问:
Sets and the principle of inclusion and exclusion
I found an interesting problem regarding the principle of inclusion and exclusion.

Given $1985$ sets, each containing $45$ items, find their total union, if the union of any two is $89$.

This is what I did:
First, I found out the intersection of any two. Using the principle of inclusion and exclusion, I managed to find out that |$A_{1}$|$\cup$|$A_{2}$|=|$A_{1}$|+|$A_{2}$|-|$A_{1}$|$\cap$|$A_{2}$|. From this, we know that $89=45+45-x$, where $x=1$. The first thing that came to my mind was that this intersection could be the common intersection of all the sets:

A solution I found to the entire problem is that there is one element that every set has in common, and 44 elements that each set does not share with no other set. To clarify, one can imagine a flower, whose petals are individual sets, their only common item being the center of the flower. Each and every pair of sets therefore contains the required 89 elements ($44$ elements on one petal $+$ $44$ elements on the other petal $+ 1$ element in the centre), each set also contains $45$ elements. Therefore, the total sum of all the elements, or the union of all the sets, is $1985\times44+1$. But I do not know how to prove this with formulas and expressions. I am also not sure whether this is correct, because of what I found when I tried to solve a similar problem but only with three sets:

The rules of this problem are exactly the same as before, only now we have $3$ sets instead of $1985$. I found out two ways to arrange the elements of the sets which satisfy the rules, each of them having a different total union.

If we had a three-set Venn diagram, one solution is this:

1 element in the intersection of them all ($A \cap B \cap C$)

44 elements in each individual set but not in any intersection (in $A$,$B$,$C$)

The other solution is this:

1 element in each intersection of every pair, but not in the intersection of them all

43 elements in each individual set.

Their unions are different, because while in the first solution, the union is $44\times3+1$, while in the second solution, the union is $43\times3+3$. These are different numbers and therefore I am unsure about my solution to the original problem. My question is whether there is an analytical way to solve this problem, or whether I am missing something.

答:
Fix set $B=\{b_1,...,b_{45}\}$ and let $d_i$ be a number of other sets element $b_i$ is in it and let $d$ be a maximal $d_i$.

Then we have: $1984 =\sum_{i=1}^{1984}|A_i\cap B| \leq 45\cdot d \implies d\geq 45$. Thus there is an element $b$ in $B$ which appears at least in $46$ sets, say $A_1,A_2,...,A_{46}$ (one of them is $B$) and suppose there is a set $A=\{a_1,...,a_{45}\}$ which does not contain $b$.

Now for each $j\leq 46$ there is $a_i$ such that $a_i\in A_j\cap A$ and $a_i\ne b$. But then some $a \in A$ would be common to some $A_p$ and $A_q$ which means they have $2$ common elements. A contradiction, so $b$ apperas in every set.

Now by PIE we have:

$$n = 1985\cdot 45-{1985\choose 2} + {1985\choose 3} - {1985\choose 4}+..+{1985\choose 1985} $$

$$ n=1985\cdot 45 -(1-1)^{1985} +1-1985 = 1985\cdot 44+1$$


问:
Prove that every positive scalar multiple of a positive definite matrix is also positive definite

Prove that if $K$ is any positive definite matrix, then every positive scalar multiple $cK$,
$c > 0$, is also positive definite.

I feel like there's an easy solution to this but I'm stuck.
I know that if $K$ is a positive definite matrix then there's some form $q(x)=x^\intercal Kx>0$. How could I show that $x^\intercal cKx>0$? Can we just say that since $c>0$ the form $cq(x)$ must also be greater than $0$? I feel like that's not a good enough proof.

答:
Yes it suffices since $x^TcKx=c(x^TKx)>0$ or equivalently we also have

$$x^T(cK)x=(\sqrt cx^T)K(\sqrt c x)>0$$

ContentLicense: CC BY-SA 4.0


问:
Equation of the line that has $x$ and $y$ intercepts at $a$ and $b$.
Please can anyone help me with proving the following problem:

Show that the line that crosses the $X$-axis at $a \neq 0$ and the $Y$-axis at $b \neq 0$ has the equation $$\dfrac{x}a + \dfrac{y}b -1=0$$

I don't have an idea of where I should start from for proving the problem.

答:
Clearly, the equation $\dfrac{x}a + \dfrac{y}b - 1 =0 $ defines a straight line.

Also, two distinct points uniquely define a straight line.

Now check that $(a,0)$ and $(0,b)$ satisfy the equation.

This would imply $(a,0)$ and $(0,b)$ lie on the straight line.

Since $a \neq 0 \neq b$, these are two distinct distinct points.

ContentLicense: CC BY-SA 3.0


问:
Deriving Welford's Corrected Sum of Squares
I am studying Welford's paper on "Note on a Method for Calculating Corrected Sums of Squares and Products" and I am stuck on the following "corrected sum of squares" derivation:

I fully understand how we moved from Equation (1) to Equation (2) but I can't seem to see how we move from Equation (2) to Equation (3). When I expand Equation (2), I get:

which looks nothing like Equation (3) from above. Can anybody spot where I'm overlooking something obvious?

答:
The last term inside the square bracket of your expansion of equation ($2$) should be $\ \displaystyle\frac{1}{n^2}\big(x_n-m_{(n-1)}\big)^2\ $.  What you're probably missing is that $\ \displaystyle m_{(n-1)}=\frac{1}{n}\sum_{i=1}^{n-1}x_i\ $, by definition (on p.419 of Welford's paper), so
$$
\sum_{i=1}^{n-1}-\frac{2}{n}\big(x_i-m_{(n-1)}\big)\big(x_n-m_{(n-1)}\big)=0\ .
$$
Also,
$$
\sum_{i=1}^{n-1}\frac{1}{n^2} \big(x_n-m_{(n-1)}\big)^2=\left(\frac{n-1}{n^2}\right) \big(x_n-m_{(n-1)}\big)^2\ .
$$
If you apply these identities to your (corrected) expansion of your equation ($2$) you'll get your equation ($3$).


问:
Range of square root of rational function
How can I determine the range of the function $$f(x)=\sqrt{\frac{-x^2+2x+3}{18x-3x^3}}$$ without using limits or derivatives? I have factorised numerator and denominator, but nothing simplifies. I tried solving $y=f(x)$ for $x$, because the domain of the inverse function is the range of the initial function, but I am somehow stuck. My guess is that the range is all non-negative real numbers. How can I be sure that we reach all of them? If we have it for the fraction, we will also have it for the whole root. Can someone help me here? Thanks!

答: First the denominator has degree 3 while the numerator has degree 2.  For very large x, this is close to 0 so the minimum is 0.  The denominator is 0 for x= 0, $\sqrt{3}$, and $-\sqrt{3}$ while the numerator is not 0 for those x values so the function value can be arbritrarily large.  (I hope I haven't come too close to "limits" for you?)


问:
Which convex deltahedra are inscribable in the sphere?
A convex deltahedron in $\mathbb{R}^3$ is a convex polyhedron whose faces are all equilateral triangles. There exist precisely 8 convex deltahedra. Some examples are the regular tetrahedron, the regular octahedron, and the regular icosahedron. As regular polyhedra, these three can be inscribed in the sphere $\mathbb{S}^2$, meaning all of their vertices lie in the sphere. Are there any other inscribable deltahedra?

答:
None of the non-regular convex deltahedra can be inscribed in a sphere.

Proofs:

Triangular bipyramid: Consider inscribing an equilateral triangle in a sphere (this will be the "equatorial triangle" consisting of the vertices of degree 4 in the bpyramid). There are many ways to do this, but up to symmetry, they only depend on how far "up" or "down" an axis of the sphere the plane of the triangle moves. Because the "polar" vertices of the bipyramid have to be on this axis at equal distances from the plane of the triangle, we know that they have to be at antipodal points of the sphere, and so the center of the sphere aligns with the center of our equatorial triangle. But then (if we normalize to unit edge length) the triangle's equatorial vertices are at distance $\sqrt{3}/3$ from the center while its polar vertices are at distance $\sqrt{6}/3$ from the center.

Pentagonal bipyramid: Exactly the same argument as above, but now the equatorial vertices are further away from the center than the polar vertices (which you can easily verify with some trigonometry or coordinate-bashing). Note that the square bipyramid, AKA the regular octahedron, hits the happy medium between these two scenarios, so you can inscribe it in the sphere.

Snub disphenoid: Consider the two most distant edges of the shape. These four vertices form a tetrahedron which can only be inscribed in the sphere in one way, forcing the center of the sphere to coincide with the center of this tetrahedron by symmetry. Then the center of the sphere must line up with the origin in the Wikipedia list of coordinates for the shape. But you can easily check that the vertices are not all at the same distance from the origin, so they do not all lie on the surface of the sphere.

Triaugmented triangular prism: This polyhedron contains square pyramids, each of which can only be inscribed in the sphere in one way up to rotation about the surface of the sphere. But we know from the octahedron (which is a union of two square pyramids) that this inscription aligns the center of the square base of the pyramid with the center of the sphere. But the triaugmented triangular prism has three such square pyramids, each of whose square bases have different centers. So there is no consistent positioning of the center of any circumscribing sphere.

Gyroelongated square bipyramid: Same reasoning as above with its two opposing square pyramids.


答:
Hint: you have two points from the problem description: $(a, 0)$, and $(0, b)$. Can you write the equation of a line connecting them?

If you are still stuck:

Using slope-intercept form, we have:

$$y = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}x + b$$ where $b$ is the $y$-intercept and $(x_{1}, y_{1}), (x_{2}, y_{2})$ are two points on the line. Evaluating, then, we see:

$$y = \frac{b - 0}{0-a}x + b = \frac{-b}{a}x + b$$
$$\implies y + \frac{b}{a}x - b = 0$$
$$\implies \frac{1}{b}y + \frac{1}{a}x - 1 = 0$$
$$\implies \frac{x}{a} + \frac{y}{b} - 1 = 0$$

Title: Equation of the line that has $x$ and $y$ intercepts at $a$ and $b$.

Body:
Please can anyone help me with proving the following problem:

Show that the line that crosses the $X$-axis at $a \neq 0$ and the $Y$-axis at $b \neq 0$ has the equation $$\dfrac{x}a + \dfrac{y}b -1=0$$

I don't have an idea of where I should start from for proving the problem.


问:
Proof that $\frac{d}{dt} \log |A(t)| = \text{Tr}\left[A(t)^{-1} \frac{d}{dt} A \right]$
For a symmetric $n\times n$  matrix $A(t)$ that depends on the variable $t$, I want to prove that
$$
\frac{d}{dt} \log |A(t)| = \text{Tr}\left[A(t)^{-1} \frac{d}{dt} A(t) \right].
$$

Here, $|A|$ denotes the determinant of $A$ and $\text{Tr}[X]$ denotes the trace of $X$. How can I prove this equality?

The reason I ask this question: The equality is presented in equation C.22 of Pattern Recognition and Machine Learning of C. Bishop and the proof is left "as an exercise to the reader".

My attempt so far (to not clutter the notation, I leave out the dependency on $t$):

We know that $|A|=\prod_{i=1}^n \lambda_i$, where $\lambda_i$ denotes the $i$-th eigenvalue.
Hence, we can write
$$
\frac{d}{dt} \log|A|
=\frac{d}{dt}\log\prod_{i=1}^n\lambda_i
=\frac{d}{dt}\sum_{i=1}^n\log(\lambda_i)
=\sum_{i=1}^n\frac{d}{dt}\log(\lambda_i)
=\sum_{i=1}^n\frac{1}{\lambda_i} \frac{d}{dt}\lambda_i.
$$

The textbook further suggests to make use of the equalities
$$
A = \sum_{i=1}^n \lambda_i u_i u_i^T
\quad \text{and} \quad
A^{-1} = \sum_{i=1}^n \frac{1}{\lambda_i} u_i u_i^T,
$$
where $u_i$ denotes the $i$-th eigenvector.

I do not know how to proceed from here. I also do not know if this is going into the right direction.

答:
$\newcommand{\Tr}{\mathrm{Tr}}$
$\newcommand{\adj}{\mathrm{adj}}$

In fact, the equality holds as long as $\det(A(t)) > 0$ (the symmetry condition is not needed).

The key identity we need is that for any $A(t)$ ($\det(A(t))$ may equal to $0$):
\begin{align*}
\frac{d(\det(A(t)))}{dt} = \sum_{i = 1}^n\sum_{j = 1}^n \frac{d a_{ij}(t)}{dt}A_{ij}(t) = \Tr\left(\frac{d(A(t))}{dt}\mathrm{adj}(A(t))\right), \tag{1}
\end{align*}
where
\begin{align*}
\adj(A(t)) = \begin{pmatrix} A_{11}(t) & A_{21}(t) & \cdots & A_{n1}(t) \\
A_{12}(t) & A_{22}(t) & \cdots & A_{n2}(t) \\
\vdots & \vdots & \ddots & \vdots \\
A_{1n}(t) & A_{2n}(t) & \cdots & A_{nn}(t)
\end{pmatrix}
\end{align*}
is the adjugate matrix of $A(t)$ thus it satisfies
\begin{align*}
A(t)\adj(A(t)) = \det(A(t))I_{(n)}.
\end{align*}

When $\det(A(t)) > 0$ hence $A(t)$ is invertible, $\adj(A(t)) = \det(A(t))A(t)^{-1}$, whence by the chain rule we have
\begin{align*}
\frac{d\log(\det(A(t)))}{dt} &= \frac{1}{\det(A(t))}\frac{d(\det(A(t)))}{dt}
= \frac{1}{\det(A(t))}\Tr\left(\frac{d(A(t))}{dt}\mathrm{adj}(A(t))\right) \\
&= \frac{1}{\det(A(t))}\Tr\left(\frac{d(A(t))}{dt}\det(A(t))A(t)^{-1}\right)
 = \Tr\left(\frac{d(A(t))}{dt}A(t)^{-1}\right).
\end{align*}

The proof of the first equality in $(1)$ is subtle and quite lengthy. If you are interested, I may post it as an appendix.


问:
Finding the inverse of $y=\mu-\sigma\log\left[\frac{e^{-x}}{1-e^{-x}}\right]$

What is the inverse of this function?
$$y=\mu-\sigma\log\left[\frac{e^{-x}}{1-e^{-x}}\right]$$

I've only calculated up to here:

$$\log\left[\frac{e^{-x}}{1-e^{-x}}\right]=(\mu-y)/\sigma$$

and I guess that $\log[e^{-x}]=-x$ but can't find something for $\log(1+e^{-x})$

Body: $$\log\left[\frac{e^{-x}}{1-e^{-x}}\right]=\log\left[\frac1{e^x-1}\right]=-\log[e^x-1]=(\mu-y)/\sigma$$which gives$$x=\ln[1+e^{(y-\mu)/\sigma}]$$


问:
Why is $\mathbb{Z}[X]$ not a euclidean domain? What goes wrong with the degree function?
I know that $K[X]$ is a Euclidean domain but $\mathbb{Z}[X]$ is not.

I understand this, if I consider the ideal $\langle X,2 \rangle$ which is a principal ideal of $K[X]$ but not of $\mathbb{Z}[X]$. So $\mathbb{Z}[X]$ isn't a principal ideal domain and therefore not an Euclidean domain.

But I don't understand this, if I consider the definition of Euclidean domains. Basically, a Euclidean domain is a ring where I can do division with remainders. For polynomial rings, the Euclidean function should be the degree of the polynomials. What's the crucial difference between $K[X]$ and $\mathbb{Z}[X]$ with respect to this?

I already did exercises involving polynomial division in $\mathbb{Z}[X]$, so clearly I must be missing something here.

答: $\mathbb{Z}[X]$ is not PID since $\langle 2,x\rangle$ is not principal. So if $\mathbb{Z}[X]$ were euclidian it would be PID.


问:
Is every proper subgroup of $(\mathbb{Z}/n\mathbb{Z})^\times$ cyclic?
I know that $(\mathbb{Z}/n\mathbb{Z})^\times$ is not necessarily cyclic. But is it true that all of its proper subgroups are cyclic? If not, for which $n$ is it true?

答:
No.  But it would be true if you take an $n$ for which $(\Bbb Z/n\Bbb Z)^\times$ is cyclic.  That would be, it turns out, any $n=1,2,4,p^k$ or $2p^k$ for $p$ an odd prime.

Those are the only $n$ that work,  with the exception, that is, of $n=8$, as @ArturoMagidin points out. Because if a finite abelian group is not cyclic, or of order $p^2$, it contains a proper subgroup of the form $\Bbb Z_p\times\Bbb Z_p$ for some $p$.


答:
Let $M \in (0, +\infty)$. $M=0$ is obviously hit, for example by $x = 3$.

We want to solve the equation $$\sqrt{\frac{x^2 - 2x - 3}{3x^3 - 18x}} = M.$$Squaring and then transforming the expression gives us $$3M^2 x^3 - x^2 + (2-18M^2)x + 3 = 0.$$For $M > 0$ this is a cubic equation, so it has at least one real root. Let $x_0$ be such a root. Obviously $x_0 \notin \{0, \sqrt{6}, -\sqrt{6}\}$, so $$\frac{x_0^2 - 2x_0 - 3}{3x_0^3 - 18x_0}$$ is a well-defined number equal to $M^2$, so it's positive. Therefore, its square root is defined and equal to $M$.

Title: Range of square root of rational function

Body: How can I determine the range of the function $$f(x)=\sqrt{\frac{-x^2+2x+3}{18x-3x^3}}$$ without using limits or derivatives? I have factorised numerator and denominator, but nothing simplifies. I tried solving $y=f(x)$ for $x$, because the domain of the inverse function is the range of the initial function, but I am somehow stuck. My guess is that the range is all non-negative real numbers. How can I be sure that we reach all of them? If we have it for the fraction, we will also have it for the whole root. Can someone help me here? Thanks!


问:
proof for that exactly one normal can be drawn to a parabola from an exterior point.
It is well known that a maximum of three normals can be drawn from the interior of a parabola. But I am interested in the proof for the fact that exactly one normal can be drawn to a parabola from an exterior point wrt to the parabola.

So far I have done this.

WLOG we consider $y^2=4ax$.

Let the exterior point be $(h,k)$.

We must have $$k^2-4ah> 0\tag1$$

Also the equation of a normal to a parabola is $$y=mx-2am-am^3$$
so $(h,k)$ satisfies $$am^3+2am-mh+k=0$$

All I have to do is show that there is only one real root possible for this cubic using $(1)$.
Can anyone complete the proof?

Alternative methods are welcome!

NoteBy a normal i mean that a segment that is external to parabola

Thank you!

答:
From the cubic

$$ax^3 +(2a-h)x +k =0$$

and using the conventions in this paper, one has

$$\begin{align*} x_N &=0 \\
\\
\delta^2 &= \dfrac{h-2a}{3a} \\
\\
y_N &= k \\
\\
h' &= 2a\left(\dfrac{h-2a}{3a}\right)^\frac{3}{2}\\
\end{align*}$$

For simplicity and WLOG, assume $a >0$.

For the case of $h \le 2a$, then $\delta^2 \le 0$ and the cubic has no real turning points, and thus only 1 real root.

For the case of $h \gt 2a$, then $\delta^2 \gt 0$ and $h' \in \mathbb{R}$ and the cubic has two turning points.  In this case, to have only 1 real root, one needs to satisfy

$$\left|\dfrac{-y_N}{h'}\right| = \left|\dfrac{-k}{2a\left(\dfrac{h-2a}{3a}\right)^\frac{3}{2}}\right| \gt 1$$

or equivalently

$$\begin{align*}\left|\dfrac{-k}{2a\left(\dfrac{h-2a}{3a}\right)^\frac{3}{2}}\right|^2 &\gt 1^2\\
\\
k^2 &\gt 4a^2\left(\dfrac{h-2a}{3a}\right)^3\\
\\
k^2 - 4a\left[a\left(\dfrac{h-2a}{3a}\right)^3\right] &\gt 0 \\
\end{align*}$$

Now, to be consistent with the constraint in (1), one needs to show that

$$\begin{align*}a\left(\dfrac{h-2a}{3a}\right)^3 &\le h \\
\\
\left(h-2a\right)^3&\le 27a^2h \\
\end{align*}$$

which is certainly not true for $h \gg 2a$.

Therefore, it appears the constraint in (1) is not sufficient to guarantee that there is only one real root for $h \gg 2a$.


答:
From $A(t)$ diagonalizable and $tr(BC)=tr(CB)$ we have $\log \det A(t)=tr \log A(t)$.

Next, $(A(t)^n)'=\sum_{k=0}^{n-1} A^k A'(t)A^{n-1-k}$ gives that for $f(z)=\sum_{n\ge 0} c_n z^n$ analytic on a disk containing $\|A(t)\|$ then $$tr(f(A(t)))'=\sum_{n=0}^\infty c_n \sum_{k=0}^{n-1} tr(A^k A'(t)A^{n-1-k})=\sum_{n=0}^\infty c_n \sum_{k=0}^{n-1} tr(A^{n-1}A'(t))$$ $$=tr( f'(A(t)) A'(t))$$
$f(z)=\log z$ is not analytic on such a disk, but by analytic continuation in $s$ of $tr(f(I+s(A(t)-I)))'$ it will still hold, ie. $$tr(\log(A(t)))' = tr(A(t)^{-1}A'(t))$$

No need that the eigenvalues of $A(t)$ are positive, just that $A(t)$ is differentiable, $\det(A(t))\ne 0$ and our choice of $tr \log A(t)$ is continuous in $t$.

Title: Proof that $\frac{d}{dt} \log |A(t)| = \text{Tr}\left[A(t)^{-1} \frac{d}{dt} A \right]$

Body:
For a symmetric $n\times n$  matrix $A(t)$ that depends on the variable $t$, I want to prove that
$$
\frac{d}{dt} \log |A(t)| = \text{Tr}\left[A(t)^{-1} \frac{d}{dt} A(t) \right].
$$

Here, $|A|$ denotes the determinant of $A$ and $\text{Tr}[X]$ denotes the trace of $X$. How can I prove this equality?

The reason I ask this question: The equality is presented in equation C.22 of Pattern Recognition and Machine Learning of C. Bishop and the proof is left "as an exercise to the reader".

My attempt so far (to not clutter the notation, I leave out the dependency on $t$):

We know that $|A|=\prod_{i=1}^n \lambda_i$, where $\lambda_i$ denotes the $i$-th eigenvalue.
Hence, we can write
$$
\frac{d}{dt} \log|A|
=\frac{d}{dt}\log\prod_{i=1}^n\lambda_i
=\frac{d}{dt}\sum_{i=1}^n\log(\lambda_i)
=\sum_{i=1}^n\frac{d}{dt}\log(\lambda_i)
=\sum_{i=1}^n\frac{1}{\lambda_i} \frac{d}{dt}\lambda_i.
$$

The textbook further suggests to make use of the equalities
$$
A = \sum_{i=1}^n \lambda_i u_i u_i^T
\quad \text{and} \quad
A^{-1} = \sum_{i=1}^n \frac{1}{\lambda_i} u_i u_i^T,
$$
where $u_i$ denotes the $i$-th eigenvector.

I do not know how to proceed from here. I also do not know if this is going into the right direction.


答:
The $x$ intercept is given by $(a, 0)$, and the $y$ intercept is given by $(0, b)$. And we know $a \neq 0, b\neq 0$, so the points are distinct.

All you need to do is substitute these values into your equation to see that equality holds: i.e., that they lie on the line defined by $$\frac xa + \frac yb - 1 = 0$$

$$(a, 0): x = a, y = 0 \implies  \frac aa + \frac 0b - 1 = 0\implies \frac aa = 1\implies 1 = 1\quad \checkmark$$

$$(0, b): x = 0, y = b \implies \frac 0a + \frac bb - 1 = 0 \implies \frac bb = 1 \implies 1 = 1 \quad \checkmark$$

Indeed: These are unique points lying on the line given by $$\frac xa + \frac yb - 1 = 0$$


问:
Why is $SL(n, \mathbb{R})$ the kernel of $\det : GL(n, \mathbb{R}) \mapsto\Bbb R^*$?
The special linear group of invertible matrices is defined as the kernel of the determinant of the map:

$$\det:GL(n,\mathbb{R}) \mapsto \mathbb{R}^*$$

In my mind the kernel of a linear map is the set of vectors that are mapped to the zero vector. So the map above would contain all the matrices that have determinant zero (which doesn't make sense since the codomain of the function excludes zero)? But isn't the special linear group made of matrices with determinant 1?

答: But $\det$ is not a linear map. It is a group homomorphism. Its kernel is$$\left\{M\in GL(n,\Bbb R)\,\middle|\,\det(M)=1\right\},$$since $1$ is the identity element of $\bigl(\Bbb R\setminus\{0\},\times\bigr)$.


答:
$\mathbb{R}^∗$ is the multiplicative group, so the identity is $1$ - in particular the kernel is the elements of the group homomorphism sent to it.

NB: the determinant map is not linear and indeed the group operation on $GL_2(\mathbb{R})$ is not abelian so it's clearly not a vector space in a naive way.

Title: Why is $SL(n, \mathbb{R})$ the kernel of $\det : GL(n, \mathbb{R}) \mapsto\Bbb R^*$?

Body:
The special linear group of invertible matrices is defined as the kernel of the determinant of the map:

$$\det:GL(n,\mathbb{R}) \mapsto \mathbb{R}^*$$

In my mind the kernel of a linear map is the set of vectors that are mapped to the zero vector. So the map above would contain all the matrices that have determinant zero (which doesn't make sense since the codomain of the function excludes zero)? But isn't the special linear group made of matrices with determinant 1?


问:
Writing an expression using logic
Write an expression using letters $\land, \lor, and$ $\neg$ which has the following truth table:

$$\begin{array}{ccc|c}
P&Q&R&???\\ \hline
T&T&T&F\\
T&T&F&T\\
T&F&T&F\\
T&F&F&T\\
F&T&T&T\\
F&T&F&F\\
F&F&T&F\\
F&F&F&F
\end{array}
$$

How would you in general solve for ??? I was told there is a mechanical way of doing it but I don't see it.

答:
Entered by truthtable:

```<code>??? = P' Q R + P Q' R' + P Q R';
</code>```

Minimized:

```<code>??? = P R' + P' Q R;
</code>```

Related Karnaugh-Veitch map as screenshot from KVD Pro:

Title: Writing an expression using logic

Body:
Entered by truthtable:

??? = P' Q R + P Q' R' + P Q R';

Minimized:

??? = P R' + P' Q R;

Related Karnaugh-Veitch map as screenshot from KVD Pro:


答:
Suppose that there is $B(t)$ such that
$$ A(t)=e^{B(t)}. $$
Then
$$ \frac{d}{dt}\log |A(t)|=\frac{d}{dt}\log |e^{B(t)}|=\frac{d}{dt}\log e^{\text{Tr}B(t)}=\frac{d}{dt}\text{Tr}B(t). $$
On the other hand,
$$ \text{Tr}\left[A(t)^{-1} \frac{d}{dt} A(t) \right]=\text{Tr}\bigg[e^{-B(t)}e^{B(t)}B'(t)\bigg]=\text{Tr}B'(t)=\frac{d}{dt}\text{Tr}B(t). $$
So
$$ \frac{d}{dt}\log |A(t)|=\text{Tr}\left[A(t)^{-1} \frac{d}{dt} A(t) \right].$$


问:
Proof that The diameter of graph $G$ is at most $2$ and $G$ doesn't have a cycle of length $4$
I am new to graph theory and I try to learn how to be better at making proofs and thinking but don't have any idea how to solve the following problem. Could you help me please?

The problem says that there is a network where $2$ users have exactly one common acquaintance. $G = (V,E)$ is the graph used for this network. (for each two users $u,v$ in $V$, $uv$ is in $E$ if $u$ and $v$ doesn't know each other).

I have to come up with a proof to show that the diameter of $G$(named $d(G)$) is at most $2$ and $G$ doesn't have a cycle of length $4$.

Thank you!

答:
Diameter is the largest distance between any two vertices. Say the diameter of your graph is the distance between vertices $v_1,v_2$. It is given that they have exactly one common neighbour, so $v_1,v_2$ are connected by a path of length $2$ that goes through the common neighbour. This means the distance between $v_1,v_2$ (length of the shortest path joining them) is not more $2$.

For the second part, suppose you have a cycle of length $4$ i.e. $v_1,\color{blue}{e_{12}},v_2,\color{blue}{e_{23}},v_3,\color{blue}{e_{34}},v_4,\color{blue}{e_{41}}$ where all vertices and edges are distinct. Then $v_1,v_3$ have two common neighbours in $v_2,v_4$ which is a contradiction.


问:
Prove that $\int_0^\infty\frac{\sin(t)}{\sinh(t)}\mathrm{d}t = \sum_{n=0}^\infty\frac{2}{(2n+1)^2+1}$
In order to prove that

$$\int_0^\infty\frac{\sin(t)}{\sinh(t)}\mathrm{d}t = \sum_{n=0}^\infty\frac{2}{(2n+1)^2+1}$$

I thought about writting that

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

$\sinh(x)=\sum_{n=0}^\infty\frac{x^{2n+1}}{(2n+1)!}$

But I don't really know what I can do then.
If you could give me a hint, I would be thankful !
(Sorry for my Eng.)

Body:
$$\begin{split}
\int_0^\infty\frac{\sin(t)}{\sinh(t)}\mathrm{d}t  &= 2 \int_0^{+\infty} \frac {\sin t}{e^t - e^{-t}}dt \\
&= 2\int_0^{+\infty}\frac{e^{-t}\sin t}{1-e^{-2t}}dt\\
&= 2\int_0^{+\infty}e^{-t}\sin t\sum_{n\geq 0}e^{-2nt}dt\\
&=2\sum_{n\geq 0} \int_0^{+\infty}e^{-(2n+1)t}\frac{e^{it}-e^{-it}}{2i} dt\\
&=-i\sum_{n\geq 0} \left ( \frac 1 {i-2n-1} +\frac 1 {2n+1+i}\right)\\
&=-i \sum_{n\geq 0} \frac{-2i}{1+(2n+1)^2}\\
&= \sum_{n\geq 0} \frac{2}{1+(2n+1)^2}
\end{split}$$


问:
A basic question about $\operatorname{supp}f$ (support of f).
Is it true that $\operatorname{supp}f$ is the complement of the biggest open set where $f=0
$?

Here $\operatorname{supp}f=$ {$x\in \Bbb R^n  ; f(x)\not=0$} and $f\in C$ (collection of continuous maps from $\Bbb R^n \to \Bbb R$)

答:
Since your question is tagged $L^p$-spaces (which are spaces of equivalence classes of functions), one should distinguish between

Support of a function $f$, which is the complement of the largest open set on which $f$ is zero identically.

Support of an equivalence class $f\in L^p$, which is  the complement of the largest open set on which (every representative of) $f$ is zero almost everywhere.

In first case, the  existence of largest such open set follows by taking the union of all open sets that qualify. In the second case one has to be more careful: assume that there is a countable base of the topology and take the union of all base elements that meet the requirement.

The second notion of support is sometimes called essential support, I think.


问:
Cut theorem for tableau?
Cut theorem is

Suppose Γ |- ϕ and Γ ∪{ϕ} |- ψ.Then Γ |- ψ.

For natural deduction, it's simply a corollary of deduction theorem.

PROOF.Suppose we have Γ |- ϕ and Γ ∪ {ϕ} |- ψ.Then by the deduc-tion theorem we have Γ |- ϕ → ψ.Thus we may combine the derivations of ϕ and ϕ → ψ from Γ to get a derivation of ψ from Γ using →-E.

However, when we want to prove it for tableau in the same way, we go wrong.

However, given that we have deduction theorem for tableau, it's strange given Γ |- ϕ and Γ ∪{ϕ} |-  ψ, i. e. Γ |- ϕ and Γ |- ϕ → ψ, we don't have Γ |-ψ.

What goes wrong here?

答:
There is a result in R.Smullyan, First-order Logic (1968):

if $\Gamma_1 \cup \{ \phi \}$ closes and $\Gamma_2 \cup \{ \lnot \phi \}$ closes, then $\Gamma_1 \cup \Gamma_2$ closes.

Apply it to $\Gamma_1=\Gamma$ and $\Gamma_2= \Gamma \cup \{ \lnot \psi \}$.

Title: Cut theorem for tableau?


问:
How to solve this congruence class equation?
how to solve for this equation in $\mathbb{Z}$12

$$
[8][x]+[3][y] = [9]\\
[6][x]+[5][y]=[2]
$$

答: Hint: if it had a solution mod $12$ then the solution persists mod $2,\,$ where it is inconsistent.


问:
Check if $\frac{x'^{2}}{2}-\sin(x)$ is a lyapunov function for $x''=\cos(x)-0.1x'$ near $(x,x')=(\frac{\pi}{2},0)$

Check if $\frac{x'^{2}}{2}-\sin(x)$ is a lyapunov function for $x''=\cos(x)-0.1x'$ near $(x,x')=(\frac{\pi}{2},0)$

I rewrote this into a system:
$x'=y$, $y'=\cos(x)-0.1y$, $F(x,y)=\frac{y^2}{2}-\sin(x)$,

And want to show that this is lyapunov function near $(x,y)=(\frac{\pi}{2},0)$

And then $F'(x,y)=yy'-\cos(x)x'=y\cos(x)-0.1y^2-y\cos(x)=-0.1y^2$

which is strictly decreasing so the sign doesn't change and $(\frac{\pi}{2},0)$ is a critical point of my system.

So do I know this is a lyapunov function?

This is my definition of Lyapunov function:

I have this lyapunov stability theorem,
Assume $U$ is an open neighbourhood of $x^*, F:U\to \mathbb{R}^n$ and $F(x^*)=0$. Further assume that there is a $C^1$-function $V:U\to\mathbb{R}$ satisfying $V(x^*)=0$, $V(x)>0$ for $x\neq x^*$. Then, V is called a lyapunov function for $x'=F(x)$ and:

a. If $V'(x)\leq 0$ for all $x\in U$, then $x^*$ is L-stable
b. If $V'(x)<0$ for all $x\in U\{x^*\}$, then $x^*$ is asymptotically stable.
c. If $V'(x)>0$ for all $x\in U\{x^*\}$ then $x^*$ is unstable.

答:
Let's transform the ODE into an SDE:

$y_1=x,y_2=x'\Rightarrow y_1'=y_2,y_2'=x''=\cos(x)-0.1x'=\cos(y_1)-0.1y_2\Rightarrow\begin{cases}y_1'=y_2\\y_2'=\cos(y_1)-0.1y_2\end{cases}$

Then the function would be $V(y_1,y_2)=\dfrac{y_2^2}{2}-\sin(y_1)$. We need points near $(\frac{\pi}{2},0)$, so we could consider points $(y_1,y_2)=(\frac{\pi}{2}+r,0)$ with $|r|>0$ as small as we want. For those points $V(y_1,y_2)=-\sin(\frac{\pi}{2}+r)<0$ for $0<r<\frac{\pi}{2}$, so $V$ can't be a lyapunov function.


问:
Intuition of generalized eigenvector.
I was trying to get an intuitive grasp about what the the generalized eigenvector intuitively is. I read this nice answer, so I understand that in the basis given by the generalized eigenvectors, a jordan block is a linear map that is the sum of a stretch by a factor $\lambda$ (eigenvalue associated to the block) and a "collapse", but I don't understand the conclusion on what these famous generalized eigenvectors actually are...

Thus the kernel of $(T−λI)k$ picks up all the Jordan blocks associated with eigenvalue $λ$ and, speaking somewhat loosely, each generalized eigenvector gets rescaled by $λ$, up to some "error" term generated by certain of the other generalized eigenvectors.

Maybe someone that actually understand the last argument can care to explain with some more detail? Thank you.

答:
First, I want to introduce the Jordan–Chevalley decomposition:

In essence it formalizes what you said: For every trigonalizable map $f \in End(V)$ with transformation Matrix $A$, there exists a diagonal Matrix $D$ (the stretching part) and an nilpotent Matrix $N$ (the “collapse” part) such that, $A = D + N$

(which makes intuitive sense: $D$ will have the same Main diagonal as $A$, so $A – D$ will be an strictly triangular matrix $\Leftrightarrow$ nilpotent).

Now consider a Matrix $B$ with just one Eigenvalue $\lambda$:

Because $\lambda$ is the sole Eigenvalue of $B$, the diagonal Matrix $D$ in the Jordan–Chevalley decomposition will just have $\lambda$ on its diagonal, so: $\lambda * \mathbb{I}_n = D$.

Now we rearrange:

$B = D + N$ to $B - D = N \Leftrightarrow (B - \lambda * \mathbb{I}_n) = N$.

Because $N$ is nilpotent, it follows that for an sufficient $k$, every Vector in the generalized Eigenspace will be in the Kernel of $(B - \lambda * \mathbb{I}_n)^k$, because for a sufficient $k$, $(B - \lambda * \mathbb{I}_n)^k$ will be the zero map.

Important: If you have more than one Eigenvalue, you need to restrict the Matrix to the respective generalized Eigenspace to reduce it to the above situation.

In this case, $(A - \lambda * \mathbb{I}_n)^k$ will not map every vector in V to zero, but just Elements of the generalized Eigenspace associated to $\lambda$
(Because you applied the Jordan–Chevalley decomposition to the generalized Eigenspace and not to V)

This is what he meant by “$(A - \lambda * \mathbb{I}_n)^k$ picks up all the Jordan blocks associated with eigenvalue Lambda“.

I think the "error" term he wrote about, are the Jordanchains / the way you calculate them:

$A * v_j = v_{j-1} + \lambda * v_j \Leftrightarrow v_j * (A - \lambda * \mathbb{I}_n) = v_{j-1}$

So every Vector not only gets scaled, but you also add a Vector. This added Vector is the “error Term".

Title: Intuition of generalized eigenvector.


问:
Find the matrix B such that it's null space is $V=\operatorname{span}(v_1,v_2,v_3)$

Find the matrix B such that it's null space is $V=\operatorname{span}\{(1,1,0),~(1,2,0),~(1,5,0)\}$.

Since this set of vectors is linearly dependent, I take $\operatorname{span}((1,1,0),~(1,2,0))$ as the linearly independent set. Then, if I think about special solutions, I'd have two, and one pivot variable. Since $(1,1,0)$ is the solution you get by considering $x_1=0,x_2=1$ and $(1,2,0)$ taking $x_1=1,x_2=0$, then the column with the only pivot of the matrix is the third one, $B^3=(0,0,1)^T$

However, I can't figure out how to complete the $3\times 3$ matrix using these special solutions.

答: The span of those three vectors is $S=\{(x,y,0)\mid x,y\in\Bbb R\}$. So, you can take $B=\left[\begin{smallmatrix}0&0&1\end{smallmatrix}\right]$. It is clear that\begin{align}B.(x,y,z)=0&\iff z=0\\&\iff(x,y,z)\in S.\end{align}


问:
Mapping from quaternion Jacobian to geometric Jacobian
I have a Jacobian $J_{A}(q)$ which maps from a robot's joint velocities to the robot's end effector time derivative:
$$\dot{x} = J_{A}(q)\cdot \dot{q}$$

$x \in \mathbb{R}^{7\times 1}$ is the end-effector representation, where the first 3 elements are Cartesian coordinates and the remaining 4 are the orientation as a quaternion. I would like to convert my analytic Jacobian to a geometric Jacobian $J_{G}(q)$, where $\dot{x_{G}} = J_{G}(q)\cdot \dot{q}$ gives a vector $\mathbb{R}^{6\times 1}$ which is the linear velocities and angular rates of the end-effector.

From what I've seen, there exists a mapping $E$ such that $\dot{x_{G}} = E\cdot J_{A}(q)\cdot \dot{q}$.

I tried to derive $E$ and got the following result. Let $\xi = [\xi_{0}, \xi_{1}, \xi_{2},  \xi_{3}]^{T}$ be the quaternion orientation of the end-effector.
$$
E = \begin{bmatrix}
I_{3\times 3} & 0 \\
0 & 2H
\end{bmatrix}\\
H = \begin{bmatrix}
-\xi_{1} & \xi_{0} & -\xi_{3} & \xi_{2} \\
-\xi_{2} & \xi_{3} & \xi_{0} & -\xi_{1} \\
-\xi_{3} & -\xi_{2} & \xi_{1} & \xi_{0}
\end{bmatrix}
$$

However, when implementing this solution and comparing the end-effector linear and angular rates against the ground truth, it is incorrect for the angular rates (linear rates are okay). Am I doing something wrong here? I am confident the analytic Jacobian is correct.

答:
Turns out the source where I got the $H$ matrix was incorrect. I worked it out by hand using this link and got that:
$$
E = \begin{bmatrix}
I_{3\times 3} & 0 \\
0 & -2H
\end{bmatrix}\\
H = \begin{bmatrix}
-\xi_{1} & \xi_{0} & \xi_{3} & -\xi_{2} \\
-\xi_{2} & -\xi_{3} & \xi_{0} & \xi_{1} \\
-\xi_{3} & \xi_{2} & -\xi_{1} & \xi_{0}
\end{bmatrix}
$$

Now the results are correct.


问:
Limit of $\mathbb{Z}/n\mathbb{Z}$
This is a super weird question coming from someone who is just starting out in mathematics.

If $n$ is a positive integer, one can notice that as $n$ grows, $\mathbb{Z}/n\mathbb{Z}$ starts to look more and more like $\mathbb{N}$. If I were a small integer $k$, the entire group would begin to look more and more indistinguishable from $\mathbb{N}$ as $n$ grows - my additive inverse is getting further and further away. Is this idea something that is studied or formalized anywhere?

答:
This can be formalized to some extent in a few different ways depending on what you want to do. In additive number theory it's common to model problems in $\mathbb{Z}$ as problems in $\mathbb{Z}/n$ for large $n$; the idea is that e.g. if all you're doing is adding up a bounded number of bounded elements it doesn't matter which of the two you do it in, if $n$ is large enough. This can be formalized using concepts like local groups or Freiman homomorphisms; this is not at all my area though so I don't know much more than that.

In the areas I'm more familiar with it is possible to rigorously define a particular kind of "limit" of the groups $\mathbb{Z}/n$, namely their categorical limit. This produces a very interesting group called the profinite integers $\widehat{\mathbb{Z}}$, which show up in various places, for example as the absolute Galois group of a finite field. These are a kind of "completion" of the integers; in particular they're uncountable, but $\mathbb{Z}$ still sits inside as a dense subgroup.

Title: Limit of $\mathbb{Z}/n\mathbb{Z}$


问:
There exist a degree $n+1$ polynomial of best approximation if there exist a degree $n$ polynomial of best approximation
This is a problem from Mathematical Analysis by Zorich, from the chapter on continuous functions.

Definition:
Let $P_n$ be a polynomial of degree $n$. For a function $f:[a,b]\to\mathbb{R}$, Let $\Delta(P_n) = \sup_{x\in[a,b]} |f(x)-P_n(x)|$. and $E_n(f) = \inf_{P_n} \Delta(P_n)$. A polynomial $P_n$ is the best approximation of degree $n$ of $f$ is $\Delta(P_n) = E_n(f)$.

I have already proved the following:

There exist a polynomial $P_0(x) = a$ of best approximation of degree 0.

If $Q_\lambda(x) = \lambda P_n(x)$ for some fixed polynomial $P_n$. Then there exist a polynomial $Q_{\lambda_0}$ such that $\Delta(Q_{\lambda_0}) = \min_{\lambda\in \mathbb{R}} \Delta(Q_\lambda)$

I'm stuck on proving the following:

If there exists a polynomial of best approximation of degree $n$, there also exists a polynomial of best approximation of degree $n+1$.

My intuition is to prove $\lambda_0 x^{n+1}+P_n$ is the $n+1$ best approximation for $f$. Where $\Delta(\lambda_0 x^{n+1}) =  \min_{\lambda\in \mathbb{R}} \Delta(\lambda x^{n+1})$ and $\Delta(P_n) = E(f(x) - \lambda_0 x^{n+1})$.

I don't know if this approach is right. I'm stuck on how to proceed.

答:
I think Zorich's original idea on induction is to establish a mapping from $(P_n,\lambda)=(a_1,...,a_n,\lambda)\mapsto \mathbb{R}$. Prove it's a continuous mapping (similar to part b) on a compact interval (using the fact from part b and our inductive assumption for $n$).

But the claim is wrong since $\lambda$ can be $0$, where we may not extend from $n$ to $n+1$. A very simple counter-example is $f(x)=0$ on $[-1,1]$. There exists a polynomial of best approximation of degree $0$, that is $P_0(x)=0$. But there is not any polynomial $P_1(x)$ of best approximation with degree $1$.


答:
If you define $T$ by $T(1,1,0)=T(1,2,0)=0$ and $T(0,0,1)=(x,y,z)$, where not all of $x,y,z$ are zero, you will have one.  Giving the values on a basis determine a linear transformation.

This $T$ will go from $\Bbb R^3$ to $\Bbb R^3$.

Title: Find the matrix B such that it's null space is $V=\operatorname{span}(v_1,v_2,v_3)$

Body:
Find the matrix B such that it's null space is $V=\operatorname{span}\{(1,1,0),~(1,2,0),~(1,5,0)\}$.

Since this set of vectors is linearly dependent, I take $\operatorname{span}((1,1,0),~(1,2,0))$ as the linearly independent set. Then, if I think about special solutions, I'd have two, and one pivot variable. Since $(1,1,0)$ is the solution you get by considering $x_1=0,x_2=1$ and $(1,2,0)$ taking $x_1=1,x_2=0$, then the column with the only pivot of the matrix is the third one, $B^3=(0,0,1)^T$

However, I can't figure out how to complete the $3\times 3$ matrix using these special solutions.


问:
Var(X_2months) = 2 Var(X_1month) with independent random variables?
My instructor insists that
$$\mathrm{Var}(X_{2\text{months}}) = 2 \mathrm{Var}(X_{1\text{month}}) $$
with independent random variables $X_{1\text{month}}$ and $X_{2\text{months}}$, but generally with a random variable $Y$
$$\mathrm{Var}(2Y) = 2^2\mathrm{Var}(Y).$$
So
$$\mathrm{Var}(X_{2\text{months}}) = \mathrm{Var}(2X_{1\text{month}}) = 2^2 \mathrm{Var}(X_{1\text{month}}).$$
Which one is right and why?</p>

答:
The assumption made here is that $X_{2\text{months}}$ is equal to the sum of two copies of $X_{1\text{month}}$, say $X_{\text{July}}$ and $X_{\text{August}}$. Then
$$ \mathrm{Var}[X_{2\text{months}}] = \mathrm{Var}[X_{\text{July}} + X_{\text{August}}] = \mathrm{Var}[X_{\text{July}}] + \mathrm{Var}[X_{\text{August}}] = 2\mathrm{Var}[X_{1\text{month}}].$$


问:
Relationship between the length of the tangent line through a point on sphere and great-circle distance
As an aviator I'm familiar with the concept of great-circle navigation because when we fly a route between 2 points on the globe we know the shortest distance between these two points is the great circle distance.

I'm developing a navigation app in Google Earth and I need to calculate the shortest distance from the surface of the "spherical" Earth to any point on the tangent line through A (origin) when flying the great circle path.

Also, I'm using a mean earth radius of 6,371.009 km for WGS84 ellipsoid.

Just to be clear, I'd like to refer to the diagram in the following link:

http://www.alaricstephen.com/main-featured/2017/5/22/the-haversine-formula

I use the Haversine formula to calculate the distance, d, between the points A and D (see diagram). What I'd like to calculate is the distance D to E as a function of d.

In the diagram this is referred to as the external secant (exsec) which is the portion DE of the secant exterior to the circle.

答:
HaverSine Formula is used routinely to compute long distances in navigation along shortest path great circles of the Earth between two points of given latitude and longitude.

First find $d$ on the Earth. Next air distance along a tangent if the flight point $B$ is above the Earth: $ t= r \tan \dfrac{d}{r}.$

$-------------------------$

After clarification the above can be ignored.

For perfect sphere model of earth it is simple trig. calculation.

Distance $ AD= r \theta =  $ the arc distance you calculated using Haversine formula along a great circle arc of earth radius $=r$ as shown. Calculate $ \theta $ in radians in the plane of kite shape $OAEB$ if we imagine $B$ on another tangent point below. We have $  \theta= \dfrac{\text{arc} AD}{r}$

Central dimension is length $OE$. From this subtract earth radius.

$$ DE = r \sec \theta - r\; = r (\sec \theta -1 )$$

This is the red height above target/destination/landing place which should vanish on landing at $D$. It is indicated by exsec in the supplied link   for unit earth radius.


问:
$\omega$-types and general $\alpha$-types, where $\alpha$ is infinite, in saturated models
Let $\mathcal{M}$ be an $\mathcal{L}$-structure and define $S_\omega^\mathcal{M}(A)$ be the set of complete $\omega$-types with parameters in $A$ (ie the set of complete types in $\omega$-many variables). Is it true that if $\mathcal{M}$ is $\kappa$-saturated for some $\kappa>|A|$ then every type in $S_\omega^\mathcal{M}(A)$ will be realized in $\mathcal{M}$?

I think the answer is yes. Since the case where $A$ is nonempty is similar, for simplicity we will just consider $A=\emptyset$, so suppose $\mathcal{M}$ is weakly saturated and $\Sigma(v_n)_{n\in\omega}\in S_\omega^\mathcal{M}(\emptyset)$. We define a sequence of elements $m_l\in M$ for all $l\in\omega$ by induction, such that each $\Sigma(m_0,...,m_l,v_n)_{l<n\in\omega}$ is consistent with $\mathcal{M}$:

Let $\Delta_0(v_0)=\exists(v_n)_{0<n\in\omega}\Sigma(v_n)_{n\in\omega}:=\{\exists v_{n_1}...v_{n_k}\varphi(v_0,v_{n_1},...,v_{n_k}):\varphi(v_0,v_{n_1},...,v_{n_k})\in\Sigma(v_n)_{n\in\omega}\text{ and each }n_i>0\}$. ($v_0$ may or may not appear in each $\varphi$, of course.) Certainly $\Delta_0(v_0)$ is finitely satisfiable, since $\Sigma(v_n)_{n\in\omega}$ is, and so by weak-saturation there is $m_0\in M$ such that $\mathcal{M}\models \Delta_0(m_0)$. But this means that  $\Sigma(m_0,v_n)_{0<n\in\omega}$ is consistent with $\mathcal{M}$, as desired.

Now suppose we have constructed $m_i$ for all $i<l$. We define $\Delta_l(v_l)=\exists(v_n)_{l<n\in\omega}\Sigma(m_0,...,m_{l-1},v_n)_{l\leqslant n\in\omega}$ in the obvious way, as $\{\exists v_{n_1}...v_{n_k}\varphi(m_0,...,m_{l-1},v_l,v_{n_1},...,v_{n_k}):\varphi(v_0,...,v_l,v_{n_1},...,v_{n_k})\in\Sigma(v_n)_{n\in\omega}\text{ and each }n_i>l\}$. (Apologies for the horrendous notation.) Now, $\Delta_l(v_l)$ is finitely satisfiable by construction of the $m_i$, else (taking conjunctions) there would be a formula $\varphi(v_0,...,v_l,v_{n_1},...,v_{n_k})\in\Sigma(v_n)_{n\in\omega}$ such that $\mathcal{M}\nvDash\exists v_l\exists v_{n_1}...v_{n_k}\varphi(m_0,...,m_{l-1},v_l,v_{n_1},...,v_{n_k})$, contradicting the induction hypothesis. Thus $\Delta_l(v_l)$ is finitely satisfiable and hence by weak saturation is realized, so there is an $m_l\in M$ such that $\mathcal{M}\models\Delta_l(m_l)$, and thus $\Sigma(m_0,...,m_l,v_n)_{l<n\in\omega}$ is consistent with $\mathcal{M}$, as desired.

Now, since any element of $\Sigma(v_n)_{n\in\omega}$ has only finitely many variables, it is clear that $\mathcal{M}\models \Sigma(m_n)_{n\in\omega}$, and so we are done.

I have two questions: first, is this proof correct? And second, if so, to what extent can we obtain similar results for general $\alpha$-types, where $\alpha$ is an infinite ordinal? Attempting to replicate the proof above for the general case using transfinite induction fails: although the successor ordinal case works fine, the limit ordinal case cannot be handled by this kind of argument.

(And this definitely makes sense, for it is clear that the result does not hold for sufficiently large $\alpha$... for instance, let $\mathcal{M}$ be any saturated structure of cardinality $\kappa$, and let $\alpha$ have cardinality strictly greater than $\kappa$. Letting $\Sigma(v_\delta)_{\delta<\alpha}=\{v_\delta\neq v_\gamma\}_{\{\delta\neq\gamma\in\alpha\}}$ we see that $\Sigma(v_\delta)_{\delta<\alpha}$ is clearly finitely satisfiable in $\mathcal{M}$ but not realizable in $\mathcal{M}$.)

So we certainly can't expect any kind of naive generalization of the above proof to work for general infinite ordinals $\alpha$, as the result does not hold in general. However, can we get any kind of analog provided that $|\alpha|\leqslant|M|$? Alternatively, can we get any kind of analog by restricting our attention to a more specific class of $\mathcal{L}$-formulas?

答:
First question. Your proof looks correct to me, except for one point: You really do need $\aleph_0$-saturation here, not just weak saturation. Weak saturation only tells you that every $n$-type over the empty set is realized, but you need to realize types over finitely many parameters.

Second question. A very natural generalization is true: If $M$ is $\kappa$-saturated and $|A|<\kappa$, then every type in $S^M_\kappa(A)$ is realized in $M$.

In fact, your proof generalizes immediately to show this. You wrote "although the successor ordinal case works fine, the limit ordinal case cannot be handled by this kind of argument." But the limit ordinal case is easy, by compactness. If $\lambda$ is a limit ordinal and $(m_\alpha)_{\alpha<\lambda}$ is a sequence from $M$ such that for all $\alpha<\lambda$, $\Sigma((m_\beta)_{\beta<\alpha},(v_\beta)_{\beta\geq \alpha})$ is consistent with the elementary diagram of $M$, then $\Sigma((m_\beta)_{\beta<\lambda},(v_\beta)_{\beta\geq \lambda})$ is consistent with the elementary diagram of $M$. The reason is that any finite subset of $\Sigma((m_\beta)_{\beta<\lambda},(v_\beta)_{\beta\geq \lambda})$ only mentions finitely many of the $(m_\beta)_{\beta<\lambda}$, and hence is contained in $\Sigma((m_\beta)_{\beta<\alpha},(v_\beta)_{\beta\geq \alpha})$ for some $\alpha<\lambda$.

The reason you can't continue the induction to arbitrary ordinals is that once you hit a sequence of length $\kappa$, you can't continue with the successor step - if $M$ is only $\kappa$-saturated, you can't realize a type over the entire sequence you've built so far in $M$.

Follow-up: Here's an alternative proof. It really has the same ideas as your proof, but it's a bit cleaner. First a lemma. Recall that for $A\subseteq M$, a map $f\colon A\to M$ is partial elementary if any tuple from $A$ has the same type in $M$ as its image under $f$. Given a type $p(x)$ over $A$, write $f_*p(x)$ (the pushforward of $p$ by $f$) for the set of formulas obtained by replacing elements in $a$ appearing in $p$ by their images under $f$. So $$\varphi(x,a_1,\dots,a_n)\in p(x) \text{ iff } \varphi(x,f(a_1),\dots,f(a_n))\in f_*p(x).$$

Lemma: Suppose $A\subseteq M$, $p(x)\in S_1^M(A)$, and $f\colon A\to M$ is a partial elementary map. Then $f_*p(x)\in S_1^M(f(A))$ (in particular, it is consistent).

The proof of this lemma is where you have to do some syntactic mucking around with existential quantifiers like in your proof. But it's a bit nicer, since we only have to deal with one free variable, not infinitely many!

Now suppose $M$ is $\kappa$-saturated and $A\subseteq M$ with $|A|<\kappa$. Let $\Sigma((x_\alpha)_{\alpha<\kappa})\in S^M_\kappa(A)$.

Since $\Sigma$ is consistent, there is some realization $(n_\alpha)_{\alpha<\kappa}$ in an elementary extension $M\preceq N$. We construct a sequence $(m_\alpha)_{\alpha<\kappa}$ in $M$ by transfinite induction. To define $m_\alpha$, we let $p_\alpha(x) = \text{tp}(n_\alpha/A(n_\beta)_{\beta<\alpha})$, define a partial elementary map $f_\alpha$ by $f_\alpha(n_\beta) = m_\beta$ for all $\beta<\alpha$, and realize $(f_\alpha)_*p_\alpha(x)$ in $M$ by saturation (since $|A\cup \{m_\beta\mid  \beta<\alpha\}|<\kappa$).

To make this construction work, we just need to prove by transfinite induction on $\alpha\leq \kappa$ that each $f_\alpha$ is partial elementary, so we can apply the lemma. The case $\alpha = \kappa$ will establish that $M\models \Sigma((m_\alpha)_{\alpha<\kappa})$.

Base case: $f_0$ is the empty map, so it is trivially partial elementary.

Successor case: If $f_\alpha$ is partial elementary, use the fact that $m_{\alpha}$ realizes $(f_{\alpha})_*\text{tp}(n_{\alpha}/A(n_\beta)_{\beta<\alpha})$ to check that $f_{\alpha+1}$ is partial elementary.

Limit case: If $\lambda$ is a limit and $f_\alpha$ is partial elementary for all $\alpha<\lambda$, then $f_\lambda$ is partial elementary, since any finite tuple from the domain of $f_\lambda$ is actually in the domain of some $f_\alpha$.


问:
Distance to the nearest integer $<r$ implication
The following is an adaptation of a result in the middle of a proof I am reading. The result is not obvious to me and I don’t know how they were able to conclude it.

I define $\langle a \rangle$ to be the distance from $a$ to the nearest integer for $a\in \mathbb{R}$.

Let $\alpha$ be irrational; $q$ be a natural number and $r \in [0,\frac{1}{2}]$. Then:

$$\langle q\alpha \rangle < r \implies \mid \alpha - \frac{p}{q} \mid < \frac{r}{q}$$ for some $p\in \mathbb{N}$

答:
Let the integer closest to $ q \alpha $ be $p$. Then the result follows.

$$ \langle q\alpha \rangle < r \Rightarrow |q\alpha - p | < r \Rightarrow |\alpha - \frac{p}{q} | < \frac{ r}{q}.$$


问:
What is an involutional automorphism?
I am reading a paper that is a little too advanced for me, and it has the following statement:

$$ \text{There is a division ring } K \text{ with involutional automorphism } *: K \rightarrow K$$

I am struggling to find a proper definition of "involutional automorphism" so any help would be really appreciated.

答:
An automorphism is an isomorphism from $K$ to itself. That is, a bijection from $K$ to itself which preserves all the relevant algebraic structure. You might imagine an invertible matrix viewed as a function from a vector space to itself.

An involution is a function which is its own inverse. You might imagine a reflection, or a rotation by $180^\circ$.

So an involutional automorphism is an isomorphism $*$ from $K$ to itself such that $**x = x$.

As a concrete example, consider $\mathbb{C}$ with the complex conjugation operation $z \mapsto \overline{z}$. I will leave it to you to check that this is

A nontrivial field isomorphism from $\mathbb{C}$ to itself (it preserves all the field structure)

Involutional, in the sense that $\overline{\overline{z}} = z$.

As another example, consider the matrix

$$A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$$

which we view as a function from $\mathbb{R}^2$ to itself. The fact that this is invertible and linear means exactly that it is an automorphism of $\mathbb{R}^2$, and the fact that it is its own inverse ($A^2 = I$) means it is an involution.

I hope this helps ^_^

Title: What is an involutional automorphism?


问:
Proving that $f$ is constant on $E$
Suppose that $f: \mathbb{R^n} \rightarrow \mathbb{R^m}$ and $a \in E \subseteq \mathbb{R^n}$, where $E$ is compact and connected. Suppose also that for each $x \in E$ there is a $\delta_x >0$ such that $f(x)=f(y)$ for all $y \in B_{\delta_x}(x)$. Prove that $f$ is constant on $E$.

The definition that my book provides for being compact is as follows:The set $E$ is said to be compact if and only if every open covering of $E$ has a finite subcovering; that is, if and only if given any open covering $\{V_α\}_{α∈A}$ of $E$, there is a finite subset $A_0 = \{α_1, . . . , α_N\}$ of $A$ such that $E \subseteq \bigcup\limits_{j=1}^N V_{a_j}$.

Similarly the definition of open covering is given as:

An open covering of $E$ is a collection of sets $\{V_α\}_{α∈A}$ such that each $V_α$ is
open and $E \subseteq \bigcup\limits_{\alpha \in A}V_{\alpha}$.

Is there any way that we can just pick our $B_{\delta_x}$ such that it is an open covering of $E$ since we know that $E$ is compact already so that $E \subseteq \bigcup\limits_{x\in E} B_{\delta_x}(x)$.

If not I am really just confused on how to start this question, if someone could just nudge me in the right direction I think I can get this problem down.

答:
Actually, the fact that $E$ is compact doesn't matter here.

Take $a\in E$, let $p=f(a)$ and define $F=\{x\in E\mid f(x)=p\}=f^{-1}\bigl(\{p\}\bigr)\cap E$. Then:

$a\in F$ and therefore $F\ne\emptyset$;

if $x\in E\setminus F$, then there is a $\delta_x>0$ such that $\bigl(\forall y\in B_{\delta_x}(x)\bigr):f(y)=f(x)$ and therefore $B_{\delta_x}(x)\subset E\setminus F$; so $E\setminus F$ is an open subset of $E$, which means that $F$ is a closed subset of $E$;

if $x\in F$, then there is a $\delta_x>0$ such that $\bigl(\forall y\in B_{\delta_x}(x)\bigr):f(y)=p$ and therefore $B_{\delta_x}(x)\subset F$; so $F$ is open.

Since $F$ is non-empty, open, and closed, the fact that it is connected implies that $F=E$. In other words, $(\forall x\in E):f(x)=p$.

Title: Proving that $f$ is constant on $E$


问:
Determine whether the following limit exists and either find the limit or prove it doesn’t exist.
The limit is that I have to prove the existence of is $\lim_{(m,n)\rightarrow(0,0)} \frac{m^3-n^3}{m^2+n^2}$. My first thought was to factor ${m^3-n^3}$ into $(m-n)(m^2+mn+n^2)$ and try to evaluate it from there. Past that I am struggling on were to take this proof. I am just for looking for help in starting this proof, but any help would be appreciated.

答:
Note that the limit along $m=0$ and the limit along $n=0$ are trvially seen to be $0$.

For $m \neq 0$ and $n \neq 0$ we have $0\leq |\frac {m^{3}-n^{3}} {m^{2}+n^{2}}|\leq \frac {|m|^{3}+|n|^{3}} {m^{2}+n^{2}}\leq  \frac {|m|^{3}} {m^{2}}+\frac {|n|^{3}} {n^{2}}\leq |m| +|n| \to 0$.  Apply Squeeze Theorem.


问:
Fixed-point theorem for a component-wise convex function.
I'm looking for a proof or counterexample of the following statement:

Conjecture: Let $\phi :[0,1]^n \to [0,1]^n $ be smooth. Suppose that each component of $\phi$ is convex, nondecreasing (in the sense that $\phi_i (x) \leq \phi_i (y)$ whenever $x_k \leq y_k$ for all $1\leq k \leq n$). Let $\mathbf{1}_n$ be the vector with all entries equal to $1$ and suppose $\phi(\mathbf{1}_n) = \mathbf{1}_n$. Assume further that the Jacobian $\mathcal{J}_\phi (\mathbf{1}_n)$ at that point has an eigenvalue $\lambda$ strictly greater than $1$. Further, suppose that $\phi(e_i)$ is a strictly positive vector for every unit vector $e_i$. Then $\phi$ has a fixed point in its domain other than $\mathbf{1}_n$.

The $n=1$ case is rather trivial, and I'm looking to generalize it. If the above doesn't work, can we add more 'niceness' assumptions to $\phi$ in order to make it work (like being a diffeomorphism)?

Any help is greatly appreciated.

答:
Here is an analysis of your problem with simplified hypotheses on derivatives of $\phi$, but without convexity.

Assume $\phi: [0, 1]^n \to [0, 1]^n$ is $C^1$, $\dfrac{\partial \phi_i}{\partial \phi_j}(1_n) > \dfrac{1}{n}$ for any $(i, j)$, $\phi$ is non-decreasing (as defined in your post), and $\phi(1_n) = 1_n$.

Because of the second hypothesis, there is for any $i$ some $\epsilon_i > 0$ such that $\phi_i((1 - \epsilon_i) 1_n) \leq 1 - \epsilon_i$.

Proof:

Assume $\forall \epsilon > 0$, $\phi_i((1 - \epsilon_i) 1_n) > 1 - \epsilon$

Using $\phi(1_n) = 1_n$, you get $\dfrac{\phi_i(1_n) - \phi_i((1-\epsilon) 1_n)}{\epsilon} < 1$

Since $\dfrac{\phi_i(1_n) - \phi_i((1-\epsilon) 1_n)}{\epsilon} \to \sum_j \dfrac{\partial \phi_i}{\partial x_j}(1_n)$ as $\epsilon \to 0$, you have $\sum_j \dfrac{\partial \phi_i}{\partial x_j}(1_n) \leq 1$, which is a contradiction.

So, $\exists \epsilon_i > 0$ such that $\phi_i((1 - \epsilon_i) 1_n) \leq 1 - \epsilon_i$

Take $\epsilon = \min_i \epsilon_i$, then for any $i$, $\phi_i((1 - \epsilon) 1_n \leq 1 - \epsilon$

Since $\phi$ is non-decreasing, then $\phi$ maps $[0, 1-\epsilon]^n$ to $[0, 1-\epsilon]^n$ (which does not contain $1_n$).

As $\phi$ is continuous, you get existence of a fixed point in $[0, 1-\epsilon]^n$ (convex compact) by the Brouwer fixed-point theorem.

Remark:

The $\dfrac{1}{n}$ in the second hypothesis is surprising for me, so if someone finds a mistake, please tell me.


问:
Probability of retirement event
This is elementary, but not clear to me.

Suppose I know that the mean age of retirement is $\mu$ and the standard deviation $\sigma$.
What is the probability that someone of age $x$, who has not yet retired,
will retire sometime in the next year,
i.e., between $x$ and $x+1$?

Clearly for small $x$, far below the mean, the probability is near zero, and for large $x$, it approaches $1$.  So it is a type of cumulative distribution...

Thanks for your help!

答: Integrate the Gaussian probability density function from x to x+1</p>

Title: Probability of retirement event


问:
A functor $\mathcal{F}$ which is not exact but it has an exact power $\mathcal{F}^n$
Is there a functor $\mathcal{F}$ on an abelian category $\mathcal{C}$ which is not exact but there is a natural number $n$ such that $\mathcal{F}^n$ is an exact functor? What about the same question when we replace exactness by half exactness? Namely a  functor which is not left exact but it has a left exact power (or the same for right case)?

Remark: We exclude the case that  $\mathcal{F}^n$ is the identity functor for some $n\in \mathbb{N}$.

答: Here's a simple example.  Let $\mathcal{C}$ be the category of $\mathbb{Z}[x]/(x^2)$-modules, and let $\mathcal{F}$ be the functor that takes a module $M$ to the submodule $xM$.  Then $\mathcal{F}$ is not half exact (consider $0\to xM\to M\to M/xM\to 0$ which $\mathcal{F}$ sends to $0\to 0\to xM \to 0\to 0$ which is not exact in the middle unless $xM=0$) but $\mathcal{F}^2=0$ is exact.


问:
How to prove that a triangle is uniquely determined by an angle, its opposite side and its perpendicular height.
I am trying to solve a homework problem and as part of this problem, I need to show that a certain situation is impossible. I have the following situation. Given some line, I have $2$ points $H$ and $I$ on a perpendicular to said line. The rays traced from $H$ and $I$ both intersect the line at the same points $K$ and $L$ as follows:

My hypothesis is that $\angle KHL = \angle KIL$. Intuitively, I know that this forces the $2$ triangles $\triangle KHL$ and $\triangle KIL$ to be equal, but I can't seem to be able to show it. I tried using the law of sines since I know the opposite angle to the side I know has equal length for both triangles, but since I don't know any other side lengths I couldn't seem to continue. I also tried using Pythagoras on triangles $\triangle KJI$ and $\triangle KJH$, but I ran into a similar problem of not knowing how to relate the same angle hypothesis into this.

Is there a simple argument as to why these two triangles have to be equal that I'm missing? Or could anyone point me in the right direction? Thank you!

答:
Hint: $\angle KIJ = \angle HKI + \angle KHJ$.

Conclude that $ \angle HKI + \angle HLI = 0 $, but these angles have the same sign, so must both be 0.


问:
When the intersection of projective varieties is finite
What I want to solve is

Let $C$ be a projective curve, $P\in C$. Then there is a binational morphism $f:C \longrightarrow C'$, $C'$ a projective plane curve, such that $f^{-1}(f(P))=\{P\}$.

Here, a projective curve means that $C\subseteq \mathbb{P}^{n+1}$ is projective variety whose dimension is 1, and projective "plane" curve is just a curve in $\mathbb{P}^2$. The outline of proof is following :

We can assume : Let $T,X_1,\cdots,X_n,Z$ be coordinates for $\mathbb{P}^{n+1}$; Then $C\cap V(T)$ is finite ; $C\cap V(T,Z)=\emptyset$; $P=[0:0:\cdots :0:1]$ and $k(C)$ (function field on C) is algebraic over $k(u)$, where $u=\bar{T}/\bar{Z}\in k(C)$.

But I can't understand this assumption, because basically I think I'm not sure when the intersection of two varieties is finite set.(In $\mathbb{P}^2$, I know this set is finite if the homogeneous polynomials have no common component, but I don't know the case for larger dimension) Can anyone explain why we can assume such things?

答:
$C\cap V(T)$ is a closed subset of $C$, which means it's either all of $C$ or a finite collection of points. This is developed in Hartshorne chapter I exercise 4.8, for instance. So if $C$ isn't contained in $V(T)$, then this intersection is just a finite collection of points; if $C\subset V(T)$, just go to the $\Bbb P^{n-1}$ which is $V(T)$, pick new coordinates, and try again.

Your larger question about when the intersection of two projective varieties is finite is more difficult. The best available thing here is that if $X,Y$ are closed subvarieties of $\Bbb P^n$ of dimension $r,s$ respectively, then every irreducible component of $X\cap Y$ has dimension at least $r+s-n$, and if $r+s-n\geq 0$ then this intersection is nonempty. This is only an existence result - you can certainly have a nonfinite intersection if $r+s-n <0$: think about two lines in $\Bbb P^3$, for instance. They can be the same line, intersect in a point, or be skew lines and have no intersection. But the intersection can only be finite if $r+s\leq n$.


答:
\begin{align}
\mathbb{P}(X \in [x,x+1) \vert X \geq x) & = \dfrac{\mathbb{P}(X \in [x,x+1))}{\mathbb{P}(X \geq x)} = \dfrac{F_X(x+1) - F_X(x)}{1-F_X(x)}\\
& = 1 - \dfrac{1-F_X(x+1)}{1-F_X(x)}
\end{align}
Since you have not clarified if you are looking for a bound on the probability, this is what we get.</p>

Title: Probability of retirement event

Body:
This is elementary, but not clear to me.

Suppose I know that the mean age of retirement is $\mu$ and the standard deviation $\sigma$.
What is the probability that someone of age $x$, who has not yet retired,
will retire sometime in the next year,
i.e., between $x$ and $x+1$?

Clearly for small $x$, far below the mean, the probability is near zero, and for large $x$, it approaches $1$.  So it is a type of cumulative distribution...

Thanks for your help!


问:
Integer programming problem, 5 pens and 10 bags
I am trying to solve an Integer Programming question (trying to come up with a model). The question goes like this: There are 5 pens a person has. He wants to have all 5 of them at the same time. And he has 10 different weighted bags. He wants to cary minimum weight as he can while having all five of the pens. Different bags carry different pens, the distribution is like this:

The weights of the bags are like this:
bag 1 --> 4
bag 2 --> 6
bag 3 --> 4
bag 4 --> 2
bag 5 --> 3
bag 6 --> 2
bag 7 --> 4
bag 8 --> 3
bag 9 --> 6
bag 10 --> 3

What I thought is this:
Let xi = 1 be that he brings the ith bag and xi = 0 be that he does not bring the ith bag. Then the obj. function is the minimum of sum of all of the x variables multiplied with their weight. And I could think of only one constraint: The sum of all the x variables i = 1..10 should be at most 10 and at least 3.

I could not think of anything else. Dont even know if this is the right path to the solution.

答:
Each bag contains just two pens or one pen, and given that you must have at least $5$ (different) pens, you must have at minimum of $3$ bags.  For minimum weight, two bags must each contain $2$ pens and the final bag can contain $1$ pen (if possible).

This is quite easy to achieve.

Pick any bag you like containing two pens but not pen 3 (call it bag A).

Pick another bag that contains the two pens not in bag A and not pen 3. (Call it bag B.)

Pick another bag that contains the only missing pen.  (Call it bag C.)

I easily found: $\{ 1,6,5 \}$.

But try others!


问:
is 1001 the only sum of two positive cubes that is the product of three consecutive odd primes?
That is $\ 10^3+1^3=7.11.13$.

I could find no other examples.
So I am looking to see if there are any more solutions to $ x^3+y^3=p.q.r$, where $ x, y$  are positive integers and $ p<q<r$ are consecutive odd primes. Now $ x^3 + y^3 =LQ$ where $ L=x+y$, and $ Q = L^2 -3xy$.

After playing with the problem for a while I thought maybe I should try and express the triple of factors in terms of the gaps between them. So $ g_1 = q-p$, and $ g_2=r-q$, and I was thinking what was the biggest possible factor that divided the resulting sum of cubes after choosing the gaps.

Using elementary arguments I found that, $ L=q$, or $ L=r$ , so dealing with the later case put $ L=r$, and $ Q=pq= L^2 -3xy$, eliminate $p$, and $q$, and put $ x = (L+d)/2$, $ y=(L-d)/2$ for $d$ an odd positive integer. Giving $ L^2 - (g_1+2g_2)L+g_1(g_1+g_2)=L^2-3(L^2-d^2)/4$. Completing the square and simplifying gives $ z^2-9d^2 = 4f$, where $ z= 3L-2(g_1+2g_2)$, and $ f = g_1^2+g_1g_2+g_2^2$, the gaps are even so suitable factors can always be found, that is $a$, $b$, such that $4f= ab$, and we get using the difference of squares $z= (a+b)/2$, and so $L=\frac{z+2(g_1+2g_2)}{3}$.
Now the largest $z$ is $f+1$, and so the largest,

$$L=\frac{g_1^2+g_1g_2+g_2^2+1+2g_1+4g_2}{3}$$
and chosing $g$ to be the bigger of the two gaps gives a largest $L=\frac{3g^2+6g+1}{3}<(g+1)^2$

so for both cases the largest possible "prime" factor in any triple of solutions was bounded above by roughly the square of the largest gap.
Looking at Wikipedia prime gaps then Dr T Nicely's site on first occurrence of prime gaps https://faculty.lynchburg.edu/~nicely/gaps/gaplist.html
I noticed that apart from a number of small cases that the primes where gaps first occurred where larger than $(g+1)^2$ suggesting that 1001 is the only case for the range of known first occurrences of prime gaps. I also saw somewhere but cannot remember at the moment something called Shank's conjecture, which is something like the first occurrence of a gap is after a prime that is about $e^\sqrt g$, which is obviously for large enough gap much bigger than $(g+1)^2$.
See section 7 Marek Wolf 'Some heuristics on the gaps between consecutive primes' https://arxiv.org/pdf/1102.0481v2.pdf.

So this is roughly where I have got with the problem. I suspect I have missed a simpler solution?

35 is the only sum of two cubes equal to the product of two consecutive odd primes, using the notation above, let $ x^3 +y^3 = pq $, and $ 8<p<q$ be consecutive odd primes then $ L< Q = L^2 -3xy $, Put $ p=L$, $ q=Q$ and $ x=(L+d)/4$, $ y = (L-d)/4$, then $ (p^2+3d^2)/4=q$, so $ (p^2+3)/4 \le q$ but for $p> 8$, $ q>2p$, which contradicts Bertrand's Postulate whereby if we have consecutive primes then $ p< q<2p$
https://en.wikipedia.org/wiki/Bertrand's_postulate

答:
Since the product of three consecutive odd primes is odd, the sum of the two cubes is odd and $a$, $b$ have opposite parity.

OP comments that $35=3^3+2^3=5\cdot7$ is the only sum of two cubes equal to the product of two consecutive odd primes.  I note that it is also the least that is the product of two distinct odd primes.  The next one is $4^3+1^3=5\cdot13$. Similarly here, besides being the product of three consecutive odd primes, $10^3+1^3=7\cdot11\cdot13$ is the least $a^3+b^3$ that is the product of three distinct primes. The next one is $12^3+1^3=7\cdot13\cdot19$.

I. In seeking a second solution for$$a^3+b^3=pqr$$supposing $a>b$ and $p$, $q$, $r$ consecutive odd primes, note that  since$$a^3+b^3=(a+b)(a^2-ab+b^2)$$then for any given $a$, $3$ will be a factor of $a^3+b^3$ for $(a+b)\equiv 0 \mod 3$, that is for every third odd $b$ if $a$ is even, or even $b$ if $a$ is odd.  Further, $5$ will be a factor when $(a+b)\equiv 0\mod5$, i.e. for every fifth odd/even $b$ when $a$ is even/odd.  And similarly for $7$, $11$, $13$, and all odd primes.

But it appears that $7$ also divides $(a^2-ab+b^2)$ twice for every seven consecutive $b$, and thus divides $(a^3+b^3)$ for three out of every seven consecutive $b$.  I gather this from inspection, but assume it can be proven.

For example, with $a=73$ and $b=2, 4, 6, 8,...,72$, then, in six rows of six, $a^3+b^3=$

$$(3^2\cdot5^2\cdot7\cdot13\cdot19), (7\cdot11\cdot31\cdot163), (13\cdot79\cdot379), (3^5\cdot7\cdot229), (37\cdot83\cdot127), (5\cdot17\cdot4597)$$$$(3^2\cdot19\cdot29\cdot79), (7\cdot89\cdot631),(7\cdot13\cdot4339), (3^2\cdot31\cdot1423), (5\cdot7\cdot19\cdot601), (97\cdot4153)$$$$(3^3\cdot11\cdot37^2), (13\cdot101\cdot313), (7\cdot103\cdot577), (3^2\cdot5\cdot7\cdot13\cdot103), (107\cdot4003), (7\cdot109\cdot571)$$$$ (3^2\cdot31\cdot37\cdot43), (19\cdot113\cdot211), (5\cdot23\cdot4027), (3^3\cdot7\cdot13\cdot193), (7\cdot17\cdot61\cdot67), (11^2\cdot4129)$$$$ (3^2\cdot7\cdot41\cdot199), (5^3\cdot19\cdot223), (13\cdot127\cdot331), (3^2\cdot43\cdot1459), (7^3\cdot13\cdot131), (7\cdot19\cdot4549)$$$$ (3^4\cdot5\cdot1549), (7^2\cdot97\cdot137), (31\cdot139\cdot157), (3^2\cdot47\cdot1663), (11\cdot13\cdot5119), (5\cdot7\cdot29\cdot751)$$

II. Since neither $3\cdot5\cdot7=105$ nor $5\cdot7\cdot11=385$ is the sum of two cubes, and hence neither $3$ nor $5$ can be among the three consecutive prime factors of any solution, then for any given $a$ we can disregard one-third of all $b$, plus two-thirds of one-fifth of all $b$ (since one in three $a+b$ divisible by $5$ is also divisible by $3$ and so $b$ for that case has already been removed):$$\frac{1}{3}+\frac{2}{3}\cdot\frac{1}{5}=\frac{5}{15}+\frac{2}{15}=\frac{7}{15}>46\%$$of all $b$ are ruled out.

And since $7$ was the least of the three consecutive odd primes in the one known solution, it cannot appear in any greater solution, and the fraction of $b$ excluded will now be$$\frac{7}{15}+\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{3}{7}=\frac{49}{105}+\frac{24}{105}=\frac{73}{105}>69\%$$

Again, since $11\cdot13\cdot17=12^3+9^3+3^3-1$ is not a solution, $11$ cannot be a factor in a second solution.

And since $13\cdot17\cdot19=14^3+11^3+5^3-1$ is not a solution, we can likewise exclude $13$ from any other solution.  And here again, as happens with $7$, and is also clear in the example above, $13$ divides $a^3+b^3$ not only for $(a+b)\equiv 0\mod13$, but also for another two of every  thirteen consecutive $b$ (i.e. when $13$ divides $a^2-ab+b^2$).

With $11$ and $13$ ruled out, the portion of $b$ excluded increases to$$\frac{73}{105}+\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{4}{7}\cdot\frac{1}{11}+\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{4}{7}\cdot\frac{10}{11}\cdot\frac{3}{13}=\frac{805}{1001}>80\%$$

Finally, if we grant that $17\cdot19\cdot23=19^3+8^3+58$ and $19\cdot23\cdot29=23^3+6^3+290$ are not sums of two cubes, then we can rule out $17$ and $19$ from any solution, and the fraction of all $b$ excluded is$$\frac{805}{1001}+\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{4}{7}\cdot\frac{10}{11}\cdot\frac{10}{13}\cdot\frac{1}{17}+\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{4}{7}\cdot\frac{10}{11}\cdot\frac{10}{13}\cdot\frac{16}{17}\cdot\frac{3}{19}=\frac{4114625}{4849845}\approx85\%$$

III. I know that taking this approach in response to OP's request for "a simpler solution," may seem like embarking on a long ground war with little prospect of a decisive victory.  But I was surprised to find how great an advance is made as we exclude primes $p=7, 13, 19,...\equiv 1\mod6$, since the above example clearly suggests (and I assume it can be proven) that these are factors of $a^3+b^3$ for three of every $p$ consecutive $b$, instead of just one as happens when $p=5, 11, 17,...\equiv 5\mod6$.  Small odd primes, especially those $\equiv 1\mod6$, are so plentiful as factors of $a^3+b^3$ that excluding just the first seven odd primes eliminates a large portion of possible solutions of$$a^3+b^3=pqr$$where $p$, $q$, $r$ are consecutive odd primes.


问:
How to find the cost to cover with a fence a side of a triangle?
The problem is as follows:

The figure from below shows a certain terrain has the shape of a triangle $\triangle ABC$ which is obtuse on $B$. Assuming the line connecting $AM$ is a bisector whose angle is $\beta$ and $\angle ACB=20^{\circ}$ and $\angle ABC= 120^{\circ}$ and $\angle CMA=90^{\circ}$ and $CM=8\sqrt{3}\,m$. Find the cost of placing a fence on $CM$ if the owner of the terrain consulted with the hardware store and the price is $\$3.0$ per meter.

The alternatives in my book are as follows:

$\begin{array}{ll}
1.&\$\,21.00\\
2.&\$\,24.00\\
3.&\$\,48.00\\
4.&\$\,32.00\\
\end{array}$

In my attempt to solve this problem the only thing which I was able to spot is that the bisector angle is $\beta= 20^{\circ}$.

This comes because of the sum of the interior angles in the triangle add up to $180^{\circ}$.

Hence on $\triangle ABC$:

$2\beta+120+20=180$

$\beta= 20^{\circ}$

Then I could also spot is that the small right triangle has the interior angles of 40^{\circ} and 50^\circ because by interior and exterior angle on the smaller triangle. But other than that I cannot spot something else.

Can someone help me?. I think in this problem is required congruence or similarity but I don't know how to use those here.

Please an answer must include a drawing Since it is not easy for me to spot the relationships and construction that would help me better to understand what's going on.

答:
Extend the line BC to the left until it comes "under" point A.  Then drop
a line from point A to the extended line BC so that the line is perpendicular
to BC.  Denote the point of intersection as E.

Then $\triangle$ AEC is a 20-70-90 right triangle.

Since triangles AEC and ACM have similar angles, and since they share the long
side of AC,

$$ \text{AE is also} ~8\sqrt{3}.$$

Since $\triangle$ AEB is a 30-60-90 right triangle,

$$ \text{AB is} ~16.$$

Since the cost is 3 dollars per meter, the cost of the fence is 48 dollars.


问:
MAT 2003 Oxford Admissions Exam Question 5 Solution Verification
As ongoing trend, I have been solving old MAT past papers that do not have published solutions. I was wondering if anybody can verify my solution to the following question.

My Solution

(i) We prove this by contradiction. Suppose that after one second there is exactly one bulb ON. This implies that only two consecutive light bulbs say $n$ and $n+1$, must have had different  lights in the beginning. This implies that $n+3$ and $n+4$ had the same state since only one is ON after 1 second. Similarly, $n+4$ and $n+5$ must have had the same state and so on until we get that $n-1$ and $n$ have had the same state. This is a contradiction as $\text{state of } n \neq \text{state of } n+!$ and hence there could not be exactly one light bulb ON after one second.

(ii) Let $m$ be an odd number. Let there be $m$ many turned on lights after one second. Therefore, out of $N$ many lights m many consecutive pairs have different initial states. Let these pairs be

$$\{m_0,m_0 +1\},\{m_1,m_1 + 1\},\, ... , \{m_m, m_{m}+1\}$$

Say the initial state of $m_0$ is ON/OFF. By construction we will have, $m_0 +1, m_0 +2,\, ... ,m_1$ have the same initial signal. This implies that $m_2$ has a signal of ON/OFF. This can be done $m$ many times to show that $m_m$ is ON/OFF which implies that that $m_n +1, \, ..., m_0 -1$ are OFF/ON since $m$ is odd. This implies that $m_0 -1$ and $m_0$ have different initial values. This is a contradiction because if true this would imply that there are $m+1$ of these pairs. Thus there can not be odd many turned ON lights after the first second.

(iii) Let $b_n$ be the nth bulb. And let $v_i (b_n)$ be the status of the nth bulb after $i$ seconds have passed. The formula for $2$ seconds passed is
$$ v_2(b_n)
 \begin{cases}
      \text{ON} & \text{if } v_1(b_n) \neq v_1(b_{n+2}) \\
     \text{OFF} & \text{if } v_1(b_n) = v_1(b_{n+2})
   \end{cases}
$$
The work of which I have done but I really can not put in LaTeX as it would take a lot of time. From the equation we see that it is only dependent on $b_n$ and $b_{n+2}$ as required.
(iv)

$$ v_4(b_n)
 \begin{cases}
      \text{ON} & \text{if } v_1(b_n) \neq v_1(b_{n+4}) \\
     \text{OFF} & \text{if } v_1(b_n) = v_1(b_{n+4})
   \end{cases}$$
again the proof of which I have on paper but I really can not write in LaTeX due to time.

(v) After 8 seconds.

答:
There's a very clean way to present the problem.

Denote a bulb to be OFF by 1.

Denote a bulb to be ON by -1.

Let $ S_t (i)$ denote the state of the bulb $i$ at time $t$.

From the definition, verify that $S_t (i) = \frac{ S_{t-1} (i)}{ S_{t-1} (i+1)} $.

i, ii) Observe that for $ t \geq 1$, $ \prod S_t (i) = \prod \frac{S_{t-1} (i) } { S_{t-1} (i+1) } = 1  = (1) ^{ \text{number of OFF} } (-1) ^ { \text{ number of ON} }$.

Hence, the number bulbs that are ON is even.

iii, iv, v) It can be easily proven (E.g. by induction) that $S_{2^t} (i)  = \frac{ S_0 (i) } { S_0 (i+2^t)}$, so the state at time $2^t$ only depends on $S_0 (i)$ and $S_0 (i+2^t)$.

Thus when $N= 2^t$, $S_N(i) = \frac{ S_0 (i) }{S_0 (i+N) } = 1$, so all bulbs are OFF.


问:
Factorial decomposition of integers?
This question might seem strange, but I had the feeling it's possible to decompose in a unique way a number as follows:

if $x < n!$, then there is a unique way to write x as:
       $$x = a_1\cdot 1! + a_2\cdot 2! + a_3\cdot3! + ... + a_{n-1}\cdot(n-1)!$$
       where $a_i \leq i$

I looked at factorial decomposition on google but I cannot find any name for such a decomposition.

example:

If I chose :

(a1,a2) =

1,0   -> 1

0,1   -> 2

1,1   -> 3

0,2   -> 4

1,2   -> 5

I get all number from $1$ to $3!-1$

ideas for a proof:

The number of elements between $1$ and $N!-1$ is equal to $N!-1$ and I have the feeling they are all different, so this decomposition should be right. But I didn't prove it properly.

Are there proofs of this decomposition? Does this decomposition as a name? And above all is this true ?

Thanks in advance

答:
It is really easy to see why the representation is unique.  It is analogous to finding the representation of a number in a different base.  For example, to find the base 3 representation of n, you do the following:

Get the next digit by finding n mod 3

Do an integer divide of n by 3

If n > 0 then go to step 1

In the case of factorial sums, do the following:

Set i = 2

The next digit is n mod i

Do an integer divide of n by i

If n > 0, set i = i + 1 and go to step 2

Title: Factorial decomposition of integers?


问:
Some Distinct Subgroups of $S_4$
Prove that there are at least 3 distinct subgroups of order 8 in $S_4$.

I have various pieces, but I'm struggling to put them together. Since $|S_4| = 4! = 24 = 2^3 \cdot 3$, we know from Sylow's theorem that there exists a subgroup of order 8 in $S_4$.

My first idea was to use the natural projection homomorphism to delineate the 3 cosets of $S_4$ mod this subgroup, and claim that the preimage of each of these cosets under the homomorphism are exactly the subgroups we need, but I'm not sure this it the way to go.

I've been given a hint that showing $S_4 / K \cong S_3$ is the big part, where $K$ is the Klein $4$-group. Assuming that I've already shown this, is there a simple way to get the solution? I've had ideas to use the lattice/fourth isomorphism theorem after this isomorphism is shown, but I need some help filling in the details.

All in all, are both/either method a viable solution to the problem, and if so could someone help describe the necessary things I'm missing?

答:
Consider the isomorphism
$$ \phi : \frac{S_4}{ K} \to S_3$$

Then, each of the three transpositions $(i,j)$ in $S_3$ defines a subgroup $H \leq S_3$ of order 2.

The isomorphism theorem tells you that for each such $H$ there exists a different $K \leq H' \leq S_4$  such that
$$\frac{H'}{K} \simeq H$$

This means that $H'/K$ has 2 elements. How many elements are in $H'$?

Title: Some Distinct Subgroups of $S_4$


问:
How to find a centroid of a quarter circle?

Find the centroid of the quarter circle $x^2+y^2 \le  10$, $y\ge |x|$ assuming the density $ {\displaystyle \delta }(x,y) = 1$. give your answer as point coordinates in the form (X,Y)

So I first convert to polar coordinates:

$$rsin(\theta)^2+rcos(\theta)^2=10$$

Then I know it's a quarter circle that's bounded by $y=|x|$

So $\theta = \frac{\pi}{4}$ to $\frac{3\pi}{4}$

So now I integrate:

$$\int _{\frac{\pi }{4}}^{\frac{3\pi }{4}}\int _0^{2.5}\:\:\:rsin\left(\theta \right)^2+rcos\left(\theta \right)^2drd\theta $$

But this is just one integral. How do I get two? What am I missing?

Thank you.

edit:

答:
Clearly $\bar{x}$ coordinate is $0$ (by symmetry).  The $\bar{y}$ coordinate is:

$$\bar{y} =
\frac{ \int\limits_{y=0}^{\sqrt{5}} y \cdot 2 y\ dy + \int\limits_{\sqrt{5}}^{\sqrt{10}} y \cdot 2 \sqrt{10 - y^2}\ dy}{\pi 10/4} = \frac{8 \sqrt{5}}{3 \pi }$$

The total mass (in the denominator) is just the density times the area of a quarter circular disk of radius $\sqrt{10}$


问:
Understanding Limit without l'Hopital
I have been trying to understand this limit:

$$\lim_{x \to 0}\frac{tan(x)-sin(x)}{x^2}$$

When aplying the l'Hopital rule I arrive to the limit being $0$ but when doing things organically I get an indetermination:

$$
\lim_{x \to 0}\frac{tan(x)-sin(x)}{x^2}=\lim_{x \to 0}\frac{tan(x)}{x^2}-\frac{sin(x)}{x^2}= \lim_{x \to 0} \frac{sin(x)}{cos(x)x^2}-\frac{sin(x)}{x^2}= \lim_{x \to 0}\frac{sin(x)}{x^2}(\frac{1}{cos(x)}-1)
$$

Clearly $\lim_{x \to 0} \frac{1}{cos(x)}=1$ hence $(\frac{1}{cos(x)}-1)=0$ and I could well aply $\lim_{x \to 0}\frac{sin(x)}{x}=1$ but that still leaves $\lim_{x \to 0}\frac{1}{x}$ which is undetermined because it has different limits on $0^-$ and $0^+$.

Is there something I'm missing?

答: Rearrange this as  $\frac{\sin x}{x} \frac{1}{\cos x}\frac{1- \cos x}{x}$ and use the standard limit $\frac{1- \cos x}{x} \to 0$ as $x \to 0$.

Title: Understanding Limit without l'Hopital


问:
For Brownian motion $B_t,$ we have $\lim_{t\to \infty}\frac{B_t}{t^{\alpha}}=0$ almost surely.
This might be a silly problem. I am trying to show that $\lim_{t\to \infty}\frac{B_t}{t^{\alpha}}=0$ almost surely if $\alpha>\frac{1}{2}.$ I can apply Chebyshev's inequality to show that $P(|B_{t}/t^{\alpha}|>\epsilon)\to 0$ as $t\to \infty.$ This tells me that $\frac{B_t}{t^{\alpha}}$ converges to $0$ in probability. But I am not sure how to argue the almost sure limit. I found at the very beginning of a book I am reading, so I am looking forward to something which does not use $0-1$ laws if that's possible. I am wondering if this can be done cheaply using the Borel-Cantelli lemma.

答:
I do not think this is easy at all from the ground up, but it can be derived from some other well known facts, which are perhaps proved in your text.

Let $W_t=tB_{1/t}$. It is well known that $W_t$ is also a Brownian motion. Then $$\lim_{t\to \infty}B_{t}/t^{\alpha}=\lim_{t\to\infty}W_{1/t}/t^{\alpha-1}=\lim_{s\to0}W_s/s^{1-\alpha}$$
The fact that the last limit is zero is the definition of $W_s$ being Hölder continuous at $0$ with exponent $1-\alpha$. The fact that $W_s$ is Hölder continuous for all exponents less than $\frac12$ is often proved early on in textbooks.


答:
Here is an alternate approach which only uses the Sylow theorems. Hopefully, I am not making any stupid mistakes.

First, note that by writing a permutation as a product of disjoint cycles that each $\sigma \in S_4$ has order $2,3,4$. Therefore, each $\sigma$ generates a $p$-group and hence belongs to some Sylow subgroup. This means that $S_4$ is the union of its $p$-Sylow subgroups.

By the Sylow theorem $n_2|\frac{24}{8}=3$ and hence $n_2=1$ or $n_2=3$.

Assume by contradiction that $n_2=1$.

Note that $n_3=1 \pmod{3}$ and $n_3|8$ thus $n_3 \leq 4$.

Now, the unique $2$-Sylow subgroup has $8$ elements and the $n_3 \leq 4$, $3-$Sylow subgroups have at most $4\cdot 3=12$ elements.

Therefore $S_4$ has at most $8+12=20$ (note that we actually count the identity $5$ times here, so this can be improved but who cares)  elements. Contradiction

Title: Some Distinct Subgroups of $S_4$

Body:
Prove that there are at least 3 distinct subgroups of order 8 in $S_4$.

I have various pieces, but I'm struggling to put them together. Since $|S_4| = 4! = 24 = 2^3 \cdot 3$, we know from Sylow's theorem that there exists a subgroup of order 8 in $S_4$.

My first idea was to use the natural projection homomorphism to delineate the 3 cosets of $S_4$ mod this subgroup, and claim that the preimage of each of these cosets under the homomorphism are exactly the subgroups we need, but I'm not sure this it the way to go.

I've been given a hint that showing $S_4 / K \cong S_3$ is the big part, where $K$ is the Klein $4$-group. Assuming that I've already shown this, is there a simple way to get the solution? I've had ideas to use the lattice/fourth isomorphism theorem after this isomorphism is shown, but I need some help filling in the details.

All in all, are both/either method a viable solution to the problem, and if so could someone help describe the necessary things I'm missing?


问:
Find $f(t,w)$ such that $\int_0^{2 \pi} f(t, w) \ln|\delta(t) - \delta(\theta)| dt = \ln|w - \delta(\theta)|$

Problem

As part of a 2d Stokes flow fluid simulation I'm working on I'm trying to find a $f(t, w)$ such that

$$\int_0^{2 \pi} f(t, w) \ln|\delta(t) - \delta(\theta)| dt = \ln|w - \delta(\theta)|, \forall \theta \in [0, 2\pi)$$

where $w \in \mathbb{C}, |w| < 1$ and $\delta(t)$ is just a complex finite fourier series: $$\delta(t) = \sum_{n=-N}^N c_n e^{int}, c_n \in \mathbb{C}$$

Partial solutution

If $\delta(t) = e^{it}$ I have that

$$f(t, w) = \frac{1}{\pi} Re\Big(\frac{1}{1 - w e^{-it}}\Big)$$

to give some flavor of what $f(t,w)$ might look like (this value for $f(t,w)$ is not unique).  I'm trying to generalize this result to a wider class of $\delta(t)$, ideally all finite Fourier series but maybe it's only possible for a certain class, I'm not sure.

Possible approach?

I thought maybe I could factor $\delta(t) - \delta(\theta)$ as $a e^{-iNt} \prod_{n=-N}^N (e^{it} - \alpha_n)$.  Then the log term is linearly seperable:

$$ \ln|a e^{iNt} \prod_{n=-N}^N (e^{it} - \alpha_n) | = \ln|a| + \sum_{n=-N}^N \ln|e^{it} - \alpha_n| $$

I can then split the integral into smaller, solvable pieces:

$$\int_0^{2 \pi} f(t, w) \ln|\delta(t) - \delta(\theta)| dt = $$
$$\ln|a| \int_0^{2 \pi} f(t, w) dt + \sum_{n=-N}^N \int_0^{2 \pi} f(t, w) \ln|e^{it} - \alpha_n| dt$$

Assuming $|\alpha_n| = 1$, (which is a big assumption!), I think I can use my solution for $\delta(t) = e^{it}$ and a little bit of linear algebra to build up a more complex solution.  But the $\theta$ term disappears in the factorization (I have to pick some arbitrary value of $\theta$ to even do the factorization).  Maybe that's okay?  My solution when $\delta(t) = e^{it}$ should hold for all $\theta$ so maybe picking a value for $\theta$ arbitrarily is okay?  It's unclear to me and I'm not sure how to justify or refute the idea.

Any help would be appreciated.  Mostly I'm out of tools in my toolbox and I'm not sure if I'm on the right track or if this is even possible in principle.

答:
I've managed to solve this for a slightly larger class of curves.  The result is interesting so I thought I'd share:

First, generalize $\delta(t) - \delta(\theta)$ to the interior of the unit disk and call it $\gamma(z)$:

$$\gamma(z) = \left( \sum_{n=-N}^N c_n z^n \right) - \delta(\theta), c_n \in \mathbb{C}.$$

Basically, $\gamma(e^{it}) = \delta(t) - \delta(\theta)$.  Crucially, assume that $\gamma(z)$ has no roots in the unit disk.  Note that $\gamma(z)$ is analytic and $\ln|\gamma(z)|$ is harmonic.

Find a value $y$ such that $\gamma(y) = w - \delta(\theta)$ and $|y| < 1$, assuming there is one (it helps here to relax the requirements on $w$ from the original question and instead place those requirements on $y$).

Then we can construct a Poisson Integral:

$$ \log|w - \delta(\theta)| = \log|\gamma(y)| = \frac{1}{2 \pi} \int_0^{2\pi} P_r(\phi - t) \log|\gamma(e^{it})|dt, \ \ \ \ y = r e^{i \phi}$$

where $P_r(\theta)$ is the Poisson Kernel and is given by

$$ P_r(\phi) = \Re\left( \frac{1 + r e^{i \phi}}{1 - r e^{i \phi}} \right), \ \ \ \ 0 \leq r < 1.$$

Therefore

$$ \boxed{f(t, w) = \frac{1}{2 \pi} \Re\left( \frac{1 + y e^{-it}}{1 - y e^{-it}} \right)}$$

Obviously there are a lot of limitations here.  $\gamma(z)$ can't have any roots in the unit disk for any value of $\theta$ (if there are roots the Poisson Integral gets some extra terms that I don't know how to account for).  Also you need to be able to find a $y$ in the unit circle, because the Poisson Kernel is only available there.  Otherwise you're free to select $w$ and $\delta(t)$, though.


问:
existence of a derivative $x\cdot f(x)$
Lets say we have a function $f(x)$ that has a derivative at point $a$.
Can we prove that the function $x\cdot f(x)$ has also a derivative at point $a$?
If this is not true, can anybody give an example that shows this ...</p>

答: Hint: Let $h(x) = x\cdot f(x)$. By the product rule, what is $h'(a)$?

Title: existence of a derivative $x\cdot f(x)$


问:
How could I solve this limit with polar coordinates?
The problem I'm working on is the following:

$$\lim _{(x,y)→(0,0)} \frac{(x^2ye^y)}{(x^4+4y^2)}$$

I don't quite understand how to separate the equation to solve it. Because if I substitute directly by polar coordinates, it gives me zero, and the real answer is that the limit doesn't exist.

答:
Along the coordinate axes we get zero, so this means that if we can find a path  for which the limit along it is not zero, we may conclude that the limit does not exist.

So let $x = t$ and $y = t^2/2$. Then $$\lim_{t \to 0} \frac{t^2(t^2/2)e^{t^2/2}}{2t^4} = \lim_{t\to 0} \frac{e^{t^2/2}}{4} = \frac{1}{4}\neq 0.$$The choice of path was made to make the denominator as simple as possible, and it worked.


答:
In the numerator, since both $\tan(x)$ and $\sin(x)$ are odd functions, the difference is also an odd function. The denominator $x^2$ is an even function. The quotient is an odd function. If the limit as $x$ approaches $0$ from positive reals exists, then it is the negative of the limit from negative reals. Thus, the bidirectional limit exists and it must be equal to its negative. That implies the limit is zero if it exists.

Your last step is not valid since the first factor goes to infinity and the second goes to zero. This is an indeterminate form and hence we can't tell what the limit is from that.

Title: Understanding Limit without l'Hopital

Body:
I have been trying to understand this limit:

$$\lim_{x \to 0}\frac{tan(x)-sin(x)}{x^2}$$

When aplying the l'Hopital rule I arrive to the limit being $0$ but when doing things organically I get an indetermination:

$$
\lim_{x \to 0}\frac{tan(x)-sin(x)}{x^2}=\lim_{x \to 0}\frac{tan(x)}{x^2}-\frac{sin(x)}{x^2}= \lim_{x \to 0} \frac{sin(x)}{cos(x)x^2}-\frac{sin(x)}{x^2}= \lim_{x \to 0}\frac{sin(x)}{x^2}(\frac{1}{cos(x)}-1)
$$

Clearly $\lim_{x \to 0} \frac{1}{cos(x)}=1$ hence $(\frac{1}{cos(x)}-1)=0$ and I could well aply $\lim_{x \to 0}\frac{sin(x)}{x}=1$ but that still leaves $\lim_{x \to 0}\frac{1}{x}$ which is undetermined because it has different limits on $0^-$ and $0^+$.

Is there something I'm missing?


问:
Bug moving on number line in random directions
A bug at the origin moves on the number line either left or right $u$ units every second, where $u$ is am arbitrary real between -1 and 1. For example, the bug could go from 0 to -0.4 in one second, then -0.4 to 0.3 in the next second, etc.. How long (in seconds) do you expect the bug to take before it is more than or at least a distance 1 from the origin (i.e. its position is less than or equal -1 or greater than equal to 1)?

EDIT #1: I tried representing the expected value as a function of the position it is currently at, but couldn't get any further

EDIT #2: I ran a computer simulation for about 100 million trials and got an answer of around $5.36457$, similar to @DreiCleaner .

@Mike Earnest had a promising approach expanding on my [EDIT #1].

His appraoch:

Let $f(x)$ be the expected value of remaining steps if you are currently at position $x$. Then it can be shown that $f(x)=1+\frac12\int_{\max(x-1,-1)}^{\min(x+1,1)} f(t)\,dt$.

I do not see why this is true yet.

答:
OK, here is what I have so far; I think this leads to a solution, but there is just some messy algebra near the end that I cannot bring myself to attempt.

As I mentioned in my comment, and you stated in your post, letting $f(x)$ be the expected number of steps starting, then for $x>0$, we have
$$
f(x)=1+\frac12\int_{x-1}^1f(t)\,dt\tag{1}
$$
Differentiating both sides with respect to $x$, you get
$$
f'(x)=-\frac12f(x-1)=-\frac12 f(1-x)\tag2
$$
where the last step follows since $f$ is an even function. Differentiating $(2)$ again, you get $f''(x)=\frac12f'(1-x)$, which combined with $(2)$ shows that
$$
f''(x)=-\frac14 f(x),\qquad 0<x<1\tag3
$$
Now, $(3)$ implies that
$$
f(x)=A\cos(x/2)+B\sin(x/2),\qquad 0<x<1
$$
for some constants $A$ and $B$. Note that this expression is only valid on the positive half of the domain, $0<x<1$. To extend this to the entire domain, you would use the fact that $f$ is even to write
$$
f(x)=A\cos(x/2)+B\sin (|x|/2)\tag 4
$$
All that remains is to determine the constants $A$ and $B$. This should be doable by substituting $(4)$ into $(1)$.