问:
Prove $D(a;R_1,R_2)$ is a connected set
I was solving problems from the start of my Complex Analysis course, and I found this one (the beggining of my course focuses a lot in topology):

Prove that $D(a; R_1,R_2)$ is a connected set.

The notation $D(a;R_1,R_2)$ refers to the open disk of center $a$ and with radius greater that $R_1$ but lower that $R_2$ (I guess, because this is the first time I see a disk notation with two different radius, tell me if I'm wrong) (Personally, I prefer the ball notation, but this is the one my course notes use). Also consider I'm working inside the metric space $(\mathbb{C},d)$, where $d$ is the euclidean metric function. To be said:

$$D(a;R_1,R_2)=\{z\in\mathbb{C} : |z-a|>R_1, |z-a|<R_2\}$$

Intuitively, I understand this set is connected, since it's like a "donut", but I have trouble proving it by using the definition of connected space, to be said, proving it isn't union of to open (or closed) sets. I'm not sure if the concept of connected by arcs is useful here, though I clearly see it in my mind I don't know how to write a consistent proof of it.

How can I proof that set is connected? Any help will be appreciated, thanks in advance.

答:
These things are often called "annuli" (in the singular, "annulus"), from the Latin for something that looks like a ring.  It makes intuitive sense that these things are arcwise connected because there is nothing stopping you from drawing a curve from any point in one of these things to any other without leaving the annulus.

If you want example formulas, I'll handle the case $a = 0$ because it's simple.  In polar form, $D(a,R_1,R_2) = \{r e^{i\theta}: R_1 < r < R_2\}$.  I'll just write $D$ for this set.

Fix any $z$ and $w$ in $D$ and choose positive $r_0, r_1$, and real $\theta_0$ and $\theta_1$ such that $z = r_0 e^{\theta_0}$ and $w = r_1 e^{\theta_1}$.  Note that because $z$ and $w$ are in $D$, $r_0$ and $r_1$ are both in the real interval $(R_1, R_2)$.

Consider the path $\gamma: [0,1] \to \mathbb{C}$ given by
$$
\gamma(t) = ((1-t) r_0 + t r_1) \exp(i((1-t) t_0 + t \theta_1)), \qquad t \in [0,1].
$$
Convince yourself that $\gamma$ is continuous and that $\gamma(0) = z$ and $\gamma(1) = w$, and it may help to think geometrically about what $\gamma$ does.  As $t$ goes from $0$ to $1$ it just moves the modulus and angle from that of $z$ to that of $w$ in a linear way; it'll look a bit like part of a spiral or something - maybe graph a few examples.

Fix $t$.  We have that
$$
|\gamma(t)| = (1-t) r_0 + t r_1
$$
Convince yourself via the fact that $r_0, r_1$ are both in $(R_1, R_2)$ that $|\gamma(t)| \in (R_1, R_2)$ also.  (In other language $|\gamma(t)|$ is a "convex combination" of $r_0$ and $r_1$ and $r_2$, so will lie in any convex set that contains $r_1$ and $r_2$.  But you don't need this language to verify the simple inequalities.)  So $|\gamma(t)| \in D$.

So $\gamma$ is an arc in $D$ connecting $z$ to $w$.  Since $z$ and $w$ were arbitrary, $D$ is arcwise connected.

The general case could be handled via a use of this case, or by just adding "$a+$" to a lot of the formulas.

Title: Prove $D(a;R_1,R_2)$ is a connected set


问:
Probability that the number of heads is coprime with the number of throws
Let's throw a coin $n$ times. Say that we have obtained heads $k$ times and tails $n-k$ times. I'm interested in the probability that the number of heads is coprime with the total number of throws. Obviously, this probability is described by the formula
$$
P=\sum_{\substack{0<k<n,\\gcd(k, n)=1}}\binom{n}{k}p^k(1-p)^{n-k},
$$
where $p\in(0, 1)$ is a probability of throwing heads.

The question.
Is it true that this sum is lower bounded by some positive constant for all integers $n > 1$?

My thoughts. I think that it should be true. Take $p=1/2$ for simplicity. Some computer calculations suggest that $P\ge 0.1875$, and this bound is achieved for $n=6$. But I don't know how to prove it.

It's also easy to show that $P\ge c n^{-1/2}$ for some constant $c$ because we can always find such number $k$, which is coprime with $n$ and close to $n/2$. But this bound is too weak.

Update.
@donaastor made some interesting comments below and convinced me that for all constant $p$ the lower limit of the probability equals 0. Also, further computer calculations for $p=1/2$ showed that the probability for $n=30030$ is lower than for n=6 and equals $\approx 0.1828$.

答:
If we define $P_d$ as the sum over $k$ multiples of $d$, one can see that

$$ P_{coprime} = \sum_{d \mid n } \mu(d) P_d $$

Where $\mu$ is the Moebius function. This is sophisticated way of expressing an inclusion-exclusion: first you put all terms via $P_1$; then you subtract all stuff that have a prime in common with $n$ via $-P_q$ for q prime; then you have to add back things that are multiple of two primes... Moebius function implements this in a neat way. Now $P_d$ is a bit easier to describe.  Let

$$F_n(z) = (pz + 1-p) ^n$$

his is a polynomial in $z$ which has exactly the terms of your summation in each degree: in particular $F_n(1) =1$ is the total probability. This is also called the generating function of the distribution.

If you average on d-th roots of unity $\zeta_d$ you will get rid of terms which are not divisible by d: this is called root of unity sieve technique. Now let $\zeta$ by a primitive $n$-th root of unity. We have that $\zeta^{nk/d} $ for $k=1, \ldots, d$ enumerates all the d-roots of unity. Thus

$$ P_d = \frac{1}{d} \left ( \sum_{k=1}^d (p \zeta^{nk/d} + 1-p) ^n \right ) $$

Plugging this into $P_{coprime}$ yields

$$ P_{coprime} = \sum_{d \mid n} \mu(d) \frac{1}{d} \left ( \sum_{k=1}^d (p \zeta^{nk/d} + 1-p) ^n \right ) $$

We want to exchange this summation in terms of when a given root of unity appears. A power $( \ldots )^n $ containing the root $\omega = \zeta^r$ appears for $r=nk/d$ with a coefficient of $\mu(d) /d$ in front. In particular, it appears exactly when $n/d \mid r$. Changing variable $d' = n/d$ we can thus write

$$ P_{coprime} = \sum_{r=1}^n (p \zeta^r + 1-p) ^n \sum_{d' \mid r} \mu(d') \frac{1}{d'}  = \sum_{r=1}^n (p\zeta^r + 1-p)^n \frac{\varphi(r)}{r} = \sum_{r=1}^n (1+ p(\zeta^r-1))^n \frac{\varphi(r)}{r} $$

For $p << 1$ we get

$$ \simeq \sum_{r=1}^n (1+np(\zeta^r-1)) \frac{\varphi(r)}{r} $$

And I think we can approximate this number by its real part, because it is close in norm to $P_{coprime}$ which is real. This we get

$$\simeq \sum_{r=1}^n (1+np(\cos(2\pi r/n) -1)) \frac{\varphi(r)}{r} = (1-np) \sum_{r=1}^n \frac{\varphi(r)}{r} +np \sum_{r=1}^n \cos(2\pi r/n) \frac{\varphi(r)}{r} $$

ContentLicense: CC BY-SA 4.0


问:
concentration inequality and how to bound densities of a sequence of random variables
Let $X_n$ be a sequence of random Variables such that $E(X_n)=0$ and $Var(X_n)=1$
I need to bound the following:

$$
P(|X_n|<\delta_n)\lesssim \delta_n
$$

for a positiv sequence $\delta_n$ tending to zero.
My question is under which asumptions this inequality holds.
Lets first assume that there is a density $f_{X_n}$ according to the lebesgue measure for each $X_n$. Then it follows
$$
P(|X_n|<\delta_n)=\int^{\delta_n}_{-\delta_n}f_{X_n}.
$$
Do i need additional asumptions to bound $f_{X_n}<c$ for all $n$ in order to bound the Integral by $2c\delta_n$ or is it somehow enough to assume that $Var(X_N)=1$?

答:
Let $X_N$ be $-N, 0$, or $N$ with probabilities $\frac{1}{2 N^2}, 1-\frac{1}{N^2}, \frac{1}{2 N^2}$.

Then, $E[X_N]=0$ and $Var[X_N]=1$.

But, for any $\delta>0$, $P[|X_N|<\delta] \ge 1-\frac{1}{N^2}\rightarrow 1$.

So, for any sequence of positive $\delta_N$ that converges to $0$, you also have $P[|X_N|<\delta_N]\rightarrow 1$

That shows $Var[X_N]=1$ is not sufficient.

You should be more precise about what the approximation means.

If $F_N$ is the cdf, and you want $$\lim_{N\rightarrow \infty}\frac{F_N(\delta_N)-F_N(-\delta_N)}{\delta_N}=1$$
then essentially this means (provided the derivative of $F_N$ exists and is denoted $f_N$):

$$1=2\lim_{\delta \rightarrow 0}\frac{F_N(\delta)-F_N(-\delta)}{2\delta}=2f_N(0)$$


问:
Does $CD$ bisect $AE$?
Construct triangle $ABC$.choose a $D$ point on the side $AB$ such that $\frac{BD}{AD}=\frac{2}{1}$.Now draw median $AE$. $AE$ intersects $CD$ at $O$ .İf area of triangle $AOD$ is 20, what is the area of triangle $ABC$? İ have tried using formula $S=\frac{1}{2}AB\cos \alpha$ for triangles $AOD$ and $AEB$ and comparing them but the side $AO$ is different than $AE$ so im so confused

答:
To answer your question: Yes, $O$ is the midpoint of $AE$, and is a consequence of Menelaus' Theorem. However you do not need this to solve your question. We just need the fact that the ratio of the area of two triangles sharing the same height is equal to the ratio of the length of their bases.

Construct the segment $OB$.

Given that $AD : DB = 1:2$, the area of $\triangle BOD$ is twice that of $\triangle AOD$, i.e. its area is $40$.

Since $BE =EC$, the areas of $\triangle OEB$ and $\triangle OEC$ are equal. The areas of $\triangle AEB$ and $\triangle AEC$ are equal too. Hence the areas of $\triangle AOB$ and $\triangle AOC$ are equal by subtraction. Therefore the area of $\triangle AOC = 60$.

Now we finish off by noting that the area of $\triangle CDB$ is twice that of $\triangle CDA$.

Hence the area of $\triangle ABC = (20 + 60)\times 3 = 240$.

We can also find that the areas of the triangles $OEB$ and $OEC$ are $60$, which is equal to that of $\triangle AOC$, and gives an alternate proof that $CD$ does indeed bisect $AE$.

Title: Does $CD$ bisect $AE$?


问:
Show that if $x\geq 0$ and $n$ is a positive integer, then $\sum_{k=0}^{n-1}\left\lfloor {x+\frac{k}{n}}\right\rfloor=\lfloor {nx}\rfloor$
I need help with this question, where $\lfloor x\rfloor$ means the floor function of $x$.

Show that if $x\geq 0$ and $n$ is a postive integer, then $$\sum_{k=0}^{n-1}\left\lfloor x+\frac{k}{n} \right\rfloor =\lfloor nx\rfloor$$

答:
For given $x\in{\mathbb R}$ write
$$x=j+{r\over n}+\xi\qquad{\rm with}\qquad j\in{\mathbb Z}, \quad 0\leq r<n, \quad 0\leq\xi<{1\over n}\ .$$
Then
$$\lfloor nx\rfloor=\lfloor nj + r+n\xi\rfloor=nj+r\ .$$
On the other hand
$$\left\lfloor x+{k\over n}\right\rfloor=\left\lfloor j+{r+k\over n}\right\rfloor=\cases{j\quad&$(0\leq k<n-r)$ \cr
j+1\quad&$(n-r\leq k<n)$\cr}$$
and therefore
$$\sum_{k=0}^{n-1}\left\lfloor x+{k\over n}\right\rfloor= nj + r\ ,$$
as above.</p>

ContentLicense: CC BY-SA 3.0


问:
Probability of stepping on n-th field in game where you move using dice

We play a game where in each round we roll a six-sided dice and move the appropriate number of squares. How likely are we to step on the $n$-th field during the game, if we started with the number $0$?

The answer is sufficient in the form of a recurrent formula, which refers to a constant number of values for the lower one. Can you use only $2$ values for lower?

I am kinda lost here. I did bit know how even approach this.

Thanks to anyone who can help me guide to right direction

I made this:

```<code>| 1 | 1                                             |
| 2 | 1+1,2                                         |
| 3 | 1+1+1, 2+1, 3                                 |
| 4 | 1+1+1+1, 1+1+2, 1+3, 2+2, 4                   |
| 5 | 1+1+1+1+1, 1+1+1+2, 1+2+2, 1+1+3, 1+4, 2+3, 5 |
</code>```

But I can not see any pattern or how I can calculate probability from it. Does order of numbers matter ?

答:
At each step you can move by $x_k = 1,2, \ldots, 6$, so after $m$ steps you have moved away from $0$ of $x_1+x_2+ \cdots + x_m$ units.

You can reach to $n$ in min $\left\lceil {n/6} \right\rceil $ and max $n$ steps.

Once you reach to it in $m$ steps you will pass beyond it in the next steps.

At each step you have $6$ equally probable outcomes, after $n$ steps they are $6^n$.

The number of outcomes that bring you to $n$ im steps are the number of
compositions of $n$ into $m$ parts with each part in the range $[1,6]$,i.e.
$$
\eqalign{
  & {\rm No}{\rm .}\,{\rm of}\,{\rm solutions}\,{\rm to}\;\left\{ \matrix{
  {\rm 1} \le {\rm integer}\;x_{\,j}  \le 6 \hfill \cr
  x_{\,1}  + x_{\,2}  + \; \cdots \; + x_{\,m}  = n \hfill \cr}  \right. =   \cr
  &  = {\rm No}{\rm .}\,{\rm of}\,{\rm solutions}\,{\rm to}\;\left\{ \matrix{
  {\rm 0} \le {\rm integer}\;y_{\,j}  \le 5 \hfill \cr
  y_{\,1}  + y_{\,2}  + \; \cdots \; + y_{\,m}  = n - m \hfill \cr}  \right. =   \cr
  &  = N_b (n - m,5,m) \cr}
$$
where $Nb$ is given by
$$
N_b (s,r,m)\quad \left| {\;0 \leqslant \text{integers  }s,m,r} \right.\quad  =
\sum\limits_{\left( {0\, \leqslant } \right)\,\,k\,\,\left( { \leqslant \,\frac{s}{r+1}\, \leqslant \,m} \right)}
{\left( { - 1} \right)^k \binom{m}{k}
 \binom
 { s + m - 1 - k\left( {r + 1} \right) }
 { s - k\left( {r + 1} \right)}\ }
$$
as explained in this related post.

Concerning the recurrence, it is clear that $N_b$ satisfies the following (among others)
$$
N_b (s,r,m) = \sum\limits_{i\, = \,s - r}^s {N_b (i,r,m - 1)}
$$

I leave to you to continue with adapting the above to your case (summing over $m$, ..).


问:
How to prove $C'$ is reflection of $C$ when we know $\triangle ABC\cong \triangle ABC'$?
We know $\triangle ABC\cong \triangle ABC'$ (the triangles are congruent). We may assume $C\neq C'$. How do I prove $C'$ is the reflection of $C$ in the line $AB$?

I know I first need to prove the fact that the orthogonal projections of $C$ and $C'$ on the line $AB$ are the same, but I really don't know how to start.

答:
ABC, ABC' are congruent mounted on the same side AB. Since CC' should make equal angles with AB and since thay are adjacent, ( total angle $180^{\circ}$) each angle is $90^{\circ}$, C reflects to C'.

ACBC' is a kite.


问:
Initial value problem $y''=e^{2y}, y(0)=0, y'(0)=1$
Supposedly, it is possible to determine information about the constants of this IVP solution, without computing the solution of the differential equation.  Here's how I solve this.

Let $z=y'$. In brief, we then have the separable equation $$z\frac{dz}{dy}=f(y,z)=e^{2y}$$

and solving for $z$ $$z=\sqrt{e^{2y}+c}$$

which can be made into a separable equation and integrated. After playing around with the initial conditions, I did find that $c=0$. But this is a contradiction with the results of the second separable equation. Can we really solve $c$ before finding the explicit form of $y$?

答:
You have $y'' = e^{2y}$. Multiplying by $y'$ you have $y'y'' = y'e^{2y}$, which you can integrate:
\begin{align*}
\int y'y''\, dt &= \int y' dy' \\
&=\tfrac{1}{2}(y')^2 + C
\end{align*}
where $C$ is a constant of integration.
On the other side you have
\begin{align*}
\int y' e^{2y}\, dt &= \int e^{2y}\, dy \\
&= \tfrac{1}{2}e^{2y} + D
\end{align*}
with $D$ also a constant of integration. Combining you end up with
\begin{equation*}
(y')^2 = e^{2y} + 2E
\end{equation*}
where $E = C + D$.
So if you happen to know $y(0)$ and $y'(0)$, you have your $E$.


问:
Question regarding fixed point (integrals)
Let $f:[0,1]\rightarrow\mathbb{R}$  be a continuous function. Show that, if $$\int\limits_0^1 f(t)\,\mathrm{d}t =\frac{1}{2},$$ then $f$ has a fixed point in $[0,1]$, i.e. $\exists x_0\in [0,1]$ such that $f(x_0)=x_0$.

Body: Apply MVT on $[0,1]$ for $$F(x)=\int_0^xf(t)\;dt-\frac{x}{2}$$


答:
Hint

Set $g(x)=f(x)-x$. It continuous and s.t. $\int_0^1 g(x)\,\mathrm d x=0$. Your problem is equivalent as proving that $g$ cancel at some point $c\in [0,1]$. Suppose by contradiction that it doesn't, and see what happen.

Title: Question regarding fixed point (integrals)

Body: Let $f:[0,1]\rightarrow\mathbb{R}$  be a continuous function. Show that, if $$\int\limits_0^1 f(t)\,\mathrm{d}t =\frac{1}{2},$$ then $f$ has a fixed point in $[0,1]$, i.e. $\exists x_0\in [0,1]$ such that $f(x_0)=x_0$.


问:
Basic ring isomorphism question
Let $R=\mathbb{Z}[x], I=(x^2+1,x+1).$ Prove that $R/I \cong \mathbb{Z}[i]/(i+1) \cong \mathbb{Z}/2\mathbb{Z}$.

I am confused with the rather messy looking of $R/I$. My first step is to define the homorphism $f: R/I \rightarrow \mathbb{Z}[i]$ but then I don't know what to do, please help!

Title: Basic ring isomorphism question

Body:
Alternatively, one may use the $\color{blue}{\rm FIT}$ = First (vs. Third) Isomorphism Theorem  as follows.

$\quad  \smash[t]{\Bbb Z\stackrel{h}{\to}}\, \Bbb Z[i\,]/(1\!+\!i)\ \ {\rm is}\ \ \color{#0b0}{\bf onto,}\ \ {\rm by\ \ mod}\,\ 1\!+\!i\!:\ i\,\equiv -1\Rightarrow\:a\!+\!b\,i\,\equiv a\!-\!b\in \Bbb Z\ \\[0.1em]
\quad  n\in \ker\ h\iff 1\!+\!i\,\mid n\iff \dfrac{n}{1\!+\!i}\, =\, \dfrac{n\,(1\!-\!i\,)}2\,\in\, \Bbb Z[i\,] \iff \color{#c00}2\mid n\\[1.2em]
\quad {\rm So} \ \ \ \Bbb Z[i\,]/(1\!+\!i\,)\, \color{#0b0}{\bf =\ Im\:h}\!\smash[t]{\stackrel{\color{blue}{\rm\ \ \ FIT_{\phantom{I^2}}}}\cong} \Bbb Z/\ker h \,=\, \Bbb Z/\color{#c00}2\,\Bbb Z$

Similarly for $\,\Bbb Z[x]/I,\,\ I = (x^2\!+\!1,x\!+\!1).\,$  $\, x\equiv -1\ $ so $\ h\,$ is onto.  $\  \ker h = \color{#c00}2\,\Bbb Z\,\ $ by

$\quad \color{#c00}2 = x^2\!+\!1-(x\!-\!1)(x\!+\!1)\in I,\ \ \ n\in I\,\Rightarrow\, n = (x^2\!+\!1)f+(x\!+\!1)g \smash[t]{\stackrel{\large\,\ x\,=\,-1\,\ }\Rightarrow} n = \color{#c00}2f(-1)$

Similarly $\ \Bbb Z[x]/(f(x),x\!-\!a)\,\cong\,\Bbb Z/\color{#c00}{f(a)}\Bbb Z\ \ $ by $\, \ f(x)\equiv \color{#c00}{f(a)}\,\pmod{\!x\!-\!a}$


问:
Possible length of median and height
In a triangle $ABC$ let $D$ be the intersection point of the side $AB$ with the angle bisector of the inner angle at $C$.

It holds that $|AC|=4$, $|CD|=3$.

Let the inner angle at $A$ be equal to $\frac{\pi}{6}$.

(a) Which length are possible for the median through $C$ and the height through $C$ of the triangle $ABC$ ?

(b) In each case calculate the legth of the radius of the circumscribed circle of $ABC$.

$$$$

I don't really see how $CD$ helps us.

For (a) do we use the apply the cosine law?

答:
If $CH$ is the height of $\triangle ABC$ through $C$, $AB = c, BC = a, AC = b$,

$CH = b \displaystyle \sin 30^0 = 2$

Applying Pythagoras,

$AH = \sqrt{AC^2 - CH^2} = 2 \sqrt3$

$DH = \sqrt{CD^2 - CH^2} = \sqrt5$

i) If $H$ is between $A$ and $D, AD = 2\sqrt3 + \sqrt5 \gt 5$.

In $\triangle ACD, AC = 4, CD = 3, AD \gt 5$, which means $\angle ACD \gt 90^0$ but $CD$ is internal angle bisector hence $\angle ACD$ must be less than $90^0$. So we conclude $H$ is not in between $A$ and $D$.

ii) If $D$ is between $A$ and $H, AD = AH - DH = 2 \sqrt3 - \sqrt5 \lt 3$.

In $\triangle ACD, AD$ is the smallest side and hence $\angle ACD \lt 30^0$ which means $\angle C \lt 60^0$ and hence $\triangle ABC$ is obtuse angled triangle with $\angle B \gt 90^0$.

If $BH = x$, applying angle bisector theorem,

$\displaystyle \frac{4}{a} = \frac{2 \sqrt3 - \sqrt5}{\sqrt5 - x}$ ...($1$)

Applying Pythagoras, we also have $x^2 + 2^2 = a^2$ ...($2$)

Solving $(1)$ and $(2)$ and I leave further workings to you, we get $a \approx 2.5, b = 4$ and $c \approx 2$. Knowing the sides, you can apply the formula for median in terms of the side lengths of $\triangle ABC$ or you can apply law of cosine.

Title: Possible length of median and height


问:
Landau big $\Omega$ notation
I have seen a table of Landau notation definitions which defines
$$f \in \Omega(g) \iff \exists k, N > 0 \colon \forall  n > N, \,f(n) \geq k \,g(n)$$

But every other definition in the table puts the $f$ and $g$ in their respective rhs analogs each in absolute values and writes the respective inequalities so that they are strict. If we did so here, then the rhs would read "$|f(n)| > k |g(n)|$" instead of the present "$f(n) \geq k \,g(n)$".

I like this more because it maintains symmetry. However, it is not technically equivalent if, for example, $\exists N > 0 \colon \forall n > N, g(n) < 0$. But I have never come across such a bound in practice. Elaborating on that particular case, any eventually-nonnegative function would be $\Omega$ of any eventually-negative function, which seems useless. But maybe I am not seeing some other situation in which this freedom is nice to have, particularly if equality is permitted.

So, my question is whether there is any practical utility in intentionally failing to require both that $g$ be nonnegative and that the inequality be strict (prohibiting the possibility of equality). Can we alter the definition in the table so that "$g(n)$" be replaced with "$|g(n)|$" and "$\geq$" be replaced with "$>$"?

答: There are several definitions of the "Landau" notations floating around. Look at how it is used, you might see that the author has only the case $f, g$ positive in mind. This definition of $\Omega$ makes little sense otherwise anyway.

Title: Landau big $\Omega$ notation


问:
number of subgroup in $S_n$ which are isomorphic to $\mathbb{Z}_n$
This is a generalization of my previous question in number of subgroups in $S_5$ which are isomorphic to $\mathbb{Z}_5$

I want to find the number of subgroups in $S_n$ that are isomorphic to $\mathbb{Z}_n$.

Inspired by @Brian M. Scott, I write my proposal as follow:

Since $\mathbb{Z}_n$ is cyclic, I guess $p=\{1,2,3,4,5,\cdots, n\}$ and $p, p^2, p^3, \cdots, p^n=e$.

Letting $p=(1,2,3,\cdots, n)$ and since it is cyclic there are $(n-1)!$ case of permutations. So my guess is $\frac{(n-1)!}{n-1} = (n-2)!$

Am I right?  Is there any other way to find the number of subgroups?

答:
No.  Powers of $n$-cycles $(123\dots n)$ are not $n$-cycles unless the power is prime to the order.  And there are more elments of order $n$ than just an $n$-cycle.

Examples:

$n=4$: There are indeed $3!=6$ 4-cycles, but each $C_4$ only contains 2 of them.

$n=6$: cycletypes (6) and (2,3).  There are $5!=120$ cycletype (6) and another $\binom{6}{3,2,1}$ cycletype (2,3).  Each $C_6$ contains $\phi(6)=2$ generators.

$n=8,9$ similar to $n=4$.

$n=10$: cycletypes (10), (2,5), (2,2,5) are all order 10.  I'll leave you to count it.

$n=12$: I'll leave you to list the cycletypes and count.


问:
Onto property of the function
How can we prove the onto property of this non-homogeneous function
$$f(x)= (|x|^{p-2} + |x|^{q-2}) x \quad \text{for} \quad x \in \mathbb{R} \quad \text{where} \quad p, q >1.$$

Any ideas (without performing any complicative calculations)?

答:
HINT

Since $f(0)=0$ and $f(x) \to \infty$ as $x \to -\infty$, IVT implies $\mathbb{R}^+ \subset \mathrm{Im}(f)$.

Can you do the other side?

Title: Onto property of the function


问:
Prove $\cos(x)$ is continuous
I'm stuck at a particular step and could use some help.

By definition, a function is continuous at $x=a$ iff $\lim_{x \to a} f(x) = f(a)$.

So I assume to prove $\cos(x)$ is continuous we must use the definition of a limit to show that:

$$\lim_{x \to a} \cos(x) = \cos(a) \iff \forall \epsilon > 0, \exists \delta>0 : 0 < |x - a| < \delta \implies |\cos(x)-\cos(a)| < \epsilon$$

From some triangle proofs it can be shown that:

$\cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha) \sin(\beta)$

$\cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha) \sin(\beta)$

Subtract the two equations:

$$\cos(\alpha + \beta) - \cos(\alpha - \beta) = -2\sin(\alpha) \sin(\beta)$$

Let $x = \alpha + \beta$ and $a = \alpha - \beta$.

Then $\alpha = \frac{x+a}{2}$ and $\beta = \frac{x-a}{2}$, implying:

$$\cos(x) - \cos(a) = -2\sin\left(\frac{x+a}{2}\right)\sin\left(\frac{x-a}{2}\right)$$

So

$$\left|\cos(x)-\cos(a)\right| = 2\left|\sin\left(\frac{x+a}{2}\right)\right| \cdot \left|\sin\left(\frac{x-a}{2}\right)\right|$$

I feel like I am close because I was able to change it so at least we have an $x-a$ term but now I'm not sure where to go from here.

答:
Note that
$$
\cos(x+h)=\cos(x)\cos(h) - \sin(x)\sin(h)
$$
and so
\begin{align}
\lim_{h \to 0}\cos(x+h) &= \lim_{h \to 0}\cos(x)\cos(h)-\lim_{h \to 0}\sin(x)\sin(h) \\
&= \cos x\lim_{h \to 0}\cos(h)-\sin(x)\lim_{h \to 0}\sin(h) \, .
\end{align}
The problem reduces to showing that $\sin$ and $\cos$ are continuous at $0$. For $\sin$, note that when $h>0$,
$$
0 < \sin(h) < h \, ,
$$
and since $\lim_{h \to 0^{+}}h=0$, by the squeeze theorem we find that $\lim_{h \to 0^+}\sin(h)=0$. The same argument can be made for $x<0$, but with the inequalities reversed. Hence, $\lim_{h \to 0}\sin(h)=0$. We can prove $\lim_{h \to 0}\cos(h)=1$ directly by the epsilon-delta definition of a limit. First, note that because $\cos$ is an even function, this problem reduces to proving that $\lim_{h \to 0^+}\cos(h)=1$. Now, if we set $\delta=\arccos(1-\varepsilon)$*, then
\begin{align}
& h < \arccos(1-\varepsilon) \\
\implies & \cos(h) > 1-\varepsilon \\
\implies & \cos(h) - 1 > -\varepsilon \\
\implies & 1 - \cos(h) < \varepsilon
\end{align}
and the result follows.

If $\varepsilon > 2$, then any value of $\delta$ suffices.

Title: Prove $\cos(x)$ is continuous


问:
$(a_n)$ diverges and $(b_n)$ converges, then $(a_n b_n)$ diverges
If $(a_n)$ diverges and $(b_n)$ converges, then $(a_n b_n)$ diverges. How can I prove that? I think this is true, but I can't prove it.

答:
As other answers have pointed out, this isn't true if $b_n \rightarrow 0$. So let's assume $b_n \rightarrow b \neq 0$. Then $1/b_n\rightarrow 1/b$.

If $a_nb_n \rightarrow c$ then $a_n \rightarrow c \times1/b = c/b$. (Since the limit of products is the product of limits.)

Therefore if $a_n$ diverges then $a_nb_n$ must diverge.


问:
Prove that $A = \{P(A)\}$
Prove that

If $A=\{B\},
B=\{\{\},A\}
$ then $A=\{P(A)\}$

$P(A)$ denotes the power set of $A$.

I have studied well founded set theory but this proof calls for Anti Foundation Axiom. I am not familiar with proving such statements. Can anyone provide a hint?

Image for reference of problem from Vicious Circles:

答: $A$ is a singleton, so $B=\{\emptyset,A\}=P(A)$. Thus, $A=\{B\}=\{P(A)\}$.

Title: Prove that $A = \{P(A)\}$


问:
Elegant proof of pre-annihilator equal to orthogonal complement
Let $X,Y$ be Hilbert spaces. I just proved that, given $A: X\rightarrow Y$ a bounded operator, one has:$$\ker(A^*)_\perp=\ker(A^t)^\perp$$

Here $U_\perp$ means pre-annihilator, $V^\perp$ means orthogonal complement, $A^*: Y^*\rightarrow X^*$ is the dual of $A$, $A^t: Y\rightarrow X$ the Hilbert space adjoint of $A$.

My proof distinguishes the two inclusions, and uses the definition of $(\cdot)^t$ and the Riesz representation theorem multiple times to compare the notions of pre-annihilator and orthogonal complement. However it is quite long winded and I feel there is a sleek way of seeing this result, which I couldn't find.

Do you have any idea?

答:
Let $\phi\leftrightarrow y_\phi$ be the Riesz correspondence on $Y^*$.

Then for all $\phi\in Y^*$, by definition of the Hilbert space adjoint, \begin{align}A^*\phi=0&\iff (\forall x\in X, \phi(Ax)=(A^*\phi) x=0)\\
&\iff (\forall x\in X, \langle y_\phi,Ax\rangle=0)\\&\iff A^ty_\phi=0.\end{align}

Hence, since $\phi\in Y^*$ correspond to $y\in Y$ by the Riesz representation, \begin{align}
x\in(\ker A^*)_\perp&\iff \forall\phi\in Y^*,(A^*\phi=0\implies\phi(x)=0)\\
&\iff \forall \phi\in Y^*, (A^ty_\phi=0\implies\langle y_\phi,x\rangle=0)\\
&\iff \forall y\in Y, (A^ty=0\implies \langle y,x\rangle=0)\\
&\iff \forall y\in Y, (y\in \ker A^t\implies y\perp x)\\
&\iff x\in(\ker A^t)^\perp\end{align}


问:
Closure of the set $\{f\in C[0,1] : f(x) >0 \ \forall x\in[0,1]\}$

Let $C[0,1]$ be defined with the max-norm and $A=\{f\in C[0,1] : f(x) >0 \ \forall x\in[0,1]\}$. Show that the closure of $A$ is the set $\{f \in C[0,1] : f(x) \geqslant 0 \ \forall x\in[0,1] \}$.

The closure of $A$ would be $A \cup \partial A$, but I'm not sure how to find the boundary here? It also seems that there would be alternative ways to show this?

答: If you look in the pointwise topology, $A$ can be seen as $(0,+\infty)^{[0,1]}$ and in this (product, essentially) topology the closure is $[0,+\infty)^{[0,1]}$ by standard facts about products (the continuous functions are dense in these product sets so that transfers to $C([0,1])$ as well). The max-topology is finer so we at least have $\overline{A} \subseteq B:=\{f \in C([0,1]]\mid \forall x: f(x) \ge 0\}$. The reverse follows as $B(f, r) \cap A \neq \emptyset$ whenever $f \in B$ and $r>0$; balls taken in the max-metric, so that $B \subseteq \overline{A}$.


问:
What is the distribution of primes modulo $n$?
Let $n\geq 2$ and let $k$ be "considerably larger" than $n$ (like some large multiple of $n$). Then for each $i$ such that $0<i<n$ and $\gcd(i,n)=1$ let's define
$$c_i=\left|\{p_j\;|\; p_j\equiv i \mod n,\;\mbox{where $p_j$ is the $j$-th prime, $1\leq j\leq k$}\}\right|$$
so $c_i$ represents how many of the first $k$ primes are congruent to $i$ modulo $n$.

What can we say about $c_i$s, that is about the distribution of the first $k$ primes modulo $n$?

I thought the distribution will be seemingly random, and that is mostly true - for example $c_i$s are always very close together. But there are observable non-random patterns. For example for $n=3$, for various $k$s I've tried  (up to $10^6$) I always got $c_2>c_1$. If this were just some kind of random discrepancy due to distribution of small primes, it would eventually vanish for large $k$s, which I don't observe.

答:
Dirichlet's theorem on primes in arithmetic progression tells us that the proportion of primes will be the same, for values of $i$ that are coprime to $n$ (and 0 otherwise).

However, Chebyshev's bias tells us that numerically there are more primes with a quadratic non-residue, than a residue, when you're counting primes up to $N$.


问:
Finding the Expected Value and Variance
The distribution function of a random variable $X$ is given by

$P (X \leq x)$ =
\begin{cases}
0 & \text{if $x<0$}, \\
x^{k} & \text{if $0 \leq x \leq 1, \,\,\,\,\,\, k\geq1$}, \\
1     & \text{if $x > 1$}.
\end{cases}

Determine the mean and variance of $X$.

Workings:

I am aware that I must start by calculating $\sum_{-\infty}^{\infty} x \,\cdot p_X(x)$, but because this is a continuous distribution, I will instead have to calculate $\int_{-\infty}^{\infty} x \, \cdot p_X(x) \,dx$. However, once I integrate it, I am confused as to what further steps as to which I am meant to take given that I get infinity as the expected value. Could someone please tell me if it is possible for the expected value to be infinite and if I have made a mistake in my workings? Thank you.

\begin{align}
\text{E(X)} & = \int_{-\infty}^{\infty} x \, \cdot p_X(x) \,dx \\
            & = \int_{-\infty}^{0} x \, \cdot 0 \,dx \: + \int_{0}^{1} x \, \cdot x^{k} \,dx \: + \int_{1}^{\infty} x \,dx \\
& = C + \bigg{[}\frac{x^{k+2}}{k+2}\bigg{]}_{0}^{1} + \bigg{[}\frac{x^{2}}{2}\bigg{]}_{1}^{\infty}
\end{align}

答:
$$ \int_{-\infty}^{0} x \, \cdot 0 \,dx+\ldots \\ =C+\ldots$$

A side note to this part: If you have a definite integral you don´t have the constant C as a part of the result.

$$\int_{a}^b f(x) \ dx=F(b)-F(a)$$

That means in the interval $(-\infty, 0)$ your integral is
$$\int_{-\infty}^0 x\cdot 0 \ dx=0\cdot \int_{-\infty}^0 x \ dx=0\cdot \left[ \frac12\cdot x^2 \right]_{-\infty}^0=0-0=0$$

The last equality holds since one factor is zero:

$"\textrm{If at least one factor of a product is zero, then the entire product is zero.}"$

Title: Finding the Expected Value and Variance


问:
Show that $f$ is Riemann Integrable on $\left[0,\pi/2\right]$
Let $f(x)=\cos^2(x)$ if $x\in \mathbb{Q}$ and $f(x)=0$ if not. Show that $f$ is Riemann Integrable on $\left[0,\pi/2\right]$.

The problem that I have, is that I don't really see why this function would be Riemann integrable. I will start the proof and explain in details what I did. So, if someone could tell me where my intuition is worng, I would appreciate it.

By definition, $f$ is integrable iff. $\exists$ a subdivision $\sigma$: $\overline{S}_{\sigma}(f)<\underline{S}_{\sigma}(f)+\epsilon \ \forall \epsilon>0$.

By density of irrational numbers, the lower Darboux sum is equal to $0$. So, we have to show that $$\overline{S}_{\sigma}(f)<\epsilon \iff \sum_{i=0}^{n}M_i(x_{i+1}-x_i)<\epsilon,$$ with $M_i=\sup\{f(x):x\in[x_{i},x_{i+1}]\}$. By density of rational numbers, we consider the following sum:

$$\sum_{i=0}^{n}\cos^2(x_i)(x_{i+1}-x_i)$$ as $\cos^2(x)$ is decreasing on $[0,\pi/2]$. But, $0\le\cos^2(x)\le 1$ on $[0,\pi/2]$. Therefore,
$$\sum_{i=0}^{n}\cos^2(x_i)(x_{i+1}-x_i)\le 1 \cdot \sum_{i=0}^{n}(x_{i+1}-x_i)=\frac{\pi}{2}-0>\epsilon$$ for every subdivision. Thus, $f$ is not intergable on $[0,\pi/2]$.

答:
Using the density of irrationals and rationals and the monotonicity of $x \mapsto \cos^2x$ on $[0,\pi/2]$, we have for any subinterval $[x_{j},x_{j+1}]$ of a partition $P$,

$$\sup_{x \in [x_{j},x_{j+1}]} f(x) = \cos ^2 x_{j}, \quad \inf_{x \in [x_{j},x_{j+1}]} f(x) = 0$$

Immediately we see that the lower Darboux sum is $L(P,f) =0$ and the upper Darboux sum is

$$U(P,f) = \sum_{j=0}^{n-1} \cos ^2 x_{j}(x_{j+1}-x_j)$$

Using the facts that $U(P,f)$ is also an upper Darboux sum for the decreasing and Riemann integrable function $x \mapsto \cos^2 x$, we get

$$ U(P,f)\geqslant \int_0^{\pi/2}\cos^2 x\, dx = \frac{\pi}{4} >0$$

This proves that $f$ is not Riemann integrable since for any $0 < \epsilon < \frac{\pi}{4}$ there can be no partition such that $U(P,f) - L(P,f) < \epsilon$.


问:
What do you take as the Particular Integral when is cancels out
I'm trying to solve the second order ODE $\ddot{x} + \omega^2x = A\cos(\omega t)$ and when I try to solve the PI it cancels out to be just $0 = A\cos(\omega t)$.  What do I take for the PI?

答:
$$\ddot{x} + \omega^2x = A\cos(\omega t)$$
For the particular solution try:
$$x_p=Ct\sin(\omega t)$$


问:
How to evaluate $\sum_{i=0}^{k-1} 2^i\cdot 4(k-i-1).$
I'm currently having trouble evaluating the following sum to get a formula in terms of $k$:
$$\sum_{i=0}^{k-1} 2^i\cdot 4(k-i-1)$$

I know that $$\sum_{i=0}^n 2^i = 2^{n+1}-1$$
but since my $2^i$ is multiplied with another term inside my summation, I can't use that formula to evaluate it. Any suggestions?

答:
Hint.

$$
\frac{d}{dx}x^{k-i-1}=(k-i-1)x^{k-i-2}
$$

now making $x=\frac 12$ we have

$$
\left(\frac12\right)^{k}2^i4(k-i-1)
$$

then think on

$$
\frac{d}{dx}\sum x^{k-i-1}
$$


答:
No reason to compute the PDF here. If $X$ is a nonnegative random variable (w.p.1) which it is in this case, you can calculate:

$$\mathbb{E}[X] = \int_0^\infty \mathbb{P}(X \geq x)\mathrm{d}x =\int_0^\infty (1- F_X(x)\mathrm{d}x = \int_0^1 1-x^k \mathrm{d}x + \int_1^\infty 1-1\mathrm{dx} = \int_0^1 (1-x^k)\mathrm{d}x$$

For the variance, we just need to compute:
$$
\mathbb{E}[X^2] = \int_0^\infty \mathbb{P}(X^2 \geq x)\mathrm{d}x = \int_0^\infty \mathbb{P}(X \geq \sqrt{x} \text{ or } X \leq -\sqrt{x} )
$$
Because $X$ is nonnegative, we have:
$$
\mathbb{E}[X^2] = \int_0^\infty \mathbb{P}(X \geq \sqrt{x})\mathrm{d} x = \int_0^\infty
(1- F_X(\sqrt x))\mathrm{d}x = \int_0^1 1- x^{k/2} \mathrm{d}x$$

Once you have this the variance is:
$$
\mathbb{E}[X^2] - \mathbb{E}[X]^2
$$

To prove the formula, let $X$ be a nonnegative random variable with density/PDF $f_X$. Note that:
$$\mathbb{P}(X \geq x) = \int_x^\infty f_X(y) \mathrm{d}y$$
then:
$$
\int_0^\infty \mathbb{P}(X \geq x) \mathrm{d}x = \int_0^\infty \int_x^\infty f_X(y) \mathrm{d}y\mathrm{d}x
$$
This is a double integral over the set in the plane given by $\{(x,y) \in \mathbb{R}^2 : x > 0 , y>x\}$ (i.e. an infinite triangle) of the function $g(x,y) = f_X(y)$ (only a function of $y$).

We can use Fubini's theorem (or more technically, Tonelli's theorem if you wish) to switch the order of integration and integrate with respect to $x$ first and $y$ second. This then becomes:
$$
\int_0^\infty \int_x^\infty f_X(y) \mathrm{d}y\mathrm{d}x = \int_0^\infty\int_0^y f_X(y)\mathrm{d}x\mathrm{d}y = \int_0^\infty yf_X(y) \mathrm{d}y
$$
which is the formula for $\mathbb{E}[X]$.

This formula holds in more generality, for example for discrete RVs and continuous RVs that don't have densities.

Title: Finding the Expected Value and Variance

Body:
The distribution function of a random variable $X$ is given by

$P (X \leq x)$ =
\begin{cases}
0 & \text{if $x<0$}, \\
x^{k} & \text{if $0 \leq x \leq 1, \,\,\,\,\,\, k\geq1$}, \\
1     & \text{if $x > 1$}.
\end{cases}

Determine the mean and variance of $X$.

Workings:

I am aware that I must start by calculating $\sum_{-\infty}^{\infty} x \,\cdot p_X(x)$, but because this is a continuous distribution, I will instead have to calculate $\int_{-\infty}^{\infty} x \, \cdot p_X(x) \,dx$. However, once I integrate it, I am confused as to what further steps as to which I am meant to take given that I get infinity as the expected value. Could someone please tell me if it is possible for the expected value to be infinite and if I have made a mistake in my workings? Thank you.

\begin{align}
\text{E(X)} & = \int_{-\infty}^{\infty} x \, \cdot p_X(x) \,dx \\
            & = \int_{-\infty}^{0} x \, \cdot 0 \,dx \: + \int_{0}^{1} x \, \cdot x^{k} \,dx \: + \int_{1}^{\infty} x \,dx \\
& = C + \bigg{[}\frac{x^{k+2}}{k+2}\bigg{]}_{0}^{1} + \bigg{[}\frac{x^{2}}{2}\bigg{]}_{1}^{\infty}
\end{align}


问:
Does an inward vector field on a boundary require a zero point or singularity in the interior?
Consider a continuous 3D vector field $\vec{V}:\mathbb{R}^n\to\mathbb{R}^n$ except that $\vec{V}$ may have isolated singularities, and an open region $\Omega\subset\mathbb{R}^n$. Suppose $\vec{V}(\vec{r})$ is nonsingular for $\vec{r}\in\partial\Omega$, and let $\vec{n}(\vec{r})$ be the outward-pointing normal at a boundary point $\vec{r}\in\partial\Omega$. If the normal component of the vector field points inward at every point on the boundary,

$$\vec{V}(\vec{r})\cdot \vec{n}(\vec{r}) < 0\ \forall\ \vec{r}\in\partial\Omega.$$

Does this imply that $\vec{V}(\vec{r})$ is either zero or singular somewhere in $\Omega$? How would it be shown? I imagine this must be a reasonably well known result but I wasn't sure how to look for it.

I did find this question, which states for $\vec{V}$ defined on the unit $n$-ball $B^n$, if $\vec{V}(\vec{r})\neq 0\ \forall\ \vec{r}\in B^n$, there must be a point $\vec{r}\in\partial B^n$ at which $\vec{V}(\vec{r})$ points outward and one at which it points inward. Logically, then, if those points don't exist on $\partial B^n$ (such as if $\vec{V}$ points inward everywhere), then there must be some point in $B^n$ at which $\vec{V}(\vec{r}) = 0$, or $\vec{V}$ has a singularity in $B^n$. Intuitively I would think this should continue to hold for any region topologically equivalent to a ball, since a homotopy between $B^n$ and $\Omega$ wouldn't change whether a given zero or singularity of $V$ is inside or outside the region or on the boundary. (Is this right so far?) But what about spaces with holes? Does anything change about the argument when $\Omega$ is not homotopic to a ball?

For the curious: I was inspired to think about this by considering whether any continuous mass distribution necessarily has a point at which the gravitational forces from all different parts of the distribution cancel out. That question is easily answered in the affirmative, because the gravitational field is a gradient of a scalar function which must have some local minimum, but then I started wondering if the condition that the vector field is a gradient was really necessary.

答:
Your instincts are right about a region equivalent to a ball. But the topology of the region definitely matters. Imagine a closed annulus (say $\{x\in \mathbb R^2: 1\le |x|\le 2\}$), with the flow spiralling in from the outside and out from the inside and a limit cycle at $|x|=3/2 cycling around. In general, see Milnor's Topology from the Differentiable Viewpoint or
http://arxiv.org/pdf/0903.0697.pdf for the Poincaré-Hopf Theorem, even on manifolds with boundary. They typically specify that the vector field be normal and outward-pointing on the boundary (so you can glue two copies of the manifold with boundary, together with the vector field, smoothly along the boundary). Inward-pointing is just as good, and your case can be smoothly deformed to that.


问:
Show that $\alpha(t)$ lies on a sphere
I'm asked to prove that the trace of the curve
$\alpha(t)=(r\sin^2(t), r \sin(t),r \cos(t))$
lies on a sphere.

What I tried so far:

Letting $\alpha(t)=(x(t),y(t),z(t))$, we have that $x^2(t)+y^2(t)+z^2(t)=r^2(\sin^4(t)+1)$ which does not satisfy the equation of a sphere. Also, I plotted the curve and I'm not sure whether the trace does actually lie on a sphere.

Any hints? Thanks!

答: There is no such sphere. If there was, let $(a,b,c)$ be its center. Then the distances from $(a,b,c)$ to $p(0)(=(0,0,1))$, to $p\left(\frac\pi2\right)(=(1,1,0))$, to $p\left(-\frac\pi2\right)(=(1,-1,0))$, and to $p(\pi)(=(0,0,-1))$ would all be equal. This means that $(a,b,c)$ is a solution of the system$$\left\{\begin{array}{l}a^2+b^2+(c-1)^2=(a-1)^2+(b-1)^2+c^2\\a^2+b^2+(c-1)^2=(a-1)^2+(b+1)^2+c^2\\a^2+b^2+(c-1)^2=a^2+b^2+(c+1)^2\end{array}\right.$$This sytem has one ond only one solution, which is $\left(\frac12,0,0\right)$. Then the radius of the sphere would be the distance from $\left(\frac12,0,0\right)$ to $(0,0,1)$, which is $\frac{\sqrt5}2$. But the distance from $p\left(\frac\pi6\right)\left(=\left(\frac34,\frac{\sqrt3}2,\frac12\right)\right)$ to $\left(\frac12,0,0\right)$ is $\frac{\sqrt{17}}4\ne\frac{\sqrt5}2$.

Title: Show that $\alpha(t)$ lies on a sphere


问:
Express $S =\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\ldots+\frac{1}{(n-2)\cdot(n-1)}+\frac{1}{n\cdot(n+1)}$ in terms of $n$.

Express $S =\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\ldots+\frac{1}{(n-2)\cdot(n-1)}+\frac{1}{n\cdot(n+1)}$ in terms of $n$.

Here's what I have done so far:

$$\begin{align*}
S &=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{n-2}-\frac{1}{n-1}\right)+\left(\frac{1}{n}-\frac{1}{n+1}\right)\\[5pt]
&= \left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+\ldots+\frac{1}{n}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\ldots+\frac{1}{n+1}\right)\\[5pt]
&= \sum_{k=1}^n \frac{1}{2k-1} - \sum_{k=1}^{n+1} \frac{1}{2k} \\[5pt]
\end{align*}$$

How do I continue the problem from here? Is it even possible?

Body: Let $ n\in\mathbb{N}^{*} $ be an odd integer, we have : \begin{aligned}\frac{1}{1\times 2}+\frac{1}{3\times 4}+\cdots +\frac{1}{n\left(n+1\right)}=\sum_{k=1}^{\frac{n+1}{2}}{\frac{1}{2k\left(2k-1\right)}}&=\sum_{k=1}^{\frac{n+1}{2}}{\frac{1}{2k-1}}-\sum_{k=1}^{\frac{n+1}{2}}{\frac{1}{2k}}\\ &=\sum_{k=1}^{n}{\frac{1}{k}}-\sum_{k=1}^{\frac{n-1}{2}}{\frac{1}{2k}}-\sum_{k=1}^{\frac{n+1}{2}}{\frac{1}{2k}}\\ \sum_{k=1}^{\frac{n+1}{2}}{\frac{1}{2k\left(2k-1\right)}}&=H_{n}-\frac{H_{\frac{n-1}{2}}+H_{\frac{n+1}{2}}}{2}\end{aligned}


问:
Is $x^x=y$ solvable for $x$?
Given that

$x^x = y$; and

given some value for $y$

is there a way to expressly solve that equation for $x$?

答:
Given y, you can solve y = x**x with a simple iteration:

Set x_0 = 2 (or anything else really)

x_(i+1) = ln y / ln x_i

Repeat 2 until abs(x_(i+1) - x_i) < threshold

It's akin to Newton's method of finding square roots.

The convergence of this process is the "natural base of y to itself", or the "exponential root" of y.

If you think about this iteration, you're iteratively performing the base-change logarithm formula. So at first you get y's logarithm (or "bit length") in base 2. Then in base of what y was in base 2, and so on. Successive values oscillate around the "natural base of y to itself" until it converges.

It's pretty interesting. I wonder if there's anything special about the value of x for each y.

As other people have said, you can take log of each side, to get:

ln y = ln x**x, which is ln y = x ln x, and the iteration

Title: Is $x^x=y$ solvable for $x$?


问:
Prove that $h(u)=\frac{2}{u^2}(u-\log(1+u))$ is decreasing on $(-1,\infty)$

Prove that $h(u)=\frac{2}{u^2}(u-\log(1+u))$ is decreasing on $(-1,\infty)- \{0\}$

This function appears on the proof of Stirling's formula in Principles of Mathematical Analysis, by Walter Rudin.

The classic way to approach the "show a function is decreasing" is to show that its derivative is negative on the given range.

$$h'(u)= -\frac{2}{u^2}+\frac{4}{u^3}\log(1+u)-\frac{2}{u^2(1+u)}$$

And after we factor out the $-\frac{2}{u^2}$ we are left with showing that

$$g(u):= 1-\frac{2}{u}\log(1+u)+\frac{1}{1+u}$$ is positive. We could try showing that $g$ has positive derivative and show that the limit of $g$ as $u$ approaches $-1$ is positive, but this seems like rhe derivative of $g$ will be something much more complicated.

答:
Another approach is to notice that \begin{align*}
\log (1 + u) = \int_0^u {\frac{{dt}}{{1 + t}}}  & = u - \int_0^u {\frac{t}{{1 + t}}dt}  = u - u^2 \int_0^1 {\frac{s}{{1 + su}}ds} \\ & =u - u^2 \int_0^1 \int_u^{ + \infty } {\frac{{s^2 }}{{(1 + st)^2 }}dt} ds
\end{align*}
for $u>-1$.


问:
Differential equation with one univariate function and several variables
I have a functional equation where I need to find one univariate function which satisfies conditions involving two variables, for example, find a function $f$ which satisfies, for all $x,y$:

$$
x (x+y) f'(x)(f(y)-f(x)) = (x-y)(f(x)-f^2(x))
$$

How are such equations called? I would like to find literature, but I don't even know where to begin, and what terms to use. This is not an ODE, but also not a PDE.

答:
The direct answer to your question is (as stated in the comments) that the proper term is "functional differential equation".  However, I would caution you about making a full frontal assault on your problem via the methods associated with FDE's.  There are many problems which can be phrased as FDE's, but trying to solve them via general FDE techniques is often very ineffective.

A somewhat general "meta-technique" for solving hard problems involving derivatives$^1$ is to reduce the problem by

Finding a necessary (but perhaps not sufficient) ODE that your solution must solve.

Determining the solution space of the ODE.

Looking within the solution space of the ODE for the solution to your original problem.

You can apply this technique to your example problem as follows: Set $y=x+\epsilon$ and take $\epsilon\rightarrow 0$ to find a related ODE $$2x^2 f'(x)^2 = f(x)^2-f(x)$$ which your unknown function $f$ must solve.  This ODE is separable and can be solved analytically!  It is first order, so there is a 1-parameter solution space.  Check through the solution space to determine which solutions satisfy the original equation involving $x$ and $y$, and then you're done.

So you can see how this line of reasoning can substantially constrain your possible solutions.  I would advise trying to make reductions like this before wandering through FDE land too much.

1. For example, the calculus of variations and the method of characteristics can be thought of this way.


问:
Geometric Tangent Vectors - looking for and understanding of and what the point is.
The problem that I am having is that I am having quite a hard time understanding the ideas of Geometric Tangent vectors and why they are even needed - I mean one already has the usual "Tangent" of a curve, why more definitions?

We let X be an n-dimensional differentiable manifold $p \in X$, and $(U,h,V)$ is a chart for X.  The set of all differentiable curves $\gamma_{1},\gamma_{2}$ through p is denoted by $K_{p}$ and $\gamma: (-\epsilon,\epsilon) \rightarrow X, \epsilon > 0$, with $\gamma(0) = p.\space$    Now $\gamma_{1} \sim \gamma_{2}$ if $\dot{\gamma_{1}}(0)_{h} = \dot{\gamma_{2}}(0)_{h}$. Where $\space \dot{\gamma}(0)_{h} \equiv d/dt(h \circ \gamma)(0) \in \mathbb{R}^{n}$

Defn:  The geometric Tangent space of a space X at p is the quotient $T_{p}^{geom}X  \equiv K_{p} /\sim$ where  is the equivalence relation of two curves $\gamma_{1},\gamma_{2}$ through point p.

The equivalence class of $\gamma \in K_{p}$ is denoted $[\gamma] \in T_{p}^{geom}X$ and is called the geometric tangent vector of X at p.

Thanks for any help at getting a better insight.

Brian

答:
Yes one have the usual definition of tangent vector of a curve but the tangent of a curve is defined if the curve lies in some $\mathbb{R}^n$ i.e. if $\gamma$ the curve then $\gamma :(0,1)\rightarrow\mathbb{R}^n $. Now when we are considering the tangent vector of a point in manifold then you have to consider the curve to be in that manifold. So my point is a priori you don't know whether the curve lies in $\mathbb{R}^n$ or not. In other words  you don't know
whether every smooth manifold can be embedded in $\mathbb{R}^n$ or not.

P.S. The above statement is true. It is called called Whitney embedding theorem.


问:
Intuitive Understanding of Independence of 3 Events
I understand what it means to say that an even E is independent of an event F, it means that knowing that F has occurred does not change the probability that E will occur. Algebraically, it implies:

$$P(E) = P\left(E \mid F\right)   \stackrel{\text{def}}{=} \frac{P\left(E \cap F\right)}{P\left(F\right)} \Leftrightarrow P\left(E \cap F \right) = P\left(E\right)  P\left(F\right)$$

Now I'm trying to tackle what it means for events E, F, and G to be independent, the definition stated in the book says:

$$\begin{gathered}
  P\left(E \cap F \cap G\right) = P\left(E\right) P\left(F\right) P\left(G\right)\\
  P\left(E \cap F\right) = P\left(E\right) P\left(F\right) \\
  P\left(E \cap G\right) = P\left(E\right) P\left(G\right) \\
  P\left(F \cap G\right) = P\left(F\right) P\left(G\right)\end{gathered}$$

Based on my previous understanding the bottom three lines would be implying that E is independent of F, E is independent of G, and that F is independent of G. But I'm not sure what the first line is saying. The way I've made sense of it so far is that:

$$ P\left(E \cap F \cap G\right)= P\left( E \cap \left( F \cap G  \right) \right) \stackrel{?}{=} P\left(E \right) P\left(F \cap G\right) \stackrel{\alpha}{=} P\left(E\right) P\left(F\right)P\left(G\right)$$

(Where $\alpha$ comes from the fact that E and F were independent.)

In order for the equality with the question mark to hold, I think we would need $E$ to be independent from $F \cap G$.

I am a little confused by their definition and tried to figure it out algebraically, but that's what I've gotten up to.

I am hoping someone can:

Give me an intuitive understanding of what it means for 3 events to be independent

Help me connect this understanding to the definition they have provided

Bonus: Help me take that understanding to the independence of $n$ events

答:
It can help to compare the definition of independent events with the concept of disjoint events.

Let's say I have some collection of $k$ events, $E_1, E_2, ..., E_k$, in a sample space of outcomes $\Omega$.

The formal definition of what it means for these events to be disjoint is that, for every subset $\{ j_1, j_2, ..., j_r \} \subseteq \{ 1, 2, ..., k \}$, we have $$ E_{j_1} \cap ... \cap E_{j_r} = \emptyset.$$ Disjointness means these events are mutually exclusive: one of these events happening excludes any of the others from happening.

In order for the whole group of $k$ events to be disjoint, it suffices for them to be pairwise disjoint: if I know $E_m \cap E_n = \emptyset$ for all $1 \leq m < n \leq k$, then I have that the whole collection of events $E_1, E_2, ..., E_k$ is disjoint as well, since any $3$-way, $4$-way, $5$-way, etc. intersection of events is contained in some $2$-way intersection by definition. For this reason, many sources will define disjointness using the condition for pairwise disjointness, since it is equivalent but easier to check.

The formal definition of what it means for these events to be independent is that, for every subset $\{ j_1, j_2, ..., j_r \} \subseteq \{ 1, 2, ..., k \}$, we have $$\Bbb{P}(E_{j_1} \cap ... \cap E_{j_r}) = \Bbb{P}(E_{j_1}) ... \Bbb{P}(E_{j_r}).$$ This does actually imply that every event is independent of any intersection of the other events, as per your question, and we want our definition of independence to have this property. That question-marked equality is a consequence of the definition, and not a thing that needs to be proven.

The reason is that, even if a group of events is pairwise independent, dependencies can come up in groups of more than $2$ events that would not be apparent from looking at pairs of events individually. If I want to be able to generalize my way of calculating probabilities of intersections of independent events to groups of more than $2$ events, I have to build it in as part of the definition, because it will not be implied by the pairwise property, as it was for disjointness.

Simple example: I flip a fair coin 3x. Every individual flip is independent--there's no possible effect they can have on each other. Let event $A$ = "first and second flips are the same," event $B$ = "second and third flips are the same", event $C$ = "first and third flips are the same". For these events, we have $$\Bbb{P}(A) = \Bbb{P}(B) = \Bbb{P}(C) = \frac{1}{2},$$ and also $$\Bbb{P}(A \cap B) = \Bbb{P}(A \cap C) = \Bbb{P}(B \cap C) = \frac{1}{4},$$ so these events are pairwise independent. However, they are not fully independent, because knowing whether both $A$ and $B$ happened tells me whether $C$ happened, since $$C = (A \cap B) \cup (A^c \cap B^c).$$ And in fact, $\Bbb{P}(A \cap B \cap C) = \frac{1}{4}$, not $\frac{1}{8}$ as it would be if we multiplied $\Bbb{P}(A)$, $\Bbb{P}(B)$, $\Bbb{P}(C)$. That's why our definition has to include the $k$-way intersections for $k > 2$ explicitly, and can't just use the two-way intersections. When one event is completely controlled or determined by a group of other events in this way, we don't want to consider the group as a whole "independent", even if the events in the group are independent in pairs.


问:
Contruction of vectors
Draw the vectors $\vec{a}$ and $\vec{b}$ with $|\vec{a}|=a=2\ cm$ and $|\vec{b}|=b=3\ cm$, these have angle $40^{\circ}$ between them.

Contruct the vectors:

(a) $\vec{c}=2\vec{a}-\vec{b}$

(b) $\vec{d}=\vec{a}+2\vec{b}$

(c) $\vec{e}$ such that $5\vec{a}-2\vec{e}=\vec{b}$

I have done the following for (a) and (b):

As for (c) do we solve for $\vec{e}$ first and then do the graphs?

答: You can proceed geometrically like you did in (a) and (b). The vector going from  the tip of $\mathbf b$ to the tip of $5\mathbf a$ is $-\mathbf b + 5\mathbf a = 2 \mathbf e$, and now you can scale this down by $1/2$.

Title: Contruction of vectors


问:
A Class of Diophantine Equations in Three Variables
This question is inspired by the problem https://projecteuler.net/problem=748

Consider the Diophantine equation
$$\frac{1}{x^2}+\frac{1}{y^2}=\frac{k}{z^2}$$
$k$ is a squarefree number. $A_k(n)$ is the number of solutions of the equation such that $1 \leq x+y+z \leq n$, $x \leq y$ and $\gcd(x,y,z)=1$. This equation has infinite solutions for $k=1$ and $k>1$ that can be expressed as sum of two perfect squares.

Let $$A_k=\lim_{n \to \infty}\frac{A_k(n)}{\sqrt{n}}$$

\begin{array}{|c|c|c|c|c|}
\hline
k& A_k\left(10^{12}\right)& A_k\left(10^{14}\right)& A_k\left(10^{16}\right)& A_k\left(10^{18}\right)& A_k \\ \hline
1& 127803& 1277995& 12779996& 127799963& 0.12779996...\\ \hline
2& 103698& 1037011& 10369954& 103699534& 0.1036995...\\ \hline
5& 129104& 1291096& 12911049& 129110713& 0.129110...\\ \hline
10& 90010& 900113& 9000661& 90006202& 0.0900062...\\ \hline
13& 103886& 1038829& 10388560& 103885465& 0.103885...\\ \hline
17& 86751& 867550& 8675250& 86752373& 0.086752...\\ \hline
\end{array}

From these data, it seems that these limits converge. I wonder if it is possible to write them in terms of known constants or some sort of infinite series. For Pythagorean triples, there are about $\frac{n}{2\pi}$ triples with hypotenuse $\leq n$.

The Python code to calculate $A_1(n)$:

```<code>def gcd(a, b):
   if b == 0:
       return a
   return gcd(b, a % b)

N = 10 ** 14

cnt = 0
for a in range(1, 22000):
   a2 = a * a
   for b in range(1, a):
       if (a + b) % 2 == 1 and gcd(a, b) == 1:
           b2 = b * b
           x = 2 * a * b * (a2 + b2)
           y = a2 * a2 - b2 * b2
           z = 2 * a * b * (a2 - b2)
           if x + y + z > N:
               continue
           cnt += 1
print(cnt)
</code>```

The Python code to calculate $A_2(n)$:

```<code>def gcd(a, b):
   if b == 0:
       return a
   return gcd(b, a % b)

N = 10 ** 14

cnt = 1     # (1, 1, 1)
for a in range(1, 22000):
   a2 = a * a
   a4 = a2 * a2
   for b in range(1, a):
       if gcd(a, b) == 1:
           b2 = b * b
           b4 = b2 * b2
           x = 2 * a * b * (a2 + b2) - (a4 - b4)
           y = 2 * a * b * (a2 + b2) + (a4 - b4)
           z = 6 * a2 * b2 - (a4 + b4)
           if x > 0 and y > 0 and z > 0 and x + y + z <= N and gcd(x, gcd(y, z)) == 1:
               cnt += 1
           x = (a4 - b4) - 2 * a * b * (a2 + b2)
           y = 2 * a * b * (a2 + b2) + (a4 - b4)
           z = (a4 + b4) - 6 * a2 * b2
           if x > 0 and y > 0 and z > 0 and x + y + z <= N and gcd(x, gcd(y, z)) == 1:
               cnt += 1
print(cnt)
</code>```

答:
This is a partial answer. Consider the case $k=1$. The parametric solutions are given by $(x,y,z)=\left(2ab(a^2+b^2),a^4-b^4,2ab(a^2-b^2)\right)$, where $\gcd(a,b)=1$, $a>b>0$ and $a+b$ is odd. In the triples $(x,y,z)$ generated by this parametrization, it is not necessarily always $x\leq y$. For example, $(20,15,12)$ is a triple generated by this parametrization, and $(15,20,12)$ will not be generated. This parametrization generates only the unique solutions. So we can have two possible cases, either $0<a^2-b^2\leq 2ab$ or $0<2ab\leq a^2-b^2$.

The sum of the triples = $x+y+z=4a^3b+a^4-b^4 \leq n$.

Let us consider the first case. Denote the number of solutions by $A_1^1(n)$ and $A_1^1=\lim_{n \to \infty}\frac{A_1^1(n)}{\sqrt{n}}$.

$$0<a^2-b^2\leq 2ab \implies \sqrt{2}-1\leq\frac{b}{a}=x<1$$
Number of pairs $(a,b)$ satisfying above constraints and also $4a^3b+a^4-b^4 \leq n$ can be approximated by
\begin{align}
\begin{split}
p(n)&\approx\int_{\frac{b}{a}=\sqrt{2}-1}^{1}\int_{4a^3b+a^4-b^4 \leq n}a \ da\ db\\
&=\int_{x=\sqrt{2}-1}^{1}\int_{0<a^4\leq \frac{n}{1+4x-x^4}}a \ da\ dx \\
&=\frac{\sqrt{n}}{2}\int_{x=\sqrt{2}-1}^{1}\frac{1}{\sqrt{1+4x-x^4}}\ dx
\end{split}
\end{align}

The natural density of coprime numbers is $\frac{6}{\pi^2}$. The condition $gcd(a,b)=1$ will contribute a multiplication factor of $\frac{6}{\pi^2}$. Among all the coprime pairs $(a,b)$ such that $a>b>0$, fraction of which have odd value of $a+b$ is $\frac{2}{3}$. We have to multiply this factor.

\begin{align}
\begin{split}
A_1^1 &=\lim_{n \to \infty}\frac{A_1^1(n)}{\sqrt{n}}\\
&= \lim_{n \to \infty} \frac{2}{3} \cdot \frac{6}{\pi^2} \cdot\frac{p(n)}{\sqrt{n}}\\
&= \frac{2}{\pi^2} \int_{x=\sqrt{2}-1}^{1}\frac{dx}{\sqrt{1+4x-x^4}} \\
\end{split}
\end{align}
Similarly, it can be shown that for the other case exactly the same limit exists. So, $$A_1=2A_1^1=\frac{4}{\pi^2} \int_{x=\sqrt{2}-1}^{1}\frac{dx}{\sqrt{1+4x-x^4}} \approx 0.1277999513464289...$$

Using a similar approach it can be shown that
\begin{align}
\begin{split}
A_2&=\frac{2}{\pi^2} \left[ \int_{x=0}^{\sqrt{2}-1}\frac{dx}{\sqrt{3-6x^2-x^4}} +\int_{x=\sqrt{2}-1}^{1}\frac{dx}{\sqrt{4x+6x^2+4x^3-1-x^4}} \right]\\
& \approx 0.1036994744684913...
\end{split}
\end{align}
\begin{align}
\begin{split}
A_{13}&=\frac{52}{7\pi^2} \int_{x=(\sqrt{13}-3)/2}^{(\sqrt{26}-1)/5}\frac{dx}{\sqrt{20x+36x^2-5-7x^4}}\approx 0.1038855856479065...
\end{split}
\end{align}


问:
Differentiable function and polynomials: Proof of $\phi (x)=ce^x + q(x)$ unclear to me.
I came across a proof which I can't quite understand:

If $p:\mathbb{R}\to\mathbb{R}$ is a polynomial of degree $n$ and $\phi:\mathbb{R}\to\mathbb{R}$ is a differentiable function with $\phi'=\phi+p$, there exists a polynomial $q$ of degree $n$ with a number $c\in\mathbb{R}$ with $\phi(x)=ce^x+q(x)$ for all $x\in\mathbb{R}$.

The proof is given as follows:

$$\phi'=\phi+p$$
$$\Leftrightarrow\phi'-\phi = p$$
$$\Leftrightarrow\phi'*e^{-x}-\phi*e^{-x}=p*e^{-x}$$
$$\Leftrightarrow\left(\frac{d}{dx}\right)[\phi*e^{-x}]=p*e^{-x}$$
$$\int\left(\frac{d}{dx}\right)[\phi*e^{-x}]=\int p*e^{-x}$$
$$\Leftrightarrow\phi*e^{-x}=q*e^{-x}+c$$
$$\Rightarrow\phi(x)=ce^x+q(x)$$

Isn't the notation (especially the integrals) missing something?

Why exactly are we multiplying by $e^{-x}$ in line 3? Is this a special technique?

How does the integral with $p$ in it suddenly yield $q$? And why is the arbitrary constant $c$ on the left hand side of the equation missing?

Also, the conclusion is unclear to me. How exactly do the $e^{-x}$ and $c$ disappear and how does the $ce^x$ appear?

Thank you very much in advance!

答:
Why exactly are we multiplying by $e^{−x}$ in line 3:

The OE $$\phi'(x)-\phi(x) = p(x)$$ is a Linear differential equation of first degree and as an method we need a magic function called Integrating factor to solve it. The method tells us to multiply this given function to the OE, both sides of it, just to make the OE an exact OE. Here we have $-1$ as the coefficient of $\phi(x)$ in OE, so according to cited approach the integrating factor is: $$\mu(x)=\exp\int(-1)dx=\text{e}^{-x}$$ Now continue from line 3 to the rest.

How does the integral with p in it suddenly yield $q$?

Do as @nullUser noted, he did the rest complete.


Title: How to evaluate $\sum_{i=0}^{k-1} 2^i\cdot 4(k-i-1).$

Body:
I'm currently having trouble evaluating the following sum to get a formula in terms of $k$:
$$\sum_{i=0}^{k-1} 2^i\cdot 4(k-i-1)$$

I know that $$\sum_{i=0}^n 2^i = 2^{n+1}-1$$
but since my $2^i$ is multiplied with another term inside my summation, I can't use that formula to evaluate it. Any suggestions?

Body: Let $ k\in\mathbb{N}^{*} $, we have :\begin{aligned}\sum_{i=0}^{k-1}{\left(k-1-i\right)2^{i}}&=\sum_{i=0}^{k-1}{\left(k-1-i\right)\left(2^{i+1}-2^{i}\right)}\\ &=\sum_{i=0}^{k-1}{\left(k-1-i\right)2^{i+1}}-\sum_{i=0}^{k-1}{\left(k-1-i\right)2^{i}}\\ &=\sum_{i=1}^{k}{\left(k-i\right)2^{i}}-\sum_{i=0}^{k-1}{\left(k-1-i\right)2^{i}}\\ &=\sum_{i=0}^{k-1}{\left(k-i\right)2^{i}}-\sum_{i=0}^{k-1}{\left(k-1-i\right)2^{i}}-k\\ &=\sum_{i=0}^{k-1}{2^{i}}-k\\ \sum_{i=0}^{k-1}{\left(k-1-i\right)2^{i}}&=2^{k}-1-k\end{aligned}


问:
Prove that the area of a right triangle is no more than the square of the hypotenuse divided by 4
I'm struggling with this proof. Let $a,b,c$ denote the base, height, and hypotenuse of a right triangle. Let $A$ denote area of this triangle

We have the following relations:
$$
a^2 + b^2 = c^2  \\
A = a*b/2
$$

and also the triangle inequality, though I'm not sure if this inequality comes into play.

Here is some of my work and I'm unsure if it's in the right direction
$$
a^2 + b^2 = c^2 \\
a^2*b + b^3 = c^2*b \\
\implies A*a + b^3 = b * c^2 \\
\implies A = b*c^2/a - b^3/a
$$

I feel that this is not in the right direction as I don't know how to show that this last expression is $\leq c^2/4$. Can anyone advise?

答: The claim is $(a^2+b^2)/2\ge ab$, which is the AM-GM inequality in $2$ positive variables. In particular, $a^2+b^2-2ab=(a-b)^2\ge0$.


问:
Integration of autocorrelation function when $\int_0^1 f(t) dt = 0$
When $\int_0^1 f(t) dt = 0$ and $f(t)=0$ for $t\in \mathbb{R} \setminus [0,1]$, my conjecture is that
$$
  \int_{\mathbb{R}} R_{ff}(\tau)\,d\tau = 0,
$$
where
$$
R_{ff}(\tau) = \int_{\mathbb{R}} f(t+\tau) f(t)\,dt.
$$
How can I prove this?

答:
Since $f(t) = 0$ for $t>1$ or $t<0$, it is enough to show
\begin{align*}
 \int_{-C}^C R_{ff}(\tau)\,d\tau,
\end{align*}
for some $C>1$.
By the integration by parts,
\begin{align*}
\def\bbR{{\mathbb{R}}}
\require{cancel}
 \int_{-C}^C R_{ff}(\tau)\,d\tau &= \cancelto{0}{\tau R_{ff}(\tau)\rvert_{-C}^C} - \int_{-C}^C \tau R'_{ff}(\tau)\,d\tau   \\
 &=- \int_{-C}^C \tau \int_{\bbR} f'(t+\tau) f(t)\,dt \,d\tau   \\
 &=- \int_{\bbR} \int_{-C}^C \tau  f'(t+\tau)\,d\tau  f(t)\,dt   \\
 &=- \int_{\bbR} \left( \cancelto{0}{\tau f(t+\tau) \rvert_{-C}^C} - \cancelto{0}{\int_{-C}^C f(t+\tau)\,d\tau} \right) f(t)\,dt  \\
 &=0.
\end{align*}


问:
Question regarding fixed point (integrals)
Let $f:[0,1]\rightarrow\mathbb{R}$  be a continuous function. Show that, if $$\int\limits_0^1 f(t)\,\mathrm{d}t =\frac{1}{2},$$ then $f$ has a fixed point in $[0,1]$, i.e. $\exists x_0\in [0,1]$ such that $f(x_0)=x_0$.

答:
If $ g:x\mapsto f\left(x\right)-x $, and $ G:x\mapsto\int_{0}^{x}{g\left(x\right)\mathrm{d}x} $, notice that : $$ G\left(1\right)-G\left(0\right)=0 $$

Now you can apply Rolle's theorem to $ G $, which is a $ \mathcal{C}^{1} $ function on $ \left[0,1\right] $.


问:
Initial value problem $y''=e^{2y}, y(0)=0, y'(0)=1$
Supposedly, it is possible to determine information about the constants of this IVP solution, without computing the solution of the differential equation.  Here's how I solve this.

Let $z=y'$. In brief, we then have the separable equation $$z\frac{dz}{dy}=f(y,z)=e^{2y}$$

and solving for $z$ $$z=\sqrt{e^{2y}+c}$$

which can be made into a separable equation and integrated. After playing around with the initial conditions, I did find that $c=0$. But this is a contradiction with the results of the second separable equation. Can we really solve $c$ before finding the explicit form of $y$?

答:
From my answer to this, we see that we need to take the $+$ of the square root and the answer will be case $3$ where  $C_1=0$. Thus the answer is

$$y=-\log(1-x)$$


答:
The notation is quite abusive. The author assumes the reader is familiar enough with differential equations to understand what the notation means, where the $dx$'s go and so on.

We multiply by $e^{-x}$ because we recognize the ODE $y'-y = p$ to be first order linear, which can always be solved by multiplying by a suitable integrating factor, which in this case turns out to be $e^{-x}$. See http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx for examples.

The statement $\int p(x) e^{-x}dx = q(x)e^{-x}+c$ follows by induction from the fact that $\int x^k e^{-x}dx$ is solved by repeated integration by parts. Differentiating powers of $x$ and integrating $e^{-x}$ always leaves you with a term of the form $x^n e^{-x}$ up to a multiplicative constant.

To get the last line from its predecessor just multiply the equation by $e^{x}$.

Title: Differentiable function and polynomials: Proof of $\phi (x)=ce^x + q(x)$ unclear to me.

Body:
I came across a proof which I can't quite understand:

If $p:\mathbb{R}\to\mathbb{R}$ is a polynomial of degree $n$ and $\phi:\mathbb{R}\to\mathbb{R}$ is a differentiable function with $\phi'=\phi+p$, there exists a polynomial $q$ of degree $n$ with a number $c\in\mathbb{R}$ with $\phi(x)=ce^x+q(x)$ for all $x\in\mathbb{R}$.

The proof is given as follows:

$$\phi'=\phi+p$$
$$\Leftrightarrow\phi'-\phi = p$$
$$\Leftrightarrow\phi'*e^{-x}-\phi*e^{-x}=p*e^{-x}$$
$$\Leftrightarrow\left(\frac{d}{dx}\right)[\phi*e^{-x}]=p*e^{-x}$$
$$\int\left(\frac{d}{dx}\right)[\phi*e^{-x}]=\int p*e^{-x}$$
$$\Leftrightarrow\phi*e^{-x}=q*e^{-x}+c$$
$$\Rightarrow\phi(x)=ce^x+q(x)$$

Isn't the notation (especially the integrals) missing something?

Why exactly are we multiplying by $e^{-x}$ in line 3? Is this a special technique?

How does the integral with $p$ in it suddenly yield $q$? And why is the arbitrary constant $c$ on the left hand side of the equation missing?

Also, the conclusion is unclear to me. How exactly do the $e^{-x}$ and $c$ disappear and how does the $ce^x$ appear?

Thank you very much in advance!


问:
Sine Fourier coefficients of derivative
Let $f$ be a $C^1$-function on $[0,\pi]$ with $f(0)=f(\pi)=0$. We know that the family $(\sin(nx))_{n\in\Bbb N}$ is a complete orthonormal system on $L^2(0,\pi)$ (maybe up to some rescaling). Let $(a_n)$ be the Fourier coefficients of $f$ with respect to this system, i.e. $$a_n=\frac{2}{\pi}\int_0^\pi f(x)\sin(nx)dx$$ Then we have by Parseval's theorem $$\frac{2}{\pi}\int_0^\pi f^2=\sum_{n=1}^\infty a_n^2$$ (for simplicity assume that $f$ only assumes real values, so that we can ignore absolute values)

My question is about this answer. There it is claimed that we also have $$\frac{2}{\pi}\int_0^\pi (f')^2=\sum_{n=1}^\infty n^2a_n^2 \tag{$*$}$$
Why is this the case? Let $b_n$ be the Fourier coefficients of $f'$. So $$b_n=\frac{2}{\pi}\int_0^\pi f'(x)\sin(nx)dx=-\frac{2n}{\pi}\int_0^\pi f(x)\cos(nx)dx$$(Since $f(0)=f(\pi)=0$.)
If we could show that $\int_0^\pi f(x)\cos(nx)dx=\int_0^\pi f(x)\sin(nx)dx$, we would have $b_n=-na_n$ and the result follows again from Parseval's identity, but some examples show that this equality doesn't hold.

The original problem was to show the inequality $\int_0^\pi f^2dx\leq\int_0^\pi (f')^2$. I think we can avoid the problem above by simply choosing the complete orthonormal system $(e^{2ikx})_{k\in\Bbb Z}$ instead of the sine functions, because in this case we get for $k\ne0$ by the same calculation as above: $$\langle f',e^{2ikx}\rangle=2ik\langle f, e^{2ikx}\rangle$$
Unfortunately there is a problem with $k=0$, see also the approach in the linked question.

So I would like to understand the equality $(*)$.

答:
I think that using that $\sin nx$ is a complete orthonormal system of $L^2(0,\pi)$ gets into subtleties that are best treated with classical Fourier series theory.

The way to proceed is using the fact that given $f \in L^2(0,\pi)$, one can extend it to an odd function $F$ on $[-\pi,\pi]$ (ends matter only for continuos functions) and then it has a sine Fourier series which gives the Parseval relation $\frac{2}{\pi}\int_0^\pi f^2=\sum_{n=1}^\infty a_n^2$.

The hypothesis that $f \in C^1, f(0)=f(\pi)=0$ implies that (actually it is equivalent to) $F$ (the odd extension) is absolutely continuous on $[-\pi,\pi]$

(by $C^1$ on $(-\pi,0) \cup (0, \pi)$ and $C^1$ from the sides at the ends, continuity at $0$ and $\pi =-\pi$ modulo circle is equivalent to absolute continuity for $F$ and that in turn is clearly equivalent with $f(0)=f(\pi)=0$).

But then by general Fourier series theory, $F'$ (an even integrable function, possibly discontinuous at $0, \pi$ as a periodic function - with left/right limits though eg take $f(x)=x(\pi-x)$ on $[0, \pi]$) has a cosine Fourier series which is the term by term derivative of the one of $F$, hence its coefficients are indeed $na_n$ and the claim follows by  Parseval and symmetry since $F'=f'$ on $(0, \pi)$

Title: Sine Fourier coefficients of derivative


问:
inverse trignometric inequality finding value for x

Find the interval of x for which the equation is satisfied

答:
Since divide by $0$ is not defined, $1 - x^2 \neq 0 $.

So $x \neq +1 \ \text{and}\  x \neq -1$.

Therefore $x \in (-\infty , +\infty) \setminus\{-1,1\}.$


问:
Using the chain rule to differentiate a multivariate function
Can anyone explain to how to do this derivative.
I am given the function

$$f(x, g(x)) = k$$

I have been told by my teacher that the derivative w.r.t $x$ is:

$$f_x + g'(x)f_y$$

He didn't explain it very well so it would be very helpful if someone could explain this to me in simple terms.

答:
This is a consequence of:

On the one hand the chain rule as the derivative of the composition of functions
$t \mapsto f(x, g(t))$ is $g^\prime(t) f_y(x,g(t))$.

On the other hand the (Fréchet) derivative of the map $f(x,y)$ is given by $f^\prime(x,y)(h,k)=f_x(x,y)h+ f_y(x,y)k$.


问:
$D=\{(a,b,c,d,e)\in\mathbb{R}^5 \mid ax^4+bx^3+cx^2+dx+e=0\}$ Prove that ${(1,2,-4,3,-2)}\in \operatorname{int}(D)$
I'm having some troubles with it:

$$D=\{(a,b,c,d,e)\in\mathbb{R}^5 \mid \exists x\in\mathbb{R} : ax^4+bx^3+cx^2+dx+e=0\}$$
Prove that ${(1,2,-4,3,-2)}\in \operatorname{int}(D)$

Might use some hints,
thanks

$x\in int(D) : if\ there\ is \ a \ r>0\in \mathbb{R} \ that\ B_r(x)\subset D $

$D$ is a subset of $\mathbb R^5$ that each vector from $D$ solves the $4$ degree poly of $x,$ $x$ is just a number in $\mathbb R$ (might means that each vector in $D$ leads to factorization? not sure), I think $\operatorname{int}(D)$ might be just all the vector in $D.$ I'm thinking that they might want me to show that $(1,2,−4,3,−2)$ is extremum of some what by using Lagrange Multipliers, not sure what are my functions $f$ and $F$

答:
So I came up with the solution eventually, hope it is right, it goes like this:
First, let us be aware that for $\overline p_0=(1,2,-4,3,-2)\in\mathbb{R^5}\ x=1$ solves the polynomial $p(x)=x^4+2x^3+-4x^2+3x-2$ thus $\overline p_0\in D$
Now lets' define a function $F:\mathbb{R^6}\rightarrow \mathbb{R}
 \\F(a,b,c,d,e,x)=ax^4+bx^3+cx^2+dx+e$

It is easy to see that $F$ is $C^1(\mathbb{R^6}\times \mathbb{R}$) and $F(\overline p_0,1)=F((1,2,-4,3,-2),1)=0$

$D(F)=(\nabla F(a,b,c,d,e,x))^t=(x^4,x^3,x^2,x,1,4ax^3+3bx^2+2cx+d)^t$ with $rank(D(F))=1$ for all $(a,b,c,d,e,x)\in \mathbb{R^6} $ including our $(\overline p_0,1)$.

now let us apply the implicit function theorem and it gives us that there exist a domain $V \times W \subset \mathbb{R^5} \times\mathbb{R}$ of $(\overline p_0,1)$ and a function $g\in C^1(V,\mathbb{R})\ ,\ g(V)\subset W $ in which that:

$g(\overline p_0)=1$

$\forall (\overline p,x)\in(V\times W) \mid F(\overline p,x)=0 \Leftrightarrow g(\overline p)=x  \Leftrightarrow F(\overline p,g(\overline p))=0$

and the second claim implies us that we've found a domain $V$ of  $ \ \overline p_0$ in $D$.


问:
why $\frac{d}{dt}U(\gamma(t))=\langle \nabla_U(\gamma(t)),\gamma'(t)\rangle$?
I do not understand why $\frac{d}{dt}U(\gamma(t))=\langle \nabla_U(\gamma(t)),\gamma'(t)\rangle$ where $U:\mathbb{R^2}\rightarrow\mathbb{R^2}$ and $\gamma:\mathbb{R}\rightarrow\mathbb{R^2}$ are $c^{\infty }$ functions.

Should not the LHS be a vector while the RHS a number/scalar?

答:
$\nabla_U (\gamma(t))$ makes little to no sense as $U$ is a vector field. The $\nabla$-notation is more common for scalar fields $f: \mathbb{R}^N \rightarrow \mathbb{R}$. You should write $DU$ instead, where $(DU(x))_{ij} = \partial_j U_i(x)$ for $i, j \in \lbrace 1, 2 \rbrace$. Then
$$
\frac{\mathrm{d}}{\mathrm{d}t} U(\gamma(t)) = DU(\gamma(t))\gamma'(t)
$$
according to the chain rule. Since we are using matrix-vector multiplication, both the LHS and the RHS are $\mathbb{R}^2$ vectors.


问:
How do you generate the numbers from an empty set?
In a short 2007 Scientific American article, the former Harvard mathematician and author Dr Robert M Kaplan stated that:

in mathematics we can generate all numbers from the empty set

I've not been able to find the proof of that statement. I assume he means the natural numbers. I can see how you can get to the empty set by removing each number from a set of numbers, therefore, reverse that process. (Though I wonder if that applies to an infinite set?) But what is the start point? Is the start related to counting the span of the empty set? I.e. by counting the span of the empty set to get the first number then a proof can be shown that involves cardinality.

What is the straightforward proof that demonstrates that all numbers can be generated from the empty set?

答:
You can do better than just all natural numbers: you can construct every ordinal!

Nickname $\varnothing$ as $0$. For any $a$, let $a+1=a\cup\{a\}$. For any limit ordinal $\delta$, let $\delta=\bigcup\limits_{\alpha<\delta}\alpha$.

Hence, we've defined the ordinals as sets, just by starting with the empty set and building upwards.

Now how do we know that this is comprehensive? All we want to show is that every well-ordered set is in bijection with one of these ordinals. But that's a classic theorem you can find here.

(I'd be remiss to not mention that this construction is a bit circular, since we are assuming the existence of ordinals while constructing them, but if you want a proper treatment of this subject, there are plenty of texts for you.)


问:
Show that every eigenvalue to a unitary operator has absolute value 1
Let $A$ be a unitary operator on hilbert space $H$, i.e.
$$(Au|Av) = (u|v)$$
for all $u,v \in D_A.$.

I'm asked to show that all eigenvalues to this unitary operator has absolute value 1.

My attempt:

Let u,v be eigenvectors to $A$ and $\lambda_1, \lambda_2$ their eigenvalues.

$$(Au|Av) = (\lambda_1u|\lambda_2v) = \lambda_1\lambda_2 (u|v) = (u|v) \implies \lambda_1\lambda_2=1.$$

For the statement to be true, $\lambda_1$ must equal $\lambda_2$ but I can't see why this is true. What am I missing?

答:
Let $\lambda$ and eigenvalue of $A$. With $u$ his eigenvector.

Then $(u,u) = (Au,Au) = (\lambda u, \lambda u) = \lambda\overline{\lambda}(u,u) = |\lambda|^{2}(u,u)$

Thus $|\lambda|^{2} = 1 \implies |\lambda| = 1$


答:
By integration from $0$ to $y$, $$2zz'=2e^{2y}\implies z^2-z_0^2=e^{2y}-1$$

and

$$z=y'=e^y$$ (the solution $z=-e^y$ is not possible).

Then integrating between $0$ and $x$,

$$-e^{-y}+e^{-y_0}=x$$

or

$$y=-\log(1-x).$$

Title: Initial value problem $y''=e^{2y}, y(0)=0, y'(0)=1$

Body:
Supposedly, it is possible to determine information about the constants of this IVP solution, without computing the solution of the differential equation.  Here's how I solve this.

Let $z=y'$. In brief, we then have the separable equation $$z\frac{dz}{dy}=f(y,z)=e^{2y}$$

and solving for $z$ $$z=\sqrt{e^{2y}+c}$$

which can be made into a separable equation and integrated. After playing around with the initial conditions, I did find that $c=0$. But this is a contradiction with the results of the second separable equation. Can we really solve $c$ before finding the explicit form of $y$?


问:
Drawing vectors in 3dimension
The vectors $\vec{a}, \vec{b}, \vec{c}$ are as follows:

with $a=2\ cm, \ b=2.5\ cm, \ c=3\ cm$.

Construct the vector

(a) $\vec{u}=2\vec{a}+\vec{b}-3\vec{c}$

(b) $\vec{v}$ such that $3\vec{a}-2\vec{b}+3\vec{c}+\vec{v}=\vec{0}$

(c) $\vec{w}$ such that $\vec{a}+3\vec{b}-4\vec{c}+2\vec{w}=\vec{0}$

For (a) I have done the following:

Is that correct?

For (b) and (c) we have to solve as for the unknown vector first, right?

答: a) is not correct; look carefully as to in which direction c is going, and mind that it says minus 3 so -3c, so the added -3c vector should be in the opposite direction of the original c.

Title: Drawing vectors in 3dimension


问:
Semisimple Lie algebra representation to reductive lie algebra
Let us work over $\Bbb C$.

If I know the classification of irreducible representations of semisimple Lie algebras, how do I classify the irreducible representations of reductive Lie algebras?

Let $\mathfrak{g}$ be reductive, then we can always write $\mathfrak{g}=\mathfrak{s}\oplus \mathfrak{a}$ for $\mathfrak{s}$ a semisimple Lie algebra and $\mathfrak{a}$ an abelian Lie algebra. From this I then just need to choose how $\mathfrak{a}$ act right? So $\mathfrak{s}$ has finite dimensional highest weight irreducible representations classified by the quotients $M(\lambda)$ from the verma modules $V(\lambda)$ where $\lambda \in P^+$ (dominant weights) how does $\mathfrak{a}$ act?

Should I be considering the correspondence between $U(\mathfrak{g})$ representations and $\mathfrak{g}$ representations, and then:
$$U(\mathfrak{s}\oplus \mathfrak{a})=U(\mathfrak{s})\otimes U(\mathfrak{a})\cong U(\mathfrak{s})\otimes S(\mathfrak{a})$$
and then how do I find what to quotient by to get all of the finite dimensional irreducible reps?

答:
If you only want irreducible representations (denoted by
$\phi: \mathfrak{g}\to \mathrm{End}(V)$) of a reductive Lie algebra $\mathfrak{g}$, then by Schur's lemma, $\phi(a)=\lambda I$ for $a\in \mathfrak{a}$, where $I$ is the identity map in $\mathrm{End}(V)$. [Note that Schur's lemma works for irreducible representations of arbitrary finite dimensional Lie algebras, provided that the field is algebraically closed.] i.e. the central elements can only act by constant multiplication in irreps.

The real subtlety when it comes to classifying all representations of reductive Lie algebras is that there exists indecomposible representations that are not irreducible [while for semisimple Lie algebras all representations are completely reducible, so indecomposible and irreducible mean the same thing]. Take $\mathfrak{gl}(2)=\mathfrak{sl}(2)\oplus\mathbb{C}$ for example, spanned by $\{x,y,h,c\}$ where $c$ is the central element. Consider the four dimensional representation
$$x=\begin{pmatrix}
0 & 1 & 0& 0\\
0 & 0 & 0& 0\\
0 & 0 & 0& 1\\
0 & 0 & 0& 0
\end{pmatrix},
y=\begin{pmatrix}
0 & 0 & 0& 0\\
1 & 0 & 0& 0\\
0 & 0 & 0& 0\\
0 & 0 & 1& 0
\end{pmatrix},\\
h=\begin{pmatrix}
1 & 0 & 0& 0\\
0 & -1 & 0& 0\\
0 & 0 & 1& 0\\
0 & 0 & 0& -1
\end{pmatrix},
c=\begin{pmatrix}
0 & 0 & 1& 0\\
0 & 0 & 0& 1\\
0 & 0 & 0& 0\\
0 & 0 & 0& 0
\end{pmatrix}.
$$
You can easily show that it is reducible, but indecomposible.


问:
Equivalent Grothendieck toposes
Let $(\mathbb{C}, J)$ and $(\mathbb{D}, K)$ be small sites whose associated Grothendieck toposes are equivalent: $\mathsf{Sh}(\mathbb{C}, J) \simeq \mathsf{Sh}(\mathbb{D}, K)$. I am aware that a given Grothendieck topos can have many "inequivalent" sites of definition, but is there any relation between the sites $(\mathbb{C}, J)$ and $(\mathbb{D}, K)$ that can be deduced from the associated Grothendieck toposes being equivalent?

答:
In general that is a very difficult question, and as far as I know there isn't yet a site characterization for the resulting topoi to be equivalent.

However, restricting to certain classes of topoi and sites there do exist characterizations.  Just to give a flavour of this:

If $X$, $Y$ are sober spaces, they can be recovered from their topoi $Sh(X)$ and $Sh(Y)$ and so $Sh(X) \simeq Sh(Y)$ iff $X \cong Y$.  More generally, if $L$, $K$ are distributive lattices then $Sh(L,J_{can}) \simeq Sh(K, J_{can})$ iff $K$ and $L$ have isomorphic co-completions.

If the topoi are presheaf topoi, then $Sets^{\mathcal{C}^{op}} \simeq Sets^{\mathcal{D}^{op}}$ iff $Ind(\mathcal{C}) \cong Ind(\mathcal{D})$ (the Ind-completion).

Theorem 3 from this paper by Moerdijk gives sufficient conditions for the topoi of sheaves on two topological groupoids to be equivalent.

Corollary 6.6 in this paper characterizes equivalence between presheaves on monoids algebraically.

Title: Equivalent Grothendieck toposes


问:
Finding unknown values in Matrix with eigenvalues
"Consider the companion matrix  $$A =  \begin{pmatrix}
    0 & 1 & 0\\
    0 & 0 & 1\\
    a & b & c
    \end{pmatrix}$$Knowing that  −1 ,  3  and  −2  are the eigenvalues of $A$ , what is $a$, $b$ and $c$?

I have tried to find $c$ with the trace of the Matrix: $0+0+c = tr(A) = -1 + 3 -2 \Rightarrow c = 0$ (sum of eigenvalues), and find $a$ using the determinant of A and by multiplying eigenvalues $det(A) = -1 \times a = (-1) \times (3) \times (-2) = 6 \Rightarrow a = -6$. But like it is wrong according to solutions.

How can I find the values for $a, b, c$?

答:
Characteristic polynomial is $-x^3+cx^2+bx+a$

Knowing that eigenvalues are $-1,3,-2$ we must have

$-(x+1)(x-3)(x+2)=-x^3+7 x+6=0$

thus $a=6,b=7,c=0$


问:
Proof of the exact sequence $0\to H^i(M)\to \operatorname{coker} d^{i-1}\to\ker d^{i+1}\to H^{i+1}(M)\to 0$.
Let $M^\bullet$ be a complex in an abelian category. By the universal property of cokernels, there is a unique induced morphism making the diagram

commute. I want to prove that its kernel is $H^i(M^\bullet)$ and that its cokernel is $H^{i+1}(M^\bullet)$.

Since $H^i(M^\bullet)$ is, by definition, the cokernel of $I^{i-1}_{M^\bullet}\to K^i_{M^\bullet}$, the universal property of cokernels induces a morphism $H^i(M^\bullet)\to C^{i-1}_{M^\bullet}$ making the diagram

commute. I see that the composition $H^i(M^\bullet)\to C^{i-1}_{M^\bullet}\to K^{i+1}_{M^\bullet}$ is zero but I don't understand why $H^i(M^\bullet)\to C^{i-1}_{M^\bullet}$ satisfies the universal property of the kernel. After all, we understand the morphisms getting out of $H^i(M^\bullet)$, not those going in.

I have the same problem showing that $H^{i+1}(M^\bullet)$ is the cokernel of $C^{i-1}_{M^\bullet}\to K^{i+1}_{M^\bullet}$.

答:
After a lot of thought, I think I solved this problem by finding a more natural candidate for the morphism on the middle. (What follows are my own notes.)

Then our desired result becomes way simpler.


问:
Using Schur Complement to prove a matrix is negative definite
Need to find range of values of $\epsilon$ for which $$\begin{bmatrix} A+\epsilon B & \epsilon C \\ \epsilon C^\top &  \epsilon D \end{bmatrix} \prec 0$$ given that $A,D \prec 0$ and $\epsilon > 0$.

The way I am going about it is using Schur Complement Lemma whereby this matrix is negative definite iff $ \epsilon D \prec 0$ and $A+\epsilon B-\epsilon CD^{-1}C^\top \prec 0$. Should the value of $\epsilon$ depend on the maximum eigenvalue of $A-\epsilon CD^{-1}C^\top$? What is the right way to go about it?

答:
One could use the generalized eigenvalue problem

$$
A\,v = \lambda\,(C\,D^{-1}C^\top - B)\,v, \tag{1}
$$

with $\lambda$ and $v$ one of the eigenvalue-eigenvector pairs associated with this.

Namely, when setting $\epsilon$ equal to one of those generalized eigenvalues one gets that

\begin{align}
\det\!\left(A + \epsilon\,(B - C\,D^{-1}C^\top)\right) &= 0, \\
A + \epsilon\,(B - C\,D^{-1}C^\top) &\nprec 0.
\end{align}

The eigenvalues of $A + \epsilon\,(B - C\,D^{-1}C^\top)$ should change continuously when changing the value of $\epsilon$. Therefore, in order for that matrix to switch from negative definite to not being negative definite one of its eigenvalues has to become equal to zero, which only happens at one of the generalized eigenvalues. Denoting $\lambda_1$ as the smallest positive generalized eigenvalue of $(1)$. From this it follows that at least for $0 < \epsilon < \lambda_1$ it holds that $A + \epsilon\,(B - C\,D^{-1}C^\top) \prec 0$.

I am not sure how one can directly find if there might be more intervals for $\epsilon$ after $\lambda_1$ where the inequality is satisfied. However, the remaining positive generalized eigenvalues should be the only candidates for the bounds of those intervals.


问:
Prove that $h(u)=\frac{2}{u^2}(u-\log(1+u))$ is decreasing on $(-1,\infty)$

Prove that $h(u)=\frac{2}{u^2}(u-\log(1+u))$ is decreasing on $(-1,\infty)- \{0\}$

This function appears on the proof of Stirling's formula in Principles of Mathematical Analysis, by Walter Rudin.

The classic way to approach the "show a function is decreasing" is to show that its derivative is negative on the given range.

$$h'(u)= -\frac{2}{u^2}+\frac{4}{u^3}\log(1+u)-\frac{2}{u^2(1+u)}$$

And after we factor out the $-\frac{2}{u^2}$ we are left with showing that

$$g(u):= 1-\frac{2}{u}\log(1+u)+\frac{1}{1+u}$$ is positive. We could try showing that $g$ has positive derivative and show that the limit of $g$ as $u$ approaches $-1$ is positive, but this seems like rhe derivative of $g$ will be something much more complicated.

答:
To show $g(u)\ge 0$ it suffices to show $$k(u)=u(1+u)-2\log (1+u)\cdot (1+u)+u\le 0\tag 1$$ for all $u\in(-1,0]$

and $$k(u)=u(1+u)-2\log (1+u)\cdot (1+u)+u\ge 0\tag 2$$ for all $u\in[0,\infty)$

Now $k'(u)=2(u-\log(1+u))\ge 0$

therfore $\forall u\in(-1,0] k(u)\le k(0)=0$ and $\forall u\ge 0,k(u)\ge k(0)=0$ So we are done!


问:
Understanding a specific process of finding the derivative of $x^TAx$
I am referring to @copper.hat's response to : Derivative of Quadratic Form. I do not have the reputation to reply directly. My goal is to find a way to better differentiate and understand these functions for the purpose of learning the gradient. I will try to write everything here to make this question independent.

Let $Q(x) = x^TAx$. $x∈R^n$ and $A∈R^{nxn}$

Fitting to $Q(x+h)−Q(x)$, $Q(x)$ expands to $(x+h)^TA(x+h)$ which expands further to $x^TAx+x^TAh+h^TAx+h^TAh-x^TAx$ simplifying to $x^TAh+h^TAx+h^TAh$.

How is it that this changes to $x^TAh+h^TAx$ in the reply on the linked post, dropping the $h^TAh$?

I see the references to $|h^TAh|≤∥A∥∥h∥^2$ however Googling "Cauchy Scwarz" gives very broad results and I'm having trouble understanding.

I've looked up many ways to find the gradient of matrix functions (ex: $b^TX^TXb$ , $1/2x^TAx+b^Tx$) and this process seems to be the most intuitive for me.

答:
By definition a derivative at a point $x$ is such a linear function $Dh$ that this holds
$$f(x+h)=f(x)+Dh+g(h),\ g(h)\in o(||h||)$$
where by $g(h)\in o(||h||)$ we mean some function $g(h)$ such that $\lim_{h\to 0}\frac{|g(h)|}{||h||}=0$.

In the case of quadratic form $f(x)=x^TAx$ we have
$$f(x+h)=\dots=x^TAx+2x^TAh+h^TAh$$
and this is exactly in the form of the first formula with
$$f(x)=x^TAx,\ Dh=2x^TAh,\ g(h)=h^TAh\in o(||h||)$$
The last statement follows indeed from Cauchy–Schwarz inequality $|a^Tb|\leq||a||||b||$ for vectors $a,b$ and also from matrix norm inequality $||Ax||\leq||A||||x||$.
$$\lim_{h\to 0}\frac{|h^TAh|}{||h||}\leq\lim_{h\to 0}\frac{||h||||Ah||}{||h||}\leq \lim_{h\to 0}\frac{||h||||A||||h||}{||h||}=\lim_{h\to 0}||h||||A||=0$$


问:
Prove that the area of a right triangle is no more than the square of the hypotenuse divided by 4
I'm struggling with this proof. Let $a,b,c$ denote the base, height, and hypotenuse of a right triangle. Let $A$ denote area of this triangle

We have the following relations:
$$
a^2 + b^2 = c^2  \\
A = a*b/2
$$

and also the triangle inequality, though I'm not sure if this inequality comes into play.

Here is some of my work and I'm unsure if it's in the right direction
$$
a^2 + b^2 = c^2 \\
a^2*b + b^3 = c^2*b \\
\implies A*a + b^3 = b * c^2 \\
\implies A = b*c^2/a - b^3/a
$$

I feel that this is not in the right direction as I don't know how to show that this last expression is $\leq c^2/4$. Can anyone advise?

答:
Here is a picture to visualize the property.

Any right triangle lies within the semicircle erected on a square with side length equal to the hypotenuse.

All such triangles have height less than or equal to the radius of semicircle i.e., half the hypotenuse length. The right triangle with maximum area can be seen as covering exactly one-fourth of the above square. Hence,
$$\text{Area of any right triangle} \le \frac{(\text{Hypotenuse})^2}{4}$$

The maximum (equality) is achieved by right isosceles triangles.


问:
The Method Of Frobenius

The ODE $xy'' + y = 0$ has a real degeneracy.  Use The Method Of
Frobenius to find a fundamental set of solutions.

Here is the procedure, as I understand it:

1)  Plug the guess $y = x^s \sum_{n = 0}^\infty a_n x^n$ into the ODE and do the algebra/calculus to separate out the indicial equation and the recurrence relation.

$xy'' + y = 0$

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

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

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

$s(s - 1) a_0 x^{s - 1} + \sum_{n = 1}^\infty (n + s)(n + s - 1) a_n x^{n + s - 1} + \sum_{n = 1}^\infty a_{n - 1} x^{n + s - 1} = 0$

$s(s - 1) a_0 x^{s - 1} + \sum_{n = 1}^\infty (n + s)(n + s - 1) a_n x^{n + s - 1} + a_{n - 1} x^{n + s - 1} = 0$

$s(s - 1) a_0 x^{s - 1} + \sum_{n = 1}^\infty [(n + s)(n + s - 1) a_n + a_{n - 1}] x^{n + s - 1} = 0$

Indicial Equation:  $s(s - 1) = 0$

Recurrence Relation:  $(n + s)(n + s - 1) a_n + a_{n - 1} = 0$, where $n \geq 1$

2)  Solve the recurrence relation, treating $s$ as a constant, but without plugging in any indicial equation solutions.

$(n + s)(n + s - 1) a_n + a_{n - 1} = 0$

$a_n = -\frac{1}{(n + s)(n + s - 1)}a_{n - 1}$

$a_n = a_0 \Pi_{m = 1}^n -\frac{1}{(m + s)(m + s - 1)}$

Here I would simplify using a formula I was given, $\Pi_{j = c}^d (j + k) = \frac{(d + k)!}{(c + k - 1)!}$.

$a_n = a_0 [\Pi_{m = 1}^n -1] [\frac{1}{\Pi_{m = 1}^n (m + s)}] [\frac{1}{\Pi_{m = 1}^n (m + s - 1)}]$

$a_n = a_0 [(-1)^n] [\frac{1}{\frac{(n + s)!}{s!}}] [\frac{1}{\frac{(n + s - 1)!}{(s - 1)!}}]$

$a_n = \frac{(-1)^n s! (s - 1)!}{(n + s)! (n + s - 1)!} a_0$

3)  To find one fundamental solution, plug the largest indicial equation solution $s_1$ and its associated recurrence relation solution into the guess, and make the particular choice $a_0 = 1$.

$s(s - 1) = 0$

$s = 0, 1$

$s_1 = 1$

$y = x^1 \sum_{n = 0}^\infty \frac{(-1)^n 1! (1 - 1)!}{(n + 1)! (n + 1 - 1)!} 1 x^n$

$y = \sum_{n = 0}^\infty \frac{(-1)^n x^{n + 1}}{n! (n + 1)!}\leftarrow$  FIRST FUNDAMENTAL SOLUTION

4)  To find another fundamental solution, plug the other indicial equation solution $s_2$ and its associated recurrence relation solution into $y = [$FIRST FUNDAMENTAL SOLUTION$]ln|x| +\ x^{s_2} \sum_{n = 0}^\infty [\frac{\partial}{\partial s} ((s - s_2)a_n)]_{s = s_2} x^n$, and make the particular choice $a_0 = 1$.

$s_2 = 0$

Here I use the earlier, less simplified expression for $a_n$ which is in $\Pi$ product notation, in anticipation of taking the logarithmic partial derivative.

$y = (\sum_{n = 0}^\infty \frac{(-1)^n x^{n + 1}}{n! (n + 1)!})ln|x| + x^0 \sum_{n = 0}^\infty [\frac{\partial}{\partial s} ((s - 0) a_0 \Pi_{m = 1}^n -\frac{1}{(m + s)(m + s - 1)})]_{s = 0} x^n$

$y = \sum_{n = 0}^\infty \frac{(-1)^n x^{n + 1} ln|x|}{n! (n + 1)!} + \sum_{n = 0}^\infty [\frac{\partial}{\partial s} (s a_0 \Pi_{m = 1}^n -\frac{1}{(m + s)(m + s - 1)})]_{s = 0} x^n$

I make the substitution $b_n = s a_0 \Pi_{m = 1}^n -\frac{1}{(m + s)(m + s - 1)}$, appearing in the solution as $y = \sum_{n = 0}^\infty \frac{(-1)^n x^{n + 1} ln|x|}{n! (n + 1)!}$
$+ \sum_{n = 0}^\infty [\frac{\partial}{\partial s} b_n]_{s = 0} x^n$.

$b_n = s a_0 \Pi_{m = 1}^n -\frac{1}{(m + s)(m + s - 1)}$

$ln|b_n| = ln|s| + ln|a_0| + \sum_{m = 1}^n ln|\frac{1}{m + s}| + ln|\frac{1}{1 - m - s}|$

$\frac{\partial}{\partial s} ln|b_n| = \frac{\partial}{\partial s} ln|s| + \frac{\partial}{\partial s} ln|a_0| + \sum_{m = 1}^n \frac{\partial}{\partial s} ln|\frac{1}{m + s}| + \frac{\partial}{\partial s} ln|\frac{1}{1 - m - s}|$

I set $\frac{\partial}{\partial s} ln|a_0|$ to $0$ because the chapter stipulates that this procedure only seeks solutions where $a_0$ is nonzero and does not depend on $s$.

$\frac{\frac{\partial b_n}{\partial s}}{b_n} = \frac{1}{s} + 0 + \sum_{m = 1}^n -\frac{1}{m + s} + \frac{1}{1 - m - s}$

$\frac{\partial b_n}{\partial s} = \frac{b_n}{s} + b_n \sum_{m = 1}^n -\frac{2m + 2s - 1}{(m + s)(m + s - 1)}$

$\frac{\partial b_n}{\partial s} = \frac{s a_0 \Pi_{m = 1}^n -\frac{1}{(m + s)(m + s - 1)}}{s} + s a_0 (\Pi_{m = 1}^n -\frac{1}{(m + s)(m + s - 1)})(\sum_{m = 1}^n -\frac{2m + 2s - 1}{(m + s)(m + s - 1)})$

Now I can simplify the $\Pi$ products using the above formula.  I also make the choice for $a_0$ here.

$\frac{\partial b_n}{\partial s} = \frac{s [\Pi_{m = 1}^n -1] [\frac{1}{\Pi_{m = 1}^n (m + s)}] [\frac{1}{\Pi_{m = 1}^n (m + s - 1)}]}{s} + s ([\Pi_{m = 1}^n -1] [\frac{1}{\Pi_{m = 1}^n (m + s)}][\frac{1}{\Pi_{m = 1}^n (m + s - 1)}])(\sum_{m = 1}^n -\frac{2m + 2s - 1}{(m + s)(m + s - 1)})$

$\frac{\partial b_n}{\partial s} = \frac{s [(-1)^n] [\frac{1}{\frac{(n + s)!}{s!}}] [\frac{1}{\frac{(n + s - 1)!}{(s - 1)!}}]}{s} + s ([(-1)^n] [\frac{1}{\frac{(n + s)!}{s!}}] [\frac{1}{\frac{(n + s - 1)!}{(s - 1)!}}])(\sum_{m = 1}^n -\frac{2m + 2s - 1}{(m + s)(m + s - 1)})$

$\frac{\partial b_n}{\partial s} = \frac{(-1)^n s! (s - 1)! + (-1)^{n + 1} s!^2 \sum_{m = 1}^n \frac{2m + 2s - 1}{(m + s)(m + s - 1)}}{(n + s)! (n + s - 1)!}$

Unwinding the substitution, I plug the expression for $\frac{\partial b_n}{\partial s}$ back into the fundamental solution.

$y = \sum_{n = 0}^\infty \frac{(-1)^n x^{n + 1} ln|x|}{n! (n + 1)!} + \sum_{n = 0}^\infty [\frac{(-1)^n s! (s - 1)! + (-1)^{n + 1} s!^2 \sum_{m = 1}^n \frac{2m + 2s - 1}{(m + s)(m + s - 1)}}{(n + s)! (n + s - 1)!}]_{s = 0} x^n$

$y = \sum_{n = 0}^\infty \frac{(-1)^n x^{n + 1} ln|x|}{n! (n + 1)!} + \sum_{n = 0}^\infty \frac{(-1)^n (-1)! x^n + (-1)^{n + 1} x^n \sum_{m = 1}^n \frac{2m - 1}{m^2 - m}}{n! (n - 1)!}$

And now we see that something is broken.  $(-1)!$ is undefined, and can't even be rescued by being generalized to the Gamma or Pi function.  What am I doing wrong, and how can I use this method to solve this ODE?

答:
Just a pictorial comment. It's always nice to see what the outcome of a differential equation looks like.

And, by the way, it does not seem much of an effort to solve it numerically, for example as in the Pascal snippet below. The solution looks like a sine with decreasing frequency, as has to be expected. The MUCH parameter has been given two different values for checking convergence.

```
procedure test(much : integer);
var
  y : array of double;
  x,dx : double;
  k : integer;
begin
  SetLength(y,much+1);
  dx := (xmax-xmin)/much;
  y[0] := 0;
  y[1] := y[0] + dx;
  for k := 1 to much-1 do
  begin
    x := k*dx;
    y[k+1] := 2*y[k]-y[k-1]-sqr(dx)/x*y[k];
  end;
end;
```

Title: The Method Of Frobenius

Body:
Just a pictorial comment. It's always nice to see what the outcome of a differential equation looks like.

And, by the way, it does not seem much of an effort to solve it numerically, for example as in the Pascal snippet below. The solution looks like a sine with decreasing frequency, as has to be expected. The MUCH parameter has been given two different values for checking convergence.

procedure test(much : integer);
var
  y : array of double;
  x,dx : double;
  k : integer;
begin
  SetLength(y,much+1);
  dx := (xmax-xmin)/much;
  y[0] := 0;
  y[1] := y[0] + dx;
  for k := 1 to much-1 do
  begin
    x := k*dx;
    y[k+1] := 2*y[k]-y[k-1]-sqr(dx)/x*y[k];
  end;
end;