问:
Gauge transformation (about mathematical definition of wikipedia, other sources)
I've had some exposure to vector bundles, but as good as none to principal bundles. I'm trying to get some orientation in the subjects concerning vector bundles and gauge theory.

On Wiki, there's a definition of a gauge transformation from a 'mathematical standpoint', namely that a gauge transformation of a vector bundle $\pi:E\rightarrow M$ is a diffeomorphism $\phi:E\rightarrow E$ commuting with $\pi$ and being a linear vector space isomorphism on each fiber.

[https://en.wikipedia.org/wiki/Gauge_theory_(mathematics)#Gauge_transformations ]

What is meant by $\varphi$ and $\pi$ to commute?

Using this definition and given some vector field, in order to show that it is gauge invariant, would I have to apply $\phi$ and show that the object remains the same?

Is there a good source for someone interested in the above, to learn more about gauge transformations?

Thanks in advance!

答:
The fact that the maps commute means that, given a point in a the fibre, $p\in E_x$ for $x\in M$, we have $\pi\circ\phi(p)=\phi\circ\pi(x)$. In other words, the gauge transformation $\phi$ preserves the fibre, it takes points in a fibre to points in the same fibre.

If you wish to read about mathematical gauge theory, there are two books that I've recently gone through that I would highly recommend: Principal Bundles by Sontz, and Mathematical Gauge Theory by Hamilton. Both can be accessed through Springer Link, so perhaps your university has given you free access.


问:
Give a combinatorial argument to evaluate sum
$$\sum_{k}^{n}\binom{n }{k}\binom{m}{k}$$

I'm not really sure how a combinatorial argument would help me evalaute this?
I was thinking maybe I'm adding up all the ways that you could divide a group of n+m people into 2 groups of size k?

I'm tempted to just plug in the fact that $$\binom{n}{k}$$ =n!/k!(n-k)! bu that's not really a combinatorial argument, right?

答:
Choose $k$ things from $n$ and choose $m-k$ things from $m$ ($k$ ranges over $0$ to $\min(n,m)$) ... This is the same as choosing $m$ things from $n+m$.

Thus
\begin{eqnarray*}
\sum_{k=0}^{\min(n,m)}\binom{n }{k}\binom{m}{k} = \binom{n+m}{m}.
\end{eqnarray*}

ContentLicense: CC BY-SA 4.0


问:
understanding a lemma in a proof that the axiom of choice implies zorn's lemma

I'm having some trouble understanding the proof of Lemma 2 on page 4 of these Axiom of Choice notes. To ensure I understand the material, I was wondering if the following justifications are correct:

$A\backslash A'$ and $A'\backslash A$ are nonempty (as otherwise the minimum would be undefined).

$A$ and $A'$ are well-ordered, so since $A\backslash A'$ is a nonempty subset of $A$, it has a minimum and similarly for $A'\backslash A.$

$C\subseteq A'$ because if not then $\exists x \in C\backslash A'\subseteq A\backslash A',$ but then $x \prec z = \min A\backslash A'$. This is a contradiction, so no such $x$ exists.

$z = \min A\backslash C$ because $A\backslash C$ is the set of all elements that are $\succ $

All elements of $C$ precede $z$, so since $A$ is a chain, every element in $A\backslash C$ is at least $z$ (or succeeds it).

Since $z = \min A\backslash C$ and every element of $A\backslash C$ is an upper bound for the elements of $C, z = f(Upp(C)) = g(C).$

However, I have a few questions:

Why are $A\backslash A'$ and $A'\backslash A$ both nonempty? Why can't $A' = A$?

Why does $A'$ being a chain imply that $x\prec z'$ for all $x\in C$?

If $x\prec z'\,\forall x \in C,$ why does this imply that $C$ is a proper initial segment of $A'$? I get that $C\subseteq A',$ but why does this show that $\forall x \in A', \forall y \in C', x \prec y \Rightarrow x \in C'$?

If $C$ is a proper initial segment of $A',$ why does this show that $g(C)\in A'$?

If $z' \not \prec z,$ doesn't this just mean $z' \succ z$ as $A$ is a chain?

答:
We are assuming that $A\nsubseteq A'$; this is exactly equivalent to saying that $A\setminus A'\ne\varnothing$. Similarly, $A'\nsubseteq A$ means that $A'\setminus A\ne\varnothing$. These set differences aren’t non-empty because otherwise their minima wouldn’t exist: their minima exist because they are non-empty.

Your next two bullet points are correct.

Your fourth bullet point is incomplete. $z\in A\setminus A'\subseteq A$, and each $x\in C$ satisfies $x\prec z$ and hence $x\ne z$, so $z\in A\setminus C$. For any $x\in A\setminus C$ we have $x\not\prec z$, and $\prec$ well-orders $A$, so $z\preceq x$, and therefore $z=\min(A\setminus C)$. This also covers your fifth bullet point.

I don’t understand your sixth bullet point: $A$ is an attempt, and $C$ is a proper initial segment of $A$, so by definition $g(C)=\min(A\setminus C)=z$.

Now for the questions. We are assuming (in order to get a contradiction) that $A\nsubseteq A'$ and $A'\nsubseteq A$; each of these alone already implies that $A\ne A'$.

In second question we are considering the case in which $C\ne A'$. Let $x\in C$. Then $x\prec z$, so if $z'\preceq x$, we have $z'\preceq x\prec z$ and hence $z'\prec z$, contradicting our assumption that $z'\not\preceq z$. Thus, we must have $z'\not\preceq x$ for every $x\in C$. Now recall that $z=\min(A\setminus A')$, and that $x\prec z$ for every $x\in C$; this means that $x\in A'$ for every $x\in C$, i.e., that $C\subseteq A'$. And $z'\in A'$, so for each $x\in C$ we must have at least one of $x\preceq z'$ or $z'\preceq x$. We’ve already ruled out the latter, so we must have $x\preceq z'$ for each $x\in C$. Finally, $C\subseteq A$, and $z'\notin A$, so $z'\notin C$, and therefore $x\prec z'$ for each $x\in C$.

Clearly $C$ is a proper subset of $A'$, since $z'\in A'\setminus C$. Suppose that $x\in C$, $y\in A'$, and $y\prec x$. Then $y\prec x\prec z'$, so $y\prec z'=\min(A'\setminus A)$, and therefore $y\in A$. Since $x\in C$, we also have $y\prec x\prec z$, so $y\prec z$. Thus, $y\in A$ and $y\prec z$, so by definition $y\in C$. It follows that $C$ is a proper initial segment of $A'$.

$A'$ is an attempt, and $C$ is a proper initial segment of $A'$, so by definition $g(C)=\min(A'\setminus C)\in A'$.

Finally, the fact that $A$ is a chain says nothing about $z'$, since $z'\notin A$. In particular, $z$ and $z'$ could be incomparable: it might be that $z'\not\preceq z$ and $z\not\preceq z'$.


问:
Converse for: If $A$ is a Normal matrix, then the field of values is the convex hull of its eigenvalues
If $A$ is  a Normal matrix, then the field of values of $A$ is the convex hull of the eigenvalues, that is $\mathcal{F}(A)=C_0(\sigma(A))$. What about the converse?

Book says the converse is not true, that is the following matrix is not normal but $\mathcal{F}(A)=C_0(\sigma(A)).$ To prove this consider $$A=\left(
\begin{array}{cccccc}
 1 & 0 & 0 & 0 & 0 & 0 \\
 0 & i & 0 & 0 & 0 & 0 \\
 0 & 0 & -1 & 0 & 0 & 0 \\
 0 & 0 & 0 & -i & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & 1 \\
 0 & 0 & 0 & 0 & 0 & 0 \\
\end{array}
\right)$$
is clear that $A$ is not normal because $AA^*\neq A^*A$ and also $\sigma(A)=\{1,i,-1,-i,0,0\}$ therefore the convex hull of the eigenvalues of $A$ is  $$C_0(\sigma(A))=\Big\{\sum_{i=1}^6\alpha_i\lambda_i:\lambda_i\in \sigma(A),\quad\sum_{i=1}^6\alpha_i=1\Big\}=\big\{\alpha_1-\alpha_2+i\alpha_3-i\alpha_4:\sum_{i=1}^6\alpha_i=1\Big\}.$$
Also if $x=(x_1,x_2,x_3,x_4,x_5,x_6)^T\in \mathbb{C}^6$ then $x^*=(\overline{x_1},\overline{x_2},\overline{x_3},\overline{x_4},\overline{x_5},\overline{x_6})$ where the overline indicates complex conjugation. The definition of the field of values is $$\mathcal{F}(A)=\Big\{x^*Ax:x\in \mathbb{C}^6,\quad x^*x=1\Big\}$$ but, we can verify that $$x^*Ax=(\overline{x_1},\overline{x_2},\overline{x_3},\overline{x_4},\overline{x_5},\overline{x_6})\left(
\begin{array}{cccccc}
 1 & 0 & 0 & 0 & 0 & 0 \\
 0 & i & 0 & 0 & 0 & 0 \\
 0 & 0 & -1 & 0 & 0 & 0 \\
 0 & 0 & 0 & -i & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & 1 \\
 0 & 0 & 0 & 0 & 0 & 0 \\
\end{array}
\right)\left(
\begin{array}{c}
 x_1  \\
 x_2  \\
 x_3  \\
 x_4   \\
 x_5   \\
 x_6   \\
\end{array}
\right)=x_1 \overline{x_1}+i x_2\overline{x_2}-x_3\overline{x_3}-i x_4\overline{x_4}+x_6 \overline{x_5}$$ that is $$x^*Ax=|x_1|^2+i|x_2|^2-|x_3|^2-i|x_4|^2+x_6 \overline{x_5}$$ so far so good except for that last term, is it illegal to say: let $x_6=0$?
Any suggestions?

答:
To prove this you arrive at $x^*Ax=(|x_1|^2-|x_3|^2-x_6\bar{x_5})+i(|x_2|^2-|x_4|^2)$

In the complex plane $(X,Y)$ the boundaries of $C_0(\sigma(A)$ are lines  $Y+X=1$, $Y+X=-1$, $Y-X=1$, $Y-X=-1$.

If $X=|x_1|^2-|x_3|^2-x_6\bar{x_5}$, $Y=|x_2|^2-|x_4|^2$ with $\sum_{i=1}^6|x_i|^2=1$ and $|x_6\bar{x_5}|\le |x_6||x_5|\le |x_6|^2+|x_5|^2$,
the combinations

$|X\pm Y|\le 1$

i let you justify this so $\mathcal{F(A)}\subset C_0(\sigma(A))$  the converse is straight if you take $x_6$ or $x_5$ equal zero.


问:
Intersecting two planes in Cartesian space
Say we have two equations,
$$
7x=9-9z
$$ and
$$
7y=17+10z
$$
if we want to find the vector form for these in a $3D~$ Cartesian plane. How would we do that, because in the answer sheet of my book, the answer would be:

$$
 x=\begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix}+t\begin{pmatrix} 9 \\ -10 \\-7 \end{pmatrix}
$$
Which I can't figure out why it is after an hour of trying to get to it.

I got this from a question to find the plane of intersection between planes

3x + 2y + z = -1 & 2x - y + 4z = 5. I don't even know if I am doing anything wrong.

答:
Rearrange both equations in terms of $z$, so: $$z=\frac19(9-7x)=\frac{1}{10}(7y-17)$$ and then pick $x$ and $y$ in the form $\alpha+t\beta$ so that $z$ cancels nicely. As long as you use the $z$ from a given $x$ to find $y$, any $x$ you pick is fine, but make sure that $|\alpha|<|\beta|$ at the end by subtracting multiples to simplify.

Since $z$ in terms of $x$ cancels if $x$ is a multiple of $9$, we pick $x=9t\implies z=1-7t$ and then $$1-7t=\frac{1}{10}(7y-17)\implies y=\frac{27}{7}-10t)$$

We then compile and see that: $$\begin{pmatrix} x \\ y \\ z \end{pmatrix}=\begin{pmatrix} 0 \\ \frac{27}{7} \\ 1 \end{pmatrix}+t\begin{pmatrix} 9 \\ -10 \\-7 \end{pmatrix}$$

which is the correct answer.


问:
Yes or No, Is the Concept of a Tangent (Tangent Line) Generally Clearly Understood?
Per an educational source, to quote:

Tangent, in geometry, straight line (or smooth curve) that touches a given curve at one point; at that point the slope of the curve is equal to that of the tangent. A tangent line may be considered the limiting position of a secant line as the two points at which it crosses the curve approach one another.

Note, the definition implies the ostensible equivalent of the tangent to a tangent line. The latter apparently is explicitly defined as a limit of a secant line, which is comprised of two points.

This inherently makes sense as two points determine a line.

So, the concept of a 'tangent' appears to be short for a tangent line, which by a limiting definition, is explicitly defined.

So, my answer is generally no, as geometry is usually taught first.

My argument is that the concept of a 'tangent' (implying a tangent line) at a point (which does not determine a line) is not clear for those not acquainted calculus-based concept of a limit. In the current context, the definition explicitly cites to quote again "limiting position of a secant line as the two points at which it crosses the curve approach one another".

I also cite a related prior question "how to draw a tangent line to a curve fit" which, by the very question, supports my answer. Note: the provided answer makes use of the explicit knowledge of the equation for the curve in question [EDIT] which is not always the case. [EDIT][EDIT] To expound on this comment, per Wikipedia on tangent, to quote:

The geometrical idea of the tangent line as the limit of secant lines serves as the motivation for analytical methods that are used to find tangent lines explicitly. The question of finding the tangent line to a graph, or the tangent line problem, was one of the central questions leading to the development of calculus in the 17th century. In the second book of his Geometry, René Descartes[8] said of the problem of constructing the tangent to a curve, "And I dare say that this is not only the most useful and most general problem in geometry that I know, but even that I have ever desired to know".[9]

答:
COMMENT

►From viewpoint of tangent as a right line touching a given curve at one point, let me give two examples of curves.
$$(1)\hspace2cmy= |x-3|+2\\(2)\hspace2cmy^2=(x-3)^3$$
The curve $(1)$ has in the point $(3,2)$ infinitely many tangents and so is for the curve $(2)$ in the point $(3,0)$.

►From viewpoint of tangent as limiting position of secants we have for the curve $(1)$ only two (distinct) tangents while for the curve $(2)$ there are just one tangent. (nevertheless the curve is singular and is not an elliptic curve).

►Is for many reasons that the definition of tangent in a point of a curve is given for differentiable curves in the considered point.


问:
$x+1/x$ an integer implies $x^n+1/x^n$ an integer
Suppose that $0\neq x\in\mathbb{R}$ and $x + \frac1x\in\mathbb{Z}$. Prove that, for all  $n\ge1$, $x^n + \frac1{x^n}\in\mathbb{Z}$.

I can't figure out and understand the question. Can you give me some hints ?

答:
The following is an extension to the hint provided by Martin Brandenburg:
expanding $(1+1/x)^n$, which is an integer by hypothesis:
$$nC0 * x^m * 1/x^0 + nC1 * x^(m-1) * 1/x + ... mC(m-1) * x * 1/x^(m-1) + mCm * x^0 * 1/x^m$$

Simplifying and rearranging, we get:
$$x^m + x^(-m) + mx^(m-2) + mx^(2-m)... $$

Clearly, this is a linear combination of terms of the for $x^a + 1/x^a$
Now assume the original hypothesis is true for all k < n. Then the terms $$mx^(m-2) + mx^(2-m)...$$ are integers, leaving $$x^m + 1/x^m$$. However, since $(1+1/x)^n$ is an integer by hypothesis, so is $$x^m + 1/x^m$$.

ContentLicense: CC BY-SA 3.0


问:
Binomial coefficients and $p$-adic limits
I was reading Schoof's Catalan's Conjecture, and on page 37, while proving Cassels' Theorem it is mentioned that by Exercise 5.7, the denominators of the binomial coefficients $\binom{\frac{p}{q}}{k}$ for $0 \leq k \leq \lfloor \frac{p}{q} \rfloor + 1$, where $p$ and $q$ are odd prime numbers, are powers of $q$.

I found a proof by Keith Conrad, but it uses $p$-adic limits, so I decided to read through Gouvea's $p$-adic Numbers: An Introduction.

I still don't know what is going on, any insight please?

答:
This was really a bit too long to put into a comment to KCd's answer, but I thought it might help to see a specific concrete example of such a limit.

We already know that $\binom{m}{n}$ is an integer when $m$ is an integer, so we could write down a specific limit of integers that approaches $\frac{a}{b}$ in the p-adic topology when $b$ and $p$ are relatively prime, because

$$p^{\varphi(b)k}(b-a) + a \equiv 0 \mod b$$
$$1*(b-a) + a \equiv 0 \mod b$$
$$b \equiv 0 \mod b$$

This means $\frac{p^{\varphi(b)k}(b-a) + a}{b}$ is an integer. Then we know that because $p^{\varphi(b)k} \to 0$ as $k \to \infty$ this means we have our sequence of integers approaching $\frac{a}{b}$,

$$\frac{p^{\varphi(b)k}(b-a) + a}{b} \to \frac{a}{b}$$.

This means our limit of integers is

$$\binom{\frac{a}{b}}{n} = \lim_{k \to \infty} \binom{\frac{p^{\varphi(b)k}(b-a) + a}{b}}{n}$$

And we have that

$$\left|\binom{\frac{a}{b}}{n}\right|_p = \lim_{k \to \infty} \left|\binom{\frac{p^{\varphi(b)k}(b-a) + a}{b}}{n}\right|_p \le 1$$


问:
Integer values of $\sqrt{n}+\sqrt{n+2005}$ where $n \in \mathbb{Z}$
Find integer values of $\sqrt{n}+\sqrt{n+2005}$ where $n$ is an integer.

So far, I have just listed some squares which are larger than $2005.$ The first few are $2025,2116, 2209,2304,$ etc. I can just plug in values of $n$ which match up the square minus $2005.$ However, this seems way too bashy. Is there any good way to do this problem? Thanks in advance.

答:
$$\begin{align}
\sqrt n+\sqrt{n+2005}=k
&\implies \sqrt{n+2005}=k-\sqrt n\\
&\implies n+2005=k^2-2k\sqrt n+n\\
&\implies \sqrt n=\left(k^2-2005\over2k\right)={1\over2}\left(k-{2005\over k}\right)\\
&\implies k\mid2005\\
&\implies k\in\{1,5,401,2005\}
\end{align}$$

But $k\ge\sqrt{n+2005}\gt\sqrt{25}$ (since $n$ must be nonnegative in order for $\sqrt n$ to be real), so we are left with $401$ and $2005$ as the only possible integer values for $\sqrt n+\sqrt{n+2005}$ with $n\in\mathbb{Z}$.

Remark: This derivation essentially contains a proof of the fact that the sum of two square roots (of integers) is an integer if and only if the two square roots are already integers.


问:
Trace norm of a complex $|B| \times|A|$ matrix $M$.
Let $A$ and $B$ be two finite-dimensional Hilbert spaces, and let $M: A \rightarrow B$ be a $|B| \times|A|$ complex matrix. Show that the trace norm of $M$ can be expressed as
$$
\|M\|_{1}=\max _{V: B \rightarrow A} \operatorname{Tr}[V M]
$$
where the maximum is over all partial isometries $V: B \rightarrow A$.

$\mathbf{NOTE:}$ $V$ is a partial isometry if $V^*V=I^B$ or $VV^*=I^A$.

$\mathbf{ATTEMPT:}$ I have two ideas, one of them is that we have to show that $\max _{V: B \rightarrow A} \operatorname{Tr}[V M] \geqslant \|M\|_{1}$ and $\max _{V: B \rightarrow A} \operatorname{Tr}[V M] \leqslant \|M\|_{1}$. So then we can conclude $\|M\|_{1}=\max _{V: B \rightarrow A} \operatorname{Tr}[V M]$.

"$\leqslant$":The singular values of $M$ are $\mu_1 \geqslant \mu_2 \geqslant \cdots \geqslant \mu_n$ and the singular values of $V$ are $\nu_1 \geqslant \nu_2 \geqslant \cdots \geqslant \nu_n$.  Now, using von Neumann's trace inequality: $$\operatorname{Tr}[V M] \leqslant \sum_{i=1}^{n} \mu_i \nu_i \leqslant \max _{i \in \{1,\cdots,n\}} \{\nu_i\}\sum_{i=1}^{n} \mu_i \leqslant \nu_{max} \|M\|_{1} \leqslant \|M\|_{1}$$

But for the other direction I have no idea and I'm not sure that the above proof is true. I should mention that I'm also struggling to understand the von Neumann's trace inequality.

Could you please help me to complete the proof or give me an alternative proof?
Thanks

答:
I think this answer is true, but not sure.

We can consider our partial isometry to be $V=\sum_{x=1}^{k} \left|u_x\right\rangle^A\left\langle v_x\right|^B $ where k is the minimum dimension between $|A|$ and $|B|$, $\{\left|u_x\right\rangle\}$  is an orthonormal set in $A$ and $\{\left|v_x\right\rangle\}$ is an orthonormal set in $B$. and we can write $M=\sum_{y=1}^{n} \left|s_y\right\rangle^B\left\langle r_y\right|^A$, where $\{\left|r_y\right\rangle\}$ and $\{\left|s_y\right\rangle\}$ are another orthonormal set in $A$ and $B$ respectively. So then $\operatorname{Tr}[VM]=\cdots=\sum_{x,y}^{} \mu_y \left\langle v_x \middle| s_y \right\rangle \left\langle r_y \middle| u_x \right\rangle$. So then, since $\left\langle v_x \middle| s_y \right\rangle , \left\langle r_y \middle| u_x \right\rangle \leqslant 1$ , then the maximum of the $\operatorname{Tr}[VM]$ occurs, if $\left\langle v_x \middle| s_y \right\rangle = \left\langle r_y \middle| u_x \right\rangle = \delta_{xy}$ and it depends on choosing the partial isometry. It means that $\max _{V: B \rightarrow A} \operatorname{Tr}[V M]=\max _{V: B \rightarrow A} \sum_{x,y}^{} \mu_y \left\langle v_x \middle| s_y \right\rangle \left\langle r_y \middle| u_x \right\rangle=\sum_{x,y}^{} \mu_y=\|M\|_1$.


问:
What is 1.999999 in 4 significant digit arithmetic?
Say you are doing some calculations in 4 significant digit arithmetic and at some point you get something like 1.9999999. You need to use this number in further calculations, also in 4 significant digit arithmetic. Do you use 2.000 or 1.999?</p>

答:
$$2.000$$

When calculating to $n$ significant figures, first write down the first $n+1$ digits of the number, and then round the first $n$ digits based on the value of the $n+1$th.

So I would write down $1.9999$ and then note that the $9$ at the end implies I round up, giving $2.000$ to $4$sf.


问:
Does "Let" imply existence?
Many years ago I kept trying to explain a proof to professor. Each time I started the proof saying "Let x be..." the professor would interrupt with "what do you mean by let?"

I came to believe what they meant is when you use "Let" you are asserting such an object exists. In this case the existence of the object had not been established so we were not free to choose it. More or less I stopped using "let".

But I notice in proofs people often say "Let" when no such object exists. So I was wondering what is the proper use of "Let"?

答: Technically it would probably be better to say Suppose that $x$ is (rather than Let $x$ be) when we don’t know for a fact that such an $x$ exists or when we intend to prove by contradiction that it does not: we can always suppose something hypothetically. In practice, however, it is common — I would even say quite normal — to use let in both contexts, and I have little if any sympathy for your professor, who seems to me to have been riding an idiosyncratic hobbyhorse.

Title: Does "Let" imply existence?


问:
Find the exact value of the trigonometric expression when $\sin(u) = − 5 /13$ and $\cos(v) = − 4/ 5$.
Find the exact value of the trigonometric expression when   $\sin(u) = − 5 /13$ and $\cos(v) = − 4/ 5$.  (Both $u$ and $v$ are anglese in III° quadrant.)

$$\cos(u + v) $$

答:
The sum formula for cosine is: $$\cos(u+v)=\cos(u)\cos(v)−\sin(u)\sin(v)$$.

In order to solve, you can visualize the triangles which define the sine and cosine.

After this, the problem is rather simple as you only need to plug in the values of cosine and sine that you get.

$$\cos(u+v) = \left(\frac{-12}{13} \cdot \frac{-4}{5}\right) - \left(\frac{-5}{13} \cdot \frac{-3}{5}\right)
= \frac{48}{65} - \frac{15}{65} = \frac{33}{65}$$

Thus, $\cos(u+v) = \frac{33}{65}$.


问:
Solving Equation Using Algebraic Method
How to solve these equations using an algebraic method?
I need to show my working, don't you do something in reverse, like 7 multiplies by something. I haven't done it in class.

$$\dfrac{5(3y-4)}{2y}=7$$

答:
We are given:

$\dfrac{5(3y-4)}{2y}=7$

We must next eliminate the $2y$ from the denominator. We can do this by multiplying $2y$ by the $7$ on the other side of the equation. What we have now is a straightforward solve for y question.

$\implies$ $5(3y-4)=7(2y)$

$\implies$ $15y-20=14y$

We must now isolate y.

$\implies$ $15y-20=14y\implies15y-14y=20\implies y=20$

We can now check our solution to see if we are correct by plugging the value of y (which is 20) back into the initial equation.

$\dfrac{5(3(20)-4)}{2(20)}=7$

$\dfrac{5(60-4)}{40}=7$

$\dfrac{280}{40}=7$

We find that 7 is indeed equal to 7 which proves our answer is correct.
$7=7$

Title: Solving Equation Using Algebraic Method


答: I don't think that "let" necessarily implies existence, it's just a symbolic assignment. For example, I think it's valid to do something like "let $x$ be something that doesn't exist" - you could show $x$ doesn't exist by showing it contradicts some property that is known to be true. Ultimately we should be careful about what words we use in what context, but I don't think using "let" in this way is abusive at all.

Title: Does "Let" imply existence?

Body:
Many years ago I kept trying to explain a proof to professor. Each time I started the proof saying "Let x be..." the professor would interrupt with "what do you mean by let?"

I came to believe what they meant is when you use "Let" you are asserting such an object exists. In this case the existence of the object had not been established so we were not free to choose it. More or less I stopped using "let".

But I notice in proofs people often say "Let" when no such object exists. So I was wondering what is the proper use of "Let"?


问:
Prove: If $x$ is a base point for a topological space $X$, then $U(x)\in B$ for every base $B$ of $X$
Prove: If point $x$ for a topological space $X$ has a minimal neighbourhood $U(x)$, then $U(x)\in B$ for every base $B$ of $X$.

答: Let $\mathcal{B}$ be any base for $X$. As $U(x)$ is open and contains $x$, for some $B\in \mathcal{B}$ be have $x \in B \subseteq U(x)$, by the definition of a base for $X$. The inclusion cannot be proper as that would contradict the minimality of $U(x)$. So $B = U(x)$ and so $U(x) \in \mathcal{B}$.


问:
Find the first 3 decimal places of a definite integral

$I(x)=\int_0^x \frac{\sin (t)}{t} \, dt$

Find $I(1)$ correct to 3 decimal places

I thought that we just put $1$ instead of $x$ and we get $\frac{\sin 1}{1} = 0.841 $ but the answer is different.
Any ideas what is the correct way to solve it?

答:
$$I(x)=\int_0^x \frac{\sin t}{t} \, dt$$

$$\sin t=t-\frac{t^3}{3!}+\frac{t^5}{5!}-\frac{t^7}{7!}+O\left(t^8\right)$$

$$\frac{\sin t}{t}=1-\frac{t^2}{3!}+\frac{t^4}{5!}-\frac{t^6}{7!}+O\left(t^7\right)$$

$$\int\frac{\sin t}{t}\,dt=t-\frac{t^3}{3\cdot 3!}+\frac{t^5}{5\cdot 5!}-\frac{t^7}{7\cdot 7!}+O(t^8)$$

Being an alternate series, the error is, in absolute value, less than the first term neglected, thus to have 3 correct decimal digits, the error must be less than $10^{-3}$.

Therefore we need to sum up to the $5$th power, because $\frac{1}{7\cdot 7!}\approx 0.00003$

So finally
$$\int_0^1\frac{\sin t}{t}\,dt=1-\frac{1}{18}+\frac{1}{600}=1703/1800\approx 0.946$$


问:
Formulas for calculating pythagorean triples
I'm looking for formulas or methods to find pythagorean triples. I only know one formula for calculating a pythagorean triple and that is euclid's which is:

$$\begin{align}
&a = m^2-n^2 \\
&b = 2mn\\
&c = m^2+n^2
\end{align}$$

With numerous parameters.

So are there other formulas/methods?

答: It doesn't seem cumbersome to me.  Just loop on $m$ starting at $2$.  Loop on $n$ starting at $1$ or $2$ depending on the parity of $m$ and going up to $m-1$.  Check the GCD using Euclid's algorithm.  If it is $1$ you have a primitive set, so calculate $a,b,c$.  If you want all sets up to some $N$, multiply $a,b,c$ by all values up to $\lfloor \frac Nc \rfloor$.  Stop the $m$ loop at $\sqrt N$</p>


问:
proof with differential equation

How to prove the last equation from the first one. If I integrate both side there is something left side in terms of arc sine and on left side there is t. then how to prove the equation that is given in terms of cos???

答:
I may have done something dumb, but the below gives almost the answer given with the exception that there is an extra square root in the solution.  I'll leave it to someone to find my error.

The left-hand side can be rewritten to be the derivative of the $\arccos()$ function.  A standard differential is:
$$d\left(\arccos\left(\frac{u}{n}\right)\right) = -\frac{du}{\sqrt{n^2 - u^2}}$$
If $E$, $m$, and $k$ are constants, then we can simply say:
$$n^2 = \frac{2E}{m}\quad\quad u = x\sqrt{\frac{k}{m}} \quad\quad du = \sqrt{\frac{k}{m}}\,dx\quad\quad dx = \frac{du}{\sqrt{\frac{k}{m}}}$$

That means:
$$
\int -\frac{\frac{du}{\sqrt{\frac{k}{m}}}}{\sqrt{n^2 - u^2}} = \pm\int dt \\
\int -\frac{du}{\sqrt{n^2 - u^2}} = \pm \sqrt{\frac{k}{m}} \int dt \\
\arccos\left(\frac{u}{n}\right) = \sqrt{\frac{k}{m}} t + K \\
\cos\left(\arccos\left(\frac{u}{n}\right)\right) = \cos\left(\sqrt{\frac{k}{m}} t + K\right) \\
\frac{u}{n} = \cos\left(\sqrt{\frac{k}{m}} t + K\right)
 $$
A few notes: (1) in the first step I added a negative sign.  Since the right was $\pm$ anyway, I could multiply both sides by $-1$.  Then, I got rid of the $\pm$ once I was multiplying by a square root, since those are $\pm$ by definition anyway.

Now we can undo our substitutions and get:
$$ x \frac{\sqrt{\frac{k}{m}}}{\sqrt{\frac{2E}{m}}} = \cos\left( \sqrt{\frac{k}{m}} t + K\right) \\
x \sqrt{\frac{\frac{k}{m}}{\frac{2E}{m}}} = \cos\left(\sqrt{\frac{k}{m}} t + K\right) \\
x \sqrt{\frac{k}{m}\frac{m}{2E}} = \cos\left(\sqrt{\frac{k}{m}} t + K\right) \\
x \sqrt{\frac{k}{2E}} = \cos\left(\sqrt{\frac{k}{m}} t + K\right) \\
x = \sqrt{\frac{2E}{k}} \cos\left(\sqrt{\frac{k}{m}} t + K\right)$$

This is the solution in the original question, but with an extra square root.  Not sure if I went wrong somewhere or the book or the questioner.

Title: proof with differential equation


问:
Complex analysis: Compute $\int_{-\infty}^\infty \frac{\cos(x)}{1+x^4} dx$.
This is a problem that I was trying to solve in preparation for an entrance exam. The first part of the problem was to solve $\int_{-\infty}^\infty \frac{1}{1+x^4} dx$ which is a fairly straight-forward application of complex analysis using a half-circle toy contour. I am not sure however, how I should proceed with this integral. One approach I thought to use was an "expanding rectangle" persae but that didn't seem to work. Any help is appreciated.

答:
As Feng Shao said, for this integral you could first consider: $$ \int_{-\infty}^{\infty}\frac{e^{ix}}{x^{4}+1}dx = \int_{-\infty}^{\infty}\frac{\cos\left(x\right)+i\sin\left(x\right)}{x^{4}+1}dx $$

And after you'd take as a result just the real part. Converting that integral into the complex world we get $$ \int_{c}^{ }\frac{e^{iz}}{z^{4}+1}dz $$

Now, let's find the poles of this expression, letting $z^4$ + 1 = 0, we get $z^4$ = -1, or ($z^2$ + i)($z^2$ - i) = 0, then let's find all roots. $$ z^{4} + 1 = (z + \sqrt{i})(z - \sqrt{i})(z + i\sqrt{i}))(z - i\sqrt{i}))$$

So, we get to know that the poles are $ ±\sqrt{i}, and ±i\sqrt{i}$. We're going to be integrating on the upper half of the complex plane, such that the poles that matter for us are the positive versions, $ \sqrt{i}\ and, i\sqrt{i}\ $. $\hspace{0cm}$

As the expression in the integral goes to 0 when the limits of the integration of the circle goes to infinity, we can conclude that. $$ \int_{-\infty}^{\infty}\frac{e^{ix}}{x^{4}+1}dx =  Resi_{z=\sqrt{i}}\frac{e^{iz}\left(z-\sqrt{i}\right)}{z^{4}+1}+Resi_{z=i\sqrt{i}}\frac{e^{iz}\left(z-i\sqrt{i}\right)}{z^{4}+1}  $$

Now let's work on the residues, simplifying both expressions, we get $$ \int_{-\infty}^{\infty}\frac{e^{ix}}{x^{4}+1}dx = Resi_{z=\sqrt{i}}\frac{e^{iz}}{\left(z^{2}+i\right)\left(z+\sqrt{i}\right)}+Resi_{z=i\sqrt{i}}\frac{e^{iz}}{\left(z^{2}-i\right)\left(z+i\sqrt{i}\right)} $$

Pluging in the values of the residues we get $$ \int_{-\infty}^{\infty}\frac{e^{ix}}{x^{4}+1}dx = Residue\frac{e^{i\sqrt{i}}}{\left(2i\right)\left(2\sqrt{i}\right)}-\frac{e^{-\sqrt{i}}}{\left(2i\right)\left(2i\sqrt{i}\right)} $$

To get the result we want, we need to multiply the residue by 2iπ. $$ \int_{-\infty}^{\infty}\frac{e^{ix}}{x^{4}+1}dx = \left(\frac{e^{i\sqrt{i}}}{\left(2i\right)\left(2\sqrt{i}\right)}-\frac{e^{-\sqrt{i}}}{\left(2i\right)\left(2i\sqrt{i}\right)}\right)2i\pi $$

$$ \int_{-\infty}^{\infty}\frac{e^{ix}}{x^{4}+1}dx = \left(\frac{ie^{i\sqrt{i}}-e^{-\sqrt{i}}}{\left(2i\sqrt{i}\right)}\right)\pi $$

Now, let's actually figure out what $\sqrt{i}$ is. We know $ i = e^{iπ/2}$, taking the square root on both sides, we get $$ \sqrt{i} = e^{iπ/4} $$ $$ \sqrt{i} = cos(π/4) + isin(π/4) $$ $$ \sqrt{i} = \frac{i+1}{\sqrt{2}} $$. Now that we've just figured out what $\sqrt{i}$ is, let's plug that in in our expression. $$ \int_{-\infty}^{\infty}\frac{e^{ix}}{x^{4}+1}dx = \left[\frac{ie^{\frac{i}{\sqrt{2}}-\frac{1}{\sqrt{2}}}-e^{-\frac{i}{\sqrt{2}}-\frac{1}{\sqrt{2}}}}{2\left(\frac{i-1}{\sqrt{2}}\right)}\right]\pi $$

$$ \int_{-\infty}^{\infty}\frac{e^{ix}}{x^{4}+1}dx = \left[\frac{ie^{\frac{i}{\sqrt{2}}}-e^{-\frac{i}{\sqrt{2}}}}{\sqrt{2}\left(i-1\right)e^{\frac{1}{\sqrt{2}}}}\right]\pi $$

Simplifying the complex e's. $$ \int_{-\infty}^{\infty}\frac{e^{ix}}{x^{4}+1}dx = \left[\frac{i\cos\left(\frac{1}{\sqrt{2}}\right)-\sin\left(\frac{1}{\sqrt{2}}\right)-\cos\left(\frac{1}{\sqrt{2}}\right)+i\sin\left(\frac{1}{\sqrt{2}}\right)}{\sqrt{2}\left(i-1\right)e^{\frac{1}{\sqrt{2}}}}\right]\pi $$

If you take a look at the expression in the numerator, we can factor out (i - 1), so we get $$ \int_{-\infty}^{\infty}\frac{e^{ix}}{x^{4}+1}dx = \left[\frac{\left(i-1\right)\left(\cos\left(\frac{1}{\sqrt{2}}\right)+\sin\left(\frac{1}{\sqrt{2}}\right)\right)}{\sqrt{2}\left(i-1\right)e^{\frac{1}{\sqrt{2}}}}\right]\pi $$

Now we can cancel out the (i-1) on the top and bottom, then we get. $$ \int_{-\infty}^{\infty}\frac{e^{ix}}{x^{4}+1}dx = \frac{\pi\left(\cos\left(\frac{1}{\sqrt{2}}\right)+\sin\left(\frac{1}{\sqrt{2}}\right)\right)}{\sqrt{2}e^{\frac{1}{\sqrt{2}}}} $$

Expanding the integral on the left, we get. $$ \int_{-\infty}^{\infty}\frac{\cos\left(x\right)+i\sin\left(x\right)}{x^{4}+1}dx = \frac{\pi\left(\cos\left(\frac{1}{\sqrt{2}}\right)+\sin\left(\frac{1}{\sqrt{2}}\right)\right)}{\sqrt{2}e^{\frac{1}{\sqrt{2}}}} $$

As $i\int_{-\infty}^{\infty}\frac{\sin\left(x\right)}{x^{4}+1}dx$ is the imaginary part, and we don't have any imaginary expression here, as a bonus we get that $$ \int_{-\infty}^{\infty}\frac{\sin\left(x\right)}{x^{4}+1}dx = 0 $$

And for the final answer, we get: $$ \int_{-\infty}^{\infty}\frac{\cos\left(x\right)}{x^{4}+1}dx = \frac{\pi\left[\cos\left(\frac{1}{\sqrt{2}}\right)+\sin\left(\frac{1}{\sqrt{2}}\right)\right]}{\sqrt{2}e^{\frac{1}{\sqrt{2}}}} ≈ 1.54427600962... $$

I hope this explanation was helpful to someone, thank you!


问:
very hard proof on binary tree on number of child's
I self study for preparing exam, and see a very difficult problem on question (2-b) on spring $2009$ at Here as attached follows.

I'm not need a very hard proof, just an intuitive step or idea about part (b). How we can deduce it?

答:
I would do parts (a) and (b) directly. We first note the following:

Claim 1: Let $y_i$ be any node in $T$ where size$(y_i) > 1$.
Then $y_i$ has a child $y_{i+1}$ where size$(y_{i+1})$ $\ge$ $\lfloor$size$(y_i)/2\rfloor$.

Indeed, as $T$ is binary, $y_i$ has at most 2 children, and at least 1 iff size$(y_i)>1$. If $y_i$ has exactly one child $y_{i+1}$ then size$(y_{i+1}) = $size$(y_i) -1 \ge \lfloor$size$(y_{i})/2\rfloor$ so Claim 1 is established for the case where $y_i$ has only one child. If $y_i$ has exactly 2 children $y^1_{i+1}$ and $y^2_{i+1}$
then size$(y^1_{i+1})$ $+$ size$(y^2_{i+1}) + 1$ $=$ size$(y_i)$. Assuming WLOG that size$(y^1_{i+1}) \ge $ size$(y^2_{i+1})$ it follows that
size$(y^1_{i+1}) \ge \lfloor$size$(y_{i})/2 \rfloor$ then, so set $y_{i+1}$ to be the vertex $y^1_{i+1}$. $\surd$

So we now show that there is a vertex $y$ such that $k \le$ size$(y) \le 2k$ for any positive integer $k<n$. Let $y_0$ be the root of the binary tree, and for each nonnegative integer $i$ set $y_{i+1}$ denote the child of $y_{i}$ of larger size. Then by Claim 1 $\lfloor$size$(y_i)/2\rfloor \le$ size$(y_{i+1})$ $<$ size$(y)$ because the tree is binary and $y-1$ has at most 2 children.

So let $l$ be the smallest integer such that size$(y_l) \le 2k$. If $l=0$ then $y_l=y_0$ the root of $T$ and size$(y_0)=n$ so $k \ge n/2$. If $k < n/2$ then $l$ is positive, and so size$(y_{l-1}) > 2k$ which implies by Claim 1 that size$(y_l) \ge k$ as well. Taking $y=\lfloor \frac{n}{2} \rfloor$ gives (b).

I think the exam writer wanted you to put $k=\lfloor n/3 \rfloor$ as in (a) and then take the edge between the node $v$ satisfying $k \le $ size$(v)  \le k$, and $v$'s parent.

We cannot always do any better $\lfloor \frac{n}{3} \rfloor$ for the smaller size w a separator that has only one edge. Let $T$ be a tree with 13 vertices: the vertices of $T$ are $v,w_1,w_2,w_3,w_4, y_1,y_2,y_3,y_4,u_1,u_2,u_3, u_4$, and the edges of $T$ are
$vw_1,vy_1, vu_1$, and then $w_1w_2, w_2w_3, w_3w_4, y_1y_2, y_2y_3, y_3y_4, u_1u_2, u_2u_3, u_3u_4$. So $T$ is a root $v$ where hanging off $v$ are 3 vertex-disjoint paths with 4 vertices each. Then every 1-edge separator of $T$ has a size with only at most $4 < 13/3$ vertices.


答:
"Let" uses an existential proposition implicitly. In formal logic, there is a rule for existence elimination. What it says is that if you have a proposition $\exists x P(x)$, where $P$ is a predicate, then you can construct a subproof with $P\left(c\right)$ as a premise through introducing a new symbol $c$. If you can derive a proposition $Q$ which is irrelevant to $c$ in your subproof, then you $Q$ can be placed outside your subproof and be a derived proposition in the main proof. So "let ..." can be seen as a combination of an existence proposition and the introduction of a subproof with a new symbol.

I strongly recommend reading the book "language, proof and logic" by Jon Barwise and John Ethemendy and getting familiar with the fitch proof style introduced in this book. Honestly, most of the difficulties in learning rigorous math like real analysis or linear algebra lie in the lack of understanding of formal logic. Universities are lacking fundamental math education for new undergraduates and expect them to understand complicated proofs without knowing thoroughly first order logic. It is very difficult for students to understand the so-called $\epsilon-\delta$ language and the implicit use of logic rules in English. The book provides you with a fundamental understanding of first order logic and its application in set theory. Even if it is unrealistic to write daily proofs in fitch style, understanding it helps you figure out the correspondence between daily language and formal language.

Title: Does "Let" imply existence?


答:
Whatever you prove after "Let $x$ be [blah]" is a theorem under the assumption that $x$ is a blah. Notice that it's not a theorem under the assumption that there exists a blah, but that $x$ is such a blah.

In particular it is a result with a free variable, but it doesn't say anything about whether such a blah exists. "Let $x$ be a real number with square $-1$. [blah] Therefore $0=1$" (I'll let you fill in the gaps of this proof !) is a perfectly valid proof, and it is not operating under the assumption that there exists a real number with square $-1$. It is operating under the assumption that $x$ is a real number with square $-1$.

In particular, if you already know that $0=1$ is false, then you can deduce the following result (with no free variable) : there is no real number with square $-1$.

Let me now state things more precisely. In most formal deduction systems, you have variables $x,y,z,x_0,...$. In a formula (not a proved formula, just a syntactically correct formula), some variables can be bound (for instance $x$ appearing within the scope of a quantifier such as $\forall x$), some can be free.

To deal with this precisely in the deduction system, one very convenient tool is the tool of contexts - this is better known in things such as type theory but it makes complete sense and is very relevant in classical logic as well. A context $\Gamma$ is a declaration of free variables, and assumptions on these variables (in type theory, you can somehow package these into just saying the type of the variable, but in classical logic this will come in the form of formulas). An example of a context $\Gamma$ is "variables : $x$, assumptions: $x$ is a real number, $x> 0$" - one variable, and two formulas.

Along a deduction, the context can change, which can make your deduction system very flexible and much closer to actual mathematical practice than other deduction systems without contexts. One example of such a change of context is the $\exists$-introduction rule.

This says, very vaguely (there are precise hypotheses that I will not spell out) that if in the context $\Gamma \cup \{x, \psi(x)\}$, where $\Gamma$ does not talk about $x$, you can deduce $\phi$, which does not talk about $x$ either, then in the context $\Gamma$, you can deduce $(\exists x, \psi(x))\implies \phi$. Informally, what this is saying is "if just by assuming that $x$ satisfies $\psi(x)$, and nothing else on $x$, I can deduce that $\phi$ is true, then just the existence of such an $x$ implies $\phi$" - it's obvious, when you say it like that, but in a formal deduction system you have to write it down.

In any case, what's the point of all of this ?

Well the sentence "Let $x$ be a [blah]" should be seen as a context declaration : you're declaring that you're working in the context $\Gamma = x$, such and such assumption on $x$. This does not, in any way, assume anything about existence. When you conclude your proof, say you prove $\phi$ which doesn't mention $x$, then you reach the state "$\Gamma \vdash \psi$" (read "$\psi$ can be proved in the context $\Gamma$")

This does not allow you to conclude that $\psi$ is correct, of course, because you're working within a context. Now, by the $\exists$-intro rule that I mentioned earlier, this does give you $\vdash (\exists x, blah)\implies \psi$ - in the empty context (or under some axioms, with no free variables, defining a global context which you don't specify at all times).

Theorems are results that are proved in the empty context. Now if you want to deduce $\psi$ from this theorem, then you have to make a statement about existence of an $x$ such that blah - this is how implications work : if I want to prove $B$ but only know $A\implies B$, I better say something about $A$ !

So in conclusion you are completely free to use "let" even when things don't exist, or when you don't know whether they exist - as long as you draw the right conclusion from what you've actually proved, and don't say something like "let $x$ be such that blah, therefore $\psi$. Therefore $\psi$" without knowing whether such an $x$ exists.

Note that it would be absurd to have this restriction on "let" : math is not decidable, and so that means that there would be sentences where it's not decidable whether or not you're allowed to write them down !!

Title: Does "Let" imply existence?


问:
Game with 6 dice, what is the chance to get a score below $10$
You are given 6 dice (with values 1 to 6) and your goal is to get a total number of pips that is less than 10.

You start by rolling all 6 dice. Say you get $(1,1,2,5,6,6)$. You may (in fact, you must) now choose one number, and put all dice with that number of pips to the side. Say you choose 1, for example. Then $(1,1)$ is put aside. You roll the remaining 4 dice again. Now you might get $(2,4,4,5)$. Again, you choose one number, say 2, and put aside all dice with that number of pips. You now have $(1,1,2)$ on the side and 3 dice left to roll. You repeat this procedure until you can throw no more dice. For instance, if you would now roll the 3 dice and get $(4,4,4)$, we can put all fours to the side and our total score will be the sum of $(1,1,2,4,4,4)$, which is $16$ in this case. The goal is to get a score $<10$.

The question is: what is the best strategy in order to have the highest change of getting below 10? And with this strategy, what is your chance to get a score $<10$?

Some quick simulations show that if your strategy is to always set aside the smallest number that appears on at least one of the rolled dice, then your chance of winning is roughly about 30%. I think there are some cases, though, in which this is not the best strategy. But I haven't seen a nice way to formulate it mathematically.

答:
You can compute your maximum chance of winning, and the corresponding optimal strategy, by using dynamic programming to work backwards from the final throw.  Let $\ G(d,t)\ $ denote the game in which you start with $\ d\ $ dice, and win whenever your ultimate total is less than $\ t\ $, and let $\ w(d,t)\ $ be the probability that you win when you play optimally in that game.  If you throw your $\ d\ $ dice in that game, the probability that you will get $\ n_1,$$n_2,$$\,\dots,$$n_6\ $ of each of the sides $\ 1,$$2,$$\,\dots,$$6\ $, respectively $\Big($with $\ \displaystyle\sum_{i=1}^6n_i=d\ \Big)$,  is
$$
p\big(n_1,n_2,\dots,n_6\big)=\frac{{d\choose n_1}{d-n_1\choose n_2}\dots{d-n_1-n_2\dots-n_4\choose n_5}}{6^d}\ .
$$
If you now choose to set aside the dice showing side $ i\ $ (where $\ n_i>0\ $), you will then be faced with trying to win the game $\ G(d-n_i,t-in_i)\ $, which you will do with probability $\ w(d-n_i,t-in_i)\ $ if you play optimally from then onwards.  If $\ n_i=d\ $ then you win immediately if $\ t-in_i>0\ $, or lose immediately if $\ t-in_i\le0\ $, so we set $\ w(0,t)=1\ $ for $\ t>0 $ and $\ w(0,t)=0\ $ for $\ t\le0 $.  Obviously, your optimal first move in $\ G(d,t)\ $ is to choose the value of $\ i\ $ with $\ n_i>0\ $ which maximises $\ w(d-n_i,t-in_i)\ $.  Let $\ c^*(\boldsymbol{n})\ $ be the optimal choice of $\ i\ $ for the throw $\ \boldsymbol{n}=\big(n_1,n_2,\dots,n_6\big)\ $. Then
$$
w(d,t)=\sum_\boldsymbol{n}p(\boldsymbol{n})w\big(d-n_{c^*(\boldsymbol{n})},c^*(\boldsymbol{n})n_{c^*(\boldsymbol{n})}\big)\ .
$$
Once $\ w\big(d',t'\big)\ $ has been tabulated for $\ d'<d\ $ and $\ t'<t\ $, this equation can be used to determine the value of $\ w(d,t)\ $.

The table below gives the values of $\ w(d,t)\ $ for $\ 1\le d\le4\ $ and $\ 2\le t\le11\ $, and approximate values (correct to $7$ decimal places) for $\ d=5,6\ $ and $\ 2\le t\le11\ $.  The entry for $\ d=6\ $ and $\ t=10\ $ indicates that your chance of winning the game $\ G(6,10)\ $ is only $17$-$18\%$, which is inconsistent  with the OP's statement that your chance of winning it is about $30\%$ if you always choose the smallest number that appears on at least one dice.  The table was computed with a script, but I have verified that the entries for $\ d=2\ $ and $\ 3\le t\le7\ $, and $\ d=3\ $ and $\ t=6,7\ $ are correct by hand, so I'm reasonably confident the numbers are all correct.  One possible explanation for the discrepancy is that the OP's quick simulations counted a win as occurring when the total was less than or equal to $10$, rather than strictly less than $10$.  The table entry for $\ d=6\ $ and $\ t=11\ $ indicates that, with best play, your chances of winning that game are $\ 27$-$28\%\ $.
$$
\begin{array}{c|c|c|c|c|c|c|}
&2&3&4&5&6&7&8&9&10&11\\
\hline
1&\frac{1}{6}&\frac{1}{3}&\frac{1}{2}&\frac{2}{3}&\frac{5}{6}&1&1&1&1&1\\
\hline
2&0&\frac{2}{27}&\frac{17}{108}&\frac{8}{27}&\frac{23}{54}&\frac{16}{27}&\frac{79}{108}&\frac{23}{27}&\frac{49}{54}&\frac{26}{27}\\
\hline
3&0&0&\frac{163}{3888}&\frac{769}{7776}&\frac{761}{3888}&\frac{2431}{7776}&\frac{875}{1944}&\frac{511}{864}&\frac{5593}{7776}&\frac{6355}{7776}\\
\hline
4&0&0&0&\frac{35387}{2^63^9}&\frac{177793}{2^73^9}&\frac{185101}{2^63^9}&\frac{25565}{2^43^8}&\frac{926741}{2^73^9}&\frac{626611}{2^63^9}&\frac{197237}{2^43^9}\\
\hline
5&0&0&0&0&\tiny0.0210745&\tiny0.0551424&\tiny0.1185422&\tiny0.2025660&\tiny0.3128971&\tiny0.4353564\\
\hline
6&0&0&0&0&0&\tiny0.0170844&\tiny0.0459158&\tiny0.1008971&\tiny0.1759971&\tiny0.2767458\\
\hline
\end{array}
$$
While there's  unlikely to be any simple formula for the optimal strategy for general $\ d\ $ and $\ t\ $, it turns out that for $\ G(6,10)\ $ the best moves are independent of your current position and can be succinctly described by ordering the non-fatal ones from best to worst:
$$
\{1,1^2,1^3,1^4,1^5,1^6\}\succ\{2,2^2\}\succ\{3\}\succ\{2^3\}\succ\{4\}\ .
$$
Any moves not in the above list will result in a sure loss from any position, and all moves in the above list except the $1$s can become non-viable as the game proceeds. The moves $\ \{2^3\}\ $ and $\ \{4\}\ $, for instance, will only remain viable as long as the dice that have so far been saved are all $1$s.


问:
$A \subset \mathbb{R}$ uncountable $\Rightarrow$ $ \exists ~I$ interval such that $A \cap I$ is dense in $I$?
If $A \subset \mathbb{R}$ is uncountable, does it follow that there exists a non-degenerate interval $I$ such that $A \cap I$ is dense in $I$?

I was thinking that Cantor set might be a counterexample, but eventually I reached the conclusion that it isn't. I'm not really sure, though.

I think that we can assume WLOG that $A \subset [0,1]$. This is because if we choose $A_k=A \cap [k,k+1]$, then $A= \bigcup\limits_{k \in \mathbb{Z}}A_k$, and since this is a countable union, it follows that there exists $n \in \mathbb{Z}$ such that $A_n$ is uncountable. We can assume WLOG that $n=0$ and then we can just take the problem from the beginning, but now with $A_0$ instead of $A$.

I tried to solve the problem by contradiction, but I didn't succeed.

答: What you are asking is whether for an uncountable set $A$ the closure of $A$ has a non-void interior. So any Cantor set (uncountable, closed, and not containing an interval) will be a counterexample. Note that the Cantor space does not have connected subsets with more than one point, so this will work in many other spaces.


问:
Norm in $C^{1}([0,T];H^{s}(\mathbb R^d))$
I am reading Evans' PDE textbook and found to prove the existence of solutions to a PDE using a priori estimates, if one chooses the space $X=C([0,T];H^{s}(\mathbb R^d))$, then the norm is defined as $$\|u\|_{X}=\sup_{0\leq t\leq T} \|u\|_{H^s},$$ where $H^s$ is just Sobolev spaces.

My question is how to define the norm for the PDE if we choose $X=C^{1}([0,T];H^{s}(\mathbb R^d))$? Is it $$\|u\|_{X}=\sup_{0\leq t\leq T} \|u\|_{H^s}+\sup_{0\leq t\leq T} \|\partial_{t}u\|_{H^s}$$ or
$$\|u\|_{X}=\sup_{0\leq t\leq T} \|u\|_{H^s}+\sup_{0\leq t\leq T} \|\nabla u\|_{H^s}?$$

答: The space $C^1([0,T],X)$ is the space of all functions from $[0,T]$ to $X$ that are continuously differentiable. The norm is the $\sup$-norm of the function and its time-derivative. So the first choice is correct. For the second one, there is the possibility of ambiguity as $\nabla$ is typically used to denote derivative with respect to spatial variables, not time.


问:
Spectrum of the follolowing operator in $C[0,1]$
I want to find a spectrum (with classification of points) of the following operator
$$
Af(x) = \frac{x-1}{x-2}f(x) + f(0),
$$
where $A:C[0,1] \to C[0,1]$.

There is only one point $\lambda = \frac{1}{2}$ with which I stuck. I give here my partial solution.

Firstly by direct calculations I proved that equation
$$
Af = \lambda f
$$
can be solved if and only if
$
\lambda = \frac{3}{2}
$.
So $$
\sigma_p(A) = \frac{3}{2}
$$
Newxt I investigate when it is possible to solve the equation
$$
(A-\lambda I)f(x) = g(x),
$$
for arbitrary functions $g(x)\in C[0,1]$. We assume that $\lambda \neq \frac{3}{2}$. Then from
$$
\left(\frac{x-1}{x-2} - \lambda\right)f(x) + f(0) = g(x)
$$
considering $x = 0$ we obtain
$$
f(0) = \frac{g(0)}{\frac{3}{2} - \lambda}
$$
So $f(0)$ is uniquely determined. If $\lambda$ belongs to the range of $\frac{x-1}{x-2}$  then from
$$
\left(\frac{x-1}{x-2} - \lambda\right)f(x) = g(x)  - f(0)
$$
we obtain that for $x_{\lambda} = \frac{1-2\lambda}{1-\lambda}$ we have
$$
0\cdot f(x_{\lambda}) = g(x_{\lambda}) - f(0)
$$
so
$$
g\left(\frac{1-2\lambda}{1-\lambda} \right) = f(0) = \frac{g(0)}{\frac{3}{2} - \lambda}
$$
So we obtain a constration on $g(x)$. But this is a non trivial constration only for $\lambda \neq \frac{1}{2}$. So we can show that
$$
\overline{Im(A - \lambda I)} \neq C[0,1]
$$
for all $\lambda \in [0, \frac{1}{2})$.

But what can we say for $\lambda = \frac{1}{2}$? I'm stuck here...
Also I want to say that this is not a duplication of the question Finding a spectrum of an operator in $C[0, 1]$
since in my question the main task is to determine the classification of spectral points (that is only one point $\lambda = \frac{1}{2}$).

答:
$A-\frac 1 2 I$ is not surjective, hence $\frac 1 2 \in \sigma (A)$. To see this take any $g \in C[0,1]$ such that $g'(0)$ does not exist. Suppose $(A-\frac 1 2 I) f=g$. This gives $\frac {x-1}{x-2}f(x)+f(0)-\frac  12 f(x)=g(x)$. This can be written as $f(x)=\frac {2(x-2)(g(x)-f(0))} x$ for $x \neq 0$. For $f$ to be continuous at $0$ it is necessary that $f(0)=g(0)$. Also, letting $x \to 0$ we see that $g'(0)$ must exist.

Let $f$ be continuously differentiable. Let $(q_n)$ be a sequence of polynomials converging uniformly to $f'$. Let $p_n(x)=\int_0^{x}q_n(t)dt+f(0)+\frac {2A} {\sqrt n} \sqrt {x+\frac 1  n}$ where $A=-\frac 1  4[f(0+4f'(0)]$. Then $p_n'(0)=q_n(0)+A \to f'(0)+A$.  Hence $4p_n'(0)+p_n(0) \to 0$. Let $c_n=4p_n'(0)+p_n(0)$ Then $c_n \to 0$ and $p_n-c_n \in M$ where $M=\{g\in C^{1}[0,1]:4g'(0)+g(0)=0\}$.  Since $p_n-c_n \to \int_0^{x} f'(t)dt +f(0)=f(x)$ uniformly  it follows that any continuously differentiable is a uniform  limit of functions from $M$. Any continuous function is a uniform limit of functions which are continuously differentiable so $M$ is dense in $C[0,1]$. But every function in $M$ is in the range of $A-\frac 1  2 I$. It follows that $A-\frac 1  2 I$ is injective with a dense range (but not surjective). Thus, $\frac1 2$ belongs to the continuous spectrum of $A$.


问:
Finding $\alpha$ such that $\alpha(f(x)-\sqrt{\pi})=\cos(2\pi x)$, where $f(x)=\sum_{n\in\mathbb{Z}}{e^{-(x-n)^2}}$
Given the function

$$f(x)=\sum_{n\in\mathbb{Z}}{e^{-(x-n)^2}}$$

find the constant $\alpha$ (independent of $x$) such that

$$\alpha(f(x)-\sqrt{\pi})=\cos(2\pi x)$$

Numerically alpha seems to be around $5453.9329935$, but that may not be perfectly accurate. I have no how to find the exact form of this number. How do you get an exact form of $\alpha$?

Here's a desmos graph if this makes it easier: https://www.desmos.com/calculator/53boxpbjnv

答:
Using the Poisson summation formula, we have $$f(x) := \sum_{n \in \mathbb{Z}}e^{-(x-n)^2} = \sum_{k \in \mathbb{Z}}\sqrt{\pi}e^{-\pi^2k^2-i2\pi kx}.$$ The $k = 0$ term is simply $\sqrt{\pi}$. The $k = \pm 1$ terms are $\sqrt{\pi}e^{-\pi^2}(\cos(2\pi x)\mp i\sin(2\pi x))$.

Meanwhile, the sum of the absolute values of the terms with $|k| \ge 2$ is bounded by $$\sum_{\substack{k \in \mathbb{Z} \\ |k| \ge 2}}\left|\sqrt{\pi}e^{-\pi^2k^2-i2\pi kx}\right| = 2\sqrt{\pi}\sum_{k = 2}^{\infty}e^{-\pi^2k^2} \le 2\sqrt{\pi}\sum_{k = 2}^{\infty}e^{-\pi^2(5k-6)} = \dfrac{2\sqrt{\pi}e^{-4\pi^2}}{1-e^{-5\pi^2}} < 3 \times 10^{-17}.$$

Hence, we have $f(x) \approx \displaystyle\sum_{k = -1}^{1}\sqrt{\pi}e^{-\pi^2k^2-i2\pi kx} = \sqrt{\pi}+2\sqrt{\pi}e^{-\pi^2}\cos(2\pi x)$. More specifically, $$\left|f(x)-\sqrt{\pi}+2\sqrt{\pi}e^{-\pi^2}\cos(2\pi x)\right| \le \dfrac{2\sqrt{\pi}e^{-4\pi^2}}{1-e^{-5\pi^2}} < 3 \times 10^{-17},$$

i.e. $$\left|\alpha(f(x)-\sqrt{\pi}) - \cos(2\pi x)\right| \le \dfrac{e^{-3\pi^2}}{1-e^{-5\pi^2}} < 2 \times 10^{-13},$$ where $\alpha = \dfrac{e^{\pi^2}}{2\sqrt{\pi}} \approx 5453.9329936539385\ldots$.

Note that we don't have $\alpha(f(x)-\sqrt{\pi}) = \cos(2\pi x)$ exactly for all $x$, but the approximation is close enough that you won't see it on a graph without zooming in really close.


问:
Expanding convolution integrals for sharply-peaked functions.
Question:

I have integrals like
$$ I(x) = \int_0^\infty K(y) F(x,y) dy $$
where $K(y)$ is some kernel function that is sharply peaked at some special value $y = y_0$ and otherwise trends toward zero away from $y_0$.

Considering $K(y) \approx \delta(y-y_0)$ it's easy to derive
$$I(x) = F(x,y_0),$$
but I would like to derive a correction term to this equation since $K(y)$ is not exactly a delta function. This would probably involve $K''(y_0).$ I am sure this is similar to a Laplace expansion but I am unclear exactly how. Any guidance for an expansion of this integral around the peak of $K(y)$ would be greatly appreciated!

Attempt:

writing
$$I(x) = \int_0^\infty e^{ \ln K(y) } F(x,y) dy $$ and expanding $K$ around $y_0$ and taking the lower limit $ y \rightarrow -\infty$ without introducing much error provides

$$I(x) \approx K(y_0) \int_{-\infty}^\infty \exp\Big[\frac{1}{2}\frac{K''(y_0)}{K(y_0)}(y-y_0)^2\Big] F(x,y) dy, $$
which is now at least a convolution integral with a sharply-peaked Gaussian kernel...
Is there a further approximation I can make to evaluate this integral as a series expansion?

答:
A method to deal with highly peaked kernels is described in this article. Only the domain near $y=y_0$ contributes significantly to the integral. Then, one may use a Talor expansion to represent $F(x,y)$ in this region:
\begin{equation}
 F(x,y)=F(x,y_0)+(y-y_0)\frac{\partial F}{\partial y}(x,y_0)+\frac{(y-y_0)^2}{2!}\frac{\partial^2 F}{\partial y^2}(x,y_0)+\cdots
\end{equation}
The integral can be written as
\begin{align}
 I(x) &= \int_0^\infty K(y) F(x,y) \,dy\\
 &=k_0F(x,y_0)+k_1\frac{\partial F}{\partial y}(x,y_0)
 +k_2\frac{\partial^2 F}{\partial y^2}(x,y_0)+\cdots
\end{align}
where the coefficients are related to the moments of the kernel:
\begin{equation}
 k_n=\frac{1}{n!}\int_0^\infty (y-y_0)^n K(y)\,dy
\end{equation}
for $n=0,1,2\cdots$. These coefficients quickly vanish if the function is highly peaked. The first term is equivalent to the delta contribution. Moreover, for a symmetric peak, the odd indices vanish.


问:
Show similarity of triangles
We first fold a square piece of paper in the middle, so that two congruent rectangles are created.

Then we fold $A$ onto midpoint $B$ of side $EC$ and mark points $D$, $F$, $G$ and $I$.

(a) Show that the triangles $\triangle IHG$, $\triangle BDC$ and $\triangle BEF$ are similar.

(b) Show that $|BD|:|BC|:|CD|=3:4:5$.

(c) Show that $|HF|=\frac{1}{3}\cdot |HE|$.

$$$$

At (a) we have that the triangles $\triangle BDC$ and $\triangle BEF$ are similar because:

The angles $\angle BEF$ and $\angle BGD$ are equal , they are both right angles.

It also holds that $|AG|=|BD|+|DG|$, $|EH|=|EF|+|FG|+|GH|=|EF|+|GI|+|GH|$ (since $|FG|=|GI|$, or not?), $|EG|=|EB|+|BG|=2|EB|=2|BG|$.

So we get that $\frac{|EB|}{|EF|}=\frac{|GB|}{|EF|}$. How can we continue?

答:
I'm not sure if showing the sides being proportional is possible (at least I haven't put much effort to it yet. I might come back to it). The angle-angle(-angle) approach seems easier.

Also, I think you've typo'd. It should be $ \lvert AC\rvert = \lvert BD\rvert + \lvert DC\rvert $, etc.

Note that $\angle DBF$ is also a right angle, which, by symmetry, is the same as the right angle at A.

$\angle EBF = 90^\circ - \angle DBC$,

and hence $\angle EFB = 90^\circ - \angle EBF = \angle DBC$.

Together with the right angles at E and C, we have $\triangle BDC \sim \triangle FBE$.

As to why $\triangle IGH$ is also similar to the two triangles mentioned, think about the small triangle that went out of the square after folding the paper, say we call it $\triangle FGX$. By symmetry, $\triangle FGX \cong \triangle IGH$. Then it should be pretty straight-forward to show that $\triangle FGX \sim \triangle FBE$.

Title: Show similarity of triangles


答:
As said in the comments, the cantor set is a counterexample, as it has empty interior, so it is nowhere dense (if $C\cap I$ was dense, then it would equal $I$, since $C$ is closed).

Similar examples show that even if you strengthen the hypothesis to say that $A$ is of positive Lebesgue measure, this is still not true. However in this case, you do get something: by the Lebesgue density theorem, you have that if $A$ has positive measure (in fact, as large as you want, up to but not including full measure), then $A$ is dense at almost every point of $A$ in the sense of approximate density.

This shows that having large cardinality and measure imply nothing about category (in the sense of Baire category).

Title: $A \subset \mathbb{R}$ uncountable $\Rightarrow$ $ \exists ~I$ interval such that $A \cap I$ is dense in $I$?

Body:
If $A \subset \mathbb{R}$ is uncountable, does it follow that there exists a non-degenerate interval $I$ such that $A \cap I$ is dense in $I$?

I was thinking that Cantor set might be a counterexample, but eventually I reached the conclusion that it isn't. I'm not really sure, though.

I think that we can assume WLOG that $A \subset [0,1]$. This is because if we choose $A_k=A \cap [k,k+1]$, then $A= \bigcup\limits_{k \in \mathbb{Z}}A_k$, and since this is a countable union, it follows that there exists $n \in \mathbb{Z}$ such that $A_n$ is uncountable. We can assume WLOG that $n=0$ and then we can just take the problem from the beginning, but now with $A_0$ instead of $A$.

I tried to solve the problem by contradiction, but I didn't succeed.


问:
Pointwise and uniform boundedness of sequence of functions
Define the sequence of functions $f_n:(0,1)\to\mathbb{R}$ by:

$$f_n(x):=1/x.$$

I believe the sequence of functions $\{f_n\}$ are pointwise bounded but not uniformly bounded. Furthermore, for example, take $M(x):=\frac{1}{x-1}$, then $\left|f_n(x)\right|\leq M(x)$ $\forall x$ and for every $n\in\mathbb{N}$. My question is, is this correct? Also, why is the sequence of functions
$f_n(x):=n$ not uniformly bounded or pointwise bounded? I would think I could simply take $M:=n+1$, then obviously $\left|f_n(x)\right|\leq M$ $\forall x$ and for every $n\in\mathbb{N}$. However, this is not the case, and so my question is why? Can neither $M$ or $M(x)$ depend on $n$?

答:
Pointwise bounded means there exists $M$ possibly depending on $x$ such that $f_n(x) \leq M$ for all $n$ and $x$. As you have observes $M(x)=\frac 1 x$ does the job.

Uniformly bounded means the bound $M$ above is not allowed to depend on $x$. Suppose $f_n(x) <M$ for all $x$ and $n$. Take any $x \in (0,\frac 1M)$ to get  a contradiction. Hence, the sequence is not uniformly bounded.

$f_n(x)=n$ is not even pointwise bounded because there is no constant $M$, not depending on $n$ but possibly depending on $x$, such that $f_n(x) <M$ for all $n$ and all $x$. .


问:
Question about limits with variable on exponent
So I have to find the following limit $$\lim_{n\to\infty}\left(1+\frac{2}{n}\right)^{1/n}.$$I said that this is $$\lim_{n\to\infty}\left[\left(1+\frac{2}{n}\right)^n\right]^{1/n^2}=\left(e^2\right)^{\lim_{n\to\infty}1/n^2}=1.$$Now I know that the final answer is correct, but my method seems to me to be wrong. Can I seperate a limit in the way I  did - i.e. is my working correct? Is there any other more elegant way to find the limit?

Thanks in advance.

答:
You don't actually need to go through this all:
$$\lim_{n\rightarrow\infty} 1+\frac{2}{n}=1$$ and
$$\lim_{n\rightarrow\infty} \frac{1}{n}=0$$
therefore
$$\lim_{n\rightarrow\infty} \left(1+\frac{2}{n}\right)^{1/n}=1^0=1$$</p>


问:
For a recursion based on $x+y+z+2\sqrt{xy+yz+zx},$ does what happens in $\mathbb Z\left[\sqrt n\right]$ stay in $\mathbb Z\left[\sqrt n\right]$?
This problem has a geometric origin which I'll outline below, but I believe the concepts and explanation are algebraic.

Given a function on triples
$$K((x,y,z))=x+y+z+2\sqrt{xy+yz+zx}$$

we build a general recursion as follows.

Start with an integer triple $X=(a,b,c)$, and a set $S=\{a,b,c\}$.

Compute $k=K(X)$ and add $k$ to $S$.  Writing $X$ as $(X_0,X_1,X_2)$, repeat step 2 for the triples $X=(k,X_1,X_2),(X_0,k,X_2),(X_0,X_1,k)$ respectively.

In general, the members of $S$ will contain nested radicals.  For example, if we start with the triple $(1,2,3)$ and use a restricted recursion  $x_{n+1}=K((1,2,x_n)),x_1=3$ we get

$$
S'=\left\{3,6+2 \sqrt{11},9+2 \sqrt{11}+2 \sqrt{2+3 \left(6+2 \sqrt{11}\right)},\\12+2
   \sqrt{11}+2 \sqrt{2+3 \left(6+2 \sqrt{11}\right)}+2 \sqrt{2+3 \left(9+2
   \sqrt{11}+2 \sqrt{2+3 \left(6+2 \sqrt{11}\right)}\right)},\dots \right\}
$$

However, Mathematica is able to denest these to

$$
S'=\left\{3,6+2 \sqrt{11},15+4 \sqrt{11},30+6 \sqrt{11},\dots\right\}
$$

suggesting the conjectures that

starting with initial triple $(1,2,3)$,  $S'\subset S\subset\mathbb Z[\sqrt {11}]$

starting with integer triple $(a,b,c)$, $S\subset \mathbb Z[\sqrt n]$, where $n=ab+bc+ca.$

Question: is the latter conjecture $S\subset \mathbb Z[\sqrt n]$ true, and if so why?

Bonus question: is there some characterization or invariant of the triples generated in the general recursion?

I've done some hand and computer calculations to understand why the radicals denest in particular cases, but I'm guessing that if the conjecture is true it will be "obvious" to someone more versed in this area.

Some background and observations:

This problem originates in playing with the construction of Apollonian gaskets and the function $K()$ comes from a theorem of Descartes concerning the curvatures of 4 mutually tangent circles.

In the case that $n=ab+bc+ca$ is a square, then all the numbers generated will be integers. And this leads to integral Apollonian circle packings.  But I have not seen any discussions of the case when curvatures are in some $\mathbb Z[\sqrt n]$.

see also this demo

答:
Please read my answer to MSE 2953315
"A Diophantine Equation Related to the Markoff Numbers" for some context for the following.

Define a symmetric homogeneous quadratic polynomial
$$ P(x,y,z,w) := x^2+y^2+z^2+w^2-2(xy+xz+yz+wx+wy+wz). \tag{1}$$
Given $\,x,y,z\,$ there are two roots of $\,P(x,y,z,w) = 0.\,$ They are
$$ w_\pm = x+y+z\pm 2\sqrt{xy+yz+zx}. \tag{2}$$
As usual, these two roots satisfy
$$ w_+w_-=x^2+y^2+z^2-2(xy+xz+yz) \tag{3}$$
and
$$ w_++w_-=2(x+y+z). \tag{4}$$

Given any commutative ring $\,R,\,$ define the set
$$ T := \{(x,y,z,w)\in R^4: P(x,y,z,w)=0\}. \tag{5}$$
If $\,(x,y,z,w)\in T\,$ then $\,(x,y,z,2(x+y+z)-w)\in T\,$
because of equation $(4)$. Now define
$$ S := \{\{x,y,z\}\; \text{ where }\; (x,y,z,w)\in T\}. \tag{6}$$
If we suppose $\,\{x,y,z\}\in S\,$ then $\,(x,y,z,w_\pm)\in T\,$
where $\,w_\pm\,$ is from equation $(2)$ and also
$$ (w_\pm,y,z,x),\,(x,w_\pm,z,y),\,(x,y,w_\pm,z)\in T \tag{7}$$
as in step 2 of the recursion since $\,P\,$ is a symmetric function.

In the question we have the special case of
$\,R=\mathbb Z[\sqrt {11}]\,$ and
$\,(1,2,3,6+2\sqrt{11})\in T.\,$


问:
How do you describe the proportionality for the function y = 1 / (1-x)?
How do I describe the proportion between x and y for this function?

$$ y=\frac{1}{1-x} $$

When x increases, y increases. But they are not directly proportional with each other.
I understand that the equation is just a rotated hyperbola, but what terminologies do you use to describe the proportions when the expressions get somewhat complicated and you can't specify it such as a direct proportion or inverse square proportion where you use adjectives related to common constructions of the function such as position of variables or the exponent of the independent variable.

How do I describe that when x increases, y increases? I thought of using the word proportional and omitting the word direct but that feels so vague and it doesn't bring out the "x increases, y increases" out that well. The closest you could get is saying is that 1 - x is inversely proportional to y but I want to describe the relationship in terms of x and y.

This question is more about the wordings. I have to describe the proportion of x and y in a sentence and want to correctly do so. I am aware that there are appropriate ways to explain things in mathematics such as the usage of terms "direct proportion" and "inverse square proportion" so I know that I have to be concise with word usage.

答: I guess you come from another area. In Mathematics there is no especial words for this relation for $x$ and $y$. To expect that every relation between two variables have a name is very very naive. There are infinitely many functions and describe them with formulas is much more practical that describe the relation with words. That's what Algebra was invented for, and it has become a very powerful tool. Trying to describe relations with words is to going many centuries backward.


Title: Prove by definition that $\frac{1-2n^3}{4n^3-2n^2-1} \to -\frac12$

Body:
$a_n = \dfrac{1-2n^3}{4n^3-2n^2-1}$

My attempt (of the scratch):

$\left| \dfrac{1-2n^3}{4n^3-2n^2-1} + \dfrac{1}{2}\right|=\left| \dfrac{2-4n^3+4n^3-2n^2-1}{8n^3-4n^2-2}\right|=\left|\dfrac{-2n^2+1}{8n^3-4n^2-2} \right|=\dfrac{2n^2-1}{8n^3-4n^2-2}<\dfrac{2n^2}{8n^3-4n^2-2}=\dfrac{n^2}{4n^3-2n^2-1}<\epsilon$

But I don't know what to do next to choose my $N \in \Bbb{N}$

Body: For large $n$, we have $4n^3 - 2n^2 - 1 > n^3$, in which case $\frac{n^2}{4n^3-2n^2-1} < \frac{n^2}{n^3} = \frac{1}{n}$.


问:
Proof that limits are unique
I have been working through a few calculus proofs, and to prove that a limit was unique i assumed the fact that if $c,d\in\mathbb{R}$ if given any $\epsilon>0$ we have $|c-d|<\epsilon$, then $c=d$.  However, is there an actual proof of this statement, or is this more of an obvious reality?

答:
It may be obvious, but in math we like to prove things.

Let $c$ and $d$ be real numbers.  Then $|c-d|\ge0$, so $|c-d|=0$ or $|c-d|>0$.

Assume, with an eye toward contradiction, that $|c-d|>0$,

and for all $\epsilon>0$ we have $|c-d|<\epsilon.$

But taking $\epsilon=|c-d|$, then we would have $\epsilon<\epsilon$, which is a contradiction.

Thus, our assumption that $|c-d|>0$ yielded a contradiction, and it must be that $|c-d|=0$.

Thus, $c=d$.

Title: Proof that limits are unique


问:
If two norm $1$ vectors have the same $L1$ norm, are they the "same" vector?
Suppose I have two vectors $u, v \in \mathbb{R}^n$ where $\|u\|_2 = \|v\|_2 = 1$. I am trying to see if these two vectors are the "same" under the following definition.

Two vectors are the "same" if you can change the sign, or shuffle the entries to make them equivalent.

For example, $[1,2]$ is the "same" as $[-1,2],[1,-2],[-1,-2],[2,1],[-2,1],[2,-1],[-2,-1]$. This is not a norm-$1$ vector but I just thought this was an easy example.

Now suppose $u,v$ are norm-$1$ such that
$$
\sum u_i^2 = \sum v_i^2 = 1
$$
and that their associated L$1$ norms are equal
$$
\sum |u_i| = \sum |v_i|.
$$

Does that mean that the two vectors must be the "same"?

In $\mathbb{R}^2$ this seems to be true just by plotting the vectors $[\sin\theta, \cos\theta]^T$, and could probably be proven pretty easily using triginometry identities. However I was wondering if anyone knew if this generalized to higher dimensions $n$. Thank you for the help!

答:
Yes, in $\mathbb{R}^2$ (as well as, trivially, in $\mathbb{R}$); no, in $\mathbb{R}^n$ for $n\geq 3$.

The "yes in $\mathbb{R}^2$" part is easily seen by, as you said, considering the vectors $\begin{bmatrix}\cos\theta\\\sin\theta\end{bmatrix}$.

A counterexample to the statement in general would be the two vectors in $\mathbb{R^3}$, $\begin{bmatrix}-1/3\\2/3\\2/3\end{bmatrix}$ and $\begin{bmatrix}1\\0\\0\end{bmatrix}$.


问:
Show $ \coth $ is a conformal mapping of the horizontal strip
I want to show that $\coth=\frac{e^{2z}+1}{e^{2z}-1}$ is a conformal mapping of the horizontal strip $S=\{z\in C: \pi/4<\text{Im}(z)<3\pi/4\}$ onto the unit disc U, but I can't seem to get the right idea?</p>

答:
First of all, notice that $\coth$ is the composition of the maps $f(z)=e^{2z}$ and $g(z)=\dfrac{z+1}{z-1}$, i.e. $\coth=g\circ f:\{z\in C: \pi/4<\text{Im}(z)<3\pi/4\}\longrightarrow \mathbb C
$

Now, where does $f$ map the strip $S=\{z\in C: \pi/4<\text{Im}(z)<3\pi/4\}
$?

First $z$ is mapped conformally to $2z$, i.e. the strip is mapped to $$S'=\{z\in C: \pi/2<\text{Im}(z)<3\pi/2\}$$
Then $S'$ is mapped via $e^z$ conformally to the left half plane $$H=\{z\in\mathbb C:\mathrm{Re}(z)<0\}\mathbf{\text { (Check it!) }}$$
Finally, where, does $g$ map the left half plane H?

$g$ is a Möbius transform, which maps conformally the boundary of $H$ to the boundary of the unit disk, and the interior of $H$ to the interior of the disk (Check it!!)


问:
Why is prime spectrum compact?
I'm doing exercise 17 in Atiyah-MacDonald, where I am proving that the prime spectrum of a ring is (quasi-)compact. I have shown that an open covering of sets $X_{f_i}$ for $i\in I$ for some index set $I$ implies that $f_i$ for $i\in I$ generates the unit ideal $(1) = A$. In the hint it is claimed that this means there is a finite combination of the generators $f_i$ which is equal to $1$, and I am having trouble proving this. Even on the Columbia stacks, this step is stated without much reason. Maybe I am just overlooking something simple.

Here is my attempt at reasoning. We can create a chain of ideals by adding more and more of the $f_i$. Supposing that no finite combination of the $f_i$ is equal to $1$, we can create an infinite chain of ideals. Now, I imagine Zorn's lemma might come in handy here? But again, I would like to think there is a simpler way to resolve this problem.

答:
The ideal generated by $S = \{f_i : i \in I\}$ is the intersection of all ideals containing $S$. But you can also show that this intersection is just the set of all finite linear combinations from $S$:

$$ (S) = \bigcap\{J \trianglelefteq A : S \subseteq J\} = \left\{ \sum_{r = 1}^n a_r f_r : a_r \in A, f_r \in S, n \in \mathbf{Z}^+ \right\}. $$

Because that set of finite linear combinations is:

an ideal containing $S$

contained in any ideal which contains $S$

Title: Why is prime spectrum compact?


答: The most straightforward way is to say "$y$ is inversely proportional to $1-x$".  If you want a more, purely verbal formulation you might say "$y$ is inversely proportional to the distance of $x$ from $1$", but this might be understood to refer to $y = \frac{1}{|1-x|}$".  Another, less ambiguous option might be "$y$ is inversely proportional to the amount by which $x$ is less than $1$".

Title: How do you describe the proportionality for the function y = 1 / (1-x)?

Body:
How do I describe the proportion between x and y for this function?

$$ y=\frac{1}{1-x} $$

When x increases, y increases. But they are not directly proportional with each other.
I understand that the equation is just a rotated hyperbola, but what terminologies do you use to describe the proportions when the expressions get somewhat complicated and you can't specify it such as a direct proportion or inverse square proportion where you use adjectives related to common constructions of the function such as position of variables or the exponent of the independent variable.

How do I describe that when x increases, y increases? I thought of using the word proportional and omitting the word direct but that feels so vague and it doesn't bring out the "x increases, y increases" out that well. The closest you could get is saying is that 1 - x is inversely proportional to y but I want to describe the relationship in terms of x and y.

This question is more about the wordings. I have to describe the proportion of x and y in a sentence and want to correctly do so. I am aware that there are appropriate ways to explain things in mathematics such as the usage of terms "direct proportion" and "inverse square proportion" so I know that I have to be concise with word usage.


答:
Here it is the proof in the context of metric spaces for the sake of curiosity.

Proposition

Let $f:(X,d_{X})\to(Y,d_{Y})$ be a function between metric spaces, $x_{0}\in X$ such that $x_{0}$ is an adherent point of $E\subseteq X$ and $L\in Y$. If $f$ converges to $L$ as $x$ approaches $x_{0}$ along $E$, then such limit is unique.

Proof

Let us assume that $f$ converges to $L$ and $L'$ as $x$ approaches $x_{0}$ along $E$, and $L \neq L'$.

Then we can take $\varepsilon = d_{Y}(L,L') > 0$.

For such value of $\varepsilon$, there corresponds $\delta^{1}_{\varepsilon} > 0$ and $\delta^{2}_{\varepsilon} > 0$ such that for every $x\in E$ one has that
\begin{align*}
\begin{cases}
\displaystyle d_{X}(x,x_{0}) \leq \delta^{1}_{\varepsilon} \Rightarrow d_{Y}(f(x),L) \leq \frac{\varepsilon}{3} = \frac{d_{Y}(L,L')}{3}\\\\
\displaystyle d_{X}(x,x_{0}) \leq \delta^{2}_{\varepsilon} \Rightarrow d_{Y}(f(x),L') \leq \frac{\varepsilon}{3} = \frac{d_{Y}(L,L')}{3}
\end{cases}
\end{align*}

Thus, for such value of $\varepsilon > 0$, there corresponds $\delta_{\varepsilon} = \min\{\delta^{1}_{\varepsilon},\delta^{2}_{\varepsilon}\}$ s.t. for every $x\in E$ one has that
\begin{align*}
d_{X}(x,x_{0}) \leq \delta_{\varepsilon} \Rightarrow d_{Y}(L,L') & \leq d_{Y}(f(x),L) + d_{Y}(f(x),L')\\\\
& \leq \frac{d_{Y}(L,L')}{3} + \frac{d_{Y}(L,L')}{3} = \frac{2d_{Y}(L,L')}{3}\\\\
& \Rightarrow d_{Y}(L,L') \leq 0 \Rightarrow L = L'
\end{align*}
which contradicts our assumption.

Thus, if the limit exists, it is unique.

Hopefully this contributes!

Title: Proof that limits are unique

Body: I have been working through a few calculus proofs, and to prove that a limit was unique i assumed the fact that if $c,d\in\mathbb{R}$ if given any $\epsilon>0$ we have $|c-d|<\epsilon$, then $c=d$.  However, is there an actual proof of this statement, or is this more of an obvious reality?


答:
Let $N \in \mathbb{N}$ s.t. $4N-2>n'+\delta$ for $n' \in \mathbb{N}$ and $\delta >0$, then $n \geq N$ implies $4n-2 \geq 4N-2 > n'+\delta$.

Let $N' \in \mathbb{N}$ s.t. $\frac{1}{N'^2}< \delta$ for $\delta >0$, then $n\geq N'$ implies $\frac{1}{n^2} \leq \frac{1}{N'^2} < \delta$.

If $n \geq \text{max}\{N, N'\}$ then $4n-2-\frac{1}{n^2} > n'+ \delta - \delta= n'$ and $\frac{n^2}{4n^3 -2n^2 -1} =\frac{1}{4n-2-\frac{1}{n^2}}<\frac{1}{n'}.$

Let $n'$ be such that $\frac{1}{n'}<\epsilon$.


问:
Probability Density Functions and Metropolis Hastings
I have a quick question regarding implementation of Metropolis-Hastings for a particular problem I'm dealing with.

Suppose that I have a probability density function $P(X)$ for a continuous random variable $X$.  In Metropolis-Hastings, I am required to compute acceptance probability $\frac{ Pr(x') Q(x_{t}|x') } {Pr(x_{t} Q(x' | x)}$.  However, the probability of any single event in a continuous space is zero.  Do  I just replace $Pr$ with $P$ and go on my merry way?

My primary concern is that the $P(x)$ is not necessarily less than or equal to 1.

答: This is usually the way to convert from discrete probabilities to continuous probabilities. You can think of this as promoting $x$ to an $\epsilon$-neighborhood of itself, whose probability is roughly $\epsilon P(x)$.</p>


问:
Right-angled isosceles triangles
If a right-angled triangle is isosceles then the other two angles must be equal to $45^\circ$ ?

Is this always the case or are there other possible right-angled isosceles triangles?

答:
Here's our isosceles right triangle.

Because it's a right triangle, we know Pythagorean theorem holds true, and so:

$$c^2 = a^2 + b^2$$

But, this is an isosceles triangle, and so by definition, $a = b$, and so:

$$c^2 = a^2 + a^2$$

$$c^2 = 2a^2$$

$$a = \frac{c}{\sqrt 2}$$

Moreover, we know that $\sin A = \frac{a}{c}$, and so:

$$\sin A = \frac{a}{c}$$

$$\sin A = \frac{\frac{c}{\sqrt 2}}{c}$$

$$\sin A = \frac{c}{c\sqrt 2}$$

$$\sin A = \frac{1}{\sqrt 2}$$

$$A = \sin^{-1}(\frac{1}{\sqrt 2})$$

$$A = 45^o$$

(Note that $A$ is technically an infinite amount of answers because of the nature of $\sin^{-1}$, but because in this case it cannot be negative, and the angles in a triangle cannot add up to more than $180^o$, this is the only possibility).

We can then find $B$, as we know all angles in a triangle in Euclidean space add up to $180^o$, and we find that $B = 45^o$ too.

Thus, all right angle isosceles triangles have two angles of $45^o$.

Title: Right-angled isosceles triangles


问:
Whether differentiable means smooth in do Carmo's Riemannian Geometry
Picture below is from the 26th page of Do Carmo's Riemannian Geometry. In my view, by (5), if $f\in C^k$, then $Xf\in C^{k-1}$. Therefore, the last red line should be wrong. But in this book, there is not $C^k$. I guess the differentiable function is smooth function (namely, $f\in C^\infty$), is it ?

答:
The index is your friend! A quick look at the index turns up the entry "Differentiable structure" pointing to page 2, where you'll find the following sentence:

Differentiable always signifies of class $C^\infty$.


问:
If a topological space $X$ has open path connected components and is connected then $X$ is path connected
Why does it hold?

I know that connected, locally path connected topological spaces are path-connected. Is it an equivalence statement?

答:
The path connected components of $X$ give a partition $\left\{C_i:i\in I\right\}$ of $X$. By hypothesis, all $C_i$ are open, so all $C_i=X\setminus\bigcup_{j\neq i}C_i$ are also closed.

Taking any $C_i$, it is a nonempty clopen set, so $X=C_i$ by connectedness.


答: Let $X = \bigcup_{i \in I} U_i$ be the path components of $X$. By hypothesis, the $U_i$ are open. Furthermore, path components are disjoint, as they are equivalence classes under the relation $x \sim y$ if $x$ and $y$ can be joined by a path. Hence, the above is a disjoint union of open subsets of $X$. As $X$ is connected, this can only occur for $|I|=1$, i.e. if $X$ equals some path component. Thus, $X$ is path connected.

Title: If a topological space $X$ has open path connected components and is connected then $X$ is path connected

Body:
Why does it hold?

I know that connected, locally path connected topological spaces are path-connected. Is it an equivalence statement?


问:
Help on typology of integrals
So I've been wrestling with understanding exactly what line integral and surface integrals represent, in both scalar and vector form. Also standard double and triple integrals can solve for different things. Having searched the internet, this seems to be a common problem, and I struggle to find really good answers.

I've created a typology of integrals. I'd love if more people could correct my entries or add anything to them. When you study these in multivariable calculus, it all becomes somewhat overwhelming. So I hope this, when rendered comprehensive and correct, can help future students.

Please add corrections in comments and I'll edit the table. Thanks in advance for any insights that can be offered.

答:
I feel like this might almost be more of a physics than a math question, but I'll try to help by stating a few different things.

Generally, in terms of dimensional analysis, an $n$-fold integral over length-like coordinates will add $n$ length dimensions to the dimension of the integrand. This is why one calls triple integrals volume integrals and double integrals area integrals. But if the integrand itself is already of dimension $L^1$ (i.e., length), then a double integral of that integrand will have the dimensions of volume, $L^3$ (how one might calculate the volume of a cylinder, for example).

This dimensional analysis applies to integration with respect to non-length variables as well, but the interpretation may not always be as straightforward. Integration with respect to time $t$, for example, will typically be done on integrands of dimension $1/t$, though that is by no means a rigorous rule—it all depends on the mathematical or physical context. When calculating trajectories, for example, the integrand may be of dimension $L/t$ (velocity) and the integral with respect to $t$, leading to a length $L$ as a result. The key point, again, is that integration with respect to some variable increases the dimension of the result with respect to that variable by one.

This applies to line and surface integrals in the same way. General line and surface integrals are really just generalizations of 1D and 2D integrals to non-Euclidean (curved) spaces. It might be a good idea to look for pedagogical introductions into the basic concept behind integration on manifolds if you want to learn more about these generalizations.

Title: Help on typology of integrals


问:
A Generalization of Linear Functions. But is it Useful?
Let $X$ be a normed vector space, and let $Y$ be another normed vector space over the same scalar field. Let $\|\cdot\|$ denote the norm of $Y$.

We say that a function $g: X \to Y$ is quasi-linear if there exists a $C>0$ such that

$\|g(x+y)\| \leq C(\|g(x)\|+\|g(y)\|)$ ${}$ and ${}$ $\|g(\alpha x)\| = |\alpha|\|g(x)\|$

for all vectors $x$ and scalars $\alpha$.

My question concerns the class of functions $g : X \to Y$ for which there exists a $C>0$ such that

$\|g(x+y)\| \leq C(\|g(x)+g(y)\|)$ ${}$ and ${}$ $\|g(\alpha x)\| = |\alpha|\|g(x)\|$

for all vectors $x$ and scalars $\alpha$ (note, the value of $C$ is allowed to depend on the particular function $g$).

First, does the latter class of functions have a name? Second, observe that we have the containments
$$
    \text{linear functions} \subseteq \text{this class of functions} \subseteq \text{quasi-linear functions}
$$
so that my other question is is either of these containments strict?

Thanks very much.

答:
The function $g\colon\mathbb{R}\to\mathbb{R}$ given by
$$g(x)=\begin{cases}x&\text{ if }x\neq 2\\-2&\text{ if }x=2\end{cases}$$
is quasi-linear (it preserves norm/absolute value), but
$$|g(1+2)|=|g(3)|=3$$
whereas
$$|g(1)+g(2)|=|1-2|=1$$
so it is not in your class of functions.


问:
Question on sums of products of divisors (cont.)
In my previous Question on sums of products of divisors, I asked the following question:

Does it exist some positive integer $n$ such that a sum of $k$ products of its divisors greater than $1$ equal $n$ itself, when $k$ is not a divisor of $n$ itself?

An example was provided, but one of the products was less than $\sqrt{n}$. Subsequently, it came to me the following subsequent question:

Does it exist some positive integer $n$ such that a sum of $k$ products of its divisors greater than $1$ equal $n$ itself, when $k$ is not a divisor of $n$ itself and all the products are equal or greater than $\sqrt{n}$?

Any positive or negative answer to the question, with a counterexample, hint or sketch of a proof in the affirmative or negative direction would be really welcomed. Thanks!

Edit

The answer to the question is yes, as perfectly showed in the answer provided. Would it change if we put the restriction that all the products must be distinct?

答:
The example you were given scales suitably.  That is, where the example had the divisors $(2,3,6,9)$ just multiply them, say, by $5$.  We get $(10,15,30,45)$ and the parallel sum is $$10\times 15+10\times 30+10\times 45+15\times 30=150+300+450+450=1350$$

Of course, $1350$ is divisible by each of $(10,15,30,45)$ and now each product is considerably greater than $\sqrt {1350}=36.742\cdots$.  And, of course, $4$ still doesn't divide $1350$.


问:
$E(\mathbb{F}_q)$ is a torsion group where $E$ is an elliptic curve?
Let $E$ be an elliptic curve defined over $\mathbb{F}_q$ with $q=p^n$, then how to deduce $E(\mathbb{F}_q)$ is a torsion group?

In other words, for any $\mathbb{F}_q$-rational point $P$, why does there exist $m\geq1$ such that $[m]P=0$?

Thanks in advance.

答: The group is necessarily finite, hence torsion.</p>


问:
MVT Differentiation
Can anyone help me with these

Suppose that $f$ is differentiable on $[4,6]$ and is such that $f(4) = 6$ and $f' \geq 2$ on
$[4,6]$. Find the lowest possible value of $f(6)$.

I think the answer for the lowest possible value of $f(6)$ is $10$. Is it right?

Give a function $f$ that is differentiable on $\Bbb R$ and which has the following 3 properties, or explain why such a function can't exist: $f(1) = 3$, $f(3) = 8$, and $f'(x)\leq 2$ for all $x$.

答:
As per mean value theorem : $f'(c) = \frac{f(b)-f(a)}{b-a}$ where c is any point in the close interval [a b]

Therefore :

$f' = \frac{f(6)-f(4)}{6-4} =2 $

$\Rightarrow  \frac{f(6)-6}{6-4} = 2$

$\Rightarrow = f(6)=6+4 =10$

Title: MVT Differentiation


问:
Formal power series with coefficient in $\{0,1 \}$ which is not rational
Recall that $f \in \mathbb{C}[[x]]$ is rational if $f = ab^{-1}$ where $a, b \in \mathbb{Q}[x]$ (the inclusion  $ \mathbb{Q}[x] \to \mathbb{C}[[x]]$ is obvious).

How do I obtain an example of  $f \in \mathbb{C}[[x]]$ with coefficients $0,1$ which is not rational?  I have an experience of proving that some series is not rational (say, $exp(x)$).I could't come up with an example myself, though.

答:
This is far from my field, so please check it.

How about $\sum_{n=0}^\infty x^{2^n}$. Then its evaluation at $1/2$ has binary expansion with $1$’s only at places not at powers of two. This expansion is not periodic, even eventually, and so is not a rational number.


问:
Divide after taking mod
There's probably a formal way to put this with rings, but suppose I have a number x and mod base M and divisor d. I know x mod M = x', and that d is divisible by x. How do I find x/d mod M?

It seems possible because if you take x = 12, M = 5, d = 3 for example, only knowing that x' = 2, you can deduce that x/d = 4.

答:
Let $\,x = dy,\,$ so $\,y = x/d.\,$ Then $\bmod m\,$ we seek to solve $\,dy\equiv \bar x\,$ for $\,y.\,$ By general theory there is a unique solution $\!\iff\! \gcd(d,m)=1.\,$ If so, $\,d^{-1}$ exists $\!\bmod m\,$ and $\,y \equiv \bar x d^{-1} \pmod{\!m}$.

e.g. $x = 14 = 2(7) = 2y$. Then $\,2y = 14\equiv 4\pmod{\!10}\!\iff$ $ 2y = 4+10n\iff$ $ y = 2+5n$ $\iff y \equiv 2,7\pmod{\!10}\,$ has $\,\color{#c00}2\,$ solutions (since $\,\gcd(2,10)= \color{#c00}2),\,$ so it is impossible to uniquely determine $\,y\bmod 10\,$ given only $\,2y\bmod 10$.

As explained in the link, we can view the solution $\,y\,$ of $\,dy\equiv \bar x\pmod{\!m}$ as a modular fraction $$y\,\equiv\, \dfrac{\bar x}{d}\!\!\!\pmod{\!m}\,\equiv\, \dfrac{\bar x\color{#c00}{/c}}{d\color{#c00}{/c}}\!\!\!\pmod{\!m\color{#c00}{/c}}\qquad $$

by cancelling $\,\color{#c00}{c} = \gcd(d,m)\,\ \rm\color{#c00}{everywhere},\,$ e.g. $\,c = \gcd(2,10) = 2\,$ in the  example above so

$$y\,\equiv\,\dfrac{4}2\!\!\!\!\pmod{\!10}\, \equiv\, \dfrac{2}1\!\!\!\!\pmod{\!5}\qquad$$

in the example above. As shown in the link, if $\,c\nmid \bar x\,$ then there are no solutions, else there are $\,c\,$ solutions $\bmod m,\,$ i.e. the map $\,y\mapsto dy\,$ is $\,c\,$ to $1\,$ when viewed modulo $\,m$.

Title: Divide after taking mod


问:
The position of a ladder leaning against a wall and touching a box under it
I was reading a newspaper and there was a little math riddle, I thought "how funny, that's gonna be easy, let's do it" and here am I...

The problem goes as follow :
in a barn, there is a 1 meter cubic box against a wall and a 4 meter ladder is leaning against the wall, touching the box at its corner. Here is a picture :

So, the big triangle has a hypotenuse $FE$ of $4$, the square $ABDC$ has sides of length 1 and is basically "insquared" at the right angle, i.e. $D\in \overline{FE}$.
The question is "what is the length of the biggest cathetus", here $AF$.

So far, no problem.

Now here are my solutions:

By Thales' intercept theorem, $\frac{FB}{FA}=\frac{BD}{AE}$, by hypothesis, $FB=FA-1$ and $BD=1$. Now by Pythagoras, $FA^2+AE^2=FE^2$; by hypothesis, $FE=4$, so we end up with a system of equations, letting $h=FA, d=AE$: $$ \begin{align} &\frac{h-1}{h}=\frac{1}{d} \\ &h^2+d^2=4^2  \end{align} $$
Which solves (removing 3 non-relevant solutions) into $d \cong 1.3622$ and $h \cong 3.76091$.

Now, if I consider the "function" of the line : $f(x)=\frac{-h}{d}x+h$, I know that $f(1)=1$ and I end up with Pythagoras with the system :$$ \begin{align} &\frac{-h}{d}+h=1 \\ &h^2+d^2=4^2  \end{align} $$
it solves again into the same, again removing 3 non-relevant solutions

Okay, this means that using Pythagoras is no good since it ends up giving a quartic equation (4 answers of which 3 are "non-relevant").

Now if I consider the length of the arc $f(x)$ between $0$ and $d$ it has to be $4$ and again $f(1)=1$ I end up with the system:
$$ \begin{align} &\frac{-h}{d}+h=1 \\ &\int_0^d \sqrt{1+(f'(x))^2} dx =\int_0^d \sqrt{1+\left(\frac{-h}{d}\right)^2} dx = d \sqrt{1+\left(\frac{-h}{d}\right)^2}   \end{align} $$
Which solves again into the same answers, but this time removing only 2 non-relevant solutions (i.e. it gives a cubic equation instead of a quartic).

I tried also using the areas and the smaller trangles $FAD$ and $AED$ for example : $\frac{h \cdot d}{2} = \frac{h\cdot 1}{2}+\frac{d\cdot 1}{2}$

Yet I wasn't able to get to any "hand solvable" solution : if I were able to bring it down to some quadratic equation, that would be nice, since it is a common assumption, here, that everybody has seen the "general formula for solving quadratic equations" in school and so would be able to solve this, I may then see how it is seen as a funny riddle in the newspaper.

My best trial, with "just" a cubic equation, is way too complicated for the normal readers of this newspaper, so it's bugging me.

What am I missing? Some basic properties maybe? It's really bugging me, not being able to solve this without Wolfram.

答:
Let $x$ be BF, then

$(x+1)^2 + (1+\frac1x)^2 = 16$

$x^2 + \frac1{x^2} + 2x + \frac2{x} + 2 = 16$

Let $a = x + \frac1x$, then $a^2 = x^2 + 2 + \frac1{x^2}$.

The equation becomes $a^2 + 2a = 16$.

Solve for $a$, then solve for $x$.


答:
A power series $f = \sum f_n z^n$ is rational over $\mathbb{C}$ (that is, contained in the image of $\mathbb{C}(x)$, not just $\mathbb{Q}(x)$; I think this is the standard meaning of "rational" in this context) iff the sequence $f_n$ of its coefficients satisfies a linear recurrence with constant coefficients (possibly after ignoring finitely many initial terms) iff $f_n$ has the form

$$f_n = \sum_{i=1}^k p_i(n) \lambda_i^n$$

where $p_i \in \mathbb{C}[x]$ are polynomials and $\lambda_i \in \mathbb{C}$, again possibly after ignoring finitely many initial terms. This can be proven in many ways, e.g. using partial fraction decomposition. ($f$ is rational over $\mathbb{Q}$ iff in addition $\lambda_i$ and the coefficients of $p_i$ have values in $\overline{\mathbb{Q}}$ and the sum above is closed under Galois conjugation.)

There are many obstructions to a sequence being of this form; among other things, because it satisfies a linear recurrence of fixed length $k$, any stretch of $k$ consecutive terms of the sequence uniquely determines all future terms (given the coefficients of the recurrence), and in particular if any stretch of $k$ consecutive terms is zero then all subsequent terms must be zero. This proves:

Proposition: If $f_n$ is not eventually zero but for every $k$ there are at least $k$ consecutive zeroes among the terms of $f_n$, then $f = \sum f_n z^n$ is irrational (over $\mathbb{C}$, and hence also over $\mathbb{Q}$).

A power series with this property is said to be lacunary. This implies, for example, that $\sum z^{n^2}$ is irrational as in Raffaele's deleted answer, that $\sum z^{2^n}$ is irrational as in Lubin's answer (but over $\mathbb{C}$ and not just over $\mathbb{Q}$), and that $\sum z^{n!}$ is irrational as Arthur speculates in the comments. A longer argument shows, IIRC, that lacunary power series are even transcendental (not-algebraic), in the sense that they don't satisfy a polynomial relation with coefficients in $\mathbb{C}[x]$.

Various other arguments for these power series being irrational are possible; for example you can show that if $f$ is rational and $f_n$ is 1) bounded and 2) consists of integers, then it must be eventually periodic, analogous to the eventually periodic decimal expansions of ordinary rational numbers. (Sketch: boundedness implies that $|\lambda_i| \le 1$ and that if $|\lambda_i| = 1$ then $p_i$ is constant. Integrality implies the $\lambda_i$ are algebraic and $|\lambda_i| = 1$, and then a cute argument involving the pigeonhole principle shows that this implies the $\lambda_i$ are roots of unity.)

Title: Formal power series with coefficient in $\{0,1 \}$ which is not rational

Body:
Recall that $f \in \mathbb{C}[[x]]$ is rational if $f = ab^{-1}$ where $a, b \in \mathbb{Q}[x]$ (the inclusion  $ \mathbb{Q}[x] \to \mathbb{C}[[x]]$ is obvious).

How do I obtain an example of  $f \in \mathbb{C}[[x]]$ with coefficients $0,1$ which is not rational?  I have an experience of proving that some series is not rational (say, $exp(x)$).I could't come up with an example myself, though.


问:
Find all triplets (x, y, z) of positive integers such that 1/x + 1/y + 1/z = 4/5
Here's what i did :-

i wrote Find all triplet $$1/x + 1/y + 1/z = 4/5$$ as $$1/x + 1/y = 4/5 - 1/z$$

$(4z-5)/5z = 1/x + 1/y$ which can be written as $1/x + 1/y = 1/(5z)/(4z-5)$

Let $(5z)/(4z-5) = n$ therefore $1/x + 1/y = 1/n$ which on factorization gives $(n-x)(n-y) = n^2$ and $n$ should be an integer , i am not sure about the proof but i got $(5z)/(4z-5)$ is never an integer except $z = <o,1>$ but z cannot be 0 .

Here's my problem :-
if we take $n = 1$ we get 0 solutions of $<x,y,-1>$
is what i did correct ?

答:
To get a pro bono Diophantine equation, multiply both sides by $5xyz$:

$5yz+5xz+5xy=4xyz$

Reduce (mod 5) to get $4xyz \equiv 0$.  Assuming $x\leq y,z$, we have $4/5\leq 3/x$, so $x <4$.  Let us take $y$ to be a variable divisible by 5.

Case $x=2$:

$y=5$ implies $z=10$.

$y=10$ implies $z=5$.

Otherwise, $z<5$.  Since $1/z\leq 4/5-1/x$, then $z>3$.

$z=4$ implies $y=20$.

Case $x=3$.

We have $\frac 1y+\frac 1z=\frac 45-\frac 1x=\frac 7{15}$.

Multiplying by $15xy$, we get

$15z+15y=7yz$.

Reduce mod 3: $0\equiv yz$ -- so, either $y$ or $z$ divisible by $3$.

Case $y$ divisible by 3: we already assumed $y$ divisible by $5$; so $y$ divisible by 15.

So, $\frac 1z=\frac 7{15} - \frac 1y\geq \frac 7{15}-\frac 1{15}=\frac 25$, or $z\leq 2.5$.  Since $z\geq x$, then $z=2$, and clearly this does not work.

Case $z$ divisible by $3$:

As above, $\frac 1z=\frac 7{15} - \frac 1y\geq \frac 7{15}-\frac 15=\frac 4{15}$, so $z\leq 15/4$.  Then $z=3$ is the only choice, and again this does not work.


问:
Clarification about proof involving product/sum manipulation
In this post, the following simplification was made:
$$2\pi i\sum_k\lim_{x\to ia_k}(x-ia_k)
\frac{x^2e^{itx}}{\prod_j(x^2+a_j^2)}\\
=-\pi\sum_ke^{-ta_k}a_k\sum_{j\neq k}\frac1{a_j^2-a_k^2}.$$

What I have done is the following, but am unsure why it is wrong:
$$
2\pi i\sum_k\lim_{x\to ia_k}(x-ia_k)
\frac{x^2e^{itx}}{\prod_k(x^2+a_k^2)}\\
=2\pi i\sum_k \frac{(ia_k)^2e^{-ta_k}}{\prod_k (ia_k+ia_k)}.
$$
Even after reduction it does not matched the stated closed form. Where does the $\sum_{i\neq j}$ come from? I am unfamiliar with this notation.

Thank you!

答:
For a given $k$, the $k$-th inner term of the sum is (before taking the limit)
$$\begin{split}
(x-ia_k)\frac{x^2e^{itx}}{\prod_j(x^2+a_j^2)} &= \frac {(x - ia_k)}{x^2+a_k^2}
\cdot\frac{x^2e^{itx}}{\prod_{j \neq k}(x^2 + a_j^2)}\\
&=\frac {1}{x+ia_k}\cdot\frac{x^2e^{itx}}{\prod_{j\neq k}(x^2+a_j^2)}\\
\end{split}$$
where we used the fact that $x^2+a_k^2=(x-ia_k)(x+ia_k)$.

This implies
$$\lim_{x\rightarrow ia_k}(x-ia_k)\frac{x^2e^{itx}}{\prod_j(x^2+a_j^2)}=-\frac 1 {2ia_k}\frac{a_k^2e^{-ta_k}}{\prod_{j\neq k}(a_j^2-a_k^2)}$$
which yields the final result.


问:
Every countably compact, scattered $T_3$-space is sequentially compact
A space $X$ is called scattered provided every non-empty subspace $Y$ has an isolated point( with respect to the subspace topology on $Y$).

How to prove that:

Every countably compact, scattered $T_3$-space is sequentially compact?

Thanks for your help.

答:
It seems the following. Let $X$ be a countably compact, scattered regular (that is $T_3$ and $T_1$) space and $S=\{x_n:n\in\omega\}\subset X$ be a sequence. If the set $\overline S\backslash S$  is empty then $S$ is closed in $X$ and we put $Y=S$. In the opposite case the subspace $\overline S\backslash S$ has an isolated point $x_0$. Therefore there exists a neighborhood $U_0$ of the point $x_0$ in the space $\overline S$ such that $U_0\subset\{x_0\}\cup S$. Since the space $\overline S$ is regular, there exists an open  neighborhood $V_0$ of the point $x_0$ in the space $\overline S$ such that $\operatorname{cl}_{\overline S} V_0\subset U_0$. Put $Y=\operatorname{cl}_{\overline S} V_0$. In both cases $Y$ is a countable countably compact space, containing  infinitely many members of the sequence $S$ (some of them may coincide as points of $X$). Because $Y$ is a regular countably compact space with $G_\delta$-diagonal, by Chaber   Theorem is a metrizable compact (see, for instance, “Handbook of Set-Theoretic Topology 1984”, p.422 of the file or p.433 of the book). Because $Y$
contains  infinitely many members of the sequence $S$, we can construct a convergent subsequence of $S$ by them.