问:
Is my intuition for ${\vdash}$ and ${\models}$ correct?
I just want to check my idea for what exactly is meant by ${\vdash}$ and ${\vDash}$ is correct (I am a noob to logic).

${S\vdash t}$ means "$t$ is provably true using finitely many elements of $S$ within that system", whereas ${S\vDash t}$ means "$t$ is true given $S$", but doesn't make any statements about whether we could prove it within that system $S$. Is this essentially correct? Thank you.

答:
I don't think you have it quite right.

"$S\models\varphi$" is defined purely semantically: it means "If $\mathcal{M}$ is a structure satisfying each sentence in $S$, then $\mathcal{M}$ satisfies $\varphi$ as well." In particular, this notion is secondary to the notion of a sentence being true in a structure (which uses the same
"$\models$" symbol, incidentally).

Meanwhile, "$S\vdash\varphi$" is not actually fully precise! It really only makes sense once we specify a proof system $\mathfrak{P}$. This is a totally different object from $S$ and $\varphi$; basically, it's just a set of rules for deriving sentences from sets of sentences. If we can derive $\varphi$ from $S$ via $\mathfrak{P}$, we write "$S\vdash_\mathfrak{P}\varphi$."

A priori these are two totally different things, and we can whip up silly "proof systems" which are obviously unrelated to the semantics: e.g. "bogus ponens," the proof system which lets you deduce anything from anything.

(Note that I haven't defined "proof system" precisely - nor am I going to. There are various definitions floating around in the literature, with no one particular one being universally accepted. And there are further subtleties which can emerge, see e.g. here.)

So how are these connected, and why do we abuse notation by writing "$\vdash$" instead of "$\vdash_\mathfrak{P}$?"

Well, it turns out that for a wide variety of $\mathfrak{P}$s we in fact have soundness and completeness of the proof system with respect to the semantics, that is, $$S\vdash_\mathfrak{P}\varphi\quad\iff\quad S\models\varphi$$ (the left-to-right direction is soundness, the right-to-left is completeness - and it's the latter which is by far the more interesting direction). Moreover, since we're specifically interested in the semantics above, proof systems which are not sound and complete aren't generally of interest.

This means that we can usually write "$\vdash$" without fear of serious confusion, since all the "reasonable" proof systems we look at will behave identically here. But in other contexts, and for certain questions (like questions about proof length), the specific proof system may matter a lot. So when first learning this stuff it's a good idea to fix a single proof system from the get-go, and only generalize later once the basics are understood.


问:
Why for a point in a given circle to be closest to the origin it must be collinear to center of circle and origin?
Why for a point in a given circle to be the closest to origin, it must be collinear to center of circle and origin?</p>

答: Does this help?  Just think about it!

ContentLicense: CC BY-SA 4.0

Body: Does this help?  Just think about it!


答: Straight line is the shortest way from the center to the origin and if you draw that line, it will intersect the circle. The point of intersection is on that line. If there is another point on the circle that's closer to the origin, it means that there is a shorter way from the origin to the center, which is impossible.</p>

Title: Why for a point in a given circle to be closest to the origin it must be collinear to center of circle and origin?

Body: Why for a point in a given circle to be the closest to origin, it must be collinear to center of circle and origin?</p>


问:
How Many Clock Hand Positions Swap to a Valid Position?
My wording will not be exactly clear, but this is what I remember.

Suppose you have a clock with minute and hour hands and you switch their places to form another correct time. How many such times can be formed in a clock?

答:
At $h$ hours and $m$ minutes, the hour hand has position $(h+m/60)/12$ and the minute hand has position $m/60$ (where positions range from $0$ to $1$). If we swap positions, the new hours $h'$ and minutes $m'$ must satisfy $(h'+m'/60)/12=m/60$ and $m'/60=(h+m/60)/12$. Thus we're looking for solutions of

\begin{align}
60h'+m'&=12m\;,\\
60h+m&=12m'
\end{align}

with $0\le h,h'\lt12$, $0\le m,m'\lt60$ and $h,h'$ integer. We can solve for $m$ and $m'$ in terms of $h$ and $h'$ by elimination:

\begin{align}
m&=\frac{60h+720h'}{143}\;,\\\\
m'&=\frac{60h'+720h}{143}\;.
\end{align}

Then demanding $m,m'\lt60$ yields

\begin{align}
h+12h'&\lt143\;,\\
h'+12h&\lt143\;.
\end{align}

This is fulfilled by all combinations of $h$ and $h'$ except $h=h'=11$. Thus there are $12\cdot12-1=143$ such times, and you get them by substituting the corresonding integer values of $h,h'$ into the equations for $m,m'$ above.

ContentLicense: CC BY-SA 3.0


问:
Is it possible to find an ellipse's axis lengths knowing the proportion and circumference?
If we have a symmetrical ellipse with, i.e. a circumference of 10 meters, and we know that the proportion between the axis a and b is of 5:3 (meaning a is 3x, and b is 5x), can we calculate the lengths?

I feel like this should be possible, but I can't figure it out.

答:
For an ellipse with semi-major axis $r_a$ and semi-minor axis $r_b$, the
circumference $L$ is given by the formula:

$$L = 4r_aE(e)$$
where $e = \sqrt{1 - \frac{r_b^2}{r_a^2}}$ is the eccentricity and
$$E(k) = \int_0^{\frac{\pi}{2}} \sqrt{1-k^2\sin^2\theta} d\theta$$
is the complete elliptic integral of the second kind.

For your case, you have $L = 10$ and $b > a$, so $b$ is the semi-major axis.
This means
$$b = \frac{L}{4E(e)} = \frac{10}{4E(\frac45)}$$

On wolfram alpha, you can compute the value of $E(k)$ using the command EllipticE[k^2]. Throwing following command
10/(4*EllipticE[(4/5)^2]) to WA, you will find
$$b\simeq 1.9587104722949813375395$$


问:
Calculate $\displaystyle \lim_{x \to 3} \frac{\sqrt{19-x} - 2\sqrt[4]{13+x}}{\sqrt[3]{11-x} - x + 1}$
Calculate:
$$\displaystyle  \lim_{x \to 3}  \frac{\sqrt{19-x} - 2\sqrt[4]{13+x}}{\sqrt[3]{11-x} - x + 1}$$

The problem with that case is that the roots are in different powers so multiplication in nominator and denominator by conjugate is not an option (at least I think it's not).

答: HINT.-Use the identity $A^3-B^3=(A-B)(A^2+AB+B^2)$ with $A=\sqrt[3]{11-x}$ and $B=x-1$ and multiplying numerator and denominator by $(A^2+AB+B^2)$ you' ll get again the undeterminated form $\dfrac{(4-4)(4+4+4)}{8-8}$. But the new denominator being equal to $11-x-(x-1)^3$ you have its derivative is equal to $-13\ne0$ so the undetermination disappear after apply Hospital's rule.


问:
Squares in a finite field $\mathbb{F}_p$
I need to find all prime $p$ s.t $n+3$ is the inverse of $n-3$ in $\mathbb{F}_p$. So obviously this means $(n+3)(n-3)=1\mod p$, meaning $n^2=10\mod p$. So the question is - for which $p$ does the equation $x^2-10=0$ has a solution in $\mathbb{F}_p$, but I'm not quite sure how to approach this.

I tried using $(\frac{10}{p})=10^{\frac{p-1}{2}}\mod p$ but this got me nowhere. Any help would be apperciated.

答:
I assume the problem is to determine all primes for which there exists an $n$ for which $n+3$ is the inverse of $n-3$ in $\mathbb{F}_p$. As you note, this requires finding an $n$ such that $n^2\equiv 10\pmod{p}$, i.e., find the primes for which $10$ is a quadratic residue. Which of course just begs for Quadratic Reciprocity.

If $p=2$, then any even $n$ will work. If $p=5$, then any multiple of $5$ will do. So we may assume that $p\neq 2,5$.

We know that, from the quadratic character of $2$, that
$$\left(\frac{2}{p}\right) = \left\{\begin{array}{ll}
1 &\text{if }p\equiv \pm1 \pmod{8}\\
-1 &\text{if }p\equiv \pm3 \pmod{8}.
\end{array}\right.$$
And from Quadratic Reciprocity, we have
$$\left(\frac{5}{p}\right) = \left(\frac{p}{5}\right)(-1)^{\frac{p-1}{2}\,\frac{5-1}{2}} = \left(\frac{p}{5}\right) = \left\{\begin{array}{ll}
1 & \text{if }p\equiv \pm 1\pmod{5},\\
-1 &\text{if }p\equiv \pm 3\pmod{5}.
\end{array}\right.$$

Since $\left(\frac{10}{p}\right)=\left(\frac{2}{p}\right)\left(\frac{5}{p}\right)$, we see that $10$ is a square modulo $p$ if and only if both $2$ and $5$ are squares modulo $p$, or neither is. So there are $8$ remainders modulo $40$ where it is a square (four for the possibilities that it be $\pm 1$ modulo $8$ and $\pm 1$ modulo $5$; and four for when it is $\pm 3$ modulo $8$ and $\pm 3$ modulo $5$), and $8$ remainders modulo $40$ where it is not a square (the remaining combinations/remainders modulo $40$ that are relatively prime to $40$).  So use the Chinese Remainder Theorem to figure out those values.


问:
Matrices manipulation
I am having difficulty with the following question

I have to determine if the following claim is true or not.

If it is true I have to proof it else I need to give an example

I believe it is not true but I do not find.

$A$ is a matrix (with scalar in real).

$A$ is symmetric.

if there exists $k>0$ so that $A^{k}=Id$ then $A^{2}=Id$

how can I proof if it is not true ?

in order to proof do I need to assume that $k=3$ and $A^{2}!=Id$ ?

although I still believe this is not true

Thanks in advance

答: Hints: a real symmetric matrix is diagonalizable (via an orthogonal matrix, but you don't even need that here) in $M_n(\mathbb{R})$ and the condition $A^k=Id$ implies that the eigenvalues of $A$, which are all real, belong to $\{\pm 1\}$. Finally, note that the result is true.

Title: Matrices manipulation


Title: Calculate $\displaystyle \lim_{x \to 3} \frac{\sqrt{19-x} - 2\sqrt[4]{13+x}}{\sqrt[3]{11-x} - x + 1}$

Body:
Calculate:
$$\displaystyle  \lim_{x \to 3}  \frac{\sqrt{19-x} - 2\sqrt[4]{13+x}}{\sqrt[3]{11-x} - x + 1}$$

The problem with that case is that the roots are in different powers so multiplication in nominator and denominator by conjugate is not an option (at least I think it's not).

Body:
$$\frac{\sqrt{19-x}-2 \sqrt[4]{x+13}}{\sqrt[3]{11-x}-x+1}$$
plug $x=y+3$ so we get
$$\underset{y\to 0}{\text{lim}}\frac{\sqrt{16-y}-2 \sqrt[4]{y+16}}{\sqrt[3]{8-y}-y-2}$$
we have the following expansions as $y\to 0$
$$
\begin{array}{rll}
 \sqrt{16-y} &\sim &4-\frac{y}{8} \\
 2 \sqrt[4]{y+16} &\sim& \frac{y}{16}+4 \\
 \sqrt[3]{8-y} &\sim& 2-\frac{y}{12} \\
\end{array}
$$
So the limit can be written as
$$\underset{y\to 0}{\text{lim}}\frac{4-\frac{y}{8}-\left(\frac{y}{16}+4\right)}{2-\frac{y}{12}-y-2}=\lim_{y\to0}\frac{-\frac{3}{16}y}{-\frac{13}{12}y}=\frac{9}{52}$$


问:
Derivative of 'inside' matrix in the squared Frobenius norm of a product of three matrices
How does one approach finding $\nabla_X\|AXB\|_F^2$?

Normally I would expand to $\|AXB\|_F^2=\text{Trace}(B^TX^TA^TAXB)$ and rearrange using the cyclic property of traces to make $X$ the first and/or last matrix in the product, but this isn't possible here.

$A$ is wide and $B$ is tall; both are full-rank. $X$ is square. All matrices are real-valued.

答:
Use a colon as a convenient product notation for the trace
$$\eqalign{
A:B &= {\rm Tr}(A^TB) \\
A:A &= \big\|A\big\|^2_F \\
}$$
and, for typing convenience, define the matrix
$\;Y = AXB$.

Write the cost function in terms of this new variable, then calculate its differential and gradient.
$$\eqalign{
\phi &= \big\|Y\big\|^2_F = Y:Y \\
d\phi &= 2Y:dY = 2Y:A\,dX\,B = 2A^TYB^T:dX \\
\frac{\partial \phi}{\partial X} &= 2A^TYB^T = 2A^TAXBB^T \\
}$$


问:
Question about Euler numbers
How to prove that

where $\ E_{2n}$ is the $2n^{th}$ euler number

and $$\frac{1}{\cosh(x)}=\sum_{n=0}^{\infty }\frac{E_n}{n!}x^n$$

Is there maybe some link with an answer, or a book in which the above is shown?

答: We have $\cosh (x) \cdot \frac{1}{\cosh (x)} = 1$ and $\cosh (x) = \sum_{k=0}^\infty \frac{1}{2k!}x^{2k}$. Multiplying and taking the coefficient of $x^{2n}$ we arrive at $E_0 = 1$ and $\sum_{k=0}^n \frac{1}{2k!}\cdot\frac{E_{2n-2k}}{(2n-2k)!} = 0$ for $n \geq 1$. From here on this is a straight forward proof by induction.</p>

Title: Question about Euler numbers


问:
Rational Numbers: Ordering equality question
Ordering of rationals:

$(b \gt 0 \land d \gt 0) \implies ((\frac a b \lt \frac c d) \iff ad \lt bc)$

Why does the inequality hold?

I understand the algebra of multiplying by $b,d$ on both sides, which equals the RHS of $\iff$.  I understand that the equality should hold where $\frac a b = \frac c d \iff ad = bc$.

答: Multiplication by a positive number preserves inequalities.</p>


问:
Algebraic expansion of $1 + (1 + x) + ... + (1 + x)^{p-1}$
I was proving that $p(x) = 1 + x + \text{ ... }  + x^{p-1}$, where $p$ is a prime number, is irreducible over the rationals, by using the translation $1 + (1 + x) + \text{ ... } + (1 + x)^{p-1}$. I managed to do it, but I would like to get a closed expression for this as a polynomial.

So far, I have:
\begin{equation}
\sum_{k = 0}^{p-1}{(1+x)^k} = \sum_{k = 0}^{p-1}{(\sum_{i = 0}^k{\binom{k}{i}x^i)}}
\end{equation}

By inspection, it looks like the final result will be:

\begin{equation}
\sum_{k = 0}^{p-1}{(1+x)^k} = \sum_{k = 0}^{p-1}{(\sum_{i=k}^{p-1}{\binom{i}{k}}} )x^k
\end{equation}

I.E.

\begin{equation}
\sum_{i=0}^{p-1}{\binom{i}{0}} + \sum_{i=1}^{p-1}{\binom{i}{1}} x + \text{ ... } + \sum_{i=p-1}^{p-1}{\binom{i}{p-1}}x^{p-1}
\end{equation}

First of all, is this correct? If not, what would be the correct expression and why? And if it is correct, how can I get from the first equation to the second?

Thanks in advance!

答:
Assuming that I have not misinterpreted your query:

A case can be made that the following hint-answer is defective, because it avoids examining your work.  Through no fault of your own, you went down a natural but bad path.

I actually don't know how to prove that any polynomial is irreducible, but I can give you a hint about how to very easily find a closed form expression for the requested polynomial.

Hint-1:

Given the geometric series,

$f(t) = 1 + t + t^2 + \cdots + t^n$, what is the result of

$f(t) \times (1-t)$?

Hint-2:

What happens if you set $t = (x+1)$?


问:
Finding the least number of plants watered by three gardeners

There are 100 plants and three gardeners A, B, and C. If A watered 68 plants, B watered 78 plants, and C watered 88 plants, at least how many plants are watered by all three gardeners?

I drew a Venn diagram and let $|A \cap B| - |A \cap B \cap C| = |A \cap C| - |A \cap B \cap C| = 34.$ Then, I let $|B \cap C| - |A \cap B \cap C| = 44,$ so that meant that $|C| - |A \cap C| - |B \cap C| = 10.$ However, this totaled to $122,$ so I concluded that the answer was $122 - 100 = 22.$ Can somebody tell me if my logic is correct or not? Thanks.

答:
Hint: 32 plants weren't watered by A, 22 plants weren't watered by B, and 12 plants weren't watered by C.  The number of plants watered by all three gardeners will be minimized when there is no overlap between those three sets of plants.

Let $A'$, $B'$, and $C'$ represent the sets of plants not watered by A, B, and C respectively.  The set of plants watered by all three is clearly the complement of $A'\cup B'\cup C'$.  Thus, to minimize the size of that set, we must strive to maximize the size of $A'\cup B'\cup C'$.  Since the sizes of $A'$, $B'$, and $C'$ are fixed, their union has maximum size when the sets are disjoint.  Therefore, the minimum number of plants watered by all three gardeners is $100-(12+22+32)=34$.


问:
Sum of square roots inequality

For all $a, b, c, d > 0$, prove that
$$2\sqrt{a+b+c+d} ≥ \sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}$$

The idea would be to use AM-GM, but $\sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}$ is hard to expand. I also tried squaring both sides, but that hasn't worked either. Using two terms at a time doesn't really work as well. How can I solve this question? Any help is appreciated.

答: You can use QM-AM for $\sqrt{a},\ \sqrt{b},\ \sqrt{c},\ \sqrt{d}$; I'm leaving the details for you.

Title: Sum of square roots inequality


问:
Relation between hemi-continuity of correspondences and semi-continuity of functions?
I have just been introduced to the concept of "hemi-continuity" (the "h" is not a typo) of correspondences.

If I understand the concept correctly then the following conjecture should be true:

A correspondence that maps $x\to \{f(x)\}$ (i.e. a single valued correspondence), is lower (or upper) Hemi-continuous iff $f$ is lower (or upper) Semi-continuous

Is my conjecture correct? This is just my first attempt at connecting hemi-continuity to what I already understand.

答: No. An upper hemi-continuous correspondence has a closed graph, but a semi-continuous doesn't. They are related in the sense that $\leq$ over real numbers is similar to $\subseteq$ over subsets.


答: If $p\in\Bbb P$ then for $0\le k\le p-1$ the $x^k$ coefficient in $\sum_{j=0}^{p-1}(1+x)^j$ is $\sum_{j=k}^{p-1}\binom{j}{k}=\binom{p}{k+1}$ by the hockey-stick identity. This is presumably how you proved irreducibility by Eisenstein's criterion, since the non-leading coefficients will be multiples of $p$. This implies the closed form$$\sum_{k=0}^{p-1}\binom{p}{k+1}x^k=\frac1x\sum_{\ell=1}^p\binom{p}{\ell}x^\ell=\frac{(1+x)^p-1}{x},$$but that's more easily obtained by @user2661923's strategy. Both approaches make obvious that this closed form is true even for composite $p$.

Title: Algebraic expansion of $1 + (1 + x) + ... + (1 + x)^{p-1}$

Body:
I was proving that $p(x) = 1 + x + \text{ ... }  + x^{p-1}$, where $p$ is a prime number, is irreducible over the rationals, by using the translation $1 + (1 + x) + \text{ ... } + (1 + x)^{p-1}$. I managed to do it, but I would like to get a closed expression for this as a polynomial.

So far, I have:
\begin{equation}
\sum_{k = 0}^{p-1}{(1+x)^k} = \sum_{k = 0}^{p-1}{(\sum_{i = 0}^k{\binom{k}{i}x^i)}}
\end{equation}

By inspection, it looks like the final result will be:

\begin{equation}
\sum_{k = 0}^{p-1}{(1+x)^k} = \sum_{k = 0}^{p-1}{(\sum_{i=k}^{p-1}{\binom{i}{k}}} )x^k
\end{equation}

I.E.

\begin{equation}
\sum_{i=0}^{p-1}{\binom{i}{0}} + \sum_{i=1}^{p-1}{\binom{i}{1}} x + \text{ ... } + \sum_{i=p-1}^{p-1}{\binom{i}{p-1}}x^{p-1}
\end{equation}

First of all, is this correct? If not, what would be the correct expression and why? And if it is correct, how can I get from the first equation to the second?

Thanks in advance!


问:
Cubic Diophantine
I cannot seem to find a solution to the following Diophantine Equation: $x^2-y^3=2$, where $x,y \in \mathbb{Z}.$ I thought that I could maybe reduce it to a simpler equation , maybe check for the extension $\mathbb{Q}\left[\sqrt{2}\right],$ but nothing that I have tried seems to work. Perhaps it is a known equation/curve I am not aware of? I was first introduced to the problem when my friend pointed out that he could not solve it. I would appreciate your help. Thanks, in advance.

答:
Let $x$ and $y$ be integers such that $x^2-y^3=2$. Then $x$ and $y$ are both odd, and
$$y^3=(x-\sqrt{2})(x+\sqrt{2}),$$
where the gcd of the two factors on the right hand side divides their sum $2\sqrt{2}=\sqrt{2}^3$, and because their product is odd we see that they are coprime. Because $\Bbb{Z}[\sqrt{2}]$ is a unique factorization domain we have
$$x+\sqrt{2}=u(a+b\sqrt{2})^3,$$
for some integers $a$ and $b$, and some unit $u\in\Bbb{Z}[\sqrt{2}]^{\times}$. Then $u=\pm(1+\sqrt{2})^k$ for some integer $k$, and without loss of generality we have the $+$-sign and $k\in\{0,1,2\}$. So we distinguish three cases:

If $k=0$ then
$$x+\sqrt{2}=(a+b\sqrt{2})^3=a(a^2+6b^2)+b(3a^2+2b^2)\sqrt{2},$$
and comparing the coefficients of $\sqrt{2}$ shows that
$$b(3a^2+2b^2)=1,$$
which clearly has no integral solutions.

If $k=1$ then in a similar way we find that
$$a^3+3a^2b+6ab^2+2b^3=1,$$
which is a(nother) cubic Thue equation. Its unique integral solution is $(a,b)=(1,0)$, corresponding to $x=1$.

If $k=2$ then in a similar way we find that
$$2a^3+9a^2b+12ab^2+6b^3=1,$$
which is another cubic Thue equation. Its unique integral solution is $(a,b)=(-1,1)$, corresponding to $x=-1$.

Title: Cubic Diophantine


问:
Rouché's Theorem
I'm trying to understand the proof of the Rouché's Theorem that my professor wrote, but I have a doubt about what he did.

Rouché's Theorem: Let $K\subseteq D \subseteq \mathbb{C}$ with $K$ compact and regular and $D$ open. Let $f,g\in \mathcal{O}(D)$ be holomorphic funtions in $D$, s.t $\forall z\in \partial K, |f(z)-g(z)|<|g(z)|$. Then $f$ and $g$ have the same number of zeros on k, counting multiplicities.

Proof: The inequality implies that $f|_{\partial K}\neq 0$ and $g|_{\partial K}\neq 0$.

By the Argument principle #{zeros of f}-#{zeros of g}$=\frac{1}{2\pi i}\int_{\partial K}(\frac{f'}{f}-\frac{g'}{g})dz$.

By the inequality $|\frac{f(z)}{g(z)}-1|<1, \forall z\in \partial K$, i.e $\forall z\in \partial K$, $\frac{f(z)}{g(z)}\in B_{1}(1)\in\mathbb{C}-\mathbb{R}_{\leq 0}$. Then $h(z):=log(\frac{f(z)}{g(z)}), \forall z\in \partial K$ is well defined and holomorphic in neighborhood of  $\partial K$.

Then $\int_{\partial K}h'dz=0$.

My doubt:

Why we have in the second line that the integral correspond to #{zeros of f}-#{zeros of g}?
I tried to apply the Argument principle and I obtained that

$\frac{1}{2\pi i}\int_{\partial K}(\frac{f'}{f}-\frac{g'}{g})dz=\frac{1}{2\pi i}\int_{\partial K}\frac{f'}{f}dz-\frac{1}{2\pi i}\int_{\partial K}\frac{g'}{g}dz$=

=(#{zeros of f}-#{poles of f})-(#{zeros of g}-#{poles of g}).

But at this point I don't understand why #{poles of f}=#{poles of g}.

答: One of your hypotheses is that $f$ and $g$ are holomorphic functions. How many poles do they have in $D$?

Title: Rouché's Theorem


问:
Implication of sequences
I would like to ask whether the following statement is true or not:

$\lim\limits_{n\to\infty}a(n)=0$ and $\lim\limits_{n\to\infty}a(n)b(n)=1$ implies that $\lim\limits_{n\to\infty}|b(n)|=\infty$

I would like to check my solution. I came to the conclusion that the implication holds.

答:
Step: Show that $b_n$ cannot be bounded. Suppose your statement is not true.
What can you say?

Step: You must indeed have $|b_n| \to \infty$ as $n\to\infty$.

I strongly encourage you not to view all steps at once but rather try for yourself.

Step:

 Suppose there is an $M > 0$ such that $|b_n| \leq M$ for every $n\in\mathbb{N}$.

 Now $|a_nb_n| \leq |a_n|M$.

 Hence $|a_n b_n| \leq |a_n| M \to 0$, meaning $a_nb_n \to 0 \neq 1$ as $n\to\infty$.

Step:

 You already know that $b_n$ cannot be bounded. All that is left to do is to show that for any possible bound $M>0$ there would be an index $N\in\mathbb{N}$ such that $|b_n| \geq M$ for all $n\geq N$. It is not possible that you have a subsequence that is bounded.

 If this was not true, you would (for some fixed $M>0$) find $n_1$ such that $b_{n_1} \leq M$. By using $N=n_1+1$ you would get an $n_2 > n_1$ such that $b_{n_2} \leq M$.

 Iterating this argument would give you $n_1 < n_2 < \ldots$ such that $|b_{n_k}| \leq M$ for all $k\in\mathbb{N}$.

 But this would mean $|a_{n_k}b_{n_k}| \leq |a_{n_k}|M \to 0$ as $n\to\infty$, contradicting $a_nb_n \to 1$.

More generally, $\lim_{n\to\infty} a_nb_n = c \neq 0$ would imply the same result by the exact same arguments.

Edit: The arguments from Step 2 should already suffice as a proof, making Step 1 superfluous but nevertheless i think it's good to think about Step 1 separately to get an intuition for the problem, since in Step 2 we refine the argument from Step 1.

Title: Implication of sequences


问:
Galois conjugation preserves valuation
Let $K$ be a complete discrete valuation field, let $L$ be an algebraic closure of $K$. Let $a\in L\setminus K$ be any element. Why the Galois conjugates of $a$ all have the same valuation?

The valuation ring $\mathcal{O}_L$ is the integral closure of $\mathcal{O}_K$ in $L$, so since Galois automorphism maps integral elements to integral elements, it preserves $\mathcal{O}_L$. But I don't know how to extend this argument further. Any hint or help is much appreciated.

答:
It has to do with the Extension Theorem, according to which (using the additive valuation $v$), we must have, for $z\in L$, $v_L(z)=\frac1{[L:K]}v_K\bigl(\mathcal N^L_K(z)\bigr)$, as long as you want the restriction of $v_L$ to $K$ to be equal to $v_K$. Here, $\mathcal N$ is the Galois-theoretic norm. Since $z$ and any of its conjugates have the same norm down to $K$, their $L$-valuations must be the same.

The above description of $v_L$ doesn’t depend on normality or separability of the extension.

Title: Galois conjugation preserves valuation


问:
If integral of vector field along closed path is zero, must it be conservative?
I know that for any conservative vector field the following is true
$$\oint_C \nabla f \  \mathrm \cdot \ d\vec r = 0$$

However, my question is: if the integral on a closed loop of a vector field evaluates to $0$, then must it be conservative?

Or algebraically put:

Suppose that $\vec F$ is a vector field (we don't know if it's conservative) and $C:\vec r(t)$ is a closed path.
Are the following statements equivalent?
$$\vec F=\nabla f\quad \Longleftrightarrow\quad \oint_C \vec F \  \mathrm \cdot \ d\vec r = 0$$
or should it be $$\vec F=\nabla f\quad \implies\quad \oint_C \vec F \  \mathrm \cdot \ d\vec r = 0$$ instead?

I would like to add the following comment to my question:

For any vector field not of the form $\vec F = \nabla f$, is it the case that $$\oint_C \vec F \  \mathrm \cdot \ d\vec r \neq 0\quad \forall C$$ Or it possible that a vector field of that same form has $$\oint_C \vec F \  \mathrm \cdot \ d\vec r = 0$$ over some $C$.

答:
The statements are indeed equivalent as long as the domain $U \subset \Bbb R^n$ of $\vec F$ is path connected (but not necessarily simply connected!). As the proof here explains: if $\vec F: U \to \Bbb R^n$ satisfies $\oint_C \vec F\cdot d \vec r = 0$ for all closed loops in the domain, then an associated potential function $f:U \to \Bbb R$ can be constructed as follows.

Select a fixed element $\vec a \in U$. For every $\vec x \in U$, there exists a path $\gamma[\vec a, \vec x]$ connecting $\vec a$ and $\vec x$. Define
$$
f(\vec x) = \int_{\gamma[\vec a, \vec x]} \vec F(\vec r) \cdot d \vec r.
$$
Because $\vec F$ is conservative (i.e. because of the integral condition), this function is well defined (i.e. we get the same result no matter which path we choose). With this $f$, we indeed find that $\nabla f = \vec F$.


问:
Probability of winning sweep stakes, put all coupons in one box or split them between boxes?

Taken from : here

A mall has sweep stakes where they give you a coupon for every X dollars you spend (amount is not important). You fill the coupon with your contact details and put it in the box. Every two weeks, they draw 10 coupons, give their owners prizes, and discard the rest of the coupons.

Now, thanks to some very big purchase we made, we have about 300 coupons (going to hurt when filling them). To get the highest probability of winning, should we put all the coupons at once, or split them evenly between the three remaining draws?

答:
Let's assume that in each box there are n other tickets besides yours. So if you put them all in the first box and 10 tickets are drawn your chance of winning something is

$$q = 1 - \prod_{k=1}^{10} \left( 1 - \frac{300}{n+301-k} \right).$$

Now if you split them, 100 each amongst three boxes (where, again, each box has n other tickets), your chance of winning something is

$$p = 1 - \prod_{k=1}^{10} \left( 1 - \frac{100}{n+101-k} \right)^3 .$$

Now p>q, but only slightly, for large n.

For example, with n=10000, $$p \approx 0.258176 \textrm{ and } q \approx 0.256004$$

So you're slightly better off if you split them up into three lots of 100. However, this assumes that the number of other tickets placed in the box every two weeks remains constant.

Note that the difference between p and q decreases with increasing n.

LastEditorDisplayName: anonymous

ContentLicense: CC BY-SA 2.5


问:
Composition factors and annihilation
Let $A$ be an associative algebra over a field $k$ and let $M$ be some $A$-module. If $\mathfrak{a}$ is an ideal of $A$ which annihilates $M$, i.e. if $\mathfrak{a}M=0$, then why must every composition factor of $M$ be a composition factor of $A/\mathfrak{a}$?</p>

答: Let E be a composition factor of M. Then E is an irreducible A-submodule of M, and since aM=0 we can consider E as an irreducible A/a-module. Since every irreducible A/a-module must occur as a composition factor in its regular representation, we are done.</p>

Title: Composition factors and annihilation


答:
If you square both sides:

$(\sqrt a + \sqrt b + \sqrt c +\sqrt d)^2 =a + b+c+d + 2(\sqrt{ab} + \sqrt{ac}+\sqrt{ad} + \sqrt{bc}+\sqrt{bd} + \sqrt{cd})$

while $(2\sqrt {a+b+c+d})^2= 4(a+b+c+d)$ so it suffices to prove

$2(\sqrt{ab} + \sqrt{ac}+\sqrt{ad} + \sqrt{bc}+\sqrt{bd} + \sqrt{cd})\le 3(a+b+c+d)$.

If we apply AM/GM though, we get $\sqrt{ab} \le \frac {a+b}2$ or in other words $2\sqrt{ab} \le a+b$.

And that does it:

$2(\sqrt{ab} + \sqrt{ac}+\sqrt{ad} + \sqrt{bc}+\sqrt{bd} + \sqrt{cd})\le$

$(a+b) + (a+c) + (a+d) + (b+c) + (b+d) = 3(a+b+c+d)$.

Title: Sum of square roots inequality

Body:
For all $a, b, c, d > 0$, prove that
$$2\sqrt{a+b+c+d} ≥ \sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}$$

The idea would be to use AM-GM, but $\sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}$ is hard to expand. I also tried squaring both sides, but that hasn't worked either. Using two terms at a time doesn't really work as well. How can I solve this question? Any help is appreciated.


问:
Evaluating $\sum_{i=0}^{30}(-1)^i {{30} \choose {i}}(30-i)^{31}$
As the title says, I've been asked to evaluate this sum: $$\sum_{i=0}^{30}(-1)^i {{30} \choose {i}}(30-i)^{31}$$ and have been unable to do so. The possible answers are $$\frac {30\cdot31!} 2, \frac {31!} 2, \frac {31!} 3, \frac {30\cdot31!} 3$$
and I don't even see where the factorial could come from. I tried expanding the inner binomial expression but that didn't yield anything helpful. Help would be appreciated.

答:
Evaluating $$\sum_{q=0}^n (-1)^q {n\choose q} (n-q)^{n+1}$$ we find

$$(n+1)! [z^{n+1}] \sum_{q=0}^n (-1)^q {n\choose q} \exp((n-q)z)
\\ = (n+1)! [z^{n+1}] (\exp(z)-1)^n
\\ = (n+1)! [z^{n+1}] (z+z^2/2+\cdots)^n
\\ = (n+1)! \frac{1}{2} {n\choose 1}.$$

With $n=30$ this yields $$31! \times \frac{30}{2}.$$

For a combinatorial interpretation we recognize the Stirling number multiple $n! \times {n+1\brace n}.$ We have ${n+1\choose 2}$ ways to partition $[n+1]$ into $n$ disjoint sets by choosing the single pair. We get $n! \times  {n+1\choose 2},$ the same as before.


问:
If $CC^*$ = $DD^*$ then C=DQ
If C and D be two m by l matrices and $CC^*$ = $DD^*$ then $C=DQ$ where $Q$ is a l by l unitary matrix .
If C and D  are  of order 1 then it is trivial  But for general case I was trying to use induction .
But I had faced slide problems . How can I do this for general case.

答: Hint: Note that $C$ spans the row space of $CC^*$.

Title: If $CC^*$ = $DD^*$ then C=DQ


问:
Nonattacking "queens" on tiled triangles
I'll start with what my question is not. My question is not Nonattacking rooks on a triangular chessboard or The number of spacing $k$ non-attacking towers on the board $\left\{(i,j):1 \le i \le j \le n \right\}$. Those questions are about taking a square grid and making a "triangular chessboard" by removing the squares of a chessboard above the diagonal.

My question is about tiled triangles of the following form (not truncated square grids):

I will define a "queen" on this tiled triangle to attack in the three directions parallel to the triangle on which the queen is located. Below, I have placed a queen on the darkest blue triangle and the light blue triangles are where the queen attacks.

I am interested in the maximal number of non-attacking queens I can place on this tiled triangle. For example, I can imagine placing down queens until every triangle is under attack:

Or, as another example,

My questions are: For a tiled equilateral triangle of side length $n$, what is the maximal number of non-attacking queens I can add to the triangle? How many configurations have the maximal number of non-attacking queens? By side length, I mean for example that the above tiled triangles have a side-length of five.

I think my trouble has been that I'm unsure of a good coordinate system for the tiled triangle. It's two dimensional, but I take out three straight lines corresponding to the horizontal, diagonal, and antidiagonal directions when I add a queen. Further, knowing which horizontal row and diagonal I am in does not typically fix the antidiagonal I am in (there are typically two choice for the antidiagonal).

From inspection, it looks like the maximum number of non-attacking queens on a tiled triangle of sidelength $n$ is $\lceil n/2 \rceil$. Maybe induction on the length of the sides could work? It looks like adding row of triangles at the base of a maximal-number-of-queens odd $n$ triangle gives a maximal number-of-queens even $n+1$ triangle, but I am unsure of how to show this.

答:
Your problem is similar to the one of placing non-attacking rooks on a triangular array of hexagons, discussed in this Math Overflow post: https://mathoverflow.net/questions/103540/hexagonal-rooks/103610. The same proof method works in your problem. I am generalizing the proof of optimality in this comment.

The maximal number of queens is $\left\lfloor 2n+1\over 3\right\rfloor$.

To place that many queens, do the following:

Letting $k=\left\lfloor 2n+1\over 3\right\rfloor$, place a queen in the leftmost triangle in the $k^{th}$ row from the bottom.

Place a queen in the rightmost triangle in the row below the previous.

Go down the rows one by one. For each row, either place a queen in the triangle below the queen in step $1$, or the queen in step $2$, alternating each time.

Here is an illustration when $n=10$:

Here is how you prove this is optimal. Each triangle has a triple of coordinates $(x,y,z)$, where $x$ is the number of triangles a queen must move to reach the left edge, and $y$ and $z$ are the corresponding distances for the right and bottom edges. Orient the entire grid so a vertex points upward, as in your pictures. You can show that:

For an upward pointing triangle, $x+y+z=2n-2$, and $x,y$ and $z$ are all even.

For downward pointing triangle, $x+y+z=2n-1$, and $x,y$ and $z$ are all odd.

For example, when $n=4$, you can verify the possible coordinates are permutations of $(0,0,6),(0,2,4),(2,2,2),(1,1,5)$ or $(1,3,3)$.

Now, for a placement of $k$ queens, let $\Sigma_x$ be the sum of the $x$-coordinates of the queens, and define $\Sigma_y$ and $\Sigma_z$ similarly. Furthermore, let $d$ be the number of queens on downward pointing triangles. The preceding two bullets imply
$$
\Sigma_x+\Sigma_y+\Sigma_z=k\cdot (2n-2)+d
$$
Therefore, at least one of the $\Sigma_x,\Sigma_y$ or $\Sigma_z$ must be at most $(k(2n-2)+d)/3$. WLOG, say $$\Sigma_x\le \frac{k(2n-2)+d}3.\tag1$$

Furthermore, as $i$ ranges over the $x$-coordinates of the queens, the values $\lfloor i/2\rfloor$ must be distinct, else the queens with the same value would attack each other in the direction parallel to the line $x=0$. If there are $k$ queens, the smallest possible value of $\Sigma_x$ for which the values $\lfloor i/2\rfloor$ are distinct is

$$
\Sigma_x \ge (0+2+4+\dots+2(k-1))+d=k(k-1)+d\tag2
$$
this is because the smallest value is attained by starting with the list $0,2,\dots,2(k-1)$, and increasing $d$ of these values by $1$, since the $d$ coordinates of the downard pointing triangles must be odd.

Combining $(1)$ and $(2)$, you get
$$
k(k-1)+d\le {k(2n-2)+d\over 3}
$$
Divide both sides by $k$ and rearrange to get
$$
k\le {2n-2\over 3}+1- {2d\over 3k}\le {2n-2\over 3}+1
$$
so $k\le \left\lfloor 2n+1\over 3\right\rfloor$.


问:
geodesics on sphere loring tu help
Loring Tu in his book Differential geometry [page 104] states (in my own words) the following on geodesics on spheres:

Consider the 2 sphere of radius $a$ in $\mathbb{R}^3$ .

Parameterise a great circle by arc length $\gamma$.

$\gamma''(t)$ is perpendicular to $\gamma'(t)$

$\gamma''(t)$ lies in the plane of the circle (does this mean $T_{\gamma(t)}M?)$

$\gamma''(t)$ is perpendicular to the tangent plane at $\gamma(t)$

I'm not sure why $(4).$ holds, why is $\gamma''(t)$ in the normal space $N_{\gamma(t)}M$?

Please do not explain using curvature.

答:
Note that the plane of the great circle contains the origin (i.e. the center of the sphere). $\gamma''$ is orthogonal to $\gamma'$, so because $\gamma'$ is parallel to the circumference, $\gamma''$ must be parallel to the radius of the great circle.

Now, note that the radius of the great circle is also a diameter of the sphere. So, $\gamma''$ is parallel to the radius of the sphere and hence orthogonal to its tangent plane.

Title: geodesics on sphere loring tu help


问:
Distance between a point inside a circle and circumference on a line along a point outside of circle
Point $A$ is located outside of a circle centered at $C$ and with radius $r$. Point $B$ is given point inside the circle. How to calculate $d$, the length of line segment between $B$ and circumference on the line $\overline{AB}$. Would there be a solution regardless of where $B$ is inside the circle?

Update: given to the problem is coordinates $A$, $B$ and $r$. I will need to extend the same problem to 3D geometry with a sphere and 3D coordinates. I would appreciate any help.

答:
A method for you to use.

Yes there is always a solution. To find it you can do the following, where the centre of the circle is taken as the origin.

You can find the equation of the line $AB$ in the form $y=mx+c$.

Then the point $P(x,y)$ where the line crosses the circle satisfies the equation
$$ x^2+(mx+c)^2=r^2.$$

Solve this equation for $x$ and then find $y$ from $y=mx+c$. There will be two points, choose the one on the same side of the circle as $A$.

Finally calculate the distance $BP$.


问:
Strict inequalities in real-valued continuous random variable
Let $X$ be a real-valued continuous random variable, such as the exponential random variable. Is the equation
$$
\mathbb{P}(X\le a) = \mathbb{P}(X < a)
$$
true? In other words, can we replace an "inequality" with its "strict" version? For instance, in the case of exponential random variable, do we have $\mathbb{P}(X\le a) = \mathbb{P}(X < a) = 1 - e^{-\lambda a}$? Intuitively, this seems correct to me (at least for non-pathological random variables) because the probability of a single point is zero, i.e., the integral of pdf over a single point is zero.
I appreciate any comment or answer to this question.

答: Yes, that's exactly right.  The definition of "continuous" for a random variable is $\mathbb P(X=a) = 0$ for all $a$.  Thus $$\mathbb P(X \le a) = \mathbb P(X < a) + \mathbb P(X = a) = \mathbb P(X < a)$$


问:
What does it mean to find something in Set theory? What is formal-language equivalent of "find"?
For example in proof that every real number $x$ has a decimal expression $x=a_0.a_1a_2a_3….$ it says:

so we can find $a_1$ between $0$ and $9$ such that...

https://math.stackexchange.com/a/2625318/360552

What if $x$ is, for example, Chaitin constant where we cannot find $a_n$ that satisfies given condition because we don't know how to compute the digits or exactly how big/small is Chaitin constant but we only know that it exists and is in some fixed interval.

Obviously, find is not the same as compute but I wonder what mathematicians mean by find. Which ZF Axiom is used to find a number that satisfies a certain condition.

Another example is in this Cantor proof that real numbers are uncountable. It says

Find the first two numbers of the given sequence that are in the open interval $(a, b)$.

The problem with this sentence is that if a sequence is a series of uncomputable numbers like Chaitin's constant, how can we "find" those first two numbers that are in the open interval $(a, b)$ if we don't know exact values of those numbers.

So which ZF axiom is used in Cantor's proof to find those first two numbers in a sequence that satisfy a given condition? What does it mean to construct an interval with those 2 numbers if we cannot find those 2 numbers and only know that they exist? It is obvious that I also do not know what a constructive proof means (I thought that I knew).

Obviously, find means something other than what I intuitively think it means so I wonder which axioms are used to find something that satisfies a certain condition like in Cantor's proof. What axioms are required for this Cantor's proof?

Summary:

How could we express find in a formal language?

Which axioms are used to find an element of a Set that satisfies a certain condition.

答:
At least in your first example, it corresponds to "there exists". Your second example is slightly different, but with some adjustment it is essentially the same. Basically, we can take it as:
"there exists at least two elements of the sequence which are in the interval $(a, b)$. Let $x_{1}, x_{2}$ be the first two."


问:
If $a_k>0\ \forall\ k\in \mathbb{N}\ $ and $r>0,\ $ prove that $\sum\frac{1}{a_k}$ converges $\iff \sum\frac{1}{a_k+r}$ converges.
If $a_k > 0\ \forall\ k \in \mathbb{N}\ $ and $r>0,\ $ prove that $\displaystyle \sum_{k=1}^\infty\frac{1}{a_k}$ converges $\iff \displaystyle \sum_{k=1}^\infty\frac{1}{a_k+r}$ converges.

$\implies\ $ is easy: $\frac{1}{a_k + r}<\frac{1}{a_k}\ \forall\ k \in \mathbb{N}.$

$\impliedby\ $ is more difficult. I provide my answer below, but certainly welcome other answers: there are often simpler solutions as I tend to over-complicate things.

答:
We are trying to prove that, if $r>0,\ $ $\large{\frac{1}{a_k-r}}$ $\ > 0\ \forall\ k \in \mathbb{N}\ $ and $\ \displaystyle \sum_{k=1}^\infty\frac{1}{a_k}\ $ converges then $\displaystyle \sum_{k=1}^\infty\frac{1}{a_k-r}\ $ also converges.

I will prove the contrapositive of $\impliedby,\ $ that is, $\displaystyle \sum_{k=1}^\infty\frac{1}{a_k}\ $ diverges $\implies\ \displaystyle \sum_{k=1}^\infty\frac{1}{a_k+r}\ $ diverges.

If $\displaystyle\lim_{k\to\infty}\left(\frac{1}{a_k}\right) \neq 0,\ $ then $\exists\ c>0\ $ such that $\large{\frac{1}{a_k}}$ $ \geq c\ $ for infinitely many $k,\ $ which implies $\large{\frac{1}{a_k+r} \geq \frac{c}{1+rc}}$ $>0\ \forall\ k \in \mathbb{N},\ $ which implies that $\displaystyle \sum_{k=1}^\infty\frac{1}{a_k+r}\ $ diverges, and we are done. So let's assume for the rest of the proof that $\displaystyle\lim_{k\to\infty}\left(\frac{1}{a_k}\right) = 0.$

Then $\exists\ j \in \mathbb{N}\ $ such that $a_k \geq r\ \forall\ k \geq j,\ $

\begin{align} \implies\ \frac{1}{2a_k} = \frac{1}{a_k + a_k} \leq \frac{1}{a_k + r}\ \forall k \geq j\\
\\
\implies \displaystyle \sum_{k=1}^\infty\frac{1}{a_k+r} = \displaystyle \sum_{k=1}^{j-1}\frac{1}{a_k+r}+\displaystyle \sum_{k=j}^\infty\frac{1}{a_k+r} > \displaystyle \sum_{k=j}^\infty\frac{1}{a_k+r} \geq \frac{1}{2}
 \displaystyle \sum_{k=j}^\infty\frac{1}{a_k} = \infty.\\
\end{align}


问:
In the lattice of ideals, what are the lowerbounds of the prime ideals?
Take the lattice of ideals in a non-commutative ring with 1. It is well known that the lowerbounds of all the maximal ideals are the superfluous ideals. Is there a similar characterization for the lowerbounds of all the prime ideals? They must be close to the sum of nilpotent ideals but I can't seem to prove it.</p>

答:
The situation for the maximal ideals is clear cut since the greatest lower bound (their intersection, the Brown-McCoy radical) happens to be the biggest superfluous ideal of the ring. Any ideal below it is also superfluous, and it contains all superfluous ideals.

Following this, we consider the intersection of all prime ideals (The "lower nilradical", "prime radical" or "Baer radical").

It's known (you can check Lambek's Lectures on rings and modules) that the prime radical of a ring consists of all its strongly nilpotent elements. (The definition of $a$ being "strongly nilpotent" is that every sequence beginning with $a_1=a$ such that $a_{i+1}\in a_nRa_n$ has a term which is zero, so that it terminates in zeros.)

If we define a "strongly nil ideal" to be one in which all elements are strongly nilpotent, then the prime radical is the largest strongly nil ideal, and it is obvious that all ideals contained within it are strongly nil ideals, and it's clear that all strongly nil ideals are contained in the prime radical.

The sum of all nilpotent ideals is obviously contained inside the prime radical, but it may be strictly smaller.


答:
$x^2-y^3=2\quad\implies\quad
 y=\sqrt[\large3]{x^2-2}\quad\lor\quad x=\sqrt{y^3+2}\quad$

Both equations are true only when $\quad (x,y)=(\pm1,-1)$

For larger values, no squares and cubes differ by only $2$.

Title: Cubic Diophantine

Body: I cannot seem to find a solution to the following Diophantine Equation: $x^2-y^3=2$, where $x,y \in \mathbb{Z}.$ I thought that I could maybe reduce it to a simpler equation , maybe check for the extension $\mathbb{Q}\left[\sqrt{2}\right],$ but nothing that I have tried seems to work. Perhaps it is a known equation/curve I am not aware of? I was first introduced to the problem when my friend pointed out that he could not solve it. I would appreciate your help. Thanks, in advance.


问:
Sum of square roots inequality

For all $a, b, c, d > 0$, prove that
$$2\sqrt{a+b+c+d} ≥ \sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}$$

The idea would be to use AM-GM, but $\sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}$ is hard to expand. I also tried squaring both sides, but that hasn't worked either. Using two terms at a time doesn't really work as well. How can I solve this question? Any help is appreciated.

答:
How about Jensen's inequality? $\sqrt x$ is concave so

$$\sqrt{\frac{a+b+c+d}{4}} \ge \frac 14 \left( \sqrt a +\sqrt b +\sqrt c +\sqrt d \right)$$

Title: Sum of square roots inequality


问:
Positive but discontinuous derivative?
Let $f: \mathbb{R} \to \mathbb{R}$ be differentiable on an open interval $I$, with $f' > 0$ on $I$. Can $f'$ be discontinuous at a point of $I$?

Title: Positive but discontinuous derivative?

Body: Yes. Try $f(x) = 2x + x^2 \sin(1/x)$ with $f(0)=0$.


答:
We may use the limit comparison test. Note that, for either series to converge, we must have
$$\frac{1}{a_k}\to 0\Leftrightarrow a_k\to\infty.$$
In particular, for all $N$, there exists some $k_0$ for which $a_k>N$ for all $k>k_0$. We claim that
$$\lim_{k\to\infty}\frac{a_k}{a_k+r}=1.$$
Indeed, for any $\epsilon$,
$$\left|1-\frac{a_k}{a_k+r}\right|<\epsilon$$
is equivalent to $a_k>N$ for some $N$, so there exists a $k_0$ for which this holds for all $k>k_0$. By the limit comparison test, the convergences of the two series are equivalent.

Title: If $a_k>0\ \forall\ k\in \mathbb{N}\ $ and $r>0,\ $ prove that $\sum\frac{1}{a_k}$ converges $\iff \sum\frac{1}{a_k+r}$ converges.

Body:
If $a_k > 0\ \forall\ k \in \mathbb{N}\ $ and $r>0,\ $ prove that $\displaystyle \sum_{k=1}^\infty\frac{1}{a_k}$ converges $\iff \displaystyle \sum_{k=1}^\infty\frac{1}{a_k+r}$ converges.

$\implies\ $ is easy: $\frac{1}{a_k + r}<\frac{1}{a_k}\ \forall\ k \in \mathbb{N}.$

$\impliedby\ $ is more difficult. I provide my answer below, but certainly welcome other answers: there are often simpler solutions as I tend to over-complicate things.


问:
Proving the remainder is $1$ if the square of a prime is divided by $12$
Given, $p$ is a prime number and $p>3$. How do we prove that the remainder $r$ is always $1$ if $p^2$ is divided by $12$?</p>

答: $ p^2 -1 =(p-1)(p+1) $. Since $p$ is odd, these are two even factors, so the product is divisible by $4$. Also, one of $p-1$, $p$, $p+1$ must be divisible by $3$, but since $p$ is prime, it is not divisible by $3$. Thus one of $p-1$, or $p+1$ must be divisible by three.


答:
You can find it, for example, on Morris H DeGroot, Mark J Schervish - Probability and statistics-Pearson Education (2012), pages 101-102 under Note: Continuous Distributions Assign Probability $0$ to Individual Values:
$$Pr(a \leqslant X \leqslant b) =\int\limits_{a}^{b}f(x)dx=Pr(a \leqslant X \lt b)=Pr(a \lt X \lt b)
$$
So if $X$ have continuous distribution, then  $Pr(X = a) = 0$ for each number $a$, however it does not imply that $X = a$ is impossible.

Title: Strict inequalities in real-valued continuous random variable

Body:
Let $X$ be a real-valued continuous random variable, such as the exponential random variable. Is the equation
$$
\mathbb{P}(X\le a) = \mathbb{P}(X < a)
$$
true? In other words, can we replace an "inequality" with its "strict" version? For instance, in the case of exponential random variable, do we have $\mathbb{P}(X\le a) = \mathbb{P}(X < a) = 1 - e^{-\lambda a}$? Intuitively, this seems correct to me (at least for non-pathological random variables) because the probability of a single point is zero, i.e., the integral of pdf over a single point is zero.
I appreciate any comment or answer to this question.


问:
If an infinite sum of positive terms $\sum x_n$ diverges, must there exist a sequence $y_n$ with $y_n/x_n \to 0$ and $\sum y_n$ divergent?
Suppose that $\sum_{n=1}^{\infty} x_n$ diverges, where $x_n>0$ $\forall n$. Must there exist a real sequence $y_n$ such that $y_n/x_n \to 0$ as $n \to \infty$ and such that $\sum_{n=1}^{\infty} y_n$ also diverges?

I've been stuck on this for quite a while, and have tried considering cases such as $x_n=1/n$ (for which this is possible e.g. with $y_n=1/n \log n$) to get an idea of in what situations this holds, but it doesn't seem very straightforward to generalize. I don't really know where to start in terms of approaching the problem abstractly.

Body:
Since $\sum_{k=1}^{\infty} x_{k}$ is not finite, it is possible to find blocks of integers, say $[n_{i},n_{l+1}]$ for which $\sum_{k=n_{l}}^{n_{l+1} -1} x_k >= 1$
and such that $n_{l+1} > n_{l}$.

For $k\in[n_{l},n_{l+1} -1]$ define $y_{k} = \frac{1}{l}*x_{k}$.
Then $$\sum_{k=1}^{\infty} y_{k} = \sum_{l}\sum_{k = n_{l}}^{n_{l+1}-1} y_{k} >= \sum_{l}\frac{1}{l}\sum_{k=n_{l}}^{n_{l+1}-1}x_{k} >= \sum_{l}\frac{1}{l}$$


答:
We should restrict to the case that $a_{k}>0$ for all $k$. Otherwise,
it is possible that $\sum_{k=1}^{\infty}\frac{1}{a_{k}+r}$ converges
but $\sum_{k=1}^{\infty}\frac{1}{a_{k}}$ is undefined (because $\frac{1}{0}$
is undefined).

Since $\frac{1}{a_{k}+r}\leq\frac{1}{a_{k}},$ we have $\sum_{k=1}^{\infty}\frac{1}{a_{k}+r}\leq\sum_{k=1}^{\infty}\frac{1}{a_{k}}$.
It follows that $\sum_{k=1}^{\infty}\frac{1}{a_{k}}$ converges $\Rightarrow$
$\sum_{k=1}^{\infty}\frac{1}{a_{k}+r}$ converges.

Conversely, assume that $\sum_{k=1}^{\infty}\frac{1}{a_{k}+r}$ converges.
In particular, $\frac{1}{a_{k}+r}\rightarrow0$ and hence $a_{k}\rightarrow\infty.$
Now, $\frac{a_{k}+r}{a_{k}}\rightarrow1$, so there exists $K$ such
that $\frac{a_{k}+r}{a_{k}}\leq2$ whenever $k\geq K$. Observe that
\begin{eqnarray*}
\sum_{k=K}^{\infty}\frac{1}{a_{k}} & = & \sum_{k=K}^{\infty}\frac{1}{a_{k}+r}\cdot\frac{a_{k}+r}{a_{k}}\\
 & \leq & \sum_{k=K}^{\infty}\frac{1}{a_{k}+r}\cdot2\\
 & < & \infty.
\end{eqnarray*}
It follows that $\sum_{k=1}^{\infty}\frac{1}{a_k}$ converges.


答:
The expression can be recognized as inclusion-exclusion count for the number of ways to place 31 balls into 30 bins in such a way that no bin is empty.

Clearly this can be performed only by filling 29 bins with one ball and one bin with two balls.
Thus the number in question can be alternatively computed as:
$$\binom{30}1\binom{31}2 29!,
$$
where the strategy to perform the task is the following: first choose a bin for two balls, then choose the two balls, and finally distribute the remaining balls between the remaining bins.

Title: Evaluating $\sum_{i=0}^{30}(-1)^i {{30} \choose {i}}(30-i)^{31}$

Body:
As the title says, I've been asked to evaluate this sum: $$\sum_{i=0}^{30}(-1)^i {{30} \choose {i}}(30-i)^{31}$$ and have been unable to do so. The possible answers are $$\frac {30\cdot31!} 2, \frac {31!} 2, \frac {31!} 3, \frac {30\cdot31!} 3$$
and I don't even see where the factorial could come from. I tried expanding the inner binomial expression but that didn't yield anything helpful. Help would be appreciated.


问:
Commutator property
Can you please show me how to prove this? If H,K,L are normal subgroups of G then $$[[H,K],L]\subseteq [[K,L],H][[L,H],K]$$ Thanks in advance!</p>

答:
Hints (Care: you need to know well commutators identities) By normality we have

$$[[H,K],L]=[H,L][K,L]\;,\;\;[[K,L],H]=[K,H][L,H]\;,\;\;[[L,H],K]=[L,K][H,K]$$

So you want to show

$$[H,L][K,L]\le[K,H][L,H][L,K][H,K] =[H,L][K,L]^{[H,K]}$$

Title: Commutator property


问:
Computing a single value within 0 and 1 from a set of features
I have a set instances with features that I would like to compare with a reference set of values. For example, I have the instances with their features below:

\begin{array}{|l|l|l|l|}\hline
Arrival Time & Departure Time & Processing Time & Wait Time \\\hline
3 & 6 & 25 & 20 \\\hline
19 & 5 & 20 & 21 \\\hline
40 & 10 & 42 & 42\\\hline
\end{array}

The reference instance is:
\begin{array}{|l|l|l|l|}\hline
Arrival Time & Departure Time & Processing Time & Wait Time \\\hline
20 & 15 & 30 & 25 \\\hline
\end{array}

I have been calculating how each instance matches to the reference features by treating each row as a vector and calculating their magnitude and dividing it with by the magnitude of the reference to give me values between 0 and 1.

I have a problem where I get values greater than 1. Is there any way to rectify that or even an alternate way of comparison? I require that the value be within that range so that I can feed it into another equation.

答: For $d$ any metric on $\mathbb{R}^4$, $\frac{d}{d + 1}$ takes values between $0$ and $1$. So if $y$ is your reference, you can take $$x \mapsto \frac{d(x, y)}{d(x, y) + 1}$$ for your favourite metric $d$. For example the Euclidean one. This has as added benefit that the farther away you are from the reference value, the closer to $1$ this comparison function becomes, and it is only $0$ for the reference value itself.


问:
How can I prove (without cardinal numbers) that $P(P(\mathbb{N}))$ is equipollent to $P(P(\mathbb{N})) \times P(P(\mathbb{N}))$
I am stuck trying to prove that $P(P(\mathbb{N}))$ is equipollent to $P(P(\mathbb{N})) \times P(P(\mathbb{N}))$. I was thinking that it is enough to prove $P(\mathbb{R}) \sim P(\mathbb{R}) \times P(\mathbb{R})$, or even $2^{\mathbb{R}} \sim 2^{\mathbb{R}} \times 2^{\mathbb{R}}$ , but I still don't know how to finish this. It would be even better if you can prove that in general for any infinite set $X$ it's true that $X \times X \sim X$.

答:
The statement that $|X|=|X\times X|$ is equivalent to the axiom of choice. You can prove it as follows. Let: $$\mathcal{H}:=\{f:B \longrightarrow B\times B : B\subset X \text{ and } f \text{ is a bijection}\}$$ with the relation order given by $g<h \Longleftrightarrow \operatorname{Dom}g\subset\operatorname{Dom} h \text{ and } h|_{\operatorname{Dom} g}=g$. It's easy to proof that for every chain $\mathcal{C}\subset\mathcal{H}$, the function defined as $f_\mathcal{C}(c):=f(c)$ if $c\in S\subset\mathcal{C}$ is well defined, is a element of $\mathcal{H}$ and is an upper bound for $\mathcal{C}$.

Hence there is a maximal $f_0\in\mathcal{H}$ (Zorn's Lemma). It can be proved (it's not an trivial proof, but I can write it down here if you want), that $\operatorname{Dom} f_0=X$, so $f_0$ is a bijection between $X$ and $X\times X$.

Let $A_0:=\operatorname{Dom} f_0$ and lets prove that there is a bijective map between $A_0$ and $X$.

For this purpose, assume that there exists a surjective aplication $g:X\setminus A_0\longrightarrow A_0$. Then, you can easily prove using axiom of choice that there exists $S\subset X\setminus A_0$ such that $\phi:=g|_{S}$ is bijective. Now note that:
$$(A_0\cup S)\times(A_0\cup S)=(A_0\times A_0)\cup(A_0\times S)\cup(S\times A_0)\cup(S\times S)$$

And also note that the function $\theta:=\phi\circ f_0\circ(\phi^{-1},\phi^{-1})$ is a bijection between $S$ and $S\times S$.

There exists a bijective map $j:S\longrightarrow(A_0\times S)\cup(S\times A_0)\cup(S\times S)$ (the existence of this map is given by the fact that $|S|=|S\times S|=|(A_0\times S)\cup(S\times A_0)\cup(S\times S)|$, because the union of disjoint sets of same cardinality has the same cardinality as any of the sets or without using cardinals at all, there is a bijection between $T$ and $T\cup P$ whenever there exists one between $T$ and $P$; again, if you want a proof of this I can write it down). Then, the map defined as:
\begin{align*}
    h_0:A_0\cup S &\longrightarrow(A_0\cup S)\times(A_0\cup S) \\
    x &\longmapsto h_0(x):=\begin{cases}f_0(x) &\text{ if } x\in A_0 \\
    j(x) &\text{ if } x\in S\setminus A_0\end{cases}
\end{align*}
is a bijection, and $h_0|_{A_0}=f_0$. Hence $f_0<h_0$, which is a contradiction. Then there doesn't exists a surjective map $g:X\setminus A_0\longrightarrow A_0$, which implies that there actually is a bijection $k:X\longrightarrow A_0$ (this is also a consecuence of the cardinality of union of disjoint sets). Finally, the map: $$k\circ f_0 \circ (k^{-1},k^{-1}): X\longrightarrow X\times X$$ is bijective, which concludes the proof.


答:
As Jacob points out, the phrase 'find an $X$ such that $Y$' is just a replacement of 'since we know there exists an $X$ such that $Y$, we may pick any such $X$'. You seem to think that in order to pick such $X$ we need an axiom that allows you to pick elements, similar to the axioms of ZFC. This is not the case: in classical logic we simply define the phrase `there exists' to be true whenever there is such $X$, and if this is the case we may pick any such $X$ without knowing a way to construct it. You're right that this is very different from a constructive proof.

To answer your question in the comments: I'm not sure whether the the diagonal argument is constructive, but at least what you can show constructively is that there are uncountably many integer sequences: for a given sequence of numbers you can constructively find a new sequence that is not in your list, therefore proving your list was incomplete. See also wiki. (On wiki it says that this proof is a proof by contradiction, but it is still a constructive proof since you really just use the definition of 'uncountable', which is 'not countable', which in turn must be proven by 'if it were countable, it would lead to contradiction'.)

Title: What does it mean to find something in Set theory? What is formal-language equivalent of "find"?

Body:
For example in proof that every real number $x$ has a decimal expression $x=a_0.a_1a_2a_3….$ it says:

so we can find $a_1$ between $0$ and $9$ such that...

https://math.stackexchange.com/a/2625318/360552

What if $x$ is, for example, Chaitin constant where we cannot find $a_n$ that satisfies given condition because we don't know how to compute the digits or exactly how big/small is Chaitin constant but we only know that it exists and is in some fixed interval.

Obviously, find is not the same as compute but I wonder what mathematicians mean by find. Which ZF Axiom is used to find a number that satisfies a certain condition.

Another example is in this Cantor proof that real numbers are uncountable. It says

Find the first two numbers of the given sequence that are in the open interval $(a, b)$.

The problem with this sentence is that if a sequence is a series of uncomputable numbers like Chaitin's constant, how can we "find" those first two numbers that are in the open interval $(a, b)$ if we don't know exact values of those numbers.

So which ZF axiom is used in Cantor's proof to find those first two numbers in a sequence that satisfy a given condition? What does it mean to construct an interval with those 2 numbers if we cannot find those 2 numbers and only know that they exist? It is obvious that I also do not know what a constructive proof means (I thought that I knew).

Obviously, find means something other than what I intuitively think it means so I wonder which axioms are used to find something that satisfies a certain condition like in Cantor's proof. What axioms are required for this Cantor's proof?

Summary:

How could we express find in a formal language?

Which axioms are used to find an element of a Set that satisfies a certain condition.


答:
It comes down to how you interpret the symbol $\exists$ in "$\exists x. Px$". If you're a constructivist, it means some variant of "I have a witness to the truth of $P$". If you're a normal mathematician, it means something more like "$\neg \forall \neg$". For a normal mathematician, we don't care that equality of reals is undecidable; we just go ahead and assert that $x = y$ or $x \ne y$ anyway.

Constructivists also can have problems even defining many real numbers (depending on quite how constructivist you go), and this is another source of the difference.

The first answer you linked to is https://math.stackexchange.com/a/2625318/360552. This is not constructive in the strictest computational sense, because it relies on being able to compute whether a given real is greater than or equal to some integer $n$. The most highly upvoted answer is https://math.stackexchange.com/a/409706/259262 which makes it more explicit where the noncomputable step is: it says "take the least such $n$", which certainly exists (in the sense that it can't not exist), but which there is no algorithm to compute in general.

The problem with many of the answers on that page is that they rely on being able to output a specific n-ary representation of a real, whereas in fact to do so is uncomputable.

A computable answer might be as follows: if reals are defined as Cauchy sequences of rationals (as they often are in constructive-land, because that definition is extremely friendly), then there is a constructive algorithm to generate a decimal expansion: simply approximate to within $\frac{1}{100}$ to obtain the first digit, then to within $\frac{1}{1000}$ to obtain the second, and so on.

However, this constructive answer is inherently limiting in practice because it requires you to know up front what sequence of rationals converges to the real in question (because a real is such a sequence of rationals - you can't specify a real without specifying that sequence!). So you've already been forced to do all the hard work of approximation already; we've just offloaded the hard work into the definition of what it means to be a real number.


问:
$f^{-1}(G\setminus H)=f^{-1}(G)\setminus f^{-1}(H)$
Show that if $f:A\to B$ and $G,H$ are subsets of $B$, then
$$f^{-1}(G\setminus H)=f^{-1}(G)\setminus f^{-1}(H)$$

My attemp:

$f^{-1}(G \setminus H)=f^{-1}(G\cap H^{c})$

$=f^{-1}(G)\cap f^{-1}(H^{c})$

$=f^{-1}(G)\cap (f^{-1}(H))^{c}$

$=f^{-1}(G)\setminus f^{-1}(H)$

Can someone kindly confirm whether this is a correct proof? :c

答:
Yes. Basically, this is because the set difference can be written as intersection and complement, and assume that you are allowed to use the fact that

if you have a function $f:A\to B$, then $f^{-1}:P(B)\to P(A)$ commutes with union, intersections, and complements.


答:
You can see that for any sets $A$ and $B$, there is a bijection from $\mathcal{P}(A)\times \mathcal{P}(B)$ to $\mathcal{P}(A\sqcup B)$, where $A\sqcup B:=\{0\}\times A\cup \{1\}\times B$ is the disjoint union of $A$ and $B$: Just consider the map that sends $(X,Y)$ to $\{0\}\times X\cup \{1\}\times Y$.

(Note: Constructing the bijection does not require any knowledge about cardinals, but is directly related to showing equalities on cardinal arithmetic.)

Hence it suffices to construct a bijection between $\mathcal{P}(\mathbb{N})\sqcup \mathcal{P}(\mathbb{N})$ and $\mathcal{P}(\mathbb{N})$. Here is a way to constructing it: for each $(i,X)$, consider $f(i,X)=\{0\mid i=1\}\cup \{n+1\mid n\in X\}$.

Title: How can I prove (without cardinal numbers) that $P(P(\mathbb{N}))$ is equipollent to $P(P(\mathbb{N})) \times P(P(\mathbb{N}))$

Body: I am stuck trying to prove that $P(P(\mathbb{N}))$ is equipollent to $P(P(\mathbb{N})) \times P(P(\mathbb{N}))$. I was thinking that it is enough to prove $P(\mathbb{R}) \sim P(\mathbb{R}) \times P(\mathbb{R})$, or even $2^{\mathbb{R}} \sim 2^{\mathbb{R}} \times 2^{\mathbb{R}}$ , but I still don't know how to finish this. It would be even better if you can prove that in general for any infinite set $X$ it's true that $X \times X \sim X$.


问:
$\log(x)$ or $\ln(x)$ when solving definite integrals
I was solving $$\int^\infty_0 \frac{dx}{e^x+1}$$ and the answer was $\log(2)$. However, I used natural logarithm when solving this integral and so I got $\ln(2)$. Would both answers be accepted?

答: In mathematics we use denote the natural logarithm by $\log(x)$, You can sometime in  old books denote $\ln(x)$ like a natural logarithm, but actually in the new science papers we use $\log(x)$ to denote the natural logarithm.


答:
Here it is a slightly different way to approach it:
\begin{align*}
x\in f^{-1}(G\backslash H) & \Longleftrightarrow f(x) \in G\backslash H\\\\
& \Longleftrightarrow (f(x)\in G)\wedge(f(x)\not\in H)\\\\
& \Longleftrightarrow (x\in f^{-1}(G))\wedge(x\not\in f^{-1}(H))\\\\
& \Longleftrightarrow x\in f^{-1}(G)\backslash f^{-1}(H)
\end{align*}
and we are done.

Hopefully this helps!

Body:
Show that if $f:A\to B$ and $G,H$ are subsets of $B$, then
$$f^{-1}(G\setminus H)=f^{-1}(G)\setminus f^{-1}(H)$$

My attemp:

$f^{-1}(G \setminus H)=f^{-1}(G\cap H^{c})$

$=f^{-1}(G)\cap f^{-1}(H^{c})$

$=f^{-1}(G)\cap (f^{-1}(H))^{c}$

$=f^{-1}(G)\setminus f^{-1}(H)$

Can someone kindly confirm whether this is a correct proof? :c


答:
You don’t need the axiom of choice for this specific case of $|X|=|X\times X|$.

Let $f:\Bbb N\to 2\Bbb N:n\mapsto 2n$ and $g:\Bbb N\to 2\Bbb N+1:n\mapsto 2n+1$. Define

$$\begin{align*}
\varphi&:\wp(\wp(\Bbb N))\times\wp(\wp(\Bbb N))\to\wp(\wp(\Bbb N))\\
&:\langle\mathscr{A},\mathscr{B}\rangle\mapsto\{f[A]\cup g[B]:\langle A,B\rangle\in\mathscr{A}\times\mathscr{B}\}\,.
\end{align*}$$

The map $\varphi$ actually recoverably encodes each member of $\mathscr{A}\times\mathscr{B}$ as a subset of $\Bbb N$.

Suppose that $\mathscr{C}\in\operatorname{ran}\varphi$. For each $C\in\mathscr{C}$ let

$$A_C=\left\{\frac{n}2:n\in C\cap 2\Bbb N\right\}$$

and

$$B_C=\left\{\frac{n-1}2:n\in C\cap(2\Bbb N+1)\right\}\,;$$

then $C=f[A_C]\cup g[B_C]$. Let $\mathscr{A}=\{A_C:C\in\mathscr{C}\}$ and $\mathscr{B}=\{B_C:C\in\mathscr{C}\}$; then $\mathscr{C}=\varphi(\langle\mathscr{A},\mathscr{B}\rangle)$. Thus, $\varphi$ is an injection. Since there is an obvious injection in the other direction, the result follows from the Cantor-Schröder-Bernstein theorem.

In the original version of this answer I thought that I had avoided having to use the Cantor-Schröder-Bernstein theorem and incorrectly said that $\varphi$ was a bijection, but of course it is not: the range of $\varphi$ is the family of $\mathscr{C}\in\wp(\wp(\Bbb N))$ such that if $\mathscr{A}$ and $\mathscr{B}$ are defined as in the previous paragraph, then $\mathscr{C}=\{f[A]\cup g[B]:\langle A,B\rangle\in\mathscr{A}\times\mathscr{B}\}$, so that $\mathscr{C}$ encodes all of $\mathscr{A}\times\mathscr{B}$, not just an arbitrary set $\mathscr{D}\subseteq\mathscr{A}\times\mathscr{B}$ big enough so that $\pi_0[\mathscr{D}]=\mathscr{A}$ and $\pi_1[\mathscr{D}]=\mathscr{B}$.


问:
Let $f(x)=g(x)$ for all $x$ in $[-\pi,\pi]$ except when $x=0$. Show that $f(x)$ and $g(x)$ have the same Fourier Coefficients?
Question:

Let $f(x)=g(x)$ for all $x$ in $[-\pi,\pi]$ except when $x=0$. Show that $f(x)$ and $g(x)$ have the same Fourier Coefficients?

My Attempt:

So I've tried to calculate the fourier coefficients but i can only show that $a_0^f=a_0^g$, not $a_n^f=a_n^g$ and $b_0^f=b_0^g$. I also tried letting $h(x)=f(x)-g(x)=0$ but I don't think that works since $h(0)=f(0)-g(0)\neq0$.

Thanks for the help.

答:
By definition, the Fourier coefficients of $f$ are given by
$$
\hat{f}(n)=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)e^{-inx}dx
$$

So changing one point of $f$ does not change the coefficients.

On the other hand, if $f$ is continuous on $[-\pi,\pi]$ and $\hat{f}(n)=0$ for all $n$, then $f\equiv 0$.


问:
Block matrices: Prove that $ \det \begin{bmatrix} A & B\\ B & A \end{bmatrix}\neq \det(A^2-B^2) $
Prove that $ \det\begin{bmatrix}
A & B\\
B & A
\end{bmatrix}\neq \det(A^2-B^2)$ using the following matrix product $ \begin{bmatrix} I & B\\
0 & A - B\end{bmatrix} \begin{bmatrix} A+B & 0\\ 0 & I \end{bmatrix} $

I don't even know where to start here. Can someone please drop some hints?

答:
First of all, the expression $A^2 - B^2$ is only interpretable if $A,B$ are both square. So, I presume that $A$ and $B$ are square matrices of the same size.

Hint: We will have $\det \pmatrix{A&B\\B&A} = \det(A^2 - B^2)$ whenever $AB = BA$. A suitable counterexample exists with $A,B$ of size $2 \times 2$.

A further hint: we may select an example where $A$ and $B$ both have $3$ zero-entries.


问:
Reciprocal inequality
Given that $a, b, c, d$ are greater than zero, prove that $$\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} ≥ \frac{64}{a+b+c+d}$$ The first thing that came to mind was QM-AM-GM-HM, but I haven't been able to create the second term with HM-AM since the geometric mean or the quadratic mean don't really make sense.
$$\frac{4}{a+b+\frac{c}{4}+\frac{d}{16}} = \frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}$$

I could probably multiply both sides by the denominator, but that seems too hard for a contest-like question. How can I solve this? Any help is appreciated.

答:
Using Titu's lemma, we have

$$\frac{1^2}{a}+\frac{1^2}{b}+\frac{2^2}{c}+\frac{4^2}{d}\ge\frac{(1+1+2+4)^2}{a+b+c+d}$$

Equality holds for
$$\frac{1}{a}=\frac{1}{b}=\frac{2}{c}=\frac{4}{d}=k$$
for some constant $k$.

Title: Reciprocal inequality


答:
Mathematical notations are mostly a matter of conventions, and understanding notations depends on context.

In mathematics literature, $\log(x)$ has been used for both $\log_{10}(x)$ or $\log_e(x)$.

In complex analysis, you will see that sometimes people use $\log$ as the complex logarithm with base $e$ and reserve $\ln$ as the real logarithm with base $e$.

If you are taking an exam, and you should follow the instructor's definitions. Your solution simply means $\log(2):=\log_e2$.

Title: $\log(x)$ or $\ln(x)$ when solving definite integrals

Body: I was solving $$\int^\infty_0 \frac{dx}{e^x+1}$$ and the answer was $\log(2)$. However, I used natural logarithm when solving this integral and so I got $\ln(2)$. Would both answers be accepted?


问:
Given that $x^TAx = 0$, can I call $A$ a null transformation?
I am wondering what to call the matrix $A$ given that $x^TAx=0$ for all $x$, which implies that $A=\textbf{0}$. Would I then simply call $A$ the zero matrix or maybe does it have a special name?

答:
No, it does not imply that $A = \mathbf 0$. Try any skew-symmetric matrix!! Indeed, working over $\Bbb R$, we can prove that this hypothesis implies that $A$ must be skew-symmetric.

Since $x^\top Ax = 0$ for all $x$, we use the standard "polarization" trick, substituting $x+y$ for $x$:
\begin{align*}
0=(x+y)^\top A(x+y) &= x^\top Ax + x^\top Ay + y^\top Ax + y^\top Ay \\ &= x^\top Ay + y^\top Ax = x^\top (A+A^\top)y.
\end{align*}
But if $x^\top By = 0$ for all $x$ and $y$, we must have $B=\mathbf 0$ (take the standard basis vectors, if you want). So we conclude that $A^\top = -A$.

(The fundamental formula here at the last step is that, identifying a scalar and a $1\times 1$ matrix, we have
$$x^\top Ay = (x^\top Ay)^\top = y^\top A^\top x = y^\top A^\top x.)$$


问:
geodesics on sphere loring tu help
Loring Tu in his book Differential geometry [page 104] states (in my own words) the following on geodesics on spheres:

Consider the 2 sphere of radius $a$ in $\mathbb{R}^3$ .

Parameterise a great circle by arc length $\gamma$.

$\gamma''(t)$ is perpendicular to $\gamma'(t)$

$\gamma''(t)$ lies in the plane of the circle (does this mean $T_{\gamma(t)}M?)$

$\gamma''(t)$ is perpendicular to the tangent plane at $\gamma(t)$

I'm not sure why $(4).$ holds, why is $\gamma''(t)$ in the normal space $N_{\gamma(t)}M$?

Please do not explain using curvature.

答:
$\gamma(t)$ lies in a plane so $\gamma'(t)$ and $\gamma''(t)$ also lie in the same plane. (If you want to be precise: planes are closed subsets, and  $\gamma'(t)$ is a limit of a difference quotients with values in the plane.)

Differentiating $\langle\gamma(t),\gamma(t)\rangle=a$ and  $\langle\gamma'(t),\gamma'(t)\rangle=1$ yields $\langle\gamma(t),\gamma'(t)\rangle=\langle\gamma(t'),\gamma''(t)\rangle=0$. Hence for each $t$ ,$\gamma(t)$ and $\gamma'(t)$ are an orthogonal basis of the plane and $\gamma''(t)$ is a scalar multiple of $\gamma(t)$.

But the normal space at a point $p$ of a sphere centered at the orgin consists exactly of the scalar multiples of $p$, so $\gamma''(t)$ lies in the normal space.

Title: geodesics on sphere loring tu help


问:
Solve PDE $\rho_t + c\rho_x +\rho^2 = 0$ using method of characteristics
$$
\frac{\partial\rho}{\partial t} + c\frac{\partial\rho}{\partial x} + \rho^2 = 0
$$

I am not sure how to begin this problem, I have looked up how to use the method of characteristics but can find no example where $\rho^2$ so I am unsure of how one would approach this.

答:
Define the characteristic curve
$$
X'_a(t)=c,\;X_a(0)=a.
$$
You have $X_a(t)=a+ct$. Now study the evolution of $\rho(X_a(t),t)$. You have
$$
\frac{d}{dt}\rho(X_a(t),t)=\rho_t+\rho_xc=-\rho^2(X_a(t),t).
$$
This ODE has an explicit solution. From here I think that you can continue :-).</p>


问:
Prove that $E[F]$ is not a field.
This is a question from Lang's algebra:

Let $E, F$ be extensions of a field $k$, contained in some bigger field $L$. We
can form the ring $E[F]$ generated by the elements of $F$ over $E$.

It is clear that the elements of $E[F]$ can be written in the form $a_1 b_1 + \dots + a_n b_n$ with $a_i \in E$ and $b_i \in F$.

How are we claiming that it is not a field? I can prove all the properties of a ring. But I can't prove/disprove that it is a not a field.

What will be the quotient field of this ring?

答:
It certainly can be a field, for example if

$E\subseteq F$ or $F\subseteq E$ or

The elements of $F$ are algebraic over $E$.

An example when $E[F]$ is not a field: Take $E=\Bbb Q(x),F=\Bbb Q(y)$ inside $L=\Bbb Q(x,y)$. Then $E[F]$ consists of those rational functions whose denominator can be written in the form $p(x)q(y)$ with $p(x)\in \Bbb Q[x], q(y)\in \Bbb Q[y]$. Hence $x+y$ doesn't have an inverse in $E[F]$, so $E[F]$ is not a field.

The field of fractions of this ring will be the compositum $EF$ of $E$ and $F$ in $L$. It consists of the elements of the form $$\frac{a_1b_1+\dots+a_nb_n}{c_1d_1+\dots+c_md_m}$$
where $a_i,c_j\in E, b_i,d_j\in F$.

Edit: Assume that the elements of $F$ are algebraic over $E$, then all elements in $E[F]$ are also algebraic over $E$, so if $0\ne x\in E[F]$, there are some $a_1,\dots,a_n$ with $$x^n+a_1x^{n-1}+\dots+a_n=0$$
We may assume $a_n\ne0$. We get: $$x^{-1}=(-a_n)^{-1}(x^{n-1}+a_1x^{n-2}+\dots+a_{n-1})\in E[F]$$ Hence $E[F]$ is a field (alternatively this follows from the fact that if $\alpha$ is algebraic over $E$, then $E[\alpha]=E(\alpha)$ is already a field).

As the elements of $E[F]$ are of the form $a_1b_1+\dots+a_nb_n$ it is clear that the field of fractions $Q$ contains the elements of the form $$\frac{a_1b_1+\dots+a_nb_n}{c_1d_1+\dots+c_md_m}$$ As the set of elements of this form is already closed under taking inverses (in $L$) it follows that this is already a field, hence it is the field  of fractions of $E[F]$. As this is a field that contains both $E$ and $F$ it contains the compositum $EF$. On the other hand, since $EF$ contains $E,F$ and is a field it also has to contain those elements of the form $\frac{a_1b_1+\dots+a_nb_n}{c_1d_1+\dots+c_md_m}$, therefore $Q$ is a subfield of $EF$. It follows $EF=Q$.

Title: Prove that $E[F]$ is not a field.


问:
Identify this graph problem.
[Edit 01/02/2021] : Now the problem has been identified, as a relaxation of "minimal spanning tree" (it is more a minimal spanning sub-graph). Are they some references I could use on this problem ?

Let consider a graph made of $N$ nodes, numeroted from $B_1$ to $B_N$.

Each nodes can or cannot be connected to the other nodes, but overall we consider a family of weighed edges : $(\nu_{ij})_{1 \leq i < j \leq N} \in \mathbb{R}^{\frac{N(N-1)}{2}}$.

The edges must be understood as followed :

$\nu_{ij}=0$ if $B_i$ and $B_j$ are not linked by an edge.

there is no edge connecting one node to himself.

I want to operate with a specific way on this graph (I want to select some specific edges), and I believe I am looking for a family of numbers $(x_{ij})_{1 \leq i < j \leq N}$ with $x_{ij}=0$ or $1$ that verifies the following discrete optimization problem :

$$\underset{(x_{ij})_{1 \leq i < j \leq N} \in (0,1)^{\frac{N(N-1)}{2}}}{\min} \sum_{i<j}\  \nu_{ij} \  x_{ij}$$
with the following constraints on the $(x_{ij})$:

$\sum_{i<j} x_{ij} = N-1$ (i.e we are only gonna keep $N-1$ edges in the graph)

$\forall i \in \{1,\dots,N\}$, we have $$\sum_{j=i+1}^N x_{ij} + \sum_{j=1}^{i-1} x_{ji}  \geq 1$$ (i.e every nodes have at least one "non-zero" edge)

My questions

Is this problem known in operation research or discrete optimization or is it closed to a known problema with a theoritical analysis ? I don't know a lot of them, except for the traveling salesman problem or the cover set problem but I am quite unfamiliar with them.

Separately from this first question, I believe (heuristically, at least) that any family $(x_{ij})_{1 \leq i < j  \leq N}$ admissible for this problem verifies the following property :

the final graph with value $(x_{ij} \nu_{ij})$ on the edge is made of cycles and of only one path

But i am not sure how to prove it. I'have been looking for counter-examples using $4/5$ nodes and cycle but couldn't find any...
Any help, references or remarks are welcomed on the problem :).

答: This is a relaxation of the minimum spanning tree (MST) problem.  The connectivity requirement of MST has been replaced with a "no isolated nodes" requirement in your problem.</p>

Title: Identify this graph problem.


问:
Properties of Simple Functions
Suppose $f$ is a non-negative measurable function. i.e $f \in L^+$.

Is it true that f is the decreasing limit of a sequence of simple functions?

I'm suspecting that it's false, since it's we don't know if $f$ is bounded.

But I couldn't find any counter examples.

is the set of real valued simple functions closed under taking the max and min of two functions?

I'm also suspecting that it's false, because the limit of simple functions may not be simple. But I'm not sure about this one.

Thanks in advance!

答:
Real-valued simple functions are bounded, because they are the measurable functions with finite range. Therefore no unbounded positive function can be smaller than a simple function. If you allow simple functions taking values in $[0,+\infty]$, then yes you can: with the convention of convenience $\frac10=+\infty$, approximate from below $\frac1f$ with a sequence of simple functions $0\le\varphi_n\nearrow \frac1f$, then consider the sequence $\phi_n=\frac1{\varphi_n}$.

Yes, because the maximum and minimum of two measurable functions with finite range are measurable functions with finite range (in fact, their ranges are subsets of the union of the ranges).

Title: Properties of Simple Functions


问:
Is there a common symbol for concatenating two (finite) sequences?
Say we have two finite sequences $X = (x_0,...,x_n)$ and $Y = (y_0,...,y_n)$.
Is there a more or less common notation for the concatenation of these sequences, like $\sum (X,Y) = (x_0,...,x_n,y_0,...,y_n)$?</p>

答:
Starting with your defined sequences $X = (x_0, \ldots, x_n)$ and $Y=(y_0,\ldots,y_n)$, you can use the commonly accepted tuple/ordered pair notation:
\begin{align}
   (X,Y) &= \left(
    \left(x_0,\ldots,x_n\right),
    \left(y_0,\ldots,y_n\right)
   \right)
   \\
   &= \left(x_0,\ldots,x_n,y_0,\ldots,y_n\right)
  \end{align}


答:
It is not true what you write. In fact, your relation holds if, and only if, the symmetric part of $A$ vanishes, i.e., if $A$ is skew-symmetric.

To see this, just compute the $i$-th derivatives. For every $x$ you have
$$
0=\partial_i(Ax\cdot x)= A e_i\cdot x+ Ax\cdot e_i=(A+A^T)x\cdot e_i \, .
$$
Just set $x=e_j$ to get that, under your assumption
$$
(A+A^T)e_i\cdot e_j=0
$$
for every $i,j$.
So to answer your question, the right name to give to those matrices is skew-symmetric.

Title: Given that $x^TAx = 0$, can I call $A$ a null transformation?

Body: I am wondering what to call the matrix $A$ given that $x^TAx=0$ for all $x$, which implies that $A=\textbf{0}$. Would I then simply call $A$ the zero matrix or maybe does it have a special name?


问:
Change of Variables and Splitting fields
A comment in Size of Galois Group based on roots and minimal polynomial says:

"By a change of variables a polynomial $f(X)$ in $\mathbb{Q}[X]$ can be made monic in $\mathbb{Z}[X]$ without changing its splitting field over $\mathbb{Q}$."

I guess the standard way to convert a polynomial $f(X)\in \mathbb{Q}[X]$ into a monic polynomial over $\mathbb{Z}$ is: First make it monic by multiplying it by a rational number and then replace it by $D^{ \, \text{deg}(f)} f(\frac{X}{D})$, where $D$ is a common denominator of the coefficients.

I don't see why the splitting field of the thus obtained polynomial is the same as that of the original polynomial? For example, the splitting of $5X^2 - 2$ is $E=\mathbb{Q}(\sqrt{\frac{2}{5}})$, the corresponding monic polynomial over $\mathbb{Z}$ is $X^2 - 10$, whose splitting field is $F=\mathbb{Q}(\sqrt{10})$. Are $E$ and $F$ isomorphic, and why so?

答:
To state everything clearly:

We have $E = \Bbb Q[X]/(5X^2 - 2)$ and $F = \Bbb Q[Y]/(Y^2 - 10)$.

We define a homomorphism $f$ from $\Bbb Q[X]$ to $F$ by sending $X$ to the class of $\frac 1 5 Y$.

It is clear that the image of $5X^2 - 2$ under $f$ is zero, i.e. the ideal $(5X^2 - 2)$ lies in the kernel of $f$.

Therefore $f$ induces a quotient homomorphism $\phi: E \rightarrow F$, sending the class of $X$ to the class of $\frac 1 5 Y$.

In the same way, we may define a homomorphism $\psi: F \rightarrow E$, sending the class of $Y$ to the class of $5X$.

We then verify that $\phi \circ \psi$ is the identity, and $\psi \circ \phi$ is the identity.

This shows that $E$ and $F$ are isomorphic to each other.


问:
Nonsingular M-matrices are nonsingular

Question: What is a good reference for a proof of Proposition 1 below?
  It is definitely a known result, appearing e.g. in
  Plemmons's M-matrix characterizations,
  but I have not managed to follow the chain of references
  to an actual proof of it (e.g., if it appears in Ostrowski's papers
  cited by Plemmons, then not in this exact form). Has anyone seen it
  appear in the literature explicitly and with a self-contained proof?
  I give such a proof further below in an answer to this very question,
  but I'd prefer to have a published reference I can cite as well.

Definition. Let $\ell$ be a nonnegative integer.

(a) In the following, $\mathbb{R}^\ell$ denotes the
$\mathbb{R}$-vector space of column vectors of size $\ell$.

(b) If $w \in \mathbb{R}^\ell$ is a column vector,
then the notation $w_i$ shall be used for
the $i$-th entry of $w$ (for each $i \in \left\{1,2,\ldots,\ell\right\}$).

(c) For two column vectors $u \in \mathbb{R}^\ell$
and $v \in \mathbb{R}^\ell$, we write $u > v$ if and
only if each $i \in \left\{1,2,\ldots,\ell\right\}$ satisfies
$u_i > v_i$.

(d) If $Q \in \mathbb{R}^{n\times m}$ is any matrix,
then $Q_{i,j}$ shall denote the $\left(i,j\right)$-th entry
of $Q$ for each $i$ and $j$.

(e) A nonsingular $M$-matrix means a matrix
$Q \in \mathbb{R}^{\ell \times \ell}$ satisfying the following
two conditions:

The off-diagonal entries of $Q$ are nonpositive.
In other words, $Q_{i,j} \leq 0$ for $i \neq j$.

There exists some $x \in \mathbb{R}^\ell$ such that
both $x>0$ and $Qx > 0$.

Proposition 1. Let $Q \in \mathbb{R}^{\ell \times \ell}$ be a
  nonsingular $M$-matrix. Then,
  $Q$ is nonsingular (i.e., we have $\ker Q = 0$).

Proposition 1 is, of course, part of the reason for the
suggestive name "nonsingular $M$-matrix". Nevertheless, it is not
really obvious. I give a proof in an answer to this question, but my real
question is: What is a good reference for Proposition 1 in published
literature that doesn't send its reader on a wild goose chase?
The notion of a nonsingular $M$-matrix is famous for its many equivalent
definitions (R.J. Plemmons, M-matrix characterizations. I
-- nonsingular M-matrices gives dozens of them), which is at the same time a blessing and a curse,
the latter because it means that a result like Proposition 1 can be
spread across several parts of the literature without ever being stated
explicitly in one of them. My impression so far is that this is what
has happened.

答:
We can as well show the stronger
$$
Qy \ge0 \quad\Rightarrow\quad y\ge0,
$$
which easily implies injectivitiy, since $Qy=0$ then implies $y\ge0$ and $-y\ge0$, i.e. $y=0$.

To see the claim, let $Qy\ge0$ and
$$
i \in \operatorname{argmin}\{\frac{y_j}{x_j}: 1\le j\le i\}
$$
Suppose that $y_i<0$; then, $\mu:= -\frac{y_i}{x_i}> 0$ so that $Q(y + \mu x) > 0$ (since $Qy \ge 0$ and $Qx > 0$). This leads to
$$
0 < Q (y + \mu x) = Q_{ii}\underbrace{(y_i+\mu x_i)}_{=0} + \sum_{j\ne i}Q_{ij}\underbrace{(y_j+\mu x_j)}_{\ge0}\le0 ,
$$
a contradiction.
So $y_i\ge0$ and then by the definition of $i$ also $y_j\ge0$ for all $j$.

Title: Nonsingular M-matrices are nonsingular


问:
Computing cross product using norm and angle
Sorry for the weird title, if someone finds a better title for my problem be my guest to edit it ;)

For $\mathbf{v,w} $ in R³ with $\mathbf{||v||=1 ;||w||=4; \theta
 =\frac{2\pi}{3}}$

Solve the following:

$\lVert (3v+w)\times(v-2w)\rVert$

$\langle w\times 5v, v-3w\rangle$

I've already calculated

$$\langle v,w\rangle = \frac{-1}{2}||v|| \cdot||w|| =-2$$

Also, I've transformed 1. into

$$\lVert-7(v\times w)\rVert$$

But I'm a bit lost here.

Next I masked $w\times 5v $ as $x$ so as 2. we now have

$$\langle x,v-3w\rangle$$

This all seems to bring me not a bit closer to the solution, can someone suggest a valid approach?

答:
Hints: First,
$$
\begin{align}
(3v+w)\times(v-2w)
&=-7v\times w
\end{align}
$$
and
$$
\|-7v\times w\|=7\|v\|\|w\|\left|\sin\left(\frac{2\pi}{3}\right)\right|
$$
For $\langle w\times 5v, v-3w\rangle$, note that $w\times v$ is perpendicular to both $v$ and $w$.


问:
$\displaystyle \lim_{x \to 0} \frac{e^{\frac{1}{x}}-1}{x^{\alpha}}$ values of $\alpha $ for which limit exists in $\mathbb{R}$
Examine if for any value $\alpha > 0$ limit in $0$ exists and is $\in \mathbb{R}$.

$$\displaystyle \lim_{x \to 0} \frac{e^{\frac{1}{x}}-1}{x^{\alpha}}$$

Now, I know that it doesn't from plotter, but how do I check that by from the expression above? When I checked for which parameters series is convergent I used tests / criteria. But here I have nothing like that. I know that the expression is a variety of a common limit:

$$\lim_{x \to 0} \frac{a^{x}-1}{x} = \ln a$$

But then again, the only value of $\alpha$ for which that would work is $-1$, and we know that $\alpha > 0$. So, my question is, how to prove formaly that there is no such $\alpha$?

答: For any $\alpha >0$ the denominator $x^{\alpha}$ of the above fraction goes to 0, whereas the numerator $e^{\frac{1}{x}}-1$ goes to infinity for $x$ going to 0; $x$ positive. So there cannot be any limit.


问:
Let $\mathbb{Z}$ be defined with the usual metric. Show that every subset of $\mathbb{Z}$ is open in $\mathbb{Z}$.

Let $\mathbb{Z}$ be defined with the usual metric. Show that every subset of $\mathbb{Z}$ is open in $\mathbb{Z}$.

I can write $\mathbb{Z}$ as $\bigcup_{n=0}^{\infty} \{n\}$ and then letting $B(n, \frac{1}{2})$ I get that $B(n, \frac12) \cap \mathbb{Z} = \{n\} $. So it seems that I can write any element in $\mathbb{Z}$ as the intersection of two open sets and thus it would be open? How can I formalize this, I seem to have a bit hard time doing that...?

答: By the usual metric on $\Bbb Z$ you seem to mean  a metric (induced by the Euclidean distance function?)  which would induce the discrete topology.  Once you get points to be open, it's discrete.  But then every single subset of $\Bbb Z$ would be open.  So you are done.


问:
Bernoulli sums as singular measure
Let $X_n$ be Ber(1/2).

To show $\sum_{n=0}^\infty\frac{X_n}{\gamma^n}$ is singular to the Lebesgue Measure for all $\gamma > 2$.

Of course for $\gamma = 2$ this decomposes into the ordinary Lebesgue Measure over $[0,1]$. This seems like an interesting problem and would seem to give better understanding of measure theoretic probability.

EDIT I think for $\gamma = 3$ or $\gamma=4$ one can use an argument from Cantor Ternary set or quaternary set. But I dont think that's going to solve the problem

答:
Let $\gamma >2$. For all $n \geq 0$,

$$0 \leq \sum_{k=n}^{+\infty} \frac{X_n}{\gamma^n} \leq \sum_{k=n}^{+\infty} \gamma^{-n} = \frac{\gamma^{-n}}{1-\gamma^{-1}}.$$

Hence, $\mu$ is supported by the union $A_n$ of $2^n$ intervals:

$$A_n := \bigcup_{(x_0, \ldots, x_{n-1}) \in \{0, 1\}^n} \left[\frac{x_k}{\gamma^k}, \frac{x_k}{\gamma^k}+\frac{\gamma^{-n}}{1-\gamma^{-1}}\right].$$

Hence, $\mu(A_n) = 1$ for all $n$. In addition, $\text{Leb} (A_n) \leq \frac{2^n \gamma^{-n}}{1-\gamma^{-1}}$.

As a consequence, $\mu \left(\bigcap_{n \geq 0} A_n\right) = 1$ and  $\text{Leb} \left(\bigcap_{n \geq 0} A_n\right) = 0$, so that $\mu$ and $\text{Leb}$ are mutually singular.

Title: Bernoulli sums as singular measure


问:
Limit of function involving natural logarithm
I was wondering why the limit of the following function is:

$$ \lim_{x\rightarrow 0^{+}}{\dfrac{\ln x}{x}} = -\infty $$

When we try to solve the limit by substituting in 0, one would expect the numerator to go to minus infinity and the denominator to 0. Thus, having an indeterminate form as a quotient, we could apply L'Hôpitals Rule. But then we would get:

$$ \lim_{x\rightarrow 0^{+}}{\dfrac{1}{x}} = +\infty $$
What's wrong with what I've written above? Thanks beforehand.

答:
The problem is that what you get is not an indeterminate form. In order to apply L'Hopital, you need "$\frac{0}{0}$" or "$\frac{\infty}{\infty}$", or other forms that are in the assumptions of the L'Hopital rule.  Your limit is none of these forms.

To see why   $ \lim_{x\rightarrow 0^{+}}{\dfrac{\ln x}{x}} = -\infty $, follow the definition.

Given any $M>0$, let $\delta=\min(\frac1e,\frac1M)$. Then
$
0<x<\delta$ implies that

$$
\frac{1}{x}>M,\quad \ln(x)<-1
$$
and thus
$$
\frac{\ln(x)}{x}<-M
$$


答:
When you write $B(n, \frac 1 2) \cap \Bbb Z$, you seem to have in mind the idea of a "ball" in $\Bbb R$.

So you should forget the fact that $\Bbb Z$ is a subspace of $\Bbb R$. It is just a set, together with a metric function $d: \Bbb Z \times \Bbb Z \rightarrow \Bbb R_{\geq 0}$, sending $(x, y)$ to $|x - y|$ (viewed as a non-negative real number).

By definition, a subset $X$ of $\Bbb Z$ is open if for every element $x \in X$, there exists a positive real number $r$ such that the set $B(x, r) = \{y \in \Bbb Z: d(x, y) < r\}$ is contained in $X$.

You can then show that every subset $X$ is open by choosing $r$ above to be $\frac 1 2$.

Body:
Let $\mathbb{Z}$ be defined with the usual metric. Show that every subset of $\mathbb{Z}$ is open in $\mathbb{Z}$.

I can write $\mathbb{Z}$ as $\bigcup_{n=0}^{\infty} \{n\}$ and then letting $B(n, \frac{1}{2})$ I get that $B(n, \frac12) \cap \mathbb{Z} = \{n\} $. So it seems that I can write any element in $\mathbb{Z}$ as the intersection of two open sets and thus it would be open? How can I formalize this, I seem to have a bit hard time doing that...?


问:
A specific type of orthogonal bases
In $\mathbb{R}^n$, can we find an orthogonal basis $v_1,v_2,\ldots,v_n$ with $\|v_i\|=1$ such that
$$\|\bar{v}_1\|=\|\bar{v}_2\|= \cdots =\|\bar{v}_n\|$$
where $\bar{v}_i\in\mathbb{R}^m$ is the vector formed by the first $m$ coordinates of $v_i$ with respect to the standard basis and $m$ is a given integer $m<n$.</p>

答:
[Edit:] I just realized that this only works for $mk\le n$, so it's also only a partial answer.

Note first that the orthonormal column vectors $v_i$ form an orthogonal matrix whose rows are also orthonormal. Since we can expand any set of $m$ orthonormal row vectors to a set of $n$ orthonormal row vectors, we only have to worry about filling in the first $m$ rows with orthonormal row vectors such that the truncated column vectors satisfy the condition.

If $m\mid n$, this is easy; just fill the $m$ rows with $n/m$ copies of the $m\times m$ identity matrix. If $m\nmid n$, write $n=qm+k$ and put $q$ copies of the identity matrix into the $m$ rows (say, beginning at the left). That produces orthogonal rows, but they don't fulfill the condition, since the norm of the first $qm$ truncated column vectors is $1$ and that of the last $k$ is $0$. To fix this, rotate the row vectors in the plane formed by the two row vectors consisting of $1$s in the first $qm$ columns and in the last $k$ columns, respectively. That leaves the row vectors orthogonal (since they're all rotated by the same isometry), but shifts the excess weight from the first $qm$ columns to the last $k$ columns. By choosing the rotation angle suitably, we can make the weights in all columns coincide. The result looks something like this:

$$
\begin{array}{l}
\overbrace{\hphantom{\begin{array}{ccccccccccc}
1+\alpha & 1+\alpha & 1+\alpha & \cdots & 1+\alpha & 1+\alpha & 1+\alpha & \cdots & \beta & \cdots & \beta
\end{array}}}^{\displaystyle n}\\
\overbrace{\hphantom{\begin{array}{ccccccccccc}
1+\alpha & 1+\alpha & 1+\alpha & \cdots & 1+\alpha & 1+\alpha & 1+\alpha&\cdots
\end{array}}}^{\displaystyle qm}
\hphantom{\cdots}
\overbrace{\hphantom{\begin{array}{ccccccccccc}
\beta & \cdots & \beta
\end{array}}}^{\displaystyle k}\\
\overbrace{\hphantom{\begin{array}{ccccccccccc}
1+\alpha & 1+\alpha & 1+\alpha & \cdots
\end{array}}}^{\displaystyle m}
\hphantom{\cdots}
\overbrace{\hphantom{\begin{array}{ccccccccccc}
1+\alpha & 1+\alpha & 1+\alpha&\cdots
\end{array}}}^{\displaystyle m}\\
\left.
\begin{array}{ccccccccccc}
1+\alpha & \alpha & \alpha & \cdots & 1+\alpha & \alpha & \alpha & \cdots & \beta & \cdots & \beta\\
\alpha & 1+\alpha & \alpha & \cdots & \alpha & 1+\alpha & \alpha & \cdots & \beta & \cdots & \beta\\
\alpha & \alpha & 1+\alpha & \cdots & \alpha & \alpha & 1+\alpha & \cdots & \beta & \cdots & \beta\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots&&\vdots
\end{array}
\;\;\;\right\}m\\
\vphantom{q}
\end{array}
$$

with $\alpha=(\sqrt{1-mk/n}-1)/m$ and $\beta=\sqrt{q/n}$.

Title: A specific type of orthogonal bases


问:
Expecation for tensor products
We are told that $$Z|0 \rangle = | 0\rangle  \\ Z | 1\rangle =-|1\rangle \\ X|0\rangle =|1\rangle \\ X|1\rangle =|0\rangle$$

and we have the state $$|\psi\rangle =|0\rangle |1\rangle +|1\rangle |0\rangle$$

The question is: Without doing an explicit calculation explain why expectation of Z*X and X*Z in the given state are the same?

My answer: I kind of done the calculation in my head and said that "For the right hand side of both inner products Z will only change the coefficient of a ket while X will switch the ket, resulting in tensor products of similar kets i.e either $|0\rangle|0\rangle$ and $|1\rangle |1\rangle$. This won't match up to the left hand side $\langle\psi|$ which means both expectations will be zero."

I don't think this is the non-calculation answer they were looking for. X and Z do not commute, my only other guess is because the state is entangled. Are there any properties linear operators can have that make their expectations the same either way round?

答:
I assume $Z*X$ means the operator that acts as $Z$ on the first tensor factor and as $X$ on the second, and analogously for $X*Z$.  If so, then $Z*X=T\cdot(X*Z)\cdot T$, where $T$ is the linear operator that interchanges the two tensor factors.  Using this and the fact that your particular state $|\psi\rangle$ satisfies $T|\psi\rangle=|\psi\rangle$ and $\langle\psi|T=\langle\psi|$, you get
$$
\langle\psi|Z*X|\psi\rangle=\langle\psi|T(X*Z)T|\psi\rangle=\langle\psi|X*Z|\psi\rangle.
$$
I'm not sure whether this counts as an explicit computation and is therefore prohibited.  If so, I recommend meditating on the equations (or on symmetry) instead of writing them.</p>

Title: Expecation for tensor products


问:
Calculate perceived angle of a ray at a known distance from origin of actual ray without distance metrics
So lets say I have a camera at origin, and the camera perceives the direction vector of angle(θ) of an object in its field of view. I have another device(lidar) at a known distance of "r" behind the camera. I need to calculate the direction/angle(Φ) perceived by the new device(lidar) without knowing how far away the object that was initially detected is actually sitting at

Diagram

"h" = Arbitrary distance of object from center line (Not Known)

"x" = Vertical distance of object from center line (Not Known)

"r" = Offset distance of new device origin point (Known)

"θ" = Direction angle of object perceived at the original point (Known)

"Φ" = Direction angle to be calculated

Since we can use traditional trigonometry to find the relation between the variables mentioned above, I was able to come to this relationship

```<code>tan(Φ) = tan(θ) * (x / (x+r))
</code>```

However here, while both θ and "r" are known values, "x" can vary from 0 to infinity. So is there a way to come up with a solution. The only hint I was given was "Taylor's series" but am not sure how to use that to approximate a good value for Φ. Any help would be much appreciated

答:
You cannot.

The camera data tells you that the object is somewhere on the green ray you have drawn at angle $\theta$ from the LIDAR/camera ray.  You have drawn the red ray from the LIDAR, but that ray is misleading : the information from the camera only restricts the object to be somewhere on the green ray, so there are a range of angles that $\phi$ may be hence there is no definite direction for the red ray from the LIDAR.

By itself, that means the angle the LIDAR will report is anywhere from $0$ (the object is touching the camera) to $\theta$ (the object is infinitely far away).  With neither the focal distance between the object and the camera nor the range from the LIDAR to the camera, the angle $\phi$ is not specified by the data given.


答:
Remember that using Taylor expansions you have

\begin{align}
   \ln x &= \int_1^x \frac{1}{t} \, dt = \int_0^{x - 1} \frac{1}{1 + u} \, du \\
   &= \int_0^{x - 1} (1 - u + u^2 - u^3 + \cdots) \, du \\
   &= (x - 1) - \frac{(x - 1)^2}{2} + \frac{(x - 1)^3}{3} - \frac{(x - 1)^4}{4} + \cdots \\
   &= \sum_{k=1}^\infty \frac{(-1)^{k-1} (x-1)^k}{k}=(x-1)+\dotsb.
 \end{align}

Hence $\ln x \approx x-1$ (first order) and

$$\frac{\ln x}{x}\approx \frac{x-1}{x}=1-\frac{1}{x}\overset{x\to 0^+}{\to} -\infty$$

Title: Limit of function involving natural logarithm

Body:
I was wondering why the limit of the following function is:

$$ \lim_{x\rightarrow 0^{+}}{\dfrac{\ln x}{x}} = -\infty $$

When we try to solve the limit by substituting in 0, one would expect the numerator to go to minus infinity and the denominator to 0. Thus, having an indeterminate form as a quotient, we could apply L'Hôpitals Rule. But then we would get:

$$ \lim_{x\rightarrow 0^{+}}{\dfrac{1}{x}} = +\infty $$
What's wrong with what I've written above? Thanks beforehand.