问:
Chern classes of tangent bundle over the Grassmannian G(2,4)
What are the Chern classes of the tangent bundle $\tau_G$ of the Grassmannian $G=G(2,4)$ of lines in $\mathbb{P}^3$? This is Exercise 5.37 on page 191 of 3264 & All That by Eisenbud and Harris.

By Theorem 3.5, the tangent bundle $\tau_G$ is isomorphic to $\mathcal{S}^{*} \otimes \mathcal{Q}$, where $\mathcal{S}$ and $\mathcal{Q}$ are the universal sub and quotient bundles of $G$. From Section 5.6.2, we have $c(\mathcal{Q})=1+\sigma_{1}+ \cdots+ \sigma_{n-k}$ and $c(\mathcal{S}^{*})=1+\sigma_1+\sigma_{1,1}+\cdots+\sigma_{1,1,\dots,1}$.

I found the following formulas here (https://stacks.math.columbia.edu/tag/02UK) for the first two Chern classes of a tensor product of vector bundles $\mathcal{E}$ and $\mathcal{F}$ which are finite locally free of ranks $r,s$.
$$ c_1(\mathcal{E} \otimes \mathcal{F})=rc_1(\mathcal{F})+sc_1(\mathcal{E})$$
$$ c_2(\mathcal{E} \otimes \mathcal{F})=r^2c_2(\mathcal{F})+rsc_1(\mathcal{F})c_1(\mathcal{E})+s^2 c_2(\mathcal{E}).$$
So I think the first two Chern classes are
$$ c_1(\tau_G)=2\sigma_1+2\sigma_1=4\sigma_1$$
$$c_2(\tau_G)=4\sigma_2+4\sigma_2\sigma_{1,1}+4\sigma_{1,1}=4(\sigma_2+\sigma_{1,1}),$$
but I haven't been able to find a formula for $c_3$.

答:
In schubert, the maple package for enumerative geometry and intersection theory,

```<code>with(SF);
with(schubert);
Gc = grass(2, 4, c, tan);
chern(tangentbundle(Gc));
</code>```

 $1+4\,{\it c1}\,t+7\,{{\it c1}}^{2}{t}^{2}+ \left( 8\,{{\it c1}}^{3}-4\,{\it c1}\,{\it c2} \right) {t}^{3}+ \left( 8\,{{\it c1}}^{4}-16\,{\it c2}\,{{\it c1}}^{2}+6\,{{\it c2}}^{2} \right) {t}^{4}$

For how the schubert package does the calculation $(Sc)^*\otimes Qc$ or Hom(4-Qc,Qc) of chern characters and how they relate to chern classes look at this:

```<code>chern(Qc)
</code>```

$1+{\it c1}\,t+{\it c2}\,{t}^{2}$

Or in terms of the chern character (elementary symmetric to power sums in the chern roots)

```<code>Qc
</code>```

$2+{\it c1}\,t+1/2\, \left( -2\,{\it c2}+{{\it c1}}^{2} \right) {t}^{2}
-1/6\, \left( 3\,{\it c1}\,{\it c2}-{{\it c1}}^{3} \right) {t}^{3}+1/
24\, \left( 2\,{{\it c2}}^{2}-4\,{\it c2}\,{{\it c1}}^{2}+{{\it c1}}^{
4} \right) {t}^{4}$

```<code>subs(t=-t,4-Qc)*Qc;
</code>```

$4+4\,{\it c1}\,t+{{\it c1}}^{2}{t}^{2}-2\,{\it c1}\,{\it c2}\,{t}^{3}+
2/3\,{{\it c1}}^{3}{t}^{3}+1/12\,{{\it c1}}^{4}{t}^{4}-{{\it c2}}^{2}{
t}^{4}$

i.e. multiply the chern character of the dual of $Sc=4-Qc$ with the chern character of $Qc$ truncated above the dimension $(4),$ then convert from power sums back to elementary symmetric:

$1+4\,{\it c1}\,t+7\,{{\it c1}}^{2}{t}^{2}+ \left( 8\,{{\it c1}}^{3}-4
\,{\it c1}\,{\it c2} \right) {t}^{3}+ \left( 8\,{{\it c1}}^{4}-16\,{
\it c2}\,{{\it c1}}^{2}+6\,{{\it c2}}^{2} \right) {t}^{4}$

and read off the $i$-th chern class of the tangent bundle as the coefficients of the $i$-th power of $t$.

Body:
In schubert, the maple package for enumerative geometry and intersection theory,

with(SF);
with(schubert);
Gc = grass(2, 4, c, tan);
chern(tangentbundle(Gc));

 $1+4\,{\it c1}\,t+7\,{{\it c1}}^{2}{t}^{2}+ \left( 8\,{{\it c1}}^{3}-4\,{\it c1}\,{\it c2} \right) {t}^{3}+ \left( 8\,{{\it c1}}^{4}-16\,{\it c2}\,{{\it c1}}^{2}+6\,{{\it c2}}^{2} \right) {t}^{4}$

For how the schubert package does the calculation $(Sc)^*\otimes Qc$ or Hom(4-Qc,Qc) of chern characters and how they relate to chern classes look at this:

chern(Qc)

$1+{\it c1}\,t+{\it c2}\,{t}^{2}$

Or in terms of the chern character (elementary symmetric to power sums in the chern roots)

Qc

$2+{\it c1}\,t+1/2\, \left( -2\,{\it c2}+{{\it c1}}^{2} \right) {t}^{2}
-1/6\, \left( 3\,{\it c1}\,{\it c2}-{{\it c1}}^{3} \right) {t}^{3}+1/
24\, \left( 2\,{{\it c2}}^{2}-4\,{\it c2}\,{{\it c1}}^{2}+{{\it c1}}^{
4} \right) {t}^{4}$

subs(t=-t,4-Qc)*Qc;

$4+4\,{\it c1}\,t+{{\it c1}}^{2}{t}^{2}-2\,{\it c1}\,{\it c2}\,{t}^{3}+
2/3\,{{\it c1}}^{3}{t}^{3}+1/12\,{{\it c1}}^{4}{t}^{4}-{{\it c2}}^{2}{
t}^{4}$

i.e. multiply the chern character of the dual of $Sc=4-Qc$ with the chern character of $Qc$ truncated above the dimension $(4),$ then convert from power sums back to elementary symmetric:

$1+4\,{\it c1}\,t+7\,{{\it c1}}^{2}{t}^{2}+ \left( 8\,{{\it c1}}^{3}-4
\,{\it c1}\,{\it c2} \right) {t}^{3}+ \left( 8\,{{\it c1}}^{4}-16\,{
\it c2}\,{{\it c1}}^{2}+6\,{{\it c2}}^{2} \right) {t}^{4}$

and read off the $i$-th chern class of the tangent bundle as the coefficients of the $i$-th power of $t$.


问:
Number of roots of $\lfloor\frac x3\rfloor=\frac x2$
How many roots does the equation $$\left\lfloor\frac x3\right\rfloor=\frac x2$$ have?

$1$

$2$

$3$

infinitely many

I checked that $x=0$ and $x=-2$ are the answers, so I think the answer is $(2)$. but I don't know how to solve the problem in general.

答:
First observe that
$$\color{blue}{\frac x3 } \ge \left\lfloor\frac x3\right\rfloor=\color{blue}{\frac x2} \implies x \le 0 \\
\color{blue}{\frac x3 - 1} \le \left\lfloor\frac x3\right\rfloor=\color{blue}{\frac x2} \implies x \ge -6$$
Now, from the equation, notice that $x/2$ is an integer which means $x$ is an even number. Therefore, the candidates are $x \in \{-6, -4,-2,0\}$.

Checking for all these four values we get $x = -4$ or $x = -2$ or $x = 0$.

So, the equation has $3$ roots.

ContentLicense: CC BY-SA 4.0


问:
Solving equations involving two sets
Out of curiosity, how would one go about solving an equation involving two sets? For example,

$$ \{1, 2, 3\} = \{a + b + c, a + b - c, a - b + c, a - b - c\} $$

An intuitive solution to this is $ \{a = 2, b = 0.5, c = 0.5\} $, but is there a specific process?

答:
Notice the set on the left has three distinct elements, but the set on the right has four representations so two of those represetations are of the same number.

So we have six options.

$a+b+c = a+b-c$ and $c = 0$ and we have the values $a+b$ and $a-b$.  But that's only two different values (or fewer) and we have exactly $3$ so that's impossible.

$a+b + c = a-b+c$  and $b =0$ and have the values $a+c$ and $a-c$ and that's the same problem.

$a+b+c = a-b-c$ and $b+c= 0$ and $b=-c$ and we have the values $a, a+2b, a-2b$.  We'll get back to this.

$a+b-c = a-b + c$ and $b-c=0$ and $b=c$ and we have the values $a+2b; a; a-2b$.  That will have the same solutions as above.

$a+b-c=a-b-c$ and $b=0$ and we have the same problem we had earlier.

Or $a-b+c = a-b-c$ adn $c=0$ and we have the same problem from the very begining.

So either way we have $a,a+2b, a-2b = 1,2,3$.

$a\pm 2b, a , a\pm 2b$ are in arithmetic progression so we must have either $a-2b < a < a+2b$ and $a-2b = 1, a=2, a+2b=3$ or $a+2b < a < a-2b$ and $a+2b = 1, a=2, a-2b =3$

So we must have $a=2$ and $b =\pm 0.5$ and $c = \pm 0.5$ (so there are four sets of solutions.)

There isn't really any way to do this in general.

Title: Solving equations involving two sets


答:
$$\left\lfloor \frac{x}{3} \right\rfloor = \frac{x}{2}$$

Hint Since $\left\lfloor \frac{x}{3} \right\rfloor$ is an integer, any solution $x$ must be a multiple of $2$. Now, for $x > 0$ we have $$\left\lfloor \frac{x}{3} \right\rfloor \leq \frac{x}{3} < \frac{x}{2} ,$$ and for $x \leq -6$ we have
$$\left\lfloor \frac{x}{3} \right\rfloor > \frac{x}{3} - 1 \geq \frac{x}{2} ,$$ leaving only a few possibilities to check manually.

Title: Number of roots of $\lfloor\frac x3\rfloor=\frac x2$

Body:
How many roots does the equation $$\left\lfloor\frac x3\right\rfloor=\frac x2$$ have?

$1$

$2$

$3$

infinitely many

I checked that $x=0$ and $x=-2$ are the answers, so I think the answer is $(2)$. but I don't know how to solve the problem in general.


问:
Entropy of extractions
A box contains $3$ white and $6$ black balls. We draw $2$ balls consequentially without replacement. Find the entropy of first and second extractions and the entropy for both of them.

答:
For the first case, when $2$ balls are drawn sequentially, the corresponding pmf is

$$P(B_1B_2)=\frac{6}{9}.\frac{5}{8}=\frac{5}{12}$$

$$P(W_1B_2)=\frac{3}{9}.\frac{6}{8}=\frac{1}{4}$$

$$P(B_1W_2)=\frac{6}{9}.\frac{3}{8}=\frac{1}{4}$$

$$P(W_1W_2)=\frac{3}{9}.\frac{2}{8}=\frac{1}{12}$$

Hence, the entropy is $$H(P) = -\sum\limits_{i}p_{i}\log_2{p_i}=1.825011$$

For the second case, when $2$ balls are drawn sequentially, the corresponding pmf is

$$P(2B) = \frac{{6 \choose 2}}{{9 \choose 2}} = \frac{5}{12}$$

$$P(2W) = \frac{{3 \choose 2}}{{9 \choose 2}} = \frac{1}{12}$$

$$P(1B1W) = \frac{{3 \choose 1}{6 \choose 1}}{{9 \choose 2}} = \frac{1}{2}$$

Hence, the entropy is $$H(P) = -\sum\limits_{i}p_{i}\log_2{p_i}=1.325011$$

Which is as per our intuition that in the second case entropy is lower because the uncertainty is lower (the order of drawing need not be considered).

Title: Entropy of extractions


问:
Convergence of Integral of gamma function
The reason for finiteness of the integral definition of the $\Gamma$-function in $\operatorname{Re}z>1$ is,

$$
\left|\int\limits_0^{\infty}t^{z-1}e^{-t}\mathrm{d}t\right|\leq \int_0^{\infty}\left|t^{z-1}e^{-t}\right|\mathrm{d}t=\int_0^{\infty}t^{\alpha}e^{-t}dt,
$$ where $\alpha>0$. Now, for lower value of $t$ close to zero, $t^{\alpha}$ is small, and as $t$ increases, we can see by L' Hospital rule that $\lim_{t\rightarrow \infty} t^{\alpha}/e^t=0$. Hence, intuitively the integral converges. But can anyone give a more formal explanation of this?

答:
Splitting the integrals up
$$\int\limits_0^\infty t^{\alpha-1}e^{-t}dt=\int\limits_0^1 t^{\alpha-1}e^{-t}dt+\int\limits_1^\infty t^{\alpha-1}e^{-t}dt$$

Now, we have that

$$0\le t\le 1\;\implies\;\; t^{\alpha-1}e^{-t}\le t^{\alpha-1}\;,\;\;\text{and}\;\;\int\limits_0^1t^{\alpha-1}dt=\left.\frac{t^\alpha}\alpha\right|_0^1=\frac1\alpha$$

and we also have that for sufficiently big $N$, with $N\leq t$

$$t^{\alpha-1}e^{-t}\stackrel{\text{}}\le e^{-t/2}\;,\;\;\text{and}\;\;\int\limits_1^\infty e^{-t/2}dt=\left.-2e^{-t/2}\right|_1^\infty=2 e^{-1/2}.$$


问:
Cluedo probabilities as information progresses
I am trying to calculate the probability that each card is inside the envelope as the information revealed progresses during the game.

The deck is composed of 21 cards, which are divided into three different classes ('Suspects', 'Weapons' and 'Rooms'). The first two classes contain 6 cards each, while the third one contains 9 cards. Then one card is removed for each class and placed in an enevelope, so that the deck now contains 18 cards in total. After shuffling, these 18 cards are dealt among 6 players (I am insterested in this case only), so that each player holds 3 cards (not necessarily one for each class).

Let's assume I am one of the 6 players. At this point, the probability that any of the cards belonging to the class 'Rooms' has been removed from the deck and placed in the envelope is 1/9, while for the other two classes is 1/6. Conversely the probabilities that these kind of cards are in the hands of the players are respectively 8/9 and 5/6.

Then let's assume that I become aware of the three cards that I have been dealt with. At this point I want to answer these two kinds of questions:

What is, for example, the probability that the card 'Kitchen' (which belongs to the 'Rooms' class) is in the enevelope conditioned to the fact that it is not among my three cards?

What is, for example, the probability that the card 'Kitchen' is in the envelope conditioned to the fact that it is not among my cards but 'Living Room' (another card from the class 'Rooms') is?

I know that we are dealing with conditional probabilities, but I cannot figure out how to calculate them, as the initial probability of the card 'Kitchen' to be removed from the deck and placed in the envelope is 1/9, which is different from the probability that this card will end in the hands of each player (which if I am not wrong should be 4/27).

答:
There is $1$ card of Rooms class in envelope and $8$ cards among players.

i) Now if you have no card of Rooms class in your hand, there are still $8$ Rooms cards out there among other players. So the probability still remains the same that the Kitchen card is in the envelope is $\frac{1}{9}$ and in hands of players is $\frac{8}{9}$.

ii) If you have one card of Rooms class in your hand but it is not Kitchen card. Now this information does change the probability (for you) that the Kitchen card is in the envelope as it is now one of the $8$ cards. That is simply $\frac{1}{8}$.

If you want to write it as conditional probability -

Event $A$ is that Kitchen card is in the envelope. Event $B$ is that it is one of the $8$ cards distributed (including one in envelope),

$P(B) = \displaystyle \frac{8 \choose 7}{9 \choose 8} = \frac{8}{9}$.

$P(A \cap B) = \frac{1}{9}$. $P(A|B) = \frac{1/9}{8/9} = \frac{1}{8}$.

Similarly,

iii) If you have two cards of Rooms class and none of them are Kitchen card, the probability of Kitchen card in envelope is $\frac{1}{7}$.

iv) If you have three non-Kitchen cards from Rooms class, the probability of Kitchen card in envelope is $\frac{1}{6}$.


问:
Continuity of a function defined differently on $\mathbb Q,\mathbb R\setminus \mathbb Q$
Let's say I define the function $f(x)=2^x$ for rational $x$, and $f(x)=1$ for irrational $x$.

My question is: is this function continuous everywhere? I think it's not, because for any $2$ irrational numbers you can find a rational in between, and for any $2$ rationals you can find an irrational in between.

My second question is: is the function continuous and differentiable at $x=0$? I think it is, and I think that $f'(0)=0$ because of the picture. Am I correct? (By the way please keep the answers at a pretty low level, I'm a calculus AB student.) Thanks!

答:
It's certainly not continuous for any $x$ such that $2^x\neq 1$, you're correct.

Also, you're right that it's continuous at $x=0$, since in any neighbourhood of 0, you're no further away from $1$ than $2^x$ is (though you might be closer). Hence the limit as you approach 0 is well-defined.

However, the function is not differentiable because you can choose a sequence of rationals $x_n\to 0$ and a sequence of irrationals $y_n\to 0$ and consider
$$\lim_{n\to\infty} \frac{f(x_n)-1}{x_n-0} \equiv \frac{2^{x_n}-1}{x_n-0} = \left[\frac {\mathrm d 2^x}{\mathrm d x}\right]_{x=0},\qquad \lim_{n\to\infty} \frac{f(y_n)-1}{y_n-0} \equiv \frac{1-1}{y_n-0} = 0$$

Then the first derivative is that of $e^{x\log 2}$ which is $\log 2\neq 0$. Hence there is no derivative.

Notice in your dots that the dots for rationals are sloped at the origin, but the irrational dots are flat. This is what the above expresses.

ContentLicense: CC BY-SA 3.0


问:
Maximum volume of a box
In his Calculus textbook (chapter 3.2, problem 23), Strang writes:

The airlines accept a box if length + width + height = $l+w+h <62''$ or 158cm. If h is fixed, show that the maximum volume $(62-w-h)wh$ is $V = h(31 - \frac{1}{2}h)^2$. Choose h to maximize V. The box with greatest volume is a _______.

Can you help me understand the problem?

The volume is $V=whl$. Since one side is fixed, we can get that side from the formula $w+h+l=62$.

How do we get to $V = h(31 - \frac{1}{2}h)^2$?

Differentiating in terms of h: $(62-h-w)(wh) = 62-2h-1$ ? I can see that h=31 is a stationary point if disregard the -1 in my differentiation, i.e. $62-2h = 0$. I'm lost as to what to do next, and why is the -1 disregarded..?

答:
The procedure they are describing goes like this.

First $l+w+h=62$, since you can certainly increase the volume by lengthening any one side.

Next, solve that for $l$: $l=62-w-h$.

Next, fix $h$; I don't know what it is going to be yet, but regardless of what it is, once I fix $h$, $V$ is now only a function of $w$. So I look at $V(w)=(62-w-h)wh$. Since $h>0$, this is a downward-opening quadratic function, so you can find its maximum. It will happen that $w=62-w-h$ (geometrically, this is because the rectangle with fixed perimeter and largest area is a square) so $w=31-h/2$.

Next you examine $V$ as a function of just $h$, no $w$ anymore: $V(h)=h(31-h/2)^2$, and you optimize that.

I don't know where this $62-2h-1$ came from, it is not what you get when differentiating $h(31-h/2)^2$. I think you got it by differentiating $(62-w-h)wh$ with respect to $h$, but if you do that then there are still $w$ terms floating around, so you would end up with $h$ as a function of $w$ (and you would still need to solve for $w$).

I'll also point out that there was nothing special about solving for $l$, nor about finding the optimal value of $w$ by fixing $h$. You could've dealt with the variables in any order.

Title: Maximum volume of a box


问:
How to show the following formula in the sense of distribution?
How to show the following formula in the sense of distribution in $\mathbb{R}^2$?
\begin{equation}(\partial_x +i\partial_y)\frac{1}{x+iy}=2\pi\delta_{(0,0)}
\end{equation}

答:
Let ${\psi} \left(x , y\right)$ be a test function and let
${\varphi} \left(r , {\theta}\right) = {\psi} \left(r \cos  {\theta} , r \sin  {\theta}\right)$.
We compute

\begin{equation}
\renewcommand{\arraystretch}{2}  \begin{array}{rcl}\displaystyle  \frac{\partial  {\varphi}}{\partial  r}&=&\displaystyle  \cos  {\theta} \frac{\partial  {\psi}}{\partial  x}+\sin  {\theta} \frac{\partial  {\psi}}{\partial  y}\\
\displaystyle  \frac{\partial  {\varphi}}{\partial  {\theta}}&=&\displaystyle -r \sin  {\theta} \frac{\partial  {\psi}}{\partial  x}+r \cos  {\theta} \frac{\partial  {\psi}}{\partial  y}
\end{array}
\end{equation}

Hence

\begin{equation}
\frac{1}{r} \frac{\partial  {\varphi}}{\partial  r}+\frac{i}{{r}^{2}} \frac{\partial  {\varphi}}{\partial  {\theta}} = \frac{1}{x+i y} \left(\frac{\partial  {\psi}}{\partial  x}+i \frac{\partial  {\psi}}{\partial  y}\right)
\end{equation}

Hence

\begin{equation}
\renewcommand{\arraystretch}{2}  \begin{array}{rcl}\displaystyle\left\langle \left(\frac{\partial  }{\partial  x}+i \frac{\partial  }{\partial  y}\right) \left(\frac{1}{x+i y}\right) , {\psi}\right\rangle &=&\displaystyle -\int_{{\mathbb{R}}^{3}}^{}\frac{1}{x+i y} \left(\frac{\partial  }{\partial  x}+i \frac{\partial  }{\partial  y}\right) \left({\psi}\right) d x d y\\
&=&-\displaystyle  \int_{{\left[0 , \infty \right[}\times{\left[0 , 2 {\pi}\right]}}^{}\left(\frac{\partial  {\varphi}}{\partial  r}+\frac{i}{r} \frac{\partial  {\varphi}}{\partial  {\theta}}\right) d r d {\theta}
\end{array}
\end{equation}

Using that $\displaystyle  \int_{0}^{2 {\pi}}\frac{\partial  {\varphi}}{\partial  {\theta}} d {\theta} = 0$, it remains

\begin{equation}
\left\langle \left(\frac{\partial  }{\partial  x}+i \frac{\partial  }{\partial  y}\right) \left(\frac{1}{x+i y}\right) , {\psi}\right\rangle  =-\int_{0}^{2 {\pi}}\int_{0}^{\infty }\frac{\partial  {\varphi}}{\partial  r} d r d {\theta} = \int_{0}^{2 {\pi}}{\psi} \left(0 , 0\right) d {\theta} = 2 {\pi} {\psi} \left(0 , 0\right)
\end{equation}


问:
Differentiating $ y= xe^{1\over x} $
Can someone please help me? I'm trying, but I really can't find the second derivative of $y= xe^{1/x}$. Thanks!</p>

Title: Differentiating $ y= xe^{1\over x} $

Body:
By the product rule:

$$y'=x'e^{1/x}+x\left(e^{1/x}\right)'=e^{1/x}-\frac{1}{x}e^{1/x}$$

I used: $y=e^{f(x)}$, then $y'=f'(x)e^{f(x)}$. Then:

$$y''=-\frac{1}{x^2}e^{1/x}+\frac{1}{x^2}e^{1/x}+\frac{1}{x^3}e^{1/x}=\frac{1}{x^3}e^{1/x}$$


问:
Related to Hom, Images (im) and Kernels (ker)
Determine whether the following statements are correct or not and prove your claims:

If f ∈ Hom(U,V) and g ∈ Hom(V,W) such that im(f) ⊂ ker(g), then g ◦ f = 0

If f ∈ Hom(U,V) and g ∈ Hom(V,W) such that im(f) ∩ ker(g)= {0}, then g ◦ f is isomorphic

I am quite puzzled about the concepts of Hom, images and kernels, so please help me. I may need some detailed explanations to understand these concepts.

答:
Let $ v \in U$

$ v \in U \Rightarrow f(v) \in im(f)$

$ im(f) ⊂ ker(g) \Rightarrow f(v) \in ker(g)$

$f(v) \in ker(g) \Rightarrow g(f(v))=0$

$g(f(v))=0 \Rightarrow g ◦ f = 0 $


问:
Alternative Methods for Trigonometry Proofs (Varying Angles)
Consider all the ways to prove $\sin(2x)=2\sin(x)\cos(x)$. There are many!

I personally would use Euler's formula involving complex numbers due to it being extremely simple and straightforward. We know $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$ and $\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$. The LH and RH sides become $\frac{e^{i2x}-e^{-i2x}}{2i}=2(\frac{e^{ix}-e^{-ix}}{2i})(\frac{e^{ix}+e^{-ix}}{2})$. And now this proof can be done with some basic algebra!

At the same time, I have always been bothered by rewriting $\sin(x)$ and $\cos(x)$ this way. To rewrite everything in terms of $e$ feels like an utter avoidance to use the words "sine" and "cosine." It also seems like an avoidance to represent such formulas along the unit circle to prove them. The proofs involving the sum and difference formulas that use the unit circle have always appeared so clean to me.

This got me thinking... is there a dependable way to prove results like these using the unit circle? Or is there a way to rewrite everything in terms of sine and cosine another way?

I have been curious about this for a long time, and I would be curious as to what other dependable ways are used to prove trigonometric results such as these without $i$ and infinite sequences. What do other people do?

答:
Let $R_{\theta}$ be a rotation matrix that rotates an arbitrary vector by an angle of $\theta$ radians anticlockwise about the origin:
$$
R_{\theta} =
\begin{bmatrix}
\cos \theta & -\sin\theta \\
\sin \theta & \cos\theta
\end{bmatrix}
$$
Clearly, $R_{\theta}R_{\theta}=R_{2\theta}$, since two rotations of size $\theta$ has the same result as a rotation of size $2\theta$:
$$
R_{2\theta}=
\begin{bmatrix}
\cos 2\theta & -\sin2\theta \\
\sin 2\theta & \cos2\theta
\end{bmatrix}
\, .
$$
On the other hand, if we multiply out the matrices the long way, we get
$$
R_{\theta}R_{\theta}=
\begin{bmatrix}
\cos \theta & -\sin\theta \\
\sin \theta & \cos\theta
\end{bmatrix}
\begin{bmatrix}
\cos \theta & -\sin\theta \\
\sin \theta & \cos\theta
\end{bmatrix}
=
\begin{bmatrix}
\cos^2\theta-\sin^2\theta & -2\sin\theta\cos\theta \\
2\sin\theta\cos\theta & \cos^2\theta - \sin^2\theta
\end{bmatrix}
\, ,
$$
and comparing the two gives us $\cos 2\theta \equiv \cos^2\theta - \sin^2\theta$ and $\sin 2\theta \equiv 2\sin\theta\cos\theta$.


问:
Let $a,b,c$ be positive integers such that $a^3+b^3=2^c.$ Show that $a=b$.

Let $a,b,c$ be positive integers such that $$a^3+b^3=2^c.$$ Show that $a=b$.

I have that $$a^3+b^3=(a+b)(a^2-ab+b^2)=2^c =2^x\cdot2^y$$ now it can only be that $a$ and $b$ are both odd or even since they sum to an even number. Thus if $a$ and $b$ are both odd I have that $a^2$ is odd, $ab$ is odd and $b^2$ is odd. This would imply that $a^2-ab+b^2 = 2^y =1$ which in turn implies that $a^3+b^3 = a+b \implies a=b.$ The problem I have is that if I would have considered that both $a$ and $b$ are even I would have gotten that $a=2t, b=2k$ from where $$8t^3+8k^3=2^c \implies t^3+k^3=2^{c-3}$$ but I couldn't deduce anything from here why cannot $a$ and $b$ be even?

答:
HINT:

For the case where $a,b$ both even, factor out the largest power of $2$ from both of them. You will either reduce to exactly one of $a'$, $b'$ odd [which you already noted is impossible] or both $a,b$ odd. To elaborate a bit more, let $j$ be the largest integer s.t. $2^j$ divides both $a$ and $b$. Then write $2^ja'=a$ and $2^jb'=b$, with at least one of $a',b'$ odd.
Then $$(a')^3+(b')^3 = 2^{3j}2^{c-3j},$$ so as at least one of $a',b'$ is odd, then by your reasoning above, both $a'$ and $b'$ must be odd.

For the case where $a,b$ both odd: You noted that this implies the equation
$$(a+b)(a^2-ab+b^2)= 2^c.$$ So then both $(a^2-ab+b^2)$ and $(a+b)$ must be powers of $2$. However, $a^2-ab+b^2$ is odd [why is this] so this gives $a^2-ab+b^2=1$, because $1$ is the only odd power of $2$. So this gives $(a+b)=(a^3+b^3)$ [why is this] which
is only possible [assuming $a,b$ positive integers] for $a,b = 1$.


答:
Write $a = 2^x \cdot m$ and $b = 2^y \cdot n$ where $m$ and $n$ are odd. WLOG $x \geq y$. Now,
$$2^{3x} \cdot m^3 + 2^{3y} \cdot n^3 = 2^{3y} \cdot (2^{3x-3y} \cdot m^3 + n^3) = 2^c$$ If $x > y$, we have a contradiction since in that case an odd number divides LHS. So, $x = y$ and so you would arrive at your argument.

Title: Let $a,b,c$ be positive integers such that $a^3+b^3=2^c.$ Show that $a=b$.

Body:
Let $a,b,c$ be positive integers such that $$a^3+b^3=2^c.$$ Show that $a=b$.

I have that $$a^3+b^3=(a+b)(a^2-ab+b^2)=2^c =2^x\cdot2^y$$ now it can only be that $a$ and $b$ are both odd or even since they sum to an even number. Thus if $a$ and $b$ are both odd I have that $a^2$ is odd, $ab$ is odd and $b^2$ is odd. This would imply that $a^2-ab+b^2 = 2^y =1$ which in turn implies that $a^3+b^3 = a+b \implies a=b.$ The problem I have is that if I would have considered that both $a$ and $b$ are even I would have gotten that $a=2t, b=2k$ from where $$8t^3+8k^3=2^c \implies t^3+k^3=2^{c-3}$$ but I couldn't deduce anything from here why cannot $a$ and $b$ be even?


问:
Integrate $\int_2^5 \dfrac{1}{1+\sqrt{x-1}}\, \mathrm{d}x $ using substitution with a fraction
Im doing this question currently

$$\int_2^5 \dfrac{1}{1+\sqrt{x-1}}\, \mathrm{d}x $$

and I set $u = x-1$ to get $\tfrac{\mathrm{d}u}{\mathrm{d}x} =1$ so $du =dx$.

I then write it out as $$\frac{1}{1+\sqrt{u}}\,\mathrm{d}u$$

But then I hit a brick wall and I'm not sure how to further progress. I know I need to replace the $u$ but I'm not sure about anything else

答:
hint

To eliminate the square root, you can put

$$\sqrt{u}=t \text{ or } u=t^2\text{ and } du=2tdt$$

the integral becomes

$$\int_1^4\frac{du}{1+\sqrt{u}}=\int_1^2\frac{2tdt}{1+t}$$

$$=2\int_1^2\frac{t+1-1}{t+1}dt$$


问:
curl of what yields $(0,s^{-1},0)$ in cylindrical coordinates?
In cylindrical coordinates $(s,\theta,z)$, what function $\mathbf{A}$ has the property $$\nabla\times \mathbf{A} = (0, \frac{1}{s} , 0) $$

I know generally that $$\nabla\times \mathbf{A} = \left(\frac{1}{s}\frac{\partial A_3}{\partial \theta}-\frac{\partial A_2}{\partial z}, \frac{\partial A_1}{\partial z}-\frac{\partial A_3}{\partial s} , \frac{1}{s} \left( \frac{\partial}{\partial z}(s\cdot A_2) - \frac{\partial A_1}{\partial \theta}\right)\right) $$

Is there some better way of solving for $\mathbf{A}$ other than a lot of ugly equations?

答:
Then you can use a Green's function to solve this PDE.

The relevant Green's function is

$$\mathbf G(\mathbf r) = \frac{\mathbf r}{4\pi r^3}$$

The function $\mathbf A$ can then be found by convolution with the Green's function.  Let $\mathbf B = s^{-1} \hat \theta$.  The solution is then

$$\mathbf A(\mathbf r) = \int_{M} \mathbf B(\mathbf r') \times \mathbf G(\mathbf r - \mathbf r') \, dV'$$

Edit: there was a surface integral here, but I realized I had it written down wrong. I'll have to find a source for this, for while the volume integral is pretty simple, the surface integral looks, to me, to be a bit complicated, and you probably don't want to do that.  I'll try to verify the form of that integral.  Choosing all space as your integration region safely ignores the surface integral, but you may not be able to do this easily with $1/s$ in your vector field.


答:
Hom(X, Y) (Homomorphism, X and Y are vector spaces) A function from X to Y which preserves scalar multiplication ($f(cx) = cf(x)$ for any c in the scalar field) and addition of vectors ($f(x+y) = f(x) + f(y)$ for any $x, y \in X$) (I assume you are working with vector spaces here because of the linear algebra tag)

Im(f) (image)  the set of all outputs of F. if f is a function from A to B then $b \in Im(B)$ if and only if there exists $a \in A$ such that f(a) = b

Ker(f) (kernel) the subset of Im(f) such that $a \in Ker(f) \Leftrightarrow f(a) = 0$

I think that Nevzat Eren Akkaya's answer answers question 1 as well as I could and if the clarification is given for Q2 then I will answer it if someone hasn't before me

Title: Related to Hom, Images (im) and Kernels (ker)

Body:
Determine whether the following statements are correct or not and prove your claims:

If f ∈ Hom(U,V) and g ∈ Hom(V,W) such that im(f) ⊂ ker(g), then g ◦ f = 0

If f ∈ Hom(U,V) and g ∈ Hom(V,W) such that im(f) ∩ ker(g)= {0}, then g ◦ f is isomorphic

I am quite puzzled about the concepts of Hom, images and kernels, so please help me. I may need some detailed explanations to understand these concepts.


问:
Are there any hidden relationships between two numbers $a$ and $b$ such that $a^a=b^b$ when $a \neq b$?
Are there any hidden relationships between two numbers $a$ and $b$ such that $a^a=b^b$ when $a \neq b$?

I am attempting to do some research into this relationship, and I came across this when looking at the graph $f(x)=x^x$, and saw that this function has a decreasing interval, before it increases again. I was then intrigued by the idea that this function decreases in the first place, and that there are numbers that can be described by $a^a=b^b$ such that $a \neq b$.

Somethings I've tried. I first looked at how this function behaves when graphed in the imaginary plane, taking ideas from the following paper: https://www.jstor.org/stable/2691469?seq=1. I wasn't really able to find anything special here, although some parts of the paper are above my current level of knowledge. Essentially it looks like the author was able to rewrite $x^x$ (through a process that would take an extra paragraph or two to explain) into this equation $$|x|^x(\cos(nπx)+i\sin(nπx))$$ where n is even for $x>0$ and $n$ is odd for $x<0$. Comparing this to $f(x)=x^x$ I didn't find any action around my original search for numbers $a^a=b^b$.

I also set up this relationship into a relation and graphed it. Graphing $y^y=x^x$ gives a graph that looks like $y=x$ for $x>0$, but it also has interesting behavior around $[0, 1]$. It creates a curve that represents all of the numbers which I am looking for. Graphing this quickly in Desmos gives a little better description, but I am not allowed to imbed pictures yet so I cannot show this.

I also tried expanding through a Taylor Series and looking at the numbers in a table directly. Nothing exciting showed up which I could see.

For context, I am a senior in highschool, and so my knowledge of math is $\leq$ Calc 2. I know very basic set notation if you want to use that in your response.

TL;DR If anyone happens to know anything intresting about $x^x$ or numbers described by $a^a=b^b$ when $a\neq b$ I would love to hear what you have to say.

答:
This can be done using the Lambert W function.  If $0<a<b<1$ and $a^a=b^b$, then
$$
a=\frac{b\ln b}{W_{-1}(b\ln b)},\qquad
b=\frac{a\ln a}{W_0(a\ln a)}
$$


答:
With $u=x-1$

\begin{align}
 \int \dfrac{dx}{1+\sqrt{x-1}}=
&\int \dfrac{du}{1+\sqrt{u}}
=\int \frac{ 1+\sqrt{u}-1}{1+\sqrt{u}} \frac{du}{\sqrt{u}}\\
= &\>2\int \left(1-\frac1{1+\sqrt{u}}\right)d(\sqrt{u})
= 2\sqrt{u}-2\ln(1+\sqrt{u})
\end{align}

Title: Integrate $\int_2^5 \dfrac{1}{1+\sqrt{x-1}}\, \mathrm{d}x $ using substitution with a fraction

Body:
Im doing this question currently

$$\int_2^5 \dfrac{1}{1+\sqrt{x-1}}\, \mathrm{d}x $$

and I set $u = x-1$ to get $\tfrac{\mathrm{d}u}{\mathrm{d}x} =1$ so $du =dx$.

I then write it out as $$\frac{1}{1+\sqrt{u}}\,\mathrm{d}u$$

But then I hit a brick wall and I'm not sure how to further progress. I know I need to replace the $u$ but I'm not sure about anything else


问:
Expressing the truth sets of $\Box\varphi$ and $\Diamond\varphi$ in terms of the truth set of $\varphi$.
Since that the answer to this question will depend on the particular semantics that I'm using, allow me to first define truth of $\Box\varphi$ and $\Diamond\varphi$ at a world $\alpha$ in a model $\mathfrak{M}=(\mathit{W},\mathit{R},\mathit{P})$, where $\mathit{R}$ is a preorder and $\mathit{P}$ is a map from the set of atomic sentences of the language to $\mathcal{P}(\mathit{W})$ (the power set of $\mathit{W}$).

Note: I'm pretty sure this it what most people would simply call a Kripke model, but I thought that to avoid misunderstandings it would be better to be completely explicit.

$\bullet$ $\vDash_{\alpha}^{\mathfrak{M}}\Box\varphi$ if and only if for all $\beta$, $\alpha\mathit{R}\beta$ implies $\vDash_{\beta}^{\mathfrak{M}}\varphi$.

$\bullet$ $\vDash_{\alpha}^{\mathfrak{M}}\Diamond\varphi$ if and only if there exists $\beta$ such that $\alpha\mathit{R}\beta$ and $\vDash_{\beta}^{\mathfrak{M}}\varphi$.

Note: Again, I think this is just Kripke semantics, but I wanted to make it explicit for clarity.

Now, if I wanted to write the truth sets corresponding to these formulae in the model $\mathfrak{M}$, I could just define

$\bullet$ $\Vert\Box\varphi\Vert^{\mathfrak{M}}=\lbrace \alpha\in\mathit{W}\mid(\forall \alpha\mathit{R}\beta)(\beta\in\Vert\varphi\Vert^{\mathfrak{M}})\rbrace$

$\bullet$ $\Vert\Diamond\varphi\Vert^{\mathfrak{M}}=\lbrace \alpha\in\mathit{W}\mid(\exists \alpha\mathit{R}\beta)(\beta\in\Vert\varphi\Vert^{\mathfrak{M}})\rbrace$

The thing is I don't find this completely satisfying, since the truth sets of the formulae $\neg\varphi$, $\varphi\wedge\psi$, $\varphi\Rightarrow\psi$, etc, can be written in terms of elementary operations involving the truth sets of $\varphi$ and $\psi$. I've been trying for a while to find a more compact way to treat the case of the modal operators, but there doesn't seem to be any way other than what I just did. I thought I could pull it off using upsets, but I'm not entirely sure about that approach.

Anyway, please let me know if there's a nicer way to describe $\Vert\Box\varphi\Vert^{\mathfrak{M}}$ and $\Vert\Diamond\varphi\Vert^{\mathfrak{M}}$ in terms of $\Vert\varphi\Vert^{\mathfrak{M}}$. Maybe I'm just wasting my time, but my gut tells me there should be a nice compact way to do it.

Thank you all in advance.

答:
This can't be done in any good sense I can think of.

The issue is that modalities take into account the relational aspect of a Kripke model in a way that connectives don't. For example, if I know the set of worlds $W$ and the truth set $\Vert\varphi\Vert$, I can figure out the truth set $\Vert\neg\varphi\Vert$: it's just $W\setminus\Vert\varphi\vert$. However, $W$ and $\Vert\varphi\Vert$ alone are not enough to determine either $\Vert\Diamond\varphi\Vert$ or $\Vert\Box\varphi\Vert$: contrast e.g. $R=W^2$ and $R=\emptyset$.

So unlike the connectives, we can't describe how modalities affect truth-sets without invoking $R$. And once we bring $R$ into the picture, the standard descriptions are really as simple as we could hope for.


问:
Prove if $n^2$ is even, then $n$ is even.
I am just learning maths, and would like someone to verify my proof.

Suppose $n$ is an integer, and that $n^2$ is even.  If we add $n$ to $n^2$, we have $n^2 + n = n(n+1)$, and it follows that $n(n+1)$ is even.  Since $n^2$ is even, $n$ is even.

Is this valid?

答:
This proof is valid.

I suggest you add the small detail that the difference of even integers is even.


问:
Prove formula equivalence without induction
Consider the following formula:

$$\frac{(n-i)!}{n!}\sum_k \frac{(k-1)!}{(k-i)!} [2 \le i \le k \le n] = \frac{1}{i}$$

It's easy to prove the equality by induction over $n$ but is there also a more direct way (a continued equality, ideally)?

答:
OPs identity is the Hockey stick identity in disguise.

We write OPs identity as
\begin{align*}
\frac{(n-i)!}{n!}\sum_{k=i}^n\frac{(k-1)!}{(k-i)!}=\frac{1}{i}\qquad\qquad 2\leq i\leq n\tag{1}
\end{align*}
Multiplying both sides of (1) with $\frac{n!}{(n-i)!(i-1)!}$ gives
\begin{align*}
\sum_{k=i}^n\binom{k-1}{i-1}=\binom{n}{i}\qquad\qquad 2\leq i\leq n\tag{2}
\end{align*}

Shifting the index $k$ by one results in
\begin{align*}
\color{blue}{\sum_{k=i-1}^{n-1}\binom{k}{i-1}=\binom{n}{i}\qquad\qquad 2\leq i\leq n}
\end{align*}
which is the Hockey stick identity.


问:
Intuition behind the Galois Group
I've just learned the definition of the Galois Group. Although I understand the formal definition, I'm having a hard time understanding the intuition behind it. So my question is: intuitively, what information about a field extension does the Galois Group capture? I'm looking for an answer along the lines of: "it measures the symmetries of the field extension", but hopefully something a little more concrete.

To be clear, I understand how the Galois Group is used in practice (e.g: it allows you to convert problems about fields into problems about groups, which are normally a lot easier to tackle). But I'm asking, aside from its practical utility, why is the Galois Group important? What information does the Galois Group give us about the "symmetries" of a field extension? (And what does "symmetry" even mean in this context?) Any clarification would be appreciated! Thanks.

答:
Try thinking about the Galois group $\operatorname{Gal}(\mathbb{C}/\mathbb{R})$ with two elements:

the identity automorphism

the complex conjugation automorphism

So the Galois group detects a mapping within elements of group that does not affect the basic group structure.

Notice that if a polynomial in $\mathbb{R}$ has a complex zero it instantly has its conjugate as another zero, but essentially we cannot really say which one is which. We do mark them as $a+bi$ and $a-bi$, but the agreement about $i$ is implicit, we could easily replace $i$ with $j=-i$ and $-j$ and start calling $j$ imaginary unit and nothing would change among polynomials in $\mathbb{R}$.

In that sense, a Galois field is talking about how much the elements within a group are indistinguishable among each other.

A Galois group contains mapping that are preserving the structure. The same idea you have all around math. Even and odd numbers, for example, you do not have to take care about the exact parity for the basic mathematical operations in order to get the parity of the entire expression, and so on. The basic idea, here, is not more complex than that.

It is somewhat unfortunate that modern math places Galois where it really does not belong, in some sort of advanced math. It is there if you take into account a general math overview, but so is multiplication and addition. It is still a simple thing.

Title: Intuition behind the Galois Group


答:
As the question fixes $h$ and seeks you to find $l$ and $w$ in terms of $h$, you should differentiate wrt. $w$ and not $h$.

$V = (62-w-h)  w  h$

$\frac{dV}{dw} = (62-w-h) h - wh = 0 \implies 62 - 2w - h = 0$

That leads to $w = 31 - \frac{h}{2}$ and $V = h(31 - \frac{h}{2})^2$.

Now to see for which $h$ the volume is max, differentiate wrt. $h$ and equate to zero.

You should get a cube with $l = w = h$. You can confirm it by AM-GM.

$V = (62-w-h) wh$

So by AM-GM,

$\frac{(62-w-h) + w + h}{3} \geq [(62-w-h) w h]^{1/3} = V^{1/3}$

$\implies V \leq (\frac{62}{3})^3$.

Equality occurs when $62-w-h = w = h$.

Title: Maximum volume of a box

Body:
In his Calculus textbook (chapter 3.2, problem 23), Strang writes:

The airlines accept a box if length + width + height = $l+w+h <62''$ or 158cm. If h is fixed, show that the maximum volume $(62-w-h)wh$ is $V = h(31 - \frac{1}{2}h)^2$. Choose h to maximize V. The box with greatest volume is a _______.

Can you help me understand the problem?

The volume is $V=whl$. Since one side is fixed, we can get that side from the formula $w+h+l=62$.

How do we get to $V = h(31 - \frac{1}{2}h)^2$?

Differentiating in terms of h: $(62-h-w)(wh) = 62-2h-1$ ? I can see that h=31 is a stationary point if disregard the -1 in my differentiation, i.e. $62-2h = 0$. I'm lost as to what to do next, and why is the -1 disregarded..?


问:
$\Delta$ operator is it as a differential operator?
Consider the second law of dynamics written in terms of both momentum, momentum of a force and angular momentum,

$$\bbox[5px,border:3px solid #F5B041]{\overline{F}=\frac{\Delta\overline{p}}{\Delta t}} \tag 1$$

$$\bbox[5px,border:3px solid #A2B151]{\overline{M}=\frac{\Delta\overline{L}}{\Delta t}} \tag 2$$

In the proof of the $(2)$ we use the concept of $\overline{L}=\overline r \times \overline p$ where $\overline r$ it is the position vector and the $\overline{p}$ it is the momentum.

$$\Delta \overline L=\Delta (\overline r \times \overline p)$$

If I consider $\Delta$ as $d$ (differential operator) I know that using the calculus I will obtain$$\bbox[lightgray,5px,border:2px solid teal]{d(\overline r \times \overline p)=(d\overline r)\times \overline p \, + \,\overline r\times (d\overline p)} $$

In my textbook of Physics I have:
$$\bbox[orange,5px,border:2px solid gray]{\Delta(\overline r \times \overline p)=(\Delta\overline r)\times \overline p \, + \,\overline r\times (\Delta\overline p)} \tag 4$$

Because the operator $\Delta$ must have the same role as the differential operator $d$? Does $\Delta$ a laplacian operator? Is it possible to proof the $(4)$ without the calculus?

PS: My students, still they not use the calculus (derivates).

答:
No - here, $\Delta$ is certainly not the "Laplacian" operator!

[In the following, I'm not decorating vectors with overlines for reasons of laziness.]

In any event:  $\Delta (r\times p)$ equals, the first line by definition, and the next line by expanding out the terms in the cross product,
$$ \begin{align}\Delta (r\times p) &= (r+\Delta r ) \times (p+\Delta p) - r\times p\\&= r\times p + (r \times \Delta p) + (\Delta r \times p) + (\Delta r \times \Delta p)- r\times p.\end{align}  $$
Canceling (subtraction) the two $r\times p$ terms, on gets

$$ \Delta (r\times p) = (r \times \Delta p) + (\Delta r \times p) + (\Delta r \times \Delta p). $$

The above is almost a tautology...

Now, if, in your case, $\Delta r \times \Delta p =0$  [which is possible!], then

$$ \Delta (r\times p) = (r \times \Delta p) + (\Delta r \times p).$$

Otherwise, under the assumption that $(\Delta r \times \Delta p)$ is small compared to the other terms, one "only" has that

$$ \Delta (r\times p) \approx (r \times \Delta p) + (\Delta r \times p).$$

In that case, your text is at best taking a liberty in writing $=$ instead of $\approx$ [unless of course $\Delta r \times \Delta p =0$ in your case]. I am not sure that I approve! On the other hand, I have read an equation of Fermi which ended up equating an irrational number ($\pi$, say) to a rational number  ($3.1$, say) - and Fermi certainly knew what he was doing...  Fermi was only interested in maintaining accuracy of to $0.1$ (say)... For instance $ 5\times 0.1 + 0.1\times 10 = 1.5 $ is much bigger than $0.1\times 0.1 = 0.01$, which one might legitimately 'neglect'. But bear in mind what you mean.

I hope this helps...


答:
I will write a group-theoretical answer since there are two correct answers before of mine.

By definition,
$$Hom(U,V) = \{f: U \to V: f\text{ is a homomorphism}\}$$
and given some $g \in Hom(V,W)$,
$$\ker g = \{v \in V: g(v) = 0\}.$$
So, if $f(u) \in \ker g$, then $g(f(u)) = 0$ and since it holds for every $u \in U$, then $g \circ f = 0$ where $0$ is the zero map.

The item $2$ is incomplete, I will try to write an answer assuming some facts.

$$\ker g \circ f = \{u \in U : (g\circ f)(u) = 0\}.$$
Since $im(f) \cap \ker g = \{0\}$, $g(f(u)) = 0$ if and only if $f(u) = 0$. In that point you need to show that $f(u) = 0$ if and only if $u = 0$, otherwise if $f(u) = 0$ for some $u \neq 0$, then $g(f(u)) = (g\circ f)(u) = 0$ for some $u \neq 0$ which implies the non injectivity of $g \circ f$. So, assuming $f$ injective, $(g \circ f)(u) = 0$ if and only if $u = 0$ so that $g \circ f$ is injective. Also, you need to show that $g \circ f$ is surjective, that is, given $w \in W$ there is some $u \in U$ such that $(g\circ f)(u) = w$. If $g$ is not surjective, there is $w \in W$ such that $g(v) \neq w$ for every $v \in V$, so $g$ must be surjective.

If $f$ is not injective:

Define $f: \Bbb{R} \to \Bbb{R}$ by $f(x) = 0$ and $g: \Bbb{R} \to \Bbb{R}$ by $g(x) = x$. Then $im(f)\cap \ker g = \{0\}$ but $(g \circ f)(x) = g(f(x)) = g(0) = 0$.

If $g$ is not surjective:

Define $f: \Bbb{Q} \to \Bbb{Q}$ by $f(x) = x$ and $g: \Bbb{Q} \to \Bbb{R}$ by $g(x) = x$. Then $im(f) \cap \ker g = \{0\}$ but there is no $x \in \Bbb{Q}$ such that $(g \circ f)(x) = \pi$.


问:
Are there stronger Axioms of Constructibility that say when $L_\beta$ starts containing $V_\alpha$?
Are there stronger alternative axioms to the axiom of constructibility that additionally tell us an upper bound on how long it can take for things in $V$ to appear in $L$?

For example, let $\kappa \leftarrow \lambda$ denote ordinal exponentiation $\kappa^\lambda$.

$$ \small\textbf{STRAWMAN:}\;\; V_{\alpha} \subset L_{\omega \leftarrow \alpha} $$

I'm curious because I was reading about the differences between $V$ and $L$ and saw a handful of results on Wikipedia relating $V_\alpha$ and $L_\alpha$.

What follows is background, motivation, and my attempt to make sense of the material.

I was reading this answer to this question, which is substantially more advanced than my background, and stumbled on the notion of a Silver indiscernible. I tried to look up what a Silver indiscernible is and ended up on this page describing $0^\#$. That led me to this discussion of the Gödel constructible universe, which is different from the von Neumann universe that I've seen before but don't understand deeply. I tried to understand what the difference between these things is, and noticed there's an alternate axiom, the Axiom of Constructibility that asserts their equality. That got me wondering if we can consistently insist on bounds for when things in $V$ appear in $L$ if we take the axiom of constructibility (and so a set in $V$ always appears in $L$ eventually).

A given "slice" of the von Neumann universe $V_\alpha$ is indexed by an ordinal $\alpha$.

$$ V_0 = \emptyset $$
$$ V_{\beta+1} = P(V_{\beta}) \;\; \text{if $\beta$ is an ordinal number} $$
$$ V_{\lambda} = \bigcup_{\alpha < \lambda} V_{\alpha} \;\; \text{if $\lambda$ is a limit ordinal} $$

And the universe as a whole is the union of all such slices.

Gödel's construction is similar, but in the $\beta+1$ case, it only takes some of the subsets, not all of them.

$$ L_0 = \emptyset $$
$$ L_{\beta+1} = \text{Def}(L_\beta) \;\; \text{if $\beta$ is an ordinal number} $$
$$ L_{\lambda} = \bigcup_{\alpha < \lambda} L_{\alpha} \;\; \text{if $\lambda$ is a limit ordinal} $$

Where $\text{Def}$ is defined as follows.

Let's define a three-argument version of $\text{Def}$ where $\Phi$ is a well-formed formula and $\vec{z}$ is a finite vector of sets.

$$ \text{Def}(X, \Phi, \vec{z}) = 1 \;\;\text{iff}\;\;\text{each element of $\vec{z}$ is in $X$ and $(X, \in)$ is a model of $\Phi(\vec{z})$ } $$

$$ \text{otherwise $\text{Def}(X, \Phi, \vec{z})$ is zero } $$

So, a triple of $X, \Phi, \vec{z}$ gets sent to one if and only if $(X, \in)$ is a model of $\Phi(\vec{z})$ and $\vec{z}$, considered as a set, is a subset of $X$.

We can then define the one-argument version of $\text{Def}$ as follows.

$$ y \in \text{Def}(X) \;\;\text{iff}\;\; \text{there exists a $\Phi$ and $\vec{z}$ so that $\vec{z}_0 = y$ and $\text{Def}(X, \Phi, \vec{z}) = 1$} $$

It's clear that $\text{Def}(X)$ is a subset of $X$ and thus each $L_\alpha$ is clearly a subset of the corresponding $V_\alpha$.

The axiom of constructibility asserts that $V = L$, but doesn't say anything about how long we need to wait in $L$ to get an element that appears in $V_\alpha$.

I'm wondering if there are stronger axioms that do insist on a bound up front.

答:
This is actually already completely determined by $V=L$.  In particular, note that $|L_\alpha|=|\alpha|$ for all infinite $\alpha$, while $|V_{\omega+\alpha}|=\beth_\alpha$ for all $\alpha$ (and this is equal to $\aleph_\alpha$ assuming $V=L$).  So, $V_{\omega+\alpha}$ cannot be contained in $L_\beta$ for any $\beta<\omega_\alpha$, simply for cardinality reasons.  Conversely, I claim that $V=L$ implies $V_{\omega+\alpha}\subseteq L_{\omega_\alpha}$.  You can prove this by induction on $\alpha$.  Limit steps are trivial; for successor steps, suppose $V_{\omega+\alpha}\subseteq L_{\omega_\alpha}$ and we wish to show that $V_{\omega+\alpha+1}\subseteq L_{\omega_{\alpha+1}}$.  Since $V_{\omega+\alpha+1}=P(V_{\omega_\alpha})$, this follows from the fact that all constructible subsets of $L_{\omega_\alpha}$ are contained in $L_{\omega_{\alpha+1}}$ (this is a consequence of Gödel's condensation lemma, by essentially the same argument as the proof that $V=L$ implies GCH).

So, $V=L$ implies that for every $\alpha$, the least $\beta$ such that $V_{\omega+\alpha}\subseteq L_\beta$ is $\beta=\omega_\alpha$.


答:
Note that $f$ is continuous only in $0$ by definition, for $x\neq 0$ is also easy to prove discontinuity using $\epsilon$-$\delta$ definition.

Note that derivative of $f$ does not exists for $x\neq 0$ since $f$ is discontinuous. For $x=0$ note that $2^x$ has derivative different of $0$, and is easy to prove that $f$ is not differentiable at $0$.

But what happens with the following function $h$ (which is also continuous only at $x=0$), defined as

$$h(x)=\begin{cases} 0\ \  \text{ if } x\in\mathbb{Q}\\
x^2 \text{ if } x\in\mathbb{Q}^c\end{cases}$$

Both pieces have the same derivatives at $x=0$ might think that here the derivative exists at a single point $ x = 0 $, and $ h '(0) = 0$.

Is the differentiability in one point true in this last case?

Title: Continuity of a function defined differently on $\mathbb Q,\mathbb R\setminus \mathbb Q$

Body:
Let's say I define the function $f(x)=2^x$ for rational $x$, and $f(x)=1$ for irrational $x$.

My question is: is this function continuous everywhere? I think it's not, because for any $2$ irrational numbers you can find a rational in between, and for any $2$ rationals you can find an irrational in between.

My second question is: is the function continuous and differentiable at $x=0$? I think it is, and I think that $f'(0)=0$ because of the picture. Am I correct? (By the way please keep the answers at a pretty low level, I'm a calculus AB student.) Thanks!


问:
Recursive Definition of the Closure of a Set in a Family of Functions
For the sake of consistency, I will define some terms below the question, however, if you are familiar with the terms, that section can be skipped.

I am attempting to prove this lemma, for my text provides none.  I have some questions of which direction I should take my proof in.

Let $\theta$ be a regular infinite cardinal, and let $\mathcal{F}$ be a family of $<\theta$-ary functions on $B$.  Fix $S \subseteq B$, and define $S_\xi$ recursively by:
$$1. \,\,S_0 = S. \hspace{3.27in}\\ 2. \,\,S_\eta = \bigcup_{\xi<\eta}S_\xi \,\, \text{when } \eta \text{ is a limit ordinal.} \hspace{1.2in}\\ 3. \,\,S_{\xi +1} = S_\xi \cup\{f(\vec{x}): f \in \mathcal{F}\wedge f \text{ is } \alpha\text{-ary} \wedge \vec{x} \in (S_\xi)^\alpha\}.$$
Then $S_\theta$ is the closure of $S$ under $\mathcal{F}$.

In other words, to get $S_{\xi+1}$, we are taking $S_\xi$ and are adding to it all the values of every $f$ at every vector from $S_\xi$.  It is clear that if $\vec{x} \in (S_\xi)^\alpha$ then $f(\vec{x}) \in S_{\xi + 1}.$  My question is how do we go about proving this from here?  Why is it important that $\theta$ must be regular?  Why do we terminate the recursion at $\theta$?  How is the ordinal we terminate at related to the arity of the functions i.e. why are they both $\theta$?

TERM DEFINITONS:

A $<\theta \text{-ary}$ function on $B$ is an function $f$ such that $f:B^\alpha \rightarrow B$ where $\alpha < \theta$.  Because functions can have an infinite arity, the input is not a tuple, instead it is itself a function from the arity $\alpha$ into $B$. We say $\vec{x} \in B^\alpha$ but $\vec{x}$ is not a vector it is a function.  So $f$ maps functions in $B^\alpha$ to elements of $B$.

A set $A \subseteq B$ is closed under $f$ if $f(A^\alpha) \subseteq A$ i.e. $f$ maps "vectors" in $A^\alpha$ to $A$.  Let $\mathcal{F}$ be a family of $<\theta$-ary functions on $B$, then $A$ is closed under $\mathcal{F}$ if $A$ is closed under every $f \in \mathcal{F}$.  And lastly, the term which we are going to be focusing on, the closure of $S \subseteq B$ under $\mathcal{F}$ is defined as

$$A = \bigcap\{X:S \subseteq X \subseteq B \wedge X \text{ is closed under } \mathcal{F}\}.$$
In other words, $A$ is the smallest closed set under $\mathcal{F}$ which contains all of $S$.

答:
Let $f\in\mathscr{F}$, and let $\alpha$ be the arity of $f$. Let $\vec x\in S_\theta^\alpha$; for each $\xi<\alpha$ there is a $\beta_\xi<\theta$ such that $x_\xi\in S_{\beta_\xi}$. Let $\beta=\sup_{\xi<\alpha}\beta_\xi$; then $\beta<\theta$, since $\alpha<\theta=\operatorname{cf}\theta$. (This is where we use the fact that $\theta$ is regular and that $\alpha<\theta$.) Thus, $\vec x\in S_\beta^\alpha$, and $f(\vec x)\in S_{\beta+1}\subseteq S_\theta$. Since $f\in\mathscr{F}$ was arbitrary, we’ve shown that $S_\theta$ is closed under $\mathscr{F}$.

Let $\overline S$ be the closure of $S$ under $\mathscr{F}$; we’ve shown that $\overline S\subseteq S_\theta$. Suppose that $S_\theta\setminus\overline S\ne\varnothing$. For each $y\in\overline S\subseteq S_\theta$ there is a minimal $\xi(y)<\theta$ such that $y\in S_{\xi(y)}$; fix $y\in\overline S\subseteq S_\theta$ for which $\xi(y)$ is minimal. The minimality of $\xi(y)$ implies that $\xi(y)$ is not a limit ordinal, so let $\xi(y)=\eta+1$. Then $y\in S_{\xi(y)}\setminus S_\eta$, so there must be an $f\in\mathscr{F}$ of some arity $\alpha$ and an $\vec x\in S_\eta^\alpha$ such that $y=f(\vec x)$. But $S_\eta\subseteq\overline S$ by the choice of $y$, so $y\in f[S_\eta^\alpha]\subseteq\overline S$, contradicting the choice of $y$. It follows that $S_\theta\setminus\overline S=\varnothing$ and hence that $S_\theta=\overline S$.


问:
Proving there do not exist natural numbers $m$ and $n$ such that $7m^2=n^2$
I'm stuck on my proof of this concept and I could use some help on understanding what to do next.

Prove there do not exist natural numbers $m$ and $n$ such that $7m^2=n^2$

Proof will be by contradiction. So, assume there does exist natural numbers such that $7m^2=n^2$ is true.

Let $m=p_1^{r_1} p_2^{r_2 }…p_k^{r_k}$ and $n=p_1^{t_1} p_2^{t_2}…p_s^{t_s}$

Then $m^2=(p_1^{r_1} p_2^{r_2 }…p_k^{2r_k})^2=p_1^{2r_1} p_2^{2r_2 }…p_k^{2r_k}$ and $n^2=(p_1^{t_1} p_2^{t_2}…p_s^{t_s})^2=p_1^{2t_1} p_2^{2t_2}…p_s^{2t_s}$

Substitute this into $7m^2=n^2$ to obtain,

$7(p_1^{2r_1} p_2^{2r_2 }…p_k^{2r_k})=p_1^{2t_1} p_2^{2t_2}…p_s^{2t_s}$

The idea is that all values of m and n chosen to satisfy $7m^2=n^2$ are natural numbers so they therefore can be expressed as a product of primes.

This is where I'm stuck. How do I go about showing the contradiction?

Thanks for the help

答: The power of $7$ in the decomposition of $7m^2$ in prime numbers is odd, whereas the power of $7$ in the decomposition of $n^2$ in prime numbers is even.


答:
Note that if $(m,n)$ solution then $(\lambda n,\lambda m)$ is also solution, so without loss of generality we can suppose $\gcd(n,m)=1$.

Now since $7\mid 7m^2\implies 7\mid n^2$ and since $7$ prime $\implies 7\mid n$

Then $n=7n'$ and $7m^2=(7n')^2\iff m^2=7n'^2$.

But with a similar process then $7\mid m$ which is a contradiction since $\gcd(m,n)=1$.

Therefore no solutions except the trivial one $(0,0)$ exists.

Title: Proving there do not exist natural numbers $m$ and $n$ such that $7m^2=n^2$

Body:
I'm stuck on my proof of this concept and I could use some help on understanding what to do next.

Prove there do not exist natural numbers $m$ and $n$ such that $7m^2=n^2$

Proof will be by contradiction. So, assume there does exist natural numbers such that $7m^2=n^2$ is true.

Let $m=p_1^{r_1} p_2^{r_2 }…p_k^{r_k}$ and $n=p_1^{t_1} p_2^{t_2}…p_s^{t_s}$

Then $m^2=(p_1^{r_1} p_2^{r_2 }…p_k^{2r_k})^2=p_1^{2r_1} p_2^{2r_2 }…p_k^{2r_k}$ and $n^2=(p_1^{t_1} p_2^{t_2}…p_s^{t_s})^2=p_1^{2t_1} p_2^{2t_2}…p_s^{2t_s}$

Substitute this into $7m^2=n^2$ to obtain,

$7(p_1^{2r_1} p_2^{2r_2 }…p_k^{2r_k})=p_1^{2t_1} p_2^{2t_2}…p_s^{2t_s}$

The idea is that all values of m and n chosen to satisfy $7m^2=n^2$ are natural numbers so they therefore can be expressed as a product of primes.

This is where I'm stuck. How do I go about showing the contradiction?

Thanks for the help


问:
Prove that $s_{n+1}=s_n^2-2$ diverges.
Let $s_1=5$ and for $n\geq 1$, $s_{n+1}=s_n^2-2$. Do the following steps to show that $(s_n)$ diverges.

1.If $\lim s_n = a$ then $a^2 -2 = a$.

$a=\lim s_{n+1}=\lim {s_n^2-2}=\lim {s_n^2}-\lim (2)=(\lim {s_n})^2-\lim (2)=a^2-2$

2. If $\lim s_n = a$ then a has to be either $−1$ or $2$.

From (1), we have

$a=a^2-2$

$a^2-a-2=0$

$a=−1, 2$

3. Show that $\lim s_n$ can neither be $−1$ nor $2$. (Hint: write down the first few terms of $(s_n)$ to get some insights.)

$s_n\leq s_{n+1}$ for all $n$

Proof, by induction:

$P_1$ is true; at $n=1$, $5=s_1\leq s_2=5^2-2=23$

Suppose that $s_n\leq s_{n+1}$ for some $n$

By the assumption we have $s_{n+1}=s_n^2-2\leq s_{n+1}^2-2=s_{n+2}$

As a result, $s_n$ is increasing, so $s_n\geq 5$ for all $n$. Now since $s_n\geq 5$, this implies that $\lim s_n \geq 5$ which implies that it cannot be $-1$ or $2$.

4. Combine these steps to show that $\lim s_n$ does not exist.

Is there any further step to do here or just combine the 2 and 3?

Are my answers correct, please? Thank you in advance.

Title: Prove that $s_{n+1}=s_n^2-2$ diverges.

Body:
Other way.

It is easy to prove by induction that
$$(\forall n\ge 0)\; \;s_n\ge 2$$

on the other hand, for $ n> 1, $

$$s_n-2=s_{n-1}^2-4$$
$$=(s_{n-1}+2)(s_{n-1}-2)$$
$$\ge 4(s_{n-1}-2)\ge 4^{n-1}(s_1-2)$$

So,
$$s_n\ge 2+3.4^{n-1}$$
and
$$\lim_{n\to+\infty}s_n=+\infty$$


问:
Proof that if $a \mid bc$ and $\gcd(a,b) = d$ then $a/d$ divides $c$

Prove that if $a \mid bc$ and $\gcd(a,b) = d$ then $a/d$ divides $c$.

Attempt at solution

\begin{align}
&a \mid bc \implies \exists \ x \in \mathbb{Z} \ \text{s.t} \quad  bc = ax \tag{1}\\
&\gcd(a,b) = d \implies a \mid b \implies \exists \ t \in \mathbb{Z} \ \text{s.t} \quad b = at\tag{2} \\
&\gcd(a,b) = d \implies a \mid d \implies \exists \ s \in \mathbb{Z} \ \text{s.t} \quad d = as\tag{3}
\end{align}
From $(2)$ , we get $a = b/t$. Substituting in $(3)$, we have $d = as = bs/t  \implies b = dt/s$. Substituting this last result in $(1)$ ;
$$ bc = ax \implies \frac{dt}{s} c = ax.$$
Since $t,s$ and $x$ are all arbitrary constants in $\mathbb{Z}$, define $sx/t := x^{\prime} \in \mathbb{Z}$. Finally,
$$ dc = ax^{\prime}\implies c= \frac{a}{d}x' \implies \frac{a}{d} \mid c  \qquad \square.$$

Can anyone confirm my proof ? I feel like there are some justifications missing. Particularly in the substitions part could there not be divisions by $0$?

答:
$$a \text{ divides }bc \quad\Longrightarrow \quad\frac{a}{d} \text{ divides }  \frac{b}{d}  c \quad\Longrightarrow \quad\frac{a}{d} \text{ divides }  c$$

the last implication being true because $a/d$ and $b/d$ are coprime.


问:
Evaluating the elliptic integral $\int^1_0 \frac{K(k)}{\sqrt{1-k^2}}\,dk$
Find the elliptic integral $$\int\limits^1_0 \frac{K(k)}{\sqrt{1-k^2}}\,\mathrm{d}k,$$
where
$$K(k) = \int\limits_{0}^{1} \frac {\mathrm dt}{\sqrt{(1 - t^2)(1 - k^2t^2)}}=\int\limits_{0}^{\pi/2} \frac{\text{d}\vartheta}{\sqrt{1 - k^2 \sin^2\vartheta}}$$

My approach: Let $k=\dfrac{\sin\theta}{\sin\phi}$

Now after some calculation, I have reached the following: $$\int\limits_{0}^{\pi/2}\int\limits_{0}^{\phi}\dfrac{d\theta d\phi}{\sqrt{\sin^2\phi-\sin^2\theta}}$$
Now how to proceed from this point..please give some idea/hint.

Definition of elliptic integral: https://en.wikipedia.org/wiki/Elliptic_integral

答:
Replace $K$ with its $_2F_1$ representation (DLMF 19.5.1).

Evaluate using the Euler integral representation of $_3F_2$ (DLMF 16.5.2).

Simplify using Dixon's well-poised $_3F_2$ sum (DLMF 16.4.4).

Simplify using the functional relations for $\Gamma$ (DLMF 5.5).

The result is

$$\frac{\Gamma(1/4)^4}{16\pi}\text{.}$$

In detail:

$\newcommand{\d}{\mathrm{d}}$

$$\begin{split}
  \int_0^1 \frac{K(k)}{\sqrt{1-k^2}}\d k
    &= \frac{\pi}{2} \int_0^1 \frac{_2F_1(\tfrac{1}{2},\tfrac{1}{2};1;k^2)} {\sqrt{1-k^2}}\d k \\
    &= \frac{\pi}{4} \int_0^1 \frac{_2F_1(\tfrac{1}{2},\tfrac{1}{2};1;x)} {\sqrt{x(1-x)}}\d x \\
    &= \frac{\pi^2}{4}\,_3F_2(\tfrac{1}{2},\tfrac{1}{2},\tfrac{1}{2};1,1;1) \\
    &= \frac{\pi^2}{4}\left(\frac{\Gamma(\tfrac{5}{4})\Gamma(1)\Gamma(1)\Gamma(\tfrac{1}{4})}{\Gamma(\tfrac{3}{2})\Gamma(\tfrac{3}{4})\Gamma(\tfrac{3}{4})\Gamma(\tfrac{1}{2})}\right) \\
    &= \frac{\Gamma(\tfrac{1}{4})^4}{16\pi}\text{.}
\end{split}$$


问:
A function vanishing at $0$ that grows faster than the exponential
Does there exist an increasing function $f$ that is continuous and vanishes at $0$ and satisfies $$f\left(\frac{1}{k}\right)e^{2k} \to 0$$ as $k \to \infty$? The common examples of modified factorials do not work since the input is supposed to take the reciprocal of $k$ instead of just $k$.

答:
Consider $f(x)=xe^{-1/x^2}$. Then, $f$ is continuous, increasing, $f(0)=0$ and

$$\lim_{k\to\infty} f\left(\frac{1}{k}\right)e^{2k}=\lim_{k\to\infty} \frac{1}{k}e^{-k^2+2k} = 0.$$


问:
Intersection of two balls in n-dimension
I've been thinking about how two prove this statement (mathematically) but cant seem to figure it out

the definition of a ball two dimensions is $B(x,r)=:\|x-y\|\leq r ~~\forall y\in\mathbb{R} $ where $x$ is the center of the ball and r is the radius.
Now define

$B_\mathbb{R^{m_1+m_2}}((x_1x_2),r)$
a ball in $\mathbb{R^{m_1+m_2}}~~~$ where $x_1 =(x1,x2,x3,\dots,xm_1)$$x_1 =(xm_1+1,xm_1+2,xm_1+3,\dots,xm_2)$
and
$B_\mathbb{R^{m_1}}(x_1,r)\times B_\mathbb{R^{m_2}}(x_2,r)$
the cartesian product of two balls in $\mathbb{R^{m_1}}$ and $\mathbb{R^{m_2}}$

prove that $B_\mathbb{R^{m_1+m_2}}((x_1,x_2),r)\subset B_\mathbb{R^{m_1}}(x_1,r) \times B_\mathbb{R^{m_2}}(x_2,r)$

in the one dimensional case a ball is just an interval so the cartesian product of two intervals is a square and its easy to see geometrically that a circle centered at $(x_1,x_2)$ and radius r is less than a square with sides of length r. One idea I have is to choose an arbitrary point $z \in B_\mathbb{R^{m_1+m_2}}((x_1,x_2),r)$ and show that it is bounded by the cartesian product of the two balls.

the first step I think would be to center the balls at the origin to make calculation simpler.

defining  $B_\mathbb{R^{m_1}}(0,r)\times B_\mathbb{R^{m_2}}(0,r)$ in n dimensions.
would it be $B_\mathbb{R^{m_1}}(0,r)\times B_\mathbb{R^{m_2}}(0,r)=:max\{\|x_1\|,\|x_2\|\}$ (i think) where do i go next?

Body:
Consider $\|\cdot\|$ the maximum norm.

If $x \in B_{\Bbb{R}^{m_1 + m_2}}((x_1,x_2);r)$, then
$$\|(x_1,x_2) - x\| \leq r.$$
Making the identification $(v_1,...,v_{m_1},\underbrace{0,...,0}_{m_2\text{ times}})$ with $(v_1,...,v_{m_1}) \in \Bbb{R}^{m_1}$, you can write $x = (y_1,y_2)$ where $y_1 \in \Bbb{R}^{m_1}$ and $y_2 \in \Bbb{R}^{m_2}$. Thus,
$$\|(x_1-y_1,x_2-y_2)\| \leq r.$$
Note that
$$\|x_1 - y_1\| \leq \max\{|x_1 - y_1|,|x_2 - y_2|\} = \|(x_1-y_1,x_2-y_2)\| \leq r$$
in the same way, $\|x_2 - y_2\| \leq r$. Therefore, $y_1 \in B_{\Bbb{R}^{m_1}}(x_1,r)$ and $y_2 \in B_{\Bbb{R}^{m_2}}(x_2,r)$.


答:
It could use a little more explanation, but yes, it works. I’d expand it to point out explicitly why $n(n+1)$ is even and that $n=(n^2+n)-n^2$ is then the difference of two even numbers and as such is even (assuming that you already have this fact available to use).

An alternative approach is to show that if $n$ is odd, then $n^2$ is odd; the desired result is the contrapositive and therefore follows at once.

Title: Prove if $n^2$ is even, then $n$ is even.

Body:
I am just learning maths, and would like someone to verify my proof.

Suppose $n$ is an integer, and that $n^2$ is even.  If we add $n$ to $n^2$, we have $n^2 + n = n(n+1)$, and it follows that $n(n+1)$ is even.  Since $n^2$ is even, $n$ is even.

Is this valid?


答:
Assuming you are referring to the standard Euclidean norm, an answer to your question "where to go next"? is to simply observe that
$$\max\{\|x_1\|,\|x_2\|\}\leq (\|x_1\|^2+\|x_2\|^2)^{1/2}$$
and the r.h.s is the norm of $B_{\mathbb{R}^{m_1+m_2}}$.

Title: Intersection of two balls in n-dimension

Body:
I've been thinking about how two prove this statement (mathematically) but cant seem to figure it out

the definition of a ball two dimensions is $B(x,r)=:\|x-y\|\leq r ~~\forall y\in\mathbb{R} $ where $x$ is the center of the ball and r is the radius.
Now define

$B_\mathbb{R^{m_1+m_2}}((x_1x_2),r)$
a ball in $\mathbb{R^{m_1+m_2}}~~~$ where $x_1 =(x1,x2,x3,\dots,xm_1)$$x_1 =(xm_1+1,xm_1+2,xm_1+3,\dots,xm_2)$
and
$B_\mathbb{R^{m_1}}(x_1,r)\times B_\mathbb{R^{m_2}}(x_2,r)$
the cartesian product of two balls in $\mathbb{R^{m_1}}$ and $\mathbb{R^{m_2}}$

prove that $B_\mathbb{R^{m_1+m_2}}((x_1,x_2),r)\subset B_\mathbb{R^{m_1}}(x_1,r) \times B_\mathbb{R^{m_2}}(x_2,r)$

in the one dimensional case a ball is just an interval so the cartesian product of two intervals is a square and its easy to see geometrically that a circle centered at $(x_1,x_2)$ and radius r is less than a square with sides of length r. One idea I have is to choose an arbitrary point $z \in B_\mathbb{R^{m_1+m_2}}((x_1,x_2),r)$ and show that it is bounded by the cartesian product of the two balls.

the first step I think would be to center the balls at the origin to make calculation simpler.

defining  $B_\mathbb{R^{m_1}}(0,r)\times B_\mathbb{R^{m_2}}(0,r)$ in n dimensions.
would it be $B_\mathbb{R^{m_1}}(0,r)\times B_\mathbb{R^{m_2}}(0,r)=:max\{\|x_1\|,\|x_2\|\}$ (i think) where do i go next?


问:
What's a formula for a graph like this?
I'm trying to find a method to turn a value (that can reach a maximum of N) into a percentage. But I don't want it to be linear, I need it to be non-linear like the in this graph:

graph

答:
Generally, any function that fits the following qualifications will be somewhat similar to the graph:

Increasing from $x = 0$ to $x = N$

Concave from $x = 0$ to $x = N$

$f(0) = 0$ and $f(N) = 100$

To keep things simple, we can use a function that is twice-differentiable. Then, we need $f'(x) > 0$ for $x\in (0,N)$, and $f''(x) < 0$ for $x\in (0,N)$.

There are a lot of options out there for functions that fit these requirements, but two easy ones to implement are

$f(x) = 100-\frac{100}{N^2}(x-N)^2$

$f(x) = 100\sqrt{\frac{x}{N}}$

Title: What's a formula for a graph like this?


问:
Is logistic regression cost function in SciKit Learn different from standard derivations?
I am trying to understand the math behind logistic regression. Going through a couple of websites, lectures and books, I tried to derive the cost function by thinking of it as the negative of the maximum likelihood. My derivation matches the cost function shown in this Wikipedia page https://en.wikipedia.org/wiki/Logistic_regression and in other places.

If the inputs are $x^{(i)}$ and outputs are $y^{(i)}$, where $(i)$ refers to the $i$th data point, then the cost as a function of weights $w$ seems to be

$$-\sum_{i=1}^m y^{(i)} \log\left(\frac{1}{1 + e^{-w^Tx^{(i)}}}\right)+\left(1-y^{(i)} \right)
\log \left( 1-\frac{1}{1 + e^{-w^Tx^{(i)}}} \right) $$

I can simplify further to $$\sum_{i=1}^m -w^Tx^{(i)} y^{(i)} +\log (1+e^{-w^Tx^{(i)}})$$
However the expression shown in the SciKit Learn guide https://scikit-learn.org/stable/modules/linear_model.html#logistic-regression
is $$\sum_{i=1}^m \log ( 1 + e^{-w^Tx^{(i)}y^{(i)}}) $$

I have tried some algebra and am not able to derive their formulation.  Am I missing something?  It is highly possible that I haven't tried all the tricks there are in simplifying

答:
From the same scikit-learn link, note that, in this notation, it’s assumed that the target $y^{(i)}$ takes values $\{-1,+1\}$ in the set at trial $i$.

To the contrary, $y^{(i)} \in \{0,1\}$ as per your initial definition of the binary cross entropy (BCE) cost function.

In the scikit-learn notation, we have

$$P(y^{(i)}=+1\mid x^{(i)})=\sigma(w^Tx^{(i)})=\frac{1}{1+e^{-w^Tx^{(i)}}}$$

$$P(y^{(i)}=-1\mid x^{(i)})=1-\sigma(w^Tx^{(i)})=\frac{1}{1+e^{w^Tx^{(i)}}},$$

so that in both cases we have

$$P(y^{(i)}\mid x^{(i)})=\frac{1}{1+e^{-w^Tx^{(i)}y^{(i)}}}$$

With independence assumption, the likelihood is $\prod\limits_{i=1}^m \frac{1}{1+e^{-w^Tx^{(i)}y^{(i)}}}$

The negative log-likelihood is $\sum\limits_{i=1}^m \log(1+e^{-w^T x^{(i)} y^{(i)}})$, which is the cost function to be minimized (for MLE), along with some ($L_2$) regularization term.


问:
Induction of a sum
I need to prove using induction
$$1 \cdot n + 2(n-1)+3(n-2)+ \cdots +n \cdot 1 = {n \over 6}(n+1)(n+2)$$

but when I put $n=k+1$ I get:
$$1 \cdot (k+1) + 2(k)+3(k-1)+ \cdots +k \cdot 2+ (k+1) \cdot 1 = {k+1 \over 6}(k+2)(k+3)$$

and I have no idea how to procceed. :(

Thank for anyone who can help!

Title: Induction of a sum

Body:
I'm only writing the inductive step as you should be able to fill in the rest.
$$1 \cdot (k+1) + 2(k)+3(k-1)+ \cdots +k \cdot 2+ (k+1) \cdot 1\\
=1 \cdot (k+1)+k+(k-1)+\cdots+2+1+{k \over 6}(k+1)(k+2)\\
= {(k+1)(k+2) \over 2}+{k \over 6}(k+1)(k+2)\\
= {3(k+1)(k+2)+k(k+1)(k+2)\over 6}\\
={k+1 \over 6}(k+2)(k+3)$$


问:
Proof of the Moreau Envelope of $l_1$ norm
Given function $| \cdot | : \mathbb{R} \rightarrow \mathbb{R}_{+}$ and $\alpha > 0$, its Moreau envelope $e_{\alpha}|\cdot|: \mathbb{R} \rightarrow \mathbb{R}_{+}$, reads:
\begin{equation}
e_{\alpha}|\cdot|\left(x\right) = \inf_{y \in \mathbb{R}}{\bigg\{|y| + \frac{1}{2\alpha} | x - y|^2\bigg\}}
\end{equation}

How would you formally prove that the Moreau Envelope is given by?
\begin{equation}
e_{\alpha}|\cdot|\left(x\right) =
\begin{cases} \frac{1}{2\alpha}x^2 & |x| \leq \alpha \\ |x| - \frac{\alpha}{2} & |x| > \alpha \end{cases}
\end{equation}

答:
Suppose $y\ge 0$. Observe that
\begin{align}
|y|+\frac{1}{2\alpha}|x-y|^2 =&\ \frac{x^2+y^2}{2\alpha}+\frac{xy}{\alpha}+y = \frac{x^2-(x+\alpha)^2}{2\alpha} + \frac{1}{2\alpha}\left(y^2+2(x+\alpha)y+(x+a)^2 \right)\\
=&\ \frac{x^2-(x+\alpha)^2}{2\alpha} +\frac{(y+x+\alpha)^2}{2\alpha}.
\end{align}
If $x+\alpha>0$, then the minimum occurs when $y = 0$, i.e.
\begin{align}
\frac{x^2-(x+\alpha)^2}{2\alpha} +\frac{(y+x+\alpha)^2}{2\alpha} \ge \frac{x^2}{2\alpha}.
\end{align}
When $x+\alpha<0$ (i.e. $-x>\alpha>0$), then the minimum occurs when $y = -(x+\alpha)$ which means
\begin{align}
\frac{x^2-(x+\alpha)^2}{2\alpha} +\frac{(y+x+\alpha)^2}{2\alpha} \ge \frac{x^2-(x+\alpha)^2}{2\alpha} = -x-\frac{\alpha}{2}=|x|-\frac{\alpha}{2}.
\end{align}

For $y<0$. Observe that
\begin{align}
|y|+\frac{1}{2\alpha}|x-y|^2 =&\ \frac{x^2+y^2}{2\alpha}+\frac{xy}{\alpha}-y = \frac{x^2-(x-\alpha)^2}{2\alpha} + \frac{1}{2\alpha}\left(y^2+2(x-\alpha)y+(x-\alpha)^2 \right)\\
=&\ \frac{x^2-(x-\alpha)^2}{2\alpha} +\frac{(y+x-\alpha)^2}{2\alpha}.
\end{align}

I will leave you to complete the argument.


问:
How to find the largest ellipsoid which fits within a convex hull?
I am looking for a mathematical realization of finding the largest ellipsoid which fits within a convex hull. Let's say the ellipsoid is defined as

$$ E = \left\{ By+d \mid \left \| y \right \|_2\ \leq 1 \right\} $$

where $d$ is the center of the ellipsoid. For a given set of points, corresponding convex hull can be calculated in H-Ref (hyperplane representation), i.e., $Ax \leq b$. Thus, the idea is to find the ellipsoid which satisfied these two type of constraints minimizing the volume of the ellipsoid, i.e., $-\log(\det(A))$.

Here the problem is I still can't understand the way constraints set should be defined?

In other words,

答:
Note: thanks to @RodrigodeAzevedo and @ Dominic for sharing valuable information.
The answer is based on this

Let's say you are given a set of inequalities that describes by a polytope:

\begin{equation}
    P = \{ x \in \mathbb{R}^n \mid a_i^Tx \leq b_i, \; i=1,...,m\}
\end{equation} So intention is to find inscribed ellipsoid with maximum volume. Let $\varepsilon$ be the ellipsoid that is to be estimated:
\begin{equation}
    E = \{x \mid x = By + d, y \in \mathbb{R}^n, \left \| y \right \|_2 \leq 1, \; B= B^T \succ 0\}.
\end{equation} In order to maximize the volume, the condition $E \subseteq P$ should be satisfied; this condition can be formed as the following way as a set of inequalities:
\begin{equation}
  \left \| Ba_i \right \| + a_i^Td \leq b_i, \; i=0,...,m
\end{equation}

Now we are ready to find the inscribed ellipsoid.
\begin{equation}\label{max_volume}
\begin{aligned}
\min_{B,d} \quad & -\log \det(B)\\
\textrm{s.t.} \quad &\left \| Ba_i \right \| + a_i^Td \leq b_i, \; i=0,...,m \\
  & B \succeq 0\\
\end{aligned}
\end{equation}

In python, this can be written as follows:

```<code>B = cp.Variable((dim,dim), PSD=True)
d = cp.Variable(dim)
constraints = [cp.norm(B@A[i],2)+A[i]@d<=b[i] for i in range(len(A))]
prob = cp.Problem(cp.Minimize(-cp.log_det(B)), constraints)
optval = prob.solve()
</code>```

For example,

Body:
Note: thanks to @RodrigodeAzevedo and @ Dominic for sharing valuable information.
The answer is based on this

Let's say you are given a set of inequalities that describes by a polytope:

\begin{equation}
    P = \{ x \in \mathbb{R}^n \mid a_i^Tx \leq b_i, \; i=1,...,m\}
\end{equation} So intention is to find inscribed ellipsoid with maximum volume. Let $\varepsilon$ be the ellipsoid that is to be estimated:
\begin{equation}
    E = \{x \mid x = By + d, y \in \mathbb{R}^n, \left \| y \right \|_2 \leq 1, \; B= B^T \succ 0\}.
\end{equation} In order to maximize the volume, the condition $E \subseteq P$ should be satisfied; this condition can be formed as the following way as a set of inequalities:
\begin{equation}
  \left \| Ba_i \right \| + a_i^Td \leq b_i, \; i=0,...,m
\end{equation}

Now we are ready to find the inscribed ellipsoid.
\begin{equation}\label{max_volume}
\begin{aligned}
\min_{B,d} \quad & -\log \det(B)\\
\textrm{s.t.} \quad &\left \| Ba_i \right \| + a_i^Td \leq b_i, \; i=0,...,m \\
  & B \succeq 0\\
\end{aligned}
\end{equation}

In python, this can be written as follows:

B = cp.Variable((dim,dim), PSD=True)
d = cp.Variable(dim)
constraints = [cp.norm(B@A[i],2)+A[i]@d<=b[i] for i in range(len(A))]
prob = cp.Problem(cp.Minimize(-cp.log_det(B)), constraints)
optval = prob.solve()

For example,


答: Incrementing $n$ from $k$ to $k+1$ increases the $i$th of the original $k$ terms by $i$; we also add an extra term equal to $k+1$. The result is$$\begin{align}\tfrac16k(k+1)(k+2)+\tfrac12k(k+1)+k+1&=\tfrac16(k+1)(k^2+2k+3k+6)\\&=\tfrac16(k+1)(k+2)(k+3).\end{align}$$

Title: Induction of a sum

Body:
I need to prove using induction
$$1 \cdot n + 2(n-1)+3(n-2)+ \cdots +n \cdot 1 = {n \over 6}(n+1)(n+2)$$

but when I put $n=k+1$ I get:
$$1 \cdot (k+1) + 2(k)+3(k-1)+ \cdots +k \cdot 2+ (k+1) \cdot 1 = {k+1 \over 6}(k+2)(k+3)$$

and I have no idea how to procceed. :(

Thank for anyone who can help!


问:
Proving stochastic dominance for hypergeometric random variables
I suspect that the following statement is true:

Statement: Let $f$ be a nondecreasing function, and let $X_k$ be a hypergeometric($2n$, $k$, $n$) random variable, where $n\ge1$ is a constant integer. Then $\mathbb{E}[f(X_k/n)]$ is nondecreasing as $k$ increases.

(Here, $X_k$ is the number of "good" balls out of $n$ balls taken uniformly at random, all at once, from a bag containing $2n$ balls, $k$ of which are "good".)

This is equivalent to verifying whether $\frac{X_{m+1}}{n}\succeq\frac{X_{m}}{n}$ (and, obviously by extension, $X_{m+1}\succeq X_{m}$) in terms of first-degree stochastic dominance (Levy 1998, Stochastic dominance). This means that the distribution function of $X_{m+1}$ (or the probability that $X_{m+1}\le j$) is less than or equal to that of $X_m$ for each $j$ in the interval [0, n].

However, I find that proving this is quite tedious, as the distribution functions have a non-trivial form. It's not enough to say that $\mathbb{E}[X_k/n]=k/(2n)$ is increasing with k, since this does not imply the statement given above as stochastic dominance does.

Is there an existing proof of this statement or a generalization of it, or existing results that could make proving it less tedious?

答:
Here is a combinatorial argument that $X_{k+1}\succeq X_{k}$:

Let's suppose your bag has $k$ good balls, $1$ intermediate ball, and  $2n-k-1$ bad balls, so $2n$ in total.  $n$ balls are taken without replacement and $X_k$ counts the number of good balls taken and $Y_{k+1}$ counts the number of good or intermediate balls taken.

Clearly $Y_{k+1} \ge X_k$, and $\mathbb P(Y_{k+1} \ge x) \ge \mathbb P(X_{k} \ge x)$ for all $x$.

Since $Y_{k+1}$ has the same distribution as $X_{k+1}$, this implies $\mathbb P(X_{k+1} \ge x) \ge \mathbb P(X_{k} \ge x)$ for all $x$ and so $X_{k+1}\succeq X_{k}$.


答:
Set $$f(x)=\frac{200 x}{N}-\frac{100 x^2}{N^2}$$
where $N$ is the maximum $x$ value.

For instance, if you want $N=150$ you set
$$f(x)=\frac{4 x}{3}-\frac{x^2}{225}$$
and get a curve like the following

Title: What's a formula for a graph like this?

Body:
I'm trying to find a method to turn a value (that can reach a maximum of N) into a percentage. But I don't want it to be linear, I need it to be non-linear like the in this graph:

graph


问:
Sufficient condition for biconnectivity
Assume we have a planar graph  $G$ with $2n+1$ vertices and $3n$ edges, such that it is bipartite $G=A\cup B$ and every vertex from $A$ is of degree $\leq 3$. Does it suffice to conclude that $G$ is a biconnected graph?

or guide me to the conditions from literature so that I can write what I have tried.

答:
No, this is not true.

Consider the graph on $7=2\cdot3+1$ vertices $\{A,b,C,D,e,f,G\}$ with edges

$A\sim b$;

$b\sim C,D$;

$C,D\sim e,f$; and

$e,f\sim G$.

This has $1+2+4+2=9=3\cdot3$ edges and a bipartition is $\{\text{upper-case}\}\sqcup\{\text{lower-case}\}$.  Moreover, every vertex has degree at most $3$.

But removing $b$ disconnects the graph into $\{A\}$ and $\{C,D,e,f,G\}$.

Title: Sufficient condition for biconnectivity


问:
Integrate $\int_2^5 \dfrac{1}{1+\sqrt{x-1}}\, \mathrm{d}x $ using substitution with a fraction
Im doing this question currently

$$\int_2^5 \dfrac{1}{1+\sqrt{x-1}}\, \mathrm{d}x $$

and I set $u = x-1$ to get $\tfrac{\mathrm{d}u}{\mathrm{d}x} =1$ so $du =dx$.

I then write it out as $$\frac{1}{1+\sqrt{u}}\,\mathrm{d}u$$

But then I hit a brick wall and I'm not sure how to further progress. I know I need to replace the $u$ but I'm not sure about anything else

Body: A single substitution will suffice:  $$u = 1 + \sqrt{x-1}$$ implies $$x = (u-1)^2 + 1 = u^2 - 2u + 2, \\ dx = 2u - 2 \, du.$$  Hence $$\int_{x=2}^5 \frac{dx}{1 + \sqrt{x-1}} = \int_{u=2}^3 \frac{2u-2}{u} \, du =2 \int_{u=2}^3 1 - \frac{1}{u} \ du = 2 \left[u - \log u \right]_{u=2}^3 = 2(1 + \log \tfrac{2}{3}).$$


问:
If $f_x(x,y)>0$, $f_{xx}(x,y)<0$, $f_y(x,y)>0$, $f_{yy}(x,y)<0$ can $f_{xy} $ change sign?
That is, suppose we have a continuous and (at least) twice differentiable function $f(x,y)$ which is

increasing but concave in each of its individual arguments$^*$,

Note that these derivatives are all assumed to be non-zero

the domain of $f$ is $[0,C]$ with $C>0$ (if changing the domain matters please feel free to let me know. I'm interested)

Is it possible for the cross partial derivative to change signs (? (i.e. $\frac{\partial f}{\partial x \partial y} >0$ at some point but $<0$ at another point)

If it is possible, what if we look along a line (i.e. fix an $x$ or $y$ value and look along the line from that point) (just answering the bold question is an acceptable answer)

If it is possible, can someone provide an example function? If not, I am looking for a proof (or hints for a proof, or some intuition)

*By increasing but concave in each argument I mean  $\frac{\partial f}{\partial x} >0$ and $\frac{\partial^2 f}{\partial x^2}<0$, and similar for $y$

Edit: I made question more slightly general. I apologize if someone was already typing an answer. (It is more general because if this edited question is true, then so is the original)

答:
For convenience, I will work on the domain $x,y > 0$. Of course you can shift this solution to get one in your desired domain.

There are no obvious relationships between the second order partials of a function in general that I'm aware of, so I'm going to guess there is an $f$ with the properties you desire.

Guess a solution of the form $f(x,y) = F(xy)$ where $F:\mathbb{R}_+ \to \mathbb{R}$. Then we compute
$$f_x = F'(xy)y,\qquad f_y=F'(xy)x,$$
$$
f_{xx} = F''(xy)y^2, \qquad f_{yy} = F''(xy)x^2
$$
$$
f_{xy} = F''(xy)xy + F'(xy).
$$
In order that $f_x, f_y > 0$ and $f_{xx}, f_{yy} < 0$, we restrict to $F$ such that $F' > 0$ and $F'' < 0$.
Note that $f_{xy}$ is a function of $xy$, so we really just need that $F''(t) t + F'(t)$ changes sign. Near $t=0$ (but positive), we have $F''(t) t + F'(t) \approx F'(t)> 0$, but when $t$ is large, we hope that $t F''(t)$ dominates, making the sum negative, which will be the case if $F'$ and $F''$ are approximately the same magnitude. At this point, a particularly easy choice stands out, $F(t) := -e^{-t}$ which has $F'(t) = e^{-t} > 0$ and $F''(t) = -e^{-t} < 0$, and $$F''(t) t + F'(t) = -e^{-t} t + e^{-t} = e^{-t}(1-t),$$
which changes sign at $t=1$.

Therefore, $f(x,y) = -e^{-xy}$ satisfies all your conditions and has $f_{xy} < 0$ if $xy > 1$, and $f_{xy} > 0$ if $xy < 1$.

By chance, this example also answers your second question. Even if you restrict to a line $x = x_0$ you can still choose $y$ with $x_0 y < 1$ or with $x_0 y > 1$ (and similar for fixed $y=y_0$).


问:
Open (closed) balls are convex set
A set $E \subset \mathbb{R}^d$ is convex if whenever $x,y \in E$ then $\lambda x + (1- \lambda) y \in E$ where $\lambda \in (0,1)$

Show that $ B(x) = \{ y \in \mathbb{R}^d : |x-y| < \epsilon \}$ is convex set.

${\bf Attempt}$

Take two elements $a,b \in B (x)$ and so $|a-x| < \epsilon / (2 \lambda + 1 )$ and $|b-x| < \epsilon / (2 \lambda + 1 )$

Now,

$$ |\lambda a + (1-\lambda) b - x | = | \lambda ( a - b) + (b-x) | \leq \lambda |a-x+x-b| + |b-x| \leq \lambda (|a-x| + |b-x| ) + |x-b| < \dfrac{2 \lambda \epsilon}{2 \lambda + 1} + \dfrac{ \epsilon }{2 \lambda + 1} = \epsilon $$

So, B(x) is convex

Is this a correct solution?

答:
You are assuming that $|a-x|$ and $|b-x|$ are smaller than what they are in general if $a,b\in B(x)$. There is no need for that; you must only assume that $a,b\in B(x)$, that is $|x-a|<\varepsilon, |b-x|<\varepsilon$. Under these assumptions only, you must take $\lambda\in(0,1)$ and show that $\lambda a+(1-\lambda)b\in B(x)$, i.e. you want to show that
$$|\lambda a+(1-\lambda)b-x|<\varepsilon.$$
This is only a matter of calculations:

$$|\lambda a+(1-\lambda)b-x|=|\lambda a+(1-\lambda)b -(\lambda+1-\lambda)x|=$$
$$|\lambda a-\lambda x+ (1-\lambda )b-(1-\lambda)x|\leq|\lambda (a-x)|+|(1-\lambda)(b-x)|=$$ $$=\lambda\cdot|x-a|+(1-\lambda)\cdot|b-x|<\lambda\cdot\varepsilon+(1-\lambda)\cdot\varepsilon=\varepsilon. $$

Title: Open (closed) balls are convex set


问:
Monic polynomial with certain properties is the characteristic polynomial?
I'm trying to determine whether this statement is true or false:

Assume that $dim V = n$ and let $T \in L(V)$. Let $f(z) \in P_n(F)$ be a
monic polynomial of degree
$n$ such that $f(T) = 0$. Then $f(z)$ is the characteristic polynomial of $T$.

I know that because $f(T) = 0$, the minimal polynomial of T divides it and that the minimal polynomial of T divides the characteristic polynomial of T as well, but I can't seem to put these together in a way that conclusively establishes the truth or falsehood of the above statement.

If anyone could shed some insight on the matter, it would be greatly appreciated. Thanks!

答:
This is not true in general. The minimum polynomial is the smallest degree polynomial that the linear map satisfies, and the characteristic polynomial is defined to be $\det(xI-T)$. If the minimum polynomial is degree $m$, we know that the characteristic polynomial will be degree $n$ so $m\leq n$. If $p(x)$ is the minimum polynomial and $g(x)$ is any other poly, then $T$ will satisfy the polynomial $p(x)g(x)$. So if $m < n$, we can multiply $p(x)$ by any polynomial of degree $(n-m)$ to get a new polynomial of degree $n$ that the linear map satisfies. This will not, in general, be the characteristic polynomial.

The above is all done thinking almost solely about polynomials. Now I'll talk about how we can understand the problem thinking more about linear maps. Working over an algebraically closed field, we know that we can pick a basis with respect to which the matrix of $T$ is in Jordan Normal Form (JNF). The multiplicities of the roots of the characteristic polynomial are the number of times that root appears on the diagonal of the matrix in JNF, but the multiplicities of the roots of the minimum polynomial tell us the size of the largest block corresponding to that root. From looking at the matrix, it is intuitively clear that the matrix will satisfy a given polynomial $f(x)$ iff $p(x)$ divides $f(x)$. Thus we can, as before, multiply $p(x)$ by any polynomial of degree $n-m$ to get one of degree $n$ that the linear map satisfies.


答:
First I would rephrase the problem so there are not so many subscripts.
Let's work with $\mathbb R^{m+n} = \mathbb R^m \times \mathbb R^n.$
And we usually use the letter $x$ to describe the coordinates of an arbitrary point in space (or in the ball), not the center of the ball.
As you suggest, the problem is simpler if we suppose that the coordinates at the center of the ball are all zero; later we might consider non-zero coordinates at the center.

Also, we know the balls we're dealing with are all in real Cartesian coordinate spaces, so it's enough to specify the dimension of the space, that is,
$B_k$ rather than $B_{\mathbb R^k}.$

So the $(m+n)$-dimensional ball of radius $r$ with center $0$ in $\mathbb R^{m+n}$ is

$$ B_{m+n}(0,r)
 = \{ (x_1,x_2,\ldots,x_{m+n})\mid x_1^2 + x_2^2 + \cdots + x_{m+n}^2 \leq r^2 \}. $$

The Cartesian product of $B_m(0,r)$ and $B_n(0,r)$ is

\begin{eqnarray}
 B_m(0,r) \times B_n(0,r)
&=& \{ (x_1,x_2,\ldots,x_m)\mid x_1^2 + x_2^2 + \cdots + x_m^2 \leq r^2 \}\\
&& {}\quad \times \{ (x_{m+1},x_{m+2},\ldots,x_{m+n})\mid x_{m+1}^2 + x_{m+2}^2
   + \cdots + x_{m+n}^2 \leq r^2 \} \\
&=& \{ (x_1,x_2,\ldots,x_{m+n})\mid x_1^2 + \cdots + x_m^2 \leq r^2 , \,
       x_{m+1}^2 + \cdots + x_{m+n}^2 \leq r^2 \} \\
\end{eqnarray}

To show that $B_{m+n}(0,r) \subseteq B_m(0,r) \times B_n(0,r),$
you merely need to show that if $(x_1, x_2, \ldots, x_{m+n})$ is a point in
$\mathbb R^{m+n}$ such that

$$ x_1^2 + x_2^2 + \cdots + x_{m+n}^2 \leq r^2 $$

then both of the following inequalities also are true:

$$
\begin{cases}
x_1^2 + x_2^2 + \cdots + x_m^2 \leq r^2,\\[0.5ex]
x_{m+1}^2 + x_{m+2}^2 + \cdots + x_{m+n}^2 \leq r^2.
\end{cases}
$$

To show this, observe that

$$
x_1^2 + x_2^2 + \cdots + x_{m+n}^2 =
 (x_1^2 + x_2^2 + \cdots + x_m^2) + (x_{m+1}^2 + x_{m+2}^2 + \cdots + x_{m+n}^2)
$$

and that

\begin{gather}
x_1^2 + x_2^2 + \cdots + x_m^2 \geq 0,\\[0.5ex]
x_{m+1}^2 + x_{m+2}^2 + \cdots + x_{m+n}^2 \geq 0.
\end{gather}

Therefore

$$ x_1^2 + x_2^2 + \cdots + x_m^2 \leq x_1^2 + x_2^2 + \cdots + x_{m+n}^2 $$

and

$$ x_{m+1}^2 + x_{m+2}^2 + \cdots + x_{m+n}^2
\leq x_1^2 + x_2^2 + \cdots + x_{m+n}^2. $$

For a ball with center somewhere other than all zero coordinates,
we can suppose we have a ball
$B_{m+n}(\mathbf a, r)$ centered at
$$ \mathbf a = (a_1, a_2, \ldots, a_{m+n}). $$

Then instead of $x_1^2 + x_2^2 + \cdots + x_{m+n}^2 \leq r^2$ we must write

$$ (x_1 - a_1)^2 + (x_2 - a_2)^2 + \cdots + (x_{m+n} - a_{m+n})^2 \leq r^2 $$

and so forth.


问:
A simple equality related to an irreducible representation of a finite group
Let $G$ be a finite group of cardinal $|G|$. Let $u \colon G \to M_n$ be an irreducible unitary representation of $G$. How to show that for any matrix $A \in M_n$ we have
$$
\frac{1}{|G|} \sum_{g \in G} u_g A u_g^*
=\frac{\mathrm{Tr(A)}}{n} \mathrm{I} \ ?
$$
It is maybe a consequence of orthogonality relations ?

答:
Let $B = \sum_{g \in G} u_g A u_g^{-1}$. This is an intertwiner for the action of $G$, since for any fixed $x \in G$ we have

$$ u_x B = \sum_{g \in G} u_{xg} A u_{g}^{-1} = \sum_{h \in G} u_hAu_{x^{-1} h}^{-1} = \sum_{h \in G} u_h A u_h^{-1} u_x = B u_x,$$
where we have re-indexed the sum using the bijection $h = xg$. Since $B$ is an intertwiner of the irreducible representation $u$, it must be equal to some scalar times the identity: we have $B = \lambda I$ for some $\lambda \in \mathbb{C}$. We can find that scalar by taking the trace:

$$ \operatorname{tr}(B) = \sum_{g \in G} \operatorname{tr}(u_g A u_g^{-1}) = \sum_{g \in G} \operatorname{tr}(u_g^{-1} u_g A) = \sum_{g \in G} \operatorname{tr}(A) = |G| \operatorname{tr}(A).$$

On the other hand since $B = \lambda I$ we have $\operatorname{tr}(B) = n \lambda$, and conclude that $\lambda= \frac{|G|}{n} \operatorname{tr}(A)$, making $B = \frac{|G|}{n} \operatorname{tr}(A) I$. In your case the identity follows because when $u$ is unitary, $u_g^{-1} = u_g^*$ for all $g \in G$.


问:
Implication of existence of non-measurable set by Terence Tao (Analysis 2)
In Real Analysis 2 by Terence Tao, the author claims that:

Lemma 7.4.5 and Proposition 7.3.3, when combined,
imply that there exist non-measurable sets

With $m^*$: outer measure, $m$: Lebesgue measure:

Lemma 7.4.5 (Finite additivity)

If $(E_j)_{j \in J}$ are a finite collection of disjoint measurable sets and any set A (not necessarily measurable), we have:

$m^*(A \, \cap \bigcup \limits_{j \in J} E_j) = \sum \limits_{j \in J} m^*(A \, \cap E_j)$

Furthermore, we have:

$m(\bigcup \limits_{j \in J} E_j) = \sum \limits_{j \in J} m(E_j)$

Proposition 7.3.3 (Failure of finite additivity)

There exists a finite collection $(A_j)_{j \in J}$ of disjoint subsets of $\mathbb{R}$, such that

$m^*(\bigcup \limits_{j \in J} A_j) \neq \sum \limits_{j \in J} m^*(A_j)$

Could you please give me some explanations why this claim is true ? (or at least some intuitive explanations).

P/s: I quote the definition of Lebesgue measure $m$ by Terence Tao for further information:

Definition 7.4.1 (Lebesgue measurability)

Let $E$ be a subset of $\mathbb{R^n}$.
We say that $E$ is Lebesgue measurable, or measurable for short, iff we have the identity:

$m^*(A) = m^*(A \cap E) + m^*(A$ \ $E)$

for every subset $A$ of $\mathbb{R^n}$.

If $E$ is measurable, we define Lebesgue measure of $E$ to be $m(E) = m^*(E)$.

答: By proposition 7.3.3 one of the sets $A_j$ of the collection must be non-measurable,because if all the sets $A_j$ where measurable,then the set $B=\bigcup_{j \in J}A_j$ would be  measurable and  we would have the equality $m^*(B)=\sum_{j \in J}m^*(A_j)$ byLemma 7.4.5


问:
Help with this exponential random variable
The question is

(a) Probability of an external hard drive lasts less than 5 years? What is the probability of it lasting between 4 and 8 years?
This being an exponential random variable my thought process is this, if it is less than 5 years, that would mean that
 $α=1−e^{−.1qα}$
 $e^{−.1qα}=1−α $
 $−.1q_{α}=  log_{e}(1−α)$
 $q_{α}=\frac{−log_{e}(1−α)}{.1}$
 I am then stuck after getting here, I am unsure how to incorporate the less than 5 years and lasting between 4 and 8. And for example, say I am paranoid, I buy two hard drives, would the probability of them lasting less than 5 years be my answer to (a) multiplied by 2? Thank you in advance!

答:
By definition of the CDF,$ F(y=5\text{ years})=P(X\leq 5\text{ years} )=1-e^{-0.1\times 5}=1-e^{-0.5}=0.39$

Then, the probability of it lasting between $4$ and $8$ years is :
$$P(4\leq X\leq 8)=F(8)-F(4)=e^{-0.1\times 4}-e^{-0.1\times 8}=0.22 $$

For your last question, if you bought 2 hard drives, the probability that both of them last less than $5$ years is the square of the first result, not the probability multiplied by $2$, so it would be $0.15 $


问:
Convex closure (need for proof explanation)
I got stuck while reading a proof regarding convex sets (Corollary 2). I was supposed to use the first statement and Corollary 1 below, if necessary.

First statement(theorem):

Let $S$ be a convex set in $R^{n}$ with a nonempty interior. Let $x_{1} \in \operatorname{cl} S$ and $x_{2} \in$ int
S. Then $\lambda x_{1}+(1-\lambda) x_{2} \in$ int $S$ for each $\lambda \in(0,1)$.

Corollary 1:

Let $S$ be a convex set. Then int $S$ is convex.

Corollary 2:

Let $S$ be a convex set with a nonempty interior. Then cl $S$ is convex.

-> Proof by the Author:

Let $x_{1}, x_{2} \in \operatorname{cl} S .$ Pick $z \in$ int $S$ (by assumption, int $\left.S \neq \varnothing\right) .$ By the
theorem, $\lambda x_{2}+(1-\lambda) z \in$ int $S$ for each $\lambda \in(0,1) .$ Now fix $\mu \in(0,1) .$ By the
theorem, $\mu \mathbf{x}_{1}+(1-\mu)\left[\lambda \mathbf{x}_{2}+(1-\lambda) \mathbf{z}\right] \in$ int $S \subset S$ for each $\lambda \in(0,1) .$ If we take
the limit as $\lambda$ approaches $1,$ it follows that $\mu \mathrm{x}_{1}+(1-\mu) \mathrm{x}_{2} \in cl(S),$ and the proof is complete.

From the book: Nonlinear Programming (Theory and Algorithms)

Where am i stuck?

I just cannot understand the part where the author says:

"If we take the limit as $\lambda$ approaches $1,$ it follows that
$\mu \mathrm{x}_{1}+(1-\mu) \mathrm{x}_{2} \in cl(S)..."$

I can see that the limit equals $\mu \mathrm{x}_{1}+(1-\mu) \mathrm{x}_{2}$. However, two things are really bothering me:

I thought that $\lambda$ was a fixed number in $(0, 1)$ and because of that i should not be able to just "consider $\lambda$ as it approaches 1". Why can he do that?

I cannot see why $\mu \mathrm{x}_{1}+(1-\mu) \mathrm{x}_{2} \in cl(S)$. It has something to do with the set inclusions $int(S) \subset S \subset cl(S)$? Or does it have something to do with the fact that $S$ is convex? I also considered the possibility that the limit points are also in $S$ since $cl(S)$ is closed.

Can someone help me? I am really struggling.

Thanks in advance, Lucas

答:
(#1) The statement is true for every choice of $\lambda \in (0,1)$. That means it's true for $\lambda=0.9, \lambda=0.99, \lambda=0.999$, etc., and you can choose these $\lambda$ such that their limit is $1$.

(#2) You have a sequence of points $\mu \mathbf{x}_{1}+(1-\mu)\left[\lambda \mathbf{x}_{2}+(1-\lambda) \mathbf{z}\right] \in S$. Taking the limit as $\lambda \to 1$, you can see the points approach $\mu \mathbf{x}_{1}+(1-\mu)\mathbf{x_2}$. In other words, this is a convergent sequence of points contained in $S$; now apply your definition of closure.


问:
Proving there do not exist natural numbers $m$ and $n$ such that $7m^2=n^2$
I'm stuck on my proof of this concept and I could use some help on understanding what to do next.

Prove there do not exist natural numbers $m$ and $n$ such that $7m^2=n^2$

Proof will be by contradiction. So, assume there does exist natural numbers such that $7m^2=n^2$ is true.

Let $m=p_1^{r_1} p_2^{r_2 }…p_k^{r_k}$ and $n=p_1^{t_1} p_2^{t_2}…p_s^{t_s}$

Then $m^2=(p_1^{r_1} p_2^{r_2 }…p_k^{2r_k})^2=p_1^{2r_1} p_2^{2r_2 }…p_k^{2r_k}$ and $n^2=(p_1^{t_1} p_2^{t_2}…p_s^{t_s})^2=p_1^{2t_1} p_2^{2t_2}…p_s^{2t_s}$

Substitute this into $7m^2=n^2$ to obtain,

$7(p_1^{2r_1} p_2^{2r_2 }…p_k^{2r_k})=p_1^{2t_1} p_2^{2t_2}…p_s^{2t_s}$

The idea is that all values of m and n chosen to satisfy $7m^2=n^2$ are natural numbers so they therefore can be expressed as a product of primes.

This is where I'm stuck. How do I go about showing the contradiction?

Thanks for the help

答:
Let $m=p_1^{r_1} p_2^{r_2 }…p_k^{r_k}$ and $n=p_1^{t_1} p_2^{t_2}…p_s^{t_s}$

Your indexes here are the same implying that $n$ and $m$ have the same prime factors.

Better to write this as $m=p_{a_1}^{r_1} p_{a_2}^{r_2 }…p_{a_k}^{r_k}$ and $n=p_{b_1}^{t_1} p_{b_2}^{t_2 }…p_{b_s}^{t_s}$.

Substitute this into $7m^2=n^2$ to obtain, $7(p_1^{2r_1} p_2^{2r_2 }…p_k^{2r_k})=p_1^{2t_1} p_2^{2t_2}…p_s^{2t_s}$

Okay that means that there is a $7$ on the LHS so there must be a $7$ on the RHS.  So  one of the $p_{b_i}$ must be equal to $7$.  Without loss of generality let's assume that $p_{b_1} = 7$ then we have

$7m^2 = 7p_{a_1}^{2r_1} p_{2a_2}^{2r_2 }…p_{a_k}^{2r_k}=n^2 = 7^{2t_1} p_{2b_2}^{2t_2 }…p_{b_s}^{2t_s}$

Now $2t_1 \ne 1$ and $2t_1 > 1$ so that $7$ on the LHS can't be the only $7$ in the factorization of $7m^2$.  So one of the $p_{a_1}$ must be $7$.  Without loss of generality let's assume that $p_{b_1} = 7$.   then we have

$7m^2 = 7\cdot 7^{2r_1}  p_{2a_2}^{2r_2 }…p_{a_k}^{2r_k}=7^{2r_1 + 1} p_{2a_2}^{2r_2 }…p_{a_k}^{2r_k}=n^2 = 7^{2t_1} p_{2b_2}^{2t_2 }…p_{b_s}^{2t_s}$

But that means $2r_1 + 1 = 2t_1$.  But that's impossible as $2r_1 + 1$ is odd and $2t_1$ is even.


问:
Prove $\int_0^1 f(g(x)) \, dx \leq \int_0^1 f(x)\,dx + \int_0^1 g(x)\,dx$
$f$ and $g$ are continuous functions on $[0, 1]$. $f$ doesn't decrease on this segment. Values of both $f$ and $g$ are in range $[0, 1]$. Prove following inequality

$$\int_0^1 f(g(x)) \, dx \leq \int_0^1 f(x)\, dx + \int_0^1g(x) \,dx$$

I couldn't think of any great idea to show that this inequality is actually satisfied by all functions with mentioned properties.

Thank you for your hints and help in advance!

Body:
Let $x_0$ be such that :
$$\forall x \in [0,1], \quad f(x) - x \leq f(x_0) - x_0.$$

Then : $$\int_0^1 [f(g(x)) - g(x) ] \,dx \leq \int_0^1 [f(x_0) - x_0] \,dx = f(x_0) - x_0 $$

And : $$\int_0^1 f(x) \,dx \geq \int_{x_0}^{1} f(x) \,dx \geq (1 - x_0) f(x_0) \geq f(x_0) - x_0$$


问:
Set Inequality Proof Check
Looking at a problem on an intro. analysis practice exam:
Prove that

$$ U\setminus(A\setminus B)\ \;=\; (U\setminus A)\setminus B $$

I don't think this is true, here is the counterexample that I cooked up:
$$ U\;=\;\{1,2,3,\dots, 9, 10\}$$
$$ A\;=\;\{2,4,6,8\} $$
$$ B\;=\; \{2,4,5,7\} $$
Considering $u=2$, we have $u\in \{1,2,3,4,5,7,9,10\}=U\setminus(A\setminus B)$, yet $u\notin \{1,3,9,10\}=(U\setminus A)\setminus B$.

Just looking for a quick verification, seeing as the awkward scenario here is an undergraduate is claiming that a tenured professor is incorrect (maybe he did it on purpose?).

答:
You can prove it is not valid through the properties of the operators involved.

Here I assume that $U$ is the universe of discourse.

On one hand, we have that

\begin{align*}
U - (A - B) & = U - (A\cap B^{c})\\\\
& = U\cap(A\cap B^{c})^{c}\\\\
& = U\cap (A^{c}\cup B)\\\\
& = A^{c}\cup B
\end{align*}

On the other hand, we have that

\begin{align*}
(U - A) - B & = (U\cap A^{c}) - B\\\\
& = (U\cap A^{c})\cap B^{c}\\\\
& = U\cap (A^{c}\cap B^{c})\\\\
& = A^{c}\cap B^{c}
\end{align*}

In general, $A^{c}\cup B \neq A^{c}\cap B^{c}$.

Hopefully this helps!

Title: Set Inequality Proof Check