问:
Inequality with the Lambert's function $x^{{\operatorname{W}(2ex)}^{2x}}+(1-x)^{{\operatorname{W}(2e(1-x))}^{2(1-x)}}\leq 1$
Let $0<x<1$ then we have :

$$x^{{\operatorname{W}(2ex)}^{2x}}+(1-x)^{{\operatorname{W}(2e(1-x))}^{2(1-x)}}\leq 1$$

The equality case is $x=0.5$.

To show it I have tried to follow the lemma 7.1 and 7.2 of this paper by Vasile Cirtoaje.The problem is that the resulting expression is awful !

I have tried also Bernoulli's inequality with any effect because it's not sharp enough.

Update 18/12/2020 :

It's a another tried .We can build an approximation like this :

Let $0<\beta<x\leq 0.5$ then we have to determine the constants such that:

$$x^{{\operatorname{W}(2ex)}^{2x}}\leq \frac{1}{2}\operatorname{W}^{\alpha}(2ex)$$

We have numerically speaking $\frac{115}{100}<\alpha<\frac{125}{100}$

To reduces the gap I have tried to introduce a linear function :
$$x^{{\operatorname{W}(2ex)}^{2x}}\leq \frac{1}{2}\operatorname{W}^{\alpha}(2ex)+ax+b$$

But again it's not enough to work so we may consider a general polynomial like :

$$x^{{\operatorname{W}(2ex)}^{2x}}\leq \frac{1}{2}\operatorname{W}^{\alpha}(2ex)+\sum_{k=0}^{n}a_nx^n$$

Well it's a first step and in the future I shall tried to find the coefficients of this general polynomial .

Update 20/12/2020 :

We can reformulate the problem as :

Let $x,y>0$ such that $ye^y+xe^x=2e$ then we have :

$$\left(\frac{xe^x}{2e}\right)^{(x)^{\frac{xe^x}{e}}}+\left(\frac{ye^y}{2e}\right)^{(y)^{\frac{ye^y}{e}}}\leq 1$$

Where I use the inverse function of the Lambert's function .

Well using the form $f(x)=\left(\frac{xe^x}{2e}\right)^{(x)^{\frac{xe^x}{e}}}=g(x)^{h(x)}$ I can show that the function $f(x)$ is convex on $(0,W(2e))$ so (I have tried) we can use Slater's inequality to find an upper bound .Like this  it doesn't work . On the other hand we can use Karamata's inequality but I have not tried !

Well If we use Karamata's I have a strategy :

We have by Karamata's inequality and $0\leq\varepsilon_n'\leq\varepsilon_n<y<x$:

$$f(x)+f(y)\leq f(x+\varepsilon_n)+f(y-\varepsilon_n')$$

With $(y-\varepsilon_n')e^{y-\varepsilon_n'}+(x+\varepsilon_n)e^{x+\varepsilon_n}\geq 2e$

Now we want to repeat the process to get a series of inequalities of the kind :

$$f(x)+f(y)\leq f(x+\varepsilon_n)+f(y-\varepsilon_n')\leq f(x+\varepsilon_{n-1})+f(y-\varepsilon_{n-1}')< 1$$

But it's very complicated.

It doesn't work for all the value but I think we have the inequality $y> 0.5 \geq x$ :

$$p(x)=(1-x^{xe^{x-1}})^2+x^{xe^{x-1}} \frac{xe^{x-1}}{2} (2-x^{xe^{x-1}})-x^{xe^{x-1}} \frac{xe^{x-1}}{2} (1-x^{xe^{x-1}}) \ln\left(\frac{xe^{x-1}}{2}\right)$$
We have :
$$f(x)+f(y)\leq p(y)+2^{-\varepsilon}p^{1+\varepsilon}(x)< 1$$

With $0\leq \varepsilon \leq\frac{1}{10}$

Where I use the Lemma 7.2 of the paper above .

The last idea :

Using the majorization theorem :

Let $a\geq b>0$ and $c\geq d >0$ and $n$ a natural number large enough such that :

$$a\geq c$$

And :

$$\left(a\frac{n}{n+1}+c\frac{1}{n+1}\right)\left(b\frac{n}{n+1}+d\frac{1}{n+1}\right)\geq cd$$

Then we have :

$$a+b\geq c+d$$

Proof:It's a direct consequence of the Karamata's inequality.

We have another theorem :

Let $2>x,y>0$ ,$n$ a natural number large enough and $\varepsilon>0 $

If we have :

$$xy<1-\varepsilon $$
$$x+y<2-\varepsilon$$
then we have :

$$\ln\left(\frac{n}{n+1}+x\frac{1}{n+1}\right)+\ln\left(\frac{n}{n+1}+y\frac{1}{n+1}\right)\leq 0$$

Example :

Using the theorem of majorization we have ($x=0.4$):

$$(1-x)^{{\operatorname{W}(2e(1-x))}^{2(1-x)}}< 1-\operatorname{W}^{1.25}(2ex)0.5$$

And :

$$\left(\frac{1}{4000}x^{{\operatorname{W}(2ex)}^{2x}}+\frac{3999}{4000}\operatorname{W}^{1.25}(2ex)0.5\right)\left(\frac{1}{4000}(1-x)^{{\operatorname{W}(2e(1-x))}^{2(1-x)}}+\frac{3999}{4000}(1-\operatorname{W}^{1.25}(2ex)0.5)\right)< (1-\operatorname{W}^{1.25}(2ex)0.5)\operatorname{W}^{1.25}(2ex)0.5$$

Dividing both side by the RHS and using the second theorem remarking that :

$$\frac{x^{{\operatorname{W}(2ex)}^{2x}}(1-x)^{{\operatorname{W}(2e(1-x))}^{2(1-x)}}}{\operatorname{W}^{1.25}(2ex)0.5(1-\operatorname{W}^{1.25}(2ex)0.5)}<1-\varepsilon$$

And :

$$\frac{x^{{\operatorname{W}(2ex)}^{2x}}}{\operatorname{W}^{1.25}(2ex)0.5}+\frac{(1-x)^{{\operatorname{W}(2e(1-x))}^{2(1-x)}}}{(1-\operatorname{W}^{1.25}(2ex)0.5)}<2-\varepsilon\quad (I)$$

Now I think it's easier because we can take the logarithm and study the behavior of the function .

To prove the $(I)$ we can use the bound :

Let $0<x<\frac{1}{100}$ :

$$e^x<(1+x)^2-x$$

Obviously if we study separately the differents elements of the LHS .

Then to study $(I)$ we have a quite good approximation :

Let $0< x \leq \frac{1}{2}$ then we have :

$${\operatorname{W}(2ex)}^{2x}\geq (2x)^{\frac{915}{1000}\left(x\right)^{\left(\frac{87}{100}\right)}}$$

In fact we have the following refinement on $(0,0.5]$ :

$$x^{{\operatorname{W}(2ex)}^{2x}}+(1-x)^{{\operatorname{W}(2e(1-x))}^{2(1-x)}}\leq x^{(2x)^{\frac{915}{1000}\left(x\right)^{\left(\frac{87}{100}\right)}}}+ (1-x)^{(2(1-x))^{\frac{915}{1000}\left((1-x)\right)^{\left(\frac{87}{100}\right)}}}\leq 1$$

Remarks : The method using the majorization theorem have two advantages . We need to choose two values of the same order with respect to the values in the LHS . One can be inferior (and the other necessary superior).On the other hand the bound with the exponential ,his accuracy depends of the initial approximation in $(I)$ . Finally if we split in two the LHS in $(I)$ and if for a one we prove a stronger result then the other element is slighlty easier to show .

I build an approximation on $(0,1)$ wich have the form :

$$x^{(2x)^{\frac{915}{1000}\left(x\right)^{\left(\frac{87}{100}\right)}}}\simeq \left(\left(2^{(2x)^{x^{1.25}}} \frac{x}{2}\right)^{0.5}0.5^{0.5}*0.5^{{(2 (1-x))}^{x^{-0.25}}}\right)^{0.5}\quad (S)$$

You can play with the coefficients $-0.25$ and $1.25$ wich are not the best (make me a comment if you have better please:-))

We can slightly improve $(S)$ in using the logarithm we have on $[0.5,1)$:

$$x^{(2x)^{\frac{915}{1000}\left(x\right)^{\left(\frac{87}{100}\right)}}}\simeq \left(\left(2^{(2x)^{x^{3}}} \frac{x}{2}\right)^{0.5}0.5^{0.5}*0.5^{{(2 (1-x))}^{x^{-0.2}}}\right)^{0.5}-0.5\ln\left(\left(\left(2^{(2x)^{x^{3}}} \frac{x}{2}\right)^{0.5}0.5^{0.5}*0.5^{{(2 (1-x))}^{x^{-0.2}}}\right)^{0.5}\right)+0.5\ln\left(x^{(2x)^{\frac{915}{1000}\left(x\right)^{\left(\frac{87}{100}\right)}}}\right)\quad (S')$$

We can replace the coefficient $\frac{915}{1000}$ by $\frac{912}{1000}$,and $3$ by $3.5$ and finally $-0.2$ by $-0.19$ and I think it's the same order so we can apply the majorization theorem .Ouf!

Any idea to solve it ?

Thanks

Body:
Some thoughts

Let me show how to use bounds for the case $0 < x < \frac{1}{10}$.

Denote $F = W(2\mathrm{e}x)^{2x}$ and $G = W(2\mathrm{e}(1-x))^{2(1-x)}$.
We need to prove that $x^F + (1-x)^G \le 1$.

Fact 1: If $u > 0$ and $0 \le v \le 1$, then $u^v \ge \frac{u}{u + v - uv}$.

(Note: By Bernoulli inequality, $(\frac{1}{u})^v=(1+\frac{1}{u}-1)^v\leq 1 + (\frac{1}{u}-1)v = \frac{u + v - uv}{u}$.)

Fact 2: $0 \le 5 - 5F \le 1$ for all $x\in (0, 1/2]$.

Fact 3: $1 \le G < 2$ for all $x\in (0, 1/2]$.

Fact 4: $W(y) \ge \frac{y}{y + 1}$ for all $y\ge 0$.

(Hint: Use $W(y)\mathrm{e}^{W(y)} = y$ for all $y\ge 0$ and that $u \mapsto u\mathrm{e}^u$ is strictly increasing on $(0, \infty)$.)

Fact 5: $F \ge \left(\frac{2\mathrm{e}x}{1 + 2\mathrm{e}x}\right)^{2x}$ for all $x > 0$. (Use Fact 4.)

Fact 6: $G = W(2\mathrm{e}(1-x))^{1 - 2x} W(2\mathrm{e}(1-x))
\ge \frac{W(2\mathrm{e}(1-x))^2}{2x W(2\mathrm{e}(1-x)) + 1 - 2x}$ for all $x \in (0, 1/2]$.

(Hint: Use Fact 1, $u = W(2\mathrm{e}(1-x))$, $v = 1-2x$.)

Fact 7: $W(2\mathrm{e}(1-x)) \ge \frac{48}{35} - \frac{3}{5}x$ for all $x$ in $(0, 1/10)$.

Fact 8: $G \ge \frac{9(16-7x)^2}{-1470x^2+910x+1225}$ for all $x$ in $(0, 1/10)$.
(Use Facts 6-7.)

Now, by Facts 1-2, we have
$$x^F = \frac{x}{x^{1-F}} = \frac{x}{\sqrt[5]{x}^{5 - 5F} } \le x + (x^{4/5} - x)(5 - 5F).$$
(Note: $u = \sqrt[5]{x}, v = 5-5F$.)

By Facts 1, 3, we have
$$(1-x)^G = \frac{(1-x)^2}{(1-x)^{2-G}} \le (1-x)^2 + x(1-x)(2-G).$$
(Note: $u = 1-x, v = 2-G$.)

It suffices to prove that
$$ x + (x^{4/5} - x)(5 - 5F) + (1-x)^2 + x(1-x)(2-G) \le 1$$
or
$$5(x^{4/5} - x)(1 - F) \le x(1-x)(G-1).$$

By Facts 5, 8, it suffices to prove that
$$5(x^{4/5} - x)\left(1 - \left(\frac{2\mathrm{e}x}{1 + 2\mathrm{e}x}\right)^{2x}\right) \le x(1-x)\left(\frac{9(16-7x)^2}{-1470x^2+910x+1225}-1\right).$$

Omitted.


问:
clarifying understanding of expectation of the absolute value of a random variable
Given pdf f(x), is $\mathbb{E}(|X|)=\sum_{-\infty}^{\infty}|x|f(|x|)$? why would it not be $\sum_{-\infty}^{\infty}xf(|x|)$ or $\sum_{-\infty}^{\infty}|x||f(x)|$?

答:
In general, if $g$ is a function and $X$ is a (real) random variable with pdf $f$
$$
E[g(X)]=\int_{\mathbb{R}}g(x)f(x)\,dx.
$$
In your case, $g(\cdot)=|\cdot|$.

ContentLicense: CC BY-SA 4.0


问:
Let $f(x) = \ln(\sqrt{1+x^2}-x)\space$ for every $x\in{R}.\space$ then $f$ is well defined and odd.
I need to prove or disprove this statement :

Let $f(x) = \ln(\sqrt{1+x^2}-x)\space$ for every $x\in{R}.\space$ then $f$ is well defined and ood.

I need to say that this is my first time working with $\ln()$ functions so sorry if I got some misunderstandings :)

It's easy to see that the function is not odd since for:

$x=1, f(1) = e^{\sqrt{2}-1} = \frac{e^{\sqrt{2}}}{e}$

$x=-1, f(-1) = e^{\sqrt{2}+1} = {e^{\sqrt{2}}}{e}$

So $f(-x)\neq -f(x)$

but I'm not sure about "well defined". I don't see why the function is not well defined since for every $x\in R$ it's valid.

I would love to hear your thoughts!.

答:
The function is no doubt well-defined for all real $x$ since $\sqrt{x^2+1}>x$ holds for all real $x$. It is odd too, since
$$f(-x)=\ln(\sqrt{x^2+1}+x)=\ln\frac{(x^2+1)-x^2}{\sqrt{x^2+1}-x}=\ln\frac1{\sqrt{x^2+1}-x}$$
$$=-\ln(\sqrt{x^2+1}-x)=-f(x)$$
In fact $f(x)=-\sinh^{-1}x$.


问:
Determine if a linear transformation exists between vectors
If I am given example vectors $ \mathbf a, \mathbf b, \mathbf c $ and $ \mathbf r, \mathbf s, \mathbf t $, how can I find if there exists a linear transformation such that

$$ \mathbf T \mathbf a = \mathbf r $$
$$ \mathbf T \mathbf b = \mathbf s $$
$$ \mathbf T \mathbf c = \mathbf t $$

The vectors can be in any dimension space $ \Bbb R ^n $ where $ n = 1,2,3,4... $

Is there a general formula/method to determine this? $ \mathbf a, \mathbf b, \mathbf c, \mathbf r, \mathbf s, \mathbf t $ is not necessarily different. For example, if $ \mathbf a = \mathbf b $ and $ \mathbf r \neq \mathbf s $, we can instantly know that $ \mathbf T $ does not exist.

Is there a general method or series of steps to determine this?

答:
Such a $T$ exists if and only if the following 3 statements are true:

If $a=0$, then $r=0$.

If $b = ka$ for some constant $k$, then $s=kr$.

If $c = ma+nb$ for some constants $m$ and $n$, then $t=mr+ns$.


问:
How do I calculate the equation of a circle given 3 complex numbers?
Given three complex values (for example, $2i, 4, i+3$), how would you calculate the equation of the circle that contains those three points?  I know it has something to do with the cross ratio of the three points and $z$ and the fact that the cross ratio is a real number, but I don't know what to do with that information.</p>

答:
"Might as well flesh out" @Dinesh's comment.

This solution uses algebraic manipulations and consideration of the geometry of the figures. [A more concise solution might leverage vector formality.]

The solution is the equation in the form (x-h)^2+(y-k)^2=r^2, where we give the values of the 3 parameters, h, k, and r.

First, we convert to Cartezian coordinates:

```<code>2i -> (0,2)
4 -> (4,0)
i+3 -> (3,1)
</code>```

1 -- To find the center of the circle (h,k), we need the equations of two lines that are perpendicular bisectors of any two of the sides of the implied triangle above. The intersection of these lines is the center of the circle.

2 -- A perpendicular bisector will have a slope, M, of -1/m, where m is the slope of the corresponding side of the triangle. We get the slope of a side, m, easily from the two given points for that side. We also can easily derive a point, P, on the perpendicular bisector: the midpoint of the side of the triangle. With a point and slope we essentially have the line equations we need.

3 -- To get the length of the radius, r, we compute the distance between the center of the circle and any of the points on the circle.

The perpendicular bisectors of the sides of a triangle intersect in the center of the circumscribing circle of that triangle. The proof is easy, as we note that every point along one of these perpendicular bisectors is equidistance from each of the two points that form the ends of the corresponding triangle side. Thus, the center of this circle (being the intersection of these perpendicular bisectors) has the property that it is equidistant from each of the 3 points, ie, it is a circumscribing circle. Note that given any 3 points, there is a single triangle implied by them (so we can use the perpendicular bisector argument to conclude that a single circle exists).

Now, for key calculations:

2:

```<code>m_1 = (0-1)/(4-3) = -1
m_2 = (2-1)/(0-3) = - 1/3
M_1 = -1/(-1) = 1
M_2 = -1/(-1/3) = 3
P_1 = ( (4+3)/2 , (0+1)/2 ) = (7/2,1/2)
P_2 = ( (0+3)/2 , (2+1)/2 ) = (3/2,3/2)
</code>```

1:

EQN_1

```<code>P_1y = (M_1)(P_1x) + B_1
1/2 = (1)(7/2) + B_1
B_1 = -3

Y = X - 3
</code>```

EQN_2

```<code>P_2y = (M_2)(P_2x) + B_2
3/2 = (3)(3/2) + B_2
B_2 = -3

Y = 3X - 3
</code>```

Intersection

```<code>X - 3 = 3X - 3
X = 0

Y = (0) - 3
Y = -3
</code>```

Thus, h = 0 and k = -3

3:

We can calculate distance from (h,k) to (0,2)

```<code>r = sqrt [ (0 - 0)^2 + (-3 - 2)^2 ]
r = 5
</code>```

More specifically, the equation of the circle is:

```<code>(x)^2 + (y + 3)^2 = 25
</code>```

ContentLicense: CC BY-SA 3.0

Body:
"Might as well flesh out" @Dinesh's comment.

This solution uses algebraic manipulations and consideration of the geometry of the figures. [A more concise solution might leverage vector formality.]

The solution is the equation in the form (x-h)^2+(y-k)^2=r^2, where we give the values of the 3 parameters, h, k, and r.

First, we convert to Cartezian coordinates:

2i -> (0,2)
4 -> (4,0)
i+3 -> (3,1)

1 -- To find the center of the circle (h,k), we need the equations of two lines that are perpendicular bisectors of any two of the sides of the implied triangle above. The intersection of these lines is the center of the circle.

2 -- A perpendicular bisector will have a slope, M, of -1/m, where m is the slope of the corresponding side of the triangle. We get the slope of a side, m, easily from the two given points for that side. We also can easily derive a point, P, on the perpendicular bisector: the midpoint of the side of the triangle. With a point and slope we essentially have the line equations we need.

3 -- To get the length of the radius, r, we compute the distance between the center of the circle and any of the points on the circle.

The perpendicular bisectors of the sides of a triangle intersect in the center of the circumscribing circle of that triangle. The proof is easy, as we note that every point along one of these perpendicular bisectors is equidistance from each of the two points that form the ends of the corresponding triangle side. Thus, the center of this circle (being the intersection of these perpendicular bisectors) has the property that it is equidistant from each of the 3 points, ie, it is a circumscribing circle. Note that given any 3 points, there is a single triangle implied by them (so we can use the perpendicular bisector argument to conclude that a single circle exists).

Now, for key calculations:

2:

m_1 = (0-1)/(4-3) = -1
m_2 = (2-1)/(0-3) = - 1/3
M_1 = -1/(-1) = 1
M_2 = -1/(-1/3) = 3
P_1 = ( (4+3)/2 , (0+1)/2 ) = (7/2,1/2)
P_2 = ( (0+3)/2 , (2+1)/2 ) = (3/2,3/2)

1:

EQN_1

P_1y = (M_1)(P_1x) + B_1
1/2 = (1)(7/2) + B_1
B_1 = -3

Y = X - 3

EQN_2

P_2y = (M_2)(P_2x) + B_2
3/2 = (3)(3/2) + B_2
B_2 = -3

Y = 3X - 3

Intersection

X - 3 = 3X - 3
X = 0

Y = (0) - 3
Y = -3

Thus, h = 0 and k = -3

3:

We can calculate distance from (h,k) to (0,2)

r = sqrt [ (0 - 0)^2 + (-3 - 2)^2 ]
r = 5

More specifically, the equation of the circle is:

(x)^2 + (y + 3)^2 = 25


问:
How to show that $x-1$ and $x^2+x+1$ are irreducible over $\mathbb R[x]?$
How to show that $x-1$ and $x^2+x+1$ are irreducible over $\mathbb R[x]?$

I can see that $x - 1$ and $x^2 + x + 1$ are nonzero and non-units (for any field $F$ the set of all units of $F[x]$ = $F-\{0\}$)?

Added: Actually I would like to show $f(x)|(x-1)(x^2+x+1)~\text{where }\deg f(x)\ge 1~\implies f(x)=(x-1)(x^2+x+1)~or~(x-1)~or~(x^2+x+1).$ Does showing $(x-1)~and~(x^2+x+1)$ irreducible help me anyhow?

答: 1 is the only root of $X-1$, and it is not a root of $X^2+X+1$, so the GCD of both polynomials is 1. That's all there is to say :) </p>


问:
Can I have a hint on this combinatorics question?
Six faces of identical cubes are painted 6 distinct colors. How many different cubes can be formed?

The answer is surprisingly small but that may be because I don't know how to express the different ways properly. I just incorrectly did $6!$ because if there are 6 ways to color the first face, there are now 5 colors left so 5 ways to color the second face. And so on. Why is this wrong?

Perhaps I need to make use of "identical" and "distinct" from the question. If the cubes are identical, I must divide the answer by something, perhaps $6!$ for the 6 identical faces? And there should be some "choosing" happening but I can see where this fits in but it must be for the "distinct" part. Choose colors as in $6C1, 5C1 ...etc$ but this is the same as $6!$

Please help

答: Two cubes that differ by a rotation are considered the same. There are $24$ ways to orient the cube, all of which are distinct because of the use of distinct colours, so $6!$ must be divided by $24$ to get the correct answer of $30$.


问:
Expectation of two cards
I have a question and I will be happy for help.
there is a normal deck of cards (ace = 1...king=13, 52 cards, 4 cards for each value)
someone takes out without return two cards.
what is the expectation of the sum of the cards?
I tried to seperate it to odd sums and even sums but it is very difficult and I think there is an easier way.
does someone have an Idea how to solve it?

答:
Suppose $X$ is random number which is drawn from the full deck, and $Y$ is random number drawn at the second step. We want to find $\mathbb{E}[X+Y] = \mathbb{E}[X] + \mathbb{E}[Y]$. Clearly, $\mathbb{E}[X] = \frac{13 \cdot 14}{2} \cdot \frac{4}{52} = 7$. It remains to find $\mathbb{E}[Y]$. Let's use the law of total expectations $$\mathbb{E}[Y] = \sum_{x=1}^{13} \mathbb{E}[Y|X=x]P(X=x)$$
$$\mathbb{E}[Y|X=x] = 1\cdot \frac{4}{51}+2\cdot \frac{4}{51}+\dots+(x-1)\cdot\frac{4}{51}+x \cdot \frac{3}{51}+(x+1)\cdot\frac{4}{51}+\dots+13\cdot\frac{4}{51} = \frac{(x-1)x}{2}\frac{4}{51}+\frac{3x}{51}+\left( \frac{(13-x)(13-x+1)}{2} +x(13-x)\right)\frac{4}{51} = \frac{364-x}{51}$$
So, $$\mathbb{E}[Y] = \sum_{x=1}^{13} \frac{364-x}{51} \frac{4}{52} = 7$$
Thus, $\mathbb{E}[X+Y] = 7+7 = 14$

Title: Expectation of two cards


问:
Cannot find limit using epsilon delta definition
Prove
$$\lim\limits_{x\to 2} x^3 = 8$$
using epsilon delta definition.

I try as below.

Let $\varepsilon>0$. We choose $\delta>0$.

Consider that
\begin{align}
\vert x^3-8\vert &= \vert (x-2) (x^2+2x+4)\vert\\
&=\vert (x-2) \vert\vert(x^2+2x+4) \vert \\
&=\vert (x-2) \vert\vert(x-2)^2+6x \vert .
\end{align}

Now I don't know how to continue this answer. I confused with $6x$.

Anyone can help me?

EDIT: I have tried as below as JC12's answer.

Let $\vert x-2\vert <1$, then $\vert x\vert -2< \vert x-2\vert <1$
then we have $$\vert x\vert -2<1 \iff \vert x\vert<3.$$

Now, $$ \vert(x-2)^2+6x \vert < \vert(3-2)^2+6\cdot 3 \vert =19. $$

Thus, \begin{align} \vert x^3-8\vert&= \vert (x-2)
 \vert\vert(x-2)^2+6x \vert < 19 \vert (x-2) \vert. \end{align}

Now choose $\delta=\min(1,\frac{\varepsilon}{19})$. We have
\begin{align} \vert x^3-8\vert < 19 \vert (x-2) \vert< 19
 \frac{\varepsilon}{19} = \varepsilon. \end{align}

So, we can conclude $\lim\limits_{x\to 2} x^3 =8$.

答:
Proposition: Let $D\subseteq\mathbb{R}$. Let $f:D\to\mathbb{R}$ and $g:D\to\mathbb{R}$ be continuous. Then $fg:D\to\mathbb{R}$ defined as $(fg)(x):=f(x)g(x)$ is continuous.

Proof: Fix $x_0\in D$. Then $f$ is continuous at $x_0$. Consider a particular vaule $\varepsilon_0>0$. Then there exists a $\delta_1>0$ such that if $x\in D$ satisfies $|x-x_0|<\delta_1$, then $$||f(x)|-|f(x_0)||\leq|f(x)-f(x_0)|<\varepsilon_0$$ which implies $|f(x)|<|f(x_0)|+\varepsilon_0$ for $x\in D$ with $|x-x_0|<\delta_1$.

(This part of the proof actually showed that continuous functions are locally bounded and can also be modified to show that if $f$ is continuous, then $|f|$ is continuous.)

Now, let $\varepsilon>0$.

Since $f$ is continuous at $x_0$, then there exists a $\delta_2>0$ such that if $|x-x_0|<\delta_2$ then $$|f(x)-f(x_0)|<\frac{\varepsilon}{2(|g(x_0)|+\varepsilon_0)}$$

Since $g$ is continuous at $x_0$, then there exists a $\delta_3>0$ such that if $|x-x_0|<\delta_3$ then $$|g(x)-g(x_0)|<\frac{\varepsilon}{2(|f(x_0)|+\varepsilon_0)}$$

Take $\delta=\min\{\delta_1,\delta_2,\delta_3\}$. Then, if $x\in D$ satisfies $|x-x_0|<\delta$, we have $$\begin{align*}|f(x)g(x)-f(x_0)g(x_0)|&=|f(x)g(x)-f(x)g(x_0)+f(x)g(x_0)-f(x_0)g(x_0)|\\&\leq |f(x)||g(x)-g(x_0)|+|g(x_0)||f(x)-f(x_0)|\\&<|f(x)|\cdot \frac{\varepsilon}{2(|f(x_0)|+\varepsilon_0)}+|g(x_0)|\cdot\frac{\varepsilon}{2(|g(x_0)|+\varepsilon_0)}\\&<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}\\&=\varepsilon \end{align*}$$

This implies that $fg$ is continuous on $D$.

Now, in your case, we set $D=\mathbb{R}$ and $f(x)=g(x)=x$. It is trivial to show that $f$ is continuous. Then we obtain $f(x)f(x)=x^2$ is continuous so $f(x)f(x)f(x)=x^3$ is continuous, which implies that $$\lim_{x\to 2}x^3=2^3=8 $$

From here it can actually be inductively shown that for all $n\in\mathbb{N}$, $f_n:\mathbb{R}\to\mathbb{R}$ defined as $f_n(x)=x^n$ is continuous.


问:
Find a rotation matrix with two constraints: Aligns two vectors AND a third vector is perpendicular to a fourth vector when transformed
I have four vectors $ d,w,c,n \in \mathscr{R}^3$.  I want to find a rotation matrix $R$ that satisfies these constraints: $w$ is aligned with $d$ after rotation  AND $c$ is perpendicular to $n$ after rotation. You can imagine it in two steps: after a first rotation that aligns $w$ with $d$, make a second rotation around $d$ so that $c$ becomes perpendicular to $n$. In a particular problem I am working on $d=[1,1,1]^T$ and $n=[0,0,1]^T$.

I believe the constraints can be expressed with these two equations

$$ d \times Rw=0$$
$$ n^T  Rc=0$$

In addition, we must impose a third condition for $R$ being a rotation matrix
$$ RR^T=I$$

Is there a nice way to solve for R using matrix algebra? Or how to formulate the two steps above?

I tried to play with Kronecker products to form a vectorized version of $R$, but did not reach a solution. With $S=skew(d)$ being the skew vector for cross product:
$$d \times Rw=SRw=0 \rightarrow (w^T \oplus S)vec(R)=0$$
$$n^T  Rc=0 \rightarrow(c^T \oplus n^T)vec(R)=0$$
but could not impose the $RR^T=I$

答:
I think this will not work unless the angle between $n$ and $w$ equals that between $n$ and $d$, which may not be true.

I think $R = R_1R_2$ works where:

$R_1$ is a rotation around $c\times n$ by an angle of $90^o + \cos^{-1} (\frac{c\cdot n}{|c||n|})$

$R_2$ is a rotation around $v\times u$ by an angle of $\cos^{-1} (\frac{u\cdot v}{|u||v|})$, where $v = (c' \times w) \times c'$ and $u
> = (c' \times d) \times c'$, where $c' = R_1c$.

The first step maps $c$ to $c'$ which is normal to n. the second step
aligns $w$ with $d$ by aligning $v$ with $u$. Intuitively, let $S$ be
the sphere with $v\times u$ (which is parallel to $c'$) at the north
pole, and let $P$ be the plane of its equator. $u$ is obtained by
sliding $d$ along $d$'s longitude until it falls on $P$, and $v$ is
similarly obtained from $w$.


问:
Proof With Cyclic Quadrilateral and Circle
Let $ABCD$ be a cyclic quadrilateral. Let $P$ be the intersection of $\overline{AD}$ and $\overline{BC}$, and let $Q$ be the intersection of $\overline{AB}$ and $\overline{CD}$. Prove that the angle bisectors of $\angle DPC$ and $\angle AQD$ are perpendicular.

All I know about cyclic quadrilaterals is that opposite angles add up to $180^\circ.$  How can I use that to finish my proof?  Thanks in advance for answering!

答: Here is another, simpler solution. Apologies for the orientation of the image.


问:
Simple geometry/trigonometry question
How to find the X coordinate of the red point if i know it's Y coordinate and the angle? Let's say the Y is 40 and the angle is 30 degrees:

答:
Suppose that the angle is between the $x$-axis and the line that goes through the origin is $\theta$. Using trigonometry, we have $$\tan\theta= \frac{y}{x}.$$

What you have is the $y$-coordinate and the value of $\theta$. So you put the values in

$$\tan (30^{\circ})= \frac{40}{x}.$$

Rearranging, we have $$x = \frac{40}{\tan (30^{\circ})}.$$

Title: Simple geometry/trigonometry question


问:
Completion of a surface along a curve.
Let $R$ be the ring $\mathbb{Z}[x,y,z]/(x(y^2-4z)-y^2+3z)$. I am trying to prove that the completion of this ring at the ideal $(x)$ is given by $\mathbb{Z}[[x]][y,z]/(x(y^2-4z)-y^2+3z)$. I am not really sure how one would go about do this or if this is even true. Any suggestions are appreciated. Thanks in advance.

From the inverse limit definition of the completion, we can get a map from $\mathbb{Z}[[x]][y,z]/(x(y^2-4z)-y^2+3z)$ to the completion. But essentially I am not sure how to show this if this is an isomorphism.

答: Let $I$ be an ideal of $\mathbb{Z}[x,y,z]$ and let $\wedge_{x}$ denote the $x$-adic completion. We have \begin{align} &(\mathbb{Z}[x,y,z]/I)^{\wedge_{x}} \stackrel{1}{\simeq} (\mathbb{Z}[x,y,z])^{\wedge_{x}} \otimes_{\mathbb{Z}[x,y,z]} \mathbb{Z}[x,y,z]/I \simeq \mathbb{Z}[y,z][[x]] \otimes_{\mathbb{Z}[x,y,z]} \mathbb{Z}[x,y,z]/I \simeq \mathbb{Z}[y,z][[x]]/I \mathbb{Z}[y,z][[x]] \end{align} where the isomorphism 1 is by part (3) of this. When mixing polynomials with power series, the order matters -- note that the natural map $\mathbb{Z}[[x]][y,z] \to \mathbb{Z}[y,z][[x]]$ is (injective but) not surjective. Note also that the leading coefficient of the polynomial (with respect to $y$) is $x-1$ which is a unit in $\mathbb{Z}[[x]]$ so the desired completion is a finite free $\mathbb{Z}[z][[x]]$-module (of rank $2$).

Title: Completion of a surface along a curve.


答:
This is perhaps a more concrete way to arrive at the answer.  Say we will color the faces with the colors, $R,O,Y,G,B,V$.  We have to color some face $R$.  Place the cube on the table, with that face down.  Now we have to color some face $O$.

We can paint the top face, or one of the lateral faces.  Suppose first that we paint the top face.  Now we have to paint one of the lateral faces $Y$.  Whichever face we choose, we can rotate the cube so that that face is facing us.  Now the cube is fixed in place.  $R$ on bottom, $O$ on top, $Y$ facing us.  We can't rotate it without changing the arrangement.   There are $3!$ ways to paint the remaining $3$ faces.

If instead, we paint one of the lateral faces $O$, we can rotate the cube so $O$ is facing us, and again we find it is fixed in place; $R$ on bottom, $O$ facing us.  We have $4!$ ways to paint the remaining $4$ faces.

You may be interpreting the question a bit too literally.  There are indeed $6!$ orders in which we can choose to color the faces of the cube, but in many cases, the results are indistinguishable.  The question is asking for the number of distinct cubes that can be made by coloring each face with one of $6$ given colors.

Title: Can I have a hint on this combinatorics question?

Body:
Six faces of identical cubes are painted 6 distinct colors. How many different cubes can be formed?

The answer is surprisingly small but that may be because I don't know how to express the different ways properly. I just incorrectly did $6!$ because if there are 6 ways to color the first face, there are now 5 colors left so 5 ways to color the second face. And so on. Why is this wrong?

Perhaps I need to make use of "identical" and "distinct" from the question. If the cubes are identical, I must divide the answer by something, perhaps $6!$ for the 6 identical faces? And there should be some "choosing" happening but I can see where this fits in but it must be for the "distinct" part. Choose colors as in $6C1, 5C1 ...etc$ but this is the same as $6!$

Please help


问:
Prove that, if L is regular, $P(L) = \{ w \in L\text{ } |\text{for any prefix } w' \text{ of w, } w' \in L\}$ is regular
I have found one specific problem about regular languages which I am not sure how to solve:

Prove that the following language $$P(L) = \{ w \in L\text{ } |\text{for any prefix } w' \text{ of w, } w' \in L\}$$ is regular, given that L is regular too.

I thought of using closure properties, but I can't think of any other language to use for that. Any tips?

答:
Your idea of using closure properties was the right one. First observe that $P(L)$ is the set of words in $L$ with no prefix in $L^c$ (the complement of $L$). Now, the set of words with no prefix in $L^c$ is the complement of the set $L^cA^*$ of words having a prefix in $L^c$. Altogether, you get
$$
  P(L) = L \cap (L^cA^*)^c
$$
Since regular languages are closed under intersection, complement and product, $P(L)$ is regular. This solution is more powerful that a proof using automata. It shows for instance, that if $L$ is star-free, then so is $P(L)$.


答:
If you label the cube counterclockwise $1$ through $4$ starting from the yellow face, the top face $5$, the bottom $6$, then if we are given the colors: yellow, blue, orange, clear, red and green, pretending the front most face is clear/see-through, a coloring of the cube would be a permutation: $\{1,2,3,4,5,6\}\rightarrow$ {yellow, blue, orange, clear, red, green}.

There are $6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1\cdot = 6!$ ways to color this cube. But now every coloring can be described the same way $6\cdot 4 =24$ 'different ways'. Looking at our cube, if we rotate it front, back, left, right, enough times we can have $6$ distinct bottom faces, the top face is automatically determined once the bottom face is. Next given a bottom face, we can rotate the cube $90^{\circ}$ counterclockwise 4 times and still have the same coloring.


答:
Let the four vectors be made unit, and let $c$ and $w$ have angle $\phi$ between them. Since orthogonal matrices preserve angles, $Rc$ must lie on the cone with axis $d=Rw$ and angle $\phi$, after rotation.

On the other hand, the vectors perpendicular to $n$ form a plane. So $c$ needs to be mapped to the intersection of the cone with this plane. Whether that is possible depends on the angle between $n$ and $d$; call it $\eta$.

If $\eta<\frac{\pi}{2}-\phi$ then the intersection is $0$, which is inadmissible. There is no solution.

If $\eta=\frac{\pi}{2}-\phi$ then the intersection is a line, so there is only one point where $Rc$ can be mapped to.

If $\eta>\frac{\pi}{2}-\phi$ then the intersection is a pair of intersecting lines, so there are two points where $Rc$ can be mapped to. To find these points, we need to solve for $\tilde{c}=Rc$, $$\tilde{c}\cdot d=\cos\phi,\qquad \tilde{c}\cdot n=0,\qquad \tilde{c}\cdot \tilde{c}=1$$
Note that $\tilde{c}\times(n\times d)=(\cos\phi) n$. Solving for $$\tilde{c}=\alpha n+\beta d+\gamma n\times d$$ gives $$\alpha=-\cos\phi\cos\eta/\sin^2\eta,\\ \beta=\cos\phi/\sin^2\eta,\\\gamma=\pm\sqrt{1-\cos^2\phi/\sin^2\eta}$$

The problem is then to find a rotation that maps $w,c$ to $d,\tilde{c}$.

The rotation that maps $(1,0,0)$, $(\cos\phi,\sin\phi,0)$ to $d,\tilde{c}$ is $$R_1=[d,\frac{\tilde{c}-(\cos\phi)d}{\sin\phi},\frac{d\times\tilde{c}}{\sin\phi}]$$ and similarly, $$R_2=[w,\frac{c-(\cos\phi) w}{\sin\phi},\frac{w\times c}{\sin\phi}]$$ so the final rotation is $$R=R_1R_2^\top$$

Title: Find a rotation matrix with two constraints: Aligns two vectors AND a third vector is perpendicular to a fourth vector when transformed

Body:
I have four vectors $ d,w,c,n \in \mathscr{R}^3$.  I want to find a rotation matrix $R$ that satisfies these constraints: $w$ is aligned with $d$ after rotation  AND $c$ is perpendicular to $n$ after rotation. You can imagine it in two steps: after a first rotation that aligns $w$ with $d$, make a second rotation around $d$ so that $c$ becomes perpendicular to $n$. In a particular problem I am working on $d=[1,1,1]^T$ and $n=[0,0,1]^T$.

I believe the constraints can be expressed with these two equations

$$ d \times Rw=0$$
$$ n^T  Rc=0$$

In addition, we must impose a third condition for $R$ being a rotation matrix
$$ RR^T=I$$

Is there a nice way to solve for R using matrix algebra? Or how to formulate the two steps above?

I tried to play with Kronecker products to form a vectorized version of $R$, but did not reach a solution. With $S=skew(d)$ being the skew vector for cross product:
$$d \times Rw=SRw=0 \rightarrow (w^T \oplus S)vec(R)=0$$
$$n^T  Rc=0 \rightarrow(c^T \oplus n^T)vec(R)=0$$
but could not impose the $RR^T=I$


问:
Proving $\sum_{cyc}\sqrt[3]{\frac{1}{a}+\frac{2}{bc}+a+2b+c}\leq\frac{6}{abc}$ for positive values such that $ab+bc+ca=3$
This problem is from my teacher.

Know that $a,b,c>0$, $ab+bc+ca=3$.

Prove that: $$\sum_{cyc}\sqrt[3]{\frac{1}{a}+\frac{2}{bc}+a+2b+c}\leq\frac{6}{abc}$$

I tried  to change the number '$6$' into $2(ab+bc+ca)$.

Then get
$$\sum_{cyc}\sqrt[3]{\frac{1}{a}+\frac{2}{bc}+a+2b+c}\leq\frac{2ab+2bc+2ca}{abc}=\frac{2}{c}+\frac{2}{b}+\frac{2}{a}$$

but i have no idea what to do next.

答:
By holder: $$\sum_{cyc}\sqrt[3]{\frac{1}{a}+\frac{2}{bc}+a+3b+c}\le \sqrt[3]{\sum_{cyc}({\frac{1}{a}+\frac{2}{bc}+a+3b+c})(1+1+1) \cdot (1+1+1)}$$ It hence suffices to prove $$\sqrt[3]{\sum_{cyc}({\frac{1}{a}+\frac{2}{bc}+a+3b+c})(1+1+1)(1+1+1)}\le \frac{6}{abc}$$ $$\sum_{cyc}({\frac{1}{a}+\frac{2}{bc}+a+3b+c})\le \frac{24}{a^3b^3c^3}\tag 1$$ Now let $p=a+b+c,q=ab+bc+ca=3,r=abc$ then we rewrite (1) as $$\frac{q}{r}+\frac{2p}{r}+5p\le \frac{24}{r^3}$$ $$\iff qr^2+2pr^2+5pr^3\le 24$$ Now as $q^2\ge 3pr \iff p\le \frac{3}{r}$  and $q=3$

it suffices to prove $$3r^2+6r+15r^2\le 24$$ which is true as $r\le 1$(by AM-GM)

Note in your original question you had $...a+2b+c$  i think its a typo and should be $a+3b+c$

all the same since $$\sum_{cyc}\sqrt[3]{({\frac{1}{a}+\frac{2}{bc}+a+2b+c})}\le \sqrt[3]{\sum_{cyc}({\frac{1}{a}+\frac{2}{bc}+a+3b+c})}$$ the same proof applies here here so we are done in either case


问:
Haar measure on compact group completely positive
Is it true that the Haar measure $\mu$ on a compact group $G$ is always completely positive, i.e. every nonempty open set has positive measure? I think I have a very simple proof of it, but honestly, the fact I tried Googling this fact and couldn’t find any mention of it, so I’m second-guessing my argument, which is as follows:

Let $G$ be a compact group, and $U \subseteq G$ a nonempty open subset of $G$. Then $\mathscr{U} = \{ g U : g \in G \}$ is an open cover of $G$, so there exist $g_1, \ldots, g_n \in G$ such that $G = \bigcup_{j = 1}^n g_j U$. Then $1 = \mu(G) \leq \sum_{j = 1}^n \mu (g_j U) = n \cdot \mu(U)$. Thus $\mu(U) \geq \frac{1}{n} >0$.

答:
The Haar measure on an arbitrary locally compact (not just compact) Hausdorff group is positive on all nonempty open subsets.  This is because Haar measure on a locally compact Hausdorff group  always satisfies two regularity conditions: (i) for each Borel set $A$, $\mu(A) = \inf_{U \supset A} \mu(U)$ where $U$ runs over open subsets that contain $A$ and (ii) for each open set $A$, $\mu(A) = \sup_{K \subset A} \mu(K)$ where $K$ runs over compact subsets of $A$.

Remark.
You can't assume (ii) holds for all Borel sets. For example, let $G = S^1 \times \mathbf R_d$, where $S^1$ has its standard compact topology, $\mathbf R_d$ is the real line with the discrete topology, and $G$ has the product topology. This is a locally compact Hausdorff since $S^1$ and $\mathbf R_d$ are both locally compact Hausdorff. The product topology on $G$ is non-discrete and makes $G$ look like an infinitely long vertical cylinder whose horizontal slices are "topologically independent" circles.
Haar measure on $G$ does not satisfy (ii) for the closed set $A = \{1\} \times [0,1]$, which is not an open set: compact subsets of $A$ are finite and thus have Haar measure $0$ (one-point sets in a nondiscrete group have Haar measure 0) and you should show open sets containing $A$ have infinite measure, so property (i) tells us $\mu(A) = \infty$ and thus (ii) isn't true for this $A$.

Returning to the general situation, how do (i) and (ii) show Haar measure is positive on all open subsets? I'll give the argument for left Haar measure (for right Haar measure replace my right translates below by left translates). Since Haar measure is not identically $0$ on the group, there is a Borel set $A_0$ such that $\mu(A_0) > 0$ (perhaps $\mu(A_0) = \infty$). By (i), $\mu(V_0) > 0$ for some open set $V_0$ containing $A$. By (ii), $\mu(K_0) > 0$ for some compact set $K_0$ contained in $V_0$. For an arbitrary nonempty open set $U$ in the group, its left translates $gU$ as $g$ runs over $G$ are an open cover of $G$, so finitely many of them cover $K_0$: $K_0 \subset \bigcup_{i=1}^n g_iU$. Then $\mu(K_0) \leq \sum_{i=1}^n \mu(g_i U) = n\mu(U)$. Thus positivity of $\mu(K_0)$ implies positivity of $\mu(U)$.


问:
Proof using the fixed point theorem

We have just started learning about continuity and this is basically the first assignment on this subject so I'm very unsure about how everything works. My attempt of a solution is as follows;

Let $g:\left[0,\frac{1}{2}\right]\ ->\left[0,\frac{1}{2}\right]\ $such that  $ g\left(x\right)\ =\ f\left(x\right)\ -\ f\left(x+\frac{1}{2}\right)$
Note that $g$ is continuous, hence by the fixed point theorem there exists a point $C$ in $\left[0,\frac{1}{2}\right]$ such that $g\left(C\right)\ =\ C$.

Therefore there exists a $C$ in $\left[0,\frac{1}{2}\right]$ such that $ g\left(C\right)\ =\ f\left(C\right)\ -\ f\left(C+\frac{1}{2}\right)$.

Now am I missing something? Because it seems just too simple to me..
also I didn't even use the fact the $f(0) = f(1)$ which is suspicious.

Edit: I'm seeing now that i didn't use the theorem correctly, since it just says that $g\left(C\right)\ =\ f\left(C\right)\ -\ f\left(C+\frac{1}{2}\right)= C $. and C is not necessarily $0$..

答:
Your function $g$ may not necessarily be a self-map of $ [0,\frac{1}{2}]$ (and thus you can’t use the fixed point theorem justifiably). Rather it is only guaranteed (as you have noted) that it is continuous on $[0,\frac{1}{2}]$ and satisfies  $$g(0)=f(0)-f(\frac{1}{2})\,~\,~\,~ and \,~\,~\,~g(\frac{1}{2})=f(\frac{1}{2})-f(1)\,.$$ Since $f(0)=f(1)$, it follows that $$g(0)=-g(\frac{1}{2})\,.$$
Thus either $g(0)=0$ —— in which case $f(0)=f(\frac{1}{2})$ —— or by the Intermediate Value Theorem there exists $x_0\in (0,\frac{1}{2})$ such that $g(x_0)=0$ — that is $f(x_0)=f(x_0+\frac{1}{2})$.

Title: Proof using the fixed point theorem


问:
Proof for a general integer solution of the equation +=
Given $a,b\in\mathbb{Z}-\{0\}$ and $N\in\mathbb{Z}$, it is easy to show that if $x_0,y_0\in\mathbb{Z}$ are a particular solution to $ax+by=N$, then $x=x_0+\frac{b}{d}t$ and $y=y_0-\frac{a}{d}t$, where $d=gcd(a,b)$ and $t\in\mathbb{Z}$, are also solution to $ax+by=N$.

But may I ask how to proof that they are actually the general solution to $ax+by=N$ if we restrict the solutions inside $\mathbb{Z}$? (i.e. all integer solutions have been counted)

Thanks!

答:
Let A be the set of all ordered pairs integer solutions. Let B be the set of all ordered pairs of integer solutions only of the form you gave. We know $B \subseteq A$

First find all the rational solutions for the equation, then restrict them.

Let

$x=x_0+bu$

for $u \in\mathbb{Q}$

This is solvable for u for any rational x.

And then using

$ax+by=N$

$a(x_0+bu)+by=N$

$y=\frac{N-a(x_0+bu)}{b}$

$y=\frac{by_0-abu}{b} = y_0-au$, which is also rational.

So every element of A can be written as $(x_0+bu,y_0-au)$ for some rational u.

So let $(x_0+bu,y_0-au) \in A$

We require

$bu \in \mathbb{Z}$

$au \in \mathbb{Z}$

write $u=\frac{m}{n}$. Assume this is in lowest terms

So

$\frac{bm}{n} \in \mathbb{Z}$

$\frac{am}{n} \in \mathbb{Z}$

So $n|b$ and $n|a$

That means $n|d$ where $d=gcd(a,b)$

We can write $rn=d$ for some integer r

So $n = \frac{d}{r}$

$\frac{bm}{n} = \frac{b}{d}(rm)$

$\frac{am}{n} = \frac{a}{d}(rm)$

So letting $t=rm$, we know that $(x_0+\frac{b}{d}t,y_0-\frac{a}{d}t) \in B$

So $A \subseteq B$ giving us $A=B$.


问:
To show that function is constant
Let $f$ be defined on $\mathbb{R}$ and suppose that |$f(x)$ - $f(y)$| $\leq$ $(x-y)^2$ $x,y \in\mathbb{R}$. Here I have to show that $f$ is a constant function.

I think I have to show that $f'(x)$ = 0 for all $x$.
But I don't know from where to start this. I tried taking it as (|$f(x)$ - $f(y)$|/|$x$-$y$|) $\leq$ |$x$ - $y$|. Am I right in doing so? Any hint or suggestion will be helpful. Thanks

答:
Consider the definition of the derivative of $f$, $$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$

So,
$$\begin{align}
|f'(x)| &= \lim_{h \to 0} \left |\frac{f(x+h)-f(x)}{h}\right|
\\
&\leq \lim_{h \to 0} \left |(x+h)-x\right|
\\
&=0
\end{align}\\
\therefore f'(x)=0 \text{ }\forall x$$

I went from the first step to the second by taking $x:=x+h$ and $y:=h$ in the given inequality. Since $f'=0$ identically, $f$ is constant.

Title: To show that function is constant


问:
Circumcircle of mid points of a triangle touching the circumcircle of the triangle
$D,E,F$ are mid points of sides $BC,CA,AB$ of triangle $ABC$ and the circumcircles of DEF,ABC touch each other then find $\sum{\cos^2A}$.

The problem is that I am able to imagine a scenario where this would happen to draw the figure and start off with solving the problem.
Will the two circles touch at any of the vertex or some other intermediate point?

Could someone please help me solve this problem?

Thanks!

答:
WLOG let $\angle A$ be the largest angle in $ABC$.

It's pretty easy to show $AFE$ and $DEF$ are congruent, furthermore $AFE$ and $ABC$ are homothetic with respect to point $A$ so the circumcircles of $AFE$ and $ABC$ touches at point $A$.

From this point it's pretty easy to see the circumcircle of $DEF$ will intersect the circumcircle of $ABC$ $0$ times when $\angle A < 90^{\circ}$, and $2$ times when $\angle A > 90^{\circ}$. When $\angle A = 90^{\circ}$ the circumcircles of $AFE$ and $DEF$ coincides and we are good.

Then the sum of cosines is just $0+1=1$


问:
Is defining set equality using $\notin$ correct?
Two sets $A$ and $B$ are defined to be equal if $\forall x \in A,x\in B$ and vice-versa

I was thinking that it would be practical to define set equality using $\notin$, and that is:

Two sets $A$ and $B$ are defined to be equal if $\forall x \notin A,x\notin B$ and vice-versa

The reason I found this to be practical is that it is the easiest way to prove that we have only one empty set $\emptyset$, however, I do acknowledge that the definition would be impractical when it comes to proving two non-empty sets are equal but is it logically correct? I am new to set-theory and I am afraid I might be missing something so I am posting to make sure my logic isn't flawed.

答:
Two sets $A,B$ are equal means

$\forall x[x\in A\Longleftrightarrow x\in B]$.

This equivalently means that

$\forall x[x\not\in A\Longleftrightarrow x\not\in B]$.


问:
Find number of skew-symmetric matrices of order $3\times 3$ in which all non-diagonal elements are different
Find number of skew-symmetric matrices of order $3\times 3$ in which all non-diagonal elements are different and belong to the set $\{-9,-8,-7,...,7,8,9\}$

My Attempt:
I did a simple calculation and obtained $$\binom{9}{3}\times(3!)\times 2^3=4032$$

But answer given is $$\frac{4032}{6}=672$$

Why has it been divided by $6$

答: The given answer would be correct if the problem specified "up to permutations of rows and columns". But it did not, and your answer is correct.</p>


答: You are on the right track. Use the definition of the derivative: $f'(y)=\lim _{x\to y}\frac{f(x)-f(y)}{x-y}$. You wish to show this limit is $0$. Now, look at what you can estimate: $|f(x)-f(y)|$. Well, it's not far off from the numerator in the definition of the derivative. Now, a limit of an expression is $0$ iff the limit of the absolute value of the expression is $0$. So look at the limit as $x\to y$ of $\frac{|f(x)-f(y)|}{|x-y|}$. Fill in something in $0\le \frac{|f(x)-f(y)|}{|x-y|}\le ... $ based on what is given. Then conclude the desired result. </p>

Title: To show that function is constant

Body:
Let $f$ be defined on $\mathbb{R}$ and suppose that |$f(x)$ - $f(y)$| $\leq$ $(x-y)^2$ $x,y \in\mathbb{R}$. Here I have to show that $f$ is a constant function.

I think I have to show that $f'(x)$ = 0 for all $x$.
But I don't know from where to start this. I tried taking it as (|$f(x)$ - $f(y)$|/|$x$-$y$|) $\leq$ |$x$ - $y$|. Am I right in doing so? Any hint or suggestion will be helpful. Thanks


问:
What is the algorithm to generate the cards in the game "Dobble" ( known as "Spot it" in the USA )?h
In the game Dobble ( known in the USA as "Spot it" ) , there is a pack of 55 playing cards, each with 8 different symbols on them. What is remarkable ( mathematically ) is that any two cards chosen at random from the pack will have one and only one matching symbol . This spurred me on to investigating the Maths behind generating such a pack of cards, starting with much more basic examples with only 2 symbols on each card and gradually working my way up to 8 .

This has been explored extensively in the linked question "What is the Math behind the game Spot it".

What has been established is that if the number of symbols on each card is N, then the maximum number of different symbols throughout the pack is C , the maximum number of cards in a pack is also C, the number of times any given symbol is repeated throughout the pack is N, and N and C are related as follows :

C = N^2 - N + 1    [ N squared minus N plus one ]

But I still do not understand the algorithm for generating the cards from a given symbol set .

I am trying to follow the matrix generated by Don Simborg , but I just can't quite follow his formula . Even for a simple matrix with N=3 and C=7, I know what the matrix should look like , but can't seem to understand his descriptive syntax .
For example in column 2, row 4, his formula suggests the symbol is the one numbered 3N-1 in the sequence of 7 symbols, but 3N-1= 8 , so which symbol should I use? Is he making an assumption that we just wrap around (subtract 7) and start counting again from the beginning of the sequence ? But this still generates the wrong symbol . I know from looking at the pattern that it should be either symbol no 4 or symbol no 5, but just can't see how this arises from his formula .
If we take the 7 symbols as being the letters "A", "B", "C", "D", "E" and "F", then the matrix should be as follows below :

A B C

A D E

A F G

B D F

B E G

C D G

C E F

Can anyone help me? I've been trying to crack how to generate the symbol arrangements on the "Dobble" cards for months, and have succeeded in generating the sequence as far as N=6, C=31  but I am stuck at N=7 .
I would welcome any assistance or enlightenment with this , thank you !

And if I have misunderstood Don Simborg's formula, then the error lies with me !

Here are the matrices I have found from my own trial and error :

For N=4, C=13, with a symbol set being A B C D E F G H I J K L M , the matrix is as follows :

A B C D

A E F G

A H I J

A K L M

B E H K

B F I L

B G J M

C E J L

C F H M

C G I K

D E I M

D F J K

D G H L

For N= 5, C = 21, with the symbol set : A B C D E F G H I J K L M N O P Q R S T U , the matrix is as follows :

A B C D E

A F G H I

A J K L M

A N O P Q

A R S T U

B F K N R

B G J O S

B H L P T

B I M Q U

C F J P U

C G K Q T

C H M N S

C I L O R

D F L Q S

D G M P R

D H K O U

D I J N T

E F M O T

E G L N U

E H J Q R

E I K P S

To state again, both the sets above have the remarkable quality that any two rows chosen at random will have one and only one matching symbol .

答:
Yesterday, I played the card game "Spot It" with my family and was facinated by the maths behind it.  After working with my wife for a couple hours, we discovered the method to generate the cards.  The method is now illustated using $8$ symbols per card.

Creating $49$ ($= 7$ sets of $7$) required cards:

Creating another set of $7$ using the matrix transpose:

the $57$th card:

The above method of construction can deal any number of symbols on the cards.  It also shows that the maximum number of cards that meet the rules of the game is $57$, i.e. the same as the number of different symbols.

The general formula for the maximum number of cards with n symbols on each card satisfying the rules of the game is $(n-1)^2 + n$.


答:
Do you know about derivatives? For a function $f$ around a point $x$, the derivative $f'(x)=\frac{df}{dx}$ is the slope of the tangent to the graph of the function at the point $(x, f(x))$.

Now, take the function $f(x)=e^x$. It goes through the point $(0, e^0)=(0,1)$. Recall that the derivative of $e^x$ is $e^x$ (again). Thus, the slope of the tangent to the graph of $f$ at the point $(0,1)$ is $f'(0)=e^0=1$. Thus, the tangent of $f(x)=e^x$ at the point $(0,1)$ is the line $y=x+1$. Now, "translate" everything "down" by $1$: the tangent to the graph of the function $y=e^x-1$, which goes through $(0,0)$, is $y=x$.

That is why the lines $y=e^x-1$ and $y=x$ are so close to each other: the latter is the tangent to the former at the point $(0,0)$.

Title: why for small value of $x$, $e^{x}-1$ is close to $x$?

Body:
why for small value of $x$, $e^{x}-1$ is very close to $x$?

Suddenly, I wrote following code and I surprised. The question is why this two  functions are close for small value of $x$.

```<code>beta1<-structure(c(0.00581299965594357, 0.00398749971786833, 0.00555549997749978,
0.0039364999682459), .Dim = 4L)
beta2<-structure(c(-1.7105e-07, -8.1175e-08, -4.7495e-07, -1.3675e-07
 ), .Dim = 4L)

>  exp(beta1)-1
[1] 0.005829928 0.003995460 0.005570960 0.003944258
>     beta1
[1] 0.0058130 0.0039875 0.0055555 0.0039365
>  exp(beta2)-1
[1] -1.710500e-07 -8.117500e-08 -4.749499e-07 -1.367500e-07
>     beta2
[1] -1.7105e-07 -8.1175e-08 -4.7495e-07 -1.3675e-07
</code>```


问:
When $F'(x) = f(x)$ where $f$ is not continuous...
This question comes from problem 2vii of Chapter 14 of Spivak's Calculus.

Problem 2vii) Let $f$ be the function shown in the link below. Let $F(x) = \int_0^x f(t) dt$. For what values of $x$ does $F'(x) = f(x)$?

<img src="https://i.stack.imgur.com/dXKUH.png" alt="Picture of $f$" />

I know $F'(x) = f(x)$ for every $x$ except $0$, because $f$ is continuous everywhere except $0$, but I don't know if $F'(0) = f(0)$ is true, false, or undecidable.

Going back to limit definitions, I need to evaluate $\lim_{h \rightarrow 0^+}(\frac{F(h)}{h})$. How to do this?

答:
Try proving that the limit does not exist.

Hint1:

 You will be done if you find two sequences $x_n$ and $y_n$ both approaching 0 such that $\lim F(x_n)/x_n \neq \lim F(y_n)/y_n$.

Hint 2:

 Pick $x_n$ So that $\lim F(x_n)/x_n$ is as large as possible and $\lim F(y_n)/y_n$ is as small as possible.

Hint 3:

 Using geometry: $\lim_{n\to \infty} F(1/2^n) = \frac{1}{2}\frac{1}{2^{n+1}} + \frac{1}{2}\frac{1}{2^{n+2}} + ... = \frac{1}{2^{n+1}}$ and $F(1.5/2^n) = F(1/2^n) + 0.25/2^n = 1.5/2^{n+1}$.


问:
Question on Euler -function
Let $\varphi$ be the Euler $\varphi$-function. We have for any $a\in\mathbb{Z}^+$:

$\varphi(p^a)=p^a-p^{a-1}$

It is clear that all the positive number of the form $np$ with $1\le n\le p^{a-1}$ are not relatively prime to $p^a$.

May I ask how to prove that all the rest of the positive number less than $p^a$, as the above formula indicates, are indeed relatively prime to $p^a$？

I've just written a trial proof:

So here I think I'm asking if there exists a positive integer $y$ such that

$y<p^a$ [given]

$y\neq np$ for any $1\le n\le p^{a-1}$ [given]

$y$ is not relatively prime to $p^a$ ($\implies$ there exists $q\in\mathbb{Z}^+$ with $q>1$ such that $q\mid y$ and $q\mid p^a$) [assumed]

If so, then there exists a prime $\alpha$ such that $\alpha\mid q$. Hence, there exists $m\in\mathbb{Z}^+$ such that $m\alpha =p^a$. Thus, $\alpha=p$ since $p$ is the only prime factor of $p^a$. Then $p\mid q$. Hence, $p\mid y$. We must have $y=np$ for some $1\le n\le p^{a-1}$ because $y<p^a$. By contradiction, such $y$ doesn't exist.

Could anyone verify my proof please?

答:
A number in $X_a=\{0,1,2,\dots,p^a-1\}$ is coprime with $p$ if and only if it is not divisible by $p$.

The map $X_{a-1}\to X_a$ defined by $x\mapsto px$ is injective and its image consists of the elements divisible by $p$.

Therefore the number of elements in $X_a$ that don't belong to the image of the map is $p^a-p^{a-1}$.

Title: Question on Euler -function


问:
Solutions that oscillate more rapidly
Which one of the following two equations has solutions that oscillate more rapidly?

$$x''+\sqrt{t^6+3t^5+1} x =0$$

$$ x''+2t^3 x =0$$

I was trying to use sturm separation theorem but not getting exact approach towards it!!

Is there any direct way to do it?

Edit: The book from where I have given this question has answer as first equation in the hints provided at the back of book.
Book name: "A Textbook  on Ordinary  Differential  Equations" by Shair Ahmad, Antonio Ambrosetti(Second Edition)-Chapter5-Exercise-D2

答:
You could just apply the Sturm-Picone comparison theorem. To refine that, consider $u(t)=x_2(ct)$ for the solution $x_2$ of the second equation. Then
$$
u''(t)=c^2x_2''(ct)=-2c^5t^3x_2(ct)=-2c^5t^3u(t).
$$
Now if we set $2c^5=1$, then $t^3$ will always be smaller than $\sqrt{t^6+3t^5+1}$, so that $x_2(t/\sqrt[5]2)$ has less roots than the solution $x_1$ of the first equation.

On the other hand, for any $c>1/\sqrt[5]2$ there is some $t_c$ so that for $t>t_c$ one also has $4c^{10}t^6>t^6+3t^5+1$, so that on $[t_c,\infty)$ the scaled function $x_2(ct)$ has more roots than the unscaled $x_1(t)$. As $c<1$ can be chosen, this implies that the root density of the unscaled $x_2$ is even greater.

Asymptotically, as a rule of thumb, whenever $x_1$ has $7$ roots on some interval, $x_2$ will have at least $8$, as $7/8>1/\sqrt[5]2$.

Title: Solutions that oscillate more rapidly


问:
Inertia degree of primes in p-adic extensions
I'm reading through some number theory and ran across a theorem where the proofs referenced were incomprehensible to me, and I was hoping there might be a simpler proof than slogging through another $\approx$100 pages of background.

The statement is that if you have a finite extension $K$ of the $p$-adic rationals $\mathbb{Q}_p$, and $\pi$ is a uniformizer of $\mathcal{O}_K$, then

$$[\mathcal{O}_K/\pi\mathcal{O}_K : \mathbb{Z}_p/p\mathbb{Z}_p] = v_p(\vert N(\pi)\vert)$$

In the non-$p$-adic case, where $K'$ is some extension of $\mathbb{Q}$ and where $\pi \vert p$, showing that $[\mathcal{O}_{K'}/\pi\mathcal{O}_{K'}:\mathbb{Z}/p\mathbb{Z}] = v_p(\vert N(\pi)\vert)$ is more straightforward to me: Since $\mathcal{O}_{K'}$ is a $\mathbb{Z}$-module, I can imagine it as $\mathbb{Z}^n$, and then the quotient by $\pi$ is like a lattice, with $\pi$ as matrix giving a basis of that lattice, and the volume of the fundamental parallelepiped is the determinant of that matrix, which is $N(\pi)$. The number of integer points in the fundamental parallelepiped will be equal to its volume, showing that $\vert \mathcal{O}_{K'}/\pi\mathcal{O}_{K'}\vert = \vert N(\pi)\vert$. Since $\vert\mathbb{Z}/p\mathbb{Z}\vert =p$, that gives the result.

But this doesn't seem to extend to the $p$-adic case. $\mathbb{Z}_p^n$ doesn't look like a lattice to me, and even if it were, I'm not sure how to relate the volume to the number of points in it.

Maybe I could argue that $\mathbb{Z}$ is a subring of $\mathbb{Z}_p$, and that $\mathbb{Z}_p/p\mathbb{Z}_p$ can take representatives of each coset from $\mathbb{Z}$, so I can treat $\mathcal{O}_K/\pi\mathcal{O}_K$ as a $\mathbb{Z}$-module. But then I can't seem to treat $\pi$ as a linear operator without causing problems with the determinant.

Or maybe, since $\mathbb{Z}_p/p\mathbb{Z}_p \cong \mathbb{Z}/p\mathbb{Z}$, I could show that $\mathcal{O}_K/\pi\mathcal{O}_K\cong \mathcal{O}_{K'}/\pi\mathcal{O}_{K'}$. Certainly I can say that every linear operator over $\mathbb{Z}_p/p\mathbb{Z}_p$ is a linear operator over $\mathbb{Z}/p\mathbb{Z}$, but I'm not sure that every element of $\mathcal{O}_K$ will form a linear operator that is equal to one from $\mathcal{O}_{K'}$. In fact this is almost certainly false, so the quotient by $\pi$ must be crucial, but I don't know how to incorporate this fact.

Is there a relatively simple proof of this, or any way to complete my attempts?

答:
Here is a solution that is similar in many ways to Alex's solution, but the (second) way I explain $n = ef$ will be quite different.

The definition of $e$ is that $p = \pi^e u$ for some $u \in \mathcal O_K^\times$.  Taking the norm from $K$ down to $\mathbf Q_p$ of both sides,
$$
p^n = {\rm N}(\pi)^e{\rm N}(u),
$$
where ${\rm N} = {\rm N}_{K/\mathbf Q_p}$. Because the $p$-adic absolute value on $K$ is defined by the rule
$$
|x| = |{\rm N}(x)|_p^{1/n}
$$
for all $x$ in $K$ (the function $|\cdot|$ is the unique extension of $|\cdot|_p$ as an absolute value from $\mathbf Q_p$ to $K$).

Let's recall why the norm maps units to units: ${\rm N}(\mathcal O_K^\times) \subset \mathbf Z_p^\times$. From $u \in \mathcal O_K$ we have $|{\rm N}(u)|_p = |u|^n = 1$. Thus
$$
p^n = {\rm N}(\pi)^e{\rm N}(u) \Longrightarrow
|p^n| = |{\rm N}(\pi)|^e \Longrightarrow  \left(\frac{1}{p}\right)^n = |{\rm N}(\pi)|^e
$$
since $|p| = 1/p$. From the fundamental identity $ef = n$, which is how $e$, $f$, and $n$ are related for $p$-adic fields,
$(1/p)^{ef} = |{\rm N}(\pi)|^e$. Take $e$th roots of both sides and we're left with $|{\rm N}(\pi)| = (1/p)^f$.

Why does $n = ef$? Surely you saw this already, since it's hard to imagine how you could be reading about extensions of $p$-adic fields without having this fact being proved somewhere in the treatment.  An argument for $n = ef$ that is different from the approach Alex describes (or at least it feels different) is to build a $\mathbf Q_p$-basis of $K$ of size $ef$, and since that dimension is $n$, we'd get $ef = n$.  This is proved in Gouvea's book "$p$-adic Numbers" and the idea is this: $\mathcal O_K/(\pi)$ is a vector space over $\mathbf F_p = \mathbf Z_p/(p)$ of dimension $f$ (that's the definition of $f$), so pick an $\mathbf F_p$-basis of $\mathcal O_K/(\pi)$ and lift up the terms in the basis of $\mathcal O_K$, say to $\omega_1, \ldots, \omega_f$. Gouvea shows that the set $\{\omega_i\pi^j\}$ for $1 \leq i \leq f$ and $0 \leq j \leq e-1$ is a $\mathbf Q_p$-basis of $K$. (He really shows first that it is a $\mathbf Z_p$-basis of $\mathcal O_K$.) So that gives a $\mathbf Q_p$-basis of $K$ of size $ef$.

An argument for $n = ef$ that in some ways feels more different from what Alex wrote can be based on measure theory. (You did express an interest in a volume-theoretic approach since that is how you think about some norm results in number fields.) Since $K$ as an additive group is locally compact and Hausdorff, it has a Haar measure $\mu$. Normalize this Haar measure to give the compact open subset $\mathcal O_K$ measure 1. (Haar measure is always nonzero on open sets and finite on compact sets, so it has a positive finite value on a compact open subset like $\mathcal O_K$.)  The size of the residue field $\mathcal O_K/(\pi)$ is $p^f$. Let $\alpha_1, \ldots,  \alpha_{p^f}$ be coset representatives for that residue field:
$$
\mathcal O_K = \bigcup_{1 \leq i \leq p^f} (\alpha_i + \pi\mathcal O_K),
$$
where the union is disjoint. Because Haar measure on the additive group $K$ is invariant under additive translations,
and that union above is disjoint,
$$
1 = \mu(\mathcal O_K) = \sum_{i=1}^{p^f} \mu(\alpha_i + \pi\mathcal O_K) = p^f\mu(\pi\mathcal O_K).
$$
Therefore $\mu(\pi\mathcal O_K) = 1/p^f$.

Now we're going to calculate the Haar measure of $\pi\mathcal O_K$ in a second way.  The function $\mu_\pi(A) = \mu(\pi A)$ for all Borel sets $A$ is a Haar measure  on $K$: it is invariant under left translation since multiplication distributes over addition, and it satisfies the regularity conditions for Haar measure since multiplication by a nonzero number is a homeomorphism of $K$.  Haar measure is determined up to scaling by a positive constant, so there is some $c > 0$ such that $\mu_\pi = c \mu$: $\mu(\pi A) = c\mu(A)$ for all Borel sets $A$. Taking $A = \mathcal O_K$, $\mu(\pi\mathcal O_K) = c\mu(\mathcal O_K) = c$.  Thus $c = \mu(\pi\mathcal O_K) = 1/p^f$. That means $\mu(\pi A) = (1/p^f)\mu(A)$ for all Borel sets $A$.

Let's compute the measure of $p\mathcal O_K$ in two ways. Since $p = \pi^e u$, $p\mathcal O_K = \pi^e\mathcal O_K$, so
$$
\mu(p\mathcal O_K) = \mu(\pi^e\mathcal O_K).
$$
We saw above that the effect on Haar measure of  multiplying by $\pi$ is to scale Haar measure by $1/p^f$. Therefore

$$
\mu(p\mathcal O_K) =
\mu(\pi^e\mathcal O_K) = \left(\frac{1}{p^f}\right)^e\mu(\mathcal O_K) = \frac{1}{p^{ef}}
$$
since $\mathcal O_K$ has Haar measure $1$.

Since $\mathcal O_K$ is a free $\mathbf Z_p$-module of rank $n$, $[\mathcal O_K:p\mathcal O_K] = p^n$. Therefore $\mathcal O_K$ is a disjoint union of $p^n$ cosets of $p\mathcal O_K$, and $\mu(\alpha + p\mathcal O_K) = \mu(p\mathcal O_K)$ for all $\alpha$, so
$$
\mu(\mathcal O_K) = p^n \mu(p\mathcal O_K).
$$
Since $\mathcal O_K$ has measure $1$, $\mu(p\mathcal O_K) = 1/p^n$. Comparing this value with what we found in the previous paragraph, $1/p^n = (1/p)^{ef}$, so $n = ef$.

Since this argument dealt a lot with cosets of $p\mathcal O_K$ and $\pi\mathcal O_K$ in $\mathcal O_K$, as Alex's argument does, his approach to $n = ef$ and the approach here with measure theory clearly have some underlying similarities even if the surface details are not the same.

It was not essential to normalize the Haar measure of $\mathcal O_K$ to be $1$: whatever value it has must be finite and positive since $\mathcal O_K$ is compact and open, and if we did not normalize the Haar measure by setting $\mu(\mathcal O_K) = 1$ then we'd be dragging a scaling factor $\mu(\mathcal O_K)$ around in many calculations, so at the end we'd have $(1/p^n)\mu(\mathcal O_K) = (1/p^{ef})\mu(\mathcal O_K)$, and then cancelling the unspecified finite positive number $\mu(\mathcal O_K)$ from both sides we'd get $1/p^n = 1/p^{ef}$, so $n = ef$.


问:
The elements of $\mathbb{Z}[\sqrt{-5}]/(2,\sqrt{-5}+1)$
I'm really confused with this one...

How can I determine the elements of the module $\mathbb{Z}[\sqrt{-5}]/(2,\sqrt{-5}+1)$? Or its cardinality?

Does $$\mathbb{Z}[\sqrt{-5}]/(2,\sqrt{-5}+1)=\{\bar{0},\bar{1},\overline{0+\sqrt{-5}}\}?$$

答:
Recall that
$$\mathbb{Z}[\sqrt{-5}]=\{a+b\sqrt{-5}\mid a,b\in\mathbb{Z}\}.$$
Therefore, as an abelian group (i.e., forgetting about the multiplicative structure for a second), we can make an isomorphism.
$$\mathbb{Z}[\sqrt{-5}]\cong\mathbb{Z}^2,\quad 1\mapsto (1,0),\quad \sqrt{-5}\mapsto (0,1).$$
Under this isomorphism, the ideal $(2)\subset\mathbb{Z}[\sqrt{-5}]$ corresponds to the subgroup of $\mathbb{Z}^2$ generated by $(2,0)$ and $(0,2)$. Therefore (again, as abelian groups)
$$\mathbb{Z}[\sqrt{-5}]/(2)\cong \mathbb{Z}^2/\langle (2,0),(0,2)\rangle\cong (\mathbb{Z}/2\mathbb{Z})^2,$$
and because representatives for $\mathbb{Z}^2/\langle (2,0),(0,2)\rangle$ are
$$\overline{(0,0)},\quad \overline{(1,0)},\quad \overline{(0,1)},\quad \overline{(1,1)},$$
we can undo our isomorphism and conclude that representatives for $\mathbb{Z}[\sqrt{-5}]/(2)$ are
$$\overline{0},\quad \overline{1},\quad \overline{\sqrt{-5}},\quad \overline{1+\sqrt{-5}}.$$
Now, $\mathbb{Z}[\sqrt{-5}]/(2,1+\sqrt{-5})$ is just $\mathbb{Z}[\sqrt{-5}]/(2)$ further quotiented by $1+\sqrt{-5}$ (or rather $\overline{1+\sqrt{-5}}$).

You could either determine which of
$$\overline{0},\quad \overline{1},\quad \overline{\sqrt{-5}},\quad \overline{1+\sqrt{-5}}$$
differ by a multiple of $1+\sqrt{-5}$ (remember, this means a $\mathbb{Z}[\sqrt{-5}]$-multiple, not a $\mathbb{Z}$-multiple), or determine what the ideal $(2,1+\sqrt{-5})$ corresponds to as a subgroup of $\mathbb{Z}^2$, or compute representatives modulo $1+\sqrt{-5}$ of these four elements.


问:
Can I have a hint on this combinatorics question?
Six faces of identical cubes are painted 6 distinct colors. How many different cubes can be formed?

The answer is surprisingly small but that may be because I don't know how to express the different ways properly. I just incorrectly did $6!$ because if there are 6 ways to color the first face, there are now 5 colors left so 5 ways to color the second face. And so on. Why is this wrong?

Perhaps I need to make use of "identical" and "distinct" from the question. If the cubes are identical, I must divide the answer by something, perhaps $6!$ for the 6 identical faces? And there should be some "choosing" happening but I can see where this fits in but it must be for the "distinct" part. Choose colors as in $6C1, 5C1 ...etc$ but this is the same as $6!$

Please help

答:
Perhaps to record a more advanced answer for those who knows Burnside lemma from group theory:

Denote $X$ to be the set of all possible ways to label each of the 6 faces of the cube with exactly one of six distinct colors. There are $6!$ ways of doing this without considering which ones are the same by rotation.

Now the rotational symmetry group for a cube is the octahedral group $O \simeq S_4$ (https://en.wikipedia.org/wiki/Octahedral_symmetry) with $|O| = 24$.

Now the group $O$ acts on the set $X$, and in each orbit it contains all the colorings that are considered the same when up to rotation. By Burnside lemma, we have the number of orbits $|X/O| = \frac{1}{|O|}\sum_{g\in O} |X^g|$, where $X^g$ is the set of colorings that are unchanged by the action of $g$. Since the cube has six different colors, whenever $g\neq e$ we have $|X^g|=0$, as any nontrivial rotation will not fix the coloring configuration. Unless $g=e$, then we have $|X^e|=|X|= 6!$, as every configuration is fixed by identity.

Hence $|X/O| = \frac{1}{|O|}\sum_{g\in O} |X^g| = \frac{6!}{24}=30$.


问:
What is: $A= \{ a \in \mathbb{N_0} ^ \mathbb{N} : \forall n\in \mathbb{N}, \sum_{i\in \{ 1,2,3,...,n\}}^{} a_i=n \} $
$\mathbb{N_0}$ is $ \mathbb{N}\cup {\{ 0\} }$.

I can't really understand what this set stands for, and I need to know if its cardinality is finite or infinite, it would be really appreciated if someone can explain how to read these type of sets.

Theres also this set, but I think I can understand it if I understand the first one.

$B= \{ a \in \mathbb{N_0} ^ \mathbb{N} : \forall n\in \mathbb{N}, \sum_{i\in \{ 1,2,3,4\}}^{}  a_i<10*\sqrt2 \} $

答:
$\mathbb N_0^{\mathbb N}$ is the set of all functions that map $\mathbb N$ into $\mathbb N_0$. In other words, this is the set of all sequences $(a_i)_{i\in\mathbb N}$ such that $a_i\in\mathbb N_0$ for all $i\in\mathbb N$.

(Generally, for two sets $X$ and $Y$, the set $X^Y$ is the set of all functions mapping $Y$ into $X$.)

This means that the set $A$ is the set of all of those sequences where $a_1+a_2+\ldots+a_n=n$ for all $n\in \mathbb N$. This implies $a_1=1, a_2=1,\ldots$, and in general $a_n=(a_1+\ldots+a_n)-(a_1+\ldots+a_{n-1})=n-(n-1)=1$ i.e. $a_i=1$ for all $i\in\mathbb N$, so the set $A$ is a $1$-element set $\{(1,1,1,\ldots)\}$.

The definition of $B$ is a bit strange as $n$ does not play any role in the criterion $a_1+a_2+a_3+a_4\lt 10\sqrt{2}$ - so that criterion is true for every $n$ if it is true at all. If there is no typo in the definition of $B$, then $B$ is simply:

$$B=\{a\in\mathbb N_0^{\mathbb N}:\sum_{i\in\{1,2,3,4\}}a_i\lt 10\sqrt{2}\}$$

i.e. set of all of those sequences where $a_1+a_2+a_3+a_4\lt 10\sqrt{2}$ - which means this is an infinite set, as $a_1,a_2,a_3,a_4$ can be chosen somehow to satisfy the above criterion (e.g. to be all zeros), and then $a_5,a_6,\ldots$ can be chosen arbitrarily.


问:
Lebesgue decomposition of a function
$F(x)=\begin{cases} x^3+5, & x\ge 1\\ x^3+2, & 0\leq x<1\\ x^3, & x<0 \end{cases}$

Let $\mu_{F}$ be the Lebesgue-Stieltjes measure associated with $F$. Find the lebesgue decomposition of $\mu_{F}$ with respect to the Lebesgue measure $m$.

$\frac{dF(x)}{dx}=3x^2$. It has jump of height 2 at $x=0$ and has jump of height 3 at $x=1$.

My answer is $\mu_{F}=\mu_{a}(E)+\mu_{d}(E)=\int_{E} 3x^2dm(x)+2\delta_{0}+3\delta_{1}$, where $\mu_{a}<<m$, and $\mu_{a}\perp\mu_{d}$.

I am asking because I do not quite understand the concept of Lebesgue decomposition. Are there any flaws in my analysis? I asked similar questions here before, but no one answers.

答:
F is an increasing function, so you have the Lebesgue decomposition for the positive Lebesgue Stieltjes measure. See for example Theorem 8 of Complex Measure, Dual Space of L
p Space, Radon-Nikodym Theorem and Riesz Representation Theorems. (https://my-calculus-web.firebaseapp.com/MA3110/Complex%20Measure_Dual_space_Riesz_Thm.pdf). The proof there is a general existence proof. For Lebesgue Stieltjes measure, specifically we have a more detail decomposition: for the function F here, we can write it as a sum of an increasing continuous function x3 and an obvious Saltus function. Note that x3 is not absolutely continuous on the set of real numbers but is locally absolutely continuous. Taking the domain as a bounded interval, from Theorem 19 of Lebesgue Stieltjes Measure, de La Vallée Poussin’s Decomposition,
Change of Variable, Integration by Parts for Lebesgue Stieltjes Integrals (https://my-calculus-web.firebaseapp.com/MA3110/Lebesgue%20Stieltjes%20Measure%20and%20de%20La%20Vallee%20Poussin%20Decomposition.pdf), appropriate interpreted, we have a Lebesgue decomposition on any bounded interval. By Passing the domain to the whole of the reals we have the Lebesgue decomposition. The explicit form of the decomposition is given by Theorem 22 of Lebesgue Stieltjes Measure, de La Vallée Poussin’s Decomposition, Change of Variable, Integration by Parts for Lebesgue Stieltjes Integrals, where we can ignore the contributions from the points having infinite derivative, since there are none. Note that the absolute part is given by the integral of the derivative of x3.  The contribution from the saltus part of the function F is given by the sum of the jumps at 0 and 1 which is the same as the sum of the jumps at all points of a Borel set and so are the sum of the function given by you, 2 times the characteristic function of {0} plus 3 times the characteristic function at {1}. Note that Theorem 19 of Lebesgue Stieltjes Measure, de La Vallée Poussin’s Decomposition,
Change of Variable, Integration by Parts for Lebesgue Stieltjes Integrals says that these two parts are mutually singular.  Proposition 12 of Lebesgue Stieltjes Measure, de La Vallée Poussin’s Decomposition,Change of Variable, Integration by Parts for Lebesgue Stieltjes Integrals affirms that the absolute part of the Lebesgue Stieltjes measure of F, is absolutely continuous with respect to the Lebesgue measure on the reals since x3 is a locally absolutely continuous function on the reals. Hope the references cited here would be useful.

Title: Lebesgue decomposition of a function


问:
Problem 25.1 (b) Kosniowski. Computing the fundamental group of a pentagonal region.

Hi. I have just calculated the fundamental group of this pentagonal region but I am not sure if it is correct. I attach my solution.

Solution:

Let $U_1=X-\left\{y\right\},\, U_2=X-(a_1\cup a_2)$ open sets and path connected.

The figure 8 is a strong deformation retract of $U_1$ then $\pi_1(U_1,x_1)=\pi_1(\text{ Fig. 8 },x_1)=\langle [\alpha_1],\, [\alpha_2];\emptyset\rangle$ with $\alpha_1,\,\alpha_2$ closed path in $x_1$.

Therefore, $\pi_1(U_1,x_0)=\langle [\delta*\alpha_1*\overline{\delta}],\,[\delta*\alpha_2*\overline{\delta}];\emptyset\rangle=\langle A_1,A_2;\emptyset\rangle$ with $A_i=[\delta*\alpha_i*\overline{\delta}],\, i=1,2.$

Now, $\pi_1(U_2,x_0)=1$ because $U_2$ is contractible and $\pi_1(U_1\cap U_2,x_0)=\langle [\gamma];\emptyset\rangle=\mathbb{Z}$ because the circle generated by $\gamma$ is a strong deformation retract of $U_1\cap U_2$.

On the other hand, $\varphi_{1*}([\gamma])=[\delta*\alpha_1*\overline{\delta}][\delta*\alpha_2*\overline{\delta}][\delta*\alpha_1*\overline{\delta}]^2[\delta*\overline{\alpha_2}*\overline{\delta}]=A_1A_2A_1^2A_2^{-1}$
and $\varphi_{2*}[\gamma]=1$

Therefore, by Seifert Van Kampen, $\pi_1(X,x_0)=\langle A_1,A_2; A_1A_2A_1^2A_2^{-1}\rangle$

This is correct?

答:
It is correct. In fact, let $Y = S^1_1 \vee S^1_2$ denote the wedge of two copies $S^1_i$ of $S^1$. This is the space you denote as "the figure $8$". You attach a two-cell to $Y$ as indicated in fig. 25.10 to obtain the space $X$. We have $\pi_1(Y) = \mathbb Z * \mathbb Z$ = free group with generators $A_1, A_2$, where $A_i$ is represented by the identity map $S^1 \to S^1_i \subset Y$.

The attaching map $\phi : S^1 \to Y$ represents the word $A_1A_2A_1^2A_2^{-1}$ which gives the presentation $\pi_1(X,x_0)=\langle A_1,A_2; A_1A_2A_1^2A_2^{-1}\rangle$.

For a more general approach to questions like that see my answer to Can this counter example disprove the statement?


问:
Finite dimensional k[x]-module is torsion module
Let $k$ be a field and $R=k[x]$. Let $M$ be a $k[x]$-module which is finite dimensional as a $k$-vector space. I want to prove $M$ is a torsion module.

I’m wondering how to build a relationship between basis and torsion, because from the question, all I know is $M$ has a basis of size $n$ ( assume the dimension is $n$).

答: Hint: Apply the Cayley-Hamilton Theorem to the $k$-endomorphism given by multiplication with $x$.


问:
A $2021$ problem: $20\sim 21$ and $43\times 47$
Notice that $2021$ is a concatenation of consecutive integers:  $20\sim 21$

Also $2021$ is a product of consecutive primes: $43\times 47$.

What is the next number with both of these properties?

$24502451$ is close, $4943\times 4957$  but $4951$ is in between.

$2484224843$ is close, $49831\times 49853$ but $49843$ is in between.

$715353612\sim 715353607$ isn't quite there but is $845785793\times 845785799$

Are there any other numbers with the $2021$ property?

答: I would have just commented but I don't have sufficient reputation. A quick brute-force approach in bases 2 to 16 is here and shows solutions where the second part of the concatenation contains leading zeros. Hence 249950~00249951 = 4999493*4999507.</p>


问:
prove that $\operatorname{Hom}_\Bbb{Z}(\Bbb{Z}/n\Bbb{Z},A)\cong A_n.$
Let $A$ be any $\Bbb{Z}$-module, let $a$ be any element of $A$ and let $n\in \Bbb{Z}^+$. Prove that the map $\psi_a\colon\Bbb{Z/nZ}\to A$ given by $\psi_a(\overline{k})=ka$ is a well-defined $\Bbb{Z}$-module homomorphism if and only if $na=0$.

Then, prove that $\text{Hom}_\Bbb{Z}(\Bbb{Z}/n\Bbb{Z},A)\cong A_n$, where $A_n=\{a\in A: na=0\}$.

I am struggling to find a map so that i can prove the isomorphism between $\text{Hom}_\Bbb{Z}(\Bbb{Z}/n\Bbb{Z},A)$ and $A_n$.

答:
Define a map $f:\text{Hom}_\Bbb{Z}(\Bbb{Z/nZ},A)\to A_n$ by $(\psi_a(\overline{k})\mapsto ka)\mapsto a$.

$f$ is a $\Bbb{Z}$-module homomorphism:

$\psi_a(\overline{x})+\psi_b(\overline{x})=xa+xb=x(a+b)=\psi_{a+b}(\overline{x})$,

$\psi_a(r\overline{x})=r(xa)=r\psi_a(\overline{x})$
Therefore,
$f(\psi_a+\psi_b)=f(a+b)=a+b=f(\psi_a)+f(\psi_b).$

$f(r\psi_a)=ra=rf(\psi_a).$

$f$ is injective:
\begin{align}f(\psi_a)=f(\psi_b)\Rightarrow a=b,
ka=kb\Rightarrow \psi_a=\psi_b.\end{align}

$f$ is surjective:
For any $a\in A_n$, $\psi_a$ is a homomorphism in $\text{Hom}_\Bbb{Z}(\Bbb{Z/nZ},A)$. Therefore $f(\psi_a)=a$.


问:
Uniform c.r.v and volume of slice (probability). two variable
I have this homework I'm struggling with.

Given Uniform
$$f_{X,Y}(x,y) = \begin{cases} \frac{1}{4} ,  \  -1\leq x\leq   1, \  -1\leq y\leq   1 \\ 0, \ \textrm{otherwise}  \\ \end{cases}$$

I need to find the probability of
$$ \mathbb{P}(X^{2}+Y^{2} \leq 1) $$

this what I'm thinking

The volume of this slice of a sphere with centrum (0,0,0) and r= 1 is the probability.

Is this assumption correct and if so how do I find the volume?
2

答:
Since X and Y are uniformly distributed, the probability of $X^2+Y^2<1$ is the ratio of the area of the circle to the area of the square, i.e., $\pi/4$. Following a Monte-Carlo approach, you may give an estimate for $\pi$, something like:
https://www.geeksforgeeks.org/estimating-value-pi-using-monte-carlo/


问:
Why is $0/\lim_{x\to0} x$ undefined, when $\lim_{x\to2} (x^2-4)/(x-2)$ is defined?
For $\lim\limits_{x\to2}\ (x^2-4)/(x-2)$,
we are able to cancel out $x-2$ and rewrite it as $\lim\limits_{x\to2} x+2 $
But in maths, we are not able to cancel out $0$ values so $x-2$ is not zero, and it totally makes sense. $x$ is aproaching to 2 but it is NEVER 2, it's extremely close to 2 but never 2.

Here is my main problem,
$0/\lim_{x\to0}x$

I am told that this is undefined but I don't understand why.
I mean, as we have done above, $x$ is something extremely close to $0$ but its never $0$ and '$0$ divided by something not $0$ ' is not undefined. It is $0$, so why isn't the answer $0$ then?

答:
This is the mathematical definition of limit- we say $ \lim_{x \to c } f (x) = L$ if for each $\epsilon > 0$ there exist a $\delta > 0$ such that $0< |x-c|< \delta \implies |f(x) - L | < \epsilon$. By this definition we can easily avoid $\frac 00 $ confusion.

The above definition is only abstraction of our intution. Suppose a function has a limit at a point it can be made arbitrarily close to a number which is called limit value of the function at the given point . In your case suppose you want make distance between $f(x) = \frac{x^2-2}{x-2}$ and $4$ less than $\frac{1}{1000}$ then if you consider $x \in (2 - 10^{-3} , 2 + 10^{-3})$ then you observe that $|f(x) - 4 | = |x-2| < \frac{1}{1000}$. How wonderful it is! Similarly you can take any number ($\epsilon$ ) and find an interval for values of $x$ ($|x-2| < \delta$) such that distance between $f(x)$ and $4$ is less than $ \epsilon$.

Thus we have avoided confusion about $0/0$ form in limits. If you consider $ \frac{0}{lim_{x \to 0 } x }$ this is little different than the above. We have $lim_{x \to 0 } x = 0$ and obviously $0/0$ is undefined (why? see here).


问:
Let there be $2n+1$ integers such that when one of them is removed, the sum can be divided into $2$ groups each with $n$ integers.
Let $a_1,a_2,\cdots,a_{2n+1}$ be integers such that, if any one of them is removed, the remaining ones can be divided into two sets of $n$ integers with equal sum. Prove $a_1 = a_2 = \cdots = a_{2n+1}$

$Note:-$
The solution is mentioned in this website - https://www.fmf.uni-lj.si/~lavric/Santos%20-%20Number%20Theory%20for%20Mathematical%20Contests.pdf.

I understood the first three lines of the solution but however from the fourth line, I couldn't understand, why are all $a_k$ congruent mod $4$ ?

THE PROOF FROM THE WEBSITE - "As the sum of the $2n$ integers remaining is always even, no matter which of the $a_k$ be taken, all the $a_k$ must have the
same parity. The property stated in the problem is now shared by $a_k/2$ or ($a_k −1$)/$2$, depending on whether they are all even, or
all odd. Thus they are all congruent mod $4$. Continuing in this manner we arrive at the conclusion that the $a_k$ are all congruent
mod $2^k$
for every $k$, and this may only happen if they are all equal."

答:
Let, for each $i$, $S_i$ be the sum of the $a_j$ over the $1\leq j \leq 2n+1$ with $j \neq i$.

Clearly, if $1 \leq i,j \leq 2n+1$, $S_i+a_i=S_j+a_j$, ie $a_j-a_i=S_i-S_j$.

Now, for each $1 \leq i \leq 2n+1$, the $a_j$, $1 \leq j \leq 2n+1$ and $j \neq i$, can be split in two subsets of same sum, so $S_i$ is even.
It follows that for each $1\leq i,j \leq 2n+1$, $a_i-a_j$ is even. Thus all the $a_i$ have the same parity.


问:
Negative Combination
If given $0<r<n$, then $C_r^n=\frac{n!}{(n-r)!r!}$. But what if $r<0$ since $r!$ will fail?

答:
If $r<0$, then the combinatorial interpretation of $C^n_r$ as the number of ways to choose $r$ items from a set of $n$ objects fails, so you must specify what you mean by this expression. Normally, we define it to be $0$. Think of this as meaning that the Pascal triangle outside the normal range consists of only entries, or to represent the fact that there's no way to choose "$-1$ elements" from any set, or just as a convention to make it convenient to allow us to write
$$(1+x)^n=\sum_k \binom{n}{k}x^k$$
and have $k$ run through all integers with no problem---this is a very useful (convenient) property when working with generating functions, for instance.

Title: Negative Combination


问:
Most elegant/simple proof of the irrationality of $\pi$
What is the most elegant/shortest proof of this? The proofs I have seen are quite long, unlike the proof of the irrationality of $e$.

thanks

答: Niven's proof. See the comments for a link. </p>


问:
If $f(x)=f(x+5)$, is $\int f(x) dx=\int f(x+5) dx$?

Given a function $f$ such that $f(x)=f(x+5)$ for all $x \in \mathbb{R}$. If $\int_1^5 f(x) dx=3$ and $\int_{-5}^{-4} f(x)dx=-2$, find $\int_5^{15} f(x)dx$.

Here's my attempt. Let $\int f(x) dx=F(x)$. Since $f(x)=f(x+5)$, then $\int f(x) dx=\int f(x+5) dx$ or we can say that $F(x)=F(x+5)$. Next, observe that

$F(-4)=F(-4+5)=F(1)$

$F(-5)=F(-5+5)=F(0)=F(0+5)=F(5)$

Then, we get $F(5)-F(1)=3$ and $F(1)-F(5)=-2$. But we will get nothing. I am worried that $F(x)\neq F(x+5)$.

答: Let $I = \int_{5}^{15}f(x)dx$. Then, do the substitution $x \mapsto x+5$, to have $I = \int_{0}^{10}f(x+5)dx=\int_{0}^{10}f(x)dx$. Do the same substitution to have $I=\int_{-5}^{5}f(x)dx$. Now, let $J = \int_{-5}^{-4}f(x)dx$. Do the substitution $x \mapsto x-5$ to have $J = \int_{0}^{1}f(x-5)dx = \int_{0}^{1}f(x)dx$ ($ \forall x (f(x)=f(x+5)) \iff  \forall x(f(x)=f(x-5))$). Now, we have, $1=J+\int_{1}^{5}f(x)dx=\int_{0}^{5}f(x)dx$. Do the substitution $x \mapsto x+5$, to have $1 = \int_{-5}^{0}f(x)dx$. Now, $I=\int_{-5}^{0}f(x)dx + \int_{0}^{5}f(x)dx=1+1=2$.


问:
Planar networks
Let $(b_1,...,b_n)$ be a sequence of positive integers. Define a $(b_1,...,b_n)$-network as the following graph $G$:

$V(G)=V_1\cup...\cup V_n$

$V_i\cap V_j=\emptyset$ ($1\leq i<j\leq n$)

$|V_i|=b_i$ ($1\leq i\leq n$)

$E(G)=\{\{u,v\}:u\in V_i,v\in V_{i+1} \text{ for some }i<n\}$

Example of the $(3,5,3)$-network:

It is straightforward that the condition $b_i\leq 2$ or $b_{i+1}\leq 2$ ($\text{for all }i<n$) is necessary for a $(b_1,...,b_n)$-network to be planar. Is it true that this is also the sufficient condition? If not, what the sufficient condition could be?

答:
Here is a sufficient condition we can show.

Proposition. Consider $n\ge 3$ many positive integers $b_1 ,\ldots ,b_n$, where $b_i \le 2$ for all $i\neq 1,n$, and $b_1,b_n$ can be any positive integer. Then the $(b_1,\ldots ,b_n)$-graph is planar.

The idea is a graph is planar if and only if it is also planar on the sphere (that the edges can be untangled on the sphere). This can be seen by taking a stereographic projection. And being planar on the sphere is the same as being planar on any surface homeomorphic to the sphere, say a cylinder with tops. So to show this $(b_1,\ldots,b_n)$-graph satisfying those our conditions is planar, we just need to draw it without the edges crossing on the cylinder.

Indeed, the two ends of the graph can be easily made planar, and the inner parts we can wrap around the cylinder with the edges not crossing. A sketch is shown below:

Occasionally we have some $b_i = 1$, where $1<i<n$, but that is also easily seen to be not a problem for planarity.

Title: Planar networks


问:
Determining whether following sequences are Cauchy
$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$
I'm self-learning mathematical analysis from Understanding Analysis by Stephen Abbott. I'd like someone to verify my solution, if the proof checks out.

Additionally, if $(a_n)$ is Cauchy, is $[[a_n]]$ also Cauchy? Because, there is no strict relationship between $\absval{[[a_n]] - [[a_m]]}$ and $\absval{a_n - a_m}$ or so I think.

Exercise 2.6.4

Let $(a_n)$ and $(b_n)$ be Cauchy sequences. Decide whether each of the following sequences is a Cauchy sequence, justifying each conclusion.

(a) $c_n = \absval{a_n - b_n}$

(b) $c_n = (-1)^n a_n$

(c) $c_n = [[a_n]]$, where $[[x]]$ refers to the greatest integer less than or equal to $x$.

Proof.

(a) The following inequalities hold for any reals $a,b \in \mathbf{R}$:
\begin{align*}
 \absval{a + b} \le \absval{a} + \absval{b}
\end{align*}
Replacing $b$ by $-b$, we have:
\begin{align*}
 \absval{a - b} \le \absval{a} + \absval{b}
\end{align*}
Writing $a = a - b + b$, we have:
\begin{align*}
 \absval{a} &= \absval{a - b + b}\\
  &\le \absval{a - b} + \absval{b}\\
 \iff \absval{a} - \absval{b} &\le \absval{a - b}
\end{align*}
Writing $b = b - a + a$, we have:
\begin{align*}
 \absval{b} &= \absval{b - a + a}\\
  &\le \absval{a - b} + \absval{a}\\
 \iff -(\absval{a} - \absval{b}) &\le \absval{a - b}
\end{align*}
Consequently,
\begin{align*}
 \absval{\absval{a} - \absval{b}} \le \absval{a - b}
\end{align*}

Now, the distance between any two terms of the sequence $(c_n)$ is given by,

\begin{align*}
 \absval{c_n - c_m} &= \absval{\absval{a_n - b_n} - \absval{a_m - b_m}}\\
 &\le \absval{{a_n - b_n} - ({a_m - b_m})} = \absval{(a_n - a_m) - (b_n - b_m)}\\
 &\le \absval{a_n - a_m} + \absval{a_m - b_m}
\end{align*}

We desire each of these distances to be smaller than $\epsilon/2$.

Since $(a_n)$ and $(b_n)$ are Cauchy sequences, given any $\epsilon > 0$, there exists $N_1,N_2$ such that for all $m,n \ge N_1$, $\absval{a_n - a_m} < \epsilon/2$ and for all $m,n \ge N_2$, $\absval{b_n - b_m} < \epsilon/2$. Altogether, define $N = \max \{ N_1,N_2 \}$, if $m,n \ge N$, the terms in the above inequality are each smaller than $\epsilon/2$, which is what we required.

(b) $c_n = (-1)^n a_n$

If $(a_n)$ is Cauchy, $((-1)^n a_n)$ is also Cauchy, because
\begin{align*}
 \absval{(-1)^n(a_n - a_m)} = \absval{a_n - a_m}
\end{align*}

答:
(a) It is correct, but you can use the inequality $\bigl||a|-|b|\bigr|\leqslant|a-b|$ without proving it. It's a standard inequality. Besides, when you wrote $|a_n-a_m|+|a_m-b_m|$, what you meant was $|a_n-a_m|+|b_n-b_m|$.

(b) That is wrong. The constant sequence $1$ is a Cauchy sequence, but $\bigl((-1)^n\bigr)_{n\in\Bbb N}$ isn't.

(c) The statement is false. For instance, the sequence $\left(\frac{(-1)^n}n\right)_{n\in\Bbb N}$ is a Cauchy sequence, but$$(\forall n\in\Bbb N):\left\lfloor\frac{(-1)^n}n\right\rfloor=\begin{cases}-1&\text{ if $n$ is odd}\\0&\text{ if $n$ is even.}\end{cases}$$and therefore $\left(\left\lfloor\frac{(-1)^n}n\right\rfloor\right)_{n\in\Bbb N}$ is not a Cauchy sequence.


问:
Find all polynomials $f(x)$ satisfying $f(x^2 ) + f(x) \cdot f(x + 1) = 0$ ∀ $x ∈ R$
Find all polynomials $f(x)$ satisfying $f(x^2 ) + f(x) \cdot f(x + 1) = 0$  ∀  $x ∈ R$

Obvious polynomials are $f(x)=0$ and $f(x) = -1$

There are no linear polynomials satisfying the equation.

I took $f(x) = a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+...a_nx^n$

For $f(x)$ to satisfy the condition

$a_0(1+a_0+a_1+a_2+...a_n) = 0$

$a_n + {a_n}^2=0$

Taking simple values of $x$ like $x=1,x=0$ did not help and I am unable to proceed further.

Can anybody help?

答: Hint: Basic Complex Analysis tells us that the given equation holds for complex $x$ also. [I am referring to the Identity Theorem].  If $f$ has  a root $x$ then the hypothesis shows that $x^{2}, x^{4},..$ are all zeros. But  a polynomial can have only finite number of zeros (unless it is the zero polynomial). So the only possible roots are roots of unity and $0$.


答:
You can't use any results involving primitive function as you are not given continuity. Rather, we make use of the following equality that always hold:

$$\int_a^b f(x) dx = \int_{a-c}^{b-c} f(x+c) dx$$

In particular, the case $c=5$ interests us because we then obtain
$$\int_a^b f(x) dx = \int_{a-5}^{b-5} f(x) dx$$
Hence, making use of this observations multiple times, we obtain:

\begin{align}\int_5^{15} f(x) dx &=\int_0^{10}f(x)dx \\
&= 2 \int_0^5 f(x) dx\\
&= 2\left(\int_0^1 f(x) dx + \int_1^4 f(x) dx\right)\\
&= 2\left(\int_{-5}^{-4}f(x)dx + \int_1^4 f(x) dx\right) \\
&= 2 (-2+3) = 2\end{align}

Title: If $f(x)=f(x+5)$, is $\int f(x) dx=\int f(x+5) dx$?

Body:
Given a function $f$ such that $f(x)=f(x+5)$ for all $x \in \mathbb{R}$. If $\int_1^5 f(x) dx=3$ and $\int_{-5}^{-4} f(x)dx=-2$, find $\int_5^{15} f(x)dx$.

Here's my attempt. Let $\int f(x) dx=F(x)$. Since $f(x)=f(x+5)$, then $\int f(x) dx=\int f(x+5) dx$ or we can say that $F(x)=F(x+5)$. Next, observe that

$F(-4)=F(-4+5)=F(1)$

$F(-5)=F(-5+5)=F(0)=F(0+5)=F(5)$

Then, we get $F(5)-F(1)=3$ and $F(1)-F(5)=-2$. But we will get nothing. I am worried that $F(x)\neq F(x+5)$.


Body:
\begin{align}
\int_{5}^{15} f(x) dx &= \int_{5}^{10} f(x) dx + \int_{10}^{15} f(x) dx \\
&= \int_{0}^{5} f(x) dx + \int_{0}^{5} f(x) dx \\
&=2 \cdot \int_{0}^{5} f(x) dx \\
&=2 \cdot \int_{0}^{1} f(x) dx + 2 \cdot \int_{1}^{5} f(x) dx \\
&= 6+ 2 \cdot \int_{0}^{1} f(x) dx \\
&= 6+ 2 \cdot \int_{-4}^{-5} f(x) dx \\
&= 6+ 2 (-2) \\
&= 2
\end{align}


问:
The elements of $\mathbb{Z}[\sqrt{-5}]/(2,\sqrt{-5}+1)$
I'm really confused with this one...

How can I determine the elements of the module $\mathbb{Z}[\sqrt{-5}]/(2,\sqrt{-5}+1)$? Or its cardinality?

Does $$\mathbb{Z}[\sqrt{-5}]/(2,\sqrt{-5}+1)=\{\bar{0},\bar{1},\overline{0+\sqrt{-5}}\}?$$

答:
Almost. Let's call $\alpha:=\sqrt{-5}$, and let $a\equiv b$ denote that $a-b\in\, (2,\,1+\alpha)$. Then, as $1\equiv -1$, we also have
$$1+\alpha\equiv 0 \ \implies\ \alpha\equiv -1\equiv 1$$
So that, the quotient has only $2$ elements.

For $0\not\equiv 1$ observe that $a+b\alpha\in (2,\,1+\alpha)\,\implies\,2\,|\,a+b$.


答:
Let $m\in M$.

Since $M$ is finite-dimensional over $k$, the infinite sequence
$$m,xm,x^2m,...$$
of elements of $M$ can't be linearly independent.

It follows that for some nonnegative integer $n$ there must exist $c_0,...,c_n\in k$, not all zero, such that
$$
\sum_{i=0}^n c_ix^im=0
$$
hence $pm=0$, where
$$p=\sum_{i=0}^n c_ix^i$$
so $m$ is a torsion element.

Title: Finite dimensional k[x]-module is torsion module

Body:
Let $k$ be a field and $R=k[x]$. Let $M$ be a $k[x]$-module which is finite dimensional as a $k$-vector space. I want to prove $M$ is a torsion module.

I’m wondering how to build a relationship between basis and torsion, because from the question, all I know is $M$ has a basis of size $n$ ( assume the dimension is $n$).


问:
Proof of theorem about sequence. Check my proof.
Prove this theorem:

Let $s_n, t_n$ be sequences on real number. If there exist $N\in \mathbb N$ such that $0\leq s_n\leq t_n$ and $\lim \limits_{n\to \infty} t_n=0$ for all $n\geq N$, then $\lim \limits_{n\to \infty} s_n=0$.

I have tried as below.

$\lim \limits_{n\to \infty} t_n=0$ means for all $\varepsilon>0$ there exist $N\in \mathbb N$ such that
$$\vert t_n\vert<\varepsilon$$

for all $n\geq N$.

Since $s_n\leq t_n$, we have $\vert s_n\vert \leq \vert t_n\vert$. So,
$$\vert s_n-0\vert=\vert s_n\vert \leq \vert t_n\vert<\varepsilon.$$
Thus,
$$\lim \limits_{n\to \infty} s_n=0.$$

I'm not sure with my answer. Anyone can check my proof? Is it correct proof?

答:
Your proof is correct, but perhaps you want to rewrie the statement as such:

Let $s_n,t_n$ be sequences of real numbers. Suppose that $\lim_{n\to\infty}t_n=0$ and there exists $N\in\mathbb N$ so that $0\leq s_n\leq t_n$ for all $n>N$. Then $\lim_{n\to\infty}s_n=0$.

The point is that having limiting value $0$ is a property of the sequence $(t_n)$ independent of $n$, so your original phrasing sounds awkward. Another minor criticism I would have is that when you reiterated the definition of $\lim t_n=0$, you implicitly assumed the $N$ in the definition is the same number as the $N$ given to you in the question, which is not necessarily always true---but in this case your proof is not affected. You also want to make explicit that the logical implication $s_n\leq t_n\implies |s_n|\leq |t_n|$ is dependent on the fact that $s_n,t_n\geq 0$, because otherwise this is not true (e.g. $-3<-1$ but $|{-}3|>|{-}1|$).


问:
Zorn's lemma: Given a non-empty, closed, convex set E in Hilbert space $\mathrm{H},$ it must obtain a unique element of smallest norm
Given a non-empty, closed, convex set E in Hilbert space $\mathrm{H},$ it must obtain a unique element of smallest norm.

A classic proof is use parallelogram law. But I want to show this using Zorn's lemma directly. Say let $P$ be set of all closed balls that intersects with $E$. Define ordering as containment of the balls. For each chain, an upper bound can be intersection of all the balls in this chain because intersection of all balls with $E$ is still non-empty (Show that the infinite intersection of nested non-empty closed subsets of a compact space is not empty). This gives existence of a maximal element: the smallest ball that has non-empty intersection with $E$. But this does not even uses completeness. Where went wrong? Is there any way I can do this?

答: Yes, but nothing ensures that $E$ is compact. For example, let $E$ be the closed unit ball of your Hilbert space. Then it is non-empty, closed and convex, but non-compact when $H$ is infinite-dimensional. Hence, your argument with the finite-intersection property is flawed.


问:
Calculating $a^p+b^p$ given $a^2+b^2=c$
If one has $a^2+b^2=c$ where $a$,$b$ and $c$ are real numbers, is there any way to calculate $a^p+b^p$ where $p$ may be any real number? If there is, would you please explain with an example with $p=3$?</p>

答:
There are times where $c$ does uniquely specify $a,b$, which is maybe worth mentioning.  Suppose $a,b,c$ are all positive integers.  Then we can look at the sum of squares function for $c$, and we know that (up to order) $a,b$ will be unique when $c$ has exactly one prime factor of the form $4k+1$, and each prime factor of the form $4k+3$ appears to an even power.

In other words, if $c$ is of that form, and $a^2+b^2=c$, $a>0$, $b>0$, then $a^p+b^p$ is completely determined for any $p$.

Title: Calculating $a^p+b^p$ given $a^2+b^2=c$


问:
To show that function is constant
Let $f$ be defined on $\mathbb{R}$ and suppose that |$f(x)$ - $f(y)$| $\leq$ $(x-y)^2$ $x,y \in\mathbb{R}$. Here I have to show that $f$ is a constant function.

I think I have to show that $f'(x)$ = 0 for all $x$.
But I don't know from where to start this. I tried taking it as (|$f(x)$ - $f(y)$|/|$x$-$y$|) $\leq$ |$x$ - $y$|. Am I right in doing so? Any hint or suggestion will be helpful. Thanks

答:
Ya it suffices to prove that $f'(x)$ = 0  $\forall x \in \mathbb{R}$. You can use $\epsilon - \delta$ approach to prove this statement.

Let $\epsilon > 0$ be given then you need to show that $\exists$ $\delta > 0$ such that

$|\frac{f(x+h) -f(x)}{h} - 0| = |\frac{f(x+h) -f(x)}{h} | < \epsilon$ when $|h|<\delta$
but, by the definition of the function given above we have that

$|\frac{f(x+h) -f(x)}{h}| \leq \frac{h^2}{|h|} = |h|$ . So choose $\delta = \epsilon$, and if

$|h|< \delta$ we have $|\frac{f(x+h) -f(x)}{h} - 0| < \epsilon$. Since $\epsilon > 0$ is arbitrary we have that $f'(x) = 0$

Title: To show that function is constant


问:
Finite dimensional k[x]-module is torsion module
Let $k$ be a field and $R=k[x]$. Let $M$ be a $k[x]$-module which is finite dimensional as a $k$-vector space. I want to prove $M$ is a torsion module.

I’m wondering how to build a relationship between basis and torsion, because from the question, all I know is $M$ has a basis of size $n$ ( assume the dimension is $n$).

答:
By contrapositive:

If $M$ has non-torsion elements as a $k[X]$-module, it can't have finite dimension as a $k$-vector space.

Indeed, let $m\in M$ such that $\operatorname{Ann}_{k[X]}=\{0\}$. Then, the submodule $k[X]m$ is  isomorphic to $k[X]$ as a $k[X]$-module, hence as a $k$-vector  space, and the latter is not finite dimensional over $k$.