答:
Look for perfect squares

$x^2 + y^2 + 1 \ge xy + x+y\iff $

$x^2 - 2xy+y^2 \ge -xy + x + y - 1\iff$

$(x-y)^2 \ge x+y - xy - 1$.

Now as the LHS is $\ge 0$ if we can show the RHS is $\le 0$ that be great.  If not... well we may hit an inspiration on the way.  We can always factor

$x+y - xy - 1 = x-xy + y-1 = x(1-y) + y-1) = x(1-y)-(1-y) = (x-1)(1-y)$.

Hmmm, no need for that to be negative but if it is positive then either $(x-1)$ and $1-y$ are both positive or both negative.

If both are positive then $x > 1$ and $y< 1$.  So $x-y > x-1$ and $x-y > 1-y$ and so $(x-y)(x-y) > (x-1)(1-y)$.

And if both are negative then $x < 1<y$ and so $(x-y)^2 = (y-x)^2$ and $y-x > y-1>0$ and $y-x > 1-x>0$ and $(y-x)^2 > (y-1)(1-x) = (x-1)(1-y)$.

Title: Prove that $\forall x,y, \, \, x^2+y^2+1 \geq xy+x+y$

Body:
Prove that  $\forall x,y\in \mathbb{R}$ the inequality  $x^2+y^2+1 \geq xy+x+y$ holds.

Attempt

First attempt: I was trying see the geometric meaning, but I´m fall.

Second attempt: Consider the equivalent inequality given by $x^2+y^2\geq (x+1)(y+1)$
and then compare $\frac{x}{y}+\frac{y}{x} \geq 2 $ and the equality $(1+\frac{1}{x}) (1+\frac{1}{y})\leq 2$ unfortunelly not is true the last inequality and hence I can´t conclude our first inequality.

Third attempt:comparing  $x^2+y^2$ and $(\sqrt{x}+\sqrt{y})^2$ but unfortunelly I don´t get bound the term $2\sqrt{xy}$ with $xy$.

Any hint or advice of how I should think the problem was very useful.


问:
A pseudofinite structure that does not have the finite model property
Is there a pseudofinite structure $M$ that does not have the finite model property? Pseudofinite means that any sentence $M$ satisfies is satisfied in a finite structure, and finite model property means, in addition, that we can take the finite structure to be a substructure of $M$.

答:
Sure. Consider any pseudofinite field of characteristic zero (e.g. take a nonprincipal ultrapower of a sequence of finite fields of increasing characteristic). The field axioms, conjoined into a single sentence, are satisfied in such a structure but no finite field embeds into any field of characteristic zero.

A simpler, if less mathematically interesting, example is the successor graph $S$ on $\mathbb{Z}$. The sentence "Every vertex has degree $2$" has as its finite models only (unions of) cycles, but $S$ is acyclic. To see that $S$ is pseudofinite, you can show that for every $m$, Duplicator has a winning strategy in the Ehrenfeucht-Fraisse game between $S$ and the graph consisting of a single cycle of length $2^m$.

ContentLicense: CC BY-SA 4.0


问:
Solving $2 + \frac 1{\cos^2 x} + \frac 5{\cos x} = 0$
Simple but confusing trig question:

$$2 + \frac 1{\cos^2 x} + \frac 5{\cos x} = 0$$

I could not find anything in my textbook similar to this....

答:
$2y^2+ 5y+ 1= 0$

By the quadratic formula

$y=\dfrac{-5\pm \sqrt{25-8}}{4}$


问:
Conditional distribution of jointly Gaussian random variables where one is degenerate
If random variables $x \in \mathbb R^n$ and $y \in \mathbb R^m$ have the joint Gaussian distribution
$$
\begin{bmatrix} x \\ y \end{bmatrix} \sim
\mathcal N \left(
\begin{bmatrix} a \\ b \end{bmatrix},
\begin{bmatrix} A & C \\ C^{\top} & B \end{bmatrix}
\right),
$$
then the conditional distributions of $x$ and $y$ are given as
$$
x \, \vert \, y \sim \mathcal N \left( a + C B^{-1} \left( y - b \right), A - C B^{-1} C^{\top} \right) \quad (*)
$$
provided that $B^{-1}$ exists.

Question: Suppose that we know two random vectors $x$ and $y$ are jointly Gaussian with mean and covariance as stated above. What can we say about the conditional distributions when $B^{-1}$ doesn't exist? I am mainly interested in the case where $B$ is restricted to be block diagonal with the submatrix in the upper-leftmost block being a $k \times k$ (with $k < m$) invertible matrix and there are zeros everywhere else (e.g. $B = \begin{bmatrix} \Sigma & \bf 0 \\ \bf 0 & \bf 0 \end{bmatrix}$ where $\Sigma^{-1}$ exists), but still would like to know if there is an answer when $B$ lacks this type of restriction. In this case, what can we say about the:

shape? Can we still say that $x \,\vert\, y$ is Gaussian?

center? Can we find an expression that represents the conditional mean $\mathbb{E} \left[ x \, \vert\, y \right]$?

spread? Can we find an expression that represents the conditional variance $\text{Var} \left[ x \, \vert \, y \right]$?

My attempt:

Shape:
I heard the phrase "conditional distributions of jointly Gaussian random variables is Gaussian", so it seems reasonable to say that the conditional distribution $x \, \vert \, y$ is Gaussian, but degenerate in the sense that it doesn't have probability density function.

The comments to the question here seems suggests this is true.

Center:
When it comes to the conditional mean, I constructed a toy example.

Suppose $x \in \mathbb{R}$ and $y = (y_1, y_2, y_3)^{\top} \in \mathbb{R}^3$ such that $x \sim \mathcal{N}(1,1)$ and $y \,\vert\, x \sim \mathcal{N} (Ax, \Sigma)$ where
$$A = \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}, \quad \Sigma = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}.$$
Then
$$\begin{bmatrix} x \\ y_1 \\ y_2 \\ y_3 \end{bmatrix}
\sim \mathcal{N} \left( \begin{bmatrix} 1 \\ 1 \\ 2 \\ 0 \end{bmatrix},
\begin{bmatrix} 1 & 1 & 2 & 0 \\ 1 & 2 & 2 & 0 \\ 2 & 2 & 5 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \right)$$

$\mathbb{E} [x \, \vert \, y ]$ should be an affine transformation of $y_1, y_2, y_3$, so
$$\mathbb{E} [x \, \vert \, y ] = a_0 + a_1 (y_1 - 1) + a_2(y_2 - 2) + a_3 y_3.$$
Now, $x - \mathbb{E} [x \, \vert \, y ]$ and $y_i$ are independent (right?) for $i = 1, 2, 3$, so then $\text{Cov} (x - \mathbb{E} [x \, \vert \, y ], y_i) = 0$ for each $i$. Also, using the tower property, it follows that $\mathbb{E} \left[ \mathbb{E} [x \, \vert \, y ]  \right] = a_0 = 1$.

Next,
\begin{align}
0 = \text{Cov} (x - \mathbb{E} [x \, \vert \, y ], y_i) &= \text{Cov} \left( x - \left( a_0 + a_1 (y_1 - 1) + a_2(y_2 - 2) + a_3 y_3 \right), y_i \right) \\
&= \text{Cov} \left( x, y_i \right) - a_0 \text{Cov} \left(1 ,y_i \right) - a_1 \text{Cov} \left( y_1, y_i \right) - a_2 \text{Cov} \left( y_2, y_i \right) - a_3 \text{Cov} \left( y_3, y_i \right) \\
&= \text{Cov} \left( x, y_i \right) - a_1 \text{Cov} \left( y_1, y_i \right) - a_2 \text{Cov} \left( y_2, y_i \right).
\end{align}

I think I can say that $y_3 = 0$, so this leads to the system
\begin{align}
1 - 2a_1 - 2a_2 &= 0 \\
2 - 2a_1 - 5a_2 &= 0
\end{align}
which has solution $(a_0, a_1, a_2)= \left( 1, \frac16, \frac13 \right)$ with $a_3$ being free. Therefore, $\mathbb{E} [x \, \vert \, y ] = \frac16 + \frac16 y_1 + \frac13 y_2$

The solution in this case is the same as $\mathbb{E} [x \,\vert\, y ]$ if $y = (y_1, y_2)^{\top}$, $A = (1, 2)^{\top}$, and $B = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.

So it seems that you'd just use $(*)$, but use the only the first $k$ columns of $C$ and the first $k \times k$ square of $B$.

Spread: To find the conditional variance in the toy example, I would need to find $\mathbb{E} \left[ x^2 \,\vert\, y \right]$, but I have not figured out a reasonable approach to compute this expectation. But, perhaps it is similar to the idea of using a restrict version of $C$ and $B$.

答: If a random vector $z$ has joint Gaussian distribution, possibly degenerate, and we partition $z$ into subvectors, say $z=\begin{pmatrix}x \\ y\end{pmatrix}$, then the conditional distribution of $x$ given $y$ also has Gaussian distribution. The conditional mean and variance are a function of the original mean vector and original covariance matrix. In the case where the original covariance matrix is singular, you will need to find a Moore-Penrose pseudoinverse. For the full derivation, please see this answer.


问:
First order logic. Describing that there exists a set with exactly 2 elements
∃∃(∈ ∧ ∈ ∧ ¬(=))

Is this a valid way to describe that? I believe it's correct but I'm not so sure

答: $\exists A \exists x \exists y ~(x \in A \land y \in A \land \lnot (x=y) \land \forall z (z \in A \Rightarrow (z=x \lor z=y)))$.


问:
Prove that $\forall x,y, \, \, x^2+y^2+1 \geq xy+x+y$
Prove that  $\forall x,y\in \mathbb{R}$ the inequality  $x^2+y^2+1 \geq xy+x+y$ holds.

Attempt

First attempt: I was trying see the geometric meaning, but I´m fall.

Second attempt: Consider the equivalent inequality given by $x^2+y^2\geq (x+1)(y+1)$
and then compare $\frac{x}{y}+\frac{y}{x} \geq 2 $ and the equality $(1+\frac{1}{x}) (1+\frac{1}{y})\leq 2$ unfortunelly not is true the last inequality and hence I can´t conclude our first inequality.

Third attempt:comparing  $x^2+y^2$ and $(\sqrt{x}+\sqrt{y})^2$ but unfortunelly I don´t get bound the term $2\sqrt{xy}$ with $xy$.

Any hint or advice of how I should think the problem was very useful.

Body:
Answer
$$
\begin{align}
0
&\le\tfrac34((x-1)-(y-1))^2+\tfrac14((x-1)+(y-1))^2\\
&=(x-1)^2+(y-1)^2-(x-1)(y-1)\\
&=\left(x^2-2x+1\right)+\left(y^2-2y+1\right)-(xy-x-y+1)\\
&=\left(x^2+y^2+1\right)-(xy+x+y)
\end{align}
$$

Motivation

To prove $x^2+y^2+1\ge xy+x+y$, look at the difference
$$
\begin{align}
\left(x^2+y^2+1\right)-(xy+x+y)
&=(x-y)^2+(x-1)(y-1)\tag1\\
&=(u-v)^2+uv\tag2\\[2pt]
&=(u-v)^2+\tfrac14\left((u+v)^2-(u-v)^2\right)\tag3\\
&=\tfrac34(u-v)^2+\tfrac14(u+v)^2\tag4\\
&=\tfrac34((x-1)-(y-1))^2+\tfrac14((x-1)+(y-1))^2\tag5\\
&\ge0\tag6
\end{align}
$$
Explanation:

$(1)$: subtract $2xy+1$ from both sides of the minus sign

$(2)$: $u=x-1$ and $v=y-1$

$(3)$: $uv=\frac14\left((u+v)^2-(u-v)^2\right)$

$(4)$: combine terms

$(5)$: $u=x-1$ and $v=y-1$

$(6)$: sum of two non-negative numbers is non-negative


问:
What can I attempt when the derivative tests fail?
The simultaneous equation I got from my Lagrange multiplier has proven itself a masochistic task. I found one trivial solution and I don't give a #### about the rest as they're too hard to find. I tried the second partial derivative test and fun fact. Three of the eigenvalues were 0.  This is the function: $f(a, b, c, d, e) = (a^2+1)(b^2+1)(c^2+1)(d^2+1)-e(ab+ac+ad+bc+bd+cd-abcd-5)$

答:
Since it isn't mentioned what you were being asked about the function, but you mention the second partial-derivative test, I am assuming you were to say something about its maximum and minimum values.

In an academic course, when students are given a problem that looks horribly complicated, they are often being asked to notice something that makes it much simpler.  (This doesn't necessarily apply to real-world problems.)

This is going to be a partial answer.  What we see in the expression for the function is that it is essentially unchanged if we permute the variables $ \ a \ , \ b \ , \ c \ , \ $ and $ \ d \ . $  In functions which have this kind of symmetry, the extrema can occur where all of these variables are equal, or where all but one of them equals zero.

In the first case, for $ \ a = b = c = d \ , $ the expression "collapses" to
$$ (a^2+1)^4 \ - \ e·(6a^2-a^4-5) \ \ = \ \ a^8 \ + \ 4a^6 \ + \ (6+e)a^4 \ + \ (4-6e)a^2 \ + (1+5e) \ \ . $$

I have left $ \ e \ $ as a "free" variable.   We see that it is an even function in $ \ a \ , $ so any solution for $ \ a \ $ has a "mirror" solution $ \ -a \ . $  What is interesting about the behavior of the function is that for $ \ e \ < \ \approx 0.7 \ $ (could it be $ \ \frac{1}{\sqrt2} \ $ ? ) , there is only an absolute minimum at $ \ a \ = \ 0 \ $ , but for $ \ e \ > \ \approx 0.7 \ , $ this changes to a local maximum at $ \ a \ = \ 0 \ $ with a pair of local minima at $ \ a \ = \ \pm a_{min} \ $ nearby.

For the other case, with a single non-zero variable, say, $ \ a \ $ , the expression is reduced to $ \ a^2 \ + \ (1+5e) \ , $ which plainly have only an absolute minimum at $ \ a \ = \ 0 \ . $  That just means there is a minimum value $ \ (1+5e) \ $  for the function at the origin for any value of $ \ e \ . $

The second-partial derivative test is not always helpful for polynomials with degrees higher than $ \ 2 \ $ ; this one counts as having degree $ \ 8 \ $ .  It looks like the only interesting case is for $ \ e \ > \ 0.7... \ , $ for which local minima occur at points with various combinations of signs for $ \ ( \pm a_{min} \ , \pm a_{min} \ , \pm a_{min} \ , \pm a_{min}) \   $  and a local maximum at the origin.

Without some further context concerning the interpretation of this function or the intention of the problem, it is difficult to say much more. (Perhaps someone else has additional thoughts.)


问:
generalized f-mean/power mean inequalities
I'd like to prove the following inequality (which seems to be true by numerics)

$$
(p-1)\frac{a^2+b^2}{2} \leq \Big(\frac{a^p+b^p}{2}\Big)^{2/p}
$$
for all $a,b\in [0,1]$ and $p\in [1,2]$. I'd eventually like a generalization when I am summing over many numbers and not just $2$, but I'd like to see how to prove this simple case first.

Body:
By applying the Holder's inequality  where $\frac{1}{m}+\frac{1}{n} = 1$
$$|x_1 y_1 +x_2y_2| \le (x_1^m +x_2^m)^{\frac{1}{m}} (y_1^m +y_2^n)^{\frac{1}{n}}$$

with $(m,n)=\left(\frac{p}{2},\frac{p}{p-2} \right)$, $(x_1,x_2) = (a^2,b^2)$  and $(y_1,y_2) = (1,1)$, we have

\begin{align}
a^2 +b^2 \le \left((a^2)^{\frac{p}{2}}+(b^2)^{\frac{p}{2}} \right)^{\frac{2}{p}} \left(1+1 \right)^{\frac{p-2}{p}} &\iff a^2 +b^2 \le \left(a^p +b^p\right)^{\frac{2}{p}} 2^{\frac{p-2}{p}} \\
&\iff \frac{a^2 +b^2}{2} \le \left(\frac{a^p +b^p}{2}\right)^{\frac{2}{p}} \tag{*}
\end{align}

And because $\frac{a^2 +b^2}{2} \ge (p-1)\frac{a^2 +b^2}{2} $ for $p \in [1,2]$, then from $(*)$ we can conclude that
$$
(p-1)\frac{a^2+b^2}{2} \leq \Big(\frac{a^p+b^p}{2}\Big)^{2/p}
$$


问:
Need help with showing divergence
I'd be glad if someone could help me once again :)

Given is the series:

$$\sum_{n=1}^\infty \frac{n+1}{n^2+n+1}$$ I need to show the divergence. Which way is the right? (Am I correct that the series $$\sum_{n=1}^\infty (-1)^n \cdot \frac{n+1}{n^2+n+1}$$ converges? Tried it using Leibnitz)

and:

$$\sum_{n=1}^\infty \frac{2^n}{2^n+3^n}$$ (again: correct that the series $$\sum_{n=1}^\infty \frac{(-2)^{n}}{2^n+3^n}$$ converges? Tried it using Leibnitz)

Thank you very much!

EDIT: The second series was the wrong one, someone already helped me :)

答:
You could compare each term with $\frac{1}{2n}$:
$$\frac{n+1}{n^2 + n + 1} \ge \frac{1}{2n} \Leftrightarrow n+1 \ge \frac{n}{2} + \frac{1}{2} + \frac{1}{2n},$$
which is clearly valid for $n \ge 1$.
It is well known that the Harmonic series diverges.</p>

ContentLicense: CC BY-SA 3.0

Title: Need help with showing divergence


问:
Zig Zags of infinity on a triangle
There's triangle below. The line is zig zags, and $P_1, P_2, P_3, ... P_n $ is where the zig zags is touching the hypotenuse.

I 2 questions:

Is the zig zags length infinite? To me, no because it's so much like the question if you cut a square in half each time, and then forever, but the length is not infinite.

If the zig zags length, is not infinite. Is the length of zig zags greater than the perimeter of the triangle? Wow, this is very tricky. Now it's using a zig zags series with infinity, like $\sum_{0}^{\infty}$. I can also find angles of each divded triangle. Finally I think the perimeter of the triangle is 3, because it's equilateral. How do I prove this, and how do I find if zig zags is greater than the perimeter?

答:
For either adding the horizontal parts or the up-left ward parts of the zig-zag, each new part is half the length of the previous. So for the horizontal parts total length is $1+1/2+1/4+\cdots=2,$ and for the up-left ward parts total length is $\sqrt{2},$ making total ziz-zag length $2+\sqrt{2}$. Note that this is equal to the perimeter of the largest triangle.

Added: A good way to see the length of the up-left ward parts of the zig-zag should be $\sqrt{2}$ is to start at the lower right (the point forming the right angle of the largest triangle), and for each new piece sliding it over to line up with the end of the previous piece. That done one gets a line segment which is the diagonal of a 1 by 1 square.

A geometric argument can be given for the total length of the horizontal parts of the zig-zag: Leave the lowest one where it is, and rotate each other smaller part (hinged at its right end) downward until it hits the vertical side of the big triangle. Then we get total length is 1 from the horizontal leg and 1 more from the set of rotated parts which exactly fill out the vertical leg of the large triangle.

Title: Zig Zags of infinity on a triangle


问:
Directional derivative Multivariate
Find the gradient, Hessian and the directional derivative of the following function at [1,1]
$$f(x)=x^T \left| \begin{array}{cc}
1 & 2  \\
4 & 8 \end{array} \right|x + x^T \left| \begin{array}{c}
3  \\
4  \end{array} \right|+6.$$
Hello, please, my problem isn't about finding the gradient or the Hessian but how to simplify this function. Thanks.</p>

答:
Actually, one does not need to rewrite $f(x)=x^TAx+x^Tb+c$ in order to find its Hessian or gradient. The linear and constant parts do not contribute to the Hessian. As for the quadratic term $x^TAx$, its Hessian matrix is nothing but $A+A^T$. This is a generalization of the one-variable formula $(ax^2+bx+c)''=2a$. </p>

<p>One can also keep the matrix notation when finding the gradient. Write $x=x_0+dx $ where $x_0=\begin{pmatrix} 1 \\ 1\end{pmatrix}$. Then
$$f(x)=f(x_0)+(dx)^TAx_0 + x_0^T A(dx) + (dx)^Tb +(dx)^TA(dx) \tag1$$
which implies
$$\nabla f(x_0)= Ax_0+(x_0^TA)^T +b = (A+A^T)x_0 +b \tag2$$
This is a generalization of the one-variable formula $(ax^2+bx+c)'=2ax+b$. </p>

Title: Directional derivative Multivariate

Body:
Actually, one does not need to rewrite $f(x)=x^TAx+x^Tb+c$ in order to find its Hessian or gradient. The linear and constant parts do not contribute to the Hessian. As for the quadratic term $x^TAx$, its Hessian matrix is nothing but $A+A^T$. This is a generalization of the one-variable formula $(ax^2+bx+c)''=2a$.

One can also keep the matrix notation when finding the gradient. Write $x=x_0+dx $ where $x_0=\begin{pmatrix} 1 \\ 1\end{pmatrix}$. Then
$$f(x)=f(x_0)+(dx)^TAx_0 + x_0^T A(dx) + (dx)^Tb +(dx)^TA(dx) \tag1$$
which implies
$$\nabla f(x_0)= Ax_0+(x_0^TA)^T +b = (A+A^T)x_0 +b \tag2$$
This is a generalization of the one-variable formula $(ax^2+bx+c)'=2ax+b$.


问:
Comparing between $\omega_2$ and $\omega_1$
My question, it seems very basic about cardinal numbers. I know $\omega_1$ is the first uncountable regular cardinal. Also, $\omega_2$ is the second uncountable regular cardinal.

I did not know how I can envision the difference between them. In other words, what are things that that might be achieved for $\omega_1$ but not for $\omega_2$ and so on.

答:
I suppose I'll give this a stab after all. For me personally, there are two differences between $\omega_1$ and $\omega_2$ which leap reflexively to mind whenever I consider them in the same thought.

At an elementary level, there is a difference in the structure of "natural" stationary sets on each cardinal. In line with Jason Zesheng Chen's comment above, $\mathsf{ZF+AD}$ proves that the club filter on $\omega_1$ is an ultrafilter. By contrast, consider $$A=\{\alpha\in\omega_2: cf(\alpha)=\omega\}\mbox{ and }B=\{\alpha\in\omega_2: cf(\alpha)=\omega_1\}.$$ As long as $\omega_1$ is regular (e.g. assuming countable choice) $A$ and $B$ are stationary but clearly $A\cap B=\emptyset$.

On a more sophisticated level, the related structures $H_{\omega_1}$ and $H_{\omega_2}$ are quite different, model-theoretically speaking. Elements of $H_{\omega_1}$ are "morally equivalent" to real numbers, and so the first-order theory of $H_{\omega_1}$ is relatively immune to set-theoretic shenanigans, especially granting large cardinals. We have no such tameness in $H_{\omega_2}$, however; for example, the continuum hypothesis can be turned on or off as desired by forcing and is equivalent to a first-order sentence in $H_{\omega_2}$. Indeed, the search for a theory to "tame" $H_{\omega_2}$ analogous to projective determinacy's impact on $H_{\omega_1}$ is a very difficult one. While much more complicated than the previous bulletpoint, this is actually more reflexive to me since I use the relative tameness of $H_{\omega_1}$ (or its moral equivalents) quite frequently in my own work.

But these are more technical differences: why should one care about natural stationary sets or model-theoretic tameness? Ultimately we're left with the bare fact that these ordinals are rather technical ($\omega_2$ especially), and ultimately a good picture of them will only emerge after (or at best, during) the process of learning how to work with them.


问:
Compute $P(B_{t}>0,B_{s}>0)$ where $B$ is a standard Brownian motion
Compute $P(B_{t}>0,B_{s}>0)$ for $t >s$ where $B$ is a standard BM

My attempt:

$P(B_{t}>0,B_{s}>0)=P(B_{t}-B_{s}+B_{s}>0,B_{s}>0)$ and now considering $X:=B_{t}-B_{s}\sim \mathcal{N}(0,t-s)$ and $Y:=B_{s}\sim\mathcal{N}(0,s)$ such that for $\xi$, $\eta$ standard normal iid random variables we have $X=\sqrt{t-s}\xi$ and $\sqrt{s}\eta$ such that we obtain:

$$ P(B_{t}-B_{s}+B_{s}>0,B_{s}>0)=P( \sqrt{t-s}\xi+\sqrt{s}\eta>0,\sqrt{s}\xi>0)=P( \sqrt{t-s}\xi+\sqrt{s}\eta>0,\xi>0)$$

Now if I want to switch the above into polar coordinates I have:

$$\{(r,\theta):\; \sqrt{t-s}r\cos(\theta)+\sqrt{s}r\sin(\theta)>0,r\cos(\theta)>0\}=\{(r,\theta):\; \sqrt{t-s}r\cos(\theta)+\sqrt{s}r\sin(\theta)>0,\theta \in (-\pi/2,\pi/2)\}\\ = \{(r,\theta):\; -\frac{\sqrt{t-s}}{\sqrt{s}}<\tan(\theta),\theta \in (-\pi/2,\pi/2)\} = \{(r,\theta):\; \arctan(-\frac{\sqrt{t-s}}{\sqrt{s}})<\theta,\theta \in (-\pi/2,\pi/2)\}$$

This looks rather ugly, am I on the right path?

答:
Use the following fact about the bivariate normal distribution: If $X$ and $Y$ are bivariate normal with mean $0$ and correlation $\rho$, then
$$P(X >0,Y>0) = \frac14 + \frac{1}{2\pi} \cdot\arcsin(\rho).$$
The link assumes $X$ and $Y$ have mean zero and standard deviation $1$, but the same result holds with arbitrary standard deviations. So your problem reduces to finding the correlation between $B_t$ and $B_s$. (Your calculation replicates the derivation of the above equation, but obscures the generality of this fact.)


问:
Is this a circular proof of Pythagorean Theorem? If not, are there better ones?
I found this proof of Pythagorean Theorem from  what 3Blue1Brown shows in his Lockdown math lecture  "Trigonometry Fundamentals":

In the following right triangle, project $\cos\alpha$ and $\sin\alpha$ back to the hypotenuse of length $1$, they become $\cos^2\alpha$ and $\sin^2\alpha$:
$\to\cos^2\alpha+\sin^2\alpha =1 $.

I thought this is what a simple, visual, and elegant proof!  My questions:

Is it a circular proof, why?

If not, are there better ones?

I want clarity about whether or not this proof is circular.  The argument only depends on the definition of the sine and cosine functions and nothing more (please correct me if I am wrong).  This is why I love it.  Many other geometry solutions involve more than three shapes.

答:
Elegance is in the eye of the beholder.

Your proof is simple.  But I think to be elegant it should also be tangible and convincing.  If we divide a hypotenuse of $1$ into two segments $x$ and $y$, it's simple clear and convincing that by similar triangles that the proportions $x:: \cos \alpha$ and $\cos\alpha:: 1$ are equal and $y :: \sin \alpha $ and $\sin\alpha :: 1$ are equal.

ANd from there the algebra that $x = \cos^2 \alpha$ and $y = \sin^2 \alpha$ is clear and simple so the result $1 = x + y = \cos^2 \alpha + \sin^2 \alpha$ is immediate.

But, at least to me, just what the algebra is saying isn't tangible or obvious and is quite abstract.  I can convince myself of it.  It means that the area $\cos \alpha \times \cos \alpha$ square has the same the area of a $x \times 1$ rectangle but... well, it's easy to loose it.  I find myself second guessing "wait, what does area mean again when we aren't talking of integers?  And I don't want to rely on limits and real analysis where are we again?"  Somehow we have turn linear measurements to area measurements being convertible and ...

I find proofs like Andrea Marino's far more tangible (and, yes, I can go through the same existential questions of what is area; but in that type of proof I'm not compelled to).  However.... maybe it's too practical and not abstract enough to be elegant.

The more I think and worry of the abstraction, the more I'm compelled to prefer a Euclid like proof of comparing two sets of rectangles, parallegrams, and squares.  .... But I can hardly call those simple or elegant.


问:
Proving certain límit exists.
I'm proving the next exercise and I'm so close to conclude but I don't know what can I do.

Let $(\varphi_n)_{n\in\mathbb{N}}$ a sequence of non-negative functions in $C([0,1])$ such that for all $k\in\omega$ the next limit exist $$\lim\limits_{n\to\infty}\int_{0}^{1}t^{k}\varphi_{n}(t)\, dt$$Prove that for all $f\in C([0,1])$ the next limit exist $$\lim\limits_{n\to\infty}\int_{0}^{1}f(t)\varphi_n(t)\, dt$$

Let $k=0$. By hypothesis $$\lim\limits_{n\to\infty}\int_{0}^{1}\varphi_n(t)\,dt$$exist. Moreover, if $p(t)$ is a polynomial then $$\lim\limits_{n\to \infty}\int_{0}^{1}p(t)\varphi_n(t)\,dt $$because $p(t)=a_0+a_1t+\cdots+a_mt^{m}$ and then $$\lim\limits_{n\to\infty}\int_{0}^{1}p(t)\varphi_n(t)\,dt=a_0\lim\limits_{n\to\infty}\int_{0}^{1} \varphi_n(t)+\cdots+a_m\lim\limits_{n\to\infty} \int_{0}^{1} t^{m}\varphi_n(t)\,dt$$Let $f\in C[0,1]$. We can take $(p_k)_{k\in\mathbb{N}}$ a sequence of polynomials such that converges uniformly to $f$. Since $$\lim\limits_{n\to\infty}\int_{0}^{1}\varphi_n(t)\,dt$$then there exist $N\in\mathbb{N}$ such that for all $n,m\geq N$  $$\left| \int_{0}^{1}\varphi_n(t)\, dt-\int_{0}^{1}\varphi_m(t)\, dt\right|=\left|\int_{0}^{1}\varphi_n(t)-\varphi_m(t)\, dt \right|<\text{apropiate $\varepsilon$}$$ If $n,n\in\mathbb{N}$ such that $n,m\geq N$ then $$\left| \int_{0}^{1} f(t)\varphi_n(t)\, dx-\int_{0}^{1} f(t)\varphi_m(t)\, dt\right|=\left|\int_{0}^{1} f(t)(\varphi_n(t)-\varphi_m(t)) \, dt\right|$$Let $k\in\mathbb{N}$ such that $\sup\{|f(t)-p_k(t)\mid t\in[0,1]\}<\text{another apropiate $\varepsilon$}$. Then $$\left|\int_{0}^{1} f(t)(\varphi_n(t)-\varphi_m(t)) \, dt\right|=\left|\int_{0}^{1} (f(t)-p_k(t)+p_k(t))(\varphi_n(t)-\varphi_m(t)) \, dt\right|$$This is the same as $$\left|\int_{0}^{1} (f(t)-p_k(t))(\varphi_n(t)-\varphi_m(t))+p_k(t)(\varphi_n(t)-\varphi_m(t))\, dt \right|\leq$$ $$\leq\int_{0}^{1}\left| (f(t)-p_k(t))(\varphi_n(t)-\varphi_m(t))+p_k(t)(\varphi_n(t)-\varphi_m(t))\right|\, dt$$ $$\leq\int_{0}^{1} \left| (f(t)-p_k(t))(\varphi_n(t)-\varphi_m(t))\right|\, dt+\int_{0}^{1}\left|p_k(t)(\varphi_n(t)-\varphi_m(t))\right| \, dt$$But from here I don't know how to conclude. It's not easy because I can bound the absolute value of integrals but not the integral of absolute value. In summary, my idea is aproximate $f$ by polinomials and use the fact that for polynomials the limit exist. Also I found the same question here Existence of Limit : but the answer is so short and I can't see really how to conclude from there.

答:
\begin{array}
\left|| \int_0^1(f(t) - p_k(t))(\phi_n(t) - \phi_m(t)) + p_k(t) (\phi_n(t) - \phi_m(t)) dt || \\
= || \int_0^1(f(t) - p_k(t))(\phi_n(t) - \phi_m(t)) dt + \int_0^1 p_k(t) (\phi_n(t) - \phi_m(t)) dt || \\
\leq || \int_0^1 (f(t) - p_k(t))(\phi_n(t) - \phi_m(t)) dt || + ||\int_0^1  p_k(t) (\phi_n(t) - \phi_m(t)) dt || \\
\leq \int_0^1 || (f(t) - p_k(t))(\phi_n(t) - \phi_m(t)) || dt + ||\int_0^1  p_k(t) (\phi_n(t) - \phi_m(t)) dt || \\
\leq \int_0^1 || \varepsilon_2 (\phi_n(t) - \phi_m(t)) || dt + || \int_0^1  p_k(t) (\phi_n(t) - \phi_m(t)) dt ||
\end{array}
and working with the second integral.

In other ways: First, split up the integral and only then use the monotony.

Please excuse the terrible typesetting, \left and \right do not react as I wish them to do.

Title: Proving certain límit exists.


问:
$\langle a\rangle \simeq \langle b \rangle$ if and only if $a$ and $b$ have the same index and period

If $a$ and $b$ are elements of finite order in the same or in different semigroups, the  $\langle a\rangle  \simeq \langle b \rangle$ if and only if $a$ and $b$ have the same index and period.

$\Leftarrow]$ we have: $\langle a\rangle=\{a, a^2, \dots, a^m, a^{m+1}, \dots, a^{m+r-1}\}$ and $\langle b\rangle =\{b, b^2, \dots , b^m, b^{m+1}, \dots, b^{m+r-1}\}$ where $m$ is the index and $r$ is the period of $a$ and $b$, which by hypothesis is the same. Let's consider the function
$f: \langle a \rangle \to \langle b \rangle$ by $f(a^{i})= b^{i}$ it is easy to verify that $f$ is an isomorphism.

I have trouble showing that if $\langle a \rangle$ and $\langle b \rangle$ are isomorphic then  the index and period of $ a$ are equal to the indexes and period of $b$.
I think that if we assume that they are not true, it may contradict the minimality of the period and the index, but I do not know if it is true, any suggestions? thanks

答:
Let $a$ have period $r$ and index $m$; let $b$ have period $s$ and index $n$. We want to show that $r=s$ and $m=n$.

Note that if $a^i=a^j$ for some $i\neq j$, then it must be the case that $i\equiv j\pmod{r}$ and $i,j\geq m$; symmetrically, if $b^u=b^w$ with $u\neq w$, then $u\equiv w\pmod {s}$ and $u,w\geq n$.

Let $f\colon \langle a\rangle \to \langle b\rangle$ be an isomorphism, and let $g$ be the inverse of $f$.

Say $f(a)=b^k$ and $g(b)=a^h$. Then
$$a=g(f(a)) = g(b^k) = (g(a))^k = a^{hk}.$$
If either $h$ or $k$ are greater than $1$, then $hk\neq 1$. Therefore, $1\geq m$ (so $m=1$), and $hk\equiv 1\pmod{r}$; symmetrically, since
$b=f(g(b)) = f(a^h) = f(a)^h = b^{kh}$, then $1\geq n$ (so $n=1$) and $kh\equiv 1\pmod{s}$. Thus we have $m=n=1$. But then $|\langle a\rangle| = r$, $|\langle b\rangle|=s$, and since the two are isomorphic, then $r=s$.

Thus we may assume that $f(a)=b$ and $g(b)=a$. Can you finish it in this case?

(Note: we need to treat the case $m=1$ separately from $m\neq 1$ (symmetrically for $n$), because whenb $m=1$ we have a group, and $\langle a\rangle$ may have more than one generator; it is a cyclic group, after all. But when $m\gt 1$ the automorphisms must send $a$ to itself, as that is the only generator.)


问:
How to solve a polar equation when $r$ is $r^2$ instead?
I have $r^2=-4\sinθ$

and I'm asked to set $r=0$, then find θ. If I just set $r^2=0$ then I'll get $\sin(2θ)=0$. That doesn't seem right.

Then I'm asked to set $θ=0$ and then find $r$. If I use the $r^2=-4\sinθ$ and set $θ=0$ then I will get "DNE". Not sure what to do instead then...

EDIT:

Sorry everyone I wrote the problem here wrong. It was supposed to be r^2=-4sin2θ

That's where the 2θ came from

答:
For any real number, $r=0$ if and only if $r^2=0$, so "set[ting] $r=0$" is the same as setting $r^2$ to zero. Equivalently: if $r=0$, then $r^2=0$, so of course you get that $r^2=0$.

However, I don't understand why you think you get $\sin(2\theta)=0$. If $r^2=0$, then $-4\sin(\theta)=0$. That means that $\sin(\theta)=0$; where did that $2$ come from?

If you set $\theta=0$ instead, then $\sin(\theta) = \sin(0)$. How much is $\sin(0)$? How much is that when multiplied by $-4$? And what is the (only) value of $r$ that will make $r^2 = -4\sin(0)$ true?

Again, I don't understand why you think you will get "Does not exist" if you plug in $\theta=0$. This is simply not the case. (Though, if you had $r^2 = -4\cos(\theta)$, and tried to find a real value of $r$ for the case $\theta=0$, then you would be unable to find one; are you sure you are computing $\sin(0)$ correctly?)


问:
Definition of $\exp(A)$ in terms of spectral decomposition.
I am read this question Plugging a matrix multiplied by an imaginary number in the exponential function. Here the question'author defined $$\exp(A) := \sum_{1\le k\le n}\exp(\lambda_k) P_k$$

What reference contain this definition "exp is defined in terms of the spectral decomposition"?

答:
Why $\exp A=P^{−1}(\exp D)P$?

Because plugging $A=P^{-1}DP$ into
$$\exp A=\sum_{n=0}^\infty \frac{1}{n!} A^n \tag1$$
and simplifying $(P^{-1}DP)^n = P^{-1}D^n P$, one obtains
$$\exp A=P^{-1}\left(\sum_{n=0}^\infty \frac{1}{n!} D^n \right) P = P^{−1}(\exp D)P \tag2$$


答:
There's a cute non-explicit argument for finiteness here. First, look at the non-dotted lines. These have lengths which add up to a finite number - namely, the length of the hypoteneuse of the big triangle. Since corresponding sides in similar triangles scale identically, the sum of the dotted lengths must also be finite.

Of course this doesn't actually compute the sum of the dotted lengths, but I find it quite nice nonetheless. In particular, it works if we replace $\pi\over 4$ with whatever other angle we want.

As to comparing the sum of the dotted lengths and the perimeter of the whole triangle, consider the following. Let $T_n$ be the right triangle in the diagram with right angle at the top right vertex and $P_n$ as the top left vertex. The top edge and (dotted) right edge have the same length, while the (dotted) hypoteneuse has the same length as the line from $P_n$ to $P_{n-1}$. This just leaves the dotted line touching $\mathcal{O}$. Putting this together, we get that the sum of the dotted lines' lengths is the same as the perimeter of the large triangle.

While more intricate than the previous part of this answer, this is again calculation-free - we just think about isosceles triangles, and draw the right ones.

Title: Zig Zags of infinity on a triangle

Body:
There's triangle below. The line is zig zags, and $P_1, P_2, P_3, ... P_n $ is where the zig zags is touching the hypotenuse.

I 2 questions:

Is the zig zags length infinite? To me, no because it's so much like the question if you cut a square in half each time, and then forever, but the length is not infinite.

If the zig zags length, is not infinite. Is the length of zig zags greater than the perimeter of the triangle? Wow, this is very tricky. Now it's using a zig zags series with infinity, like $\sum_{0}^{\infty}$. I can also find angles of each divded triangle. Finally I think the perimeter of the triangle is 3, because it's equilateral. How do I prove this, and how do I find if zig zags is greater than the perimeter?


问:
If $P$ be the orthogonal projection of Hilbert space $X$, then $\langle Px,x \rangle = \|Px\|^2, \forall x \in X $
Let $P$ be the orthonormal projection of a Hilbert space $X$ onto a closed subpace $Y$. Also, assume that $Y$ is the linear span of the orthonormal set $[y_1, \cdots, y_n].$ Then I need to show that $\langle Px,x \rangle = \|Px\|^2, \forall x \in X.$

I began by the very definition: $$Px = \sum_{i=1}^n \langle x,y_i \rangle y_i$$ Thus, $$\langle Px,x \rangle = \Big \langle \sum_{i=1}^n \langle x,y_i \rangle y_i,x \Big \rangle = \sum_{i=1}^n \langle x,y_i \rangle \langle y_i,x \rangle. \quad (I)$$

On the other hand, $$ \|Px\|^2 = \langle Px,Px \rangle = \Big \langle \sum_{i=1}^n \langle x,y_i \rangle y_i , \sum_{j=1}^n \langle x,y_j \rangle y_j  \Big \rangle = \sum_{i=1}^n \sum_{j=1}^n \langle x,y_i \rangle \langle x,y_j \rangle \delta_{ij} = \sum_{i=1}^n \langle x,y_i \rangle \langle x,y_i \rangle. \quad (II) $$
Now, I need to show that (I) and (II) are the same. But are they in general? Knowing that generally, $\langle x,y\rangle = \overline{\langle y,x\rangle}$ and not necessarily $\langle x,y\rangle = {\langle y,x\rangle}.$ Am I right so far? If so, how to continue from this point onward?

答:
If $P$ is a projection than $P=P^2$ and if $P$ is an orthogonal projection then $P^*=P$. Hence
$\langle Px , x \rangle = \langle P^2x , x \rangle = \langle P^*Px , x \rangle = \langle Px , Px \rangle $.


问:
Prove that $\sup\{f(x) : x < s\} ≤ \inf\{f(y) : y > s\}$ and both values are real numbers given an increasing function
Suppose there's an increasing function $f : R → R,$ indicating that $f(x) ≤ f(y)$ whenever $x ≤ y.$

My question is: given $s ∈ R,$ how would I show that $\sup\{f(x) : x < s\} ≤ \inf\{f(y) : y > s\}$ and also that both values are real numbers ?

答: You have the inequality $f(x) \leq f(y)$ for all $x \leq y$, so take $x_n$ strictly increasing to $s$ and $y_n$ strictly decreasing to $s$. Necessarily $x_n < s < y_n$ for each $n$, so $f(x_n) \leq f(y_n)$, so $\sup\{f(x): x < s\} = \lim_n f(x_n) \leq \lim_n f(y_n) = \inf\{f(y): y > s\}$.


问:
$a \circ' b = b^{-1} \circ a \circ b^{2}$ Show that for every $a, b \in G$, the equations $x \circ' a = b$ and $a \circ' x =b $ has unique solutions.
Let $(G, \circ)$ be a group. Define a new binary operation $\circ'$ on $G$ as follows, $$a \circ' b = b^{-1} \circ a \circ b^{2}$$ Show that for every $a, b \in G$, the equations $x \circ' a = b$ and $a \circ' x =b $ has unique solutions.

First part is easy since $a^{-1}xa^{2}=b$, so $x = aba^{-2}$

But I am stuck in second part, here is what I tried.

We want to find $x$ such that $x^{-1}ax^{2}= b$, Firstly I just considered $f_{x} = x^{-1}ax^{2}$, and saw what is happening with it for different $x$, I found if $x = a^{n}$ then $f_x = a^{n+1}$, also clearly $x = a^{-1}$ we get identity element and also tried some other things, which seems irrelevant. So, it did not helped much, Then I considered $ax^{2} = xb$ and tried to simplify this, for this firstly I replaced $x$ by $xb$, which gave $axbxb = xb^{2}$ or $axbx = xb$ and I tried to manipulate this in more ways, but did not get anything, finally I just started guessing and putting, which did not worked.

Also is there a general way to solve equations in group.

Thanks in Advance.

答:
This as written above is not always true though.

Let us consider $S_3$ acting on $\{1,2,3\}$, and let $a$ be the following element:
$a(1)=3$; $a(3)=1$; $a(2)=2$. Then as $e^{-1}ae^2 =a$ already, by the Pigeonhole Principle, there is a $b \in S_3$ such that there is no $x \in S_3$ satisfying $x^{-1}ax^2 =b$ if there is a $y \in S_3; y \not = e$ that satisfies $y^{-1}ay^2=a$.

Take $y$ as follows: $y(1)=2$; $y(2)=3$; $y(3)=1$. Then $y^{-1}ay^2=a$. [Indeed, first note the equation $y^2=y^{-1}$,
and the note $y^{-1}ay^2(1)= y^{-1}a(3) = y^{-1}(1) = 3$;
$y^{-1}ay^2(3) = y^{-1}a(2) = y^{-1}(2) = 1$; and so $y^{-1}ay^2(2) = 2$.]


答:
As nicely compact as the cited diagram is, I might be inclined ---as a matter of diagrammatic pedagogy--- to render things thusly:

This arrangement can help make clear(er) that

"$\sin\theta$" and "$\cos\theta$" are effectively definitions (of particular ratios in a right triangle)

"$\cos^2\theta$" and "$\sin^2\theta$" are consequences (via proportions in similar right triangles)

so that

"$\cos^2\theta+\sin^2\theta=1$" is a deduction, not an assumption.

As a result, some (most? all?) of the confusion about the logical dynamics is mitigated. There's no circular argument here.

(Plus, as a matter of diagrammatic style, I find it's often preferable to avoid overlapping elements.)

The contention (dating back at least as far as to Loomis in the 1940s) that "no trigonometric proof [of the Pythagorean Proposition] is possible" seems to manifest in knee-jerk rejections of any argument that makes any reference whatsoever to sine or cosine. Granted, caution is advised when considering such things: once trig relations become second-nature to us all, it's all-too-easy for an author to inadvertently invoke a Pythagorean identity ... and for a reader to overlook the mistake. But, when caution is applied, it's certainly possible to construct a viable proof.

Title: Is this a circular proof of Pythagorean Theorem? If not, are there better ones?

Body:
I found this proof of Pythagorean Theorem from  what 3Blue1Brown shows in his Lockdown math lecture  "Trigonometry Fundamentals":

In the following right triangle, project $\cos\alpha$ and $\sin\alpha$ back to the hypotenuse of length $1$, they become $\cos^2\alpha$ and $\sin^2\alpha$:
$\to\cos^2\alpha+\sin^2\alpha =1 $.

I thought this is what a simple, visual, and elegant proof!  My questions:

Is it a circular proof, why?

If not, are there better ones?

I want clarity about whether or not this proof is circular.  The argument only depends on the definition of the sine and cosine functions and nothing more (please correct me if I am wrong).  This is why I love it.  Many other geometry solutions involve more than three shapes.


问:
Is there a way to generalize the connected sets in $\mathbb{R}^2$?
How do I generalize the connected sets in $\mathbb{R}^2$ if there is a way. I know how connected sets in $\mathbb{R}$ look like, so is there a way to say something about the connected sets in $\mathbb{R}^2$?

答:
There is no good way to classify the connected sets in $\mathbb{R}^2$. Even the planar continua - that is, connected subsets of $\mathbb{R}^2$ which are closed and bounded - are extremely varied (and have a very interesting theory!), see e.g. here.

Now granted, since there isn't a good rigorous notion of "classification" it's hard to turn this into a precise theorem. However, I think the following pretty is pretty definitive:

$(*)\quad$ There are $2^{2^{\aleph_0}}$-many homeomorphism types of connected subsets of the plane. In fact, this is true even for simply connected and path connected subsets!

That is, there are as many distinct "topological types" of connected sets in the plane as there are sets in the plane in the first place! Contrast this with the situation for $\mathbb{R}$, where there are only four.

The proof of $(*)$ is a bit lengthy, but ultimately elementary. Here's one way to do this:

Basically, we want to associate to each set $A$ of paths through the infinite binary tree (there are $2^{2^{\aleph_0}}$-many sets of such paths) an $X_A\subseteq\mathbb{R}^2$ which is connected and such that $A\not=B$ implies $X_A\not\cong X_B$. The first step to doing this is to build a copy of the complete binary tree inside $\mathbb{R}^2$, in a "topologically-decodable" way. For example:

Here the root of the tree is recognizable as the only degree-five point, and we can distinguish left and right via the degree-four points (the little hash marks in the middle of the left-pointing edges). Of course I've only drawn the first few branchings of the tree.

Now we attach, to the "end" of each path through this infinite tree, a copy of $[0,1]$ or $[0,1)$ depending on whether the infinite binary sequence representing that path is in $A$

This is all a little vague; it's a good exercise to fill in the details. Note that the sets so constructed are topologically quite nice in may ways (e.g. contractible, and locally homeomorphic to $\mathbb{R}$ except on a nowhere-dense set).


问:
Counting numbers smaller than $N$ with exactly $k$ *distinct* prime factors
Using common notation, $\omega(n)$ is the number of distinct prime factors on $n$. Similiarly, $\Omega(n)$ is the number of prime factors of $n$, not necessarily distinct: $120=2^{3}\cdot 3 \cdot 5$ , therefore $\omega(120)=3$ , $\Omega(120) = 5$.

If we let $W_k(N)$ be the count of numbers not exceeding $N$ with exactly $k$ distinct prime factors {$i | i \leq N , \omega (i) = k$}, is there any way or formulation to calculate it? Or at least a relatively coincise algorithm for it, which does not include factoring each number and checking for $\omega$ individually?

For example, one can easily find that $W_3(100)=8$ , $W_3(1000000)=379720$ , $W_4(1000000)=208034$.

Please note here I am not looking for an asymptotic approximation but rather an exact count. And $N$ can be large, with order of magnitude of $10^{12}$ and over.

I have found that a solution exists for the $\Omega$ counterpart, known as k-Almost Primes, which admits closed-form formulas, but as for $\omega$, I have found non.

答:
The system asked if I really wanted to post another answer, but I really do feel like this belongs separately. Two things, actually. First, a simple (thought not algebraic) expression for $W_1(N)$, in terms of the prime counting function:

$$W_1(N) = \sum_{i = 1}^{\lfloor \log_2 N \rfloor} \pi \left(N^{\frac{1}{i}} \right)$$

$W_1(N)$ is just all the primes and their prime powers. Primes $\sqrt{N} \leq p$ will only have their first power on the list. Primes $\sqrt[3]{N} \leq p \leq \sqrt{N}$ will also have their second power on the list, and are counted twice, and so on to $i = \lfloor \log_2 N \rfloor$, for which $\pi(N^{\frac1i}) = 1$, that is, only $2$.

Second, a different take on algorithmic sieving. We start with:

$$|\{i | i \leq N , \omega (i) = k\}| = |\{i | i \leq N , \omega (i) \geq k\}| - |\{i | i \leq N , \omega (i) \geq k+1\}|$$

That is, for say $k=4$, if we find the set of integers where $\omega(i) \geq 4$, then remove everything that has $\omega(i) \geq 5$, what remains must have $\omega(i) = 4$. We do this because it means we can just use multiples without worrying about whether the multiples have specific factors. So for instance, $2\cdot3\cdot5\cdot7 = 210$ has four factors. Every multiple of it must have at least as many factors, in some cases many more.

I first thought "Hey we can just divide to find how many multiples there are," but that doesn't work--we end up counting the same thing multiple times. But then it struck me that the only numbers we'll count multiple times are the ones that have at least five distinct factors. So we only need to run this enumeration once, then remove all copies of any values that appear more than once in the list.

There are almost certainly optimizations to be done once you get to larger primes, see the above discussion. Right now my issue seems to be the time used to find and remove the multiples is dominating the other calculations. Weird. Will comment later with Sage code.


答:
No, they're not the same. You have an error in version II. Since $\langle\cdot,\cdot\rangle$ is conjugate-linear in the second variable, you should have $\overline{\langle x,y_j\rangle}$, and then the two expressions do in fact become equal.

Note that it should have been suspicious that your formula for $\|Px\|^2$ did not necessarily give a real answer.

Title: If $P$ be the orthogonal projection of Hilbert space $X$, then $\langle Px,x \rangle = \|Px\|^2, \forall x \in X $

Body:
Let $P$ be the orthonormal projection of a Hilbert space $X$ onto a closed subpace $Y$. Also, assume that $Y$ is the linear span of the orthonormal set $[y_1, \cdots, y_n].$ Then I need to show that $\langle Px,x \rangle = \|Px\|^2, \forall x \in X.$

I began by the very definition: $$Px = \sum_{i=1}^n \langle x,y_i \rangle y_i$$ Thus, $$\langle Px,x \rangle = \Big \langle \sum_{i=1}^n \langle x,y_i \rangle y_i,x \Big \rangle = \sum_{i=1}^n \langle x,y_i \rangle \langle y_i,x \rangle. \quad (I)$$

On the other hand, $$ \|Px\|^2 = \langle Px,Px \rangle = \Big \langle \sum_{i=1}^n \langle x,y_i \rangle y_i , \sum_{j=1}^n \langle x,y_j \rangle y_j  \Big \rangle = \sum_{i=1}^n \sum_{j=1}^n \langle x,y_i \rangle \langle x,y_j \rangle \delta_{ij} = \sum_{i=1}^n \langle x,y_i \rangle \langle x,y_i \rangle. \quad (II) $$
Now, I need to show that (I) and (II) are the same. But are they in general? Knowing that generally, $\langle x,y\rangle = \overline{\langle y,x\rangle}$ and not necessarily $\langle x,y\rangle = {\langle y,x\rangle}.$ Am I right so far? If so, how to continue from this point onward?


问:
Convolution of regulated functions
In his Analysis 1, Konrad Königsberger defines a convolution of two $2\pi$ periodic regulated functions (that have left and right limits at every point ) $f$ and $g$ as:

$$
(f*g)(x):=\frac{1}{2\pi}\int_{\mathbb{T}}f(t)g(x-t)dt
$$

where $\mathbb{T}$ is an arbitrary $2\pi$ long interval. Right after this, he states that $f*g$ is clearly $2\pi$ periodic and regulated. To me, it is not obvious that it's regulated.

I've managed to prove that if the functions are continuous, then their convolution is also continuous, but I can't use the same argument to prove that it's regulated.

What seemed as a promising attempt was to prove that the convolution of two step (piecewise-continuous) functions is a piecewise linear function. The problem is that I cant find a nice way to show this. Knowing that this holds, one could approximate $f(t)g(x-t)$ with $\varphi(t)\psi(x-t)$ such that $|f(t)g(x-t)-\varphi(t)\psi(x-t)|<\frac{\varepsilon}{2(b-a)}$ for all $t$, and $\varphi*\psi$ with a step function $\rho$ such that $||\varphi*\psi-\rho|| < \frac{\varepsilon}{2}$.

Edit:

I managed to prove it using piecewise linear and step functions: First we must prove that the convolution of two step functions with at most one point of discontinuities is a piecewise linear function. To do this, we can let:

\begin{equation*}
\begin{split}
&\varphi(t)=
\begin{cases}
\alpha & x\in[0,a)\\
\beta & x\in[a,2\pi)
\end{cases}\\
&\psi(t)=
\begin{cases}
\gamma & x\in[0,b)\\
\delta & x\in[b,2\pi)
\end{cases}
\end{split}
\end{equation*}

They're both extended $2\pi$ periodically. We can assume that $a\leq b$. Let $x\in[0,2\pi)$

\begin{equation*}
\begin{split}
&2\pi(\varphi*\psi)(x) = \int_0^{2\pi}\varphi(t)\psi(x-t)dt = \alpha\int_0^a\psi(x-t)dt + \beta\int_a^{2\pi}\psi(x-t)dt
&
\end{split}
\end{equation*}

Now you simply look at cases $x\in[0,a);x\in[a,b);x\in[b,2\pi)$ and notice, after a few substitutions, that the function is linear on all of them.

When $\varphi$ and $\psi$ are arbitrary step functions they can be written as sums of step functions with one point of discontinuity. Using the bilinearity of the convolution and the fact that the sum of piecewise linear functions is piecewise linear, we get that convolution of any two step functions is piecewise linear, thus obviously regulated.

Finally, if we let $f,g$ be $2\pi$ periodic regulated functions, we pick step functions $\varphi,\psi,\rho$ such that:

\begin{equation*}
\begin{split}
&1.||g-\psi|| < \frac{\varepsilon}{3||f||}\\
&2.||f-\varphi|| < \frac{\varepsilon}{3||\psi||}\\
&3.||\varphi*\psi-\rho||<\frac{\varepsilon}{3}
\end{split}
\end{equation*}

From here, we get that for all $x\in[0,2\pi)$

\begin{equation*}
\begin{split}
|(f*g)(x)-\rho(x)|\leq|(f*g)(x)-(f*\psi)(x)|+|(f*\psi)(x)-(\varphi*\psi)(x)| +
|(\varphi*\psi(x)-\rho(x))|<\varepsilon
\end{split}
\end{equation*}

答: Note that any regulated function on a compact interval $[a, b]$ is bounded. If it were unbounded, one could choose $x_n \in [a,b]$ with $|f(x_n)| > n$ for each $n$, and these $x_n$ would admit an accumulation point $x$. The left/right limits at this $x$ cannot exist in $\mathbb{R}$, giving a contradiction. Thus $f,g$ are bounded functions, and so the integrand is bounded. Since $\mathbb{T}$ has finite measure and $f,g$ are bounded we may apply the dominated convergence theorem to pass limits inside the integral, from which it immediately follows that $f * g$ is regulated by passing the limits inside the integral and using that $f,g$ are regulated.

Title: Convolution of regulated functions


问:
Limits in functions from $\mathbb R^2 \to \mathbb R$
Good evening,
I would like tips / methods for calculating limits of functions of $\mathbb R ^ 2 \in \mathbb R$.

The squeeze theorem is sometimes a problem for me to set up because I do not know how to put in order this new type of functions in the good way, it is necessary that $f (x, y) <g (x, y) <h (x, y)$ but I must make sure this is true for all $x$ and all $y$?

Concerning the formal definition, making $\lvert \lvert x - x_0 \rvert \rvert$ (where $x$ and $x_0$ are in $\mathbb R^2$) appear in the expression $\lvert f(x) -l \rvert$ is sometimes difficult. Obviously, my teacher has the bad habit of giving the simplest examples in the world by looking for example at the limit when $(x, y)$ tends to $(1,0)$ of a function where there is already $\sqrt {(x-1) ^ 2 + y ^ 2}$ which is not very beneficial for the students.

I have heard of switching to polar. The course on these functions being discovered today, I have not yet seen this method but if it is really essential and useful I would like to know what it consists of.

As an example I am interested in the limit when (x, y) goes to $(0,1)$ of $f (x, y) = \frac {y} {x ^ 2} e ^ {\frac {- y} {x ^ 2}}$ for $x \neq 0$ and $0$ when $x = 0$.

答:
If you want to treat it as a single limit (as opposed to $x\rightarrow x_0$ and $y\rightarrow y_0$ independently) then you need to account for all possible sequences that converge to $(x_0,y_0)$. The limit of $f$ at $(x_0,y_0)$ only exists if the images of all such sequences converge to the same value.

This is because function limits are defined in terms of universal qualification of points arbitrarily close to the limit point. This doesn't make much of a difference in $\mathbb{R}$ because you can only approach a point from 2 different directions, but in higher dimensions there are infinite directions of approach.

If you know that the limit exists and you'd just like to compute it, then you can use any parametric function $z(n)\rightarrow(x_0,y_0)$, as long as $f$ is defined for every $z(n)$. Note that it's up to you how to use $n$; you could do $n\rightarrow\infty$ or $n\rightarrow0$. It just needs to match how you use $n$ in $z(n)$. ($n\rightarrow\infty$ is more common, such as in real-analysis proofs that use convergent sequences.)

You can also use this to prove that the limit doesn't exist, by creating a sequence where $f$ doesn't converge, or two sequences where $f$ converges to different values.

In your example, fixing $x=0$ is only one possible way to approach $(0,1)$. You also need to try other sequences as a sanity check, such as $\left(\frac{1}{n}\mathrm{cos}(n),1+\frac{1}{n}\mathrm{sin}(n)\right)$.

Note that there's nothing special about choosing the sequence, as long as it converges to $(x_0,y_0)$. The goal is to find a sequence that makes the computation of (or the proof of non-existence of) the limit convenient.

You also mentioned polar coordinates. If you used something like $\left(\frac{1}{n}\mathrm{cos}(\theta),\frac{1}{n}\mathrm{sin}(\theta)\right)$, then you can leave $\theta$ unspecified to show that the angle of approach doesn't affect the result.

Here is a specific example of a limit that appears to exist if you only look at one sequence approaching the limit point.

Let $f(x,y)=\mathrm{tanh}\left(\frac{1}{x+y}\right)$ and $(x_0,y_0)=(0,0)$.

With $z(n)=\left(0,\frac{1}{n}\right)$, $f(z(n))=\mathrm{tanh}\left(n\right)\rightarrow 1$.

With $z(n)=\left(\frac{1}{n},-\frac{2}{n}\right)$, $f(z(n))=\mathrm{tanh}\left(-n\right)\rightarrow -1$.

Since the images of the two sequences don't converge to the same value, that means that the limit at $(0,0)$ doesn't exist.


问:
Proof that $f(x) = \frac{2}{1 - |x|}$ is surjective on $\mathbb{R}→(0,2]$
How can I prove this function is surjective ? I'm not sure how to start.
$$f(x) = \frac{2}{1 - |x|}$$
on $\mathbb{R}→(0,2]$

Thank you!

答: It seems there is an error or so in your post because $f(\frac{1}{2}) = 4 \notin (0,2]$. If your function is $f(x) = \dfrac{2}{1+|x|}$, then yes, it is surjective, i.e $f(\mathbb{R}) = (0,2]$. Observe that if $r \in (0,2]\implies f(d) = \dfrac{2}{1+|d|} = r\implies |d| = \dfrac{2-r}{r}\implies d = \pm \dfrac{2-r}{r} \implies \text{surjectivity of f}$ .


问:
Does the $\sum\limits_{n=1}^{\infty}(\sqrt[n]{4}-1)^{\ln(n)}$ converges?
To check for the convergence, I used most of the common tests like, Ratio test, Root test, Divergence test, integral test and also tried to use a comparison. But none of them gave a satisfactory result.

Appreciate your help of this

答:
$4^{1/n}$ is approximately $1 + c/n$ (where $c = \log 4,$ but it is irrelevant.

So, the term is approximately $\frac{1}{n^{\log n}},$ which goes to $0$ faster than any power of $n.$ By the comparison test with $\sum \frac1{n^2}$ the series converges.


问:
Type of distribution and mgf
A tutoring class consists of 5 students and meets for a total of $11$ times. After each lesson, students take a quiz. The score of each of the $5$ students is equally likely to be $1,2,3, 4$ independently of other students.  Let $X$ be the sum of the quiz scores of the students in today’s lesson. What is the moment generating function of $X$?

I am very confused about what type of distribution $X$ follows. I figured I would assign a single distribution to each of the students, $X_k$, for the student $k$. Thus far, my guess is that $X$ follows a binomial distribution but the parameters are not clear. Any assistance is much appreciated.

答:
One approach for this specific situation is to consider each student's score to be the result of two coin flips. So it is two sets of $5$ coin flips; the first set can be zero or one, the second set can be the "2's digit" and would add $2$ to the score if it is a $1$ and $0$ if it is a zero.

Suppose the first student gets a $1$. That would correspond to $X_1 = 0$ and $Y_1 = 0$ and the first score would be $2X_1 + Y_1 + 1$. If the second student got a $4$, that would correspond to $X_2 = 1$ and $Y_2 = 1$ for a score of $2X_2 + Y_2 + 1$.  The total score would be $\sum_{i = 1}^{5} \big(2X_{i} + Y_{i} + 1\big)$.

So you would have a sum of $5$ Bernoulli random variables (a binomial with $n=5$ and $p = 1/2$), and $2$x a sum of $5$ other Bernoulli random variables.
From there you can use theorems about moment generating functions.

Title: Type of distribution and mgf


问:
Maximum number of edges in minimally $k$-edge-connected multigraph
A graph or multigraph is $k$-edge-connected if it cannot be disconnected by deleting fewer than $k$ edges. It is minimally $k$-edge-connected if it loses this property when any edges are deleted. (Equivalently, if any edge of the graph is part of a $k$-edge cut).

According to this paper,

It is easy to show that a minimally $k$-edge-connected multigraph on $n$ vertices has at most $k(n-1)$ edges, and that this value is best possible for all values of $n$ and $k$.

I can see how this is best possible: if you take any $n$-vertex tree, and replace each edge of the tree by $k$ parallel edges, then we get a minimally $k$-edge-connected graph.

(To disconnect it, we need to destroy an entire edge of the original tree, which requires deleting $k$ edges. Every edge is part of such a $k$-edge cut, together with its parallel copies.)

How do we show this?

Notes:

This came up in my answer to this question, where I gave a proof that for simple graphs, the maximum number of edges is smaller.

A possibly-useful lemma (see this question) is that any minimally $k$-edge-connected graph, or multigraph, has a vertex of degree exactly $k$.

答:
We induct on $k$; for $k=1$, this statement is just the claim that every minimally connected $n$-vertex graph is a tree (with $n-1$ edges).

Now let's take a minimally $k$-edge-connected $n$-vertex graph $G$. Inside it, let $H$ be any minimally $(k-1)$-edge-connected spanning subgraph, and let $F$ be the subgraph consisting of all edges of $G$ not in $H$. By induction, $H$ has at most $(k-1)(n-1)$ edges, and if $F$ has at most $n-1$ edges, we're done.

To prove this, we prove that $F$ is a forest. Let $e$ be any edge of $F$; by the minimality of $G$, it is part of a $k$-edge cut. In other words, there is a partition $V(G) = S \cup T$ such that $G$ has only $k$ edges between $S$ and $T$, and one of them is $e$. Well, because $H$ is $(k-1)$-edge-connected, $H$ has at least $k-1$ edges between $S$ and $T$. That and $e$ already gets us up to $k$ edges. Therefore $F$ can have no other edges between $S$ and $T$.

This proves that $e$ is a bridge of $F$, so it is not part of any cycle in $F$. Because $e$ was arbitrary, we conclude that no edges of $F$ are part of any cycles in $F$: $F$ is a forest. Therefore $|E(F)| \le n-1$, and $|E(G)| = |E(H)| + |E(F)| \le k(n-1)$, as desired.


问:
Help with probability mass function, specifically understanding sample mean
The question is

Let $X$ be a discrete random variable with probability mass function

(a) Find $E(X)$
I did $E(X) = -2*.3+ -1*6 + 12*.1 = -5.4$

(b) Find $Var(X)$
 I did $E(x^{2})-(E(x))^{2}= (-2^{2} * .3 + -1*.6 + 12*.1) - (-5.4)^{2} = -29.76$

(c) Find expected value of X(bar) (the sample mean)
 Would this just be the same thing as $E(X)$?
(d) if n = 100, what is the variance of X(bar)?
 would I just multiply the variance by 100?

Last question is if I wanted to put this all in R, how would I input my CDF into R and find the expected value and variance, etc. I understand it on paper (most of the time) but I am unsure how I would put it in R, any help is appreciated, thank you in advance.

答:
You forgot to square the x's before multiplying and then adding.
$$E[X^2] = (-2)^2\cdot 0.3 + (-1)^2 \cdot 0.6 + (12)^2\cdot 0.1 = 1.2 + 0.6 + 14.4 = 16.2$$
When you multiply a variable by $c$, you multiply the mean of it by $c$ and the standard deviation by $c$, which means the variance is multiplied by $c^2$.  The expected value of a sum is the sum of the expected values.

In R, you could construct a vector of $X$ values and a vector of $p$ values

$x \leftarrow c(-2,-1,12)$

$p \leftarrow c(0.3, 0.6, 0.1)$

I'm afraid I don't know the sampling functions in R very well yet. I have used the ones in the rethinking module.


问:
Covering map with a right inverse is a homeomorphism
I am stuck on the following problem. I believe I am very close.

Let $p:E\to B$ be a covering map. Suppose that there exists a continuous $f: B\to E$ so that $p\circ f = id_B$ (that is, $p$ has a right inverse).
Prove that if $E$ is path connected, then $p$ is a homeomorphism.

Since $p$ is covering map, I know that it suffices to show $p$ is injective. I am guessing I am to show that for any $b\in B$, $|p^{-1}(\{ b\} )|=1$.

Let $b\in B$. Let $e\in p^{-1}(\{ b\} ) $. I know that the monodromy action is transitive since $E$ is path connected. This, and the fact that $p$ is surjective, gives $1 \leq |p^{-1}(b)| \leq | \pi _1 (B,b)| $. (This comes from the correspondence $\pi _1(B,b) \to p^{-1}(b) $ via $[\gamma ] \mapsto e \cdot [\gamma ] $ being surjective.)

From this, I predict that I will need to show that $\pi _1(B,b)$ is trivial (i.e. $B$ is simply connected). However I have no idea how to show this.

How do we use the right inverse? The most I can get out of it is that $f$ is a lift of the identity function on $B$. This gives the induced homomorphism $p_*: \pi _1 (E,e) \to \pi _1 (B,b)$ is an isomorphism. Does this help me?

Any suggestion would be great.

答:
You are correct that it suffices to show that $p$ is injective. This is because $p \circ f \circ p = p \circ id_E$, so if $p$ is injective then $f \circ p = id_E$ since injective maps can be cancelled from the left.

To show that it is injective, consider $a, b \in E$ such that $p(a) = p(b)$. Since $E$ is path-connected, let $\gamma : [0, 1] \to E$ be a path such that $\gamma(0) = a$ and $\gamma(1) = b$. Now consider the map $p \circ \gamma : [0, 1] \to B$. Note that $p \circ (f \circ p \circ \gamma) = (p \circ f) \circ (p \circ \gamma) = id_B \circ (p \circ \gamma) = p \circ \gamma$.

By the path lifting property of covering spaces, this implies that $(f \circ p \circ \gamma) = \gamma$. In that case, $a = \gamma(0) = f(p(\gamma(0))) = f(p(a)) = f(p(b)) = f(p(\gamma(1))) = \gamma(1) = b$.

Therefore, $a = b$.


问:
Prove ${\rm Hom}_\mathbb{Z}(\mathbb{Z}_n,\mathbb{Z}_m)\cong \mathbb{Z}_{m,n}$
Prove ${\rm Hom}_\mathbb{Z}(\mathbb{Z}_n,\mathbb{Z}_m)\cong \mathbb{Z}_{m,n}$

I think we can definine a map

$$\phi: \mathbb{Z}_n \to \mathbb{Z}_m,$$

where $\phi \in {\rm Hom}_\mathbb{Z}(\mathbb{Z}_n,\mathbb{Z}_m)$

Then $\pi :{\rm Hom}_\mathbb{Z}(\mathbb{Z}_n,\mathbb{Z}_m) \to \mathbb{Z}_{m,n}$

$\phi \mapsto t$ which is the lowest common denominator of $m,n$.

Then we show this is an bijective and a ring homomorphism.

First, I'm still working through this notation and I'm not sure if things above make sense. Second defining my map $\pi$ is weird because I'm not sure how we express an element in $\mathbb{Z}_{m,n}$

答:
One can show more generally that for a one-dimensional principal ideal domain $R$ and any elements $a$ and $b$ of $R,$ we have that $$\operatorname{Hom}_R\!{\left(\frac R {aR}, \frac R {bR}\right)} \cong \frac R {\gcd(a, b) R} \cong \frac R {aR} \otimes_R \frac R {bR}.$$ Convince yourself first that $\operatorname{Hom}_R(R/I, M) \cong (0 :_M I) = \{x \in I \,|\, x \cdot m = 0\}$ for any commutative unital ring $R,$ any ideal $I$ of $R,$ and any $R$-module $M.$ Considering that $R / I$ is cyclic, a well-defined $R$-linear map $\varphi : R / I \to M$ is uniquely determined by $m = \varphi(1_R + I).$ Consequently, for any element $x$ of $I,$ it follows that $$x \cdot m = x \cdot \varphi(1_R + I) = \varphi(x + I) = \varphi(0_R + I) = 0.$$ Can you construct an isomorphism $\psi : \operatorname{Hom}_R(R/I, M) \to (0 :_M I)$? Once you have established this, apply it to the case that $I = aR$ and $M = R / bR$ to show that $\operatorname{Hom}_R(R/aR, R/bR) \cong (0 + bR :_{R/bR} a),$ where $$(0 :_{R/bR} a) = \{r + bR \,|\, ra + bR = 0 + bR\} = \{r + bR \,|\, b \text{ divides } ra\} = (b :_R a)/bR.$$ Last, you must prove that $(b :_R a)/bR \cong R / dR,$ where we denote $d = \gcd(a, b).$ Consider the map $$\gamma : R \ni r \mapsto \frac{\operatorname{lcm}(a, b)} a r + bR \in \frac{(b :_R a)}{bR}.$$ Of course, we must convince ourselves that $\gamma$ is well-defined and surjective with $\ker \gamma = dR.$ But this can be achieved by manipulating the identity $ab = \gcd(a, b) \operatorname{lcm}(a, b).$

Ultimately, the question you have asked is a special case of this with $R = \mathbb Z.$


问:
Definition of set containment
Let $X$ and $Y$ be sets. Does "$X$ contains $Y$" mean $Y \subseteq X$ or $Y\in X$, or is it ambiguous?

答:
The symbols $\subset$ and $\subseteq$ are used for sets which are subsets of another, which can also be described as containment. The former is used when the sets are not equal, the latter when they can be.
The symbol $\in$ is used for elements of a set, rather than whole sets. For example, you could say: $$y\in Y\subseteq X \implies y\in X$$

Title: Definition of set containment


问:
Problem related with solving ODE
I was solving old exam papers and am stuck on the following problem:

Consider the system of ODE

  $\frac {d}{dx}Y=AY,Y(0)=\begin{pmatrix}
2\\
-1
\end{pmatrix}$ where $A=\begin{pmatrix}
1 &2 \\
 0&-1
\end{pmatrix},Y=\begin{pmatrix}
y_1(x)\\
y_2(x)
\end{pmatrix}$.  Then I have to determine which of the following options hold good.

$y_1(x) \to \infty,y_2(x) \to 0\,\,$ as $\,\,x \to \infty$

$y_1(x) \to 0,y_2(x) \to 0\,\,$ as $\,\,x \to \infty$

$y_1(x) \to \infty,y_2(x) \to -\infty \,\,$ as $\,\,x \to -\infty$

$y_1(x),y_2(x) \to -\infty \,\,$ as $\,\,x \to -\infty$ .

My Attempt: Using the eigenvalues of $A$,I get the ODE of the form:

$y(x)=c_1e^x+c_2e^{-x},$ where $c_1,c_2$ being arbitrary constants to be determined .
Here,I am stuck.I could not use the fact that $Y(0)=\begin{pmatrix}
2\\
-1
\end{pmatrix}$ for finding $c_1,c_2$.

Can someone point me in the right direction with some explanation? Thanks in advance for your time.

答:
If you want to use the eigenvalue approach, you would generate eigenvectors with them.

These are shown below as column vectors (is it clear how to generate the eigenvectors). You would write:

$$y(x) = c_1 e^x \begin{pmatrix}1 \\ 0 \end{pmatrix} + c_2 e^{-x}\begin{pmatrix}-1 \\ 1 \end{pmatrix}$$

From this, we have two solutions, $y_1(x)$ and $y_2(x)$.

Now, you have two equations, two unknowns and two initial conditions to use for finding $c_1$ and $c_2$.

So, we have:

$$y_1(x) = c_1e^x - c_2e^{-x}$$

$$y_2(x) = c_2e^{-x} $$

From the given ICs, we have:

$$y_1(0) = c_1 - c_2 = 2$$

$$y_2(0) = c_2 = -1$$

So, $c_2 = -1$ and $c_1 = 1$, thus:

$$y_1(x) = e^x + e^{-x}$$

$$y_2(x) = - e^{-x} $$

Can you now answer the four questions?

Title: Problem related with solving ODE


问:
Do exists a mathematical operation that reflects a matrix along an axis?
Frequently I need to mirror a matrix along an horizontal or vertical axis, in Excel.

I ask if there is a matrix operation that does it. For example, a product with some rotation matrix.

For example:

```<code>1 0 3
0 0 0
7 0 8
</code>```

Mirrored along a vertical axis:

```<code>3 0 1
0 0 0
8 0 7
</code>```

Or along an horizontal axis

```<code>7 0 8
0 0 0
1 0 3
</code>```

I'm not particularly asking about Excel, although if it is easier to implement in Excel, or with non rectangular matrices, it would be a plus.

Since a matrix can rotate vectors, and reflections are rotations on higher dimensions (I don't know how), I hope there is an easy matrix product capable of doing it.

答: You can use the square matrix with $1$'s along the anti-diagonal. Right multiply for the vertical axis and left multiply for the horizontal axis.


答:
Yes, a matrix can be viewed as a vector in $\mathbb{R}^{n^2}$, and because you are permuting indices there is a linear map $L: \mathbb{R}^{n^2} \to \mathbb{R}^{n^2}$ which does this. As a matrix it will be $n^2 \times n^2.$ Note that if you want to swap columns and rows, then

$$\begin{pmatrix} 0 & 1 & 0 \\
1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} A$$

Will swap the first two rows of $A$, and
$$A \begin{pmatrix} 0 & 1 & 0 \\
1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} $$
will swap the first two columns of $A$ (and similar matrices for swapping the other row/columns with each other).

However, perhaps what you want is
$$\begin{pmatrix} a_1 & a_2 & a_3 \\
a_4 & a_5 & a_6 \\ a_7 & a_8 & a_9 \end{pmatrix} \mapsto \begin{pmatrix} a_2 & a_3 & a_6 \\
a_1 & a_5 & a_9 \\ a_4 & a_7 & a_8 \end{pmatrix} $$
when you say 'rotation?

Title: Do exists a mathematical operation that reflects a matrix along an axis?

Body:
Frequently I need to mirror a matrix along an horizontal or vertical axis, in Excel.

I ask if there is a matrix operation that does it. For example, a product with some rotation matrix.

For example:

```<code>1 0 3
0 0 0
7 0 8
</code>```

Mirrored along a vertical axis:

```<code>3 0 1
0 0 0
8 0 7
</code>```

Or along an horizontal axis

```<code>7 0 8
0 0 0
1 0 3
</code>```

I'm not particularly asking about Excel, although if it is easier to implement in Excel, or with non rectangular matrices, it would be a plus.

Since a matrix can rotate vectors, and reflections are rotations on higher dimensions (I don't know how), I hope there is an easy matrix product capable of doing it.


答: In general it is ambiguous, and the sense must be determined from context. In this case the fact that $X$ and $Y$ are the same kind of letter, both being upper case Latin letters, suggests that they represent similar kinds of mathematical objects, or at least ones at the same level of complexity, and hence that the intended sense is $Y\subseteq X$: if $Y$ were an element of $X$, one would be somewhat more likely to see $y\in X$. (This is by no means a hard and fast rule, however.)

Title: Definition of set containment

Body: Let $X$ and $Y$ be sets. Does "$X$ contains $Y$" mean $Y \subseteq X$ or $Y\in X$, or is it ambiguous?


答:
Multiplication on the right by a permutation matrix permutes the columns, so, for example,

```<code>a b   0 1     b a
    x      =
c d   1 0     d c
</code>```

Multiplication on the left permutes rows.

The matrix

```<code>0 0 1
0 1 0
1 0 0
</code>```

will manage the reflections you need.

Body:
Multiplication on the right by a permutation matrix permutes the columns, so, for example,

a b   0 1     b a
    x      =
c d   1 0     d c

Multiplication on the left permutes rows.

The matrix

0 0 1
0 1 0
1 0 0

will manage the reflections you need.


答: It is ambiguous. For example, I can say the real numbers contain the rationals and this would mean $\mathbb{Q} \subseteq \mathbb{R}$. I can also say the real numbers contains $3$ and this would mean $3 \in \mathbb{R}$.

Title: Definition of set containment


问:
Applied Algebra Proof
No answers please, hints only.

I must prove, by induction, that $$(1-a_1)(1-a_2)\cdots >(1-a_n) > 1-a_1-a_2-\cdots-a_n$$ for real numbers $0<a_i<1$ and integers $n\geq 2$.

My attempt:

Proof. Induction
Basis case: \begin{align} (1-a_1)(1-a_2)&>1-a_1-a_2  \\ 1-a_2-a_1+a_1a_2 &> 1-a_1-a_2 \\ a_1a_2 &> \text{ since } 0<a_i<10 \end{align}
Induction: Assume $P(n)$ holds. Then $$(1-a_1)\cdots(1-a_n)(1-a_{n+1})>1-a_1-a_2-\cdots - a_n-a_{n+1}$$

I'm unable to find a way to use $P(n)$ to show that $P(n+1)$ is true. At this point I am stuck.

Title: Applied Algebra Proof

Body: Hint: $$(1-a_1) \cdots (1-a_n)(1-a_{n+1}) > (1-a_1- \cdots - a_n)(1-a_{n+1})$$ We used $P(n)$ to claim $(1-a_1) \cdots (1-a_n) > 1-a_1- \cdots - a_n$.


答:
You know that $1-a_{n+1}>0$, so the induction hypothesis tells you that

$$(1-a_1)\ldots(1-a_{n+1})>(1-a_1-\ldots-a_n)(1-a_{n+1})\,,$$

and all that remains is to show that

$$(1-a_1-\ldots-a_n)(1-a_{n+1})\ge 1-a_1-\ldots-a_n-a_{n+1}\,.$$

Try multiplying out the lefthand side of that last inequality.

Title: Applied Algebra Proof

Body:
No answers please, hints only.

I must prove, by induction, that $$(1-a_1)(1-a_2)\cdots >(1-a_n) > 1-a_1-a_2-\cdots-a_n$$ for real numbers $0<a_i<1$ and integers $n\geq 2$.

My attempt:

Proof. Induction
Basis case: \begin{align} (1-a_1)(1-a_2)&>1-a_1-a_2  \\ 1-a_2-a_1+a_1a_2 &> 1-a_1-a_2 \\ a_1a_2 &> \text{ since } 0<a_i<10 \end{align}
Induction: Assume $P(n)$ holds. Then $$(1-a_1)\cdots(1-a_n)(1-a_{n+1})>1-a_1-a_2-\cdots - a_n-a_{n+1}$$

I'm unable to find a way to use $P(n)$ to show that $P(n+1)$ is true. At this point I am stuck.


问:
Question of rank of a matrix $4\times 6$
Let $a,b,c,d$ be complex numbers. Find the rank of the following matrix
\begin{align*}
    \begin{pmatrix}
    2& -3& 6& 0& -6& a\\
    -1& 2& -4& 1& 8& b\\
    1& 0& 0& 1& 6& c\\
    1& -1& 2& 0& -1& d\\
    \end{pmatrix}
\end{align*}
Idea: Proceed the elementary row operations and reduce to the raw Echelon form then the number of nonzero rows are the rank of the matrix.

Solution
\begin{align*}
&\begin{pmatrix}
    2& -3& 6& 0& -6& a\\
    -1& 2& -4& 1& 8& b\\
    1& 0& 0& 1& 6& c\\
    1& -1& 2& 0& -1& d\\
    \end{pmatrix}\overset{r_1  \xrightarrow{} r_4} {}
    \begin{pmatrix}
    1& -1& 2& 0& -1& d\\
    -1& 2& -4& 1& 8& b\\
    1& 0& 0& 1& 6& c\\
    2& -3& 6& 0& -6& a\\
    \end{pmatrix} \underset{\overset{r_1(-2) + r_4 \xrightarrow{}r_4{}}{}}{\overset{r_1 + r_2 \xrightarrow{}r_2,\, (-1)r_1 + r_3 \xrightarrow{}r_3}{}}\\
    &\begin{pmatrix}
    1& -1& 2& 0& -1& d\\
    0& 1& -2& 1& 7& b+d\\
    0& 1& -2& 1& 7& c-d\\
    0& -1& 2& 0& -4& a-2d\\
    \end{pmatrix}  \underset{\overset{r_2 + r_4 \xrightarrow{}r_4{}}{}}{\overset{r_2(-1) + r_3 \xrightarrow{}r_3}{}}\begin{pmatrix}
    1& -1& 2& 0& -1& d\\
    0& 1& -2& 1& 7& b+d\\
    0& 0& 0& 0& 0& c-b-2d\\
    0& 0& 0& 1& 3& a+b-d\\
    \end{pmatrix}\overset{r_3  \xrightarrow{}\, r_4} {}\\
    & \begin{pmatrix}
    1& -1& 2& 0& -1& d\\
    0& 1& -2& 1& 7& b+d\\
    0& 0& 0& 1& 3& a+b-d\\
    0& 0& 0& 0& 0& c-b-2d\\
    \end{pmatrix}
\end{align*}

Answer: If $c-b-2d = 0$ the rank of the matrix is 3, If $c-b-2d \neq 0$ the rank of the matrix is 4

This is my answer, am I correct here? I feel it is not enough for an answer. Is there anything I have to do? Any advice would be appreciated.

答:
Guide:

Your idea is fine.

The question is what is the rank of the matrix, the answer should be a number, of which you have not provided.

Consider two cases:

$c-b-2d=0$.

$c-b-2d \ne 0$.


问:
intuition for the closed form of the fibonacci sequence
I'm trying to picture this closed form from Wikipedia visually:

The idea is, if you take $\phi^n / \sqrt{5}$ and round it to the nearest integer, you'll get the $n$th Fibonacci number. I see how it works out on paper, but is there an intuitive way to understand this? Specifically I'm trying to wrap my brain around how $\phi^n$ would lead to the correct answer... if I want the 8th Fibonacci number, for example, how does it help to multiply $\phi$ by itself 8 times?

答:
The fact that $F_n$ is the integer nearest to $\dfrac{\varphi^n}{\sqrt5}$ follows from the closed form for the Fibonacci numbers known as the Binet formula:

$$F_n=\frac{\varphi^n-\widehat\varphi^n}{\sqrt5}=\frac{\varphi^n}{\sqrt5}-\frac{\widehat\varphi^n}{\sqrt5}\;,$$

where $\widehat\varphi=\dfrac{1-\sqrt5}2$. Note that $\widehat\varphi\approx-0.618$, so $|\widehat\varphi|<1$, and $|\widehat\varphi^n|$ decreases rapidly as $n$ increases. It turns out that even for small $n$ the correction $\dfrac{\widehat\varphi^n}{\sqrt5}$ is small enough so that $F_n$ is the integer nearest to $\dfrac{\varphi^n}{\sqrt5}$.

The $\sqrt5$ in the Binet formula ultimately comes from the initial conditions $F_0=0$ and $F_1=1$; a sequence with the same recurrence $x_n=x_{n-1}+x_{n-2}$ but different initial values would still grow approximately proportionally to $\varphi^n$ (or in exceptional cases $\widehat\varphi^n$), but the coefficient of approximate proportionality would be different.

Added: Specifically, each such sequence has a closed form $x_n=\alpha\varphi^n+\beta\widehat\varphi^n$, where $\alpha$ and $\beta$ are chosen to give the correct values when $n=0$ and $n=1$. Suppose that $x_0=a$ and $x_1=b$. Then from $n=0$ we must have $\alpha+\beta=a$, and from $n=1$ we must have $\alpha\varphi+\beta\widehat\varphi=b$. This pair of linear equations can then be solved for $\alpha$ and $\beta$, and provided that $\alpha\ne0$, we’ll have $x_n\approx\alpha\varphi^n$ for sufficiently large $n$. (Just how large sufficiently large actually is will depend on $\alpha$ and $\beta$.)


问:
Showing that average velocity equals change in position over change in time
It is frequently stated that average velocity equals the displacement divided by the time increment over which this displacement occurred. I have two questions:

How is this quantity the average? How would one go about showing that the displacement divided by the time increment over which this displacement occurred is the same as summing the constituent velocities and dividing by the number of different velocities (i.e. just taking the arithmetic mean)?

If I say velocity changes over time according to $v(t)=t^2$ and position changes over time according to $x(t)=t^3/3$, and I want the average velocity over the time interval $[0,2]$, using the average velocity formula I would think average velocity = $((2^3/3)-0)/2-0 = 4/3$. Then I go to the velocity function and see that the velocity starts at $0$ at $t=0$ and is $4$ at $t=2$. How does it make sense that the average velocity here is $4/3$?

答:
If we consider the change in position as $\Delta x$, then it is true that the average velocity would be your total distance traveled $\Delta x$ over time $t$ just by definitions. However I see a calculus tag so I am going to assume you are familiar with integrals. Recall that average value of a function $f$ over an interval $[a,b]$ is given by $$f_{avg} = \frac{1}{b-a}\int\limits_a^b f(x)dx.$$ Now when you want to find the velocity given a distance function $s(t)$, we take the derivative and thus $v(t)=s'(t)$. Moreover, the average velocity over the time interval $[a,b]$ is as follows:
$$V_{avg}=\frac{1}{b-a}\int\limits_a^b v(x)dx\\
=\frac{1}{b-a}[s(b)-s(a)].$$

I hope this helps and encourage you to work out any details you are missing.


问:
Two player game with marbles using optimal strategies
Consider the two-player game that consists of a single bowl of $n$ marbles. The players alternate turns. During each turn, a player can remove $2^k$ marbles, for any $k \ge 0$ of his or her choice. The winner is the person who removes the last marble. Determine all $n \ge 0$ for which the first player wins, supposing both players use optimal strategies.

Here is what I have done so far: n=1 is a winner. n=5 is a winner.

This is a problem I am doing for fun, it would not harm me to show me the answer and work

答:
You can try and think about this problem inductively. Note that every single position is either a winning position or a losing position assuming both players play the optimal strategy. Furthermore, a position is winning if and only if you can put your opponent in a losing position by taking out $2^k$ marbles for some $k \geq 0$. Using this idea:

1 is winning because player one can simply take all the marbles

2 is winning for the same reason

3 is losing because player one can only put player two in a winning position

4 is winning

5 is winning because player one can take 2 marbles and put player two in a losing position (3)

Now we can continue this process for each $n$: $n$ is winning if and only if $n$ is a power of 2 or $n$ is $2^k$ more than one of the previous losing positions for some $k \geq 0$.

(Though this idea is assuming every position is either winning or losing, which probably requires more thought.)


问:
Probability that the product of two numbers in $\{0,1,...,10^{n - 1}\}$ contains $2n - k$ digits, $k = 0,1,...$
How to find the probability that the product of two numbers in ${0,1,...,10^n - 1}$ contains $2n-k$ digits, $k = 0,1,...$? As I understand, the sum of all the combinations of two numbers is equal to $10^{2n}$. But I can't understand how to find the sum of all the positive combinations. I understand that the product of these numbers is in $10^{2n-k-1} \le x*y < 10^{2n-k}$. But it doesn't help me. Explain me please, how I can find the sum of all the positive combinations.

答:
This will give an approximate answer, which will be very close.  The region of interest is the square $[0,10^n-1] \times [0,10^n-1]$ in the $xy$ plane.  The numbers with $2n-k$ digits are the numbers from $10^{2n-k-1}$ to $10^{2n-k}-1$.  Instead of counting lattice points, we can view $x$ and $y$ as continuous variables and find the area that corresponds to this range of products.  This is where the approximation comes from.  It is the area between the two hyperbolas $xy=10^{2n-k-1}$ and $10^{2n-k}$.  Do the integral and you are done.

Added:  let us do an example with $n=7, k=4$ so we are looking for factors less than or equal to $10^6$ that multiply to an eight digit number.  Such numbers are in the range $[10^7,10^8)$  Our square is $10^6 \times 10^6$ for a total area of $10^{12}$.  The lower boundary of the area we are looking at is $xy=10^7$ and the upper boundary is $xy=10^8$.  The lower hyperbola hits the edges of the square at $(10,10^6)$ and $(10^6,10)$.  The area below it and inside the square is $$10 \cdot 10^6+\int_{10}^{10^6}\frac {10^7}x\ dx=10^7+10^7 \log(10^5) \approx 1.2513\cdot 10^8$$
where the first term is the rectangle to the left of the intersection with the top of the square.  Similarly the upper hyperbola hits the square at $(100,10^6)$ and $(10^6,100)$ so the area below it is
$$100 \cdot 10^6+\int_{100}^{10^6}\frac {10^8}x\ dx=10^8+10^8 \log(10^4) \approx 10.2103\cdot 10^8$$
The difference of areas is then $8.959 \cdot 10^8$  That is approximately the number of pairs that have a product with eight digits.


答:
You’ve discovered that the multiples of $3$ appear to be the losing positions, the other non-negative integers being the winning positions. There is a standard way to prove this kind of result (when it’s true!).

Let $W$ be the set of winning positions, and let $L$ be the set of losing positions. As you may have realized in analyzing the smaller positions, a winning position is one from which you can put your opponent in a losing position, and a losing position is one from which every legal move leaves your opponent in a winning position. In the case of this game that means:

if $n\in W$, then there is some $k\ge 0$ such that $n-2^k\in L$, and

if $n\in L$, then for every $k\ge 0$ either $2^k>n$, or $n-2^k\in W$.

In fact, the cases

$$n-2^k\in L\text{ for some }k\ge 0$$

and

$$\text{for each }k\ge 0\text{ either }2^k>n,\text{ or }n-2^k\in W$$

are mutually exclusive, so we can go further:

$n\in W$ if and only if there is some $k\ge 0$ such that $n-2^k\in L$, and

$n\in L$ if and only if for every $k\ge 0$ either $2^k>n$, or $n-2^k\in W$.

You’ve conjectured that $L=\{3n:n\ge 0\}$, and $W=\Bbb N\setminus L$. You can prove this by showing that two bulleted statements above are true for this choice of $W$ and $L$.

Title: Two player game with marbles using optimal strategies

Body:
Consider the two-player game that consists of a single bowl of $n$ marbles. The players alternate turns. During each turn, a player can remove $2^k$ marbles, for any $k \ge 0$ of his or her choice. The winner is the person who removes the last marble. Determine all $n \ge 0$ for which the first player wins, supposing both players use optimal strategies.

Here is what I have done so far: n=1 is a winner. n=5 is a winner.

This is a problem I am doing for fun, it would not harm me to show me the answer and work


问:
Is my reasoning sound for proving that the intersection over a family of sets is empty?
So as a homework problem I was asked to find the intersection of the following family of sets. I was told that my reasoning was invalid as I did not correctly prove that the intersection is the empty set. Is the fact that $\displaystyle \frac{1}{n} \lt \frac{1}{n+1}$ not sufficient to show that there are no common elements over the entire family of sets and thus the intersection contains no elements?

1. Find the union and intersection of each of the following families.

(g) $\displaystyle \mathscr{A} =\{A_{n} :n\in \mathbb{N}\}$, where $\displaystyle A_{n} =\left( 0,\frac{1}{n}\right)$ for each natural number $\displaystyle n$.

Solution.

Since $\displaystyle n\in \mathbb{N}$, and $\displaystyle 0$ is not in any set because it is an open interval, it follows that the quantity $\displaystyle \frac{1}{n} \leq 1$. Therefore, the union of the family of sets $\displaystyle \mathscr{A}$ is $\displaystyle \bigcup _{n\in \mathbb{N}} A_{n} =( 0,1)$.

Since for each set $\displaystyle A_{n}$ the quantity $\displaystyle \frac{1}{n} \lt \frac{1}{n+1}$, and $\displaystyle 0$ is not in any set because it is an open interval, it follows that the intersection of the family of sets is $\displaystyle \bigcap _{n\in \mathbb{N}} A_{n} =\emptyset $, since no two sets have a common element.

答:
The point you're missing is that for any $x > 0$ there is some $n$ such that $\dfrac{1}{n} \le x$.  The mere fact that $\dfrac{1}{n} > \dfrac{1}{n+1}$ (btw, you wrote $<$, but I guess that was just a typo) is not sufficient.  For example,
you could also say $1 + \dfrac{1}{n} > 1 + \dfrac{1}{n+1}$, but it's not true
that $\displaystyle \bigcap_{n \in N} \left(0, 1 + \frac{1}{n}\right) = \emptyset$.


答:
For starters, $\frac1n$ is larger than $\frac1{n+1}$: you have the inequality backwards. But the argument is inadequate even after you correct the inequality. Consider instead the family of intervals $\mathscr{C}=\{C_n:n\in\Bbb Z^+\}$, where

$$C_n=\left(0,\frac1{10}+\frac1n\right)\,.$$

If your argument were valid, it would apply equally well to $\mathscr{C}$, since $$\frac1{10}+\frac1{n+1}<\frac1{10}+\frac1n$$ for each $n\in\Bbb Z^+$, yet

$$\bigcap\mathscr{C}=\bigcap_{n\in\Bbb Z^+}\left(0,\frac1{10}+\frac1n\right)=\left(0,\frac1{10}\right]\ne\varnothing\,.$$

Title: Is my reasoning sound for proving that the intersection over a family of sets is empty?

Body:
So as a homework problem I was asked to find the intersection of the following family of sets. I was told that my reasoning was invalid as I did not correctly prove that the intersection is the empty set. Is the fact that $\displaystyle \frac{1}{n} \lt \frac{1}{n+1}$ not sufficient to show that there are no common elements over the entire family of sets and thus the intersection contains no elements?

1. Find the union and intersection of each of the following families.

(g) $\displaystyle \mathscr{A} =\{A_{n} :n\in \mathbb{N}\}$, where $\displaystyle A_{n} =\left( 0,\frac{1}{n}\right)$ for each natural number $\displaystyle n$.

Solution.

Since $\displaystyle n\in \mathbb{N}$, and $\displaystyle 0$ is not in any set because it is an open interval, it follows that the quantity $\displaystyle \frac{1}{n} \leq 1$. Therefore, the union of the family of sets $\displaystyle \mathscr{A}$ is $\displaystyle \bigcup _{n\in \mathbb{N}} A_{n} =( 0,1)$.

Since for each set $\displaystyle A_{n}$ the quantity $\displaystyle \frac{1}{n} \lt \frac{1}{n+1}$, and $\displaystyle 0$ is not in any set because it is an open interval, it follows that the intersection of the family of sets is $\displaystyle \bigcap _{n\in \mathbb{N}} A_{n} =\emptyset $, since no two sets have a common element.


答:
Your argument is not valid because it could just as easily be applied to the family of sets $A_n = (0, 1 + 1/n)$.

Here is a correct argument:

$\bigcap\limits_{n \in \mathbb{N}} A_n$ is empty for the following reason: suppose we have some $x \in \bigcap\limits_{n \in \mathbb{N}} A_n$. Then $x \in A_1 = (0, 1/1)$; then $x > 0$. Therefore, $1/x$ is defined. By the Archimedean property of real numbers, there exists a natural number $n > 1/x$. In this case, we see that $1/n < x$, since the reciprocal function is decreasing on $(0, \infty)$. Because $x \in \bigcap\limits_{n \in \mathbb{N}} A_n $, it follows that $x \in A_n = (0, 1/n)$ and therefore $x < 1/n < x$. This is a contradiction.


答:
(a) Your calculation was off.  It seems that you dropped the decimal.$$\mathsf E(X)=−2\cdot 0.3+−1\cdot \underbrace{0.6}+12\cdot0.1=0.0$$

(b) A negative variance is an error warning.   You need to square each supported value when you multiply by the probability mass.$$\mathsf{Var}(X)=((−2)^2\cdot 0.3+(−1)^2\cdot 0.6+(12)^2\cdot0.1) -(0.0)^2= 16.2$$

(c) The expected sample mean is indeed the expected value of the distribution.   We just apply the Linearity of expectation, and identical distribution of the samples: $$\begin{align}\mathsf E(\bar X)&=\mathsf E\left(\dfrac{\sum_{i=1}^nX_i}{n}\right)\\[1ex]&=\dfrac{\sum_{i=1}^n\mathsf E(X_i)}{n}\\[1ex]&=\mathsf E(X_1)\\[1ex]&=\mathsf E(X)\end{align}$$

(d) For the variance of the sample mean, you must apply the Bilinearity of Covariance, and the independence and identical distribution of the samples.

$$\begin{align}\mathsf {Var}(\overline X)&=\mathsf{Var}\left(\frac{\sum_{i=1}^nX_i}{n}\right)\\[1ex]&=\frac 1{n^2}\mathsf {Cov}(\sum_{i=1}^n X_i,\sum_{j=1}^n X_j)\\[1ex]&~~\vdots\end{align}$$

Title: Help with probability mass function, specifically understanding sample mean

Body:
The question is

Let $X$ be a discrete random variable with probability mass function

(a) Find $E(X)$
I did $E(X) = -2*.3+ -1*6 + 12*.1 = -5.4$

(b) Find $Var(X)$
 I did $E(x^{2})-(E(x))^{2}= (-2^{2} * .3 + -1*.6 + 12*.1) - (-5.4)^{2} = -29.76$

(c) Find expected value of X(bar) (the sample mean)
 Would this just be the same thing as $E(X)$?
(d) if n = 100, what is the variance of X(bar)?
 would I just multiply the variance by 100?

Last question is if I wanted to put this all in R, how would I input my CDF into R and find the expected value and variance, etc. I understand it on paper (most of the time) but I am unsure how I would put it in R, any help is appreciated, thank you in advance.


问:
Tent & Ziegler Model Theory Exercise 6.2.2 about Morley rank and Morley degree
I am trying to solve Exercise 6.2.2 in Tent & Ziegler's A Course in Model Theory. The exercise is:

Let $\varphi$ be a formula of Morley rank $\alpha<\infty$ and $\psi_0,\psi_1, \psi_2, ...$ an infinite sequence of formulas. Assume that there is a number $k$ such that the conjunction of any $k$ of the $\psi_i$ has Morley rank smaller than $\alpha$. Then $\text{MR}(\varphi\wedge\psi_i)<\alpha$ for almost all $i$.

There is a hint at the end of the book, which says:

Let $I$ be the set of all $i$ such that $\text{MR}(\varphi\wedge\psi_i)=\alpha$. The hypothesis implies that all $k$-element subsets of $I$ contain two indices $i,j$ such that $\varphi\wedge\psi_i\not\sim_\alpha\varphi\wedge\psi_j$. So $|I|\leq (k-1)MD(\varphi)$.

I am lost on how the hypothesis implies the statement in the hint and also how that statement implies $|I|\leq (k-1)MD(\varphi)$. It seems like I have been going in circles.

My attempt:

Let $i_1,...,i_k$ be $k$ elements from $I$. Assume for contradiction that $\varphi\wedge\psi_{i_1}\sim_\alpha...\sim_\alpha\varphi\wedge\psi_{i_k}$. By definition, $\text{MR}(\varphi\wedge \psi_{i_m}\triangle \varphi\wedge \psi_{i_n})<\alpha$ for any $m,n\leq k$. Because $\wedge$ distributes over $\triangle$, we have $\text{MR}(\varphi\wedge (\psi_{i_m}\triangle\psi_{i_n}))<\alpha$. At this point, write $\varphi$ as the disjunction witnessing its Morley degree, $\varphi\equiv \varphi_1\vee...\vee\varphi_d$ so $\text{MR}(\varphi\wedge (\psi_{i_m}\triangle\psi_{i_n}))=\text{MR}(\varphi_l\wedge (\psi_{i_m}\triangle\psi_{i_n}))$ for some $\varphi_l$ that is $\alpha$ strongly minimal. Then I tried writing out $(\psi_{i_m}\triangle\psi_{i_n})$, but I think I am moving further and further away from being able to apply the assumption.

答:
Let $d = \mathrm{MD}(\varphi)$, and let $p_1,\dots,p_d$ enumerate the complete types of Morley rank $\alpha$ containing $\varphi$.

For each $1\leq i\leq d$, $p_i$ contains at most $(k-1)$ of the formulas $\psi_n$. Indeed, if $p_i$ contains $\psi_{n_1},\dots,\psi_{n_k}$, then $\bigwedge_{j=1}^k \psi_{n_j}\in p_i$, contradicting the hypothesis that $\mathrm{MR}(\bigwedge_{j=1}^k \psi_{n_j})<\alpha$.

It follows that only finitely many (at most $d(k-1)$) of the formulas $\psi_n$ are contained in any of the types $p_i$ for $1\leq i \leq d$. And if $\psi_n$ is a formula which is not contained in any of the types $p_i$, then $\varphi\land \psi_n$ is not contained in any type of Morley rank $\geq \alpha$, so $\mathrm{MR}(\varphi\land \psi_n)<\alpha$.


问:
How to solve a polar equation when $r$ is $r^2$ instead?
I have $r^2=-4\sinθ$

and I'm asked to set $r=0$, then find θ. If I just set $r^2=0$ then I'll get $\sin(2θ)=0$. That doesn't seem right.

Then I'm asked to set $θ=0$ and then find $r$. If I use the $r^2=-4\sinθ$ and set $θ=0$ then I will get "DNE". Not sure what to do instead then...

EDIT:

Sorry everyone I wrote the problem here wrong. It was supposed to be r^2=-4sin2θ

That's where the 2θ came from

答:
1) It is right.

2) You do not get DNE, you get 0... (Sqrt[-4*0] = 0)

It comes out as a very nice figure 8 loop.
You seem to have understood the question, but are lacking some basic algebra/trig that the problem was not designed to test.


问:
Alternating Sequence Limit

Suppose $a_{n} a_{n+1}<0$ for all $n \in \mathbb{N},$ and suppose $\lim a_{n}=L$. What is $L ?$ Prove your claim.

I've encountered this weird question with limits. I feel like the answer is L=0, but I don't know how to show this.

答: If $a_n a_{n + 1} < 0$ for all $n$, and if we have $\lim\limits_{n \to \infty} a_n = L$, then we see we have $\lim\limits_{n \to \infty} a_n a_{n + 1} = \lim\limits_{n \to \infty} a_{n} \lim\limits_{n \to \infty} a_{n + 1} = L^2 \leq 0$. Then $L = 0$.

Title: Alternating Sequence Limit


问:
Solving Equations Regarding Modular
146x = 12 (mod 421)

I found out that 421 is a prime number and I still did not know how to start after going over the notes

I feel like it is similar to solve 7x = 1 (mod 180) if I know how to solve one of them, any tips? Thank you.

答:
Hints:

Do you know how to use the
Euclidean Algorithm
to find a positive integer $r$ such that $(146)r \equiv 1\pmod{421}?$
If so, given that $(146)x \equiv 12 \pmod{421},$ what could you say about
$(146)(r)(x)$, which must be congruent to $(12)r \pmod{421}?$

Title: Solving Equations Regarding Modular


问:
What is the correct notation for writing events in terms of other events?
In my probability and statistics class, I am confused about the notation for describing events. I'd like to use the following problem as an example:

I know what the answers are but am confused about the notation I need to use to describe them. For example, part a is $A_{ab} = [(A_1, A_2, A_3)]$ while part b is $A_{ab} = [(A_1, A_2, A_3), (A1, A2), (A1, A3)]$

答:
Events are sets of outcomes, so we therefore use set notation to describe their combinations.

So, the event that the three switches are all closed (ie they are in series) would be the intersection: $(A_1\cap A_2\cap A_3)$ , while the event that at least one among the three is closed (ie they are in parallel) would be the union: $(A_1\cup A_2\cup A_3)$.


问:
Question on usage of Löwenheim–Skolem in Beginner Logic
I'm taking a beginner course in mathematical logic, and I was struggling to understand some applications of the downward Löwenheim–Skolem theorem. The version of Löwenheim–Skolem theorem that was presented in the lectures is:

Suppose $L$ is countable first order language, and $\mathcal{A}$ is an $L$-structure. Let $T$ be the theory of $\mathcal{A}$, i.e. $T = Th(\mathcal{A}) =\{ \theta: \mathcal{A} \models \theta \text{ where } \theta \text{ is an L-sentence}\}$. Then, there is a countable model $\mathcal{B}$ such that $\mathcal{B}\models T$.

$\Delta$ was defined to be the sentences which describe a dense linear order with no end points, i.e. $\Delta = \{ \varphi_1, \dots , \varphi_6\}$ where for example $\varphi_1 = \forall x_1, \exists x_2, \  x_2 < x_1$. The proof of the "Los-Vaught test" was given.

(Los-Vaught Test) For every $L$-sentence $\theta$, $\Delta \vdash \theta$ or $\Delta \vdash \lnot \theta$.

The proof starts: Suppose not. Then, $\Delta \cup \{\theta\}$ and $\Delta \cup \{ \lnot\theta\}$ are both consistent. Godel's completeness theorem tells us (the model existence lemma tells us?) that if it is consistent, then it should have a model. Then Löwenheim–Skolem theorem tells us, that we should have a countable model.

I don't understand the last sentence, namely, why the Löwenheim–Skolem theorem tells us that. How can I know that $\Delta \cup \{\theta\}$ and $\Delta \cup \{ \lnot\theta\}$ are theories of some $L$-structure?

答: If $M$ is a model of $\Delta\cup\{\theta\}$, then $\Delta\cup\{\theta\}\subseteq\mathrm{Th}(M)$. So any model of $\mathrm{Th}(M)$ is a model of $\Delta\cup\{\theta\}$.


问:
Two rings with the same multiplicative structure but non-isomorphic underlying Abelian groups
I am giving a series of lectures where I introduce some undergraduates to basic ideas from category theory. One of the things I would like to show them is how category theory could be used to make precise some of the intuitive ideas they have been exposed to (e.g., what "natural" means when we speak about the natural isomorphism of a finite dimensional vector space with its double dual, what "universal" really means when we speak of a universal construction, etc.).

One of the subtleties I would like to illustrate with explicit examples is the difference between "property" and "structure". I would really like to show them that "being a preadditive category" (by preadditive here I just mean $Ab$-enriched category, I do not assume existence of a $0$-object) is not a property, so it does not make sense to give me an abstract category and ask me if it is preadditive or not (I can ask whether it can be made into a preadditive category, but even in that case the preadditive structure may not be unique).

To illustrate this, I would like to find some minimal example of two different $Ab$-enrichments of the same abstract category. A good way to do that seems that of finding two (unitary and associative) rings $(R,+,.)$ and $(S,+',*)$, such that $(R,.)\cong (S,*)$ as monoids (i.e., as one-object categories) but $(R,+)\not\cong (S,+')$ as Abelian groups.

Do you have some easy example of two rings with the properties described above?

答: A small finite example is $R=\mathbb{Z}/4\mathbb{Z}$, $S=\mathbb{F}_2[x]/(x^2)$. The multiplicative monoids are isomorphic by $0\mapsto0$, $1\mapsto1$, $2\mapsto x$, $3\mapsto1+x$.