问:
Proving the Alternate Series Test
I am self-learning Real Analysis from Understanding Analysis by Stephen Abbott. I am finding it difficult to come up with a proof for the Alternating Series Test for an infinite series. I only have some initial ideas, so I'll put those down.

I'd like to ask if somebody could help, by providing any hints (please do not write the complete proof), on how to think about this.

Exercise 2.7.1

Proving the Alternating Series Test amounts to showing that the sequence of partial sums
\begin{align*}
 s_n = a_1 - a_2 + a_3 - \ldots \pm a_n
\end{align*}
converges. Different characterizations of completeness lead to different proofs.

(a) Produce the Alternating Series Test by showing that $(s_n)$ is a Cauchy sequence.

(b) Supply another proof for this result using the Nested Interval Property.

(c) Consider the sequences $(s_{2n})$ and $(s_{2n+1})$, and show how the Monotone Convergence Theorem leads to a third proof for the Alternating Series Test.

Proof.

(a)Let $(a_n)$ be a sequence satisfying

(i) $a_1 \ge a_2 \ge a_3 \ge \ldots \ge a_n \ge a_{n+1} \ge$

(ii) $(a_n) \to 0$

Our claim is that, the alternating series $\sum_{n=1}^{\infty}(-1)^n a_n$ converges.

Let $(s_n)$ be sequence of the partial sums, where
\begin{align*}
s_n = a_1 - a_2 + a_3 - a_4 + a_5 - a_6 + \ldots + (-1)^{n+1}a_n
\end{align*}

We are interested to show that $(s_n)$ is a Cauchy sequence. Let's see if can find a simple expression for $\lvert s_n - s_m \vert$. If $n > m$,

\begin{align*}
\lvert (s_n - s_m) \rvert &= \lvert (-1)^{m+2}a_{m+1} + (-1)^{m+3}a_{m+2} + (-1)^{n+1}a_{n} \rvert\\
&= \lvert {a_{m+1} - a_{m+2} + a_{m+3} - \ldots + (-1)^{n-m+1}a_n} \rvert
\end{align*}

A consequence of condition(i) is that $(a_n)$ is a decreasing sequence and has a lower bound zero, so $a_n \ge 0$ for all $n \in \mathbf{N}$. Since $a_{m+2} \ge 0$, $a_{m+4} \ge 0$, $a_{m+6} \ge 0, \ldots$ we can write,

\begin{align*}
\lvert (s_n - s_m) \rvert &= \lvert {a_{m+1} - a_{m+2} + a_{m+3} - \ldots + (-1)^{n-m+1}a_n} \rvert\\
&\le \lvert {a_{m+1} + a_{m+3} + a_{m+5} + \ldots + a_{n}} \rvert \\
&\le \vert {a_{m+1} + a_{m+1} + a_{m+1} + \ldots + a_{m+1}} \rvert
\end{align*}

But this gives me a variable number of terms (whereas I need to prove that the distance $\lvert s_n - s_m \rvert$ becomes smaller than fixed $\epsilon$ or $k\epsilon$. I am not quite sure, where to go from here. I know that the original sequence $(a_n)$ converges, so given any $\epsilon > 0$, $\exists N \in \mathbf{N}$, such that $\lvert a_n \rvert < \epsilon$ for all $n \ge N$.

(b) I am thinking of constructing a sequence of closed nested intervals $I_1 \supseteq I_2 \supseteq I_3 \supseteq \ldots \supseteq I_n \supseteq I_{n+1} \supseteq \ldots$, in such a way that, $I_1=\{s_1,s_2,s_3,\ldots\}$, $I_2 = I_1 - \{s_1\}$ and in general $I_{k+1} = I_k - \{s_k\}$, as I am interested in the tail of the sequence of partial sums. Then, I can apply the nested interval property to show that $\bigcup_{k\ge 1}I_k$ is not empty.

The one thing I don't know, is whether $(s_n)$ is bounded. If and only if this set is bounded, one may construct a closed interval. I think this is important for the construction.

(c) Again, I need some bounds for $(s_n)$, I guess.

答:
$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$
For completeness, I add the proof using NIP(nested interval property).

Given that $a_1 \ge a_2 \ge a_3 \ge \ldots $

\begin{align*}
 s_n &= a_1 - a_2 + a_3 - a_4 + a_5 - a_6 + \ldots + (-1)^{n+1}a_n\\
  &= a_1 - (a_2 - a_3) - (a_4 - a_5) - (a_6 - a_7) - \ldots \\
  &\le a_1 = s_1
\end{align*}

And, moreover,

\begin{align*}
 s_n &= (a_1 - a_2) + (a_3 - a_4) + (a_5 - a_6) + \ldots + (-1)^{n+1}a_n\\
  &\ge 0
\end{align*}

So, $0 \le s_n \le s_1$ for all $n \in \mathbf{N}$.

Also, consider the subsequences,
\begin{align*}
 (s_{2n}) &= (s_2,s_4,s_6,s_8,\ldots,s_{2n-2},s_{2n})   \\
 (s_{2n-1}) &= (s_1,s_3,s_5,s_7,\ldots,s_{2n-3},s_{2n-1})
\end{align*}

We observe:
\begin{align*}
 s_{2} &= a_1 - a_2\\
 s_{4} &= (a_1 - a_2) + (a_3 - a_4) \\
 s_{6} &= (a_1 - a_2) + (a_3 - a_4) + (a_5 - a_6)
\end{align*}
Thus, $(s_{2n})$ is a monotonically increasing sequence.
\begin{align*}
 s_{1} &= a_1\\
 s_{3} &= a_1 - (a_2 - a_3)\\
 s_{6} &= a_1 - (a_2 - a_3) - (a_4 - a_5)
\end{align*}
Thus, $(s_{2n-1})$ is a monotonically decreasing sequence.

Let $S := [0,s_1]$. Next, let $I_1$ be the closed interval that contains $s_2$ and $s_3$. In general,
\begin{align*}
 I_{k} = [s_{2k},s_{2k+1}]
\end{align*}

By the nested interval property, $\bigcap_{k\ge1} I_k \ne \emptyset$ and contains atleast one element $x$.

Define $A :=\{s_2,s_4,s_6,\ldots\}$ and $B := \{s_1,s_3,s_5,s_7,\ldots\}$. Every $s_{2n+1} \in B$ is an upper bound for $A$ and every $s_{2n} \in A$ is a lower bound for $B$. Since, these are bounded subsets of $\mathbf{R}$, we are justified in setting $x = \sup A = \inf B$. Thus, we have a candidate for the limit of the alternating series. Our claim then is, $\lim s_{2n} = x = \lim s_{2n+1}$.

Since $x = \sup A$, for any $\epsilon > 0$, $x - \epsilon$ is not an upper bound for A. Thus, there exists a point $s_{2N} \in A$, such that $x - \epsilon < s_{2N}$. Therefore, $\absval{s_{2N} - x} < \epsilon$. A similar argument can be made for $B$.

Title: Proving the Alternate Series Test


问:
Issues with integration using Dirac-$\delta$ function. What is $f(0)?$
This integral arose in a problem where I needed to apply Fourier transform of the impulse response. I have
$$\int\limits_{-\infty}^{\infty}u(t)e^{-(b+j\omega)t} \ dt, \quad u(t)=1 \ \text{if} \ t>0 \ \text{and}\  0 \ \text{otherwise.}\tag1$$

Using integration by parts I then get

$$u(t)\int\limits_{-\infty}^{\infty}e^{-(b+j\omega)t} \ dt - \int\limits_{-\infty}^{\infty}\delta(t)\underbrace{\left[\int\limits_{-\infty}^{\infty}e^{-(b+j\omega)t} \ dt\right]}_{=f(t)} \ dt. \tag2$$

Where $u'(t)=\delta(t).$ Now, in the first integral, $u(t)=0$ for $t<0$ so the contributions are only from $t>0$ and we can adjust the limits accordingly. So $(2)$ becomes

$$\int\limits_{0}^{\infty}e^{-(b+j\omega)t} \ dt-\int\limits_{-\infty}^{\infty}\delta(t)f(t) \ dt = \left[-\frac{e^{-(b+j\omega)t}}{b+j\omega}\right]^{\infty}_{0}-f(0) = \frac{1}{b+j\omega}-f(0).\tag3$$

What is $f(0)$? Plugging in $t=0$ I just get $f(0)=\infty$. According to my book the correct answer is just $1/(b+j\omega).$

Note that to evaluate the second integral in $(2)$ I wanted to use the well known fact that
$$\int\limits_{-\infty}^{\infty}\delta(t)f(t) \ dt = f(0).$$

答: I think you’ve mistakenly calculated $e^0$ as 0 instead of 1 with regards to $f(0)$.

ContentLicense: CC BY-SA 4.0


问:
Dividing a polynomial with integer coefficients by a quadratic
In the following question, one of the coefficients of the divisor quadratic polynomial is unknown; so, it cant be factored for performing synthetic division. I tried to perform long division hoping to get some clues but ran into a conflict.

Appreciate your help. Thanks!!

Here is the question:
$x^2 -x +a$  evenly divides $x^8 + 5x^6 + 13x^4 + 20x^2 +36$. Determine positive integer $a$.

A few things I noted are:

$a$ must be a factor of 36 so, I am thinking $a$ would be from $\{1, 2, 3, 4, 6, 9, 12, 18, 36\}$

All the coefficients of the dividend polynomial are integers and degrees of all the terms are even. Not sure how to use this info.

$P(x) = Q(x)(x^2 -x +a) + mx + n$.  Since the remainder is zero, $m$ and $n$ are zero. When I tried long division, I got two equations; but the value of $a$ would come as imaginary which conflicts with the first observation and the question asks for real integer $a$.

答:
Let the $q(x)$ be the quotient when f(x) is divided by $x^2-x+a$. (where $f(x)$ be the given 8 degree polynomial). $f(x) = q(x)(x^2-x+a)$. $ f(0)= 36= a*q(0)$ , $f(1)=75=a*q(1)$ . As $ q(0), q(1) $ are integers (Since $q(x) $ is a polynomial with integer coefficients as coefficients of divisor are integers check that) ,$a|36, a|75$ this implies $a=1 $ or $3$ or no integer solution.

If $a=1$ then roots of $x^2-x+1=0$ i.e $ -w, -w^2$ are also the roots of $ f(x)$,  where $1+w +w^2=0$ and $w^3=1$ check that. You can check the case of $a=3$  using synthetic division or any other process.


问:
Count the numbers in which the product of the digits is 48 (1 not allowed)

Count the numbers in which the product of the digits is 48 (1 not allowed as digit).

Proposed solution: I addressed the problem by brute force, first noticing that
$$48=2^4\times 3.$$

Then I found the possible groups of digits to be permuted for each number: (1) 2,2,2,2,3; (2) 2,2,2,6; (3) 4,2,2,3; (4) 4,2,6; (5) 4,4,3; (6) 8,2,3; (7) 8,6.

By adding the permutations with repetition for each case, I've found 38 possible numbers. Which agrees with the answer provided in the source.

My problem: I found this solution is not elegant and I'm wondering if there is a better way to address problems like these. By brute force it is easy to undercount terms (as I did as first solving the problem). Any help?

答:
There isn't a much better way, no. There are better ways for calculating the number of ordered factorizations but those will include factorizations with factors that aren't single digits.

If $a_n = $ the number of ordered factorizations of $n$ then we have two identities:

$a_n = \sum a_d$ over all factors $d$ of $n$ not including $d = n$.

$a_{p^k} = 2^{k - 1}$ when $p$ is a prime.

See http://oeis.org/A074206 for references/additional formulae/etc.

So using this we have

\begin{align}
a_{48} &= a_1 + a_2 + a_3 + a_4 + a_6 + a_8 + a_{12} + a_{16} + a_{24} \\
&= 1 + 1 + 1 + 2 + a_6 + 4 + a_{12} + 8 + a_{24} \\
&= 17 + a_6 + a_{12} + a_{24} \\
&= 17 + (a_1 + a_2 + a_3) + ([a_1 + a_2 + a_3] + a_4 + a_6) \\
&\quad\qquad+ ([a_1 + a_2 + a_3 + a_4 + a_6] + a_8 + a_{12}) \\
&= 17 + 3 + (3 + 2 + 3) + (8 + 4 + 8) \\
&= 48
\end{align}

So there are 48 total ordered factorizations. But we know this is overcounting because it includes factorizations which are permutations of $2,24$ and $2,2,12$ and $4,12$ and $3,16$. So that's an additional $2 + 3 + 2 + 2 = 10$.

So there are $48 - 10 = 38$ ordered factorizations where no factor is larger than $9$.

Was this easier than calculating the number $38$ directly? Probably not. But hopefully it's still useful to someone.


问:
Calculating the probability of total payout for an insurance company

Automobile losses reported to an insurance company are independent and uniformly distributed between $0$ and $20,000$. The company covers each loss subject to a deductible of $5,000$.

Calculate the probability that the total payout on 200 reported losses is between $1,000,000$ and $1,200,000$.

My attempt:

Let $S$ denote the random variable for all losses reported to the insurance company. Then, using CLT, $S \sim N\left(200 \times \dfrac{20,000}{2}, 200 \times \dfrac{20,000}{12}\right) = (2\times 10^6, 577.35^2)$. Now let $Y$ be the random variable for the payout on all claims. Then, $Y=200(S-5000)$ if $10^6\le s<\infty$ and $Y=0$ otherwise.

The desired probability is given by
$$P[10^6<Y<1.2\times 10^6] = P[10^6<200(S-5000)<1.2\times 10^6] = P[10^4<S<1.1\times 10^4]$$

Now, I computed $$P\left[Z\le\frac{(1.1\times 10^4) - 2\times 10^6}{577.35}\right] - P\left[Z\le\frac{10^4 - (2\times 10^6)}{577.35}\right]$$

but this didn't lead me to the correct answer. Can someone please suggest what I did wrong in this question? Thanks.

答:
Not sure, because you give few reasons along with
your computations, but it seems you're deducting \$5000 from all claims, including those for less than \$5000. [About $1/4$ of claims result in no payout because of the deductible; then payout is \$0, not -\$5000.]

Here is a simulation (of 100,000 lots of 200 claims)
that may be helpful: Simulation uses thousands of dollars.

If $T$ is the total payout for 200 claims, then the simulation illustrates that
$E(T)\approx \$1,125,000$ and $P(\$1\,000\,000 < T < \$1\,200\,000) \approx 0.82.$ With 100,000 iterations it is reasonable to expect about
2-place accuracy.

```<code>set.seed(2020)
m = 10^5;  tot.pay = numeric(m)
for(i in 1:m){
 loss = runif(200, 0, 20)
 tot,pay[i] = sum(pmax(0, loss-5))
}
mean(tot,pay)
[1] 1125.617
mean(tot.pay > 1000 & pay < 1200)
[1] 0.81652  <span class="math-container">$ aprx P(Total pay btw $</span>1m & 1.2m)

summary(pay)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  825.1  1077.9  1125.5  1125.6  1173.0  1428.6

hdr = "Payouts for 200 Claims"
hist(pay, prob=T, col="skyblue", xlab="$1000s", main=hdr)
 abline(v = c(1000, 1200), col="red")
</code>```

Body:
Not sure, because you give few reasons along with
your computations, but it seems you're deducting \$5000 from all claims, including those for less than \$5000. [About $1/4$ of claims result in no payout because of the deductible; then payout is \$0, not -\$5000.]

Here is a simulation (of 100,000 lots of 200 claims)
that may be helpful: Simulation uses thousands of dollars.

If $T$ is the total payout for 200 claims, then the simulation illustrates that
$E(T)\approx \$1,125,000$ and $P(\$1\,000\,000 < T < \$1\,200\,000) \approx 0.82.$ With 100,000 iterations it is reasonable to expect about
2-place accuracy.

set.seed(2020)
m = 10^5;  tot.pay = numeric(m)
for(i in 1:m){
 loss = runif(200, 0, 20)
 tot,pay[i] = sum(pmax(0, loss-5))
}
mean(tot,pay)
[1] 1125.617
mean(tot.pay > 1000 & pay < 1200)
[1] 0.81652  $ aprx P(Total pay btw $1m & 1.2m)

summary(pay)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  825.1  1077.9  1125.5  1125.6  1173.0  1428.6

hdr = "Payouts for 200 Claims"
hist(pay, prob=T, col="skyblue", xlab="$1000s", main=hdr)
 abline(v = c(1000, 1200), col="red")


问:
Two ramified covers $\Rightarrow$ reducible ramification divisor?
Let $X,X',X''$ be algebraic varieties and let $X''\to X'$ and $X'\to X$ be two ramified covers. Is the ramification divisor of the composition $X''\to X'\to X$ reducible?</p>

答: Let $X=X'=X''=\Bbb{A}^1$ be the affine line, and let the two covers be $z \mapsto z^2$. Their composite is the (ramified) cover $z \mapsto z^4$, and the ramification divisor of this morphism is supported at $0$ only. This shows in particular that the ramification divisor can be irreducible.</p>

ContentLicense: CC BY-SA 3.0


问:
A 2-digit number is one more than 6 times the sum of its digits. If the digits are reversed, the new number is 9 less than the original number.
How to teach kids to solve this problem?

A 2-digit number is one more than 6 times the sum of its digits. If the digits are reversed, the new number is 9 less than the original number. Find the original number.

Body:
$$10 a + b - 1 = 6 (a + b)$$

$$10 b + a + 9 = 10 a + b$$

Solve those.


问:
Line of Intersection of 2 Planes
I'm trying to understand how to find line of intersection between 2 planes. In almost every explanation I've seen, it involves removing one of the unknowns by substituting it with 0.

For example in this video https://www.youtube.com/watch?v=q2FO675dmyM

```<code>For the planes:
x + 2y - 4z = 16
2x - y + 3z = 6

Substituting x = 0,
We get the equations
2y - 4z = 16
2y + 6z = 12
</code>```

And eventually we can solve the equation.

But what if we cannot substitute x = 0?

```<code>Consider the planes:
x + y = 8
x = 5
</code>```

In this case, the line of intersection will never touch x = 0 or y = 0.
I know that with this equation, we can already easily solve this question as there is only 2 unknowns (technically 1 unknown since x = 5).

But I'm just curious if it is possible for such a case to occur with 3 unknowns.
Where we have 3 unknowns but the line of intersection will never pass through one of the axis.

答:
We are trying to find nonzero $x_1, x_2, y_1, y_2, z_1, z_2$ and real $c_1, c_2$ such that for the system of equations

$$\begin{equation}\begin{cases}
x_1x+y_1y+z_1z=c_1 \\
x_2x+y_2y+z_2z=c_2
\end{cases}\end{equation}$$

such that this is equivalent to $x=a$ for some nonzero $a$.

This notation is rather clunky, so we can use matrices to represent this:

$$\left(\begin{array}{rrr|r}
x_1 & y_1 & z_1 & c_1 \\
x_2 & y_2 & z_2 & c_2
\end{array}\right)$$

Now we can begin using Gaussian Elimination to solve this system of equations.

We first divide the first row by $x_1$:

$$\left(\begin{array}{rrr|r}
1 & \frac{y_1}{x_1} & \frac{z_1}{x_1} & \frac{c_1}{x_1} \\
x_2 & y_2 & z_2 & c_2
\end{array}\right)$$

Now we get rid of the $x_2$ term by subtracting $x_2$ copies of row 1 from row 2:

$$\left(\begin{array}{rrr|r}
1 & \frac{y_1}{x_1} & \frac{z_1}{x_1} & \frac{c_1}{x_1} \\
0 & y_2-\frac{y_1x_2}{x_1} & z_2-\frac{z_1x_2}{x_1} & c_2-\frac{c_1x_2}{x_1}
\end{array}\right)$$

Now divide row 2 by $y_2-\frac{y_1x_2}{x_1}$:

$$\left(\begin{array}{rrr|r}
1 & \frac{y_1}{x_1} & \frac{z_1}{x_1} & \frac{c_1}{x_1} \\
0 & 1 & \frac{z_2-\frac{z_1x_2}{x_1}}{y_2-\frac{y_1x_2}{x_1}} & \frac{c_2-\frac{c_1x_2}{x_1}}{y_2-\frac{y_1x_2}{x_1}}
\end{array}\right)$$

Finally subtract row 1 by $\frac{y_1}{x_1}$ copies of row $2$:

$$\left(\begin{array}{rrr|r}
1 & 0 & \frac{z_1}{x_1}-\frac{y_1\left(z_2-\frac{z_1x_2}{x_1}\right)}{x_1\left(y_2-\frac{y_1x_2}{x_1}\right)} & \frac{c_1}{x_1}-\frac{y_1\left(c_2-\frac{c_1x_2}{x_1}\right)}{x_1\left(y_2-\frac{y_1x_2}{x_1}\right)} \\
0 & 1 & \frac{z_2-\frac{z_1x_2}{x_1}}{y_2-\frac{y_1x_2}{x_1}} & \frac{c_2-\frac{c_1x_2}{x_1}}{y_2-\frac{y_1x_2}{x_1}}
\end{array}\right)$$

Let's look at the first row, which equivalently is:

$$x+\left(\frac{z_1}{x_1}-\frac{y_1\left(z_2-\frac{z_1x_2}{x_1}\right)}{x_1\left(y_2-\frac{y_1x_2}{x_1}\right)}\right)z=\frac{c_1}{x_1}-\frac{y_1\left(c_2-\frac{c_1x_2}{x_1}\right)}{x_1\left(y_2-\frac{y_1x_2}{x_1}\right)}$$

Recall that we want $x=a$ for nonzero $a$, so we need:

$$\frac{z_1}{x_1}-\frac{y_1\left(z_2-\frac{z_1x_2}{x_1}\right)}{x_1\left(y_2-\frac{y_1x_2}{x_1}\right)}=0$$
$$\frac{c_1}{x_1}-\frac{y_1\left(c_2-\frac{c_1x_2}{x_1}\right)}{x_1\left(y_2-\frac{y_1x_2}{x_1}\right)}\neq0$$

Both look awfully complicated so we can clean up both equations.

Working on the first equation:

$$\frac{z_1}{x_1}=\frac{y_1\left(z_2-\frac{z_1x_2}{x_1}\right)}{x_1\left(y_2-\frac{y_1x_2}{x_1}\right)}$$
$$z_1=\frac{y_1\left(z_2-\frac{z_1x_2}{x_1}\right)}{y_2-\frac{y_1x_2}{x_1}}$$
$$z_1\left(y_2-\frac{y_1x_2}{x_1}\right)=y_1\left(z_2-\frac{z_1x_2}{x_1}\right)$$
$$z_1(y_2x_1-y_1x_2)=y_1(z_2x_1-z_1x_2)$$

As for the second equation:

$$\frac{c_1}{x_1}\neq\frac{y_1\left(c_2-\frac{c_1x_2}{x_1}\right)}{x_1\left(y_2-\frac{y_1x_2}{x_1}\right)}$$
$$c_1\neq\frac{y_1\left(c_2-\frac{c_1x_2}{x_1}\right)}{y_2-\frac{y_1x_2}{x_1}}$$
$$c_1\left(y_2-\frac{y_1x_2}{x_1}\right)\neq y_1\left(c_2-\frac{c_1x_2}{x_1}\right)$$
$$c_1(y_2x_1-y_1x_2)\neq y_1(c_2x_1-c_1x_2)$$

Combining,

$$\begin{equation}\begin{cases}
z_1(y_2x_1-y_1x_2)=y_1(z_2x_1-z_1x_2) \\
c_1(y_2x_1-y_1x_2)\neq y_1(c_2x_1-c_1x_2)
\end{cases}\end{equation}$$

Considering there are $8$ different constants and $2$ equations, there should be infinitely many $x_1,x_2,y_1,y_2,z_1,z_2,c_1,c_2$ satisfying both equations.

An example of such a system of equations is:

$$\begin{equation}\begin{cases}
x+y+z=2 \\
x+2y+2z=3
\end{cases}\end{equation}$$

Which is equivalent to

$$\begin{equation}\begin{cases}
x=1 \\
y+z=1
\end{cases}\end{equation}$$

Title: Line of Intersection of 2 Planes


问:
Finding the “root” of a monotone function (in the sense of composition)
Let $f:[0,\infty)\rightarrow [0,\infty) $ be a smooth and monotone function s.t $f(0)=0$. Let $N\in\mathbb{N}$. Can we find a function $g: [0,\infty) \rightarrow [0,\infty) $ s.t $g\circ\cdots\circ g$ ($g$ composed with itself $N$ times) equals $f$?

Can we say something about $g$‘s monotonicity? Its smoothness? I cannot come up with any basic answers. Thanks in advance to the helpers.

答:
Thank you everyone, @Hanno posted a collection of references with a paper which addresses this problem:

http://pldml.icm.edu.pl/pldml/element/bwmeta1.element.desklight-3881187f-67b6-4d8e-a977-31e52dd4e414/c/apm22_2_09.pdf


问:
Defect version of Fatou's Lemma (repost)
Someone recently posted the following question, but it was deleted.  Since it was an interesting question and I didn't know the answer, I'm reposting it here.

The question: is the following limit theorem true?

Suppose $(X,\mathcal{A},\mu)$ is a measure space, $f \in L^{1}(X)$, and $\{f_{n}\}_{n \in \mathbb{N}} \subseteq L^{1}(X)$ with $f_{n} \geq 0$ a.e. for all $n \in \mathbb{N}$ and $f_{n} \to f$ pointwise a.e.  If $\{f_{n}\}_{n \in \mathbb{N}}$ is bounded in $L^{1}(X)$, then $f_{n} \to f$ in $L^{1}(X)$ if and only if $\int_{X} f_{n} \, d \mu \to \int_{X} f \, d \mu$.

The answer turns out to be yes.  This isn't completely obvious, and it's interesting since I'm aware of at least one very closely related result in the literature.  (Namely, if $\{\mu_{n}\}_{n \in \mathbb{N}}$ are finite signed Borel measures on a compact metric space $X$ and $\mu_{n} \overset{*}{\rightharpoonup} \mu$, then $\|\mu_{n} - \mu\| \overset{*}{\rightharpoonup} 0$ if and only if $\|\mu_{n}\|(X) \to \|\mu\|(X)$.)

答:
The "only if" direction is immediate so I'll only treat the "if" direction.

A sequence in $L^{1}(X)$ converges if and only if it converges locally in measure, it's uniformly integrable, and it's tight.  This follows from the Vitali Convergence Theorem, the Arzela-Ascoli Theorem analogue for $L^{1}$.

Local convergence in measure: $\{f_{n}\}_{n \in \mathbb{N}}$ converges locally in measure (i.e. it converges in measure on finite measure sets) since it's pointwise convergent.

Tightness: $\{f_{n}\}_{n \in \mathbb{N}}$ is tight if and only if, for each $\epsilon > 0$, there is a $N_{\epsilon} \in \mathbb{N}$ and a measurable $A_{\epsilon}$ with $\mu(A_{\epsilon}) < \infty$ such that
\begin{equation*}
\int_{X \setminus A_{\epsilon}} f_{n}(x) \, \mu(dx) < \epsilon.
\end{equation*}

To see this is so, recall that $\{f\}$ is tight so there is an $A_{\epsilon}$ with $\mu(A_{\epsilon}) < \infty$ such that
\begin{equation*}
\int_{X \setminus A_{\epsilon}} f(x) \, \mu(dx) < \epsilon/2.
\end{equation*}
Fatou's Lemma thus gives
\begin{equation*}
\int_{X} f(x) \, \mu(dx) - \epsilon/2 \leq \int_{A_{\epsilon}} f(x) \, \mu(dx) \leq \liminf_{n \to \infty} \int_{X \setminus A_{\epsilon}} f_{n}(x) \, \mu(dx)
\end{equation*}
and so
\begin{equation*}
\liminf_{n \to \infty} \int_{A_{\epsilon}} f_{n}(x) \, \mu(dx) \leq \int_{X} f(x) \, \mu(dx) - \limsup_{n \to \infty} \int_{X \setminus A_{\epsilon}} f_{n}(x) \, \mu(dx) \leq \epsilon/2.
\end{equation*}
Therefore, there is a $N_{\epsilon} \in \mathbb{N}$ with the desired property.

Uniform integrability:  We proceed by contradiction.  $\{f_{n}\}_{n \in \mathbb{N}}$ is not tight if and only if there is a sequence of measurable sets $\{A_{j}\}_{j \in \mathbb{N}}$, a subsequence $(n_{j})_{j \in \mathbb{N}} \subseteq \mathbb{N}$, and a $\nu > 0$ such that
\begin{equation*}
\inf \left\{ \int_{A_{j}} f_{n_{j}}(x) \, \mu(dx) \, \mid \, j \in \mathbb{N} \right\} \geq \nu, \quad \lim_{j \to \infty} \mu(A_{j}) = 0.
\end{equation*}
Passing to a further subsequence, we see that we can assume that $\mu(A_{j}) \leq 2^{-j}$ for all $j \in \mathbb{N}$, hence $\sum_{j = 1}^{\infty} \mu(A_{j}) < \infty$.  Therefore, by the Borel-Cantelli Lemma, $\mu(\limsup_{j \to \infty} A_{j}) = 0$ (def. of lim-sup of sets can be found here), and an immediate consequence is
\begin{equation*}
\lim_{j \to \infty} f_{n_{j}} \chi_{X \setminus A_{j}} = f \quad \text{a.e.}
\end{equation*}
From this, we see that
\begin{equation*}
\int_{X} f(x) \, \mu(dx) \leq \liminf_{j \to \infty} \int_{X \setminus A_{j}} f_{n_{j}}(x) \, \mu(dx) \leq \int_{X} f(x) \, \mu(dx) - \nu.
\end{equation*}
This contradicts the choice of $\nu > 0$.

(As Ramiro pointed out, the non-negativity of $\{f_{n}\}_{n \in \mathbb{N}}$ isn't necessary.  The proof above is readily adapted to that setting.)


问:
Prove $\lim_{n\to\infty}\int_0^1\sqrt{x^n+1}~dx=1$
I was messing around on my calulator and I noticed that for large values of $n$, $\int_0^1\sqrt{x^n+1}~dx\to1$. I would like to prove this. So, this is my question:

Prove $$\lim_{n\to\infty}\int_0^1\sqrt{x^n+1}~dx=1$$
Geometrically this seems to make sense; for all values of $x$ in the interval $[0,1)$ it is obvious that as as $n$ increases then $x^n+1$ is very close to $1$, but I would like a formal proof.

Many thanks for your help.

答:
The point is we can permute intergral and limit (see below):

$$\lim_{n\to\infty}\int_0^1\sqrt{x^n+1}~dx=\int_0^1\lim_{n\to\infty}\sqrt{x^n+1}~dx = \int_0^1 1 dx = 1$$

Justification of the permutation limit - integral :

The dominated convergence theorem states that :

If $(f_n)$ are functions such that :

$f_n$ are piecewise continuous functions

$(f_n)$ simply converges to $f$ which is piecewise continuous

$\exists \varphi : I\mapsto \mathbb{R}^{+}$ piecewise continuous and integrable such that $\forall n\in \mathbb{N}, |f_n|\leqslant \varphi$

Then :
The $f_n$ and $f$ are integrable on $I$ and $$\int_I f_n \rightarrow \int_I f$$

Here you can use $\varphi :x\mapsto 2$, which is of course integrable.


问:
Standard equation of an ellipse given both foci
I want to find the standard equation of an ellipse $E$ given its two foci points $p_1=(x_1, y_1)$ and $p_2=(x_2,y_2)$. My ellipse is shifted in the x and y-direction to a new center point $(x_e,y_e)$. Note that the points $p_1$ and $p_2$ are aligned so that the ellipse will be rotated by some angle $\theta$. I know the standard equation for an ellipse that is aligned with the x axis and is centered at $(0,0)$ is
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$

I searched the internet for a broader definition that does include shifting, rotation, and does only use the coordinates of the points $p_1$ and $p_2$ and some distance $c$ so that every point $q$ on the border of the ellipse is of distance $c=|p_1q| + |p_2q|$ to $q$. I was surprised that I did not find any. Can you help me out?

Edit

From what I am understanding I can calculate the major/minor axis $a,b$ as follows:

$$a=\frac{1}{2}c$$
$$b=\sqrt{a^2-f^2}$$

Where $f$ is the distance from $(x_e,y_e)$ to the focus point $p_1$:
$$f=\sqrt{(p_1^x-x_e)^2+(p_1^y-y_e)^2}$$

I also found that the standard form of an ellipse that is rotated by an angle $\theta$ from the x-axis is

$$\frac{(x\cos{\theta}+y\sin{\theta})^2}{a^2}+\frac{(x\sin{\theta}-y\cos{\theta})^2}{b^2}=1$$

However, this representation does not include the shift to the new center point $(x_e,y_e)$ and I am not sure if the equations for $a,b$ hold for the rotated ellipse case.

答:
There is a general equation of an ellipse so that every point $(x,y)$ implicitly satisfies a quadric as was mentioned in the comments. However, in my specific use case, I can substitute $(x-x_e)$ and $(y-y_e)$ in the formula that I posted in the original question. Again, thanks for the useful comments on this. I conclude the following general form of an ellipse using only the points $p_1,p_2$ the angle $\theta$ (which can be derived from $p_1p_2$) and the distance $c$.

Start by calculating the minor/major as already suggested in the question
$$a=\frac{1}{2}c$$
$$b=\sqrt{a^2-f^2}$$

Where $f$ is the distance from $(x_e,y_e)$ to the focus point $p_1$:
$$f=\sqrt{(p_1^x-x_e)^2+(p_1^y-y_e)^2}$$

Now the general form for all points $(x,y)$ on the ellipse satisfies the following equation
$$\dfrac {((x-x_e)\cos(\theta)+(y-y_e)\sin(\theta))^2}{a^2}+\dfrac{((x-x_e) \sin(\theta)-(y-y_e) \cos(\theta))^2}{b^2}=1$$


问:
Diameter of a compact in a metric space : is it reached?
Context :

In a metric space $(E,d)$, we define the diameter of a compact party $K$ as being $\delta =\sup \{d(x,y); x\in K, y\in K\}$.

Question :
Is this $\sup$ a $\max$, since $K$ is compact?

I think yes.

My proof :

$\delta =\sup \{d(x,y); x\in K, y\in K\}$ so there exists two sequences $(x_n)$ and $(y_n)$ of elements of $K$ such that $d(x_n,y_n) \rightarrow \delta$.

As elements are in $K$ which is compact, we can extract $(u_{\varphi(n)})$ from $(u_n)$ which converges to $x\in K$. And as $(y_{\varphi(n)})$ is a sequence of elements of $K$, we can extract $(y_{\varphi(\psi(n))})$ which converges to $y\in K$. So $(x_{\varphi(\psi(n))})$ also converges to $x\in K$.

By continuity of the distance, we get that $d(x_{\varphi(\psi(n))}, y_{\varphi(\psi(n))}) \rightarrow d(x,y)$. By unicity of the limit, we have that $\delta = d(x,y)$.

But I'm not sure of what I've done.

Is is right?

Thanks by advance.

答:
The map

$$\begin{array}{l|rcl}
d : & E \times E & \longrightarrow & \mathbb R \\
    & (x,y) & \longmapsto & d(x,y) \end{array}$$

is continuous when $E \times E$ is endowed with the distance
$$D((x_1,y_1), (x_2,y_2)) = d(x_1,x_2) + d(y_1,y_2).$$

$K \times K$ is compact as it is the product of two compact subsets of $E$. Therefore $f$ attains its supremum on $K \times K$, which proves the desired result.


问:
isometry group of $\mathbb{R}^n$
How to calculate the isometry group of $\mathbb{R}^n$? And show that $G$ is generated by rotations, reflections and translations?

Can someone help me?

答:
Here's a guideline:

Show that $f:\Bbb R^n \longrightarrow \Bbb R^n$ is an isometry if and only if for all translations $T:\Bbb R^n \longrightarrow \Bbb R^n$, $T\circ f$ is an isometry.

Show that if $f:\Bbb R^n \longrightarrow \Bbb R^n$ is an isometry with $f(0) = 0$, then $f$ is linear.

What can you say about linear isometries of $\Bbb R^n$? What do you know about these transformations and the matrices that represent them?

Title: isometry group of $\mathbb{R}^n$


答:
I think I came in 5 minutes too late,  but since my answer is already typed up,  here it goes:

As it is clear by  the interesting  comments,  this question is not a walk in the park,  so let me offer my two cents by
addressing the simpler problem in which $f$ is assumed to be strictly increasing and onto $[0,\infty )$.

I will  ignore
the smoothness of $f$, hence also giving up on the question of the smoothness of $g$, and concentrate instead on an
increasing homeomorphism
$$
  f: [0,\infty )\to   [0,\infty ).
  $$

Hopefully the argument below can be refined by bringing the smoothness aspects back in to the picture.

Considering any  increasing homeomorphism  $h:{\mathbb  R}\to (0,\infty )$, we may replace $f$ by $h^{-1}fh$, and hence think of $f$ as
an increasing homeomorphism
$$
  f:{\mathbb  R}\to{\mathbb  R}.
  $$

Case 1: If $f(x)>x$, for all $x\in  {\mathbb  R}$, then there exists a homeomorphism $\varphi :{\mathbb  R}\to {\mathbb  R}$, such that
$$
  \varphi f\varphi ^{-1}(x) = x+1, \quad\forall x\in  R,
  $$
and therefore  $f$ has an increasing  $n^{\text{th}}$ root $g$, given by
$$
  g(x) = \varphi ^{-1}\big (\varphi (x)+1/n\big ).
  $$

Proof:  Choose any $x_0$ in ${\mathbb  R}$, and observe that for every $x$ in ${\mathbb  R}$, there exists a unique $n=n_x\in {\mathbb  Z}$, such that
$$
  f^n(x)\in \big[x_0,f(x_0)\big).
  $$
Letting $L$ denote the length of the interval $[x_0,f(x_0)\big)$,
we may then write $f^n(x)=x_0+aL$, for a unique $a=a_x\in [0,1)$.  The desired homeomorphism can then be taken to be
$\varphi (x) = n_x+a_x$.  QED

Case 2: $f(x)<x$, for all $x\in  {\mathbb  R}$.  This follows as an easy generalization of  case (1) with the obvious modifications (e.g.,
replacing $x+1$ by $x-1$).

To treat the general case,  let
$$
  F=\{x\in {\mathbb  R}: f(x)=x\},
  $$
and let the complement of $F$ be written as the disjoint union of open intervals
$$
  {\mathbb  R}\setminus F = \bigcup_{k=1}^\infty  J_k.
  $$

Clearly each $J_k$ is invariant under $f$, and since $J_k$ is order-homeomrphic to ${\mathbb  R}$, we may apply either case (1)
or (2) to produce an $n^{\text{th}}$ root $g_k$, on each open interval,  and then patch them together to obtain a global $n^{\text{th}}$ root $g$.

Title: Finding the “root” of a monotone function (in the sense of composition)

Body:
Let $f:[0,\infty)\rightarrow [0,\infty) $ be a smooth and monotone function s.t $f(0)=0$. Let $N\in\mathbb{N}$. Can we find a function $g: [0,\infty) \rightarrow [0,\infty) $ s.t $g\circ\cdots\circ g$ ($g$ composed with itself $N$ times) equals $f$?

Can we say something about $g$‘s monotonicity? Its smoothness? I cannot come up with any basic answers. Thanks in advance to the helpers.


答:
Since this tools may be advanced for you (as you wrote in the comment) I just would like to give you a counterexample that rises the main problem here: it is not true that you can, in general, exchange limit and integral, that formally means: $$\lim _{n \rightarrow \infty} \int_{a}^{b} f_{n}(x) d x \neq \int_{a}^{b} \lim_{n \rightarrow \infty}f_n(x) d x$$

Just as an example try to compute the followings: $$\lim _{n \rightarrow \infty} \int_{-n}^{n} \frac{1}{2n} d x =1$$ and $$\int_{-n}^{n} \lim _{n \rightarrow \infty} \frac{1}{2n} d x =0$$ Clearly you get they are not equal.

In general, to have such limit theorems either you need uniform convergence of a sequence of functions if you are dealing with Riemann integrable functions, or you need more powerful tools like Lebesgue integral and the Lebesgue theory of integration.

What you are applying is what is known as the Lebesgue Dominated Convergence Theorem. All what you need is to consider the sequence of functions (with values in $[0,1]$) given by $$f_n(x)=\sqrt{x^n+1}$$ and noticing that we have the following pointwise convergence where $f(x)=0$: $$f_n(x) \rightarrow f(x) $$ and the following integrability condition $$|f_n| \le g \text{  for all  } n \in \mathbb{N}$$ where $g$ is an integrable function (in this case just pick $g$ equal the constant function 2 on $[0,1]$, that has clearly finite integral and dominates all the $|f_n|$). Then, since all this holds in your example, you can exchange limit ans integral and conclude the result.

You could also apply what is known as the Monotone Convergence Theorem (a bit modified) in this case for monotone decreasing sequences.

Title: Prove $\lim_{n\to\infty}\int_0^1\sqrt{x^n+1}~dx=1$

Body:
I was messing around on my calulator and I noticed that for large values of $n$, $\int_0^1\sqrt{x^n+1}~dx\to1$. I would like to prove this. So, this is my question:

Prove $$\lim_{n\to\infty}\int_0^1\sqrt{x^n+1}~dx=1$$
Geometrically this seems to make sense; for all values of $x$ in the interval $[0,1)$ it is obvious that as as $n$ increases then $x^n+1$ is very close to $1$, but I would like a formal proof.

Many thanks for your help.


问:
Point-wise and Norm Convergence of Vectors in a finite dimensional space
I'm trying to prove the theorem, that states, that if I have a normed vector space with a finite dimension (so that each vector I can express as a linear combination $$ \vec{v}=a\vec{e}_1+b\vec{e}_2+c\vec{e}_3+\cdots+q\vec{e}_n $$ where the vectors $\{\vec{e_k}\}_{k=1}^{n} $ are linearly independent), and I have some norm defined there, and I have a sequence of vectors $$ \{\vec{v_k}\}_{k=1}^{\infty} $$ that converges in norm to some vector $\vec{V}$, then it also converges to it Point-wisely, that is, if
$$\vec{v}_k=a_k\vec{e}_1+b_k\vec{e}_2+c_k\vec{e}_3+\cdots+q_k\vec{e}_n, $$ and
$$\vec{V}=a\vec{e}_1+b\vec{e}_2+c\vec{e}_3+\cdots+q\vec{e}_n, $$ and
$$\lim_{k\rightarrow\infty} \lVert\vec{v}_k-\vec{V}\rVert=0,$$ then also
$$\begin{align*}
\lim_{k\rightarrow\infty} |a_k-a|&=0,\\
\lim_{k\rightarrow\infty} |b_k-b|&=0,\\
&\,\vdots\\
\lim_{k\rightarrow\infty} |q_k-q|&=0.
\end{align*}$$

Thanks.

答:
You have
$$
\| \vec{v_k} - \vec V \|^2 = |a_k - a|^2 + |b_k - b|^2 + ... + |q_k - q|^2 \ge |a_k - a|^2
$$
for instance, so that $0 \le |a_k - a| \le \| \vec{v_k} - V \| \to 0$, so that you can use a so-called sandwich theorem to conclude that your limit is zero. With a better notation for the components of $v_k$, something like $v_{k_i}, 1 \le i \le n$, the proof would be more elegant though.

Note that this statement is not just an implication, it's an equivalence : a vector sequence in a finite dimensional space whose components converge is always converging to the vector with components being the limits of the components of the sequence, i.e.
$$
a_k \to a, b_k \to b, \dots, q_k \to q \quad \Longleftrightarrow \quad v_k \to V.
$$

EDIT : Oh well. This isn't much general, I admit, 'cause I tend to spare the details when it's getting late, so I didn't notice you were in a context of a general norm. The only way to get an upper bound function of $\|v_k - V\|$ is to use norm equivalence on finite-dimension space, so I guess I can't make any better than what's been said.


问:
Find an orthogonal matrix $M$ and diagonal matrix $D$ such that $ M^{T}AM=D$
Hello can you help me with that. I tried all my best but I don't know how to find $M$ orthogonal matrix
For the matrix $A=
\begin{bmatrix}
2 & 2 & -2 \\
2 & -1 & 4 \\
-2 & 4 & -1
\end{bmatrix}$
Find an orthogonal matrix $M$ and diagonal matrix $D$ such that $ M^{T}AM=D$

At first I tried to find Diagonal matrix  using eigenvalues

$
\begin{bmatrix}
2 - \lambda & 2 & -2 \\
2 & -1- \lambda & 4 \\
-2 & 4 & -1- \lambda
\end{bmatrix}$

$-\lambda^3 +27\lambda -54 = -(\lambda-3)(\lambda+6)(\lambda-3)$

$\lambda_1 = 3$ $\lambda_2=3$ $\lambda_3 = 6$

$D=
\begin{bmatrix}
3 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & -6
\end{bmatrix}$

Okay I found diagonal matrix and I also know that $A^{T} =
\begin{bmatrix}
2 & 2 & -2 \\
2 & -1 & 4 \\
-2 & 4 & -1
\end{bmatrix}$

$A^{-1}=
\begin{bmatrix}
\frac{5}{18} & \frac{1}{9} & -\frac{1}{9} \\
\frac{1}{9} & \frac{1}{9} & \frac{2}{9} \\
\frac{1}{9} & \frac{2}{9} & \frac{1}{9}
\end{bmatrix}$

答: Look for the eigenvectors corresponding to the eigenvalue $3$. You will have to solve the system$$\left\{\begin{array}{l}2x+2y-2z=3x\\2x-y+4z=3y\\-2x+4y-z=3z,\end{array}\right.$$which is equivalent to$$\left\{\begin{array}{l}-x+2y-2z=0\\2x-4y+4z=0\\-2x+4y-4z=0.\end{array}\right.$$Each of these equations is a multiple of the other two. So, you really have only one equation to deal with, say, $-x+2y-2z=0$. Take two unit vectors $(x,y,z)$ which satisfy this condition and which are orthogonal, such as $u=\left(\frac2{\sqrt5},\frac1{\sqrt5},0\right)$ and $v=\left(\frac2{3\sqrt5},-\frac4{3\sqrt5},-\frac5{3\sqrt5}\right)$. Now, take an eigenvector corresponding to the eigenvalue $-6$, which is also an unit vector; you can take, say, $w=\left(\frac13\,-\frac23,\frac23\right)$. Then you can take the matrix$$M=\begin{bmatrix}\frac2{\sqrt5}&\frac2{3\sqrt5}&\frac13\\\frac1{\sqrt5}&-\frac4{3\sqrt5}&-\frac23\\0&-\frac5{3\sqrt5}&\frac23\end{bmatrix},$$whose colmuns are the vectors $u$, $v$; and $w$.


答:
Some quick observations : if $x$ is a root of $P(x)$ then so is $-x$. Thus if $(x^2-x+a)$ is a factor of $P(x)$, so is $(-x)^2-(-x)+a=x^2+x+a.$

Then $(x^2-x+a)(x^2+x+a)=x^4+(2a-1)x^2+a^2$ is a factor of $P(x)$. $\, a^2|36 \,$ $\Rightarrow a \in \{1,2,3,6\}$. (Here @Anwesha1729 observe in their answer that as $P(x)$ takes odd values also for integer $x$, $a$ must be odd.)

Let $P(x)=(x^4+(2a-1)x^2+a^2)(x^4+bx^2+c)$.

Comparing coefficients :
$$b+(2a-1)=5$$
$$a^2+c+(2a-1)b=13$$
$$a^2b+(2a-1)c=20$$
$$a^2c=36$$

It can be readily checked $a=3$ $\, (b=0,c=4)$

Title: Dividing a polynomial with integer coefficients by a quadratic

Body:
In the following question, one of the coefficients of the divisor quadratic polynomial is unknown; so, it cant be factored for performing synthetic division. I tried to perform long division hoping to get some clues but ran into a conflict.

Appreciate your help. Thanks!!

Here is the question:
$x^2 -x +a$  evenly divides $x^8 + 5x^6 + 13x^4 + 20x^2 +36$. Determine positive integer $a$.

A few things I noted are:

$a$ must be a factor of 36 so, I am thinking $a$ would be from $\{1, 2, 3, 4, 6, 9, 12, 18, 36\}$

All the coefficients of the dividend polynomial are integers and degrees of all the terms are even. Not sure how to use this info.

$P(x) = Q(x)(x^2 -x +a) + mx + n$.  Since the remainder is zero, $m$ and $n$ are zero. When I tried long division, I got two equations; but the value of $a$ would come as imaginary which conflicts with the first observation and the question asks for real integer $a$.


问:
Mistake when substituting constraint $4x^2+y^2=1$ into a function $f(x,y)=x^2+y^2$ in extrema problem
Consider the function $f(x,y)=x^2+y^2$ under the constraint $4x^2+y^2=1$. The extrema of $f$ under that constraint can be easily found with Lagrange multipliers, and they are attained for $(0,1)$, $(0,-1)$ for maximum and $(1/2,0)$, $(-1/2,0)$ for minimum; however, if we isolate $y^2=1-4x^2$ and we substitute it in the function, we get a wrong result (in particular, the one variable function $g(x)=1-3x^2$ obtained has only a maximum for $x=0$ and Weierstrass theorem assures that the are both maximum and minimum for $f$). Can someone explain me why this fails?

答:
Solving

$$
\min_{x,y}(\max_{x,y})(x^2+y^2)\ \ \text{s. t.}\ \ \ 4x^2+y^2=1
$$

using the Lagrange multipliers method, reduces to determine the stationary points for

$$
\nabla(x^2+y^2)+\lambda\nabla(4x^2+y^2-1)=0
$$

by solving for $x,y,\lambda$

$$
\cases{
2x+8x\lambda=0\\
2y+2y\lambda=0\\
4x^2+y^2-1=0
}
$$

giving as solutions the four tangency points between the level curves for $z=x^2+y^2$ and the ellipse $4x^2+y^2-1=0$. Those points can be depicted in the attached plot.

Now by making the substitution $y = 1-4x^2$ into $z = x^2+y^2$ giving $z=1-3x^2$ we are searching extrema along the $x$ axis at $x^*=0$ with value $1$ and the corresponding $y^*$ we obtain from the restriction $4(x^*)+(y^*)^2=1$.   Also with the substitution $x^2=\frac{1-y^2}{4}$ we are searching extrema along the $y$ axis with value $\frac 14$.


问:
When proving hyperbolic identities why do we add one or two
Example:

$\cosh^2x+\sinh^2x=\cosh2x$

Proof:
$$\frac{1}{2}(e^x+e^{-x})^2+\frac{1}{2}(e^x-e^{-x})^2$$
Where does this two's come from?
$$\frac{1}{4}(e^{2x}+e^{-2x}+2)+\frac{1}{4}(e^{2x}+e^{-2x}-2)$$

Body:
The two's come from expanding the brackets, i.e $(x+y)^2=x^2+y^2+\color{red}{2}xy$.

Since by definition we have

$$\cosh(x)=\frac{e^x+e^{-x}}{2}\quad\text{and}\quad\sinh(x)=\frac{e^{x}-e^{-x}}{2}$$

Then

$$\cosh^2(x)+\sinh^2(x)=\big(\frac{e^x+e^{-x}}{2}\big)^2+\big(\frac{e^x-e^{-x}}{2}\big)^2$$
$$ =\frac{1}{4}(e^x+e^{-x})^2+\frac{1}{4}(e^x-e^{-x})^2$$
$$=\frac{1}{4}(e^{x+x}+e^{-x-x}+2e^{x-x})+\frac{1}{4}(e^{x+x}+e^{-x-x}-2e^{x-x})$$
$$=\frac{1}{4}(e^{2x}+e^{-2x}+2)+\frac{1}{4}(e^{2x}+e^{-2x}-2)$$
$$=\frac{1}{4}(2e^{2x}+2e^{-2x})$$
$$=\frac{e^{2x}+e^{-2x}}{2}=\cosh(2x)$$


问:
$(\sqrt{n}\cos(nx))_{n\in\mathbb{N}}$ is a divergent sequence
I want to prove that $(\sqrt{n}\cos(nx))_{n\in\mathbb{N}}$ is a divergent sequence for all $x\in\mathbb{R}$. I would like to know if there are any mistakes or ways to improve my proof. Thank you.
(Note: I use the fact that $\lim_{n\to\infty}\cos(nx_0)\neq 0$ as a given lemma).

My attempt:

Let $x_0\in\mathbb{R}$: I already know that $\lim_{n\to\infty}\cos(nx_0)\neq 0$ so that means there exists $\bar{\varepsilon}>0$ such that for all $N\in\mathbb{N}$ there is $n\geq N$ such that $|\cos(nx_0)-0|=|\cos(nx_0)|\geq\bar{\varepsilon}$ so the set $A:=\{n\in\mathbb{N}:|\cos(nx_0)|\geq\bar{\varepsilon}\}$ is infinite and we can build an increasing sequence $(n_j)_{j=0}^{\infty}$ by setting $n_j:=\min\{n\in A:n\neq n_i, i<j\}$ and we have $|\sqrt{n_j}\cos(n_jx_0)|\geq\sqrt{n_j}\bar{\varepsilon}>0$ for all $j\in\mathbb{N}$. Now if we let $M>0$ we know there will be some $J\in\mathbb{N}$ such that $\sqrt{n_J}>\frac{M}{\bar{\varepsilon}}$ so $\sqrt{n_J}\bar{\varepsilon}>M$ which implies $|\sqrt{n_J}\cos(n_Jx_0)|>M$ and this is true for all $j\geq J$ so $|\sqrt{n_j}\cos(n_jx_0)|\xrightarrow[]{n\to\infty}+\infty$ which in turn implies that $\sqrt{n_j}\cos(n_jx_0)$ cannot converge too (i.e. it either goes to $\pm\infty$ or it does not have a limit).

答: If you already know that for every $x$, the sequence $\cos(nx)$ does not tend to $0$, then the proof is over. Indeed, the fact that $(\sqrt{n}\cos(nx))$ converges directly implies that $\cos(nx)$ must tend to $0$...


问:
Showing that there exists a certain function
We have the following functions:

$f(x) = x-\sin(x); \ \ \ \ g(x) = 1-\cos(x)$

I have to prove that there exists a function h such that h(f(x)) = g(x). But I don't exactly how to show that there exists such a function

It looks like I don't need to find the exact function $h$, only to show the existence of one. I already proved that $f(x)$ is bijective. But what about that function $h$ which should exists? What can I do? Can I use the fact that $f$ is bijective?

I also trying to find the derivative of $f(x)$. Yes, I am able to do that with my own knowledge of derivatives (and it is equal to $g(x)$), but I am also trying with the definition of the derivative. I tried the following:
$f(x+h) = x+h - \sin(x+h)$, but then I have write is as $f(x) + Df_x + r(h)$ where $r(h) \to 0 \text{ as } h \to 0$. But how can I rewrite $f(x+h)$ to get in into those terms?

答:
I will try to answer your second query about proving from first principles that
$$\frac{d}{dx}(x-\sin x)=1-\cos x$$
Recall that the derivative, $f'(x)$, of a function $f(x)$, is defined as
$$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$
For $f(x)=x-\sin x$
$$\frac{f(x+h)-f(x)}{h}=\frac{((x+h)-\sin(x+h))-(x-\sin x)}{h}=\frac{h}{h}-\frac{\sin(x+h)-\sin x}{h}$$
$$=1-\frac{\sin(x+h)-\sin x}{h}$$
Hence,
$$f'(x)=\lim_{h\to0}\left(1-\frac{\sin(x+h)-\sin x}{h}\right)=1-\lim_{h\to0}\frac{\sin(x+h)-\sin x}{h}$$
Now to focus on $\lim_{h\to0}\frac{\sin(x+h)-\sin x}{h}$:
$$\lim_{h\to0}\frac{\sin(x+h)-\sin x}{h}=\lim_{h\to0}\frac{\sin x \cos h+\sin h \cos x-\sin x}{h}$$
$$=\lim_{h\to0}\left(\sin x\frac{\cos h-1}{h}+\cos x\frac{\sin h}{h}\right)$$
Now we use the facts that
$$\lim_{h\to0}\frac{\cos h-1}{h}=0,~~~\lim_{h\to0}\frac{\sin h}{h}=1$$
Hence we are left with
$$\lim_{h\to0}\frac{\sin(x+h)-\sin x}{h}=0+\cos x \times 1=\cos x$$
So finally,
$$f'(x)=x-\cos x$$ as required.

For a proof for the $2$ standard limits used above, see the Wikipedia article on the squeeze theorem (the second example): https://en.wikipedia.org/wiki/Squeeze_theorem


问:
Find a function $g$ such that $\int_0^{x^2}tg(t)dt = x + x^2$
This problem $5b$ from Chapter $14$ of Spivak's Calculus:

Find a function $g$ such that $\int_0^{x^2}tg(t)dt = x + x^2$ (Note $g$ is not assumed to be continuous at $0$).

I think this question is impossible? First of all there's no restrictions on $x$, so letting $x = 1$ and $x = -1$, we get $0 = 2$.

So maybe we have to restrict to $x \geq 0$? In that case, we can take the square root:

Find $g$ such that $\forall x \geq 0 : \int_0^{x}tg(t)dt = x + \sqrt{x}$

Let $f(t) = tg(t)$, so we need $\int_0^{x}f(t)dt = x + \sqrt{x}$

Now $\frac{d}{dx} (x + \sqrt{x}) = 1 + \frac{1}{2\sqrt{x}}$. But $f(t) = 1 + \frac{1}{2\sqrt{t}}$ won't work because $1 + \frac{1}{2\sqrt{t}}$ isn't bounded as $t \rightarrow 0$, so the integral won't be defined.

Am I missing something?

答:
If we let:
$$F(x)=\int tg(t)dt$$
then we have:
$$F(x^2)-F(0)=x+x^2$$
now try differentiating both sides:
$$2xF'(x^2)=1+2x$$
but we know that $F'(x)=xg(x)$ and so:
$$2x\left[x^2g(x^2)\right]=1+2x$$
$$2x^3g(x^2)=1+2x$$
$$g(x^2)=\frac{1}{2x^3}+\frac{1}{x^2}$$
now find $g(x)$


问:
Is $\mathcal{N}(\mathcal{N}(x, \Sigma), cI) = \mathcal{N}(x, \Sigma + cI)$?
Is a multi-variate normal distribution with the mean defined by a normal distribution again a normal distribution where the covariance matrices are summed up?

That is, is
$\mathcal{N}(\mathcal{N}(x, \Sigma), cI) = \mathcal{N}(x, \Sigma + cI)$?

Or more general, is
$\mathcal{N}(\mathcal{N}(x, A), B) = \mathcal{N}(x, A+B)$?

答:
If you recognize that $\mathcal{N}(\mu, B)$ has the same distribution as $\mu + Z$ where $Z \sim \mathcal{N}(0, B)$, then it seems that you are asking for the distribution of $Z+X$ where $X \sim \mathcal{N}(x, A)$ and where $Z$ and $X$ are independent. (Otherwise, you should clarify what $\mathcal{N}(\mathcal{N}(x, A), B)$ means.)

Then $Z+X$ is jointly normal, as it is the sum of two independent Gaussian vectors. Its mean is $E[Z+X] = E[Z] + E[X] = x$ and its covariance is $A+B$ because
$$\text{Cov}(Z_i + X_i, Z_j + X_j)
= \text{Cov}(X_i, X_j) + \text{Cov}(Z_i, Z_j) = A_{i,j} + B_{i,j}.$$


问:
An edge that connects 3 vertices
Are there any practical situations that can be represented using a modified definition of an edge - where each edge connects three vertices instead of two?</p>

答:
You're talking about $3$-uniform hypergraphs.  They will arise in a range of contexts, some of which are:

Steiner triple systems are $3$-uniform hypergraphs in which each pair of vertices belongs to exactly one hyperedge.

Triangulations of surfaces (e.g. the plane).

The problem of completing a partial triangulation of the complete tripartite graph $K_{n,n,n}$ has been shown to be NP-complete.  Charlie Colbourn gave a polynomial time reduction of this problem to the problem of completing partial Latin squares, thereby showing that this problem is also NP-complete (ref.).

These contexts are more areas of pure mathematics, and can be rephrased to not be $3$-uniform hypergraphs (however, I'm guessing all $3$-uniform hypergraph problems can be thus rephrased).

Combinatorial designs, such as Steiner triple systems, have applications in software testing.  Hypergraphs in general arise in the study of complex networks, but I haven't seen the particular case of $3$-uniform hypergraphs arise.

However, we don't need applications for mathematics to be interesting; the following quote applies to mathematics as well as physics.

Physics is like sex. Sure you can get some interesting results, but that's not why we do it.  -- Richard Feynman (disputed).

Title: An edge that connects 3 vertices


答:
Note that $\sqrt{1+x^n} \le \sqrt{1+x^n+x^{2n}/4}=1+x^n/2$ for $x \ge 0$; so
$$
1=\int_{0}^{1}dx\le \int_{0}^{1}\sqrt{1+x^n}\;dx \le \int_{0}^{1}\left(1+x^n/2\right)dx=x+\frac{x^{n+1}}{2(n+1)}\bigg\vert_{0}^{1}=1+\frac{1}{2(n+1)}.
$$
Since the endpoints of this interval $\rightarrow 1$ as $n\rightarrow \infty$, the value of the integral must also approach $1$ in this limit.


问:
How many functions $ f: \{1, 2, 3, \dots, 10\} \to \{0,1\}$ satisfy $f(1) + f(2) + \dots + f(10) = 2$?
How many functions $ f: \{1, 2, 3, \dots, 10\} \to \{0,1\}$ have this property: $$f(1) + f(2) + \dots + f(10) = 2.$$

I understand just $2$ functions can be $1$, the rest have to be $0$, in total there are $2^{10}$ functions, but how can I find out how many of them have this property?

答: HINT: As you said in the question, exactly two of the functions values must be $1$, and the other eight must be $0$. How many ways are there to choose $2$ things from a set of $10$ things (i.e., to choose which two function values are to be $1$)?</p>


问:
The hypotenuse of a triangle is shortest when the other two legs are equal. Prove this using the Calculus of Variations.
I started off my proof by of course stating that the different right triangles I would be comparing should have the same area A. I was able to show that what the question is asking is true visually and computationally using the Pythagorean Theorem, and even using the triangle inequality, but I don't really know how to set it up in such a way that I can use the Euler-Lagrange equation in order to prove that this is true.

Equations:

$$\frac{1}{2}xy=A$$

So

$$y(x) = 2A/x \Longrightarrow y'(x) = -\frac{2A}{x^2}$$

答:
$$\min_{xy=2A}\sqrt{x^2+y^2} = \sqrt{\min_{xy=2A}x^2+y^2} $$
and $(x-y)^2\geq 0$ is equivalent to $x^2+y^2\geq 2xy$, with equality attained only at $x=y$. This implies

$$\min_{xy=2A}\sqrt{x^2+y^2} = \sqrt{4A}. $$


问:
A 2-digit number is one more than 6 times the sum of its digits. If the digits are reversed, the new number is 9 less than the original number.
How to teach kids to solve this problem?

A 2-digit number is one more than 6 times the sum of its digits. If the digits are reversed, the new number is 9 less than the original number. Find the original number.

答:
If the kids are too young to be familiar with solving linear equations, one can motivate them to look for pattern.

Adding $9$ to a two-digit number reduces its units digit by $1$. This new units digit was tens digit of original. The required number is of the form $$\overline{(d-1)d}$$

where $d \in \{1,2,\ldots, 9\}$. There is only a few of them and one of them satisfies the other condition.


问:
Finding a closed form for the following integral: $\int_0^1\sqrt{1+x^k}dx$
Is there a nice closed form, for the following integral:

$$\int_0^1\sqrt{1+x^k}dx$$

And how can I derive it?

I have no idea how to get started, thanks for any help. This problem came up when I was working on a number theory problem in my university for some values of $k$.

答:
Well, using IBP you can show that:

$$\mathscr{S}_\text{n}:=\int_0^1\sqrt{1+x^\text{n}}\space\text{d}x=\sqrt{2}-\frac{\text{n}}{2}\int_0^1\frac{x^\text{n}}{\sqrt{1+x^\text{n}}}\space\text{d}x\tag1$$

Using the fact that:

$$\frac{x^\text{n}}{\sqrt{1+x^\text{n}}}=\sqrt{1+x^\text{n}}-\frac{1}{\sqrt{1+x^\text{n}}}\tag2$$

We can write:

$$\mathscr{S}_\text{n}=\sqrt{2}-\frac{\text{n}}{2}\cdot\left\{\underbrace{\int_0^1\sqrt{1+x^\text{n}}\space\text{d}x}_{=\space\mathscr{S}_\text{n}}-\int_0^1\frac{1}{\sqrt{1+x^\text{n}}}\space\text{d}x\right\}\tag3$$

Using the Binomial series we know that we can write:

$$\mathscr{S}_\text{n}=\sqrt{2}-\frac{\text{n}}{2}\cdot\left\{\mathscr{S}_\text{n}-\int_0^1\sum_{\text{k}\ge0}\frac{\left(-1\right)^\text{k}\left(2\text{k}\right)!}{2^{2\text{k}}\left(\text{k}!\right)^2}\cdot x^{\text{nk}}\space\text{d}x\right\}=$$
$$\sqrt{2}-\frac{\text{n}}{2}\cdot\left\{\mathscr{S}_\text{n}-\sum_{\text{k}\ge0}\frac{\left(-1\right)^\text{k}\left(2\text{k}\right)!}{2^{2\text{k}}\left(\text{k}!\right)^2}\int_0^1x^{\text{nk}}\space\text{d}x\right\}=$$
$$\sqrt{2}-\frac{\text{n}}{2}\cdot\left\{\mathscr{S}_\text{n}-\underbrace{\sum_{\text{k}\ge0}\frac{\left(-1\right)^\text{k}\left(2\text{k}\right)!}{2^{2\text{k}}\left(\text{k}!\right)^2}\cdot\frac{1}{1+\text{nk}}}_{:=\space\mathcal{I}_\text{n}}\right\}\tag4$$

Solving for $\mathscr{S}_\text{n}$, gives:

$$\mathscr{S}_\text{n}=\frac{2\sqrt{2}+\text{n}\cdot\mathcal{I}_\text{n}}{2+\text{n}}\tag5$$


问:
Find constants in system of equations
In the below system of equations, find the values of constants $q_1, q_2, q_3 \in \mathbb R$ so that the system has no solution.

$$\begin{cases}
2x&+y-z&+2s &= q_1\\
3x&+y-2z&+s &= q_2\\
x&+y+3z&-s &= q_3
\end{cases}$$

In order to eliminate some of the unknowns, we add the 1st with the 3rd and subtract the 2nd, so we get:
$y+4z = q1+q3-q2$
By similar more ways (multiplying, adding and subtracting) we get some more such equations.

I don't see how this system can have no solution; plus that we are one equation short (3 equations - 4 unknowns).

Any assistance is much appreciated.

EDIT:

(Apologies but I am not familiar with linear algebra!!)

What I believe I must do is the following:

Reduce the unknowns by one (don't see how), so as to have a 3x3 system. Calculate the determinant, for which it must be $D \neq 0$.
Then calculate $D_x, D_y, D_z$ assuming we have eliminated s.
Then, for the system NOT to have a solution, it must be $D=0$ and $D_x, D,y, D_z \neq 0$.

Any further help?

答:
Thanks to Rouché-Capelli theorem we can state that the given system has always solutions.

Indeed the rank of the matrix $A$ of the coefficients is $3$ and the augmented matrix $A|B$ can't have a different rank, even if $q_i$ were all zero.

Title: Find constants in system of equations


问:
Find and classify the spectrum of the operator: $T(x_1, x_2, x_3 \ldots) = (2x_2 - 3x_1, 2x_3 - 3x_2, 2x_4 - 3x_3, \ldots)$
Let $T:l_p \to l_p$ be an operator defined as:
$$T(x_1, x_2, x_3 \ldots) = (2x_2 - 3x_1, 2x_3 - 3x_2, 2x_4 - 3x_3, \ldots)$$

I need to find and classify its spectrum.

First, I noticed that $T = 2S_l - 3I$, when $S_l$ is the shift left operator.

Using this I can say that $||T|| \le 2||S_l|| + 3||I|| = 2 + 3 = 5$. Which means that for $| \lambda | > 5$, $\lambda$ is a regular point.

Then, I tried to find its point spectrum. If $Tx = \lambda x$, then:
$$(2x_2 - 3x_1, 2x_3 - 3x_2, 2x_4 - 3x_3, \ldots) = (\lambda x_1, \lambda x_2, \lambda x_3, \ldots)$$
which means that: $x_2 = \frac{3 + \lambda}{2} x_1, x_3 = \frac{3 + \lambda}{2} x_2, \ldots$

This gives $x = (\frac{3 + \lambda}{2} x_1, (\frac{3 + \lambda}{2})^2 x_1, (\frac{3 + \lambda}{2})^3 x_1, \ldots)$

This vector is in $l_p$ if $\frac{|3+\lambda|}{2} < 1$.

From here, I am not sure how to continue, and I still need to find the continuous spectrum and residual spectrum of $T$.

Help would be appreciated.

答:
It's easier to concentrate on $S_l$ first.

You've found that it has eigenvectors $(1,\lambda,\lambda^2,\ldots)$ with eigenvalue $\lambda$, for any $|\lambda|<1$. Moreover its norm is $1$, so any $|\lambda|>1$ is a regular point.

This means that its point spectrum $\sigma_p(S_l)$ is the open unit ball, while $\sigma(S_l)$ is a subset of the closed unit ball. Moreover, the whole spectrum $\sigma(S_l)$ is a closed set bounded by the norm $\|S_l\|=1$. This implies that $\sigma(S_l)$ is precisely the closed unit ball.

By the spectral mapping theorem, the spectrum of $T=2S_l-3I$ is "$\sigma(T)=2\sigma(S_l)-3$", that is, it is the closed disk of radius $2$ centered at $-3$.

To classify the spectrum requires one more step. The adjoint of $S_l$ is $S_r$ so $$T^*=2S_r-3I$$
Repeating your exercise for the right-shift operator shows that it has no eigenvectors at all.

Now for any operator $A$, $$\sigma_r(A)\subseteq\sigma_p(A^*)$$
For $S=S_l$, this gives $\sigma_r(S_l)=\emptyset$. Hence the remainder of the spectrum of $S_l$ must be the continuous spectrum.

These results pass on to $T$ since it is just a translation of $S_l$, i.e., $\sigma_r(T)\subseteq\sigma_p(T^*)=\sigma_p(2S_r-3I)=\emptyset$. (If $\lambda\in\sigma_p(2S_r-3I)$ then $2S_r-(3+\lambda)I$ is not injective, so $(3+\lambda)/2$ would be an eigenvalue of $S_r$.)


Title: How many functions $ f: \{1, 2, 3, \dots, 10\} \to \{0,1\}$ satisfy $f(1) + f(2) + \dots + f(10) = 2$?

Body:
How many functions $ f: \{1, 2, 3, \dots, 10\} \to \{0,1\}$ have this property: $$f(1) + f(2) + \dots + f(10) = 2.$$

I understand just $2$ functions can be $1$, the rest have to be $0$, in total there are $2^{10}$ functions, but how can I find out how many of them have this property?

Body: $$ $$ $$ 10 \choose 2$$ $$ $$ $ $


问:
What Is Wrong Being Compactness of $[0,1]$ in $\mathbb R$ and Is $(0,1]$ Compact in $\mathbb R^{+}$?
I know that any [a,b] are compact. However, if it is compact, then $f(\,[0,1]\,)$ with the continuous function $f(x)=\ln(x)$ would be compact too, but $f(\,[0,1]\,)=(-\infty,0]$ which is not compact. What is wrong with that?

For second question, I think $(0,1]=[0,1]$ and it is compact in $\mathbb R^{+}$ considering $f$ is defined on $\mathbb R^+$. But with the same logic in first question, it seems not compact. What is wrong with that?

答:
The problem is that $\ln$ is only defined on $(0,\infty)$, thus it does not make sense to consider $f([0,1])$.

Moreover, we have $\lim_{x \to 0} \ln(x) = -\infty$, therefore there is no continuous extension of $\ln$ to a map $[0,\infty) \to \mathbb R$. But we can work with the extended real line $\bar {\mathbb R} = \mathbb R \cup \{-\infty, +\infty \}$ and topologize it in the well-known way. Note that $\bar {\mathbb R}$ is compact. Then $\ln$ extends continuously to $\ln : [0,\infty) \to  \bar {\mathbb R}$ by defining $\ln(0) = -\infty$. Now $\ln([0,1]) = [-\infty,0]$ which is compact.


问:
Why does a constant positive Gaussian curvature imply a sphere?
The Gaussian curvature value $G$ is the product of the two principle curvatures, which are the largest, $\kappa_1$, and the smallest (most downwards-curving), $\kappa_2$, normal curvatures through a point):
$$G=\kappa_1\kappa_2$$

For a mountain top as well as a valley, we have a positive $G>0$, because both principle curvatures are either upwards or downwards. The same is the case for a sphere (rightmost figure below). A saddle point, on the other hand, will have negative $G<0$ (leftmost figure). And a flat surface as well as a surface which is 'flat' in one dimension, such as a cylinder (middle figure), will have $G=0$.

Image from Wikipedia

My question is regarding a specific claim from the article on Gaussian curvatures on Wikipedia ("Relation to geometries" headline, middle sentence):

When a surface has a constant zero Gaussian curvature, then it is a developable surface and the geometry of the surface is Euclidean geometry.

When a surface has a constant positive Gaussian curvature, then it is a sphere and the geometry of the surface is spherical geometry.

When a surface has a constant negative Gaussian curvature, then it is a pseudospherical surface and the geometry of the surface is hyperbolic geometry.

These three sentences seemingly cover all types of surfaces. But the middle sentence (I have emphasized it) tells that for a positive $G>0$, the curvature is a sphere. How can that be true? Don't many mountain top surfaces have a positive $G>0$ all the way down from the top to forever? Such as the function $f=x^2+y^2$ (graph below)?

from WolframAlpha

答:
This is a non-trivial fact. It is possible to create surfaces of revolution of the form

$$(\phi(v) \cos(u), \phi(v) \sin(u), \psi(v)),\quad \phi \ne 0$$

If $v$ is arc length we have $(\phi')^2 + (\psi')^2 = 1$. Having constant Gaussian curvature, K, then implies $\phi''+K\phi = 0$ and $\psi = \int \sqrt{1-(\psi')^2}\ dv$.

With $K=1$ and the curve intersection the X-Y plane perpendicularly we have

$$\phi(v) = C \cos v,\quad \psi(v) = \int_0^v \sqrt{1-C^2 \sin^2 v}\  dv.$$

So $\psi(v)$ is an Incomplete elliptic integral of the second kind. (Reference Do Carmo)

Now these surfaces have constant positive Gaussian curvature, if $C=1$, it gives a sphere, if $C\ne 1$, you have surface which have two singular points on the rotation axis. Examples of such surfaces can be seen at Wolfram demonstrations.

One of the comments above points to a looseness in Wikipedia's statement. The sphere is the only compact, simply-connected, Riemannian surface of constant positive curvature. The above examples, either fails to be compact if we exclude the singular points, or if they are left it fails to be smooth, hence not Riemannian.

Wikipedia does give a hint to a proof of the fact, it references the Liebmann's theorem (1900), and a brief comment that

A standard proof uses Hilbert's lemma that non-umbilical points of extreme principal curvature have non-positive Gaussian curvature.

This may have come from Hilbert and Cohn Vossen (p228). They first show that surface of constant positive Gaussian curvature, without boundary or singularities, must be a closed surface. Apart from the sphere there are no surfaces where both principal curvatures are constant. So we just need to consider cases like the above where the two principle curvatures vary, but their products are constant. As the surface is closed there must be a point where one of the principle curvature obtains its maximum value. But it can be proved analytically that such points can't exist, except on the boundary, on surfaces with constant positive Gaussian curvature.

do Carmo, Manfredo Perdigão, Differential geometry of curves and surfaces. Mineola, NY: Dover Publications (ISBN 978-0-486-80699-0/pbk). xvi, 510 p. (2016). ZBL1352.53002.


答:
You cannot take $\log([0,1]$, since it is not well defined in zero.

But if you take $\log([\varepsilon,1]$, with $\epsilon>0$, then you get a compact interval as image since logarithm is continuous in its domain.

Title: What Is Wrong Being Compactness of $[0,1]$ in $\mathbb R$ and Is $(0,1]$ Compact in $\mathbb R^{+}$?

Body:
I know that any [a,b] are compact. However, if it is compact, then $f(\,[0,1]\,)$ with the continuous function $f(x)=\ln(x)$ would be compact too, but $f(\,[0,1]\,)=(-\infty,0]$ which is not compact. What is wrong with that?

For second question, I think $(0,1]=[0,1]$ and it is compact in $\mathbb R^{+}$ considering $f$ is defined on $\mathbb R^+$. But with the same logic in first question, it seems not compact. What is wrong with that?


问:
Calculating the probability of total payout for an insurance company

Automobile losses reported to an insurance company are independent and uniformly distributed between $0$ and $20,000$. The company covers each loss subject to a deductible of $5,000$.

Calculate the probability that the total payout on 200 reported losses is between $1,000,000$ and $1,200,000$.

My attempt:

Let $S$ denote the random variable for all losses reported to the insurance company. Then, using CLT, $S \sim N\left(200 \times \dfrac{20,000}{2}, 200 \times \dfrac{20,000}{12}\right) = (2\times 10^6, 577.35^2)$. Now let $Y$ be the random variable for the payout on all claims. Then, $Y=200(S-5000)$ if $10^6\le s<\infty$ and $Y=0$ otherwise.

The desired probability is given by
$$P[10^6<Y<1.2\times 10^6] = P[10^6<200(S-5000)<1.2\times 10^6] = P[10^4<S<1.1\times 10^4]$$

Now, I computed $$P\left[Z\le\frac{(1.1\times 10^4) - 2\times 10^6}{577.35}\right] - P\left[Z\le\frac{10^4 - (2\times 10^6)}{577.35}\right]$$

but this didn't lead me to the correct answer. Can someone please suggest what I did wrong in this question? Thanks.

答:
Your calculation is incorrect because you are applying an aggregate deductible when in fact the deductible is applied on an individual basis.  This means that if a ground-up loss is below $5000$, no claim payment is made.  This is not adequately expressed in your assumption about your aggregate claim random variable $Y$.

To understand why your approach is flawed, consider the toy model where you have $10$ random ground-up losses, each uniformly distributed on $[0,1]$.  The deductible on each loss is $0.5$.  Then according to your logic, the aggregate claim amount on the $10$ losses is $Y = 10(S - 0.5)$ if $S > 5$, and $0$ otherwise.  But this does not work because the losses could look like this:
$$\{0.2, 0.7, 0.8, 0.9, 0.7, 0.8, 0.9, 0.7, 0.8, 0.9\}$$ and only the first loss is less than the deductible.  The sum of these losses is greater than $5$ but the aggregate claim amount is not $$\{-0.3, 0.2, 0.3, 0.4, 0.2, 0.3, 0.4, 0.2, 0.3, 0.4\}$$ as your model would suggest:  the first claim amount is $0$.

Rather, the model behaves like this.  Let all numbers be expressed in units of thousands.  $X_i \sim \operatorname{Uniform}(0,20)$ is the ground-up loss for the $i^{\rm th}$ reported loss.  Then $$Y_i = \max(X_i - 5, 0)$$ is the claim size for the $i^{\rm th}$ reported loss.  The aggregate claim on $n = 200$ reported losses is $$S = \sum_{i=1}^n Y_i.$$  It is on $S$ that the central limit theorem may be used; to this end, we would need to compute $\operatorname{E}[Y_i]$ and $\operatorname{Var}[Y_i]$.  This is not difficult when $X_i$ is uniform.  We have $$\begin{align}
\operatorname{E}[Y_i] &= \operatorname{E}[X_i - 5 \mid X_i > 5]\Pr[X_i > 5] + \operatorname{E}[0 \mid X_i \le 5]\Pr[X_i \le 5] \\
&= \left(\frac{20+5}{2} - 5 \right)\frac{3}{4} + (0)\frac{1}{4} \\
&= \frac{45}{8}.
\end{align}$$
The variance calculation I leave to you as an exercise (hint: calculate $\operatorname{E}[(X_i - 5)^2 \mid X_i > 5]$); the result is $$\operatorname{Var}[Y_i] = \frac{1575}{64}.$$
Now we can use a normal asymptotic approximation:  $S$ is approximately normal with mean $$\mu = n \operatorname{E}[Y_i] = 1125,$$ and variance $$\sigma^2 = n \operatorname{Var}[Y_i] = \frac{39375}{8}.$$  Then the desired probability is $$\begin{align}
\Pr[1000 \le S \le 1200]
&= \Pr\left[\frac{1000 - 1125}{70.1561} \le \frac{S - \mu}{\sigma} \le \frac{1200 - 1125}{70.1561}\right] \\
&\approx \Pr\left[-1.78174 \le Z \le 1.06904\right] \\
&\approx 0.82008. \end{align}$$


问:
Number of ordered pair $(a,b)$ such that ${ab\over a+b}=n$ ...

Number of ordered pair $(a,b)$ of natural numbers such that ${ab\over a+b}=n$ (n $\in \mathbb{N}$) is $a_n$ then find , $\lim_{n \to \infty} {a_n\over n^2}$.

MY attempt :
Suppose the case where $n=2^k$ for some $k \in \mathbb{N}$ . Then ,  ${ab\over a+b}=n$ is equivalent to $(a-2^k)(b-2^k)=2^{2k}$ . In this case $a_n=4k$  and thus $\lim_{n \to \infty} {a_n\over n^2}=0$

Now I have an intuition (from observations) that ${a_n\over n^2}$ would be maximum when $a_n$ is a power of 2 , and if bigger factors (like $3,5,etc$) are included then the ratio ${a_n\over n^2}$ would decrease. And thus $\lim_{n \to \infty} {a_n\over n^2}=0$ for any $n$

But I am not able to think of a solid mathematical proof for this .

Could someone please help me solve this ?

Thanks!

答:
As already remarked in the comments, the integer solutions of $\frac{ab}{a+b}=n$ are the integer solutions of $(a-n)(b-n)=n^2$, so they are given by $(a,b)=\left(d+n,\frac{n^2}{d}+n\right)$ for any integer divisor $d$ of $n^2$.

The wanted limit is zero since the following Dirichlet series is clearly convergent:
$$ \sum_{n\geq 1}\frac{\tau(n^2)}{n^2}=\prod_p\left(1+\frac{\tau(p^2)}{p^2}+\frac{\tau(p^4)}{p^4}+\ldots\right)=\prod_{p}\left(1+\frac{3}{p^2}+\frac{5}{p^4}+\ldots\right) $$
$$ \sum_{n\geq 1}\frac{\tau(n^2)}{n^2}=\prod_{p}\frac{p^2(p^4-1)}{(p^2-1)^3} =\frac{\zeta(2)^3}{\zeta(4)}=\frac{5\pi^2}{12}. $$


问:
Morphism into Dedekind scheme is an open immersion?
I was working on exercise 4.1.14 from Liu's 'Algebraic Geometry and Arithmetic Curves' book:

Let $X$ be an integral scheme of dimension 1, and let $f:X\to Y$ be a separated birational morphism of finite type from $X$ to a Dedekind scheme. Show that $f$ is an open immersion.

I had some problems at a certain point in the proof, so I decided to take a look at an example that 'shows' these problems:
$$X=\operatorname{Spec}\mathbb{Z}[3/2]$$
$$Y=\operatorname{Spec}\mathbb{Z}$$
and $f$ is simply the morphism induced by the inclusion $\mathbb{Z}\to\mathbb{Z}[3/2]$. First of all $X$ is an integral scheme of dimension 1 and $Y$ is a Dedekind scheme. $f$ is obviously separated and of finite type and it is birational as it induces the identity on the function fields (which are just $\mathbb{Q}$). However, $f$ is not an open immersion for the following reason: Since $2$ is not a unit in $\mathbb{Z}[3/2]$ there is some maximal ideal $\mathfrak{m}$ of $\mathbb{Z}[3/2]$ with $2\in\mathfrak{m}$. Then $$\mathfrak{m}\cap\mathbb{Z}=2\mathbb{Z}=:\mathfrak{p}$$
The morphism $$f^\#_{\mathfrak{m}}:\mathcal{O}_{Y,\mathfrak{p}}\to\mathcal{O}_{X,\mathfrak{m}}$$
is not surjective as $3/2$ is not in $\mathbb{Z}_{2\mathbb{Z}}=\mathcal{O}_{Y,\mathfrak{p}}$. Therefore $f$ is not an open immersion.

What concerns me about this is that Liu writes in that exercise that this statement is a special case of "Zariski's Main Theorem", so there is probably some mistake somewhere in this 'counterexample'. Can someone tell me where I am wrong?

答: $2$ is a unit in $\Bbb Z[3/2]$ with inverse $\frac{3}{2}-1$.


答:
As pointed out by BruceET, you deduce $5000$ from every claim, including those who where valued less than $5000$.  This fact changes the total payout.

Here is an example for why it didn't work. Let's say that we have $100$ claim of $20\ 000$ and $100$ claim of $0$. The total loss is $2\cdot10^6$. According to your reasonning, we deduce $5000$ for every claim, since it is a total loss of over $10^6$. After the deduction, we have a total payout of $10^6$. $100$ people recieved $15\ 000$ for the insurance company while the other $100$ people will give back $5000$ to the insurance company! In reality, the insurance company will pay $15\ 000$ to a $100$ people, and $0$ to the other, for a total of $1.5\cdot10^6$.

Let $X_i$ be an automobile loss. Then $X_i\sim \text{Unif}(0, 20\ 000)$. Let $P_i$ be the payout from the insurance company, then
$$P_i = \begin{cases}0 &\text{if }X_i \leq 5000\\X_i-5000 &\text{if } 5000<X_i\leq 20\ 000\end{cases}$$

Since it is not an uniform distribution anymore, we have to compute the mean and variance.
$$E[P_i] = \int_0^{15\ 000} P_i\times \frac{1}{20\ 000}\ \mathrm{d}P_i = 5625$$
$$E[P_i^2]\int_0^{15\ 000} P_i^2\times \frac{1}{20\ 000}\ \mathrm{d}P_i = 5.625\times10^7$$
$$Var[X_i] = 24\ 609\ 375$$

Now, let $Y$ be the total payout by the insurance company, for $200$ claims, then
$$Y\sim N(200\times 5625, 200\times 24\ 609\ 375)$$
$$Y\sim N(1\ 125\ 000, 70\ 156.076^2)$$
All is left if to evaluate the probability as you did.

Title: Calculating the probability of total payout for an insurance company

Body:
Automobile losses reported to an insurance company are independent and uniformly distributed between $0$ and $20,000$. The company covers each loss subject to a deductible of $5,000$.

Calculate the probability that the total payout on 200 reported losses is between $1,000,000$ and $1,200,000$.

My attempt:

Let $S$ denote the random variable for all losses reported to the insurance company. Then, using CLT, $S \sim N\left(200 \times \dfrac{20,000}{2}, 200 \times \dfrac{20,000}{12}\right) = (2\times 10^6, 577.35^2)$. Now let $Y$ be the random variable for the payout on all claims. Then, $Y=200(S-5000)$ if $10^6\le s<\infty$ and $Y=0$ otherwise.

The desired probability is given by
$$P[10^6<Y<1.2\times 10^6] = P[10^6<200(S-5000)<1.2\times 10^6] = P[10^4<S<1.1\times 10^4]$$

Now, I computed $$P\left[Z\le\frac{(1.1\times 10^4) - 2\times 10^6}{577.35}\right] - P\left[Z\le\frac{10^4 - (2\times 10^6)}{577.35}\right]$$

but this didn't lead me to the correct answer. Can someone please suggest what I did wrong in this question? Thanks.


问:
Minimum value for A+B, given rounded percentages
The percentage of boys in a Math circle, rounded to an integer is equal to 51%. The percentage of girls in this Math circle, rounded to an integer is equal to 49%. What is the minimal possible number of participants in the circle?
(Source: "Formula of unity" contest, 2020 qualification rounds, R7, problem 2 - https://www.formulo.org/en/olymp/2020-math-en/)

What is a principled way to tackle such a problem?

Here was my sketch of solution:
It is easy to see 51 - 49 is a solution, although not minimal.

Assuming the girls percentage can be at least 48.501 and boys percentage 51.499, and that the difference between the two is at least 1 -- I could guess the solution is somewhere around 33 children in the class.
Then, 35 (17 + 18) is the smallest solution.

答:
Alternative proof (advantage is no need for trial-and-error):

Using Calvin Lin's notation, we have $\frac{1}{100} < \frac{2B}{A} - 1 < \frac{3}{100}$

If $A=2m$ then $B=m+d, d\ge1$, $$\frac{1}{100} < \frac{2(m+d)}{2m} - 1 = \frac dm < \frac{3}{100} \\\implies m > \frac{100d}{3} > 33, m\ge 34, A=2m \ge 68.$$

If $A=2m+1$ then $B=m+d, d\ge1$,
$$\frac{1}{100} < \frac{2(m+d)}{2m+1}-1 = \frac{2d-1}{2m+1} < \frac{3}{100}\\
 \implies 2m+1 > 100(2d-1)/3>33 \text{ and } 2m+1 < 100(2d-1) \\ \implies 2m+1 \ge 35.$$

BTW: a quick and dirty trick is to look at the continued fraction of $0.515$ which turns out to be $\{0; 1, 1, 16, \ldots,\}$ and the first fraction between $0.505$ and $0.515$ is $\frac{1}{1+\frac{1}{1+\frac{1}{16+1}}}=\frac{18}{35}.$


问:
For linear operator $T(t)$, what is the difference between "$T(t)$ is continuous", "$t\to T(t)$ is continuous", and "$t\mapsto T(t)$"?
Let $T(t)$ be a family of linear operators defined on a vector space $X$, $t\geq 0$. What is the difference between the following statements:

$T(t)$ is continuous

$t\rightarrow T(t)$ is continuous

$t\mapsto T(t)$

答: Just saying “$T(t)$ is continuous” might not make sense unless $T(t)$ is a continuous function. However, saying “$T : X \to X$ is continuous” or “$t \mapsto T(t)$ is continuous makes sense. Keep in mind the arrow $\to$ Is used to show the domain and codomain of $T$, while $\mapsto$ is used to show how $T$ is defined.


问:
When is the number of Divisors of a Number equivalent to one of its Factors?
My math teacher asked me this problem for homework and I am unsure how to solve it.

Which numbers contain a number of factors equivalent to the value of one of their divisors?

I found that 8 works, because it has 4 factors and 4 divides 8, but I did that by manually computing the number of factors of each number. I am unsure how to develop a  general process of proof to find all such numbers.

答: You're asked to solve $\sigma_0(n)|n$, with $\sigma_0$ the divisor function. Luckily, $\sigma_0$ is a multiplicative function. Writing an arbitrary $n\in\Bbb N$ as $\prod_ip_i^{a_i}$, with the $p_i$ finitely many distinct primes and the $a_i$ positive integers, $\sigma_0(n)=\prod_i(a_i+1)$, so $n$ qualifies iff each $a_i+1$ is a product of powers of the $p_i$. Since once you choose which prime factors you want the legal choices of the $a_i$ are thereby specified, that's about as elegant an all-solutions characterization as you'll get.


问:
Different method to find fundamental unit
Consider the quadratic number field $K=\mathbb{Q}(\sqrt{61})$. By Dirichlets unit theorem we know that the unit group has rank $1$. I now want to find a unit of infinite order.

I know that one way to do this is by finding the smallest non-negative integer $y$ for which
$x^2+xy-15y^2=N_{K/\mathbb{Q}}(x+y(\frac{1+\sqrt{61}}{2}))=\pm 1$ has an integer solution for $x$. After trying one can show that the smallest one is for $y=5$ and consequently $x=17$. This gives us the fundamental unit $17+5\alpha$, where $\alpha=\frac{1+\sqrt{61}}{2}$.

The problem with this method is that it is a trial and error method and that the first $y$ could be very big. So computational wise this method can be very inefficient.

So I wondered if there is another method for finding the fundamental unit for quadratic fields or if there is an easier way to find at least one unit of infinite order? Any help would be appreciated.

答:
The usual method is factoring elements of small norm. In your case,
the ideals generated by $\alpha = 8 + \sqrt{61}$ and $\beta = (7 - \sqrt{61})/2$ have norm $3$, and the quotient is a nontrivial unit, namely $\frac{39 - 5 \sqrt{61}}2$.

Afterwards you have to show that $\frac{39 + 5 \sqrt{61}}2$ is not a power, which is not hard since the fundamental unit is $\ge \frac{1 + \sqrt{61}}2$, and this bounds possible exponents.


问:
Is magnitude of vector cross product always equal to dot product?
Is magnitude of vector cross product always equal to dot product ?

What I mean to say is that

|vector cross product|(magnitude) = ABcos$\theta$ or magnitude of dot product.
^^
This statement up there means that magnitude of cross product  = magnitude of dot product

is $$|A\times B|=|A||B|\sin(\theta)$$ $$A\cdot B=|A||B|\cos(\theta).$$

答: Your equation is incorrect. What is true is $$|A\times B|=|A||B|\sin(\theta)$$ $$A\cdot B=|A||B|\cos(\theta).$$


答:
If we expand $\sqrt{1+x^k}$ in series and then intergrate  then we get a close representation in terms of hypergeometric function
$$
\int_0^1 \sqrt{1+x^k}={}_2F_1\left(-\frac{1}{2},\frac{1}{k};\frac{k+1}{k};-1 \right)
$$

Title: Finding a closed form for the following integral: $\int_0^1\sqrt{1+x^k}dx$

Body:
Is there a nice closed form, for the following integral:

$$\int_0^1\sqrt{1+x^k}dx$$

And how can I derive it?

I have no idea how to get started, thanks for any help. This problem came up when I was working on a number theory problem in my university for some values of $k$.


答:
In order for this to be the case, two elements in the domain must map to 1, and the rest map to 0:

$\binom{10}{2}$


问:
Number of homeomorphism types of separable closed subspaces of $\beta \mathbb N$.
When thinking about subalgebras of $\ell_\infty$, the algebra of bounded, scalar-valued, I came across the following question related to counting subalgebras with weak*-separable dual ball obtainable by restricting to closed subsets of the spectrum of $\ell_\infty$.

Let $\beta \mathbb N$ be the Čech–Stone compactification of the discrete space of natural numbers. How many pairwise non-homeomorphic closed and separable subspaces does $\beta \mathbb N$ have?

Such subspaces cannot be second-countable unless finite, but there are many separable closed subspaces (take the closure of any countable subset) and there is no obvious way to classify them up to homeomorphism.

答:
There are exactly $2^c$ pairwise nonhomeomorphic separable closed subspaces in $\beta \mathbb N$.

The proof is essentially just a reference to a great paper:

Since $|\beta \mathbb N| = 2^c$, there are $2^c$ countable subsets in $\beta \mathbb N$, hence $\beta \mathbb N$ has only $2^c$ separable subspaces.

The other, non-trivial, direction is mainly the Main Theorem in
A. Dow, A.V. Gubbi, A. Szymanski, "Rigid Stone Spaces within ZFC":

There exist $2^c$ pairwise nonhomeomorphic rigid separable Stone spaces,

where Stone space is (in this paper) an extremally disconnected compact Hausdorff space.
As it is well-known, each separable extremally disconnected space can be embedded into $\beta \mathbb N$
(see, for instance, Corollary 3.2 here).

Remark: Perhaps there might be much easier constructions, since here we don't need that the spaces are rigid
(= the only autohomeomorphism is the identity).


答: Since$$\begin{align}(A\cdot B)^2+|A\times B|^2&=A_iB_iA_jB_j+\epsilon_{ijk}\epsilon_{ilm}A_jA_lB_kB_m\\&=(\delta_{jk}\delta_{lm}+\epsilon_{ijk}\epsilon_{ilm})A_jA_lB_kB_m\\&=\delta_{jl}\delta_{km}A_jA_lB_kB_m\\&=A^2B^2,\end{align}$$the correct result is $|A\times B|=\sqrt{A^2B^2-(AB\cos\theta)}=AB|\sin\theta|$.This equals $AB\cos\theta$ iff $\cos\theta=\tfrac{1}{\sqrt{2}}$, i.e. $A,\,B$ are separated by an angle of $\pi/4$.

Title: Is magnitude of vector cross product always equal to dot product?

Body:
Is magnitude of vector cross product always equal to dot product ?

What I mean to say is that

|vector cross product|(magnitude) = ABcos$\theta$ or magnitude of dot product.
^^
This statement up there means that magnitude of cross product  = magnitude of dot product

is $$|A\times B|=|A||B|\sin(\theta)$$ $$A\cdot B=|A||B|\cos(\theta).$$


问:
Why is 5.93e-56 considered to be close to zero?
I am very poor in math but have recently been studying some math courses, unfortunately they do not explain things well.

I have heard multiple times now that values like 5.93e-56 is very close to 0.

I do not know how this is. Is this not a 5.something value? And thus is it not supposed to be higher than 0?

Can anyone explain? I am specifically talking about when this value shows up as F-value in linear regression results.

答:
$5.93$e$-56$ usually means the number $5.93 \times 10^{-56} = 0.0000000000000000000000000000000000000000000000000000000593,$ which is a positive number and in most every-day contexts is considered to be quite close to zero.

Addendum:

I always find it ironic when an entire question is asked asking about the meaning of a notation which is designed to abbreviate terms. It would have been quicker if the "abbreviated notation" wasn't invented at all - i.e. scientists just used  $5.93 \times 10^{-56}$ as opposed to $5.93$e$-56$, rather than us all having to spend time explaining to someone that this is what the notation means...


答: I would say that "$T(t)$ is continuous" is an abuse of notation as $T(t)$ is the value of the operator $T$ in the point $t\in X$ (unless $T(t)$ is an element in a space of maps). On the other hand, "$t\mapsto T(t)$" can be read as "the map who sends $t$ to $T(t)$". Finally, I wouldn't use the arrow "$\rightarrow$" in the way you have done; in general, "$A\to B$" means "a maps which send each element in the set $A$ to an element of $B$.

Title: For linear operator $T(t)$, what is the difference between "$T(t)$ is continuous", "$t\to T(t)$ is continuous", and "$t\mapsto T(t)$"?

Body:
Let $T(t)$ be a family of linear operators defined on a vector space $X$, $t\geq 0$. What is the difference between the following statements:

$T(t)$ is continuous

$t\rightarrow T(t)$ is continuous

$t\mapsto T(t)$


问:
Definition of equation vs. expression vs. polynomial in the context of zeros vs. roots
I was trying to figure out the distinction of a root and a zero and found people in such discussions make distinctions between equation vs. expression vs. polynomial without defining them.  What is the difference exactly?

I'm especially confused by function.  Likely I've forgotten many details from my high school math such as this definition.  I can tell you about functions in several programming languages, but that's not the same thing at all.

So are root and zero the same things or are they applied to different things?   Interestingly two of the suggested links for this question were "Roots of Polynomial Equation?" and "Finding the zeros of a polynomial equation."

答:
A polynomial is a linear combination of natural powers of the variable. E.g.

$$ax^2+bx+c.$$

An expression is a series of symbols that can be evaluated (this is a very general term). E.g.

$$\sin(x).$$

(A polynomial is an expression; the converse is not necessarily true.)

An equation is an equality relation between two expressions. E.g.

$$ax^2+bx+c=\sin(x).$$
An equation can be polynomial or not.

A zero is a value of an unknown that makes an expression vanish, i.e. it is a solution of an equation with a zero RHS. E.g. $x$ such that

$$ax^2+bx+c=0.$$

A root is a value that satisfies an equation. E.g. $x$ such that

$$ax^2+bx+c=\sin(x).$$


问:
Solving for $y$, given $y = 0.8(0.78 - 0.15 - 0.15 - x)$ and $x = \frac1y$
I have the below equation:

$$y = 0.8(0.78 - 0.15 - 0.15 - x) \tag1$$

$x$ relates to $y$ by the below:

$$x = \frac1y \tag2$$

This would give:

$$y = 0.8\left(0.78 - 0.15 - 0.15 - \frac1y\right) \tag3$$

But how do I solve for $y$?

答:
$\begin{align*} y &= 0.8(0.48 - x) \\
xy &= 0.8(0.48x - x^2) \\
1 &= 0.8(0.48x - x^2) \end{align*}$

What was said in the comments basically.


问:
characterization of quotient rings
In the ring $\mathbb{Q}{[x]}$ consider the following subset

$I = \{f(x) (x^2 + 3) | f(x) \in \mathbb{Q}[x]\}$

Prove that $I$ is an ideal in $\mathbb{Q}[x]$ and express the quotient ring $\mathbb{Q}[x]/I$.

First I'm not sure of all the conditions to verify to be sure in an ideal.

I verified $I$ is an additive subgroup since $f(x)(x^2 + 3) + g(x)(x^2 + 3) = h(x) (x^2 + 3)$ with $h = g + f$.
Then it is also a two-sided ideal $g(x) \cdot f(x)(x^2 + 3)\in I$ with $g \in \mathbb{Q}{[x]}, f(x)(x^2 + 3) \in I$ and $f(x)(x^2 + 3) \cdot g(x) \in I$ as well.

Is it enough, or are there other conditions to be satisfy?

Regarding the characterization of $\mathbb{Q}[x]/I$, I'm uncertain. This should be made of all congruence classes with regards to the equivalence relation ~ defined as:

$g \sim f \quad \iff f - g \in I \quad for \quad g,f \in \mathbb{Q}[x]$

my manual express it as $R/I := \{a+I | a \in \mathbb{Q}[x]\}$

with

$a+I := \{a + i | i \in I \} = \{x \in \mathbb{Q}[x] | x \sim a\}$

答: By the division algorithm, for every $p(x) \in \mathbb Q[x]$ there exists $q(x),r(x) \in \mathbb Q[x]$ such that $$p(x) = (x^2+3)q(x)+r(x)$$ and that $\deg r(x) < \deg(x^2+3) = 2$ whenever $r(x) \neq 0$; put in another way: for every $p(x) \in \mathbb Q[x]$ there exists $q(x) \in \mathbb Q[x]$ and $a,b \in \mathbb Q$ such that $p(x) = (x^2+3)q(x)+(a+bx)$, which is the same as saying $$p(x)+I = (a+bx)+I.$$ Thus $\mathbb Q[x]/I = \{(a+bx)+I : a,b \in \mathbb Q\}$.

Title: characterization of quotient rings


问:
Showing that the sequence is bounded without finding its limit
I am given a sequence ${x_n}=1+\frac{1}{2}+...+\frac{1}{n}-\ln n$ and I have to show that it's bounded.

I tried doing this: $$1\leq 1+\frac{1}{2}+...+\frac{1}{n}\leq n$$
$$1-\ln n\leq 1+\frac{1}{2}+...+\frac{1}{n}-\ln n\leq n-\ln n$$
$$1-\ln n\leq{x_n}\leq n-\ln n$$
But I don't think I am allowed to do this since the boundaries depend on n.
Is there maybe another way to solve this? and could you please help me?
Thanks a lot!
I can't use integrals, derivatives or limits

答:
Hint for an elementary solution, without integral or asymptotic approximation :

Define the sequences $(x_n)_{n \geq 1}$ and $(y_n)_{n \geq 1}$ by
$$x_n = 1 + \frac{1}{2} + ... + \frac{1}{n} - \ln(n) \quad \quad \quad \text{and} \quad \quad \quad y_n = 1 + \frac{1}{2} + ... + \frac{1}{n} - \ln(n+1)$$

Show that $(x_n)$ is decreasing.

Show that $(y_n)$ is increasing.

Show that $x_n - y_n \longrightarrow 0$ when $n$ tends to $+\infty$.

(under these three conditions, we say that $(x_n)$ and $(y_n)$ are adjacent sequences).

Deduce that both $(x_n)$ and $(y_n)$ converge to the same limit. In particular they are bounded.


问:
Express $\operatorname{Log}(i-z)$ as a Taylor series centered at $z=1$
How can I express $\operatorname{Log}(i-z)$ as a Taylor series centered at $z=1$ ???

I have tried to first derivate the function, for then integrating it. But I think that I'm doing it wrong... I'm having problems with the $i$

How can I do it?

*Edit:

With the suggestion given, I've done the following:$\space$

$Log(i-z)=Log((i-1)(1-(\frac{z-1}{i-1})))=Log(i-1)+Log(1-(\frac{z-1}{i-1}))$$\space$

We know that $\space$ $Log(i-1)=ln(\sqrt{2})+\frac{3\pi}{4}i$$\space$

And we also know that $\space$ $Log(1-(\frac{z-1}{i-1}))=-\sum_{n=0}^{\infty}\frac{(\frac{z-1}{i-1})^{n+1}}{n+1}$

But now I don't know how to continue...

答:
Suggestion: $$i - z = (i-1) - (z-1)= (i-1) \left( 1 -\left(z-1\over i-1 \right) \right).
$$


答:
The magnitude of the cross product equals the dot product when $$|A| |B| \sin \theta = |A| |B| \cos \theta \\ \iff |A|=0, \ \text{or} \ |B|=0, \ \text{or} \ \sin\theta =\cos \theta \iff \theta =\frac{\pi}{4}$$

The magnitude of the cross product equals the magnitude of the dot product when
$$|A| |B| \sin \theta = |A| |B| | \cos \theta| \\ \iff |\tan \theta| =1 \iff \theta =\frac{\pi}{4} , \frac{3\pi}{4} $$

or if one of $|A|, |B|$ is zero.

So no, neither of these is always the case.

Here, it is assumed that $\theta \in [0, \pi)$.


问:
Understanding when convergence implies uniform convergence for sequences of non-continuous functions
I am working on the following problem:

Let $(f_n)$ be a sequence of functions $[a,b] \rightarrow \mathbb{R}$ such that: (i) $f_n(x)≤0$ if $n$ is even, $f_n(x)≥0$ if $n$ is odd; (ii) $|f_n(x)|≥|f_{n+1}(x)|$ for all $x$; (iii) $f_n$ converges to $0$ uniformly. Prove that then $\sum_n f_n$ is uniformly convergent.

I take as given the following theorem (3.43 in Rudin's Principles of Mathematical Analysis):

Suppose (a) $|c_1|≥|c_2|≥|c_3|≥...$; (b) $c_{2m-1}≥0$, $c_{2m}≤0$ ($m=1,2,3,...$); (c) $lim_{n\rightarrow\infty} c_n=0$ Then $\sum_n c_n$ converges.

Now, it follows immediately that for given $x$, $\sum_n f_n(x)$ converges, say, to $F_n(x)$, a constant function.

I am wondering how to show rigorously that this convergence is in fact uniform. We are dealing with a compact set, but the $\sum_n f_n(x)$ are not continuous (are they?), so we can't use Dini's theorem.

I have tried saying something along these lines:

Choose $\epsilon>0$. Then $\forall x \in [a,b]$, $\exists N_x$ s.t. $n≥N_x \implies |\sum_nf_n(x)-F_n(x)|<\epsilon.$

Put $M=sup_{x\in [a,b]}N_x$. Then $n≥M \implies|\sum_nf_n(x)-F_n(x)|<\epsilon$ $ \forall x \in [a,b]$.

But I worry that the reasoning is faulty. ($\forall x \exists N$ does not usually imply $\exists N \forall x$ - but I wonder if it does in this case because we're working on a compact set?) If anyone has any thoughts I'd be very grateful. Thanks!

答:
Following David's hint, we have:

We are given that $f_n \rightarrow 0$ uniformly, so for any $\epsilon > 0$ $\exists N$ s.t. $n≥N \implies |f_n(x)|<\epsilon$ $\forall x$.

We have, then, for $m>n>N$, $|\sum_1^n f_n(x) - \sum_1^m f_n(x)|=|\sum_{n+1}^m f_n(x)|≤|f_n(x)|≤\epsilon$ $\forall x$, where the penultimate inequality follows immediately from conditions (i) and (ii).

Thus, $\sum_n f_n$ is uniformly convergent.


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

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

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

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

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

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

Thanks in advance

答:
This is asking you to first find the level surface of the function $f(x,y,z)$ that contains the point $(\pi/2,\pi,\pi)$. This would just be the surface implicitly defined by $0=\cos(x+2y+3z)$. The correct equation for the tangent plane of an implicitly defined surface is $$0=f_x(a,b,c)(x-a)+f_y(a,b,c)(y-b)+f_z(a,b,c)(z-c).$$

The equation you have in the question is a combination of this equation and the tangent plane of an explicitly defined surface $z=f(x,y)$, which is $$z=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)$$


问:
How to prove $\int _0^\infty \operatorname{si}(x) \operatorname{Ci}(x) \, dx=\ln 2$
How to prove the integral
$$\begin{align}&\int _0^\infty \operatorname{si}(x) \operatorname{Ci}(x) dx\\=&\int_0^\infty\left (\int_x^\infty\frac{\sin t}{t}dt \int_x^\infty\frac{\cos t}{t}dt\right)dx\\=&\ln 2\end{align}$$

I try to use integration by parts
$$\begin{align}
&\int _0^\infty \operatorname{si}(x) \operatorname{Ci}(x) dx\\
=&\int_0^\infty \int_x^\infty\frac{\sin t}{t}dt \int_x^\infty\frac{\cos t}{t}dt\\
=&x\left.\int_x^\infty\frac{\sin t}{t}dt \int_x^\infty\frac{\cos t}{t}dt\right|_0^\infty+\int_0^\infty \cos x \int_x^\infty\frac{\sin t}{t}dx+\int_0^\infty \sin x  \int_x^\infty\frac{\cos t}{t}dx
\end{align}$$
But the last two integrals are divergent.

答:
Integration by parts shows that $$ \int \operatorname{Ci}(x) \, \mathrm dx =  x \operatorname{Ci}(x) - \int \cos (x) \, \mathrm dx  = x \operatorname{Ci}(x) - \sin(x) +C_{1} .$$

Then again using integration by parts, we have

$$ \begin{align} \int \operatorname{si}(x) \operatorname{Ci}(x) \, \mathrm dx &= \int \underbrace{\left(\operatorname{Si}(x)- \frac{\pi}{2} \right) }_{u} \underbrace{\operatorname{Ci}(x) \, \mathrm dx}_{dv} \\ &= \left(\operatorname{Si} (x)- \frac{\pi}{2} \right) \left( x\operatorname{Ci}(x) - \sin (x)\right)  - \int \frac{\sin (x)}{x}\left(x\operatorname{Ci}(x) - \sin (x) \right) \, \mathrm dx, \end{align}$$

where $$ \begin{align} \int \underbrace{\operatorname{Ci}(x))}_{u} \underbrace{\sin(x)\, \mathrm dx}_{dv} &= - \cos(x) \operatorname{Ci}(x) + \int \frac{\cos^{2}(x)}{x} \, \mathrm dx  \\ &= - \cos(x) \operatorname{Ci}(x) + \frac{1}{2} \int \left(\frac{1}{x} + \frac{\cos (2x)}{x} \right) \, \mathrm dx \\ &=- \cos(x) \operatorname{Ci}(x) + \frac{\ln (x)}{2} + \frac{\operatorname{Ci}(2x)}{2} + C_{2} \end{align}$$

and $$ \int \frac{\sin^{2}(x)}{x} \, \mathrm dx = \int \frac{1}{2} \left(\frac{1}{x}- \frac{\cos(2x)}{x} \right) \, \mathrm dx = \frac{\ln (x)}{2} -
\frac{\operatorname{Ci}(2x)}{2} + C_{3}.$$

Therefore,

$$\int \operatorname{si}(x) \operatorname{Ci}(x) \, \mathrm dx =  \left(\operatorname{Si} (x)- \frac{\pi}{2} \right) \left( x\operatorname{Ci}(x) - \sin (x)\right)+ \cos(x) \operatorname{Ci}(x) - \operatorname{Ci}(2x)   + C_{4} \tag{1}. $$

You can use the asymptotic expansion $$\operatorname{Ci}(x) \sim \frac{\sin x}{x} + \mathcal{O} \left(\frac{\cos(x)}{x^{2}} \right) $$ to show that the limit of the right side of $(1)$ as $x \to + \infty$ is $C_{4}$, and then use use the series expansion $$\operatorname{Ci}(x) =  \gamma + \ln(x) + \mathcal{O}(x^{2})$$ to show that limit of the right side of $(1)$ as $x \to 0^{+}$ is $-\ln(2) + C_{4}$.

Since $\int_{0}^{\infty} \operatorname{Ci}(x) \, \mathrm dx = 0$, it turns out that $$ \int_{0}^{\infty} \operatorname{si}(x) \operatorname{Ci}(x) \, \mathrm dx = \int_{0}^{\infty} \operatorname{Si}(x) \operatorname{Ci}(x) \, \mathrm dx.  $$


问:
Show that $\vec{F}(x,y,z) = y\vec{i}-x\vec{j}$ is not a gradient field.
By definition, for this to be a gradient field, it must satisfy the condition that for a function $\phi$ : $\Bbb{R}^3 \to \Bbb{R}$, $F_1(x,y,z) = \frac{d\phi}{dx}$ $F_2(x,y,z) = \frac{d\phi}{dy}$ and $F_3(x,y,z) = \frac{d\phi}{dz}$

Or equivalently $\vec{F}(x,y,z)=\operatorname{grad}(\phi)(x,y,z) = \frac{d\phi}{dx}(x,y,z)\vec{\mathrm{i}}+\frac{d\phi}{dy}(x,y,z)\vec{\mathrm{j}}+\frac{d\phi}{dz}(x,y,z)\vec{\mathrm{k}}$

For my function, $\vec{F}(x,y,z)$ = $y\vec{\mathrm{i}}-x\vec{\mathrm{j}} +0\vec{\mathrm{k}}$, these conditions are satisfied, since $\frac{d\phi}{dx}(x,y,z) = y$ and $\frac{d\phi}{dy}(x,y,z) = -x$ and $\frac{d\phi}{dz} =0$

So therefore it is a gradient field. But I'm supposed to show that it is not a gradient field. How can I do so, if the conditions for it being a gradient field are satisfied?

答:
The curl of a gradient field vanishes.

If

$\vec F = y \vec i - x \vec j \tag 1$

were a gradient field, then we would have

$\nabla \times \vec F = 0; \tag 2$

but

$\nabla \times \vec F$
$= (\vec F_{z,y} - \vec F_{y, z})\vec i +  (\vec F_{x, z} - \vec F_{z, x})\vec j +  (\vec F_{y, x} - \vec F_{x, y})\vec k = -2\vec k \ne 0; \tag 3$

thus $\vec F$ is not a gradient vector field.


问:
Deciding Binomial or Poisson Distribution for a Particular Scenario
I am writing a program and am confused with deciding on whether to use a Binomial or Poisson distribution. I will give an example of an identical problem.

Let's say I have a deck of 52 cards, and our random variable X = the # of times (with replacement of course) I choose a King of Hearts.

Let's also say that each day I take note of how many times I managed to draw a King of Hearts (lets just assume that I recorded the occurrences for 3 hours each day).

If I were to choose the Poisson distribution, assume that after 3 days I summed a total of 30 occurrences (1st day = 10, 2nd day = 5, 3rd day = 15) where $\lambda$ = 30/3 = 10 occurrences per day. Now I want to find P(X=1) or the probability of choosing a single King of Hearts on any given day.

Given this context, would it make sense to use this distribution to compute for P(X = 1)? With the way I have explained it, it seems it would be correct but I am not sure.

Please note: I am by no means an expert when it comes to statistics concepts, so I may have not been able to explain this the best; if you are confused or if there exists an error in the logic provided please let me know.

答: Maybe see here: Difference between Poisson and Binomial distributions.. In general, if $n p_n \to \lambda$ as $n \to\infty$, then the distribution $\text{Bin}(n,p_n)$ converges to a Poisson distribution with mean $\lambda$ as $n\to\infty$. Since the probability of drawing a King of Hearts is relatively low ($1/52$) and your number of draws is relatively high, the Poisson approximation should be reasonable.