问:
Expectation of a martingale
I have a stochastic variable $Z$ and know that that $Z(t)/Z(0)$ is a martingale. Now, my book claims that this means that it has mean 1. I assume this is because of the martingale-property:

$$
E[Z(t)/Z(0)] = Z(0)/Z(0) = 1
$$

However, the above I guess only makes sense if we take the expectation value up to and include time $0$. But I would think that when we refer to the mean of a variable, then we are taking the expectation up to infinity, and not 0?

答:
As @charlus said, for every $t \in \mathbb R^+$, $Z(t)$ is a random variable. So when we take the expectation $\mathbb E[Z(t)]$, we integrate with respect to $\omega$ and not $t$. In other words, $$\mathbb E[Z(t)] := \int_{\omega \in \Omega} Z(t,\omega) d\mathbb P(\omega) \quad \forall t\in\mathbb R^+ $$
So there isn't any notion of "taking expectation up to infinity", we are simply computing the expectation of every random variable $Z(t)$ in the family $\{Z(t),t\in\mathbb R^+\}$.

As for the result you want to prove, it is indeed a consequence of the martingale property :

I assume that $Z(0)$ is a constant almost surely, and that you are working with the canonical filtration $(\mathcal F_t)_{t\ge0}$ for which the process $Z(t)/Z(0)$ is a martingale.

The martingale property tells you that for every $s\le t$, $\mathbb E[Z(t)/Z(0)\ |\ \mathcal F_s] = Z(s)/Z(0)$

Now noticing that $\mathcal F_0 = \{\varnothing, \Omega\}$, what your book says is that

$$\begin{align}\mathbb E[Z(t)/Z(0)] &= \mathbb E [Z(t)/Z(0)\  |\ \Omega] \\
&=\mathbb E [Z(t)/Z(0) \ |\ \mathcal F_0] \\
&= Z(0) / Z(0) \ \text{ by the martingale property} \\
&= 1 \end{align} $$

Edit :
As suggested by @Brian Moehring I'll do the proof without the assumption of almost surely constant $Z(0)$, which is unnecessary. So now we just assume that $Z(0) \ne 0$ almost surely :
$$\begin{align}\mathbb E[Z(t)/Z(0)] &= \mathbb E[\mathbb E [Z(t)/Z(0)\  |\ \mathcal F_0]] \text{ (law of total expectation)} \\
&=\mathbb E [Z(0)/Z(0)] \text{ by the martingale property}  \\
&= 1 \end{align} $$

Title: Expectation of a martingale


问:
F[x] is Principal Ideal Domain implies F is a Field. Note: This is the converse of a known theorem.
F is a Field implies F[X] is a Principal Ideal Domain.
But is the converse true?
Can we also always say that if F[x] is a PID then F is a Field?

I follow Gallian's Contemporary Abstract Algebra wherein is a Theorem stating F is a Field implies F[x] is a PID. The converse isn't given.
I had come across the converse in a few answers to the question asked here: Is $\mathbb{Z}[x]$ a principal ideal domain?.
Hope you understand the context.

答:
Yes.

Take $\alpha\in F\setminus\{0\}$ and take the ideal $\langle \alpha, x\rangle$ of $F[x]$. Being a principal ideal, it is generated by a single polynomial $f\in F[x]$, which divides both $\alpha$ and $x$. $f$ dividing $\alpha\ne 0$ means that $f$ is of degree $0$, i.e. $f\in F$, and $f$ dividing $x$ means that its "leading coefficient" (i.e. $f$ itself) is a unit. This means that, WLOG, we can assume that $f=1$. Thus $1\in\langle\alpha, x\rangle$, i.e. $1=\alpha p(x)+xq(x)$ for some polynomials $p,q\in F[x]$. Now for the final blow: set $x=0$ to obtain $1=\alpha p(0)$, i.e. $\alpha$ is invertible.

ContentLicense: CC BY-SA 4.0


答: If $F[x]$ is a PID then so is F.  Consider the ideal $(a, x)$ where $a\in F$ and $a\neq 0$.  By hypothesis there is a polynomial $g$ that generates $(a, x)$.  Thus for some $f\in F[x]$ we have $g f(x)= a$.  The only way this can happen is if g and f are in $F$.  Thus $g \in F$.  Since $x \in (g)$ then there is a polynomial $h(x)$ such that $g h(x) = x$.  Thus $h(x) = c x$ for some $c \in F$ and $g c x = x$. Hence $g$ is a unit and thus $a$ is a unit since the ideal $(a, x) = F[x]$.  Therefore $F$ is a field.

Title: F[x] is Principal Ideal Domain implies F is a Field. Note: This is the converse of a known theorem.

Body:
F is a Field implies F[X] is a Principal Ideal Domain.
But is the converse true?
Can we also always say that if F[x] is a PID then F is a Field?

I follow Gallian's Contemporary Abstract Algebra wherein is a Theorem stating F is a Field implies F[x] is a PID. The converse isn't given.
I had come across the converse in a few answers to the question asked here: Is $\mathbb{Z}[x]$ a principal ideal domain?.
Hope you understand the context.


问:
struggling with derivative of natural log of sin x
I was hoping someone could help me understand my struggle to derive the $y(x) = \ln (\sin x)$

My two assumptions

the derivative of $\ln(g(x)) = \frac{g(x)'}{g(x)}$.

And deriving the original equation $\ln(\sin(x))$ will require an application of the chain rule.

So if:

$$g(x) = \sin(x)$$

Then the derivative using the chain rule would be:

$$y(x)' = \frac{g(x)'}{g(x)} g(x)'$$

Thus I thought the solution would be the following:

$$y(x)' = \frac{\cos(x)}{\sin(x)}\cos(x) = \frac{\cos^2(x)}{\sin(x)} = \cos(x)\cot(x)$$

But the reported answer (for the online course I'm following) is:

$$y(x)' = \frac{1}{\sin(x)}\cos(x) = \cot(x)$$

So I'm pretty close, but it seems to be the $\frac{1}{\sin(x)}$ that is alluding me. I thought this would be $\frac{\cos(x)}{\sin(x)}$ $\left(\text{the result of } \frac{g(x)'}{g(x)}\right)$ but I'm clearly missing something.

Any advice for an amateur but enthusiastic learner is much appreciated.

答: The chain rule is already taken into account in $g^\prime/g$, so you don't need a second $g^\prime$ factor. The correct result is $\cot x$.


问:
Big-O: Prove $2^n$ is $O(n!)$
I am a little stuck trying to prove that $2^n$ is $O(n!)$. Obviously, I can tell in a few ways that this is the case. For one, based on Big-$O$ hierarchy, the exponential is beneath the factorial in terms of complexity. I also substituted $C = 2$ and tried several values of $n$, all following the Big-$O$ definition of $f(n) \le cg(n)$.

However, my professor says neither of these is a sufficient justification and I should be able to use the definition alone.

How can I do this? I know in the past it has been advantageous for me to compare the functions by converting one into a form that resembles the other, but I just can't figure out how to relate the factorial and $2^n$ functions.

Can anyone point me in the right direction?

答:
We have
$$
\sum_{n=0}^{\infty} \frac{2^n}{n!} = e^2 < \infty,
$$
hence $2^n/n! \to 0$. This means $2^n = O(n!)$.

Hope that helps,

ContentLicense: CC BY-SA 3.0

Title: Big-O: Prove $2^n$ is $O(n!)$


问:
Probability of $3$ of $4$ numbers matching numbers in same position on license plate
License plates in our area end in a 4-digit number $0000$ through $9999$. What is the probability of my second car's plate matching $3$ of the $4$ digits in the identical position as my first car's plate? This actually happened and it seems the probability would be fairly low. I think the odds are $1:10,000$ that all four numbers would match, so it must be somewhat higher probability that any three would match in the same position, e.g., $0123$ and $0523$.

What is the probability, and what is the equation to obtain it?

答:
We need to calculate in how many ways can three digits match (but not the last one). Let me give you an example - suppose your plate was previously $0000$. The "matching at three positions" license plates are:

$$1000,2000,\ldots,9000$$
$$0100,0200,\ldots,0900$$
$$0010,0020,\ldots,0090$$
$$0001,0002,\ldots,0009$$

In other words, pick the "mismatching" position (in $4$ ways) and then pick the mismatching digit (in $9$ ways). The total number of ways to get a mismatch of that sort is $4\times 9=36$. Thus, the probability for this to happen (if all the combinations of digits are equally probable) is $\frac{36}{10,000}=0.0036=0.36\%$.


答:
Look the difference :
Assume that
$$y(x)=\ln(g(h(x)))=\ln(G(x))$$

then

$$y'(x)=\frac{g'(h(x))}{g(h(x))}h'(x)$$

In your case $ h(x)=x$ and $ h'(x)=1$.

For example, if $$y(x)=\ln(\sin(3x))$$
then
$$y'(x)=3\frac{\cos(3x)}{\sin(3x)}$$

You have also
$$y'(x)=\frac{G'(x)}{G(x)}$$

Title: struggling with derivative of natural log of sin x

Body:
I was hoping someone could help me understand my struggle to derive the $y(x) = \ln (\sin x)$

My two assumptions

the derivative of $\ln(g(x)) = \frac{g(x)'}{g(x)}$.

And deriving the original equation $\ln(\sin(x))$ will require an application of the chain rule.

So if:

$$g(x) = \sin(x)$$

Then the derivative using the chain rule would be:

$$y(x)' = \frac{g(x)'}{g(x)} g(x)'$$

Thus I thought the solution would be the following:

$$y(x)' = \frac{\cos(x)}{\sin(x)}\cos(x) = \frac{\cos^2(x)}{\sin(x)} = \cos(x)\cot(x)$$

But the reported answer (for the online course I'm following) is:

$$y(x)' = \frac{1}{\sin(x)}\cos(x) = \cot(x)$$

So I'm pretty close, but it seems to be the $\frac{1}{\sin(x)}$ that is alluding me. I thought this would be $\frac{\cos(x)}{\sin(x)}$ $\left(\text{the result of } \frac{g(x)'}{g(x)}\right)$ but I'm clearly missing something.

Any advice for an amateur but enthusiastic learner is much appreciated.


答:
Suppose one license plate is fixed. Choose one of the four digits that will not match. Once that is chosen, there are nine digits you can change it to. So, there are a total of 36 license plates that will match exactly three digits out of all possible license plates. That is:

$$\dfrac{4\cdot 9}{10^4} = \dfrac{9}{2500} = 0.0036$$

Title: Probability of $3$ of $4$ numbers matching numbers in same position on license plate

Body:
License plates in our area end in a 4-digit number $0000$ through $9999$. What is the probability of my second car's plate matching $3$ of the $4$ digits in the identical position as my first car's plate? This actually happened and it seems the probability would be fairly low. I think the odds are $1:10,000$ that all four numbers would match, so it must be somewhat higher probability that any three would match in the same position, e.g., $0123$ and $0523$.

What is the probability, and what is the equation to obtain it?


答:
\begin{align}
\frac d {dx} \ln x & = \frac 1 x \\[10pt]
\frac d {dx} \ln g(x) & = \big(\ln'g(x)\big)\cdot g'(x) = \frac 1 {g(x)} \cdot g'(x) \\[10pt]
\frac d {dx} \ln \sin x & = \big(\ln'\sin x\big)\cdot \sin'(x) \\[10pt]
& = \frac 1 {\sin x} \cdot \cos x.
\end{align}


问:
Real part of an analytic function
I'd like to know how to take the real part (of the power series representation) of an analytic function with the goal of showing the (real part of the analytic) function is real analytic.

I know this may seem trivial but I haven't been able to convince myself that that's the case.

答:
I will show an example with the power series for $e^z,\,z\in\mathbb{C}$. The power series is
$$\sum_{n=0}^{\infty}\frac{z^n}{n!}=\sum_{n=0}^{\infty}\frac{(re^{i\varphi})^n}{n!}.$$
Here $r$ is the magnitude of $z$ and $\varphi$ the phase angle. Note that $\operatorname{Re}(z_1+z_2)=\operatorname{Re}(z_1)+\operatorname{Re}(z_2),$ so
$$\operatorname{Re}\sum_{n=0}^{\infty}\frac{r^ne^{i\varphi n}}{n!}=\sum_{n=0}^{\infty}\operatorname{Re}\frac{r^ne^{i\varphi n}}{n!}.$$
Eulers formula tell us that  $e^{ix}=\cos x+i\sin x,\,x\in\mathbb{R}.$ From now on you should be able to calculate the last steps. Hint: $r^n/n!$ is real so you can first take the real part of $e^{i\varphi n}$ and then multiply it.

Title: Real part of an analytic function


答: When you fix one (of four) position and want to count how many combinations are there that match the given number at three other positions but the one you fixed you get $\frac{1}{10}\cdot\frac{1}{10}\cdot\frac{1}{10}\cdot\frac{9}{10}$ out of the total number of combinations. The last factor is that what it is because you allow all digits to happen on that place but the one that is written on the original four-digit number. That leaves $9$ possibilities. Now circle through all four positions, fixing them one at the time, and add up the combinations that match at all but fixed place. There are four identical summands. So the total sum is $4\cdot\frac{1}{10}\cdot\frac{1}{10}\cdot\frac{1}{10}\cdot\frac{9}{10}$, which is $\frac{36}{10000}=\frac{9}{2500}$.


答:
Let $u_n=\frac{2^n}{n!}$ then by ratio test we have
$$\frac{u_{n+1}}{u_n}=\frac{2}{n+1}\to_\infty 0$$
hence the sequence $(u_n)$ is convergent to $0$ and then we find that $2^n=O(n!)$ and more precisely $2^n=o(n!)$</p>

Title: Big-O: Prove $2^n$ is $O(n!)$

Body:
I am a little stuck trying to prove that $2^n$ is $O(n!)$. Obviously, I can tell in a few ways that this is the case. For one, based on Big-$O$ hierarchy, the exponential is beneath the factorial in terms of complexity. I also substituted $C = 2$ and tried several values of $n$, all following the Big-$O$ definition of $f(n) \le cg(n)$.

However, my professor says neither of these is a sufficient justification and I should be able to use the definition alone.

How can I do this? I know in the past it has been advantageous for me to compare the functions by converting one into a form that resembles the other, but I just can't figure out how to relate the factorial and $2^n$ functions.

Can anyone point me in the right direction?


问:
If $X_n \to X $ in probability then $f(X_n) \to f(X)$ in probability for a Borel function $f$
$(X_n)_n$ is a sequence of identically distributed random variables, $f:\mathbb{R} \to \mathbb{R}$ a Borel function.

Prove that if $X_n$ converges in probability to $X,$ then $f(X_n)$ converges in probability to $f(X).$

答:
Following up on a line of reasoning suggested by MikeG and  Danny Pak-Keung Chan:

Let $\mu$ be the law of the $X_n$ (which is also the law of $X$).  Fix $\epsilon, \eta > 0$.  By Lusin's theorem applied to the measure $\mu$, there is a compact set $K \subset \mathbb{R}$ with $\mu(K) > 1-\eta$ on which which $f|_K$ is continuous.  It is thus even uniformly continuous, so there exists $\delta > 0$ such that for every $x,y \in K$ with $|x-y| < \delta$, we have $|f(x)-f(y)| < \epsilon$.

As such, for any given $n$, if we are to have $|f(X_n) - f(X)| \ge \epsilon$, then we must either have $X_n \notin K$, $X \notin K$, or $|X_n - X| \ge \delta$.  Thus by union bound
$$\begin{align*}P(|f(X_n) - f(X)| \ge \epsilon) &\le P(X_n \notin K) + P(X \notin K) + P(|X_n - X| \ge \delta)\\ &\le 2 \eta + P(|X_n - X| \ge \delta).\end{align*}$$
Since $X_n \to X$ in probability, letting $n \to \infty$ we conclude $$\limsup_{n \to \infty} P(|f(X_n) - f(X)| \ge \epsilon) \le 2\eta$$
and $\eta > 0$ was arbitrary.

A previous of the version asked whether, if $X_n \to X$ almost surely, we can conclude $f(X_n) \to f(X)$ almost surely.  The answer is no.  For simplicity, work on the group $(S^1,+)$.  You can take it back to $\mathbb{R}$ if you like by identifying $S^1$ with $[0,1)$ and $+$ with addition mod 1.

Let $E \subset S^1$ be a closed nowhere dense set of positive Lebesgue measure, e.g. a generalized Cantor set, and $f = 1_E$.   I claim there exists a sequence $t_n \in S^1$ with $t_n \to e$, the identity element, for which $\liminf_{n \to \infty} 1_E(x+t_n) = 0$ for every $x$.  Once this is shown, we can let $X$ be uniformly distributed on $S^1$ and let $X_n = X + t_n$.  Clearly every $X_n$ is uniformly distributed on $S^1$ and $X_n \to X$ a.s., but $\liminf f(X_n) = 0$ a.s., so $f(X_n) \not\to f(X)$ on the event $\{X \in E\}$, which has positive probability.

To prove the claim, note that $U=E^c$ is open and dense.  So for any integer $m>0$, the sets $U-s$ where $d(s,e) < 1/m$ form an open cover of $S^1$.  By compactness, there is a finite subcover $U-s_{m,1}, \dots, U-s_{m, k_m}$.  This means that for every $x$ and every $m$, we have $x + s_{m,i} \in U$ for some $i$.  Let $t_n$ be the sequence $$(s_{1,1}, \dots, s_{1, k_1}, s_{2,1},\dots, s_{2, k_2}, \dots).$$

Since $d(s_{m,i}-e) < 1/m$, we have $t_n \to e$.  And by construction, for every $x \in S^1$, there are infinitely many $t_n$ such that $x + t_n \in U$.  This means that $\liminf_{n \to \infty} 1_E(x+t_n) = 0$.


问:
Restricting Domain and Range in Inverse Trigonometric Function
After an explanation of the restricted domains and ranges of inverse trigonometric functions, I.M. Gelfand's Trigonometry gives the following exercise:

Show that $$\sin(\arccos b) = \pm \sqrt{1-b^2}$$  What determines whether we should choose the positive sign or the negative sign?

It is my understanding that:

$b^2 + \sin^2 x = 1$

So $\sin x = \pm \sqrt{1-b^2}$.

But  $0\leq \arccos b \leq \pi$

So $-1\leq b \leq 1$

So $\sin 0 \leq \sin x \leq \sin \pi $

And $0 \leq \sin x \leq 1$.

I must be missing something because it seems to me that we would not need the ambiguous sign and the original expression should be $$\sin(\arccos b) = \sqrt{1-b^2}$$ because $\sin x$ where $x = \arccos b$ will always be positive.

What am I missing?

答:
Your answer is correct. We can verify it in a slightly different way by noting that
$$
\sin^2 x + \cos^2x = 1
$$
and so
$$
\left|\sin x\right| = \sqrt{1-\cos^2x} \, .
$$
Hence,
\begin{align}
\left|\sin(\arccos(b))\right| &= \sqrt{1-\cos^2(\arccos(b))} \\
&= \sqrt{1-b^2} \, .
\end{align}
Because $\arccos(b) \in [0,\pi]$, we know that $\sin(\arccos b) \geq 0$. Hence, we are left with
$$
\sin(\arccos(b))=\sqrt{1-b^2} \, .
$$
The crux of solving this problem is understanding the range of the $\arccos$ function, so that we can draw this conclusion. Strictly speaking, $\arccos$ is not the inverse of $\cos$—rather, it is the inverse of $\operatorname{Cos}$, which is the function $f:[0,\pi] \mapsto \mathbb{R}$ defined by $f(x)=\cos(x)$. On the other hand, $\cos$ is the inverse of $\arccos$, since $\arccos$ is one-to-one. It is because of this that $\cos(\arccos(x))$ is always equal to $x$, but $\arccos(\cos(x))$ might not be equal to $x$.


问:
Homomorphic images of $\mathbb{Z}$
Suppose there is a homomorphism $f:\mathbb{Z}\to R$  where $R$ is a non-zero ring.
By the definition of homomorphism $f(n)=f(1+1+1+1.....+1)=f(1)+f(1)+f(1)+....+f(1)$
Now suppose $f(1)=x$ then $f(n)=nx$ by definition. By varying x we can say that homomorphic images of $\mathbb{Z}$ is $0,\mathbb{Z},\mathbb{2Z},\mathbb{3Z}$...etc.

Am I right? Please provide some deep intuitions if any.

答:
Hint : if $\phi:\mathbb{Z} \to R $ is a homorphism then $ \mathbb{Z}/ker(\phi) \cong \phi(\mathbb{Z}) $

now, ker($\phi$) is an ideal of $\mathbb{Z}$ and any ideal of $\mathbb{Z}$ is of the form n$\mathbb{Z}$

Title: Homomorphic images of $\mathbb{Z}$


问:
Subset permutations
Given that $B = (a_1, a_2 \cdots, a_{12})$ is a permutation of the set $(1, 2, \cdots, 12)$ such that $a_1>a_2>a_3>a_4>a_5>a_6$ and  $a_6<a_7<a_8<a_9<a_{10}<a_{11}<a_{12}$, then how many possible permutations can B represent?

I know of the permutation formula $P(n, r) = \frac{n!}{(n-r)!}$, but I don't know how that can be used here. How can I use $P(n, r)$?

答: Firstly, you can see that $a_6$ is the minimum since it is smaller than any other nubmer in $B$. So, $a_6=1$. Now, choose sets $\{a_1,a_2,a_3,a_4,a_5\}$ and $\{a_7,a_8,a_9,a_{10},a_{11},a_{12}\}$. This you can do in $\binom{11}{5}=\binom{11}{6}$ ways. Once you have defined these two sets there is a unique way to put them in the array $B$ and that is to sort them appropriately.

Title: Subset permutations


问:
Why is the bridge index of the trefoil equal to 2?
It seems to me, all three 3 bridges are needed?</p>

答: The bridge index of a knot $K$ is the minimum number of local maxima (or equivalently, minimum number of local minima) of the height function on all possible knot diagrams of $K$. So from the standard picture of the trefoil, you can see that its bridge index is at most $2$. It is clear that the only knot with bridge index $1$ is the unknot, so in fact the bridge index of the trefoil $T_{2,3}$ is $2$. Here's a picture of $T_{2,3}$ with its standard $2$-bridge presentation:


问:
How do I prove that $\lim_{x\to \infty}(\sqrt[4]{x})$ is not Cauchy by using the definition of the Cauchy sequence
The Cauchy Sequence is that: $\forall \varepsilon >0 \, \exists n_0 \in \mathbb ℕ$ such that $\forall m > n \geq n_0, \, |x_m−x_n|≤\varepsilon.$ Let's negate it, we will get $∃\varepsilon> \, ∀ \in \mathbb ℕ$ such that $∃>≥_0 \, |_−_|≥\varepsilon.$
Need to show that $|\sqrt[4]{_} - \sqrt[4]{_}| ≥\varepsilon $

答: You mean $x_n = \sqrt[4]{n}$, then on $[m, 2m], m > 1, f(x) = \sqrt[4]{x}$   is differentiable thus using MVT yields: $\left|x_m - x_{2m}\right| = \left|\sqrt[4]{m} - \sqrt[4]{2m}\right|  \ge |2m - m|\cdot \dfrac{1}{4\sqrt{m}} = \dfrac{\sqrt{m}}{4}  >  \dfrac{1}{4} = \epsilon$. Thus it’s not Cauchy ( sequence ).


问:
Moment generating function, probability problem
Given: $M_x(t) = \frac{1}{6} + \frac{1}{3} \exp(t) + \frac{1}{6} \exp(2t) + \frac{1}{3} \exp(3t)$.

What will be $P(X>2)$?

I couldn’t identify this moment generating function.

I guess if I could identify the mgf this will become easier.

答: As @StubbornAtom notes, you can read the probability distribution off viz.$$P(X=0)=\tfrac16,\,P(X=1)=\tfrac13,\,P(X=2)=\tfrac16,\,\color{blue}{P(X=3)=\tfrac13}.$$


问:
Proof for vectors
I am struggling with solving tasks on vectors. The task:
Prove that the end of vector $\vec c$ lies on the segment joining the ends of vectors $\vec a$ and $\vec b$ if and only if the equality

$$\vec c = m\vec a+n\vec b \, , \quad m \geq0 \, , \quad n \geq0 \, , \quad m+n=1$$

is satisfied. In what ratio does the end of vector $\vec c$ divide this segment?

The statement that $\vec c$ is decomposed by vectors $\vec a$ and $\vec b$ is quite intuitive since $\vec a$, $\vec b$ and $\vec c$ vectors are coplanar vectors. But I couldn't prove equality $m+n=1$. I really appreciate any help you can provide.

答: Note $m+n=1$ is one of the defining conditions of the locus you're asked to prove is the specified line segment; it's not something you have to "prove" separately. Since $m\vec{a}+n\vec{b}=\vec{a}+n(\vec{b}-\vec{a})$, the given parameterization is precisely of the points a proportion $n\in[0,\,1]$ of the way from $\vec{a}$ to $\vec{b}$.

Title: Proof for vectors


问:
Minimize $\quad z=4\left(x_{1}-1\right)^{2}+\left(x_{2}-2\right)^{2}$ s.t $\quad x_{1}+x_{2} \leq 2 \ ,x_{1}, x_{2} \geq-1$ without KKT conditions
The problem is to minimize $\quad z=4\left(x_{1}-1\right)^{2}+\left(x_{2}-2\right)^{2}$
Subject to the following constraints.
$$\quad x_{1}+x_{2} \leq 2$$$$ x_{1}, x_{2} \geq-1$$
I don't want to solve this problem with the KKT conditions. I am looking for a more intuitive way, maybe using geometry or inequalities. a simple approach but I can't seem to think of anything outside KKT which I can't use.

Any suggestions??

答:
You can use the ubiquitous Lagrange multipliers with the help of some slack variables to handle the inequalities.

Calling

$$
\cases{
f(x_1,x_2)=4\left(x_{1}-1\right)^{2}+\left(x_{2}-2\right)^{2}\\
g_1(x_1,x_2) = 2- x_{1}-x_{2}\\
g_2(x_1,x_2) = x_{1} + 1\\
g_3(x_1,x_2) = x_{2} + 1
}
$$

you can formulate

$$
L(x_1,x_2,\lambda,s) = f(x_1,x_2)+\sum_{k=1}^3\lambda_k(g_k(x_1,x_2)-s_k^2)
$$

The stationary points are the solutions for

$$
\nabla L = 0 = \left\{
\begin{array}{l}
 8 (x_1-1) -\lambda_1+\lambda_2\\
 2 (x_2-2)-\lambda_1+\lambda_3 \\
 -x_1-x_2+2 -s_1^2\\
 x_1+1 -s_2^2\\
 x_2+1-s_3^2 \\
 \lambda_1 s_1\\
 \lambda_2 s_2 \\
 \lambda_3 s_3 \\
\end{array}
\right.
$$

Pay attention to the complementary relations $\lambda_ks_k = 0$. They simplify a lot the calculations.
$$
\left[
\begin{array}{ccccccccc}
f& x_1 & x_2 & \lambda_1 &\lambda_2 & \lambda_3 & s_1^2&s_2^2& s_3^2\\
 \frac{4}{5} & \frac{4}{5} & \frac{6}{5} & -\frac{8}{5} & 0 & 0 & 0 & \frac{9}{5} & \frac{11}{5} \\
 9 & 1 & -1 & 0 & 0 & 6 & 2 & 2 & 0 \\
 16 & -1 & 2 & 0 & 16 & 0 & 1 & 0 & 3 \\
 17 & -1 & 3 & 2 & 18 & 0 & 0 & 0 & 4 \\
 25 & -1 & -1 & 0 & 16 & 6 & 4 & 0 & 0 \\
 25 & 3 & -1 & 16 & 0 & 22 & 0 & 4 & 0 \\
\end{array}
\right]
$$

Here $s_k = 0$ indicates that the corresponding restriction is active.

Follows a level curve plot showing in red the stationary points. Finally follows the point qualification.


问:
What does it mean to "embedding of a sequence into an sequence"? Can you show it with real-world examples?
Please refer to the screenshot below from a research article:

Screenshot showing embedding of sequences.

What is the meaning of embedding here, could you point out specific resources/sources to learn more about this. This concept is crucial for understanding the relevant concepts (i.e., interleaving of two or more sequences) in the research paper. An explanation with suitable example(s) would be highly appreciated.

I have tried to search this concept online, but have not been able to figure out. While the following definition defines the term embedding, it is not making sense in the context (see the screenshot below).

答:
Don't they just mean something like embedding of $a=(1,2,3,4)$ in the sequence $b=(0,0,0,1,0,0,2,0,0,0,3,4,0,\ldots)$?

In other words, you have some strictly increasing re-mapping of indices $emb$ such that $b_{emb(n)}=a_n$ for all $n$ in the domain of $a$? In the example above, $emb=\begin{pmatrix}0&1&2&3\\3&6&10&11\end{pmatrix}$.


问:
How to prove (p → q) ∨ (q → r) ≡ (p ∧ q) → r
My attempt at this mathematical demonstration was as follows:

(p → q) ∨ (q → r) ≡ (~p ∧ q) ∨ (q → r)
≡ q ∨ (p → r)
≡ (q ∨ p) → r
≡ (p ∨ q) → r

答:
Your second step seems off, you went from $p\rightarrow r$ to $\lnot p\land q$, it is not clear how q appeared from r. Instead, try this, using the fact that $a\rightarrow\ b$ means $\lnot a \lor b$:

$$\begin{split}(p\rightarrow r)\lor (q\rightarrow r)\\
(\lnot p \lor r)\lor(\lnot q \lor r)\\
(\lnot p \lor \lnot q)\lor (r\lor r)\\
(\lnot p \lor \lnot q)\lor r\\
\lnot(p\land q)\lor r\\
(p\land q)\rightarrow r\end{split}$$


问:
Example of homomorphism $\underset{\alpha \in A}{\ast} \pi_1(U_\alpha) \to \pi_1(X)$, where $U_\alpha \ni x_0$ are an open cover of $(X,x_0)$.
I’m reading a passage leading up to the Van Kampen’s theorem, and I have trouble understanding the settings. Here’s the materials.

Let us consider a pointed space $\left(X, x_{0}\right)$ with an open cover $\left\{U_{\alpha}: \alpha \in A\right\}$ by path-connected open sets $U_{\alpha} \ni x_{0}$ containing the basepoint. We fix the basepoint $x_{0}$ throughout, using it for $X$ and for each $U_{\alpha},$ as well as for their intersections; we omit it from the notation for fundamental groups. The inclusions $U_{\alpha} \hookrightarrow X$ induce homomorphisms $j_{\alpha}: \pi_{1}\left(U_{\alpha}\right) \rightarrow \pi_{1}(X) .$ By the universal property of the free product, these combine to give a homomorphism
$$\varphi: F:=\underset{\alpha \in A}{*} \pi_{1}\left(U_{\alpha}\right) \rightarrow \pi_{1}(X)$$
Consider now, the inclusions $U_{\alpha} \cap U_{\beta} \hookrightarrow U_{\alpha}($ for $\alpha, \beta \in A)$. These induce homomorphisms
$$
i_{\alpha \beta}: \pi_{1}\left(U_{\alpha} \cap U_{\beta}\right) \rightarrow \pi_{1}\left(U_{\alpha}\right)
$$

The inclusions of $U_{\alpha} \cap U_{\beta}$ into $X$ via $U_{\alpha}$ or $U_{\beta}$ commute:

By functoriality, the same is true of the induced homomorphisms:

That is, for all $\alpha, \beta \in A$ we have
$$
j_{\alpha} \circ i_{\alpha \beta}=j_{\beta} \circ i_{\beta \alpha}: \pi_{1}\left(U_{\alpha} \cap U_{\beta}\right) \rightarrow \pi_{1}(X)
$$
This implies that for any loop $\sigma \in \pi_{1}\left(U_{\alpha} \cap U_{\beta}\right)$ we have
$$
\varphi\left(i_{\alpha \beta}(\sigma)\right)=\varphi\left(i_{\beta \alpha}(\sigma)\right) \in \pi_{1}(X)
$$
Thus the two-letter word $i_{\alpha \beta}(\sigma) i_{\beta \alpha}(\sigma)^{-1}=i_{\alpha \beta}(\sigma) i_{\beta \alpha}\left(\sigma^{-1}\right)$
in the free group $F= \underset{\alpha}{\ast} \pi_{1}\left(U_{\alpha}\right)$ is in the kernel of $\varphi$.

My questions are as follows:

Just to be sure, are the maps $i$ and $j$ just the inclusions of homotopy classes of loops from one fundamental group to another?

What does the map $\varphi$ do exactly? I have trouble understanding how the maps $j_\alpha$ and $j_\beta$ “combine” to give the map $\varphi$.

Perhaps an example would make the idea clearer. Please use the following setting if it makes sense: say $U_\alpha$ and $U_\beta$ are both homeomorphic to $S^1$, and $a$ and $b$ are the generators of their respective fundamental groups. A word in $F$ could be something like $a^3b^{-5}$. What are the remaining components of the second diagram above, and the corresponding image of $\varphi$? If this doesn’t make sense, please let me know why and offer another example if possible.

答:
Yes

$\phi$ takes a free product (or word) $\Pi_\alpha \ \gamma_\alpha$ of loops $\gamma_\alpha \in \pi_1(U_\alpha)$ to the product $\Pi_\alpha \ j_\alpha(\gamma_\alpha) \in \pi_1(X)$ where the product in $\pi_1(X)$ is concatenation of loops. It's really only the maps $j$ which combine to give $\phi$.

This question doesn't really make sense, it's impossible to say what the space $X$ is from this situation. If $X = U_0 = U_1 = S^1$ then we have that $\phi(a^3b^{-5}) = a^{3-5} = a^{-2}$ assuming $a = b$.

More clearly, $\phi: \mathbb Z* \mathbb Z \rightarrow \mathbb Z$ in this situation is the identity on each of the copies of $\mathbb Z$ so takes a word of integers to its product (really sum) in $\mathbb Z$.

If you don't understand something I would be happy to elaborate.


问:
Probability question about putting X items in K things

The question is: "8 employees want to book meetings with their manager on randomly chosen days of the work week, from Monday till Friday. What's the probability that the manager has at least one day without a booking?"

I've tried it in two ways but it hasn't worked:

I said that the probability of an employee booking a specific day is $1/5$, so the probability of all $8$ booking on any day except one is $(4/5)^8$. Then, I multiplied that by $5$ to accomodate for all days, but didn't work.

I said the total number of ways of employees booking a meeting is $(8+5-1)C(5-1)$, which is $495$. Then, I said the number of ways employees can book where there's at least one meeting a day is $(3+5-1)C(5-1)$, which is $35$. Then, I said $35/495$ is the probability that there's at least one meeting every day, so the probability that there's at least one day off should be that subtracted from $1$, and it didn't work either.

答:
Using Inclusion/Exclusion, the solution is

$$\dbinom{5}{4}\left(\dfrac{4}{5}\right)^8-\dbinom{5}{3}\left(\dfrac{3}{5}\right)^8+\dbinom{5}{2}\left(\dfrac{2}{5}\right)^8-\dbinom{5}{1}\left(\dfrac{1}{5}\right)^8 = \dfrac{2117}{3125} = 0.67744$$


Title: Probability question about putting X items in K things

Body:
The question is: "8 employees want to book meetings with their manager on randomly chosen days of the work week, from Monday till Friday. What's the probability that the manager has at least one day without a booking?"

I've tried it in two ways but it hasn't worked:

I said that the probability of an employee booking a specific day is $1/5$, so the probability of all $8$ booking on any day except one is $(4/5)^8$. Then, I multiplied that by $5$ to accomodate for all days, but didn't work.

I said the total number of ways of employees booking a meeting is $(8+5-1)C(5-1)$, which is $495$. Then, I said the number of ways employees can book where there's at least one meeting a day is $(3+5-1)C(5-1)$, which is $35$. Then, I said $35/495$ is the probability that there's at least one meeting every day, so the probability that there's at least one day off should be that subtracted from $1$, and it didn't work either.

Body:
One approach is to set up a recursion: the probability the manager has $d$ days booked is by $n$ employees satisfies $$p_n(d) = \tfrac{d}{5}p_{n-1}(d) + \tfrac{5-(d-1)}{5}p_{n-1}(d-1)$$ starting at $p_0(0)=1$. You get a table like this

    d   0    1           2           3           4          5
n
0       1   0           0           0           0           0
1       0   1           0           0           0           0
2       0   0.2         0.8         0           0           0
3       0   0.04        0.48        0.48        0           0
4       0   0.008       0.224       0.576       0.192       0
5       0   0.0016      0.0960      0.4800      0.3840      0.0384
6       0   0.00032     0.03968     0.34560     0.49920     0.11520
7       0   0.000064    0.016128    0.231168    0.537600    0.215040
8       0   0.0000128   0.0065024   0.1483776   0.5225472   0.3225600

and you want $1-p_8(5) = 0.67764 = \frac{2117}{3125}=\frac{264625}{5^8}$ where $5^8$ is the total number of ways the meetings could be booked


答:
If $A$ is the set of all permutations where manager has at least one day free from one to one meetings with employees, then by principle of inclusion-exclusion,

$|A| = \displaystyle {5 \choose 4}4^8 - {5 \choose 3}3^8 + {5 \choose 2}2^8 - {5 \choose 1} = 264625$

You can alternatively compute this by using Stirling Number of the second kind,

$|A| = 5^8 - 5! \ S2[8,5] = 264625$.

$S2[8,5]$ returns number of permutations where manager has meetings all $5$ days without any distinction between days. Multiplying by $5!$ makes days distinct. We then subtract it from unrestricted permutations $5^8$ and that gives us all permutations where manager has at least one of the days free.


答:
Here's one way to prove it that's nonstandard, it uses $[ a \, b \, c \,\cdots]$ as a form of nand that takes an arbitrary number of arguments. This is a linear notation for an existential graph and leverages the fact that nand is a sole sufficient operator for classical logic.

$$ [ a \, b \, c \, \cdots ] \;\;\text{is equivalent to}\;\; \lnot(a \land b \land c \land \cdots) $$

Repetition of the arguments and order of the arguments to a single $[\cdots]$ does not affect the truth value of the expression.

Inserting or removing a double negation also does not affect the truth value of an expression.

For example,

$$ [a \, b] \;\; \text{is equivalent to} \;\; [a \, [[b]]] $$

Note that

$$ p \to q \;\;\text{is equivalent to}\;\; [p \, [q]] $$

$$ p \lor q \;\;\text{is equivalent to}\;\; [[p]\,[q]] $$

And the definition of and, by way of double negation elimination

$$ p \land q \;\;\text{is equivalent to}\;\; [[p \, q]] $$

The first expression is equivalent to

$$ [[[p \, [r]]] \, [[q \, [r]]]] $$

We begin by removing the double negation

$$ [ p \, [r] \, q \, [r] ] $$

Remove the repeated argument

$$ [p \, q \, [r]] $$

Insert a double negation

$$ [[[p \, q]] \, [r]] $$

Apply the definition of $\land$ and the definition of $\to$

$$ (p \land q) \to r $$

Title: How to prove (p → q) ∨ (q → r) ≡ (p ∧ q) → r

Body:
My attempt at this mathematical demonstration was as follows:

(p → q) ∨ (q → r) ≡ (~p ∧ q) ∨ (q → r)
≡ q ∨ (p → r)
≡ (q ∨ p) → r
≡ (p ∨ q) → r


问:
Determine whether it is homomorphism:
Determine whether it is homomorphism:

$\phi: (\mathbb{Q}, \cdot) \rightarrow (\mathbb{R}, \cdot)$, where

$\phi(x) =  \begin{cases}
      \sqrt x, & \text{if}\ x \geq 0 \\
      0, & \text{otherwise.}
    \end{cases} $

I have never seen this kind of problem, so thank you for helping.

答:
EDIT: As brought up in the comments to my answer, I shouldn't have mentioned the identity element at all. I've edited to remove that.

As discussed in the comments you would be well served by explicitly stating what sort of structure you want your homomorphism to preserve. Given that you write these as $(\mathbb Q, \cdot), (\mathbb R, \cdot)$ I will assume you mean to see if $\phi$ is a homomorphism of semigroups, i.e. it preserves the multiplication operation $\phi(xy) = \phi(x) \phi(y)$ for all $x, y \in \mathbb Q$. Unfortunately, this condition fails. Consider indeed $\phi((-1)(-1)) = \phi(1) = \sqrt{1} = 1$. On the other hand, $\phi(-1) \phi(-1) = 0*0 = 0$. As $1 \neq 0$ in $\mathbb R$, $\phi((-1)(-1)) \neq \phi(-1) \phi(-1)$. Hence, $\phi$ is not a homomorphism of semigroups.

Title: Determine whether it is homomorphism:


问:
Dependance of variables in variational calculus
I stumbled across the following functional for which a stationary path should be found:
$$  L[y]=\int\limits_a^b x[y‘(x)]^4+ ... \,\mathrm{d}x$$
Where the dots indicate the other terms that were included in the original problem but dont cause any confusion. The first question that arises is how to make any sense of this expression since both $y$ and $y‘$ are functions of $x$ but then $x$ itself happens to depend on $y‘$ which then again is a function of itself (?). I hope someone can elude me, or at least confirm that it really does not.
The next step would then be to differentiate this partially with respect to $y$ and $y‘$ in order to be able to use the Euler-Lagrange equation. If the first term that causes the problem does not reduce to $0$ then how could one possibly find a solution to a differential equation where the variable depends on the function itself?

答: I think that you just kind of misread. Isn't it $x \cdot [y'(x)]^4$, i.e. $x$ times $y'^4$?


问:
Prove $ n \left(\frac{|z_2|}{|z_1|}\right)^{n-1} < \frac{|z_1|}{|z_1|-|z_2|}$ for $|z_1| > |z_2|$
Let $z_1 , z_2 \in \mathbb{C}$ and $|z_1|>|z_2|$. Prove that $\forall n\ge 2$ we have  $ n \left(\frac{|z_2|}{|z_1|}\right)^{n-1} < \frac{|z_1|}{|z_1|-|z_2|}$.

I started with induction:

For $n=2$ this inequality is ok because $2\frac{|z_2|}{|z_1|} <2 <  \frac{|z_1|}{|z_1|-|z_2|}$.

Now we check for $n+1$ so: $(n+1){(\frac{|z_2|}{|z_1|})}^n < \frac{|z_1|}{|z_1|-|z_2|} $ but I cannot prove that such inequality exists ..

答:
The inequality can be proved directly, without induction: With $c = |z_1|/|z_2| > 1$ it is equivalent to
$$
 \frac{n}{c^{n-1}} < \frac{c}{c-1}
$$
or
$$
 c^n > n(c-1)
$$
and that follows from Bernoulli's inequality:
$$
 c^n = (1 + (c-1))^n \ge 1 + n(c-1) > n(c-1) \, .
$$


问:
Applied PDE's book suggestion
I am searching for a book on applied PDEs which do not focus on how to solve them, but instead different areas where they appear and what can we do with them. Maybe a book in which each chapter shows applications of pdes in areas like economics, finance, physics, biology, social sciences, etc... .I also would be nice if it contains some numerical ways to use them. Can you suggest one book like that to me?</p>

答: It depends on which level you are aiming for. For undergraduate level, the book "Partial Differential Equations An Introduction" by Walter A. Strauss is decent (but it does not contain too much material on numerical PDE methods). For graduate level PDE, the classical book "Partial Equation Equations" by L.C.Evans seems quite popular.</p>

Title: Applied PDE's book suggestion


问:
Why does an exponential function eventually get bigger than a quadratic
I have seen the answer to this question and this one.

My $7$th grade son has this question on his homework:

How do you know an exponential expression will eventually be larger than any quadratic expression?

I can explain to him for any particular example such as $3^x$ vs. $10 x^2$ that he can just try different integer values of $x$ until he finds one, e.g. $x=6$. But, how can a $7$th grader understand that it will always be true, even $1.0001^x$ will eventually by greater than $1000 x^2$? They obviously do not know the Binomial Theorem, derivatives, Taylor series, L'Hopital's rule, Limits, etc,

Note: that is the way the problem is stated, it does not say that the base of the exponential expression has to be greater than $1$. Although for base between $0$ and $1$, it is still true that there exists some $x$ where the exponential is larger than the quadratic, the phrase "eventually" makes it sound like there is some $M$ where it is larger for all $x>M$. So, I don't like the way the question is written.

答:
Perhaps this is above the level of a 7th-grader, but a series expansion gives:

$$e^x > 1+x+\frac{x^2}{2}+\frac{x^3}{6}>\frac{x^3}{6}$$

and thus the cubic (and higher) terms will always dominate a quadratic at large $x$.


问:
A product of a symmetric and an antisymmetric tensor
Let $g^{ij}$ be a symmetric $(2,0)$ tensor and $F_{ij}=F_{[ij]}$, then what can we say about the product $g^{ij}F_{ij}$? Does it always have the same sign, is it always zero or it depends on the specific $g$ and $F$? (About $F_{[ij]}$ : Any rank $2$ tensor can be written as $F_{ij}=\frac{F_{(ij)}+F_{[ij]}}{2}$, where $F_{(ij)}$ is a symmetric tensor and $F_{[ij]}$ is an antisymmetric tensor.)

答: While $g^{ji}F_{ji}=g^{ij}F_{ij}$ by relabelling, on your assumptions $g^{ji}F_{ji}=-g^{ij}F_{ij}$ by exchanging indices. Combining these, $2g^{ij}F_{ij}=0$. The only way to avoid $g^{ij}F_{ij}=0$ is to work in characteristic $2$.


问:
Is $5^{1/5} - 3\cdot i$ algebraic?
I am studying the book Complex Variables with Applications written by Herb Silverman. In this book, problem number 8 in Question 1.7 is as in the following.

Is $5^{1/5} - 3\cdot i$ algebraic? (i.e, Is the $5$th-root of $5$ minus $3\cdot i$ algebraic?)

Can you help me to solve this?

答:
Multinomial Combiniations of Algebraic Integers

Given a set of monic polynomials defining a set of algebraic integers, how can we derive a polynomial for a multinomial combination of those algebraic integers?

For example, given $\alpha^2-3=0$ and $\beta^3-2=0$, how do we derive a polynomial for $\alpha+\beta$?

First, create the basis for all multinomial combinations of $\alpha$ and $\beta$:
$$
B=\begin{bmatrix}
1\\
\alpha\\
\beta\\
\alpha\beta\\
\beta^2\\
\alpha\beta^2
\end{bmatrix}
$$
Next compute $(\alpha+\beta)B$:
$$
\begin{align}
(\alpha+\beta)B
&=(\alpha+\beta)
\begin{bmatrix}
1\\
\alpha\\
\beta\\
\alpha\beta\\
\beta^2\\
\alpha\beta^2
\end{bmatrix}
=\begin{bmatrix}
\alpha+\beta\\
\alpha^2+\alpha\beta\\
\alpha\beta+\beta^2\\
\alpha^2\beta+\alpha\beta^2\\
\alpha\beta^2+\beta^3\\
\alpha^2\beta^2+\alpha\beta^3
\end{bmatrix}
=\begin{bmatrix}
\alpha+\beta\\
3+\alpha\beta\\
\alpha\beta+\beta^2\\
3\beta+\alpha\beta^2\\
\alpha\beta^2+2\\
3\beta^2+2\alpha
\end{bmatrix}\\[6pt]
&=\begin{bmatrix}
0&1&1&0&0&0\\
3&0&0&1&0&0\\
0&0&0&1&1&0\\
0&0&3&0&0&1\\
2&0&0&0&0&1\\
0&2&0&0&3&0\\
\end{bmatrix}
\begin{bmatrix}
1\\
\alpha\\
\beta\\
\alpha\beta\\
\beta^2\\
\alpha\beta^2
\end{bmatrix}
=MB
\end{align}
$$
$M$ is an integer matrix since the polynomials for $\alpha$ and $\beta$ are monic.

Since multiplication by a scalar commutes with matrix multiplication, induction gives
$$
(\alpha+\beta)^kB=M^kB
$$
SInce $M$ satisfies its own characteristic polynomial,
$$
0=P(M)B=P(\alpha+\beta)B
$$
therefore, $\alpha+\beta$ also satisfies the characteristic polynomial of $M$:
$$
P(x)=x^6-9x^4-4x^3+27x^2-36x-23
$$
Since the characteristic polynomial, $P(x)=\det(M-Ix)$, for an integer matrix is monic, $\alpha+\beta$ is also an algebraic integer.

Application to the Question

The example above should be general enough to show how to apply the process to any set of monic polynomials, a cubic and quadratic were used to reduce the footprint of the example.

To show that $5^{1/5}-3i$ is an algebraic integer, apply the program above to $\alpha^5-5=0$ and $\beta^2+9=0$ to produce a monic polynomial for $\alpha-\beta$:
$$
x^{10}+45x^8+810x^6-10x^5+7290x^4+900x^3+32805x^2-4050x+59074
$$

Afterword

I had thought I had written something along these lines before. I finally came across this answer that I wrote a while back. Hopefully, between this post and that one, the ideas are sufficiently clear.

Title: Is $5^{1/5} - 3\cdot i$ algebraic?


问:
Tutte Berge Formula Simple Proof Clarification
I am trying to understand the one line proof of Tutte-Berge formula here in 24.1 (just the $\le$ side proof), i.e. $$ν(G) ≤ |U| + ν(G − U) ≤ |U| + \frac 1 2 (|V \setminus U| − o(G − U)) = \frac 1 2 (|V | + |U| − o(G − U))$$

I do not fully understand where the first two inequalities and the third equality come from. Any explanation would be appreciated, thanks.

答:
For the first inequality $$\nu(G)\leq |U|+ \nu(G-U),$$ take any matching in $G$ and split it into edges that contain an element of $U$ and edges that do not.  There are at most $|U|$ edges of the first type, and at most $\nu(G-U)$ edges of the second type since they form a matching in $G-U$.

The second inequality $$\nu(G-U)\leq \frac{1}{2}(|V\setminus U|-o(G-U))$$ is because any matching in a graph must omit a vertex from each odd component.  So, a matching in $G-U$ can involve at most $|V\setminus U|-o(G-U)$ vertices.

The last equality is just algebra since $|V\setminus U|=|V|-|U|$.


问:
Suppose $F ⊂ K$ are fields. Let $f(x) ∈ F[x] ⊂ K[x]$. Suppose that $f(x)$ is irreducible in $K[x]$. Prove that $f(x)$ is also irreducible in $F[x]$.
Suppose $F ⊂ K$ are both fields. Let $f(x) ∈ F[x] ⊂ K[x]$. Suppose that $f(x)$ is irreducible in $K[x]$.

$a)$ Prove that $f(x)$ is also irreducible in $F[x]$.

$b)$ Is it true that if $f(x)$ is irreducible in $F[x]$, then it is irreducible in $K[x]$, if not, give an example.

My attempt:

$a)$ Since $f(x)$ is irreducible over $K$, then $K[x]/(f(x))$ is a field (I have previously proven this).

But since $F ⊂ K$, then $F[x] ⊂ K[x]$ and thus, $F[x]/(f(x))⊂K[x]/(f(x))$.

Since $F[x]/(f(x))$ is a subfield, then it is a field, and so $f(x)$ is irreducible over $F$.

Is my attempt correct?

And I don't think this still holds if $f(x)$ is irreducible over $F$. Can someone please clarify part $(b)$ and give an example? Thank you

答:
Your proof for (a) isn't quite there, although the result is true. You correctly have an inclusion $F[x]/(f) \subseteq K[x]/(f)$ but $K[x]/(f)$ being a field does not a priori mean that $F[x]/(f)$ is a field. You have shown that it is a subring, but it is not true that all subrings of fields are field. For instance, $\mathbb Z \subseteq \mathbb Q$. This sort of thing cannot happen in this case, however, but you have not proven that it will not.

By the way, you can actually just stop here without doing all the work to show that $F[x]/(f)$ is a field. It is a subring of the field $K[x]/(f)$ so it is a domain. Thus, $(f) \subseteq F[x]$ is a prime ideal, so $f$ is prime. Hence, it is irreducible. I'd still like to show you that you can prove that $F[x]/(f)$ is a field, however.

The key is that $F[x]/(f)$ is not just some random subring of $K[x]/(f)$ - it contains $F$, i.e. it is an $F$-algebra. This will let us show that $F[x]/(f)$ must be a field. Indeed, we have inclusions $F \subseteq F[x]/(f) \subseteq K[x]/(f)$. In particular, we know that $F[x]/(f)$ is a domain. Furthermore, by polynomial long division, $F[x]/(f)$ is a vector space over $F$ of finite dimension. Indeed, let $n = \deg f$ and let $\overline g$ denote the image of $g \in F[x] \longrightarrow F[x]/(f)$. By polynomial long division, we can write $g = qf + r$ where $\deg r < n$ (we take the convention that $\deg 0 = -\infty$). Hence, $\overline{g} = \overline{r}$ and $\overline{r} = \sum_{i=0}^n r_i \overline{x^i}$. Thus, $1, \overline{x}, \dots, \overline{x^n}$ spans $F[x]/(f)$ as an $F$ vector space.

This puts us in the following general situation: let $F$ be a field and $D \supseteq F$ be an $F$ algebra that is also a domain. Suppose additionally that $D$ is a finite dimensional vector space over $F$. Then $D$ is a field. Indeed, take any $0 \neq d \in D$ and consider the map $D \longrightarrow D$ via $a \mapsto ad$. This map is injective as $D$ is a domain. Furthermore, it is $F$-linear: for $\lambda \in F$, $(\lambda a) \mapsto (\lambda a) d = \lambda (ad)$. Thus, this map $D \longrightarrow D$ is an injective linear transformation of finite dimensional $F$-vector spaces. By linear algebra, it is a bijection. In particular, it is onto so $1$ is in the image. That is, there exists some $a \in D$ such that $ad = 1$. Hence, $d$ is a unit in $D$ so $D$ is a field.

For part (b), here's an explicit counterexample. Let $F = \mathbb R$ and $K = \mathbb C$. Then the polynomial $x^2 + 1$ is irreducible in $\mathbb R[x]$, as if it were reducible it'd split into linear factors but you can show that this has no real roots. However, it is reducible in $\mathbb C[x]$ as $(x - i)(x + i)$. More generally, if $f \in F[x]$ is irreducible then $F \subseteq F[x]/(f)$ is a field and $f$ is reducible in $F[x]/(f)$. In fact, it has a root $\overline{x}$.


答:
Perhaps still the best to work on a concrete example: $1.0001^x$ vs. $1000x^2$ is a good one.

Can you prove that you can find big enough $x_0$ so that the ratio of $\frac{1000(x+1)^2}{1000x^2}$ becomes smaller than, say, $1.00005$ for all $x\ge x_0$? This ratio is $(1+\frac{1}{x})^2$ and one could perhaps try to solve the inequality $1+\frac{1}{x}\le\sqrt{1.00005}$. (Use the calculator to calculate the latter - you get that $x\ge x_0\approx 40,000.5$ would work.)

Now, from that point on, $1.0001^x$ may still be tiny compared to $1000x^2$, but whenever $x$ grows by $1$, $1.0001^x$ grows by a factor $1.0001$, while $1000x^2$ grows by a factor at most $1.00005$. Thus their quotient grows by a factor at least $\frac{1.0001}{1.00005}\approx 1.00005$. Even if it is very small to start with (near $x=x_0$), that quotient grows exponentially and will soon become bigger than $1$.

(Left as exercise: why does $1.00005^x$ eventually becomes bigger than any constant? Could use $1.00005^x=(1+0.00005)^x\ge 1+0.00005x$, the latter easily proven at least for natural $x$  by reasoning in an inductive way, even if mathematical induction has not been formally introduced.)

I don't know of a simpler proof, and my honest opinion is that this is probably not the topic suitable for introduction at that level. Maybe they have a very ambitious teacher who sadly cannot identify with the ways of thinking of the students at that level, and believes that, however amazing the mere fact is (that the exponential always grows faster than a polynomial), they can provoke more amazement by introducing the proof of that fact. I expect the effect to be quite the opposite - the kids may be put off maths for life. (Majestic however it is, the middle of Pacific ocean is not the place to drop a beginner swimmer to enjoy the beauty of the ocean, so to speak.)

So your best strategy may be to not do the assignment, and wait to see what the teacher would say. At least you spare your kid from the hours of frustration of them working with you, and the teacher may end up producing some handwavy argument in the end, which we will all accept as "good enough at this stage" - and move on.

Title: Why does an exponential function eventually get bigger than a quadratic

Body:
I have seen the answer to this question and this one.

My $7$th grade son has this question on his homework:

How do you know an exponential expression will eventually be larger than any quadratic expression?

I can explain to him for any particular example such as $3^x$ vs. $10 x^2$ that he can just try different integer values of $x$ until he finds one, e.g. $x=6$. But, how can a $7$th grader understand that it will always be true, even $1.0001^x$ will eventually by greater than $1000 x^2$? They obviously do not know the Binomial Theorem, derivatives, Taylor series, L'Hopital's rule, Limits, etc,

Note: that is the way the problem is stated, it does not say that the base of the exponential expression has to be greater than $1$. Although for base between $0$ and $1$, it is still true that there exists some $x$ where the exponential is larger than the quadratic, the phrase "eventually" makes it sound like there is some $M$ where it is larger for all $x>M$. So, I don't like the way the question is written.


问:
Prove that the least upper bound of the sequence $\left\{ \frac{2^n - 1}{2^{n-1}} \right\}$ is $2$
It can be easily verified that the sequence $\left\{ \dfrac{2^n -1} {2^{n-1}} \right\}$ is strictly positive, the task now is to prove that 2 is an upper bound.

Proof:

Suppose for the sake of contradiction that 2 is not the least upper bound of the sequence. Let $x_n$ be an element of the sequence. Using the density of the real numbers in combination with the fact that the sequence is strictly increasing we can write the following:

$$ x_i < 2 < x_j , \quad j>i  $$

That is, we can find a $j \in \mathbb{N}$ such that:

\begin{align}
0 &< x_j - 2 \\
  &<\dfrac{2^j - 1}{2^{j-1}} - 2 \\
&< \dfrac{2^j - 1 - 2(2^{j-1})}{2^{j-1}} \\
&< (-1)\dfrac{1}{2^{j-1}} \\
\end{align}

So we arrive at a contradiction.

I just want to check to see if my approach makes sense to this problem.

答:
Your reasoning is incomplete, because it only establishes that there is no $j$ such that $x_j > 2$.  It does not demonstrate that $2$ is the least such bound; e.g., if you replace $2$ with $4$, your reasoning would still apply.  But $4$ is not the least upper bound.

In order to show that $2$ is the least upper bound, one way is to establish that if $B$ is the set of all upper bounds for $x_n$, that $2$ is the smallest member of $B$.


问:
Finding the common modulus given many congruences
Consider the following relationships:

$a_1 \equiv b_1 \mod{g}$

$a_2 \equiv b_2 \mod{g}$

$\cdots $

$a_n \equiv b_n \mod{g}$

Where:

All values $a_1, a_2, ... , a_n$ are known

All values $b_1, b_2, ... , b_n$ are unknown

$g$ is an unknown modulus

$n$ is an unknown limit

It is known that such a relationship exist for all values $a_1, a_2, ... , a_n$

All the numbers are generated randomly (AKA: $g$ is not intentionally prime and there is no intentional coprime relationships), so an RSA common modulus attack is not intentional)

Question is: what would be the fastest way to find $g$ and/or all of $b_1, b_2, ... , b_n$ values?

I do realize that a brute-force attack would definitely reveal it, especially if the bit space is not high, but I'm looking for an easier and less noisy way. Does such a way exist?

答:
You do not have enough information to find anything, because you introduce a new unknown with every added equation.

When you start off with $a_1 \equiv b_1 \pmod g$, then any value of $g$ is possible, and $b_1$ can be any of the infinitely many values $a_1 + k_1 g$ for any $k_1 \in \mathbb Z$.

For each equation $a_i \equiv b_i \pmod g$ you add, any existing hypothesis $(b_1, b_2, \dots, b_{i-1}, g)$ can be extended to solve this equation as well, in infinitely many ways, by setting $b_i = a_i + k_i g$ for any $k_i \in \mathbb Z$.

You might be able to make guesses about which values of $b_1, b_2, \dots, b_n, g$ are more likely than others, if you know the law by which they are randomly chosen (and how you receive the values $a_i$ - for example, is it true that each $a_i$ is between $0$ and $g-1$?)


答:
These statements are not equivalent to one another. In fact, the LHS expression is a tautology! To see this, note that if $q$ is true, then $p \to q$ is true (anything implies a true statement) and if $q$ is false, then $q \to r$ is true (a false statement implies anything).

However, the RHS is not a tautology. Specifically, it’s false if $p$ and $q$ are true and $r$ is false.

Hope this helps!


答:
$$\frac{2^n-1}{2^{n-1}}<2$$
multiply both sides by $2^{n-1}$
$$2^n-1<2^n$$
which is true for any $n$.

Furthermore $$\frac{2^n-1}{2^{n-1}}=\frac{2^n}{2^{n-1}}-\frac{1}{2^{n-1}}=2-\frac{1}{2^{n-1}}$$

For any $\varepsilon >0$ there exists $N$ such that if $n>N$ and $n\in\mathbb {N}$ then
$$\frac{1}{2^{n-1}}<\varepsilon$$
we take $N=1-\log _2 \varepsilon $.

Therefore $2$ is the least upper bound of the sequence $\{\frac{2^n-1}{2^{n-1}}\}$

Title: Prove that the least upper bound of the sequence $\left\{ \frac{2^n - 1}{2^{n-1}} \right\}$ is $2$

Body:
It can be easily verified that the sequence $\left\{ \dfrac{2^n -1} {2^{n-1}} \right\}$ is strictly positive, the task now is to prove that 2 is an upper bound.

Proof:

Suppose for the sake of contradiction that 2 is not the least upper bound of the sequence. Let $x_n$ be an element of the sequence. Using the density of the real numbers in combination with the fact that the sequence is strictly increasing we can write the following:

$$ x_i < 2 < x_j , \quad j>i  $$

That is, we can find a $j \in \mathbb{N}$ such that:

\begin{align}
0 &< x_j - 2 \\
  &<\dfrac{2^j - 1}{2^{j-1}} - 2 \\
&< \dfrac{2^j - 1 - 2(2^{j-1})}{2^{j-1}} \\
&< (-1)\dfrac{1}{2^{j-1}} \\
\end{align}

So we arrive at a contradiction.

I just want to check to see if my approach makes sense to this problem.


问:
Greater precision with regression?
I used my TI-83 to find the quadratic regression of two data columns. The accuracy wasn't close at all. So I tried cubic and then finally quartic regression. The accuracy still isn't close enough. Is there any way I can improve the accuracy? I can't add any more data, unfortunetly.

Edit 1:

To address Henry's comment, my plotted data looks similar to y=x^(1/3)

答: If, as you say, it looks like $y=x^\frac{1}{3}$, then you might try a power regression, which will give you a formula of the form $y=ax^b$.</p>

Title: Greater precision with regression?


问:
Show $z^4 + e^z = 0$ has a solution in $\{z \in \mathbb{C} : |z| \leq 2\}$
Show $z^4 + e^z = 0$ has a solution in $\{z \in \mathbb{C} : |z|  \leq 2\}$.

I would like if in the proof the tools of algebraic topology were preferred over the other tools of analysis, complex analysis, algebra etc.

答: The tools are the same. Let $f(z)=z^4+e^z$ and let $g(z)=z^4$. Show that $\dfrac{f}{|f|}$ and $\dfrac{g}{|g|}$ are homotopic as maps $\{z: |z|=2\} \to S^1$. Note that if $h\colon S^1\to S^1$ has nonzero degree, then $h$ has a root in $D^2$. (Why?)</p>


问:
How to derive a model for a recursive situation
An animal use 50% of its fat every night, but is able to add $4$ grams to its fat reserve every day. This same animal begin with 3 grams of fat on the first night.

Derive a model for its fat reserve at the beginning of $(t-1)$ night.

I tried to write down first couple nights and find a relation.

Night 1: $3$

Night 2: $3*0.5+4$

Night 3: $(3*0.5+4)*0.5 + 4$

Night 4: $((3*0.5+4)*0.5 + 4)*0.5 + 4$

I feel like I am onto something, but I can't quite get it. Maybe it's my lack of understanding of recursive relationship. I need some help on this.

答: Since you're trying to generalize a pattern, it'll help to have notation for the items in the sequence. For every natural number $t$, let's define $x_t$ to be the fat amount on the beginning of the $t$th night. You've calculated $x_1,x_2,x_3,x_4$. Could you write a formula for $x_{100}$ in terms of $x_{99}$? A formula for $x_{t+1}$ in terms of $x_t$? Then you're done. (It looks like you're not required to solve the recurrence relation, just write it down.)


问:
Second-order Eulerian numbers, Lambert's W function, and Schröder's fourth problem
Recently, some of the remarkable properties of second-order
Eulerian numbers $ \left\langle\!\!\left\langle n\atop k\right\rangle\!\!\right\rangle$  A340556
have been proved on MSE [  a ,
 b ,  c  ].
A special property unfolds when these numbers are interpreted as coefficients of polynomials:

$$ \left\langle\!\left\langle x \right\rangle\!\right\rangle_n =
\sum_{k=0}^n \left\langle\!\!\left\langle n\atop k \right\rangle\!\!\right\rangle \, x^k $$
If we evaluate these polynomials at the point $ x = 1/2 $, we get
a sequence whose exponential generating function is based on
the amazing Lambert W function.

$$ 2^n \left\langle\!\!\left\langle  \frac{1}{2}  \right\rangle\!\!\right\rangle_n
= n! \, [x^n]\, ({2 + 2\operatorname{W}(- \exp((x-1)\,/\,2)\,/\,2))^{-1}} $$

Not only that: the sequence represents the solution of a
famous combinatorial problem, Schröder's fourth problem
(see A000311)!
It would be great if someone could show us a proof of this formula.

答:
We seek to show that the following identity holds:

$$2^{n+1} \sum_{k=0}^n
\left\langle\!\! \left\langle n \atop k
\right\rangle\!\! \right\rangle
\frac{1}{2^k} =
n! [x^n] \frac{1}{1 + W(-\exp((x-1)/2)/2)}.$$

We will be using data from Wikipedia on Lambert
W and work with the
combinatorial branch which is  $W_0(z)$.

Recall that

$$W'(z) \frac{z}{W(z)} = \frac{1}{1+W(z)}.$$

We obtain

$$[z^m] \frac{1}{1+W(z)} =
\frac{1}{2\pi i}
\int_{|z|=\varepsilon} \frac{1}{z^{m}}
\frac{1}{W(z)} W'(z) \; dz.$$

Putting $W(z) = v$ we find

$$\frac{1}{2\pi i}
\int_{|v|=\gamma} \frac{1}{v^{m} \exp(mv)}
\frac{1}{v} \; dv
= \frac{1}{2\pi i}
\int_{|v|=\gamma} \frac{1}{v^{m+1}} \exp(-mv) \; dv
= \frac{(-1)^m m^m}{m!}.$$

so that

$$\frac{1}{1+W(z)} = \sum_{m\ge 0} (-1)^m m^m \frac{z^m}{m!}.$$

We get for the original RHS

$$n! [x^n] \sum_{m\ge 0} \frac{m^m}{m!}
\exp(m(x-1)/2) \frac{1}{2^m}
\\ = n! [x^n]
\sum_{m\ge 0} \frac{m^m}{m!} \frac{\exp(-m/2)}{2^m} \exp(mx/2)
\\ = \sum_{m\ge 0} \frac{m^{m+n}}{m!}
\frac{\exp(-m/2)}{2^{m+n}}.$$

First part.
Introduce the tree function $T(z)$  from combinatorics where $T(z) = z
\exp  T(z)$ and  $T(z) =  - W_0(-z).$  Note that  we have  by Cayley's
theorem that $T(z)  = \sum_{m\ge 1} m^{m-1}  \frac{z^m}{m!}.$ We claim
that with $n\ge 1$

$$Q_n(z) = \sum_{m\ge 0} m^{m+n} \frac{z^m}{m!} =
\frac{1}{(1-T(z))^{2n+1}}
\sum_{k=1}^n
\left\langle\!\! \left\langle n \atop k
\right\rangle\!\! \right\rangle
T(z)^k.$$

This  means the  RHS is  $\frac{1}{2^n} Q_n(\exp(-1/2)/2).$  To verify
this last identity  note that $Q_{n+1}(z) = z  \frac{d}{dz} Q_n(z)$ so
we may prove it by induction.

We get for the RHS of the series identity on differentiating and
multiplying by $z$

$$\frac{(2n+1) z T'(z)}{(1-T(z))^{2n+2}}
\sum_{k=1}^n
\left\langle\!\! \left\langle n \atop k
\right\rangle\!\! \right\rangle
T(z)^k +
\frac{z}{(1-T(z))^{2n+1}}
\sum_{k=1}^n
\left\langle\!\! \left\langle n \atop k
\right\rangle\!\! \right\rangle
k T(z)^{k-1} T'(z)$$

Extracting the term $z T'(z)/(1-T(z))^{2n+2}$ in front leaves us with

$$(2n+1) \sum_{k=1}^n
\left\langle\!\! \left\langle n \atop k
\right\rangle\!\! \right\rangle
T(z)^k +
(1-T(z))
\sum_{k=1}^n
\left\langle\!\! \left\langle n \atop k
\right\rangle\!\! \right\rangle
k T(z)^{k-1}
\\ = (2n+1) \sum_{k=1}^n
\left\langle\!\! \left\langle n \atop k
\right\rangle\!\! \right\rangle
T(z)^k +
\sum_{k=0}^{n-1}
\left\langle\!\! \left\langle n \atop k+1
\right\rangle\!\! \right\rangle
(k+1) T(z)^{k}
- \sum_{k=1}^n
\left\langle\!\! \left\langle n \atop k
\right\rangle\!\! \right\rangle
k T(z)^{k}
\\ = \sum_{k=1}^n
\left\langle\!\! \left\langle n \atop k
\right\rangle\!\! \right\rangle
(2n+2-(k+1)) T(z)^k +
\sum_{k=0}^{n-1}
\left\langle\!\! \left\langle n \atop k+1
\right\rangle\!\! \right\rangle
(k+1) T(z)^{k}.$$

We may include $k=0$ in the first sum and $k=n$ in the second.
Now the Eulerian number recurrence (second order) according to OEIS
A349556 is

$$\left\langle\!\! \left\langle n \atop k
\right\rangle\!\! \right\rangle
=
\left\langle\!\! \left\langle n-1 \atop k
\right\rangle\!\! \right\rangle k
+ \left\langle\!\! \left\langle n-1 \atop k-1
\right\rangle\!\! \right\rangle (2n-k)$$

We have shown that

$$Q_{n+1}(z) = \frac{z T'(z)}{(1-T(z))^{2n+2}}
\sum_{k=0}^n \left\langle\!\! \left\langle n+1 \atop k+1
\right\rangle\!\! \right\rangle T(z)^k
\\ = \frac{z T'(z)}{T(z) (1-T(z))^{2n+2}}
\sum_{k=1}^{n+1} \left\langle\!\! \left\langle n+1 \atop k
\right\rangle\!\! \right\rangle T(z)^k.$$

Now we just have to verify that

$$\frac{z T'(z)}{T(z) (1-T(z))^{2n+2}}
= \frac{1}{(1-T(z))^{2n+3}}
\quad\text{or}\quad
z T'(z) (1-T(z)) = T(z).$$

The functional equation tells us that $T'(z) = \exp T(z) + z \exp T(z)
T'(z)$ so that $T'(z) (1 - T(z)) = \exp T(z) = T(z) / z$ which is just
what we need. It remains to verify the base case so the induction
starts properly. We seek

$$Q_1(z) = \sum_{m\ge 0} m^{m+1} \frac{z^m}{m!} =
\frac{T(z)}{(1-T(z))^3}.$$

We verify this by coefficient extraction. We get

$$m! [z^m] Q_1(z) =
\frac{m!}{2\pi i}
\int_{|z|=\varepsilon}
\frac{1}{z^{m+1}} \frac{T(z)}{(1-T(z))^3} \; dz.$$

With $T(z) = z + \cdots$ this integral will produce the correct value
zero for $m=0.$ For $m\ge 1$,  we put $T(z) = w$ so that $z = w \exp(-w)$
and $dz = \exp(-w) (1-w) \;  dw$ and obtain

$$\frac{m!}{2\pi i}
\int_{|w|=\gamma}
\frac{\exp((m+1)w)}{w^{m+1}} \frac{w}{(1-w)^3}
\exp(-w) (1-w) \; dw
\\ = \frac{m!}{2\pi i}
\int_{|w|=\gamma}
\frac{\exp(mw)}{w^{m}} \frac{1}{(1-w)^2}
\; dw.$$

This is

$$m! \sum_{q=0}^{m-1} \frac{m^q}{q!} (m-q)
= m! \sum_{q=0}^{m-1} \frac{m^{q+1}}{q!}
- m! \sum_{q=1}^{m-1} \frac{m^q}{(q-1)!}
\\ = m! \sum_{q=0}^{m-1} \frac{m^{q+1}}{q!}
- m! \sum_{q=0}^{m-2} \frac{m^{q+1}}{q!}
= m! \frac{m^m}{(m-1)!} = m^{m+1}$$

as desired.

Sequel. Note  that in  the identity  for $Q_n(z)$  we have  by the
definition of the Eulerian numbers that $\left\langle\!\! \left\langle
n  \atop 0  \right\rangle\!\! \right\rangle$  is zero  when $n\ge  1.$
Therefore we  may extend $k$  to include zero  (with $n\ge 1$  for the
moment) which yields

$$\bbox[5px,border:2px solid #00A000]{
Q_n(z) = \sum_{m\ge 0} m^{m+n} \frac{z^m}{m!} =
\frac{1}{(1-T(z))^{2n+1}}
\sum_{k=0}^n
\left\langle\!\! \left\langle n \atop k
\right\rangle\!\! \right\rangle
T(z)^k.}$$

Now  observe  that this  will  produce  $Q_0(z)  = \sum_{m\ge  0}  m^m
\frac{z^m}{m!}    =   \frac{1}{1-T(z)}$   due   to   $\left\langle\!\!
\left\langle 0 \atop  0 \right\rangle\!\! \right\rangle =  1$ which is
in fact correct because unlike $Q_n(z)$  with $n\ge 1$, $Q_0(z)$ has a
constant term, which  is one (this is because $m^{m+n}  = 0$ for $m=0$
and  $n\ge 1$  and  $m^{m+n} =  1$ for  $m=0$  and $n=0$).   Therefore
$$Q_0(z) = 1 + z T'(z)  = 1 + \frac{T(z)}{1-T(z)} = \frac{1}{1-T(z)}$$
as obtained from the boxed version of the main identity, which is seen
to hold for all $n\ge 0.$

Conclusion. We are now ready  to answer the original question.  We
have shown that  the RHS is $\frac{1}{2^n}  Q_n(\exp(-1/2)/2).$ By our
formula for  $Q_n(z)$ in  terms of  the tree  function we  obtain with
$T(\exp(-1/2)/2) = \frac{1}{2}$ at last the closed form

$$\frac{1}{2^n} \frac{1}{(1-1/2)^{2n+1}}
\sum_{k=0}^n
\left\langle\!\! \left\langle n \atop k
\right\rangle\!\! \right\rangle \frac{1}{2^k}
= 2^{n+1}
\sum_{k=0}^n
\left\langle\!\! \left\langle n \atop k
\right\rangle\!\! \right\rangle \frac{1}{2^k}$$

which is the LHS and hence the claim.


Body: Your solution is fine, you could also notice that $\frac{2^n-1}{2^{n-1}}=\frac{2^n}{2^{n-1}}-\frac{1}{2^{n-1}}=2-\frac{1}{2^{n-1}}\leq2$ as $\frac{1}{2^{n-1}}\geq0$


问:
Express $\operatorname{sech}^{-1}(x)$ in terms of logarithms
I'm trying to express the following  $\operatorname{sech}^{-1}(x)$ in terms of logarithms, and would warmly appreciate feedback towards my approach. The solution should be :

$$\ln\left(\dfrac{1+\sqrt{(1-x^2)}}{x}\right)$$ when $x > 0.$ However, I cannot seem to get this. My working out:

$y = \operatorname{sech}^{-1}(x)\implies \operatorname{sech}(y) = x$

$$-\operatorname{sech}(y)\operatorname{tan}(y) = x; $$

Once I got to here, I decided to transform the $\operatorname{sech}$ and $\tan$ into their hyperbolic identities.

$$-\dfrac{\operatorname{sinh}(y)}{\operatorname{cosh}^2(y)}=\dfrac{e^y-e^{-y}}{2}(\dfrac{e^y+e^{-y}}{2})^2 = x$$

Though I've tried reworking this into a quadratic form $\dfrac{b \space \pm \space \sqrt{b^2-4ac}}{2a}$, however, I couldn't manage to get the right form. I would greatly appreciate some help on the next steps towards this.

答:
By definition $\operatorname{sech}(x)$ is equal to
$$y = \frac{2}{e^x+e^{-x}}.$$
What we now want to do is solve for $x$. First, divide both sides by $2$ and then take the multiplicative inverse of both sides to get $$\frac{2}{y}=e^x+e^{-x}$$
Now substitute $u=e^x.$ We will get a new and easy equation.
$$u+\frac{1}{u}=\frac{2}{y}.$$ You should be able to solve this equation for $u$
$$u=\pm\sqrt{y^{-2}-1}+\frac{1}{y}.$$
If you are done, don't forget to substitute back $e^x=u$.


问:
Master Theorem change of variables with root other than 2
I'm working on this:

$$T(n) = 12T(n^{1/3}) + \log(n)^{2}$$

Using change of variables, and substituting $m = \log n$, I get as far as:

$$S(m) = 12S(m/3) + m^{2}$$

I see how a square root would work but with a cube root I'm not sure that $\Theta(m \log m)$ makes sense since the convention seems to be that this means $\log_2 n$ and I'm not seeing how that accounts for a base $3$ $\log$.

答:
Let $n=3^m$ (that is, let $m=\log_3n$). Then by a change of variables, we have:
$$
T(3^m) = 12T((3^m)^{1/3}) + (\log(3^m))^{2} = 12T(3^{m/3}) + (m\log3)^{2} = 12T(3^{m/3}) + (\log3)^{2}m^2
$$
Renaming $S(m)=T(3^m)$ yields: $\boxed{S(m)=12S(m/3)+(\log3)^{2}m^2}$

Since $\log_3{12}>\log_3{9}=2$, it follows that $(\log3)^{2}m^2=O(n^{\log_3{12}-\epsilon})$ if  $0<\epsilon\le\log_3{12}-2$. Thus, by Case 1 of the Master Theorem, we have $\boxed{S(m)=\Theta(m^{\log_3{12}})}$. Changing variables back to the original recurrence yields:
$$
T(n)=T(3^m)=S(m)=\Theta(m^{\log_3{12}})=\Theta((\log_3n)^{\log_3{12}})=\boxed{\Theta((\log n)^{\log_3{12}})}
$$
Using the identity $x^{\log_b{y}}=y^{\log_b{x}}$, we can alternatively write this as:
$$
T(n)=\boxed{\Theta(12^{\log_3{\log n}})}
$$


问:
How to determine the center of symmetry of a function?
I'm learning about functions and I came to a question that I'm struggling at:

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

The question is:

Show that the graph of $f$, has a center of symmetry to be determined in an orthonormal system.

What is the best approach to solve this question?

Body: Since $f=x-1-\frac{1}{x+1}$, rotation around $(a,\,b)$ gives$$y=2b-f(2a-x)=2b-2a+x+1-\frac{1}{x-2a-1}.$$This is the original curve for $2b-2a+1=-1,\,-2a-1=1$, i.e. $a=-1,\,b=-2$.


答: Once you have $\operatorname{sech}(y) = x$ it implies $$\dfrac{e^y+e^{-y}}2=\dfrac1x$$ which simplifies to the quadratic $$xe^{2y}-2e^y+x=0$$ of $e^y.$ Now solve for $y$ as you discribed in the last sentence.

Title: Express $\operatorname{sech}^{-1}(x)$ in terms of logarithms

Body:
I'm trying to express the following  $\operatorname{sech}^{-1}(x)$ in terms of logarithms, and would warmly appreciate feedback towards my approach. The solution should be :

$$\ln\left(\dfrac{1+\sqrt{(1-x^2)}}{x}\right)$$ when $x > 0.$ However, I cannot seem to get this. My working out:

$y = \operatorname{sech}^{-1}(x)\implies \operatorname{sech}(y) = x$

$$-\operatorname{sech}(y)\operatorname{tan}(y) = x; $$

Once I got to here, I decided to transform the $\operatorname{sech}$ and $\tan$ into their hyperbolic identities.

$$-\dfrac{\operatorname{sinh}(y)}{\operatorname{cosh}^2(y)}=\dfrac{e^y-e^{-y}}{2}(\dfrac{e^y+e^{-y}}{2})^2 = x$$

Though I've tried reworking this into a quadratic form $\dfrac{b \space \pm \space \sqrt{b^2-4ac}}{2a}$, however, I couldn't manage to get the right form. I would greatly appreciate some help on the next steps towards this.


问:
Upper bound on difference between Gaussian CDFs
May I know if there are any non-trivial upper bounds $f$ on the following:

$$\Phi(a + \Delta) - \Phi(a) \leq f(a, \Delta)$$

for $\Phi$ the CDF of a standard normal and all $a, \Delta > 0$.

Thanks!

答:
Let $X$ be a standard normal random variable. Then you want an upper bound of

$$
P(X>a) - P(X>a+\Delta).
$$

Now this post gives $P(X>a) \leq \frac{1}{a\sqrt{2\pi}}\exp(-a^2/2)$ and this other post gives $$P(X>a+\Delta) \geq \frac{e^{-1/2}(1- e^{-(a+\Delta)-(a+\Delta)^2/2})}{\sqrt{2\pi}(a+\Delta)}.$$ Therefore a choice of $f(a,\Delta)$ would be

$$
f(a,\Delta) := \frac{1}{a\sqrt{2\pi}}\exp(-a^2/2) - \frac{e^{-1/2}(1- e^{-(a+\Delta)-(a+\Delta)^2/2})}{\sqrt{2\pi}(a+\Delta)}.
$$


问:
What makes $5$ and $6$ so special that taking powers doesn't change the last digit?
Why are $5$ and $6$ (and numbers ending with these respective last digits) the only (nonzero, non-one) numbers such that no matter what (integer) power you raise them to, the last digit stays the same? (by the way please avoid modular arithmetic) Thanks!</p>

答:
A small piece that's missing out of the other answers: once you know that $5\times5$ ends in a $5$, and that $6\times 6$ ends in a $6$, then you can quickly show it'll be the case for the rest of the powers.  For instance, consider $5^3$; then this is $5\times(5^2)=5\times25=5\times(20+5)=(5\times20)+(5\times5)$.  Since $20$ ends in $0$ then multiplying it by anything will lead to another number ending in $0$ (i.e., another multiple of $10$), and we've already shown that $5\times5$ ends in $5$, so their sum will end in $0+5=5$.  Likewise, $5^4=5^3\times5=$somethingsomethingsomething$5\times5=($somethingsomethingsomething$0+5)\times5$, etc.

Once you've showed that $5^n$ ends in $5$, then you can use the result for $5\times5$ to show that $5^{(n+1)}$ ends in $5$, and then use the principle of mathematical induction to show that it holds for all powers of $5$.  The exact same approach works for the powers of $6$, using the fact that $6\times6$ ends in $6$.


问:
How to prove this propositional logic equivalence?
Problem : $\lnot[(\lnot p∧q)∨\lnot q]∨[\lnot(p∧\lnot q)]≡\lnot p∨q$  Prove this equility without using truth table.

I have managed to solve this question until this part :

$$\begin{align}\lnot[(\lnot p∧q)∨[\lnot(p∧\lnot q)] &≡ [\lnot(\lnot p∧q)∧q]∨(\lnot p∨q)\\& ≡ [(p∧q)∨(\lnot q∧q)]∨(\lnot p∨q)]\\&  ≡ (p∧q)∨(\lnot p∨q)\end{align}$$

I stucked here. I tried to $(p∧q)∨\lnot(p∧\lnot q)$ like this for de morgan rule but it doesnt work. What do I need to do?

答:
Start by opening everything up.
=> ~[(~p v ~q) ^ (q v ~q)] v (~p v q)
=> ~[(~p v ~q) ^ T] v (~p v q)
=> ~(~p v ~q) v (~p v q)
=> (p ^ q) v ~p v q
=> (p v ~p) ^ (q v ~p) v q
=> T ^ (~p v q) v q
=> ~p v q</p>


问:
$\int \sin( \pi x) \cos( \pi x)\ dx$
I would like some help computing this integral:

$$\int \sin(\pi t)\cos(\pi t)dt$$

Apparently, the answer should be $\frac{\sin^2(\pi t)}{2\pi}+C$ but I get $\frac{-\cos(2 \pi t)}{4 \pi}$+C instead. Here's what I did:

I rewrote the integrand as $\frac{1}{2}\sin(2\pi t)$.

I took out the constant $\frac{1}{2}$ out of the integral, so my expression became $\frac{1}{2} \int \sin(2\pi t)dt$.

I evaluated the antiderivative of $\sin(2\pi t)$, which I believe is $-\frac{\cos(2\pi t)}{2 \pi}+C$.

Lastly, I multiplied the antiderivative with $\frac{1}{2}$, yielding my final answer.

Would it be wrong to assume that both answers are equivalent, since the integral can either be rewritten using the half-angle formula or solved by U-substitution? If not, could you please explain why?

答:
Yes, both answers are equivalent. Using the identity $\cos(2x)=1-2\sin^2(x)$ we have

$$-\frac{\cos(2\pi t)}{4\pi}+C_1=\frac{2\sin^2(\pi t)-1}{4\pi}+C_1=\frac{\sin^2(\pi t)}{2\pi}+C_2$$

where $C_2=C_1-\frac{1}{4\pi}$ (for constants $C_1$ and $C_2$).

Let $u=\sin(\pi t)$, then $\frac{du}{dt}=\pi\cos(\pi t)$ and we have

$$\int\sin(\pi t)\cos(\pi t)dt=\frac{1}{\pi}\int udu=\frac{u^2}{2\pi}+C=\frac{\sin^2(\pi t)}{2\pi}+C$$

Title: $\int \sin( \pi x) \cos( \pi x)\ dx$


答:
Your answer is okay but it's ludicrously complicated.

Why not just say $\frac {2^n -1}{2^{n-1}} < \frac {2^n}{2^{n-1}} = 2$?

That's it.  You are done.

But is that all of the question?

Are you sure you weren't also being as to prove that $2$ is the least upper bound.

In that case if $x < 2$ you must prove there exists a $x < \frac {2^n-1}{2^{n-1} } < 2$ no matter how close $x$ is to $2$.

$x < \frac {2^n-1}{2^{n-1} } = \frac {2^n}{2^{n-1}} - \frac 1{2^{n-1}} = 2-\frac 1{2^{n-1}} < 2 \iff$

$x-2 <  -\frac 1{2^{n-1}} < 0\iff$

$0 < \frac 1{2^{n-1}} < 2-x \iff$

$x < 2$ and $2^{n-1} > \frac 1{2-x}$.

If we know that $2^m$ is unbounded then we know that there is an $m = n-1$ so that $2^{n-1} > \frac 1{2-x} $ for any $x < 2$.

If $\{2^m\}$ were bounded above then $\sup 2^m$ would exist.  And $\sup 2^m -1$ wouldn't be an upper bound so there would be a $k$ so that $\sup 2^m -1 < 2^k < \sup 2^m$  So $2\sup 2^m - 2 < 2^{k+1}$.  But as $\sup 2^m  > 2=2^2$ as must have $2\sup 2^m - 2 > \sup 2^m$ and so $\sup 2^m < 2^{k+1}$ which is a contradiction.


问:
For symmetric real matrices, each $L^2$ circle contains at least an eigenvalue?
Let $A=(a_{ij})_{ij}$ be a symmetric real matrix. I know by the Gershgorin circle theorem, every eigenvalue $\lambda$ of $A$ must lie in a closed circle defined as
$$D_1(A,i):=\left\{x\in\mathbb{R}\left||x-a_{ii}|\le\sum_{j\neq i}|a_{ij}|\right.\right\}$$
for some $i$. However, the converse does not necessarily hold, which means for each index $i$, it is possible that $D_1(A,i)$ does not contain any eigenvalue of $A$.

Here is my question, if we choose the $L^2$ norm rather than the $L^1$ norm to define the circles
$$D_2(A,i):=\left\{x\in\mathbb{R}\left||x-a_{ii}|^2\le\sum_{j\neq i}|a_{ij}|^2\right.\right\},$$
then for each index $i$, $D_2(A,i)$ must contain at least one eigenvalue of $A$. I have tried to search for it online, and it turns out a more general version of the conclusion can be found in this paper (by setting $x_i=y_i=\left(\sum_{j\neq i}|a_{ij}|^2\right)^{1/2}$ in Lemma 3.1). Of course I can follow the proof of the Lemma directly, but I am wondering if there exists a simpler and easier approach since the proposition itself is not complex at all.

Any help would be appreciated.

答:
A word of caution: the lemma of Barnes and Hoffman is not a generalisation of Gerschgorin disc theorem, because the union of all Gerschgorin discs always covers the spectrum of a matrix, but this is not the case in Barnes and Hoffman's lemma. E.g. when $A$ is the $3\times3$ matrix of ones and $x_i=y_i=\left(\sum_{j\ne i}|a_{ij}|^2\right)^{1/2}$, the eigenvalue $3$ lies outside $D_2(A,i)$ for all $i$.

However, since the Euclidean norm is bounded above by the $\ell_1$ norm, the lemma of Barnes and Hoffman implies that each Gerschgorin disc must contain some eigenvalue of $A$ when $A$ is real symmetric. (This in general is false when $A$ is not Hermitian.)

Anyway, to prove the lemma in the special case where $x_i=y_i=\left(\sum_{j\ne i}|a_{ij}|^2\right)^{1/2}$, we suppose that $i=1$ (the proof is similar for other values of $i$). Let $A=\pmatrix{a&\mathbf v^T\\ \mathbf v&S}$ and $B=A-aI$.
Since $B$ is symmetric, its singular values are the absolute values of its eigenvalues. As the smallest singular value of $B$ is bounded above by the norm of the first column of $B$, we obtain $|\lambda(B)|=\sigma_\min(B)\le\|\mathbf v\|_2$ for the smallest-sized eigenvalue $\lambda(B)$ of $B$. It follows that $|\lambda(A)-a|^2\le\|\mathbf v\|_2^2$ for the eigenvalue $\lambda(A)$ of $A$ that is the closest to $a$.


问:
Constants in Laplace's equation for a cube
I'm working a Laplace's equation $\Delta F=0$  for a cube in Cartesian coordinates $((0,0,0),(a,a,a))$ and after separation I have $$\frac{X''(x)}{X(x)} +\frac{Y''(y)}{Y(y)}+\frac{Z''(z)}{Z(z)} = 0\tag{1}$$

with boundary conditions

$$ x=0 \Rightarrow F=F_0$$
$$ x=a \Rightarrow F=-F_0$$
$$ y=0 \Rightarrow F=0$$
$$ y=a \Rightarrow F=0$$
$$ z=0 \Rightarrow F=0$$
$$ z=a \Rightarrow F=0$$

Thinking about my boundary condition, I figured I'd want the Y and Z terms in $(1)$ to be equal to negative constants $-k^2$ and $-\ell^2$ so I could get a $\sin{\pi y\over a}\sin{\pi z\over a}$ out of them, and the X term to equal a negative constant $-j^2$ as well whence I'd take a $F_0\cos{2\pi x\over a}$. Problem is, if they are all negative constants, they can't add up to zero as in $(1)$! What other options do I have here? I don't want to take a positive $-j^2$ because

答:
Scale the variable such that $x' = \pi x/a$, $y' = \pi y/a$, $z' = \pi z/a$ first, we the equation still holds
$$
\frac{X''(x')}{X(x')} +\frac{Y''(y')}{Y(y')}+\frac{Z''(z')}{Z(z')} = 0
$$
and we can still get what you mentioned in your question:
$$
\begin{cases}
Y'' = -k^2 Y
\\
Y(0) = Y(\pi) = 0
\end{cases},
\quad
\begin{cases}
Z'' = -l^2 Z
\\
Z(0) = Z(\pi) = 0
\end{cases}.
$$
Slightly abuse of notation, these two lead to
$$
Y_k(y) = \sin(ky),\quad\text{and}\quad Z_l(z) = \sin(lz).
$$
Now $X$ satisfies:
$$
X'' = (k^2 + l^2)X,
$$
hence let $m = \sqrt{k^2+l^2}$:
$$
X = a_{kl} e^{mx} + b_{kl} e^{-mx},
$$
where the coefficient $a_{kl}, b_{kl}$ depend on $k$ and $l$.

When $x = 0$:
$$
F\big|_{x=0} =F_0  = \sum_{l=1}^{\infty}\sum_{k=1}^{\infty} (a_{kl}+b_{kl}) \sin(ky)\sin(lz)\tag{1}
$$
When $x=\pi$:
$$
F\big|_{x=\pi} =-F_0  = \sum_{l=1}^{\infty}\sum_{k=1}^{\infty} (a_{kl} e^{m\pi} + b_{kl} e^{-m\pi}) \sin(ky)\sin(lz) \tag{2}
$$
For $\{ \sin(ky)\sin(lz)\}_{k,l=1}^{\infty}$ are orthogonal pairwise, multiply (1) by $\sin(ky)\sin(lz)$ and integrate on $(0,\pi)$ for $y$ and $z$:
$$
F_0 \int^{\pi}_0\int^{\pi}_0 \sin(ky)\sin(lz)\,dydz =\int^{\pi}_0\int^{\pi}_0  (a_{kl}+b_{kl}) \sin^2(ky)\sin^2(lz)\,dydz, \tag{$1'$}
$$
The right side only have one term for $k' \neq k$ or $l' \neq l$
$$
\int^{\pi}_0\int^{\pi}_0 \sin(ky)\sin(lz) \sin(k'y)\sin(l'z) \,dydz= 0,
$$
this is the orthogonality I mentioned above. Now $(1')$ gives:
$$
F_0\frac{1-(-1)^k}{k}\frac{1-(-1)^l}{l} = (a_{kl}+b_{kl})\frac{\pi^2}{4}.
$$
Doing the same thing for (2) yields:
$$
-F_0\frac{1-(-1)^k}{k}\frac{1-(-1)^l}{l} =  (a_{kl} e^{m\pi} + b_{kl} e^{-m\pi})\frac{\pi^2}{4}.
$$
When $k$ and $l$ are both odd, we have:
$$
a_{kl} = - \frac{16F_0 (e^{-m\pi}+1)}{\pi^2 kl(e^{m\pi} - e^{-m\pi})},\text{ and }
b_{kl} = \frac{16F_0 (e^{m\pi}+1)}{\pi^2 kl(e^{m\pi} - e^{-m\pi})},
$$
When one of $k$ and $l$ is even, we have:
$$
a_{kl} = b_{kl} = 0.
$$

Now the final solution can be obtained just by scaling back by the factor of $\pi/a$, also setting $k = 2p-1$, $l = 2q-1$, for $p,q \in \mathbb{N}$:
$$
F = \frac{16F_0}{\pi^2}\sum_{p=1}^\infty\sum_{q=1}^{\infty} X_{pq}(x)\sin\frac{(2p-1)\pi y}{a} \sin\frac{(2q-1)\pi z}{a}
$$
where
$$X_{pq}(x) = - \frac{\cosh\frac{m\pi}{2} e^{m(x-1/2)\pi/a}}{(2p-1)(2q-1)\sinh(m\pi)}  +\frac{\cosh\frac{m\pi}{2}e^{-m(x-1/2)\pi/a}}{(2p-1)(2q-1)\sinh(m\pi)} $$
and
$$
 m =\sqrt{(2p-1)^2+(2q-1)^2}.
$$


问:
Understanding pseudovectors and pseudoscalars
In doing an exercise about pseudovectors

How do the components of a cross product transform under inversion?

I got stuck. I didn't know how to compute the inversion of a cross product.

My naive approach was as follows. Say we we invert the cross product $\mathbf a = \mathbf b \times \mathbf c$ about the origin. We then have $\mathrm{inv} (\mathbf a) = \mathrm{inv} (a_x,a_y,a_z)=(-a_x,-a_y,-a_z)$, and we're done.

This means $\mathrm{inv}(\mathbf a) = (-)\mathbf a = - \mathbf b \times \mathbf c$.

At this point I have no means of checking whether this is correct or not.

Hence, I now choose $\mathbf b = \hat{\mathbf{x}}$ and $\mathbf c =\hat{\mathbf{y}}$, so that $\mathbf a = \hat{\mathbf z}$.

Now, $\mathrm{inv}(\hat{\mathbf z}) = -\hat{\mathbf z} = - \hat{\mathbf x} \times \hat{\mathbf y} $.

This I can check, using the right-hand rule.

$ - \hat{\mathbf x} \times \hat{\mathbf y}  = \hat{\mathbf y} \times \hat{\mathbf x} = - \hat{\mathbf z} .$

Hence, this seems to check out, in keeping with my naive calculations. Why is this wrong?

Reading this question and it's answer, I seem to have to calculate the inversion of $\mathbf a$ about the origin by inverting its components seperately: $\mathrm{inv}(\mathbf a) = \mathrm{inv}(\mathbf b) \times \mathrm{inv}(\mathbf c)$. How could I have derived that you should invert the components $\mathbf b, \mathbf c$ seperately?

In the answer, the argument goes that we simply define the action of the linear operator $\mathrm{inv}$ on a bivector (?) that way. But how does this --- $\mathrm{inv}(\mathbf a) = \mathrm{inv}(\mathbf b) \times \mathrm{inv}(\mathbf c)$--- follow naturally?

Also, bivectors are associative. But cross products aren't. How is this relevant to the discussion?

If I would accept this rule, then my answer to my question is readily given: we simply have $$\mathrm{inv}(\mathbf b) \times \mathrm{inv}(\mathbf c) = (-\mathbf b) \times (-\mathbf c) = \mathbf a$$ hence the cross product is invariant under inversions.

My post relates directly to this question.

This answer, to that question question attempts to answer my question by saying the laws of physics are converved under coordinate inversion, which I find unalluminating. I read this same sentence

One of the most powerful ideas in physics is that physical laws do not change when one changes the coordinate system used to describe
these laws.

in the Wikipedia article on pseudoscalars.

By using the same rule --- inverting the components seperately --- on the scalar triple product, the answer follows readily: $$\mathrm{inv}(\mathbf d \cdot \mathbf e \times \mathbf f) = \mathrm{inv}(\mathbf d) \cdot \mathrm{inv}(\mathbf e) \times \mathrm{inv}(\mathbf f)  = - \mathbf d \cdot \mathbf e \times \mathbf f \ .$$

My same question in this case: how does this rule of inverting the seperate components follow naturally?

I want to deeply understand what relates pseudovectors and pseudoscalars to bivectors and trivectors respectively, in a clear way. This answer, atttempts to relate them to the exterior and Clifford algebra, which I don't follow. It says that inversion changes the handedness of a trivector:

Trivectors, too, are different from ordinary scalars.  All trivectors can be written as a scalar multiple of $\epsilon \equiv \hat x \wedge \hat y \wedge \hat z$, and we can interpret this as an oriented volume.  If some trivector $\tau = \alpha \epsilon$ for some positive $\alpha$, then $\tau$ is right-handed.  If $\alpha < 0$, then $\tau$ is left-handed.

What do we mean by changing the handedness?

I decided to ask this question on math.stackexchange instead of physics.stackexchange because the question relates to exterior- and Clifford algebra. I also haven't used any applications in physics in the post.

答:
You should imagine that when you take $\mathbf x, \mathbf y \in \mathbb R^3$ and compute $\mathbf x \times \mathbf y$, the result lives in a "different $\mathbb R^3$". We call elements of the first $\mathbb R^3$ "vectors", and elements of the second $\mathbb R^3$ "pseudovectors". It is legal to add two vectors together; it is legal to add two pseudovectors together; it is illegal to add a vector and a pseudovector.

So the question is:

If we reflect the first $\mathbb R^3$ through the origin, what happens to the second $\mathbb R^3$?"

We have a transformation $\text{inv}$ that takes a vector $(x,y,z)$ to a vector $\text{inv}(x,y,z) = (-x,-y,-z)$. We also have a corresponding transformation, which you're also calling $\text{inv}$, but which I'll call for clarity $\text{inv}^{\times}$. It operates on the second $\mathbb R^3$, where pseudovectors live. We don't know what it does yet, but we'd like it to do something that plays nicely with cross product: we'd like to have
$$
    \text{inv}^\times(\mathbf x \times \mathbf y) = \text{inv}(\mathbf x) \times \text{inv}(\mathbf y).
$$
Now, if we apply the definition of cross product, we get that
$$
    \text{inv}(\mathbf x) \times \text{inv}(\mathbf y) = \det\begin{bmatrix} \mathbf i & \mathbf j & \mathbf k \\ -x_1 & -x_2 & -x_3 \\ -y_1 & -y_2 & -y_3\end{bmatrix} = (-1) \det\begin{bmatrix} \mathbf i & \mathbf j & \mathbf k \\ x_1 & x_2 & x_3 \\ -y_1 & -y_2 & -y_3\end{bmatrix} = \\ = (-1)^2 \det\begin{bmatrix} \mathbf i & \mathbf j & \mathbf k \\ x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3\end{bmatrix} = \mathbf x \times \mathbf y.
$$
We conclude that $\text{inv}^\times$ must be the identity: $\text{inv}^\times(\mathbf z) = \mathbf z$ for all $\mathbf z$ in the "second $\mathbb R^3$".


问:
Solving a quintic function for zero
I got this  question on my homework and I cannot for the life of me figure out how to solve for $0$.

$$x^5+2x-10=0$$

I have tried this every which way and this is my last resort. Thanks in advanced.

答:
Just for completeness. The real solution of this equation can be expressed in terms of a generalized hypergeometric function

$$
5 \, {}_4F_3\!\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{2},\frac{3}{4},\frac{5}{4};-\frac{1953125}{512}\right).$$

Title: Solving a quintic function for zero


问:
Show that no number of the form $2^a3^b$ is $3$-perfect.
Show that no number of the form $2^a3^b$ is $3$-perfect.

A number is $3$-perfect if $\sigma(n)=3n$. I assumed that there exists a number $n=2^a3^b$ such that $\sigma(n)=\sigma(2^a3^b)=\sigma(2^a)\sigma(3^b)=3\cdot2^a3^b=3n$. Then I had $(1+2+\cdots +2^a)(1+3+\cdots+3^b)=2^a3^{b+1}$ and so $(2^{a+1}-1)({1\over 2}(3^{b+1}-1))=2^a3^{b+1}$. and through some algebra on the LHS I convinced myself that these two things are not equal to each other, contradiction.

I'm confident there is a better and more elementary way to do this problem. Any hints or solutions are greatly appreciated.

Body:
Multiplying both sides of $(2^{a+1}-1)(\frac{3^{b+1}-1}{2})=2^a3^{b+1}$ by $2$, we get $(2^{a+1} - 1)(3^{b+1} - 1) = 2^{a+1} 3^{b+1}$.

But this is impossible, because $2^{a+1}-1 < 2^{a+1}$ and $3^{b+1}-1 < 3^{b+1}$.


问:
How can we prove the following product-sum inequality involving e?
Take any sequence $x_1, \dots, x_k$ of positive reals such that $\sum_{i=1}^{k-1} x_i < 1$ and $\sum_{i=1}^k x_i \ge 1$. I suspect that the following inequality holds:
$$
\prod_{i=1}^k (1 + x_i) \le e \sum_{i=1}^k x_k.
$$

This didn't seem very hard at first, but I can't quite get the desired inequality.  Here's my best attempt so far. By AM-GM inequality, we have
$$
\prod_{i=1}^k (1 + x_i) = (1 + x_k) \prod_{i=1}^{k-1} (1 + x_i) \le (1 + x_k) \left( 1 + \frac{1}{k-1} \sum_{i=1}^{k-1} x_i \right)^{k-1}.
$$
By a standard exponential function inequality, we then have
$$
(1 + x_k) \left( 1 + \frac{1}{k-1} \sum_{i=1}^{k-1} x_i \right)^{k-1} < (1 + x_k) \exp \left( \sum_{i=1}^{k-1} x_i \right).
$$
Finally, by the first constraint on the sequence, we get
$$
(1 + x_k) \exp \left( \sum_{i=1}^{k-1} x_i \right) \le e (1 + x_k).
$$

This is very close to what I want, but not quite there. I'm assuming there's a trick somewhere that I'm missing. Any help would be appreciated.

答:
Starting with the second step above, we want to show that
$$
(1 + x_k) \exp \left( \sum_{i=1}^{k-1} x_i \right) \le e \sum_{i=1}^k x_i,
$$
which is equivalent to
$$
(1 + a) \exp(b) \le e (a + b),
$$
for $0 \le b < 1 \le a + b$.

I tried using polynomial upper bounds on $\exp(b)$ for $0 \le b < 1$, but none of the bounds I tried were tight enough. I suspect that no finite polynomial upper bound on $\exp(b)$ is sufficient to get the desired inequality.

Ultimately, I overlooked a simple solution. First, rewriting it as $f(a, b) = e(a + b) - (1 + a) \exp(b)$, we want to show that $f(a, b) \ge 0$ for all $0 \le b < 1 \le a + b$. This function is smooth and has a saddle point at $b = 1$, $a = 0$ for which $f(a, b) = 0$. Then it's enough to show that $f$ is weakly increasing in $a$ and in $a - b$ for all $a, b$ in the domain. We can check that this is true by differentiating.


问:
how to prove that $\lim_{j \rightarrow \infty} \ln\left(\frac{j^2+1}{j+1}\right) = \infty$?
How can i see that beacuse the logarithm is strictly increasing and not bounded and the input $\frac{j^2+1}{j+1}$ converges to $\infty$, $\lim_{j \rightarrow \infty} \ln \frac{j^2+1}{j+1} = \infty$ ??

答:
For a really simple proof, @saulspatz gave you a nice little observation. We have
$$
j^2 + 1 > j^2 - 1
$$
and so
$$
\frac{j^2 + 1}{j+1} > \frac{j^2-1}{j+1} = \frac{(j+1)(j-1)}{(j+1)} = j-1
$$
so
$$
\frac{j^2+1}{j+1} > j-1
$$
since ${\log}$ is increasing, we have
$$
\log\left(\frac{j^2+1}{j+1}\right) \geq \log\left(j-1\right)
$$
can you finish it from here?

If you can't:

 We know $\log$ is unbounded, i.e. ${\lim_{j\to\infty}\log(j)=\infty}$. This means that for any ${R>0}$, we can find a corresponding ${j^*}$ such that ${\log(j^*)>R}$. We can do the same with ${\log(j-1)}$, just choose ${(j^* + 1)}$. This shows ${\log(j-1)\to \infty}$, and since the original expression ${\log\left(\frac{j^2 + 1}{j+1}\right)}$ is greater than this, it too must go to infinity.


答:
You could do the following:

$$\frac{j^2 +1}{j+1} \enspace = \enspace \frac{j + \tfrac{1}{j}}{1 + \tfrac{1}{j}}$$

And therefore

$$\lim_{j\rightarrow \infty} \ln \Big( \frac{j + \tfrac{1}{j}}{1 + \tfrac{1}{j}} \Big) \enspace = \enspace \lim_{j\rightarrow \infty} \ln \Big( j + \tfrac{1}{j} \Big) - \underbrace{\lim_{j\rightarrow \infty} \ln \Big( 1 + \tfrac{1}{j} \Big)}_{= 0} \\ = \enspace \lim_{j\rightarrow \infty} \ln \Big(j + \tfrac{1}{j} \Big)  \\ = \enspace \lim_{j\rightarrow \infty} \ln j (1+\tfrac{1}{j^2}) \\ = \enspace  \lim_{j\rightarrow \infty} \ln j + \underbrace{\lim_{j\rightarrow \infty} \ln(1+\tfrac{1}{j^2})}_{=0} \\ = \enspace \lim_{j\rightarrow \infty} \ln j \\ = \enspace \infty$$

You would still have to show that it is strictly increasing and not bounded though.

EDIT: Thanks to J.G. for contributing within the last steps.

Title: how to prove that $\lim_{j \rightarrow \infty} \ln\left(\frac{j^2+1}{j+1}\right) = \infty$?

Body: How can i see that beacuse the logarithm is strictly increasing and not bounded and the input $\frac{j^2+1}{j+1}$ converges to $\infty$, $\lim_{j \rightarrow \infty} \ln \frac{j^2+1}{j+1} = \infty$ ??


答:
According to the Mean value theorem: if $f(x): [a, b] \rightarrow \mathbb{R}$ is a continuous function on a closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, where $a < b$, there exists some $c \in (a, b)$ such that

$$
f(b)-f(a) = f'(c)(b-a).
$$

Applying this theorem to our example ($f(x) = \Phi(x)$, $a = a$, $b = a + \Delta$),

$$
\Phi(a + \Delta) - \Phi(a) = \Phi'(c)\left((a + \Delta) - a\right), \text{ where } c \in (a, a+\Delta).
$$

$$
\Phi(a + \Delta) - \Phi(a) = \varphi(c)\Delta, \text{ where } \varphi(c) = \frac{1}{\sqrt{2\pi}}e^{-\frac{-c^2}{2}} \text{ is a p.d.f. of the standard normal distribution} \Rightarrow.
$$

$$
\Phi(a + \Delta) - \Phi(a) = \varphi(c)\Delta < \left|\text{ since } 0 < a < c \right| < \varphi(a)\Delta.
$$

The choice of $f(a, \Delta)$ is

$$
f(a, \Delta) := \frac{\Delta}{\sqrt{2\pi}}e^{-\frac{a^2}{2}}
$$

Title: Upper bound on difference between Gaussian CDFs

Body:
May I know if there are any non-trivial upper bounds $f$ on the following:

$$\Phi(a + \Delta) - \Phi(a) \leq f(a, \Delta)$$

for $\Phi$ the CDF of a standard normal and all $a, \Delta > 0$.

Thanks!


问:
Does the series converge $\sum_{n=2}^{\infty}\frac{(-1)^n}{\cos{\frac{\pi}{2n}}}$
Does the series converge?
$$\sum_{n=2}^{\infty}\frac{(-1)^n}{\cos{\frac{\pi}{2n}}}$$

I tried to solve it with alternating series test and got $\lim_{n\rightarrow\infty}a_n=1\neq0\implies\text{??Divergence?? or how to prove that}$

Idea: To show that the limit of $(-1)^n$ doesnt exist, thus the series is divergent. Is this correct?

答: Series diverges, as general member doesn't tends to zero.</p>


答: You're correct, this series is not convergent because sequence $$ \frac{(-1)^n}{\cos\frac{\pi}{2n}}$$ is not convergent. For a series $\sum_n {a_n}$ to be convergent, it is necessary (but not yet sufficient) that $\lim_{n\to\infty} a_n$ exists and is equal to $0$.

Title: Does the series converge $\sum_{n=2}^{\infty}\frac{(-1)^n}{\cos{\frac{\pi}{2n}}}$

Body:
Does the series converge?
$$\sum_{n=2}^{\infty}\frac{(-1)^n}{\cos{\frac{\pi}{2n}}}$$

I tried to solve it with alternating series test and got $\lim_{n\rightarrow\infty}a_n=1\neq0\implies\text{??Divergence?? or how to prove that}$

Idea: To show that the limit of $(-1)^n$ doesnt exist, thus the series is divergent. Is this correct?


答:
Think about the behavior of $\cos\frac{\pi}{2n}$ for many $n$'s. Then evaluate the partial sums as $n$ increases.

It's a bit of a trick question.


问:
Solving an inequality with a constraint
I have the following inequality:

$|a\frac{e^{-ax}}{(e^{-ax}+1)^2}| \geq 1$

The solution for this can only exist if $|a| \geq 4$. I want to find the range of values for $x$ where the above inequality is true. I think if I can isolate $x$ in this equation, I would have solved it. Unfortunately, I can't find a way to do that. I begin by moving the denominator towards the right.

$a e^{-ax} \geq (e^{-ax}+1)^2$

This leads me to

$\ln a -ax \geq \ln (e^{-ax}+1)^2$

At this stage I am stuck. I can't figure out how to isolate $x$ from the RHS. I have tried other approaches as well - such as expanding the square term before bringing in the log or partially moving out one term from the denomination towards the right - they all leads to an expression where I can't figure out how to isolate $x$

答:
WLOG, we can assume $ a>0$.

Put $$ t=e^{-ax}$$

The inequation becomes
$$f(t)=\frac{t}{(1+t)^2}\ge \frac 1a$$

but

$$f'(t)=\frac{1-t^2}{(1+t)^4}$$
So
$$\max f(t)=f(1)=\frac 14$$

and necessarily,
$$\frac 1a\le \frac 14$$
or
$$a\ge 4$$

You need to solve the inequation above for $ t $ and then for $ x$.

We have
$$t^2+(2-a)t+1\le 0$$

then

$$t\in[\frac{a-2-\sqrt{a(a-4)}}{2},\frac{a-2+\sqrt{a(a-4)}}{2}]\cap (0,+\infty)$$

Title: Solving an inequality with a constraint


问:
Disprove that for all integers $a,b$ that if $d=gcd(a,b)$ then $d=gcd(a^2,b)$ or $d=gcd(a,b^2)$
Question:
Disprove that for all integers $a,b$ that if $d=gcd(a,b)$ then $d=gcd(a^2,b)$ or $d=gcd(a,b^2)$

Attempts.

I thought to first try $a=b$, then $d=a$ and $gcd(a^2,b)=b=d$

Ok, how about $a=0, b \neq 0$, then $d=b$ and $gcd(a^2,b)=b=d$

Ok, if $a,b$ are co-prime, then the $d=1$ and $gcd(a^2,b)=1$

Ok, maybe $a=b^2$, then $d=b$ and $gcd(a^2,b)=b$

Not sure what else to try. Maybe the problem is wrong? I copied it down carefully. Hints appreciated.

Body:
$a=12$, $b=18$.

$gcd(a,b)=6$, but $gdc(a,b^2)=gdc(12,18^2)=12$ and $gdc(a^2,b)=gdc(12^2,18)=18$


答:
$\ln$ being unbounded means that for all $M\in\mathbb{R}$ we can find $x_M\in\mathbb{R}$ such that $\ln(x_M)>M$. Being increasing implies on the previous that for every $x\geq x_M$, $\ln(x)\geq\ln(x_M)>M$.

Plug your sequence $(x_n)_{n\in\mathbb{N}}=\left(\dfrac{n^2+1}{n+1}\right)_{n\in\mathbb{N}}$ and pick $M\in\mathbb{R}$. You know that $x_n\to\infty$, so there exists $n(M)\in\mathbb{N}$ such that $x_n>x_M$ for all $n\geq n(M)$. This means that $\ln(x_n)\geq\ln(x_M)>M$, which is exactly the definition of $\lim_{n\to\infty}\ln(x_n)$ being infinity.


问:
finding the derivative $\frac{dy}{dx}$ for $y=(e^{x^{2+1}})^{sin(x)}$
I'm calculating for the derivative on this exercise, and would kindly appreciate feedback on my solution and especially on whether my solution can be simplified further if correct.

$$y =(e^{x^{2+1}})^{sin(x)}$$

My calculation:

Take the natural log of both sides:

$$\ln(y)=\sin(x)\ln(e^{x^{2+1}})$$

Then taking the derivative of both sides and applying the product rule $fg'+f'g$ in respect to $x$.

$$\frac{1}{y}\cdot\frac{dy}{dx}=\sin(x)\frac{2x}{e^{x^{2+1}}}+\cos(x)\ln(e^{x^{2+1}})$$

transporting $y$ onto the right side to find $\frac{dy}{dx}$ and substituting in the value $(e^{x^{2+1}})^{sin(x)}$ for $y$

$$(e^{x^{2+1}})^{sin(x)}[\sin(x)\frac{2x}{e^{x^{2+1}}}+\cos(x)\ln(e^{x^{2+1}})] = \frac{2x\sin(x)(e^{x^{2+1}})^{sin(x)}}{e^{x^{2+1}}}+\cos(x)(e^{x^{2+1}})^{sin(x)}\ln(e^{x^{2+1}})=\frac{(e^{x^{2+1}})^{sin(x)}[2x\sin(x)]+(e^{x^{2+1}})^{sin(x)}[\cos(x)(e^{x^{2+1}})\ln(e^{x^{2+1}})]}{e^{x^{2+1}}}$$

答:
In order to ease derivation it is convenient to simplify $y=y(x)$ at first.

We obtain
\begin{align*}
\color{blue}{\frac{dy}{dx}}&=\frac{d}{dx}\left(e^{x^3}\right)^{\sin(x)}\\
&=\frac{d}{dx}e^{x^3\sin(x)}\tag{1}\\
&=\left(\frac{d}{dx}\left(\sin(x)x^3\right)\right)e^{x^3\sin(x)}\tag{2}\\
&=\left(3x^2\sin(x)+x^3\cos(x)\right)e^{x^3\sin(x)}\\
&\,\,\color{blue}{=x^2\left(3\sin(x)+x\cos(x)\right)e^{x^3\sin(x)}}
\end{align*}

Comment:

In (1) we apply the rule $\left(p^{q^r}\right)^s=\left(p^{\left(q^r\right)}\right)^s=p^{\left(sq^r\right)}$.

In (2) we apply $\left(e^{f(x)}\right)^{\prime}=f^{\prime}(x)e^{f(x)}$.