问:
when does $\det(AB^T+BA^T)\le \det(AA^T+BB^T)$ hold?
When does the following matrix inequality hold?

$$\det(AB+B^TA^T)\le \det(AA^T+BB^T)$$

$A$ and $B$ are any real matrices.

My reply gives a counter example. The question is under what condition does that hold?

[Update]

The inequality actually I would like to prove is not the above one. It should be

$$\det(AB^T+BA^T)\le \det(AA^T+BB^T)$$

I also changed the title. Thanks julien.

答:
I assume that $A,B$ are real $n\times m$ matrices. But the same proof gives the same result as below for complex matrices with the adjoint instead of the transpose.

For every real $n\times m$ matrices $A,B$, we have
  $$
|\det (AB^T+BA^T)|\leq \det(AA^T+BB^T).
$$

Proof: if $x\in\ker (AA^T+BB^T)$, then $\|A^Tx\|^2+\|B^Tx\|^2=0$, whence $A^Tx=B^Tx=0$ and so $x\in \ker(AB^T+BA^T)$. In other words, if $\det(AA^T+BB^T)=0$, we do have $\det(AB^T+BA^T)=0$ and the inequality holds. So we will assume that $AA^T+BB^T$ is invertible from now on.

The matrix $C=(A-B)(A-B)^T$ is positive semidefinite. So
$$
0\leq (Cx,x)=((AA^T+BB^T)x,x)-((AB^T+BA^T)x,x)
$$
whence
$$
((AB^T+BA^T)x,x)\leq ((AA^T+BB^T)x,x).
$$
Changing $B$ for $-B$, we get altogether

$$
|((AB^T+BA^T)x,x)|\leq ((AA^T+BB^T)x,x).
$$

Denote $C:=AB^T+BA^T$ which is symmetric, and $D:=AA^T+BB^T$ which is positive definite and has a unique positive definite square root $D^{1/2}$. Now the inequality above entails, with $x=D^{-1/2}y$:
$$
|(D^{-1/2}CD^{-1/2}y,y)|\leq(y,y)=\|y\|^2\qquad\forall y\in\mathbb{R}^n.
$$
Since $T=D^{-1/2}CD^{-1/2}$ is symmetric, it is diagonalizable in an orthonormal basis and the above shows that its spectrum is contained in $[-1,1]$. A fortiori
$$
|\det (D^{-1/2}CD^{-1/2})|\leq 1\qquad \Rightarrow \qquad |\det C|\leq \det D.
$$
QED.

Note: the inequality you wanted initially requires that $A,B$ be square matrices of the same size. As you observed, the inequality is not true in general. The transpose was not at the right place.  It follows in particular from the inequality above that the inequality you first wanted holds when $A,B$ are both symmetric, or both skew-symmetric. Of course, it is true also if $AB=0$. One can add some more solutions. All $(A,B)$ such that the symmetric matrix $AB+B^TA^T$ is singular or has an odd number of negative eigenvalues. But I doubt there is a nice general characterization. Although I might be wrong.


问:
which of these is a Hausdorff Space?
I have found the topologies on the set $\{a, b\}$:

$\{\emptyset, \{a, b\}\}$,

$\{\emptyset, \{a\}, \{a, b\}\}$,

$\{\emptyset, \{b\}, \{a, b\}\}$

and $\{\emptyset, \{a\}, \{b\}, \{a, b\}\}$,

and now want to determine which of these is Hausdorff/separated.

I have the definition that in order of for this each distinct point of the space must have a disjoint neighbourhood.

My thinking is that the only separated space in the list is $\{\emptyset, \{a, b\}\}$, because it is the only one that has no two elements with both $a$ and $b$?

答:
A space $X$ is Hausdorff if, given $x,y\in X$, with $x\ne y$, there are open sets $U,V$ such that $x\in U,y\notin U$ and $x\notin V,y\in V$.

You need a pair of open sets for any pair of distinct points.

By the way, a finite set is Hausdorff if and only if the topology is discrete, because in a Hausdorff space singleton sets are closed.

ContentLicense: CC BY-SA 4.0

Title: which of these is a Hausdorff Space?


答:
Not right. The reduced row echelon form turns into two equations:
\begin{align*}
x_1 + x_3 &= 0 \\
x_2 + x_3 &= 1
\end{align*}
(not pictured: $x_4$). We indeed have $x_3$ and $x_4$ as free variables, but something seems to have gone wrong when you figured out $x_1$ and $x_2$ based on these free variables.

The first equation rearranges to $x_1 = x_3$, i.e. $x_1$ must be the same as $x_3$. On the other hand, $x_2 + x_3 = 1$ if and only if $x_2 \neq x_3$ That is, $x_2$ must be the opposite value to both $x_1$ and $x_3$. On the other hand, $x_4$ is totally free, not affecting any other variable. So, our solutions are:
$$(x_1, x_2, x_3, x_4) = (1, 0, 1, 0), (1, 0, 1, 1), (0, 1, 0, 0), (0, 1, 0, 1).$$

EDIT: If you specifically wish to write it as a particular solution plus a homogeneous solution, you can follow the usual method. The variable $x_4$ is completely free, so we shall call it $t$. The variable $x_3$ is also free, but $x_1$ and $x_2$ depend on it. Let's call it $s$, and note $x_1 = -s = s$ and $x_2 = 1 - s = 1 + s$. Thus,
$$(x_1, x_2, x_3, x_4) = (s, s + 1, s, t) = (0, 1, 0, 0) + s(1, 1, 1, 0) + t(0, 0, 0, 1).$$
The first term is a particular solution, while the other two terms form the full homogeneous solution.

Title: $\left( {\begin{array}{cccc|c} 1 & 0 & 1 & 0 & 0\\ 1 & 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 & 1\\ \end{array} } \right) $ in $\mathbb{Z}_{2}$

Body:
\begin{align}
&\left( {\begin{array}{cccc|c}
1 & 0 & 1 & 0 & 0\\
1 & 1 & 0 & 0 & 1\\
0 & 1 & 1 & 0 & 1\\
\end{array} } \right)
\text{in}\ \mathbb{Z}_{2} \\
&=\left( {\begin{array}{cccc|c}
1 & 0 & 1 & 0 & 0\\
0 & 1 & 1 & 0 & 1\\
0 & 0 & 0 & 0 & 0\\
\end{array} } \right)
\end{align}

I set $z=t=0    (x,y,z,t)$ and I got a partial solution $(0,1,0,0).$

I solved two homogeneous matrices once for $z = 1$ and $t = 0$, then for $z = 0$ and $t = 1$ and I got two solutions $(1,1,1,0)$ and $(1,1,0,1).$

Then, I got $(0,1,0,0) + a*(1,1,1,0)+b*(1,1,0,1$)

Therefore, all possible results are $(0,1,0,0),(1,0,1,0),(1,0,0,1),(0,1,1,1)$

Would this be correct?


问:
Find the area of the shaded region between $r=e^{\theta/2}$ and $r=θ$ .

That's the picture of the shaded region I have to find the area of. I'm totally stuck on this problem mainly because these two curves don't intersect so I'm not sure how to find the bounds of integration. Once I find those my guess would be to the outside curve - the inside curve and just do the integration from there.

答:
We have to find the area of the region $R$ between $r=\theta $ and $r=e^{\theta/2}$, and between $\theta =0$ and $\theta =\pi /2$. Since the Jacobian of the transformation from Cartesian to polar coordinates is $r$, the area $A$ is given by the double integral (see Explain $\iint \mathrm dx\,\mathrm dy = \iint r \,\mathrm \,d\alpha\,\mathrm dr$):
$$
\begin{eqnarray*}
A &=&\iint_{R}dA=\int_{0}^{\pi /2}\left( \int_{\theta }^{e^{\theta
/2}}r\;dr\right)\ d\theta  \\
&=&\int_{0}^{\pi /2}\big( \frac{e^{\theta }}{2}-\frac{\theta ^{2}}{2}
\big)\ d\theta  \\
&=&\frac{1}{2}\int_{0}^{\pi /2}e^{\theta }d\theta -\frac{1}{2}\int_{0}^{\pi
/2}\theta ^{2}\ d\theta  \\
&=&\big( \frac{e^{\pi /2}}{2}-\frac{1}{2}\big) -\frac{\pi ^{3}}{48}.
\end{eqnarray*}
$$</p>

ContentLicense: CC BY-SA 3.0


问:
Is it mathematically correct to write the primitive of a function in this way : $F(x)=\int f(x)dx$
Is it mathematically correct to write the primitive of a function in this way :
$F(x)=\int f(x)dx$

or should we absolutely change the variable name : as in $F(t)=\int f(x)dx$ ?
(that is : not use the same variable inside the integral)

答:
$F(x)=\int f(x)dx$ is very common.  So you will never convince everyone not to do it.

However $F(t)=\int f(x)dx$ is worse.  The RHS does not reveal what the variable is.

Some peculiar people may write $F = \int f$, which is OK as far as variables are concerned.

Related:  Do not write $F(x)=\int_0^x f(x)dx$. That has $x$ used for two different things in the same formula.

But it is OK to write $F(t)=\int_0^t f(x)dx$.  This time the RHS does reveal what the variable is.


问:
Finding the determinant of a $n\times n$ matrix of rational functions
Here is my matrix. How do I find the determinant of this one?

$$ \begin{pmatrix} 1 & 1 & 1 & \ldots & 1\\ \dfrac{z^{2}+1}{z}& \dfrac{z^{4}+1}{z^2}& \dfrac{z^{6}+1}{z^3}& \ldots & \dfrac{z^{2n}+1}{z^n}\\ \dfrac{z^{4}+1}{z^2}& \dfrac{z^{8}+1}{z^4}& \dfrac{z^{12}+1}{z^6}&  \ldots &\dfrac{z^{4n}+1}{z^{2n}}\\ \vdots & \vdots & \vdots & \ddots & \vdots &\\ \dfrac{z^{2(n-1)}+1}{z^{n-1}}& \dfrac{z^{4(n-1)}+1}{z^{2(n-1)}} & \dfrac{z^{6(n-1)}+1}{z^{3(n-1)}}& \ldots & \dfrac{z^{2n(n-1)}+1}{z^{n(n-1)}}&\end{pmatrix} $$

I'm really trying to solve it but I can't think of anything. Does it have anything to do with Vandermonde?

答:
As you suggest, we can reformulate this determinant ($D$) as the determinant of a Vandermonde matrix. Until I saw this question I had never even heard of a Vandermonde matrix, so there are probably easier ways of handling this.

First we note that the elements of the matrix can be written as
$$p_k=z^k+z^{-k}$$

Then
$$ D=\begin{vmatrix} 1 & 1 & 1 & \ldots & 1\\ p_1&p_2 & p_3 & \ldots & p_n\\ p_2 & p_4 & p_6 &  \ldots &p_{2n}\\ p_3 & p_6 & p_9 &  \ldots &p_{3n}\\ \vdots & \vdots & \vdots & \ddots & \vdots &\\ p_r & p_{2r} & p_{3r} &  \ldots &p_{rn}\\ \vdots & \vdots & \vdots & \ddots & \vdots &\\ p_{n-1} &  p_{2(n-1)}  &  p_{3(n-1)}  & \ldots &   p_{n(n-1)} &\end{vmatrix} $$

This looks a bit more like a Vandermonde determinant except that (a) it is transposed from the way the Wikipedia page lays it out; and (b) we want things like $p^k$ rather than $p_k$.

(a) is irrelevant because taking the transpose of a matrix doesn't affect its determinant.
(b) can be addressed with some row manipulations.
We note that
$$p_r^2 = (z^r+z^{-r})^2=p_{2r}+2 $$
$$p_r^3 = (z^r+z^{-r})^3=p_{3r}+3p_r $$
$$p_r^4 = (z^r+z^{-r})^4=p_{4r}+4p_{2r}+6 $$
$$p_r^5 = (z^r+z^{-r})^5=p_{5r}+5p_{3r}+10p_{r} $$

and in general

$$p_r^k = p_{kr} + a_{k-2}p_{(k-2)r} + a_{k-4}p_{(k-4)r}+\cdots $$
for some constants $a$. So we can change $p_{kr}$ to $p_r^k$ with the only changes being combinations of the previous rows.

For example we can rewrite the third row ($p_2,p_4,p_6,\dots$) as ($p_1^2,p_2^2,p_3^2\dots$) by subtracting twice the first row from it. Similarly we can rewrite the fourth row ($p_3,p_6,p_9,\dots$) as ($p_1^3,p_2^3,p_3^3\dots$) by subtracting three times the second row from it. In general, we can rewrite all of the rows and get:
$$ D=\begin{vmatrix} 1 & 1 & 1 & \ldots & 1\\ p_1&p_2 & p_3 & \ldots & p_n\\ p_1^2 & p_2^2 & p_3^2 &  \ldots &p_{n}^2\\ p_1^3 & p_2^3 & p_3^3 &  \ldots &p_{n}^3\\ \vdots & \vdots & \vdots & \ddots & \vdots &\\ p_1^r & p_{2}^r & p_{3}^r &  \ldots &p_{n}^r\\ \vdots & \vdots & \vdots & \ddots & \vdots &\\ p_1^{n-1} &  p_2^{n-1}  &  p_3^{n-1}  & \ldots &   p_n^{n-1} &\end{vmatrix} $$

This is exactly a Vandermonde determinant and so, as shown on https://en.wikipedia.org/wiki/Vandermonde_matrix
, we have
$$D= \prod_{1 \le i < j \le n} (p_j - p_i)$$

It is helpful to write
$$q_k=z^k-z^{-k}$$
Which leads to
$$p_j-p_i=q_{\frac12(j+i)}q_{\frac12(j-i)}$$
and so
$$D= \prod_{1 \le i < j \le n} q_{\frac12 (j+i)} q_{\frac12 (j-i)} $$

Writing $r_k=q_{\frac12 k}$ gives the slightly simpler looking
$$D= \prod_{1 \le i < j \le n} r_{j+i} \,  r_{j-i} $$
$$= \prod_{j=2}^n \prod_{i=1}^{j-1}  r_{j+i} \, r_{j-i} $$
$$= \prod_{j=2}^n \frac{\prod_{k=1}^{2j-1}  r_{k} }{r_j} $$
$$=   \frac{r_1 \, r_3}{r_2}  \frac{r_1 \, r_3 \, r_5}{r_3}\cdots  \frac{r_1 \, r_3 \, r_5 \dots r_{2n-1}}{r_n}  $$
$$= \frac{r_1^n \, r_3^{n-1} \,  r_5^{n-2} \dots r_{2n-1}^{1} }{r_1 \, r_2\,  r_3\dots r_n}   $$
$$=  \frac{\prod_{k=1}^{n}  r_{2k-1}^{n+1-k} }{\prod_{k=1}^{n}  r_{k} } $$

where
$$r_k=z^{\frac12 k}-z^{-\frac12 k}$$


问:
Eigenvector of $T$ with $T : V \rightarrow V$

Show that for each $0 \neq v \in V$ either $v$ is an eigenvector of $T$ corresponding to $λ = 1$ or $T(v) − v$ is an eigenvector of $T$ corresponding to $λ = 0$

I would just like some guidance on where to start and how to approach the question.
Thank you.

Let $T : V \rightarrow V$ be a linear operator on a finite-dimensional vector space $V$ over $C$ such that $T^2 = T$

Show that for each $0 \neq  v \in V$ either $v$ is an eigenvector of $T$ corresponding to $λ = 1$ or $T(v) − v$ is an eigenvector of $T$ corresponding to $λ = 0$

答:
The hint tells the whole story. Because $T^2 = T$, for any vector $v$ we have
$$
T(Tv - v) = T^2 v - Tv = Tv - Tv = 0.
$$
This means that for any vector $v$ whatsoever, $Tv - v$ is in the nullspace of $T$.  Nonzero vectors in the nullspace of an operator are eigenvectors corresponding to the eigenvalue zero.  So $Tv - v$ will be an eigenvector for the eigenvalue $0$ whenever $Tv - v$ is nonzero.  And if $Tv - v$ happens to be zero?  That means that $Tv = v$, which means (if $v$ is nonzero, as is assumed here) that $v$ is an eigenvector for the eigenvalue $1$.

Note that there is no use of the assumption that $V$ is finite dimensional or that $\mathbb{C}$ is the field.


问:
Adding "C" to an absolute value equation
Does anybody know how the following equation:

$$|a|+|b| \ge |a+b|$$ changes when a $c$ is added? This way:

$$|a| + |b| + |c| \ge |a+b| + |c| \ge |a+b+c|$$

How should I expand this? What happens when a $c$ is added to the equation?

I'd be grateful for help!

答:
You can prove your version with the extra term involving $c$ by using your first inequality a few times with substitution. First to make things clear re-write the first inequality as
$$|x|+|y| \ge |x+y|. \tag{1} $$
Now use (1) with $x=a,y=b$ to get $|a|+|b| \ge |a+b|$ (actually your first ineuality) to which you can add $|c|$ to both sides giving the first of the two inequalities you want, namely
$$|a|+|b|+|c| \ge |a+b|+|c|. \tag{2}$$
Now in (1) put $x=(a+b),y=c$ to get
$$|(a+b)|+[c| \ge |(a+b)+c|=|a+b+c|, \tag{3}$$ the last equality since we don't need the parentheses on the first term on the right of (3).

Finally putting (2) and (3) together gives your desired inequality.


问:
How to find the class of a function
Let $$f(x) = \begin{cases}
 x^3\sin(\frac{1}{x}); & \text{ if, } x \neq 0 \\
 0; & \text{ $x=0$ }
\end{cases}$$

How do I find the greatest value of n such that $f \in C^n ([-1,1])$?

I know to find n, I must find the greatest n such that $f^{(n)}(x)$ is differentiable in the interval $[-1,1]$. But how do I prove its differentiability?

答:
Check that $f'(0)=0$ and $f'(x)=3x^2\sin (1/x) - x\cos (1/x)$ if $x\ne 0.$ It follows that $f'$ is continuous everywhere.

However, for $x\ne 0,$

$$\frac{f'(x)-f'(0)}{x-0} = 3x\sin (1/x) - \cos (1/x).$$

As $x\to 0,$ the first term on the right $\to 0.$ But $\cos(1/x)$ has no limit as $x\to 0.$ Therefore $f''(0)$ does not exist. It follows that $n=1$ is the answer to the problem.

Title: How to find the class of a function


答:
By equalizing, see Calculating basis of null space, one gets immediately the solutions without any further calculations.  From
$$\left( {\begin{array}{cccc|c}
1 & 0 & 1 & 0 & 0\\
0 & 1 & 1 & 0 & 1\\
\end{array} } \right)
$$
equalizing gives
$$\left( {\begin{array}{cccc|c}
1 & 0 & 1 & 0 & 0\\
0 & 1 & 1 & 0 & 1\\
0 & 0 & -1 & 0 & 0\\
0 & 0 & 0 & -1 & 0
\end{array} } \right),
$$
hence the solutions are given by
$$\begin{pmatrix} 0\\1\\0\\0\end{pmatrix}
+r\begin{pmatrix}1\\1\\-1\\0\end{pmatrix}
+s\begin{pmatrix}0\\0\\0\\-1\end{pmatrix}.
$$
Remember that $-1=1$.


问:
Simple Algorithm for Finding Clusters in Matrix
Let's say I have a simple 4x4 matrix which describes a set of clusters. If $A_{ij}=0$ then node $i$ and node $j$ are in the same cluster. For example:
$$
A =
\left(\begin{array}{cccc} 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 1\\ 1 & 1 & 1 & 0 \end{array}\right)
$$

would imply that nodes $1$, $2$, and $3$ are in a cluster together and node $4$ is in a cluster by itself.

Now lets say there is a slight mistake in the matrix where:

$$
A =
\left(\begin{array}{cccc} 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 1\\ 1 & 1 & 1 & 0 \end{array}\right)
$$

Now we have that node $1$ and $2$ should be in the same cluster, node $2$ and $3$ should be in the same cluster, but not node $1$ and $3$.

In the first case, I would just compare equivalency of the rows, but in the latter, now I need something a bit smarter. Any simple solutions in MatLab to make sure that nodes $1$, $2$, and $3$ all make it in the same cluster?

答:
The zero entries of given matrix $A$ form (we assume) a symmetric relation, but not necessarily reflexive or transitive one.  Those three properties are required of an equivalence relation, or what amounts to the same thing, a partition of the nodes (a family of disjoint subsets whose union is all of those nodes, which you called *clusters^).

A natural approach would be to set additional nodes of $A$ to zero in order to achieve the reflexive and transitive properties without losing the symmetric property.  The reflexive property, that every individual node is related to itself, is easily imposed.  Set all the diagonal entries of $A$ to zero.  (Clearly this does not change the symmetric character of $A$, and we shall assume hereafter that it has already been done.)

That leaves us with just the problem of imposing the transitive property.  This can usually be done in more than one way, such as the trivial approach of setting all entries of $A$ to zero.  The (unique) solution which sets the fewest additional entries of $A$ is called the transitive closure, and this may well be the "something a bit smarter" that you want.

Computationally an easy way to do this is by matrix multiplication, which accords well with your desire to do a Matlab implementation.  First take the complement of each entry in $A$, mapping one entries to zeros and zero entries to ones. We will call this matrix $B$.

Now we perform repeated multiplications of $B$, together with setting any nonzero entries that appear to one.  That is one matrix multiplication per step, followed by truncating the entries:

$$ B_1 = B\, ; B_{k+1} = f(B \times B_k) $$

where $f$ is the operation of setting nonzero entries to $1$.

$B_k$ is the identity matrix plus the adjacency matrix of a simple graph, namely the graph whose edges connect a pair of distinct nodes which can be reached in $k$ or fewer steps using the original relation between nodes.

The sequence of matrices $B_k$ eventually stabilizes, and actually $B_n = B_{n-1}$ where $n$ is the matrix size.  So we check when computing the next matrix to see if any additional nonzero entries have been introduced, and we stop whenever there are no new $1$ entries.

It will likely be expeditious to take steps by squaring $B_{2k} = f(B_k^2)$ the matrices.  The termination criterion then becomes $B_{2k} = B_k$.  This still requires only one matrix multiplication per step, and typically only logarithmically as many steps.  So the computational cost is reduced.

Finally one complements the converged matrix $B_k$ by turning $1$ entries to zeros and $0$ entries to ones, and that gives us the "corrected" matrix $A$ you want.


问:
Help on a parsing algorithm for FOL
In Enderton, Section 2.3, to prepare to prove Unique Readability for the wffs in FOL he proves a series of Lemmas involving a function K. On pg. 105 says the following about $K$:

"Recall that the terms are built up from the variables and constant symbols by operations corresponding to the function symbols. We now define a function $K$ on the symbols involved such that for a symbol $s$, $K(s)=1-n$, where $n$ is the number of terms that must follow $s$ to obtain a term:

$K(x)=1-0=1$ for a variable $x$;

$K(c)=1-0=1$ for a constant symbol $c$;

$K(f)=1-n$   for an $n$-place function symbol $f$."

He then sketches inductive proofs for general claims like: For any term $t$, $K(t)=1$.

Before attempting to flesh out this proof, I am having trouble understanding $K$. Take for example a 1-place function like the successor function $S$. Since $S$ is 1-place, by the above definition isn't $K(S)=1-1=0$. In contrast, take a 2-place function like $+$. Again, since $+$ is 2-place, by the above definition, $K(+)=1-2=-1$. What information do I get from the values of $K$?

Is it that by solving for $n$, I am told how many terms need to follow $s$ in order to get a term, i.e. so that $K(s)=1$?

答:
Intuitively, $K(f)=1-n$ signals a need for $n$ terms, whereas $K(x)=1$ signals a supply of one term. In the end, you want to reach the number $1$, meaning that all demands have been met.

More formally, given a sequence
$$
s_1 s_2 \ldots s_n
$$
of symbols, parse the sequence from left to right while computing partial sums $$K(s_1)\;,\; K(s_1)+K(s_2)\;,\;\ldots\;,\;\sum_{i=1}^n K(s_i).$$ Then $s_1 s_2\ldots s_n$ is a well-formed term iff the lower sequence ends with $1$ and all previous numbers are $<1$.

For $m<n$, the number $\sum_{i=1}^m K(s_i)=1-k$ tells you that $k$ further terms have to be supplied as arguments to the function symbols in $s_1\ldots s_m$ until you obtain a well-formed expression.

E.g. given a binary function symbol $f$ ($K(f)=-1$) and a variable $x$ ($K(x)=1$), parsing the sequence
$$
ffxxx
$$
of symbols yields the sequence of partial sums
$$
-1,-2,-1,0,1
$$
and so you do have a well-formed term, namely $f(f(x,x),x)$. The third number $-1 = 1-2$ in the list says that in $ffx$, there are still $2$ terms missing (namely, both the outer and the inner $f$ lack their second argument).

Title: Help on a parsing algorithm for FOL


问:
Generic smoothness in the analytic setting.
If $X, Y$ are non-singular algebraic varieties over an algebraically closed field of characteristic 0, and $f: X \to Y$ is a morphism, then there exists a Zariski-open, non-empty subset $U \subset Y$, such that $f^{-1}(U) \to U$ is a smooth morphism.

I wonder: Is the same true if $X, Y$ are non-singular complex analytic spaces?

I know that Sard's theorem shows that "most" fibers are smooth, in the sense that $f(\operatorname{Sing}(f)) \subset Y$ has Lebesgue measure $0$. But is $f(\operatorname{Sing}(f))$ also contained in a Zariski-closed subset?

If this is true, what is a good reference?

答:
Here is a reference. I have translated the statement and changed the wording slightly.

Theorem [Bănică, Théorème, (ii)]. Let $f\colon X \to Y$ be a morphism of complex analytic spaces. If $f$ is flat, then the set
$$U = \bigl\{ x \in X \bigm\vert X_{f(x)}\ \text{is regular at}\ x\bigr\}$$
is open in $X$, and the complement of $U$ is analytic in $X$.

For proper flat morphisms, we have:

Corollary [Bănică, Corollaire, (ii)]. Let $f\colon X \to Y$ be a proper morphism of complex analytic spaces. If $f$ is flat, then the set
$$V = \bigl\{ y \in Y \bigm\vert X_{y}\ \text{is a manifold}\bigr\}$$
is open in $Y$, and the complement of $V$ is analytic in $Y$.

As Moishe Kohan suggests, the Corollary follows from the Theorem using Remmert's theorem.

To obtain a statement that does not mention flatness, you can use Frisch's theorem [Frisch, Théorème (IV, 9)] (see [Kiehl, Satz 4] for an alternative proof), which says that the locus in $X$ at which $f$ is not flat is closed and analytic. For a textbook account of Frisch's theorem, see [Bănică–Stănăşilă, Theorem 4.5] (their proof follows [Kiehl]).

Finally, you can see [Bingener and Flenner, Corollary 2.1] for an English reference proving variants of these results (they also prove a version for real analytic spaces).


问:
Prove $\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}≥\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
Question : Prove $$\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$
$(a, b, c \in \mathbb{R}^+)$

I tried to solve it like this :
$$\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 2 \; (\frac{a}{b^2}+\frac{b}{c^2}+\frac{c}{a^2})$$

Am I doing this right? How can I finish this problem?

答:
We can rewrite LHS as:

$$3[\frac 1b(\frac{a^2}{b^2})+\frac 1c(\frac{b^2}{c^2})+\frac 1a(\frac{c^2}{a^2})]\geq(\frac 1c+\frac 1b+\frac 1a)(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2})$$

This holds , due to Tchebychev inequality, if $a<b<c$, such that both factors on RHS are in ascending order of magnitude. Also we apply this inequality:

$$a^2c+b^2a+c^2b\geq3abc$$

Dividing both sides by $abc$ we get:

$$\frac ab+\frac bc+\frac ca\geq 3$$

Therefore:

$$\frac {a^2}{b^2}+\frac {b^2}{c^2}+\frac {c^2}{a^2}\geq 3$$

Therefore:

$$\frac 1b(\frac{a^2}{b^2})+\frac 1c(\frac{b^2}{c^2})+\frac 1a(\frac{c^2}{a^2})\geq(\frac 1c+\frac 1b+\frac 1a)$$


问:
Queueing model with two servers
I have a two-server queue with Poisson arrival rate and $\lambda$ exponential services with $\mu$ ( first server service rate) and 2$\mu$ ( 2nd server service rate). Capacity is infinite.

Then why is the number of customers in the queue at time $t$ not a Markov Process?

Can you please help me out?

Thank You.

答: You need extra information to make the process a Markov process. The Markov property requires that the future depend only on the current state, but suppose you have only a single customer in the system. The future behaviour of the system depends on the history of the process (namely which server that job started service with) not just the current state. (It could experience service at rate either $\mu$ or $2\mu$.)</p>

Title: Queueing model with two servers


问:
Classification of the Finite Abelian Groups
I am studying "Abstract Algebra" by Gregory T. Lee. I am going through chapter 5, "Direct Products and the Classification of Finite Abelian Groups" at the moment, but it has been rather difficult for me. I would like to know if you know of any books, papers or online videos where this topic is discussed in detail and with examples, in order to understand a little better. Also, it would help me if you told me here what are the main ideas of this subject, what I should focus on to overcome it. In this book the subject is developed using p-groups. Thanks in advance</p>

答: I found Gallian's, "Contemporary Abstract Algebra" to be very clear on these topics. It comes with plenty of exercises, half of which with solutions.


问:
Let $f:\mathbb{R} \to \mathbb{R}$ continuous with $f(f(x))=e^x$, show that $\lim_{x\to \infty } \frac{f(x)}{x^n}=\infty$ (Brazilian Olympiad)
Given $f:\mathbb{R} \to \mathbb{R}$ continuous with $f(f(x))=e^x$ for all $x\in\mathbb{R}$. Show that $\lim_{x\to \infty } \frac{f(x)}{x^n}=\infty$ for all $n\in\mathbb{N}$.

I appreciate any help!

Edit 1: It is easy to see that $f$ is injective. So, if $f$ increases anywhere, $f$ will be an increasing function.

Edit 2: As said in the comments, $f$ must indeed be increasing, since
$f(1)=f(e^0)=f(f(f(0)))=e^{f(0)}>f(0)$.

答:
remark

It is not enough to assume (for example) that $f(f(x)) = e^x$ for all $x>1$.  You must assume (as the OP says) $f(f(x))=e^x$ for all $x$.  This is used in Edit 2.

The following function $f : \mathbb R \to \mathbb R$
is continuous and $f(f(x)) = e^x$ for all $x > 1$ but fails the required
conclusion.

$$
f(x) = \begin{cases}
-2x,\quad &x\ge T
\\
e^{-x/2},\quad &x < T
\end{cases}
$$
Here $T \approx -0.7148$ is one of the points where the two graphs $-2x,e^{-x/2}$ cross.


问:
If $a,b\in \mathbb R^+$, $ |a-2b|\leq\frac {1}{\sqrt{a}}$, $|b-2a|\leq\frac {1}{\sqrt{b}}$ Prove $a+b\leq 2$

Question:  If $a,b\in \mathbb R^+$, $|a-2b|\leq\frac {1}{\sqrt{a}}$,  $|b-2a|\leq\frac {1}{\sqrt{b}},$ prove $a+b≤2$.

I figured out that $a+b\leq \frac {1}{\sqrt{a}}+\frac {1}{\sqrt{b}}$, but I am not sure how to prove that $a+b\leq 2$ after doing this.

Can anyone help me?

答: Note by given condition$$a{(a-2b)}^2+b{(b-2a)}^2\le 1+1 $$ $$\iff a^3+b^3\le 2$$ Now use by power-mean inequality that $$\frac{{(a+b)}^3}{4}\le a^3+b^3$$


问:
Homeomorphism between spiral and positive real axis
I have given the following two subsets of $\mathbb{R^2}$:
$$ M = \lbrace{(t,0) \in \mathbb{R}^2 \rbrace | t >0 \rbrace},~~ N = \lbrace{ ( t \cos(1/t),t \sin(1/t)) \in \mathbb{R}^2 | t > 0\rbrace}.$$
The exercise asks me to show that there is a homeomorphism $\psi \colon \mathbb{R}^2 \to \mathbb{R}^2$ that satisfies $\psi(M) = N$.
My approach was to identify $\mathbb{R}^2$ with the complex plane $\mathbb{C}$ using the map $(x,y) \mapsto z = x+ iy$, then the map $z \mapsto w = z \exp(i/z)$ and finally map back to $\mathbb{R}^2$ with $w \mapsto (\text{Re}(w),\text{Im}(w))$. This however does not lead to a continuous function on $\mathbb{R}^2$ and I don't see what function I could take in stead of $ z \exp(i/z)$ to satisfy $\psi(M) = N$. I would be thankful for any hints as to how I can find this homeomorphism.

答:
Identifying $\mathbb R^2$ with $\mathbb C$, define $$f(z)=ze^{-i/|z|}$$
In polar cohordinates this is $f(\rho,\theta)=(\rho,\theta-1/\rho)$. Then check that $f$ extends contintuously in $z=0$ by $f(0)=0$ and  that the result is indeed a homeo.

Your spiral $N$ in polar cohordinate is $\{(\rho,1/\rho)|\rho>0\}$, so $f(N)$ becomes $\{(\rho,0)|\rho>0\}=M$.


问:
Is this problem in P or NP-Complete?
I need to determine if the following problem is in P or NP-Complete:

$\mathrm{2-IS} = \{\left< G, k\right > | G \text{ is a graph which every node in it has a degree 2 AND there is an independent set of size $k$ in $G$}\}$

My intuition says it's in NP-Complete but I can't find another NP-Complete problem to reduce it to... any help?

答:
Here are a few ingredients you might find useful for cooking the polynomial algorithm for 2-IS (no, it's not NP-complete; unless P=NP):

Checking if every node in a graph has degree $2$ is easy.

A graph in which every node has degree $2$ is a disjoint union of one or more cycles.

The size of maximum independent set of a cycle on $n$ nodes is $\lfloor\frac{n}{2}\rfloor$.

Finding all the cycles can be performed by a depth-first search, for example.

Title: Is this problem in P or NP-Complete?


问:
Am I missing something in the definition of universal generalization?
Today I came across the definition of universal generalization in first-order logic.

The definition in my script goes as follows (translated):

If we know, that for any element $a \in X: P(a)$ is true, we can conclude, that $\forall x \in X:P(x)$ is true.

Now for my question: Is it legitimate when I just write: $\forall a \in X:P(a)\rightarrow \forall x \in X: P(x)$

This definition of mine seems redundant to me, and therefore I have doubts that it is right. Is it the same, or did I get something wrong?

答:
The meaning of $\forall$ is given in natural language. "$\forall$" and "$\text{for all/any/each}$" live in different realms. Natural language has meaning before we start constructing the formalism, and so it can be used to give the semantics of "$\forall$" without being circular or redundant.

Your translated bit of text is saying the following, written as an inference rule

$$ \frac{\text{For all elements $a$ in $X$, $P(a)$ holds}}{\forall x \in X \mathop. P(x)} $$

By writing down $\forall a \in X \mathop. P(a) \implies \forall x \in X \mathop. P(x)$ and concluding that it holds, you have used the above inference rule.


问:
$N(X)$ denotes collection of all nets of $X$. $T:N(X)\to \mathcal{P}(X)$ be a map such that $T\{x_\delta\}=T\{y_\gamma\}$ if one is subnet of other.

Let $X$ be a non-empty set. $N(X)$ denotes the collection of all nets on $X$ and $\mathcal{P}(X)$ denotes the collection of all subsets of $X$. Let $T:N(X)\to \mathcal{P}(X)$ be a map satisfying the condition $T\{x_\lambda\}_{\Lambda}=T\{y_\gamma\}_{\Gamma}$ if one of $\{x_\lambda\}_{\Lambda}$ and $\{y_\gamma\}_{\Gamma}$ is a subnet of the other.

Define $\tau=\{A\subset X|\ T\{x_\lambda\}_{\Lambda}\subset A^c\text{ for every net }\{x_\lambda\}_{\Lambda}\subset A^c\}$

We have to prove that $\tau$ is a topology on $X$. I have tried the problem in the following manner-

$\emptyset\in \tau$ since given any net $\{x_\lambda\}_{\Lambda}\subset (\emptyset)^c=X$, we always have $T\{x_\lambda\}_{\Lambda}\subset (\emptyset)^c=X$ . Again, $X\in \tau$ since $X^c=\emptyset$

Let $\{A_i|\ i\in I\}\subset \tau$. Then let $\{x_\lambda\}_{\Lambda}\subset \left(\bigcup A_i\right)^c=\bigcap A_i^c\implies \{x_\lambda\}_{\Lambda}\subset A_i^c\ \forall i\implies T\{x_\lambda\}_{\Lambda}\subset A_i^c\ \forall i\implies T\{x_\lambda\}_{\Lambda}\subset \bigcap A_i^c=\left(\bigcup A_i\right)^c$. Hence, $\bigcup A_i\in \tau$.

Let $A,B\in \tau$ and $\{x_\lambda\}_{\Lambda}\in (A\cap B)^c=A^c\cup B^c$. Now we have to prove $T\{x_\lambda\}_{\Lambda}\subset A^c\cup B^c$.

Now I define $\Gamma=\{\lambda\in\Lambda|\ x_\lambda\in A^c\}, \Sigma=\{\lambda\in\Lambda|\ x_\lambda\in B^c\}$. Now if I can prove at least one of $\Lambda_1,\Lambda_2$ is directed set then I will construct a subnet of $\{x_\lambda\}_{\Lambda}$, and use the property of $T$ to achieve our goal. But I'm to prove that at least one of $\Gamma,\Sigma$ is directed set, although I have shown that both $\Gamma, \Sigma$ satisfies reflexivity and transitivity condition. But we have to show the third condition which is $$\text{If }\lambda_1,\lambda_2\in \Gamma\text{ (or }\Sigma\text{)}, \exists \lambda_3\in \Gamma\text{ (or }\Sigma\text{) such that } \lambda_3\ge\lambda_1,\lambda_2$$

Can anyone help me with this regard? Thanks for your help in advance.

答: $\Gamma \cup \Sigma = \Lambda$ so one of them is cofinal in $\Lambda$ ( if neither, then the directedness gives an immediate contradiction ), so say that $\Gamma$ is cofinal. But then $\Gamma$ defines a subnet of $x$ living on $A^c$ (the cofinality ensures its also directed as a suborder) and so has the same $T$ set as $x$, and as $A\in \tau$ that $T(x) \subseteq A^c$ and we’re done.


问:
Prove by induction: $9\mid (4^{3n} + 8)$ for all integers $n \geq 0$.
I cannot seem to figure this one out.

My though process so far is the following:

$\textit{Proof}$: For the base case n = 0, we have $9 \mid (4^0 + 8)$. This is true.

$\textit{Inductive step:}$ Suppose for $k \geq 0$ we have that $9\mid (4^{3k} + 8)$. We will show that $9 \mid (4^{3(k + 1)} + 8)$. The statement $9 \mid (4^{3k} + 8)$ can be expressed as $4^{3k} + 8 = 9a$ for integer $a$. Observe that
$$\begin{equation}\begin{split} 4^{3(k + 1)} + 8 &= 4^{3k}4^3 + 8 \\ &=
 \dots \\ &\vdots \\ &= 9 (z)\end{split}\end{equation}$$
Where $z$ is some integer. I feel I am forgetting some algebraic properties or a manipulation of the equation above of some sort to help in this prove.

Any help is great! Thanks!

答:
Hint: Induction is not necessary. What is the order of $2\bmod 8$?

It you really need induction, you can rewrite
\begin{align}
4^{3(k + 1)} + 8&= 4^{3k} \,4^3 + 8\cdot4^3 -8\cdot4^3+8 \\
&=4^3\bigl(4^{3k}+8\bigr)+8(1-4^3).
\end{align}


Title: Prove by induction: $9\mid (4^{3n} + 8)$ for all integers $n \geq 0$.

Body:
I cannot seem to figure this one out.

My though process so far is the following:

$\textit{Proof}$: For the base case n = 0, we have $9 \mid (4^0 + 8)$. This is true.

$\textit{Inductive step:}$ Suppose for $k \geq 0$ we have that $9\mid (4^{3k} + 8)$. We will show that $9 \mid (4^{3(k + 1)} + 8)$. The statement $9 \mid (4^{3k} + 8)$ can be expressed as $4^{3k} + 8 = 9a$ for integer $a$. Observe that
$$\begin{equation}\begin{split} 4^{3(k + 1)} + 8 &= 4^{3k}4^3 + 8 \\ &=
 \dots \\ &\vdots \\ &= 9 (z)\end{split}\end{equation}$$
Where $z$ is some integer. I feel I am forgetting some algebraic properties or a manipulation of the equation above of some sort to help in this prove.

Any help is great! Thanks!

Body:
$4^{3k}+8=9a\to 4^{3k}=9a-8$

$4^{3k+3}+8=4^{3k}\cdot 4^3+8=(9a-8)\cdot 64+8=576 a-504=9(64a-56)=9z$


问:
Is $u(t)=u(2t)$ true?
Since $u(t)=1$ for $t>0$ and $0$ otherwise, this means that $u(t)=u(2t)$ for all $t$. Also, they both have identical graphs.

But when obtaining the derivative of both of these functions they give different results.
$$\frac{du(t)}{dt} = \delta(t)$$
$$\frac{du(2t)}{dt} = 2\delta(2t)$$
Where the second derivative is obtained using the chain rule.

So does this mean that $u(t)\neq u(2t)$? or the second derivative is wrong?

答:
Hint

We want derivative in the sense of distributions.  So to check equality, you need to check equality in the sense of distributions.  What distribution is $2 \delta(2t)$?  That is, if $\varphi$ is a test function, what is $\langle 2\delta(2t), \varphi\rangle$? Is that the same distribution as $\delta(t)$?  Answer: Yes.

But: if you do not know how to compute in distributions, then this question is too advanced for you.

Title: Is $u(t)=u(2t)$ true?


问:
Find $a_n =1/\pi\int_{-\pi}^{\pi}\cos(x)\cos(nx)dx$
I am trying to compute Fourier coefficients for the function $f(x) = \cos x$. I have managed to solve that the coefficients "$c$" and "$b$" for the function are $0$. However, I am struggling to solve the following integral (in order to compute coefficient "a")

$$a_n =\frac{1}{π}\int_{-\pi}^{\pi}\cos(x)\cos(nx)dx$$

where n is a positive integer.

I have tried to solve the problem by using integration by parts but this gets me going around in circles. I have also tried to solve the integral using the cosine product-to-sum formula.

For context, I am facing this problem in a calculus textbook, not in the context of Fourier analysis . The purpose of the exercise is solely to practice integration techniques (I am not yet familiar with Fourier series or analysis).

Edit: here is what I get when using the product-to-sum formula:
$$\frac{1}{2π}\frac{2\sin(\pi+\pi n)}{(1+n)}+\frac{1}{2π}\frac{2\sin(\pi-\pi n)}{(1-n)}$$

答:
Recall that
$$\cos(\theta+\varphi)=\cos\theta\cos\varphi-\sin\theta\sin\varphi$$
Therefore, we can show that
$$\cos\theta\cos\varphi=\frac 12\bigr(\cos(\theta+\varphi)+\cos(\theta-\varphi)\bigr)$$
Setting $\theta=x$ and $\varphi=nx$ gives
$$\int\limits_{-\pi}^{\pi}\mathrm dx\,\cos x\cos nx=\frac 12\int\limits_{-\pi}^{\pi}\mathrm dx\,\cos x(1+n)+\frac 12\int\limits_{-\pi}^{\pi}\mathrm dx\,\cos x(1-n)$$
Can you finish the rest?


问:
Tips on how to start calculating a definite integral
i would appreciate if you could help me with this problem.
I don't want the answer only how should i proceed with solving an integral like this
$$I=\int_{0}^{\pi}\frac{x \sin^{2018}(x)}{\cos^{2018}(x)+\sin^{2018}(x)}dx$$
I am completely overwhelmed on how to proceed with this and i am stuck. so i would appreciate some tips to start.

after reading the tips i have done this
$$\int_{0}^{\pi}\frac{(\pi-x) \sin^{2018}(\pi-x)}{\cos^{2018}(\pi-x)+\sin^{2018}(\pi-x)}dx$$

after that i distributed it
$$\int_{0}^{\pi}\frac{\pi\sin^{2018}(\pi-x)-x\sin^{2018}(\pi-x)}{\cos^{2018}(\pi-x)+\sin^{2018}(\pi-x)}=\int_{0}^{\pi}\frac{\pi\sin^{2018}(x)}{\cos^{2018}(x)+\sin^{2018}(x)}-I$$
But after that i'm lost again, i know i should be look for a way to make some terms cancel but that's as far as i got

please don't give me the direct solution, i would like to solve it by myself

Thanks in advance!

答:
Start with $$\int_a^b f(x)\,\mathrm dx=\int_a^b f(a+b-x)\,\mathrm dx$$

Edit: Since you are still stuck after performing the above step, you get $$I=\int_{0}^{\pi}\frac{(\pi-x) \sin^{2018}(x)}{\cos^{2018}(x)+\sin^{2018}(x)}\, \,\mathrm dx$$
and add this to $I$ to get $$2I=\pi \int_{0}^{\pi}\frac{ \sin^{2018}(x)}{\cos^{2018}(x)+\sin^{2018}(x)}\,\mathrm dx $$

If you are still stuck, use the property

$\displaystyle \int_0^{2a}f(x)\,\mathrm dx=2\int_0^a f(x)\,\mathrm dx$, when $f(2a-x)=f(x)$. If you are unaware of this property, you may see it's proof in this answer.

I hope, now you can take it from here.


问:
Relation between hypergeometric functions $_2F_1$ of $z$ and $\frac{1}{1+z}$.
What are the relation between hypergeometric functions $_2F_1$ of $z$ and $\frac{1}{1+z}$.

Specifically, I need a transformation that transforms:

$_2F_1\left(a,b;c; -\sinh^2(x)\right)$
to $_2F_1\left(a',b';c'; \cosh^{-2}(x)\right)$

Thanks in advance.

答:
As pointed out in the comments, if we use DLMF 15.8.3 and $a\neq b$ then
$$
F\left({a,b\atop c};-\sinh^2x\right)=\frac{(b)_{-a}}{(c)_{-a}}\cosh^{-2a}x\ F\left({a,c-b\atop a-b+1};\cosh^{-2}x\right)
+\frac{(a)_{-b}}{(c)_{-b}}\cosh^{-2b}x\ F\left({b,c-a\atop b-a+1};\cosh^{-2}x\right).
$$
If $a=b$ we end up with an indeterminate form that will require L'Hôpital.


问:
What does the dash mean in "$v \mapsto \langle v, -\rangle$"?
This quote is taken from a math.SE answer (https://math.stackexchange.com/a/622695/816960):

By the way, for Euclidean vector spaces, there is a canonical isomorphism, given by $V \to V^*$, $v \mapsto \langle v, -\rangle$.

What does the dash in this inner product mean? I've never seen this notation before? I assume it's some kind of placeholder, but for what I can't guess.

答: An element of $V^*$ is a linear map on $V$, ultimately a function $V \to F$. The notation $v \mapsto \langle v, -\rangle$ is a shorthand for $v \mapsto (w \mapsto \langle v, w \rangle)$, producing a member $w \mapsto \langle v, w \rangle$ of $V^*$ given a member $v$ of $V$.


问:
If $a$ is a positive constant no matter how small, is $x^a$ eventually larger than $\log x$?
If $a$ is a positive constant no matter how small, is $x^a$ eventually larger than $\log x$?

Specifically I tried plotting $f(x)=x^a$ for $a = 0.07$ and $g(x)=\log x$ but it showed $f(x) \leq g(x)$ for large $x$ on a large interval.

答: Yes. This follows from L'Hopital's rule:$$\lim_{x\to\infty}\frac{x^a}{\log x}=\lim_{x\to\infty}\frac{ax^{a-1}}{1/x}=\lim_{x\to\infty}ax^a=\infty,$$since $a>0$.


答:
One way to see this is to find the limit of the ratio of the functions at infinity:
$$
\lim_{x\to\infty}\frac{\log x}{x^a}=\lim_{x\to\infty}\frac{(\log x)'}{(x^a)'}=\lim_{x\to\infty}\frac{\frac1x}{ax^{a-1}}
=\lim_{x\to\infty}\frac{1}{ax^a}=0
$$
for any positive $a$.

Body:
If $a$ is a positive constant no matter how small, is $x^a$ eventually larger than $\log x$?

Specifically I tried plotting $f(x)=x^a$ for $a = 0.07$ and $g(x)=\log x$ but it showed $f(x) \leq g(x)$ for large $x$ on a large interval.


问:
A cubic polynomial over $\mathbb{Q}$ with a degree three splitting field?
I'm wondering if it's possible to have a cubic polynomial $f(x)\in\mathbb{Q}$ with three distinct roots (in $\mathbb{C}$) that has a degree three splitting field?

If $f(x)=(x-a)(x-b)(x-c)$, with $a,b,c$ distinct, then I believe they must all be real, because otherwise we would get an automorphism of order two (complex conjugation) in $\operatorname{Gal}(f)\cong \mathbb{Z}/3\mathbb{Z}$, contradicting Lagrange's theorem. Now, I can't cook up an example that works, but I'm not sure how to prove nothing does work either. What am I missing?

答:
All you need is a cubic polynomial whose discriminant is a square.

E.g. $f(x) = x^3 - 3x + 1$.

Reference: the wiki page on cubic equation


问:
Is a linear map of norm $1$ always an isometry?
Let $(E,\|.\|_{E})$ and $(F,\|.\|_{F})$ be two normed spaces and let $f:E\longrightarrow F$ a linear map such that $\|f\|=1$. I don't know if this means that $f$ is an isometry $(\|f(x)\|_{F}=\|x\|_{E}, \,\, \forall x\in E)$. Thanks </p>

答: It does not. Consider the function $f : \mathbb{R}^2 \rightarrow \mathbb{R}$ defined by $f(x,y)=x$, with both $\mathbb{R}^2$ and $\mathbb{R}$ endowed with the $\| \|_\infty$ norm. Then clearly $\|f\|=1$. However $f$ is not an isometry (it is not even invertible : $f(0,y)=0$ for all $y$).</p>


答:
Such fields are known as cyclic cubic fields, because their Galois group is cyclic of order 3. Here's the LMFDB's list of cyclic cubic fields, which includes over four thousand examples.

One easy example (and the first entry in the list) is the cubic subfield of the cyclotomic field $\mathbb{Q}(e^{2\pi i/7})$, which has Galois group $\mathbb{Z}/6\mathbb{Z}$ and therefore has a unique cubic subfield.

Title: A cubic polynomial over $\mathbb{Q}$ with a degree three splitting field?

Body:
I'm wondering if it's possible to have a cubic polynomial $f(x)\in\mathbb{Q}$ with three distinct roots (in $\mathbb{C}$) that has a degree three splitting field?

If $f(x)=(x-a)(x-b)(x-c)$, with $a,b,c$ distinct, then I believe they must all be real, because otherwise we would get an automorphism of order two (complex conjugation) in $\operatorname{Gal}(f)\cong \mathbb{Z}/3\mathbb{Z}$, contradicting Lagrange's theorem. Now, I can't cook up an example that works, but I'm not sure how to prove nothing does work either. What am I missing?


问:
Calculating the integral of $\int_0^\infty \int _{-\infty}^\infty \frac{x e^{-x}}{0.2(\sin(1.5x)+y)^2+0.1}dy dx$
I was wondering if this has a name (I was taking the Gamma distribution and tried to create a 2D with "wiggly-tail" version of it). And also if there is a way to solve it? I used numerical methods to estimate that the integral (i.e. normalizing-constant) should be around $20, 21$. Wolfram alpha puts it at $5\sqrt 2 \pi\approx 22.2144$.

$$\int_0^\infty \int _{-\infty}^\infty \frac{x e^{-x}}{0.2(\sin(1.5x)+y)^2+0.1}dy dx$$

答:
I'll just add that for the more general case:

$$ \frac{x^ae^{-bx}}{c (\sin(dx)+ey)^2 + f}
$$
The integral (normalizing constant) turns to be:
$$\frac{\Gamma(a+1)}{b^{a+1}}\frac{\pi}{e\sqrt{cf}}
$$

Hence the PDF of this function could be written as:

$$f(x,y) = \frac{b^{a+1}}{\Gamma(a+1)}\frac{e\sqrt{cf}}{\pi}\frac{x^ae^{-bx}}{c (\sin(dx)+ey)^2 + f}
$$


问:
Set defined for proving existence of $ r > 0 $ s.t. $ r^2 = 2 $ is problematic
Theorem: There exists a positive real number $ r $ s.t. $ r^2 = 2 $

The proof in my book begins with defining a set $ L = \{ x \in \mathbb{R} | x \geq 0 , x^2 \leq 2 \} $  and then proceeds to show that $ r^2 = 2 $ is the maximum of this set. However, I don't understand the following:

Since the set L contains real numbers $ x $ such that $x \geq 0, x^2 \leq 2 $, then specifically, it contains a number $ x>0 $ s.t. $ x^2 = 2 $ . How am I supposed to show the existence of such a number ( $ r^2 = 2 $ ) if it already exists  in the set I've defined? ( In which case, the set $ L $ is problematic since it already contains the number which I need to prove exists )

答:
We don't know much about real numbers at this stage.  We know that every rational number is a real.  And we know that every set of real numbers that is bounded above will have a least upper bound.  But that's about all. (Okay, we know it's an ordered fields and... more stuff)

We do not know that there actually is any number $r$ so that $r^2 = 2$?  Why should there be?  Sure,  if there were such a number than $r\in L$ but we don't know any number exists.

(Here's an analogy.  Suppose I asked you to prove that something that doesn't exist does exist.  Suppose I asked you to prove that there is a smallest positive number $s$.  That is a number $s$ so that $s > 0$ and for any $t>0$ then $s \le t$.  There is no such number.  But I'm asking you to prove it.  So you say let $R= \{x| x > 0\}$.  But then you declare that because $s> 0$ then $s \in R$ so what's left to be done.....)

(Or maybe if we defined $S = \{x| 5< x \le \infty\}$ and we argue well if $w = \infty$ then $w \in S$......  Just because the set is defined to allow $w = \infty$ doesn't mean that it does have it.)

So we don't know that there is an $r$ so that $r^2=2$.

But we do know that $s=\sup L$ exists. (That because $L$ is not empty [$1\in L$] and $L$ is bounded above [If $s\ge 2$ then $s^2 \ge 4>2$ so $s\not \in L$ so $2$ is an upper bound] and the reals have the least upper bound property.)

So how does $s^2$ compare to $2$?  Either $s^2 < 2$ or $s^2 > 2$ or $s^2 = 2$.  If we can prove $s^2 < 2$ and $s^2 > 2$ are both impossible than we'd have to conclude $s^2 \ge 2$ and $s^2 \le 2$ and we must conclude $s^2 = 2$

=======

 if $s^2 < 2$ then if we let $\epsilon = \frac {2-s^2}3$.  As $1$ and $1\in L$ we know that $\epsilon \le \frac 13$ so $\epsilon^2 < \epsilon$.  And so $(s+ \epsilon)^2 = s^2 + 2\epsilon + \epsilon^2< s^2 + 3\epsilon =s^2-(2-s^2) = 2$.  So $s < s+\epsilon \in L$ contradicting that $s$ is an upper bound of $L$.

 So $s^2 \ge 2$.

 Do something similar if we assume $s^2 > 2$.  Then let $\delta = \frac{s^2 -2}2$.  As $y \ge 1.5\implies y^2 \ge 2.25$ we know $\delta < 1$ so $\delta^2 < \delta$ and .... same stuff.   $(s-\delta)^2 = s^2 - 2\delta + \delta^2> s^2 -2\delta = s^2 -(s^2-2) = 2$.  (And if $t\ge  (s-\delta)$ then $t^2 \ge (s-\delta)^2 > 2$.) So $s-\delta$ is an upper bound contradicting that $s$ is the least upper bound of $L$.

So $s^2 \le 2$.

Well if $s$ exists, and $s^2 \ge 2$ and $s^2 \le 2$ then we must have $s^2 = 2$.

And that's that.

And, yes, $s \in L$.

(Actually the proof would have worked just as well if we had defined $L$ as $L = \{x| x\ge 0; x^2 < 2\}$.  ANd just as an aside there was no need to specify $x \ge 0$--- it doesn't matter that negative values are in $L$ because the $\sup$ would still be positive.)


问:
Why is the Maclaurin series for $\cos(x^2)$ simply the Maclaurin series for $\cos(x)$ with $x^2$ substituted for $x$?
So I'm trying to understand why we can use expansions of Maclaurin series in this form.

If I try to convert the following into a Maclaurin Series
$$f(x) = x^3 \cos(x^2)$$
Using the following:
$$\cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}$$
I get
$$x^3 \cos(x^2) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+3}}{(2n)!}$$

Why can we plug in $(x^2)$ in in the place of $\cos(x)$? Isn't there a chain rule to consider, since the Maclaurin/Taylor Series have something to do with derivatives? Why is it that we can just plug in $x^2$ and it still works?

答: Since it is true that, for any real number $x$, you have$$\cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots,$$then, in particular, if $x\in\Bbb R$, then you also have $x^2\in\Bbb R$, and therefore\begin{align}\cos(x^2)&=1-\frac{(x^2)^2}{2!}+\frac{(x^2)^4}{4!}-\cdots\\&=1-\frac{x^4}{2!}+\frac{x^8}{4!}-\cdots\end{align}And now you multiply both sides of this equality by $x^3$, in order to get that$$x^3\cos(x^2)=x^3-\frac{x^7}{2!}+\frac{x^{11}}{4!}-\cdots.$$


问:
How would one set up this differential equation?
I have just started learning about differential equations, and I would like to know to set up this differential equation, and I would like to see if I am right.

QUESTION: The population of a certain mining town is known to increase at a rate proportional to itself. After 2 years the population doubled, and after 1 more year, the population was 10,000. What was the original population?

My solution: Denote $P(t)$ as the population of the town at year $t$. We then have $\frac{dP}{dt}$=$kP$, for some constant $k$. Then we have (switching signs and so forth), that $\frac{dP}{P}$=$k dt$. So integrating both sides yeilds $\log P$=$kt+C$. So raising both sides by $e$, we have $P$=$e^{kt+C}$=$C_1e^{kt}$. So we then have that at year $3$, the population is $10000$. In particular, $P(3)$=$10000$, and so solving for $k$ yields $k$=$\frac{\log 10000}{3}$. So this was the initial population.

But my dilemma is that this does not seem intuitive. $\frac{\log 10000}{3}$ is about $3$, which sounds quite absurd. So here are my final questions: Did I use the right method? Is my answer correct? If not correct, what would be the correct answer, and please show all possible details. (I am a beginner. Sorry).

Body:
$$p(t)= c e^{k t}$$
$$p(3)=10000\to ce^{3k}=10000$$
$$p(2)=2p(0)\to  c e^{2k}=2c\to e^{2k}=2\to k=\frac{\ln 2}{2}$$

$$p(t)=c 2^{t/2}$$
$$p(3)=10000\to c 2^{3/2}=10000\to c=3536$$
so finally $$p(t)=3536\cdot 2^{t/2}$$
and we get $p(0)=3536$


问:
Computing associated primes of $\mathbb Z/n\mathbb Z$ over $\mathbb Z$.
Let $n = p_1^{\alpha_1} \cdots p_r^{\alpha_r}$.

Modulo $n$, each $p_i$ annihilates $p_1^{\alpha_1} \cdots p_i^{\alpha_i -1} \cdots p_r^{\alpha_r}$, so each $(p_i)$ is an associated prime of $\mathbb Z/n\mathbb Z$.

Here is where I'm stuck.

Now suppose $(q)$ is an associated prime. Then there exists $x \notin (n)$ such that $qx \in (n)$.

So, $n \mid qx$ and so some $p_i \mid qx$. Then $p_i \mid q$ or $p_i \mid x$.

If for some $i$ we have $p_i \mid q$, then $p_i =q$.

On the contrary, if no $p_i$ divides $q$ then $q$ is not among the $p_i$ and thus every $p_i \mid x$.

I believe we want to derive a contradiction by showing $x \in (n)$.

答:
Let $A$ be a general (commutative) ring and $I\subset A$ be an ideal whose primary decomposition exists. Then the primes associated with $I$ are precisely those primes which occur in $r(I:x)$ as $x$ ranges over all the elements of the ring. Here $r(B)$ denotes the radical of the ideal $B$.

In your case, if $(q)$ is a prime associated with $(n)$ then $(q) = r((n) : x)$ for some $x$. Obviously $x$ should not be inside $(n)$ otherwise $((n):x) = \mathbb{Z}$. Now
$$((n):x) = \left(\frac{n}{\gcd(n,x)}\right).$$

Radical of an arbitrary ideal $(b)$, where $b=p_1^{\alpha_1}p_2^{\alpha_2}\dots p_k^{\alpha_k}$, is simply $(p_1\dots p_k)$. Hence for $r((n):x)$ to be prime we must have
$$\frac{n}{\gcd(n,x)} = p_i^{m}.$$
I think I have already over-explained.

Where you went wrong is the following. If $(q)$ is a prime associated with $(n)$ then there exists an $x \notin (n)$ such that $q^mx \in (n)$ for some integer $m$.


问:
Comparing $2^n$ to $n!$
Comparing $\;2^n\;$ and $\;n!\;$

I need to rewrite one or the other so they can be more easily compared (i.e. have similar form). Because of the factorial, I'm a little lost as to how to compare the two functions. Normally, I would take the logarithm when trying to re-express a power like $2^n$, but I don't think this will work, since I don't know how to take the logarithm of a factorial, if at all.

Might the series expansions of either functions be the right approach? If not could, someone point me in the right direction?

答:
Note that if $n\ge 2$ then
$$\frac{2^n}{n!} =\left(\frac{2}{1}\cdot\frac{2}{2}\cdot\frac{2}{3}\cdots+\frac{2}{n-1}\right)\cdot\left(\frac{2}{n}\right)=\left(\frac{2}{2}\cdot\frac{2}{2}\cdot\frac{2}{3}\cdots+\frac{2}{n-1}\right)\cdot\left(\frac{4}{n}\right).$$
The product $\frac{2}{2}\cdot\frac{2}{2}\cdot\frac{2}{3}\cdots\frac{2}{n-1}$ is $\le 1$. It follows that
$$\frac{2^n}{n!}\le \frac{4}{n}$$
when $n\ge 1$,
and therefore $\frac{2^n}{n!}\to 0$ as $n\to\infty$.</p>

Title: Comparing $2^n$ to $n!$


问:
Transposing Lemma for n x n determinants
Given real numbers $\,x_1,\dots,x_n,\,$
real functions $\,\phi_0,\dots,\phi_{n-1},\,$
and $\,\varphi_0,\dots,\varphi_{n-1},\,$
I have to prove the following lemma:

$$\det[\phi_{j-1}(x_k)]\det[\varphi_{j-1}(x_k)] = \det[\sum_{m=1}^n \phi_{m-1}(x_j)\varphi_{m-1}(x_k)].$$
The RHS is determinant of the sum of rank $1$ matrices.

Can someone help me ?

答:
You ask about the identity
$$\det[\phi_{j-1}(x_k)]\det[\varphi_{j-1}(x_k)] = \det[\sum_{m=1}^n \phi_{m-1}(x_j)\varphi_{m-1}(x_k)] .$$
Define the $\,n\times n\,$matrices
$$ A := \{ \phi_{j-1}(x_k)\}_{j,k=1}^n,\quad
B := \{ \varphi_{j-1}(x_k)\}_{j,k=1}^n,\\
C := \{ \sum_{m=1}^n \phi_{m-1}(x_j)\varphi_{m-1}(x_k)\}_{j,k=1}^n. $$
Since $\, C = A^T B \,$ by definition of matrix
multiplication, using properties of determinants
we get
$$\det[A]\det[B] = \det[A^T]\det[B] =
\det[A^T B] = \det[C] $$ which is the identity in question.


答:
Let $I_n = \pi a_n$, just for convenience. For $n = \pm1$, you can use the closed-form of the anti-derivative, which is $\frac{x}{2}+\frac{\sin(2x)}{4}$ to get $I_1=\pi$. For other $n$, integrate by parts to get $$I_n = -\frac{2}{n}\sin(n\pi)+\frac{1}{n}\int_{-\pi}^{\pi}\sin(x)\sin(nx)dx$$

Integrate by parts again to get $$I_n = -\frac{2}{n}\sin(n\pi) + \frac{1}{n^2}\underbrace{\int_{-\pi}^{\pi}\cos(x)\cos(nx)dx}_{I_n}$$

Which means that $$I_n = -\frac{\frac{2}{n}\sin(n\pi)}{1-\frac{1}{n^2}}=-\frac{2\sin(n\pi)}{n-\frac{1}{n}}$$

For integer $n$ greater than $1$, this is simply $0$.

Title: Find $a_n =1/\pi\int_{-\pi}^{\pi}\cos(x)\cos(nx)dx$

Body:
I am trying to compute Fourier coefficients for the function $f(x) = \cos x$. I have managed to solve that the coefficients "$c$" and "$b$" for the function are $0$. However, I am struggling to solve the following integral (in order to compute coefficient "a")

$$a_n =\frac{1}{π}\int_{-\pi}^{\pi}\cos(x)\cos(nx)dx$$

where n is a positive integer.

I have tried to solve the problem by using integration by parts but this gets me going around in circles. I have also tried to solve the integral using the cosine product-to-sum formula.

For context, I am facing this problem in a calculus textbook, not in the context of Fourier analysis . The purpose of the exercise is solely to practice integration techniques (I am not yet familiar with Fourier series or analysis).

Edit: here is what I get when using the product-to-sum formula:
$$\frac{1}{2π}\frac{2\sin(\pi+\pi n)}{(1+n)}+\frac{1}{2π}\frac{2\sin(\pi-\pi n)}{(1-n)}$$


答: It turns out that inside their respective radii of convergence you can manipulate power series "like polynomials". That isn't too hard to prove, makes a nice exercise. In your case the power series are $x^3$ (clearly everywhere convergent) and the Maclaurin series for $\cos x$, also everywhere convergent. Thus the resulting series is everywhere convergent, and as power series are unique, is the (Maclaurin) power series of the function in question.

Title: Why is the Maclaurin series for $\cos(x^2)$ simply the Maclaurin series for $\cos(x)$ with $x^2$ substituted for $x$?

Body:
So I'm trying to understand why we can use expansions of Maclaurin series in this form.

If I try to convert the following into a Maclaurin Series
$$f(x) = x^3 \cos(x^2)$$
Using the following:
$$\cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}$$
I get
$$x^3 \cos(x^2) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+3}}{(2n)!}$$

Why can we plug in $(x^2)$ in in the place of $\cos(x)$? Isn't there a chain rule to consider, since the Maclaurin/Taylor Series have something to do with derivatives? Why is it that we can just plug in $x^2$ and it still works?


问:
Reasoning about Binomial distribution?
A binomial distribution is a sequence of $n$ Bernoulli experiments - each with a probability $p$ of success. Then it is said in the book that I read that to derive the PMF for binomial distributions, we consider the probability of any specific sequence of k successes and n − k failures, which is is $p^k (1 − p)^{n−k}$.

So here is my question: Why don't we just consider the probability of $k$ successes? I mean, if you have $k$ successes, doesn't that automatically imply $n-k$ failures? So for example, I would just have used $p^k$ as the formula for a sequence of $k$ successes and $n-k$ failures.

答:
To complete my comment: as pointed out by @saulspatz, the key issue is that in probability we prefer talking about the exact number of successes, because otherwise it brings a whole bunch of scenario that are difficult to sum up with one number.

For that reason, the binomial distribution is useful because it is elementary: it corresponds to counting the exact amount of successes when we repeat the exact same experience.

The probability for $k$ successes obviously depends heavily on the number of attempts, it is quite rare to get 100 tails in 100 attempts with a fair coin, while it is completely normal if you toss it 10 000 times. For that reason it is quite clear that the corresponding probability should depend on $p,k$ but also $n$.

Finally, you can look at this sequentially: to have 3 successes among three attempts you need first two successes, which corresponds to $p^2$ and then one failure with probability $1-p$. You can not leave $p^2$ here as it would correspond to the case where you have at least two successes since
$$ p^2 = p^3 + (1-p)p^2$$
(the first term corresponds to exactly 3 successes, the second one to exactly two successes).
Of course you also need to count the number of positions available for the two successes (which is where the binomial coefficient plays a role).

Title: Reasoning about Binomial distribution?


问:
Potential function of $F(x, y)=\left(\lambda x-x^3, y\right)$
I have the following ODE: $$x''=\lambda x-x^3$$
To work with it, I turned it into a system of equations:
$$\begin{matrix}
x' & = & y \\
y' & = & \lambda x -x^3
\end{matrix}$$
Now I want to find a first integral for this system, as well as a potential function for $F(x, y)=\left(\lambda x-x^3, y\right)$. To find the first integral I've done:
$$\frac{dy}{dx}=\frac{\lambda x-x^3}{y} \quad \Longleftrightarrow \quad \frac{y^2}{2}=\lambda \frac{x^2}{2}-\frac{x^4}{4} +C$$
Therefore the first integral is:
$$E(x,y)=y^2-\lambda x^2+\frac{x^4}{2}$$
I've tried calculating the potential function but I got the exact same result, and I don't think this is correct. Can someone help me with the potential? Thank you in advance

答:
If you have a separable situation like the present, then the connection between first integral/Hamiltonian/energy and the potential is
$$
H(x,y)=\frac12y^2+P(x)
$$
So in this case you can chose $P(x)=\frac14(x^2-λ)^2$.


问:
What is the topology defined on $[0, \omega_1)$ it and how do we define its basis?
The set $[0, \omega_1)$ is a countably compact topological space. How do we prove it to be a topological space. How does its basis elements look like?

答:
The set is the first uncountable ordinal which is a linearly ordered set, of size $\aleph_1$, and whose elements are all countable ordinals (so it starts with $0,1,2,\ldots,\omega, \omega+1, \omega+2, \ldots, \omega+\omega,\ldots$ and the order is a well-order (every non-empty subset has a minimum). The topology on this set is the standard order topology, which in this case is: $0$ and every successor ordinal $\alpha+1$ is an isolated point, and if $\alpha$ is a limit ordinal, a local base are all sets $(\beta,\alpha]$ where $\beta < \alpha$. All subsets of the form $(\alpha, \rightarrow)$ are also open.

A typical feature of $\omega_1$ is that every countable sequence $(\alpha_n)_n$ has a supremum in $[0,\omega_1)$. This is what causes it to be countably compact, but not compact. It's a commonly used counterexample, not only in topology but also in measure theory e.g.


问:
Solving a linear system over $\mathbb{Z}_{5}$
I would like to ask how to solve this matrix, if I substitute for z (x, y, z) z = 4, I can't calculate the matrix. I tried to adjust the matrix to this shape, but I don't know how to proceed.

$\left( {\begin{array}{ccc|c}
1 & 2 & 1 & 3\\
1 & 0 & 0 & 4\\
1 & 3 & 4 & 0
\end{array} } \right)$ in $\mathbb{Z}_{5}$

$\left( {\begin{array}{ccc|c}
1 & 2 & 1 & 3\\
0 & 1 & 3 & 2\\
0 & 0 & 0 & 0
\end{array} } \right)$

答:
This is the linear system $Ax=b\;$ with
$$
A=\pmatrix{1 & 2 & 1 \\1 & 0 & 0 \\1 & 3 & 4}\qquad b = \pmatrix{3\\4\\0}
$$
Using ${\mathbb Z}_{5}$ arithmetic and examining the powers of $A^k$ one finds that
$$\eqalign{
A^5 &= A
 \quad&\big({\rm periodic}\big) \\
A^4 &\ne I
 \quad&\big({\rm not\,invertible}\big) \\
A^3 &= A^+
 \quad&\big({\rm generalized\,inverse}\big) \\
}$$
Using the generalized inverse, one can write the solution of the system as
$$\eqalign{
x &= A^+b + \left(I-A^+A\right)z \\
  &= A^3b + \left(I-A^4\right)z \\
}$$
where $z$ is an arbitrary vector and $(I-A^4)$ is the nullspace projector.

In terms of components, this becomes
$$\eqalign{
x &= \pmatrix{4&3&4\\4&0&0\\4&2&1}\pmatrix{3\\4\\0}
   + \pmatrix{0&0&0\\1&3&1\\3&4&3}\pmatrix{z_1\\z_2\\z_3} \\
  &= \pmatrix{4\\2\\0} + \pmatrix{0\\z_1+3z_2+z_3\\3z_1+4z_2+3z_3} \\
}$$
Check the result by setting $z_k=0\,$ to obtain
$$\eqalign{
Ax &= \pmatrix{1&2&1\\1&0&0\\1&3&4}\pmatrix{4\\2\\0}
   &= \pmatrix{3\\4\\0}
   &= b \\
}$$
To connect this with Henno's result note that
$$\eqalign{
\lambda &= 3z_1+4z_2+3z_3 \\
2\lambda &= 6z_1+8z_2+6z_3 \;\overset{{\mathbb Z}_{5}}{\longrightarrow}\; z_1+3z_2+z_3 \\
}$$


问:
Prove by induction: $9\mid (4^{3n} + 8)$ for all integers $n \geq 0$.
I cannot seem to figure this one out.

My though process so far is the following:

$\textit{Proof}$: For the base case n = 0, we have $9 \mid (4^0 + 8)$. This is true.

$\textit{Inductive step:}$ Suppose for $k \geq 0$ we have that $9\mid (4^{3k} + 8)$. We will show that $9 \mid (4^{3(k + 1)} + 8)$. The statement $9 \mid (4^{3k} + 8)$ can be expressed as $4^{3k} + 8 = 9a$ for integer $a$. Observe that
$$\begin{equation}\begin{split} 4^{3(k + 1)} + 8 &= 4^{3k}4^3 + 8 \\ &=
 \dots \\ &\vdots \\ &= 9 (z)\end{split}\end{equation}$$
Where $z$ is some integer. I feel I am forgetting some algebraic properties or a manipulation of the equation above of some sort to help in this prove.

Any help is great! Thanks!

答:
Well, we need to use $4^{3k} +8$ is divisible by $9$.

SO we have somehow isolate $4^{3k} + 8$ from $4^{3k}4^3 + 8$.

Only way I see to do that is $4^{3k}4^3 + 8= (4^3 - 1)4^{3k} + \color{blue}{(4^{3k} + 8)}$

Now we know $9$ divides $\color{blue}{(4^{3k} + 8)}$ so it suffices to show $9$ divides $(4^3-1)4^{3k}$ as well.

As the only prime factors of $4^{3k}$ is $2$ at suffices to show $9$ divides $4^3-1$.

And that's a single constant value.

  Oh... come on! Did you really need to peek at the spoiler to figure out how to prove $9$ divides into $4^3 -1$?


问:
Hypergeometric $\lim_{z\to 1}{}_3F_2(a,b,c;d,e;z)$ with $a+b+c-d-e=0$?
Is there anything special one can say about a hypergeometric function

$$\lim_{z\to 1}{}_3F_2\left({{a,b,c}\atop{d,e}};z\right)$$

in the case when $a+b+c-d-e=0$?

答:
If $a+b+c=d+e$ then ${_3F}_2(\mathbf a;\mathbf b;z)$ is said to be zero-balanced which results in a logarithmic singularity at $z=1$. Specifically, as $z\to 1$:
$$
{_3F}_2\left({a,b,c \atop d,e};z\right)\sim-\frac{\Gamma(d)\Gamma(e)}{\Gamma(a)\Gamma(b)\Gamma(c)}\log(1-z).
$$
Hence
$$
\lim_{z\to 1}{_3F}_2\left({a,b,c \atop d,e};z\right)=\infty.
$$

See here for more details.


问:
The incongruent solutions of a linear congruence
My question is to do with the incongruent solutions of a linear congruence.
This is the problem:

Find all integer solutions to the linear congruence $15x \equiv 36 \mod 57$.

I'm able to use Euclid's algorithm, the gcd etc to solve the linear Diophantine equation and get a general solution for $x$.
I get $x=48+19t$ with $t\in\mathbb{Z}$.

Now I am required to express my answer as a linear congruence:
So from the above it follows that $x \equiv 48 \mod 19 $.

However, I don't understand the next steps and would appreciate an explanation.

Notes then go on to say "now express your answer in the same modulus as the question (i.e., $57$). If we vary $t (=-2,-1,0,1,2)$ we find solutions $10,29,48,67$. But $67\equiv10 \mod 57$ and thus after $10,29,48$ we get no new solutions mod 57."

My questions are to do with the statement in bold:

Why does $67\equiv 10 \mod 57$ imply that we would get no new solutions?
Also, why are there only $3$ incongruent solutions?

(I cooked up a sort of rough explanation, but it doesn't exactly satisfy me: $x=10,29,48,67,86$ etc depending on the value if $t$ we choose.
But as $19(3)$ every $3$ solutions from $10,29,48$ will be equivalent to adding $3(19)=57$ (or a multiple of $57$) to one of $10,29,48$ and thus all the 'new' solutions will be equivalent to the original three solutions mod $57$.)

答: I'd say your 'rough explanation' is pretty good. When one talks about congruence $\pmod {57}$, there are only $57$ possibilities, represented by $0, 1, 2,..., 56$.  Anything outside that range of values is congruent to one of the values in that range. So once you get outside that range of values, you don't get anything new.


问:
If $a$ is a positive constant no matter how small, is $x^a$ eventually larger than $\log x$?
If $a$ is a positive constant no matter how small, is $x^a$ eventually larger than $\log x$?

Specifically I tried plotting $f(x)=x^a$ for $a = 0.07$ and $g(x)=\log x$ but it showed $f(x) \leq g(x)$ for large $x$ on a large interval.

答:
Using the multi-branched Lambert W function, we can explicitly solve for $x$ in
$$x^a = \log x$$

Let $x = e^y$
$$e^{ay}=y$$
$$a = \frac{\log y}y$$
Let $y=e^{-u}$
$$-a=ue^u$$
$$W(-a)=u$$
Thus
$$y=\exp(-W(-a))$$
and
$$x=\exp(\exp(-W(-a)))$$

Clearly, at $x=1$, $x^a > \log x$. The $W_0$ branch tells us where $x^a$ goes below $\log x$, and the $W_{-1}$ branch shows us where $x^a$ overtakes $\log x$.

For $a = 0.07$, these solutions are (approximately)
$$x=2.94000454319605, \log x=1.07841112665417$$
and
$$x=1.55857105089153\times 10^{25}, \log x=58.0083967333164$$

Here is the Sage / Python code I used for those calculations.


问:
Combinatorics: precomputed all combination without repetition, need fast indexing
I've asked a similar (but not identical) question on SO here https://stackoverflow.com/questions/5984102 but the answers aren't that helpful for I need very fast indexing (and maybe that my question wasn't clear enough).

I precomputed a lot of values that took days to compute (with the load spread on several computers).  Now I've got a huge lookup table that I need to frequently access and that access has to be as fast as possible.

Basically, I've got a problem like this: I precomputed C(38,6) values (i.e. 2 760 681 values)  that are now store in a big lookup table.  Memory ain't an issue.  Precomputing ain't an issue.  The "order" ain't an issue and I can re-order in any way I like.

Also, C(38,6) is just an example, I could be needing C(42,7) as well, etc.

When I did the precomputing, I basically looped like this (the actual code is more complicated for the load needed to be spread across several machines and for the actual computation aren't here but here's the meat of it):

```<code>for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
        for (int k = j + 1; k < n; k++) {
            for (int l = k + 1; l < n; l++) {
                for (int m = l + 1; m < n; m++) {
                    for (int o = m + 1; o < n; o++) {
                       ...
                    }
                }
            }
        }
    }
}
</code>```

The only thing that matters is to have the fastest lookup time possible.

I've got i,j,k,l,m and n all in [0..37] (or [1..38], doesn't matter) and I need to access the corresponding precomputed element in my huge lookup array as fast as possible.

Is there a closed formula for this that is fast to compute?

What I've got now is a binary search: my i shall always be lower than j, j lower than k etc.  I then created an arbitrary index doing:

```<code>(i + 1) * (n exp 5) + (j + 1) * (n exp 4) + (k + 1) * (n exp 3) + (l + 1) * (n exp 2) + (m + 1) * n + o
</code>```

This is probably overkill but guarantees no collisions. I then do a binary search on that index and find where the precomputed value is stored.

This is kinda fast but I was wondering if there was a faster way to "reach" my precomputed index?

Basically because I don't know enumerative combinatorics well enough I'm precomputing my own index and using that index to do a binary search.  This is fast but I'm looking for something faster, like a closed formula (I doubt anything doing "enumerations" and i++ or things like that would be faster than those few multiplication to compute that index + the binary search).

I think the complexity of my method is log(n) but I'm just not sure.

答:
The squashed order is a nice order for this sort of thing.

Clarification: This algorithm takes a $k$-subset $0\leq a_1<a_2<...<a_k<n$ to a unique index number from $0$ to ${n \choose k}-1$.

Given $k$ integers, $0\leq a_1<a_2<...<a_k$, the squashed order index of this sequence is ${a_1\choose 1} + {a_2\choose 2} + {a_3 \choose 3} ... + {a_k \choose k}$.  If your maximum value for $a_k$ is $n$, you can precompute $m \choose i$ for $m\leq n$ and $i\leq k$ to increase performance of this computation.

If $A$ and $B$ are sets of $k$ natural numbers, then $A$ comes before $B$ in the squashed order if the largest number that is in the symmetric difference of $A$ and $B$ is in $B$.

Aside from the ease of computing the index of a set, the other nice thing about the order is that the index does not depend on $n$.

See, for example: Combinatorics of finite sets, by Ian Anderson, p. 113.


问:
Show that the sequence $X_n$ converges to a limit $Y$
Let $A$ be a positive definite $n\times n$ matrix. We use the iteration of the mapping $$f(X)=0.5(X^2+B), \ X\in \mathbb{R}^{n\times n}$$ where $B=I-A$.

Show that the sequence $X_0:=0$ (zero matrix), $X_1=f(X_0), X_2=f(f(X_0)), \ldots $ converges to a limit $Y$ if $\|B\|<1$ and show that $(I-Y)(I-Y)=A$.

We want to show that for every $\epsilon>0$ there is a natural number $N$ such that $|X_n-Y|<\epsilon$, right? How could we do that?

答:
By mathematical induction, we see that each $X_k$ is a polynomial in $A$. Since $A$ is diagonalisable, it suffices to prove the problem statement in the scalar case.

Presumably, the matrix norm in question is submultiplicative. Hence $\rho(B)\le\|B\|<1$ and in the scalar case, we may assume that $|B|<1$. From $X_0\le|B|<1$, we can prove by mathematical induction that $X_k=\frac12(X_{k-1}^2+B)\le|B|$. Clearly we also have $\frac12(X_{k-1}^2+B)\ge\frac12 B\ge-|B|$. Therefore $|X_k|\le|B|$ for every $k$ and
$$
|X_{k+1}-X_k|
=\left|\frac{X_k^2-X_{k-1}^2}{2}\right|
=\left|\frac{X_k+X_{k-1}}{2}\right|\left|X_k-X_{k-1}\right|
\le|B|\left|X_k-X_{k-1}\right|.
$$
Hence $\{X_k\}$ is a Cauchy sequence and it converges to some $Y\in\mathbb R$. By passing both sides of $X_k=\frac12(X_{k-1}^2+B)$ to the limit, we get $Y=\frac12(Y^2+B)$, which can be rewritten as $(1-Y)^2=A$.


问:
Difference between projection matrix and vector projection
I am trying to resolve my confusion regarding vector projection and projected vectors with projection matrices.

Suppose the column vectors of $\mathbf{G}$ are orthonormal. To my understanding the vector projection of vector $\mathbf{a}$ onto vector $\mathbf{b}$ is $\hat{\mathbf{a}}(\frac{\mathbf{a}\cdot\mathbf{b}}{||\mathbf{a}||||\mathbf{b}||}$). Then how come in, say, PCA the projection of a data matrix $\mathbf{X}$ onto the new orthonormal basis vectors of $\mathbf{G}$ is given by $\mathbf{G}^T(\mathbf{X} - \mathbb{E}(\mathbf{X}))$? Specifically, the first column vector of $\mathbf{X}$ ends up being $\mathbf{G}^T(\mathbf{x}_1 - \mathbb{E}(\mathbf{x}_1)) = \sum_i\mathbf{g}_i^T(\mathbf{x}_1 - \mathbb{E}(\mathbf{x}_1))$, where the $i$th component of the projected version of $(\mathbf{x}_1 - \mathbb{E}(\mathbf{x}_1)$ is given by the dot product with the $i$th column vector of $\mathbf{G}$. But at what point are we dividing with the length of the vector $(\mathbf{x}_1 - \mathbb{E}(\mathbf{x}_1)$ like in the vector projection? Or is there a reason why this is not necessary?

答:
Suppose that $\mathbf G$ has orthonormal columns $\mathbf g_1,\dots,\mathbf g_k$. Because the columns are orthonormal, we can get the projection onto the span of these vectors by adding up individual projections. That is, the projection a vector $\mathbf a$ is given by
$$
\operatorname{proj}_{\mathbf G}(\mathbf a) = \frac{\mathbf g_1\cdot \mathbf a}{\|\mathbf g_1\|^2} \mathbf g_1 + \cdots + \frac{\mathbf g_k \cdot \mathbf a}{\|\mathbf g_k\|^2}\mathbf g_k
\\ =
(\mathbf g_1 \cdot \mathbf a)\mathbf g_1 + \cdots + (\mathbf g_k \cdot \mathbf a)\mathbf g_k.
$$
The coefficients of this sum are $\mathbf g_1\cdot \mathbf a, \dots, \mathbf g_k \cdot \mathbf a$, which are the entries of the product $\mathbf G^T\mathbf a.$ To get the corresponding projected vector, take the corresponding linear combination of the columns of $\mathbf G$. In other words, we have
$$
\operatorname{proj}_{\mathbf G}(\mathbf a) = \mathbf G(\mathbf G^T \mathbf a).
$$
With that in mind, we can find the coefficients of the projection for each column of the matrix $\mathbf M = \mathbf X - \mathbb E(\mathbf X)$ by computing $\mathbf G^T\mathbf M$ and obtain the vectors corresponding to these columns of coefficients with the product $\mathbf G\mathbf G^T\mathbf M$.


问:
Determining sample proportion probability using population proportion and a sample of size
Suppose 40% of all seniors have a computer at home and a sample of 64 is taken. What is the probability that more than 30 of those in the sample have a computer at home?"

My attempt:

n=64

0.4x64=25.6

p=?

x=??

A>30=??

Don't have an idea of what equation would be appropriate to determine proportion so as to use s=root p(1−p)/n

答:
The number $X$ of seniors in a random sample of size $n = 64,$ who will have computers at home has $X \sim \mathsf{Binom}(n = 64, p = 0.4).$ You seek
$P(X > 30) = 1 - P(X\le 30).$ You may be expected to
solve this problem using software to get an exact answer
or using a normal approximation to get a useful approximate answer.

Using R statistical software, where pbinom is a binomial DCF and dbinom is a binomial PDF, you could obtain the answer $0.1062776$ in either of two ways, as shown below. [The second method is a software implementation of the sum displayed in @user's (+1) Answer.]

```<code>1 - pbinom(30, 64, 0.4)
[1] 0.1062776
sum(dbinom(31:64, 64, 0.4))
[1] 0.1062776
</code>```

Normal approximation begins by finding $ \mu = E(X) = np = 64(.4) = 25.6$ and $\sigma = \sqrt{64(.4)(.6)} = 3.9192.$ Then $X \stackrel{aprx}{\sim}
\mathsf{Norm}(\mu=25.6, \sigma = 3.9192).$
Using a continuity correction, you would find
$P(X > 30.5) = 1 - P(X < 30.5) \approx 0.1056.$ This could be evaluated in R, where pnorm is a normal CDF:

```<code>1 - pnorm(30.5, 25.6, 3.9192)
[1] 0.1056032
</code>```

Often, a normal approximation of a binomial probability
is accurate to two decimal places, provided $n$ is large
enough that $np$ and $n(1-p)$ both exceed $5.$ (Roughly speaking, this ensures that there is very little normal
probability below $9$ or above $n.)$

Standardizing and using a printed standard normal table
is done as follows:

$$P(X > 30.5) =
P\left(\frac{X=\mu}{\sigma} > \frac{30.5 - 25.6}{3.9192}\right)\\
\approx P(Z > 1.25) =  0.1056,$$
where $Z$ is standard normal and the final answer
is obtained from a printed table. Sometimes this method
is a little less precise because printed table give
z-scores only to two decimal places.

The plot below shows the good fit of the normal distribution to this particular binomial distribution.

```<code>x = 10:40; PDF= dbinom(x, 64, .4)
hdr = "BINOM(64, 0.4) with Normal Approx."
plot(x, PDF, type="h", lwd=2, main=hdr)
 curve(dnorm(x, 25.5, 3.9192), add=T, col="red", lwd=2)
 abline(v=30.5, col="blue", lty="dotted", lwd=2)
 abline(h=0, col="green2")
</code>```

Body:
The number $X$ of seniors in a random sample of size $n = 64,$ who will have computers at home has $X \sim \mathsf{Binom}(n = 64, p = 0.4).$ You seek
$P(X > 30) = 1 - P(X\le 30).$ You may be expected to
solve this problem using software to get an exact answer
or using a normal approximation to get a useful approximate answer.

Using R statistical software, where pbinom is a binomial DCF and dbinom is a binomial PDF, you could obtain the answer $0.1062776$ in either of two ways, as shown below. [The second method is a software implementation of the sum displayed in @user's (+1) Answer.]

1 - pbinom(30, 64, 0.4)
[1] 0.1062776
sum(dbinom(31:64, 64, 0.4))
[1] 0.1062776

Normal approximation begins by finding $ \mu = E(X) = np = 64(.4) = 25.6$ and $\sigma = \sqrt{64(.4)(.6)} = 3.9192.$ Then $X \stackrel{aprx}{\sim}
\mathsf{Norm}(\mu=25.6, \sigma = 3.9192).$
Using a continuity correction, you would find
$P(X > 30.5) = 1 - P(X < 30.5) \approx 0.1056.$ This could be evaluated in R, where pnorm is a normal CDF:

1 - pnorm(30.5, 25.6, 3.9192)
[1] 0.1056032

Often, a normal approximation of a binomial probability
is accurate to two decimal places, provided $n$ is large
enough that $np$ and $n(1-p)$ both exceed $5.$ (Roughly speaking, this ensures that there is very little normal
probability below $9$ or above $n.)$

Standardizing and using a printed standard normal table
is done as follows:

$$P(X > 30.5) =
P\left(\frac{X=\mu}{\sigma} > \frac{30.5 - 25.6}{3.9192}\right)\\
\approx P(Z > 1.25) =  0.1056,$$
where $Z$ is standard normal and the final answer
is obtained from a printed table. Sometimes this method
is a little less precise because printed table give
z-scores only to two decimal places.

The plot below shows the good fit of the normal distribution to this particular binomial distribution.

x = 10:40; PDF= dbinom(x, 64, .4)
hdr = "BINOM(64, 0.4) with Normal Approx."
plot(x, PDF, type="h", lwd=2, main=hdr)
 curve(dnorm(x, 25.5, 3.9192), add=T, col="red", lwd=2)
 abline(v=30.5, col="blue", lty="dotted", lwd=2)
 abline(h=0, col="green2")


问:
sufficient and necessary conditions for some sign function based on inner product
Given a vector $x \in \mathbb{R}^n$ and a finite collection of vectors $ G \subset\mathbb{R}^n $.

are the two conditions sufficient and necessary for each other?

A: $\forall g \in G,\langle x, g \rangle >0 $

and B: $\forall S \subset G, \sum_{i\in S} \langle i, x \rangle > 0$

It's rather easy to prove $A \to B$, since $S \subset G, i \in G$ any finite sum of positive is still positive.

And I'm looking for help to prove $B \to A$.

答:
actually, the proof of $B \to A$ is also easy.

Since $\forall S \subset G$, we have a positive result.

We can construct a set $\{\{g\}|\forall g \in G\}$, which would then prove $A$


问:
Can you tile a heart with dominoes?
For a positive integer $n$, let $R_n$ be the set of integer lattice points $(x, y)$ such that

$0 \leq x < 2n$

$0 \leq y < 4n$

$x \leq y$

$y \leq 5n - x$

$y \leq x + 3n$,

and let $L_n = \{(-x, y) \mid (x, y) \in R_n\}$. Define the $n$th heart board to be $L_n \cup R_n$. The first four heart boards look like this:

Can heart boards ever be tiled with dominoes?

The $n$th heart board seems to have size $10n^2 + 3n - 3$. This is even iff $n$ is odd, so only the odd heart boards have any hope to be tiled. With a computer search I've verified that no heart board with $n \leq 20$ can be tiled using dominoes. I don't see how to extend any "forcing" argument to arbitrary integers, but I suspect that there is some nice pattern. (Any possible tiling seems to only use "upright" dominoes, for example.)

答:
No heart boards can be tiled with dominoes.

Gregory Puleo's suggestion turns into an excellent proof.

"Color" the heart with $+1$ and $-1$ so that the spaces alternate signs. Every domino then covers exactly one $+1$ and one $-1$, so the heart board will not be tileable if there are more $+1$'s than $-1$'s, or vice versa. It suffices to consider the heart a single column at a time.

Fix some $x \in \{0, 1, 2, \dots, 2n - 1\}$. What is the height of the $x$th column? Using the given constraints, it is exactly

$$\min(5n - x, x + 3n, 4n) - x + \delta,$$

where $\delta = 0$ if the minimum is $4n$, and $1$ otherwise. In the first case for the minimum, we get

$$5n - x - x + 1 = 5n - 2x + 1.$$

This is even if $n$ is odd, so the "color difference" is $0$. In the second case for the minimum, we get

$$x + 3n - x + 1 = 3n + 1,$$

which is even if $n$ is odd, so the color difference is again $0$. In the third case we get

$$4n - x,$$

but the conditions $4n \leq 5n - x$ and $4n \leq x + 3n$ happen to imply $x = n$, so the color difference is actually $3n$, and this occurs in only one column. This is odd if $n$ is odd, so there is exactly one more plus sign than minus sign in this column (or the other way around).

This all means that the odd heart boards have a color difference of exactly two, one from column $x = n$, and one from column $x = -n$, so they cannot be tiled by dominoes. The even heart boards have an odd number of spaces, so they obviously cannot be tiled with dominoes.

Title: Can you tile a heart with dominoes?


问:
Derivative of Gauss hypergeometric function ${}_{2}F_{1} (a,b,b+1,z)$
Given the Gauss hypergeometric function ${}_{2}F_{1} (a,b,c,z)$, its $n$th derivative can be written as

$\frac{\mathrm{d}^{n}}{\mathrm{d} z^{n}} {}_{2}F_{1} (a,b,c,z) = \frac{(a)_{n} (b)_{n}}{(c)_{n}} {}_{2}F_{1} (a+n,b+n,c+n,z)$

However, when $c = b+1$, it appears that one can write the $n$th derivative differently. For instance, with Mathematica, I obtain

$\frac{\mathrm{d}}{\mathrm{d} z} {}_{2}F_{1} (a,b,b+1,z) = \frac{b \big( (1-z)^{-a} - {}_{2}F_{1} (a,b,b+1,z) \big)}{z}$

$\frac{\mathrm{d}^{2}}{\mathrm{d} z^{2}} {}_{2}F_{1} (a,b,b+1,z) = \frac{b \left(a (1-z)^{-a-1}-\frac{b \left((1-z)^{-a}-\, _2F_1(a,b;b+1;z)\right)}{z}\right)}{z}-\frac{b \left((1-z)^{-a}-\, _2F_1(a,b;b+1;z)\right)}{z^2}$

$\frac{\mathrm{d}^{3}}{\mathrm{d} z^{3}} {}_{2}F_{1} (a,b,b+1,z) = \ldots$

Although less compact, the interesting thing of this form is that all Gauss hypergeometric functions appearing in the derivatives are the same as the original function. What is the property behind this result? I searched all the properties of the Gauss hypergeometric function in http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/ and in the book "Table of integrals, series and products" but with no luck.

答:
This behavior of $F(a,b;b+1;z)$ can be seen by working with the contiguous relations of the hypergeometric function. Using the notation in the linked paper we may write the derivative of the hypergeometric function as
$$
\partial_zF\left({a_1,a_2\atop a_3};z\right)=\frac{a_1a_2}{a_3}\mathcal A_1\mathcal A_2\mathcal A_3F\left({a_1,a_2\atop a_3};z\right)=\frac{a_2}{z}(\mathcal A_2-\mathcal I)F\left({a_1,a_2\atop a_3};z\right),
$$
which is equivalent to stating
$$
\partial_zF\left({a_1,a_2\atop a_3};z\right)=\frac{a_2}{z}\left(F\left({a_1,a_2+1\atop a_3};z\right)-F\left({a_1,a_2\atop a_3};z\right)\right).
$$
For your specific case, $a_2+1=a_3$ and so this relationship reduces to
$$
\partial_zF\left({a_1,a_2\atop a_2+1};z\right)=\frac{a_2}{z}\left((1-z)^{-a_1}-F\left({a_1,a_2\atop a_2+1};z\right)\right).
$$
Note that we can iterate this process and write higher order derivatives of $F(a,b;b+1;z)$ w.r.t. $z$ in terms of combinations of powers of $z$, $(1-z)$, and $F(a,b;b+1;z)$.


问:
Given examples of complete non-Archimedean field of characteristic $p$ s.t. $K$ is not $F$-finite
Let $K$ be a complete non-Archimedean field of characteristic $p$ with valuation ring $A$. We say a ring $R$ is $F$-finite if it is of characteristic $p$ and the absolute Frobenius map $F_R$ is finite.

We know if $A$ is $F$-finite then $K$ is $F$-finite as a localization of $F$-finite map. So the case $A$ is $F$-finite and $K$ is not $F$-finite is not possible. But I can't deduce about other situations. So I would like to know examples for the other three cases:

$A$ is not $F$-finite, $K$ is $F$-finite.

$A$ is not $F$-finite, $K$ is not $F$-finite.

$A$ is $F$-finite, $K$ is $F$-finite.

The case 3 is probably the most simple one, $K=\mathbb{F}_p((t)),A=\mathbb{F}_p[[t]]$ should do the job. For case 1 and 2, $A$ is better to be non-Noetherian because $F$-finite ring is excellent (by Kunz) so Noetherian. Thanks in advance!

答:
We want to know if $K/K^p$ is a finite extension and if $A$ is a finitely generated $A^p$-module.

With $A= \Bbb{F}_p[[t]]$ both are.

With $A= \Bbb{F}_p(x_1,x_2,x_3,\ldots)[[t]]$ neither are.

Let $K$ be the completion of $\Bbb{F}_p[x,y,x^{-1},y^{-1}]$ for the valuation $v(x^ny^m) = n+\sqrt2 m$ ie. $$K = \{ \sum_{j\ge 1} c_j t^{b_j},c_j\in \Bbb{F}_p, b_j\in \Bbb{Z+\sqrt2\ Z}, \lim_{j\to \infty} b_j=\infty\}$$ $v(\sum_{j\ge 1} c_j t^{b_j})=\inf \{ b_j,c_j\ne 0\}$.

Then $[K:K^p]=p^2$, while $A$ is not a finitely generated $A^p$-module:

If $A=A^p+\sum_{m=1}^M a_m A^p$ then wlog $v(a_m)>0$ so that for $v(\alpha)<\inf v(a_m))$ we'd have $v(\alpha)\in \Bbb{pZ+\sqrt2\ p Z}$, contradicting that the elements in $(\Bbb{Z+\sqrt2\ Z})_{\ge 0}, \not \in (\Bbb{pZ+\sqrt2\ p Z})_{\ge 0}$ can be arbitrary small.


问:
What does it mean for a set to exist?
Is there a precise meaning of the word 'exist', what does it mean for a set to exist?

And what does it mean for a set to 'not exist' ?

And what is a set, what is the precise definition of a set?

答:
Is there a precise meaning of the word 'exist', what does it mean for a set to exist?

In many contexts, the word 'exist' has a precise meaning. If the context is omitted, its meaning becomes vague. The situation is more complicated for existence of a set, because its existence ("supposedly") requires the existence of its elements, but the existence of its elements is not always enough to ensure existence of the set itself.

And what does it mean for a set to 'not exist' ?

One common situation for a set to 'not exist' is that the "abstract collection" is a proper class instead of a set. Slightly less common is the situation that the "abstract collection" contains some "provably" non-existent elements.

And what is a set, what is the precise definition of a set?

I would say that a 'set' should "model" an "extensional collection of existing objects". This is to be contrasted with an "intensional collection of potential objects", which we might call a 'proper intensional class'. The collection of all sets with two elements is a typical example of such a 'proper intensional class'. On the other hand, every specific two element set in this collection is a perfectly valid existing 'set'. However, I should add that there is a grey zone between these two extremes, for example an "intensional collection of existing objects" will often be a set, but I'm not sure that this is always the case.

Note: By "extensional", I mean that a set is (or can be) defined by the elements it contains. By "intensional", I mean the intention behind the definition of a collection. I'm thinking of polymorphic functions in modern programming languages (for example "templates" in C++) when I write this.

Title: What does it mean for a set to exist?


问:
Find $\mathbb{P}\left( \sum_{k=1}^n X_{k} < a \right)$

Let $X_1, X_2, \dots, X_n \sim U([1,3])$. Write a code in $\mathcal{R}$ to find an approximate value of  $$\mathbb{P}\left(
 \sum_{k=1}^n X_k< a \right)$$

I have seen multiple questions here where people asked about the sum of two or three uniformly distributed variables. There was one more that suggested to use Irwin-Hall distribution, but this distribution only works for $X_k \sim U([0,1])$.

My attempt so far was this (an example with $a=3$) :

```<code>k = 5
for(i in 1:k){
  measurements <- runif(10, min = 1, max = 3) # take 10,000 measurements
  above.threshold <- sum((measurements) < 3) #count the number of values < 4
  rez <- above.threshold/length(measurements) # calculate the proportion of values < 4
}
</code>```

This code doesn't work properly. I have calculated a couple of probabilities by hand:

$$\begin{equation}
\begin{aligned}
&\mathbb{P}(X_1+X_2<4)=\frac{1}{2}; \\
&\mathbb{P}(X_1+X_2+X_3<4)=\frac{1}{48}.
\end{aligned}
\end{equation}$$

Maybe there is someone who's familiar with $\mathcal{R}$ programming and has already faced this problem ? Thank you !

答:
Here is how you can run the simulation with R:

```<code>simulate.sum.uniform.iid <- function(k, a, min=1, max=3, N=10000) {
  set.seed(111)     # for reproducibility
  mean(replicate(N, # take N measurements
                 {
                   X <- runif(k, min = 1, max = 3)
                   sum(X) < a
                 }))
}
</code>```

$P(X_1 + X_2 < 4)$, where $X_1, X_2 \sim U(1, 3)$, prob. is $=\frac{1}{2}=0.5$

```<code>simulate.sum.uniform.iid(k = 2, a = 4)
# 0.5019
</code>```

$P(X_1 + X_2 +X_3 < 4)$, where $X_1, X_2, X_3 \sim U(1, 3)$, prob. is $=\frac{1}{48}=0.02083333$

```<code>simulate.sum.uniform.iid(k = 3, a = 4)
# [1] 0.021
</code>```

Body:
Here is how you can run the simulation with R:

simulate.sum.uniform.iid <- function(k, a, min=1, max=3, N=10000) {
  set.seed(111)     # for reproducibility
  mean(replicate(N, # take N measurements
                 {
                   X <- runif(k, min = 1, max = 3)
                   sum(X) < a
                 }))
}

$P(X_1 + X_2 < 4)$, where $X_1, X_2 \sim U(1, 3)$, prob. is $=\frac{1}{2}=0.5$

simulate.sum.uniform.iid(k = 2, a = 4)
# 0.5019

$P(X_1 + X_2 +X_3 < 4)$, where $X_1, X_2, X_3 \sim U(1, 3)$, prob. is $=\frac{1}{48}=0.02083333$

simulate.sum.uniform.iid(k = 3, a = 4)
# [1] 0.021


问:
Find $a_n =1/\pi\int_{-\pi}^{\pi}\cos(x)\cos(nx)dx$
I am trying to compute Fourier coefficients for the function $f(x) = \cos x$. I have managed to solve that the coefficients "$c$" and "$b$" for the function are $0$. However, I am struggling to solve the following integral (in order to compute coefficient "a")

$$a_n =\frac{1}{π}\int_{-\pi}^{\pi}\cos(x)\cos(nx)dx$$

where n is a positive integer.

I have tried to solve the problem by using integration by parts but this gets me going around in circles. I have also tried to solve the integral using the cosine product-to-sum formula.

For context, I am facing this problem in a calculus textbook, not in the context of Fourier analysis . The purpose of the exercise is solely to practice integration techniques (I am not yet familiar with Fourier series or analysis).

Edit: here is what I get when using the product-to-sum formula:
$$\frac{1}{2π}\frac{2\sin(\pi+\pi n)}{(1+n)}+\frac{1}{2π}\frac{2\sin(\pi-\pi n)}{(1-n)}$$

答:
I would go straightforward here: For $n=1$
$$a_1=\frac{1}{\pi}\int_{-\pi}^{\pi}\cos(x)\cos(x)dx=\frac{1}{\pi}\int_{-\pi}^{\pi}\sin(x)\sin(x)dx=\frac{1}{\pi}\int_{-\pi}^{\pi}[1-\cos(x)^2]dx=2-a_1$$
which results in $a_1=1$. For $n>1$
$$a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}\cos(x)\cos(nx)dx=\frac{n}{\pi}\int_{-\pi}^{\pi}\sin(x)\sin(nx)dx=\frac{n^2}{\pi}\int_{-\pi}^{\pi}\cos(x)\cos(nx)dx=n^2a_n$$
Since $n^2-1>0$, the equation $(n^2-1)a_n=0$ is only satisfied when $a_n=0$.
Thus
$$a_n=\begin{cases}
1, & n=1\\
0, & n>1
\end{cases}$$


问:
Making Sense of the weighted Least Squares method from ordinary Least Squares

It is stated that the equation for the best $\hat{x}$ uses the covariance matrix $V$ such that
$$
A^TV^{-1}A\hat{x}=A^TV^{-1}b
$$
and the error vector is $\vec{e}=(b-Ax)^TV^{-1}(b-Ax)$ and we need to minimize $e=\sum_{i=1}^m\dfrac{(b-Ax)_i^2}{\sigma_i^2}$

How do I make sense of this idea of weighted least squares in comparison with ordinary least square method ?

And how does the inverse of variances become the respective weights ?

My Understanding

Least square

We have a system $Ax=b$ and $b$ need not have to be in the column space of $A$.

We need the vector $A\hat{x}$ in $C(A)$ thats best fit to $b$, ie., closest to $b$.

So we need to minimize the error vector $\vec{e}=b-Ax$, ie., we need to minimize $e=e_1^2+\cdots+e_m^2=(b-Ax)^T(b-Ax)=||b-Ax||^2=\sum_{i=1}^m(b-Ax)_i^2$

Weighted Least square

I think here we are assigning weights for each errors such that each error is not equally likely, say $w_i$

Then we need to minimize the error $e=\sum_{i=1}^m w_ie_i=(b-Ax)^TW(b-Ax)=\sum_{i=1}^m w_i(b-Ax)_i^2$.

But how does the matrix for the weights $W$ is the inverse of the covariance matrix $V^{-1}$ ?

Reference: Page 557, Introduction to Linear Algebra, Gilbert Strang

Note: I am actually looking for an exaplanation based on linear algebra, by not going deep into statistics

答:
Here is a derivation of the formula (in terms of the weight matrix $W$).

Note that the matrix $W$ is necessarily positive definite. Let $M$ be a matrix such that $W = M^TM$ (such a matrix can be found using the Cholesky decomposition, for instance). Note that for any vector $x$, the weighted norm satisfies
$$
x^TWx = (Mx)^T(Mx).
$$
With that in mind, we can reframe our original problem as follows:
$$
\min_{x \in \Bbb R^n} (Ax - b)^TW(Ax - b) = \\
\min_{x \in \Bbb R^n} [M(Ax - b)]^T[M(Ax - b)] = \\
\min_{x \in \Bbb R^n} [(MA)x - (Mb)]^T[(MA)x - (Mb)].
$$
In other words, the solution that we're looking for is the least squares solution to the equation $(MA) x = Mb$. Using the ordinary least-squares, the solution satisfies
$$
(MA)^T(MA)\hat x = (MA)^T(Mb) \implies\\
A^T(M^TM) A \hat x = A^T(M^TM)b \implies\\
A^TWA\hat x = A^TWb.
$$

Now, a question remains: why is $W = V^{-1}$ the correct choice of weight matrix? Once answer is that even if entries of $b$ are not uncorrelated with identical variance, then entries of $Mb$ will be, where $M$ is chosen such that $W = V^{-1} = M^TM$.

Indeed: we note that the covariance matrix of $b$ is given by
$$
\Bbb E(bb^T) - \Bbb E(b)\Bbb E(b)^T.
$$
We note that the covariance of the new variable $Mb$ is given by
$$
\begin{align}
\Bbb E((Mb)(Mb)^T) - \Bbb E(Mb) \Bbb E(Mb)^T &=
\Bbb E(Mbb^TM^T) - \Bbb E(Mb)\Bbb E(b^TM^T)
\\ & =
M\Bbb E(bb^T)M^T - M\Bbb E(b)\Bbb E(b)^T M^T
\\ & =
M[\Bbb E(bb^T) - \Bbb E(b)\Bbb E(b)^T]M^T
\\ & =
MVM^T =
M(M^TM)^{-1}M^T
\\ & =
MM^{-1}M^{-T}M^T = I.
\end{align}
$$
In other words, the covariance of each pair of components of $Mb$ is zero and the variance of each component is $1$.

Because the equation $(MA)x = (Mb)$ is such that the right side has entries that are uncorrelated with identical variance, it is more appropriate to compute its least squares solution (in the usual sense). As I explain in the first part of this answer, finding this least squares solution is equivalent to finding the weighted least squares solution for the original problem.


问:
Pseudovectors in Geometric / Clifford Algebra
I am studying geometric algebra and I am confused about why pseudovectors are written as single vectors with an $i$ in front.  In other words, for basis vectors $\gamma_{\mu}$ where $\mu = 1,2,3,4$, there are 4 pseudovectors, written as $i \gamma_{\mu}$.

I know that in general, for a space of dimension $N$, pseudovectors are elements of the set of $(N-1)$ fold wedge products, namely $\wedge^{N-1}\mathcal{C}_N$ where $\mathcal{C}$ denotes a Clifford algebra from dimension $N$.

So from this, for a 4 dimensional space, the pseudovectors are trivectors, for example $(\gamma_1\wedge \gamma_2 \wedge \gamma_3)$.  Where does the $i$ come in?  I know that the pseudoscalar is written as an $N$ fold wedge product, so in this case would be a quad-vector, but I don't see how that can be worked in.

I can't seem to find any literature that explains this directly.  I did find something that says there is a one-to-one mapping between $(N-1)$ fold wedge products, and single vectors, but it didn't go much into exactly how this mapping works.

答: The pseudoscalar for an N=4 dimensional algebra is $I=\gamma_1\gamma_2\gamma_3\gamma_4$, so if you multiply $I$ by any vector you get a grade 3 object (trivector). For example $I\gamma_4=\gamma_1\gamma_2\gamma_3\gamma_4\gamma_4=\gamma_1\gamma_2\gamma_3$ (assuming each basis vector $\gamma_\mu$ squares to +1). I think this is the type of calculation you were looking for.


问:
Asumptions required for Inverse Limit on Infinite Galois Group
I am currently trying to get some knowledge on Infinite Galois Theory.

During the construction of the Galois Group of an infinite Galois extension through an inverse limit of Galois group of intermediary finite Galois extensions, there is one point that puzzles me.

Some authors require the inverse system of the inverse limit to be based on a directed set (for example Milne), while some authors only require the weaker asumption of a partially ordered set.

So for example in Milne's approach we have to check that the compositum EF of two finite galois extensions E and F is still a finite galois extension before we are able to talk about the inverse limit.

Do we gain anything from this extra property of the inverse limit in this context, or is it just a difference of convention used by different authors ?

答: Doesn't matter. Define a profinite group to be a projective limit of finite groups, each with the discrete topology. Whether or not you require the projective limit to be over a directed set, you find that a profinite group is compact and totally disconnected. But a compact totally disconnected group is a projective limit of finite groups over a directed set, namely, over the set of its open normal subgroups, which is directed because the intersection of two open normal subgroups is open normal. See Serre, Galois Cohomology, first page.</p>