问:
Integration using digamma functions
I'm trying to prove that for $x,y,r>0$ the following identity (gotten via CAS) holds:

$$\int_0^x \frac{x^r}{x^r+y^r}\,dy=\frac{x}{2r}\left(\psi\left(\frac{1+r}{2r}\right)-\psi\left(\frac{1}{2r}\right)\right),$$

where $\psi(\cdot)=\frac{\Gamma'(\cdot)}{\Gamma(\cdot)}$ denotes the Digamma function. Does anyone has a clue, how integration can be done this way?

As an additional information following identity, holds also to be true:
$$\int_0^x \frac{y^r}{y^r+x^r}\,dy=\frac{x}{2r}\left(\psi\left(\frac{1}{2r}+1\right)-\psi\left(\frac{1}{2r}+\frac{1}{2}\right)\right),$$
Which was obtained using:
$$\psi(x)=H_{x-1} - \gamma,$$
where $\gamma$ denotes the Euler-Mascheroni constant and $H_x$ denotes the $x^{th}$ harmonic number.

答:
Just to add a bit to the evaluation provided by @giobrach and @Roman - another way to get Digamma function explicitly.
$$I=\int_0^x \frac{x^r}{x^r+y^r}\,dy\overset{z=\frac{y}{x}}=x\int_0^1 \frac{1}{1+z^r}\,dz=x\int_0^1 \frac{1-z^r}{(1+z^{r})(1-z^r)}\,dz=x\int_0^1 \frac{1-z^r}{1-z^{2r}}\,dz$$
Next we make regularisation:
$$I=lim_{\epsilon\to0} \,I_\epsilon $$ where

$$I_\epsilon=x\int_0^1 \frac{1-z^r}{(1-z^{2r})^{1-\epsilon}}\,dz\overset{t=z^{2r}}=\frac{x}{2r}\int_0^1 \frac{t^{(\frac{1}{2r}-1)}-t^{(\frac{1}{2r}-\frac{1}{2})}}{(1-t)^{1-\epsilon}}\,dt=\frac{x}{2r}\Bigr(B(\frac{1}{2r},\epsilon)-B(\frac{1}{2r}+\frac{1}{2},\epsilon)\Bigr)$$ $B(a,b)$ - Beta function: $B(a,b)=\int_0^1t^{a-1}(1-t)^{b-1}=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$
$$I_\epsilon=\frac{x}{2r}\biggl(\frac{\Gamma(\epsilon)\Gamma(\frac{1}{2r})}{\Gamma(\frac{1}{2r}+\epsilon)}-\frac{\Gamma(\epsilon)\Gamma(\frac{1}{2r}+\frac{1}{2})}{\Gamma(\frac{1}{2r}+\frac{1}{2}+\epsilon)}\biggr)$$

Taking into consideration that $\Gamma(\epsilon)=\frac{1}{\epsilon}+O(1)$ at $\epsilon\to0$ we get
$$I=lim_{\epsilon\to0} \,I_\epsilon = lim_{\epsilon\to0}\,\frac{x}{2r}\,\frac{1}{\epsilon}\biggl(\frac{\Gamma(\frac{1}{2r})}{\Gamma(\frac{1}{2r}+\epsilon)}-\frac{\Gamma(\frac{1}{2r}+\frac{1}{2})}{\Gamma(\frac{1}{2r}+\frac{1}{2}+\epsilon)}\biggr)$$ $$=lim_{\epsilon\to0}\,\frac{x}{2r}\,\frac{1}{\epsilon}\biggl(\frac{\Gamma(\frac{1}{2r})}{\Gamma(\frac{1}{2r})+\Gamma \,'(\frac{1}{2r})\epsilon + O(\epsilon^2)}-\frac{\Gamma(\frac{1}{2r}+\frac{1}{2})}{\Gamma(\frac{1}{2r}+\frac{1}{2})+\Gamma\,'(\frac{1}{2r}+\frac{1}{2})\epsilon)+O(\epsilon^2)}\biggr)$$
$$=lim_{\epsilon\to0}\,\frac{x}{2r}\,\frac{1}{\epsilon}\biggl(\frac{1}{1+\log'(\Gamma(\frac{1}{2r}))\epsilon + O(\epsilon^2)}-\frac{1}{1+\log'(\Gamma(\frac{1}{2r}+\frac{1}{2}))\epsilon+O(\epsilon^2)}\biggr)$$
$$=\frac{x}{2r}\,\biggl(\log'(\Gamma(\frac{1}{2r}+\frac{1}{2}))-\log'(\Gamma(\frac{1}{2r}))\biggr)$$
Finally

$$I=\frac{x}{2r}\,\biggl(\Psi(\frac{1}{2r}+\frac{1}{2})-\Psi(\frac{1}{2r})\biggr)$$

were $\Psi(t)=\frac{d}{dt}\log(\Gamma(t))$ - Digamma function

Title: Integration using digamma functions


问:
Differential Operators
I was given some lecture notes on invariant differential operators. In the notes, there was something that I did not quite understand (link to original image)

and recall the matrix exponential function
$$ \exp(X) = \sum\limits_{k=0}^\infty \frac{X^k}{k!}, $$
which satisfies $\det\exp(X) = e^{\text{trace}(X)}$. So, if $\text{trace}(X)=0$, we can define a differential operator on the manifold $G = \text{SL}(2,\Bbb R)$ via
$$ \underline X.f(g) := \frac{d}{dt}\biggr\rvert_{t=0}f(g\exp(tX)) $$

I don't understand the correlation of having a matrix $X$ having a zero trace and being able to define a differential operator on $SL_2(\mathbb{R})$. Here $f$ is any differentiable function on $SL_2(\mathbb{R})$.

答:
Closed subgroups $G$ of $Gl(n,k)$ ($k=\mathbb{R}$ or $\mathbb{C}$) are defined by a system of equations (they are called matrix Lie groups).
A tangent vector $X$ at $1_G$ (the unit of $G$) is described by the one-parameter group with $X$ as infinitesimal generator (in fact, any path $\gamma$ such that $\gamma(0)=1$ and $\gamma'(0)=X$), this group is $\{exp(tX)\}_{t\in \mathbb{R}}$ in general. Here for
all $X\in \frak{sl}(2,\mathbb{R})$ (i.e. with $tr(X)=0$) the path $\{exp(tX)\}_{t\in \mathbb{R}}$ is drawn on $SL(2,\mathbb{R})$ and satisfies the two preceding conditions ($\gamma(0)=1$ and $\gamma'(0)=X$), whence this description.

ContentLicense: CC BY-SA 4.0

Title: Differential Operators


问:
Why do I have an equality sign instead of a proportionality sign in the solution of the heat equation using Fourier Transform?
I am trying to answer the following question:

Consider an infinitely long metal rod. The one dimensional heat equation is $\frac{\partial u}{\partial t}=a^2\frac{\partial^2 u}{\partial x^2}$. Let the temperature at $t=0$ be the delta function. Show that $u(x,t)\propto \frac{1}{a\sqrt{2t}}\exp({\frac{-x^2}{4a^2t}})$.

In lectures we were shown that $u(x,t)$ has the following solutions (we are using the convention that the forward Fourier transform uses $e^{+ixy}$ and the inverse uses $e^{-ixy}$).

I applied these solutions to and got:

Which is the correct formula, but I am getting and equality rather than a proportionaliy. I think the error must be somewhere in the inverse fourier transform, but I just can't figrue out where.

答:
I think you just have a careless mistake in your last line.  You have $\sqrt{2\pi}^2$ in the denominator of the first equality, but in the second you just have $\sqrt{2}\sqrt{ \pi}$, with the square having been dropped.  If you fix this you find your answer ought to have a constant factor of $\frac{1}{\sqrt{2 \pi}}$ in front, which would agree with the formula for the standard heat kernel when $a= \frac{1}{\sqrt{2}}$.

(The minus sign in front of the second equality also seems to be a mistake.)


问:
Find length BC of triangle with incircle and circumcircle

Some thoughts... some chord theorem to get the angle between PQ and BC... AB and AC are tangent to the circle, there has to be another theorem about that, perhaps ADE is isosceles and that helps looking at the angled PDE and AEQ, and with all this one should be able to figure the angle at A, and once I have that probably I can use yet another theorem on chords to determine the length BC from the length PQ...idk

答:
Here are 2 solutions 1) an analytical solution 2) a solution inspired by the solution of @Quanto.

First solution; Let us take coordinate axes as displayed in Fig. 1 below.

Line $PQ$ is taken as the $x$-axis, with origin and unit such that $P,D,E,Q$ have resp. abscissas $-3,-1,1,2$.

Triangle $ADE$ is isosceles, due to the fact that $A$ is the intersection point of tangents in $D$ and $E$ to the incircle ; therefore $A$ belongs to the ordinate axis with coordinates $(0,\tan a)$, where  $a$:=angle($ODA$).

Fig. 1.

As a consequence:

$$AD=AE=\dfrac{1}{\cos a} \ \implies \ \vec{AD}=\binom{-1}{-\tan a}  \ \text{and} \ \vec{AE}=\binom{1}{-\tan a}\tag{1}$$

A classical theorem on intersecting chords in a circle gives:

$$AD.DB=PD.PQ=6 \ \implies \ DB=\dfrac{6}{AD}=6 \cos a\tag{2}$$

For a similar reason:

$$EC=4 \cos a\tag{3}$$

Therefore, using (1) and (2) and the trigonometric formula $1+\tan^2 a=1/\cos^2 a$:

$$\vec{AB}=\dfrac{\dfrac{1}{\cos a}+6 \cos a}{\dfrac{1}{\cos a}} \vec{AD}=(1+6 \cos^2 a)\binom{-1}{-\tan a}$$

implying

$$B=\binom{-1-6 \cos^2 a}{- 6 \sin a\cos a}\tag{4}$$

For completely similar reasons, using (3):

$$C=\binom{1+4 \cos^2 a}{- 4\sin a \cos a}\tag{5}$$

As a consequence of (4) and (5):

$$\vec{BC}=\binom{2+10 \cos^2 a}{2 \cos a \sin a}\tag{6}$$

It remains to express that line segment $BC$ must be tangent to the incircle. This will be done by the following condition:

$$BC=BD+CE\tag{7}$$

Squaring (7) and using (2), (3) and (6):

$$\underbrace{(2+10 \cos^2 a)^2+(2 \cos a \sin a)^2}_{BC^2}
=\underbrace{(10 \cos a)^2}_{(BD+CE)^2}\tag{8}$$

Setting $x=\cos^2 a$ in (8), we get the following quadratic equation:

$$(1+5x)^2+x(1-x)=25x \ \ \iff \ \ 24 x^2-14x+1=0\tag{9}$$

whose roots are

$$x_1=(\cos a_1)^2=1/2 \ \ \text{and} \ \ x_2=(\cos a_2)^2=1/12\tag{10}$$

Therefore, as we are looking for positive values of $\cos a$:

$$\cos a_1=\dfrac{\sqrt{2}}{2} \ \ \text{and} \ \ \cos a_2=\dfrac{1}{2\sqrt{3}}\tag{11}$$

Consequently (see (8)), the two solutions are:

$$BC=10 \cos a_1=5 \sqrt{2}  \ \ \text{and} \ \ BC=10 \cos a_2=\dfrac{5}{\sqrt{3}}\tag{12}$$

Remarks :

a) Figure 1 corresponds to the second case in (12).

b) This solution doesn't use any specific triangle formula.

Second solution:

Let $a:=AD$; due to chord theorem in circles, we have :

$$DB=6/a \ \ \ \ \& \ \ \ \ EC=4/a\tag{1}$$

Consider Fig. 2.

Fig. 2.

In rectangular triangles $AHD$ and $ADI$, we have :

$$\sin(\tfrac12 A) = \dfrac 1a \ \ \ \ \& \ \ \ \ \tan(\tfrac12 A) = \dfrac{r}{a}\tag{2}$$

Using relationship $1+\dfrac{1}{\tan^2 x}=\dfrac{1}{\sin^2x}$, we deduce from (2) that:

$$r^2=\dfrac{a^2}{a^2-1}\tag{3}$$

Besides, we have this triangle formula:

$$\tan(\tfrac12 A)\tan(\tfrac12 B)\tan(\tfrac12 C)=\dfrac{r}{p}\tag{4}$$

where $p$ is the semiperimeter (half of perimeter).

Relationship (4) can be written:

$$\dfrac{r}{a}\dfrac{r}{(6/a)}\dfrac{r}{(4/a)}=\dfrac{r}{a+\tfrac6a+\tfrac4a}\tag{5}$$

that can be transformed into

$$r^2=\dfrac{24}{a^2+10}\tag{6}$$

Equating (3) and (6) gives a second degree polynomial equation in $x=a^2$:

$$x(x+10)=24(x-1)$$

whose two real positive roots are $x_1=2=a_1^2$ and $x_2=12=a_2^2$.

And we find back the two solutions for $BC=\dfrac{10}{a}$:

$$BC=\dfrac{10}{a_1}=\dfrac{10}{\sqrt{2}} \ \ \ \text{and} \ \ \ BC=\dfrac{10}{a_2}=\dfrac{5}{\sqrt{3}} $$

Remark: relationship (4) could have been obtained using Heron's formula.


问:
Counting tuples which satisfy certain additive conditions
I have $4$-tuples

$(a_1, a_2, a_3, a_4) \in \mathbb{N}_0$ such that $ a1 + a2 + a3 + a4 = n$ and $2 | a_2$ and $3 | a_3$ and $4 | a_4$

For example, if $n = 2,$ then we could have $(2,0,0,0)$ or $(0,2,0,0)$. If $n = 3,$ then we have $(0,0,3,0)$ and $(3,0,0,0)$ and $(1,2,0,0)$.

How can I count how many tuples I have for each $n \in \mathbb{N}$? It looks like it really blows up after $n = 5$.

Body:
You are asking to count how many ways there are to write
$$
n=k_1+2k_2+3k_3+4k_4
$$
with nonnegative integers $k_1,...,k_4$.

This is the same as the coefficient of $x^n$ in
$$
(1+x+x^2+..)(1+x^2+x^4+..)
(1+x^3+x^6+..)
(1+x^4+x^8+..)=
\frac 1{1-x}
\frac 1{1-x^2}
\frac 1{1-x^3}
\frac 1{1-x^4}
$$
$$
=1 + x + 2 x^2 + 3 x^3 + 5 x^4 + 6 x^5 + 9 x^6 + 11 x^7 + 15 x^{8} +
 18 x^9 + 23 x^{10} + 27 x^{11} + 34 x^{12} + 39 x^{13} + 47 x^{14} + 54 x^{15} +
 64 x^{16} + 72 x^{17} + 84 x^{18} + 94 x^{19} + 108 x^{20} +\mathcal O(x^{21}).
$$
It is the number of partitions of $n$ with parts at most equal to $4$.


问:
Concise proof that $\lvert a-b\rvert < \frac{\lvert b \rvert}{2} \implies \lvert a\rvert > \frac{\lvert b \rvert}{2}$
I am only considering $\mathbb{R}$ in this context. I understand why the statement holds and can prove it geometrically, but I can’t think of a concise way to derive the consequent from the precedent in a purely algebraic way.

答:
Assume that $|a-b|\lt\frac12|b|$ and $|a|\le\tfrac12|b|$.

The triangle inequality says that $|b|\le|a-b|+|a|$, but the assumption implies that $|a-b|+|a|\lt|b|$. Thus, the assumption is false.

One negation of the assumption is
$$
|a-b|\lt\frac12|b|\implies|a|\gt\tfrac12|b|
$$


问:
Do smooth, injective maps preserve linear independence?
Let $\phi: U \to V$ be a smooth map, where $U \subset \mathbb{R}^2$ and $V \subset \mathbb{R}^3$. Suppose $\phi$ is injective. Then, $\phi$ maps linearly independent vectors in $\mathbb{R}^2$ contained within $U$ to linearly independent vectors in $\mathbb{R}^3$. Is this true? I think this property is used in one of my lectures, but I can't find a proof anywhere.

(Context: Although this doesn't have much to do with the question at this point, it might be helpful. This question comes up when defining curvilinear triangles for surfaces. A curvilinear triangle is defined as some set in the plane satisfying several conditions, and then a triangle for a surface is defined as the image of that triangle under a chart $\phi$. It seems to be the case that $\phi$ maps linearly independent sets to linearly independent sets, though I don't know why. This is from a course in elementary differential geometry, though the question is just basic linear algebra.)

答:
The conjeture that a smooth injective map might send linearly independent sets to linearly dependent sets is immediately disproved, because as soon as $\phi(v)=0$ for some $v\ne0$, we have that $\{v\}$ is linearly independent and $\phi[\{v\}]$ isn't. Nor, in fact, is $\phi[\{v,w\}]$ for any $w$.

For another example, we can devise a function in the form $\phi(x,y)=(e^{ax}-b,e^{x+cy}-d,0)$ such that $\phi(1,2)=(1,0,0)$ and $\phi(-1,-1)=(-1,0,0)$. Namely, solve for the coefficients $$\begin{cases}e^a=1+b\\ e^{-a}=b-1\\ e^{1+2c}-d=0\\ e^{-1-c}-d=0\end{cases}\Leftrightarrow \begin{cases}b=\sqrt2\\ a=\ln(\sqrt2+1)\\ c=-\frac23\\ d=e^{-1/3}\end{cases}$$

Since $a,c\ne 0$, this $\phi$ is injective.


问:
Convergence in distribution $\sqrt{n}X_n \xrightarrow{n \to +\infty} \mathcal{N}(0,1)$ implies $\sup_{n\in \Bbb N^*}E(|X_n|)<+\infty$
Given the random variables $(X_n)_{n\in \Bbb N^*}$ such that $\sqrt{n}X_n$ converges in distribution to the standard normal distribution
$$\sqrt{n}X_n \xrightarrow[n \to +\infty]{\mathcal{D}} \mathcal{N}(0,1)$$

Suppose $E(|X_n|) <+\infty$ for all $n\in \Bbb N^*$.

Prove that
$$\sup_{n\in \Bbb N^*}E(|X_n|)<+\infty$$

In other words, there exist an $M <+\infty$ such that $E(|X_n|)<M$ for all $n\in \Bbb N^*$.

My attempt:

Let's write
$$E(|X_n|) =  E(|X_n|\mathbb{I}_{\{|X_n|\le M_1\}})+ E(|X_n|\mathbb{I}_{\{|X_n|>M_1\}}) $$
with $M_1>0$

The first term is bounded by $M_1$.

About the second term, from an $N$ sufficiently large, we have intuitively $X_n \sim\frac{1} {\sqrt{n}}\mathcal{N}(0,1) $ then $f_{X_n}(x) ~\approx \frac{\sqrt{n}}{\sqrt{2\pi}}\exp\left(-\frac{x^2}{2n} \right)$ hence the linear increasing of $x$ is much less than the decreasing of $f_{X_n}(x)$ for $x>M_1$. Hence,  $E(|X_n|\mathbb{I}_{\{|X_n|\le M_1\}})$ will be bounded by a constant $M_2$ for all $n> N$.

But I haven't been able to formalize this intuition.

答: Let $X\sim N(0,1)$. Let  $Y_n$ take the values $0$ and $n^{2}$ with probabilities $1-\frac 1 {\sqrt n}$ and $\frac 1 {\sqrt n}$ respectively. Let $X_n=\frac X {\sqrt n}+\frac {Y_n} {\sqrt n}$. Since $P(|Y_n| >\epsilon)=\frac 1 {\sqrt n} \to 0$ it follows that $\sqrt n X_n \to X$ in distribution. Howvever, $EX_n \geq 0+\frac {n^{2}} {\sqrt n}\frac 1 {\sqrt n} \to \infty$.


问:
Find all natural numbers such that $\sqrt{n+2020}+\sqrt{n} \in \mathbb{N}$
My attempt:

$\sqrt{n+2020}+\sqrt{n} \in \mathbb{N} $$\Rightarrow $$\frac{2020}{\sqrt{n+2020}-\sqrt{n}} \in \mathbb{N} $

$\Rightarrow $$\sqrt{n+2020}-\sqrt{n}=w \in \mathbb{N} $in which w is a factor of$ 2020$

Note( w is even, you can chek this by separating the cases)

$2020=5×2^2×101$,

so the factor of $2020$ is $5^{i}×2^{j}×101^{k}$ which ($0\leq i\leq 1 $ and $ 0\leq k\leq 1 $ and $0 \leq j\leq2$)

So $i=k={0,1}$ and $j={0,1,2}$

So the numbers of factor of 2020 is $j×k×i=3×2×2=12$

After that we can see $(i, k, j)$ maybe equal this values

$(i, k, j) ={(0,0,0),(0,0,1),(0,0,2),(0,1,0),(0,1,1),(0,1,2),(1,0,0) ,(1,0,1),(1,0,2),(1,1,0),(1,1,1),(1,1,2)}$

If $(i, k, j) =(0, 0,0)$that implie $w=1 $ that is impossible because we know w is even

If $(i, k, j) =(0, 0,1)$that implie $w=2 $

If $(i, k, j) =(0, 0,2)$that implie $w=4 $

If $(i, k, j) =(0, 1,0)$that implie $w=101 $ that is impossible because we know w is even

So  i  check the  all above values, and in final i  find just 8 value of w and this value is $w={2,4,202,404,10,20,1010,2020}$

After that  suppose $n+2020=x^2$ and $n=y^2$ wich$ x, y \in \mathbb{N} $

After the last assumation we can see Two results, the first is $x+y$ is even is easy the chek that
And the seconde results is $x^2-y^2=(x-y)(x+y)=2020$

And $x-y=w$

So if $x-y=w=2$$\Rightarrow$ $x+y=1010 $ So$ x+y+x-y=2x=1012$

So $x=506$

if $x-y=w=4$$\Rightarrow $$x+y=505$ that is impossible because $x+y$ is even

if$x-y=w=202$$\Rightarrow $$x+y=10$

So $x=106$

And i am continued like, and in final, I find just $x=106$ and $x=506$

So if $x=506 $ that implie $n=506^2-2020=254016$

if $x=106 $ that implie $n=106^2-2020=9216$

So finally:

$n=9216$

And

$n=254016$

So what are you think about my attempt, is true or false?

Any comment is acceptable

答:
As I pointed out in the comments, the first step is false.

(Though the subsequent conclusions are true, so we just need to fix this first step.)

Hint: Show that if $ k \in \mathbb{N} $ and $ \sqrt{k} \in \mathbb{N}$, then $ k = a^2$ for some $ a \in \mathbb{N}$.

After proving the hint, apply it to show that $n$ and $n+2020$ are both perfect squares.

 Show that if $ \sqrt{n} + \sqrt{ n+2020} = k \in \mathbb{N}$, then consider $ n+2020 = k^2 - 2k\sqrt{n} + n $ to prove that $(4k^2n) = a^2$ for some $a\in\mathbb{N}$, and so $ n = b^2$ for some $b\in \mathbb{N}$.

 Likewise, $ n+2020 = c^2$ for some $ c \in \mathbb{N}$.

The rest of your proof then follows.


问:
the nonexistence of a division algebra on $\mathbb R^{2n+1}$

Prove that there is no division algebra structure on $\mathbb R^{2n+1}$.

The hint says

Suppose that there is such a structure on $\mathbb R^{2n+1}$. Take a nonzero $a \in \mathbb R^{2n+1}$. Consider $f: S^{2n} \rightarrow S^{2n}, x \mapsto \frac{ax}{|ax|}$. Prove that $f$ and $-f$ are homotopic.

But how is this homotopy constructed? Thanks for the help.

答: Hint: Find a path between $a$ and $-a$ that does not go through $0$.</p>

ContentLicense: CC BY-SA 3.0


问:
Prove that the acceleration is either zero or parallel to the velocity given the curve lies on a plane
I'm stuck on this problem (section 14.9 exercise 13 of Apostol Calculus) :

A particle moves along a curve in such a way that the velocity vector makes a constant angle with a
given unit vector $\mathbf{c}$ . If the curve lies in a plane containing $\mathbf{c}$, prove that the acceleration vector  is either zero or parallel to the velocity

My attempt :

Define a unit vector $\displaystyle  \mathbf{T} = \frac{\mathbf{v}}{\|\mathbf{v}\|}$,  so $\displaystyle \mathbf{T} \ ' =  \frac{\mathbf{a}\|\mathbf{v}\|-\mathbf{v}\|\mathbf{v}\|'}{\|\mathbf{v}\|^2}$.

Then, because of the angle between the velocity $v$ and $c$ is constant we have

$$\mathbf{T} \cdot \mathbf{c} = \cos \alpha \implies \mathbf{T}\ ' \cdot \mathbf{c} = 0 \implies \mathbf{T} \ '=\mathbf{0} \  \lor \  \mathbf{T}\ ' \bot \ \mathbf{c}  $$

Also,

$$\mathbf{T} \cdot \mathbf{T}\ = 1 \implies \mathbf{T} \ ' \cdot v = 0 \implies    \mathbf{T} \ '=\mathbf{0} \  \lor \  \mathbf{T}\ ' \bot \ \mathbf{v}$$

If it is true that $ \mathbf{T}\ '=\mathbf{0} $ then $\displaystyle \mathbf{a}=\frac{\|\mathbf{v}\|'}{\|\mathbf{v}\|}\mathbf{v}  $ ,  so we have done.

Edit:

But as @user pointed out, $\mathbf{T} '$ cannot be perpendicular to both $\mathbf{v}$ and $\mathbf{c}$ because $\mathbf{v}$ and $\mathbf{a}$ have to lie on the same plane as $\mathbf{c}$ so does $\mathbf{T} '$, but this means it must be $\mathbf{0}$ because there isn't a non zero vector on a plane perpendicular to $\mathbf{v}$ and $\mathbf{c}$.

I think all of my problem was that I was not aware of the definition of vector lying on a plane, i thought just the spike of it was included in the plane.

答:
Since the curve and unit vector are in the same plane, let that plane be xy plane.Set original velocity  $\hat v =<x,y,0 >$, unit vector = $\hat u$=<0,1,0>,  $\theta$ is the angle between $\hat v$ and $\hat u$ . After the acceleration, velocity vector becomes  $\hat v$=<x',y',z'>.

Since the angle between  $\hat v'$ and $\hat u$ need to be $\theta$, $\hat v $ and $\hat v'$ are parallel to each other, they are scaled by a scalar value m,   or

$\hat v' = m\hat v$ =$<mx,my,0>$,

the velocity change is

$\Delta \hat v$=$<mx-m,my-y,0>$ =(m-1)$\hat v$.

Acceleration $\hat a$ =  $\frac{\Delta v}{\Delta t}$=$\frac{(m-1) \hat v}{\Delta t}$,
$\hat a$  is either parallel to  $\hat v$ (m$\neq 1$)or $\hat 0$ ($ m=1$).


答: Basically, you want: $n = p^2, n+2020 = q^2\implies q^2 -p^2 = 2020\implies (q-p)(q+p)=2020$. Observe that both $q-p, q+p$ must be even. Thus looking at the prime factorization of $2020 = 2^2\cdot 5\cdot 101$. Thus it must be that $q-p = 2, q+p = 1010\implies q = 506, p = 504\implies n = 504^2 $, or $q-p = 10, q+p = 202 \implies q = 106, p = 96$ which gives $n = 96^2$. Thus there are two solutions: $n = 96^2, 504^2$.

Title: Find all natural numbers such that $\sqrt{n+2020}+\sqrt{n} \in \mathbb{N}$

Body:
My attempt:

$\sqrt{n+2020}+\sqrt{n} \in \mathbb{N} $$\Rightarrow $$\frac{2020}{\sqrt{n+2020}-\sqrt{n}} \in \mathbb{N} $

$\Rightarrow $$\sqrt{n+2020}-\sqrt{n}=w \in \mathbb{N} $in which w is a factor of$ 2020$

Note( w is even, you can chek this by separating the cases)

$2020=5×2^2×101$,

so the factor of $2020$ is $5^{i}×2^{j}×101^{k}$ which ($0\leq i\leq 1 $ and $ 0\leq k\leq 1 $ and $0 \leq j\leq2$)

So $i=k={0,1}$ and $j={0,1,2}$

So the numbers of factor of 2020 is $j×k×i=3×2×2=12$

After that we can see $(i, k, j)$ maybe equal this values

$(i, k, j) ={(0,0,0),(0,0,1),(0,0,2),(0,1,0),(0,1,1),(0,1,2),(1,0,0) ,(1,0,1),(1,0,2),(1,1,0),(1,1,1),(1,1,2)}$

If $(i, k, j) =(0, 0,0)$that implie $w=1 $ that is impossible because we know w is even

If $(i, k, j) =(0, 0,1)$that implie $w=2 $

If $(i, k, j) =(0, 0,2)$that implie $w=4 $

If $(i, k, j) =(0, 1,0)$that implie $w=101 $ that is impossible because we know w is even

So  i  check the  all above values, and in final i  find just 8 value of w and this value is $w={2,4,202,404,10,20,1010,2020}$

After that  suppose $n+2020=x^2$ and $n=y^2$ wich$ x, y \in \mathbb{N} $

After the last assumation we can see Two results, the first is $x+y$ is even is easy the chek that
And the seconde results is $x^2-y^2=(x-y)(x+y)=2020$

And $x-y=w$

So if $x-y=w=2$$\Rightarrow$ $x+y=1010 $ So$ x+y+x-y=2x=1012$

So $x=506$

if $x-y=w=4$$\Rightarrow $$x+y=505$ that is impossible because $x+y$ is even

if$x-y=w=202$$\Rightarrow $$x+y=10$

So $x=106$

And i am continued like, and in final, I find just $x=106$ and $x=506$

So if $x=506 $ that implie $n=506^2-2020=254016$

if $x=106 $ that implie $n=106^2-2020=9216$

So finally:

$n=9216$

And

$n=254016$

So what are you think about my attempt, is true or false?

Any comment is acceptable


答:
I believe the answer (valid for $n \geq 3$) is the closest integer to
$$
\frac{1}{288}(2n^3 + 30n^2 + 135n + 175) + \frac{(-1)^n n}{32}.
$$
This seems to agree with Giulio's answer.

The answer will be the coefficient of degree $n$ in the product series of
$$
(1 + x + x^2 + \dots)(1 + x^2 + x^4 + \dots)(1 + x^3 + x^6 + \dots)(1 + x^4 + x^8 + \dots) = \frac{1}{(1-x)(1-x^2)(1-x^3)(1-x^4)}.
$$

According to Wolfram Alpha, the partial fractions decomposition of this rational function is
$$
\frac{x + 1}{9(x^2 + x + 1)} + \frac{1}{8 (x^2 + 1)} - \frac{68 x^3 - 263 x^2 + 358 x - 175}{288 (x - 1)^4} + \frac{4 x + 5}{32 (x + 1)^2}.
$$
The general terms of the series corresponding to the last three terms can be determined using the binomial series. Letting $\omega$ be a complex cube root of unity, the first term can be rewritten as $-\frac{1}{18}\frac{\omega}{x - \omega} -\frac{1}{18}\frac{\omega^2}{x - \omega^2}$, and then the same method can be applied to that term. Presumably, there will be distinct formulae according to $n \mod 6$.

Using complex numbers, the answer won't depend formally on $n \mod 6$. We get
$$
\frac{1}{18}(\omega^n + \omega^{-n})
+ \frac{1}{16}(i^n + i^{-n}) -\frac{68}{288}(-1)^{n-3}\binom{-4}{n-3} + \frac{263}{288}(-1)^{n-2}\binom{-4}{n-2} - \frac{358}{288}(-1)^{n-1}\binom{-4}{n-1} + \frac{175}{288}(-1)^n\binom{-4}{n} + \frac{4}{32}\binom{-2}{n-1} + \frac{5}{32}\binom{-2}{n},
$$
where $\omega^n + \omega^{-n}$ is $2$ or $-1$ according as whether $n$ divides $3$ or not, $i^n + i^{-n}$ is $2(-1)^{n/2}$ or $0$ according as whether $n$ is even or odd, and
$$
\binom{-2}{n} = (-1)^n(n+1), \quad \binom{-4}{n} = \frac{(-1)^n}{6}(n+3)(n+2)(n+1),$$
except when $n < 0$, in which case $\binom{-2}{n}$ and $\binom{-4}{n}$ are zero. The first two terms being much smaller than $1$ in absolute value, we can say in short that the answer for $n \geq 3$ is the closest integer to
$$\frac{1}{288}(2n^3 + 30n^2 + 135n + 175) + \frac{(-1)^n}{32}(n+5).$$

Since $1/9 + 1/8 + 5/32 < 1/2$, we can also ignore the last constant term.

A way to just get the answer from Wolfram Alpha or some similar system would be to take the $n$th derivative of $\frac{1}{(1-x)(1-x^2)(1-x^3)(1-x^4)}$ evaluated at $x = 0$ and divide the answer by $n!$. The free online version of Wolfram Alpha is telling me its computation time is too long for this request, so I can't actually do this for you.

Title: Counting tuples which satisfy certain additive conditions

Body:
I have $4$-tuples

$(a_1, a_2, a_3, a_4) \in \mathbb{N}_0$ such that $ a1 + a2 + a3 + a4 = n$ and $2 | a_2$ and $3 | a_3$ and $4 | a_4$

For example, if $n = 2,$ then we could have $(2,0,0,0)$ or $(0,2,0,0)$. If $n = 3,$ then we have $(0,0,3,0)$ and $(3,0,0,0)$ and $(1,2,0,0)$.

How can I count how many tuples I have for each $n \in \mathbb{N}$? It looks like it really blows up after $n = 5$.


问:
Value of $x$ in a failure rate problem
An electronic component is known to have a useful life which follows a probability
distribution with failure rate of $10^{-5}$
failures per hour. What is the probability that the
component would fail before the mean life of expected life?

This is Poisson Distribution right? Here the mean is $10^{-5}$ right but what is the value of $x$?

答:
@Heropup (+1) has already given you the correct distribution.
I think you're dealing with an exponential distribution with failure rate $\lambda,$ mean $\mu = 1/\lambda.$ and
CDF $F_X(x) = 1 - e^{\lambda x}.$ for $x > 0.$

So
$$P(X \le \mu) = 1 - e^{-\lambda\mu} = 1 - e^{-1} =  0.6321206.$$

In R, where pexp is a CDF:

```<code>lam = 10;  pexp(1/lam, lam)
[1] 0.6321206
lam = 1; pexp(1/lam, lam)
[1] 0.6321206
lam = .02; pexp(1/lam, lam)
[1] 0.6321206
lam = 10^-5; pexp(1/lam, lam)
[1] 0.6321206
1 - exp(-1)
[1] 0.6321206
</code>```

Graph for one example. The relevant area is under the density curve to
the left of the vertical dotted line.

Title: Value of $x$ in a failure rate problem

Body:
@Heropup (+1) has already given you the correct distribution.
I think you're dealing with an exponential distribution with failure rate $\lambda,$ mean $\mu = 1/\lambda.$ and
CDF $F_X(x) = 1 - e^{\lambda x}.$ for $x > 0.$

So
$$P(X \le \mu) = 1 - e^{-\lambda\mu} = 1 - e^{-1} =  0.6321206.$$

In R, where pexp is a CDF:

lam = 10;  pexp(1/lam, lam)
[1] 0.6321206
lam = 1; pexp(1/lam, lam)
[1] 0.6321206
lam = .02; pexp(1/lam, lam)
[1] 0.6321206
lam = 10^-5; pexp(1/lam, lam)
[1] 0.6321206
1 - exp(-1)
[1] 0.6321206

Graph for one example. The relevant area is under the density curve to
the left of the vertical dotted line.


问:
Does mean/expectation affect the variance of a sample?
Let's say I have an app $A$ that has a mean of $4$ stars and an app $B$ that has a mean of $3$ stars. Both have the same sample size. Would the variance stay the same? I thought it would because what's up top (of the unbiased variance formula) would stay the same, since it's $X$ minus the mean and the $X$ would change too along with the mean. I am kinda confused on it though.

答: Typically, when you have a bounded random variable, the closer the mean is to one of the extremes, the lower the variance. Consider an app with a 5-star mean. It will necessarily have 0 variance as if an outcome other than a 5-star rating were possible, this would imply the mean is less than 5. Examples of bounded families of random variables are the beta distribution and the binomial distribution, which exhibit this property. Though an app rating would not necessarily follow any family of distribution other than the general multinomial distribution.</p>


问:
Bounding functions involving integrals (PDEs by Walter Strauss)
Multiple times in Partial Differential Equations by Walter Strauss, e.g. in Ch 3.4 and 3.5, the author bounds the absolute value of a function involving integrals by using the maximum of the absolute value of the functions being integrated. For example they bound the following function:

$u(x,t)=\frac{1}{2}[\phi(x+ct)+\phi(x-ct)]+\frac{1}{2c}\int_{x-ct}^{x+ct}\psi+\frac{1}{2c}\int\int_{\Delta}f$

By

$|u(x,t)|\le max|\phi|+\frac{1}{2c}\cdot max|\psi|\cdot 2ct+\frac{1}{2c} \cdot max|f| \cdot ct^2$

What allows them to do this?

It seems like they multiply the area of the region being integrated over by the maximum value of the absolute value of the function.

They do a similar thing in Ch 3.5 and mention the inequality is related to the maximum principle.

答:
For a bounded area $\Omega$ and a continuous function $f$:
\begin{align*}
\left |\int_{\Omega} f(x) dx\right| &\leq \int_{\Omega} |f(x)| dx \\
&\leq \max_{\overline{\Omega}} |f| \int_{\Omega} dx\\
&=\left(\max_{\overline{\Omega}} |f|\right)  |\Omega|,
\end{align*}
where $|\Omega|$ denotes the (Lebesgue) measure of $\Omega$.


问:
"Dual" definition of 1-connectivity in finite simplicial complex?
I am studying the concepts of finite simplicial complexes (so please forgive for any mistakes :)) and, in particular, trying to grasp the concept of higher-order connectedness.

I have the following intuition from graph theory. We can define connected graphs in the following two equivalent ways (see Is this an equivalent definition for connectivity in graphs? )

$G$ is connected iff for all pairs of vertices $u,v \in V$, there exists a path between $u$ and $v$.

$G$ is connected iff for all partitions $\{A,B\}$ of $V(G)$ in two non-empty parts, there exists an edge with one endpoint in $A$ and the other in $B$.

My impression is that, using a topological point of view, we can view a graph $G$ as a $1$-dimensional simplicial complex, and "graph connectivity" is equivalent to the $0$-connectivity (or path connectivity) of this complex, i.e. for each continuous map $f: S^0 \rightarrow G$ can be extended to a continuous map $\overline{f}: B^1 \rightarrow G$. The parallel is clear with the first definition of graph connectivity.

Now, considering $1$-connectivity in a finite simplicial complex $X$ (i.e. each continuous map $f: S^1 \rightarrow X$ can be extended to a continuous map $\overline{f}: B^2 \rightarrow X$), I wonder if there is "more combinatorial" way of expressing that property in such a way that is more similar to the definitions for graph-connectivity I gave earlier.

For instance, I kind of visualize this property as (being vague here) saying that any "cycle" $C$ in $X$ we can find some "surface" in $X$ with $C$ as its border. But at the moment I don't imagine something which is analogous to the definition of connectivity in graphs in terms of its "cuts", that is, a partition of the vertices which "separate" the graph somehow.

So my question is, does such a "dual" definition of $1$-connectivity exist? If so, how it is formalised? Can it be understood in combinatorial terms like the cuts in graphs?

答:
The usual "$1$-connectedness" analogue of your property 1 is that any cycle of edges in the 1-skeleton extends to a Van Kampen diagram.

But I don't know of an analogue of your property 2.


问:
Is there a solution for following integral?
Is it possible to solve the following integral?

$$\int\frac{\exp(-ax^2)}{x}\,dx.$$

Body:
Substitute $u=ax^2$:
$$\int\frac{\exp(-ax^2)}{x}\,dx=-\frac{1}{2}\int-\frac{\exp(-u)}{u}\,dx=-\frac{1}{2}\operatorname{E}_1(u)+C=-\frac{1}{2}\operatorname{E}_1(ax^2)+C$$


问:
How many sequences of length 10 with elements $\{a, b, c, d\}$ have exactly $3$ out of $4$ elements?
My logic is since $3$ out of $4$ elements are chosen, each element would appear once.
So a sequence would look like: $a\,b\,c\,x\,x\,x\,x\,x\,x\,x$

We have $7$ spots $x$ that can be whatever elements so: $3^7$

Choosing $3$ out of $4$ elements:  $4\choose 3$$ = 4$
So in total, there are $4\cdot(3^7)$ sequences

Can someone tell me if this is right or not? Thanks

答:
Use $\binom{4}{3}$ to make the selection of $3$ elements to show up in the sequence.

Now you have to ask, "How many of each should there be?" Since there needs to be at least one of each, we are looking for how many ways there are to partition $10$ into the sum of $3$ positive integers. There are probably fast ways to do that, but I just wrote them all down by hand:

$\{8,1,1\}$, $\{7,2,1\}$, $\{6,2,2\}$, $\{6,3,1\}$, $\{5,4,1\}$, $\{5,3,2\}$, $\{4,4,2\}$, $\{4,3,3\}$.

For each of these, you then compute the number of ways to arrange the elements.

e.g. for $\{8,1,1\}$, there are $\binom{10}{8}\cdot\binom{2}{1}\cdot\binom{1}{1}$ to arrange, e.g. eight a's, one b, and one c. You would need to multiply that result by $3$ to account for the fact that you could have eight of the other letters instead.

Similarly for $\{7,2,1\}$, you would do $\binom{10}{7}\cdot\binom{3}{2}\cdot\binom{1}{1}$, but this you would multiply by $3!=6$ because there are that many ways to determine the quantities of each letter you have.

And so on for each partition of $10$. It seems a bit painstaking to do all of it, so I won't, but hopefully you can take it from here, and maybe someone will come along with an easier method.


答:
If you choose $10$ items from a pool of $3$ elements, you of course have $3^{10}$ possibilities, although that will include some where not all are used.

If you choose $10$ items from a pool of $2$ elements, you have $2^{10}$ outcomes, although again not every option will have both elements present.

Finally if there is only one element to choose from, there's no choice and a single outcome. $1^{10}=1$.

To get to exactly three elements present, we can choose the elements $\binom 43 = 4$ ways and then use inclusion-exclusion to eliminate the deficient options:

$$\binom 43 \left[ 3^{10} - \binom 32 2^{10} + \binom 31 1^{10} \right]$$

Title: How many sequences of length 10 with elements $\{a, b, c, d\}$ have exactly $3$ out of $4$ elements?

Body:
My logic is since $3$ out of $4$ elements are chosen, each element would appear once.
So a sequence would look like: $a\,b\,c\,x\,x\,x\,x\,x\,x\,x$

We have $7$ spots $x$ that can be whatever elements so: $3^7$

Choosing $3$ out of $4$ elements:  $4\choose 3$$ = 4$
So in total, there are $4\cdot(3^7)$ sequences

Can someone tell me if this is right or not? Thanks


问:
Poisson Process: Event arrival times
I am starting to learn about Poisson processes and I am struggling to put the wording of a question into a probability statement. The questions states that 12 tour boats come per hour and follow a homogenous Poisson process and then asks if you have already been waiting 6 minutes for a tour boat what is the probability you will have to wait another 6 mins for one to arrive.

My initial thought was P(T1 = 12|T1>6) -- T1 being the event time before the first event -- but when going through my notes couldn't find any examples of P(T1= "n") so I then ended up with P(T1<=12|T1>6) however this does not look intuitive to me. If it is correct could you explain why P(T1 <= 12) is correct and if not it would be much appreciated if you could provide some insight on where I'm going wrong.

答:
My understanding is $\mathbb{P}(T\geq 12\mid T \gt 6) = \mathbb{P}(T \gt 6)$
due to the memory-less property.

Title: Poisson Process: Event arrival times


问:
Analytic operator semigroup bounds
Suppose that $A: D(A) \subset X \to X $ with $X$ Banach is the infinitesimal generator of an analytic semigroup. Then how do you show that

$$|(-A)^{\gamma}A^k e^{At}|_{L(X)}\leq M t^{-k-\gamma}$$
for every $t \geq 0$, for $\gamma \in (0,1), K=0,1$ ?

This could be found in [Da Prato - stochastic differential equations in infinite dimensions - p 131]

Anyway shouldn't the estimate hold for every $t \leq T$ with $T$ fixed?

答:
See the proof of Theorem 6.13, Chapter 2 in Pazy's book:

A. Pazy, Semigroups of Linear Operators and Applications to Partial Differential Equations (Berlin: Springer), 1983.

Take $\alpha=\gamma+k$.

Title: Analytic operator semigroup bounds


问:
Prove the following about an eigenvector, $w$

If $w$ is an eigenvector of the $n \times n$ matrix $A$ with a corresponding eigenvalue $\lambda$, prove that $w$ is also an eigenvector of the $n \times n$ matrix $B = A^2$. Find the eigenvalue $\mu$ corresponding to this eigenvector of $B$.

Linear algebra really trips me up, but we know that $w \lambda = wA$ but how does that help prove anything about the eigenvectors and eigenvalues of $B$? Is there a trick I'm not using to manipulate $w \lambda = wA$ to get $w\mu = wB$?

答: If $Aw = \lambda w$, then $Bw = A^2 w = A(Aw) = A(\lambda w) = \lambda A w = \lambda^2 w$


问:
Solving $2^k +k \equiv 0 \pmod {323}$
Find all $k$ such that
$$2^k + k \equiv 0 \pmod{323}.$$

I noticed that $323 = 17\cdot 19$ so I thought about using the Chinese Remainder theorem by considering $2^k+k$ modulo 17 and 19. I got $k \equiv 16 \pmod{17}$ and $k\equiv 18 \pmod{19}$, which gives $k\equiv 322 \pmod{323}$. However, after trying this, the given solution was not valid : $2^{322} + 322 \equiv 156 \not \equiv 0 \mod{323}$.

Can someone explain why this didn't work, and what I can do to solve this problem? Thanks.

答:
As explained in the comments, it didn't work because CRT wasn't applied correctly.

In particular, the solution to $ 2^k + k \equiv 0 \pmod{17}$ has a cycle length of $ 16 \times 17$, because $2^k$ has a cycle length of 16 and $k$ has a cycle length of 17.

So, the (theoretical) approach is to find all the solutions to $2^k+k \equiv 0 \pmod{17} $ working in $\pmod{17 \times 16}$, and likewise for the other equation working in $\pmod{19 \times 18}$, and finally combine them via CRT in $ \pmod{17 \times 19 \times 144}$.

If you understand the above logic, I do not recommend actually finding all the solutions, as it's just a very tedious process that's best left to the computer.

Title: Solving $2^k +k \equiv 0 \pmod {323}$


问:
Prove that there is no positive rational number a such that $a^2 = 3$.
I solved a similar question $a^4 = 2$ but I didn't find it hard. This one is taking a lot of my time and I still don't understand how to solve this.

Here's what I have tried so far.

edit:

\begin{align}
a^2 &= 3 \\
a &= \left(\frac mn\right) \\
a^2 &= \left(\frac mn\right)^2 = 3 \\
m^2 &= 3n^2 \\
m &= 3p \\
m^2 &= (3p)^2 \\
9p^2 &= 3n^2
\end{align}

In another similar question $a^3 = 2$, I had to divide a common number from each side to finally solve the problem. But with this problem, I don't know how to continue.

答: The classic method known as "infinite descent". Assume from where you had : $m^2 = 3n^2 \implies m = 3k \implies n^2 = 3k^2 \implies n = 3p$, and it keeps going indefinitely with new pairs $(k,p)$ with $k < m, p < n$ and both are multiple of $3$ which is impossible. Thus there is no $(m,n)$ which satisfies the equation. So $\sqrt{3}$ must be irrational.


答:
Alright, make the straightforward assumption that $a= \frac mn$ with $m,n$ in lowest terms and they are integers. Thus, we get $(\frac mn)^2 = 3$, which soon becomes $m^2 = 3n^2$. Thus, $m^2$ is a multiple of $3$.

However, for $m^2$ to be a multiple of $3$, it needs to have an even number of factors of $3$. If it had an odd number of factors of $3$, you wouldn't get an integer for $m$ when you take the square root of $m^2$ because squares come in factors of even powers like $3^2$ or $2^2$. Thus, you must have an even number of $3$s in $m^2$, for it to be divisible by $3$. Now, when we take the square root of $m^2$ to get $m$, half of the $3$s disappear, but not all of them, so $m$ is divisible by $3$. Since $m$ is divisible by $3$, it can be expressed as $m=3p$. Thus, we get $(3p)^2 = 3n^2 \implies 3p^2 = n^2 $. Using the same logic as before, we can show that $n$ is divisible by $3$. This means that both $m,n$ are divisible by $3$, contradicting our initial assumption that $\frac mn$ is in lowest terms and shouldn't have any common factors besides $1$. Thus, there is no such $a = \frac mn$ with $m,n$ in lowest terms and integers such that $a^2=3$.

Here is a bonus easy proof. Rearrange $a^2=3$ to get $a^2-3=0$. By the rational root theorem, all the POSSIBLE rational roots of this polynomial are $a=-1,1,-3,3$. Let's see if any of these $a$ make $a^2=3$. Testing all these $a$, we see that none of these work, thus there is no RATIONAL $a$ such that $a^2=3$.

Title: Prove that there is no positive rational number a such that $a^2 = 3$.

Body:
I solved a similar question $a^4 = 2$ but I didn't find it hard. This one is taking a lot of my time and I still don't understand how to solve this.

Here's what I have tried so far.

edit:

\begin{align}
a^2 &= 3 \\
a &= \left(\frac mn\right) \\
a^2 &= \left(\frac mn\right)^2 = 3 \\
m^2 &= 3n^2 \\
m &= 3p \\
m^2 &= (3p)^2 \\
9p^2 &= 3n^2
\end{align}

In another similar question $a^3 = 2$, I had to divide a common number from each side to finally solve the problem. But with this problem, I don't know how to continue.


答:
From
$$ m^2 = 3n^2  \text{,}  $$ we know $3$ divides $m$, so $m = 3 m_1$
$$  9 m_1^2 = 3 n^2  $$
$$  3 m_1^2 = n^2  $$
This tells us $3$ divides $n$.  We have found that $3$ is a common factor of $m$ and $n$.

You failed to introduce $m$ or $n$ in your derivation and you are missing any words that would make this common divisor lead to a contradiction.  Perhaps you believe that a proper proof is just a barrage of equations.  It is not.  You need to introduce new symbols and explain how they relate.  For instance, a reader is right to ask "Why did you set $a = m/n$?  What is $m$?  What is $n$?  Are there any conditions among $m$ and $n$?"  Your demonstration is not a proof -- it is missing all the words that explain what you are doing and why.


问:
Can you determine if a measure is finite by looking at its Fourier-Stieltjes transform?
Let $\hat{\mu}:\mathbb{Z}\rightarrow\mathbb{C}$ be an arbitrary function. Then, letting $\mathbb{C}\left[e^{2\pi it},e^{-2\pi it}\right]$ denote the vector space space of all trigonometric polynomials (linear combinations of the form $\sum_{n=-N}^{N}a_{n}e^{2n\pi it}$), we can define a linear functional $d\mu:\mathbb{C}\left[e^{2\pi it},e^{-2\pi it}\right]\rightarrow\mathbb{C}$ by way of the formula:$$\int_{0}^{1}e^{-2n\pi it}d\mu\left(t\right)\overset{\textrm{def}}{=}\hat{\mu}\left(n\right),\textrm{ }\forall n\in\mathbb{Z}$$ which we then extend by linearity. My question is: are there any useful criteria on $\hat{\mu}$ to determine whether or not the $d\mu$ thus defined is in fact a finite Borel measure on $\mathbb{R}/\mathbb{Z}$? In particular, I would like to know if there are criteria that are easier to check than the following:

It is well-known that linear combinations of the form $\sum_{n=-N}^{N}a_{n}e^{2n\pi it}$ are dense in the Banach space $C\left(\mathbb{R}/\mathbb{Z}\right)$ of all continuous, $1$-periodic complex-valued functions on the unit interval, with respect to that space's norm (supremum). By standard functional analysis, it is known that every continuous linear functional on $C\left(\mathbb{R}/\mathbb{Z}\right)$ is a finite Borel measure on $\mathbb{R}/\mathbb{Z}$; as such, in order for $d\mu$ to be finite, it is sufficient that:$$\lim_{N\rightarrow\infty}\int_{0}^{1}\left(\sum_{n=-N}^{N}a_{n}e^{-2n\pi it}\right)d\mu\left(t\right)=\lim_{N\rightarrow\infty}\sum_{n=-N}^{N}a_{n}\hat{\mu}\left(n\right)$$ converges to a finite value whenever:$$\sum_{n=-N}^{N}a_{n}e^{-2n\pi it}$$ converges in supremum norm to a function in $C\left(\mathbb{R}/\mathbb{Z}\right)$ as $N\rightarrow\infty$. The reason I say this sufficient condition is “not easy to check” because, for example, you can't assume that the $a_{n}$s are in $\ell^{1}\left(\mathbb{Z}\right)$—in that case, $\sum_{n=-\infty}^{\infty}a_{n}e^{-2n\pi it}$ would be an element of the Wiener algebra on $\mathbb{R}/\mathbb{Z}$, and it is known that there are continuous functions on $\mathbb{R}/\mathbb{Z}$ which are not in the Wiener algebra.

答:
The Fourier-Stieltjes algebra of a discrete group $G$, usually denoted by $B(G)$, was defined  by  P. Eymard in 1964 to
be the linear span,  within the space of complex valued functions on  $G$,  of the positive type functions,
namely the functions
$f:G\to {\mathbb C}$,  such that
$$
  \sum_{i,j=1}^nc_i\overline {c_j}f(g_j^{-1} g_i)\geq 0,
  $$
whenever $g_1, g_2, \ldots , g_n\in  G$, and $c_1, c_2, \ldots , c_n\in  {\mathbb C}$.

As the name suggests,  $B(G)$ is indeed an algebra under pointwise multiplicaion of functions.

In the case of an abelian group $G$, such as the one of interest for the OP, the Fourier-Stieltjes algebra is precisely
formed by the Fourier transforms of the finite measures  on the dual group $\widehat G$.

Summarizing,  a function $\hat \mu :{\mathbb Z}\to {\mathbb C}$ satisfies the condition mentioned in the question if and only if it is a linear
combination of positive type functions on ${\mathbb Z}$.


问:
Showing that $E(\mathbb Q)_{\text{tors}} \cong \left\{\mathcal O\right\}$?
Suppose you have an elliptic curve $E/\mathbb{Q}: y^2= x^3+ax+b$. Suppose that $E$ has good reduction at primes $p$ and $q$, and suppose that you want to show $E(\mathbb Q)_{\text{tors}} \cong \left\{\mathcal O\right\}.$ Further more, suppose that $|E(\mathbb F_p)| = m$ and $|E(\mathbb F_q)| = n$. Then $|E(\mathbb Q)_{\text{tors}}| $ divides $\gcd(m, n)$. If $\gcd(m, n) = 1$ then $E(\mathbb Q)_{\text{tors}} \cong \left\{\mathcal O\right\}$. But say $\gcd(m, n)=2$. How do you rule out the curve having torsion subgroup of size $2$? And the same question for $\gcd(m, n)=3$. What's the process in general? Normally I would think of checking the order of non-trivial points over $E(\mathbb Q)$, but in this case there are no such points and this is what we're trying to prove.

My other question is if you have a set of points $\left\{a_0 \cdots, a_m\right\}$ torsion points over $E(\mathbb{F}_p)$ and $\left\{a_0 \cdots, a_n\right\}$ torsion points over $E(\mathbb{F}_q)$, can this give me explicit solutions over $\mathbb{Q}$? I know that you can have solutions over $\mathbb F_p$ but none over $\mathbb{Q}$, but when solutions exist over $\mathbb Q$ is there a way to determine these from the solutions over $\mathbb F_p$?

I'm new to elliptic curves, so I apologise if these are not good questions.

答:
A point $P = (x_0,y_0)$ is two torsion if and only if $y_0= 0$, you can see this from the definition of doubling on an elliptic curve, the tangent line goes to infinity only for a vertical tangent line, which happens only for points with $y$-coordinate zero (take the partial derivatives of the defining equation).
So to find the two torsion points we simply have to factor $x^3 + ax + b$ over the rationals.

More generally however this sort of analysis of tangent lines to find torsion points gets harder, but there is a theorem by Lutz-Nagell (https://en.wikipedia.org/wiki/Nagell%E2%80%93Lutz_theorem) that classifies torsion points in a very explicit way for elliptic curves over the rationals.
This theorem always gives us a finite list of possible points that we can check if they are torsion (if they even define points on the curve) by repeatedly adding one point to itself. Note that as soon as we get non-integer coordinates for a multiple of our point we know it cannot have been torsion so we can stop.


问:
Continuous semigroup
Let $X$ Banach, suppose that $A: D(A) \subset X \to X$ is the infinitesimal generator of a strongly continuous semigroup. $e^{At}, t \geq 0$. Then is the following number finite?

$$M=\sup_{0 \leq t \leq T} |{e^{At}}|_{L(X)}$$

答:
It is obvious from the exponential bound for $C_0$-semigroups: there exist always $M\ge 1$ and $\omega \in \mathbb{R}$ such that
$$\|e^{tA}\| \leq M e^{\omega t} \quad \text{for all } t\ge 0.$$
The proof is basically the same as in the answer by @MaoWao.

Title: Continuous semigroup


答:
Okay to solve $2^k \equiv \pmod {17}$ we know by FLT than $2^{16}\equiv 1 \pmod {17}$ and as $2^4 =16 \equiv -1$ that $2^8\equiv 1 \pmod {17}$.

So if $k \equiv 0, 1,2,3,4....,7\pmod {16}$ we have $2^k\equiv 1,2,4,8,-1,-2,-4, -8\pmod {17}$ and by CRT will be a unique solution for each pair. $k\equiv a \pmod {16}$ and $k \equiv 2^a \pmod {16}$. So that is eight solutions $\pmod {16*17}$.

(exmaple:  If $k\equiv 0\pmod {16}$ and $k \equiv 2^0=1\pmod {17}$ then $k\equiv -16 \equiv 256 \pmod {272}$.  Or if $k \equiv 1 \pmod 16$ and $k \equiv 2^1 \equiv 2\pmod {17}$ the solution $\pmod{272}$ is .... whatever)

Then we can do the same thing for $k\equiv b\pmod {18}$ and $k\equiv 2^b\pmod {19}$

So there are $18$ solutions $\pmod{18\cdot 19}$ for those.

However as $\gcd(16,18)=2\ne 1$ we must have $a,b$ either both even or both odd.

Still that is $4$ solutions $\pmod {16*17}$ with $a$ odd, and $4$ \solutions $\pmod{16*17}$ with $a$ even.  And $9$ solutions $\pmod{18*19}$ with $b$ even or odd.

So that is  $4*9=36$ solutions $\pmod{\frac {16*17*18*19}2}$ for $a,b$ odd and $36$ for $a,b$ even.

So there are $72$ solutions.

And it's way to much work do to.

But to find one solution we can have $k\equiv 256 \pmod {272}$

And $k \equiv 0 \pmod {18}$ and $k\equiv 2^0=1 \pmod{19}$ so $k\equiv -18\equiv 18^2 \mod {18*18}$ or $k \equiv 324 \pmod {342}$.

So if we solve $k\equiv 256\pmod {272}$ and $k \equiv 324{342}$ that solution ought to give us $2^k \equiv k\pmod {323}$ and ... oh, that $2^k - k\equiv 0$.... oh well the same ideaa will hold.

Title: Solving $2^k +k \equiv 0 \pmod {323}$

Body:
Find all $k$ such that
$$2^k + k \equiv 0 \pmod{323}.$$

I noticed that $323 = 17\cdot 19$ so I thought about using the Chinese Remainder theorem by considering $2^k+k$ modulo 17 and 19. I got $k \equiv 16 \pmod{17}$ and $k\equiv 18 \pmod{19}$, which gives $k\equiv 322 \pmod{323}$. However, after trying this, the given solution was not valid : $2^{322} + 322 \equiv 156 \not \equiv 0 \mod{323}$.

Can someone explain why this didn't work, and what I can do to solve this problem? Thanks.


问:
Young's inequality for three variables
Let $x, y, z \geqslant 0$ and let $p, q, r > 1$ be such that
$$
\frac{1}{p} + \frac{1}{q} + \frac{1}{r} = 1.
$$
How can one show that under these hypotheses we have
$$
xyz \leqslant \frac{x^p}{p} + \frac{y^q}{q} + \frac{z^r}{r}
$$
with equality if and only if $x^p = y^q = z^r$, using twice the standard two-parameters Young's inequality which says that for all $x, y \geq 0$ and for all $p, q > 1$ for which $\frac{1}{p} + \frac{1}{q} = 1$ we have
$$
xy \leqslant \frac{x^p}{p} + \frac{y^q}{q}
$$
with equality if and only if $x^p = y^q$ ?

I've tried to apply it twice directly, to multiply two inequalities and to add two inequalities, but in each case it gets quite messy and I can't get the desired result, even though I'm sure it should be quite simple.

答:
The function $\ln$ is concave, so if $\sum_n \lambda_n =1$, with $\lambda_n \ge 0$, then $\ln ( \sum_n \lambda_n x_n ) \ge \sum_n \lambda_n \ln x_n$ (with $x_n >0$, of course).

Hence $\ln ( \frac{x^p}{p} + \frac{y^q}{q} + \frac{z^r}{r} ) \ge \frac{1}{p} \ln x^p + \frac{1}{q} \ln y^q + \frac{1}{r} \ln x^r = \ln (x y z)$.

Taking exponents yields the desired result.

Since $\ln$ is strictly concave, we have equality iff $x_i = x_j$ in the first inequality, which corresponds to $x^p = y^q = z^r$.

Title: Young's inequality for three variables


Title: Solving $2^k +k \equiv 0 \pmod {323}$

Body:
values mod 323 and 46512 . The earlier post was in error because the final line of four numbers mod 46512 was printed below a calendar line and I did not notice them when pasting into this window...

Fri Mar 19 17:38:28 PDT 2021

    9   13   18   21   25   26   33   36   42   43
   49   50   52   66   67   69   72   77   83   84
   86   87   89   93   98  100  103  104  111  121
  132  134  135  138  144  154  161  166  168  172
  174  178  183  186  195  196  200  203  205  206
  208  213  217  222  229  242  253  259  263  264
  268  270  276  288  291  293  307  308  315  319
  321  322

    43    52   582   671  1820  2758  3438  3452  3574  3579
  3721  3846  4265  4405  4949  5431  5625  5975  6309  7335
  7496  8067  8345  8448  9029  9286  9563  9890 10055 10624
 10657 10842 11054 11304 11341 11440 11712 12367 12686 12697
 13018 13381 13652 13698 13834 14106 14323 14407 14604 14867
 16227 16233 16780 16917 17152 17460 17596 17868 17943 18920
 18947 19546 20552 20721 21259 21314 21405 22217 22431 22886
 22901 22946 23299 23308 23838 23927 25076 26014 26694 26708
 26830 26835 26977 27102 27521 27661 28205 28687 28881 29231
 29565 30591 30752 31323 31601 31704 32285 32542 32819 33146
 33311 33880 33913 34098 34310 34560 34597 34696 34968 35623
 35942 35953 36274 36637 36908 36954 37090 37362 37579 37663
 37860 38123 39483 39489 40036 40173 40408 40716 40852 41124
 41199 42176 42203 42802 43808 43977 44515 44570 44661 45473
 45687 46142 46157 46202


问:
Equilateral Triangles On Sides of a Parallelogram
On the sides of a parallelogram $ABCD$, construct similarly oriented equilateral triangles $ABC_1, BCA_1, CDD_1$, and $DAB_1$. Let $N,P,Q$ and $M$ be the centers of the triangles, respectively. Show that the quadrilateral $MNPQ$ happens to be a parallelogram.

I think this is a really interesting problem. I found it in the geometric transformations unit in my textbook. The given hint is to use the (central) symmetry with center the intersection of the diagonals of the given parallelogram ($ABCD$). I am not sure I see it though. Any help would be appreciated. Thank you in advance!

答:
Here is a proof via the complex geometry. Given the vertexes of the parallelogram at $z_1$, $z_2$, $-z_1$ and $-z_2$, the third vertexes of the equilateral triangles are

$$A = z_1e^{-i\frac\pi3} - z_2e^{i\frac\pi3}, \>\>\>
B = z_2e^{-i\frac\pi3} + z_1e^{i\frac\pi3}, \>\>\>
C = -z_1e^{-i\frac\pi3} + z_2e^{i\frac\pi3}, \>\>\>
D = -z_2e^{-i\frac\pi3} - z_1e^{i\frac\pi3}
$$

and their centers
$$X = \frac13(A+ z_1 -z_2) = \frac13 z_1(1+e^{-i\frac\pi3}) - \frac13z_2(1+e^{i\frac\pi3})$$
$$Y = \frac13(B+ z_2 +z_1) = \frac13 z_2(1+e^{-i\frac\pi3}) + \frac13z_1(1+e^{i\frac\pi3})$$
$$Z = \frac13(C- z_1 +z_2) = -\frac13 z_1(1+e^{-i\frac\pi3}) + \frac13z_2(1+e^{i\frac\pi3})$$
$$W = \frac13(D- z_2 -z_1) = -\frac13 z_2(1+e^{-i\frac\pi3}) - \frac13z_1(1+e^{i\frac\pi3})$$

Then, it is straightforward to verify

$$Y-X=Z-W,\>\>\>\>\>Z-Y = W-X$$

Thus, $XYZW$ is a parallelogram .


问:
If $\mathbb P(A)=0.61, \mathbb P(B)=0.56$ and $\mathbb P(A\cup B)=0.81$. Find $\mathbb P(A \mid A\cap B)$
I’ve figured out what $(A\cap B)$ is, which is $0.36.$

I tried putting 0.61 over 0.36 but the answer is over 1 which is obviously incorrect. How do I answer this question?

答: Obverse that $A \cap B \subseteq A\implies P(A|A\cap B) = 1$. Or maybe you want to find $P(A|A \cup B)$ ?


答: Given that you are in $A\cap B$ you are in $A$ with probability 1

Title: If $\mathbb P(A)=0.61, \mathbb P(B)=0.56$ and $\mathbb P(A\cup B)=0.81$. Find $\mathbb P(A \mid A\cap B)$

Body:
I’ve figured out what $(A\cap B)$ is, which is $0.36.$

I tried putting 0.61 over 0.36 but the answer is over 1 which is obviously incorrect. How do I answer this question?


答:
WLOG pick one element you choose to omit while creating that sequence. The number of possible sequences times 4 gives the final result because ${4\choose 1}={4\choose 3}=4$. Then consider that to generate a sequence containing each element at least once you can first fixate those 3 choices, you have ${10\choose 3}$ ways of doing so; for each of those ways you are left with $3^7$ choices to complete the series, because any of the 3 elements will be valid for any of the remaining 7 positions.

The total number of ways thus equals

${4\choose 1}{10\choose 3}3^7$

Edit: obviously the restraint of containing each element at least once leads to less ways than arbitrary sequences of length 10, which make for $3^{10}$, though obviously ${10\choose 3} > 3^3$ and thus I was mistaken for a moment. Instead, use inclusion-exclusion at that part of the proof.


问:
Counting 11 letter words with subword "FRED" - check my answer
I'm preparing for an exam and just want to make sure I am getting things correctly.

How many 11-letter 'words' constructed from the 26 letters of the english alphabet contain the subword "FRED" Where 'words' are just any string of 11 letter words (does not have to be a word in the dictionary)

a subword is a word inside another word. So an 11 letter word must have the word "FRED" in it, so FREDAABBCDEF is allowed, or XYZFREDABBC

My answer:
FRED contains 4 letters, so:

Choose the 4 positions to put FRED in: $c(11,4)$

Choose the remaining 7 letters (can be any repeatable letters): $p(26,7)$

So we have $c(11,4) * p(26,7)$

答:
The answer that you suggest is not right, unfortunately in more than one way. Here is a concrete analysis of the problem, lengthy but I hope fairly straightforward.  Your mistake was to try to use a simple formula, before carrying out an analysis of what is really going on.

How many $11$ letter words have FRED as a subword, with the letters FRED, in that order, occupying the first $4$ positions in the word? The remaining letters can be filled in $26^7$ ways, so the answer is $26^7$.

How many words have FRED as a subword, with the letters FRED, in that order, occupying positions $2$ to $5$ in the word? The remaining letters can be filled in $26^7$ ways, so the answer is $26^7$.

The same is true for the words that have FRED occupying positions $3$ to $6$, $4$ to $7$, $5$ to $8$, and so on up to $8$ to $11$.

Now it is tempting to add up, getting a total of $(8)(26^7)$.  However, this will double count, for example, the words that start with FRED, then some useless letter, then FRED again,  with $2$  letters added at the end.  There are several other types of words that have two FRED's in them that will be double-counted.

Luckily, we don't have to worry about words that have $3$ or more occurrences of FRED.  (If we were dealing with $47$-letter words, things could get very unpleasant.)

Let us find out how many double-counted words there are.  We will do it slowly, and then in a slicker way.

How many "two FRED" words have the first one at the beginning?  The second one can then start in any of positions $5$ to $8$, so $4$ positions in all.  For each of thse choices, the remaining "useless" letters can be placed in $26^3$ ways, for a total of $(4)(26^3)$.

How many "two FRED" words have the first FRED starting at position $2$?  The same reasoning as in the paragraph above shows that there are $(3)(26^3)$ such words.

How many have the first FRED starting at position $3$? Clearly $(2)(26^3)$. How many with the first FRED starting at position $4$? Clearly $(1)(26^3)$.  And that's all.

So in total the number of "two FRED" words is $(10)(26^3)$.

Recall that with some double-counting, we got a total of $(8)(26^7)$.  Get rid of the double counting by subtracting $(10)(26^3)$ and we get
$$(8)(26^7) -(10)(26^3)$$

A slicker way: We describe another way of counting the "two FRED" words.  Take some Scrabble tiles, and glue them together to form two copies of the word "FRED."  Think of these as two special mega letters.  To make a "two FRED" word, we take our $2$ mega letters, and $3$ ordinary letter, and put them in a row.  So we are making a weird $5$-letter word.  The positions of the two mega letters can be chosen in $\binom{5}{2}$ ways.  For each such choice, the other letters can be filled in $26^3$ ways, for a total of $\binom{5}{2}(26^3)$.  This is exactly the count of $(10)(26^3)$ that we got in a slower way earlier.

Cute, but maybe more than a little dangerous.  The more plodding approach probably lets us retain greater control over what is going on.

Comment: It is all too easy to count in a plausible way that turns out to be incorrect.  As a partial check, one might try to use the same reasoning on say $5$-letter words, and MO instead of FRED.  Use the kind of "general" reasoning mentioned above, and compare with a detailed hand-count.


问:
Intersection of two lines in vector form
Sorry about formatting, this is my first question here, and I have no idea how to do it properly.

Assuming I have two lines

$\displaystyle l_1 = \binom{x_1}{y_1} + a\binom{u_1}{v_1}$

$\displaystyle l_2 = \binom{x_2}{y_2} + b\binom{u_2}{v_2}$

Thus the intersection holds

$\displaystyle \binom{x_1}{y_1} + a\binom{u_1}{v_1} = \binom{x_2}{y_2} + b\binom{u_2}{v_2}$

and have made sure those two lines will intercept (so $u_1,v_1$ and $u_2,v_2$ aren't parallel), how do I find the intersection point elegantly?

Currently (I am using this in a C++ code), I need to do a lot of if() cases, since I cannot divide by 0, but in order to find the point, I need to substitute b in the formula

I get

$a = \dfrac{x_2-x_1+bu_2}{u_1}$

so here I have to check for $u_1\neq0$, and then substitute in the second line, where I then need to check for $v_2\neq0$, and this creates one hell a lot of code.

Is there a better way, am I just missing something easy?

答: You could use the elimination method to solve the simultaneous equations rather than the substitution method. If the two lines aren't parallel then there should be no cases when you could accidentally divide by $0$.</p>


问:
Does mean/expectation affect the variance of a sample?
Let's say I have an app $A$ that has a mean of $4$ stars and an app $B$ that has a mean of $3$ stars. Both have the same sample size. Would the variance stay the same? I thought it would because what's up top (of the unbiased variance formula) would stay the same, since it's $X$ minus the mean and the $X$ would change too along with the mean. I am kinda confused on it though.

答:
The mean of a sample is not informative of the variance of that sample.  For instance, both of your samples could have zero variance--all of your app A ratings are exactly $4$ stars, and all of your app B ratings are exactly $3$ stars.  Then the mean ratings are $4$ and $3$ respectively, and each has variance $0$ because all the ratings are identical within each type of app.

On the other hand, app A ratings could have an average of $4$ stars, half of them are $3$ star ratings and half are $5$ star ratings.  Now the variance of app A ratings is positive.  Or, you can apply the same reasoning to app B ratings.

If you don't specify any other information about how individual ratings are distributed on the rating scale, then you cannot answer the question about how the variance of the rating might or might not change with a changing mean.  And even if you did supply some underlying parametric model, it may not be possible to relate mean and variance because the answer depends on the choice of model and other assumptions.


问:
Prove that $F$ is non-singular and find the orthogonal projection of $ v$
Let be the follow bilinear form on $\mathbb{R}^4$ given by the matrix:

\begin{pmatrix}
    1 & 1 & 1 & 1 \\
    1 & 0 & 0 & 0 \\
    1 & 0 & 1 & 1 \\
    1 & 0 & 1 & 0
  \end{pmatrix}

Consider the subspace $F$, generated by $ w_1 = (1, 1, 0, 1) $ and $ w_2 = (0, −1, 1, 1)$. Prove that $F$ is non-singular and find a basis for $F^{\perp}$. Find the orthogonal projection of $ v = (1, 0, −1, 1) $ over $F$.

Is there an easier way to show that $F$ is non-singular? My attemp:

$F$ non-singular if $rad(F) = \{ {0_{\mathbb{R}^4}} \} $.

We calculate $F^{\perp}$ = $\langle (-1/2,1,1/4,0),(0,0,-1/2,1) \rangle $

$ rad(F) = F \cap  F^{\perp} = \{ {0_{\mathbb{R}^4}} \}  $

Then, I have several problems to find the projection of $v$ over $F$. I know that I need an orthogonal basis for $F$, let be $W=(w_1, w_2)$, and then we can do something like:

$ p^r_F (v) = \frac{\langle v,w_1 \rangle}{\langle w_1,w_1 \rangle } w_1 + \frac{\langle v,w_2 \rangle}{\langle w_2,w_2 \rangle } w_2 $

But, well, I'm not sure at all about that last step... Any help for that questions?? Thanks!!

答:
$$  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc     \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc $$

$$ P^T H P = D  $$

$$\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
 - 1 & 1 & 0 & 0 \\
0 &  - 1 & 1 & 0 \\
0 & 0 &  - 1 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
1 & 1 & 1 & 1 \\
1 & 0 & 0 & 0 \\
1 & 0 & 1 & 1 \\
1 & 0 & 1 & 0 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
1 &  - 1 & 0 & 0 \\
0 & 1 &  - 1 & 0 \\
0 & 0 & 1 &  - 1 \\
0 & 0 & 0 & 1 \\
\end{array}
\right)
 = \left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
0 &  - 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 &  - 1 \\
\end{array}
\right)
$$
$$  $$

$$ Q^T D Q = H  $$

$$\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
1 & 1 & 0 & 0 \\
1 & 1 & 1 & 0 \\
1 & 1 & 1 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
0 &  - 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 &  - 1 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
1 & 1 & 1 & 1 \\
0 & 1 & 1 & 1 \\
0 & 0 & 1 & 1 \\
0 & 0 & 0 & 1 \\
\end{array}
\right)
 = \left(
\begin{array}{rrrr}
1 & 1 & 1 & 1 \\
1 & 0 & 0 & 0 \\
1 & 0 & 1 & 1 \\
1 & 0 & 1 & 0 \\
\end{array}
\right)
$$

$$  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc     \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc $$

...


问:
I don't understand how implication works in logic
The statement $(p \wedge q ) \Rightarrow (p \Rightarrow q)$ is a tautology, i.e., always true.

Now, imagine I am writing a proof and I want to show that $p \Rightarrow q$. I show that statement $p$ is always true, then I show that statement $q$ is always true, then I can end my proof and claim $p \Rightarrow q.$

This doesn't make sense to me. If I prove that $\sqrt2$ is irrational and then I prove that $7$ is odd, I don't understand how "$\sqrt2$ is irrational" $\Rightarrow$ "$7$ is odd".

Can someone help this disconnect I am having? Perhaps, you can provide information about what these logical statements actually mean and how they are used.

答: Here is an example, outside of mathematics, that I think shows how material implication is accepted and understood in ordinary human discourse. There was a court case, in which the wife of a serial killer contested claims, made in the satirical magazine Private Eye, that she had tried to profit on her husband's crimes. When, she won the case, Ian Hislop, the editor of the the magazine stated that "if that's justice, I am a banana". He was quite right and she ended up the poorer for it.


答:
If $A$ and $B$ both occur, then certainly $A$ occurs, so the probability is $1$.  If you want to use the formula,

$$\Pr(A|A\cap B)=\frac{\Pr(A\cap(A\cap B))}{\Pr(A\cap B)}=\frac{\Pr(A\cap B)}{\Pr(A\cap B)}=1$$


问:
How would I prove this statement using an indirect proof?
So I am new to discrete math and I am learning about proofs. I have learned a little about contrapositive and contradiction proofs.

This is the example I have trouble with $$\forall k, l \in \Bbb{Z} \, , \, kl \text{ is even} \implies k \text{ is even } \vee l \text{ is even}.$$

Suppose $k$ is odd and $l$ is odd, then there exist integers $a$ and $b$ such that $k = 2a + 1$, so $kl = (ab??)$

What I have?

答:
For an indirect proof, we assume the opposite to be true and find a contradiction. This means we assume

$\exists k,l \in \mathbb{Z} : ( 2\mid kl \land 2\nmid k,l)$

Notice that then $k=2a+1$ and $l=2b+1$, therefore $kl=(2a+1)(2b+1)=2(2ab+a+b)+1$ which clearly contradicts the assumption $2\mid kl$


问:
How many sequences of length 10 with elements $\{a, b, c, d\}$ have exactly $3$ out of $4$ elements?
My logic is since $3$ out of $4$ elements are chosen, each element would appear once.
So a sequence would look like: $a\,b\,c\,x\,x\,x\,x\,x\,x\,x$

We have $7$ spots $x$ that can be whatever elements so: $3^7$

Choosing $3$ out of $4$ elements:  $4\choose 3$$ = 4$
So in total, there are $4\cdot(3^7)$ sequences

Can someone tell me if this is right or not? Thanks

答:
First, select the $3$ elements to show up, in $\binom{4}{3}$ ways.

In all, there are $3^{10}$ sequences of $3$ elements, of which $\binom{3}{2} \cdot 2^{10}$ are of at most two symbols. Thus:

$$
   \binom{4}{3} \left( 3^{10} - \binom{3}{2} \cdot 2^{10} \right)
      = 223908
$$


问:
Finitely generated $R$-module as a sum of cyclic submodules
I know that  $M$ is finitely generated if there exist $a_1, ..., a_n$ in $M$ such that for any $x$ in $M$, there exist $r_1, ..., r_n$ in $R$ such that $x = r_1a_1 + \dots + r_na_n.$

But if I want to express this by its subsets is it correct to say, let $M$ be a finitely generated $R$-module, then $M = \sum_{i \in I}M_i$ where $I$ is a finite set of integers and $M_i \subseteq M$ for every $i$?

答: This would be correct if you require each $M_i$ to be the submodule of $M$ generated by a single element $a_i$ of $M$.


问:
Sequence of elements representing a Hamiltonian cycle and the generating element of a subgroup
Let $S$ be a subset of a finite group $G$. The Cayley graph $Cay(G,S)$ can be defined as the graph whose vertices are the elements of $G$, with an edge joining $g$ and $gs$, for every $g \in G$ and $s \in S$.

For $s_1,s_2,\cdots,s_n \in S \cup S^{-1}$, $(s_1,s_2, \cdots, s_n)$ is used to denote a walk in $Cay(G,S)$ that visits (in order) the vertices, $e, s_1, s_1 s_2, \cdots, s_1 s_2 s_3 \cdots s_n$.

Lemma: Suppose,

$S$ is a generating set of $G$,

$H$  is a cyclic subgroup of $G$, with index $|G:H|=n$,

$s_1,s_2,\cdots,s_n$ is a sequence of $n$ elements of $S \cup S^{-1}$ such that,

i) the elements $e, s_1, s_1 s_2, \cdots, s_1 s_2 s_3 \cdots s_n$ are all in different right cosets of $H$,

ii) the product $s_1 s_2 \cdots s_n$ is a generator of $H$.

Then, $(s_1, s_2, \cdots , s_n)^{|H|}$ is a Hamiltonian cycle in $Cay(G,S)$.

In this lemma, what is the importance of having $s_1 s_2 \cdots s_n$ as a generator of $H$?

I have seen other lemmas and theorems too where similar condition that such a product should be a generating element of some subgroup is mentioned. But I'm unable to see why?

Also, suppose a case where $S=\{u,t\}$. Then a Hamiltonian cycle in a Cayley graph with respect to $S$ will be expressed in terms of $u,t$. Then, if the Hamiltonian cycle is $(u,t,t^{-1},t, \cdots, t)^{|H|}$, I thought probably the above definition of notations mean to take $s_1=s, s_2=t, s_3=t^{-1}, s_4=t, \cdots, s_n=t$ since they have not mentioned that $s_1, s_2, s_3, \cdots, s_n$ are distinct elements. Am I right?.

Can some one help me with the above questions.

Thanks a lot in advance.

答:
In the walk $(s_1, s_2, \dots , s_n)^{|H|}$, we visit $H$ every $n$ elements, so after $n|H|$ steps, the elements of $H$ we see are exactly $(s_1 s_2 \cdots s_n)^k$ for $0 \le k \le |H|-1$. If $s_1 s_2 \cdots s_n$ does not generate $H$, then we won't get a Hamiltonian cycle, because we won't visit all the different elements of $H$.

And yes, it's fine for some of the elements $s_1, s_2, \dots, s_n$ to coincide in the definition of $(s_1, s_2, \dots, s_n)$.


问:
Is my solution to this homework problem in geometry correct? Can we avoid trigonometry in solving this?
I'm trying to solve this nice homework problem in (Euclidean) geometry - https://snipboard.io/oVp3Ic.jpg, and I wrote down a detailed solution with step-by-step explanation at: https://snipboard.io/ZzdsA0.jpg (click on the + sign on the image to make it bigger). Note that in the picture of the solution, DE=7cm. Also, note that, in the picture of the problem, the drawings aren't correct, because some sides AB an CE supposed to be parallel, but they're not drawn to be so, which could be misleading.

Some steps in the solution I gave:

i) I first solved for the angle ABC

ii) I expressed BC and CD as a trigonometric functions of the angle ABC (called $\theta.)$

iii) Observed $x=BC + CD.$

I'd greatly appreciate if you could please the take a bit of time to see if my solution is correct, and if we can avoid the use of trigonometry in solving this problem? I think in the picture of the solution, if you consider the similar right-angles triangles $CDE$ and $CGB,$ then we can write:

$$\frac{DE}{CE}=\frac{CG}{BG}$$

which is, algbraically ( where $y=AG$ in the picture of the solution.):

$$\frac{7}{6+y}=\frac{3}{\sqrt{(9-y)^2 + 3^2}}.$$

答:
HINT.-

$(1)$ Plot in the cartesian plane the red broken line of vertices $A=(0,0),B=(0,9),C=(3,9),D=(3,15)$ as in the attached figure. This broken line satisfies the wanted condition of perpendicularity.

$(2)$ Find the tangent from the point $A$ to the circle of equation $(x-3)^2+(y-15)^2=7^2$.

You must find $x\approx13.60120$.


问:
Does strict convexity everywhere implies strong convexity on a neighborhood of almost every point?
Prove or disprove the following:

For every strictly convex $f : \Bbb R^n \to \Bbb R$, for almost every $x \in \Bbb R^n$ there exists a neighborhood $N(x)$ of $x$ such that $f$ is strongly convex in $N(x)$.

The usual examples of strictly convex functions that are not strongly convex, for instance $x \mapsto x^4$, do satisfy this property.

答:
A counterexample for $n=1$: Let $h: \Bbb R \to \Bbb R$ be a function with the following properties:

$h$ is continuous,

$h$ is strictly increasing,

$h'(x) = 0$ almost everywhere.

Such functions do exist, see for example Showing the existence of a continuous, strictly increasing function $f$ on $\mathbb{R}$ such that $f'(x) = 0$ almost everywhere or Existence of a Strictly Increasing, Continuous Function whose Derivative is 0 a.e. on $\mathbb{R}$.

Then $f: \Bbb R \to \Bbb R$ defined by $f(x) = \int_0^x h(t) \, dt$ is strictly convex, because $f'=h$ is strictly increasing.

But $f$ is not strongly convex on any non-empty open interval $I$:  Assume that there is a constant $m > 0$ such that
$$
 h(y) - h(x) =  f'(y) - f'(x) \ge m (y-x)
$$
for all $x, y \in I$ with $x< y$. Then $h'(x) \ge m > 0$ a.e. in $I$, in contradiction to the construction of $h$.

Therefore no $x \in \Bbb R$ has a neighborhood on which $f$ is strongly convex.

This can be extended to a counterexample $F: \Bbb R^n \to \Bbb R$ for any $n$ by defining
$$
F(x_1, \ldots, x_n) = f(x_1) +  \cdots + f(x_n) \, .
$$


问:
How can I interpret the slice category of a Monoid under its uneque element?
Let $M$ be a monoid, regarded as a category with one single object, $\bullet$. Obviously, the objects of the slice category $\bullet/M$ are all morphisms of $M$, and its morphisms $h:(f: \bullet\rightarrow \bullet) \rightarrow (g:\bullet \rightarrow \bullet)$ are all morphims of $M$ that satisfy $g=hf$. With what can I identify $\bullet/M$ within Abstract Algebra?  For instance, we say that the slice category $c/P$ of a Poset $P$ is the up-set $\uparrow(c)$; I'm trying to do something similar for $M$.

答:
In general, we can't do better than what you obtained.

The slice category $\bullet/M$ is a category, its objects are the arrows $\bullet\to\bullet$, that is, the elements of $M$, and the arrows $f:a\to b$ are also (labelled by) elements of $M$, namely which satisfy $fa=b$.

If $M$ is right cancelable, $xa=ya\implies x=y$ (e.g. if it is a submonoid of a group), then $\bullet/M$ becomes a posetal category by the order $a\le b$ if $xa=b$ for some (hence a unique) $x$.

As in the comments, it gives the usual ordering in the case $M$ is the additive monoid of nonnegative (integer/real) numbers, and it gives the divisibility relation in the case $M$ is a multiplicative submonoid of an integral domain.


问:
Difference between Bi-annual and semi-annual in Financial Maths
I was wandering what the difference was between compounding interest when they use bi-annual and semi-annual and hence how to change your value of i

I think semi-annual means twice in 1 year so your i would be i/2? and then you would multiply your years by two as well

however im not sure how to deal with bi-annual, i think it means once every 2 years so would you take you i and divide it by 0.5 as well as your number of years?

答:
I assume you are using the formula for compound interest: $$A = P \left(1 + \frac{i}{n}\right)^{nt}$$ where $A$ is the future value, $P$ is the present value, $i$ is the annual interest rate (as a decimal), $n$ is the number of times compounded per year and $t$ is the length of time in years. It is very important here that the question states interest as the annual interest rate.

Semi-annual means twice in one year. Therefore, your $n$ will equal 2. Hence, your formula becomes $$A = P \left(1 + \frac{i}{2}\right)^{2t}.$$

You are correct that bi-annual means once every two years. Therefore, the interest is compounded "half" a time per year (1 compounding every 2 years for $\frac{1}{2}$). Now we have $n = \frac{1}{2}$ and $$A = P \left(1 + \frac{i}{\frac{1}{2}}\right)^{\frac{1}{2}t}$$


问:
Obtaining a subcanonical topology
Given a small site $(\mathbb{C}, \mathcal{J})$, it is known that one can devise a subcanonical site $(\mathbb{D}, \mathcal{K})$ so that $\mathsf{Sh}(\mathbb{C}, \mathcal{J}) \simeq \mathsf{Sh}(\mathbb{D}, \mathcal{K})$. Is it possible in general to choose $\mathbb{D} := \mathbb{C}$, i.e. is it possible to keep the same base category and just modify the topology? If not, then is there a general construction of $\mathbb{D}$ from $(\mathbb{C}, \mathcal{J})$ that is "close" to the category $\mathbb{C}$ (which is an admittedly vague question)?

答:
No, it is not always possible to choose $\mathcal{D} = \mathcal{C}$.
For example, suppose $J$ is the topology where every sieve – even the empty sieve – covers; then the topos is trivial, so any subcanonical site for it must have the property that there is exactly one morphism between any two objects.
But $\mathcal{C}$ could be an arbitrary category.

In general, the way to obtain a subcanonical site from an arbitrary site is to take the full subcategory $\mathcal{D}$ of the topos spanned by the sheafification of the representable presheaves.
The topology $K$ has as its covering sieves the sieves in $\mathcal{D}$ that are jointly epimorphic in the topos.
By construction there is a bijective on objects functor $\mathcal{C} \to \mathcal{D}$ that sends $J$-covering sieves to $K$-covering sieves.

In fact, $J$ is subcanonical if and only if this $\mathcal{C} \to \mathcal{D}$ is fully faithful; and if $J$ is subcanonical then $J = K$, modulo the identification of $\mathcal{C}$ and $\mathcal{D}$.
So you could think of this construction as "the best" subcanonical site related to the site you start with.

Title: Obtaining a subcanonical topology


问:
calculating a probability from a T-table

This problem comes from Probability and Statistics for Engineers by Scheaffer, Mulekar, McClave

Can someone please lead me in the right direction here? My understanding is that I cannot use a Z-score here because I am not being given the population standard deviation. Thus I have

$t = ((x̅ - u)$ $\sqrt{n}$)) / $s$

$u = 200, n = 15, df = 14, s = 10$

Of course, one confusing part of the question is that it states there are 15 resistors but the standard deviation is of "12" measurements. I'm just assuming this should be "15" measurements, thus s = 10

a) plugging in the values of $x̅ = 199$ and $x̅ = 202$ I get $-0.3873$ and $0.7746$

$P(-0.3873 \leq t \leq 0.7746)$ with $df = 14$ using a calculator I get $0.422$

The book has an answer key for odd numbers in the back, and the only thing they have is 0.3224 which makes even less sense because that's one answer for a two-part question.

b) 5,100 ohms among 15 resistors is an average of $340$.

Plugging this into $t = ((x̅ - u)$ $\sqrt{n}$)) / $s$ I get that $t = 54.22$ which is a ridiculous number because to my understanding, this is supposed to represent a deviation from the mean. Yet it doesn't seem unreasonable, because if the expected value is 200, 15 resistors at random should be about 3000 ohms. The likelihood of them adding up to a number greater than 5,100 seems like it should be virtually impossible (and that is reflected in the numbers)

Even if I did use Z-Scores, it wouldn't give me a more rational answer. Can someone please give me a sanity check here? Am I doing something extraordinarily wrong or is this question just complete nonsense? These aren't the first errors I've found in this book. They're not even the 10th errors I've found in this book. The authors of this book should be embarrassed for publishing this text, and my university should be equally embarassed for teaching from it. Excuse my rant, It's more frustrating than I can put into words trying to learn content that I personally find challenging compounded with dozens of careless mistakes.

答:
This is a very poorly worded question and I don't blame you for being confused.  And the claimed answer is wrong.

The intent of the question is that the standard deviation of the resistance is estimated as $\hat \sigma = 10$, based on $12$ measurements, and that this estimate is assumed to represent the true standard deviation $\sigma$.  Therefore, no $t$-distribution is used for this question at all.  Everything is calculated using a normal distribution.  Don't ask me how I know; it's just experience from reading too many badly written statistics textbook questions.

So what you're trying to do--using a $t$-distribution to model the sample mean--won't work.  To understand why, let's suppose $X_1, X_2, \ldots, X_n$ represents an iid sample of resistances.  With $n = 15$ such observations, the sample mean is $$\bar X = \frac{1}{n} \sum_{i=1}^n X_i.$$  You are asked to calculate the probability $$\Pr[199 \le \bar X \le 202]$$ under the assumption that $$X_i \sim \operatorname{Normal}(\mu = 200, \sigma = \sigma_x),$$ which is equivalent to $$\bar X \sim \operatorname{Normal}(\mu = 200, \sigma_{\bar x} = \sigma_x/\sqrt{15}).$$  Note here that I have left $\sigma_x$ unknown.  Now, you could standardize: $$\Pr[199 \le \bar X \le 202] = \Pr\left[\frac{199 - 200}{\sigma_{\bar x}} \le \frac{\bar X - \mu}{\sigma_{\bar x}} \le \frac{202 - 200}{\sigma_{\bar x}}\right],$$ and the quantity in the center of the inequality is standard normal, but you don't know $\sigma_{\bar x}$ unless you use the estimate $\sigma_x = \hat \sigma = 10.$  However, if you studentize, you are even worse off:  $$\Pr[199 \le \bar X \le 202] = \Pr\left[\frac{199 - 200}{s/\sqrt{n}} \le \frac{\bar X - \mu}{s/\sqrt{n}} \le \frac{202 - 200}{s/\sqrt{n}}\right]$$ where $s$ is the sample standard deviation $$s = \sqrt{ \frac{1}{n-1} \sum_{i=1}^n (X_i - \bar X)^2 }$$ estimated from the same sample, and now the quantity in the center is student $t$ with $n - 1$ degrees of freedom.  But even if you knew the value of $s$ (which you don't because it is a random variable), you cannot use it to solve for $\bar X$ because you don't know the individual values of the $X_i$.  This approach is impossible.  So the only feasible interpretation here is to use the normal model and assume that $\sigma_x = \hat \sigma$.  Under this assumption, the answer should be $0.43144330780783071032\ldots$.

In any case, for the second part of the question, the answer is effectively $1$, no matter what model you use.  This is because the mean of the total of $15$ resistors is going to be $200(15) = 3000$ ohms, well below $5100$ ohms--in fact, it is over $54$ standard deviations above the mean.  The resulting probability is less than $1$ by about $2 \times 10^{-641}$, and even if you assume a standard deviation that is $10$ times larger, the probability is still around $0.99999997$.  Such questions are not interesting to ask.

I would also like to add in closing that sometimes we learn more from badly written textbooks than we do from a good one, because it keeps us vigilant.  If you feel like you can't trust everything in the book to be accurate, you're more likely to work harder to check every claim in it.  In a perverse way, this can lead to a better understanding of the material, especially if you have access to a better text where you can compare the two.  When working on poorly written questions such as these, your goal shouldn't be to replicate the answers that are provided.  Sometimes, it won't even be possible to give any meaningful answer to a bad question.  Instead, the goal should be to formulate a complete response to the question by addressing the following:

In what ways is the question deficient/defective, and why?

How would you rewrite or correct the question so that it can be answered?

How would you write up a complete, step-by-step solution to such a revised question?

In essence, this is the approach I have taken above.


问:
Is $\sqrt{x} = \sqrt{y} \iff x = y$ true?
Is the following statement true for $x \geq 0,y \geq 0$?
$$
\sqrt{x} = \sqrt{y} \iff x = y
$$
Starting with the statement on the left, multiplying both sides by $\sqrt{x}$:
$$
\sqrt{x} \cdot \sqrt{x} = \sqrt{y} \cdot \sqrt{x}
$$
Since it is assumed that $\sqrt{x} = \sqrt{y}$, then:
$$
\sqrt{x} \cdot \sqrt{x} = \sqrt{y} \cdot \sqrt{x} = \sqrt{y} \cdot \sqrt{y} \\
\sqrt{x^2} = \sqrt{y^2}
$$
Not sure how to proceed from here.

答:
HINT

If $x\geq 0$ and $y\geq 0$, then it makes sense to take square roots.

Consequently, we deduce that
\begin{align*}
x = y & \Longleftrightarrow x - y = 0\\\\
& \Longleftrightarrow (\sqrt{x})^{2} - (\sqrt{y})^{2} = 0\\\\
& \Longleftrightarrow (\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y}) = 0\\\\
\end{align*}

Can you take it from here?


答:
Your last step unnecessarily complicates things. Just go straight to $\sqrt{x} \sqrt{x} =x$ and $\sqrt{y} \sqrt{y} =y$ by definition.

The other direction is basically tautologically true. If $x=y$ then their image under $f(x)=\sqrt{x}$ must not be ambiguous, hence $\sqrt{x} =\sqrt{y}$

(Note that here f can without problems be defined such that $f: \mathbb{R}_{+}\rightarrow \mathcal{P}(\mathbb{R}), x\mapsto \{-\sqrt{x},\sqrt{x}\}$, not neglecting the fact that the square root technically produces 2 outputs)

Title: Is $\sqrt{x} = \sqrt{y} \iff x = y$ true?

Body:
Is the following statement true for $x \geq 0,y \geq 0$?
$$
\sqrt{x} = \sqrt{y} \iff x = y
$$
Starting with the statement on the left, multiplying both sides by $\sqrt{x}$:
$$
\sqrt{x} \cdot \sqrt{x} = \sqrt{y} \cdot \sqrt{x}
$$
Since it is assumed that $\sqrt{x} = \sqrt{y}$, then:
$$
\sqrt{x} \cdot \sqrt{x} = \sqrt{y} \cdot \sqrt{x} = \sqrt{y} \cdot \sqrt{y} \\
\sqrt{x^2} = \sqrt{y^2}
$$
Not sure how to proceed from here.


问:
Can this sum be written as a matrix product?
Given matrices $D,C$ and vector $\vec{\lambda}$, I am wondering if the sum  $\sum_kc_{jk}\lambda_kd_{kl}$ can be written as some sort of matrix product. The context comes from the following scenario:

Say we have some operator $A$ and want to act it on an expansion over some basis $\{\phi_j\}$, ie, we want to calculate $A\sum\alpha_j\phi_j$. We can do this by expanding each $\phi_j=\sum_kc_{jk}\chi_k$ where the $\chi_k$ are the eigenvectors of $A$ with corresponding eigenvalues $\lambda_k$. The calculation now becomes

$$
A\sum_j\alpha_j\sum_kc_{jk}\chi_k=\sum_{j,k}\alpha_jc_{jk}\lambda_k\chi_k
$$

Now given the previously mentioned change of basis matrix $C=(c_{jk})$, we can use its inverse $C^{-1}=D=(d_{kj})$ to expand the eigenbasis for $A$ back into our basis $\{\phi_j\}$: $\chi_k=\sum d_{kj}\phi_j$ and we can expand our result from earlier:

$$
\sum_{j,k}\alpha_jc_{jk}\lambda_k\chi_k=\sum_{j,k,l}\alpha_jc_{jk}\lambda_kd_{kl}\phi_l
$$

the sum of interest from earlier is seen here as $e_{jl}=\sum_kc_{jk}\lambda_kd_{kl}$. If I consider the matrix $E=(e_{jl})$, can $E$ be written in simple terms with respect to $C,D$ and $\vec{\lambda}$?

答:
The only way I can think of is if we let
$$\Lambda = \text{diag}( \lambda_1, \ldots, \lambda_n ),$$that is, if it has matrix elements $$\Lambda_{kl} = \delta_{k,l} \lambda_k,$$ then using your notation we would have
$$ E = C \Lambda D.$$


问:
Calculate the measure in degrees of the acute angle formed by $ NMA $?
Consider the $PAT$ triangle such that the angle $ \angle P = 36 $ degrees, $ \angle A = 56 $ degrees and $ PA = 10 $. Knowing that the points $ U $ and $ G $ belong, respectively, to the sides $ TP $ and $ TA $, so that $ PU = AG = 1 $. If $ M $ and $ N $ are the midpoints of the $ PA $ and $ UG $ segments, respectively. What is the measure in degree of the acute angle formed by $ NMA $?

Can anyone give me any tips?

答:
If you construct $P^*$ and $A^*$ so that $$\,\,\,PP^* \, || \, UN \, , \,\,\,PP^* = UN = \frac{1}{2} UG\,\,\,v\text{ and }\,\,\,PP^* \, || \, UN\, , \,\, PP^* = UN = \frac{1}{2} UG\,\,\,$$
then $UPP^*N$ and $GAA^*N$ are parallelograms and so
$$\,\,\,NP^* \, || \, UP \, , \,\,\,NP^* = UP \,\,\, \text{and} \,\,\,NA^* \, || \, GA\, , \,\, NA^* = GA\,\,\,$$ which yields $NP^* = NA^*$ and thus triangle $\Delta \, NP^*A^* $ is isosceles. Furthermore, since $NP^* \, || \, TP$ and $NA^* \, || \, TA$
$$\angle \, P^*NA^* = \angle \, PTA = 88^{\circ}$$
Observe that by construction, $PP^* \, || \, UG\,, \,\,\,AA^* \, || \, UG\,\,$ and $\,\, PP^* = \frac{1}{2}UG = AA^*$, which implies that $PP^*AA^*$ is a parallelogram. By the fact that the diagonals of a parallelogram share a common midpoint, since $M$ is the midpoint of $PA$, it follows that $M$ is also the midpoint of $P^*A^*$. That implies that $NM$ is a median in the isosceles triangle $\Delta \, NP^*A^*$ which means that $MN$ is also the angle bisector of $\angle \, P^*NA^*$ so $$\angle \, MNA^* = \frac{1}{2} \angle \, P^*NA^* = 44^{\circ}$$ Set $L = NA^* \,\cap \, PA$ and since $NA^* \, || \, TA$
$$\angle \, NLM = \angle \, TAP = 56^{\circ}$$ We know two of the three angles od the triangle $\Delta \, NML$ so the thrid angle is

$$\angle \, NML = 180^{\circ} - 56^{\circ} - 44^{\circ} = 80^{\circ}$$


答: If $\sqrt{x} = \sqrt{y}$, then you can square both sides to obtain x =y. But the other way is tricky, if x =y and you are taking square root on both sides, you have option to take either positive or negative square root. If you take both of them to be positive, then they are equal or if you take both of them to be negative, they are equal. For example, if x=4, y= 4; when you take square root if the sign of both square roots are same, only then they will be equal.


问:
How does one show that if $B \subset A$ (where $A$ is a directed set) is cofinal in $(A, \leq)$ then $(B, \leq)$ is a directed set?
First two axioms of directed sets readily follow for $(B,\leq)$ by the virtue of being a subset of $(A,\leq)$ but I don't see how the third one follows.

Directed set $(A,\leq)$ is a set with the order relation $\leq$, where the order relation $\leq$ is reflexive, transitive and that every pair of elements has an upper bound in $(A,\leq)$.

By "$B$ is cofinal in $A$.", I mean, that each element of $A$ is bounded above by some element of $B$.

答:
Since $\leq$ is a partial order on $A$, it is a partial order, by restriction on any subset; this is a property of partial orders, directed or not.

Let $b_1,b_2\in B$. Since $A$ is directed, there exists $a\in A$ such that $b_1\leq a$ and $b_2\leq a$. Since $B$ is cofinal in $A$, there exists $b\in B$ such that $a\leq b$. Thus, there exists $b\in B$ such that $b_1\leq b$ and $b_2\leq b$. Hence, $B$ is directed.


答:
Direct proof:

Strategy: assume $kl$ is even.  Prove that either $k$ is even or $l$ is even.

Proof by contradiction:

Strategy: assume $k$ is odd, $l$ is odd and $kl$ is even.  Get a contradiction (Possibly that $kl =2m$ and $m$ is not an integer.)

Proof by contrapositive

Strategy: assume $k$ is odd, $l$ is odd.  Show that this must mean $kl$ is odd.

.....

The contrapositive proof is (ironically) the most direct and easy.

If $k,l$ are odd there are integers $a,b$ so that $k = 2a + 1$ and $l = 2b+1$.

So $kl = (2a+1)(2b+1) = 4ab + 2a + 2b + 1=2(2ab + a+b) + 1$ is odd.

....

The contradiction proof is similar.

Let $k=2a+1$ and $l = 2b + 1$ and $kl = 2m$ so $2m = (2a+1)(2b+1)$.  The contradiction is RHS is odd and LHS is even.

But it was really unnecessary to assume $kl$ was even.  It added something unnecessary we never really worked with.

....

And direct... not very direct.

Let $kl = 2m$ be even.

The $l = \frac {2m}k=2\frac mk$.

Now either $k$ divides evenly into $m$ or it doesn't.  If it does, then $l$ is even and we are done.

If it doesn't then $\frac mk$ is not an integer but $2\frac mk$ is so $\frac mk = w + \frac 12$ for some integer $w$.  So $m = (w+ \frac 12)k$.  But $m$ is an integer and that can only happen if $k$ is even.

So either $l$ or $k$ is even (or both).

Title: How would I prove this statement using an indirect proof?

Body:
So I am new to discrete math and I am learning about proofs. I have learned a little about contrapositive and contradiction proofs.

This is the example I have trouble with $$\forall k, l \in \Bbb{Z} \, , \, kl \text{ is even} \implies k \text{ is even } \vee l \text{ is even}.$$

Suppose $k$ is odd and $l$ is odd, then there exist integers $a$ and $b$ such that $k = 2a + 1$, so $kl = (ab??)$

What I have?


问:
Intersection of two lines in vector form
Sorry about formatting, this is my first question here, and I have no idea how to do it properly.

Assuming I have two lines

$\displaystyle l_1 = \binom{x_1}{y_1} + a\binom{u_1}{v_1}$

$\displaystyle l_2 = \binom{x_2}{y_2} + b\binom{u_2}{v_2}$

Thus the intersection holds

$\displaystyle \binom{x_1}{y_1} + a\binom{u_1}{v_1} = \binom{x_2}{y_2} + b\binom{u_2}{v_2}$

and have made sure those two lines will intercept (so $u_1,v_1$ and $u_2,v_2$ aren't parallel), how do I find the intersection point elegantly?

Currently (I am using this in a C++ code), I need to do a lot of if() cases, since I cannot divide by 0, but in order to find the point, I need to substitute b in the formula

I get

$a = \dfrac{x_2-x_1+bu_2}{u_1}$

so here I have to check for $u_1\neq0$, and then substitute in the second line, where I then need to check for $v_2\neq0$, and this creates one hell a lot of code.

Is there a better way, am I just missing something easy?

答:
The most elegant formulation (IMO) uses Cramer's Rule. You can reformulate your two equations into $$\begin{bmatrix}
u_1 & -u_2\\
v_1 & -v_2
\end{bmatrix} \cdot \begin{bmatrix}
a \\
b\end{bmatrix} = \begin{bmatrix}
x_2-x_1 \\
y_2 -y_1
\end{bmatrix}$$
and use determinants to solve for $a$.

Another nice way to think about it is areas: For segments AB and CD, the fraction along AB at which the intersection occurs is equal to the area of the triangle ACD divided by the area of the quadrilateral ACBD.


问:
An estimate for required $n^{\alpha}\ge \ln(n^2)$
I am trying to find an estimate in terms of $\alpha\in(0,1)$ of how large $n$ should be so that the below inequality is satisfied for all large $n$.
$$ n^{\alpha}\ge \ln(n^2)$$
Is there a good way to go about this?

答:
One approach is graphically.  $\ln x$ is negative for $x < 1$, and climbs slowly after that. A power function $x^\alpha$ passes through the origin for $\alpha > 0$ and rises fairly rapidly.  For example, if $\alpha = \frac{1}{2}$, then I don't believe the curves ever intersect.  You might use desmos.com to let $\alpha$ vary.

Another approach is calculating specific examples.  If $n = 1, \ln n = 0$ and $n^\alpha = 1$ for any \alpha.  If $n = 2$, then $2^\alpha \geq \ln 2$ can be solved for \alpha.  I expect any intersections will happen at low values of $n$.


问:
In how many ways can a team be formed when some players can play multiple positions?
I want to form a lineup where we have 1 goalkeeper, 2 defenders and 3 forwards. We have 3 goalkeepers, 7 defenders and 10 forwards available, as well as 4 players who can play as defender or forward. How many possible lineups are there?

My approach was to separately consider the cases where $n$ of the 4 players (lets call them hybrid players) who can play as defender or forward are first chosen as defenders or forwards. If we first pick the 2 defenders, we can pick 0, 1 or 2 of the hybrid players as defenders, and then either 4, 3 or 2 of these players will be left available for forward. I tried calculating each case separately, and summing the combinations for each case together. However, if I calculate it such that we first choose the forwards, and we can choose 3, 2, 1 or 0 of the hybrid players to be forwards, I get a different result. I assume there is a smarter way to do this, but I can't seem to figure it out.

答:
Partitioning the $4$ hybrid players into $d$ defenders, $f$ forwards, and $4-d-f$ unused players yields
$$
\binom{3}{1}\sum_{d=0}^2\sum_{f=0}^{\min(4-d,3)}\binom{4}{d,f,4-d-f}\binom{7}{2-d}\binom{10}{3-f} = 50916
$$


问:
roots of $z^4-4.z^3+6.z^2-4.z-2$
What would be the roots of $z^4-4.z^3+6.z^2-4.z-2$, including the complex ones? I don't know how to discover them and in those sites they just give the result.

答:
HINT

The given expression can be rewritten as:
\begin{align*}
z^{4} - 4z^{3} + 6z^{2} - 4z - 2 = (z^{4} - 4z^{3} + 6z^{2} - 4z + 1) - 3
\end{align*}

where the first part can be expressed as
\begin{align*}
{4\choose 0}z^{4}(-1)^{0} + {4\choose 1}z^{3}(-1)^{1} + {4\choose 2}z^{2}(-1)^{2} + {4\choose 3}z^{1}(-1)^{3} + {4\choose 4}z^{0}(-1)^{4}
\end{align*}

which equals $(z-1)^{4}$ due to the binomial theorem.

Can you take it from here?

Title: roots of $z^4-4.z^3+6.z^2-4.z-2$


答:
\begin{gather*}
Expanding\ out\ ( z-1)^{4} \ using\ the\ binomial\ theorem,\\
( z-1)^{4} =z^{4} -4z^{3} +6z^{2} -4z-1\\
\Longrightarrow z^{4} -4z^{3} +6z^{2} -4z-2=( z-1)^{4} -3\\
The\ equation\ boils\ down\ to\ finding\ the\ roots\ of\\
( z-1)^{4} =3\\
\Longrightarrow z-1=3^{\frac{1}{4}}\left( e^{\frac{2\pi ki}{4}}\right)\\
where\ k=0,1,2,3
\end{gather*}
Hope this helps!

Title: roots of $z^4-4.z^3+6.z^2-4.z-2$

Body: What would be the roots of $z^4-4.z^3+6.z^2-4.z-2$, including the complex ones? I don't know how to discover them and in those sites they just give the result.