问:
Question on M/M/2 queue variation
I have the following question:

Two workers handle three machines(i.e. we can at most repair two machines at a time). The time until the machine breaks down is exponential distributed with expectation value $1/2$ and independent of other repair and break down times. Every repair time is exponential distributed with expectation value $1/3$ and independent of other repair and break down times. Let $X_t$ be the amount of broken down machines at time t.

a) Determine the birth and death frequencies in the birth and death chain $(X_t)_{t \geq 0}$.

b) Determine $\lim_{t \to \infty}\mathbf{P}_i(X_t =0)$ for $i=0,1,2,3$.

Attempt

a) The jump times are minimum of the exponential times which we know have the parameter the sum of each of them. Thus I got the Q-matrix (I assumed that only one worker could repair a machine):

$$
Q =
 \begin{pmatrix}
 -9 & 9 & 0 & 0 \\
  2 & -8 & 6 & 0 \\
  0  & 4  & -7 & 3  \\
  0 & 0 & 4 & -4
 \end{pmatrix}$$

So the frequency of going from 0 to 1 is 9, from 1 to 2 is 6, 2 to 3 is 3. To the opposite direction I get from 1 to 0 is 2, 2 to 1 is 4 and from 3 to 2 is 4.

b) Since this is irreducible, aperiodic and finite we know that this limit is just $\pi_0$ where $\pi$ is the invariant distribution. From this I get $\pi_0=16/277$ (With my $\pi$,  $\pi Q=0$) .

Is this this right or am I making some wrong conclusions in my calculations?

Regards, Raxel.

答:
$X_t$ is the number of broken machines. I think your $Q$ matrix should be
$$Q=\begin{pmatrix}
-6 & 6 & 0 & 0 \\
3 & -7 & 4 & 0 \\
0 & 6 & -8 & 2 \\
0 & 0 & 6 & -6
\end{pmatrix}$$

because

repairs happen at rate 3

failures happen at rate 2

and in state

0: all machines are working and can independently fail at rate 2 (so total failure rate 6)

1: two machines can fail (total failure rate 4) and a single machine can be repaired (at rate 3)

2: the one remaining machine can fail (rate 2) or the two broken machines can be repaired. They are being repaired in parallel so repairs happen at total rate 6.

3: All machines are broken, but only two are being repaired (there are only two workers) so the total repair rate remains at 6.

Computing the stationary distribution using $\pi Q = 0$ we get
$$ \pi = \left(\frac{9}{43},\frac{18}{43},\frac{12}{43},\frac{4}{43}\right)$$
so $\pi_0 = \frac{9}{43}$ (which you have correctly identified is the answer irrespective of starting state $i$).

Title: Question on M/M/2 queue variation


问:
Does there exist two infinite subsets of naturals $A, B$ such that each $n \in A + B$ uniquely determines $a \in A, b \in B$ such that $a + b = n$?
Does there exist two infinite sets of naturals $A, B$ such that each $n \in A + B = \{ a + b : a\in A, b\in B\}$ has a unique solution $a + b = n$ in which $a \in A, b \in B$?

For example:

$$
A = \{0,3,6\} \\
B = \{0,2,7\} \\ \implies \\
0 = 0 + 0 \text{ is unique } \\
0 = 0 + 2 \text{ is unique } \\
\vdots \\
6 + 7 = 13 \text{ is unique }
$$

But I need the sets $A, B$ to be infinite.

I need it so that when multiplying two numbers in binary:

$$
(\sum_{i \in A}^{\infty} a_i 2^i)(\sum_{j \in B}^{\infty} b_j 2^j)
$$

Each coefficient is of the form $a_i b_j$ for some $i, j$ and so that the position $i$ that $a_i$ came from and the position $j$ that $b_j$ came from is uniquely determined by the value $i + j$.

I only need to always be able to determine two finite sets $A, B$ of roughly the same size such that $|A||B| = |A + B| = N^2$ for any $N \in \Bbb{N}$, which is why infinity came to mind, but an algorithm that returns the two finite sets given $N$ is sufficient for my purposes.

答:
Sure. Let $A$ be the powers of 10, and $B$ be $\{2, 22, 222, 2222, \dotsc\}$.

Numbers in $A+B$ are either of a form like 1000022222, or like 2223222, and in either case you can determine what $a$ and $b$ are.

If you want more numbers in the sets, we can generalize quasi's answer:
Let $A$ be the sums of distinct odd powers of two, and $B$ be the sums of distinct even powers of two. Then each set has 32 elements below 1000, which is better than the 3 each in my answer, or the 5 each in quasi's answer. $A$ starts with $\{0, 2, 8, 10, 32, 34, \dotsc\}$ and $B$ starts with $\{0, 1, 4, 5, 16, 17, 20, \dotsc\}$.

Then the number of elements in $A$ or $B$ less than any given $N$
is about $2^{\log_2(N)/2}$: for half the available bit positions (either the even or odd ones), you make a choice between putting a 0 or 1 there.
This is about $\sqrt{N}$.
By the pigeonhole principle, the best you can do is less than $\sqrt{2N}$, so this is close to optimal.

ContentLicense: CC BY-SA 4.0


问:
Should I prove AGM inequality with 4 variables with Induction or Cauchy-Schwarz?
I'm trying to prove the following: $\sqrt[4]{abcd} \le \frac{a+b+c+d}{4}$.

While I managed to prove base case $n=1$ and $n=2$;

$(\sqrt{a}-\sqrt{b})^2 \geqslant 0$

$a-2\sqrt{ab}+b \geqslant 0$

$a+b \geqslant 2\sqrt{ab}$

$\sqrt{ab} \le \frac{a+b}{2}$

I am unsure how to proceed for $n=4$. Is there a way to continue proving 4 variables using induction, or is it better to use Cauchy-Schwarz Inequality to go about the proof?

答:
You can prove AM-GM using Jensen's inequality which states that if $f$ is continuous and convex (or equivalently, concave up), then we know for $x_1, x_2, ..., x_n$ and $a_1, a_2, ..., a_n$ that $f(\frac{a_1x_1 + ... + a_nx_n}{a_1 + ... + a_n}) \leq \frac{a_1 f(x_1) + ... + a_n f(x_n)}{a_1 + ... + a_n}$. This may seem complicated, but we can think of the $a_i$ as weights being applied before or after $f$. Since $f$ is convex, the point $\frac{a_1x_1 + ... + a_nx_n}{a_1 + ... + a_n}$ will be somewhere in the middle of the $x_i$. Then, when $f$ is applied to this value, we get something less than a value which is somewhere in the middle of the $f(x_i)$. It can be shown that Cauchy-Schwarz is actually a specific case of Jensen's inequality.

We actually apply this as follows: Consider $x_1, x_2, ..., x_n \in \mathbb{R}$. Let $y_i = \text{ln}(x_i)$, let $a_i = \frac{1}{n}$, and let $f(x) = e^x$, which is convex/concave up. Thus, by Jenson's inequality, $e^{\frac{1}{n}(y_1 + ... + y_n)} \leq \frac{1}{n}(e^{y_1} + ... + e^{y_n})$. By the construction of $y_i$, we have that $e^{y_i} = x_i$. Additionally, $y_1 + ... + y_n = \text{ln}(x_1) + ... + \text{ln}(x_n) = \text{ln}(x_1 x_2 ... x_n)$. So plugging everything in and simplifying, we get $e^{\frac{1}{n}(y_1 + ... + y_n)} = (x_1 ... x_n)^\frac{1}{n} \leq \frac{1}{n}(x_1 + ... + x_n)$. At this point, we can let $n = 4$, to get that $n = 4$ to get your result.


问:
Prove that $S^n$ is $(n-1)$-connected.
I heard someone mention that since $S^n$ is $(n-1)$-connected (i.e. $\pi_k (S^n) = 0$ for $k<n$), $\pi_2 (S^3)=0$. However, I can't seem to imagine how this is the case. The person who said it made it seem trivial, so I've come to the conclusion that this is probably a stupid question, but are there any easy/intuitive ways to understand this?

答: The only way I am aware of for proving this is the Freudenthal suspension theorem, which is not very hard.</p>

Title: Prove that $S^n$ is $(n-1)$-connected.


问:
How to find pdf of $max|X_i|$
Let $X_1,...,X_n$ be a random sample from a population with pdf

$$f(x|\theta)=\frac{1}{2\theta}, -\theta<x<\theta,\theta>0$$.

I want to know the pdf of $max|X_i|$. How to find?

答:
From @runr's Comment, $P(\max_i|X_i|\le x) = P(-x \le X_1\le x)^n = (x/\theta)^n.$ for $0 < x <\theta.$

For $n = 10, \theta=3,$ the following simulation in R illustrates the density
function.

```<code>set.seed(317)
n = 10;  th = 3
abs.max = replicate(10^5, max(abs(runif(n,-th,th))))
summary(abs.max)

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
 0.8256  2.6093  2.7973  2.7261  2.9139  3.0000

hist(abs.max, prob=T, br=50, col="skyblue2")
 curve(10*x^9/3^10, 0, 3, add=T, col="orange", lwd=2)
</code>```

Title: How to find pdf of $max|X_i|$

Body:
From @runr's Comment, $P(\max_i|X_i|\le x) = P(-x \le X_1\le x)^n = (x/\theta)^n.$ for $0 < x <\theta.$

For $n = 10, \theta=3,$ the following simulation in R illustrates the density
function.

set.seed(317)
n = 10;  th = 3
abs.max = replicate(10^5, max(abs(runif(n,-th,th))))
summary(abs.max)

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
 0.8256  2.6093  2.7973  2.7261  2.9139  3.0000

hist(abs.max, prob=T, br=50, col="skyblue2")
 curve(10*x^9/3^10, 0, 3, add=T, col="orange", lwd=2)


问:
How to prove the following evaluation map is surjective?
Let $P_{1}, P_{2},\cdots, P_{k}\in \Bbb{C}^n$. Now define a homomorphism
$\phi: \Bbb{C}[x_{1},x_{2},\cdots,x_{n}]\rightarrow \Bbb{C}^k$ which sends $f$ to $(f(P_{1}),f(P_{2}),\cdots,f(P_{k})).$ Then how to show that $\phi$ is surjective?

I saw it here, but don't know how to prove it.

答:
In general proving that functions are surjective is hard but inyectivity is much easier because we have tools, for example the Jacobian that tell us when a function is inyective.

Fortunely for us, the Ax's thoerem states that any polinomial map that is inyective is also surjective, precisley

Let $f:\mathbb{C}^n\rightarrow \mathbb{C}^n$, $f=(f_1,...,f_n), f_i\in \mathbb{C}[x_1,...,x_n]$ if $f$ is inyective then, it also is surjective. (Hils, M., & Loeser, F. (2019). A first journey through logic. Theorem 3.6.3)

I think that Ax's theorem also generalizes to any algebraic variety.


答: Continuous maps are homotopic to smooth maps. Smooth maps $S^k \to S^n$, for $k<n$, are for dimensional reasons not surjective due to Sard's theorem. By removing a point from $S^n$ you get $\mathbb{R}^n$, and thus such maps are homotopic to a constant.

Title: Prove that $S^n$ is $(n-1)$-connected.

Body: I heard someone mention that since $S^n$ is $(n-1)$-connected (i.e. $\pi_k (S^n) = 0$ for $k<n$), $\pi_2 (S^3)=0$. However, I can't seem to imagine how this is the case. The person who said it made it seem trivial, so I've come to the conclusion that this is probably a stupid question, but are there any easy/intuitive ways to understand this?


答:
The argument goes that when $0\lt i\lt n $, any continuous map from $S^i $ to $S^n $ is homotopy equivalent to a non-surjective map.  Then since $S^n\setminus\{\text {pt}\}\cong\mathbb R^n $ is contractible,  the map is homtopically trivial.

To make the first part rigorous,  see the cellular approximation theorem.  This is somewhat difficult because of the existence of space-filling curves.  When $i=1$ and $n=2$, there's the intuition that a rubber band wrapped around a sphere can be deformed to a point.

Title: Prove that $S^n$ is $(n-1)$-connected.


问:
Why does $(e^{-\frac{3}{4}t}-e^{it})(\frac{3}{4}-i) = e^{-\frac{3}{4}t}-\cos t+ \frac{3}{4}\sin t$
I read this while going through some derivations in my mathematical physics class. I thought of multiplying everything and applying Euler's formula but I don't understand how the $ie^{-\frac{3}{4}t}$ could be turned into something real.

A little more context: this was after doing convolution with the Laplace transform and then doing something with real numbers to solve this deceptively easy-looking problem $y''+y'+\frac{3}{16}y(t) = \sin (t),\quad y(0) = y'(0) = 0$.

答:
Your simplification is not correct. use $e^{it}=\cos t+ i\sin t$, then
$$(e^{-3t/4}-\cos t-i \sin t)(\frac{3}{4}-i)=\frac{3}{4} e^{-3t/4}-\frac{3}{4} \cos t- \frac{3i}{4}\sin t-i 3^{-3t/4}-i \cos t  -\sin t.$$


问:
Is the direct sum decomposition of formal power series provable without choice?
If we view the ring of formal power series $F[[x]]$ as a vector space over $F$, and we view the polynomial ring $F[x]$ as a subspace of $F[[x]]$, then the axioms of choice implies that
$$F[[x]]=F[x]\oplus W$$ for some subspace $W$ of $F[[x]]$; we just need to extend $\{1,x,x^2,...\}$ into an (uncountable) Hamel basis for $F[[x]]$, and then take the span of the other basis elements.

My question is, is the existence of $W$ provable in $ZF$, or do we require the Axiom of Choice?  And if it is provable without Choice, can we explicitly construct such a $W$?

答:
Even for $F=\mathbb{Z}/2\mathbb{Z}$ this is not provable in $\mathsf{ZF}$. Note that the formal power series language is unnecessarily complicated, given that we're just thinking about vector spaces; really we just have $A=\bigoplus_{i\in\mathbb{N}}F$ sitting inside $B=\prod_{i\in\mathbb{N}}F$ in the obvious way and we're looking for a way to write $B=A\oplus W$ for some $W$.

A key point here is that $B$ is extremely "flexible" compared to $A$: if we pass to a symmetric extension $V'\supset V$ of the universe we have $A^{V'}=A$ but $B^{V'}\not=B$ in general.

In particular, fix any field $F\in V$ and consider forcing with finite sequences of elements of $F$ (intuitively we're building a Cohen-generic element of $B$) and look at the finite supports symmetric submodel corresponding to this forcing. If memory serves, in this model we wind up killing off any possible summand $W$, although the details are a bit annoying to check.


答:
Define $A,B$ as follows . . .

Let $A$ be the set of positive integers of the form $2^s$ where $s$ is even.

Let $B$ be the set of positive integers of the form $2^t$ where $t$ is odd.

If for $a\in A$ and $b\in B$, we have
$$n=a+b=2^s+2^t$$
where $s$ is even and $t$ is odd, then $s,t$ are uniquely determined by $n$, hence so are $a,b$.

Title: Does there exist two infinite subsets of naturals $A, B$ such that each $n \in A + B$ uniquely determines $a \in A, b \in B$ such that $a + b = n$?

Body:
Does there exist two infinite sets of naturals $A, B$ such that each $n \in A + B = \{ a + b : a\in A, b\in B\}$ has a unique solution $a + b = n$ in which $a \in A, b \in B$?

For example:

$$
A = \{0,3,6\} \\
B = \{0,2,7\} \\ \implies \\
0 = 0 + 0 \text{ is unique } \\
0 = 0 + 2 \text{ is unique } \\
\vdots \\
6 + 7 = 13 \text{ is unique }
$$

But I need the sets $A, B$ to be infinite.

I need it so that when multiplying two numbers in binary:

$$
(\sum_{i \in A}^{\infty} a_i 2^i)(\sum_{j \in B}^{\infty} b_j 2^j)
$$

Each coefficient is of the form $a_i b_j$ for some $i, j$ and so that the position $i$ that $a_i$ came from and the position $j$ that $b_j$ came from is uniquely determined by the value $i + j$.

I only need to always be able to determine two finite sets $A, B$ of roughly the same size such that $|A||B| = |A + B| = N^2$ for any $N \in \Bbb{N}$, which is why infinity came to mind, but an algorithm that returns the two finite sets given $N$ is sufficient for my purposes.


问:
Prove that $f(x)=0$.
If $ f: [a, b] \to \Bbb {R} $ is continuous in $ [a, b] $, and there is a number $ \alpha \neq \beta $ such that $$ \alpha \int_a^c f ( x) \, dx + \beta \int_c^b f (x) \, dx = 0,$$ for all $ c \in (a, b) $. Prove that $f(x)=0$, for all $x\in[a,b]$.

For absurdity, suppose that $ f (x) \neq 0 $. So $ \alpha \int_a^c f (x) \, dx \neq 0 $, likewise with $ \beta \int_c^b f (x) \, dx \neq 0 $. Therefore, the sum of the integrals will be different from $ 0 $, but that contradicts the general hypothesis.

答:
The question as phrased prior to the edit has a counterexample. For the corrected question, the answer is positive. The argument given by OP in the comments to this answer is correct. Basically, consider the sum of two integrals as a function of $c$ and use the fundamental theorem of calculus.

The previous version of the question said $$\alpha \int_a^c f(x) dx + \beta \int_{b}^c f(x)dx \equiv 0.$$
Note the order of the variables for the integration domains. For this we have a counterexample:
Note that for all $c\in(0,2\pi)$, $$0=\int_0^{2\pi} \sin xdx=\int_0^{c}  \sin xdx + \int_c^{2\pi}  \sin xdx = \int_0^c \sin x dx - \int_{2\pi}^c  \sin xdx $$
Therefore $f(x)=\sin(x)$ is a  continuous nonzero function on $[a,b]=[0,2\pi]$ such that $\alpha=1,\beta=-1\neq \alpha$ such that for all $c\in(0,2\pi)$,
$$\alpha \int_0^c f(x) dx + \beta \int_{2\pi}^c f(x)dx \equiv 0.$$

Title: Prove that $f(x)=0$.


问:
Continuous Probability Density Function: question on definition
I have seen this with multiple definitions.  It also appears the Fundamental Theorem of Calculus applies.  Does anyone know the correct definition?  I think my confusion is between $\le$ and $=$. I list some definitions below to show my confusion.

$f_X(z) = \frac {d F_{\le X}(z)} {dz}$, $f_X : \mathbb{R} \rightarrow [0,1]$ is the PDF and $F_{\le X} : \mathbb{R} \rightarrow [0,1]$ is the CDF.

Wiki definition:

a)
$$
P(a \le X \le b) = \int_a^b f_X(z) \, dz = F_{\le X}(b) - F_{\le X}(a)
$$

I think this should really be two different CDFs, $\le$ and $\lt$ as follows.  Otherwise the value at $a$ is excluded.:

b)
$$
P(a \le X \le b) = \int_a^b f_X(z) \, dz = F_{\le X}(b) - F_{\lt X}(a)
$$

this definition is from my notes (which I think is the correct one):

c)
$$
P(a \lt X \le b) = \int_a^b f_X(z) \, dz = F_{\le X}(b) - F_{\le X}(a)
$$

I think whats confusing me is application of the Fundamental Theorem of Calculus.  It doesn't agree with equality in form (c) above:

d)
$$
\int_a^b f(t) \, dt = F(b) - F(a)
$$

答: They are all equivalent. I will illustrate by showing that a) gives c): If $f_X$ is an integrable function and, $$P(a \leq X \leq b)=\int_a^b f_X(x)\,dx$$ then we can put $a=b$ to see that for all $a$,$$P(X=a)=\int_a^a f_X(x) \, dx=0$$Therefore, $$P(a<x\leq b)=P(a\leq X \leq b)-P(X=a)=\int_a^b f_X(x) \, dx$$


问:
Change of variables using local approximation
Let $$ T(u, v, w)=x(u, v, w) \hat{i}  +y(u, v, w)\hat{j} + z(u, v, w)\hat{k}$$
be a continuously differentiable bijection.

Then $$  \int \int \int_D f(x, y, z) dx dy dz=\int \int \int _D f \circ T(u, v, w) \det{J} du dv dw, $$
where $$ J=\left[\begin{matrix}
\frac{\partial x}{\partial u}&\frac{\partial x}{\partial v}&\frac{\partial x}{\partial w}\\
\frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}&\frac{\partial y}{\partial w}\\
\frac{\partial z}{\partial u}&\frac{\partial z}{\partial v}&\frac{\partial z}{\partial w}
\end{matrix}\right]. $$

The main goal is to show that $$ \Delta V= \Delta x \Delta y \Delta z = \det{J} \Delta u \Delta v \Delta w .$$

Attempt of proof:

Using local approximation we get
$$\Delta T \approx J \left[\begin{matrix}
\Delta u\\
\Delta v\\
\Delta w
\end{matrix}\right]  \\
\left[\begin{matrix}
\Delta x\left(u{,}\ v{,}\ w\right)\\
\Delta y\left(u{,}\ v{,}\ w\right)\\
\Delta z\left(u{,}\ v{,}\ w\right)
\end{matrix}\right]\approx\left[\begin{matrix}
\frac{\partial x}{\partial u}&\frac{\partial x}{\partial v}&\frac{\partial x}{\partial w}\\
\frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}&\frac{\partial y}{\partial w}\\
\frac{\partial z}{\partial u}&\frac{\partial z}{\partial v}&\frac{\partial z}{\partial w}
\end{matrix}\right]\left[\begin{matrix}
\Delta u\\
\Delta v\\
\Delta w
\end{matrix}\right] \\
\left[\begin{matrix}
\Delta x\left(u{,}\ v{,}\ w\right)\\
\Delta y\left(u{,}\ v{,}\ w\right)\\
\Delta z\left(u{,}\ v{,}\ w\right)
\end{matrix}\right]\approx\left[\begin{matrix}
\frac{\partial x}{\partial u}\Delta u+\frac{\partial x}{\partial v}\Delta v+\frac{\partial x}{\partial w}\Delta w\\
\frac{\partial y}{\partial u}\Delta u+\frac{\partial y}{\partial v}\Delta v+\frac{\partial y}{\partial w}\Delta w\\
\frac{\partial z}{\partial u}\Delta u+\frac{\partial z}{\partial v}\Delta v+\frac{\partial z}{\partial w}\Delta w
\end{matrix}\right]$$

So now if we read row by row $$ \Delta x= \dots, \Delta y= \dots, \ \mathrm{and \ so \ on}, $$ and calculate $$  \Delta V= \Delta x \Delta y \Delta z,$$ we end up in a different result. So why doesn't it work using this approach?

答:
I think your attempt is fine, the only thing that you need to formalize is what "$\approx$" means, and what it means is that you choose on Taylor series up to first derivative.

Here is a simpler way to not make all the computations.
It should be easier (and equivalent) to see $T$ one component at a time, that is, verify this three cases: $$T(u,v,w)=x(u,v,w)\hat{i}+u\hat{j}+w\hat{k}$$
$$T(u,v,w)=u\hat{i}+y(u,v,w)\hat{j}+w\hat{k}$$
$$T(u,v,w)=u\hat{i}+v\hat{j}+z(u,v,w)\hat{k}$$
And by symmetry, it is enough to show only one, choose the first one.

In this case
$$J=\left[\begin{matrix}\frac{\partial x}{\partial u}&\frac{\partial x}{\partial v}&\frac{\partial x}{\partial w}\\
\frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}&\frac{\partial y}{\partial w}\\
\frac{\partial z}{\partial u}&\frac{\partial z}{\partial v}&\frac{\partial z}{\partial w}\end{matrix}\right]=
\left[\begin{matrix}\frac{\partial x}{\partial u}&\frac{\partial x}{\partial v}&\frac{\partial x}{\partial w}\\
0&1&0\\
0&0&1\end{matrix}\right]$$
Hence, $$\Delta x =\frac{\partial x}{\partial u}\Delta u +\frac{\partial x}{\partial u} \Delta v +\frac{\partial x}{\partial u} \Delta w, \quad \Delta y=\Delta v,\quad \Delta z=\Delta w $$

Finally, $$\Delta V=\Delta x \Delta y \Delta z\approx \left(\frac{\partial x}{\partial u}\Delta u +\frac{\partial x}{\partial u} \Delta v +\frac{\partial x}{\partial u} \Delta w\right)\Delta v \Delta w \approx \frac{\partial x}{\partial u}\Delta u\Delta v \Delta w =\operatorname{det}J \Delta u \Delta v \Delta w$$

Where terms of second order ($\Delta v^2$ and $\Delta w^2$) vanish in our approximation.


问:
Statistic - Population and Sample Mean Switching
I am figuring out a method to calculate the population mean via sample info. Is that possible?

For example, sample info sample size n=100, ̅x=53, σ=7.7.
The population size is 6700.
Can I calculate the population mean when P(x ̅≥53 hours)=0.5,

Can anyone give me hints and tell me is there any trading between population and sample mean? Thanks a lot

答:
With a sample of size $n=100,$ the sample mean $\bar X$ will tend to be very close to the population mean $\mu.$

Furthermore, by the Central Limit Theorem,
$\bar X$ will have very nearly the distribution
$\mathsf{Norm}(\mu, \sigma/\sqrt{n} = 7.7/10 = 0.77).$

So you can say
$$P(\bar X \ge 53) \approx P\left(Z = \frac{\bar X - \mu}{\sigma/\sqrt{10}}
\ge
\frac{53-\mu}{0.77} = 0\right) = 0.5,$$
where $Z$ is a standard normal random variable. Solve for $\mu.$


问:
Smallest Lipschitz constant of $x \mapsto A_0 + \sum_{i=1}^n x_i A_i$
Let $\mathcal{S}_m (\mathbb{R})$ denote the space of symmetric $m \times m$ matrices.

Consider the matrix-valued affine-linear function $A: \mathbb{R}^n \to \mathcal{S}_m (\mathbb{R})$ given by

$$ A(x) := A_0 + \sum_{i=1}^{n} x_i A_i  $$

with $A_0, A_1, \dots , A_n \in \mathcal{S}_m (\mathbb{R})$.

Function  $A: \mathbb{R}^n \to \mathcal{S}_m (\mathbb{R})$ is affine-linear and thus (globally) Lipschitz-continuous.

Question: What is the  (smallest) Lipschitz-constant $L>0$ of $A: \mathbb{R}^n \to \mathcal{S}_m (\mathbb{R})$ such that

$$ \| A(y) - A(x) \|_2 \leq L \| y -x \|_2 \qquad \forall x,y \in \mathbb{R}^n$$

where $\| z \|_2 = \sqrt{z^{\top}z}$ for any vector $z \in \mathbb{R}^n$ and

$$\| B \|_2 = \sup_{\| x\| =1} \| Bx \|_{2} \qquad \forall B \in \mathbb{R}^{m \times m} $$

denotes the induced matrix norm.

答:
The following is a very rough sketch written in haste. It may be totally wrong. Please check every step very carefully.

$$ {\rm A} ({\rm y}) - {\rm A} ({\rm x}) = \sum_{k=1}^n \left( y_k - x_k \right) {\rm A}_k = \begin{bmatrix} {\rm A}_1 & {\rm A}_2 & \cdots & {\rm A}_n\end{bmatrix} \left( ({\rm y} - {\rm x}) \otimes {\rm I}_m \right) $$

where $\otimes$ denotes the Kronecker product. Since the spectral norm is submultiplicative,

$$\begin{aligned} \left\| {\rm A} ({\rm y}) - {\rm A} ({\rm x}) \right\|_2 &= \left\| \begin{bmatrix} {\rm A}_1 & {\rm A}_2 & \cdots & {\rm A}_n\end{bmatrix} \left( ({\rm y} - {\rm x}) \otimes {\rm I}_m \right) \right\|_2 \\\\ &\leq \left\| \begin{bmatrix} {\rm A}_1 & {\rm A}_2 & \cdots & {\rm A}_n\end{bmatrix} \right\|_2 \underbrace{\left\| ({\rm y} - {\rm x}) \otimes {\rm I}_m \right\|_2}_{=: \| {\rm y} - {\rm x} \|_2 }\end{aligned}$$


问:
Proving an inequality with $6$ variables and a constant $C$
Find the greatest constant $C$ so that for all real numbers $x_1, x_2,\ldots,x_6$ the inequality
$$(x_1 + x_2 + ...+ x_6)^2 \geq C \cdot (x_1 (x_2 + x_3) + x_2 (x_3 + x_4) + \ldots+ x_6 (x_1 + x_2))$$ applies.

Can someone give me a hint as to how to solve this?

答:
Hint: It is well known that $(a+b+c)^2 \geq 3 (ab+bc+ca)$, with equality when $ a = b = c$.

Further hint, (actually the crux in this approach, so I encourage you to think about how the previous line could be a hint. If you don't know what to do, I suggest to skip this part first and read the rest of the solution below the separator):

 Set $ a = x_1 + x_4, b = x_2 + x_5, c = x_3 + x_6$.

Hence $ C = 3$.

Steps that one could take to approach the inequality directly:

Setting $ x_i = 1$ gives $ C \leq 3$. A reasonable guess is that $ C = 3$ is the answer (though this need not be the case).

Notice that with $C=3$, equality holds when $ x_1 = x_3 = x_5 , x_2 = x_4 = x_6$, so it's not a standard "all terms equal" scenario.

Calling the term on the RHS $X$ (without the constant), show that  $$X = (x_1 + x_4) ( x_2 + x_5) + (x_2 + x_5) ( x_3 + x_6) + (x_3 + x_6) ( x_1 + x_4).$$ Arguably, this is a "nicer" expression.

Show that

$$(x_1 + x_2 + x_3 + x_4+ x_5 + x_6) ^2  - 2X = (\sum x_i^2) + 2 ( x_1x_4 + x_2x_5 + x_3x_6) \geq X.$$

Equality holds when $x_1 + x_4 = x_2 + x_5 = x_3 + x_6$.

Hence $(x_1 + x_2 + x_3 + x_4+ x_5 + x_6) ^2 \geq 3X$

Hence, the answer is $C = 3$.

From here, it's easy to motivate the original substitution.


问:
Proof that one of three terms with 3 variables is always smaller or equal to 1
Let $x, y$ and $z$ positive real numbers with $x + y + z = 3$. Prove that at least one of the three numbers

$x (x + y − z), y (y + z − x)$ or $z (z + x − y)$

is less than or equal to $1$.

Can someone give me a hint where to start with this?

答:
If any of (cyclic) $x+y - z $ is negative, then we are done.

Otherwise, show that $0 < xyz \leq 1$ and $ 0 \leq \prod (x+y-z) \leq 1$.

Hence, the product of those 3 numbers is between 0 and 1, hence at least one of them is $ \leq 1$.

Note: Try showing that at least one of them is $ \geq 1$.


问:
Conditional probability of scoring more than $8$ questions correctly given at least $6$ is correct.

Question: In a multiple choice test, there are 20 questions, and each question has four choices. Jane is clueless and answers by purely guessing. She is equally likely to pick any of the possible choices, and each such choice is independent of the other ones. If it is known that Jane guessed correctly strictly more than six questions correctly, what is the chance that she scored strictly more than eight questions correctly?

I have attempted it and my work is shown below for this practice problem. However, I got it wrong and I am not sure why is the answer correct. Could anyone please tell me what I did wrong and how should I solve it?

My answer:

$$P(X > 8| X> 6) = \frac{P(X > 8)}{P(X>6)}=\frac{\sum_{i=9}^{20}\binom{20}{i}(0.75)^{20-i}(0.25)^i}{\sum_{i=7}^{20}\binom{20}{i}(0.75)^{20-i}(0.25)^i}$$

But answer provided was:

$$\frac{\sum_{i=7}^{20}\binom{20}{i}3^{20-i}}{\sum_{i=9}^{20}\binom{20}{i}3^{20-i}}$$

答:
The "actual answer" can't be correct, since the numerator is greater than the denominator.

Probability can't exceed $1$.


问:
What value of theta we need to take after the substitution
Suppose we have the def integral of $\frac{1}{√[(x-a)(b-x)] }dx$ from $a$ to $b$ , where $b>a$ , there is a special substitution which can be elegant only for def integral of $1/√$(quadratic having real  roots) that is $x = asin^2\theta + bcos^2\theta$ , but when we now change the limits , what should be those $\theta s$ we will take, I mean as for lower one will it be $\pi/2$ or $3\pi/2$ .. which one and similarily for above will it be $0$ or $\pi$ or $2\pi$ ..etc ?

[ I know one thing that for other types of problems where we substitute something the substituted thing from the integrand must be bijective , but here what we r substituting does the substituing thing too need to bijective , as "x" itself is bijective which i dont have problem]. For substituent thing i dont know what conditions should be imposed in that .

答:
The question is, if we are dealing with $\int_a^bf(x)\,dx$, and we find it useful to make the substitution $x=a\sin^2\theta+b\cos^2\theta$ to get $\int_?^?g(\theta)\,d\theta$, then what should the new limits of integration be?

We need a value of $\theta$, call it $\theta_1$, that leads to $x=a$, and a value of $\theta$, call it $\theta_2$, that leads to $x=b$; we need $\theta_1<\theta_2$; and we need $x$ to take on every value between $a$ and $b$ exactly once as $\theta$ goes from $\theta_1$ to $\theta_2$.

To get $x=a$, we need $\sin^2\theta_1=1$ and $\cos^2\theta_1=0$. This will happen precisely when $\theta_1$ is an odd multiple of $\pi/2$. [That is, when $\theta_1=k\pi/2$ for some odd integer $k$.]

To get $x=b$, we need $\sin^2\theta_2=0$ and $\cos^2\theta_2=1$. This will happen precisely when $\theta_2$ is a multiple of $\pi$.

That leaves us with many options. E.g., we might try $\theta_1=\pi/2,\theta_2=2\pi$. But this doesn't work, since we get $x=b$ twice, at $\theta=\pi$ and again at $\theta=2\pi$. So we see we have to make our interval of $\theta$-values short, so as to get $x=a$ and $x=b$ only once each.

So, let's try $\theta_1=\pi/2$, $\theta_2=\pi$. Now we get $x=a$ exactly once, and $x=b$ exactly once. But how do we know we get every $x$ between $a$ and $b$ exactly once, and that we never get $x<a$ or $x>b$?

One way to do this is to rewrite $$x=a\sin^2\theta+b\cos^2\theta=a(1-\cos^2\theta)+b\cos^2\theta=a+(b-a)\cos^2\theta$$ Now, we know that $\cos^2\theta$ is continuous and monotone increasing from zero to one as $\theta$ goes from $\pi/2$ to $\pi$, so $x=a+(b-a)\cos^2\theta$ will be continuous and monotone increasing from $a$ to $b$ over that same range of values of $\theta$, and thus will take on every value in that interval exactly once. We're done.


问:
How do you determine generalized eigenvectors using Jordan form?
I have the following matrix

$$
A= \begin{bmatrix}
0 & 0 & 0\\
0 & i & 1\\
0 & 0 & i
\end{bmatrix}
$$

I can see that it is in Jordan normal form. I read that you can find the generalized eigenvector just by looking at the matrix but I don't know how. The answer is supposed to be $v=(0,0,1)^T$.

答:
All three vectors in the standard basis are generalized eigenvectors of the matrix since this is the basis in which it takes Jordan normal form. You can read their associated eigenvalue from the corresponding diagonal element of the matrix and you can find their rank by counting their position within the respective Jordan block starting from the left.

For example, in the case of the matrix in the question, we see that

$(1, 0, 0)^T$ is a generalized eigenvector of rank $1$ (ordinarily called an eigenvector) associated with eigenvalue $0$,

$(0, 1, 0)^T$ is a generalized eigenvector of rank $1$ associated with eigenvalue $i$ and

$(0, 0, 1)^T$ is a generalized eigenvector of rank $2$ associated with eigenvalue $i$.

You can confirm this directly from definition by checking that rank $r$ generalized eigenvector $v$ associated with eigenvalue $\lambda$ satisfies $(A-\lambda I)^r v = 0$ and $(A-\lambda I)^{r-1} v \ne 0$ where $A$ is the matrix in question.


问:
What's the purpose of the word $w$ in the proof for $ L \in \mathsf{REG} \implies L(\varphi) \in \mathsf{MSO}$ with $L(\varphi) = L$?
I'm taking a lecture where we proved $ L \subseteq \mathsf{REG(\Sigma^*)} \iff \exists \phi \in \mathsf{MSO}$ with $L(\phi) = L$. For the first direction $L \subseteq \mathsf{REG(\Sigma^*)} \implies \exists \phi \in \mathsf{MSO}, L(\phi) = L$ we were shown the following:

Proof:

We define $\mathsf{MSO}$ for directed graphs of the form $G=(V, E ,\lambda)$ where $V$ is the set of vertices, $E$ the set of edges and $E \subseteq V \times V$ and a labeling function $\lambda : V \to \Sigma$, with $\Sigma$ being an finite alphabet.

So we label every vertice with a letter $a\in \Sigma$.
We use variables $x$ for elements from $V$ and $X$ for subsets of $V$ (monadic pradicates).

Let $L= L(A)$ for some NFA $A= (Q, \Sigma, \delta, I, F)$. We now construct $\phi \in \mathsf{MSO}$ with: $$w \in L(A) \iff w \models \phi$$

Therefore let $w = a_1 \ldots a_n$ with $a_i \in \Sigma$ for $1\leq i \leq n$. Then also $V = \{1, \ldots, n\}$ and $E=\{(i, i+1) \mid 1 \leq i <n\}$, and $\lambda(i) = a_i$.

Let $Q = \{0, \ldots, m\}$. Then there exists set variables $X_0, \ldots, X_m$, such that $\phi$ can be set to:

$$\phi = \Bigl\lbrack(\exists X_q : 1 \in X_q \land q \in I) \land (\exists X_q : n \in X_q \land q \in F) \land  (\bigwedge_{p, q \in Q} (p \not = q \implies X_p \cap X_q = \emptyset)) \, \,\land $$

$$ \forall x \forall y ( \bigwedge_{p, q, a} (x \in X_p \land \lambda(x)= a \land (x,y \in E)) \implies  \bigvee_{(p,q,a) \in \delta} y \in X_q)\Bigr\rbrack$$

Questions:

1.) Why is it enough to use only one word for constructing $\phi$? I thought $\phi$ should describe all words from the language $L$. If that isn't the case, and we construct a new formulae for every $w\in L$, how is that done for infinte languages like for example $ L = a^*$?

2.) What exactly is the purpose of the graph? In "Finite Model Theory" by Leonid Libkin (page 124), Theorem 7.21 there is a similar proof, but I still don't understand why and how we use graphs with the formulaes.

答:
Instead of using graphs, it is easier to interpret a word $u = a_1 \dotsm a_n$ as a finite structure ${\cal M}_u =\{1, \ldots, n\}$
equipped with the natural order $<$ on ${\cal M}_u$ and, for each letter $a$, a predicate $R_a$ giving the positions of the letter $a$ in $u$. For isntance, if $u = ababba$, then ${\cal M}_u = \{1, \ldots, 6\}$, $R_a = \{1,3,6\}$ and $R_b = \{3, 4, 5\}$.

If you use finite graphs, you simply identify a word $a_1 \cdots a_n$ with the graph $G_u$
$$
(0) \xrightarrow{a_1} (1) \xrightarrow{a_2} (2) \rightarrow \cdots \rightarrow (n-1) \xrightarrow{a_n} (n)
$$
So your set $V$ should rather be $\{0, 1, \dots, n\}$.

Now you have
\begin{align}
L(\phi) &= \{ u \in \Sigma^+ \mid u \text{ satisfies } \phi \} \\
&= \{ u \in \Sigma^+ \mid {\cal M}_u \models \phi \} \quad &&\text{(ordered structure version)}\\
&= \{u \in \Sigma^+ \mid G_u \models \phi \} &&\text{(graph version)}
\end{align}
The proposed formula $\phi$ just encodes the behaviour of the NFA $\cal A$ accepting $L$. If
$$
q_0 \xrightarrow{a_1} q_1 \xrightarrow{a_2} q_2 \rightarrow \cdots \rightarrow q_{n-1} \xrightarrow{a_n} q_n
$$
is a path in $\cal A$ and $q \in Q$, let $X_q = \bigl\{i \in \{0, \ldots, n\} \mid q_i = q\bigr\}$, then $\phi$ (slightly modified) can be interpreted as follows:
\begin{align}
&\exists X_q \quad 0 \in X_q \land q \in I && \text{$0 \in X_q$ for some initial state $q$}\\
{} \land {} &\exists X_q \quad n \in X_q \land q \in F && \text{$n \in X_q$ for some final state $q$}\\
{} \land {} &\bigwedge_{p, q \in Q} \bigl(p \not = q \implies X_p \cap X_q = \emptyset\bigr) && \text{the sets $X_q$ are pairwise disjoint}\\
 {}\land{} & \forall x \forall y\  \bigwedge_{p, q, a} \Bigl( \bigl(x \in X_p \land \lambda(x)= a \land (x,y \in E)\bigr) && \text{if $i \in X_p$ and if $a$ is a letter,} \\
&\implies  \bigvee_{(p,q,a) \in \delta} y \in X_q\Bigr) && \text{then $i+1 \in X_q$ for some state $q$ such that $p \xrightarrow{a} q$ is a transition}
\end{align}


问:
How can $4^x = 4^{400}+4^{400}+4^{400}+4^{400}$ have the solution $x=401$?
How can $4^x = 4^{400} + 4^{400} + 4^{400} + 4^{400}$ have the solution $x = 401$?

Can someone explain to me how this works in a simple way?

ContentLicense: CC BY-SA 3.0

Body: $4^{401} = 4\times 4^{400} = 4^{400} + 4^{400} + 4^{400} + 4^{400}$


问:
Reference request: Where is this exercise concerning the exponential distribution found in textbooks?
\begin{align}
& T_1,T_2,T_3,\ldots\sim\text{i.i.d.} \\
& \text{with distribution } e^{-t/\mu} (dt/\mu)\text{ for } t>0. \\[8pt]
& \Pr(N=n) = p^{n-1}q \text{ for } n = 1,2,3,\ldots \\
& \text{where $p+q=1,$ $p,q>0$} \\
& \text{and $N$ is independent of } T_1,T_2,T_3,\ldots \\
& \text{so that } \operatorname E(N) = 1/q. \\[8pt]
& T= \sum_{n\,=\,1}^N T_n \\
& \text{so } \operatorname E(T) = \mu/q.
\end{align}
An exercise is to show that the distribution of $T$ is exponential: $e^{-qt/\mu} (q\,dt/\mu) \text{ for } t>0.$

MY QUESTION IS whether this is somehow a standard exercise known to everyone except me, or more literally: Did I just somehow miss that this one is in various textbooks? And while I'm at it, do others solve this differently from the way I did (I'll post something below)?

答: This is fairly standard. One easy way to prove it is to use Laplace transforms. The Laplace transform of $\exp(\lambda)$ is $\frac {\lambda} {\lambda+t}$. You can compute the Laplace transform of $T$ by conditioning on $N$ and then compute the geometric sum you get. You will get $\frac {q/\mu} {q/m+t}$ which is the Laplace transform of $\exp(q/\mu)$.


答:
This will answer the exercise but not the main question.

A standard exercise posits that $M\sim\operatorname{Poisson}(\lambda)$ and $X\mid M \sim \operatorname{Binomial}(M,q),$ and asks for proof of the following:

$X\sim\operatorname{Poisson}(q\lambda),$

$M-X\sim\operatorname{Poisson}((1-q)\lambda),$

$X$ and $M-X$ are independent.

Then one has the "thinned Poisson process": The number of arrivals in any time interval is $\lambda$ times the length of the interval, and the numbers of arrivals in disjoint intervals are independent, and one "thins" the process by accepting each arrival with probability $q$ independently of the Poisson process. The exercise above shows that the "thinned" process must also be a Poisson process, so its interarrival times are exponentially distributed. Finally, one observes that those exponentially distributed interarrival times are just what is described in the question above.

So the next question is: What is a more direct way to do it? Without mentioning the Poisson distribution. I expect this will be easy, but I'm about to crash and I won't think about that further tonight.

This does not actually answer the main question above.

Title: Reference request: Where is this exercise concerning the exponential distribution found in textbooks?

Body:
\begin{align}
& T_1,T_2,T_3,\ldots\sim\text{i.i.d.} \\
& \text{with distribution } e^{-t/\mu} (dt/\mu)\text{ for } t>0. \\[8pt]
& \Pr(N=n) = p^{n-1}q \text{ for } n = 1,2,3,\ldots \\
& \text{where $p+q=1,$ $p,q>0$} \\
& \text{and $N$ is independent of } T_1,T_2,T_3,\ldots \\
& \text{so that } \operatorname E(N) = 1/q. \\[8pt]
& T= \sum_{n\,=\,1}^N T_n \\
& \text{so } \operatorname E(T) = \mu/q.
\end{align}
An exercise is to show that the distribution of $T$ is exponential: $e^{-qt/\mu} (q\,dt/\mu) \text{ for } t>0.$

MY QUESTION IS whether this is somehow a standard exercise known to everyone except me, or more literally: Did I just somehow miss that this one is in various textbooks? And while I'm at it, do others solve this differently from the way I did (I'll post something below)?


问:
How to correctly understand a Corollary of the Metatheorem on Constants in predicate logic
In Lectures in Logic and Set Theory(Vol.2) by George Tourlakis, there is a metatheorem as follows:

I cannot see intuitively why the $x_1$ that occurs free in $A[e_1,...,e_n]$ does not matter. For example, let $A[e_1,...,e_n] \equiv x_1=e_1\rightarrow P(e_1, e_2)$, then $A[y_1,...,y_n] \equiv x_1=y_1\rightarrow P(y_1, y_2)$ and $A[x_1,...,x_n] \equiv x_1=x_1\rightarrow P(x_1, x_2)$. Doesn't the $x_1$ introduced in the final step interfere with the original one?

答:
The issue is that there are no issues with the "identification" of free variables.

Consider the following simple example: $\Gamma \vdash \mathscr A(x,y)$.

Using Coroll.I.4.9 we have that $\Gamma \vdash \forall x \forall y \mathscr A(x,y)$, and thus (instantiating): $\Gamma \vdash \mathscr A(z,z)$.

Now, by a pair of similar moves, we get:

$\Gamma \vdash \mathscr A(x,x)$.

The key-point is convincing yourself (maybe semantically) that the above Corollary is sound.


问:
Proving that $7n^{2}\log_{6}n + 12n +7$ is Big O of $3n^{3} - 4n^{2} + 12$
Been banging my head against this problem for an hour now. The best I've been able to come up with is
this, but I feel like I'm still missing something. Am I on the right track? How should I approach this question?</p>

答: You only need concern yourself with the dominant terms.  For large $n$, $$7n^{2}\log_{6}n + 12n +7\leq 8n^3\leq 3(3n^{3} - 4n^{2} + 12)$$


问:
The sums of squares from standard deviation
I am trying to figure out how to get a sum of squares from a standard deviation. The standard deviations I have are $11.04$, $9.91$ and $9.43$. How do I calculate the sum of squares associated with these standard deviation values? Thanks!

答:
It holds that
$$
\sigma^2 = \frac{1}{n}\sum_{i=1}^n (x_i-\bar{x})^2 = \Big[\frac{1}{n}\sum_{i=1}^n x_i^2\Big] - (\bar{x})^2
$$

hence we have
$$
\sum_{i=1}^n x_i^2\ = n[\sigma^2+(\bar{x})^2].
$$

Note that $\bar{x}$ is arithmetic mean and $n$ is number of observation.


问:
Equation for area of ellipse given length of major axis and position of focus along the major axis?
I'm solving a physics question where the nearest and furthest distance of a satellite from Earth are given, along with the satellite's speed at the nearest point, and it asks to find the period of the satellite's orbit. In solving the question, the answer uses the following equation for the area of an ellipse:

$$S=\frac{\pi}{2}(\overline{F_1V_1}+\overline{F_1V_2})\sqrt{\overline{F_1V_1}\cdot\overline{F_1V_2}}$$

I have never seen this equation before, and searches did not return any results.

Does anyone know a derivation for this equation?

答:
Your description of $a,b$ doesn't match the diagram.  Anyway, I think you don't want $F_1,F_2$ (which would make your $a=b$) but a single focus $F$ and measure the distances $V_1,V_2$ from it, say $\ell_1,\ell_2$.

Then $\ell_1+\ell_2=2a$ and $\ell_1\ell_2=(a+OF)(a-OF)=b^2$ and we recover the usual formula for the area $\pi ab$ of an ellipse with semimajor and semiminor axes $a,b$.


Title: How can $4^x = 4^{400}+4^{400}+4^{400}+4^{400}$ have the solution $x=401$?

Body:
How can $4^x = 4^{400} + 4^{400} + 4^{400} + 4^{400}$ have the solution $x = 401$?

Can someone explain to me how this works in a simple way?

Body: $$4^{401}=4^{400+1}=4^{400}\cdot 4^1=4\cdot 4^{400}$$


答: When $n$ is sufficiently large, we have $2n^3\leq 3n^3-4n^2+12\leq 4n^3$, and $6n^2\log_6n\leq 7n^2\log_6n+12n+7\leq 8n^2\log_6n$. So we just need to verify that $n^2\log_6n$ is Big O of $n^3$. Given that $\log n$ is Big O of $n$, this is true.

Title: Proving that $7n^{2}\log_{6}n + 12n +7$ is Big O of $3n^{3} - 4n^{2} + 12$

Body:
Been banging my head against this problem for an hour now. The best I've been able to come up with is
this, but I feel like I'm still missing something. Am I on the right track? How should I approach this question?</p>


问:
Given a collection of curves, find the differential equation associated with them
The question is as follows:

$$
\text { 1) } y=\frac{\sec x}{c_{1}+c_{2} \tan x} \quad ; \quad \underline{\text { The Answer is  }}: y y^{\prime \prime}-2 y^{\prime 2}-y^{2}=0
$$

My main problem with this question is the long algebraic process. Is there any fast way to solve this question?

One possible way of solving this is to form a matrix involving y', y'', and y and try to make a system of equations with that matrix. Then because the equation should have a non-zero answer, we set the determinant to zero.

This is algebraically time consuming and provides the possibility for mistakes.

答:
Write it as $C_1+C_2 \tan x=\frac{\sec x}{y}$. D. W.r. t. $x$
to get $$C_2 \sec^2 x=\frac{y \sec x \tan x-\sec x ~~y'}{y^2}$$
$$C_2=\frac{y \sin x-y'\cos x }{y^2}$$
Again D.w.t. $x$, we get
$$0=\frac{y^2(y' \sin x +y \cos x-y'' \cos x+y' \sin x)-(y
 \sin x-y' \cos x)~2yy'}{y^4}$$
$$\implies \frac{\cos x(y^2-yy''+2y'^2)}{y^3}=0 \implies y^2-yy''+2y'^2=0$$

One should actually be thankful that one gets it by hand irrespective of the length of the calculation.


问:
Show triangle equality implies colinearly
Let $X$ be a normed vector space and suppose the following holds;

If $f \in X^*$ is a nonzero bounded linear functional then there exists
at most one element $x \in X$ such that $||x|| = 1$ and $f(x)=||f||$ (This is supposed to be equivalent to strict convexity).

I want to show that if $x \neq 0 \neq y$ and $||x+y||=||x||+||y||$ then $x$ and $y$ are colinear.

It feels like I'm supposed to use Hahn-Banach here (how else would I get a functional?).

I tried the obvious things like $M=\mathbb{F}y$ or $\mathbb{F}x+\mathbb{F}y$ and defining some functional like $f(cy)=c||y||$ but it doesn't work out.

EDIT:
Since it hasn't been commented on yet I guess its not trivial (or maybe so trivial that its not interesting). Ill share some more details;

The most naïve approach is to define $f:\mathbb{F}y \rightarrow \mathbb{F}$ by $f(cy)=c||y||$ which meets H.B conditions (with $p(x)=||x||\geq f(x)$) so we can extend it to a functional on X, call the extension $F$.

set $v=y/||y||$ and we see $||F(x)||<||x||$ (since this was our seminorm) so $||F|| \leq 1$  and $|F(v)|=f(v)=1$ so $||v||=||F||=1$ and we are done if we can show $F(\frac{x}{||x||})=1$ somehow using $||x||+||y||=||x+y||$ (this is where I"m stuck).

The problem is from https://people.math.ethz.ch/~salamon/PREPRINTS/funcana-ams.pdf on page 106 (118 of the PDF).

答:
First, we show that $X$ is strictly convex, that is, for any $x\neq y \in X$ with $||x||=||y||=1$, we get that $||\frac{x+y}{2}|| <1$. Indeed, suppose  that $X$ is not strictly convex and so there exist $x \neq y \in X$ with $||x||=||y||=1 $ and $||\frac{x+y}{2}|| =1$. By the Hahn-Banach theorem, we can find $f \in X^*$ such that $f(\frac{x+y}{2}) = ||\frac{x+y}{2}|| =1 $ and $||f||=1$. Since $f$ is linear, we infer that
$$\frac{f(x)+f(y)}{2} =1 .$$
Since $f(x) \leq 1 $ and $f(y) \leq 1$, it must be true that $f(x)=1$ and $f(y)=1$. By the hypothesis, we get that $ x=y$ which is impossible.

We now show that if $X$ is strictly convex, then for any $x_1,x_2 \in X$, if $||x_1+x_2||=||x_1||+||x_2||$, then $x_1=ax_0$ for some $a>0$. Let $x_1,x_2 \in X$ with $||x_1|| +||x_2||=||x_1+x_2||$ . Without loss of generality, we may assume that $ ||x_1||=1$. Let $ y= x_2/||x_2||$. Then,
\begin{align*}
    2 & \geq  ||{x_1+y}||   \\
    &= ||x_1+x_2 - (1- ||x_2||^{-1} ) x_2|| \\
    &\geq ||x_1+x_2|| - (1- ||x_2||^{-1}) ||x_2|| \\
    &= ||{x_1}||+||{x_2}||-||{x_2}||+1\\
    &=2
   \end{align*}
and so $|| \frac{x_1+y}{2}=1|| $. Since $X$ is strictly convex, we infer that $x_1=y_1= ||x_2||^{-1} x_2$.


问:
Convergent sequence/series proof
Let $\{x_n\}$ for $n \geq 1$ be a sequence of positive real numbers and let $s_n = \frac{x_n}{n}$. Let $\{y_n\}$ for $n \geq 1$ be another sequence such that $y_n = n^{-1} \sum_{i = 1}^{n} x_i$. Prove that if $\sum_{j = 1}^{\infty} s_j$ converges, then $\{y_n\}$ converges to $0$.

A proof I was shown:

Let $S_n = \sum_{j=1}^{n} s_j = \sum_{j=1}^n \frac{x_j}{j}$. Because it converges there is some $S$ such that $\lim_{n \rightarrow \infty} S_n = \sum_{j=1}^{\infty} \frac{x_j}{j} = S$.

We can also write:

$$y_n  =n^{-1}\sum_{k=1}^n x_k = n^{-1} \sum_{k=1}^nk \frac{x_k}{k} = n^{-1} \sum_{k=1}^nk s_k$$

Using summation by parts, with $S_0 := 0$,

$$\begin{align*}y_n &= n^{-1} \sum_{k=1}^nk(S_k- S_{k-1})\\
&= n^{-1}\left(nS_n + \sum_{k=1}^{n-1}S_k (k - (k+1)) \right)\\
&= S_n - n^{-1}\sum_{k=1}^{n-1}S_k
\end{align*}$$

We need to show $n^{-1}\sum_{k=1}^{n-1}S_k \to S$ as $n \to \infty$ when $S_n \to S$.

Since $S_n$ converges to $S$, there is some $N \in \mathbb{N}$ such that for $n \geq N$

$$\begin{align*}
\left|n^{-1} \sum_{k=1}^n (S_k- S_{k-1})\right| &\leq n^{-1} \sum_{k=1}^N |S_k- S_{k-1}| \\
&\qquad+ n^{-1} \sum_{k=N + 1}^n |S_k- S_{k-1}|\\
&= n^{-1} \sum_{k=1}^N |S_k- S_{k-1}| + \epsilon\left(1 - \frac{N}{n}\right)
\end{align*}$$

Indeed by taking the limsup for both sides we see that for any $\epsilon > 0$

$$0 \leq \limsup_{n \rightarrow \infty} \left|n^{-1} \sum_{k=1}^n (S_k- S_{k-1})\right| \leq \epsilon$$

By squeeze theorem,

$$\lim_{n \rightarrow \infty} \left|n^{-1} \sum_{k=1}^n (S_k- S_{k-1})\right| = \limsup_{n \rightarrow \infty} \left|n^{-1} \sum_{k=1}^n (S_k- S_{k-1})\right| = 0$$

It then follows that $y_n \to S-S=0$ as $n \to \infty$.

QED.

Is this proof correct? If so, could somebody provide some intuition about it. Any insight much appreciated.

答:
It’s not quite correct, but it can be fixed. The first part is correct, but in case you’re not accustomed to summation by parts, I’ll note that you don’t need that technique:

$$\begin{align*}
\sum_{k=1}^nks_k&=\sum_{k=1}^n\sum_{\ell=1}^ks_k=\sum_{\ell=1}^n\sum_{k=\ell}^ns_k\\
&=\sum_{\ell=1}^n(S_n-S_{\ell-1})=nS_n-\sum_{\ell=1}^nS_{\ell-1}\tag{1}\\
&=nS_n-\sum_{k=1}^{n-1}S_k\,,
\end{align*}$$

since $S_0=0$, and that gives us

$$y_n=S_n-n^{-1}\sum\limits_{k=1}^{n-1}S_k\,.\tag{2}$$

The one sequence whose convergence we do know is the sequence of partial sums $S_n$ of the series $\sum_{k\ge 1}s_k$, so expressing the numbers $y_n$ in terms of those partial sums is a natural thing to try, and $(2)$ accomplishes it. Once we get that, it’s clear that we want $S_n-n^{-1}\sum_{k=1}^{n-1}S_k$ to approach $0$ as $n\to\infty$.

We don’t know much about $\sum_{k=1}^{n-1}S_k$, but we do know that when $n$ and $k$ are large, $|S_n-S_k|$ is small, and the first summation in line $(1)$ shows that

$$y_n=n^{-1}\sum_{k=1}^n(S_n-S_k)\,,$$

and if $n$ is large enough, ‘most’ of the terms in the summation ought to be small. Let $\epsilon>0$. There is an $N\in\Bbb N$ such that $|S_n-S_k|<\epsilon$ whenever $n,k\ge N$, so for $n\ge N$ we have

$$\begin{align*}
|y_n|&=\left|n^{-1}\sum_{k=1}^n(S_n-S_k)\right|\\
&\le n^{-1}\sum_{k=1}^N|S_n-S_k|+n^{-1}\sum_{k=N+1}^n|S_n-S_k|\\
&<n^{-1}\sum_{k=1}^N|S_n-S_k|+\left(1-\frac{N}n\right)\epsilon\,.
\end{align*}$$

It follows that $\lim\limits_{n\to\infty}|y_n|\le\epsilon$ for all $\epsilon>0$ and hence that $\lim\limits_{n\to\infty}y_n=0$.

Title: Convergent sequence/series proof


问:
Does this graph already have a name?
I am studying the Cartesian product of such a graph with other graphs. I want to ask if it has a known name?

By the way, for some special graphs, is there a good website to check?

答: Apparently (according to this website) this graph is called "X84", or "g6: ElD?" (whatever that means). Some details about this nomenclature have been helpfully provided in David Scholz's answer.</p>

Title: Does this graph already have a name?


Body:
It only works because $x=401$.

$4^{401}$ means multiply $4$ by itself 401 times.  So $$\begin{array}{rl}
4^{401} &= 4\times 4^{400} \\ &= 4^{400} +4^{400} +4^{400} +4^{400} \end{array}$$


问:
Is this sketch work correct way to prove?
If $$S\subseteq T\subseteq R$$ where S is non empty, then show that : if T is bounded above, then $$SupS\leqslant SupT$$.

Sketch: Assume t to be supremum of T. Then $$x\leqslant t$$ for all x belonging to T.

Let p be supremum of S.
Then $$s\leqslant p$$ for all s belonging to S.

Since s,p belongs to S, they also belong to T.
Now if p>t, then t is not supremum of T(by definition).
This implies p<t. If p=t( trivial).
Thus $$s\leqslant p\leqslant t$$

答:
You are assuming that the supremum of a set must belong to that set. That is not true in general.

If $s\in S$, then $s\in T$, and therefore $s\leqslant\sup T$. So, $\sup T$ is an upper bound of $S$. Since $\sup S$ is, by definition, the least upper bound of $S$, it follows that $\sup S\leqslant\sup T$.


答: From $x^2 + px + q = a(x+1)(x-4) = x^2 -3x -4$ and from comparison of the coefficients follows $a = 1$, $p = -3$ and $q = -4$.

Title: Given $x^2 +px + q$ has roots -1 & 4, find the values for p & q.

Body:
Given $x^2 +px + q$ has roots $-1$ & $4$, find the values for $p$ & $q$.

Attempt:
$$
x = \frac{-p\pm\sqrt{p^2-4q}}{2}\\(p + 2x)^2 = p^2 - 4q\\
p^2 + 4px + 4x^2 = p^2 -4q
\\
q = -x^2 - px
\\
q = -(-1)^2 -p(-1)
\\
q = -1 - p
\\
q = -(4)^2-p(4)
\\
q = -16 - 4p
\\
-1 - p = - 16 - 4p
\\
3p = -15
\\
p = -5
\\
q = -1 - (-5)
\\
q = 4
$$

According to the book the answer should be $-3$ & $-4$


问:
Prove the existence of the limit using fixed point theorem
Consider the sequence $x_{n+1}=x_n-\frac{x_n^2-a}{2}$, prove that $\forall a \in (0,1) $ the sequence converges to $\sqrt{a}$.

Considering a function $f(x)=x-\frac{x^2-a}{2}$ we have $$f'(x) = 1 - x,$$ which means the function is increasing $\forall x < 1$.
Next if we look at what happnes if $x<1$, we have $$f(x) \le f(1) = 1 - \frac{1-a}{2} = \frac{1}{2} +\frac{a}{2} <1.$$
But this is where I got stuck. I can't figure out such $Q$ that $f:Q \rightarrow Q$ and moreover I can't show that $f(x)$ is a contractor since Lagrange's theorem yields nothing useful:
$$|f(x) - f(y)| = |1-\xi |* |x-y|.$$
But I don't see a way of bounding $|1-\xi|$ above especially not knowing the domain $Q$.
Without those 2 conditions I can't use the fixed point theorem which allows me to find the limit by simply solving the following equation:
$$x=x-\frac{x^2-a}{2}$$ which sure enough yields $x^*=\sqrt{a}$.

How would one approach this ?

答:
To get a contraction mapping, what is required is $$|f(x)-f(y)|=|x-y-\frac{x^2-y^2}{2}|=\left|1-\frac{x+y}{2}\right||x-y|\le c|x-y|$$ with $0\le c<1$. So one condition is $$1-c\le\frac{x+y}{2}\le 1+c\tag{1}$$ Let $Q=[\alpha,\beta]$, then the contraction condition would be ensured if $1-c\le\alpha$ and $\beta\le 1+c$.

The other condition is $f:Q\to Q$, that is $$\alpha\le x-\frac{x^2-a}{2}\le \beta\tag{2}$$ Given that $a\in(0,1)$ and $x-\frac{x^2}{2}\le\frac{1}{2}$, one can pick $\alpha=a/2$, $\beta=1$, to ensure both $(1)$ and $(2)$.


答: First remember that $4^x$ means you multiply $4$ together with itself $x$ times. For example $4^2 = 4*4= 16.$ Note that $4^{400} + 4^{400}+4^{400}+4^{400} = 4^{400}(1+1+1+1) = 4^{400}*4$. And $4^{400}*4$ is just $4$ multiplied together with itself 401 times, which is the same as writing $4^{401}$.</p>


问:
factor $1/(x\pm y)$ as $f(x)g(y)$
Is it possible to rewrite
$$\frac{1}{x\pm y}$$
as multiplication of an expression containing only $x$ and another containing only $y$ ($x$ and $y$ are real independent variables), i.e.
$$\frac{1}{x\pm y}\stackrel{?}{=}f(x)g(y)$$
If it is possible, what are $f(x)$ and $g(y)$?

答:
Sure it's possible. Take $A=\frac{a}{a+b}$ and take $B=0$. Then it should be trivial to show that $$\frac Aa + \frac Bb = \frac{\frac{a}{a+b}}{a} + \frac0b = \frac{1}{a+b}.$$

In fact, taking any value of $B$, you can set $A=a\cdot\left(\frac{1}{a+b} - \frac Bb\right)$ and get the equality
$$\frac{A}{a} + \frac Bb = \frac1{a+b}.$$

However, if you demand that $a,b, A, B$ are all integers, then the answer is no. A simple counterexample can be found by taking $a=b=1$. Then, $\frac{1}{a+b}=\frac12$, while $\frac Aa + \frac Bb = \frac A1 + \frac B1 = A + B$, and since $A$ and $B$ are integers, this means $\frac Aa + \frac Bb$ must also be an integer.

Since $\frac12$ is not an integer, it's clear that $\frac Aa + \frac Bb$ cannot be equal to $\frac1{a+b}$.

Title: factor $1/(x\pm y)$ as $f(x)g(y)$


问:
Completeness of $(C_c^k(\mathbb R^n),\left\|\cdot\right\|_{k,\mathbb R^n})$.
It's well-known that $(C^k(\overline{\Omega}),\left\|\cdot\right\|_{k,\overline{\Omega}})$ is a Banach space if $\Omega\subset \mathbb R^n$ is open, connected and bounded.
I want to study some properties of the unbounded situation $(C_c^k(\mathbb R^n),\left\|\cdot\right\|_{k,\mathbb R^n})$.
Here is my problem (I'm not sure whether this is true):

Define the $k$-norm $\left\|\cdot\right\|_{k,\mathbb R^n}$ for $C_c^k(\mathbb R^n)$ by
\begin{align*}
  \left\|u\right\|_{k,\mathbb R^n} =\max_{|\alpha|\le k}\max_{x\in\mathbb R^n} |\partial^\alpha u(x)|
  \quad \forall u\in C_c^k(\mathbb R^n)
 \end{align*}
and we extend $\left\|\cdot\right\|_{k,\mathbb R^n}$ to $C^k(\mathbb R^n)$ by
\begin{align*}
  \left\|\cdot\right\|_{k,\mathbb R^n}: C^k(\mathbb R^n)&\to [0,+\infty],
  \quad u\mapsto \max_{|\alpha|\le k}\sup_{x\in\mathbb R^n} |\partial^\alpha u(x)|
 \end{align*}
Let $(f_i)_{i\ge 1}$ be a Cauchy sequence of functions in $(C_c^k(\mathbb R^n),\left\|\cdot\right\|_{k,\mathbb R^n})$.
Then there exists $f\in C^k(\mathbb R^n)$ such that
\begin{align*}
  \lim_{n\to\infty} f_n(x)=f(x) \quad \forall x\in \mathbb R^n
 \end{align*}
Furthermore,
\begin{align}
  \lim_{n\to\infty}\left\|f_n-f\right\|_{k,\mathbb R^n}=0 \tag{1}
 \end{align}

My attempy

It's obvious that $\lim_{n\to\infty} f_n(x)$ exists for each $x\in \mathbb R^n$, and we put
\begin{align*}
  f(x)=\lim_{n\to\infty} f_n(x)\quad \forall x\in \mathbb R^n
 \end{align*}
In the next we prove that $f\in C^k(\mathbb R^n)$.
It suffices to prove that for each ball $B\subset \mathbb R^n$, $f\in C^k(\overline B)$.
Similarly, we define the $k$-norm $\left\|\cdot\right\|_{k,\overline{B} }$ for $C^k(\overline B)$ by
\begin{align*}
  \left\|u\right\|_{k,\overline{B} } =\max_{|\alpha|\le k}\max_{x\in \overline{B}} |\partial^\alpha u(x)|
  \quad \forall u\in C^k(\overline B)
 \end{align*}
Note that $(f_i|_{\overline B})$ is a Cauchy sequence in $(C^k(\overline{B}),\left\|\cdot\right\|_{k,\overline{B}})$, and
hence via the completeness we know that $(f_i|_{\overline B})$ converges to $f|_{\overline{B}}$ in
$(C^k(\overline{B}),\left\|\cdot\right\|_{k,\overline{B}})$ and that $f\in C^k(\overline B)$.

It seems well to prove the conclusion except (1). But I got stuck on (1), and I'm not sure whether (1) is true.

I would highly appreciate it if you could share any thoughts on how to solve this problem. Thanks in advance!

答:
The proof of (1) does not differ from proofs of completeness of $C_0(X)$.
Take $\epsilon>0$. Then there is $N$ such that for all $m,n>N$
$$
|f_n(x)-f_m(x)|\le \epsilon.
$$
Now let $m\to\infty$. This implies
$$
|f(x)-f_m(x)|\le \epsilon
$$
for all $m>N$.


问:
Is there a tight bound on following binomial summations involving squares on arithmetic progressions?
The summations of interest is following:

$$\sum_{i=0}^{\lfloor\sqrt n\rfloor}\binom{n}{i^2}$$
$$\sum_{i\in\{a,q+a,2q+a\dots,\lfloor\sqrt n\rfloor\}}\binom{n}{i^2}$$
where $q<n$ and $a\in\{0,1,\dots,q-1\}.$

Is there a tight asymptotic bound on these?

答:
As stated under the OEIS entry $\texttt{A003099}$ mentioned in comments, the quantity
$$f(n)=\frac{\sqrt n}{2^n}\sum_{k\geqslant 0}\binom{n}{k^2}$$

stays bounded, in $(\approx 0.587,\approx 0.827)$ for large $n$. More precisely, one obtains

For any $x\in\mathbb{R}$ we have $\displaystyle\color{blue}{\lim_{n\to\infty}f\big(\lfloor2(n+x)^2\rfloor\big)=\sqrt\frac2\pi\sum_{m\in\mathbb{Z}}e^{-4(m-x)^2}}$.

The idea is basically to put $k=n+m$ above, and use the following consequence $$\lim_{n\to\infty}\frac{\sqrt{2n(n+a)+b}}{2^{2n(n+a)+b}}\binom{2n(n+a)+b}{(n+m)^2}=\sqrt\frac2\pi e^{-(2m-a)^2}$$ of Stirling's formula. Thus, the exact values of the bounds are
\begin{align*}\limsup_{n\to\infty}f(n)&=\sqrt\frac2\pi\sum_{m\in\mathbb{Z}}e^{-(2m)^2}&={0.827112271364145742192103543257}\cdots\\
\liminf_{n\to\infty}f(n)&=\sqrt\frac2\pi\sum_{m\in\mathbb{Z}}e^{-(2m-1)^2}&={0.587247586271786487501375771221}\cdots\end{align*}

I'm sure that the second (more general) sum in the question can be handled similarly.


问:
Find the $n$th derivative for the function $y=\sin(kx)$, $n\in\mathbb{R}$.
I created a question for a high school calculus exam:

Find the $n$th derivative, $\dfrac{d^ny}{dx^n}$, for the function $y=\sin(kx)$, $n\in\mathbb{N}$.

The solution:
\begin{align}
\dfrac{d}{dx}\sin(kx)&=k\cos(kx) =k\sin k(x+\frac{\pi}{2})\\
\dfrac{d^2}{dx^2}\sin(kx)&=\dfrac{d}{dx}k\sin k(x+\frac{\pi}{2})=k^2\cos k(x+\frac{\pi}{2})=k^2\sin k(x+\pi)\\
\vdots \\
\dfrac{d^n}{dx^n}\sin(kx)&=k^n\sin k(x+\frac{n\pi}{2})
\end{align}

One of my students had asked me "Why did you include $n\in\mathbb{N}$ in the question?  Does that mean it's possible to be asked for $n\in\mathbb{R}$ later on?"

I'm aware of the existence of the fractional derivative, but I'm unable to give it any meaning or explain it.  So the problem I want to find a solution for is (and I'm totally aware that this is well beyond any high school calculus content):

Find the $n$th derivative for the function $y=\sin(kx)$, $n\in\mathbb{R}$.

BONUS:  Find the $n$th derivative for the function $y=x^2e^x$, $n\in\mathbb{R}$.

答:
There is no reason whatsoever to use a complicated formulas instantly. Especially not from the educational point of view. Fractional derivative is a natural extension of integer derivative, and there is not much to say there.

If $f^{(n)}(x)=g(x,n), n \in \mathbb{N}$ is $n^{th}$ derivative of $f(x)$, then the fractional derivative of $f(x)$, in various notations, is $ \left ( \mathbb{D}^r \right ) f(x)=\frac{d^r}{dx^r}f(x)=f^{(r)}(x)=g^*(x,r), r \in \mathbb{R}$. where $g^*$ is an extension of $g$ from integers to reals.

(We could use the same and define other derivatives like complex ones.)

The only purpose of deriving a more general formula is to have a proof that such derivative is mathematically sounded object. But for ordinary functions, it is as simple as the above statement. This is because the fractional derivative is defined as an extension itself, albeit a more generic one that uses the least possible amount of ammunition.

(The only question here is which $g^*$ to use and if it is sufficiently generic.)

From the educational point of view, this shows the direction we want to go, and it is very logical and substantial. After this explanation, it is much more clear why we need to introduce a more generic way of treating all functions the same way, rather than trying to guess in each particular situation, for more complicated functions, what this extending function to reals contains.

Therefore

$$\left( \mathbb{D}^r \right ) \sin(kx)=k^r\sin(kx+\frac{r\pi}{2})$$

$$\left( \mathbb{D}^r \right ) x^2e^x=e^x((x+r)^2-r)$$

In-depth reasoning: in more generic formulae, fractional derivative, in general, is continuous over reals, there is nothing in particular that makes integers different from reals and the only needed extension is, for example, a trivial one, like Gamma function, i.e. factorial.

$$\left( \mathbb{D}^r \right ) f( x )  = \frac1{ \Gamma ( r ) } \int_0^x \left(x-t\right)^{r-1} f(t) \, dt$$

Regarding generality we talked about, this apparently trivial one is a good illustration.

It is not true that $\left( \mathbb{D}^r \right ) 1 = 0$. Meaning, we need to start from a more generic form before we can define the situation.

$$ \left( \mathbb{D}^r \right ) x^k = \frac{d^r}{dx^r}x^k=\frac{k!}{(k-r)!}x^{k-r} $$

and now for $k=0$

$$ \frac{d^r}{dx^r}x^0=\frac{d^r}{dx^r}1=\frac{0!}{(0-r)!}x^{0-r}=\frac1{(-r)!}x^{-r}=\frac1{\Gamma(1-r)}x^{-r} $$

(There is some strange belief that we are allowed to introduce a "simple" mathematical object in a simple way, yet more intricate in a complex way. The truth is that all mathematical objects are complex if we want them to be complex, and if humanity needed to learn these starting from simpler notions, why would a 21st century kid have to be smarter than geniuses of old and digest all that at once? No, the fact that we know it does not make it easier to digest. As a student, I would not say a word if you would introduce fractional derivatives as above, but I would object a direct introduction into formulas I can barely decipher after first view.)


问:
Sum of a p-series with coefficients
I am trying t find an expression for the partial sum of a p-like-series.  The problem comes from trying to sum the elements of a matrix whose entries are inversely related to their distance from the diagonal.

$$
\begin{matrix}
k^{-\alpha} & (k+1)^{-\alpha} & (k+2)^{-\alpha} & (k+3)^{-\alpha} & (k+4)^{-\alpha} & (k+5)^{-\alpha} & (k+6)^{-\alpha} & (k+7)^{-\alpha}\\
  & k^{-\alpha} & (k+1)^{-\alpha} & (k+2)^{-\alpha} & (k+3)^{-\alpha} & (k+4)^{-\alpha} & (k+5)^{-\alpha} & (k+6)^{-\alpha} \\
 & & k^{-\alpha} & (k+1)^{-\alpha} & (k+2)^{-\alpha} & (k+3)^{-\alpha} & (k+4)^{-\alpha} & (k+5)^{-\alpha} \\
& & & k^{-\alpha} & (k+1)^{-\alpha} & (k+2)^{-\alpha} & (k+3)^{-\alpha} & (k+4)^{-\alpha} & \\
& & & & k^{-\alpha} & (k+1)^{-\alpha} & (k+2)^{-\alpha} & (k+3)^{-\alpha} \\
& & & & & k^{-\alpha} & (k+1)^{-\alpha} & (k+2)^{-\alpha} \\
& & & & & & k^{-\alpha} & (k+1)^{-\alpha} \\
& & & & & & & k^{-\alpha}
\end{matrix}
$$

Now, if I divide this matrix into 2 $\times$ 2 squares and sum the elements, I get

$$
\tiny
\begin{matrix}
2k^{-\alpha} + (k+1)^{-\alpha} & (k+1)^{-\alpha} + 2(k+2)^{-\alpha} + (k+3)^{-\alpha} & (k+3)^{-\alpha} + 2(k+4)^{-\alpha} + (k+5)^{-\alpha} & (k+5)^{-\alpha} + 2(k+6)^{-\alpha} + (k+7)^{-\alpha} \\
& 2k^{-\alpha} + (k+1)^{-\alpha} & (k+1)^{-\alpha} + 2(k+2)^{-\alpha} + (k+3)^{-\alpha} & (k+3)^{-\alpha} + 2(k+4)^{-\alpha} + (k+5)^{-\alpha} \\
& & 2k^{-\alpha} + (k+1)^{-\alpha} & (k+1)^{-\alpha} + 2(k+2)^{-\alpha} + (k+3)^{-\alpha} \\
& & & 2k^{-\alpha} + (k+1)^{-\alpha}
\end{matrix}
\normalsize
$$

If I repeat the process one more time we get:

$$
\tiny
\begin{matrix}
4k^{-\alpha} + 3(k+1)^{-\alpha} + 2(k+2)^{-\alpha} + (k+3)^{-\alpha} & (k+1)^{-\alpha} + 2(k+2)^{-\alpha} + 3(k+3)^{-\alpha} + 4(k+4)^{-\alpha} + 3(k+5)^{-\alpha} + 2(k+6)^{-\alpha} + (k+7)^{-\alpha}  \\
 & 4k^{-\alpha} + 3(k+1)^{-\alpha} + 2(k+2)^{-\alpha} + (k+3)^{-\alpha}
\end{matrix}
\normalsize
$$

The pattern which emerges is some sort of weighted p-series with strictly deceasing coefficients on the diagonal, and coefficients that increase and decrease on the off diagonals, starting at 1, then increasing to $\frac{n(n+1)}{2}$, then decreasing and returning to 1 again.

MY QUESTIONS:

Is there a name for such a series?

Is there an approximation for such a sum? (I found this post about approximating p-series sums, but the addition of the above coefficients complicate the issue)

Things to note:

$k$ is likely to be set to $0$

$\alpha \gt 1$

The above process is likely to be repeated up to 18 times on a very large matrix, so approximations to a large sum are welcome.

TO SUMMARISE

Let's set k = 0.

We have a square matrix with dimension $\propto 2^J$.

Let $r$ and $c$ represent the row and column, respectively.

Each cell equals $(c - r)^{-\alpha} * I(r < c)$

I have a square, upper triangular Toeplitz matrix

$$
\begin{matrix}
0 & 1^{-\alpha} & 2^{-\alpha} & 3^{-\alpha} & 4^{-\alpha} & 5^{-\alpha} & 6^{-\alpha} & 7^{-\alpha} & ...\\
  & 0 & 1^{-\alpha} & 2^{-\alpha} & 3^{-\alpha} & 4^{-\alpha} & 5^{-\alpha} & 6^{-\alpha} & ...\\
 & & 0 & 1^{-\alpha} & 2^{-\alpha} & 3^{-\alpha} & 4^{-\alpha} & 5^{-\alpha} & ... \\
& & & 0 & 1^{-\alpha} & 2^{-\alpha} & 3^{-\alpha} & 4^{-\alpha} & ...\\
& & & & 0 & 1^{-\alpha} & 2^{-\alpha} & 3^{-\alpha} & ...\\
& & & & & 0 & 1^{-\alpha} & 2^{-\alpha} & ...\\
& & & & & & 0 & 1^{-\alpha} & ...\\
& & & & & & & 0 & ... \\
& & & & & & & & \ddots
\end{matrix}
$$

I break the matrix into square bins of size $2^j$, $j \in {1,2,3,...., J}$

The number of diagonals within each square bin (and therefore the number of unique terms) is 2 x dimension - 1 = $2^{j+1} - 1$.

Since the coefficients start at 1, ascend to the midpoint, and then descend back to 1, the middle (and therefore maximum) coefficient is $\frac{(2^{j+1} - 1)+1}{2} = 2^j$.

The summands within each square bin change as they move further away from the diagonal.  If we consider the $b^{th}$ bin from the diagonal, $b \in {1,2,3,..., 2^{J-j}}$ then the maximum summand is $(b2^j - 1)$

The minimum summand is the maximum summand - # of diagonal elements + 1 = $(b2^j - 1) - (2^{j+1} - 1) + 1 = b2^j - 2^{j+1} + 1$

Therefore, the midpoint summand will be $\frac{max + min}{2} = \frac{(b2^j - 1)+(b2^j - 2^{j+1} + 1)}{2} = \frac{b2^{j+1} - 2^{j+1}}{2} = (b-1)*2^j$

Bringing this all together, we get:  (diagonal term) + (lower triangle terms) + (Upper triangle terms).

$$2^j((b-1)2^j)^{-\alpha} + \sum_{i=1}^{2^j - 1}(2^j -i)((b-1)2^j - i)^{-\alpha} + \sum_{i=1}^{2^j - 1}(2^j -i)((b-1)2^j + i)^{-\alpha}$$

Note that if we are on one of the "triangular" diagonal bins, the first two terms evaluate to ZERO, since $r \ge c$

$$= 2^j((b-1)2^j)^{-\alpha} + \sum_{i=1}^{2^j - 1}(2^j -i)((b-1)2^j - i)^{-\alpha} + \sum_{i=1}^{2^j - 1}(2^j -i)((b-1)2^j + i)^{-\alpha}$$

In order to make the above expression conform with Dr. Wolfgang Hintze's solution below, I simply need to change the index.

$\ell = (2^j - i) \rightarrow (i = 2^j - k)$

$$
\begin{align}
& 2^j((b-1)2^j)^{-\alpha} + \sum_{i=1}^{2^j - 1}(2^j -i)((b-1)2^j - i)^{-\alpha} + \sum_{i=1}^{2^j - 1}(2^j -i)((b-1)2^j + i)^{-\alpha} \\
& = 2^j((b-1)2^j)^{-\alpha} + \sum_{\ell=1}^{2^j - 1}\ell((b-1)2^j - (2^j - \ell))^{-\alpha} + \sum_{\ell=1}^{2^j - 1}\ell((b-1)2^j + (2^j - \ell))^{-\alpha} \\
& = 2^j((b-1)2^j)^{-\alpha} + \sum_{\ell=1}^{2^j - 1}\ell((b-2)2^j + \ell)^{-\alpha} + \sum_{\ell=1}^{2^j - 1}\ell(b2^j - \ell)^{-\alpha} \\
& = 2^j((b-1)2^j)^{-\alpha} + \sum_{\ell=1}^{2^j - 1}\ell(\ell + (b-2)2^j)^{-\alpha} + \sum_{\ell=1}^{2^j - 1}\ell(- \ell + b2^j)^{-\alpha} \\
\end{align}
$$

As a final pass, I also came up with the following expression by not changing the term in the final sum, instead simply distributing the power summand across the coefficient term:

$$
\begin{align}
2^j((b-1)2^j)^{-\alpha} &+ \sum_{\ell=1}^{2^j - 1}\ell(\ell + (b-2)2^j)^{-\alpha} \\
& + 2^j\sum_{i=1}^{2^j - 1}(i + (b-1)2^j)^{-\alpha} \\
& - \sum_{i=1}^{2^j - 1}i(i + (b-1)2^j)^{-\alpha}
\end{align}
$$

$$
\begin{align}
2^j((b-1)2^j)^{-\alpha} &+ s\left((b-2)2^j, \alpha, 1, 2^j - 1\right) \\
& + 2^j\left\{\zeta(\alpha, (b-1)2^j +1) - \zeta(\alpha, (b-2)2^j -1)\right\} \\
& - s\left((b-1)2^j, \alpha, 1, 2^j - 1\right)
\end{align}
$$

答:
I'm not sure if I understand your problem correctly but it seems that you are interested in sums of the type

$$s(k,\alpha,m,n)=\sum _{i=0}^n i^m (i+k)^{-\alpha }\tag{1}$$

For $m=0$ we have

$$s(k,\alpha,0,n)=\zeta (-a,k)-\zeta (-a,k+n+1)\tag{2}$$

where $\zeta (r,n)$ is the Hurwitz Zeta function.

For general integer $m$ it is useful to start from the genrating function

$$\begin{align}g(z)= & \sum _{i=0}^n z^i (i+k)^{-\alpha }\\=& \Phi (z,\alpha ,k)-z^{n+1} \Phi (z,\alpha ,k+n+1)
\end{align}\tag{3}$$

where $\Phi (z,\alpha ,k)$ is the Lerch $\Phi$ function.

You can generate the $m$-th power of $i$ by applying the operation $(z \frac{\partial}{\partial z})^m$ to $g(z)$ and letting $z\to 1$.

Since the derivative

$$z \frac{\partial \Phi (z,\alpha ,k)}{\partial z} = \Phi (z,\alpha -1,k)-k \Phi (z,\alpha ,k)\tag{4}$$

can also be expressed in terms of the Lerch $\Phi$-funcition we conclude that, the sum $(1)$ can be expressed in terms of the Lerch function $\Phi (z,\alpha ,k)$.

Title: Sum of a p-series with coefficients


问:
Number Theory $a$ coprime with 10
How do I prove that

$$a^{a^{a}} \equiv a^a \pmod{10} \quad \text{and}\quad a^{a^{a^a}} \equiv a^{a^a} \pmod{100}$$

where $a \in \mathbb{N}$ such that $a$ is coprime with 10. Thanks.

答:
By Euler's theorem,

$a^{4} \equiv 1 \pmod{10}$ since $\varphi(10)=4$

which will give you

$\left(a^{a}\right)^4=a^{4a}=\left(a^{4}\right)^a \equiv 1 \pmod{10}$

$\left(a^{a^a}\right)^{4}=a^{4a^a}=\left(a^{4}\right)^{a^a} \equiv 1 \pmod{10}$

$\Rightarrow \left(a^{a^a}\right)^{4} - \left(a^{a}\right)^4\equiv 0 \pmod{10}$

$\Rightarrow \left(\left(a^{a^a}\right)^{2} + \left(a^{a}\right)^2\right)\left(a^{a^a} + a^{a}\right)\left(a^{a^a} - a^{a}\right)\equiv 0 \pmod{10}$

From here can you argue that $\left(a^{a^a} - a^{a}\right)\equiv 0 \pmod{10}$
$\Rightarrow a^{a^a} \equiv a^{a} \pmod{10}$ ?

e.g., since $a$ is odd, we must have $\left(a^{a^a} - a^{a}\right)\equiv 0 \pmod{2}$, then all we have to show is $\left(a^{a^a} - a^{a}\right)\equiv 0 \pmod{5}$, in other words, we need to show $5 \nmid \left(\left(a^{a^a}\right)^{2} + \left(a^{a}\right)^2\right)\left(a^{a^a} + a^{a}\right)$.

also, since $(a,10) = 1$,

$a \equiv \pm1 \pmod{10}$ or

$a \equiv \pm 3 \pmod{10}$

for the first case, it's trivial to show that both $\left(a^{a^a}\right)^{2} + \left(a^{a}\right)^2 \not\equiv 0 \pmod{5}$ and $a^{a^a} + a^{a} \not\equiv 0 \pmod{5}$.

for the second case, let $a \equiv 3 \pmod{10}$

$\Rightarrow \left(a^{a^a}\right)^{2} + \left(a^{a}\right)^2 \equiv (9^{a^{a}} + 9^{a}) \pmod{5}\not \equiv 0 \pmod{5}$ since for an odd $a$, $9^a$ has the last digit as $9$ (so will have $9^{a^a}$, s.t. the last digit of $9^{a^{a}} + 9^{a}$ is $8$).

Also, note that $a \equiv \pm 1 \pmod{4}$, $a$ being odd.

Now $a \equiv 1 \pmod{4} \Rightarrow a^a \equiv 1 \pmod 4$ which implies that the last digit of both $3^{a^a}$ and $3^a$ is $3 \Rightarrow 3^{a^a} + 3^a \not \equiv 0 \pmod 5$ (since last digit of $3^{a^a}+3^a$ will be $6$).

Similarly, $a \equiv -1 \pmod{4}$ implies that last digit of both $3^{a^a}$ and $3^a$ is 7, i.e., the last digit of $3^{a^a}+3^a$ will be $4$, i.e., $3^{a^a}+3^a \not \equiv 0 \pmod 5$.

Combining the above two, we have for any odd $a$ with $a \equiv 3 \pmod {10}$, we have $a^{a^a}+a^a \equiv 3^{a^a}+3^a \pmod 5 \not \equiv 0 \pmod 5$

$\implies \left(\left(a^{a^a}\right)^{2} + \left(a^{a}\right)^2\right)\left(a^{a^a} + a^{a}\right) \not \equiv 0 \pmod 5$.

Hence, in both the cases, we have $5 \nmid \left(\left(a^{a^a}\right)^{2} + \left(a^{a}\right)^2\right)\left(a^{a^a} + a^{a}\right) $

$\Rightarrow \left(a^{a^a} - a^{a}\right)\equiv 0 \pmod{5}$.

EDIT

We could also start from Fermat's Little Theorem:

$a^{4} \equiv 1 \pmod{5}$

Title: Number Theory $a$ coprime with 10


问:
Should I prove AGM inequality with 4 variables with Induction or Cauchy-Schwarz?
I'm trying to prove the following: $\sqrt[4]{abcd} \le \frac{a+b+c+d}{4}$.

While I managed to prove base case $n=1$ and $n=2$;

$(\sqrt{a}-\sqrt{b})^2 \geqslant 0$

$a-2\sqrt{ab}+b \geqslant 0$

$a+b \geqslant 2\sqrt{ab}$

$\sqrt{ab} \le \frac{a+b}{2}$

I am unsure how to proceed for $n=4$. Is there a way to continue proving 4 variables using induction, or is it better to use Cauchy-Schwarz Inequality to go about the proof?

答:
Alternative approach.

Lemma-1

Given:

$x_1, x_2, \cdots, x_n \in \mathbb{R^+},~:~n > 1$.

It is not the case that the $x_i$ are all equal.

$\prod_{i=1}^n x_i = 1.$

To Prove: $\sum_{i=1}^n x_i > n.$

Proof by Induction

$n = 2:$

Without Loss of Generality (WLOG), $x_1 = 1 - k, ~0 < k < 1.$

Then $\displaystyle x_2 = \frac{1}{x_1}
= \frac{1}{1 - k} = 1 + k + \frac{k^2}{1 - k} \implies $
$\displaystyle x_1 + x_2 = 2 + \frac{k^2}{1 - k} > 2.$

$n = (N+1):$

Assume conjecture true when $n = N$.

Given $x_1, \cdots, x_N, x_{(N+1)} \in \mathbb{R+},~$
such that they are not all equal and $\prod_{i=1}^{(N+1)} x_i = 1.$

WLOG $~x_1 < 1 < x_{(N+1)}.$

Let $b = x_1 \times x_{(N+1)}.$

By induction hypothesis,

since $\displaystyle b \times \prod_{i=2}^N x_n = 1~$,
$~~b + \sum_{i=2}^N x_i > N.$

Edit

Minor flaw in the above analysis, which (fortunately) doesn't invalidate the analysis at the end of the proof to Lemma 1.

As $b, x_2, x_3, \cdots, x_N$ are defined, it is possible that

these $N$ elements are all equal to $1$.

Therefore, the correct intermediate conclusion is that

$b + \sum_{i=2}^N x_i \geq N.$

Since $~x_1 < 1 < x_{(N+1)},~~$
$0 > [x_1 - 1]~[x_{(N+1)} - 1] = -x_1 - x_{(N+1)} + 1 + b \implies$

$x_1 + x_{(N+1)} > 1 + b \implies \sum_{i=1}^{N+1} x_i > (N+1).$

Notes:

When proving that the geometric mean is greater than or equal to the arithmetic mean, you can assume that each of the numbers is non-negative.  Otherwise, the conjecture is false.  For example: the geometric mean of $(-4)$ and $(-9)$ will be construed to be either  $\pm 6$, both of which are greater than $\frac{(-4) + (-9)}{2}.$

Consequently, you can also assume that each of the numbers is non-zero, because otherwise the geometric mean would be $0$.  Therefore, by presumption, each of the numbers is positive.

The situation is trivial when all of the numbers are equal, since then the geometric mean and arithmetic mean are then clearly equal.  Therefore, WLOG, the numbers are all positive, and not all equal.

Given $\displaystyle x_1,x_2, \cdots, x_n, ~\text{let}~ G = \left(\prod_{i=1}^n x_i\right)^{(1/n)}.$

For $i \in \{1,2,\cdots,n\},~$ let
$\displaystyle a_i = \frac{x_i}{G}.$

Then, $\displaystyle\prod_{i=1}^n a_i = 1.$

Therefore, by Lemma-1, $\sum_{i=1}^n a_i > n.$

Thus $\displaystyle
\frac{x_1}{G} + \cdots + \frac{x_n}{G} > n \implies
\frac{x_1 + \cdots + x_n}{n} > G.$


问:
Can all elements of all rings be multiplied by positive integers?
Let's say we have an arbitrary ring $R$ and an element $a \in R$, and we have a positive integer $n \in ℤ^{+}$. What can we say about the multiplication operations $an$ and $na$? Do they exist in ring $R$? What is the proof? Are there any other properties of this operation? Since multiplication within a ring is associative, is multiplication of an arbitrary ring with members of $ℤ^{+}$ also associative? What about distributivity? Or commutativity if $R$ has commutative multiplication?

Context: I'm being asked to prove $m(na) = (mn)a$ where $a \in R$ and $m, n \in ℤ^{+}$, and I assume that both sides are $\in R$, but I don't know what the proof is for the first step of $na$ (and for the subsequent steps). I don't know what the rules are for operations that have one member inside a ring and one member outside of the ring.

答:
Here is how one can define a $\mathbb{Z}$-action on any ring $R$ (note that I included the negative integers) :
$$
\begin{aligned}
0\cdot a &= 0\\
n\cdot a &= \underbrace{a+\ldots+a}_n\quad n>0\\
n\cdot a &= \underbrace{(-a)+\ldots+(-a)}_{-n}\quad n<0
\end{aligned}$$
Note @saulspatz's alternative inductive definition in the comments. Then all your questions basically follow from the same arguments one could do about the multiplication in $\mathbb{Z}$ :

it is associative because addition in the ring is associative

it is distributive because addition in the ring is commutative

it is commutative (follow from the definition)

Note that these properties never use the multiplicative structure of the ring: the only thing needed is the abelian group structure. Actually, the above is exactly saying that an abelian group is the same as a $\mathbb{Z}$-module.


问:
How can $4^x = 4^{400}+4^{400}+4^{400}+4^{400}$ have the solution $x=401$?
How can $4^x = 4^{400} + 4^{400} + 4^{400} + 4^{400}$ have the solution $x = 401$?

Can someone explain to me how this works in a simple way?

Body:
$a=4^{400}$

$4^x=a+a+a+a$
$$4^x=4\cdot a$$
$$4^x=4\cdot 4^{400}$$
$$u^m\times u^n= u^{m+n}\implies 4^1\times 4^{400}=4^{401}$$
$$4^x=4^{400+1}$$
$$a^m=a^n \implies m=n$$
$$x=401$$


答:
By Euler's theorem,  you can reduce  the exponent $a^a\bmod4 $.  By Euler again, you can reduce the new exponent,  $a$, $\bmod2$.  But $a\equiv1\bmod2$.

Now for part two, apply Euler and you can reduce the exponent  $\bmod40$.  To do that, apply CRT.  So $\bmod5$ you reduce the new exponent, $a^a$, $\bmod4$.  And $\bmod8$, by Euler you reduce that exponent  $\bmod\phi (8)=4$.  In both cases you get $a $, as we saw in part one.

Title: Number Theory $a$ coprime with 10

Body:
How do I prove that

$$a^{a^{a}} \equiv a^a \pmod{10} \quad \text{and}\quad a^{a^{a^a}} \equiv a^{a^a} \pmod{100}$$

where $a \in \mathbb{N}$ such that $a$ is coprime with 10. Thanks.


问:
Find $\sum_{j=0}^{n}\sum_{i=j}^{n} {n \choose i}{i \choose j}$.
Find $\sum_{j=0}^{n}\sum_{i=j}^{n} {n \choose i}{i \choose j}$.

I don't know how to double summations like this very well. Can someone expand this to show how the $i=j$ thing works?

I tried the following:
${n \choose i}{i \choose j}={n \choose j}{n-j \choose i-j}$

$\sum_{j=0}^{n}\sum_{i=j}^{n} {n \choose i}{i \choose j}=\sum_{j=0}^{n}\sum_{i=j}^{n} {n \choose j}{n-j \choose i-j}=\sum_{j=0}^{n} {n \choose j}{n-j \choose 0}=2^n$

Where am I going wrong?

答:
$$
\begin{align}
\sum\limits_{i=j}^n\binom{n-j}{i-j}
&=\sum\limits_{i=0}^{n-j}\binom{n-j}{i}\tag1\\[3pt]
&=2^{n-j}\tag2
\end{align}
$$
Explanation:

$(1)$: subsitute $i\mapsto i+j$

$(2)$: evaluate $(1+1)^{n-j}$ with the Binomial Theorem

If step $(1)$ is confusing, break it into two steps:

$i\mapsto k+j$: since $i$ ranges from $j$ to $n$, $k=i-j$ ranges from $0$ to $n-j$

$k\mapsto i$: simply change the variable of summation back

Thus, this sum is $2^{n-j}$, not $\binom{n-j}{0}$.

Now, the rest is evaluating either
$$
2^n\sum_{j=0}^n\color{#C00}{\binom{n}{j}}2^{-j}=2^n\left(1+2^{-1}\right)^n\tag{3a}
$$
or
$$
\sum_{j=0}^n\color{#C00}{\binom{n}{n-j}}2^{n-j}=(1+2)^n\tag{3b}
$$
with the Binomial Theorem (the parts in red are equal).


答:
Change the order of summation (overall there are only finitely many summands). For that note that

\begin{align*}
M:&=\{ (i,j)\in\mathbb{N}^2~:~j=0,...,n~~i=j,...,n\}\\
 &= \{(i,j)\in \mathbb{N}^2~:~i=0,...,n~~j=0,...,i\}.
\end{align*}

Therefore we can change the representation of $M$ and obtain

\begin{align*}
\sum_{j=0}^n \sum_{i=j}^n {n\choose i}{i\choose j} &= \sum_{(i,j)\in M} {n \choose i}{i \choose j}= \sum_{i=0}^n \sum_{j=0}^i {n \choose i}{i \choose j}\\
&=  \sum_{i=0}^n {n \choose i} \sum_{j=0}^i {i\choose j} = \sum_{i=0}^n {n\choose i} 2^i = (1+2)^n = 3^n.
\end{align*}

Body:
Find $\sum_{j=0}^{n}\sum_{i=j}^{n} {n \choose i}{i \choose j}$.

I don't know how to double summations like this very well. Can someone expand this to show how the $i=j$ thing works?

I tried the following:
${n \choose i}{i \choose j}={n \choose j}{n-j \choose i-j}$

$\sum_{j=0}^{n}\sum_{i=j}^{n} {n \choose i}{i \choose j}=\sum_{j=0}^{n}\sum_{i=j}^{n} {n \choose j}{n-j \choose i-j}=\sum_{j=0}^{n} {n \choose j}{n-j \choose 0}=2^n$

Where am I going wrong?


答:
See robjohn's answer for the problem in your attempt.

However, a good way to start these problems is to stop and think what the sum actually represents. $\binom ni\binom ij$ is the number of ways to choose a committee of $i$ people out of $n$, and then a subcommittee of $j$ people from those $i$. If you sum this over all possible values $i\geq j$ you get the number of ways to divide $n$ people into three groups (those on the subcommittee, those on the main committee only, those on neither) with groups allowed to be any size (including empty). What is a simpler way to calculate that?

Title: Find $\sum_{j=0}^{n}\sum_{i=j}^{n} {n \choose i}{i \choose j}$.


问:
How do you solve this problem from Kangaroo Math Competition 2016 Junior?
Dates can be written in the form DD.MM.YY. For example, today's date is 17.03.2016. A date is called "surprising" if all 8 digits in its written form are different. In what month will the next surprising date occur?

From what I understand, I think we need to make sure that the year does not have a 1 digit, a 0-digit, or a 2-digit, in order for us to be able to maximize the probability that the month is a number underneath 10. However, I do not know where to go from there. Any help? The answer is June, but I do not know why.

答:
The day starts with 0,1,2 or 3, and the month starts with 0 or 1. So the year can contain at most two of the digits 0 to 3.

Using this fact, I'll try to find the first year that could have a surprising date.

 If the year contains 2 and 0 then 1 and 3 must be used in the day and month. So 3 is used to start the day number and 1 starts the month. A day starting with 3 must end with 0 or 1, which will be a repeat. So the year cannot contain both 2 and 0.

 The same argument works if the year contains 2 and 1.

 Therefore the year cannot start 20 or 21, but maybe could start with 23, as long as it does not contain 0 or 1. The first year this happens is 2345.

Now I'll try to find the first month in that year that could have a surprising date:

 The day and month must use 0 and 1. The first month that uses exactly one of 0,1 and not any digits of the year is 06. The first day that works is then 17. The first surprising date is therefore 17-06-2345.

So the answer is June.


问:
Prove or disprove that if V is a finite set over F with a dimension 1, then F is a finite field.
I think it is a True statement, and my proof is as follows:

let $\theta $ be the additive identity in F

Given any element v in V, since v + v = $\theta$ v + $\theta$ v = 2$\theta$ v, then 2$\theta$ $\in$ F, and $\frac{\theta}{2} \in$ F, hence $\frac{\theta}{2}v \in$ V, by this argument we can construct infinite $\{2v, \frac{\theta}{2}v , 3v,  \frac{\theta}{3}v...\}$

Hence V cannot be finite, and we are done.

答:
I think your proof doesn't work like this because:

For a general field, you don't necessarily have that the set $\{ 2 \theta, 3 \theta, 4 \theta, 5 \theta, ..., \theta (2 \theta )^{-1}, \theta (3 \theta)^{-1}, ... \}$ is infinite! Take $\mathbb{F}_2 =  \mathbb{Z} / 2 \mathbb{Z}$ as an example...The set you use is only infinite, if the characteristic of the field is not finite.

There are infinite fields with finite characteristic (see for example here or consider the algebraic closure of a finite field).
Your proof only shows that the characteristic has to be finite, but that's not the same as showing that the field is a finite field.

My attempt for proving the statement would be this. (Please be doubtful, I might be wrong!)

Don't read further if you want to find a proof yourself.

Let $v \in V \setminus \{ 0 \}$ (this exists since $\mathrm{dim}(V) > 0$).
We consider the map
$$
M: F \to V: \lambda \mapsto \lambda v.
$$

We show by contradiction that the map $M$ is injective. Let $\lambda, \mu \in F$ with $\lambda \neq \mu$ and $\lambda v = \mu v$, then
$$
(\lambda - \mu) v = 0
$$
but
$$
(\lambda - \mu)^{-1} (\lambda - \mu) v = v \neq 0
$$
which is a contradiction to the vector space axioms (a zero vector multiplied with anything has to be the zero vector again). Hence there are no such elements such that $\lambda \neq \mu$ and $\lambda v = \mu v$.

Since $M$ is injective, we get
$$
\vert F \vert \leq \vert V \vert < \infty.
$$


问:
Where did the word "logarithm" come from?
Where did the word logarithm come from? Any relation to the word algorithm?

答:
"There is no relation between the two words." feels wrong. The spelling of  algorithm was influenced by the Greek árithmos, which is the second compound in logarithm.

I quote from Word Origins (2005 2e) by John Ayto. p 16 Left column.

algorithm [13]

Algorithm comes from the name
of a Persian mathematician, in full Abu Ja far Mohammed ibn-Musa al-Khwarizmi (c. 780–c.
850), who lived and taught in Baghdad and
whose works in translation introduced Arabic
numerals to the West. The last part of his name
means literally ‘man from Khwarizm’, a town on
the borders of Turkmenistan, now called Khiva.

      The Arabic system of numeration and
calculation, based on 10, of which he was the
chief exponent, became known in Arabic by his
name – al-khwarizmi. This was borrowed into
medieval Latin as algorismus (with the Arabic
-izmi transformed into the Latin suffix -ismus
‘-ism’). In Old French algorismus became
augorime, which was the basis of the earliest
English form of the word, augrim. From the 14th
century onwards, Latin influence gradually led
to the adoption of the spelling algorism in
English. This remains the standard form of the
word when referring to the Arabic number
system; but in the late 17th century an alternative
version, algorithm, arose owing to association
with Greek árithmos ‘number’ (source of
arithmetic [13]), and this became established
from the 1930s onwards as the term for a step-by-
step mathematical procedure, as used in
computing.

      Algol, the name of a computer programming
language, was coined in the late 1950s from
‘algorithmic language’.

Op cit. p 314 Right column.

logarithm [17]

Greek lógos had a remarkably
wide spread of meanings, ranging from ‘speech,
saying’ to ‘reason, reckoning, calculation’, and
‘ratio’. The more ‘verbal’ end of its spectrum has
given English the suffixes -logue and -logy (as in
dialogue, tautology, etc), while the ‘reasoning’
component has contributed logic [14] (from the
Greek derivative logiké), logistic [17] (from the
Greek derivative logistikós ‘of calculation’), and
logarithm, coined in the early 17th century by
the English mathematician John Napier from
Greek logós ‘ratio’ and arithmós ‘number’
(source of English arithmetic [13]).


问:
Why is every complex manifold not Kähler?
I'm playing around with what happens when we have a short exact sequence
$$
0 \longrightarrow T_X \stackrel{j}{\longrightarrow} E \stackrel{g}{\longrightarrow} Q \longrightarrow 0
$$
of holomorphic vector bundles over a complex manifold $X$. The question I'd like to answer is: Suppose we put a Hermitian metric on the middle bundle $E$. When is the induced metric on $X$ Kähler?

This can happen; submanifolds of Kähler manifolds are Kähler, and the induced metric arises in this way. This can also not happen; take a non-Kähler manifold and consider it as a submanifold in itself times a curve. No matter what metric we put on $E$, we won't get a Kahler metric on $X$.

Luckily this situation has been studied before, in particular by Demailly in his book (Theorem 14.3, page 273). Using that reference, we argue as follows:

The torsion tensor $\tau$ of the induced metric on $X$ is the exterior covariant derivative of the identity morphism on $T_X$. We have $\operatorname{id}_{T_X} = j^\dagger \circ j$, where $j^\dagger : E \to T_X$ is the adjoint to $j$. Then
$$
\tau
= D_{\operatorname{End}(T_X)} \operatorname{id}
= D_{\operatorname{Hom}(E,T_X)}j^\dagger \circ j
+ j^\dagger \circ D_{\operatorname{Hom}(T_X,E)}j.
$$
By Demailly, we have
$$
D_{\operatorname{Hom}(E,T_X)}j^\dagger
= \beta^\dagger \circ g,
$$
where $\beta$ is the second fundamental form of $T_X$, and
$$
D_{\operatorname{Hom}(T_X,E)}j
= g^\dagger \circ \beta.
$$
This gives
$$
\tau
= \beta^\dagger \circ g \circ j + j^\dagger \circ g^\dagger \circ \beta
= 0
$$
because $g \circ j = 0$ and $j^\dagger \circ g^\dagger = 0$.

Thus the torsion tensor of the Chern connection of the induced metric on $X$ is zero, so that metric is Kähler. As the tangent bundle of any manifold trivially fits into a short exact sequence
$$
0 \to T_X \to T_X \oplus \mathcal O_X \to \mathcal O_X \to 0
$$
we conclude that all complex manifolds are Kähler. Hurray!

Clearly this is not true, but I can't find the mistake in this argument. The induced connection is really the Chern connection, and as far as I can tell this calculates its torsion. What is the mistake here?

答:
I think I've figured out where this goes wrong (nothing like asking people in public and then sleeping on the problem).

The problem with this "proof" is that what we actually calculate is the covariant derivative of the identity, and not the covariant exterior derivative. If $D$ is a connection on a vector bundle $E$, we have
$$
D_E s
= D_E(\operatorname{id}_{E} s)
= (D_{\operatorname{End} E}\operatorname{id}_E)(s) + \operatorname{id}_E(D_Es)
= (D_{\operatorname{End} E}\operatorname{id}_E)(s) + D_Es
$$
for any section $s$ of $E$, so $D_{\operatorname{End} E}\operatorname{id}_E = 0$.

Now, the torsion tensor is the covariant exterior derivative $d^D$ of the identity morphism considered as a one-form with values in $E = T_X$, and the two notions are not the same. If $f \in \operatorname{End} E$ then $D_{\operatorname{End} E}f \in \mathcal C^{\infty}(X, T_X^* \otimes \operatorname{End} E)$. For $T_X$, this is $\mathcal C^\infty(X,T_X^* \otimes T_X^* \otimes T_X)$. Meanwhile the torsion tensor $\tau = d^D \operatorname{id}_{T_X}$ is antisymmetric, and thus an element of $\mathcal C^\infty(X,\bigwedge^2 T_X^* \otimes T_X)$. What the above argument establishes at some length is that $D_{\operatorname{End} T_X} \operatorname{id}_{T_X} = 0$, which is true for any connection as we saw above.


问:
For which values of $a$ and $b$ does the equation $\sin(\sqrt{x})=\frac{a}{x}+b$ have no solution?
I‘m trying to find for what values of $a$ and $b$ there is no solutions to this equation:

$$\sin(\sqrt{x})=\frac{a}{x}+b$$

I would like to get some advice.

Thanks

答:
Let $k=\sqrt{x}$. Reformat the equation as $sin(k)-b=\frac{a}{k^2}$, with the condition that $k>0$.

We consider three cases: $a>0$, $a<0$, and $a=0$.

First, assume $a>0$. Then, The graph of $y=\frac{a}{k^2}$ in respect to $k$ has a vertical asymptote at $k=0$, a horizontal asymptote at $y=0$, and is located in the first quadrant. A visual example can be seen here: http://www.wolframalpha.com/input/?i=1%2Fx%5E2+for+x+from+0+to+4.

We see that any $sin(k)$ graph translated in the positive y-direction inevitably intersects with $y=\frac{a}{k^2}$. In order for $y=sin(k)-b$  to not intersect this graph, the maximum value of the sine curve must lie on the horizontal asymptote $y=0$ or below. Thus, we see that the two graphs do not intersect for $a>0$ and $b\geq1$.

Similarly, for the parallel case of $a<0$ (in which the graph is reflected across the x-axis), we see that the two graphs do not intersect for $a<0$ and $b\leq1$.

For the case of $a=0$, we require that $sin(k)\neq b$. This is satisfied when $b>1$ or $b<1$.


问:
Given a collection of curves, find the differential equation associated with them
The question is as follows:

$$
\text { 1) } y=\frac{\sec x}{c_{1}+c_{2} \tan x} \quad ; \quad \underline{\text { The Answer is  }}: y y^{\prime \prime}-2 y^{\prime 2}-y^{2}=0
$$

My main problem with this question is the long algebraic process. Is there any fast way to solve this question?

One possible way of solving this is to form a matrix involving y', y'', and y and try to make a system of equations with that matrix. Then because the equation should have a non-zero answer, we set the determinant to zero.

This is algebraically time consuming and provides the possibility for mistakes.

答:
One could observe that
$$
y^{-1}=c_1\cos x+c_2\sin x
$$
which is the general solution of the harmonic oscillator. Thus
$$
(y^{-1})''+(y^{-1})=0
$$
Now $(y^{-1})'=-y^{-2}y'$ and $(y^{-1})''=-y^{-2}y''+2y^{-3}y'^2$ so that inserting gives
$$
y^{-3}(-yy''+2y'^2+y^2)=0.
$$


问:
Reference Request: Parameter Dependent Center Manifold Theorem for ODEs
Suppose we have an $n$-dimensional first order ODE of the form $\frac{dx}{dt}= f_{\mu}(x)$ with $\mu \in \mathbb{R}^k$ a parameter and with an equilibrium at $x=0$ $(f_{\mu}(0) =0)$.

For fixed $\mu$ I am able to apply the center manifold theorem which will give me information about the stable, unstable and center manifolds.

I am curious to understand how these manifolds depend on the parameter $\mu$.

Does anyone know a good reference for the corresponding theorem?

Thanks in advance.

答:
So you need extend the system as pointed out by John:
\begin{cases}
x'=f_\mu(x),\\
\mu'=0
\end{cases}
and then you apply the center-unstable, center-stable or center manifold theorem depending on which manifold's smoothness you want to show.

Kelley, Al. "Stability of the center-stable manifold." Journal of Mathematical Analysis and Applications 18.2 (1967): 336-344.


问:
An orthogonal matrix the given power of which is the identity matrix
What is the general form of a real orthogonal matrix of $n$th order that satisfies the condition $A^p=E$ where $p$ is a given positive integer, and $E$ is the identity matrix?

As I see, in the case of $p=2$ we have $A^2=E$ whence $A^{-1}=A$, then by the definition of orthogonality $A^T=A^{-1}$, whence $A^T=A$, that is $A$ is a symmetric matrix.

And what can one say about the general view of such matrices, for example for $p=5$ or $p=8$? Is it possible to display this general view?

答:
According to wikipedia , any Special Orthogonal matrix $R$ can be written as the exponential of a skew symmetric matrix $A$ , $R=e^A$. So if you can find an $A$ such that $I=e^A$ then $R = e^{{1\over p}A}$ is the $p$'th root of the identity matrix. Furthermore, $A$ can be written as
$Q\Sigma Q^T$ with $Q$ orthogonal and $\Sigma$ skew symmetric and block diagonal with blocks of order 1 or 2. selecting each block to be $2\pi \begin{bmatrix}0 & 1\\-1 & 0\end{bmatrix}$  or $\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}$  will ensure that indeed $I=e^A$.


问:
Intepretation of generated $\sigma$-algebra
I'm readying Probability with Martingales by David Williams, and I encountered the following discussion in section 3.13. It says that if we have a collection of random variables $(Y_\gamma:\gamma \in C)$ on a probability triple $(\Omega, \mathcal{F}, \mathbb{P})$, then the $\sigma$-algebra generated by these random variables are precisely the events $F$ such that for every $\omega$, one can decide if $\omega\in F$ based on $Y_\gamma(\omega), \gamma \in C$. I'm trying to prove this but I can't see how it can be done. Can anyone please help?

答: Let $x:=Y(\omega)$ for some chosen $\omega \in \Omega $, then as every $F\in \sigma (Y)$ is of the form $F=Y^{-1}(A)$ for some Borel set $A$, then if $x\in A$ you knows that $\omega \in F$. And viceversa: if $x\notin A$ then necessarily $\omega \notin F$.


问:
Boltzmann distribution as generalization of Gaussian
I'm trying to gain some intuition on the Boltzmann distribution (also called Gibbs canonical distribution).  The definition is very broad: it's an exponential probability distribution that is a function of an energy and temperature, $$ P(x) \propto e^{-\epsilon(x)/kT(x)} $$ where $\epsilon$ and $T$ are the energy and temperature of the state $x$, and $k$ is the Boltzmann constant.

An example is the 1D Gaussian distribution, which has pdf $$P(x) \propto e^{-\frac{(x-\mu)^2}{2\sigma^2}}.$$ Apparently this is the Boltzmann distribution corresponding to the potential energy $$\epsilon(x) = \frac{(x-\mu)^2}{2\sigma^2}.$$

How do I interpret this? I can see that if a point $x$ is far from the mean, then it has high potential energy.

In statistics, the idea of "entropy" is related to the spread of a distribution. Is this saying that a high entropy distribution has high potential energy?

答: The potential energy ${1\over2}k(x-\mu)^2$ is exactly the well known potential energy of a  harmonic spring with offset $\mu$. A spring in a thermal bath where the kinetic energy can be neglected, or can be integrated over will have a position distribution proportional to $e^{-{1\over 2}\beta k (x-\mu)^2}$  where $\beta = {1\over \kappa T}$


问:
How to test the convergence of the series $\sum_{n=1}^\infty n^{-1-1/n}$?

How to test the convergence of the series $\displaystyle\sum_{n=1}^\infty\frac{1}{n^{1+1/n}}?$

Help me. I'm clueless.

答:
You for $a_n = \frac1n$ and $b_n = \frac{1}{n^{1+1/n}}$ that
$$
  \lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{n}{n^{1+1/n}} = \lim_{n\to\infty}n^{-1/n} = 1.
$$
Since $\sum a_n$ diverges, so does $\sum b_n$.</p>