问:
Solving a written problem involving system of linear equation
It is about a loan of 25.000.000.

Some of it was borrowed with 8% interest.

Some was borrowed with 10% interest.

And the rest with 9% interest.

We know that the sum that got 10% interest is 2.000.000 plus half of the 8% sum.

We also know that the total interest paid for the first year was 2.220.000.

I need to calculate the amount for each borrowed part using a system of linear equations

This is the system i came up with:

```<code>x + y + z = 25000000  all 3 parts should equal 25000000
y - 0.04x = 2000000   the sum that got 10% interest is 2.000.000 plus half of the 8% sum
0.08x + 0.1y + 0.09z = 2220000 the total interest paid is 2.220.000
</code>```

When i solve it i get these values:

```<code>x = 15625000/3
y = 6625000/3
z = 52750000/3
</code>```

I am not asking for the answer to this problem. Can someone tell me what i am doing wrong/point me in the right direction

Thanks in advance :)

答:
Some of it was borrowed with 8% interest.

Call that $x$.

Some was borrowed with 10% interest.

Call that $y$

And the rest with 9% interest.

Call that $z$.

It is about a loan of 25.000.000.

"x + y + z = 25000000  all 3 parts should equal 25000000"

Correct.  $x + y + z = 25000000$.

We know that the sum that got 10% interest is 2.000.000 plus half of the 8% sum.

"y - 0.04x = 2000000   the sum that got 10% interest is 2.000.000 plus half of the 8% sum"

Wait, what?  The "8% sum" refer to $x$.  We don't take half of 8% of $x$.  The "8%" just refers to what was borrowed at 8%.

So $y = 2000000 + \frac 12 x$.  That's what you should have written.

We also know that the total interest paid for the first year was 2.220.000.

"0.08x + 0.1y + 0.09z = 2220000 the total interest paid is 2.220.000"

Correct. $0.08 x + 0.1y + 0.09z = 2220000$

SO you have the three equations:

$x + y + z = 25000000$

$y = 2000000 + \frac 12 x$

$0.08 x + 0.1y + 0.09z = 2220000$

Solve them.


问:
Show that $\text {Hom}_{\mathbb R[x]}(M,N)=\{0\}$ where $M,N=\mathbb R^2$ are $\mathbb R[x]$-modules where $X$ acts as $A$ in $M$ and $B$ in $N$
Let $A=\begin{pmatrix}2&0\\0&3\end{pmatrix}$ and $B=\begin{pmatrix}1&0\\0&0\end{pmatrix}$

Show that $\text {Hom}_{\mathbb R[x]}(M,N)=\{0\}$ where $M,N=\mathbb R^2$ are $\mathbb R[X]$-modules and $X$ acts as $A$ in $M$ and $B$ in $N$.

I've completed the first part of this question which asked to calculate $(1-3X+X^2)\cdot\begin{pmatrix}1\\2\end{pmatrix}$ in both $M$ and $N$ for which I got $\begin{pmatrix}-1\\2\end{pmatrix}$ in both $M$ and $N$ (which is probably important in showing the part above that I can't do.)

However I'm struggling to use the fact that both answers were the same (if that's even relevant) to arrive at $\text {Hom}_{\mathbb R[x]}(M,N)=\{0\}$. I've tried fiddling around with the properties of homomorphisms of modules like $\varphi(rm)=r\varphi(m)$ for $r\in\mathbb R[X]$ and $m\in M$. I also don't really understand how we can define homomorphisms like these when $M$ and $N$ are essentially modules over different rings $\mathbb R[A]$ and $\mathbb R[B]$. I know the ring they're over is $\mathbb R[X]$ but I hope you get what I mean.

I couldn't find much about modules over polynomial rings involving a matrix online so I'm hoping someone here can help me. Thanks

EDIT: Corrected wrong entry in $B$

答:
Use the fact that $\varphi$ is a homomorphism and your first part to deduce
$$ \varphi((1,2))-2\varphi(e_1) = \varphi((-1,2)) = \varphi((1-3A+A^2) (1,2)) = (1-3B+B^2) \varphi((1,2)). $$
This implies
$$ 2\varphi(e_1) = (3B-B^2) \varphi((1,2)). $$
We write now $\varphi((1,2)) = (a,b)$ and obtain
$$ 2\varphi(e_1) = (2a,0). $$
Hence,
$$ \varphi(e_1) = (a,0). $$
Then we obtain
$$ \varphi((0,2)) = \varphi((1,2)-e_1) = (a,b) - (a,0) = (0,b). $$
Then we get
$$ (0,0) = B(0,b) = B \varphi(e_2) = \varphi(Ae_2) = \varphi(3e_2), $$
which implies $b=0$. On the other hand we have
$$ (a,0) = B(a,0) = B \varphi(e_1)= \varphi(Ae_1) = \varphi(2e_1) = (2a, 0). $$
Thus, $a=0$ and so we get $\varphi(e_1) = (0,0) = \varphi(e_2)$. Hence, we have $\varphi=0$.

ContentLicense: CC BY-SA 4.0


问:
Convergence of infinite series $\sum (-1)^{n+1}\frac{1}{n!}$
I have this question: Do the series converge absolutely or conditionally?

$$
\sum (-1)^{n+1}\frac{1}{n!}
$$
I would say it does not converge absolutely, since I suggest, by using the ratio test, that
$$
\frac{a_{n+1}}{a_n}
$$
does not approach a limit
$$
L<1
$$

but rather
$$
L=1
$$

However, I can see from the result list that I am wrong, and it does converge absolutely, can someone please explain to me why?

答:
The Ratio Test for absolute convergence should give you

$$\lim_{n \rightarrow \infty} \frac{1/(n+1)!}{1/n!} \ = \ \lim_{n \rightarrow \infty} \frac{n!}{(n+1)!} \ . $$

Be careful in setting up the ratio, since you have a compound fraction here...

EDIT: I might add that, after you've worked with infinite series for a while, you will develop a "feel" for what sorts of series are likely to be absolutely convergent.  We can make the comparison test  $  \frac{1}{n!} < \frac{1}{2^n} \ , $ for $ \ n > 3 \ $ , so the "infinite tail" of $ \ \Sigma_{n=1}^{\infty} \frac{1}{n!} \ $ is smaller than a convergent geometric series.  So it is reasonable to expect that this series should be absolutely convergent and "pass" the Ratio Test.

ContentLicense: CC BY-SA 3.0


问:
What is the result of a 180 degree rotation of a cube around the line joining the midpoints of opposite edges?
This is supposed to be an isomorphism, but I can't visualize it.  Suppose that there is a 180 degree rotation around the line joining the midpoint of the left edge of the top face to the midpoint of the right edge of the bottom face.  Where do the top and bottom faces end up?</p>

答: Top face ends up left, bottom face ends up right. As I convinced myself by picking up the nearest cube lying around (a Rubik's cube), holding it at those midpoints, and doing the rotation. The top and left faces swap places, as do back and front, and bottom and right.</p>


问:
Counting problem assigning student to groups
Say I want to assign $4$ students $a,b,c,d$, to three teams, and each team needs at least one student in it. Only one issue, $a$ and $b$ cannot be in the same team. How many ways are there to assign teams these teams?

The way I went about this problem is lets say we have the set of students $\{a,c,d \}$ and we assign these students to a group using the permutation $P(3,3)$, then we can assign $b$ to a team, $P(1,1)$. Together we get $6 \cdot 1$ and there is $2!$ ways to do this because we do the first set mentioned or take another set $\{b,c,d \}$ and do the same thing we did above.

All together we get $36$ ways of arranging these teams.

Is this a correct way of going about this question?

答:
Select $2$ out of $4$ students $\{a \ b \ c \ d \}$ who will be together in a group. That is $\displaystyle {4 \choose 2} = 6$ ways. One of these selections will have $a$ and $b$ together. Taking that out, we have $5$ acceptable ways to make three groups with $2$ students in one and remaining $2$ automatically in two individual groups.

Now if they are being assigned to distinct teams, we can permute the groups in $3!$ ways.

So the answer is $5 \cdot 3! = 30$ ways.


问:
Proof explanation for a sum of sines
Looking through proofs and there's one part I'm confused about, that if solved will get me straight to the answer.

I need to show that $\sum_{m=0}^{N-1} \sin((m+\frac{1}{2})x) = \frac{\sin(Nx/2)^2}{\sin(x/2)}$

How should I go about this?

答:
This is quite a simple solution based on telescoping – see e.g. the Wikpiedia article.

Let's multiply both sides by the denominator first:
$$\sin(x/2) \sum\limits_{m=0}^{N-1} \sin((m+\tfrac{1}{2})x) = \sin(Nx/2)^2$$

Then we include the leading term into the sum and utilize a well-known trigonometric identity $\sin(a)\sin(b)=\tfrac 12(\cos(a-b)-\cos(a+b))$ to transform products into sums:

$$LHS = \sum\limits_{m=0}^{N-1} \left(\sin(x/2) \sin((m+\tfrac{1}{2})x)\right) = $$
$$\frac 12\,\sum\limits_{m=0}^{N-1} \left(\cos(x/2 - (m+\tfrac{1}{2})x) - \cos(x/2 + (m+\tfrac{1}{2})x)\right) = $$
$$\frac 12\,\sum\limits_{m=0}^{N-1} \left(\cos(-mx) - \cos((m+1)x)\right) =$$
$$\frac 12\,\sum\limits_{m=0}^{N-1} \left(\cos(mx) - \cos((m+1)x)\right)$$

The second term of each sum cancels the first term of the next sum, and the whole sum telescopes so that only the first and last terms remain:

$$= \frac 12\, \left(\cos(0x) - \cos(Nx)\right) = $$

which we transform back into a product by the same identity mentioned above:

$$= \sin(Nx/2)\sin(Nx/2) = \sin^2(Nx/2) = RHS.$$

Q.E.D.

Title: Proof explanation for a sum of sines


问:
Equation for cloud-like curve
It might be a bit naive question, but I'm looking for an equation to describe a curve that looks like a typical cloud clipart icon. This is an example (not sure how to make it smaller):

I think it could be something like epicycloid (particularly, ranunculoid), but I'm not sure how to make it more oblong.

答: To get those images, you could freely choose finitely many connection points forming a convex set and connect these connection points with smooth curves with positive (or negative) curvature.</p>

Title: Equation for cloud-like curve


问:
A binomial inequality with factorial fractions: $\left(1+\frac{1}{n}\right)^n<\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{n!}$
Prove that $$\left(1+\frac{1}{n}\right)^n<\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{n!}$$ for $n>1 , n \in  \mathbb{N}$.</p>

答:
We have by the binomial identity that
\begin{align*}
  \left(1 + \frac 1n \right)^n &= \sum_{k=0}^n \binom nk \frac 1{n^k}\\
     &= \sum_{k=0}^n \frac{n!}{(n-k)! n^k} \cdot \frac 1{k!}\\
     &= \sum_{k=0}^n \frac{n \cdot (n-1) \cdots (n-k+1)}{n \cdot n \cdots n} \cdot \frac 1{k!}\\
     &\text{now the first factor is $<1$ for $k\ge 2$}\\
     &< \sum_{k=0}^n \frac 1{k!}
\end{align*}
for $n \ge 2$.</p>


问:
Avoiding brute force: determining when a specific polynomial in $\mathbb{Q}[x]$ is an integer for any integer $x$
I have to prove that $\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n$ is an integer for any $n$. I solved this by brute-force, exhausting all the possibilities methods. I was wondering if there was a way to solve this at-a-glance with some sort of theory? Because

Although my answer is correct, I was apparently supposed to use some theory to answer this question.

I started by putting everything on common denominator and factoring out $n$:

$$\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n = \frac{n(3n^4+5n^2+7)}{15}$$

then I proceeded to plug in $\pm1 ,\pm2, \pm3 ,..., \pm7$ into $3n^4+5n^2+7 \pmod {3,5,\text{ or }15}$

$$3(\pm 1)^4 +5(\pm 1)^2 +7 \equiv 15 \equiv 0 \pmod{15}$$

$$3(\pm 2)^4 +5(\pm 2)^2 +7 \equiv 75 \equiv 5\cdot 15 \equiv 0 \pmod{15}$$

$$3(\pm 3)^4 +5(\pm 3)^2 +7 \equiv 295 \equiv 0 \pmod{5} \text{ while } n \equiv 0 \pmod 3$$

$$3(\pm 4)^4 +5(\pm 4)^2 +7 \equiv 855 \equiv 57\cdot 15\equiv 0 \pmod{15}$$

$$3(\pm 5)^4 +5(\pm 5)^2 +7 \equiv 2007 \equiv 0 \pmod{3}\text{ while } n \equiv 0 \pmod 5$$

$$3(\pm 6)^4 +5(\pm 6)^2 +7 \equiv 4075 \equiv 0 \pmod{5}\text{ while } n \equiv 0 \pmod 3$$

$$3(\pm 7)^4 +5(\pm 7)^2 +7 \equiv 7455 \equiv 497 \cdot 15\equiv 0 \pmod{15}$$

to conclude that if $n\equiv 0 \pmod {15}$ then $\frac{n}{15}$ is an integer from which $\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n$ is an integer,

and if $n \not\equiv 0 \pmod {15}$ then either $3n^4+5n^2+7 \equiv 0 \pmod{15}$ or $n \equiv 0 \pmod 3$ and $ 3n^4+5n^2+7 \equiv 0 \pmod{5}$ or $n \equiv 0 \pmod 5$ and $ 3n^4+5n^2+7 \equiv 0 \pmod{3}$ from which the statement is clearly true.

Is there a less brute-force-ish way of concluding this? Is there some theory I should be using that would cause me to not be excessively lengthy in calculation if I were given different, larger numbers than $15$?

答:
$$ \begin{align}f(n+1)-f(n)&=\frac{(n+1)^5-n^5}5+\frac{(n+1)^3-n^3}3+\frac7{15}
\\&=\frac{5n^4+10n^3+10n^2+5n+1}5+\frac{3n^2+3n+1}3+\frac7{15}\\&=\text{integer}+\frac15+\frac13+\frac 7{15}\end{align}$$
is an integer for all integers $n$ and so is $f(0)$. The claim follows by induction.

Here's a generalization for you: Suppose $p_1,p_2,\ldots,p_m$  are primes, $a_1,a_2,\ldots,a_m$ are integers and $c$ is real. Let
$$ f(n)=\frac{a_1n^{p_1}}{p_1}+\frac{a_2n^{p_2}}{p_2}+\cdots +\frac{a_mn^{p_m}}{p_m}+cn.$$
If $f(1)$ is an integer, then $f(n)$ is an integer for all integers $n$.


问:
Towards a proof of global convergence for Newton-like methods
Assume $B_k$ are positive definite with a uniformly bounded condition number, i.e., there is an $M$ such that
$$
||B_k||\cdot ||B_k^{-1}|| \leq M \quad \forall k\in\mathbb{N},
$$
$\nabla f(x_k) := \nabla f_k$, and
$p_k := -B_k^{-1} \nabla f_k$ is a descent direction.
The book (Numerical Optimization by Nocedal & Wright, 2nd ed, page 40) says it is easy to show
$$
\cos \theta_k := \frac{-\nabla f_k^T p_k}{||\nabla f_k || \cdot ||p_k||}
\geq \frac{1}{M}.
$$
How can I prove this?

I tried approaches using the Cholesky decomposition $B_k =L L^T$ and using
$||x|| \leq ||B_k^{-1}||\cdot ||B_k x||$, but to no avail. I think it should be a simple one liner that I'm not seeing? Any help is much appreciated!

答:
I'm working really hard on the same problem. The Cholesky decomposition approach you mentioned really inspired me and I found a stronger inequality
$$
\cos(\theta_k) \geq \frac{1}{\sqrt{M}}.
$$
Here is my proof. Let $B_k^{-1} = R_k^T R_k$, then we get
$$
\nabla f_k^T B_k^{-1} \nabla f_k = \nabla f_k^T R_k^T R_k \nabla f_k = ||R_k \nabla f_k||^2. \qquad \qquad \qquad \qquad (1)
$$
Furthermore, since
$$
||\nabla f_k|| \leq ||R_k \nabla f_k||\ ||R_k^{-1}||
$$
and
$$
\ ||B_k^{-1} \nabla f_k|| = ||R_k^T R_k \nabla f_k|| \leq ||R_k^T||\ ||R_k \nabla f_k|| = ||R_k||\ ||R_k \nabla f_k||,
$$
we get
$$
||\nabla f_k ||\ ||B_k^{-1}\nabla f_k|| \leq ||R_k^{-1}|| \ ||R_k|| \ ||R_k \nabla f_k||^2. \qquad \qquad \qquad \qquad (2)
$$

Finally, combining Eqs. (1) and (2), we get
$$
\frac{\nabla f_k^T B_k^{-1} \nabla f_k }{||\nabla f_k ||\ ||B_k^{-1}\nabla f_k||} \geq
\frac{||R_k \nabla f_k||^2}{||R_k^{-1}|| \ ||R_k|| \ ||R_k \nabla f_k||^2} = \frac{1}{|| R_k||\ ||R_k^{-1}||} = \frac{1}{\sqrt{M}}.
$$


问:
Find all $m,n \in \mathbb N$ Such that $3(2^m)+4=n^2$.

Find All $n,m\in \mathbb N$ Such that $3(2^m)+4=n^2$

I’ve tried to plug some values of $m$, and it turns up that the only valid values is $$m\in \{2,5,6\}.$$

So once i saw that i tried to prove that it doesn’t exist a solution to the equation $\forall m\geq7$. But i have no idea how to prove this.

Something i’m not sure whether it’s true:

$$3(2^m)+4=n^2 \iff3(2^m+1)=n^2-1 \\ \iff 3(2^m+1)=(n+1)(n-1)$$

Thus $3\mid (n+1)(n-1) $, But since $3$ is a prime and $\gcd(n+1,n-1)=1$ or $2 \implies$
$3\mid n+1$ or $3\mid n-1$ What’s next?

答:
(Here's a suggested approach. If you're stuck, explain where you're stuck and what you've tried.)

Factor it as $ 3 \times 2^m = (n-2) ( n+2)$.

Hence $ \{ n-2, n+2 \} = \{ 3 \times 2^a, 2^b \}$.

Hence $  3 \times 2^a = 2^b \pm 4 $.

If $ b = 0, 1, 2$, what can we say? What are the solutions with these small cases?

 $3 \times 2^a = 1+4, 2+4, 4+4$ (cannot take negative branch).

 So we get $ b=1, a = 1 \Rightarrow 2 \times 6 = 3\times 2^2  $.

If $ b \geq 3$, what can we say?

 RHS is a multiple of 4 but not 8, so $ a = 2$.

 $3 \times 4 = 2^b \pm 4$, which gives us $b=3, 4$.

 $a = 2, b = 3 \Rightarrow 8\times 12 = 3\times 2^5 .$

 $a = 2, b = 4 \Rightarrow 12 \times 16 = 3\times 2^6 . $

In conclusion, we have solutions only when $ m = 2, 5, 6$.


问:
Degree vs Valence of a vertex in a graph
To my understanding, "degree" and "valence" mean the same thing: the number of edges incident to a vertex (including multiplicity for loops).

Is there a difference in the contexts in which they are used?

答:
Both terms are equivalent. However, the term "degree" is used more often these days. In old publications, such as the ones of Erdős (c.f. "On a valence problem in extremal graph theory", Erdős and Simonovits, 1972), the term valence is ocurring more frequently.

In today's graph theoretical literature, however, it has almost disappeared (standard works such as Bondy's and Murty's "Graph Theory", already no longer mention it, and it is not appearing in their older work called "Graph Theory with applications" (1972), either). Diestel's "Graph Theory" uses both terms equivalently ("The degree (or valency) [...] of a vertex $v$ is the number of edges at $v$", p.14). However, he mostly uses the term "degree".

In chemical graph theory, one often tries to strictly separate the terms in order to make a clear distinction between the valence of chemical bonds and an abstract graph theoretic model (see for example "A review on molecular topology: applying graph theory to drug discovery and design" by Amigó et. al.).

Whether there are other applications in which the term "valence" takes on a specifically different meaning than the term "degree", I can't say for sure. In pure graph theory it does not.


答:
hint

It is easy to check that $ m $ satisfies
$$m\ge 2.$$

but

$$3.2^m=(n+2)(n-2)$$
with $ n $ even.

So
$$(n+2)|3.2^m $$
and by Gauss Theorem
$$(n+2)|2^m$$

Title: Find all $m,n \in \mathbb N$ Such that $3(2^m)+4=n^2$.

Body:
Find All $n,m\in \mathbb N$ Such that $3(2^m)+4=n^2$

I’ve tried to plug some values of $m$, and it turns up that the only valid values is $$m\in \{2,5,6\}.$$

So once i saw that i tried to prove that it doesn’t exist a solution to the equation $\forall m\geq7$. But i have no idea how to prove this.

Something i’m not sure whether it’s true:

$$3(2^m)+4=n^2 \iff3(2^m+1)=n^2-1 \\ \iff 3(2^m+1)=(n+1)(n-1)$$

Thus $3\mid (n+1)(n-1) $, But since $3$ is a prime and $\gcd(n+1,n-1)=1$ or $2 \implies$
$3\mid n+1$ or $3\mid n-1$ What’s next?


答:
The sequence that you are looking at is similar to the Fibonacci-Narayana sequence, which is given by $f_n=f_{n-1}+f_{n-1-m}, m\ge 1$. The generating function is given by

$$
\frac{1}{1-x-x^{n+1}}
$$

and the limiting ratio is given by

$$\lim_{n\to\infty}\frac{f_{n+1}}{f_{n}}=\chi\\
\chi-1=\chi^{-n}
$$

This sequence can be found in OEIS A000930 and related. $n=1, 2, 4$ are the golden ratio, supergolden ratio, and the plastic number, respectively.

Additional properties of this sequence are

$$
\sum_{n=0}^{\infty}\frac{1}{\chi^n}=\chi^{n+1}\\
S_n=\sum_{j=1}^{n}f_j \quad \text{(cumulative sum)}\\
\lim_{n\to\infty}\frac{S_{n}}{S_{n+1}}=\chi
$$

You might also be interested in the Fibonacci-Padovan series, $f_n=f_{n-m+1}+f_{n-m}$ with a limiting value given by $X+1=X^m$.

Title: Is there a general formula for the ratio limit of a(n) = a(n-1) + a(n-m).?

Body:
I am exploring the properties of a generalization of Fibonacci numbers and have been encountered an issue.

Specifically I am trying to find the limit of the ratio of the the sequence:

```<code>A(n) = A(n-1) + A(n-m)
A(n) = n+1 for 0 <= k < m
</code>```

when m = 2 this is the Fibonacci sequence.

It's clear that the solution is the real root of the polynomial:

x^m - x^(m-1) - 1 = 0

For n=1 the ratio is 2, n=2 is the golden ratio and n=3 is the supergolden ratio. But beyond that I haven't made much progress. I have written a script to generate the sequences and approximate the ratio which has given the following for the for 1 <= m <= 10:

```<code>1: 2.0: [1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536, 131072, 262144, 524288]
2: 1.618033988749895: [1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946]
3: 1.465571231876768: [1, 2, 3, 4, 6, 9, 13, 19, 28, 41, 60, 88, 129, 189, 277, 406, 595, 872, 1278, 1873]
4: 1.3802775690976141: [1, 2, 3, 4, 5, 7, 10, 14, 19, 26, 36, 50, 69, 95, 131, 181, 250, 345, 476, 657]
5: 1.324717957244746: [1, 2, 3, 4, 5, 6, 8, 11, 15, 20, 26, 34, 45, 60, 80, 106, 140, 185, 245, 325]
6: 1.2851990332453493: [1, 2, 3, 4, 5, 6, 7, 9, 12, 16, 21, 27, 34, 43, 55, 71, 92, 119, 153, 196]
7: 1.2554228710768465: [1, 2, 3, 4, 5, 6, 7, 8, 10, 13, 17, 22, 28, 35, 43, 53, 66, 83, 105, 133]
8: 1.2320546314285723: [1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 14, 18, 23, 29, 36, 44, 53, 64, 78, 96]
9: 1.21314972305964: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 15, 19, 24, 30, 37, 45, 54, 64, 76]
</code>```

Is there a known solution for this? Or a proof there isn't one?

Thank you


答:
You do indeed have absolute convergence, and by the ratio test:

$$\lim_{n\to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to \infty} \frac{\frac{1}{(n+1)!}}{\frac 1{n!}} = \lim_{n\to \infty} \frac{n!}{(n+1)!} = \lim_{n\to \infty} \frac 1{n+1} = 0 < 1$$

Added: recall that the factorial is a product of factors, and so $$\frac{n!}{(n+1)!} = \frac{(n!)}{(n+1)(n!)} = \frac 1{n+1}$$
The factors given by $n! = n(n-1)(n-2)\cdots (2) \cdot (1)$ are common in both the numerator and denominator, and hence cancel, leaving only the factor of $1$ in the numerator, and $(n+1)$ in the denominator.

Title: Convergence of infinite series $\sum (-1)^{n+1}\frac{1}{n!}$

Body:
I have this question: Do the series converge absolutely or conditionally?

$$
\sum (-1)^{n+1}\frac{1}{n!}
$$
I would say it does not converge absolutely, since I suggest, by using the ratio test, that
$$
\frac{a_{n+1}}{a_n}
$$
does not approach a limit
$$
L<1
$$

but rather
$$
L=1
$$

However, I can see from the result list that I am wrong, and it does converge absolutely, can someone please explain to me why?


问:
Lower Bounds on Markov's Inequality

Problem 20.3
If $R$ is a nonnegative random variable, then Markov’s Theorem gives an upper
bound on $Pr[R \geq x]$ for any real number $x \gt Ex[R]$. If b is a lower bound on R, then Markov’s Theorem can also be applied to $R - b$ to obtain a possibly different bound on $Pr[R \geq x]$.

a)  Show that if $b > 0$, applying Markov’s Theorem to $R - b$ gives a smaller
upper bound on $Pr[R \geq x]$ than simply applying Markov’s Theorem directly to R.

b) What value of $b > 0$ in part (a) gives the best bound?

This question is from Mathematics for Computer Science text book (MIT text for discrete math course). I would like to receive some help with this problem and if possible to send any resources to check if my solution is right or not. I am using the textbook but the problem is that I don't have a way to verify my solutions.

答:
For part a:

$$\mathbb P (R - b \geq t) \leq \frac{\mathbb E R - b}{t} \leq \frac{\mathbb E R}{t}$$

where the second inequality follows when $b \geq 0$

For part b:

The RHS of the inequality given is linear in $b$ and thus is maximized at the constraint: $ X-b \geq 0$ with probability $1$. So the best value of $b$ is the largest lower bound of $X$ also known at the essential infimum of $X$.

Title: Lower Bounds on Markov's Inequality


问:
Looking for an other method to compare $100!$ and $50^{100}$
There is a question asked me to compare $100! \ and \ 50^{100}$. I think more than 2 hours to solve it.
I show my work below, but I looking for other Ideas to prove the inequality.  Thanks in advance for any hints or new Ideas.

my work :
$$50 \times 50 >  1 \times 99 \times 2\\
50 \times 50 > \ 2 \times 98\\
50 \times 50 > \ 3 \times 97\\
50 \times 50 > \ 4 \times 96\\ \vdots \\
50 \times 50 > \ 48 \times 52=(50+2)(50-2)=50^2-4\\
50 \times 50 > \ 49 \times 51=(50+1)(50-1)=50^2-1\\
50 \times 50 \geq 50 \times 50 $$ then multiply them
$$50^{49}\times 50^{49}\times 50^2 >1.2.3...50....99.(2.50)\\50^{100}>100!$$

Remark: I use numerical approximation for $50^{100}\sim 7.8886\times 10^{169} $ and $100!=9.33236\times 10^{157}$ But is not interesting( Ithink).

答:
Well they both have $100$ factor components.

The LHS the factor components run from $1$ to $100$.  The RHS the factors are a constant coomponent of $50$.

$50$ is almost the arithmetic average of $1$ to $100$.  By the A.M. G.M inequality that is greater than the geometric mean of $1$ to $100$ which is  $\sqrt[100]{1*2*3*....*100}$.

And so we ought to have $\sqrt[100]{100!} < 50$ and the $100! < 50^{100}$.

.... but the arithmetic average is actually $50.5$ so we have $\sqrt[100]{100!} < 50.1$.

But we can probably tweak it not to make such a difference.

.....

And indeed you did tweak it!

$2\times 4 \times ....... \times 98 =98!$ and $\sqrt[97]{98!}$ is the geometry mean of $2,....,98$.  And the arithmetic mean of $2,...., 98$ is $50$.

So we have $\sqrt[97]{98!} < 50$ and so $98! < 50^{97}$

Now the tweak: $99\times 100 < 100^2 = 4\times 50^2 < 50^3$ so $100! = 98!\cdot (99\times 100) < 50^{97} \times 50^3= 50^{100}$

The tweak makes sense to look at the edge the difference between geometric mean and arithmetic mean is greatest when the numbers are far apart.  In this case we managed to be a product of three terms which wasn't just less than the cube of the AM.  It was barely significantly more than the square of the AM.


答:
Another way is to first use exclusion-inclusion principle to get the number of classes with all groups being non empty. This is
$$
3^4-3 \cdot 2^4 + 3 =36
$$
Now you need to exclude cases where and b are together. There are $2+2+2=6$ such cases. Hence the solution is $30$.

Title: Counting problem assigning student to groups

Body:
Say I want to assign $4$ students $a,b,c,d$, to three teams, and each team needs at least one student in it. Only one issue, $a$ and $b$ cannot be in the same team. How many ways are there to assign teams these teams?

The way I went about this problem is lets say we have the set of students $\{a,c,d \}$ and we assign these students to a group using the permutation $P(3,3)$, then we can assign $b$ to a team, $P(1,1)$. Together we get $6 \cdot 1$ and there is $2!$ ways to do this because we do the first set mentioned or take another set $\{b,c,d \}$ and do the same thing we did above.

All together we get $36$ ways of arranging these teams.

Is this a correct way of going about this question?


答:
In respect of the stated comment, I add our second case (and so the whole solution) I left for OP to solve.

First, check $m=1$.

If $m≥2, m-2=k$ and $n=2n_1$, then

$$3×2^{k}+1=n_1^2, n_1≥2$$

Case $-1.$

$n_1=3a+1$, where $a≥1.$

$$3×2^k=9a^2+6a$$

$$2^k=3a^2+2a$$

$$a≥1 \Longrightarrow \begin{cases} a=2b \\ k≥3 \end{cases}$$

$$2^k=12b^2+4b$$

$$2^{k-2}=3b^2+b=b(3b+1)$$

$$b=2^x, 3b+1=2^y$$

If $x≥1$, then we get $3b+1≥7$ and $3b+1$ is an odd number, which gives a contradiction.

So, we deduce that $x=0.$

Then, we can check $n_1=3a+2$ by the same way.

Of course, $n_1=3a$ is impossible.

The method of looking at our second case is the same as the first one, as I mentioned earlier.

Case $-2.$

$n_1=3a+2$, where $a≥0.$

First, check $a=0$. If, $a≥1$ then

$$3×2^k+1=9a^2+12a+4$$

$$2^k=3a^2+4a+1, a≥1$$

$$a≥1 \Longrightarrow \begin{cases} a=2b-1 \\ k≥3 \end{cases}$$

We have,

$$2^k=3(2b-1)^2+4(2b-1)+1, b≥1$$

$$2^k=12b^2-4b$$

$$2^{k-2}=3b^2-b=b(3b-1)$$

$$b=2^x, 3b-1=2^y$$

If $x≥1$, then we get $3b-1≥5$ and $3b-1$ is an odd number, which gives a contradiction.

So, we deduce that $x=0.$

Finally, backward all the steps and you will find all required integer values of $m$ and $n.$


问:
Coordinate geometry: calculating the height of an equilateral triangle
If I have equilateral $\Delta ABC$ with A being $(-x,0)$ and B being $(x,0)$, how can I solve for the coordinates of C in terms of $x$?

I tried the following:

$2x^2 = x^2 + b^2 $ -- pythagorean thm, since we know that one side of the triangle is two times the length of half of the base.

$3x^2 = b^2$ -- simplification

$ x\sqrt3 = b $ -- so this says that the height of point C must be x\sqrt3 units.

This gives me an end result with the coordinates $(0, 2)$. However, frankly, this doesn't seem logical to me...what I'm basically saying with that result is that all equilateral triangles are two units tall...what?! This sounds completely incorrect.

Any help would be great; thanks.

edited: I actually had the right idea, I just made a stupid mistake with my simplification. ._.

答:
Since the triangle is equilateral, the $x$-coordinate of $C$ must be $x=0$, so that it falls in the middle of the $x$-coordinates of $A$ and $B$.

The base has length $|AB|=2x$ and so all of the sides have length $2x$. The point $C$ will lie on the circle, centre $A$ and radius $2x$. The whole circle is parametrized by $(-x,0) + 2x(\cos\theta,\sin\theta)$, where $\theta = 0$ corresponds to the base. Since we have an equilateral triangle, we have $\theta = 60^{\circ}$:

\begin{array}{ccc}
(-x,0)+2x(\cos 60, \sin 60) &=& (-x,0) + 2x(\tfrac{1}{2},\tfrac{\sqrt{3}}{2}) \\ \\
&=& (0,x\sqrt{3})
\end{array}

If you want an "inverted" triangle then let $C=(0,-x\sqrt{3})$.


问:
Use compactness to show that $f(x_0)=\inf_{x\in K}f(x)$
Let $f: \mathbb{R} \to \mathbb{R}$ be a function bounded below and lower semi-continuous. Show that for every compact $K \subset \mathbb{R}$, there exists $x_0 \in K$ such that
$$
f(x_0)=\inf_{x\in K}f(x)
$$

I've read a proof for a similar theorem here, but in that proof $f$ is a continuous coercive function. That makes the proof pretty straightforward.

答:
A characterization of lower semi continuity is that for any $a \in \mathbb{R}$, the set $f^{-1}((-\infty, a])$ is closed in $\mathbb{R}$ (I don't know what definition of lower semi continuity you're using becasue there are a few, but showing that the definition is equivalent to this is a good exercise, show it!)

Let $m = \inf_{x \in K} f(x)$ (finite by assumption). Let $A_{n} = \{x \in K: f(x) \leq m + \frac{1}{n}$}. By the characterization of lower semi continuity above, each $A_{n}$ is closed. Any closed subset of a compact set is compact, so each $A_{n}$ is compact. Also, clearly $A_{n} \subset A_{n+1}$ for each $n \in \mathbb{N}$ -- by definition of $A_n$. Thus,
$$
A := \bigcap_{n \in \mathbb{N}} A_{n} \neq \emptyset
$$
since nested sequences of compact sets in $\mathbb{R}$ (or any complete metric space, actually) have a non-empty intersection. Pick any $x_{0} \in A$. Check that $x_{0}$ is a minimizer like you want.


问:
Vakil 14.2 isomorphism of the pair (invertible sheaf, rational section)
In Vakil's chapter 14 of "The Rising Sea", he talks about a map from invertible sheaves to divisors. Specifically, for a Noetherian irreducible (can be relaxed apparently) scheme regular in codimension one, it makes sense to talk about the divisor of a rational section $s$ of an invertible sheaf $\mathcal{L}$. He then has a map:
$$\{(\mathcal{L}, s)\}/ \text{isomorphism} \rightarrow \operatorname{Weil} X$$

I am not sure what this isomorphism of a pair $(\mathcal{L}, s)$ is supposed to mean. One way to interpret it is that two such pairs are isomorphic if there is sheaf isomorphism which takes one section to another.  But I don't think that's all there is to it, because we also have the case when we can extend a rational section to a bigger open set and in that case it is still supposed to be from the same isomorphism class. So what is it?

Another way to think of it was along the lines of rational functions.  And I haven't studied the non-irreducible case but in case of irreducible scheme rational functions have a natural notion of "isomorphism" in the sense that they are same if they map to the same element in the function field.  Is there a similar notion for invertible sheaves?  Do we similarly have an injective map from regular sections on open sets to the stalk of the invertible sheaf at the generic point?

Edit: In the last paragraph, I mean function field of an integral scheme, not just irreducible.

答:
You're close - the correct notion of isomorphism of a pair $(\mathcal{L},s)$ here is that there's an isomorphism $\varphi:\mathcal{L}_1\to\mathcal{L}_2$ under which $\varphi(s_1)$ and $s_2$ agree on a dense open subset (sometimes we say that "they're equal as rational sections"), which is equivalent to agreeing at the generic point.

For your final paragraph, you're not too far away either. Suppose $\eta$ is the generic point of $X$ and $\mathcal{L}$ is an invertible sheaf. Then $\mathcal{L}_\eta \cong \mathcal{O}_{X,\eta}$: since stalks can be calculated from any open neighborhood, simply pick an open set where $\mathcal{L}\cong\mathcal{O}_X$ as it's locally free. On the other hand, it's not true that $\mathcal{O}_X(U)\to\mathcal{O}_{X,\eta}$ is an injection without the condition that $X$ is reduced - think about adding some nilpotent fuzz on a closed subset. For instance, $\operatorname{Spec} k[x,y]/(x^2,xy)$ is $\operatorname{Spec} k[y]$ with fuzz at the origin, and indeed $x\neq 0$ is in the kernel of the map from the global sections to the stalk at the generic point.


问:
How to show vector field is conservative?
Conditions for Conservative Vector Field?

Is it sufficient for a vector field to be conservative, to say that it’s image is simply connected (no holes) and the partials are equivalent?? If not then what can you do if you have a situation where you can’t test every possible loop or paths? Thanks!

答: One way is to find a function $f:\mathbb{R}^n\rightarrow \mathbb{R}$ whose gradient is that vector field. A vector field is conservative if and only if it is a gradient.


问:
Decreasing function
Given two increasing function $f(x)$ and $g(x)$ with all $x \ge 0$. Moreover, $f(x) > g(x)
$ for all $x \ge 0$ and $$\mathop {\lim }\limits_{x \to \infty } \left( {f(x) - g(x)} \right) = 0.$$
Is $d(x) = f(x)-g(x)$ a decreasing function ?</p>

答:
It is not even eventually decreasing. Consider
$$
  f(x)
= e^x \ \
  \mbox{ and } \ \
  g(x)
= \begin{cases}
  x                    & \mbox{ if } x \leq 1 \\
  e^x - \frac{1}{x}    & \mbox{ if } x \in (2n+1,2n+2], n \in \mathbb{N} \\
  e^x - \frac{1}{2n+2} & \mbox{ if } x \in [2n, 2n+1], n \in \mathbb{N}; \\
  \end{cases}
$$
they satisfy your conditions but on intervals of the form $[2n,2n+1]$ $f(x) - g(x)$ is constant and therefore not decreasing. There are infinitely many such intervals. </p>

Title: Decreasing function


问:
$L^p$ is reflexive for $1<p<2$, Brezis proof
Theorem 4.10 [pages 95-97] of Brezis proves that $L^p$ is reflexive for $p\in(1,\infty)$.

The last statement of the proof says that since $T(L^p)$ is reflexive, with $T$ being an isometry, so is $L^p$. Why is it true?

答:
In general, if $E, \ F$ are isomorphic normed vector spaces, then
$F$ is reflexive $\Leftrightarrow$ $E$ is reflexive. Since $T$ is an isomorphism (onto $T(L^p)$), it follows that $L^p$ is reflexive.


问:
Ratio of area of triangle and hexagon
I am looking for the proof of the following claim:

Claim. If the sides of the triangle are  partitioned into $n$ equal segments for $n$ an even integer and each division point adjacent to the corresponding central division point  is connected to the opposite vertex, a central hexagon which has an area of $\frac{32}{(3n-2)(3n+2)}$ of the area of the original triangle  is still formed.

GeoGebra applet that demonstrates this claim can be found here.

I tried to adapt the proof of the Marion's theorem given on this page. The problem I encountered is the following: How to find a trilinear coordinates of the division points : $D,E,F,G,H,I$ . The only formula I am aware of is the formula for the first n-multisection.

答:
I will provide a picture and go into detail, but need my notations first - as in the following picture:

Let $\Delta ABC$ be the given triangle. Let $A'$, $B'$, $C'$ be the mid points of the sides respectively opposed to $A$, $B$, $C$. We denote by $A_-$ and $A_+$ the two division points ($\ne A'$) to the left and to the right of $A'$ on the side $BC$, then use similar notations for the other sides,
so that we have
$$
\begin{aligned}
A_-A'&=A'A_+=xBC=xa\ ,\\
B_-B'&=B'B_+=xCA=xb\ ,\\
C_-C'&=C'C_+=xAB=xc\ ,
\end{aligned}
$$
where $x$ is one and the same proportion. We can let it be general,
in the OP it is $\displaystyle \color{blue}{\frac1n}$.

Let $A^*$ be the intersection of the cevians $BB_+$, $CC_-$, and the median $AA'$.

Let $A^!$ be the intersection of the cevians $BB_-$, $CC_+$, and the median $AA'$.

Similarly we construct $B^*$, $B^!$; $C^*$, $C^!$.

The OP is immediately solved if we answer the following questions:

Which is the position of $A^*$ on the segment $GA$?

Which is the position of $A^!$ on the segment $GA'$?

(Try to get it on your own.) Applying Menelaus in $\Delta AGB'$ w.r.t. the line $BA^*B+$, and w.r.t. the line $BA^!B_-$ we have:
$$
\begin{aligned}
1 &=
\frac {A^*G}{A^*A}
\cdot
\frac {B_+A}{B_+B'}
\cdot
\frac {BB'}{BG}
=
\frac {A^*G}{A^*A}
\cdot
\frac {\frac 12-x}{x}
\cdot
\frac 32\ ,
&\text{ so }\qquad
\frac {GA^*}{GA}
&=
\frac {4x}{3-2x}\ ,
\\
1 &=
\frac {A^!G}{A^!A}
\cdot
\frac {B_-A}{B_-B'}
\cdot
\frac {BB'}{BG}
=
\frac {A^!G}{A^!A}
\cdot
\frac {\frac 12+x}{x}
\cdot
\frac 32\ ,
&\text{ so }\qquad
\frac {GA^!}{GA}
&=
\frac {4x}{3+2x}\ ,
\\
&
&\text{ so }\qquad
\frac {GA^!}{GA'}
&=
\frac {8x}{3+2x}\ .
\end{aligned}
$$
Similar computations can be done for $B^*$, $B^!$, and $C^*$, $C^!$.
This gives the proportions of areas:
$$
\frac{[\color{magenta}G\color{blue}{A^*B^!}]}{[\color{magenta}{G}AB']}
=
\frac {\color{magenta}{G}\color{blue}{A^*}}{\color{magenta}{G}A}\cdot
\frac {\color{magenta}{G}\color{blue}{B^!}}{\color{magenta}{G}B'}
=
\frac {4x}{3-2x}\cdot
\frac {8x}{3+2x}
=
\color{blue}{
\frac{32}{(3n-2)(3n+2)}}
\ .
$$
This is also the final answer since the medians cut the triangle $\Delta ABC$ in six smaller triangles of same area. This matches the claimed formula.

$\square$

Here is also explicitly the situation $n=4$:

We have placed

on $GA$, $GB$, $GC$ points to illustrate $GA^*:GA=2:5$ (and similar proportions for the other two segments),
and

on $GA'$, $GB'$, $GC'$ points to illustrate
$GA^!:GA'=4:7$ (and similar proportions for the other two segments).

Title: Ratio of area of triangle and hexagon


问:
Which is larger? $20!$ or $2^{40}$?
Someone asked me this question, and it bothers the hell out of me that I can't prove either way.

I've sort of come to the conclusion that 20! must be larger, because it has 36 prime factors, some of which are significantly larger than 2, whereas $2^{40}$ has only factors of 2.

Is there a way for me to formulate a proper, definitive answer from this?

Thanks in advance for any tips. I'm not really a huge proof-monster.

Title: Which is larger? $20!$ or $2^{40}$?

Body:
$\log _{10}\left( {2^{40} } \right)=40 \log _{10}\left( {2} \right)= 12,04$ so the numbers of digits of $2^{40}$=13

$\log _{10}\left( {20! } \right)=\sum _{1}^{20} \log_{10} \left({x})\right)=18,38$ so the numbers of digits of 20! Is $19$

So $
20!>2^{40} $


问:
What would be the logical equivalence of not P or T?
I'm finishing my prove but is the logical equivalence of not P OR T equal to True?
This is what I'm trying to prove $$(p \land q) \implies q $$

答: Hint:$(p \land q) \implies q$ is same with $\neg p \lor \neg q \lor q$


问:
Differential Equations - How to solve a logistic growth model with the added mean
For the general logistic growth model that can be applied to biology and economics

$\frac{dP}{dt} = gP(1-\frac{P}{K})$

I know to separate the variables and divide by $P(1-\frac{P}{K})$ resulting in $ln(p) - ln(1-\frac{p}{k}) = gt + A$. Then just rearranging to find p.

That I understand fine, however when another term is added, uP or anything involving P, I do not know how to separate the variables.

For example how do I separate
$\frac{dP}{dt} = gP(1-\frac{P}{K}) + uP$

I cannot seem to find a way to get it to result in all Ps on one side  without horrendous mess and trying to integrate

$\frac{1}{(gP(1-\frac{P}{K}) +uP)}$

答:
Both DE are Bernoulli's differential equations:
$$\frac{dP}{dt} = gP(1-\frac{P}{K}) + uP$$
$$P'-P(g+u) = -g\frac{P^2}{K}$$
By integrating factor method we get:
$$(Pe^{-(g+u)t})' = -g\frac{P^2}{K}e^{-(g+u)t}$$
$$\int \dfrac {d(Pe^{-(g+u)t})}{P^2e^{-2(g+u)t}} = - \frac{g}{K}\int e^{(g+u)t}dt$$
It's easy to integrate now.
$$ \dfrac 1{Pe^{-(g+u)t}} =  \frac{ge^{(g+u)t}}{K(g+u)} +C$$
$$ \dfrac 1{P} =  \frac{g}{K(g+u)} +Ce^{-(g+u)t}$$

With fractions decomposition method:
$$\frac{dP}{dt} = gP(1-\frac{P}{K}) + uP$$
$$\frac{dP}{dt} = gP(1+\dfrac ug-\frac{P}{K}) $$
$$\frac{dP}{dt} = gP(B-\frac{P}{K}) $$
$$\frac{dP}{dt} = \dfrac g KP(KB-P) $$
$$\frac{dP}{dt} = \dfrac g KP(A-P) $$
$$\int \frac{dP}{P(A-P)} = \int \dfrac g K dt$$
Where $A=BK=K(1+\dfrac ug)$.
$$\dfrac 1 A \left(\int \frac{dP}P-\int \dfrac {dP}{(P-A)}\right) = \int \dfrac g Kdt$$
Now integrate.


问:
Why is this a smooth manifold?
I read that the space of quantum states, i.e. the space of density operators

$\mathcal{S}_n = \{ \rho \in H_n : \rho \geq 0, \, \, Tr[\rho]=1\}$

is a smooth manifold of dimension $n^2-1$, without further explanation. Could anyone help me understand why this is the case?

Here $H_n$ denotes the space of hermitian matrices: $H_n = \{ A \in \mathbb{C}^{n \times n} : \, A^* =A  \}$, and $\rho \geq 0$ means that the matrix $\rho$ is positive semi-definite.

答: This is false, since the set has a boundary, but is not a even a manifold with boundary, since the boundary (the set of singular matrices with trace equal to one) is very singular. However, the set of positive definite hermitian matrices with trace equal to one is, indeed, an open manifold.

Title: Why is this a smooth manifold?


问:
Does this discrete curve have a name?
Generated using the following logic.

Begin at a center starting point and draw two one-unit line segments from the center point in opposite directions.

Draw two one-unit line segments that begin at the end of each of the previous step's endpoint, perpendicular to the previous segment and opposite one another. Continue doing this while a given path does not terminate.

A path terminates if both new segments would touch another path from this or a previous step.

Here is a visual of the first few generations, center point and terminal points marked.

答:
I believe this is equivalent to the toothpick sequence, OEIS A139250.

There are many comments, connections, and conjectures listed there.

This image shows the construction of the toothpick sequence:

Each "toothpick" is the same as two of your line segments, and is added with its midpoint at the end of a previous toothpick. If you delete the halves of toothpicks that run into previously or simultaneously placed toothpicks, it should look the same as your picture.

The OEIS also has code to generate animations such as this one:

Title: Does this discrete curve have a name?


答:
A vector field $\mathbf F\in \mathcal C^1$ is said to be conservative if exists a scalar field
$\varphi$ such that:

$$\mathbf{F}=\boldsymbol{\nabla}\varphi$$
$\varphi$ it is called a scalar potential for the field $\mathbf{F}$.

In general, a vector field does not always admit a scalar potential. A necessary condition for a field to be conservative is that the equalities are satisfied:

$${\frac {\partial F_{x}}{\partial y}(x,y,z)=\frac {\partial F_{y}}{\partial x}}(x,y,z),$$$$ {{\frac {\partial F_{y}}{\partial z}}(x, y,z)={\frac {\partial F_{z}}{\partial y}}(x,y,z),\, {\frac {\partial F_{x}}{\partial y}}(x, y,z)=\frac {\partial F_{y}}{\partial x}}(x,y,z)$$
or

$$\boldsymbol \nabla \times \mathbf F = \mathbf 0$$

Title: How to show vector field is conservative?

Body:
Conditions for Conservative Vector Field?

Is it sufficient for a vector field to be conservative, to say that it’s image is simply connected (no holes) and the partials are equivalent?? If not then what can you do if you have a situation where you can’t test every possible loop or paths? Thanks!


问:
Find dominant eigenvector

Consider the $5x5$ tridiagonal matrix with entries $a_{i,i} = 2$,
$i=1,2,...,5$ and $a_{i,i+1}= a_{i+1, i} = 1$ , $i=1,...,4$. Knowing
that the initial approximation $x_0  =(0.5,0.8,1,0.8,0.5)$  is almost
aligned with the dominant eigenvector of the matrix, determine until
the fifth iteration an approximation of the dominant eigenvalue.

So, the first thing I did was to construct the matrix:
$A = \begin{bmatrix}
2 & 1 & 0 & 0 & 0\\
1 & 2 & 1 & 0 & 0\\
0 & 1 & 2 & 1 & 0\\
0 & 0 & 1 & 2 & 1\\
0 & 0 & 0 & 1 & 2\\ \end{bmatrix}$

Then I calculated $A^5x_0 = \begin{bmatrix} 348.2\\603.1\\696.4\\603.1\\348.2\end{bmatrix} = \lambda^5x_0$. But from now on I'm stuck. How can I calculate the dominant eigenvector from here?

答: The question only asks you to compute the dominant eigenvalue, which is approximated by the fifth root of 696.4.

Title: Find dominant eigenvector


问:
Hyperbola parametrics
I have the following problem.

A particle moves on the top part of hyperbola  $$y^2 - x^2 = 9$$ in such a way that the horizontal component of the velocity vector is $3$.

Determine vectors velocity and acceleration when $t = 2$ and the particle is at point $P(4,5)$.

First try:

$$\vec{r} (t)= (3 \tan(t), 3 \sec(t))$$

$$\vec{v} (t) = (3 \sec^2(t), 3 \sec(t) \tan(t))$$

$$\vec{a} (t) = (6 \sin(t) / \cos^3(t), 3 (\sin(t) + 1) / \cos^3(t))$$

Second try:

$$\vec{r} (t) = ( t , \sqrt{9+t^2})$$

$$\vec{v} (t) = ( 1 , t / \sqrt{9+t^2} )$$

$$\vec{a} (t) = ( 0 ,  9 / (9+t^2)\sqrt{9+t^2} )$$

Is there something wrong with what I am doing? The problem says the horizontal component of the velocity vector is $3$, but I get different results, and I am not sure what to do with $P(4,5)$.

Any tips on how to solve that problem?

Please excuse my english. Thanks.

答:
As horizontal component of velocity is $3$, $x = 3t + c$

When $t = 2$, particle is at $P(4,5) $ i.e $x = 4$

So, $4 = 3 \cdot 2 + c \implies c = -2$

Now $y^2-x^2 = 9 \implies y^2 = 9 + (3t-2)^2 = 9t^2 - 12t + 13$

As $y \geq 0, r(t) = (3t-2, \sqrt{9t^2-12t+13})$

Now you can differentiate and find $v(t)$ and $a(t)$ vectors at $t=2$.

Title: Hyperbola parametrics


问:
Simplifying the sum of a product of multinomial coefficients
From the multinomial theorem the following holds

$$
  \sum_{k_1 + k_2 + \ldots + k_m=n} {n \choose k_1, k_2, \ldots, k_m} = m^n
$$

I have the following sum
$$
 \sum_{\substack{k_1 + k_2 + \ldots + k_m &=& B \\ g_1 + g_2 + \ldots + g_m &=& W}} {B \choose k_1, k_2, \ldots k_m} \cdot {W \choose g_1, g_2, \ldots, g_m}
$$
where $B + W = n$, and $\forall_i: g_i = x - k_i$, where $x = \frac{n}{m}$, and $m\mid n$ and $k_i \geq 1$

Can this sum be simplified like the original one?

答:
We assume positive integers $m,n\geq 2$. Since $m\mid n$ we know that $n=Nm$ is a multiple of $m$ with $N\geq 1$. This way we can write $W=Nm-B$ and $x=N$. We show by induction of $m$ that for all $N\geq 1, m\leq B\leq Nm$ the following is valid:

\begin{align*}
\color{blue}{\sum_{\substack{k_1+\cdots+k_m=B\\k_1,\ldots,k_m\geq 0}}
\binom{B}{k_1,\ldots,k_m}\binom{mN-B}{N-k_1,\ldots,N-k_m}=\prod_{q=2}^m\binom{qN}{N}}\tag{1}
\end{align*}

Base step: $m=2$

\begin{align*}
\color{blue}{\sum_{{k_1+k_2=B}\atop{k_1,k_2\geq 0}}}
&\color{blue}{\binom{B}{k_1,k_2}\binom{2N-B}{N-k_1,N-k_2}}\\
&=\sum_{k_1=0}^B\binom{B}{k_1}\binom{2N-B}{N-k_1}\tag{2}\\
&=\sum_{k_1=0}^B\binom{B}{k_1}[z^{N-k_1}](1+z)^{2N-B}\tag{3}\\
&=[z^N](1+z)^{2N-B}\sum_{k_1=0}^B\binom{B}{k_1}z^{k_1}\tag{4}\\
&=[z^N](1+z)^{2N-B}(1+z)^{B}\\
&=[z^N](1+z)^{2N}\\
&\,\,\color{blue}{=\binom{2N}{N}}\tag{5}
\end{align*}
in accordance with (1).

Comment:

In (2) we write the multinomial as binomial coefficients and substitute $k_2 = B-k_1$.

In (3) we use the coefficient of operator $[z^N]$ to denote the coefficient of $z^N$ of a series.

In (4) we apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.

In (5) we select the coefficient of $z^{N}$.

Induction hypothesis: $m=M$

We assume the validity of
\begin{align*}
\color{blue}{\sum_{{k_1+\cdots+k_M=B}\atop{k_1,\ldots,k_M\geq 0}}
\binom{B}{k_1,\ldots,k_M}\binom{MN-B}{N-k_1,\ldots,N-k_M}=\prod_{q=2}^M\binom{qN}{N}}\tag{6}
\end{align*}

Induction step: $m=M+1$

We have to show
\begin{align*}
\color{blue}{\sum_{\substack{k_1+\cdots+k_{M+1}=B\\k_1,\ldots,k_{M+1}\geq 0}}
\binom{B}{k_1,\ldots,k_{M+1}}\binom{(M+1)N-B}{N-k_1,\ldots,N-k_{M+1}}=\prod_{q=2}^{M+1}\binom{qN}{N}}\tag{7}
\end{align*}

We obtain for $M\geq 2$:
\begin{align*}
&\color{blue}{\sum_{\substack{k_1+\cdots+k_{M+1}=B\\k_1,\ldots,k_{M+1}\geq 0}}}\color{blue}{\binom{B}{k_1,\ldots,k_{M+1}}\binom{(M+1)N-B}{N-k_1,\ldots,n-k_{M+1}}}\\
&\qquad=\sum_{{k_1+\cdots+k_{M+1}=B}\atop{k_1,\ldots,k_{M+1}\geq 0}}\frac{B!}{k_1!\cdots k_{M+1}!}\,\frac{\left((M+1)N-B\right)!}{\left(N-k_1\right)!\cdots\left(N-k_{M+1}\right)!}\tag{8}\\
&\qquad=\sum_{k_{M+1}=0}^B\frac{B!}{k_{M+1}!\left(B-k_{M+1}\right)!}\,
\frac{\left((M+1)N-B\right)!}{\left(N-k_{M+1}\right)!\left(MN-\left(B-k_{M+1}\right)\right)!}\\
&\qquad\quad\cdot\sum_{{k_1+\cdots+k_M=B-k_{M+1}}\atop{k_1,\ldots,k_{M}\geq 0}}\frac{\left(B-k_{M+1}\right)!}{k_1!\cdots k_{M}!}\,\frac{\left(MN-\left(B-k_{M+1}\right)\right)!}{\left(N-k_1\right)!\cdots\left(N-k_{M}\right)!}\tag{9}\\
&\qquad=\sum_{k_{M+1}=0}^B\binom{B}{k_{M+1}}\,
\binom{(M+1)N-B}{N-k_{M+1}}\\
&\qquad\quad\cdot\sum_{{k_1+\cdots+k_M=B-k_{M+1}}\atop{k_1,\ldots,k_{M}\geq 0}}
\binom{B-k_{M+1}}{k_1,\ldots, k_{M}}\,\binom{MN-\left(B-k_{M+1}\right)}{N-k_1,\ldots,N-k_{M}}\tag{10}\\
&\qquad=\prod_{q=2}^M\binom{qN}{N}\sum_{k_{M+1}=0}^B\binom{B}{k_{M+1}}
[z^{N-k_{M+1}}](1+z)^{(M+1)N-B}\tag{11}\\
&\qquad=[z^{N}](1+z)^{(M+1)N-B}\prod_{q=2}^M\binom{qN}{N}\sum_{k_{M+1}=0}^B\binom{B}{k_{M+1}}z^{k_{M+1}}\\
&\qquad=[z^{N}](1+z)^{(M+1)N-B}\prod_{q=2}^M\binom{qN}{N}(1+z)^B\\
&\qquad=[z^{N}](1+z)^{(M+1)N}\prod_{q=2}^M\binom{qN}{N}\\
&\qquad=\binom{(M+1)N}{N}\prod_{q=2}^M\binom{qN}{N}\\
&\qquad\,\,\color{blue}{=\prod_{q=2}^{M+1}\binom{qN}{N}}
\end{align*}
in accordance with (7) and the claim follows.

Comment:

In (8) we write the multinomial coefficients using factorials.

In (9) we separate the summation of the summand $k_{M+1}$ as preparation to apply the induction hypothesis. We expand with $\left(B-k_{M+1}\right)!$ and $\left(MN-\left(B-k_{M+1}\right)\right)!$ to write the expression with binomial and multinomial coefficients.

In (10) we write the expression using binomial and multinomial coefficients.

In (11) we apply the induction hypothesis. We also use the coefficient of operator of $[z^N]$ to denote the coefficient of $z^N$ of a series.

In the following lines we use the same techniques as in the base step.


问:
Programming connectedness of a graph
I have to model the following problem as an optimization problem with constant objecive function and I am stuck at one of the restrictions.

Say that we have an $n\times n$ table for a fixed $n$. We have to allocate a list of nodes on this table. The nodes are labelled with natural numbers (greater or equal to 1) that can be repeated. The restriction to the position (cell) of this nodes are the following:

There can only be one node on each cell

Every time a node is allocated in a cell, it is connected with an edge to all nodes in adjacent cells (horizontal, vertical or diagonal).

The degree of each node  (number of other nodes it is connected to) coincides with its label.

The graph must be connected.

I have been able to model all the conditions except for the last one. The way I have done this is by defining variables $x_{ij}^k$ where $0\leq i,j\leq n+1$ and $1\leq k\leq n$ such that $x_{ij}^k=1$ if there is a node with label $k$ on the $(i,j)$ cell and $0$ otherwise. I have also added $x_{i0}^k=x_{0j}^k=x^k_{i,n+1}=x^k_{n+1,j}=0$ for convenience in the restriction to the degree of the nodes.

How can I specify the restrictions for the graph to be connected?

If I new in advance the number of nodes, say $m$, I know that if it is not connected, then there should be a connected component of at most $\lfloor \frac{m}{2}\rfloor$ nodes and I could write equations to detect whether there is a connected component with $l$ verticies for each $1\leq l\leq \lfloor \frac{m}{2}\rfloor$. However, this is not the case, I should be able to write all the restrictions before I know the input. But even if I new the amount of vertices, the number of equations for this approach would be huge, so it would be highly inefficient.

If it helps, and example where the set of solutions provides only provide disconnected graphs is an input of 4 nodes with label 1, as they would be paired in two connected components of 2 vertices each one. So, for this input, the problem should be infeasible.

答: This type of constraint appears in political districting problems.  See Imposing contiguity constraints in political districting models for several formulations.</p>

Title: Programming connectedness of a graph


答: $x_0$ is close to the dominant eigenspace. $A^4x_0$ is even closer to the dominant eigenspace. Choose an entry (say, the first entry), and calculate the ratio between that entry in $A^5x_0$ and $A^4x_0$.

Title: Find dominant eigenvector

Body:
Consider the $5x5$ tridiagonal matrix with entries $a_{i,i} = 2$,
$i=1,2,...,5$ and $a_{i,i+1}= a_{i+1, i} = 1$ , $i=1,...,4$. Knowing
that the initial approximation $x_0  =(0.5,0.8,1,0.8,0.5)$  is almost
aligned with the dominant eigenvector of the matrix, determine until
the fifth iteration an approximation of the dominant eigenvalue.

So, the first thing I did was to construct the matrix:
$A = \begin{bmatrix}
2 & 1 & 0 & 0 & 0\\
1 & 2 & 1 & 0 & 0\\
0 & 1 & 2 & 1 & 0\\
0 & 0 & 1 & 2 & 1\\
0 & 0 & 0 & 1 & 2\\ \end{bmatrix}$

Then I calculated $A^5x_0 = \begin{bmatrix} 348.2\\603.1\\696.4\\603.1\\348.2\end{bmatrix} = \lambda^5x_0$. But from now on I'm stuck. How can I calculate the dominant eigenvector from here?


问:
A countable compact set has infinitely many isolated points
I tried to prove that if $(X, d)$ is a complete metric space and $K$ a compact countable subset of $X$, then $K$ has infinite many isolated points. I tried to prove it with Baire's Theorem but I can't.
My try:

By reduction to the absurd, suppose that $ K $ has finitely many isolated points, that is, $ K \setminus K '=\{x_0, x_1, \ldots, x_n \} $. Note that,
\begin{align*}
    K \setminus \{x_0, x_1, \ldots, x_n \} & = K \setminus (K \setminus K ') \\
    & = K \cap (K^c \cup K') = K \cap K'
\end{align*}
On the other hand, we have that $ K $ and $ K'$ are closed, therefore, $ K \cap K' $ also is, and therefore this set is complete. For all $ i \in \mathbb{N} $ such that $ 0 \leq i \leq n $, we have that $ M_i = K \setminus \{x_i\} $ is an open set for the induced topology on $ K $, moreover, $ \overline{M_i} \cap K = K $, that is, $ M_i $ is dense in $ K $. With this, $ \{M_k \}_{k = 0} ^ n $ is a family of open and dense sets in $ K $, since $ K $ is closed, $ K $ is complete. Thus, using the Baire Category Theorem, we have that $ \overline{\displaystyle \bigcap_{k = 0}^n M_k} = K $, but $ \displaystyle \bigcap_{k = 0}^n M_k =\emptyset $, which is a contradiction.

I note that $ \overline{M_i} \cap K = K $ this could be false and I do not use that $K$ is countable. Any help will be of use, thank you very much.

答:
Will you accept a proof that does use (at least, not explicitly)  Baire's Theorem? If so, suppose the claim is false.  From the set $K=(x_n)_{n\in \mathbb N}$, extract a subsequence which converges to some $x\in K$, and which we still call $(x_n)$ for convenience. Of course, then $x$ is not isolated.

But now, if $\{y_1,\cdots, y_k\}$ are the isolated points in $K$ then consider $B:=B_r(x)$ where $0<r<\min\{d(x,y_i):1\le i\le k\}$. If $x\neq y\in \overline{B_{r/2}(x)}\cap K$ then, $y\in K'$ because it is not isolated. Then, $\overline {B_{r/2}(x)}\cap K'$ is perfect set and so uncountable. But this is impossible, so $B_{r/2}(x)\cap K= \{x\}$. But now we have a contradiction because this means that $x$ is isolated.


答: Since $K$ is compact, $K$ is closed, and therefore $K\cap K'=K'$. $K'$ is closed, so you can apply the Baire category theorem to it: each of the sets $K'\setminus\{x\}$ for $x\in K'$ is a dense open subset of $K'$, and there are only countably many of these sets, but their intersection is empty and therefore not dense in $K'$.

Title: A countable compact set has infinitely many isolated points

Body:
I tried to prove that if $(X, d)$ is a complete metric space and $K$ a compact countable subset of $X$, then $K$ has infinite many isolated points. I tried to prove it with Baire's Theorem but I can't.
My try:

By reduction to the absurd, suppose that $ K $ has finitely many isolated points, that is, $ K \setminus K '=\{x_0, x_1, \ldots, x_n \} $. Note that,
\begin{align*}
    K \setminus \{x_0, x_1, \ldots, x_n \} & = K \setminus (K \setminus K ') \\
    & = K \cap (K^c \cup K') = K \cap K'
\end{align*}
On the other hand, we have that $ K $ and $ K'$ are closed, therefore, $ K \cap K' $ also is, and therefore this set is complete. For all $ i \in \mathbb{N} $ such that $ 0 \leq i \leq n $, we have that $ M_i = K \setminus \{x_i\} $ is an open set for the induced topology on $ K $, moreover, $ \overline{M_i} \cap K = K $, that is, $ M_i $ is dense in $ K $. With this, $ \{M_k \}_{k = 0} ^ n $ is a family of open and dense sets in $ K $, since $ K $ is closed, $ K $ is complete. Thus, using the Baire Category Theorem, we have that $ \overline{\displaystyle \bigcap_{k = 0}^n M_k} = K $, but $ \displaystyle \bigcap_{k = 0}^n M_k =\emptyset $, which is a contradiction.

I note that $ \overline{M_i} \cap K = K $ this could be false and I do not use that $K$ is countable. Any help will be of use, thank you very much.


问:
Avoiding brute force: determining when a specific polynomial in $\mathbb{Q}[x]$ is an integer for any integer $x$
I have to prove that $\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n$ is an integer for any $n$. I solved this by brute-force, exhausting all the possibilities methods. I was wondering if there was a way to solve this at-a-glance with some sort of theory? Because

Although my answer is correct, I was apparently supposed to use some theory to answer this question.

I started by putting everything on common denominator and factoring out $n$:

$$\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n = \frac{n(3n^4+5n^2+7)}{15}$$

then I proceeded to plug in $\pm1 ,\pm2, \pm3 ,..., \pm7$ into $3n^4+5n^2+7 \pmod {3,5,\text{ or }15}$

$$3(\pm 1)^4 +5(\pm 1)^2 +7 \equiv 15 \equiv 0 \pmod{15}$$

$$3(\pm 2)^4 +5(\pm 2)^2 +7 \equiv 75 \equiv 5\cdot 15 \equiv 0 \pmod{15}$$

$$3(\pm 3)^4 +5(\pm 3)^2 +7 \equiv 295 \equiv 0 \pmod{5} \text{ while } n \equiv 0 \pmod 3$$

$$3(\pm 4)^4 +5(\pm 4)^2 +7 \equiv 855 \equiv 57\cdot 15\equiv 0 \pmod{15}$$

$$3(\pm 5)^4 +5(\pm 5)^2 +7 \equiv 2007 \equiv 0 \pmod{3}\text{ while } n \equiv 0 \pmod 5$$

$$3(\pm 6)^4 +5(\pm 6)^2 +7 \equiv 4075 \equiv 0 \pmod{5}\text{ while } n \equiv 0 \pmod 3$$

$$3(\pm 7)^4 +5(\pm 7)^2 +7 \equiv 7455 \equiv 497 \cdot 15\equiv 0 \pmod{15}$$

to conclude that if $n\equiv 0 \pmod {15}$ then $\frac{n}{15}$ is an integer from which $\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n$ is an integer,

and if $n \not\equiv 0 \pmod {15}$ then either $3n^4+5n^2+7 \equiv 0 \pmod{15}$ or $n \equiv 0 \pmod 3$ and $ 3n^4+5n^2+7 \equiv 0 \pmod{5}$ or $n \equiv 0 \pmod 5$ and $ 3n^4+5n^2+7 \equiv 0 \pmod{3}$ from which the statement is clearly true.

Is there a less brute-force-ish way of concluding this? Is there some theory I should be using that would cause me to not be excessively lengthy in calculation if I were given different, larger numbers than $15$?

答:
You can cut some work by just looking at the remainder mod $3$.

Let $n=3q+r$. Easy calculation gives
$$\frac15(3q+r)^5+\frac13(3q+r)^3+\frac{7}{15}(3q+r) = \text{some integer} + \frac{243q^5+7q}{5} + \frac{r^5}5+\frac{r^3}3+\frac{7r}{15}.$$

By Little Fermat we have $q^5 \equiv q \pmod{5}$ so
$$243q^5+7q \equiv 250q \equiv 0\pmod{5}.$$
The second term
$$\frac{r^5}5+\frac{r^3}3+\frac{7r}{15}$$
is the same as the original expression but you only have to check $r=0,1,2$ which is almost trivial.


答:
$X$ is called a Baire space if the countable intersection of open and dense subsets $D_n$ has a dense intersection. These include (this is what the Baire category theorem says) all complete metric spaces and also all locally compact Hausdorff spaces.

Lemma:

If $X$ is a Baire space that is $T_1$ without isolated points (a crowded space, this is called) then $X$ is uncountable: If not, write $X=\{x_m\mid n \in \Bbb N\}$ and note that $D_n = X\setminus \{x_n\}$ is open and dense (open as $\{x_n\}$ is closed by $X$ being $T_1$, and dense as $\{x\}$ is not open, so $D_n$ is not closed so the closure can only be $X$). But then $\bigcap_n D_n = \emptyset$ (by de Morgan, essentially) contradicting Baire-ness.

Now apply this to your $K \subseteq (X,d)$ situation: $K$ is Baire (for two reasons: it's locally compact Hausdorff, and also complete metric, being closed in $X$), so $K$ has isolated points $I(K) = K\setminus K'$. Note that $I(K)$ is open in $K$, and if it were finite (!) it would be closed too and $L = K\setminus I(K)$ would be open (and closed) in $K$ so wouldn't have isolated points ($L$ open in $K$ implies $I(L) \subseteq I(K)$, and $I(L) \cap I(K)=\emptyset$ by construction) and still be countable, Baire (still compact Hausdorff or complete) and crowded, contradicting the lemma. So $I(K)$ cannot be finite.


问:
$\operatorname{\mathcal{Jac}}\left( \mathbb{Q}[x] / (x^8-1) \right)$
$\DeclareMathOperator{\Jac}{\mathcal{Jac}}$
Using the fact that $R := \mathbb{Q}[x]/(x^8-1)$ is a Jacobson ring and thus its Jacobson radical is equal to its Nilradical, I already computed that $\Jac \left( \mathbb{Q}[x] / (x^8-1) \right) = \{0\}$. But I'd like to calculate this in an elementary way, preferably using nothing beyond the very basic properties of the Jacobson radical, quotients of rings and maximal ideals. And here starts my struggle.</p>

答: Hint: $\,{\rm max}\ M\subset K[x]/(p_1\cdots p_n)\! \iff\! M=(p_i)\,\Rightarrow\,  \cap\, M_ i =\, \cap\, (p_i) = (p_1\cdots p_n) = (0)$</p>


问:
Counterexample? Equality holds in the Cauchy-Schwarz Inequality iff $\textbf{x} = \alpha\textbf{y}$ for some $\alpha \in \mathbb{R}$
I found this way of expressing that equality holds in Cauchy Schwarz inequality iff $\textbf{x},\textbf{y}$ and are linearly dependent in a book (as an exercise), but I think it is wrong, so there's a counterexample for it. I'm not sure if it is a valid counterexample.

Statement:

$| \langle\textbf{x} ,\textbf{y}\rangle| = \| \textbf{x}\|\| \textbf{y}\|$ iff $\textbf{x} = \alpha\textbf{y}$ for some $\alpha \in \mathbb{R}$

Counterexample:

Let $\textbf{y} = (0,0) \in \mathbb{R^2}$, $\textbf{x} = (1,1) \in \mathbb{R^2}$. Then we have:

$$|\langle \textbf{x},\textbf{y}\rangle| = |\langle \textbf{x},\textbf{0}\rangle| = 0 = 1\cdot0 = \| \textbf{x}\|\| \textbf{y}\|$$
But
$$\textbf{x} = (1,1) \neq (\alpha\cdot0,\alpha\cdot0) = \alpha(0,0) = \alpha\textbf{y}, \forall \alpha \in \mathbb{R}$$

I know that if we let $\textbf{y} = (1,1)$ and $\textbf{x} = (0,0)$ the statement is true, but if the equality must work for all values of $\textbf{y}$ and $\textbf{x}$, then the instance gives is a counterexample, right?

答:
The statement

$\textbf{x} = \alpha\textbf{y}$ for some $\alpha \in \mathbb{R}$

does not cover all possibilities for the two vectors to be linearly dependent. It fails precisely when $\mathbf y$ is the zero vector but $\mathbf x$ isn't. So for such a case, the Caychy-Schwarz equality will hold, but the above property doesn't.

Instead, the true characterisation of linear dependence is usually phrased as

$\alpha\textbf{x} + \beta\textbf{y} = \mathbf 0$ for some $\alpha, \beta \in \mathbb{R}$, not both $0$

This version allows for either of the vectors to be the zero without the other one. And the Cauchy-Schwarz equality will hold iff this holds.


答:
The question asks how to prove a polynomial
$\,f(n)\,$ takes only integer values for any
integer $\,n\,.$ If the polynomial is of
degree $0$, then it is a constant and if
that constant is an integer we are done.
If the polynomial is of degree $1$, then if
any two consecutive values,
$\,f(k), f(k+1)\,$ are integers,
then it is an arithmetic progression and
we are done. In general, for any degree
$\,d\,$ polynomial $\,f(n),\,$ it is
sufficient to verify that
$$f(k),f(k+1), \dots,f(k+d-1) $$ are all integers for some
integer $\,k,\,$ which implies that the polynomial takes on only integer values for all integer $\,n.\,$

A proof of this can be done by using a
difference table to find an explicit
expression for the polynomial as a sum of
binomial coefficients. For example, using
the particular polynomial in the question,
$$ f(n) := \frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n $$
the forward difference table is:

$$ \begin{matrix}
\Delta^6f(n)  &&&&&&& 0 \\
\Delta^5f(n)  &&&&&& 24 && 24 \\
\Delta^4f(n)  &&&&& 48 && 72 && 96 \\
\Delta^3f(n)    &&&& 32 && 80 && 152 && 248 \\
\Delta^2f(n)    &&& 8 && 40 && 120 && 272 && 520\\
\Delta f(n)   && 1 && 9 && 49 && 169 && 441 && 961  \\
f(n)    & 0 && 1 && 10 && 59 && 228 && 669 && 1630 \\
n      & 0 && 1 && 2 && 3 && 4 && 5 && 6
\end{matrix} $$
where the 5th differences are constant as expected for a
5th degree polynomial. Thus,
$$ f(n) = 0{n \choose 0} +1{n \choose 1} + 8{n \choose 2}
 + 32{n \choose 3}  + 48{n \choose 4}  + 24{n \choose 5}$$
where the coefficient of $\,{n\choose k}=\Delta^kf(0),\,$
the $k$th difference of $\,f\,$ at zero.
The Wikipedia article finite difference
explains some of this  theory. The forward differences of $\,f\,$ are
$$ \Delta f(n) := f(n+1)-f(n),
\:\: \Delta^2 f(n) := \Delta f(n+1) - \Delta f(n), \:\: \dots. $$

In this particular case, the 1st backward difference is the
OEIS sequence A058031
$$ a_n := (n^2-n+1)^2 = \nabla f(n) := f(n)-f(n-1).$$
Thus the polynomial is the partial sums of an integer sequence
and therefore is an integer sequence itself.

Title: Avoiding brute force: determining when a specific polynomial in $\mathbb{Q}[x]$ is an integer for any integer $x$

Body:
I have to prove that $\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n$ is an integer for any $n$. I solved this by brute-force, exhausting all the possibilities methods. I was wondering if there was a way to solve this at-a-glance with some sort of theory? Because

Although my answer is correct, I was apparently supposed to use some theory to answer this question.

I started by putting everything on common denominator and factoring out $n$:

$$\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n = \frac{n(3n^4+5n^2+7)}{15}$$

then I proceeded to plug in $\pm1 ,\pm2, \pm3 ,..., \pm7$ into $3n^4+5n^2+7 \pmod {3,5,\text{ or }15}$

$$3(\pm 1)^4 +5(\pm 1)^2 +7 \equiv 15 \equiv 0 \pmod{15}$$

$$3(\pm 2)^4 +5(\pm 2)^2 +7 \equiv 75 \equiv 5\cdot 15 \equiv 0 \pmod{15}$$

$$3(\pm 3)^4 +5(\pm 3)^2 +7 \equiv 295 \equiv 0 \pmod{5} \text{ while } n \equiv 0 \pmod 3$$

$$3(\pm 4)^4 +5(\pm 4)^2 +7 \equiv 855 \equiv 57\cdot 15\equiv 0 \pmod{15}$$

$$3(\pm 5)^4 +5(\pm 5)^2 +7 \equiv 2007 \equiv 0 \pmod{3}\text{ while } n \equiv 0 \pmod 5$$

$$3(\pm 6)^4 +5(\pm 6)^2 +7 \equiv 4075 \equiv 0 \pmod{5}\text{ while } n \equiv 0 \pmod 3$$

$$3(\pm 7)^4 +5(\pm 7)^2 +7 \equiv 7455 \equiv 497 \cdot 15\equiv 0 \pmod{15}$$

to conclude that if $n\equiv 0 \pmod {15}$ then $\frac{n}{15}$ is an integer from which $\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n$ is an integer,

and if $n \not\equiv 0 \pmod {15}$ then either $3n^4+5n^2+7 \equiv 0 \pmod{15}$ or $n \equiv 0 \pmod 3$ and $ 3n^4+5n^2+7 \equiv 0 \pmod{5}$ or $n \equiv 0 \pmod 5$ and $ 3n^4+5n^2+7 \equiv 0 \pmod{3}$ from which the statement is clearly true.

Is there a less brute-force-ish way of concluding this? Is there some theory I should be using that would cause me to not be excessively lengthy in calculation if I were given different, larger numbers than $15$?


问:
If $A \cong I$ in a monoidal category, then any arrow $X \otimes A \to Y \otimes A$ decompose into $(X \to Y) \otimes 1$
The question is originated from the answer of this post. Let me make it precise first. Suppose $(\mathcal{A}, \otimes, I)$ is a monoidal category and $A\in \mathcal{A}$ is an object isomorphic to $I$. Then for any morphism $g:X \otimes A \to Y \otimes A$ in $\mathcal{A}$ there is a unique morphism $f:X\to Y$ such that $g = f \otimes 1_A$.

$\textbf{Attempt}$
Let $k:A \to I$ be an isomorphism. Given any such $g$, there is an arrow
$$X \xrightarrow{\rho^{-1}} X \otimes I \xrightarrow{1 \otimes k^{-1}} X \otimes A \xrightarrow{g} Y \otimes A \xrightarrow{1 \otimes k} Y \otimes I \xrightarrow{\rho} Y$$
(and I think that's the only way to construct a morphism $X \to Y$). But I don't see how this map tensoring with the identity gives you $g$. I believe it is just some sort of combining different diagrams together + the interchange law, as you normally do proofs in monoidal category. I have no clue on the uniqueness part of the morphism either. Any help is appreciate.

答:
You can show that if $g$ is of the form $f \otimes 1_A$ then your composite returns $f$: use functoriality of the tensor product to cancel out the $1 \otimes k$'s and then use naturality of $\rho$.

At the same time, your construction is clearly invertible, because it's given by composing with the invertible morphisms $1 \otimes k$ and $\rho$. Therefore if we define $f$ to be your composite

$$X \xrightarrow{\rho^{-1}} X \otimes I \xrightarrow{1 \otimes k^{-1}} X \otimes A \xrightarrow{g} Y \otimes A \xrightarrow{1 \otimes k} Y \otimes I \xrightarrow{\rho} Y$$

we can look at applying the inverse of your construction to get

$$X \otimes A \xrightarrow{1 \otimes k} X \otimes I \xrightarrow{\rho} X \xrightarrow{f} Y \xrightarrow{\rho^{-1}} Y \otimes I \xrightarrow{1 \otimes k^{-1}} Y \otimes A$$

This will give both $g$ (because that's what we started with) and $f \otimes 1_A$ (because as we saw in the first part we could have started with that and got the same $f$). Therefore $g$ and $f \otimes 1_A$ are equal, and this establishes your construction as being the inverse of the function given by tensoring by $1_A$.


问:
$f$ is continuous at $a$ iff for each subset $A$ of $X$ with $a\in \bar A$, $f(a)\in \overline{ f(A)}$.
Definiton.

$f$ is continuous at $a$ provided that for each open set $V$ in $Y$ containing $f(a)$ there is an open set $U$ in $X$ containing $a$ such that $f(U) \subset V$.

Problem.

$f$ is continuous at $a$ iff for each subset $A$ of $X$ with $a\in \overline{A}$, $f(a)\in \overline{ f(A)}$.

答:
Suppose $f$ is continuous at $a$, according to the definition.

Let $A$ be a subset with $a \in \overline{A}$, we want to show that $f(a) \in \overline{f[A]}$.

We will use that a point is in the closure of a set iff every open neighbourhood of the point intersects that set.

So let $V$ be an open neighbourhood of $f(a)$. Using the definition, we find a neighbourhood $U$ of $a$ such that $f[U] \subset V$. As $a \in \overline{A}$ we know that $V$ intersects $A$, say $a' \in A \cap V$. But then $f(a') \in V \cap f[A]$ (as $f[U] \subset V$), so indeed $V$ intersects $f[A]$, and as $V$ is arbitrary, $f(a) \in \overline{f[A]}$. This shows one implication.

Now let $f$ fulfill the closure condition at $a$; we want to see that $f$ is continuous at $a$. So let $V$ be an open neighbourhood of $f(a)$. Suppose now (striving for a contradiction) that for all open neighbourhoods $U$ of $a$ we have that $f[U] \not\subset V$. This means that every neighbourhood of $a$ contains points of $X \setminus f^{-1}[V]$, so $a \in \overline{X \setminus f^{-1}[V]}$, and so, by the closure condition, $f(a) \in \overline{f[X \setminus f^{-1}[V]} \subset \overline{Y \setminus V}$, but this is false, as $V$ does not intersect $Y \setminus V$. This contradiction shows the continuity of $f$ at $a$.


问:
Prove subset sum of squares errors are less than sum of squares error of the original set
As part of a proof I'd like to show if $R_1 \cup R_2 \subseteq R_3$, then $SSE(R_1) + SSE(R_2) \leq SSE(R_3)$ where SSE is the sum of squared deviations from the mean of each group.  I'm assuming the values of the group are real or rational numbers if that makes a difference.

I know that since the mean minimizes the sum of squared errors, then each group must be minimized, I'm stuck on how to leverage the fact that the groups are subsets to show that $SSE(R_1 \cup R_2) \geq SSE(R_1) + SSE(R_2)$.  Is there a basic piece of theory I am missing?  Any pointers would be appreciated!

答:
Using your insight about the mean being the estimator that minimizes the sum of squared errors we can write:

$$SSE(R_1 \cup R_2) = \min_\gamma \sum_{R_1 \cup R_2}(x_i - \gamma)^2$$

Now we can split this sum into two components:

$$= \min_\gamma \sum_{R_1}(x_i - \gamma)^2 + \sum_{R_2}(x_i - \gamma)^2$$

But of course, minimizing each sum individually will always give a value at least as small as minimizes the two sums together:

$$\geq \min_\alpha \sum_{R_1}(x_i - \alpha)^2 + \min_\beta\sum_{R_2}(x_i - \beta)^2$$

But this is exactly $SSE(R_1) + SSE(R_2)$


问:
Why I get $e^0=-\inf$ !?
I am trying to understand why my calculation for result of $$e^{0}=1$$ is not the same of others like posted here;

My calculation is based of using the $\ln$ and power actions which is shown below:
$$$$

$$\text{If}\, e^{0}=y\overset{\ln}{\rightarrow} \ln(e^{0})=\ln(y)$$

$${\rightarrow} 0\times \ln(e)=\ln(y)\overset{}{\rightarrow} 0\times 1=\ln(y) \rightarrow 0=\ln(y)\rightarrow y =- \infty !$$

I guess my calculation s not correct by If seems correct, i guess it shows some strange aspect of e and power (^) by the blow

Thanks for your consideration.

答: You obviously confused logarithm and exponential functions. $\log(y)=0$ means $y=1$.

Title: Why I get $e^0=-\inf$ !?


答:
From your calculations

$\ln y=0 \Longrightarrow y=e^0=1$. This is a circular trivial manipulations. Because, you get $y=e^0$ again. I could not see why $y=- \infty?$.

Here, $y=e^0$ by using only the definition of logarithm. Just definition. Also, $e^0=1$ is based on exponential definition.

Title: Why I get $e^0=-\inf$ !?

Body:
I am trying to understand why my calculation for result of $$e^{0}=1$$ is not the same of others like posted here;

My calculation is based of using the $\ln$ and power actions which is shown below:
$$$$

$$\text{If}\, e^{0}=y\overset{\ln}{\rightarrow} \ln(e^{0})=\ln(y)$$

$${\rightarrow} 0\times \ln(e)=\ln(y)\overset{}{\rightarrow} 0\times 1=\ln(y) \rightarrow 0=\ln(y)\rightarrow y =- \infty !$$

I guess my calculation s not correct by If seems correct, i guess it shows some strange aspect of e and power (^) by the blow

Thanks for your consideration.


问:
Calculate all possible combinations of 1, 2 & 3 which equal 67
My son (a third grader) was given this problem to solve. Is states that:

In a basketball game, one of the teams scored 67 points. If baskets can be worth 1 point for a free throw, 2 points for a field shot or 3 points for a 3 pointer, how did the team reach 67 points? How can you know that you have found all possibilities.

We can easily start out by doing 1 x 67 times, and then adding a 2 into the mix until we had mostly 2 and then starting with 3 and rounding that pattern back to introducing 1. We got up to 77 possible combinations when applying this simple pattern based solution. But that doesn't account for all the other combinations where any of the 3 numbers could be used which still equal 67.

It seems stars and bars may be a good way to deal with this problem, but I'm unclear on how to apply it. Would anyone be able to explain this? I'm not looking for the answer, but keen to understand how to calculate how many combinations there could be for any 3 sets of numbers to equal any given number.

Much appreciated!

答:
Here's an start (assuming order doesn’t matter).

Imagine that you want to know how many ways there are to buy a $67 candy bar from a vending machine that only accepts \$3, \$2, and \$1 coins. Since order doesn’t matter, you can count the ways you can pay using \$3 coins first, then \$2, then \$1 coins. (It’s the same problem, but sometimes making a problem involve money makes it easier to think about.)

You can use anywhere from $0$ and $22$ \$3 coins. If you use, say, exactly $7$ \$3 coins, then you have to pay the remaining \$67 - \$21 = \$46 dollars with \$2 and \$1 coins. You can do this using $0$, $1$, $2$, ... or $23$ \$2 coins (and the rest \$1 coins), for $24$ possible ways.

Can you figure out how many ways there are to finish paying after using a given number of \$3 coins, say $c$ of them (not necessarily $7$), and then (by hand is fine) add up your answers for $c=0$ through $c=22$?


答: You are correct, that's a counterexample. I don't know the exact statement of the book, but the statement can be fixed replacing it by "One of them is a scalar multiple of the other" (may be that's what the book says?). That way, if $\textbf{y}=\textbf{0}$, then $\textbf{y} = 0 \textbf{x}$.

Title: Counterexample? Equality holds in the Cauchy-Schwarz Inequality iff $\textbf{x} = \alpha\textbf{y}$ for some $\alpha \in \mathbb{R}$

Body:
I found this way of expressing that equality holds in Cauchy Schwarz inequality iff $\textbf{x},\textbf{y}$ and are linearly dependent in a book (as an exercise), but I think it is wrong, so there's a counterexample for it. I'm not sure if it is a valid counterexample.

Statement:

$| \langle\textbf{x} ,\textbf{y}\rangle| = \| \textbf{x}\|\| \textbf{y}\|$ iff $\textbf{x} = \alpha\textbf{y}$ for some $\alpha \in \mathbb{R}$

Counterexample:

Let $\textbf{y} = (0,0) \in \mathbb{R^2}$, $\textbf{x} = (1,1) \in \mathbb{R^2}$. Then we have:

$$|\langle \textbf{x},\textbf{y}\rangle| = |\langle \textbf{x},\textbf{0}\rangle| = 0 = 1\cdot0 = \| \textbf{x}\|\| \textbf{y}\|$$
But
$$\textbf{x} = (1,1) \neq (\alpha\cdot0,\alpha\cdot0) = \alpha(0,0) = \alpha\textbf{y}, \forall \alpha \in \mathbb{R}$$

I know that if we let $\textbf{y} = (1,1)$ and $\textbf{x} = (0,0)$ the statement is true, but if the equality must work for all values of $\textbf{y}$ and $\textbf{x}$, then the instance gives is a counterexample, right?


问:
What's wrong with my Surface Area of a solid of revolution formula?
When I learnt about the derivation for the formula $$V=\pi\int_{x_1}^{x_2} y^2~dx$$
where $V$ is volume of the solid generated when $y=f(x)$ is rotated about the $x$ axis by $2\pi$ radians between $x_1$ and $x_2$, my teacher showed that this works by approximating the volume by splitting up the solid generated into very thin cylinders. Each cylinder would have width $\delta x$ and radius $y$, so we'd have
$$V=\lim_{\delta x\to0}\sum_{x=x_1}^{x=x_2}\pi y^2\delta x=\pi\int_{x_1}^{x_2} y^2~dx$$

I thought of doing the same thing to find the surface area of thus solid that we have produced. Once more, consider the very thin cylinders that we used before. The surface area of the outer strip of each cylinder is $2\pi y\delta x $. Hence, I thought that the surface area should be equal to
$$SA=\lim_{\delta x\to0}\sum_{x=x_1}^{x=x_2}2\pi y\delta x=2\pi\int_{x_1}^{x_2}y~dx$$
But this is wrong! Why is that? What is wrong my reasoning?

Thank you for your help.

答: Your formula accounts for the fact that the surface gets larger (in terms of its circumference) as the curve gets further from the axis of revolution, but it does not account for the slope of that curve.  Therefore, your integrand has to be multiplied by the infinitesimal arclength, which accounts for this slope.


问:
Banach Space Condition help
Is $X$ a Banach space, $b\in x$ a vector and $A\in L\left ( x,x \right ) $ an application that checks the condition $\left |A ^{n} \right |< 1$, for all $n\in \mathbb{N}$.

(a) show that the equation $x=Ax+b$ has the only solution $x= \sum_{k=0}^{\infty }\left (A^{k} \right )b$.

thanks for all help

答:
First note, that $\|A \| < 1$ implies that $$\sum_{k=0}^\infty \|A^k\| \le \sum_{k=0}^\infty \|A\|^k = (1- \|A\|)^{-1} < \infty $$ converges. Hence, as $L(X,X)$ is a Banach space, the series $\sum_{k=0}^\infty A^k$ converges in $L(X,X)$. Let's call its limit $B$. Then
\begin{align*}
  B(\mathrm{Id} - A) &= \sum_{k=0}^\infty A^k(\mathrm{Id} - A)\\
          &= \sum_{k=0}^\infty A^k - \sum_{k=0}^\infty A^{k+1}\\
          &= \mathrm{Id}.
\end{align*}
Along the same lines, one sees $(\mathrm{Id} - A)B = \mathrm{Id}$, so $B = (\mathrm{Id} - A)^{-1}$. Now note that
$$ x = Ax + b \iff (\mathrm{Id} - A)x = b \iff x = (\mathrm{Id} - A)^{-1}b. $$

EDIT: Note that it is sufficient to have $\|A^n\| < 1$ for some $n$, in view of your other question this seems to be meant instead of all $n$. As: If $\|A^n\| < 1$, and $m \in \mathbb N$ is given, we may write $m = kn + \mu$ with $0 \le \mu < n$ and $k \in \mathbb N$, giving
$$ \|A^m\| = \|A^{kn + \mu}\| \le \|A^n\|^k \max_{0 \le \mu < n} \|A^\mu\| =: C \|A^n\|^k $$
One now has
$$ \sum_{m=0}^\infty \|A^m\| \le \sum_{k=0}^\infty n\|A^n\|^kC = Cn(1 - \|A^n\|)^{-1} < \infty $$
Now continue as above.

Title: Banach Space Condition help


问:
Is cocountable topology generally separable.What else is wrong with this table?
Is the cocountable complement topology always separable? Also are there Really any other issues with this table I created?

$$\begin{array}{cc|c}
 & \mbox{T1} & \mbox{Hausdorff}&\mbox{Regular} & \mbox{Normal}&\mbox{Separable}\\
\hline
\mathbb{R}&Y & Y & Y&Y&Y\\
\mathbb{R}^n&Y & Y & Y&Y&Y\\
\mbox{indiscrete}&N & N & Y&Y&Y\\
\mbox{discrete}&Y & Y & Y&Y&Y\\
\mbox{Cofinite}& Y& N&N &N&Y \\
\mbox{Cocountable}&Y&N&N&N&Y\\
\mathbb{R}_l&Y &Y &Y &Y&Y\\
\mbox{line w  2 origins} &Y &N &N &N&Y\\
\mbox{ordered  square} & Y&Y &Y &Y&N \\
\mathbb{R}_k &Y &Y &N &N&Y\\
\{0,1\}^A &Y &Y &Y &Y&Y
\end{array}$$

I am going to need to know these.Any help will be greatly appreciated.

Here is a link to all the discrepancies that were cleared up.

Table of separation properties of various topological spaces

答:
As said before: "Cocountable" means nothing. What set ? If the set is uncountable then the cocountable topology is not separable, because all countable sets are closed and so the closure is never $X$.

If $X$ is countable, then any topology on it is separable. So you do need to specify the underlying set in order to be able to fill in the table at all.

For "discrete" the same holds: not separable if $X$ is uncountable, and separable otherwise. Specify the set! Also for "indiscrete" you need to be precise. And what is the size of $A$ in $\{0,1\}^A$?

The table is incomplete and partially false as it stands, and the spaces are underspecified.


答: You are making a fundamental error.  The logarithm of 0 is negative infinity, the logarithm of 1 is 0.  From ln(y)= 0 you should conclude that y= 1, not negative infinity.</p>

Title: Why I get $e^0=-\inf$ !?


问:
Proof that $\sqrt{\det(XY)^2} \leq \sqrt{\det(Y^TY)\det(XX^T)}$?
Proof that $\sqrt{\det(XY)^2} \leq \sqrt{\det(Y^TY)\det(XX^T)}$? I feel like I could use Cauchy-Binet here but I'm unsure of how to do so. Any thoughts?

答:
The only non-trivial case where the products are well defined is where $X$ is $m \times n$ and $Y$ is $n \times m$ with $m \leq n$. We can prove that this formula holds by combining the Cauchy-Binet formula with the Cauchy-Schwarz inequality. In particular, in the notation of the CB formula wiki page, we have
$$
\begin{align}
\det(XY)^2 &= \left(\sum_{S \in \binom{[n]}{m}}\det(X_{[m],S})\det(Y_{S,[m]})\right)^2
\\ & \leq \left(\sum_{S \in \binom{[n]}{m}}\det(X_{[m],S})^2\right)\left(\sum_{S \in \binom{[n]}{m}}\det(Y_{S,[m]})^2\right)
\\ & = \left(\sum_{S \in \binom{[n]}{m}}\det(X_{[m],S})\det(X^T_{S,[m]})\right)\left(\sum_{S \in \binom{[n]}{m}}\det(Y^T_{[m],S})\det(Y_{S,[m]})\right)
\\ &= \det(XX^T) \det(Y^TY),
\end{align}
$$
which was what we wanted.


问:
How do I find the foci of the hyperbola $x^2 + \frac{2}{\sqrt{3}} x y - y^2 = 1$?
I have the hyperbola
$$x^2 + \frac{2}{\sqrt{3}} x y - y^2 = 1$$ and I want to find the foci, but the only resources I can find that talk about finding the foci require the formula to be in standard form,
$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$

I don't think I can convert my hyperbola to standard form because it isn't purely horizontal or vertical. How do I go about finding the foci in this case?

答:
Notice first of all that your hyperbola is centred at $O=(0,0)$. If $P=(x,y)$ is a vertex of the hyperbola, then the tangent at $P$ is perpendicular to $PO$, that is $y'=-x/y$. You can compute $y'$ by differentiating the equation of the ellipse:
$$
y'={x+y/\sqrt3\over y-x/\sqrt3}.
$$
Equating that to $-x/y$ gives:
$$
x^2-y^2-2\sqrt3 xy=0,
$$
which can be combined with the equation of the hyperbola to obtain
$$
P=\left(\pm\sqrt{2\sqrt3+3\over8}, \pm\sqrt{2\sqrt3-3\over8}\right).
$$

But yours is a right hyperbola (coefficients of $x^2$ and $y^2$ are opposite),
hence we immediately obtain the foci as:
$$
F=\sqrt2 P=\left(\pm\sqrt{2\sqrt3+3\over4}, \pm\sqrt{2\sqrt3-3\over4}\right).
$$


问:
How do you solve an equation in relation to curvature?
Image of the the points and lines. Show that
$$  \frac{MN}{KL} = 1 + \Big( \frac{LN - KM}{KL} \Big).  $$

I am unsure how to even start this problem.  Any help would be appreciated.

答: Just observe that $KL+LN=KN$ and $KN-KM=MN$. Therefore $KL+LN-KM=MN$. Now divide by $KL$ throughout.


问:
Geometry of Set of States of Random Process
Consider the following random process in the unit circle. Starting at $(0,0)$, move up, down, left, right a distance $1/2$ with equal probability.

At the next step there are three possibilities: stay where you are with probability $1/2$, go a distance $1/4$ in a direction orthogonal to the previous step with probability $1/4$ (*).

Keep iterating, at step $k$ stay where you are with probability $1/2$, go a distance $1/2^k$ in a direction orthogonal to the previous step with probability $1/4$.

Can we describe the set of points this process can visit?

It seems to have a fractal nature, and I think the closest it gets to the circle is $(2/3,1/3)$.

(*) In context this probability might not be $1/4$ and there are some absorbing states. The idea is to encode products of non commuting orthogonal projections $p, 1-p, q, 1-q$ as points in the unit circle. $p$ would be up, $1-p$ down and similar for $q$ and $1-q$.

答: Well, a bit of simulation shows there isn't anything particular amazing going on:


答: If $K$ has a finite number of isolated points, then their complement $K^\prime$ is a perfect Polish space (perfect means that every point is a limit point, Polish means complete metrizable (which a compact set in $\mathbb{R}^n$ obviously is). Then, it is a theorem (Theorem 6.2 in Kechris' "Classical descriptive set theory") that such a space contains a copy of the Cantor set, and is thus uncountable. The proof is simple: pick a subset $U_0$, then two disjoint subsets of $U_0$  of at most half the diameter of $U_0$ then for each of those two disjoint subsets of at most half the diameter of, and so on. The uncountability then follows by the Cantor (natch) diagonal argument.


问:
Is the zero locus of a sheaf homomorphism closed subscheme?
Let $\phi:\mathcal F\rightarrow \mathcal G$ be a morphism between quasicoherent sheaves over a scheme $S$. Then is the zero locus of $\phi$ is a closed subscheme?

In the language of representable functor, let $F:(Sch/S)\rightarrow (Sets)$ be the functor $F(T) = \{f:T\rightarrow S|f^*(\phi)= 0\}$. It is equivalent to say $F$ is represented by a closed subscheme of $S$.

I already know it is true when $\mathcal G$ is finite locally free. I'm wondering if there is any mild(for example, $\mathcal G$ is flat, sheaves are coherent, $S$ is Noetherian, etc) condition guarantees this property.

Thank you.

答:
Not in general. Let $Z\subset X$ be a closed subscheme, and consider the identity map $\mathcal{O}_Z\to\mathcal{O}_Z$. The zero locus is $X\setminus U$, which is open and not generally closed.

Off the top of my head, I do not know any condition more general than what you've already indicated which imply the result you seek. I suspect $\mathcal{G}$ flat is necessary if you're asking about all morphisms, but I doubt that's sufficient. It also seems likely that some mild finiteness hypothesis (either on the sheaves or on $S$) will be necessary, and you may be curious to know that this is already somewhat close to finite locally free, as seen in Stacks 053N.


问:
Simplifying an algorithm with matrix algebra.
Sorry in advances for my lack of knowledge on this topic.

Given a $p$ by $p$ positive semidefinite matrix $S$ from which, I want to compute this algorithm

```<code>for (i in 1:p){
for (j in 1:p){
Target[i,j] = S[i,j]*S[j,i]+S[i,i]*S[j,j]
}
}
</code>```

is there a better efficient way using matrix algebra?

答:
In terms of the Hadamard product, we could write
$$
T = S \circ S^T + \operatorname{diag}(S)\operatorname{diag}(S)^T,
$$
where $\operatorname{diag}(S)$ is the column-vector $(S_{1,1},\dots,S_{p,p})$.


问:
Is there by chance an analytic solution of the following non-linear ODE: $x'(t) = a\, x(t) + b\, x^3 (t)$?
I'm in fact interested in a PDE for which I try to get some intuition (roughly, I interpret my PDE as a function of time with value in a space of functions of space variables). Does someone by luck know a solution to the ODE in the title (wandering around this site I did see some incredible analytic solutions to some ODEs).

As for my line of thoughts so far, I see that $\enspace x:0< t\mapsto \frac{C}{\sqrt{t}}$ satisfies $x' = -\frac{1}{2}\, x^3$. The ODE being non-linear, I guess it is not very helpful to look at solutions of $x' = b x^3$  alone. I also don't think that Laplace or Fourier transfo will work. Series solution (haven't checked yet...)? My last idea is to use the trick of writing the equation in integral form and insert it within itself...

Follow up question: the next step is to consider a system of ODE of the form
$$\dot{\mathbf{X}} = A\cdot \mathbf{X} + B\cdot \mathbf{X}^3 $$
where here $\mathbf{X}^3$ denotes a vector with $i^{\text{th}}$-components $x_i^3$. Let us assume also that $A$ is diagonalizable but not $B$ (or at least no in the same basis). The problem boils down for each component of $\mathbf{X}$ to solving
$$ x_i'(t) = a\, x_i(t) + b\, x_i^3 (t) + f(t)$$

答:
Other ideas for the system of equations: for the one dimension case, the crux of the proof is the relation $\left(- \frac{1}{2\ x^2(t)} \right)'= \frac{x'(t)}{x^3(t)}$. In higher dimension, there is a priori no multiplication of vectors (although such an operation is just a linear map $\mu:V \otimes V \to V $ ). In fact, multiplication components by components does define a product for which the previous relation is satisfied and such a product can be recasted as multiplication of diagonal matrices after identification of vectors with diagonal matrices: ($V$ vector space with basis $\mathcal{B}:=(\mathbf{e}_i)_{i\in I}$, vectors are mapped to "multiplication" operators)
$$\mathrm{Identif}_{\mathcal{B}}: \left\lbrace
\begin{aligned} V &\to \mathrm{End}(V) \\ \mathbf{X} & \mapsto \mathrm{Diag}(\mathbf{X}):= \mathrm{Diag}(x_i, i\in I) \end{aligned}\right.\qquad \text{e.g. dimension 2}\quad \begin{pmatrix}x_1 \\ x_2 \end{pmatrix}\mapsto \begin{pmatrix}x_1 & 0 \\ 0 & x_2 \end{pmatrix}$$
The problem is now that the action of a matrix $A\in \mathrm{End}(V)$ on a vector $\mathbf{X}\in V$ cannot be written with a matrix of the same size on $\mathrm{Diag}(x_i) $, although it is still a linear transformation. (Linear map from a $n\times n$ matrix space to itself would in general be written with matrices of size $n^2\times n^2$). So let us introduce the notation $\mathcal{A}\big(\mathrm{Diag}(\mathbf{X})\big)= \mathrm{Diag}(A\cdot \mathbf{X}) $. The matrix equation can thus be rewritten
$$\dot{\mathbf{X}} = A\cdot \mathbf{X} + B\cdot \mathbf{X}^3 \quad \Longleftrightarrow \quad \mathrm{Diag}(\dot{\mathbf{X}}) = \mathcal{A}\big(\mathrm{Diag}(\mathbf{X})\big) + \mathcal{B}\big(\mathrm{Diag}(\mathbf{X})^3\big) $$
The second equality is that of diagonal matrices. I thought about multiplying on the right by $\mathrm{Diag}(\mathbf{X})^{-3}$ but although $\mathcal{B}$ is linear, one does not have equality
$$\mathcal{B}\big(\mathrm{Diag}(\mathbf{X})^3\big)\cdot \mathrm{Diag}(\mathbf{X})^{-3} \neq \mathcal{B}\big(\mathrm{Diag}(\mathbf{X})^3\cdot \mathrm{Diag}(\mathbf{X})^{-3}\big) = \mathcal{B}(\mathrm{Id})$$
unless the matrix $B$ is diagonal...

Finally it seems that $\mathcal{A}, \mathcal{B}$ can actually be written with matrices: let us stick to a 2-d example as below (\ref{G}). We want to write
$$\begin{pmatrix}\dot{x_1} \\ \dot{x_2} \end{pmatrix} = \begin{pmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{pmatrix} \cdot \begin{pmatrix}x_1 \\ x_2 \end{pmatrix} + \begin{pmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{pmatrix} \cdot \begin{pmatrix}x_1^3 \\ x_2^3 \end{pmatrix}$$
but with the vector replaced by diagonal matrices:

$$\begin{split}
\begin{pmatrix}\dot{x_1} & 0 \\0 & \dot{x_2} \end{pmatrix} &= \mathcal{A} \left( \begin{pmatrix}x_1 & 0\\ 0 & x_2 \end{pmatrix} \right) + \mathcal{B} \left( \begin{pmatrix}x_1^3 & 0 \\ 0 & x_2^3 \end{pmatrix} \right) \\
&= \begin{pmatrix} a_{11} & 0 \\ 0 & a_{22} \end{pmatrix} \cdot \begin{pmatrix}x_1 & 0\\ 0 & x_2 \end{pmatrix} + \begin{pmatrix} 0 & a_{12}\\ a_{21} & 0 \end{pmatrix} \cdot \begin{pmatrix}x_1 & 0 \\ 0 & x_2 \end{pmatrix} \cdot \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}  \\
&\enspace + \begin{pmatrix} b_{11} & 0 \\ 0 & b_{22} \end{pmatrix} \cdot \begin{pmatrix}x_1^3 & 0\\ 0 & x_2^3 \end{pmatrix} + \begin{pmatrix} 0 & b_{12}\\ b_{21} & 0 \end{pmatrix} \cdot \begin{pmatrix}x_1^3 & 0 \\ 0 & x_2^3 \end{pmatrix} \cdot \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \\
&= \begin{pmatrix} a_{11} & 0 \\ 0 & a_{22} \end{pmatrix} \cdot \begin{pmatrix}x_1 & 0\\ 0 & x_2 \end{pmatrix} + \begin{pmatrix} a_{12} & 0\\ 0 & a_{21} \end{pmatrix} \cdot \begin{pmatrix}x_2 & 0 \\ 0 & x_1 \end{pmatrix} \\
&\enspace + \begin{pmatrix} b_{11} & 0 \\ 0 & b_{22} \end{pmatrix} \cdot \begin{pmatrix}x_1^3 & 0\\ 0 & x_2^3 \end{pmatrix} + \begin{pmatrix} b_{12} & 0\\ 0 & b_{21} \end{pmatrix} \cdot \begin{pmatrix}x_2^3 & 0 \\ 0 & x_1^3 \end{pmatrix}
\end{split}$$

The equation can also be rewritten as follows (example in dimension $2$)
$$ \begin{split}
\begin{pmatrix}\dot{x_1} \\ \dot{x_2} \end{pmatrix} &= \begin{pmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{pmatrix} \cdot \begin{pmatrix}x_1 \\ x_2 \end{pmatrix} + \begin{pmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{pmatrix} \cdot \begin{pmatrix} x_{1}^2 & 0 \\ 0 & x_{2}^2 \end{pmatrix} \cdot \begin{pmatrix}x_1 \\ x_2 \end{pmatrix} \\
&= \left[ \begin{pmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{pmatrix} + \begin{pmatrix} b_{11}\, x_{1}^2 & b_{12}\, x_{2}^2 \\ b_{21}\, x_{1}^2 & b_{22}\, x_{2}^2 \end{pmatrix}  \right] \cdot \begin{pmatrix}x_1 \\ x_2 \end{pmatrix}\\
&= \begin{pmatrix} a_{11}+ b_{11}\, x_{1}^2 & a_{12} + b_{12}\, x_{2}^2 \\ a_{21} + b_{21}\, x_{1}^2 & a_{22} + b_{22}\, x_{2}^2 \end{pmatrix}  \cdot \begin{pmatrix}x_1 \\ x_2 \end{pmatrix}
\end{split} \label{G}\tag{G}$$

A dead end: If solutions existed for all times the resolvant would define a one parameter group (of $2\times 2$ matrices in example (\ref{G})). I thought that such groups were known and of the form $e^{tH}$ where $H\in M_2(\mathbb{K})$ is the so-called generator. BUT this holds for groups of linear transformations, and our resolvant is looking at (\ref{G}) probably not a linear transformation...

My last idea was to use tensor products. Assume again for simplicity of notation that $\mathbf{X}$ lives in a 2 dimensional vector space $V$ with basis $\mathcal{B}:=(\mathbf{e}_1, \mathbf{e}_2)$. Define
$$ M_{\mathcal{B}}: \left\lbrace
\begin{aligned} V \otimes V \otimes V &\longrightarrow \quad V\\
\mathbf{X}\otimes \mathbf{Y}\otimes \mathbf{Z} &\longmapsto \begin{pmatrix} x_1 \cdot y_1 \cdot z_1 \\ x_2 \cdot y_2 \cdot z_2\end{pmatrix} \end{aligned}
\right.$$
or by its action on basis vectors $M_{\mathcal{B}}(\mathbf{e}_i \otimes \mathbf{e}_j \otimes \mathbf{e}_k) = \mathbf{0}$ as soon as one of the indice is different from the other and $M_{\mathcal{B}}(\mathbf{e}_i \otimes \mathbf{e}_i\otimes \mathbf{e}_i)= \mathbf{e}_i $. Introducing $\mathcal{u}=\mathbf{e}_1 + \mathbf{e}_2$ the equation reads
$$\dot{\mathbf{X}} = M_{\mathcal{B}}\big( (A\cdot \mathbf{X})\otimes \mathbf{u} \otimes \mathbf{u} \big) + B\cdot M_{\mathcal{B}}\big( \mathbf{X}\otimes \mathbf{X} \otimes \mathbf{X}\big) \label{T}\tag{T} $$

Let $\displaystyle A= \bigoplus_{k=0}^{+\infty} V^{\otimes k} = \mathbb{K} \oplus V \oplus \big(V \otimes V \big) \oplus \cdots $ be the freely generated associative algebra built out of $V$, a little bit like the Fock space in quantum mechanics (without the Hilbert space structure...) and $\Delta: V \to A,\ \mathbf{X} \mapsto \epsilon(\mathbf{X}) \oplus \mathbf{X} \oplus \big(\mathbf{X}\otimes \mathbf{X} \big) \oplus \cdots$ the "diagonal map" ($\epsilon$ some linear form; $\Delta$ is not the usual linear injection of $V \hookrightarrow A,\, \mathbf{X} \mapsto \mathbf{X}$). Let us indicate with indices $(n,p)$ a linear map from a "homogeneous" subspace $L^{(n,p)}: V^{\otimes n} \to V^{\otimes p}$, e.g. $M_{\mathcal{B}}^{(3,1)}: V^{\otimes 3} \to V $. Introducing also the projections $\Pi_k: A \to V^{\otimes k}$, (\ref{T}) can be written:

$$\begin{split}
\dot{\mathbf{X}} &= A^{(1,1)}(\mathbf{X})+ B^{(1,1)} \circ M_{\mathcal{B}}^{(3,1)}\big( \mathbf{X}\otimes \mathbf{X} \otimes \mathbf{X}\big) \\
&= \left( A^{(1,1)}\circ \Pi_1 + B^{(1,1)} \circ M_{\mathcal{B}}^{(3,1)}\circ \Pi_3 \right) \Delta(\mathbf{X})
\end{split} \label{T2}\tag{T2} $$
Can one find a solution of the form $ \left( e^{\left(A^{(1,1)} + B^{(1,1)} \circ M_{\mathcal{B}}^{(3,1)} \right) t} \Delta(\mathbf{X})\right)$? We have manage to get a linear operator withing the exponential but $\Delta(\mathbf{X})$ is not linear, and it seems that this doesn't have the correct behavior under derivation...


问:
Avoiding brute force: determining when a specific polynomial in $\mathbb{Q}[x]$ is an integer for any integer $x$
I have to prove that $\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n$ is an integer for any $n$. I solved this by brute-force, exhausting all the possibilities methods. I was wondering if there was a way to solve this at-a-glance with some sort of theory? Because

Although my answer is correct, I was apparently supposed to use some theory to answer this question.

I started by putting everything on common denominator and factoring out $n$:

$$\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n = \frac{n(3n^4+5n^2+7)}{15}$$

then I proceeded to plug in $\pm1 ,\pm2, \pm3 ,..., \pm7$ into $3n^4+5n^2+7 \pmod {3,5,\text{ or }15}$

$$3(\pm 1)^4 +5(\pm 1)^2 +7 \equiv 15 \equiv 0 \pmod{15}$$

$$3(\pm 2)^4 +5(\pm 2)^2 +7 \equiv 75 \equiv 5\cdot 15 \equiv 0 \pmod{15}$$

$$3(\pm 3)^4 +5(\pm 3)^2 +7 \equiv 295 \equiv 0 \pmod{5} \text{ while } n \equiv 0 \pmod 3$$

$$3(\pm 4)^4 +5(\pm 4)^2 +7 \equiv 855 \equiv 57\cdot 15\equiv 0 \pmod{15}$$

$$3(\pm 5)^4 +5(\pm 5)^2 +7 \equiv 2007 \equiv 0 \pmod{3}\text{ while } n \equiv 0 \pmod 5$$

$$3(\pm 6)^4 +5(\pm 6)^2 +7 \equiv 4075 \equiv 0 \pmod{5}\text{ while } n \equiv 0 \pmod 3$$

$$3(\pm 7)^4 +5(\pm 7)^2 +7 \equiv 7455 \equiv 497 \cdot 15\equiv 0 \pmod{15}$$

to conclude that if $n\equiv 0 \pmod {15}$ then $\frac{n}{15}$ is an integer from which $\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n$ is an integer,

and if $n \not\equiv 0 \pmod {15}$ then either $3n^4+5n^2+7 \equiv 0 \pmod{15}$ or $n \equiv 0 \pmod 3$ and $ 3n^4+5n^2+7 \equiv 0 \pmod{5}$ or $n \equiv 0 \pmod 5$ and $ 3n^4+5n^2+7 \equiv 0 \pmod{3}$ from which the statement is clearly true.

Is there a less brute-force-ish way of concluding this? Is there some theory I should be using that would cause me to not be excessively lengthy in calculation if I were given different, larger numbers than $15$?

答:
To show that $15$ divides $n(3n^4+5n^2+7)$, show that $3$ and $5$ do.

$3$ does because either $3$ divides $n$ or $n^2\equiv1\pmod3$ by Fermat's little theorem,

in which case $3n^4+5n^2+7\equiv5n^2+7\equiv5+7=12\equiv0\pmod3.$

$5$ does because either $5$ divides $n$ or $n^4\equiv1\pmod5$ by Fermat's little theorem,

in which case $3n^4+5n^2+7\equiv3n^4+7\equiv3+7=10\equiv0\pmod5$.