答:
Here's one way of visualizing it.

In the same way that $S^2$ can be mapped to the extended plane $\mathbb R^2+\infty$ via the stereoscopic projection, use stereoscopic projection to map $S^3$ to $\mathbb R^3+\infty$.

Now imagine a solid torus $T$ and its complement $T'$ in $\mathbb R^3$. $T$ is evidently $S^1\times D^2$.  But so is $T'$, because it can be decomposed into circles that go through the disk spanned by the smallest circle in $T$ (i.e. the circle that travels around the "hole" in the torus) and travel around $T$. This decomposition will be preserved topologically when the inverse stereographic projection takes it back to $S^3$.

(Lest the verbal description above doesn't make sense, a cross section would look like the following picture from wikipedia)

Title: $S^3$ as union of 2 solid tori

Body:
I am working through Hatcher's Algebraic Topology and don't get what he means by "standard decomposition of $S^3$ into two solid tori",

$$S^3= S^1 \times D^2 \cup D^2 \times S^1$$

Can someone explain (or visualize) what the union looks like?
First, I thought it looked like a genus-$2$-torus:

but I guess that's not what he means, actually...
I have basic knowledge in algebraic topology and I am not familiar with Heegaard Splittings.

To obtain an embedding of $X$ in $S^3-K$ as a deformation retract we will use the standard decomposition of $S^3$ into two solid tori $S^1\times D^2$ and $D^2\times S^1$, the results of regarding $S^2$ as $\delta D^4=\delta(D^2\times D^2)=\delta D^2\times D^2\cup D^2\times\delta D^2$. Geometrically, the first solid torus $S^1\times D^2$ can be identified with the compact region in $\Bbb R^3$ bounded by the standard torus $S^1\times S^1$ containing $K$, and the second solid torus $D^2\times S^1$ is then the closure of the complement of the first solid torus, together with the compactification point at infinity. Notice that meridional circles in $S^1\times S^1$ bound disks in the first solid torus, while it is longitudinal circles that bound disks in the second solid torus.

I have read many similar questions and answers, but none I found could help me out, so I started a new one.


问:
Atiyah-MacDonald Ex 3.3, trouble with well definedness

Let $A$ be a ring with $S,T$ multiplicative closed sets. Let $U$ be the image of $T$ under $f:A\rightarrow S^{-1}A$. Show that $U^{-1}(S^{-1}A)\cong (ST)^{-1}A$.

There is a slick answer to this question using universal properties of the inclusion into the localization (for example Isomorphism of ring localized twice - Atiyah Macdonald Exercise 3.3). However I'd like to get my hands dirty and work the elements.

Our relevant sets are the following:

$S^{-1}{A}=\{(a,s)\mid a \in A \quad s \in S\}/\sim$

$U^{-1}S^{-1}{A}=\{([(a,s)],(t,1))\mid a \in A \quad s \in S \quad t \in T\}/\sim$

$(ST)^{-1}{A}=\{(a,p)\mid a \in A \quad p \in ST\}/\sim$

The trouble I'm encountering is if we try to define a map from $(ST)^{-1}{A}$ to $U^{-1}S^{-1}{A}$ we get in a little bit of trouble as we must chose a lift $(s,t)\in S\times T$ of $p\in ST$. If we try to define a map the other way I haven't been able to show well-definedness

答:
The map $U^{-1}(S^{-1}A)\to (ST)^{-1}A$ is the obvious one: given an element of the form
$$\frac{\quad\frac{a}{s}\quad}{\frac{t}{s'}}$$
send it to $\frac{as'}{st}$.

To verify that this is well-defined, assume first that
$$\frac{\quad\frac{a}{s}\quad}{\frac{t}{s'}} = \frac{\quad\frac{\alpha}{\sigma}\quad}{\frac{\tau}{\sigma'}}.$$
That means that there exists $\frac{t''}{s''}\in U$ such that
$$\frac{t''}{s''}\left(\frac{a\tau}{s\sigma'} - \frac{\alpha t}{s'\sigma}\right) = 0.$$
This is equivalent to asking that
$$\frac{t''}{s''}\left(\frac{a\tau s'\sigma - \alpha ts\sigma'}{ss'\sigma\sigma'}\right) = 0,$$
or that
$$\frac{t''(a\tau s'\sigma - \alpha ts\sigma')}{s''ss'\sigma\sigma;} = 0.$$
This in turn means that there exists $\sigma''\in S$ such that
$$\sigma''t'' (a\tau s'\sigma - \alpha ts\sigma') = 0.$$

On the other hand, to verify that $\frac{as'}{st} = \frac{\alpha\sigma'}{\sigma\tau}$, you would need the existence of some $s''\in ST$ such that
$$s''(as'\sigma\tau - \alpha\sigma'st) = 0.$$
But clearly you can take $s''=\sigma''t''$ to make this hold. So the map is well-defined.

ContentLicense: CC BY-SA 4.0


问:
Combinatorics on divisors
Consider the set $D$ of all divisors of $10000$. What is the number of subsets $H \subseteq D$ containing at least two elements such that for every $a, b \in H$, $a \mid b$ or $b \mid a$?
I tried translating the problem into one about integer grid, which makes it ask for the number of ways to choose at least two points in the square with vertices $(0, 0)$ and $(4,4)$ ordered in such a way that they can be reached sequentially by only moving up and right. But I fail to see a way to calculate it besides setting up this recurrence: let $a_{m,n}$ be the number of subsets of the square between $(0,0)$ and $(m,n)$ that satisfy the above property. We can form such a subset in a number of ways: it could be a subset already counted in $a_{m-1,n}$, or in $a_{m,n-1}$, or one of those same subsets with the addition of $(m,n)$, or be composed of $(m,n)$ and a point in the rectangle with vertices $(0,0$) and $(m, n)$ (there are $(m+1)(n+1)-1$ such points). If we simply added $a_{m-1,n}$ and $a_{m,n-1}$ we would be counting twice the subsets already counted in $a_{m-1, n-1}$.

The recurrence is thus: $\begin{cases} a_{0, 0} = 0 \\ a_{n,0} = a_{0,n} = 2a_{n-1, 0} + n \\ a_{m,n} = 2(a_{m-1,n} + a_{m,n-1} - a_{m-1,n-1}) + (m+1)(n+1)-1\end{cases}$.

And the answer we are looking for is $a_{4,4}$. Is there a more elegant way to answer this however?

答:
I'm not sure if you'll think this is elegant but here goes:-

For convenience, temporarily drop the restriction that $H$ must contain $2$ elements. We can form a variant on Pascal's triangle as follows.

Write $1$ in the top right square i.e. $(4,4)$. This represents the number of sets $H$ with $(4,4)$ as the 'least' point.

Now consider the squares in the next diagonal down i.e. $(3,4)$ and $(4,3)$. For each of these squares we calculate $1$ more than the sum of the numbers above and to the right of it i.e. a total of $2$. For $(3,4)$, the first $1$ represents the number of sets $H$ which only contain  $(3,4)$. The sum of the other numbers adds up all those sets $H$ containing  $(3,4)$ and at least one other point.

Subsequent diagonals contain the numbers
$$4,6,4$$
$$8,16,16,8$$
$$16,40,52,40,16$$
$$96,152,152,96$$
$$416,504,416$$
$$1536,1536$$
$$5136$$

So, I obtain the number $5136$ in square $(0,0)$ i.e. as the number of sets $H$ with $(0,0)$ as the 'least' point.

That means that the sum of all the numbers in the $5\times 5$ block is $ 5136+5135=10271$. We must now exclude the $5^2$ cases where we have only $1$ element in $H$ giving a total of $10246$.

Title: Combinatorics on divisors


问:
Weak Induction implies Strong Induction Proof Inquiry
I am trying to improve my ability to think formally and write public posts informally. This requires a sort of "toggling" between one and the other. In the following question, I would like to confirm that I am correctly producing this conversion.

I read a post (What exactly is the difference between weak and strong induction?) about the distinctions (really, semantic equivalence) between weak induction and strong induction.

The accepted and heavily thumbed-up answer shows that weak induction implies strong induction and that strong induction implies weak induction.

The following definitions were provided for weak induction and strong induction:

Weak Induction:

"Let $S(n)$ denote a statement regarding an integer $n$, and let $k\in\mathbb{Z}$ be fixed. If

(i) $S(k)$ holds, and

(ii) for every $m\geq k, S(m)\to S(m+1)$,

then for every $n\geq k$, the statement $S(n)$ holds."

Strong Induction:
"Let $S(n)$ denote a statement regarding an integer $n$. If

(i) $S(k)$ is true and

(ii) for every $m\geq k, [S(k)\land S(k+1)\land\cdots\land S(m)]\to S(m+1)$,

then for every $n\geq k$, the statement $S(n)$ is true. "

Here is the proof offered for Weak Induction Implies Strong Induction:

Assume that for some $k$, the statement $S(k)$ is true and for every $m\geq k, [S(k)\land S(k+1)\land\cdot\land S(m)]\to S(m+1)$. Let $B$ be the set of all $n>m$ for which $S(n)$ is false. If $B\neq\varnothing, B\subset\mathbb{N}$ and so by well-ordering, $B$ has a least element, say $\ell$. By the definition of $B$, for every $k\leq t<\ell, S(t)$ is true. The premise of the inductive hypothesis is true, and so $S(\ell)$ is true, contradicting that $\ell\in B$. Hence $B=\varnothing$. $\blacksquare$

Here is my conversion to a formal FOL expression (step by step).

The claim to prove:

$\forall k \Big ( \Big [ S(k) \land \forall m \geq k \big (S(m) \rightarrow S(m+1) \big) \Big] \rightarrow \Big[S(k) \land \forall m \geq k \big(S(k) \land S(k+1) \land ... \land S(m) \rightarrow S(m+1) \Big] \Big ) $

The author proceeds with a proof by contradiction which amounts to negating the previous statement and arriving at a contradiction. The negation of $P \rightarrow Q$ is $P \land \neg Q$. And the negation of $R \land S$ is $\neg R \lor \neg S$.  Using this, the negated statement is:

$\exists k \Big ( \Big [ S(k) \land \forall m \geq k \big (S(m) \rightarrow S(m+1) \big) \Big] \land\Big[\neg S(k) \lor \neg [\forall m \geq k \big(S(k) \land S(k+1) \land ... \land S(m) \rightarrow S(m+1)] \Big] \Big ) $

In order for this statement to be true, it is clear that $S(k)$ is true, which means that $\neg S(k)$ cannot also so be true. Therefore, in the disjunction, it must be the formula  $\neg [\forall m \geq k \big(S(k) \land S(k+1) \land ... \land S(m) \rightarrow S(m+1)]$ that is true.

This is equivalent to $\exists m \geq k \Big ( S(k) \land S(k+1) \land ... \land S (m) \land \neg S(m+1) \Big )$.

It seems like we could really just stop right here...because $\neg S(m+1)$ would contradict the fact that the weak induction hypothesis gives us $\forall n \geq k \  S(n)$. If someone could explain why the author continues with his set $B$ construction, I would greatly appreciate it. Even if we were to construct a set $B:=\{n \in \mathbb N| n \gt m\ \land \neg S(n)\}$, it is still the case that $\neg S(m+1)$ is a contradictory statement. The construction of $B$ seems unnecessary.

Thank you!

答:
The author was simply giving an ordinary mathematical proof. He didn’t continue with his set $B$ construction: he started with it. The FOL manipulations with which you started probably never even entered his mind; they certainly would not have entered mine, and they strike me as completely unnecessary. Given the equivalence between weak induction and well-ordering of $\Bbb N$, this ‘least counterexample’ approach is a standard, natural way to use induction.

One can, however, also prove the result directly from the statement of weak induction. Assume the hypothesis of strong induction. For $m\ge k$ let $P(m)$ be the statement $\bigwedge_{i=k}^mS(i)$. Then $P(k)=S(k)$ is true, and $P(m)\to P(m+1)$ for each $m\ge k$, so by weak induction $P(n)$ is true for each $n\ge k$, and therefore $S(n)$ is true for each $n\ge k$.


问:
Expressions are not perfect square for any prime number p
For what values of $n \in \mathbb{N}$ does the two expressions $n^2 + 4n + 1 - p$ and $2n^2 + n + 2 - p$ ARE NOT perfect squares for any prime number $p$.

some progress - Manually I found out one such instance, where $n = 21$ there doesn't exist any prime number $p$ such that those two expressions could be perfect square.

We might solve each expression individually or both simultaneously.

EDIT : so peter pointed out that - For $n=21$, the second expression is a perfect square , for example , for $p=761$

Now adding the main background of this problem -

Conjecture 1 : For every $2n = a + b$, where $a, b, n \in \mathbb{N}$ there always exists minimum one pair $(a,b)$ such that $2(a+n) + ab$ is prime.

Conjecture 2 : For every $2n = a + b$, where $a, b, n \in \mathbb{N}$ there always exists minimum one pair $(a,b)$ with $gcd(a,b) = 1$ such that $n^2 + n + (ab + 2)$ is prime.

At this point I manually found the Exception to both the conjectures for $2n = 42$ implies $n = 21$. Even further supported by some numerical evidences.

Now lets substitute $b = 2n - a$ in the above expressions, with them resulting in some prime number say $p$

From first expression reduced to $a^2 -a(2n + 2) -2n + p$ we have its discriminant $\Delta = n^2 + 4n +1 -p$ which should must be perfect square.

Similarly for the second expression reduced to $a^2 -2an +p -n^2 -n -2$ we have its discriminant $\Delta = 2n^2 +n +2 -p$ which should must be a perfect square.

NOTE : Above both are quadratic expressions in $``a"$

so meanwhile translating the main problem to this very reduced format, I missed very essential constraint of Conjecture 2 , that is - $gcd(a,b) = 1$ and now when peter pointed out that it fails for $p = 761$ ,surprisingly there are other primes as well...

so lets check if our $gcd(a,b) = 1$ criteria is met or not,
After plugging $n = 21$ and $p = 761$ in the expression $a^2 -2an +p -n^2 -n -2$ it boils down to solve for a quadratic equation in $``a"$ hereby -
$a^2 -42a +297 = 0$ having roots as $a = 9, 33$

Finally $a = 9$ and $b = 33$ for $n = 21$ in the second expression, here clearly $gcd(a,b) = 3$

Hence my bad I couldn't induce this constraint into the problem you faced in the very beginning.

How should we approach this now, to conclude a rigorous solution.

答:
For $\space n=21\space $, $2n^2 + n + 2 - 5=900=30^2.\space $ Also, $p=229\rightarrow 26^2\quad p=421\rightarrow 22^2\quad p=709\rightarrow 14^2\quad p=761\rightarrow12^2 \quad 941\rightarrow 6^2\quad $
There are also five "square" solutions to $n^2 + 4n + 1 - p,n=21$

The following is not proof. The following numbers work for $\space 2\le p \le 997\space $ but this simply indicates that the "square" solutions get rarer with altitude. For any of these, there may exist a larger $\space p\space$ for which either function will yield a perfect square.

$\space n\in\{59,71,75,96,98,101,120,121,122,124,126,129,133,134,136,138,140,145,\\146,149,156,164,170,171,173,174176,179,180,181,184,185,191,194,196,197,\\198,204\cdots\} \space$

I'm suggesting that there may be no value of $n$ meets your criteria.


问:
Regression coefficient DiD
Following the discussion on: https://www.publichealth.columbia.edu/research/population-health-methods/difference-difference-estimation
If we specify the regression equation to estimate the DiD as:
$$Y= β_0 + β_1*[Time]  + β_3*[Time*Intervention] + β_4*[Covariates]+ε$$

instead of :

$$Y= β_0 + β_1*[Time] + β_2*[Intervention] + β_3*[Time*Intervention] + β_4*[Covariates]+ε$$

Our $β_3$ coefficient would still be yielding the DiD estimator right?

答:
The estimator from the first equation would be biased. One of the assumptions of difference-in-difference estimations is that potential variations between the treatment and control group are accounted for and constant over time (parallel trends assumption). That's the "constant difference in outcome" in Figure 1 from your link. If you don't control for Intervention, then that constant difference assumption is no longer part of your parametric equation.

Similarly, you must control for Time because of the parallel trends assumption - both groups are expected to have evolved in the same way over time in the counterfactual that neither received a treatment.

Without either assumption, you no longer have a quasi-experimental design and an argument for the DiD estimator being an estimate of the causal effect of the treatment.

Title: Regression coefficient DiD


问:
Loss function gradiant - why the transpose?
I have a product of two matrices, defined as $Y=AB$. I need to find $\nabla_A J$ where $J(Y)$ is a scalar-valued loss function dependent on $Y$. As I understand, this gradient can be written as $\nabla_A J = \nabla_A Y \cdot B^T$. I am having some difficulty proving this to myself.

Any element in $Y$ can be written down as $y_{ij} = \sum_{k=1}^M a_{ik}b_{kj}$ and it's derivative can be written as $\frac{\partial y_{ij}}{\partial a_{ik}} = b_{kj}$.

Some intense googling led me to this page which uses concrete matrices to demonstrate. They use the chain rule element by element like so:

\begin{equation}
  \label{eq:dLdX11-chain}
  \frac{\partial L}{x _ {1,1}}
  = \sum _ {i=1}^N\sum _ {j=1}^M \frac{\partial L}{\partial y _ {i,j}}\frac{\partial y _ {i,j}}{\partial x _ {1,1}}
  = \frac{\partial L}{\partial Y} \cdot \frac{\partial Y}{x _ {1,1}}
\end{equation}

However, I do not know how to extend this generally to an entire matrix.

答:
$\def\p#1#2{\frac{\partial #1}{\partial #2}}$There is a problem with your first derivative
$$\eqalign{
y_{ij} &= a_{ik}b_{kj} \\
\p{y_{ij}}{a_{mn}}
 &= \left(\p{a_{ik}}{a_{mn}}\right)b_{kj} \\
 &= \delta_{im}\delta_{kn}\;b_{kj} \\
 &= \delta_{im}\,b_{nj} \\
}$$
If you switch the indices to $\Big((m,n)=(i,k)\Big)$, then you obtain
$$\eqalign{
\p{y_{ij}}{a_{ik}} &= \delta_{ii}\,b_{kj} = N\,b_{kj} \\
}$$
The first rule of index notation is that repeated indexes indicate summation. Whenever you violate this rule you obtain invalid results. You did not intend to sum over the $i$ index on the LHS, but that is what happened.

Now let's to the whole calculation in index notation.
$$\eqalign{
\p{L}{a_{mn}}
 &= \left(\p{L}{y_{ij}}\right) \left(\p{y_{ij}}{a_{mn}}\right) \\
 &= \left(\p{L}{y_{ij}}\right) \left( \delta_{im}\,b_{nj} \right) \\
 &= \left(\p{L}{y_{mj}}\right)  b_{nj} \\
 &= \left(\p{L}{y_{mj}}\right)  b^T_{jn} \\
\p{L}{A}
 &= \left(\p{L}{Y}\right) B^T \\
}$$

In the above, $\delta_{im}$ denotes a Kronecker delta symbol. It has a special property under summation. It substitutes its indices onto adjacent terms in the expression.


问:
Matrices Multiplication $AB + BC = D$
Suppose I have the following equation $AB + BC=D$, where $A$, $C$, and $D$ are known, how can I derive the solution for $B$?

答:
This is a Sylvester equation. Suppose that $B$ has shape $m \times n$. In terms of vectorization and the Kronecker product, we have
$$
(I_n \otimes A + C^T \otimes I_m)\operatorname{vec}(B) = \operatorname{vec}(D).
$$
If the coefficient matrix is invertible, this can be solved as usual with
$$
\operatorname{vec}(B) = (I_n \otimes A + C^T \otimes I_m)^{-1}\operatorname{vec}(D).
$$

Title: Matrices Multiplication $AB + BC = D$


问:
Show that the equation has a periodic orbit
I've been learning differential equations and need help with this exercise:

$$
x^{\prime \prime}+\left(5 x^{4}-9 x^{2}\right) x^{\prime}+x^{5}=0
$$
I've tried to write it as a system:
$$
\begin{array}{l}
x^{\prime}=y \\
y^{\prime}=\left(9 x^{2}-5x^4 \right) y-x^{5}
\end{array}
$$
any help would be appreciated

答:
The main idea is that, when you consider a small neighborhood of zero, $x^2$ is the most powerful and $y’$ has the same sign as $y$. So the zero point is unstable. But when $x$ is big enough, higher powers dominate, so trajectories become attracted to zero. The only consensus between those two phenomena is a cycle somewhere in the middle.

Try to find an appropriate Lyapunov function to make it formal. Also, I’m not sure, but such problem should have a known general solution.


答:
It is often helpful to combine the first two terms in
$$
x''+c(x)x'+g(x)=0
$$
to form the second variable $y=x'+C(x)$, $C'(x)=c(x)$, in the first order system, which now is
\begin{align}
x'&=y-C(x)\\
y'&=-g(x)
\end{align}
Then you get a phase portrait like

where the blue line is the null-cline $y=C(x)=x^5-3x^3$. This has the typical almost rectangular or trapezoidal structure of a fast-slow system, a Lienard dynamic, ... which is commonly demonstrated on the van der Pol oscillator $\ddot x+\mu(x^2-1)\dot x+x=0$. It is clearly visible that a limit cycle has to exist, however to find a Lyapunov function or to demonstrate that in other ways seems not so easy.

See Trajectory in the Lienard system on how  one can start to show that solutions starting "outwards" have to spiral inwards. Such a spiral then also provides a trapping region.

Title: Show that the equation has a periodic orbit

Body:
I've been learning differential equations and need help with this exercise:

$$
x^{\prime \prime}+\left(5 x^{4}-9 x^{2}\right) x^{\prime}+x^{5}=0
$$
I've tried to write it as a system:
$$
\begin{array}{l}
x^{\prime}=y \\
y^{\prime}=\left(9 x^{2}-5x^4 \right) y-x^{5}
\end{array}
$$
any help would be appreciated


问:
When does the distance traveled in vertical and horizontal lines become the hypotenuse?

I am looking for a formal proof for the following problem:

You travel from point A to point B on a right triangle only along its legs. For a 3 4 5 right triangle you would travel a distance of seven. Now imagine going in horizontal and vertical lines to point B while remaining within the right triangle (basically steps going from point A to B). You would again travel a distance of seven: Figure. My question is once you start decreasing the size of your vertical and horizontal lines until they are infinitely small, when do you travel the length of your hypotenuse (for this example five) instead of the length of the legs (for this example seven)? Does this also mean at some point you travel a distance of in between the length of both legs and hypotenuse? Figure 2

答:
This example illustrates that limits$^1$ need not commute, specifically here:

The limit of the lengths is not the length of the limit.

$^1$ Note that length of a curve is itself a limit


问:
using variable in a function
my question is that I have a function and a variable and I want the result of function in that point:

```<code>f := exp(x)*sin(x):
m:=5:
n:=evalf(f, x:=m);
</code>```

but it doesn't work.

in description of f we have:

description(exp(__SELECTION(1, 2)) sin(__SELECTION(1, 2)))

but I couldn't figure out what is the problem.
thanks for your help.

答:
Questions primarily about Maple programming rather than mathematics are off-topic in this forum. (More suitable are stackoverflow.com or www.mapleprimes.com)</p>
```<code>f := exp(x)*sin(x):
m:=5:

eval(f, x=m);

         exp(5) sin(5)

evalf(eval(f, x=m));

         -142.3169809
</code>```

Title: using variable in a function

Body:
Questions primarily about Maple programming rather than mathematics are off-topic in this forum. (More suitable are stackoverflow.com or www.mapleprimes.com)

f := exp(x)*sin(x):
m:=5:

eval(f, x=m);

         exp(5) sin(5)

evalf(eval(f, x=m));

         -142.3169809


问:
R is not Artinian but R/J is semisimple
Can there be a ring R (with 1) such that R is not Artinian but R/J is semisimple where J is the Jacobson radical?

I know if R is Artinian, then R/J is semisimple. That's why I was wondering about the converse.

I've tried working with some non Artinian rings like $\mathbb Z$ and $\mathbb Z[x]$ but it didn't solve my question.

答: As an example, any local domain satisfies the desired criterion. For instance, consider the power series ring $F[[x]]$ over your favorite field $F$. This ring is not Artinian, since $$(x)\supset(x^2)\supset(x^3)\supset\dots$$ is an infinite descending chain. But the ring is local, with maximal ideal and thus Jacobson radical $(x)$, and $F[[x]]\big/(x)\cong F$ is simple, hence semisimple.

Title: R is not Artinian but R/J is semisimple


问:
Infinite sum convergence $ \sum_{i\geq 1}\frac{1}{x^i-y^i}$
For certain values of x and y, the sum $$\sum_{i=1}^{\infty}{\frac{1}{x^i-y^i}}$$ converges...is there a way to get the exact value, given x and y?</p>

答:
This is not precisely what you asked.  (It goes from $-\infty$ to $\infty$ and must have $u \ne 1$.) But there is this very interesting series in terms of Jacobi theta functions...
$$
\sum _{n=-\infty }^{\infty } \frac{1}{ {\alpha}^{n}-u{\beta}^{n} }
=\frac{-i}{2}\,\frac{{ \theta_1}
 \left( \frac{1}{2}\,i\ln  \left( {\frac {
\alpha}{u}} \right) ,\sqrt {{\frac {\beta}{\alpha}}} \right) {
\theta_2} \left( 0,\sqrt {{\frac {\beta}{\alpha}}} \right) {
\theta_3} \left( 0,\sqrt {{\frac {\beta}{\alpha}}} \right) {
\theta_4} \left( 0,\sqrt {{\frac {\beta}{\alpha}}} \right) }{
  { \theta_1} \left( \frac{1}{2}\,i\ln  \left( \alpha \right) ,
\sqrt {{\frac {\beta}{\alpha}}} \right)   {
\theta_1} \left( \frac{1}{2}\,i\ln  \left( u \right) ,\sqrt {{\frac {\beta
}{\alpha}}} \right) }$$</p>

ContentLicense: CC BY-SA 3.0


问:
Where am I going wrong in my tensor notation?
Consider the following expression for order-2 tensor $T$ and vector $v$, in Cartesian coordinates, where $\nabla$ is the gradient operator, and $e_i$ is a unit vector in the $i$ direction.

$$T:\nabla v = T_{ij} e_i e_j : \frac{\partial v_m}{\partial x_n} e_n e_m = T_{ij}\frac{\partial v_m}{\partial x_n} e_i\cdot e_m e_j \cdot e_n=T_{ij}\frac{\partial v_m}{\partial x_n}\delta_{im}\delta_{jn}=T_{ij}\frac{\partial v_i}{\partial x_j}$$

However, I am told that that the above expression should yield $$T_{ij}\frac{\partial v_j}{\partial x_i}$$

Where am I going wrong here?

For reference, $T = T_{ij} e_i e_j, v = v_k e_k$, and so $\nabla v = \frac{\partial v_m}{\partial x_n} e_n e_m$.

答:
It seems to me that you should use the uniform approach to the tensors index order.
If $T:= T_{ij} e_i e_j$, in this case $\nabla v = A_{nm}e_ne_m=\frac{\partial v_m}{\partial x_n} e_n e_m$
$T:\nabla v = T_{ij} e_i e_j : \frac{\partial v_m}{\partial x_n} e_n e_m = T_{ij}\frac{\partial v_m}{\partial x_n} (e_i\circ e_j),(e_n \circ e_m)=T_{ij}\frac{\partial v_m}{\partial x_n} (e_i,e_n)\circ(e_j, e_m)=T_{ij}\frac{\partial v_m}{\partial x_n}\delta_{in}\delta_{jm}=T_{ij}\frac{\partial v_j}{\partial x_i}$


问:
Heron's Formula; an Intuitive or Visual Proof
I've found several proofs for Heron's formula for the area of a triangle in term of its sides, but none of them is simple and intuitive enough to show WHY the formula works.

Do you know an intuitive or visual proof for it?

Thanks.

答:
I noticed that many proofs for Herons formula use the properties of angles, which is why I wanted to show a proof that doesn't involve any,  and seems to be simpler than the usual methods that are applied.

A triangle with side lengths $a, b, c,$ where $c$ is the greatest side length and an altitude($h$) that intercepts $c$ such that $c$ is the  sum of two side lengths ,
$c = m + n$, using Pythagorean theorem and the area formula of a triangle,  we can prove Herons formula.

First we find $a$ and $b$ in terms of $m$ and $n$ using Pythagorean theorem:

$a^2 - m^2 = h^2$

$b^2 - n^2 = h^2$

$a^2 - m^2 = b^2 - n^2$

$a^2 - b^2 = m^2 - n^2$

Then using $c = m + n$, eliminate $m$:

$a^2 - b^2 = (c-n)^2 - n^2$

$a^2 - b^2 = c^2 - 2cn$

$n = \frac{1}{2c}(c^2 + b^2 - a^2)$

We can Eliminate $n$ from $b^2 - n^2 = h^2$ to find $h$:

$h = \sqrt{b^2 - (\frac{1}{2c}(c^2 + b^2 - a^2))^2}$

Plugging this into the area formula ($A = \frac{1}{2}ch$) gives:

$A = \frac{1}{2}c\sqrt{ b^2 - (\frac{1}{2c}(c^2 + b^2 - a^2))^2} $

$A = \frac{1}{2}c\sqrt{ \frac{1}{4c^2}(2a^2b^2+2a^2c^2-a^4-b^4+2b^2c^2-c^4)} $

$A = \sqrt{\frac{1}{16}(c^2 - (a - b)^2)(( a + b)^2 - c^2)} $

$A = \sqrt{\frac{1}{16}(a + b - c)( a + b + c)( b + c - a)(a + c - b)} $

$A = \sqrt{s(s - a)(s- b)(s- c)}$

Q.E.D.


答: I already addressed in a comment below my answer to your other question:$$e_i:e_j=e_i^Te_j=\delta_{ij},\,e_ie_j:e_ne_m=(e_ie_j)^Te_ne_m=e_j^Te_i^Te_ne_m=e_j^T\delta_{in}e_m=\delta_{in}\delta_{jm},$$in accordance with @Svyatoslav's answer, which uses a different, equivalent method.

Title: Where am I going wrong in my tensor notation?

Body:
Consider the following expression for order-2 tensor $T$ and vector $v$, in Cartesian coordinates, where $\nabla$ is the gradient operator, and $e_i$ is a unit vector in the $i$ direction.

$$T:\nabla v = T_{ij} e_i e_j : \frac{\partial v_m}{\partial x_n} e_n e_m = T_{ij}\frac{\partial v_m}{\partial x_n} e_i\cdot e_m e_j \cdot e_n=T_{ij}\frac{\partial v_m}{\partial x_n}\delta_{im}\delta_{jn}=T_{ij}\frac{\partial v_i}{\partial x_j}$$

However, I am told that that the above expression should yield $$T_{ij}\frac{\partial v_j}{\partial x_i}$$

Where am I going wrong here?

For reference, $T = T_{ij} e_i e_j, v = v_k e_k$, and so $\nabla v = \frac{\partial v_m}{\partial x_n} e_n e_m$.


问:
recurrence relation using master method
I know that the Master theorem is used for the recurrence relations of the form:
$$T(n)=aT(n/b)+f(n)$$
In my question, I am supposed to solve the following recurrence relation by using Master theorem:
$$T(n)=T(n/4)+n^{1/2}+5$$
$$T(1)=3$$
What should I do? Thanks.</p>

答:
I think a nice way to deal with such a recurrence is to define a new sequence $U_k$:

$$T\left(4^k\right) = U_k$$

that satisfies

$$U_k - U_{k-1} = 2^k + 5$$

$$U_0=3$$

This is a linear, inhomogeneous recurrence relation with general solution

$$U_k = A \cdot 2^k + B k +C$$

Plugging back into the equation and applying the initial condition, we find that

$$U_k = 2^{k+1} + 5 k + 1$$

which means that, using $T(n) = U_{(\log_2{n})/2}$

$$T(n) = 2 \sqrt{n} + \frac{5}{2} \log_2{n} + 1$$

Title: recurrence relation using master method


问:
Question regarding the proof of $\pi$ is transcendental.
I was reading a proof of the fact that $\pi$ is transcendental over $\mathbb{Q}$, and the author mentioned that if $\pi$ is algebraic then so is $i \pi$, but I'm not particular sure why. Can anybody explain this to me? Thank you.

答: Suppose that $i\pi$ is algebraic. Then there are rational numbers $a_0,a_1,\ldots,a_n$, not all of which are $0$, such that $i\pi$ is a root of $P(x)=a_0+a_1x+\cdots+a_nx^n$. Now, let$$Q(x)=P(ix)=a_0-a_2x^2+\cdots+(a_1x-a_3x^3+\cdots)i.$$Then $Q(\pi)=P(i\pi)=0$. So, since $\pi\in\Bbb R$, $\pi$ is a root of both polynomials$$Q_1(x)=a_0-a_2x^2+\cdots\quad\text{and}\quad Q_2(x)=a_1x-a_3x^3+\cdots,$$and at least one of them is not the null polynomial. But then $\pi$ is a root of $Q_1(x)^2+Q_2(x)^2$.


问:
Quartic function increasing , decreasing , global and local minima maxima problem
Given this Quartic function $p(x)=ax^4+3x^3+6x+2$ , we need to solve according to the information given about $'a'$ and about the polynomial attribute(sorry if the word is not correct) in the interval $(-\infty,+\infty)$. We need to choose 1 correct answer
A. The polynomial decreases in the interval

B.No local maxima or minima

C.The polynomial increases in the interval

D. there is no global maxima

E. there is no global minima

The information about $a$:

$'a'$ is negative

$'a'$ is positive

My try for the first part($a$ is negative):

I checked the limits when ${x \to +\infty }$ and ${x \to -\infty }$ because if we have $(-\infty,+\infty)$ and the limits are $\lim \limits_{x \to +\infty  }p(x)=-\infty$ and $\lim \limits_{x \to -\infty  }p(x)=-\infty$ then it means there exists a maxima in $(-\infty,+\infty)$ .

$\lim \limits_{x \to +\infty  }ax^4+3x^3+6x+2=-\infty$ (since we only deal with the highest power and $a$ is negative
$\lim \limits_{x \to -\infty  }ax^4+3x^3+6x+2=-\infty$( for the same reason and here the even power make it positive)

according to this we have a maxima , and since $a$ is negative we know that our function will look similar in a way to a parabola with a frown which is our global maxima. and of course the derivative would be $p'(x)=4ax^3+9x^2+6$ but atleast for me I could not see a way that the derivative would help me.
my conclusions were that there is no local maxima or minima(B) , there is no global minima(E) , there is a global mxima(D) , for A and C I do not know.

for the second part ($a$ is positive):
According to the derivative $p'(x)=4ax^3+9x^2+6$ the function increases for every x.

so I will also check to see if there is a minimum or maximum according to a similar thing but this time if $\lim \limits_{x \to +\infty  }p(x)=+\infty$ and $\lim \limits_{x \to -\infty  }p(x)=+\infty$ then it means we have a minima but no maxima in the interval.

$\lim \limits_{x \to +\infty  }ax^4+3x^3+6x+2=+\infty$ (since again we only deal with the highest power and $a$ is positive this time)
$\lim \limits_{x \to -\infty  }ax^4+3x^3+6x+2=+\infty$(for the same reason)
so according to this we have a minima , no maxima , the function increases to every x but I still do not know the correct answers.

Thank you for the help , I would like to know if my ways or at least the thoughts were correct in some way And I would appreciate any tips and advice on how to approach this better and how to solve $A$ and $C$.
Sorry for the English mistakes.

答:
Problem can be attacked by meta-cheating based on the idea that there is only 1 correct answer.  Clearly $ax^4$ is the dominant term, so if $a$ is positive, there is no global maxima and if $a$ is negative, there is no global minima.

It remains to

Discuss the other options, for completeness

Discuss your work.

Options A and C are easily dismissed, because for the generic cubic represented by $p'(x)$, how in the world could it be guaranteed that the cubic is always less than zero or always greater than zero.

Option B is similarly dismissed.  How in the world could one guarantee that the pertinent cubic has no real root?

From what I can see, your overall analysis seems sound: I saw no obvious mistakes.

Exception

for the second part ($a$ is positive):
According to the derivative $p'(x)=4ax^3+9x^2+6$ the function increases for every x.

The issue is (for example) whether it can be guaranteed that $p'(x)$ is always positive, regardless of the value of $a$.  If $a$ is positive (non-zero), then for large negative $x$, $p'(x)$ will be negative and for large positive $x$, $p'(x)$ will be positive.

By focusing on whether $p'(x)$ is increasing or decreasing, you are mistakenly focusing on whether $p''(x)$ is positive or negative, rather than focusing on whether $p'(x)$ is positive or negative.


问:
An example of $p$-divisible group
In page 76 of Demazure's book "Lectures on $p$-divisible groups". The formal group $G^{\lambda}$ defined by the exact sequence
$$ 0\to G^{\lambda} \to W(p) \xrightarrow{F^r-V^s} W(p)  $$
is claimed to be a $p$-divisible group over $\mathbb{F}_p$, where $W(p)=\varinjlim(\mathbf{Ker}p^n: W_{\mathbb{F}_p}\to W_{\mathbb{F}_p})$.

Question 1. Why this is a $p$-divisible group?

Question 2. It is claimed that its Dieudonné module is $\mathbb{Z}_p[F]/(F^{r-s}-V^s)$, how should I understand this module (the "$V$" is not well-defined inside  $\mathbb{Z}_p[F]$? ) and why this fact is true?

ps. As I understand, $W_{\mathbb{F}_p}$ can be written (as a group scheme): $$W_{\mathbb{F}_p}=\prod_{n\geq 0}\mathbb{A}^1_{\mathbb{F}_p}=\mathbf{Spec}  \mathbb{F}_p[X_0, X_1, \cdots].$$ (This question is part of the following question---not answered yet: An example of $p$-divisible group from the book of Demazure.)

答:
I think you have to first check the claim about Dieudenne module. for that you have to understand for each n $M(G[p^n])$ which you can easily compute from the exact sequence and the fact that you now the Dieudenne module of $ker(p^n:W_{F_p}\to W_{F_p})$.and then take the limit.$V$ is coming from here because it is defined on each $M(G[p^n])$ and hence on the limit.

now you can use the equivalence between p-divisible groups and (M, F, V) modules with $M$ free stated at the page 71 of the book.

Title: An example of $p$-divisible group


问:
Is "element of" an a Relation with two Parameters?
Is "element of" a relation with two Parameters in propositional logic like for example father(X,Y) ?</p>

答:
Yes.  A relation on $n$ sets is ANY subset of the Cartesian product (set of all $n$-tuples) of those sets.  Frequently for a relation on 2 sets the sets will be the same ($<$, $\subseteq$, etc), but for $\in$ the first set is whatever domain the elements of the second set come from (such as the integers) and the second set is the powerset (set of sets) of that domain.

Symbolically, $"\in"\subseteq A\times\mathcal{P}(A)$ where $A$ is the domain and $\mathcal{P}(A)$ the powerset.


问:
divide the distance between two points
I have two points(in two dimension) and I want to divide the distance between them to n part with same length, I was thinking about doing this manually by using for, but because it is numerical it wont be precise, so is there any way to do this?
any help is appreciated.</p>

答:
In your original question you wrote, "n part with same length", which implies n+1 equispaced points and n equal segments between them.

In your followup comment you wrote "n points", which implies n-1 equal spaces between them.

It's now unclear which you actually want.

```<code>p1:=[2,4]:
p2:=[6,1]:

n:=7:

# n equal distances
S1:=Matrix([seq(p1+(p2-p1)*(i-1)/(n), i=1..n+1)]);

plot(S1, style=pointline,
     symbolsize=20, symbol=solidcircle);

# n equispaced points
S2:=Matrix([seq(p1+(p2-p1)*(i-1)/(n-1), i=1..n)]);

plot(S2, style=pointline,
     symbolsize=20, symbol=solidcircle);
</code>```

Title: divide the distance between two points

Body:
In your original question you wrote, "n part with same length", which implies n+1 equispaced points and n equal segments between them.

In your followup comment you wrote "n points", which implies n-1 equal spaces between them.

It's now unclear which you actually want.

p1:=[2,4]:
p2:=[6,1]:

n:=7:

# n equal distances
S1:=Matrix([seq(p1+(p2-p1)*(i-1)/(n), i=1..n+1)]);

plot(S1, style=pointline,
     symbolsize=20, symbol=solidcircle);

# n equispaced points
S2:=Matrix([seq(p1+(p2-p1)*(i-1)/(n-1), i=1..n)]);

plot(S2, style=pointline,
     symbolsize=20, symbol=solidcircle);


问:
What is the variable(s) or such written before the integration symbol?
From 'Distance measures (cosmology)' on Wikipedia:

Cosmologists commonly use the following measures for distances from the observer to an object at redshift $z$ along the line of sight:

Comoving distance:
$$d_C(z)=d_H\int_0^z\frac{dz'}{E(z')}.$$

Transverse comoving distance:
$$d_M(z)=\left\{\begin{array}{ll}
\frac{d_H}{\sqrt{\Omega_k}}\sinh\left(\frac{\sqrt{\Omega_k}d_C(z)}{d_H}\right) & \Omega_k>0,\\
d_C(z) & \Omega_k=0,\\
\frac{d_H}{\sqrt{|\Omega_k|}}\sin\left(\frac{\sqrt{|\Omega_k|}d_C(z)}{d_H}\right) & \Omega_k<0.
\end{array}\right.$$

Angular diameter distance:
$$d_A(z)=\frac{d_M(z)}{1+z}.$$

Luminosity distance:
$$d_L(z)=(1+z)d_M(z).$$

Light-travel distance:
$$d_T(z)=d_H\int_0^z\frac{dz'}{(1+z')E(z')}.$$

Note that the comoving distance is recovered from the transverse comoving distance by taking the limit $\Omega\to0$, such that the two distance measures are equivalent in a flat universe.

What do you do with the dH that comes up before the integration symbol under 'comoving distance' and 'light-travel distance'?

Not just here, but in general...

Also, the z is a constant in astronomy, so how do you take the derivatives of z in those same two equations?  If the derivative of a constant is always 0?  It doesn't make sense to me....

P.S.  I posted something similar in Astronomy S.E., I hope that is okay... Here, I wanted to understand the general math aspect...

答:
We can rewrite the equations with every "calculus $\mathrm d$" deitalicised, so$$d_C(z)=d_H\int_0^z\tfrac{\mathrm dz^\prime}{E(z^\prime)},\,d_T(z)=d_H\int_0^z\tfrac{\mathrm dz^\prime}{(1+z^\prime)E(z^\prime)}.$$The other equations don't contain any such $\mathrm d$, so I won't reproduce those. For all the difference it makes, $d_A,\,d_C,\,d_H,\,d_L,\,d_M,\,d_T$ might as well have been called $r_A,\,r_C,\,r_H,\,r_L,\,r_M,\,r_T$.

It helps to understand why these definite integrals start at $0$. The value of $a:=1+z$ is proportional to the width of the universe, so is a function of time. The usual convention is to take $a=1$, or equivalently $z=0$, in the present day. Then $d_C,\,d_T$ are distances from our present-day location to something observed by dint of light it emitted in the past: in particular, these distances are $0$ if $z=0$.

Of course, $z=0\implies d_C=d_T=0\implies d_A=d_L=d_M=0$. But $d_H$ is approximately proportional to $a$.


问:
Why is the conditional expectation $E[x|y]$ a function of the conditioning argument $y$? How about the pdf $p(x|y)$?
For a random variable $x$ and $y$, consider the conditional pdf $p(x|y)$ and the conditional expectation $E[x|y]$.

I previously thought that the conditional pdf or conditional expectation were functions of $x$, not $y$.

But I read that the $E[x|y] = \int_{-\infty}^{\infty}x\cdot p(x|y)dx$ is a function of $y$. Hence, when taking the expectation again on $E[x|y]$, we write:

$$
E[E[x|y]] = \int_{y = -\infty}^{\infty}\left[\int_{-\infty}^{\infty}x\cdot p(x|y)dx\right]p(y)dy.
$$

I do not understand this first because I thought $p(x|y)$ is a function of $x$, and therefore $E[x|y]$ also a function of $x$. Am I wrong in the first place?

Is $p(x|y)$ actually a function of $y$? Why isn't it a function of $x$?

Why isn't $E[x|y] = \int_{-\infty}^{\infty}x\cdot p(x|y)dx$ a function of $x$ when it clearly has $x$ in the expression? Is it a function of both $x$ and $y$?

答:
$$\int_{-\infty}^{\infty} x f(x|y)dx$$

is clearly a function of $y$ because $x$ is integrated

For the same reason

$$\int_{-\infty}^{\infty} x f(x)dx= E(X)$$

is a constant...when the integral exists

$p(x|y)$ is surely a function of $y$ but it can also be a function of both $x,y$, i.e.

$$f(x|y)=y e ^{-xy}$$

$x>0$; $y \sim  U[1;2]$

Which is a Negative exponential density with a parameter that is uniformly distributed in $[1;2]$


答:
It was discussed in the comments that $g(z)=(2-3i)(e^{1/z}-1)$ is holomorphic in the complex plane excluding the origin.  The function $e^{1/z}$ is a composition of holomorphic functions, thus holomorphic.

The residue is $2-3i$ so the integral evaluates to $2 \pi i \cdot(2-3i)$.

Title: Holomorphic Function $g(z)=\sum_{n=1}^\infty (2-3i)\cdot \dfrac{1}{n! z^n}$

Body:
Let $g(z)=\sum_{n=1}^\infty (2-3i)\cdot \dfrac{1}{n! z^n}$

Prove that this function is holomorphic in $\mathbb{C} \setminus \{0\}$ and calculate $$\int_{|z|=2} g(z)\,\mathrm{d}z$$

Edit:

What  I have done:
\begin{align}
\begin{split}
g(z) & = \sum_{n=1}^\infty (2-3i)\cdot \dfrac{1}{n! z^n}\\
     & = (2-3i)\cdot \sum_{n=1}^\infty \dfrac{1}{n! z^n}\\
     & = (2-3i)\cdot \sum_{n=1}^{\infty} \dfrac{z^{-n}}{n!}\\
     & = (2-3i)\cdot \sum_{n=0}^{\infty} \dfrac{z^{-n}}{n!} - \dfrac{z^{-0}}{0!}\\
     & = (2-3i)\cdot  (e^{z^{-1}} -1)\\
     & = (2-3i)  e^{z^{-1}}  -(2-3i) \\
     & = 2e^{z^{-1}} - 3ie^{z^{-1}} +3i - 2
\end{split}
\end{align}

let $z^{-1}=z^*$ for a $z^* = x+ iy$

Then let's see $2e^{z^{*}} - 3ie^{z^{*}} +3i - 2$ is holomorphic

Proof:

$ 2e^{z^{*}} - 3ie^{z^{*}} +3i - 2$

$ = 2e^{x + iy} - 3ie^{x + iy} +3i - 2 $

$ = 2e^{x} e^{iy} - 3ie^{x}e^{iy} +3i - 2 $

$ = 2e^{x} [ \cos(y) + i \sin(y) ] - 3 ie^{x} [ \cos(y) + i \sin(y) ] +3i - 2 $

$ = 2e^{x}\cos(y) + 2 i e^{x} \sin(y) - 3 ie^{x} \cos(y) - 3 i^2 e^{x} \sin(y)+3i-2 $

$ = 2e^{x}\cos(y) + 2 i e^{x} \sin(y) - 3 ie^{x} \cos(y) + 3 e^{x} \sin(y) +3i - 2 $

$ = 2e^{x}\cos(y) + 3e^{x} \sin(y) - 2 + 2 i e^{x} \sin(y) - 3 i e^{x} \cos(y) +3i $

$ = 2e^{x}\cos(y) + 3e^{x} \sin(y) - 2 + i[2 e^{x} \sin(y) - 3 e^{x} \cos(y) +3] $

Now let $u(x,y) = 2e^{x}\cos(y) + 3e^{x} \sin(y) - 2$

And let $v(x,y) = 2 e^{x} \sin(y) - 3 e^{x} \cos(y) +3$


问:
Cauchy-Binet Formula Proof Intuition (Determinants)

I am having trouble proving the Cauchy-Binet Theorem. I jotted down how far I got in the proof, but I just find myself stuck. Any guidance would be greatly appreciated!

I understand that

$\begin{align*}\det(AB)
&=\sum_{j_1,j_2, ...,j_k=1}^n b_{j_1,1}b_{j_2,2}...b_{j_k,k}\det(A(J)) \\
&=\sum_{j_1,j_2,...,j_k\in \{1, 2, ..., n\} \text{ and all distinct}} b_{j_1,1}b_{j_2,2}...b_{j_k,k}\det(A(J)) \\
\end{align*}$

The last equations work as for any $J$, we will only consider the $j's$ to be all distinct (otherwise the determinant would be zero) and be integers that are between $1$ and $n$. Now, fix $J'=(j_1', j_2', ..., j_k')$ which organizes these $j's$ from least to greatest. Now, let $\sigma\in S_k$ and have $j'_i=j_{\sigma(i)}$ for $i=1, 2, ...,k$.

I'm not sure why $\sigma$ is a permutation of $[n]$ here instead of being in $S_k$ like how I defined it above? I thought $\sigma$ was defined here by looking at the index of $j$ and not by $j$ itself (so it isn't associated with n).

So, then I continue to get $\operatorname{sgn}(\sigma)\det(J')=\det(J)$. Thus, $j_i=j_{\sigma(\underbrace{\sigma^{-1}(i)}_{\in \{ 1, 2, ..., k\}})}=j'_{\sigma^{-1}(i)}$.

Thus, continuing our equation where we left off, we know
$\begin{align*}
\det(AB)&=\sum_{j_1,j_2,...,j_k\in \{1, 2, ..., n\} \text{ and all distinct}} b_{j_1,1}b_{j_2,2}...b_{j_k,k}\det(A(J)) \\
&=\sum_{j_1,j_2,...,j_k\in \{1, 2, ..., n\} \text{ and all distinct}}  b_{j_1,1}b_{j_2,2}...b_{j_k,k}\operatorname{sgn}(\sigma)\det(A(J'))\\
&=\sum_{j_1,j_2,...,j_k\in \{1, 2, ..., n\} \text{ and all distinct}}\operatorname{sgn}(\sigma^{-1}) b_{j_1,1}b_{j_2,2}...b_{j_k,k}\det(A(J'))\\
&=\sum_{j_1,j_2,...,j_k\in \{1, 2, ..., n\} \text{ and all distinct}}\operatorname{sgn}(\sigma^{-1}) b_{j'_{\sigma^{-1}(1),1}}b_{j'_{\sigma^{-1}(2),2}}...b_{j'_{\sigma^{-1}(k),k}}\det(A(J'))\\
&= \text{and then I get confused here to show} = \sum_{J'}\det(A(J')\det(B(J'))
\end{align*}$

答:
It is helpful to define some notation to aid visualizing the various summations :

$[n]$ = the set $\{1, 2, \ldots, n\}$

$\binom{[n]}{k}$ = set of all subsets of $[n]$ of size $k$

$P_k$ = set of all bijections ('permutations') $: [k] \rightarrow [k]$

$Q_k$ = set of all mappings $: [k] \rightarrow [k]$

$P_{k, n}$ = set of all injections $: [k] \rightarrow [n]$, ($k \leq n$)

$Q_{k, n}$ = set of all mappings $: [k] \rightarrow [n]$

$R_{k, n}$ = set of all strictly increasing injections $: [k] \rightarrow [n]$, ($k \leq n$)

Then :

$\left|\binom{[n]}{k}\right|$ = ${}^nC_k = \binom{n}{k} = \frac{n!}{k!(n - k)!} $

$|P_k|$ = $k!$

$|Q_k|$ = $k^k$

$|P_{k, n}|$ = ${}^n\!P_k = \frac{n!}{(n - k)!}$

$|Q_{k, n}|$ = $n^k$

$|R_{k, n}|$ = ${}^nC_k = \left|\binom{[n]}{k}\right|$

There is a one-to-one correspondence between the set $R_{k, n}$ and the set $\binom{[n]}{k}$ - for, each strictly increasing injection $: [k] \rightarrow [n]$ defines and is defined by a unique $k$-element subset of $[n]$. The set $P_{k, n}$ includes $R_{k, n}$ as a subset but is $k!$ times larger since for every strictly increasing injection $f \in R_{k, n}$ there are $k!$ reorderings of $f$, thus producing $k!$ distinct injection mappings in $P_{k, n}$.

We can denote a mapping by a list notation, eg $(i_1, \ldots, i_k) \in Q_k$ denotes a mapping taking $r$ to $i_r \in [k]$ for each $r \in \{1, 2, \ldots, k\}$. And $(j_1, \ldots, j_k) \in Q_{k, n}$ denotes a mapping taking $r$ to $j_r \in [n]$ for each $r \in \{1, 2, \ldots, k\}$.

The summation notation $\displaystyle \sum_{i_1, i_2, \ldots, i_k = 1}^{k}$ which denotes the iterated ordered sum $\displaystyle \sum_{i_1 = 1}^{k} \, \sum_{i_2 = 1}^{k} \ldots \sum_{i_k = 1}^{k} $ can also be written as the unordered associative/commutative sum $\displaystyle \sum_{(i_1, i_2, \ldots, i_k) \in Q_k}$, since the order of adding up the summands and the bracketing does not matter. And because the $i_r$ independently take on all the values from $1$ to $k$, the mappings $(i_1, \ldots, i_k)$ are not in general injective, so that they range over $Q_k$ rather than $P_k$.

By contrast the summation $\displaystyle \sum_{j_1, j_2, \ldots, j_k \in \{1, 2, \ldots, n\}\; \text{and all distinct}}$ can be written as the associative/commutative sum $\displaystyle \sum_{(j_1, j_2, \ldots, j_k) \in P_{k, n}}$ with $P_{k, n}$ as the indexing set, since we are now confined to mappings $(j_1, j_2, \ldots, j_k)$ from $[k]$ to $[n]$ which are injective.

Assume we have $A \in F_{k \times n}$ and $B \in F_{n \times k}$, where $F$ is a field of scalars, and $k < n$. (The case $k =n$ of the Cauchy-Binet formula reduces to the product formula for determinants, ie $\det AB = \det A \cdot \det B$, and the case $k > n$ implies $A$, and hence $AB \in F_{k \times k}$, has rank $\leq n < k$, so $AB$ must be singular $\therefore$ LHS of formula is zero, and RHS is also zero as it is an empty sum).

For any $J = (j_1, j_2, \ldots, j_k) \in Q_{k, n}$, we define square matrix $A(J) \in F_{k \times k}$ as having $r^{\text{th}}$ column equal to column $j_r$ of $A$ $(r = 1, 2, \ldots, k)$, and square matrix $B(J) \in F_{k \times k}$ as having $r^{\text{th}}$ row equal to row $j_r$ of $A$ $(r = 1, 2, \ldots, k)$. This means columns of $A(J)$ are picked from $A$, but not necessarily in the order they appear in $A$, and a column may be picked more than once, producing an $A(J)$ with $\det A(J) = 0$, and similarly with $B$.

Thus from

\begin{equation}
\det AB = \sum_{(j_1, j_2, \ldots, j_k) \in Q_{k, n}} (b_{j_1, 1} b_{j_2, 2} \cdots b_{j_k, k}) \cdot \det (A(J))
\tag{1}
\label{eq:detAB-Qkn}
\end{equation}

we obtain

\begin{equation}
\det AB = \sum_{(j_1, j_2, \ldots, j_k) \in P_{k, n}} (b_{j_1, 1} b_{j_2, 2} \cdots b_{j_k, k}) \cdot \det(A(J))
\tag{2}
\label{eq:detAB-Pkn}
\end{equation}

by removing all the non-injective mappings. Now define a mapping $J = (j_1, j_2, \ldots, j_k) \rightarrow \psi(J) = J' = (j'_1, j'_2, \ldots, j'_k)$ from $P_{k, n} \rightarrow R_{k, n}$, where $(j'_1, j'_2, \ldots, j'_k)$ is the ordered version of $(j_1, j_2, \ldots, j_k)$. For each $J \in P_{k, n}$, let $\sigma_J \in P_k$ be the permutation that maps $(j'_1, j'_2, \ldots, j'_k)$ back to $(j_1, j_2, \ldots, j_k)$, ie

\begin{equation}
\sigma_J =
\left(
\begin{array}{llll}
j'_1 & j'_2 & \cdots & j'_k \\
j_1 & j_2 & \cdots & j_k \\
\end{array}
\right)
\tag{3}
\label{eq:perm-sigma}
\end{equation}

(ie $\sigma_J$ maps the original position of an object in the list to its new position in the permuted list), so that $j_{\sigma_J(r)} = j'_r, \forall r \in \{1, 2, \ldots, k\}$. Note the numbers that are being permuted are numbers in the range $\{1, 2, \ldots, n\}$ but $\sigma_J \in P_k$ acts on their positions which are numbers in the range $\{1, 2, \ldots, k\}$.

If $\sigma_J$ is expressed as a composition of $t$ swap permutations (ie transpositions) then $\operatorname{sgn}(\sigma_J) = (-1)^{t}$ and if the corresponding sequence of $t$ column swaps is applied to the column selections for $A(J')$ then the matrix $A(J)$ is produced. But since each column swap multiplies the determinant by $-1$, we then have

\begin{eqnarray}
\det A(J) & = & (-1)^{t} \det A(J') \tag{4} \label{eq:JtoJ'-perm-t} \\
\mbox{ie.} \hspace{1em} \det A(J) & = & \operatorname{sgn}(\sigma_J) \det A(J') \tag{5} \label{eq:JtoJ'-perm-sigma} \\
\therefore\  \hspace{1em} \det AB & = & \sum_{J = (j_1, j_2, \ldots, j_k) \in P_{k, n}} (b_{j_1, 1} b_{j_2, 2} \cdots b_{j_k, k}) \cdot \operatorname{sgn}(\sigma_J) \cdot \det(A(J')) \tag{6} \label{eq:detAB-sigma}
\end{eqnarray}

Now gather together the terms in the summation (\ref{eq:detAB-sigma}) which have a common $J'$. The set of all possible $J'$ is $R_{k, n}$ and for each $J' \in R_{k, n}$ there will be $k!$ terms $J \in P_{k, n}$ that map to that same $J'$ under $\psi$, ie all the injective maps which when placed in order are the same as $J'$. Thus the above sum (\ref{eq:detAB-sigma}) partitions into the double sum :

\begin{equation}
\det AB = \sum_{J' = (j'_1, j'_2, \ldots, j'_k) \in R_{k, n}} \left( \sum_{\substack{J = (j_1, j_2, \ldots, j_k) \in P_{k, n}, \\ \psi(J) = J' }} (b_{j_1, 1} b_{j_2, 2} \cdots b_{j_k, k}) \cdot \operatorname{sgn}(\sigma_J) \right) \cdot \det(A(J'))
\tag{7} \label{eq:detAB-double-sum}
\end{equation}

Writing $j_r = j_{\sigma_J(\sigma_J^{-1}(r))}$ we have $j_r = j'_{\sigma_J^{-1}(r)}$, and so as the inverse of a permutation has the same signature as the permutation itself, the summand in the inner sum of (\ref{eq:detAB-double-sum}) is :

\begin{equation}
(b_{j'_{\sigma_J^{-1}(1)}, 1} b_{j'_{\sigma_J^{-1}(2)}, 2} \cdots b_{j'_{\sigma_J^{-1}(k)}, k}) \cdot \operatorname{sgn}(\sigma_J^{-1})
\tag{8} \label{eq:detAB-inner-sum-sigma-inverse}
\end{equation}

and this ranges over all $J \in P_{k, n}$ satisfying $\psi(J) = J'$. But such a range of $J$ corresponds with all $k!$ permutations of the in-order $J'$ (which is fixed in the inner sum), and thus $\sigma_J$ over this range of $J$ covers all $k!$ elements of the set $P_k$ of permutations of $[k]$ - hence $\sigma_J^{-1}$ over this range of $J$ covers the latter set of permutations also. Thus with the substitution $\tau = \sigma_J^{-1}$ the inner sum in (\ref{eq:detAB-double-sum}) equals (for a given fixed $J'$) :

\begin{equation}
\sum_{\tau \in P_k} (b_{j'_{\tau(1)}, 1} b_{j'_{\tau(2)}, 2} \cdots b_{j'_{\tau(k)}, k}) \cdot \operatorname{sgn}(\tau)
\tag{9} \label{eq:detAB-inner-sum-tau}
\end{equation}

But from the definition of $B(J')$, $B(J')_{s,r} = b_{j'_{s},r} \forall\:s, r$, and so $B(J')_{\tau(r),r} = b_{j'_{\tau(r)},r}$ for $r \in \{1, 2, \ldots, k\}$, and thus the inner sum (\ref{eq:detAB-inner-sum-tau}) equals :

\begin{equation}
\sum_{\tau \in P_{k}} (B(J')_{\tau(1), 1} B(J')_{\tau(2), 2} \cdots B(J')_{\tau(k), k}) \cdot \operatorname{sgn}(\tau)
\tag{10} \label{eq:detAB-inner-sum-BJ'}
\end{equation}

which is the column-wise Leibniz formula for $\det B(J')$. Thus from (\ref{eq:detAB-double-sum}) :

\begin{eqnarray}
\det AB & = & \sum_{J' \in R_{k,n}} \det B(J') \cdot \det A(J') \tag{11} \label{eq:detAB-Rkn} \\
& = & \sum_{S \in \binom{[n]}{k}} \det A_S \cdot \det B_S \tag{12} \label{eq:detAB-S}
\end{eqnarray}

using the one-to-one correspondence between sets $R_{k, n}$ and $\binom{[n]}{k}$, where $A_S$ denotes the $k \times k$ matrix obtained from $A$ by selecting the column numbers in $A$ from the set $S$, and $B_S$ denotes the $k \times k$ matrix obtained from $B$ by selecting the row numbers in $B$ from the set $S$.


问:
Harmonic functions in $\mathbb{R}^d$
I want to establish the equivalence of the 3 standard definitions, and that harmonic functions are $C^\infty$.  The 3 definitions are:

Mean value property and continuous.

$C^2$ and $0$ Laplacian.

Mean value property on arbitrarily small balls and continuous.

The only help I think I need here is proof of the existence and uniqueness of solution for the Dirichlet problem on the ball. (That would help me to get 3 implies 2, along with the maximal property for harmonic functions of type 3, which should be easy to prove.)  2 implies 1 by using arguments to take a derivative under the integral, and then using gauss' divergence theorem.  1 implies 3 obviously.  I also need help seeing that harmonic functions must be $C^\infty$.  I am not used to methods that don't involve complex analysis, as must be used here.

I know that there is a theory of plurisubharmonic functions.  As a bonus question, do those tend to be useful outside of complex analysis, and are they ever discussed for odd dimension?  For example, I have never seen them discussed in harmonic analysis, nor do they seem to be useful in relation to Brownian motions, which is why I am learning the d-dimensional version of harmonic function theory now.

Edit: Come to think of it, I'd also like some help proving the open mapping property and maximal properties for harmonic functions.  Please only assume definition 3 here, because I will use it and a connectedness argument to establish 3 implies 2.  The precise statement of definition 3 is as in Greene and Krantz but for d dimensions:

$f$ is "harmonic (3)" if $f$ is continuous and $\forall x \in U$ the domain of $f$ there exists $\epsilon>0$ such that all balls of radius $\epsilon$ or less centered at $x$ are contained in U, and $f$ satisfies the MVP for that ball/spherical shell.

答:
For the Dirichlet problem on a Ball in $\mathbb R^n$ of radius $r$:

$$\begin{cases}\Delta u=0&\text{ in }B^\circ (0,r)\\u=g&\text{ in }\partial B(0,r)\end{cases}$$ assuming $g$ is continuous on $\partial B(0,r)$.

The Poisson Kernel on this Ball is $$K(x,y)=\dfrac {1}{na(n)r}\cdot\dfrac {r^2-|x|^2}{|x-y|^n}$$ where $a(n)$ is the volume of the unitary ball.

Let $$u(x)=\dfrac{r^2-|x|^2}{na(n)r}\int_{\partial B(0,r)}\dfrac{g(y)}{|x-y|^n}dy, x\in B^\circ (0,r)$$
Then you have to prove that:

$1.u\in C^\infty (B^\circ (0,r))$

$2.\Delta u=0,\text{ in } B^\circ (0,r)$

$3.\lim\limits_{x\to x_0}u(x)=g(x_0),\forall x_0\in\partial B(0,r)$

Title: Harmonic functions in $\mathbb{R}^d$


答:
The definition of conditional density is f(x|y) = f(x $\cap$ y)/f(y).  Plus it is right in the name: the probability density depends on the condition, the value of y, so it has to be a function of y.  p(x|y) is generally a function of both x and y. If you integrate it over all y then you end up with a function that depends on x, called the marginal probability, f(x).  E[x|y] is not a function of x because it is an expectation--when you integrate over a variable between constant limits, the variable goes away.  It still depends on y though.  E[x|y] says "the average value of x given this value of y."
When you do a double integral to get a number, the outer integral has to have numbers for limits, but the inner integral can have functions of the outer variable for limits, because it will still be a function of the outer variable once you have done the inner integral, but it won't be a function of the inner variable any more because you "totaled it all up."  It can't depend on a specific value of inner variable because it already is built out of ALL the values of the inner variable.  Try writing out a couple of easy double integrals for yourself; find the area of a triangle, for example.  Watch when you have functions of what, and integrate in both orders so you can see the difference.

Title: Why is the conditional expectation $E[x|y]$ a function of the conditioning argument $y$? How about the pdf $p(x|y)$?

Body:
For a random variable $x$ and $y$, consider the conditional pdf $p(x|y)$ and the conditional expectation $E[x|y]$.

I previously thought that the conditional pdf or conditional expectation were functions of $x$, not $y$.

But I read that the $E[x|y] = \int_{-\infty}^{\infty}x\cdot p(x|y)dx$ is a function of $y$. Hence, when taking the expectation again on $E[x|y]$, we write:

$$
E[E[x|y]] = \int_{y = -\infty}^{\infty}\left[\int_{-\infty}^{\infty}x\cdot p(x|y)dx\right]p(y)dy.
$$

I do not understand this first because I thought $p(x|y)$ is a function of $x$, and therefore $E[x|y]$ also a function of $x$. Am I wrong in the first place?

Is $p(x|y)$ actually a function of $y$? Why isn't it a function of $x$?

Why isn't $E[x|y] = \int_{-\infty}^{\infty}x\cdot p(x|y)dx$ a function of $x$ when it clearly has $x$ in the expression? Is it a function of both $x$ and $y$?


答:
Possibly same trick: $x^4+y^4 \ge 2x^2y^2$

You end up bounding the entire sum by $2\times 2=4$, which is achievable at $x=y=1$, so this bound is sharp.

Title: $xy=1$ find the minimum value of $x^{6}+x^{4}y^{2}+x^{2}y^{4}+y^{6} $

Body:
$x,y \in \mathbb{R}$ and this is a math Olympiad problem.

My Attempt :

Note:$a^{2}+b^{2} \geq 2ab$ For all $a,b \in \mathbb{R}$.
$$x^{6}+x^{4}y^{2}+x^{2}y^{4}+y^{6} =(x^{2}+y^{2})(x^{4}+y^{4})$$
So we get:
$$(x^{2}+y^{2})(x^{4}+y^{4}) \geq 2xy(x^{4}+y^{4})$$
Since $xy=1$:
$$(x^{2}+y^{2})(x^{4}+y^{4}) \geq 2(x^{4}+y^{4})  $$
So the min of $x^{6}+x^{4}y^{2}+x^{2}y^{4}+y^{6}$ is $2(x^{4}+y^{4})$.

What can i do to get rid of this $x^{4}+y^{4} $?


答:
AM-GM says
$$
\frac{x^6+x^4y^2+x^2y^4+y^6}{4}\geq\sqrt[4]{x^6\cdot x^4y^2\cdot x^2y^4\cdot y^6}=1
$$
where equality is obtained iff all the four terms in the numerator are equal. And it is possible to make them equal, so we can obtain equality.


Body:
Hint

Replace $y$ with $\frac{1}{x}$ to find

$$f(x)=x^6+\frac{1}{x^6}+x^2+\frac{1}{x^2}$$

$$=(x^3-\frac{1}{x^3})^2+2+(x-\frac{1}{x})^2+2$$

$$\ge2+2=4$$
The equality occurs if $x-1/x=0$

Alternatively if $y=x^2+1/x^2=(x-1/x)^2+2\ge2,$

$$f(x)=y^3-3y+y=y(y^2-2)\ge2(2^2-2)=?$$


问:
Prove or disprove, limits of antiderivatives
If $g(x), f(x)$ are real and continuous, and $g(x)\leq$ $f(x)\leq$$g(x)+\frac{1}{x}$ for every $x>1$.
The following limit is equal to $0$:
$$\lim\limits_{x \to \infty} \int_x^{x+\sqrt{x}}(f(t)-g(t))\,dt =0$$

答:
If:
$$g(x)\le f(x)\le g(x)+\frac1x$$
then:
$$0\le f(t)-g(t)\le\frac1t$$ so:
$$\int_x^{x+\sqrt{x}}0\,dt\le\int_x^{x+\sqrt{x}}f(t)-g(t)dt\le\int_x^{x+\sqrt{x}}\frac{dt}t$$
now integrate all parts and you get:
$$0\le\int_x^{x+\sqrt{x}}f(t)-g(t)dt\le\ln(x+\sqrt{x})-\ln(x)=\ln\left(1+\frac{1}{\sqrt{x}}\right)$$
now take the limit...


答:
As $y=\frac{1}{x}$, $x^6+x^4y^2+x^2y^4+y^6=x^6+x^2=x^{-2}+x^{-6}$.

We know that, for $t\gt 0$, we have $t+\frac{1}{t}\ge 2$ with the equality satisfied if and only if $t=1$. Taking $t=x^2$ and $t=x^6$ you conclude that both $x^6+x^{-6}\ge 2$ and $x^2+x^{-2}\ge 2$, with equality satisfied if and only if $x=\pm 1$. Thus:

$$x^6+x^4y^2+x^2y^4+y^6\ge 4$$

with the equation satisfied if and only if $x=y=\pm 1$.


问:
What's so well in having a least element in a set?
What's so well about the principle of well-ordering? Why is this pattern of order named like this? I am unable to trace a connection between well-ordering and a set that has a least element so I presume that there must be some mathematical becakground I'm unaware of. I've searched some wikipedia links but they describe only what the principle is, not the reason why it's named like this.</p>

答: Remember that Cantor was led to introduce the basic concepts of set theory (especially ordinal an cardinal numbers) by the need for transfinite inductive definitions in treating some aspects of trigonometric series.  So it's not surprising that he should call a set well-ordered (wohlgeordnet) when it serves his purpose of supporting transfinite inductions.</p>


问:
Show $\log\left(\frac{x-y}{x}\right)-2\sqrt{\frac{x-y}{x}}<\log\left(\frac{x+y}{x}\right)-2\sqrt{\frac{x+y}{x}}$ if $x\geq 5$ and $1\leq y\leq x-2$
Assume that all logarithms are natural.

Let $x$ and $y$ be integers that satisfy $x \geq 5$ and $1 \leq y \leq x-2$. I am trying to show that $$\log\left( \frac{x-y}{x}\right)-2\sqrt{\frac{x-y}{x}} < \log\left( \frac{x+y}{x}\right)-2\sqrt{\frac{x+y}{x}}.$$

This is equivalent to showing that $$\log\left( \frac{x-y}{x+y}\right)<2\left(\sqrt{\frac{x-y}{x}}-\sqrt{\frac{x+y}{x}}\right).$$

My first inclination was to apply the well known inequality, $\log(z)\leq 2(\sqrt{z}-1)$ if $z>0$, to the left side by letting $z=\frac{x-y}{x+y}$, but this did not help me.

I also tried double induction on $(x,y)$ (since $x$ and $y$ are integers) but I could not finish it.

I appreciate any help, thank you.

答:
The inequality
$$
\log\left( \frac{x-y}{x+y}\right)<2\left(\sqrt{\frac{x-y}{x}}-\sqrt{\frac{x+y}{x}}\right)
$$
holds for all real numbers $x, y$ satisfying $0 < y < x$.

With the substitution $t=y/x$ this is equivalent to
$$ \tag{*}
\log\left( \frac{1+t}{1-t}\right)>2\left(\sqrt{1+t}-\sqrt{1-t}\right)
$$
for $0 < t < 1$. This suggests to investigate the function
$$
 f(t) = \log\left( \frac{1+t}{1-t}\right) -2\left(\sqrt{1+t}-\sqrt{1-t}\right) \, .
$$
Then $f(0) = 0$, and for $0 < t < 1$ is
$$
 f'(t) = \frac{2}{1-t^2} - \left( \frac{1}{\sqrt{1+t}} + \frac{1}{\sqrt{1-t}}  \right) \\
= \frac{2 - \sqrt{1-t^2}\left(\sqrt{1+t} + \sqrt{1-t}\right)}{1-t^2}
$$
and that is positive, because
$$
\sqrt{1-t^2} \cdot (\sqrt{1+t} + \sqrt{1-t}) < \sqrt{1+t} + \sqrt{1-t} \le 2
$$
as can be verified easily.

So $f$ is strictly increasing on $[0, 1)$, and that proves the inequality $(*)$.


问:
Combinatorics on divisors
Consider the set $D$ of all divisors of $10000$. What is the number of subsets $H \subseteq D$ containing at least two elements such that for every $a, b \in H$, $a \mid b$ or $b \mid a$?
I tried translating the problem into one about integer grid, which makes it ask for the number of ways to choose at least two points in the square with vertices $(0, 0)$ and $(4,4)$ ordered in such a way that they can be reached sequentially by only moving up and right. But I fail to see a way to calculate it besides setting up this recurrence: let $a_{m,n}$ be the number of subsets of the square between $(0,0)$ and $(m,n)$ that satisfy the above property. We can form such a subset in a number of ways: it could be a subset already counted in $a_{m-1,n}$, or in $a_{m,n-1}$, or one of those same subsets with the addition of $(m,n)$, or be composed of $(m,n)$ and a point in the rectangle with vertices $(0,0$) and $(m, n)$ (there are $(m+1)(n+1)-1$ such points). If we simply added $a_{m-1,n}$ and $a_{m,n-1}$ we would be counting twice the subsets already counted in $a_{m-1, n-1}$.

The recurrence is thus: $\begin{cases} a_{0, 0} = 0 \\ a_{n,0} = a_{0,n} = 2a_{n-1, 0} + n \\ a_{m,n} = 2(a_{m-1,n} + a_{m,n-1} - a_{m-1,n-1}) + (m+1)(n+1)-1\end{cases}$.

And the answer we are looking for is $a_{4,4}$. Is there a more elegant way to answer this however?

答:
I think that $a(5)$ in this OEIS sequence might contain the number you’re looking for, if you subtract from it the $26$ subsets containing fewer than two divisors that satisfy your condition. At least it has the right number, seems related, and seems to be correct for divisors of $2^1 5^1 = 10$ as well. (The number $a(2)$ counts the $12$ subsets of $\{1, 2, 5, 10\}$ not containing both $2$ and $5$.)

The entry for this OEIS sequence, whose terms are one-half the values of the one above, gives a generating function and other information you might find useful.

Title: Combinatorics on divisors


问:
Matrix Determinant Rules
During my preparation for my final exam in Linear Algebra I, I am currently investigating Determinants, and its uses.

I wanted to calculate $\det(2A)$, and I have proved to myself that:

$$\det(\alpha A)= \alpha^n \det(A)$$ for some $\alpha \in \mathbb{R}$ and for $A \in \mathbb{M}_{n\times n}(\mathbb{R})$.

Is that correct? If not, is there any other method to calculate it?

Also, if any of you have some tips and tricks that he/she would like to share before my exam, I will very appreciate it!

Thanks a lot!

答:
It's correct.

Just learn all basic properties of determinant. Determinant of a product, Laplace theorem, connection with eigenvalues, standard methods of calculation.

Title: Matrix Determinant Rules


问:
Bilinear forms and polarization identity
Let $H$ be a Hilbert space and $\sigma: H \times H \to \mathbb{C}$ a sesquilinear form (linear in the first variable, anti-linear in the second variable). Then it is well-known that the following polarization identity holds:

$$\sigma(x,y)=\frac{1}{4} \sum_{k=0}^3 i^k \sigma(x+i^k y, x + i^ky)$$

Is there a similar result when $\sigma: H \times H \to \mathbb{C}$ is bilinear (linear in both variables)? I.e. is there a way to express $\sigma(x,y)$ as a linear combination of elements of the form $\sigma(z,z)$?

答:
It's even easier:
\begin{align*}
\sigma(x,y) &= \frac{\sigma(x+y,x+y)-\sigma(x,x)-\sigma(y,y)}{2} \\ &=\frac{\sigma(x+y,x+y)-\sigma(x-y,x-y)}{4}.
\end{align*}


答:
Try working with the inequalities on $x$ and $y$.  We are given $x\in[5,\infty)$ and $y\in[1,x-2]$.  First, we find $x-y\in[2,x-1]$ and $x+y\in[6,2x-2]$.  For a fraction, we can find the minimum by maximizing the denominator and minimizing the numerator, for instance $\frac{x-y}{x+y}\in[\frac{2}{2x-2},\frac{x-1}{6}]$.

Then, $\log(z)$ and $\sqrt(z)$ are strictly increasing functions so we can plug in the minimum and maximum values for their arguments.  So it will suffice to show
$$\log\left(\frac{x-1}{6}\right)<2\left(\sqrt{\frac{x-1}{5}}-\sqrt{\frac{6}{x}}\right)$$

We can split up the inequality to find the inequality you mentioned:

$$\log\left(\frac{x-1}{5}\cdot\frac{5}{6}\right)<2\left(\sqrt{\frac{x-1}{5}}-1\right)+2\left(1-\sqrt{\frac{6}{x}}\right)$$
$$\log\left(\frac{x-1}{5}\right)+\log\left(\frac{5}{6}\right)<2\left(\sqrt{\frac{x-1}{5}}-1\right)+2\left(1-\sqrt{\frac{6}{x}}\right)$$

And extracting the $\log(z)\le 2(\sqrt(z)-1)$ identity we need to show:

$$\log\left(\frac{5}{6}\right)<2\left(1-\sqrt{\frac{6}{x}}\right)$$

Which is clearly true because the lhs is negative and the rhs is positive for the allowed values of $x$.

Title: Show $\log\left(\frac{x-y}{x}\right)-2\sqrt{\frac{x-y}{x}}<\log\left(\frac{x+y}{x}\right)-2\sqrt{\frac{x+y}{x}}$ if $x\geq 5$ and $1\leq y\leq x-2$

Body:
Assume that all logarithms are natural.

Let $x$ and $y$ be integers that satisfy $x \geq 5$ and $1 \leq y \leq x-2$. I am trying to show that $$\log\left( \frac{x-y}{x}\right)-2\sqrt{\frac{x-y}{x}} < \log\left( \frac{x+y}{x}\right)-2\sqrt{\frac{x+y}{x}}.$$

This is equivalent to showing that $$\log\left( \frac{x-y}{x+y}\right)<2\left(\sqrt{\frac{x-y}{x}}-\sqrt{\frac{x+y}{x}}\right).$$

My first inclination was to apply the well known inequality, $\log(z)\leq 2(\sqrt{z}-1)$ if $z>0$, to the left side by letting $z=\frac{x-y}{x+y}$, but this did not help me.

I also tried double induction on $(x,y)$ (since $x$ and $y$ are integers) but I could not finish it.

I appreciate any help, thank you.


答:
Starting with @labbattercarjee's substitution, call it $f$, and take the derivative with respect to $x$ and get:

$$\frac{df}{dx} = -\frac{6}{x^7}+6 x^5-\frac{2}{x^3}+2 x$$

Set it to $0$ to find the only real solutions are $x \pm 1$.


问:
I found the solution of Ax=b in the least squares but I don't understand how to use it
This is probably a very obvious question, but I don't understand it.

I know how to use the least squares method, which I did, and verified using an online calculator. My system is

("e" = and)

I solved x = $[1 , -1]^T$

I don't understand how this solves the problem. I know that if I multiply A by x I get b. What I don't understand is that, I know this method is applied when you have a bunch of points and want to find a line or function that goes through them, in an approximate way. How do I get a line from x? I know x has to do with the coefficients of a polynomial, and that A and b are the x and y coordinates of the points, so why do A has two columns? It's probably a really silly question but I'm very confused.

Thanks.

答:
The least squares problem is $y=X\beta+\epsilon$, and the solution is given by $\beta=(X^TX)^{-1}X^Ty$.

Here, we want to use the least squares technique to find the solution to $b=Ax$. However, least squares is not needed here because the system is consistent, i.e. it has a solution at $(1,-1)$. You can still get the solution by solving $\operatorname{arg min}\limits_x\|b-Ax\|^2=(1,-1)$ and the error is $\|b-A(1,-1)\|^2=0$ because it's a solution.

It becomes more interesting if the equation is inconsistent, i.e. say $b=\left(\begin{matrix}2\\-\sqrt 2\\-3\end{matrix}\right)$. Then the least squares solution will be given by $x=(A^TA)A^Tb=(\frac 3 2,-1)$. This minimizes the Euclidean difference between b and Ax in $R^3$. Here, the difference is not 0, in fact it is $\|b-A(\frac 3 2,-1)\|^2=\|(1/2,0,1/2)\|^2=\frac 1 2$.

There are two ways to think about this

Way 1

Geometrically, the column space of A is a plane given by $\left(\begin{matrix}1\\-\sqrt 0\\-1\end{matrix}\right)x+\left(\begin{matrix}-1\\\sqrt 2\\1\end{matrix}\right)y$, which is a plane in $\mathbb R^3$. You seek a point (x, y) such that the perpendicular (Euclidean) distance between b and Ax is minimum. It's given by (3/2, -1) and the distnace is $1/2$.

Way 2

You've got three planes in $R^3$, $z_1=x-y, z_2=\sqrt 2 y, z_3=-x+y$. It looks like this where the origin is at the center:

The least squares solution is found at the (x,y) coordinate (i.e. moving around in the x-y plane) such that the difference between x-y and the plane z=2, plus the squared difference between $\sqrt 2 y$ and the plane z=$-\sqrt 2$, and the squared  difference between -x+y and the plane z=-3 is minimized. You must imagine that there's a "pin" perpendicular to the xy plane and the position of the pin where these vertical distances are minimized is the least squares solution.

Conversely, if the system were consistent, in which case you wouldn't use least squares, the pin would precisely land where x-y=2, $\sqrt 2 y=-\sqrt 2$, and -x+y=-3.


问:
Correlation between multiplied numbers?
I do not have a strong math background, but I'm curious as to what this pattern is from a mathematical standpoint.

I was curious how many minutes there were in a day, so I said "24*6=144, add a 0, 1440."  Then it immediately struck me that 12*12=144, 6 is half of 12, and 12 is half of 24.  So, I checked to make sure it worked in other circumstances:

4*16=64
8*8=64

11*44=484
22*22=484

9*36=324
18*18=324

So what exactly is going on here from a logical standpoint to create that pattern?  Thanks in advance for satiating my curiosity!

答:
I'm not sure exactly what you are asking, but I can't think of a more logic pattern than:

There are $24$ hours to a day, and $60$ minutes to an hour, thus

$$\text{minutes/day} = 24*60$$

Since $24*6 = (24/2)*(6*2)$, your logic was faulty.

Title: Correlation between multiplied numbers?


Body:
As $x^6 + x^4 y^2 + x^2 y^4 + y^6=(x^2+y^2)(x^4+y^4)$

$$
\frac 12(x^2+y^2)\frac 12(x^4+y^4)\ge \sqrt{x^2y^2}\sqrt{x^4y^4}=1
$$


问:
Definition of Inner Product boolean function
I want to define a boolean function $f: \{-1,1\}^{2n} \rightarrow \{-1,1\}$ which gives the inner product in the given range and defined as :

$$f(x_1,x_2,.......,x_n,x_{n+1},....,x_{2n})= x_1x_{n+1}+x_2x_{n+2}+.....+x_nx_{2n}$$
Here, value of $x_1x_{n+1}+x_2x_{n+2}+.....+x_nx_{2n}$ should be either $+1$ or $-1$, So it means I have to use some operation to get values as $\{\pm 1\}$ but I am not getting what operation should I have to do so that it will give values either $+1$ or $-1$.

For example, if we have $n=2$ then for $(x_1,x_2,x_3,x_4) = (1,1,1,1)$, So, $f(1,1,1,1)= 2$. Now, which operation should I have to perform so that value 2 will be converted to $+1$ or $-1$.

One way I am thinking about to do the mapping from $\mathbb{F}_2^{2n} \rightarrow \mathbb{F}$ and then encode it as $0 \mapsto 1$ and $1 \mapsto -1$ like for $n=1$ in truth table, I can replace $f(0,0)=0$ to $f(1,1)=1$ and $f(1,1)=1$ to  $f(-1,-1)=-1$ etc.

If I have a function $g: \mathbb{F}_2^{2n} \rightarrow \{-1,1 \}$ then I can define the inner product mod 2 boolean function as

$$g(x_1,x_2,.....,x_n,y_1,y_2,....,y_n) = (-1)^{x.y}$$
where $x.y$ is the dot product of $x=(x_1,x_2,....,x_n)$ and $y=(y_1,y_2,....,y_n)$ in vector space $\mathbb{F}_2^n.$

But how to define it on $f: \{-1,1\}^{2n} \rightarrow \{-1,1\}$. If I do the same mapping as defined for $\mathbb{F}_2^{2n} \rightarrow \{-1,1 \}$ i.e. $(-1)^{x.y}$ then for $n=2$, I am getting all the values of $+1$ and for $n=3$, it gives $-1$. Here, $(-1)^{odd}=-1$ and $(-1)^{even}=+1$  So, I can use this mapping i.e.$(-1)^{x.y}$  but Is there any other way to define mapping for $f$ because I have to find the Fourier expansion of inner product boolean function which is defined on $\{-1,1\}^{2n} \rightarrow \{-1,1\}.$

Any help would be appreciated.

答:
$n=1$ is trivial and uninformative. So look at the case for $n=2$.

By commutivity of addition, you want

$f(x_1, x_2, y_1, y_2) = f(x_2, x_1, y_2, y_1)$

By distributivity of multiplication over addition, you want

$f(x_1, x_2, -y_1, -y_2) = -f(x_1, x_2, y_1, y_2)$

Here in lies a problem.

By 1, $f(1,1,1,-1) = f(1,1,-1,1)$

By 2, $f(1,1,1,-1) = -f(1,1,-1,1)$

So $f(1,1,1,-1) = -f(1,1,1,-1)$. But the only values allowed are $1$ and $-1$, neither of which are its own opposite.

You can't do this without either giving up commutivity of addition, or distributivity.


问:
Equality of products of reduced residue classes
I am working on the following problem:

Suppose $n>1$. Prove that
$$
\prod_{\substack{1\leq a\leq n\\(a,n)=1}}a\equiv\prod_{\substack{1\leq b\leq n\\(b,n)=1\\b^2\equiv1\pmod n}}b\pmod n.
$$

A computer program seems to show that for each $n$, the product either equals $1\pmod n$ or $n-1\pmod n$. Why is this the case, and how may I prove the result?

Looking at the product on the right, if $(b,n)=1$ and $b^2\equiv1\pmod n$ then I am inclined to think $b=1$, however clearly this is not the case. What is faulty with my reasoning? Any help would be appreciated!

答:
First of all, name the multiplicative inverse of any $t$ (mod $n$ clearly) as $t^*$. Now trivially the elements of the reduced residue class fall into two disjoint sets :

$i)$ Set $A$ consisting of all elements whose multiplicative inverse is equal to themselves.

$ii)$ Set $B$ consisting of all elements whose multiplicative inverse is not equal to themselves.

Now one can say that $$\displaystyle\prod_{1\leqslant a\leqslant n\\(a,n)=1}a=\displaystyle\prod_{1\leqslant a\leqslant n\\(a,n)=1\\{a\in A}}a\displaystyle\prod_{1\leq a\leq n\\(a,n)=1\\a\in B}a$$

Now note that $A$ is made of a pair of residues that in each pair the two numbers are multiplicative inverses of each other mod $n$ so $\displaystyle\prod_{1\leqslant a\leqslant n\\(a,n)\\a\in A}\overset{n}{\equiv}1$

Hence:$$\displaystyle\prod_{1\leqslant a\leqslant n\\(a,n)=1}a=\displaystyle\prod_{1\leqslant a\leqslant n\\(a,n)=1\\{a\in A}}a\displaystyle\prod_{1\leq a\leq n\\(a,n)=1\\a\in B}a=\displaystyle\prod_{1\leqslant a\leqslant n\\(a,n)\\a\in B}$$

But $B$ is the set of all residues whose multiplicative inverse is equal to themselves so for each element $b\in B$, we have $b^2\overset{n}{\equiv}1$ and hence: $$\prod_{\substack{1\leq a\leq n\\(a,n)=1}}a\equiv\prod_{\substack{1\leq b\leq n\\(b,n)=1\\b^2\overset{n}{\equiv}1}}b$$


问:
Sum of a dyadic sequence satisfying $\delta\omega_\lambda \leq \omega_{2\lambda}\leq \kappa\omega_\lambda$ with $1<\delta<\kappa$
I was reading the proof of a lemma and I found something that seemed to be a "trivial inequality" but I haven't been able to prove it, so I was wondering if anyone has a good answer for it.

Lets fix two real numbers $1<\delta<\kappa$. Now, consider a sequence $(\omega_{\lambda})_{\lambda}\subset\mathbb{R}_+$ with $\lambda$ running over the integer dyadic numbers and where $\mathbb{R}_+=(0,+\infty)$. Suposse that $(\omega_\lambda)_\lambda$ satisfies the following inequalities: $$
\delta\omega_\lambda \leq \omega_{2\lambda}\leq \kappa\omega_\lambda.
$$
Then, there exists a constant $c>0$ such that $$
\sum_{\mu\geq \lambda/8}\omega_\mu^{-1}\leq c\omega_{\lambda}^{-1}.
$$
Here, notation $\sum_\lambda \omega_\lambda$ means $\sum_{n=0}^{+\infty}\omega_{2^n}$. I think it should be a silly trick using power series but I am getting confuse since I am not very used to dyadic stuffs.

答:
As Steven Stadnicki suggested in the comments, if 'integer dyadics' means numbers of the form $2^n$, we can rewrite the problem with $\mu_i := \omega_{2^i}$. The assumption is now :
$$\forall i \geq 0, \quad \delta \mu_i \leq \mu_{i+1} \leq \kappa \mu_i$$

We want to prove that there exists $c > 0$ such that :

$$\forall n \geq 0,\quad \sum_{i \geq ~\max(n - 3, 0) } \frac1{\mu_i} \leq c \frac1{\mu_n}$$

First, we can suppose $n \geq 3$ because for $0 \leq n \leq 2$ the inequality is true with $c_1 = \mu_2 \times (\sum_{i \geq 0 } \frac1{\mu_i})$.

Heuristically, $\mu_i$ should behave as a geometric sequence. Remember that, to compute the sum $\sum_{i \geq n} q^i$ (with $0 < q < 1$) we use a telescopic sum :

$$\sum_{i \geq n} q^i (1 - q) = \sum_{i \geq n} q^i - q^{i+1} =  q^n $$

In fact, we can do the same with our problem :

$$\sum_{i \geq n-3} \frac1{\mu_i} (1 - \frac1{\delta}) = \sum_{i \geq n-3} \frac1{\mu_i} - \frac1{\mu_i \delta} \leq  \sum_{i \geq n-3} \frac1{\mu_i} - \frac1{\mu_{i+1}} = \frac1{\mu_{n-3}} \leq \kappa^3 \frac1{\mu_{n}} $$

We used the assumption in the two inequalities.

We conclude by taking $c = \max(c_1, \frac{\kappa^3}{(1 - \frac1{\delta})})$


答:
Is it possible to interpret the example you gave in terms of the p(x,y) and p(y)?

Of course it is! There are several possible interpretation. Here is one...

$$f(x|y)=\frac{f(x,y)}{f(y)}=\frac{2}{2y}$$

where

$$f(x,y)=2$$

is the joint uniform distribution on the following triangle

and $f(y)=2y$ is the density of the variable $Y$ where $y \in (0;1)$


问:
exponential between 2 points
I haven't done some math in a long long time... I've been trying to find an exponentially decreasing function (instead of a linear one that's easy) bound by 2 known points (x1,y1), (x2,y2), as below:

I've been trying to play with functions like f(x)=a*b^(k*x)+c, unsuccesfully. It's easy to guess that b effects the depth of the curve. But that leaves a, k and c to find.

答:
If we consider $f(x)=ae^{kx}+b$, (where $e$ is Euler's Constant), then we have 3 free parameters and we are solving for two points, so there are actually infinitely many solutions. We can let the function go through the first point by setting

$$b=y_1-ae^{kx_1}$$

for any $a,k$. It remains to find values such that

$$y_2=ae^{kx_2}+b=ae^{kx_2}+(y_1-ae^{kx_1})$$

and so

$$y_2-y_1=a(e^{kx_2}-e^{kx_1})$$

hence, we can choose

$$a=\frac{y_2-y_1}{e^{kx_2}-e^{kx_1}}$$

and $k$ can still be whichever value we like.

Title: exponential between 2 points


问:
Indeterminate limit algebraic method
Friends, I am stuck with this problem: what is the limit of

$$\lim_{x\to-2}\frac{x^2-2}{x^3+8}$$

答:
The proposed limit does not exist. Indeed, on one hand we have

\begin{align*}
\lim_{x\to-2^{+}}\frac{x^{2} - 2}{x^{3} + 8} = +\infty
\end{align*}

On the other hand, we have

\begin{align*}
\lim_{x\to-2^{-}}\frac{x^{2} - 2}{x^{3} + 8} = -\infty
\end{align*}

Title: Indeterminate limit algebraic method


问:
Fun calculus problem I can't seem to solve
I've recently picked up a math book I haven't read since college (highly recommended reading by the way!). I was reviewing multi-dimentional derivatives and such, and I stumbled upon a problem I've been trying to solve for two days, and I can't get it out of my head, so please help me out! = )

Problem (from memory):

There is a rabbit that runs in a perfect circle of radius $r$ with a constant speed $v$. A fox chases the rabbit, starting from the center of the circle and also moves with a constant speed $v$ such that it is always between the center of the circle and the rabbit. How long will it take for the fox to catch the rabbit?

I tried using the fact that $|x'(t)| = |y'(t)| = v$ where $x(t)$ is fox's position and $y(t)$ is rabbit's position, and that $x'(t)x''(t) = 0$ because of constant speed restriction, but I'm still failing to find a solution.

Anyone feel like attacking this one?

答: Hint:  the rabbit moves at constant angular velocity of $\frac{v}{r}$.  If we let the fox's radius be R, we have that $(R')^2+(\frac{Rv}{r})^2=v^2$, decomposing the fox's speed into radial and tangential pieces.</p>


答:
Can't resist jumping on the bandwagon:

Find minimum of $x^{6}+x^{4}y^{2}+x^{2}y^{4}+y^{6} $

Let $$f(x,y) = x^{6}+x^{4}y^{2}+x^{2}y^{4}+y^{6}. \tag{1}$$

Here, it is assumed that the domain of $f$ is restricted by the constraint that $(xy) = 1.$

Let $$g(x) = x^6 + x^2 + x^{(-2)} + x^{(-6)}$$

and restrict the domain of $g$ to $x > 0.$

Since $xy=1,$ the domain of $f$ is restricted to $x,y$ either both being positive or both being negative.  Also, for $(x > 0), f(x,y) = f(x,[1/x]) = g(x).$

From the definition of $g$:

$$g'(x) = 6x^5 + 2x -2x^{(-3)} - 6x^{(-7)}$$

and

$$g''(x) = 30x^4 + 2 + 6x^{(-4)} + 42x^{(-8)}.$$

At this point, there are two immediate conclusions:

$g'(x) = 0$ has a root at $x=1$.

$\forall $x > 0$, g''(x) > 0.$

Therefore, $g(x)$ must have a local minimum at $x=1, g(x)$ can not have a local minimum anywhere else, and the local minimum at $(x=1)$ must be a global minimum, with respect to the domain of $g$ being restricted to $(x>0)$.

Remember that the domain of $f(x,y)$ is restricted to $x,y$ either being both positive or both negative.

For $(x < 0)$, by equation (1) above, $f(x,y) = f([-x],[-y])$, so the behavior of $g(x)$ for $(x>0)$ is conclusive.


问:
How to find all third roots of $-i$?
How to find all third roots of $-i$?
I have no idea.
Could you give me some hint?

答:
As $i^3=-i,$

If $x^3=-i,$

$$x^3-i^3=(x-i)(x^2+ix+i^2)=0$$

Can you solve the quadratic equation?

Title: How to find all third roots of $-i$?


问:
Algebraic integer definition (basic question)
The notes of a module I'm doing have the following definition:

Let $\alpha \in \mathbb{C}$. We say that $\alpha$ is an algebraic integer if $\alpha$ is the root of a monic polynomial with integer coefficients, that is, $p(\alpha)=0$ for some $p \in \mathbb{Z}[t]$.

My problem here is with what goes after the "that is" above. Is it not possible for $p(\alpha)=0$ for some $p \in \mathbb{Z}[t]$, where $p$ is not a monic polynomial? I thought $\mathbb{Z}[t] = a + bt$ with $a, b \in \mathbb{Z}$, so $b$ does not need to be 1 for $p \in \mathbb{Z}[t]$ to hold, or did I misunderstand something?

答: Your doubt is natural. The part that comes after the “that is” doesn't make sense. It should be “that is, $p(\alpha)=0$ for some monic polynomial $p(x)\in\Bbb Z[x]$”.


问:
Find the GCD Product of a number and its power
Let $n, k \in \mathbb{N}$.

Find $\gcd(n, 2^n)\cdot\gcd(n+1, 2^{n+1}) \cdots \gcd(n+k, 2^{n+k-1})$.

I've been thinking about this for a while. I tried to plug in $n=1$, but didn't find patterns. The answer seems to be $2^{k-1}$. Not sure if this has anything to do with the Euler Phi function.

Thanks!

答:
Note that $2^k>k$ holds for all $k\in\mathbb{N} \implies V_{2}(2^k)>V_{2}(k) \implies gcd(k,2^k)=V_2(k)$

So what you're truly seeking is the value of $$\displaystyle\prod_{t=n}^{n+k}{V_{2}(t)}=V_2\Bigg({\displaystyle\prod_{t=n}^{n+k}}t\Bigg)$$

Which is equal to $$V_2\Bigg(\frac{(n+k)!}{(n-1)!}\Bigg)=V_2\bigg((n+k)!\bigg)-V_2\bigg((n-1)!\bigg)$$

But it's well known that $V_2(k!)=\displaystyle\sum_{i=1}^{\infty}{\bigg\lfloor\frac{k}{2^i}}\bigg\rfloor$ so finally we deduce that :
$$\displaystyle\prod_{t=n}^{n+k}{gcd(t,2^t)}=2^s$$ Where $s$ is equal to $${\displaystyle\sum_{i=1}^{\infty}{\bigg\lfloor\frac{n+k}{2^i}}\bigg\rfloor-\displaystyle\sum_{i=1}^{\infty}{\bigg\lfloor\frac{n-1}{2^i}}\bigg\rfloor}={\displaystyle\sum_{i=1}^{\infty}{\Bigg(\scriptsize{\bigg\lfloor\frac{n+k}{2^i}}\bigg\rfloor-\bigg\lfloor\frac{n-1}{2^i}}\bigg\rfloor\Bigg)}$$


问:
What notation to use for case distinction in a set
I'm currently in the following framework: $n \in \mathbb{N}$, $I \subseteq \{1, \dots, n\}$ and $X \subseteq \mathbb{R}$. Now, I need to define a set which contains all vectors $v \in \mathbb{R}^n$ such that the components of $v$ with indices in $I$ are elements of $X$ and the other components are zero. My naive approach to define this set looks like this

$$\{(x_1, \dots, x_n) \in \mathbb{R}^n: x_i \in X, i \in I, x_j = 0, j \in \{1, \dots, n\} \setminus I\}.$$

I don't like this definition as it contains two indices $i, j$ for the components. Also it is not immediately clear that the cases are well defined (each index is covered by a case and no index appears in both terms). In a normal cases environment one could write "else" for the second case. However a cases environment does not look nice within the braces of the surrounding set.

Is there some notation I am missing? How can I write this set more clearly? Of course I could use a description of the set as I did in my introduction of this question.

答:
You can describe it very simply if you do a little preliminary setting up. Let $$Y_i=\begin{cases}X,&\text{if }i\in I\\\{0\},&\text{otherwise;}\end{cases}$$ the desired set is then $\prod\limits_{i=1}^nY_i$.

If you want everything in the curly braces, it’s going to be a bit ugly. One fairly readable version is

$$\{\langle x_1,\ldots,x_n\rangle\in\Bbb R^n:x_i\in X\text{ for all }i\in I,\text{ and }x_i=0\text{ otherwise}\}\,.$$


答:
Think about multiplication as having piles of rocks. $6*24$ represents 6 piles of 24 rocks.

Now what happens if you split each pile in two halves? The number of rocks in your pile half, but the number of piles double. Thus, you get 12 piles of 12 rocks, or $12*12$.

You didn't change the number of rocks, you only rearrange them in a different way.
Thus you have $6*24=12*12$ rocks.

And the same holds with any numbers. If you split each pile in halves, the number of rocks in piles half, and the number of piles double. But in total you have the same number of rocks... Thus, if you multiply two numbers, half one and double the other, you get the same product. [Note that this intuitive explanation works for whole numbers, but can also be made to work easily for fractions].

Title: Correlation between multiplied numbers?

Body:
I do not have a strong math background, but I'm curious as to what this pattern is from a mathematical standpoint.

I was curious how many minutes there were in a day, so I said "24*6=144, add a 0, 1440."  Then it immediately struck me that 12*12=144, 6 is half of 12, and 12 is half of 24.  So, I checked to make sure it worked in other circumstances:

4*16=64
8*8=64

11*44=484
22*22=484

9*36=324
18*18=324

So what exactly is going on here from a logical standpoint to create that pattern?  Thanks in advance for satiating my curiosity!


答:
If you don't know polar form you can do $(a+bi)^3 = a^3 + 3a^2bi + 3ab^2i^2 + b^3i^3 = a^3 + 3a^2bi -3ab^2 -b^3i = (a^3-3ab^2) +(3a^2b -b^3)i = -i$.  So $a^3-3ab^2 = 0; 3a^2b-b^3 = -1$.

Solve for real $a,b$.

If you do know polar form then $-i = e^{\frac {3\pi}2i}$. so $(-i)^{\frac 13} = e^{\frac {\frac {3\pi}2+2\pi}3 i}$.

Also $i^3 = (i^2)i = -i$ so $i$ is one of the cube roots.

Title: How to find all third roots of $-i$?

Body:
How to find all third roots of $-i$?
I have no idea.
Could you give me some hint?


问:
On an Isomorphism of Semigroup Rings via Congruence Classes
Let $\mathbb Z_{\geq 0}$ denote the set of non-negative integers. Let $\mathbb Z_{\geq 0}^n$ denote the set of $n$-tuples of non-negative integers.

(Theorem 2.1.5, Herzog, 1969) Given a finitely generated commutative semigroup $S$ of rank $n$ and any surjective semigroup homomorphism $\varphi : \mathbb Z_{\geq 0}^n \to S,$ the congruence $\sim_\varphi \, : \mathbb Z_{\geq 0}^n \times \mathbb Z_{\geq 0}^n \to \mathbb Z_{\geq 0}^n$ defined by $\mathbf a \sim_\varphi \mathbf b$ if and only if $\varphi(\mathbf a) = \varphi(\mathbf b)$ induces an isomorphism of semigroup rings $$R[S] \cong R[\mathbb Z_{\geq 0}^n / \sim_\varphi] \cong R[x_1, \dots, x_n] / (x^{\mathbf a} - x^{\mathbf b} \,|\, \mathbf a \sim_\varphi \mathbf b).$$

One can establish by the First Isomorphism Theorem that there is an isomorphism $k[x, y] / (x^3 - y^2) \cong k[t^2, t^3]$; however, I am interested in using the above theorem of Herzog to prove the same result. For this purpose, we may consider the numerical semigroup $\mathbb Z_{\geq 0} \langle 2, 3 \rangle = \{2r + 3s \,|\, r, s \in \mathbb Z_{\geq 0}\}.$ Of course, there is a natural surjection $\varphi : \mathbb Z_{\geq 0}^2 \to \mathbb Z_{\geq 0} \langle 2, 3 \rangle$ defined by $\varphi(r, s) = 2r + 3s,$ hence we have a congruence $\sim_\varphi$ on $\mathbb Z_{\geq 0}^2 \times \mathbb Z_{\geq 0}^2$ defined by $\mathbf a = (r_1, s_1) \sim_\varphi (r_2, s_2) = \mathbf b$ if and only if $\varphi(r_1, s_1) = \varphi(r_2, s_2)$ if and only if $2r_1 + 3s_1 = 2r_2 + 3s_2$ if and only if $r_1 \equiv r_2 \text{ (mod } 3)$ and $s_1 \equiv s_2 \text{ (mod } 2).$

From here, I am not sure how to conclude that $(\mathbf x^{\mathbf a} - \mathbf x^{\mathbf b} \,|\, \mathbf a \sim_\varphi \mathbf b) = (x^3 - y^2).$ I would be very gratefully for any feedback, comments, or suggestions about how to prove this equality.

Body:
Let $2r_1+3s_1=2r_2+3s_2$ and assume that $r_1 > r_2$ and $s_1 < s_2$. It follows that $\frac{s_1-s_2}{2}=\frac{r_2-r_1}{3}$ and we denote this number by $N$. We have
$$
x^{r_1}y^{s_1}-x^{r_2}y^{s_2}=x^{r_2}y^{s_1} (x^{r_1-r_2}-y^{s_2-s_1})=x^{r_2}y^{s_1} (x^{3N}-y^{2N})= \\
=x^{r_2}y^{s_1} (x^{3}-y^{2})(x^{3(N-1)}+x^{3(N-2)}y^2+x^{3(N-3)}y^4+ \dots + x^3 y^{2(N-2)}+y^{2(N-1)})
$$
and, then, $x^{r_1}y^{s_1}-x^{r_2}y^{s_2} \in (x^3-y^2)$.
Now it is enough to note that (3,0) and (0,2) are in the same congruence class.


问:
What's the motivation for the term "compact" operator?
I am self-studying functional analysis and have just learned the definition of compact operators. However, it isn't clear to me why the name "compact operator" was chosen.

The operator itself could be considered a one-point subset of a topological vector space of operators, but the "compactness" doesn't seem to refer to any such topology (every operator would constitute a compact subset of such a space, I think).

The definition involves compactness, but only in an indirect way (bounded sets get mapped to relatively compact subsets). Thus it is unclear to me if there is a good reason for the term.

答:
Well, as my professor says, compact operators are small and Fredholm operators are big since cokernel of a Fredholm operator should be finite dimensional while the image of a compact operator between Banach spaces can not contain an infinite dimensional closed subspace.

There is also a deeper connection between two of those types of operators which is a subject of Riesz--Schauder theory.


问:
Help with calculus of variations question Chiang
I am trying to solve the following problem from Chiang's Fundamental Methods of Dynamic Optimisation:
$$
V \left[y \right] = \int_{0}^{2} \left( 2ye^t + y^2 +y'^2\right) dt
$$

with boundary conditions, $y(0) = 2$ and $y(2) = 2e^2+e^{-2}$

I can get to the general solution, but the definitive solution escapes me. The general solution (I think) is as follows:

Using Euler's equation
$$
F_{y'y'} y''(t) + F_{yy'}y'(t)+F_{ty'}-F_y = 0
$$

We have

$$
2y''(t) -(2e^t+2y) =0 \\
y''(t) =e^t+y
$$

Integrating with respect to time twice gives the general solution

$$
y{*}(t) = e^t +\frac{1}{2}yt^2 +c_1t +c_2
$$

Solving for the constants $c1$ and $c2$ using the boundary conditions yeilds

$$
y(t=0) = e^{(0)} + \frac{1}{2}y(0)^2+c_1(0)+c_2 = 2 \\
c_2 = 2-1 - 0-0 \\
c_2 = 1
$$

And

$$
y(t=2) = e^{(2)} + \frac{1}{2}y(2)^2+c_1(2)+c_2 = 2e^2+e^{-2} \\
$$

Substituting $c_2 = 2$
$$
y(t=2) = e^{2} + 2y+2c_1+2 = 2e^2+e^{-2} \\
y(t=2) =  2c_1 = 2e^2-e^{2}+e^{-2}-2y-2 \\
y(t=2) =  2c_1 = e^2+e^{-2}-2y-2 \\
y(t=2) =  c_1 = \frac{1}{2}(e^2+e^{-2})-y-1 \\
$$

Substituting these back into the general solution

$$
y^{*}(t) = e^t +\frac{1}{2}yt^2 +(\frac{1}{2}(e^2+e^{-2})-y-1)t +2
$$

Which I don't think is correct. The answer given in the solutions appendix is which I am miles away from and can't see how to get there

$$
y*(t) = e^t +e^{-t}+\frac{1}{2}te^t
$$

答:
You want to express $y$ purely in terms of $t$ to substitute the boundary conditions. Since $y$ depends on $t$, you cannot integrate with respect time twice without finding an explicit expression for $y(t)$ first. So use the method of undetermined coefficients.

We have $$y''-y=e^t$$ which is a second order inhomogeneous differential equation. Trying a complementary function of the form $y=e^{kt}$, we obtain the auxiliary equation $k^2=1$ and the roots $k=\pm 1$. Thus the complementary function is $y_{c}(x)=Ae^{t}+Be^{-t}.$ Then for the particular solution we try $y_{p}(x)=cte^{t}$ and substituting gives $c=\frac{1}{2}$.

Thus the general solution is $$y(t)=y_{c}(x)+y_p(x)=Ae^{t}+Be^{-t}+\frac{1}{2}te^t.$$

Now substitute the boundary conditions to find $A$ and $B$.


问:
Prove divergence for improper integral of $y \sin(2/y)$
How can I prove that the following improper integral diverge?

$$\int_e^\infty y \sin \left( \frac{2}{y} \right) \ dx$$

I think I'm supposed to separate it into two parts, but I don't know how to do it. Any input?

答:
Put $$t=\frac 1x$$

Your integral has the same nature than
$$\int_0^1\frac 1t\sin(t)\frac{dt}{t^2}$$

$$=\int_0^1\frac{\sin(t)dt}{t^3}$$

but, near $ 0^+ $,

$$\frac{\sin(t)}{t^3}\sim \frac{1}{t^2}$$

and
$$\int_0^1\frac{dt}{t^2} \text{ diverges}$$


答:
The key word(s) is(are) not "that is", but the following word "some". The idea is: $\alpha$ is an algebraic integer if there exists a monic polynomial $p\in\mathbb Z[x]$ such that $p(\alpha)=0$.

It doesn't mean that, if $\alpha$ is already an algebraic integer, $p(\alpha)$ cannot not be additionally true for some other polynomials, including those in $\mathbb{Z}[x]$ that are not monic, or even some that are not in $\mathbb Z[x]$... However, at least one of polynomials $p$ satisfying $p(\alpha)=0$ must be monic, with integer coefficients.

Example: $\alpha=i$ is an algebraic integer because we can take the monic polynomial $p(x)=x^2+1$ such that $p(i)=0$.

Example 2: $\alpha=\frac{1}{2}$ is not an algebraic integer, because no polynomial $p(x)=a_0+a_1x+\ldots+a_{n-1}x^{n-1}+x^n$ (with $a_i\in\mathbb Z$) satisfies $p(\frac{1}{2})=0$. Namely, if it was satisfied for some polynomial $p$, then the multiple: $2^{n-1}p(\frac{1}{2})=a_02^{n-1}+a_12^{n-2}+\ldots+a_{n-1}+\frac{1}{2}$ would also be zero, which it isn't, as it isn't even an integer.

(In fact, it can be shown that a number from $\mathbb Q$ is an algebraic integer if and only if it is an integer in the ordinary sense, i.e. if it is in $\mathbb Z$. This justifies the terminology "algebraic integer" - it is a generalisation of the notion of an "integer".)

Also, a correction in your question: $\mathbb Z[x]=\{a_0+a_1x+\ldots+a_nx^n\mid n\in\mathbb N, a_0,a_1,\ldots,a_n\in\mathbb Z\}$ i.e. it contains polynomials of arbitrary degrees.

Title: Algebraic integer definition (basic question)

Body:
The notes of a module I'm doing have the following definition:

Let $\alpha \in \mathbb{C}$. We say that $\alpha$ is an algebraic integer if $\alpha$ is the root of a monic polynomial with integer coefficients, that is, $p(\alpha)=0$ for some $p \in \mathbb{Z}[t]$.

My problem here is with what goes after the "that is" above. Is it not possible for $p(\alpha)=0$ for some $p \in \mathbb{Z}[t]$, where $p$ is not a monic polynomial? I thought $\mathbb{Z}[t] = a + bt$ with $a, b \in \mathbb{Z}$, so $b$ does not need to be 1 for $p \in \mathbb{Z}[t]$ to hold, or did I misunderstand something?


问:
What's the word for being able to "pull something out of a function"?
If I have something like $(5x + 7x) = x(5 + 7),$ I've "factored" out the $x$.

With linear functions you can say $f(a \cdot x) = a \cdot f(x)$ "by linearity", you've (for lack of better phrasing) "pulled out the multiplication".

Or for preimages of sets, the preimage of union = union of preimage $f^{-1}(\cup_i A_i) = \cup_i f^{-1}(A_i)$.

Is there a general term for saying some operation inside a function works in a similar way? Something like

$$f(A \odot B) = f(A) \odot f(B)$$

where $\odot$ is some general operator. I guess we could say "linearity in $\odot$", is that standard?

答: In the context of abstract algebra, where the domain and range of $f$ may not always be the same set, the term "homomorphism" can often be seen.


答:
"Linearity" is a notion for vector spaces.

The term you are looking for is called "homomorphism".

A homomorphism is a map between two algebraic structures of the same type (that is of the same name), that preserves the operations of the structures.

One says often that the homomorphism $f$ preserves the operation $\odot$ or is compatible with the operation.

Title: What's the word for being able to "pull something out of a function"?

Body:
If I have something like $(5x + 7x) = x(5 + 7),$ I've "factored" out the $x$.

With linear functions you can say $f(a \cdot x) = a \cdot f(x)$ "by linearity", you've (for lack of better phrasing) "pulled out the multiplication".

Or for preimages of sets, the preimage of union = union of preimage $f^{-1}(\cup_i A_i) = \cup_i f^{-1}(A_i)$.

Is there a general term for saying some operation inside a function works in a similar way? Something like

$$f(A \odot B) = f(A) \odot f(B)$$

where $\odot$ is some general operator. I guess we could say "linearity in $\odot$", is that standard?