thanks in advance.

答:
Here's an outline of an induction proof  for a fixed $k$ :

$k(k-1)$ cents can be paid with $(k-1)$  k-cent coins.

$k(k-1) +1$ cents can be paid with $(k-2)$  k¢ coins plus $1$ k+1 cent coin.

$k(k-1) +2$ cents can be paid with $(k-3)$  k¢ coins plus $2$ k+1 cent coins.

$$\vdots$$

$k(k-1) + k-2$ cents can be paid with $1$  k¢ coin plus $(k-2)$ k+1¢ coins.

$k(k-1) + k-1$ cents can be paid with $(k-1)$ k+1 cents coins.

i.e. $(k^2-1)$ cents can be paid with $(k-1)$  k+1-cent coins.

But then $k^2$ cents can be paid with $k$ k-cent coins, and from there the cycle repeats.

Title: Proof by induction on coin problem


问:
Given $(X,Y)$ bivariate normal, find a linear combination independent of $Y$
I have a question about the answer to this question. At the start of the proof we assume that $$\frac{X-\mu_1}{\sigma_1} = \rho \frac{Y-\mu_2}{\sigma_2} + \sqrt{1-\rho^2} Z \tag{$*$}$$
But I can't figure out where this assumption comes from and how it is valid.

答:
The claim is:

If $(X,Y)$ have bivariate normal distribution, where $X\sim N(\mu_1,\sigma_1^2)$ and $Y\sim N(\mu_2,\sigma_2^2)$ with correlation $\rho$, then
$$ Z:=\frac1{\sqrt{1-\rho^2}}\left(\frac{X-\mu_1}{\sigma_1}\right)-\frac\rho{\sqrt{1-\rho^2}}\left(\frac{Y-\mu_2}{\sigma_2}\right)
$$ has standard normal distribution and is independent of $Y$.

To prove this, write $X':=\frac{X-\mu_1}{\sigma_1}$ and $Y':=\frac{Y-\mu_2}{\sigma_2}$, so $X'$ and $Y'$ each have standard normal distribution, and their covariance is $\rho$. Define
$$Z:=aX'+bY'$$ for constants $a,b$ to be determined. Then $(Z,Y')$ are jointly normal since we can write
$$
\begin{pmatrix}Z\\Y'\end{pmatrix}=\begin{pmatrix}a&b\\0&1\end{pmatrix}\begin{pmatrix}X'\\Y'\end{pmatrix}
$$ as a linear transformation of the jointly normal $(X',Y')$. Check that the mean of $Z$ is zero and its variance is
$$\operatorname{Var}(Z)=
\operatorname{Var}(aX'+bY')=a^2+b^2+2ab\rho.\tag a$$ Moreover, the covariance between $Z$ and $Y'$ is
$$\operatorname{Cov}(Z,Y')=
\operatorname{Cov}(aX'+bY',Y')=a\operatorname{Cov}(X',Y')+b\operatorname{Var}(Y')=a\rho+b.\tag b
$$ We now find $a$ and $b$ such that (a) equals $1$ and (b) equals $0$ . Solving the equations $$a^2+b^2+2ab\rho=1\qquad{\rm and}\qquad a\rho+b=0$$ yields
$$a=\frac1{\sqrt{1-\rho^2}}\qquad{\rm and}\qquad b=\frac{-\rho}{\sqrt{1-\rho^2}}.$$ With these choices for $a$, $b$ we have that $Z$ has standard normal distribution, and $Z$ is independent of $Y'$ (since zero covariance implies independence for jointly normal variables), and therefore $Z$ is independent of $Y$.


问:
Probablity of all different socks pulled out.
There were 5 pairs of socks in a pile of clothes. On pulling out one by one and after pulling out 5 socks, I found they were all different. What is the probability of such an event?</p>

答: For randomly choosing $5$ socks from $10$ socks, we have $\displaystyle \binom{10}5$ ways. For choosing $5$ different socks, we have $2^5$ ways. Thus the probability is $\displaystyle \dfrac{2^5}{\binom{10}{5}}=\dfrac{8}{63}$


问:
Find condtion of A is a bijective
Let $(X,\langle\cdot,\cdot\rangle)$ be a Hilbert space with orthonormal system $\{e_n\}$ and $\{\lambda_k\}$ is a bounded sequence. We define
$A(x)=\displaystyle\sum_{k=1}^\infty \lambda_k \langle x,e_k\rangle e_k$ and $A_n(x)=\displaystyle\sum_{k=1}^n \lambda_k \langle x,e_k\rangle e_k$, $x \in X$, $n \in \mathbb{N}$.

I have already proved $A: X \rightarrow X$ is linear and continuous and $\Vert A \Vert =\displaystyle\sup_{k \in \mathbb{N}} \vert \lambda_k \vert $.

I tried to find a necessary and sufficent condtion of A is a bijective, but I don't have any idea.
Any help is appreciated. Thank you.

答:
If $A$ is bijective then it has  a continuous inverse by Open Mapping Theorem. Since $Ae_k=\lambda_k e_k$ it follows that $\frac 1 {\lambda_k}e_k=A^{-1} e_k$ is bounded. So $\inf \{|\lambda_k|: k \geq 1 \} >0$.

Conversely, suppose $\inf \{|\lambda_k|: k \geq 1 \} >0$.  If $Ax=Ay$ then $\lambda_k \langle x, e_k \rangle=\lambda_k \langle y, e_k \rangle$ for all $k$ which gives $x=y$. Any $y =\sum a_ke_k \in X$ (with $\sum |a_k|^{2}< \infty$)  is the image of $\sum \frac {a_k} {\lambda_k} e_k$ so $A$ is also surjective.

Title: Find condtion of A is a bijective


问:
On integral of a function over a simplex
Help w/the following general calculation and references would be appreciated.

Let $ABC$ be a triangle in the plane.
 Then for any linear function of two variables $u$.
$$
\int_{\triangle}|\nabla u|^2=\gamma_{AB}(u(A)-u(B))^2+
\gamma_{AC}(u(A)-u(C))^2+\gamma_{BC}(u(B)-u(C))^2,
$$
 where
$$
\gamma_{AB}=\frac{1}{2}\cot(\angle C),
\gamma_{AC}=\frac{1}{2}\cot(\angle B),
\gamma_{BC}=\frac{1}{2}\cot(\angle A).
$$

What is a good reference for the formula? Is it due to R. Duffin?

Is there generalization to linear functions of three variables? The number of parameters fits (in any $nD$), and it seems like one needs to calculate 6x6 Cayley-Menger like determinants, but it's difficult and the geometric interpretation is not clear.
Thank you.

答:
Let $(x_k, y_k,z_k)$ be four points in 3D. Then for a linear function

$$u(x,y,z)=Ax+By+Cz+D,$$
and
$$ x_{kl}=x_k-x_l, \dots $$
matching coefficients in the identity in question, one gets the following system of equations:

$$
\begin{pmatrix}
x_{12}^2 & x_{13}^2 & \dots & x_{34}^2 \\
y_{12}^2 & y_{13}^2 & \dots & y_{34}^2 \\
z_{12}^2 & \dots & \dots & \dots \\
x_{12}y_{12} & x_{13}y_{13} & \dots & \dots \\
x_{12}z_{12} & \dots & \dots & \dots \\
y_{12}z_{12} & y_{13}z_{13} & \dots & y_{34}z_{34} \\
\end{pmatrix}
\begin{pmatrix}
\gamma_{12} \\ \gamma_{13} \\ \dots \\ \dots \\ \dots \\ \gamma_{34}
\end{pmatrix} =
Vol_{\triangle}
\begin{pmatrix}
1 \\ 1 \\ 1  \\ 0 \\ 0 \\ 0
\end{pmatrix},
$$
where
$$ x_{kl}+x_{lm}+x_{mk}=0, \dots $$
 Can these expression be simplified? Is there geometric interpretation of the solution?


问:
How to find a and b estimates in regression from least square error formula
In regression, we want to minimize $1/n \sum_1^n(y_i-(a+bx_i))^2 $. To minimize, we use gradients:
$$\begin{align} \partial/\partial b = 0 \Rightarrow \sum_i^n x_i(y_i-(a+bx_i)) =0 \Rightarrow
\sum_1^nx_iy_i-a\sum x_i-b\sum x_i^2=0 \end{align}$$
$$\partial/\partial a = 0 \Rightarrow
\sum_i^n (y_i-(a+bx_i))=0 \Rightarrow \\
\sum y_i - \sum a -b\sum x_i = 0  \Rightarrow \\
 \sum y_i - na -b\sum x_i = 0 $$

which eventually in the second equation we have $na$ which I can't find a way to properly get rid of to get to the formalization of $\hat b = \frac{\overline {XY} -\overline X \overline Y}{\overline {X^2} - \overline X^2}$ where $\overline {XY} ,$ $\overline X,$ and $\overline Y $ represent averages. Can someone explain where I am going wrong?

答: your first equation can be written as $\overline{XY}-a\overline{X}-b\overline{X^2}=0$ and then second can be written as $-a=b\overline{X}-\overline{Y}$ plugging $-a$ into the first $\overline{XY}-(b\overline{X}-\overline{Y})\overline{X}-b\overline{X^2}=0\iff b(\overline{X}^2-\overline{X^2})=\overline{X}\overline{Y}-\overline{XY}$ as desired


答:
As of there are 5 pairs of socks or 10 socks so pulling out any 5 can be found out from the formula.

$C(10,5) = 252$

And the favourable outcomes will be only $2^5$ as of

Every pair has two socks and there are five pairs so you have select different so at first pull you have two choices of any pair and then again at the second pull it goes same.
So the probability becomes

$\displaystyle \dfrac{2^5}{252}=\dfrac{8}{63}$

Title: Probablity of all different socks pulled out.

Body: There were 5 pairs of socks in a pile of clothes. On pulling out one by one and after pulling out 5 socks, I found they were all different. What is the probability of such an event?</p>