问:
Find the generating function for $c_n = na_{n-1}$
Let $A(x)$ be the generating function of the sequence $(a_n)$. What is the generating function of the sequence:
$$
c_n = na_{n-1}
$$

In principle I know these rules:

$$
GF(a_{n-1}) = xA(x)\\
GF(na_n) = xA'(x)
$$

I am not sure how to combine these two though, my idea would have been to basically first compute the generating function for $a_{n-1}$ and then simply apply the second rule (i.e. differentiate the first line) to get something like this:

$$
C(x) = x*(xA(x))' = x(A(x) + xA'(x))
$$

Can someone tell me if that is a valid approach or otherwise point me in the right direction?

Body:
\begin{align}
\sum_{n \ge 1} n a_{n-1} x^n
&= \sum_{n \ge 1} (n-1) a_{n-1} x^n + \sum_{n \ge 1} a_{n-1} x^n \\
&= x \sum_{n \ge 1} (n-1) a_{n-1} x^{n-1} + x \sum_{n \ge 1} a_{n-1} x^{n-1} \\
&= x \sum_{n \ge 0} n a_n x^n + x \sum_{n \ge 0} a_n x^n \\
&= x^2 A'(x) + x A(x)
\end{align}


问:
If $a \in \Bbb R$ and $W=\{p \in P_n \mid p(a)=0\}$, show that $W$ is a subspace of $P_n$.
Let $P_n$ be a vector space of all polynomials with real coefficients and of degree $n$ or less. If $a \in \Bbb R$ and $W=\{p \in P_n \mid p(a)=0\}$, show that $W$ is a subspace of $P_n$.

Edit:

Attempt:
Denote $0_p$ be the zero polynomial.

Since $0_p(a)=0 \in W$, then
$0_p \in W$ and thus, $W \ne \emptyset$.

Let $p,q \in W$. Then,
$p(a)=0=q(a)$ for some $a \in \Bbb R$.
Now, $(p+q)(a) = p(a)+q(a)=0+0=0$. Hence, $p+q \in W$.

Let $p \in W$. Let $c$ be an arbitrary scalar. Then, $p(a)=0$ for some $a \in \Bbb R$. Now, $(cp)(a)=cp(a)=c0=0$.
Hence, $cp \in W$.

Thus, $W$ is a subspace of $P_n$.

Am I true?

答: This is almost right. The train of thought and the important algebraic steps are right. But throughout your argument you refer to "some $a$". That's not  correct. The $a$ is fixed from the start. The space you are interested in might be called $W_a$. It's a different space for each fixed $a$.

ContentLicense: CC BY-SA 4.0


问:
Finding squares that add to a certain sum
I have been looking for an algebraic way to solve for 2 squares when given a sum, but I got nothing so far.

For example:

$$y^2 + x^2 = 9797$$

I thought that the solution would have to do something like this, but I want a method which is certain to give me a solution without trial and error.

$$f(x) = \sqrt{(9797 - x^2)}$$

or this where I would just increase the variable a by one until x^2 is a perfect square.

$$9797 - y^2 = x^2$$

I came across the sum of two squares theorem, as well Fermat's theorem on sums of two squares. However, they state that a number can be written as $x^2 + y^2 = z$ but from my research, I believe they did not state a solution to solve the two squares.

Edit: I wanted to add that I know there can be multiple solutions to problem, as the example I gave has $4$ solutions.

答:
$$x^2+y^2=9797\implies y=\sqrt{9797-x^2}\\
\implies\bigg\lceil\sqrt{9797-\big(\big\lfloor\sqrt{9797}\;\big\rfloor\big)^2}\space \bigg\rceil=14\le
 x\le \big\lfloor\sqrt{9797}\;\big\rfloor=98$$
This shows a finite search and any $\space x \space $ that yields an integer $\space y\space $ is a solution.
Since $\space x,y\space $ are interchangeable, this means
$\qquad|x|,|y|\in\big\{31,49,86,94\big\}\qquad$ with the valid solutions being combinations of the pairs $\space 31,94\space $ and $\space 49,86\space $ for a total of $\space  16\space $ solutions.

Note: The reason that the ends and middles of the set are paired is because the farthest and nearest members must be paired in the same way that $\space 100,1\space $ and $\space 51,50\space $ would be paired to sum integers. If paired otherwise, their squares summed would be smaller or larger than $\space 9797.\space $


问:
Question about Conditional Expectation from One Variable
I'm having a bit of trouble understanding how to solve this question in my textbook.

It says to allow X to have the following distribution:

Suppose you are then told that $X \geq 2$. Record a prediction of a future value of X
that uses this information.

I know that I need the expectation, but I was struggling a bit to find $\text{E}(X|X \geq 2)$. $$\text{E}(X|X \geq 2) = \sum_x{x\Pr(X=x|X \geq 2)}$$ and $$\Pr(X=x|X \geq 2) = \frac{\Pr(X=x \cap X\geq2)}{\Pr(X \geq 2)} \\ \Pr(X=x \cap X\geq2) = \frac{1}{2} \\ \Pr(X \geq 2) = \frac{1}{2}$$ so $$\text{E}(X|X \geq 2)= 8$$ which doesn't seem right at all.

edit: I'm really not sure how I got 8 lol!

答:
The probability that $X=x$ and $X\geq 2$ is not a flat $1/2$.   It depends on what value of $x$ we have.

Clearly $\mathsf P(X=1\cap X\geq 2)=0$ but for all $x\geq 2$ there $\mathsf P(X=x\cap X\geq 2)=\mathsf P(X=x)$

$$\begin{array}{|c|c|}\hline x & 1 & 2 & 3 & 4 \\\hline\mathsf P(X=x\cap X\geq 2) & 0 & 1/4 & 1/8 & 1/8 \\\hline \end{array}$$

$$\mathsf E(X\mid X\geq 2) =\dfrac{2\cdot 1/4+3\cdot 1/8+4\cdot 1/8}{1/2}=\dfrac{11}{4}$$


问:
How can I find the value of x given multiple Parallel Lines and multiple Transversals?
Ive been given the following problem:

I was unable to figure out how to solve the problem, and I was told the answer was the following;
However, the program has failed to explain to me exactly how it got the answer... Looking online, I can only find solutions for single transversal, 2 line problems, which aren't necessarily difficult..

My question is, how exactly, piece by piece, do I find X given multiple transversals or parallel lines?

答: Just reduce the problem to the problem with two lines by passing a line parallel to one of your lines - for example (sorry for sloppy drawing):


问:
Probability question involving colored balls
Probability question.  I’ve been trying to figure this out for a while and need a little help.  This is not a homework problem just something I made up but can’t figure it out on my own.

I have a total of 30 balls(10 red, 10 green, 10 blue) I dump all 30 balls into a large bin.

I have 4 smaller containers.  I randomly take 2 balls out and put them in each of the 4 smaller containers (I remove 8 balls total).  What are my chances of having 2 bins or more of just red balls?  Can you please show the math behind the answer.

答:
Let $E$ be the event of interest ( at least 2 bins have 2 red balls). Let $R$ be the amount of red balls extracted (out of 8). Then

$$ P(E) = \sum_{R=0}^8 P(E \mid R) P(R)$$

Now,

$$P(R)= \frac{\binom{10}{R} \binom{20}{8-R}}{\binom{30}{8}}$$

And $P(E \mid R)=0$ for $R\le 3$ and $P(E \mid R)=1$ for $R\ge 6$.

You need only to compute $P(E \mid R=4)$ and $(E \mid R=5)$... can you go on from here?


问:
Percentage question
Certain alloy contains 20% of copper and 5% of tin. How much of copper and how much tin must be added to 100kg of this alloy to get another with 30% of copper and 10% of tin?

I can't figure how to start it

答:
Given your problem statement, you have the following quantities: 20kg of copper, 5kg of tin
and 75kg of the base metal, 100kg total.

You want to add an unknown amount $x$ to the copper such that the new percentage of copper in the alloy is 30%, thus  $(20 + x) / (100 + x + y) = 0.3$. Likewise, you are looking for an amount $y$ to add to the amount of tin such that the new percent is 10%, thus $(5 + y) / (100 + x + y) = 0.1$. Likewise, you known that the base percentage is now going to be 60%, thus $75 / (100 + x + y) = 0.6$.

With some algebra you get:

$$20 + x = 0.3 (100 + x + y)$$
$$20 + x = 30 + 0.3x + 0.3y$$
$$-10 = -0.7x + 0.3y$$

$$5 + y = 0.1 (100 + x + y)$$
$$5 + y = 10 + 0.1x + 0.1y$$
$$-5 = 0.1x - 0.9y$$

Which will give you the following system:

$$\pmatrix{ -10 \\ -5 } = \pmatrix{ -0.7 & 0.3 \\ 0.1 & -0.9} \pmatrix { x \\ y }$$

Solving it will yield:

$$\pmatrix{x \\ y } = \pmatrix{ 17.5 \\ 7.5 }$$

Thus, you would need to add 17.5kg of copper and 7.5kg of tin to get a new alloy with 30% copper, 10% tin and 60% base.

ContentLicense: CC BY-SA 3.0

Title: Percentage question


问:
Probability: determine if compliment, mutually exclusive, and independent
A certain illness has two symptoms associated with it - a fever and fatigue. There is a 90% probability that at least one of the two symptoms occurs for a randomly selected person with the illness. There is an 80% probability that a randomly selected person with the illness will come down with a fever and there is a 50% probability that a randomly selected person with the illness will feel fatigued.

Note: For simplicity, we will refer to the event of "coming down with a fever" simply as "fever" and the event of "feeling fatigued" simply as "fatigue". You may do the same.

Please answer the following three questions:

Are the events of "fever" and "fatigue" complementary?

Are the events of "fever" and "fatigue" mutually exclusive (or disjoint)?

Are the events of "fever" and "fatigue" independent?

My answers:

P(Fever) = 0.9 * 0.8 P(Fatigue) = 0.9 * 0.5 (I am thinking the answer is no, but don't know why)

I think this answer is no because we are told there is a 90% chance that "at least one" of the two symptoms are shown, so there could be both. But I don't know how to show this using math.

Yes they are independent

I am not sure how to approach this problem because the two probabilities given add up to more than 100%. Would I have to multiply 0.9 * 0.8 to figure out the probability of just getting a fever, and likewise for feeling fatigued? I need help showing the math behind these answers if I am even right.

答:
It's important to know what the different terms complement, mutually exclusive and independent mean mathematically.

Two events $A, B$ are complementary if $P(A) +P(B)=1$.

Two events are mutually exclusive if they cannot both happen together. In other words, $P(A\cap B)=0$

Two events are independent if they don't affect each other. This means that $P(A\cap B) =P(A)P(B)$

Lets denote event $A$ as fever and $B$ as fatigue.

The question gives you: $$P(A\cup B) =0.9$$$$ P(A) =0.8$$$$P(B) =0.5$$

Clearly they are not complementary as $P(A)+P(B)=1.3>1$. Can you find $P(A\cap B)$ to determine if they are mutually exclusive, independent, or neither?


问:
Convergence of a filter defined in a subspace
I am reading about convergence of filters in topological spaces and cluster points. My doubt is the following: let $X$ be a topological space, $A\subset X$ and $\mathcal F$ a filter on $A$. Given $x \in \overline A$, what is the meaning of $\mathcal F \longrightarrow x$?

Bourbaki defines the convergence of a filter (page 68 - topology 1) which is defined in the whole space $X$. Of course that is not the case of $\mathcal F$ if $A\ne X$. But in Proposition 6 of page 70 they use this kind of convergence. I am assuming that this means that  $\mathcal F$ is finer than $\mathcal B(x)_A$, which is the trace of the filter of the neighborhoods of $x$. Is that correct? I found it very strange since Bourbaki's book (as far as I know) does not make this kind of imprecision (to work with a not given definition).

In Dugundji's book only the definition of convergence of a filterbase is given (Def. 2.1 of Chapter 10). So his Theorem 4.1 (that is the cited proposition of Bourbaki's book) is coherent.

答:
I agree with you that a filter on $A$ is a filter of subsets of $A$. However, if we denote by $\Bbb F(Y)$ the set of filters on a set $Y$, then $\Bbb F(A)=\{\mathscr{F}\cap\wp(A):A\in\mathscr{F}\in\Bbb F(X)\}$, so there is a natural correspondence between filters on $A$ and filters on $X$ that contain $A$.

But you don’t actually need to make this explicit for Prop. $6$. The first part of the proposition is clear: if $\mathfrak{B}$ is a filter base on $A$, then it is automatically a filter base on $X$ for a filter that contains $A$, so by the remark following Definition $2$ every cluster point of $\mathfrak{B}$ is in $\overline{A}$. Conversely, if $x\in\overline{A}$, then every nbhd of $x$ in $X$ meets $A$, so the trace on $A$ of the nbhd filter of $x$ in $X$ is a filter on $A$ and therefore a filter base on $X$; clearly every nbhd of $x$ in $X$ contains a member of this filter base, so by Prop. $1$ of $7.1$ this filter base converges to $x$ in $X$.


问:
What is the limit $x\to\infty$?
I'm working on some exercises about bases of filters and in one of them they want me to compare the bases $x\to+\infty, x\to-\infty$ and $x\to\infty$. I know that I could take the base $\{(x,+\infty):x\in\mathbb{R}\}$ for $x\to+\infty$ and in an analogous manner for $x\to-\infty$. My question is what means $x\to\infty$?

答: Base(bases of filter) for $x \to \infty$ is $U(\infty) = \{ x \in \mathbb{R}\colon \delta \lt |x| \}$ where $\delta \in \mathbb{R}$ - Vladimir A. Zorich - Mathematical Analysis I-Springer (2016), page 127.

Title: What is the limit $x\to\infty$?


问:
Given the sequence $(x_n)$ is unbounded, show that there exist a subsequence $(x_{n_k})$ such that $\lim(1/x_n)=0$.

Given the sequence $(x_n)$ is unbounded, show that there exist a subsequence $(x_{n_k})$ such that $\lim(1/x_{n_k})=0$.

I guess I have to prove that $(x_{n_k})$ diverge, but I don't know how to carry on.

Thanks.

答: Hint: If $(x_n)$ is unbounded, then for each $N$ there are infinitely many $n$ with $|x_n| \ge N$. Now choose inductively $n_k$ such that $n_1 < n_2 < \cdots$ and $|x_{n_k}| > k$.


问:
express prime as sum of squares, $p = a^2 + b^2$
Espress $2017$ as sum of two squares.

attempt: by Fermat's Theorem on sums of squares, the prime $p = 2017$ is the sum of two squares $2017 = a^2 + b^2$ , $a,b \in \mathbb{Z}$, if and only if $p \equiv 1 mod 4$.

And The irreducible elements in the Gaussian integers $\mathbb{Z[i]}$ are as follows
$(a + bi)(a - bi) $ for primes $p\in \mathbb{Z}$ with $p \equiv 1 mod 4$ (both of which have norm $p$).

Then since $2017 \equiv 1 (mod 4)$ Then $2017 = a^2 + b^2$ .

Notice that $\sqrt2017 $ is approximately $44.91$. So $a^2, b^2 $
will be between values $1,2^2,....,44^2$ .

Then plugging different values from the above squares in  $2017 - a^2 = b^2$

we find $2017 - 44^2 = 81 = 9^2$

So $2017 = 44^2 + 9^2$.

However, I found them using that approach.  But is there a way to find them without doing this approach?

I dont' know how to use $p = a^2 + b^2 = (a + bi)(a - bi) $ for primes $p\in \mathbb{Z}$ with $p \equiv 1 mod 4$ (both of which have norm $p$).

So $2017 = a^2 + b^2 = (a+ bi)(a - bi) $. I don't' know how I would proceed assuming I would not have found the values .
Any feedback or better approach would be appreciated it.
Thank you!

答:
It would be easy if $\space 2017\space $ were a perfect square but there are no better approaches that i know of except for limiting the search.
$$x^2+y^2=2017\implies y=\sqrt{2017-x^2}\\
\implies\bigg\lceil\sqrt{2017-\big(\big\lfloor\sqrt{2017}\;\big\rfloor\big)^2}\space \bigg\rceil=9\le
 x\le \big\lfloor\sqrt{2017}\;\big\rfloor=44$$
This shows a finite search and any $\space x \space $ that yields an integer $\space y\space $ is a solution.

Of these $\space 36\space $ values to test, the only $y$-integers found are for $\space x=9\space$ and $\space x=44.\quad$ Since $\space x,y\space $ are interchangeable, this means
$\quad|x|,|y|\in\big\{9,44\big\}\quad$ with the valid solutions being all combinations for a total of $\space  8\space $ solutions.


问:
Localization of a principal ring is principal?
I'm trying to prove that the localization $S^{-1}A$ is a principal ring. Here, $A$ is a principal ring and $S$ a multiplicative subset of $A$. Furthermore we assume $0 \notin S$.

The term principal ring is used in Lang's algebra, referring to a nontrivial commutative ring whose all ideals are principal. It allows zero divisors, which is the only difference form PID.

Anyway I start with an ideal $\mathfrak a'$ in $S^{-1}A$. Let $f:A \to S^{-1}A$ be the canonical map. Then $f^{-1}(\mathfrak a')$ is a principal ideal, say $aA$, in $A$.

Now $\mathfrak a'$ seems to be the principal ideal $(a/1)(S^{-1}A)$. Showing that this ideal is contained in $\mathfrak a'$ is easy, but I got stuck to show the converse.

Any braekthrough?

答:
Take an ideal $\mathfrak a\leqslant S^{-1}A$ and consider the underlying ideal in $A$, i.e. $\mathfrak a\cap A$. By assumption, this ideal is principal and hence $\mathfrak a\cap A=aA$ for some $a\in A$. Let $x\in\mathfrak a$. Then there is some $s\in S$ such that $sx\in A$ by construction and hence $sx\in \mathfrak a\cap A=aA$. So $x\in a(S^{-1}A)$ lifting the principal ideal to the localization. This shows at once that $\mathfrak a$ is principal with generator $a$.

The map $\mathfrak a\mapsto\mathfrak a\cap A$ is a left inverse of your canonical map. One can show that composing these the other way around is the identity on prime ideals disjoint from $S$, which might be an interesting fact. See Milne's Algebraic Number Theory $§1$ for the details.


问:
Show that a $2$-connected graph with minimum degree greater than or equal to $3$, has a vertex $v$ such that $G-v$ is $2$-connected
Assume that $G$ is a $2$-connected graph with $δ(G)≥3$. Show that there is a vertex $v\in G$ such that $G−v$ is $2$-connected.

This question is a duplicate, I'm aware. However as the author of that other question said when he posted an answer, his proof is not based on well known theorem from textbooks. I'd appreciate if someone could provide a proof based on such theorems, or maybe point me to the right direction on how I could prove this myself.

答:
All we need is the result that a graph is $2$-connected if and only if it has an (open/proper) ear decomposition. This is a way of building a graph, starting with a cycle. We pick two distinct vertices in the graph built so far, and add a path between them, of any length. If the path has any internal vertices, then those are completely new to the graph.

A graph can have many ear decompositions. Most of those will not help. We'd like to find one which ends by adding a length-$2$ ear (with one extra vertex $v$) and then some length-$1$ ears (edges) with one endpoint at $v$. If we find such an ear decomposition, we can delete $v$. $G-v$ is $2$-connected: its ear decomposition is just the ear decomposition of $G$, stopped before we add $v$.

To find such an ear decomposition, out of all the ear decompositions $G$ might have, pick one such that:

The last ear added which has length $>1$, call it $E^*$, is as short as possible;

Subject to 1, the number of ears of length $1$ added after $E^*$ is as small as possible.

I claim that this ear decomposition has the property we want.

Suppose that $E^*$ has more than one vertex: it is a path $v_0, v_1, \dots, v_k$ where $v_1, \dots, v_{k-1}$ are new. Right after we added $E^*$, the vertices $v_1, \dots, v_{k-1}$ have degree $2$, and $G$ has minimum degree $3$, so those vertices are going to get extra edges added to them later. There's two cases:

We add an edge $v_i w$, where $w$ is not on $E^*$. Then we could have added the vertices of $E^*$ in two adding-an-ear steps. We could have added the ear $v_0, v_1, \dots, v_i, w$ and then $v_i, v_{i+1}, \dots, v_k$, which is a better ear decomposition by our measure unless $i=k-1$ and that last ear has no internal vertices. In that case, we could also have added the ear $w, v_i = v_{k-1}, v_k$ and then the ear $v_0, v_1, \dots, v_{k-1}$, which is a better ear decomposition by our measure.

We add an edge $v_i v_j$, where $j > i+1$. Then instead of adding ear $E^*$, we could have added the ear $v_0, v_1, \dots, v_i, v_j, \dots, v_k$ followed by the ear $v_i, v_{i+1}, \dots, v_j$. This is a better ear decomposition by our measure.

Either way, we contradict the extremal property of the ear decomposition we chose. So $E^*$ just has one vertex $v$.

After we add $E^*$, all ears of length $1$ we add must be incident to $v$. Otherwise, we could have added them earlier, and had a better ear decomposition. Therefore, $G-v$ has an ear decomposition: the ear decomposition we had before we added the ear $E^*$. Therefore, $G-v$ is $2$-connected.


问:
Establish rules
I think the above constraint when the absolute value of the HW for 2 strings is less than the HD between those strings. Any thoughts on the proof, please?

Also, how do I make it a constraint if it's provable? It's not necessary that the explanation has to be in terms of HD and/or HW.

答:
Let $x_i$ be an indicator variable that can only take the value $1$ if bit $i$ flips from $0$ to $1$. You can model that with the constraints $x_i \leq b_i$ and $x_i \leq 1 - a_i$. To ensure at least one flip occurs, you add $\sum_i x_i \geq 1$.

You can do the same with a variable $y_i$ that can only take the value $1$ if bit $i$ flips from $1$ to $0$.

Title: Establish rules


问:
For number fields $L/K$ the ring of integers $O_L$ is finite over $O_K$ as an immediate consequence of a Lemma from Neukirch
Let $K \subset L$ a finite extension of number fields. Let $O_K$ and $O_L$ be
the rings of integers of $K$ and $L$. The goal of this question was
to elaborate that $O_L$ is finite $O_K$-module. Although the question is
answered, Jeroen remarked that this claim could alternatively proved using
following Lemma from Neukirch's Algebraic Number Theory:

Let be a basis of finite extension $L/K$. Multiplying them by suitable
elements from $O_K$ ('clearing the denominators') we can assume that all
$\alpha_i$ lie in $O_L$. Then

Lemma (cf. Neukirch, I$.2.9$). Let $d \in K$ be the discriminant of
$\alpha_1,\ldots,\alpha_n$. Then
$d\mathcal{O}_L \subseteq \mathcal{O}_K \alpha_1 + \cdots + \mathcal{O}_K \alpha_n$.

Then we have a chain of inclusions
$$d\mathcal{O}_L \subseteq \mathcal{O}_K \alpha_1 + \cdots + \mathcal{O}_K \alpha_n
\subseteq \mathcal{O}_L.$$

But I not know how from this we can conclude
$O_L = \mathcal{O}_K \alpha_1 + \cdots + \mathcal{O}_K \alpha_n$. Equivalently
how can I show that for any $a \in O_L$ which by the Lemma can be written as
$a=\frac{ad}{d}=  \sum \frac{c_i}{d} \alpha_i$ we can conclude that
$c_i/d \in O_K$.

答:
Note that in this case

$$\mathcal O_L\subseteq d^{-1}\alpha_1\mathcal O_K+\cdots+d^{-1}\alpha_n\mathcal O_K$$

and the latter is a free $\mathcal O_K$-module and in particular finitely generated. We may consider the special case $K=\Bbb Q$ and $\mathcal O_K=\Bbb Z$. Then, $\mathcal O_L$ is as a submodule of a finitely generated module over a Noetherian ring is itself finitely generated. This by passes the usual trace argument in so far as this argument is hidden in the lemma. Using this on $\mathcal O_K\subseteq\mathcal O_L$ for a finite extension of number fields $L/K$ then gives your result.

However, I am not completely sure if the $\alpha_i$ do form an integral basis necessarily. The argument I presented only proves that $\mathcal O_L$ is itself finitely generated and hence finite over $\mathcal O_K$ containing the latter (which is the question asked in the title and what the linked answer is concerned with; please correct me if your questions was something else).


答:
Hopefully I didn't make any mistake.

Note that the squares modulo $9$ are $0,1,4,7$.

Now
$$2017 = a^2 + b^2 \Rightarrow \\
1 \equiv a^2+b^2 \pmod{9}$$

This gives that, WLOG we have
$$
a^2 = 0 \pmod{9} \\
b^2= 1 \pmod{9}
$$
and hence $a=3k$ and $b=9l \pm 1$.

Then,
$$
2017 = 9k^2+ 81l^2 \pm 18l +1 \Rightarrow \\
224 = k^2+ 9l^2 \pm 2l
$$
Finally
$$
224 \geq 9l^2 \pm 2l  \geq 9l^2-2l \leq 7l^2 \Rightarrow \\
l \leq 5
$$

A case by case analysis is now very fast.

Title: express prime as sum of squares, $p = a^2 + b^2$

Body:
Espress $2017$ as sum of two squares.

attempt: by Fermat's Theorem on sums of squares, the prime $p = 2017$ is the sum of two squares $2017 = a^2 + b^2$ , $a,b \in \mathbb{Z}$, if and only if $p \equiv 1 mod 4$.

And The irreducible elements in the Gaussian integers $\mathbb{Z[i]}$ are as follows
$(a + bi)(a - bi) $ for primes $p\in \mathbb{Z}$ with $p \equiv 1 mod 4$ (both of which have norm $p$).

Then since $2017 \equiv 1 (mod 4)$ Then $2017 = a^2 + b^2$ .

Notice that $\sqrt2017 $ is approximately $44.91$. So $a^2, b^2 $
will be between values $1,2^2,....,44^2$ .

Then plugging different values from the above squares in  $2017 - a^2 = b^2$

we find $2017 - 44^2 = 81 = 9^2$

So $2017 = 44^2 + 9^2$.

However, I found them using that approach.  But is there a way to find them without doing this approach?

I dont' know how to use $p = a^2 + b^2 = (a + bi)(a - bi) $ for primes $p\in \mathbb{Z}$ with $p \equiv 1 mod 4$ (both of which have norm $p$).

So $2017 = a^2 + b^2 = (a+ bi)(a - bi) $. I don't' know how I would proceed assuming I would not have found the values .
Any feedback or better approach would be appreciated it.
Thank you!


问:
Why does doubling a covector make the "ruler markings" closer together?
I was watching this video about how to visualize covectors as "ruler markings" or "topographic maps" with which to measure vectors. If $\alpha$ is a covector and $v$ is a vector, my confusion starts when we ask what $(2\alpha)(v)$ is and how to visualize it. I understand that $(2\alpha)(v) = 2(\alpha(v))$ and that the ruler markings get denser (so that the vector $v$ pierces more lines). But we could also have defined things so that $(2\alpha)(v) = \frac12(\alpha(v))$ and where the ruler markings get farther apart.

Here's another way I am thinking about this. We have objects in the real world whose lengths we can measure, and we have units with which to measure these objects. If "mouse" is an object and "cm" is a unit, we can write (mouse,cm) = 8 to mean that the mouse has length 8 cm. Now we can have different units, and in particular we want to do some arithmetic with units and define e.g. what "2cm" means as a unit. We could write "2cm" to mean "measure the object using units twice as long as cm" so that (mouse,2cm) = 4, or we could write "2cm" to mean "twice the number you get if you measure the object using cm" so that (mouse,2cm) = 16. As far as I know, this is just a convention. The convention has the nice property that $(\cdot,\cdot)$ is linear in each argument. (I am using this awkward notation because the units here are like inner products and we can think of this as asking whether $\langle v,2u\rangle$ should equal $2\langle v,u\rangle$ or $\frac12\langle v,u\rangle$.)

So is there a deeper reason (than just "inner products/covectors should be linear") for defining things this way?

答:
I think that by using an inner product, you have introduced too many rulers to the party, since an inner product already defines its own notion of length: $\lVert v \rVert := \sqrt{(v, v)}$. Furthermore, there is no reason why $(\text{mouse}, \text{cm})$ is the correct way of measuring the length of a mouse in centimetres (and in fact, it is not).

Let's say I have a one-dimensional vector space $L$. I have a yardstick $y \in L$ which I wish to use to measure a mouse $m \in L$. The length of the mouse $\ell \in \mathbb{R}$ in the solution to the equation $\ell y = m$. You can see here clearly why doubling my yardstick halves the length, since if $\ell y = m$ then $\frac{\ell}{2} (2y) = m$ also. You can also see that $\ell$ has a linear relationship to $m$, for example $(2 \ell) y = 2m$. Measuring is linear, but changing the side of the yardstick is not linear (it is the reciprocal).

Getting more abstract, to any nonzero yardstick $y \in L$ we can define the measure-with-respect-to-$y$ functional $\ell_y \colon L \to \mathbb{R}$, which is the unique functional such that $\ell_y(m) y = m$ for all $m \in L$. (A simpler definition is the unique functional such that $\ell_y(y) = 1$). The question now is: how is $\ell_{2y}$ related to $\ell_y$? We can find that
$$ 1 = \ell_{2y}(2y) = (2 \ell_{2y})(y)$$
which shows that $2 \ell_{2y} = \ell_y$, or that $\ell_{2y} = \ell_y / 2$. So the assignment $y \mapsto \ell_y$ is nonlinear. This is important: the functional $2 \ell_y$ corresponds to the yardstick $y/2$, not to $2y$.

Finally, what happens if $L$ already comes with an inner product? There must exist a vector $z_y \in L$ such that $\ell_y(m) = (m, z_y)$ for all $m \in L$. It is not hard to see that $z_y = \frac{y}{(y, y)}$ works, since then we have $(y, z_y) = 1$ as required. Again, we can see that $z_{2y} = z_y / 2$. You might also recognise $(m, z_y)$ as the formula for the orthogonal projection of $m$ onto $y$.

Long answer short the length of a mouse in centimetres is $\frac{(\text{mouse}, \text{cm})}{(\text{cm}, \text{cm})}$, not $(\text{mouse}, \text{cm})$, and converting yardsticks to rulers works like reciprocals, not like a linear map.


问:
Noncompact surface with finite fundamental group
Let $M$ be a noncompact surface with finite fundamental group. Can we prove that $M$ must be homeomorphic to $\mathbb R^2$?

答:
Let $M$ be a non-compact manifold with finite fundamental group. The universal cover of $M$ is either $S^2$ or $\mathbb{R}^2$, but $M$ is non-compact, so it must be the latter; in particular, $M$ is aspherical. As the fundamental group of an aspherical manifold is torsion-free (see below), we see that $M$ is simply connected and hence homeomorphic to its universal cover, namely $\mathbb{R}^2$.

Proposition: Let $X$ be a finite-dimensional aspherical CW complex. Then $\pi_1(X)$ is torsion-free.

Proof: If $\alpha$ is a torsion element of $\pi_1(X)$, then $\langle \alpha\rangle \cong \mathbb{Z}_k$ for some $k$; let $X' \to X$ be the corresponding covering. Then $X'$ is a finite-dimensional aspherical CW complex with $\pi_1(X') \cong \mathbb{Z}_k$, so $X'$ is a $K(\mathbb{Z}_k, 1)$. As Eilenberg-MacLane spaces are unique up to homotopy, we see that $X'$ is homotopy equivalent to the infinite-dimensional lens space $S^{\infty}/\mathbb{Z}_k$. Therefore $H^n(X'; \mathbb{Z}_k) \cong H^n(S^{\infty}/\mathbb{Z}_k; \mathbb{Z}_k) \cong \mathbb{Z}_k$ for every $n$, but this is impossible as $X'$ is finite-dimensional.


问:
Limit of a sequence defined by a recurrence
Let $x_1$ and $x_2$ be positive real numbers and define, for $n>2$:$$\displaystyle x_{n+1}=\sum_{k=1}^{n}\sqrt[n]{x_k}$$

Evaluate $\displaystyle \lim_{n\to\infty}\frac{x_n-n}{\ln n}$

答:
I am absolutely not sure the answer I will give is correct, so please pay attention to details!

Let us assume $x_1, x_2 >1$, now $$\lim_{n\to\infty}\frac{x_n-n}{\ln n}=\lim_{n\to\infty}\frac{\sum_{k=1}^{n}\sqrt[n]{x_k}-1}{\ln n}$$Now we can observe that the letter series diverges by ratio test with the harmonic series. In particular we have that $\forall \epsilon ,\exists \nu_\epsilon: \forall n> \nu_\epsilon, (1+\epsilon)\frac{1}{k}>\sqrt[n]{x_k}-1>(1-\epsilon)\frac{1}{k}.$
Let us consider $c=\sum^{\nu_\epsilon}\sqrt[n]{x_k}-1$, now we know that: $$ \frac{c+(1-\epsilon)\sum_{\nu_\epsilon}\frac{1}{k}}{\ln(n)} \leq \frac{\sum_{k=1}^{n}\sqrt[n]{x_k}-1}{\ln (n)} \leq \frac{c+(1+\epsilon)\sum_{\nu_\epsilon}\frac{1}{k}}{\ln(n)} $$ Now we use that $\sum_0^n \frac{1}{k} - ln(n+1)\rightarrow \gamma$, the Euler-Mascheroni constant, which means that $\sum_{\nu_\epsilon}^n \frac{1}{k} - ln(n+1)\rightarrow c'=\gamma-\sum_0^{\nu_\epsilon} \frac{1}{k}$, so we get: $$\frac{c+(1-\epsilon) \left( \sum_{\nu_\epsilon}\frac{1}{k}-\ln(n+1) +\ln(n+1)\right)}{\ln(n)}\leq \frac{\sum_{k=1}^{n}\sqrt[n]{x_k}-1}{\ln (n)} \leq \frac{c +(1+\epsilon) \left( \sum_{\nu_\epsilon}\frac{1}{k}-\ln(n+1) +\ln(n+1)\right)}{\ln(n)},$$ which in terms of limits means: $$1-\epsilon \leq \lim_{n\to\infty}\frac{x_n-n}{\ln n} \leq 1+\epsilon,$$ $\forall \epsilon >0$. For $x_1, x_2 <1$ there should be only some sign swaps when doing the ratio test.


问:
Solving functional equation $f(x)+f(1/x)=f(x)\cdot f(1/x)$
Let a function $f(x)$, and $x$ not equal to zero be such that:
$f(x)+f(1/x)=f(x)\cdot f(1/x)$ then $f(x)$ is ?

I tried differentiating it but did not find any useful outcome.
answer given at back of book is

A)$1+x^n$, $n$ belongs to $\mathbb R$

B)$1-x^n$, $n$ belongs to $\mathbb R$

C)$\pi/(2\arctan|x|)$

D)$2/(1+k\ln|x|) $

答:
The given functional equation factored as $$(f(x)-1)(f(1/x)-1)=1.$$ Now if we let $f(x)=g(x)+1$ for some other function $g,$ then the only constrain on $g$ is $$g(x)g(1/x)=1.$$ Note that $g$ can never be zero and $(g(\pm1))^2=1.$ Also, if we know $g$ on $[-1,0)\cup(0,1],$ then this last condition determine it everywhere on on $\mathbb{R}_{\neq0}.$ Hence the general solution to this functional equation is
$$f(x)=\left\{ \begin{array}{ll} 1+g(x), &\text{if }0<|x|\lt 1\,,\\
0\text{ or }2, &\text{if }|x|=1\,,\\
1+\dfrac{1}{g\left(1/x\right)},&\text{if }|x|>1\,. \end{array} \right.$$ as mentioned in a comment. However this doesn't mean that $f$ is always piecewise or non-smooth. The condition $g(x)g(1/x)=1$ on $g$ is equivalent to say that the composition $$h(x)=\ln |g(\exp(x))|$$ is an odd-function. So, always we can take any odd function $h$ and create a nice solution to this functional equation via $f(x)=\pm\exp( h(\ln |x|))+1$ on $\mathbb{R}_{\neq0}.$ For some examples:

\begin{array}{|c|c|}
h(x) & f(x) \\
\hline
nx & \pm x^n+1 \\
n\ln\left(\dfrac{\cot^{-1}(e^x)}{\tan^{-1}(e^x)}\right) & \left(\dfrac{\pi}{2\tan^{-1}(x)}-1\right)^n+1\\
-2\tanh^{-1}(nx) & \dfrac{2}{1+n\ln |x|}, \dfrac{2n\ln |x|}{1+n\ln |x|}\\
n\sinh^{-1}(x) & \pm(\ln|x|+\sqrt{(\ln |x|)^2+1})^n+1\\
\tanh (nx) & \exp\left(\dfrac{x^{2n}-1}{x^{2n}+1}\right)+1\\
\sinh (nx) & \exp\left(\dfrac{x^{2n}-1}{2x^n}\right)+1\\
\end{array}

for any $n\in\mathbb{R}.$


问:
Proof a subset is valid
No answers, please, hints only.

Given that $A=\{p+q\sqrt2: p,q\text{ are rational}\}, B=\{r+s\sqrt3: r, s\text{ are rational}\}, Q=\mathbb{Q}$.
Show that $Q\subseteq A\cap B$.

I tried using a direct proof. I have a feeling that since $\sqrt2, \sqrt3$ are irrational then $A$ and $B$ cannot be rational, so their intersections cannot contain the elements of $Q$, but I'm not getting anywhere with this

答: Hint- $\mathbb Q\subset A$ and $\mathbb Q\subset B$ so that $A\cap B=...$

Title: Proof a subset is valid


问:
Using properties of inverse functions to prove a derivative exists.
A quick differentiation question that I'm caught up on:

Let $n \in \mathbb{N}$ and let $A=[0,\infty)$ if $n$ is even, and $A=\mathbb{R}$ if $n$ is odd. Let $g: A\rightarrow\mathbb{R}$ be $g(y)=y^{1/n}.$

Prove: For $y \neq 0$, $g'(y)$ exists and $g'(y)=\frac{1}{n}y^{\frac{1}{n}-1}$.*

Thus far, I've tried to define the inverse function $f(y)=y^n$, then using the fact that if $g(f(y))=y$ then $g'(f(y))=\frac{1}{f'(x)}$.

Though I think I have to show that $g$ is continuous, and $f$ differentiable to be able to use this result.

If we show this, do we then get that
$$g'(y)= \lim_{f(x) \to f(y)} \frac{g(f(x))-g(f(y))}{f(x)-f(y)} = \lim_{f(x) \rightarrow f(y)}\frac{x-y}{f(x)-f(y)}=\lim_{f(x) \rightarrow f(y)}\frac{x-y}{x^n-y^n}?$$

I'm not quite sure how to approach this question. Any help is appreciated.

答:
If we want to talk about inverse functions, then we need to restrict the codomain so as to make the function invertible: Instead of $g : A \to \Bbb R$, we need $g : A \to A$. Then by the definition of $y^{1/n}, g : A \to A$ is the inverse of $f : A \to A$.

Though I think I have to show that $g$ is continuous, and $f$ differentiable to be able to use this result.

There are many variants of the theorem. I don't know which one you are referring to. But you should look it up and make sure you know its requirements.

That $f$ is differentiable is surely something you've already seen in your course. If nothing else, it follows by induction from the Leibnitz rule $(uv)' = u'v + v'u$.

That $g$ is continuous follows from it being the inverse of a continuous function on an interval. Because $f$ is both continuous and invertible, it has to be strictly monotone. (Otherwise it would violate the intermediate value theorem - do you see why?) Therefore $g$ is also strictly monotone and can only have jump discontinuities. Such a discontinuity would mean that $f$ was not defined in the jumped region, but $f$ is defined everywhere in the interval.

If we show this, do we then get that
$$g'(y)=\lim_{f(x) \to f(y)} \frac{g(f(x))-g(f(y))}{f(x)-f(y)} =\lim_{f(x) \to f(y)}\frac{x-y}{f(x)-f(y)}=\lim_{f(x) \to f(y)}\frac{x-y}{x^n-y^n}$$

No. The definition of $g'(y)$ is $g'(y) = \lim_{x \to y}\frac{g(x) - g(y)}{x-y}$. If you already knew that $g'(y)$ existed, then you could claim that $\lim_{x \to y} \frac{g(f(x))-g(f(y))}{f(x)-f(y)} = g'(y)$. But the existence of $\lim_{x \to y} \frac{g(f(x))-g(f(y))}{f(x)-f(y)}$ does not in general imply the existence of $\lim_{x \to y}\frac{g(x) - g(y)}{x-y}$, since $f$ might be sidestepping where $g$ misbehaves.

And note that I am still taking the limit as $x \to y$. You would have to define what the limit as "$f(x) \to f(y)$" even means. Because this particular $f$ is well-behaved, you could get away with doing so, but in general "$\lim_{f(x) \to f(y)}$" is not a real thing.

However, you do get
$$\lim_{x\to y} \frac{g(x) - g(y)}{x-y}=\lim_{x\to y}\dfrac1{\frac{f(g(x)) - f(g(y))}{g(x) - g(y)}}= \dfrac1{\lim_{x\to y}\frac{f(g(x)) - f(g(y))}{g(x) - g(y)}}$$
And in this case, you do already know that $f$ is differentiable, and since $g$ is continuous $\lim_{x \to y} g(x) - g(y) = 0$.


问:
How can I prove that for any sets A, B, C, D, that $(A \cap B) \cup (C \cap D) \subseteq (A \cup C) \cap (B \cup D)$
I tested a few cases and it seems to check out, but I'm not sure how to prove that this statement is true.

$(A \cap B) \cup (C \cap D) \subseteq (A \cup C) \cap (B \cup D)$

答:
If you already know that, for any sets $X,Y,Z$,
$$
X\subseteq Y\cap Z\iff (X\subseteq Y \text{ and } X\subseteq Z)\qquad(1)
$$
and
$$
(X\cup Y)\subseteq Z\iff (X\subseteq Z\text{ and }Y\subseteq Z)\qquad (2)
$$
then this problem becomes easy, as follows. The formula you want,
$$
(A\cap B)\cup(C\cap D) \subseteq (A\cup C)\cap (B\cup D),
$$
is equivalent by (1) to
$$
(A\cap B)\cup(C\cap D) \subseteq A\cup C \quad\text{ and }\quad
(A\cap B)\cup(C\cap D) \subseteq B\cup D,
$$
which is in turn equivalent by (2) to
$$
A\cap B\subseteq A\cup C\quad\text{ and }\quad
C\cap D\subseteq A\cup C\quad\text{ and }\quad
A\cap B\subseteq B\cup D\quad\text{ and }\quad
C\cap D\subseteq B\cup D.
$$
The four statements on the last line are all obvious; for example $A\cap B\subseteq A\subseteq A\cup C$.

(If you don't already know (1) and (2), it would be a good idea to prove them. They're easy and often useful.)


问:
if I have a length AB, defined by points A and B, how do I construct a circle with circumference AB with a compass?
I want to construct a circle with a given linear circumference with compass and straightedge. I know that you cant construct a length ¨pi¨ if the linear distance is ¨1¨, but can you find the radius?

I want to do this so I can possibly take the relationships of a square, and its divisibility into n equal parts, and copy it onto a circle, and vice versa

答: You'd need to construct $\frac1{2\pi}$, which is also impossible.


问:
If the true population mean is 43 and the significance level is 5%, the probability of a Type II error is closest to:
A normal population has a population standard deviation of $12$. The hypotheses about the population mean, $H_0: \mu = 40$ versus $H_1: \mu > 40$ are to be tested. A random sample of $100$ observations will be selected and the sample mean is denoted by $\bar x$.

If the true population mean is $43$ and the significance level is $5\%$, the probability of a Type II error is closest to:

A. $0.305$

B. $0.273$

C. $0.229$

D. $0.195$

E. $0.187$

I know that the $z$ score is $(43-40)/1.2 = 2.50$, but am lost on what to do afterwards

答:
The Type II error $\beta$ is the probability of failing to reject the null hypothesis $H_0$ when the alternative $H_1 : \mu = 43$ is true.  In other words, given that the true sampling distribution of the sample mean is normal with mean $\mu = 43$ and standard error $\sigma/\sqrt{n} = 12/\sqrt{100} = 1.2$, then $H_0$ is not rejected if it is "too small."  But what does this mean?

Specifically, suppose we observe a value of the test statistic $$Z \mid H_0 = \frac{\bar x - \mu_0}{\sigma/\sqrt{n}} = \frac{\bar x - 40}{1.2}$$ such that it is less than the critical value for rejection at the $\alpha = 0.05$ level; i.e., $$\frac{\bar x - 40}{1.2} \le z^*_\alpha \approx 1.64485.$$  This means the sample mean would need to be at most $$\bar x \le 41.9738,$$ otherwise you would reject.  So we need to compute $$\beta = \Pr[\bar x \le 41.9738 \mid H_1] = \Pr\left[\frac{\bar x - 43}{1.2} \le \frac{41.9738 - 43}{1.2}\right] = \Pr[Z \le -0.855146] \approx 0.196235.$$  Therefore, the Type II error probability is just shy of $20\%$.

The thinking is like this:  The test statistic assumes the null is true.  Under this assumption, what is the complement of the rejection region; i.e., for what values of the sample mean would we not reject the null?  Then, assuming the alternative hypothesis is true, what is the chance of observing a sample mean that lies in the complement of the rejection region?  This gives you the probability of committing a Type II error--failure to reject when we should reject.


问:
One of $2^1-1,2^2-1,...,2^n-1$ is divisible by $n$ for odd $n$

Let $n$ be an odd integer greater than 1. Show that one of the numbers $2^1-1,2^2-1,...,2^n-1$ is divisible by $n$.

I know that pigeonhole principle would be helpful, but how should I apply it? Thanks.

答:
Hint: We know that $\gcd(2,n)=1$

$2^{\phi(n)} \equiv 1\pmod n$

And $1<\phi(n)<n$ and we already have $2^1-1 \equiv 1 \pmod n$

You can use pigeon-hole principle after this.

Another Approach:

Actually, the $\phi(n)$ or $\lambda(n)$ occurs just between $1$ and $n$, therefore,

$2^{\lambda(n)} \equiv 1 \pmod n  \implies 2^{\lambda(n)} -1 \equiv 0 \pmod n$.

This involves no Pigeons or Pigeon-holes.:)

Here $\lambda(n)$ is smallest number which occurs before $\phi(n)$ such that $a^{\lambda(n)} \equiv 1 \pmod n$. Also known as Carmichael Number.


问:
Find a general method to find particular solutions where the sum of the squares of two consecutive integers is equal to the square of another integer
Question:  If the sum  of the squares of two consecutive integers is equal to the square of another integer, then find a general method to find particular solutions.  E.g., $27304196^2+27304197^2=38613965^2$.

I tried to figure out other solutions such as $3^2+4^2=5^2$ and $20^2+21^2=29^2$, but I don't know how to find a method for general solutions.

Can anybody help me

Body:
For $A^2+B^2=C^2:\space B=A\pm1\quad$ the $C$-values are the odd Pell numbers  $(5,29,169,985\cdots)$ as shown in A000129
and the triples $T_n$ that contain them can be generate sequentially by the formula below with a seed of $\quad T_0=(0,0,1)$

$$A_{n+1}=3A_n+2C_n+1\qquad B_{n+1}=3A_n+2C_n+2\qquad C_{n+1}=4A_n+3C_n+2$$

For example, it generates the following
$$T_1=(3,4,5)\quad T_2=(20,21,29)\quad T_3=(119,120,169)\quad T_4=(697,696,985)$$

They may also be generated directly using Euclid's formula

$$\text{where} \qquad A=m^2-k^2 \qquad B=2mk \qquad C=m^2+k^2\quad \text{and}$$

\begin{equation}
 m_n= \frac{(1 + \sqrt{2})^{n+1} - (1 - \sqrt{2})^{n+1}}{2\sqrt{2}}\qquad \qquad\qquad
 k_n= \frac{(1 + \sqrt{2})^n - (1 - \sqrt{2})^n}{2\sqrt{2}}
 \end{equation}

For example
\begin{align*}
 \frac{(1 + \sqrt{2})^{2} - (1 - \sqrt{2})^{2}}{2\sqrt{2}}=2 \qquad &
     \frac{(1 + \sqrt{2})^1 - (1 - \sqrt{2})^1}{2\sqrt{2}}=1 \\
&\qquad\qquad   F(2,1)=(3,4,5)\\
 \frac{(1 + \sqrt{2})^{3} - (1 - \sqrt{2})^{3}}{2\sqrt{2}}=5 \qquad &
     \frac{(1 + \sqrt{2})^2 - (1 - \sqrt{2})^2}{2\sqrt{2}}=2 \\
& \qquad\qquad  F(5,2)=(21,20,29)\\
  \frac{(1 + \sqrt{2})^{4} - (1 - \sqrt{2})^{4}}{2\sqrt{2}}=12 \qquad &
      \frac{(1 + \sqrt{2})^3 - (1 - \sqrt{2})^3}{2\sqrt{2}}=5 \\
&\qquad\qquad  F(12,5)=(119,120,169)\\
 \frac{(1 + \sqrt{2})^{5} - (1 - \sqrt{2})^{5}}{2\sqrt{2}}=29 \qquad &
     \frac{(1 + \sqrt{2})^4 - (1 - \sqrt{2})^4}{2\sqrt{2}}=12 \\
&\qquad\qquad  F(29,12)=(697,696,985)
 \end{align*}


问:
What is the probability that at least one of the blocks you pulled out is Red, Orange or Yellow when you pull out 2 blocks?
Consider a bag of 7 blocks. Each block has a different color. The colors are Red, Orange, Yellow, Green, Blue, Indigo and Violet.

You reach into the bag of 7 blocks and pull out 2 blocks. What is the probability that at least one of the blocks you pulled out is Red, Orange or Yellow?

答:
We'll take Graham's advice and think about the complements - that's a slightly easier calculation.

We want to know the probability that one of the two blocks is one of red, orange, or yellow. The complement of this event is that neither of the first two blocks is one of these three colors.

There are four blocks in the bag that are not one of the desired colors, and seven total blocks. So, the probability that the first block is not red, orange, or yellow is $\frac{4}{7}$.

If the first block is not one of our colors, then the block that was removed first was one of the four we did not want: green, blue, indigo, or violet. So, three of these remain in the bag. We removed one of the seven, so six total remain in the bag. This means that the probability that the second block is also not red, orange or yellow (given that the first one was not) is $\frac{3}{6}$.

Then, the probability that the first block was not one of the three colors and the second block was not is

$$ \frac{4}{7}*\frac{3}{6} = \frac{2}{7} . $$

We want the probability of the complement, so the probability we're looking for is

$$ 1 - \frac{2}{7} = \frac{5}{7} . $$


问:
Can all positive real polynomials be expressed as a sum of polynomial squares?
For $P(x)=y$,   where $P(x)$ is some real polynomial strictly above the $x$-axis, can it be expressed as the sum of polynomial squares?

答:
Yes. We induct on the degree. If $P$ is constant we are done. Otherwise we must have the degree of $P$ be even and the limit of $P$ is infinity in both $\pm \infty$ dirrections. Hence there exists an $a \in \mathbb R$ such that $P$ obtains it's minimum value at $a$.  But then $(x - a)^2$ divides $P(x) - P(a)$. Thus we can write

$$P(x) = (x-a)^2 \frac{P(x) - P(a)}{(x-a)^2} + \sqrt{P(a)}^2$$ and by inductive hypothesis $\frac{P(x) - P(a)}{(x-a)^2}$ can be written as the sum of polynomial squares so we are done.


问:
Find the perimeter of Triangle ABC
This practice engineering board exam which I've tried seems very tricky (question no.2). The inscribed circle of $\triangle ABC$ is tangent to $AB$ at $P$ . If the radius is $21$, $AP=23$ and $PB=27$, find the perimeter of $\triangle ABC$ .I've attempted to use the Pythagorean Theorem and Heron's Formula but the problem is that I've missed the values of the two sides $BC$ and $CA$ .Can you please provide solutions by two methods(geometry and trigonometry)?

答:
Let the circle touch side $BC$ and $CA$ at points $Q$ and $R$ respectively. Let $CQ=CR=m$.

$AR=AP=23$ and $BQ=BP=27$.

$AB=50, BC=27+m, CA=23+m$

Now, apply Heron's Formula and then use $\Delta=rs$.

Title: Find the perimeter of Triangle ABC


问:
Is the kth central moment less than the kth raw moment for even k?
If $X$ is a real-valued random variable, then the $k$th raw moment is $\mathbb{E}[X^k]$, while the $k$th central moment is $\mathbb{E}[(X-\mathbb{E}[X])^k]$.  If $k$ is even, is the $k$th central moment always bounded above the $k$th raw moment?

When $k = 2$, then $\mathbb{E}[(X-\mathbb{E}[X])^2] = \mathbb{E}[X^2]-\mathbb{E}[X]^2$, and because $\mathbb{E}[X]^2$ is always positive, it follows that this is less than or equal to $\mathbb{E}[X^2]$.  But I'm having trouble extending this to larger moments.

答:
Roberto Rastapopoulos' proof could be simplified by firstly proving the inequality holds when $a\ge 0$:
$$a^m-(a-1)^m\ge a-1;$$

Then similarly for $X\ge0$ with $E(X)=1$,
$$E(X^r)-E[(X-1)^r] = \int_{X\ge 0}[x^r-(x-1)^r]dP(\omega)\ge \int_{X\ge 0}(x-1)dP(\omega)=0.$$


问:
radius of convergence and the interval of convergence for $\sum_{n=0}^{\infty}(\frac{\pi}{2} +sin(n))x^n$
I'm having trouble finding the radius of convergence and the interval of convergence of the following sequence:

$\sum_{n=0}^{\infty}(\frac{\pi}{2} +sin(n))x^n$

I tried using the ratio test, but got stuck when trying to solve the limit.

Then went on to try the root test, thought that I could use the sandwich therom and therefore the limit is 1, which means that the radius is 1? Not sure (about the limit).
If so, I need to check the ends of the interval:  $x=1, x=(-1)$

for $x=1$
$\sum_{n=0}^{\infty}(\frac{\pi}{2} +sin(n))1^n$ = $\sum_{n=0}^{\infty}(\frac{\pi}{2} +sin(n))$

for $x=(-1)$
$\sum_{n=0}^{\infty}(\frac{\pi}{2} +sin(n))(-1)^n$

Both diverge (how do I prove that? and is that even true?), therefore the interval of convergence is $(-1,1)$?

Could really use help here.

Thanks!

答:
It seems like you are confused about the boundary of your interval of convergence. Keep in mind that if $\sum^{\infty}_{n=1}a_{n}$ converges then $a_{n} \to 0$. This should remove all doubts. Since

{$(-1)^{n}(\frac{\pi}{2}+sin(n))$} and {$(1)^{n}(\frac{\pi}{2}+sin(n))$} are both bounded away from 0.


问:
How can 1+1=3 be possible?
I've been looking at some proofs but none make any sense to me, can someone prove this is true?

One of them...
Suppose $a=b$

$$ \Rightarrow a=b $$

Multiply each side by $b$

$$ \Rightarrow ab=b^2 $$

Subtract a^2 from each side

$$ \Rightarrow ab-a^2=b^2-a^2 $$

Simplify
$$ \Rightarrow a=a+b $$
$$ \Rightarrow a+1=a+b+1 $$

Because a=b
$$ \Rightarrow a+1=2a+1 $$

Suppose a=1
$$ \Rightarrow 1+1=2+1 $$

NOTE: I'm a 10th grader

答:
OK. Let's walk through this, somewhat more nicely formatted:

Suppose $a=b$. Then
\begin{align}
a&=b \tag{1}
\end{align}
Multiply each side by $b^2$
\begin{align}
 ab&=b^2
\tag{2}
\end{align}

Subtract $a^2$ from each side
\begin{align}
ab-a^2& =b^2-a^2 \tag{3}
\end{align}

Simplify
\begin{align}
a&=a+b  \tag{4}\\
a+1&=a+b+1  \tag{5}
\end{align}
Because $a=b$
\begin{align}
a+1=2a+1  \tag{6}
\end{align}

Suppose $a=1$
\begin{align}
1+1&=2+1  \tag{7}
\end{align}

Where to begin? Well, right at the start, I think you're saying "Let $a$ be any real number, and let $b = a$." So let me just run with that, writing notes to the right explaining each step:

\begin{align}
a & = b & \text{by assumption} \tag{1} \\
ab & = b^2 & \text{multiply both sides by $b$} \tag{2} \\
ab-a^2 & = b^2 -a ^2 & \text{subtract $a^2$ from both sides} \tag{3} \\
a(b-a) & = (b+a)(b-a)& \text{factoring, difference of squares} \tag{4} \\
a & = (b+a)& \text{Divide through by $b-a$} \tag{5} \\
a+1 & = b+a + 1& \text{Add 1 to each side} \tag{6} \\
a+1 & = a+a + 1& \text{Replace $b$ by $a$ using Eq. 1} \tag{7} \\
1+1 & = 1+1 + 1& \text{Because $a$ was arbitrary,}\\
& & \text{we can apply equation 7 to $a = 1$} \tag{8} \\
2 & = 3& \text{Simplification} \tag{8} \\
\end{align}

So that's the "proof". The failure is at step 5, where you say that $pq = pr$ implies that $q = r$; that's true only if you know $p$ is nonzero. For instance $0 \cdot 5 = 0 \cdot 7$, but that doesn't let you conclude that $5 = 7$, because $0$ has no multiplicative inverse. In your case, the number $b-a$ is playing the role of $p$, and because $b$ and $a$ are the same, you're really "dividing through by $0$", and that's the error.

Title: How can 1+1=3 be possible?


问:
Why is the distribution of the x coordinate of a point randomly selected from the circumference the unit cricle not uniform?
I am currently attempting to solve a question which deals with the distribution of the $x$ and $y$ coordinates of a randomly chosen point from the circumference  of the unit circle. When I first attempted the question I thought that since point is chosen at random, it would follow that it is uniformly distributed on the circumference of the circle, and moreover, its $x$ and $y$ coordinates will be uniformly distributed on the region $[-1,1]$. However, this is wrong.

I have found a couple of  threads in regards to this issue, where the answers suggest one uses the fact that $x=\cos \theta$ and start from there. My issue is that this was not my original thought and that without going through the algebra with $x=\cos \theta$ I wouldn't be able to say why the distribution of $x$ (and $y$ for that mater) are not uniform.

Could someone explain to me how can I intuitively rebutle the idea that $x$ (and $y$) are uniformly distributed, without involving calculation?

答:
It is because the circle is more steep on its left and right edges so a small interval there will contain a lot of arc length. But as you go toward the center of the circle, a small interval will be smaller. Consider an interval of [-1,-.99] vs an interval of [0,.01]. The slope is much higher in the first interval than the second, which is mostly flat, so the first interval will contain more “points,” and hence probability.

Also, Compare the interval [-1,-1/2] vs [-1/2, 0], dividing the left half in two equal interval widths. The first interval contains more arc length than the second. To convince yourself, draw it out. It looks like the first contains about twice as much as the second.

The more arc length, the more “points” there are to choose from.

Two intervals of the same width won’t contain the same arc length unless they are mirrors across the y axis.


问:
for which n is $(3+ i\sqrt{3})^{n} $a real number
I understand complex numbers and their exponents, but I'm not sure how to go about this problem or where to start it.</p>

答:
Write $3+i\sqrt{3}$ with the $re^{i\theta}$ form.

$$3+i\sqrt{3} = \sqrt{12} e^{i\tfrac{\pi}{6}}$$

$$(3+i\sqrt{3})^{n} = \sqrt{12} e^{i\tfrac{n\pi}{6}}$$

Which is real when $n=6k$ for some integer $k$.


问:
Methods to solve this differential equation
How can I solve this differential equation?

$$
A\;\frac{d^2\theta}{dt^2} + B\cos(\theta) +C\sin(\theta) = 0
$$

答:
First note that
\begin{align}
B\cos\theta + C\sin\theta =&\ \sqrt{B^2+C^2}\left(\frac{B}{\sqrt{B^2+C^2}}\cos\theta + \frac{C}{\sqrt{B^2+C^2}}\sin\theta \right)\\
=&\ \sqrt{B^2+C^2} \sin(\theta + \delta) = D \sin(\theta+\delta)
\end{align}
where
\begin{align}
\cos\delta = \frac{C}{\sqrt{B^2+C^2}} \ \text{ and } \ \sin\delta = \frac{B}{\sqrt{B^2+C^2}}.
\end{align}

Hence, your differential equation becomes
\begin{align}
A \frac{d^2\theta}{dt^2}+ D \sin(\theta+\delta)=0.
\end{align}
In particular, if you define a new function $\phi(t) = \theta(t) + \delta$, i.e. shift by a constant, then we see that $\phi$ satisfies
\begin{align}
A\frac{d^2\phi}{dt^2} = A\frac{d^2\theta}{dt^2} = - D\sin(\theta +\delta) = -D\sin(\phi).
\end{align}
Hence you have the equation
\begin{align}
\frac{d^2\phi}{dt^2}+\frac{D}{A}\sin(\phi)=0.
\end{align}
This is just the standard pendulum equation which can be "readily solved".


答:
$$(3+ i\sqrt{3})^{n}=({\sqrt{3}})^{n}\Bigl(\sqrt{3}+i\Bigr)^n$$
$$({\sqrt{3}})^{n}\Bigl(\sqrt{3}+i\Bigr)^n=(2\sqrt3)^n\Bigl(\frac{\sqrt3}{2}+\frac{1.i}{2}\Bigr)^n$$
$$(2\sqrt3)^n\Bigl(\frac{\sqrt3}{2}+\frac{1.i}{2}\Bigr)^n=(2\sqrt3)^n\Bigl(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\Bigr)^n$$

I'm not sure if you are aware of this identity: $\cos\theta+i\sin\theta=e^{i\theta}$. You can find proofs for this online.
$$(2\sqrt3)^n\Bigl(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\Bigr)^n=(2\sqrt3)^n\Bigl(e^{i\frac{\pi}{6}}\Bigr)^n$$

Going back to the stated identity:$\cos\theta+i\sin\theta=e^{i\theta}$, we can observe the following:
$$\cos\pi+i\sin\pi=e^{i\pi}=-1$$
$$\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}=e^{i\frac{\pi}{2}}=i$$

$e^{i\theta}$ is after all, a complex number which is written as $\cos\theta + i\sin\theta$. For it to be completely real, it shouldn't have an imaginary part. So, $\sin\theta=0$. And this is true for all  $\theta \in n\pi$, where $n=0,1,2,....$. Thus, $e^0, e^{i\pi},e^{2i\pi},e^{3i\pi}.. $ are all real numbers.

For our question to be real:
$$(2\sqrt3)^n\Bigl(e^{i\frac{\pi}{6}}\Bigr)^n=(2\sqrt3)^ne^{ik\pi}$$

Thus for all $n=6k$, the given complex number is real

Title: for which n is $(3+ i\sqrt{3})^{n} $a real number

Body: I understand complex numbers and their exponents, but I'm not sure how to go about this problem or where to start it.</p>


问:
What is the locus of $\vec r$ if $(\vec a - \vec b)\cdot\vec r=\frac {1}{2}({\vert\vec a\vert}^2-{\vert\vec b\vert}^2)$?
Question. I want to find out the locus of point whose position vector is $\vec r$, in three-dimensional space. Vectors $\vec a$ and $\vec b$ are two fixed position vectors. The following relationship between the three vectors is satisfied.$$(\vec a - \vec b)\cdot\vec r=\frac {1}{2}({\vert\vec a\vert}^2-{\vert\vec b\vert}^2)$$

My Attempt. I tried to rearrange the relationship equation by "multiply out" the dot product and turning the moduli squared into dot products. Finally I got the following equation but still stuck.$$\vec a\cdot (2\vec r - \vec a)=\vec b\cdot(2\vec r - \vec b)$$

Comment. Please do not tell me the answer directly. What I need is some hint with which I could make further progress on my own. Thanks a lot!

答:
$\textbf{Hint}$: $$\vec{n}\cdot(\vec{r}-\vec{r}_0)=0$$

is the equation of a plane in $3$D, that passes through the point $\vec{r}_0$ with normal vector $\vec{n}$.


答: Transform to polar coordinates $r^2=9+3=12$, so  $3+i\sqrt{3}=2\sqrt{3}(\frac{\sqrt{3}}{2}+i\frac{1}{2})=2\sqrt{3}e^{i\frac{\pi}{6}}$ To get n to make it real, the exponent has to be an integer multiple of $\pi$ so $\frac{n}{6}$ must be an integer.


问:
Intuition for Brownian motion time-inversion formula
Background: I am studying a course in Stochastic Financial Models, and have been introduced to Brownian motions (BMs) and Gaussian processes.

The BM time-inversion formula states that if $W_t$ is a BM, then $$B_t := tW_{1/t}$$ is also a BM.

I know how to prove the result, through Gaussian processes, similar to the answer here.

However, the proof is unenlightening and does not offer any intuition. In fact, my course lecturer does not offer any intuition either.

Question: Please provide some intuition why the formula “should be” true.

Example of my intuitions for another formula:

“Scaling” formula: if $W_t$ is a BM, then $cW_{t/c^2}$ is also a BM.

Intuition: picture time on the $x$ axis and the motion on the $y$ axis. BMs “don’t go that far up or down”, i.e. their standard deviation is “only” sd$(W_t)=\sqrt t$, so to compensate for the factor of $c$ stretch in the $y$ axis, we have to stretch by a factor of $c^2$ in the $x$ axis.

答:
This explanation is intuitive but very far from stringent. I think one issue with building an intuition for time inversion scaling based on previously obtained intuition for the regular scaling formula $c W_{t/c^2}$ is that the regular scaling formula is a kind of "uniform global scaling" whereas the time inversion scaling can be seen as more of a "local scaling". Assume that we have built an intuition for how $c W_{t/c^2}$ is BM. I wanted to show how one can "build a bridge" to time inversion scaling.

Consider the time interval $t_1-t_0$, where we will later set $t_0=0$. If we now take a large step back so we can see a big chunk, the interval $t-t_0$, of the random walk for which the interval $t_1-t_0$ is relatively tiny, $t_1\ll t$, then we can consider the BM to be fairly constant (in some sense) in the $t_1-t_0$ interval compared to the much larger $t-t_0$ interval.

Let's now use the regular scaling formula on the little interval, with $c=t_1-t_0$:
$$c W_{t/c^2}=(t_1-t_0) W_{t/(t_1-t_0)^2}$$

Here we need  $c=t_1-t_0$ and $W_{t_1-t_0}$  to be "small" compared to the expected variation of $W_t$.

Set $t_0=0$ to get:
$$c W_{t/c^2}=t_1 W_{t/t_1^2}$$
If we let $t_1$ grow and catch up with $t$ (or $t$ shrink down to $t_1$) we get
$$t_1 W_{t/t_1^2} \rightarrow t W_{t/t^2}=t W_{1/t}$$
and this provides a link between the regular scaling formula and the time inversion formula. Apologies for the utter and complete lack of stringency.
Several things are problematic here. For example, what do we mean with the BM to be relatively (fairly) constant? How can we be allowed to let $t_1$ approach $t$? However, I suspect making it stringent would be a larger undertaking and not give the wanted intuition. The key, I think, is to realize that any subinterval $t_1-t_0$ will look like a regularly scaled BM when compared to a much larger interval. For stringent proofs, see:
Prove the time inversion formula is brownian motion
or
http://www.math.uchicago.edu/~may/VIGRE/VIGRE2011/REUPapers/Stolarski.pdf

It is very difficult to get a nice computer plot that shows time inversion. This can be understood, for example, as you needing an extremely large number of terms in a, say, Fourier series or Karhunen–Loève representation (https://en.wikipedia.org/wiki/Wiener_process). And if you try this in the obvious naïve way by just summing terms you will inevitably get a BM that is highly resolved close to $t=0$ but poorly resolved for larger value of $t$. In a way you get penalized for not being able to represent this as a true infinite series.


问:
Does the integral converge $\int_{-\infty}^{+\infty}\frac{\exp(ibx)}{1+e^x}\,dx$ for $b>0$?
I try to calculate integral $$\int_{-\infty}^{+\infty}\frac{\exp(ibx)}{1+e^x}\,dx.$$ Does it converge for $b>0$? Wolfram Mathematica shows contradictory results: for $b=2$ it converges, for $b=1$ and $b=3$ it does not. I need to calculate it.

答:
The integral $I^+(b)=\int_{0}^\infty \frac{e^{ibx}}{1+e^x}\,dx$ exists, both as an improper Riemann integral or a Lebesgue integral.  In fact, by expanding $\frac{1}{1+e^x}=\frac{e^{-x}}{1+e^{-x}}$ in a geometric series, we can write $I^+(b)$ in series form as

$$I^+(b)=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n-ib}\tag1$$

But the integral $I^-(b)=\int_{-\infty}^0 \frac{e^{ibx}}{1+e^x}\,dx$ fails to exist.  As has already been discussed by others on this page, we may interpret $I^-(b)$ as a distribution.

So, I thought it might be instructive to present an approach to evaluating the distribution $I^-(b)$ that is distinct to the other posted answers.  It is to that end that we now proceed.

Let $R>0$ and denote $I^-_R(b)=\int_{-R}^0 \frac{e^{ibx}}{1+e^x}\,dx$ as the integral.  Note that $\lim_{R\to \infty}I^-_R(b)=I^-(b)$ in disribution.  Now, enforcing the substitution $x\mapsto -x$ and expanding $\frac{1}{1+e^{-x}}$ in a geometric series

$$\begin{align}
I^-_R(b)&=\sum_{n=0}^\infty \frac{(-1)^n(1-e^{-(n+ib)R})}{n+ib} \\\\
&=-\frac ib (1-e^{-ibR})-\sum_{n=1}^\infty \frac{(-1)^{n-1}(1-e^{-(n+ib)R})}{n+ib}\\\\
\end{align}$$

In the sense of distributions, $\lim_{R\to\infty}\frac{1-e^{-ibR}}b=\text{PV}\left(\frac1b\right)+i\pi\delta(b)$.  (See the Proof in the Appendix herein)

Therefore, in the sense of distributions

$$I^-(b)=-i\text{PV}\left(\frac1b\right)+\pi \delta(b)-\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n+ib}\tag2$$

Putting $(1)$ and $(2)$ together, the Fourier Transform of $f(x)=\frac1{1+e^x}$ is

$$\mathscr{F}\{f\}(b)=\pi\delta(b)-i\text{PV}\left(\frac1b\right)+i2b\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^2+b^2}\tag3$$

The series in $(3)$ can be found using the Residue Theorem by integrating $\frac{\csc(\pi z)}{z^2+b^2}$ over a circular contour with radius $N+1/2$, $N\in \mathbb{N}$ and center at $z=0$ and then letting $N\to \infty$.  Proceeding reveals

$$\sum_{n=-\infty}^\infty \frac{(-1)^n}{\pi(n^2+b^2)}+\frac{1}{ib\sin(i\pi b)}=0$$

from which we find that

$$i2b\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^2+b^2}=\frac ib-i\pi \text{csch}(\pi b)\tag4$$

Substituting $(4)$ in $(3)$ yields the coveted result

$$\bbox[5px,border:2px solid #C0A000]{\mathscr{F}\{f\}(b)=\pi\delta(b)-i\text{PV}\left(\frac1b\right)+\frac ib-i\pi \text{csch}(\pi b)}$$

APPENDIX:  Proof that $\displaystyle \lim_{R\to\infty}\frac{1-e^{-ibR}}{b}=\text{PV}\left(\frac1b\right)+i\pi \delta(b)$

STEP $1$

Let $\phi(b)$ be a Schwartz function.  We will show that

$$\lim_{R\to \infty}\int_{-\infty}^\infty \phi(b) \frac{\sin(bR)}{b}\,db=\pi \phi(0)$$

Integrating by parts with $u=\phi(b)$ and $v=\int_{-\infty}^b \frac{\sin(kR)}{k}\,dk$, and applying the Dominated Convergence Theorem, we find that

$$\begin{align}
\lim_{R\to\infty}\int_{-\infty}^\infty \phi(b) \frac{\sin(bR)}{b}\,db&=-\lim_{R\to\infty}\int_{-\infty}^\infty \phi'(b) \int_{-\infty}^b \frac{\sin(kR)}{k}\,dk\,db\\\\
&=-\lim_{R\to\infty}\int_{-\infty}^\infty \phi'(b) \int_{-\infty}^{Rb} \frac{\sin(k)}{k}\,dk\,db\\\\
&=-\pi \int_{-\infty}^\infty \phi'(b)H(b)\,db\\\\
&=\pi \phi(0)
\end{align}$$

And we are done.

STEP $2$

Let $\phi(b)$ be a Schwartz function.  We will show that

$$\lim_{R\to \infty}\int_{-\infty}^\infty \phi(b) \frac{1-\cos(bR)}{b}\,db=\lim_{\varepsilon\to 0^+}\int_{|x|\ge\varepsilon}\frac{\phi(b)}{b}\,db$$

Fix $\varepsilon>0$.  Then, applying the Riemann-Lebesgue Lemma reveals

$$\begin{align}
\lim_{R\to\infty}\int_{-\infty}^\infty \phi(b) \frac{1-\cos(bR)}{b}\,db&=\lim_{R\to\infty}\int_{|b|<\varepsilon} \phi(b) \frac{1-\cos(bR)}{b}\,db+\lim_{R\to\infty}\int_{|b|>\varepsilon} \phi(b) \frac{1-\cos(bR)}{b}\,db\\\\
&=\lim_{R\to\infty}\int_{|b|<\varepsilon} \phi(b) \frac{1-\cos(bR)}{b}\,db+\int_{|b|>\varepsilon} \frac{\phi(b)}{b}\,db
\end{align}$$

Next, writing $\phi(b)=\phi(0)+\phi'(0)b+O(b^2)$ for $|b|<\varepsilon$ and noting that $\frac{1-\cos(bR)}{b}$ is integrable and an odd function of $b$, we find that

$$\int_{|b|<\varepsilon} \phi(b) \frac{1-\cos(bR)}{b}\,db=O(\varepsilon)$$

Letting $\varepsilon \to 0^+$ we conclude

$$\begin{align}
\lim_{R\to\infty}\int_{-\infty}^\infty \phi(b)\frac{1-\cos(bR)}{b}\,db&=\lim_{\varepsilon\to0^+}\int_{|x|\ge \varepsilon}\frac{\phi(b)}{b}\,db\\\\
&=\text{PV}\int_{-\infty }^\infty \frac{\phi(b)}{b}\,db
\end{align}$$

PUTTING STEPS $1$ AND $2$ TOGETHER

Using the results in Steps $1$ and $2$ yields

$$\lim_{R\to \infty }\left(\frac{1-e^{-ibR}}{b}\right)=\text{PV}\left(\frac1b\right)+i\pi \delta(b)$$


问:
Probability question involving colored balls
Probability question.  I’ve been trying to figure this out for a while and need a little help.  This is not a homework problem just something I made up but can’t figure it out on my own.

I have a total of 30 balls(10 red, 10 green, 10 blue) I dump all 30 balls into a large bin.

I have 4 smaller containers.  I randomly take 2 balls out and put them in each of the 4 smaller containers (I remove 8 balls total).  What are my chances of having 2 bins or more of just red balls?  Can you please show the math behind the answer.

答:
You could use an inclusion-exclusion strategy:

Call the bins A,B,C,D, and let AB represent the event that both A and B are all red (with no requirements on the other bins); ABC the event that A,B,C are all red (with no requirements on D); etc.

First, let's find P(AB). The order in which we fill the bins doesn't matter - we might as well have put the first four removed balls into bins A and B, and the probability these are all red is $\frac{10}{30}\frac{9}{29}\frac{8}{28}\frac{7}{27}$. The same goes for any other pair of bins, so P(AB)+P(AC)+P(AD)+P(BC)+P(BD)+P(CD) $= 6P(AB) =6\cdot\frac{10}{30}\frac{9}{29}\frac{8}{28}\frac{7}{27}$.

This sum does some over-counting. Any outcome where exactly two bins are all red falls into just one of the six cases, so we've counted it correctly, but any outcome in which three bins are all red appears in three of the two-bin events (e.g. ABC is counted in each of AB,AC,BC), so by adding the six probabilities, we've over-counted these outcomes by a factor of 3. Let's correct that by un-counting them twice, by subtracting 2(P(ABC)+P(ABD)+P(ACD)+P(BCD)) $=2\cdot 4\cdot P(ABC)=8 \cdot \frac{10}{30}\frac{9}{29}\frac{8}{28}\frac{7}{27}\frac{6}{26}\frac{5}{25}$ (because P(ABC) equals the probability that the first six balls we look at are all red).

Now our adjusted sum correctly accounts for the exactly-two-bin and exactly-three-bin outcomes, and we just need to fix it for ABCD, which was "accidentally" included in all the events considered above. We added its probability 6 times and subtracted it 8 times, so we should add it back 3 times to get its count back to 1.

So the probability that at least two bins are all red is
$$6\cdot\frac{10}{30}\frac{9}{29}\frac{8}{28}\frac{7}{27}-8\cdot \frac{10}{30}\frac{9}{29}\frac{8}{28}\frac{7}{27}\frac{6}{26}\frac{5}{25}+3\cdot \frac{10}{30}\frac{9}{29}\frac{8}{28}\frac{7}{27}\frac{6}{26}\frac{5}{25}\frac{4}{24}\frac{3}{23}.$$


问:
Floor property In general, ⌊ n x ⌋ = n ⌊ x ⌋ .
I am trying to proof the above statement. I have concluded this statement is true if x and n is an integer and real. For example if I take n=3 and x =2 then ⌊ 3. 2 ⌋ = 3 ⌊ 2 ⌋ which is ⌊6⌋ = (3 . 2)  which is 6=6 which proves the statement. But how do I prove this statement without taking n=3 or n=4 etc . ? Any advice will be appreciated</p>

Body:
Let $n=3$ and $x=\frac{1}{3}$

floor(NX)= floor(1) = 1 $\neq$ 0 = 3(0) = 3 (floor($\frac{1}{3}$)) = N*floor(X).


问:
Find the value of angle x (Langley's Adventitious Angles variant)
This problem had came to a Facebook post. Given that the left angles are given (with only 40° angle on the upper right triangle,find the angle x . What I did us to make equilateral triangles but it seems I still don't get two angles to get the vale of x . Can anyone give me a solution ???

答:
This is a different problem, but it follows the same principles with this problem. In these problems there are two methods to pick from: Synthetic geometry or trigonometry. Although synthetic solutions are more beautiful, they are harder to come up with. To obtain a synthetic solution you need to create isosceles triangles until you end up with an equilateral triangle that can be used to solve the problem or can be used to obtain a congruency.

A more practical theorem is Trigonometric Ceva's Theorem. (The proof of this theorem is done by applying the law of sines to the three triangles , and multiplying all the obtained expressions.) To apply the theorem just multiply the ratios of the sines of the angles at each vertex respectively:
$$\frac{\sin(40°)}{\sin(70°)}×\frac{\sin(10°)}{\sin(10°)}×\frac{\sin(x)}{\sin(50° - x)}= 1$$
Rearranging:
$$\sin(40)×\sin(x)=\sin(70°)×\sin(50° - x)= 1$$
From here, the answer turns out to be,
$$x=30°.$$


问:
Stopping times and $\sigma$-algebras
We have the usual $(\Omega, \mathcal{F}, P)$ stochastic basis. Let $\rho, \tau: \Omega \to T \cup \{+\infty\}$ be stopping times and $\mathcal{F}_{\rho}, \mathcal{F}_{\tau}$ their respective $\sigma$-algebras. Prove that $[\rho \le \tau ] \in \mathcal{F}_{\rho} \cap \mathcal{F}_{\tau}$.

Here is what I tried:

Let's first prove that $[\rho \le \tau ] \in \mathcal{F}_t$: $[\rho \le \tau ] = [\rho \le t] \cap [\tau \le t] \in \mathcal{F}_t \space$. Is this correct?

But now I'm having a hard time proving that it lies in the intersection of the $\sigma$-algebras.

Any hints?

答:
To show that $[\rho\leqslant \tau]$ is in $\mathcal F_\rho$, use the identity
$$
[\rho\gt\tau]\cap[\rho\leqslant t]=[\rho\leqslant t]\cap\bigcup_{s\in\mathbb Q,s\lt t}[\rho\gt s]\cap[\tau\leqslant s].
$$
Likewise, $[\rho\leqslant \tau]$ is in $\mathcal F_\tau$ since
$$
[\rho\gt\tau]\cap[\tau\leqslant t]=[\tau\leqslant t]\cap\bigcup_{s\in\mathbb Q,s\leqslant t}[\rho\gt s]\cap[\tau\leqslant s].
$$</p>

Title: Stopping times and $\sigma$-algebras


问:
Solving a linear system of the form C .* (BX) = AX
I have formed the above problem, which involves a mix between element-wise (Hadamard) multiplication (indicated by ".*") and ordinary matrix multiplication. In this problem I know all of A, B and C but I still have no idea how to solve for X.

Dimensions, if this helps:

A = N x N (sparse)

B = N x N (sparse)

C = N x 1 (not sparse)

X = N x 1 (not sparse)

The sparse matrices above are populated only with either 0 (of course), +1, or -1.

Alternatively, the problem can also be stated in the form:

log(C) + log(BX) = log(AX)

In a simpler form of the problem, A and B were so sparse that they had only one entry (+1) per row. I found (clumsily) that I was able to go like this:

log(C) + B * log(X) = A * log(X)

... which is easy enough to solve, and gave the correct solution (I am just doing some modelling, so can check against whatever X I use). However, in the "real" problem, my clumsy attempt to pull A and B outside the log operator does not work out.

Does anyone have any ideas on how I can go about solving this problem? It seems I need either

Some fancy method to combine Hadamard and ordinary matrix multiplication

Some fancy method to express the logarithm of ordinary matrix multiplication

Something else I have no idea about (likely)

Many thanks in advance!

答:
The Hadamard product of two vectors is the same as the matrix product of one vector with the diagonal matrix of the other vector. Namely
$$
C.*(BX) = D_CBX
$$
where
$$
D_C = \begin{bmatrix}
c_1&0&\dots&0\\
0&c_2&&\vdots\\
\vdots&&\ddots&0\\
0&\dots&0&c_N
\end{bmatrix}
$$

Then the equation reduces to
$$(D_CB-A)X = 0$$


问:
Proof about the dimension of a vector space
In Gilbert Strang's Linear Algebra and Its Applications (4th edition), he states:

If $v_1,...,v_m$ and $w_1,...,w_n$ are both bases for the same vector space, then m=n. The number of vectors is the same

He then goes onto the proof of this theorem:

He expresses the vectors $w_1,...,w_n$ as a combination of $v_1,...,v_n$:

$W=\begin{bmatrix} &&&\\w_1 & w_2 & ... & w_n\\&&&\end{bmatrix}=\begin{bmatrix}&&\\v_1 & ... & v_m\\&&\end{bmatrix}\begin{bmatrix}a_{11}&&&&&\\\vdots&&&&&\\a_{m1}&&&&& \end{bmatrix}=VA$

Here, $A$ is a $m$ by $n$ matrix, so it must have a nonzero solution to $Ax=0$. Therefore, $W$ and $VA$ is multiplied by this nonzero solution:

$VAx=0=Wx$

Hence, the vectors $w_1,...,w_n$ are dependent and cannot be a basis.

Similarly when $m>n$, the same steps can be applied and there will also be a contradiction. So $m=n$ to avoid any contradiction.

However, when $A$ is a non-invertible matrix (where $A$ is $m$ by $n, \ m=n$), that also means that there will be a nonzero solution.

Do we just assume that $A$ will always be invertible when it is a square matrix?

答:
It's not an assumption. You've proved that $A$ is square; you could go on and prove it's invertible. The same argument applies: if $Ax=0$ for some nonzero $x$ then $Wx=0$, which is a contradiction. So $Ax$ must be nonzero for all nonzero $x$, which means $A$ must be invertible.

This wasn't relevant to the statement being proved, which is why it wasn't mentioned, but it's certainly a worthwhile thing to know.


问:
Independent and Identically Distributed Random Variables, with 0.01 and 0.99
good day

Can anybody help me? thanks

Let ${X_i}$ Independent and Identically Distributed Random Variables as normals with mean $μ$ and standard deviation 2, What is the minimum value of n such that $P(|\dfrac{S_n}{n}-μ|<.01) \geq .99$

Thanks

Independent and Identically Distributed Random Variables

答: $\frac{S_n}{n}$ is normal with mean $\mu$ and standard devation $\frac{2}{\sqrt{n}}$.  To get your answer look up table of normal distribution.


答: For any integer $N$, we have $floor(N)=N$. Also if $x$ and $n$ are integers $nx$ is an integer, since integers are closed with respect to multiplication. So $$floor(nx) = nx = n(x) = n (floor(x)).$$

Title: Floor property In general, ⌊ n x ⌋ = n ⌊ x ⌋ .

Body: I am trying to proof the above statement. I have concluded this statement is true if x and n is an integer and real. For example if I take n=3 and x =2 then ⌊ 3. 2 ⌋ = 3 ⌊ 2 ⌋ which is ⌊6⌋ = (3 . 2)  which is 6=6 which proves the statement. But how do I prove this statement without taking n=3 or n=4 etc . ? Any advice will be appreciated</p>


问:
Evaluation of Complex number.
I want to prove that
\begin{equation}
|e^{irz}-1|\leqq re^{|y|r} \text{ for } z=x+iy, |z|=1, r>0.
\end{equation}

My attempt

\begin{align}
|e^{irz}-1|
&=|e^{-ry} \cos(rx) -1 + ie^{-ry} \sin(rx)| \\
&=\sqrt{(e^{-ry} \cos(rx) -1)^2+(e^{-ry} \sin(rx))^2} \\
&=\sqrt{e^{-2ry}-2e^{-ry}\cos(rx)+1} \\
&\leqq \sqrt{e^{-2ry}+2e^{-ry}+1} \\
&=\sqrt{(e^{-ry}+1)^2} \\
&=|e^{-ry}+1|.
\end{align}

But this doesn't seem to work.
I would like you to give me some ideas.

答:
Proof: Please note that, by $\textbf{Newton-Leibniz formula}$,

You can refer https://www.oulu.fi/sites/default/files/151/complex_book.pdf    Page73

$$e^{irz}-1=\int_{0}^{r}ize^{iuz}du.$$
So
\begin{align*}
|e^{irz}-1|
&\leq |iz|r\cdot\max_{u\in[0,r]}|e^{iuz}|\\
&=|iz|r\cdot\max_{u\in[0,r]}e^{\text{Re}(iuz)}\\
&=r\cdot\max_{u\in[0,r]}e^{-uy}\\
&\leq r\cdot\max_{u\in[0,r]}e^{u|y|}\\
&=re^{r|y|}.
\end{align*}

Autually, there is a lemma like this:

If $a,b$ are real numbers and $s=\sigma+it(\sigma>0)$,
then
$$e^{bs}-e^{as}\leq |s||b-a| e^{\max\{a,b\}\cdot\sigma}.$$
$$e^{bs}-e^{as}=\int_{a}^{b}se^{xs}dx.$$
So (W.L.G we suppose $a<b$)
\begin{align*}
|e^{bs}-e^{as}|
&\leq |s||b-a|\max_{x\in[a,b]}|e^{xs}|\\
&=|s||b-a|\max_{x\in[a,b]}e^{\text{Re}(xs)}\\
&=|s||b-a|\max_{x\in[a,b]}e^{x\sigma}\\
&=|s||b-a| e^{\max\{a,b\}\cdot\sigma}.
\end{align*}

Title: Evaluation of Complex number.


问:
Interesting BMO Final Selection Test Problem 1998 Q3

An infinite arithmetic progression, whose terms are all positive integers, contains

(i) a perfect square which is not a perfect cube, and

(ii) a perfect cube which is not a perfect square.

Prove that the arithmetic progression contains a perfect sixth power.

Finding examples where the base is the same is trivial but to find a general proof I tried looking at examples where square and cube are co-prime (hence common difference d is also co-prime with these) and using mod d and Euler's theorem to come up with a relatively short algebraic proof however, nothing has immediately opened up.

答:
We know that there exist $x$, $y$ with $x^2\equiv y^3\equiv a\pmod n$. Let $d$ be the order of $a$ mod $n$.

If $d=1$, we are done as $x^6\equiv1\pmod n$. If $n=2$, then $(xy)^6\equiv a^5\equiv a\pmod n$, and again we are done.

So assume $d>2$. For any $i,j$ we have $(x^iy^j)^6\equiv a^{3i+2j}\pmod n$. By the Chicken McNugget theorem, there exist $i,j$ such that $3i+2j=m$ for all $m>6-2-3=1$. Since $d-1>1$, we can find $i,j$ such that $3i+2j=d-1$, so
$$(x^iy^j)^6\equiv a^{d-1}\pmod n\implies\left(x^{i(d-1)}y^{j(d-1)}\right)^6\equiv a^{(d-1)^2}\equiv a\pmod n,$$
and we are done!


问:
A directed inverse limit of finite connected spaces is connected
Apparently, a directed inverse limit of finite connected spaces is connected. Does anyone have a reference?</p>

答:
Suppose $X=\lim_i X_i$ is a directed limit of finite connected spaces.

Proof — As $X_i$ are finite connected spaces, by the Mittag-Leffler condition we can replace the directed system of $X_i$ by one in which all the transition maps are surjective. (Each $X_i$, being the image of a connected space, is still connected.) As a directed limit of nonempty finite sets is nonempty, the projections $p_i : X \to X_i$ are then surjective. If $U$ and $V$ are complementary nonempty clopens in $X$, they are quasi-compact since $X$ is. Therefore since the system is directed there is an $i$ so that $U = p_i^{-1}(A)$ and $V = p_i^{-1}(B)$ for opens $A,B$ in $X_i$. The opens $A$ and $B$ are disjoint since $p_i$ is surjective and $U$ and $V$ are disjoint. Also $A$ and $B$ partition $X_i$ for the same reason. QED


问:
How to express the series of $[\tan^{-1}(x)][\tanh^{-1}(x)]$ as $x^2+\left(1-\frac{1}{3}+\frac{1}{5}\right)\frac{x^6}{3}...$
How to express the series of $[\tan^{-1}(x)][\tanh^{-1}(x)]$ as

$x^2+\left(1-\dfrac{1}{3}+\dfrac{1}{5}\right)\dfrac{x^6}{3}+\left(1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}\right)\dfrac{x^{10}}{5}+\left(1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{13}\right)\dfrac{x^{14}}{7}+...$

I have tried to use Cauchy product or long multiplication, but the result, although of course, are equivalent, is not as elegantly expressed as this expression. This is taken from Ferrar "A textbook of convergence" page 123.

For example, for $x^6$, I have $\dfrac{2}{5}-\dfrac{1}{9}$, while for $x^{10}$, I have $\dfrac{2}{9}+\dfrac{1}{25}-\dfrac{2}{21}$. I don't know how to make it pop out the partial Leibnitz series for the coefficients.

答:
Since,
for $|x| < 1$,
$\arctan(x)
=\sum_{n=0}^{\infty} \dfrac{(-1)^nx^{2n+1}}{2n+1}
$,
so
$\arctan(\sqrt{x})
=\sum_{n=0}^{\infty} \dfrac{(-1)^nx^{n+\frac12}}{2n+1}
=\sqrt{x}\sum_{n=0}^{\infty} \dfrac{(-1)^nx^n}{2n+1}
$.

Similarly,
for $|x| < 1$,
$arctanh(x)
=\sum_{n=0}^{\infty} \dfrac{x^{2n+1}}{2n+1}
$,
so
$arctanh(\sqrt{x})
=\sum_{n=0}^{\infty} \dfrac{x^{n+\frac12}}{2n+1}
=\sqrt{x}\sum_{n=0}^{\infty} \dfrac{x^n}{2n+1}
$.

Therefore
for $|x| < 1$,
$\arctan(\sqrt{x})arctanh(\sqrt{x})
=x\sum_{n=0}^{\infty}
x^n\sum_{k=0}^n \dfrac{(-1)^k}{2k+1}\dfrac1{2(n-k)+1}
$
so

$\begin{array}\\
\arctan(x)arctanh(x)
&=x^2\sum_{n=0}^{\infty}
x^{2n}\sum_{k=0}^n \dfrac{(-1)^k}{2k+1}\dfrac1{2(n-k)+1}\\
&=\sum_{n=0}^{\infty}
x^{2n+2}\sum_{k=0}^n \dfrac{(-1)^k}{2k+1}\dfrac1{2(n-k)+1}\\
&=\sum_{n=1}^{\infty}
x^{2n}\sum_{k=0}^{n-1} \dfrac{(-1)^k}{2k+1}\dfrac1{2(n-1-k)+1}\\
&=\sum_{n=1}^{\infty}
x^{2n}c_n\\
\end{array}
$

The coefficients are thus
$c_n
=\sum_{k=0}^{n-1} \dfrac{(-1)^k}{2k+1}\dfrac1{2(n-1-k)+1}
$.

According to Wolfy,
the first few terms are
$x^2 + \dfrac{13 x^6}{45} + \dfrac{263 x^{10}}{1575} +
\dfrac{36979x^{14}}{315315}+O(x^{18})
$.

It looks like
$c_{2n} = 0$
and
$c_{2n+1} \ne 0$.
The first should be
straightforward to prove
by reversing the order of summation.

A similar technique
should prove that
$c_{2n+1} > 0$.

As to a closed form
for the sum,
I don't know.


问:
How can I find the value of x given multiple Parallel Lines and multiple Transversals?
Ive been given the following problem:

I was unable to figure out how to solve the problem, and I was told the answer was the following;
However, the program has failed to explain to me exactly how it got the answer... Looking online, I can only find solutions for single transversal, 2 line problems, which aren't necessarily difficult..

My question is, how exactly, piece by piece, do I find X given multiple transversals or parallel lines?

答:
With more parallel lines or transversals you just apply the "Parallel lines cut transversals into proportional segments." theorem for more than once. (This theorem is essentially Thales' Theorem.)

I created an example in the figure below. Since you had successfully obtained $x=5$ from the lower trapezoid, we can write the following for the upper trapezoid.
$$\frac{y}{3}=\frac{10}{x}=\frac{10}{5}\implies y=6$$


答:
In general, this property does not hold for an integer $x$ and a real number $n$. For example, If I chose $x=2$ and $n=0.6$, I would have $n \lfloor x \rfloor$ = $0.6\lfloor2\rfloor = 1.2$. However, $\lfloor nx \rfloor = \lfloor0.6*2 \rfloor = \lfloor1.2\rfloor = 1$. Thus, this is not true in general.

This property does not hold true for any REAL $x$ and an INTEGER $n$, either. For example, if I chose $x=0.5$ and $n=2$, I would have $n \lfloor x \rfloor$ = $2\lfloor0.5\rfloor = 0$. However, $\lfloor nx \rfloor = \lfloor2*0.5 \rfloor = \lfloor1\rfloor = 1$. Thus, this is not true in general, either.


问:
Can someone explain to me what are these 2 statements talking about?
I have to prove that these 2 statements are equivalent, but I can't even understand them.

There exist $\epsilon_0>0$ such that for all $k\in\mathbb N$, there exist $n_k\in\mathbb N$ such that $n_k\geq k$ and $|x_{n_k}-x|\geq\epsilon_0$?

There exist $\epsilon_0>0$ and a subsequence $(x_{n_k})$ of sequence $(x_n)$ such that $|x_{n_k}-x|\geq\epsilon_0$ for all $k\in\mathbb N$.

Thank you.

答:
They both say

There are infinitely many $n$, such that $x_n$ is not an element of $(x-\epsilon_0, x + \epsilon_0)$, that is $I := \{n \in \mathbb N \mid x_n \not\in (x - \epsilon_0, x + \epsilon_0)\}$ is infinite.

In the first formulation, the point that $I$ is infinite is formulated as $I$ being unbounded in $\mathbb N$, that is for each $k$ there is some element (called $n_k$) of $I$ being larger than $k$. In the second one, the point is that with $I$ being infinite, we can write $I = \{n_1, n_2, \ldots\}$, $n_1 < n_2 < \cdots$ and obtain a subsequence $(x_{n_k})$ by using the elements of $I$ as indices.


问:
How to prove that if $S(S,f,\{p_i\})=\sum f(p_j)A(D_j)$ then $|\iint_{R}f-S|(S,f,\{p_i\})|<\varepsilon$
Let $$s(S,f,\{p_i\})=\sum_{j=1}^m f(p_j)A(D_j)$$ where $p_j\in D_j$ and $S$ is a general division of a close rectangle $R$, i.e., a finite number of domains $D_j$ each having area, and together covers $R$ and such that no pair have common interior points. The norm $d(S)$ is the largest of the diameters of the set $D_j$.  How can I prove that if $f$ is continuous on $R$, then for $\varepsilon>0$ there is $\delta>0$ such that $$\left| \iint_R f-s(S,f,\{p_j\}) \right| $$ whenever $d(S)<\delta$ ??

I saw this problem from Buck's Advanced Calculus, and I'm wondering how can I prove it.

Attempt: When one integrates over domains $D_j$ which are not rectangles, then we choose a rectangle $R_j$ which contains $D_j$ and defined $F=f$ on $D_j$ and $0$ outside $D_j$, so that $$\iint_{D_j}f=\iint_RF $$ The associated Riemann sum, is given by $$ S(N,f,\{p_{ij}\}) = \sum_{i,j}f(p_{i,j})A(R_{ij}) $$ where $A(R_{ij})$ denotes the area of rectangles and $N$ is a grid. One can think that if the difference of this two sums are equal, then we are done, I mean making small the difference
$$\left| \sum_{i,j}f(p_{ij})A(R_{ij})-\sum_{j}f(p_j)A(D_j)\right|<\varepsilon$$

The uniform continuity of $f$ can make $|f(p_j)-f(p_{i,j})|$ small but how to deal with $|A(R_{ij})-A(D_j))|$ and also is there a way of operating with only one sum in $\sum_{i,j}f(p_{i,j})A(R_{ij})$??

Here there is the same question: but in a different complicated approach (uses lower and upper bounds, I'm wondering if using the uniform continuity of $f$ could be done easier) Riemann Integrability in $\Bbb R^2$

答:
We can show that under the given hypotheses that

$$\int_R f = \int_{\cup_j D_j} f = \sum_j \int_{D_j} f$$

We will use the fact that if $f_1$ and $f_2$ are Riemann integrable on $Q$ then $\int_Q f_1 \pm f_2 = \int_Qf_1 \pm \int_Q f_2$.

First assume that $f$ is both continuous and nonnegative and consider two sets $D_1$ and $D_2$.  Let $f_{D_j}(x) = f(x)$ for $x \in D_j$ and $f_{D_j}(x) = 0$ for $x \in D_j^c$.

We have

$$f_{D_1\cup D_2}(x) = \max(f_{D_1}(x), f_{D_2}(x))= \frac{1}{2}\left(f_{D_1}(x)+f_{D_2}(x)+|f_{D_1}(x)-f_{D_1}(x)| \right),\\ \quad f_{D_1\cap D_2}(x) = \min(f_{D_1}(x), f_{D_2}(x))= \frac{1}{2}\left(f_{D_1}(x)+f_{D_2}(x)-|f_{D_1}(x)-f_{D_1}(x)| \right),$$

and, both $f_{D_1\cup D_2}$ and $f_{D_1\cap D_2}$ are Riemann integrable.

Since, $f_{D_1\cup D_2}  = f_{D_1}+ f_{D_2} - f_{D_1\cap D_2}$, it follows that if $m(D_1\cap D_2) = 0$, then

$$\int_{D_1\cup D_2} f = \int_{D_1} f +\int_{D_2} f - \int_{D_1\cap D_2} f= \int_{D_1} f +\int_{D_2} f, $$

By induction this result generalizes to

$$\int_{\cup_j D_j} f = \sum_j \int_{D_j} f,$$

We can easily prove this for the general case where $f$ is not nonnegative or nonpositive, by applying  the previous result to the positive and negative parts, $f_+ = \max (f,0)$ and $f_- = \max (-f,0)$.


问:
In a binomial experiment, which will produce a probability distribution with the smallest standard deviation, given n is the same for each situation?
A. P = 0.5

B. P = 0.4

C. P = 0.3

D. P = 0.9

E. (1-P) =0.8

No idea where to even start

答: Binomial distributions have two parameters, and are given by  $Binomial(np, np(1-p))$ where the first is the mean and the second is the variance. A has a variance of n(.5)(.5)=.25n, the largest, while D has a variance of n(.9)(.1)=.09n. Thus since n is fixed, D is the answer. Intuitively, a binomial distribution is the number of successes in n trials with probability p of success. What you found is there is less variability in the number of successes when p is large (or small) as opposed to being about even chance of success and failure.


Title: How to express the series of $[\tan^{-1}(x)][\tanh^{-1}(x)]$ as $x^2+\left(1-\frac{1}{3}+\frac{1}{5}\right)\frac{x^6}{3}...$

Body:
How to express the series of $[\tan^{-1}(x)][\tanh^{-1}(x)]$ as

$x^2+\left(1-\dfrac{1}{3}+\dfrac{1}{5}\right)\dfrac{x^6}{3}+\left(1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}\right)\dfrac{x^{10}}{5}+\left(1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{13}\right)\dfrac{x^{14}}{7}+...$

I have tried to use Cauchy product or long multiplication, but the result, although of course, are equivalent, is not as elegantly expressed as this expression. This is taken from Ferrar "A textbook of convergence" page 123.

For example, for $x^6$, I have $\dfrac{2}{5}-\dfrac{1}{9}$, while for $x^{10}$, I have $\dfrac{2}{9}+\dfrac{1}{25}-\dfrac{2}{21}$. I don't know how to make it pop out the partial Leibnitz series for the coefficients.

Body:
Hint:

$f(x)=[\tan^{-1}(x)][\tanh^{-1}(x)]$

$f'(x)=\dfrac{ \tan^{-1}(x)}{1-x^2}+\dfrac{ \tanh^{-1}(x)}{1+x^2} $

If $|x|<1$:

$\displaystyle f'(x)=\left(\sum_{n=0}^{+\infty}  \dfrac{(-1)^n}{2n+1} x^{2n+1}   \right) \left(\sum_{n=0}^{+\infty}   x^{2n}  \right)   +  \left(\sum_{n=0}^{+\infty}  \dfrac{1}{2n+1} x^{2n+1}   \right) \left(\sum_{n=0}^{+\infty} (-1)^n  x^{2n}  \right)    $

Edit to show the idea

$f′(x)=\left(x−\dfrac{x^3}{3}+\dfrac{x^5}{5}−\dfrac{x^7}{7}+\dfrac{x^9}{9}...\right)(1+x^2+x^4+x^6+x^8...)+\left(x+\dfrac{x^3}{3}+\dfrac{x^5}{5}+\dfrac{x^7}{7}+\dfrac{x^9}{9}...\right)(1−x^2+x^4−x^6+x^8...) $

By using Cauchy product:

$f′(x)=2x+ 2\left(1−\dfrac{1}{3}+\dfrac{1}{5}\right)x^5+2\left(1 −\dfrac{1}{3}+\dfrac{1}{5}−\dfrac{1}{7}+\dfrac{1}{9}\right)x^9+...$

Then we integrate:

$f(x)=x^2+\left(1−\dfrac{1}{3}+\dfrac{1}{5}\right)\dfrac{x^6}{3} +\left(1 −\dfrac{1}{3}+\dfrac{1}{5}−\dfrac{1}{7}+\dfrac{1}{9}\right)\dfrac{x^{10}}{5} +...  $


问:
Prove that for all natural numbers $x$, $y$, $z$, and $w$ that $gcd(x,y)gcd(z,w) | gcd(xz,yw)$.
Prove that for all natural numbers $x$, $y$, $z$, and $w$ that  $gcd(x,y)gcd(z,w) | gcd(xz,yw)$. I established a prime factorization product for each variable and I've tried using GCD's via prime factorizations to prove this but I get stuck when I multiply $gcd(x,y) \times gcd(z,w)$ and I'm not sure what to do with $\prod_{j=1}^{\infty}{p_i^{min(a_i, b_i) + min(c_i, d_i)}}$. Any help is appreciated!

答:
There are three key facts:

First, $\gcd(a,b)$ divides both $a$ and $b$ as it is a common divisor.

Second, if $a|b$ and $a|c$ then $a|\gcd(b,c)$, by definition of the greatest common divisor.

Third, if $a|b$ and $c|d$ then $ac|bd$, for if $b=ax$ and $d=cy$ , we have $bd=ac(xy).$

Notice that $\gcd(x,y)$ divides $x$ and $y$.

Similarly, $\gcd(z,w)$ divides $z$ and $w$.

Hence, $\gcd(x,y)\gcd(z,w)$ divides $xz$ and $yw$. This implies $\gcd(x,y)\gcd(z,w)|\gcd(xz,yw).$

As for your particular approach, it suffices to observe that the exponent of $p_i$ in the prime factorization of $xz$ is $a_i+c_i$. In particular, this is at least $\min(a_i,b_i)+\min(c_i,d_i)$. This implies $$ \gcd(x,y)\gcd(z,w)=\prod_{j=1}^\infty p_i^{\min(a_i,b_i)+\min(c_i,d_i)}|\prod_{j=1}^\infty p_i^{a_i+c_i}=xz.$$
Similarly, $\gcd(x,y)\gcd(z,w)$ divides $yw$.


答:
The variance for a binomial distribution is: $np(1-p)$. Therefore the standard deviation will be $SD = \sqrt{np(1-p)}$. The question asks you to minimize $SD$, given that $n$ is fixed, i.e. the only variable is $p$.
Let $$f(p) = \sqrt{np(1-p)} = \sqrt{np-np^2}$$
Now in order to minimize the function $f$, you can take the first-order derivative and equate it to $0$, thereby getting a condition on $p$.

Title: In a binomial experiment, which will produce a probability distribution with the smallest standard deviation, given n is the same for each situation?

Body:
A. P = 0.5

B. P = 0.4

C. P = 0.3

D. P = 0.9

E. (1-P) =0.8

No idea where to even start


问:
Is a class of sets a set if its union is a set?
I am wondering if a class of sets X is a set given that $ \cup X $ is a set.

Going in the other direction is trivial -- if X is a set then $\cup X$ is a set by the axiom of union. However, how would one prove it in this direction?

My initial thought is that if $\cup X$ is a set, we can construct its power set $P(\cup X)$. Then, to show X itself (the class of sets) is a set, it would suffice if it were a member of $P(\cup X)$. I'm not sure if this is true or if it is, how I would show that it is?

答:
The real question is how you can define the class of sets in the first place without knowing that it is a set. Presumably you have some formula $\varphi$ that defines the members of the class, and you can prove

$$\exists x\forall y\big(y\in x\leftrightarrow\exists z(\varphi(z)\land y\in z)\big)\,;$$

that is, you can prove that there is a set that behaves like the union of the class of sets satisfying $\varphi$. In that case you can let $a$ be such a set $x$, so that

$$\forall y\big(y\in a\leftrightarrow\exists z(\varphi(z)\land y\in z)\big)\,,\tag{1}$$

and define a set

$$X=\{y\in\wp(a):\varphi(a)\}\,;$$

Clearly every member of $X$ satisfies $\varphi$. Now suppose that $z$ is a set, and $\varphi(z)$ holds. Then $(1)$ implies that $\forall y\in z\,(y\in a)$, i.e., that $z\in\wp(a)$, and hence that $z\in X$, so $X$ is indeed the set of sets satisfying $\varphi$.


问:
Probability interpretation
Supose we have a mouse on a labyrinth and we associate $p$ as the probability of the mouse escaping this labyrinth this turn, and $(1-p)$ for the probability of him not escaping that turn. Call $X$ the random variable associated with the number of turns that the mouse takes in order to escape.
If I want him to take, at best, $4$ turns to escape the labyrinth then I would consider the possibilities of:

$$P(X \leq 4)= P(X = 4)+P(X = 3)+P(X = 2)+P(X = 1)$$

After some calculations I arrive at the probability:

$$ P(X \leq 4)= 1-(1-p)^4 $$

On the other hand:

$$ P(X \leq 4)=1-P(X > 4)$$

Which leads to:

$$ P(X > 4) = (1-p)^4$$

This gives us the probability of him failling more then $4$ times, given that it is the sum:

$$ P(X > 4) = P(X=5)+P(X = 6)+...=p(1-p)^4+p(1-p)^5+...=p \sum_{i=4}^n(1-p)^{i}=(1-p)^4$$

Can someone give me a intuitive interpretation for this? It seems a little odd that the summation would tend to $(1-p)^4$ when starting at $4$.

答:
You have that $\mathsf P(X=n)= p(1-p)^{n-1}$ and so $\mathsf P(X\gt n)=p\sum_{k=n}^\infty(1-p)^k$.

Thus $\mathsf P(X>0)= p\sum_{k=0}^\infty(1-p)^k$.  However, this almost surely equals $1$.  Theoretically, the mouse will escape sometime after the experiment starts.

$$\begin{align}\mathsf P(X>n)&=p\sum_{k=n}^n(1-p)^k\\&=(1-p)^n\cdot p\sum_{k=n}^\infty(1-p)^{k-n}\\&=(1-p)^n\cdot p\sum_{j=0}^\infty (1-p)^j\\&=(1-p)^n\cdot\mathsf P(X>0)\\&=(1-p)^n\end{align}$$

In short...

The probability that the mouse escapes after $n$ failures is the probability of having $n$ consecutive failures times the probability the probability that it escapes sometime after that.

And to check this intuition mathematically, we use the derivation for the closed form of a Geometric Series. $$\begin{align}\sum_{k=0}^\infty (1-p)^k &= (1-p)^0+(1-p)\sum_{k=1}^\infty (1-p)^{k-1}\\ &=1+(1-p)\sum_{k=0}^\infty(1-p)^k \\[2ex]\therefore\quad p\sum_{k=0}^\infty (1-p)^k &= 1 \end{align}$$

Title: Probability interpretation


问:
Numerical boundary-finding algorithms
I'm very familiar with numerical root-finding algorithms like Brent's method (scipy.optimize.brentq) or Chandrupatla's method. What if I am looking for a boundary numerically?

Suppose for example that I have an unknown monotonically increasing function $y = g(x)$ between $x=a$ and $x=b$ as represented by the boundary of some region $f(x,y) \ge 0$.

I cannot evaluate $g(x)$ directly; instead, I can evaluate $f(x,y)$ and compare it to zero. For any fixed value of $x=x_0$ I can use a root-finding method to determine $y$ such that $f(x_0,y)=0$.

Aside from sweeping across the interval $x\in[a,b]$ and repeating a root-finding method for each value of $x$, is there a more efficient way? (Suppose I know that $g(x)$ is continuous, for example, and that its derivative $dg/dx$ is "mostly" continuous; then the solution to $y=g(x_0+\Delta x)$ is close to $y=g(x_0)$ .)

答:
If you know that $g$ is Lipschitz continuous with Lipschitz constant $K$, then

$$g(x)-K\Delta\le g(x+\Delta)\le g(x)+K\Delta x\tag1$$

will provide reasonably tight bounds if $\Delta$ is sufficiently small.

If you know that $g$ is differentiable, then the first order Taylor

$$g(x+\Delta)=g(x)+g'(x)\Delta+o(\Delta)\tag2$$

is an even better estimate. This may be supplemented into the above bounds to give even faster convergence, or extended further if more derivatives are known.

If $K$ is unknown, then it may suffice to use an initial estimate $\tilde K>0$. Whenever $(1)$ fails to hold, double $\tilde K$ and try again. Another simple alternative strategy if $\Delta x$ is always very small would be to use $\tilde K=k|g'(x)|$ for some $k>1$.

If $g'(x)$ is unknown, then it may suffice to instead use a finite difference estimate in its place from the last two points. Higher order estimates can of course be derived by using more points too.

If $g$ is continuously twice differentiable then we actually know the error of $(2)$ to be bounded by $L\Delta^2$ for some $L>0$, even if you replace $g'(x)$ by an estimate with $\mathcal O(\Delta)$ error. Following a similar situation as with $K$ and $(1)$, you could extend $(2)$ to give the bounds

$$g(x)+g'(x)\Delta x-L\Delta^2\le g(x+\Delta)\le g(x)+g'(x)\Delta x+L\Delta^2\tag3$$

and so on for higher order derivatives. Analogously, such an $L$ may be estimated in the same ways as $K$.

Title: Numerical boundary-finding algorithms


问:
Random walk - Solution verification
A particle moves randomly 2n times. Each time it chooses one direction: north, south, west, east with even probability and without effect from previous choices, it then moves one step in that direction. What is the probability that the particle will end up in the same spot it started?

What I did:

There are $4$ choices for each of $2n$ steps. Therefore $|\Omega|=4^{2n}$.

We'll mark $N=|steps\space north|,\space S=|steps\space south|,\space W=|steps\space west|,\space E=|steps\space east|$.
To end up in the same spot we started, we need $N=S$ and $W=E$. Therefore $N+W=n$ and $S+E=n$. Therefore if we choose $n$ of the $2n$ steps $[\binom {2n}{n}]$ to only be composed of steps $north\cup west$ ($2$ choices for each of $n$ steps $[2^n]$). Out of the $n$ "leftover" steps, $S$ are south $[\binom {n}{S}]$. Therefore:
$$P(\text{The particle ends up where it started})=\frac{ \sum\limits_{S=0}^n\binom {2n} {n} 2^n \binom {n}{S}}{4^{2n}}$$

Thank you for your time and effort.

答:
Assume that $n=2N$. There are ${2N\choose N}$ ways to choose the times of the N-or-W steps. For each $0\leqslant k\leqslant N$, there are ${N\choose k}$ ways of choosing $k$ times of N-steps amongst the $N$ times of the N-or-W steps. Then one must also choose $k$ times of S-steps amongst the $N$ times of the S-or-E steps. There are ${N\choose k}$ ways of doing that. Hence the number of paths of length $2N$ ending at $(0,0)$ is
$$
P_N={2N\choose N}\sum_{k=0}^N{N\choose k}^2,
$$
and the probability you are after is $P_N/4^{2N}$. (Of course, the sum over $k$ in $P_N$ has a simpler expression and the resulting formula for $P_N$ can be recovered more directly but I am not sure you are after this...)</p>

Title: Random walk - Solution verification