问:
Understanding the Dirac delta function for impulses
I am looking at a very basic impulse system of a ball bouncing on the ground at time $T$. It experiences force for a short time when it hits the ground. Using Newton's 2nd Law:

$ m \ddot{x} = F(t) - mg$

$\displaystyle \int_{T - \varepsilon}^{T + \varepsilon} \! m \frac{\mathrm{d}^2x}{\mathrm{d}t^2} \, \mathrm{d}t = \int_{T - \varepsilon}^{T + \varepsilon} \! F(t) \, \mathrm{d}t \; - \int_{T - \varepsilon} ^{T + \varepsilon} \! mg \, \mathrm{d}t$

$\bigg[ m \frac{\mathrm{d}x}{\mathrm{d}t} \bigg]_{T - \varepsilon}^{T + \varepsilon} = I - 2\varepsilon mg $

Where impulse $I$ is the area under the force curve.

I can see that this can be rewritten as something of the form

$\Delta p = I - O(\varepsilon)$

And as $\varepsilon$ goes to $0$ we get

$\Delta p = I$

However, I have seen the result that the differential equation can be rewritten using the Delta function as

$ m \ddot{x} = -mg +I \delta (t - T)$

I do not understand how this result came about using the Delta function. My question is how can we rewrite the above differential equation ($ m \ddot{x} = F(t) - mg$) using the Dirac Delta function and the impulse $I$?

答:
I think it's important to understand the full implications of what it means when you write $$ \lim_{\epsilon \to 0} \int_{T - \epsilon}^{T + \epsilon} F(t) dt = I,$$
where $I \ne 0$.

When you write this, you are basically assuming that $F(t)$ has a Dirac delta function in it.  If not, then the integral would be $O(\epsilon)$ and $I$ would be $0$.

(As an aside, you might think that you can get a nonzero result by allowing a singularity in the force at time $T$, or something like that.  This may be possible, but it would probably be physically indistinguishable from a Dirac delta.  The obvious approach, saying $F(t)=(t-T)^\alpha$ for some power $\alpha$, does not work.  $I$ will turn out to be 0, $\infty$, or ill-defined.)

So far, I have argued that $I \ne 0$ is more or less equivalent to saying
$$ F(t) = I \delta(t - T) + \dots $$
where $\dots$ could represent any other terms that do not contribute a finite impulse at time $T$.  Now try plugging this $F(t)$ into
$$ m \ddot{x} = F(t) - mg $$
You get the desired result,
$$ m \ddot{x} = -mg + I \delta(t - T) $$


问:
Range and inverse of $f(x)=\sqrt{4x^2 + 1}$ on $[0,\infty)$

Consider the function $f(x) = \sqrt{4x^2 + 1}$ for $x \in [0,\infty)$. Denote the range of $f$ by $Y$. There is a function $g$ with domain $Y$ so that $g(f(x)) = x$
for every $x \in [0, \infty)$ and  $f(g(y)) = y$ for every $y \in Y$. Find an expression for $g(y)$

Also, how can I go about finding the range of $f$?

Any pointers would be much appreciated :).

答:
Note that if $x \in [0,\infty)$ then $x \ge 0$.

So $x^2 \ge 0$.  And $4x^2 \ge 0$ and $4x^2 + 1 \ge 1$ and $\sqrt{4x^2 + 1} \ge 1$.

So $Y \subset [1,\infty)$.

The only issue is can any $y \in [1,\infty)$ be in the range of $f$?

SO here are three questions for you to think about

If $y = \sqrt{4x^2 + 1}$ and $y \ge 1$ is the any limit on how big $y$ can be?

If we can have $y$ be as small as $1$ and we can have $y$ be as large as we like (can we? see question 1) does that mean we can have $y$ be any value in between?  Is $f(x) = \sqrt{4x^2 + 1}$ continuous?

If we have any $k \in [1,\infty)$ can we always find an $x$ so that $k = \sqrt{4x^2 + 1}$?  How would we know?  Are there any $k$ *without solutions?

If after you answer all those questions, if every $k \ge 1$ will have an $x$ so that $\sqrt{4x^2 + 1} =k$ is possible, then that would mean the range of $f$ is $[1,\infty)$.

Is it?

ContentLicense: CC BY-SA 4.0


问:
How to show that (G,•) is an abelian group?

Let $(G,\cdot)$ be a group such that $\forall x,y \in G$, $x\neq y$ there exists $H_1,H_2$ subgroups of $G$,
such that $x\in H_1$, $y\in H_2$ and $H_1\cap H_2$={e} then show that $(G,\cdot)$ is an abelian group.

If I assume the contrary that $ab\neq ba$ then there exist subgroups $H_1,H_2$ where $ab$ is in $H_1$ and $ba$ is in $H_2$. But now I am confused I am going the right way. Can someone provide me some hint what I have to do?

答:
We prove at first that every non-identical element of this group is of order 2.
Assume the converse; then there exists $a\in G$ of order $>2$.
Let $x = a$ and $y = a^2\neq e$. By assumption, there exist $H_1$ and $H_2$
such that $x\in H_1$ and $y\in H_2$ and $H_1\cap H_2=\{e\}$.
On the other hand, since $y=x^2$, then $y\in H_1$ and $y\in H_1\cap H_2$.
Contradiction; statement is proved.

If the order of every element of $G$ is 2, then $G$ is an Abelian group.
Indeed:
$(xy)^2=e\Rightarrow xy=(xy)^{-1}\Rightarrow xy=y^{-1}x^{-1}=yx.$


问:
Roll 2 dice. Let A = ‘the first die is odd’, B = ‘the second die is odd’, and C = ‘the sum is odd’
I am currently working on a probability problem. I am new to probability (just now studying it). I came across this question just working on some problems.

The question asks, are A and B independent? By intuition I would say so.

A = {1,3,5},
B = {1,3,5}

(Alternatively, I think I could also represent them as pairs with size sets of 36; the probabilities would
still be 1/2).

$P(A) = 1/2$

$P(B) = 1/2 $

However, whenever I find the $P(A \cap B)$, it does not
equal $P(A)P(B)$.

I get that $$P(A \cap B)=1/2$$ $$|A \cap B|=3,|\Omega|=6$$ where $\Omega$ is the total number of outcomes.

Thus, $$P(A \cap B)=|A \cap B|/|\Omega|=3/6=1/2$$ However, $$P(A)P(B)=1/4$$
Therefore, $$P(A \cap B)=1/2 \neq P(A)P(B)=1/4$$.

This leads to a rejection of
independence.

Have I made a mistake?

答:
They are independent.  There are $3$ ways for each die to be odd.  Therefore there are $3^2 = 9$ ways for both dice in a pair to be odd.  There are 6 results for each die.  Therefore there are $6^2 = 36$ possible results for a pair of dice.  The probability for a pair of dice to both be odd is then
$$ P(A \cap B) = \frac{9}{36} = \frac{1}{4} = \frac{1}{2} \cdot \frac{1}{2} = P(A) P(B) $$

The mistake you made is when you wrote $|A \cap B| = 3$.  It is actually 9 (and the full space has size 36).


问:
An application of Weierstrass Approximation Theorem to polynomial
The question is

Let $f:[0,1]\to\mathbb{R}$ be a continuous function. Let $\varepsilon>0$, show that there is a polynomial $p:[0,1]\to\mathbb{R}$ with $\vert f(x)-p(x)\vert <\varepsilon$ for all $x\in[0,1]$ and with $p(0)=f(0),p\left(\frac{1}{2}\right)=f\left(\frac{1}{2}\right), p(1)=f(1)$

What I have in my mind is that using the Weierstrass approximation theorem, let $q:[0,1]\to \mathbb{R}$ such that
$$ \forall \varepsilon>0, \vert f(x)-q(x)\vert <\frac{\varepsilon}{4} $$
Then $\vert f(x)-p(x) \vert \leq \vert f(x)-p(x) \vert+\vert f(0)-p(0) \vert + \vert f\left(\frac{1}{2}\right)-p\left(\frac{1}{2}\right)\vert + \vert f(1)-p(1) \vert <\frac{\varepsilon}{4}+\frac{\varepsilon}{4}+\frac{\varepsilon}{4}=\varepsilon$.
However, the equality conditions are not satisfied.

For only one point in the equality conditions, it is easy to see that the above argument works, changing the denominator of $\varepsilon$ to 2 is enough. But I got stuck in this problem. Do I have to consider some fancy linear combination of the three points? Or is there any other ways to show it? Many thanks!

答:
Hint: Let $q$ be a polynomial such that $|f(x)-q(x)| <\epsilon $ for all $x$. Let $p(x)=q(x)+a+bx+cx^{2}$. We can choose $a,b,c$ so that $f(0)=p(0),f(\frac  1 2)=p(\frac  1 2)$ and $f(1)=p(1)$. Now  use the fact that $|f(x)-q(x)| <\epsilon $ for $x=0,\frac 1 2$ and $x=1$ to check that  $|a|<\epsilon, |a+\frac b 2 +\frac  c  4| <\epsilon$ and $|a+b+c| <\epsilon$. Conclude that $|a|, |b|, |c|$ are all 'small' to finish the proof.

[You should check that $|a|<\epsilon,|b| <10\epsilon $ and $|c| <12 \epsilon$. Finally you get $|f(x)-p(x)| <36 \epsilon$ for all $x$].


问:
Integral of Gaussian Distribution
I'm trying to solve the below integral and my idea is to substitute $x$ by $\sqrt[]{t}x$. And my final answer is $\sqrt[]{t}$. But my textbook shows that the result should be $t$. Could someone explain where I did wrong? Thank you very much.

$$\int_{-\infty}^{\infty}\frac{x^2}{\sqrt[]{2\pi}*t}exp(-\frac{x^2}{2t})dx$$

I think the original statement is like this: For a random variable with probability density for x at time t give that $f(x,t) = $
$$\frac{1}{\sqrt[]{2\pi}*t}exp(-\frac{x^2}{2t})$$
And it says that $E(x(t)^2) = t$. I'm not sure whether I misunderstand something.

答:
$$\begin{split}\frac{1}{\sqrt t} \int_{-\infty}^{\infty}x^2\frac{1}{\sqrt {2\pi}\sqrt t}\exp{-\frac{x^2}{2t}}dx&=\frac{1}{\sqrt t} \int_{-\infty}^{\infty}x^2f_{N(0,t)}(x)dx\\
&=\frac{1}{\sqrt t}E(X^2)\end{split}$$

with $X\sim N(0, t)$

$E(X^2)=var(X)+[E(X)]^2=t+0=t$

Thus

$$\frac{1}{\sqrt t}(t)=\sqrt t$$

Title: Integral of Gaussian Distribution


问:
Proving $\sin^2x<\sin x$ for $x\in(0,\pi)$
Actually this is a calculus problem. I'm required to compare a given integral with a simpler one and check if the inequality holds 2. (Question b)

This is a part of an answer in my textbook:

$$0 < \sin x < 1 \;\;\text{on}\;\; (0,\pi) \quad\implies \sin^2x<\sin x \;\;\text{on}\;\; (0,\pi)$$
Then it says, Therefore:$$\int ^{\pi }_{0}\sin^{2} xdx< \int ^{\pi }_{0}\sin xdx$$

But how can prove this inequality? $$\sin^{2} x< \sin x\ (x\in \ ( 0,\ \pi ))$$ if $$0 < \sin x < 1 \;\;\text{on}\;\; (0,\pi)$$

Edit: Thank you for warning me about the MSE format. I'll be more careful in the future.

答: Hint: $\sin x$ is a positive number on this interval.  Try dividing both sides by $\sin x$.


问:
Gateux differential question
I am going over the book "Optimization by Vector Space Methods" by Luenberger, and I found a statement that I'm not sure I understand. I am attaching the relevant page and I drew a box around the question in hand. I am not sure why the Gateux differential equals $\frac{d f(x+\alpha h)}{d \alpha}$ at $\alpha=0$ in the case the functional is real.

Thanks in advance

答:
By definition,
$$
\delta f(x,h)=\lim\limits_{\alpha\to 0}\frac {f(x+\alpha h)-f(x)}{\alpha}=\lim\limits_{\alpha\to 0}\frac {f(x+\alpha h)-f(x+0h)}{\alpha}
$$
which is by definition the derivative of the function $\alpha\to f(x+\alpha h)$ at $\alpha=0$.

Title: Gateux differential question


Title: Roll 2 dice. Let A = ‘the first die is odd’, B = ‘the second die is odd’, and C = ‘the sum is odd’

Body:
I am currently working on a probability problem. I am new to probability (just now studying it). I came across this question just working on some problems.

The question asks, are A and B independent? By intuition I would say so.

A = {1,3,5},
B = {1,3,5}

(Alternatively, I think I could also represent them as pairs with size sets of 36; the probabilities would
still be 1/2).

$P(A) = 1/2$

$P(B) = 1/2 $

However, whenever I find the $P(A \cap B)$, it does not
equal $P(A)P(B)$.

I get that $$P(A \cap B)=1/2$$ $$|A \cap B|=3,|\Omega|=6$$ where $\Omega$ is the total number of outcomes.

Thus, $$P(A \cap B)=|A \cap B|/|\Omega|=3/6=1/2$$ However, $$P(A)P(B)=1/4$$
Therefore, $$P(A \cap B)=1/2 \neq P(A)P(B)=1/4$$.

This leads to a rejection of
independence.

Have I made a mistake?

Body:
You wrote that $|A \cap B| = 3, |\Omega| = 6$. This is not correct.

$A = \{1,3,5\}, B = \{1,3,5\}$

For throw of two dice, $|\Omega| = 36$

$A \cap B$ is the event of both dice turning up odd. That is,

$A \cap B = \{1,1\}, \{1,3\}, \{1,5\}, \{3,1\}, \{3,3\}, \{3,5\}, \{5,1\}, \{5,3\}, \{5,5\}$

$|A \cap B| = 9$

$\displaystyle |\frac{A \cap B|}{|\Omega|} = \frac{9}{36} = \frac{1}{4}$


问:
Calculate the limit: $\lim_{n\to+\infty}\sum_{k=1}^n\frac{\sin{\frac{k\pi}n}}{n}$ Using definite integral between the interval $[0,1]$.
Calculate the limit: $\lim_{n\to+\infty}\sum_{k=1}^n\frac{\sin{\frac{k\pi}n}}{n}$ Using definite integral between the interval $[0,1]$.

It seems to me like a Riemann integral definition:

$\sum_{k=1}^n\frac{\sin{\frac{k\pi}{n}}}{n}=\frac1n(\sin{\frac{\pi}n}+...+\sin\pi)$

So $\Delta x=\frac 1n$ and $f(x)=\sin(\frac{\pi x}n)$ (Not sure about $f(x))$

How do i proceed from this point?

答:
Remember that if $f: [a,b] \to \mathbb{R}$ is some function we can define the Riemann Sum of $f$ in the interval by choosing a partition $P = \{t_0, \dots, t_n\}$ of $[a,b]$ in a very specific manner: we simply divide the interval $[a,b]$ equally calculating it's length $b-a$ and then dividing by the number of subintervals that we want. So a partition with $n$ subintervals will have $t_i = a + i\Delta t$ where $\Delta t = (b-a)/n$ is the size of each interval. In that case the integral of $f$ on the interval $[a,b]$ will be the limit:

$$\int_a^b f(t)dt=\lim_{n\to \infty} \sum_{i=1}^n f(t_i)\Delta t$$

Look that in your case we are working on the interval $[0,1]$ so for a partition of $n$ subintervals we'll have $\Delta t = 1/n$. Now we want to compute the limit of the following sum:

$$\sum_{i=1}^n \frac{1}{n}\sin{\left(\frac{i\pi}{n}\right)}$$

Note that with this we have:

$$\sum_{i=1}^n \Delta t\sin{\left(i\Delta t \pi\right)}$$

However, since the interval starts on $0$ and since $t_i = a + i\Delta t$, we have that $t_i = i\Delta t$, so we have:

$$\sum_{i=1}^n \sin{\left(t_i \pi\right)}\Delta t$$

Now on the limit this will become obviously the following integral:

$$\lim_{n\to \infty} \sum_{i=1}^n \sin{\left(t_i \pi\right)}\Delta t=\int_0^1 \sin(\pi t)dt$$

This integral is computed trivially by setting $u=\pi t$, $dt=du/\pi$ and so $u$ ranges from $0$ to $\pi$ and we have:

$$\lim_{n\to \infty} \sum_{i=1}^n \sin{\left(t_i \pi\right)}\Delta t=\frac{1}{\pi}\int_0^\pi \sin(u)du=\frac{2}{\pi}$$

ContentLicense: CC BY-SA 3.0


问:
Derivative of {f(x)}^2
$h(x)= f^2(x)$

Find $h'(x)$.

I initially thought that $h'(x)=2f(x)$ but it is wrong.
When I googled it, I found it to be $h'(x) = 0$ but I'm not sure if that's true.

答: If $u(x) = x^2$, then $h(x) = u(f(x))$, and by the chain rule $h'(x) = u'(f(x)) f'(x) = 2 f(x) f'(x)$.

Title: Derivative of {f(x)}^2


答:
Consider the 1D Newton's equation
\begin{align}
m\ddot x = F(t).
\end{align}
This says that the force experienced by the particle varies as a function of time (in general, the force could also depend on the position $x$).

Suppose $F = 0$, then the particle is experiencing no external forces. Hence, it's either traveling at a constant velocity or it's not moving.

Now, suppose you apply a constant force on the particle over a time interval $[T, T+\Delta t]$, say
\begin{align}
F(t) =
\begin{cases}
C & \text{ it } T\le t \le T+\Delta t\\
0 & \text{ otherwise}
\end{cases}
\end{align}
i.e. you are pushing the particle with constant force from time $T$ to $T+\Delta t$. Then physics says that the particle experiences an impulse of size $I=C\Delta t$.

But you could have applied the following force
\begin{align}
F(t) =
\begin{cases}
2C & \text{ it } T\le t \le T+\frac{1}{2}\Delta t\\
0 & \text{ otherwise}
\end{cases}
\end{align}
which is twice the magnitude but half the time of the original force. The impulse experienced by the particle is the same. What this is modeling is the fact that you are pushing harder but for less time.

In theory, you could increase the magnitude of the force that you are applying but maintain the same impulse by applying force for a shorter period of time. Furthermore, if you consider the limit of this procedure, i.e. you smack the particle at time $T$ with magnitude $I=C\Delta t$. This is modeled by the delta function, i.e.
\begin{align}
m\ddot x = I\delta(t-T).
\end{align}

Title: Understanding the Dirac delta function for impulses

Body:
I am looking at a very basic impulse system of a ball bouncing on the ground at time $T$. It experiences force for a short time when it hits the ground. Using Newton's 2nd Law:

$ m \ddot{x} = F(t) - mg$

$\displaystyle \int_{T - \varepsilon}^{T + \varepsilon} \! m \frac{\mathrm{d}^2x}{\mathrm{d}t^2} \, \mathrm{d}t = \int_{T - \varepsilon}^{T + \varepsilon} \! F(t) \, \mathrm{d}t \; - \int_{T - \varepsilon} ^{T + \varepsilon} \! mg \, \mathrm{d}t$

$\bigg[ m \frac{\mathrm{d}x}{\mathrm{d}t} \bigg]_{T - \varepsilon}^{T + \varepsilon} = I - 2\varepsilon mg $

Where impulse $I$ is the area under the force curve.

I can see that this can be rewritten as something of the form

$\Delta p = I - O(\varepsilon)$

And as $\varepsilon$ goes to $0$ we get

$\Delta p = I$

However, I have seen the result that the differential equation can be rewritten using the Delta function as

$ m \ddot{x} = -mg +I \delta (t - T)$

I do not understand how this result came about using the Delta function. My question is how can we rewrite the above differential equation ($ m \ddot{x} = F(t) - mg$) using the Dirac Delta function and the impulse $I$?


答:
Well, if $$h(x)=\left(f(x)\right)^2$$ then using the chain rule we get $$h'(x)=2f(x)f'(x)$$

So, I'm not sure how you're getting $h'(x)$ to be $0$, the derivative is $0$ only when the function is a constant so $h'(x)$ being $0$ means that $h(x)=c$ where c is some constant.

Now if that's the case $f(x)$ would be the square root of $c$ so $f(x)=\sqrt{c}$ and this would make $h'(x)=0$

Title: Derivative of {f(x)}^2

Body:
$h(x)= f^2(x)$

Find $h'(x)$.

I initially thought that $h'(x)=2f(x)$ but it is wrong.
When I googled it, I found it to be $h'(x) = 0$ but I'm not sure if that's true.


问:
comparing a regression model to actual data
I built a regression model estimating y(x),

I found that $α=8.2$ and $β=7.5$ and am trying to compare my results to the actual $y(x)$ that is given in the data to drow a conclusion regarding the error in my regression model.

I tried building a graph with a trendline that compares the two but I'm having a difficulty drowing any conclusions.

the equations for the trendlines are as follows,

$y=0.5x+340$ for the actual y(x)

$y=1.55x+350$ for the regression model

How can I compare my regression model with the actual data to estimate the error?

答:
Short answer: there is no single way of looking at error - it all depends on your application. But there are a few popular methods of quantifying error, the most of which is Mean Square Error

Let $\bar{y}(x_i)$ be the prediction and $y(x_i)$ be the actual value

$$\text{MSE} = \frac{1}{N}\sum_{i=1}^N (\bar{y}_i - y_i)^2$$

Another way is to find the absolute mean error

$$\text{Absolute Error} = \frac{1}{N}\sum_{i=1}^N |\bar{y}_i - y|$$

You can search online for other error functions


答:
In this interval $\sin(x)$ is positive. Also $0\leq\sin(x)\leq 1$, so it is a fraction like:

$\sin(x)= \frac 1a$

$\sin^2(x)=\frac 1{a^2}$

where $ a \geq 1$

If $a>1$ then $a^2>a$ , therefore:

$ \frac 1{a^2}<\frac 1a$

$\Rightarrow\sin^2(x)<\sin(x)$

$\sin^2(x)=\sin(x)$ if $a=1$

Title: Proving $\sin^2x<\sin x$ for $x\in(0,\pi)$

Body:
Actually this is a calculus problem. I'm required to compare a given integral with a simpler one and check if the inequality holds 2. (Question b)

This is a part of an answer in my textbook:

$$0 < \sin x < 1 \;\;\text{on}\;\; (0,\pi) \quad\implies \sin^2x<\sin x \;\;\text{on}\;\; (0,\pi)$$
Then it says, Therefore:$$\int ^{\pi }_{0}\sin^{2} xdx< \int ^{\pi }_{0}\sin xdx$$

But how can prove this inequality? $$\sin^{2} x< \sin x\ (x\in \ ( 0,\ \pi ))$$ if $$0 < \sin x < 1 \;\;\text{on}\;\; (0,\pi)$$

Edit: Thank you for warning me about the MSE format. I'll be more careful in the future.


答:
$\sin(x)$ has the range of $(0, 1]$ when $x\in (0, \pi)$

Now you can see that for any number in $(0, 1]$, taking the square means multiplying by a number which is smaller than or equal to 1, therefore the square will be a number smaller than the original number.

However, the question you asked doesn't hold true as adam pointed out, $\sin^2(\frac{\pi}{2})=\sin(\frac{\pi}{2})$ therefore $\sin^2(x)\le \sin(x)$ would be true, not $\sin^2(x)<\sin(x)$


问:
Symmetric Grothendieck inequality
Grothendieck's inequality states that for all $n \times n$ matrices $(a_{ij})$ such that

$$\max_{x \in \{\pm 1\}^n,\, y \in \{\pm 1\}^n} \left|\sum_{ij} a_{ij}\, x_i\, y_j\right| \leq 1,$$

there exists a universal constant $K$, such that for $u_i, v_j$ in any Hilbert space,
$$
\max_{x \in \{\pm 1\}^n,\, y \in \{\pm 1\}^n} \left|\sum_{ij} a_{ij} \langle u_i , v_j \rangle \right| \leq K.
$$
I would like to prove the symmetric statement.  For all symmetric matrices $(a_{ij})$ such that
$$
\max_{x \in \{\pm 1\}^n} \left|\sum_{ij} a_{ij} \,x_i\, x_j\right| \leq 1,
$$
there exists a universal constant such that
$$
\max_{x \in \{\pm 1\}^n} \left|\sum_{i,j} a_{ij} \langle u_i , v_j \rangle \right| \leq 2K
$$
for $u_i, v_j$ in any Hilbert space.  This should be a consequence of the original inequality.  I tried to use the polarization identity
$$
\langle Ax, y\rangle = \langle Au, u\rangle - \langle Av, v \rangle
$$
where $u = (x+y)/2$ and $ v = (x-y)/2$.  However, as $x$ and $y$ vary over $\pm 1$ vectors, $u$ and $v$ can be vectors in $\{\pm 1, 0\}$.

答:
I would like to point you to this paper, and I feel there should be a satisfactory answer there.
https://arxiv.org/abs/2003.07345</p>

Title: Symmetric Grothendieck inequality


问:
What am I doing wrong; evaluating integrals?
$$f(x) = \dfrac{2x}{2-3x^2}$$

$$- \dfrac{1}{3} d (2-3x^2) = 2x$$

$$ \dfrac{2x}{2-3x^2} dx = -\dfrac{1}{3} d (2-3x^2) \cdot \dfrac{1}{2-3x^2} = - \dfrac{1}{3} du \cdot u^{-1} = -\dfrac{1}{3} u^{-1} \cdot du$$

And here I'm stuck, since you can't divide by 0.. What am I doing incorrectly here? I've applied the method my book gave me perfectly as far as I know.

答:
$$f(x) = \dfrac{2x}{2-3x^2}$$

$$2-3x^2=t\implies-6x=\dfrac{dt}{dx}\implies dx=\dfrac{dt}{-6x}$$

$$\int \dfrac {2x}{2-3x^2}\,dx$$
$$\int \dfrac {2x}{t}\,\dfrac{dt}{-6x}$$
$$-\int \dfrac {1}{3t}\,dt$$
$$-\dfrac13\cdot \log t+C$$
$$-\dfrac13\cdot \log (2-3x^2)+C$$

when you have situation like this : $\int \dfrac 1x \,dx$ then don't write it as $x^{-1}$.

$\int x^n\, dx=\dfrac {x^{n+1}}{n+1}$ this is not applicable on $n=-1$.

$\int x^{-1}\, dx$ solve as $\int \dfrac 1x\, dx=\log |x|$


问:
Proving $R/I \otimes_R M \cong_R M/IM$ using UMP and a biadditive map not with exact sequences.
I want to Understand a paragraph of the proof of $R/I \otimes_R M \cong_R M/IM.$ in example$(8)$ on pg. 370 of Dummit and Foote (third edition)

Here is the example from Dummit & Foote:

My questions are:

1- I do not understand exactly why we need the previous observation (as stated in the book) to say that the map $N \rightarrow (R/I) \otimes_R N$ defined by $n \mapsto 1 \otimes n$ is surjective?

2-why we need $IN$ to be in the kernel?

3- why $1 \otimes a_i n_i = a_i \otimes n_i$?

4- Why the author defined $f$ by that definition?

Could anyone help me answer those questions please?

答:
Quick answer that could be expanded if more details are provided

You need the previous observation to prove that elements $1\otimes n$ generate the tensor product as an $R$-module, since all of them are in the image and the map is $R$ linear the claim follows.

You need that to define a map from a quotient.

By definition of tensor product over $R$, $R$-scalars can "jump" from one side to the other of the product.

It was probably easier to show the inverse than to prove than the original map is injective.


问:
If $1^3 = 1^2$, and take the$\ log_1$ on both sides, $3 = 2$?
If $$\boxed{1^3=1^2}$$ and I take the $$\log_1$$ on both sides, $$ 3=2,\ \quad ?$$

答: Well, $1^2=1^3$ does hold true but if you take the log of both the sides, you get $$2\log(1)=3\log(1)$$now, $\log(1)$ is equal to $0$ for any base except 1, and log base 1 is not defined, so you can't cancel the log from both the sides as that would be the same as dividing $0$ by $0$ which you can't do, that is why what you said does not hold true


答: There is no $\log_1$ function and, if any function were defined with such name, it would not satisfy $\log_1(1^a)=a$ for all $a$.

Title: If $1^3 = 1^2$, and take the$\ log_1$ on both sides, $3 = 2$?

Body: If $$\boxed{1^3=1^2}$$ and I take the $$\log_1$$ on both sides, $$ 3=2,\ \quad ?$$


答:
Well, first of all.  $0 < \sin x < 1$ on $(0,\pi)$ is not true because $\sin \frac {\pi} 2 = 1$.

But we do have $0 < \sin x \le 1$ on $(0,\pi)$ with $\sin x = 1 \iff x = \frac \pi 2$.

And we do have $0< \sin^2 x < \sin x < 1$ for all $x\in (0,1); x\ne \frac \pi 2$ (And $\sin^2 x = \sin x = 1$ if $x =\frac \pi 2$)

But that has Nothing to do with trigonometry!

Notice for any  $0 < w < 1$ then if we multiply each term by $w$ we ALWAYS get:

$0*w < w*w < 1*w$ and $0 < w^2 < w$.

And so $0 < \sin x < 1$ if $x\ne \frac {\pi}2$ we will hgave $0 < \sin^2 x < \sin x$ when $x\ne \frac {\pi}2 $.


问:
Prove the curve is a part of circle
Given a curve $y$ having constant curvature with $y=f(x)$ with $f''(x) > 0 $. I have to prove that this curve is a part of a circle.

My Try

Assume curvature of the curve as $c$. Then, I can write $$c = \frac{f''(x)}{(1+f'(x)^2)^{\frac 32}}$$
After this, I am unable to solve the non-linear differential equation.

Is my approach wrong?
Thanks in advance !

答:
If you separate the variables in the equation for $g=f'$ you obtain
$$
\frac{dg}{(1+g^2)^{3/2}}=cdx
$$
With the substitution $g=\tan\theta$ the integral on the left is just $\sin\theta+const.$ so
$$
\sin\theta=cx+D
$$
If we choose as our initial direction $\theta=0$ (the curve starts horizontally at $x=0$) you get $D=0$. Since $\sin\theta=\tan\theta/\sqrt{1+\tan^2\theta}=g/\sqrt{1+g^2}$ we have
$$
\frac{g}{\sqrt{1+g^2}}=cx.
$$
Solving for $g$:
$$
g(x)=\frac{cx}{\sqrt{1-c^2x^2}},
$$
for $|x|\in[0,1/c)$. Integrating once again
$$
f(x)=-\frac{\sqrt{1-c^2x^2}}{c}+E
$$
Choosing the initial point of the curve at the origin, we find $E=\frac 1{c}$. Finally,
$$
f(x)=\frac 1{c}-\frac{\sqrt{1-c^2x^2}}{c}
$$
which is the lower half of the circle centered at $(0,1/c)$ with radius $1/c$, as expected (the radius is the reciprocal of the curvature).

Title: Prove the curve is a part of circle


问:
Commutation relation for Lie algebras using the root string
Let $\alpha, \beta \in \Phi$ be roots of $\mathfrak{g}$ (relative to some choice of $\mathfrak{h}$) for which $\alpha + \beta \in \Phi$, and let $\mathfrak{g}_\alpha$ denote the $\alpha$ weight space (similarly for $\beta$).
The $\alpha$-root string through $\beta$ is the set $$R(\alpha, \beta) = \{ \beta + k \alpha : k \in \mathbb{Z}\} \cap \Phi.$$
Define $V_{\alpha, \beta} = \sum_{\gamma \in R(\alpha, \beta)} \mathfrak{g}_\gamma$.
I know that there exists an isomorphic copy of $\mathfrak{sl}_2 \subset \mathfrak{g}$ generated by $\mathfrak{g}_\alpha$ and $\mathfrak{g}_{-\alpha}$ (specifically a choice of $e_\alpha$ and $f_\alpha$) - denote this algebra by $\mathfrak{s}(\alpha)$.
I also know that if $\alpha + \beta \in \Phi$ then the space $V_{\alpha,\beta}$ is an invariant irreducible representation of $\mathfrak{s}(\alpha)$ (by way of restricting the adjoint representation of $\mathfrak{g}$).

I want to show that

if $\alpha + \beta \in \Phi$ then $[\mathfrak{g}_\alpha, \mathfrak{g}_\beta] = \mathfrak{g}_{\alpha + \beta}.$

The proof in my textbook is summarised below, I have boldfaced the conclusion that I have tried to fill in the gaps for.

First, we can check directly using the Lie bracket that $[\mathfrak{g}_\alpha, \mathfrak{g}_\beta] \subset \mathfrak{g}_{\alpha + \beta}$.
Since $\alpha + \beta \in \Phi$, then $V_{\alpha,\beta}$ is irreducible under $\mathfrak{s}(\alpha)$, and by definition $\mathfrak{g}_{\alpha + \beta} \subset V_{\alpha, \beta}$.
Then $e_\alpha$ maps $\mathfrak{g}_\beta$ onto $\mathfrak{g}_{\alpha + \beta}$.

I first assume (for a contradiction) that there exists $X \in \mathfrak{g}_{\alpha + \beta}$ which does not come from the action of $e_\alpha$ on $\mathfrak{g}_\beta$.
Since each weight space has dimension 1, I know that $X$ spans $\mathfrak{g}_{\alpha + \beta}$, which means that $e_\alpha$ sends $\mathfrak{g}_\beta$ to 0.
This in turn implies that $[\mathfrak{g}_\alpha, \mathfrak{g}_\beta] = 0$, since $e_\alpha \in \mathfrak{g}_\alpha$.
Finally, I say that this shows that the action of $\mathfrak{s}(\alpha)$ on $V_{\alpha, \beta}$ is zero, contradicting the irreducibility of the representation.

It is this final statement that I am having the most trouble accepting, even though it seems correct to me.
Any input on if this is correct or not, and/or if there is an easier way to see the result, would be an ideal answer.

答: Your argument works, but this is easier: $V_{\alpha,\beta}$ is an irreducible representation of $\mathfrak{s}(\alpha)$, which is isomorphic to $\mathfrak{sl}(2)$. It follows that $e_\alpha$ only kills the highest weight vector, whose weight cannot be $\beta$, since $\alpha+\beta$ also occurs in the string. Because weight spaces are $1$-dimensional, the image of $\mathfrak{g}_\beta$ under $e_\alpha$ is precisely $\mathfrak{g}_{\alpha+\beta}$.


问:
4D Left Isoclinic Rotations as double rotations
According to this page, 4D left-isoclinic rotation matrices are double rotations of the same angle (and same sign). But in the same page, in the context of quaternions, it is mentioned that left-isoclinic matrices have the form:
$$R=\begin{bmatrix}a&-b&-c&-d \\ b&a&-d&c\\c&d&a&-b\\d&-c&b&a\end{bmatrix}$$ where $a,b,c,d\in\mathbb{R}$ such that $$a^2+b^2+c^2+d^2=1.$$ My question is, how can we view $R$ as a double rotation? I mean, how can we find the common angle? If two of $b,c,d$ are $0$ then it is easy to see. But what if this is not the case?

答:
For this matrix, $a$ is the cosine of the rotation angle.

Proof sketch: First show that $R$ is an orthogonal matrix, so that $|Ru| = |u|$ for all $u$. Now recall that $u^T v =|u| |v|\cos\theta$ where $\theta$ angle between two vectors $u$ and $v$. So then, to compute the angle between $u$ and $Ru$, show that $u^T R u = a u^T u$.


问:
Confusion with independence relations in probability
What events would need to be independent for $P(A,B,C) = P(A)P(B|C)P(C)$ to be true?

What about $P(A,B,C) = P(A|B)P(B|C)P(C)$?

I know that we can use bayes rule to expand it, but with that I end up getting $P(A,B,C) = P(A,B,C)$

答:
For the first one, as mentioned in the comments,

$$\begin{split}P(A\cap B\cap C)&=P(A)P(B|C)P(C)\\
&=P(A)P(B\cap C)\end{split}$$

Hence  $A$ and $B\cap C$ are independent (because $P(B\cap C|A)=P(B\cap C)$).

For the second one,

$$\begin{split}P(A\cap B\cap C)&=P(A|B)P(B|C)P(C)\end{split}$$

we clearly have

$$P(A|B)P(B|C)=P(A\cap B|C)$$

we know

$$P(A|B,C)P(B|C)=P(A\cap B|C)$$

therefore

$$P(A|B)=P(A|B,C)$$

so $A|B$ is independent of C.


问:
Weak * convergence in $\ell^{\infty}$
Let $x=(x_1, x_2, \cdots) \in \ell^{\infty} = (\ell^1)^*$.

Let $s_n = \sum_{k=1}^n x_k e_k$ be the $n$-th partial sum.

(1) Show that $s_n \stackrel{\ast}{\rightharpoonup}x$ (i.e. show the the weak $^*$ convergence).

(2) Does the net $s_F= \sum_{k\in F} x_k e_k$ of finite partial sums converge int the weak$^*$ topology to $x$?

Things I know:

For (1):

We have that $\ell^1 = c_0^*$.

Since $(e_n) \subset \ell^1$, $(e_n)$ converges to $0$ in the weak $^*$ topology.

Since $s_n$ is finite, we know that each $s_n \in \ell^1.$.

So we want to show that $\lim_{n\rightarrow \infty} s_n= x$.

But this above seems obvious. I'm sure I missed something important here, since I haven't really used the the definitions of $\ell^1$ and $\ell^{\infty}$...

So please let me know how to proceed from here.

For (2),

I'm not sure about this one. My guess is yes. But I don't have any clues on how to prove that.

Any help is appreciated!

Thank you!

答: Let $(y_k) \in \ell^{1}$. Then $s_n((y_k))= \sum\limits_{k=1}^{n}x_ky_k \to \sum\limits_{k=1}^{\infty}x_ky_k$, Note that the series $\sum\limits_{k=1}^{\infty}x_ky_k$ converges absolutely. This proves the first part. The second part follows from the fact that for any $\epsilon >0$ we can find $N$ with $\sum\limits_{k=N}^{\infty}|x_k||y_k| <\epsilon$.

Title: Weak * convergence in $\ell^{\infty}$


答:
There are always many ways to split $\Bbb{R}^4$ into a direct sum of two planes, stable under $R$ (and hence both rotated by the same angle).

The eigenvalues of $R$ are complex, $\lambda_{\pm}=a\pm i\sqrt{b^2+c^2+d^2}$, both with multiplicity two. Let $\vec{v}_1, \vec{v}_2\in\Bbb{C}^4$ be two linearly independent eigenvectors belonging to $\lambda_+$. Then their componentwise complex conjugates
$\vec{v}_1^*$ and $\vec{v}_2^*$ belong to $\lambda_-$. Consider the spaces
$$
T_j=\{z\vec{v}_j+z^*\vec{v}_j^*\mid z\in\Bbb{C}\},
$$
$j=1,2$. Clearly $T_1$ and $T_2$ are both subsets of $\Bbb{R}^4$ and $2$-dimensional vector spaces over $\Bbb{R}$. Furthermore
$$
R(z\vec{v}_j+z^*\vec{v}_j^*)=\lambda_+z\vec{v}_j+\lambda_-z^*\vec{v}_j^*\in T_j,
$$
so the two planes are stable under $R$.

Replacing $\vec{v}_1,\vec{v}_2$ with a different pair will produce different pairs of planes.

The common angle of rotation (assuming normalized to a unit quaternion) is $\arccos a$.

Title: 4D Left Isoclinic Rotations as double rotations

Body:
According to this page, 4D left-isoclinic rotation matrices are double rotations of the same angle (and same sign). But in the same page, in the context of quaternions, it is mentioned that left-isoclinic matrices have the form:
$$R=\begin{bmatrix}a&-b&-c&-d \\ b&a&-d&c\\c&d&a&-b\\d&-c&b&a\end{bmatrix}$$ where $a,b,c,d\in\mathbb{R}$ such that $$a^2+b^2+c^2+d^2=1.$$ My question is, how can we view $R$ as a double rotation? I mean, how can we find the common angle? If two of $b,c,d$ are $0$ then it is easy to see. But what if this is not the case?


答:
There's no such thing as $\log_1$.

And there's no such thing as $\log_0$ or $\log_{b}$ if $b < 0$.

The definition of $\log_b m = k\iff b^k = m$ relies upon the assumption that $b^k = b^j \iff k = j$ (and that for any $m > 0$ that there is a $k$ [and only on $k$] where $b^k =m$).

As this assumption is not true for $b = 0, 1$ or $b < 0$ we simply do not have any such $\log_1, \log_0$ or $\log_{b;b< 0}$

...

Also not if $m \ne 1$ the $\log_1 m $ makes no sense as we can never have $1^k = m$.  And $\log_1 1$ makes no sense because $1^k = 1$ for ALL $k$.  So $\log_1$ just makes no sense at all.


答:
This is a problem that requires proper handling of the concept of Riemann sum and its relationship to definite integral. Consider the following partition of the interval $[0,1]$:
$$
\mathscr{P}\Big( [0,1]\Big)=\left \{ 0=x_0<x_1=\frac{1}{n}<\ldots < x_k=k\cdot \frac{1}{n}<\ldots<x_{n}=n\cdot\frac{1}{n} \right\}
$$
chose $x^*_k\in [x_{k-1},x_{k}]$ equal to $k\cdot \frac{1}{n}$ and set $\Delta x_k=x_k-x_{k-1}$ .
Then we have
$$
\sum_{k=1}^{n}\inf_{\xi\in[x_k,x_{k-1}]}\sin( \xi\pi)\cdot\Delta x_k
\leq
 \sum_{k=1}^{n}\sin( x_k^*\pi)\cdot\Delta x_k
\leq
\sum_{k=1}^{n}\sup_{\xi\in[x_k,x_{k-1}]}\sin( \xi\pi)\cdot\Delta x_k
$$
that is
$$
\sum_{k=1}^{n}\inf_{\xi\in[x_k,x_{k-1}]}\sin( \xi\pi)\cdot\Delta x_k
\leq
\sum_{k=1}^n\sin{\left(\frac{k\pi}{n}\right )}\frac{1}{n}
\leq
\sum_{k=1}^{n}\sup_{\xi\in[x_k,x_{k-1}]}\sin( \xi \pi)\cdot\Delta x_k.
$$
If we have
$$
\lim_{n \to \infty}\sum_{k=1}^{n}\inf_{\xi\in[x_k,x_{k-1}]}\sin( \xi \pi)\cdot\Delta x_k
=
 \int_0^1\sin( x\pi)dx
=
\lim_{n \to \infty}\sum_{k=1}^{n}\sup_{\xi\in[x_k,x_{k-1}]}\sin( \xi\pi)\cdot\Delta x_k
$$
then
\begin{align}
\lim_{n\to+\infty}\sum_{k=1}^n\sin{\left(\frac{k\pi}{n}\right )}\frac{1}{n}=
 \int_0^1\sin( x\pi)dx
\end{align}</p>

Body:
Calculate the limit: $\lim_{n\to+\infty}\sum_{k=1}^n\frac{\sin{\frac{k\pi}n}}{n}$ Using definite integral between the interval $[0,1]$.

It seems to me like a Riemann integral definition:

$\sum_{k=1}^n\frac{\sin{\frac{k\pi}{n}}}{n}=\frac1n(\sin{\frac{\pi}n}+...+\sin\pi)$

So $\Delta x=\frac 1n$ and $f(x)=\sin(\frac{\pi x}n)$ (Not sure about $f(x))$

How do i proceed from this point?


答:
The function $\log_b(x)$ is only defined for $x,b \in \mathbb{R} :b>0, b\neq 1, x>0$

Note: Please look up Wikipedia (or other online sources that have already answered your question) before asking questions here.


答:
Or more straightforward

$$\log_1{1}=\dfrac {\ln 1}{\ln 1}=\dfrac {0}{0}=\text{undefined}.$$


问:
If $PP^*=P^*P$ and $P^2=P$ then show that $P^*=P$
To show an idempotent normal matrix is an $\textbf{orthogonal}$ projection.

Proof: If we use, spectral decomposition of normal matrices we can write, $P=U\Lambda U^* \implies P^2=U\Lambda^2U^*$. Idempotence gives $\lambda(P)\in \{0,1\}$, so that  $P^*=U\Lambda^* U^*=U\Lambda U^*=P$.

Can we do this proof without using spectral demcomposition?

答:
The main argument supposes that you know that the trace of an idempotent matrix (in characteristic 0) gives you said matrix's rank-- this result does not require understanding of spectral theorem / normality or even eigenvalues (see post-script).

main argument

$(P^*P)^2 = P^*PP^*P= P^*P^*PP = (P^*P)$

hence $(P^*P)$ is idempotent.  Since we work over $\mathbb C$, we know

$\text{rank}\Big(P^*P\Big) =\text{rank}\Big(P\Big)\implies \text{trace}\Big(P^*P\Big) =\text{trace}\Big(P\Big)=\text{trace}\Big(P^2\Big)$

but
$\text{trace}\Big(P^*P\Big) \geq \text{trace}\Big(P^2\Big)$
by Cauchy-Schwarz (met with equality)  $\implies P^*=P$

optional post-script:

'eigenvalue-free' proof that an idempotent matrix has rank given by its trace (and that said matrix must be diagonalizable).

Suppose idempotent $A$ has rank r, then apply rank nullity, and build a basis for its kernel -- call these vectors $\big\{\mathbf q_{r+1},..., \mathbf q_{n}\big\}$.  Further we have $\text{rank}\big(A^2\big)=\text{rank}\big(A\big)$ so $V=\text{image}\big(A\big)\oplus \text{ker}\big(A\big)$.  Extend the $\mathbf q_j$ to a basis for the vector space by appending $r$ linearly independent vectors from the image of $A$, and collect these vectors in a matrix.

$Q:=\bigg[\begin{array}{c|c|c|c|c}
\mathbf q_1  &\cdots & \mathbf q_{r} & \mathbf q_{r+1} &\cdots & \mathbf q_{n}\end{array}\bigg]$

$ AQ  = QD$

where $D=\begin{bmatrix} B_r &\mathbf {0}\\ * &\mathbf {0}\end{bmatrix}=\begin{bmatrix} I_r &\mathbf {0}\\ \mathbf {0} &\mathbf {0}\end{bmatrix}$

because

1.) $A$ sends vectors $\mathbf q_j$ for $r+1\leq j\leq n$ to zero

2.) $*$ is seen to be zero because e.g. $A\mathbf q_1 \in \text{image}\big(A\big)$ hence $A\mathbf q_1$ is uniquely written as a linear combination of $\mathbf q_k$ for $1\leq k\leq r$ (i.e. basis vectors for the image of $A$).

3.)  $QD^2 = A^2Q =AQ=QD\implies B_r^2 = B_r\implies B_r =I_r$
because $\text{rank}\big(A\big)= r= \text{rank}\big(B_r\big)$ and the only invertible idempotent element is the identity (a one line proof).  Since $Q$ is invertible we have  $A=QDQ^{-1}$

$\implies \text{rank}\Big(A\Big)=r = \text{trace}\Big(D\Big)= \text{trace}\Big(QDQ^{-1}\Big)= \text{trace}\Big(A\Big)$


问:
On 'proving' that $\langle x,y_1,\ldots,y_n | [x, y_1 \cdots y_n] = 1 \rangle \simeq \mathbb Z^2$.
As the title says, I want to prove that $G:= \langle x,y_1,\ldots,y_n \mid [x, y_1 \cdots y_n] = 1 \rangle \simeq \mathbb Z^2$ algebraically.

Edit: well, this is embarrassing. That is not what I want, because it is not true (and that is why I asked this in the first place, it seemed quite suspicious). So:

Where have I messed up in my proof?

Edit': I have found the error in what follows, it is now added as an answer.

Consider a regular $2(n+1)$-gon $P$ whose edges are labeled  $x,y_1,\ldots,y_n,x^{-1},y_n^{-1},\ldots,y_1^{-1}$, say, clockwise. The quotient $X$ of $P$ that identifies the edges of $\partial P$ according to this labeling gives a torus: we can "fold" $P$ so that each edge $y_i$ matches the edge labeled with its inverse. This produces a tube whose boundary corresponds to the edges labeled $x$ and $x^{-1}$ respectively, and the orientation is coherent with the torus (i.e. we do not get a Klein bottle). Hence $\pi_1(X) \simeq \mathbb Z^2$. On the other hand, a straightforward analysis via van Kampen with $U = X \setminus \{[p]\}, p \not \in \partial P$ and and $V$ (the projection of) a disk around $p$ yields $ \pi_1(X) \simeq G$.

答: Doesn’t $x\mapsto 1$ give an epinorphism from $ G$ to the free group on $n$ generators $y_1,\ldots,y_n$?


问:
How do I find the limit $\lim_{x \to 1} \frac{1}{1 - x} - \frac{3}{1 - x^3}$
How do I calculate the following limit? $$\lim_{x \to 1} \frac{1}{1 - x} - \frac{3}{1 - x^3}$$

I already transformed this into

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

but this still has zeroes up- and downstairs. L'Hospitals rule is not allowed. Any advice?

EDIT: Sorry, I was doing confused. There is a 3 in the second numerator.

EDIT 2: Wolfram Alpha says the limit is $-1$

Body:
Hints:

$$\frac1{1-x}-\frac1{1-x^3}=\frac{1+x+x^2-1}{1-x^3}=\ldots\ldots$$

Since

$$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\ldots+ab^{n-2}+b^{n-1})\;,\;\;n\in\Bbb N$$


问:
H is finite dimensional
Let $H$ be a Hilbert space and $\lbrace x_ i \rbrace_ i \subset H$ a countable set of vectors that algebraically spans $H$, i.e. every vector in $H$ is a linear combination of some finite subset of $\lbrace x _i \rbrace_ i$. Show that $H$ is finite dimensional.

My attempt:

In this set of vectors I can find a minimal spanning set, i.e basis of $H$, say $\lbrace y_i\rbrace_i$. suppose the basis is infinite. then take a non-zero element $x\in H$ then $x=\sum_{i\geq1}a_iy_i$. but $x$ can be written as linear combination of finitely many elements, say $x=\sum_{k=1}^{n}b_ky_k$. Therefore, $0=\sum_{k=1}^{n}(a_k-b_k)y_k+\sum_{i\geq n+1}a_iy_i$. So that implies the tail of the series converges to $0\Longrightarrow x=0$ (Contradiction.)

If this prove is correct then where are we really using the inner product of Hilbert space and the fact that Hilbert spaces are complete.

答:
Here is a proof, which looks like the one you had in mind. The contradiction is again in finite versus infinite sums.

Let $(y_i)$ be a countable basis of $H$. Then using the Gram-Schmidt orthonormalization process we can construct an orthonormal basis $(z_i)$ of $H$:
$(z_i)$ is a basis and $\langle z_i,z_j\rangle = \delta_{i,j}$ for all $i,j\in \mathbb N$.

Define $u_n:=\sum_{k=1}^n \frac1k z_k$. This is a Cauchy sequence in $H$. Hence converging to $u$. One can verify $u=\sum_{k=1}^\infty \frac1kz_k$ by looking at $\langle u-u_n,z_k\rangle$. Let $v\in span(z_1\dots z_n)$ with $v=\sum_{k=1}^n v_kz_k$. Then
$$
\|u-v\|_H^2 = \|\sum_{k=1}^n(\frac1k - v_k)z_k\|_H^2 +  \|\sum_{k=n+1}^\infty \frac1kz_k\|_H^2 \ge \sum_{k=n+1}^\infty \frac1{k^2} >0.
$$
Hence, $u$ cannot be a finite linear combination of the $(z_i)$'s. Contradiction.

Title: H is finite dimensional


问:
Idempotents in $n \times n$ matrices with entries in $Z_2$
OEIS has the following sequence and formula http://oeis.org/A132186 for counting how many $n \times n$ idempotents there are over $GF(2)$.

But how does one find/generate them? I haven't been able to find any resource for this.

I assume that this is solved for $GF(2)$, is it also solved in general?

答:
An idempotent matrix $X$ satisfies $X^2-X=0$, therefore it has minimal polynomial either $t$, $t-1$ or $t^2-t$. The first two cases give the zero and identity matrices. The second one decomposes into linear factors so $X$ must be diagonalizable, with eigenvalues some number of zeros and some number of ones. It follows that all the idempotents are the conjugacy classes of these matrices.

Now by the orbit stabilizer theorem, the size of the conjugacy class of $X$ is precisely $|G|/|C_G(X) |$, where $G=GL_n(\mathbb{F}_2)$ and $C_G(X) $ is the centralizer of $X$. Applying this to the diagonal matrix $D_k$ with $k$ ones and $n-k$ zeros on the diagonal (these are representatives for each conjugacy class we're interested in), we see that the centralizer consists of matrices that are block diagonal $A\oplus B$, with $A$ in $GL_k(\mathbb{F}_2)$ and $B$ in $GL_{n-k} (\mathbb{F}_2)$.

Since there is a formula for the order of $GL_n$ over a finite field, putting all together you should get a formula for the total number of idempotents.

To generate the matrices, you need coset representatives $g_i $ for $G/C_G(X)$ (computer programs give you these), and then the idempotent matrices will be precisely the $g_iXg_i^{-1}$ (for each $X=D_k$).


问:
Show sin($\bar{z}$) = $\overline{\sin(z)}$
Let $z = x + iy$ where $x$ and $y$ are real. Show that

$$\sin(\bar{z})=\overline{\sin(z)}.$$

I started by trying to take the conjugate of $z$ which was ok but I didn't know how to take the conjugate of $\sin(z)$.

答:
Sine in the complex variable is an entire function given by the series

$$
\sin z = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}z^{2n+1}
$$

convergent everywhere in the complex plane. We can compute its complex conjugate  term by term

$$
\overline{\sin z} = \sum_{n=0}^\infty \frac{\overline{(-1)^n}}{\overline{(2n+1)!}}\overline{z^{2n+1}} = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}\overline{z}^{2n+1} = \sin\overline{z}.
$$

Alternatively, as suggested in the comments, the equality follows from the Schwarz reflection principle. In both cases, we exploit the fact that sine is complex analytic.


问:
Why is $e^π$ transcendental?
the title and the tags might be correct for the question.

In the wikipedia page about gelfond's constant I saw that $e^{π}$ is transcendental and the proof was

$e^{π}=(e^{iπ})^{-i}=(-1)^{-i}$

Since $-1$ and $-i$ is algebraic and $-i$ is not rational, by the Gelfond Schneider Theorem $e^{π}$ is transcendental.

Now let's take the lindemann weierstrass theorem. According to the theorem, $e^a$ is transcendental if a is algebraic

In this case where $π$ is transcendental so by the lindemann weierstrass theorem $e^π$ cannot be transcendental so it is algebraic.

Which contradicts our first result.

I just wanna know that what am I missing(I am quite sure about it) or did I just found a contradiction in maths(I don't think so)?

答:
The Gelfond-Schneider Theorem actually says:

If:

$\alpha$ and $\beta$ are algebraic numbers such that $\alpha \notin \{0,1\}$

$\beta$ is either irrational or not wholly real

then $\alpha^{\beta}$ is transcendental.

Just thought I'd put that in there because technically speaking $-i$ is not actually "irrational" as such.

Apart from that, yes, from your argument $e^\pi$ is transcendental.

Lindemann-Weierstrass Theorem says:

Let $a$ be a non-zero algebraic number.

Then $e^a$ is transcendental.

It does not say that if $a$ is transcendental, then $e^a$ is not transcendental.

That would be like saying:

"If $a$ is even then $2 a$ is even. Therefore if $a$ is odd then $2 a$ is odd."

Title: Why is $e^π$ transcendental?


Title: How do I find the limit $\lim_{x \to 1} \frac{1}{1 - x} - \frac{3}{1 - x^3}$

Body:
How do I calculate the following limit? $$\lim_{x \to 1} \frac{1}{1 - x} - \frac{3}{1 - x^3}$$

I already transformed this into

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

but this still has zeroes up- and downstairs. L'Hospitals rule is not allowed. Any advice?

EDIT: Sorry, I was doing confused. There is a 3 in the second numerator.

EDIT 2: Wolfram Alpha says the limit is $-1$

Body:
Answer to the original question

The limit doesn't exist: $(1 - x^3) = (1 - x)(1 + x^2 + x)$, therefore
$$\frac{1}{1 - x} - \frac{1}{1 - x^3} = \frac{1}{1 - x}\Big(1 - \frac{1}{1 + x^2 + x}\Big).$$From this you get $$\lim_{x \to 1^-} \frac{1}{1 - x}\Big(1 - \frac{1}{1 + x^2 + x}\Big) = \infty$$
$$\lim_{x \to 1^+} \frac{1}{1 - x}\Big(1 - \frac{1}{1 + x^2 + x}\Big) = -\infty$$

Answer to the edited question
$$\frac{1}{1 - x} - \frac{3}{1 - x^3} = \frac{1}{1 - x}\Big(1 - \frac{3}{1 + x^2 + x}\Big) = \frac{1}{1 - x}\frac{1 + x^2 + x - 3}{1 + x^2 + x} = \frac{1}{1 - x}\frac{(x - 1)(x + 2)}{1 + x^2 + x} = \frac{-x - 2}{1 + x^2 + x} \to \frac{-3}{3} = -1$$


问:
How to proof such propositions
let $P(x,y): x - 7y = 5$ and $Q(x,y): x-3 \geq y + 11$

$\exists x \forall y, Q(x,y)$

$\exists x \exists y, P(x,y) \land \neg Q(x,y)$

$\forall x \exists y, P(x,y) \geq Q(x,y)$

How de we prove if these propositions are true or false, I know when a ($\forall x, P$) proposition is false you can prove it by showing that ($\exists x, \neg P$) is true, and to prove ($\forall x, P$) is true you have to use a detailed demonstration, and to prove ($\exists x, P$) you just show one example where the equation is true.
However in these cases I really have no idea how to prove such propositions.

答:
To disprove the first one set y = x.

To determine the second one, use the equation in P to reduce Q to a single variable and make your decision.

Explain why the third one is nonsense.

Title: How to proof such propositions


问:
How to calculate the absolute extrema for this equation
Basically, I am trying to find the absolute extrema for this problem, but I am stuck after a few steps.

$$y=\frac{4}{x}+\tan\left(\frac{\pi x}{8}\right),\quad x\in[1,2].$$

I have figured out the derivative, which is

$$y'=-\frac{4}{x^2}+\sec^2\left(\frac{\pi x}{8}\right)\cdot\frac{\pi}{8}.$$

However, after reaching this step, I am stuck. I have tried setting the derivative equal to zero, but I do not get anywhere as I don't know what to do next. How can I solve this problem? Any help is appreciated. Thanks.

答:
Since you already received good answers, let me show something amazing.

With the derivative you wrote, you cannot find its zero except using numerical methods.

To simplify the problem, let $x=\frac{8 t}{\pi }$ and the function becomes
$$f(t)=\frac{\pi }{2 t}+\tan (t)$$
$$f'(t)=\sec ^2(t)-\frac{\pi }{2 t^2}$$ Now, use the $1,400$ years old approximation
$$\cos(t) \simeq\frac{\pi ^2-4t^2}{\pi ^2+t^2}\qquad \text{for} \qquad -\frac \pi 2 \leq x\leq\frac \pi 2$$  which makes
$$f'(t) \sim \frac{\left(\pi ^2+t^2\right)^2}{\left(\pi ^2-4 t^2\right)^2}-\frac{\pi }{2 t^2}$$ Let $t=\sqrt y$ and, after simplifications, the derivative write
$$\frac {2 y^3+4 (\pi -4) \pi  y^2+2 \pi ^3 (4+\pi ) y-\pi ^5 } {2 \left(\pi ^2-4 y\right)^2 y }$$ The cubic in numerator shows only one real root since
$$\Delta=-100 \pi^9 (8 \pi ^2+83 \pi-128) <0$$ Now, using the hyperbolic method, this root is
$$y=\frac{2\pi}3\Bigg[4-\pi+\sqrt{44\pi- \pi^2-64} \sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{2048-2112 \pi +411 \pi ^2+4 \pi ^3}{4(44\pi- \pi^2-64)^{3/2}}\right)\right)  \Bigg]$$ which is $y\sim 0.701420$; then $t\sim 0.837508$ and finally $x\sim 2.13270$ while the solution given by Newton method is $x=2.13448$.

Using the approximate solution, the minimum of the function would be $2.98561$ while its "exact" value is $2.98561$ !!


问:
Silverman's theorem 3.1 on Weierstrass p function convergence.
In the Arithmetic of Elliptic curves, VI.2 Theorem 3.1 (a), Silverman states the Eisenstein series $G_{2k}(\Lambda) = \sum_{\omega \in \Lambda \backslash 0} \frac{1}{\omega^{2k}}$ converges absolutely for all $k > 1$. In Theorem 3.1(b) he says part (a) implies that the Weierstrass $\wp$ function is also absolutely convergent. But in the proof of this he only bounds each term of the Weierstrass function by the following:
$$ \left | \frac{1}{(z-\omega)^2} - \frac{1}{\omega^2} \right | \leq \frac{10 |z| }{ |\omega|^2}$$ (where $\omega$ is the lattice point). Isn't this insufficient because $G_{2k}(\Lambda)$ does not converge for $k=1$? I see other sources (e.g. https://math.berkeley.edu/~dcorwin/math185S20/SS-elliptic.pdf) bound each term by $O(\omega^{-3})$, and that argument I do follow.

Am I missing something here in Silverman's proof?

答: This mistake is noted on p. 18 of the errata on Silverman's webpage.</p>


答:
Facts:
$$sin(a+b) = sinacosb+cosasinb$$
$$\begin{align}
\sin{ix}&=\frac{1}{2i}\left(e^{i^2x}-e^{-i^2x}\right)\\
&=\frac{1}{2i}\left(e^{-x}-e^x\right)\\
\end{align}$$

$$\begin{align}
\cos{ix}&=\frac{1}{2}\left(e^{i^2x}-e^{-i^2x}\right)\\
&=\frac{1}{2}\left(e^{-x}-e^x\right)\\
\end{align}$$

NOW:

$sin(z) = sin(x+iy) = sin(x)cos(iy)+cos(x)sin(iy) = sin(x)\frac{1}{2}\left(e^{-y}-e^y\right)+cos(x)\frac{1}{2i}\left(e^{-y}-e^y\right)$

$sin( \bar {z}) = sin(x-iy) = sin(x)cos(iy)-cos(x)sin(iy) = sin(x)\frac{1}{2}\left(e^{-y}-e^y\right)-cos(x)\frac{1}{2i}\left(e^{-y}-e^y\right)$

you can take it from here, i.e. find $\bar{sin(z)}$

Title: Show sin($\bar{z}$) = $\overline{\sin(z)}$

Body:
Let $z = x + iy$ where $x$ and $y$ are real. Show that

$$\sin(\bar{z})=\overline{\sin(z)}.$$

I started by trying to take the conjugate of $z$ which was ok but I didn't know how to take the conjugate of $\sin(z)$.


问:
Finding the number of integer solutions, why is this wrong?
The question is to find the number of solutions such that $(x, y)$ are integers: $(x-8)(x-10)=2^y$. Here's what I did: $u(u-2)=2^y$. From the quadratic formula, $u=1+\sqrt{1+2^y}$. This is where I think this question was meant to be solved by modular arithmetic here, but I don't know it so please try not to use it in your answers. I now constructed two right angle triangles. Triangle 1 has sides $1$, $2^{y/2}$,a hypotenuse of $\sqrt{1+2^y}$, and an angle $t$ such that $\sin(t)=\dfrac {1}{\sqrt{1+2^y}}$. Triangle 2 has sides of  $1$, $w$, a hypotenuse of $u$, and an angle $t$ such that $\cos t=\dfrac{1}{u}$. Now we want $\cot(t)+\dfrac1u=1$, or $\cot(t)+\cos(t)=1$. The problem now just reduces to finding the number solutions to that equation. However, this gives me an infinite number of solutions. Why is this wrong? Thanks!

答:
Hint:

$(x-8)(x-10)=2^y \implies x-8=2^m$ and $x-10=2^k$. Only possible way such that both are powers of two and differ by $2$?

$2^m-2^k=2$

$2(2^{m-1}-2^{k-1})=2 \implies 2^{m-1}-1=1 \implies 2^{m-1}=2$, $m=2$ and $k=1$


问:
Complex Analysis Definition
Link to textbook: https://math.unice.fr/~nivoche/pdf/Brown-Churchill-Complex%20Variables%20and%20Application%208th%20edition.pdf

For statement 3, I am not sure why the RHS is equivalent to the LHS. What would $-arg(z_{2})$ indicate in this instance if

$z_{2}^{-1} = \frac{1}{r_2}e^{-i\theta_2}$

I think $-\arg(z_{2})$ is:

$-r_2e^{-i\theta_2}$

答: The expression $\operatorname{arg}(z)$ denotes an argument of $z$. So, if$$z_2=r_2e^{i\theta_2},\tag1$$you can write $\theta_2=\operatorname{arg}(z_2)$ since $\theta_2$ is an argument of $z_2$. On the other hand, it follows from $(1)$ that $\frac1{z_2}=\frac1{r_2}e^{-i\theta_2}$ and that therefore $-\theta_2$ is an argument of $\frac1{z_2}$; in other words, $-\theta_2=\operatorname{arg}\left(\frac1{z_2}\right)$.

Title: Complex Analysis Definition


问:
Prove that if $f$ is a continuous map which maps every open set to an open set then $f$ is monotonic.
My attempt:If the function is not monotonous then we can consider points $a < b$ and $b <c$ such that $f(a)<f(b)$ and $f(b)>f(c)$.My intuition behind this was to consider a subset $[a,c]$ of the domain where we can consider two intervals $[a,b]$ and $[b,c]$ such that in the first interval the function is montonically increasing and in the second interval it is monotonically decreasing . We can get such an element $b$ in the domain because the function is not monotone.

Since the function maps every open interval into an open interval then we can consider $(a,b)$ as the first open set which is being mapped to an open set $(f(a),f(b))$ and another open set $(b,c)$ such that $(f(b),f(c))$.[In the first case I am considering the open sets as the open interval]

case $1$: $f(a)<f(b)$ and $f(b)>f(c)$

subcase $i$ : $f(a)<f(c)$ then the interval $(a,c)$ will be mapped to the interval $(f(a),f(b)) \cup (f(b),f(c))$.Also , we see that since $f(c)$ is contained in the interval $(f(a),f(b))$ .Now, I am trying to arrive at a contradiction by concluding that since $b \in (a,c)$ so it is seen that the image of $f$ becomes $(f(a),f(c)) \cup (f(c),f(b)]$ where $f(b)$ is not an interior point of the image of $f((a,c))$ .

subcase $ii$:$f(c)<f(a)$ then the interval $(a,c)$ will be mapped to the interval $(f(c),f(a)) \cup (f(a),f(b)]$.So we can arrive at the same contradiction.

case$2$:We can assume that the function is continuous in the interval $[a,b]$ and $[b,c]$.

Now, we try to consider the case where $(f(a),f(b))$ is the union of countable disjoint intervals.

case $1$: We try to proceed with same approach that $f(a)<f(b)$ and $f(b)>f(c)$ .Then the $(f(a),f(b)) = (f(a),f(a_1)) \cup .........\cup (f(a_k),f(b)) $ .Also,  $(f(b),f(c))= (f(b),f(b_1)) \cup ...\cup (f(b_k),f(c))$. $f(a)<f(b)$ and also $f(b)>f(c)$ .Then,

subcase $1$:$f(a)<f(c)$ then the interval $(a,c)$ will be mapped to the all the disjoint  intervals between $(f(a),f(c)) \cup (f(c),f(b)]$ .Now, I can conclude by the same logic.

This is the summary of how I want to proceed with the sum but I needed some suggestions on where I am gong wrong and what more cases I am missing out in the proceeed.Hints will really be appreciated instead of complete answers.

答:
Without further hypothesis, this is wrong. It exists real functions that map any open interval onto $\mathbb R$. Those maps are obviously not monotonic.

This is however true for continuous maps. The proof can be performed by contradiction as you did. It can be simplified using the extreme value theorem.

If $f$ is not constant on $(a,b)$, $f$ attains its extreme values on that interval. By changing $f$ into $-f$ if necessary, we can suppose that the maximum of $f$ is attained at $a\lt c \lt b$. But then $f[(a,b)]$ is an interval that is not open: it contains its upper value.


问:
Visualization of length and orthogonality under non-standard inner product
In  under the standard inner product, the length of

\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}

is 1. However, under the innner product

where a>0, the length becomes

It is obvious to me that under the standard inner product the unit vector has length of 1; also I know the above calculation of the length under the non-standard inner product is following the definition of length. However, how would one visualize the length of the unit vector being

under the non-standard inner product defined above ?

Similarly, I have problem visualizing the fact that

\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}

and

\begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix}

are not orthogonal to each other under the non-standard inner product.

答:
While vectors under norm generated by standard inner product can be visualized on $n$ dimensional sphere, the non-standard products leads to $n$ dimensional ellipsoids.

You can rewrite the inner product as $x^TAy$ where $A$ is positive definite matrix. For setting $A = I$ you have standard inner product. An expression $x^TAy$ is so-called quadratic form which can be visualised as an ellipsoid (under assumption that $A$ is positive definite which is also necessary to fulfil axioms of inner product).

In your example (I switched to 2D only but this can be generalized for $n$ dimensions) vector $\begin{pmatrix} 1 & 0  \end {pmatrix}$ is unit vector which lays on x axis, vector $\begin{pmatrix} 0 & 1 \end {pmatrix}$ layes on y axis and ends of these vectors lays on unit sphere with centre in point (0,0). Under your non-standard inner product, vector $\begin{pmatrix} 1 & 0  \end {pmatrix}$ lays again on x axis and also on major axis of ellipse, vector $\begin{pmatrix} 0 & 1  \end {pmatrix}$ lays again on y axis and also on minor axis of ellipse. The ellipse is centered in point (0,0). A length of the ellipse major axis is $\sqrt{2}$ and minor axis has length 1.


答:
Remember that $$\sin(x+iy)=\cosh(y)\sin(x)+i\sinh(y)\cos(x)$$Now if we take the conjugate, we get that $$\sin(x-iy)=\cosh(-y)\sin(x)+i\sinh(-y)\cos(x)$$As $\sinh(-y)=-\sinh(y)$ and $\cosh(-y)=\cosh(y)$, we can get $$\sin(x-iy)=\cosh(y)\sin(x)-i\sinh(y)\cos(x)$$
which you can see to be the conjugate of $\sin(x+iy)$


答:
Recall that $1-x^3=(1-x)(1+x+x^2)$, so
\begin{align*}
\frac1{1-x}-\frac3{1-x^3} &= \frac1{1-x}\left[1-\frac3{1+x+x^2}\right] = \frac1{1-x}\cdot\frac{1+x+x^2-3}{1+x+x^2} \\ &= \frac1{1-x}\cdot\frac{(x+2)(x-1)}{1+x+x^2} =
\frac{x-1}{1-x}\cdot\frac{x+2}{1+x+x^2} = -\frac{x+2}{1+x+x^2}
\,.
\end{align*}
From this we see that $\lim\limits_{x\to 1} = -1$.

P.S. You have it from where you were if you use the factoring on my first line to simplify your fraction.


问:
Counting all possible arrangements
I try to solve the following counting problem:

We consider a new type of gamble. In this one there are 7 red balls, 5 blue balls and 3 purple balls. The presenter now rolls a 4-sided die for each red ball, an 8-sided die for each blue ball, and a 12-sided die for each purple ball, and writes the result on the ball. The 15 balls are then placed together in a box. How many possible arrangements of the balls can be created in this way when drawing? Assume that the balls cannot be distinguished by handwriting.

What I did so far:
First, I counted the number of possibilities in which the balls can be numbered. I think there are $12^3\cdot 8^5\cdot 4^7$ possibilities. Now it only remains to count the possible arrangements of the balls.

If any two balls would always be distinguishable, we would have $15!$ possibilities. Thus the answer would be $n:=15!\cdot 12^3\cdot 8^5\cdot 4^7$. But unfortunately it may happen that the balls are indistinguishable (when their color and number coincide). So the answer must be less than $n$, because we effectively counted some events multiple times. But in my head everything becomes so foggy at this point. How to deal with those cases?

答:
Maybe this: Let's handle each color separately-
Look on a random permutation of the $15$ balls we have:
3 balls are going to be purple, 5 are going to be blue and 7 are going to be red,
hence there are
$${15\choose3}{12\choose5}$$
possibilities to select places for the balls, by color.
Then, you would like to use
$$Select(n,k) ={k+n-1 \choose k}$$ that is the number to choose $k$ items out of $n$ with the possibility to choose again, and with no importance of order
That gives us the number of possibilities to select each group of balls, and now we only need to plug everything together to get:
$${15\choose3}{12\choose5}\cdot{3+12-1\choose3}{5+8-1\choose5}{4+7-1\choose4} \\
={15 \choose3}{12\choose5}\cdot{14\choose3}{12\choose5}{10\choose4}$$

Title: Counting all possible arrangements


问:
Statistics problem about computing UMVUE(Uniformly Minimum Variance Unbiased Estimator)
Let $\underline{X}$ ~ Be(p) of size n. Obtain UMVUE for p.

My idea is the MLE of that problem is the UMVUE but i think its wrong. Can someone help me?

答:
There are several ways to show that $T=\overline{X}_n$ is the UMVUE for $p$

This is one; I chose this because it allow us to do further considerations about that important estimator:

$$\mathbb{E}[\overline{X}_n]=p$$

Thus $T$ is  unbiased and its variance is

$$\mathbb{V}[\overline{X}_n]=\frac{p(1-p)}{n}$$

Now let's calculate the lower bound for the variance of unbiased estimator for $p$ using Cramér-Rao inequality;

$$\mathbb{V}[T]\geq \frac{1}{-n\mathbb{E}\left\{\frac{\partial^2}{\partial p^2}\log f(x|p)  \right\}}=\dots=\frac{p(1-p)}{n}$$

Thus $T=\overline{X}_n$ is UMVUE for $p$ AND its variance also attains the Cramér Rao lower bound....and this does not happen so frequently among UMVU Estimators

here are the calculations for CR lower bound, as requested.

$$f=p^x(1-p)^{1-x}$$

$$\log f=x\log p+(1-x)\log(1-p)$$

$$\frac{\partial}{\partial p}\log f=\frac{x}{p}-\frac{1-x}{1-p}$$

$$\frac{\partial^2}{\partial p^2}\log f=-\frac{x}{p^2}+\frac{1-x}{(1-p)^2}$$

Now observe that $\mathbb{E}[X]=p$ and $\mathbb{E}[1-X]=1-p$ thus substitute and immediately conclude


问:
Find the number of roots of $z^2 - \cos z=0$ for $|z| < 2$ using Rouche's theorem

Find the number of roots for the following equation for $|z| < 2$, $z\in \Bbb C$:
$$
z^2 - \cos z=0
$$

The reasoning below is based on using Rouche's theorem. So basically I picked two functions $f(z)$ and $\phi(z)$ such that:
$$
f(z) = z^2\\
\phi(z) = \cos z
$$

Now I need to show that $|f(z)| > |\phi(z)|$ for $|z| = 2$. My main issue here is proving that statement.

Consider $|z^2|$, clearly the absolute value is $4$. For $|\cos z|$ and $y\in[-2,2]$:
$$
\begin{align}
|\cos z| &= \left|\frac{e^{iz} + e^{-iz}}{2}\right| \\
&= {1\over 2}\left|e^{ix - y} + e^{-ix+y}\right| \\
&\le {1\over 2}\left(\left|e^{ix}\right|\left|e^{-y}\right| + \left|e^{-ix}\right|\left|e^{y}\right|\right)\\
&={1\over 2}\left(e^{-y}+e^y\right)
\end{align}
$$

Now $|\cos z|$ is symmetric with respect to $x = 0$ so we might consider only one case: $y \in [0;2]$. Take a look at the following equation:
$$
\begin{align}
{1\over 2}\left(e^{-y} + e^y\right) &= 4 \\
e^{-y} + e^y &= 8 \ \ | \times e^y \\
e^{2y} - 8e^y + 1 &= 0
\end{align}
$$
Solve for $e^y$:
$$
e^y = 4\pm \sqrt{15}
$$

Approximate calculations show that:
$$
\begin{align}
y = \ln(4+\sqrt{15}) &\approx 2,06343... > 2\\
y = \ln(4-\sqrt{15}) &\approx -2,06343... < -2
\end{align}
$$

Finally since ${1\over 2}\left(e^{-y}+e^y\right)$ is incresing for $y \in [0,2]$ we have that:
$$
|\cos z| = {1\over 2}\left(e^{-y}+e^y\right) < 4, \forall y\in[0,2]
$$

This means the equation has two roots (with multiplicities) in $|z| < 2$.

The question here is how do I show
$$
\ln(4+\sqrt{15}) > 2\\
\ln(4-\sqrt{15}) < -2
$$
Also, I would appreciate it if someone could show a simpler solution.

答:
You already figured out that for $z=x+iy$
$$
 |\cos(z)| \le \cosh(y)
$$
and that $\cosh(y)$ is increasing in $y$ for $y \ge 0$. Therefore
$$
 |\cos(z)| \le  \cosh(2) = \frac{e^2+e^{-2}}{2}
$$
for $|z| =2$, and it suffices to show that $e^2+e^{-2} < 8$:
$$
e^2+e^{-2} < 2.8^2 + \frac 1{2.5^2} = 7.84 + 0.16 = 8 \, .
$$

It follows that $z^2 - \cos(z)$ and $z^2$ have the same number of zeros in $|z| < 2$ (counting multiplicities), that is $2$.


问:
Showing that the set $\mathbb{R^2} \setminus \{(n,0) \mid n \in \mathbb{N} \}$ is an open subset of the plane.

For every $n \in \mathbb{N}$ denote $x_n=(n,0) \in \mathbb{R^2}.$ Show that the set $\mathbb{R^2} \setminus \{x_n \mid n \in \mathbb{N} \}$ is an open subset of the plane.

The set $\mathbb{R^2} \setminus \{x_n \mid n \in \mathbb{N} \}$ is the plane excluding the positive $x-$axis? It seems that I cannot use the definition of a ball here to conclude that the set would be open. Other definition that I know states that the union of open sets is open also, but it doesn't seem applicable here also. What other definitions might I use here?

答: As $\Bbb N$ and $\{0\}$ are closed in $\Bbb R$, $\Bbb N \times \{0\}$ is closed in $\Bbb R^2$ and so your set, which is the complement of that set, is open.


问:
Confusion on why 2 equivalence classes are either equal or disjoint
So I've just started trying to teach myself some topology and in the book I'm reading there is a proof that 2 equivalence classes are either equal or disjoint.

However I'm a bit confused on why the author just randomly states any 2 equivalence classes surely it should be any 2 equivalence classes defined by the same equivalence relation or is it that any 2 equivalence classes regardless whether or not the equivalence relations on them is equal or not are either equal or disjoint?

Thanks in advance.

答:
Yes, the discussion is about one fixed equivalence relation $R$ on $X$, say.

If two classes $[x]_R$ and $[x']_R$ of $R$ intersect, $xRx'$ (transitivity via the intersection element) and so the classes are actually equal to each other.

We rarely mix equivalence relations; there is usually one under consideration, and we discuss its classes.


问:
Kinematics stone thrown upwards past a point, show the following.
I know I should be able to do this, but I have tried for 3 hours and can't do it. I know its simple but it's driving me mad.

A particle is projected vertically upwards with speed $ u_{0}$ and passes through a point
that is a distance $ h $ above the point of projection at time $t_{1}$ going up and $t_{2}$ coming down. Show that $g t_{1} t_{2} = 2 h$.

I am assuming the time taken for the stone to go from point $h$ up to the max height is equal to the time taken for the stone to fall from the max height down to the point $h$. This time being $\frac{t_{2}-t_{1}}{2}$. I've used the SUVAT equations ... many times ,and the answer won't deliver. Any help appreciated.

答:
The kinetic and potential energies are given by
$$T=\frac 12 m \dot h^2$$
$$V=m\,g\,h$$
By applying Lagrangian mechanics the equation of motion can be found as
$$\ddot h=-g\Rightarrow h=-\frac g2 t^2+a\,t+b$$
By initial conditions $\dot h(0)=u_0$ and $h(0)=0$
$$h=-\frac g2 t^2+u_0\,t$$
It can be described as a quadratic equation
$$\frac g2 t^2-u_0\,t+h=0$$
which has the following roots
$$t_1=\frac {u_0+\sqrt{u_0^2-2\,h\,g}}{g}\qquad t_2=\frac {u_0-\sqrt{u_0^2-2\,h\,g}}{g}$$
and
$$t_1 t_2=\frac {u_0+\sqrt{u_0^2-2\,h\,g}}{g}\frac {u_0-\sqrt{u_0^2-2\,h\,g}}{g}=\frac{u_0^2-u_0^2+2\,h\,g}{g^2}$$
$$t_1\,t_2=\frac{2h}g\Rightarrow g\,t_1\,t_2=2h$$</p>


问:
closed immersion in $\mathbb{A}_A^n$
Let $A$ be a ring, $X$ an $A$-scheme. The scheme $X$ is told projective if there is some closed immersion $X\to\mathbb{P}_A^n$ over $\mathrm{Spec}(A)$. For proper schemes these embeddings are described by the ample invertible sheaves which gives multiple possible embeddings. Why is there not equivalent definition for the affine case?

I mean for $A$-scheme with some closed immersion $X\to\mathbb{A}_A^n
$. I guess for this case it's trivial but can't see why (or the affines made it the things more difficult?). Can't wee have multiple possible closed immersions? These immersions give $X\simeq\mathrm{Spec}(A[x_1,\ldots,x_n]/I)$ so we should have $A[x_1,\ldots,x_n]/I\simeq A[x_1,\ldots,x_k]/J$. We cannot conclude that $n=k$ and $I=J$ because we can add superfluous variables $A[x_1,\ldots,x_n]/I\simeq A[x_1,\ldots,x_{n+1}]/(I+(x_{n+1}))$ but appart from that is there some easy description?

答: Since any closed subscheme of an affine scheme is affine, a scheme which admits a closed embedding in to $\Bbb A^n_A$ is again affine. Thus your $X$ is the spectrum of some ring $R$, and the closed immersion $X\to \Bbb A^n_A$ is given by a surjective morphism of rings $A[x_1,\cdots,x_n]\to R$. This means that closed immersions in to affine space over $A$ are in bijection with (ordered) generating sets of $R$ as an $A$-algebra.

Title: closed immersion in $\mathbb{A}_A^n$


问:
How to conclude truth value of elements from true propositions
suppose that these propositions are true  $\qquad \neg (\neg d \lor a) \ , \quad (\neg a \land c) \lor d \ , \quad a \lor \neg b \ , \quad c \rightarrow b $

What are the value(s) of Truth of a, b, c and d?

I got an exercice like that and I'm not sure if my method is correct mainly because the question states that an element could have more than one value of truth

Answer...  $a$: False, $b$: False, $c$: False, $d$: True

We have: $ \neg (\neg d \lor a)  \Leftrightarrow d \land \neg a $ (Morgan law)

Therefore $a$ is strictely false and $d$ is  strictly True so that the $\neg (\neg d \lor a)$ proposition is True.

Also in: $a\lor \neg b$
We conclude that $\neg b$ is True so $b$ is False (knowing that $a$ is False) so that the proposition $a \lor \neg b$ is True.

In: $c \rightarrow b$.
We have $b$ False, so for the $c \rightarrow b$ proposal to be True $c$ must be strictly False to satisfy the proposal.

The statement $(\neg a \land c) \lor d$ is also True with all our results.

答:
Your reasoning is quite correct.

That evaluation of: $a$, $b$, and $c$ are false, and $d$ it true, satisfies all of the propositions.

Now, while the question says there could be more answers, ... in actuality that is the only one.


问:
Derive the Bayes estimator under the loss > $L(\theta,\delta)=\frac{(\theta-\delta)^2}{\theta^k}$

Let $X_i,...,X_n$ be iid $Poisson(\theta)$ with $Gamma(a, \frac 1 b)$
prior and a posterior $Gamma\left(a+n\bar x,\frac{1+nb}{b}\right)$.
Derive the Bayes estimator under the loss
$L(\theta,\delta)=\frac{(\theta-\delta)^2}{\theta^k}$ for some $k>0$,
also assume $a-k>0$.

I think it has to do with finding a function $\delta*$ that minimizes $E\left(\frac{(\theta-\delta)^2}{\theta^k}\right)$, but not sure how to go about it. Is $\delta$ distributed as the posterior Gamma, or are those separate things?

I know the Bayes estimator when the loss is $L(\theta,\delta)=(\theta-\delta)^2$ is the mean of the posterior distribution. But I don't know what to do when it's divided by $\theta^k$.

答:
Your loss function IS quadratic in $\delta=\hat{\theta}$, the only difference is that the loss function is

$$l(\theta;\hat{\theta})=C(\theta)(\theta-\hat{\theta})^2$$

thus doing the same passages you surely know to prove that with a quadratic loss function Bayes MMSE is the posterior mean you will easy find that in your case you have that your MMSE is

$$\hat{\theta}=\frac{\mathbb{E}[\theta C(\theta)|\mathbf{x}]}{\mathbb{E}[C(\theta)|\mathbf{x}]}$$

Using a Gamma posterior it is not difficutl to to the calculations. but observe that when posting Gamma distributions without writing their density you must explicitate if one of the two parameter is the "scale" or "rate" parameter or saying something about their expectations...oherwise we cannot do the calculations.

Obseve that if $k=1$ then

$$\hat{\theta}_{\text{MMSE}}=\frac{1}{\mathbb{E}\left[\frac{1}{\theta}|\mathbf{x}\right]}$$

which is the armonic mean of the posterior distribution


问:
Constructing a point D on AB such that PD is perpendicular to AB
I need to write the steps of construction from this exercise.

Construct a line segment $AB$, and a point $P$, not on $AB$. Use the theorem on the previous page or one of its alternative statements to construct a point $D$ on $AB$ such that $PD ⊥ AB$. Write down your steps of construction, and explain how you used the theorem to do the construction.

This is the previous theorem: In an isosceles triangle, the median to the base is the perpendicular bisector (of the base) and also the angle bisector for the angle opposite to the base.

So far I have learned how to create an equilateral triangle with side $AB$ by using the intersections of circle $(A,AB)$ and circle $(B,BA)$.

I am unsure how to start this. I think I need to somehow create a isosceles triangle that would be triangle $PAB$ but not sure how to construct it?

答:
Draw a line $\mathcal{L}_a$ through the points $A$ and $B$.

Draw a line $\mathcal{L}_p$ through the points $A$ and $P$.

Draw a circle $\mathcal{C}_a(A,|AP|)$.

Draw a circle $\mathcal{C}_p(P,|AP|)$.
Call the intersection points of
$\mathcal{C}_a$ and $\mathcal{C}_p$ as $E$ and $F$.

Draw a line $\mathcal{L}_o$ through the points $E$ and $F$.
Let $O=\mathcal{L}_a\cap\mathcal{L}_o$.

Draw a circle $\mathcal{C}_o(O,|AO|)$.
The sought point is $D=\mathcal{C}_o\cap\mathcal{L}_a$.

Steps 2-5 are used to find the midpoint $O$ of $AP$, so
that $|OA|=|OP|=\tfrac12|AP|$ is the diameter of $\mathcal{C}_o$.


问:
How to prove this lower bound on $\log(1+x)$ for $x \geq 0$? How about $x > -1$?
I want to show that for all real numbers $x \geq 0$:

$$
\log(1+x) \geq \frac{x(5x+6)}{2(x+3)(x+1)}.
$$

I'd like to understand each step necessary to prove this so that I can apply it to future problems. Amazingly, a popular online calculator immediately spits out that this inequality is true for all $x > -1$. Is there an easy way to prove this for $x \geq 0$? How about $x >-1$?

Attempt (and more info on how I'm stuck):
My guess is to use the Taylor series expansion of both sides at $x=0$ (but why is that sufficient for all $x \geq 0$?). Expanding both sides:

$$
\log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots \geq x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{5x^4}{18} + \frac{7x^5}{27} - \ldots = \frac{x(5x+6)}{2(x+3)(x+1)}.
$$

I see that the early order terms match and then they don't. Not sure how to proceed from there. The signs are alternating on both sides, which I'm not sure how to deal with. Though, it seems the coefficients are converging to $0$ on both sides. It also seems the magnitude of the coefficients on the right-hand side is larger than the left-hand side (if you go term-wise), but how do I prove that? And do I need that fact to prove the inequality?

答:
Tbe problem looks complex  and encourages us to use approximations and taylor series but its infact very easy

Consider $$f(x)=\ln(1+x)-\frac{x(5x+6)}{2(x+3)(x+1)}$$ Indeed (after some hardwork)$$f'(x)=\frac{x^3}{{(x+3)}^2{(x+1)}^2}\ge 0$$

Thus $$f(x)\ge f(0)=0$$ Done


问:
Set of prime numbers $q$ such that $\sum\limits_{p\leq\sqrt{q}}p=\pi(q)$
If we define the sum of consecutive prime numbers $p$ up to the square root of a given positive integer $n$ as $S(n)=\sum\limits_{p\leq\sqrt{n}}p$, and the prime counting function up to a given positive integer $n$ as $\pi(n)=\#\left\{p\leq n\right\}$, it can be showed that $S(n)\sim\pi(n)$ (for a proof, you can check http://vixra.org/abs/1911.0316). This led me to study the positive integers for which $S(n)=\pi(n)$.

If we define set $Q$ as the set of values of $n$ such that $\pi\left(n\right)=S\left(n\right)$, the first values are $Q=\{11,12,29,30,59,60,179,180,...\}$, for $\pi(n)=5,5,10,10,17,17,41,41,...$

It can be conjectured that the first value of $n$ with a concrete $\pi\left(n\right)=S\left(n\right)$ will be always a prime number. This conjecture is a corollary of the following

Conjecture 1.

It does not exist any squared prime number $p^{2}$ such that $\pi\left(p^{2}\right)=S\left(p^{2}\right)$ except of $p_{1}=2$. That is, $$\pi\left(p_{n}^{2}\right)\neq\sum_{k=1}^{n}p_{k}$$

If the Conjecture were false, we could have that, being $p_m$ and $p_{m+1}$ prime numbers such that $p_m<p_n^2<p_{m+1}$, then $S\left(p_{m}\right)=S\left(p_{m}+1\right)=S\left(p_{m}+2\right)=...=S\left(p_n^{2}-1\right)=S\left(p_n^{2}\right)-p_n$; and as  $\pi\left(p_n^{2}\right)=S\left(p_n^{2}\right)$, then $p_n^{2}\in Q$,and $p_n^{2}$ would be the first of a series of consecutive elements of $Q$ until $p_{m+1}$.

The conjecture has been tested and found to be true for the first thousands of primes.

If we define set $M$ as the set of values of $n$ such that $\pi\left(n\right)=S\left(n\right)$ and $n$ is some prime number, we find that $M=\{11,29,59,179,389,541,...\}$. You can see more terms in https://oeis.org/A329403. And if we define $p_{k}$ as the last prime number which is a term of $S\left(n\right)$, we can see that the first values of $k$ are $2,3,4,6,8,9,...$, which led me to formulate the following

Conjecture 2

Set $M$ has infinitely many elements

If the conjecture were true, as $M\subset Q$, the Conjecture implies that $\pi\left(n\right)$ intersects $S\left(n\right)$ infinitely many times, so $S\left(n\right)$ is not only asymptotically equivalent to $\pi\left(n\right)$: it is infinitely many times equal to $\pi\left(n\right)$. And it would show that the number of primes between $p_{n}^{2}$ and $p_{n+1}^{2}$ , on average, do not differ much from $p_{n+1}$.

Questions

(1) Any idea of how could be proved Conjecture 1, if it is already proved? Or maybe, with more computational power than mine, if anyone could verify if the conjecture supports itself also for big numbers, it would be great.

(2) With my computational power, I was able only to identify the first 19 prime numbers and $k$ values of the set (last one was $1756397$, for $k=216$). If anyone could calculate more of them to see if the conjecture supports itself also for big numbers, it would be great.

(3) Any idea of how to show that Conjecture 2 is true? I guess that it could be done with some theorem of the kind of Littlewood's oscillation theorem, but as an amateur mathematician obtaining such a result is beyond my actual knowledge.

Thanks in advance!

答: Conjecture 2 has finally been proved as correct here , thanks to GH from MO. Conjecture 1 still remains open.</p>


问:
Can't tell if the function is differentiable at a point?
I was trying to evaluate if the below function is differentiable at $(0,0),$

$$f(x,y) = \begin{cases}\frac{x^3}{x^2+y^2} & \textrm{for} (x,y) \ne (0,0)\\ 0 &\textrm{at }(x,y) = (0,0)\end{cases}$$

and what I did was that I used the definition for differentiability of a function in two variables which gives,

$$f'(0,0) = \lim_{(h,k)\to (0,0)} (-hk^2/\sqrt{h^2+k^2})$$

and here, I assumed $h = r\cos\theta$ and $k =r\sin\theta$ and limit $ r \to 0$, which gives,

$$f'(0,0) = \lim_{r\to 0} (-\sin^2\theta \cos\theta),$$

And thus, we can see that the limit is a finite value, and therefore, the limit exists, and the function is differentiable at $(0,0),$ but it is given otherwise in the answer sheet.

Please help me solve this.

答: The limit has to be the same for all $\theta$ for the derivative to exist. Since the limit in this case varies with $\theta$ it follows that $f$ is not differentiable at $(0,0)$.


问:
Another pigeonhole principle question
Have another question for you today:

A course has seven elective topics, and students must complete exactly three of them in order to pass the course. If 200 students passed the course, show that at least 6 of them must have completed the same electives as each other.

Now I know this is related to counting and the pigeonhole principle, and there are a couple of other related questions already asked but I couldn't apply them to my quuestion.

I know that the (informal) pigeonhole principle states that if you have $n$ boxes, and you have more than $n$ pigeons to distribute between those boxes, then at least one of the boxes will contain more than one pigeons, but I'm not sure what my boxes and what my pigeons are in this problem

答:
Often it's useful in pigeonhole-type problems to try to make them not work, and when you see that you cannot, you often see why you cannot.

For example, if I say there are 2 electives and 9 students, can you see that at least 5 students must select the same elective?  You can, because if you try to avoid putting 5 students in an elective, you end up with 4 in each and 1 left over.

Now apply this same approach to your problem.  First ask what are the "pigeons" and what are the "holes"?  The trick here is that you need to show that at least 6 students take exactly the same set of electives.  How many ways can you enumerate 3 from 7?  Those are your holes.  Now you have 200 pigeons to place.  Go.

Title: Another pigeonhole principle question


答:
You must keep in mind that, if $f'(0,0)$ exists, then it is a linear map, not a vector.

You have, for each $x\in\Bbb R$, that $f(x,0)=x$, and therefore $\frac{\partial f}{\partial x}(0,0)=1$. By a similar argument, $\frac{\partial f}{\partial y}(0,0)=0$. Therefore, if $f$ was differentiable at $(0,0)$, $f'(0,0)(x,y)=x$. That means that$$\lim_{(x,y)\to(0,0)}\frac{f(x,y)-x}{\sqrt{x^2+y^2}}=0.$$In other words,$$\lim_{(x,y)\to(0,0)}\frac{xy^2}{(x^2+y^2)^{3/2}}=0.$$However, this is not true, because, if $x=y>0$, $\frac{xy^2}{(x^2+y^2)^{3/2}}=1$.

Title: Can't tell if the function is differentiable at a point?

Body:
I was trying to evaluate if the below function is differentiable at $(0,0),$

$$f(x,y) = \begin{cases}\frac{x^3}{x^2+y^2} & \textrm{for} (x,y) \ne (0,0)\\ 0 &\textrm{at }(x,y) = (0,0)\end{cases}$$

and what I did was that I used the definition for differentiability of a function in two variables which gives,

$$f'(0,0) = \lim_{(h,k)\to (0,0)} (-hk^2/\sqrt{h^2+k^2})$$

and here, I assumed $h = r\cos\theta$ and $k =r\sin\theta$ and limit $ r \to 0$, which gives,

$$f'(0,0) = \lim_{r\to 0} (-\sin^2\theta \cos\theta),$$

And thus, we can see that the limit is a finite value, and therefore, the limit exists, and the function is differentiable at $(0,0),$ but it is given otherwise in the answer sheet.

Please help me solve this.


答:
Basically, the same proof as the great @AlbusDumbledore except (1) it more closely follows @RobertShore's suggestion and (2) it's not as nice. It might be easier to follow step-by-step though.

Let $f(x) = \log(1+x)$ and $g(x) = \frac{x(5x + 6)}{2(x+3)(x+1)}$. We want to show that $f(x) \geq g(x)$ for all $x \geq 0$.
Note that $f(0) = g(0) = 0$. Thus, it suffices to show that $g^\prime(x) \leq f^\prime(x)$ for $x \geq 0$, i.e., that $f$ grows faster than $g$. Also, note that $f^\prime(x) = \frac{1}{1+x}$ and $g^\prime(x) = \frac{7x^2 + 15x + 9}{(x+3)^2(1+x)^2}$. Then,
\begin{align}
&\frac{7x^2 + 15x + 9}{(x+3)^2(1+x)^2} \leq \frac{1}{1+x} \\
&\iff \frac{7x^2 + 15x + 9}{(x+3)^2} \leq 1+x \\
%&\iff \frac{7x^2 + 15x + 9}{(x+3)^2} -1 \leq x \\
&\iff \frac{3x}{(x+3)}  \leq x \\
%&\iff \frac{3x}{(x+3)} - x  \leq 0 \\
&\iff \frac{-x^2}{(x+3)}  \leq 0 \\
&\iff \frac{x^2}{(x+3)}  \geq 0,
\end{align}
which is clearly true for $x \geq 0$.

Title: How to prove this lower bound on $\log(1+x)$ for $x \geq 0$? How about $x > -1$?

Body:
I want to show that for all real numbers $x \geq 0$:

$$
\log(1+x) \geq \frac{x(5x+6)}{2(x+3)(x+1)}.
$$

I'd like to understand each step necessary to prove this so that I can apply it to future problems. Amazingly, a popular online calculator immediately spits out that this inequality is true for all $x > -1$. Is there an easy way to prove this for $x \geq 0$? How about $x >-1$?

Attempt (and more info on how I'm stuck):
My guess is to use the Taylor series expansion of both sides at $x=0$ (but why is that sufficient for all $x \geq 0$?). Expanding both sides:

$$
\log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots \geq x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{5x^4}{18} + \frac{7x^5}{27} - \ldots = \frac{x(5x+6)}{2(x+3)(x+1)}.
$$

I see that the early order terms match and then they don't. Not sure how to proceed from there. The signs are alternating on both sides, which I'm not sure how to deal with. Though, it seems the coefficients are converging to $0$ on both sides. It also seems the magnitude of the coefficients on the right-hand side is larger than the left-hand side (if you go term-wise), but how do I prove that? And do I need that fact to prove the inequality?


问:
Calculation of probability for random variable
The number of the daily car accidents in Luna-land is a random variable with mean 50 and standard deviation 5. Find the probability for the total accidents in the following 25 days to be $\leq 1300$.

I am only familiar with combinatorics, not distributions.
I did a bit of reading around and found that probably this must be approached by the normal distribution.
So we want $P (X\leq 1300)$?
(this is not homework!)

Thank you!

EDIT: I found the formula for the central limit theorem.

So, it seems I must calculate $\frac {(1300-25*50)}{\frac{5}{\sqrt25}}$?

答:
Suppose $X_i$ denotes the number of accidents on day $i$ and suppose you have $\mathbb{E}[X_i]=\mu$ and $\text{Var}(X_i)=\sigma^2<\infty$. What you are interested in is $S_{25}=\sum_{i=1}^{25}X_i$

By the CLT you have that $\frac{S_{25}-25\mu}{\sqrt{25}\sigma}$ has approximately $\mathcal{N}(0,1)$ distribution (of course, if you are willing to assume that $25$ is a reasonably large enough number for your purposes) so you have that

$$\mathbb{P}(S_{25}\leq1300)=\mathbb{P}\Big(\frac{S_{25}-25\mu}{\sqrt{25}\sigma} \leq \frac{1300-25\mu}{\sqrt{25}\sigma}\Big)=\mathbb{P}\Big(Z \leq \frac{1300-25\mu}{\sqrt{25}\sigma}\Big)$$

for $Z\sim\mathcal{N}(0,1)$. What's left is to plug in the values given and use software to get the exact probability.

If you do that, you end up with the expression $\mathbb{P}(Z \leq 2)$ which should be approximately $0.975$.


答:
Assuming that you can use CLT (25 is a little borderline) or Assuming normality in the daily car accidents, the correct way to calculate this probability is

$$\mathbb{P}[X\leq 1300]=\mathbb{P}\left[Z\leq \frac{1300-1250}{\sqrt{625}}\right]=\Phi(2)=97.725\%$$

Title: Calculation of probability for random variable

Body:
The number of the daily car accidents in Luna-land is a random variable with mean 50 and standard deviation 5. Find the probability for the total accidents in the following 25 days to be $\leq 1300$.

I am only familiar with combinatorics, not distributions.
I did a bit of reading around and found that probably this must be approached by the normal distribution.
So we want $P (X\leq 1300)$?
(this is not homework!)

Thank you!

EDIT: I found the formula for the central limit theorem.

So, it seems I must calculate $\frac {(1300-25*50)}{\frac{5}{\sqrt25}}$?


问:
How much classical algorithmics and probability theory is necessary for quantum algorithmics?
I would like to learn more about quantum algorithmics in next months/years.

I wonder - do I need to be very very good in "classical" algorithmics to be good in quantum algorithmics, or are they almost "independent" and I don't need more than basics or things I will learn "ad-hoc" when I will be needing them?

The same about probability - does quantum algorithmic uses very advanced probability theory, like more than 1 semester course of probability on Math studies? If Yes, does it use only discrete probability or also continuous probability?

Thanks in advance for answers.
Stay cool!

答:
Well, classical quantum algorithm as the Grover algorithm require knowledge of quantum states (q-bits) and operations on quantum states.

For both, you need complex numbers, linear algebra, tensors, functional calculus (but only for finite dimensional spaces which makes life easier), and basic probability.


问:
What am I doing wrong; evaluating integrals?
$$f(x) = \dfrac{2x}{2-3x^2}$$

$$- \dfrac{1}{3} d (2-3x^2) = 2x$$

$$ \dfrac{2x}{2-3x^2} dx = -\dfrac{1}{3} d (2-3x^2) \cdot \dfrac{1}{2-3x^2} = - \dfrac{1}{3} du \cdot u^{-1} = -\dfrac{1}{3} u^{-1} \cdot du$$

And here I'm stuck, since you can't divide by 0.. What am I doing incorrectly here? I've applied the method my book gave me perfectly as far as I know.

答:
All your steps are correct. But what it seems you're not aware of is that

$$\int{\frac{1}{x}}\space{dx}=\ln|x|+C$$

This is an exception to the power rule.


问:
Understanding proof of 1-Dimension Vector space is multiplication by scalar
This problem is from the book Linear Algebra Done Right. The question is :

If $T: \mathbb F \to \mathbb F$ is linear where $\mathbb F \in
 \{\mathbb R, \mathbb C \}$ then $Tv = \lambda v$ for some scalar.

One solution given in this post is the following:

Let $T:V \to V$ for V a 1-dimensional F-vector space. Let $0 \neq v_0
\epsilon V$, so that $\{v_0\}$ is a basis for V, and hence, in
particular, $Tv_0=\lambda v_0$ for some $\lambda \in F$. Then,
for any $v = \alpha v_0 \in V$,$$Tv=T(\alpha v_0)=\alpha
T(v_0)=\alpha(\lambda v_0)=\lambda(\alpha v_0)=\lambda v,$$so that
$T=\lambda\operatorname{Id}|_V$, as required. Note that the result is entirely
coordinate-independent, even though the proof does require a choice of
basis.

My question is, how this "$Tv_0=\lambda v_0$" step is coming? And if this step is true, then why we are doing rest of the prove?

Thanks!

答: We have $Tv_0 \in V= span\{v_0\}$, thus there is a scalar $ \lambda$ such that $Tv_0= \lambda v_0.$


答:
Since $V$ is $1$-dimensional, then every vector is a multiple of every non-zero vector. So, either $v_0=0$, in which case $T(v_0)=0=\lambda v_0$ for every scalar $\lambda$, or $v_0\ne0$, in which case $T(v_0)$ is a multiple of $v_0$.

But this only proves that, for a vector $v_0$, there is some $\lambda$ such that the equality $T(v_0)=\lambda v_0$ holds for that $v_0$. It still remains to be proved that there is some scalar $\lambda$ such that you have $T(v)=\lambda$ for every vector $v$.

Title: Understanding proof of 1-Dimension Vector space is multiplication by scalar

Body:
This problem is from the book Linear Algebra Done Right. The question is :

If $T: \mathbb F \to \mathbb F$ is linear where $\mathbb F \in
 \{\mathbb R, \mathbb C \}$ then $Tv = \lambda v$ for some scalar.

One solution given in this post is the following:

Let $T:V \to V$ for V a 1-dimensional F-vector space. Let $0 \neq v_0
\epsilon V$, so that $\{v_0\}$ is a basis for V, and hence, in
particular, $Tv_0=\lambda v_0$ for some $\lambda \in F$. Then,
for any $v = \alpha v_0 \in V$,$$Tv=T(\alpha v_0)=\alpha
T(v_0)=\alpha(\lambda v_0)=\lambda(\alpha v_0)=\lambda v,$$so that
$T=\lambda\operatorname{Id}|_V$, as required. Note that the result is entirely
coordinate-independent, even though the proof does require a choice of
basis.

My question is, how this "$Tv_0=\lambda v_0$" step is coming? And if this step is true, then why we are doing rest of the prove?

Thanks!


问:
Continuous injection of $[a,b]$ is strictly monotonic from Zorich's book
I am reading Zorich's book "Analysis I" and ran into the following statement:

Let $f:[a,b]\to \mathbb{R}$ and $f$ is continuous on $[a,b]$. Then $f$ is injective if and only if $f$ is strictly monotonic.

Proof: $\Leftarrow$ is quite obvious.

The most interesting part is $\Rightarrow$. Suppose that $f$ is not strictly monotonic. Then we can find three points $x_1<x_2<x_3$ from $[a,b]$ such that $f(x_2)$ does not lie in between $f(x_1)$ and $f(x_3)$. In this case $f(x_3)$ lies in between $f(x_1)$ and $f(x_2)$ or $f(x_1)$ lies in between $f(x_2)$ and $f(x_3)$.

Question: Suppose $f$ is not strictly monotonic, i.e. $f$ is not strictly increasing and is not strictly decreasing. Then we can find points $x_1,x_2,x_3,x_4\in [a,b]$ such that $x_1<x_2$ with $f(x_1)\geq f(x_2)$ and $x_3<x_4$ with $f(x_3)\leq f(x_4)$.

I have spent smth like couple of hours trying to derive that exists three points with the above properties as in Zorich's book but I failed.

I'd be thankful if you can help me, please!

答: Zorich seemingly uses the definition that $f$ is strictly monotonic that whenever $x_2$ lies between $x_1$ and $x_3$ then $f(x_2)$ lies between $f(x_1)$ and $f(x_3)$.


问:
Totient-like function
I have number written as factors for instance: n = 2 * 3 * 3 * 5. What I have to do is find how many numbers between <1, n) are co-prime to n, which means GCD = 1. It can simply be done using Euler's Totient. But what if GCD = 2 or more? Is there any totient-like function?

UPDATE:

I seeking how many numbers between ai = <1, n) will return GCD(ai,n) = 2. For GCD(ai, n) = 1. It's Euler Totient, what about higher GCD's?

答:
Let $n > 1$, and let $d < n$ be a positive divisor of $n$. You want to count the number of elements of the set
$$
A = \{ a : 0 \le a < n, \gcd(a, n) = d \}.
$$
Note that if $a \in A$, then $\gcd\left(\dfrac{a}{d}, \dfrac{n}{d}\right) = 1$, so $\dfrac{a}{d} \in B$, where
$$
B = \left\{ b : 0 \le b < \frac{n}{d}, \gcd\left(b, \frac{n}{d}\right) = 1 \right\},
$$
and $B$ has $\varphi(n/d)$ elements. Conversely, if $b \in B$, then $b d \in A$, as
$$
\gcd(b d, n) = \gcd \left(b d, \frac{n}{d} d \right) = \gcd \left( b, \frac{n}{d} \right) d = d.
$$

So $A$ has $\varphi(n/d)$ elements.

Title: Totient-like function


问:
Proof-check: Ring Isomorphism of $\Bbb Z[x]/\left< x^3-x\right>$
Claim: $\Bbb Z[x]\ /\left< x^3-x\right> \simeq \Bbb Z[x]/\left< x\right>  \times\  \Bbb Z[x]\ /\left< x^2-1\right> $

My attempt:

Here $\left< x^2-1\right> $ and $\left< x\right> $ are two ideals, where $1$ belongs to $\big(\left< x^2-1\right> +\left< x\right> \big)$, since $1=x^2-(x^2-1)$, so my claim holds true by Chinese Remainder Theorem.

Now I am checking, whether it can be further reducible or not.

Again we can show $\Bbb Z[x]\ /\left< x\right> \simeq \Bbb Z$ by taking the surjective ring homomorphism $\phi\colon \Bbb Z[x]\to \Bbb Z$, such that $\phi(f(x))=f(0)$. Here $\text{ker}(\phi)=\left< x\right>$, so by the first ring isomorphism theorem $\Bbb Z[x]\ /\left< x\right> \simeq \Bbb Z$, and $\Bbb Z[x]\ /\left< x\right>  \times\  \Bbb Z[x]\ /\left< x^2-1\right> \simeq\Bbb Z \times\  \Bbb Z[x]\ /\left< x^2-1\right> $.

Now, since $1$ does not belong to $\big(\left< x-1\right> +\left< x+1\right> \big)$, where factorization of $(x^2-1)=(x+1)\cdot(x-1)$, we can't use CRT further on this or we can't break this quotient ring further.

So my ANSWER (after reducing the claim also) to this ring isomorphism is  $\Bbb Z[x]\ /\left< x^3-x\right> $  $\simeq$  $\Bbb Z \times\  \Bbb Z[x]\ /\left< x^2-1\right> $. Is my explanation correct? Is it the most reduced version in isomorphism(even after reducing the claim)?

答:
Define ring epimorphisms $\pi_1:\Bbb Z[x]\to\Bbb Z$ and $\pi_2:\Bbb Z[x]\to\Bbb Z\times\frac{\Bbb Z[x]}{\langle x^2-1\rangle}$ by
$$\pi_1:f(x)\mapsto f(0)\;\;\;\;\;\pi_2:f(x)\mapsto f(x)+\langle x^2-1\rangle$$ so that $\Psi:\Bbb Z[x]\to\Bbb Z\times\frac{\Bbb Z[x]}{\langle x^2-1\rangle}$ is also a ring homomorphism where $$\Psi:f(x)\mapsto(\pi_1(f(x)),\pi_2(f(x)))$$
$$f(x)\in\ker(\Psi)\iff f(0)=0\;\land\;x^2-1\;|\;f(x)\iff x,x+1,x-1\;|\;f(x)$$
$$\therefore\;f(x)\in\ker(\Psi)\iff x^3-x\;|\;f(x)\;\;\;\;\;\therefore\;\ker(\Psi)=\langle x^3-x\rangle$$
Moreover, $(c,f(x)+\langle x^2-1\rangle)=\Psi(f(x)+(x^2-1)(f(0)-c))$ and thus $\Psi$ is an epimorphism.$$\therefore\;\frac{\Bbb Z[x]}{\langle x^3-x\rangle}\approx\Bbb Z\times\frac{\Bbb Z[x]}{\langle x^2-1\rangle}$$


问:
Advantages of cubic spline interpolation over cubic Hermite interpolation?
So far I have studied cubic Hermite interpolation has a less computational cost and does not have issues like undershoot and overshoot as compared to cubic spline interpolation. My question is what are the advantages of cubic spline interpolation over cubic Hermite interpolation? And in what situation/data application of cubic spline interpolation will be more suitable?</p>

答:
Cubic spline interpolation of a function $f : [a,b] \rightarrow \mathbb{R}$ produces an approximation $p : [a,b] \rightarrow \mathbb{R}$ which is two times differentiable and $p''$ is continuous. This is particularly useful when you need to fool the human eye. In the cinema the audience expects the spaceship's position, velocity and acceleration to wary continuously as it moves across the screen.

Cubic Hermite interpolation produces an approximation $q : [a,b] \rightarrow \mathbb{R}$ that is differentiable with a continuous derivative. Cubic Hermite interpolation is useful when generating continuous output when solving ordinary differential equations $$x'(t) = g(t,x(t)).$$ It uses the local position ($x$) and velocity data ($x'$) that is naturally available in this context and it matches the accuracy delivered by the classical 4th order Runge Kutta method. Cubic Hermite interpolation allows us to rapidly solve event equations of the type $$h(x(t),x'(t)) = 0.$$ This allows us to determine when, say, an artillery shell reaches the apex of its trajectory or impacts the ground.