问:
The subspace $C$ of the space $(X,d)$ is compact
Let:

$(X,d)$ - complete metric space.

$K_1,K_2,\dots $ - compact sets in the space $(X,d)$ (sequence $K_1,K_2,\dots$ need not be descending)

$C_n=\{x\in X: \text{dist} (x,K_n)\le \frac 1n \}$ for  $n=1,2,\dots$

$C=\bigcap\limits_{n=1}^{\infty} C_{n}\subset X$

Claim: The subspace $C$ of the space  $(X,d)$ is compact

My attempts:

We want to show that $\bigcap\limits_{n=1}^{\infty} C_{n}$ is compact. Let's establish $V=\{ \bigcup\limits_{\alpha}:\alpha \in A\}$ - cover of $\bigcap\limits_{n=1}^{\infty} C_{n}$. From this $V$ we want to choose finite subcover for $\bigcap\limits_{n=1}^{\infty} C_{n}$. I think that we can use the total limitation of space $C$ with metric $d$ truncated to $C$. However I don't have idea how I can do it.

答:
Recall the Heine-Borel theorem says that a set in a metric space is compact if and only if it is complete and totally bounded. It is easy to see that $C_n$ are closed, so $C$ is closed and therefore complete. It remains to show that $C$ is totally bounded.

Let $\epsilon > 0$ and choose $m \in \mathbb N$ sufficiently large such that $1/ m < \epsilon$. Since $K_m$ is compact, it is totally bounded. Therefore, we can find a finite collection of points $x_k \in X$ such that
$$ K_m \subseteq \bigcup_{k} B(x_k, \epsilon). $$
We claim that
$$ C_m \subseteq \bigcup_k B(x_k, 2\epsilon). $$
Indeed, suppose $x \in C_m$, i.e. $d(x, K_m) \leq 1/m$. By compactness, there exists a witness $y \in K_m$, that is $d(x, y) \leq 1/m$. There is some index $k$ such that $d(x_k, y) < \epsilon$. Thus by the triangle inequality
$$ d(x_k, x) \leq d(x_k, y) + d(x, y) < 2 \epsilon.  $$
Then
$$ \bigcap_n C_n \subseteq C_m \subseteq \bigcup_k B(x_k, 2\epsilon). $$
As $\epsilon$ was arbitrary, this completes the proof. The key ingredient is scaling the radius of the balls $B(x_k, \epsilon)$, which you'll find is a very common argument in the proofs of covering lemmas, which are especially useful in analysis.


问:
How to calculate $\int_{C}{}\frac{\bar{z}}{z+i} dz$ if $C$ is a circle: $|z+i|=3$?
The task:
$$
\text{Calculate } \int_{C}{}f(z)dz\text{, where } f(z)=\frac{\bar{z}}{z+i}\text{, and } C \text{ is a circle } |z+i|=3\text{.}
$$
Finding the circle's center and radius:
$$
|z+i|=|x+yi+i|=|x+(y+1)i|=3
\\
x^2+(y+1)^2=3^2
$$
Parametrizing the circle:
$$
z(t)=i+3e^{-2\pi i t}
$$

Now I need to calculate this integral:
$$
\int_{0}^{1}{f(z(t))z'(t)dt}=
2\pi\int_{0}^{1}{
\frac{1-3ie^{-2\pi i t}}{e^{4 \pi i t}}dt
}
$$

Unfortunately I calculated this integral, and it's equal to $0$. Is this correct? I don't think so. Where did I go wrong? Maybe I made a mistake when calculating the integral - what would be the best way to calculate it?

答:
Note that on the circle $C$, given by $|z+i|=3$, we have $\overline{z+i}=\frac9{z+i}$.  Hence, on $C$

$$f(z)=\frac9{(z+i)^2}+i\frac{1}{z+i}$$

Then,

$$\begin{align}
\oint_{|z+i|=3}\frac{\overline{z}}{z+i}\,dz&=\color{red}{\oint_{|z+i|=3}\frac{9}{(z+i)^2}\,dz}+\color{blue}{i\oint_{|z+i|=3}\frac{1}{z+i}\,dz}\\\\
&=\color{red}{0}\color{blue}{-2\pi}
\end{align}$$

As an alternative approach, we will evaluate the integral using parameterization.  On $C$, $z=-i+3e^{i\theta}$ where $\theta \in [0,2\pi]$.  Furthermore, on $C$C we have $f(z)=\frac{\bar z}{z+i}=\frac{i+3e^{-i\theta}}{3e^{i\theta}}$ and $dz=i3e^{i\theta}\,d\theta$.  Hence,

$$\begin{align}
\oint_C f(z)\,dz&=\int_0^{2\pi} \left(-1+i3e^{-i\theta}\right)\,d\theta\\\\
&=-2\pi
\end{align}$$

as expected!

ContentLicense: CC BY-SA 4.0


问:
Quadratic Formula problem?
There is a right triangle. The hypotenuse is 17 units. The sum of the other two sides is 23. Find the length of the two other sides.

Thanks for everyone's help in advance!

答:
$a+b=23 \implies b=23-a$.

Note that $a^{2}+b^{2}=17^{2}$ and that $a+b=23$.

Substituting $b=23-a$ in $a^{2}+b^{2}=17^{2}$ gets you:

$a^{2}+(a-23)^{2}=289 \implies 2a^{2}-46a+240=0$

Then apply the quadratic formula:

$a=\frac{46 \pm \sqrt {46^{2}-(4)(2)(240)}}{(2)(2)} =$ $\frac{46 \pm \sqrt {46^{2}-(1920)}}{(4)}\implies$$ a=15$ or $a=8$. Then you plug the a value into the first equation and you get that one of the sides is 15 units and the other side is 8 units.

ContentLicense: CC BY-SA 3.0

Title: Quadratic Formula problem?


问:
What does it mean by "depends non-linearly upon the highest order derivatives" mean?
PDE is classified as linear and non-linear ones.
Non-linear PDE is further classified as semi-linear, quasi-linear, and fully non-linear. In Evan's book, fully non-linear PDE is PDE that "depends non-linearly upon the highest order derivatives." What does it exactly mean? If I have, say, $x^2f_{xx}=0$ then since $x^2$ is non-linear, is it fully nonlinear? What if we have two highest order derivative terms, for example, $x^2f_{xx}+y^2 f_{xy}=1$?

答:
In differential equations, a non-linear DE means that the unknown function, not the independent variables, is nonlinear. So your second example,
\begin{align}
x^2f_{xx}+y^2f_{xy}=1,
\end{align}
is linear as the unknown function $f(x,y)$ has no non-linear terms.

So a fully non-linear PDE would be something like
\begin{align}
x^2(f_{xx})^2+y^2f_{xy}=1.
\end{align}
Here's a list of non-linear PDEs from wiki: https://en.wikipedia.org/wiki/List_of_nonlinear_partial_differential_equations


问:
Intersection of a downward directed family of sets
Let $X$ be a set and $Y\subseteq X$ a nonempty subset. Let $R$ be an equivalence relation on $X$ and let $K=\{K_i: i\in I\}$ be a directed family of subsets of $X$ containing $Y$. That is, for any $i,j\in I$ there exist $l\in I$ such that $K_l\subseteq K_i\cap K_j$.

I may add a comment to clarify my notation: For a subset $H\subseteq X$, $H/R=\{ [x]_R: x\in H\}$, that is, the set of $R$-classes of elements in $H$. It is also important that I am not assuming that any of the $K_i$ is the union of $R$-classes, that is, it may happen that $[x]_R=[y]_R$ and $x\in K_i$ but $y\notin K_i$. This is the real problem because otherwise the answer to the question would be trivial.

My question is the following: Does
$(\bigcap\limits_{i\in I}K_i)/R=\bigcap\limits_{i\in I}(K_i/R)$? (and does the cardinality of $I$ matter?)

For me it is clear that $(\bigcap\limits_{i\in I}K_i)/R\subseteq\bigcap\limits_{i\in I}(K_i/R)$. However, I am not able to prove the other inclusion. I can find examples where the equality does not hold if the family $K$ is not directed even for finite $I$

Edit 2: I posted this question thinking that it might be a set-theoretic property, in the context I was working $R$ is a $\emptyset$-type-definable equivalence relation and all $K_i$ are $\emptyset$-type-definable sets. In that case, one can use compactness theorem to prove the equality.

答:
Let $X=\mathbb{N}$, $R$ the equivalence relation with two classes: $\{0\}$ and $\{n\mid n>0\}$, $Y=\{0\}$, and $K_i = \{0\}\cup \{ n\in \mathbb{N}\mid n> i\}$ for all $i\in\mathbb{N}$.

Then $\left(\bigcap_{i\in\mathbb{N}} K_i\right)/R=\{0\}/R=\{\{0\}\}$ but $\bigcap_{i\in\mathbb{N}} (K_i/R)$ contains both $R$-classes, since both are represented in each $K_i$.


问:
Proving differentiability at x0 using continuity at x0

This is a question I haven't been able to complete for some time now. I have shown below what I have tried so far and I need to know the following,

1.Am I on the right track?

2.How do I show that f' is differentiable at x0?

Also I would appreciate some hints on how I could complete part (b) of the question.Thanks in advance.

答:
Consider any $\epsilon > 0$. Then, there exists a $\delta > 0$ such that if $0 < |x-x_0| < \delta$, then
$$\frac{|f(x)-a_0 - a_1(x-x_0)|}{|x-x_0|} < \epsilon$$
i.e.
$$|f(x)-a_0 - a_1(x-x_0)| < \epsilon |x-x_0| < \epsilon \delta$$
Then
$$|f(x)-a_0| - |a_1(x-x_0)| < |f(x)-a_0 - a_1(x-x_0)| < \epsilon\delta$$
So
$$|f(x)-a_0| < \epsilon \delta + |a_1| \delta$$
Now we may take $\delta$ to be smaller than $\frac{\epsilon}{|a_1|}$, so we see that
$$|f(x)-a_0| < 2\epsilon$$
That is, $\lim_{x \to x_0} f(x)-a_0 = 0$. Thus $f(x_0) = a_0$ by continuity.

Now that we have this, we see that
$$\lim_{x \to x_0} \frac{f(x)-a_0 - a_1(x-x_0)}{x-x_0} = \lim_{x \to x_0} \frac{f(x)-f(x_0)}{x-x_0} - a_1 = 0$$
i.e.
$$\lim_{x \to x_0} \frac{f(x)-f(x_0)}{x-x_0} = a_1$$
The left-hand side is $f'(x_0)$, which exists and is equal to $a_1$.


问:
Ternary and Binary representations to Prove Cantor set is uncountable (questions)
I am trying to prove that the Cantor set is uncountable, but I am very much a novice in working with different bases of numbers. I've never had to do it in any of my classes until now.

Where I'm at is this:
It is trivial that $C \underline{\subset} [0,1]$. Therefore, $|C| \leq |[0,1]|$.

Now, I want to get $|[0,1]| \leq |C|$ to apply Cantor-Bernstein-Schroeder, and I have a few questions. I am mainly looking at this link here https://sciences.csuohio.edu/sites/csuohio.edu.sciences/files/Cantor%20and%20infinite%20sets.pdf number 5 on page 2.

First, I have read that ternary numbers could have multiple binary representations. (Or maybe it's the other way around, I'm extremely new to this stuff). Is this true? If so, how would it be possible to get an injection since an input could produce multiple outputs?

Second, from looking at this link here For any two sets A and B, if $f: A \rightarrow B$ is injective, then if A is countable, B must be countable. it seems like an injection (1-1) from $C$ to $[0,1]$ (in our case) means that "$C$ has at least as many elements as $[0,1]$." So, $|[0,1]| \leq |C|$. The surjection (onto) is the other way around. With that being said, at the first link I sent, why are they giving $f(x)$? Since $f(x)$ is a surjection, doesn't that simply assert that $|C| \leq |[0,1]|$? They already said that fact was found from number 4, so I'm confused as to the importance of listing it?

Third, following on my second question, it looks like $g(y)$ is all that is needed. If $g(y)$ is injective, like they say, then we'd have a 1-1 map from $[0,1] \to C$ meaning "$[0,1]$ has at least as many elements as $C$." So, We would have $|C| \leq [0,1]$. But this goes back to my first question: if there are multiple representations, how do we know this map is indeed 1-1? Can we prove that this map is 1-1? And if so, how?

Note for readers: I know there are plenty of posts about this subject. However, I am still not fully sure how to answer the question from reading them. Due to my very limited knowledge of other bases and never having worked with the countability of sets, I'm having trouble following them. So that is why I posted this, I want some clarification.

答:
First, I have read that ternary numbers could have multiple binary representations. (Or maybe it's the other way around, I'm extremely new to this stuff). Is this true? If so, how would it be possible to get an injection since an input could produce multiple outputs?

Suppose we have two decimal expansions $x = 0.x_1x_2x_3\dots, y = 0.y_1y_2y_3\dots$ for numbers in $0$ and $1$, labelled so that $x \ge y$. Then
$$x - y = \sum_{n=1}^\infty(x_n - y_n)10^{-n}$$ where for all $n, x_n - y_n \le 9$. Therefore $$x - y \le 9\sum_{n=1}^\infty 10^{-n} = 1$$
If the first $k$ decimals agree - that is, $x_n = y_n$ for all $n \le k$ - then we can sharpen that estimate
$$x - y \le 9\sum_{n=k+1}^\infty 10^{-n} = 10^{-k}$$
But further, to get $10^{-k}, x_n - y_n =9$ for all $n > k$. The only way to accomplish this is for all the $x_n = 9, y_n = 0$. In this case, $x = y + 10^{-k}$. But $y+ 10^{-k}$ is also what you get by increasing $y_k$ by $1$ (unless $y_k = 9$, in which case, you set $y_k = 0$ and increase $y_{k-1}$ by $1$, etc.) So the original expansion $0.x_1x_2x_3\dots$ ending in repeated $9$s, and the terminating expansion from incrementing $y_k$ are two different representations of $x$.

On the other hand, if there are some $n > k$ with $x_n \ne 9$, then $x - y < 10^{-k}$. No matter how $y$ varies from $x$ beyond the first $k$ digits, it cannot be as much as a change by $1$ in the $k^\text{th}$ digit. Thus any alternative expansion for $x$ cannot make up a difference in the $k^\text{th}$ digit by the remaining digits. I.e., it must agree up to digit $k$. Similarly, if $y$ has non-zero digits beyond $y_k$, then no alternative expansion for it can disagree in the first $k$ digits.

Thus real numbers with a terminating decimal epression (i.e., all numbers of the form $\frac n{10^m}$ for $n, m \in \Bbb Z$), also have a second decimal expression agreeing up to but not including the last non-zero digit of the terminating expression. That last digit is $1$ lower, and is followed by repeating $9$. Any non-terminating decimal expression that also does not end in repeating $9$ is unique. The number it represents has no other decimal expansion.

In the the above, if you replace "$10$" with any integer base $b > 1$ and "$9$" with "$b-1$", the argument is still valid. For any base, the terminating expansions have two forms, while all other expansions are unique.

The answer to your question of how is it possible is that you take greater care than this, and pick specifically which output you want. I'll get to that later.

Second, from looking at this link here For any two sets A and B, if $f: A \rightarrow B$ is injective, then if A is countable, B must be countable. it seems like an injection (1-1) from $C$ to $[0,1]$ (in our case) means that "$C$ has at least as many elements as $[0,1]$." So, $|[0,1]| \leq |C|$.

No. The set $\{1\}$ has a injection into $\Bbb R$ (namely the map carrying $1$ to itself), but I can assure you that $\{1\}$ does not have as many elements of $\Bbb R$. The questioner failed to realize they needed to disprove statement (A), not prove it. That is why a similar, if not quite so blatant, counterexample was given in both the comments and both answers.

If you have an injection from $A$ into $B$, then that injection matches up $1-1$ every element of $A$ with an element of $B$. Every element of $A$ is matched, but $B$ can have some elements left over. If you match up the the fingers of your right and left hands, and (perhaps due to some unfortunate incident in your past), one of the hands has an unmatched finger, which hand has more fingers? The one with the left-over finger. So an injection from $A$ to $B$ means $B$ has at least as many elements as $A$, not the other way around.

Conversely, if $A$ has a surjection $g$ onto $B$, intuitively, you can create an "semi-inverse" mapping $h$ from $B$ into $A$ by picking for each element $b \in B$, one of its pre-images in $g^{-1}(b) = \{a \in A : g(a) = b\}$ (there are a lot complications if you want to prove it constructively, but this is the general idea). Since $g(h(b)) = b$, $h$ has to be injective (do you see why?). So if there is a surjection from $A$ to $B$, then $A$ has at least as many elements as $B$.

With that being said, at the first link I sent, why are they giving $f(x)$? Since $f(x)$ is a surjection, doesn't that simply assert that $|C| \leq |[0,1]|$? They already said that fact was found from number 4, so I'm confused as to the importance of listing it?

Third, following on my second question, it looks like $g(y)$ is all that is needed. If $g(y)$ is injective, like they say, then we'd have a 1-1 map from $[0,1] \to C$ meaning "$[0,1]$ has at least as many elements as $C$." So, We would have $|C| \leq [0,1]$. But this goes back to my first question: if there are multiple representations, how do we know this map is indeed 1-1? Can we prove that this map is 1-1? And if so, how?

As I've already noted, the surjectivity of $f$ does not show the same inequality already known, but the opposite inequality. But that is not why they discuss $f$. In your text, they have not (yet) discussed the relationship between surjectivity and comparison of cardinals. Yes, (under the axiom of choice) surjectivity does show the cardinality of the domain is $\ge$ the cardinality of the codomain. But that is not how they defined $\ge$ between cardinals, nor have they discussed it, so it is not something they can use.

So why talk about $f$? You'll have to ask them to be sure. $f$ and $g$ are "semi-inverses" as discussed above. $f$ is surjective, and $g$ picks one of the pre-images of $f$. I think they include $f$ as part of the general background about the Cantor set and function (if you restrict the Cantor function (often called the Cantor-Legesgue function) they define later to just the Cantor set, you get $f$.

So it is by $g$ that they show $|[0,1]| \le |\frak C|$.

But this goes back to my first question: if there are multiple representations, how do we know this map is indeed 1-1? Can we prove that this map is 1-1? And if so, how?

By specifically deciding which representation will be used:

If $y\in[0,1]$ we may write it in terms of binary expansion $y=.b_1b_2b_3\dots$, choosing one with an infinite number of zeros (in case of ambiguity).

$g$ is defined by this procedure

Expand $y$ in binary $y = 0.b_1b_2b_3\dots$. If $y$ has a terminating expansion (i.e., repeating $0$s), that one is to be used, not the expansion with repeating $1$s.

Set $g(y) = \sum_n \dfrac{2b_n}{3^n}$.

The binary expansion of $y$ is used to define the $b_n$. So that is where there is ambiguity. But there are multiple expansions only when the number has a terminating expansion, so we require that the terminating expansion always be used in that case. That removes the ambiguity. The value of $g$ is defined by a well-defined infinite sum, so there is no ambiguity in it.


问:
How to prove the existence of exactly one root of a function in between two consecutive roots of another function
I have the following question before me:
$f$ and $g$ are two functions derivable in $(a,b)$ such that $f(x) g'(x)-f'(x)g(x)>0$ for all $x$ in $(a,b)$ then prove that there lies exactly one root of $g(x)=0$ in between two consecutive roots of $f(x)=0$.

I assumed the existence of two roots of $g(x)=0$ in between two consecutive roots of $f(x)=0$ in order to reach a contradiction.
Here is how I attempted: Let $c$ and $d$ be the two consecutive roots of $f(x)=0$. Now for all $x$ in between $c$ and $d$, $f(x)$ is non-zero. Consider the function $g(x)/f(x)$ in the interval $(c,d)$. The derivative of this function is $ (f(x)g'(x)-g(x)f'(x))/((f(x))^2$.

Due to the given condition, this derivative is always positive which implies that the function is strictly increasing in nature and thus can't have equal values at any two different points in its domain. Now I consider the existence of two consecutive roots of $g(x)=0$ in $(c,d)$ which makes the value of the function $g(x)/f(x)$ $0$ at two different points which is impermissible. Thus I am able to arrive at a contradiction. From here I can conclude that there lies at most one root of $g(x)=0$ in between two consecutive roots of $f(x)=0$.

But the question asks me to show the existence of exactly one such root. How do I proceed from here to prove this? Please suggest.

答:
Assume that $g$ has no zeros in between $c$ and $d$. Consider the function $f/g$ . It has two roots at $c$,$d$ thus it's derivative has a root in $(c,d)$ but you can see that derivative of $f/g$ is always negative. Contradiction so no. of roots is greater or equal to 1 and by your previous argument no. of roots is exactly 1.

Or you can argue as follows
Since its derivative is negative so $f/g$ is strictly decreasing and now $f/g$ has a zero at $c$ then how can $f/g$ have a zero at $d$ ? as it has to increase somewhere to come back to 0.


问:
Probability that a 2 was rolled, given an even number has been rolled
I've been learning conditional probability and encountered the problem statement given in the title. We roll the die only a single time:

My method of solving the question:

A: Event that the die rolled a 2

B: Event that the die rolled an even number

P(A) = 1/6

P(B) = 1/2

Then we have P(A|B) = P(A,B)/P(B) = (1/6)*(1/2)/(1/6) = 1/2. In some places, the answer was given as 1/3. Is this method correct? Also how can we find P(A,B) when we only have a single throw of a die.

答:
Had you not been provided with the knowledge that the die rolled an even number i.e. in case of simple probability, the probability of rolling 2 on single throw of die would have been $1/6$ as there are $6$ possible outcomes. But in this case of conditional probability, there are only three possible outcomes viz. $2,4,6$ once you are told that the die rolled an even number. This makes the required probability $1/3$.

Also the formula for conditional probability $ P(A/B)=P(A\cap B)/P(B)$ gives $P(A/B)= (1/6)/(1/2)=1/3$ as the probability of rolling an even number is $1/2$.

My earlier answer had $P(A)$ in the numerator which was erroneous although that yielded the correct answer as $P(B)$ and $P(A\cap B)$ were same.


问:
How to add a non-zero mean to the equation of state of Kalman filter
The measurement data of the laser gyro is used to establish the noise random process model, and then the Kalman filter state equation is established through the model parameters. First, remove the average value of the measured data and then build the ARIMA model. The output data of the laser gyro is relatively stable (DW0 = 2.0007：Durbin-Watson statistic), and the data stationarity test is already met without difference. The established random process noise model ARIMA (1,0,0) is as follows：

${Y_t} = \mu  + {\phi _1}{Y_{t - 1}} + {\varepsilon _t}$

This formula removes the mean value and is the state equation (prediction equation) in Kalman filtering. Because if the mean value is included, it will continue to be added to the predicted value, resulting in divergence.
The parameters of ARIMA (1,0,0) are as follows：$\mu {\rm{ = }}-0.00011775$,${\phi _1} = -0.050254$,$\sigma \left( {{\varepsilon _t}} \right){\rm{ = }}2272.2$.
The figure below is the result of the measurement data after Kalman filtering:

Figure 1: The Kalman filter method is used to process the data of laser gyro without the mean value

It can be seen from the data in the figure that the center value of the filtered data (blue) and the center average of the original measurement data (black) coincide. However, if the measurement data used to establish the ARIMA model does not subtract the average value, the two center values will not coincide. The following obtains the ARIMA (1,0,0) model parameters without deducting the central value:$\mu {\rm{ = }}612.24$,${\phi _1} = -0.050254$,$\sigma \left( {{\varepsilon _t}} \right){\rm{ = }}2272.2$. And the filtered result is shown in the figure：

Figure 2: Kalman filter method is used to process the data of laser gyro without removing the mean value

The biggest difference between these two results is that the second model has a large mean $\mu {\rm{ = }}612.24$. But the mean value in the ARIMA model cannot be added to the Kalman filter's state equation. I have consulted many variants of Kalman filter algorithm, but none of them seem to mention this problem.
I tried to build a more complex ARIMA model to study this problem, and found that if the mean is small, After a limited number of iterations, the filtered mean can quickly catch up with the mean of the original measurement data. If the model mean is large, there will be a fixed deviation（as shown in Figure 2）.

How much non-zero mean value can be ignored in the state equation of Kalman filter, and the mean value of the original data can still be tracked stably？ In actual navigation applications, real-time data changes drastically, and the filtered data must track the measured data in real time. It cannot be a post-processing method that subtracts the average value and then filters. Is it that the prediction equation in the Kalman filter equation does not allow adding a mean constant? How does Kalman filter ensure that the data before and after filtering does not deviate as shown in Figure 2 in practical applications? This problem should involve the problem of stable solutions determined by the parameters of the Kalman filter equation. There should be researchers who have discussed this issue, but I have not found the relevant literature yet, I hope you can give some suggestions.

答:
One can solve this problem by adding the bias to the state vector. So one could define the state space model with state $x_t = \begin{bmatrix}Y_t & \mu_t\end{bmatrix}^\top$ as

\begin{align}
x_t &=
\begin{bmatrix}
\phi_1 & 1 \\ 0 & 1
\end{bmatrix} x_{t-1} + w_{t-1}, \\
y_t &=
\begin{bmatrix}
1 & 0
\end{bmatrix} x_t + v_t,
\end{align}

with $w_t$ and $v_t$ zero mean (possibly multivariate) normally distributed with a given covariance. This approach also allows one to estimate $\mu$ online. But in order for the state covariance of the Kalman filter using this model to remain bounded it is required that covariance of $v_t$ is non-zero. This would also allow for better state estimation if $\mu$ would drift over time.

Another possibility of incorporating $\mu$ into the state space model used by the Kalman filter is to view it as an known input. But this doesn't allow for any compensation for drifting values for $\mu$ over time, and $\mu$ would have to be estimated offline before using the Kalman filter.


问:
Sequence Induction Proof
Q: Let $(x_n)$ be a real sequence satisfying that every subsequence of $(x_n)$ does not converge in $\Bbb{R}$. Prove that $\vert{x_n}\vert$ $\implies$ $\infty$ as $n$ $\implies$ $\infty$.

My lecturer left this as an exercise that he didn't give the answer to, he set up the proof as such.

Proof: By contradiction, if $\vert{x_n}\vert$ does not $\implies$ $\infty$ (negation of $\vert{x_n}\vert$ $\implies$ $\infty$). Then $\exists$$M$ $>$ $0$, such that $\forall$ $N$ $\in$ $\Bbb{N}$, such that $\vert{x_{n^{'}}}\vert$ $\leq$ $M$

His words are "From this, there is some more work by induction, say; this will imply that there is a subsequence $(x_{n_{k}})$ using this method we can deduce a subsequence such that $(x_{n_{k}})$ $\leq$ $M$

He then continues, as $(x_{n_{k}})$ is a bounded sequence. By Bolzano-Weierstrass Theorem, there's a convergent subsequence of $(x_{n_{k}})$, call it $(x_{n_{k_{j}}})$. This is a contradiction as $(x_{n_{k_{j}}})$ is a subsequence of the original sequence $(x_n)$ which converges in $\Bbb{R}$

I just want help with the induction proof for the implication that the subsequence $x_{n_{k}}$ is bounded from the middle. How would I do this? How would this be done? I'm not sure.

答: Taking $N=1$ we see that there exist $n_1$ such that $|x_{n_1}| \leq M$. Now, taking $N=n_1$, there exist $n_2 >n_1$ such that $|x_{n_1}| \leq M$. Having found $n_1,n_2,...,n_p$ take  $N=n_p$ to find $n_{p+1}>n_p$ such that $|x_{n_{p+1}}| \leq M$.

Title: Sequence Induction Proof


问:
Evaluate $ \int_0^{2π} e^{\cos (x)}\cos(\sin x) \, \mathrm{d}x $
I have again a doubt regarding an exercise of differentiation under the integral sign. In this case, it concerns the integral:

$$ \int_0^{2π} e^{\cos(x)}\cos(\sin x) \, \mathrm{d}x $$

I tried the substitution:

$f(a,b) =  \int_0^{2π} e^{acos(x)}\cos(b\sin x) \, \mathrm{d}x  $

So then:

$\frac {\partial f} {\partial a} = \int_0^{2π} e^{a\cos(x)}\cos(b\sin x)\cos x \, \mathrm{d}x  $

and:

$\frac {\partial f} {\partial b} = -\int_0^{2π} e^{a\cos(x)}\sin(b\sin x)\sin x \, \mathrm{d}x $

Then what i did was try to integrate $\frac{\partial f}{\partial b}$ by parts, making:

$dv = -e^{a\cos x}\sin x\, \mathrm{d}x$

so:

$v = \frac{1}{a}e^{a\cos x}$

and:

$u = \sin(b\sin x)$

so:

$du = \cos(b\sin x)b\cos x$

Then we got that:

$\frac{\partial f}{\partial b} = - \frac{b}{a} \int_0^{2π} e^{a\cos(x)}\cos(b\sin x)\cos x \, \mathrm{d}x = -\frac{b}{a} \frac{\partial f}{\partial a} $

(Because evaluating $uv$ from $0$ to $2π$ gives us $0$)

So we get the partial differential equation:

$\frac{1}{b} \frac{\partial f}{\partial b} = - \frac{1}{a} \frac{\partial f}{\partial a}$

I solved this by separation of variables, making $f(a,b) = A(a)B(b)$. The end result was:

$f(a,b) = Ce^{\frac{h^2}{2}(a^2 - b^2)}$

Where $C$ is an integration constant and $h$ comes from solving the ODE associated with each $A$ and $B$.

So, my problem begins with this, because I am not sure of my result, this is mostly because of the $h$, should I pick an specific value? Or my result is just wrong? If this is the case, where did I made the mistake? Because evaluating $f(0,0)$ we get: $f(0,0) = C = 2π$, and so: $f(1,1) = 2π$, that is the correct result evaluating in wolfram alfa. I just want to know if there is a way in wich I should get the value.

答:
Let $I(a)=\int_{0}^{2\pi}e^{a\cos x}\cos(a\sin x)dx
$.
Then
\begin{align}
I’(a) =\frac1a\int_{0}^{2\pi}d(e^{a\cos x}\sin(a\sin x))=0\\
\end{align}
and $\int_{0}^{2\pi}e^{\cos x}\cos(\sin x)dx
= I(1) = I(0)=2\pi$.


问:
Find the product of the factors of $6^{12}$ that are congruent to $1 \pmod 7$

If the product of the factors of $6^{12}$ that are congruent to $1 \pmod 7$ can be expressed as $2^m \cdot 3^n$, prove that $m=n$ and find $m$.

This may be considered a sub-problem from another problem. Although I have greatly reduced calculation in my solution, if we can give a simple proof that $m=n$ before we compute them, we'd further cut calculation in my answer in half.

My efforts so far (adapted from my solution to the linked problem):

Case 1: $2^x \equiv  3^y \equiv 1$.

There are $3$ cases when $3^y\equiv 1$ for each $x \in \{0,3,6,9,12\}$ so the sum of all these $x$'s is $3 \cdot (3+6+9+12)=3 \cdot 30=90$;

There are $5$ cases when $2^x\equiv 1$ for each $y\in\{0,6,12\}$ so the sum of all these $y$'s is $5\cdot(0+6+12)=90$;

Case 2: $2^x \not \equiv 1$. Note that we must have $3^y \equiv 2,4$ otherwise we can't have $3^y 2^x \equiv 1$.

There are $2$ cases when $3^y \equiv 2^{-x}$ for each $x\in\{1,2,4,5,7,8,10,11\}$ and the sum of all these $x$'s is $2\cdot(78-30)=96$, where $78=\sum_{i=0}^{12} i$;

There are $4$ cases when $2^x\equiv 3^{-y}$ for each $y \in \{2,4,8,10\}$ so the sum of all these $y$'s is $4\cdot (2+4+8+10)=96$;

Therefore $m=n=186$.

However I feel that there is a simple argument to just prove $m=n$ so that we can cut our calculation in half. One thing I came up with to prove the sums of $x$'s and $y$'s are equal in Case 1 is as follows:

Denote $o_2, o_3$ as the orders of the elements $2$, $3$, $\pmod 7$, respectively, and $M=12$. Then in Case 1 we need to show $$3\cdot (3+6+9+12) = 5 \cdot (0+6+12)\\ \iff \color{blue}{3} \cdot \color{red}{3} \cdot \color{brown}{\frac{4(4+1)}{2}}  = \color{green}{5} \cdot \color{purple}{6} \cdot \color{red}{\frac{2(2+1)}{2}} $$ or equivalently $$\color{blue}{\left(\frac{M}{o_3}+1\right)} \cdot \color{red}{o_2} \cdot \color{brown}{\frac{\frac{M}{o_2} \left(\frac{M}{o_2}+1\right)}{2}}  = \color{green}{\left(\frac{M}{o_2} + 1\right)} \cdot \color{purple}{o_3} \cdot \color{red}{\frac{\frac{M}{o_3} \left(\frac{M}{o_3}+1\right)}{2}} \tag 1$$ which is indeed true. And for Case 2 we could do something similar but it would be lengthy and I might as well compute them directly even though there's unnecessary repetition.

My questions:

Is there any better way, e.g., via some simple argument, and better yet combining the two cases together (instead of treating them separately), that $m=n$, $\color{blue}{\it{\text{even before}}}$ we compute them? If it turns out lengthy then it's not worth typing the extra words to replace the few lines of calculation in Case 1 and Case 2 I presented earlier.

Is there a simpler way to solve the original problem by showing $a=b=c$ before computing them?

If the product of the factors of $30^{12}$ that are congruent to 1 mod 7 can be expressed as $2^{a} \cdot 3^{b} \cdot 5^{c},$ find $a+b+c$.

答: If $a$ is a factor of $6^{12}$ and $a \equiv 1 \pmod 7$, then $\frac{6^{12}}{a} \equiv \frac{(-1)^{12}}{1} \equiv 1 \pmod 7$, and these pairs multiply to $6^{12}$.  In the special case where $a = 6^6$ and $a = \frac{6^{12}}{a}$, verify that $a \equiv 1 \pmod 7$. Therefore $m = 12k + 6, k \in \mathbb N$.


问:
Electric field and curvature
My physic teacher said that

In a conductor the electric field, which is non-zero only on the surface, is stronger where the curvature is bigger*.

But he did not provide a mathematical proof for this. Furthermore, I don't know the correct statement of this proposition: does curvature mean Gaussian curvature? And which are the hypotheses? Is the statement true for every equipotential surface?

I am asking for

The correct and complete statement;

The proof (or a reference for the proof).

Thanks in advance.

EDIT

That doesn't mean that the field is proportional to curvature. It means: if we pick two points A and B, and in A the curvature is bigger than in B, then the electric field in A will be stronger than in B.

答: It is an old question, but considering the importance of the subject, I would encourage future readers to take a look at the paper: arXiv:1509.09252.</p>

Title: Electric field and curvature


问:
Finding values of $\text{Log}$ of $i$

Find all values of $\log(\sqrt 3 -i)$.

Solution:

$\log z = \log(√3 - i) = \ln(r + iθ) = \ln|z| + iθ$

It is clear that $\ln|z| = \ln\sqrt{3 + 1} = \ln|2|$

Hence $θ = \ln(√3/2) - \pi/2$.

I do not understand the last line how the principal angle is derived; I assume the principal angle would be $\tan^{-1} (1/3)$, which would make $\arg(z)=\tan^{-1} (1/3)+2n\pi$.

答: The given working is dead wrong. $\sqrt3-i=2e^{-i\pi/6}$, so $\log(\sqrt3-i)=\log2+(-\pi/6+2k\pi)i$ for $k\in\mathbb Z$.

Title: Finding values of $\text{Log}$ of $i$


问:
Evaluate $\sum\limits_{r=1}^\infty(-1)^{r+1}\frac{\cos(2r-1)x}{2r-1}$
I would like to know how to evaluate $$\sum\limits_{r=1}^\infty(-1)^{r+1}\frac{\cos(2r-1)x}{2r-1}$$

There are a couple of issues I have with this. Firstly, depending on the value of $x$, it seems, at least numerically, that this value is always
$$\pm\sqrt{\frac{\sqrt5-1}{2}}$$
If this is true, how would I prove this?

Secondly, I tried using complex numbers to evalaute this, as shown briefly below:

$$C=\sum\limits_{r=1}^\infty(-1)^{r+1}\frac{\cos(2r-1)x}{2r-1}$$
$$S=\sum\limits_{r=1}^\infty(-1)^{r+1}\frac{\sin(2r-1)x}{2r-1}$$
$$\implies C+iS=\arctan e^{ix}$$
on using the power series of $\arctan x$. Differentiating yields a completely imaginary number:
$$\frac{i}{2}\sec x$$
which proves that the value of $C$ is a constant. However, this gives me no information whatsoever on the value of $C$.

Thank you for your help.

答:
The Series

This series only converges for $x\in\mathbb{R}$. If $x\not\in\mathbb{R}$, the terms do not go to $0$.$\newcommand{\sgn}{\operatorname{sgn}}\newcommand{\Re}{\operatorname{Re}}\newcommand{\Im}{\operatorname{Im}}$
$$
\begin{align}
&\sum_{r=1}^\infty(-1)^{r+1}\frac{\cos((2r-1)x)}{2r-1}\tag{1a}\\
&=\frac12\sum_{r=1}^\infty(-1)^{r+1}\frac{e^{i(2r-1)x}}{2r-1}+\frac12\sum_{r=1}^\infty(-1)^{r+1}\frac{e^{-i(2r-1)x}}{2r-1}\tag{1b}\\
&=\frac12\tan^{-1}\left(e^{ix}\right)+\frac12\tan^{-1}\left(e^{-ix}\right)\tag{1c}\\[9pt]
&=\Re\left(\tan^{-1}\left(e^{ix}\right)\right)\tag{1d}\\[12pt]
&=\frac\pi4\,\sgn(\cos(x))\tag{1e}
\end{align}
$$
Explanation:

$\text{(1b)}$: $\cos(x)=\frac12\left(e^{ix}+e^{-ix}\right)$

$\text{(1c)}$: power series for $\tan^{-1}(z)$

$\text{(1d)}$: $\tan(\bar z)=\overline{\tan(z)}$

$\text{(1e)}$: $|\tan(z)\,|=1$ when $\Re(z)=\pm\frac\pi4$

$\phantom{\text{(1e):}}$ and $\Re(\tan(z))$ has the same sign as $\Re(z)$

$\phantom{\text{(1e):}}$ also, if $\cos(x)=0$, all the terms of the series are $0$

Concerning $\boldsymbol{\tan(x+iy)}$

Note that
$$
\begin{align}
\tan(x+iy)&=\frac{\sin(x)\cosh(y)+i\cos(x)\sinh(y)}{\cos(x)\cosh(y)-i\sin(x)\sinh(y)}\tag{2a}\\
&=\frac{\sin(x)\cos(x)+i\sinh(y)\cosh(y)}{\cos^2(x)\cosh^2(y)+\sin^2(x)\sinh^2(y)}\tag{2b}
\end{align}
$$
is on the unit circle when the absolute values of the numerator and denominator are equal. Since $\cosh^2(y)-\sinh^2(y)=1$, equation $\text{(2a)}$ says that is when
$$
\begin{align}
\sin^2(x)\cosh^2(y)+\cos^2(x)\sinh^2(y)&=\cos^2(x)\cosh^2(y)+\sin^2(x)\sinh^2(y)\tag{3a}\\[6pt]
\sin^2(x)&=\cos^2(x)\tag{3b}
\end{align}
$$
Thus, $|\tan(z)\,|=1$ when $\Re(z)=\pm\frac\pi4$

Furthermore, equation $\text{(2b)}$ says not only that $\tan(\bar z)=\overline{\tan(z)}$, but also that, when $\Re(z)=\pm\frac\pi4$, $\Re(\tan(z))$ has the same sign as $\Re(z)$.


问:
Evaluating $\int_{0}^{\pi} \frac{1}{\sqrt{u^2+2u\cos x +1}} \mathrm{d}x$

Does the integral $$\int_0^{\pi} \frac{1}{\sqrt{u^2+2u\cos x +1}} \text d x \hspace{30pt} (u \le 1) $$ has a closed form? If it has, how do we evaluate it?

I was solving a physics problem which I have asked on physics SE as well (here), and this integral popped out. How do I solve it?

答:
If you are not too familiar with elliptic integrals or have problems to compute them, you could use
$$2\,K(u^2)=\pi\, \sum_{n=0}^\infty a_n u^{2n}$$ with

$$a_n=\Bigg[\frac{ (2 n)!}{2^{2 n}\,(n!)^2}\Bigg]^2$$ the computation being simple since
$$a_{n+1}=\frac{(2 n+1)^2}{4 (n+1)^2}\, a_n$$

For the case where $u$ is close to $1$, it is much better to use
$$2\,K(u^2)=\log \left(\frac{8}{1-u}\right)+\frac{1}{2} (u-1) \left(\log
   \left(\frac{1-u}{8}\right)+1\right)+\frac{1}{16} (u-1)^2 (-5 \log (1-u)-7+15
   \log (2))+\frac{1}{96} (u-1)^3 (21 \log (1-u)+34-63 \log (2))+\frac{(u-1)^4
   (-1014 \log (1-u)-1775+3042 \log (2))}{6144}+O\left((u-1)^5\right)$$ which is extremely good for $u \geq \frac 34$.


问:
Finding integers $s$ and $t$ such that $s+t=\frac{\alpha}{b}+b$ and $st=\frac{\beta}{b}+b$ for integers $\alpha$, $\beta$, $b$
In my research I need to find integers $s$ and $t$ with the following properties (respect to $\alpha, \beta, b$) :
\begin{align}
s+t=\frac{\alpha}{b}+b\\
st=\frac{\beta}{b}+b,
\end{align}
where $\alpha, \beta, b\in\mathbb{Z}$ and $b\neq0$. Do there exist any integers $s$ and $t$ with the above properties?

Anyone can help me? Thanks in advance.

Body:
$$s+t=\frac{\alpha}{b}+b \qquad st=\frac{\beta}{b}+b\\
\implies {α + b^2 = b (s + t) \qquad b^2 + β = b s t}\\
\implies α = -b^2 + b s + b t \qquad  β = b s t - b^2\qquad  b\ne0
$$

We can see infinite solutions in terms of $\quad b, s, t\quad$ but it appears $\quad \alpha,\beta\subset\mathbb{Z}.$

Example of $(b,s,t,\alpha, \beta)$
$$(1,1,1,3,0)\quad
(2,1,1,8,-2)\quad
(3,1,1,15,-6)\quad
(4,1,1,24,-12)\\
(1,1,2,4,1)\quad
(2,1,2,10,0)\quad
(3,1,2,18,-3)\quad
(4,1,2,28,-8)\\
(1,2,2,5,3)\quad
(2,2,2,12,4)\quad
(3,2,2,21,3)\quad
(4,2,2,32,0)\\ $$

For $\quad s,t \quad\text{in terms of}\quad b, \alpha, \beta\quad$ as in Hagen von Eitzen's answer, we an immediately see that few of the lower number combinations yield integers. Some $\quad (b, \alpha, \beta )\quad$ values are

$$(1,3,2,3,1)\quad (1,3,3,2,2)\quad (1,4,3,4,1)\quad (1,4,5,3,2)\quad (1,5,4,5,1)\\$$

$$(2,4,2,3,1)\quad (2,4,4,2,2)\quad (2,6,4,4,1)\quad (2,6,8,3,2)\quad (2,8,6,5,1)\\ $$

$$(3,3,3,2,2)\quad (3,6,3,4,1)\quad (3,6,9,3,2)\quad (3,9,6,5,1)\quad (3,9,15,4,2)\\$$


问:
Is $\lim_{n \to \infty}\frac{1}{n\pi+(-1)^n\frac{\pi}{2}}=0$?
When I am trying to prove $\sin(\frac{1}{x})$ has no limit at $x=0$, I came up with this problem.  Is $$\lim_{n \to \infty}\frac{1}{n\pi+(-1)^n\frac{\pi}{2}}=0 \,\,?$$

My problem is there is $(-1)^n$ so I can't understand but I think it is convergence to $0$

Can someone help me to understand?

答:
Heuristically, $n\pi$ grows like $n$ to $\infty$, and the $(-1)^n\frac{\pi}{2}$ term is insignificant because it's just adding or subtracting a constant $\frac{\pi}{2}$.

More formally, you could use the squeeze theorem with $$(n-1)\pi\leq n\pi+(-1)^n\frac{\pi}{2}\leq (n+1)\pi.$$


答:
We always have
$$\frac1{\pi(n+1/2)}\le\frac1{n\pi+(-1)^n\pi/2}\le\frac1{\pi(n-1/2)}$$
and so by the squeeze theorem the limit is $0$.

Title: Is $\lim_{n \to \infty}\frac{1}{n\pi+(-1)^n\frac{\pi}{2}}=0$?

Body:
When I am trying to prove $\sin(\frac{1}{x})$ has no limit at $x=0$, I came up with this problem.  Is $$\lim_{n \to \infty}\frac{1}{n\pi+(-1)^n\frac{\pi}{2}}=0 \,\,?$$

My problem is there is $(-1)^n$ so I can't understand but I think it is convergence to $0$

Can someone help me to understand?


问:
How to get the coefficients of Taylor series in several variables when sums are grouped by order of partial derivatives
Assume $f$ is real analytic in several variables. The Taylor series of $f$ can be represented in two forms (cf. e.g., wikipedia). The first form is power series form: $$T(x_1,\dots,x_d)=\sum\limits_{(n_1,\dots,n_d)\in\Bbb N^d}\frac{1}{n_1!\cdots n_d!}\frac{\partial^{n_1+\cdots+n_d}f(a_1,\dots,a_d)}{\partial x_1^{n_1}\cdots\partial x_d^{n_d}}(x_1-a_1)^{n_1}\cdots(x_d-a_d)^{n_d}.\tag{1}$$
The second form is grouping summands by the order (number of variables) of partial derivatives: $$T(x_1,\dots,x_d)=f(a_1,\dots,a_d)+\sum\limits_{j=1}^d\frac{\partial f(a_1,\dots,a_d)}{\partial x_j}(x_j-a_j)\\+\frac{1}{2!}\sum\limits_{j=1}^d\sum\limits_{k=1}^d\frac{\partial^2f(a_1,\dots,a_d)}{\partial x_j\partial x_k}(x_j-a_j)(x_k-a_k)+\cdots\\+\frac{1}{n!}\sum\limits_{i_1=1}^d\cdots\sum\limits_{i_n=1}^d\frac{\partial^n f(a_1,\dots,a_d)}{\partial x_{i_1}\cdots\partial x_{i_n}}(x_{i_1}-a_{i_1})\cdots(x_{i_n}-a_{i_n})+\cdots.\tag{2}$$

I can derive the coefficients in power series form $(1)$ by differentiating term-by-term. But I encountered difficulty in deriving coefficients in the second form. Following is my attempt. The general $n$-th term in the second form is like this (please correct me if I am wrong): $$\sum\limits_{i_1=1}^d\cdots\sum\limits_{i_n=1}^d c_{(i_1,\cdots,i_n)}(x_{i_1}-a_{i_1})\cdots(x_{i_n}-a_{i_n}).\tag{3}$$ $i_1,\dots,i_n$ can take duplicated values in $1..d$. Assuming in $n$ indices $i_1,\dots,i_n$, there are $n_1\ 1$'s, $n_2\ 2$'s, $\cdots$, $n_d\ d$'s, we have actually constructed a histogram mapping $\varphi: (i_1,\cdots,i_n)\to (n_1,\dots,n_d)$ with the restriction that $n_1+\cdots+n_d=n$. Based on different histogram $d$-tuple $(n_1,\dots,n_d)$, we can further subgroup the summands in sum $(3)$ as follows. For a specific histogram distribution $(n_1,\dots,n_d)$, by merging same factors, the polynomial part of $(3)$ becomes $(x_1-a_1)^{n_1}\cdots(x_d-a_d)^{n_d}$. So, $(3)$ can be rewritten in accordance with this subgrouping as $$\sum\limits_{\begin{array}{c}(n_1,\dots,n_d)\in\Bbb N^d\\n_1+\cdots+n_d=n\end{array}}c_{(n_1,\dots,n_d)}(x_1-a_1)^{n_1}\cdots(x_d-a_d)^{n_d}.$$ This looks much like the power series form. If we take partial derivative $\frac{\partial^{n}f}{\partial x_1^{n_1}\cdots\partial x_d^{n_d}}$ and plugging in $(a_1,\dots,a_d)$, all other terms not corresponding to this $(n_1,\dots,n_d)$ tuple will vanish either because the exponent is less or because the factor is evaluated to zero, and not surprisingly we get the coefficient in exactly the same form as ones in the power series form: $c_{(n_1,\dots,n_d)}=\frac{1}{n_1!\cdots n_d!}\frac{\partial^n f(a_1,\dots,a_d)}{\partial x_1^{n_1}\cdots\partial x_d^{n_d}}=\frac{1}{n_1!\cdots n_d!}\frac{\partial^n f(a_1,\dots,a_d)}{\partial x_{i_1}\cdots\partial x_{i_n}}$ since $f$ is $C^\infty$ and therefore changing the order in which variables are taken does not change the result of mixed partial derivative.

But the denominator before the mixed partial derivative is supposed to be $n!$ from the ground truth in $(2)$ as opposed to $n_1!\cdots n_d!$. I know I must have committed some error by messing up something during the derivation, but I cannot spot it. So, could you please help me figure out the error so that I can correct it and eventually derive the coefficients in the second form having expected $n!$ in the denominator. Thank you.

答:
For any $ (n_1,\dots,n_d)\in\Bbb N^d $ let us count the coefficient at $$\frac{\partial^{n_1+\cdots+n_d}f(a_1,\dots,a_d)}{\partial x_1^{n_1}\cdots\partial x_d^{n_d}}(x_1-a_1)^{n_1}\cdots(x_d-a_d)^{n_d}$$ in the series (1) and (2). In the series (1) the coefficient is
$\frac{1}{n_1!\cdots n_d!}.$ In the series (2) it is $\frac{1}{(n_1+\dots n_d)!}$ multiplied by the multinomial coefficient ${n_1+\dots n_d\choose n_1\dots n_k}$, that is the same as in (1).


问:
Let S be a subset of $\mathbb{C}$. If $z,w \in S$, define $z ∼ w$ if and only if there is a path from z to w. Show that ∼ is an equivalence relation.
Let S be a subset of $\mathbb{C}$. If $z,w \in S$, define $z ∼ w$ if and only if there is a path from z to w. Show that ∼ is an equivalence relation.

Definition of path: A path in $\mathbb{C}$ is a continuous function $\gamma:[a,b] \longrightarrow \mathbb{C}$, $\gamma(a)$ = initial point, $\gamma(b) =$ end point.

From my understanding, I need to show 3 components: reflective, symmetric, and transitive.

For Reflective, I have:
Suppose $z_1 = x_1 +iy_1 \in \mathbb{C},$ then
$x_1 +iy_1 - (x_1 +iy_1) = 0$ is a path

Symmetric:
Suppose $z_1=x_1 +iy_1 \in \mathbb{C}$ and $w_1 = x_2+i+y_2 \in \mathbb{C}$ and $z_1Rz_2$, then there is a path from z to w. Since $\gamma:[x_2,y_2]\longrightarrow \mathbb{C}$, there is a path from w to z.

Transitive:
Suppose $z,w \in \mathbb{C}$ and z~w and w~y, then there is a path from z to w and from w to $y=x_3+iy_3$. Thus $\gamma:[x_1,y_3]\longrightarrow \mathbb{C}$ so there is a path from z to y.

Am I going about these the right way?

答:
As I said in my comment, $[a, b]$ is an interval in $\mathbb{R}$. For reflexivity, let $x \in S$ and let $\gamma:[0, 1] \rightarrow \mathbb{C}$ be the constant map given by $\gamma(t) = x$ for all $t \in [0,1]$. Then $\gamma$ is clearly continuous, and $\gamma(0) = \gamma(1) = x$. So that's a path from $x$ to $x$.

Now suppose there is a path $\gamma : [a, b] \rightarrow \mathbb{C}$ from $x$ to $y$, i.e. $\gamma(a) = x$ and $\gamma(b) = y$. Then let $\tilde{\gamma} : [a, b] \rightarrow \mathbb{C}$ be defined by
$$\tilde{\gamma}(t) = \gamma(b+a-t)$$
Then $\tilde{\gamma}$ is continuous because $\gamma$ is, and $\tilde{\gamma}(a) = \gamma(b+a-a) = \gamma(b) = y$, and $\tilde{\gamma}(b) = \gamma(b+a-b) = \gamma(a) = x$. So $\tilde{\gamma}$ is a path from $y$ to $x$.

For transitivity, suppose $\gamma:[0, 1] \rightarrow \mathbb{C}$ is a path from $x$ to $y$ and $\theta: [0, 1] \rightarrow \mathbb{C}$ is a path from $y$ to $z$. Define $\delta: [0, 1] \rightarrow \mathbb{C}$ by
$$\delta(t) = \gamma(2t), 0 \leq t \leq \frac{1}{2}$$
$$\delta(t) = \theta(2t-1), \frac{1}{2} < t \leq 1 $$
Note that $\delta$ is continuous at $t = \frac{1}{2}$. This is a path from $x$ to $z$.


问:
Quadratic Formula problem?
There is a right triangle. The hypotenuse is 17 units. The sum of the other two sides is 23. Find the length of the two other sides.

Thanks for everyone's help in advance!

答:
There are only a few sets of small integers that form right triangles.  These are:
$$(3,4,5)\\
(5,12,13)\\
(7,24,25)\\
(8,15,17)
$$

and multiples of these, for example $2\cdot(3,4,5) = (6,8,10)$.  You should memorize these, because teachers like to use them over and over again.

In this case the problem mentions that the hypotenuse is 17, so you should immediately ask if this is the $(8,15,17)$ triangle.  Since $8+15=23$, it is and you have the answer and can go on to the next problem faster than the person at the next desk.

Title: Quadratic Formula problem?


问:
This set is an ideal of $\mathbb{Z}$?
The subset $I = \{n ∈ \mathbb{Z}; \exists \,\,k \mbox{ such that }
16|n^k\}$ of $\mathbb{Z}$ is an ideal of $\mathbb{Z}$.

答:
We can prove the following more general fact: if $N>0$, $\{n\in\mathbb Z:\exists k\ge1\ N\mid n^k\}=\operatorname{rad}(N)\mathbb Z=$ $\text{A007947}(N)\mathbb Z$.

For a prospective element of the set $n$, if $n$ does not have a prime factor of $N$, all its powers clearly cannot divide $n$. Conversely, if it has all prime factors, a high enough power of $n$ will have sufficiently many prime factors (with multiplicity) to match all of $N$'s prime factors and thus be a multiple of $N$. Since $N\mid 0$, the set described above is $\operatorname{rad}(N)\mathbb Z$.

The set in the question is equivalent to $2\mathbb Z$.

Title: This set is an ideal of $\mathbb{Z}$?


问:
Implicit differentiation derivative not as expected
I am trying to solve the problem: $x^{2}+xy+y^{3}=0$ using implicit differentiation.

My workings:

$(1)\,\,\,{d\over dx}{[x^2]}\,+{d\over dx}{[xy]}\,+\,{d\over dx}{[y^3]}={d\over dx}{[0]}$

$(2)\,\,\,2x+y+{{dy^3}\over{dy}}{{dy}\over{dx}}=0$

$(3)\,\,\,2x+y+3y^2\,({dy\over dx})=0$

$(4)\,\,\,{dy\over dx}=\boxed{-{{2x+y}\over{3y^2}}}$

But the answer says it should be:

$(3)\,\,\,2x+y+{{dxy}\over{dy}}{{dy}\over{dx}} + 3y^2\,({dy\over dx})=0$

$(4)\,\,\,2x+y+{dy\over dx}(x+3y^2)=0$

$(5)\,\,\,{dy\over dx}=-{2x\,+\,y\over{x\,+\,3y^2}}$

Why?

答:
You can't conclude that $\frac{d}{dx}(xy)=y$, since this is effectively the product of the functions $x$ and $y$. Hence, using the product rule gives
$$\frac{d}{dx}(xy)=y\frac{d}{dx}(x)+x\frac{d}{dx}(y)=y+x\frac{dy}{dx}$$
The rest should work out as expected.

edited to reflect Vegeta's comment


问:
In homotopy type theory, prove that law of excluded middle implies reduction ad absurdum
It's about Excercise 2 from here:

While the principle of excluded middle $P\vee\neg P$ ( tertium non datur)
is not provable, prove its double negation using the propositions as types translation:

$\neg\neg (P\vee \neg P)$.

If for a particular proposition $P$ we can establish $P\vee\neg P$ then we can also derive the principle of indirect proof (reduction ad absurdo) for the same proposition $\neg\neg P\implies P$.
Hence show:

$(P\vee\neg P)\implies (\neg\neg P \implies P)$

However, the converse does not hold (what would be a counterexample?). However, use the two tautologies to show that the two principles are equivalent.

For the first part, $\neg\neg (P\vee \neg P)$ unfolds for me as

$
f:((P+(P\to 0))\to 0)\to 0
$

and I'm not sure how to proceed, i.e. how to construct witness $f$ (or should it ever be constructed since we are aiming at empty type).

For the second part- $(P\vee\neg P)\implies (\neg\neg P \implies P)$- I derive

$
f:(P+(P\to 0))\to (((P\to 0)\to 0)\to P)
$

and assuming $h:(P\to 0)\to 0$ and $a:P\to 0$ I'm trying something like that:

$
(f\;left\;p)h=\;...\;?
$

$
(f\;right\;a)h=h(a)\;...\;?
$

I also don't understand last paragraph of the exercise, namely that converse does not hold, but nevertheless the principles are equivalent. My understanding is that equivalence($\iff$) would mean that the converse ($\impliedby $) follows.

By the way, do you have some materials on the subject to recommend, exercises as the above, from simple to harder ones, with a lot of examples or explanation so I don't get stuck again and can develop more intuition? The materials I have found so far are vestigial in amount of such basic examples/exercises or feel like pro level for me with little or no explanation... Maybe my problem is that I skipped some prerequisites on which homotopy type theory is based and those topics would be richer in what I expect. Nonetheless I'm not sure what to google to find those topics...

答:
For the first part, you don't have a lot of choices in your strategy. You want: $$f : ¬(P ∨ ¬P) → 0$$ so your only option is to define a function: $$f\ (e : (P ∨ ¬P) → 0) = ...$$

Now you have no way of producing a value of $0$ but calling $e$: $$f\ e = e (...)$$ and you must supply a value of $P ∨ ¬P$, but you have no value of $P$, so you must try the right option: $$f\ e = e(\mathsf{right} (λ p →\ ...))$$

Now you are again in a situation where you must provide a value of $0$, but this time you have a value of $P$.

The second problem doesn't have many degrees of freedom either. One thing to keep in mind is that there is a function with type $0 → P$

The bit about the converse is perhaps a bit cryptic. When it says that the principles are equivalent, it means that e.g. assuming the axiom $$\mathsf{lem} : \prod_{P : \mathsf{Prop}_\mathcal{U}}  P ∨ ¬ P$$ is equivalent to assuming the axiom $$\mathsf{dne} : \prod_{P : \mathsf{Prop}_\mathcal{U}} ¬¬P → P$$

The reason for this is that above we exhibited $$¬¬(P ∨ ¬P)$$ for arbitrary $P$, and $\mathsf{dne}$ allows the double negation to be eliminated. However, that uses double negation elimination for $$P ∨ ¬P : \mathsf{Prop}_\mathcal{U}$$ not for $P$. I think you'll find that double negation elimination at $P$ is insufficient to prove excluded middle at $P$, which is the converse formula they were referring to. I believe they are asking you (parenthetically) to think of an example of a type $P$ for which $\mathsf{dne}\ P$ holds, but $\mathsf{lem}\ P$ does not.

As for resources, I don't have any books or anything to recommend. One thing you might want to look into, though, is trying out a computerized proof assistant like Agda (or Lean, or ...) if you haven't. Using something like that, it is very natural to have the system guide you through the sort of steps I went through in the first part. You can incrementally refine your definition, and it will show you your options at each step, and there is even an analogue of the '$...$' written in each step.

Working in such an environment can make it easier to be confident that an approach is correct or not for these sorts of problems, and aid building intuition that is necessary for similar confidence in a 'pen and paper' scenario (where you must check things that the computer would check).


问:
Rank of a matrix
If a $3 \times 3$ matrix has determinant zero, then is it possible that its rank could be $3$? I think it only could be $2$ or less. I am right or wrong? Please explain.</p>

答:
You are correct; a singular matrix (square matrix with determinant zero) cannot be full rank.

To see this, think about the effect that elementary row operations have on the determinant:

If $B$ is obtained from $A$ by exchanging two rows, then $\det(B) = -\det(A)$.

If $B$ is obtained from $A$ by multiplying a row by a $\lambda$, then $\det(B) = \lambda\det(A)$. Remember that for this to be an elementary row operation, you must have $\lambda\neq 0$.

If $B$ is obtained from $A$ by adding a multiple of one row to another row, then $\det(B)=\det(A)$.

This means that if $B$ is obtained from $A$ by a sequence of elementary row operations, then $\det(B)=0$ if and only if $\det(A)=0$; that is, you cannot go from a matrix with zero determinant to a matrix with nonzero determinant by doing elementary row operations; and neither can you go from a matrix with nonzero determinant to a matrix with zero determinant. (Though the value of the determinant can change from some nonzero number to another, because of the second type of elementary row operation mentioned above, it cannot go from zero to nonzero or from nonzero to zero).

Now suppose that $A$ is $n\times n$ and has rank $n$. Then we know that by doing elementary row operations we can transform $A$ into the $n\times n$ identity matrix. The $n\times n$ matrix has determinant $1$; so if we can transform $A$ into the identity matrix with elementary row operations, then $\det(A)\neq 0$ (because $\det(I_n)\neq 0$). So a matrix of rank $n$ has nonzero determinant. This is logically equivalent to the contrapositive: if $\det(A)=0$, then $A$ does not have rank $n$ (and so has rank $n-1$ or less).

Conversely, if the rank of $A$ is strictly less than $n$, then with elementary row operations we can transform $A$ into a matrix that has at least one row of zeros. The determinant of a matrix with one row of zeros is $0$, so that means that $\det(A)=0$.

In summary, if $A$ is $n\times n$, then $\mathrm{rank}(A)=n$ if and only if $\det(A)\neq 0$.

Title: Rank of a matrix


问:
Eigenvalue of Bessel function
Finding the eigenvalue of the Bessel function.
By applying the right boundary condition, we have:
$$
A J_{0} (\lambda a) = 0
$$
Here, we require that $A \neq 0$ to avoid a trivial solution and we let $J_{0} (\lambda a) = 0$.
How do you derive the value of $\lambda$ from this equality. In the heat equation with planar symmetry, we use the fact that $\sin (n \pi ) = 0$ and equate it with for example $\sin (\lambda x/L)$ to derive the eigenvalue $\lambda = \frac{n \pi L}{x}$.

Could someone show a resource such as a table containing the roots of $J_{0} (\lambda a)$?

Title: Eigenvalue of Bessel function

Body:
I give you below the list of the first zero's of $J_0(x)$
$$\{2.404825558,5.520078112,8.653727913,11.79153444,14.93091771,18.07106397,21.211636
   63,24.35247153,27.49347913,30.63460647,33.77582021,36.91709835,40.05842576,43.19
   979171,46.34118837,49.48260990,52.62405184,55.76551076,58.90698393,62.04846920,6
   5.18996481,68.33146933,71.47298161,74.61450065,77.75602563,80.89755587,84.039090
   78,87.18062985,90.32217264,93.46371878,96.60526795,99.74681986,102.8883743,106.0
   299309,109.1714897,112.3130503,115.4546127,118.5961766,121.7377421,124.8793089,1
   28.0208770,131.1624463,134.3040166,137.4455880,140.5871604,143.7287336,146.87030
   76,150.0118825,153.1534580,156.2950343,159.4366112,162.5781887,165.7197667,168.8
   613454\}$$

If you need more than these $54$, just tell.

For large values on $n$, you can use
$$t+\frac{1}{8 t}-\frac{31}{384 t^3}+\frac{3779}{15360 t^5}-\frac{6277237}{3440640
   t^7}+\frac{2092163573}{82575360 t^9}+O\left(\frac{1}{t^{11}}\right)$$ where $t=\pi  \left(n-\frac{1}{4}\right)$.

For $n=50$, this would give $$156.29503426853352381954991$$ while the exact value is

$$156.29503426853352381954950$$


问:
limit of $x^2e^{-{x^{-3}}}$ doesn't exist
I am trying to show that the limit when $x$ goes to zero of the function $f(x)=x^2e^{-(x^{-3})}$ doesn't exist. My intuition says that

$$\lim_{x\to 0^+}x^2e^{-(x^{-3})}=0 $$
and
$$\lim_{x\to 0^-}x^2e^{-(x^{-3})}=\infty$$

but I am not sure how to prove it. I've tried L'hopitals rule for the last one but it gets really messy.

答:
Hint:

Set $u=1/x$. This is a relatively common trick to keep in mind. Then $$\lim_{x\to 0^+}x^2e^{-(x^{-3})}=\lim_{u\to\infty}\frac{e^{-u^3}}{u^2}$$
$$\lim_{x\to 0^-}x^2e^{-(x^{-3})}=\lim_{u\to-\infty}\frac{e^{-u^3}}{u^2}$$

I wouldn't try L'Hopital's on either of these either, but perhaps these are easier to analyze.


问:
What is the mathematical definition of index set?
I find some descriptions http://en.wikipedia.org/wiki/Index_set and http://mathworld.wolfram.com/IndexSet.html .
But can't find any definition.</p>

答:
An index set is just the domain $I$ of some function $f:I\to X$. It's just a notational distinction between a function domain and an index set - when we think if it as an index set, we write $f_i$ rather than $f(i)$.

Both the Wikipedia and Wolfram links you provide indicate that the function $f$ should be $1-1$ and onto, but I don't actually think that is necessary.  For example, if we have a sequence $a_1,\dots,a_n,\dots$ then the index set is $\mathbb N$ whether or not the $a_i$ are distinct.


答:
Just use the replacement $t = \frac{-1}{x}$, then you have
$$\lim_{x\to 0^{-}} x^{2}e^{-x^{-3}} =
\lim_{t \to \infty} \frac{e^{t^{3}}}{t^{2}} =
\lim_{t \to \infty} \frac{3t^{2}e^{t^{3}}}{2t} = \lim_{t \to \infty} \frac{3te^{t^{3}}}{2} = \infty$$

Your intuition about $\lim_{x\to0^{+}}$ is of course correct because in this case both $x^{2}$ and $e^{-x^{-3}}$ both go to 0.

Title: limit of $x^2e^{-{x^{-3}}}$ doesn't exist

Body:
I am trying to show that the limit when $x$ goes to zero of the function $f(x)=x^2e^{-(x^{-3})}$ doesn't exist. My intuition says that

$$\lim_{x\to 0^+}x^2e^{-(x^{-3})}=0 $$
and
$$\lim_{x\to 0^-}x^2e^{-(x^{-3})}=\infty$$

but I am not sure how to prove it. I've tried L'hopitals rule for the last one but it gets really messy.


问:
Given 2 overlapping circles, and a given line, how can i find a pair of points such that the line fits between them?
This is hard to explain so ill attach visuals,

I have the the general equations 2 circles, i.e

$$
(x-x_1)^2 + (y-y_1)^2 = r_1^2
$$
$$
(x-x_2)^2 + (y-y_2)^2 = r_2^2
$$

and it's given that they overlap.

How can i find the pair of points (one on either circle) such that a line of some specified length $d$  and some specified angle of rotation $R$ passes through these points?

The example image depicts the 2 circles, and the black line connecting them. To clarify, the length of the black line and its angle to the X axis are to be the given variables, how can i find the pairs of points on the red and blue circles that fits the defined line?

答:
Let the two circles be $\Gamma_1$ and $\Gamma_2$. The given length and angle define up to sign a vector $v$ pointing from $p_1\in\Gamma_1$ to $p_2\in\Gamma_2$. It is clear upon translating $\Gamma_2$ by $-v_2$ that the desired pairs of points $(p_1,p_2)$ correspond to intersections of $\Gamma_1$ and $\Gamma_2-v_2$.

Thus, to find all pairs of points, translate $\Gamma_2$ by $\pm v$, find the intersections $p$ of $\Gamma_1$ and the translated $\Gamma_2$ and return for each $p$ $(p,p\mp v)$.


答: Differentiating $f$ and substituting zero, you find that you only need to find the value of $d$ from your linear system.

Body:
$f$ is a 4th degree polynomial and $f(-0.2)=2.226, f(-0.1)=2.460, f(1)=3.004, f(0.2)=3.320$. Find $f'(0)$

My attempt:

Let $$f(x)=ax^4+bx^3+cx^2+dx+e$$
then
\begin{align*}
2.226&=a(0.2)^4+b(-0.2)^3+c(0.2)^2+d(-0.2)+e\\
2.460&=a(-0.1)^4+b(-0.1)^3+c(-0.1)^2+d(-0.1)+e\\
3.004&=a+b+c+d+e\\
3.320&=a(0.2)^4+b(0.2)^3+c(0.2)^2+d(0.2)+e\\
\end{align*}
Is there any good ways to approach these equation?


问:
PDE initial value problem
Show that the solution of the initial value problem for
$u_t+u_x=\cos ^2 u$
is given by
$u(x,t)=\tan^{-1} \{ \tan [u_o(x-t)]+t\}$,
where $u_0(x)$ is the initial condition.

My attempts at a solution:
I first tried directly taking the partial derivatives of $u(x,t)$ to plug them in and verify, but I got stuck on how I would do that with the $u_o(x-t)$ part of $u(x,t)$.

I then tried separation of variables, but I couldn't successfully separate the variables $x$ and $t$.

My most recent attempt involved Laplace transform. I got to this: $U_x(x,s)+sU(x,s)=\frac{2}{s(s^2+4)}+u_0(x)$ but didn't know how to proceed (I was trying to teach myself).

Which method would you use? Also, could you please show the first few steps of your method, especially if it's one that I tried and got stuck on?

Thanks!

答:
Here is a solution by the method of characteristics. So, we assume that the equation is
$$
u_t+u_x=\cos^2 u,\quad u(x,0)=u_0(x).
$$
Consider the characteristics defined by
$$
\frac{dx}{ds}=1,\quad x(0)=\tau,\\
\frac{dt}{ds}=1,\quad t(0)=0,\\
\frac{du}{ds}=\cos^2 u,\quad u(0)=u_0(\tau).
$$
Obviously, from first two equations
$$
x=s+\tau,\quad t=s\implies \tau=x-t.
$$
From the third equation
$$
\tan u=s+\tan u_0(\tau),
$$
or, finally,
$$
u=\tan^{-1}(\tan u_0(x-t)+t)
$$
as required.</p>

Title: PDE initial value problem


问:
A set of transitive sets has an $\in$-minimal element
In "Fast Track to Forcing" Mirna Dzamonja, I have read that the Axiom of Foundation implies that

$\in$ well-orders the universe of all transitive sets. $\quad(*)$

I'm now trying to check this claim. Transitivity and asymmetry are obvious. Next, the goal is to prove that any set of transitive sets has an $\in$-minimal element. Is this even true? Am I misunderstanding $(*)$?

答:
Regularity (Foundation),

$$\forall x\big(x\ne\varnothing\to\exists y\in x(y\cap x=\varnothing)\big)\,,$$

says that every non-empty set has an $\in$-minimal element: if $y\in x$, and $y\cap x=\varnothing$, then there is no $z\in x$ such that $z\in y$, and that makes $y$ an $\in$-minimal element of $x$ by definition.


问:
Geometric expectation & perpetuity discount rate
Two observations:

If $X \sim \text{Geom}(p)$ then $E(X) = \frac{1}{p}$ (using the "first success" definition, i.e., X is supported on the set $\{1, 2, 3, ...\}$).

The present value of a perpetuity, $\text{PV}$, is the constant cashflow, $\text{CF}$, multiplied by this expectation (when the discount rate is $p$): $\text{PV} = \frac{\text{CF}}{p}$.

I'm trying to connect these concepts in an intuitive way and try to explain what this expectation is in the context of a never-ending stream of cashflows. Two questions:

What is the random variable $X$ in the context of a perpetuity?

How is $E(X)$ interpreted (as an average)?

I want to fill in the blank for the the statement:

A perpetuity is a never-ending stream of future, periodic cashflows (each cashflow in the amount of $\text{CF}$) discounted at the rate, $p$. It can be calculated by multiplying the cashflow amount by the average _____________.

答:
There is no connection in the two formulas.

$$\mathbb{E}[X]=\Sigma_{x=1}^{\infty}x\cdot p(1-p)^{x-1}=\dots=\frac{1}{p}$$

is the result of the definition of the expectation of a discrete random variable

$$\mathbb{E}[X]=\Sigma_x x\cdot p(x)$$

Where

$p$ is a probability: the probability of getting a success

$p(1-p)^{x-1}$ is a Probability Mass Function

The proof of the result of these sum is quite simple and I leave you it as an exercise

The present value of a perpetuity is the sum of a deterministic, periodic and constant amount $CF$ shifted to the present value using a fixed interest rate $p$. Just to avoid misunderstanding let's set the interest rate as $i$

Thus the PV is

$$PV=CF[(1+i)^{-1}+(1+i)^{-2}+\dots +]$$

Immediately, setting $(1+i)^{-1}=c$ you get

$$PV=CF[c+c^2+c^3+\dots +]=\frac{CF c}{1-c}=\frac{CF}{1+i}\frac{1}{1-\frac{1}{1+i}}=\frac{CF}{i}$$


问:
Let $G$ be a group with $|a_i|=n_i$ for all $1\le i \le m$. Show that $|x|=n_1n_2\ldots n_m$ where $x$ defined as follows
Let $(G,*)$ be a group. Let $a_i \in G, |a_i|=n_i, 1 \le i \le m$. Suppose $\gcd(n_i,n_j)=1$ and $a_ia_j=a_ja_i$, for all $i$ and $j$. Let $x=a_1*a_2*\ldots*a_m$. Show that
$|x| = n_1n_2\ldots n_m$.

Does it could be proved using induction on $m$? If yes, what about the base case? If I let $P(m)$ be the statement, then the base case is: $P(1)$ true
since $|x| = |a_1| = n_1$, isn't it?

答:
You need the base case to be $P(2)$ because your inductive step will look something like:

...Assuming P(n), consider $$x_1\cdot ... \cdot x_n\cdot x_{n+1} = (x_1\cdot ... \cdot x_n )\cdot x_{n+1} $$
Applying $P(2)$, this has order $|x_1\cdot ... \cdot x_n|| x_{n+1}|$...


问:
A farmer is growing tomatoes
Say a tomato farmer receives four bags of 50 seeds each of 4 types of heirloom tomatoes, let’s say Hillbillys, Mortgage Lifters, Purple Cherokees, and Brandywine reds.

However the sender warns the farmer that the chances of successfully getting a tomato plant from seed is 50%.

Given that these are specialty tomatoes, the farmer is growing them for his/her own personal consumption and enjoyment, he/she would like to grow and try at least one tomato of each variety. For greater chances of success he/she plants 7 seeds in a row of each variety, because he/she doesn’t have room to plant 4 rows of 50 seeds.

What are the chances that at least one tomato plant of each variety sprouts successfully from seed?

I tried going row by row with S for success and F for failure, thus all elements of the form SSSFSFF, counting them and then dividing to get the probability for one row $\frac{2^7 -1}{2^7}$, but other than that I don’t know where to start for the whole system.

Also, in general, what if the probability of success was $0<c<1$ and failure $1-c$ and each variety had it’s own probability of success, $c_1,c_2,c_3,c_4$?

Any help would be appreciated, thanks in advance.

答:
So in the end, it seems like you have $4$ rows of $7$ seeds each and each row of a different variety. The probability that a seed germinates is $p = \frac{1}{2}$ and that it does not germinate is $1 - p = \frac{1}{2}$.

If $X$ is the event of at least one seed germinating and considering that germination of each seed is an independent event

$P(X_1) = P(X_2) = P(X_3) = P(X_4) = 1 - \big(\frac{1}{2}\big)^7$

where $(\frac{1}{2})^7$ is the probability that no seed in a given row germinates.

Now the probability that we have at least one seed of each variety germinate,

$P = \displaystyle \big(1 - \frac{1}{2^7}\big)^4$

Title: A farmer is growing tomatoes


问:
what is the distribution for a chi-squired variable divided by its degrees of freedom?
Assume that Xi~N(0,1) are the standard normal variables that are mutually independent.

We know that Y=X1^2+...+Xm^2 is a chi-squired random variable with m degrees of freedom.

Let Z= Y/m=(X1^2+...+Xm^2)/m

what is the distribution for Z? Does Z have a chi-squared distribution with 1 degree of freedom?

答:
No, it does not. I do not know why you think so.

As you know

$$Y\sim \chi_{(m)}^2=Gamma\Bigg(\frac{m}{2};\frac{1}{2}\Bigg)$$

Thus

$$\frac{1}{m}Y\sim Gamma\Bigg(\frac{m}{2};\frac{m}{2}\Bigg)$$

The proofs of all these result are quite easy...if you are interested in show your attempts

Proof:

$$f_Y(y)=\frac{\Big(\frac{1}{2}\Big)^{m/2}}{\Gamma\Big(\frac{m}{2}\Big)}y^{m/2-1}e^{-y/2}$$

You have to calculate the density of

$$Z=\frac{Y}{m}$$

It is immediate using the Fundamental Transformation theorem:

$$f_Z(z)=f_Y[g^{-1}(z)]\Bigg|\frac{d}{dz}g^{-1}(z)\Bigg|$$

being $y=zm$ and $y'=m$


问:
Is the distribution of $f\circ X$ determined by the partial integrals over $\sigma(X)$?
This is a follow up question to this one.

Let $(\Omega_i, \mathcal{F}_i, \mathbb{P}_i)$ be a probability space for $i \in \{1,2\}$, let $X_i:\Omega_i\rightarrow[-\infty, \infty]$ be a random variable, and let $f_i:[-\infty,\infty]\rightarrow[-\infty,\infty]$ be a Borel-measurable function such that $f_i\circ X_i$ is integrable, respectively. If $\int_{\{X_1 \in A\}} f_1\circ X_1\ d\mathbb{P}_1 = \int_{\{X_2 \in A\}} f_2\circ X_2\ d\mathbb{P}_2$ for every Borel set $A$, does it follow that $f_1\circ X_1 \overset{d}{=} f_2\circ X_2$? (Note that I did not write $\int_{\{f_i\circ X_i \in A\}} f_i\circ X_i\ d\mathbb{P}_i$.)

答:
No.  Here's a simple counterexample (although there are many).  Let $X_1$ be distributed uniformly on $[0,1]$ and let $X_2$ be distributed according to the pdf $\pi(x) = 2x$ on $[0,1]$.  Then let $f_1(x) = 2x$ and $f_2(x) = 1$.  Calculate
\begin{align*}
\int_{X_1 \in A} f_1 \circ X_1 \,d\mathbb{P} &= \int_{X_1 \in A} 2X_1 \,d\mathbb{P} = \int_A 2x \,dx \\
\int_{X_2 \in A} f_2 \circ X_2 \,d\mathbb{P} &= \int_{X_2 \in A} 1\,d\mathbb{P} = \mathbb{P}(X_2 \in A) = \int_A 2x \,dx.
\end{align*}
But the distribution of $f_1 \circ X_1$ is normalized Lebesgue measure on $[0,2]$ and the distribution of $f_2 \circ X_2$ is a point mass at $1$.


Body:
Let $a_0,a_1,\dots,a_n$ and $b_0,b_1,\dots,b_n$ be sequences of real number. Suppose $a_0=b_{2020}$ and $b_0=a_{2020}$ and that $\displaystyle a_n=\frac{\sqrt{2n-1}}{2}+a_{n-1}$ and $\displaystyle b_n=\frac{\sqrt{2n-1}}{1010}-b_{n-1}$ for $n=1,2,\dots,2020$. Find $\displaystyle\sum_{k=1}^{2020}(a_kb_{k-1}-a_{k-1}b_k)$.

My attempt:

Let $$S_{2020}=\sum_{k=1}^{2020}(a_kb_{k-1}-a_{k-1}b_k)$$
then
\begin{align*}
S_{2020}&=\sum_{k=1}^{2020}\left(\frac{a_{k-1}\sqrt{2k-3}}{1010}-\frac{b_{k-2}{\sqrt{2k-1}}}{2}-\frac{a_{k-2}\sqrt{2k-1}}{1010}-\frac{b_{k-1}{\sqrt{2k-3}}}{2}\right)-S_{2019}+(a_1b_0-a_0b_1)
\end{align*}

Body:
Given two relations can be rearranged and multiplied to give $$(a_k-a_{k-1})(b_k+b_{k-1})=\frac{2k-1}{2020}$$

$$(a_kb_k-a_{k-1}b_{k-1}) + (a_kb_{k-1}-a_{k-1}b_k)=\frac{2k-1}{2020}$$

Now summing over,

$$\sum_{k=1}^{2020} (a_kb_k-a_{k-1}b_{k-1}) + \sum_{k=1}^{2020} (a_kb_{k-1}-a_{k-1}b_k) = \sum_{k=1}^{2020} \frac{2k-1}{2020}$$

The first sum telescopes. The second term is our desired sum, $S$. RHS is a multiple of sum of first $2020$ odd numbers, which is simply $2020^2$.

$$(a_{2020}b_{2020}-a_0b_0) + S =2020$$

Since $a_{2020}=b_0$ and $b_{2020}=a_0$, $$\boxed{S=2020}$$


问:
Help calculate the probablility of multiple second place lottery winners
I'm trying to calculate the probability of multiple second place lottery winners. For example, on January 13th there were 10 second place winners (https://www.beatlottery.net/powerball/results/draw_date/2021-01-13). Is having ten second place winners more or less likely than having one jackpot winner?

For reference, I am referring to the Powerball lottery (https://en.wikipedia.org/wiki/Powerball)

答:
I'll try to answer. :)

Is having ten second place winners more or less likely than having one
jackpot winner?

Going from the wikipedia on the US Powerball Odds from here:
https://en.wikipedia.org/wiki/Powerball#Prizes_and_odds

The chance of getting first place, i.e. 5 numbers plus the Powerball, is
1 in 292,201,338.

The chance of getting second place, i.e. five numbers without the Powerball, is
1 in 11,688,053.52.

So, and I'm no mathematician, by the way...
I'd hazard a guess that the chance of 10 people getting 2nd division would be more likely than 1 person getting first division...

eg. 1 in 11,688,053.52 x 10 = 1 in 116,880,535.20.

and I'd take that further and say its more likely that 20+ people would get 2nd division than 1 person getting first division...
ie. above x2...or 1 in 233,761,070.40, which is still better than the chance of 1 person winning 1st division.

If you are trying to estimate the actual number of 2nd place winners, you'd need to know, for a given draw, the number of entrants....and then divide that number by the chance of them winning 2nd place.

Obviously, for each 11,688,053.52 entrants, there would be on average 1 2nd Place winner - assuming that each entrant only bet on one group of 6 numbers, of course, which is probably not the case.


问:
Does $\int_0^{\frac {\pi}{2}}{\sqrt {\sin x\tan x}}\space dx $ diverge or converge?
I am trying to decide if this integral converges or diverges, I got stuck calculating, so I tried to use the comparison test, but I'm not sure of which function to choose, I would appreciate any hints and help in how to choose which function to compare my integral with, or any tips in general about how to approach these questions.
 Thanks in advance.

Body: Hint:\begin{align}\lim_{x\to\pi/2^-}\frac{\sqrt{\sin(x)\tan(x)}}{\frac1{\sqrt{\pi/2-x}}}&=\lim_{x\to\pi/2^-}\frac{\sin(x)}{\sqrt{-\frac{\cos(x)}{x-\pi/2}}}\\&=1.\end{align}


问:
Are Noetherian commutative rings finitely generated algebras?
I wonder if every Noetherian commutative ring is a finitely generated algebra over some PID.

Has this problem been proved or disproved?

Thank you for your answer.

答: Counterexamples exist. A finitely generated algebra over a PID has finite Krull dimension, but there are Noetherian rings with infinite Krull dimension.</p>


问:
Solving homogeneous exponential system
Suppose we want to minimize the following energy function
\begin{equation}
\min_{\boldsymbol{x}, \boldsymbol{y}}  \|\boldsymbol{Ax} - e^{\boldsymbol{By}} \|_2^2
\end{equation}
where $\boldsymbol{A} \in \mathbb{R}^{m, n}, \boldsymbol{B} \in \mathbb{R}^{m, k}$, with $n$ and $k$ less than $m$.

Questions:

Is the above optimization convex?

Is there a closed-form solution for the above problem? I know $\|\boldsymbol{Ax} - \boldsymbol{By}\|$ can be solved up to a scale ambiguity via SVD, I have no idea whether the above minimization can also be solved.

答:
Is the above optimization convex ?

Yes, it is, as squared-l2-norm is a convex function by the definition.

Is there a closed-form solution for the above problem ?

The problem seems to be not well conditioned, i.e. we could solve it by making $x$ exponentially small positive value, and $y$ exponentially large negative value, will do the job.

e.x. $ \forall x_i = 10^{-6}, \ \forall y_i = -10^{6}$

\begin{equation}
\min_{\boldsymbol{x}, \boldsymbol{y}}  \|\boldsymbol{Ax} - e^{\boldsymbol{By}} \|_2^2 \approx 0
\end{equation}

So there's no need to a closed-form solution.

Moreover $B$ is not a square matrix, which implies there's no analytical solution for this system.

Title: Solving homogeneous exponential system


问:
Find the smallest natural number that satisfy $13^N = 1 \pmod {2013}$

Moderator Note: This is a current contest question on Brilliant.org.

Find the smallest natural number that satisfy:

$$13^N = 1  \pmod {2013}$$

My idea is to use the Fermat's little theorem for prime numbers. First I factorize 2013 as 3*11*61

Then i get using the theorem i get:

$$13^{61-1} = 1 \pmod{ 61}\iff 13^{60} = 1 \pmod {61}$$
$$13^{11-1} = 1 \pmod{11}\iff13^{10} = 1 \pmod{11}$$
$$13^{3-1} = 1 \pmod 3\iff 13^{2} = 1 \pmod{3}$$

But because the modulo bases are different I doesn't know how to continue. And even if they were same, the multiplication won't change the modulo base. Is there any way to achieve this?

Actually my idea is to find the smallest natural nubmer that satisfies

$$13^N = 1 \pmod{ 3\cdot 11\cdot 61}$$

答:
Using Fermat's Little Theorem, $a^{p-1}\equiv1\pmod p$ where $p$ is prime and $a$ is any integer relatively prime to $p$

If modulo/Multiplicative  order of $a$ is $ord_pa,$ "as a consequence of Lagrange's theorem,  $ord_pa$ always divides  $\phi(p)=p-1$", we need to test for those powers of $a$ which divides $p-1$

Now,
$$13^1\equiv1\pmod 3\implies ord_3{(13)}=1$$
$$13\equiv2\pmod {11}\implies 11^2\equiv2^2=4,11^5\equiv2^5\equiv-1\implies ord_{11}{(13)}=10$$
$$13^2=169\equiv-14\pmod {61},13^3\equiv(-14)\cdot13\equiv1\implies ord_{61}{(13)}=3$$
So, $ord_{3\cdot11\cdot61}(13)=$lcm$(1,10,3)=30$

Related : Carmichael Function


问:
Does there exists any non-trivial group homomorphism from $\mathbb{R}$ to $\mathbb{Q}$
If $f:\mathbb{R}\longrightarrow \mathbb{Q}$ then it should be of the form $f(q)=qf(1)$, then how it will work for irrationals?

答:
The answer depends on the axioms of set theory. Using the axiom of choice, $\mathbb{R}$ has a basis as a vector space over $\mathbb{Q}$, so you can write down examples of nontrivial linear maps $f : \mathbb{R} \to \mathbb{Q}$ by defining then arbitrarily on such a basis.

On the other hand, it's consistent with the negation of the axiom of choice that every homomorphism $\mathbb{R} \to \mathbb{Q}$ is continuous, which means it is necessarily zero since its image must be connected. See this MO thread for some details.


问:
integral closure of $k[[t]]$ in algebraic closure of $k((t))$ is not Noether
Let $k$ be a field.
I want to prove that the integral closure of $k[[t]]$ in algebraic closure of $k((t))$ (we call it $A$) is not a Noetherian ring.

Thank you for your help. Reference (a webpage, pdf, etc.) is also appreciated.

P.S

From the theory of extension of valuation, we can say $\operatorname{Spec} A$ is Noetherian as topological space.
Thus we can construct an example of $\operatorname{Spec} A$ which is not noetherian as a scheme but Noetherian as a topological space.

答:
If $k$ has characteristic $0$ then $$A=\bigcup_{n\ge 1} \overline{k}[[t^{1/n!}]], \qquad \overline{k((t))} = A[t^{-1}]$$
and
$(t^{1/n!})$ is a strictly increasing sequence of ideals.

$Spec(A) = \{0\} \cup (\bigcup_{n\ge 1}(t^{1/n!}))$ is a 'Noetherian topological space'.

When $char(k)=p$ then $t^{1/(n+1)!-1/n!}$ is not integral over $k[[t]]$ thus $$(t^{1/n!})\subsetneq (t^{1/(n+1)!})\subsetneq A$$


问:
Find $f:[0, 1]\to \mathbb R$ that maximizes $I(f)-J(f)$, where $I(f)=\int_0^1 {x^2 f(x)dx}$, $J(f)=\int_0^1{x\left(f(x)\right)^2 dx}$
I've never seen this kind of problems - finding a function with almost no conditions which maximizes the integral - so I'm asking for a hint. The problem is as follows.

Find a continuous function $f:[0, 1]\to \mathbb R$ that maximizes $I(f)-J(f)$, where $$I(f)=\int_0^1 {x^2 f(x)dx},\ \ \ \ \ \  J(f)=\int_0^1{x\left(f(x)\right)^2 dx}$$

I tried to select candidates of types of functions that can possibly make the given integration maximum. Assuming that $f$ is a function that $\exists I(f), J(f)$, I considered a function $g(x):=xf(x)\{x-f(x)\}$, but still don't have a idea. Any helps will be very appreciated. Thanks.

答:
Hint: For any $x \in [0,1]$, we have $xy(x-y) \le \dfrac{x^3}{4}$ for all $y \in \mathbb{R}$, with equality if $y = \dfrac{x}{2}$.

You can prove this using basic calculus. Can you use this to find the maximum of $I(f)-J(f) = \displaystyle\int_{0}^{1}xf(x)(x-f(x))\,dx$?


问:
parametrization of surfaces and area
The question is:

Paraboloid $z=x^2+y^2 $ divides the sphere $x^2+y^2+z^2=1$  into two parts, calculate the area
of each of these surfaces.

I know that i need to use $\iint |n| dS$ but i need to first find the normal to this surfaces and for that i need to parameterize each part, i do not know much about parametrization of part of surfaces, how can i use $\theta,\phi$ for this parametrization or should i use something else?

Any suggestion would be great, thanks

答:
The intersection of the paraboloid and the sphere will be a circle, which occurs at

$$z+z^2=1 \implies z = \frac{\sqrt{5}-1}{2} \equiv z_0$$

From here there are many avenues you can take to calculate the area of the top cap (the bottom one is just $4\pi$ minus this area).

$\textbf{Option 1}$: Directly

Parametrize with spherical coordinates, $r = 1$ and $J = \frac{\frac{\partial(x,y,z)}{\partial r}}{\left|\frac{\partial(x,y,z)}{\partial r}\right|}\Biggr|_{r=1} = \sin\theta$

$$\iint\limits_{x^2+y^2+z^2=1\cap z\geq z_0} dS = \int_0^{2\pi}\int_0^{\cos^{-1}(z_0)}\sin\theta\;d\theta d\phi = 2\pi(1-z_0) = (3-\sqrt{5})\pi$$

$\textbf{Option 2}$: Surface area formula

Plug in $z = \sqrt{1-x^2-y^2}$

$$\iint_{x^2+y^2\leq z_0} \sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2}dA = \int_0^{2\pi}\int_0^{\sqrt{z_0}} \frac{r}{\sqrt{1-r^2}}drd\theta = 2\pi(1-\sqrt{1-z_0}) = 2\pi(1-z_0)$$

since $z_0^2 = 1-z_0$

$\textbf{Option 3}$: Divergence Theorem

Consider the vector field $F(x,y,z) = \left(0,0,\frac{1}{z}\right)$ and close the top spherical cap with the plane $z=z_0$. Using $n = (x,y,z)$ for the spherical cap, we have that

$$\iint\limits_{\text{cap}}dS = \iint_{x^2+y^2+z^2=1\cap z\geq z_0}\left(0,0,\frac{1}{z}\right)\cdot(x,y,z)dS = \iiint\limits_{x^2+y^2+z^2\leq1 \cap z\geq z_0}-\frac{1}{z^2}dV + \iint_{x^2+y^2+z^2\leq 1 \cap z=z_0}\frac{1}{z_0}dS$$

$$= \int_0^{2\pi}\int_{z_0}^1\int_0^{\cos^{-1}\left(\frac{z_0}{r}\right)}-\frac{\sin\theta}{\cos^2\theta}\:d\theta dr d\phi + \frac{1}{z_0}\iint_{x^2+y^2\leq z_0} dA = 2\pi \int_{z_0}^1 1 - \frac{r}{z_0}\:dr+\pi$$

which will simplify to the other options.

Title: parametrization of surfaces and area


问:
Show that $d/dx (a^x) = a^x\ln a$.
Show that
$$
\frac{d}{dx} a^x = a^x \ln a.
$$

How would I do a proof for this.  I can't seem to get it to work anyway I try.

I know that
$$
\frac{d}{dx} e^x = e^x.
$$
Does that help me here?

答:
Let $f : \mathbb{R} \to \mathbb{R}$ be given by $f(x)=a^x$ and consider the $\ln$ function. We can take the composition so that we have:

$$(\ln\circ f)(x)=\ln (a^x)=x\ln a$$

Now, if we take the derivative, on the left hand side we use the chain rule and on the right hand side we differentiate as usual so that we have:

$$\frac{f'(x)}{f(x)}=\ln a$$

Now solving for $f'(x)$ gives $f'(x) = f(x) \ln a$ so that $f'(x) = a^x \ln a$. This useful technique can be used to take derivatives of other functions: we compose the original function with the inverse and then differentiate on both sides and use the same idea we've used here, this technique can simplify many derivatives and save a lot of time in some situations.

Title: Show that $d/dx (a^x) = a^x\ln a$.


问:
How to solve the Cauchy problem?
I have a task that looks like
$$ \frac{\partial u}{\partial t} = 2\frac{\partial^2 u}{\partial x^2} + \frac{\partial u}{\partial x} + u + 3e^{t},\, u\left(x, 0\right)=\sin{x}$$

Please help, thanks :)

答:
We continue Matthew Pilling solution:
Using Laplace transform method we get ode
$$2v_{xx}+v_{x}+(1-s)v=-\sin(x)+\frac{3}{1-s}$$
Particular solution is
$$v=\frac{s\sin x+\sin x+\cos x}{s^2+2s+2}+\frac{3}{(s-1)^2}$$
Final solution of Cauchy problem is
$$u(x,t)=\mathcal{L}^{-1}v=e^{-t}\sin(x+t)+3te^t$$

Title: How to solve the Cauchy problem?


问:
Every Archimedean Ordered Field is field isomorphic to a subfield of $\mathbb{R}$.
Let $\mathbb{F}$ be an archimedean ordered field. Then $\mathbb{F}$ is isomorphic to a subfield of $\mathbb{R}$.

Let $x \in \mathbb{F}$, then define $S_x:=\{\frac{m}{n} \in \mathbb{Q}: \frac{m_{\mathbb{F}}}{n_{\mathbb{F}}}<x\}$.$\mathbb{F}$ being archimedean makes $S_x$ bounded in $\mathbb{R}$.

Define $\phi:\mathbb{F} \to \mathbb{R}$ such that $\phi(x)=\operatorname{lub}S_x$

Claim 1: $\phi$ is injective.

Let $x_1, x_2 \in \mathbb{F}$ with $x_1<x_2$ one can obtain an element of the form $\frac{m_{\mathbb{F}}}{n_\mathbb{F}}$ such that $x_1<\frac{m_{\mathbb{F}}}{n_\mathbb{F}}<x_2$.

Now how can I show that $\phi(x_1) \le \frac{m}{n}$?

Claim 2: $\phi(x+y)=\phi(x)+\phi(y)$ i.e $\operatorname{lub}S_{x+y}=\operatorname{lub}S_x+\operatorname{lub}S_y$

From the construction of $S_{a}$ it follows $S_x+S_y \subseteq S_{x+y}$ $\implies$
$\phi(x)+\phi(y) \le \phi(x+y)$

How do I show the reverse inequality?

Claim 3: $\phi(xy)=\phi(x)\phi(y)$ i.e $\operatorname{lub}S_{xy}=\operatorname{lub}
S_{x}\operatorname{lub}S_y$

How do I do this?

答:
I am writing @Paul K comment as an answer:

Claim 2: $\Phi(x+y)=\Phi(x)+\Phi(y)$ $\implies$
$\operatorname{lub}S_{x+y}=\operatorname{lub}S_x+\operatorname{lub}S_y$

By construction:

$S_x+S_y \subseteq S_{x+y}
\Rightarrow \operatorname{lub}(S_x+S_y) \le \operatorname{lub}S_{x+y}$

$\Rightarrow \operatorname{lub}S_x+\operatorname{lub}S_y \le \operatorname{lub}S_{x+y}$
$\Rightarrow \phi(x)+\phi(y) \le \phi(x+y)$.

Let $q \in S_{x+y}$

$\Rightarrow$ $q_{\mathbb{F}}<x+y$.

Define $h=\left(x+y-q_{\mathbb{F}}\right)$.

Choose $p \in S_x$ and $t \in S_y$ with $x-h/2<p_\mathbb{F}<x$ and $y-h/2<t_{\mathbb{F}}<y$.

$\Rightarrow$ $x+y-h<p_{\mathbb{F}}+t_{\mathbb{F}}$ $\Rightarrow$ $q_{\mathbb{F}}<p_{\mathbb{F}}+t_{\mathbb{F}}=(p+t)_{\mathbb{F}}$.

So $q<p+t\le \Phi(x)+\Phi(y)$.

$\Phi(x)+\Phi(y)$ is an upperbound of $S_{x+y}$
$\Rightarrow \Phi(x+y)\le \Phi(x)+\Phi(y)$

Hence $\Phi(x+y)=\Phi(x)+\Phi(y)$.


问:
Analytic function in domain D and local max

For $f$ that is analytic on domain D, $|f(z)|$ has no local maximum within its domain of analyticity.

I have proven the result that $|f(z)|\leq$ max $|f(z+re^{it})|$ ($0\leq t\leq 2\pi$), and am trying to use it show that $|f|$ has no strict local maximum within its domain of analyticity by contradiction.

Is it just assuming that if there is a maximum say $f(z_0)$ then we can find a small enough r s.t. $f(z_0+re^{it})$ is still within the domain but bigger than $f(z_0)$? If so, I was wondering how can I write it down in a more rigorous way.

答: By Maximum Modulus Principle the fact that $f$ attains its maximum at an interior point of  some closed disk around $z$ implies that $f$ is a constant on the disk. Since the domain of $f$ is connected this implies that $f$ is  a constant throughout $D$.


问:
Given two identical groups with X distinct entities, and randomly selecting Y entities. What are the chances each selection contains at least 1 match?
My wife asked a version of this question because she was curious about the probability of getting matching villagers in animal crossing. I was curious to know the generalized method of solving this problem so I can answer her question. In Animal Crossing, there are 397 villagers, but you are only allowed to have 10 at a time. So, knowing that she's limited to 10/397, what's the chances that at least 1/10 is matched with one of my 10?

To give a much smaller example that can be brute forced by hand, consider a pool of 5 Villagers and you can only have 2 on your island. There are 10 combinations of 2 villagers:

AB / AC / AD / AE / BC / BD / BE / CD / CE / DE

From this we can determine that there's a 70% chance that at least one of my 2 would match with one of her 2.

Say I have villagers "B" and "E", I have at least 1 match with 7 of the permutations (and this is true with any group of 2):

AB / AE / BC / BD / BE / CE / DE

答:
As Gribouillis says in the comments the key to doing this computation is to express it as $1$ minus the probability that there are no matches. If you fix your wife's set of $10$ villagers without loss of generality, you can have ${397 \choose 10}$ possible sets of villagers, and in order to have no matches you have to choose your villagers from the ones your wife doesn't have, which can be done in ${387 \choose 10}$ ways. So the probability is

$$1 - \frac{ {387 \choose 10} }{ {397 \choose 10} } \approx 0.227 \dots $$

or about $\boxed{ 23 \%}$.

An easier and rougher version of this calculation which doesn't tell you the exact probability but gives you an upper bound on it, and which you can more easily do in your head on the fly, is the following. There are $10^2$ different pairs (your wife's villager, your villager), and each of them has a probability of $\frac{1}{397}$ of being the same villager. This means the expected number of same villagers is

$$\frac{10^2}{397} \approx 0.252 \dots$$

which is both an upper bound and not that far off from the probability.

In general if there are $n$ possible villagers and you pick $k$ of them the exact answer is

$$1 - \frac{ {n-k \choose k} }{ {n \choose k} } = 1 - \prod_{i=0}^{k-1} \left( 1 -  \frac{k}{n-i} \right)$$

which is a little tricky to analyze in detail, although it can be done. But the expected value computation easily gives $\frac{k^2}{n}$, which is an upper bound, and can be shown to be a pretty acccurate upper bound as long as $k$ isn't too big compared to $n$.


问:
Proof Solver Geometry
Are there any programs that can solve Geometry Problems?

An example of such a problem would be:
The centroid of a triangle always divides its medians into two sections with a 1:2 ratio.

While that problem is very easy to solve by hand, I am wondering whether it can be automated in any way.
Thanks in advance!

答:
There are several such programs:

Geometry Expert: http://www.mmrc.iss.ac.cn/gex/

Java Geometry Expert: https://github.com/yezheng1981/Java-Geometry-Expert

GCLC: http://poincare.matf.bg.ac.rs/~janicic/gclc/

GeoTher: https://www-polsys.lip6.fr/~wang/

Implementation of theorem proving inside GeoGebra (http://www.geogebra.org) and its experimental version GeoGebra Discovery (https://github.com/kovzol/geogebra-discovery)

Implementation of the area method in Coq: https://github.com/coq-contribs/area-method

A library of examples with automatic proofs: http://hilbert.mat.uc.pt/TGTP/index.php, http://hilbert.mat.uc.pt/GeoThms/

An overview of the different methods and implementations can be found in our chapter:

Julien Narboux, Predrag Janičić and Jacques Fleuriot. Computer-assisted Theorem Proving in Synthetic Geometry. Meera Sitharam; Audrey St. John; Jessica Sidman. Handbook of Geometric Constraint Systems Principles, Chapman and Hall/CRC, In press, Discrete Mathematics and Its Applications, 1498738915. ⟨hal-01779452⟩

An overview of the implementation in GeoGebra can be found here:

Francisco Botana, Markus Hohenwarter, Predrag Janičić, Zoltán Kovács, Ivan Petrović, Tomás Recio and Simon Weitzhofer.
Automated Theorem Proving in GeoGebra: Current Achievements.
Journal of Automated Reasoning volume 55, pages 39–59 (2015)

and a more recent publication on the same program is here:

Z. Kovács, Tomás Recio, M. Pilar Vélez: Using Automated Reasoning Tools in GeoGebra in the Teaching and Learning of Proving in Geometry. International Journal of Technology in Mathematic Education. Vol. 25, no. 2. pp. 33-50. 2018.  (https://core.ac.uk/download/pdf/188016037.pdf)

or here:

Francisco Botana, Zoltán Kovács, Tomás Recio: A mechanical geometer. Mathematics in Computer Science, online Nov. 20, 2020.  https://doi.org/10.1007/s11786-020-00497-7

Title: Proof Solver Geometry


问:
Do there exists any Order Complete Field at all?
It is known that any two Order-Complete Fields are isomorphic. So there can exist at most 1 Order-Complete Field up to isomorphism.

Is there a way to tell that there exists one Order Complete Field?

答: The reals field $\mathbb R$ is one.

Body: The reals field $\mathbb R$ is one.


问:
Question about $\mathbb{Z}$
I was reading about minimal and maximal subgroups and in one example it is mentioned that the additive group $\mathbb{Z} $ does not have minimal subgroups. I tried to prove this by assuming that $ \mathbb{Z} $ does have minimal subgroups to arrive at a contradiction, starting from the fact that the subgroups of $\mathbb{Z} $ are cyclic but it doesn't get anywhere.

答: So, as pointed out already by @ArturoMagidin, any subgroup is of the form $\langle n\rangle$.  But there are always multiples of a given $n$ (in fact infinitely many: $n,2n,3n,\dots$).  Now the (cyclic) subgroup $\langle kn\rangle\lt\langle n\rangle$ for any $\Bbb Z\ni k\gt1$.  Thus $\langle n\rangle$ is not minimal.

Title: Question about $\mathbb{Z}$


问:
number of zeros of function $\prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2}\right)-1$
$$f(z)=\prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2}\right)-1$$

How many zeros does the above function have in $\Bbb{C}$?

答: As noted in a comment, this is essentially the problem of determining the number of solutions to $\sin(z)=z$.  There are infinitely many. It turns out more generally, as I learned from Aryabhata's answer at this link, with reference to an example in Markuševič's book at this link, that for every complex number $A$, the equation $\sin(z)=Az$ has infinitely many solutions in the complex plane.</p>


问:
Encoding 2 numbers into 1
Say we have two integers, $a$ and $b$. I need a way to combine these numbers into one unique number $x$, such that they can both be recovered from $x$ and no other numbers can be recovered from $x$
There are plenty of ways to do this, but the problem is that I am using this for programming and I only have the following operators:

$+$ $-$ $\times$ $\div$ $a^x$ $\sqrt{}$

I also have sin and cos but I doubt those will be needed.
Is it possible to encode and decode $a$ and $b$ using just these operators?

答:
You didn't specify whether $x$ needed to be an integer. In order to avoid massive numbers in exponentation, we could take
$$x=a+\frac{1}{b}$$
Since $a,b$ are integers, $a$ will just be the integer part of $x$ and $1/b$ will be the reciprocal part, which you should be able to computationally reciprocate and round to the nearest integer (again, if $b$ is too big this may lead to floating point errors but this should be much better than calculating $2^a$ or similar).

Title: Encoding 2 numbers into 1


答:
If you are using parametrization,

$f(x,y,z) = (x, y, g(x,y))$ where $g(x,y) = \sqrt{1-x^2-y^2}$

You can check that $|f_x \times f_y| = \sqrt{{1 + (\frac{\partial g}{\partial x}})^2 + (\frac{\partial g}{\partial y})^2} = \frac{1}{\sqrt{1-x^2-y^2}}$

Now the intersection of both surfaces is given by $z^2 = 1 - x^2 - y^2 = 1 - z$

$\implies z = \frac{\sqrt5 - 1}{2}$ (you have a negative value of $z$ from the quadratic that you ignore given the surfaces intersect above $z = 0$).

So you are now look at projection in $XY-$plane of a circle given by $x^2+y^2 = \frac{\sqrt5 - 1}{2}$.

So your integral becomes $\displaystyle \int_0^{2\pi} \int_0^{a}\frac{r}{\sqrt{1-r^2}} \ dr \ d\theta$

where $a = \sqrt{\frac{\sqrt5 - 1}{2}}$

If you are parametrizing in spherical coordinates,

$r(\theta, \phi) = (\cos \theta \sin \phi, \sin \theta \sin \phi, \cos \phi)$

Again you can check that $|r_{\phi} \times r_{\theta}| = \sin \phi$

Also you can check that paraboloid surface is given by $\rho = \frac{\cos \phi}{\sin^2\phi}$

At intersection of both surfaces, $\frac{\cos \phi}{\sin^2\phi} = 1 \implies \cos^2\phi+\cos\phi-1= 0$

$\implies \phi = \arccos(\frac{\sqrt5 - 1}{2})$

So the integral becomes

$\displaystyle \int_0^{2\pi} \int_0^{\arccos(\frac{\sqrt5 - 1}{2})} \sin\phi \ d\phi \ d\theta$

Title: parametrization of surfaces and area

Body:
The question is:

Paraboloid $z=x^2+y^2 $ divides the sphere $x^2+y^2+z^2=1$  into two parts, calculate the area
of each of these surfaces.

I know that i need to use $\iint |n| dS$ but i need to first find the normal to this surfaces and for that i need to parameterize each part, i do not know much about parametrization of part of surfaces, how can i use $\theta,\phi$ for this parametrization or should i use something else?

Any suggestion would be great, thanks


问:
Dependence of zeroes of a polynomial on polynomial's coefficients

What I have done so far:

Suppose $ \lambda $ is a root. Then,

$ \Lambda(p,q) = \lambda^2+p \lambda+q $. To prove that the zeros are $ C^1 $ functions, we need to prove that $ \frac{\partial \Lambda}{\partial p} $ and $ \frac{\partial \Lambda}{\partial q} $ are continuous.

This is where it gets confusing.

$ \frac{\partial \Lambda}{\partial q} = \lambda $ and $ \frac{\partial \Lambda}{\partial q} = 1 $.

These are constant functions so they are continuous. What am I missing here?

Also can someone give me hints for parts (b) and (c)? Any help is greatly appreciated.

答:
Part (a) You need to prove that $\frac{\partial \lambda}{\partial p}$ and $\frac{\partial \lambda}{\partial q}$ exist and are continuous i.e. $\lambda$ not $\Lambda$.

Note that, since $\frac{\partial \lambda}{\partial p}=-\frac{\lambda}{2\lambda+ p}$ and $2\lambda\ne -p$, the continuity is obvious.

Part (b) Hint

This is the same as part (a) for a polnomial of degree $n$ rather than degree $2$ and can be answered in the same way.

Part (c) Hint

Express the eigenvalue as a root of a polynomial equation.


问:
Evaluating $\int_{0}^{\pi} \frac{1}{\sqrt{u^2+2u\cos x +1}} \mathrm{d}x$

Does the integral $$\int_0^{\pi} \frac{1}{\sqrt{u^2+2u\cos x +1}} \text d x \hspace{30pt} (u \le 1) $$ has a closed form? If it has, how do we evaluate it?

I was solving a physics problem which I have asked on physics SE as well (here), and this integral popped out. How do I solve it?

Body:
I prefer to write another answer for the case where $u>1$.

In the most general case,
$$\int_{0}^{\pi} \frac{1}{\sqrt{u^2+2u\cos x +1}} \,dx=\frac{2 }{u-1}K\left(-\frac{4 u}{(u-1)^2}\right)$$ and we can expand the rhs as a series built around $u=1$. This would give
$$\frac{2 }{u-1}K\left(-\frac{4 u}{(u-1)^2}\right)=\sum_{n=0}^\infty a_n \, u^n$$ where the first $a_n$'s are
$$\left\{d,\frac{1}{2}-\frac{d}{2},\frac{5 d}{16}-\frac{7}{16},\frac{17}{48}-\frac{7
   d}{32},\frac{169 d}{1024}-\frac{1775}{6144},\frac{14789}{61440}-\frac{269
   d}{2048},\frac{1781
   d}{16384}-\frac{100741}{491520},\frac{244835}{1376256}-\frac{3035
   d}{32768},\frac{338377
   d}{4194304}-\frac{276441853}{1761607680},\frac{1482803963}{10569646080}-\frac{59
   9569 d}{8388608}\right\}$$ where $d=\log \left(\frac{8}{u-1}\right)$.

Uisng the above terms, the result is quite decent up to $u=2$. For this value, the exact result is
$2 K(-8) \sim 1.68575$ while the expansion gives $1.68208$


问:
Tangent space of algebraic variety at a point.
Could you please help me with this problem?

[Springer, Linear algebraic group, 4.1.9]Assume that $X$ is a closed subvariety of $\mathbb{A}^n$, $k$ is an algebraically closed field,
Describe the tangent space $T_{x} X$ in the following cases:

(a) $X$ is a point;

(b) $X=\mathbb{A}^{n}$, $x$ is an arbitrary point in $X$;

(c) $X=\left\{(a, b) \in \mathbb{A}^{2} \mid a b=0\right\}$, $x=(0,0)$;

(d) Assume that $\operatorname{char}(k) \neq 2,3, X=\left\{(a, b) \in \mathbb{A}^{2} \mid a^{2}=b^{3}\right\}$, $x=(0,0)$.

We have $T_x X \cong (M_x / M_x^2)^*$, where $M_x\subset k[X]$ is the maximal ideal of functions vanishing in $x$. So we need to compute $M_x$?

Thank you in advance.

答:
I can give an example computation for one of them.

Take $c$, for example. Here the coordinate ring of the variety is $R=k[a,b]/(ab)$, and the point has maximal ideal $M_x=(a,b)$, since it is $(0,0)$.

We want to first compute $M_x/M_x^2$. So let us see what is inside $M_x^2$. Since $M_x$ is generated by $a,b$, $M_x^2$ is generated by $a^2$, ab and $b^2$. But $ab=0$ is zero in $R$, so $M_x^2$ is generated by $a^2$ and $b^2$.

Now, what does elements of $M_x$ look like? As a $k$-vector space it is spanned by $a, b, a^2, b^2, a^3, ...$. Elements of $M_x^2$ on the other hand are combinations of $a^2, b^2, a^3, ...$

So modding out by $M_x^2$, we are left with $a, b$ inside $M_x/M_x^2$.

We conclude that $T_x \simeq k^2$ in this case.