问:
Square roots of $j$ and $ε$
I know how to find the square root of the imaginary unit $i$, but I'm still learning about split-complex and dual numbers. I can't find any info anywhere about the square roots of $j$ and $ε$, if they have them.

答:
Square root of $j$ is not a split-complex or complex number, it is a tessarine, a combination of complex and split-complex numbers. Actually, there are 4 roots, according the fundamental theorem of tessarine algebra:

$$z_1=\left(\frac{1}{2}+\frac{i}{2}\right)+\left(\frac{1}{2}-\frac{i}{2}\right)j$$

$$z_2=\left(\frac{1}{2}-\frac{i}{2}\right)+\left(\frac{1}{2}+\frac{i}{2}\right)j$$

$$z_3=-\left(\frac{1}{2}+\frac{i}{2}\right)-\left(\frac{1}{2}-\frac{i}{2}\right)j$$

$$z_4=-\left(\frac{1}{2}-\frac{i}{2}\right)-\left(\frac{1}{2}+\frac{i}{2}\right)j$$

You can square these expressions by hand so to verify the roots.

Title: Square roots of $j$ and $ε$


问:
Some homework questions about a Lipschitz function (cauchy sequence)
Do you want to help me with my homework? The exercise is as follows:

Consider a Lipschitz function, $h:\mathbb{R}\rightarrow\mathbb{R}$, satisfying for every $x, y$:
  $$\left| h(x)-h(y) \right| \leq \alpha \left| x-y \right|$$
  with $0 \lt \alpha \lt 1$

Show that there exists at most one $x$, with $h(x)=x$.

Prove that $h$ is uniformly continuous.

Take some $x_1\in\mathbb{R}$, and define inductively the sequence $(x_n)$ as $$ x_{n+1}=h(x_n), \quad  n= 1, 2, \cdots$$ Show that for every $x_1$ the sequence $(x_n)$ is a Cauchy sequence.

Take some $x_1$. Define $x= \lim x_n$. Show that  $x=\lim{x_n}$ satisfies $h(x)=x$

Show (using the first part), that the limits of the sequences $(x_n)$ for all choices of $x_1$ are all the same.

My work until now:

Part 1

Suppose $h(x_1)=x_1, h(x_2)=x_2$, and $x_1 \not = x_2$. Then, following the condition,
$$\left| h(x_1)-h(x_2) \right|= \left| x_1-x_2 \right| \le \alpha \left|x_1-x_2\right|.$$
This means that $\frac{\left|x_1-x_2\right|}{\left|x_1-x_2\right|}=1 \le \alpha$. But $\alpha <1$ was given, so $x_1 = x_2$. There exists at most one $x$ with $h(x)=x$

Part 2

Let $\epsilon>0$ be given and choose $\delta=\frac{\epsilon}{\alpha}$. Then, for any $x, y$ with $\left|x-y\right|<\delta = \frac{\epsilon}{\alpha}$, I have
$$\left| h(x)-h(y) \right| \le \alpha \left|x-y\right| \lt \alpha\left(\frac{\epsilon}{\alpha}\right)=\epsilon,$$
which shows that $f$ is uniformly continuous.

Part 3

I don't know where to start.

Part 4

答:
The only difficult part is Part 3.

Let $p$ be a positive integer. Then $|x_{k+1}-x_k|\leq\alpha|x_k-x_{k-1}|$ and therefore
$$\eqalign{(1-\alpha)|x_p-x_0|&\leq(1-\alpha)\sum_{k=1}^p |x_k-x_{k-1}|\leq \sum_{k=1}^p |x_k-x_{k-1}|-\sum_{k=1}^p |x_{k+1}-x_k|\cr &\leq|x_1-x_0|=:\delta\ .\cr}$$
It follows that
$$|x_p-x_0|\leq{\delta \over1-\alpha}\qquad(p\geq1)\ ;$$
whence we obtain that
$$|x_{n+p}-x_n|\leq {\delta\over1-\alpha} \ \alpha^n\qquad(n\geq0,\ p\geq1)\ .$$
Assume now that an $\epsilon>0$ is given. There is an $n_0$ such that
$${\delta\over1-\alpha}\ \alpha^n<\epsilon\qquad(n>n_0)\ .$$
It follows that $|x_m-x_n|<\epsilon$ as soon as $m>n>n_0$.

ContentLicense: CC BY-SA 3.0


问:
$(x^2-9)^{(3x+5)}=(x-3)^{(x-1)}(x+3)^{(x-1)}$ What is $x$?
$(x^2-9)^{(3x+5)}=(x-3)^{(x-1)}(x+3)^{(x-1)}$

$(x^2-9)^{(3x+5)}=(x^2-9)^{(x-1)}$

$3x+5=x-1$

$x=-3$

But when I try to use WolfarmAlpha the integer solution is $3$ instead of $-3$. The numerical solution is $x\approx 3.1622776600...$

So I tried to do other things, such as making sure the base is larger than $0$.

$x^2-9>0$

$x<-3$ or $x>3$

But I still didn't obtain the $x$.
So what is $x$?

答:
The domain is $x\ge3$ due to the power functions involved, and it may easier to examine the equation with $x=t+3$, which leads to

$$(t^2+6t)^{t+2}[( t^2+6t)^{2(t+6)} -1]=0$$

The first factor produces the solution $t=0$, or $x=3$, and the second factor reduces to $t^2+6t=1$,
which yields $t=\sqrt{10}-3$, or $x=\sqrt{10}$.

ContentLicense: CC BY-SA 4.0


问:
Finding the mean function of a time series. Question 1.6 in Shumway Stoffer Time Series Analysis.
I am self studying time series analysis with Shumway Stoffer text.  I am stuck on Question 1.6.

Consider the time series

$x_t = \beta_1 + \beta_2 t + w_t$

where $\beta_1$ and $\beta_2$ are known constants and $w_t$ is a white noise process with variance $\sigma^2_w$.

Show that the mean of the moving average

$v_t=\frac{1}{(2q+1)} \sum_{j=-q}^{q} x_{t-j}$

is $\beta_1+\beta_2t$, and give a simplified expression for the autocovariance function.

What I have is

$\mu_{t-j} = E(x_{t-j}) = \beta_1 + \beta_2 t - \beta_2 j$

$E[v_t] =\frac{1}{(2q+1)} \sum_{j=-q}^{q} E(x_{t-j})=\frac{1}{(2q+1)} [2q (\beta_1 + \beta_2 t) + \sum_{j=-q}^{q} \beta_2 j]$

seems like it should be easy from here, but I just can't see it.

Body:
$\mu_{t-j} = E(x_{t-j}) = \beta_1 + \beta_2 t - \beta_2 j$

$\sum_{j=-q}^{q} \beta_2 j = 0$ because of symmetry

And note there are $2q + 1$ terms in series $j = -q,...,q$

So we have

$E[v_t] =\frac{1}{(2q+1)} \sum_{j=-q}^{q} E(x_{t-j})$

$=\frac{1}{(2q+1)} [(2q +1) (\beta_1 + \beta_2 t) + \sum_{j=-q}^{q} \beta_2 j]$

$= \frac{2q + 1}{2q+1} \beta_1+\beta_2 t$

$=  \beta_1+\beta_2 t$


问:
What's a set of sentential formulas satisfied by $\aleph_0$ assignments in $\{0,1\}$-logic but $2^{\aleph_0}$ assignments in $\{0,1/3,2/3,1\}$-logic?
$\{0,1/3,2/3,1\}$ logic has the following connectives:

$\neg r= 1-r$

$r\wedge s=min\{r,s\}$

$r\vee s=max\{r,s\}$

$r\rightarrow s=min\{1,1+s-r\}$

$r\oplus s=min\{r+s,2-r-s\}$

$r\leftrightarrow s=min\{1+r-s,1+s-r\}$

Assume there are countably many atoms. All connectives are allowed in constructing the set. But truth constants are not allowed.

I know that sets like $\{A_{k+1}\rightarrow A_k: k \in \mathbb{N}\}$ or $\{A_i \vee A_j: i,j\in \mathbb{N}, i\neq j\}$ have $\aleph_0$ truth assignments satisfying them in $\{0,1\}$ logic, but I cannot think of an example that furthermore has $2^{\aleph_0}$ truth assignments satisfying it in $\{0,1/3,2/3,1\}$ logic. Any hints?

答:
I assume throughout that "satisfies" means "gives truth value $1$" in your four-valued logic.

Given an atom $A_i$, consider the sentence $$(*)_i:\quad (A_i\vee\neg A_i)\rightarrow A_i.$$ The truth assignments satisfying $(*)_i$ in two-valued logic are exactly those giving $A_i$ truth value $1$, while the truth assignments satisfying $(*)_i$ in your four-valued logic are more complicated. There are two key points:

Truth assignments giving $A_i$ value $0$ or $1$ give $(A_i\vee \neg A_i)$ value $1$, while truth assignments giving $A_i$ value $1\over 3$ or ${2\over 3}$ give $(A_i\vee \neg A_i)$ value $2\over 3$.

A truth assignment satisfies an implication $X\rightarrow Y$ iff it gives $Y$ a truth value at least as large as the truth value it gives $X$.

Putting these together, we get a characterization of those truth assignments satisfying $(*)_i$:

 They are exactly the truth assignments giving $A_i$ value either ${2\over 3}$ or $1$.

This gives rise to a theory which behaves very differently in the two logics in consideration, namely $$\Theta=\{(*)_i: i\in\mathbb{N}\}.$$ There are continuum-many truth assignments satisfying $\Theta$ in your four-valued logic, but only one truth assignment satisfying it in two-valued logic.

Of course this doesn't answer your question yet since $1\not=\aleph_0$, but it does give an important first step. In particular, by combining $\Theta$ with a theory which has $\aleph_0$-many satisfying truth assignments in each logic you'll get what you want:

 Consider for example $$\{(*)_{2i}: i\in\mathbb{N}\}\cup\{A_{2i+3}\rightarrow A_{2i+1}: i\in\mathbb{N}\}.$$


问:
How is it proved that 99.7% of all samples are in 6 SDs in Normal distribution?

The property of 68.2%, 95.4%, and 99.7% are amazing. But how do we prove that it always works?

答:
It is not right to say that such percentage of the sample will always be exactly such multiple of the standard deviation. (In reality, it will never be.)

In fact, the number of samples that fall in the prescribed range is a random variable that follows a Binomial law (the probability $p$ that a single sample falls in the range is given by the Normal law, and we have a Bernouilli scheme, i.e. repeating $n$ times a yes/no drawing).

For instance, for a normal sample of size $100$, the distribution of the number of samples that fall within two sigmas of the mean is given by the histogram below:

As you see, the most probable value is $96\%$, in about every fifth case (though the average value is indeed $95.4\%$.)


问:
Why do this series converge to 2 different values?
Firstly, I consider the Fourier series of the $2\pi$-periodic function $f=\cos{\frac{3}{2}\pi}$ if $-\pi\leq x\leq \pi.$ And I get $$\cos\frac{3}{2}t \sim -\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{n+1} }{(\frac{9}{4}-n^2)\pi}  \cos nt.$$ Since I find that $f\in C^2(T),$ i.e. $f,f’$ and $f’'$ are continuous on $T,$ then $$\cos\frac{3}{2}t = -\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{n+1} }{(\frac{9}{4}-n^2)\pi}  \cos nt \text{ (converge uniformly)}.$$
Let $x=\frac{\pi}{2},$ then
\begin{align}
 \cos\frac{3}{2}(\frac{\pi}{2})&=-\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{n+1} }{(\frac{9}{4}-n^2)\pi}  \cos nx=-\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{n+1} }{(\frac{9}{4}-n^2)\pi}  \cos n\frac{\pi}{2}\notag\\
 &=-\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{2n+1} }{(\frac{9}{4}-(2n)^2)\pi}(-1)^n=-\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{n+1} }{(\frac{9}{4}-(2n)^2)\pi}.\notag
\end{align}
Let $x=\frac{3}{2}\pi,$ then
\begin{align}
 \cos\frac{3}{2}(\frac{3}{2}\pi)&=-\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{n+1} }{(\frac{9}{4}-n^2)\pi}  \cos nx=-\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{n+1} }{(\frac{9}{4}-n^2)\pi}  \cos n\frac{3}{2}\pi\notag\\
 &=-\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{2n+1} }{(\frac{9}{4}-(2n)^2)\pi}(-1)^n=-\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{n+1} }{(\frac{9}{4}-(2n)^2)\pi}.\notag
\end{align}
I find that these two same series $$-\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{n+1} }{(\frac{9}{4}-(2n)^2)\pi} $$ converge to 2 different values, which are $$\cos\frac{3}{2}(\frac{\pi}{2}) \text{ and } \cos\frac{3}{2}(\frac{3}{2}\pi).$$ Did I make any mistake? Thanks~

答: No, but your function is not $cos(\frac{3x}{2})$ as such, it is a periodic extension of that defined in $-\pi$ to $\pi$. If we call this function $f(x)$ then you can see in the graph below that $f(\frac{\pi}{2})$ and $f(\frac{3 \pi}{2})$ are the same.


问:
If an earthquake has magnitude 3.9 with mms formula $M=\frac{2}{3}\log(\frac{S}{S_o})$ then what is the magnitude of an earthquake 750 larger?
$M=\frac{2}{3}\log(\frac{S}{S_o})$

From my online textbook:

Recall the formula for calculating the magnitude of an earthquake,
$M=\frac{2}{3}\log(\frac{S}{S_o})$. One earthquake has magnitude 3.9 on the MMS scale. If a
second earthquake has 750 times as much energy as the first, find the
magnitude of the second quake. Round to the nearest hundredth.

I'm unsure how to approach this problem. I could not find anything about mms scale in the book and am unsure what $S$ and $S_o$ are supposed to be.

I'm guessing that my first step would be:

$3.9=\frac{2}{3}\log(\frac{S}{S_o})$

From here, I really don't know where to go. I could try to represent $S$ and $S_o$ but am unsure if that's even needed, e.g.

$$3.9=\frac{2}{3}\log(\frac{S}{S_o})$$
$$3.9=\frac{2}{3}\log S-\frac{2}{3}\log(S_o)$$
$$\frac{3.9}{\frac{2}{3}}=\log S-\log S_o$$
$$5.85=\log S-\log S_o$$

(From this point I'm pretty lost already and not sure if below or above are even correct or the right path)

$$\log S=5.85+\log S_o$$
$$10^x=S=5.85+\log S_o$$

How can I calculate the magnitude of the second earthquake if I know that it was 750 times more powerful than the one at 3.9 magnitude ?

Body:
$\frac{2}{3}log(\frac{S}{S_0})=3.9$

$\frac{2}{3}log(\frac{750 S}{S_0})=\frac{2}{3}log(\frac{S}{S_0}) + \frac{2}{3}log(750)$

$\frac{2}{3}log(\frac{750 S}{S_0})= 3.9 + 1.92 = 5.82$

edit: where $S$ is the power and $S_0$ is the reference power.


问:
Math notation, conditional sum
I calculate an index; the length of a vector given a conditional statement (mean of the vector) in relation to the unconditional length of the vector. My question is how does one write this in proper math notation?

I've thought that something like this may be close to appropriate, but really this is just a wild guess. I can't seem to find some good guidance for such a specific question.

\begin{eqnarray}
 W = \dfrac{\sum_{i=1}^{N} X_{i}\vert X_{i} < \bar{X}}{\sum_{i=1}^{N} X_{i}}
\end{eqnarray}\

Is anyone comfortable guiding me with this problem?

答:
I don't think there is standard notation for your calculation. What you wrote is clear enough. I suggest you invent a notation for it, define it once at the start of your document, along with an example, and then use it. Perhaps

$$ W(X, \bar X) = \ldots $$ is the sum of the entries in vector $X$
less than  $\bar X$ divided by the sum of the entries in $X$. For
example ...

Note: This is not an "index". I think you should call this "weight", or "relative weight" not "length". Both "index" and "length" have other meanings that your reader will confuse with yours.

Title: Math notation, conditional sum


问:
Pigeon Hole Principle; 3 know each other, or 3 don't know each other
I found another question in my text book, it seems simple, but the hardest part is to prove it. Here the question

There are six persons in a party. Prove that either 3 of them recognize each other or 3 of them don't recognize each other.

I heard the answer use Pigeon Hole Principle, but i have no idea in using it. Could somebody please tell me the way to solve it? Thanks for the attention and sorry for the bad English and my messy post

答:
NOTE: in my answer I assume that the relation "know someone" is symmetric (i.e., A knows B if and only if B knows A). If this relation is not symmetric for you then, I did not really check it but I believe the statement is not true.\\\

Choose a person A at the party. The following two situations are possible:

(CASE 1) A knows at least three people, say B, C and D, at the party;

(CASE 2) A doesn't know at least three people at the party.

In (CASE 1), if at least a pair among {B,C}, {C,D} or {D,B} is formed by people that know each other, then you have three people that know each other. If there is no such pair among these three, then B, C and D are three people that do not know each other.

In (CASE 2) proceed similarly.


问:
What's the structure that's a combination of "Time" and "Time differences" called?
Let's say we call 2021/03/11 12:34:56 a Time

And let's call 30 minutes a TimeDiff

Then we have the following operations.

```<code>Time - Time -> TimeDiff
Time + TimeDiff -> Time
Time - TimeDiff -> Time
TimeDiff + TimeDiff -> TimeDiff
TimeDiff - TimeDiff -> TimeDiff
TimeDiff * Scalar -> TimeDiff
</code>```

Note, we for example, can't add Times. Nor can we multiply TimeDiffs with each other but we can multiply TimeDiffs by scalars.

Is there a general name for this sort of structure?

TimeDiff on it's own I think would just be called a Vector, as you can add, subtract and multiply by scalars.

But what would the combination of Time+TimeDiff be called? I feel like this should have a name, as it's a combination that might come up in other places (although I am finding it hard to find as good an example as "time").

答: It is called an affine space.</p>

Body: It is called an affine space.


答: The space of all different possible Time is called an affine space, and the operation of adding a TimeDiff to a Time is called a translation of (more generally a vector space action on) the affine space.

Title: What's the structure that's a combination of "Time" and "Time differences" called?

Body:
Let's say we call 2021/03/11 12:34:56 a Time

And let's call 30 minutes a TimeDiff

Then we have the following operations.

```<code>Time - Time -> TimeDiff
Time + TimeDiff -> Time
Time - TimeDiff -> Time
TimeDiff + TimeDiff -> TimeDiff
TimeDiff - TimeDiff -> TimeDiff
TimeDiff * Scalar -> TimeDiff
</code>```

Note, we for example, can't add Times. Nor can we multiply TimeDiffs with each other but we can multiply TimeDiffs by scalars.

Is there a general name for this sort of structure?

TimeDiff on it's own I think would just be called a Vector, as you can add, subtract and multiply by scalars.

But what would the combination of Time+TimeDiff be called? I feel like this should have a name, as it's a combination that might come up in other places (although I am finding it hard to find as good an example as "time").


问:
Prove that the function $\cos(x)$ is continuous using the topological definition
We know that $f: (\mathbb{R}, \mathcal{T_{\mathbb{R}}}) \to (\mathbb{R}, \mathcal{T_{\mathbb{R}}})$ is continuous $\iff$ $\forall a< b \in \mathbb{R}, f^{-1}(]a,b[)\in \mathcal{T_{\mathbb{R}}} $

I'm trying to apply this to the function : $f: (\mathbb{R}, \mathcal{T_{\mathbb{R}}}) \to (\mathbb{R},\mathcal{T_{\mathbb{R}}}), x \mapsto \cos(x)$.
Let $a\lt b\in\mathbb{R}$.

if $f^{-1}(]a,b[)=\phi$ then $f^{-1}(]a,b[)\in\mathcal{T_{\mathbb{R}}}$ an open subset.

if $f^{-1}(]a,b[)\not=\phi$ then $x\in f^{-1}(]a,b[)$$\iff$$\cos(x)\in]a,b[ \in \mathcal{T_{\mathbb{R}}}$$\implies$$\exists c\lt d\in \mathbb{R}$ : $\cos(x)\in ]c,d[\subseteq ]a,b[$.

Any help Please.

答:
OK I finally figured it out .
We continue the previous proof by letting $\varepsilon = min\{\cos(x) -c, d-\cos(x)\}$ which is positive $\implies$ ]$\cos(x)-\varepsilon,\cos(x)+\varepsilon[\subseteq]a,b[$.
Using the fact that $\cos(x)$ is 1-Lipschits, Let $r=\varepsilon$. Then $\forall y\in $ $]x-r, x+r[$ $\iff$ $|x-y|\lt r $ $\implies$ $|\cos(x) - \cos(y)|\lt r=\varepsilon$ $\iff$ $\cos(y) \in ]\cos(x)-\varepsilon, \cos(x)+\varepsilon[$ $\implies $ $\cos(]x-r, x+r[)\subseteq ]\cos(x)-\varepsilon,\cos(x)+\varepsilon[\subseteq ]a,b[$ $\iff$ $]x-r, x+r[\subseteq cos^{-1}(]\cos(x)-\varepsilon,\cos(x)+\varepsilon[)\subseteq \cos^{-1}(]a,b[)$.
Conclusion: $\forall y \in f^{-1}(]a,b[) , \exists r>0 : ]x-r, x+r[\subseteq cos^{-1}(]a,b[)$ $\implies$ $\cos^{-1}(]a,b[) \in \mathcal{T_\mathbb{R}}$ $\iff$ $\cos(x)$ is continuous $\forall x \in \mathbb{R}$


问:
Does pointwise convergence against a continuous function imply uniform convergence?
Let $(f_n)_{n\in\mathbb{N}}$ be a sequence of functions with $f_n:\mathbb{R}\rightarrow[0,1]$ for all $n\in\mathbb{N}$ and $f:\mathbb{R}\rightarrow[0,1]$ continuous such that $\lim_{n\rightarrow\infty}f_n(x)=f(x)$ for all $x\in\mathbb{R}$.
Is it now possible to show that $f_n$ converges uniformly to $f$?</p>

答:
That's not necessary true, standard example is $f_n(x) = x^n$ for $x\in[0,1)$ and $f_n(x) = 0$ otherwise. They converge point-wise to $f\equiv 0$ but not uniformly. What you have instead is Dini's theorem:

If $(f_n)$ are continuous functions, and they converge to a continuous function $f$ point-wise and monotonically on a compact set, then they converge uniformly to that function.


问:
How to prove that the functions which map from real numbers to rational numbers is not possibly continuous without using the IVT?

I can make up a non-constant function f(x) like this: its domain is an interval, for example [0,1], and its range is totally composed of many rational numbers.
The problem is: If I pick the range very carefully, it seems that it is possible that every  $\lim_{t\rightarrow\ x}f(t)$could equal to some rational number for every x, I mean for any sequence {tn} converging to x, just make sure that f(tn) would not converge to a irrational number, which means that this is a continuous function!
I know by the Intermediate Value Theorem, it is impossible. But how can I prove this impossiblity without using the IVT, or say how can I prove that the idea picking range process which seeems really feasible is impossible?

Edit: The intuition of this problem comes from this: when we first learn the continuity of functions, we always relate it with a non-liftinghand curve which looks very compact. But if I use so big enough amount of rational numbers that the graphy looks almost the same with the one of real numbers because there is an infinity of rational numbers (less than real numbers though).
If I want to prove this statement is wrong by using Reductio ad Absurdum, namely asume that there exist a function like that, I mean every $\lim_{t\rightarrow\ x}f(t)$ equals to some rational number, how can I find the contradiction from the perspective of limits rather than the macro things like the IVT?

答: I do not think that the statement you are trying to prove is correct. Consider the function $f(x)=r$ where $r$ is a rational number. This is a function from the reals to the rationals and it is certainly continuous. Do you mean to ask about a function from the reals onto the rationals - in which case, my example would fall short?


问:
Hermitian Matrix properties
Given
$ \begin{equation*}
Y =
\begin{pmatrix}
y & 0 & 0 \\
0 & 0 & -iy \\
0 & iy & 0
\end{pmatrix}
\end{equation*}$

Given another matrix $X$ which has two degenerate eigenvalues and commutes with $Y$, that is $XY = YX$. Can we claim anything about $X$ in general, more specifically if $X$ is Hermitian or not. I used commutative property to find $X$ as

$ \begin{equation*}
X =
\begin{pmatrix}
a & ci & c \\
-mi & t & -n \\
m & n & t
\end{pmatrix}
\end{equation*}$

But I can' use the property that $X$ has two degenerate eigenvalues.
Help would be appreciated.

Thanks in advance.

答:
No $X$ need not be Hermitian. Given any Hermitian matrix $X$ satisfying the constraints in the question, we define $X' = i X$. Then $X'$ also satisfies all constraints in the question but is not Hermitian.

But let's investigate $X$ further. Working in the orthonormal basis in which $Y$ is diagonal we have
$$
Y = \begin{pmatrix} y & 0 & 0 \\
0 & y & 0 \\
0 & 0 & -y
\end{pmatrix}
$$
Then by the commutation constraint $X$ needs to be of the form
$$
X = \begin{pmatrix}
x_{00} & x_{01} & 0 \\
x_{10} & x_{11} & 0 \\
0 & 0 & x_{22}
\end{pmatrix}
$$

But now take the top left block to be any $2\times 2$ matrix that has two distinct eigenvalues. If we pick $x_{22}$ equal to one of those eigenvalues then we satisfy the constraints in the question. This shows that $X$ need not even be normal as we can pick the top left block to be a non-normal matrix.

Title: Hermitian Matrix properties


问:
A nontrivial everywhere continuous function with uncountably many roots?
This is my first post on SE, forgive any blunders.

I am looking for an example of a function $f:\mathbb{R} \to \mathbb{R}$ which is continuous everywhere but has uncountably many roots ($x$ such that $f(x) = 0$). I am not looking for trivial examples such as $f = 0$ for all $x$. This is not a homework problem. I'd prefer a nudge in the right direction rather than an explicit example.

Thanks!

Edit: Thanks all! I've constructed my example with your help.

答:
If you just want a continuous function, then let $f(x) = 0$ over an interval say $[a,b]$ where $a0\newline 0&x\leq 0\end{cases}$$

And if you want an infinitely smooth function with a compact support

$$
f(x) =
\begin{cases}
\exp(- \frac{1}{1 - x^2}), \quad  |x| < 1 \\
0 , \quad |x| \geq 1

\end{cases}

$$

Body:
If you just want a continuous function, then let $f(x) = 0$ over an interval say $[a,b]$ where $a<b$ and over the rest of the real line find functions $g(x)$ and $h(x)$ such that $g(a) = 0$ and $h(b) = 0$ and define $f(x) = g(x), \forall x \leq a$ and $f(x) = h(x), \forall x \geq b$.

You could also have infinitely smooth functions with zero on an uncountable set as for instance
$$f(x)=\begin{cases}e^{-1/x}&x>0\newline 0&x\leq 0\end{cases}$$

And if you want an infinitely smooth function with a compact support

$$
f(x) =
\begin{cases}
\exp(- \frac{1}{1 - x^2}), \quad  |x| < 1 \\
0 , \quad |x| \geq 1

\end{cases}

$$


问:
maps forming a basis for integer function as $\mathbb{Z}$-module
so I'm currently struggling witht the following problem:

Show that the maps
$$e_k: \mathbb{N}_0 \rightarrow \mathbb{Z}; x \mapsto \frac{x(x-1) \cdots (x-k+1)}{k!}$$
for $k \in \mathbb{N}_0$ (with $e_0$ identically $1$) form a basis for Int$(\mathbb{Z})$ as a $\mathbb{Z}$-module.

In case that this is not a widespread notation, Int$(\mathbb{Z})$ refers to the functions that map integers to integers.

To be quite honest, I am quite confused by the question, I don't understand the maps. I don't understand why one has a normal arrow and one has an arrow with a little base. what are the maps supposed to mean? Any help would be really appreciated!

答:
The MSE question 4049277 "Avoiding brute
force: determining when a specific polynomial in $\mathbb{Q}[x]$ is an integer
for any integer $x$" asks essentially the same question. Some of the answers are for
the general case. My answer used the technique of a difference table to solve the
general problem of expressing an integer valued polynomial as a $\mathbb{Z}$-linear
combination of binomial coefficients. The formula is
$$ f(x) = \sum_{k=0}^\infty \Delta^k\!f(0) {x \choose k} $$
where the sum has only a finite number of non-zero terms since $\,f(x)\,$ is a
polynomial function.


问:
How do I compute this expectation of sum of product of random variables squared?
If $h_{1}, h_{2}, \dots, h_{N}$ are zero mean i.i.d Gaussian random variables with variance $\sigma_{h}^{2}$, how do I calculate the value of

$$\mathbb{E}\left[\left(h_{1}h_{2} + h_{2}h_{3} + \dots, h_{N-1}h_{N} \right)^{2}\right]$$

答:
We have
$$ \mathbb{E}\Big[\Big(\sum_{i=1}^{N-1}h_ih_{i+1}\Big)^2\Big] = \sum_{i=1}^{N-1} \sum_{j=1}^{N-1} \mathbb{E}\big[h_ih_{i+1}h_jh_{j+1} \big]$$
Assuming that the variables are independent, we have
$$ \mathbb{E}\big[h_ih_{i+1}h_jh_{j+1} \big] = \left\{\begin{array}{ll}\mathbb{E}\big[h_i^2 \big]\mathbb{E}\big[h_{i+1}^2 \big] = \sigma_h^4 & \text{for }j=i \\ \mathbb{E}\big[h_{i-1} \big]\mathbb{E}\big[h_i^2 \big]\mathbb{E}\big[h_{i+1} \big] = 0 & \text{for }j=i-1 \\ \mathbb{E}\big[h_{i} \big]\mathbb{E}\big[h_{i+1}^2 \big]\mathbb{E}\big[h_{i+2} \big] = 0 & \text{for }j=i+1 \\ \mathbb{E}\big[h_{i} \big]\mathbb{E}\big[h_{i+1} \big]\mathbb{E}\big[h_{j} \big]\mathbb{E}\big[h_{j+1} \big] = 0 & \text{else } \end{array}\right.$$
In the whole sum the only terms that give non-zero contributions are those with $j=i$, therefore
$$ \mathbb{E}\Big[\Big(\sum_{i=1}^{N-1}h_ih_{i+1}\Big)^2\Big] = (N-1)\sigma_h^4$$


问:
Is the following an example of an exact sequence?
So,in Algebra 1, I have to give an example of an exact sequence of homomorphisms in an example of vector spaces and linear maps between them.

I have an ideas. I don't know if any of them work and the main problem at both of them is that I am not able of finding the linear maps.
We learned about the connection between these exact sequences and direct sums.
The first example:

$0\rightarrow ker(f) \rightarrow V \rightarrow [v]\rightarrow0 $ where V is finite dimensional, f is a linear functional in the dualraum of V and $v \notin kerf $.

This is the first time ever working with exact sequences, therefore I have problems. I have seen that above the arrows there must be functions, I don't know which could be the functions in my example.

I would be really helpful for help. Also If there is any example which might be easier, I would be thankful to know that too in order to get a better idea.

Thanks in advance.

答:
I don't quite understand your sequence, but it looks exact, but only by definition. Like if you have a surjective linear functional $f \colon V \to F$ where $F$ is the field you're working over, then sure you have an exact sequence
$$\ker(f) \hookrightarrow V \twoheadrightarrow [v]$$
where you identify $F$ with $[v]$ for some $v \notin \ker(f)$. An important heuristic to keep in mind is that we must have $\dim V = \dim\ker(f) + \dim[v]$ for this to be an exact sequence, so the dimension of the kernel must be one less than $\dim[v]$. As for finding the linear maps, well that depends on what $f$ is. But the cheekiest way I can think to do this would be to choose a basis for $V$ containing $v$ and a basis for $\ker(f)$. So say $\ker(f)$ has basis $\{a,b,\dotsc,z\}$. Then take $V$ to have basis $\{v, a, b, \dotsc, z\}$ then $f$ can be written as

\begin{pmatrix}
 1&0&\dotsb&0
\end{pmatrix}

will map nothing to $v$, but be the identity map on the rest of the basis for $\ker(f)$.


问:
Find $E[(x+2)^2]$ Given $E[x]$
Find
$
E[(x+2)^2]
$
given that $
E[x] = 5, Var(X) = 2
$.

I'm not sure if I'm making this more complicated than it should be. However, this is what I did.

$$
E[(x+2)^2] =
E[x^2+2x+4] =
E[x^2]+E[2x]+4 = E[x^2]+2E[x]+4 = 25+10+4 = 39.
$$
I'm unsure is if $E[x^2]$ is 25. And furthermore, if there's an easier method to solving this problem if $x+2$ was raised to a higher power

答:
No, $\displaystyle E\left[ X^{2}\right] \neq 25$.

But, you can probabily recal that: $\displaystyle Var( X) =E\left[ X^{2}\right] -( E[ X])^{2}$.

From here:

$\displaystyle E\left[ X^{2}\right] =Var( X) +( E[ X])^{2} =2+25=27$.

Now,

$\displaystyle  \begin{array}{{>{\displaystyle}l}}
E\left[( X+2)^{2}\right] =E\left[ x^{2} +4X+4\right] =E\left[ X^{2}\right] +E[ 4X] +E[ 4] =E\left[ X^{2}\right] +4E[ X] +4\\
=27+4\cdotp 5+4=51
\end{array}$

Title: Find $E[(x+2)^2]$ Given $E[x]$


问:
Is there a meaningful way to define $ij$, where $i$ is the imaginary unit and $j$ is the split-complex unit?
At the moment I'm treating $ij=ji$ as its own quantity and reducing it where possible, as in $(ij)^2 =-1$ and $e^{ij} = \cos(j) + i\sin(j) = \cosh(i) + j \sinh(i)$.

From the first identity seems like you can say $ij=\sqrt{-1}=i$, and this even agrees with  $\cos(j) + i\sin(j)= \cos(1) + i\sin(1)$ and  $\cosh(i) + j \sinh(i) = \cosh(i) + 1\sinh(i)$ unless I made a mistake, but it still seems like a strange definition.

So, Is there a way to represent $ij$ as a linear combination of $1$, $i$, and $j$, and does the definition I came up with here break anything I haven't noticed?

答:
No. What you are talking about is tessarines, a 4-dimensional algebra that combines complex and split-complex numbers.

In that system $ij$ is a separate unit vector, similar to $i$ in that $(ij)^2=-1$. But not equal to $i$. It is irreducible.

The algebra of tessarines is commutative and associative.


问:
How we can write 36 in base 37?
i've been imagining that in the base 16 the number 10 is A and 11 is B so in the base 37, 35 will be Z so how we can write 36 ?</p>

答:
You could represent 36 in base 37 by (36). That could stand for a symbol that means 36. Or if you want to indicate the base with a subscript: $(36)_{37}$. This could be used with all bases no matter how big.

Or you could use an actual (Unicode) symbol that is '36' inside a circle, or inside a square, or rectangle or something like that. Or a tiny '36'. Thus you have an actual symbol (numeral) that means 36 that is 'transparent', by which I mean that what it stands for is obvious from the appearance.

This approach could be good when you want the layperson or absolute beginner to be able to read the numbers. Also if you are experimenting with other bases, or using a very large base.

Title: How we can write 36 in base 37?


问:
Not able to calculate this expectation
If $h_{1}, h_{2}, \dots, h_{N}$ are zero mean i.i.d Gaussian random variables with variance $\sigma_{h}^{2}$, then how do I calculate the value of

$$\mathbb{E}\left[\left(h_{1}^{2} + h_{2}^{2} + \dots, h_{N}^{2} \right)^{2}\right]$$

The only clue I have is that

$$\mathbb{E}\left[h_{i}^{2}\right] = \sigma_{h}^{2}$$

I do not know how to compute $\mathbb{E}\left[h_{i}^{2} h_{j}^{2}\right]$ (for all $i \neq j$). Please help.

答:
Answer for the follow-up question in the comments

When  you expand the square you will have a matrix of $(N-1)\times(N-1)$ elements.

The $N-1$ elements in the principal diagonal are in the form $E(h_i^2h_{i+1}^2)=\sigma^4$

All the other elements contains at least on element  $E(h_i)=0$ thus they are all zero.

Thus the result is $(N-1)\sigma^4$

Title: Not able to calculate this expectation


问:
Equilibrium of the following system of non linear ODE's
How would find the equilibriums of the following system:

$$
\frac{d I}{dt}=-\frac{\beta S I}{N}   \qquad (1)$$

$$\frac{d S}{dt}=\frac{\beta S I}{N} -\gamma I   \qquad (2)$$

$$\frac{d R}{dt}=\gamma I   \qquad (3)
$$

where $S+I+R=N$ and $\beta,\gamma >0$.

I know one such equilibrium is
\begin{align}
e_1 : \left( S_1^*, I_1^*, R_1^*\right)&= \left(N, 0, 0\right), \\[2ex]
\end{align}

How do I find the other(if it exists)? I believe it should be similar to the problem below unless I'm mistaken.

I have solved the more complicated case:

$$
\frac{d S}{dt}=\mu N -\frac{\beta S I}{N} - \nu S  \qquad (1)$$

$$\frac{d I}{dt}=\frac{\beta S I}{N} -\gamma I - \nu I  \qquad (2)$$

$$\frac{d R}{dt}=\gamma I - \nu R  \qquad (3)
$$

where $S+I+R=N$, $\quad$ $\mu =\nu$, $\quad$ $\mu, \beta, \nu,\gamma >0$.

Solutions are:

\begin{align}
e_1 : \left( S_1^*, I_1^*, R_1^*\right)&= \left(N, 0, 0\right), \\[2ex]
e_2 : \left( S_2^*, I_2^*, R_2^*\right)&= \left(\frac{N\left(\gamma+\nu\right)}{\beta}, N\nu\left(\frac{1}{\gamma +\nu}-\frac{1}{\beta}\right), N\gamma\left(\frac{1}{\gamma +\nu}-\frac{1}{\beta}\right)\right)
\end{align}

答:
Notice that a necessary condition for an equilibrium is given by:
$$\frac{dR}{dt} = 0 \implies I = 0.$$
Since the condition $I=0$ is enough to have $\frac{dI}{dt} = \frac{dS}{dt} = 0$, you have a sufficient condition. Thus the equilibriums are given by:
$$(S^*,I^*,R^*) = (K, 0, N-K),$$
for any $0\leq K \leq N$.


问:
Inequality proof for all $x,y,z \gt 0$

$$\frac{xy}{z}+\frac{xz}{y}+\frac{zy}{x}\ge x+y+z$$

I want to prove this inequality for all $x,y,z \gt 0$

I started by multiply both sides by $2$, but the factorisation on the left-side was confusing.

答:
Use Am-Gm
$$\frac{xy}{z}+\frac{xz}{y}\geq 2x$$
$$\frac{yz}{x}+\frac{xz}{y}\geq 2z$$
$$\frac{xy}{z}+\frac{yz}{x}\geq 2y$$
Q.E.D

Title: Inequality proof for all $x,y,z \gt 0$


问:
What's the Derivative?
What is the derivative (gradient) of $$\operatorname{vec}(W)^\top
\left[\sum_i (ue_i^\top)\otimes(e_ie_i^\top) + \sum_j(e_je_j^\top)\otimes(ue_j^\top) - 2I\otimes I\right]
\operatorname{vec}(FF^\top),\tag{3}
$$ where $e_i$ denotes the $i$-th vector in the canonical basis and $u=\sum_ie_i=(1,1,\ldots,1)^\top$
where "$\otimes$" denotes the Kronecker product (tensor product) with respect to the rectangular matrix $F$? and $W$ is a constant matrix. The above term can also be simplified as $\sum_{ijl}(F_{il}-F_{jl})^2W_{ij}$ which is a much simpler expression. What would the derivative be? All the above entries are real while $F$ is a rectangular matrix. Also please refer to : Represent in a matrix form: $\sum_{ijl}(F_{il}-F_{jl})^2W_{ij}$ for the details on simplification/algebra.</p>

答:
To find the derivative, it is much easier to work with your original expression $S=\sum_{ijl}(F_{il}-F_{jl})^2W_{ij}$. Note that, for fixed indices $I$ and $J$,
\begin{align*}
S = &\text{ terms that do not involve } F_{IJ}\\
&+\ (F_{IJ} - F_{IJ})^2 W_{II}
+ \sum_{j\neq I} (F_{IJ} - F_{jJ})^2 W_{Ij}
+ \sum_{i\neq I} (F_{iJ} - F_{IJ})^2 W_{iI}\\
= &\text{ terms that do not involve } F_{IJ}\\
&+\ \sum_{k\neq I} (F_{IJ} - F_{kJ})^2 W_{Ik}
+ \sum_{k\neq I} (F_{kJ} - F_{IJ})^2 W_{kI}.
\end{align*}
Let $K = W-W^T$. Then
\begin{align*}
\frac{\partial S}{\partial F_{IJ}}
=&2\sum_{k\neq I} (F_{IJ} - F_{kJ}) W_{Ik}
+ 2\sum_{k\neq I} (F_{kJ} - F_{IJ}) W_{kI}\\
=&2\sum_{k\neq I} (F_{IJ} - F_{kJ}) (W_{Ik}-W_{kI})\\
=&2\sum_{k\neq I} (F_{IJ} - F_{kJ}) K_{Ik}\\
=&2\sum_k (F_{IJ} - F_{kJ}) K_{Ik}\\
=&2F_{IJ}\sum_k K_{Ik} - 2\sum_k K_{Ik}F_{kJ}.
\end{align*}
Therefore
$$
\left(\frac{\partial S}{\partial F_{ij}}\right)
=2F\circ(KE) - 2KF
$$
where $E$ is the matrix with the same size as $F$ and all entries equal to $1$.</p>

Title: What's the Derivative?


问:
$F(E)$ is proper iff $E$ has the finite intersection property (Blackburn, 2.5.1(c))

Blackburn's Modal Logic, Ex 2.5.1 (c): Let $E\subseteq \mathcal P(W)$, the power set of $W$. Let $F = F(E)$ be the filter generated by $E$. Prove that $F$ is proper iff $E$ has the finite intersection property (FIP).

My work:

$[\Leftarrow]$: Suppose $E$ has the finite intersection property. Then the intersection of any finite number of elements of $E$ is non-empty. Suppose $\varnothing\in F$, i.e. $F$ is not proper. Then, can we conclude that $\varnothing\in E$ as well? If yes, then we have our contradiction - since the intersection of any finite number of sets with $\varnothing$ is $\varnothing$, violating FIP.

$[\Rightarrow]$: Suppose $F$ is proper. This is the same as saying $\varnothing\notin F$. This means $\varnothing\notin E$, because if it were, we'd be in trouble. I'm using that $$F = \bigcap \left\{X\subseteq \mathcal P(W): X \text{ is a filter}, E\subseteq X   \right\}$$
If FIP doesn't hold, then there exist a finite number of sets $Y_1,...,Y_n$, $Y_i \in E$ such that $\bigcap_i Y_i = \varnothing$. However, $Y_i\in E \implies Y_i \in F$. $F$ is a filter, and so is closed under finite intersections. We must have $\bigcap_i Y_i \in F$, but $\bigcap_i Y_i = \varnothing$ which is a contradiction since $F$ is proper and $\varnothing\notin F$.

I believe the second part is fine, but the first part isn't. Could someone check and help me refine this proof? Thanks a lot.

答:
For the implication $(\Leftarrow)$:

If $E$ has the finite intersection property, then any nonempty subset $A$ of $W$ which contains an intersection of a finite collection of elements of $E$ must be in $\mathcal{F}(E)$, because it is a filter. Actually, it can be easily checked that the collections of such $A$'s is a filter which contains $E$, and therefore we have

$$ \mathcal{F}(E)=\{A\in \mathcal{P}(W): \exists \text{ a finite collection }U_{1},\dots, U_{n}\in E\text{ such that } \bigcap\limits_{i=1}^{n}U_{i}\subseteq A\}$$

From this, we deduce that $\mathcal{F}(E)$ is a proper filter


问:
Reduction formula for $\int_0^1 x^m(1-x)^{n-1}\,\mathrm{d}x$?
So I am trying to find the reduction formula for $\int_0^1 x^m(1-x)^{n-1}\,\mathrm{d}x$.

My attempt: Taking $x^m$ as the second function and integrating by parts, I got,

$I_n=\frac{(n-1)}{m+1}I_{n-2,m+1}$. But the given answer is $\frac{(n-1)!m!}{(m+n)}$. How do I go from my answer to the one given in the textbook? I tried manipulating my answer, but it isn't the same as the one in the textbook. Thanks in advance.

答: Let me take $$J_{m,n}=\displaystyle\int_0^1x^m(1-x)^n\,dx.$$ This is really your $I_{m, n+1},$ but this renaming make it symmetric and easy to work with. For example, the substitution $x=1-t$ gives us $$J_{m,n}=J_{n, m}\tag1$$ which will transform as $I_{m, n+1}=I_{n, m+1}$ in to old notation. Note that $J_{m,0}=\dfrac{1}{m+1}$ for all $m\in\mathbb{N}$ (including zero). Letting $u=x^m$ and $dv=(1-x)^ndx$ we get $$J_{m,n}=\dfrac{m}{n+1}J_{m-1, n+1}\tag2$$ via integration by parts. Also because of $(1)$ identity, we have $$J_{m,n}=\dfrac{n}{m+1}J_{m+1, n-1}\tag3$$ for free. In fact, this is exactly what you get after integrating by parts with $u=(1-x)^n$ and $dv=x^mdx.$ Lets apply this third identity repeatedly as: $$J_{m,n}=\dfrac{n}{m+1}J_{m+1, n-1}=\dfrac{n}{m+1}\dfrac{n-1}{m+2}J_{m+2, n-2}$$ to derive $$J_{m,n}=\dfrac{n}{m+1}\dfrac{n-1}{m+2}\cdots\dfrac{1}{m+n}J_{m+n, 0}=\color{Green}{\dfrac{m!\,n!}{(m+n+1)!}}.$$ Now it is not difficult to rewrite this in your old notation.


问:
The graph of $f(x)$ lies below this line for small values of $x$
Suppose that there exists a function $f:\mathbb R_{+} \to \mathbb R_{+}$ with only the following properties  :

$f(x)>0$ for every $x>0$

$\lim_{x \to 0} f(x) =0$

My question: Are $(1),(2)$ sufficient in order to deduce that $f(x) \leq 2x\;$ if $x$ is sufficiently small?

My approach: I tried with the standard $\epsilon$-$\delta$ definition, i.e
$\begin{align}
\forall \epsilon>0 \quad \exists \delta>0 \;\text{ such that } \; 0<x<\delta \implies f(x)<\epsilon\;.
\end{align}$ Then, choosing $\delta=\frac{\epsilon}{2}$, I obtain that $f(x) <2\delta\;$ which then contradicts what I need to show. Does this mean that my question has a negative answer or am I missing something here?

Any help is much appreciated! Thanks in advance.

答:
There is the counter example $f(x)=3x$, which satisfies both properties ($3x>0$ for every $x>0$ and $\lim_{x \to 0} 3x =0$, because $f(0)=0$). It is also always bigger than $2x$.

The answer is that those properties aren‘t sufficient to deduce that $f(x) < 2x$ for sufficiently small values of x.


问:
Projective general linear groups of order 2 = projective special linear group of order 2
If $q\geq 5$ is a prime and we consider the group PGL(2,q) the projective general linear group of order 2 and the group PSL(2,q) the projective special linear group of order 2.

Can I then conclude PSL(2,q) = PGL(2,q)?

答: The cokernel of the embedding $\mathrm{PSL}(2,q) \to \mathrm{PGL}(2,q)$ is isomorphic to $\mathbb F_q^*/(\mathbb F_q^*)^2$, that is you will have equality exactly iff every element of $\mathbb F_q$ is a square. For $q = 5$ for example, we have $\mathbb F_5^* = \{1,2,3,4\}$ and $(\mathbb F_5^*)^2 = \{1,4\}$, so $\mathrm{PSL}(2,5) < \mathrm{PGL}(2,5)$.</p>


问:
Is there a base (besides base 0) that cannot represent all rational numbers?
My son is learning about different bases for numbers and he is wondering if there is any base for a number system that cannot represent all rational numbers?

I'm thinking specifically of irrational bases, like base $\pi$ or negative bases, like $-2$.

I am not a mathematician and so do not know how to prove this one way or the other.

答:
Base one can only represent integers, and some rational numbers are not integers. So, yes, there is indeed a base (besides base O) that cannot represent all rational numbers, and it is base one.

What I call 'base one', Wikipedia calls 'unary' and has this to say:
"The unary numeral system is the simplest numeral system to represent natural numbers:[1] to represent a number N, a symbol representing 1 is repeated N times.[2]

In the unary system, the number 0 (zero) is represented by the empty string, that is, the absence of a symbol. Numbers 1, 2, 3, 4, 5, 6, ... are represented in unary as 1, 11, 111, 1111, 11111, 111111, ...[3]

In the positional notation framework, the unary is the bijective base-1 numeral system. However, because the value of a digit does not depend on its position, one can argue that unary is not a positional system.[citation needed] "


问:
Finding a formula for the kth partial sum
I need help finding the kth partial sum of the question below:

Consider the series $\sum_{n=1}^{\infty}a_n$ where $a_n=\ln\left(\frac{6n-1}{6n+5}\right)$. Find a formula, in closed form, for the $k^{th}$ partial sum $s_k$.

I think this is a telescoping series but I'm sorta lost on how to answer the question. Any help would be appreciated

Picture of Question

Body:
$$\begin{aligned}a_n \;=\;& \ln\left(\frac{6n-1}{6n+5}\right) \;=\; \ln(6n-1) \;-\; \ln(6n+5) \\[5mm] =\;& \ln(6n-1) \;-\; \ln(6(n+1)-1)\end{aligned}$$

Which means that $\sum a_n$ is a telescoping sum, giving us:

$$\sum^n_{k=1}a_k \;=\; \ln(5) \;-\; \ln(6(n+1)-1)$$


问:
Why doesn't the inverse of this matrix always exist?
Take $A$ square, non-negative, irreducible matrix, with Perron root $\rho(A)$.
Let $Q = \rho(A)I - A$. Then it is argued (e.g. in the Handbook of Linear Algebra, Chapter 9) that the group inverse of $Q$ always exists.
But I fail to see why the actual inverse doesn't always exist (and must then be the group inverse)
using the following argument:
By the Perron-Frobenius theorem, $\rho(A)$ distinct from the other eigenvalues,
$0$ cannot be in the spectrum of $Q$, and so $Q$ must be invertible.

答:
$Q$ must be singular. By Perron-Frobenius theorem, $\rho(A)$ is an eigenvalue of $A$. Therefore $0=\rho(A)-\rho(A)$ is an eigenvalue of $Q=\rho(A)I-A$ and the usual inverse of $Q$ doesn't exist.

However, since $A$ is irreducible, $\rho(A)$ is a simple eigenvalue of $A$. Therefore $0$ is a simple eigenvalue of $Q$ and $\operatorname{rank}(Q^2)=\operatorname{rank}(Q)$. Hence the group inverse (here I suppose that the author means Drazin inverse) of $Q$ exists.


问:
Why are two identical lines in this context here non-parallel?
I came across this question on Linear Algebra: concepts and techniques on euclidean spaces (second edition)

This is the answer given for the question:

I do not understand why did they mention x + y - z =1 twice and state that they are two non-parallel line? With my understanding, shouldn't two identical line be parallel to each other?

答:
It's just a copy-paste misprint. It should read:

Since $x+y-z=1$ and $x-2y+z=0$ are two non-parallel planes, $\ldots$


问:
How to prove the set of group axioms is independent?
Hey I am having a hard time solving a logic problem. It one of the exercises in Ebbinghaus, Mathematical Logic (exercise 4.14, p. 39):

A set $\phi$ of S-formulas is independent if there is no $\varphi \in \phi$ s.t. $\phi  $/$  \{\varphi\} \vDash \varphi$. Let $\mathcal{S}=\{\circ, e\}$ and $$\phi := \{ \forall x \forall y \forall z(x \circ (y \circ z) \equiv (x \circ y) \circ z), \forall x \ x \circ e \equiv x, \forall x \exists y \ x \circ y \equiv e \}\text{.}$$ Show that  $\phi$ is independent.

My thoughts: I need to find three models $M1$, $M2$, $M3$ s.t. $M1$  satisfies the second and third formulas, but cannot satisfy the first formula $\forall x \forall y \forall z(x \circ (y \circ z) \equiv (x \circ y) \circ z)$, so the first formula is independent.

$M2$, $M3$ are similar cases. i,e., $M2$ can satisfy the first formula and the third formula, but cannot satisfy the second formula, so the second one is independent. $M3$ can satisfy the first formula and the second formula, but cannot satisfy the third formula, so the third one is independent.

Like, consider M3, let the domain of M3 is $\mathbb{Z}$, $\varphi(\circ)$ $= +$ on $\mathbb{Z}$，and $\varphi(e) = 0$, so $M3$ satisfies the first formula (associative law) and the second $\mathbb{Z}+0=0$. But $M3$ cannot satisfy the third one, so the third formula is independent.

But I'm wondering :

(1) Could this kind of specific model works? I mean I just find a "specific model" to satisfy two formulas in the set. I just feel I need to prove independence in a more general way, right? But I have no idea how to do.

(2) Could anyone think examples of $M1$ and $M2$?

Any ideas please? I appreciate your help.

答:
For $M_1$ I propose the following minimal possible model. An operation on a set of two elements $\{e,a\}$ is given by the Cayley table
$$
\begin{array}{c|cc}
  \circ  & e & a\\
 \hline
  e & e & e \\
  a & a & e \\
\end{array}
$$
We have $(a\circ a)\circ a=e\circ a=e$ and $a\circ(a\circ a)=a\circ e=a$.


问:
Proving vector calculus identity $\nabla \times (\mathbf a\times \mathbf b) =\cdots$ using Levi-Civita symbol
I want to prove the following relation

$$\nabla \times (\mathbf a\times \mathbf b) = \mathbf a\nabla \cdot \mathbf  b + \mathbf  b \cdot \nabla \mathbf  a - \mathbf  b \nabla \cdot \mathbf  a - \mathbf a \cdot \nabla \mathbf  b$$

using the following Levi-Civita definition of cross product
$$\mathbf{a} \times \mathbf{b} =\mathbf{e}_i \epsilon_{ijk}a_ib_j$$
where $\epsilon_{ijk} =\begin{cases}
+1 & \text{if } i,j,k \text{ are in clockwise permutation}, \\
-1 & \text{if } i,j,k \text{ are in counterclockwise permutation, and} \\
\;\;\,0 & \text{if }i=j \text{ or } j=k \text{ or } k=i.
\end{cases}$

答:
Just figured out one quick but tentative proof using skew-symmetric matrices and expecting to be verified.
The product of two skew-symmetric matrices is
$\mathbf{S}_\vec{A}\mathbf{S}_\vec{B}\vec{x}=\left[\vec{B}\vec{A}^T-(\vec{A}\vec{B}^T)\mathbb{I}\right]\vec{x}$
or in dyadic direct multiplication:
$$\begin{align}
(\vec{a}\times\vec{b})\times \mathbb{I}&=\mathbb{I}\times (\vec{a}\times\vec{b})=\vec{b}\vec{a}-\vec{a}\vec{b}.\\
{(\vec{a}\times\vec{b})\times \mathbb{I}}\cdot\vec{x}&=(\vec{a}\times \vec{b})\times \vec{x}=(\vec{b}\vec{a}-\vec{a}\vec{b})\cdot\vec{x}\\
\vec{x}\times (\vec{a}\times \vec{b})&=\vec{x}\cdot(\vec{b}\vec{a}-\vec{a}\vec{b}).
\end{align}$$
where $\mathbf{S}_\vec{A}=\vec{A}\times$ is the skew-symmetric matrix about vector $\vec{A}$, likely for $\mathbf{S}_\vec{B}$, and $\mathbb{I}$ is the identity matrix which can also be omitted.
Let $\vec{e}$ be unit vector, so
$$\begin{align}
[\nabla \times (\vec{A} \times \vec{B})]\cdot\vec{e}&=\nabla\cdot\left[ (\vec{A} \times \vec{B})]\times\vec{e}\right]=\nabla\cdot\left[\vec{B}\vec{A}^T-(\vec{A}\vec{B}^T)\mathbb{I}\right]\vec{e}\\
&=\left[\nabla {\cdot }\left(\vec{B}\vec{A}^T\right)-\nabla {\cdot }\left(\vec{A}\vec{B}^T\right)\right]\vec{e}
\end{align}$$


问:
How to prove that the functions which map from real numbers to rational numbers is not possibly continuous without using the IVT?

I can make up a non-constant function f(x) like this: its domain is an interval, for example [0,1], and its range is totally composed of many rational numbers.
The problem is: If I pick the range very carefully, it seems that it is possible that every  $\lim_{t\rightarrow\ x}f(t)$could equal to some rational number for every x, I mean for any sequence {tn} converging to x, just make sure that f(tn) would not converge to a irrational number, which means that this is a continuous function!
I know by the Intermediate Value Theorem, it is impossible. But how can I prove this impossiblity without using the IVT, or say how can I prove that the idea picking range process which seeems really feasible is impossible?

Edit: The intuition of this problem comes from this: when we first learn the continuity of functions, we always relate it with a non-liftinghand curve which looks very compact. But if I use so big enough amount of rational numbers that the graphy looks almost the same with the one of real numbers because there is an infinity of rational numbers (less than real numbers though).
If I want to prove this statement is wrong by using Reductio ad Absurdum, namely asume that there exist a function like that, I mean every $\lim_{t\rightarrow\ x}f(t)$ equals to some rational number, how can I find the contradiction from the perspective of limits rather than the macro things like the IVT?

答: You're going to need to use the connectedness of $[0,1]$ here, which is equivalent to IVT. After all, there is a nonconstant continuous function from e.g. $[0,1]\setminus\{{1\over 2}\}\rightarrow\mathbb{Q}$, so as soon as we drop connectedness we can't prove the statement. We could of course try to rephrase the standard argument to avoid explicit appeal to IVT, but this wouldn't change its mathematical content.


问:
Solve transcendental equation / maximum likelihood estimation
Consider $(X_1,\dots X_n)$ with $$\mathbb P[X=k]=\begin{cases}p_0,&k=0\\\frac{(1-p_0)\lambda^k}{k!(e^\lambda)-1},&k\geq1\end{cases}$$

I want to determine the maximum likelihood estimator of $\lambda$

I arrive at $$\frac{\partial}{\partial \lambda}\log L((p_0,\lambda))=\sum_{i=1}^{n}1_{\{X_i\geq 1\}}(X_i)\big(\frac{X_i}{\lambda}-\frac{e^\lambda}{e^\lambda-1}\big)\stackrel{!}{=}0$$

Which is equivalent to
$$\frac{1}{\lambda}\frac{e^\lambda-1}{e^\lambda}=\frac{\sum_{i=1}^{n}1_{\{X_i\geq 1\}}(X_i)}{\sum_{i=1}^{n}1_{\{X_i\geq 1\}}(X_i)X_i}$$

How do I solve this equation? When can it be solved? (If I didn't make any mistakes this is a transcendental equation)

答:
The zero-modified Poisson PMF should be $$\Pr[X = k] = \begin{cases} p_0, & k = 0 \\ \frac{(1-p_0) \lambda^k}{k! \color{red}{(e^\lambda - 1)}}, & k \ge 1. \end{cases}$$

For convenience, denote $$w = \sum_{i=1}^n \mathbb 1(X_i = 0),$$ the number of observations in the sample that equal $0$, and $$s = \sum_{i=1}^n X_i$$ be the sample total.  It is worth noting that the sample total also happens to equal the total of nonzero observations, since zero-valued observations do not contribute to the sum in either case.  Then likelihood is $$\mathcal L(\lambda, p_0 \mid \boldsymbol x) \propto p_0^w (1-p_0)^{n-w} \frac{\lambda^s}{(e^\lambda - 1)^{n-w}.}$$  Moreover, because $\mathcal L(\lambda, p_0 \mid \boldsymbol x) = g(p_0 \mid \boldsymbol x) h(\lambda \mid \boldsymbol x)$ for suitable functions $g, h$, the joint MLE equals the marginal MLEs.  So we may ignore $p_0$ to search for the MLE of $\lambda$:  the marginal log-likelihood with respect to $\lambda$ is
$$\ell(\lambda \mid \boldsymbol x) \propto s \log \lambda - (n-w) \log(e^\lambda - 1).$$  Its critical points satisfy $$0 = \frac{s}{\lambda} - \frac{(n-w) e^\lambda}{e^\lambda - 1}$$ or equivalently $$\frac{1-e^{-\lambda}}{\lambda} = \frac{n-w}{s}  = \frac{\sum_{i=1}^n \mathbb 1(X_i \ge 1)}{\sum_{i=1}^n X_i}.$$  This agrees with your computation.  Its reciprocal can be interpreted as $$\frac{\lambda}{1-e^{-\lambda}} = \bar X^+,$$ where $\bar X^+$ is the sample mean of the nonzero observations only, and in light of the discussion so far, makes perfect intuitive sense:  since $p_0$ and $\lambda$ are in a sense independently estimable from the likelihood, the information contained in the zero observations is informative only of $p_0$, and the information contained in the nonzero observations is informative only of $\lambda$.  Unfortunately, this equation has no elementary closed-form solution.  We can employ Newton's method to numerically search for the MLE; specifically, $$\lambda_{m+1} = \frac{\lambda_m^2 - 2\bar X^+ (1 - \cosh \lambda_m)}{1 + \lambda_m - e^{\lambda_m}}, \quad \lambda_0 = \bar X^+,$$ should have good convergence characteristics.  For instance, $\bar X^+ = 3$ rapidly converges to the MLE estimate $$\hat \lambda = \lim_{m \to \infty} \lambda_m \approx 2.82144.$$


问:
Questions about proof for $\int_{E}f\,dm=0 \implies m(E)=0$
I have been working on reproducing the following proof for this theorem and have come up with a few questions I cannot seem to answer:

$(1)$ Given that $f$ is a measurable function such that $f >0$ a.e. Prove that, if $\displaystyle\int_{E}fdm=0$ for some measurable set $E$, then $m(E) = 0$. (where $m$ is the Lebesgue measure on $\Bbb{R}$)

My original attempt depended on a problem I did before this one in which I showed that for a Lebesgue measurable function $f$ on $\mathbb{R}$, $f$ is finite almost everywhere. In the proof of this Here, for $n \in \Bbb{N}$ define the set $E := \{x \in \Bbb{R}: f(x) \geq n\}$ such that $E := \bigcap_{n=1}^{\infty}E_n$, I came up with the following inequaility $$\int_{\mathbb{R}}fdm \geq \int_{E_n}fdm \geq n\cdot m(E_n) \geq n \cdot m(E).$$ I figured for $(1)$ I could use this inequality (and the set $E$) to my advantage and say: since $m(E) \leq \frac{1}{n} \int_{E}fdm$, we are given that $\int_{E}fdm=0$ so therefore $m(E) =0$. But, it seems there is an obvious counterexample here to my work. With that said, I started reading this proof and had some questions (I would comment and ask the author, but it seems they haven't been on this site for awhile):

$\bullet$ I'm really not seeing how $\mathbb R^n \setminus F$ has measure zero and why we need to know this for the rest of the proof?

$\bullet$ Why is this line correct: "if $m(E) \neq 0$ then we must have $m(F_n \cap E) >0$ for some $n$ and this would immediately imply that $\int_E f(x)dx>0$"? I don't think I understand why  $m(F_n \cap E) >0$ implies $\int_E f(x)dx>0$.

答:
Let $A=\{x\in\mathbb{R}\mid f(x)>0\}$, then $m(A^{c})=0$ because
$f>0$ a.e.. For each $n$, define $A_{n}=\{x\in\mathbb{R}\mid f(x)>\frac{1}{n}\}$.
We have that $\frac{1}{n}m(A_{n}\cap E)\leq\int_{A_{n}\cap E}f\leq\int_{E}f=0$
and hence $m(A_{n}\cap E)=0$. Note that $A=\cup_{n}A_{n}$, so $A\cap E=\cup_{n}(A_{n}\cap E)$.
Therefore, $m(A\cap E)\leq\sum_{n=1}^{\infty}m(A_{n}\cap E)=0$. Finally,
$m(E)=m(E\cap A)+m(E\cap A^{c})=0$.


问:
Minimize difference between 3 positive variables
I have a very simple problem to solve, but I'm trying to find the best way to do it. Suppose we have variables $x, y, z$ (all positive and real). I basically want to set up an optimization where all three variables should be as close as possible to each other (but not exactly equal). One approach I'm taking is to construct the following vector
$$P(x,y,z) = \begin{bmatrix}x - y \\ x - z \\ y-z \end{bmatrix}$$

and then $$\min_{ \{x,y,z\} \in \mathbb{R}^+} \| P(x,y,z)\|_2$$
With no other constraints, this optimization will obviously yield $x=y=z$. Lets suppose that we have other constraints such that $x\neq y \neq z$. Then what would be the best way to formulate a problem such that $x, y, z$ are all as close as possible to each other?

Your help is much appreciated!

答:
Your quest is for minimization under constraints. Suppose that you want to minimize $\|P(z,y,z)\|^2$ subject to $a x+b y + c \ge d$. A good way to handle this kind of problems is the lagrangian formulation introducing a slack variable to transform the inequality into an equation, as follows

$$
L(x,y,z,\lambda,s) = \|P(z,y,z)\|^2+\lambda(a x + b y + c z - d-s^2)
$$

The minima/maxima points are included into the set of $L(x,y,z,\lambda,s)$ stationary points. Those stationary points are determined by solving

$$
\nabla L = 0 = \left\{
\begin{array}{l}
 a \lambda +2 (x-y)+2 (x-z) \\
 b \lambda -2 (x-y)+2 (y-z) \\
 c \lambda -2 (x-z)-2 (y-z) \\
 a x+b y+c z-d-s^2 \\
 -2 \lambda  s \\
\end{array}
\right.
$$

and then we can find

$$
x = y = z = \frac{d}{a+b+c},\ \ s = 0
$$

$s=0$ meaning that the restriction is active.


问:
Find derivatives of functions with respect to $ x$
Can someone help me with these.

Find the derivatives of the following functions with respect to $x$: here $a$ is an arbitrary
(fixed) real number.

$(a)$ $\displaystyle\int_{a}^{x^4} t^3\ \mathrm dt$

$(b)$ $\displaystyle \int_{-x^3}^{a^3} \dfrac{\mathrm dt}{1+t^2}$

答:
For example, let $\,F(x)\,$ be a function s.t. $\,F'(x)=x^3\,$ (why such a a function exists?), so that (assuming you've a typo in the first integral's upper limit):

$$g(x):=\int\limits_a^{x^4}t^3\,dt=F(x^4)-F(a)\implies g'(x)\stackrel{\text{chain rule}}=4x^3F'(x^4)=4x^3(x^3)^4=4x^{15}$$

Now you try in a similar fashion the other one.


问:
Showing $1+\cos(\theta)+\dots+\cos (n\theta)=\frac12+\frac{\sin((n+\frac12)\theta)}{2\sin(\frac\theta2)}$

Suppose that $n \geq 1$ is a natural number.

(i) Show that for $z$ in $\mathbb{C}$, provided $z \neq 1$,

$$ \frac{1-z^{n+1}}{1-z} = 1+z+...+z^n.$$

(ii) Using de Moivre’s Theorem, part (b)(i) and standard trigonometric identities, show that for $\theta$ in $(0,2\pi)$,

$$ 1+\cos(\theta) + \dots + \cos (n\theta) = \frac{1}{2} + \frac{\sin((n+\frac12)\theta)}{2\sin(\frac\theta2)}$$

I have set $z=a+bi$ and subbed this into the top equation but dont seem to be getting anwhere with this problem. Any guidance will be great!

答:
Part (i) should be straightforward if you are familiar with finding the sum of a geometric series:
\begin{align}
S &= 1+z+z^2+\ldots+z^{n-1}+z^n \\
zS &= z+z^2+z^3+\ldots + z^n+z^{n+1} \\
zS - S &= z^{n+1}-1 \\
S(z-1) &= z^{n+1}-1 \\
S &= \frac{z^{n+1}-1}{z-1} = \frac{1-z^{n+1}}{1-z} \, . \\
\end{align}
For part (ii), you should first consider the geometric series
$$
1+e^{i\theta}+e^{2i\theta}+\ldots+e^{ni\theta} \, .
$$
Since $e^{ik\theta}=\cos k\theta+i\sin k\theta$ for $k \in \mathbb{Z}$, we know that
\begin{align}
&1+e^{i\theta}+e^{2i\theta}+\ldots+e^{ni\theta}\\=&1+(\cos\theta+i\sin\theta)+(\cos 2\theta+i\sin 2\theta)+\ldots+(\cos n\theta+i\sin n\theta)
\end{align}
Hence,
$$
1+\cos\theta+\cos 2\theta+\ldots+\cos n \theta=\Re(1+e^{i\theta}+e^{2i\theta}+\ldots+e^{ni\theta}) \, .
$$


问:
computing $\int_{-\infty}^{\infty} x e^{itx} dx$ using Cauchy Theorem and gamma function rapresentation
I want to solve the above integral with $t \in \mathbb{R}$; using a software (Wolfram) to verify the result I obtain the following result:

$I(t) \equiv \int_{-\infty}^{\infty} x e^{itx} dx = -i \frac{d}{dt}\int_{-\infty}^{\infty} e^{itx} dx = -i2\pi \frac{d \delta(t)}{dt}$

To solve it I start rewritting $I(t)$ as:

$\int_{-\infty}^{\infty} x e^{itx} dx = \int_{-\infty}^{0} x e^{itx} dx + \int_{0}^{\infty} x e^{itx} dx = -\int_{0}^{\infty} w e^{-itw} dw + \int_{0}^{\infty} y e^{ity} dy $

I tried to solve both the last integrals using Cauchy's integral theorem; In my case I have (following the notation of the last link) $\Re(\beta) = 0$, but this shouldn't be a problem, In fact I can write $t = lim_{h \rightarrow 0} t \pm ih$ and interchange the limit with the integral (using DCT).

With this approach I obtain the following result:

$I(t) = -\frac{\Gamma(2)}{(it)^2} + \frac{\Gamma(2)}{(-it)^2}$

which is 0 for all $t$ except for $t=0$ where I have an indeterminate form $-\infty + \infty$.

Is it correct this solution?

答:
No.

$$I(t) = \int_{-\infty}^{\infty} x e^{itx} dx$$
diverges for all $t$. It converges only in the sense of distributions.

$$I_N(t)=\int_{-N}^N x e^{itx} dx=-2i \frac{d}{dt}\frac{\sin(Nt)}{t}$$
$$h(t)=\int_{-\infty}^t 2\frac{\sin(u)}{u}du, \qquad h(Nt)\to 2\pi 1_{t> 0}$$
For all $\phi \in C^\infty_c$
$$\lim_{N\to \infty} \int_{-\infty}^\infty I_N(t)\phi(t)dt=\lim_{N\to \infty} \int_{-\infty}^\infty i\, h(Nt)\phi''(t)dt= \int_{-\infty}^\infty 2i\pi 1_{t>0}\phi''(t)dt=2i\pi\phi'(0)$$
This is the definition of $\int_{-\infty}^{\infty} x e^{itx} dx$ converges to $-2i\pi\delta'(t)$ in the sense of distributions.

To use the Gamma function you need to do the same kind of idea, saying that $\lim_{a\to 0}\int_{-\infty}^{\infty} x e^{itx}e^{-a|x|} dx$ converges to $\int_{-\infty}^{\infty} x e^{itx} dx$ in the sense of distributions.

You'll get (something like) $\lim_{a\to 0} \frac1{(it+a)^2}-\frac1{(it-a)^2}$, the second derivative of $\lim_{a\to 0} \log(it-a)-\log(it+a)=C\ sign(t)$


问:
Matrix operations - equivalent operation for a given operation
This is the given problem, I need to write a code for this:

$(M*Q) \circ (N*Q) $

where $M,Q,N$ are known matrices, "$\ast$" denotes matrix multiplication and "$\circ$" denotes elementwise division.

Dimensions of the matrices:
\begin{align*}
M: &a\times l,\\
N: &a\times l,\\
Q: &l\times1.
\end{align*}

By element-wise division, I mean this:

$A \circ B = C$ where $C_{ij} = A_{ij}/B_{ij} \ \ \forall\ i,j$.

I want to convert this above problem to this:

Find $K$ such that $K*Q = (M*Q) \circ (N*Q)$.

答: In general, this is impossible because $(M*Q) \circ (N*Q)$ is a rational function in the entries of $Q$ but $K*Q$ is linear. For instance, consider $M=(1,0),N=(0,1)$ and $Q=\pmatrix{x\\ y}$. Then $(M*Q) \circ (N*Q)=\frac xy$. Surely it is not a linear function in $x$ and $y$.</p>


问:
Question about a worked out problem regarding epsilon-delta Proof
my question is regarding the last part since $|x+3|<6$ I thought it must be that $$
|x-2|<\frac{\epsilon}{6}<\frac{\epsilon}{|x+3|}
$$
and not $$
|x-2|<\frac{\epsilon}{|x+3|}<\frac{\epsilon}{6}
$$ I think I'm missing something, maybe someone can help me with this problem. the whole proof is in this paper it is the 4th proof.
$$
\lim _{x \rightarrow 2} x^{2}+x-2=4
$$
We start the same way we always do. By now this part should be automatic.
$$
\begin{aligned}
|f(x)-L|<\epsilon & \Longrightarrow\left|\left(x^{2}+x-2\right)-4\right|<\epsilon \\
& \Longrightarrow\left|\left(x^{2}+x-6\right)\right|<\epsilon \\
& \Longrightarrow|(x+3)(x-2)|<\epsilon \\
& \Longrightarrow|x+3||x-2|<\epsilon \\
& \Longrightarrow|x-2|<\frac{\epsilon}{|x+3|}
\end{aligned}
$$
Once again we have the left side in the form $|x-a| .$ Now we can let $\delta$ equal $\frac{\epsilon}{|x+3|}$. But what do we do about the $|x+3| .$ In general $\delta$ must be in terms of $\epsilon$ only, without any extra variables.
So how we can remove this $x+3$ term?
First we need to simplify the problem a little bit. Since the concept of limit only applies when $x$ is close to $a$, we will first restrict $x$ so that it is at most 1 away from $a$, or, mathematically, $|x-a|<1$ (in our case $|x-2|<1$. Then, this means, $1<x<3$, or $4<x+3<6$.
Now consider the original inequality
$$
\begin{array}{c}
|x-2|<\frac{\epsilon}{|x+3|}
\end{array}
$$
Notice that the right hand side is at the minimum when $x+3$ is at its maximum. Since the maximum of $x+3$ is $6,$ we know that
$$
|x-2|<\frac{\epsilon}{|x+3|}<\frac{\epsilon}{6}
$$

答:
It seems that your logic is going in the wrong direction. While it is often best to start with $|f(x) - L| < \varepsilon$ when figuring out the proof, ultimately your logical argument has to start with assuming $0 < |x - a| < \delta$ (where $\delta$ is the function of $\varepsilon$ you've chosen), and it has to end by concluding $|f(x) - L| < \varepsilon$.

Using your work so far, it looks like we need $\delta = \min\{1, \varepsilon / 6\}$. We then have
\begin{align*}
0 < |x - 2| < \min\left\{1, \frac{\varepsilon}{6}\right\} &\implies |x - 2| < 1 \text{ and } 6|x - 2| < \varepsilon \\
&\implies |x + 3| < 6 \text{ and } 6|x - 2| < \varepsilon \\
&\implies |x + 3||x - 2| < 6|x - 2| < \varepsilon \\
&\implies |x^2 + x - 2 - 4| < \varepsilon,
\end{align*}
completing the proof. Note that the logic makes perfect sense, when written in order!

EDIT: Let me see if I can explain a bit better. I still think this is a misconception about the direction of the flow of logic, but we can make this clearer, I think. You are absolutely right when you say it should be
$$|x-2|<\frac{\varepsilon}{6}<\frac{\varepsilon}{|x+3|}, \tag{1}$$
not
$$|x-2|<\frac{\varepsilon}{|x+3|}<\frac{\varepsilon}{6}. \tag{2}$$
And it is. Really, we are using $(1)$, not $(2)$. We want to put the step $|x - 2| < \varepsilon/6$ (where the $\varepsilon/6$ is derived from our choice of $\delta$) before the step $|x - 2| < \varepsilon / |x + 3|$. We want the logic to go:
$$|x - 2| < \frac{\varepsilon}{6} \implies |x - 2| < \frac{\varepsilon}{|x+3|},$$
which is what $(1)$, not $(2)$, delivers to us. This is because, for our choice of $\delta$, $|x - 2| < \varepsilon/6$ follows easily from $0 < |x - 2| < \delta$, and $|x^2 + x - 2 - 4| < \varepsilon$ follows easily from $|x - 2| < \varepsilon / |x + 3|$.

As I said, I think you were viewing this from the wrong direction. In your working, you very explicitly started with $|x^2 + x - 2 - 4| < \varepsilon$, and worked backwards. You reached the step $|x - 2| < \varepsilon / |x + 3|$ first, and started looking for ways to show that this implied $|x - 2| < \varepsilon/6$.

This implication is false. In order for something like that to be true, you'd want $(2)$ to be true, but it isn't. That way, you could then conclude that $|x - 2| < \varepsilon / 6$, and then end with $|x - 2| < \delta$. But this is simply not what the definition of a limit requires. You don't need to show
$$0 < |x - a| < \delta \impliedby |f(x) - L| < \varepsilon,$$
you need to show the converse instead.


问:
How to find the middle point of an arc slice?
I need to find centre of different arc in the Pie Chart. I have following details -

Similar question is here -
How to find the middle point of an arc slice?

here is the reference image from above post -

My Scenario. When I move to different slice I need to calculate centre of that slice. This is the piechart centre coordinates -

```<code>cx = 330.5;
cy = 140;
</code>```

First (Green) slice details -

```<code>{
    "cx": 330.5,
    "cy": 140,
    "innerRadius": 0,
    "middleRadius": 54,
    "outerRadius": 108,
    "maxRadius": 352.3850876526985,
    "startAngle": 0,
    "endAngle": 17.358490566037737,
    "midAngle": 8.679245283018869,
    "paddingAngle": 0,
    "percent": 0.04821802935010482,
}
</code>```

Second (Orange) slice details -

```<code>{
    "cx": 330.5,
    "cy": 140,
    "innerRadius": 0,
    "middleRadius": 54,
    "outerRadius": 108,
    "maxRadius": 352.3850876526985,
    "percent": 0.36547868623340324,
    "startAngle": 17.358490566037737,
    "midAngle": 83.14465408805033,
    "endAngle": 148.93081761006292,
    "paddingAngle": 0,
    "percent": 0.36547868623340324,
}
</code>```

Third (Gray) slice details -

```<code>{
    "cx": 330.5,
    "cy": 140,
    "innerRadius": 0,
    "middleRadius": 54,
    "outerRadius": 108,
    "maxRadius": 352.3850876526985,
    "startAngle": 148.93081761006292,
    "midAngle": 254.46540880503147,
    "endAngle": 360,
    "paddingAngle": 0,
    "percent": 0.5863032844164919,
}
</code>```

I tried following formula but it did not work and not getting the expected result -

```<code>x = cx + outerRadius * Math.cos(midAngle),
y = cy + outerRadius * Math.sin(midAngle),
</code>```

Update

After changing the formula to

```<code> x = cx + outerRadius * Math.cos(midAngle * 3.14159265358979323846 / 180.0);
 y = cy + outerRadius * Math.sin(midAngle * 3.14159265358979323846 / 180.0);
</code>```

I am getting following output -
Completed -

```<code>{x: 437.2632503544796, y: 156.29749593485772}
</code>```

In progress -

```<code>{x: 343.3912133107331, y: 247.2278723997505}
</code>```

Not started -

```<code>{x: 301.5754291795817, y: 35.9453547271694}
</code>```

You can see in all above three images tooltip position is not correct.

答:
So you want the point on the midangle of the sector and halfway from the center to the circumference:

$$
x = x_c + \frac{1}{2}R\cos(\theta)
$$
$$
y = y_c + \frac{1}{2}R\sin(\theta)
$$
where $\theta$ is the midangle in radians. As the comments note, you convert the midangle in degrees (denoted by $\theta_{mid}^\circ$ below) to radians by dividing by 360 and multiplying by $2\pi$: $\theta = \frac{2\pi}{360}\theta_{mid}^\circ$.

In terms of your programming variables:

```<code>x = cx + 0.5 * outerRadius * Math.cos(midAngle * 3.14159265358979323846 / 180.0);
y = cy + 0.5 * outerRadius * Math.sin(midAngle * 3.14159265358979323846 / 180.0);
</code>```

although I would at least  define a constant PI instead of using 3.14.... in two places:

```<code>PI =  3.14159265358979323846
x = cx + 0.5 * outerRadius * Math.cos(midAngle * PI / 180.0)
y = cy + 0.5 * outerRadius * Math.sin(midAngle * PI / 180.0)
</code>```

But as you can see the only difference with your formula is the halving of the radius.

Body:
So you want the point on the midangle of the sector and halfway from the center to the circumference:

$$
x = x_c + \frac{1}{2}R\cos(\theta)
$$
$$
y = y_c + \frac{1}{2}R\sin(\theta)
$$
where $\theta$ is the midangle in radians. As the comments note, you convert the midangle in degrees (denoted by $\theta_{mid}^\circ$ below) to radians by dividing by 360 and multiplying by $2\pi$: $\theta = \frac{2\pi}{360}\theta_{mid}^\circ$.

In terms of your programming variables:

x = cx + 0.5 * outerRadius * Math.cos(midAngle * 3.14159265358979323846 / 180.0);
y = cy + 0.5 * outerRadius * Math.sin(midAngle * 3.14159265358979323846 / 180.0);

although I would at least  define a constant PI instead of using 3.14.... in two places:

PI =  3.14159265358979323846
x = cx + 0.5 * outerRadius * Math.cos(midAngle * PI / 180.0)
y = cy + 0.5 * outerRadius * Math.sin(midAngle * PI / 180.0)

But as you can see the only difference with your formula is the halving of the radius.


问:
Definite integral of [{Cos^-1(x)}]/√(1-x^2) from 0 to 1
If we substitute $cos^{-1}(x)$ with $z$ then,corresponding values of $x=1$ are $z=0,2\pi,4\pi...$
And correspond values of $x=0$ are $z=\pi/2,3\pi/2,5\pi/2...$

Now if for $x$ belongs to $[0,1]$ corresponding value is taken $z$ belongs to $[\pi/2,0]$ then the value of integral comes out to be $(\pi^2)/8$ but for $z$ belongs to $[5\pi/2,2\pi]$ the value of integral comes out to be $9(\pi^2)/8$

Which interval should be taken?

I apologise for stating anything wrong if I have done any.

Edit: previously the question was written about {cos^-1(x)}^2 by mistake

答: As @Gae.S. notes, you should take $\arccos0=\pi/2,\,\arccos1=0$ so$$\int_0^1\frac{\arccos^2xdx}{\sqrt{1-x^2}}=\left[-\frac13\arccos^3x\right]_0^1=\frac{\pi^3}{24}.$$


问:
Proof of the Lebesgue-Radon-Nikodym Theorem
Theorem: Let $\lambda, \mu$ be $\sigma$-finite measures defined on the $\sigma$-algebra $\mathcal{A}$ of the space $X$. Then,

a) Lebesgue decomposition: $\lambda=\lambda_a+\lambda_s$ where $\lambda_a \ll \mu$ and $\lambda_s \perp \mu$

b) $\exists h \in L^1(\mu)$ such that $\lambda_a = \int h d\mu$

Proof:

Define $\phi=\lambda + \mu \space \phi(X)<\infty, \space f \in L^2(\phi) $. Then $f \to \int f d\lambda $ is a continuous linear functional on $L^2(\phi)$ Because:
$$\int_X |f| d\lambda \le \int |f| d\phi \le \Bigg( \int_X |f|^2 d\phi \Bigg)^{\frac12}\Bigg( \int_X d\phi \Bigg)^{\frac12} $$

Q1: How does this prove the aforementioned claim? Don't we already have $f \to \int f d\lambda $ is a continuous linear functional on $L^2(\phi)$  from the fact that $f \in L^2(\phi) $?

Moving on, from the Riesz Representation theorem we know, that $\exists ! \space g \in L^2(\phi) $ such that $$\int_X f d\lambda = \int_X fg d\phi $$.

Choosing $f=\chi_E, \space E \in \mathcal{A}, \space \phi(E)<\infty$ we obtain $\lambda(E)=\int_E g d\phi$ and I'm told, that

$$\lambda(E)=\int_E g d\phi \Rightarrow 0 \le \frac{1}{\phi(E)}\int_E g d\phi = \frac{\lambda(E)}{\phi(E)} \le 1$$

Q2 Why does the above implication hold?

From the above equality we know, that $g(x) \in [0,1]$, an absolutely crucial fact for the rest of the proof. We define $A=\{x:g(x)<1\} \text{ and } B=\{x: g(x)=1\}$. Now,
$$\lambda(E) = \lambda(E\cap A) + \lambda(E \cap B) = \lambda_a (E)+\lambda_s(E)$$

$$\int_X f d\lambda = \int_X fg d (\lambda+\mu) \Rightarrow \int_X f(1-g) d \lambda = \int_X fg d \mu $$.

Choosing $f=\chi_B$ we obtain $\mu(B)=0 \Rightarrow \mu \perp \lambda_s$

Choosing $f=\chi_E \sum_{k=0}^n g^k$ we obtain

$$\int_E (1-g^{n+1}) d \lambda = \int_E \sum_{k=1}^n g^k d \mu \\ \lim_{n \to \infty} \int_E (1-g^{n+1}) d \lambda = \lim_{n \to \infty} \int_E \sum_{k=1}^n g^k d \mu$$.

and since  $(1-g^{n+1}) \nearrow 1$ and $\sum_{k=1}^n g^k \nearrow h$ we can use the monotone convergence theorem to get

$$\lambda(E\cap A)=\lambda_a(E) = \int_E h d \mu $$

Remark: for $x \in B$ we get 0 measure because $g(x)=1$.

答:
Question 1: We define a functional $\ell\colon L^2(\phi) \to \mathbb K$ by
$$ \ell(f) = \int_X f \, d\lambda, \quad f \in L^2(\phi). $$
So to each $f \in L^2(\phi)$ we associate a number, namely the integral over $X$ with respect to $\lambda$. To know that $f$ is linear and continuous, we have to prove it. Linearity is easy, so we concentrate on continuity, we have for all $f \in L^2(\phi)\def\abs#1{\left|#1\right|}\def\norm#1{\left\|#1\right\|}$
\begin{align*}
  \abs{\ell(f)} &= \abs{\int_X f\, d\lambda}\\
                &\le \int_X \abs f\, d\lambda\\
                &\le \int_X \abs f \, d(\lambda + \mu) \text{   note that $\int_X \abs f \, d\mu \ge 0$}\\
                &= \int_X \abs f\cdot 1 \, d\phi\\
                &\le \left(\int_X \abs{f}^2\, d\phi\right)^{1/2} \left(\int_X 1\, d\phi\right)^{1/2} \text{  this is Cauchy-Schwarz}\\
                &= \phi(X)^{1/2} \cdot \norm f_{L^2(\phi)}
\end{align*}
 So $\ell$ is a bounded, hence continuous linear functional on $L^2(\phi)$.

Question 2: As $\lambda$ and $\phi$ are positive measures, we obviously have $\lambda(E)$, $\phi(E)\ge 0$, moreover note, that by defintion
$$ \phi(E) = \lambda(E) + \mu(E) \ge \lambda(E) $$
So for $\phi(E) > 0$, we have
$$ 0 \le \frac{\lambda(E)}{\phi(E)}  = \frac{\lambda(E)}{\lambda(E) + \mu(E)} \le 1. $$


问:
If $X$ Has CDF F, What Is The CDF Of $Y=e^X$?
If $X$ Has CDF F, What Is The CDF of $Y=e^X$?

Frankly, I'm not even sure where to start. Any pointers would be very helpful.

答:
I'm not even sure where to start

If I was in you I would start with the definition of CDF

$$F_Y(y)=\mathbb{P}(Y\leq y)=\mathbb{P}(e^X\leq y)=\mathbb{P}(X\leq \log y)=F_X(\log y)$$


答:
Start with the definition.  The CDF of $e^X$ is the function
$F_{e^X}(t) = \mathbb P(e^X \le t)$.  So what does $e^X \le t$ tell you about $X$?

Title: If $X$ Has CDF F, What Is The CDF Of $Y=e^X$?

Body:
If $X$ Has CDF F, What Is The CDF of $Y=e^X$?

Frankly, I'm not even sure where to start. Any pointers would be very helpful.


问:
Maximum possible edge betweenness of a graph?
Betweenness, roughly speaking, measures how many times a particular node or edge in a graph is part of the shortest path, when counting paths among all pairs of nodes.  For nodes, it's easy to see that a star graph forces the central node to have a maximum betweenness as it lies between every pair of nodes.

I'm trying to determine if there exists a graph such that a particular edge is on every shortest path (and therefore has maximum edge betweenness).  Obviously, for an undirected graph, there are a maximum of $\frac{n(n-1)}{2}$ shortest paths.  But considering some small examples, I was unable to design a graph such that any edge actually achieved this max.  Is such a (weighted) graph possible (by proof or construction), or can we find the actual maximum betweenness?

My use case is normalizing these edge betweenness scores.  I could just use the number of shortest paths (which is what networkx and I suspect other implementations do), but if we can figure out what the actual max is (or some demonstrable estimates), I'd like to use that.

Edit: To clarify, I'm considering a weighted graph (possibly with zero weights allowed, though practically that's less interesting).  But analyzing an unweighted graph may of course still be helpful.

答:
Suppose that a shortest path from vertex $v$ to vertex $w$ in the graph uses edge $xy$: let's say the path goes $(v, \dots, x, y, \dots, w)$. Then distances from $v$ are strictly increasing along the path, and distances from $w$ are strictly decreasing: in particular, $d(v,x) < d(v,y)$ and $d(w,x) > d(w,y)$.

This means that all shortest paths using edge $xy$ must start at a vertex closer to $x$ than to $y$, and end at a vertex closer to $y$ than to $x$. If there are $n_x$ vertices of the first type, and $n_y$ of the second, then at most $n_x \cdot n_y$ pairs of vertices have a shortest path using edge $xy$.

With the constraint $n_x + n_y \le n$ and $n_x, n_y \ge 0$, $n_x \cdot n_y$ is maximized at $\lfloor \frac{n^2}{4}\rfloor$ when $n_x = n_y = \frac n2$ (for even $n$) or when $\{n_x, n_y\} = \{\frac{n-1}{2}, \frac{n+1}{2}\}$ (for odd $n$).

This bound is achievable for all $n$, by letting $xy$ be the only edge between two connected subgraphs, one with $n_x$ vertices and one with $n_y$ vertices.


Title: Definite integral of [{Cos^-1(x)}]/√(1-x^2) from 0 to 1

Body:
If we substitute $cos^{-1}(x)$ with $z$ then,corresponding values of $x=1$ are $z=0,2\pi,4\pi...$
And correspond values of $x=0$ are $z=\pi/2,3\pi/2,5\pi/2...$

Now if for $x$ belongs to $[0,1]$ corresponding value is taken $z$ belongs to $[\pi/2,0]$ then the value of integral comes out to be $(\pi^2)/8$ but for $z$ belongs to $[5\pi/2,2\pi]$ the value of integral comes out to be $9(\pi^2)/8$

Which interval should be taken?

I apologise for stating anything wrong if I have done any.

Edit: previously the question was written about {cos^-1(x)}^2 by mistake

Body:
$$I=\int_{0}^{1} \frac{(\cos^{-1}x)^2}{\sqrt{1-x^2}} dx$$
Let $\cos^{-1} x=t \implies -\frac{dx}{\sqrt{1-x^2}} =dt$
So $$=\int_{0}^{\pi/2} t^2 dt= \frac{t^3}{3}|_{0}^{\pi/2}=\frac{\pi^3}{24}.$$


问:
Spliting Field over $\mathbb{F}_3$
How to find the splitting field of $f(x)=x^3-x+1$ and $g(x)=x^3-x-1$ over $\mathbb{F}_3$ and how to construct a isomorphism between them?</p>

答:
let $a$ is a root of $f(x)$ in the splitting field,

then $f(a)=0=a^3-a+1=(a+1)^3-(a+1)+1=(a-1)^3-(a-1)+1$

so, $(x-a)(x-(a+1))(x-(a-1))=f(x)$

splitting field of $f(x)$ is $F_3[x]/<f(x)>=F(a)$ where $a$ is a root of $f$

similarly you can do for $g(x)$

Title: Spliting Field over $\mathbb{F}_3$


问:
Computing integral $\int_0^1 |x-a| \, {\rm d} x$
I have the following question for a seminar and cannot find the relevant information to solve it.

Solve the integral for all $a \in \mathbb R$ $$I(a)=\int_0^1 |x-a| \, {\rm d} x$$

I understand that I have to treat the absolute value of the function $f(x)=|x-a|$ depending on the when $x$ is less or more then $a$, if $a$ had a defined value I could have solved this, but now im stuck because I dont know how to relate its unspecified value to find a solution for the integral.

Any help is appreciated that'll point me in the right direction.

答:
The integrand is an absolute value function and changing $a$ values correspond to the horizontal translation of the function $|x|$. It is best explained visually:

I think it's quite practical to visualize the integral for a variety of $a$ values. For all real values of $a$ in $(-\infty,0)\cup (1,\infty)$, the shape of the area given by the integral is trapezoidal. On the other hand, for all real values of $a$ in the closed interval $[0,1]$, the area is made up of triangles. In the light of these, we can evaluate the integral using area formulas.

For the values of $a \in (-\infty,0)$, the area bounded by the integrand function and the vertical line $x=1$ is trapezoidal. It is given by: $$\begin{align}\frac{|-a|+|1-a|}{2}
&=\frac{(-a)+(1-a)}{2} \\
&=\frac{1-2a}{2} \\
&=\frac{1}{2}-a \\ \end{align}$$

For the values of $a \in (1,\infty)$, the area bounded by the integrand function and the vertical line $x=1$ is again trapezoidal. It is given by: $$\begin{align} \frac{|-a|+|1-a|}{2}
&=\frac{(a)+(-1+a)}{2} \\
&=\frac{2a-1}{2} \\
&=a-\frac{1}{2} \\ \end{align}$$

For the values of $a \in [0,1]$, the area is given by the sum of areas of triangles: $$\begin{align} \frac{|-a|a}{2}+\frac{(1-a)|1-a|}{2}
&=\frac{a^2}{2}+\frac{1-2a+a^2}{2} \\
&=\frac{1}{2}-a+a^2 \\ \end{align}$$
Finally, the resultant integral is expressed as:

$$I(a)=\int_{0}^{1} |x-a|\space dx= \begin{cases} a-\frac{1}{2},\space \space \forall a\gt 1 \\
\frac{1}{2}-a+a^2,\space \space \forall a \in [0,1] \\
\frac{1}{2}-a, \space \space \forall a\lt 0
\end{cases}$$


问:
How to tell if a series converges/diverges based on if the sequence converges?
Maybe a stupid question, but if a sequence converges, can it be predicted that the series will also converge?</p>

答:
Here are a couple of examples that serve to illustrate that it difficult to make general statements about whether or not a series converges, based solely on the convergence of the terms of the series:

The sequence converges but the series diverges

$$
1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots
$$

(If a series is convergent, then its terms must approach $0$. However, the converse is not true: if the terms approach $0$, then the series is not necessarily convergent, as shown by the example above.)

The sequence and the series both converge

$$
1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots
$$

The sequence and the series both diverge

$$
1-1+1-1+\ldots
$$

The sequence diverges but the series converges

(No examples here. The divergence test says that if the limit of the sequence is non-zero, or does not exist, then the series diverges.)


问:
Proving a condition for when 2 bases are linearly independent
I'm currently attempting a problem in a linear algebra textbook. In the solution sheet this statement is said to be a necessary and sufficient condition for the union of 2 bases to be a basis but i have no clue how to prove it.

$$B_1 \cup B_2 \text{ is linearly independent} \iff span(B_1 \cap B_2) = span(B_1) \cap span(B_2)$$
Where $B_1$ and $B_2$ are bases

I think maybe in the $\Longrightarrow $ direction the union of $B_1$ and $B_2$ being linearly independent implies $span(B_1) \cap span(B_2) \subseteq span(B_1 \cap B_2)$ ? but i'm not sure where to go from here.
I don't know where to start in the $\Longleftarrow$ direction.

答:
Let $sp(B_1)=W_1,sp(B_2)=W_2$(sp stands for span)

$B_1 \cup B_2$ spans $W_1+W_2$. Let $w \in W_1+W_2$. Then we have $w=w_1+w_2, w_1 \in W_1, w_2 \in W_2$ by definition. $w_1$ is a linear combination of the vectors in $B_1$. Similarly $w_2$ is a linear combination of vectors in $B_2$. So, we have $w$ is a linear combination of $B_1 \cup B_2$.

$B_1 \cap B_2$ is linearly independent in $W_1 \cap W_2$(straightforward from the condition that the vectors in a basis are linearly independent).

Now,

$$B_1 \cup B_2 \text{ is linearly independant} \iff B_1 \cup B_2$$ form a basis of $W_1+W_2$
$$\iff \dim(W_1+W_2)=|B_1 \cup B_2|=\dim(W_1)+\dim(W_2)-\dim(W_1 \cap W_2)$$(well known result)
$$\iff |B_1 \cup B_2| = |B_1| + |B_2| - \dim(W_1 \cap W_2)$$
$$\iff \dim(W_1 \cap W_2) = |B_1 \cap B_2|$$
$$\iff B_1 \cap B_2$$ form a basis of $\dim(W_1 \cap W_2)$
$$\iff sp(B_1 \cap B_2) = W_1 \cap W_2 = sp(B_1) \cap sp(B_2)$$


问:
Calculating Wavelength Required To Break Bonds Using The Photon Energy Equation: ${λ = \frac{hc}{E}}$
I'm asking this here because it's a maths type question

I seem to get a much warmer response here despite making silly mistakes:

My friend told me he is going to use 365nm light to irradiate Ibotenic acid -> Muscazone

I mentioned to him that this is barely below the visible spectrum and that you need a specific amount of energy to break bonds using photons, so I went on a little adventure:

First I had to calculate the bond energy of each species which was very tedious, I wonder if there is a better way:

```<code>Ibotenic acid:
O-H  = 2*463
C--N = 615
C-C  = 3*346
C--C = 602
C-O  = 3*358
N-O  = 201
C-N  = 2*305
C--O = 799
N-H  = 2*391
{Total bond energy = 6037 kj/mol}

Muscazone:
O-H  = 463
C-O  = 3*358
C-C  = 3*346
C--C = 602
N-O  = 201
C-N  = 4*305
C--O = 799
N-H  = 3*391

Total Bond energy = 6570 kj/mol
</code>```

$$
 E=6037000-6570000=-533000j\cdot mol^{-1}
$$

planck equation:

$$
\lambda = \frac{hc}{E}
$$

Dimensional analysis:

$$
\lambda = \frac{j\cdot s\cdot m}{j\cdot s\cdot n} \Rightarrow \lambda = \frac{m}{n}
$$

Seems that I can't get rid of n or mol so I used avogadros number:

$$
\lambda = \frac{(6.63\cdot 10^{-34})  (2.99\cdot 10^{8})}{(-533000)
(6.02\cdot 10^{23})}
$$

$$
\lambda = 224nm
$$

Is my reasoning, maths and answer correct?

答:
Here are just a couple of notes that might help you understand the problem better:

There is no better way to calculate the total bond energy $E_{mole}$

Since the final energy is larger, it means that the molecules must absorb energy to transform into the final product. That's the meaning of the $-$ sign.

One photon is absorbed by one molecule, so you need to calculate the bond energy per molecule $E_{molecule}=E_{mole}/N_A$

Transform this energy to wavelength $$\lambda=\frac{hc}{E_{molecule}}=\frac{hcN_A}{E_{mole}}$$


问:
Prove uniform equicontinuity and boundness
Suppose
$K = \{x(t) \in C[a,b], \forall x \  x(t) = 0 \ has \ a \ solution \}$.

Prove that $K$ is uniformly bounded and uniformly equicontinuous.

Could anyone spare a hint on how to use the fact every function in the set has at least one root?

答:
In fact, the set $K$ is not uniformly bounded or equicontinuous.

Consider the family $x_n \in C[a,b]$ defined by $$x_n(t) = n\left(t - \tfrac{a+b}2 \right), \,\,\,\,\,\,\,\,\, t \in [a,b].$$ These are just increasingly steep linear functions which are zero at the midpoint $t = \frac{a+b}2$. The maximum then occurs at $t =a$ or $t = b$ , and has a value $$\lvert x_n(a)\rvert = \lvert x_n(b) \rvert = \frac{n(b-a)}{2}$$ which is unbounded as $n$ gets large. Therefore $K$ is not uniformly bounded. Likewise, since each function $x_n$ has increasingly large slope, the family is not equicontinuous.


问:
Extrapolating an abstract algebra proof, arriving upon an incorrect conclusion.
Could you kindly point out what is wrong with my reasoning?

EDIT: What I have unintendedly proven through my reasoning is that every field can only have one automorphism- the identity mapping. Hope this helps in navigating the mess below

Let $k_{1},k_{2},\dots k_{n}$ be an arbitrary number of distinct, non-zero elements $\in\mathbb{K}$. $\mathbb{K}$ is a field of $0$ characteristic.

Also, $\sigma_{1},\sigma_{2},\dots \sigma_{m}$ are distinct automorphisms of $\mathbb{K}$.

I will show that $m\leq n$.

Let us assume that $m>n$. We can form the following system of linear equations:

$$\sigma_{1}(k_{1})x_{1}+\sigma_{2}(k_{1})x_{2}+\dots \sigma_{m}(k_{1})x_{m}=0$$ $$\sigma_{1}(k_{2})x_{1}+\sigma_{2}(k_{2})x_{2}+\dots \sigma_{m}(k_{2})x_{m}=0$$ $$\dots$$$$\dots$$$$\sigma_{1}(k_{n})x_{1}+\sigma_{2}(k_{n})x_{2}+\dots \sigma_{m}(k_{n})x_{m}=0$$

As $m>n$, we are assured a non-zero solution for $(x_{1},x_{2},\dots x_{m})$. Let it be $(a_{1}, a_{2},\dots a_{m})$, not all zero.

Then, taking the first linear equation into consideration, $$\sigma_{1}(k_{1})a_{1}+\sigma_{2}(k_{1})a_{2}+\dots \sigma_{m}(k_{1})a_{m}=0, a_{i}\in\mathbb{K}$$

This is not possible (I can give a proof, but it would unnecessarily lengthen the discussion).

Hence, on selecting an arbitrary ($n$) number of elements from $\mathbb{K}$, this result shows that the number of automorphisms $\mathbb{K}$ can have $\leq n$. This directly implies that the number of automorphisms $\mathbb{K}$ has is $1$- the identity automorphism.

It is quite obvious that this conclusion is wrong. What is erroneous in my thought process though?  Thanks in advance!

答:
The mistake must lie in the assertion that

$$\sigma_1(k_1)a_1 + \cdots +\sigma_m(k_1)a_m = 0$$

is impossible.

Namely, if we let $k_1 = 1$, it is trivial to see that there will be $a_1\ldots a_m$ making this true: just let $a_m = -(a_1+\cdots + a_{m-1})$. This is in direct contradiction with your observation.

Without your purported proof of the above statement, not much more can be said.


问:
Find permutation matrix that minimizes a Frobenius norm
Is there an efficient way to solve (maybe approximately) the following optimization problem

$$\left\| A - X B X^T \right \|_F \xrightarrow[]{X}\min$$

where $X$ is a permutation matrix and $A$, $B$ are dense square symmetric real matrices?

答:
The problem is equivalent to:

$$\text{minimize} \quad  \|A-XBX^T\|_F^2=\|A\|_F^2-2\langle A,XBX^T\rangle+\|XBX^T\|_F^2$$

which is equivalent to:

$$\text{minimize} \quad  -\langle A,XBX^T\rangle = -\text{trace}(X^TA^TXB)$$

The latter is due to $\|A\|_F^2$ and $\|XBX^T\|_F^2$ are constants.

This problem is the quadratic assignment problem and is NP-hard, with a special case being the travelling salesman problem.


问:
Investigate convergence of a sequence of functions $(f_n)_{n \in \mathbb{N}}$ with $ f_n(x)= \frac{x^{2n}}{1+x^{2n}}$

Let $R \in \mathbb{R}$ with $R>1$. Investigate the sequence of functions $ (f_n)_{n \in \mathbb{N}}$ with $$
f_n(x)= \frac{x^{2n}}{1+x^{2n}}, x\in [R, \infty)$$
with regard to uniform convergence.

I know the definition of pointwise and uniform convergence but never seen an example how to investigate uniform convergence, therefore I have no clue how to approach such problems. I know that pointwise convergence is necessary for uniform convergence but that's it. Can someone show me how to tackle such problems?

答:
Necessary and sufficient condition for uniform convergence on set $A$ is
$$\lim\limits_{n\to\infty}\sup_{x \in A}|f_n(x)-f(x)|=0$$
As in your case $A=[R, \infty)$ with $R \gt 1$, then $f(x)=1 $ and $|f_n(x)-f(x)|=\left|\frac{x^{2n}}{1+x^{2n}}-1  \right|=\frac{1}{1+x^{2n}}$. Now having supremum $\frac{1}{1+R^{2n}}$ we can obtain uniform convergence.


问:
Showing $1+\cos(\theta)+\dots+\cos (n\theta)=\frac12+\frac{\sin((n+\frac12)\theta)}{2\sin(\frac\theta2)}$

Suppose that $n \geq 1$ is a natural number.

(i) Show that for $z$ in $\mathbb{C}$, provided $z \neq 1$,

$$ \frac{1-z^{n+1}}{1-z} = 1+z+...+z^n.$$

(ii) Using de Moivre’s Theorem, part (b)(i) and standard trigonometric identities, show that for $\theta$ in $(0,2\pi)$,

$$ 1+\cos(\theta) + \dots + \cos (n\theta) = \frac{1}{2} + \frac{\sin((n+\frac12)\theta)}{2\sin(\frac\theta2)}$$

I have set $z=a+bi$ and subbed this into the top equation but dont seem to be getting anwhere with this problem. Any guidance will be great!

答:
Hints guides

Breaking $z$ into $(x + iy)$ is bad for these particular problems.

Part (i)

multiply the denominator by the RHS.

Part (ii)

This one, which happens to be part of my Complex Analysis textbook, is really no walk in the park.  Need introductory identities.

$\cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b).$

$\sin(a+b) = \sin(a)\cos(b) + \sin(b)\cos(a).$

$\sin \,a  ~-~ \sin \,b ~=~
2\,\sin\left(\frac{a - b}{2}\right) \times
\cos\left(\frac{a + b}{2}\right) \implies $

$\sin [(n + 1)\theta] ~-~ \sin \,(n\theta)
~=~
2 \,\sin \left(\frac{\theta}{2}\right) \times
\cos \left[\frac{(2n + 1)\theta}{2}\right].$

$\cos \,a  ~-~ \cos \,b ~=~
- 2 \,\sin\left(\frac{a - b}{2}\right) \times
\sin\left(\frac{a + b}{2}\right) \implies $

$\cos (n\theta) ~-~ \cos \,[(n + 1)\theta] ~=~
2 \,\sin \left(\frac{\theta}{2}\right) \times
\sin \left[\frac{(2n + 1)\theta}{2}\right].$

Let $\;A ~\equiv ~ \sum_{k = 0}^n
        (\cos \,k\theta ~+~ i \,\sin \,k\theta).$

Let $\displaystyle\;B ~\equiv ~ \left(\frac{1}{2}\right) ~+~
\left\{\frac{\sin \,\left[\frac{(2n + 1)\theta}{2}\right]}
{2 \,\sin \,(\theta/2)}\right\}.$

Let $\displaystyle\;C ~\equiv ~
\left(\frac{\cot (\theta/2)}{2}\right) ~-~
\left\{\frac{\cos \,\left[\frac{(2n + 1)\theta}{2}\right]}
{2 \,\sin \,(\theta/2)}\right\}.$

To Show: $\,A ~=~ B + iC.$

It is assumed that $(2\pi)$ does not divide $(\theta)$.

Use part (1), substituting $e^{i\theta}$ for $z$, to
re-express $A$.

Simplify $A$ to $\frac{R + iS}{T}$ and then prove that
$B = \frac{R}{T}, C = \frac{S}{T}.$

If you wish to use this answer, and encounter questions or roadblocks, edit your query to show all of your work,
and leave very explicit questions as a comment following this
answer.

I will automatically be flagged.  For your comment/questions,
try to have the detailed preliminary work done in your query, and
then ask very specific questions, like:

"In my query, expression (1) signifies $\frac{R}{T}$ and expression (2) signifies $B$.

How do I show that
expression (1) = expression (2)"?


问:
Are orthogonal matrices unitary?
If we have an orthogonal square matrix, A, and some c in the complex numbers with |c| = 1, how do we know cA is unitary?</p>

答: We know that $AA^T = I $ and $A^TA = I$ (since $A$ is orthogonal). Then, $cAcA^T = c^2 AA^T = c^2 I = I$ (since $|c| = 1$), same for $A^TA$ (assuming $A$ is real). As also mentioned in the comments, one just has to use the definition of a unitary/orthogonal matrix.

Title: Are orthogonal matrices unitary?


问:
How to determine the existence of this function
Does exist $f:\mathbb{R}^2 \to \mathbb{R}$ for $(x_0,x_0)$ such that $$\left(\frac{\partial f}{\partial v}\right)_{(x_0,y_0)}>0$$ for every direction $v$?

答:
No. If $v = \langle a,b \rangle$ is a unit vector and $w = \langle -a,-b \rangle$ then
\begin{align*}
\frac{\partial f}{\partial v}(x_0,y_0) &= \lim_{h \to 0} \frac{f(x_0 + ah,y_0+bh)}{h} \\
&= \lim_{h \to 0^+}\frac{f(x_0 + ah,y_0+bh)}{h} \\
&= \lim_{h \to 0^-}\frac{f(x_0 - ah,y_0-bh)}{-h} \\
&= -  \lim_{h \to 0^-}\frac{f(x_0 - ah,y_0-bh)}{h} \\
&= - \lim_{h \to 0}\frac{f(x_0 - ah,y_0-bh)}{h} \\
&= - \frac{\partial f}{\partial w}(x_0,y_0)
\end{align*}
because both directional derivatives are assumed to exist. More specifically if $f$ is differentiable at $(x_0,y_0)$ then
$$\frac{\partial f}{\partial w}(x_0,y_0) = \nabla f(x_0,y_0) \cdot w =  - \nabla f(x_0,y_0) \cdot v = - \frac{\partial f}{\partial v}(x_0,y_0).$$

In any event, $\dfrac{\partial f}{\partial v}(x_0,y_0)$ and $\dfrac{\partial f}{\partial w}(x_0,y_0)$ can't both be positive.


问:
Unique decomposition of $c$ sums of products of $k$ prime numbers, allowing duplicates?
Suppose that there are $n$ different prime numbers. Define procedure a) as following ($k \leq n$ and $k$ fixed): procedure a) for each time, we select one number out of $n$ possible cases and multiply it to the product (number may be repeated). For the first time, we choose a number and that is the product. We do this $k$ times.

Then, we do the following for $c \leq n^n$ times where $c$ is fixed: procedure b) we define variable $z$. For the first time procedure b) is started, $z = 0$. Now call procedure a) and get a result into variable $y$. Then we add $y$ to $z$ - and the sum replaces the original value of $z$ into $z$. We repeat this $c$ times.

So, step is: we call procedure b) (which then recursively calls procedure a) ) and see the result $z$.

Now the question is, is this $z$ unique in its decomposition, given $k$ and $c$? (So, there would be $c$ sums of product of $k$ prime numbers, allowing duplicates.)

Or if not, what would be the general rule when this holds?

For every step, the set of $n$ prime numbers remains the same - that is, it does not change.

答:
If $c=1$ then the decomposition is unique (up to ordering), as the selection of primes gives the prime factorization of $z$.

If $c>1$ then in general the decomposition is not unique. In the simplest case $k=1,c=2$ you consider the number of ways $z$ can be written as the sum of two primes (Goldbach partitions), e.g. let your primes be $\{5,11,13,19\}$ then $z=24=5+19=11+13$.

In general procedure a) has $N=\binom{n+k-1}{k}\simeq n^k/k!$ possible results (by stars-and-bars), and as multisets procedure b) has $$M=\binom{N+c-1}{c}\simeq \frac{n^{kc}}{(k!)^cc!}$$
If $L$ and $U$ are your smallest and largest primes, then
$$
cL^k\le z \le cU^k
$$
and a counting argument shows in general they can't all be distinct. For example, let your primes be $\{29,31,37,41,43,47,53,59,61,67,71\}$, $n=10,k=5,c=5$.
Then
$$
N=\binom{14}{5}=2002 \\
M=\binom{2006}{5}=2.69\times 10^{14} \\
cL^k=5\cdot 29^5 = 1.026\times 10^8, \quad cU^k=5\cdot 71^5 = 9.02\times 10^9
$$
So procedure b) has $>10^{14}$ possible selections from procedure a), but can only output $<10^{10}$ possible values, so some values must have a large number of decompositions.


问:
Infinite product of Zetafunctions
It is well-known (I learned about this first in a video by Papa Flammy) that $\sum_{n=2}^\infty (\zeta(n)-1) = 1$. This result on its own is quite remarkable, but it also implies convergence of the infinite product
$$P = \prod_{n=2}^\infty \zeta(n).$$
I somehow got invested in this. Upon trying to work out the exact value of $P$, I found that
$$P = \sum_{n=1}^\infty \frac{c_2(n)}n,$$
where $c_2$ is the multiplicative continuation of $c_2(p^e) := b_2(e),$ where $b_2(e)$ is the amount of partitions of $e$ not containing $1$, i.e.
$$b_2(n) = \# \{(a_2, a_3, a_4, \dots ) \in \mathbb N_{\geq 0}^* | 2a_2 + 3a_3 + \dots = n\}.$$

Furthermore, the function
$$Q(s) = \prod_{n=1}^\infty \zeta(ns),$$
converges compactly on $\Re s > 1,$ and one has
$Q(s) = \sum_{n=1}^\infty c_1(n) n^{-s}.$ Here again $c_1(p^e) = b_1(e),$ the amount of partitions of $e$ (this time including 1). Also, $Q$ has Residue
$$\text{Res}_{s=1} (Q(s)) = P.$$

Doing some numerical analysis with python I found that $P \approx 2.29$, but i would love to know if there is some exact form. Does anyone know about such a result?

答:
Yes, for $\Re(s) > 1/2$, $$F(s)=\prod_{n\ge 2} \zeta(ns)= \prod_p \prod_{n\ge 2} \frac1{1-p^{-sn}}=(\prod_p 1+ \sum_{k\ge 1} b_2(k)p^{-sk})=\sum_{m\ge 1} c_2(m)m^{-s}$$
where $b_2(k)$ is the number of partitions in integers $\ge 2$ and $c_2(m)=\prod_{p^k\| m} b_2(k)$.

The product for $F(s)$ converges locally uniformly for $\Re(s)>0, 1/s\not \in \Bbb{Z}_{\ge 2}$, so it is meromorphic for $\Re(s)>0$ with simple poles at $1/s \in \Bbb{Z}_{\ge 2}$.

$Res(F(s),1/N)=\frac1N \prod_{n\ne N} \zeta(n/N)$

$F(s)$ has a naturaly boundary at $\Re(s)=0$ (it cannot be analytically continued further) because the non-trivial zeros of each $\zeta(ns)$ accumulate to the right of $\Re(s)=0$ making $F(s)$ not analytic anywhere on $\Re(s)=0$.

It is mostly obvious that nothing will have a closed-form here.

Title: Infinite product of Zetafunctions