问:
Are singletons open sets?
I've recently been learning about metric spaces and one of the very important definitions are of open sets and open balls.
I have a question that was raised when reading on a website that a singleton in a metric space $(X,d)$ such as $[p]$ is considered to be an open set for example let the  set be $[0,1]$ and the metric be the standard metric is it true to say that the singleton $[1]$ is an open set? Since the $B_r$(1)=$[1]$ and this open ball is contained in the set.
However is this true for all singletons in the metric space $(\mathbb{R},d)$.
I'm new to metric spaces and I'm sure this is a very trivial question for most of you but its really confusing me.

Thanks in advance.

答: Singleton sets on the real line are not open with the standard metric. To see this let $\{a\}$ be your singleton set, and construct a ball such that $B(a,r)\subseteq\{a\}$ or in other words with the metric we are working with the set of points $x$ such that $a-r<x<a+r\subseteq\{a\}$ but this is a contradiction since $r>0$

Title: Are singletons open sets?


问:
Some horrid integrals I am confused with
$$(1) ~~~~ \int_{1/e}^{\infty} \sqrt{\frac{\ln{x} + 1}{x^3 + 1}}~~dx$$

Speaking as an example on (1).

The only thing I could do from here is only to do a u-sub:

$$ u = \ln(x) + 1 \\ x = e^{u-1} \\ du = \frac{1}{x} dx \\ dx = x (du) \Rightarrow dx = e^{u-1} (du) $$

And so this becomes:

$$ \int \sqrt{ \frac{u}{e^{3u-3} + 1} } ~~ du \\ $$

And from here it's basically a deadend.. I would appreciate if you could give some insights..

Interestingly, these similar integrals have interesting solutions:

$$(2) ~~~~ \int_{1/e}^{\infty} \sqrt{\frac{\ln{x} +1}{x^3}}~~dx =  \sqrt{2 e \pi}$$

$$(3) ~~~~ \int_{1/e}^{\infty} \sqrt{\frac{\ln{x}}{x^3 + 1}}~~dx = \text{a complex number ?!}$$

(2) - where and why does the pi comes into play? this looks like an interesting solution..

(3) - why is the solution a complex number if the area is right infront my eyes, and it is real just like every other areas? What is happening here?

Mystery solved about the complex number - the lower bound should be $1$ and not $1/e$ my bad! But this is not the main mystery ;)

Thank you :-)

答:
$$I=\int_{1/e}^\infty\sqrt{\frac{\ln x+1}{x^3+1}}dx$$
notice that: $\ln x+1=\ln x+\ln e=\ln ex$ if we try $u=ex\Rightarrow dx=du/e$ and so:
$$I=\frac1e\int_1^\infty\sqrt{\frac{\ln u}{(u/e)^3+1}}du=\sqrt{e}\int_1^\infty\sqrt{\frac{\ln u}{u^3+e^3}}du$$
now this is quite a tough integral

$$J=\int_{1/e}^\infty\sqrt{\frac{\ln x+1}{x^3}}dx=\sqrt{e}\int_1^\infty u^{-3/2}\sqrt{\ln u}\,du$$
if we now let $v=\ln u\Rightarrow du=e^vdv$ so:
$$J=\sqrt{e}\int_0^\infty e^{-1/2v}v^{1/2}dv$$
now let $w=\frac12v,dv=2dw$
$$J=2\sqrt{2e}\int_0^\infty e^{-w}w^{1/2}dw=2\sqrt{2e}\Gamma\left(\frac32\right)=2\sqrt{2e}\times\frac{\sqrt{\pi}}{2}=\sqrt{2e\pi}$$

ContentLicense: CC BY-SA 4.0


问:
Is frational ideal finitely generated?
Let $R$ be Noetherian.

I know finitely generated $R$-submodule $I$ of $K$ is a fractional ideal. It is easy to see.

But why if $R$-submodule $I$ of $K$ is a fractional ideal then it is finitely generated? hard to understand...

答: If $I$ is a fractional ideal, then there is some non-zero $r\in R$ with $rI\subset R$. But as $R$ is noetherian we have $rI=(f_1,\dots,f_n)$ for some $f_i\in R$, hence $I=(\frac{f_1}{r},\dots,\frac{f_n}{r})$ is finitely generated.

Title: Is frational ideal finitely generated?


问:
Exponential formula to assign list items an equally distributed percentage
I am going to explain this as best as I can:

We have a number of list items, that dynamically changes (sometimes it has 5 items sometimes it has 7 items or 43 items, etc.) We are trying to find a relevant prioritypercentage to assign to each list item (starting from bottom to top) but cannot figure out a formula to do so based on the number of items.

For example (percentages are COMPLETE guesses):

Is there a formula that will take a number and distribute 100 points exponentially or any other way as shown below?

A list item with 2 items:

List item 1 (75%)

List item 2 (25%)

A list item with 3 items:

List item 1 (50%)

List item 2 (30%)

List item 3 (20%)

答:
Here's one way: if you have $n$ items, assign the 1st one ${2n\over n(n+1)}$, the 2nd one $2(n-1)\over n(n+1)$, then $2(n-2)\over n(n+1)$, etc, down to the last getting $2\over n(n+1)$.

For 2 items, this gives 2/3, 1/3.

For 3 items, 1/2, 1/3, 1/6.

For 4 items, 40%, 30%, 20%, 10%.

It's not "exponential", but it's easy....

ContentLicense: CC BY-SA 3.0


答: As some people have already commented, the inequality isn't true, because the left hand side is exactly equal to the right hand side. Note that the sum on the left-hand side is simply a telescoping sum; with the observation that $1/(n(n-1)) = 1/(n-1) - 1/n$, we have $$1 + \frac{1}{1 \cdot 2} + \cdots + \frac{1}{n (n-1)} = 1 + \bigg(\sum_{k=2}^n \frac{1}{k - 1} - \frac{1}{k} \bigg) =   1 + \bigg( 1 - \frac 1n \bigg) = 2 - \frac 1n.$$ Note that the second equality holds because the middle terms all cancel out (i.e. it's a telescoping sum), leaving only the first half of the first term and the second half of the last term.

Title: Why is $1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{n(n-1)}=2-\frac{1}{n}$ true?

Body: Why is this equality true? $$1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{n(n-1)}=2-\frac{1}{n}$$


答:
${1\over n(n-1)}={n-(n-1)\over n(n-1)}=\frac{1}{n-1}-\frac 1n$

Thus your expression simplifies to:
$1+(1-\frac 12)+(\frac 12-\frac 13)+...+(\frac{1}{n-1}-\frac 1n)=2-\frac 1n$


问:
What are the relations between pushdown automata and parsers?
For a CFG, there is a pushdown automaton; vice versa.

For a CFG, there is a parser; and I guess vice versa?

What are the relations between pushdown automata and parsers? Are they  directly related? Can they be derived from each other?

What are the relations between  pushdown automata and  parser generators (e.g. Yacc)?

Thanks.

答:
There is usually a pushdown automata at the core of every programming language parser, at least implicitly (e.g parsers based on recursive-descent do not have an explicitly-coded pushdown automaton).  YACC is one tool for generating such parsers from grammar definitions.  The pushdown automaton is used to verify that a given program is syntactically correct.

However, it should be noted that parsers for real programming languages do more than check syntax.  For example, verifying the internal datatype consistency of expressions is outside the scope of what can be expressed (easily) with syntax alone.  In that sense, parsers generally exceed pushdown automata.


问:
Solve limit using definite integration
I have to solve this limit using definite integral

$${\lim_{n\to \infty} \frac{1}{n} \cdot \sum_{i=1}^n \frac{1}{1+(\frac{i}{n})^i}}$$

Well, my development was:

I tried to give the Riemann integral definition form, using a regular partition such that:

$$\int_a^bf(x)dx=\lim_{n\to\infty}\frac{b-a}{n}\cdot\sum_{i=1}f\left(a+i\frac{b-a}{n}\right)$$

So, i set $b=1, a=0$ for simplicity, then we have $f(\frac{i}{n})$ must be equal to ${\frac{1}{1+(\frac{i}{n})^i}}$.

Let ${x=\frac{i}{n}}$, then $${f(x)=\frac{1}{1+x^{nx}}}$$

The problem is that $n$ is a dummy variable that does not make sense outside the limit, that is, it does not make sense for the function $f$

So, I need to do some kind of algebraic transformation or variable change for $n$, in such a way that the function $f$ remains only in terms of $x$ and thus I can use it in the definite integral. However, I have not been able to find such a magical algebraic transformation or variable change for $n$.

答:
By proceeding through your method, we have obtained f(x) as $\frac{1}{1+x^{nx}}$, as mentioned in the question.

So, when n approaches infinity,

$f(x)=0 $$ $ if  $ x $ $ ϵ$ $ (0,1),$

and at $x=1$,
$lim $ $n->∞, f(x)=2 $

So, $ f(x)=1 $ at all but finite points and hence,

Given Sum= $$\int_{0}^{1}1dx$$

$$=1$$

(We can further write the integral as $\int_{0}^{1}1dx$ + $\int_{1}^{1}2dx $ = $1+0$ but that is understood.)

Title: Solve limit using definite integration


问:
Chi Squared Clarification
Suppose you have $Z_1, Z_2, Z_3$ which are all independent standard Gaussian variables.

Suppose you have
$$A=\frac{(Z_1-2Z_2+Z_3)^2}{12}+\frac{(Z_1-Z_3)^2}{4}+\frac{(Z_1-Z_2)^2}{4}+\frac{(Z_1+Z_2-2Z_3)^2}{12}.$$

$A$ is $\chi_2^2$, but I don't see how this is the case.
Particularly since $\operatorname{Var}\left(\frac{Z_1-2Z_2+Z_3}{\sqrt{12}}\right)$${}=\frac{1}{2}$, which is the same issue with the other terms, i.e since none of the terms, by themselves, are standard normal before you square them.
However, I do see how:
$$B=\frac{(Z_1-2Z_2+Z_3)^2}{6}+\frac{(Z_1-Z_3)^2}{2}.$$ follows the $\chi_2^2$ distribution. I can also see that the first two terms of $A$ multiplied by $2$ is equal to $B.$

Any feedback would be appreciated.

答:
If you expand the quadratic form, you end up with

\begin{align}
A&=\frac23(Z_1^2+Z_2^2+Z_3^2-Z_1Z_2-Z_2Z_3-Z_1Z_3)
\\&=Z_1^2+Z_2^2+Z_3^2-\frac23(Z_1Z_2+Z_2Z_3+Z_1Z_3)-\frac13(Z_1^2+Z_2^2+Z_3^2)
\\&=Z_1^2+Z_2^2+Z_3^2-\frac13(Z_1+Z_2+Z_3)^2
\end{align}

In other words, if $\overline Z=\frac13\sum\limits_{i=1}^3 Z_i$, then  $$A=\sum_{i=1}^3Z_i^2-3\overline Z^2 =\sum_{i=1}^3 (Z_i-\overline Z)^2$$

So $A$ is the corrected sum of squares for a sample $Z_1,Z_2,Z_3$ drawn from a standard normal population. This is known to have a $\chi^2_{3-1}$ distribution.

One can also note that $A=Z^TPZ$ where $Z=(Z_1,Z_2,Z_3)^T$ and $P=I_3-\frac13\mathbf1\mathbf1^T$, with $\mathbf1$ being a $3\times 1$ vector of all ones. As $P$ is symmetric and idempotent, $A$ has a $\chi^2$ distribution with degrees of freedom $\operatorname{rank}(P)=\operatorname{tr}(P)=3-1$.

Title: Chi Squared Clarification


问:
$X_1,X_2$ distributed $U(0,1)$ and independent, find the distribution of $D=|X_2-X_1|$
I know it shouldn't be this hard but I am having trouble anyway.

I tried using $F_D(t)=\mathbb{P}(|X_2-X_1|\leq t)=\mathbb{P}(-t\leq X_1-X_2\leq t)$ and sketching the unit square $[0,1]^2$ but to no avail, I can't seem to find a formula that fits.

答:
Sketch for the solution: setting $Z:=(X,Y)$ you will have that
$$
\begin{align*}
\Pr [|X-Y|\leqslant c]&=\Pr [Z\in A_c]\\&=\int_{\mathbb{R}^2}\mathbf{1}_{A_c}(x,y)f_X(x)f_Y(y)\mathop{}\!d (x,y)\\
&=\int_{[0,1]^2}\mathbf{1}_{A_c}(x,y)\mathop{}\!d (x,y)
\end{align*}
$$
where $A_c:=\{(x,y)\in \mathbb{R}^2:|x-y|\leqslant c\}$ and $f_X(t)=f_Y(t)=\mathbf{1}_{[0,1]}(t)$ are the densities of $X$ and $Y$. Now observe that

$$
\mathbf{1}_{A_c}(x,y)=1 \iff |x-y|\leqslant c \iff x-y\in[-c,c]\iff y\in \mathbb{R}\,\land\, x\in [y-c,y+c]
$$

Therefore $\mathbf{1}_{A_c}(x,y)=\mathbf{1}_{\mathbb{R}}(y)\mathbf{1}_{[y-c,y+c]}(x)$, and so substituting in the last integral we find that

$$
Pr[|X-Y|\leqslant c]=\int_0^{1}\int_0^{1}\mathbf{1}_{[y-c,y+c]}(x)\mathop{}\!d x\mathop{}\!d y
$$

Now the inner integral is just the Lebesgue measure of the set $[0,1]\cap [y-c,y+c]$, this is $1-(y-c)$ when $y-c\in[0,1]$ or $y+c$ when $y+c\in[0,1]$, therefore

$$
Pr[|X-Y|\leqslant c]=\int_0^{1}((1-y+c)\mathbf{1}_{[c,c+1]}(y)+(y+c)\mathbf{1}_{[-c,1-c]}(y))\mathop{}\!d y
$$


答:
Generally
$$\frac{1}{n(n+k)}=\frac{1}{k}\left( \frac{1}{n}-\frac{1}{n+k}\right)$$
So we can calculate the exact value of left hand side as other solutions above.


问:
What is the name of a game that cannot be won until it is over?
Consider the following game:

The game is to keep a friend's secret. If you ever tell the secret, you lose. As long as you don't you are winning. Clearly, it's a game that takes a lifetime to win.

Another example would be honor. Let's say, for our purposes, once you lose your honor you can't get it back. Call it a game. Then the only way to win is to never lose your honor.

Both of these games are highly unfavorable for the player. Because they are games that can never be won, until they are over. Is there a mathematical name for such a game?

答:
In some contexts, such a game is called a closed game.  To understand the terminology, let I and II denote two players, and suppose that the players alternate indefinitely playing elements $m$ from some set $M$ of possible moves.
If we imagine that the game is played for infinitely many moves, we obtain an infinite sequence of moves $(m_1,m_2,m_3,\ldots)$.  Let $M^{\mathbb{N}}$ denote the space of such sequences.

This space $M^{\mathbb{N}}$ has a natural topology, namely the product of the discrete topologies on $M$.  Suppose that the payoff set (set of move sequences that count as a "win") for player I is closed in this topology. This means that whenever player I loses, it is because he or she made a mistake (e.g. telling the secret, in your example) at some finite stage of the game, or because the game was simply unwinnable. On the other hand, if player I wins it is because he or she has forever avoided making mistakes.

I am stretching your example here by assuming that the players are immortal, in which case the only way to win is by keeping the secret literally forever.
In this situation, the game is "over" when infinitely many years have passed.  I think that this is the most interesting interpretation of the question; otherwise one could simply consider "taking the secret to the grave" as one of the possible moves $m \in M$, and this move has the property that playing it results in an immediate win for player I, which is not very interesting.

The next few paragraphs do not directly address the question, but help to explain why some people (e.g. set theorists) consider closed games to be interesting:

A nice feature of closed games (relative to infinite games in general) is
that they are determined. This is known as the Gale–Stewart theorem. To say that a game is determined means that one player or the other has a winning strategy that prescribes the moves to make in any given situation in order to win.  If player I (the player with the closed payoff set) has a winning strategy, that strategy can be described simply as "don't make any mistakes" where a mistake is defined as a move after which player II can force a win in finitely many moves.

The more interesting case of the Gale–Stewart theorem is when player I does not have a winning strategy.
If the set $M$ of moves is finite, as in chess, then the lack of a winning strategy for player I means that there is some fixed finite number $n$ such that player II can force a win in $n$ moves (e.g. perhaps the starting position in chess is a mate in 73 moves for black, although this seems very unlikely.)  Then a winning strategy for player II proceeds by always playing so as to reduce this number (e.g. mate in 72, then on the next move mate in 71, and so on until he or she inevitably wins.)

On the other hand, if the set of moves is infinite then the construction of a winning strategy for player II in the case that player I has no winning strategy is more complicated and involves transfinite ordinals.  The issue is that, for example, if the possible moves for player I are $m_1,m_2,m_3,\ldots,$ it could be that making any of these moves leads to a win for player II, but making move $m_i$ allows player I to survive for $i$ many moves, so there is no fixed upper bound on how long it takes player II to win.  In this case we would take the supremum and say that player II can force a win in $\omega$ moves.  By always choosing moves that decrease this ordinal rank, player II always wins, because there is no infinite decreasing sequence of ordinals.


答:
No, the singleton set $\{1\}$, as a subset of the real line, is not an open set. Formally, this is because any open ball centered at $1$ must contain some number greater than and less than $1$.

When considering subsets of the real line, we can think of open sets as "having no borders;" they're the "opposite" of closed sets, which have a clearly defined boundary. It's the difference between the open interval $(a, b)$ and the closed interval $[a, b]$. One has endpoints, one doesn't.

A singleton set $\{x\}$ has boundaries, namely itself. It's the same as the closed interval $[x, x]$.

However, just note that this is not true in general metric spaces, namely the discrete metric space. In fact, a topology on any set $X$ is equal to the discrete topology on $X$ if and only if every singleton set of $X$ is open.

Title: Are singletons open sets?

Body:
I've recently been learning about metric spaces and one of the very important definitions are of open sets and open balls.
I have a question that was raised when reading on a website that a singleton in a metric space $(X,d)$ such as $[p]$ is considered to be an open set for example let the  set be $[0,1]$ and the metric be the standard metric is it true to say that the singleton $[1]$ is an open set? Since the $B_r$(1)=$[1]$ and this open ball is contained in the set.
However is this true for all singletons in the metric space $(\mathbb{R},d)$.
I'm new to metric spaces and I'm sure this is a very trivial question for most of you but its really confusing me.

Thanks in advance.


问:
Uniqueness of the topology on $\mathbb{R}$
For any non-empty set $X$, we can think of multiple topologies on $X$. That is, topologies are not uniquely determined by $X$. For example, the metric induced by $p$-norm and discrete metric yield different topologies on the same set $X$.

But when we talk about the topology on the set of real numbers (especially in analysis), it seems that the metric used for the construction of the topology is tacitly fixed to $p$-norm. For example, the topology on real numbers is induced by the absolute function (as a $p$-norm for scalars).

My questions are 1) whether I correctly understand the meaning of topologies, and 2) when we deal with real numbers, do we use only one topology that is induced by $p$-norm?

答:
Yes, you do understand correctly that there can be several topologies on the same set. That being said, a lot of them are rarely used, as they are either boring or too special. This leads to the second point.

There are loads of other topologies on $\mathbb{R}$. For example, in algebraic geometry people will mostly use the Zariski-topology. Also the discrete topology exists on $\mathbb{R}$, just as on any other set. But in differential geometry for example you will almost always encounter the "standard-topology" on $\mathbb{R}$. Same is true for most of applied mathematics. So to answer your second question: No, we do not use only the "standard-topology" on $\mathbb{R}$. It depends largely on who "we" is.


问:
Terence Tao's definition of subtraction operation to build integers
I am reading the book Analysis 1 by Terence Tao in which he discussed the construction of integers from natural numbers. The integers are constructed from the natural numbers by the operation $subtraction$.

To be able to define the $subtraction$ operation, he started with the "$--$" symbol as a placeholder and then he defined the subtraction operation based on that symbol.

I interpret his definition of integer (shown in the picture) like this: We have the expression $a$ $--$ $b$, then we call that expression an integer. By saying "we have" means we know for sure the expression has already existed.

However, the symbol "$--$" has never been defined. We don't even know whether if exists or not and don't know what it means to be.

So my question is: How can we define the new term "integer" based on something which is also "new" (in this case, the symbol "$--$")

[image]

答:
Define the relation

$(a,b)\sim (c,d) :\Longleftrightarrow a+d=b+d$.

This relation is an equivalence relation on $\Bbb N_0^2$, so we can form equivalence classes and the quotient set.

The equivalence class of $(a,b)$ denoted $a--b$ is the set of all pairs that are equivalent to $(a,b)$.

The set of equivalence classes $\{a--b\mid a,b\in\Bbb N_0\}$
is called $\Bbb Z$ and defines the set of integers.


问:
Normal distribution - What should be value of $x$ if the company wants no more than $0.5\%$ of the packets to be underweight?

A cereal machine is sending $x$ grams of cereal into each packet. The value of
$x$ is programmed into the machine and the packets are then filled
with the weight of cereal in each being normally distributed with:

mean of $x$ grams and

standard deviation of $1.8$ grams

The machine is used to fill packets that will be labelled as
containing $500$ grams. What should be the value of $x$ if the company
wants no more than $0.5\%$ of the packets to be underweight?

I'm not quite sure how to start this question but here's what I am thinking so far. Do we have to set up some sort of simultaneous equations here since we are only given an unknown mean value of $x$? Does this involve using the standard deviation of $1.8$ grams? Any starting hints would be much appreciated.

答: You are expected to use the z-score table.  $x$ is the mean of the distribution.  You want less than $5\%$ of the area to be less than $500$ grams.  How many standard deviations does that mean $x$ has to be above $500$ grams?


答: This is a purely syntactic definition.  First, '--' is introduced, then (I suspect this step is done implicitly) purely syntactic expressions of the form 'a--b' (for a and b being natural numbers) are defined based on the existing definition of '--', then the above definition is introduced.</p>

Title: Terence Tao's definition of subtraction operation to build integers

Body:
I am reading the book Analysis 1 by Terence Tao in which he discussed the construction of integers from natural numbers. The integers are constructed from the natural numbers by the operation $subtraction$.

To be able to define the $subtraction$ operation, he started with the "$--$" symbol as a placeholder and then he defined the subtraction operation based on that symbol.

I interpret his definition of integer (shown in the picture) like this: We have the expression $a$ $--$ $b$, then we call that expression an integer. By saying "we have" means we know for sure the expression has already existed.

However, the symbol "$--$" has never been defined. We don't even know whether if exists or not and don't know what it means to be.

So my question is: How can we define the new term "integer" based on something which is also "new" (in this case, the symbol "$--$")

[image]


答: You're correct. Technically, the use of a placeholder symbol like this is somewhat of a loose construction of the integers. What Tao means, formally, is that we are defining an equivalence relation on the Cartesian product $\mathbb{N} \times \mathbb{N}$, where two elements $(a, b)$ and $(c, d)$ are considered to be equal if and only if $a + d = c + b$. In other words, an integer is simply a pair of natural numbers. We compare integers and consider them to be equal to each other using the rule $a + d = c + b$. The use of the symbol $a--b$ is simply a shorthand for $(a, b)$, as he (probably) is just trying to get his readers to understand the motivation behind such a construction i.e. it's to imitate the rules of subtraction.


问:
Evaluate $\int^{2\pi}_{0}\frac{\mathrm{d}x}{(\cos x+\sin x+\sqrt{3})^2}=2\sqrt{3}\pi.$
The problem is to evaluate the following definite integral:
$$\int^{2\pi}_{0}\frac{\mathrm{d}x}{(\cos x+\sin x+\sqrt{3})^2}=2\sqrt{3}\pi.$$

I have failed in almost every way I tried, with the exception of substitution by $\tan(\frac{x}{2})$ which leads to the integration of a rational function with annoying coefficients.

Is there any "good" way to address the problem, given that the result doesn't seem that complicated? I've also tried this website, but the result it gave was far from satisfactory. Please help.

答:
I would use the following identities:
$$\sin 2\theta\equiv\frac{2\tan \theta}{1+\tan^2 \theta}$$
$$\cos 2\theta\equiv\frac{1-\tan^2 \theta}{1+\tan^2 \theta}$$
And in your question then use the substitution $t=\tan \frac{x}{2}$, as you did.

Persevere and it should work.


Title: Evaluate $\int^{2\pi}_{0}\frac{\mathrm{d}x}{(\cos x+\sin x+\sqrt{3})^2}=2\sqrt{3}\pi.$

Body:
The problem is to evaluate the following definite integral:
$$\int^{2\pi}_{0}\frac{\mathrm{d}x}{(\cos x+\sin x+\sqrt{3})^2}=2\sqrt{3}\pi.$$

I have failed in almost every way I tried, with the exception of substitution by $\tan(\frac{x}{2})$ which leads to the integration of a rational function with annoying coefficients.

Is there any "good" way to address the problem, given that the result doesn't seem that complicated? I've also tried this website, but the result it gave was far from satisfactory. Please help.


问:
Solving $\sqrt{a-\sqrt{a+x}} = x$
Solve the equation
$$\sqrt{a-\sqrt{a+x}} = x$$

My approach:

Tried shifting the variables into different options, but couldn't get anything out of it. So, please help.

答:
Let $t=x+a$ to rewrite the nested radical equation $\sqrt{a-\sqrt{a+x}} = x $ as
$$t-a- \sqrt{a-\sqrt t}=
0$$

Avoid extraneous solutions from squaring the equation and instead  factorize it as

$$(\sqrt{a-\sqrt t}+\sqrt t)(\sqrt{a-\sqrt t}+1-\sqrt t)=0$$
Note that $ \sqrt{a-\sqrt t}+\sqrt t>0$ and any solutions result from
$$ \sqrt{a-\sqrt t}+1-\sqrt t =0$$
which factorizes as
$$\left(\sqrt{a-\sqrt t}+\frac{1+\sqrt{4a-3}}2\right)
 \left(\sqrt{a-\sqrt t}+\frac{1-\sqrt{4a-3}}2\right)=0
$$
Again, the first factor is positive, permitting no solutions, and the second factor yields the sole solution

$$ x=t-a= \frac12(\sqrt{4a-3}-1)$$
(Note: the solution exists for $a\ge 1$.)

Title: Solving $\sqrt{a-\sqrt{a+x}} = x$


答:
You define a metric space by $(X,d)$ where $X$ is a non-empty set and $d$ is the distance function.  In the metric $(X,d),\,\, X$ is the universal set. So $X$ is always an open set. Now if you take $X$ as a singleton set then $X$ is always open.

Consider the Discrete metric space(trivial metric space) with $X=\mathbb Z$ or any subset of $\mathbb  Z$. If you take any $0<r<1$ then every singleton set consisting a single integer is open in $X$.

In the usual metric, (Euclidean metric of degree 1) $(\mathbb R,d)$ no singleton set is open.

Title: Are singletons open sets?


问:
Is there a similarity between $\Bbb Z \times \Bbb Q$ and $\Bbb Z \times \Bbb N$?
Is there a similarity between $\Bbb Z \times \Bbb Q$ and $\Bbb Z \times \Bbb N$? (EDITED: Both of these sets have anti lexicographic order, that is $(a,b)<(c,d)⟺b<d$ or $(b=d$ and $a<c)$).

Definition of similarity: Let $(A, <)$ and $(B,≺)$ be partially ordered sets. Then $f:A\rightarrow B$ is a similarity between $A$ and $B$ if $f$ is bijective and $\forall x,y \in A, x<y \Rightarrow f(x)≺f(y)$ and $\forall m,n \in B, m≺n \Rightarrow f^{-1}(m)<f^{-1}(n)$. We write $A \simeq B$ if $A$ and $B$ are similar.

We usually check if the sets are (not) similar by checking for possible similarity invariants. (Let $A,B$ be posets. Property P is similarity invariant if $A \simeq B \Rightarrow (P(A) \iff P(B))$. That is, if $P(A)$ and $\lnot P(B)$, then $A \not \simeq B$.)

I have tried to find an invariant that one set has and the other doesn't, but haven't been able to find it. For example, they are both not separable (they both don't have a countable subset that's dense in them), they both aren't complete (there exists a nonempty subset that is bounded from above but doesn't have a supremum ( $\Bbb Z \times \{0\}$ is example for both), they both aren't locally finite, they are both unbounded on both sides, they both aren't dense in themselves.

On other hand, I can't find a similarity between them but I also don't know how to prove that there is no similarity between them.

(Also, if $(A, <)$ and $(B,≺)$ are both totally ordered (and in my case, they are), then $f:A\rightarrow B$ is similarity between $A$ and $B$ if $f$ is surjective and $\forall x,y \in A, x<y \Rightarrow f(x)≺f(y)$.)

答:
Your sets are not just partially ordered, of course, they are totally ordered (and total ordering is an invariant of partially ordered sets).

In any totally ordered set $(S,<)$, for any two elements $r,s$, let's say that $s$ is a successor of $r$ if $r < s$ and there does not exist $t \in S$ such that $r < t < s$.

In $\mathbb Z \times \mathbb N$, every element $(m,n)$ has a successor, namely $(m+1,n)$. And the property "every element has a successor" is an invariant property. But in $\mathbb Z \times \mathbb Q$ it is not true that every element has a successor, in fact no element has a successor.


Title: Normal distribution - What should be value of $x$ if the company wants no more than $0.5\%$ of the packets to be underweight?

Body:
A cereal machine is sending $x$ grams of cereal into each packet. The value of
$x$ is programmed into the machine and the packets are then filled
with the weight of cereal in each being normally distributed with:

mean of $x$ grams and

standard deviation of $1.8$ grams

The machine is used to fill packets that will be labelled as
containing $500$ grams. What should be the value of $x$ if the company
wants no more than $0.5\%$ of the packets to be underweight?

I'm not quite sure how to start this question but here's what I am thinking so far. Do we have to set up some sort of simultaneous equations here since we are only given an unknown mean value of $x$? Does this involve using the standard deviation of $1.8$ grams? Any starting hints would be much appreciated.

Body:
Set

$$\mathbb{P}[Y<500]\leq 0.5\%$$

$$\mathbb{P}\Bigg[Z<\frac{500-x}{1.8}\Bigg]\leq 0.5\%$$

$$\frac{500-x}{1.8}=-2.58$$

$$x=504.64$$


问:
Extension of valuation to the algebraic extension of a number field.
I am trying to get the idea how we can extend the $p$-adic valuation on $\mathbb Q$ to an algebraic extension. In particular, how to extend the $p$-adic valuation for $p = 5$ from $\mathbb Q$ to $\mathbb Q(5^{1/3})$.  Thank you.</p>

答:
The analogue of the $p$-adic valuation of a number field depends upon the prime ideals of its ring of integers. Let $K$ be a number field with ring of integers $\mathcal{O}_K$, and let $\mathfrak{p}\lhd\mathcal{O}_K$ be a non-zero prime. Then for any non-zero element $x\in K$, we can factor the fractional ideal $x\mathcal{O}_K\subset K$ uniquely into a product of prime ideals, say $x\mathcal{O}_K=\mathfrak{p}_1^{e_1}\mathfrak{p}_2^{e_2}\cdots$, where the $e_i\in\mathbb{Z}$ and each $\mathfrak{p}_i\lhd\mathcal{O}_K$ is prime. The $\mathfrak{p}_i$-adic valuation $\text{ord}_{\mathfrak{p}_i}(x)$ is then the exponent $e_i$ in the prime factorisation of $x\mathcal{O}_K$.

In your particular example, $K=\mathbb{Q}(\sqrt[3]{5})$, the rational prime $5$ ramifies as $(5)=\mathfrak{p}^3$, where $\mathfrak{p}=(5,\sqrt[3]{5})$, so you could consider the $\mathfrak{p}$-adic valuation on $K$.


问:
$\int_{0}^{\pi} D_{n}(y)dy=\frac{1}{2}$ Dirichlet
I need to calculate that $\int_{0}^{\pi} D_{n}(y)dy=\frac{1}{2}$ with $D_{n}(y)= \frac{1}{2\pi}\frac{\sin((n+\frac{1}{2})y)}{\sin(\frac{y}{2})}$ from Dirichlet.

Now I tried to do this with the known that $D_{n}(y)=\frac{1}{2\pi}\sum_{k=-1}^{n}e^{iky}$.

I tried to things but I never become 1/2.

First I tried to solve the somation and then integrate: $\int_{0}^{\pi} D_{n}(y)dy=\int_{0}^{\pi} \frac{1}{2\pi}\sum_{k=-1}^{n}e^{iky}dy=\int_{0}^{\pi} \frac{1}{2\pi}(\frac{e^{-niy}-e^{iy(n+1)}}{1-e^{iy}})dy$. This integral doesn't converges so I did something wrong but I don't know what?

Second I tried to switch the integral and the sommetion and then solve it: $\int_{0}^{\pi} D_{n}(y)dy=\int_{0}^{\pi} \frac{1}{2\pi}\sum_{k=-1}^{n}e^{iky}dy=\int_{0}^{\pi} D_{n}(y)dy=\frac{1}{2\pi}\sum_{k=-1}^{n}\int_{0}^{\pi}e^{iky}dy=\frac{1}{2\pi}\sum_{k=-1}^{n}\frac{1}{ik}(1-e^{-\pi i k})$. But this sometion will give $i\infty -i \infty$ so this doesn't work eather.

Can someone eplain me what I do wrong and how I can fix it?

答:
You're using a wrong definition of the Dirichlet kernel. The proper one is

$$D_n(x) = \frac{1}{2\pi}\sum_{k=-n}^n e^{ikx} = \frac{1}{2\pi}\frac{\sin \left(\frac{(n+1)x}{2}\right)}{\sin \left(\frac{x}{2}\right)}.$$

Based on that you have by switching $\int$ and $\sum$

$$\int_0^{2\pi}D_n(x) = \int_0^{2\pi}\frac{1}{2\pi}\sum_{k=-n}^n e^{ikx} = 1.$$

as $\int_{0}^{2\pi}e^{ikx}=0 $ for $k\neq 0$ integer.

And as $D_n$ is even

$$\int_0^{\pi}D_n(x) = \frac{1}{2}.$$


问:
If A is a large number and B is a larger number than A, find C which is the nearest number to A that goes into B x times.
```<code>A = 2548788451252366445785585474844745854781554558547847844

B = 50135928365972237608005364328936872115615930177019003323815

x is unknown
</code>```

Given A and B find C, where C must be the nearest integer to A that goes into B x times.

Note: C must be smaller than A or equal with A.

Calculating C manually will involve subtracting or dividing A from B x times until we get near A and therefore to C.

Is there a formula or maybe a SageMath or PARI method to calculate C?

Thanks.

答:
You can just divide $B$ by $x$, then round down to the next whole number.  Any language that lets you handle integers this large will do that.  $A$ does not figure in this at all.

Based on the comment you are just asking for the remainder of the division of B by A.  Many languages have a modulo operator which will do this.  Often it is the percent sign, so you would write $C=B\%A$  You can do division with remainder in any language that can handle large integers.  If you can't do the modulo operation, you can do integer division and $C=B-\lfloor B/A\rfloor\cdot A$


问:
Solving modular equations that gives GCD = 1
I have problems with understanding modular equations that gives GCD = 1. For example:

$$3x \equiv 59 \mod 100$$

So I'm getting $GCD(3, 100) = 1$. Now:

$1 = -33*3 + 100$

That's where the first question appears - I always guess those -33 and 1 (here) numbers...is there a way to solve them?

And the second question - the answer to that equation, at least according to the book, is:

{$-33*59 + 100k$} (k - integer) - and why is that so? Where did the 59 came from?

答:
$$\begin{eqnarray} \text{Note} &&\ 1 &=&\ 3\, (-33)\ \ \ \, +\ \ \ \, 100\, (1)  \\
 \stackrel{\times\, 59\ }\Rightarrow && 59\, &=&\ 3\, (-33\cdot 59) + 100\, (59)\end{eqnarray}$$

Hence $\ 59 = 3x+100y \ $ has a particular solution $\rm\:(x,y) = (-33\cdot 59, 59).\, $ By linearity, the general solution is the sum of this and the solution of the associated homogeneous equation $\ 0 = 3x+100y,\,$ so $\,
\frac{y}x = \frac{-3}{100},\ $ so $\ x = 100k,\ y = -3k .\,$ So the general solution is their sum  $$(-33\cdot 59,\,59)\, +\, (100k, -3k)\, =\, (-33\cdot 59+100k,\, 59-3k)$$


问:
Let $A$ be a square matrix with all its diagonal and anti-diagonal elements equal to zero; is $A$ not invertible?
Example for $3\times 3$: $A=\begin{bmatrix}0&3&0\\1&0&7\\0&1&0\end{bmatrix}$

I tried out some examples and it seems right, but I honestly have no idea how to prove that it is not invertible in general; any help is really appreciated.

Body: How about $\pmatrix{0&1&0&0\\1&0&0&0\\0&0&0&1\\0&0&1&0}$?


问:
Evaluating a sum without using a program
$$ \sum_{k=1}^{\infty} \frac{e^k}{k^k} $$

The solution is about $\approx {5.5804}$

But I don't know how to calculate this sum, I tried using the squeeze theorem but I couldn't find $2$ series that converge such that $$a_n \leq \frac{e^k}{k^k} \leq b_n$$

The only thing I did is set $a_n = \frac{1}{k^k}$ but it didn't really help because it's way less than the original question's series, by a couple of magnitudes... ($e^k$ to be exact)

I would highly appreciate your help :-) Thanks!

答:
If you expand out $e^k$ you should find  term in $k^k/k!$ with all terms positive, so I shouldn't think it will converge.

Edit

I hadn't considered Japp's thought below, seems I could be wrong, you have no idea how hard it is to type this :-)


问:
If $A$ is a $k$-algebra for an algebraically closed field $k$, then $A/M=k$ for any maximal $M$ of $A$.
I am trying to understand why if a commutative ring $A$ is a finitely generated $k$-algebra for an algebraically closed field $k$, then $A/M=k$ for any maximal $M$ of $A$.

I know that by the Nullstellensatz that $A/M$ is a finite field extension of $k$, and that we have the homomorphism $k\rightarrow A\rightarrow A/M$. However, how can I use the algebraic closure of $k$ to see that this is indeed an isomorphism?

答: $A/M$ is a finite field extension of $k$, and is in particular algebraic over $k$.  But $k$ is algebraically closed, so $A/M = k$.


问:
Find the first term in a geometric progression
The exercise reads as follows:

The sum of the first 5 terms in a geometric progression is 62. The 5th, 8th and 11th term of this geometric sequence are also the 1st, 2nd and 10th term of an arithmetic sequence. Find the first number of the geometric sequence.

This was a rough translation so my attempt at explaining the exercise-question better is:

GP $a,b,c,d,e,f,g,h,i,j,k$

$a+b+c+d+e=62$

AP $e,h,x_3,x_4,x_5,x_6,x_7,x_8,x_9,k$

Find the number $a$.

GP is geometric progression, AP is arithmetic progression. $x_3,x_4,x_5,x_6,x_7,x_8,x_9$ are the other terms of the arithmetic progression which we know nothing of (and they probably do not matter, not sure).

Any hints? I thought about using the ratio from the geometric progression in two different ways: either starting from $e$ (therefore $d=e/r$ and $h=e*r^3$) or starting from $a$ (and we use $e=a*r^4$). The information about the sum is probably valuable but I might not know how to use it properly.

Thanks to the suggestions, I've made it a bit further:

$a=a$

$b=a*q$

$c=a*q^2$

$d=a*q^3$

$e=a*q^4$

$h=a*q^7$

$k=a*q^{10}$

I've wrote the important terms using $a$ and the geometric progression ratio. We have the following equations:

$a*(1+q+q^2+q^3+q^4)=62$ (sum)

$a*q^7=a*q^4+r$ (h)

$a*q^{10}=a*q^4+9r$ (k)

$q$ is the ratio from the geometric progression, $r$ is the ratio from the arithmetic progression

EDIT: I've managed to find some answers with the help of Ross. The arithmetic progression has the ratio equal to 224, the geometric progression has the ratio 2. The first term is equal to 2. Sadly the book has no answers and I can't double-check, they were discovered by giving values and seeing if it works, assuming all the numbers are integers (which they probably are but the exercise doesn't specify).

If anyone is still willing to help with a solution that doesn't require "trial and error" so to say, feel free.

答:
We have sum of the first $5$ terms in a geometric progression. sum of first $n$ terms can be calculated by $S_n=\frac{a(1-q^n)}{1-q}$. for $n=5$:

$$S_5=\frac{a(1-q^5)}{1-q}=62\tag1$$

The $5$th, $8$th and $11$th term of this geometric sequence are also the $1$st, $2$nd and $10$th term of an arithmetic sequence:$$\underbrace{{aq^4}}_{m},\underbrace{aq^7}_{m+d},\cdots,\underbrace{aq^{10}}_{m+9d}$$

By extracting $d$ we get:

$$d=a(q^7-q^4)=\frac{a(q^{10}-q^{7})}{8}$$
$$8(q^7-q^4)=q^3(q^7-q^4)$$

Therefor $q=2$. substitute this in the equation $(1)$ to find $a$.


问:
Time changed gaussian process
I have a question regarding a time changed Gaussian process. Suppose that $\{X_t\}_{t\in \mathbb{R}}$ is a Gaussian process with $0$ mean and covariance function $K$.

Suppose that $\phi$ is a deterministic function i.e. $\phi:\mathbb R \to \mathbb R$. What conditions should we have on the function $\phi$ such that $\{X_{\phi(t)}\}_{t\in \mathbb{R}}$ is still a Gaussian process ?

答:
As you said $\phi$ is a deterministic function, there should be no further condtions be necessary.

Because a time change is a nondecreasing cadlag function and Gaussian process is defined as (thats the definition I learned) a stochastic process $X_t$, so that for every $n\in\mathbb N$, every $t_1<...<t_n$ and every linear mapping $$l:\mathbb R^n\to \mathbb R$$ holds, that the real RV $l(X_{t_1},...,X_{t_n})$ is normal distributed. A deterministic change of time $\phi$ does not change anything of this property, since you can just choose new times $t'_1=\phi(t_1),...,t'_n=\phi(t_n)$ of which you already know, that $l(X_{t'_1},...,X_{t'_n})$ ist normal distributed.

If your function is not deterministic, this would not be the case in general as this post shows.

Title: Time changed gaussian process


问:
Evaluate $\exp\bigg(-\dfrac{x^2}{2} + 4x \bigg) \text{ as } x \to \infty.$

Compute limit of $$\exp\bigg(-\dfrac{x^2}{2} + 4x \bigg) \text{ as } x \to \infty.$$

Why is the above limit $0$?

答: Hint: $e^{-x}\rightarrow 0$ as $x\rightarrow\infty.$


问:
minmod slope total variation diminishing
This is exercise 6.5 of the book Finite Volume Methods for Hyperbolic Problems by R.J. LeVeque (2002).

Show that the minmod slope guarantees that $$
TV(q^n(·, t_n)) ≤ TV(Q^n) \tag{6.23}
$$ will be satisfied in general, and hence the minmod method is TVD.

with the minmod slope defined as

$$minmod(a,b) =\left\{\begin{array}{lll}a&\text{if } |a|<|b|&\text{and } ab>0\\b&\text{if } |b|<|a|&\text{and }ab>0\\0&\text{if } ab\leq 0&\end{array} \right.$$

$$\sigma_i^n= minmod\left(\frac{Q_i^n−Q_{i-1}^n}{\Delta x},\frac{Q_{i+1}^n−Q_i^n}{\Delta x}\right).$$

and the total variation diminution (TVD) property defined by $TV(Q_{n+1}) \leq TV(Q_n)$.

It is known that

$TV(Q)=\sum_i|Q_i-Q_{i-1}|$ (definition of total variation)

$TV(Q^{n+1})\leq TV(q^n\left(\cdot,t_{n+1})\right)$ (given in exercise 6.4)

$TV(q^n\left(\cdot,t_{n+1})\right)= TV(q^n\left(\cdot,t_{n})\right)$ (follows from REA algorithm 2nd step)

$TV(q^n\left(\cdot,t_{n})\right)=\sup\sum_i|q^n(x_i)-q^n(x_{i-1})|$ (defitinion)

My idea was

$$\begin{align}
TV(Q^{n+1})&\leq TV(q^n\left(\cdot,t_{n+1})\right)\\
&=TV(q^n\left(\cdot,t_{n})\right)\\
&=\sup\sum_i|q^n(x_i)-q^n(x_{i-1})|\\
&\leq \sup\sum_i|Q_i^n-Q_{i-1}^n|\\
&=TV(Q^n)
\end{align}$$

But nowhere I've used something from minmod. What am I missing?

Edit

The REA algorithm is the following:

Reconstruct $q^n(\cdot,t_n)=Q_i^n+\sigma_i^n(x-x_i)$, $\forall x\in C_i$, where $q^n$ is a piecewise polynomial function, $Q_i^n$ the cell average and $\sigma_i^n$ the slope.

Evolve the hyperbolic equation exactly with this initial data to obtain $q^n(x,t_{n+1})$ a time $\Delta t$ later.

Average this function over each grid cell to obtain new cell averages $Q_i^{n+1}=\frac{1}{\Delta x}\int\limits_{C_{i}}q^n(\cdot,t_{n+1})dx$

答:
We consider the advection equation $q_t + cq_x = 0$ with $c>0$. For the REA slope-limiter method with minmod slope $\sigma_i^n$ specified in OP, the reconstructed piecewise linear data of the book's Eq. (6.11) reads \begin{aligned}
q^n(x, t_n) &= \sum_{i=-\infty}^\infty \left(Q_i^n + \sigma_i^n (x-x_i)\right) \Bbb I_i(x) \\
&= Q_{-\infty}^n + \sum_{i=-\infty}^\infty \left(Q_{i+1}^n - \sigma_{i+1}^n x_{i+1} - Q_i^n + \sigma_i^n x_i\right) H(x-x_{i+1/2})  \\ &\; + x \sum_{i=-\infty}^\infty (\sigma_{i+1}^n - \sigma_i^n)\, H(x-x_{i+1/2}) \, ,
\end{aligned}
where $\Bbb I_i(x)$ is the indicator function of the $i$th finite volume $[x_{i-1/2}, x_{i+1/2}[$, and $H$ is the Heaviside step function. In the case of zero slope $\sigma_i^n \equiv 0$, a straightforward computation of the total variation yields $$
TV(q^n(\cdot, t_n)) = \sum_{i=-\infty}^\infty |Q_{i+1}^n - Q_{i}^n| = TV(Q^n) \, ,
$$ cf. Eq. (6.21) of the book. The present exercise consists in proving that the above equality becomes an inequality of the form $TV(q^n(\cdot, t_n)) \leq TV(Q^n)$ in the case of minmod slope $\sigma_i^n \not\equiv 0$ defined in OP --- in fact, the evaluation of $TV(q^n(\cdot, t_n))$ in OP is incorrect. To understand how things work, one could start with the computation of $TV(q^n(\cdot, t_n))$ for very simple piecewise linear functions $q^n$, e.g. the case of one single nonzero state $Q_0^n \neq 0$, then the case of two successive nonzero states $Q_0^n, Q_1^n \neq 0$, etc.
This way, you'll be able to tackle the case of arbitrary data $Q^n$.
From the above inequality (Eq. (6.23) of the book) and Eqs. (6.24)-(6.25), one shows that the minmod slope-limiter method is TVD.


答:
In the arithmetic progression you need one step to go from $e=aq^4$ to $h=aq^7$, but then eight steps to go from $h$ to $k=aq^{10}$.  So we must have

$aq^7-aq^4=d,aq^{10}-aq^7=8d$

$\color{blue}{8}d=aq^{10}-aq^7=q^3(aq^7-aq^4)=\color{blue}{q^3}d$

Can you find $q$ (and verify that there is only one real root for that variable) and continue?

Title: Find the first term in a geometric progression

Body:
The exercise reads as follows:

The sum of the first 5 terms in a geometric progression is 62. The 5th, 8th and 11th term of this geometric sequence are also the 1st, 2nd and 10th term of an arithmetic sequence. Find the first number of the geometric sequence.

This was a rough translation so my attempt at explaining the exercise-question better is:

GP $a,b,c,d,e,f,g,h,i,j,k$

$a+b+c+d+e=62$

AP $e,h,x_3,x_4,x_5,x_6,x_7,x_8,x_9,k$

Find the number $a$.

GP is geometric progression, AP is arithmetic progression. $x_3,x_4,x_5,x_6,x_7,x_8,x_9$ are the other terms of the arithmetic progression which we know nothing of (and they probably do not matter, not sure).

Any hints? I thought about using the ratio from the geometric progression in two different ways: either starting from $e$ (therefore $d=e/r$ and $h=e*r^3$) or starting from $a$ (and we use $e=a*r^4$). The information about the sum is probably valuable but I might not know how to use it properly.

Thanks to the suggestions, I've made it a bit further:

$a=a$

$b=a*q$

$c=a*q^2$

$d=a*q^3$

$e=a*q^4$

$h=a*q^7$

$k=a*q^{10}$

I've wrote the important terms using $a$ and the geometric progression ratio. We have the following equations:

$a*(1+q+q^2+q^3+q^4)=62$ (sum)

$a*q^7=a*q^4+r$ (h)

$a*q^{10}=a*q^4+9r$ (k)

$q$ is the ratio from the geometric progression, $r$ is the ratio from the arithmetic progression

EDIT: I've managed to find some answers with the help of Ross. The arithmetic progression has the ratio equal to 224, the geometric progression has the ratio 2. The first term is equal to 2. Sadly the book has no answers and I can't double-check, they were discovered by giving values and seeing if it works, assuming all the numbers are integers (which they probably are but the exercise doesn't specify).

If anyone is still willing to help with a solution that doesn't require "trial and error" so to say, feel free.


问:
Why do we need polar coordinates?
So, I haven't formally started learning about polar coordinates or anything yet, I'm just curious about them. I'm saying this to convey that I might not understand many technical terms.

So, I want to know how exactly polar coordinates came into existence i.e. what gave rise to the idea of polar coordinates. I would also like to know how they found their purpose/applications later on and some examples of things that they can do but Cartesian coordinates can not.

This is my first question here. Sorry if it's off topic.

Thank you!

答:
The first thing that comes to mind is for aircraft navigation. The location of a plane in the Earth's atmosphere is described by a height above land ($r$ in polar coordinates) and latitude/longitude which are two angles, $\theta$ and $\phi$. Using $x$, $y$ and $z$ would be needlessly complicated for describing where a plane is.

This is one example of where polar coordinates are beneficial over cartesian. All coordinate systems "do" the same things, but some simplify problems more than others.

Title: Why do we need polar coordinates?


问:
Since dividing $x=x^6$ by $x$ gives $1=x^5$, how can I get to $x=0$ as a root?
this might sound like a stupid question,
bear with me it probably is.

I know the solutions for $x=x^6$ are 1 and 0.

Now, since $1 \cdot x = 1 \cdot x^6 $ and it follows $ 1 \cdot x = 1 \cdot x \cdot x^5$ and I divide both sides by $ 1 \cdot x$ to get $ 1=x^5 $, how can I find $x=0$?

Thank you so much in advance.

答:
Whenever you divide by a variable quantity, you have to pause and think, "Either this is zero or not.  If it's not, I can divide by it.  If it is, that's another case."

So if $x$ is not $0$, you can divide by it to get $1=x^5$.  If it is zero, then, well, $x=0$.

But most folks would proceed like this:  $0 = x^6-x = x(x^5-1)$  and one of the last factors equals $0$.  So either $x=0$  or $x^5-1 = 0$.


答:
Don't divide. Factorise:

$x=x^6$

$x-x^6=0$

$x(1-x^5)=0$

Either $x=0$ or $1-x^5=0 \Rightarrow x=1$

Title: Since dividing $x=x^6$ by $x$ gives $1=x^5$, how can I get to $x=0$ as a root?

Body:
this might sound like a stupid question,
bear with me it probably is.

I know the solutions for $x=x^6$ are 1 and 0.

Now, since $1 \cdot x = 1 \cdot x^6 $ and it follows $ 1 \cdot x = 1 \cdot x \cdot x^5$ and I divide both sides by $ 1 \cdot x$ to get $ 1=x^5 $, how can I find $x=0$?

Thank you so much in advance.


答:
I would not like to use the definition for illustration. Instead, I am going to use the limit of a composite function for illustration. The function you have can be viewed as a composite function $h\left(x\right) = f\left(g\left(x\right)\right)$, where $g\left(x\right) = -\frac{x^{2}}{2} + 4x$ and $f\left(x\right) = e^{x}$. As
\begin{equation}
\lim_{x\to \infty}-\frac{x^{2}}{2} + 4x = -\infty
\end{equation}
and
\begin{equation}
\lim_{x\to -\infty}e^{x} = 0,
\end{equation}
the solution of your problem is
\begin{equation}
\lim_{x\to\infty}e^{-x^{2}/2+4x} = 0.
\end{equation}
Strictly, you should use the $\epsilon-\delta$ language to prove it, as the type of notation $\lim_{x\to\infty}f\left(x\right) = -\infty$ is not rigorous.

Title: Evaluate $\exp\bigg(-\dfrac{x^2}{2} + 4x \bigg) \text{ as } x \to \infty.$

Body:
Compute limit of $$\exp\bigg(-\dfrac{x^2}{2} + 4x \bigg) \text{ as } x \to \infty.$$

Why is the above limit $0$?


答:
You find $x=0$ by using the property that if $A\cdot B=0$, and $A,B$ are real numbers (or elements of a field), then $A=0$ or $B=0$.

Write $x=x^6\iff x^6-x=0\iff x(x^5-1)=0$

Either $x^5-1=0$ or $x=0$. The last thing shows that $x=0$ is a solution.


问:
Why does reindexing fail when trying to solve this infinite sum?
The infinite sum in question is $\sum_{1\leq n\leq\infty} ln\left(\frac{n}{n+1}\right)$ (I'm writing my sums in general Sigma-notation) which I broke down into $\sum_{1\leq n\leq\infty} ln(n) - \sum_{1\leq n\leq\infty} ln(n + 1)$. Writing out the partial sum and seeing the pattern, you can prove that it obviously diverges. However, when solving telescopic series, I like to reindex because it's easier, but when I reindex, I get that it converges as shown below.

$$\sum_{1\leq n\leq\infty} ln(n) - \sum_{1\leq n\leq\infty} ln(n + 1)$$
$$\sum_{1\leq n\leq\infty} ln(n) - \sum_{1\leq n - 1\leq\infty} ln((n - 1) + 1 )$$
$$\sum_{1\leq n\leq\infty} ln(n) - \sum_{1 + 1\leq n\leq\infty + 1} ln(n)$$
$$\sum_{1\leq n\leq\infty} ln(n) - \sum_{2\leq n\leq\infty} ln(n)$$
$$\sum_{1\leq n\leq\infty} ln(n) - \left(\sum_{1\leq n\leq\infty} ln(n) - ln(1)\right)$$
$$\sum_{1\leq n\leq\infty} ln(n) - \sum_{1\leq n\leq\infty} ln(n) + 0$$
$$0 \rightarrow converges$$

I've looked over this over and over and feel as if I'm doing something dumb or missing something. Someone explained to me that in the partial sum $S_n = ln(1) + (ln(2) - ln(2)) + (ln(3) - ln(3))... - ln(n+1)$ that because there will always be that last term which doesn't have a pair, that it will approach negative infinity and therefore diverge, but I don't get this because of the fact that there are an infinite amount of terms means that the last $ln(n + 1)$ term will always have a pair to cancel it out which is the next $ln(n)$ term in the series. I see how the limit of the partial sum diverges, but by my reasoning, it has to converge at infinity.
So there are two questions here, why does reindexing fail in this scenario and what is wrong about my thought process with the partial sum?

答:
Infinite series are odd beasts and if you rearrange terms you can get different convergence properties. Redo your algebra but replacing infinity with a large number $N$. You'll get, as the final result,

$$
\sum_{1\leq n \leq N} \ln(n) - \sum_{2 \leq n \leq N+1} \ln(n) = \ln(1) - \ln(N+1)
$$

This $-\ln(N+1)$ term that appears in every partial sum is swept under the rug by your telescoping manipulations that allow cancellation of terms all the way up to $N \to \infty$. Qualitatively, its as though you've staggered the two series and ignored the place where one "overhangs" the other at infinity.


答:
Curvilinear coordinate systems, like polar coordinates, came into being because they simplify computations in certain situations. For example think of a body moving with constant angular velocity $\omega$ around a circle of radius one centered at the origin. Using cartesian coordinates $x, y$ its motion is described by
$$
x = \cos\omega t,\
y = \sin\omega t
$$
in polar coordinates $r, \theta$ it becomes much nicer
$$
r = 1,
\theta = \omega t
$$
this can simplify other computations you might want to do that are related to this kind of motion.

Title: Why do we need polar coordinates?

Body:
So, I haven't formally started learning about polar coordinates or anything yet, I'm just curious about them. I'm saying this to convey that I might not understand many technical terms.

So, I want to know how exactly polar coordinates came into existence i.e. what gave rise to the idea of polar coordinates. I would also like to know how they found their purpose/applications later on and some examples of things that they can do but Cartesian coordinates can not.

This is my first question here. Sorry if it's off topic.

Thank you!


Body:
\begin{align}
-\frac{x^2}{2} + 4x &= -\frac{1}{2}\left( x^2-8x \right) \\
&= -\frac{1}{2}\left((x-4)^2-16\right) \\
&= -\frac{1}{2}(x-4)^2+8
\end{align}


答:
For $x \ge 16$
$$
 - \frac{{x^2 }}{2} + 4x =  - x^2 \left( {\frac{1}{2} - \frac{4}{x}} \right) \le  - \frac{{x^2 }}{4}.
$$
Thus, by the positivity and monotonicity of the exponential function
$$
0 < \exp \left( { - \frac{{x^2 }}{2} + 4x} \right) \le \exp \left( { - \frac{{x^2 }}{4}} \right)
$$
for $x\geq 16$. Can you see why does this imply the claim?


问:
Calculating $\lim_{n \to \infty} n^\alpha \int^{\frac{1}{n^\beta}}_0 f(x)dx$ where $\beta > 0$ and $\beta > \alpha$
An exercise in the lecture notes asks to calculate

$$\lim_{n \to \infty} n^\alpha \int^{\frac{1}{n^\beta}}_0 f(x)dx$$

where $\beta$ and $\alpha$ are real numbers with $\beta > 0$ and $\beta > \alpha$ for a function $f$ integrable on $[0,1]$.

But I am confused about how to do this for any arbitrary function $f$.

To get a sense, I tried to put $f(x)=1$ and I got $\lim_{n \to \infty} n^{\alpha-\beta}$ which evaluates to $0$ since $\alpha - \beta < 0$. Then I tried $f(x)=2x$ and got $\lim_{n \to \infty} n^{\alpha-2\beta}$ which again evaluates to $0$.

Is this the case for all such $f$, can you explain how to deal with this question?

答:
By definition if $f$ is (Riemann) integrable then $f$ is bounded; say $|f(t)|\le M$ for all $t\in[0,1]$. Now $$\left|n^\alpha\int_0^{1/n^\beta}f(x)\,dx\right|
\le n^\alpha\int_0^{1/n^\beta}|f(x)|\,dx\le n^\alpha\int_0^{1/n^\beta}M\,dx\dots .$$


问:
Seating four people next to each other probability problem
The following problem is from Carol Ash's The Probability Tutoring Book

If $4$ people are assigned seats at random in a $7$ seat row, what is the probability that they are seated together?

My approach:

We can choose $4$ from $7$ seats: $_{7}C_4$ ways,

We can permute the people in $4!$ ways.

So, my answer was $\frac{4!}{_{7}C_4}$

However the answer in the book is $\frac{4}{_{7}C_4}$. did they miss the factorial? Or is this the correct answer?

Why is the numerator just $4$?

Is it because if I permute the numerator, I also have to permute the denominator? Can someone explain why this is logically so?

答:
You have to be clear whether you are using permutations or combinations.

If you want to use permutations, $Pr =\dfrac{4!\cdot4}{7\cdot6\cdot5\cdot4}$

whereas if you want to use combinations, $Pr = \dfrac{4}{\binom7 4}$

Whenever probability is asked for, it is generally simpler to use combinations


答:
Denote $$ a=\sum_{k=1}^{\infty} \frac{e^k}{k^k}, \, a_n=\sum_{k=1}^n \frac{e^k}{k^k}.$$

We have for $n \in \mathbb N$ $$0 \le a - a_n = \sum_{k=n+1}^\infty \frac{e^k}{k^k} \le \sum_{k=n+1}^\infty \frac{e^k}{(n+1)^k} = \left(\frac{e}{n+1}\right)^{n+1}\frac{n+1}{n+1-e}.$$

If I understand well the question, you expect to compute an approximation of $a$ at $10^{-4}$. Therefore you need

$$\left(\frac{e}{n+1}\right)^{n+1}\frac{n+1}{n+1-e} \lt 10^{-4}$$ which is the case for $n=8$.

$a_8$ is an approximation of $a$ at $10^{-4}$ and you have

$$a_8 \approx 5.580377.$$ With the same process, you can compute the value of $a$ (with a paper, a pencil and some sweat...) at the desired precision.

Note: this supposes that an approximation of $e$ is already known. If not, you can find one using $e= \sum_{k=0}^\infty \frac{1}{n!}$.

Title: Evaluating a sum without using a program

Body:
$$ \sum_{k=1}^{\infty} \frac{e^k}{k^k} $$

The solution is about $\approx {5.5804}$

But I don't know how to calculate this sum, I tried using the squeeze theorem but I couldn't find $2$ series that converge such that $$a_n \leq \frac{e^k}{k^k} \leq b_n$$

The only thing I did is set $a_n = \frac{1}{k^k}$ but it didn't really help because it's way less than the original question's series, by a couple of magnitudes... ($e^k$ to be exact)

I would highly appreciate your help :-) Thanks!


答: Polar coordinates are useful in differential equations, mainly when solving problems related to variable separable form and sometimes useful in finding out wronskian</p>

Title: Why do we need polar coordinates?


问:
Non-integer powers of complex number
When introducing branch points and branch cuts I had a lecture on non-integer powers of complex number. To explore it I google it and get a paper where it was explained using polar representation of complex number. Here is the snippet,

I didn't understand how they get the set of values and why this set of values is independent of the numerator $P$? It will be great help if anyone explain it.

答:
The quantities in the curly brackets are called the roots of unity. In the complex plane,
they sit at $Q$ evenly-spaced points on the unit circle, with $1$ as one of the values.

Title: Non-integer powers of complex number


问:
Substitution with limits
If we have a limit to solve, doesn't matter which, but $\lim_{x\to 0}\ln(\frac{1}{x})$ for example, is the substitution $t=\frac{1}{x}$ allowed? Because as $x\to 0$, $t\to DNE$? Because $\lim_{x\to 0^+}\frac{1}{x}=+\infty$ and $\lim_{x\to 0^-}\frac{1}{x}=-\infty$? Hence non sided limit is DNE

答:
Note that
$$\lim_{x\to0^+}f(x) = \lim_{t\to +\infty}f\left(\frac{1}{t}\right) \\ \lim_{x\to0^-}f(x) = \lim_{t\to -\infty}f\left(\frac{1}{t}\right)$$
Usually, the "infinity" without a sign means that the "infinities" with both signs are being dealt simultaneously; that is, $\infty = \pm \infty$. So, you can say $$x\to 0 \implies \frac{1}{x} \to \infty$$

Title: Substitution with limits


问:
Show that the vertical component of the reaction is constant
A smooth wire in the shape of the helix $\vec{r} = a (cos θ)\vec{i}+ a (sin θ)\vec{j}+ cθ\vec{k}$, where $a$ and $c$ are positive constants is fixed with the $z$-axis pointing vertically downwards. A bead of mass $m$, free to slide along the wire, is released from rest
at the point A where $θ = 0$. Show that the vertical component of the reaction is constant.

I applied $\vec{f}=m\vec{a}$ to the particle vertically so I got

$$f=ma=R_{k}-mg=mc\ddot{\theta}$$
My textbooks solution is $R_{k}=mc\ddot{\theta}$

Is there anything wrong what I did?

答:
There is a laborious way to show that. The movement equations can be obtained from energy conservation considerations.

Calling  $p = (x(t),y(t),z(t))$ the movement lagrangian is

$$
L = \frac 12 m (\dot p\cdot \dot p) - m g z(t) + \lambda(x^2(t)+y^2(t)-a^2)+\mu\left(z(t)-c \arctan\left(\frac{y(t)}{x(t)}\right)\right)
$$

so the movement equations are given by

$$
\left\{
\begin{array}{rcl}
 2 \lambda  x(t)-m x''(t)+c \frac{\mu  y(t)}{x(t)^2+y(t)^2} &=&0\\
 2 \lambda  y(t)-m y''(t)-c \frac{\mu  x(t)}{x(t)^2+y(t)^2} &=&0\\
 -g m-m z''(t)+\mu  &=&0\\
z(t)-\arctan\left(\frac{y(t)}{x(t)}\right)&=&0\\
x^2(t)+y^2(t)-a^2&=&0\\
\end{array}
\right.
$$

Deriving twice regarding $t$ the last two equations and solving for $z''(t)$ over the full set of equations, we obtain

$$
z''(t) = -\frac{c \left(x(t)^2 \left(c g+2 x'(t) y'(t)\right)+y(t)^2 \left(c g-2 x'(t) y'(t)\right)+2 x(t) y(t)
   \left(y'(t)^2-x'(t)^2\right)\right)}{\left(x(t)^2+y(t)^2\right) \left(c^2+x(t)^2+y(t)^2\right)}
$$

now substituting

$$
\cases{
x(t) = a\cos(\theta)\\
y(t) = a\sin(\theta)
}
$$

we obtain

$$
z''(t) = -\frac{c^2g}{a^2+c^2}
$$


问:
Is it possible to rearrange this equation to solve for x?
$$0 = -i+(iq-ix+1)\ln(1+i)-(1-d)^xln(1+i)-(1-d)^xln(1-d)+t (1-d)^xln(1+i)+t \ln(1-d)^x$$

Variables $i, q, d, t, x$

Wanting to rearrange equation for $x$, online calculators don't seem to like it.

答: You can't solve for $x$ with the usual functions, as $x$ appears both as a linear term and an exponential.


答:
It's not the reindexing what fails. The problem is that you split a convergent series as a difference of two divergent series.

In general, $\sum_{n=0}^\infty (a_n + b_n) \ne \sum_{n=0}^\infty a_n  + \sum_{n=0}^\infty b_n $. If you consult this theorems in a textbook you'll find that you need the hypothesis of both series of the RHS to be convergent.

Title: Why does reindexing fail when trying to solve this infinite sum?

Body:
The infinite sum in question is $\sum_{1\leq n\leq\infty} ln\left(\frac{n}{n+1}\right)$ (I'm writing my sums in general Sigma-notation) which I broke down into $\sum_{1\leq n\leq\infty} ln(n) - \sum_{1\leq n\leq\infty} ln(n + 1)$. Writing out the partial sum and seeing the pattern, you can prove that it obviously diverges. However, when solving telescopic series, I like to reindex because it's easier, but when I reindex, I get that it converges as shown below.

$$\sum_{1\leq n\leq\infty} ln(n) - \sum_{1\leq n\leq\infty} ln(n + 1)$$
$$\sum_{1\leq n\leq\infty} ln(n) - \sum_{1\leq n - 1\leq\infty} ln((n - 1) + 1 )$$
$$\sum_{1\leq n\leq\infty} ln(n) - \sum_{1 + 1\leq n\leq\infty + 1} ln(n)$$
$$\sum_{1\leq n\leq\infty} ln(n) - \sum_{2\leq n\leq\infty} ln(n)$$
$$\sum_{1\leq n\leq\infty} ln(n) - \left(\sum_{1\leq n\leq\infty} ln(n) - ln(1)\right)$$
$$\sum_{1\leq n\leq\infty} ln(n) - \sum_{1\leq n\leq\infty} ln(n) + 0$$
$$0 \rightarrow converges$$

I've looked over this over and over and feel as if I'm doing something dumb or missing something. Someone explained to me that in the partial sum $S_n = ln(1) + (ln(2) - ln(2)) + (ln(3) - ln(3))... - ln(n+1)$ that because there will always be that last term which doesn't have a pair, that it will approach negative infinity and therefore diverge, but I don't get this because of the fact that there are an infinite amount of terms means that the last $ln(n + 1)$ term will always have a pair to cancel it out which is the next $ln(n)$ term in the series. I see how the limit of the partial sum diverges, but by my reasoning, it has to converge at infinity.
So there are two questions here, why does reindexing fail in this scenario and what is wrong about my thought process with the partial sum?


问:
Problem with joint PDF of two random variables
I'm going through probability problems for an upcoming exam and came across this problem:

The PDF of the random vector $(X,Y)$ is $f(x,y)=3\min\left\{x,y\right\}, 0<x,y<1$. If the random variable $Z$ is defined as $Z=\max\{X,Y\}$, calculate the coefficient of correlation between $X$ and $Z$.

Most of the stuff needed is easy enough to work out, but to get $E[XZ]$ the approach I took is to find the joint CDF of $X$ and $Z$, which I worked out to be $F_{X,Z}(x,z)=\frac{3}{2}x^2z-\frac{1}{2}x^3$ for $0<x<z<1$ and functions of only one variable in other cases. This would mean that if there's a PDF, it would have to be $\frac{\partial}{\partial x}\frac{\partial}{\partial z}F_{X,Z}(x,z)$, i.e., $f_{X,Z}(x,z)=3x$ for $0<x<z<1$ and $0$ otherwise. However, this function can't be a PDF since its integral over $\mathbb{R}^2$ is not $1$.

So my questions are: Is this correct? If so, what other approach can I take to find $E[XZ]$?

答:
There is no pdf of $(X,Z)$ under Lebesgue measure since $X=Z$ holds with positive probability.

You have $$XZ=\begin{cases} X^2 &,\text{ if }X\ge Y \\ XY &, \text{ if }X<Y \end{cases}$$

If $\mathbf1$ is an indicator variable, then by law of total expectation $$\operatorname E\left[XZ\right]=\operatorname E\left[X^2\mathbf1_{X\ge Y}\right]+\operatorname E\left[XY\mathbf1_{X<Y}\right]$$

To calculate $\operatorname E\left[g(X,Y)\right]$ for any (measurable) function $g(\cdot)$, use this theorem:

$$\operatorname E\left[g(X,Y)\right]=\iint g(x,y)f(x,y)\,\mathrm{d}x\,\mathrm{d}y$$

So for example,

\begin{align}
\operatorname E\left[X^2\mathbf1_{X\ge Y}\right]&=3\iint x^2\mathbf1_{x\ge y}\min(x,y)\mathbf1_{0<x,y<1}\,\mathrm{d}x\,\mathrm{d}y
\\&=3\iint x^2y\,\mathbf1_{0<y\le x<1}\,\mathrm{d}x\,\mathrm{d}y
\\&=3\int_0^1 y\int_y^1 x^2\,\mathrm{d}x\,\mathrm{d}y
\end{align}


问:
Interior of $A=\{x\in \Bbb{R^n} : \vert x \vert \leq 1\}$
What is the interior of $A=\{x\in \Bbb{R^n} : \vert x \vert \leq 1\}$? I claim it is the set of $\{y\in \Bbb{R^n}:\vert y\vert<1\}.$ This is because

$\exists h\in\Bbb{R}$ such that $\vert y \vert=1-h.$ Also, $\exists z\in \Bbb{R^n}$ s.t $\vert y-z\vert < h$. Therefore $\vert z-0\vert=\vert z \vert\leq\vert y-0\vert+\vert y-z\vert<1-h+h=1$. Therefore $z\in A$ and $y$ is an interior point of $A$. What's wrong with this proof? I don't believe it's correct.

答:
Proof 1:

$\newcommand{\R}{\mathbb{R}^n} \newcommand{\eps}{\epsilon} \newcommand{\ball}[1]{B_{\eps}(#1)}$
The interior of $A$ is the largest open set contained in $A$.

Let $B = \{x \in\R~|~|x| <1\}$.

Claim 1: $B$ is open.

Let $a \in B$ be arbitrary. Then, $|a| < 1$ and let $\eps = (1-|a|)/2$. For all $y \in \R$, $|y-a| < \eps$ implies $|y| < 1$, that is, $B$ is open. Note that $|y-a| \geq |y| - |a|$ from triangle inequality.

Claim 2: $B$ is the largest open set contained in $A$.

Suppose $B$ is not the largest open subset of $A$. Then, there is a non-empty subset $C$ of $A$ that is disjoint from $B$ and $B \cup C$ is an open subset of $A$. By the conditions, $C \subseteq \{x\in\R~|~|x| = 1\}$. The disjoint union of $B$ and $C$ is not open, which is a contradiction.

Therefore, $B$ is the interior of $A$.

Proof 2:

Here is another proof using a different definition you would like:

If $B$ satisfies
$$
\forall x \in \R: x \in B \Leftrightarrow \exists \eps >0 :\ball{x} \subseteq A
$$

$B$ is the interior of $A$.

We can easily show the $\Rightarrow$ direction as we did previously. We now only need to show the necessity, whose contrapositive is

$$
\forall x \in \R: x \not\in B \Rightarrow \forall \eps >0 : \exists y \in \ball{x} : y \not\in A
$$

Let $a \not\in B$. Then, $|a| \geq 1$. For each $\eps > 0$, let $y = a(1+\frac{\eps}{2|a|})$. Then, $|y| = |a|(1+\frac{\eps}{2|a|}) = |a| +\eps/2 > 1$, which means $y \not\in A$. Furthermore, $|y - a| = |\frac{\eps a}{2|a|}| = \eps/2 < \eps$, that is, $y \in \ball{x}$.

This proof looks more useful.


问:
Given a finite number of stones. We place every stone on an integer. Prove that, given different movements, we can only make finite number of moves
Given a finite number of stones. We place every stone on an integer (a number $x$ where $x\in Z$) and maybe multiple stones on an integer.

On each move we can make one of the following movements:

Remove a stone from numbers $n-1$ and $n$ and place one stone in integer $n+1$.

Remove $2$ stones from $n$ and place them in integers $n+1$ and $n-2$.

Prove that we can only make finite number moves.

I attempted to do it as follows:

On each stone we give it the weight $x^n$ where $x^2-x-1=0$, hence $x=\frac{1+\sqrt{5}}{2}$.

Hence we have that $x^{n-1}(x^2-x-1)=0$ and $x^{n+1}+x^{n-2}-2x^n=x^{n-2}(x^3+1-2x^2)=x^{n-2}(x^2+x+1-2x^2)=x^{n-2}(x+1-x^2)=0$.

Which proves that the total weight of the stones is an invariant.

This is as far as I got. I do not know how to finish it off although I do believe that this question is logically solved like this. Could you please explain to me how to solve this question?

答:
This proof is rather semantic, since my technical notation is rather poor, but it works nonetheless. I've switched move 2 around, 1 to the left, and 2 to the right. The symmetry of the line shouldn't make it any different:

Assume you can do it indefinitely. With $x$  stones, you cannot do the first move indefinitely due to the fact that you lose 2 stones and gain 1, meaning a net loss of 1. More importantly, the number of stones before and after move 2 does not change it. Hence, we can take the 'start' position of the stones to be the point at which we stop doing move 1, and continue doing move 2 (still on the assumption that doing it indefinitely is possible)

Now all that's left is to prove that no matter how you do move 2, you'll end up with no moves left (we don't need to consider if two stones are in a row since we've safely assumed that all of those moves are done). To do this we just have to prove that the number of spaces occupied is equal to the number of stones. We can show this by pointing out that in a finite time, the number of spaces occupied will grow.

Consider the stone in the highest position $c$. There's no guarantee we'll perform the second move on it, so let's consider the highest position $Y$ that will be a move, that initially has at least 1 stone on it. I.e consider a set $X$ of positions with a stone on them at the beginning (I refer to the highest position with the highest ordinality that will eventually make a move).

When that position makes its move, it will

A) Donate a stone either to an empty position ($+1$
or $0$ [if the position we chose loses all its stones] positions occupied in total)

B) Donate to a position that has already has a stone on it (that will never be activated thereafter).

This same argument can be applied on the side of less stones.

Now we just need to disprove BB  (which implies that every move will continue within finite bounds), and AB and BA (which implies that the rest of the stones will move to the left or to the right or otherwise donating stones indefinitely to a position that will grow but never be activated). We also need to disprove that A can occur forever with a net gain of occupied positions of 0

BB:

Consider the highest activated stone, same thing we did earlier. When that makes it's move it will place a stone a further distance, thus disproving BB

AB, BA:

Same argument as BB, just take it to the B side.

AA (0):

Consider below to be what happens directly after moving stones on $c$ whereby c is no empty:

```<code>0000010??????
</code>```

That (previously) empty position was not in set $X$. A stone can be added to it. Since we pointed out that moves cannot have bounds, inevitably we'll get:

```<code>0000002???????
</code>```

Again, moves cannot have bounds; that 2 will be activated. Therefore:

```<code>0000010???????
</code>```

A $-1$ is, inevitable.

Body:
This proof is rather semantic, since my technical notation is rather poor, but it works nonetheless. I've switched move 2 around, 1 to the left, and 2 to the right. The symmetry of the line shouldn't make it any different:

Assume you can do it indefinitely. With $x$  stones, you cannot do the first move indefinitely due to the fact that you lose 2 stones and gain 1, meaning a net loss of 1. More importantly, the number of stones before and after move 2 does not change it. Hence, we can take the 'start' position of the stones to be the point at which we stop doing move 1, and continue doing move 2 (still on the assumption that doing it indefinitely is possible)

Now all that's left is to prove that no matter how you do move 2, you'll end up with no moves left (we don't need to consider if two stones are in a row since we've safely assumed that all of those moves are done). To do this we just have to prove that the number of spaces occupied is equal to the number of stones. We can show this by pointing out that in a finite time, the number of spaces occupied will grow.

Consider the stone in the highest position $c$. There's no guarantee we'll perform the second move on it, so let's consider the highest position $Y$ that will be a move, that initially has at least 1 stone on it. I.e consider a set $X$ of positions with a stone on them at the beginning (I refer to the highest position with the highest ordinality that will eventually make a move).

When that position makes its move, it will

A) Donate a stone either to an empty position ($+1$
or $0$ [if the position we chose loses all its stones] positions occupied in total)

B) Donate to a position that has already has a stone on it (that will never be activated thereafter).

This same argument can be applied on the side of less stones.

Now we just need to disprove BB  (which implies that every move will continue within finite bounds), and AB and BA (which implies that the rest of the stones will move to the left or to the right or otherwise donating stones indefinitely to a position that will grow but never be activated). We also need to disprove that A can occur forever with a net gain of occupied positions of 0

BB:

Consider the highest activated stone, same thing we did earlier. When that makes it's move it will place a stone a further distance, thus disproving BB

AB, BA:

Same argument as BB, just take it to the B side.

AA (0):

Consider below to be what happens directly after moving stones on $c$ whereby c is no empty:

0000010??????

That (previously) empty position was not in set $X$. A stone can be added to it. Since we pointed out that moves cannot have bounds, inevitably we'll get:

0000002???????

Again, moves cannot have bounds; that 2 will be activated. Therefore:

0000010???????

A $-1$ is, inevitable.


问:
Calculate $\lim_{n\to\infty}\int_0^nf(\frac{x}{n})^ndx$.
Consider the following problem:

Suppose $f:[0,1] \to \mathbb{R}$ is a differentiable function with $f'(0)<0$, $f(0)=1$ and for all $x\in(0,1]:0<f(x)<1$. Calculate $$\lim_{n\to\infty}\int_0^nf\left(\frac{x}{n}\right)^ndx.$$

My attempt so far:

Let $$f_n:x\mapsto\chi_{[0,n]}(x)\cdot f\left(\dfrac{x}{n}\right)^n,$$ where $\chi_{[0,n]}$ is the indicator function on $[0,n]$. We wish to do two things:

Show that $f_n\to h$ pointwise for some integrable function $h:\mathbb{R}^+\to\mathbb{R}$ and

show that $|f_n|\leq k$ for some positively integrable $k:[0,1]\to \mathbb{R}^+$ so that we can use the dominated convergence theorem to show that $$\lim_{n\to\infty}\int_{[0,\infty)}f_n(x) \,dx = \int_{[0,\infty)}h(x)dx.$$

To do this, we first construct the function $$g:[0,1]\to\mathbb{R}:y\mapsto\begin{cases}\dfrac{\ln f(y)}{y}& y\in(0,1],\\ f'(0) & y=0.\end{cases}$$
This is a continuous extention of $\ln f(y)/y$ since $$\lim_{y\to0}\dfrac{\ln f(y)}{y}=f'(0)$$
by l'Hôpital's rule. The idea is to show that  $$f\left(\dfrac{x}{n}\right)^n\leq e^{f'(0)x}$$ for all $x\in[0,n]$. To do this we can use the function $g$ as constructed above. Indeed, taking the natural log of this inequality gives us

$$n\ln f\left(\dfrac{x}{n}\right)\leq f'(0)x$$
which is equivalent to $$g(y)\leq f'(0),$$ for all $y\in[0,1]$. The problem is that I have trouble showing that this last inequality is true. Although I'm convinced it is. I've tried calculating $g'$ to see if it is negative everywhere, but have failed to do so.

Suppose it is true. Then, since $f'(0)<0$, the function $x\mapsto e^{f'(0)x}$ is integrable on $x>0$, so we have found a dominating function for the $f_n$.

Furthermore, we need to show that there is an integrable function $h$ to which the $f_n$ converge. I cannot see how to show this pointwise convergence.

Edit: I have found conclusive proofs for my two problems and have added them as an answer.

To conclude the problem, the limit is $$\lim_{n\to\infty}\int_0^nf\left(\frac{x}{n}\right)^ndx = \int_0^\infty e^{-|f'(0)|x}dx = \dfrac{1}{|f'(0)|}.$$

答:
Here is an alternative proof that doesn't require convergence theorems for integrals. Write $\ell=|f'(0)|$ and take $0<\epsilon<\frac12\ell$: the factor $1/2$ will come into play at the end and is purely here for convenience. Since $f$ is differentiable at $0$ there exists $\eta\in(0,1]$ such that
$$
\forall x\in [0,\eta],
\quad
|f(x)-1+\ell x|\leq\epsilon x
$$
Without loss of generality we further assume $\eta < \frac{1}{2\ell}$. Then for all $x\in[0,\eta]$,
$$
0<1-\ell x-\epsilon x\leq f(x)\leq 1-\ell x+\epsilon x
$$
and so
$$
\begin{array}{rcl}
\displaystyle
\int_0^nf(x/n)^n~dx
&=&
\displaystyle
n\int_0^1f(x)^n~dx\\
&=&
\displaystyle
\underbrace{n\int_0^\eta f(x)^n~dx}_{I_n}
+
\underbrace{n\int_\eta^1f(x)^n~dx}_{II_n}
\end{array}
$$
We show that the first term $I_n$ converges to $\frac1{f'(0)}$ and the second term $II_n$ tends to $0$.

The second term. This is swiftly dealt with: by continuity of $f$ and $0<f<1$ on $(0,1]$ it follows that $M=\max_{[\eta,1]}f<1$. Then
$$
0\leq II_n\leq nM^n
$$
so that $II$ indeed (converges and) tends to $0$.

The first term. Integrating the inequality from above yields
$$
-n\bigg[\frac{(1-(\ell+\epsilon)x)^{n+1}}{(n+1)(\ell + \epsilon)}\bigg]^\eta_0
\leq I_n\leq
-n\bigg[\frac{(1-(\ell - \epsilon)x)^{n+1}}{(n+1)(\ell - \epsilon)}\bigg]^\eta_0
$$
The LHS equals
$$
\frac{n}{n+1}\frac{1}{\ell+\epsilon} - C_nq_1^n=\frac{1}{\ell+\epsilon}+o(1)
$$
for $C_n=\frac{n}{(n+1)(\ell + \epsilon)}$ and $q_1=1-(\ell + \epsilon)\eta\in(0,1)$.

Similarly, the RHS equals
$$
\frac{n}{n+1}\frac{1}{\ell-\epsilon} - D_nq_2^n=\frac{1}{\ell-\epsilon}+o(1)
$$
for $D_n=\frac{n}{(n+1)(\ell - \epsilon)}$ and $q_2=1-(\ell - \epsilon)\eta\in(0,1)$.

Since $\frac{1}{\ell+2\epsilon}<\frac{1}{\ell+\epsilon}$ and $\frac{1}{\ell-\epsilon}<\frac{1}{\ell-2\epsilon}$, there exists an integer $n_0$ such that for all $n\geq n_0$
$$
\frac{1}{\ell+2\epsilon}
\leq
I_n
\leq
\frac{1}{\ell-2\epsilon}
$$
This establishes the convergence of $I_n$ to $\frac1\ell=\frac1{|f'(0)|}$ and finishes the proof.


问:
Count 0,1 codes with specific conditions
how many $0,1$ codes of size $n$ that each $1$ is between two $0$s?

$0100010010$ is acceptable but $010110$ is not.

i know it should be solved with some recursive function.

答:
obviously each code(I call string) should end with $0$.
second last number can be $0$ or $1$. if it is zero then the sub-string omitting last element is valid too. and if it is one then the sub-string omitting last two element is valid too. so we Have:

$A_n=A_{n-1}+A_{n-2}$

$A_1=A_2=1$

it is just like Fibonacci.


问:
Why is my final vector not!= B when using slerp to interpolate from A to B
I think this is just a simple case of my misunderstanding.

I have succesfully been using linear interpolation to interpolate between two many dimensional vectors, A and B, where elements of A and B are -1 .. 1

I then learned that using slerp can give better results for my task, and decided to implement it. From examples I saw, it worked as a straight drop-in for the linear interpolation method I was using.

However,  I find that where element wise pairs in the two vectors A and B have a different sign,  the sign of the corresponding element in the final (ie, interpolation distance is 1) vector is flipped.

I was expecting, when interpolating from A to B,  that the result of the interpolation when the interpolation distance is 1.0, should be B.  This is the case when using simple linear interpolation.  Why is not the case when using slerp?

The task at hand is interpolation between two latent-space vectors to generate a morphing sequence of outputs from a trained GAN model (wavegan) --  Simple examples use linear interpolation and I have been able to do so succesfully,  but the discussion here : https://github.com/soumith/dcgan.torch/issues/14  suggested using spherical linear interpolation.  The implementation I am using of slerp is not from that discussion, but instead the python implementation from the wiki page https://en.wikipedia.org/wiki/Slerp

答:
I found my answer;

It is not noted in the wikipedia page's implementation of slerp, but it flips the direction on one vector when the interpolation would not take the shortest path. I believe this is desirable when computing interpolations of quaternions for animation purposes, but was not desirable in my case.

This commented section is notable in the c++ implementation, while the comment is removed fro the python implementation;

```<code>// If the dot product is negative, slerp won't take
// the shorter path. Note that v1 and -v1 are equivalent when
// the negation is applied to all four components. Fix by
// reversing one quaternion.
if (dot < 0.0f) {
    v1 = -v1;
    dot = -dot;
}
</code>```

By removing this section I achieved the results I expected.

Body:
I found my answer;

It is not noted in the wikipedia page's implementation of slerp, but it flips the direction on one vector when the interpolation would not take the shortest path. I believe this is desirable when computing interpolations of quaternions for animation purposes, but was not desirable in my case.

This commented section is notable in the c++ implementation, while the comment is removed fro the python implementation;

// If the dot product is negative, slerp won't take
// the shorter path. Note that v1 and -v1 are equivalent when
// the negation is applied to all four components. Fix by
// reversing one quaternion.
if (dot < 0.0f) {
    v1 = -v1;
    dot = -dot;
}

By removing this section I achieved the results I expected.


问:
Decryption Problem
The following message was posted in our math department and I wouldn't mind some help into getting started at cracking it:

gectl atnoy danwm etaim oroni snair ohass wveno faome nceto kils

Any ideas?

答: It's a transposition cipher.  Write the ciphertext as 3 rows of 18, then read it vertically instead of horizontally.</p>

Title: Decryption Problem


问:
How to prove the concurrence of three lines?

Let quadrilateral $ABCD$ inscribe in circle whose center is $O$ and let $AC$ and $BD$ intersect at $P$,while $Q$ be a point on the straight line $OP$.The circumcenters of $\Delta QAB$, $\Delta QBC$, $\Delta QCD$, $\Delta QDA$ are $O_1$, $O_2$, $O_3$, $O_4$ respectively.Then how to prove the concurrence of the straight lines $O_1 O_3$, $O_2 O_4$, $OP$?($O$ is not coincide with $P$)

I'm sorry that I can't provide any useful ideas.(I have tried to violently calculate through trigonometric functions,but it only made me crazy)

And I'm very sorry for my possible wrong grammars and strange expressions because I'm still an English learner and my first language is very different from the English system.

I'd appreciate it if someone could share his ideas about the question.

答:
My friend has given his answer.I think I should present it.

Let $E=O_1O_3\cap O_2O_4$,and $F=AB\cap CD$,$G=AD∩BC$.Let $\Gamma$ be a circle whose centre is $E$,radius is $EQ$.(If $E$ is coincide with $Q$,then let $\Gamma$ be a null circle.)And $\Gamma$ meets $\bigcirc O_1$ again at $K$.

Then $FB\cdot FA=FC\cdot FD$,which implies that  $F$ is  on the radical axis of $\bigcirc O_1$ and $\bigcirc O_3$.So $FQ\perp O_1O_3$,which means $FQ\perp O_1E$.So $F$ is on the radical axis of $\bigcirc O_1$ and $\Gamma$.Then we have that $F,K,Q$ are on the same line,which implies that $FK\cdot FQ=FB\cdot FA$.Thus $F$ is on the radical axis of $\bigcirc O$ and $\Gamma$.Similarly,$G$ is on the radical axis of $\bigcirc O$ and $\Gamma$.So the straight line $FG$ is the radical axis of $\bigcirc O$ and $\Gamma$.Therefore,$OE\perp FG$.

And by Brokard's theorem,we know that $OP\perp FG$.Thus,$O,E,P$ are collinear.

So,the staight lines $O_1O_3$,$O_2O_4$ and $OP$ are concurrent at $E$.


问:
Where does the identity $p^{-ns}=s\int_{p^n}^\infty x^{-s-1}dx$ come from?
In this paper, equation 45 (page 11) gives the identity

$$p^{-ns}=s\int_{p^n}^\infty x^{-s-1}dx$$

Can someone tell me where this comes from, and how it is derived?

答: Try to see it follows from identity 41,42,43 I see that it can be connected to the mellin transform of the function</p>


答:
This statement holds only for $s>0$.

For $s\ne0$ the anti-derivative of $x^{-s-1}$ is $-\frac{x^{-s}}s$. Applying the limits,$$\frac s{-s}[x^{-s}]_{p^n}^\infty=p^{-ns}$$since $\infty^{-s}=1/\infty^s=0$ for positive $s$.

Title: Where does the identity $p^{-ns}=s\int_{p^n}^\infty x^{-s-1}dx$ come from?

Body:
In this paper, equation 45 (page 11) gives the identity

$$p^{-ns}=s\int_{p^n}^\infty x^{-s-1}dx$$

Can someone tell me where this comes from, and how it is derived?


问:
Some questions about variations of fixed point method
I'm doing some excercises in Fixed Point Iteration methods with Matlab. I have to find roots for $f(x)=e^x -x -1.9\cos x$ by using $x_{n+1}=g(x_n)$. I know how to choose $g(x)$ such that I can find both roots. The following part of the exercise is as follows:

In general we can rewrite $f(x)=0$ to $x=g(x)$ with $g(x)=x+cf(x)$, where $c\neq 0$ is a constant.

I am given the function $f(x)=e^x-x-1.9\cos x$ , and I have to determine values for $x_0$ and $c$ such that the sequences obtained for $x_{n+1}=g(x_n)$ both converge quickly, to  roots $x_{r1}$ and $x_{r_2}$.

I'm told that this can be obtained by picking a point $x_0$ close to the zero and setting $g'(x_0)=0$ by an appropriate choice of $c$.  My question is, what is this all about? What should I do?

Then, another part says

Increase the order of convergence of the fixed point method $x_{n+1}=x_n+cf(x_n)$ by changing $c$ each iteration instead of using a constant value.

How can I do that? What do they mean?

答:
A central tenet in calculus is that we can understand the local behavior of a function near a point by examining the linear approximation of the function at that point.  Thus, let's begin by exploring the behavior of iteration near fixed points of linear functions.

To this end, consider the function $\ell(x) = x_f + m(x-x_f)$.  This clearly has a fixed point at $x_f$.  Now, suppose that we iterate this function starting from some point $x_0$.  Note that
$$\ell(\ell(x_0)) = x_f + m((x_f + m(x_0-x_f)) - x_f) = x_f + m^2 (x_0 - x_f).$$

More generally,
$$\ell^p(x_0) = x_f + m^p (x_0 - x_f).$$

As a result, we see quite easily that the orbit of $x_0$ tends towards $x_f$ iff $|m|<1$.  Furthermore, the smaller the value of $m$, the faster the convergence.  The convergence is fastest (instantaneous, in fact) when $m=0$.

Now, let's consider the iteration of a differentiable function $g$ near a fixed point $x_f$, thus $g(x_f)=x_f$. Suppose we iterate $g$ starting from some point $x_0$ near $x_f$.  By analogy with the linear example we just examined, we might expect $x_f$ to be attractive if the slope of $g$ is less than $1$ in absolute value at the point $x_f$.  More precisely, we need $|g'(x_f)|<1$.

Now, in your example, you have a given function $f(x)=e^x-x-1.9\cos(x)$ and you'll generate $g_c$ by setting $g_c(x)=x+c\,f(x)$.  Since you have a computational tool in your hands, I suggest that you examine the graph of $g_c$ together with the graph of $y=x$ to help you choose a good value of $c$.  For example, here's the graph of $y=g_{-1/2}(x)$ together with the graph of $y=x$.

The points of intersection correspond to the roots of $f$.  It appears that one such point is around $0.7$ or so and that iteration of $g_{-1/2}$ starting near there should converge to that fixed point.  It does appear, though, that you could pick a value of $c$ that is better than $-1/2$.  Note that it is important to plot the graph in it's correct aspect ratio to get a good feel for the magnitude of the derivative.

Finally, if you start with a good $c$ value, then refinement should not be necessary.  I guess the idea, though, is choose $c$ at the $i^{\text{th}}$ step so that $g_c'(x_i)=0$.  Since $x_i$ is close to $x_f$, we'd then expect $g_c'(x_f)$ to be close to zero.


问:
How to prove this question related to multi-times composite relation $\mathcal{R}^{n}$
I found a question somewhat difficult, on An Invitation to Discrete Mathematics, by Matousek et al., which goes as below.
$\newcommand{\op}[2]{ \left\langle #1 ,\, #2 \right\rangle }$
$\newcommand{\rel}[1]{ \mathcal{#1} }$

Question:

For a relation $\rel{R}$ on a set $X$ we define the symbol $\rel{R}^n$ by induction:
$$\rel{R}^1 = \rel{R},\, \rel{R}^{n+1}=\rel{R} \circ \rel{R}^n$$

(a) Prove that if $X$ is finite and $\rel{R}$ is a relation on it, then there exist $r, s \in \mathbb{N}$, $r < s$, such that $\rel{R}^r = \rel{R}^s$.

(b) Find a relation $\rel{R}$ on a finite set such that $\rel{R}^n \neq \rel{R}^{n+1}$ for every $n \in \mathbb{N}$.

(c) Show that if $X$ is infinite, the claim (a) need not hold (i.e. a relation $\rel{R}$ may exist such that all the relations $\rel{R}^n$, $n \in \mathbb{N}$, are distinct).

My Attempt:

I had some trouble to define a finite set rigorously. Later, I decided not to struggle with the definition of the finite set. Instead, I chose to use a concept, Adjacency Matrix, to solve the problem.

First, I start with observing some examples.
Let $\rel{R} = \left\{ \op{x}{y} \mid x < y \right\} \subseteq A \times A$, where the set $A = \left\{ 1,\, 2,\, 3,\, 4,\, 5 \right\}$, while $\op{x}{y}$ is an ordered pair: $\op{x}{y} = \left\{ \left\{ x \right\},\, \left\{ x,\,y \right\} \right\}$. Now it is possible to make a graphic presentation of the relation $\rel{R}$ on the set $A$ as followed:

Furthurmore, the adjacency matrix of the relation $\rel{R}$ is:

\begin{align}
J_{\rel{R}^1} =
\begin{bmatrix}
0 & 1 & 1 & 1 & 1 \\
0 & 0 & 1 & 1 & 1 \\
0 & 0 & 0 & 1 & 1 \\
0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
\end{align}

Similarly, we can get the graphic presentation and the adjacency matrix of the composite relation $\rel{R}^{2} = \rel{R}\circ\rel{R}$.

\begin{align}
J_{\rel{R}^2} =
\begin{bmatrix}
0 & 0 & 1 & 1 & 1 \\
0 & 0 & 0 & 1 & 1 \\
0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
\end{align}

And so forth, the adjacency matrix of the composite relation $\rel{R}^{3}$, $\rel{R}^{4}$ and $\rel{R}^{5}$ are:

\begin{align}
J_{\rel{R}^3} =
\begin{bmatrix}
0 & 0 & 0 & 1 & 1 \\
0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
\end{align}

\begin{align}
J_{\rel{R}^4} =
\begin{bmatrix}
0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
\end{align}

\begin{align}
J_{\rel{R}^5} =
\begin{bmatrix}
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
\end{align}

It's obvious that $\forall i \in \mathbb{N},\, i > 5 :\: J_{\rel{R}^i} = J_{\rel{R}^5} = 0$, i.e. $\forall i \in \mathbb{N},\, i > 5 :\: \rel{R}^i = \rel{R}^5 = \varnothing$.

Meanwhile, let $\rel{S} = \left\{ \op{x}{y} \mid x = y \right\} \subseteq A \times A$. We can get $\rel{S} = \left\{ \op{1}{1},\, \op{2}{2},\, \op{3}{3},\, \op{4}{4},\, \op{5}{5} \right\}$. it's obvious that $\forall i \in \mathbb{N} :\: J_{\rel{S}^i} = I \;\land\; \rel{S}^{i} = \rel{S}$.

After the observation of some special cases as listed above, I started to prove the first sub-question.

Suppose $A$ is a finite set with n elements.

$$ A = \left\{a_1,\,a_2,\,\ldots{},\,a_n\right\} $$

Suppose $\rel{R}$ is a relation on $A$. It can be completely described by an $n \times n$ matrix $M = \left(m_{ij}\right)$, $i,\,j \in \left[1, n\right] \cap \mathbb{N}$, where
\begin{align}
m_{ij} = \left\{ \begin{array}{c}
  1,\ \text{if}\ \op{a_i}{a_j} \in R \\
  0,\ \text{if}\ \op{a_i}{a_j} \notin R
\end{array} \right.
\end{align}

I've got so far, but I cannot figure it out how to prove any of three sub-questions at all. Could you please help me? Thx in advance.

答:
Question a)
For $X$ is finite, the set of relations on $X$ is also finite, as a relation is a an element of the powerset $\mathcal P(X \times X)$. Therefore $\mathcal R^r = \mathcal R^s$ for some $r \lt s$.

Question b)
Take $X = \{1,2\}$ and $\mathcal R = \{(1,2), (2,1)\}$.

Question c)
Take $X = \mathbb N$ and $\mathcal R = \{(0,1), (1,2), (2,3), \dots\}$.


问:
Largest possible common difference of an arithmetic progression given its three terms (not necessarily adjacent)
How can I find the largest possible common difference of an arithmetic progression given that its three terms (not necessarily adjacent) are $0.37$, $9$ and $\frac{71}{7}$?

答: The common difference must divide the difference of any two terms of an AP. So it must divide $9-0.37=863/100$ and $71/7-9=8/7$. You want the largest common difference, i.e. the HCF of $8.63$ and $8/7$. Now use the following link: Rational Numbers - LCM and HCF.


问:
Continuity and differentiability of $h=\sum_{n=1}^{\infty} \frac{1}{x^2+n^2}$
I have solved the following exercise and I would like to know if I have made any mistakes:

Let $h(x)=\sum_{n=1}^{\infty}\frac{1}{x^2+n^2}$.

(a) Show that $h$ is a continuous function defined on all of $\mathbb{R}$;

(b) Is $h$ differentiable? If so, is the derivative function $h'$ continuous?

My solution:

(a) $|h_n(x)|=\frac{1}{x^2+n^2}\leq\frac{1}{n^2}$ for all $x\in\mathbb{R}$ and $\sum_{n=1}^{\infty}\frac{1}{n^2}<\infty$ so by Weierstrass M-Test $\sum_{n=1}^{\infty}\frac{1}{x^2+n^2}$ converges uniformly on $\mathbb{R}$ and since $h_n(x)$ is continuous for all $n\geq 1$, by Term-by-term Continuity Theorem the limit function $h$ is continuous too.

(b) for $x\in [-M,M]\ (M\in\mathbb{R}^+)$, $|h'_n(x)|=|-\frac{2x}{(x^2+n^2)^2}|=\frac{2}{(x^2+n^2)^2}|x|\leq \frac{2M}{n^4}$ and $\sum_{n=1}^{\infty}\frac{2M}{n^4}<\infty$ so by Weierstrass M-Test $\sum_{n=1}^{\infty}h'_n(x)$ converges uniformly on each interval $[-M,M]$ to a function $h$ and since $0$ belongs to every such interval and $h(0)=\sum_{n=1}^{\infty}\frac{1}{n^2}<\infty$ by Term-by-Term Differentiability Theorem $h$ is differentiable; $h'$ is also continuous by Term-by-Term Continuity Theorem since each $h'_n$ is continuous and the convergence of the series of derivatives is uniform.

答:
$$h(x)=\sum_{k=1}^{\infty} \frac{1}{k^2+x^2}=\frac{-1+ \pi x\coth^2 \pi x}{2x^2}$$
Use $\cot z \sim \frac{1}{z}+\frac{z}{3}-\frac{z^3}{45}$

So $$h(x)=\frac{\pi^2}{6}-\frac{\pi^4x}{90}+O(x^3)$$ is comtinuous and differentiable at $x=0$ so it is differentiable everywhere.


问:
What is the MLE $\theta^*$ of $\theta$?
I have that $x_1, x_2,...,x_n$ are from a rv $X$ that has the density function $f_X(x)=\frac{2x}{\theta^2} \quad$ for $0 \le x \le \theta  \quad$ and $f_X(x)=0 \quad$ otherwise. Ihave to determine the MLE of $\theta^*$ of $\theta$

Here is how I have done it:

$L(\theta)= \frac{2}{\theta^{2n}}\prod_{i=1}^nx_i$

$\frac{\partial L(\theta)}{\partial \theta} =...=\frac{-4n}{\theta^{2n+1}}\prod_{i=1}^nx_i + \frac{2}{\theta^{2n}}\frac{\partial(\prod_{i=1}^nx_i)}{\partial \theta}$

Is this correct? and also how do I calculate the CDF $F_{\theta^*}$, the pdf $f_{\theta^*}$ and the expectation $E[\theta^*]$ of the maximum likelihood estimator $\theta^*$?

答:
The likelihood you wrote is wrong. You did not consider that the support of $X$ depends on $\theta$

The correct likelihood is

$$L(\theta)\propto \frac{1}{\theta^{2n}}\cdot\mathbb{1}_{[x_{(n)};\infty)}(\theta)$$

now, looking at your L, it is evident that it is strictly decreasing, thus

$$\hat{\theta}_{ML}=X_{(n)}$$

where $X_{(n)}$ is the max of the observations

Distribution of Max

Set $Z=max(X_i)$

$$F_Z(z)=P(Z\leq z)=P(X_1\leq z,\dots X_n\leq z)=P(X_1\leq z)\dots P(X_n \leq z)=$$

$$=[P(X_1\leq z)]^ n$$

To find the distribution of $\hat{\theta}_{ML}=X_{(n)}$ consider that the CDF of Max is the product of the single CDF's thus

$$F_X(x)=\int_{-\infty}^x f(t) dt=\frac{1}{\theta^2}\int_0^x 2t dt=\frac{x^2}{\theta^2}$$

Thus, setting $Z=\hat{\theta}$ you get

$$F_Z(z)=\frac{z^{2n}}{\theta^{2n}}$$

derivating you get the density and then calculate the expectation with the usual definition

$$f_Z(z)=\frac{d}{dz}F_Z=\frac{2n z^{2n-1}}{\theta^{2n}}$$

$$E(Z)=\int_0^\theta z f_Z(z)dz$$

Title: What is the MLE $\theta^*$ of $\theta$?


问:
Prove that the following map is bounded from the given weighted $L^p$ space.
Show that the linear map  $T:L^1\left(\mathbb{R},(1+\lambda^2)^{-1}d\lambda\right)\to L^1\left(\mathbb{R},(1+\lambda^2)^{-2}d\lambda\right)$, given by  $T(f)(\lambda)=(1+\lambda^2)f(\lambda)-\int_{0}^{\lambda}xf(x)dx$, is bounded.

答:
Well, $T=T_1-T_2$, where $T_1f(\lambda)=(1+\lambda^2)f(\lambda)$. It's trivial that $T_1$ is bounded, and it seems like if $T_2$ is bounded this "must" be just Fubini, or more properly Tonelli:

Wlog $f\ge0$. Now $$\begin{align}\int_0^\infty(1+\lambda^2)^{-2} T_2f(\lambda)&=\int_0^\infty(1+\lambda^2)^{-2}
\int_0^\lambda xf(x)\,dxd\lambda
\\&=\int_0^\infty\int_x^\infty(1+\lambda^2)^{-2}xf(x)\,d\lambda dx
\\&\le\int_0^\infty\int_x^\infty(1+\lambda^2)^{-2}\lambda f(x)\,d\lambda dx
\\&=\frac12\int_0^\infty(1+x^2)^{-1}f(x)\,dx.\end{align}$$

Similarly for $\int_{-\infty}^0$; there  probably you want to start by rewriting $\int_0^\lambda=-\int_\lambda^0$.


问:
How to show $d(x,y)= \sqrt{d_1(x_1,y_1)^2+d_2(x_2,y_2)^2}$ is a metric?
$d_1(x_1,y_1)$ and  $d_2(x_2,y_2)$ are metric on $X$ and $d(x,y)$ is defined as:

$$d(x,y)= \sqrt{d_1(x_1,y_1)^2+d_2(x_2,y_2)^2}.$$
I am trying to show this is a metric. Can you give me some clue about proving the triangle inequality for $d$ ? Thank you for your help.

Body:
\begin{align*}
d(x,z)&=\left(\sum d_i(x_i,z_i)^2\right)^{1/2}\leq\left(\sum(d_i(x_i,y_i)+d_i(y_i,z_i))^2\right)^{1/2}\\
&=\left(\sum d_i(x_i,y_i)^2+2\sum d_i(x_i,y_i)d_i(y_i,z_i)+\sum d_i(y_i,z_i)^2\right)^{1/2}\\
&\leq\left[\sum d_i(x_i,y_i)^2+2\left(\sum d_i(x_i,y_i)^2\right)^{1/2}\left(\sum d_i(y_i,z_i)^2\right)^{1/2}+\sum d_i(y_i,z_i)^2\right]^{1/2}\\
&=\left\{\left[\left(\sum d_i(x_i,y_i)^2\right)^{1/2}+\left(\sum d_i(y_i,z_i)^2\right)^{1/2}\right]^2\right\}^{1/2}=\left(\sum d_i(x_i,y_i)^2\right)^{1/2}+\left(\sum d_i(y_i,z_i)^2\right)^{1/2}\\
&\leq d(x,y)+d(y,z)
\end{align*}


答:
What you wrote is correct but it does not show that $B:= \{ y\in\mathbb{R}^n : |y| < 1\}$ is the interior of $A$.

To prove the claim, first of all recall that the interior of $A$ is the set of all interior points of $A$, i.e. of all the points that have a open neighborhood contained in $A$. If $|z|=1-h<1$ then the ball with center $z$ and radius $h/2$ is contained in $A$, hence $z$ is an interior point. If $|z|=1$ then every open ball around $z$ will contain a point with norm greater than $1$, hence $z$ does not have a open neighborhood contained in $A$ and therefore is not an interior point. This shows that the set of interior points is exactly $B$.

Title: Interior of $A=\{x\in \Bbb{R^n} : \vert x \vert \leq 1\}$

Body:
What is the interior of $A=\{x\in \Bbb{R^n} : \vert x \vert \leq 1\}$? I claim it is the set of $\{y\in \Bbb{R^n}:\vert y\vert<1\}.$ This is because

$\exists h\in\Bbb{R}$ such that $\vert y \vert=1-h.$ Also, $\exists z\in \Bbb{R^n}$ s.t $\vert y-z\vert < h$. Therefore $\vert z-0\vert=\vert z \vert\leq\vert y-0\vert+\vert y-z\vert<1-h+h=1$. Therefore $z\in A$ and $y$ is an interior point of $A$. What's wrong with this proof? I don't believe it's correct.