答:
Let's say we have an inner product $\langle \cdot, \cdot \rangle$ on $\mathbb{R}^n$. We can then use the Graham-Schmidt process to find an orthonormal basis $\{v_1, \dots, v_n\}$ of $\mathbb{R}^n$ with respect to this inner product. Then, if $v = \sum x_iv_i$ and $w = \sum y_i v_i$, then by the bilinear properties of the inner product we find that $$\langle v, w \rangle = \sum_{i = 1}^n \sum_{i = 1}^n x_i y_j \langle v_i, v_j \rangle = \sum_{i = 1}^n x_i y_i \langle v_i, v_i \rangle = \sum_{i = 1}^n x_i y_i = \begin{bmatrix}x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} \cdot \begin{bmatrix}y_1 \\ y_2 \\ \vdots \\ y_n \end{bmatrix} $$ since $\langle v_i , v_j \rangle$ is $1$ if $i = j$ and $0$ if $i \neq j$. This is why inner products give way to dot products in the usual sense.

For linear combinations, you have to be a bit careful. Usually things work pretty nicely with products of vector spaces but they get a bit weird when the dimension is infinite (ie. there is no basis).

Title: Are all product topologies/spaces over real numbers Euclidean spaces?

Body:
Until earlier today, when I thought of "real numbers" I thought of only the pure set of real numbers.
It seems I was mistaken to do so, as this seems to be the set-theoretical Baire space.
Instead, real numbers seem to be the set and what you can do with it, namely their field operations (addition and multiplication, their algebraic inverses, namely subtraction and division, and all the rules regarding associativity, commutativity and so on) and ordering.
So the real numbers can be written as $(R, +, \cdot, <)$. Usually we write them as $\mathbb R$ though and know that we can do these things with the real numbers.

Next, we can build a product space (not inner product space!), which

is the Cartesian product of a family of topological spaces equipped with a natural topology called the product space,

from a finite number ($n\in\mathbb N$) of real numbers, which would be $\mathbb R^n$.
A couple of people in How do I formally write down a Euclidean space with symbols? state that $\mathbb R^n$ is already the Euclidean space.
In other words, every finite dimensional product space of the real numbers is a Euclidean space.

But I do not see what that should be the case.
For a Euclidean space I would also need the

completeness

linear combination

Euclidean distance as metric,

Euclidean norm

the "normal" dot product

Now it seems to me that

completeness is inherited by every product space over the real numbers from the completeness of the real numbers

linear combination can be derived using the direct product. Even though I am not sure if the direct product is always implied for product topologies!

the Euclidean distance is induced by the Euclidean norm

the Euclidean norm is induced by the "normal" dot product, or according to J.W.Tanner in How do I formally write down a Euclidean space with symbols?, the dot product is induced by the Euclidean norm. However, one of them must be derived from somewhere else.

So, it seems to me that many of the properties of the Euclidean space are properties every product space over the real values has.
However, I do not see why every product space over the real values should necessarily have the dot product defined as the Euclidean space has.
For example I could choose to use a different inner product, which is not the Euclidean dot product, without having the dot product induced

So, are all product spaces over real numbers Euclidean spaces? If so, how is the dot product necessarily induced?


问:
How to prove that $n!>2^n$ holds for all $n>3$?
I suspect something with sets since one with cardinality $n$ has $n!$ permutations and its powerset contains $2^n$ elements. It could also involve binomial coefficients because of$$\begin{pmatrix}n\\0\end{pmatrix}+\begin{pmatrix}n\\1\end{pmatrix}+\begin{pmatrix}n\\2\end{pmatrix}+\cdot\cdot\cdot+\begin{pmatrix}n\\n\end{pmatrix}=2^n$$

答: The number $n!$ is the product of the $n-1$ numbers $2,3,\ldots,n$, each of which is greater than or equal to $2$ (and all of them but $2$ is actually greater than $2$) and, since $n>3$, at least one of them is greater than or equal to $4$. So, $n!>2^{n-2}\times4=2^n$

ContentLicense: CC BY-SA 4.0


问:
Approximation of alternating series $\sum_{n=1}^\infty a_n = 0.55 - (0.55)^3/3! + (0.55)^5/5! - (0.55)^7/7! + ...$
$\sum_{n=1}^\infty a_n = 0.55 - (0.55)^3/3! + (0.55)^5/5! - (0.55)^7/7! + ...$

I am asked to find the no. of terms needed to approximate the partial sum to be within  0.0000001 from the convergent value of that series. Hence, I tried find the remainder, $R_n = S-S_n$ by the improper integral

$\int_{n}^{\infty} \frac{ (-1)^n(0.55)^{2n-1}} {(2n-1)!} dn $

However, I don't know how to integrate this improper integral so is there other method to solve this problem?

答:
Your general sum,
$$
\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!}
$$
is the power series of a trigonometric function. Note that I've corrected the limits a little bit so that the correct terms come out negative. Do you know what that function is? Given that, can you evaluate the sum?</p>

ContentLicense: CC BY-SA 3.0


答: I think it's simpler than you're suggesting. Are you familiar with proof by induction? Here's a hint: once you know that $4! > 2^4$, look at how you get from $4!$ to $5!$ and from $2^4$ to $2^5$. If you can prove it for $n = 5$ given that it's true for $n = 4$, can you continue that process?

Title: How to prove that $n!>2^n$ holds for all $n>3$?

Body: I suspect something with sets since one with cardinality $n$ has $n!$ permutations and its powerset contains $2^n$ elements. It could also involve binomial coefficients because of$$\begin{pmatrix}n\\0\end{pmatrix}+\begin{pmatrix}n\\1\end{pmatrix}+\begin{pmatrix}n\\2\end{pmatrix}+\cdot\cdot\cdot+\begin{pmatrix}n\\n\end{pmatrix}=2^n$$


答:
multiply both sides by n+1
Then prove that $(n+1)!2^n$ is greater than $2^{n+1}$


答:
If $n!>2^n$ multiply both sides by $(n+1)$ to get
$$
(n+1)!>(n+1)2^n> 2 \times 2^n > 2^{n+1}
$$
with minor details to be filled.


问:
Uniqueness of congruences proof
I want to prove a statement, let $p_{1}$ and $p_{2}$ be distinct odd primes such that $(p_{1}-1)|(p_{2}-1)$. Let $m=p_{1}p_{2}$.

Prove the equation $[x]^{p_{2}}+[x]-[1]=[0]$ has a unique solution in $\mathbb{Z}_{m}$. [x] is a congruence class.

I think I can use the Chinese Remainder Theorem, but how do I prove uniqueness?

答:
I assume $\mathbb{Z}_m=\mathbb{Z}/m\mathbb{Z}$.

By the chinese remainder theorem, $\mathbb{Z}_m=\mathbb{Z}_{p_1}\times\mathbb{Z}_{p_2}$.

So the problem boils down to proving that there is a unique solution to $x^{p_2}+x-1=0$ on each $\mathbb{Z}_{p_i}$.

On $\mathbb{Z}_{p_2}$, note that $x^{p_2}=x$ for each $x$ (Fermat's little theorem) and solve the equation.

On $\mathbb{Z}_{p_1}$, note that $x=0$ is not a solution. For $x\neq 0$, use the hypothesis: $p_2-1=k(p_1-1)$ for some $k$. Then
$$x^{p_2}=x^{k(p_1-1)+1}=(x^{p_1})^kx^{-k}x$$
Similarly to the previous case, $x^{p_1}=x$. So
$$x^{p_2}=x$$
(in fact, this shows that $x^{p_2}=x$ for each $x\in\mathbb{Z}_{p_1}\setminus\left\{0\right\}$, and the same equation holds trivially for $x=0$). Solve the equation again.

Title: Uniqueness of congruences proof


问:
Evaluating $\sum_{n=0}^\infty \frac 1 {(2n+1)^4}$ using Mittag-Leffler's expansion
I am trying to evaluate the following series using Mittag-Leffler's expansion theorem. What function would be useful?
$$\sum_{n=0}^\infty \frac 1 {(2n+1)^4} = \frac{\pi^4}{96}$$

I considered differentiating the following relation twice but it did not help. And I managed to prove $0 = 0$ :D
$$\text{tanh}(z) = 2z \sum_{n=0}^\infty \frac{1}{z^2 + ((2n+1)\pi/2)^2}$$
The above expansion gives $\displaystyle \sum_{n=0}^\infty \frac{1}{(2n+1)^2}$. I need something of $4-th$ order. Which function's expansion will give me something like $\displaystyle F(z) = \sum_{n=0}^\infty \frac{1}{(2n+1)^4 + f(z)}$

\begin{align*}
\frac 1z \frac{d}{dz}\left( \frac{\text{tanh}(z)}{z} \right )
&= \frac{z \; \text{sech}(z)^2 -\text{tanh}(z)}{z^3} \\
&= \frac{z (1 - z^2 + O(z^3)) - (z  - z^3/3 + O(z^2))}{z^3} \\
&= \frac{-\frac 2 3 z^3 + O(z^5)}{z^3 } \\
&= -2/3  \\
&= \frac 1z \frac{d}{dz}\left(  2 \sum_{n=0}^\infty \frac{1}{z^2 + ((2n+1)\pi/2)^2}\right )\\
&= \frac 1z 2 \sum_{n=0}^\infty - \frac{2z}{(z^2 + ((2n+1)\pi/2)^2)^2} \\
&= - 4\sum_{n=0}^\infty \frac{1}{(z^2 + ((2n+1)\pi/2)^2)^2}
\end{align*}

答:
You should evaluate:
$$
     \lim_{z \to 0} -\frac{(\pi/2)^4}{2 z} \frac{\mathrm{d}}{\mathrm{d}z} \frac{\tanh(z)}{2z}
$$</p>


问:
Why is "If every proper subgraph of G is bipartite, then G is bipartite." False
"If every proper subgraph of G is bipartite, then G is bipartite" is apparently false.

I can't think of a justification in my head. I've been going over this, sketching out multiple non-bipartite graphs with all proper subgraphs being Bipartite, but the main graph not. I haven't come up with anything. Can someone help me?

答: So I had complicated things too much and forgotten to build up from the basics. A triangle graph would have a bipartite proper subgraph but is itself not bipartite.</p>


问:
Dual with piecewise constraint
Given the following minimization problem:

$$\begin{aligned}\min \quad & x \\ \mathrm{s.t.} \quad & f_1(x) \leq 0\end{aligned}$$

Where

$$f_1 = \begin{cases} -x + 2 &x\geq 1 \\ x  & -1 \leq x \leq 1 \\ -x -2 & x \leq - 1 \end{cases}$$

How would I go about finding the dual problem? I have been working at this for hours and cannot come up with anything.

Note: I came across this post but I don't think the solution applies here as $\max(-x + 2, x, -x -2) \neq f_1(x)$. If that were the case the solution would be trivial using epigraph form.

答:
Just use the definition:
\begin{equation}
L(x,\lambda) = x + \lambda f_1(x) =
\begin{cases}
(1-\lambda)x + 2\lambda &\mbox{if } x\geq 1 \\
(1+\lambda)x  &\mbox{if } -1 \leq x \leq 1 \\
(1-\lambda)x -2\lambda &\mbox{if } x \leq - 1.
\end{cases}\end{equation}
The dual function is $g(\lambda) = \inf_x L(x,\lambda) \ \forall \lambda \ge 0$. Notice that if $\lambda \neq 1$ then let $x\to+\infty$ or $x\to-\infty$ we obtain $L(x,\lambda) \to -\infty$. When $\lambda=1$ it is easy to check that $\min_x L(x,1) = -2$, attained at any $x\le -1$. Therefore the dual function is
\begin{equation}
g(\lambda) =
\begin{cases}
-2 &\mbox{if } \lambda = 1 \\
-\infty  &\mbox{otherwise}.
\end{cases}\end{equation}
Obviously the maximum of $g(\lambda)$ is $-2$, attained at $\lambda = 1$.

Note that the problem is not convex because $f_1(x)$ is not convex (although it can be reformulated as a convex program), but it turns out in this case that strong duality holds.

Title: Dual with piecewise constraint


问:
Why to use dummy variables in definite integrals?
I'm interested in knowing what is the utility of using dummy variables in definite integrals; that is, why use $\int_0^xf(x')dx'$ instead of $\int_0^xf(x)dx$.

In case they have no real utility and it is only a convention used for historical reasons, it would be interesting to know what were the reasons that originally led to the introduction of this concept.

Note: I have seen this question or a similar one posted here, but I'm not sure the OP was asking the question in the sense I ask, and I don't think any of the answers given explains the matter...

答:
It is a matter of correctness and to ease readability. In fact the term bound variable is much more appropriate than the term dummy variable since it  addresses the scope of variables which is relevant here.

A discrete pendant might help to better see the situation:

Consider the formulas
\begin{align*}
\sum_{k=1}^n k=\frac{n(n+1)}{2}\qquad\qquad\qquad \sum_{k=1}^k k\stackrel{???}{=}\frac{k(k+1)}{2}
\end{align*}

The index $k$ in the left sum is a bound variable with scope given by the sigma symbol. The variable $n$ on the other hand is a free variable whose existence is not bound by the sigma symbol. The right sum is nonsense. Free and bound variables are the main theme here and also in OPs example.


问:
A Semi-infinite right circular cylinder problem
A semi-infinite right circular cylinder whose axis lie along the $z$ axis, has its base on the $x$-$y$ plane.The base is maintained at a constant potential $V_0$  and the side of the cylinder is maintained at zero potential. Write down the Laplace equation to determine the potential inside the cylinder. Separate the PDE into a set of ODEs and write down the boundary conditions.

Please help....

答:
Clearly, there is no angular dependence, so the potential will be a function of $r$ and $z$.  We may separate variables by letting the potential $V(r,z)= R(r)Z(z)$  The equation for the potential is then

$$\frac{1}{rR} \frac{dR}{dr} + \frac{1}{R} \frac{d^2 R}{d r^2} + \frac{1}{Z} \frac{d^2 Z}{d z^2} = 0$$

We deal with the $R$ equation first:

$$\frac{1}{rR} \frac{dR}{dr} + \frac{1}{R} \frac{d^2 R}{d r^2} = -\lambda^2$$

where $\lambda$ is some separation constant.  Then

$$r R'' + R' + \lambda^2 r R = 0$$

where, for $z>0$, $R(a) = 0$.  The solution to this equation is a Bessel function of the first kind, of order zero: $R(r) = A \, J_0(\lambda r)$.  By requiring that the potential be zero at $r=a$ leads to $\lambda = j_{0,n}/a$, where $j_{0,n}$ is the $n$th zero of the Bessel function.

Because we have a semi-infinite cylinder with a boundary condition at $z=0$, then the solution for $Z(z)$ must take the form

$$Z(z) = B_n \, e^{- j_{0,n} z/a}$$

The general solution is

$$V(r,z) = \sum_{n=1}^{\infty} B_n \, J_0\left ( j_{0,n} \frac{r}{a} \right ) \, e^{- j_{0,n} z/a}$$

Because $V(r,0) = V_0$, and the fact that the $J_0\left ( j_{0,n} \frac{r}{a} \right )$ form an orthogonal basis set over $L^2[0,a]$, we may determine the $B_n$ from

$$B_n = \frac{\displaystyle V_0 \int_0^a dr \, r \, J_0\left ( j_{0,n} \frac{r}{a} \right )}{\displaystyle\int_0^a dr \, r \, J_0\left ( j_{0,n} \frac{r}{a} \right )^2} = \frac{2 V_0}{j_{0,n} J_1(j_{0,n})} $$


问:
How to calculate this alternating sum?
I was wondering how (and if it's even possible) to calculate the following series:
$$
\sum_{k=0}^\infty (-1)^{k}\frac{1}{(k+1)^{k+1}} = \frac{1}{1^1} - \frac{1}{2^2} + \frac{1}{3^3} - \frac{1}{4^4} + \cdots
$$
I know it's convergent, of course, but I don't know how to get the value. I've managed an expression for the denominator of the partial sums, but not one for the numerator.

答: Alpha tells us it is about $0.783431$ and will give many more digits if you ask.  It does not offer an exact form.  The alternating series theorem says it takes few terms to get many decimals-$10$ terms gives $10$ decimals and $100$ gives $200$.

Title: How to calculate this alternating sum?


问:
Lower bounding the Bhattacharyya distance with the Chernoff distance
Consider the Bhattacharyya distance

$$B = -\ln \sum_k \sqrt{p_k q_k},$$

where $p$ and $q$ are probability distributions, $\sum_k p_k = \sum_k q_k = 1$. I know that the Bhattacharyya distance is upper bounded by the Chernoff distance, i.e., $B \leq C$ with

$$C = \sup_{t\in[0,1]}\left\{-\ln\sum_k p_k^t q_k^{1-t}\right\}.$$

The equality occurs for symmetric distributions $p$ and $q$, in which case the supremum occurs at $t=1/2$.

By messing around, I found that $C$ also allows to lower bound the Bhattacharyya distance:

$$B \geq C/2.$$

A sketch of a proof is included below. Feel free to double check.

What I'd like to know is whether this inequality has a name and if I can have some reference for it. Maybe one of you knows. Or maybe it's just too trivial for that.

答:
Proof: Consider the function $c(t) = -\ln\sum_k p_k^t q_k^{1-t}$. First note the $c(t)$ is a concave function of $t$ that is positive in $[0,1]$. Moreover, it vanishes at $t = 0$ and at $t=1$. Therefore, the supremum of $c(t)$ occurs at some $t^* \in [0,1]$. The value of $c(t)$ at $t^*$ is the Chernoff distance:

$$c(t^*) = C.$$

Similarly, the Bhattacharyya distance is the value of $c(t)$ at $t = 1/2$:

$$c(1/2) = B.$$

Wihtout loss of generality, focus on the case $t^* < 1/2$. The proof for $t^* > 1/2$ is similar. Consider the line that crosses $c(t)$ at the points $(1,0)$ and $(1/2,B)$. The equation of this line is $y(t) = 2B(1-t)$. Because $c(t)$ is concave and because $t^* < 1/2$, the supremum $c(t^*)$ must occur below this line:

$$c(t^*) \leq y(t^*) \Rightarrow C \leq 2B(1-t^*) \Rightarrow C \leq 2B$$

This proves the theorem.


问:
Are all product topologies/spaces over real numbers Euclidean spaces?
Until earlier today, when I thought of "real numbers" I thought of only the pure set of real numbers.
It seems I was mistaken to do so, as this seems to be the set-theoretical Baire space.
Instead, real numbers seem to be the set and what you can do with it, namely their field operations (addition and multiplication, their algebraic inverses, namely subtraction and division, and all the rules regarding associativity, commutativity and so on) and ordering.
So the real numbers can be written as $(R, +, \cdot, <)$. Usually we write them as $\mathbb R$ though and know that we can do these things with the real numbers.

Next, we can build a product space (not inner product space!), which

is the Cartesian product of a family of topological spaces equipped with a natural topology called the product space,

from a finite number ($n\in\mathbb N$) of real numbers, which would be $\mathbb R^n$.
A couple of people in How do I formally write down a Euclidean space with symbols? state that $\mathbb R^n$ is already the Euclidean space.
In other words, every finite dimensional product space of the real numbers is a Euclidean space.

But I do not see what that should be the case.
For a Euclidean space I would also need the

completeness

linear combination

Euclidean distance as metric,

Euclidean norm

the "normal" dot product

Now it seems to me that

completeness is inherited by every product space over the real numbers from the completeness of the real numbers

linear combination can be derived using the direct product. Even though I am not sure if the direct product is always implied for product topologies!

the Euclidean distance is induced by the Euclidean norm

the Euclidean norm is induced by the "normal" dot product, or according to J.W.Tanner in How do I formally write down a Euclidean space with symbols?, the dot product is induced by the Euclidean norm. However, one of them must be derived from somewhere else.

So, it seems to me that many of the properties of the Euclidean space are properties every product space over the real values has.
However, I do not see why every product space over the real values should necessarily have the dot product defined as the Euclidean space has.
For example I could choose to use a different inner product, which is not the Euclidean dot product, without having the dot product induced

So, are all product spaces over real numbers Euclidean spaces? If so, how is the dot product necessarily induced?

答:
$\mathbb R$ is a lot of things. At its most basic it is just a set, but there are a lot of additional structure that you can tack onto the base set to make $\mathbb R$ a mathematical object of a different category, such as an ordered set, a group, a field, a vector space, a metric space, a topological space, a smooth manifold or an algebraic variety, just to name a few. Since there's usually only a single sensible/common way to do this, and since it is usually clear from the context which kind of object one is referring to, the resulting objects are also just called $\mathbb R$. Take note that all these objects differ in the additional structure they are given.

To a slightly lesser extent, this also applies to $\mathbb R^n$: it's (usually) the n-fold product of $\mathbb R$ with itself in whichever category we're currently looking at, whether it be sets, vector spaces, topological spaces, or lots of other things.

So you're talking about euclidean space, which itself is a name shared by several different types of object: when you say "euclidean space" without context, you could mean for example a metric space, a metric incidence geometry, or a Riemannian manifold.

Let's see how $\mathbb R^n$ is a metric space. A metric space is just a set X with a function $d : X \times X \to [0,\infty)$ that satisfies the properties of a distance function (symmetry, positivity, triangle inequality). We know the set, it's the usual cartesian product, now let's define the distance function:

$$d \colon \mathbb R^n \times \mathbb R^n \to [0, \infty)\\$$
$$d(x, y) = \sqrt{(x_1 - y_1)^2 + \ldots + (x_n - y_n)^2}$$

Simple enough. Here we're doing a bunch of arithmetic on real numbers, which we borrow from the usual structure of $\mathbb R$ as an ordered field, and we can use the theory of $\mathbb R$ as an ordered field to prove that this definition does indeed satisfy the of a metric space.

The definitions of the other meanings of "Euclidean space" will be very different, since they require a different type of additional structure. It turns out that these structures are very similar though, in that you can reconstruct each type of structure given only another type of structure.

Now let's talk a bit about cartesian products. In various categories, it is possible define a process that takes two objects of that category and defines, in a canonical and sensible manner, a new object of that category whose base set is exactly the cartesian product of the base sets of the two input objects. The product of two sets is a set, the product of two topological spaces is a topological space, the product of two metric spaces is another metric space.

Concrete example in a different category, this time the category of inner product spaces: given two inner product spaces $X$ and $Y$, with inner products $g_x$ and $g_y$ respectively, we can define their cartesian product like so:

An inner product space is a vector space with a dot product. For the vector space we pick $X \times Y$ (this is just a product of vector spaces), and we can define an inner product $g$ on it like so:

$$g((x_1, y_1), (x_2, y_2)) = g_x(x_1, x_2) + g_y(y_1, y_2).$$

You can check that this new space satisfies the definition of an inner product space. Indeed, using this definition of product it is now actually a mathematically provable statement that $\mathbb R^n \times \mathbb R^m \cong \mathbb R^{n+m}$, where the $\cong$ means that these two objects might be differently defined but they are indistinguishable within the category. It is in fact the case that $\mathbb R^n \times \mathbb R^m \cong \mathbb R^{n+m}$ in almost every category where you have both $\mathbb R^n$ and $\times$, and in other categories they still satisfy a weaker form of equivalence.

To conclude: there are many things called $\mathbb R$, many things called $\mathbb R^n$, many things called $\times$, because they lie in different categories. To get a Riemannian manifold called $\mathbb R^n$, it is not enough to take the product of $n$ copies of the set called $\mathbb R$, or the ordered field called $\mathbb R$, but you actually need to take the product of $n$ copies of the Riemannian manifold called $\mathbb R$.


Title: Uniqueness of congruences proof

Body:
I want to prove a statement, let $p_{1}$ and $p_{2}$ be distinct odd primes such that $(p_{1}-1)|(p_{2}-1)$. Let $m=p_{1}p_{2}$.

Prove the equation $[x]^{p_{2}}+[x]-[1]=[0]$ has a unique solution in $\mathbb{Z}_{m}$. [x] is a congruence class.

I think I can use the Chinese Remainder Theorem, but how do I prove uniqueness?

Body:
Hint: the polynomial $\,f(x)\,$ is congruent to $\,2x-1\,$ both mod $p_1$ and $q_1$ by little Fermat, so by CRT it boils down to computing its root $\,x\equiv 1/2\equiv (1\!+\!n)/2\pmod{\!n}\,$ for odd $\,n = p_1 p_2$.

i.e. let  $\,p_1,p_2 = p,q.\,$ $\,f(x) = x(x^{q-1}\!-\!1)+\!2x\!-\!1\,$ so by Fermat & $\,p\!-\!1\mid q\!-\!1^{\phantom{|^|}}\!\!$ & $ $ CRT: $\bmod p\ \&\ q\!:\ 0\equiv f(x)\equiv 2x\!-\!1^{\phantom{|^|}}\!\!\!$ $\iff \bmod{ pq}\!:\ 2x\equiv 1^{\phantom{|^|}}\!\!\!\equiv 1\!+\!pq\iff x\equiv (1\!+\!pq)/2$


问:
Checking if polynomials are chromatic polynomials
Consider the following polynomials.

$r^2(r-1)(r-2)$

$r(r-1)^2(r-2)$

$r(r-1)(r-2)^2$

$r(r-1)(r^2-2r+2)$

$r(r-1)(r^2-3r+3)$

$r(r-1)(r-2)(r-3)$

Check if these polynomials are chromatic polynomials of some simple graph.

Let $G = (X, U)$ be a simple graph, with $n$ vertices and $m$ edges and let $P_G(r)=a_n r^n +a_{n-1} r^{n-1}+\ldots+a_1 r+a_0$ the respective chromatic polynomial. Then we have that the coefficients in $P_G(r)$ alternate in sign and $m=-a_{n-1}$. In addition, the sum of the coefficients of $P_G(r)$ is zero, $\displaystyle\sum_{i=0}^n a_i=0$, if and only if $G$ has at least one edge.

Developing the notable products we have:

$r^2(r-1)(r-2)=r^4-3r^3+2r^2$

$r(r-1)^2(r-2)=r^4-4r^3+5r^2-2r$

$(r-1)(r-2)^2=r^4-5r^3+8r^2-4r$

$r(r-1)(r^2-2r+2)=r^4-3r^3+4r^2-2r$

$r(r-1)(r^2-3r+3)=r^4-4r^3+6r^2-3r$

$r(r-1)(r-2)(r-3)=r^4-6r^3+11r^2-6r$

I verified the properties described above for all polynomials and all polynomials satisfy these properties. But my professor told me that one of these polynomials that not is a chromatic polynomial.

I'm not sure if I'm in the correct way, can anyone help me?

答:
Note that the degree of a chromatic polynomial $n = |V(G)|$ and the coefficient of $r^{n-1} = -|E(G)|$. With these facts, one can find graphs with 1, 2, 3, and 6 as their chromatic polynomial fairly quickly due to their nice form. For example

$K_3$ and a singleton. (The chromatic polynomial is the product of the chromatic polynomial of each component.)

$K_3$ with an appended leaf.

$K_4$ minus an edge.

$P_G(2) = 4$, that is $G$ must be bipartite with $3$ edges and $4$ vertices. Furthermore, it must have $2$ connected components (why?).

$P_G(2)= 2$ so $G$ must be bipartite with $4$ edges and $4$ vertices, with one connected component (why?).

$K_4$.

The remaining, 4. and 5., require a bit more thinking. I hopefully have given helpful hints.


问:
Ultraproduct of pairwise non-equivalent structures
It is easy to see that the ultraproduct of a family of structures over a principal ultrafilter is elementarily equivalent to the structure whose index generates the ultrafilter, i.e. if $U$ is generated by $k \in I$, then $\prod_{i \in I} M_i / U \equiv M_k$. My question is, does the inverse also hold? That is, if $\prod_{i \in I} M_i / U \equiv M_k$ does it follow that $U$ is a non-principal ultrafilter that focuses on k?

Well, in general the answer is no, because we can simply consider the ultrapower of any structure over a non-principal ultrafilter. Clearly that's gonna be elementarily equivalent to that structure but the ultrafilter can still be non-principal.

Does the same hold if we consider a family of pairwise elementarily inequivalent structures? So, if $M_i, i \in I$ is an infinite family of structures such that $M_i \not\equiv M_j$ for $i\neq j$, does there exist a non-principal ultrafilter $U$ such that $\prod_{i \in I} M_i / U \equiv M_k$ for some $k$, or does $U$ have to necessarily be principal for this to hold?

答:
We can perform the following sort of silly trick. If $M_i$ is a family of pairwise inequivalent structures ranging over an index set $I$ and $U$ is a non-principal ultrafilter on $I$, write $M_{\infty} = \prod_{i \in I} M_i/U$ for the ultraproduct. There are two cases:

Case 1: $M_{\infty}$ is elementary equivalent to some $M_i$. In this case we are done.

Case 2: $M_{\infty}$ is not elementary equivalent to any of the $M_i$. Then we can extend $M_i$ to a family of structures ranging over the index set $I_{\infty} = I \cup \{ \infty \}$ which is still pairwise inequivalent. $U$ induces a unique non-principal ultrafilter $U_{\infty}$ on $I_{\infty}$ given by taking all the subsets in $U$ and adding to them all the subsets in $U$ together with $\infty$. Now consider the ultraproduct

$$M_{\infty + 1} = \prod_{i \in I_{\infty}} M_i/U_{\infty}.$$

By Łoś's theorem, $M_{\infty+1} \models \phi$ iff $X_{\infty} = \{ i \in I_{\infty} : M_i \models \phi \} \in U_{\infty}$. By construction, this is true iff $X = \{ i \in I : M_i \models \phi \} \in U$, which is true iff $M_{\infty} \models \phi$ by a second application of Łoś's theorem. So $M_{\infty+1}$ is elementary equivalent to $M_{\infty}$.


问:
Function $F$ that $F(x)+F(-x)=\lim_{y\rightarrow\infty}F(y), \forall x \in \mathbb{R}$
I was wondering if there is a name for function $F: \mathbb{R} \rightarrow \mathbb{R}$ with property $$F(x)+F(-x)=\lim_{y\rightarrow\infty}F(y), \forall x \in \mathbb{R}?$$

An example is when $F$ is the probability distribution function of a symmetric density function. A random variable with such probability distribution function is called symmetric.

I would like to know if properties of such kind of functions have been studied both generally and especially for probability theory i.e. when $F$ is a probability distribution function.

Added: Also I wonder in Kai Lai Chung's A Course in Probability Theory where he wrote for the probability distribution function of a symmetric distribution
$$F(x)=1-F(-x-), \forall x \in \mathbb{R},$$
what is the meaning of $-x-$? I guess that because a probability distribution function is a right-continuous function, maybe Chung wanted to emphasize that it is the left limit that is used to define symmetric distribution.

Thanks and regards!

答:
If we assume that the limit on the right hand side is zero, then $F$ is called antisymmetric or odd, that is, it satisfies $F(x) = -F(-x)$.

Now let's assume that the limit $c = \lim_{x \to \infty} F(x) + F(-x)$ exists. Then $G(x) = F(x) - \frac{c}{2}$ satisfies $G(x) + G(-x) = F(x) + F(-x) - c$, hence $\lim_{y \to \infty} G(y) + G(-y) = 0$, so $G$ is odd by the previous reasoning, hence $F(x) = G(x) + c/2$ is the sum of an odd function and the scalar $c/2$. Odd functions arise quite frequently and a bunch of their properties can be found on the Wikipedia page linked to above.


问:
Parabolic PDE existence/uniqueness
Consider the parabolic PDE:
$$\frac{\partial u}{\partial t} = u^2\frac{\partial^2u}{\partial x^2} + u^3$$
with some initial condition.

Apparently this is a straightforward parabolic PDE in which I can apply standard results to prove short term existence and uniquness. Can someone tell/refer me to these standard results please? The equation is non-linear and I haven't seen any theory for non-linear results.

Thanks

答:
I will assume that the initial data is smooth. First we define the appropriate energy (this depends on oyur problem, but usually it is enough to take some high order Sobolev norm). For instance we take $E[u]=\|u\|_{H^1}$. Now we obtain an a priori bound for this energy. Multiplying the equation by u and integrating by parts we get
$$
\frac{1}{2}\frac{d}{dt}\|u(t)\|_{L^2}^2=\int u u_t=\int u^3 u_{xx}+u^4=\int -3(u_x)^2u^2+u^4
$$
$$
\leq \|u\|_{L^2}^2\|u\|_{L^\infty}^2\leq C\|u\|_{H^1}^4 \text{ (by Sobolev embedding)}.
$$
Now we take one derivative and follow the same way:
$$
\frac{1}{2}\frac{d}{dt}\|u_x(t)\|_{L^2}^2=\int u_x u_{xt}=\int 2u(u_x)^2u_{xx}+u^2u_xu_{xxx}+3u^2(u_x)^2=
$$
We integrate by parts in the second term and we get
$$
\int u^2u_xu_{xxx}=-\int u^2(u_{xx})^2-2u(u_x)^2u_{xx}.
$$
Inserting this in the previous equation we get
$$
\frac{1}{2}\frac{d}{dt}\|u_x(t)\|_{L^2}^2\leq3\|u\|_{L^\infty}^2\|u_x\|_{L^2}^2\leq C\|u\|_{H^1}^4.
$$
Collecting both estimates, we get
$$
\frac{d}{dt}\|u\|_{H^1}\leq C\|u\|_{H^1}^3,
$$
and we can use Gronwall inequality to get
$$
\sup_{t\in[0,T)}\|u(t)\|_{H^1}\leq C(u_0),
$$
for an explicit time $T=T(u_0)$.

To conclude the argument you should regularize the system. Typically you should take some convolutions with the usual mollifier to have that the derivative operator is bounded in this Sobolev spaces. If you take the correct regularized system the same bounds hold and you have enough compactness to ensure the existence of a smooth limit which is also a solution.

All these techniques are quite standart and I refer you the book of A.Majda and A. Bertozzi "Vorticity and incompressible flow" (Chapter 3).

I hope that this is useful for you.

Title: Parabolic PDE existence/uniqueness


问:
Can we show $\frac{\nu\left(B_\varepsilon(x)\right)}{\mu\left(B_\varepsilon(x)\right)}\xrightarrow{\varepsilon\to0+}\frac{{\rm d}\nu}{{\rm d}\mu}(x)$?
Let $(E,\mathcal E,\mu)$ be a finite measure space and $\nu$ be a finite signed measure on $(E,\mathcal E)$ with $\nu\ll\mu$. By the Radon-Nikodým theorem, $$\nu=f\mu\tag1$$ for some $f\in L^1(\mu)$.

If $(E,d)$ is a metric space and $\mathcal E=\mathcal B(E)$, are we able to show that $$g_\varepsilon(x):=\left.\begin{cases}\displaystyle\frac{\nu\left(B_\varepsilon(x)\right)}{\mu\left(B_\varepsilon(x)\right)}&\text{, if }\mu\left(B_\varepsilon(x)\right)>0\\0&\text{, otherwise}\end{cases}\right\}\;\;\;\text{for }x\in E$$ is convergent in a suitable mode of convergence (e.g. in $\mu$-almost everywhere or in $L^1(\mu)$ as $\varepsilon\to0+$ and the limit $g$ is a version of $f$?

If someone isn't aware of a rigorous proof, but can give an intuitive explanation of the basic idea, I'd appreciate that as well.

答:
Note that $\frac{\nu(B_{\epsilon}(x))}{\mu(B_{\epsilon}(x))}=\frac{1}{\mu(B_{\epsilon}(x))}\int_{B_{\epsilon}(x)} f(y) \, d\mu(y).$

By the Lebesgue differentiation theorem, we know that $\frac{1}{\mu(B_{\epsilon}(x))}\int_{B_{\epsilon}(x)} f(y) \, d\mu(y)\to f(x)$ for $\mu$ a.e. $x,$ as $\epsilon\to 0.$

This completes the proof.


问:
If A and B are row equivalent, and A$\mathbf x$ = $\mathbf b$ has a solution, then B$\mathbf x$ = $\mathbf b$ has a solution
Assume towards contradiction that B$\mathbf x$ = $\mathbf b$ has no solutions. Therefore the augmented matrix is inconsistent.

Let $E_1$, $E_2$,..., $E_p$ be a sequence of elementary matrices representing row operations that render B row equivalent to some matrix A.

Thus we have:

A$\mathbf x$ = $E_1 \cdot E_2 \cdot ... E_p \cdot$ B = $\mathbf b$

Since we know that all elementary matrices are invertible, it follows that:

B = $(E_1 \cdot E_2 \cdot ... E_p)^{-1} \mathbf b$

At this point I feel that perhaps the proposition might be false

答:
Your feeling is correct.

For instance:

If $A=\left[\begin{array}{cc}1&1\\1&1 \end{array}\right]$, then $Ax=\left[\begin{array}{c}1\\1 \end{array}\right]$ has a solution.

Now $B=\left[\begin{array}{cc}1&1\\0&0 \end{array}\right]$ is row equivalent to $A$, but there is no solution to $Bx=\left[\begin{array}{c}1\\1 \end{array}\right]$


问:
Is $\mathbb{Z}[√13] $ a Unique factorization domain?
I think it is not so as $12 $ can be written in two ways $12=2.6=(1+\sqrt{13})(-1+\sqrt{13})$. Are these two factorization unique upto irreducibles? Please help.

答:
$\mathbf{2}$ is Irreducible in $\boldsymbol{\mathbb{Z}\!\left[\sqrt{13}\right]}$

Suppose that
$$
\left(a+b\sqrt{13}\right)\left(c+d\sqrt{13}\right)=2\tag1
$$
where $2\nmid\left(a+b\sqrt{13}\right)$ and $2\nmid\left(c+d\sqrt{13}\right)$. Then
$$
ac+13bd=2\tag2
$$
and
$$
ad+bc=0\tag3
$$
If $2\mid a$, then since $2\nmid\left(a+b\sqrt{13}\right)$, we must have $2\nmid b$. Then $(2)$ says that $2\mid d$ and $(3)$ says that $2\mid c$. However, this means that $2\mid\left(c+d\sqrt{13}\right)$, which contradicts the hypotheses. Therefore, $2\nmid a$. Similarly, $2\nmid c$.

Multiplying $(2)$ by $ac$ and applying $(3)$ to eliminate $d$ gives
$$
a^2c^2-13b^2c^2=2ac\tag4
$$
and therefore,
$$
(ac-1)^2-1\equiv b^2c^2\pmod4\tag5
$$
However, since $a,c\equiv1\pmod2$, we must have $(ac-1)^2-1\equiv3\pmod4$. Since $3$ is not a quadratic residue mod $4$, $(5)$ can have no solutions.

Thus, $2$ is irreducible in $\mathbb{Z}\!\left[\sqrt{13}\right]$.

$\boldsymbol{\mathbb{Z}\!\left[\sqrt{13}\right]}$ is not a UFD
$$
\left(\sqrt{13}-3\right)\left(\sqrt{13}+3\right)=2\cdot2
$$


问:
Approximation of alternating series $\sum_{n=1}^\infty a_n = 0.55 - (0.55)^3/3! + (0.55)^5/5! - (0.55)^7/7! + ...$
$\sum_{n=1}^\infty a_n = 0.55 - (0.55)^3/3! + (0.55)^5/5! - (0.55)^7/7! + ...$

I am asked to find the no. of terms needed to approximate the partial sum to be within  0.0000001 from the convergent value of that series. Hence, I tried find the remainder, $R_n = S-S_n$ by the improper integral

$\int_{n}^{\infty} \frac{ (-1)^n(0.55)^{2n-1}} {(2n-1)!} dn $

However, I don't know how to integrate this improper integral so is there other method to solve this problem?

答: For an alternating series the error is less than the first neglected term, so just keep calculating until the terms get smaller than $10^{-7}$</p>


问:
How to prove that $df_x$ is well defined with respect to different chart $\psi^{\prime}_{V^{\prime}}$
I am studying Moroianu book lectures on Kähler geometry. I am  trying to show that the map $df_x$ is well defined. Choose charts $\phi^{\prime}_{U^{\prime}}$ and $\psi^{\prime}_{V^{\prime}}$ around x and f(x). It follows that $[U,u] = [U^{\prime},u^{\prime}]$ and $[V,v] = [V^{\prime},v^{\prime}]$. It follows
$$u = d\phi_{{UU^{\prime}}_{\phi_U^{\prime}(x)}} (u^{\prime})$$
$$v  = d\psi_{{VV^{\prime}}_{\psi_V^{\prime}(x)}} (v^{\prime})$$
Therefore
$df([U,x]) =
[V,d(\psi_V \circ f \circ \phi^{-1}(U) \circ d\phi_{{UU^{\prime}}_{\phi_U^{\prime}(x)}} (u^{\prime})] = [V, d(\psi_V \circ f \circ \phi_U^{\prime})x]$

I am missing $\psi_V^{\prime}$. I am not sure how to get rid of $\psi_V$? In order to show well-definedness.

答:
Extending my comment, you have by definition
$$
\begin{align}
  \mathrm{d}f_x([U,u]) = [V, \mathrm{d}(\psi_{V}\circ f \circ \phi_{U}^{-1})_{\phi_U(x)}(u)]. \tag{1}
\end{align}
$$
The definition of tangent vectors as equivalence classes means that for tangent vectors at $x\in M$
$$
\begin{align}
  [U^{\prime},u^{\prime}]=[U,\mathrm{d}(\phi_{UU^{\prime}})_{\phi_{U^{\prime}}(x)}(u^{\prime})] \tag{2}
\end{align}
$$
and for tangent vectors at $f(x)\in N$
$$
\begin{align}
  [V^{\prime},v^{\prime}]=[V,\mathrm{d}(\psi_{VV^{\prime}})_{\psi_{V^{\prime}}(f(x))}(v^{\prime})] \tag{3}
\end{align}
$$
where $\phi_{U^{\prime}U}=\phi_{U^{\prime}}\circ \phi_{U}^{-1}$ and $\psi_{V^{\prime}V}=\psi_{V^{\prime}}\circ \psi_{V}^{-1}$ are the transition maps between two charts around $x\in M$ and $f(x)\in N$.

Carefully unraveling all the definitions you now get
$$
\begin{align}
    \mathrm{d}f_x([U^{\prime},u^{\prime}])
        &= \mathrm{d}f_x\big([U,\mathrm{d}(\phi_{UU^{\prime}})_{\phi_{U^{\prime}}(x)}(u^{\prime})]\big)  \\
        &= \big[V, \mathrm{d}(\psi_{V}\circ f \circ \phi_{U}^{-1})_{\phi_{U}(x)}\circ \mathrm{d}(\phi_{UU^{\prime}})_{\phi_{U^{\prime}}(x)}(u^{\prime})\big]  \\
        &= \big[V, \mathrm{d}\big(\psi_{V}\circ f \circ \phi_{U}^{-1} \circ \phi_{UU^{\prime}}\big)_{\phi_{U^{\prime}}(x)}(u^{\prime})\big]  \\
        &= \big[V, \mathrm{d}\big(\psi_{V} \circ \psi_{V^{\prime}}^{-1} \circ \psi_{V^{\prime}} \circ f \circ \phi_{U}^{-1} \circ \phi_{U} \circ \phi_{U^{\prime}}^{-1}\big)_{\phi_{U^{\prime}}(x)}(u^{\prime})\big]  \\
        &= \big[V, \mathrm{d}(\psi_{VV^{\prime}})_{\psi_{V^{\prime}}(f(x))} \Big(\mathrm{d}\big(\psi_{V^{\prime}} \circ f \circ \phi_{U^{\prime}}^{-1}\big)_{\phi_{U^{\prime}}(x)}(u^{\prime})\Big)\big]  \\
        &= \big[V^{\prime}, \mathrm{d}\big(\psi_{V^{\prime}} \circ f \circ \phi_{U^{\prime}}^{-1}\big)_{\phi_{U^{\prime}}(x)}(u^{\prime})\big]
\end{align}
$$
where we used in order of appearance eq. (2), eq. (1), chain rule, definition of transition maps, chain rule, eq. (3).


问:
$\{ x,y \} \times \mathbb{R}$ not connected in the standard topology why?
I know that singletons like $$\{x \}$$ is a connected set in the standard topology. Also $\{x\} \times \mathbb{R}$ is connected.

But why isn't $\{ x,y \} \times \mathbb{R}$ connected? What are the two disjoint open sets such that the union is $\{ x,y \} \times \mathbb{R}$ ? I'd say that they are precisely $U = \{x\} \times \mathbb{R}$ and $V=\{y\} \times \mathbb{R}$. Is it correct?

答:
There is a single topology on $\{x\}$, so there can be no chance to get it wrong. If $B$ is any topological (or metric, if you stick with them) space, then the map
$$
f\colon B\to \{x\}\times B,\qquad b\mapsto (x,b)
$$
is easily seen to be a homeomorphism, that is, bijective, continuous and with continuous inverse. So, if $B$ is connected, also $\{x\}\times B$ is connected.

Suppose now $A$ and $B$ are nonempty topological spaces, with $A\times B$ connected. Since the map $A\times B\to A$, $(a,b)\mapsto a$ is continuous and surjective, also $A$ is connected.

Is $A=\{x,y\}$ with the discrete topology connected?

In the case of $\{x,y\}\times\mathbb{R}$, a disconnection is indeed $\{x\}\times\mathbb{R}\cup\{y\}\times\mathbb{R}$. But having a more general argument is more helpful, in my opinion.


问:
Prove that if $z^n$ is a palindrome for some $n>0$, then $z$ is also a palindrome for any alphabet E.
Prove that if $z^n$ is a palindrome for some $n>0$, then $z$ is also a palindrome for any alphabet E.

Here's a proof of the statement above:

Let $w = x^n$.  If $n = 1$, then the result is trivial.

Suppose $n > 1$. Then $$x^n = reverse (x^n) = (reverse  (x))^n.$$
Write this as $$x x \dots x = reverse (x) reverse (x) \dots reverse (x).$$ But $x$ and $reverse (x)$ have the same length, so that $x = reverse (x)$.

Can anyone justify this line in the proof?
$$reverse (x^n) = (reverse  (x))^n$$

Or provide an alternative proof.

答:
$$\begin{array}{rcl} z^k \in \text{Palindrome} &\implies& z^k = \text{rev}(z^k) \\
&\implies& z^k{}_1 = z^k{}_{kn}~,~z^k{}_2 = z^k{}_{kn-1}~,~z^k{}_3 = z^k{}_{kn-2}~,~\dots \\
&\implies& z_1 = z_n~,~z_2 = z_{n-1}~,~z_3 = z_{n-2}~,~\dots \\
&\implies& z \in \text{Palindrome}
\end{array}$$

Third line follows from $z^k{}_{a} = z_{a ~\rm{mod}~ n}$


答: I don't believe there is a closed form. It's well known that this value is $\int_0^1x^xdx$, which is called Sophomore's Dream.

Title: How to calculate this alternating sum?

Body:
I was wondering how (and if it's even possible) to calculate the following series:
$$
\sum_{k=0}^\infty (-1)^{k}\frac{1}{(k+1)^{k+1}} = \frac{1}{1^1} - \frac{1}{2^2} + \frac{1}{3^3} - \frac{1}{4^4} + \cdots
$$
I know it's convergent, of course, but I don't know how to get the value. I've managed an expression for the denominator of the partial sums, but not one for the numerator.


问:
Calculate two vectors given their norms and angle

For two vectors $\mathbf{u,v}$ in $\mathbb{R}^n$ euclidean space, given:

$\|\mathbf{u}\| = 3$

$\|\mathbf{v}\| = 5$

$\angle (\mathbf{u,v})=\frac{2\pi}{3}$

Calculate the length of the vectors

$4\mathbf{u}-\mathbf{v}$

$2\mathbf{u}-7\mathbf{v}$

I'm not sure how to approach this with the given information

With the formula for the angle between the two vectors being

$$\cos \theta=\frac{\mathbf{u\cdot v}}{\|\mathbf{u}\|\cdot\|\mathbf{v}\|}$$

I already have the denominator, but how do I get the point product of u and v in this case?

A point to start would be most appreciated

答:
You have the angle so it's trivial to rearrange for $\mathbf {u\cdot v}$, by multiplying by the denominator. Then you know all combinations of dot products of the two vectors.

Using the linearity (distributivity) of the dot product allows you to now calculate $$ \lVert a\mathbf u + b\mathbf v\rVert^2 = (  a\mathbf u + b\mathbf v)\cdot (  a\mathbf u + b\mathbf v) = a^2  \mathbf {u\cdot u} + \cdots$$


问:
proof verification on compact metric with $d(A,B)$
Let $(X,d)$ be a metric space and $A,B$ non empty subsets. If $A$ is compact show that $d(A,B)=0\Leftrightarrow A\cap \overline{B}\neq \varnothing  $

$$  d(A,B)=\inf\left \{ d(a,b):a\in A ,b\in B  \right \}$$

$\Rightarrow $

let's suppose  $A\cap \overline{B}= \varnothing  $ now because $A$ is compact and $\overline{B}$ is closed then it's been proved that $d(A,B)>0$ , contradiction

$\Leftarrow $

let $A\cap \overline{B}\neq \varnothing  $, I will prove that $d(A,B)=0$, because   $A\cap \overline{B}\neq \varnothing  $, I can take $(x_n)\subseteq A   $ and $z_n \in A\cap \overline{B}$ so $z_n\in B\left(x_n,\frac{1}{n}\right)$

$d(x_n,z_n)< \dfrac{1}{n}\rightarrow 0$

$y_n\in B $ and $z_n\in B\left(y_n,\frac{1}{n}\right)$ we can approach $d(A,B)$ with a sequence $d(x_n,y_n)\rightarrow d(A,B)$ $d(x_n,y_n)\leq d(x_n,z_n)+ d(z_n,y_n)\rightarrow 0$ as $n\rightarrow \infty$

$d(A,B)=0 $

Is this ok?

答:
It's a good start.

$\Rightarrow$: as @José Carlos Santos has pointed out, there's a gap here. Hint: each point in $A$ has a neighborhood that's disjoint from $B$, which gives an open cover of $A$, so you can use compactness.

$\Leftarrow$: I don't understand where your $z_n$ are coming from. I suggest you start with a point $x \in A \cap \overline{B}$, and then a sequence $(x_n)$ in $B$ with $x_n \rightarrow x$, and then the rest of your argument should work.


答:
Well first thing I've thaught of was using the fact that : $$ \left(\forall k\in\mathbb{N}\right)\left(\forall x>0\right),\ \frac{1}{\left(k+1\right)^{x}}=\frac{1}{\Gamma\left(x\right)}\int_{0}^{+\infty}{y^{x-1}\,\mathrm{e}^{-\left(k+1\right)y}\,\mathrm{d}y} $$

Thus, for any $ k\in\mathbb{N} $, we have : $$ \frac{1}{\left(k+1\right)^{k+1}}=\frac{1}{k!} \int_{0}^{+\infty}{y^{k}\,\mathrm{e}^{-\left(k+1\right)y}\,\mathrm{d}y}$$

Let's now try to check if we could swith the sum with the integral :

Let $ n\in\mathbb{N} $, we have :

\begin{aligned} \left|\sum_{k=0}^{n}{\frac{\left(-1\right)^{k}}{k!}\int_{0}^{+\infty}{y^{k}\,\mathrm{e}^{-\left(k+1\right)y}\,\mathrm{d}y}}-\int_{0}^{+\infty}{\sum_{k=0}^{+\infty}{\frac{\left(-1\right)^{k}}{k!}y^{k}\,\mathrm{e}^{-\left(k+1\right)y}}\,\mathrm{d}y}\right|&=\left|\int_{0}^{+\infty}{\sum_{k=n+1}^{+\infty}{\left(-1\right)^{k}\frac{y^{k}\,\mathrm{e}^{-\left(k+1\right)y}}{k!}}\,\mathrm{d}y}\right|\\ &\leq\int_{0}^{+\infty}{\left|\sum_{k=n+1}^{+\infty}{\left(-1\right)^{k}\frac{y^{k}\,\mathrm{e}^{-\left(k+1\right)y}}{k!}}\right|\mathrm{d}y} \\ &\leq\frac{1}{\left(n+1\right)!}\int_{0}^{+\infty}{y^{n+1}\,\mathrm{e}^{-\left(n+2\right)y}\,\mathrm{d}y}\\ &\leq \frac{1}{\left(n+2\right)^{n+2}}\underset{n\to +\infty}{\longrightarrow}0\end{aligned}

Which means : \begin{aligned} \sum_{n=0}^{+\infty}{\frac{\left(-1\right)^{n}}{n!}\int_{0}^{+\infty}{y^{n}\,\mathrm{e}^{-\left(n+1\right)y}\,\mathrm{d}y}} &= \lim_{n\to +\infty}{\sum_{k=0}^{n}{\frac{\left(-1\right)^{k}}{k!}\int_{0}^{+\infty}{y^{k}\,\mathrm{e}^{-\left(k+1\right)y}\,\mathrm{d}y}}}\\ &=\int_{0}^{+\infty}{\sum_{k=0}^{+\infty}{\frac{\left(-1\right)^{k}}{k!}y^{k}\,\mathrm{e}^{-\left(k+1\right)y}}\,\mathrm{d}y}\\ &=\int_{0}^{+\infty}{\mathrm{e}^{-y\left(\mathrm{e}^{-y}+1\right)}\,\mathrm{d}y}\\ &=\int_{0}^{+\infty}{\left(\mathrm{e}^{-y}\right)^{\mathrm{e}^{-y}}\,\mathrm{e}^{-y}\,\mathrm{d}y}\\ &=\int_{0}^{1}{x^{x}\,\mathrm{d}x} \end{aligned}

Thus : $$ \sum_{n=0}^{+\infty}{\frac{\left(-1\right)^{n}}{\left(n+1\right)^{n+1}}}=\int_{0}^{1}{x^{x}\,\mathrm{d}x} $$

Title: How to calculate this alternating sum?


问:
Find power series to determine infinite sum convergence
Find an appropriate power series and apply the ratio test to show the following infinite sum converges:
$$
  \sum_{n=0}^\infty \frac{n^2}{5^n}
$$

I'd simply do it like this:
$$
  \text{Let } a_n = \frac{n^2}{5^n}. \text{Then by the ratio test we have}\\
  \begin{align*}
    \left|{\frac{a_{n+1}}{a_n}}\right|= \left|\frac{(n+1)^2}{5^{n+1}}\cdot \frac{5^n}{n^2}\right| = \left|\frac{n^2+2n+1}{5n^2}\right|
  \end{align*}\\
  \implies \lim_{n\to\infty} \frac{n^2+2n+1}{5n^2} = \lim_{n\to\infty} \frac{1+\frac{2}{n}+\frac{1}{n^2}}{5} = \frac{1}{5} < 1
$$

I'm confused since I thought I could directly apply the ratio test on this series. (without finding a power series)

I guess my professor wants me to solve it like that, but I have no idea what kind of convergent power series I could (upper-)bound this with.

答:
Hint:

Work with $\sum\limits_{n=0}^\infty\ x^n\;,\;\;|x|<1\;$ , and then differentiatie twice...In fact, you should also find out what this series converges to.


答:
Hint

Consider the geometric power series
$$\sum x^n$$
with $ R=1 $ as radius of convergence.

its derivatives at $ x\in(-1,0)\cup(0,1)$
$$\frac 1x\sum nx^n$$
and
$$\sum n^2x^n$$
have the same Radius.

Title: Find power series to determine infinite sum convergence

Body:
Find an appropriate power series and apply the ratio test to show the following infinite sum converges:
$$
  \sum_{n=0}^\infty \frac{n^2}{5^n}
$$

I'd simply do it like this:
$$
  \text{Let } a_n = \frac{n^2}{5^n}. \text{Then by the ratio test we have}\\
  \begin{align*}
    \left|{\frac{a_{n+1}}{a_n}}\right|= \left|\frac{(n+1)^2}{5^{n+1}}\cdot \frac{5^n}{n^2}\right| = \left|\frac{n^2+2n+1}{5n^2}\right|
  \end{align*}\\
  \implies \lim_{n\to\infty} \frac{n^2+2n+1}{5n^2} = \lim_{n\to\infty} \frac{1+\frac{2}{n}+\frac{1}{n^2}}{5} = \frac{1}{5} < 1
$$

I'm confused since I thought I could directly apply the ratio test on this series. (without finding a power series)

I guess my professor wants me to solve it like that, but I have no idea what kind of convergent power series I could (upper-)bound this with.


答: Your approach is fine. The approach that the question might have been looking for is to consider the power series $\sum_{n=0}^\infty a_n x^n$ where $a_n=n^2$ and check that $x=1/5$ is within the radius of convergence.


问:
Bessel function and its derivative in a boundary condition
I have a function $u_{nj}=u_{nj}(r)$ of the radius variable $r$ (cylindrical coordinates) that satisfy the following Bessel equation

$$ r^{2} u''_{nj}+ r u'_{nj} +(k_{nj}^{2}-n^{2})u_{nj}=0, $$

where $n$ is integer, $j$ is the index denoting the $j$-th zero, and the solution must have finite (physical) amplitude everywhere within the domain inside a cylinder of radius $r=a$ (so typically this excludes second-kind Bessel functions and only allows the first kind, $J_{n}$).

This function ($u_{nj}$) also needs to satisfy the following boundary condition at $r=a$

$$ \left[ u_{nj} +(1+i)\beta_{0} a M u'_{nj} \right]_{r=a}=0,  $$

where $\beta_{0}, M, a$ are all constants, and we have $M<<1$ (small).

Somehow the result that satisfies both the equation and boundary condition above turns out to be (after taking first-order approximation in the small parameter $M$) given by

$$ u_{nj}=J_{n}(k_{nj} r), \ \ \ \ \text{where}  \ \ \ k_{nj}=\frac{V_{nj}}{a} [1-(1+i)\beta_{0} M],$$

and $V_{nj}$ denotes the $j$-th zero of the $J_{n}$ function.

I am having difficulty seeing how this result was reached. In particular, how can one attempt to solve an equation of the form

$$\left[ J_{nj} + (\text{constant}) J'_{nj} \right]_{r=a}=0,$$

to reach such result? Am I missing something trivial here?

Any help would be appreciated.

答:
From the condition
\begin{equation}
 u_{nj}(a) +(1+i)\beta_{0} a M u'_{nj}(a)=0
\end{equation}
with $u_{nj}=J_{n}(k_{nj} r)$ we must have
\begin{equation}
 J_{n}(k_{nj}a) +(1+i)\beta_{0} k_{nj}a M J'_{n}(k_{nj}a)=0 \tag{1}\label{1}
\end{equation}
For $M\ll1$, the first term dominates and thus $k_{nj}a\simeq V_{nj}$, where $V_{nj}$ is a zero of $J_n$. To go a step forward, we assume that
\begin{equation}
 k_{nj}a= V_{nj}+\kappa M+o(M)
\end{equation}
This expression and a first order Taylor expansion
\begin{align}
 J_{n}(k_{nj}a)&\sim J_n(V_{nj})+ \kappa M J'_n(V_{nj})\\
 &\sim\kappa M J'_n(V_{nj})
\end{align}
plugged into the condition (\ref{1}) gives,
\begin{equation}
 \kappa M J'_n(V_{nj})+(1+i)\beta_{0} \left(V_{nj}+\kappa M  \right) M J'_{n}(V_{nj}+\kappa M)=0
\end{equation}
or, by keeping the terms up to the first order in $M$,
\begin{equation}
 \kappa M J'_n(V_{nj})+(1+i)\beta_{0} V_{nj} M J'_{n}(V_{nj})=0
\end{equation}
which reads
\begin{equation}
  M J'_n(V_{nj})\left[\kappa+(1+i)\beta_{0}V_{nj}\right]=0
\end{equation}
This condition is fulfilled if
\begin{equation}
 \kappa=-(1+i)\beta_{0} V_{nj}
\end{equation}
We deduce
\begin{equation}
 k_{nj}a\simeq V_{nj}\left( 1- (1+i)\beta_{0}M\right)
\end{equation}


问:
Given a complex number set, how do you prove which set is it in the complex plane?
Let $a,b \in \mathbb{C}$ is it true that:
$|a| < |b|$ implies $|a|^{2} < |b|^{2}$
I want to use this to build sets of points in the complex plane since I have things like:
$$|z-1| < |z|$$
So I want to apply power to both sides and then change $z = x+iy$ to discover what the set is in terms of the real coordinates.

Is it useful?

Which other strategies do you recommend in order to discover what the set is in terms of the real coordinates?

答:
The map $x\mapsto x^2$ is strictly increasing on $[0,\infty)$: since $\lvert z\rvert\in[0,\infty)$, we indeed have that $\lvert a\rvert<\lvert b\rvert$ if and only if $\lvert a\rvert^2<\lvert b\rvert^2$.

This strategy is fine by me.


问:
Number of ways to represent any N as sum of odd numbers?
I was solving some basic Math Coding Problem and found that For any number $N$, the number of ways to express $N$ as sum of Odd Numbers is $Fib[N]$ where $Fib$ is Fibonnaci , I don't have a valid proof for this and didnot understand that how this can be solved using recurrences Can someone provide with it ?

If you are not getting it Suppose for N=4 number of ways to write it as sum of Odd Numbers is 3 which is Fibonnaci at $3$

$4=> 1+1+1+1$

$4=> 1+3$

$4=> 3+1$

NOTE-> the composition is ordered $( 1+3)$ and $(3+1)$ are different .
UPD -> I do not claim that I observed it myself but in the problem solution I found it , I asked to just find some valid proof / reason to it

答:
Let's say $S(n)$ is the set of ways to write $n$ as a sum of odd numbers.

We can partition this set into two subsets: $A(n)$ and $B(n)$, where $A(n)$ is the set of sums where the last summand is a $1$, and $B(n)$ is the set of all other sums.

Can you see why $A(n)$ has the same size as $S(n-1)$? Can you see why $B(n)$ has the same size as $S(n-2)$?

If you prove this, you find that $|S(n)| = |A(n)| + |B(n)| = |S(n-1)| + |S(n-2)|$, which is the Fibonacci recurrence relation. You can then prove by induction that your sequence is equal to the Fibonacci sequence.


问:
Does the product rule for exponents apply to a power of a power?
Does the product rule for exponents apply to a power of a power?
For example: If I have 3^(4^5), will that be equal to 3^(4^2)*3^(4^3)?</p>

答:
It applies in these terms: $$3^{4^2}\times 3^{4^3}=3^{4^2+4^3}\stackrel{\text{assuming you like this}}=3^{4^2(4^{3-2}+1)}=3^{5\cdot4^2}$$

And I don't think you want claim that $3^{4^5}=3^{80}$. I sure don't.


Body:
Does the product rule for exponents apply to a power of a power?
For example: If I have 3^(4^5), will that be equal to 3^(4^2)*3^(4^3)?</p>

Body:
It depends on how you write it.

$$3^{(4^5)}\ne 3^{(4^2)}.3^{(4^3)}$$

but

$$(3^4)^5=(3^4)^2.(3^4)^3$$


答:
A short answer is no. Using the product rule that works we have
$$3^{4^2}3^{4^3} = 3^{4^2+4^3}\neq 3^{4^5} $$

Title: Does the product rule for exponents apply to a power of a power?


问:
Continuous periodic functions not approximable by a Fourier series
What is an example of a continuous periodic function that is not the limit of any Fourier series? If not, is there an more or less elementary proof?</p>

答:
In part, there is some potential imprecision of language that confuses beginners here. I know it confused me. Plus, a surprising fact:

First, it has been known since at least since Fejer that every continuous function is a uniform-pointwise limit of finite Fourier series $\sum_{|m|\le N} c_{N,n} e^{2\pi inx}$.

More specifically, Fejer gave a formula for $c_{N,n}$ in terms of the Fourier coefficients $\widehat{f}(n)$ of the function $f$ itself...

The seeming paradox is that $c_{N,n}$ definitely cannot be $\widehat{f}(n)$ in general! That is, "the Fourier series" of $f$'s finite subsums are in general not the sequence of Fourier sums that converge uniformly to $f$. Crazy, seems to me.


答:
We have from first principles that the number of compositions into odd parts is given by

$$[z^N] \sum_{q\ge 1} \left(\frac{z}{1-z^2}\right)^q.$$

This simplifies to

$$[z^N] \frac{z/(1-z^2)}{1-z/(1-z^2)}
= [z^N] \frac{z}{1-z-z^2}.$$

Now $$F(z) = \frac{z}{1-z-z^2}$$ is the OGF of the Fibonacci numbers and we have the claim.

Title: Number of ways to represent any N as sum of odd numbers?

Body:
I was solving some basic Math Coding Problem and found that For any number $N$, the number of ways to express $N$ as sum of Odd Numbers is $Fib[N]$ where $Fib$ is Fibonnaci , I don't have a valid proof for this and didnot understand that how this can be solved using recurrences Can someone provide with it ?

If you are not getting it Suppose for N=4 number of ways to write it as sum of Odd Numbers is 3 which is Fibonnaci at $3$

$4=> 1+1+1+1$

$4=> 1+3$

$4=> 3+1$

NOTE-> the composition is ordered $( 1+3)$ and $(3+1)$ are different .
UPD -> I do not claim that I observed it myself but in the problem solution I found it , I asked to just find some valid proof / reason to it


问:
Equivalent metrics generate the same topology
Let $X$ be a set. Two metrics $d_1, d_2: X\times X\to\mathbb{R}$ are equivalent, if constants $\alpha,\beta > 0$ exist such that for all $x,y\in X$ holds:

$\alpha d_1(x,y)\leq d_2(x,y)\leq \beta d_1(x,y)$

Show that equivalent metrics generate the same topology

Proof:

Let $d_1, d_2: X\times X\to\mathbb{R}$ be metrics and $\tau_1, \tau_2$ the induced topolgies.

We have to show, that $\tau_1=\tau_2$.

$\tau_1\subseteq\tau_2$.

Let $U\in\tau_1$ open. Then exists for every $x\in U$ a $\epsilon >0$ such that $B_{d_1}(x,\epsilon)\subseteq U$.

It is $B_{d_1}(x,\epsilon)=\{y\in X|d_1(x,y)<\epsilon\}$.

Since $d_1$ and $d_2$ are equivalent, there are constants $\alpha,\beta >0$ such that $\alpha d_1(x,y)\leq d_2(x,y)\leq \beta d_1(x,y)$.
Take $\epsilon':=\beta^{-1}\epsilon >0$.

We get $d_2(x,y)\leq \beta d_1(x,y)<\beta\cdot \beta^{-1}\cdot\epsilon=\epsilon$.

Hence $U\in\tau_2\checkmark$.

The other inclusion $\tau_1\supseteq\tau_2$ works analogously.

Is this proof correct?
Thanks in advance.

答:
For completeness I also wish to add the other inclusion on this same page.

Just as above we have the same $\alpha, \beta>0$ such that for all $x,y \in X$ we have $$\alpha d_1(x,y) \le d_2(x,y) \le \beta d_1(x,y)$$

Let $U \in \tau_2$. We will show $U \in \tau_1$. We take $x \in U$. As $U \in \tau_2$ there exists an $\varepsilon > 0$ such that $B_{d_2}(x,\varepsilon) \subseteq U$. Then I claim that:
$$B_{d_1}(x,\varepsilon / \beta) \subseteq B_{d_2}(x,\varepsilon)$$

Suppose $y \in B_{d_1}(x,\varepsilon/\beta)$, then $d_1(x,y) < \varepsilon/\beta$ and so  by the equivalence of metrics we have $d_2(x,y) \le \beta d_1(x,y) < \beta \cdot\varepsilon/\beta = \varepsilon$ and so $y \in B_{d_2}(x,\varepsilon)$. We see that  $B_{d_1}(x,\varepsilon / \beta) \subseteq B_{d_2}(x,\varepsilon) \subseteq U$.

We deduce that $x$ is an interior point of $U$ with respect to $d_1$. As $x \in U$ was arbitrary, $U \in \tau_1$ and so $\tau_2 \subseteq \tau_1$.


问:
How to prove that $n!>2^n$ holds for all $n>3$?
I suspect something with sets since one with cardinality $n$ has $n!$ permutations and its powerset contains $2^n$ elements. It could also involve binomial coefficients because of$$\begin{pmatrix}n\\0\end{pmatrix}+\begin{pmatrix}n\\1\end{pmatrix}+\begin{pmatrix}n\\2\end{pmatrix}+\cdot\cdot\cdot+\begin{pmatrix}n\\n\end{pmatrix}=2^n$$

答:
The OP was not very clear if them were looking for a combinatorial interpretation, in such case:

Consider the following function $Fix : S_n\longrightarrow P([n]),$
called the set of fixed points of a permutation, where $S_n$ is the set of permutations on $[n]$ and $P([n])$ is the powerset of $[n],$ defined as
$$Fix(\pi)=\{i: \pi(i)= i \},$$ for some permutation $\pi ,$ meaning we are recording the fixed points of the permutation.

Notice that if we fix $n-1$ points in the permutation, we are fixing $n$ because there is no space to make the$ n-$th element not fixed. So any set $A\subseteq [n]$ such that $|A|=n-1$ is not the set of fixed points of any permutation. So, apriori, $$n!=|S_n|\geq 2^n-n+1.$$
We actually can say more, using the Derangements(permutations without fixed points, denoted by $D_m$) we can decompose $S_n$ in the statistic of how many fixed points the permutation has i.e.,
$$n!=\sum _{k=0}^n\binom{n}{k}D_{n-k}=\sum _{k=0}^{n-2}\binom{n}{k}D_{n-k} +1,$$
where the last equality is by the argument above on how there are no permutations that fix $n-1$ points exactly. Now, notice that $D_{n-1}>2$ iff $n>3,$ because $D_k>D_{k-1}$ for $k>1$ (why? ). So we can split this as $\binom{n}{1}D_{n-1}=\binom{n}{1}+\binom{n}{1}(D_{n-1}-1)=\binom{n}{n-1}+\binom{n}{1}(D_{n-1}-1).$ So we have that
$$n!=\sum _{k=0}^n\binom{n}{k}D_{n-k}=\sum _{k=0}^{n-2}\binom{n}{k}D_{n-k} +1=D_n+\binom{n}{1}(D_{n-1}-1)+\sum _{k=2}^{n-2}\binom{n}{k}D_{n-k}+\binom{n}{n-1} +1>\sum _{k=0}^n\binom{n}{k}=2^n.$$


问:
How to supplement a relation?
Relation $\;\rho\;$ is given on the power set of $\;\mathbb{N}\;$ and is defined with $A\rho B\Leftrightarrow A\cap B\ne\emptyset \land A\cup B=\mathbb{N}\;$.

$\rho\;$ is a symmetric relation. How do I find the smallest equivalence relation that contains $\rho$ and how do I find the largest equivalence relation that is contained in $\rho$?

答:
HINT: The relation $\rho$ isn’t reflexive, so it does not contain any equivalence relation on $\wp(\Bbb N)$. There is exactly one subset of $\wp(\Bbb N)$ on which it is reflexive, and it does contain an equivalence relation on that subset; what is that subset, and what is the equivalence relation in question?

For the first part of the question note that you will definitely have to add a lot of ordered pairs just to get a reflexive relation containing $\rho$. Then you’ll have to worry about transitivity. Note that if $A$ and $B$ are any non-empty subsets of $\Bbb N$, then $A\mathrel{\rho}\Bbb N$ and $B\mathrel{\rho}\Bbb N$, so if $\hat\rho$ is an equivalence relation containing $\rho$, you will have to have $\langle A,B\rangle\in\hat\rho$.

Title: How to supplement a relation?


问:
Find borel measure for on $X=[1,\infty)$ such that for every $s\geq 1$ the following holds $\mu(T_sA)=\mu(A)$ for every measureable $A$.
For every $s\geq1$ we define the transformation $T_s(x)=s\cdot x$. Find a borel measure which is not the zero measure on $X=[1,\infty)$ such that for every $s\geq 1$ the following holds $\mu(T_sA)=\mu(A)$ for every measureable $A$.

Hint: Log(ab)=log(a)+log(b) and use lebesgue stieltjes measure.

I feel really lost here any hint or more would really appreciated.

答:
Define $\mu$ by
$$
\mu(A) = \int_A \frac1x \,dx.
$$
It suffices to check the condition for intervals $[a,b)$.  We can calculate
$$
\mu[sa,sb) = \int_{sa}^{sb} \frac1x \,dx = \log(sb) - \log(sa) = \log(sb/sa) = \log(b/a) = \log(b) - \log(a) = \int_{a}^{b} \frac1x \,dx = \mu[a,b).
$$


问:
Totally busted euler trail.
New to graph theory proofs would like some help.

Let $G_{1}$ and $G_{2}$ be Eulerian graphs . let $v_{1}$ be a vertex in $G_{1}$ and $v_{2}$ be a vertex in $G_{2}$  jion $v_{1}$ and $v_{2}$ by a single edge.

1)Does the graph have a Euler trail?
EDIT:

I think i finally get this part start at $v_{1}$ make a Euler circiut back to $v_{1}$ ignoring the new path to $v_{2}$ now delete these edges and cross form $v_{1}$ to $v_{2}$ now complete that euler circiut jion the ends together of these 2. Now as shown this is actually a Eulerian Trail. Which is where my confusion arose form.

2) is the resulting graph eulerian? why or why not?

The requirements for a graph to be Eulerian is that every vertex is of even degree, and clearly this was true in $G_{1}$ and $G_{2}$ as they where Eulerian thus all vertex of $G_{1}$ and $G_{2}$ where even. even +1 is not even hence there The new Graph is not Eulerian. ($ v_{1}$ and $v_{2}$ must both be of odd degree)

答:
It is important that the definition of an Eulerian graph here includes the property "is connected".  Without this property, we can have the following situation where $G_1$ is the blue graph and $G_2$ is the red graph, and we add in the orange edge:

Both $G_1$ and $G_2$ contain an Eulerian cycle, but the resultant graph does not contain an Eulerian trail.

We conclude that we must use the assumption that $G_1$ and $G_2$ are connected, otherwise the claim is not true due to the above counterexample.

For 1., you seem to be confusing Eulerian trail with Eulerian circuit.  Assuming $G_1$ and $G_2$ are connected, there is an Eulerian trail (and even an Eulerian circuit) starting at any vertex, and we can join them together as described.

You can also approach 1. theoretically by using the following result:

An undirected graph has an Eulerian trail if and only if at most two vertices have odd degree, and if all of its vertices with nonzero degree belong to a single connected component.  (Source: Wikipedia)

Since the vertices of $G_1$ and $G_2$ have even degrees, the resultant graph has two odd degree vertices.  It is also connected since $G_1$ and $G_2$ are connected.  So the resultant graph satisfies the above condition, and thus contains an Eulerian trial.

Your approach to 2. seems natural, but I don't think the conditions are right, e.g. this non-Eulerian graph has every vertex of even degree:

This is what it should be:

An undirected graph has an Eulerian cycle if and only if every vertex has even degree, and all of its vertices with nonzero degree belong to a single connected component.  (Source Wikipedia, op. cit.)

Title: Totally busted euler trail.


问:
If L1 is decidable and if L2 is included in L1, is L2 also decidable?
Is this statement true or false and why ?

If $L_{1}$ is decidable and if $L_{2}$$\subseteq$$L_{1}$ then $L_{2}$ is also decidable.

I would be tempted to say yes, but i am really not sure. I am also not sure of the reasoning behind.

Thanks for your help.

答:
Hint:

Let $L_2$ be any language.

Notice $L_2 \subseteq \Sigma^*$ (the language of all words) which is decidable!

I hope this helps ^_^


问:
Are all product topologies/spaces over real numbers Euclidean spaces?
Until earlier today, when I thought of "real numbers" I thought of only the pure set of real numbers.
It seems I was mistaken to do so, as this seems to be the set-theoretical Baire space.
Instead, real numbers seem to be the set and what you can do with it, namely their field operations (addition and multiplication, their algebraic inverses, namely subtraction and division, and all the rules regarding associativity, commutativity and so on) and ordering.
So the real numbers can be written as $(R, +, \cdot, <)$. Usually we write them as $\mathbb R$ though and know that we can do these things with the real numbers.

Next, we can build a product space (not inner product space!), which

is the Cartesian product of a family of topological spaces equipped with a natural topology called the product space,

from a finite number ($n\in\mathbb N$) of real numbers, which would be $\mathbb R^n$.
A couple of people in How do I formally write down a Euclidean space with symbols? state that $\mathbb R^n$ is already the Euclidean space.
In other words, every finite dimensional product space of the real numbers is a Euclidean space.

But I do not see what that should be the case.
For a Euclidean space I would also need the

completeness

linear combination

Euclidean distance as metric,

Euclidean norm

the "normal" dot product

Now it seems to me that

completeness is inherited by every product space over the real numbers from the completeness of the real numbers

linear combination can be derived using the direct product. Even though I am not sure if the direct product is always implied for product topologies!

the Euclidean distance is induced by the Euclidean norm

the Euclidean norm is induced by the "normal" dot product, or according to J.W.Tanner in How do I formally write down a Euclidean space with symbols?, the dot product is induced by the Euclidean norm. However, one of them must be derived from somewhere else.

So, it seems to me that many of the properties of the Euclidean space are properties every product space over the real values has.
However, I do not see why every product space over the real values should necessarily have the dot product defined as the Euclidean space has.
For example I could choose to use a different inner product, which is not the Euclidean dot product, without having the dot product induced

So, are all product spaces over real numbers Euclidean spaces? If so, how is the dot product necessarily induced?

答:
Euclidean Affine Spaces

A Euclidean space, $E$, refers to an affine space; think of points and coordinates. We have a Euclidean distance, determined by the Pythagorean Theorem. The Euclidean distance, $d$, with the space $E$ makes the metric space $(E,d)$

Euclidean Vector Spaces $(+,-,*)$

These are different from Euclidean spaces in that we are no longer using points. These two concepts are often used in conjunction, and it's easy to forget that they are actually two different structures.

The n-dimensional Euclidean vector space, $\overrightarrow{E}$, is an inner product space. We have vector addition, an inner product and  and the Euclidean norm is analogous to the distance between two points as defined for $E$. The Euclidean norm is clearly a metric for our vector space. $\overrightarrow{E}$ is nearly identical to $\mathbb{R^n}$, so why do we bother distinguishing the two?

Coordinate Spaces of Real Numbers/Affine Real Spaces

When we mention the real vector space $\mathbb{R}^n$ we are actually referring to some n-dimensional Euclidean vector space with the properties of an affine space included. Once we assign a coordinate system to $\mathbb{R}^n$ we have a "coordinate space". The Cartesian coordinate system is certainly the most familiar to impress on such a space, but I'm sure you can think of some other interesting ones.


问:
Determine the location and order of the zeros
I am trying to find the zeroes and the order of them for $$sin^4{\frac{1}{2}}z$$

So $f(z) = 0$ when $sin{\frac{1}{2}}z = 0$ so $z = 0 $ when $z = 2 \pi k$ when k is an integer. Is that right?

What about the order? So do I keep finding derivatives of f(z) until $f^{n}(z) \ne 0$? when  $f(z) = 0$

Is that right? How do I notate this better? How do I write that z is a zero of the function?

Anyway:

$f'(z) = 4 \sin^{3}{\frac{1}{2}z} \cdot cos \frac{1}{2}z \cdot \frac{1}{2}$

by laws of chain rules

Is that right? Do I keep going? My book says it's 4th order but I'm not sure how to get there.

答:
This looks good so far - although finding the derivatives will get progressively harder.
Perhaps a quicker way is to note that $f(z)$ is analytic and consider its power series. Starting with the zero at 0, since $\sin (\frac{z}{2})=\frac{z}{2}+O(z^3)$, raising this to the power of 4 we see that $f(z)=\frac{z^4}{16} +O(z^6)$ and so has a zero of order 4. The other zeros occur at $z=\pi k$ for k an integer, and these are also of order 4 by the periodicity properties of sine.


问:
How to find the B(x2, y2) poin on a chart knowing the angle, the segment length and point A(x1, y1)
How I find the B(x2, y2) in the below chart? I want to draw the Fibonacci Fan lines from point A, knowing the x1 and y1, to point B. The line can be a segment with a length of 200 or a ray line. After the first line, I would like to draw several other lines at different angles from point A(x1, y1) at 38.2°, 50°, and so on. Any help is appreciated, thank you.

答:
A consistent convention for bearing
angle should be adopted.

With respect to North or East. The latter is used here. From x-axis positive CCW.

$$ x2= x1+ 200 \cos 315^{\circ}$$
$$ y2= y1+ 200 \sin 315 ^{\circ}$$


问:
Algorithm to transform symmetric shapes/matrices into a standart one
Take the letter F. By turning it 90 degrees and mirroring top to bottom you can create $8$ F shapes.
Let's say you represent those letters F by matrices:

$$F_0 =
\begin{pmatrix}
0 & 1 & 1\\
0 & 1 & 1\\
0 & 1 & 0
\end{pmatrix},\quad F_1 =
\begin{pmatrix}
1 & 1 & 0\\
1 & 1 & 0\\
0 & 1 & 0
\end{pmatrix}$$
$$ F_2 =
\begin{pmatrix}
0 & 0 & 0\\
1 & 1 & 1\\
0 & 1 & 1
\end{pmatrix},\quad F_3 =
\begin{pmatrix}
0 & 0 & 0\\
1 & 1 & 1\\
1 & 1 & 0
\end{pmatrix}$$
$$F_4 =
\begin{pmatrix}
0 & 1 & 0\\
1 & 1 & 0\\
1 & 1 & 0
\end{pmatrix} \quad F_5 =
\begin{pmatrix}
0 & 1 & 0\\
0 & 1 & 1\\
0 & 1 & 1
\end{pmatrix}$$
$$ F_6 =
\begin{pmatrix}
1 & 1 & 0\\
1 & 1 & 1\\
0 & 0 & 0
\end{pmatrix} \quad F_7 =
\begin{pmatrix}
0 & 1 & 1\\
1 & 1 & 1\\
0 & 0 & 0
\end{pmatrix}$$

Is there an algorithm that when applied to any of those matrices returns the same matrix?

My gut says to transform F into all $8$ shapes, calculate the center of mass from the matrix coordinates and then pick the one with the lowest center of mass to be the output.

Sadly, I have no idea how to prove that this works and it seems like there might be a better way. Also the algorithm should work for any $A\times A$ matrix and not just with F shapes.

This is my first post on Math.StackExchange, I am thankful for any feedback.

答:
Essentially you need a way to choose a representative from class of equivalent matrices.

Center of mass will work if it's different for all $8$ transformations - it's equal to it been not on horizontal or vertical line through center.

However, if you just need any algorithm, it's probably simpler to choose say lexicographically smallest matrix - equivalently, write all elements from matrix in line in some fixed order (for example, left-to-right, top-to-bottom), interpret it as binary number and choose matrix corresponding to the smallest number.


问:
Equation of latus rectum of parabola
If parabola $\left ( y+1 \right )^{2}=k\left ( x-2 \right )$ passes through a point (1,-2), then equation of its latus rectum and directrix ?
My attempt

Is this correct answer?
because this answer is not mention in given options.

答:
Vertices of parabola are (-1,2)》(h,p)
Find the k value by substituting the given point..
Now we get k=-1
General equation of parabola( Y-p)^2=-4a(X -h)
4a=k  .. a=1/4
Negative sign Denotes that parabola opens towards negative X axis

Now focus 》(2-1/4,-1)》(7/4,-1)》(h-a,p)
Which means equation of lateus rectum is x=7/4 since it passes through focus perpendicular to axis of parabola..
Now directrix passes through》(2+1/4,-1)》(9/4,-1)》(h+a,p)
Means direct passes through (9/4,-1) perpendicular to axis of parabola then its equation is x=9/4

Title: Equation of latus rectum of parabola


问:
On the eigenvalues of a linear transformation $\tau$ such that $\tau^3 = \mathrm{id}$
I am reading  the book on representation theory by Fulton and Harris in GTM.  I came across this paragraph.

[..] we will start our analysis of an arbitrary representation $W$ of $S_3$ by looking just at the action of the abelian subgroup $\mathfrak A_3 = \mathbb Z/3 \subset \mathfrak S_3$ on $W$. This yields a very simple decomposition: if we take $\tau$ to be any generator of $\mathfrak A_3$ (that is, any three-cycle), the space $W$ is spanned by eigenvectors $v_i$ for the action of $\tau$, whose eivenvalues are of course all powers of a cube root of unity $\omega = e^{2\pi i/3}.$

I know the eigenvalues of $\tau$ are some cube root of unity, but I don't know why all of the cube roots of unity are eigenvalues of $\tau$.

I tried to calculate the minimal polynomial $f$ of $\tau$.  It is easily seen that $f$ divides $X^3 - 1$, but I cannot go further in this regard, since I do not know anything about $W$ except the fact that it is finite-dimensional.

I would be most grateful if you could help me understand the paragraph.

答:
The last sentence is remarkably ambiguous: I can stare at it and watch my perception of its meaning flicker back and forth between "Every eigenvalue of $\tau$ is a power of $\omega = e^{2 \pi i/3}$ and "The eigenvalues of $\tau$ give all of the powers of $\omega$."

Mathematically it is the former which is true -- as the authors very well know -- so that is the intended meaning.  It is enough to observe that a complex representation of the cyclic group $C_3 = \langle \tau \ | \ \tau^3 = 1 \rangle$ need not have $\omega$ as an eigenvalue.  For instance, there is always the trivial representation, whose only eigenvalue is $1$.  More generally $C_3$ has precisely $3$ irreducible complex representations: they are all one-dimensional and consist of scaling by $1$, $\omega$ or $\omega^2$.  By taking direct sums of these -- or, in simpler terms, taking a diagonal matrix with whatever combination of $1,\omega,\omega^2$ you want on the diagonal -- you can get whatever eigenvalues you want (among cube roots of $1$!) with whatever multiplicities.

That's over $\mathbb{C}$.  If we are working over a subfield $K$ of $\mathbb{C}$ which doesn't contain a primitive cube root of unity -- e.g. $\mathbb{R}$ or any of its subfields, e.g. $\mathbb{Q}$ -- then in any representation in which $\omega$ appears as an eigenvalue, so also must $\omega^2$.  There are lots of ways to see this, but perhaps the most purely representation-theoretic is to notice that instead of three one-dimensional irreducible representations of $C_3$, in this case there is the trivial (one-dimensional) representation and an irreducible two-dimensional representation with irreducible minimal polynomial $t^2+t+1$.  And this comes, for instance, from decomposing the group ring $K[C_3]$ into a product of simple rings:

$K[C_3] \cong K[t]/(t^3-1) = K[t]/((t-1)(t^2+t+1))$

$ \cong K[t]/(t-1) \oplus K[t]/(t^2+t+1) \cong K \oplus K[\omega]$.


问:
Is it true that $|x-y|^n \leq n(|x|^n+ |y|^n)$?
In a lecture note, the author states an inequality without proof. Now, to prove the inequality in his note, one could just prove that

$$|x-y|^n \leq n(|x|^n+ |y|^n)$$

My question is then, is this inequality true? And if so, how to prove it?

The lecture note presented the following inequality
The inequality is the following for $+\infty>p\geq 1$:
$$
d^p_W(\mu, \nu) = \inf_{\gamma \in \prod(\mu,\nu)} \int_{X \times X} |x-y|^pd\gamma(x,y)
\leq p \inf_{\gamma \prod(\mu,\nu)}\int_{X \times X} |x|^p + |y|^p d\gamma (x,y)
$$

If the first inequality is not true, than how come is the inequality above true?

答:
The inequality is not true.

Take $ x = 1, y = -1, n = 3$ then we have $ 2^3 \leq 3(1+1)$.

It might help for you to state the original inequality.


问:
A special trick needs a special name
What do we call the process $(-1)^n$ that we use to alternate $+$ and $-$? Do Mathematicians call it something like "the alternator"?

答: I assume you mean for when $n$ is a real integer, so it is equal to either $1$ or $-1$. We have the sign function but this is different. I am not sure if there is a specific name for this, more when considering a series of these terms we talk about their value and absolute value.

Title: A special trick needs a special name


问:
Equation with Logarithm
I want to solve the following equation:

$$3^x3^{x-1} = 243.$$

My approach is the following:

$3^{2x-1} = 243$ then:

$(2x-1)\cdot\log3 = \log 243$ and then:

$x = (\frac{\log243}{\log3}+1)/2$

Is this correct?

答:
don't need to take log cause RHS also in same base .$$3^{2x-1}=243$$
$$3^{2x-1}=3^5\implies 2x-1=5\implies x=3$$.By the way you were going on right track</p>

Title: Equation with Logarithm


问:
Can we assign a measure on sequences $\epsilon_n\in\{-1,1\}$ such that $\sum_{n\ge 1}a_n \epsilon_n $ converges? If so, what is the image?
Let $(a_n)_n$ be a sequence of real numbers, let $(\epsilon_n)_n$ be a sequence of i.i.d random variables with $P(\epsilon_n=1)=P(\epsilon_n=-1)=1/2$, and let $m$ denote Lebesgue measure. Consider the spaces
$$
\mathcal{A} = \{(a_n)_n: a_n\in\mathbb{R}\};\qquad \mathcal{E} = \{(\epsilon_n)_n: \epsilon_n = \pm 1\}
$$
$\mathcal{E}$ can be considered a "random choice of sign" in the probabilistic context, for example. Lastly, define a map
$$
r:\mathcal{E}\to[0,1], (\epsilon_n)_n\mapsto \sum_{n\ge1}\frac{1+\epsilon_n}{2^{n+1}}
$$

My question: is the following map
$$
f: \mathcal{A} \to \left[0,1\right],\,
(a_n)_n \mapsto m\left(\left\{x\in[0,1]:\sum_{n\ge 1}{r^{-1}(x)} a_n \text{ converges}\right\}\right)
$$well-defined? If so, is it surjective; if not, what is its image?

For example, if $(a_n)_n=1/n$, by Kolmogorov's Three-Series theorem, $\sum_{n\ge 1} \epsilon_n/n$ converges almost surely, so $f((1/n)_n)=1$; likewise, $f((1/\sqrt{n})_n)=0$.

Many sequences (in particular, those that whose series are absolutely convergent) map to $1$ and many sequences map to $0$ (such as sequences with $\lim_{n\to\infty} a_n\ne 0$), but I struggled to find a sequence $(b_n)_n$ with $0<f((b_n)_n)<1$. I've read a bit about random harmonic series and the Three-Series theorem but haven't made it much past that.

答:
Convergence of a series such as this one an example of a tail event and by the Kolmogorov Zero-One Law must have probability $0$ or $1$. This is a special case of a slightly more general result:

Suppose that $X_1,X_2,X_3,X_4,\ldots$ is a sequence of independent random variables. The probability that $\sum X_i$ converges is either $0$ or $1$.

Your question asks about the special case where we have fixed some sequence of real numbers $a_n$ and define the $X_n$ to be independent random variables taking the values $a_n$ or $-a_n$ with equal probability.

We can apply the zero-one law as follows. First, you ask if the probability is well-defined, so let's note that the event that $\sum X_i$ converges is indeed measurable. This may be observed by noting that, for real numbers $s_i$, the convergence of $\sum s_i$ can be stated in terms that translate easily to a countable intersection of countable unions of countable intersections of measurable sets:

$\sum s_i$ converges if and only if for every $\ell \in\mathbb N$ there exists some rational $p,q\in\mathbb Q$ and some $N\in\mathbb N$ such that $|p-q| < 1/\ell$ and for every $n\geq N$ we have $p <\sum_{i=1}^n s_i < q$.

There's a bit of analytical trickery here - we've replaced "convergence" by asking that the partial sums get restricted to arbitrarily small intervals eventually. Showing that convergence implies this statement requires knowing, essentially, that the open intervals with rational endpoints form a base for the topology on $\mathbb R$. The converse direction requires completeness of $\mathbb R$. This said, one may easily translate the resulting definition to operations that play well with the measure, hence the convergence of $\sum X_i$ does have a well-defined probability.

A tail event is any event that is independent of any finite collection of the sequence of random variables. Convergence is a tail event because $\sum_{i=1}^{\infty} X_i$ converges exactly when $\sum_{i=N}^{\infty} X_i$ converges for some larger $N$ - and the latter event is clearly independent of the variables $X_1,\ldots,X_{N-1}$ given that it is some condition involving only $X_N,X_{N+1},\ldots$ (which are, by assumption, each independent of the earlier values).

This suffices to show the conditions of the zero one law - hence we can immediately apply it to see that the probability of convergence can only be $0$ or $1$.


问:
Evaluate a Triple Integral enclosed by a cylinder and two planes
I'm having trouble setting the bounds for this:

Evaluate the triple integral $\displaystyle \iiint x \, dV$, where the solid is enclosed by the cylinder $x^2 + z^2 = 4$ and the planes $y = 0$ and $y + z = 4$.

I came up with: $0 \leq y \leq 4-z$ , $-\sqrt{4-x^2} \leq z \leq \sqrt{4-x^2}$, $-2 \leq x \leq 2$.
Is this right?

答:
Please note the given function $f(x,y,z) = x$ is negative to the left of $YZ$ plane and positive to the right of it. Given the symmetry of the cylinder and the plane $y+z = 4$ wrt $YZ$ plane, your integral will most likely be zero. Here are the bounds -

$0 \leq y \leq 4-z$ (cut by the planes $y = 0, y + z = 4)$

$x^2 + z^2 = 4 \implies -\sqrt{4-x^2} \leq z \leq \sqrt{4-x^2}$

And lastly $-2 \leq x \leq 2$.

Your integral order should be $\,dy \, dz \, dx$

Or use polar coordinates in $XZ$ plane,

$x = r \cos \theta, z = r \sin \theta, y$.

$0 \leq y \leq 4-r\sin\theta$

$0 \leq r \leq 2$

$0 \leq \theta \leq 2\pi$.

Order of integral should be $\, dy \, dr \, d\theta$.


问:
Does the product rule for exponents apply to a power of a power?
Does the product rule for exponents apply to a power of a power?
For example: If I have 3^(4^5), will that be equal to 3^(4^2)*3^(4^3)?</p>

答:
First, note that exponents are evaluated right-to-left. What I mean by this is that
$$
{a^b}^c
$$
by convention means $a^{(b^c)}$. It is important to have such a convention, since
$$
a^{(b^c)} \neq (a^b)^c
$$
in general. For example,
$$
2^{(3^4)} \neq (2^3)^4  \, .
$$
Because of this, your question as it stands is a little ambiguous. When you write 'a^b^c', do you mean $${(a^b)}^c$$ or $$a^{(b^c)} \, ?$$ If you mean the former, then the product rule for exponents does hold:
$$
(a^b)^c \times (a^b)^d = (a^b)^{c+d} \, .
$$
To explain why, try setting $k=a^b$. Then we have
$$
k^c \times k^d = k^{c+d} \, ,
$$
which is the familiar product formula. However, if by a^b^c you meant
$$
a^{(b^c)} \, ,
$$
then the product formula does not apply in the way you suggested. Instead,
$$
a^{b^c} \times a^{b^d} = a^{b^c+b^d} \, .
$$
Try working out why this is the case, and ask me if you have any questions.


问:
When can player $A$ win this game
There are $n$ positive integers on the blackboard.

For each round, player $A$ splits the integers into two groups, $X$ and $Y$. One of the groups can be empty. Player $B$ erases one group of integers, and reduces each remaining integers by one.

$A$ wins when one of the integers becomes $0$. B wins when every integer is erased.

Under what circumstances does $A$ have a must-win strategy?

I can think of the case where there are two $1$'s. $A$ just needs to keep these two $1$'s in separate groups, then he will win. But what kind of configurations can be reduced into this case?

答:
(Fill in the gaps as needed. If you're stuck, show your work and explain why you're stuck.)

Let the integers be $ a_i$.

Claim: $A$ wins iff $\sum \frac{1}{2^{a_i}} \geq 1$

Proof: Suppose the condition holds.

Show that $A$ can split into 2 groups such that no matter what group $B$ chooses, the condition holds again at the next step. (See Useful lemma below.)

Then, because the game is finite, eventually we must have one term equal to 0, so A wins.

Suppose the condition doesn't holds.

Show that no matter how $A$ splits the 2 groups, B can find a group so that the condition doesn't hold at the next step.

Thus, B will eventually erase every number, so B wins.

Notes:

Useful lemma: If we have positive integers $a_i$ such that $ \sum \frac{1}{2^{a_i}} \geq 1$, then we can find a subsequence such that $ \sum_{j \in J } \frac{1}{ 2^{a_j} } = 1 $.

This lemma, suitably applied (IE needs 1 more step), ensures that we can split into 2 groups such that $\sum_{j\in J } \frac{2}{ 2^{a_j} } = 1 \geq 1, \sum_{j \not \in J }     \frac{2}{ 2^{a_j} } \geq 1$.

Note: The useful lemma isn't true if we merely have $ \sum b_i \geq 1$. We can't always find $ \sum_{j\in J} 2b_j \geq 1, \sum_{j \not \in J} 2b_j \geq 1$.

Coming up with the invariance is pretty natural from studying the winning conditions, and how we go from one to the next.

Hint to useful lemma: Greedy Algorithm. Arrange $a_i$ in non-decreasing order. Show that if $ \sum_{i=1}^T \frac{1}{ 2^{a_i} }  = \frac{ b_T} { 2^{c_T}} < 1$, then $ a_{T+1} \leq \frac{1}{ 2^{c_T}}$, so $\sum_{i=1}^{T+1} \frac{1}{ 2^{a_i} } \leq 1 $. Since we eventually go over 1, hence at some point we must hit exactly 1.

Title: When can player $A$ win this game


问:
$x^4+y^4 \geq \frac{(x^2+y^2)^2}{2}$
I'm doing some exercise to prepare for my multivariable analysis exam. I didn't understand the second part of this question.

Given the function

$$f(x,y)=(x^2+y^2+1)^2 - 2(x^2+y^2) +4\cos(xy)$$

Prove that the taylor polynomial of degree $4$ of $f$ is equal to
  $5+x^4+y^4$.

First, $4\cos(xy) = 4 - 2(xy)^2 + 4R_3 $

$(x^2+y^2+1)^2=x^4+2 x^2 y^2+2 x^2+y^4+2 y^2+1$

Therefore: $(x^2+y^2+1)^2 - 2(x^2+y^2)=x^4+2 x^2 y^2+y^4+1$

Therefore: $f(x,y)=x^4+y^4+5+4R_3$

I don't know exactly why I can now conclude that Taylor Polynomial of degree 4 must be $5+x^4+y^4$, but I don't know exactly why.

Now the second question is: $x^4+y^4 \geq \frac{(x^2+y^2)^2}{2}$

New edit

I understand this now thanks to hint of Hagen von Eitzen, thanks !

The third question is:

Determine what kind of stationary point you have in $(0,0)$.

答: Hint: The difference left minus right is $\frac12(x^4-2x^2y^2+y^4)$

Title: $x^4+y^4 \geq \frac{(x^2+y^2)^2}{2}$


问:
Prove $f:[a,b) \longrightarrow (a,b)$ cannot be bijective and continuous (verification)
I would just like to make sure my proof for this is correct. If I made any wrong assumptions, please do let me know so I can correct them or even try a different approach. Thank you!

Let $f(a)=c$, by the bijectivity of $f$. Now, consider the case where we remove $c$ from $(a,b)$ and what we are left with is a disconnected set which is the union of two connected sets, namely $(a,c) \cup (c,b)$. We now impose a restriction on $f$ defined by $$g: (a,b) \longrightarrow (a,c) \cup (c,b), \ g(x) = f(x)$$This restriction is continuous (by Theorem proved in class). The image of $g$ is the union of two connected components. Since $(a,b)$ is connected, its image must also be connected, which means it must lie entirely within one of the two components. But then this would imply that the original function, $f$, is a proper subset of $(a,b)$ which contradicts the bijectivity of it. As such, it cannot be bijective and continuous.

The intervals are regarded with the subset topology induced by the standard topology of $\mathbb R$

答:
You should specify what assumptions you use, for example that $f$ is bijective and continuous. Therefore, write "Assume that $f$ is bijective and continuous" in the beginning of the proof.

It can also be good to tell the audience that you will use a "proof by contradiction". That will make it clear why you make that assumptions on $f$.


问:
Fundamentals of Truth, Provability and Axioms by means of the Continuum Hypothesis
Let S be the structure/language of ZFC (including PL 1). Let CH refer to the well-known continuum hypothesis. My claims are as follows and could u just say if it's true or wrong and why?

In S neither CH is true nor false because in S only tautologies and contradictions are already true/false and CH is not such.

Let's say I assume only one axiom in S that says: there exists an empty set. Now in this scenario CH again is neither true nor false because we still can't talk about cardinalities of sets at all (and so we can't talk about CH).

Let's say I assume ZFC in S. Now we can talk about cardinalities of sets. That means that here CH is a wff and so it is either true or false. But we cannot prove which one it is (Gödel, Cohen). But it means: CH is true xor false in ZFC in this very moment, we just don't know and we just will never know!

If we just brutally assume CH to be true in ZFC (ZFC + CH), then there's no inconsistency (proof by Gödel), but if we take ZFC + ~CH we can prove there's no inconsistency either (Cohen), so ZFC is - loosely spoken - too general to catch CH's truth/falseness properly, just like a fisherman's net is sometimes too big to catch certain fishes.

答:
There are several issues here, which may not feel important at first but over time will cloud the (already quite nuanced) picture.

First of all, you're conflating structures, theories, and languages. In increasing order of complexity:

A language (also called a signature or vocabulary) is a set of non-logical symbols, such as $\{\in\}$ or $\{+,\times,0,1,<\}$.

A theory is a set of first-order sentences, and for a language $\Sigma$ a $\Sigma$-theory is a theory consisting of sentences in the language $\Sigma$ - e.g. $\mathsf{ZFC}$ is an $\{\in\}$-theory and first-order $\mathsf{PA}$ is an $\{+,\times,0,1,<\}$-theory.

A structure in a given language is a set together with an interpretation of the various symbols in that language in a precise sense.

Whether or not a particular string of symbols is a wff depends only on the language involved, not on what axioms we're considering nor on what structure (if any) we're specifically focusing on. $\mathsf{CH}$ is a wff in the language $\{\in\}$. What the empty $\{\in\}$-theory (your "$S$") can't do is prove basic things about $\mathsf{CH}$ and related sentences. So $S$ can talk about $\mathsf{CH}$, it just doesn't have much to say. This issue is implicit in $(1)$ and $(2)$, and explicit in $(3)$.

Now on to the more subtle point: truth and falsity. The satisfaction relation $\models$ connects structures and sentences/theories, with "$\mathcal{A}\models\varphi$" (resp. "$\mathcal{A}\models\Gamma$") being read as "$\varphi$ is true in $\mathcal{A}$" (resp. "Every sentence in $\Gamma$ is true in $\mathcal{A}$"). But we use the term "true" only in this context; when talking about theories, the relevant term is provable.

The main reason for reserving terms like "true" and "false" for structures as opposed to theories is that the standard properties of truth such as bivalence only hold of truth-in-a-structure, not provability-in-a-theory. By separating the terms we make it easier to be precise and avoid subtle errors. This is an issue in your point $(3)$, where truth and provability get mixed up. In particular, the statement

CH is true xor false in ZFC in this very moment, we just don't know and we just will never know

doesn't parse.

OK, unfortunately you will find people say that things are true/false in $\mathsf{ZFC}$. The connection is that a sentence is provable in a theory $T$ iff it is true in all models of $T$, so this isn't totally unjustified. But this is an abuse of terminology, and should be avoided until the fundamentals of the topic are mastered.

After shifting from truth to provability, point $(4)$ then is correct with one slight additional hypothesis: assuming $\mathsf{ZFC}$ is consistent in the first place, both $\mathsf{ZFC+CH}$ and $\mathsf{ZFC+\neg CH}$ are consistent.


问:
Fermat numbers question
I am looking for direction in my question.
The question is to prove, by long polynomial division, that:
\begin{equation}
2^{2^{n+1}}-1\ |\ 2^{F_n-1}-1
\end{equation}

I have tried a lot of things, like calling:\begin{equation}  t=2^{2^n} \end{equation} and trying long polynomial division right away, but none of this worked successfully.

答:
If you want to see how to use Polynomial long division as described in here https://en.wikipedia.org/wiki/Polynomial_long_division to show that $t-1 \mid t^{2^k}-1$ let's do it.

Here the dividend is $t^{2^{k}}-1$ and the divisor is $t-1$. The first step of Polynomial long division is to divide the first term of the dividend ($t^{2^{k}}$) by the highest term of the divisor ($t$)
$$
\frac{t^{2^{k}}}{t}=t^{2^{k}-1}.
$$
Now we multiply the divisor ($t-1$) by the result just obtained ($t^{2^{k}-1}$) to get
$$
t^{2^{k}-1}(t-1)=t^{2^{k}}-t^{2^{k}-1}.
$$
Now subtract the product just obtained ($t^{2^{k}}-t^{2^{k}-1}$) from the appropriate terms of the original dividend ($t^{2^{k}}-1$) and we get $t^{2^{k}-1}-1$. Now we repeat the process with $t^{2^{k}-1}-1$ as the dividend. Repeating the process one gets $t^{(2^{k}-1)-1}-1=t^{2^{k}-2}-1$. And in the $r-1$-th step one gets $t^{2^{k}-r}-1$(prove it). Suppose WLOG that we must stop the process at this step. This means that the degree of the polynomial $t^{2^{k}-r}-1$ is smaller then the degree of the polynomial $t-1$. Since the degree of $t-1$ is $1$ we know that the degree of $t^{2^{k}-r}-1$ must be $0$. Hence we must have that $r=2^k$ and that $t^{2^{k}-r}-1=t^0-1=0$. So we have that the remainder of the long division is $0$. And we are done.

Title: Fermat numbers question


答:
Any time there are at least $2^k$ copies of $k$ $A$ can win.  The approach is like your two $1$s.  Split the copies evenly.  $B$ will erase one half and leave $2^{k-1}$ copies of $k-1$.

If there is a $1$ and two $2$s $A$ can win.  Put the $1$ in one group and the two $2$s in the other.  If $B$ erases the group with the $1$, there will be two $1$s left and $A$ will win.

If there is a $1$, a $2$, and two $3$s $A$ can win.  Put the $1$ in one group and the others in the second group.

This suggests a general criterion.  For all the numbers $i$ on the board, compute $2^{-i}$ and add them up.  If the sum is $1$ or greater $A$ can win.  $A$ should divide the numbers into two groups that each have the sum of $2^{-1}$ at least $\frac 12$.  Then when $B$ erases one group and subtracts $1$ from each of the remaining numbers, the sum will be at least $1$ again.  There must then be a time that $0$ appears.

If the sum of the $2^i$ is less than $1$, one of the two groups will have a sum less than $\frac 12$ and the sum will be less than $1$ when each number is decreased.  If $B$ always selects the group with sum less than $\frac 12$, no $0$ can then appear.

Title: When can player $A$ win this game

Body:
There are $n$ positive integers on the blackboard.

For each round, player $A$ splits the integers into two groups, $X$ and $Y$. One of the groups can be empty. Player $B$ erases one group of integers, and reduces each remaining integers by one.

$A$ wins when one of the integers becomes $0$. B wins when every integer is erased.

Under what circumstances does $A$ have a must-win strategy?

I can think of the case where there are two $1$'s. $A$ just needs to keep these two $1$'s in separate groups, then he will win. But what kind of configurations can be reduced into this case?


问:
Given a complex number set, how do you prove which set is it in the complex plane?
Let $a,b \in \mathbb{C}$ is it true that:
$|a| < |b|$ implies $|a|^{2} < |b|^{2}$
I want to use this to build sets of points in the complex plane since I have things like:
$$|z-1| < |z|$$
So I want to apply power to both sides and then change $z = x+iy$ to discover what the set is in terms of the real coordinates.

Is it useful?

Which other strategies do you recommend in order to discover what the set is in terms of the real coordinates?

答:
It is true that, as both are nonnegative real numbers, $\:|z-1|<|z|\iff |z-1|^2<|z|^2$.

However it is much simpler to solve  than what you want to undertake,  as $|z|^2=z\bar z$: you obtain instantly that
$$|z-1|<|z|\iff\iff(z-1)(\bar z -1)<z\bar z\iff z+\bar z=2\operatorname{Re}z>1$$
so the set is simply the half-plane $\operatorname{Re}z>\frac 12$.

Note: this answer was geometrically obvious, as the inequation $|z-1|>|1|$ means the image of $z$ in the complex plane is closer from the origin than from the real point $1$ on the real axis, so by a med-school argument, the image of $z$ belongs to the half-plane on the right of the perpendicular bisector of the segment $[0,1]$ on the real axis.