问:
How to show that $f : {}^\ast \Bbb R \to {}^\ast \Bbb R$ is bounded, if it obtains a limited value everywhere?
Let $f :  {}^\ast \Bbb R \to  {}^\ast \Bbb R$. For all $x \in  {}^\ast \Bbb R $, there exists $y \in \Bbb R $ such that $f(x) \leq y$. How to show that there exists  $r \in \Bbb R$, such that for all $z \in  {}^\ast \Bbb R$  we have $f(z) \leq r$?

I know this can be shown by a direct application of the dual form of idealization principle. Is there any other way to show this, say, using ultrapower construction?

Moreover, if $f$ is continuous, is it true that $f$ necessarily obtain a maximal value?

答:
The set of upper bounds for the function contains all unlimited numbers.  Therefore by the underspill principle, it must contain a finite number, as well.

Of course, this is in the assumption that the function is internal.  Otherwise there are obvious counterexamples.


问:
IMO long list spatial Combinatorial problem
Can anyone send a solution or hint to these problem? I tried a greedy algorithm but couldn’t make it work

答: Form the convex hull of the $2n$ points, and let $x$ and $y$ be adjacent extreme points. Then you can join $x$ with $y$, and pair up the remaining $2n-2$ points by induction.

ContentLicense: CC BY-SA 4.0


问:
If $\gcd(a,x)=1$, then $a\mathbb N+b, x\mathbb N+y$ intersects.
I have tried to think as intersections of two lines but for $2x+3$ and $5x+7$ is failed to have integer intersections (but not sure it is enough counterexample).

Is the above generalisation correct? Can you give a counter-example, or a hint for how to prove it?

答: This is true, and in fact you only need the first condition $\gcd(a,x)=1$. This follows from the Chinese remainder theorem. For example, $2\mathbb N+3$ and $5\mathbb N+7$ both include all points in $10\mathbb N+7$.


答: Hint: Find integers $A, X$ such that $ a  A +xX = y - b$.

Title: If $\gcd(a,x)=1$, then $a\mathbb N+b, x\mathbb N+y$ intersects.

Body:
I have tried to think as intersections of two lines but for $2x+3$ and $5x+7$ is failed to have integer intersections (but not sure it is enough counterexample).

Is the above generalisation correct? Can you give a counter-example, or a hint for how to prove it?


问:
Local property of localization and tensor product
Let $M$ be an $R$-module, and let $U\subset R$ be a multiplicatively closed subset. Let $M_U$ be the localization of $M$ at $U$, and let $R_U$ be the localization of $R$ at $U$. Then we have the "well known" isomorphism
$$
M_U\cong M\otimes_R R_U.
$$

I could prove this by constructing concrete isomorphism with both direction.

But I heard that this isomorphism can be shown by universal property of localization and tensor product.

How can I do that? Thank you for your help.

Local property which I want to use is,

The ring homomorphism j : R → R maps every element of S to a unit in R* = S^ −1R. The universal property is that if f : R → T is some other ring homomorphism into another ring T which maps every element of S to a unit in T, then there exists a unique ring homomorphism g : R* → T such that f = g∘j.*

How can I show

$$
 M\otimes_R R_U→M_U
$$

and

$$
M×R_U→M\otimes_R R_U
$$
preserves units in $U$?

答:
You can't show this with this universal property, as it only refers to rings.

However there is a universal property of $M_U$ which is:

The map $M\to M_U$ induces an isomorphism $\hom_R(M_U,N)\to \hom_R(M,N)$ for any $N$ such that every element of $U$ acts invertibly on $N$ (and of course every element of $U$ acts invertibly on $M_U$).

In other words, if $M\to N$ is a morphism, and every element of $U$ acts invertibly on $N$, then it extends uniquely to a morphism $M_U\to N$.

Now you can use this universal property to show that $M\to M\otimes_R R_U$ satisfies the same one. You'll have to use :

the universal property of the tensor product;

the universal property of $R_U$ as an $R$-module, not as a ring;

the universal property of $R$ as an $R$-module;

basic properties of the tensor product.


问:
Differentiation of Matrix Norm
I would like to differentiate the induced norm $\|I - \gamma P\|_2$ with respect to $\gamma$, where $I$ and $P$ are $n\times n$ matrices and $\gamma$ is a scalar. How do I proceed?

The Matrix Cookbook doesn't seem to address this. This question seems related, but I think it's unlikely that the result generalizes to matrices.

Thanks in advance.

答:
$\def\p#1#2{\frac{\partial #1}{\partial #2}}\def\pp#1#2{\frac{d#1}{d#2}}$
The induced ${\tt2}$-norm is identical to the Schatten ${\infty}$-norm (also known as the spectral norm).

The spectral norm of $A$ can be written in terms of its SVD
$$\eqalign{
A &= USV^T = \sum_{k=1}^{rank(A)} \sigma_k u_k v_k^T\\
\|A\| &= \sigma_1 \qquad\big(\sigma_1\ge\sigma_2\ge\ldots\ge\sigma_n\ge 0\big) \\
}$$
and its differential and gradient can be written using the first columns of $U$ and $V$
$$\eqalign{
d\,\|A\| &= u_1^T\,dA\;v_1 \\
\p{\|A\|}{A} &= u_1v_1^T \\
}$$
Setting $\,A=\gamma P-I\;$ yields the desired derivative
$$\eqalign{
d\,\|A\| &= u_1^T\big(P\,d\gamma\big)\,v_1 \\
\pp{\|A\|}{\gamma} &= u_1^TPv_1 \\
}$$

Title: Differentiation of Matrix Norm


答:
First since $gcd(x,a)=1$, there exist $\alpha$, $\beta$, such that
(Bézout's identity)$$\alpha x - \beta a = 1$$ Now multiply both sides by $k = b-y$.

$$(k \alpha ) x - (k \beta)a = b-y$$
$$(k \alpha ) x + y = (k \beta)a +b$$


问:
sequence of monotone functions on a compact interval converges uniformly to a *monotone* function?

Let $(f_n)$ be a sequence of monotone functions on a compact interval $I$ which converges pointwise to a continuous function $f$. Show that $f$ is monotone and that $(f_n)$ converges uniformly to $f$.

I am unable to show that $f$ is monotone, considering that each $f_n$ can be either increasing or decreasing and I don't know if this is even true.

I know that if all of them are increasing/decreasing then it is true.

答:
It is true. If $f$ is constant then there is nothing to show. Otherwise, all but finitely many $f_n$ have the same monotony direction:

If $f(a) < f(b)$ for some $a<b$ then the pointwise convergence of $f_n$ to $f$ implies that $f_n(a) < f_n(b)$ for all sufficiently large $n$, so that all but finitely $f_n$ are (weakly) increasing, which implies that $f$ is (weakly) increasing.

Similarly if $f(a) > f(b)$ for some $a < b$ then all but finitely $f_n$ are (weakly) decreasing, which implies that $f$ is (weakly) decreasing.


问:
Counting the number of good functions
Hy, happy new year everyone.

I have been stuck on the following problem for a while now, so I am posting it here in other to discuss it.

A function $g: [[1,n]] \to [[0,n]]$ is called good if :

$$\forall j \in[[1,n]] , \exists i~ \text{integer} \geq 0 , g^{i} (j)=0 $$

where $g^{i}=g\circ \dots \circ g ~~(i ~~\text{times})$

How many  such good functions  there is?

答:
First $g(x)\ne x$ for all $x$, and $g^i(x)=0 \Rightarrow i\leq n$ and there exist a number x such that $g(x)=0$. Let $k_j$ denote number of numbers $x$ s.t. $g^j(x)=0$. So to construct such function, we first select $k_1$ numbers $x$ and force $g(x)=0$ in $\binom{n}{k_1}$ modes. Then we choose $k_2$ numbers and force $g(x)\in g^{-1}(0)$ in $\binom{n-k_1}{k_2}\times k_1^{k_2}$ modes and so on... . Finally the number of functions is:
$$\sum_{k_1\geq 1, k_j\geq0 , j=2,3,\cdots n}\binom{n}{k_1}\times \binom{n-k_1}{k_2}\times k_1^{k_2}\times \binom{n-k_1-k_2}{k_3}\times k_2^{k_3}\cdots \binom{n-k_1-k_2-\cdots k_{n-1}}{k_n}\times k_{n-1}^{k_n}=\sum_{k_1\geq 1, k_j\geq0 , j=2,3,\cdots n}\binom{n}{k_1,k_2,\cdots ,k_{n}}\times  k_1^{k_2}\times k_2^{k_3}\cdots \times k_{n-1}^{k_n}$$

Title: Counting the number of good functions


问:
When is an initial topology pseudocompact?
In my previous question I asked if there are well-known topologies on $[0,1]$ that make it pseudocompact, but not compact. For that I was now considering initial topologies on $[0,1]$. Suppose I have the functions $f_i:[0,1]\rightarrow\mathbb{R}$ and I consider the smallest topology on $[0,1]$ with respect to which each $f_i$ is continuous. For this topology to be necessarily pseudocompact, I take each $f_i$ to be bounded. Is there any way to tell whether each function continuous in this topology will be bounded?

答:
Take the topology generated by $\{(1/n,1]:n\in\mathbb Z^+\}\cup\{0\}$, under this topology $[0,1]$ is not compact and it is pseudo compact:

Take a continuous function (under the new topology) $f:[0,1]\to \mathbb R$ and two points $x\in(0,1]$ then if $f(x)\neq f(1)$ there would be an open set $B$ containing $f(x)$ but not $f(1)$ so $x\in f^{-1}(B)$ but $1\notin f^{-1}(B)$ which is impossible so $f|_{(0,1]}$ is constant so $\text{im}(f)=\{f(0),f(1)\}$.


问:
Infinite Non-abelian group with every non-identity has infinite order.
Is there any infinite Non-abelian group whose every non-identity element has infinite order?</p>

答:
Functions from reals to reals of the form $f(x)=mx+b$ with $m>0$.

$f(g(x))=m(nx+c)+b = mnx + (mc+b)$ and $mn > 0$ and the other requirements are equally simple to check.


问:
Fractal Surface Area of Lung
I'm preparing for an upcoming exam and one of the questions is the following;

"A typical human lung has a volume of approximately 5 litres, and a fractal dimension of
approximately 2.97. One way of measuring the surface area $S$ of the alveoli (the cavities
where the gas exchange takes place) would be to wrap the lungs in a sheet of paper of
some thickness $\varepsilon$ and measure how much paper is needed to completely cover the lungs.
Derive an expression relating the surface area $S$ to the thickness of the paper $\varepsilon$, and hence
show that the maximum surface area is approximately $80m^2$
, once the size of the paper
approaches that of the diameter of a lung cell (approximately $50\mu m$).
"

We have done stuff in class to do with box counting dimensions and was wondering if this question had anything to do with that? I have tried doing this myself but I am finding this very difficult to make any progress. Any suggestions, or a solution to the problem would be greatly appreciated!

答:
By the definition of fractal dimension (via "shrinking rulers", aka "how long is the coastline of britain"), the surface area of the lung with an $\epsilon$-length ruler is approximately $S = S_0 \left(\frac{1}{\epsilon}\right)^{2.97}$.

A sphere of volume $5$ liters ($0.005 m^3$) has a radius approximately $0.106m$ (almost $11cm$, which is the right order of magnitude for an average adult human) and surface area approximately $0.141m^2$

Let $\epsilon = 50 \mu m = 0.00005 m$ with $S = 80m^2$, then compute $S_0 = 80m \times (0.00005m)^{2.97} = 1.35\times 10^{-11} m^{4.97}$.

Substituting the volume of the sphere $S = 0.141m^2$ then gives $\epsilon_0 = \left(\frac{1.35\times10^{-11}m^{4.97}}{0.141m^2}\right)^\frac{1}{2.97} = 0.000422m = 0.422mm$ as the initial thickness of the paper, which is the right order of magnitude for regular paper.

But if you started with paper twice as thick (I even tried starting with paper 10cm thick...) you'd end up with an area $2^{2.97}$ times bigger than $80m^2$, so I think this is a bad question with insufficient information.

Title: Fractal Surface Area of Lung


问:
Book recommendation for a third course in abstract algebra
I completed an abstract algebra course on groups, and a second course on rings and fields a few semesters back and I want to take a third course next semester. For the previous two courses, I followed mainly I. N. Herstein and J. A. Gallian books (and also looked into Dummit & Foote and J. Rotman, but not in detail).

For the third course, I'm a little confused and I need some book recommendations. The syllabus covers these topics :

Rings, ideals, quotient rings, homomorphisms and isomorphisms, prime ideals, maximal ideals, integral domains, field of fractions. (recap)

Modules, submodules, quotient modules, morphisms, tensor product, flat modules.

Chain condition.

Completion, localization, valuation.

Regularity, introduction to dimension theory.

答: Dummit and Foote will serve you in good stead for a long time in algebra, although no book is certainly perfect. With the exception of their treatment of tensor products which I am not a fan of, I think the rest of their explanations are quite nice and good for self studying. Ultimately, assuming cost is not an object (or if you are getting these from a library or libgen) it is helpful to have a few different books available so you can look to a different exposition of a topic if you get stuck. Having multiple books will also help expose you to a greater variety of notation, which while confusing at first will be helpful in the long run.</p>


问:
What is the probability of rolling at most three sixes in 5 independent casts of this die?
A die is loaded in such a way that the probability of the face with j dots
turning up is proportional to j2 for j = 1, 2, 3, 4, 5, 6. What is the probability
of rolling at most three sixes in 5 independent casts of this die?

I have solved for $p(j)$:

$p(j)=1/91*j^2$

I don't know how to proceed further? I think binomial distribution will be used to solve it, but I don't know how to proceed. Any hint or help?

答: Note that the probability of getting a six is $\frac{6^2}{91}=\frac{36}{91}$. If $X$ is the number of sixes in the 5 independent casts, then $X\sim B\left(5,\frac{36}{91}\right)$. The probability you want can be found by $$\mathbb P(0\le X\le 3)=\sum_{k=0}^3\binom{5}{k}\left(\frac{36}{91}\right)^k\left(1-\frac{36}{91}\right)^{5-k}\approx 0.9163$$


问:
Wondering why the induction method of proof is meant to prove something.
One thing I find weird is induction proof, I understand how to use it fully and it's not that it is hard, it's just that it doesn't come across as actual "proof".
Let me give you an example:

I was trying to prove the binomial theorem today,

$$
(a+b)^n=
\begin{equation}
\sum_{j=0}^{n} \binom{n}{j}a^{n-j}b^j
\end{equation}
$$

In a nutshell, I spent $30$ minutes trying to find a (satisfactory - at least to me if that makes sense) proof without realizing that the question was asking for an induction proof which when I found out, I easily came to the answer, but that's not the point, the point is that the answer (that I and the book came to) was

For ($a+b)^1$ or $n=1$, the binomial theorem is clear.  No question there.

For $(a+b)^{n+1}$, we get

$$
(a+b)^{n+1}=
\begin{equation}
\sum_{j=0}^{n+1} \binom{n+1}{j}a^{(n+1)-j}b^j
\end{equation}
$$

My question is, from this proof, how can we still be satisfied that $n=n+1$ clears the theorem? Thank you.

P.S. I am not sure if I made it very obvious, but $\binom{n}{j}$ is a binomial coefficient.

答:
Comments have clarified that the question is how we separate true consequences of induction from false ones. A valid proof needs two ingredients: the base step must be shown to work, and so must the inductive step.

The base step here is $(a+b)^0=\sum_{j=0}^0\binom{0}{j}a^{n-j}b^j$, which simplifies to $1=1$ so is true. With the inductive step, it helps to use slightly different notation: show if the case $n=k$ works so does the case $n=k+1$. In this problem, this looks like$$\begin{align}(a+b)^{k+1}&=(a+b)(a+b)^k\\&=(a+b)\sum_{j=0}^k\binom{k}{j}a^{k-j}b^j\\&=\sum_{j=0}^k\binom{k}{j}a^{k+1-j}b^j+\sum_{j=0}^k\binom{k}{j}a^{k-j}b^{j+1}\\&=\sum_{j=0}^k\binom{k}{j}a^{k+1-j}b^j+\sum_{j=0}^k\binom{k}{j}a^{k+1-(j+1)}b^{j+1}\\&=\sum_{j=0}^k\binom{k}{j}a^{k+1-j}b^j+\sum_{j=1}^{k+1}\binom{k}{j-1}a^{k+1-j}b^j\\&=\sum_{j=0}^{k+1}\binom{k}{j}a^{k+1-j}b^j+\sum_{j=0}^{k+1}\binom{k}{j-1}a^{k+1-j}b^j\\&=\sum_{j=0}^{k+1}\left(\binom{k}{j}+\binom{k}{j-1}\right)a^{k+1-j}b^j\\&=\sum_{j=0}^{k+1}\binom{k+1}{j}a^{k+1-j}b^j,\end{align}$$where the second $=$ assumes the case $n=k$.

OwnerDisplayName: anon

LastEditorDisplayName: anon


问:
Origin of Yor's formula
For any two semimartingales $X$ and $Y$ it is true that
$$ \mathscr{E}(X)\mathscr{E}(Y) = \mathscr{E}(X+Y+[X,Y]);\tag{1}\label{1}$$
see for example II.8.19 in Jacod, Jean; Shiryaev, Albert N., Limit theorems for stochastic processes., Grundlehren der Mathematischen Wissenschaften. 288. Berlin: Springer. xx, 661 p. (2003). ZBL1018.60002.
Some authors call \eqref{1} the Yor formula. What is the original source of this formula and who has given it its name?

答:
I have found it: In
Jacod (1979) Calcul stochastique et problèmes de martingales, the formula is credited to Proposition 4 in  Yor (1976) Sur les intégrales stochastiques optionnelles et une suite remarquable de formules exponentielles.

Title: Origin of Yor's formula


答:
The "proof" of the inductive step is either incomplete or plain wrong. In the correct proof, we should assume:

$$(a+b)^n=\sum_{j=0}^n{n\choose j}a^{n-j}b^j$$

and prove the same for $n+1$, i.e.

$$(a+b)^{n+1}=\sum_{j=0}^{n+1}{{n+1}\choose j}a^{(n+1)-j}b^j$$

This is normally done this way:

$$\begin{array}{rcl}(a+b)^{n+1}&=&(a+b)^n\cdot (a+b)\\&=&\left(\sum_{j=0}^n{n\choose j}a^{n-j}b^j\right)\cdot(a+b)\quad\text{(here we used the inductive hypothesis)}\\&=&\sum_{j=0}^n{n\choose j}a^{n-j+1}b^j+\sum_{j=0}^n{n\choose j}a^{n-j}b^{j+1}\\&=&a^{n+1}+\sum_{j=1}^{n}{n\choose j}a^{n-j+1}b^j+\sum_{j=0}^{n-1}{n\choose j}a^{n-j}b^{j+1}+b^{n+1}\\&=&a^{n+1}+\sum_{j=1}^{n}{n\choose j}a^{n-j+1}b^j+\sum_{j=1}^n{n\choose {j-1}}a^{n-(j-1)}b^j+b^{n+1}\\&=&a^{n+1}+\sum_{j=1}^{n}\left[{n\choose j}+{n\choose {j-1}}\right]a^{(n+1)-j}b^j+b^{n+1}\\&=&a^{n+1}+\sum_{j=1}^{n}{{n+1}\choose j}a^{(n+1)-j}b^j+b^{n+1}\quad\text{(see Note below)}\\&=&\sum_{j=0}^{n+1}{{n+1}\choose j}a^{(n+1)-j}b^j\end{array}$$

Note: At the critical place, we used the equality ${n\choose{j-1}}+{n\choose j}={{n+1}\choose j}$ known from combinatorics. We also used the fact that ${n\choose 0}={{n+1}\choose 0}={n\choose n}={{n+1}\choose{n+1}}=1$ liberally throughout the proof.

What seems to me to have been done in their solution is: just replace $n$ with $n+1$ and - voila, job done. It does not work as simply as that. Replacing $n$ with $n+1$ just converts your assumption into your (hopefully) conclusion, but you have to connect the two with a sound proof.

Title: Wondering why the induction method of proof is meant to prove something.

Body:
One thing I find weird is induction proof, I understand how to use it fully and it's not that it is hard, it's just that it doesn't come across as actual "proof".
Let me give you an example:

I was trying to prove the binomial theorem today,

$$
(a+b)^n=
\begin{equation}
\sum_{j=0}^{n} \binom{n}{j}a^{n-j}b^j
\end{equation}
$$

In a nutshell, I spent $30$ minutes trying to find a (satisfactory - at least to me if that makes sense) proof without realizing that the question was asking for an induction proof which when I found out, I easily came to the answer, but that's not the point, the point is that the answer (that I and the book came to) was

For ($a+b)^1$ or $n=1$, the binomial theorem is clear.  No question there.

For $(a+b)^{n+1}$, we get

$$
(a+b)^{n+1}=
\begin{equation}
\sum_{j=0}^{n+1} \binom{n+1}{j}a^{(n+1)-j}b^j
\end{equation}
$$

My question is, from this proof, how can we still be satisfied that $n=n+1$ clears the theorem? Thank you.

P.S. I am not sure if I made it very obvious, but $\binom{n}{j}$ is a binomial coefficient.


问:
Equivalent condition for a presheaf to satisfy the gluing axiom using the etale space
Let $F$ be a presheaf over $X$. I'm trying to prove that if $F$ satisfies the first sheaf axiom (locality in the wikipedia terminology), then it satisfies the gluing axiom iff the morphisms $\varepsilon_U: F(U) \rightarrow \hat{F}(U)$ defined by $\varepsilon_U (f)(x):= f_x$ ($f_x$ being the stalk of $f$ at $x$) are surjective for all $U$ open in $X$.

For context, the definition of the sheafification I'm working with $\hat{F}(U) := \left\{ \alpha: U \rightarrow Et_F, \alpha \textrm{ continuous and } \alpha(x) \in F_x \forall x \right\}$, where the topology on $Et_F$ is such that the projection to $X$ as well as all functions of the form $[x \mapsto f_x]$ where $f \in F(U)$ are continuous.

I am basically having trouble using the definition of the topology on $Et_F$ to rigorously prove the claim. I argued as follows:
Let $\alpha \in \hat{F}(U)$. Then $\exists f \in F(U), \varepsilon_U(f)=\alpha$ iff $f_x = \alpha(x)$ for any $x$ iff $\forall x \in U: \exists W_x$ a nbhd. of $x$ such that $f|_{W_x} = g^{\alpha, x}$, where $g^{\alpha,x} \in F(W_x)$ and implicitly then $g^{\alpha, x}_x = \alpha(x) = f_x$ at the level of stalks. This is almost what I want, except I don't know how to show that $g^{\alpha, x}$ agree on intersections (this would show that if $F$ has the gluing axiom, then indeed the wanted $f$ exists, it is just the result of glueing all $g^{\alpha,x}$). Conversely, I  would have to show that any local $g \in F(W_x)$ are of the form $g^{\alpha,x}$ for some (same) $\alpha \in F(U)$. This, I think, is where I should use the topology on $Et_F$, but I can't figure out the technical details.

答:
Suppose that for a covering $\{U_i\}$ you have a set of compatible elements $f^i \in F(U_i).$ You wish to show that they come from one global element $f \in F(X).$ To that end it is enough to construct a global section $s \in \hat{F}(X)$ such that $s$ is $x \mapsto f^i_x$ on each $U_i$. It will define a unique $f$ such that $f_x = f^i_x$ for $x \in U_i,$ so $f$ and $f^i$ will coincide in a vicinity of each $x$, thus $f_{U_i}=f^i$.

So, for $x \in U_i$ let's just set $s(x) := f^i_x.$ This is what we look for, as long as $s$ is continuous. But the topology on Et is defined so that taking germs is a continuous map, which you wrote at the end of the second paragraph.


答: You must have a subsequence of $f_n$ that consists entirely of either monotonically increasing functions or of monotonically decreasing functions.

Title: sequence of monotone functions on a compact interval converges uniformly to a *monotone* function?

Body:
Let $(f_n)$ be a sequence of monotone functions on a compact interval $I$ which converges pointwise to a continuous function $f$. Show that $f$ is monotone and that $(f_n)$ converges uniformly to $f$.

I am unable to show that $f$ is monotone, considering that each $f_n$ can be either increasing or decreasing and I don't know if this is even true.

I know that if all of them are increasing/decreasing then it is true.


问:
There is concept of finite sets that can have only one "interpretation"?
In our mind we have a naive idea of what a set is, and in nature we can only observe something that behave like a finite set, ZFC (or set theories in general) tries to catch these properties in axioms. Like Peano axioms describe the nature of natural numbers, and all the "things" that behave in that way are "isomorphic to the naturals, that property seems is called categoricity (in my naive understanding is that all models of the naturals are isomorphic).

Well, I think that all agree on how sets behave, in the sense that in our experience of the word, we can see only a kind of finite set, or like a bag full of something (or empty) or as a property (comprension axiom), and all sets are finite.

My question is there is a concept of set (axioms), that have a "unique interpretation" and are exactly the naive concept of set?

答:
If the "naive concept of a set" allows you to represent natural numbers (by, say, representing $0$ as the empty set, and $n+1$ as $\{N,\{N\}\}$ where $N$ is the set that represents $n$), and allows you to prove that the peano axioms hold for this representation, then this "naive concept of a set" won't have a unique interpretation.

This is because that concept of a set is then powerfull enough to apply Gödel's incompleteness theorem to it, and from that it follows that there are statements $\phi$ for which neither $\phi$ nor $\lnot\phi$ is provable. That allows you to add either $\phi$ or $\lnot\phi$ as a new axiom without causing inconsistencies, which yields two different models of the original set of axioms.

This only applies if you formalize your naive concept of a set in first-order logic, though.

ContentLicense: CC BY-SA 3.0


问:
Number Theory: Problems with Integer Squares
I have these problems assigned for homework, I was able to get through the rest of the homework fine but I'm stuck on these two. Prove that

If $n\equiv 6\pmod9$, then $n$ cannot be a sum of two integer squares.

If $n\in\mathbb{N}$ is not the sum of two integer squares, then it is not the sum of two rational squares.

I've been trying to use these theorems from class but I haven't gotten far:

Theorem 8.2: For $p$ an odd prime, there exists $a\in\mathbb{N}$ so $a^2\equiv -1\pmod{p}$.

Theorem 8.3: For $p$ an odd prime, there exist $a,b\in\mathbb{N}$ so $p=a^2+b^2\iff p\equiv 1\pmod4$.

Theorem 8.4: $n\in\mathbb{N}$ is a sum of two squares in $\mathbb{Z}\iff$ every prime divisor $p$ of $n$ with $p\equiv 3\pmod4$ occurs an even number of times in the prime factorization of $n$.

Body:
Theorem 8.2
Among primitive Pythagorean triples, for $X=\pm (B-A),\qquad X\equiv X^2\equiv \pm1 \mod 8.\qquad $ under 100,  $X\in \{1,7, 17, 23, 31, 41, 47, 49, 71, 73, 79, 89, 97\}$.

Theorem 8.3
$\qquad 17=4^2+1^2\qquad 73=8^2+3^2\qquad 97=9^2+4^2$


问:
Constructing an isomorphism between two finite fields of order 25.
The fields in question are
\begin{equation*}
\mathbb{F}_5[x]/(x^2+x+1),\ \mathbb{F}_5(\sqrt{2}).
\end{equation*}
I know that there's an isomorphism between the above fields as they are finite fields of the same order. My idea was to find a generator of the group of units of each field, and construct an isomorphism by mapping one generator to the other.

I found that $x+2$ generates $(\mathbb{F}_5[x]/(x^2+x+1))^{\times}$ and $1+\sqrt{2}$ generates $\mathbb{F}_5(\sqrt{2})^{\times}.$ Then, calling the map $\varphi$, I send $x+2$ to $1+\sqrt{2}$ which gives, after rearranging, $\varphi(x)=\sqrt{2}+4$ where I also used that any isomorphism shall fix the base field $\mathbb{F}_5$. The problem is that the map
\begin{align*}
\varphi:&\mathbb{F}_5[x]/(x^2+x+1)\longrightarrow \mathbb{F}_5(\sqrt{2})\\
&a+bx \mapsto a+4b+b\sqrt{2}
\end{align*}
doesn't satisfy $\varphi(fg)=\varphi(f)\varphi(g)$ for all $f,g \in \mathbb{F}_5[x]/(x^2+x+1).$ Is this down to the general approach being incorrect?

答:
We notice that $\omega$, a primitive third root of unity, has as minimum polynomial $f(x)=x^2+x+1 \in \mathbb{F}_5[x]$. As $\omega=\frac{-1+\sqrt{-3}}{2},$ this gives the following isomorphism $\varphi:$
\begin{align*}
\varphi: \mathbb{F}_5[x]/(x^2+x+1) &\longrightarrow \mathbb{F}_5(\frac{-1+\sqrt{-3}}{2})\\
g(x)&\longmapsto g(\frac{-1+\sqrt{-3}}{2}).
\end{align*}
However, $-3=2 \in \mathbb{F}_5$ and $\mathbb{F}_5(\frac{-1+\sqrt{-3}}{2})=\mathbb{F}_5(\sqrt{-3})$ so
\begin{equation*}
\mathbb{F}_5[x]/(x^2+x+1) \cong \mathbb{F}_5(\frac{-1+\sqrt{-3}}{2}) = \mathbb{F}_5(\sqrt{2}).
\end{equation*}


问:
Problem calculating a complex integral
I'm trying to calculate the integral from $(0,0)$ to $(1,1)$ of the following function: $z\operatorname{Im}(z) + \operatorname{Re}(z) + 2z\overline z$ on the parabola $y=x^2$.

The problem is that when I switch to $y$, I end up with the integral $d(x+ix^2)$...and I don't know what to do with the $x^2$.

Any help would be appreciated, thanks!

答:
The problem is that when I switch to $y$, I end up with the integral $d(x+ix^2)$...and I don't know what to do with the $x^2$.

Hint: $d( x+ix^2) =dx+d(ix^2)=(1+i\,2x)\, dx$

Title: Problem calculating a complex integral


问:
Finding the percentage of observations that differ from the mean by some amount for the normal distribution.
I was trying to solve the following problem:

If a set of observations is normally distributed, what percent of these differ from the mean by:

a) more than $1.3\sigma $

b) less than $0.52\sigma $

I wrote $P(X-\mu <1.3\sigma \cup  \mu -X <1.3\sigma)$ to solve  a) however, trying to do the same for b):
$$P(X-\mu >0.52\sigma \cup  \mu -X >0.52\sigma)$$
yields $1$, which makes sense when graphing both inequalities on the same line, but this doesn't agree with the actual solution to problem and I can't figure out what I'm missing.

答: Taking (a) as an example, the proportion can be written as $$\mathbb P(X-\mu >1.3\sigma )+\mathbb P(X-\mu <-1.3\sigma)=1-\mathbb P(-1.3\sigma <X-\mu <1.3\sigma)$$ $$=1-\mathbb P(\mu -1.3\sigma <X<\mu +1.3\sigma)=1-\mathbb P\left(\frac{\mu -1.3\sigma -\mu}{\sigma}<Z<\frac{\mu +1.3\sigma-\mu}{\sigma}\right)$$ $$=1-\mathbb P(-1.3<Z<1.3)$$ where $Z\sim N(0,1)$. Using a normal distribution table or a calculator yields a probability of $1-0.8064=0.1936$.


问:
Understanding a proof regarding the relationship between rational generating functions and linear recurrence relations
I'm trying to understand this lemma from my course notes:

It's my understanding that a linear recurrence relation is one of the form $x_n = p(n, x_{n - 1}, \dots, x_{n - k}) = q(n) + b_1 x_{n - 1} + b_2 x_{n - 2} + \dots + b_k x_{n - k}$ for $n \geq k$. This recurrence relation is homogeneous if $q = 0$.

I find the above proof very terse, so below I give my understanding along with comments/questions where I don't understand things.

First part: Rational generating function defines a homogeneous linear recurrence relation

Start with $f(x) = a_0 + a_1 x + \dots = \frac{g(x)}{h(x)}$ where $g(x) = b_l x^l + \dots + b_0$ and $h(x) = c_k x^k + \dots + c_0$. Bring the denominator over to the left so you get $(c_k x^k + \dots + c_i x_i) (a_0 + a_1 x + \dots ) = b_l x^l + \dots + b_0$. We want to set up something like $a_j = \alpha_1 a_{j - 1} + \dots + \alpha_n a_{j - n}$. Because we are trying to set up a linear homogeneous recurrence relation, we clearly need to set up something that will work for all $j > \text{[something]}$, but it's not clear at this stage what that something should be.

Clearly based on the product  $(c_k x^k + \dots + c_i x_i) (a_0 + a_1 x + \dots )$ we are going to get products of $a$ coefficients with $c$ coefficients. This suggests that we might have to divide an expression involving $a_j c_{\text{something}}$ by  $c_{\text{something}}$ in order to isolate $a_j$. The only two $c_{\text{something}}$s that we know are not equal to zero are $c_i$ and $c_k$ (which may or may not be distinct).

If we try to work with $c_k$ then we would be looking at the coefficient for the $x^{k + j}$ term, which would be something like $c_k a_j + c_{k - 1} a_{j + 1} + \dots + c_i a_{j + (k - i + 1)}$. This doesn't look like the type of expression we're looking for ($a_j = \alpha_1 a_{j - 1} + \dots + \alpha_n a_{j - n}$) so we'll try using $c_i$.

If we use $c_i$ then we are looking at the coefficient for the $x^{i + j}$ term, which would be something like $c_i a_j + c_{i + 1} a_{j - 1} + \dots + c_k a_{j - (k - i)}$. Comparing both sides of $(c_k x^k + \dots + c_i x_i) (a_0 + a_1 x + \dots ) = b_l x^l + \dots + b_0$, we see that $c_i a_j + c_{i + 1} a_{j - 1} + \dots + c_k a_{j - (k - i)} = b_{i + j}$. We want $b_{i + j} = 0$ so that we can just divide by $c_i$ and move terms over and get our desired recurrence relation. This implies a restriction $i + j > l$. We also see that we need $j - (k - i) \geq 0$, so our restrictions on $j$ are that $j > l - i$ and $j \geq k - i$.

Given this, we can divide both sides of $c_i a_j + c_{i + 1} a_{j - 1} + \dots + c_k a_{j - (k - i)} = 0$ by $c_i$ and we get that $a_j = - \frac{1}{c_i} ( c_{i + 1} a_{j - 1} + \dots + c_k a_{j - (k - i)})$, which is our desired recurrence relation.

Why is my bound for $j$ different from that given in the proof ($j > \max \{ k, l \}$)?

Second part: linear recurrence relation (having function $q(n)$ that has its own rational generating function) has a rational generating function

This part doesn't make much sense to me at all. I see that $q$ essentially defines a sequence starting at index $k$, as opposed to zero, so I can see that $q$ could have an associated generating function. However the generating function $g(x)$ given for $q$ is not even a polynomial, much less rational. Is the assumption that $g(x)$ can be expressed in a rational form?

I guess $f(x)$ is supposed to be the generating function for the linear recurrence relation? I suppose a linear recurrence relation defines a sequence so there is guaranteed to be an associated generating function.

We define the function $h$ in such a way that all its coefficients beyond $k - 1$ are zero. Since $g(x)$ is presumed to be rational, we can write $g(x) = \frac{r(x)}{s(x)}$ where $r$ and $s$ are polynomials, therefore to be more thorough we can write $f(x) = \frac{h(x)s(x) + r(x)}{s(x)(1 - b_1 x - b_2 x^2 - \dots - b_k x^k)}$.

Does what I'm saying about this second part make sense? I am trying to talk myself through it. What is the intuition behind the definition of $h(x)$? I guess we're trying to make some kind of finite polynomial and that is the definition that works?

I appreciate any help.

答:
First part:
We have a generating function $f(x)=a_0+a_1x+a_2x^2+\cdots$ which is a rational function
\begin{align*}
f(x)&=\frac{g(x)}{h(x)}\\
a_0+a_1x+a_2x^2+\cdots&=\frac{b_0+b_1x+\cdots+b_lx^l}{c_0+c_1x+\cdots+c_kx^k}\qquad\qquad b_l\ne 0, c_k\ne 0
\end{align*}
We consider $h(x)f(x)=g(x)$ and obtain
\begin{align*}
&\left(c_ix^i+c_{i+1}x^{i+1}+\cdots+c_kx^k\right)\left(a_0+a_1x+a_2x^2+\cdots\right)\\
&\qquad=b_0+b_1x+\cdots+b_lx^l\tag{1}
\end{align*}
with $c_i\ne 0$ the coefficient of the smallest power of $x$ of $h(x)$ not equal to zero.

In the following we use the coefficient of operator $[x^q]$ to denote the coefficient of $x^q$ of a series. We want to find a recurrence relation in terms of $a_j,$. We take $j>\max\{k,l\}$ which is convenient, since the coefficient of $[x^j]g(x)=0$ and $[x^j]h(x)f(x)$ has to respect all coefficients of $h(x)$.

We calculate $[x^{j+i}]h(x)f(x)$ and obtain from (1)
\begin{align*}
\color{blue}{[x^{j+i}]h(x)f(x)}&=[x^{j+i}]\left(c_ix^i+c_{i+1}x^{i+1}+\cdots+c_kx^k\right)\\
&\qquad\qquad\qquad\cdot \left(a_0+a_1x+a_2x^2+\cdots\right)\\
&=\left(c_i[x^{j+i}]x^i+c_{i+1}[x^{j+i}]x^{i+1}+\cdots+c_k[x^{j+i}]x^k\right)\tag{2}\\
&\qquad\qquad\qquad\cdot \left(a_0+a_1x+a_2x^2+\cdots\right)\\&=\left(c_i[x^{j}]+c_{i+1}[x^{j-1}]+\cdots+c_k[x^{j+i-k}]\right)\tag{3}\\
&\qquad\qquad\qquad\cdot \left(a_0+a_1x+a_2x^2+\cdots\right)\\
&\,\,\color{blue}{=c_ia_j+c_{i+1}a_{j-1}+\cdots+c_{k}a_{j+i-k}}\tag{4}\\
\end{align*}

Comment:

In (2) we use the linearity of the coefficient of operator.

In (3) we apply the rule $[x^p]x^qA(x)=[x^{p-q}]A(x)$.

In (4) we select the coefficient of $[x^q],  j+i-k\leq q \leq j$.

Since $j>\max\{k,l\}$ we obtain from (1)
\begin{align*}
\color{blue}{[x^{j+i}]g(x)}=[x^{j+i}]\left(b_0+b_1x+\cdots+b_lx^l\right)\color{blue}{=0}\tag{5}
\end{align*}
and we obtain from (4) and (5)
\begin{align*}
\color{blue}{a_j=-\frac{c_{i+1}}{c_i}a_{j-1}-\cdots-\frac{c_{k}}{c_i}a_{j+i-k}}
\end{align*}
in accordance with the proof of the lemma.

Second part: Note the Lemma states:

... Furthermore if ... the function $q(n)$ has a rational generating function ...

It follows a representation of the generating function $g(x)=\frac{r(x)}{s(x)}$, with $r, s$ polynomials, wlog relatively prime and $\deg{r}<\deg{s}$ is admissible.


问:
Minimizing $|\frac{f'}{f}|$ and $|\frac{f''}{f}|$ under constraints.
Let's give constraints first.

$f$ is a $C^2$ function on $[0,K]$, and we require that $f(0)=f(K)=1$, $f'(0)=C$, $C>0$ is a given constant. The goal is to minimize the quantities $$\left|\frac{f'}{f}\right|$$ and $$\left|\frac{f''}{f}\right|.$$

Here, the absolute value means the uniform norm on the interval $[0,K].$

I wonder if there's an obstruction to minimize them, like having an lower of bound of them in terms of $C,K$. Otherwise, we would be able to minimize them as much as we want.

Or, are there any useful integral inequalities related to this? Or, any deep theory explaining it, like calculus of variations? If so, please guide me to references.

Thanks in advance.

答:
Set the value of $|\frac{f''}f|$ to a constant Q.  Then put $f$ on the other side:

$$f'' = Q f$$  From this it is relatively obvious that $e^x$ will come in to play.  With that and the other constraints you can find the exact function.


问:
Landmarks of subjects of mathematics
In order to have a good view of the whole mathematical landscape one might want to know a deep theorem from the main subjects (I think my view is too narrow so I want to extend it).

For example in natural number theory it is good to know quadratic reciprocity and in linear algebra it's good to know the Cayley-Hamilton theorem (to give two examples).

So, what is one (per post) deep and representative theorem of each subject that one can spend a couple of months or so to learn about? (In Combinatorics, Graph theory, Real Analysis, Logic, Differential Geometry, etc.)

答: Primary decomposition theorem in linear algebra.

Title: Landmarks of subjects of mathematics


问:
Elevator stops in building
Bill is working in the 38th floor of an 100-floors building. This building has a strange elevator which only has 2 buttons: the green one which takes you to the next floor every time you press it (and if you are at the last floor, it takes you to the 1st) and a red button which takes you to a random floor every time you press it. Once you enter the elevator, the first stop is selected at random, without pressing any button. Bill wants to eventually arrive at his office at the 38th floor. How is he going to do it, with the least number of steps?

I am thinking of a very simplistic approach: Since the first stop is decided at random, depending on the number of the floor where the elevator stops, Bill will decide which button to use: If he is near the 38th and before it, he will keep pressing the green button until he reaches the 38th. If it stops in a floor $>38$, he will press the red button until the elevator stops at some floor $<38$, but I assume the problem involves the expected number of stops etc probabilities, which I am not very familiar with!
Any help is really appreciated!

答:
Actually, we have better to consider a circular unidirectional shuttle line, connecting stations 0 to 99.

Bill enters at a random station and takes a shuttle to go to his destination at station $0$.

He is manouvring the shuttle himself with the random (R, red) and progressive (G, green) buttons.

It is asked which is the driving stategy that minimizes the number of button pressings.

Clearly, by pressing R it is like as Bill is restarting the journey, apart from adding one button pressing.

Now, when Bill enters the line he has $p=1/100$ chances of finding himself right at station $0$,
$p$ to enter at station $99$ and thus to be at distance (pressing of G) $1$ from the goal, $p$ to be at st. $98$
and thus at distance $2$, and so on down to st. $1$ at distance $99$.

Let's put it cumulatively, so at first entrance Bill has $(m+1)p$ chances to be at no more than  distance $m$,
and the complement $1-(m+1)p$ of being farther than $m$, which is the first row in the table below.

If Bill retains that he entered too far from the goal, he will press the R button and re-enter in the 2nd
row, which is shifted vs the precedent by 1 cell, since now one button pressing has been added, and the row
expresses the chances that it might be within a total distance from the goal.

The strategy could be to fix a threshold ($m$), below which he is going to complete the run by only
pressing G. If he has entered above $m$, he will press R and restart.

Clearly, we have reached to a variant of the famous secretary's problem,
and in particular to the cardinal payoff variant,
so I am not going further to expose it here.

Title: Elevator stops in building


答:
I suspect that you have misunderstood what is meant by the induction step.  You seem to think that, in this step, you assume the statement in question, which I'll abbreviate as $P(n)$, for all values of $n$ and then prove it fo $n+1$. That would always work, trivially, by just substituting $n+1$ for $n$ in your assumption.  And it wouldn't actually prove anything about $P(n)$.

But that's not what the induction step is.

The actual induction step assumes $P(n)$ for a single natural number $n$ and tries to prove $P(n+1)$.

Sometimes this is stated, confusingly, as "assume $P(n)$ for an arbitrary $n$", and the word "arbitrary" makes it sound as if you're allowed to assume $P(n)$ for all $n$ at once.  That is not the intended meaning of "arbitrary". The intended meaning is that you're not allowed to assume anything about $n$ beyond its being a natural number satisfying $P(n)$.

So the induction step goes like this. Consider one  natural number $n$ about which you assume that $P(n)$ and nothing more. Then try to prove $P(n+1)$.

[To state it more formally, the induction step consists of proving "for all natural numbers $n$, if $P(n)$ then $P(n+1)$."  It does not consist of proving the (trivial but similar-looking) statement "if for all natural numbers $n$ $P(n)$ then $P(n+1)$."  Switching the order of "if" and "for all $n$" makes a huge difference.]


问:
Is this a new Primality Test?
Every $n \in \mathbb{Z}$ is prime if all lattice points on $x+y=n$ are visible from the origin.

Graphed points on $x+y=n$ not visible from the origin for potential primes.

```<code>// Primality Test
// Every n is prime if all lattice points on x+y=n are visible from the origin.

#include <stdio.h>
#include <stdint.h>
#include <math.h>

uint64_t gcd(uint64_t a, uint64_t b)
{
    return (b != 0) ? gcd(b, a % b) : a;
}

int isPrimeNumber(uint64_t n)
{
    if (n == 1) return 0;
    if (n == 2 || n == 3) return 1;
    if (n % 2 == 0) return 0;

    // Start near line x=y.
    uint64_t x = (n / 2) + 2;
    uint64_t y = n - x;
    uint64_t count = sqrt(n) / 2;

    for (uint64_t i = 0; i < count; ++i) {
        // Check lattice point visibility...
        if (gcd(x, y) != 1) return 0;
        x++; y--;
    }

    return 1;
}

int main(int argc, char* argv)
{
    uint64_t n = 1000000007;

    if (isPrimeNumber(n) == 1)
    {
        printf("%llu prime.", n);
    }
    else
    {
        printf("%llu not prime.", n);
    }

    return 0;
}
</code>```

答:
If $n=ab$ with $a$ and $b$ greater than $1$ then
$$
n = a + a(b-1)
$$
so $(a, a(b-1))$ is a point on the line.

That says that there are invisible points when $n$ is composite.

Conversely, if $n$ is prime and $n=x+y$ then any common divisor of $x$ and $y$ will divide $n$, so $(x,y)$ is visible.

So this is a valid primality test.

It's not an efficient test.

Title: Is this a new Primality Test?


问:
Evaluate $\int \arctan({\frac{x-2}{x+2}})\,\text dx$
$$
\int \arctan\left({\frac{x-2}{x+2}}\right)\,\text dx
$$

It seems to be solved in just one line:
$$
x \arctan\left({\frac{x-2}{x+2}}\right)-\log|x^2+4|+c
$$

but how?
I have tried with integration by parts (by picturing out $1$ multiplying arctan), but I don't think is the right way.

答:
We have using first integration by parts:
$$
\begin{aligned}
\int \arctan\frac{x-2}{x+2}\; dx
&=
\int x'\cdot\arctan\frac{x-2}{x+2}\; dx
\\
&=
x\arctan\frac{x-2}{x+2}
-
\int x\cdot\underbrace{\left(\arctan\frac{x-2}{x+2}\right)'}_{2/(x^2+4)}\; dx
\\
&=
x\arctan\frac{x-2}{x+2}
-
\int \frac{d(x^2+4)}{x^2+4}\\
&=
x\arctan\frac{x-2}{x+2}
-
\log(x^2+4)+ c\ .
\end{aligned}
$$


问:
Fibonacci numbers in two sets
Is there a way to split all natural numbers (without zero) into two sets, so that no two elements in the same set add up to a Fibonacci number? I made a c++ program and it worked for the first 100'000 natural numbers. So there has to be a proof for that...

答:
I really love the elementary argument presented in the paper of Erdos, Alladi and Hoggatt, which proves inductively that the Fibonacci numbers, or infact any linear recurrence of the Fibonacci type, provides a partition of the integers which is sum-free with respect to that subset, and in fact the split is unique. What I will do is to present the result and proof of Erdos where I outline key steps and leave the details hidden for interested readers. The argument is elementary and really nice as a read-through.

Result : Let $u_n$ be a sequence given by $u_1 = 1$, $u_2  = b>1$ and $u_{n+2} = u_{n+1} + u_n$, $n >0$. There exist unique subsets $A_1,A_2 \subset \mathbb N$ such that

$A_1 \cap A_2 = \emptyset$ and $A_1 \cup A_2 = \mathbb N$.

For $i=1,2$ and $a , b\in A_i$, $i=1,2$, we have $a+b \notin \{u_j\}$.

Erdos' proof is an existential one. Turns out we have explicit descriptions for the Fibonacci sequence, as per the sequences A005652 and A005653 of the OEIS.

Anyway, the proof starts with constructing the $A_i$, and will show by induction that no $u_i$ is the sum of two elements from the same $A_i$.

For starters, why must each of $u_1,u_3,...$ and $u_2,u_4,...$ be parts of different subsets?

 Of course : if $u_1$ is in one of the sets, then $u_2$ can't be in there because $u_1+u_2 = u_3$ will be a contradiction. Then $u_3$ can't be in the same subset as $u_2$ otherwise $u_2 + u_3 = u_4$ is a contradiction, so $u_1$ and $u_3$ are in the same subset. The argument continues.

Now we assume that $b$ is even and $b = 2a$.

All of $1,2,...,a-1,a$ must lie in the same set (taken as $A_1$ wlog). More precisely, if $c <a$ then $c,c+1$ must lie in the same set. (Hint : Consider which subset $d = b-c$ would go in!)

 Which would it go in? We have $d+c = u_2$ and $d+(c+1) = u_2 + 1 = u_3$ so $d$ can't be in the same subset as either $c$ or $c+1$. If $c,c+1$ themselves belonged to different subsets, we'd have nowhere to fit $d$!

Of course, then $\{a+1,...,2a\} \subset A_2$.

 $b = (a-x) + (a+x)$ for any $x<a$ so we can't have $a-x,a+x$ in the same subset.

Ok, now let's look at $b = 2a+1$ odd.

Use a similar reasoning to above to get that $\{1,...,a-1\}$ belong in the same subset , wlog $A_1$.

$a \in A_1$. (Hint : study where $a+1$ and $a+2$ can go)

 If $a \in A_2$, we must have $a+1 \in A_2$ as $a+a+1 = b$. But now $a+2 + a = b+1$ and $a+2 + a-1 =b$ means $a+2$ can go nowhere!

Now by a similar argument to earlier, $\{a+1,a+2,...,2a+1\} \subset A_2$.

This sorts out everything until $u_2$ for the odd and even cases. Now for the part after $u_2$, the following is proposed as a construction of the $A_i$. It is inductive in nature.

Pick an $n > u_2$. If $n$ is one of the $u_i$ we already know where it is. Otherwise, there is a unique $u_m < n < u_{m+1}$. Let $i = n - u_m$. Why is $i < u_{m-1}< n$?

 The second is obvious, the first from $u_{m-1} + u_m = u_{m+1} > n$ and subtract $u_m$ from both sides.

Now $i$ has already been assigned, as per the inductive build up. We have two parts to this :

If $i$ is never half any element of $\{u_i\}$ then put $n$ in the same subset as $i$.

If $i = \frac{1}{2} u_{k-1}$ for some $k>1$, then $ m> k-1$ (why?) so if $m \neq k$, put $n$ and $i$ in the same subset, else put them in different subsets.

Finally, we need to show that this works! Each positive integer has been put in exactly one subset, so that is not a problem. What we need to look at is the second property.

Assume that $\{1,2,...,n-1\}$ have been sorted into two subsets $A_1$ and $A_2$ in the fashion above such that no two elements of the same subset add up to some $u_i$. We prove that when $n$ is inserted according to our rule then the property remains.

There is a unique $m$ with $u_m \leq n < u_{m+1}$.

$n = u_m$ is not a problem. (Hint : how can $u_m$ be made to add up to some $u_i$ with only elements smaller than $u_m$?)

 There's only $u_m + u_{m-1} =u_{m+1}$ as a possibility, but these already belong in different subsets.

We then have $u_m < n < u_{m+1}$. Let $a<n$. Then the only possibilities for $a+n$ being some $u_i$, are $a+n= u_{m+1}$ and $a+n = u_{m+2}$.

 Of course, $a+n > u_m$ but $a+n < 2n < u_{m+1} + u_{m+2} = u_{m+3}$ so we can get only two values for $a+n$.

We split into the cases $a+n = u_{m+1}$ and $a+n = u_{m+2}$, and want to show that for any $a$ in the same subset as $n$ neither of these will happen. Basically, the point is : for $n$, neither $u_{m+1}-n$ or $u_{m+2} -n$ should be in the same subset as $n$.

Let $a+n = u_{m+1}$ and $i = n-u_m < n$.Then $a+i = u_{m-1}$.

If $a = i$ then $n$ and $a$ are in different subsets.

 Then $a=i = \frac 12 u_{m-1}$ are in the same subset but according to the rules, $n$ and $i$ are in different subsets.

If $a=i$ then $a,n$ are in different subsets.

 Well, in that case $i \neq \frac 12u_{m-1}$ so whatever happens, the rules say that $n$ and $i$ are in the same subset. $a$ and $i$ can't be since $a+i = u_{m-1}$ and they are distinct.

So either way, this case is done.

Let $a+n = u_{m+2}$. Let's use case $1$ : $n$ and $u_{m+1} - n$ belong in different subsets from there.

$u_{m+1} - n$ and $u_{m} + (u_{m+1}-n)$ belong in the same subset. (Hint : let $I = u_{m+1} - n$ , then $N= u_{m} + (u_{m+1}-n) = u_m + i$ so $N- u_m = I$. Our rules say that $N$ and $I$ belong in the same subset unless something happens, but that doesn't happen actually!)

 Yes, $N$ and $I$ can belong in different subsets , but for that to happen we must have $I = \frac 12 u_{m-1}$. But then $u_{m+1} - n = \frac 12 u_{m-1}$ and multiply by two and rearrange to get $2n= u_{m+2} = a+n$ so $a = n$, a problem.

But we just proved that $u_{m+1} - n$ and $u_{m} + (u_{m+1}-n) = u_{m+2}-n$ belong in the same subset. $n$ doesn't belong in the same subset as the first one by case $1$, so it doesn't belong in the same subset as $u_{m+2} - n = a$ either!

Consequently, we are done!

Uniqueness :  We prove by induction that this holds. We can skip some numbers, though.

The base case is made clear in the proof, there's only one way to assign everything up till $u_2$. Also, the $u_i$ themselves are assigned uniquely as well from the proof again.

For any $n$, let $u_m < n < u_{m+1}$. Then $j = u_{m+1} - n < n$ and $n$ are not in the same subset, so as $j$ is uniquely assigned, $n$ is uniquely assigned as well.

This completes the proof of the result.

Title: Fibonacci numbers in two sets


问:
If $\gcd(a,x)=1$, then $a\mathbb N+b, x\mathbb N+y$ intersects.
I have tried to think as intersections of two lines but for $2x+3$ and $5x+7$ is failed to have integer intersections (but not sure it is enough counterexample).

Is the above generalisation correct? Can you give a counter-example, or a hint for how to prove it?

答:
Thinking in terms of straight lines and looking for points of intersection where both lines have the same output for the same input is not the best analogy.  It's better to think of these as unstructured unordered sets which are like bags of groceries that they have a bunch of stuff dumped in them in no particular order and an "intersection" simply means the two bags both have the same item somewhere. ("You bought mushrooms and tampons too!  So did I!")

$2\mathbb N + 3 = \{5,7,9,11,13,15,17,19,21,23,25,27,29......\}$ and $5\mathbb N + 7= \{12,17,22,27,32,....\}$

Notice they both have $17$ and $27$ and $37$ etc in them.  Note those numbers aren't in both for the same reason... If $17 \in 2\mathbb N+3 = \{2\color{blue}n + 3|\color{blue}nn\in \mathbb N\}$ it is because if $\color{blue}{n= 7}$ we get $2\cdot\color{blue}{7} +3=17$ which $17 \in 5\mathbb N + 7=\{5\color{red}n + 7|\color{red}n \in \mathbb N\}$ because if $\color{red}n = 2$ we get $5\cdot\color{red}2+7=17$... but we don't care why the items are in the bag, just that they are both in each of the bags.

("I bought the mushrooms as the fifth item right after the milk and a picked up the tampons at the check out counter as my last item"  "Oh, not me.  I went shopping just because I needed tampons so I bought the first thing.  Then I thought about dinner so I bought pasta, mushrooms, and onions.  Mushrooms were the third thing I bought.")

..........

So this is actually much looser.

You mus show that there exist natural number $M,K$ so that $aM + b= xK +y$ where $M,K$ can be any natural numbers and need not have any thing in common.

That would mean $aM - xK = (y-b)$.  By Bezout's lemma you know we can find integers$M', K'$ so that $aM' + xK' = 1$ so if we let $M = (y-b)M'$ and $K = -(y-b)K'$ we get $aM - xK = y-b$.

Or only concern is whether or not $M,K$ are both positive.  I'll let you work the details.

(Note: if $a,b \ge 0$ then $aM' + bK' = 1$ would mean only one of $M'$ or $K'$ is positive and the other is either negative or $0$.)

=====

Postsript:  For the record the two sets $a\mathbb N + b$ and $x \mathbb N + y$ don't just intersect and have one item in common.  They have an infinite number of items in common.

.....

Notice that every $10K + 7 = 2(5K + 2) + 3= 5(2K) + 7 \in (2\mathbb N + 3)\cap( 5\mathbb N + 7)$.  So for all $K$ we have $10K + 7$ is in the intersection.

That may give you a hint to a more direct way (but not necessarily easier) way to solve it than my Bezout solution.


问:
A linear order that every proper initial segment is $\operatorname{seg}(x)$ for some $x$ is a well-order
Can someone help with the following problem:
Let $\langle A, R\rangle$  is a linear ordered set. Let every time when $w$ is an initial segment of $\langle A, R\rangle$, to be true that $w=A$ or $(∃x ∈ A)(w = \operatorname{seg}(x))$. Prove that $\langle A,R \rangle$ is a well ordered set.

Any help is appreciated, thanks in advance!

答: Given a nonempty subset $S$ of $A$, Apply the given assumption to the set $w$ of all strict lower bounds of $S$.


答:
Well, I would say that instead of computing how many observations differ from the mean by more than $1.3 \sigma$ you are computing how many differ by less than that value, that is, you are counting the internal values instead of the external ones.

In other words, if you look at this image:

you are integrating the darker part, instead of integrating outside it.

I would rewrite your expression as $P\Big(\frac{|X-\mu|}{\sigma} > 1.3 \sigma\Big)$ and than use the normal table to compute it.

In general, letting $\phi(z)$ be the standard normal probability function (i.e., it gives you the integral of the pdf from $- \infty$ to $z$) and $Z$ a normalized gaussian distribution (in your case $Z=\frac{X-\mu}{\sigma}$) you can compute your probabilities as $P(|Z|>z) = 2(1-\phi(z))$.

If I'm not wrong in substituting the values, I obtain for:

more than $1.3\sigma$, 0.1936

more than $0.52\sigma$, 0.603

If instead we want to compute $P(|Z| < z) = \phi(z) - \phi(-z) = 2\phi(z) - 1$ where the last comes from the fact that $\phi(-z) = 1 - \phi(z)$.

Using this formula I get for:

less than $0.52\sigma$, 0.397

The best way to understand these calculations is to visualize what you are doing: the probabilities are the areas that you consider and, then, you are just summing and subtracting among areas to get what you want. To be more specific, in this case I get the whole area of the pdf from $-\infty$ to $z$ using $\phi(z)$ and than subtract the area from $-\infty$ to $-z$ using $\phi(-z)$ so that you are keeping only values between $-z$ and $z$.

Maybe this picture can clear your ideas (the integration of the whole curve of course lead to 1, so you can use it to make additions/subtractions between areas):

Title: Finding the percentage of observations that differ from the mean by some amount for the normal distribution.

Body:
I was trying to solve the following problem:

If a set of observations is normally distributed, what percent of these differ from the mean by:

a) more than $1.3\sigma $

b) less than $0.52\sigma $

I wrote $P(X-\mu <1.3\sigma \cup  \mu -X <1.3\sigma)$ to solve  a) however, trying to do the same for b):
$$P(X-\mu >0.52\sigma \cup  \mu -X >0.52\sigma)$$
yields $1$, which makes sense when graphing both inequalities on the same line, but this doesn't agree with the actual solution to problem and I can't figure out what I'm missing.


问:
Every Ring is Isomorphic to a Subring of an Endomorphism Ring of an Abelian Group
Show that for every ring $(R,+,\cdot)$, there is an abelian group, $(A,+)$, such that $R$ is isomorphic to a subring of $(\operatorname{End}(A),+,\circ)$.

$(\operatorname{End}(A),+,\circ)$ is the set of homomorphisms of $A$ that form a ring under function addition and composition.

I am thinking to let $\operatorname{End}(A)$ be the group such that $A$ is the abelian group $(R,+)$ and create a ring homomorphism from $(R,+,\cdot)$ into $\operatorname{End}((R,+))$.

Thoughts?

答:
You can consider $(R,+)$, the underlying abelian group. Then, an injective morphism is
\begin{align*}
\lambda:R&\to \text{End}_{\textsf{Ab}}(R)\\
r&\mapsto \lambda_r
\end{align*}
where $\lambda_r$ is left-multiplication by $r$.


问:
Solve the system of equations for $\sqrt{xy}$
$$x + y\sqrt{x} = \frac{95}{8}$$
$$y + x\sqrt{y} = \frac{93}{8}$$
$$x, y \in \mathbb{R}$$

I can't solve this system of equations I got asked in a group. I added and substracted them to find the following equations but I don't really know what to do with them.

$$\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}-\sqrt{xy}\right) = \frac{1}{4}$$

$$\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}+\sqrt{xy}\right) - 2\sqrt{xy} = \frac{47}{2}$$

This is all I have. I checked the answer using Wolframalpha and it was $\frac{15}{4}$ for $\sqrt{xy}$.

I have been looking for this question's solution but didn't find any related question/solution on Stack Exchange or Web. Sorry if it is duplicate, thank you in advance.

答:
Let's do the substitution that Will suggests, $x=s^2$,$y=t^2$ and let's save writing those fractions $a=95/8,b=93/8$. Thus we have
\begin{eqnarray*}
s^2+st^2=a \\ t^2+ts^2=b.
\end{eqnarray*}
How do we solve these (for $(s,t)$ in terms of $a,b$) ? Rearrange the first equation to $s^2-a=s t^2$ and square this
\begin{eqnarray*}
s^4-2as^2+a^2=t^4s^2.
\end{eqnarray*}
Now multiply this by $t^2$ and use the second equation
\begin{eqnarray*}
(b-t^2)^2-2at(b-t^2)+a^2t^2=t^5(b-t^2).
\end{eqnarray*}
This shows how to solve these equations (in principle). So we just need to solve an equation of order $7$ ... Let's see what CAS says ...

So this gives the solution that Will mentions ... $t=5/2$ ... What about the order $6$ part ? Well have a look at the graph, https://www.desmos.com/calculator/xsj6mrglur there are two real roots but they give negative values for $t^2$ or $s^2$.


问:
Occurrence of 5 consecutive tails before occurrence of 2 consecutive heads
In repeated tossing of a fair coin find the probability that $5$ consecutive tails occur before occurrence of $2$ consecutive heads.

My attempt:
I tried to find the probability of non-occurrence of two consecutive heads in $n$ throws.

Let $a_{n}$ be the number of possibilities in which $2$ consecutive heads do not occur in $n$ throws.

I managed to find the recursion formula.

$a_{1}=2$

$a_{2}=3$

$a_{n}=a_{n-1}+a_{n-2}$

But I am not able to get a closed form of $a_{n}$.

Once $a_{n}$ gets determined it may be possible then to find probability of occurrence of $5$ consecutive heads

答:
There is an intuitive way to think about this.

Let $E$ be the probability of success. Let $\mathbb{T}$ denote the probability of success starting with a T first; Let $\mathbb{H}$ denote the probability of success starting with an H first.

To illustrate, the event space can be partitioned into $\mathbb{T}$ and $\mathbb{H}$:

\begin{align}
\mathbb{H}&: &H \\
\mathbb{T}&: &TH \\
& &TTH \\
& &TTTH \\
& &TTTTH \\
& &TTTTT
\end{align}

If we get H on the first flip, to succeed requires getting a T next. If we get T on the first flip, the series of outcomes listed above under $\mathbb{T}$ can lead to success. With these, we can derive the following equations.
$$
P(E|\mathbb{H}) = \frac{1}{2}P(E|\mathbb{T}) \\
P(E|\mathbb{T}) = \left ( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} \right )  P(E|\mathbb{H}) + \frac{1}{16}
$$

Finally, $P(E) = \frac{1}{2} P(E|\mathbb{H}) +  \frac{1}{2} P(E|\mathbb{T}) = \frac{3}{34}$.

In general,
$$
P(E|\mathbb{H}) = \left( 1 - \frac{1}{2^{h-1}} \right) P(E|\mathbb{T}) \\
P(E|\mathbb{T}) = \left( 1 - \frac{1}{2^{t-1}} \right) P(E|\mathbb{H}) + \frac{1}{2^{t-1}}
$$

Solving them gives
$$
P(E) = \frac{2^h - 1}{2^h + 2^t - 2}
$$


问:
Variational formulation of a Elliptique -PDE and prove that the problem resultating is well-posed

Determine the variational formulation of
\begin{cases}
-\Delta u(\mathbf{x})=1, & \mathbf{x}\in (0,1)\times (0,1) \\
-\partial_{x}u(\mathbf{x})+c(\mathbf{x})u(\mathbf{x})=0, & \mathbf{x}\in \{0\}\times (0,1)\\
\partial_{y}u(\mathbf{x})=0,  & \mathbf{x} \in (0,1)\times \{0\}\cup  (0,1)\times \{1\}\\
\partial_{x}u(\mathbf{x})=0,  & \mathbf{x} \in\{1\}\times (0,1)
\end{cases}

My approach:  Let $v\in H^{1}(\Omega)$ and assuming that $u\in H^{2}(\Omega)$, so we have that
\begin{eqnarray*}
    -\Delta u(\mathbf{x})&=&1\\
    -\Delta u(\mathbf{x})v(\mathbf{x})&=& v(\mathbf{x})\\
    \int_{\Omega}-\Delta u(\mathbf{x})v(\mathbf{x})d\mathbf{x}&=& \int_{\Omega}v(\mathbf{x})d\mathbf{x}
    \end{eqnarray*}
Using Green's formula, we obtain that
\begin{eqnarray*}
    \int_{\Omega}-\Delta u(\mathbf{x})v(\mathbf{x})d\mathbf{x}&=& \int_{\Omega}v(\mathbf{x})d\mathbf{x}\\
    \int_{\Omega}\mathbf{\nabla u(x)}\cdot \mathbf{\nabla v(x)}d\mathbf{x}-\int_{\partial\Omega}\gamma_{1}u(\mathbf{x})\gamma_{0}v(\mathbf{x})ds_{\mathbf{x}}&=&\int_{\Omega}v(\mathbf{x})d\mathbf{x}
    \end{eqnarray*}

Thank you @VoB using your answer I was able to advance from where I was stuck: The cantity $\displaystyle  \int_{\partial\Omega}\gamma_{1}u(\mathbf{x})\gamma_{0}v(\mathbf{x})ds_{\mathbf{x}}$ is equal to $\displaystyle \int_{\partial \Omega} v (\mathbf{\nabla u(\mathbf{x}) \cdot n})ds_\mathbf{x}$
we could find the  last cantity in terms of the (BC), so the variational formulations is given by $$ \quad u\in H^{1}(\Omega): \quad a(u,v)=\ell(v), \quad \forall v\in V$$
where $$a(u,v):=\int_{\Omega} \mathbf{\nabla u(x)}\cdot \mathbf{\nabla v(x)}d\mathbf{x}-\int_{\partial \Omega} v (\mathbf{\nabla u(\mathbf{x}) \cdot n})ds_\mathbf{x}$$ and $$\ell(v):=\int_{\Omega} v(\mathbf{x})d\mathbf{x}$$

Green's formula: Let $\Omega$ a bounded open set in $\mathbb{R}^{n}$ with border continous-Lipschitz $\partial \Omega$. Then,  $\forall v\in H^{1}(\Omega)$ and $u\in H^{2}(\Omega)$ we have $$\int_{\Omega} v \Delta u=-\int_{\Omega}\mathbf{\nabla u\cdot \nabla v}+\int_{\partial \Omega}\gamma_{0}(v)\gamma_{0}(\mathbf{\nabla u)\cdot \nu}$$
where $\gamma_{0}(\mathbf{\nabla u})$ is the vector that is result of apply $\gamma_{0}$ to the components of $\mathbf{\nabla u}$.

答:
After taking $v \in H^1(\Omega)$:

$$\int_{\Omega} \nabla u \cdot \nabla v dx- \int_{\partial \Omega} v (\nabla u \cdot n)ds_x = \int_{\Omega} vdx$$

Now, using the additivity of the integral on $\partial \Omega$ and the boundary conditions on the unit square:

$$\int_{\partial \Omega} v (\nabla u \cdot n)ds_x = \int_{(0,1) \times \{0\} \cup (0,1) \times \{1\}} v \partial_x u \cdot n ds_x + \int_{\{1\} \times(0,1)}v \partial_y u \cdot n_2 ds_x + \int_{\{0\} \times (0,1)} v (c(x) u(x) + \partial_y u \cdot n_2 )ds_x$$


问:
Let $f:[0,\frac{\pi}{2}] \to \mathbb R$ is defined as $f(x):=\text{max}\left\{x^2,\cos(x)\right\}$ for all $x \in [0,\frac{\pi}{2}] $
Let $f:[0,\frac{\pi}{2}] \to \mathbb R$ is defined as $f(x):=\text{max}\left\{x^2,\cos(x)\right\}$ for all $x \in [0,\frac{\pi}{2}] $,show that an absolute minimum $x_0 \in [0,\frac{\pi}{2}]$ for $f$ on $[0,\frac{\pi}{2}]$ and $x_0$ is the solution to the equation $\cos(x)=x^2$.

Let $x_0$ be the point where the two functions intersect,on $[0,x_0]$ we have that $\cos(x) \ge x^2$ and on $[x_0,\frac{\pi}{2}]$ we have that $x^2 \ge \cos(x)$,from here we see that $$f(x)=\color{blue}{\cos(x)\ge \cos(x_0)}=f(x_0) \tag{$x \in [0,x_0]$}$$
$$f(x)=x^2 \ge x_0^2=f(x_0) \tag{$x \in [x_0,\frac{\pi}{2}]$}$$

So for all $x \in [0,\frac{\pi}{2}]$ we see that $f(x) \ge f(x_0)$ and this implies $x_0$ is an absolute minimum of $f$ over $[0,\frac{\pi}{2}]$,moreover we know that $x_0$ is the solution to the given equation .

I have two questions,how without using the concept of derivative we are able to show that the blue part does hold?

And how do we know that "on $[0,x_0]$ we have that $\cos(x) \ge x^2$ and on $[x_0,\frac{\pi}{2}]$ we have that $x^2 \ge \cos(x)$"?

答:
Hint

The blue inequality immediately follows from the fact that $\cos x$ is decreasing on $[0, \frac{\pi}{2}]$.

Also the map $g(x) = \cos x - x^2$ is the difference of a decreasing map and an increasing map on $[0, \frac{\pi}{2}]$. Therefore it is decreasing on that interval. The fact that $g(0)=1$ and $g(\pi/2) = -\pi^2/4$ provides the answer to your second question.


问:
Prove combinatorially $\sum_{k=0}^{\left \lceil \frac{n}{2} \right \rceil}\binom{n}{2k}=2^{n-1}$

$$\sum_{k=0}^{\left \lceil \frac{n}{2} \right \rceil}\binom{n}{2k}=2^{n-1}$$

I have no idea how to start, can i get hints?

答:
Maybe you can try by induction?

As a starting point, compute it for $n=0, 1, 2$, just to verify than the relation is true for some base cases.

Than, you can assume the relation to be true for $n$ and show that, with this assumption, then, the relation holds also for $n+1$.

And you're done!

I would try this way! Let me know ;)


问:
Proving that a continuous map is surjective

Let $f:\mathbb{R}\to \mathbb{R}$ be a continuous map such that $\overline{f(\mathbb{N})} = \mathbb{R}$. Prove that $f$ is surjective.

I tried to use the definition of the continuity and $\overline{f(\mathbb{N})} = \mathbb{R}$ to show that $f$ is surjective but it didn't work. Since $f$ is continuous, the preimage of every open set is open. Also it results in $f(\overline{A})\subseteq \overline{f(A)}$ for every set $A$. How this couples with $\overline{f(\mathbb{N})} = \mathbb{R}$ to show surjectivity?

答: Hint: How do you prove that $f$ takes all values between $0$ and $1$ if it takes the value $0$ and $1$?


答:
Hint: Using the binomial theorem, what is $ ( 1 + 1) ^n$?

What is $ ( 1 - 1) ^n$?

Title: Prove combinatorially $\sum_{k=0}^{\left \lceil \frac{n}{2} \right \rceil}\binom{n}{2k}=2^{n-1}$

Body:
$$\sum_{k=0}^{\left \lceil \frac{n}{2} \right \rceil}\binom{n}{2k}=2^{n-1}$$

I have no idea how to start, can i get hints?


答:
The LHS gives the number of subsets of even size of a set with cardinality $n$.

Now, there are a total of $2^n$ subsets, and clearly exactly half of them will have even cardinality, i.e. $2^n/2 = 2^{n-1}$

Since both sides count the same thing, we are done.


答:
WLOG $t=\arctan\dfrac x2,x=2\tan t,dx=2\sec^2t\ dt$

$$\dfrac{x-2}{x+2}=\dfrac{\tan t-1}{\tan t+1}=\tan\left(t-\dfrac\pi4\right)$$

Using Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$,

$$\arctan\dfrac x2+\arctan(-1)=\begin{cases} \arctan\frac{\dfrac x2-1}{1+\dfrac x2} &\mbox{if } -\dfrac x2<1\\ \pi+\arctan\frac{\dfrac x2-1}{1+\dfrac x2} & \mbox{if } -\dfrac x2>1\\-\dfrac\pi2 & \mbox{if } x=-2\end{cases} $$

Now for $-\dfrac x2<1,$ $$I=\int\arctan\frac{\dfrac x2-1}{1+\dfrac x2} \ dx=2\int\left(t-1\right)\sec^2t\ dt$$

Integrating by parts,

$$I=2t\int\sec^2t\ dt-\int\left(\int\dfrac{d(2t)}{dt}\int\sec^2t\ dt\right)\ dt-2\sec^2t\ dt$$

Can you take it from here?

Title: Evaluate $\int \arctan({\frac{x-2}{x+2}})\,\text dx$

Body:
$$
\int \arctan\left({\frac{x-2}{x+2}}\right)\,\text dx
$$

It seems to be solved in just one line:
$$
x \arctan\left({\frac{x-2}{x+2}}\right)-\log|x^2+4|+c
$$

but how?
I have tried with integration by parts (by picturing out $1$ multiplying arctan), but I don't think is the right way.


答:
Hint: Consider the following bijection
$$\varphi:\mathcal{P}([n])\longrightarrow \mathcal{P}([n]),$$ given by $$\varphi (X)=\begin{cases}X\cup \{n\}&\text{if }n\not \in X\\
X\setminus \{n\}& \text{Otherwise}\end{cases}.$$
Show that this is a bijection that maps even sized to odd sized sets and viceversa. Use binomial theorem to conclude(or involution principle if you know it).


答: Take $x\in\Bbb R$; you want to prove that $x=f(y)$, for some $y\in\Bbb R$. Since $x+1\in\overline{f(\Bbb N)}$, $x+1$ is the limit of a sequence of numbers of the form $f(n_k)_{k\in\Bbb N})$, where each $n_k$ is a natural number. In particular, there is some $n\in\Bbb N$ such that $f(n)>x$. By the same argument, there is some $m\in\Bbb N$ such that $f(m)<x$. So, by the intermediate value theorem, there is some $y$ between $m$ and $n$ such that $f(y)=x$.

Title: Proving that a continuous map is surjective

Body:
Let $f:\mathbb{R}\to \mathbb{R}$ be a continuous map such that $\overline{f(\mathbb{N})} = \mathbb{R}$. Prove that $f$ is surjective.

I tried to use the definition of the continuity and $\overline{f(\mathbb{N})} = \mathbb{R}$ to show that $f$ is surjective but it didn't work. Since $f$ is continuous, the preimage of every open set is open. Also it results in $f(\overline{A})\subseteq \overline{f(A)}$ for every set $A$. How this couples with $\overline{f(\mathbb{N})} = \mathbb{R}$ to show surjectivity?


问:
How to show that $f : {}^\ast \Bbb R \to {}^\ast \Bbb R$ is bounded, if it obtains a limited value everywhere?
Let $f :  {}^\ast \Bbb R \to  {}^\ast \Bbb R$. For all $x \in  {}^\ast \Bbb R $, there exists $y \in \Bbb R $ such that $f(x) \leq y$. How to show that there exists  $r \in \Bbb R$, such that for all $z \in  {}^\ast \Bbb R$  we have $f(z) \leq r$?

I know this can be shown by a direct application of the dual form of idealization principle. Is there any other way to show this, say, using ultrapower construction?

Moreover, if $f$ is continuous, is it true that $f$ necessarily obtain a maximal value?

答:
Our goal is to prove the statement

$$ (\forall z \in {}^{*}\Bbb{R} ) (\exists r \in \Bbb{R}) ( f(z) \leq r ) \quad \Longrightarrow \quad (\exists r \in \Bbb{R})(\forall z\in {}^{*} \Bbb{R})( f(z) \leq r). $$

We prove the contrapositive:

$$ (\forall r \in \Bbb{R})(\exists z\in {}^{*} \Bbb{R})( f(z) > r) \quad \Longrightarrow \quad (\exists z \in {}^{*}\Bbb{R} ) (\forall r \in \Bbb{R}) ( f(z) > r ). $$

Assume that for any $r \in \Bbb{R}$, there exists $z \in {}^{*}\Bbb{R} $ such that $f(z) > r$. For each fixed $r \in \Bbb{R}$, applying the transfer principle, we have

$$(\exists z \in {}^{*}\Bbb{R})( f(z) > r ) \quad \Longrightarrow \quad (\exists z \in \Bbb{R})( f(z) > r ). $$

Thus we have

$$ (\forall r \in \Bbb{R})(\exists z \in \Bbb{R})( f(z) > r ). $$

Passing to the transfer principle, we obtain

$$ (\forall r \in {}^{*}\Bbb{R})(\exists z \in {}^{*}\Bbb{R})( f(z) > r ). $$

In particular, for a given infinitely large $R \in {}^{*}\Bbb{R}$, we have $f(z) > R$ for some $z \in {}^{*}\Bbb{R}$. This can be rephrased as

$$ (\exists z \in {}^{*}\Bbb{R} ) (\forall r \in \Bbb{R}) ( f(z) > r ). $$

This is exactly what we wanted, and hence the proof is completed.

To show that $f$ may fail to achieve its maximum, we consider the function

$$ f(x) = \frac{|x| + 1}{|x| + 2}. $$

It is clear that for all $x \in \Bbb{R}$ we have $f(x) < 1$ while for each $r < 1$ there exists $x \in \Bbb{R}$ such that $f(x) > r$. This can be transferred into the corresponding statement in the hyperreal field, and thus $f(x)$ has no maximum on $\Bbb{R}^{*}$.


问:
Famous deaf mathematicians?
There are some really inspiring examples of blind mathematicians. However in my experience I also think problems inside my head using words. So I was wondering if there are some examples of deaf mathematicians who nonetheless where able to make advancements in mathematics.

Regards.

答: Answering "I wonder if there are any Deaf professors at Gallaudet who do active research? They seem to offer courses in complex analysis and abstract algebra there, according to their website at least." ... there is at least one that I know of at this time.  Over the years there have been a number of deaf math professors at Gallaudet, several with PhDs.  But admittedly few continued to do serious research in their fields.</p>

Title: Famous deaf mathematicians?


答:
Take $b \in \mathbb R$. By hypothesis, it exists $a_1,a_2 \in \mathbb N$ such that
$f(a_1) \in (b-3/2, b-1/2)$ and $f(a_2) \in (b+1/2, b+3/2)$.

Then by the intermediate value theorem, it exists $a \in (a_1, a_2)$ such that $f(a)=b$.


问:
Correlation coefficient and regression, Check my answers
I am working on a question. I made more than the half of the question but stuck with the last steps, I do not have answers for standard error and part C. Could you please help me to finish?

x explanatory variable;

23.1, 32.8, 31.8, 32, 30.4, 24, 39.5, 24.2, 52.5, 37.9, 30.5, 25.1, 12.4, 35.1, 31.5, 21.1, 27.6

y dependent variable;

10.5, 16.7, 18.2, 17, 16.3, 10.5, 23.1, 12.4, 24.9, 22.8, 14.1, 12.9, 8.8, 17.4, 14.9, 10.5, 16.1

A) Estimate the sample correlation coefficient between X and Y .

B) Fit a simple linear regression model to this data and calculate the regression coefficients
$ \widehat{\beta_0}\ $ and $ \widehat{\beta_1}\ $ and their standard errors. Write the fitted regression equation. Interpret the coefficients.

C) Test the significance of the model vith controlling the significance  level at $0.01.$ State the hypotheses, the decision rule, the value of the test statistic, and the conclusion. What does the conclusion imply about the relationship between $Y$ and $X$?

My work;

A)

Sum of; X = 511.5 , Y = 267.1 , XY = 8653.45 , XX 16628.65 , Y*Y
4549.43

${\rm SS}_{XY}\ =\ 8653.45\ -\ \frac{1}{17}(511.5\ \times\ 267.1)\ =\ 616.882$

${\rm SS}_{XX}\ =\ 16628.65\ -\ \frac{1}{17}(511.5\ \times\ 511.5)\ =\ 1238.518$

${\rm SS}_{YY}\ =\ 4549.43\ -\ \frac{1}{17}(267.1\ \times\ 267.1)\ =\ 352.818$

${r}\ =\ \frac{\mathbf{616}.\mathbf{882}}{\sqrt{\mathbf{1238}.\mathbf{518}\ \times\ \mathbf{352}.\mathbf{818}}}\ =\ \mathbf{0}.\mathbf{933} $

B)

$\bar{X}\ =\ 30.09\ ,\ \ \ \ \bar{Y}\ =\ 15.71$

$\widehat{\beta_1}\ =\ \frac{{\rm SS}_{XY}}{{\rm SS}_{XX}}\ =\ \frac{616.882}{1238.518}\ =\ 0.5$

$\widehat{\beta_0}\ =\ \bar{Y}\ -\ \widehat{\beta_1}\ \bar{X}    =    15.71 – (0.5\times30.09) = 0.72538$

${\hat{y}}_i\ =\ 0.725\ +\ 0.5X_i$

答:
Model: $Y_i = \beta_1 X_i + \beta_0 + \varepsilon_i$ with i.i.d. $\varepsilon_i \sim N(0, \sigma^2)$. Here $X_i$ are fixed numbers and $Y_i$ are independent r.v.

Thus $\hat{\beta}_1 = \frac{\bar{Y}}{s_X^2}$,  $\hat{\beta}_0 = \frac{\overline{XY}}{s_X^2}$, where $s_X^2 = \overline{X^2} - (\overline{X})^2$.

$r = \frac{\overline{XY} }{s_X s_Y}$.

$H_0$:  $\{X_1, \ldots, X_n\}$ are independent of $\{Y_1, \ldots, Y_n \}$.

In case when $H_0$ is true we have

$$\frac{r\sqrt{n-2}}{\sqrt{1-r^2}} \sim T_{n-2}$$
i.e. it has a Student's t-distribution with parameter $n-2$.

Denote by $u_{c}$ the $c$-quantile of $T_{n-2}$.

Test have the next form: we accept $H_0$ iff $$ u_{\frac{\alpha}2} < \frac{r\sqrt{n-2}}{\sqrt{1-r^2}} < u_{1-\frac{\alpha}2}  $$

Addition:
$$\frac{\hat{\beta}_0 - \beta_0}{\sqrt{a_{\beta_0} \hat{\sigma}^2}} \sim T_{n-k},$$
$$\frac{\hat{\beta}_1 - \beta_1}{\sqrt{a_{\beta_1} \hat{\sigma}^2}} \sim T_{n-k},$$
where $\hat{\sigma}^2  = \frac{\sum_{i=1}^n (Y_i - \hat{\beta}_1 X_i - \hat{\beta}_0)^2}{n-2}$, $a_{\beta_0} = \frac{\overline{X^2}}{ns^2}$, $a_{\beta_1} = \frac{1}{ns^2}$.

It follows that:

Standard errof of $\beta_0$ is equal to $\sqrt{a_{\beta_0} \hat{\sigma}^2} = \frac{\hat{\sigma}}{\sqrt{n}}\sqrt{\frac{\overline{X^2}}{s^2}}$,

Standard errof of $\beta_1$ is equal to $\sqrt{a_{\beta_1} \hat{\sigma}^2} =  \frac{\hat{\sigma}}{\sqrt{n}}\sqrt{\frac1{s^2}}$.

This formulas coinside with formulas from people.duke.edu/~rnau/mathreg.htm


问:
How to check whether this transformation is canonical?
$Q=\ln(p),\; P=p(1-q+\ln(p))$ was given.

I read that one needs to check whether the Poisson brackets $\{Q,P\}=1$ and $\{Q,Q\}=0=\{P,P\}$, but we haven't discussed that kind of method for approval. We did discuss the poisson bracket, and how a transformation is canonical, if a function $K(Q,P,t)$ exists which satisfies both equations:

$\dot{Q_i}=\frac{\partial K}{\partial p_i}$ and $\dot{P_i}=-\frac{\partial K}{\partial Q_i}$ , where $K$ is the Hamiltonian with the new generalised momenta $P$ and new generalised coordinates $Q$. Note that $p,q$ are the old momenta and coordinates. We have also discussed the 4 types of generating functions for a transformation,
but I don't know how to solve the exercise with the methods presented in lecture. I don't know how to solve the exercise using these methods. I don't even know how to find out $K$ and all the more the generating function.

I would be very grateful if someone could help me somehow!

答: Hint: Your transformation may be written as $p=e^Q$, $P=e^Q (1-q+Q)$. This suggests one of the four types of generating functions. Use this to deduce the appropriate generating function and so verify that this is a canonical transformation.


问:
$3^{1234}$ can be written as $abcdef...qr$. What is the value of $q+r$?
It was possible to find $3^{15} ≡ 7\pmod{100}$. Knowing that $7^{4k}≡1\pmod{100}, (3^{15})^{80}≡3^{1200}≡(7)^{80}≡1\pmod{100}$.
$(3^{15})(3^{15})3^{1200}≡1\cdot 7\cdot 7\pmod{100}, 3^{1230}\cdot 3^{4}≡49\cdot 81\pmod{100}≡69\pmod{100}$ to get $6+9=15$. I felt that there must be a better way to solve this problem. What ideas should I keep in mind?

答:
You can directly find the last $2$ digits of $3^{1234}$ by writing it as $3^{1232}\cdot 9=(81)^{308}\cdot 9$.

Last two digits of $(81)^{308}=(80+1)^{308}\equiv 41 \pmod{100} $. Hence $41\cdot 9 \equiv 69 \pmod{100}$.


答:
Here is some information to achieve an alternative solution.

$3^4\equiv1\bmod4$.  $\;3^{20}\equiv1\bmod25,$ by Euler's theorem.  Therefore, $3^{20}\equiv1\bmod100$.

Therefore, $3^{1234}\equiv3^{14}\equiv9^7=(10-1)^7\equiv7\times10-1\bmod100$.

Title: $3^{1234}$ can be written as $abcdef...qr$. What is the value of $q+r$?

Body:
It was possible to find $3^{15} ≡ 7\pmod{100}$. Knowing that $7^{4k}≡1\pmod{100}, (3^{15})^{80}≡3^{1200}≡(7)^{80}≡1\pmod{100}$.
$(3^{15})(3^{15})3^{1200}≡1\cdot 7\cdot 7\pmod{100}, 3^{1230}\cdot 3^{4}≡49\cdot 81\pmod{100}≡69\pmod{100}$ to get $6+9=15$. I felt that there must be a better way to solve this problem. What ideas should I keep in mind?


问:
Showing the diagonal in an isosceles trapezoid equals the sum of the product of the two parallel sides plus the square on another side.

I drew this picture to answer the following question:

Show directly, without the use of Ptolmey's theorem, that in an isosceles trapezoid, the square on a diagonal is equal to the sum of the product of the two parallel sides plus the square on one of the other sides.

What I am trying to show is that $(DB)^2=(DC)(AB)+(AD)^2$

I can show easily that $(DB)^2=(FE)^2+(DF)^2$, but I don't know how to proceed from here. I know that $AF=EB, DF=CE$ and $AB-AF-EB=DC$, but I don't know if any of this helps me.

Any guidance to help me find the right track is helpful!

答:
Let $AB = a$, $CD = b$, $DF = h$.

Then, $BD^2 = (\frac{a+b}{2} ) ^2 + h^2$

$DC \times AB = ab$

$ AD^2 = (\frac{a-b}{2} ) ^2 + h^2$

We can now verify that the desired equality holds.


答:
$$\binom{n}{0}+\binom{n}{2}+\binom{n}{4} \ldots$$

counts the number of even-sized subsets of $\{1 \ldots n\}$. I assume that you know the total number of subsets of $\{1\ldots n\}$ is $2^n$. To show the even-sized subsets number $2^{n-1}$, a direct way would be to show that it's the same as the number of subsets of $\{1 \ldots n-1\}$.

So: lay out all of your subsets of even size. For each set $S$, consider whether it contains $n$ or not. If it does not contain $n$, then it is an even sized subset of $\{1 \ldots n-1\}$. If it does contain $n$, then $S -\{n\}$ is an odd-sized subset of $\{1 \ldots n-1\}$. Clearly this is a reversible function - given any $T \subset \{1 \ldots n-1\}$, I can tell you the unique even subset of $\{1 \ldots n\}$ that maps to it (depending on whether it has even or odd size itself). It follows that the two sets have the same size.


问:
Ray-Lens Intersection
So imagine that I have a ray parameterized as $\vec{R} = \vec{O} + t\vec{D}$, where $\vec{O}$ = origin, $t$ = parameter and $\vec{D}$ = direction vector.

I also have a spherical lens with aperture $A$, radius of curvature $r$, and center of curvature $C$.

Assume that the lens is centered on the $z$-axis and that the ray always comes in from "one side" of the lens. So if it's a convex lens, the rays will always come from its convex direction and not from it's backside.

So this is what I'm doing to solve this:

1) Treat the lens as a full sphere with center $C$ and radius $r$.

2) Do a normal ray-sphere intersection algorithm and return the closest of the two points.

3) Once I have an intersection point $(x,y,z)$, I check to see if $ |x| \le A/2$ and $|y| \le A/2$.

The reason for step 3 is that the aperture $A$ refers to the width of the lens, so if it's centered on the $z$-axis, the range of $x$ values on the lens' surface range from $-A/2 \le x \le A/2$ (and the same for $y$) even though the range of $x,y$ values for the sphere will be from $-r \le x,y \le r$.

So my question is: Is this a reasonable way to go about solving this problem? I've sort of convinced myself in my mind that this is correct, but perhaps if there is a flaw in my reasoning I would really appreciate the feedback!

答:
It looks OK to me, almost.

One remark about step #3. I don't really know what "aperture" means, but it sounds like the diameter of a circular "hole" in front of the lens. If so, I think the check in step #3 should be $\sqrt{x^2 + y^2} < A/2$. The test you're doing would correspond to a square aperture, rather than a circular one.

Title: Ray-Lens Intersection


答:
\begin{align}
\sum_{k=0}^{\left \lceil \frac{n}{2} \right \rceil}\binom{n}{2k}
&=\sum_{k=0, k\text{ even}}^{n}\binom{n}{k}
=\sum_{k=0}^{n}\binom{n}{k}-\sum_{k=0, k\text{ odd}}^{n}\binom{n}{k}\\
&=2^n+\sum_{k=0, k\text{ odd}}^{n}(-1)^k\binom{n}{k}\\
&=2^n+\sum_{k=0}^{n}(-1)^k\binom{n}{k}-\sum_{k=0, k\text{ even}}^{n}(-1)^k\binom{n}{k}\\
&=2^n+(1-1)^n-\sum_{k=0}^{\left \lceil \frac{n}{2} \right \rceil}\binom{n}{2k}\\
\implies 2\sum_{k=0}^{\left \lceil \frac{n}{2} \right \rceil}\binom{n}{2k}
&= 2^n
\implies \sum_{k=0}^{\left \lceil \frac{n}{2} \right \rceil}\binom{n}{2k} = 2^{n-1}
\end{align}


问:
How to prove that all the eigenvalues of the matrix satisfy $0\leq\lambda\leq1$？
Assume that $D\in{\bf R}^{n\times n}$ is a symmetric positive definite matrix, $B\in{\bf R}^{m\times n}$ is an arbitrary matrix, and $C\in{\bf R}^{m\times m}$ is a symmetric positive semidefinite matrix, such that the matrix $S=(C+BD^{-1}B^T)$ is nonsingular. Let $X=D^{-1}-D^{-1}B^TS^{-1}BD^{-1}$. Prove that all the eigenvalues of the matrix $D^{1/2}XD^{1/2}$ satisfy $0\leq\lambda\leq1$？

I try to utilize Woodbury formula to prove the above assertion, but it requires $C$ to be nonsingular which is not guaranteed. How can I prove the above assertion? Is there any hint related?

答:
As usual, write $A\le B$ to mean that the spectrum of $B-A$ is positive.

Then $S=C+BD^{-1}B^T\ge C$ since $D^{-1}>0$.

Hence $$0\le S^{-1}C\le 1\\
I=S^{-1}(C+BD^{-1}B^T)\\
\therefore\ 0\le I-S^{-1}BD^{-1}B^T\le 1\\
\therefore\ 0\le S^{-1}BD^{-1}B^T\le 1$$
But in general $AB$ and $BA$ have the same eigenvalues, except possibly for $0$. In this case, this does not matter, so $$0\le B^TS^{-1}BD^{-1}\le1\\
\therefore\ 0\le I-B^TS^{-1}BD^{-1}\le 1\\
\therefore\ 0\le DX\le 1\\
\therefore\ 0\le D^{1/2}XD^{1/2}\le 1$$ by conjugating with $D^{1/2}$.


问:
What's the most elementary way to solve this trigonometric problem?

A bead is threaded onto a light, inextensible string of length $4m$. One end of the string is fixed to a point, $A$, on a (vertical) wall. The other end of the string is attached to a point $B$ on the wall exactly $2m$ directly below $A.$ The bead is held in place so that it is at a distance of $1m$ from the wall, such that the string is taut and the plane the string is in is perpendicular to the plane the wall is in. Find the two possible vertical components of the displacement from $B$ to the bead.

This is a question I came up with, and I thought there should be some relatively simple trigonometric methods to get to the answer. I tried lots of stuff but nothing seemed to work.

Of course, we could find the equation of the ellipse that corresponds to the locus of points the bead could be when the string is taut, and then find the two values of $y$ when $x = 1$. But I'm looking for more elementary methods involving only trigonometry and Pythagoras. This is because I want the answer to be aimed at secondary school students who know elementary trigonometry only (Pythagoras, addition angle formulae, R addition formulae etc).

Thanks in advance.

答:
Here's an alternative trigonometric approach that doesn't assume knowledge of the divided triangle formulae, but at the expense of additional work algebraically.

Half the battle with trigonometry is getting started: which formulae capture the key details of your set-up? If we want to do trigonometry separately on the smaller triangles $\triangle ABC$ and $\triangle BCD$, the following points are essential:

$ABD$ is a straight line, so $\angle ABC$ and $\angle CBD$ are supplementary,

The two triangles share the length $BC = x$.

So we expect to need these angles and sides in our equations. For $\triangle ABC$ we also know $AB = 2$ and $AC = 4 - x$. A way to incorporate all three sides and  $\angle ABC$ would be the law of cosines:

\begin{align}
 \cos(\angle ABC ) &= \frac{2^2 + x^2 - (4-x)^2}{2 \cdot 2 \cdot x} \\
\implies -\cos(\angle CBD ) &= \frac{8x - 12}{4x} \\
\implies \cos(\angle CBD) &= \frac{3}{x} - 2 \tag{1} \\
\end{align}

where we used the fact $\angle ABC$ and $\angle CBD$ are supplementary, because we want to relate this finding to the other triangle.

Since $\triangle BCD$ is right-angled, we can stick to basic "SOHCAHTOA" rather than anything fancy, and as discussed above we certainly want to make use of $BC = x$ and $\angle CBD$. There are two choices which both give the solution, though in very different ways.

Take the sine of $\angle CBD$; combine with cosine by harmonic addition

Since $\sin (\angle CBD) = \frac{1}{x}$ then $(1)$ implies that

$$ \cos \theta - 3 \sin \theta = -2$$

This can be solved by the harmonic addition ("R") formula to deduce an acute angle of approximately 57.7°; using $\tan (\angle CBD) = \frac{1}{y}$ we have

$$y = \frac{1}{\tan (\angle CBD) } \approx 0.63$$

Note that the harmonic addition formula also produces an obtuse solution around 159.2°, for which we obtain $y \approx -2.63$, corresponding to the bead lying 0.63 vertically above $A$ instead of 0.63 vertically below $B$.

Take the cosine of $\angle CBD$; compare results and form a quadratic

From $\triangle ABC$ we had deduced $\cos(\angle CBD) = \frac{3}{x} - 2$.

From $\triangle BCD$ we have $\cos(\angle CBD) = \frac{y}{x}$.

Comparing results we have:

\begin{align}
\frac{y}{x} &= \frac{3}{x} - 2 \\
\implies y &= 3 - 2x
\end{align}

Filling that into our diagram we see $AD = y+2 = 5-2x$. We now have all the lengths of the right-angled triangle $\triangle ADC$ in terms of $x$, so we can form a quadratic using Pythagoras. Note we could instead write the sides in terms of $y$, but there would be fractional coefficients which make unnecessarily hard work. It's easier to find $x$ then use $y = 3-2x$ to convert back to $y$ at the end.

\begin{align}
(5 - 2x)^2 + 1^2 &= (4-x)^2 \\
\implies 3x^2 -12x + 10 &= 0\\
\implies 3(x-2)^2 &= 2 \\
\implies x = 2 \pm \sqrt {\frac 2 3} &= 2 \pm \frac{\sqrt 6 }{3}
\end{align}

Note that it's the lower of these two values which yields the positive root for $y$,

$$y = 3 - 2x = -1 + \frac{2\sqrt 6 }{3} \approx 0.63$$

while the negative root

$$y= -1 - \frac{2\sqrt 6 }{3} \approx -2.63$$

corresponds to the solution above $A$, as before.

Alternative: $\angle BAC = \angle DAC$ is common to $\triangle ABC$ and $\triangle ACD$

This works like the above, but has the slight advantage it doesn't require students to know $\cos (\pi - \theta) = - \cos (\theta)$. It makes use of the fact $ABD$ is a straight line by looking at how a small triangle lies inside the larger one. Using the law of cosines in $\triangle ABC$:

\begin{align}
 \cos(\angle BAC) &= \frac{2^2 + (4-x)^2 - x^2}{2 \cdot 2 \cdot (4-x)} \\
\implies \cos(\angle BAC) &= \frac{20 - 8x}{4(4-x)} \\
\implies \cos(\angle BAC) &= \frac{5-2x}{4-x} \tag{2}
\end{align}

Then in the right-angled triangle $\triangle ACD$, we have

\begin{align}
\cos (\angle BAC) &= \frac{y+2}{4-x} \tag{3} \\
\sin (\angle BAC) &= \frac{1}{4-x} \tag{4}
\end{align}

We can use $(2)$ and $(3)$ together to write $y$ in terms of $x$.

\begin{align}
\frac{5-2x}{4-x} &= \frac{y+2}{4-x} \\
\implies 5-2x &= y + 2 \\
\implies y = 3 - 2x
\end{align}

We can apply Pythagoras to $\triangle ACD$ and solve for $x$ and $y$, as above.

Alternatively, we can use $(2)$ and $(4)$ together to relate the sine and cosine:

\begin{align}
(2) \implies \cos(\angle BAC) &= \frac{5-2x}{4-x} \\
 &= \frac{8-2x-3}{4-x} \\
 &= 2 - 3 \cdot \frac{1}{4-x}\\
(4) \implies \cos(\angle BAC) &=  2 - 3 \sin (\angle BAC) \\
\implies \cos(\angle BAC) + 3 \sin (\angle BAC) &=  2
\end{align}

Which can then be solved e.g. by harmonic addition to produce an acute angle of 20.8° and obtuse angle of 122.3°. We can then solve for $y$ e.g. by using $\tan(\angle BAC) = \frac{1}{2+y}$.


问:
Given certain conditions, prove a function G(x) is always equal to 4.
Theres a question I've been having trouble with:

$G(x)$ is the function $|x+2|+|x-2|$. Show that if $-2<x<2$ then $G(x) = 4$.

Any help would be greatly appreciated. No calculus please :D

答: The graph of $x \mapsto |x + 2|$ is a vee-shape that touches the $x$ axis at $x = -2$ and then goes up with a gradient of 1 from there. The graph of $x \mapsto |x - 2|$ is a vee-shape that touches the $x$ axis at $x = 2$ and to the left of that has a gradient of $-1$. So when you add them together, the gradients (between $-2$ and $2$) will cancel out and you'll get a flat line, i.e. a constant function. It's now only necessary to work out how high up the flat line is – but just work out what it is for any value of $x$ ($x = 2$ is good) and the flat line must be at that height all the way across.</p>


答:
Building and expanding on Zaind and Lars Smith's answers, we start by substituting

$$ \sqrt{x} = u \tag 1 $$
$$ \sqrt{y} = v  \tag 2 $$

to give

$$ u^2+u v^2=\frac{95}{8} \tag 3 $$
$$ u^2 v+v^2=\frac{93}{8} \tag 4 $$

Note that if we have $ x, y \in \mathbb{R} $, we also have $ x \ge 0 $ and $ y \ge 0 $, so $ u \ge 0 $ and $ v \ge 0 $ as well.

Solving $ (3) $ for $ v^2 $:

$$ v^2 = \frac{95-8 u^2}{8 u} $$

This can be substituted into $(4)$ to give

$$ u^2 v+\frac{95-8 u^2}{8 u}=\frac{93}{8} \tag 5 $$

which allows solution for $ v $:

$$ v = \frac{8 u^2+93 u-95}{8 u^3} \tag 6 $$

We close the circle by substituting $ (6) $ into $ (3) $ to get:

$$ u^2+\frac{\left(8 u^2+93 u-95\right)^2}{64 u^5}=\frac{95}{8} \tag 7 $$

Multiply by $ 64 u^5 $ and expand to get:

$$ 64 u^7-760 u^5+64 u^4+1488 u^3+7129 u^2-17670 u+9025 = 0  $$

Recall we are only interested in positive roots of $ u $. Now let's see what we can find out about those roots. Wikipedia has a useful page on this subject, and gives the following upper bound on the absolute value of roots of a polynomial $ a_n u^n + a_{n-1} u^{n-1} + \ldots + a_0 $:

$$ 2 * \max \left\{ \left| \frac{a_{n-1}}{a_n} \right|, \left| \frac{a_{n-2}}{a_n} \right|^{1/2}, \ldots, \left| \frac{a_{1}}{a_n} \right|^{1/(n-1)}, \left| \frac{a_{n-1}}{a_n} \right|^{1/n} \right\} $$

In fact, if we're only interested in positive real roots, we can replace all the $ a_j $s which are positive with 0, which means we have to select the maximum value out of:

$$ \left\{\left(\frac{8835}{32}\right)^{1/6},\sqrt{\frac{95}{8}}\right\} $$

A minute with a calculator indicates that that the largest of these is $ \sqrt{\frac{95}{8}} \approx 3.446 $

Let's take a look at a plot of our polynomial from $ 0 $ to $ 3.5 $  to get a sense of where the roots might be:

It looks like one of those roots is very close to $ 5/2 $. Plugging $ 5/2 $ into the left hand side of $ (7) $ (much easier than doing all the terms in the polynomial) gives exactly $ 95/8 $, verifying that it is a solution. Plugging into $ (6) $ gives $ v = 3/2 $, and together we have $ \sqrt{ x y } = u v = 15/4 $.

Now, however, we still have the possibility of a root near $ u = 0.8 $, but substituting $ 0.8 $ into $ (6) $ gives $ v \approx  -3.78 $, which contradicts the assumption that $ v > 0 $, and we can discard it.

Title: Solve the system of equations for $\sqrt{xy}$

Body:
$$x + y\sqrt{x} = \frac{95}{8}$$
$$y + x\sqrt{y} = \frac{93}{8}$$
$$x, y \in \mathbb{R}$$

I can't solve this system of equations I got asked in a group. I added and substracted them to find the following equations but I don't really know what to do with them.

$$\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}-\sqrt{xy}\right) = \frac{1}{4}$$

$$\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}+\sqrt{xy}\right) - 2\sqrt{xy} = \frac{47}{2}$$

This is all I have. I checked the answer using Wolframalpha and it was $\frac{15}{4}$ for $\sqrt{xy}$.

I have been looking for this question's solution but didn't find any related question/solution on Stack Exchange or Web. Sorry if it is duplicate, thank you in advance.


问:
Geometric proof of angle sum identities for sine and cosine
Erik Jacobsen's YouTube video "Sine and Cosine Addition Formula Proof" features a geometric proof of the formula.

I have two questions:

What is the reason intuitively to place the hypotenuse of the "upper triangle" on top of the hypotenuse of the lower triangle? (Procedurally I can 'see' why, but I cannot put my finger on what is the reason behind it)

How does this proof generalize to all angles? To my mind it seems that this setting up of the triangles is very specific, I can't see how this proves that the conclusion holds true for all angles.

答:
First of all, never expect there to be an intuitively obvious path to a proof.
In many cases proof involves trying a few things to see what works and what does not work.

In this case, however, it does seem intuitive to try a proof with two triangles, one with angle $A$ at one vertex and one with angle $B$ at one vertex.
Then it seems intuitive that we might try to put the angle $A$ of one triangle at the same point as the angle $B$ of the other triangle and to make the two angles share a common side so that we get an angle of $A + B$.
This means either the hypotenuses of both triangles are touching, or the adjacent legs of both triangles are touching, or the hypotenuse of one triangle touches the adjacent leg of the other triangle.

Once we choose two triangle sides that should touch,
we have to decide how long each side should be.
Should one be longer than the other? If so, by how much?
Or should they be the same length?

In this proof the choice was to touch the hypotenuse of one triangle to the adjacent leg of the other triangle and to make those two sides the same length.
It works.
But perhaps some other choice would also have worked or might even have been simpler. You don't really know until  you try it.

As for generalizing to more angles, if the proof had used a ruler to measure the sides to scale then it would apply only to the exact angles shown (and would not be an exact proof, since ruler measurements are only as good as your ability to see how close a point is to the marks on the ruler).
But that is not what the proof does. It draws the triangles in schematic form
but never uses the actual scale of the figure to conclude anything.
You could draw the triangles with different proportions and all the labels
on all the sides would still be correct exactly as shown in the proof,
as long as $A$ and $B$ are both positive and $A+B$ is less than a right angle.

The proof does not generalize (at least not easily)
to angles $A$ and $B$ such that
$A+B$ is greater than a right angle,
because if $A+B$ is greater than a right angle the figure will
be structured differently than shown in this proof.
So it is not a proof of the most general form of the angle-sum identity for the sine function.