问:
Spherical coordinate integration help -- spheres.

$$L = \iiint_{D} \frac{1}{(x^2+y^2+z^2)^\frac{3}{2}}dxdydz $$
where D is a part of a sphere $x^2+y^2+z^2 \leq 1 $ that lies in the first octant between the planes $z=\frac{1}{2}$ and $z=\frac{1}{\sqrt 2}$

We are asked to find the solution to this, in spherical coordinates and I have found a expression for $L $ but I'm not sure how to go about completing the solution.

My thoughts were to decompose the integral and use spherical coordinates and then express $L$ as the following.

$$L = \int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{4}} \int_{\frac{1}{2 \cos \phi}}^{\frac{1}{\sqrt 2 \cos \phi}} \rho^{1/2} \sin  \phi d \rho d\phi d \theta + \int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{4}} \int_{\frac{1}{2 \cos \phi}}^{1} \rho^{1/2} \sin  \phi d \rho d\phi d \theta$$

Using geometry and my question is, is this correct? - I based it off a sketch off $z$ against $r$ and then saw we have to decompose this accordingly ^ and then I don't know how to complete the integral i.e. find out what the value of $L$ is.

Could someone help me with this?

答:
Asserting that $(x,y,z)$ belongs to the first octant means that $x,y,z\geqslant0$. Actually, asserting that $z\geqslant0$ is useless here, since we know that $z\in\left[\frac12,\frac1{\sqrt2}\right]$. So, we have

$x=\rho\cos(\theta)\sin(\phi)\geqslant0$, meaning that $\theta\in\left[-\frac\pi2,\frac\pi2\right]$;

$y=\rho\sin(\theta)\sin(\phi)\geqslant0$, meaning that $\theta\in[0,\pi]$;

$z=\rho\cos(\phi)\in\left[\frac12,\frac1{\sqrt2}\right]$, meaning that $\phi\in\left[0,\frac\pi2\right]$, and that $\frac1{2\cos(\phi)}\leqslant\rho\leqslant\frac1{\sqrt2\cos(\phi)}$. But we also want to have $\rho\leqslant1$. So, if $\phi\in\left[0,\frac\pi4\right]$, $\frac1{2\cos(\phi)}\leqslant\rho\leqslant\frac1{\sqrt2\cos(\phi)}$, and if $\phi\in\left[\frac\pi4,\frac\pi3\right]$, $\frac1{2\cos(\phi)}\leqslant\rho\leqslant1$

So, the integral should be\begin{multline}\int_0^{\pi/2}\int_0^{\pi/4}\int_{1/(2\cos\phi)}^{1/\left(\sqrt2\cos\phi\right)}\rho^{1/2}\sin(\phi)\,\mathrm d\rho\,\mathrm d\phi\,\mathrm d\theta+\\+\int_0^{\pi/2}\int_{\pi/4}^{\pi/3}\int_{1/(2\cos\phi)}^1\rho^{1/2}\sin(\phi)\,\mathrm d\rho\,\mathrm d\phi\,\mathrm d\theta.\end{multline}


问:
Let $X,X_i$ random variables such that $(X_n-X)^2\underset{p}{\to}0$ prove $X_i^2$ converges to $X^2$

Let $X,X_i$ random variables such that $(X_n-X)^2\underset{p}{\to}0$

prove $X_i^2$ converges to in probability $X^2$

I've tried using the definition of converges in probability but didn't get too far

any hint?

答:
Let $\eta>0$. I guess that $X$ is finite a.s. (otherwise I don't think that it's true). Therefore, there is $K>0$ s.t. $$\mathbb P\{|X|>K\}\leq \eta.$$

Let $\varepsilon >0$.
Then,
\begin{align*}
\mathbb P\{|X_n^2-X^2|>\varepsilon \}&=\mathbb P\{|X_n^2-X^2|>\varepsilon ,|X|\leq K\}\\
&\ \ \ \ \ \ +\mathbb P\{|X_n^2-X^2|>\varepsilon ,|X|>K\}\\
&\leq \mathbb P\{|X_n^2-X^2|>\varepsilon ,|X|\leq K\}+\eta.
\end{align*}

The function $f(x):=x^2$ is uniformly continuous on $[-2K,2K]$. Therefore, there is $\delta \in(0,K)$ s.t. for all $n\in\mathbb N$,  $$|f(X_n)-f(X)|>\varepsilon \implies |X_n-X|>\delta.$$

Therefore
$$\mathbb P\{|X_n^2-X^2|>\varepsilon ,|X|\leq K\}\leq \mathbb P\{(X_n-X)^2>\delta ^2\}\underset{n\to \infty }{\longrightarrow }0.$$

Therefore, $$\lim_{n\to \infty }\mathbb P\{|X_n^2-X^2|>\varepsilon \}<\eta,$$

for all $\eta>0$, and since $\eta>0$ is unspecified, we conclude that $$  \lim_{n\to \infty }\mathbb P\{|X_n^2-X^2|>\varepsilon \}=0,$$
as wished.

ContentLicense: CC BY-SA 4.0


问:
Prove that if $\lim_{n\to\infty}n a_n=L$, then $\lim_{n\to\infty} a_n =0$
I am having trouble with the proof of this question. Any help would be appreciated. Thank you!

Question

Assume that $L \in \mathbb{R}$ and $(a_n) \subset \mathbb{R}$. If $\lim_{n\to\infty}na_n = L$, prove that $\lim_{n\to\infty}a_n = 0$.

答:
Let $L \in \mathbb{R}$ and $(a_n) \subset \mathbb{R}$ and assume that $\lim_{n\to\infty} na_n = L$.

Claim: $\lim_{n\to\infty}a_n = 0$.

Proof

Fix $\epsilon > 0$, then since $\lim_{n\to\infty} na_n = L$, there exists $N \in \mathbb{N}$ such that
$$
|na_n - L| < \epsilon \quad \forall\ n \geq N.
$$
Reformatting this, we have that the above is equivalent to the statement
$$
-2\frac{\max\{|L|,\epsilon\}}{n} < \frac{1}{n}(L - \epsilon) < a_n <  \frac{1}{n}(L + \epsilon) < 2\frac{\max\{|L|,\epsilon\}}{n} , \quad \forall\ n\geq N.
$$
That is
$$
0 \leq |a_n| < 2\frac{\max\{|L|,\epsilon\}}{n}, \quad \forall\ n \geq N.
$$
Applying the squeeze/sandwich theorem (or whichever name you attribute to this theorem), since $A/n \to 0$ as $n \to \infty$ for any finite $A \in \mathbb{R}$ (independent of $n$), then $a_n \to 0$ as $n \to \infty$.


问:
Matrix with positive and negative eigenvalues (proof/saddle point)
Let $f:\mathbb{R}^n \to \mathbb{R}$ with $f(\textbf{x})=\frac{1}{2}\textbf{x}^T\textbf{B}\textbf{x}$ where $\textbf{B} \in \mathbb{R}^{2\times 2}$ has one positive and one negative eigenvalue for the existence of a saddle point.

My question is: Why does $\textbf{B}$ has to have one positive and one negative eigenvalue for the existence of a saddle point?

My idea was:

It follows from the Hessian matrix $\textbf{H}f(\textbf{x})$, since $\textbf{H}f(\textbf{x})$ is indefinite if it has positive and negative eigenvalues. So $\textbf{x}$ is a saddle point. Is this correct?

I know the proof for showing that if $\textbf{H}f(\textbf{x})$ is indefinite then $\textbf{x}$ is a saddle point.

But does it follow from this that if $\textbf{B}$ is indefinite then $\textbf{H}f(\textbf{x})$ is indefinite or how can it be shown correctly?

答: We have $\textbf{B}= \textbf{H}f(x)$ for all $x$.


答:
You can just use the operating properties of limits:

$$
\lim a_n = \lim \frac{n a_n}{n} = \lim n a_n \cdot \lim \frac 1n  = L \cdot 0 = 0
$$

Title: Prove that if $\lim_{n\to\infty}n a_n=L$, then $\lim_{n\to\infty} a_n =0$

Body:
I am having trouble with the proof of this question. Any help would be appreciated. Thank you!

Question

Assume that $L \in \mathbb{R}$ and $(a_n) \subset \mathbb{R}$. If $\lim_{n\to\infty}na_n = L$, prove that $\lim_{n\to\infty}a_n = 0$.


问:
Justify that T is compact
I have troubles solving the following problem:

Consider the Hilbert space $H=L_2([0,1],m)$ where m is the Lebesgue measure. Define $K:[0,1]\times [0,1]\to \mathbb{R}$ by:

$K(s,t)=\begin{cases}(1-s)t\ \text{if}\ 0\leq t\leq s\leq 1 \\ (1-t)s\ \text{if}\ 0\leq s<t\leq 1\end{cases}$.

And consider $T\in \mathcal{L}(H,H)$ defined by

$(Tf)(s)=\int_{[0,1]}K(s,t)f(t) dm(t),\ \ \ \ \ s\in[0,1],\ f\in H$

Justify that T is compact.

I am thinking that if I can show that T is of finite rank, then I'd have that T is compact, but I am not sure how to show that T is of finite rank.

答:
Hints: First write down $Tf(s)$ and check that it is a continuous function. Next observe that $K$ is uniformly continuous on $[0,1]\times [0,1]$. By Holder's inequality we have $|Tf(s)-Tf(s')| \le \|f\|_2 (\int_0^{1} |K(s,t)-K(s',t)|^{2}dt)^{1/2}$. Conclude that if $(f_n)$ is a bounded sequence in $L^{2}$ then $(Tf_n)$ is an equi-continuous and uniformly bounded sequence of continuous functions on $[0,1]$. By Arzela -Ascoli Theorem it follows that $(Tf_n)$ has  subsequence which converges uniformly. But then the subsequence also converges in $L^{2}$.

Additional information: It is known that if $K$ is replaced by any square integrable function on the unit square then $T$ is compact.

Title: Justify that T is compact


问:
$f(x)=|\cos x|+|\sin(2-x)|$ at which of the following point $f$ is not differentiable?
$f(x)=|\cos x|+|\sin(2-x)|$ at which of the following point $f$ is not differentiable?

1.$\{(2n+1){\pi\over2}\}$

2.$\{n\pi\}$

3.$\{{n\pi\over 2}\}$

4.$\{n\pi+2\}$

in all cases $n\in\mathbb{Z}$

well, is there any easy trick to solve this type of probelem in an competitive examination? Do I need to pick $\pi \over 2$ from $1$ and cheked that $f$ is differentiable and same trick from each set? in some what I am lost.

答:
$f(x)=|x|$ then $f'(x)=\text{sign}{(x)}$

Here $f(x)=|\cos x|+|\sin(2-x)|$

Using the result mentioned at the beginning we get

$f'(x)=\text{sign}{(\cos x)}\times (-\sin{x})+\text{sign}{(\sin(2-x))}\times \cos{\left(2-x\right)}$

Now all you have to do is to check the continuity of the function $f'(x)$ at the point mentioned .

Note:
 $g(x)=\text{sign}{(t(x))}=\begin{cases}{ -1 \quad \text{if } f(x)<0\\+1 \quad \text{if } f(x)>0 } \end{cases}$ .

ContentLicense: CC BY-SA 3.0


问:
Find the CDF of $\ X $ where $\ X = \sqrt{A^2 + B^2} $
Say $\ A,B \sim U(0,1) $ and $\ X = \sqrt{A^2 + B^2} $

and I want to find the the value $\ t$ of $\ X $ for which $\ P(X \le t) = 0.5 $

I understand I need to find the CDF of $\ X $ first and that $\ A, B $ distributed uniformly in the unit square. But I can't see how to find the CDF of $\ X $ ?

答:
It is very easy. Do a drawing and calculate the area $\sqrt{a^2+b^2}<r$ in the unit square $[0;1]\times [0;1]$ then set this probability as 0.5 and solve in $r$

... of course you have to assume independence between $A,B$

Thus the solution is

$$\frac{\pi r^2}{4}=\frac{1}{2}$$

$$r=\sqrt{\frac{2}{\pi}}$$


问:
Can this special case of a Theorem by Bonsall be generalised slightly?
When choosing $\Delta := \mathbb N$, $M_1 = M_2 = 1$ in Theorem 1 in this paper by F.F. Bonsall, we obtain the following variant:

Theorem 1 (Variant).
Let $(X, \| \cdot \|)$ be a Banach space, $(u_k)_{k \in \mathbb N} \subset X$ such that $\| u_k \| \le 1$ for all $k \in \mathbb N$ and
$$
\sup_{k \in \mathbb N} | \psi(u_k) | = \| \psi \|_{X^*}
$$
for all $\psi \in X^*$ and for $f \in X$
$$
L(f, (u_k)_{k \in \mathbb N})
:= \left\{ \lambda = (\lambda_k)_{k \in \mathbb N} \in \ell^1(\mathbb N): f = \sum_{k \in \mathbb N} \lambda_k u_k \right\}.
$$

Then $L(f, u) \ne \emptyset$ for all $f \in X$ and $\| f \| = \inf_{\lambda \in L(f, u)} \| \lambda \|_{\ell^1(\mathbb N)}$.

My Question.
If I take a set $E \subset X$ such that $\| u \| \le 1$ for all $u \in E$ and $\sup_{u \in E} | \psi(u) | = \| \psi \|_{X^*}$ for all $\psi \in X^*$ instead of the sequence $(u_k)_{k \in \mathbb N} \subset X$ with those properties, can I still conclude that the set
$$
L(f, E) := \left\{ \lambda = (\lambda_k)_{k \in \mathbb N} \in \ell^1(\mathbb N): f = \sum_{k \in \mathbb N} \lambda_k u_k, \ (u_k)_{k \in \mathbb N} \subset E \right\}
$$
is nonempty for each $f \in X$ and that $\| f \| = \inf_{c \in L(f, E)} \| c \|_{\ell^1(\mathbb N)}$?

I hope this is true as one can write $E$ as a subset of the union of all sequences $u$ considered in the variant of theorem 1.

My try.
Nonemptyness:
By the properties of $E$, there is a sequence $(u_k)_{k \in \mathbb N} \subset E$ fulfilling the conditions in the variant of Theorem 1.
Let $f \in X$.
There there exists a $\tilde{\lambda} \in L(f, (u_k)_{k \in \mathbb N}) \subset L(f, E)$ by the variant of Theorem 1, so $L(f, E) \ne \emptyset$.

Is the norm condition true, as well?

Note.
I am aware that the result I seek is true, as it is proven in this later paper by the same author. But that proof is rather cumbersome and unintuitive, as it strictly avoids the use of Banach's closed range theorem (because later in the paper the author demonstrates an easy proof of the closed range theorem with the help of the above result). The proof in the first paper uses the closed range theorem and thus is much shorter.

答:
From
$$
|\psi(u)| \le \|u\|_X \, \|\psi\|_{X^*}
$$
you still have $\sup_{u \in E} |\psi(u)| = \|\psi\|_{X^*}$,
i.e. there is no difference in this assumption.

As you said, there is a sequence $(u_k) \subset E$ such that theorem 1 can be applied. Hence,
$$
\inf_{c \in L(f,E)} \|c\| \le \inf_{c \in L(f, (u_k))} \|c\| = \|f\|.$$
To the contrary,
for every $c \in L(f,E)$, there is a sequence $(u_k^{(c)}) \subset E$ (take the above sequence amended by the terms appearing in the statement of $c \in L(f,E)$), such that
it satisfies the assumptions of theorem 1 and
$c \in L(f,(u_k^{(c)}))$.
Hence,
$$\|c\| \ge \inf_{d \in L(f,(u_k^{(c)}))}\|d\| = \|f\|.$$

This shows the norm statement.


问:
Which of the following are true?.
Suppose $\mathfrak{T}_1$ and $\mathfrak{T}_2$ are topologies on $X$ induced by metric $d_1$ and $d_2$, respectively, such that $\mathfrak{T}_1\subseteq \mathfrak{T}_2$. Then, which of the following statement is true?

(a)If a sequence converges in $(X, d_2)$, then it converges in $(X, d_1)$

(b)If a sequence converges in $(X, d_1)$, then it converges in $(X, d_2)$

(c) Every open ball in $(X, d_1)$ is an open ball in $(X, d_2)$

(d)The map $x\mapsto x$ from $(X, d_1)$ to $(X, d_2)$ is continuous

Intuitively, if a sequence convergent in weaker topology $\mathfrak{T}_1$, then it also convergent in $\mathfrak{T}_2$. So I think option (a) is correct.
(b) is false with the above idea.

I am confused with (c) and (d). Also, I am looking for a formal proof of statement (a).

答:
b has a simple counterexample ($d_1$ usual metric on $\Bbb R$, $d_2$ the discrete metric on that same set).

a is indeed true: if every open set under $d_2$ contains a tail of the sequence this certainly holds for all $d_1$-open sets, which are a subset. Note that the reaosn you give hasstronger/weaker in the wrong order: if you converge in the stronger, than also in the weaker. The weaker has fewer sets so fewer constraints.

c is false, same counterexample, or use even two equivalent metrics on the plane, like the taxicab and Euclidean one.

d is false, same counterexample, but true if you interchange domain and codomain.

The only place the metric places any role is in c.

Title: Which of the following are true?.


答: A function isn't differentiable where it has sharp corners since the tangent line at that point is not well-defined. In this case, it fails to be differentiable when $\cos(x)$ and $\sin(2-x)$ change sign since the absolute value of a function has a sharp cusp when its argument changes sign.

Body:
$f(x)=|\cos x|+|\sin(2-x)|$ at which of the following point $f$ is not differentiable?

1.$\{(2n+1){\pi\over2}\}$

2.$\{n\pi\}$

3.$\{{n\pi\over 2}\}$

4.$\{n\pi+2\}$

in all cases $n\in\mathbb{Z}$

well, is there any easy trick to solve this type of probelem in an competitive examination? Do I need to pick $\pi \over 2$ from $1$ and cheked that $f$ is differentiable and same trick from each set? in some what I am lost.


答:
In fact,
\begin{align*}
(Tf)(s)&=\int_0^s(1-s)tf(t)\mathrm{d}t+\int_s^1(1-t)sf(t)\mathrm{d}t
\end{align*}
so
\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}s}(Tf)(s)=(1-s)sf(s)+(1-s)sf(s)-\int_0^stf(t)\mathrm{d}t-\int_s^1tf(t)\mathrm{d}t=\int_0^1tf(t)\mathrm{d}t.
\end{align*}
The above computation first works for smooth test functions then extends to $L^2([0,1],\mathrm{d}m)$ and yields (using Cauchy-Schwarz inequality):
\begin{align*}
\left|\frac{\mathrm{d}}{\mathrm{d}s}(Tf)(s)\right|&\leq\int_0^1t|f(t)|\mathrm{d}t\leq\|t\|_{L^2([0,1],\mathrm{d}m)}\|f\|_{L^2([0,1],\mathrm{d}m)}=\frac{1}{3}\|f\|_{L^2([0,1],\mathrm{d}m)}.
\end{align*}
So
\begin{align*}
\left\|\frac{\mathrm{d}}{\mathrm{d}s}(Tf)\right\|_{L^2([0,1],\mathrm{d}m)}&\leq\frac{1}{3}\|f\|_{L^2([0,1],\mathrm{d}m)}.
\end{align*}
This show that $Tf\in H^1([0,1],\mathrm{d}m)$. As the embedding $H^1([0,1],\mathrm{d}m)\subset H^{1-\epsilon}([0,1],\mathrm{d}m)$ is compact for all $\epsilon>0$, we conclude that $T:L^2([0,1],\mathrm{d}m)\to H^{1-\epsilon}([0,1],\mathrm{d}m)$ is compact. In particular, $T:L^2([0,1],\mathrm{d}m)\to L^{2}([0,1],\mathrm{d}m)$ is compact. Actually, a boot-strap argument shows that $T$ is regularizing.

Title: Justify that T is compact

Body:
I have troubles solving the following problem:

Consider the Hilbert space $H=L_2([0,1],m)$ where m is the Lebesgue measure. Define $K:[0,1]\times [0,1]\to \mathbb{R}$ by:

$K(s,t)=\begin{cases}(1-s)t\ \text{if}\ 0\leq t\leq s\leq 1 \\ (1-t)s\ \text{if}\ 0\leq s<t\leq 1\end{cases}$.

And consider $T\in \mathcal{L}(H,H)$ defined by

$(Tf)(s)=\int_{[0,1]}K(s,t)f(t) dm(t),\ \ \ \ \ s\in[0,1],\ f\in H$

Justify that T is compact.

I am thinking that if I can show that T is of finite rank, then I'd have that T is compact, but I am not sure how to show that T is of finite rank.


问:
Mathematical definition vs Geometric Explanation of Homeomorphism.
How can we connect the mathematical definition of Homeomorphism with its geometric interpretation?

Lets say if I give a Cut what properties of Homeomorphism will be violated?

答: If a function introduces a cut, it is not continuous: continuous functions preserve connectedness. If a function glues, then its inverse is not continuous: functions with continuous inverses are open, meaning that they map open sets to open sets. Intuitively, if you glue two open sets together, there is a kind of boundary between the glued sets where they were glued together. Before gluing, these points weren't boundary points (since open sets dont contain their boundary points) while after gluing they are. For instance, $0$ is not a boundary point of $[0,1/2)$ as a subset of the topological space $[0,1)$, but $1=f(0)$ is a boundary point of $f([0,1/2))$, where $f(t)=\exp(2\pi\mathrm it)$ is a map to the unit circle in the complex plane which glues the ends of the line together. So this map is not open, since sets containing their own boundary points aren't open. And finally, if a function squishes points into each other, it is not injective, and thus not invertible.


问:
Prove that the set {¬,∧,∨} is functionally complete.
So far I have been using the given set to prove the functional completeness of other sets, but I don't know how to prove this one. That seems to be the case for similar questions too.Do I need to construct the truth table with every connective and build their equivalent with 3 given connectives?</p>

答:
There are only $16$ possible boolean functions of two arguments. They have the outputs

$$0000,0001,0010,0011,0100,0101,0110,0111,1000,1001,1010,1011,1100,1101,1110,1111.$$

From these, we can withdraw those that are independent of one or two arguments. Remain

$$0001,0010,0100,0110,0111,1000,1001,1011,1101,1110.$$

We can also withdraw those obtained by negation.

$$0001,0010,0100,0110,1000.$$

These can be synthetized as

$$a\land b,a\land\lnot b,\lnot a\land b,(a\land\lnot b)\lor(\lnot a\land b),\lnot a\land \lnot b.$$


问:
Is there a difference between the number most closest to zero and zero
Let's call the number most closest to zero x. Hence if any number subtracted  by 0 is equal to the number itself..

x - 0 = x
meaning that the answer would be the number that's most closest to zero..

Im confused with that if there's no difference between the number that's most closest to zero and 0 , how can there be an accumulation in the number system as the difference would be 0

答: There is no number closest to zero (if we are only considering non-zero numbers) If we had a number that was "closest", call it $x$, then $x/2$ is even closer to $0$, a contradiction.


问:
$X=(-\infty,0]\cup\left\{{1\over n}:n\in\mathbb N\right\}$ with subspace topology Then
$X=(-\infty,0]\cup\left\{{1\over n}:n\in\mathbb N\right\}$ with subspace topology. Then

$0$ is an isolated point

$(-2,0]$ is an open set

$0$ is a limit point of the subset $\left\{{1\over n}:n\in\mathbb N\right\}$

$(-2,0)$ is open set.

$0$ is not an isolated point of $X$ as every nbd of $0$ contains a point of the form $1\over n$ so  $3$ is true and $1$ is false, $2$ is false as $0$ is not interior point, $4$ is true as every point is interior point. Am i right?

答:
You are right.

3) is true as every nbd of 0 contains a point of the form $\frac 1n$.

4) is true as for any points $x\in (0,2)$, there exists an open set $U$ of $\mathbb R$ such that $x \in U \cap X \subseteq (0,2)$.


问:
ways to walk on a hexagon to return to same point with n steps
So I have an hexagon and I start at point 1 and I can only move right or left, I need to find the formula to reach the same point within n steps,
Assuming I have an odd number of steps allowed, I can't ever return to the point.

But assuming I have even steps, I'm not sure how do I even think about it, I divided by two because I can go from 1>>2 or from 1>>6 so it's easier to just look at it as if i started with 1 step right.

This means that if the sequence is $ a_n $and I took a step right now i have $ a_{n-1}$ steps to come back.
how do I go on from here?

答:
Another approach:

While the situation is simple enough to describe it the way you did, I think the context of a walk on a graph naturally arises here: your graph is the (undirected) $6$-cycle. Let $A$ be its adjacency matrix, finding the number of ways to start from a given point $i$ and end at $i$ in $n$ steps is just the number of paths with length $n$ from $i$ to $i$. That number is given by computing $B = A^n$ and looking at $b_{ii}$.

Here $A$ is known:
$$A=\begin{bmatrix}
0  &    1    &     0   &    0    &     0   &  1  \\
1   & 0 &  1 &  0 &    0    &  0 \\
0  & 1  &      0 &  1 & 0  &  0  \\
 0  &  0 & 1  &      0 & 1  &   0 \\
0   &    0    & 0  & 1 & 0 &  1 \\
1  &    0    &      0  &     0  &     1   & 0
\end{bmatrix}$$

$A$ is diagonalisable, making it possible to find a "simple" closed form for $A^n$. In the end, the result you are looking for is:
$$\frac{1}{6}\left (2(1+(-1)^n) + 2^n + (-2)^n \right)$$

However, as you noticed that for odd $n$, the result is always $0$, you can spare yourself some work and look directly at $A^2$ and find that the formula is $\frac{2^n + 2}{3}$ for even $n$.


问:
simplify $c=\sqrt{290-143\sqrt2}$
I am trying to simplify $c=\sqrt{290-143\sqrt2}.$ I am solving a triangle and I got that $c^2=290-143\sqrt2.$ I have tried to use the formula for $\sqrt{a\pm\sqrt{b}}$ but it seemed useless at the end. Can you give me a hint? I want to remove the square root.

答: It is impossible to remove the square root and get something of the form $a+b\sqrt c$ where $a,b,c$ are rational numbers. If that were possible, the minimal polynomial of the given number would be quadratic. But that of $\sqrt{290-143\sqrt2}$ is a quartic $x^4-580x^2+43202=0$.

Title: simplify $c=\sqrt{290-143\sqrt2}$


答:
Hint

$$(\sqrt{a}-\sqrt{b})^2=290-143\sqrt{2}$$
$$a+b-2\sqrt{ab}=290-143\sqrt{2}$$
so,
$$a+b=290 \quad \text{and} \quad 4ab=2.143^2$$
which is impossible to solve for $a$ and $b$ rational numbers.

Title: simplify $c=\sqrt{290-143\sqrt2}$

Body: I am trying to simplify $c=\sqrt{290-143\sqrt2}.$ I am solving a triangle and I got that $c^2=290-143\sqrt2.$ I have tried to use the formula for $\sqrt{a\pm\sqrt{b}}$ but it seemed useless at the end. Can you give me a hint? I want to remove the square root.


问:
Taylor series of $f(x)=\frac {e^x-1}{x}$
I am asked to expand $f(x)=\frac {e^x-1}{x}$ centered at 0 using the known Talyor series of functions.
How to simplify the function so that it can be expanded more easily?</p>

答:
In order to expand this into a Taylor series. Do the following. Convert $e^x$ into it's taylor series which is $$\sum_{n=0}^\infty \frac{x^n}{n!}$$

Therefore for $e^x$ you have:

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

If you subtract from it $1$ then you will have:

$$x + \frac{x^2}{2} + \frac{x^3}{6} \ldots$$

Now you can divide that by $x$.

$$1 + \frac{x}{2} + \frac{x^2}{6} \ldots$$

Then it is easy to see that the answer is:

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


问:
A few questions on an application of the residue theorem
I have got the following closed curve $\gamma$

and I am supposed to calculate
\begin{equation}
\int_\gamma\frac{\cos(z)}{z^3(z^2 +1)}dz
\end{equation}

with the help of the residue theorem.
Now, I've got a few questions

$0$, i and -i are the isolated singularities of our given function. As far as I understand it, the winding number of $0$ is $-2$, the winding number of $-$i is $-1$ and the winding number of i is $0$. Is this correct?

How can I determine the residue in this case? For a start, I tried finding the Laurent expansion for $z_0=0$, but I have no idea how to rearrange the equation further. I only got as far as
\begin{equation}
\frac{\cos(z)}{z^3(z^2 +1)}=\frac{1}{z^2+1}\sum\limits_{n=0}^{\infty}(-1)^n\cdot\frac{z^{2n-3}}{(2n)!}.
\end{equation}
Could someone give me a hint on how to find the Laurent expansion?

Thanks in advance!

答:
Your conclusions about the winding numbers and singularities are correct.

At $z=0$, $\cos(z)=1-\frac12z^2++O\!\left(z^4\right)$ and $\frac1{1+z^2}=1-z^2+O\!\left(z^4\right)$. Thus,
$$
\frac{\cos(z)}{z^3(1+z^2)}=\frac1{z^3}-\frac32\frac1z+O(z)
$$
At $z=\omega$, where $\omega^2+1=0$, the residue is
$$
\begin{align}
\lim_{z\to\omega}(z-\omega)\frac{\cos(z)}{z^3(z^2+1)}
&=\frac{\cos(\omega)}{\omega^3}\lim_{z\to\omega}\frac{z-\omega}{z^2+1}\\
&=\frac{\cos(\omega)}{\omega^3}\frac1{2\omega}\\[3pt]
&=\frac{\cosh(1)}2
\end{align}
$$


问:
Function $\sqrt[5]{x}$
Wolfram alpha tell me that $\lim_{x\to 0-}\frac{\sqrt[5]{x}}{x}$ does not exists and in particular the domain of $\sqrt[5]{x}$ is given by all $x\geq 0$...why? I should say that $\lim_{x\to 0-}\frac{\sqrt[5]{x}}{x}=+\infty$ and the domain is all $\mathbb{R}$

答:
For $x^{1/5}$, Wolfram|Alpha will take the principal root by default but you can 'force' it to take the real-valued root instead: it then correctly reports the domain to be $\mathbb{R}$.

The same goes for the limit you are trying to calculate: principal (default) vs. real-valued with the correct result:
$$\lim_{x \to 0^-}\frac{\sqrt[5]{x}}{x}=+\infty$$

I think it mainly illustrates how you should be very careful with interpreting what mathematical computer software tells you.

See also: Principal value.

Title: Function $\sqrt[5]{x}$


问:
Conditional Probability with uncertainty
My friend and I have been debating over this for hours and seem to not get to a conclusion nor wrap our heads around. Anyone has an indication for what we need to look in literature to get a proper explanation?

You have two cups and are 80% sure that the marble is under one of
them. 40% that it is under the left and 40% that it is under the right
cup. You take a look under the left cup and find out that the marble
is not there (condition). What is then the conditional probability
then that the marble is under the right cup?

We've defined the events as follows:

$A$: marble is under the left cup on the first look $\longrightarrow \mathbb P(A)=40\%$

$B$: marble is under the right cup on the second look

And came to the conclusion:

$\mathbb P(B|\neg A)$: marble is under the right cup if it was not left on the first look: $80\%$?

答:
Assuming there is only one marble, $$P(B| \bar A ) = \frac{P(B \cap \bar A)}{P(\bar A)} =\frac 23$$

$P(B\cap \bar A)=40\% $ as this event is equivalent to the marble being in the right cup.

$P(\bar A)= (100-40)\%=60\% $ as this event is equivalent to the marble just not being in the first cup.


问:
Approximation of a function in $L^2$ by functions with bounded reciprocals
Let $f \in L^2(0,1)$.
Can one construct a sequence $\{f_\epsilon\}_{\epsilon>0} \subset L^2(0,1)$ such that

$f_\epsilon \longrightarrow f$ strongly in $L^2(0,1)$ as $\epsilon \longrightarrow 0$;

$\frac{1}{\epsilon} f_\epsilon$ is bounded in $L^2(0,1)$ for all $\epsilon>0$?

答: If $(\frac1\epsilon f_\epsilon)$ is bounded in $L^2$, i.e., there exists $M>0$ such that $\epsilon^{-1} \|f_\epsilon\|_{L^2} \le M$, then necessarily $f_\epsilon \to0$ in $L^2$.


问:
indirectly convex
Let g:$\mathbb{R^2}$$\rightarrow$$\mathbb{R}$ be defined by g(x)=Max{$x_1$,$x_2$} at each x=($x_1$,$x_2$)$\in$ $\mathbb{R}$. Determine whether or not g is indirectly convex on $\mathbb{R^2}$.
definition:
let f:$\mathbb{R^n}$$\rightarrow$$\mathbb{R}$ be a function. We say that f is indirectly convex if there exist F:$\mathbb{R^n}$$\rightarrow$$\mathbb{R}$ and h:$\mathbb{R}$$\rightarrow$$\mathbb{R}$ such that F is convex, h is increasing with $f=h \circ F$.</p>

答: Indirect convex implies convex sub-level sets $V_a = \lbrace x | f(x) \leq a \rbrace$ and that higher level curves contain lower ones.  This could be the exact condition, and it is true for $g(x)$.</p>

Title: indirectly convex


问:
Solving a congruence system (which has other solutions)
I am trying to solve this congruence system :
$$\begin{cases}
3x+7y\equiv 5[11]\\
8x+4y\equiv6[11]
\end{cases}
$$
I multiply first by $8$ and $3$ both equations resp.
$$\begin{cases}
24x+56y\equiv40[11]\equiv7 [11]\\
24x+12y\equiv18[11]\equiv7 [11]
\end{cases}
$$
We then substract the second equation from the first one.
$$44y\equiv0[11],$$
Since $gcd(11,44)=11$, then previous equation is equivalent to : $4y\equiv0[11]$, and then we find that :
$$y\equiv0[11],$$
and now I replace it in the following equation : $3x+7\times 0\equiv5[11]$ which is equivalent to :
$3x\equiv5[11]$, hence
$$x\equiv 9[11]$$
As a result $(9+11k,0+11k')$ with $k,k' \in \mathbb{Z}$ are solutions of the system.

Now my remark is the following : I noticed that also $(0+11k,7+11k')$ with $k,k' \in \mathbb{Z}$ and also $(7+11k,4+11k')$ with $k,k' \in \mathbb{Z}$ are also solutions of the system which I don't understand.

答:
Since upon elimination the two equations results in $44y \equiv 0 \pmod {11}$ which is true for all integers $y$, the two equations are not independent. As a result, we must express the values of $x$ in terms of $y$ using one of the equations. For example:

$$3x+7y\equiv 5 \pmod {11} \implies 12x + 28y \equiv 20 \pmod {11} \implies x\equiv 9+5y\pmod {11}$$

To verify that $(9+5y,y)$ gives all solutions, we substitute this to the second equation, and we have:

$$8x+4y \equiv 72+44y\equiv 6\pmod {11}$$

so both equations are satisfied. $(9,0), (0,7), (7,4)$ are particular solutions, modulo $11$.


问:
Find the liar and the reliable
I have two categories of people, the reliable and the liars.The reliable telling always the truth,the liars always telling lies. One visitor meets both of them:

A declares: "I and B are the same."
B declares : "From us, only one is reliable."

What situation is true?

```<code>a)both are liars
b) the A is liar and the B reliable
c)both reliable
d)A and B reliable
</code>```

答:
A and B cannot be in the same category, since -

if A and B are both liars, then A's statement ('I and B are the same') is true. But A is a liar, so A will always speak falsehoods, and not true statements, thus A and B cannot be both liars.

if A and B are both reliable, then B's statement ('From us, only one is reliable') is false. But B is reliable, therefore B cannot speak falsehoods, only truths, thus A and B cannot be both reliable.

That eliminates options a, c, d from your list, leaving option b.

If A is liar and B is reliable - then A's statement ('I and B are the same') is false, and B's statement ('From us, only one is reliable') is true. Since A can only speak lies and B can speak only truths, this is a valid situation.

Title: Find the liar and the reliable


答:
If A is a liar then A and B are not the same hence B is a reliable. Then B says one one of us is reliable hence pointing to himself. Therefore the statement is true.

However if A is reliable then both B and A the same contradicting B's statement of only one reliable between both of them. Contradiction.

Hence A is a liar and B is reliable B is the answer. Talk about 'there is one imposter among us.'

Title: Find the liar and the reliable

Body:
I have two categories of people, the reliable and the liars.The reliable telling always the truth,the liars always telling lies. One visitor meets both of them:

A declares: "I and B are the same."
B declares : "From us, only one is reliable."

What situation is true?

```<code>a)both are liars
b) the A is liar and the B reliable
c)both reliable
d)A and B reliable
</code>```


问:
How to find the minimum of $xy$ from the equation $2x+y+6=xy$ , $x,y>0$
I have tried doing factoring and completing the square and also AM-GM

$(2-y)(x-1)=-8$

$xy=2x+y+6≥\sqrt{12xy} $

答:
Hint:

With $p:=xy$, after elimination of $y$ the equation can be written

$$p=\frac{2x^2+6x}{x-1},$$ with $x,p>0$.

Note that

$$\frac{2x^2+6x}{x-1}=2x+8+\frac 8{x-1},$$ and finding the extrema is easy by differentiation.


问:
What's the equation for a line segment?
I already know that the standard equation for a line is $y=mx+b$, but what if I want the line to have specific endpoints and not go on forever? For example, the equation for a line beginning at $(3, 1)$ and ending at $(7, 2)$. Can you help me?

What's the standard equation for this?

答:
In your example, the equation of a line passing from $(3,1)$ and $(7,2)$ is $$y=\frac{1}{4}x+\frac{1}{4}$$
defined for $3<x<7$ and $1<y<2$

Now let's take $3<x<7$  and split it into two inequalities $$3<x$$ and $$x<7$$, or equivalently $$x-3>0$$ and $$7-x>0$$

Now write $$\sqrt{\frac{|7-x|}{7-x}}$$ and  $$\sqrt{\frac{|x-3|}{x-3}}$$
Notice that in $$\sqrt{\frac{|7-x|}{7-x}}$$ if I plug in a value greater than $7$ the I get a complex number, this is why I included the square root, similarly   $$\sqrt{\frac{|x-3|}{x-3}}$$ becomes complex number for values of $x$ less than $3$ (note that the purpose of the absolute values is to "normalize" the term so that the only possible values that I can get are $-1$ or $1$)

Next multiply  $$\sqrt{\frac{|7-x|}{7-x}}$$ and  $$\sqrt{\frac{|x-3|}{x-3}}$$ together and multiply with $x$ in the equation of the line to get $$y=\frac{1}{4} x \sqrt{\frac{|7-x|}{7-x}} \sqrt{\frac{|x-3|}{x-3}}+\frac{1}{4}$$
and this equation describes a line segment if you include the points $(3,1)$ and $(7,2)$

This method will work for any curve.

Title: What's the equation for a line segment?


问:
Showing this integral is always greater than $\pi/2$
Define $$I(b)= \int_1^\infty {(u+b)^{1/2} du\over (u^2+b)}$$

I want to show that $\mathit I(b)$ is greater than $\pi/2$, for every $\mathit b >0. $

答:
The following is a more-or-less elementary approach:

Changing variables $u = bv$ in the original $I(b)$ integral, we get

$$ I(b) = \int_{1/b}^{\infty} \frac{b^{1/2} \sqrt{v+1}}{(b^{1/2}v)^2 +1} \, dv = \int_{1/b}^{\infty} \sqrt{v+1} \, \partial_v(\arctan(b^{1/2}v)-\pi/2) \, dv.$$

Integrating by parts, we get

$$I(b) = \sqrt{1/b + 1} (\pi/2 - \arctan(1/b^{1/2})) + \int_{1/b}^{\infty}\frac{(\pi/2 - \arctan(b^{1/2}v)}{2\sqrt{v+1}} \, dv.$$

As mentioned before, $\limsup_{b \to \infty} I(b) \ge \pi/2,$ which also follows from this last identity. In order to prove the result, it suffices to show that $I(b)$ is nonincreasing in $b>0.$ In order to do so, it suffices to prove that the function

$$g(t) = \sqrt{\frac{1}{t^2} + 1} (\pi/2 - \arctan(1/t)) + \int_{1/t^2}^{\infty} \frac{\pi/2 - \arctan(tv)}{2\sqrt{v+1}} \, dv $$

is nonincreasing. Before we do that, we remark that a function

$$\eta(t) = \int_{a(t)}^{\infty} F(x,t) \, dx$$

is, given $a,F$ are smooth, is differentiable, and its derivative is

$$\eta'(t) = -a'(t) F(a(t),t) + \int_{a(t)}^{\infty} \partial_t F(x,t) \, dx.$$

The derivative of $\int_{1/t^2}^{\infty} \frac{\pi/2 - \arctan(tx)}{2\sqrt{x+1}} \, dx$ is then

$$ 2 t^{-3} \frac{\pi/2 - \arctan(1/t)}{2\sqrt{\frac{1}{t^2} + 1}} - \int_{1/t^2}^{\infty} \frac{x}{2\sqrt{x+1}} \frac{1}{(xt)^2+1} \, dx,$$

while that of $\sqrt{\frac{1}{t^2}+1}(\pi/2 - \arctan(1/t))$ is

$$ -2 t^{-3} \frac{1}{2\sqrt{\frac{1}{t^2} + 1}} (\pi/2 - \arctan(1/t)) + \frac{1}{t(1+t^2)^{1/2}}.$$

Therefore,

$$g'(t) = \frac{1}{t} \frac{1}{(1+t^2)^{1/2}}  -\int_{1/t^2}^{\infty} \frac{x}{2\sqrt{x+1}} \frac{1}{(xt)^2+1} \, dx. $$

Change once more variables $x = v/t^2$ in the integral on the right. This yields

$$g'(t) = \frac{1}{t(1+t^2)^{1/2}} - \frac{1}{t} \int_1^{\infty} \frac{v}{2\sqrt{v+t^2}} \frac{1}{v^2 + t^2} \, dv.$$

We then use the Cauchy-Schwarz inequality $\sqrt{v+t^2} \le (v^2 + t^2)^{1/4} (1+t^2)^{1/4}$ in the integral, which yields

$$g'(t) \le \frac{1}{t} \left( \frac{1}{(1+t^2)^{1/2}} -\frac{1}{(1+t^2)^{1/4}} \int_1^{\infty} \frac{v}{2(v^2+t^2)^{5/4}} \right) \, dv.$$

But then $-\frac{v}{2(v^2 + t^2)^{5/4}} = \partial_v ((v^2+t^2)^{-1/4}),$ so the integral evaluates explicitly to $\frac{1}{(1+t^2)^{1/4}}.$ Thus $g'(t) \le 0.$ Notice, moreover, that the inequality is strict (due to the Cauchy-Schwartz inequality) unless $t=0.$ Therefore, $g$ is strictly decreasing on $t>0,$ and so is $I(b),$ finishing the problem.


问:
Critiquing the notion of a tangent as a line that "just touches" a curve.
The most common answer to this question "What is tangent to any curve?" is as follows:

"Tangent to a plane curve at a given point is the straight line that
"just touches" the curve at that point. "

But this definition have two problems.

First is what do you mean by "just touches"? How can I know if a line "just  touches"  a curve or not?

For example: How do you know that the red line in the following image "just touches" the curve while green line doesn't?

Second problem is that this definition doesn't generalize to the "straight-line curve" .

A tangent to a straight line is the straight line itself. But this can't be possible under the "just touches" definition. (This is just what I think, if you think my reasoning is incorrect then please correct me.)

So what is tangent to any curve?  And also if my objections are correct then why is this definition so famous?

答: I think that the "just touches" means that "it doesn't go through". Maybe by looking to the contrapositive (the one I just mentioned) you will be "happier". And by the "just touches" in the sense that "it doesn't go through", a straight line tangent to another "just touches it". But yes, it is not the most rigorous approach and definitely not the one I would use, but it can be useful for the abstraction of this concept, if it satisfies you.</p>


问:
Evaluating the integral with help of Laplace transforms
I came across an integral that is to be solved by the Laplace transform method.
$$ \int_0^{\infty} \frac{\sin^2{x}}{x^2}$$
My Approch
let $ f(t) = \sin^{2}t$ and $$\sin^{2}t = \frac{1-\cos2t}{2}$$
also we know from the properties of Laplace transforms
$$\frac{\mathcal{L}{f(t)}}{t} = \int_s^\infty F(s) ds $$
so $$ \mathcal{L}\left[\frac{1-\cos2t}{2}\right] = \frac{1}{2} \left[ \frac{1}{s} - \frac{s}{s^2+4} \right]$$
now, $$\mathcal{L}\left[\frac{1-\cos2t}{2t}\right] = \frac{1}{2}\int_s^\infty \left(\frac{1}{s} -\frac{s}{s^2+4} \right) ds$$
after integrating
$$ \mathcal{L} \left[\frac{1-cos2t}{2t}\right]= \frac{1}{2}\left[\ln(s) - \frac{\ln(s^2+4)}{2}\right]\big{|}_s^\infty$$
$$ \mathcal{L}\left[\frac{1-\cos2t}{2t}\right] =\frac{1}{2} \left[\ln(\frac{s}{(s^2+4)^{1/2}})\right] \big{|}_s^\infty $$
but at this point the integral becomes not-defined,
How can I approach this question as it has to be solved by the method of Laplace transforms
Hints are appreciated

答:
First, there's a missing $2$ in the denominator of $\frac{1-\cos 2t}{t}$. I suppose it's a typo?
Next,
$$\mathcal{L}\left[\frac{1-\cos2t}{2t}\right] = \frac{1}{2}\int_0^\infty \left(\frac{1}{s} -\frac{s}{s^2+4} \right) ds $$
can't be true because the RHS is possibly a number (the integral actually diverges) while the LHS is a function. What we have instead is
$$\mathcal{L}\left[\frac{\sin^2 t}{t}\right]=\mathcal{L}\left[\frac{1-\cos2t}{2t}\right]=\frac 12\int_s^{\infty}\left(\frac 1y-\frac{y}{y^2+4}\right)dy=\frac 14\log\frac{s^2+4}{s^2} $$
Now, if we let $g(t)=\sin^2 t/t$, we've found $\mathcal{L}[g(t)]$. So
$$\int_0^{\infty}\frac{\sin^2 x}{x^2}dx=\int_0^{\infty}\frac{g(x)}{x}dx=\frac 14\int_0^{\infty}\log\frac{p^2+4}{p^2}dp $$
Can you continue from here? Use integration by parts.


问:
Open Loop Transfer Function and Stability?
Given the following open loop transfer function
$$G_{OL} = \frac{k}{(s+1)^2(0.5s + 1)^2}e^{-\tau s}$$

Let $k = k_{max}/2$. Let $\tau_{max}$ denote the value of the parameter $\tau$ such that the system in a unity negative feedback loop is stable for $\tau < \tau_{max}$. Find $\tau_{max}$

I know $k_{max} = 9/2$ from applying the Routh Hurwitz stability criterion with $\tau = 0$, so $k = 9/4$ in this case.

Questions:

How do I determine $\tau_{max}$? I can expand $G_{OL}$ find the magnitude and phase, approximate a Bode plot or make a Nyquist plot, but how does this help? I should also be able to find the closed-loop transfer function since its unity negative feedback. I guess I just don't see the connection between the tools (Bode, root locus, Nyquist) used to analyze stability and actually finding $\tau_{max}$.

Any help would be appreciated!

答: You can find $\tau_{max}$ as the delay margin. Set $\tau=0$, find the phase margin and the corresponding frequency. Then compute for which $\tau$ the phase shift induced by the delay will be equal to the phase margin.


问:
Intuitive proof for the Schwarz inequality (Complex Analysis)
I have been following Serge Leng's Introduction to Complex Analysis at a graduate level and I feel a bit stuck on one of the exercises. The exercise in question is

I can't seem to prove N3 without assuming the Schwarz inequality. On the other hand, I am able to prove the Schwarz inequality but I would have to assume N3.

There have been published solutions for these questions:

however, this is not the slitest bit intuitive (atleast to me). Can anybody help me/explain to me an intuitive way of looking at the problem?

答:
The intuition behind the Schwarz inequality is perhaps easily seen for an inner product over a real vector space;  for any vectors $u, v$,
\begin{align}
\langle u \pm v, u \pm v\rangle \geqslant 0.
\end{align}
Now take $u, v$ to have unit length and expand the product to obtain,
\begin{align}
2 \geqslant \pm 2 \langle u, v \rangle
\end{align}
so that $\lvert \langle u,v \rangle \rvert \leqslant 1$.  This is extended to general vectors $u$ and $v$ by dividing each by its norm and using the bi-linearity of the inner product.

The argument is also extended to a complex linear space by taking again unit vectors $u$ and $v$ but now multiply $u$ by $e^{i\theta}$ where $\theta$ is chosen to make  $\langle u,v \rangle$  real and positive.  This is always possible.

Then
\begin{align}
\langle e^{i\theta}u - v, e^{i\theta} u - v \rangle &= \langle e^{i\theta} u, e^{i \theta}u\rangle - e^{i\theta} \langle u, v\rangle
 - e^{-i\theta}\langle v,u\rangle  +\langle v,v \rangle \\
&=\langle u,u \rangle - 2 \Re \big( e^{i\theta}\langle u,v \rangle\big) + \langle v,v \rangle
\end{align}
Now the choice of $\theta$ comes into play. We have made $e^{i\theta}\langle u,v\rangle$  real and positive, so as before we now obtain
\begin{align}
\Re \big( e^{i\theta} \langle u,v \rangle \big) &= e^{i\theta} \langle u,v \rangle \\
&= \lvert e^{i\theta} \langle u,v \rangle \rvert \\
&= \lvert \langle u,v \rangle \rvert
\end{align}
leading neatly to $\lvert \langle u,v \rangle \rvert \leqslant 1$.  This is extended to general $u$ and $v$ as before.


答:
Tavish' answer provides a formal proof.  The informal approach is:

There are two possibilities:

The marble is under the first cup

The marble is not under the first cup and the ratio of the probability that the marble is under the 2nd cup vs the probability that the marble is not under the 2nd cup is 40 to 20.

Once you determine that the first case above is impossible, you are left with the second case.

Title: Conditional Probability with uncertainty

Body:
My friend and I have been debating over this for hours and seem to not get to a conclusion nor wrap our heads around. Anyone has an indication for what we need to look in literature to get a proper explanation?

You have two cups and are 80% sure that the marble is under one of
them. 40% that it is under the left and 40% that it is under the right
cup. You take a look under the left cup and find out that the marble
is not there (condition). What is then the conditional probability
then that the marble is under the right cup?

We've defined the events as follows:

$A$: marble is under the left cup on the first look $\longrightarrow \mathbb P(A)=40\%$

$B$: marble is under the right cup on the second look

And came to the conclusion:

$\mathbb P(B|\neg A)$: marble is under the right cup if it was not left on the first look: $80\%$?


问:
Finding unique number and value of p.
Function = f(x) = $\frac{x^2+px+1}{x^2+x+1}.$

Q 1For any value of p , a unique number will be there in the range . Find it.

Q 2 If nine integers are in the range of f(x) , then solve for p.

Q3If  range of function is non negative , then find maximum no of integers in the range can be .

For the 1st one , i thought unique number as  neither a prime number nor a composite number. It has only one factor, i.e, the number itself. Is it correct ?

For the 2nd , I am not getting

For 3rd , no of integers will change as value of p will change . Then how to find the amswer

答:
Hint:

The function can be written

$$\frac{x^2+px+1}{x^2+x+1}=1+(p-1)\frac1{x+1+\dfrac1x}.$$

It is an easy matter to establish the range of $x+\dfrac1x$, which is $(-\infty,-2]\cup[2,\infty)$. From this, the range of the function is

$$1+(p-1)\left[-1,\frac13\right].$$

(For $p<1$, the bounds of the interval need to be swapped.)

$1$ is always in the range.

the reduced range has length $\dfrac43$. To cover $9$ integers, you need to multiply by at least $6$ ($p=7$). The integers will be $-5$ to $3$.

for $p\ge1$, the range is $\left[2-p,\dfrac{3p+2}3\right]$ and for $p=2$, it contains $0,1,2$.

For $p<1$, the range is $\left[\dfrac{3p+2}3,2-p\right]$ and the smallest $p$ is $-\dfrac23$, giving $0,1,2$ as well.

Title: Finding unique number and value of p.


问:
How to show that $(\Bbb Q, <)$ contains an order isomorphic copy of $\epsilon_0$ which is the least ordinal satisfies $\omega^\epsilon = \epsilon$?
We define $\epsilon_0 = \sup\{\omega, \omega^\omega,\omega^{\omega^\omega},\omega^{\omega^{\omega^\omega}},\ldots\}$. How to find a subset of $ \Bbb Q$, $(A, <_{\Bbb Q})$ that is isomorphic to $\epsilon_0$?

So far, I'm only able to show that there's an order isomophic copy of $\omega^\omega$ in $\Bbb Q$ and don't know how to proceed.

答:
A different proof of the fact cited by mercio (every countable linear ordering embeds into ${\bf Q}$) can be formulated as follows:

${\bf Q}$ is the unique (up to isomorphism) countable dense linear ordering without endpoints (this is a very well-known fact, can be shown using back-and-forth method).

Any linear order $L$ can be extended to a dense linear ordering without endpoints of cardinality $\aleph_0+\lvert L\rvert$ by recursively adding a single point at the end, at the beginning, and between each pair of points which have nothing between. At each step of the construction we add at most $\lvert L\rvert+\aleph_0$ points, so after countably many steps we have added at most $\aleph_0+ \lvert L\rvert$ points, and it's easy to see that we obtain a dense linear ordering.

Therefore, if $L$ was countable to begin with, the dense linear ordering we obtain in the previous point is countable and hence isomorphic to ${\bf Q}$.


问:
Dimensions of subspaces and generating sets.
Say I have two subspaces $U$, $V$ of some finite-dimensional vector space with generating sets $A$ and $B$.

I can't seem to find a nice and concise proof of the following (also implying I'm not even sure it is true):

If and only if none of the elements in $B$ can be expressed as a linear combination of vectors in $A$ and vice versa, then $dim(U+V)=dim(U)+dim(V)$.

From a different perspective, I could also ask: If I want to check if $U\cap V=\{0\}$, is it sufficient to check if I can make vectors in $B$ with vectors in $A$ and the other way around?

I would be happy if someone could help with that.

答: Suppose $v\in U\cap V$. Then $v$ can be written as a linear combination of elements of $V$ and also as a linear combination of elements of $U$. By hypothesis, this is only possible for $v=0$.


答:
A note on reducing $\quad ax\equiv b\pmod m\quad$ and a divider $\quad d$

if $\gcd(m,d)=1$ then you can divide $a,b$ by $d$ and keep the same modulo

e.g. $\quad 6x\equiv 9\pmod{11}\iff 2x\equiv 3\pmod{11}$

if $\gcd(m,d)\neq 1$ and $d$ divides $a,b$ then you can reduce the equation to a lower modulo

e.g. $\quad 6x\equiv 9\pmod{15}\iff 2x\equiv 3\pmod{5}$

In all other cases, try multiplying by $a^{-1}$ or $b^{-1}$ depending on which is relatively prime with the modulo.

e.g. $\, 2^{-1}\equiv 6\pmod{11}\, $ thus $\, 2x\equiv 3\pmod{11}\iff 12x\equiv x\equiv 18\equiv 7\pmod {11}$

In the present case $44y\equiv 0\pmod{11}$, you get in first case for $d=4$ and second case for $d=11$.

$44y\equiv 0\pmod{11}\iff 11y\equiv 0\pmod{11}\iff y\equiv 0\pmod 1\quad$ which is always true for a random integer $y$.

Note that since $44$ is a multiple of $11$ the equation is also simply $0y\equiv 0\pmod{11}$, and this is just some $0=0$ equation, bringing no fuel to the problem (or rather that $y$ is a free variable).

Title: Solving a congruence system (which has other solutions)

Body:
I am trying to solve this congruence system :
$$\begin{cases}
3x+7y\equiv 5[11]\\
8x+4y\equiv6[11]
\end{cases}
$$
I multiply first by $8$ and $3$ both equations resp.
$$\begin{cases}
24x+56y\equiv40[11]\equiv7 [11]\\
24x+12y\equiv18[11]\equiv7 [11]
\end{cases}
$$
We then substract the second equation from the first one.
$$44y\equiv0[11],$$
Since $gcd(11,44)=11$, then previous equation is equivalent to : $4y\equiv0[11]$, and then we find that :
$$y\equiv0[11],$$
and now I replace it in the following equation : $3x+7\times 0\equiv5[11]$ which is equivalent to :
$3x\equiv5[11]$, hence
$$x\equiv 9[11]$$
As a result $(9+11k,0+11k')$ with $k,k' \in \mathbb{Z}$ are solutions of the system.

Now my remark is the following : I noticed that also $(0+11k,7+11k')$ with $k,k' \in \mathbb{Z}$ and also $(7+11k,4+11k')$ with $k,k' \in \mathbb{Z}$ are also solutions of the system which I don't understand.


问:
Fourier cosine transform of $\arctan(1/x)/x$
Consider
$$
\int_{-\infty}^{\infty}\cos(\omega x)\frac{\arctan\left(\frac{1}{x}\right)}{x} dx, \quad \text{for } \omega > 0,
$$
although for $\omega \in \mathbb{C}$ would be interesting. What is the above cosine transform? Does it exist?

My attempt so far: for $x \in \mathbb{R}$ we know that
$$
\arctan(x) + \arctan(1/x) = \text{sign}(x)\pi/2,
$$
and we know that
$$
\int_{-\infty}^{\infty}\cos(\omega x)\frac{\arctan(x)}{x} dx = \pi \Gamma(0,|\omega|),
$$
where $\Gamma$ is the upper incomplete Gamma function. Using this, then
$$
\begin{align}
\int_{-\infty}^{\infty}\cos(\omega x)\frac{\arctan\left(\frac{1}{x}\right)}{x} dx
&= \int_{0}^{\infty}\cos(\omega x)\frac{\arctan\left(\frac{1}{x}\right)}{x} dx + \int_{-\infty}^{0}\cos(\omega x)\frac{\arctan\left(\frac{1}{x}\right)}{x} dx \\
&= -\int_{-\infty}^{\infty}\cos(\omega x)\frac{\arctan\left(x\right)}{x} dx -\frac{\pi}{2}\int_{-\infty}^{0}\cos(\omega x)\frac{1}{x} dx + \frac{\pi}{2}\int_{0}^{\infty}\cos(\omega x)\frac{1}{x} dx \\
&= -\pi\Gamma(0,|\omega|) - \pi\int_{0}^{\infty}\cos(\omega x)\frac{1}{x} dx.
\end{align}
$$
But the integral on the right does not make sense. I assume this is not the correct approach.

答:
I fact the initial integral $g(\omega)=\frac{1}{2}\int_{-\infty}^{\infty}\cos(\omega x)\frac{\arctan\left(\frac{1}{x}\right)}{x} dx=\int_{0}^{\infty}\cos(\omega x)\frac{\arctan\left(\frac{1}{x}\right)}{x} dx$ diverges at a point $х=0$ logarithmically - even in a principal value sense.

Indeed, the integrand is even, and at $x\to0$ behaves as $\sim 1*\frac{\pi}{2}*\frac{1}{x}$. To investigate the integral in more details let's make a regularisation at $x\to0$, for example by adding the second term.

Let's consider $$f(\omega) =g(\omega)-\frac{\pi}{2}\int_{0}^{\infty}\cos(\omega x)\frac{\exp(-ax)}{x} dx$$ This integrand does not have a singularity at $x=0$ and, as soon as $a\to{\infty}$ $f(\omega)\to{g(\omega)}$.
Differentiating by $\omega$  we get:
$\frac{df}{d\omega}=-\int_{0}^{\infty}\sin(\omega x)\arctan(\frac{1}{x}) dx+\frac{\pi}{2}\int_{0}^{\infty}\cos(\omega x)\exp(-ax) dx$.$$\frac{df}{d\omega}=\frac{\pi}{2}\left(\frac{\exp(-\omega)-1}{\omega}+\frac{\omega}{a^2+\omega^2}\right)$$
$f(\omega=0)=\frac{\pi}{2}\int_{0}^{\infty}\frac{\arctan(\frac{1}{x})}{x}dx-\frac{\pi}{2}\int_{0}^{\infty}\frac{\exp(-ax)}{x}dx$.

Integrating by parts we get: $f(0)=\int_{0}^{\infty}\frac{\log(x)}{1+x^2}dx-\frac{\pi}{2}a\int_{0}^{\infty}\log(x)\exp(-ax)dx=\frac{\pi}{2}\log(a)-\frac{\pi}{2}\int_{0}^{\infty}\log(x)\exp(-x)dx$

$$f(0)=\frac{\pi}{2}\log(a)+\frac{\pi}{2}\gamma$$ where  $\gamma=-\int_{0}^{\infty}\log(x)\exp(-x)dx$ - Euler constant.

$f(\omega)=f(0)+\int_{0}^{\omega}\frac{df}{dx}dx=\frac{\pi}{2}\log(a)+\frac{\pi}{2}\gamma+\frac{\pi}{2}\int_{0}^{\omega}\left(\frac{\exp(-x)-1}{x}+\frac{x}{a^2+x^2}\right)dx$.

Finally we get:
$$f(\omega)=\frac{\pi}{2}\left(\log(\omega)(\exp(-\omega)-1)+\frac{1}{2}\log(a^2+\omega^2)-\int_{\omega}^{\infty}\log(x)\exp(-x)dx\right)$$

$a\to{\infty}$ (we get the initial integral without regularisation):
$$f(\omega)\to{g(\omega)}\to\frac{\pi}{2}\log(a)\to\infty$$ as expected

$\omega\to{\infty}$ and $ (a\lt{\infty})$ $$f(\omega)\to-\frac{\pi}{2}\frac{\exp(-\omega)}{\omega}\to{0}$$


问:
Find $\det (A+B)$ where $\det A = 0$, and $A$ and $B$ are square matrices of order $n$ and $x\in\mathbb{R}.$
Well considering two $n\times n$ matrices what is the general formula of
$\det(A+xB)$ ?
What can be said about $\det(A+B)$?

By the way $A$ is not invertible. $B$ as well.
I tried expansion but there is too much calculus.

Help me to find an easier way and a concrete general formula.

答:
What can be said about det(A+B)?

From just knowing that $A$ and $B$ are non-invertible, nothing. Literally.

For every $d\in\mathbb R$ and every $n\geq 2$, there exists a pair of non-invertible matrices such that $\det(A+B)=d$.

For example, take the matrix $E$ the matrix with all elements except $E_{1,1}$ set to zero, with $E_{1,1}=1$, in other words $$E=\begin{bmatrix}1 &0& \dots&0\\0 & 0 & \dots &0\\\vdots&\vdots&\ddots&\vdots\\0&0&\dots&0\end{bmatrix}$$

Now take $A=d\cdot E$ and $B=I-E$. Then, $A$ is non-invertible, $B$ is non-invertible, and $\det(A+B)=d$.


问:
Combinatorics: Number of ways to make 1500 dollars
I have 50 bills of 20\$ and 20 bills of 50\$ and 10 bills of 100\$. With how many ways a bank machine could make the 1500\$ ?

what I think is :as the cofficient of $x^{1500}$
in  $(1+x+x^2+\ldots+x^{1000})^3$

答:
To keep numbers small let's divide the note values by $10$ so we have $50$ of $2$, $20$ of $5$ and $10$ of $10$ and want to make $150$. Because the other two denominations and the target amount are divisible by $5$, in all ways of making $150$ the number of $2$ notes must be a multiple of $5$, so effectively we have $10$ notes of $10$ but in a different kind – think different series of notes. Then, by the same logic, the number of $5$ notes used must be even, so it is like $10$ notes of $10$ in a third kind. Now we can divide note values by $10$ again, leading to the ultimate reduced question:

What is the number of integer solutions to $a+b+c=15$ where $0\le a,b,c\le10$?

We count the number of ways with no upper limit on the variables, which is $\binom{15+3-1}{3-1}=136$, then subtract the number of ways where one of the variables is at least $10$: three times the number of nonnegative integer solutions to $a+b+c=4$, equalling $45$. Thus the final answer is $91$: there are $91$ ways for the ATM to dispense \$1500.


问:
Show that $e^{ix}=\cos(x)+i\sin(x)$ using the Fourier Series only
As stated in the title.

Any arbitrary function can be expressed as
$$f(x)=\frac{a_0}{2}+\sum^{\infty}_{n=1}(a_n\cos(nx)+b_n\sin(nx)) \tag{1}$$
where
$$a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(nx)dx \tag{2}$$
$$b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(nx)dx \tag{3}$$

I am trying to show that (1)-(3) are consistent with $f(x)=e^{ix}=\cos(x)+i\sin(x)$.

However, for $a_n=\frac{1}{\pi}\int^{\pi}_{-\pi}e^{ix}\cos(nx)dx$ I got $a_n=0$, similarly, $b_n=0$.

What's gone wrong? Are the integral limits $\int_{-\pi}^{\pi}$ not valid in general, does it require modifications for non-periodic functions?

答:
Let's try this definition: $f(x) = e^{ix}$ is the unique differentiable function $f : \mathbb R \to \mathbb C$ that satisfies
$$
f'(x) = if(x),\qquad f(0)=1.
\tag4$$
Then let's try to evaluate the required integrals $(2),(3)$ using integration by parts.

We will do this only using $(4)$.  No complex exponentials appear in what I do below, so I will not accidentally use other properties of $e^{ix}$

Start with $(2)$, integrate by parts, use $(4)$ on the result.  We get
$$
a_n = \frac{-i}{n}\;b_n
\tag5$$
Start with $(3)$ in the same way.  We get
$$
b_n = \frac{i}{n}\;a_n + \frac{(-1)^n}{\pi n}(f(-\pi)-f(\pi))
\tag6$$
Solving the system $(5),(6)$, we get (in case $n \ne 1$)
$$
a_n = \frac{i n (-1)^n}{\pi(n^2-1)}\;(f(\pi)-f(-\pi))
\\
b_n = \frac{n(-1)^n}{\pi(n^2-1)}\;(f(-\pi)-f(\pi))
$$

Unless we can prove $f(\pi) = f(-\pi)$ from the definition $(4)$, it seems we will not be able to complete this.

Now consider $n=1$.  We get equations
$$
b_1 = ia_1
\\
a_1 = a_1 +\frac{i}{\pi}\;(f(\pi)-f(-\pi))
$$
so we conclude $f(\pi)-f(-\pi) = 0$.  This tells us $a_n = b_n = 0$ for $n \ne 1$.  Write $A = a_1$.  Putting this all into $(1)$ we have
$$
f(x) = A\big(\cos(x) + i\sin(x)\big)
\tag7$$
Finally, we evaluate the constant $A$.  For that we use $f(0)=1$
in $(7)$ to get $A=1$.


答:
First note that you can't have a lone \$50 bill - they must always be paired with another to make a multiple of \$100. Then the same applies to \$20 bills, except they need to appear in packs of $5$, again to make a \$100 bundle.

This simplifies the issue considerably, since now you have 3 stacks of $10$ items: \$100 bills, \$50 pairs, \$20 bundles, with each item worth \$100.

So now it's a division with limits. We can enumerate picking $15$ items from $3$ classes with $\binom {15+2}2$. Then we can remove all cases when we pick more of than one class than the $10$ allowed by allocating $11$ items to each class in turn and counting how we can divide the remaining $4$ items.

So the result is $\binom {17}{2}-3\binom {6}{2}$.

Title: Combinatorics: Number of ways to make 1500 dollars

Body:
I have 50 bills of 20\$ and 20 bills of 50\$ and 10 bills of 100\$. With how many ways a bank machine could make the 1500\$ ?

what I think is :as the cofficient of $x^{1500}$
in  $(1+x+x^2+\ldots+x^{1000})^3$


问:
"Almost" ring homomorphism
This is an exercise out of Herstein which seems pretty straightforward but is eluding me.

Let $R,R'$ be rings and let $\phi:R\to R'$ be a mapping such that, for every $x,y\in R$:
$${\rm 1.}\qquad\phi(x+y)=\phi(x)+\phi(y)$$
$${\rm 2.}\qquad\phi(xy)=\phi(x)\phi(y)\qquad{\rm or}\qquad\phi(xy)=\phi(y)\phi(x)$$
I'm to show that one of the conditions holds uniformly. That is, for all $x,y\in R$, $\phi(xy)=\phi(x)\phi(y)$ or, for all $x,y\in R$, $\phi(xy)=\phi(y)\phi(x)$. Of course, this doesn't exclude the possibility that both conditions hold uniformly (which happens when e.g. $R'$ is commutative).

Herstein hints to fix $a\in R$ and to consider the sets
$$W_a=\{x\in R\mid\phi(ax)=\phi(a)\phi(x)\},$$
$$V_a=\{x\in R\mid\phi(ax)=\phi(x)\phi(a)\}.$$
I have shown that one of $W_a$ or $V_a$ must be the entire ring $R$. I think I'm missing something simple to complete the argument. Any hints?

Edit

I forgot to mention that in my copy of the text, the condition to be shown is stated as "$\phi(xy)=\phi(x)\phi(y)$ or $\phi(x)=\phi(y)\phi(x)$". I am pretty confident that it's a typo, but maybe not?

答:
Fact. If a group $G$ is a union of two subgroups $G = G_1 \cup G_2$, then $G=G_1$ or $G=G_2$.

We use this two times. Let $a \in R$. Then $W_a$ and $V_a$ are additive subgroups of $R$ with union $R$. Hence, $W_a=R$ or $V_a=R$. Now, $\{a \in R : W_a=R\}$ and $\{a \in R : V_a=R\}$ are additive subgroups of $R$ with union $R$. Hence, $\{a \in R : W_a=R\}$, i.e. $\phi$ is a homomorphism, or $\{a \in R :V_a=R\}$, i.e. $\phi$ is an anti-homomorphism.

Title: "Almost" ring homomorphism


问:
How to compare log and exponential

$x^{1/3}$

$　 (\log_{2}{x})^3$

How to compare them?

答:
Have you tried this? If not, take a look. Yes, there is no method to exactly compare them, unless you consider this special function $W_z(k)$, as a known function.

https://www.wolframalpha.com/input/?i=x%5E%281%2F3%29-log2%28x%29%5E3

If you refer to the behavior of log/power to infinity (i.e. for x sufficiently large), always the power $x^n, n>0$ goes faster to $\infty$ than $(log_mx)^p$, where $p,m>1$.

Title: How to compare log and exponential


问:
Function $\sqrt[5]{x}$
Wolfram alpha tell me that $\lim_{x\to 0-}\frac{\sqrt[5]{x}}{x}$ does not exists and in particular the domain of $\sqrt[5]{x}$ is given by all $x\geq 0$...why? I should say that $\lim_{x\to 0-}\frac{\sqrt[5]{x}}{x}=+\infty$ and the domain is all $\mathbb{R}$

答:
Extending StackTD's answer, the first thing to notice is that in the expression below, since $x$ is approaching $0$ from below, both the numerator and the denominator will be negative.  Therefore, the fraction is positive.

$$\lim_{x \to 0^-}\frac{\sqrt[5]{x}}{x}.$$

For simplicity, you can therefore convert the problem to evaluating:

$$\lim_{x \to 0^+}\frac{\sqrt[5]{x}}{x}.$$

At this point there are three ways to complete the evaluation, two formal and one informal:

divide both the numerator and the denominator by $\sqrt[5]{x}$, which converts the fraction to $\frac{1}{x^{(4/5)}}.$  Here, since the denominator is approaching $0$ from above, the fraction is clearly going to $+\infty.$

Apply L'Hopital's Rule, which transforms the fraction into
$\frac{[1/5]x^{(-4/5)}}{1}$, which, as in the first approach, clearly evaluates to $+\infty.$

Rely solely on intuition.  Consider what happens as $x$ progressively takes on the values $0.9, 0.8, 0.7, ...$ so that $x^2$ is simultaneously taking on the values $0.81, 0.64, 0.49, ...$.  Clearly as $x$ goes to $0^+$, $x^2$ goes to $0^+$ more quickly (told you this was informal).  Therefore, as $x$ goes to $0^+, \sqrt[5]{x}$ goes to $0^+$ more slowly.  This insight permits you to reach the accurate conclusion without (much) math.

Title: Function $\sqrt[5]{x}$


问:
Finding volume via integration
When we are asked to find the volume for example of a sphere,we slice it into numerous disks with small depth $dr$ and get the infinitesimal volume $dV$.And then we sum those slices together by putting an integral symbol.But my question is the basic integration we learnt was to find area under the curve.In that case we proved that the area is simply the anti derivative of the function.But here how do we use integration by putting an integral symbol before $Adr$ without proving that the derivative of the volume is the area of the disks?I am so disturbed by this as I didn't see any author giving the proof with proper intuition.

答:
What the Fundamental Theorem of Calculus actually tells us is that in general,
$$\int_a^b g(x)~dx=\lim_{\delta x\to0}\sum_{x=a}^{x=b}g(x)\delta x$$
Applying this to your example gives us the required formula.

This is how we can apply it volume. Suppose we have a function $f(x)$, and we want to revolve around the $x$ axis the area bounded by $x=a$, $x=b$, the $x$ axis and $y=f(x)$ to form a volume, $V$.

Consider a very small strip of the volume that is created, at a point $x$. This will be a cylinder with width $\delta x$ and its radius of its $2$ flat surfaces will be $f(x)$. So the volume of this cylinder is $$\pi r^2h= \pi [f(x)]^2\delta x$$

Hence, using the fact that for any function $g(x)$ we have
$$\int_a^b g(x)~dx=\lim_{\delta x\to0}\sum_{x=a}^{x=b}g(x)\delta x$$ we see that the total volume, $V$, that we are interested in is equal to
$$\lim_{\delta x\to0}\sum_{x=a}^{x=b}\pi[f(x)]^2\delta x=\pi\lim_{\delta x\to0}\sum_{x=a}^{x=b}[f(x)]^2\delta x$$
which is equal to
$$\pi\int_a^b[f(x)]^2~dx$$
as required. We can apply this very easily to a sphere by considering a circle's Cartesian equation: $x^2+y^2=r^2$

Title: Finding volume via integration


问:
How can I see a visual representation of the difference between Euclidean and Affine geometry?
I have just started studying different types of geometry with the first to I have covered being Euclidean geometry and Affine geometry. I am aware that Euclidean geometry is what we have used the most up until degree-level maths and Affine geometry is to do with the preservation of parallel lines but beyond this I am struggling to visualise the differences between these geometries. I was wondering if anybody knew a good way of explaining this or some resources to aid in my understanding.</p>

答:
First of all, it can be very useful to know that there is an even more general geometry including Euclidean and affine geometry: Projective geometry (shortly said: the geometry of perspective).

This said, I think that there at least two ways to consider this issue.

First point of view attached to figures :

Euclidean geometry: the image of a square can be a/any square.

Affine geometry: the image of a square is a/any parallelogram.

Projective geometry: the image of a square is a/any quadrilateral.

Second point of view attached to group of transformations (The "modern" point of view developed by Klein in the so-called Erlangen program):

Euclidean geometry is characterized by the group of transformations generated by rotations, symmetries, translations, all of them preserving (Eucledean) norm:

$$\begin{cases}x'&= &a x - \varepsilon b y + e\\y'&=& b x + \varepsilon a y +f\ \end{cases} \ \text{with} \ a^2+b^2=1 \ \ \text{or} \ \ \begin{cases}x'&=&x+ e\\y'&=&y +f\ \end{cases}$$

(with $\varepsilon = +1$ for a rotation, $\varepsilon = -1$ for a symmetry with respect to a straight line ; the second case deals with pure translations).

Affine geometry is characterized by any transformation of the form :

$$\begin{cases}x'&=&ax+by+e\\y'&=&cx+dy+f \end{cases}$$

Projective geometry is characterized by any transformation of the form :

$$\begin{cases}x'&=&(ax+by+e)/(gx+hy+i)\\y'&=&(cx+dy+f)/(gx+hy+i) \end{cases}$$

(please note that if, in the common denominator, we take $f=h=0$ and $i=1$, we are back to an affine transformation).


问:
Spherical coordinate integration help -- spheres.

$$L = \iiint_{D} \frac{1}{(x^2+y^2+z^2)^\frac{3}{2}}dxdydz $$
where D is a part of a sphere $x^2+y^2+z^2 \leq 1 $ that lies in the first octant between the planes $z=\frac{1}{2}$ and $z=\frac{1}{\sqrt 2}$

We are asked to find the solution to this, in spherical coordinates and I have found a expression for $L $ but I'm not sure how to go about completing the solution.

My thoughts were to decompose the integral and use spherical coordinates and then express $L$ as the following.

$$L = \int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{4}} \int_{\frac{1}{2 \cos \phi}}^{\frac{1}{\sqrt 2 \cos \phi}} \rho^{1/2} \sin  \phi d \rho d\phi d \theta + \int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{4}} \int_{\frac{1}{2 \cos \phi}}^{1} \rho^{1/2} \sin  \phi d \rho d\phi d \theta$$

Using geometry and my question is, is this correct? - I based it off a sketch off $z$ against $r$ and then saw we have to decompose this accordingly ^ and then I don't know how to complete the integral i.e. find out what the value of $L$ is.

Could someone help me with this?

答:
I see two errors.

One error is that you let $\phi$ cover the same range of angles in both integrals. You have one integral that covers all the region in the first octant within the solid cone $z^2 \geq x^2 + y^2$ and between the planes $z=\frac12$ and $z =\frac1{\sqrt2},$
which covers part of the region you are trying to integrate.
This is fine.
But the second integral covers the region in the first octant within the same cone, above the plane $z=\frac12$ and within the sphere.
This provides double coverage on the region you already integrated and includes a region above the plane $z =\frac1{\sqrt2},$ which should have been excluded.

The other error appears to be in converting the expression
$(x^2 + y^2 + z^2)^{3/2}.$
Remember that in the conversion to spherical coordinates,
$\rho = (x^2 + y^2 + z^2)^{1/2}.$
You seem to have taken $\rho$ to be equal to $x^2 + y^2 + z^2$ instead.


问:
Must the vector multiplied by an orthogonal matrix be unique in order to get a unique resultant vector?
Let's say I have a Hadamard matrix $W$ and a vector $y$ such that $W*x = y$ where $y$ is fixed and supposed to be known, and $x$ is a vector too. my question, how many possible vectors of $x$ can be exist in order to get the same result $y$?

For example, let's have $W = [1, -1; -1, 1]$ the vector $x$ can be either [1; 1] or [-1; -1] and the result of both multiplication is unique which is $y = 0$.

Is there a general rule for that ?

答: The kernel $\ker(W) := \{x \in \mathbb{R}^n : Wx = 0\}$ is always a vector space. That means, if $Wx = 0$ has a solution that is different from $x = 0$, you will automatically have infinitely many solutions. Indeed, If $x$ solves $Wx = 0$, then also $W(\lambda x) = \lambda Wx = 0$ for all $\lambda \in \mathbb{R}$. Likewise, if $x$ is a solution of $Wx = y$, then $x' = x+v$ is also a solution for all $v \in \ker(W)$. So it all depends on the dimension of $\ker(W)$. If $\ker(W) = \{0\}$, then the solution is unique.


问:
Writing as integral
I wanted to write $$\prod_{x=3}^9 (x)$$ so I wrote it as $$\frac{9!}{2!}$$ which I further wrote as $$\frac{\int_0 ^\infty t^9 e^{-t} dt}{\int_0 ^\infty z^2 e^{-z} dz}$$

Is this way of writing correct?

Please Help

答:
If you write out the product $\prod_{x=3}^9(x)$ in full, you should get $3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9$ which is equivalent to $\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9}{1\cdot2}$ or $\frac{9!}{2!}$ not $\frac{9!}{3!}$ so the correct use of the Pi/Gamma function would be $\frac{\int^\infty_0t^9e^{-t}dt}{\int^\infty_0z^2e^{-z}dz}$.
Other than the small error in the original, this is all equivalent.

Title: Writing as integral


问:
$X$ and $Y$ be i.i.d $U(0,1)$ random variables. Then $E(X|X>Y)$

Let $X$ and $Y$ be i.i.d $U(0,1)$ random variables. Then $E(X|X>Y)$
equals?

The way I interpreted it is as follows:

$E(X|X>Y)$ is $E(X)$ realizing $X>Y$

so $E(X)=\int_{0}^{1}\int_{y}^{1}xdxdy=\int_{0}^{1}(\frac{x^2}{2})_{y}^{1}dy=\frac{1}{2}\int_{0}^{1}(1-y^2)dy=\frac{1}{2}(y-\frac{y^3}{3})_{0}^{1}=\frac{1}{2}\cdot\frac{2}{3}=\frac{1}{3}$

What can be a better alternative to solve this problem? I tried so hard but couldn't get the answer using the proper definition of expectation, can anyone help me?And can we use law of total expectation here?

答:
The cdf of $\mathcal U(0,1)$ is $U(x) = x$, and the the pdf $u(x) = 1$.

Note that $\mathbb E[X|X>Y] = 2\mathbb E[X1_{\{X>Y\}}]$ since $$\mathbb E[X|X>Y] = \frac{\mathbb E[X1_{\{X>Y\}}]}{\mathbb P(X>Y)},$$ and $\mathbb P(X>Y) = \frac{1}{2}$. Moreover, $$\mathbb E[X1_{\{X>Y\}}] = \mathbb E[\mathbb E[X1_{\{X>Y\}}]|X] = \mathbb E[X\mathbb P(X>Y|X)]$$ using the law of iterated expectations. Since $\mathbb P(X>Y|X) =U(X)$, we obtain $$\mathbb E[X1_{\{X>Y\}}] = \mathbb E[XU(X)] = \int_0^1xU(x)u(x)\,\mathrm dx = \int_0^1x^2\,\mathrm dx = \frac{1}{3}$$


问:
Boy's surface and normal vectors
Denote by $X$ the image under the immersion $RP^2 \to R^3$ (that is, the Boy's surface). At each point $x \in X$ consider a small orthogonal segment that is centered at the point $x$. Denote by $S$ the set of all endpoints of these segments. Thus, we obtain a new surface $S$. What is this?
I think it should be $S^2$ but I cannot prove this. This construction looks like orientable double cover but there is a problem: the Boy's surface has self-intersection points (the whole curve of such points), therefore for the natural map $p: S \to X$ the preimage $p^{-1} (x)$ for any self-intersection point $x$ consists of 4 points instead of 2. So, it is not a covering or even not a ramified covering with finite number of ramification points, so we cannot use Riemann-Hurwitz formula to find the Euler characteristic of $S$. Am I missing something? How can it be done?

答:
Say your surface is parametrized by some smooth function $f(x,y)$ (this can be said because Boy's surface is a smooth immersion of $\mathbb RP^2$ into $\mathbb R^3$). Then you consider
$$
g_\varepsilon(x,y) = f(x,y) + \varepsilon \cdot \frac{\partial_xf(x,y)\times \partial_yf(x,y)}{\|\partial_xf(x,y)\times \partial_yf(x,y)\|}=:f(x,y) + \varepsilon \cdot n(x,y),
$$
so $g_\varepsilon$ is just a normal shift of $f$. Since $f$ is an immersion, the vector $n(x,y)$ is always defined and its length equals $1$ (and moreover, $n$ is smooth as well). Therefore you conclude that
$
g_\varepsilon - f = \varepsilon n
$
becomes arbitrarily small if $\varepsilon \to 0$. Regarding first derivatives you see that first derivatives of the difference between $g_\varepsilon - f$ are given by
$\varepsilon\cdot \partial_xn$ and $\varepsilon \cdot \partial_yn$ respectively. Again they become arbitrarily small, provided $\varepsilon \to 0$ and you conclude that $g_\varepsilon$ and $f$ are $C^1$-close. From there you get that $g$ is again an immersion since immersions form an open set in the $C^1$-topology.

Title: Boy's surface and normal vectors


问:
Applying Urysohn Lemma on $\mathbb{R^2}$
$A_1=\text{ Closed Unit Disk}$, $A_2=\{(1,y):y\in\mathbb{R}\}$, $A_3=\{(0,2)\}$.
Then there always exists a real-valued continuous function on $\mathbb{R}^2$ such that

$f(x)=a_j$ for $x\in A_j$, $j=1,2,3$

Iff at least two of the numbers $a_1,a_2,a_3$ are equal.

if $a_1=a_2=a_3$

$\forall a_i\in\mathbb{R}$

iff $a_1=a_2$

As all three sets are closed and we are in normal space, we can apply Urysohn's lemma to ensure the existence of such continuous function, so I see that as $A_1$ and $A_2$  are not disjoint, so $4$ is true, am I right?

答: You are right. Note that 2. is also true. </p>


问:
Sequence of non-constant integer polynomials whose limit is constant
Does there exist a sequence of non-constant polynomials $\big \{p_{n} \big \}_{n \geq 1}$, each $p_n$ in $\mathbb{Z}[X]$, such that $$ \lim_{n \to \infty} p_n =c $$ pointwise on $[0,1]$, where $c$ is a constant?

We can find such polynomials easily on the open interval $(0,1)$, such as $p_n =x^n$, however the closed interval does not seem as simple. The answer seems to be no, but I wouldn't know where to start.

答: Yes. Consider $p_n(x) = x^{2n} - x^n + 1$. Note that all we need to do is find a sequence of polynomials such that $p_n (1) = p_n (0)$ for all sufficiently large $n$, and with the degrees of all monomial terms going off to infinity.


问:
Integral : $\int_0^{\infty} \frac{1-\cos(ax)}{x^2} dx. (a>0) $
I have to calculate $\displaystyle\int_0^{\infty} \dfrac{1-\cos(ax)}{x^2} dx. (a>0) $

My attempt:

Let $f(z):= \dfrac{1-e^{iaz}}{z^2}.$

$C_1 : z=t, t:0 \to R$

$C_R : z=Re^{i\theta}, \theta : 0 \to 2\pi $

$C_2 : z=te^{2\pi i}, t:R \to 0$

$C_r : z=re^{i\theta}, \theta : 2\pi \to 0$

$(R>r)$

$f(z)$ is a regular analytic function, so I can use Cauchy's Integral Theorem.

$\displaystyle\int_{C_1} f(z)dz
+\displaystyle\int_{C_R} f(z)dz
+\displaystyle\int_{C_2} f(z)dz
+\displaystyle\int_{C_r} f(z)dz
=0.
$

But this didn't work because $\lim_{r\to 0} \displaystyle\int_{C_r} f(z)dz \neq 0$.

How should I set $f(z)$ and Integral route?

Body:
First note that
$$
I(a) = \int_0^{ + \infty } {\frac{{1 - \cos (ax)}}{x^2}dx}  = a\int_0^{ + \infty } {\frac{{1 - \cos t}}{t^2}dt}  = aI(1).
$$
Now
$$
I(1) = I'(a) = \int_0^{ + \infty } {\frac{{\sin (ax)}}{x}dx}  = \int_0^{ + \infty } {\frac{{\sin t}}{t}dt}  = \frac{\pi }{2}.
$$
Hence, $I(a)=\frac{\pi }{2} a$.


问:
Convex hull and number of vertices
Let $S_N \subseteq \mathbb{R}^n$ be the convex hull of a finite set of $N$ points in $\mathbb{R}^n$, i.e., $S_N := \mathrm{conv}(\{x_1, \ldots, x_N\})$, and let $V_N$ be the associated set of extreme points (vertices).
If one randomly adds an additional point (different from the previous ones), say $x_{N+1} \in \mathbb{R}^n$, can it happen that $V_{N+1} \subset V_{N} \cup \{x_{N+1}\}$, where $V_{N+1}$ is the set of vertices associated to $S_{N+1} := \mathrm{conv}(S_{N} \cup \{x_{N+1}\})$?
In summary, is it true that, in adding a new point, what can only happen is that either $V_{N+1} = V_{N}$, or $V_{N+1} = V_{N} \cup \{x_{N+1}\}$, and therefore the new point does not cut-off any vertex of $S_N$?

I know the answer is trivial, but I'm not able how to work this out.

答:
If $x_N\in\text{conv}(V_n)$, then $V_{N+1}=V_N$.

If $x_N\notin\text{conv}(V_n)$, then $x_N$ may eliminate from $0$ to $N-n+1$ points (worst case when it reduces the hull to a simplex).

Title: Convex hull and number of vertices


问:
Expression of the partial sum of a sequence
I was working on the following sequence
$$
u_n=\sum_{k=0}^{n}\frac{1}{q_0 \times q_1 \times \dots \times q_k}
$$
where $\left(q_n\right)$ is an increasing sequence with $q_0 \geq 2$. I've shown that $\left(u_n\right)$ converges.

By trying $q_n=2^{n+1}$ for example, we would have
$$
u_n=\sum_{k=0}^{n}\frac{1}{2 \times 2^2 \times 2^{k+1}}=\sum_{k=0}^{n}\frac{1}{2^{1+2+\dots+\left(k+1\right)}}=\sum_{k=0}^{n}\frac{1}{2^{\left(k+1\right)\left(k+2\right)/2}}
$$

Is there a way to compute this for all $n$, or to find its limit, apparently it approaches $0,641633$.

答:
You have
$$u_n=\frac{\left\lfloor c\,\, 2^{\frac{1}{2} (n+1) (n+2)}\right\rfloor } {2^{\frac{1}{2} (n+1) (n+2)} }$$ where $c$ is given in $A190405 $ in $OEIS$ as already commented by @player3236.

If you want to make $c$ rational

$$c \sim \frac{436459}{680232}$$ is in error of $6.23 \times 10^{-13}$


问:
Proof for the Ratio Test using the squeeze theorem
Excerpt from my University's lecture notes

The assumption in the excerpt above is that $a_n >0$.

In the above image, at the bottom it mentions "so $a_n \rightarrow 0$ by the sandwich theorem" but I don't see anywhere that I could apply the squeeze theorem. Is the squeeze theorem being applied to this set of inequality?
$$
0\leq a_{N+s} \leq y^{N+s}\left(\frac{a_N}{y^N}\right)
$$

If I'm missing out any piece of information, please let me know. I've been trying to understand this for hours and I've looked at many resources but I do not understand where or how the squeeze theorem is being used. Thank you in advance.

答:
Yes, the squeeze theorem implies that $\lim_{s \to \infty} a_{N+s} = 0$ since the we can 'squeeze it' between two sequences which converge to zero as $s \to \infty.$
These are the constant sequence $0$ on the left, and $y^sa_N =y^{N+s}\left(\frac{a_N}{y^N}\right)$ on the right.

From $\lim_{s \to \infty} a_{N+s} = 0$ it easily follows that $\lim_{n \to \infty} a_n=0$, as you can check from the definition of the limit of a sequence.


问:
Let $K=\mathbb{Z}_2[x]/\langle x^4+x^2+x \rangle$ Find a $g\in K$ such that $g^2=x^3+x+1$
Let $K=\mathbb{Z}_2[x]/\langle x^4+x^2+x \rangle$. Find a $g\in K$ such that $g^2=x^3+x+1.$
I tried $x^3+x+1$ itself but unfortunately the degree is only $2$. I don't know how I can multiply something and get a polynomial of degree $3$.

答:
A nice trick to know.

$\newcommand{\d}{\frac{\mathbb Z_2[x]}{\langle x^4+x^2+x\rangle}}$In $\mathbb Z_2[x]$, we have the nice identity that $(a+b)^2 = a^2+2ab+b^2 = a^2+b^2$, because $2ab \equiv 0$ in $\mathbb Z_2[x]$! This naturally becomes $(a+b+c)^2 = a^2+b^2+c^2$ in $\mathbb Z_2[x]$ as well.

We can use this to great advantage as follows : let $p$ be an element satisfying $p^2 = x$ in the ring $\d$. Then $(p^3)^2 = x^3$ and $1^2 = 1$ therefore $(p^3+p+1)^2 = p^6+p^2+1 = x^3+x+1$ in $\d$.

So let's search for such a $p$ : $p^2 = x + h(x)(x^4+x^2+x)$ for some polynomial $h \in \mathbb Z_2[x]$. We start with the simplest $h$ (note $h = 0$ is not useful) , i.e. $h \equiv 1$. For this , we get the RHS as $$x^4+x^2+2x = x^4+x^2 = x^2(x^2+1) = x^2(x+1)^2 = [x(x+1)]^2$$

(The step $x^2+1 = (x+1)^2$ follows from the same $\mathbb Z_2$ identity I discussed earlier)

Et voilà! So $p = x(x+1)$, and $p^3+p+1$ provides the required square root.

Confirmation :
$$
p^3+p+1 = x^3(x+1)^3 + x(x+1)+1 = x^6 + x^5+x^4+x^3+x^2+x+1
$$

modulo $x^4+x^2 + x$ this reduces to $x^3+x+1$, whose square is in fact $(x^3+x+1)^2 = x^3+x+1$.


问:
Some questions about determinants
I'm studying some Linear Algebra and determinants are being a real struggle for me. I can't understand them at all, but my biggest problem is with all the equivalent definitions and why they are in fact equivalent.

Given $V, W$ vector spaces, we define the tensor product $\otimes$ of $f \colon V^k \to \mathbb{R}$ and $g \colon W^l \to \mathbb{R}$ as $(f \otimes g )(v_1, \dots, v_{k+l})$ $= f(v_1, \dots, v_k)g(v_{k+1}, \dots, v_{k+l})$. Also, we define $V \otimes W = \langle v \otimes w \colon v \in V, w \in W \rangle$ (note that the tensor product of vectors is defined, since $V = V^{**}$, which is a space of functions).

Given $V$ a vector space, we define $X = V^{\otimes k} = V \otimes V \otimes \dots \otimes V$ ($k$ times) and $Y =$ $\langle v_1 \otimes \dots \otimes v_k \colon \exists i,j \textrm{ } v_i = v_j \rangle$ (i.e. the set generated by the tensor product of $k$ vectors in which at least two are equal). We define $\wedge^k V = X/Y$ (call it exterior $k-$power of $V$) and denote $\overline{u_1 \otimes \dots \otimes u_k} = u_1 \wedge \dots \wedge u_k$.

Given those important definitions of tensor product and exterior power of a space, let's introduce the definitions of determinant.

Given $V$ a vector space with $\dim V = n$ and $T\colon V \to V$ a linear transformation, note that $\dim \wedge^n V = 1$. Hence, the linear transformation $\wedge^n T \colon \wedge^n V \to \wedge^n V$ must be of the form $\wedge^n T(v) = cv$ for some constant $c$. This constant is called the determinant of $T$ and denoted $c = \det T$.

The first thing that bothers me here is the assumption that $\wedge^n T(v) = cv$ for some constant $c$, why is that? It seems trivial but I'm not getting it.

The usual way we see determinants (in high school) is by beeing introduced to determinants of matrices, not of linear operators, but we can translate this definition to matrix just by saying that the determinant of $T$ is the determinant of it's matrix in some basis. My first question is: why is this definition independent of the chosen basis?

Given a matrix $M = [m_{ij}]_{n \times n}$ we define its determinant $$\det T = \sum_{\sigma \in S_n} \textrm{sgn}(\sigma) m_{i\sigma(1)}\cdots m_{i\sigma(n)}.$$

Now given both definitions, why are those equivalent?

Thanks in advance and if there is anything I can do to make this question better, let me know please!

答:
I'll sketch how you can get from the exterior power perspective to the sum over permutations perspective, but the treatment is rather rushed and may be hard to follow or appreciate without proper study of the subject. The key idea at work is that, working concretely with the vector space $\mathbf{R}^n$, the determinant is the unique alternating multilinear map taking $n$ vectors in $\mathbf{R}^n$ to a real number in $\mathbf{R}$ that sends the standard basis $(e_1,\dots,e_n)$ to $1$. I'll explain those terms as we go along. Also, it's very much worth noting that the determinant can be understood as the signed volume of a parallelepiped. (For intuition for the geometric perspective of linear algebra I recommend 3Blue1Brown's video series Essence of Linear Algebra.)

It is straightforward to verify that the tensor product is multilinear — that is, when all inputs but one are fixed, we get a linear map. So $v\otimes(u+w)=v\otimes u+v\otimes w$ and $(cv)\otimes w=c(v\otimes w)$. (In fact, the tensor product is in some sense the 'freest' such product.) Consider $(v+w)\otimes(v+w)$ under the $\wedge$ map. We find $$(v+w)\wedge(v+w)=v\wedge v+v\wedge w+w\wedge v+w\wedge w$$ by multilinearity. By definition of $\wedge$, the terms $v\wedge v$, $w\wedge w$ and $(v+w)\wedge(v+w)$ are all equal to zero, and so we have $$v\wedge w=-w\wedge v$$ in the exterior algebra. In fact, for the same reason, whenever we swap two terms in some wedge of vectors, a negative sign appears. So we can write things like $u\wedge v\wedge w=-u\wedge w\wedge v=w\wedge u\wedge v$. Roughly speaking, a map that has this property of picking up a minus sign every time we swap elements is said to be alternating. Notice that if we have a permutation of $n$ letters $\sigma\in S_n$ and $n$ vectors $v_1$, ..., $v_n$, we have the formula
$$v_{\sigma(1)}\wedge\dots\wedge v_{\sigma(n)}=\operatorname{sgn}(\sigma)v_1\wedge\dots\wedge v_n,$$
where $\operatorname{sgn}(\sigma)$ is the sign of the permutation $\sigma$, since we can think of $\operatorname{sgn}(\sigma)$ as $(-1)^{T(\sigma)}$, where $T(\sigma)$ counts the number of transpositions of $\sigma$; that is, the number of swaps made.

A brief digression. You wrote:

It seems trivial but I'm not getting it.

I think this tensor products and exterior powers stuff is not very trivial, and is best appreciated within a larger framework (say, as a tool for differential geometry; see the references I provide at the end), as opposed to being contrived into a short two-paragraph definition that cuts through the theory to produce a definition of the determinant that I find rather unenlightening to people who do not already know the theory of tensor products and exterior powers. Okay, back to math.

Now if $V$ is an $n$-dimensional vector space, then the $n$-th exterior power of $V$ is denoted $\Lambda^nV$ and is by definition a vector space consisting of elements that are linear combinations of objects like $v_1\wedge\dots\wedge v_n$. If $\{e_1,\dots,e_n\}$ is a basis for $V$, then we may expand vectors like $v_j$ in the basis, so that $v_j=\sum_{1\le i\le n}v_{i,j}e_i$. By multilinearity we can now compute
\begin{align*}
v_1\wedge\dots\wedge v_n
&=\left(\sum_{1\le i_1\le n}v_{i_1,1}e_{i_1}\right)
\wedge\dots\wedge
\left(\sum_{1\le i_n\le n}v_{i_n,n}e_{i_n}\right)\\
&=\sum_{1\le i_1\le n}\dots\sum_{1\le i_n\le n}v_{i_1,1}\dots v_{i_n,n}
e_{i_1}\wedge\dots\wedge e_{i_n}.
\end{align*}
This is quite a big sum, with $n^n$ terms. But remember that the wedge product $\wedge$ is defined so that, whenever we have a repeated vector, we must get zero. (So things like $v\wedge w\wedge v=0$ since $v$ is repeated.) And so most of the terms $e_{i_1}\wedge\dots\wedge e_{i_n}$ vanish, except when $i_1$, ..., $i_n$ is a permutation of $1$, ..., $n$. So we have
\begin{align*}
&\sum_{1\le i_1\le n}\dots\sum_{1\le i_n\le n}v_{i_1,1}\dots v_{i_n,n}
e_{i_1}\wedge\dots\wedge e_{i_n}\\
&\quad=\sum_{\sigma\in S_n}v_{\sigma(1),1}\dots v_{\sigma(n),n}e_{\sigma(1)}\wedge\dots\wedge e_{\sigma(n)}\\
&=\left(\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)v_{\sigma(1),1}\dots v_{\sigma(n),n}\right)e_1\wedge\dots\wedge e_n.
\end{align*}
Interesting. This large computation revealed to us that every element of $\Lambda^nV$ is a constant multiple of $e_1\wedge\dots\wedge e_n$; that is, $\dim(\Lambda^nV)=1$.
And so if we have a linear map $T\colon V\to V$, and if $$\Lambda^nT:\Lambda^nV\to\Lambda^nV$$ is defined by sending $v_1\wedge\dots\wedge v_n$ to $T(v_1)\wedge\dots\wedge T(v_n)$, then it is just multiplication by a scalar. We call that scalar the determinant of $T$; that is, $T(e_1)\wedge\dots\wedge T(e_n)=\det(T)e_1\wedge\dots\wedge e_n$, by definition. But as we computed earlier, we already know what that scalar is:

$$\det(T)=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)T_{\sigma(1),1}\dots T_{\sigma(n),n}$$

(The fact that we get $T_{\sigma(j),j}$ instead of $T_{j,\sigma(j)}$ is inconsequential — we may multiply by $\sigma^{-1}$ and everything works out since $S_n$ is a group. This also uses the fact that a matrix can be thought of as representing some linear map with respect to chosen bases, and understanding the interplay between the abstract linear maps between vector spaces perspective and the concrete matrices-as-arrays-of-numbers perspective is very useful. It is important to know that matrices represent linear maps, though there are also many other useful interpretations of matrices, fit for different occasions, like as the adjacency matrix of a graph, or as encoding weights in a weighted directed graph, in which case the determinant can be understood combinatorially as counting non-intersecting systems of paths via the Lindström–Gessel–Viennot Lemma! But I am digressing...)

One last thing. We did all those computations in terms of this abstract construction of a quotient of tensor spaces and whatnot, but these ideas can be dealt with more concretely, thinking about areas and volumes of parallelograms and parallelepipeds in 2D and 3D space. Let's return to the idea of an alternating multilinear map
$$\det\colon\underbrace{\mathbf{R}^n\times\dots\times\mathbf{R}^n}_{\hbox{$n$ copies}}\to\mathbf{R}$$
that sends $(e_1,\dots,e_n)$ to $1$. You'll find that the requirements of alternation and multilinearity are rather natural requirements for any map that takes $n$ vectors in $n$-dimensional space and returns the hypervolume of a parallelotope. (For example, in $\mathbf{R}^2$, this is a map that takes two vectors $v$ and $w$, and returns the area of the parallelogram that they span. If you stretch one of the vectors by a factor of $c$, you would expect the area of the parallelogram they span to increase by that same factor; so $\det(cv,w)=c\det(v,w)$. If a vector is repeated, like in $\det(v,v)$, then we just get a line with no area and so $\det(v,v)=0$. Similarly, you would expect the area of the unit square to be $1$; that is, $\det(e_1,e_2)=1$. And so all the relevant axioms are motivated geometrically and concretely.) From here we may then repeat what is really the same computation as above, just dressed up differently:
\begin{align*}
\det(v_1,\dots, v_n)
&=\det\left(\sum_{1\le i_1\le n}v_{i_1,1}e_{i_1},\dots,
\sum_{1\le i_n\le n}v_{i_n,n}e_{i_n}\right)\\
&=\sum_{1\le i_1\le n}\dots\sum_{1\le i_n\le n}v_{i_1,1}\dots v_{i_n,n}
\det(e_{i_1},\dots, e_{i_n})\\
&=\dots
\end{align*}
And so the whole abstract wedge product business is really just the essence of a determinant, abstracted out into its own algebraic thing.

Just for the fun of it, and to end off, let's work out the computation of the determinant in the plane. Let $e_1=\begin{pmatrix}1\\0\end{pmatrix}$ and $e_2=\begin{pmatrix}0\\1\end{pmatrix}$. Then
\begin{align*}
\det\left(\begin{pmatrix}a&b\\c&d\end{pmatrix}\right)
&=\det\left(\begin{pmatrix}a\\c\end{pmatrix},
\begin{pmatrix}b\\d\end{pmatrix}\right)\\
&=\det(ae_1+ce_2,be_1+de_2)\\
&=\det(ae_1,be_1)+\det(ae_1,de_2)+\det(ce_2,be_1)+\det(ce_2,de_2)\\
&=ab\det(e_1,e_1)+ad\det(e_1,e_2)+bc\det(e_2,e_1)+cd\det(e_2,e_2)\\
&=ad\det(e_1,e_2)+bc\det(e_2,e_1)\\
&=ad-bc,
\end{align*}
just as we learned in high school. (The fact that $\det(e_2,e_1)=-1$ can be interpreted as the idea that the square going from $e_2$ to $e_1$ has orientation opposite to that of the unit square which goes from $e_1$ to $e_2$. But that's a story for another day.)

In conclusion, I think an ideal place to learn these ideas would be a text on manifolds, since the tools of tensors and exterior powers are part of the underlying machinery for geometry.
See John Lee's Introduction to Smooth Manifolds or Loring Tu's An Introduction to Manifolds. See also Keith Conrad's notes on Tensor Products and Exterior Powers.

Title: Some questions about determinants


问:
Show that $e^{ix}=\cos(x)+i\sin(x)$ using the Fourier Series only
As stated in the title.

Any arbitrary function can be expressed as
$$f(x)=\frac{a_0}{2}+\sum^{\infty}_{n=1}(a_n\cos(nx)+b_n\sin(nx)) \tag{1}$$
where
$$a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(nx)dx \tag{2}$$
$$b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(nx)dx \tag{3}$$

I am trying to show that (1)-(3) are consistent with $f(x)=e^{ix}=\cos(x)+i\sin(x)$.

However, for $a_n=\frac{1}{\pi}\int^{\pi}_{-\pi}e^{ix}\cos(nx)dx$ I got $a_n=0$, similarly, $b_n=0$.

What's gone wrong? Are the integral limits $\int_{-\pi}^{\pi}$ not valid in general, does it require modifications for non-periodic functions?

答:
If you allow for complex coefficients, then the Fourier-series of $x\mapsto \cos (x) + \mathrm i \sin(x)$ is just itself (by uniqueness of the Fourier-series).
You can formally compute the Fourier-coefficients of $x\mapsto \mathrm e^{\mathrm i x}$ using integration by parts twice and you obtain (for $k>1$)
$$
a_k = -\frac{2k \sin(k \pi)}{\pi(k^2-1)}=0,
$$
since $k\in\mathbb N$ and $a_0 = 0$ and $a_1 = 1$. The completely similar computation (again for $k>1$) shows
$$
b_k = -\frac{2\mathrm i \sin(k\pi)}{\pi(k^2-1)}=0
$$
and $b_0=0$ and $b_1 =\mathrm i$. Therefore the Forier-series of the two functions agree and they are therefore the same.

EDIT: GEdgar made a very valid point. In order to evaluate the integrals, one can use the ODE she or he had given as mentioned in the comments but one gets another proof of the identity along the way. So I propose another way of doing it:

We use that $f(x) = \mathrm e^{\mathrm i x}$ satisfies the ODE $f' = \mathrm i f$ with the initial datum $f(0)=1$. If $f$ has the Fourier-series
$$
f(x) = \frac{a_0}{2}+\sum_{n=1}^\infty (a_n\cos(nx) + b_n\sin(nx))
$$
you can take a formal derivative and see that
$$
f'(x) = \sum_{n=1}^\infty (-na_n\sin(nx) + nb_n\cos(nx))
$$
Using the ODE above you get $a_0=0$ and the relations (for $n>0$): $- n a_n = \mathrm ib_n$ and $nb_n = \mathrm i a_n$.
From there you get $\mathrm i n^2b_n=\mathrm i b_n$ and therefore $b_n=0$ for $n>1$. Similarly you get $a_n=0$ for $n>1$. If $n=1$ one finds $-a_1=\mathrm ib_1$ and $b_1 = \mathrm i a_1$. But now we know that $f(0) =a_1 = 1$ and therefore $b_1 = \mathrm i$.


答:
Consider a sphere, centred at the origin, with a radius of $R$. Now think of slicing the sphere into discs of thickness $dr$. The disc has a surface area of $\pi r_i^2$, where $r_i$ is the radius of a single disc, so has volume $\pi r_i^2 dr$.

What is $r_i$? It is found by realising that $r_i$ is the horizontal side of a right-angled triangle with $R$ being the hypotenuse and $r$ being the vertical side. Draw it out to convince yourself.

To find the volume of the sphere, we have to add up all of the $\pi r_i^2dr$. This is an integral over $r$, which ranges from $-R$ to $R$:
$$
V = \int_{-R}^R \pi (R^2 - r^2)dr.
$$
Can you take it from here?

Title: Finding volume via integration

Body: When we are asked to find the volume for example of a sphere,we slice it into numerous disks with small depth $dr$ and get the infinitesimal volume $dV$.And then we sum those slices together by putting an integral symbol.But my question is the basic integration we learnt was to find area under the curve.In that case we proved that the area is simply the anti derivative of the function.But here how do we use integration by putting an integral symbol before $Adr$ without proving that the derivative of the volume is the area of the disks?I am so disturbed by this as I didn't see any author giving the proof with proper intuition.


问:
Calculate this limit without using L'Hôpital rule
the limit is, I can't find where to start

$$\lim_{x\rightarrow 0} \frac{\arcsin\frac{x}{\sqrt{1-x^{2}}} }{\ln (1-x)}$$

your help is appreciated

答:
HINT

Note that the Taylor expansion for both sides of the fraction looks like
$$
\begin{split}
\ln(1-x) &= -x -\frac{x^2}{2} - \frac{x^3}{3} - \ldots\\
\arcsin(u) &= u + \frac{u^3}{6} + \ldots
\end{split}
$$


问:
Describe the open sets in the topology $(\mathbb{R}^2,\tau)$
From Topology Without Tears:

Let $S$ be the collection of all the circles in the plane.If $S$ is a subbasis for  a topology $\tau$ on $\mathbb{R}^2$.Describe the open sets in $(\mathbb{R}^2,\tau)$.

For two different circles,their intersection will be either one point or two points.So,considering intersection of all two possible circles that intersect at a single point,the basis will include singleton sets $\{r\} ,\forall r\in \mathbb{R}^2$

So,our topology is discrete topology.so,all subsets of $\mathbb{R}^2$ will be open.Am I correct?

Thanks in advance!

答: Any point in the plane is the kissing point of two circles, i.e. their intersection.  Looks good!  Since the topology generated by a sub-basis is all unions of finite intersections, we are done.</p>


问:
Unique attracting set is the whole space?
I'm looking for a simple example in which $\Omega(f) \neq X$ but the unique attracting set is $X$.

$(X,f)$ is a Dynamical System if $f:X \to X$ is a homeomorphism and $X$ is a compact space. $\Omega(f)$ is the set of non wandering points, i.e. all $x$ such that $\forall$ U open containing $x$ and $\forall$ $N>0$ there exists some $n>N$ such that $f^n(U) \cap U \ne \emptyset$.

An open set $U \subseteq X$ is called a trapping region of $f$ if $f(\overline{U}) \subseteq U$. A compact invariant set $A \subseteq X$ is said to be attracting of $f$ if it has a neighborhood $U$ which is a trapping region such that $A = \bigcap_{n \geq 0}f^{n}(\overline{U})  $.

答:
I think you can take $x \mapsto x^2$ on the torus $\mathbb{R}/\mathbb{Z}$.

The only point in $\Omega(f)$ is $0$. Indeed, every fixed point is in the non-wandering set, and $0$ is a fixed point. For every over point, $x \in ]0;1[$ you can take a small intervall $I=]x- \epsilon; x+ \epsilon[$ aroud it, such that $0 \not\in I$. Then for $N$ big enough, $(x+\epsilon)^N \leq x-\epsilon$ so $N \geq \frac{\ln(x-\epsilon)}{\ln(x+\epsilon)}$ and for all $n \geq N$, $f^n(I) \subset[0;x-\epsilon[$ which has empty intersection with $I$. In conclusion $\Omega(f)=x$

Now if $U$ is atrapping region, then we have two case

$0 \in U$, then there is a $\epsilon$ such that $]1- \epsilon;1] \subset U$, taking image with $f$ you find that $U = X$

$0 \not\in U$ in this case you can do the same argument as above and this case is not

So the only trapping region is $X$ and then the only attractor is $X$.


问:
indirectly convex
Let g:$\mathbb{R^2}$$\rightarrow$$\mathbb{R}$ be defined by g(x)=Max{$x_1$,$x_2$} at each x=($x_1$,$x_2$)$\in$ $\mathbb{R}$. Determine whether or not g is indirectly convex on $\mathbb{R^2}$.
definition:
let f:$\mathbb{R^n}$$\rightarrow$$\mathbb{R}$ be a function. We say that f is indirectly convex if there exist F:$\mathbb{R^n}$$\rightarrow$$\mathbb{R}$ and h:$\mathbb{R}$$\rightarrow$$\mathbb{R}$ such that F is convex, h is increasing with $f=h \circ F$.</p>

答:
Convexity implies direct convexity, and $g$ is convex. So simply choosing $F=g$ and $h(x)=x$ will suffice.

EDIT: coffeemath makes a fair point: what if you don't know that $\max$ is convex; that is, what if you have to prove it? I omitted that because in my experience, $\max$ is one of the first functions you get to put in your "atomic convex function toolbox." But if you can't do that yet, his approach is simple: $\max\{a,b\}=\tfrac{1}{2}(a+b+|a-b|)$, which is the sum of a linear function $a+b$ and a convex function $|a-b|$, scaled by a positive constant. Hence it is convex.

Title: indirectly convex


问:
Generator find the coefficient
I have this : $$6a(n)=-2^n+a(n+1), a(1)=4$$

what is the coefficient  of this ? I beleive 1) 5*6/4  what do you think?

1)$ (5*6)/4 $

2)$ (17*6)/4 $

3)$  (15*6)/4 $

4)$  (31*6)/4 $

答:
So let's solve the recurrence $a_{n+1}-6a_n = 2^n$ with the initial condition $a_1=4$.

The homogeneous recurrence $a_{n+1}-6a_n=0$ has solution $a_n = A 6^n$ and imposing the inhomogeneous RHS you should end up with $a_n = A\cdot 6^n + B\cdot 2^n$, and now imposing $a_1=4$ can solve for $B$.

Then you have the whole sequence, and the answer will depend on what you mean by "coefficient."

UPDATE

I am getting the same output from Wolfram Alpha, indeed $a_n = 2^{n-2}\left(3^{n+1}-1\right)$. I still don't understand what is meant here by "coefficient." Perhaps the intent is to rewrite
$$
a_n = 2^{n-2}\left(3^{n+1}-1\right) = \frac{3 \cdot 6^n - 2^n}{4},
$$
then the coefficient of $6^n$ is $3/4$ and the coefficient of $2^n$ is $-1/4$?

Title: Generator find the coefficient