问:
Which of these subsequences converge, and why?
Let $a_n$ = $\frac{n}{n+1}$$(-1)^{\frac{n(n+1)}{2}}$

I've been asked which of the following converge, ($a_{2k}$), ($a_{2k+1}$), ($a_{4k+1}$), or ($a_{6k}$)

Now, for $a_{2k}$ gives $\frac{2k(2k+1)}{2}$ which is even, thus $a_{2k}$ = $\frac{2k}{2k+1}$

In fact, for all of these examples of subsequences $\frac{n(n+1)}{2}$ is even, thus

$a_{2k+1}$ = $\frac{2k+1}{2k+2}$

$a_{4k+1}$ = $\frac{4k+1}{2k+2}$

and $a_{6k}$ = $\frac{6k}{6k+1}$

Now, how do I see which subsequence is convergent? Because it seems to me that they all converge to 1

答:
Your claim isn't all that true, in fact for example if $k=5$, then
$$\frac{k(k+1)}{2}= \frac{5 \cdot 6}{2}=15.$$

For $a_{2k}$, we know it is
$$a_{2k}=\frac{2k}{2k+1}(-1)^{k(2k+1)},$$
so if we pick the subsequence with even $k$, we get
$$a_{2k} = \frac{2k}{2k+1} \to 1;$$
instead, if we pick the subsequence with odd $k$, we get
$$a_{2k} \to -1,$$
in fact we have that $k(2k+1)$ is product of two odds (since $2k$ is even $\Rightarrow 2k+1$ is odd).

This shows that $a_{2k}$ does not converge.

For $a_{2k+1}$ we have that
$$a_{2k+1}=\frac{2k+1}{2k+2}(-1)^{(2k+1)(k+1)},$$
so, proceeding just as before, we observe that the subsequence of odd $k$ gives us
$$a_{2k+1} = \frac{2k+1}{2k+2} \to 1;$$
instead the subsequence of even $k$ gives us
$$a_{2k+1} \to -1$$
since  both $(2k+1)$ and $(k+1)$ are odd.

This proves that also $a_{2k+1}$ does not converge.

For $a_{4k+1}$, we have that
$$a_{4k+1}=\frac{4k+1}{4k+2}(-1)^{(4k+1)(2k+1)},$$
where both $(4k+1)$ and $(2k+1)$ are odd, so we always have
$$a_{4k+1}=\frac{4k+1}{4k+2}(-1) \to -1.$$
Proving that $a_{4k+1}$ converges.

Lastly, for $a_{6k}$, we have
$$a_{6k}= \frac{6k}{6k+1}(-1)^{3k(6k+1)},$$
where $(6k+1)$ is always odd, meanwhile $3k$ is even when $k$ is even, so once again we have that there is the subsequence of even $k$ that converges to $1$ and the subsequence of odd $k$ that converges to $-1$.
This shows that also this last subsequence doesn't converge.


问:
If two topologies have the same continuous maps into and out of them, must they be the same?
Let $X$ be a set and let $T_1,T_2$ be topologies on $X$. Assume that for any topological space $Y$, the following conditions hold:

(1) $f:Y\rightarrow X$ is continuous wrt $T_1$ iff $f:Y\rightarrow X$ is continuous wrt $T_2$

(2) $f:X\rightarrow Y$ is continuous wrt $T_2$ iff $f:X\rightarrow Y$ is continuous wrt $T_1$

Must $T_1,T_2$ be the same topologies?

答: There is one, special topological space $Y$ so that for any set $X$ and for any two topologies $T_1, T_2$ on $X$, we have $T_1=T_2$ if and only if the continuous functions from $X$ to $Y$ are the same for $T_1$ and $T_2$.  Can you find such a topological space $Y$?

ContentLicense: CC BY-SA 4.0


问:
If $y_{n}=\frac{x_{1}+\ldots+x_{n}}{n}$ converges, the sequence $x_n$ must converge?
Given  $\left(x_{n}\right)$ a sequence, we define  $y_{n}=\frac{x_{1}+\ldots+x_{n}}{n}$ for all $n \in \mathbb{N} .$ If we know $\left(y_{n}\right)$ converges. the sequence  $\left(x_{n}\right)$ must converge ?

I think yes because $y_n=\frac{1}{n}\
\sum_{i=1}^{n}x_{i}
$

then by Necessary condition of series convergence $x_n \rightarrow 0$ is right?

答: No, take $x_n = (-1)^n$.  Then $x_1 + \dots + x_n$ is always equal to either $0$ or $-1$, so $y_n \to 0$, but $(x_n)$ doesn't converge.


问:
Cardinality of intersection of sets
Consider the following problem: find $n(A \cap B)$ if $n(A)=10$, $n(B)=13$ and $n(A \cup B) = 15$.

I know if I want to find the union I use the Cardinal Number formula:

$$n(A\cup B) = n(A) + n(B) - n(A\cap B)$$

But how do I do it the other way: to find $n(A\cap B)$?

Would it be $n(A\cap B) = n(A) + n(B) - n(A\cup B)$ ?

答:
We know that: $n(A)=10$, $n(B)=13$ and $n(A \cup B) = 15$.

Using the Cardinal Number formula:

$$\underbrace{n(A\cup B)}_{15} = \underbrace{n(A)}_{10} + \underbrace{n(B)}_{13} - n(A\cap B)\tag{1}$$

So indeed, we can write the equivalent to $(1)$: $$n(A\cap B) = \underbrace{n(A)}_{10} + \underbrace{n(B)}_{13} - \underbrace{n(A\cup B)}_{15}\tag{2}$$

$$\text{So, given what we know}:\;\; n(A\cap B) = 10 + 13 - 15 = 8$$

ContentLicense: CC BY-SA 3.0

Title: Cardinality of intersection of sets


问:
Probability of cards using law of total probability
Given two decks shuffled decks of 54 cards, split each in half (27 cards), then take one half of each deck and form a new deck of 54 cards. Then shuffle it and pick first two cards. Whats the probability of a match?

I already know how to solve this problem using conditional probability, which is indeed simpler.

However, I want to develop some intuition on how to solve this problem using law of total probability. I added below the solution to this problem using conditional probability, but I tried to also solve using law of total probability but was unable to came up with a full solution.

The real question here is: how to solve this using law of total probability?

My attempts so far use the following logic: the probability of a match depend on how many pairs of cards are formed. So I tried to calculate each probability of exact number of pairs for each k=1, 2, ..., n/2 pairs, and then for each of those probabilities I multiply by the probability of picking a matching pair having k pairs in the deck. However the final probability does not seem to agree with my simulation.

The solution of this problem is easier if we use conditional probability as follows:

Start by observing that a match is only possible if the both come from different decks, as one deck has no repeated cards.

Define de event ${A: \text{both pick cards came from different decks}}$

Then find the probability of both cards coming from different decks. Notice that there is probability of 1/2 of taking the card from a specific half deck, since a card can come from each of the halves with equal probability.

Below $V_1$ means the first valued picked, and $V_2$ means second value picked.

$P(A) = P(V_1 \text{from deck 1})P(V_2 \text{from deck 2}) + P(V_2 \text{from deck 1})P(V_1 \text{from deck 2})  = 2 \frac{1}{4} = \frac{1}{2}$

Now $P(V_1 = V_2| A) = 1/54$, then,

$P(V_2 = V_2) = P(V_1 = V_2|A)P(A) = 1/108$

How can we go about solving this without conditional probability?

Thanks

Edit: turns out this calculation using conditional probability is wrong, as the probability of cards coming from different decks is not 1/2. The probability of taking a card from, say, deck 1, changes when you pick a card from deck 2.

Correct probability is 1/106.

答:
Here's how you can solve this using the total law of probability.

Let $E$ be the event of a match, and let $X$ denote the number of matches in your newly formed deck of $54$ cards.

For $n\in \{0,1,2,\dots ,27\}$ we have $$P(X=n)=\frac{{54 \choose n}{54 - n \choose 27 - n} {27 \choose 27-n} }{{54 \choose 27}^2}$$ $$P(E|X=n)=\frac{n}{{54 \choose 2}}$$ From the total law of probability, $$P(E)=\sum_{n=0}^{27}P(E|X=n)P(X=n)=\frac{1}{106}$$


答: Suppose $\tau_1$ and $\tau_2$ are topologies on a set $X$. Let $X_1=Y_1=(X,\tau_1)$ and $X_2=Y_2=(X,\tau_2)$, and let $f$ be the identity function on $X$. Then $f:X_1\to Y_1$ and $f:X_2\to Y_2$ are continuous. If $f:X_2\to Y_1$ and $f:X_1\to Y_2$ are also continuous, then $X_1$ and $X_2$ are homeomorphic.

Title: If two topologies have the same continuous maps into and out of them, must they be the same?

Body:
Let $X$ be a set and let $T_1,T_2$ be topologies on $X$. Assume that for any topological space $Y$, the following conditions hold:

(1) $f:Y\rightarrow X$ is continuous wrt $T_1$ iff $f:Y\rightarrow X$ is continuous wrt $T_2$

(2) $f:X\rightarrow Y$ is continuous wrt $T_2$ iff $f:X\rightarrow Y$ is continuous wrt $T_1$

Must $T_1,T_2$ be the same topologies?


问:
Proving that quotient space is pre-Hilbert and not separable
I was having some trouble with the following problem from Luenberger's Optimization by Vector Space Methods (p. 72).

Update (2/7/2021): I attempted (a) based on a hint from GReyes and gave another shot at (b), updating the solutions below. Please comment if you see any mistakes!

Consider the set $X$ of all real functions $x$ defined on $(-\infty, \infty)$ for which $$\lim_{T\to\infty} \frac{1}{2T} \int_{-T}^T |x(t)|^2 dt < \infty$$
Let $M$ be the subspace on which the indicated limit is zero.

a) Show that the space $H = X/M$ becomes a pre-Hilbert space when the inner product is defined as $$([x]|[y]|) = \lim_{T\to\infty} \frac{1}{2T} \int_{-T}^T x(t)y(t) dt$$
b) Show that $H$ is not separable.

a) I first show that $\langle[x],[y]\rangle$ has the same value regardless of which function is chosen from each coset. Let $x$ and $\tilde{x}$ be arbitrary functions in $[x]$ and $y$ and $\tilde{y}$ arbitrary functions in $[y]$ and consider the difference between each pair's inner products $\langle x, y \rangle - \langle \tilde{x}, \tilde{y}\rangle$. Using the facts that $xy - \tilde{x}\tilde{y} \leq |x||y - \tilde{y}|  + |x-\tilde{x}||\tilde{y}|$ and that the integral and limit operations are linear, we obtain the inequality

\begin{gather}
\langle x, y \rangle - \langle \tilde{x}, \tilde{y}\rangle \leq \langle |x|, |y-\tilde{y}|\rangle + \langle |x-\tilde{x}|, |\tilde{y}|\rangle
\text{ and by Cauchy Schwarz, }\\
\langle |x|, |y-\tilde{y}|\rangle + \langle |x-\tilde{x}|, |\tilde{y}|\rangle \leq \lvert|x\rvert| \lvert|y-\tilde{y}\rvert| + \lvert|x - \tilde{x}\rvert| \lvert|y\rvert|
\end{gather}

Because $y - \tilde{y}$ and $x - \tilde{x}$ are in the subspace $M$, $\lvert|y-\tilde{y}\rvert| = 0$ and $\lvert |x - \tilde{x}\rvert | = 0$; and because $\lvert |x \rvert|$ and $\lvert |y \rvert|$ are finite, $\lvert|x\rvert| \lvert|y-\tilde{y}\rvert| + \lvert|x - \tilde{x}\rvert| \lvert|y\rvert| = 0$ and so $\langle x, y \rangle - \langle \tilde{x}, \tilde{y} \rangle = 0$. The elements in the coset were chosen arbitrarily, and so the inequality holds in both directions, meaning that $\langle x, y \rangle - \langle \tilde{x}, \tilde{y} \rangle = 0$ and so $\langle[x],[y]\rangle = \langle[\tilde{x}],[\tilde{y}]\rangle$.

To check whether this is a pre-Hilbert space, I check whether the inner-product satisfies necessary properties when applied to arbitrary functions $x\in [x]$ and $y\in [y]$ from the associated cosets. By the commutativity of multiplication it is clear that $\langle[x],[y]\rangle = \langle[y],[x]\rangle$. From the distributive property of multiplication and linearity of the integral operator, $\langle[x]+[y],[z]\rangle = \langle[x],[z]\rangle + \langle[y],[z]\rangle$ and $\langle\lambda[x],[y]\rangle = \lambda\langle[x],[y]\rangle$. Finally, because $|x(t)|^2 \geq 0$ and $T\to \infty$, the inner product $\langle[x],[x]\rangle \geq 0$. If $\langle[x],[x]\rangle = 0$ then $x\in M$ by construction. If $x\in M$ then its coset includes all $\tilde{x}$ satisfying $\tilde{x} - x \in M$, and so $\tilde{x} \in M$ by the closure of $M$. Satisfying the four necessary conditions, the inner product defines a pre-Hilbert space on $X/M$.

Update (2/7/2021): Here's my attempt at showing that the space is not separable. I assume that $H$ is uncountable. Is this ok? I started writing a proof by trying to show that the inner product is a surjective mapping from any two cosets $[x], [y] \in H$ to $\mathbb{R}$, but I keep having trouble coming up with examples.

b) Proof: If $H$ is separable, it contains a countable dense subset $Z$ and because $H$ is uncountable, $Z$ must be a proper subset. Choose an arbitrary $[g] \in H\setminus Z$ and use $N([g], \delta)$ to denote the neighborhood of radius $\delta$ around $[g]$. Because $Z$ is dense in $H$, for each $\delta > 0$ the intersection $Z \cap N([g], \delta) \neq \emptyset$. But, fixing $[g]$, we can choose $[z] \in N([g], \epsilon/3)$, $\tilde{z}\in [z]$, and $\tilde{g}\in [g]$ so that
\begin{align}
\lim_{T\to\infty} \frac{1}{2T} \int_{-T}^T |\tilde{z} -  \tilde{g}|^2 dt &\leq \lim_{T\to\infty} \frac{1}{2T} \int_{-T}^T |\tilde{z} - z|^2dt + \lim_{T\to\infty} \frac{1}{2T} \int_{-T}^T |g - \tilde{g}|^2 dt + \epsilon/3
\end{align}
Because $z-\tilde{z}$ and $g-\tilde{g} \in M$, there exists some $T'$ such that for all $T \geq T'$
\begin{align}
\frac{1}{2T} \max\Big(\int_{-T}^T |\tilde{z} - z|^2dt, \int_{-T}^T |g - \tilde{g}|^2 dt\Big) < \epsilon/3
\end{align}
and so
\begin{equation}
\lim_{T\to\infty} \frac{1}{2T} \int_{-T}^T |\tilde{z} -  \tilde{g}|^2 dt < \epsilon.
\end{equation}
But then, $\tilde{z} - \tilde{g} \in M$ and so $\tilde{z}$ and $\tilde{g}$ are in the same coset, contradicting the assumption that $[g] \in H\setminus Z$.

答:
To finish your computation you can use Cauchy-Schwartz inequality for both integrals:
$$
(\frac 1{2T}\int_{-T}^T|x(t)||\tilde y(t)-y(t)|dt)^2\le \frac 1{2T}\int_{-T}^T|x(t)|^2 dt\cdot\frac 1{2T}\int_{-T}^T|\tilde y(t)-y(t)|^2 dt
$$
And a similar computation for the second one. The right hand side goes to zero because one of the limits is zero and the other one is finite by the definitions of the quotient and the space itself.

Checking the Hilbertian structure is straightforward after this.


问:
Closed-form solution to finite series involving cos and sin
I have the following problem: Find the closed-form solution of:
$$\sum_{k=-m}^m(-1)^k(\cos(k\omega)-\sin(k\omega))$$
Since $\sin(x)=-sin(-x)$ and $\sin(0)=0$, the sine term will cancel. Furthermore, cosine is even, hence, the summation becomes $1+2\sum_{k=1}^m(-1)^k\cos(k\omega)$

My attempt to compute that sum is as follows:
$$\sum_{k=1}^m(-1)^k\cos(k\omega)=-\cos(\omega)-(-1)^{m+1}\cos((m+1)\omega)+\sum_{k=2}^{m+1}(-1)^k\cos(k\omega)$$
Then
$$\sum_{k=1}^m(-1)^k\cos(k\omega)\Big(1-A\Big)=-\cos(\omega)-(-1)^{m+1}\cos((m+1)\omega)$$
My problem is that I don't know what $A$ should be.

I use this approach, because it usually does the trick for this kind of series. However, maybe here a completely different approach is required. I'd appreciate if someone can help me figure out what $A$ should be, or a different approach to find the closed-form solution.

Thanks

答:
Hint

As $k$ is an integer, $(-1)^k=\cos k\pi$

Now use https://mathworld.wolfram.com/WernerFormulas.html

and finally follow How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?


问:
Line integral calculation for a line segment

Compute the line integral $\int_y e^z dz$ where $y$ is the line segment from $0$ to $z_0$.

Since there is only one variable z here, I can directly compute $\int_y e^z dz=\int_0^{z_0} e^z dz=e^{z_0}-1$. I assume $z_0$ means a fixed point. Am I missing something or is this it?

答:
It is not that "there is only one variable $z$ there" you can directly do so, but that when $f$ has a primitive $F$, i.e., $F'(z)=f(z)$ in the domain (that contains the path $\gamma$),
$$
\int_\gamma f(z)=F(B)-F(A)
$$
where $A$ is the initial of $\gamma$ and $B$ the final point of $\gamma$.

In general, you can't calculate $\int_\gamma f(z)dx$ in this way. Consider for instance $f(z)=\frac{1}{z}$.


答:
Remark that it does not even imply that $x_n$ is bounded.

Indeed you can take $x_i=0$ for almost all indices except for $x_{n^2}=n$ then $y_n\to \frac 12$ but $x_n$ unbounded.

Title: If $y_{n}=\frac{x_{1}+\ldots+x_{n}}{n}$ converges, the sequence $x_n$ must converge?

Body:
Given  $\left(x_{n}\right)$ a sequence, we define  $y_{n}=\frac{x_{1}+\ldots+x_{n}}{n}$ for all $n \in \mathbb{N} .$ If we know $\left(y_{n}\right)$ converges. the sequence  $\left(x_{n}\right)$ must converge ?

I think yes because $y_n=\frac{1}{n}\
\sum_{i=1}^{n}x_{i}
$

then by Necessary condition of series convergence $x_n \rightarrow 0$ is right?


问:
Is a curve homologous to zero according to Ahlfors actually homologous to zero?
The presentation of the homology version of Cauchy's theorem in Ahlfors is slick, but sweeps a lot of the topology under the rug using clever arguments. This question is an attempt to reconcile Ahlfors' analytic notion of a curve being homologous to zero (presented in his book Complex Analysis and originally due to E. Artin, I believe) with the standard definition in homology as found in Hatcher.

We work in the complex plane and fix $a\in \mathbb C$. Let $\gamma$ be a continuous map $[0,1]\rightarrow C\backslash \{ a\}$. Following Munkres in his book Topology, we define the winding number of $\gamma$ with respect to the point $a$ by considering

$$g(t)=\frac{\gamma(t)-a}{|\gamma(t)-a|}.$$

This is clearly a loop in $S^1$ and corresponds to some multiple of of the generator of the fundamental group of $S^1$. If the generator is $\tau$ and $g(t)$ corresponds to $m\tau$, $m\in\mathbb Z$, we define the winding number $n(\gamma, a)$ to be $m$. This is the definition of winding number I will use in this question, but in case $\gamma$ is piecewise differentiable, it corresponds to the analytic definition by integration given in Ahlfors.

Ahlfors calls a curve contained in an open region $\Omega$ homologous to zero if $n(\gamma,a)=0$ for all $a\in \Omega^c$.

In homology theory, as I understand it, we would call a curve homologous to zero if it represents the zero element in $H_1(\Omega, \mathbb Z)$. That is, $\gamma$ is the boundary of some singular $2$-chain.

My question: Why do all of these notions agree? Why is Ahlfors' definition of being homologous to zero (using Munkres' definition of winding number) agree with the usual homological one?

I would like to use Munkres' definition because it works for continuous curves, not just piecewise differentiable ones, and it seems to me the equivalence should hold in this generality.

Edit: This result appears as proposition 1.9.13 in Berenstein and Gay's book on complex analysis.

答: Imagine the curve as made of segments parallel to the coordinate axes. Create a grid on a large rectangle containing the curve in its boundary and interior so that  every segment of our curve is an edge of some subrectangle. Now define a singular $2$-chain by using each subrectangle with the coefficient given by the winding number of our curve around the center of the subrectangle. Check that the boundary of this $2$-chain is our original curve. </p>


问:
How long on average to brute force a probability?
Imagine someone is making a 4 digit pin for their computer. 10 possible numbers for each character in the pin (0-9), so the total number of combinations is $10^4$ = 10,000. If someone were to guess through every pin until they got the right one, on average, would it take 5,000 guesses? Logically this makes sense, however, how would a proof be constructed for this? Thanks.

答:
Suppose there are $n$ possible pins and each are equally likely to occur in the brute force method. Imagine randomly permuting all possible pins from $1, \dots, n$. Then the correct pin can be found with equal probability at any particular site. Let this random variable be denoted by $k$. Then $\mathbb P(k) = 1/n$ for all $k$. Then,
$$ \mathbb E(k) = \sum_{k=1}^n \frac{k}{n} = \frac{n(n+1)}{2n} = \frac{n+1}{2}. $$


问:
what are all functions with $x>1$ and $y>1$ $\rho$ that follows $\rho(xy)=\frac{1}{\frac{y}{\rho(x)}+\frac{x}{\rho(y)}}$ and is continuous
What are all functions with $x>1$ and $y>1$ $\rho$ that follows $$\rho(xy)=\frac{1}{\frac{y}{\rho(x)}+\frac{x}{\rho(y)}}$$ and is continuous

If this doesn't have any solutions then prove no such solution exists.

I would like to understand how to solve a problem like this

答:
Rewrite the property as
\begin{equation*}
\rho(xy) (\rho(x)x + \rho(y)y) = \rho(x)\rho(y)
\end{equation*}
If $\rho$ has that property, it follows that $\rho(1) = 0$. See this by setting $x = y = 1$ and getting $2\rho^2(1) = \rho^2(1)$.

Now look at $y = 1/x$ for any $x$. Using the above we get $\rho(x)\rho(1/x) = 0$, so $\rho = 0$ for the formulation above.


问:
surjective group homomorphism between $\mathbb{Z}^2$ and $\mathbb{Z}_{30}$

Let $H = \langle (6,2), (3,6)\rangle$, which is a subgroup of $\mathbb{Z}^2$ (denoted $H\leq \mathbb{Z}^2$). Show that $|\mathbb{Z}^2/H| = 30$ and that $\mathbb{Z}^2/ H$ is cyclic, and then find a surjective group homomorphism $\phi : \mathbb{Z}^2 \to \mathbb{Z}_{30}$ with $Ker(\phi) = H.$

I think that $\mathbb{Z}^2/ H = \{(r,s) + H : 0\leq r < 15, 0\leq s < 2\} =: T$ and $T = \langle (2,1) + H\rangle$. Indeed $(2,1) + H \in T$ and $T\leq \mathbb{Z}^2/H$ so $\langle (2,1) + H\rangle \subseteq T.$ Also, every element in $T$ is a multiple of $(2,1) + H$ so $T\subseteq \langle (2,1) + H\rangle$ (one can show this using the fact that $(6,2), (15,0)\in H$). Also, by repeated use of the division algorithm, one can show that every coset is in $T.$ To show they're distinct one can obtain a contradiction from assuming $(r_1, s_1) + H = (r_2, s_2) + H$ if $(r_1, s_1)\neq (r_2, s_2) $ . One can show $s_1 = s_2$ and if $r_1 \neq r_2$ then they must differ by a multiple of $15.$ Also, $\phi : \mathbb{Z}^2/ H \to \mathbb{Z}_{30}, \phi((r,s) + H) = s \cdot 15 + r$ is a group isomorphism, which shows $\mathbb{Z}^2/ H\cong Z_{30}$ and hence $|\mathbb{Z}^2/ H| = 30$.

However, I’m not sure how to find a surjective group homomorphism $f: \mathbb{Z}^2 \to \mathbb{Z}_{30}$ with $Ker(\phi) = H.$ I tried determining the values of  $f(1,0)$ and $f(0,1)$ but apparently I seem to get a contradiction.

答:
Consider instead of the given generators
(colomns in $A$ below)
the ones in the columns of the following matrix $B$:
$$
\underbrace{
\begin{bmatrix}
0 & 3 \\ -10 & 6
\end{bmatrix}}_{B}
=
\underbrace{
\begin{bmatrix}
6 & 3 \\ 2 & 6
\end{bmatrix}}_{\text{Given $A$}}
\underbrace{
\begin{bmatrix}
1 & 0 \\ -2 & 1
\end{bmatrix}}_T
\ .
$$
And note that $TAT=TB=\begin{bmatrix}
0 & 3 \\ -10 & 0
\end{bmatrix}$.

This suggest to define a surjective group morphism from the column vector with components $a,b\in\Bbb Z$ as follows (so that the columns of $B$, thus also those of $A$ are mapped to "zero"):
$$
\begin{bmatrix}
a\\ b
\end{bmatrix}
\to
T
\begin{bmatrix}
a\\ b
\end{bmatrix}
=
\begin{bmatrix}
a\\ b-2a
\end{bmatrix}
=:
\begin{bmatrix}
a'\\ b'
\end{bmatrix}
\to
\begin{bmatrix}
a'\mod 3\\ b'\mod 10
\end{bmatrix}
\in
\overset{\displaystyle\Bbb Z/3}{\underset{\displaystyle\Bbb Z/10}\oplus}
\cong \Bbb Z/30\ .
$$


问:
Show that $\lim_{y \to 0} f(x+iy)$ exists for almost all $x \in \mathbb{R}$
This is part of a question in Rudin's RCA. It says that if $f$ is holomorphic in the upper half plane and satisfies
\begin{equation*}
\sup_{0<y<\infty}\int_{-\infty}^\infty |f(x+iy)|^2 dx<\infty,
\end{equation*} then
\begin{equation*}
f(z)=\frac{1}{2\pi i} \int_{-\infty}^\infty \frac{f(\xi+i\epsilon)}{\xi+i\epsilon-z}d\xi
\end{equation*} holds for $0<\epsilon<y$. I can prove this part by using paley and wiener with a semicircle in the upper half plane. However, I got stuck with the second question of showing that
\begin{equation*}
\lim_{y \to 0} f(x+iy)
\end{equation*} exists for almost all $x \in \mathbb{R}$. Any suggestions or hints?

答:
Using a linear fractional transformation $\phi$ to map the unit disk $D$ conformally onto the upper half plane. Consider $F=f \circ \phi$ which is analytic in $D$. Also it is clear that $F$ is bounded a.e. in $D$ so that $\|F\|_\infty<\infty$. Now we may apply Fatou's Theorem to conclude that $F(re^{it})$ exists a.e. as $r \to 1$. By transforming back to $f$, we can show that the limit
\begin{equation}
\lim_{y \to 0} f(x+iy)
\end{equation} exists a.e. on $\mathbb{R}$.


问:
Does $\csc(x)$ violate the Intermediate Value Theorem?
I have been asked a question to explain as to why $f(x) = \csc(x)$ does not "break" the IVP between
$\left(\dfrac{-\pi}2,\dfrac{\pi}2\right)$.

Considering that the function is discontinuous, it already breaks the IVP. In addition, there is no $f(x)$ value that can be found between $f\left(\dfrac{\pi}2\right)$ and $f\left(-\dfrac{\pi}2\right)$ which makes the question even more confusing for me.

答: As @J. W. Tanner pointed out, the Intermediate Value Theorem requires that $\csc(x)$ be continuous on the given interval, but $\csc(x)$ is not defined at $x=0$. Therefore, the Intermediate Value Theorem does not apply because one of its pre-conditions was not met.


问:
Is the "cyclotomic diagonalization" always squarefree?
For every integer $n\ge 2$ , define $$f(n):=\Phi_n(n)$$ where $\Phi(n)$ is the $n$ th cyclotomic polynomial. This can be considered as the "cyclotomic diagonalization"

Prove or disprove the conjecture that $f(n)$ is squarefree for every integer $n\ge 2$

The partial factorizations I determined with PARI/GP (usually, the factors upto $12-13$ digits are found) gave no exponent greater than $1$ upto $n=200$. A rigorous test upto $n=46$ revealed that upto this limit there are only squarefree values.

If it cannot be proven, a good heuristic would also be welcome.

答:
We want to show that the problem is equivalent to

For any number $n$ and any prime $q \mid n^n-1$, then $ord_{q^2}(n) \neq n$

We will see at the end why we expect this to be false, but also that counterexamples are very rare and are likely to have more than forty digits.

Recall the Moebius function defined here, and the notation $v_p(k)$ for the maximum exponent $\alpha$ such that $p^{\alpha} \mid k$. Notice that $v_p(xy) = v_p(x)+v_p(y)$ behaves like a logarithm. The key observation is the following:

Lemma. It holds
$$ (**) \ \ \ \ v_p(\Phi_n(n) ) = \sum_{d \mid n} \mu \left (\frac{n}{d} \right ) v_p(n^d-1) $$

Proof. It is well known that

$$ \prod_{ d \mid n} \Phi_d(x) = x^n -1 $$

Now for any natural number $x$, taking $v_p$ yields

$$ \sum_{d \mid n} v_p \Phi_d(x) = v_p(x^n-1) $$

Applying the Moebius Inversion Formula we get
$$ v_p \Phi_n(x) = \sum_{d \mid n} \mu \left ( \frac{n}{d} \right ) v_p(x^d-1)$$
and the thesis follow by specializing at $x=n$.

Main Proof. Now that we have a formula for the exponent, let's compute it. Set $r:= ord_p(n)$. Note that if $r$ does not divide $n$, it does not divide any of its divisors $d \mid n$; it follows that $n^d \neq 1 \pmod{p}$ for all $d$. Thus equation (**) yields $v_p(\Phi_n(n) ) = 0$. Let us suppose for the rest that $r \mid n$ and set $n'=n/r$.

Case 1. $n'=1$, that is $ord_p(n) = n$. In this case the only summand that contributes to the summation in (**) is $v_p(n^n-1)$. This is $\le 1$ if and only if $p^2 \nmid n^n - 1$. Note that $n = ord_p(n) \mid ord_{p^2}(n) $:

If the $ord_{p^2}(n)$ is equal to $n$, then $p^2 \mid n^n -1 $ and we found a counterexample;

If instead $ord_{p^2}(n)>n$, we must have that $n^n \neq 1 \pmod{p^2}$ and we are happy.

Case 2 $n' >1$. This part can seem heavy, but if you are familiar with Moebius Function you will recognize that this is just ordinary Yoga.

The summands that contribute to (**) are all the $d$ that are divisible by $r$. Note that any number $d$ such that $r \mid d \mid n$ can be written as $rd'$, where $d' \mid n'=n/r $. With this observation in mind we write

$$ v_p(\Phi_n(n) ) = \sum_{r \mid d \mid n} \mu \left (\frac{n}{d} \right ) v_p(n^d-1) =
\sum_{d' \mid n'} \mu \left (\frac{r n'}{r d'} \right ) v_p(n^{rd'}-1) = \sum_{d' \mid n'} \mu \left (\frac{n'}{d'} \right ) v_p((n^r)^{d'}-1)
$$

Now we use the Lifting The Exponent Lemma to split the $v_p$ factor. We get
$$ \sum_{d' \mid n'} \mu \left (\frac{n'}{d'} \right )( v_p(n^r-1) + v_p(d') )= v_p(n^r-1) \sum_{d' \mid n'} \mu \left (\frac{n'}{d'} \right ) + \sum_{d' \mid n'} \mu \left (\frac{n'}{d'} \right ) v_p(d') $$

Since $n'>1$, the first summand is zero: indeed $\sum_{k \mid h} \mu(h/k) = 0$ for $h> 1$. The second factor is more interesting. Suppose $n'= p^{\alpha} m$ with $(m,p) =1$. Every divisor of $n'$ will uniquely write as $p^{\beta} m'$, where $\beta \le \alpha$ and $m' \mid m$. Using the fact that $\mu(ab) = \mu(a)\mu(b)$ for $a,b$ coprimes we get

$$\sum_{d' \mid n'} \mu \left (\frac{n'}{d'} \right ) v_p(d') = \sum_{\beta=0}^{\alpha} \sum_{m' \mid m} \mu(p^{\alpha-\beta}) \mu(m/m') \beta = \sum_{\beta= 0}^{\alpha} \mu(p^{\alpha-\beta})\beta \sum_{m' \mid m} \mu(m/m')$$

We have to do a little distinction here. If $m>1$ this is zero again by the "summation moebius" property. If $m=1$, we are basically saying that $n'=p^{\alpha}$. In this case the formula reduce to

$$ \sum_{\beta=0}^{\alpha} \mu(p^{\alpha-\beta}) \beta = \beta - (\beta-1) = 1$$

And... Voilà!

Heuristics. We want to show that it is almost certain that a counterexample exist, but it is very likely to have hundreds of digits. Define for $d \mid p(p-1)$:
$$ X(p) = \{ 1 \le z \le p^2: \ \ (z,p) =1 \} $$
$$ X_d(p) := \{1 \le z \le p^2 : \ \ (z,p) = 1 \text{ and } ord_{p^2}(z) = d \} $$
Note that
$$\# X(p) = \phi(p^2) = p(p-1) \ \ \ \ \#X_d(p) = \#\{\text{roots of } x^d-1 \text{ in } \mathbb{F}_p\} = d$$
It is reasonable to assume that $X_d(p)$ is a random set in $X(p)$; indeed, if I am not wrong, there are some results of Davenport, Erdos, Vinogradov showing that quadratic residues, cubic residues ecc. are equidistributed among $X(p)$, and you can say that $ord_{p^2}(z) = k$ if and only if $z$ is a $\phi(p)/k$ residue but not a $\phi(p)/kq$ residue for all $q \mid \phi(p)/k$. However, I am not expert enough to produce a real proof.

We say that $p$ is a counterexample to the cdp (cyclotomic diagonalization problem) if $p^2$ appears as a factor of some $\Phi_n(n)$. We have shown above that this can happen iff $ord_{p^2}(n) = n$. Note that such an $n$ should be a divisor of $\phi(p) = p(p-1)$, because $n = ord_{p^2}(n) \mid p(p-1)$ and must be also coprime with $p$ (otherwise $p \nmid n^n-1$). On balance, the $n$ that make $p$ a counterexample to the cdp must be searched among the divisors of $(p-1)$.

For any such divisor $b$, we must hope that $b \in X_b(p)$. Since $X_b(p)$ is a random set in $X(p)$, this happens with probability $\frac{b}{p(p-1)}$. So the probability that this does not happen for any divisor $b$ is

$$ \prod_{b \mid (p-1) } \left ( 1- \frac{b}{p(p-1)} \right ) $$
The logarithm of this sum can be estimated as
$$ - \sum_{b \mid (p-1) } \frac{b}{p(p-1)} = - \frac{1}{p(p-1)} \sum_{b \mid q-1} b \le - \frac{1}{p} $$
Using the fact that $\sum_{b \mid q-1} b \ge q-1$. So the probability that $p$ is a counterexample is $\ge $

$$ 1- e^{-1/p} \simeq \frac{1}{p} $$

We use this logarithm-product strategy to estimate the probability $Q_x$ of finding a counterexample $\le x$. The probability of not having counterexamples up to $x$ is

$$ 1- Q_x = \prod_{p \le x} \left ( 1 - \frac{1}{p} \right ) \simeq exp(-\sum_{p \le x} \frac{1}{p} )\simeq exp(-\log \log x) = \frac{1}{\log(x)}$$

From which we get $Q_x \simeq 1- \frac{1}{\log x} $. This tends to $1$, so it is almost certain that a counterexample exists. On the other side, it goes to zero very slowly.

Example. Suppose you want to be $99\%$ sure to find a counterexample up to $x$. Then
$$\frac{99}{100} \simeq 1- \frac{1}{\log x} \ \ \Leftrightarrow \ \ \ \log x \simeq 100 $$

Which implies $x \simeq e^{100} \simeq 10^{43}$.

P.S. I am missing the case $p=2$, in which the Lifting the Exponent Lemma is slightly different; it should work the same way, though (the only different passage is the last, I which I think it is enough to verify the case $\Phi_{2^n}(x) = x^{2^{n-1}}+1$ :) )


问:
Statistics how to calculate likelihood over several draws
I am working on a spreadsheet and am lousy at statistics. I am working on what is essentially a lottery to draw a winning number

I have a likelihood of drawing a win in percent, and number of attempts. What would I do to calculate it across $x$ number of tries?

I did number of tries * percent and I got numbers that seemed right. I wanted to make sure I was correct

答:
Suppose the probability of winning in one attempt is $p$.

Across $n$ attempts, the probability that you lose in all $n$ is $(1-p)^n$, assuming that each attempt is independent. So the probability that you win at least once is $\boxed{1-(1-p)^n}$.

When $p$ is very small, and so $np\ll1$, we have by the binomial approximation
$$1-(1-p)^n\approx1-(1-np)=np,$$
so your formula of number of tries x winning probability is a decent approximation.


问:
Theorem 6.10 Rudin PMA, Partition
The third paragraph of the proof begins,

'Now form a partition $P = \{x_0,x_1,...,x_n\}$ of $[a,b]$, as follows: Each $u_j$ occurs in $P$. Each $v_j$ occurs in $P$. No point of any segment $(u_j,v_j)$ occurs in $P$. If $x_{i-1}$ is not one of the $u_j$, then $\Delta x_i < \delta$.'

I don't understand the final sentence, 'If $x_{i-1}$ is not one of the $u_j$, then $\Delta x_i < \delta$.' I interpret this as meaning that $|v_j - u_{j+1}| < \delta$ for each $j$. Is my interpretation correct? If so, I don't know what allows him to say this--we don't know apriori where the points of discontinuity are within $[a,b]$, so how can rudin assert that the distance between the intervals $[u_j,v_j]$ and $[u_{j+1},v_{j+1}]$ is less than $\delta$?

答: I assume this is what it means: Take all the points $u_j$ and $v_j$ and throw in to $P$. Then in $P$, there are end points of intervals $(u_j,v_j)$. However outside of all the intervals $(u_j,v_j)$ are the complement intervals $(v_j,u_{j+1})$. In these complement intervals, chop each interval into many shorter intervals, each of length $<\delta$. Now take all the points where you chopped up these intervals $(v_j,u_{j+1})$, and add them into $P$. Hopefully this makes sense.

Title: Theorem 6.10 Rudin PMA, Partition


问:
A problem of systems of linear equations (very basic).

The exercise is:

Wheat flour is packaged in a supermarket in bags of 2 kg, 5 kg and 12 kg. In September, 250 bags were used and 5500 kg of wheat flour were packed. In October, due to problems with the 12 kg bags, 50 more bags of 2 kg and 5 kg were used, so only 4250 kg of flour were packaged.

The goals to solve:

a) How many bags of each type were used in September?

b) How many percent of the total flour packaged between September and October was made in 2 kg bags?

*Doubt

I know how to do the process of reduction method but I do not know how to transform each sentence into equations to solve the problem.

1- "In a supermarket, wheat flour is packaged in bags of 2 kg, 5 kg and 12 kg." I guess it is: (2x + 5y + 12z = 0), but I'm not sure.

2- "In the month of September, 250 bags were used and 5500 kg of wheat flour were packed." I guess it is: (250 = 5500), but I'm not sure.

3- "In October due to problems with the 12 kg bags, 50 more 2 kg and 5 kg bags were used, so only 4250 kg of flour were packed." I guess it is: (50(2x) + 50(5y) = 4250), but I'm not sure.

Please help me, thank you.

答:
You have the right idea about writing out equations in terms of $x,y,z$. What will help is being very specific about what each of these quantities are.

In the first part of your answer, you let $x$ be the number of 2kg bags, $y$ the number of 5kg bags and $z$ the number of 12kg bags. You want the total number of bags to be 250, which gives you the equation $x+y+z=250$.

When you are finding the total weight, you need to multiply the number of bags by the weight of each type, so in September you get $2x+5y+12z=5500$. See if you can make a similar equation for October and then use linear reduction to find the solutions.

P.S. As a general hint, always make sure any equation you want to solve makes sense. For example, $2x+5y+12z=0$ is not realistic here as that implies that you have zero of each bag. Similarly, you will never be able to solve the equation $250=5500$, so you need to think through how you approach the problem again.


问:
Tangent space of a product of algebraic group.
I need help with this problem from Shafarevich Basic Algebraic Geometry.

Let $G$ be an algebraic group and $\Psi:G\times G\rightarrow G$ the
  regular map defined by the group law. Let $T_{e}G$ and be
  $T_{e'}(G\times G)$ the tangent spaces to $G$ and $G\times G$ at their
  respective identity elements. Prove that $T_{e'}(G\times G)=T_{e}G\oplus T_{e}G$ and that $d_e \Psi:T_{e}G\oplus
 T_{e}G\rightarrow T_{e}G$ is given by addition of vectors.

I tried this way but I'm not acquainted with the tensor product. Let $T_{e'}(G\times G)=\left( \frac{M_{e'}}{M_{e'}^2} \right)^{*}$ where $M_{e'}=\{ f\in k[G\times G]\mid f(e')=0 \}$. But $k[G\times G]=k[G]\otimes_k k[G]$ so $M_{e'}=M_{e}\otimes M_{e}$.
Then
$$T_{e'}(G\times G)=\left( \frac{M_{e'}}{M_{e'}^2} \right)^{*} = \left(\frac{M_{e}}{M_{e}^2} \otimes \frac{M_{e}}{M_{e}^2}\right)^{*}=T_e G\oplus T_e G$$.

I'm not sure about the second and the third equalities. Could you help me, please? Thanks!

答:
You're correct that $k[G\times G]\cong k[G]\otimes_k k[G]$, but the next assertion that you make, $M_{e'}=M_e\otimes M_e$, is not correct (this unfortunately fouls up everything afterwards). For example, if you're working with $\Bbb G_a=\operatorname{Spec} k[t]$, then $k[G\times G]=k[t_1,t_2]$ with maximal ideal at $e'$ $(t_1,t_2)$, while $(t_1)\otimes_k (t_2) = (t_1t_2)$. Instead, you should have $$M_{e'}=(M\otimes k[G]) + (k[G]\otimes M)\subset k[G]\otimes_k k[G].$$

From here, we compute: $M_{e'}^2=M^2\otimes k[G] + M\otimes M + k[G]\otimes M^2$, so applying the third isomorphism theorem to $M\otimes M \subset M_{e'}^2\subset M_{e'}$, we see that $M_{e'}/M_{e'}^2$ is isomorphic to the quotient of $(M\otimes_k k[G]/M) \oplus  (k[G]/M\otimes_k M)$ by $(M^2\otimes_k k[G]/M) \oplus  (k[G]/M\otimes_k M^2)$, where we've upgraded from a sum to a direct sum because the intersection of those submodules is zero. But as $k[G]/M\cong k$ and $M\otimes_k k\cong M$, this is exactly $(M\oplus M)/(M^2\oplus M^2)\cong (M/M^2)\oplus (M/M^2)$. Dualizing, we have the result.


答: To solve $$\frac{1}{xy\rho(xy)}=\frac{1}{x\rho(x)}+\frac{1}{y\rho(y)}$$ consider $f(t):=1/(t\rho(t))$. Then it is well known that the only continuous functions satisfying $$f(xy)=f(x)+f(y)$$ are $$f(x)=A\log x$$ Hence $$\rho(x)=\frac{a}{x\log x}$$ where $a\ne0$.

Title: what are all functions with $x>1$ and $y>1$ $\rho$ that follows $\rho(xy)=\frac{1}{\frac{y}{\rho(x)}+\frac{x}{\rho(y)}}$ and is continuous

Body:
What are all functions with $x>1$ and $y>1$ $\rho$ that follows $$\rho(xy)=\frac{1}{\frac{y}{\rho(x)}+\frac{x}{\rho(y)}}$$ and is continuous

If this doesn't have any solutions then prove no such solution exists.

I would like to understand how to solve a problem like this


问:
Probability of picking a complete set of x items from an indefinitely large field after y tries
I'm looking for an equation that will help me calculate probabilities for a project at work. I've been Googling and reading posts here trying to figure out what I need but my unskilled brain is running into roadblocks. I hope that you can point me in the right direction.

In simplest terms, we have a promotion running that involves users finding 7 different virtual items evenly distributed among a field of essentially unlimited locations. In order to determine if the promotion is encouraging participation (i.e. visiting locations), we want to see if the number of people obtaining the complete set of 7 items is greater or lesser than what would be expected by chance given x number of location visits. I hope that makes sense.

I am looking for an equation that I can plug into Excel or SQL that returns this probability. I have found questions similar to this one here, but I can't seem to get over the hump of converting things into the usable equation I need (mostly due to my limited knowledge of statistics and notation).

Thank you in advance for your help,

Jon

答:
As I take your question, you have a large number of locations (say $N$), of which $7$ of these contain a particular prize. You want to determine the probability that if a user randomly selects $x$ of these locations, then they will receive all $7$ prizes.

There is a special notation in mathematics called the binomial coefficient $\binom{n}{k}$, which gives the total number of ways to choose $k$ objects out of a total $n$. In this case, the total number of locations a user can choose is
$$\binom{N}{x}$$
And if we require that $7$ of these choices are the 'prizes', then since the user must visit these locations, the remaining number of choices is
$$\binom{N-7}{x-7}$$
and hence the probability is simply
$$\frac{\binom{N-7}{x-7}}{\binom{N}{x}}$$
or $0$ if $x<7$.

In Excel, we can find the value $\binom{n}{k}$ with COMBIN(n,k), and hence you would want COMBIN(N,x)/COMBIN(N-7,x-7). Note that for this to work you require $x$ to be between $7$ and $N$, otherwise Excel may return an error.

UPDATE (in order to reflect the correction given in the comments)

Consider calculating the probability that a user receives all the prizes on the $n^\text{th}$ visit to a location, which we call $P(n)$. Since there are 7 possible prizes, the number of 'prize sequences' a user could encounter in $n$ locations is $7\times7\times\cdots\times 7=7^n$.

Now, the hard part is determining how many of these sequences involve the user finding the final prize at the $n^\text{th}$ location. To find this, we choose one of the 7 prizes to be the 'last discovered', and then count the number of weighs a user can find all remaining $6$ prizes in the first $n-1$ locations. Mathematically, this is equivalent to the number of surjections from $[n-1]$ to $[6]$ (don't worry about this if you don't understand it). In any case, the probability will be

$$P(n)=\frac{7\times 6!S(n-1,6)}{7^n}=\frac{6!S(n-1,6)}{7^{n-1}}$$

where $S(a,b)$ refers to the Stirling Numbers of the Second Kind. Since a user must visit at least $7$ locations, the average number of visits for a user will be

$$\mu=\sum_{n=7}^{\infty}nP(n)=\sum_{n=7}^{\infty}n\frac{6!S(n-1,6)}{7^{n-1}}$$

Doing some numerical calculation, I find this average to be $\approx 18.15$. I'm not sure Excel has a native option for finding these Stirling Numbers, but an explicit formula is given on the Wikipedia page.


问:
Let f be a function from $A$ to $B$. Let $S$ be a subset of $B$. Show that $f^{−1}( \overline{ S}) =(\overline{f^{−1}(S)}) $.
As you saw in the question, assume $S \subseteq B$, show that $f^{−1}( \overline{ S}) =(\overline{f^{−1}(S)})$.

I just need to show the first part $f^{−1}( \overline{ S}) \subseteq (\overline{f^{−1}(S)})$.  without proving the other part $ (\overline{f^{−1}(S)}) \subseteq f^{−1}( \overline{ S}) $ to prove they are equal.

Proof:

To prove the first part $f^{−1}( \overline{ S}) \subseteq (\overline{f^{−1}(S)})$, we will let $x \in f^{-1}(\overline{S})$ and show in a sequence of steps that  $x \in \overline{f^{-1}(S)}$ based on my understanding eventually. Then $ \exists y \in \overline{S} $ such that $f(x)=y$.

The proof proceeded then by saying that, based on the definition of the complement, $ \exists y \in \overline{S} $ is equivalent to,

$$ \neg (y\in S)$$

Since $f(x) = y$, then,
$$ \neg (f(x)\in S)$$

Up to now, the proof makes sense, but I don't understand the following steps that lead to the final conclusion:

$f^{-1}(S)$ contains all elements whose image is in $S$, $$ \neg
    (f(x)\in S)$$, then
$$\neg(x \in f^{-1}(S))$$

By the definition of the complement,

$$x\in \overline{f^{-1}(S)}$$

By the definition of the subset,

$$f^{−1}( \overline{ S})\subseteq (\overline{f^{−1}(S)})$$

Can you please clarify point 1 above as I don't get how it was brought above in the proof.

答:
By stating the definition of preimmage
$$f^{-1}(S) = \{ x\in A : f(x)\in S \} ,$$
so, by definition, if $f(x) \not \in S \Longrightarrow x\not \in f^{-1}(S)$.


问:
$L^1$ norm of a function is greater than the norm of its root
Edit: My friend helped me to solve this, and now it is solved for me. We must first separate the simple cases, the only serious case is when $\sum_{i=0}^n |f_i|$ and $\alpha$ are both strictly greater than $1$.

Let $L/K$ be an extension of local fields. and let $|.|_L$ be the absolute value on $L$. Suppose we have $K[x] \ni f(x)=\prod_{i=1}^n(x-\alpha_i)$ for $\alpha_i$'s in $L$. I want to show that $|\alpha| < \sum_{i=0}^n |f_i|,$ for every root $\alpha$ of $f$, where $f(x)=\sum_{i=0}^n f_ix^i$. It is about an hour that I stuck in it without no progress, what's make it worser is that I don't have even a raw idea.

答:
Let $m := \max \lvert \alpha_i \rvert$, and let $ r \in \{1, .., n\}$ be the number of $\alpha_i$'s such that $\lvert \alpha_i \rvert = m$. Then note that by the ultrametric maximum principle, $\lvert f_{n-r} \rvert = m^r$. Hence

$$ \sum \lvert f_i \rvert \ge \lvert f_n \rvert + \lvert f_{n-r} \rvert = 1+m^r > m $$

(the last inequality is true for any real $m \ge 0$), which by definition is $\ge \lvert \alpha_i \rvert$ for all $i$.


答:
Let $x\in f^{-1}(\overline{S})$ then by definition $f(x)\in \overline{S}$ it is
$f(x)\not \in S$ then$f(x) \not \in f^{-1}(S)$ and therefore
$f(x)\in \overline{f^{-1}(S)}$  therefore $x\in \overline{f^{-1}(S)}$.

And $f^{-1}(\overline{S}) \subset \overline{f^{-1}(S)}$ holds.

Now let $y\in \overline{f^{-1}(S)}$ then $f(y)\not \in S$ and therefore
$f(y)\in f^{-1}(\overline{S})$.

Therefore $f^{-1}(\overline{S}) \supset \overline{f^{-1}(S)}$.

And then $f^{-1}(\overline{S}) = \overline{f^{-1}(S)}$ .

Title: Let f be a function from $A$ to $B$. Let $S$ be a subset of $B$. Show that $f^{−1}( \overline{ S}) =(\overline{f^{−1}(S)}) $.

Body:
As you saw in the question, assume $S \subseteq B$, show that $f^{−1}( \overline{ S}) =(\overline{f^{−1}(S)})$.

I just need to show the first part $f^{−1}( \overline{ S}) \subseteq (\overline{f^{−1}(S)})$.  without proving the other part $ (\overline{f^{−1}(S)}) \subseteq f^{−1}( \overline{ S}) $ to prove they are equal.

Proof:

To prove the first part $f^{−1}( \overline{ S}) \subseteq (\overline{f^{−1}(S)})$, we will let $x \in f^{-1}(\overline{S})$ and show in a sequence of steps that  $x \in \overline{f^{-1}(S)}$ based on my understanding eventually. Then $ \exists y \in \overline{S} $ such that $f(x)=y$.

The proof proceeded then by saying that, based on the definition of the complement, $ \exists y \in \overline{S} $ is equivalent to,

$$ \neg (y\in S)$$

Since $f(x) = y$, then,
$$ \neg (f(x)\in S)$$

Up to now, the proof makes sense, but I don't understand the following steps that lead to the final conclusion:

$f^{-1}(S)$ contains all elements whose image is in $S$, $$ \neg
    (f(x)\in S)$$, then
$$\neg(x \in f^{-1}(S))$$

By the definition of the complement,

$$x\in \overline{f^{-1}(S)}$$

By the definition of the subset,

$$f^{−1}( \overline{ S})\subseteq (\overline{f^{−1}(S)})$$

Can you please clarify point 1 above as I don't get how it was brought above in the proof.


问:
Differential Equation Solution By Power Series

Solve $(1 + x)y' = py;\ \ \ y(0) = 1$, where $p$ is an arbitrary constant.

First I plugged in the guess $y = \sum_{n = 0}^\infty a_n x^n$:

$(1 + x)(\sum_{n = 0}^\infty a_n x^n)' = p\sum_{n = 0}^\infty a_n x^n$

Then I expanded the derivative and multiplication:

$\sum_{n = 0}^\infty n a_n x^{n - 1} + \sum_{n = 0}^\infty n a_n x^n = p\sum_{n = 0}^\infty a_n x^n$

Then I shifted the left index (the first term yielding $0$ allows the lower bound to remain $0$) and algebraically combined the summations:

$\sum_{n = 0}^\infty (n + 1)a_{n + 1} x^n + (n - p)a_n x^n = 0$

This leads to the following recurrence relation:

$a_{n + 1} = \frac{p - n}{n + 1}a_n$

Thus for various values of $n$:

$a_1 = p a_0$, $a_2 = \frac{p(p - 1)}{2}a_0$, $a_3 = \frac{p(p - 1)(p - 2)}{6} a_0$, etc.

So applying definitions for the exponential taylor series and falling factorial, the guessed solution would be:

$y = \sum_{n = 0}^\infty \frac{p! a_0 x^n}{n! (p - n)!} = \sum_{n = 0}^\infty a_0 e^x p^{\underline n}$

Solving the initial value problem:

$1 = \sum_{n = 0}^\infty a_0 e^0 p^{\underline n} \implies a_0 = \frac{1}{\sum_{n = 0}^\infty p^{\underline n}}$

My final solution is:

$y = \frac{\sum_{n = 0}^\infty e^x p^{\underline n}}{\sum_{n = 0}^\infty p^{\underline n}}$

However, the answer is supposed to be $y = (1 + x)^p$.  Are these identical, or did I make an error somewhere?

答:
Your solution
$$
y = \sum_{n = 0}^\infty \frac{p! a_0 x^n}{n! (p - n)!}
$$
is mostly the binomial series for $a_0\,(1+x)^p$, assuming you know what you are doing (it looks like you are treatng $p$ as an integer, which it isn't necessarily according to the question as you posted it).

The "equality"
$$
\sum_{n = 0}^\infty \frac{p! a_0 x^n}{n! (p - n)!}= \sum_{n = 0}^\infty a_0 e^x p^{\underline  n}
$$
is not true; it's missing the $n!$ in the denominator on the right.


问:
Is a Radon measure always positive on non-empty open sets? The same question about the Haar measure.
A measure is locally finite if it is finite on all compact sets from the underlying $\sigma$-algebra.

A measure is regular if every measurable set $A\in\Sigma$ can be approximated from above by open measurable sets:
$$
 \mu(A)\,=\,\operatorname{inf}\,\left\{\,\mu(O)\;\Big{|}\;A\subseteq O\;,\;\;O\in\Sigma\,,\;\,O\;\,\mbox{open}   \,\right\}\;\;,\;\;
 $$
and from below by compact measurable sets:
$$
 \mu(A)\,=\,\operatorname{sup}\,\left\{\,\mu(C)\;\Big{|}\;C\subseteq A\,,\;\,C\in\Sigma\,,\;\,C\;\,\mbox{compact}\,\right\}\;\;.
 $$

A Radon measure is a Borel measure that is introduced on the Borel algebra of a Hausdorff topological space, and is regular and locally finite.

Question 1: $\;\;$is a non-zero Radon measure always positive on non-empty open sets?

Question 2: $\;\;$the same for a non-zero Haar measure

答: Unless I'm overlooking something trivial, the answer is no. Look at the "restriction" of Lebesgue measure $\lambda$ on $\Bbb{R}$ to $[0,1]$, by which I mean $\mu(E):= \lambda(E\cap [0,1])$. Then, $\mu$ is a finite Borel measure on $\Bbb{R}$, hence is regular, but the open set $(10,11)$ has measure $0$ with respect to $\mu$.


问:
Proving uniqueness of intersection multiplicities
I'm trying to understand a proof from my Algebraic curves course. Additional context is provided Geometric Interpretation of Intersection Multiplicities, but I'll copy the relevant details here. Partial answers (e.g. answering only one of the questions) are very welcome, and I'll upvote them! This looks like a long question, but only because I have provided unnecessary details for the sake of completeness.

(Axioms of intersection multiplicity)

Symmetry: $\mathbf{I}(p, P, Q)=\mathbf{I}(p, Q, P)$

Detects intersection points: $\mathbf{I}(p, P, Q) \neq 0$ if and only if $P(p)=Q(p)=0$.

Detects common components: $\mathbf{I}(p, P, Q)=\infty$ if and only if $P$ and $Q$ have a common irreducible factor that vanishes at $p$.

Transversality: suppose that $P$ and $Q$ both have degree one and that $Q$ is not of the form $\lambda P$ for some $\lambda \in \mathbb{C}^{*} .$ If $p$ is the unique point in $\mathbb{P}_{\mathbb{C}}^{2}$ such that $P(p)=Q(p)=0,$ then $\mathbf{I}(p, P, Q)=1$

Additivity: if $R$ is another non-zero homogeneous polynomial in $\mathbb{C}\left[x_{0}, x_{1}, x_{2}\right],$ then $\mathbf{I}(p, P, Q R)=$ $\mathbf{I}(p, P, Q)+\mathbf{I}(p, P, R)$

Deformation: assume that $\operatorname{deg}(Q) \geq \operatorname{deg}(P),$ and let $R$ be a homogeneous polynomial in $\mathbb{C}\left[x_{0}, x_{1}, x_{2}\right]$ of degree $\operatorname{deg}(Q)-\operatorname{deg}(P)$ such that $Q+P R \neq 0 .$ Then $\mathbf{I}(p, P, Q)=\mathbf{I}(p, P, Q+P R)$
Moreover, if $P$ and $Q$ are non-constant and have no repeated factors, and we denote by $C$ and $D$ their zero sets in $\mathbb{P}_{\mathbb{C}}^{2},$ then $\mathbf{I}(p, P, Q)=\mathbf{I}(p, C, D)$

Proposition:

There is at most one invariant $\mathbf{I}(p, P, Q)$ that satisfies all the axioms above.

Proof. Since the axioms are coordinate-independent, we may assume that $p=[1,0,0] .$ Indeed, if we could find two distinct invariants $\mathbf{I}(q, P, Q)$ and $\mathbf{I}^{\prime}(q, P, Q)$ at another point $q$ in $\mathbb{P}_{\mathbb{C}}^{2}$ satisfying all the axioms, then we could pick an invertible rank 3 matrix $A$ over $\mathbb{C}$ such that the projective coordinate transformation $\Phi_{A}$ maps $q$ to $p=[0,0,1],$ and produce two distinct invariants $\mathbf{I}(p, P, Q)=\mathbf{I}\left(q, P\left(\left(x_{0}, x_{1}, x_{2}\right) A^{t}\right), Q\left(\left(x_{0}, x_{1}, x_{2}\right) A^{t}\right)\right)$
and $\mathbf{I}^{\prime}(p, P, Q)=\mathbf{I}^{\prime}\left(q, P\left(\left(x_{0}, x_{1}, x_{2}\right) A^{t}\right), Q\left(\left(x_{0}, x_{1}, x_{2}\right) A^{t}\right)\right)$ satisfying all the axioms with respect to the point $p$.

Question 1: We know that $\mathbf{I}^{\prime}(q, P, Q)$ satisfy the axioms, but surely this doesn't imply that $\mathbf{I}^{\prime}\left(q, P\left(\left(x_{0}, x_{1}, x_{2}\right) A^{t}\right), Q\left(\left(x_{0}, x_{1}, x_{2}\right) A^{t}\right)\right)$ satisfies the axioms as well? (Because the polynomials with respect to which the intersection multiplicity is being measured has changed?) Then, we cannot claim that $\mathbf{I}^{\prime}(p, P, Q)$ satisfies the axioms? I suppose $\mathbf{I}(p, P, Q)=\mathbf{I}\left(q, P\left(\left(x_{0}, x_{1}, x_{2}\right) A^{t}\right), Q\left(\left(x_{0}, x_{1}, x_{2}\right) A^{t}\right)\right)$
and $\mathbf{I}^{\prime}(p, P, Q)=\mathbf{I}^{\prime}\left(q, P\left(\left(x_{0}, x_{1}, x_{2}\right) A^{t}\right), Q\left(\left(x_{0}, x_{1}, x_{2}\right) A^{t}\right)\right)$ holds because intersection multiplicities do not change under projective transformations.

So, suppose that we have two invariants $\mathbf{I}(p, \cdot, \cdot)$ and $\mathbf{I}^{\prime}(p, \cdot, \cdot)$ that satisfy all the axioms. We will show that $\mathbf{I}(p, P, Q)=\mathbf{I}^{\prime}(p, P, Q)$ for all non-zero homogeneous polynomials $P$ and $Q$ in $\mathbb{C}\left[x_{0}, x_{1}, x_{2}\right] .$ Denote by $r$ and $s$ the degrees of the polynomials $P(1,0, x)$ and $Q(1,0, x)$ in $\mathbb{C}[x] .$ By symmetry, we may assume that $r \leq s .$ We will prove the result by induction on $r .$

First, suppose that $r \leq 0$. By the detection of intersection points, the equality $\mathbf{I}(p, P, Q)=\mathbf{I}^{\prime}(p, P, Q)$ is satisfied when $P$ or $Q$ does not vanish at $p$. Thus, by additivity and symmetry, we can reduce to the case where $P(p)=Q(p)=0$ and $P$ is irreducible. By the detection of common components, we can exclude the case where $Q$ is divisible by $P$.

Question 2: If $P$ has repeated factors, i.e. $P_1^2 \mid P$, then $\mathbf{I}(p,P,Q) = \mathbf{I}(p,P_1,Q) + \mathbf{I}(p,P_1,Q) + \dots$ so I don't think we can reduce to the case $\mathbf{I}(p,P,Q)$ where $P$ is irreducible? (As there may be more than one such irreducible $P_1$ in the factorization of the original $P$, in the sense that $P_1^2 \mid P$)

(The rest of the proof is provided for completeness, but is probably irrelevant to the above two questions)

Now $P(1,0, x)$ is a constant polynomial that vanishes at $x=0,$ so that $P(1,0, x)=0 .$ This implies that $P\left(x_{0}, x_{1}, x_{2}\right)$ is divisible by $x_{1} .$ Since we are assuming that $P$ is irreducible, it follows that $P=a x_{1}$ for some $a$ in $\mathbb{C}^{*}$. Now we can use the deformation axiom to remove all the terms in $Q$ that are divisible by $x_{1} ;$ this brings us to the case where $Q$ is a non-zero homogeneous polynomial in $\mathbb{C}\left[x_{0}, x_{2}\right] .$ Factoring $Q$ into linear homogeneous polynomials, we deduce from the additivity and transversality axioms that $\mathbf{I}(p, P, Q)=$ $\mathbf{I}^{\prime}(p, P, Q)=\operatorname{mult}_{0} Q(1,0, x):$ this is the number of linear factors in $Q$ that vanish at the point $p$.

Therefore, we may assume that $s \geq r \geq 1,$ and that $\mathbf{I}\left(p, P^{\prime}, Q^{\prime}\right)=\mathbf{I}^{\prime}\left(p, P^{\prime}, Q^{\prime}\right)$ for all non-zero homogeneous polynomials $P^{\prime}$ and $Q^{\prime}$ in $\mathbb{C}\left[x_{0}, x_{1}, x_{2}\right]$ such that the degree of $P^{\prime}(1,0, x)$ or $Q^{\prime}(1,0, x)$ is strictly less than $r$. We may again assume that $P(p)=Q(p)=0$ and that $P$ is irreducible and does not divide $Q$. Dividing $P$ and $Q$ by suitable elements in $\mathbb{C}^{*},$ we can reduce to the case where $P(1,0, x)$ and $Q(1,0, x)$ are monic. We set
$$
R\left(x_{0}, x_{1}, x_{2}\right)=x_{0}^{\operatorname{deg}(P)+s-r} Q\left(x_{0}, x_{1}, x_{2}\right)-x_{0}^{\operatorname{deg}(Q)} x_{2}^{s-r} P\left(x_{0}, x_{1}, x_{2}\right)
$$
This is a homogeneous polynomial such that the degree of $R(1,0, x)=Q(1,0, x)-x^{s-r} P(1,0, x)$ is at most $s-1 .$ Moreover, $R$ is non-zero because $P$ is irreducible and does not divide $x_{0}$ or $Q .$ By detection of intersection points, additivity and deformation, we have
$$
\mathbf{I}(p, P, Q)=\mathbf{I}\left(p, P, x_{0}^{\operatorname{deg}(P)+s-r} Q\left(x_{0}, x_{1}, x_{2}\right)\right)=\mathbf{I}(p, P, R)
$$
and the same equalities hold for $\mathbf{I}^{\prime}$. Replacing $Q$ by $R$ and repeating the argument, we arrive at a situation where the degree of $Q$ becomes strictly less than $r,$ where we can invoke the induction hypothesis.

答:
Q1: projective transformations do indeed preserve intersection multiplicity. On one hand, this should be geometrically clear: the axioms are coordinate-independent, so why should our specific choice of coordinates on $\Bbb P^2$ matter? If you want to prove it explicitly for yourself, the big idea of the proof is that we can build up any $I(q,P,Q)$ for any $P,Q$ via our axioms by starting with $I(q,\lambda_1,\lambda_2)$ for $\lambda_1,\lambda_2$ linear forms where we know the value of $I$ via axioms 2, 3, and 4 by combining them via the additivity and deformation axioms.

Q2: I don't understand your objection. Additivity says that if $P$ factors as $P_1P_2\ldots P_n$, then $I(q,P,Q)=I(q,P_1,Q)+I(q,P_2,Q)+\ldots+I(q,P_n,Q)$, and it places no restriction on the factorization - there is no requirement that any two $P_i$ be distinct, coprime, or anything. As $k[x,y,z]$ is a UFD, we can always break $P$ down uniquely (up to rearranging and multiplication by constants) in to a product of irreducible $P_i$, apply this result, and then we're dealing with a sum of $I(q,P_i,Q)$ where every $P_i$ is irreducible.


答:
Hint: You know that the solution (by separation of variables) is:

$y(x)=(x+1)^p$

Now try to use the binomial theorem, expand the above, and equate coefficients. You will see that:

$a_0=1$

$a_1=pa_0=p= \binom{p}{1}$

$a_2=\frac{p(p-1)}{2}a_0= \binom{p}{2}$

.
.
.

$a_k=\binom{p}{k}$

As the others have pointed out, $a_k=0$ for $k\geq p+1$. This is because the solution is a polynomial.

So your recursion is correct, but after you applied the exponential Taylor series, things took a different turn.

Moral of the story: If you have an easier method, use it to double check your work :)

Edit: The above assumes that $p$ is an integer. If $p$ is an arbitrary scalar value (i.e. real, or even complex) it turns out we can in fact use a similar formula, we just have to be careful of how we are computing the binomial coefficients.

Indeed, we have that for arbitrary $p$,

$$y(x)= \sum_{k=0}^{\infty}\binom{p}{k}x^k$$ for $|x|<1$ and $x,p \in \mathbb{R}$

I did a little investigating, and found this to be helpful:
Generalized Binomial Expansion of $\left(1+x \right )^{y}$

Edit: Looks like there are multiple formulas for the "generalized binomial coefficients", but digging into Wolfram Mathworld, I came across:

$$\binom{p}{k}:= \frac{(p)_k}{k!}=\frac{p(p-1)(p-2)\cdots(p-k+1)}{k(k-1)\cdots 2 \cdot 1}$$

Look at formulas $(3),(4)$ in:

https://mathworld.wolfram.com/BinomialCoefficient.html

Title: Differential Equation Solution By Power Series

Body:
Solve $(1 + x)y' = py;\ \ \ y(0) = 1$, where $p$ is an arbitrary constant.

First I plugged in the guess $y = \sum_{n = 0}^\infty a_n x^n$:

$(1 + x)(\sum_{n = 0}^\infty a_n x^n)' = p\sum_{n = 0}^\infty a_n x^n$

Then I expanded the derivative and multiplication:

$\sum_{n = 0}^\infty n a_n x^{n - 1} + \sum_{n = 0}^\infty n a_n x^n = p\sum_{n = 0}^\infty a_n x^n$

Then I shifted the left index (the first term yielding $0$ allows the lower bound to remain $0$) and algebraically combined the summations:

$\sum_{n = 0}^\infty (n + 1)a_{n + 1} x^n + (n - p)a_n x^n = 0$

This leads to the following recurrence relation:

$a_{n + 1} = \frac{p - n}{n + 1}a_n$

Thus for various values of $n$:

$a_1 = p a_0$, $a_2 = \frac{p(p - 1)}{2}a_0$, $a_3 = \frac{p(p - 1)(p - 2)}{6} a_0$, etc.

So applying definitions for the exponential taylor series and falling factorial, the guessed solution would be:

$y = \sum_{n = 0}^\infty \frac{p! a_0 x^n}{n! (p - n)!} = \sum_{n = 0}^\infty a_0 e^x p^{\underline n}$

Solving the initial value problem:

$1 = \sum_{n = 0}^\infty a_0 e^0 p^{\underline n} \implies a_0 = \frac{1}{\sum_{n = 0}^\infty p^{\underline n}}$

My final solution is:

$y = \frac{\sum_{n = 0}^\infty e^x p^{\underline n}}{\sum_{n = 0}^\infty p^{\underline n}}$

However, the answer is supposed to be $y = (1 + x)^p$.  Are these identical, or did I make an error somewhere?


问:
How to prove this inequality $\sqrt{\prod\limits_{cyc}(a+\sqrt{\frac{bcd}{a}})}+2\sqrt{abcd} \ge ab+bc+cd+da+ac+bd$
Let $a,b,c,d>0$. Show that
$$\sqrt{\left(a+\sqrt{\dfrac{bcd}{a}}\right)\left(b+\sqrt{\dfrac{acd}{b}}\right)\left(c+\sqrt{\dfrac{abd}{c}}\right)\left(d+\sqrt{\dfrac{abc}{d}}\right)}+2\sqrt{abcd}\ge ab+bc+cd+da+ac+bd$$

My idea: make use of
$$a+\sqrt{\dfrac{bcd}{a}}=a+\dfrac{\sqrt{abcd}}{a}\ge 2\sqrt{\sqrt{abcd}}$$

It is said this inequality can be proved using AM-GM( Cauchy-Schwarz) inequality. But I can't.

Thank you for you help in proving it.

答:
WLOG, assume $abcd = 1$. The desired inequality is written as
$$\sqrt{(a^2+1)(b^2+1)(c^2+1)(d^2+1)} + 2 \ge ab + bc + cd + da + ac + bd.$$
Since $ab + bc + cd + da + ac + bd \ge 6$, it suffices to prove that
$$(a^2+1)(b^2+1)(c^2+1)(d^2+1) - (ab + bc + cd + da + ac + bd - 2)^2 \ge 0.$$
Using $d = \frac{1}{abc}$, we have
$$\mathrm{LHS} = \frac{(ab - 1)^2(bc-1)^2(ca-1)^2}{a^2b^2c^2}.$$
We are done.


问:
Show that $d/dx (a^x) = a^x\ln a$.
Show that
$$
\frac{d}{dx} a^x = a^x \ln a.
$$

How would I do a proof for this.  I can't seem to get it to work anyway I try.

I know that
$$
\frac{d}{dx} e^x = e^x.
$$
Does that help me here?

答: Hint: Write $y = a^x$ or equivalently $\ln y = x \ln a$ and use implicit differentiation.</p>

Title: Show that $d/dx (a^x) = a^x\ln a$.


问:
epsilon delta limit proof of $\lim_{x \to 0} \arctan{(x)}=0$

Let $\lim x→0 \arctan{(x)}=0$  then there exists $\delta>0$ such that if $0<|x|<\delta$ then $|\arctan{(x)}|< \varepsilon $

I know that,

$|\arctan{(x)}|<\pi/2$

But how do I relate this to $|x|$ so I can find a suitable $\delta$? Because I don't even know how to check if $|\arctan{(x)}|\leq|x|$ is true (for those $x$ near zero).

答:
By deriving $f(x)=\arctan(x)$ with respect to $x$ you get
$$f^\prime (x) =\frac{1}{1+x^2} ,$$
so
$$0 < f^\prime (x) \le 1$$
and, finally, this means
$$|f(x)| = |\arctan(x)| \le |x|.$$
Knowing this, now let $\epsilon >0$, then for all $x$ such that $|x|<\epsilon$, $|\arctan(x)| \le |x| < \epsilon$ just as we wanted; in particular we choose $\delta = \epsilon$.


问:
What are the steps in breaking down the exponent in this limit analysis?
I'm trying to understand the reasoning in the following step of a limit analysis:

$$\lim_{n \to \infty} n\left(\left[1- \frac{1+c}{\frac{n}{\ln(n)}} \right]^{\frac{n}{\ln (n)}}\right)^{(n-1)\ln n/n} = \lim_{n \to \infty} ne^{-[(1+c)\ln(n)]}$$

I understand the "inner" part; $\lim_{n \to \infty} [1-\frac{1+c}{\frac{n}{\ln (n)}}]^{\frac{n}{\ln n}} = e^{-(1+c)}.$ And I sort of see that outer exponent $((n-1)\ln n) /n = (\ln n) - (\ln n / n)$ and the second part goes to 0, but it's not clear to me what rules actually justify "bringing the limit to the exponent". What are the actual steps involved in deducing this limit?

More generally, these types of asymptotic analysis show up in comp sci all the time and I feel there is a bag of tricks that I am missing.

答:
You can't in general replace a sub-expression by its limit while evaluating the limit of a complicated expression. There are ways to do so in very specific circumstances when the sub-expression is a term or a factor in the whole expression, but otherwise this is not allowed.

A better strategy is to take logarithm and thus if $L$ is the desired limit then we have by continuity of logarithm $$\log L=\lim_{n\to\infty} \log n+(n-1) \log\left(1-\frac{(1+c)\log n} {n} \right) $$ and the expression under limit above can be written as $$\log n+n\log\left (1-\frac{(1+c)\log n} {n} \right) - \log\left(1-\frac{(1+c)\log n} {n} \right) $$ Clearly the last term tends to $0$ and hence we get $$\log L=\lim_{n\to\infty} \log n+n\log\left(1-\frac{(1+c)\log n} {n} \right) \tag{1}$$ If $t=(\log n)/n$ then we can write $$\log L=\lim_{n\to \infty} \log n\cdot\frac{t+\log(1-(1+c)t)}{t}$$ The second factor above tends to $-c$ and the fraction can thus be replaced by its limit $-c$ as long as $c\neq 0$.

Thus if $c\neq 0$ we get $$\log L=-\lim_{n\to\infty} c\log n$$ which means that $L=0$ if $c>0$ and $L=\infty$ if $c<0$.

If $c=0$ then we have more work to do. In that case $$\log L=\lim_{n\to\infty} \log n\cdot t\cdot\frac{t+\log(1-t)}{t^2}$$ and last factor tends to $-1/2$ and then we have $$\log L=-\frac{1}{2}\lim_{n\to\infty }  \frac{(\log n) ^2}{n}=0$$ so that $L=1$ in this case.

Most cases of asymptotic analysis during evaluation of limit are nothing more than omission of valid steps (on the assumption that they are obvious to the reader) and may appear as tricks. But in reality they are nothing more than application of limit laws.

The specific treatment in your question is nothing more than applying Taylor series in equation $(1)$ and discarding all terms except the first because all the other terms tend to $0$. Thus we get from $(1)$ $$\log L=\lim_{n\to\infty } \log n-n\cdot(1+c)\frac{\log n} {n} $$ and thus $$L=\lim_{n\to\infty} ne^{-(1+c)\log n} =\lim_{n\to\infty} n^{-c} $$


问:
Coding $\omega$-sequences of 0,1 with $\mathbb{Z}$-sequences of 0,1
This is a question from the talk here on page 115. Define an equivalence relation $\sim$ on ${}^\mathbb{Z}2$ by $f\sim g$ iff $f$ is a constant shift of $g$ (i.e. we are forgetting the origin). What is an example of an injection from ${}^\omega 2$ to ${}^\mathbb{Z}2/\sim$, the equivalence classes of ${}^\mathbb{Z}2$ under $\sim$? Do we need the axiom of choice for this?

答:
Choice is not needed here.

One easy approach is the following: given a $\mathbb{N}$-indexed sequence $\alpha=(a_i)_{i\in\mathbb{N}}$, let $\hat{\alpha}=(\hat{a}_i)_{i\in\mathbb{Z}}$ be defined by $$\hat{a}_i=\begin{cases}
0 & \mbox{ if } i<-1\\
1 & \mbox{ if } i=-1\\
a_i & \mbox{ if } i\ge 0.
\end{cases}$$

The point is that the "$...0001$"-part tells us exactly where the origin is, and so if $\alpha\not=\beta$ we get $\hat{\alpha}\not\sim\hat{\beta}$. (Note that the $i=-1$ clause is important: if we set $\hat{a}_i=0$ for all $i<0$, we would lose injectivity since e.g. any two elements of $^\mathbb{N}2$ with exactly one $1$ would be sent to $\sim$-equivalent sequences.)

Choice is however needed for an injection $^\mathbb{Z}2/{\sim}\rightarrow{}^\mathbb{N}2$.


问:
Evaluating $\lim_{x\to\infty}(\sqrt{x^2+1} - x)(x+1) $
The question is $$\lim_{x\to\infty}(\sqrt{x^2+1} - x)(x+1) $$ I know the answer is $\frac{1}{2}$ and I found it using this equality :

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

But is there any other way to solve this? Any hints would be appreciated.

答:
Just for fun, try letting $x=\tan\theta=\sin\theta/\cos\theta$ with $\theta\to\pi/2^-$, and use

$$\sqrt{\tan^2\theta+1}=\sqrt{\sec^2\theta}=\sec\theta={1\over\cos\theta}$$

so that

$$\begin{align}
(\sqrt{x^2+1}-x)(x+1)
&=\left({1\over\cos\theta}-{\sin\theta\over\cos\theta}\right)\left({\sin\theta\over\cos\theta}+1\right)\\
&={(1-\sin\theta)(\sin\theta+\cos\theta)\over\cos^2\theta}\\
&={(1-\sin\theta)(\sin\theta+\cos\theta)\over1-\sin^2\theta}\\
&={\sin\theta+\cos\theta\over1+\sin\theta}
\end{align}$$

We get

$$\lim_{x\to\infty}(\sqrt{x^2+1}-x)(x+1)=\lim_{\theta\to\pi/2^-}{\sin\theta+\cos\theta\over1+\sin\theta}={1+0\over1+1}={1\over2}$$


问:
Probability that one randomly chosen subset of a finite set is a subset of another
If $α$ and $β$ are subsets of $\{1,\dots,n\}$, what is the probability that $α \subseteq β$, given that $α$ and $β$ are chosen independently of each other and with the same probability.

I know that there are $2^n$ possible subsets of $\{1,\dots,n\}$ (including the empty set and the whole $\{1,\dots,n\}$) and if $|β| = k$ then there are $2^k$ possible subsets of $β$, which can be our $α$.

I suspect (but I'm not 100% sure) that probability space $\Omega$ has a size of $2^{2n}$. If this is true I need to find the number of ways in which $α$ can be a subset of $β$. I'm not really sure how to do this since I don't know the number of elements in $β$.

Can someone perhaps give me a little hint?

I'm so sorry if this problem is too trivial, I'm very much a beginner and I find this to be pretty hard.

答:
Here is a different approach:

Take $A,B$ independent and uniformly random subsets of $\{1,..,n\}$. Since $|A\cup B|$ follows a Binomial$(n,3/4)$ distribution,${}^\dagger$ we have
$$
\Pr[A\cup B = \{1,...,n\}] = (3/4)^n. \tag{1}
$$
(the Binomial statement is even a bit overkill, but true).

Now, set $A=\{1,...,n\}\setminus \alpha$ and $B=\beta$. Then
$
\alpha\subseteq \beta$ if, and only if, $A\cup B = \{1,\dots,n\}$, so
$$
\Pr[\alpha\subseteq \beta] = \Pr[A\cup B = \{1,\dots,n\}] = (3/4)^n. \tag{2}
$$

${}^\dagger$ As every element $k$ independently belongs to $A\cup B$ with probability $1-\Pr[k\notin A]\Pr[k\notin B] = 1-1/2^2$.


问:
Suppose that f is meromorphic in $\mathbb{C}$ such that $|f(z)|< |z|^{k}, \ |z|> R.$ Show that f is rational.
Suppose that f is meromorphic in $\mathbb{C}$ and that there are $ K \in \mathbb{N}$ e positive numbers C,K such that $|f(z)|< |z|^{k}, \ |z|> R.$ Show that f is rational.

To make an argument that has a finite number of points, I was thinking of making a polynomial estimate for the f-module, conclude that it has a finite number of poles inside the closed and compact R-ray ball.
Then multiply the meroforphic function by a polynomial function so it will be correct analytic ? and without poles.
I think it has to extend to every complex plane there will be no poles and it will be limited there uses Liouvile's theorem ... I wonder if my thinking is correct.

答: After multiplying by a polynomial to get rid of the poles, you have an analytic function which is again bounded by some $|z|^m$ for $|z|>R$.  Use Cauchy's formula on circles of large radius to bound the Maclaurin series coefficient of $z^n$ for $n > m$.


问:
Value of $i^\sqrt3$

Find all values of $i^\sqrt3$.

I am trying to apply de Moivre's formula here but cannot find a way to do so. I am not sure if i am approaching this wrong.

答: Hint: by definition, $b^c = \exp(c \log b)$.  Use all values of $\log b$.

Title: Value of $i^\sqrt3$


问:
Concrete Mathematics Exercise 3.22 Solution Explanations
Evaluate the sums $$S_n=\sum_{k\geqslant1}\lfloor n/2^k+1/2\rfloor$$

The answer key stated that for n and n-1 all the terms would be equal except for k where $$n=2^{k-1}q$$ and q is odd. This seems to me it comes from nowhere.

From high level I understand that there should be some critical $k$ value that will cause $S_n$ and $S_{n-1}$ different and that will be the only differences as well between adjacent terms. But I can get the details sorted. Could someone give me a hint here? Thanks.

答:
Assuming $n$ is an integer.

$n = 2^{k-1}q$ means that, in $S_n$ the $k$th term is

$$\left\lfloor\frac{n}{2^k} + \frac12\right\rfloor = \left\lfloor\frac{q}{2} + \frac12\right\rfloor = \left\lfloor\frac{q+1}{2}\right\rfloor $$

while in $S_{n-1}$, the corresponding term is

$$\left\lfloor\frac{n-1}{2^k} + \frac12\right\rfloor  = \left\lfloor\frac{q+1}{2} - \frac 1{2^k}\right\rfloor $$

$\frac{q+1}2$ is an integer, so these two floors are different.

For terms before the $k$th one, call that term $1\le k'<k$. In $S_n$,

$$\left\lfloor\frac{n}{2^{k'}} + \frac12\right\rfloor  = \left\lfloor2^{k-1-k'}q + \frac 1{2}\right\rfloor$$

In $S_{n-1}$,

$$\left\lfloor\frac{n-1}{2^{k'}} + \frac12\right\rfloor  = \left\lfloor2^{k-1-k'}q + \frac 1{2}-\frac{1}{2^{k'}}\right\rfloor$$

$2^{k-1-k'}q$ is an integer, and $\frac12 \ge \frac{1}{2^{k'}}$, so the two floors are the same.

For terms after the $k$th one, call that term $k'' > k$. In $S_n$,

$$\left\lfloor\frac{n}{2^{k''}} + \frac12\right\rfloor  = \left\lfloor\frac{q}{2^{k''-k+1}} + \frac 1{2}\right\rfloor = \left\lfloor\frac{q+2^{k''-k}}{2^{k''-k+1}}\right\rfloor$$

The numerator is odd and the denominator is even, so the fraction inside the floor is not an integer. Its fractional part is at least $\frac{1}{2^{k''-k+1}}$:

$$\begin{align*}
\frac{q+2^{k''-k}}{2^{k''-k+1}} &> a&a\in\mathbb Z\\
q+2^{k''-k} &> 2^{k''-k+1}a\\
q+2^{k''-k} &\ge 2^{k''-k+1}a + 1\\
\frac{q+2^{k''-k}}{2^{k''-k+1}} -a&\ge \frac{1}{2^{k''-k+1}}\\
\end{align*}$$

While in $S_{n-1}$,

$$\left\lfloor\frac{n-1}{2^{k''}} + \frac12\right\rfloor
 = \left\lfloor\frac{q}{2^{k''-k+1}} + \frac 1{2}-\frac{1}{2^{k''}}\right\rfloor
 = \left\lfloor\frac{q+2^{k''-k}}{2^{k''-k+1}}-\frac{1}{2^{k''}}\right\rfloor$$

The least fractional part mentioned above $\frac{1}{2^{k''-k+1}}\ge \frac1{2^{k''}}$, so the two floors are the same.


问:
Given a string of length $n$, how many substrings of length $k$ of consecutive items exist?
You have a string of characters {ABCDE}.

You want to know how many substrings of consecutive characters that are three digits long exist.

So you have:

{ABC},{BCD},{CDE}

So my question is: how do I expand this to any arbitrary arguments for the string length and the length of each substring?

答:
Thanks @QiaochuYuan for giving me the answer. After thinking about it for a bit I've come to the proof:

The number of possible substrings is equal to the number of starting locations:

Possible substrings = Number of starting locations

The number of starting locations is equal to the total number of locations ($n$) minus the number of impossible starting locations.

Number of starting locations = $n$ - Number of impossible starting locations

The number of impossible starting locations is $k-1$. This is because the last possible substring occurs at the $(n-k)$th digit, because $(n-k) + k = n$, and you cannot exceed the value of $n$.

So the total number of possible substrings of length $k$ formed of consecutive characters of a string of length $n$ is:

$n - (k -1) = n - k + 1$


问:
Is there a way to characterize all nowhere dense sets of $\mathbb{R}$?
I'm reading Functional Analysis from Joseph Muscat's book. Muscat asks the following question in an exercise:

What are the nowhere dense sets in $\mathbb{R}$?

He gives the hint that every open set in $\mathbb{R}$ is a countable union of disjoint open intervals.

By definition, closure of nowhere dense sets cannot contain a open ball which is an open interval in $\mathbb{R}$. Now, how can I use the hint further characterize nowhere dense sets in $\mathbb{R}$?

I do not need a complete answer. A nudge in the right direction will be appreciated!

The question as given in the book:

答:
Frankly, this is bizarre to me. The reference to the prior exercise showing that open sets in $\mathbb{R}$ are countable disjoint unions open intervals suggests that the desired classification is the following:

In $\mathbb{R}$, $A$ is nowhere dense iff $\overline{A}$ does not contain any nonempty open interval.

This is a trivial application of the prior exercise. However, the author has marked this exercise as particularly difficult (the "$*$" preceding it). Moreover, the "result" above doesn't even use the interesting part of the prior exercise! All it uses is that every open set contains a nonempty open interval, which is basically trivial. So this feels wrong. On the other hand, I see no other interpretation of the problem which makes sense.

For what it's worth, in my opinion there is no good classification of nowhere dense sets in $\mathbb{R}$. There are simply too many of them: since every subset of a nowhere dense set is nowhere dense, and there are nowhere dense sets of size continuum (e.g. the Cantor set), the number of nowhere dense subsets of $\mathbb{R}$ is as large as it could be: $2^{2^{\aleph_0}}$. By contrast, there are only $2^{\aleph_0}$-many open sets in $\mathbb{R}$. (This is a good exercise; the first step is to show that every open set in $\mathbb{R}$ can be written as a not-necessarily-disjoint union of countably many open intervals with rational endpoints, and that's worth doing even if you're not interested in cardinality.)

Moreover, nowhere dense sets need not be small in all senses. In particular, there are nowhere dense sets which have positive Lebesgue measure (e.g. the "fat Cantor sets"). In general, in contrast with null sets nowhere dense sets are not "ignorable" in analysis.


问:
Does a function exist that maps the natural numbers to the natural numbers excluding multiples of 2 and 3 and starting at an arbitrary point?
This question is a generalization of this other question. To recap, I was looking for a function that mapped the set of natural numbers to the set of natural numbers excluding multiples of 2 or 3 or both. So a function $f: x \mapsto y$ where $x = \{0, 1, 2, 3,...,m\}$ and $y=\{1,5,7,11,13,17,...,n\}$ for some arbitrary limit $m$. @A.J. gave a great solution

$y=\qquad 6\left \lceil \dfrac{x}{2} \right \rceil + (-1)^x$

(Erratum: the independent set $y$ should have been $y=\{5,7,11,13,...,n\}$ however since an answer was given for $y=\{1,5,...,n\}$ I didn't want to change the original and the difference is trivial.)

Now I'm wondering about a generalization that allows for the set $y$ to start at an arbitrary point (that is not a multiple of 2 nor 3). For example instead of $x=0$ mapping to $y=1$ it would instead map to $y=5$ and the formula would then be $y=6\left \lceil \dfrac{x+3}{2} \right \rceil + (-1)^{x+3}$ in other words translated 3 units. Or if we wanted $x=0$ to may to $y=41$ then it would be $y=6\left \lceil \dfrac{x+13}{2} \right \rceil + (-1)^{x+13}$. But I only know this because I precomputed the set starting at 1 and translated the number per the number of skipped members. If the number is bigger this would be non-trivial. For example, $101$ is not a multiple of 2 nor 3 so what function would start with $x=0$ mapping to $y=101$ and then continue by skipping multiples of 2 and 3 e.g. $103, 107, 109$.

Is there a generalization for a function that given the "starting point" $a$ that is not a multiple of 2 nor 3, and the parameter $x$, it maps $x=\{0,1,2,3,...,m\}$ to $y$ where each member of $y$ is $\ge$ $a$ and each subsequent term is the next non-multiple of 2 nor 3?

答:
The additional difficulty in starting at an arbitrary $a$ which is not a multiple of $2$ or $3$ is that we have to consider two cases: either $a \equiv 1 \bmod 6$ or $a \equiv 5 \bmod 6$. The first case is just a translation of the solution from the earlier question; to handle the second case, we first consider the base case $\{0,1,2,3,4,...,m\} \longrightarrow \{-1,1,5,7,11,\ldots,n \}$.

[Strictly speaking our base case should use $\{y\} = \{5,7,11,13,17,...,n\}$ since you want $\{y\}$ to be a subset of the natural numbers, but starting at $-1$ will give a slightly simpler formula.]

Using the same approach as last time, we note the set $\{y\}$ can be written as

$$y=\{6(0)-1,6(0)+1,6(1)-1,6(1)+1,6(2)-1,6(2)+1,...\}$$

We first need a function to map $\{0,1,2,3,4,5,...\} \longrightarrow \{0,0,1,1,2,2,...\}$, which could be

$$\left \lfloor \dfrac{x}{2} \right \rfloor \quad \text{or} \quad \left \lceil \dfrac{x-1}{2} \right \rceil$$

Then we need to map $\{0,1,2,3,4,5,6,...\} \longrightarrow \{-1,1,-1,1,-1,1,-1,...\}$, so we use

$$(-1)^{x-1}$$

giving us the functions

$$y=6\left \lfloor \dfrac{x}{2} \right \rfloor + (-1)^{x-1} \qquad \text{or} \qquad 6\left \lceil \dfrac{x-1}{2} \right \rceil + (-1)^{x-1}$$

Now we can find a function for any starting value $a$ by a suitable translation. If $a \equiv 1 \bmod 6$, we can use the function from the earlier question and translate by $a-1$; if $a \equiv 5 \bmod 6$, we use one of the above functions and translate by $a+1$. Thus we have

$$f(x) = \begin{cases}
6\left \lceil \dfrac{x}{2} \right \rceil + (-1)^{x} + a-1& \quad a \equiv 1 \bmod 6 \\
\\
6\left \lfloor \dfrac{x}{2} \right \rfloor + (-1)^{x-1} + a+1& \quad a \equiv 5 \bmod 6
\end{cases}$$


问:
$\lim_{x \to \infty} \frac {\log(|\sin(x)|)}{\log(x)}$
Does $\lim_{x \to \infty} \frac {\log(|\sin(x)|)}{\log(x)}$ exist considering $\sin(x) = 0$ When $x=n\pi$?

答: The limit inf is $-\infty$ and obtained when $\sin=0$, while the limit sup is $0$ and is obtained when $\sin=1$. The limit does not exist because the two limits are not equal.


问:
An example where great-circle distance gives a different answer than $L_2$-norm
Given three 3D vectors A B C, when does great-circle distance tell us A and B are closer, but $L_2$ norm tells us A and C are closer?

Also, any geometric intuition would be great.

答:
No such example can be found. Consider that if we're on the unit circle, the great-circle distance between unit vectors $U$, $V$ is just the angle between them in radians: $$0 \leq d_{\bigodot}(U, V) = \theta(U, V) \leq \pi.$$

The $L^2$-distance between them, $d_{L^2}(U, V) = \| U - V \|$,  can be found using elementary trig as $$0 \leq d_{L^2}(U, V) = 2 \sin \left( \frac{\theta(U, V)}{2} \right) = 2\sin \left( \frac{d_{\bigodot}(U, V)}{2} \right) \leq 2.$$

On  the interval $[0, \pi]$, the function $f(\theta) = 2\sin(\theta/2)$ is an increasing bijection onto its image $[0, 2]$,  so if $d(A, B) > d(A, C)$ in one metric, it's also bigger in the other metric as well. In fact, these metrics are strongly equivalent, since $$\|U - V \| \leq \theta(U, V) \leq \frac{\pi}{2} \|U - V\|$$ for all unit vectors $U, V \in S^2$ (i.e. $x \leq 2 \arcsin(x/2) \leq \pi x/2$ for all $x \in [0, 2]$).


答:
The de Moivre's formula is useful when you deal with a complex number of the form $e^{ix}$ ($x\in\mathbb{R}$).

When you have exponentiation of a complex number, what you should look for first is the definition, particularly when you have $z^w$ where $z$ is a complex number and $w$ is not an integer.

In your case,
$$
i^{\sqrt{3}}:=\exp(\sqrt{3}\log(i))
$$
where $\log$ denotes the complex logarithm, which is a multi-valued function.

Title: Value of $i^\sqrt3$

Body:
Find all values of $i^\sqrt3$.

I am trying to apply de Moivre's formula here but cannot find a way to do so. I am not sure if i am approaching this wrong.


问:
Evaluate $\int_0^{\pi} \frac{\sin^2 \theta}{(1-2a\cos\theta+a^2)(1-2b\cos\theta+b^2)}\mathrm{d\theta}, \space 0<a<b<1$
Evaluate by complex methods

$$\int_0^{\pi} \frac{\sin^2 \theta}{(1-2a\cos\theta+a^2)(1-2b\cos\theta+b^2)}\mathrm{d\theta}, \space 0<a<b<1$$

Sis.

答:
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
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 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
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\begin{align}
&\bbox[5px,#ffd]{\left.\int_{0}^{\pi}{\sin^{2}\pars{\theta} \over \bracks{1 - 2a\cos\pars{\theta} + a^{2}}
\bracks{1 - 2b\cos\pars{\theta} + b^{2}}}\,\dd\theta
\,\right\vert_{\, 0\ <\ a\ <\ b\ <\ 1}}
\\[5mm] = &\
{1 \over 2}
\int_{-\pi}^{\pi}{\pars{\expo{\ic\theta} - \expo{-\ic\theta}}^{2}/\pars{-4} \over
\pars{a - \expo{\ic\theta}}\pars{a - \expo{-\ic\theta}}
\pars{b - \expo{\ic\theta}}
\pars{b - \expo{-\ic\theta}}}\,\dd\theta
\\[5mm] = &\
-{1 \over 8}\,\Re
\int_{-\pi}^{\pi}{\pars{z - z^{-1}}^{2} \over
\pars{a - z}\pars{a - z^{-1}}
\pars{b - z}
\pars{b - z^{-1}}}\,{\dd z \over \ic z}
\\[5mm] = &\
-{1 \over 8ab}\,\Im
\int_{-\pi}^{\pi}{\pars{z^{2} - 1}^{2} \over
z\pars{z - a}\pars{z - a^{-1}}
\pars{z - b}
\pars{z - b^{-1}}}\,\dd z
\end{align}
The integrand has simple poles at $\ds{z \in \braces{0,a,b}}$ with
$$
\left\{\begin{array}{rcl}
\ds{\on{Res}_{\, z\ =\ 0}} & \ds{=} & \ds{\phantom{-}1}
\\[1mm]
\ds{\on{Res}_{\, z\ =\ a}} & \ds{=} &
\ds{-{\pars{1 - a^{2}}b \over \pars{b - a}\pars{1 - ab}}}
\\[1mm]
\ds{\on{Res}_{\, z\ =\ b}} & \ds{=} &
\ds{\phantom{-}
{a\pars{1 - b^{2}} \over \pars{b - a}\pars{1 - ab}}}
\end{array}\right.
$$
Then,
\begin{align}
&\bbox[5px,#ffd]{\left.\int_{0}^{\pi}{\sin^{2}\pars{\theta} \over \bracks{1 - 2a\cos\pars{\theta} + a^{2}}
\bracks{1 - 2b\cos\pars{\theta} + b^{2}}}\,\dd\theta
\,\right\vert_{\, 0\ <\ a\ <\ b\ <\ 1}}
\\[5mm] = &\
-{1 \over 8ab}\Im\bracks{2\pi\ic\pars{\on{Res}_{\, z\ =\ 0}\ +\
\on{Res}_{\, z\ =\ a}\ +\ \on{Res}_{\, z\ =\ b}}}
\\[5mm] = &\
\bbx{\pi/2 \over 1 - ab} \\ &
\end{align}


问:
Monotone Convergence Property $\iff$ Least Upper Bound Property
Let $\mathbb{F}$ be an arbitrary ordered field that satisfies Monotone Convergence Property then $\mathbb{F}$ is Order Complete.

Order Completeness Property:

If $S \subset \mathbb{F}$ is bounded above, then $\exists \ c \in \mathbb{F}$ that is an upperbound of S and for every upperbound $b$ of S, we have $c \leq b.$

Monotone Convergence Property

Every monotone bounded sequence in $\mathbb{F}$ is convergent in $\mathbb{F}$.

I am not posting anything new. The answers were given in parts
here
and
here.

I have just brought it together for my own future reference.

答:
Monotone Convergence Property $\implies$ Archimedean Property.

Let's assume $\mathbb{F}$ is not archimedean,  $\exists$ $c \in \mathbb{F}$ with $n<c$
for all $n \in \mathbb{N}_{\mathbb{F}}$.

By assumption $(1,2,...)$ must converge,say to $r \in \mathbb{F}$.
$\implies$ $(0,1,2,...)$ also converges to $r$.

Subtracting the two sequence we find (1,1,1,...) converges to 0, which is absurd.

Therefore $\mathbb{F}$ must be Archimedean.

Let $(\phi \ne)A \subseteq \mathbb{F}$ which is bounded above in $\mathbb{F}$, with $U$ being the set
of upperbounds of $A$ in $\mathbb{F}$.

Want: To find an lub for $A$.

We now move on to build a few tools in order to prove the existence of $\operatorname{lub} A$.

Claim 1: $\{u-\varepsilon: u \in U\}=:U- \varepsilon \nsubseteq U$,
for all $\varepsilon (>0) \in \mathbb{F}$.

Let $\varepsilon>0$.

Let $U-\varepsilon \subseteq U$, Now we use induction on $n \in \mathbb{N}_{\mathbb{F}}$.

If $U-n\varepsilon \subseteq U$ for some $n \in \mathbb{N}_{\mathbb{F}}$.

then $U-(n+1)\varepsilon=(U-\varepsilon)-n\varepsilon$ $\subseteq U-n\varepsilon$ $\subseteq U$.
$\implies$ $U-n\varepsilon \subseteq U$ for all $n \in \mathbb{N}_{\mathbb{F}}$.

Hence by Archimedean Property we have $\mathbb{F}= \bigcup_{n=1}^\infty U-n\varepsilon \subseteq U$,

which contradicts $A \neq \phi$.

Claim 2: $\bigcap_{n=1}^\infty U-\frac{1}{n} \subseteq U$.

Let $x \in \bigcap_{n=1}^\infty U-\frac{1}{n}$ and let $(x<)y \in \mathbb{F}$.

By archimedean property $\exists$ $n \in \mathbb{N}_{\mathbb{F}}$ such that $x+\frac{1}{n} <y$.

$x \in U-\frac{1}{n}$\pause $\Rightarrow$ $x+\frac{1}{n} \in U$\pause $\Rightarrow$ $(x+\frac{1}{n}<)y
   \notin A$.

$y$ is chosen arbitrarily, so $x$ is an upperbound of A, i.e $x \in U$.

Claim 3: $U-\frac1n$ $\subseteq$ $U-\frac1m$ for all $m \le n$.

$m \le n \Rightarrow \frac1n \le \frac1m$.

$x \in U-\frac1n$ $\Rightarrow$ $x+\frac1n \le x+\frac1m$ $\Rightarrow x+\frac1m \in U$ $\Rightarrow x
   \in U-\frac1m$.

Corollary:

Let $n_k$ be any increasing sequence in $\mathbb{N}_{\mathbb{F}}$ then
$\bigcap_{k=1}^\infty \left(U-\frac{1}{n_k}\right)$ =
$\bigcap_{n=1}^\infty \left(U-\frac{1}{n}\right) \subseteq U.$

Claim 4: If $x \in \mathbb{F}$ with $x \notin U-\frac1n$, then
$x<y$ for all $y \in U-\frac1n$.

Let $u-\frac1n<x$ for some $u \in U.$

Then $x-(u-\frac1n)>0$ $\implies u+x-(u-\frac1n) \in U$
$\implies x \in U-\frac1n$, contradiction.

Hence $x$ is less than every member of $U-\frac1n$.

Now we proceed in creating an increasing sequence so that we can apply Monotone Convergence Property.

Let $x_1\in (U-1) \setminus U$ ($\ne \phi$ by Claim 1)

$x_1\notin U$ $\Rightarrow$ $x_1 \notin \bigcap_{n=1}^\infty U-\frac{1}{n}$.(from Claim 2)

so $\exists$ $n_1>1$ such that $x_1 \notin U-\frac{1}{n_1}$.

Let $x_2 \in \left(U-\frac{1}{n_1}\right) \setminus U$.

Then $\exists$ $n_2>n_1$(from Claim 3)

such that $x_2 \notin U - \frac{1}{n_2}$.

Again consider $x_3 \in \left(U-\frac{1}{n_2} \setminus
   U\right)$ and so on.

This yields an increasing sequence $(1,n_1,n_2,...)$ in $\mathbb{N}_{\mathbb{F}}$(from Claim 3)
and an increasing sequence $(x_k)$ in $\mathbb{F}$

(from Claim 4,3) such that

$x_k \in \left(U-\frac{1}{n_{k-1}}\right) \setminus U.$

Then $(x_k)$ is bounded above by each element of U.

$\implies$ $(x_k)\rightarrow x \in \mathbb{F}$.(M.C.P)

$\implies$ $x \le u$ for all $u \in U$.

Remains to show that $x \in U$

Let $x_k \leq x$ for all k and $x_k \in \left(U-\frac{1}{n_k}\right)$

$\implies x \in \bigcap_{k=1}^\infty \left(U-\frac{1}{n_k}\right) = \bigcap_{n=1}^\infty \left(U-
   \frac{1}{n}\right) \subseteq U$, i.e $x \in U$.

Therefore $x$ is the smallest element in $U$, which means $x=\operatorname{lub}A$.\

Hence for any nonempty bounded subset of $\mathbb{F}$, $\operatorname{lub} A \in \mathbb{F}$

i.e $\mathbb{F}$ is Order Complete.


问:
Difference of two powers where a term is multiplied by some constant

We know that the difference of two powers $x^n-y^n$ over its linear factor $x-y$ is

$$\frac{x^n-y^n}{x-y}=\sum_{j=0}^{n-1}(x^{n-j-1}y^{j}),$$

where $n\in\mathbb{N}$ and all $x,y\in\mathbb{F}, \ \mathbb{F}$ being one of the standard number systems (i.e. $\mathbb{Z},\mathbb{Q},\mathbb{R},\text{and} \ \mathbb{C}$).

I was wondering if there is a similar theorem for when one of the terms is multiplied by some constant $a$. Is there a theorem that states $ax^n-y^n$ over $ax-y$ is equal to some summation along the lines of

$$ax^n-y^n=(ax-y)\sum_{j=0}^{n-1}???$$

I have tried figuring this out myself, but I have no formal training in mathematics beyond a high school level and am only in the beginning stages of learning number theory.

Without knowing a better route to take, I attempted to get the first few terms of the summation by long division:

$$\frac{ax^n-y^n}{ax-y}=x^{n-1}+\frac{x^{n-2}y}{a}+\frac{x^{n-3}y^2}{a^2}+\ .\ .\ .\ $$

from which I wrote a summation that seems to match:

$$\frac{ax^n-y^n}{ax-y}=\sum_{j=0}^{n-1}\frac{x^{n-j-1}y^{j}}{a^{j}}.$$

However, I am unable to prove this statement to be true or false.

答:
$$ax^n-y^n=(ax-y)\sum_{j=0}^{n-1}???$$

If such identity holds, then when $y=ax$, the expression vanishes.

That is we would require $$ax^n-(ax)^n=0$$

and we have $$a=a^n$$

This can't hold true for all the $a$.

We can however view $ax^n-y^n=(a^\frac1n x)^n -y^n$.


问:
Show that $d/dx (a^x) = a^x\ln a$.
Show that
$$
\frac{d}{dx} a^x = a^x \ln a.
$$

How would I do a proof for this.  I can't seem to get it to work anyway I try.

I know that
$$
\frac{d}{dx} e^x = e^x.
$$
Does that help me here?

答: Hint: $a^x=e^{\ln a^x}=e^{x\ln a}$.

Title: Show that $d/dx (a^x) = a^x\ln a$.

Body: Hint: $a^x=e^{\ln a^x}=e^{x\ln a}$.


问:
How does this sum go to $0$?
http://www.math.chalmers.se/Math/Grundutb/CTH/tma401/0304/handinsolutions.pdf

In problem (2), at the very end it says

$$\left(\sum_{k = n+1}^{\infty} \frac{1}{k^2}\right)^{1/2} \to 0$$

I don't see how that is accomplished. I understand the sequence might, but how does the sum $$\left ( \frac{1}{(n+1)^2}+ \frac{1}{(n+2)^2} + \frac{1}{(n+3)^2} + \dots \right)^{1/2} \to 0$$?

Is one allowed to do the following?

$$\lim_{n \to \infty}\left ( \frac{1}{(n+1)^2}+ \frac{1}{(n+2)^2} + \frac{1}{(n+3)^2} + \dots \right)^{1/2} \to 0$$

$$=\left (\lim_{n \to \infty}\frac{1}{(n+1)^2}+ \lim_{n \to \infty}\frac{1}{(n+2)^2} + \lim_{n \to \infty}\frac{1}{(n+3)^2} + \dots \right)^{1/2} \to 0$$

$$(0 + 0 + 0 + \dots)^{1/2} = 0$$

答:
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 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{%
\pars{\sum_{k = n + 1}^{\infty}\ {1 \over k^{2}}}^{1/2}} =
\pars{\sum_{k = 1}^{\infty}\ {1 \over k^{2}} -
\sum_{k = 1}^{n}\ {1 \over k^{2}}}^{1/2}
\\[5mm] = &\
\braces{\zeta\pars{2} -
\bracks{-{1 \over n} + \zeta\pars{2} + 2\int_{n}^{\infty}{\braces{x} \over x^{3}}\dd x}}^{1/2}
\end{align}
where I used a zeta function identity. Moreover,
$$
0 < 2\int_{n}^{\infty}{\braces{x} \over x^{3}}\dd x <
2\int_{n}^{\infty}{\dd x \over x^{3}} = {1 \over n^{2}}
$$
Then,
\begin{align}
&\bbox[5px,#ffd]{%
\pars{\sum_{k = n + 1}^{\infty}\ {1 \over k^{2}}}^{1/2}} =
\bracks{{1 \over n} - 2\int_{n}^{\infty}{\braces{x} \over x^{3}}\dd x}^{1/2}
\\[5mm] &\
\bbx{\to \color{red}{\large 0}\ \mbox{as}\ n \to \infty}
\\ &
\end{align}

Title: How does this sum go to $0$?


问:
Cantor diagonal process to extract a subsequence of continuously differentiable functions
Let $\left(f_{h}\right)_{h} \subset \mathcal{C}^{1}([a, b]) . $ Suppose we have by assumption the sets of functions
$$
\left\{f_{h}: h \in \mathbb{N}\right\} \quad and \quad\left\{f_{h}^{\prime}: h \in \mathbb{N}\right\}
$$
relatively compact in $\left(\mathcal{C}^{0}([a, b]),\|\cdot\|_{\infty}\right)$ Then $\exists\left(h_k\right)_{k}$ increasing sequence of integers and $\exists g_{1}, g_{2} \in \mathcal{C}^{0}([a, b])$ such that
$$
f_{h_{k}} \longrightarrow g_{1} \text { for } k \rightarrow+\infty \text { in }\left(\mathcal{C}^{0}([a, b]),\|\cdot\|_{\infty}\right)
$$
and
$$
f_{h_{k}}^{\prime} \longrightarrow g_{2} \text { for } k \rightarrow+\infty \text { in }\left(\mathcal{C}^{0}([a, b]),\|\cdot\|_{\infty}\right)
$$

What I have is that the $
f_{h_{k}}^{\prime}$ now are actually not the derivatives of the $
f_{h_{k}}$ and my goal is to be able to extract from the sequences $
f_{h_{k}}$ a subsequence such that the $
f_{h_{k}}^{\prime}$ of my new subsequence are actually their derivatives. My goal is to apply the Cantor diagonal procedure on a dense set of $[a,b]$, but I have difficulties in formalizing it. Thanks in advance!

答:
No need for diagonal procedure. First pick a  subsequence along which $(f_h)$ converges in the norm. Look at the derivatives along this subsequence. There is  a further subsequence along which $(f_h')$ converges. You now have one subsequence along which both $(f_h)$ and $(f_h')$ converge.

Now use the fact that $f_{h_k} (x)=f_{h_k} (c)+\int_c^{x} f_{h_k}' (y)dy$ to conclude that $g_1$ is differentiable and its derivative is $g_2$.


问:
Iteratively solving (sparse) homogeneous linear systems
Solving (sparse) non-homogeneous linear systems can be done iteratively, by using the LSQR algorithm or similar. However, in the homogeneous case, we have $Ax =0$, where we typically want to find the non-zero solutions which is generally done by demanding a linear constraint on $x$, e.g. $|x|=1$. The method usually used to solve this constrained homogeneous system is by performing a singular value decomposition, and choosing the vector corresponding to the smallest singular value.

I'm looking for an iterative equivalent of LSQR that may be applied to sparse rectangular systems. LSQR does not seem appropriate, as it's stages implicitly assume $|b|\neq0$. Is anyone aware of such a method or can think of a method to modify LSQR appropriately?

答:
You could use LSQR, by setting $b$, very close to zero (not exactly zero), which prevents, the trivial solution $(x = 0)$, In practice, it's a quite good approximation.

$Ax = \varepsilon$ ,

where $\varepsilon \approx 0$

ex. $\varepsilon = 10^{-7}$

In case of your application, requires an exact zero solution, then you might need to involve $m$ inequality constraints, $x > 0$, which could be solved easily using Lagrangian multipliers.


问:
Showing the solution to a differential equation is periodic
I have an initial value problem for which I have to show that the the function is periodic. The definition and lemmas I can use is:

We say a function $f$ is periodic if $f(t) = f(t + T)$ for some period $T$

Given $f$ which is continuous and periodic with period $T$. Then for any $a \in \mathbb{R}$, $\int_a^{a+T} f(t) dt = \int_0^T f(t) dt$

The given differential equation is $y' + cy = cos(t)$ with inital condition $y(0) = b$. Im asked to find its solution and show that $y(t)$ is periodic if and only if $b = \frac{c}{c^2 + 1}$. I got $y(t) = \frac{sin (t) + c cos (t)}{c^2 + 1} + \frac{d}{e^{ct}}$ where $d$ is an arbitrary constant. Solving at $y(0) = b$ I get $b = \frac{c}{c^2 + 1} + d$. How do I go about showing its periodic? Im almost there but i dont quite know what to do next.

答:
The proof of this fact only requires that you recognize that $\sin(t)=\sin(t+T)$ and $\cos(t)=\cos(t+T)$, i.e. they're periodic.

The $\Leftarrow$ direction is like DuduBob is saying, if $d=0$ then $b=\cfrac{c}{c^2+1}$ and so $y(t)=e^{-ct}(b-b)+\cfrac{1}{c^2+1}(\sin(t)+c\cos(t))=\cfrac{1}{c^2+1}(\sin(t)+c\cos(t))$. The solution is periodic since $\sin(t)$ and $\cos(t)$ are periodic on their own. The sum of periodic functions is again periodic.

Edit: As @Ian mentions, this is only true when the periods are commensurable, that is the ratio is a rational number. But in this case, $\sin(t)$ and $\cos(t)$ share the same period $T$, so the ratio is in fact $1$.

The $\Rightarrow$ direction is obtained by subtracting $y(t+T)-y(t):$

$y(t+T)-y(t)=(e^{-c(t+T)}-e^{-ct})d=0$. It equals $0$ since by assumption, $y(t)=y(t+T)$. Since $e^{-c(t+T)}-e^{-ct}\neq0$ (it is an exponential), then it follows that $d=0$