问:
In $\Delta ABC$ , $\angle A = 20^\circ, \angle C = 90^\circ$. $O$ is a point on $AB$ and $D$ is the midpoint of $OB$ . Find $\angle BCD$ .

In $\Delta ABC$ , $\angle A = 20^\circ, \angle C = 90^\circ$. $O$ is a  point on $AB$ and $D$ is the midpoint of $OB$ . The circle centered at $O$ with radius $OD$ intersects $AC$ at $T$. Find $\angle BCD$ .

What I Tried: Here is a picture :-

You can see what I tried. The first part is to do angle-chasing and find the respective angles in the picture. I also find that $OT \parallel BC$, since $OBCT$ is a trapezium . Now I think that to find $\angle BCD$ , somehow I have to use the fact that $OD = DB$ (which I have not used) and maybe the parallel lines, but I seem to be stuck here.

Can anyone help me figure this out? Thank You.

答:
Given you already have a geometric solution, here is a trigonometric one.

Say $\angle BCD = x, $ radius of the circle = $r$. Portion of $BA$ on the left outside circle is say, $a$.

Using Sine law in $\triangle BCD$,

$\displaystyle \frac{r}{\sin x} = \frac{CD}{\sin 70^0}$ ...(i)

(as $BD = r$)

Applying Sine law in $\triangle AOT$,

$\displaystyle \frac{a+r}{\sin 90^0} = \frac{r}{\sin 20^0} \implies a = \frac{r(1-sin 20^0)}{\sin 20^0}$ ...(ii)

(as $AO = a + r, OT = r$)

Applying Sine Law in $\triangle ACD$,

$\displaystyle \frac{a+2r}{\sin (90^0-x)} = \frac{CD}{\sin 20^0}$

(as $AD = a + 2r, \angle ACD = (90^0 - x)$)

So, $\displaystyle \frac{a+2r}{\cos x} = \frac{CD}{\sin 20^0} \implies CD = r\frac{1 + \sin20^0}{\cos x}$ (substituting $a$ from (ii))

Now substituting $CD$ in (i),

$\tan x = \displaystyle \frac{\sin 70^0}{1 + \sin 20^0} = \frac{\sin 70^0}{1 + \cos 70^0}$

i.ie $\tan x = \displaystyle \frac{2 \sin 35^0 \cos 35^0}{2 \cos^2 35^0} = \tan 35^0$


问:
Question about randomly selection and selected uniformly at random?
I am starting a probability class and I am a bit confused looking at my textbook. An exercice is mentionning selecting cards randomly when another one says picking balls uniformly at random
I have seen that a lot of posts are dealing about this but all of them have not made any sense to me unfortunately...

Could you explain me the difference between the two ? (if there is)

Complementary question :

If an exercice is saying: picking balls uniformly at random

How do I know if we take the four balls at the same time or the balls one by one ?

答: For the second question, you need to know if there is replacement, and when. Each ball replaced is different than all four replaced before the next operation, which is different than no replacement. Combinatorics with replacement is trickier, but Professor Woo on Youtube has excellent videos from 2013.</p>

ContentLicense: CC BY-SA 4.0


问:
When $f(x+1)-f(x)=f'(x)$, what are the solutions for $f(x)$?
The question is:

When $f(x+1)-f(x)=f'(x)$, what are the solutions for $f(x)$?

The most obvious solution is a linear function of the form $f(x)=ax+b$. Is this the only solution?

Edit

I should add that $f:\mathbb R\to\mathbb R$ to the question.

答:
Let $\lambda = \alpha + \beta i \in \mathbb{C}$ be any root of the equation $\lambda = e^{\lambda} - 1$ subject to $\beta \ge 0$ and $\delta$ any real number, then
$$f(x) = \Re( e^{\lambda x + i\delta} ) = e^{\alpha x} \cos(\beta x + \delta)\tag{*}$$
will be a $C^{\infty}$ solution for the functional equation $f'(x) = f(x+1) - f(x)$.

For example, $\lambda = e^{\lambda} -1 $ has a root at $\sim 2.088843015613044 + 7.461489285654254 i$.

This gives us following $C^{\infty}$ solution:

$$f(x) \sim e^{ 2.088843015613044\,x}\,\cos(7.461489285654254\,x)$$

Other solutions can be constructed by forming linear combination of solutions of the form in $(*)$. Please note that $\lambda = e^{\lambda} - 1$ has a double root at $\lambda = 0$,
there are two solutions at $\lambda = 0$ which one can use:

$$ \Re( e^{\lambda x} )\Big|_{\lambda = 0} = 1
\quad\quad\text{ and }\quad\quad
\frac{\partial}{\partial \lambda} \Re( e^{\lambda x} )\Big|_{\lambda = 0} = x
$$

ContentLicense: CC BY-SA 3.0


问:
How can I solve $x^2y''+2xy'+2y=x$ using VoP?
Is it possible to solve this differential equation by variation of parameters? If so, how can I compute the complementary solution? Otherwise, how should approach this problem?
$$x^2y''+2xy'+2y=x$$

My Attempt:

When I try to substitute $y=x^m$ into the equation I get the characteristic eq. as $m^2+m+2=0$ which gives $m=(−1−\sqrt 7)/2$ and $m=(−1+ \sqrt 7)/2$. Is this correct ? Thanks in advance!

答:
For the homogeneous differential equation:
$$x^2y''+2xy'+2y=0$$
Substitute $y=x^m$:
$$m(m-1)+2m+2=0$$
$$m^2+m+2=0$$
it should give you complex solutions:
$$\Delta =1-8=-7$$
$$\implies m=\dfrac {-1\pm i\sqrt 7}2$$
$$y(x)=\dfrac {\sqrt x}{x}\left(A \cos \left( \dfrac {\sqrt 7}2 \ln x\right)+B
\sin \left (\dfrac {\sqrt 7}2\ln x \right)\right)$$

Note that you have:
$$x^{a+bi}=x^ax^{bi}$$
Now use Euler's formula:
$$=x^a(e^{bi \ln x})=x^a(\cos( b \ln x)+i\sin (b\ln x))$$
And you also have that:
$$x^{a-bi}=x^a(\cos( b \ln x)-i\sin (b\ln x))$$
Since $\sin (-b)=-\sin (b)$.


问:
Probabilities of nested intervals tending to single point
Say I have a probability measure $\mathbb{P}$ over $\mathbb{R}$ without any atoms, by which I mean $\mathbb{P}(\{x\}) = 0 \; \forall x \in \mathbb{R}$. Does it follow that a sequence of nested intervals tending to a point will have probabilities which tend to 0?

For instance, does it follow that $\mathbb{P}([x - \frac1n, x + \frac1n]) \to 0$ as $n \to \infty$? The limit itself should exist, by monotone convergence, but it's not clear to me why it would need to be 0 (if this is indeed true). I'm looking for a result about how probability measures interact with infinite intersections/chains of sets, or a counter-example.

答:
The result you are looking is known as "continuity of the probability measure" and it can be found (i guess) in every introductory probability textbook. This says that whenever you have a decreasing (there is also a corresponding result for increasing sequences) sequence of measurable sets $A_n$, then $$\mathbb{P}\bigl(\bigcap_{n=1}^{\infty}A_n\bigr)=\lim_{n\to \infty}\mathbb{P}(A_n)$$

In your case the $A_n's$ are the intervals $[x-\frac{1}{n},x+\frac{1}{n}]$. Here, its easy to check that $\bigcap_{n=1}^{\infty}[x-\frac{1}{n},x+\frac{1}{n}]=\{x\}$. Which means that
$$\lim_{n\to \infty}\mathbb{P}([x-\frac{1}{n},x+\frac{1}{n}])=\mathbb{P}(\{x\})$$

If you are dealing with a measure with no atoms, then $\mathbb{P}(\{x\})=0$.


问:
How to calculate the derivative of Fourier transform of a matrix?
$X$ is a complex matrix and $Y = FT(|X|^2)$, $|X|^2$ stands for elementwise modulus. I want to calculate the derivative of $Y$ to X. Could anyone tell me how to do this?

答:
$\def\p{\partial}$
Your function is
$$Y=F(X^*\circ X)$$
where $X^*$ is the complex conjugate of $X,\,$ and
$(\circ)$ denotes the elementwise/Hadamard product.

Calculate the gradient wrt $X$ in the CR-Calculus sense, i.e. by treating $X^*$ as independent of $X$
$$\eqalign{
dY &= F(X^*\circ dX) \\
{\rm vec}(dY) &= (I\otimes F)\;{\rm vec}(X^*\circ dX) \\
  &= (I\otimes F)\;\Big({\rm vec}(X^*)\circ{\rm vec}(dX)\Big) \\
  &= (I\otimes F)\cdot{\rm Diag}\Big({\rm vec}(X^*)\Big)\,{\rm vec}(dX) \\
\frac{\p\,{\rm vec}(Y)}{\p\,{\rm vec}(X)}
  &= (I\otimes F)\cdot{\rm Diag}\Big({\rm vec}(X^*)\Big) \\\\
}$$
where $(\otimes)$ denotes the Kronecker product
and the gradient is a second-order tensor (aka a matrix).

The alternative to vectorizing the matrices is to replace the Hadamard product with a sixth-order tensor
$$\eqalign{
dY &= F\big(X^*:{\mathbb M}:dX\big) \\
\frac{\p Y}{\p X}
 &= F\big(X^*:{\mathbb M}\big) \\
}$$
where $(:)$ denotes the double-dot product and the gradient is a fourth-order tensor.


问:
Computation of PDF $f_{Z}(z)$, where $Z=XY^{-1}$ and $X$ and $Y > 0$ are independent [I am primarily seeking a hint]
The distributions of $X$ and $Y$ are respectively unknown, hence I am trying to derive a (preferably) compact expression of e.g. $f_X$ etc., if possible. Furthermore it is given that $Y:\Omega \rightarrow \mathbb{R}_+$, as presented in the title.

My assertion is as follows,

$F_Z(z)\triangleq \mathbb{P}\text{r}(Z \leq z)=\mathbb{P}\text{r}(XY^{-1}\leq z)$,

thereby I am seeking for an alternative (suitable) formulation of $\mathbb{P}\text{r}(XY^{-1}\leq z)$ and furthermore differentiate to obtain the sought for PDF,

$f_Z(z)=\dfrac{\partial \mathbb{P}\text{r}(Z\leq z)}{\partial z}=\dfrac{\partial \mathbb{P}\text{r}(XY^{-1}\leq z)}{\partial z}$.

I have also derived $f_{Z \mid Y}(z)=yf_{X}(yz)$, $f_{Y\mid Z}(y)=yf_X(yz)$ and $f_{Z\mid X}(z)=\dfrac{x}{z^2}f_X\left(\dfrac{x}{z}\right)$ etc. if someone perhaps sees an application.

Any hints are highly appreciated!

答:
Since $Y > 0$, you could equivalently write
$$ \mathbb P(X/Y \le z) = \mathbb P(X \le z Y)$$
If $z < 0$ this is
$$ \int_{-\infty}^0 f_X(x) (1 - F_Y(x/z))\; dx $$
while if $z > 0$ it is
$$ F_X(0) + \int_0^\infty F_Y(x/z)\; dx$$


问:
Prove that a closed rectangle in $\mathbb{R}^n$ is a closed set
Here's what I'm trying to prove:

A closed rectangle in $\mathbb{R}^n$ is a closed set.

This was a statement in Michael Spivak's Calculus on Manifolds and I'm trying to write a proof for it.

Let $([a_i,b_i])_{i=1}^{n}$ be a finite sequence of closed intervals and let $C = \prod_{i=1}^{n}[a_i,b_i]$ be a closed rectangle in $\mathbb{R}^n$. To prove that this is closed, we just have to show that $\mathbb{R}^n \setminus C$ is open. This amounts to showing that for every $p \in \mathbb{R}^n \setminus C$, there is an open rectangle $A$ such that $p \in A \subseteq \mathbb{R}^n \setminus C$.

Let $p \in \mathbb{R}^n \setminus C$. Then, $p = (p_i)_{i=1}^{n}$. Since $p \notin C$, there is an $i \in \{1,2,\ldots,n\}$ such that $p_i \notin [a_i,b_i]$. So, $p_i \in (-\infty,a_i) \cup (b_i,+\infty)$.

Construct open intervals $(c_i,d_i)$ as follows:

If $p_i \in [a_i,b_i]$, let $c_i = p_i-\epsilon$ and $d_i = p_i+\epsilon$, with $\epsilon > 0$ large enough so that $c_i < a_i$ and $d_i > b_i$.

If $p_i \notin [a_i,b_i]$, let $c_i = p_i-\epsilon$ and $d_i = p_i+\epsilon$, with $\epsilon > 0$ small enough so that $(c_i,d_i)$ is contained in either $(-\infty,a_i)$ or in $(b_i,+\infty)$, depending on whether $p_i < a_i$ or $p_i > b_i$.

By definition, $p \in \prod_{i=1}^{n} (c_i,d_i)$ and clearly, this open rectangle is a subset of $\mathbb{R}^n \setminus C$. Since $p$ was arbitrary, $\mathbb{R}^n \setminus C$ is open so $C$ is closed. $\Box$

Does the argument above work? If it doesn't, then why? How can I fix it?

答: Your proof is correct. However, you can simplify your argument. You have $p_i \in U_i = (-\infty,a_i) \cup (b_i,\infty)$ which is open in $\mathbb R$. Hence $\pi_i^{-1}(U_i)$ is open in $\mathbb R^n$, where $\pi_i : \mathbb R^n \to \mathbb R$ denotes the projection onto the $i$-th coordinate. Clearly $p \in \pi_i^{-1}(U_i) \subset \mathbb R^n \setminus C$.


问:
Show that the polynomial $x^{2}+1$ has no fixed prime divisor
An integer-valued polynomial $Q(x)$ has a fixed divisor $m$ if there is an integer $m>1$ such that
$\frac {Q(x)}{m}$ is also an integer-valued polynomial. See https://en.wikipedia.org/wiki/Schinzel%27s_hypothesis_H#Fixed_divisors. If $m$ is a prime then we use the term fixed prime divisor.

I am interested in this problem:

Show that the polynomial $x^{2}+1$ has no fixed prime divisor.

答:
$P(0)=1$ is enough to say that there are no fixed divisor, prime or not.

If an integer-valued polynomial has $1$ in its range then it has no fixed divisors.

Edit

Generalization: If an integer-valued polynomial has a prime number in its range, then it has no fixed number, excluded trivial cases like $P(x)=7x^2+7$


问:
Core of the graph $G_1\cup G_2$, where $G_1$ and $G_2$ are disjoint unions of complete graphs, is a complete graph
Question

I'm looking for an answer to the following question from Hell and Nesetril's Graphs and Homomorphisms:

Let $G_1$ and $G_2$ be graphs such that $V = V(G_1) = V(G_2)$, and both $G_1$ and $G_2$ are disjoint unions of complete graphs (so each component of $G_i$ is complete). Let $G$ be the graph $(V, E(G_1)\cup E(G_2))$. Show that the core of $G$ is a complete graph.

Definitions

A retract from a graph $G$ onto a subgraph $H$ is a graph homomorphism that fixes $H$.
I.e., For $H\leq G$, a retract is a map $f:G\rightarrow H$ such that $f(u) = u$ for $u\in V(H)$, and such that $uv\in E(G)$ implies $f(u)f(v) \in E(H)$.
A core is a graph that does not retract onto any proper subgraph. The core of a graph $G$ is the unique-up-to-isomorphism subgraph of $G$ that is itself a core.

Commentary

It should be sufficient, and probably easier, to prove that the chromatic number of $G$ is equal to the size of the largest clique in $G$ (which in turn will be the size of the largest connected component of $G_1$ or $G_2$). So I'll be happy with any answer that shows $\chi(G) = \omega(G)$ instead.

I do not want an answer that makes use of a heavy literature result like the strong perfect graph theorem.

答:
Consider the bipartite multigraph $H$ with:

a vertex corresponding to each clique of $G_1$ on one side;

a vertex corresponding to each clique of $G_2$ on the other side;

for each $v \in V(G)$, an edge from the vertex representing $v$'s clique in $G_1$ to the vertex representing $v$'s clique in $G_2$.

By König's line coloring theorem, $H$ has an edge coloring with $\Delta(H)$ colors. This means $G$ has a vertex coloring with $\omega(G)$ colors. (The degree of a vertex in $H$ is the number of vertices in the corresponding clique of $G$.)

If you're uncertain about using König's line coloring theorem, we can prove it from Hall's theorem without too much effort. By adding dummy edges and vertices, we can make $H$ be $\Delta(H)$-regular. By applying Hall's theorem, we can find a perfect matching in the resulting multigraph; color the edges of that matching one color, remove them, and repeat on the $(\Delta(H)-1)$-regular multigraph remaining.


问:
Jacobian of a Change of Variable
I am studying this article for my PhD, where every thing was quite clear except the Jacobian claimed in the proof of proposition 5 (which is not related to the topic of the article).

In particular, I am confused about the Jacobian of the change of variable
$$S^2\times\mathbb{R}^3\mapsto\mathbb{R}^3\times\{{p}\}^\perp$$
$$(w,V_*)\mapsto(p,q),$$
where $p=(V_*.w)w$ and $q=V_*-p$, which turned out to be
$$dV_*dw=\frac{2}{p^2\sin(p,p+q)}dpdq.$$

I found an answer on stack exchange which was to view the Jacobian the infinitesimal area of the paralellogram of sides $|p|dV_*$ and $|p|dw$. I didn't understand what does the side $|p|dV_*$ have to do here!I hope someone could give me a formal explanation, and I'm very grateful.

答:
Here is the begining of an answer. I did not make all the computations (especially in the end) but hope it will help you answer your own question.

Let $f$ be the differentiable function
\begin{align}
f : \mathbb{R}^3 \times \mathbb{R}^3 & \longrightarrow \mathbb{R}^3 \times \mathbb{R}^3 \\
(x,v) & \longmapsto \left(\langle x,v\rangle x, v-\langle x,v\rangle x \right)
\end{align}
Where $\langle \cdot,\cdot\rangle$ stands for the euclidean structure. It is differentiable as for $(h_1,h_2)\in \mathbb{R}^3\times\mathbb{R}^3$, one has
\begin{align}
f(x+h_1,v+h_2) = f(x,v)+ L_{x,v}(h_1,h_2) + o\left(\|h_1,h_2\|\right)
\end{align}
where
\begin{align}
L_{x,v}(h_1,h_2) = \left(\left(\langle h_1,v\rangle + \langle x,h_2 \rangle \right) x + \langle x,v\rangle h_1, h_2 -\left(\langle h_1,v\rangle + \langle x,h_2 \rangle \right) x - \langle x,v\rangle h_1  \right)
\end{align}

By restriction, $f$ is differentiable on $\mathbb{S}^2\times \mathbb{R}^3 \subset \mathbb{R}^3\times\mathbb{R}^3$ and the differentiable is just the restriction of $L$ on the tangent spaces.

I do not show that $f$ is injective and that $L_{x,v}$ is invertible on $T_{x,v}\left(\mathbb{S}^2\times \mathbb{R}^3 \right)$ as this is due to easy but messy computations. This shows that $f$ is a diffeomorphism from $\mathbb{S}^2\times \mathbb{R}^3$ to its image. Moreover, the image of $f$ is
\begin{align}
f\left(\mathbb{S}^2\times \mathbb{R}^3\right) = \bigcup_{x \in \mathbb{S}^2} \mathbb{R}^3\times \{x\}^{\perp}
\end{align}

We now have a clean writing of $f$, and the jacobian of $f$ is the determinant of $L_{x,v}$ in orthonormal bases.
Moreover, in the article they are using $f^{-1}$. Maybe the inverse function $f^{-1}$ has a nicer form and can help you. For the $\sin(p,p+q)$ part, I feel like, as $p+q = v$ and $\langle x,v\rangle = \langle p,v\rangle$ (easy computations), we can hope that using $f$ above show a formula like $f^*\mathrm{dVol}_{RHS} = \langle x,v\rangle^2 \frac{\sin\left(x,v\right)}{2}\mathrm{dVol}_{LHS}$ (for volume on the left hand side and right hand side) and thus just use $f$.

Title: Jacobian of a Change of Variable


问:
Estimation of Survival Probability
I am reading a book which gives the way to calculate the Survival and fault probabilities as follows. Below table is given an reference -

```<code>+------+-------------+----------------+
| Year | Death Prob. | Survival Prob. |
+------+-------------+----------------+
|    1 |        0.02 |           0.98 |
|    2 |      0.0196 |         0.9604 |
|    3 |      0.0192 |         0.9412 |
|    4 |      0.0188 |         0.9224 |
|    5 |      0.0184 |         0.9039 |
+------+-------------+----------------+
</code>```

The calculation of the elements are mentioned as below -

Suppose that the probability of an object is damaged during a year conditional
on no earlier damage is 2%. The table shows survival probabilities and unconditional
probabilities of being damaged for each of the 5 years. The probability that an object will get damaged during the first year is 0.02 and the probability that it will survive until the end of the first year is 0.98. The probability of being damaged during the second year is 0:02 *  0:98 = 0:0196 and the probability of survival until the end of the second year is 0:98* 0:98 = 0:9604.

I failed to understand the reasoning for the calculation that probability of being damaged during the second year is 0:02 *  0:98 = 0:0196.

Assuming I am given the column for the total Survival probability (the last column), how should I get the Death Probability column except the trivial first case?

Any insight is very helpful.

答:
Let $N$ be the random variable denoting the year of death, and for $n = 1,2, \ldots$ we denote
$$D_n = \mathbf P[N = n], \qquad S_n = \mathbf P[N > n]$$
respectively the probability of death in year n (the left hand column in your table), and the probability of survival beyond the $n$-th year (the right hand column).

The probability you have been given is the probability of death in a given year, conditioned on having survived to the previous year. I.e. the condition

$$ \mathbf P[N = n + 1 | N > n] = 0.02 = \rho.$$

In the first year we have:
$$D_1 = \rho, \qquad S_1 = (1 - \rho).$$

For the second year we make use of the law of total probability:

$$
\begin{align}
D_2 & = \mathbf P[N = 2 | N = 1 ] \mathbf P[N = 1] +
 \mathbf P[N = 2 | N > 1 ]\mathbf P[N > 1] \\
& = \big( 0 \times D_1 \big) + \mathbf P[N = 2 | N > 1 ]S_1 \\
& = 0 + \rho S_1 \\
& = \rho \times (1 - \rho).
\end{align}
$$
The survival probability is then

$$
\begin{align}
S_2 & = S_1 - D_2\\
& = (1-\rho) - (1-\rho)\rho \\
& = (1- \rho)^2
\end{align}
$$

More generally you can derive the probabilities for the higher terms by induction.

We suppose that $D_n, \, S_n$ follow the equations

$$D_n = \rho(1-\rho)^{n-1}\, \qquad S_n = (1- \rho)^n$$

then the same approach as for the case $n = 2$ can be applied
$$
\begin{align}
D_{n+1} & = \mathbf P[N = n+1 | N \leq n] \mathbf P[N \leq n ] + \mathbf P[N = n +1 | N > n ] \mathbf P[N > n ] \\
& = \mathbf P[N = n +1 | N > n ] \mathbf P[N > n ]  \\
& = \rho S_n
\end{align}
$$
and
$$
\begin{align}
S_{n+1} & = S_n - D_{n+1} \\
& = S_n - \rho S_n \\
& = (1 - \rho)S_n
\end{align}
$$

Title: Estimation of Survival Probability


问:
Homeomorphism between $\mathbb{R}^2$ and the open unit disc
Given the function $f(x,y)=(\frac{x}{1+\sqrt{x^2+y^2}},\frac{y}{1+\sqrt{x^2+y^2}})$, I have to prove $f$ is a homemorphism between $\mathbb{R}^2$ and the open unit disc. Proving this function is conitunuous is trivial, proving it's a bijection is a bit harder (I wasn't able to find it's inverse). I did find the same function in a different notation - $f(z)=\frac{z}{1+\|z\|}$ where $z\in\mathbb{R}^2$, but since it's my first time dealing with two variable functions and homemorphisms, I can't see why $f(z)=f(x,y)$, other than $\|z\|=\|(x,y)\|$.
What is the best way to approach this with given notation ($f(x,y)$) - more generally how can I prove this function is a bijection and find its inverse.

(I was able to show that $lim_{(x,y)\rightarrow(\infty,\infty)}f(x,y)$ is (1,1) which I guess shows its a surjection but otherwise had no idea).

答:
We have $f(x,y)=(\frac{x}{1+\sqrt{x^2+y^2}},\frac{y}{1+\sqrt{x^2+y^2}})$.

It's easier to see in polar coordinates: $\vec x=(x,y)=(r\cos \theta,r\sin \theta);\ f(x,y)=(r'\sin \varphi,r'\cos \varphi)$. Then,

$r'(\cos \theta,\sin \theta)=\frac{r}{1+r}(\cos \varphi,\sin \varphi)$  so it is clear that $f$ maps $(x,y)$ into the disk along the radial line it makes with the origin. That is, $(x,y)\mapsto \frac{\|\vec x\|}{1+\|\vec x\|}(x,y).$ $f$ is clearly bijective.

It is also open: take a basic open set

$\vec x_0\in (x_0-\delta)\times ,x_0+\delta)\times (y_0-\eta,y_0+\eta):=U.$ If $\vec x\in U$ then, wlog $x_0,y_0\ge 0$, we have then

$\sqrt{(x_0-\delta)^2+(y_0-\eta)^2}< \|\vec x\|< \sqrt{(x_0+\delta)^2+(y_0+\eta)^2}$

and so

$\frac{\sqrt{(x_0+\delta)^2+(y_0+\eta)^2}}{1+\sqrt{(x_0+\delta)^2+(y_0+\eta)^2}}<\|f(x,y)\|<(x_0+\delta)^2+(y_0+\eta)^2.$

It follows that $f$ is a homeomorphism.


问:
if $a,b,c,d$ be the first positive solutions of $\sin x=\frac{1}{4}$ then...

if $a,b,c,d$ be the first positive solutions of $\sin x=\frac{1}{4}$ then find $$\sin (d/2)+2\sin (c/2)+3\sin (b/2)+4\sin (a/2)$$.$(a<b<c<d)$

Attempts
$$\sin x=1/4$$,$$4\sin^2(x/2)(1-\sin^2(x/2))=\frac{1}{16}$$
Let $\sin^2 (x/2)=t$:it remains to solve $$t-t^2=\frac{1}{64}$$ .The rest is just brute force i.e we calculate  the 4 values $\sin(x/2)$ can take , make sure the angles are arranged in ascending order and substitute the roots.As far as i can see i a getting irrational roots.

My question is does anyone spot a trick, that can avoid all this work.!

答:
Hint: Since $\sin a$ takes a positive value and $a$ is the smallest solution of $\sin x = \frac{1}{4}$, it follows that angle $a$ is between $0$ and $\frac{π}{2}$. This leads to the conclusion that $b=π−a, c=2π+a$ and $d=3π−a$.

Substituting these values in the original equation and using some basic trigonometric identities gives $2\sqrt{1+\sin a}.$


问:
Using FFT to calculate derivatives
I know that there are several posts about FFT and derivatives, but i dont get it.

The formula is:

$$
f(t) \to \hat{f}(\xi), f'(t) \to  2\pi i \xi\hat{f}(\xi)
$$

First, i dont understand what is the $\xi$, when i learned FFT as much as i poorly understand it, we compute it at different complex roots.

Let's take an example, i want to calculate the derivative of the polynomial:

$$
P(x) = -1 + 2x + 5x^2 - 4x^3
$$

The analytic solution will be:

$$
P'(x) = 0 + 2 + 10x - 12x^2
$$

Now, i take the FFT of $P(x)$ and get:

What should i multiply this by in order to get the values that by using FFT$^{-1}$  will get me to $P'(x)$?

答:
Suppose you have an arbitrary function $f\equiv \hat{f}$. The function can be represented in physical space as $f(x)$ or in frequency space as $\hat{f}(\xi)$. Here $\xi$ is the frequency or mode. You can derive the spectral derivative in three steps:

Apply the Fourier transform on $f(x)$ to derive $\hat{f}(\xi)=\mathcal{F}(f(x))$.

Multiply the $k$-th mode with $i \xi$ where $\xi=0,1,2,3,...$ to derive $\frac{\partial \hat{f}(\xi)}{\partial \xi}= i \xi \hat{f}(\xi)$.

Apply the inverse Fourier transform on $\frac{\partial \hat{f}(\xi)}{\partial \xi}$ to derive $\frac{\partial f(x)}{\partial x} = \mathcal{F}^{-1}(\frac{\partial \hat{f}(\xi)}{\partial \xi})$.

Note that here $\omega= \frac{2\pi}{L} = 1$ or $L=2 \pi$. Otherwise you have to scale the $i$-th mode with $\omega$.

Summarizing:

$\frac{\partial f(x)}{\partial x} = \mathcal{F}^{-1}(\frac{\partial \hat{f}(\xi)}{\partial \xi})=\mathcal{F}^{-1}( i \xi \mathcal{F}(f(x))$.

Matlab code for even $N$:

```<code>N=64;
L=2*pi;
dx=L/N;
x=linspace(0,L-dx,N);
f=sin(x);
k(1:N/2)=1i*(0:N/2-1);
k(N/2+2:N)=1i*(-N/2+1:-1);
% Consider the Nyquist mode my young Padawan
k(N/2+1)=0;
fhat=fft(f);
fhatx=k.*fhat;
fx=ifft(fhatx);
</code>```

Title: Using FFT to calculate derivatives

Body:
Suppose you have an arbitrary function $f\equiv \hat{f}$. The function can be represented in physical space as $f(x)$ or in frequency space as $\hat{f}(\xi)$. Here $\xi$ is the frequency or mode. You can derive the spectral derivative in three steps:

Apply the Fourier transform on $f(x)$ to derive $\hat{f}(\xi)=\mathcal{F}(f(x))$.

Multiply the $k$-th mode with $i \xi$ where $\xi=0,1,2,3,...$ to derive $\frac{\partial \hat{f}(\xi)}{\partial \xi}= i \xi \hat{f}(\xi)$.

Apply the inverse Fourier transform on $\frac{\partial \hat{f}(\xi)}{\partial \xi}$ to derive $\frac{\partial f(x)}{\partial x} = \mathcal{F}^{-1}(\frac{\partial \hat{f}(\xi)}{\partial \xi})$.

Note that here $\omega= \frac{2\pi}{L} = 1$ or $L=2 \pi$. Otherwise you have to scale the $i$-th mode with $\omega$.

Summarizing:

$\frac{\partial f(x)}{\partial x} = \mathcal{F}^{-1}(\frac{\partial \hat{f}(\xi)}{\partial \xi})=\mathcal{F}^{-1}( i \xi \mathcal{F}(f(x))$.

Matlab code for even $N$:

N=64;
L=2*pi;
dx=L/N;
x=linspace(0,L-dx,N);
f=sin(x);
k(1:N/2)=1i*(0:N/2-1);
k(N/2+2:N)=1i*(-N/2+1:-1);
% Consider the Nyquist mode my young Padawan
k(N/2+1)=0;
fhat=fft(f);
fhatx=k.*fhat;
fx=ifft(fhatx);


问:
Prove that this set is a borel set
I have difficulties proving the following :

Let A be the set of all real x between [0 , 1], such that x has a decimal representation and each digit has infinite occurence. Now, what i do understand is, that a borel set is a set that can be constructed using countable union or intersection. But i dont really see a way to construct this set, since there are uncountably many of such x.
Any guesses?

答:
First note that the numbers with non-unique decimal representation are not in $A$, and neither of their two representations (one ending in $000\ldots$ and one in $999\ldots$ has infinitely many of any digit other than $0$ or $9$.
So we may restrict our attention to the representation that, in case of non-uniqueness, ends in $0$'s.  Let $D_n(x)$ be the $n$'th digit of $x$ in this representation. We have $A = A_0 \cap \ldots \cap A_9$, where $A_d$ is the set of numbers whose decimal representation has infinitely many $d$'s.
Now $x \in A_d$ iff for every positive integer $m$ there is integer $n > m$ such that the $n$'th digit of $x$ is $d$.  That says
$$ A_d = \bigcap_{m=1}^\infty \bigcup_{n=m+1}^\infty \{x\in (0,1): D_n(x) = d\}$$
And finally, write $\{x \in (0,1): D_n(x) = d\}$ as a finite union of intervals.

Title: Prove that this set is a borel set


问:
Why do we study prime ideals?
I hope this isn't an inappropriate question here!

I'd like to ask the following (perhaps slightly ill-posed) question: why do we study prime ideals in general (commutative or non-commutative) rings? The situation is somewhat clear to me in some nice cases such as Dedekind rings, but even in slightly more general settings (e.g. general commutative Noetherian rings) the theory becomes rather horrible. What benefit do we gain from an understanding of the prime ideals of a ring?

Thank you!

答:
An answer from algebraic geometry is that the set of prime ideals of a ring satisfy the axioms for a closed set in a topology. My personal elaboration on this answer is the following.

To interpret a ring $R$ "geometrically" is to interpret the ring $R$ as a ring of "functions" on a "space" $X$ such that the collections of points of $X$ on which "functions" of $R$ "vanish" form interesting "geometric" objects that we thinkof as living in the "space" $X$.

Making this precise amounts to the following:

The ring $R$ consists of functions on $X$ means that there are (surjective) evaluation maps $R\to R_x$ notated as $f\mapsto f(x)$, where $R_x$ is the set of values that the elements of $R$ can take when evaluated at the point $x\in X$. Note that for the sake of generality, we may have different sets of values possible at different points.

We say that A function $f\in R$ vanishes at the point $x\in X$ if its value $f(x)$ at that point is the same as the value $0(x)$ of the zero element of $R$ function (this is the same as requiring that vanishing is a point-wise condition, and that $0$ vanishes everywhere).

Further, we wish vanishing to satisfy nice properties: if a function $f$ vanishes at $x$, so does any multiple $f\cdot g$ of it, and two functions $f$ and $g$ have the same value at $x$ if their difference vanishes at $x$.

These three basic properties ensure that the evaluation maps $f\mapsto f(x)$ give each $R_x$ the structure of a quotient ring, and hence each evaluation map corresponds to an ideal of $R$. I like to call this ideal $\ker x$ because I like to think of the points on the space as the evaluation maps. Clearly, from the point of view of the ring $R$, points with the same kernels are indistinguishable, so we can in fact identify the "space" $X$ on which the ring $R$ is a ring of "functions" as simply a collection of ideals of $R$, which we designate as special point ideals.

What is interesting is that now vanishing can be easily interpreted as follows: $f$ vanishes at $x$ if and only if $f\in\ker x$ if and only if $\left<f\right>\subset\ker x$. Pushing this further, a collection of functions vanishes at a point $x$ if and only if $I=\left<f_i\right>\subset\ker x$, so a "geometric object" defined as where a collection of functions vanishes consists of the point ideals of $R$ contained in an ideal: geometric objects are specified by ideals of $R$!

Now, the geometric objects behave nicely: arbitrary intersections of geometric objects are geometric objects. Unfortunately, they are not nice enough without an extra assumption: if we want the union of two geometric objects (as collections of point ideals) to be a geometric object, then it is in fact necessary that every point ideal be prime. Once this assumption is made, we can talk about finite unions of geometric objects and arbitrary intersections of them, which allows us to talk about the geometric objects "locally": by investigating small manageable pieces of them, which we would then patch back together.

Title: Why do we study prime ideals?


问:
"Exactly one event of A_1,...,A_n occurs" using set operations?
I'm supposed to describe the situation using set theoretical operators $(\cup, \cap,\setminus,$ etc.). So we have $A_1,..,A_n$ events and only one of them occurs. I know that $A^c_n$ means $A_n$ does not occur. So I'm supposed to combine, let's say, $A_1$ with $n$ events that do not occur, correct? I'm not so sure about how to express that. I know for example, $(A_1 \cup ... \cup A_n)$ means at least one event occurs.

Does that mean the solution here is $(A_1 \cup A^c_2\cup ... \cup A^c_n )$?

答: Hint: you want $A_1 \cup \ldots \cup A_n$, but you do not want $(A_1 \cap A_2) \cup \ldots $


问:
Triangular Factorials
I came across a statement online and have been looking for a proof :

It states that 1, 6 and 120 are the only numbers which are both triangular and factorials.

Is there any way I can prove this? This claim looks too 'big' and I've tried to prove it but

I couldn't. Can anyone help me to prove this?

答:
I thought I give it a go to search for solutions of $x(x+1) = 2(n!)$ since it seemed it was checked "only" for numbers
up to $10^{77337}$. I guess this was n!, can't imagine that someone could calculate = $10^{77337}$!. After one night my
Java programm is at n=70000 and n! = $10^{309,000}$. And of course no further solution other than 1, 6 and 120.

Following the argumentation of user $zyx$ above, I try to estimate the probability that a solution will come up after
n=70000 (or whatever number n).
Let's write $2(n!) = 2^a 3^b 5^c 7^d ... p^z$ (prime factorisation). I do not know which $u$ I should take for the cutoff $n^u$
of my primes (user $zyx$ says "suitable $u<1$"), but if I go for all primes then the "probability" that a given 2(n!)
can be written as x(x+1) is:
$$P = \frac{2}{2^a} \cdot \frac{2}{3^b} \cdot \frac{2}{5^c} \cdot \; \; ... = \frac{2^{n_p}}{2^a 3^b 5^c 7^d ... p^z} = \frac{2^{(n_p-1)}}{n!}$$
where $n_p$ is the number of primes up to $n$.
If we approximate $n_p$ by $\frac{n}{ln(n)}$ we end up with a probability of $2^{(\frac{n}{ln(n)} -1)}$/$n!$

For n=70000, this expression is equal to about $10^{-307,000}$. And the sum of the probabilities from n=70001 to infinity is
even smaller. So with these odds I should better stop my Java program and save the energy for more useful calculations.
Could someone confirm my estimations?

Title: Triangular Factorials


答:
You can read $\cup$ like "or", $\cap$ like "and" and $\cdot^c$ like "not".

So what you want to say is "$A_1$ happens and none of $A_2,\dots, A_n$ happens or $A_2$ happens and none of $A_1, A_3,\dots, A_n$ happens or $\dots$ or $A_n$ happens and none of $A_1,\dots, A_{n-1}$ happens."

Translate this into the set-theoretical writing and you obtain $$(A_1\cap (A_2\cup \dots \cup A_n)^c) \cup (A_2\cap (A_1\cup A_3\cup \dots \cup A_n)^c \cup\dots \cup (A_n\cap (A_1\cup \dots \cup A_{n-1})^c).$$

You can of course rewrite this using the usual rules of set-theory.

Title: "Exactly one event of A_1,...,A_n occurs" using set operations?

Body:
I'm supposed to describe the situation using set theoretical operators $(\cup, \cap,\setminus,$ etc.). So we have $A_1,..,A_n$ events and only one of them occurs. I know that $A^c_n$ means $A_n$ does not occur. So I'm supposed to combine, let's say, $A_1$ with $n$ events that do not occur, correct? I'm not so sure about how to express that. I know for example, $(A_1 \cup ... \cup A_n)$ means at least one event occurs.

Does that mean the solution here is $(A_1 \cup A^c_2\cup ... \cup A^c_n )$?


问:
About definition of $f_n \rightarrow f$ in measure

Definition: We say that $f_n \rightarrow f$ in measure to mean that for any $\epsilon >0, \mu(\{x: |f_n(x)-f(x)|> \epsilon \}) \rightarrow 0$ as $n \rightarrow \infty$.

My question for this is why is $\mu(\{x: |f_n(x)-f(x)|> \epsilon \})$ "$>$" not $<$? Thanks in advance.

答:
The concept of convergence of functions is that for every $x$ of interest, $|f_n(x) -f(x)|$ should go to $0$ as $n$ increases. That is, for every $\epsilon > 0$, $|f_n(x) - f(x)| < \epsilon$ for sufficiently large $n$. And it makes no difference it we replace that with $|f_n(x) - f(x)| \le \epsilon$, since a slightly smaller $\epsilon$ will make the stricter condition true.

For given $n, \epsilon$, the set $F_n^\epsilon = \{x : |f_n(x) - f(x)| > \epsilon\}$ is the set of all points where the condition $|f_n(x) - f(x)| \le \epsilon$ fails to hold. If $f_n \to f$ pointwise, then $F_n^\epsilon$ must eventually get smaller as $n$ increases, since for each $x$, there is some $N$ such that for $n > N, x \notin F_n^\epsilon$.

This is where the concept of convergence in measure comes from. Instead of demanding that at every point $f_n$ eventually gets close to $f$, it instead demands that the set of points where $f_n$ fails to be close to $f$ should shrink to insignficance as $n$ increases. For measure, "insignificance" means "measure $0$".

If you wanted to talk about $C_n^\epsilon = \{x : |f_n(x) - f(x)| < \epsilon\}$, the set where $f_n$ is close to $f$, the corresponding choice would be that $C_n^\epsilon$ should grow as $n$ increases until it is practically everything. So you might think about requiring $\mu\left(C_n^\epsilon\right) \to \mu(D)$, where $D$ is the domain of $f$ and the $f_n$. But there are pitfalls with this concept. In particular, it breaks down when $\mu(D) = \infty$. As then you can have $\mu\left(C_n^\epsilon\right) \to \mu(D)$, but still have the set where they fail also be infinitely large in measure.

So instead we define convergence in measure as we do, as it contains what is essential - that $f_n(x) \to f(x)$ except on insignificant subsets - without those problems.


问:
show that a language is context-free by giving a pushdown automaton that accepts the language
Given the language $K$ : the words x$y where x and y have odd lengths on {a, b} and the median letter of x is equal to the median letter of y.

Example:

abb$b and

aaaaa$bab are in K

aba$abaab isn't in K

So i am kinda lost here, i am trying to figure out how to do this problem. I struggle to understand how will it recognize that the numbers are odd and how can it know that the median of both words are the same letter.

Could anyone help please, thanks a lot.

I started this drawing but i am so unsure on what to put on state 2 and 3 to make it work.
( I used : http://madebyevan.com/fsm/ if u wanna draw).

EDIT
I wrote a new solution this seem ok, if someone could confirm it would be nice.

答:
A CFG would be much easier , then it can be converted to a PDA :

The $ was replaced by #

S --> A # A | B # B

A --> ΣAΣ | a

B --> ΣBΣ | b

Σ --> a | b

Hopefully you can see that A generates all strings of odd length with a in middle and B generates all strings of odd length with b in middle ,

And so S generates the strings correctly


问:
Evaluating $\iiint_B (x^2+y^2+z^2)dV$ where $B$ is the ball of radius $5$ centered at the origin.
The question asks to use spherical coords. My answer is coming out wrong and symbolab is saying I'm evaluating the integrals correctly so my set up must be wrong.

Since $\rho$ is the distance from the origin to a point on it, and it's a sphere, I got $0 \le \rho \le 5$

Since it's a sphere I did $\theta$ from $0$ to $2\pi$. And then for $\phi$ I have from $0$ to $\pi$.

From an example problem, $x^2+y^2+z^2=\rho^2$

Thus

$$\int^\pi_0\int^{2\pi}_0\int^5_0 [( \rho^2) \rho^2 \sin(\phi)]\,d\rho \,d\theta \,d\phi$$

The answer is $\frac{312,500\pi}{7}$ and I'm getting $\frac{-1250}{\pi}$.

答:
Partition your ball $B$ into spherical shells of radius $\rho$ $(0\leq\rho\leq5)$ and thickness $d\rho$. The volume of such a shell is $dV=4\pi \rho^2\,d\rho$. It follows that
$$\int_B \rho^2\>dV=\int_0^5\rho^2\>4\pi\rho^2\>d\rho=2500\pi\ .$$


答:
So a is in the first quadrant

$b = \pi-a$

$c = a + 2\pi$

$d = 3\pi-a$

$\sin(\frac{b}{2}) = \sin(\frac{\pi}{2}-\frac{a}{2}) = \sin(\frac{\pi}{2})\cos(-\frac{a}{2})+\cos(\frac{\pi}{2})\sin(-\frac{a}{2}) = \cos(\frac{a}{2})$

$\sin(\frac{c}{2}) = \sin(\frac{a}{2}+\pi) = \sin(\frac{a}{2})\cos(\pi)+\cos(\frac{a}{2})\sin(\pi) = -\sin(\frac{a}{2})$

$\sin(\frac{d}{2}) = \sin(\frac{3\pi}{2}-\frac{a}{2}) = \sin(\frac{3\pi}{2})\cos(-\frac{a}{2})+\cos(\frac{3\pi}{2})\sin(-\frac{a}{2}) = -\cos(\frac{a}{2})$

So $\sin(\frac{d}{2}) + 2\sin(\frac{c}{2}) + 3\sin(\frac{b}{2}) + 4\sin(\frac{a}{2})$

$=2\sin(\frac{a}{2}) + 2\cos(\frac{a}{2})$

We know $\cos(a) = \sqrt{1-\frac{1}{16}} = \frac{\sqrt{15}}{4}$

Using the half-angle formla

$\sin(\frac{a}{2}) = \sqrt{0.5(1-\cos(a))}$

$\sin(\frac{a}{2}) = \sqrt{\frac{4-\sqrt{15}}{8}}$

similarly

$\cos(\frac{a}{2}) = \sqrt{\frac{4+\sqrt{15}}{8}}$

So

$2\sin(\frac{a}{2}) + 2\cos(\frac{a}{2}) = \sqrt{\frac{4-\sqrt{15}}{2}}+\sqrt{\frac{4+\sqrt{15}}{2}}$

ok. simplifying the right side by squaring (comes out to 5) and then taking the square root, I get $\sqrt{5}$.

Title: if $a,b,c,d$ be the first positive solutions of $\sin x=\frac{1}{4}$ then...

Body:
if $a,b,c,d$ be the first positive solutions of $\sin x=\frac{1}{4}$ then find $$\sin (d/2)+2\sin (c/2)+3\sin (b/2)+4\sin (a/2)$$.$(a<b<c<d)$

Attempts
$$\sin x=1/4$$,$$4\sin^2(x/2)(1-\sin^2(x/2))=\frac{1}{16}$$
Let $\sin^2 (x/2)=t$:it remains to solve $$t-t^2=\frac{1}{64}$$ .The rest is just brute force i.e we calculate  the 4 values $\sin(x/2)$ can take , make sure the angles are arranged in ascending order and substitute the roots.As far as i can see i a getting irrational roots.

My question is does anyone spot a trick, that can avoid all this work.!


问:
About the ideal of the product of rings
If $R_1, \cdots,R_n$ are rings with identities and $I$  is an ideal of $R_1\times\cdots\times R_n$, then $I=A_1\times\cdots\times A_n$ where each $A_i$ is an ideal of $R_i$ ($1\le i\le n$).

What I want to know is that how does the condition "$R_1,\cdots,R_n$ have identities" works in the conclusion? Is the condition necessary?

Is the conclusion necessarily true if the rings don't have identities?

答:
The assumption is needed (it could be relaxed, though). Let $R$ be a nontrivial (additive) abelian group and consider the trivial ring structure on $R$, where $rs=0$ for every $r,s\in R$; then every subgroup of $R\times R$ (which is a trivial ring as well) is an ideal; the subgroup $\{(x,x):x\in R\}$ is an ideal that's not the product of two ideals.

If we want to plunge more deeply in the question, it is sufficient to analyze the case of two rings. Suppose $I$ is an ideal of $R\times S$. Then $A=\{x\in R:(x,0)\in I\}$ is an ideal of $R$ and $B=\{y\in S:(0,y)\in I\{$ is an ideal of $S$. The verification is elementary.

It is also clear that $A\times B\subseteq I$, because $(x,y)=(x,0)+(0,y)$.

If $R$ has an identity, then $A\times\{0\}=I(1,0)=(1,0)I$; indeed, if $x\in A$, then $(x,0)\in I$ and $(x,0)=(x,0)(1,0)$. Similarly if $S$ has an identity. As a consequence, $I=A\times B$.

However, if the rings are not supposed to have an identity, the statement can be false and the counterexample above proves this.

A sufficient condition is that $I(R\times S)=I$.

Title: About the ideal of the product of rings


答: Suppose $I$ is an ideal of $R_1×\cdots ×R_n$ (assume commutative rings for convenience). Define $A_i:=\{a_i|(0,...,0,a_i,0,...,0)\in I\}$. Then $I=A_1×\cdots× A_n$. This is because if $(a_1,...,a_n)\in I$ then $(1,0,...,0)(a_1,...,a_n)=(a_1,0,...,0)\in I$ (this is where identity is required) so $a_1\in A_1$ and similarly $a_i\in A_i$ for each $i$. Conversely if $(a_1,..,a_n)\in A_1\times\cdots \times A_n$ then by definition of $A_i$'s $(0,...,0,a_i,0,...,0)\in I$, adding these we get $(a_1,...,a_n)\in I$. Now $I$ is a group with the additive structure hence each of the $A_i$ 's are closed under addition and similarly closed under taking inverses and contains the zero of each ring, $R_i$. Take an $A_i$ and consider the element $(0,...,0,a_i,0,...,0)\in I=A_1×\cdots ×A_n$ with $a_i\in A_i$. Then $(0,...,0,r_i,0,..,0)(0,...,0,a_i,0,...,0)=(0,...,0,r_ia_i,0,...,0)\in I$ hence $r_ia_i\in A_i$ for all $r_i\in R_i$, showing $A_i$ is an ideal.

Title: About the ideal of the product of rings

Body:
If $R_1, \cdots,R_n$ are rings with identities and $I$  is an ideal of $R_1\times\cdots\times R_n$, then $I=A_1\times\cdots\times A_n$ where each $A_i$ is an ideal of $R_i$ ($1\le i\le n$).

What I want to know is that how does the condition "$R_1,\cdots,R_n$ have identities" works in the conclusion? Is the condition necessary?

Is the conclusion necessarily true if the rings don't have identities?


问:
Calculate the volume of the region bounded by the coordinate planes and the surface $z=4-x^2-y^2$
Calculate the volume of the region bounded by the coordinate planes and the surface $z=4-x^2-y^2$

Then I think its $x=0, y=0$ and $z=0$ and the surface $z=4-x^2-y^2$

I think the integral is:

$\int_0^{2}\int_0^{\sqrt{{-x^2}+4}}4-x^2-y^2dydx$

is right?

答:
Your integral approach is correct. An easy way to do it is using polar coordinates. @Rezha Adrian Tanuharja said cylindrical coordinates, but you use them for triple integrals which is not the case (you can also solve this problem with $\displaystyle\iiint_E dV$).

Let $x=rcos(\theta)$, $y=rsin(\theta)$.

Your integration region is the circle in the first quadrant:

Then, we can see that $\theta$ should go from 0 (starting from point $C$) to B, and that angle is $90$ degrees. Then $0\leq\theta\leq \frac{\pi}{2}$. It is also easy to see that the radius of the cirlce goes from $0$ to $2$ (because $2^2=x^2+y^2$).

You should also remember that the jacobian when using polar coordinates is $r$, and we set up the following integral:

$\displaystyle\int_0^{{\pi/}{2}}\int_0^2(4-r^2)rdrd\theta=2\pi$ (this is an easy to solve integral, just a polynomial).


问:
Give a context-free grammar that generate this language
Language :

I would like to know if my approach is right or am i missing anything.

Thanks a lot :)

My context-free grammar:

$S \Rightarrow aSb $

$S \Rightarrow aaSb $

$S \Rightarrow aab $

$S \Rightarrow ab $

答:
One of the words in the language generated by your CFG, $aab$, is not in $K$. I'd do the following:
$$S\to aSC,aC,\varepsilon$$
$$C\to b,bb$$
Starting from $S$, the $S$ rules can be applied $n$ times where $n$ is any natural number (excluding zero), giving $a^nC^n$. The $C$ rules turn the $C$ symbols into $b$ terminals, the latter of which may eventually be any number $m$ from $n$ (if $C\to b$ is always applied) to $2n$ (if $C\to bb$ is always applied). Thus the description of words in $K$ is satisfied exactly.


问:
Calculating an expected value of a sum of random variables
I'm trying to see that I have understand the expected value definition correctly. However, I think that my intuition in the following exercise is wrong.

Definitions:

We define a set of $n$ jobs and $m$ machines, where each of the jobs is denoted by $W_j$ and takes $0 \leq W_j \leq 1$ seconds.
The assignment of the jobs is random.

We denote ${M_i}_j$ as the $j$ th job that was assigned, and ${M_i}_j$ = $W_j$. If $j$ wasn't assigned to the $M$ machine, ${M_i}_j$ = 0.

Also, we define ${M_i}$ = $\sum_{j=1}^n{M_i}_j$ (the sum of seconds that it will take the machine to run the jobs).

How do I calculate the expected value of $E[{M_i}]$?

My proposed solution:

Using the definition of expected value and ${M_i}$:

$E[{M_i}]={\frac 1m}\sum_{j=1}^n{M_i}_j$ $={\frac {M_i}m}$

I'm using ${\frac 1m}$ because the probability of each ${W_j}$ to be assigned to some machine ${M_i}$ is the same (since its random). I even did some tests using python to check my intuition, and it seems that it is ${\frac 1m}$.

答:
The expected Wj is (.5) and with any combination of machines, the total machine time is n(.5) and dividing by m machines, E[Mi] = (1/m)(n)(.5)
For Expected Total Run Time, you would break each n,m scenario into combinations, multiplying each way by the maximum jobs on any one machine, and dividing by total combinations, then  summing (left for others to formulize, good luck).</p>


问:
Closed Immersions are open in the strong topology $C_S^r(M,N)$
I am trying to show that closed immersions are open in the strong topology $C_S^r(M,N)$. I know that a proper map $f$ into a locally compact haussdorf space, which $N$ is , will be closed . So now I am just trying to see that if I have a closed immersion between manifolds then it will be a proper map. If I have this then I have that the set of closed Immersions is just the intersection of the open set of Immersions with the one of proper maps and so it is open.

So does anyone know why a closed immersion is a proper map ? I have tried seeing this using subsequences but I got nowhere, so any hints are aprecciated.

Thanks in advance.

答:
After thinking about this for a while I think I have come up with a solution:

So let's suppose that we have a closed immersion $f: M\rightarrow N$, that is not a proper map, where $M$ and $N$ are manifolds. So there exists a compact set $K\subset N$ such that $f^{-1}(K)$ is not compact,and so there exists a sequence $\{x_n\}_{n=1}^{\infty}$ that has no convergent subsequences. Now since $f$ is an immersion there will exists neighborhoods of $x_n$ such that $f|_{V_n}$  is injective. Since $K$ is compact there exists a subsequence $f(x_{n_k})\rightarrow y\in N$. Now using the fact $f|_{V_n}$ is injective we can choose $z_k$ such that $f(z_k)\neq y$, $d(z_k,x_{n_k})<\frac{1}{k}$ and $f(z_k)\rightarrow y$. Now since we choose $\{z_k\}$ in a way that $d(z_k,x_{n_k})<\frac{1}{k}$ we have that $B=\{z_k\}_{k=1}^{\infty}$ is a closed set, since it has no convergent subsequences. Also note that $y\in \bar f(B)-f(B)\neq \emptyset$ and so we obtain a contradiction with the fact that $f$ is a closed map, hence proving the dersired result.


问:
If $f$ is differentiable on $[x_1,x_2]$, $x_1x_2>0$, prove that $\frac{x_1f(x_2)-x_2f(x_1)}{x_1-x_2}=f(c)-cf'(c)$, where $c\in(x_1,x_2)$
So I thought of making up a function $\phi(x)$ (just as how the mean value theorem was proven) and by using rolle's theorem($\phi(x_1)=\phi(x_2)=0$), I could get the answer-- $f(c)-cf'(c)$ reminds me of....something related to $e^{-x}f(x)$? but is that how its done? At least, I couldn't devise the function; couldn't link the dots together... any help would be appreciated! Thanks!

答:
With
$$
 \phi(x) = x f\left(\frac{x_1 x_2}{x}\right)
$$
we have
$$
\frac{x_1 f(x_2) - x_2 f(x_1)}{x_1 -x_2} = \frac{\phi(x_1)-\phi(x_2)}{x_1 - x_2} \, .
$$
According to the mean-value theorem, this is equal to
$$
\phi'(d) = f\left(\frac{x_1 x_2}{d}\right)  - \frac{x_1 x_2}{d}f'\left(\frac{x_1 x_2}{d}\right)
$$
for some $d \in (x_1, x_2)$. Then $c = x_1x_2/d$ is also in the interval $(x_1, x_2)$ and satisfies
$$
\frac{x_1 f(x_2) - x_2 f(x_1)}{x_1 -x_2} = f(c) - cf'(c) \, .
$$


问:
Show that sequence approaches fixed point of a function
Problem
Let $f(x)$ be a differentiable function on $\Bbb R$ with $\left|\,f ' (x)\right| \leq r < 1$, where $r$ is constant.  Then consider the sequence $\{x_n\}$ such that $x_1 = 0$, $x_{n+1} = f(x_n)$, $n\geq1$.  Show that $x_n \to x^*$ as $n$ approaches infinity.  Moreover, $x^* = f(x^*)$.

Attempt
I tried showing that when $f ' (x) = r$, then eventually you end up with $x_{n+1} = \sum_{n=1}^{\infty}r^n$.  That sum converges by the $p$ test, so then $f(x)$ should converge to some number $x^*$.  Then for $\epsilon>0$ there exists $N$ such that $n\geq N$ such that
$$
\left|\,f(x^*)-x^*\right| \leq \left|x_{n+1} - x_n\right| \leq \epsilon.
$$
Then to show for $f ' (x) < r$ use direct comparison test to say the bigger series converges so the smaller one must converge as well.

I'm not sure if I can just do this or not, can someone let me know if I'm using the correct method or not?

答:
The trick is to show the sequence is Cauchy, i.e. given any $\varepsilon >0$ there is some $N$ s.t.

$$|x_n-x_m|<\varepsilon$$

whenever $n,m>N$. The trick here is that you know that $\sum_i r^i$ converges, so you end up with an upper bound which is the tail $\sum_{i=N}^\infty r^i$. This converges to zero as $N\to \infty$. This should be enough hints to let you fill in the details.


问:
The strength of a flow equals the total flow entering the sink?
The strength of the flow in a transportation network is defined as the sum of the flow of each edge leaving the source.

My textbook remarked without proof that this is the same as the sum of the flow of each edge entering the sink, but I'm not sure how to prove this formally.

Terminology:

A transportation network is a finite directed graph with 2 distinguished vertices the source and the sink.

The flow is a nonnegative real valued function on the edges of the graph satisfying the following conservation criterion:

for each vertex x that is not the source nor the sink, the sum of flow on edges entering x is equal to the sum of flow on edges leaving x.

答:
To deduce this equality, just add up the flow balance constraints for $i \in N \setminus \{s,t\}$:
$$\sum_{(i,j)\in A} x_{i,j} - \sum_{(j,i)\in A} x_{j,i} = 0$$

Almost everything cancels out in pairs, leaving:
$$\sum_{(s,j)\in A} x_{s,j} - \sum_{(j,t)\in A} x_{j,t} = 0$$


问:
Does the starting index affect the convergence/divergence of two equal-term series?
Say I have a convergent series $\sum_{m}^∞ a_i$. Does $\sum_{n}^∞ a_i$ converge when $m>n$? Similarly, does it converge when $m<n$?

答:
Convergence or divergence is unchanged, since the difference between the series is a finite sum of finite terms, so it is finite.

I am assuming that each $a_i$ is finite.


问:
Proof regarding families of sets and Cartesian products
I have to prove the following and I believe that the choose an element method is likely the best. I gave an attempt but it was lackluster, I don’t know that I fully understand families of sets. Any help at all in starting would be greatly appreciated.

\begin{equation*}
    \bigcup_{t\in (0,1)}\bigl( (t,1) \times (0,t^2) \bigr)
= \{\, (x,y) \in \mathbb{R}^2 \mid 0<x<1 \text{ and } 0<y<x^2 \,\}.
\end{equation*}

答:
Denote:

$$\begin{cases}
A & = \bigcup_{t\in (0,1)}\bigl( (t,1) \times (0,t^2) \bigr)\\
B &= \{\, (x,y) \in \mathbb{R}^2 \mid 0<x<1 \text{ and } 0<y<x^2 \,\}
\end{cases}$$

You have to prove that the two sets are equal. A usual way to do so is to prove that $A \subseteq B$ and $B \subseteq A$. Also notice that $A,B$ are subsets of $\mathbb R^2$. Notice that $A$ is a union of open rectangles.

Proof that $A \subseteq B$

Take $(x,y) \in A$. By definition, it exists $t \in (0,1)$ such that $(x,y) \in (t,1) \times (0,t^2)$. Note that the notations are unfortunately confusing... $(x,y)$ is here an ordered pair while $(t,1)$ is an interval. That's our poor life of mathematicians!

Anyhow, that means that $0 \lt t \lt x \lt 1$ and $0 \lt y \lt t^2 \lt x^2$. So $(x,y) \in B$: we've proven first inclusion. Let's prove the reverse one.

Proof that $B \subseteq A$

So take $(x,y) \in B$. We have to prove that $(x,y) \in A$, i.e. find a rectangle $(t,1) \times(0,t^2)$ in which $(x,y)$ lie. You'll verify that

$$(x,y) \in (t,1) \times (0,t^2)$$ where $ t = \frac{x + \sqrt y}{2}$. A drawing will help to understand why... and computations validate it.

We're done!


问:
Basic Partial Fractions
I feel super foolish asking this, but I've reached a mental block.

I'm trying to find the inverse laplace transform of:

$$\frac{s+3}{(s + 1)^2 (s-2)}$$

but, when I expand it into partial fractions:-

$$A(s+1)(s-2) + B(s-2) + C(s+1)^2$$

I get

$$B = -2/3$$
$$C = 5/9$$
$$A = -5/9$$

This is wrong, I've checked on wolfram alpha and it says the answer is

$$\\1/9{}{ e^{-t} (-6 t+5 e^{3t}-5)}$$

Please can you tell me where I'm going wrong?

答:
The partial fraction expansion is:

$$\displaystyle \frac{s+3}{(s-2) (s+1)^2} = -\frac{5}{9(s+1)}-\frac{2}{3(s+1)^2}+ \frac{5}{9(s-2)}$$

The inverse Laplace Transform can now be done for each of those terms. We get:

$$\displaystyle \frac{1}{9} e^{-t} (-6 t+5 e^{3 t}-5)$$

Note that this is a somewhat easier way to display it. Of course, if you want to line things up with the partial fraction fractions, this is identically (just multiply each term in the above expression):

$$\displaystyle -\frac{5}{9} e^{-t} -\frac{2}{3} t e^{-t} + \frac{5}{9} e^{2 t}$$

Do you see it now? I think you may have done everything okay as your constants look correct.

Title: Basic Partial Fractions


问:
proof using Brownian motion scaling property
I am trying to prove $A_t^+$ has the same distribution as $tA_1^+$ using the scaling property, where
$$
A_t^+ = \int_0^t\mathcal 1_{[0,\infty)}(W_s)\,ds
$$
I tried to scale the Brownian motion but it cannot lead to the right conclusion. I wonder which step is wrong. Here is what I attempted.
$$
\begin{aligned}
tA_1^+ & = t\int_0^1\mathcal 1_{[0,\infty)}(W_s)\,ds \\
& = \sqrt{t}\int_0^1\mathcal 1_{[0,\infty)}(\sqrt{t} W_{s})\,ds \\
& = \frac{1}{\sqrt{t}}\int_0^1\mathcal 1_{[0,\infty)}(W_{ts}) \,d(ts) \\
& = \frac{1}{\sqrt{t}}\int_0^t\mathcal 1_{[0,\infty)}(W_u) \,du \\
& = \frac{1}{\sqrt{t}}A_t^+
\end{aligned}
$$

答:
For $t>0$.
\begin{align}
A_t^+ &= \int_0^t\mathbb{1}({W_s \geq 0)} ds\\
&=^{u = s/t} t\int_0^1\mathbb{1}({W_{ut} \geq 0})du\\
&= t\int_0^1{\mathbb{1}\left(\frac{W_{ut}}{\sqrt{t}}\geq 0\right)}du\\
&=^{(d)} t\int_0^1{\mathbb{1}(\tilde{W}_{u} \geq 0)}du\\
\end{align}
In the last equality, we use the scaling property of the BM i.e. for $u\geq0$, $\tilde{W}_u = \frac{1}{\sqrt{t}}W_{ut}$ is again a BM.


问:
Martingale whose distribution is independent of n
Claim: A one-point stationary sequence $(X_n)_{n\in\mathbb{N}}$ of square integrable r.v's is a martingale w.r.t the natural filtration iff it is surely constant in time.

I am convinced that this is a false statement, and therefore I'd like to provide a counterexample. As a simple counterexample I wanted to take a sequence of $i.i.d. ~ Bernoulli(p)$.

Then $E[X_{n+1} | \mathcal{F}_n$] = $E[X_{n+1}] = E[X_n] = p$, independent of $n$, and therefore one-point stationary. On the other hand, $P(X_{n+1} = X_{n}) \neq 1$.

As I'm not very familiar yet with martingales, I am wondering if the sequence of Bernoulli i.i.d. r.v's is indeed a martingale, and if it suffices as a counterexample to disprove the claim.

EDIT: $E[X_{n+1} | \mathcal{F}_n] = E[X_n] = p \neq X_n$. Therefore, this sequence is not a martingale. What is an appropriate counterexample?

答:
You have
$$
\eqalign{
E[(X_{n+1}-X_n)^2]
&= E[X_{n+1}^2]+E[X_n^2]-2E[X_{n+1}X_n]\cr
&=2E[X_n^2]-2E[X_{n+1}X_n]\cr
&=2E[X_n^2]-2E[X_nX_n]=0,\cr
}
$$
the second equality because of 1-point stationarity and the third because $X_n$ is a martingale.


问:
Prove the inequality $x-4x \frac{x}{\sqrt{x}+x} +3x \left( \frac{x}{\sqrt{x}+x} \right)^2 -3 \left( \frac{x}{\sqrt{x}+x} \right)^2 <0$.
I want to prove $$x-4x \frac{x}{\sqrt{x}+x} +3x  \left( \frac{x}{\sqrt{x}+x} \right)^2 -3 \left( \frac{x}{\sqrt{x}+x} \right)^2 <0$$
for $x>0$ real number.
I tried supposing that  $$x-4x \frac{x}{\sqrt{x}+x} +3x  \left( \frac{x}{\sqrt{x}+x} \right)^2 -3 \left( \frac{x}{\sqrt{x}+x} \right)^2 \geq 0$$ holds to get a contradiction but I couldn't find the solution.
Can you help me proceed?

Body:
Answer :

$x-4x\frac{x}{\sqrt{x} +x} +3x(\frac{x}{\sqrt{x} +x})^2 - 3(\frac{x}{\sqrt{x} +x})^2 =1-4\frac{x}{\sqrt{x} +x} +3(\frac{x}{\sqrt{x} +x})^2 - 3\frac{x}{(\sqrt{x} +x)^2 }=  \frac{(\sqrt{x} +x)^2 - 4x(\sqrt{x} +x) +3x^2 - 3x}{(\sqrt{x} +x)^2} =\frac{x+2x\sqrt{x} +x^2 - 4x\sqrt{x} - 4x^2 +3x^2 - 3x}{(\sqrt{x} +x)^2} =\frac{-2x-2x\sqrt{x}}{(\sqrt{x} +x)^2}=\frac{-2\sqrt{x}}{\sqrt{x} +x} <0$

Finally :

$x-4x\frac{x}{\sqrt{x} +x} +3x(\frac{x}{\sqrt{x} +x})^2 - 3(\frac{x}{\sqrt{x} +x})^2 <0$


问:
Given a particular matrix $A \in \operatorname{GL}_n(F)$, what is an easy way to determine $A^n$ $\forall n \in \Bbb Z$?
$
\newcommand{\GL}{\operatorname{GL}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\m}[1]{\left( \begin{matrix} #1 \end{matrix} \right)}
\newcommand{\ds}{\displaystyle}
$
Original Problem: Consider the matrix $A \in \GL_3(\R)$ given by, for $x > 0$,

$$A = \m{-1 & x & -x \\ 1 & -1 & 0 \\ 1 & 0 & -1 }$$

What is the general form of $A^n$ for any $n \in \Z$?

My Work So Far: By doing some particular cases by hand and doing some pattern matching, I was able to observe that

$$A^n= (-1)^n \m{ 1 & -nx & nx \\
-n &\ds \frac{n(n-1)}{2} x + 1 &\ds  -\frac{n(n-1)}{2} x \\
-n & \ds \frac{n(n-1)}{2} x &\ds 1 - \frac{n(n-1)}{2} x }$$

Proving this through a sort of "bidirectional induction" (i.e. show it with $A$ as the base case for $n \in \Z_{>0}$, then with $A^{-1}$ as the base case for $\Z_{<0}$, and then $A^0 = I_3$ trivially) seems more than doable, if a little tedious.

However, I don't think that this is the way it was intended to be solved. This comes as a homework problem and was tied to some others I was given. Namely:

Find the characteristic polynomial $\mu_A$ of $A$

Deduce $A$ is not diagonalizable

Find vectors $U_i \in \R^3$ such that
$$AU_1 = -U_1 \qquad AU_2 = U_1 - U_2 \qquad AU_3 = U_1 + U_2 - U_3$$

Show $A$ is similar to the matrix
$$B = \m{ -1 & 1 & 1 \\ 0 & -1 & 1 \\ 0 & 0 & -1}$$
(I did so by showing $\mu_A = \mu_B$.)

I've done all of these, so I don't really need help there. It would not be unusual for these to tie into the problem somehow, based on how this professor's homework tends to go, but I'm not sure where the connection lies.

So I'm left wondering: based on this information, what is some more elegant way to calculate $A^n$ or prove its general form as given, if any? Be it for this specific matrix tied to the information given, or a more general case of $A \in \GL_n(F)$.

答:
First find the Jordan matrix of $A$. The eigenvalues of $A$ are found from the characteristic equation $$0=\det\begin{pmatrix}-1-\lambda & x & -x \\ 1 & -1-\lambda & 0 \\ 1 & 0 & -1-\lambda \end{pmatrix}=-(1+\lambda)^3$$ to be $\lambda=-1$ thrice. There is only one associated eigenvector $(0,1,1)$.

A generalized eigenvector is found from $(A+I)^2v=0$, $$\begin{pmatrix}0&0&0\\0&x&-x\\0&x&-x\end{pmatrix}v=0\Longleftarrow (A+I)v=\begin{pmatrix}0\\1\\1\end{pmatrix}\Longleftarrow v=\begin{pmatrix}1\\0\\0\end{pmatrix}$$ The third, a solution of $(A+I)^3v=0$, satisfies $(A+I)v=\begin{pmatrix}1\\0\\0\end{pmatrix}$, namely $\begin{pmatrix}0\\1/x\\0\end{pmatrix}$. So letting $P=\begin{pmatrix}0&1&0\\1&0&1/x\\1&0&0\end{pmatrix}$ gives $$A=PJP^{-1}=P\begin{pmatrix}-1&1&0\\0&-1&1\\0&0&-1\end{pmatrix}P^{-1}$$

Hence $A^n=PJ^nP^{-1}$ where $$J^n=(-1)^n\begin{pmatrix}1&(-1)^nn&\binom{n}{2}\\0&1&(-1)^nn\\0&0&1\end{pmatrix}$$ Multiplying out $PJ^nP^{-1}$ should give your answer.


问:
$k$-th derivative of an exponantial function
How do I derive this $k$-th derivative:
$\dfrac{d^k}{dt^k}e^{tx}=x^ke^{tx}?$

I believe that induction is to be used.

答:
Induction will work great, but I'll mention that you might be getting a bit confused or distracted by the variables they chose to use. I'm curious if you find

$$\dfrac{d^n}{dx^n}e^{ax}=a^n e^{ax}$$

a bit more obvious. Because it's the exact same thing as your question.


问:
Nearly, but not almost, continuous
Lusin's Theorem asserts that a measurable function f is nearly continuous in the sense that for all $\epsilon>0$ there is a set S of measure less than $\epsilon$ such that f is continuous on the complement of S. </p>

<p>I am looking for an example of such a function which is not continuous on the complement of any measure zero set. For instance, something like the characteristic function of the rationals won't do, since it is continuous on the complement of the rationals, which is a zero set. </p>

答: The characteristic function of a fat Cantor set gives an example (or any closed set with positive measure and empty interior).  If $C$ is such a set, and $A$ is any null set, then $\mathbb{R}\setminus A$ contains an element $x$ of $C$.  Every neighborhood of $x$ intersects $\mathbb{R}\setminus C$ in a set of positive measure (because nonempty open sets have positive measure), so for all $\delta>0$, $(x-\delta,x+\delta)\setminus A$ contains elements of $\mathbb R\setminus C$.  This implies that the restriction of $\chi_C$ to $\mathbb{R}\setminus A$ is discontinuous at $x$. </p>

Title: Nearly, but not almost, continuous


问:
What does it mean when a function has an accent over it? Eg, $\widehat d(x, y) = \min\bigl(d(x, y), 1\bigr)$
Take the following function:
$$\widehat d(x, y) = \min\bigl(d(x, y), 1\bigr)$$
What does that mean, is it just a function name? Or does it have a mathematical meaning?

答:
It's just an adjective to change the name slightly.
To write it in mathjax use $\hat d$:  $\hat d$.

It's often used for the Fourier transform, but not in this example.


问:
Derivative of an inverse map from $\mathbb{R}^2 \rightarrow \mathbb{R}^2$
Let $f$ be a map from $\mathbb{R}^2 \rightarrow \mathbb{R}^2$ be defined as:

$$ f: \left( \begin{matrix}x\\y\end{matrix} \right) \mapsto \left( \begin{matrix}x+y-xy\\x^2\end{matrix} \right)$$

How do you determine the set of points in $\mathbb{R^2}$ such that $f$ is invertible near those points, and compute the derivative of the inverse map. I believe this is done using inverse and/or implicit function theorem. I get the derivative:

$$Df(x,y) = \left( \begin{matrix}1-y&1-x\\2x&0\end{matrix} \right)$$

But according to the inverse function theorem since $f$ continuously differentiable we have:

$$Df^{-1}(x,y) = \left[ Df(f^{-1}(x, y)) \right]^{-1}$$

and I computed $f^{-1}(x, y) = \left( \begin{matrix}\pm \sqrt x\\ \frac{x \mp \sqrt y}{1 \mp \sqrt y}\end{matrix} \right)$ But this inverse mapping sends one point to two so I'm not sure if I'm on the right track.

答: Take $(a,b)\in\Bbb R^2$. Then $f$ is differentable at $(a,b)$ and the Jacobian of $Df_{(a,b)}$ is$$\begin{bmatrix}1-b&1-a\\2a&0\end{bmatrix}.\tag1$$If $a=0$ or $a=1$, then $(1)$ is not invertible. At all other points of $\Bbb R^2$, $(1)$ is invertible and its inverse is$$\begin{bmatrix}0&-\frac1{2a}\\\frac1{1-a}&\frac{1-b}{2a(a-1)}\end{bmatrix}.\tag2$$So, at those points $f$ is locally invertible, and the Jacobian of $Df^{-1}$ at $(a+b-ab,a^2)$ is $(2)$.


问:
How to prove that the remainder of $15^{2098}$ divided by $14$ is $1$?
I have to compute the remainder of $15^{2098}$ divided by $14$, I know that the answer will be $1$ because $15$ is $14+1$ and the remainder of every $15$'s power will be $1$ when divided by $14$, the problem is that I can't write that as answer because it is not too mathematical, so I want know how can I prove this without having to do all the long computation.

Body:
$$15\equiv 1 \pmod{14}$$

$$15^{2098}\equiv 1^{2098}\equiv 1 \pmod{14}$$


问:
Finding a basis for $V$ such that the matrix of $T:V\to V$ is strictly upper triangular

Let $T:V\to V$ be a linear map on a vector space $V$ of dimension $n$. Assume that there exists $k\in \mathbb{N}$, such that for each $i$, $0\leq i\leq k$, there exists a subspace $V_i$ of $V$ (denoted $V_i \leq V$) with the following properties:

$V_i \subset V_{i+1}$ for $0\leq i\leq k-1$

$V_i \neq V_{i+1}$ for $0\leq i\leq k-1$

$V_0 = \{0\}$ and $V_k = V$

$TV_i \subseteq V_{i-1}$ for $1\leq i\leq k$

Then there exists a basis for $V$ such that the matrix of $T$ with respect to this basis is strictly upper triangular (i.e. all non-zero entries lie above the diagonal only)

So, I'm in the search of a basis $\mathcal{B} = \{v_1,...,v_n\}$ which makes $[T]_{\mathcal{B}}$ strictly upper triangular. This would mean that $Tv_j = \text{span}(v_1,v_2,...,v_{j-1})$. Intuitively, it makes sense to think about considering the basis of $V_i$ and extending it to a basis for $V_{i+1}$ (until we reach $V_n$), but I'm not sure how to put this down formally. Moreover, it's a little bothering that we don't know what $k$ is. Is that a typographical error in the question? Possibly not.

Please help me with hints so that I can work further in the right direction!

答:
Hint

Take $\{e_1, e_2, \dots , e_{i_1}\}$ as a basis of $V_1$.

And by induction if $\{e_1, e_2, \dots, e_{i_1}, e_{i_1+1}, e_{i_1+2}, \dots, e_{i_{l-1}}\}$ is a basis of $V_{l-1}$, complete it into a basis $\{e_1, e_2, \dots, e_{i_1}, e_{i_1+1}, e_{i_1+2}, \dots, e_{i_{l-1}}, e_{i_{l-1}+1},e_{i_{l-1}+2},\dots, e_{i_l}\}$ of $V_l$.

At the last step and as $V_k = V$, you'll ensure that you get a basis of $V$.

Then look at $T(e_i)$ for $1 \le i \le n$ to prove that the matrix of $T$ in the built basis is strictly upper diagonal.


问:
$\det A = a_{1 , 1}A_{1 , 1} - a_{1 , 2}A_{1 , 2} + \cdots + (-1)^{n+1} a_{1 , n}A_{1 , n}$.
$A = (a_{ij})$ is a square matrix  of order $n$ . It's determinant will be $a_{1 , 1}A_{1 , 1} - a_{1 , 2}A_{1 , 2} + \cdots + (-1)^{n+1} a_{1 , n}A_{1 , n}$. Here $A_{i , j}$ is the co-factor of $a_{i , j}$.

The definition of determinant is following.

I was trying to prove it only from the definition of determinant . I was not allowed to use any other result than the definition of determinant.

But I am not being able to do that. Can anyone please help me to do this ?

答:
Hint: Consider a permutation $\sigma$ on $n$ elements. Iterate over the value of $\sigma (1).$

You will start something like
$$det(A)=\sum _{\sigma \in S_n}(-1)^{|inv(\sigma )|}\prod _{i=1}^{n}A_{i,\sigma (i)}=\sum _{k=1}^n(-1)^{k-1}\sum _{\substack{\tau \in S_{n-1}\\\sigma = \sigma(1)\tau}}(-1)^{|inv(\tau)|}A_{1,k}\prod _{i=2}^{n}A_{i,\sigma(i)}.$$
In the formula, $S_{n-1}$ is used instead of the permutations of $[n]\setminus \{\sigma (1)\}$ but they are in bijection.

It is important to notice that the sign of a permutation is $(-1)^{|inv(\sigma)|},$ where $inv(\sigma)=\{(i,j):i<j \text{ and }\sigma _i>\sigma _j\}.$ and if you fix $k=\sigma(1)$ then you already know that the inversion contains, at least, $k-1$ elements i.e., $(1,\sigma(1)),(2,\sigma(1))\cdots (\sigma(1)-1,\sigma(1)).$

Which part of the corresponds to the co-factor and how is the inversion set that defines the sign gets modified by fixing $\sigma(1)=k$?

As an example, take $\sigma = 31254,$ then $k=\sigma(1)=3$ and when you take out $3$ you end up with the sequence $1254$ that can be though as a permutation $\tau = 1243$ that respects the order of this. Show that $|inv(\sigma)|=k-1+|inv(\tau)|.$


答: Hint: Your attempt will work: just fix any basis $v_1,\dots,v_{d_1}$ of $V_1$, then extend it to a basis of $V_2$ by $v_{d_1+1},\dots,v_{d_2}$, and so on.. Here $d_i=\dim V_i$.

Title: Finding a basis for $V$ such that the matrix of $T:V\to V$ is strictly upper triangular

Body:
Let $T:V\to V$ be a linear map on a vector space $V$ of dimension $n$. Assume that there exists $k\in \mathbb{N}$, such that for each $i$, $0\leq i\leq k$, there exists a subspace $V_i$ of $V$ (denoted $V_i \leq V$) with the following properties:

$V_i \subset V_{i+1}$ for $0\leq i\leq k-1$

$V_i \neq V_{i+1}$ for $0\leq i\leq k-1$

$V_0 = \{0\}$ and $V_k = V$

$TV_i \subseteq V_{i-1}$ for $1\leq i\leq k$

Then there exists a basis for $V$ such that the matrix of $T$ with respect to this basis is strictly upper triangular (i.e. all non-zero entries lie above the diagonal only)

So, I'm in the search of a basis $\mathcal{B} = \{v_1,...,v_n\}$ which makes $[T]_{\mathcal{B}}$ strictly upper triangular. This would mean that $Tv_j = \text{span}(v_1,v_2,...,v_{j-1})$. Intuitively, it makes sense to think about considering the basis of $V_i$ and extending it to a basis for $V_{i+1}$ (until we reach $V_n$), but I'm not sure how to put this down formally. Moreover, it's a little bothering that we don't know what $k$ is. Is that a typographical error in the question? Possibly not.

Please help me with hints so that I can work further in the right direction!


问:
In $\Delta ABC , D$ and $E$ are midpoints of the sides $AB$ and $AC$ respectively, $CD$ and $BE$ intersect at $P$ with $\angle BPC = 90^\circ$.

In $\Delta ABC , D$ and $E$ are midpoints of the sides $AB$ and $AC$ respectively, $CD$ and $BE$ intersect at $P$ with $\angle BPC = 90^\circ$. Suppose $BD = 8\sqrt{5} , CE = 6\sqrt{5}$ . Find $BC$ .

What I Tried: Here is a picture :-

It was not long that I made some wonderful observations. For example, I see that $P$ is the median of this triangle. So probably I can assume $DP = x$ and $PE = y$ , and $PC = 2x$ and $PB = 2y$ , and they would work fine as we got a lot of right angles too?

Also from Midpoint Theorem, $DE$ is half of $BC$. So finding $DE$ alone, will be enough, but I have not worked out a way for that.

Another thing I thought of is let $[\Delta DBP] = M$ , $[\Delta PBC] = N$ , $[\Delta PCE] = O$ , $[DPEA] = K$ . Then I get :-
$$\rightarrow M + N = O + K$$
$$\rightarrow N + O = M + K$$

I think I have a lot of information, but I cannot combine all of these together (or maybe some of them), to get a good result. I need a bit of help with that.

Can anyone help me? Thank You.

答: Hint:By Pythagoras in $\Delta PDB,\Delta PEC$ $$x^2+4y^2=320$$,$$4x^2+y^2=180$$


问:
Could I define the supremum of a function, given that that function is not continuous?
Starting from a function $f(x)$ which is not continuous on $\mathbb{R}$, it that possible to "define" objects like $\sup\limits_x f(x)$, $x\in\mathbb{R}$?

答:
Yes of course. Let $f(x)=1$ if $x\in\Bbb Q$ and $f(x)=0$ elsewhere. Then $\sup_x f(x)=1$.

To be more precise, $Im(f)=\{f(x):\ x\in\Bbb R\}$ is a subset of $\Bbb R$ assuming that $f:\Bbb R\to\Bbb R$.

If $f$ is a bounded (from above) function (not necessarily continuous), then $Im(f)$ wolud be a bounded set. So there exists $\sup Im(f)\in\Bbb R$.

If $f$ is (upper) unbounded, then $Im(f)$ is unbounded and you can set that $\sup Im(f)=+\infty$.


问:
Confusion about a Measure Theory problem
I am trying to solve the following problem (1.76)

This problem is taken from Junghenn's Principles of Analysis book.

According to the mentioned book, we have the following.

If $\mu^*$ is an outer measure in $[0,1]$, then $\mathcal{M}(\mu^*)$ denote the set of all $\mu^*$-measurable sets.

Recall that $E\subseteq [0,1]$ is a $\mu^*$-measurable set if for all $C\subseteq [0,1]$ we have $\mu^*(C)=\mu^* (C \cap E)+ \mu^* (C \cap E^c).$

In the other hand, the function $\lambda$ just assigns the lenght to a left-open interval $(a,b];$ that is, $\lambda ((a,b])=b-a.$ Note that $\lambda$ is not an outer measure since it is not defined in all $\mathcal{P}(\mathbb{R}).$

So my first confusion is here: The set $\mathcal{M}(\mu^*)$ is defined if $\mu^*$ is an outer measure, right? But the problem states that $B\in \mathcal{M}(\lambda),$ but again, $\lambda$ is not an outer measure. So, what does it means $\mathcal{M}(\lambda)?$ Is this a possible errata?

My second confusion is here: Since $\lambda (B)=1,$ and $B\subseteq [0,1]$, then does it follows that $B=[0,1)?$ I conclude that since first of all, $\lambda$ is defined for the left open intervals of $\mathbb{R}$ according to the book mentioned. So $B$ must be a left open interval of lenght $1$ that is contained in $[0,1]$, so $B$ must be $(0,1]$, isn't it?

If you have the chance, it would be helpful to check the chapter 1 of the Junghenn's book Principles of Analysis to get a better backround about what am I talking about.

Thanks in advance!

答:
Regarding your second question, No.  $B$ can be $[0,1]$ minus any measure zero set.  For instance, $[0,1/2) \cup (1/2,1]$.  In fact, $B$ can be any (up to countable) union of intervals whose sum of lengths is $1$.

Think carefully about what exercise 1.71 is telling you: that you do not care whether the endpoints are included in any of these intervals.


问:
Interesting counter examples to "If a sequence tends to zero, must the series converge?"
We know that the sequence

$ 1, 1/2, 1/3, 1/4 \cdots $

converges to zero, yet most of us have seen that the "Harmonic series"

$ \sum_{n=1} \frac{1}{n}  $

is unbounded above and does not converge

I'm looking for more interesting counterexamples other than this one in general - so other sequences that tend to zero, yet their sum are unbounded.

答: any sequence that is larger than the harmonic series for $n$ large. For example every series of the form $\frac{1}{a_n}$ where $a_n<n$ for large $n$. Examples are $\frac{1}{\sqrt{n}}, \frac{1}{\ln(n)}$ etc.


问:
improper integral question - $\int_{1}^{\infty}\!e^{-x}\ln x\,dx$
I ran into this integral question:

does this integral converge:

$$\int_{1}^{\infty}\!e^{-x}\ln x\,dx$$
?

Thank you very much in advance,

Yaron

答:
Hints: Use integration by part by letting $u=\ln(x)$ and then use the exponential integral $E_1(x)$

$$ E_{1}(x) = \int_{x}^{\infty}\frac{e^{-t}}{t}dt. $$

The final answer should be

$$E_1(1).$$

Added: For convergence test, you can use the fact that $\ln(x)<x$ which implies

$$ \int_{1}^{\infty} \ln(x) e^{-x}dx < \int_{1}^{\infty} x e^{-x}dx < \infty. $$


答: If you define $\sup_xf(x)$ as $\sup\{f(x)\mid x\in\Bbb R\}$, then this has nothing to do with the continuity of $f$. If, for instance, $f(x)=x$, then $\sup_xf(x)$ doesn't exist. And if$$f(x)=\begin{cases}1&\text{ if }x\in\Bbb Q\\0&\text{ otherwise,}\end{cases}$$then $f$ is discontinuous everywhere, but $\sup_xf(x)=1$.

Title: Could I define the supremum of a function, given that that function is not continuous?

Body: Starting from a function $f(x)$ which is not continuous on $\mathbb{R}$, it that possible to "define" objects like $\sup\limits_x f(x)$, $x\in\mathbb{R}$?


问:
Finding the supremum and infimum of this set
The set:
$$\Big\{\frac{1}{n} + (-1)^{n} \mid n \in \mathbb{N}\Big\}$$
So I know that if I plotted this graph it would be full of discontinuities as n changes from odd to even but I'm not sure how I can find the supremum and infimum without using any limits (or calculus but I don't think that's important here anyways, correct me if I'm wrong). I think that the supremum is going to occur when $n$ is even and n is as small as possible. So in this case I think the supremum would be $\frac{3}{2}$, when $n$ is equal to $2$. I think the infimum would be $-1$ as when $n$ tends to infinity and is odd it will output $-1$.

I'd really appreciate if someone could tell me how to do this sort of question more rigorously or if I've made a mistake please point it out. I would greatly appreciate it. Thank you

答:
Consider the case where $n$ is odd. Then, $\frac{1}{n}+(-1)^{n} = \frac{1}{n}-1$. Thus, the infimum for the sequence of odd $n$ is $-1$ as $n\rightarrow \infty$, and the supremum is $0$ at $n=1$.

Now, consider the case where $n$ is even. Then, $\frac{1}{n}+(-1)^{n} = \frac{1}{n} + 1$. Thus, the infimum for the sequence of even $n$ is $1$ as $n\rightarrow\infty$, and the supremum is $\frac{3}{2}$ at $n=2$.

Thus, the infimum for the entire set is $-1$, and the supremum is $\frac{3}{2}$. $\blacksquare$


问:
Dieudonné's proof of Taylor's formula
I'm having trouble understanding a lemma which Dieudonné in "Foundations of Modern Analysis", chapter 8.14, pp. 185-186, uses to proof Taylor's formula, and then the proof of the formula itself with this lemma. First, for proving the lemma, Dieudonné uses the following theorem for the derivative of a continuous bilinear map:

Theorem 8.1.4: Let E, F, G be Banach spaces, and $ E \times F \rightarrow G: (x,y) \mapsto [x \cdot y] $ a continuous bilinear map. Then this mapping is differentiable at every point (x, y), and for $ (s, t) \in E \times F $ the derivative is the linear mapping

$ D[x \cdot y](s,t) = [x \cdot t] + [s \cdot y]. \qquad \qquad $(1)

The lemma is then:

Lemma 8.14.1: Let $ I \subset \mathbb{R} $ be an open interval, with two functions $ f \in C_E^p(I) $ and $ g \in C_F^p(I) $, and the continuous bilinear map $ E \times F \rightarrow G: (x,y) \mapsto [x \cdot y] $. Then $ [f \cdot g] $ belongs to $ C_G^p(I) $, and

$ [f \cdot D^pg] - (-1)^p [D^pf \cdot g] = D \big( [f \cdot D^{p-1}g] +(-1)^1 [Df \cdot D^{p-2}g] + \cdots + (-1)^{p-1} [D^{p-1}f \cdot g] \big) $.

And finally Taylor's formula:

Theorem 8.14.2: Let $ I \subset \mathbb{R} $ be an open interval, and a function $ f \in C_E^p(I) $. Then for any pair of points $ \alpha, \xi \in I $

$ f(\xi) = f(\alpha) + \frac{\xi - \alpha}{1!} f'(\alpha) + \frac{(\xi - \alpha)^2}{2!} f''(\alpha) + \cdots + \frac{(\xi - \alpha)^{p-1}}{(p-1)!} f^{(p-1)}(\alpha) + \int_{\alpha}^{\xi} \frac{(\xi - \zeta)^{p-1}}{(p-1)!} f^{(p)}(\zeta) \, d\zeta $.

Dieudonné says, tersely, that the lemma is immediately verified by applying theorem 8.1.4. And that the formula itself is proved by applying 8.14.1 to the bilinear mapping $ (\lambda,x) \mapsto \lambda x $ and to the function $ g(\zeta) = \frac{(\xi - \zeta)^{p-1}}{(p-1)!} $, and then by integrating both sides between $ \alpha $ and $ \xi $.

Thinking about this for quite some time and trying around didn't bring me any closer to understanding these proofs. I have no idea how to apply 8.1.4 to prove 8.14.1. If the arguments of the bilinear map are functions, then what are the arguments s and t? What would be the meaning of the terms on the rhs of an equality like

$ D[f \cdot D^{p-1}g](s,t)= [f \cdot t]+[s \cdot D^{p-1}g] $?

The only idea I have here is that (1) can turn into something like a product rule if the derivatives of the arguments x, y of the bilinear map are applied as (s, t):

$ D[f \cdot g](Df,Dg) = [f \cdot Dg] + [Df \cdot g]. \qquad \qquad $(2)

And I don't understand how to apply 8.14.1 to Taylor's formula, either.

Can anyone help me with this? Thanks a lot in advance.

答:
Well, the original idea has turned out to be valid, at least for proving the lemma. The decisive point is that $ D[f \cdot g] $ always has to be understood as $ D[f \cdot g](Df,Dg) $; it is a shortcut which Dieudonné uses. This is because we have to apply the chain rule. This becomes clearer if we use the notation $ \lambda(x,y) $ for the continuous bilinear map as Serge Lang does. Then we see immediately that

$ D \big( \lambda(f,g) \big) = \lambda'(f,g) \circ (Df,Dg) $.

With this we can write lemma 8.14.1 as

$
 \begin{align*}
  &[f \cdot D^pg] - (-1)^p [D^pf \cdot g] = D \big( [f \cdot D^{p-1}g] +(-1)^1 [Df \cdot D^{p-2}g] + \cdots + (-1)^{p-1} [D^{p-1}f \cdot g] \big) \\
  &= \sum_{i=0}^{p-1} (-1)^i D[D^if \cdot D^{p-1-i}g] = \sum_{i=0}^{p-1} (-1)^i D[D^if \cdot D^{p-1-i}g] (D^{i+1}f, D^{p-i}g) \\
&= \sum_{i=0}^{p-1} (-1)^i [D^if \cdot D^{p-i}g] + [D^{i+1}f \cdot D^{p-1-i}g].
 \end{align*} (3)
$

The proof of the lemma can then be done by induction over p. The beginning is done by (2). The induction step is achieved by differentiating the induction premise. Later in the process, a substitution $ Df \rightarrow f $ has to be made, i.e., the induction premise is applied again (here in the original, non-differentiated shape), this time to the bilinear map $ [Df \cdot g] $.

For anyone interested, here are the details:

The induction claim, under the assumption that $ f, g \in C^{p+1} $, is

$
 \begin{align*}
  &[f \cdot D^{p+1}g] - (-1)^{p+1} [D^{p+1}f \cdot g] = \sum_{i=0}^p (-1)^i D[D^if \cdot D^{p-i}g] (D^{i+1}f, D^{p+1-i}g) \\
  &= \sum_{i=0}^p (-1)^i [D^if \cdot D^{p+1-i}g] + [D^{i+1}f \cdot D^{p-i}g].
 \end{align*} (4)
$

We differentiate (3)

$
 \begin{align*}
  &[f \cdot D^{p+1}g] + [Df \cdot D^pg] - (-1)^p \big( [D^pf \cdot Dg]+[D^{p+1}f \cdot g] \big) \\
  &= \sum_{i=0}^{p-1} (-1)^i \big( [D^if \cdot D^{p+1-i}g] + [D^{i+1}f \cdot
  D^{p-i}g] + [D^{i+1}f \cdot
  D^{p-i}g] + [D^{i+2}f \cdot D^{p-1-i}g] \big).
 \end{align*}
$

Rewriting the result gives

$
 \begin{align*}
  &[f \cdot D^{p+1}g] + (-1)^{p+1} [D^{p+1}f \cdot g] + [Df \cdot D^pg] - (-1)^p [D^pf \cdot Dg] \\
  &= \sum_{i=0}^{p-1} (-1)^i \big( [D^if \cdot D^{p+1-i}g] + [D^{i+1}f \cdot D^{p-i}g]  \big)+ \sum_{i=0}^{p-1} (-1)^i \big( [D^{i+1}f \cdot D^{p-i}g] + [D^{i+2}f \cdot D^{p-1-i}g] \big) \\
  &= \underbrace{\sum_{i=0}^p (-1)^i \big( [D^if \cdot D^{p+1-i}g] + [D^{i+1}f \cdot D^{p-i}g]  \big)}_{(*)} -(-1)^p [D^pf \cdot Dg] + (-1)^{p+1} [D^{p+1}f \cdot g] \\
  &+ \sum_{i=0}^{p-1} (-1)^i \big( [D^{i+1}f \cdot D^{p-i}g] + [D^{i+2}f \cdot D^{p-1-i}g] \big).
 \end{align*}
$

In addition to some further clean-up we substitute (*) by the lhs of (4) in order to see, whether we get a valid equation

$
 \begin{align*}
  &[f \cdot D^{p+1}g] + [Df \cdot D^pg] \\
  &=! \quad [f \cdot D^{p+1}g] - (-1)^{p+1} [D^{p+1}f \cdot g] + \sum_{i=0}^{p-1} (-1)^i \big( [D^{i+1}f \cdot D^{p-i}g] + [D^{i+2}f \cdot D^{p-1-i}g] \big).
 \end{align*}
$

Further rearrangements deliver

$
 \begin{align*}
  [Df \cdot D^pg] - (-1)^p [D^{p+1}f \cdot g] \quad =! \quad \sum_{i=0}^{p-1} (-1)^i D[D^{i+1}f \cdot D^{p-1-i}g].
 \end{align*}
$

Now we substitute $ Df \rightarrow f $ and get again (3)

$
 \begin{align*}
  [(Df) \cdot D^pg] - (-1)^p [D^p(Df) \cdot g] = \sum_{i=0}^{p-1} (-1)^i D[D^i(Df) \cdot D^{p-1-i}g].
 \end{align*}
$

The only thing I cannot confirm ist that one verifies all this immediately.

Concerning the proof of Taylor's formula itself, this is explained in @blargoner's excellent notes. Thanks a lot!

Title: Dieudonné's proof of Taylor's formula


答: A close relative to the harmonic series, but a "strictly stronger" result, is $$\sum_{p\text{ prime}}\frac1p=\infty$$

Title: Interesting counter examples to "If a sequence tends to zero, must the series converge?"

Body:
We know that the sequence

$ 1, 1/2, 1/3, 1/4 \cdots $

converges to zero, yet most of us have seen that the "Harmonic series"

$ \sum_{n=1} \frac{1}{n}  $

is unbounded above and does not converge

I'm looking for more interesting counterexamples other than this one in general - so other sequences that tend to zero, yet their sum are unbounded.


答: If you have enough information about congruences, then use the answer by Lion Heart. Otherwise, notice that $(1+14)^{2098}$, when multiplied out, gives 1 followed by a lot of further terms all of which have $14$ as a factor.

Title: How to prove that the remainder of $15^{2098}$ divided by $14$ is $1$?

Body: I have to compute the remainder of $15^{2098}$ divided by $14$, I know that the answer will be $1$ because $15$ is $14+1$ and the remainder of every $15$'s power will be $1$ when divided by $14$, the problem is that I can't write that as answer because it is not too mathematical, so I want know how can I prove this without having to do all the long computation.


答:
Using integration by parts, we let:

$$u = \ln x; du = \frac{1}{x}$$
$$dv = e^{-x}dx; v = -e^{-x}$$

Then the integral becomes:

$$\int_{1}^{\infty}e^{-x}\ln x dx = -e^{-x}\ln x|^{\infty}_{1} + \int_{1}^{\infty}\frac{e^{-x}}{x}dx$$

The integral $$\int_{1}^{\infty}\frac{e^{-x}}{x}dx$$

is special, and is denoted the exponential integral (Ei$(-1)$).

Via wolfram, we have that $$\lim_{A\rightarrow -\infty} \text{Ei}(A) = 0$$
Furthermore, have $$\text{Ei}(-1) = .2193$$

to four decimal places.

Hence, our integral is:

$$\int_{1}^{\infty}e^{-x}\ln x dx = -e^{-x}\ln x|^{\infty}_{1} + \int_{1}^{\infty}\frac{e^{-x}}{x}dx = (0 - 0) + (\lim_{A\rightarrow -\infty} \text{Ei}(A) - \text{Ei}(-1)) = -.2193$$

Title: improper integral question - $\int_{1}^{\infty}\!e^{-x}\ln x\,dx$

Body:
I ran into this integral question:

does this integral converge:

$$\int_{1}^{\infty}\!e^{-x}\ln x\,dx$$
?

Thank you very much in advance,

Yaron


问:
How can I show that the rank of a matrix is not altered if a column is multiplied by a non-zero scalar?
How can I show that the rank of a matrix is not altered if a column is multiplied by a non-zero scalar? I don't know where to start.</p>

答: The dimension of the space spanned by the row vectors is the same as that spanned by its column vectors.  Let $C = \{c_1, \dots, c_n\}$ be the collection of $n$ column vectors.  Its span is unchanged if we multiply one of the $c_i$ by a nonzero scalar.  So whether you define rank using rows or columns, you have not changed the rank.


答:
If $A \subseteq \mathbb{N}$ is any set of positive upper density, which is to say that
$$
\limsup_{n \to \infty} \frac{|A \cap \{1, \dots, n\}|}{n} > 0,
$$
then $\sum_{a \in A} 1/a = \infty$.
This gives many examples of series which answer your question.
However, it is not a biconditional, as exemplified by e.g. the set of prime numbers as Arthur mentioned.
(There is probably some weaker asymptotic than positive upper density which guarantees divergence of the sum of reciprocals that would have the sum of reciprocals of primes as a special case, but I'm not aware of any.)

However, the set of integers lacking a $9$ in the decimal expansion (and any similar construction) does have convergent reciprocal sum — this is the Kempner series.


问:
Joint entropy of cards color
If we have 52 cards, 26 red and 26 black, and we take out one card at a time (and don't return it).
Xi is the color of the card i.
What is H(X1,X2,X3...X52)?

I tried with H(X1,X2,X3) or I(X1,X2,X3) to see if any parameters gets canceled by one another, but couldn't find any combinations that's "easy to calculate".

答: A simple argument should allow you to assert that $(X_1,X_2,\dots,X_{52})$ is unifom on $\binom{52}{26}$ issues, therefore its entropy is easy to compute.

Title: Joint entropy of cards color


问:
Prove that if $x\in \mathbb{R}$ and $x\in[m,M]$, then $x$ can be written as $m+\mu (M-m)$ where $\mu\in [0,1]$.
Prove that if $x\in \mathbb{R}$ and $x\in[m,M]$, then $x$ can be written as $m+\mu (M-m)$ where $\mu\in [0,1]$.

This is obviously a very intuitive result but I struggle to write a formal proof for it. I feel like I need to use completeness of $\mathbb{R}$ here or the continuity of linear function. Graphically, I suppose you can construct an algorithm to locate $\mu$ by keep replacing the right and left interval.

It would be great if I could see as many alternative proofs as I can in this thread.

答:
$m\le x\le M\Rightarrow x=(1-\mu)m+\mu M$ for some $\mu\in [0,1].$ Then, $x=m+\mu(M-m).$

Remark: the first inplication follows by considering the function $\mu\mapsto (1-\mu)m+\mu M$ and using the intermediate value theorem for continuous real-valued functions on intervals.


问:
Evaluate Integral using Laplace transform
I want to find the integral of the following function using Laplace Transform
$$
\int_0^{\infty}\frac{\cos(6t)-\cos(4t)}{t}\,dt
$$

答:
Hint : write it as $$\int _0^{\infty} \frac {cos 6t- cos 4t}{t} e ^{-st} dt $$

Where s= 0

Now it become Laplace transformation

$$L(\frac {cos 6t- cos 4t}{t}) $$

We have  property $$L ( \frac{ f(t) }{t}) = \int_s^{\infty}F(s) ds $$
Here $F(s) = L(f(t)) $

Using above property

$L(\frac {cos 6t- cos 4t}{t}) = \int_s^{\infty} \left(\frac{s}{s^2+6^2}-\frac{s}{s^2+4^2} \right) ds$

Integrate you will get
$$=\frac{1}{2}\ln(\frac{s^2+4^2}{s^2+6^2}) $$

Now put s=0

= $$=\frac{1}{2}\ln(\frac{4^2}{6^2}) $$

$$= \ln ( 2/3) $$


问:
Prove the inequality $\|x\|_p \le N^{1 - 1/q}\|x\|_q$.
In my functional analysis course, we define
$$\|x\|_p = \left(\sum_{k = 1}^N |x_i|^p\right)^{1/p}$$
for $x \in \mathbb{R}^N$ and $p \in [1, +\infty)$. Then the teacher told us that, for $p \le q$, it was obvious that
$$\|x\|_q \le \|x\|_p \le N^{1 - 1/q}\|x\|_q.$$
The first inequality is not too hard, because
$$\left|\frac{x_j}{\|x\|_q}\right|^q \le \left|\frac{x_j}{\|x\|_q}\right|^p \quad \forall j \in \{1, \ldots, N\},$$
and then, after summation $1 \le (\|x\|_p/\|x\|_q)^p$. But I don't really see how can I prove the second inequality. Any idea?

答:
Observe that
$$
\|x\|_p^q = N^{q/p}\left(\frac 1N\sum_{k = 1}^N |x_i|^p\right)^{q/p}.
$$
Then use Jensen's inequality $\varphi\left(\sum_{k=1}^N\alpha_kt_k\right)\leqslant \sum_{k=1}^N\alpha_k\varphi\left(t_k\right)$ where $\alpha_k\geqslant 0$ and $\sum_{k=1}^N\alpha_k=1$, here with $\alpha_k=1/N$ and $\varphi(t)=t^{q/p}$.


问:
Integral of a two dimensional function
A function is defined by
$$f(x,y) = \int_{x^2}^{xy} e^{t^2} \,dt$$
We are to decide the partial derivatives of the function.

I am quite unsure whether this is a trick question because we could do an approximation of the integrand with the Taylor series, but I can't see how that would simplify the work.

Could I just get a direction to as how I should approach this problem? I would very much not like to just get the solution.

Note that this is one of the easiest questions on the paper.

答:
So basically the fundamental theorem of calculus gives us
$$f(x,y) = \int_{x^2}^{xy} e^{t^2} \,dt = \int_{0}^{xy} e^{t^2} \,dt - \int_{0}^{x^2} e^{t^2} \,dt$$
$$\implies$$
$$\dfrac{\partial f}{\partial x} = \dfrac{\partial}{\partial x} (F(xy) - F(x^2)) =yF'(xy) - 2xF'(x^2) = ye^{x^2 y^2}-2xe^{x^4}$$

$$\dfrac{\partial f}{\partial y} = xe^{x^2y^2}$$

Thanks for the help, I knew this was supposed to be easy. Just looking at it from a different perspective!

Title: Integral of a two dimensional function


问:
There exist infinitely many prime numbers $p≥2$ and there exist positive integers $k,m$ such that $k^2+1=p+m$
I am asking about the validity of the following conjecture:

Conjecture: There exist infinitely many prime numbers $p≥2$ and there exist positive integers $k,m$ such that $k^2+1=p+m$ with $1≤m≤p-1$.

However, I am not able to prove or disprove it.

答:
Let's note (usual notation) by $\pi(n)$ - prime counting function and $p_n$-$n$'th prime number. We have
$$p_{\pi (k)} \leq k < p_{\pi (k)+1}$$
or
$$p_{\pi (k^2)} < k^2 < p_{\pi (k^2)+1}$$
We can add $m-1\geq2$ such that
$$p_{\pi (k^2)} <p_{\pi (k^2)} +m-1 =k^2 < p_{\pi (k^2)+1}$$
It has to be $m-1\geq2$, because $m-1=1$ would lead to $p_{\pi (k^2)}=k^2-1=(k-1)(k+1)$ (not a prime for large $k$'s).
Finally, from the strict version of the Bertrand's postulate
$$2\leq m-1 < p_{\pi (k^2)+1}-p_{\pi (k^2)}\leq p_{\pi (k^2)}-2$$

Summary: we have found a subsequence of prime numbers $\left(p_{\pi (k^2)}\right)_{k\in\mathbb{N}}$, $k>1$ satisfying:
$$k^2 +1=p_{\pi (k^2)} +m$$
with $3\leq m\leq p_{\pi (k^2)}-1$. It yields infinite different primes because:

Legendre's conjecture or

from Bertrand's postulate $$\pi((2\cdot k)^2)=\pi(4\cdot k^2)>\pi(2\cdot k^2)>\pi(k^2) \Rightarrow p_{\pi((2\cdot k)^2)} > p_{\pi(k^2)}$$

whichever version you prefer.


问:
if $a,b,c,d$ be the first positive solutions of $\sin x=\frac{1}{4}$ then...

if $a,b,c,d$ be the first positive solutions of $\sin x=\frac{1}{4}$ then find $$\sin (d/2)+2\sin (c/2)+3\sin (b/2)+4\sin (a/2)$$.$(a<b<c<d)$

Attempts
$$\sin x=1/4$$,$$4\sin^2(x/2)(1-\sin^2(x/2))=\frac{1}{16}$$
Let $\sin^2 (x/2)=t$:it remains to solve $$t-t^2=\frac{1}{64}$$ .The rest is just brute force i.e we calculate  the 4 values $\sin(x/2)$ can take , make sure the angles are arranged in ascending order and substitute the roots.As far as i can see i a getting irrational roots.

My question is does anyone spot a trick, that can avoid all this work.!

答:
The problem can be simplified somewhat by looking at the geometry of the exercise with respect to the unit circle. The four half angles are distributed as in the diagram:

We see that $\sin(a/2)+sin(c/2)=0$ and $\sin(b/2)+\sin(d/2)=0$ which results in

$$\sin (d/2)+2\sin (c/2)+3\sin (b/2)+4\sin (a/2)=2\sin(a/2)+2\sin(b/2)$$

with much less hassle.


问:
how do I prove that in a jump discontinuity, the size of the jump is the oscillation
for $f:A \rightarrow \mathbb{R}$ and $x\in A$, the oscillation of $f$ at $x$ is $$\operatorname{Osc}_x(f):=\lim_{r \to 0^+} \text{diam}(f([x-r, x+r]\cap A))$$
where, diam$(S) := \sup \{|x-y|: x,y \in S\}$

Now, given that $f$ is an increasing function that has a jump discontinuity at $x$,  how do I prove that $\operatorname{Osc}_x(f) = f(x^+) - f(x^-)$?

what I have tried is this:

since f is a jump discontinuity, hence $f(x^-) \neq f(x^+)$

since f is increasing, hence $f(x^-) < f(x^+)$ and $f(x) \in [f(x^-) , f(x^+)]$

now drawing a picture makes it obvious that  $\operatorname{Osc}_x(f) = f(x^+) - f(x^-)$, but how do I rigorously prove that?

EDIT:-
A part of @EuYu 's answer wasn't quite clear to me, so I have added a clarification for that part. Could someone verify if this clarification is correct or not?

set $\epsilon_0>0$ such that $f(x_0^-) + \epsilon_0 < f(x_0^+) - \epsilon_0 $, this can be done since, $f(x_0^-) < f(x_0^+)$

Now, for any $\epsilon \in (0,\epsilon_0)$, there exists some $\delta > 0 $ such that
$$|f(x) - f(x_0^-)| < \epsilon,\qquad \text{and}\qquad |f(x)-f(x_0^+)|<\epsilon$$
for all $x \in (x_0-\delta,x_0)$ and $x \in (x_0, x_0 + \delta)$, respectively.

This implies that
$$\quad f(x_0^-) - \epsilon < f(x) < f(x_0^-) + \epsilon,\quad \forall x\in (x_0-\delta,x_0)$$ and
$$ f(x_0^+) - \epsilon <f(x) < f(x_0^+) + \epsilon,\quad \forall x\in (x_0,x_0+\delta)$$
and
$$ f(x_0^-) + \epsilon < f(x_0^+) - \epsilon $$

This implies that, $\forall \epsilon \in (0,\epsilon_0)$, there exists some $\delta > 0 $ such that
$$ f(x_0^-) - \epsilon < \sup_{x\in (x_0-\delta,x_0)}f(x) \le f(x_0^-) + \epsilon < f(x_0^+) - \epsilon < \sup_{x\in (x_0,x_0+\delta)}f(x) \leq f(x_0^+) + \epsilon,$$
This implies that $\forall \epsilon \in (0,\epsilon_0)$ , there exists some $\delta > 0 $ such that
$$ f(x_0^+) - \epsilon < \sup_{x\in (x_0-\delta,x_0+\delta)}f(x) \le f(x_0^+) + \epsilon $$
This implies that $\forall \epsilon>0$ , there exists some $\delta > 0 $ such that
$$ f(x_0^+) - \epsilon < \sup_{x\in (x_0-\delta,x_0+\delta)}f(x) \le f(x_0^+) + \epsilon $$
similarly $\forall \epsilon>0$, there exists some $\delta > 0 $ such that
$$ f(x_0^-) - \epsilon < \inf_{x\in(x_0-\delta,x_0+\delta)}f(x) \le f(x_0^-) + \epsilon$$

答:
Let me denote the oscillation of the function $f$ on a set $S$ by
$$\omega_f(S) = \sup_{x\in S}f(x) - \inf_{x\in S}f(x).$$
Suppose that $f$ has a jump discontinuity at $x_0$. You don't need the function to be increasing, just for the value of $f(x_0)$ to be between its one-sided limits $f(x_0^-)$ and $f(x_0^+)$. Without loss of generality, let's assume that $f(x_0^-) \le f(x_0) \le f(x_0^+)$.

Since the left and right limits exist by assumption, for any $\epsilon > 0$, there exists some $\delta > 0 $ such that
$$|f(x) - f(x_0^-)| < \epsilon,\qquad \text{and}\qquad |f(x)-f(x_0^+)|<\epsilon$$
for all $x \in (x_0-\delta,x_0)$ and $x \in (x_0, x_0 + \delta)$, respectively. This implies that
$$f(x_0^+) - \epsilon \le \sup_{x\in (x_0-\delta,x_0+\delta)}f(x) \le f(x_0^+) + \epsilon,$$and$$\quad f(x_0^-) - \epsilon \le \inf_{x\in(x_0-\delta,x_0+\delta)}f(x) \le f(x_0^-) + \epsilon.$$
Therefore the oscillation of $f$ on $(x_0-\delta,x_0+\delta)$ is bounded by
$$f(x_0^+)-f(x_0^-) - 2\epsilon \le \omega_f((x_0-\delta,x_0+\delta)) \le f(x_0^+) - f(x_0^-) + 2\epsilon.$$
Since this holds for arbitrary $\epsilon > 0$, it follows that we must have
$$\omega_f(x_0) = f(x_0^+) - f(x_0^-).$$


答:
Here are some counterexamples I can think of. The rough rule of thumb is that any series $\sum_{n = 1}^\infty a_n$ where $a_n$ is bigger than $n^{-(1 + \epsilon)}$ for every $\epsilon > 0$ could potentially diverge. Of course, you have to be careful and you should use the various convergence/divergence tests to test your hypothesis. Here are some examples:

The boring examples: $a_n = 1/\sqrt n$, $a_n = 1/ n^{1/3}$, $a_n = 1 / \log n$, $a_n = 1 / \log \log n$, etc.

Some examples whose terms are smaller than the respective terms in the harmonic sequence: $a_n = 1/(n (\log n)^k)$ for any positive number $k$, $a_n = 1 / (n \log ^{(k)}n)$, where $\log^{(k)}$ denotes the iterated logarithm (i.e. apply it $k$ times).

$a_n = (n\text{th prime number})^{-1}$. This series converges because we know from the prime number theorem that $a_n \sim n \log n$.

Let $\chi(n)$ denote the expected chromatic number of the random graph $G(n, 1/2)$ (in a nutshell, this is a graph on $n$ vertices where each edge is included with probability $1/2$). Then choosing $a_n = \chi(n)^{-1}$ gives a divergent series, because we know that $\chi(n) \sim n / \log n$.

To answer your last question, actually this would yield a convergent series. As jakehuryn mentioned, this is the Kempner series. As Wikipedia writes:

Heuristically, this series converges because most large integers contain every digit. For example, a random $100$-digit integer is very likely to contain at least one $9$, causing it to be excluded from the above sum.


答:
Following from the comment by Mike Daas that you are asking whether a number $k^2$ falls among $p,\dots 2p-2$: For any suitably large square number $k^2$, we can always pick the largest prime number $p_k<k^2$.

Assume that $k^2>2p_k-2$. Bertrand's postulate assures us that for suitably large $p$, there is at least one prime number in any interval $(p,2p-2)$. So if $k^2$ is not in the interval $(p_k,2p_k-2)$, then there is another  prime number in that interval that is smaller than $k^2$, and $p_k$ is not the largest prime smaller than $k^2$. In other words, every suitably large $k^2$ falls in an interval $(p,2p-2)$.

Body:
I am asking about the validity of the following conjecture:

Conjecture: There exist infinitely many prime numbers $p≥2$ and there exist positive integers $k,m$ such that $k^2+1=p+m$ with $1≤m≤p-1$.

However, I am not able to prove or disprove it.