问:
markov chain application
Two workers handle three machines(i.e. we can at most repair two machines at a time). The time until the machine breaks down is exponentially distributed with expectation value 1/2 and independent of other repair and break down times. Every repair time is exponentially distributed with expectation value 1/3 and independent of other repair and break down times. Let $X_t$ be the amount of broken down machines at time $t$.

Determine the birth and death frequencies in the birth and death chain $(X_t)_{t\geq 0}$.

Determine $\lim_{t \to \infty} \mathbb{P}_i (X_t=0)$ for i=0,1,2,3.

I need some explain aboute the question like this for compute the birth and death rate
as a markov chain in continues time

答:
The system can be in three states $\{0,1,2,3\}$ which correspond to the number of working machines. A single machine is repaired at rate 3, so when both workers are repairing they repair at rate 6 (but only when two or more machines are broken). Machines break at rate 2 independently so when 2 or 3 are active the failure rates are 4 and 6 respectively.

Therefore the $Q$ matrix is
$$ Q = \begin{pmatrix}
 -6 & 6 & 0 & 0 \\
 2 & -8 & 6 & 0 \\
 0 & 4 & -7 & 3 \\
 0 & 0 & 6 & -6
\end{pmatrix}$$

and the stationary distribution (the limiting distribution as $t\rightarrow \infty$) is
$$ \pi = \begin{pmatrix} \frac{4}{43} & \frac{12}{43} & \frac{18}{43} & \frac{9}{43} \end{pmatrix}.$$

Title: markov chain application


问:
Morphism of admissible $1$-dimensional $(\varphi,N)$-module
In these notes in section 7.3  admissible one-dimensional $(\varphi,N)$-modules are classified. Let $K$ be a $p$-adic field. Then for any $\lambda\in K_0$, they give an element of $MF_K^{ad}(\varphi,N)$ called $D_\lambda$, whose Frobenius is $\varphi=\lambda \sigma$, the monodromy operator is $N=0$, and $$Fil^r(D_{\lambda}\otimes_{K_0} K)=\begin{cases} D_{\lambda}\otimes_{K_0} K&, r\leq v_p(\lambda)\\
0&, r>v_p(\lambda)\end{cases}.$$
Then a morphism $D_\lambda\to D_\mu$ is given by a $K_0$-linear morphism given by multypling $u\in K_0^\times$ such that
$$\lambda=\frac{\sigma(u)}{u}\mu.$$
However, in the notes it is claimed that $u$ actually has to lie in $\mathcal{O}_{K_0}^{\times}$ and I don't understand why. Clearly I'm missing something and I'd appreciate any help.

答:
I admit the way it is phrased in the notes seems a bit ambiguous / confusing, but the actual statement there is that there is an isomorphism $D_\lambda \simeq D_\mu$ if and only if there exists a $u \in \mathcal{O}_{K_0}^\times$ such that $\mu = \dfrac{\sigma(u)}{u} \lambda$.

Now if $D_\lambda \simeq D_\mu$, then indeed a priori there is just a $\tilde u \in K_0^\times$ such that the iso is given by multiplication with $\tilde u$. But now note that $p$ is a prime element of $\mathcal{O}_{K_0}$ (which is totally unramified), i.e. $\tilde u = u \cdot p^n$ for some $n \in \mathbb Z$ and $u \in \mathcal{O}_{K_0}^\times$, and because $p \in \mathbb Z$ is fixed by $\sigma$ we have$\dfrac{\sigma( \tilde u)}{\tilde u} = \dfrac{\sigma(u)}{u}$.

So indeed there exists a $u \in \mathcal{O}_{K_0}^\times$ such that ..., it's just that the iso is not necessarily given by multiplication with that $u$, but by multiplication with some $u \cdot p^n$.

ContentLicense: CC BY-SA 4.0


问:
$\mathrm{int}( \mathrm{c\ell}(U)) = U$ implies every open set is closed
Can someone provide a proof or counterexample to the following conjecture?

Conjecture: if $X$ is a topological space such that every open set $U \subseteq X$ satisfies $\mathrm{int}( \mathrm{c\ell}(U)) = U$, then every open set is closed (i.e. clopen).

Context: for any topological space $X$, its lattice of open sets forms a "model" of intuitionistic propositional logic. We interpret propositions as open sets $U$, $\land$ as intersection, $\lor$ as union, $\varnothing$ as $\bot$ and $X$ as $\top$. Implication is defined $U \to V = \mathrm{int}( U^\complement \cup V)$ - see this nLab article - and negation is defined $\lnot U = U \to \bot = \mathrm{ext}(U)$.

It is known that, over intuitionistic logic, the law of excluded middle $P \lor \lnot P$ and double negation elimination $\lnot \lnot P \to P$ are equivalent. Saying $U \cup \lnot U = X$ for an open set $U \subseteq X$ is equivalent to saying $U$ is closed (clopen). Meanwhile, saying that $\lnot \lnot U = U$ is equivalent to saying $\mathrm{int}( \mathrm{c\ell}(U)) = U$.

Thus, I was trying to prove the equivalence of $\mathsf{LEM}$ and $\mathsf{DNE}$ using this topological interpretation. The forward direction is easy enough, but I couldn't prove the reverse easily, which is the conjecture above.

答:
As mentioned in my comment above, it is perhaps worth posting here a very succinct argument which establishes the OP's conjecture.

Proposition. If all the open subsets of the space $(X, \mathscr{T})$ are regular then they are all also closed (the topology is clopen).

Proof. For every $x \in X$ let us write $Z_x=\mathring{\overline{\{x\}}} \in \mathscr{T}$. By dualising via complementation the given hypothesis of regularity -- according to which $\mathring{\overline{U}}=U$ for any open subset $U \in \mathscr{T}$ -- we gather that $\overline{\mathring{F}}=F$ for any closed subset $F \subseteq X$. Applying this to the particular case of the closed subset $\overline{\{x\}}$ we infer that $\overline{Z_x}=\overline{\{x\}}$.

In order to proceed with this argument it is also essential that we establish the nontrivial relation $x \in Z_x$. This we achieve by considering an arbitrary point $x \in X$ and assuming the contrary, $x \in X \setminus Z_x$. For ease of notation let us write $\overset{\smile}{M}\colon=X \setminus \overline{M}=\left(X \setminus M\right)^{\circ}$ for the exterior of arbitrary subset $M \subseteq X$. Our assumption then means that $x \in X \setminus Z_x=\overline{\left(X \setminus \{x\}\right)^{\circ}}=\overline{\overset{\smile}{\{x\}}}$, whence we infer that $\overline{\overset{\smile}{\{x\}}} \supseteq \overset{\smile}{\{x\}} \cup \{x\}$. Since it is in general (in any topological space) true that $\overline{M \cup \overset{\smile}{M}}=X$ for any subset $M \subseteq X$, we gather in particular that $\overline{\overset{\smile}{\{x\}}}\supseteq \overline{\overset{\smile}{\{x\}} \cup \{x\}}=X$, which simply means that $\overline{\overset{\smile}{\{x\}}}=X$. We now apply the hypothesis of regularity by considering interiors in this last relation in order to obtain $X=\mathring{\overline{\overset{\smile}{\{x\}}}}=\overset{\smile}{\{x\}}$. This is an obvious contradiction as $\overset{\smile}{\{x\}} \subseteq X \setminus \{x\} \subset X$ (where $\subset$ denotes strict inclusion). In other words, it is in general true that exteriors of nonempty subsets are proper subsets and that under our particular assumption of regularity the adherence of a proper open subset must remain a proper subset.

Considering now an arbitrary closed subset $F \subseteq X$ and an arbitrary point $x \in F$ we clearly have $\overline{\{x\}} \subseteq F$. This however leads to $Z_x \subseteq \overline{Z_x}=\overline{\{x\}} \subseteq F$ and as $x \in Z_x \in \mathscr{T}$ the point $x$ is established to be an interior point of $F$. By virtue of the arbitrariness of $x$ this entails $F \subseteq \mathring{F}$, which means that $F=\mathring{F}$ and hence any closed subset is equally open. The topology is thus seen to consist of clopen subsets. $\Box$


问:
Help in proving relations involving finite sums containing factorials, coming from an odd card problem
Let's say I have a deck of $2P$ cards, with $P$ of them red and $P$ black ($P$ is a "large" integer, such as 26). I then randomly form $P$  pairs of cards. I was able to show that the probability of having exactly $c$ red-red ($RR$) pairs, which is equal to the number of $BB$ pairs, is given by
$$
    \text{Pr}(P,\,c) = \frac{P!\,P!}{(2P)!} \times
    \frac{P!}{c!\,c!\,(P-2c)!} \times 2^{P-2c} \qquad (0 \le c \le P/2)
$$
But I'm having trouble trying to prove that these probabilities are normalized, that is
$$
\sum_{c=0}^{P/2} \text{Pr}(P,\,c) = 1 \,,
$$
and also that the expected $c$ value is
$$
\langle c \rangle = \sum_{c=0}^{P/2} c \, \text{Pr}(P,\,c) =  \frac{P(P-1)}{2 (2P-1)} \,.
$$
By the way, I checked the normalization and calculated the last expression using  Mathematica. But I have no idea how the software arrived at the results...

My question is: How to prove the last two relations? I don't mean a proof by induction. I mean a strategy to arrive at the expressions.

Thanks for the feedback, all help will be greatly appreciated!

答:
For starters, I would say that your argument constitutes a full proof that that complicated summation is equal to $1$. You have started with a partition of the set of possible pairings into several classes, and proven an enumeration for each class. It follows the sum of the sizes is equal to the total number of pairings. However, we can provide a proof by induction as well.

Mathematica likely simplifies these summations using Zeilberger's algorithm. Using the same algorithm, we can generate a human readable proof. Let $$F(n,k)=\frac{n!^3}{(2n)!k!^2(n-2k)!}2^{n-2k},\qquad S_n=\sum_{k=0}^{\lfloor n/2\rfloor} F(n,k)$$
so you want to prove $S_n=1$ for all $n\ge 1$. I will adopt the convention $F(n,k)=0$ when $2k>n$, so $S_n=\sum_{k=0}^\infty F(n,k)$ as well.

Now, let $$
G(n,k)=F(n,k)\cdot\color{blue}{\frac{4k^2}{(n-2k+1)(2n+1)}}=\frac{n!^3k^2}{(2n+1)!k!^2(n-2k+1)!}2^{n-2k+2}
$$
with the convention that $G(n,k)=0$ when $2k>n+1$. You can then show that
$$
F(n,k)-F(n+1,k)=G(n,k+1)-G(n,k)\tag1
$$
Proving this just takes a lot of tedious algebraic manipulation, no creativity at all. Start by canceling common factorials until all that remains is ratio of polynomials in $k$ and $n$ on both sides, get a common denominator, and then simplify.

Now, sum both sides of $(1)$ from $k=0$ to $\infty$. We get
$$
\sum_{k=0}^\infty F(n,k)-\sum_{k=0}^\infty F(n+1,k)=\sum_{k=0}^\infty (G(n,k+1)-G(n,k)
$$
The summations on the left are $S_n$ and $S_{n+1}$, while the right hand side is a telescoping series. We therefore get
$$
S_n-S_{n+1}=-G(n,0)=0,
$$
(to see why $G(n,0)=0$, note the $k^2$ in the numerator of $G(n,k)$), so that $S_{n+1}=S_n$ for all $n$. Therefore, to verify $S_n=1$, you need only check that $S_1=1$.

A similar method can be used to prove $\sum c\text{Pr}(P,c)=P(P-1)/2(2P-1)$, but the linearity of expectation argument in the comments is the nicer way. (In case it gets deleted: for each card, the probability it is paired with a same-colored card is $\frac{P-1}{2P-1}$. Summing over call cards, you get $2P\cdot \frac{P-1}{2P-1}$, and this will count each pair of cards twice, so divide by $2$.)


问:
Why do you need three perpendicular bisectors to circumscribe a triangle?
All the perpendicular bisectors of the sides of a triangle intersect at one point -- the circumcenter. So, why do we typically find all three perpendicular bisectors when trying to circumscribe a triangle in a circle? Wouldn't two perpendicular bisectors be sufficient to find the radius of the circle since the third perpendicular bisector would intersect in the same place anyway?</p>

答:
First, even though this isn't what you asked, I wanted to explain why the 3 perpendicular bisectors are concurrent in the first place, and why they intersect at the circumcenter.

Basically, in order to circumscribe a triangle using a circle, you need to find the point that is equidistant from each of the vertices in the triangle. A perpendicular bisector of a segment AB is actually the set of all points that are equidistant from A and B. You can see this by picking an arbitrary point C on the line, and the intersection of the perpendicular bisector of AB with AB D. It is easy to see that AD=DB, CD=CD, and m<ADC=<BDC=90º. Therefore, triangles ADC and BDC are congruent by SAS, so AC=BC.

Now let's focus on the triangle in question. Call it XYZ. Call line z the perpendicular bisector of XY, y the bisector of XZ, and x the bisector of YZ. The intersection of x and z is the point equidistant from X, Y, and Z, and since two lines can only intersect at one point, there is only one point with this property, This means that, since x and y also intersect at a point with this property, and the only point with that property is the intersection of x and z, we know that x, y, and z must be concurrent at the point equidistant from X, Y, and Z, called the circumcenter of triangle XYZ.

So, yes, you can find the circumcenter using only two perpendicular bisectors, but using 3 is usually a good thing to do since if they don't concur at the same point you know you made a mistake :)


问:
How to convert from a power of base two to a power of base 10?
I might have an extremely silly question:
If I have a number, say $2^{32}$ and I need to convert to base 10, how should I do it? I know it should be $4 * 10^9$, but I do not know how did we get it.

I understand that $10$ is $2^3 + 2$, but I cannot understand how to proceed further in my reasoning...

Thanks!

答:
To brush up on algebra rules for exponents, the relevant rule here is that $x^{A*B}=(x^A)^B$. And as the others have already mentioned, $2^{10}=1024≈10^3=1000$. If we pick an example number that is so horrendously gigantic where this small inaccuracy doesn't matter, something like $2^{(2^{20})}=2^{1,048,576}$, you can factor out the exponent and replace it as follows:
$$2^{1,048,576}=2^{10*104,857}=(2^{10})^{104,857}=(10^3)^{104,857}=10^{3*104,857}=10^{314,571}$$
Extremely simple, as long as you remember the algebra fundamentals :)

ContentLicense: CC BY-SA 3.0


问:
How to determine the leading order asymptotic behaviour of this integral?
I tried to put it in the form f(t) exp(tx) to apply watson's lemma but I couldn't. Is there a trick?

$$\int_{0}^{1} e^{t}\left ( \frac{t}{1+t^{2}} \right )^{x} dt$$  as $ x \rightarrow {\infty} $

答:
Let
$$S\left( x \right)=\int\limits_{0}^{1}{{{e}^{t}}{{e}^{x\log \left( \frac{t}{1+{{t}^{2}}} \right)}}dt}=\int\limits_{0}^{1}{{{e}^{t}}{{e}^{xh\left( t \right)}}dt}$$
Note that
$$h'\left( {{t}_{0}} \right)=0\Rightarrow t=1$$
And so the end point is a stationary point of the argument against x, therefore consider

$$S\left( x \right)=\int\limits_{0}^{1}{{{e}^{t}}{{e}^{x\left( h\left( 1 \right)+\frac{1}{2}h''\left( 1 \right){{\left( t-1 \right)}^{2}}+... \right)}}dt}$$
Taking an approximation about the maximum and keeping only the first term of ${e}^{t}$ and the quadratic term in $h\left(t\right)$

$$S\left( x \right)\simeq {{e}^{xh\left( 1 \right)+1}}\int\limits_{1-\delta }^{1}{{{e}^{\tfrac{1}{2}xh''\left( 1 \right){{\left( t-1 \right)}^{2}}}}dt}\simeq {{e}^{xh\left( 1 \right)+1}}\int\limits_{-\delta }^{0}{{{e}^{\tfrac{1}{2}xh''\left( 1 \right){{u}^{2}}}}du}\simeq {{e}^{-x\log \left( 2 \right)+1}}\int\limits_{-\infty }^{0}{{{e}^{-\tfrac{1}{2}x{{u}^{2}}}}du}$$
Completing the integral we have (effectively by Watson’s lemma)
$$S\left( x \right)\sim \sqrt{\frac{\pi }{2x}}{{e}^{-x\log \left( 2 \right)+1}}=\sqrt{\frac{\pi }{2x}}\frac{e}{{{2}^{x}}}$$


问:
What is the number of solution of the equation $\ x^3 - \lfloor x\rfloor = 3$?
What is the number of solution of the equation $\ x^3 - \lfloor x\rfloor = 3$ ? (where $\lfloor x \rfloor\ $ is the greatest integer $\le x$)

I tried plotting the graphs of these equations on Desmos graph calculator and that they intersect each other in between x = 1 and 2 but I couldn't figure out a way to get to this conclusion on my own.

Is there any way by which I can determine where these functions intersect?

答:
First of all $x > 0$.

If $x \le 0$ then $x^3 < 0$ so $[x] = x^3 -3 \le -3$ so $[x] \le -3$ and $x \le -2$.  (Obviously I'm being overly cautious. $[x]\ne x$ unless $x$ is on integer so $x \le -3$ and obviously $x$ can't be $0$ so $[x]\le -4$ and ... so on but... this just stick with $x \le -2$.)

This means that $|x^3| > |x|$ and that $x^3 - x < 0$.  So $x^3 - [x] \ge x^3 -x > x^3 -[x] -1$ so $x^3 -[x] < 1< 3$.

So $x > 0$.

....

Now $[x] \le x$ so $[x]^3 - [x] \le x^3-[x] = 3$.

We must have $[x]^3 -[x] = [x]([x]^2 - 1) = [x]([x]-1)([x]+1) \le 3$.

If $[x] \ge 2$ then $[x]([x]-1)([x]+1)\ge 2\cdot 1\cdot 3 > 3$ so we must have

$[x]=1$

Okay.  So $1\le x < 2$.  Let $r = x-1$ and $x = 1+ r$.

Then $x^3 -[x] = (1+r)^3 =1 = 3r + 3r^2 + r^3 = 3$.

Now $r$ is monotonically increasing on $r\in [0,1)$ so there is at most one solution.

And $3\cdot 0 + 3\cdot 0^2 + 0^3 < 3$ and $\lim_{r\to 1} 3r + 3r^2 + r^3 = 7 > 3$.

So by continuity there is one solution.

We might be able to solve for $r$ but we don't have to.  $r$ is the unique and only solution to $3r + 3r^2 + r^3 = 3$ and $x = 1+r$.

And... oh, googly moogly......

If $[x]=1$ then $x^3 - 1 =3$ and..... $x =\sqrt[3] 4$ and $r = \sqrt[3] 4 -1$  is the unique solution to $3r+3r^2 +r^3 = 3$ and ....

Ah... geeze.... that was trivial to solvi

$3r + 3r^2 + r^3 = 3$

$1+3r + 3r^2 + r^3 = 3+1 = 4$.

$(1+r)^3 = 4$.

.... forest for the trees, forest for the trees.


问:
Schmidt Decomposition of a Matrix
Let $H,K$ be Hilbert spaces. Let $T \in B(H,K)$. By Schmidt Theorem, we know that there exists countable collections of orthonormal vectors $(\phi_n)_n$ and $ (\psi_n)_n$ such that $T(x) = \sum_n s_n(T) \langle \phi_n|x \rangle \psi_n$.

Let $T=
    \begin{pmatrix}
    1 & 1  \\
    0 & 0  \\
    \end{pmatrix}
$.

I'm trying to write the Schmidt Decomposition of $T$.

I found that we have eigenvalues $\lambda_1=2$, $\lambda_2=0$, and eigenvectors $v_1 = (1,1)$ and $v_2= (-1,1)$.

From here, I'm not sure how to find the orthonormal vectors.

I'm very new to this. I wonder if someone could walk me through the steps.

Thank you for your help.

答:
Note that the Schmidt decomposition is essentially the same as SVD with $s_n$ being the singular values (not eigenvalues of $T$), and $\psi_n$, $\phi_n$ being the orthonormal columns of $U,V$ (not the eigenvectors of $T$).

In this case $$T=UDV^*=\begin{pmatrix}1&1\\0&0\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}\begin{pmatrix}\sqrt2&0\\0&0\end{pmatrix}\frac{1}{\sqrt2}\begin{pmatrix}1&1\\-1&1\end{pmatrix}$$

The $s_n$ are $\sqrt2$ and $0$, while $\phi_1=v_1=(1,1)/\sqrt2$, $\phi_2=v_2=(-1,1)/\sqrt2$, $\psi_1=e_1=(1,0)$, $\psi_2=e_2=(0,1)$.

So \begin{align}Tx=UDV^*x&=\frac{1}{\sqrt2}\begin{pmatrix}\sqrt2&0\\0&0\end{pmatrix}\begin{pmatrix}1&1\\-1&1\end{pmatrix}x=\begin{pmatrix}\sqrt2&0\\0&0\end{pmatrix}\begin{pmatrix}\langle v_1,x\rangle\\\langle v_2,x\rangle\end{pmatrix}\\
&=\sqrt2\langle v_1,x\rangle e_1+0\langle v_2,x\rangle e_2\end{align}

Title: Schmidt Decomposition of a Matrix


问:
Trefoil knot cannot be injectively projected to a plane?
I'm trying to do the following geometry qual problem from my university:

Give an example of a smooth embedding $f:S^1\rightarrow\mathbb{R}^3$, such that for each plane $P$ in $\mathbb{R}^3$, $\pi_P\circ f$ is not injective; here $\pi_P$ is the orthogonal projection of $\mathbb{R}^3$ onto $P$.

My guess is that the trefoil knot will work, but since I have zero knowledge of knot theory, can someone please let me know if there is a simple argument to see if (or why) this works? Alternatively, is there a simple example that I missed? Thank you!

Edit: Now I'm sort of confident that this is due to $R^3$-trefoil knot and $R^3-S^1$ have different fundamental groups. I'm still not sure about the details here. Can someone please help me finish the argument? Thank you!

答: Your guess is correct. Assume $\pi_P\circ f$ is injective. Then, the image of $\pi_P\circ f$  is a simple closed curve, injective image of $S^1$, contained in a plane. The Jordan curve theorem tells us that this s.c.c. bounds a disk, i.e. represents the unknot. A knot is the unknot if there exists an unknotted diagram in some planar projection. Therefore, if $f$ is knotted, then there exists no injective projection.


问:
When to use Inclusion Exclusion principle?
I'm a bit confused as to when you should use the Incl/Excl principle. Take for example the following problem:

How many 13-card hands can be selected from the standard 52-card deck with exactly four spades or exactly four diamonds?

My university tutor solved this problem by using the principle. My theory would be that you need to use it in this case because a 13-card hand with exactly four spades can also contain four diamonds, and a hand with exactly four diamond can also contain four spades - so you end up counting too many times.

Is there a trick or a general rule for when you should use it?

Sub-question: My tutor wrote this for choosing exactly four spades, and I believe there might be a mistake:

$c(13,4) * c(9,39)$

$c(13,4)$ is the number of ways to choose four positions for the spades

$c(9,39)$ is the number of ways to choose the remaining 9 cards that are not spades

But isn't it missing another $c(13,4)$ to choose the four spades out of 13 spades?

答:
First and foremost, your notation $c(9,39)$ should read $c(39,9)$.  That is presumably a typo.  Now:

The inclusion-exclusion principle helps you find the cardinality of the sets $A \cup B$, $A \cup B \cup C$, $A \cup B \cup C \cup D$, etc.  In your case, you have to find the cardinality of the set $A \cup B$, where $A$ is the event that you draw exactly four spades and $B$ is the event that you draw exactly four diamonds.  If $|A|$ denotes the cardinality (or number of elements) in the set $A$, then

$$
|A \cup B| = |A| + |B| - |A \cap B|.
$$

This makes intuitive sense, since if you want to count the number of elements in the set $A \cup B$, you count the number of elements in $A$, you count the number of elements in $B$ and you subtract the number of elements in $A \cap B$, since they were counted twice.

To find $|A|$ or the number of $13$ card hands which have exactly four spades, you need to choose $4$ spades from the possible $13$ and $9$ non spades from the rest of the deck.  The number of ways of doing this is

$$
c(13,4) \cdot c(39,9) = 151519319380.
$$


问:
Ordinal arithmetic $3+\omega^2 = \omega^2$
Which one of the following equalities is false:

a) $2\cdot \omega = 3\cdot\omega$

b) $3+\omega+\omega^2 = \omega+3+\omega^2$

c) $\omega^2 + 3 = 3+\omega^2$

d) $12\cdot(5+\omega)=60 \cdot\omega$

I think the wrong one is c):

In a) and d) both sides are $\omega$. And for b) I found that:

$\omega^2$ can be viewed as $\{0,1,2,\dots,\omega,\omega+1,\dots,\omega\cdot 2,\omega\cdot 2+1,\omega\cdot 2+2,\dots,\omega\cdot 3, \omega\cdot 3+1,\dots\}$.

Thus in b) LHS we have $3+\omega = \omega$ and RHS $3+\omega^2=\omega^2$.

But in c) we get $\omega^2+3\not=\omega^2$.

Is that right?

答:
In c) you have limit ordinal on one side of the equation and non-limit on the other side.  The ordinal sum $\alpha+\beta$ is limit if and only if $\beta$ is limit. The same is true for non-limit (successor) ordinals.

So if you have been told that exactly one of them is incorrect, this must be the one.

The other computations you have given seem to be correct.

a) You have $n\cdot \omega=\omega$ for every $n<\omega$. (This can be visualized as follows: If we replace in the usual ordering of $\omega=\{0,1,2,3,\dots\}$ each number by finitely many copies, we get an order-isomorphic set.)

b) You are correct in saying that $3+\omega=\omega$ and $3+\omega^2=\omega^2$. Then you have $\omega+\omega^2$ on both sides. This can be further simplified as
$\omega+\omega^2=1\cdot\omega+\omega\cdot \omega=(1+\omega)\cdot\omega=\omega\cdot\omega=\omega^2$.

d) $5+\omega=\omega$; so you get $12\cdot\omega=60\cdot\omega=\omega$ for the same reasons as in a).


问:
Number of measurable maps $f: \{1,2,3\} \rightarrow \{1,2,3\}$
I'm asked to find the number of measurable maps $f:\Omega_1 \rightarrow \Omega_2$,  where $\Omega_1 = \Omega_2 = \{1,2,3\}$.

Furthermore, we have that $\mathcal{F_1} = \mathcal{F_2} =\{\emptyset, \{1\}, \{2,3\}, \{1,2,3\}\}$  and that $(\Omega_1,\mathcal{F_1} ), (\Omega_2, \mathcal{F_2})$ are measurable spaces.

How can I proceed to find how many such maps exists? I'm not very good at measure theory since (by now) I've only seen a little and I don't know where I should start from.
Any hints?

答:
The number of measurable such maps is 15.

This is due to the fact that the preimage of any set in $\mathcal{F_2}$ must be in $\mathcal{F_1}$.
Let's analyse which cases satisfy such condition:

We fix $f(1)=1$, then

$f(2)=2$ & $f(3)=3$

$f(2)=1$ & $f(3)=1$

$f(2)=3$ & $f(3)=2$

$f(2)=2$ & $f(3)=2$

$f(2)=3$ & $f(3)=3$

These are the only maps that satisfy the above condition when $ f(1)=1$, hence they are measurable.
Similarly:

$f(1)=2$, then

$f(2)=2$ & $f(3)=3$

$f(2)=1$ & $f(3)=1$

$f(2)=3$ & $f(3)=2$

$f(2)=2$ & $f(3)=2$

$f(2)=3$ & $f(3)=3$

And

$f(1)=3$

$f(2)=2$ & $f(3)=3$

$f(2)=1$ & $f(3)=1$

$f(2)=3$ & $f(3)=2$

$f(2)=2$ & $f(3)=2$

$f(2)=3$ & $f(3)=3$

The condition is always satisfied, therefore we have 3 situations having 5 measurable maps each, giving us a total of $$3\cdot5=15$$
measurable maps.


问:
Am trying to fit a gas strut to some brackets. Think I have an equation to work it out, but not sure where to next.
I am trying to work out where to fit the pins on a gas strut for a 90 degree opening hinge. BTW - I am a farmer, not a maths person and this is my first post.

Strut details: Open length is $325mm$. Closed length is $205mm$. Hinge goes from $0^\circ$ degrees (fully closed) to $90^\circ$ degrees (fully open).  The problem is where to drill the pins on each side of the hinge so it opens to the maximum length and closes to the shortest.

It seems such a simple problem, but I just cannot see how to solve it. The formula I have worked out is this (based on the squared rule for a right angle triangle):

$(x+y)^2 + y^2 = z^2$

So for my case, it is $(205 + y)^2 + y^2 = 325^2$

By substituting a few real-life values I can see the value for $y$ is about $103mm$, but that took about $4$ or $5$ high/low guesses to get to that.

Is there a mathematical way to work it out, or am I heading into integration/differentiation maths to get an answer?

Thanks

David

答:
Welcome to MSE!

you can solve the equation $(205+y)^2 +y^2=325^2$ through the quadratic equation formula or also, probably through pythagorean triples in some weird way.

I’de say maybe the most straight forward way would be to expand everything,

$$2y^2+410y-63600 =0$$

and then use $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ which will give you both solutions, where you have a polynomial of the form $ax^2+bx+c=0$

WolframAlpha gives $y=-\frac{205}{2}\pm\frac{5\sqrt{6769}}{2}$ with positive $y\approx 103.18$ and negative $y\approx -308.18$, but I’m pretty sure the positive $y$ is only useful to you.

It turns out you were pretty close to the right answer, only off by $0.18$


问:
$f$ is continuous on $\mathbb{R}$ and $f(x) = f(x+2\pi)$ for all $x\in\mathbb{R}$. Then $f$ is a uniformly continuous function on $\mathbb{R}$.
I am reading "A Course in Analysis vol. 4" (in Japanese) by Kazuo Matsuzaka.

The author uses the following proposition without a proof in this book.

I think this proposition is intuitively obvious.

And I proved this proposition as follows.

This proposition is very simple, but my proof is not elegant.

Please give me a simple and natural and elegant proof of this proposition.

Proposition:

Let $f$ be a continuous function defined on $\mathbb{R}$ such that $f(x) = f(x+2\pi)$ for all $x\in\mathbb{R}$.

Then, $f$ is a uniformly continuous function on $\mathbb{R}$.

Proof:

$f$ is continuous on $[-\pi,3\pi]$.

So, $f$ is uniformly continuous on $[-\pi,3\pi]$.

So, if $\epsilon$ is an arbitrary positive real number, then there exists a positive real number $\delta$ such that
$$x,y\in[-\pi,3\pi]\text{ and }|x-y|<\delta \implies |f(x)-f(y)|<\epsilon.$$

(1) In the case $2\pi<\delta$.

Suppose that $x,y\in\mathbb{R}$ and $|x-y|<\delta$.

There exists an integer $m$ such that $x-2m\pi\in[-\pi,\pi]$.

There exists an integer $n$ such that $y-2n\pi\in[-\pi,\pi]$.

Since $x-2m\pi,y-2n\pi\in[-\pi,3\pi]$ and $|(x-2m\pi)-(y-2n\pi)|\leq 2\pi<\delta$, $|f(x)-f(y)|=|f(x-2m\pi)-f(y-2n\pi)|<\epsilon$.

(2) In the case $\delta\leq 2\pi$.

Suppose that $x,y\in\mathbb{R}$ and $|x-y|<\delta$.

If $x=y$, then $|f(x)-f(y)|=0<\epsilon$.

If $x\ne y$, we can assume that $x<y$ without loss of generality.

Then, $0<y-x<\delta\leq 2\pi$.

There exists an integer $n$ such that $x-2n\pi\in[-\pi,\pi]$.

$-\pi\leq x-2n\pi<y-2n\pi<(x+2\pi)-2n\pi=(x-2n\pi)+2\pi\leq 3\pi$.

Since $x-2n\pi,y-2n\pi\in[-\pi,3\pi]$ and $|(x-2n\pi)-(y-2n\pi)|=|x-y|<\delta$, $|f(x)-f(y)|=|f(x-2n\pi)-f(y-2n\pi)|<\epsilon$.

答:
$f$ factorises as $\mathbb{R}\overset{\pi}\rightarrow \mathbb{R}/2\pi \mathbb{Z}
\overset{\tilde {f}}{\rightarrow}\mathbb{R}$. Note that$\mathbb{R}/2\pi \mathbb{Z}$ can be given a metric structure compatibile with the quotient structure and that with such a metric $\pi$ is uniformly continuous. Since $\mathbb{R}/2\pi \mathbb{Z}$ is compact and $\tilde f$ is continuous, $\tilde f$ is uniformly continuous. The result then follows since $f=\tilde f\circ \pi$ is a composition of uniformly continuous functions.


问:
Section of a set with measure 0 in the product space is measurable almost everywhere for a complete space
Let $(X,\mathcal S, \mu)$ and $(Y,\mathcal T, \nu)$ be complete positive measure spaces, and let $(X\times Y,\mathcal V, w)$ be the product space. And let $N$ be a set such that $w(N)=0$.

How can I prove that:
$$N_x := \{y\in Y \mid (x, y)\in N\}$$
is measurable for almost all $x\in X$ in the measure space $(Y,\mathcal T, \nu)$?

答:
It is clear that $\mathbb{1}_N$ is lebesgue measurable in the product space $(X\times Y,\mathcal V, w)$.
Additionally, $$\int\limits_{X \times Y} \, \mathbb{1}_N (x,y) \, \mathrm{d} w = w(N)= 0$$ so $\mathbb{1}_N$ is integrable $\mathrm{d} w$.
Now notice that both $(X,\mathcal S, \mu)$ and $(Y,\mathcal T, \nu)$ are complete positive measure spaces, so we may apply fubini's theorem:
$$\int\limits_{X} \, \left(\int\limits_{Y} \, \mathbb{1}_{N_x }(x,y) \, \nu(\mathrm{d} y)\right)\, \mu(\mathrm{d} x) = \int\limits_{X} \, \nu \left(N_x (x,y) \right)\, \mu(\mathrm{d} x )= 0$$
But $\nu \left(N_x (x,y) \right) \geq 0 \; a.e$ and $\int\limits_X \nu \left(N_x (x,y) \right)\, \mathrm{d} x = 0$ and thus it follows that $\nu \left(N_x (x,y) \right) = 0 \Rightarrow N_x$  is measurable for almost all $x\in X$ in the measure space $(Y,\mathcal T, \nu)$

Edit:
This may also be proven without the use of fubini’s theorem.

To do so define $\Sigma := \{E \in \mathcal S \times \mathcal T \mid \forall x \in X:E_x \in \mathcal T\}$. Notice that if $E=A\times B$ then $\forall x \in A: E_x=B$ and $\forall x \in A^c: E_x=\emptyset$.
Therefore every measurable rectangle in the product space belongs to $\Sigma$.
Now since $\mathcal T$ is a sigma-algebra it is simple to prove that $\Sigma$ is also a sigma-algebra in the product space $X \times Y \Rightarrow \Sigma = \mathcal S \times \mathcal T$.

Now we know that $w(N) = 0$ and so $N$ is measurable in the product space. But then according to according to what was proven above: $\forall x \in X: N_x \in \mathcal T$

q.e.d.


问:
How much would you pay to play this game?
There is a game in which you have a $\frac{1}{3}$ chance of winning 100 dollars and getting the fee you paid to play the game back. There is a $\frac{2}{3}$ of losing the fee you payed to play the game. How much would you pay to play this game?

My approach:

$\frac{1}{3}(100+x)+\frac{2}{3}(-x)=x$

Solving this equation we get $x=25$

This means that, on average, we break even if we pay $25$ dollars.

So, I should be willing to pay any amount under $25$ dollars to play this game.

Is this the correct thinking?

答:
The house (whoever runs this game) puts 100 in the pot (that usually means on the table between you). You pay the buy-in of $x$, which is to say, you put $x$ in the pot. Then you play.

I see two ways of thinking about this:

$\frac{1}{3}(100)+\frac{2}{3}(-x)$ is the expected net change in your wallet contents after playing the game one time (including the action of paying the buy-in). For a fair game, you want that to be $0$.

$\frac{1}{3}(100+x)+\frac{2}{3}(0)$ is how much money you're expected to take from the pot after playing the game one time. For a fair game, you want that to be equal to the buy-in $x$.

Either approach works, but mixing them like you did gives the wrong result.

Your analysis after the equation solving is correct, though. Any buy-in below what gives a fair game will tilt the game in your favour.


问:
Metric space in $\mathbb{R}^{2}$ and its non-emptiness.
I will start my question by giving Wikipedia information of Sierpinski Triangle first. https://en.wikipedia.org/wiki/Sierpi%C5%84ski_triangle

I was proving why the Sierpinski triangle is compact in metric space $\mathbb{R}^{2}$, and came up with the idea that

"since the area of $F_{n}$, which is the set of remaining triangles after $n^{th}$ process, is greater than $0$ $\forall n\in \mathbb{N}$, $F_{n}$ is non-empty.
If we call the Sierpinski triangle $F=\bigcap_{n=1}^{\infty }F_{n}$, then its area is still greater than 0 because of 'the area is greater than $0$ property.' Thus, $F$ is non-empty."

My concern is this: can we really say the set $F$ is non-empty in $\mathbb{R}^{2}$ just because the set's area is not $0$? Basically, can we ever claim that there is an area of some closed and bounded subset of Metric Space $\mathbb{R}^{2}$?

I know the proof using subsequences and sets, but I just wonder if this idea works.
Or, since this is about the metric space, we are only allowed to use 'elements, sequence, sets, distance, etc' terms?

Thanks a lot for your help.

答:
If the area (or measure) of $F$ were positive, you could. This is so simply because the area of the empty set id $0$ and hence would be different from that of $F$.
However, the area of $\bigcap F_n$ cannot be larger than the area of any of the $F_n$, and these get arbitrarily small. Thus the area of $F$ is in fact $0$.

To show that $\bigcap F_n$ is non-empty, it suffices to exhibit a single common element of all $F_n$, and at least the three vertices come to mind as easy candidates.


问:
How do I rewrite the vector equation of a line in symmetric form?
The problem given was "Given a the vector equation $r(t)=(−4−1t)i+(−3+1t)j+(3+5t)k$, rewrite this in terms of the symmetric equations for the line."

(quotient involving $x$)$=\quad -x- 4$

(quotient involving $y$)$=\quad\frac{y+3}{1}$

(quotient involving $z$) ?

I am having trouble figuring out the quotient involving $z$. I intially got $\frac{z-3}{5}$ but that was wrong. I googled how to do problems similiar to this one and it said to write them as parametric equations and then solve for $t$ which worked for the other two. I don't know what I am doing wrong and any help would be appreciated. Thank you !

答:
In 3-dimensional space, a line's vector equation has the following form, right?
$$\begin{pmatrix} x \cr y \cr z \end{pmatrix}=\begin{pmatrix} a_x \cr a_y \cr a_z \end{pmatrix}+t\begin{pmatrix} u_x \cr u_y \cr u_z \end{pmatrix}$$

Rearranging:
$$\begin{pmatrix} x-a_x  \cr y-a_y  \cr z-a_z \end{pmatrix}=t\begin{pmatrix} u_x \cr u_y \cr u_z \end{pmatrix}.$$

Dividing and multiplying:
$$\begin{pmatrix} \frac{x-a_x}{u_x}  \cr \frac{y-a_y}{u_y}  \cr \frac{z-a_z}{u_z}  \end{pmatrix}=\begin{pmatrix}  t \cr t \cr t \end{pmatrix}.$$

Eliminating $t$ to obtain the symmetric form:
$$\frac{x-a_x}{u_x} = \frac{y-a_y}{u_y} = \frac{z-a_z}{u_z}.$$

Therefore, in your example,
$$\mathbf{r}(t)=(−4−1t)\mathbf{i}+(−3+1t)\mathbf{j}+(3+5t)\mathbf{k} \\
\implies \frac{x+4}{-1} = \frac{y+3}{1} = \frac{z-3}{5}.$$


问:
Probability that the lot contains defective articles.
I tried much in the problem but I didn't get my answer correct.

The question is---

A lot contains 20 articles.the probability that the lot contains exactly 2 defective articles is 0.4 and that the lot contains exactly 3 defective articles is 0.6.articles are drawn from the lot at random one by one without replacement and are tested till all defective articles are found.then the probability that the testing procedure ends at the twelfth testing is------

My attempt -----

let E1 be the event that the lot contains exactly 2 defective articles and E2 be the event that the lot contains exactly 3 defective articles.

I noticed that $P(E1)+P(E2)=1 $and also it is easy to see E1 and E2 ate mutually exclusive. So that implies that the lot contains exactly 3 or 2 defective articles.

Hence my approach was $$\frac{\binom{12}{2} \times 0.4 +\binom{12}{4} \times 0.6}{\binom{20}{12}}$$
But I am not getting the answer.please help me in this regard.

Thanks.

答:
There are two cases

2 defective items

3 defective items

Let us see one by one

Case-1: 2 defective items

Total sample space = $_{}^{20}\textrm{P}_{12}$

Because we can select any 12 items and arrange them in 12! ways.

Now the number of ways the 12$^{th}$ item turns to be our last defective item = ( ways to select 1 of the 2 defective objects ) * ( ways to select 10 of the 18 non-defective objects ) * ( ways to arrange these 11 objects ) * ( selecting the last left defective item and only one way to arrange this, i.e. at the 12$^{th}$ position ) = $\binom{2}{1} * \binom{18}{10} * 11! * \binom{1}{1}$

Probabililty(case-1) = $\frac{\binom{2}{1} * \binom{18}{10} * 11! * \binom{1}{1}}{_{}^{20}\textrm{P}_{12}} = \frac{11}{190}$

Case-2: 3 defective items

Similarly here,

Probabililty(case-2) = $\frac{\binom{3}{2} * \binom{17}{9} * 11! * \binom{1}{1}}{_{}^{20}\textrm{P}_{12}} = \frac{11}{228}$

Final Answer:

$0.4 * Probabililty\:(case-1)  +  0.6 * Probabililty\:(case-2) = 0.052$


问:
Looking for some Abstract Proof help about proving a number group is commutative
I'm a bit confused on where to begin on this proof.  I understand that in order to prove a group, $G$, is commutative, I'd say $a,b \in G$, then show that $ab = ba$.

Here is the theorem: "If $g^{-1}=g$ for all $g\in G$, then $G$ is commutative."

However, I'm not entirely sure how I can use the inverse property provided in this theorem to show the commutative property.

Would my proof utilize the idea that $a * a^{-1} = 1$ and simply show that $a^{-1} * a = 1$ and then show the same for $b$ or am I looking at this the wrong way?

Thank you in advance for any insight you can provide! :)

Here is what I had submitted to my professor, which he quickly sent back for me to rework and resubmit:
Assume $g^{-1}=g$ for all $g \in G$. Then, we must show that $g(g^{-1}) = (g^{-1})g$.  From the definition of an inverse, we know that $g * g^{-1} = 1$. We also know that the inverse of $g^{-1}$ is $g$ and $g^{-1} * g  = 1$.  Hence, by substitution, we can express $g(g^{-1}) = (g^{-1})g$. Thus, we conclude that $G$ is commutative.

Body: If $a,b\in G$, then $a^{-1},b^{-1}\in G$, and $(a^{-1}b^{-1})^{-1}=a^{-1}b^{-1}$. In other words, $ba=a^{-1}b^{-1}=ab$.


答: You need to prove $ab=ba$. You know $a=a^{-1}$ and $b=b^{-1}$. Then $ab=a^{-1}b^{-1}=(ba)^{-1}=ba$, which concludes the proof.

Title: Looking for some Abstract Proof help about proving a number group is commutative

Body:
I'm a bit confused on where to begin on this proof.  I understand that in order to prove a group, $G$, is commutative, I'd say $a,b \in G$, then show that $ab = ba$.

Here is the theorem: "If $g^{-1}=g$ for all $g\in G$, then $G$ is commutative."

However, I'm not entirely sure how I can use the inverse property provided in this theorem to show the commutative property.

Would my proof utilize the idea that $a * a^{-1} = 1$ and simply show that $a^{-1} * a = 1$ and then show the same for $b$ or am I looking at this the wrong way?

Thank you in advance for any insight you can provide! :)

Here is what I had submitted to my professor, which he quickly sent back for me to rework and resubmit:
Assume $g^{-1}=g$ for all $g \in G$. Then, we must show that $g(g^{-1}) = (g^{-1})g$.  From the definition of an inverse, we know that $g * g^{-1} = 1$. We also know that the inverse of $g^{-1}$ is $g$ and $g^{-1} * g  = 1$.  Hence, by substitution, we can express $g(g^{-1}) = (g^{-1})g$. Thus, we conclude that $G$ is commutative.


问:
Why is $\sigma(B_{s}-s: 0\leq s\leq t)$ strictly included in $\mathcal{F}_{t}^{B}$
Why is $\sigma(B_{s}-s: 0\leq s\leq t)$ strictly included in $\mathcal{F}_{t}^{B}$ , i.e. the completed natural filtration of the standard Brownian motion $B$?

It is clear to me that $B_{s}-s$ for $0\leq s\leq t$ is contained in $\mathcal{F}_{t}^{B}$, since $f_{s}(B_{s})=B_{s}-s$ is a composition of continuous map $f_{s}$ and $B_{s}$, hence $B_{s}$ measurable. Why is the inclusion strict?

I am struggling to find sets that are $\mathcal{F}^{B}_{t}$- measurable but not in $\sigma(B_{s}-s: 0\leq s\leq t)$.

答:
Let $\mathcal G^B_t$ be the usual incomplete filtration generated by $(B_s)_{0 \leq s \leq t}$. Then the claim is that $\mathcal G^B_t = \sigma(B_s -s : 0 \leq s \leq t)$.

Indeed, recall that $$\sigma(B_s - s : 0 \leq s \leq t) = \sigma\left(\bigcup_{n=1}^\infty \bigcup_{0 \leq t_1 \leq ... \leq t_n \leq t} \sigma(B_{t_1}-t_1,...,B_{t_n} - t_n)\right)$$

and similarly for $\mathcal G^B_t$. It is sufficient to prove that for all $n$ and $0 \leq t_1 \leq ... \leq t_n \leq t$ we have $\sigma(B_{t_1}-t_1,...,B_{t_n} - t_n) = \sigma(B_{t_1},...,B_{t_n})$.

But now we use a result : a function is measurable w.r.t a sigma-algebra generated by finitely many random variables, if and only if it is a Borel function in those variables.

That is, let $f$ be measurable w.r.t. $\sigma(B_{t_1}-t_1,...,B_{t_n} - t_n)$. Then $$f = g(B_{t_1}- t_1, B_{t_2} - t_2,....,B_{t_n}-t_n)$$ for a measurable function $g$. Define $g'(x_1,...,x_n) = g(x_1-t_1,...,x_n - t_n)$. Then note that $g'$ is measurable and $f = g(B_{t_1},...,B_{t_n})$.

Therefore $f$ is measurable w.r.t $\sigma(B_{t_1},...,B_{t_n})$. I am sure you can go the other way as well.

Therefore, the strictness of the containment comes from the fact that one is complete while the other is apparently not. If they are both complete, then the filtrations will be the same as the null sets of both filtrations are the same.

However, we can show that the natural filtration is not complete, using a characterisation of the sets that are measurable w.r.t  the sigma algebras in the natural filtration.

Recall that the one-dimensional Brownian motion on $[0,t]$ sits on the probability space
$\Omega = \{f : [0,t] \to \mathbb R\}$, the sigma algebra being $\mathcal B^{[0,t]}(\mathbb R)$ i.e. the sigma-algebra generated by $\{\pi_s^{-1}(B) : B \in \mathcal B(\mathbb R) , 0 \leq s \leq t\}$ where $\pi_s (w) = w(s)$ is the $s$th coordinate projection , and the measure being the Wiener measure i.e. that whose finite dimensional distributions are given by $\mathbb P_{t_1,...,t_n}(C_1 \times ... \times C_n) = \mathbb P(B_{t_i} \in C_i , 1 \leq i \leq n)$ (which has an explicit formula using the normal densities and independent increments assumption).

Note that $\mathcal B^{[0,t]}(\mathbb R)$ is the same as $\mathcal F^B_t$.

Using this, we can prove the following result on sets which are in $\mathcal B^{[0,t]}(\mathbb R)$.

Let $\Gamma \in \mathcal B^{[0,t]}(\mathbb R)$. Then there exists a countable set $S$ such that if $f \in \Gamma$, $w \in \Omega$ and $w|_S = f|_S$ then $w \in \Gamma$.

In other words, every measurable subset of functions  w.r.t the usual sigma-algebra is determined by countably many coordinates.

To prove this, let $$\Sigma = \{\Gamma \in \mathcal B^{[0,t]}(\mathbb R) : \exists S_{\Gamma} \textrm{ countable } , \forall w \in \Omega,f \in \Gamma, w|_{S_\Gamma} = f|_{S_\Gamma}
 \implies w \in \Gamma\}$$

i.e. those $\Gamma$ which do satisfy what we need. We note that if $C \in \mathcal B(\mathbb R)$ and $0\leq s \leq t$ then $\{w : \pi_s(w) \in C\}$ belong in $\Sigma$ because you can take $S_{\Gamma} = \{s\}$, it is the single coordinate at which information will tell you if a function should belong in this set or not.

On the other hand, one can show that $\Sigma$ is a sigma-algebra (left as an exercise, you can see the hidden block for hints).

 $\phi, \Omega \in \Sigma$ obviously. For $\Gamma \in \Sigma$, we can show that $S_{\Gamma^c} = S_{\Gamma}$ works out so $\Gamma^c \in \Sigma$. For the infinite union, just take the infinite union of the $S_i$ as well.

Therefore $\Sigma$ is a sigma-algebra which contains the generating set for $\mathcal B^{[0,t]}(\mathbb R)$. It follows that $\Sigma = \mathcal B^{[0,t]}(\mathbb R)$.

This can be used to find sets which are not elements of $\mathcal B^{[0,t]}(\mathbb R)$ but are contained in elements of $\mathcal B^{[0,t]}(\mathbb R)$, which show incompleteness of the natural filtration.

Indeed, let $C$ denote the Cantor set and let $E = \{w : w_{i} \in [0,1] \text{ for all } i \in C$. Then I claim that $E$ is not an event which is measurable with respect to $\mathcal F^B_t$ for $t \geq 1$.

If not, let $S$ be a set such that if $f \in E$ and $w|_S = f|_S$ then $w \in E$. Clearly the Cantor function belongs in $E$ (I don't actually need continuity of the function so I could have gone with just the indicator of the Cantor set. I'm just doing this because usually we associate Brownian paths with continuity so I thought the function could be continuous as well!).

Now, because $S$ is countable and $C$ is not, we know that there is a point in $C \setminus S$, call it $s'$. Define $w(s) = f(s)$ except at $s'$ where $w(s') = 2$. Then $f|_{S} = w|_{S}$ and yet $w \notin E$ because $w(s') \notin [0,1]$.

Thus, no such countable set can exist, so $E$ is not in $\mathcal F^B_t$. However, you can try to find a measurable $\mathbb P$-null set by yourself such that $E$ is contained in that null set.

Thus, the natural filtration is very much extended by completion. Completion of the filtration allows us to do one very important thing : it allows us to ensure that we can modify processes to retain measurability and adaptedness. For example, constructing continuous modifications of a stochastic process like a martingale will require us to modify the martingale on sets which are not null sets but are clearly contained in them. Thus we require completeness to ensure measurability and adaptability are not affected as we make these changes.


问:
Cardinality of a power set (cartesian product)
$A = \{0,1,2\}$ and $C = \{1,2\}$

$|P(A \times C)| = ?$

The answer states $|P(A \times C)| = 2^{3×2} = 2^6 = 64$

What formula/logic is used to obtain this answer please?

答:
There are two theorems at play here:

Theorem 1 If $|A|=n$ and $|B|=m$ then $|A \times B|= n\cdot m$.

Theorem 2 If $|C|=n$ then $|\mathcal{P}(C)| = 2^n$.

In your particular example, as $|A|=3$ and $|C|=2$, then by Theorem 1 we have $|A \times C| = 6$. Then, by Theorem 2, we have that $|\mathcal{P}(A \times C)| = 2^6=64.$


问:
Does $\sum^{\infty}_1 k \tan(1/k^3)$ converge?
I'm trying to determine if $$\sum^{\infty}_1 k \tan(1/k^3)$$ converges.

Since $1/k^3 \to 0$ as $k \to \infty$, I'm thinking that all the terms will be positive, so I'm thinking the alternating series test is out of the question. I'm trying to think of an appropriate series to compare it to, but I am not sure what to compare a tan-series to! Do I compare it to another tan-series or something else? Thanks!

答:
Using the Taylor Series for $Tan(x)$, substituting $\frac{1}{k^3}$ for $x$, multiplying by k, and then summing yields the equivalent series:

$\sum_{k=1}^{\infty} \frac{2 \cdot (2^{2k}-1)}{\pi^{2k}} \zeta(2k)
\zeta(6k-4) = \sum_{k=1}^{\infty} k \cdot Tan(\frac{1}{k^3}) $

Using the ratio test for the left hand side yields $\frac{4}{\pi^2}$ which is less than one. Therefore the series converges absolutely.


问:
If the roots of the equation $x^n-1=0$ are $1,\alpha,\beta,\gamma,\cdots$, prove that $(1-\alpha)(1-\beta)(1-\gamma)\cdots= n$
My first post here

So I was doing some past year questions and this one popped out and I haven't been able to progress much upon it and I believe there's a trick which will get it done in no time.

If the roots of the equation
$x^n-1=0$ are $1,\alpha,\beta,\gamma,\cdots$

we need to show that $$(1-\alpha)(1-\beta)(1-\gamma)\cdots= n$$

Can I be helped with some hints to push me in the right direction to solve it?

Thanks in advance.

答:
Denote the roots by $x_0 = 1, x_1, x_2, x_3, \dots, x_{n-1}$

(roots of unity, usual indexing in counter-clockwise direction)

Then $x_k = x_1^k$

So this whole product is an expression involving just $x_1$

And from the roots of unity theory we know that

$$x_1 = \cos\frac{2\pi}{n} + i \cdot \sin\frac{2\pi}{n}$$

Proceeding from here is a matter of doing some simple algebra.


问:
Prove that if $ST=I$. Then $\text{dim}(U)\le\text{dim}(V)$
I'm hoping someone could help me with the following problem. Let $S\in\mathcal{L}(V,U)$ and $T\in\mathcal{L}(U,V)$. Prove that if $ST=I$. Then $\text{dim}(U)\le\text{dim}(V)$

答:
I think that we have $W=U,$ otherwise $ST=I$ makes no sense. If we assume that $U$ and $V$ are finite-dimensional, we get

$$ \dim V = \dim ker(S)+ \dim im(S).$$

From $U= im (ST) \subseteq im(S) \subseteq U,$ we see that $im(S) =U.$

Hence we derive

$$ \dim V = \dim ker(S)+ \dim U \ge \dim U.$$


问:
About Method of Moments and Maximum Likelihood
\begin{aligned}
&X_{1}, X_{2}, \ldots, X_{n} \text { are n samples from a population with density }\\
&f(x)=\left\{\begin{array}{ll}
(\theta+1) x^{\theta} & 0<x<1 ; \quad \theta>0 \\
0 \text { otherwise }
\end{array}\right.
\end{aligned}

I have found that method of moment for this is $\frac{2 \sum_{i=1}^{n} x_{i}-n}{n-\sum_{i=1}^{n} x_{i}}$ and MLE is $\frac{-\left(n+\sum_{i=1}^{n} \ln \left(x_{i}\right)\right)}{\sum_{i=1}^{n} \ln \left(x_{i}\right)}$ .

Now I have this question ;  Suppose $\mathrm{n}=2,$ and $X_{1}=.125, X_{2}=.375,$ what is your (moment) esti-
mated value of $\theta$ ?

The value turns out to be negative but parmeter $\theta > 0$ so what is method of moment estimate in this case ?

答:
The fact that the MoM estimate can give an estimation that is not acceptable, out of the parameter support, is one of the reason why it is not always a good estimator, moreover if you compare it with the ML one.

Usually your MoM estimator can be expressed in a more suitable way

$$\hat{\theta}_{MM}=\frac{2\overline{X}_n-1}{1-\overline{X}_n}$$

EDIT: in this case, both the estimators are evidently out of the specified parameter support.

Actually your $f(x)$ is a beta distribution, $Beta(\theta+1;1)$ thus it is well defined for any $\theta>-1$.

Your exercise states $\theta>0$  and states a particular pair of observation to get you in trouble... thus the solution is the following

$$\hat{\theta}_{MM}=\max\left\{\frac{2\overline{X}_n-1}{1-\overline{X}_n};0^+\right\}$$

Actually, setting $\theta>0$ or $\theta\geq 0$ the only change is that if $\theta=0$ $X\sim U(0;1)$ thus I will write

$$\hat{\theta}_{MM}=\max\left\{\frac{2\overline{X}_n-1}{1-\overline{X}_n};0\right\}$$

Same story for MLE


问:
Integral test for convergence
Show that
$$\sum_{n=3}^\infty \frac1{n \log n(\log \log n)^p}$$
converges if and only if $ p > 1$.

I know the hint for this: we have to use integral test so we know that the sum of $f(n)$ converges $\iff$  integral of $f(x)$ converges.

Anyone know how exactly to do this?

答:
Note that
$$
\int\limits_3^{+\infty}\frac{dx}{x\log x(\log\log x)^p}
=\int\limits_3^{+\infty}\frac{d(\log x)}{\log x(\log\log x)^p}
=\int\limits_3^{+\infty}\frac{d(\log\log x)}{(\log\log x)^p}
=\frac{(\log\log x)^{-p-1}}{-p-1}\Biggl|_3^{+\infty}
=\lim\limits_{x\to+\infty}\frac{(\log\log x)^{-p-1}}{-p-1}-\frac{(\log\log 3)^{-p-1}}{-p-1}=+\infty
$$
Hence by integral test the series
$$
\sum\limits_{n=3}^{+\infty}\frac{1}{n\log n(\log\log n)^p}
$$
diverges</p>

Title: Integral test for convergence


答:
Apparently

$
x^n-1 = (x-1)(x-\alpha)(x-\beta)(x-\gamma)...
$

setting $P(x) = \tfrac{x^n-1}{x-1}$

$
P(x)= (x-\alpha)(x-\beta)(x-\gamma)...
$

Now just evaluate at x = 1.

Title: If the roots of the equation $x^n-1=0$ are $1,\alpha,\beta,\gamma,\cdots$, prove that $(1-\alpha)(1-\beta)(1-\gamma)\cdots= n$

Body:
My first post here

So I was doing some past year questions and this one popped out and I haven't been able to progress much upon it and I believe there's a trick which will get it done in no time.

If the roots of the equation
$x^n-1=0$ are $1,\alpha,\beta,\gamma,\cdots$

we need to show that $$(1-\alpha)(1-\beta)(1-\gamma)\cdots= n$$

Can I be helped with some hints to push me in the right direction to solve it?

Thanks in advance.


问:
Finding spectrum of $M_f.$

Let $f \in C(\Bbb R)$ be a bounded continuous function on $\Bbb R.$ Consider the multiplication operator $M_f : L^2 (\Bbb R) \longrightarrow L^2 (\Bbb R)$ defined by $M_f (g) = fg, g \in L^2 (\Bbb R).$ What will be $\sigma (M_f),$ the spectrum of the operator $M_f\ $?

By definition, $$\sigma (M_f) = \left \{\lambda \in \Bbb C\ |\ (M_f - \lambda I) \notin GL \left (L^2(\Bbb R) \right ) \right \}.$$ Now let $\lambda \in \rho (M_f),$ the resolvent of the operator $M_f.$ That implies $(M_f - \lambda I) \in GL \left (L^2 (\Bbb R) \right ).$ When can it happen? If $\lambda \notin \text {ran}\ f$ then clearly it will be the case because then $M_{(f - \lambda)^{-1}}$ will serve as the inverse of $(M_f - \lambda I).$ What about the converse? If $(M_f - \lambda I) \in GL \left (L^2 ( \Bbb R ) \right )$ can we always say that $\lambda \notin \text {ran}\ f\ $? I think it is not necessarily the case for otherwise $\sigma (M_f) = \left (\rho (M_f) \right )^c = \text {ran}\ f,$ which is not necessarily closed. So I guess that $\sigma (M_f) = \overline {\text {ran}\ f}.$ But how to prove it rigorously?

Any help in this regard will be warmly appreciated. Thanks for your time.

答:
Your guess is correct. The spectrum is the closure of the range $R$ of $f$.

If $\lambda \notin \overline R$ then the equation $(M_f-\lambda I)g=h$ has  a unique solution $g=\frac h {f-\lambda}$ for every $h \in L^{2}(\mathbb R)$ and $\|g\|_2 \leq M\|h\|_2$ where $M=\frac  1 {\inf \{|f(x)-\lambda|: 0 \leq x \leq 1\}}$. Hence $\lambda \notin \sigma (M_f)$.

Now suppose $M_f-\lambda I$ has a bounded inverse. Then $h (\frac 1 {f-\lambda}) \in L^{2}$ for every $h \in L^{2}$. By a well known result in Functional Analysis (which has appered many times on this site) this implies that $\frac  1{f-\lambda}$ is essentially bounded. By continuity this implies that $\lambda \notin \overline R$

Title: Finding spectrum of $M_f.$


问:
Find the equation of a plane given a vector parallel to the plane and a line lying in the plane
Im struggling to understand the marking scheme's answer to this particular question.

The points A, B and C have position vectors, relative to the origin O, given by

$\\\vec{OA} = \mathbf{i +2j +3k}$

$\vec{OB} = \mathbf{4j +k}$

$\vec{OC} = \mathbf{2i+5j -k}$

The plane $p$ is parallel to $OA$ and the line $BC$ lies in $p$. Find the equation of $p$, giving your answer
in the form $ax + by + cz = d.$

The marking scheme gave 4 possible methods to solve the question.

Either

Obtain two equations $a+2b+3c=0$ or $2a+b-2c=0$ and solve for the ratio of $a$,$b$, and $c$ and substitute a relevant point in the plane equation. Or,

Calculate the vector product of relevant vectors and substitute a relevant point in the plane equation.

Form a 2-parameter equation with relevant vectors

State a correct equation. e.g $\mathbf{r=2i+5j-k} + \lambda(\mathbf{i +2j+3k}) + \mu(\mathbf{2i +j -2k})$

State 3 equations in $x$, $y$, $z$, $\lambda$, and $\mu$

Eliminate $\lambda$ and $\mu$

Obtain answer $-7x+8y-3z=29$ or equivalent

Using a relevant point and relevant direction vectors, form a determinant
equation for the plane

State a correct equation, e.g
$$\begin{bmatrix}
x-2 & y-5 & z-1 \\
1 & 2 & 3 \\
2 & 1 & -2
\end{bmatrix}=0$$

Attempt to expand the determinant

Obtain correct values of two cofactors

Obtain answer $-7x+8y-3z=29$ or equivalent

I understand the first two methods, but im struggling to understand the third and fourth method. i know that $\mathbf{r=2i+5j-k} + \lambda(\mathbf{i +2j +3k}) + \mu(\mathbf{2i +j -2k})$ is a vector equation but i dont know how its obtained and how its related to the plane.

It will be really helpful if the fourth method is explained as well, but I have no idea what are determinant equations and cofactors

the original question (number 9) and the marking scheme

答:
For the third method, we can think of a plane by starting at an initial point.$$\mathbf{r}=2\mathbf{i}+5\mathbf{j}−\mathbf{k}+\lambda(\mathbf{i}+2\mathbf{j}+3\mathbf{k})+\mu(2\mathbf{i}+\mathbf{j}−2\mathbf{k}),$$ Imagine you're a bug, starting at the point $(2,5,-1)$, which is on the plane. You can first travel as far as you want in the direction of $\langle 1,2,3\rangle$ or opposite this direction, which produces a line. Then you can also travel as much as you want in the direction of $\langle 2,1,-2\rangle$. Now you have a plane.

If you take the parametric equations that this equation gives, you'd get $$x=2+\lambda+2\mu,\quad y=5+2\lambda+\mu,\text{ and } z=-1+3\lambda-2\mu,$$
and you can eliminate $\lambda$ and $\mu$ to get a single equation involving just $x$, $y$, and $z$.

For the fourth method, the determinant of a matrix is zero precisely when one vector can be written as a combination of the others. So this is just giving a slick way to find when $\langle x-2,y-5,z-1\rangle$ is in the span of $\langle 1,2,3\rangle$ and $\langle 2,1,-2\rangle$.


问:
How do I incorporate initial conditions when given a transfer function?
I'm trying to understand how to incorporate a set of initial conditions when starting from a transfer function, i.e. I know the general response of my system, and I want to reach a time-domain representation where the initial state is nonzero.  I am familiar with this process for polynomial functions: take the inverse Laplace transform, then take the Laplace transform with the initial conditions included, and then take the inverse Laplace transform of the results.  However, it is not clear how to do so when the impulse response is not a polynomial function.

For example, given y(0) = 1, y'(0) = 0, and the transfer function
$$
H(s) = \frac{K_0}{(s-a)(s-b)}
$$

I can decompose this into first order components to get the impulse response:
$$
h(t) = u(t)[ K_1e^{at} + K_2e^{bt} ]
$$

How do I incorporate my initial conditions?  It is not clear from the definition of the transform how initial conditions would be captured for non-polynomial expressions.  My first guess was to treat the input as a homogeneous solution, and then use my initial conditions and solve for a particular solution, and since this is a LTI system, I can sum the results.  However, I couldn't find any text or discussion that supports this as valid.

答:
The equation you are solving with $\frac{Y(s)}{X(s)}=H(s)$ is
$$
y''-(a+b)y'+aby=K_0x(t),
$$
$x(t)=y(t)=0$ for $t<0$.

Now translate back the homogeneous equation incorporating arbitrary initial conditions
$$
[s^2Y(s)-y(0)s-y'(0)]-(a+b)[sY(s)-y(0)]+Y(s)=0
$$
so that
$$
Y(s)=\frac{(s-a-b)y(0)+y'(0)}{(s-a)(s-b)}=H(s)\frac{(s-a-b)y(0)+y'(0)}{K_0}.
$$
Which tells you that the input to realize wanted initial conditions is
$$
x(t)=\frac{y(0)u(t)+\bigl[y'(0)-(a+b)y(0)\bigr]\delta_0(t)}{K_0}.
$$

So with $y(0)=1$, $y'(0)=0$ one gets
$$
Y(s)=\frac{ s-a-b }{(s-a)(s-b)}=-\frac{a}{(b-a)(s-b)}+\frac{b}{(b-a)(s-a)}
\\
 \implies y(t)=u(t)\frac{be^{at}-ae^{bt}}{b-a}
$$


问:
Closed form of $\sum_{m=1}^{\infty} \frac{(-1)^mH_{\frac{3m}{4}}}{3m}$
I've been working on an integral, namely: $${\displaystyle \int_0^1 \frac{x^2}{1 + x^3}\ln(1 - x^4)dx}$$

Which I managed to narrow down to the following expression: $$\sum_{m=1}^{\infty} \frac{(-1)^mH_{\frac{3m}{4}}}{3m}$$

Where $H_n$ is the n-th harmonic number. I managed to get here after converting the integrand into a double summation, recognizing the digamma function hidden inside, and summing back to a harmonic number, but couldn't get past that.

Is there another way to solve the integral, or to solve the summation? I've tried breaking the sum down to its parts, but the best I could do was to find a few terms and was left with
$$\sum_{m=1}^{\infty} \frac{(-1)^m\psi(\frac{3m}{4})}{3m}$$
Where $\psi(x)$ is the digamma function, and still couldn't solve that.

Body:
Too long for comments

I do not know how you arrived at
$$S_p=\sum _{m=1}^p \frac{(-1)^m }{3 m}\psi \left(\frac{3 m}{4}\right)$$ with $p\to \infty$ because this does not converge; for example
$$S_{10000}\sim 0.341858 \qquad \text{and} \qquad S_{10001}\sim 0.341561$$ and I do not see how this relates to the value of the integral which is $-0.157197$.

Assuming that you wrote
$$\frac{x^2}{1+x^3}=\sum _{n=0}^\infty (-1)^n x^{3 n+2}$$ you should have for the integral
$${\displaystyle \int_0^1 \frac{x^2}{1 + x^3}\log(1 - x^4)\,dx}=\sum _{n=0}^\infty \frac{(-1)^{n+1}}{3(n+1) } H_{\frac{3 (n+1)}{4}}$$ which converges to the correct value.

Consider numerically
$$T_k=\sum _{n=0}^{10^k} \frac{(-1)^{n+1}}{3(n+1) } H_{\frac{3 (n+1)}{4}}\qquad \text{and} \qquad
U_k=\sum _{n=0}^{10^k+1} \frac{(-1)^{n+1}}{3(n+1) } H_{\frac{3 (n+1)}{4}}$$
$$\left(
\begin{array}{cccc}
 k & T_k & U_k  & T_k - U_k\\
 1 & -0.197575 & -0.118992 & -0.0785825 \\
 2 & -0.165269 & -0.149187 & -0.0160818 \\
 3 & -0.158395 & -0.155999 & -0.0023952 \\
 4 & -0.157355 & -0.157038 & -0.0003166 \\
 5 & -0.157216 & -0.157177 & -0.0000393 \\
 6 & -0.157199 & -0.157194 & -0.0000047 \\
 7 & -0.157197 & -0.157196 & -0.0000005
\end{array}
\right)$$


答:
If the roots are $r_i$, $i=1\ldots n$, with $r_1=1$, we have

$\prod\limits_{i=1}^{n}r_i= -(-1)^n$

$\sum\limits_{i=1}^{n}r_i = \sum\limits_{i<j}^{n}r_ir_j = \sum\limits_{i<j<k}^{n}r_ir_jr_k =\ldots= 0$ (till taking $n-1$ terms at a time)

Now, $\sum\limits_{i=2}^{n}r_i=\sum\limits_{i=1}^{n}r_i-1=0-1=-1$

Also, $\sum\limits_{i,j=2,i<j}^{n}r_ir_j=\sum\limits_{i,j=1,i<j}^{n}r_ir_j-\sum\limits_{i=2}^{n}r_i=0-(-1)=1$

We can show that the terms $\sum\limits_{i=2}^{n}r_i, \sum\limits_{i,j=2,i<j}^{n}r_ir_j,\sum\limits_{i,j,k=2,i<j<k}^{n}r_ir_jr_k, \ldots$
have alternating values $-1$ and $1$

$\therefore (1-\alpha)(1-\beta)(1-\gamma)\ldots$

$=\prod\limits_{i=2}^{n}(1-r_i)$

$=1-\sum\limits_{i=2}^{n}r_i+\sum\limits_{i,j=2}^{n}r_ir_j-\ldots$

$=1-(-1)+1-(-1)+\ldots=n$


问:
Does a monotone function on an arbitrary subset of $\mathbb R$ always have at most countable number of discontinuity?
I know a monotone function of a closed and bounded interval can have at most countably many point of discontinuity. And hence a monotone function on $\mathbb R$ can have at most countably many point of discontinuity since $\mathbb R=[0,1]\cup[1,2]\cup[2,3]\cup...\cup[-2,-1]\cup[-3,-2]\cup...$

Now my question is

Does a monotone function on an arbitrary subset of $\mathbb R$ always have at most countable number of discontinuity?

Does the funtion $f:(\mathbb R-\mathbb Q)\to\mathbb R:x\mapsto x$ serve our purpose?

答:
First, let me give some terminology that I will use here:

Suppose that $E\subseteq\Bbb R,$ $x_0\in\Bbb R,$ and $f:E\to\Bbb R$ are given.

We will say that $x_0$ is $E$-supremal if $$x_0=\sup\{x\in E:x<x_0\},$$ and $E$-infimal if $$x_0=\inf\{x\in E:x>x_0\}.$$ Equivalently, $x_0$ is $E$-supremal if $E\cap(x_0-\delta,x_0)$ is non-empty for all $\delta>0,$ and $E$-infimal if $E\cap(x_0,x_0+\delta)$ is non-empty for all $\delta>0$. Note that $x_0$ is a limit point of $E$ iff $x_0$ is $E$-supremal and/or $E$-infimal. If $x_0\in E$, then $x_0$ is isolated in $E$ iff $x_0$ is neither $E$-supremal nor $E$-infimal.

If $x_0$ is a limit point of $E$ and $L\in\Bbb R$, we say that $L$ is the limit of $f(x)$ as $x$ approaches $x_0$--written $L=\lim_{x\to x_0}f(x)$--if for all $\epsilon>0$ there is some $\delta_\epsilon>0$ such that $|f(x)-L|<\epsilon$ whenever $x\in E$ and $0<|x-x_0|<\delta$. We say that $\lim_{x\to x_0}f(x)$ fails to exist if there is no such $L$, or if $x_0$ is not a limit point of $E$.

If $x_0$ is $E$-supremal and $L\in\Bbb R$, we say that $L$ is the limit of $f(x)$ as $x$ approaches $x_0$ from below--written $L=\lim_{x\to x_0^-}f(x)$--if for all $\epsilon>0$ there is some $\delta_\epsilon>0$ such that $|f(x)-L|<\epsilon$ whenever $x\in E\cap(x_0-\delta_\epsilon,x_0).$ We say that $\lim_{x\to x_0^-}f(x)$ fails to exist if there is no such $L$, or if $x_0$ is not $E$-supremal.

If $x_0$ is $E$-infimal and $L\in\Bbb R$, we say that $L$ is the limit of $f(x)$ as $x$ approaches $x_0$ from above--written $L=\lim_{x\to x_0^+}f(x)$--if for all $\epsilon>0$ there is some $\delta_\epsilon>0$ such that $|f(x)-L|<\epsilon$ whenever $x\in E\cap(x_0,x_0+\delta_\epsilon).$ We say that $\lim_{x\to x_0^+}f(x)$ fails to exist if there is no such $L$, or if $x_0$ is not $E$-infimal.

Now, suppose $x_0\in E$. Recall that $f$ is said to be continuous at $x_0$ iff for all $\epsilon>0$ there is some $\delta>0$ such that $|f(x)-f(x_0)|<\epsilon$ whenever $|x-x_0|<\delta.$ It can be shown that $f$ is continuous at $x_0$ iff either $x_0$ is isolated in $E,$ or $x_0$ is a limit point of $E$ and $f(x_0)=\lim_{x\to x_0}f(x).$ $f$ is said to be discontinuous at $x_0$ iff it is not continuous there--that is, iff there is some $\epsilon_0>0$ such that for all $\delta>0$ we have $|f(x)-f(x_0)|\ge\epsilon_0$ for some $x\in E$ with $0<|x-x_0|<\delta.$ Note that if $f$ is discontinuous at $x_0$, then $x_0$ is $E$-supremal and/or $E$-infimal (why?).

You should be able to prove the following:

Lemma: Suppose $E\subseteq\Bbb R,$ $x_0\in E,$ and $f:E\to\Bbb R$ are given, with $f$ discontinous at $x_0.$ Then exactly one of the following $5$ situations occurs:

(i) $x_0$ is $E$-supremal and $E$-infimal. There is some $L\in\Bbb R$ with $L\ne f(x_0)$ and $L=\lim_{x\to x_0}f(x).$

(ii) $x_0$ is $E$-supremal and not $E$-infimal. There is some $L\in\Bbb R$ with $L\ne f(x_0)$ and $L=\lim_{x\to x_0^-}f(x).$

(iii) $x_0$ is $E$-infimal and not $E$-supremal. There is some $L\in\Bbb R$ with $L\ne f(x_0)$ and $L=\lim_{x\to x_0^+}f(x).$

(iv) $x_0$ is $E$-supremal and $E$-infimal. There are some $L_1,L_2\in\Bbb R$ with $L_1\ne L_2$, $L_1=\lim_{x\to x_0^-}f(x),$ and $L_2=\lim_{x\to x_0^+}f(x).$

(v) $x_0$ is $E$-supremal but $\lim_{x\to x_0^-}f(x)$ fails to exist, and/or $x_0$ is $E$-infimal but $\lim_{x\to x_0^+}f(x)$ fails to exist.

We say that $f$ has a removable discontinuity at $x_0$ if (i) holds; that $f$ has an essential discontinuity at $x_0$ if (v) holds; that $f$ has a jump discontinuity at $x_0$ in the other three situations.

Now for the kicker: real-valued monotone functions on subsets of the reals can have only jump discontinuities, and can only have at most countably many of those.

Let me give you an idea of how to prove this. (Toward the end, I will use the fact that the rationals are countable and dense in the reals.)

Proof Idea: Suppose $E\subseteq\Bbb R,$ that $f:E\to\Bbb R$ is monotone function, and that $f$ is discontinuous at some point $x_0\in E$. (WLOG, suppose $f$ is nondecreasing. The proof when $f$ is nonincreasing will be similar.) By definition, then, there is some $\epsilon_0>0$ such that for all $\delta>0,$ there exists $x\in E$ where $0<|x-x_0|<\delta$ and $|f(x)-f(x_0)|\ge\epsilon_0.$ We must address three cases.

(Case 1): If $x_0$ is not $E$-supremal, then $x_0$ must be $E$-infimal since $f$ is discontinuous at $x_0$. Moreover, since $f$ is non-decreasing, we must have that $f(x)\ge f(x_0)+\epsilon_0$ for all $x\in E\cap(x_0,\infty),$ for otherwise, this would violate our choice of $\epsilon_0$. (Why?) Then $\{f(x):x\in E,x>x_0\}$ is a non-empty set bounded below by $f(x_0)+\epsilon_0,$ so since $f$ is nondecreasing then we can conclude (why?) that $$\lim_{x\to x_0^+}f(x)=\inf\{f(x):x\in E,x>x_0\}\ge f(x_0)+\epsilon_0>f(x_0).$$ Thus, we have a type (iii) discontinuity.

(Case 2): If $x_0$ is not $E$-infimal, then in a similar fashion to Case 1, we conclude that $x_0$ is $E$-supremal, that $\lim_{x\to x_0^-}f(x)$ exists and is at most $f(x_0)-\epsilon_0,$ meaning we have a type (ii) discontinuity.

(Case 3): If $x_0$ is both $E$-supremal and $E$-infimal, then since $f$ is nondecreasing, we can conclude that $$\lim_{x\to x_0^+}f(x)=\inf\{f(x):x\in E,x>x_0\}\ge f(x_0)$$ and likewise that $\lim_{x\to x_0^-}f(x)\le f(x_0).$ By our choice of $\epsilon_0$, we can show that these two limits cannot be equal, so we have a type (iv) discontinuity.

Thus, $f$ must have a jump discontinuity at $x_0.$

Now, let $Y$ be the union of the "gaps" in $\text{range}(f)$--more precisely: $$Y=\{y\in\Bbb R:\exists x_1,x_2\in E\text{ s.t. }f(x_1)<y<f(x_2),\text{ and }\exists a<y<b\text{ s.t. }(a,b)\cap\text{range}(f)=\emptyset\}.$$ Note that $Y$ is a union of pairwise disjoint open intervals, and is disjoint from $\text{range}(f)$. (Why?) Note further that if $f$ has a jump discontinuity at any point $x_0\in E$, then at least one (and at most two) of the component intervals of $Y$ has an endpoint at $f(x_0)$. (Why?) Thus, we readily have a one-to-one function from the set of discontinuities of $f$ into the set of component intervals of $Y$--in particular, say that for any given point of discontinuity $x_0$, we take $x_0$ to the component interval of $Y$ with $f(x_0)$ for an endpoint, and if we have a choice, take $x_0$ to the component interval of $Y$ whose lower endpoint is $f(x_0)$.

Now, let $\{q_n\}_{n\in\Bbb N}$ be any enumeration of the rationals, and note that for each component interval $J$ of $Y,$ there is some least $n$ such that $q_n\in J$--call this $n_J$ for each such $J$. Different components of $Y$ correspond to different $n_J$ (why?), so the function $J\mapsto n_J$ is a one-to-one function from the set of component intervals of $Y$ into the naturals, meaning that $Y$ has at most countably-many component intervals. Since we also have a one-to-one function from the set of discontinuities of $f$ into the set of component intervals of $Y$, then $f$ has at most countably-many discontinuities.


问:
Modulus not Riemann integrable
Suppose a function $f\colon I \rightarrow \mathbb{C}$ is (complex) Riemann integrable (cf. https://proofwiki.org/wiki/Definition:Complex_Riemann_Integral).

Is it possible for $\vert f \vert$ (modulus) not to be integrable? I can't think of an example as by definition both the real and imaginary part have to be integrable.

答:
Suppose that $f=u+iv$ with $u,v:I \to \mathbb R.$

If $f$ is complex Riemann integrable then $u$ and $v$ are Riemann integrable, hence $u^2+v^2$ is Riemann integrable, therefore $|f|= \sqrt{u^2+v^2}$ is Riemann integrable

Title: Modulus not Riemann integrable


问:
What is the group $G=\langle a,b\mid a^2=b^2=1\rangle$?
I am trying to describe all the elements of $G=\langle a,b\mid a^2=b^2=1\rangle$, and I am really struggling with where to start.

I feel like the group could be infinite, with each element in the group, aside from the identity, having order $2$ (I have since learnt that these are called Boolean, or Abelian-$2$, Groups - again, not sure if this is the right track to go down). I would like to go on to build two subgroups, $S_1$ and $S_2$ that are isomorphic to $\mathbb{Z} /2\mathbb{Z}$ and $\mathbb{Z}$, but obviously to do that I need to understand what $G$ actually looks like!

Thank you for any help.

答:
So, how $G$ looks like. Elements of $G$ are words with letters $a,b,a^{-1},b^{-1}$ where you have to substitute $a^2$ and $b^2$ with $1$. So some elements are

$$a,b,ab,aba,abab,ababa,ababab,...$$

Note that I don't use $a^{-1}$ and $b^{-1}$ since $a=a^{-1}$ and $b=b^{-1}$

It is clear that $ab$ has infinite order.


问:
Finding square roots of an unknown
$p$ and $q$ are rational numbers, find the values of p and q given that
$2p = q \sqrt {76} + \sqrt {19}$

attempt:
$$2p = 2q \sqrt{19} + \sqrt{19}$$
$$2p = \sqrt{19}(2q+1) $$

How to continue or how should I do?

答:
Observe that the left side is rational, but the right side will be irrational unless one of the numbers is zero. This comes from the fact that the product of an irrational number and a nonzero rational number is always irrational.

Thus one of the brackets on the RHS must be zero, which implies $2q + 1  =0$. So we must have $p = 0, q = -\frac{1}{2}$.

Title: Finding square roots of an unknown


问:
How to convert from a power of base two to a power of base 10?
I might have an extremely silly question:
If I have a number, say $2^{32}$ and I need to convert to base 10, how should I do it? I know it should be $4 * 10^9$, but I do not know how did we get it.

I understand that $10$ is $2^3 + 2$, but I cannot understand how to proceed further in my reasoning...

Thanks!

Body:
$2^{32} = (2^{10})^3\cdot 2^2=$

$1024^3\times 4 = $

I guess this is why you said $2^{32} = 4\times 10^9$.  Of course there is rounding and it isn't equal to that. The trick is to know that $2^{10}=1024 \approx 10^3$ so $2^{32} = 2^{30 + 2}=1024^3\times 4 \approx 10^9 \times 4$. But that's an under approximation.

$(10^3 + 24)^3 \times 4 =$

$(10^9 + 3\cdot 24 \cdot 10^6 + 3\cdot 24^2\cdot 10^3 + 24^3)\times 4$.

Okay.....

$24^2 = (25-1)^2 = 5^4 - 50 + 1 = 625-50-1 = 576$.

And $24^3 = (25-1)^3 = 5^6 - 3\cdot 5^4 + 3\cdot 25 -1 = 25*625 - 1875 + 75 -1= 5*3125 -1801 = 15625-1801 = 13824$

So $2^{32} = 4\times 10^9 + 12\cdot 24\cdot 10^6 + 12\cdot 576\cdot 10^3 + 4\cdot 13824 =$

And $12\cdot 24 = 2\cdot 12^2 = 288$

$12\cdot 576 = 5760 + 1152=6912$

$4\cdot 13824 = 5\times 13824 - 13824 = \frac {138240}2-13824 = 69120-13824 = 55296$

So $2^{32} =$

$4,000,000,000 + 288,000,000 + 6,912,000 + 55,296 =$

$4,288,000,000 + 6,967,296=$

$4,294,967,296$

..... or......

we could figure $2^{10} = 10^3 + \frac 14 10^2 - 1$ so

$2^{30 + 2} = (10^3 + \frac 14 10^2 - 1)^3 \times (5-1)$

Which could get interesting.

$= (10^3 + \frac 1{2^6}10^6 - 1 + 3(10^3\cdot \frac 1{2^4}10^4 + 10^3(-1)^2 +\frac 14\times 10^2\times 10^6 + \frac 1410^2 - 10^6 -\frac 1{2^4}10^4 + 2\cdot 10^3\cdot \frac 14\cdot 10^2 \cdot(-1))(5-1)=$

$= (10^3 + \frac 1{2^6}10^6 - 1 + (5-2)(10^3\cdot \frac 1{2^4}10^4 + 10^3(-1)^2 +\frac 14\times 10^2\times 10^6 + \frac 1410^2 - 10^6 -\frac 1{2^4}10^4 + 2\cdot 10^3\cdot \frac 14\cdot 10^2 \cdot(-1))(5-1)=$

etc.....

Which I really don't recommend unless you are morbidly curious and have lots of paper.  But its a goo exercise to deal with only the components $5$ and $2$ and seeing how all digits are a matter of cascade.


问:
Representations of an integer as the sum of other integers
Given a finite set $S$ of (distinct) integers $s_1, \dots, s_n$ and an integer $x$, I'm looking for all representations (where order is important)
$$
x=\sum_{i=1}^ks_{t_i} (t_i\in\{1,\dots,n\})
$$

For example

$x = 4, S = \{1,2\}$: Five solutions.
$$4=1+1+1+1=1+1+2=1+2+1=2+1+1=2+2$$

$x = 7, S = \{3,5\}$: No solution.

$x$ odd, all $s_i$ even: No solution.

This is for a photo gallery on a website and can easily be solved programmatically using recursion. However, it left me thinking about the theory, which may or may not help optimize the algorithm.

Now, as a generalization of 3., it can be easily seen that there are no solutions when $$\gcd(s_1, \dots, s_n)\ne \gcd(x, s_1, \dots, s_n)$$

What other conclusions can we draw, particularly on the existence of solutions?

Is this problem known by a specific name? (Edit: yes, I'm looking for restricted partitions; related: Coin problem, Subset sum problem, Knapsack problem)

答:
The problem of deciding whether there is a solution for $x$ and $S$ is difficult to characterize in general, but I do know a useful result in the special case where $|S| = 2$.

Theorem.  If $S = \{m,n\}$ where $m,n$ relatively prime, then there is a solution for $x$ if and only if there is not a solution for $mn - m - n - x$.

Note that this easily generalizes to when $m,n$ are not relatively prime, because if $(m,n) = d > 1$, you can just consider $\frac{x}{d}$ and $S = \{\frac{m}{d},\frac{n}{d}\}$.

Proof.
Note that there is a solution for an integer $y$ if and only if we can write $y$ as a sum of $m$s and $n$s, i.e. $y = am + bn$, where $a,b \ge 0$.

First, suppose towards contradiction that there is a solution for both $x$ and $mn - m - n - x$.
Then adding the two solutions together gives a solution for $mn - m - n$.
Thus we can write $mn - m - n = am + bn$.
But this implies $a \equiv -1 \pmod n$ and $b \equiv -1 \mod m$,
so $a \ge n-1$ and $b \ge m - 1$, so
$$
mn - m - n = am + bn \ge (n-1)m + (m-1)n = 2mn - m - n,
$$
a contradiction.

Therefore, it suffices to show that there is a solution for at least one of $x$ and $mn - m - n - x$.
By the fact that $m$ and $n$ are relatively prime, we can find $a, b \in \mathbb{Z}$ such that
$$
am + bn = x
$$
Let $a \equiv a' \pmod n$ and $b \equiv b' \pmod m$, where $0 \le a' < n$ and $0 \le b' < m$.  Then
$$
a'm + b'n \equiv x \pmod {mn}
$$
Let $a'm + b'n = x + kmn$, where $k \in \mathbb{Z}$.
If $k \le 0$ we are done because $x = a'm + b'n + (-k)mn$ is a sum of $m$s and $n$s.
Conversely, if $k > 0$, we write
\begin{align*}
mn - m - n - x
&= mn - m - n - a'm - b'n + kmn \\
&= (n - 1 - a')m + (m - 1 - b')n + (k-1)mn
\end{align*}
and since $n -1 - a', m - a - b', k-1 \ge 0$, there is a solution for $mn - m - n - x$.

$$
* \quad * \quad *
$$

A well-known corollary to this theorem is that there is always a solution for $x$ whenever $x > mn - m - n$, but not when $x = mn - m - n$.

$$
* \quad * \quad *
$$

EDIT: The above theorem only applies when $m$ and $n$ are nonnegative.

If exactly one of $m$ and $n$ is negative, then clearly there is a solution for any $x$ since there is a solution for $1$ and $-1$.
I.e., write $am + bn = 1$ and $cm + dn = -1$ for $a,b,c,d \ge 0$, and then we have
$(a|x|)m + (b|x|)n = |x|$ and $(c|x|)m + (d|x|)n = -|x|$.  (This suffices since $x$ is equal to either $|x|$ or -|x|).

If both $m$ and $n$ are negative, then the situation is basically the same as when they are positive.  In this case, there is a solution for $x$ if and only if there is a solution for $-x$ using $S = \{|m|,|n|\}$.


问:
Help understanding two theorems on primes and residue systems
My textbook lists two theorems and I'm not sure how I'm supposed to interpret them. I don't need a proof; I'm only trying to figure out what information I'm being told by each theorem.

Let $p$ be a prime and let $a$ be an integer not divisible by $p$; that is, $gcd(a,p)=1$. Then $\{a,2a,3a,...,pa\}$ is a complete residue system modulo $p$

So for this first theorem, I believe the canonical complete residue system $p$ would be $\{0,1,2,...,p-1\}$. So the set $\{a,2a,3a,...,pa\}$ would be the integers that satifsy the congruences, $a\equiv0 \space (mod \space p), 2a\equiv1 \space (mod \space p), 3a\equiv2 \space (mod \space p),...,pa\equiv (p-1) \space (mod \space p)$.

Is my interpretation correct?

Let $p$ be a prime and let $a$ be an integer not divisible by $p$. Then,

$a\cdot2a\cdot3a\cdot...\cdot(p-1)a\equiv1\cdot 2 \cdot 3 \cdot...\cdot(p-1) \space (mod \space n)$

For this second theorem, is this theorem telling me that I can decompose congruence integers into products of primes?

Thanks for any help.

答:
No, you have misunderstood both. Let's consider a simple special case:  $\,a=2,\, p=5.\,$ i.e. we consider the doubling map $\,n\to 2n\bmod 5\,$ whose action is as follows.

$\qquad\qquad\bmod 5\!:\,\ \begin{align}&2\{\color{#c00}1,2,3,\color{#0a0}4\} \\
                       \equiv\, &\ \ \{\color{#c00}2,4,1,\color{#0a0}3\}\end{align} $

i.e. $\ \ \ \quad\qquad\qquad\qquad\begin{align}2\cdot \color{#c00}1&\equiv\color{#c00} 2\\ 2\cdot 2&\equiv 4\\ 2\cdot 3&\equiv 1\\ 2\cdot \color{#0a0}4&\equiv\color{#0a0}3  \end{align}\,$

Multiplying them all yields product of all on LHS  is congruent to product of all om RHS, i.e.

$\quad\qquad\qquad\begin{align}&(2\cdot\color{#c00} 1)(2\cdot 2)(2\cdot 3)(2\cdot\color{#0a0} 4)\\[.1em]
\equiv &\ \ \, (\color{#c00}2)\, \cdot\,  (4)\,\cdot\, (1)\,\cdot\, (\color{#0a0}3)\\[.1em] \equiv&\ \ \,4!\,\equiv -1\end{align}\,$ by the Congruence Product Rule.

That's what the theorem is telling you (and how it is proved).

Remark $ $ The key idea is: $\,\gcd(a,p)\!=\!1\Rightarrow$ the map $\,x\to ax\,$ is invertible $\!\bmod p\,$ so it is a bijection, so it simply permutes the complete set of nonzero residues $\,1,2,\ldots,p\!-\!1,\,$ which does not alter their product $\,(p\!-\!1)!,\,$ e.g. above the doubling map permuted $\,\color{#c00}1,2,3,\color{#0a0}4\ $ to $\ \color{#c00}2,3,1,\color{#0a0}3$.

The multuiplication is  a congruence generalization of the fact that equal finite sets of integers have equal products, by replacing "equal" by "congruent" (congruence can be viewed as a generalized equality - in fact it becomes an equality if we work in the quotient ring $\,\Bbb Z_p\,$ of cosets $\,[k] = k+p\Bbb Z)$

For another common example of this idea see the Wilson reflection formula where the permutation arises from an additive shift by $\,a\,$ (vs. multiplicative scaling by $\,a)$.


问:
How would I solve these equations of motion? (Coupled Second Order PDEs or ODEs?)
$\ddot{x}(t)=\frac{x}{x^2 + y^2}$

$\ddot{y}(t)=\frac{y}{x^2 + y^2}$

I have no idea where to start with something like this.

答:
Consider polar coordinates, $x+iy=re^{i\phi}$. Then
$$
\ddot r+2i\dot r\dot\phi+ir\ddot\phi-r\dot\phi^2=\frac1r
$$
so that, like in any radial field, $r^2\dot\phi=L$ with a constant angular moment $L$, and
$$
\ddot r=\frac1r+\frac{L^2}{r^3}=0.
$$
This can be integrated once again to
$$
E=\frac12\dot r^2-\ln(r)+\frac{L^2}{2r^2}
$$
with a constant energy $E$.

So one gets an acceleration away from the origin, which is higher with higher initial angular speed. The acceleration and angular speed go to zero the farther away the particle gets from the origin.


问:
Sum of Random Variables...
Imagine we repeat the following loop some thousands of times:

$$
\begin{align}
& \text{array} = []\\
& \text{for n} = 1: 10 000 \\
& k = 0 \\
& \text{while unifrnd}(0,1) < 0.3 \\
& k = k + 1 \\
& \text{end} \\
& \text{if k} \neq 0 \\
& \text{array} = [\text{array,k}] \\
& \text{end} \\
\end{align}
$$

whereas "unifrnd(0,1)" means a random number drawn from the uniform distribution on the unit interval.

My question is then: What is then the value of k, which is the most often observed - except for k = 0? And is that the expectation of k?

Thanks very much

答:
You can model your process as a random walk with unit steps in the positive direction. In other words, define independent random variables $Z_i$, $i \in 1, 2, n$ such that:

$P(Z_i = 1) = \frac{1}{3}$ and

$P(Z_i = 0) = \frac{2}{3}$

Set the initial position of the walk as $S_0 = 0$ and define:

$S_i = S_0 + \sum_{j=1}^iZ_j$

Your question then reduces to investigating the behavior of the random variable $S_n$. Specifically, you are asking for the mode of $S_n$ and $E(S_n)$.

By linearity of expectations, it can be seen that:

$E(S_n) = \frac{n}{3}$

My intuition suggests that the mode will coincide with the mean.

Title: Sum of Random Variables...


问:
Why is log-of-sum-of-exponentials $f(x)=\log\left(\sum_{i=1}^n e^ {x_i}\right)$ a convex function for $x \in\mathbb R^n$?
How to prove $f(x)=\log\left(\displaystyle\sum_{i=1}^n e^{x_i}\right)$ is a convex function?

EDIT1: for above function $f(x)$, following inequalities hold:

$$\max\{x_1,x_2,\ldots,x_n\}\leqslant f(x)\leqslant\max\{x_1,x_2,\ldots,x_n\}+\log n$$

and I have tried proving its convexity via definition of a convex function with above inequalities, but that didn't work.

EDIT2: I have posted my answers below.

答:
I have a preference for proving that the Hessian matrix is non negative as fully developed by @littleO.
Once you have shown that the gradient is the softmax function $S(x)$ and the Hessian matrix has the expression
$$
\nabla^2 f(x) = \operatorname{diag}(S(x)) - S(x)S(x)^T,
$$
you can simply use the convexity of the real map $t\mapsto t^2$ to conclude that
$\nabla^2 f(x)$ is non negative: Since $S(x)^T v$ is a convex linear combination of the coordinates of the vector $v$ (that is $S(x)_k\geq 0$ and $\sum_{k=1}^n S(x)_k =1$), one has
$$
v^T S(x) S(x)^T v
= (S(x)^T v)^2
= \left(\sum_{k=1}^n S(x)_k v_k\right)^2
\leq \sum_{k=1}^n S(x)_k v_k^2 = v^T \operatorname{diag}(S(x)) v.
$$


问:
Convergence of positive series $\sum_{n=1}^{\infty}\frac{1}{a_n}$.
Suppose $a_n>0$ for all $n\in \mathbb{N}$, and consider two series
$$\sum_{n=1}^{\infty}\frac{1}{a_n},\qquad \sum_{n=1}^{\infty}\frac{n}{a_1+a_2+\cdots+a_n}.$$

Due to the inequality from this question
$$\sum_{k=1}^{n}\frac{k}{a_1+a_2+\cdots+a_k}<2\sum_{k=1}^{n}\frac{1}{a_k},\qquad n\in\mathbb{N},$$
we can conclude that
$$\sum_{n=1}^{\infty}\frac{1}{a_n}<\infty\implies\sum_{n=1}^{\infty}\frac{n}{a_1+a_2+\cdots+a_n}<\infty.$$
$\textbf{What I want to do is to prove or disprove}$:
$$\sum_{n=1}^{\infty}\frac{n}{a_1+a_2+\cdots+a_n}<\infty\implies\sum_{n=1}^{\infty}\frac{1}{a_n}<\infty.$$
Some progress:
When $a_n<a_{n+1}$, it is easy to see
$$\frac{1}{a_n}<\frac{n}{a_1+a_2+\cdots+a_n},$$
which implies that
$$\sum_{n=1}^{\infty}\frac{n}{a_1+a_2+\cdots+a_n}<\infty\implies\sum_{n=1}^{\infty}\frac{1}{a_n}<\infty.$$

How about the general case? Any help and hint will welcome!

答: This isn't true. Start from any sequence where $\sum\frac{n}{a_1+\cdots+a_n}$ converges. If you choose any value $k$ and replace $a_k$ by $a_k+a_{k+1}-1$ and $a_{k+1}$ by $1$, you decrease the value of $\frac{k}{a_1+\cdots+a_k}$ without changing any other term, so this still converges. Do this for infinitely many different (non-consecutive) values for $k$ and the sequence $a_n$ has infinitely many terms equal to $1$, so $\sum\frac{1}{a_n}$ doesn't converge.


问:
derive general solution to first-order inhomogeneous pde
Given $u_t+cu_x=b(x,t)$ with initial value $u(x,0)=\phi(x)$, I am asked to derive the following general solution
$$u=\phi(x-ct)+\int_{0}^t b(x-c(t-\tau),\tau)d\tau$$

I am not sure how to proceed with this $b(x,t)$. I have solved such PDE where the explicit form of $b(x,t)$ given, but not in the general case. I tried the method of characteristics,
$$\frac{dt}{1}=\frac{dx}{c}=\frac{du}{b(x,t)}$$
but got stuck. Hope someone could help! Thanks.

答:
The general solution of equation $u_t+cu_x=b(x,t)$ is sum of general solution of homogenous equation $u_t+cu_x=0$ and the particular solution of inhomogenous equation $u_t+cu_x=b(x,t)$.

General solution of homogenous equation $u_t+cu_x=0$ can be obtained using method of characteristics. We compose the equation of characteristics $\frac{dt}{1}=\frac{dx}{c}$, and obtain its the first integral $x - c t$. So, general solution of homogenous equation is $u(x, t) = \psi(x - c t)$, where $\psi$ is a continuous-differentiable function.

The particular solution of equation $u_t+cu_x=b(x,t)$. We build it using extra condition $u(0, t) = 0$. It can be obtained using Fourier transform: $u(x, t) \overset{\mathcal{F}}{\rightarrow} \hat{u}(\omega,t)$
$$
\hat{u}(\omega,t) = \frac{1}{\sqrt{2 \pi}}\int\limits_{-\infty}^{+\infty} \exp(-i\omega x) u(x, t) dx,
$$
$$
u(x,t) = \frac{1}{\sqrt{2 \pi}}\int\limits_{-\infty}^{+\infty} \exp(i\omega x)\hat{u}(\omega, t) d\omega.
$$
For function $\hat{u}$ we get the Cauchy problem $\hat{u}_t+i \omega c \hat{u}=\hat{b}(\omega,t)$, $\hat{u}|_{t=0} = 0$. After solving it, we obtain formula
$$
\hat{u}(\omega,t) = \int\limits_0^t \exp(i \omega c (\tau - t)) \hat{b}(\omega,\tau) d\tau.
$$
So,
$$
u(x,t) = \frac{1}{\sqrt{2 \pi}} \int\limits_{-\infty}^{+\infty} \left(\int\limits_0^t \exp(i \omega c (\tau - t)) \hat{b}(\omega,\tau) d\tau \right) \exp(i\omega x) d\omega,
$$
or
$$
u(x,t) = \int\limits_0^t \frac{d\tau}{\sqrt{2 \pi}} \int\limits_{-\infty}^{+\infty}  \exp(i \omega c (\tau - t)) \hat{b}(\omega,\tau) \exp(i\omega x) d\omega.
$$
After calculating it using $\exp(i \omega c (\tau - t)) \overset{\mathcal{F}^{-1}}{\rightarrow} \sqrt{2 \pi} \delta (x - c (t -\tau))$ and convolution theorem, we obtain

$$u(x, t)=\int_{0}^t b(x-c(t-\tau),\tau)d\tau.$$

So, the general solution of equation $u_t+cu_x=b(x,t)$ is
$$
u(x, t)=\psi(x - c t) + \int_{0}^t b(x-c(t-\tau),\tau)d\tau,
$$
where $\psi$ is a continuous-differentiable function. Substituting it into Cauchy condition $u(x, 0)=\phi(x)$ leads to $u(x, 0)=\psi(x)=\phi(x)$.

Thus,
$$
u(x, t)=\phi(x - c t) + \int_{0}^t b(x-c(t-\tau),\tau)d\tau,
$$
is the solution of your Cauchy problem.


问:
Cute lemma on reduced words on basis elements of free group
In an answer to a 2014 MSE post here, user DonAntonio states the following "cute" fact, which I've restated here in simpler terms for ease-of-reading:

Lemma: Let $F_n$ be the free group of rank $n$ with basis $\{ w_1 \dots w_n\}$. Then $\prod w_{a_i}^{b_i} \in [F,F] \iff$ for each $w_i$, the sum of the powers to which $w_i$ is raised to in this word equals zero.

So for example, $w_1^{-2}w_2w_3w_1^2w_3^{-1}w_2^{-1} \in [F_3,F_3]$ but $w_1^4w_2^{-3}w_1^3w_2^3 \not \in [F_3,F_3]$.

I'm trying to prove this, but I've only started getting into group theory this semester and am a bit uneasy with all...the words.

(Originally, I had a proof attempt here. It was nowhere near complete nor insightful. I like shorter posts, so I removed it.)

答:
Denote the right-hand side property by (P). Then it is easy to see that

(1) if two elements of $F_n$ satisfy (P), then so does their product,

(2) each element of the form $[x,y]$ (i.e. each generator of $[F_n,F_n]$) satisfies (P).

This establishes "$\implies$".

For the other direction, your attempt at induction over the word length is a good idea:

The start of the induction is trivial.

For the induction step, use the (likewise trivial) formula
$$w_k^b\,u\,w_k^c\,x=[w_k^{-b},u^{-1}]\,u\,w_k^{b+c}\,x$$
and the fact that if $w_k^b\,u\,w_k^c\,x$ satisfies (P), then so does $u\,w_k^{b+c}\,x$, and the length of the latter is one less than that of the former.


问:
Will the "worst case scenario" work for all pigeon hole questions?
I am looking for a generalized method to solve simple to advanced pigeonhole principle questions (that are PURELY pigeon-hole ie not in combination with any other topics like probability)

The definition:

Suppose that $n$ pigeons fly into $d$ pigeonholes. Performing the division, let $n = qd + r$, where the remainder is $r = 0, 1, 2, . . . , d − 1.$

If $r = 0$, there must be at least one pigeonhole with $q$ pigeons.

If $r > 0$, there must be at least one pigeonhole with $q + 1$ pigeons.

The least number of pigeons that will guarantee that there is at least one pigeonhole with at least
$q + 1$ pigeons is $qd + 1$

Normally I don't use the above definition because I dont really understand it but I can solve wuestion anyway by intuition. So I am asking: If $r=0$, why is $n = q$ and not $qp$? If $r > 0$, why is the remainder $q+1$? Since the remainder can be up to $d-1$, I don't understand where the $q+1$ comes from. And From the third statement, I have realised that I am probably interpreting the law incorrectly: it seems that I should be ignoring $d$ for some reason or maybe I am just supposed to set it equal to $1$?

My second question is: will considering the "worst case scenario" work for every question? This method has not failed me yet but if I search online, there is hardly any information on it (and on the PHP in general).

Typical questions involve something along the lines of "how many socks/ties/hats would I have to pick to ensure two of the same color" and I just consider the worst case scenario where I delay the desired outcome until the very end. Then I just add one to force the ceiling function to go to the next positive integer, which basically ensures the "same color" pair. HOWEVER, when the "worst case scenario" gets more abstract eg $$\text{Prove by using PHP that there are two powers of $2$ that differ by a multiple of 8479}$$

I cannot identify the worst case scenario. Basically,should I be able to use it in all PHP questions?

I have a statement which I think is what I need to prove:

RTP $2^a - 2^b = 8479k$, where $a,b,k$ are integers.

But I'm not sure how to proceed from here USING MY METHOD. I know there is another method that consists of considering a set of powers of $2$ and dividing by $8480$ to produce $8479$ remainders. But in this case, is there a "worst case scenario"?

答:
The version of pigeonhole principle that you cited is ridiculous. Not because it is wrong, but because it is NOT the right way to learn at all. In fact, it is completely pointless to learn the name "pigeonhole principle". Instead, just think as follows:

If you have $d$ groups, each having at most $k$ objects, then in total you have at most $d·k$ objects. That is it. Now think what this implies for the applications you want.

No need for your vague intuition about 'worst-case scenario' either. If every power of two is different modulo $m$, then obviously there are at most $m$ powers of two.......


问:
Prove the following statement using the contrapositive method. If $ n^2 $ is divisible by 3 then n is divisible by 3.
Prove the following statement using the contrapositive method. If $ n^2 $ is divisible by 3 then n is divisible by 3.

I tried using contrapositive method by assuming if not b then not a, i.e., if n is not divisible by 3 then $ n^2$ is not divisible by 3. and prove it.

Is this the correct way to do it or am I doing it wrong?

If anyone could layout an explanation could be helpful. Thanks!

答:
With contrapositive method：

If n is not divisible by 3, let n = 3m + k, where m, k are integers and k=1,2 only.
好
Then
$n^2$  = $(3m+k)^2$=9$m^2$+6mk+$k^2$.

If k=1, $n^2$  = $(3m+k)^2$=9$m^2$+6mk+1=3(3$m^2$+2mk)+1, not divisible by 3.

If k=2, $n^2$  = $(3m+k)^2$=9$m^2$+6mk+3+1= 3(3$m^2$+2mk+1)+1,  not divisible by 3.

This is in contradictory with the assumption $n^2$ is divisible by 3. So n must be divisible by 3.


问:
Minimize $(x+y)(x+z)$ with constraint without calculus

Let $x,y,z \in \mathbb R^+$ such that $xyz\cdot(x+y+z) = 1$

Find $\min\{(x+y)(x+z)\}.$

Using calculus, and Lagrange multipliers, I get:

$(x+y)(x+z) \ge2$ (with the equality occurring if and only if $y=z=1,\ x=\sqrt{2} - 1$).

But I want to solve it in an easy way, which doesn't need calculus.

How can I do it?
Thanks.

答: Hint: Your expression is $x(x+y+z)+yz= 1/yz+yz$</p>


问:
If $E=X-x$ find $\sqrt{X}-\sqrt{x}$
In the footnotes of my old (numerical methods and types of errors) lecture notes there is a question to find $\sqrt{X}-\sqrt{x}$ given that $E=X-x$. I am assuming that $E$ represents the error difference. The notes are old and not very clear but it mentions that
$\sqrt{X}-\sqrt{x}$ is proportional to $E/2\sqrt{X}$. Is there a way to show that?

答: Since $E=X-x=(\sqrt X-\sqrt x)(\sqrt X+\sqrt x)$, when the error is small you can approximate $(\sqrt X+\sqrt x)$ by $2\sqrt X$. Hence $\sqrt X-\sqrt x$ is around $E/2\sqrt X$.

Title: If $E=X-x$ find $\sqrt{X}-\sqrt{x}$


答:
For an arbitrary group $G$, we have $g \in [G, G]$ if and only if $\bar{g} = 1$ in the abelianization $G/[G, G]$. For the free group, $F_n/[F_n, F_n]$ is just $\mathbb{Z}^n$ (take your pick of arguments). The stated property follows immediately.

(Edit: Hah! I just noticed my link equals yours. Well, it's another approach, anyway.)

Title: Cute lemma on reduced words on basis elements of free group

Body:
In an answer to a 2014 MSE post here, user DonAntonio states the following "cute" fact, which I've restated here in simpler terms for ease-of-reading:

Lemma: Let $F_n$ be the free group of rank $n$ with basis $\{ w_1 \dots w_n\}$. Then $\prod w_{a_i}^{b_i} \in [F,F] \iff$ for each $w_i$, the sum of the powers to which $w_i$ is raised to in this word equals zero.

So for example, $w_1^{-2}w_2w_3w_1^2w_3^{-1}w_2^{-1} \in [F_3,F_3]$ but $w_1^4w_2^{-3}w_1^3w_2^3 \not \in [F_3,F_3]$.

I'm trying to prove this, but I've only started getting into group theory this semester and am a bit uneasy with all...the words.

(Originally, I had a proof attempt here. It was nowhere near complete nor insightful. I like shorter posts, so I removed it.)


问:
Characterization of continuity in terms of closure
Let $f:X \to Y$ be a continuous function between topological spaces then $f(\bar A) \subseteq \overline{f(A)}$. I know that for every $A \subset X$ the subset $f^{-1}( \overline{f(A)})$ is a closed one that containing $A$. Now how can I deduce that $\bar A \subseteq f^{-1}( \overline{f(A)})$?

答:
Indeed $A \subseteq f^{-1}[f[A]] \subseteq f^{-1}[\overline{f[A]}]$, the first by definition of $f^{-1}$ and $f[A]$ the second as $f[A] \subseteq \overline{f[A]}$. And by continuity of $f$, the right hand set is closed, as the inverse image of a closed set.

Now, two ways to proceed: always $B \subseteq C$ implies $\overline{B} \subseteq \overline{C}$ and so $\overline{A} \subseteq \overline{f^{-1}[\overline{f[A]}]} = f^{-1}[\overline{f[A]}]$ (as $\overline{B}=B$ iff $B$ is closed). Or just use that the closure of $A$ is the smallest closed set that contains $A$ and $f^{-1}[\overline{f[A]}]$ is a closed subset that contains $A$, so the minimality of closure can be invoked.

Either way $\overline{A} \subseteq f^{-1}[\overline{f[A]}]$ which implies that if $x \in \overline{A}$, $f(x) \in \overline{f[A]}$ so that $f[\overline{A}] \subseteq \overline{f[A]}$, for all $A \subseteq X$ and $f$ continuous.

For the reverse see this question, which shows that this property could be used as an alternative definition of continuity.

A "pointwise" way to see the inclusion: let $x \in \overline{A}$ and let $V$ be any neighbourhood of $f(x)$, as $f$ is continuous at $x$, there is some neighbourhood $U$ of $x$ so that $f[U] \subseteq V$. Because $x \in \overline{A}$, there is some $a \in A \cap U$ and then $f(a) \in f[A] \cap f[U] \subseteq f[A] \cap V$. So $V$ intersects $f[A]$ and as $V$ was arbitrary, we conclude that indeed $f(x) \in \overline{f[A]}$.


答:
$$u_t+cu_x=b(x,t)$$
You have correctly written the Charpit-Lagrange system of ODEs :
$$\frac{dt}{1}=\frac{dx}{c}=\frac{du}{b(x,t)}$$
A first characteristic equation comes from solving $\frac{dt}{1}=\frac{dx}{c}$ :
$$x-ct=C_1\quad\implies\quad x=C_1+ct$$
A second characteristic equation comes from solving $dt=\frac{du}{b(x,t)}=\frac{du}{b\left(C_1+ct\:,\:t\right)}$
$$du-b\left(C_1+ct\:,\:t\right)dt=0$$
$$u-\int_0^t b\left(C_1+c\tau\:,\:\tau\right)d\tau=C_2$$

The general solution of the PDE $\quad C_2=\phi(C_1)\quad$ is :
$$u-\int_0^t b\left((x-ct)+c\tau\:,\:\tau\right)d\tau=\phi(x-ct)$$

$\phi$ is an arbitrary function (to be determined according to some initial condition).
$$u=\phi(x-ct)+\int_{0}^t b(x-c(t-\tau),\tau)d\tau$$

Title: derive general solution to first-order inhomogeneous pde

Body:
Given $u_t+cu_x=b(x,t)$ with initial value $u(x,0)=\phi(x)$, I am asked to derive the following general solution
$$u=\phi(x-ct)+\int_{0}^t b(x-c(t-\tau),\tau)d\tau$$

I am not sure how to proceed with this $b(x,t)$. I have solved such PDE where the explicit form of $b(x,t)$ given, but not in the general case. I tried the method of characteristics,
$$\frac{dt}{1}=\frac{dx}{c}=\frac{du}{b(x,t)}$$
but got stuck. Hope someone could help! Thanks.


答:
Two facts to note:

If $n$ is not divisible by 3, then $n\equiv1 \ \text{mod} \ 3$ or $n\equiv2 \ \text{mod} \ 3.$

If $a\equiv b \ \text{mod} \ n$, then $a^2 \equiv b^2 \ \text{mod} \ n$.

Body:
Prove the following statement using the contrapositive method. If $ n^2 $ is divisible by 3 then n is divisible by 3.

I tried using contrapositive method by assuming if not b then not a, i.e., if n is not divisible by 3 then $ n^2$ is not divisible by 3. and prove it.

Is this the correct way to do it or am I doing it wrong?

If anyone could layout an explanation could be helpful. Thanks!