问:
If $R$ is a binary relation over $A$ where $I_A \subseteq R $ and $R \circ R \subseteq R$ then $R$ is reflexive and transitive
My proof:

Lets assume $A$ is an arbitrary set, where $R$ is a relation over the set $A$. Consider $I_A$ to be an arbitrary identity relation. We will prove if $I_A \subseteq R$ and $R \circ R \subseteq R$ then $R$ is a preorder relation.

Consequently we know that $I_A \subseteq R$, therefore for every $x$, we know $(x,x) \in R$. Therefore we can conlude that $R$ is reflexive. We need to prove that $R$ is transitive.

To prove $R$ is transitive, we know $R \circ R \subseteq R$. This tells us that when we combined R with itself, no new relations were created. Because we know no new relations were created,lets assume, a,b and c are arbitrary elements, we can conclude if aRb and bRc then aRc must be true. Therefore the we can conclude $R$ is transitive. This implies that $R$ is reflexive and transitive. $\blacksquare$

My problem:

I'm not sure if i've done the proof right, so if anyone can check it. That would be of great help

When I tried writing the proof, I felt like i didn't go into too much mathematical detail regarding the composition. To check that if $R \circ R \subseteq$ is really transitive, I did some examples and checked if that was true. Then i did the same thing but with untransitive relations, which resulted in new relations. Which is what i sort of expected, but I couldn't really mathematically explain what $R \circ R$ was actually doing and the mathematical explanation behind why $R \circ R \subseteq R$ is transitive. And like i'm still really confused when I read my proof .

答:
Your proof that $R$ is reflexive is correct, it's still missing some important details though. First of all, we need to show what we prove and how we are going to prove this:

First, we prove that $R$ is reflexive, by showing that for any $x \in A$, that $xRx$.

This helps a reader of your proof a lot, so you and he/she are on the same page. Then you need to pick up an arbitrary $x$ and conclude somehow that $(x, x) \in R$ or $xRx$.

So, consider any $x \in A$. Since $I_a \subseteq R$ and $x I_a x$, we know that $xRx$, thus $R$ is reflexive.

See how I write how I got to that conclusion that $xRx$.

The same here: first of all state what you're going to prove and how. We can always reach out to the definition of the transitivity. This helps us to understand what we can assume and what exactly we need to prove.

Now we prove that $R$ is transitive, so consider any $x, y, z \in A$, where $x R y$ and $y R z$. We will show that $x R z$.

Now see carefully at the definition of composition of relations, which states that:

Given two binary relations $R_1$ and $R_2$ over some set $A$, the composition of $R_1$ and $R_2$, denoted $R_1 \circ R_2$, is the binary relation over $A$ defined as $$x(R \circ R)y \quad \text{if} \quad \exists z\in A.(x R_1 z \land z R_2 x )  $$

So go back to our proof. Let $R_1 = R$ and $R_2 = R$. Then we see that $x R y$ and $y R z$, or $x R_1 y$ and $y R_2 z$. So as it turned out we have such $z \in A$ (namely $y$), where $x R_1 z$ and $z R_2 y$. Therefore we can conclude that $x(R_1 \circ R_2)y$.

Since $x R y$ and $y R z$ and $y \in A$ we know that $x(R \circ R)z$ by definition of composition of relations. Since $x(R \circ R)z$ and $R \circ R \subseteq R$ we know that $xRz$, thus R is transitive.


问:
Hypothesis Testing for Poisson rvs
Let $x_1,\dots,x_m$ be iid Poisson$(\lambda_1)$ and $y_1,\dots,y_m$ be iid Poisson$(\lambda_2)$ and independent of $x_1,\dots,x_m$. $S_1=\displaystyle\sum_{i=1}^mx_i$ and $S_2=\displaystyle\sum_{i=1}^ny_i$ are independent Poissons with parameters $m\lambda_1$ and $n\lambda_2$.

a) Show that the conditional distribution of $S_1$, given that $S_1+S_2=S$ is binomial with parameters $(s,p)$ where $$p=\frac{m\lambda_1}{m\lambda_1+n\lambda_2}$$b) Test the null hypothesis $H_0:\theta=1$ versus $H_1:\theta<1$, where $\theta=\frac{\lambda_1}{\lambda_2}$. What is $p$ under $H_0$? Should you reject the null hypothesis for large or small values of $S_1$?
c) Suppose that $m=n=25$ and $S_1=6$ and $S_2=18$. Compute the one-sided $p$-value for testing $H_0:\theta=1$ versus $H_1:\theta<1$ using the conditional binomial test suggested above.

I was able to figure out the first part, but need help figuring out parts b and c. Any help will be appreciated.

答:
First of all

$X_1,...,X_n$ are rv's thus Capital letter is required and obviously same observation for $Y_i$

Second:

you literally wrote $y_1,...,y_m$ are iid poisson with parameter $\lambda_2$; this means that the two random sample have the same size $m$ but then you show the sum of $y_i$ going to $n$, that is a different size from $X-$ sample

(clarify if the two random samples have the same size or not)

(b) it is evident that $(p|\mathcal{H}_0)=\frac{m}{m+n}$

(c) under the given data verify that $\mathcal{H}_0:\theta=1$ against the alternative $\theta<1$ means to verify that, in a Binomial

$$B(24;p)$$

the parameter $p=0.5$ against the alternative that $p<0.5$

Using the conditional distribution, as suggested, this is the drawing

As you can see, for $S_1\leq 6$ the pvalue is less than 2% (it's about $1.133\%$) thus, i.e. you can reject the null hypothesis with any significance level $\alpha > p_{value}$

In other words, you reject the null hypothesis with a significance level $\alpha=5\%$ but you cannot reject it with a significance level $\alpha=1\%$

This also answer to the second (b) question

ContentLicense: CC BY-SA 4.0

Title: Hypothesis Testing for Poisson rvs


问:
How far can you get in math by doing random problems and generalizing them in as many possible ways as you can imagine?
Off course 1) the notion of 'far' would have to be defined. As well as 2) what counts as a 'random problem'. And of course 3) different people can think of different generalizations and can do the problems at different rates.

Let's say 1) An undergraduate degree in mathematics, just to have some milestone.
2) Let's say you take the standard undergraduate mathematics books used in universities and you
pile all the problems on a big heap, you randomly take a problem one after the other. If you come across a problem that is included in some generalization of another one, you skip it.
3) Let's say we're talking about an average undergraduate in mathematics who is beginning his
studies.

答:
Not far at all. On top of that undergraduate math you mention, thousands over thousands of mathematicians, decade after decade, have put a lot of study, thinking, re-thinking, picking the best ideas, re-writing, refining.

By skipping all that knowledge, even if you were the new Gauss (or Cauchy, or Euler, or take your pick) you would be starting decades (if not centuries) behind current knowledge, so at best you could aspire to re-discover a tiny part of what is already in some graduate textbook in your library. And you would be not even aware of most modern math.


问:
Relation between $n$ and $m$ with $\mathfrak{g}^{(n)} = \mathfrak{g}^{(n+1)}$ and $\mathrm{rad}(\mathfrak{g})^{(m)} = 0$
Let $\mathfrak{g}$ be a finite-dimensional Lie algebra, let $\mathfrak{g}^{(i)}$ with $i \geq 0$ denote the derived series of $\mathfrak{g}$ (with $\mathfrak{g}^{(0)} = \mathfrak{g}$) and let $\operatorname{rad}(\mathfrak{g})$ be the solvable radical of $\mathfrak{g}$, i.e. the unique maxmimal solvable ideal of $\mathfrak{g}$.
There exist minimal natural numbers $n$ and $m$ with
$$
  \mathfrak{g}^{(n)} = \mathfrak{g}^{(n+1)} \,
  \qquad\text{and}\qquad
  \operatorname{rad}(\mathfrak{g})^{(m)} = 0 \,.
$$

Question.
Is there some relation between the numbers $n$ and $m$?

Some thoughts:

If $\mathfrak{g}$ is solvable, then $\operatorname{rad}(\mathfrak{g}) = \mathfrak{g}$ and $n = m$.

If $\mathfrak{g}$ is semisimple, then $n = 0 = m$.

There exist perfect, non-semisimple Lie algebras. It may therefore happen that $n = 0 < m$.
But the only examples I know have $m = 1$.

答:
I assume some field of characteristic zero is fixed. The answer is then: the possible $(n,m)$ are exactly those $(n,m)$ with $0\le n\le m$.

Proof:

All $(n,m)$ with $0\le n\le m$ can be achieved.

Indeed, $(0,0)$ is clear (it means semisimple), so suppose $m\ge 1$ Let $\mathfrak{v}_m$ be the free $2^m$-step nilpotent Lie algebra on $m$-generators, and consider $\mathfrak{g}_m=\mathfrak{sl}_m\ltimes\mathfrak{v}_m$. It's perfect ans has $m(\mathfrak{g}_m)=m$. Hence for $\mathfrak{g}_{n,m}=\mathfrak{v}_n\times\mathfrak{g}_m$, for $n\le m$, achieves $(n,m)$.

(The dimension is clearly not optimal but I don't really pay attention. The smallest achieving $(0,1)$ is the perfect 5-dimensional Lie algebra $\mathfrak{sl}_2(K)\ltimes K^2$ and the smallest achieving $(0,2)$ is the 6-dimensional $\mathfrak{sl}_2(K)\ltimes \mathfrak{hei}_3(K)$.)

Conversely, $n\le m$ is a necessary condition.

Indeed, let $\mathfrak{r}_2$ be the radical of $\mathfrak{g}$ and $\mathfrak{r}_1$ the quotient of $\mathfrak{g}$ by the intersection of its derived series (this is the largest solvable quotient of $\mathfrak{g}$). So by definition $n(\mathfrak{g})$ is the solvability length of $\mathfrak{r}_1$ and $m(\mathfrak{g})$ is the solvability length of $\mathfrak{r}_2$.

The inequality then follows from the observation that $\mathfrak{r}_1$ is quotient of $\mathfrak{r}_2$. Indeed, write $\mathfrak{r}=\mathfrak{s}\ltimes\mathfrak{r}_2$. Then the quotient map $\mathfrak{g}\to\mathfrak{r}_1$ is trivial on $\mathfrak{s}$, hence is surjective in restrictio to the radical $\mathfrak{r}_2$.


问:
How to find the basis of the given vector space.
Let $V=\begin{cases}\begin{bmatrix}x\\y\end{bmatrix}:x\in \mathbb{R}^+, y\in \mathbb{R}\end{cases}\bigg\}.$
Then it can be proved that under the operations $$\alpha \cdot \begin{bmatrix}x \\  y\end{bmatrix}=\begin{bmatrix}x^{\alpha}\\ \alpha y\end{bmatrix}$$

$$\begin{bmatrix}x_1 \\  y_1\end{bmatrix} \oplus \begin{bmatrix}x_2 \\  y_2\end{bmatrix}=\begin{bmatrix}x_1x_2 \\  y_1+y_2\end{bmatrix}$$
$V$ is a vector space over $\mathbb{R}$. How to find the basis of $V$?

I know the meaning of basis and how to obtain. But the normal method doesn't work here. Please help.

答:
You mean find a basis, as it's not unique.

We seek $\begin{bmatrix}a\\b\end{bmatrix},\begin{bmatrix}c\\d\end{bmatrix}$ such that $\begin{bmatrix}a^\alpha c^\beta\\\alpha b+\beta d\end{bmatrix}$ can be any element of $V$, so e.g. $a=e,\,b=0,\,\,c=1,\,d=1$ would work, reducing the above general form to $\begin{bmatrix}e^\alpha\\\beta\end{bmatrix}$.

Edot: as @Christoph notes, it's worth saying why a basis would be of cardinality $2$. Certainly this one is, so they all are. Even if we don't notice the isomorphism in @user126154's answer, that there are $2$ components makes the dimension being $2$ our only reasonable guess.


问:
Efficient product of vector with inverse of positive definite circulant matrix plus real diagonal matrix?

Question:

Hi! Would anyone happen to know an efficient way to compute:

$$
u^\top [C + D]^{-1},
$$

where:

$C$ is a square matrix, and is real, positive definite, and circulant

$C$ is too large to fit in computer memory, and is instead defined by a single row $c$ (or its Fourier transform $\tilde c$)

$D$ is a real-valued diagonal matrix (and is not proportional to the identity matrix)

$u$ is a vector

(If $D$ is also circulant, then there is a fast solution via the fast Fourier transform. But, $D$ is not circulant my use case.)

"Efficient" here is deliberately vague, meaning "something I could actually compute on a laptop". Something less than $\mathcal O(n^3)$ would be nice. But, more important is the memory constraint: for the problem I'm working on it's just not possible to explicitly construct a matrix with size $C$, so $\mathcal O(n^2)$ space requirement just isn't feasible. I'd settle for an algorithm that only needs $\mathcal O(n)$ or $\mathcal O(n \log(n))$ memory ($n$ here is $\approx 10,000$), and takes at most a few minutes to run.

Context:

I'm working with a spatially-extended Gaussian process on a regular square grid. I have observations $y_i(\mathbf x_i)$ at each grid point $\mathbf x_i$, and I evaluate the posterior mean $\mu$ at these same grid points. The prior kernel is translationally invariant on a periodic domain. This kernel is evaluated on a uniformly-spaced grid. The prior covariance $C$ is therefore a real positive circulant matrix. The observation/measurement noise model consists of independent observations at each grid point $\mathbf x_i$ with noise $\sigma^2_i$ (i.e. the measurement noise is inhomogeneous). The observation covariance $D$ is therefore a diagonal real-valued matrix. In this case, the posterior covariance $\Sigma$ is given by

$$
\begin{aligned}
\Sigma &= C - C [C+D]^{-1} C
= [C^{-1} + D^{-1} ]^{-1}
\end{aligned}
$$

I want the marginal variance (or covariance) of the posterior along the vector(s) $a_1$, $a_2$. This is

$$
\begin{aligned}
\sigma^2_{a_1,a_2} = a_1^\top \Sigma a_2
\end{aligned}
$$

The grid is large, so $C$ (and therefore $C+D$) can't fit in memory. Thankfully, matrix-vector products can be quickly computed using the FFT if the matrix is circulant, so terms like $u^\top C$ are fast. If there were also a fast way to compute $u^\top [C + D]^{-1}$, then $\sigma^2_{a_1,a_2}$ could be computed quickly.

Conclusion:

I suspect that there might be no easy/fast way to compute this via ordinary matrix operations and the FFT, but perhaps there is some clever iterative algorithm that converges toward the right answer with acceptable time and space complexity.

Alternatively, since the underlying problem is related to a spatial Gaussian process, if the spatial correlations decay to zero at a certain distance, I could approximate the solution by working on smaller, localized domains. But something more formal (and maybe even exact) would be welcome.

答:
Calculating $[C + D]^{-1} u$ can be re-cast as a solving for $x$ in the linear system $[C + D] x = u$. Although the matrix $C + D$ is too large to construct, one can define a quadratic error function with $x$ as the solution:

$$
\epsilon = \tfrac 1 2 \| [C+D] x - u \|^2
$$

We can find $x$ by minimizing this error via gradient descent. The gradient in $x$ is:

$$
\nabla_x \epsilon = [C+D]^\top \left( [C+D] x - u \right)
$$

This can be computed using the FFT with relatively limited memory requirements. If $\tilde x = \mathcal F x$ is the Fourier transform of $x$, and $\tilde c$ is the Fourier transform of the kernel defining $C$, then:

$$
[C+D] x = Cx + Dx = \mathcal F^{-1} [\tilde c \circ \tilde x] + \operatorname{diag}[D]\circ x,
$$

where $\circ$ denotes element-wise multiplication. Define the vector $\nu = [C+D] x - u$. The gradient is then $\nabla_x \epsilon = [C+D]^\top \nu = [C+D] \nu$. This matrix vector product can also be computed via the FFT as above.

This might not be especially fast, but in cases where $\Sigma$ can be decomposed into a sum special matrices with compact representations (e.g. Toeplitz, circulant, diagonal, sparse, etc), it provides a space-efficient alternative that does not require writing down a large linear system.


答: The function $f(x,y)=(\log x,y)$ is an isomorphism between $V$ and $\mathbb R^2$. So a basis is formed by $v_1=f^{-1}(1,0)=(e,0)$ and $v_2=f^{-1}(0,1)=(1,1)$.

Title: How to find the basis of the given vector space.

Body:
Let $V=\begin{cases}\begin{bmatrix}x\\y\end{bmatrix}:x\in \mathbb{R}^+, y\in \mathbb{R}\end{cases}\bigg\}.$
Then it can be proved that under the operations $$\alpha \cdot \begin{bmatrix}x \\  y\end{bmatrix}=\begin{bmatrix}x^{\alpha}\\ \alpha y\end{bmatrix}$$

$$\begin{bmatrix}x_1 \\  y_1\end{bmatrix} \oplus \begin{bmatrix}x_2 \\  y_2\end{bmatrix}=\begin{bmatrix}x_1x_2 \\  y_1+y_2\end{bmatrix}$$
$V$ is a vector space over $\mathbb{R}$. How to find the basis of $V$?

I know the meaning of basis and how to obtain. But the normal method doesn't work here. Please help.


问:
Double integral of a product in calculus of variations
Let's say I have an integral of the form
$$
  V(u) = \iint\limits_{[0,T]^2}f(x,y)u(x)u(y)\mathrm dx\mathrm dy
$$
which I would like to optimize over smooth functions $u$. For the variation I get
$$
  \delta V(u) \approx \iint\limits_{[0,T]^2}f(x,y)[u(x)\delta u(y) + u(y)\delta u(x)]\mathrm dx\mathrm dy
$$
since I hope the term $\delta u(x)\delta u(y)$ can be disregarded. How can I find the optimal $u$ now? In the 1d case I'd have
$$
  F(u) = \int_0^T G(x,u(x))\mathrm dx \quad \implies \quad \delta F(u) \approx \int_0^TG_u(x, u(x))\delta u(x)\mathrm dx
$$
and by fundamental lemma of calculus of variations I would get $G_u(x, u(x)) = 0$. However, I am not sure whether this passage is extendable to my case. Fwiw, my original case is a bit more complicated: I deal with an integral
$$
  W(u) = V(u) + F(u),
$$
and since I know how to variate the latter part, I hope it is enough to know how to variate the former. Would be happy to hear any suggestions. I tag it as LA, since the $V$ functional is a bilinear form, so maybe there are some results in LA known for that case.</p>

答:
I wanted to provide a little more detail on the answer given by Siminore.  I don't have a reference, as I worked this out on my own, but I think it is correct.

From the original problem statement, we may take the variation of $W(u)$ (I am using the more general problem), yielding
\begin{align*}
\delta W(u) &= \delta V(u) + \delta F(u) \\
&= \int_0^T\int_0^Tf(x,y)\left[ u(x)\delta u(y) + u(y)\delta u(x) \right]\ dx\ dy + \int_0^T G_u(x,u)\delta u(x)\ dx.
\end{align*}
Distribute the $f(x,y)$ to the terms in the brackets and break the integrals into two:
\begin{equation*}
\delta W(u) = \int_0^T\int_0^Tf(x,y)u(x)\delta u(y) \ dx\ dy + \int_0^T\int_0^Tf(x,y)u(y)\delta u(x) \ dx\ dy + \int_0^T G_u(x,u)\delta u(x)\ dx.
\end{equation*}
In the first set of integrals, we note that the integration variables are just dummy indices, and may be swapped [e.g., $(x,y)\rightarrow (y,x)$] without changing the result.  Furthermore, the limits of integration are the same for both integrals and are constants, and so we may arbitrarily change the order of integration.  Making use of these observations, we may write
\begin{align*}
\delta W(u) &= \int_0^T\int_0^Tf(y,x)u(y)\delta u(x) \ dx\ dy + \int_0^T\int_0^Tf(x,y)u(y)\delta u(x) \ dx\ dy \\
&\hspace{10mm}+ \int_0^T G_u(x,u)\delta u(x)\ dx \\
&= \int_0^T\int_0^T\left[f(y,x) + f(x,y)\right]u(y)\delta u(x) \ dx\ dy + \int_0^T G_u(x,u)\delta u(x)\ dx \\
&= \int_0^T\left( \int_0^T\left[f(x,y) + f(y,x)\right]u(y)\ dy + G_u(x,u) \right)\delta u(x)\ dx. \tag{1}
\end{align*}
Then, by the fundamental lemma of variational calculus you would conclude
\begin{equation*}
\int_0^T\left[f(x,y) + f(y,x)\right]u(y)\ dy + G_u(x,u) = 0,
\end{equation*}
which is an integral equation that may be solved for $u$.  If we make the additional assumption made by Siminore, i.e. that $f(x,y)=f(y,x)$, and let $G=0$ then the equation simplifies to
\begin{equation*}
\int_0^Tf(x,y)u(y)\ dy = 0,
\end{equation*}
which is what Siminore obtained.

As a special case that I was interested in, consider $f(x,y)=h(x)h(y)$ and $G=0$.  In this case we may write $W(u)=V(u)=H^2(u)$ where
\begin{equation*}
H(u) = \int_0^Th(x)u(x)\ dx.
\end{equation*}
Note that the variation of $H$ may be written as
\begin{equation*}
\delta H(u) = \int_0^Th(x)\delta u(x)\ dx.
\end{equation*}
From Eq. (1) above, we may write
\begin{align*}
\delta W(u) = \delta H^2(u) &= \int_0^T\int_0^T h(x)h(y)u(y)\delta u(x)\ dx\ dy \\
&= \int_0^Th(y)u(y)\ dy \int_0^T h(x)\delta u(x)\ dx \\
&= 2H(u)\delta H(u).
\end{align*}
Thus, the simple differentiation rule applies to these variations.

ContentLicense: CC BY-SA 3.0


问:
prove that $1^3+2^3+\ldots+n^3= \left(\frac{n(n+1)}{2}\right)^2$ using geometrically
For all integers $n\geq1$ prove the following  identity using geometrically.

$$1^3+2^3+\ldots+n^3= \left(\frac{n(n+1)}{2}\right)^2$$

I can prove the statement using mathematical induction. I want to find a geometrically proof

Thanks!

答:
The usual approach here is to slice the cubes into layers of n(n-1), except for the last, which is n².

So you put the 1.1.1 cube, and around it two rectangles 2.1.1, and the square 2.2.1.  This makes a square whose sides are (1+2).

The 3-cube gives 3 slices of 3.3.1.  Since the side of the square is now 3, we can place these squares to double the existing square.

The 4.4.4 cube is cut into a 4.4.1 layer, and the remainder is cut 4.3.1 four times.  We can put two of these 4.3.1 so that the 3 side is against the existing square, and the 4 side enongens the edge, a pair of these will fill the inner edge, and the 4.4.1 array goes in the corner far from 1.

Repeat steps 3, 4 for odd and even numbers, as far as necessary.  The shape formed will be a square, of the sum of numbers.


问:
calculate location on an isometric world map
I want to display a point with a latitude and a longitude value in an isometric world map using JavaScript and jQuery.

For a rectangular map, I use the following code to calculate the x and y position of the marker:

```<code>marker.y = map_range(position.lat, 90, -90, 0, $('#map').height());
marker.x = map_range(position.lng, -180, 180, 0, $('#map').width());

function map_range(value, low1, high1, low2, high2)
{
    return low2 + (high2 - low2) * (value - low1) / (high1 - low1);
}
</code>```

How do I transfer this formula to a pseudo-isometric map? (skewed 45 degrees, height = .5 * width)?

P.S. I first posted this question on StackOverflow, and people directed me here to ask for an answer. For those not familiar with JavaScript or jQuery: $('#map').height() represents the height of the map, the same goes for the width. If you have any other questions about the code, please ask.

答:
Having previously done this kind of isometric transforming in my own programming, I think what will be useful to you is a function that converts from rectangular to isometric. I believe something like this is what you're looking for (feel free to tweak as needed).

(Let $x$ and $y$ be the original rectangular coordinates and $u$ and $v$ be the transformed isometric coordinates.)

$u = \displaystyle \frac{(2x+2y)(x+3y)}{\sqrt{2}}$

$v = \displaystyle \frac{(x-y)(3x-y)}{\sqrt{2}}$

I derived this by transforming from rectangular into isometric ($x \mapsto 2x+2y$ and $y \mapsto x-y$) and then rotating by 45 degrees around the origin (using sum and difference identities to simplify).

Is that sufficient for your needs?


问:
find and prove probability of P({D1,D2,D3,...}=N)
Let $D_1, D_2,\cdots$ be independent with $D_n\sim Unif\{1,2,\cdots,n\}$. Find and prove tthe probability of $P(\{D_1,D_2,\cdots\}=\mathbb{N})$. I am a bit lost as to how to prove this. My intuition tells it should be 1. Hope someone could help! Thanks!

答:
The probability is indeed $1$.

For any $k \le n$ the probability of the event
$$k \notin \{ D_k , \dots , D_n\}$$
is equal to
$$\frac{k-1}{k} \cdot \frac{k}{k+1} \cdots \frac{n-1}{n} = \frac{k-1}{n}$$
As $n \to \infty$ this probability tends to $0$.

Hence the event
$$k \notin \{ D_k , \dots \}$$
has probability $0$.

This means that $k$ belongs almost certainly to the set $\{ D_n : n \in \Bbb N\}$.

Since $\Bbb N$ is countable, and countable intersection of almost certain events is almost certain,
the probability of the event
$$\Bbb N \subseteq \{ D_n : n \in \Bbb N\}$$
is $1$.


问:
derivatives of function f: [0; + ∞)
ok, this exercise is quite difficult, at least (for me).
f: [0; + ∞) → R satisfies
$$
f'(x) \left( f(x)+1 \right)^2 = \frac{1}{1+6x}
$$

for $x>0,  f(0)=2$ . So $f(x)$= ?

I'm really quite confused where to start. I tried to do something but it gave me 3√(1/2 ln(1+6x) +27 -1) but I don't think it's okay

Title: derivatives of function f: [0; + ∞)

Body:
Let $g(x):=\frac{1}{3}(f(x)+1)^3.$ Then we have

$$g'(x)=\frac{1}{1+6x},$$

Hence

$$g(x)= \frac{1}{6} \ln (1+6x)+c.$$

Can you proceed ?


问:
Using the Hessian, prove that affine scalar field is convex

Using the Hessian, prove that the affine scalar field $f : \Bbb R^n \to \Bbb R$ defined by $$f(x) = A^T x + b$$ where $A \in \Bbb R^n$ and $b \in \Bbb R$, is convex.

My intuition on this problem is to apply the theorem that if $f(x)$ is twice continuously differentiable then $f(x)$ is convex if and only if the Hessian is positive semidefinite, but I don't know how to find the hessian of this function and show that it is positive semidefinite.

答:
If $A^T=(a_1,a_2,...,a_n)$ and $x=(x_1,...,x_n)^T$, then

$$f(x)=a_1x_1+...+a_nx_n+b.$$

Hence

$$f_{x_j}(x)=a_j$$

for $j=1,...,n.$

Conclusion: the hessian of this function $=?$


答:
set $y=f(x)\to \frac{dy}{dx}=f'(x)$
$$
f'(x) \left( y+1 \right)^2 = \frac{1}{1+6x}
$$
means
$$\frac{dy}{dx}\left( y+1 \right)^2 =\frac{1}{1+6x}$$
and then
$$\left( y+1 \right)^2\,dy=\frac{dx}{1+6x}$$
now integrate both sides. Don't forget the constant of integration which can be found with the initial condition.

Title: derivatives of function f: [0; + ∞)

Body:
ok, this exercise is quite difficult, at least (for me).
f: [0; + ∞) → R satisfies
$$
f'(x) \left( f(x)+1 \right)^2 = \frac{1}{1+6x}
$$

for $x>0,  f(0)=2$ . So $f(x)$= ?

I'm really quite confused where to start. I tried to do something but it gave me 3√(1/2 ln(1+6x) +27 -1) but I don't think it's okay


问:
De Rham Theorem
I was just reading about De Rham theorem :

Let $\Omega$ be an open set of $\mathbb R^n$.
Let pose $\mathcal V := \{u=(u_1,...,u_N)\in  \mathcal D^N (\Omega)\}$, And $f=(f_1,...,f_N) \in (\mathcal D'  (\Omega))^N $; So The two properties are equivalent:

1) $\exists P \in \mathcal D' (\Omega)$ such that: $f = \nabla  P . \\ $

2) $\left \langle f,v \right \rangle = 0$, $\forall v \in \mathcal V .$

I want to prove that $2\Rightarrow 1 $, ($1\Rightarrow 2$  is OKEY).

I need a reference for learning the proof. I will be very happy if someone could help me to prove it.

Thanks!

答:
I know it's a bit late, but you can find the original proof in

De   Rham,     G.:   Differentiable Manifolds  1984.

at page 114, theorem 17.

Title: De Rham Theorem


问:
How to infer about the distribution of $Y$ from the given
Let $X\sim\mathsf{Ber}\left(p\right)$. Let $Y$ be random variable such that $\left(Y|X=k\right)\sim\mathsf{Poi}(k+1)$ for $k\in\{0,1\}$. find the distribution of $Y$

my attempt:

$$P(Y|X=k)=\frac{P(Y=t,X=k)}{P(X=k)} $$ $$P(Y|X=k)=\frac{P(Y=t,X=k)}{p(1-p)}$$

using the given about $P(Y|X=k)=$ I think I can write $$P(Y|X=k)=\frac{(k+1)^n}{n!}e^{-(k+1)}$$

from here I dont know how to procced. I am not sure if I wrote the expressions for $P(Y|X=k)$ and $P(X=k)$ correctly

Body: $$\mathbb{P}(Y=l) = \mathbb{P}(Y=l,X=0)+\mathbb{P}(Y=l,X=1) = \mathbb{P}(Y=l|X=0)\mathbb{P}(X=0)+\mathbb{P}(Y=l|X=1)\mathbb{P}(X=1) = (1-p)\cdot \frac{1^l}{l!}e^{-1} + p\frac{2^l}{l!}e^{-2} = \frac{e^{-1}}{l!}(p2^le^{-1}+(1-p))$$


问:
Find the number x such that the point $C=(x,0)$ achieves the minimum of the function $|AC|+|BC|$
Consider two points $A=(0,5)$ and $B=(10,3)$ in the $xy$-plane. Find the number $x$ such that the point $C=(x,0)$ achieves the minimum of the function $|AC|+|BC|$. Here $|AC|$ denotes the distance between the two points $A$ and $C$, and $|BC|$ denotes the distance between the two points $B$ and $C$.

I don't know how to approach this question... I would assume you have to use the distance formula... but not too sure what to do after that part... Would appreciate it if you could help me out with this question!

答:
Consider the point $B'(10,-3)$ symmetric of $B$ wrt $x$-axis.

The point $C$ where $AB'$ intersects the $x$-axis is the wanted point.

Indeed any other point $D$ has distances $AD+DB>AB$ for the triangular inequality

Hope this helps
$$...$$


问:
How to show that these integrals converge?
What test do I use to show that the following integral converges?
If you could provide me with the process that leads to the answer that would really help.

$\displaystyle  \int_{0}^{1}\frac{x^n}{\sqrt{1-x^4}}\,dx$

$\displaystyle \int_{0}^{\pi /2}\frac{\ln(\sin(x))}{\sqrt{x}}\,dx$

Thanks

答:
To check convergence at $x=0$, check that the integrand behaves as $x^a$, where $a>-1$.  If there is something that looks like a singularity away from $x=0$, say at $x=x_0$, then substitute $y=x_0-x$.  Further, $\ln{x}$ represents an integrable singularity.

For the first integral, you need to check the that singularity at $x=1$ is integrable; i.e., that, when you substitute $y=1-x$, then the integral behaves as $y^a$, where $a>-1$ at $y=0$.  That is, check the behavior of

$$[1-(1-y)^4]^{-1/2}$$

at $y=0$.

For the second integral, use the fact that $x^a \ln{x}$ is integrable at $x=0$ when $a>-1$.  How does $\sin{x}$ behave near $x=0$?


问:
How to prove that $\mathbb{N}$ is a set?
We all know that the collection of all natural numbers $\mathbb{N}$ is a set. But before that how do we prove it really is a set, by using only Zermelo-Fraenkel (ZF) axioms.

I tried using the axiom of infinity, but could not come up with a satisfactory proof ( or you can say construction). Im now wondering can we actually prove that $\mathbb{N}$ is a set using only ZF axioms? If yes how?. Any help is appreciated.

Edit: Here the collection of natural numbers is already assume to exist by Peano's axiom. The collection of natural numbers is define to be a collection $\mathbb{N}$, that satisfies the following:

Peano's Axioms

(i)  $1\in \mathbb{N}$.

(ii) For each $a\in \mathbb{N}$, there exists a unique $a'\in \mathbb{N}$, called the successor of $a$.

(iii) $a\neq a'$ for any $a\in \mathbb{N}$.

(iv) For any $a,b \in \mathbb{N}$, $a'=b' \implies a=b$.

(v) If $S$ is a subset $\big($I should say $S$ is a subcollection, because we didn't know yet whether $\mathbb{N}$ is a set$\big)$of $\mathbb{N}$ such that : (a) $1\in S$, and (b) if $n\in S $ then $ n'\in S$. Then $S= \mathbb{N}$.

Now I want to prove that the above collection $\mathbb{N}$ is indeed a set using the following ZF axioms.

ZF Axioms

(i)(Extension Axiom.) Any two sets are equal if and only if they have the same elements.

(ii)(Empty Set Axiom.) There exist a set with no elements.(It is called the empty set and denoted by $\emptyset$.)

(ii)(Pairing Axiom) For any two sets $A$ and $B$, there exist a set $\{A,B\}$ consisting of exactly $A$ and $B$ as elements.

(iv)(Union Axiom.) let $\mathcal{F}$ be a set. Then there exist a set $S$ whose elements belong to atleast one element of $\mathcal{F}$ .

(v)(Subset Axiom.) Let $\phi(x)$ be a formula. If $A$ is a set then the collection of all $x\in A$ such that $\phi(x)$ is a set. ($S$ is called a subset of $A$.)

(Vi)(Power Set Axiom.) Let $A$ be a set then there exist a set $P(A)$ consisting of all subsets of $A$.($P(A)$ is called the power set of $A$)

(vii)(Replacement Axiom). Let $\phi(x,y)$ be a formula in two variables. Let $A$ be a set. If for each $x\in A$ there exist a unique $y$ such that $\phi(x,y)$, then the collection of all $y$ in which there exist atleast one $x\in A$ such that $\phi(x,y)$ is a set.

(viii)(Infinity Axiom.) There exist a set $I$ such that $\emptyset\in I$ and for each $x\in I$, $x\cup\{x\}\in I.$

(ix)(Regularity Axiom.) For every non-empty set $A$ there exist atleast one element $x\in A$ that $x$ is disjoint from $A$.(i.e., there exist $x\in A $ such that $x\cap A = \emptyset$.)

答:
There are various ways to prove the existence of $\mathbb{N}$. I assume that $\mathbb{N}$ is the least inductive set (i.e., a set $I$ such that $0\in I$ and $x\in I\to S(x)\in I$, where $S(x)=x\cup\{x\}$ is a successor of $x$) and the axiom of infinity is the statement there is an inductive set.

In your case (i.e., working with $\mathsf{ZF}$), relying on the axiom of separation suffices. Observe that $\mathbb{N}$ is the intersection of all inductive sets. Hence the following equality holds:
$$\mathbb{N} = \{x\in I \mid \forall J (\text{$J$ is inductive}\to x\in J)\}.$$

Some textbooks state the axiom of infinity as there is an infinite set, and do not mention anything about inductive sets. Even in that case, we can define what finite sets and infinite sets are. (For example, we may define natural numbers as ordinals whose elements are 0 or successor ordinals, and finite sets as an image of a natural number.)

Moreover, we can still prove $\mathbb{N}$ exists: let $I$ be an infinite set. For each natural number $n$, we can find $x\subseteq I$ whose cardinality is $n$. Hence

$$\{x\in\mathcal{P}(I): |x|=n\}$$

is a non-empty set. (where $\mathcal{P}(X)$ is the power set of $X$.) Now consider the set

$$A = \{x\in\mathcal{P}(I) \mid \text{$x$ is finite}\}$$
and a map $F$ of domain $A$, defined by $F(x)=|x|$. By the axiom of replacement, the image of $F$ exists, and it is $\mathbb{N}$.


问:
Can anything be said about the average value of X using its CDF?
Marginal CDF of X:

$$F(x)=\begin{cases} 0 & \text{if }x\in(-\infty,-3)\\
\frac{2x+11}{25}&x \in [-3,2)\\
1&x\in[2,\infty)
\end{cases}$$

Can we say anything about $X$’s average value?

I don’t think we can find the expected value of $X$, because $X$ is not absolutely continuous so we can’t find $X$’s pdf. Also, I said that $X$ is not a non-negative random variable so we can’t do that method either. When I think of saying something about $X$’s average value, I think of the expected value of $X$ or finding a bound of the average value. The latter I don’t think we can do either.

Am I missing something? Can we say anything about X’s average value?

答:
Your rv has 2 discontinuity points

$X=-3$ with probability $\frac{1}{5}$

$X=2$ with probability $\frac{2}{5}$

Thus the expectation is

$$\mathbb{E}[X]=-3\times\frac{1}{5}+\int_{-3}^{2}\frac{2}{25}xdx+2\times\frac{2}{5}=0$$

so we can’t find X’s pdf.

Actually there is the possibility to derive a "mixed PDF" with two discrete points with probability $>0$

This is often used in Telecommunication Enigineering


问:
Prove that a metric space $M$ is compact iff every fully ordered family of closed and non-empty subsets of $M$ has a non-empty intersection.
We say that a family of sets $\{A_\lambda\}_{λ\in\Lambda}$ is fully ordered, if given $\lambda,\mu \in \Lambda$, one has $A_\lambda \subset A_\mu$ or $A_\mu \subset A_\lambda$. Prove that a metric space $M$ is compact if, and only if, every fully ordered family of closed and non-empty subsets of $M$ has a non-empty intersection.

Could someone help me?

I tried use Bolzano-Weierstrass property, or more simply, using a theorem relating the compactness and the finite intersection property. But I am not sure how to prove.

答:
Assume that $M$ is compact and let $(A_\lambda)_{\lambda\in\Lambda}$ be a fully ordered family of closed and non-empty subsets of $M$. Suppose that $\bigcap_{\lambda\in\Lambda}A_\lambda=\emptyset$. Then $\left(A_\lambda^{\,\complement}\right)_{\lambda\in\Lambda}$ is an open cover of $M$ and therefore it has a finite subcover $\left(A_\lambda^{\,\complement}\right)_{\lambda\in F}$, where $F$ is a finite subset of $\Lambda$. But there is a $\lambda_0\in F$ such that $(\forall\lambda\in F):A_\lambda\supset A_{\lambda_0}$ and therefore $\bigcap_{\lambda\in F}A_\lambda\supset A_{\lambda_0}$. In particular, $\bigcap_{\lambda\in F}A_\lambda\ne\emptyset$, which is impossible, since$$\left(\bigcap_{\lambda\in F}A_\lambda\right)^\complement=\bigcup_{\lambda\in F}A_\lambda^{\,\complement}=M.$$

And if $M$ is not compact, there is some sequence $(x_n)_{n\in\Bbb N}$ of elements of $M$ without any convergent sequence. Let $A_n=\overline{\{x_n,x_{n+1},x_{n+2},\ldots\}}$. Then each $A_n$ is closed and non-empty, and $(A_n)_{n\in\Bbb N}$ is fully ordered, but $\bigcap_{n\in\Bbb N}A_n=\emptyset$, since any element of $\bigcap_{n\in\Bbb N}A_n$ would be the limit of some subsequence of $(x_n)_{n\in\Bbb N}$.


问:
Simplify $(1+i)^{2017} + (i-1)^{2017}$
I had to simplify complex number $(1+i)^{2017} + (i-1)^{2017}$ where $i= \sqrt {-1} $ on a test but I could not do it as the exponent is so large.

Correct answer was $2^{1009}i$

I tried to convert it to trigonometric form and got

$(\sqrt 2)^{2017}(\cos(\frac{2017\pi}{4}) +i\sin(\frac{2017\pi}{4})) \quad + \quad (\sqrt 2)^{2017}(\cos(\frac{2017\cdot 3\pi}{4}) +i\sin(\frac{2017\cdot 3\pi}{4})) $

But I need help going on from here

Any help is appreciated!

Title: Simplify $(1+i)^{2017} + (i-1)^{2017}$

Body:
Hint: $$(1+i)^2 = 1+2i+i^2 = 2i$$

So $$(1+i)^{2n+1} = (1+i)(1+i)^{2n}= (1+i)2^ni^n$$

and $$(i-1)^{2n+1} = -(1-i)^{2n+1}$$


问:
An exercise about field extension
Let $K/F$ be a field extension, $L$ an intermediate field, $\alpha \in K$ is algebraic over $F$ with minimal polynomial $p(x) \in F[x] $. If $p(x)$ is irreducible in $L[x]$, then $F(\alpha)\cap L=F$.
Could you please help me this exercise? Thank you in advance.

答:
Take an element $\beta \in F(\alpha) \cap L$. Then you have that by the Tower Law

$$\deg p(x) = [F(\alpha) : F] = [F(\alpha) : F(\beta)][F(\beta) : F]$$

They key thing here is that $\beta \in L$ so $F(\beta) \subseteq L$. $p$ is irreducible in $L$ so it must be irreducible in $F(\beta)$ and $[F(\alpha) : F(\beta)] = \deg p(x)$.

From the above equality, $[F(\beta) : F] = 1$ so $\beta \in F$. This shows $F(\alpha) \cap L \subseteq F$. But clearly it contains $F$ so equality follows.

Title: An exercise about field extension


答:
Continuing from your trigonometric form, remember that $\cos(x+2\pi)=\cos(x) \implies \cos(\frac{2017\cdot 3}{4}\pi)=\cos(\frac{6051}{4}\pi)=\cos(1512\pi+\frac{3}{4}\pi)=\cos(\frac{3}{4}\pi)$

I believe you can take it from here :)

Title: Simplify $(1+i)^{2017} + (i-1)^{2017}$

Body:
I had to simplify complex number $(1+i)^{2017} + (i-1)^{2017}$ where $i= \sqrt {-1} $ on a test but I could not do it as the exponent is so large.

Correct answer was $2^{1009}i$

I tried to convert it to trigonometric form and got

$(\sqrt 2)^{2017}(\cos(\frac{2017\pi}{4}) +i\sin(\frac{2017\pi}{4})) \quad + \quad (\sqrt 2)^{2017}(\cos(\frac{2017\cdot 3\pi}{4}) +i\sin(\frac{2017\cdot 3\pi}{4})) $

But I need help going on from here

Any help is appreciated!


问:
Simple closed curve in $\mathbb C$
A simple closed curve is a continuous function from [a b] to $\mathbb C$. Does a pair of concentric circles represent a simple closed curve? What about 3 concentric circles?

My thinking is that a pair of concentric circles is not the continuous image of a closed and bounded interval in $\mathbb R$. Since continuity fails it can not be a simple closed curve. Similarly any collection of n concentric circles. Can someone throw light on the concept.

答:
If $f:[a,b] \to \mathbb C$ is continuous, then $f([a,b])$ is  connected, since $[a,b]$ is connected.

Any  collection of $n$ concentric circles is  not connected !

Conclusion ?

Title: Simple closed curve in $\mathbb C$


问:
Optimal Stock Selection Algorithm (Python)
Given a total amount of available capital, the values of stocks at the current date and the values of the same stocks at a later date, I need to find a way to calculate the optimal return if I can buy at most one share of any stock.

For example:

Given

```<code>input:
capital = 10
cur_vals = [3,4,5,6]
fut_vals = [6,7,5,3]

output:
6
</code>```

So far I've come up with the following:

```<code>def optimal_select(spend, cur_value, fut_value):

    #create list of (returns,current_vals)

    diffs = sorted([(fut_value[i] - cur_value[i],cur_value[i]) for i in range(len(cur_value))], reverse = True)

    values = []

    for k in range(len(diffs)):

        #initial balance:
        balance = spend - diffs[k][1]

        #initial returns
        returns = diffs[k][0]

        #attempt to add each additional stock by highest return

        for j in range(k+1,len(diffs)):

            if diffs[j][1] <= balance and diffs[j][1] > 0:

                balance += -diffs[j][1]
                returns += diffs[j][0]

        values.append(returns)

    return max(values)
</code>```

Although, I know this won't give me the correct answer all the time because it doesn't account for the case where it makes sense to forgo the highest return selection at a given iteration for multiple less costly stocks that sum to higher returns.
For example:

```<code>optimal_select(8, [4,4,2,2],[10,9,5,5])

output: 11
</code>```

Where the correct output in this case should be 12.

Is there a particular algorithm or area of optimization that I can reference that might help me better structure my approach?

答:
Thanks to the comment, I was able to work through the correct solution with the 0-1 knapsack problem and the following code:</p>
```<code>def optimal_selection(w, cost, returns):

    m = [[None for i in range(len(cost)+1)] for j in range(w + 1)]

    for j in range(w+1):
        m[j][0] = 0

    for i in range(1,len(cost)+1):

        for j in range(w+1):

            if cost[i-1] > j:

                m[j][i] = m[j][i-1]

            else:

                m[j][i] = max(m[j][i-1], m[j-cost[i-1]][i-1] + returns[i-1])

    return max([i for sub in m for i in sub])

<span class="math-container">```</span>
</code>```

Body:
Thanks to the comment, I was able to work through the correct solution with the 0-1 knapsack problem and the following code:

def optimal_selection(w, cost, returns):

    m = [[None for i in range(len(cost)+1)] for j in range(w + 1)]

    for j in range(w+1):
        m[j][0] = 0

    for i in range(1,len(cost)+1):

        for j in range(w+1):

            if cost[i-1] > j:

                m[j][i] = m[j][i-1]

            else:

                m[j][i] = max(m[j][i-1], m[j-cost[i-1]][i-1] + returns[i-1])

    return max([i for sub in m for i in sub])

```


问:
Another approach for calculating the sum
$$
\sum_{n=0}^{\infty}(-\cos \beta)^{n} \frac{\sqrt{\pi} \Gamma\left(\frac{n}{2}+\frac{1}{2}\right)}{2 \Gamma\left(\frac{n}{2}+1\right)}, \beta\in(o,\pi)
$$

Answer:
$$\frac{\pi-2\arcsin\cos\beta}{2\sin\beta}=\frac{\beta}{\sin\beta}$$

I have an indirect proof， but it is more complicated. I believe there is a more elegant way to do this.

My complicated answer:
$$\int_{0}^{\frac{\pi}{2}} \frac{\mathrm{d} \alpha}{1+\cos \alpha \cos \beta} =\int_{0}^{\frac{\pi}{2}} \sum_{n=0}^{\infty}(-\cos \alpha \cos \beta)^{n} \mathrm{~d} \alpha \\
  =\sum_{n=0}^{\infty}(-\cos \beta)^{n} \int_{0}^{\frac{\pi}{2}} \cos ^{n} \alpha \mathrm{d} \alpha \\
  =\sum_{n=0}^{\infty}(-\cos \beta)^{n} \frac{\sqrt{\pi} \Gamma\left(\frac{n}{2}+\frac{1}{2}\right)}{2 \Gamma\left(\frac{n}{2}+1\right)} \\
$$
On the other hand:
$$\int \frac{d x}{1+\varepsilon \cos x}=\frac{1}{\sqrt{1-\epsilon^{2}}} \operatorname{arctan}\left(\sqrt{\frac{1-\varepsilon}{1+\varepsilon}} \operatorname{tan} \frac{x}{2}\right)+C, \varepsilon\in(-1,1)$$
In this way, I can get the answer. But it is complicate. So I need help.

答:
Using Legendre's duplication formula, $\Gamma(z)\Gamma(z+1/2)=2^{1-2z}\pi^{1/2}\Gamma(2z)$, your sum
$$\frac{\pi}{2}+\frac{\sqrt{\pi}}{2}\sum_{n=1}^{\infty}(-\cos \beta)^{n} \frac{\Gamma\left(\frac{n}{2}+\frac{1}{2}\right)}{ \Gamma\left(\frac{n}{2}+1\right)}$$
can be rewritten
$$
\frac{\pi}{2}+\frac{\sqrt{\pi}}{2}\sum_{n=1}^{\infty}(-\cos \beta)^{n}\frac{2^{1-n}\pi^{1/2}\Gamma(n)}{\Gamma(n/2)}\frac{2}{n\Gamma(n/2)}
=\frac{\pi}{2}+2\pi\sum_{n=1}^{\infty}x^n\frac{(n-1)!}{n\Gamma^2(n/2)}.
$$
where $x=-1/2\cos\beta$. Splitting into $n$ even/odd and applying the duplication formula again gives,
$$\frac{\pi}{2}+\left(2\pi\sum_{n=1}^{\infty}x^{2n}\frac{(2n-1)!}{2n(n-1)!^2}\right)+\left(2x+\frac{1}{8}\sum_{n=1}^{\infty}(4x)^{2n+1}\frac{(2n)!}{(2n+1)}\frac{(n-1)!^2}{(2n-1)!^2}\right).$$
The coefficients of powers of $x$ in the left hand sum appear to be A00170 and those of the right, $1/n$ multiples of reciprocals of Apéry numbers A005430, so we have
$$\frac{\pi}{2}+2x+\pi\sum_{n=1}^{\infty}{2n-1\choose n-1}x^{2n}+\sum_{n=1}^{\infty}\frac{1}{(n+1){2n+2\choose n+1}}(4x)^{2n+1}.$$

Mathematica has no problem evaluating these sums to obtain
$$\frac{\pi -\cos ^{-1}(2 x)}{\sqrt{1-4 x^2}}=\frac{\beta}{\sin\beta},$$
for $\beta\in(0,\pi)$. Check out this Binomial Sums link to see sums very similar to those here.

This might not be as elegant as you're looking for (evaluating the sums might not be considered elegant) but I spent some time on it so thought I'd show it anyway.


问:
How to prove $\int_{0}^{2\pi } \frac{ \sin(\theta - \phi) } { R^2 - 2rR \cos(\theta -\phi) +r^2 } d{\phi} =0$
This question was part of mock test of masters exam for which I am preparing and I am unable to solve it.

Show that  $\int_{0}^{2\pi }  \frac{ \sin(\theta - \phi) } { R^2 - 2rR \cos(\theta -\phi) +r^2 } d{\phi} =0$ .

The part ${ R^2 - 2rR \cos(\theta -\phi) +r^2 }$ seems to be from Poisson's integral fomula for harmonic functions butI am not sure how to use it here. (Due to this reason i am using harmonic functions tag )

If u(z) is harmonic in a domain consisting of disc $|z|\leq R$ . Then for $z=r e^{i\theta}$ we have ,

$u(r e^{i \theta})=\frac{1}{2\pi}\int_{0}^{2\pi} \frac{R^2 -r^2} { R^2 - 2rR \cos(\theta - \phi) +r^2 } u(R e^{i \phi})d {\phi}$.

But $\sin(\theta -\phi)$ is an implicit function.So, I am not able to find u(z) and $R^2 -r^2$.

Can you please help me in proving it?

答: Since the function that you are integrating is periodic with period $2\pi$, your integral is equal to$$\int_{\theta-\pi}^{\theta+\pi}\frac{\sin(\theta-\phi)}{R^2-2rR\cos(\theta-\phi)+r^2}\,\mathrm d\phi$$andd now the substitution $\theta-\phi=t$ and $\mathrm d\phi=-\mathrm dt$ leads to$$\int_{-\pi}^\pi\frac{\sin t}{R^2-2rR\cos(t)+r^2}\,\mathrm dt,$$which is $0$, since you are integrating an odd function.


Title: Find the number x such that the point $C=(x,0)$ achieves the minimum of the function $|AC|+|BC|$

Body:
Consider two points $A=(0,5)$ and $B=(10,3)$ in the $xy$-plane. Find the number $x$ such that the point $C=(x,0)$ achieves the minimum of the function $|AC|+|BC|$. Here $|AC|$ denotes the distance between the two points $A$ and $C$, and $|BC|$ denotes the distance between the two points $B$ and $C$.

I don't know how to approach this question... I would assume you have to use the distance formula... but not too sure what to do after that part... Would appreciate it if you could help me out with this question!

Body:
Using the distance formula, $dist(A,B) = \sqrt{(X_B-X_A)^2+(Y_B-Y_A)^2}$:

$$|AC| = dist(A,C) = \sqrt{(X_C-X_A)^2+(Y_C-Y_A)^2}=\sqrt{(x-0)^2+(0-5)^2} = \sqrt{x^2+25}=(x^2+25)^{0.5}$$

$$|BC| = dist(B,C) = \sqrt{(X_C-X_B)^2+(Y_C-Y_B)^2}=\sqrt{(x-10)^2+(0-3)^2} = (x^2-20x+109)^{0.5}$$

Let $f(x)=|AC|+|BC|= (x^2+25)^{0.5} + (x^2-20x+109)^{0.5}$.

Now, in order to find the minimum of the function, we need to find $f'(x)$ and equate it to $0$.

Using the chain rule:

$$f'(x) = 0.5(x^2+25)^{-0.5}\cdot2x+0.5(x^2-20x+109)^{-0.5}\cdot(2x-20) = \frac{2x}{2\sqrt{x^2+25}}+\frac{2x-20}{2\sqrt{x^2-20x+109}} = \frac{x}{\sqrt{x^2+25}}+\frac{x-10}{\sqrt{x^2-20x+109}}$$

Now, equate the derivative to $0$ and solve for $x$:

$$\frac{x}{\sqrt{x^2+25}}+\frac{x-10}{\sqrt{x^2-20x+109}}=0$$

The equation yields two possible $x$ values: $x_1=6.25$ and $x_2=25$.
Note that:

$f(x_1)=f(6.25)=12.806$

$f(x_2)=f(25)=40.792$

Therefore, the correct answer is $x=6.25$.


问:
Calculate bounds of the inner rectangle of the polygon based on it's constant bounds along with border
Yesterday I asked this question seems not clear. So, I am writing new question with mathematical terms only.

Here is some points about shapes I used:

Square: Rectangle that all angles are 90° & all sides are equal length. So, consider Square also a Rectangle.

Triangle: I used an equilateral triangle is a triangle in which all three sides have the same length.

Polygon: A finite number of straight line segments connected to form a closed.

So, Here after I use the term Polygon to describe triangle & rectangle shapes when it's required to mention consider this shape as polygone. Becuase, I used triangle (regular polygon) to simplfy the work to find out the formula.

Variables used in this problem:

b = Border width

Bounds of Outer Rectangle (r) that enclosing the polygon(p):

x = Origin x

y = Origin y

w = Width

h = Height

So, If r = (x,y,w,h) then, Bounds of Inner Rectangle that encloses the inner polygon(p1) (r1) = (x1,y1,w1,h1)

Here, inner polygon = Polygon by subtracting the border width.

What is my final goal:

I should able to calculate inner rectangle boundary of any polygon like below image, I have angles, outer rectangle bounds & border width. The problem is to find out the inner rectangle bounds across to the outer rectangle based on the polygon at that enclosed by inner rectangle.

Problem 1:

Consider a Polygon(p) as a shape of rectangle(r) which has width(w) is 100 & height(h) is 100. The border width of the polygon is 10. The what is the inner boundary of the rectangle that encloses the inner polygon?

Note: I solved it by mind calculation(I didn't took angles in the calculation). But we can't apply this same
formula for triangle. So, this is not solution for my problem.

Problem 2:

Consider the polygon (don't consider regulard) with three edges and three vertices & all angles are equal in measure.

If the rectangle(r) boundary is zero & size(w,h) is (100,87) that encloses the polygon(p) along with polygon border width as 10, then

What is the bounds of the inner rectangle(r2) that encloses the inner polygon(p2) ?

Hope I clearly explained the problem.  If anything I am missed then please let me know. Thanks in advance.

答:
Inner triangle is the image of outer triangle under a homothetic transformation centred at the centre of the triangle.

The centre of the triangle is located on the altitude, at $1/3$ of its length starting from the base.

The ratio $r$ of the homothety can be computed as the ratio of the distances from the centre to the midpoint of the bases:
$$
r={{1\over3}{\sqrt3\over2}100-10\over {1\over3}{\sqrt3\over2}100}=
1-{\sqrt3\over 5}.
$$

Inner rectangle is then the image of outer rectangle under the same homothety.


Title: Simplify $(1+i)^{2017} + (i-1)^{2017}$

Body:
$e^{i\theta}=\cos \theta+i\sin \theta$.

$(i+1)=\sqrt{2}\cdot e^{\frac{\pi i}{4}}$,$\qquad $ $(i-1)=\sqrt{2}\cdot e^{\frac{3\pi i}
{4}}$.

$(i+1)^n+(i-1)^n=2^{\frac{n}{2}}\cdot (e^{\frac{n\pi i}{4}}+e^{\frac{3n\pi i}{4}})$.

$2017\equiv 1\pmod 8$.

$(i+1)^{2017}+(i-1)^{2017}=2^{\frac{2017}{2}}\cdot (e^{\frac{\pi i}{4}}+e^{\frac{3\pi i}{4}})=2^{\frac{2016}{2}}\cdot 2i=2^{1009} \cdot i$.


问:
Does the equation $k^k+1=a^2$ have any positive integer solutions?
As the title of the question says, I want to know if the equation
$k^k+1=a^2$
has any integer positive solutions, (which means k and a are both positive integers.)

I'd really appreciate it if someone could provide the following:

If the given solutions exists, please provide the minimal solution. Are the solutions infinite or finite?

If the given solutions doesn't exist, please provide your proof.

答:
There are no positive integer solutions by Mihailescu's theorem.

With a less powerful hammer, Ke Zhao proved the nonexistence of solutions to $x^y+1=a^2$ in 1964. See the last reference in this answer.


问:
Calculate time to fill an empty tank

An empty tank can be filled with water in 20 minutes by using Pipe A
  or in 30 minutes by Pipe B, and the tank filled with water can be
  emptied of water in 40 minutes by using Pipe C. When the three pipes
  A, B, and C work together, approximately how long (in minutes) does it
  take to fill the empty tank with water?

I would love to know the method to solve this kind of math. Thanks lot!

答:
Efficiency method

Pipe                   A            ,                B                      ,          -C
Time                 20               ,           30                   ,            40

Lcm : 120

Efficiency     6 (120/20) ,          4(120/30)          ,      -3 (120/40)

Total capacity of tank = 120 litre (lcm)

A= fill 6 part per min,
B = fill 4 part per min,
C = empty 3 part per min

A,B,C  opened together = A+B-C =6+4-3=  fill 7 part per min..
So, time taken to fill the tank = 120/7 min or 17.14 min

Title: Calculate time to fill an empty tank


问:
Marble Possibility P(At least one yellow)
There are $2$ black and $3$ yellow marbles in a bag. $2$ marbles are drawn randomly without replacement. What is the possibility that at least $1$ yellow marble is selected. </p>

答:
Imagine drawing the balls one at a time. The probability the first ball is black is $\frac{2}{5}$. Given that the first ball was black, the probability the second is black is $\frac{1}{4}$ (there is only $1$ black left, out of $4$ balls). So the probability both are black is $\frac{2}{5}\cdot\frac{1}{4}$.

It follows that the probability that there is at least one yellow is $1-\frac{2}{5}\cdot\frac{1}{4}$.

There are many other ways to solve the problem. Imagine the balls have ID numbers. There are $\binom{5}{2}$ equally likely ways to choose $2$ balls. There is only $1$ way to choose $2$ black, so the probability of $2$ black is $\frac{1}{\binom{5}{2}}$. It follows that the probability of at least one yellow is $1-\frac{1}{\binom{5}{2}}$.

In each of these solutions, we worked with the blacks, because it is easier. But we could work with the yellows. Either of the two ways we solved the problem can be adapted.

For example, we get at least one yellow if we draw BY (black, then yellow) or YB or YY.

The probability of BY is $\frac{2}{5}\cdot\frac{3}{4}$. In a similar way, find the probability of YB and of YY, and add up.

We can also do a counting argument. There are $\binom{5}{2}$ ways to choose $2$ balls. How many result in at least one yellow? We could have a yellow and a black, which can be chosen in $\binom{3}{1}\binom{2}{1}$ ways. Or we can have two yellow, which can be chosen in $\binom{3}{2}\binom{2}{0}$ ways. So our probability is $\frac{\binom{3}{1}\binom{2}{1}+  \binom{3}{2}\binom{2}{0}}{\binom{5}{2}}$.


问:
Prove that volume of a cylinder section cut off by a plane is $\frac16$ of circumscribing parallelepipe
I was given the following question:

Given a cylinder circumscribed within a parallelepiped with a square base that has a plane going through the center of the base circle and through one side of the square on the top of the parallelepiped. Use calculus to prove that the volume of the segment of the is $\frac16$ the volume of the rectangular parallelepiped circumscribing the cylinder.

Here is what I've done:

I created a diagram of the described cylinder on GeoGebra (see below). Given that the base is a square, the volume of the parallelepiped can be given as $a^2b$ (if necessary this can be shown using calculus by doing $\int^a_0 ab\ dx$). The radius of the cylinder is $\frac a2$ and the height is $b$. The volume would therefore be: $\frac{\pi a^2}{4}b$. The only problem is that I need to factor in the plane. I would take the integral of the equation of the plane minus the area of each cross section of the cylinder ($\frac{\pi a^2}{4}$) but I cannot figure out the equation of the plane! I know it goes through the point $(\frac a2,\frac a2,0)$ and the line $z=b,x=a$ (I'm not sure if I wrote that line right). But I don't know how to get the equation of that plane.

Am I going about this problem in the right way? How can I find that equation?

答:
Let the cylinder be $x^2+ y^2=\frac{a^2}4$.  Then, the corresponding cutting plane is $z=\frac {2b}a x$ and the section volume is

$$\int_ {-\frac a2} ^{\frac a2} \int_0^{\sqrt{\frac{a^2}4-y^2} }z \>dxdy
= \int_{-\frac a2} ^{\frac a2} \int_0^{\sqrt{\frac{a^2}4-y^2} }\frac {2b}a x\> dxdy
= \frac16 a^2b$$
which is $\frac16$ that of the rectangular parallelepipe.


问:
homeomorphism on the sphere
I want to prove that for all homeomorphism (or diffeo) $g:S^2\to S^2$ on the sphere, it doesn't exists $\lambda\in (0,1)$ with
$$d(g^n(x),g^n(y))\leq \lambda^nd(x,y)$$
for every pair of points $x,y\in S^2$ and $n>0$. Here the distance $d(x,y)$ is the the infimum of the lengths of differentiable curves in $S^2$ connecting the two points.

I know that if the above relation holds for some $\lambda\in (0,1)$, then every two points will be iteratively brought closer to each other. So, every open set will be condenced to compact sets or points when $n\to \infty$. This should mean that $f$ is not even bijective, but I don't know how to formalize it properly. Any help?

Maybe there is also an alternative proof by the Brouwer fixed point but I'm not sure either.

答: Hint: Such an homeomorphism has a fixed point $x$, (for every $z$, $g^n(z)$ is Cauchy)  if $y$ is another point, $d(x,g(x))< d(x,y)\leq 1$ implies that $g$ is not surjective. Contradiction.

Title: homeomorphism on the sphere


问:
Affine subspace generated by inner product
Let $v$ be a vector of $\Bbb{R}^n$ and $c \in \Bbb{R}$. Let $A$ be a point of the affine space $\Bbb{R}^n$.

Show that $\mathcal{E} = \{ B \in \Bbb{R}^n | \langle \overrightarrow{AB},v \rangle = c \}$ is an affine subspace and give its direction and dimension.

When I write down the inner product $\langle \overrightarrow{AB},v \rangle = c$ it gives me :
$$ v_1 (b_1 - a_1) + \dots + v_n (b_n - a_n ) = c $$
$$ v_1 b_1 + \dots + v_n b_n = c + v_1 a_1 + \dots v_n a_n $$
This instantaneously show that $\mathcal{E}$ is an affine subspace because of the form of the equation.
So the director $\overrightarrow{\mathcal{E}}$ is defined by the homogeneous equation
$$ v_1 w_1 + \dots + v_n w_n = 0, ~ w \in \Bbb{R}$$
which is a vector subspace of dimension $n-1$. So the dimension of $\mathcal{E}$ is $n-1$.

Does it seem correct ?

答:
You are correct. To be more precise: the direction is

$$\{v\}^{\perp}.$$


问:
How to show that these integrals converge?
What test do I use to show that the following integral converges?
If you could provide me with the process that leads to the answer that would really help.

$\displaystyle  \int_{0}^{1}\frac{x^n}{\sqrt{1-x^4}}\,dx$

$\displaystyle \int_{0}^{\pi /2}\frac{\ln(\sin(x))}{\sqrt{x}}\,dx$

Thanks

答:
I like to use the comparison tests for these kind of problems.

For the first integral, use the fact that $x^n \leq x$ on $x \in [0,1]$ $\forall n \in \mathbb{N}$, and that $x^4<x^2$ on $[0,1]$, giving
$$\frac{x^n}{\sqrt{1-x^4}} \leq \frac{x}{\sqrt{1-x^2}}$$
and use the fact that $f\leq g \implies \int f \leq \int g$ on any bounded interval.

Hence,

$$\begin{align}
\int_0^1 \frac{x^n}{\sqrt{1-x^4}} \mathrm{d}x &\leq \int_0^1 \frac{x}{\sqrt{1-x^2}} \mathrm{d}x
\\
&=\left. -\sqrt{1-x^2} +c \right|_0^1
\\
&= 1
\end{align}$$

For the second use a similar idea, except here note that $\ln x < x$ $\forall x>0$ and that $\sin x< x$ for all $x$.

We do all this to get the integral of $g$ on the right side to be one which we can compute in closed form, which gives a bound on the integral of $f$ which you want to show converges, which is of course the standard comparison test.


问:
Specify the maximum interval for solving the Cauchy problem depending on the initial data
I have a differential equation: $$y'+ \frac{2y}{xln(x)}+ y^2(ln(x))^2=0, x=e, y = y_0$$
I need to find the maximum interval for solving the Cauchy problem depending on the initial data.

So, I found a solution to the differential equation:
$$y=\frac{1}{(x+C)(ln(x))^2}$$
How can I find the maximum interval for my case ($x =e, y=y_0)$?

P.S. I don't know what I was doing, but I'll write it down just in case.
$$y_0=\frac{1}{C+e}$$
$$C=\frac{1-y_0e}{y_0}$$

答:
Putting $C$ in the solution gives
$$y=\frac{1}{(x+C)(\ln(x))^2}=\frac{1}{(x+\frac{1-ey_0}{y_0})(\ln(x))^2}.$$
If $\frac{1-ey_0}{y_0}\ge 0$, then the maximal interval is $(0,\infty)$. If $\frac{1-ey_0}{y_0}< 0$, then the maximal interval is $(0,-\frac{1-ey_0}{y_0})$.


问:
How to find the limit of $f(x)$.
Let $$f(x) = {\frac{e^{e^x - 1} - \frac{1}{1 - x}}{\ln(\frac{1 + x}{1-x}) - 2\sin x}}$$

I need to find a $ \lim_{x \to 0}f(x)$. As I understand, we have to do some Taylor series decomposition and problem should be pretty easy. But I have had some troubles with it.

P.S : I can decompose $\ln(\frac{1 + x}{1-x})$ as $\ln(1 + \frac{2x}{1-x})$ to use some standard Taylor formulas. I know $\sin x$ Taylor series, but I've got some troubles with $e^{e^x-1}$ and some further transformations.

Title: How to find the limit of $f(x)$.

Body:
I think that Taylor series decomposition is good.
But you can also use L'Hôpital's rule.

\begin{align}
&f(x)
=\dfrac{e^{e^x-1}-(1-x)^{-1}}{\ln|1+x|-\ln|1-x|-2\text{sin}x}.\\
\lim_{x\to 0} f(x)
&=\lim_{x\to 0}\dfrac{e^{e^x-1}-(1-x)^{-1}}{\ln|1+x|-\ln|1-x|-2\text{sin}x}\\
&=_{\text{L'Hôpital's rule}}\lim_{x\to 0} \dfrac{e^{e^x-1}e^x-(1-x)^{-2}}{(1+x)^{-1}+(1-x)^{-1}-2\text{cos}x}\\
&=_{\text{L'Hôpital's rule}}\lim_{x\to 0} \dfrac{e^{e^x-1}e^{2x}+e^{e^x-1}e^x-2(1-x)^{-3}}{-(1+x)^{-2}+(1-x)^{-2}+2\text{sin}x}\\
&=_{\text{L'Hôpital's rule}}\lim_{x\to 0} \dfrac{e^{e^x-1}e^xe^{2x}+e^{e^x-1}・2e^{2x}+e^{e^x-1}e^xe^x+e^{e^x-1}e^x-6(1-x)^{-4}}{2(1+x)^{-3}+2(1-x)^{-3}+2\text{cos}x}\\
&=\dfrac{1+2+1+1-6}{2+2+2}\\
&=-\dfrac{1}{6}.
\end{align}


问:
Counting the number of possible strings given some constraints
So this is more of a statement of an interesting approach I found (of solving a particular type of problem) than a question, but I would like to read up more on the subject, if there even is a subject here. So maybe that would be my question - where to investigate this topic further?

The question is about finding the number "string" representations given certain conditions. For example, consider the question: How many three digit numbers contain exactly one '3'? Now, I suppose you could write out the combinations manually, but what I found was a more general way of solving these kinds of problems.

The main part of the approach is to reword the problem in terms of a finite-state automaton, that is finding the possible states and the transitions between them, deriving a recurrence relation for the $n$-th case (where $n$ represents the string length), solving it and finding the desired value for a particular length. These states represent the mental model of when you try to solve such problems manually. For the example above, imagine trying to determine the possible values for the first digit, based on those the possible values of the second and lastly, based on those the ones for the third. You could argue that for the first digit you can have any digit other than '0' (that would make the number a two digit number). The possible digits in the second position depend on the digit in the first - if it was a '3', you can have a digit from '0'-'9' except '3', and if it wasn't you can have any. For the third digit, if you had a digit '3' in either the first or second place, you can have but '3', but if you haven't you can only have the digit '3'. So, as you can see, this gets complicated quickly.

In my approach, you recognize there are two states - one where the digit '3' was not yet included (let's call that state $N$), and one where it was (let's call that state $Y$). Now, you can see that you can stay in $N$ by choosing any digit but the digit '3', you can go from $N$ to $Y$ by choosing the digit '3' and you stay in $Y$ by choosing any digit but the digit '3'. Also, you can start in $N$ in 8 possible ways (any digit but '0' and '3') and you can start in $Y$ in one possible way (first digit is '3'). Here's a visual representation of that.

Now, let $N_n$ denote the number of possible combinations generated after $n$ steps and ending in state $N$, and $Y_n$ the same, but when ending in state $Y$. Now, looking at the diagram, we see that this must hold:

$N_n=9 N_{n-1}$ with $N_1=8$

$Y_n=N_{n-1} + 9 Y_{n-1}$ with $Y_1=1$

The reasoning behind this is that the number of combinations after $n$ steps that end in state $X$ is equal to the number of combinations after $n-1$ steps that end in any of the possible states times the number of ways we can get from a particular state to state $X$.

Now we just need to solve this, but this is relatively simple when written in matrix form like this:

$\begin{pmatrix}N_n \\ Y_n\end{pmatrix}=\begin{pmatrix}9 & 0 \\ 1 & 9\end{pmatrix}\begin{pmatrix}N_{n-1} \\ Y_{n-1}\end{pmatrix}=\begin{pmatrix}9 & 0 \\ 1 & 9\end{pmatrix}^{n-1}\begin{pmatrix}N_{1} \\ Y_{1}\end{pmatrix}=\begin{pmatrix}9 & 0 \\ 1 & 9\end{pmatrix}^{n-1}\begin{pmatrix}8 \\ 1\end{pmatrix}$

So, for the example above, when $n=3$ we have $\begin{pmatrix}N_3 \\ Y_3\end{pmatrix}=\begin{pmatrix}9 & 0 \\ 1 & 9\end{pmatrix}^{2}\begin{pmatrix}8 \\ 1\end{pmatrix}=\begin{pmatrix}648 \\ 225\end{pmatrix}$.

And since the valid end-state for this example is only the $Y$ state, we get the answer to be $Y_3=225$.

Now, for a simple example such as tone above, there doesn't seem to be much benefit in this over-the-top approach. However, I believe it does for examples that are a bit more complicated. An example to prove my point: What is the number of possible 4 digit numbers where the digits '0' and '1' are never next to each other? So $6013$ and $1210$ are note allowed, but $1304$ and $7007$ are. Doing this manually would be a chore (for me), but doing it like I've described is manageable. Here's the finite-state machine describing this problem. From this we have:

$\begin{pmatrix}N_n \\ Z_n \\ Y_n\end{pmatrix}=\begin{pmatrix}8 & 8 & 8 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix}^{n-1}\begin{pmatrix}8 \\ 0 \\ 1\end{pmatrix}$

Plugging in $n=4$ gives

$\begin{pmatrix}N_4 \\ Z_4 \\ Y_4\end{pmatrix}=\begin{pmatrix}6984 \\ 792 \\ 793\end{pmatrix}$

And since all of the states are valid final states, the total number of combinations is $N_4+Z_4+Y_4=8569$.

Note that it would probably be simpler to solve this example by calculating the number of valid strings as the number of all strings minus the number of invalid ones, but even that approach has it's downfalls.

So, that's it. Now - how could I expand my knowledge on this?

I hope this made a somewhat interesting read...

EDIT: Just to add one more example I find interesting: What's the number of possible binary strings (strings containing only digits '0' and '1') of length $n$ where the digit '1' is never in succession? Turns our the resulting matrix is the Fibonacci number matrix.

答:
Much of what you discuss can be simplified.

Three digit numbers with exactly one $3$:  if you allow the number to start with $0$, then you have to place the $3$ and then the others can be anything (other than $3$), hence $3\times 9^2=243$.   If you don't allow the first digit to be $0$ then we split into cases.  There are $9^2$ with first digit $3$ and then there are $2\times 8\times 9=144$ where the $3$ is not first.  Thus, in that case there are $81+144=225$.   I gather from your result that you are interested in the second case.

And your final problem can be addressed recursively, which makes the connection to Fibonacci more obvious.  Indeed, a good string of length at least $2$ must end in either $0$ or $01$, so $A_n=A_{n-1}+A_{n-2}$  (where, of course, $A_n$ is the number of good strings of length $n$).


问:
Suppose that the error, in grams, of a balance is a random variable with density $f(x) = \frac{1}{4}e^{−\frac{|x−1|}{2}}$...

Question:

Suppose that the error, in grammes, of a balance is a random variable with density $f(x) = \frac{1}{4}e^{−\frac{|x−1|}{2}}$, for $−\infty < x < \infty.$

Suppose that 100 items are weighed, and the errors of measurements are independent.

a) Find expectation and variance of the error in a single measurement.

b) Use the central limit theorem to find approximate value of the probability that absolute value of the total error is less than 50 grammes.

c) Use Chebyshev’s inequality to lower bound the probability that the total error is
between 50 and 150 grammes.

My Answer:

(a)
$\mathbb E(X)=\int_{-\infty}^{\infty}xf(x)dx=\frac{1}{4}\int_{-\infty}^{\infty}xe^{−\frac{|x−1|}{2}}dx=\frac{1}{4}\int_{-\infty}^{1}xe^{\frac{(x−1)}{2}}dx+\frac{1}{4}\int_{1}^{\infty}xe^{\frac{-(x−1)}{2}}dx=1$ (after integrating by parts a few times and using limits)

Similarly $Var(X)=\mathbb E(X^2)-(\mathbb E(x))^2=\int_{-\infty}^{\infty}\frac{x^2}{4}e^{\frac{-|x−1|}{2}}dx -1=9-1=8$ (after similar computations)

(b)
I've used my pervious answers $\mu =\mathbb E(X)=1$ and $Var(X)=\sigma^2=8$. To give $\overline{X_n}\sim N(1,\frac{8}{100})$

$\Rightarrow \mathbb P(\overline{X_n}<50)=\Phi(\frac{50-1}{{\sqrt\frac{8}{100}}})=1$ Makes me think I've made a mistake as I dont think that makes sense :)

(c)

Think I need to be confident with (b) to tackle (c)

Comments

I'm unsure of all questions really, I dont think I fully understand how to use the central limit theorem.
Any clarification, alternative solutions or even solution verification would be greatly appreciated

答: I'm doing a similar question for my homework, I found that E[X$^2$]=11, apologies, due to my reputation I am not able to comment.


问:
Lipschitz continuous
Let $\delta$ be an interval in $\mathbb{R}$. Recall that a function $f$ is called Lipschitz continuous on $\delta$ with Lipschitz constant $L$ if there holds
 $|f(x) - f(y)| \leq L|x-y|$ for all $x,y$ in $\delta$.

(a) Show that the composition of Lipschitz continuous functions is again Lipschitz continuous.

(b) Is the pointwise maximum of two Lipschitz continuous functions necessarily Lipschitz
continuous?

答:
For part b, I think YES.

Because,

$|\max(f(x),g(x))-\max(f(y),g(y))|$

can take $4$ values:

$|f(x)-f(y)|$,

$|g(x)-g(y)|$,

$|f(x)-g(y)|$ and

$|g(x)-f(y)|$.

For the first $2$ cases, $L1$ and $L2$ are the required Lipschitz constants, where

$|f(x)-f(y)| \le L1 |x-y|$

and

$|g(x)-g(y)|\le L2 |x-y|$

Now, for the $3$rd case, wlog, assume that $x < y$.

Then we must have $f(x)\ge g(x)$, but $f(y) \le  g(y)$, so by the intermediate value theorem for the continuous function $f-g$, we get a point t such that $x < t < y$, satisfying: $f(t)=g(t)$.

Now,

$$|f(x)-g(y)| \le |f(x)-f(t)|+|f(t)-g(t)|+|g(t)-g(y)|\le
> L1(t-x)+L2(y-t) \le (\max(L1,L2))*(y-x)$$

The $4$th case is similarly dealt with and $\max(L1,L2)$ is your required Lipschitz constant for part b. Part a is all right, $L1*L2$ being the Lipschitz constant. Please correct me if I'm mistaken anywhere.

Title: Lipschitz continuous


答:
For $0 \le r < R$ and $z = re ^{i \theta}$ we have (from Cauchy's integral formula or the Residue theorem or the definition of the winding number):
$$
 1 = \frac{1}{2 \pi i } \int_{|\zeta| = R} \frac{d\zeta}{\zeta - z}
= \frac{1}{2 \pi } \int_0^{2 \pi} \frac{R}{R - r e ^{i ( \theta - \phi)}} d\phi \\
= \frac{1}{2 \pi } \int_0^{2 \pi} \frac{R(R-r e ^{-i ( \theta - \phi)})}{R^2 - 2rR \cos(\theta-\phi) + r^2} d\phi \, .
$$
Taking the imaginary part gives
$$
0 = \frac{1}{2 \pi } \int_0^{2 \pi} \frac{rR (\sin\theta - \phi)}{R^2 - 2rR \cos(\theta-\phi) + r^2} d\phi \, .
$$

The connection to the Poisson formula is that
$$
\frac{R^2-r^2}{{R^2 - 2rR \cos(\theta-\phi) + r^2} } = \operatorname{Re}\left( \frac{R+r e ^{i ( \theta - \phi)}}{R+r e ^{i ( \theta - \phi)}}\right)
$$
and
$$
\frac{2Rr \sin(\theta - \phi)}{{R^2 - 2rR \cos(\theta-\phi) + r^2} } = \operatorname{Im}\left( \frac{R+r e ^{i ( \theta - \phi)}}{R+r e ^{i ( \theta - \phi)}}\right)
$$

Title: How to prove $\int_{0}^{2\pi } \frac{ \sin(\theta - \phi) } { R^2 - 2rR \cos(\theta -\phi) +r^2 } d{\phi} =0$

Body:
This question was part of mock test of masters exam for which I am preparing and I am unable to solve it.

Show that  $\int_{0}^{2\pi }  \frac{ \sin(\theta - \phi) } { R^2 - 2rR \cos(\theta -\phi) +r^2 } d{\phi} =0$ .

The part ${ R^2 - 2rR \cos(\theta -\phi) +r^2 }$ seems to be from Poisson's integral fomula for harmonic functions butI am not sure how to use it here. (Due to this reason i am using harmonic functions tag )

If u(z) is harmonic in a domain consisting of disc $|z|\leq R$ . Then for $z=r e^{i\theta}$ we have ,

$u(r e^{i \theta})=\frac{1}{2\pi}\int_{0}^{2\pi} \frac{R^2 -r^2} { R^2 - 2rR \cos(\theta - \phi) +r^2 } u(R e^{i \phi})d {\phi}$.

But $\sin(\theta -\phi)$ is an implicit function.So, I am not able to find u(z) and $R^2 -r^2$.

Can you please help me in proving it?


答:
When you want to use Taylor series of complex expressions, do it for one piece at the time
$$y=e^{e^x-1}\implies \log(y)=e^x-1=x+\frac{x^2}{2}+\frac{x^3}{6}+O\left(x^4\right)$$
Now
$$y=e^{\log(y)}=1+x+x^2+\frac{5 x^3}{6}+O\left(x^4\right)$$

Title: How to find the limit of $f(x)$.

Body:
Let $$f(x) = {\frac{e^{e^x - 1} - \frac{1}{1 - x}}{\ln(\frac{1 + x}{1-x}) - 2\sin x}}$$

I need to find a $ \lim_{x \to 0}f(x)$. As I understand, we have to do some Taylor series decomposition and problem should be pretty easy. But I have had some troubles with it.

P.S : I can decompose $\ln(\frac{1 + x}{1-x})$ as $\ln(1 + \frac{2x}{1-x})$ to use some standard Taylor formulas. I know $\sin x$ Taylor series, but I've got some troubles with $e^{e^x-1}$ and some further transformations.


答:
If you want, you can directly calculate the integral. In fact,
\begin{eqnarray}
&&\int_{0}^{2\pi }  \frac{ \sin(\theta - \phi) } { R^2 - 2rR \cos(\theta -\phi) +r^2 } d{\phi} \\
&=&-\frac{1}{2rR}\int_{0}^{2\pi } \frac{1} { R^2 - 2rR \cos(\theta -\phi) +r^2 } d\bigg[R^2 - 2rR \cos(\theta -\phi) +r^2 \bigg] \\
&=&-\frac{1}{2rR}\ln\bigg[ R^2 - 2rR \cos(\theta -\phi) +r^2 \bigg]\bigg|_{\phi=0}^{2\pi}\\
&=&0.
\end{eqnarray}


问:
Find a set $A$ in $\mathbb{R}$ such that the interior of $A$, the closure and the set itself all have a different number of connected components
I thought that $X=\mathbb{R}$ and $A=\mathbb{R}\setminus \{ 1\}$ at first but now I’m second guessing myself and I think it's wrong. Help would be much appreciated

答:
Consider $\{ -1 \} \cup [0,1) \cup (1,2] \cup \{5\}$.

The set has 4 components, the closure - 3, the interior - 2.


答: Hint: Take $A=\Bbb Q$ and check that it works.

Title: Find a set $A$ in $\mathbb{R}$ such that the interior of $A$, the closure and the set itself all have a different number of connected components

Body: I thought that $X=\mathbb{R}$ and $A=\mathbb{R}\setminus \{ 1\}$ at first but now I’m second guessing myself and I think it's wrong. Help would be much appreciated


问:
Infinite products of a (finite) group
So I'm having a little trouble understanding the concept of infinite (cartesian) products of a group -- specifically, my notes (and, of course, homework questions) have concepts of, say $S_3^\mathbb{Z}$ and $S_3^\mathbb{R}$ (where $S_3$ is the permutation group of 3 elements, of course), but I'm having a hard time conceptualising the difference between these.

If I assume that we're taking what seems to be the canonical approach: $G^I = \Pi_{i \in I} G$, how do I begin to understand the notion of, say, 3.141519 copies of $G$ (or in my case, $S_3$)? Is there a distinction between $\mathbb{Z}$ and $\mathbb{N}$ copies of a group? Or $\mathbb{Z}$ and $\mathbb{R}$?

It feels like there's something obvious I'm missing, but, of course, I don't know it!

答:
The direct product of copies of a single group $G$ over an indexing set $I$ can be thought of as the space of functions $f:I\to G$ with componentwise multiplication (i.e. $(f_1f_2)(i)=f_1(i)f_2(i)$ for all $i\in I$). This makes sense; any coordinate vector $(a_1,a_2,\cdots)$ can be thought of as a function from $\bf N$ into whatever, as $i\mapsto a_i$, and conversely any function $f:{\bf N}\to{\rm blah}$ can be thought of as a coordinate vector via $(f(1),f(2),f(3),\cdots)$. The utility of the functional approach is that it's kind of hard to write a "vector" with uncountably many coordinates (there are too many coordinates to enumerate), but functions are perfectly acceptable to talk about.

If $I$ and $J$ have the same size, then $\prod_IG\cong\prod_JG$. This can be accomplished by any bijection given by $\varphi:I\xrightarrow{\sim}J$; given $x\in\prod_IG$, define $\bar{x}\in\prod_JG$ to be the element whose $\varphi(i)$th coordinate is equal to the $i$th coordinate of $x$. So for example, let $\varphi:1,2,3,\cdots\mapsto0,1,-1,2,-2,\cdots$. Then

$$(a_1,a_2,a_3,\cdots)\mapsto (\cdots,a_5,a_3,a_1,a_2,a_4,\cdots)$$

is an isomorphism $\prod_{\bf N}G\cong\prod_{\bf Z}G$ for any group $G$. In the language of functions this means that the map $f\mapsto f\circ \varphi^{-1}$ is an isomorphism $\prod_IG\to\prod_JG$.

If $I,J$ are different sizes, then $G^I\not\cong G^J$ since in the first place they do not have the same sizes as sets. If we drop the condition that $G$ is finite though then I'm not so sure what can be said.

It does not make sense to take the direct product of a non-integer number of copies of $G$, any more than it makes sense to have a set of non-integer size. (Makes even less sense; you can define "fuzzy" sets, but I wouldn't know how to begin defining a direct product indexed by a fuzzy set.)

It's noteworthy that when working with an infinite number of nontrivial groups, direct sums and direct products are two different things. The direct product $\prod_I G_i$ can be formalized as the space of functions $f:I\to\bigcup_I G_i$ such that $f(i)\in G_i$ for all $i\in I$, with pointwise multiplication. The direct sum $\bigoplus_IG_i$ is a subgroup of the direct product consisting of those functions with "finite support," i.e. $f(i)$ is not an identity element of $G_i$ for only finitely many $i\in I$.

Title: Infinite products of a (finite) group


问:
Weak limit of invariant measures
Let $([0,1],F)$ be a dynamical system. Which hypotheses do I need on $F$ to be sure that a measure which is weak limit of a sequence of finitely supported $F$-invariant measures is an $F$-invariant measure? In particular, is Baire 2 sufficient? Is Baire 1 sufficient? Is left-continuous sufficient?

答:
All Baire function are measurable function.

Then if you take $A$ a measurable set, and if we name $\mu_n \to^* \mu$ the sequence of measure and its weak limit we have : $$
\mu_n(F^{-1}A)=\mu_n(A) \to \mu(A) = \mu(F^{-1}A)
$$
So $\mu$ is also $F$-invariant. In conclusion to be measurable is enough (and a fortiori any Baire).

Title: Weak limit of invariant measures


问:
Question about $(n^2+1)*(\dfrac{1}{n}+o(\dfrac{1}{n}))$
In the limit $\lim\limits_{n \rightarrow +\infty} \dfrac{ (n^2+1)\log(1+\dfrac{1}{n}) }{ n+n^x }$

Can you explain me why the product

$(n^2+1)*(\dfrac{1}{n}+o(\dfrac{1}{n}))$

is equal to $n$ ?

Body: $(n^2+1)*(\dfrac{1}{n}+o(\dfrac{1}{n}))= n+n^2o(\dfrac{1}{n})=n$


答:
Just compute term by term to get
$$n + o(n) + 1/n + o(1/n).$$
Basically the last two terms go to 0, so they don’t interfere with the asymptotic behaviour. The term o(n) is the result of $n^2o(1/n)$ which is for sure something that asymptotically grows slower than n, whence you have the result that asymptotically all that expression behaves like $n$.

Title: Question about $(n^2+1)*(\dfrac{1}{n}+o(\dfrac{1}{n}))$

Body:
In the limit $\lim\limits_{n \rightarrow +\infty} \dfrac{ (n^2+1)\log(1+\dfrac{1}{n}) }{ n+n^x }$

Can you explain me why the product

$(n^2+1)*(\dfrac{1}{n}+o(\dfrac{1}{n}))$

is equal to $n$ ?


答:
It is not:
\begin{align*}
(n^2+1)\left(\frac{1}{n}+o_{n\to+\infty}\left(\frac{1}{n}\right)\right)&=n+\frac{1}{n}+o_{n\to+\infty}\left(n\right)+o_{n\to+\infty}\left(\frac{1}{n}\right).
\end{align*}
The following equivalence thus holds
\begin{align*}
(n^2+1)\left(\frac{1}{n}+o_{n\to+\infty}\left(\frac{1}{n}\right)\right)&\underset{n\to+\infty}{\sim}n
\end{align*}
since
\begin{align*}
\frac{1}{n}(n^2+1)\left(\frac{1}{n}+o_{n\to+\infty}\left(\frac{1}{n}\right)\right)&=1+\frac{1}{n^2}+o_{n\to+\infty}\left(1\right)+o_{n\to+\infty}\left(\frac{1}{n^2}\right)\underset{n\to+\infty}{\longrightarrow}1.
\end{align*}
The equivalence is not an equality but rather an asymptotic behavior.


问:
alternating series test for $\sum_{n=1}^{\infty}(-1)^n\frac{\sqrt{n}}{n+4}$
I must determine if this series converges (using specifically the alternating series test) $$\sum_{n=4}^{\infty}(-1)^n\frac{\sqrt{n}}{n+4}.$$

I know the necessary and sufficient conditions are:
The series $\displaystyle\sum_{n=1}^{\infty}(-1)^na_n$ converges if $a_{(n+1)} \leq a_n$ (or if $f'(x) < 0$) and $\displaystyle\lim_{n\to\infty}a_n = 0.$

So first I tried to show the derivative, $\frac{d}{dx}(\frac{\sqrt{x}}{x+4}) < 0$ and after a bunch of algebra got $\displaystyle\frac{2-x}{\sqrt{x^5} + \sqrt{x^3} + 16 \sqrt{x}} < 0$, but I don't know how to demonstrate if that's really true for all $x \geq 1$.

Next I tried to show $a_{n+1} \leq a_n\  \forall \ n \geq 1$, or
$\frac{\sqrt{n+1}}{4(n+1)^2 + 1} \leq \frac{\sqrt{n}}{4n^2+1}. \ $ After cross multiplying and some rearranging I get $(4n^2 + 1)\sqrt{n+1} \leq 4n^{\frac{5}{2}} + 8n^\frac{3}{2} + 2\sqrt{n}$, and I don't know how to tell if that's true for all $n \geq 1$.

So I'm stumped.  I know there are other tests.  But this is in the alternating series section so either the alternating series test can show it is convergent, or if not then I know it's divergent.
But the answer in the back of the book says it's convergent.  I just don't know how to prove it.

Any suggestions?

n.b.  The original series goes from $n=4\to\infty$ but I don't know if the 4 is a typo or is intended.

答:
Your test of whether $a_n \gt a_{n+1}$ looks odd. Since $a_n, a_{n+1} \gt 0$, you can test whether $a_n^2\gt a_{n+1}^2$ and this comes out as $$\frac{n}{(n+4)^2}\gt \frac {n+1}{(n+5)^2}$$

It is then simple to check (clear fractions, since you are multiplying by positive numbers) that this is true for $n \ge 4$, but not for $n=3$ - which may be why the series starts at $n=4$ in the original question from which yours is derived. Note that adding the extra terms does not affect the convergence of the series.


问:
two vectors ∩S in linear transformation
i have de next problem
Let  be a basis of R3 and . If v and w∈R3 satisfy , then ⟨v, w⟩∩S is equal to

and they give me the following options

I'll tell you what I did (I'm sure I did it wrong haha).
try looking for (v + w) b = (α, β, ɣ)
so
v + w = 1 (1,0,0) +3 (-1,1,0) -1 (-1, -1,1)
which gave me that v + w = (- 1,4, -1)
I did the same with v-w and it gave me (-10, -2,5)
then assume that v + w + v-w = (- 1,4, -1) + (- 10, -2,5)
from there take v = (- 11 / 2,1,2)
and w = (9 / 2,3, -3)
but I'm very far from all the results (from the options they give me to choose from)

答:
We know that $S=\langle \begin{pmatrix} 2\\1\\0 \end{pmatrix},\begin{pmatrix} 1\\0\\1 \end{pmatrix} \rangle$ and $\langle v,w\rangle=\langle \begin{pmatrix} -11\\2\\4 \end{pmatrix},\begin{pmatrix} 9\\6\\-6 \end{pmatrix} \rangle$, so, if we want to find the generator of $S \cap\langle v,w\rangle$ we need to solve the system $$\alpha\begin{pmatrix} 2\\1\\0 \end{pmatrix}+\beta\begin{pmatrix} 1\\0\\1 \end{pmatrix}=\gamma\begin{pmatrix} -11\\2\\4 \end{pmatrix}+\delta\begin{pmatrix} 9\\6\\-6 \end{pmatrix}\text{, with }\alpha,\beta,\gamma,\delta\in \mathbb R.$$
Starting from the matrix associated to the homogeneous system $\begin{pmatrix} 2&&1&&11&&-9\\1&&0&&-2&&-6\\0&&1&&-4&&6 \end{pmatrix}$

we get $\gamma=-\dfrac{20}{17}\delta$ $\implies S\cap\langle v,w \rangle$, whose dimension is $1$, is given by $$\langle -\dfrac{20}{17}\begin{pmatrix} -11\\2\\4 \end{pmatrix}+\begin{pmatrix} 9\\6\\-6 \end{pmatrix} \rangle=\langle\begin{pmatrix} -\dfrac{23}{17}\\-\dfrac{20}{17}\\1 \end{pmatrix}\rangle=\langle \begin{pmatrix} 23\\20\\-17 \end{pmatrix}\rangle.$$

Title: two vectors ∩S in linear transformation


问:
Counter-example concerning isometry.

Can $(\Bbb R^2 , \|\cdot\|_1)$ and $(\Bbb R^2, \|\cdot\|_2)$ be isometric?

I think that these two spaces can't be isometric. But I can't able to show that. Would anybody please help me in this regard?

Thanks in advance.

答: (Apologies, I cannot comment). This question is a duplicate, and already has answers here.</p>

Title: Counter-example concerning isometry.


问:
Expand $\ln(\sqrt[3]{x^2})$
I am to expand $\ln(\sqrt[3]{x^2})$. The textbook solutions page say it should be $\frac{2}{3}\ln(x)$.

I gave it a try but suspect I went wrong further up in my calculations. I got $\frac{1}{2}\ln(x)$.

My working:
$$\ln(\sqrt[3]{x^2})$$
$$\ln(x^\frac{1}{2})$$

(Is there a rule for the nth mth root? I forget? Can I simplify $\sqrt[3]{x^2}$?. Here I thought that if it's a 3rd root of $x^2$ then I still have one more root to go, assuming they cancel out to two powers?)

And then
$$\frac{1}{2}\ln(x)$$

How can I arrive at $\frac{2}{3}\ln(x)$?

答: Your suspicion is correct.  That is the third root of $x^2$, which is $(x^2)^\frac13=x^\frac23$.  So that's how it turns to $\frac23\ln x$.

Title: Expand $\ln(\sqrt[3]{x^2})$


答:
You just need to know that $\sqrt[n]{x} = x^{1/n}.$  This makes sense because

$$x^{1/3}x^{1/3}x^{1/3} = x^{1/3+1/3+1/3} = x^1.$$

Title: Expand $\ln(\sqrt[3]{x^2})$

Body:
I am to expand $\ln(\sqrt[3]{x^2})$. The textbook solutions page say it should be $\frac{2}{3}\ln(x)$.

I gave it a try but suspect I went wrong further up in my calculations. I got $\frac{1}{2}\ln(x)$.

My working:
$$\ln(\sqrt[3]{x^2})$$
$$\ln(x^\frac{1}{2})$$

(Is there a rule for the nth mth root? I forget? Can I simplify $\sqrt[3]{x^2}$?. Here I thought that if it's a 3rd root of $x^2$ then I still have one more root to go, assuming they cancel out to two powers?)

And then
$$\frac{1}{2}\ln(x)$$

How can I arrive at $\frac{2}{3}\ln(x)$?


问:
Tensor power- Notation question
I have a notational question, which is written usually in papers, but I can not figure it out what could be. Let $M$ be an $A$-module. I have seen this notation
$$M^{\otimes -n}$$

I think this would mean $Hom_A(M^{\otimes n},A)$, but in other way this usually is denoted by $(M^{\otimes n})^*$. Am I right?

答:
Perhaps the other answer was describing the Picard group of a commutative ring A.  If M is a finitely generated projective A-module of rank 1, then the tensor product M ⊗A HomA(M, A) is isomorphic to A.  In particular, the set of isomorphism classes of finitely generated projective modules of rank 1 forms a group under tensor product.

If A is an integral domain, then the M we consider are just the isomorphism classes of projective modules that are isomorphic to ideals of A.  This forms the ideal class group.

In particular for principal ideal domains or fields, this is not a useful concept.

Also, none of this works for general modules, not even finitely generated projective modules of rank larger than 1, so I would be very cautious of assuming this is true for a negative tensor power of some random module.

Title: Tensor power- Notation question


答: In general, $$\sqrt[n]{x^m} = x^{m/n},$$ and hence $$\ln\left(\sqrt[n]{x^m}\right) = \ln x^{m/n} = \frac{m}{n}\ln x.$$

Title: Expand $\ln(\sqrt[3]{x^2})$


答: You may unwrap it as $\ln\sqrt[3]{x^2}=\tfrac13\ln(x^2)=\tfrac132\ln x=\tfrac23\ln |x|$ for $x\in\Bbb R\setminus\{0\}$. For $x>0$, we can drop the mod signs. More generally, $\ln\sqrt[N]{x^m}=\tfrac1N\ln(x^M)=\tfrac{M}{N}\ln x$ works for $x>0$, while we get $\tfrac{M}{N}\ln|x|$ for $x\ne0$ if $M$ is even.

Title: Expand $\ln(\sqrt[3]{x^2})$


答:
Assume that $U:(\mathbb{R}^d,\|\cdot\|_1)\to(\mathbb{R}^d,\|\cdot\|_2)$ is an isometry. Then
\begin{align*}
\|Ux+Uy\|_2^2+\|Ux-Uy\|_2^2&=2\|Ux\|_2^2+2\|Uy\|_2^2
\end{align*}
by the parallelogram law. On the other hand, as $U$ is an isometry,\begin{align*}
\|x+y\|_1^2+\|x-y\|_1^2&=2\|x\|_1^2+2\|y\|_1^2.
\end{align*}
Thus $\|\cdot\|_1$ also satisfies the parallelogram law, thus comes from a scalar product, a contradiction. The conclusion is the same if we replace $\|\cdot\|_1$ by any $\|\cdot\|_p$ with $p\neq2$.

Title: Counter-example concerning isometry.

Body:
Can $(\Bbb R^2 , \|\cdot\|_1)$ and $(\Bbb R^2, \|\cdot\|_2)$ be isometric?

I think that these two spaces can't be isometric. But I can't able to show that. Would anybody please help me in this regard?

Thanks in advance.


问:
Relative distance to prime numbers
Let $p_{n,1}$ be the next biggest prime for $n \in \mathbb{N}$, $p_{n,2}$ the next biggest but one and so on. (e.g. $p_{4,1} = 5$, $p_{4,2} = 7$, ...)

Are there $n,m \in \mathbb{N}, n\neq m$ so that
$$p_{n,j}-n=p_{m,j}-m$$
for each $j\in\mathbb{N}$?

答: There are prime gaps of arbitrary size (e.g. $k!+2,...,k!+k$ are all composite). If there were such a pair $m,n$ it would mean that the pattern of prime gaps repeats with a period of $|m-n|$, so arbitrarily large prime gaps would not be possible.

Title: Relative distance to prime numbers


问:
Why do diagonal entries of odd potencies of adjacency matrices of forests sum up to zero?
I came across the statement that the trace (sum of diagonal entries) of an odd potency of an adjacency matrix of a forest is always zero. I have tried out a few options and it seems to be correct for forests or trees but not for other graphs, e.g.:

$A = \begin{bmatrix} 0 & 1 & 0 & 1 & 1 \\ 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \end{bmatrix}$ and $A^3 = \begin{bmatrix} 0 & 4 & 0 & 3 & 3 \\ 4 & 0 & 2 & 0 & 0 \\ 0 & 2 & 0 & 1 & 1 \\ 3 & 0 & 1 & 0 & 0 \\ 3 & 0 & 1 & 0 & 0 \end{bmatrix}$

While if I add one edge to create a circle:

$B = \begin{bmatrix} 0 & 1 & 1 & 1 & 0 \\ 1 & 0 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 \\1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$ and $B^3 = \begin{bmatrix} 2 & 4 & 4 & 3 & 0 \\ 4 & 2 & 3 & 1 & 0 \\ 4 & 3 & 2 & 1 & 0 \\ 3 & 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$

Can someone explain to me why this is the case?

答: If $A$ is the adjacency matrix of a graph, then $A^n_{ij}$ counts the number of paths of length $n$ from vertex $i$ to vertex $j$.  If you are in a forest, because there are no cycles, every step along a path is either one step closer or one step father away from any particular vertex.  Because of this, if we start at vertex $i$, we will alternate between an even distance away and an odd distance away.  In particular, at odd times, we are an odd distance away, and therefore NOT back at our starting point.  So all the diagonal entries of $A^n$ are $0$ when $n$ is odd.


问:
Weakness of the test
By performing a hypothesis test on a sample from a normal population, you find that the hypothesis $\mu = 20$ cannot be rejected at $10\%$ risk. A colleague obtains the same conclusion by testing $\mu= 22$.

I think this is the weakness of the test: several hypotheses may not be rejected. Am I wrong?

答:
Verifying a null hypothesis without expliciting an alternative does not make sense.

Then, to compare your test with your friend's one you can calculate the two $p_{value}$'s.

A test is measured by its $p_{value}$.

If the alternative is a simple hypothesis too, you can also calculate the power of both tests to compare them

Title: Weakness of the test


问:
Solving a summation problem - I know the answer I think, but not why it works
This problem I intuitively know the answer, (I think) but I am trying to get a handle on why it works.

$$\sum_{n=0}^x n=\frac{\sum_\limits{n=0}^x n^3}{\sum_\limits{n=0}^x n^2} $$

We want to know what x is.
I figured the answer was x = 1, because the only number I can come up with where $\frac{n^3 + (n+1)^3 + ... }{n^2 + (n+1)^2 + ...}=n$ is one, when I did a few sums and realized that anything other than n=1 doesn't work. And I know it's because if I add a number I am not getting a cube over a square of n anymore, $\frac{(1^3 + 2^3)}{(1^2+2^2)}=\frac{9}{4}$ for example. But I feel like there is some basic piece of a proof/procedure I am missing, even though this intuition is correct. Anyhow, I am curious what the "real" reason might be -- I suspect this is some relatively simple algebraic principle that makes this work, I feel like I am just not quite seeing it. Or maybe I got it right and just overthought this.

答:
The sum of cube is the square of the triangular numbers, viz $(n^2+n)^2/4$,

So the equation is correct if you drop the square from the $n^2$ im the denominator.


答:
These are well-known sums: $$\sum_{0}^x n = \frac{x(x+1)}{2} \\ \sum_0^x n^2 = \frac{x(x+1)(2x+1)}{6} \\ \sum_0^x n^3 = \frac{x^2(x+1)^2}{4} $$ Substituting in gives $$\frac{x(x+1)}{2} = \frac{\frac{x^2(x+1)^2}{4}}{\frac{x(x+1)(2x+1)}{6}} \\ \implies \frac{2x+1}{3} =1 \implies x=1 $$ is the only solution.

See here for the proofs of these identities.

Title: Solving a summation problem - I know the answer I think, but not why it works

Body:
This problem I intuitively know the answer, (I think) but I am trying to get a handle on why it works.

$$\sum_{n=0}^x n=\frac{\sum_\limits{n=0}^x n^3}{\sum_\limits{n=0}^x n^2} $$

We want to know what x is.
I figured the answer was x = 1, because the only number I can come up with where $\frac{n^3 + (n+1)^3 + ... }{n^2 + (n+1)^2 + ...}=n$ is one, when I did a few sums and realized that anything other than n=1 doesn't work. And I know it's because if I add a number I am not getting a cube over a square of n anymore, $\frac{(1^3 + 2^3)}{(1^2+2^2)}=\frac{9}{4}$ for example. But I feel like there is some basic piece of a proof/procedure I am missing, even though this intuition is correct. Anyhow, I am curious what the "real" reason might be -- I suspect this is some relatively simple algebraic principle that makes this work, I feel like I am just not quite seeing it. Or maybe I got it right and just overthought this.


答:
Let's say the initial equation is: $$ S_1 = \frac{S_2}{S_3} $$

It is well known that:

$$S_1 = \frac{x(x+1)}{2}$$

$$S_2 = {\frac{x^2(x+1)^2}{4}}$$

$$S_3 = \frac{x(x+1)(2x+1)}{6}$$

Plug these into the original equation and solve for $x$.


问:
How many people must be present at a party to guarantee that at least three of their birthdays are in the same month?
I keep getting 36 as an answer since if 36 people were present at a party, 36/12 (for the months in a given year) = 3. But this is incorrect. Any help would be appreciated.</p>

答:
If you had $24$ people you could have every month with two people having their birthdays then

But if you had $25$ then at least one month must have at least three people having their birthdays then, since the average number of birthdays per month would be $\frac{25}{12}>2$


问:
Hearts are open sets in the standard topology in $\mathbb{R}^2$?
In Adams and Franzosa's Introduction to Topology, there are two ways that the standard topology on $\mathbb{R}^2$ is defined.

Definition 1. Let $d: \mathbb{R}^2 \times \mathbb{R}^2 \to \mathbb{R}$ be Euclidean distance, and for each $x \in \mathbb{R}^2$ and $\epsilon > 0$, let
$$B(x, \epsilon) = \{p \in \mathbb{R}^2: d(x, p) < \epsilon\}\text{.}$$
Let $$\mathcal{B} = \{B(x, \epsilon): x \in \mathbb{R}^2, \epsilon > 0\}\text{.}$$
Then $\mathcal{B}$ is a basis for the standard topology on $\mathbb{R}^2$.

Theorem. Let $$\mathcal{B}^{\prime} = \{(a, b) \times (c, d) \subset \mathbb{R}^2: a < b, c < d\}\text{.}$$
Then $\mathcal{B}^{\prime}$ is a basis for the standard topology on $\mathbb{R}^2$.

In Figure 1.10 (p. 35), there is a drawing of a heart-shape object with dotted lines outlining it in $\mathbb{R}^2$. In Example 1.12 (p. 36), they aim to show such heart-shape objects are a basis for the standard topology on $\mathbb{R}^2$, and in showing this, they state

the hearts themselves are open sets in the standard topology.

Why is this the case? I suspect it's because hearts (as described above) can be written as a union of open rectangles, but I'm not certain.

If it helps, they say the same is also true for diamonds.

答:
Thank you to Shubham Johri for pointing me in the right direction.

Theorem 1.4 of Adams and Franzosa states:

Let $X$ be a topological space and $A \subset X$. Then $A$ is open in $X$ if and only if for each $x \in A$, there is a neighborhood $U$ [i.e., an open set] of $x$ such that $x \in U \subset A$.

Under the basis $\mathcal{B}$ provided in the question, given any point in the heart, we may find a ball (which is open) which is centered at the point with radius $\epsilon > 0$ which is still contained in the heart, thus the hearts themselves are open in the standard topology.

It is important to note that the boundary outlining the heart is excluded from points that we look at, otherwise this would not hold.


答:
Yes, your proof is correct.

A note on terminology: What you call the "external direct sum" is what I would call the "sum" (of submodules of a fixed module $M$). For me, the external direct sum of $M_1,\dots,M_n$ is the set of ordered $n$-tuples $$\{(m_1,\dots,m_n)\mid m_i\in M_i \text{ for  all  }1\leq i \leq n\},$$ with the component-wise addition and scalar multiplication. Note that in the external direct sum, we don't need to know that the modules $M_i$ are submodules of some module $M$ - that's what makes it "external", rather than internal to $M$.

Title: Associativity of the internal direct sum

Body:
Let ${{M}_{1}},{{M}_{2}},...,{{M}_{n}}$ be submodules of an $R$-module $M$. We define the external direct sum as${{M}_{1}}+{{M}_{2}}+...+{{M}_{n}}=\left\{ {{m}_{1}}+{{m}_{2}}+...+{{m}_{n}}\left| \forall i\in \left\{ 1,2,...,n \right\}:{{m}_{i}}\in {{M}_{i}} \right. \right\}$. On the other hand, the internal direct sum has an additional condition:

${{M}_{1}}\oplus {{M}_{2}}\oplus ...\oplus {{M}_{n}}=\forall j\in \left\{ 1,2,...,n \right\}:\left( {{M}_{1}}+{{M}_{2}}+...+{{M}_{j-1}}+{{M}_{j+1}}+...+{{M}_{n}} \right)\bigcap {{M}_{j}}=\left\{ 0 \right\}$

This condition ensures that, for instance, every element of the set ${{M}_{1}}\oplus {{M}_{2}}\oplus ...\oplus {{M}_{n}}$ can be expressed as a sum ${{m}_{1}}+{{m}_{2}}+...+{{m}_{n}}\left| \forall i\in \left\{ 1,2,...,n \right\}:{{m}_{i}}\in {{M}_{i}} \right.$ in a unique way.

My question is whether this proof is correct. Let ${{M}_{1}},{{M}_{2}},{{M}_{3}}$ be arbitrary submodules of an $R$-module $M$. Obviously,

${{M}_{1}}+{{M}_{2}}=\left\{ {{m}_{1}}+{{m}_{2}}\left| {{m}_{1}}\in {{M}_{1}}\And {{m}_{2}}\in {{M}_{2}} \right. \right\}$

and

$\left( {{M}_{1}}+{{M}_{2}} \right)+{{M}_{3}}=\left\{ {{m}_{s}}+{{m}_{3}}\left| {{m}_{s}}\in {{M}_{1}}+{{M}_{2}}\And {{m}_{3}}\in {{M}_{3}} \right. \right\}=\left\{ \left( {{m}_{1}}+{{m}_{2}} \right)+{{m}_{3}}\left| \left( {{m}_{1}}\in {{M}_{1}}\And {{m}_{2}}\in {{M}_{2}} \right)\And {{m}_{3}}\in {{M}_{3}} \right. \right\}$

Applying the same method, we obtain ${{M}_{1}}+\left( {{M}_{2}}+{{M}_{3}} \right)=\left\{ {{m}_{1}}+{{\left( {{m}_{2}}+m \right)}_{3}}\left| {{m}_{1}}\in {{M}_{1}}\And \left( {{m}_{2}}\in {{M}_{2}}\And {{m}_{3}}\in {{M}_{3}} \right) \right. \right\}$.

Therefore, $\left( {{M}_{1}}+{{M}_{2}} \right)+{{M}_{3}}={{M}_{1}}+\left( {{M}_{2}}+{{M}_{3}} \right)$. The internal direct sum is just a special case of the external direct sum, so $\left( {{M}_{1}}\oplus {{M}_{2}} \right)\oplus {{M}_{3}}={{M}_{1}}\oplus \left( {{M}_{2}}\oplus {{M}_{3}} \right)$. Could this be considered as a valid proof?