问:
Prove that all members of a recursive sequence are coprime
The problem is:

$k$ is a positive integer. Sequence $(a_n)$ is defined by: $$a_1=k+1 \\ a_{n+1}=a_n^2-a_n k + k. $$
Prove that every pair of members of that sequence is coprime.

By induction I got that $a_{n+1}=a_1 a_2 ... a_n + k $. A special case of that sequence is when $k=1$, it's called Sylvester's sequence. In that case, it's pretty easy to solve ($a_n\equiv 1\pmod b, b=a_m, m>n$ .) Otherwise, I'm completely stuck.

答: This is reminiscent to the trick used by Euclid to prove the infinitude of primes.  See, none of $a_1,\dots,a_n$ can divide $a_1a_2\cdots a_n+k$, because none of them divides $k$.


问:
Unicity of solution of $\left(1+x\right)^\alpha = 1+\alpha x$
Consider the two Bernoulli (thank you @ErikSatie) inequalities for $x>-1$:
$$
0<\alpha<1 \quad \to\quad \left(1+x\right)^\alpha\le 1+\alpha x,\tag{1}
$$
$$
\alpha<0\lor 1<\alpha  \quad \to\quad \left(1+x\right)^\alpha\ge 1+\alpha x.\tag{2}
$$

It remains me to prove that the inequality is strict for every $x\ne 0$ (keeping $x>-1$)

It seams to me that this point is more difficult than the inequalities themselves.
I can't use derivation.

In the following  picture:

I plotted different cases of $\alpha\in\{-1,-0.1,0.1,0.5,0.9,1.1,2\}$, and with a different color the asymptotic cases $\alpha=0$ (yellow) and $\alpha=1$ (green).

What I have to do is to prove that in $(-1,\infty)$ there is only a single solution for
$$
\left(1+x\right)^\alpha = 1+\alpha x,\quad \alpha\in (-\infty,0)\cup(0,1)\cup (1,\infty).\tag{3}
$$

EDIT 1 The case (1) derives directly for the corresponding aspect for the arithmetic-geometric inequality (which is used to prove it).

答:
Consider the function $f(x) = (1+x)^{\alpha}-1-\alpha x$. We know that $f(0)=0$, and we essentially wish to show that for $\alpha\neq 0,1$, $x=0$ is the only zero of $f(x)$.

Firstly, we can easily see that $f(x)$ is twice differentiable on $\mathbb{R}/\{0\}$ with first derivative $f'(x) = \alpha(1+x)^{\alpha-1}-\alpha$ and second derivative $f''(x)=\alpha(\alpha-1)(1+x)^{\alpha-2}$.

From this, we can see that $f'(0)=0$, and $f''(x)\neq0$ except at $\alpha=0,1$ where it is zero throughout.

Now, if there exists another $x_0\neq 0$ at which $f(x_0)=0$, then by Rolle's theorem, we would have a $c$ between $x_0$ and $0$ such that $f'(c)=0$.

But, then as $f'(0)=0$, we can again apply Rolle's theorem to get a $d$ between $c$ and $0$ at which $f''(d)=0$. This is impossible, unless $\alpha=0,1$.

Hence, the Bernoulli inequality is strict for every $x\neq 0$ for $\alpha\neq 0,1$

ContentLicense: CC BY-SA 4.0


答:
Let $m > n$. Suppose there exists a prime $p$ such that $p|a_{m}$ and $p |a_{n}$.

Then using your induction result, $p|a_{1}a_{2}\cdot\cdot\cdot a_{m-1}+k$, and $p|a_{1}a_{2}\cdot\cdot\cdot a_{n-1}+k$.

Hence, $p|(a_{1}a_{2}\cdot\cdot\cdot a_{m-1})-(a_{1}a_{2}\cdot\cdot\cdot a_{n-1})=(a_{1}a_{2}\cdot\cdot\cdot a_{n-1})(a_{n}a_{n+1}\cdot\cdot\cdot a_{m-1}-1)$.

If $p|(a_{1}a_{2}\cdot\cdot\cdot a_{n-1})$, then using the above, $p|k$ which implies for any $s$ that $p|a_{s+1} \implies p|a_{s}$ since $a_{s+1}=a_s^2-a_s k + k.$

By repeating this process, $p|a_{1} = k+1 \implies p|1$, a contradiction.

If $p|(a_{n}a_{n+1}\cdot\cdot\cdot a_{m-1}-1)$, then since $p|a_{n}, p|(a_{n}a_{n+1}\cdot\cdot\cdot a_{m-1}) \implies p|1$, a contradiction.

Hence there does not exist such a prime and all pairs are coprime.

Title: Prove that all members of a recursive sequence are coprime

Body:
The problem is:

$k$ is a positive integer. Sequence $(a_n)$ is defined by: $$a_1=k+1 \\ a_{n+1}=a_n^2-a_n k + k. $$
Prove that every pair of members of that sequence is coprime.

By induction I got that $a_{n+1}=a_1 a_2 ... a_n + k $. A special case of that sequence is when $k=1$, it's called Sylvester's sequence. In that case, it's pretty easy to solve ($a_n\equiv 1\pmod b, b=a_m, m>n$ .) Otherwise, I'm completely stuck.


问:
Show that the SU(2) group is a Lie group
How can I prove that the SU(2) is a Lie group with the Pauli matrices as generators?</p>

答:
First study the group $U(2)$. This is the set of matrices $A$ satisfying
$$A \overline{A}^t=1,$$ where the overline denotes complex conjugation of the entries and the $t$ is the transpose. The map $A \mapsto A \overline{A}^t$ is a differentiable map from the set of all $2$ by $2$ complex matrices to the set of all matrices $C$ satisfying $\overline{C}^t=C$. Its derivative at a matrix $A$ is the linear map
$$B \mapsto A \overline{B}^t+B \overline{A}^t.$$ One verifies (do it first for $A=1$ where it is trivial and then derive the general case from this) that this is surjective at all points $A$ of $U(2)$, which is therefore an embedded real submanifold of $\mathbb{C}^4$ by the implicit function theorem. It is immediate that the group operations are smooth: they are given by rational functions of the entries of the matrix. This proves that $U(2)$ is a Lie group.

Now $SU(2)$ is the kernel of the Lie group homomorphism
$$\mathrm{det}: U(2) \rightarrow S^1.$$ It is therefore an embedded subgroup.

I don't believe $SU(2)$ is generated as an abstract group by the Pauli matrices (at least, not by the Pauli matrices I know, which correspond via the isomorphism between the unit quaternions and $SU(2)$ to the quaternions $i$, $j$, and $k$). You can modify the Pauli matrices to obtain generators of the Lie algebra of $SU(2)$, though, and the one-parameter subgroups corresponding to these do generate $SU(2)$. If you clarify (give a precise definition of Pauli matrices) and it's still useful to you, I will give more details.

ContentLicense: CC BY-SA 3.0


问:
Banach spaces, $\lVert\cdot\rVert_X$ and $\lVert\cdot\rVert_{X+X}$ are equivalent.
Let $(X,\lVert\cdot\rVert_X)$ and $(Y,\lVert\cdot\rVert_Y)$ be Banach spaces. Then as I understand, $X+Y$ endowed with $$\lVert v\rVert_{X+Y}=\inf\limits_{a+b=v\\ a\in X\\b\in Y}\lVert a\rVert_X+\lVert b\rVert_Y$$ is also a Banach space. Now I want to show that if $X=Y$ than the two norms are equivalent. Taking $a = v, b = 0,$ I have $\lVert v\rVert_{X+X}\le \lVert v\rVert_X$ but how to show $\lVert v\rVert_X\le C\lVert v\rVert_{X+X}$?

答:
In general your notations don't make sense. You cannot add elments of two Banach spaces in general. But if $X$ and $Y$ are subspaces of one big vector space then this question makes sense.

$\|v\|_X \leq \|a\|_X+\|b\|_X$ for all choices of $a$ and $b$ with $v=a+b$. Taking infimum over all choices of $a$ and $b$ we get $\|v\|_X \leq \|v\|_{X+X}$.


问:
Simple Logical Simplification - Distributive Law
I can't seem to understand the hopefully simple step of getting from :

$(a \land b \land c) \lor (a \land b \land \lnot c)$

to:

$(a \land b) \land (c \lor \lnot c)$

This step is in the answers and only has '(distribution)' written next to it, however even when breaking this down into several parts with Distributive Laws, I got nothing close to the expected solution (most likely my fault rather than teacher's). So any help explaining this step would very much be appreciated. Thank you!

答: Apply the distributive law to $d \wedge (c\vee \neg c)$ where $d=a\wedge b$. Then you get $$\begin{array}{rcl}d\wedge (c\vee \neg c)&=&(d\wedge c )\vee (d\wedge \neg c)\\&=&((a\wedge b)\wedge c)\vee ((a \wedge b)\wedge \neg c)\\&=&(a\wedge b\wedge c)\vee (a\wedge b \wedge \neg c). \end{array}$$Hope this helps!


答:
It might help to put it into words:

I will use $TRUE/FALSE$ notation.

$A,B,C$ are $TRUE$ or $A,B$, NOT $C$ are $TRUE$. In either case $A$ and $B$ are $TRUE$.

Then either $C$ is $TRUE$ or NOT $C$ is $TRUE$ since this is an or statement and may be either one.

Title: Simple Logical Simplification - Distributive Law

Body:
I can't seem to understand the hopefully simple step of getting from :

$(a \land b \land c) \lor (a \land b \land \lnot c)$

to:

$(a \land b) \land (c \lor \lnot c)$

This step is in the answers and only has '(distribution)' written next to it, however even when breaking this down into several parts with Distributive Laws, I got nothing close to the expected solution (most likely my fault rather than teacher's). So any help explaining this step would very much be appreciated. Thank you!


问:
determining the value of e by using mathematical induction
using the fact that $n!>2^{n}$ $\forall n\ge 4$ conclude that $e<\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\displaystyle\sum_{n=4}^{\infty}\frac{1}{2^{n}}$

where $e:=\displaystyle\sum_{n=0}^{\infty}\frac{1}{n!}=1+\frac{1}{1!}+....$

should I inverse the inequality

$\frac{1}{n!}<\frac{1}{2^{n}}$ $\forall n\ge 4$ ??

Please help

答: Write $e=1+1/1!+1/2!+1/3!+\sum_{k=4}^\infty 1/n!$ and use your inequality for the summands of the summation. </p>


问:
Fuctions: into, onto or both
Determine whether each of the following functions, defined from $\mathbb{Z}\times\mathbb{Z}\to\mathbb{Z},$ is one-to-one, onto, or both. Explain your answers.

i) $f(x, y) = x^2 + 1$

ii) $g(x, y) = x + y + 2$

Having a lot of confusion with this question, any help will be appreciated, Thank you!

答:
Here $f:\mathbb{Z}\times\mathbb{Z}\to\mathbb{Z}$ defined by $f(x,y)=x^2 + 1$ is neither one one nor onto.

Hint: $f(1,0)=2$ and $f(-1,0)=2$ but $(1,0)\neq(-1,0).$
Also $0\in\mathbb{Z},$ but there does not exist any $(x,y)\in\mathbb{Z}\times\mathbb{Z}$ such that $f(x,y)=0.$

Similarly, apply definitions to check for the second function.

Title: Fuctions: into, onto or both


问:
Polynomial $p\in\mathbb Q[x,y]$ such that $p(e,\pi)=0$.
Is there any polynomial $p\in\mathbb Q[x,y]$ that $p(e,\pi)=0$?

I was inspired by the fact that $\pi^4+\pi^5-e^6\approx 0$ and this lead me to the aforementioned question. Since there is no $p\in\mathbb Q[x]$ that $p(\pi),p(e)=0$ and the non-polynomial $f(e,\pi)=e^{i\pi}+1=0$. Is the statement in question a well-known result from somewhere? Any help would be appreciated, but please keep it as a (little as possible) hint so that I can study it further. Thank you in advance.

答:
In the comments already a few easy examples are mentioned. Just to list them in an answer here as well.

You could take $p(x, y) = xy - 1$ and then pick any transcendental $x$ and set $y = 1/x$ (as suggested by Pedro Tamaroff).

A similar strategy: take $p(x, y) = x - y^2$ and again pick any transcendental $x$ and now set $y = \sqrt{x}$.

You did also ask about a polynomial $p(x, y)$ such that $p(e, \pi) = 0$. This is still an open problem (as mentioned in the comments as well). It is conjectured that there is no such $p$. That is, that $e$ and $\pi$ are algebraically independent.

You might be interested in Schanuel's conjecture, which is the following much more general statement.

Let $z_1, \ldots, z_n \in \mathbb{C}$ be linearly independent over $\mathbb{Q}$ then the transcendence degree of $(z_1, \ldots, z_n, e^{z_1}, \ldots, e^{z_n})$ (over $\mathbb{Q}$) is at least $n$.

So to see how this implies that $e$ and $\pi$ are algebraically independent we can consider $z_1 = 1$ and $z_2 = \pi i$. Then by Schanuel's conjecture we would have that $(\pi i, 1, -1, e)$ has transcendence degree at least 2. Since $1$ and $-1$ are algebraic we have that $\pi i$ and $e$ must be algebraically independent, and hence $\pi$ and $e$ must be algebraically independent.

The conjecture is much more general than that and is studied a lot in model theory. This is also mentioned in the MO post that leoli1 links to. One nice thing to highlight there is that Zilber proved that there is indeed an algebraically closed field $\mathbb{B}$ with an exponential function (i.e. a group homomorphism from the additive group of $\mathbb{B}$ to the multiplicative group of $\mathbb{B}$), which satisfies Schanuel's conjecture.


问:
Please explain how to derive the PGF of a 6 sided dice
I understand the PGF is
$$ P(s) = \frac{s}{6} \frac{1-s^6}{1-s} $$
Can someone explain all the steps to get there

Body:
We have $p(1)=...=p(6)=\frac16$, $p(0)=0$ and for $k>6$, $p(k)=0$

The PGF is

$G(s)=\sum_{k=0}^{+\infty} p(k) s^k=\sum_{k=1}^{6} p(k) s^k$

So $G(s)=\frac16 \sum_{k=1}^{6} s^k$

We recognize a geometric sum:

$G(s)=\frac{s}{6} \frac{1-s^6}{1-s}$


问:
optimizing 3 variable function
I have the following function:

m = f(x,y,z) = (a * (1 + x/100) * b - d * (1 + y/100)) / (h * (1 + z/100))
where a, d, h, b are positive constants and the variables are x,z,y (positive too)

My question is: how can I find for given x (between 1 and 1000), the optimal y,z so that m is as low as possible?

It seems that for different values x, some times y is more "important" (in lowering m) and others z is more important. I also don't have a clue how to visualize this.

Thank you in advance.

答:
if a,d, h,b are positive and x,y,z are also positive and m is not constrained than you need to make (a * (1 + x/100) * b as small as possible meaning x should be very neare to zero...
and you need to try to make d * (1 + y/100)) / (h * (1 + z/100)) as large as possible..meaning y/z should be as large as it could be...y/z  goes to infinity
a*(1+0)- infinity
your answer should be infinite...</p>

Title: optimizing 3 variable function


问:
Why does the diagonal morphism of a map between affine schemes correspond to the following morphism of rings?
Suppose I have a morphism of affine schemes $(f, f^\sharp) : \operatorname{Spec} A \to \operatorname{Spec} B$, then my question is - why does the diagonal morphism $(\Delta_f, \Delta_f^\sharp) : \operatorname{Spec} A \to \operatorname{Spec} A \times_{\operatorname{Spec} B} \operatorname{Spec} A = \operatorname{Spec} (A \otimes_B A)$ correspond the following morphism of rings $\phi : A \otimes_B A \to A$ given by $\phi(a \otimes a') = a\cdot a'$?

This was my attempt to try and asnwer that.
I know that we have the result that $$\operatorname{Hom}_{\textsf{Sch}}\left(\operatorname{Spec} A, \operatorname{Spec} (A \otimes_B A)\right) \cong \operatorname{Hom}_{\textsf{Ring}}(A \otimes_B A, A)$$
with the bijection given by $(g, g^\sharp) \mapsto g^\sharp_{\operatorname{Spec} A}$ so one possible way to show that the diagonal morphism does correspond to the claimed morphism of rings would be to show that ${\Delta_f}_{_{\operatorname{Spec} A}}^\sharp = \phi$ but I was unable to actually verify this.

Another thing I was thinking of was the following. If I let $(p_1, p_1^\sharp) : \operatorname{Spec} (A \otimes_B A) \to \operatorname{Spec} A$ denote the projection map that comes with the fibered product, then we know that $p_1 \circ \Delta_f = 1_{\operatorname{Spec} A}$ the identity map on $\operatorname{Spec} A$. Now I guess, (but I cannot show), that the morphism $(p_1, p_1^\sharp)$ corresponds to the morphism of rings $\iota : A \to A \otimes_B A$ given by $\iota(a) = a \otimes_B 1_A$, because then we see that $\phi \circ \iota = 1_A$ the identity map on $A$.

If it were the case that the morphism $(p_1, p_1^\sharp)$ corresponds to $\iota$, then since taking global sections is a contravariant functor (and really what that bijection above is) I would see that ${\Delta_f}_{_{\operatorname{Spec} A}}^\sharp\circ \iota = 1_A$ as well, but this isn't enough to prove that ${\Delta_f}_{_{\operatorname{Spec} A}}^\sharp = \phi$.

So I have in addition two further follow up questions in addition to the original one:

How does one prove that the morphism $(p_1, p_1^\sharp)$ corresponds to $\iota$?

Is there a more efficient way to prove that a morphism of schemes corresponds to a morphism of rings?

If it is relevant, you can assume that the construction of the fibered product I am working with is the same one given in Theorem 3.3 in Hartshorne's book.

答:
Question: "How does one prove that the morphism (p1,p♯1) corresponds to ι?"

Answer:  By the universal property of the fiber product, it follows the diagonal map $\Delta: X\rightarrow X\times_S X$ at the level of schemes, is the unique map commuting with the projection maps $p,q:X\times_S X \rightarrow X$. The multiplication map has the same property: The map $m:A\otimes_k A\rightarrow A$ defined by $m(a\otimes b):=ab$ is the unique map commuting with the maps $p,q:A \rightarrow A\otimes_k A$ defined by $p(a):=a\otimes 1, q(a):=1\otimes a$. Hence it follows the maps $\Delta$ and $m$ determine each other when $X:=Spec(A)$.


答: If you start from $n! \gt 2^n$ and divide both sides by $2^nn!$ you get the inequality you want.</p>

Title: determining the value of e by using mathematical induction

Body:
using the fact that $n!>2^{n}$ $\forall n\ge 4$ conclude that $e<\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\displaystyle\sum_{n=4}^{\infty}\frac{1}{2^{n}}$

where $e:=\displaystyle\sum_{n=0}^{\infty}\frac{1}{n!}=1+\frac{1}{1!}+....$

should I inverse the inequality

$\frac{1}{n!}<\frac{1}{2^{n}}$ $\forall n\ge 4$ ??

Please help


问:
$2020$ AIME I Problem $13$ Identifying Cyclic Quadrilaterals
Link: https://artofproblemsolving.com/wiki/index.php/2020_AIME_I_Problems/Problem_13#Solution_1

I will be referring to Solution 1.

I understand the solution completely, except for the part where it says

$GB=HC=1$

I don't understand how this is true. I tried to prove it using Power of a Point, angle chasing, and Law of Sines/Cosines, but none of these steps got me far.

Can someone please explain why it is true?

Thank you.

答:
$\triangle ABD : $ the angle bisector of $\angle B$ meets perpendicular bisector of opposite side $AD$, on its circumcircle. This point is $E$. So $ABDE$ is cyclic as is $ACDF$.

$$\angle AHB = \dfrac{1}{2}\angle AED = 90-\dfrac{B}{2}=\angle HAB$$

Conclude $AB=BH$ and similarly $CG=AC$ from which $GB=HC=1$ follows.


问:
If $S$ be any set of intervals I in $\mathbb R$ and $J=\bigcap_{I \in S}I$ show $(\alpha,\beta)\subset J\subset [\alpha,\beta]$
Let $S$ be a set of intervals $I$ in $\mathbb R$ s.t. $\exists I_1,I_2\in S$ where $I_1$ is bounded above and $I_2$ is bounded below.

If $J=\bigcap_{I\in S}I$ then show that $\exists \alpha,\beta \in \mathbb R$ with $\alpha\le \beta$ s.t. $(\alpha,\beta)\subset J \subset [\alpha,\beta]$

答:
Let $b$ be the right end-point of $I_1$ and $a$ be the left end-point of $I_2$. Clearly $I_1\cap I_2\subset [a,b]$ if $a\le b$. When $a>b,I_1\cap I_2=\emptyset$. Now $J=\bigcap_{I\in S}I\subset I_1\cap I_2$. Thus $J\subset [a,b]$ if $a\le b$ and $J=\emptyset$ if $a>b$. In any case $J$ is a bounded set in $\mathbb R$.

If $J=\emptyset$ then $(c,c)\subset J\subset [c,c] \,\,\forall c \in \mathbb R$. If $J$ is a singleton set, say $\{x\}$ then $(x,x)\subset J\subset [x,x]$.If $J$ contains more than one element, then $\exists x,y\in J$ s.t. $x<y$.  $J\subset I\,\,\forall I \in S \Rightarrow x,y\in I \,\,\forall I\in S$. Therefore $[x,y]\subset I\,\,\forall I\in S$ Hence $[x,y]\subset \bigcap_{ I\in S}I=J$.

By characterization theorem of intervals,  $J$ is an interval. As $J$ is a bounded interval it has a left end-point,say $\alpha$ and right end-point, say $\beta$ in $\mathbb R$ s.t. $(\alpha,\beta)\subset J\subset [\alpha, \beta]$.


答:
We want

$$f(x)=x+x^2+x^3+\cdots +x^6.$$

By long division we can prove that

$$\frac{1}{1-x}=1+x+x^2+x^3+\cdots$$

If we multiply the above expression by $1-x^7$ we are going to see a pattern where all the $x^n$ exponents will cancel out after $x^6$:

$$\begin{align}&(1-\color{red}{x^7})+(x-\color{blue}{x^8}) +(x^2-x^9)+\cdots+(\color{red}{x^7}-x^{14})+(\color{blue}{x^8}-x^{15})+\cdots\\[2ex]&=1+x+x^2+\cdots+x^6
\end{align}$$

Now we have to get rid of the $1$ in front by multiplying the series by $x.$ That will require reducing the exponent in the numerator exponent from $1-x^7$ to $1-x^6.$

This leaves us with $x\frac{1-x^6}{1-x}.$ And since the probability of each outcome is $1/6,$ the expression is derived:

$$f(x)=\frac 1 6 \frac{x(1-x^6)}{1-x}$$

Title: Please explain how to derive the PGF of a 6 sided dice

Body:
I understand the PGF is
$$ P(s) = \frac{s}{6} \frac{1-s^6}{1-s} $$
Can someone explain all the steps to get there


问:
Cat and Mice puzzle!
I was studying an Iranian olympiad book problems then I faced a very intersting puzzle, I wanted to solve this but faced lots of problems. I'd like to share this with you and know your mind.

We have divided the following figure garden into two areas with fences.We have drawn the fences on a simple and closed curve.
Only three parts of the garden are known and the rest of the garden is covered with trees.
Three cats are waiting for the mice.
All three are in the outer area and can not cross the fence. How many mice might be trapped? (Iranan Math Olympiad).

答:
Hint:
The walls form a closed curve, so every time you cross a wall you go from the outside area to the inside area, or vice versa.</p>

Title: Cat and Mice puzzle!


答:
Since the fence is a simple closed curve, and the cats are on the outside, we can tell which  mice are on the inside by counting how fences a nearby cat has to cross to reach a particular mouse.

Every time you cross a fence, you go from the inside of the curve to the outside, or the reverse. If a cat and a mouse are separated by an odd number of fence sections, the mouse is safe; if by an even number the mouse may be caught.

By inspection, only one mouse is vulnerable.

Title: Cat and Mice puzzle!

Body:
I was studying an Iranian olympiad book problems then I faced a very intersting puzzle, I wanted to solve this but faced lots of problems. I'd like to share this with you and know your mind.

We have divided the following figure garden into two areas with fences.We have drawn the fences on a simple and closed curve.
Only three parts of the garden are known and the rest of the garden is covered with trees.
Three cats are waiting for the mice.
All three are in the outer area and can not cross the fence. How many mice might be trapped? (Iranan Math Olympiad).


问:
Can you give an example of a continuous function with this property?
Can you give an example of a continuous function $f:[0, \infty)$ $\to$ $\Bbb R$
such as $\lim_{b\to \infty}$ $\int_0^b{f(x)dx}$ exists but $f$$\notin$ $L^{1}$ ([0, $\infty$)) ?

($f$$\notin$ $L^{1}$ ([0, $\infty$)) means that $f$ is not Lebesgue integrable in that interval)

答:
Assuming that you mean finitely exists when you say limit exists (otherwise constant function works), an example is
\begin{equation}
f(x)=\begin{cases}
1&\text{if $x=0$}\\
\dfrac{\sin x}{x}&\text{if }x>0
\end{cases}
\end{equation}
This is a continuous function which is conditionally integrable on $\Bbb R^+$.


问:
What is the last non-zero digit of $((\dots(((1!)!+2!)!+3!)!+\dots)!+1992!)!$?
What is the last non-zero digit of $((\dots(((1!)!+2!)!+3!)!+\dots)!+1992!)!$?

Clarification of the given expression:

Let $A_1=(1!)!$

To get $A_2$, we add $2!$ to $A_1$ then we take the factorial of that expression. So, $A_2=((1!)!+2!)!$

To get $A_3$, we add $3!$ to $A_2$ then we take the factorial of that expression. So, $A_3=(((1!)!+2!)!+3!)!$

To get $A_4$, we add $4!$ to $A_3$ then we take the factorial of that expression. So, $A_4=((((1!)!+2!)!+3!)!+4!)!$

and so on.

Now we need to find the last non-zero digit of $A_{1992}$. I am not asking for a solution, but asking for some hints how shall I begin.

Your help would be appreciated. THANKS.

答:
Let $a_n$ denote the last nonzero digit of $n!$; we need only consider $n\ge5$.

Prove $a_n\in S:=\{2,\,4,\,6,\,8\}$.

Prove $a_{10k+5}=a_{10k+14}$.

Determine the order of the permutation $\sigma$ of $S$ for which $a_{10k+15}=\sigma(a_{10k+14})$.

Find a fairly small upper bound $p$ on the period of $a_n$.

Convince yourself $a_p,\,\cdots,\,a_{2p-1}$ is a period.

Explain why the original problem's answer is $a_p$, then compute it.


答:
Let $f$ be the piecewise linear interpolation between the points $(n,0)$ and  $(n+\frac12,(-1)^n)$, both with $n\in\Bbb N_0$.
Then $\int_0^bf(x)\,\mathrm dx$ will behave like the alternating harmonic series and converge, whereas $\int |f|$ is like the normal harmonic series and diverges (both when taking the limit of Riemann over finite intervals and when taking the Lebesgue integral  directly).

Title: Can you give an example of a continuous function with this property?

Body:
Can you give an example of a continuous function $f:[0, \infty)$ $\to$ $\Bbb R$
such as $\lim_{b\to \infty}$ $\int_0^b{f(x)dx}$ exists but $f$$\notin$ $L^{1}$ ([0, $\infty$)) ?

($f$$\notin$ $L^{1}$ ([0, $\infty$)) means that $f$ is not Lebesgue integrable in that interval)


问:
$P(X_1 < X_2 < X_3)$ for Exponential Random Variables
Let $X_1 \sim \exp(\lambda_1)$, $X_2 \sim \exp(\lambda_2)$ and $X_3 \sim \exp(\lambda_3)$ be independent random variables, where the exponential density I'm using is
$$
f_X(x) = \frac{1}{\lambda}e^{-x/\lambda}
$$

I'm trying to calculate $P(X_1 < X_2 < X_3)$ but am stuck.  So far, I've rewritten this as
$$
P(X_1 < X_2 < X_3) = P(X_1 < X_2 \cap X_2 < X_3) = P(Y_1 < 0 \cap Y_2 < 0) \tag{$*$}
$$
where $Y_1 = X_1 - X_2$ and $Y_2 = X_2 - X_3$.  I've also worked out the marginal distributions for $Y_1$ and $Y_2$.  Here's the pdf for $Y_1$:
$$
f_{Y_1}(y) = \begin{cases}
\dfrac{e^{-y/\lambda_1}}{\lambda_1 + \lambda_2}, \qquad y > 0 \\[6pt]
\dfrac{e^{y/\lambda_2}}{\lambda_1 + \lambda_2}, \qquad y \leq 0
\end{cases}
$$

But, I need the joint pdf of $(Y_1, Y_2)$ to calculate $(*)$ since the two are not independent.  I can write the vector $(Y_1, Y_2)$ as
$$
\begin{pmatrix}
Y_1 \\
Y_2
\end{pmatrix}
=
\begin{pmatrix}
1 & -1 & 0 \\
0 & 1 & -1
\end{pmatrix}
\begin{pmatrix}
X_1 \\
X_2 \\
X_3
\end{pmatrix}
$$
but this doesn't seem to help me work out the joint distribution as it does for linear combinations of normals.

答:
To simplify notation, I'll use the parameters $a_k:=1/\lambda_k$ to express the exponential densities. Since the r.v.s are independent, their joint pdf is simply the product of the individual pdfs. Hence we obtain the relevant probability by integrating over the region $x\leq y\leq z$:

$$
P(X_1<X_2<X_3)
=\int_{x_3=0}^\infty \int_{x_2=0}^{x_3}\int_{x_1=0}^{x_2} f_{X_1}(x_1)f_{X_3}(x_3)f_{X_3}(x_3)\,dx_1\,dx_2\,dx_3
$$
(Note that $x\geq 0$ by the nature of the exponential pdf.) We now compute these integrals inside-out:
\begin{align}
\int_{x_1=0}^{x_2}f_{X_1}(x_1)\,dx_1&=\int_0^{x_2}a_1 e^{-a_1 x_1}\,dx_1\\&=1-e^{-a_1 x_2},\\\\
\int_{x_2=0}^{x_3}f_{X_2}(x_2)(1-e^{-a_1 x_2})\,dx_2
&=\int_{x_2=0}^{x_3}a_2 e^{-a_2 x_2/\lambda_2}(1-e^{-a_1 x_2})\,dx_1\\
&=1-e^{-a_2 x_3}-\frac{a_2}{a_1+a_2}\left(1-e^{-(a_1+a_2)x_3}\right),\\\\
\int_{x_3=0}^{\infty }f_{X_3}(x_3)\Big[1-e^{-a_2 x_3}-&\frac{a_2}{a_1+a_2}\left(1-e^{-(a_1+a_2)x_3}\right)\Big]\,dx_3\\&=\int_{x_3=0}^{\infty }a_3 e^{-a_3 x_3}\Big[1-e^{-a_2 x_3}-\frac{a_2}{a_1+a_2}\left(1-e^{-(a_1+a_2)x_3}\right)\Big]\,dx_3\\
&=1-\frac{a_3}{a_2+a_3}-\frac{a_2}{a_1+a_2}\left(1-\frac{a_3}{a_1+a_2+a_3}\right)\\
&=\frac{a_2}{a_2+a_3}-\frac{a_2 }{a_1+a_2+a_3}\\
&=\frac{a_1 a_2}{(a_2+a_3)(a_1+a_2+a_3)}
\end{align}

Returning to the original notation, we have
$$P(X_1<X_2<X_3)=\frac{\lambda_1^{-1} \lambda_2^{-1}}{(\lambda_2^{-1}+\lambda_3^{-1})(\lambda_1^{-1}+\lambda_2^{-1}+\lambda_3^{-1})}=\frac{\lambda_2 \lambda_3^2}{(\lambda_2+\lambda_3)(\lambda_1\lambda_2+\lambda_1\lambda_3+\lambda_2\lambda_3)}.$$


问:
How to prove that propositions are not logically entailed?
Let's consider the next proposition: A⇔B⊨A∨B

I used a truth table to show that there exists model1={A=False, B=False} where (A⇔B) is True but (A∨B) is False. It means A⇔B does Not entail A∨B.

Can you please suggest how to use deduction rules and what is a general mechanism to show that Statement1 does not entail Statement2?

Should I prove that (Statement1 ⊨ ¬Statement2)?
Or prove that (Statement1 ∧ ¬Statement2)? And this will show that (Statement1 ⊨ Statement2) is a contradiction.

答: Deduction rules are used to generate formal proofs $\varphi\vdash\psi$. The existence of a formal proof implies the corresponding  entailment $\varphi\models\psi$. Deduction rules cannot be used to demonstrate non-entailments. For this, you need countermodels (like the row of your truth table).


问:
Number of solutions for $x_1 + x_2 + x_3 + x_4 = 30$ with $(x_1,x_2,x_3,x_4) \in \{-10,...,19,20\}^4$
I have to solve the following problem: calculate the number of solutions for $$x_1 + x_2 + x_3 + x_4 = 30$$ with $$(x_1,x_2,x_3,x_4) \in \{-10,...,19,20\}^4$$.

I know that if the domain is $$\mathbb{N}$$ the solution is as simple as
$$
\binom{n - 1}{k - 1}=\binom{30 - 1}{4 - 1}=\binom{29}{3}
$$

But the negative values are confusing me.
I wrote a little Java program to calculate the result, it displays 17266.
Could somebody please help me?

答:
For any formal Laurent series $\sum\limits_{k=-\infty}^\infty \alpha_k t^k$ in indeterminate $t$, let $[t^n] \sum\limits_{k=-\infty}^\infty \alpha_k t^k$ be a short hand for the coefficient $\alpha_n$ associated with the monomial $t^n$.

If one expand following product of sums,

$$\sum_{x_1=-10}^{20} (tz_1)^{x_1}
\sum_{x_2=-10}^{20} (tz_2)^{x_2}
\sum_{x_3=-10}^{20} (tz_3)^{x_3}
\sum_{x_4=-10}^{20} (tz_4)^{x_4}
$$
and look at coefficients of $t^{30}$ as a finite Laurent series in $z_1,z_2,z_3,z_4$, one will notice
there is an one-one correspondence between solutions of $x_1 + x_2 + x_3 + x_4 = 30$ and monomials $z_1^{x_1} z_2^{x_2} z_3^{x_3} z_4^{x_4}$ with matching exponents.

This means the number of solutions of $x_1 + x_2 + x_3 + x_4 = 30$, let's call it $\mathcal{N}$, can be computed by first setting $z_1, z_2, z_3, z_4$ to $1$ and then extract the coefficients of $t^{30}$ in above expression. In other words,

$$\begin{align}
\mathcal{N} &= [t^{30}] \left( \sum_{k=-10}^{20} t^k\right)^4
= [t^{30}] \left(t^{-40} \sum_{k=0}^{30} t^k \right)^4
= [t^{70}] \left(\frac{1 - t^{31}}{1-t}\right)^4\\
&= [t^{70}] \frac{1 - 4 t^{31} + 6 t^{62} - 4t^{93} + t^{124} }{(1-t)^4}\\
&= [t^{70}]\frac{1}{(1-t)^4}
 -4[t^{39}]\frac{1}{(1-t)^4}
+ 6[t^8]\frac{1}{(1-t)^4}
\end{align}$$
Notice for any positive integer $\ell$, we have  following power series expansion in $t$:
$$\frac{1}{(1-t)^\ell} = \sum_{n=0}^\infty \binom{n+\ell-1}{\ell-1} t^n$$

This implies $\displaystyle\;[t^n]\frac{1}{(1-t)^4} = \binom{n+3}{3}$. As a result,
$$\begin{align}
\mathcal{N}
&= \binom{70+3}{3}
- 4\binom{39+3}{3}
+ 6\binom{8+3}{3}\\
&= \binom{73}{3} - 4\binom{42}{3} + 6\binom{11}{3}\\
&= 17266
\end{align}
$$


问:
Show that if $f'$ is increasing on $(a,b)$, then $f'$ is continuous on $(a,b)$

Let $a < b$ and $f : [a, b] \to \mathbb R$ be differentiable on $[a, b]$. Show that,

(i) if $f'$ is increasing on $(a,b)$, then $f'$ is continuous on $(a,b)$.

I have seen this exercise question in the lecture notes, but I am confused. The statement we want to prove only has the derivative $f'$ but no $f$. Am I supposed to use a property of derivatives?

答:
I'm going to justify @DavidCUllrich's suggestion for a right-neighborhood $(\alpha, \alpha+\delta)$ of any $\alpha \in (a,b)$.  As the collection of such neighborhood $\{(\alpha,\alpha+\delta)\}_{\alpha\in(a,b)}$ covers the open interval $(a,b)$, then we're done.

Apply MVT on the interval $[\alpha+\delta,\alpha+2\delta]$ to establish a point $\xi \in (\alpha+\delta,\alpha+2\delta)$ such that $$f'(\xi) = \frac{f(\alpha+2\delta)-f(\alpha+\delta)}{\delta}.$$

Use the given condition that $f'$ is increasing to conclude that
$$f'(\alpha) \le f'(\alpha+\delta) \le f'(\xi).$$
Combine the result of step 1 to see that we need to establish a relationship between the difference quotient in step 1 and $f'(\alpha)$.  Once that's done, we may apply the Sandwich Theorem to conclude.

Rewrite the difference quotient in step 1 as
\begin{align}
f'(\xi) &= \frac{f(\alpha+2\delta)-f(\alpha+\delta)}{\delta} \\
&= 2\;\frac{f(\alpha+2\delta)-f(\alpha)}{2\delta} - \frac{f(\alpha+\delta) - f(\alpha)}{\delta}.
\end{align}

Then take limit as $\delta \to 0$.  This is possible as $f$ is differentiable on $[a,b]$.  It's not hard to see that the RHS tends to $2f'(\alpha) - f'(\alpha) = f'(\alpha)$.

Apply the Sandwich Theorem to conclude.


问:
ring homomorphism and cardinality
suppose $(F,+,\cdot)$ is a field and $(R,\oplus,*) $ is a ring then there is no ring homomorphism from $F$ to $R$ when $|F|>|R|$

(here all rings assumed commutative and non zero i.e multiplicative identity is different from additive identity)

答:
Here I am not proving it completely but here are the propositions(which you can prove yourself) with which the given statement can be claimed.

s0: $(F,+,\cdot)$ is a field and $(R,\oplus,*) $ with ring homorphism $\phi: F \rightarrow R$

p1: Fields has only nontrivial ideals.

p2: If $\psi$ is a ring homomorphism then $Ker(\psi)$ is Ideal of a domain ring.

p3: If $\psi:R_1 \rightarrow R_2$ is a ring homomorphism then $Ker(\psi) \neq R_1 $.

now from s0, p1, p2, and p3;

s4: $Ker(\phi)=\{0\}$ for s0

p5: if $\psi$ is a ring homomorphism and $Ker(\psi)=\{0\}$ then $\psi$ is injection.

p6: if $\psi: A \rightarrow B $ is injection then $|A|\leq|B|$.

now from s4, p5 and p6,

s7: $|F|\leq|R|$ for s0.

done.

Here s is used for statement and p is used for the proposition.

Title: ring homomorphism and cardinality


问:
How many entries in a bisymmetric matrix can be chosen independently?
A bisymmetric matrix is symmetric along both diagonals (it is both symmetric and persymmetric). A simple geometric argument (filling in by rows, in which for every new iteration, the outer two entries are non-unique) shows that for a bysymmetric n x n matrix, the answer is n + (n-2) + (n-4) + ... + 4 + 2 (for n even) or n + (n-2) + (n-4) + ... + 3 + 1 (for n odd). In other words, like n!! (but with addition rather than multiplication). Is there a way to simplify this?</p>

答: Thanks to Joppy's comment I figured it out: $n(n+2)/4$ for even n, and ($n+1)^2/4$ for odd $n$.


问:
If $\left.\lambda\right|_r$ is the restriction of $λ$ to the ball of radius $r$, can we show $\left.λ\right|_1\ast\left.λ\right|_1=\left.λ\right|_2$?
This shouldn't be to difficult, but I struggle to obtain the following result:

Let $\lambda$ be a finite nonnegative measure on the Borel $\sigma$-algebra $\mathcal B(E)$ of a normed $\mathbb R$-vector space $E$, $$\overline B_r:=\left\{x\in E:\left\|x\right\|_E\le r\right\}$$ and $$\left.\lambda\right|_r(A):=\lambda\left(A\cap\overline B_r\right)\;\;\;\text{for }A\in\mathcal B(E)$$ for $r>0$.

Are we able to show $\left.\lambda\right|_1\ast\left.\lambda\right|_1=\left.\lambda\right|_2$?

As usual, if $\mu_i$ is a finite signed measure on $\mathcal B(E)$, then $$(\mu_1\ast\mu_2)(A):=(\mu_1\ast\mu_2)\left(\left\{(x,y)\in E^2:x+y\in A\right\}\right)=\int\mu_1({\rm d}x)\mu_2(A-x)\tag1$$ for $A\in\mathcal B(E)$.

Intuitively, we may clearly write $$\overline B_2=\left\{x+y:(x,y)\in\overline B_1\times\overline B_1\right\}\tag2,$$ which should somehow immediately yield the result.

So, how can we prove the desired claim?

答:
The claim is false. Here is a simple counter-example:

Consider $\mathbb R$. Then $\overline B_1 =[-1,1]$ and  $\overline B_2 =[-2,2]$. It is easy to see that
$$\overline B_2=\left\{x+y:(x,y)\in\overline B_1\times\overline B_1\right\}$$
However, take $A=[0,1]$. The
$$ (\lambda |_1 * \lambda |_1)(A)= (\lambda \otimes \lambda)(\{(x,y) \in [-1,1]^2: 0 \leqslant x+y \leqslant 1 \}) = \frac{3} {2}$$
On the other hand,
$$ \lambda |_2 (A) = 1 $$


问:
Examples of nonlinear ordinary differential equations with elementary solutions.
I am looking for nice examples of nonlinear ordinary differential equations that have simple solutions in terms of elementary functions. (But are not trivial to find, like, for example, with separation of variables).

A perfect example of what I am looking for is the Lane-Emden equation of index 5:
$$y''+\frac{2}{x}y'+y^5=0,\qquad y(0)=1,y'(0)=0$$
which admits the solution
$$y(x)=\frac{1}{\sqrt{1+x^2/3}}$$

Another very good example is
$$4y^2y'''-18yy'y''+15y'^3=0$$
from which we find
$$y(x)=\frac{1}{(ax^2+bx+c)^2}$$
for some constants $a,b,c$.

Do you have more examples like these? The higher the order, the better. (But I consider, for example, $H(x,y'',y''')=0$ to be of order 1). And it is a plus if it comes from a physics problem.

答:
A very simple non-linear system to analyze is what I like to call the "Parachute Equation" which is essentially

$$\ddot{x}+k\dot{x}^2-g=0 \tag{1}$$

With initial conditions $x(0)=0$ and $\dot{x}(0)=0$.

where $\displaystyle k=\frac{\pi \rho C_d D^2}{8m}$

such that:

$m$ is the mass of the body and parachute,

$\rho$ is the density of the fluid in which the body moves,

$C_d$ is the drag coefficient for the parachute ($1.5$ for parabolic profile and $0.75$ for flat),

$D$ is the effective diameter of the parachute.

$(1)$ admits the solution:

$$x=\frac{1}{k}\left(\log\left(\frac{e^{2\sqrt{gk}t}+1}{2}\right)-\sqrt{gk}{t}\right)$$

$(1)$ can also be converted into velocity form considering $\dot{x}=v$ into

$$\dot{v}+k{v}^2-g=0 \tag{2}$$

Solution of $(2)$ is

$$ v=\sqrt{\frac{g}{k}}\left(1-\frac{2}{e^{2\sqrt{gk}t}+1}\right) $$


答:
If you have two dices and you throw them you have a probability of

1/36 -> sum is 2

2/36 -> sum is 3

3/36 -> sum is 4

...

6/36 -> sum is 7

...

1/36 -> sum is 12

if you do it  with 3 dices:
1/216, 3/216, 6/216, ... (Tartaglia)

For 4 dices is the next diagonal (1, 4, 10, 20, 35)

$$x_1 + x_2 + x_3 + x_4 = 30$$ with $$(x_1,x_2,x_3,x_4) \in \{-10,...,19,20\}^4$$

Is equivalent to:

$$x_1 + x_2 + x_3 + x_4 = 70$$ with $$(x_1,x_2,x_3,x_4) \in \{0,...,30\}^4$$

out of $31*4$, yours is the 70th element, since the mid element is 31*4/2=62, so 124-70=54, where the first is $x_1+x_2+x_3+x_4=0$ so it is $\binom{54+3}{3}=55*56*57/6=29260$

Which makes me think your little java program might have a mistake.

Title: Number of solutions for $x_1 + x_2 + x_3 + x_4 = 30$ with $(x_1,x_2,x_3,x_4) \in \{-10,...,19,20\}^4$

Body:
I have to solve the following problem: calculate the number of solutions for $$x_1 + x_2 + x_3 + x_4 = 30$$ with $$(x_1,x_2,x_3,x_4) \in \{-10,...,19,20\}^4$$.

I know that if the domain is $$\mathbb{N}$$ the solution is as simple as
$$
\binom{n - 1}{k - 1}=\binom{30 - 1}{4 - 1}=\binom{29}{3}
$$

But the negative values are confusing me.
I wrote a little Java program to calculate the result, it displays 17266.
Could somebody please help me?


答:
For the first order non linear OE, you certainly are aware of Clairaut's equation. I see you are searching for higher order, so considering the following: $$y^{'''}=(x-1)^2+y^2+y'-2\\\\ y(1)=1,~y'(1)=0,~y''(1)=2$$
We can find a particular solution, of course by using series method and the undetermined coefficient method simultaneously. W e will see that $$y_p(x)=1+(x-1)^2-1/6(x-1)^3+1/12(x-1)^4+...$$</p>

Title: Examples of nonlinear ordinary differential equations with elementary solutions.

Body:
I am looking for nice examples of nonlinear ordinary differential equations that have simple solutions in terms of elementary functions. (But are not trivial to find, like, for example, with separation of variables).

A perfect example of what I am looking for is the Lane-Emden equation of index 5:
$$y''+\frac{2}{x}y'+y^5=0,\qquad y(0)=1,y'(0)=0$$
which admits the solution
$$y(x)=\frac{1}{\sqrt{1+x^2/3}}$$

Another very good example is
$$4y^2y'''-18yy'y''+15y'^3=0$$
from which we find
$$y(x)=\frac{1}{(ax^2+bx+c)^2}$$
for some constants $a,b,c$.

Do you have more examples like these? The higher the order, the better. (But I consider, for example, $H(x,y'',y''')=0$ to be of order 1). And it is a plus if it comes from a physics problem.


问:
unitary decomposition of diagonal matrices
Let $D \in \mathbb{R^{n,n}}$ be a diagonal Matrix with diagonal entries $d_{ii} \neq 0$.
Let $\tilde{D} \in \mathbb{R}^{n,n}$ be a diagonal matrix with the same diagonal entries that could be transposed in any way (for example $\tilde{d}_{11}=d_{22}$). Does there exist a unitary matrix $U$ such that $$UDU^H=\tilde{D}$$?

答:
I am not sure if I understand your condition. It seems that your are looking for something like
$$
\begin{pmatrix}
0&1\\1&0
\end{pmatrix}
\begin{pmatrix}
a&0\\0&b
\end{pmatrix}
\begin{pmatrix}
0&1\\1&0
\end{pmatrix}^{-1}
=
\begin{pmatrix}
b&0\\0&a
\end{pmatrix},
$$
where $\begin{pmatrix}
0&1\\1&0
\end{pmatrix}$ is unitary. For general $n\times n$ matrices we can use matrix blocks to construct such examples.


问:
When do intermediate fields being Galois imply entire extension is Galois?
Let $k \subset K$ be Galois. The Fundamental Theorem of Galois Theory gives us a criteria for when any intermediate field $k \subset E \subset K$ is Galois over our base field $k$. Also, it is in general not true that if $k \subset E$ and $E \subset K$ are both Galois, then $k \subset K$ will be Galois. For example, take $k = \mathbb{Q}, E = \mathbb{Q}[\sqrt{2}]$, and $K = \mathbb{Q}[\sqrt[4]{2}]$.

However, are there any minimal restrictions we can put on on the extensions $k \subset E \subset K$ such that when $E$ is Galois over $k$ and $K$ is Galois over $E$ that guarantees us $K$ being Galois over $k$?

答:
The terms of your question are too vague. In the tower of (finite) extensions $k < E < K $, with $E/k$ and $K/E$ galois, you need a minimal amount of information relative to the two intermediate extensions. Suppose e.g. that $G=Gal(E/k)$ is given, and $K$ is determined by a set of generators over $E$, say $K=E(A), A$ finite $\subset K$. Assume that separability is automatic. Then, checking normality, it is straightforward to see that $K/k$ is galois iff for all $\sigma \in G$, all the prolongations $\tau$ of $\sigma$ to $k$-embeddings of $E$ into a separable closure of $k$ stabilize $E$, i.e. $\tau (A) \subset E$. The problem is thus reduced to a convenient expression of this last property.

A simple instructive illustration comes from Kummer's theory. Suppose e.g. that $k$ has characteristic $0$ and contains the group $\mu_n$ of $n$-th roots of unity, and moreover that $H=Gal(K/E)$ is abelian of exponent $n$. By Kummer theory, there exists a finite subgroup $B$ of $E^*/{E^*}^n$, the Kummer radical of $K/E$, such that $B\cong Hom(H,\mu_n)$ and $K=E(B^{1/n})$ in the usual notations. The criterion for the normality of $K/k$ is then the $G$-stability of $B$. Things become even more visible when, in addition, $n$ is a prime $p$, $G$ is a cyclic $p$-group of order $p^n$ and $B$ is cyclic of order $p$, in which case $G$ acts trivially on $B$. If $Gal(K/k)$ is abelian, it's isomorphic to either  $C_p \times C_{p^n}$ or $C_{p^{n+1}}$; in general, $Gal(K/k)$ is characterized up to isomorphism by a cohomology class of $H^2(G, C_p)$; an istructive example is when $Gal(K/k)$ has order $p^3$. Other analogous variations can be considered in the kummerian situation. In the case of number fields, with $K/E$ abelian, a $G$-module description of $H$ in terms of class-field theory can be used, but of course things get more complicated.


问:
Show that f is bounded below and assumes its infimum
I got a metric space $(M,d)$ and the function $f:M\to \mathbb{R}$ with the following property: $\forall r\in$$\mathbb{R},$ $\{x\in M:f(x)>r\}$ is an open set. Now I need to show that this function in an compact metric space is bounded below and assumes its infimum. I thought that I only need to show that the function is continuous, but I don't really know how.
I hope someone can help me with this.

答:
Hint

You won't be able to prove that $f$ is continuous as it may not be. However your $f$ is lower semi-continuous. For example

$$f= \begin{cases}
1 &\text{if } x \lt 1\\
0 &\text{if } x = 1\\
1/2 &\text{if } x \gt 1
\end{cases}$$

is lower semi-continuous but not continuous.

A lower semi-continuous map is attaining its minimum on a compact set. To prove it consider a sequence $\{x_n \}$ in the compact $K$ such that $\{f(x_n)\}$ converges to the infimum $m$ of $f$ in $K$. As $K$ is compact, there is a subsequence $\{x_{\varphi(n)}\}$ of $\{x_n \}$ that converges. Let say to $k \in K$. Then prove that $f(k) = m$.


问:
Proof of $BAC-CAB$ identity missing step
I'm stuck on one step of the proof for the identity: $$ \vec{A}\times(\vec{B}\times\vec{C}) = \vec{B}(\vec{A}\cdot\vec{C}) - \vec{C}(\vec{A}\cdot\vec{B})$$

So far, the proof follows as:

We know that $\vec{B}\times\vec{C}$ gives a vector perpendicular to both $\vec{B}$ & $\vec{C}$, and that $ \vec{A}\times(\vec{B}\times\vec{C})$ gives a vector perpendicular to both $\vec{A}$ & $(\vec{B}\times\vec{C})$. Therefore, the vector $\vec{A}\times(\vec{B}\times\vec{C})$ must lie in the plane containing both $\vec{B}$ & $\vec{C}$.

Provided $\vec{B}$ & $\vec{C}$ are not parallel (if they were, $\vec{A}\times(\vec{B}\times\vec{C}) = 0$ regardless), vectors $\vec{B}$ & $\vec{C}$ span the 2D plane containing them both.

Therefore, we can  express any vector in the plane as a linear combination of both $\vec{B}$ and $\vec{C}$ and so we can write: $$\vec{A}\times(\vec{B}\times\vec{C}) = \alpha\vec{B} + \beta\vec{C} \tag{1}$$

Taking the scalar product of both sides with $\vec{A}$: $$\vec{A} \cdot (\vec{A}\times(\vec{B}\times\vec{C})) = \vec{A} \cdot (\alpha\vec{B} + \beta\vec{C}) = 0$$ So, $$\alpha(\vec{A} \cdot \vec{B}) + \beta(\vec{A} \cdot\vec{C}) = 0$$

Now writing, $$\lambda = \frac{\alpha}{\vec{A} \cdot\vec{C}} = -\frac{\beta}{\vec{A} \cdot \vec{B}}$$
and substituting $\alpha$ and $\beta$ back into (1) we get:
$$ \vec{A}\times(\vec{B}\times\vec{C}) = \lambda(\vec{B}(\vec{A}\cdot\vec{C}) - \vec{C}(\vec{A}\cdot\vec{B}))$$
I am able to show $\lambda = 1$ with particular choices of unit vectors for $\vec{A}, \vec{B}, \vec{C}$ but I am unable to prove that $\lambda$ is independent of the magnitude of vectors (i.e. $\lambda = 1$ for all choices of $\vec{A}, \vec{B}, \vec{C}$). This is the step that I am struggling with. Any suggestions?

答:
Observe that both sides of the equation to be proved are linear in each of the three variables (when the other two are fixed).

Based on this, it's enough to prove the statement for all possible combinations of the standard basis, say.


问:
Given X is Uni, Prove -X is Uni?
I had this problem from my last question:

Given that:
$X \sim \text{Uni}(a,b)$.

Then why does this say that: $-X \sim \text{Uni}(-a,-b)$

and in general how can I prove such claims?

Please show a formal explanation to help me understand every step.

My book only has this definition of the density function:

$$f_X(x) = \begin{cases} \dfrac{1}{b-a} &x \in [a,b] \\ 0 & \text{otherwise}. \end{cases}$$

答:
In order to find the distribution of a random variable $Y$, one determines for example the socalled CDF
$$
F_Y(x)=\mathbb{P}[Y\leq x].
$$
The function you are given in your book is called PDF and in most applications and simple cases is related via
$$
F_Y(x)=\int_{-\infty}^xf_Y(y)dy
$$
and
$$
f_Y(x)=\frac{d}{dx}F_Y(x)
$$
to the CDF which allows quick calculations forth and back.

Can you determine the CDF of $-X$ and relate it to the CDF of a $\text{Uni}(-b,-a)$ variable?

Title: Given X is Uni, Prove -X is Uni?


问:
General formula to stay inside the boundaries?
I have the following inequality:

x<y<z.

I want to write a general formula that is going to help me to always stay between x and z. Is there any formula that includes all variables x,y & z for that?

For example, something like the following:

a = z*x/y.

No matter what the x,y,z values, it always give a value between the boundries.

3<4<5

a=15/4, a is between x and z.

or

1<49<50

a=50/49, a is between x and z.

I guess this holds when we have the nonnegativity constraint. But in our problem x<y<z is the only constraint.

答: We have $x<\frac{x+y}{2}<y<z$ and $x<y<\frac{y+z}{2}<z$ hence $x<\frac{\frac{x+y}{2}+\frac{y+z}{2}}{2}<z$.


答:
First we prove that the function is bounded from below.

Note that we have an open cover
$$\bigcup_{n=0}^\infty \{f>-n\}$$
of $X.$ By the compactness of $X,$ it admits a finite open cover. Thus we know
$$X=\{f>-n\}$$
for some $n\in\mathbb N,$ so $f$ is bounded below by $-n.$

Next, we let $m:=\inf_X f.$ By definition, we can find a sequence $x_n$ in $X$ such that $f(x_n)\searrow m$ as $n\to \infty.$ The compactness of $X$ implies that it admits a convergent subsequence $x_{i_n},$ say converging to $x.$ By construction, given any $\delta>0,$ $x_{i_n}$ belongs to $\{f\le m+\delta\}$ for any large enough $n.$ Since $\{f\le m+\delta\}$ is closed, we know the limit $x$ also belongs to this set. Thus we derive
$$x\in\bigcap_{\delta>0}\{f\le m+\delta\}=\{f\le m\},$$
so $f(x)=m.$

Title: Show that f is bounded below and assumes its infimum

Body:
I got a metric space $(M,d)$ and the function $f:M\to \mathbb{R}$ with the following property: $\forall r\in$$\mathbb{R},$ $\{x\in M:f(x)>r\}$ is an open set. Now I need to show that this function in an compact metric space is bounded below and assumes its infimum. I thought that I only need to show that the function is continuous, but I don't really know how.
I hope someone can help me with this.


问:
An inequality involving summation and products
I am trying to prove the following inequality (all the variables, ie, all $z_{i,j}$, are positive):

\begin{equation}
\left(\frac{1}{mk}\sum_{i=1}^{m}\sum_{j=1}^{k} z_{i, j}\right)^{k} \geq \frac{1}{m}\sum_{i=1}^{m}\prod_{j=1}^{k}z_{i,j}
\end{equation}

I checked that it holds for some trivial cases (taking, for example, $m=k=2$), but how do I generalize it?

For convenience (and if the inequality does not hold in general), you can take all the variables to be less than or equal to $1$).

答:
This is false. If $z_{i,j}=i$, $m=k=2$, LHS = $(\frac 64)^2 = 2.25$, RHS=$\frac 12 (1+4)=2.5 >$ LHS.

In fact, if you write $a_i, g_i$ as the arithmetic mean and geometric mean of $z_{i,j}$'s, respectively, your inequality becomes

$$\left( \frac{a_1+\cdots+a_m}{m} \right)^k \ge \frac 1m \sum_{i=1}^m g_i^k \tag 1$$

If $a_i=g_i$, but different across $i$'s then $(1)$ is reversed via the power mean inequality.


问:
How can I prove that $aN\cap bN=cN$?
Is there any general method for showing that $a\mathbb{N}\cap b\mathbb{N}=c\mathbb{N}$ in set builder form? It can be done by showing that $a\mathbb{N} \cap b\mathbb{N}\subset c\mathbb{N}$ and $c\mathbb{N} \subset a\mathbb{N} \cap b\mathbb{N}$, which will prove that $a\mathbb{N}\cap b\mathbb{N}=c\mathbb{N}$, but the question goes like this, for instance, what is $3\mathbb{N}\cap7\mathbb{N}=?$, the answer is known here, that it is $21\mathbb{N}$ but what will be the method to do that if $a\in \mathbb{N}$ such that $a\mathbb{N}=\{ax:x\in\mathbb{N}\}$?

答:
Let $a,b\in \Bbb N$ and let $c=lcm (a,b).$

(1). Now $c/a,c/b\in \Bbb N,$  so for any $x$ we have $$x\in c\Bbb N \implies [\,x/a\in (c/a)\Bbb N\subset \Bbb N \,\land \,x/b\in (c/b)\Bbb N\subset \Bbb N\,]\implies$$ $$\implies [\,x/a\in \Bbb N\land x/b\in \Bbb N\,]\implies$$ $$\implies [\,x=(x/a)\cdot a\in a\Bbb N \land x=(x/b)\cdot b\in b\Bbb N\,]\implies$$ $$\implies [x\in a\Bbb N\land x\in b\Bbb N\,]\implies$$ $$\implies x\in a\Bbb N\cap b\Bbb N.$$

(2). We need to know that $\gcd (c/a,c/b)=1$  and hence that if $n\in \Bbb N$ and $(c/b) |(c/a)n$ then $(c/b)|n.\;$(see Appendix).

Now $c/a,c/b\in \Bbb N,$  so for any $x$ we have $$x\in a\Bbb N\cap b\Bbb N\implies\exists m,n \in \Bbb N(x=am=bn)\implies$$ $$\implies x(c/ab)=(c/b)m=(c/a)n\in \Bbb N\implies$$ $$\implies (c/b)|(c/a)n\implies (c/b)|n\implies $$ $$\implies bn/c=n/(c/b)\in \Bbb N\implies  x=bn=(bn/c)c\in c\Bbb N.$$ I have deleted the "$\exists m,n \in \Bbb N$" from the last 3 lines above for easier reading.

Appendix. By definition of $c=lcm (a,b)$ we have $c=\min (a\Bbb N\cap b\Bbb N).$ If $p\in \Bbb N$ and $p$ divides both of $c/a, c/b$ then we have $$[\,(c/a)/p\in \Bbb N\land (c/b)p\in \Bbb N \,] \implies$$ $$\implies [(c/p)=a\cdot (c/a)/p\in a\Bbb N\land (c/p)=b\cdot (c/b)/p\in b\Bbb N ]\implies$$ $$\implies (c/p)\in a\Bbb N\cap b\Bbb N\implies $$ $$\implies(c/p)\ge\min (a\Bbb N\cap b\Bbb N)=c\implies$$ $$ \implies (c/p)\ge c\implies 1\ge p.$$ So $\gcd (c/a,c/b)=1 .$

It is a fundamental theorem that if $u,v,n\in \Bbb N$ with $\gcd (u,v)=1$ and $u|vn$ then $u|n.$ In particular for this Q, when $u=(c/b)$ and $v=(c/a).$

Title: How can I prove that $aN\cap bN=cN$?


答:
You can easily prove this using Moment Generating Function.

We know

$M_{aX}(t)=M_X(at)$

$a$ in our case is $-1$

$M_X(t)=\dfrac{e^{bt}-e^{at}}{t(b-a)}$

$M_{(-X)}(t)=M_{X}(-t)=\dfrac{e^{-bt}-e^{-at}}{-t(b-a)}=\dfrac{e^{(-b)t}-e^{(-a)t}}{t(a-b)}$

You can see that it is MGF of uniform distribution but parameters are $-a$ and $-b$.

Therefor $-X \sim U(-b,-a)$

Title: Given X is Uni, Prove -X is Uni?

Body:
I had this problem from my last question:

Given that:
$X \sim \text{Uni}(a,b)$.

Then why does this say that: $-X \sim \text{Uni}(-a,-b)$

and in general how can I prove such claims?

Please show a formal explanation to help me understand every step.

My book only has this definition of the density function:

$$f_X(x) = \begin{cases} \dfrac{1}{b-a} &x \in [a,b] \\ 0 & \text{otherwise}. \end{cases}$$


答:
The same expression reached by achille-hui can also be achieved through some inclusion-exclusion reasoning on the limits imposed, using the transformed form of $$y_1 + y_2 + y_3 + y_4 = 70$$ with $$(y_1,y_2,y_3,y_4) \in \{0,...,29,30\}^4$$.

Without the constraint, this is a straight stars-and-bars evaluation of dividing 70 items into four groups, $\binom {73}{3}$.

We can then find the constraint-breaking cases for each variable in turn by pre-allocating one more than the limit ($31$) to each variable in  turn and repeating the division on the remaining items, giving four cases of $\binom {42}{3}$ of above-constraint cases to remove.

Finally we can find the double-counted cases of two constraints being simultaneously broken (the maximum possible), again preallocating $31$ to each of two chosen variables in each combination, $\binom 4 2=6$ cases, and dividing to give $\binom {11}{3}$.

The inclusion-exclusion result is then $$\binom {73}{3} -4\binom {42}{3} +6\binom {11}{3}$$ as before.


问:
Solve $3+e^{2t}=7e^{2t}$
I am to solve $3+e^{2t}=7e^{2t}$. My textbook solution says it's $t=\ln\left(\frac{1}{\sqrt{2}}\right)=-\frac{1}{2}\ln(2)$ whereas I got $t=\frac{1}{2} \ln(\frac{1}{2})$

My working:

$$3+e^{2t}=7e^{2t}$$
$$3=6e^{2t}$$
$$\frac{1}{2}=e^{2t}$$
$$\ln\left(\frac{1}{2}\right)=2t$$
$$t=\frac{1}{2} \ln\left(\frac{1}{2}\right)$$

Where did I go wrong and how can I arrive at $t=\ln\left(\frac{1}{\sqrt{2}}\right)=-\frac{1}{2}\ln(2)$

答: Nothing is wrong. Use the fact that$$\log\left(\frac1{\sqrt2}\right)=\log\left(\left(\frac12\right)^{1/2}\right)=\frac12\log\left(2^{-1}\right)=-\frac12\log(2).$$

Title: Solve $3+e^{2t}=7e^{2t}$


Title: Solve $3+e^{2t}=7e^{2t}$

Body:
I am to solve $3+e^{2t}=7e^{2t}$. My textbook solution says it's $t=\ln\left(\frac{1}{\sqrt{2}}\right)=-\frac{1}{2}\ln(2)$ whereas I got $t=\frac{1}{2} \ln(\frac{1}{2})$

My working:

$$3+e^{2t}=7e^{2t}$$
$$3=6e^{2t}$$
$$\frac{1}{2}=e^{2t}$$
$$\ln\left(\frac{1}{2}\right)=2t$$
$$t=\frac{1}{2} \ln\left(\frac{1}{2}\right)$$

Where did I go wrong and how can I arrive at $t=\ln\left(\frac{1}{\sqrt{2}}\right)=-\frac{1}{2}\ln(2)$

Body:
Just note that
$$t_{\text{yours}}=\frac{1}{2}\ln(\frac{1}{2})=\color{blue}{\frac{1}{2}\ln(2^{-1}) = \frac{1}{2}(-1)\ln(2)} = -\frac{1}{2}\ln(2) = t_{\text{book's}}$$


答:
Using the logarithm power rule:

The logarithm of $x$ raised to the power of $y$ is $y$ times the logarithm of $x$.

$$\log(x^ y) = y \cdot \log(x)$$

In your case is:
$$y=-0.5,
x=2$$

Title: Solve $3+e^{2t}=7e^{2t}$


问:
Is the set $J^n := \{ (x,y)\in \mathbb{R}^2: |x| > m_n, y = h(x)\}$ measurable?
Im currently working on a proof regarding APARCH-models, where the unction $h$ in the following context comes from:

Given is the set $J^n := \{ (x,y)\in \mathbb{R}^2: |x| > m_n, y = h(x)\}$, where $m_n$ is a positive, monotonely decreasing sequence with $m_n \rightarrow 0$ and $h(x) = (|x| - \gamma x)^\delta$ for some $|\gamma| < 1$ and $\delta > 0$.

To use a nice theorem, I need (Borel-)measurability of the set $J^n$.
I thought about writing the set as $J^n = (-\infty,m_n)  \cup (m_n,\infty) \times h((-\infty,m_n)  \cup (m_n,\infty))$, where $(-\infty,m_n)$ and $(m_n,\infty)$ as open interval are clearly measurable, hence their union is measurable too. Since $h$ is continuous, I thought that $h((m_n,\infty))$ is also open and hence measurable, but continuity only means that the preimage of an open set is open, not the image of an open set. Moreover, I don't know whether the image then is an open interval.

Moreover, I need that $\bigcup_{n=1}^\infty J^n = \mathbb{R}^2 \setminus \{0\}$, which is (clearly) not the case since $h$ is nonnegative and hence the second component of the elements in $J^n$ will never be less than $0$, so I will never reach the whole $\mathbb{R}^2$. But just out of curiosity I would like to know how I can show that the set $J^n$ is measurable (if it is at all).

答:
Since $(x,y)\to |x|$ is continuous, the set
$$A_n=\{(x,y)\in\mathbb{R}^2: |x|>m_n\}$$
is measurable for each natural number $n$.

The set
$$B=\{(x,y)\in\mathbb{R}^2: y=h(x)\}$$
is a graph of a continuous function, hence it is measurable. In fact, the graph of any measurable function is measurable.

It follows that for every $n$, the set
$$J^n=A_n\cap B=\{(x,y)\in\mathbb{R}^2: |x|>m_n, y=h(x)\}$$
is measurable, because it is an intersection of two measurable sets.


问:
Proving a function has no extreme points.

Is there any possible way I can make this proof shorter?

Question 5

a.

Consider the function $f:\ R\ ➜\ R,\ f\left(x\right)\ =\ \frac{x}{\left|x\right|+1}$

Note that by the algebra of continuous functions f is continuous.

We show the limits of f as x tends to $∞,-∞$ exist.

$\lim _{x\to \infty }\left(\frac{x}{\left|x\right|+1}\right)=\lim _{x\to \infty }\left(\frac{x}{x+1}\right)=\lim _{x\to \infty }\left(\frac{x}{x\left(1+\frac{1}{x}\right)}\right)=\lim \:_{x\to \infty \:}\left(\frac{1}{1+\frac{1}{x}}\right)=\left(\frac{1}{1+0}\right)=1$
$\lim \:\:_{x\to \:-\infty \:\:}\left(\frac{x}{\left|x\right|+1}\right)=\lim \:_{x\to -\infty \:}\left(\frac{x}{-x+1}\right)=\lim _{x\to -\infty }\left(\frac{x}{x\left(-1-\frac{1}{x}\right)}\right)=\lim \:_{x\to -\infty \:}\left(\frac{1}{-1-\frac{1}{x}}\right)=\left(\frac{1}{-1-0}\right)=-1\:$

Now we show that for every $x ∈ R \ \ $ we have $-1 < f(x) < 1$.

Assume toward contradiction that there exist $x ∈ R$ such that $f(x) ≥ 1$ or $f(x) ≤ -1$

Case 1: $x = 0$. In this case $f$ will equal to $0$ which is obviously in $(-1,1)$, so we have a contradiction.

Case 2: $x > 0$.

$\frac{x}{\left|x\right|+1}\ge1 \ ➜ \ \frac{x}{x+1}\ge1 \ ➜ \ x\ge x+1 \ ➜ \ 0\ge1$ Contradiction!

$\frac{x}{\left|x\right|+1}\le-1 \ ➜ \ \frac{x}{x+1}\le-1 \ ➜ \ x\le-x-1 \ ➜ \ 2x\le-1$ Contradiction! Since $2x > 0$.

Case 3: $x < 0$.

$\frac{x}{\left|x\right|+1}\ge1 \ ➜ \ \frac{x}{-x+1}\ge1 \ ➜ \ x\ge-x+1 \ ➜ \ 2x\ge1$ Contradiction! since $2x < 0$.

$\frac{x}{\left|x\right|+1}\le-1 \ ➜ \ \frac{x}{-x+1}\le-1 \ ➜ \ x\le x-1 \ ➜ \ 0\le-1$ Contradiction!

Therefore -1 < f(x) < 1 for every $x ∈ R$.

Now we show f(x) is strictly increasing.
Assume toward contradiction that f(a) ≥ f(b) for every a,b ∈ R such that a < b.

$\frac{a}{\left|a\right|+1}\ge\frac{b}{\left|b\right|+1}\ \ ➜\ \ a\left(\left|b\right|+1\right)\ge b\left(\left|a\right|+1\right)\ $

Case 1: b = 0, 0 > a.

$a\left(\left|0\right|+1\right)\ge0\left(\left|a\right|+1\right)\ \ ➜\ \ a\ \ge0$ Contradiction!

Case 2: b > 0, a = 0.

$0\left(\left|b\right|+1\right)\ge b\left(\left|0\right|+1\right)\ \ ➜\ \ 0\ \ge b$ Contradiction!

Case 3: b > 0, a > 0

$a\left(b+1\right)\ge b\left(a+1\right)\ \ ➜\ \ ab\ +a\ge ab+b\ \ \ ➜\ \ a\ \ge b\ $ Contradiction!

Case 4: 0 > b, 0 > a.

$a\left(-b+1\right)\ge b\left(-a+1\right)\ \ ➜\ \ -ab\ +a\ge-ab+b\ \ \ ➜\ \ a\ \ge b\ $ Contradiction!

Case 5: b > 0, 0 > a.

$a\left(b+1\right)\ge b\left(-a+1\right)\ \ ➜\ \ ab\ +a\ge-ab+b\ \ \ ➜\ \ 2ab\ \ge b-a$ Contradiction! Since $2ab < 0$ and $b - a > 0$

We've reached a contradiction for every possible case of $a,b$, therefore $f(a) < f(b)$ for every $a,b ∈ R$, such that $a < b$ which means $f$ is strictly increasing.

Now we show $f$ has no extreme points.

Assume toward contradiction that $f$ takes a maximum at a point, we will denote it $a$.

Since $f$ is strictly increasing, there exist a point $b ∈ R$ such that $a < b$ for which $f(a) < f(b)$,

in contradiction to the minimality of $a$.

Assume toward contradiction that $f$ takes a minimum at a point, we will denote it $c$.

Since $f$ is strictly increasing, there exist a point $d ∈ R$ such that $d < c$ for which $f(d) < f(c)$,

in contradiction to the minimality of $c$.

Therefore we've shown that $f$ satisfies all requirements.

答:
That's some serious struggle you got there. here's a simpler example:

Consider the function $f\left(x\right)=\frac{e^{x}}{1+e^{x}}$

$f$ is continuous. Also We have:

$$\underset{x\rightarrow\infty}{\lim}\frac{e^{x}}{1+e^{x}}=1$$

$$\underset{x\rightarrow-\infty}{\lim}\frac{e^{x}}{1+e^{x}}=0$$

$f$ doesn't have a max because for every $x_1\in\mathbb{R}$ we can take $x_1+1$  such that:

$$\underbrace{f\left(x_1+1\right)}=\frac{e^{x_{1}+1}}{1+e^{x_{1}+1}}=\frac{e\cdot e^{x_{1}}}{1+e\cdot e^{x_{1}}}=\frac{e^{x_{1}}}{\frac{1}{e}+e^{x_{1}}}\underset{\frac{1}{e}<1}{\underbrace >}\frac{e^{x_{1}}}{1+e^{x_{1}}}=\underbrace{f\left(x_{1}\right)}$$

Also, $f$ doesn't have a min because for every $x_2\in\mathbb{R}$ we can take $x_2-1$  such that:

$$\underbrace{f\left(x_2-1\right)}=\frac{e^{x_{2}-1}}{1+e^{x_{2}-1}}=\frac{\frac{e^{x_{2}}}{e}}{1+\frac{e^{x_{2}}}{e}}=\frac{e^{x_{2}}}{e+e^{x_{2}}}\underset{e>1}{\underbrace <}\frac{e^{x_{2}}}{1+e^{x_{2}}}=\underbrace{f\left(x_{2}\right)}$$


问:
In the given figure find $\angle FBC - \angle BED$
In $\Delta ABC$, $\angle A$ = $24°$, $\angle B$ = $78°$, $D$ and $E$ are points on $AB$ and $AC$ respectively in such a way that $\angle BCD$ = $24°$, $\angle CBE$ = $51°$. $AC$ is extended to a point $F$ such that $C$ is the mid-point of $EF$. Find $
|\angle FBC – \angle BED|$.

My Work :-

Well I tried angle chasing(in figure) but to no avail.

答:
There are some calculation mistakes. $\angle ACB = 180 - 24- 78= 78=\angle ABC$. So $AB=AC$.

Also $\angle BDC = 78$. On more angle chasing, it will follow $$BC=CD=CE=CF$$

Thus $C$  is center of circle passing through $B,D,E,F$. From this it will eventually follow that $\angle FBC - \angle BED = 27$.


问:
Getting rid of a floor function in the next expression:$\left\lfloor\frac{(x-2)^2}{4}\right\rfloor $, It is known x is odd.
I was wondering if there's a way in which you can get rid of a floor function in the next expression:$$\left\lfloor\frac{(x-2)^2}{4}\right\rfloor $$ It is known x is odd.</p>

答:
You can use:$$\left\lfloor\frac{(x-2)^2}{4}\right\rfloor=\frac{(x-2)^2}4-0.25$$

More generally:
 $$\left\lfloor x\right\rfloor=x-(x \mod 1)$$


问:
Analytic continuation of a Dirichlet series
Suppose we have a Dirichlet series

$$
D(s) = \sum_{n=1}^\infty \frac{a(n)}{n^s}
$$

which we know is absolutely convergent for $Re(s)>1$. Suppose that we prove that $\lim_{s\to 1^+}D(s) < \infty$. Does this imply that $D(s)$ continues analytically to some half plane to the left of $s=1$, that is to the region $Re(s) > 1-\epsilon$ for some positive $\epsilon$?

Remark: The case where $a(n)>0$ for all $n$ (or equivalently finitely many $n$) is a classical theorem of Landau. I am interested when this is not the case.

答:
This is just a comment that seems important....

Note that $$\frac{\zeta(2s)}{\zeta(s)}=\sum_{n=1}^{\infty}\frac{\lambda(n)}{n^s},$$
for Re$(s)>1$ and that $\frac{\zeta(2s)}{\zeta(s)}=\sum_{n=1}^{\infty}\frac{\lambda(n)}{n^s}=0$ at $s=1$ (this is equivalent to the prime number theorem). So, your question includes the question of whether $\frac{\zeta(2s)}{\zeta(s)}$ can be analytic continued to a half-plane Re$(s)>1-\epsilon$, That is, Is there  $\epsilon>0$ such that $\zeta(s)\neq 0$ for Re$(s)>1-\epsilon$? This is an open problem, about the zeros of $\zeta(s)$ in the critical strip $0<$Re$(s)<1$.


问:
Number of solutions for $x_1 + x_2 + x_3 + x_4 = 30$ with $(x_1,x_2,x_3,x_4) \in \{-10,...,19,20\}^4$
I have to solve the following problem: calculate the number of solutions for $$x_1 + x_2 + x_3 + x_4 = 30$$ with $$(x_1,x_2,x_3,x_4) \in \{-10,...,19,20\}^4$$.

I know that if the domain is $$\mathbb{N}$$ the solution is as simple as
$$
\binom{n - 1}{k - 1}=\binom{30 - 1}{4 - 1}=\binom{29}{3}
$$

But the negative values are confusing me.
I wrote a little Java program to calculate the result, it displays 17266.
Could somebody please help me?

答:
Assuming $0$ is included in $\Bbb {N}$, let $Y_i = X_i +11$,
then the problem transforms to

$Y_1 + Y_2 +Y_3+Y_4 = 74,\;$subject to$\,\; 1\leq{Y_i}<31$

Applying "stars and bars" formula and inclusion-exclusion, we get

(All ways) - (at least one variable breaks upper limit) + (at least two variables break upper limit)

$\binom{74-1}{4-1} - \binom 4 1\binom{74-1-31}{4-1} + \binom 4 2\binom{74-1-62}{4-1}$

= $\binom{73}3 - \binom 4 1 \binom{42}3 + \binom4 2\binom{11}3 = 17266$

Further simplification

I added $11$ rather than $10$ to transform the problem as you wanted, but just realized that it also had the advantage of putting it into a larger class of familiar problems  regarding sum obtained from rolling dice.

The problem is to obtain a sum of $74$ rolling a $31$ faced die $4$ times. Now the median value of the sum $= 4*16 = 64$, and the required sum is $74$ which is $10$ above the median, hence by symmetry, will be identical to the number of ways to get $10$ below the median, viz. $54$

The answer is then obtained more simply as $\binom{53}3-\binom4 1\binom{22}3 = 17266$

In this formulation, only one variable can break the upper limit, in some others, you might not need to apply inclusion-exclusion at all !

Transforming a problem into one of a familiar class of problems often leads to new insights !

We can even encapsulate the whole thing in a formula (ignoring the final twist) as

$$W(n,k,D,j)= \sum_{j=0}^{\lfloor{\frac{n-k} D}\rfloor}(-1)^j \binom{k} {j}\binom{n-1-Dj}{k-1}$$

where $W$ = answer, $n$ = desired sum, $k$ = number of throws, $D$ is number of faces on die, and $j$ is the counter


问:
A question about bipartite graphs and maximal independent (stable) sets
Here's an excerpt from an article:

My question is: why are the conjectures 1 and 2 equivalent for non-connected bipartite graphs? For example, if there are only 2 vertices connected by an edge in a graph, the first conjecture implies those are the ones that belong to at most half maximal stable sets, whereas second permits (in theory) for other vertices to hold that property.

答:
For non-connected bipartite graphs, the two bipartition classes aren't uniquely determined, so it's less clear how to interpret the conjecture.

But the second conjecture is equivalent to the first if we say:

Conjecture 2'. Let $G$ be a finite bipartite graph with bipartition classes $X,Y$ and at least one edge. Then each of $X$ and $Y$ contains a vertex belonging to at most half of the maximal stable sets.

Suppose that $G$ is a bipartite graph such that in each connected component, only one or possibly no bipartition class contains an at-most-half vertex. Then define $X$ by picking a side with no at-most-half vertices from each connected component, and let $Y$ be its complement. This choice of $X$ and $Y$ will contradict Conjecture 2'.

Therefore there must be a connected component $C$ such that both $X \cap C$ and $Y \cap C$ contain an at-most-half vertex. By looking at a path between these vertices, we can find an edge whose endpoints are both at-most-half vertices, and obtain Conjecture 1.

We can even reduce to the case of

Conjecture 3. Let $G$ be a finite connected bipartite graph with at least two vertices. Then each bipartition class contains a vertex belonging to at most half of the maximal stable sets.

To get conjecture 2' from conjecture 3, let $G$ be any finite bipartite graph with at least one edge. Begin by applying conjecture 3 to any connected component $C$ of $G$, finding vertices $x$ and $y$.

Then note that any maximal stable set of $G$ has the following structure: a maximal stable set in $C$, and a maximal stable set in the rest of the graph. So when we go from $C$ to $G$, all counts of maximal stable sets (containing $x$, containing $y$, and with no restrictions) are simply multiplied by the number of maximal stable sets in the rest of the graph. This does not affect the at-least-half property, so $x$ and $y$ are also the vertices that satisfy Conjecture 2'.


答:
Another way to look at it:

If is $x$ is odd, then $x-2$ is odd as well. We have that $x-2 \equiv 1$ (mod 4) or $ x-2 \equiv 3$ (mod 4). In either case we have that $(x-2)^2 \equiv 1$ (mod 4).
Hence 4 divides $(x-2)^2-1$, therefore:

$$\left\lfloor\frac{(x-2)^2}4\right\rfloor= \frac{(x-2)^2-1}4$$

If you expand $(x-2)^2-1$, you get the same answer as the one given by Brian M. Scott.

Title: Getting rid of a floor function in the next expression:$\left\lfloor\frac{(x-2)^2}{4}\right\rfloor $, It is known x is odd.

Body: I was wondering if there's a way in which you can get rid of a floor function in the next expression:$$\left\lfloor\frac{(x-2)^2}{4}\right\rfloor $$ It is known x is odd.</p>


问:
Prove or disprove this statement: For any $y \in \mathbb{Z}$ and $y \neq \pm 1$, $(y+1)^{2}$ is not divisible by $y$.
I want to prove or disprove this statement:

For any $y \in \mathbb{Z}$ and $y \neq \pm 1$, $(y+1)^{2}$ is not divisible by $y$.

The case where $y$ is even can be easily proved. However, I am stuck at the case where $y$ is odd. I try to start with the fact that "$y+1$ is not divisible by $y$" and expect to use something like the Euclid's Lemma. Any idea on how to proceed?

答: Numbers $y$ and $y+1$ are consecutive and thus relatively prime and thus $y\ne \pm1$ can not divide $y+1$ and so it can not divide $(y+1)^2$.


问:
If $a_1=1,a_{n+1}^2-2a_na_{n+1}-a_n=0$ then sum of the series $\sum {a_n\over 3^n}$
Let $\{a_n\}$ be a sequence of positive real numbers such that $a_1=1,a_{n+1}^2-2a_na_{n+1}-a_n=0\,\,\forall n\ge 1$. Then show that $$1<\sum_{n=1}^\infty {a_n\over 3^n}<2$$.
I got this problem from a question paper. Though it didn't asked for exact sum of the series , if you also manage to find that then please post it.

答:
(Fill in the gaps as needed. If you're stuck, show your work and explain what you've tried.)

Hints:

After calculating the first few terms, the ratio of the terms is very similar. If we estimate $ a_n \approx a_1 r^{n-1}$, this suggests we should solve for $ \frac{ 1/3} { 1 - (r_1/3) } = 1$ and $ \frac{1/3} { 1 - (r_2/3)} = 2$ to give us an idea of how to bound the sequence. This gives us $ r_1  = 2, r_2 = 5/2$, so we want to show that $ 2 a_n < a_{n+1} < \frac{5}{2} a_n$ (with some flexibility if this doesn't immediately work out).

Show that $ a_{n+1} = \frac{ 2a_n + \sqrt{ 4a_n^2 + 4a_n  } } { 2} $. In particular, reject the negative root.

Show that $ a_n \geq 1$.

Show that $ 2 a_n < a_{n+1} < \frac{5}{2} a_n$.

Hence, show that $1=\frac{ 1/3 } { 1 - 2/3} < \sum \frac{a_n}{3^n} < \frac{ 1 / 3 } { 1 - 5/6}=2 $

Note:

The LHS is true by calculating the first 5 terms.

Of course, we can't prove the RHS just by calculating enough terms.

In fact, the bounding inequality $ 2a_n < a_{n+1} < 2a_n + \frac{1}{2}$, so the ratio $ a_{n+1} / a_n$ is very close to 2, esp at (slightly) larger values of $n$.

Not surprisingly, the value of the summation is much much closer to the 1. Using the first ~10 terms to get a better approximation, you can in fact show that the summation is between 1.2 and 1.25.


问:
IMU Change of reference frame
Im trying to rotate one accelerometer vector from body frame to ned frame, but i cant found what im doing wrong. For now, im using an online dataset that provides me roll, pitch and heading information, where heading is 0 rad when alligned with east axis, not north axis, the roll is positive with left-side up and the pitch is positive with the front of the vehicle down. The dataset also provides accelereration in x, y, z. For making the rotation im using a direct cossine matrix, like this below:
\begin{bmatrix}
CP*CY & -CR*SY + SR*SP*CY & SR*SY + CR*SP*CY  \\
CP*SY & CR*CY + SR*SP*SY & -SR*CY + CR*SP*SY\\
-SP &  SR*CP & CR*CP
\end{bmatrix}

The x axis of the imu is alligned with the movement of the car, the y axis with the lateral acceleration, and z axis with gravity. What im doing is first change the heading angle to get an orientation with respect of north, after this, multiply the DCM with the acceleration [Ax,-Ay,Az] in this order, Ay is negated because the dataset gives me acceleration positive in the left axis.  Im expecting to have the NED acceleration. But the answer seems wrong. Is the process for obtaining the NED acceleration correct? Or am I  missing something ?

答: The output of the accelerometer is given in the body-fixed coordinate system, that is, $a = {}^{i}\dot{v}^b_{C/i} - g^b$ where ${}^{i}\dot{v}^b_{C/i}$ is the spatial acceleration of the sensor and $g^b$ is the gravitational acceleration. The DCM you wrote seems to be formed from Euler angles under 3-2-1 rotation; from body to NED. If you want to get an acceleration in the NED coordinate system, multiply the DCM with the sensor output.

Title: IMU Change of reference frame


答:
I think I'd go back to the definition here.  If $y\mid (y+1)^2$ then there is an integer $k$ such that

$$yk = (y+1)^2 = y^2+2y+1$$

so

$$1 = yk - y^2 -2y = y(k-y-2)$$

Therefore $y$ is a divisor of $1$, which is disallowed.

Title: Prove or disprove this statement: For any $y \in \mathbb{Z}$ and $y \neq \pm 1$, $(y+1)^{2}$ is not divisible by $y$.

Body:
I want to prove or disprove this statement:

For any $y \in \mathbb{Z}$ and $y \neq \pm 1$, $(y+1)^{2}$ is not divisible by $y$.

The case where $y$ is even can be easily proved. However, I am stuck at the case where $y$ is odd. I try to start with the fact that "$y+1$ is not divisible by $y$" and expect to use something like the Euclid's Lemma. Any idea on how to proceed?


问:
Given $f(i, j) = f(i − 1, j − 1) + f(i, j − 1)$, find the value of $f(1009, 2019)$

Let $f:(\mathbb{N}\ \cup \{0\})^2$ $\rightarrow\mathbb{N}\ \cup\{0\}$ be such that $f(0, x) = f(x, 0) = 1$ for all x $\in$ $\mathbb{N}\ \cup$ {$0$}.
Also for all $i, j \in\ \mathbb{N}$ we have $f(i, j) = f(i − 1, j − 1) + f(i, j − 1)$. Find $f(1009, 2019)$.

What I've tried:

I took two cases: $i>=j$ and $i<j$.

As $j$ is reduced by $1$ in both $f(i-1,j-1)$ and $f(i,j-1)$, if $j<=i$, then $j$ will reach $0$ first irrespective of $i$. So the value of $f(i,j)$ is $2^{j}$. Here's a branch diagram to illustrate what I meant

So $f(3,2)=2^2=4$

However I'm stuck at finding the pattern for the other case ($j>i$). Example for $f(i,j)$ where $i<j$:

$f(2,3)=7$

I do recognize that I can stop evaluating once I reach a point where $i=j$ but I can't find at which places and how many times they reach that point. It would be highly helpful If you would help me find the pattern. Also, am I going in the right direction? Is there any other way to solve this problem(a formula or something)? Help would be highly appreciable!

答:
Here is a generating function approach. We consider the recurrence relation
\begin{align*}
f(i,j)&=f(i-1,j-1)+f(i,j-1)\qquad\qquad i,j\geq 1\tag{1}\\
f(i,0)&=f(0,j)=1\qquad\qquad\qquad\qquad\qquad\quad i,j\geq 0
\end{align*}

We set
\begin{align*}
\color{blue}{F(x,y)=\sum_{i=0}^\infty\sum_{j=0}^\infty f(i,j)x^iy^j}
\end{align*}
and use guided from (1) the Ansatz
\begin{align*}
\sum_{i=1}^\infty\sum_{j=1}^\infty f(i,j) x^iy^j=\sum_{i=1}^\infty\sum_{j=1}^\infty f(i-1,j-1) x^iy^j+\sum_{i=1}^\infty\sum_{j=1}^\infty f(i,j-1) x^iy^j\tag{2}
\end{align*}

Left-hand side of (2):

We obtain
\begin{align*}
\sum_{i=1}^\infty&\sum_{j=1}^\infty f(i,j) x^iy^j\\
&=\sum_{i=0}^\infty\sum_{j=0}^\infty f(i,j) x^iy^j-\sum_{i=0}^\infty f(i,0)x^i-\sum_{j=0}^\infty f(0,j)y^j+f(0,0)\\
&=F(x,y)-\frac{1}{1-x}-\frac{1}{1-y}+1\tag{3}
\end{align*}

Right-hand side of (2):

We obtain
\begin{align*}
\sum_{i=1}^\infty&\sum_{j=1}^\infty f(i-1,j-1) x^iy^j+\sum_{i=1}^\infty\sum_{j=1}^\infty f(i,j-1) x^iy^j\\
&=\sum_{i=0}^\infty\sum_{j=0}^\infty f(i,j) x^{i+1}y^{j+1}+\sum_{i=1}^\infty\sum_{j=0}^\infty f(i,j) x^{i}y^{j+1}\\
&=xyF(x,y)+yF(x,y)-\sum_{j=0}^\infty f(0,j)y^{y+1}\\
&=xyF(x,y)+yF(x,y)-\frac{y}{1-y}\tag{4}
\end{align*}

From (3) and (4) we get:

\begin{align*}
F(x,y)-\frac{1}{1-x}-\frac{1}{1-y}+1&=xyF(x,y)+yF(x,y)-\frac{y}{1-y}\\
F(x,y)(1-y(1+x))&=\frac{1}{1-x}\\
\color{blue}{F(x,y)}&\color{blue}{=\frac{1}{1-x}\,\frac{1}{1-y(1+x)}}\tag{5}
\end{align*}

The final step is to extract $f(i,j)=[x^iy^j]F(x,y)$ where we use the coefficient of operator $[x^i]$ to denote the coefficient of $x^i$ of a series.

We obtain from (5) for $i,j\geq 0$:
\begin{align*}
\color{blue}{f(i,j)}&=[x^iy^j]\frac{1}{1-x}\,\frac{1}{1-y(1+x)}\\
&=[x^iy^j]\sum_{k=0}^\infty y^k(1+x)^k\,\frac{1}{1-x}\tag{6}\\
&=[x^i](1+x)^j\,\frac{1}{1-x}\tag{7}\\
&=[x^i](1+x)^j\sum_{k=0}^\infty x^k\tag{8}\\
&=\sum_{k=0}^i[x^{i-k}](1+x)^j\tag{9}\\
&=\sum_{k=0}^i\binom{j}{i-k}\tag{10}\\
&\color{blue}{=\begin{cases}
\sum_{k=0}^i\binom{j}{k}&\qquad j>i\\
2^j&\qquad j\leq i
\end{cases}}\tag{11}
\end{align*}

Comment:

In (6) we use the geometric series expansion.

In (7) we select the coefficient of $y^j$.

In (8) we expand $\frac{1}{1-x}$.

In (9) we use the linearity of the coefficient of operator and apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$. We also set the upper limit of the series to $i$ since other terms do not contribute.

In (10) we selet the coefficient of $x^{i-k}$.

In (11) we change the order of summation $k\to i-k$ and we use the binomial identity $\sum_{k=0}^j\binom{j}{k}=2^j$ as well as $\binom{j}{k}=0$ if $k>j$.


问:
Calculate an integral solving a differential equation
I've completely stuck with this problem:

Calculate:

$$ \int_0^{0.1} e^{-x^2} \,dx $$

solving the problem:

$$ y'(x)= e^{-x^2}, x \in [0, 0.1]$$
$$ y(0) = 0 $$

My doubt is just how to do it. I'm used to use aproximations (with Taylor, Euler, Runge-Kutta methods), but the statement always says "use X method to calculate an aproximate the value of an equation", but I have no idea about how to calculate the value of an integral.

Please, anybody could help me?
Thanks in advance.

UPDATE:

The complete statement says:

i)

We can see that numerical methods which are used to solve initial value problems, are also used to calculate integrals:

Demostrate that all initial value problem of the form:

$$ y'(x)= f(t,y(t)) , t \in [t_{0}, T]$$
$$ y(t_{0}) = y_{0} $$

is equivalent to the integral equation:

$$ y(t) = y_{0} + \int_{t_0}^{T} f(s,y(s)) \,ds $$

The second section ii) is the question I ask in the beggining.

I understand the second section is an application of the first section but the integral of the problem has no anylitic solution, so that...how to solve it?

I still don't see the solution even with your help and tips..

Thanks a lot.

Body:
Hint:

$$0.1-\frac{0.1^3}{3}+\frac{0.1^5}{5\cdot2!}-\frac{0.1^7}{7\cdot3!}+\cdots$$


问:
Proof of $BAC-CAB$ identity missing step
I'm stuck on one step of the proof for the identity: $$ \vec{A}\times(\vec{B}\times\vec{C}) = \vec{B}(\vec{A}\cdot\vec{C}) - \vec{C}(\vec{A}\cdot\vec{B})$$

So far, the proof follows as:

We know that $\vec{B}\times\vec{C}$ gives a vector perpendicular to both $\vec{B}$ & $\vec{C}$, and that $ \vec{A}\times(\vec{B}\times\vec{C})$ gives a vector perpendicular to both $\vec{A}$ & $(\vec{B}\times\vec{C})$. Therefore, the vector $\vec{A}\times(\vec{B}\times\vec{C})$ must lie in the plane containing both $\vec{B}$ & $\vec{C}$.

Provided $\vec{B}$ & $\vec{C}$ are not parallel (if they were, $\vec{A}\times(\vec{B}\times\vec{C}) = 0$ regardless), vectors $\vec{B}$ & $\vec{C}$ span the 2D plane containing them both.

Therefore, we can  express any vector in the plane as a linear combination of both $\vec{B}$ and $\vec{C}$ and so we can write: $$\vec{A}\times(\vec{B}\times\vec{C}) = \alpha\vec{B} + \beta\vec{C} \tag{1}$$

Taking the scalar product of both sides with $\vec{A}$: $$\vec{A} \cdot (\vec{A}\times(\vec{B}\times\vec{C})) = \vec{A} \cdot (\alpha\vec{B} + \beta\vec{C}) = 0$$ So, $$\alpha(\vec{A} \cdot \vec{B}) + \beta(\vec{A} \cdot\vec{C}) = 0$$

Now writing, $$\lambda = \frac{\alpha}{\vec{A} \cdot\vec{C}} = -\frac{\beta}{\vec{A} \cdot \vec{B}}$$
and substituting $\alpha$ and $\beta$ back into (1) we get:
$$ \vec{A}\times(\vec{B}\times\vec{C}) = \lambda(\vec{B}(\vec{A}\cdot\vec{C}) - \vec{C}(\vec{A}\cdot\vec{B}))$$
I am able to show $\lambda = 1$ with particular choices of unit vectors for $\vec{A}, \vec{B}, \vec{C}$ but I am unable to prove that $\lambda$ is independent of the magnitude of vectors (i.e. $\lambda = 1$ for all choices of $\vec{A}, \vec{B}, \vec{C}$). This is the step that I am struggling with. Any suggestions?

答:
(This is a long comment.) This is probably not the simplest way to derive the identity, but the identity follows quite quickly from a special case of Cauchy-Binet formula:
\begin{aligned}
\left(A\times(B\times C)\right)\cdot x
&=\det(A,B\times C,x)\quad\text{(defintion of cross product)}\\
&=\det(x,A,B\times C)\\
&=(x\times A)\cdot(B\times C)\quad\text{(defintion of cross product)}\\
&=(x\cdot B)(A\cdot C)-(x\cdot C)(A\cdot B)\quad\text{(Cauchy-Binet formula)}\\
&=\left(B(A\cdot C)-C(A\cdot B)\right)\cdot x.\\
\end{aligned}
Since $x$ is arbitrary, the result follows.


问:
Probability question regarding mars, venus, and pluto
Background:

Venus cycle is $584$ days

Mars cycle is  $780$ days

Pluto cycle, in this case, is 245.5 YEARS

Question:

What are the odds of all three converging at the same degree point on the same day?

This came from an astronomer friend of mine but essentially he was interested to know what is the probability that all three planets converge at the same degree point. I was thinking this may be somewhat complicated if we take into consideration the rotations and speed of which each planet rotates around the sun. On the other hand, wouldn't it just be

$$\left(\frac{1}{360}\right)^3$$

since all planets rotate in a circular sort of shape around the sun hence the $360$. Any suggestions or thoughts on calculating this would be greatly appreciated.

答:
You speak about convergence "at the same degree point", but do we imply that we are watching from the Earth or from the Sun?

Consider we are "staying" at the Sun. As Pluto has rather big inclination: 17,14°, and  two other planets have almost zero inclination, then formally  it's possible that such event never occurs.

If we do not consider inclination and if we start looking at random moment of time, then the answer is $\frac1{360^2}$.


问:
Lebesgue Measure of Disk is $\pi r^2$ using Fubinis Theorem
Sorry for being unhelpful with this question, I normally do try to have a good stab at questions before posting them here but I seem to have mathematicians block today! Could anyone just give me a very brief outline of how I can find show the Lebesgue Measure of Disk is $\pi r^2$ using Fubinis Theorem?

Thanks a lot and sorry for not putting a lot of effort into this question.

EDIT

I have decided to have a stab. Notice that $\lambda^2=\lambda^1 \times \lambda^1$ and a similar result for Borel Sigma Algebras.

Hence $\lambda^2(D)=\lambda^1\times\lambda^1(D):=\int_{\mathbb{R}}\lambda^1(D_x)d\lambda^1(x)$

Where $E_x:=\{y \in \mathbb{R} : (x,y)\in E\}$ for any $E \in \mathcal{B}(\mathbb{R}^2)$

Here we have $D_x= \chi_{[-r,r]}\cdot\chi_{[-\sqrt{r^2-x^2},\sqrt{r^2-x^2}]} $  (from simple geometry)

And hence $\lambda^2(D)= \int_{[-r,r]}\lambda^1(\chi_{[-\sqrt{r^2-x^2},\sqrt{r^2-x^2}]})d\lambda^1(x)= \int_{[-r,r]}2\sqrt{r^2-x^2}d\lambda^1(x) $

Which I can then solve. But I dont think I have used Fubinis theorem?

答:
With your notations, we must calculate $\int_{\mathbb{R}^2} \chi_{D}(x,y) \lambda^2(x,y)$

For all $x\in \mathbb{R}$, let $D_x=\{y\in\mathbb{R}| (x,y)\in D\}$

We can write $\chi_{D}(x,y) = \chi_{[-r,r]}(x).\chi_{D_x}(y)$

From Fubini

$\int_{\mathbb{R}^2} \chi_{D}(x,y) \lambda^2(x,y) = \int_{\mathbb{R}}\left(\int_{\mathbb{R}} \chi_{[-r,r]}(x).\chi_{D_x}(y)d\lambda^1(y)\right)d\lambda^1(x)$

$= \int_{\mathbb{R}}\chi_{[-r,r]}(x) \left(\int_{\mathbb{R}} \chi_{D_x}(y)d\lambda^1(y)\right)d\lambda^1(x)$

$= \int_{\mathbb{R}}\chi_{[-r,r]}(x) \lambda^1(D_x)d\lambda^1(x)$


答: If $y$ divides $(y+1)^2=y^2+2y+1$, knowing that $y$ already divides the terms $y^2$ and $2y$, we conclude that $y\mid 1$. Thus, $y=\pm 1$.


问:
Is there a direct proof for $\int_0^{2\pi}\frac{r^2+r(\cos t-\sin t)}{1+2r\cos t+r^2}dt=2\pi$
Is there a direct method to prove that, if $r>1$
$$I=\int_0^{2\pi}\frac{r^2+r(\cos(t)-\sin(t))}{1+2r\cos(t)+r^2}dt=2\pi$$

I ask this question because this integral can be thought as an integral of a 1-form along a circle:
$\omega=\frac{x-y}{x^2+y^2}dx+\frac{x+y}{x^2+y^2}dy$
and $\gamma=(1+r\cos(t),r\sin(t))$ for $t\in[0,2\pi]$ and if I'm not mistaken this curve is homotopic to the unit circle centered in $(0,0)$ if $r>1$, then the integral is equal to the integral along this curve that is $2\pi.$ Is my argument correct?

答:
We can reduce the problem a little bit. By elementary algebra and symmetry you'll see that it suffices to show that
$$
\int_0^{2 \pi} \dfrac{1+r \cos t}{1+2 r \cos t + r^2} \,dt = 0.
$$

Under suitable transformation we can evaluate this integral as
$$
\left[ -\frac{1}{2} \arctan \left(\frac{(r+1) \cos \left(\frac{t}{2}\right)}{\sin
   \left(\frac{t}{2}\right)-r \sin \left(\frac{t}{2}\right)}\right)+\frac{1}{2} \arctan \left(\frac{(r+1) \cos \left(\frac{t}{2}\right)}{r \sin
   \left(\frac{t}{2}\right)-\sin \left(\frac{t}{2}\right)}\right)+\frac{t}{2}\right]_0^{2 \pi} = 0
$$

note: Out of laziness I computed the primitive with Wolfram Mathematica, but the transformation is standard.


问:
Confusing math problem
How would I solve this question? I came across it and is really confused.

The payment of Jon was bigger by $960$ than the payment of David.
After the payment of David got increased by $10\%$, Jon and David got the same payment amount.

$\text{A}$. We will mark with $X$ the payment of David in the beginning and using the $X$ find the payment of Jon.

$\text{B}$. Find out the payment of David in the beginning.

How can I solve this? Any hints?

答:
We know that: (Jon's Payment) = (David's Payment) + $960$

But, we also know that: (Jon's Payment) = (David's Payment) + $0.1$ (David's Payment)

Let Jon's Payment be deonted by $J$ and David's Payment by $X$. Thus, we have:

$J=X+960$ and

$J = X + 0.1 X$

Does that help?

Title: Confusing math problem


问:
Show that if $A\in S_n^{++}(\Bbb R)$ and $B \in A_n(\Bbb R)$ therefore $AB$ is diagonalizable in $\Bbb M_n(\Bbb C)$.

Let $S_n^{++}(\Bbb R)$ be the set of positive definite symmetric matrices with real coefficients, $A_n(\Bbb R)$ the set of skew-symmetric matrices with real coefficients. Show that if $A\in S_n^{++}(\Bbb R)$ and $B \in A_n(\Bbb R)$ therefore $AB$ is diagonalizable in $\Bbb M_n(\Bbb C)$.

If $AB=BA$ I know how to get the result. But I'm stuck if $AB \neq BA$... Does someone have a hint?

答: Hint: prove that $AB$ is similar to a skew-symmetric matrix with real coefficients.


问:
How would one solve the following: $\sin(\frac{\pi x}{30})+\sin(\frac{\pi x}{360})-\cos(\frac{\pi x}{30})-\cos(\frac{\pi x}{360})=0$
I want to use the idea of vectors to solve this problem: "How many times a day do the hour and minute hands of a clock face the opposite of each other?"
We know that the minute hand turns $6^{\circ}$ every minute. If I assume my starting point is 12 o'clock, then my angle (with respect to the positive x-axis, of course), $\theta_m(t)=90^{\circ}-6t$. Similarly, the angle between the hour-hand and the positive x-axis at any given time (in minutes), t, is $\theta_h= 90^{\circ}- \frac{1}{2}t$.
If we think of the hour and minute hands as two dimensional vectors whose tails are on the origin, then the vector of the minute hand can be specified as $\textbf{m}$, where:
\begin{align*}
 \textbf{m} &= \begin{bmatrix}
  \sin(6t) \\
  \cos(6t) \\
 \end{bmatrix}
\end{align*}
In the same way, the vector of the hour hand, $\textbf{h}$= \begin{bmatrix}
 \sin( \frac{1}{2}t) \\
 \cos( \frac{1}{2}t) \\
\end{bmatrix}

We want the instances where:
\begin{align*}
  \textbf{m} &= -\textbf{h} \\
  \textbf{m}+\textbf{h} &=\textbf{0} \\
  \begin{bmatrix}
   \sin(6t)+\sin\left( \frac{1}{2}t\right) \\
   \cos(6t)+\cos\left( \frac{1}{2}t\right) \\
  \end{bmatrix} &= \textbf{0}
 \end{align*}
So we have:
$$\sin(6t)+\sin\left( \frac{1}{2}t\right) -\cos(6t)-\cos\left( \frac{1}{2}t\right)=0$$
In radians:
$$\sin\left( \frac{\pi t}{30}\right)+\sin\left( \frac{\pi t}{360}\right) -\cos\left( \frac{\pi t}{30}\right)-\cos\left( \frac{\pi t}{360}\right)=0$$
My question is, can I find out the roots of that by hand (for my puposes, $0<t \le 1440$)? I plotted it and the period seems to be changing.

Edit: So technically from the vector equation I get this:
$$sin(\frac{\pi t}{30})+sin(\frac{\pi t}{360})=0$$
$$cos(\frac{\pi t}{30})+cos(\frac{\pi t}{360})=0$$

It turns solving these get me the true answers, while solving the one in the title gives me unnecessary ones (I'm saying all of this by looking at the plots). So, how do I solve these 2 sets by hand?

答:
Ok, we have $$sin( \frac{\pi t}{30})+sin( \frac{\pi t}{360})=0$$
Let $ \frac{\pi t}{360}=y$. So we have: $$sin(12y)+sin(y)=0$$
As lan bhattacharjee pointed out, we could use Simpson's formulas here:
\begin{align*}
 sin(12y)+sin(y) &= 2sin[ \frac{1}{2}(12y+y)]cos[ \frac{1}{2}(12y-y)] \\
   &= 2sin( \frac{13y}{2})cos( \frac{11y}{2})
\end{align*}
So we have:
\begin{equation}
 2sin( \frac{13y}{2})cos( \frac{11y}{2})=0
\end{equation}

From the main post it is also given:
$$cos( \frac{\pi t}{30})+cos( \frac{\pi t}{360})=0$$
From this we get:
\begin{equation}
 2cos( \frac{13y}{2})cos( \frac{11y}{2})=0
\end{equation}

Comparing (1) & (2), it is apparent that $cos( \frac{11y}{2})$ has to be zero, because if it isn't zero, then we get $sin( \frac{13y}{2})=cos( \frac{13}{y})$. But we know that $sinx=cosx$ only when $sinx=cosx= \frac{1}{\sqrt{2}}$. Yet, we assumed that $cos( \frac{11y}{2})$ doesn't equal zero. That would leave us with $sin( \frac{13y}{2})cos( \frac{11y}{2}) \neq 0$, which is not true.
So the only way for equations 1 and 2 to be true is if:
\begin{align*}
 cos( \frac{11y}{2}) &= 0 \\
 \frac{11y}{2} &= \frac{\pi}{2}(2n+1), n \in \mathbb{Z} \\
 y &= \frac{\pi}{11}(2n+1) \\
 \frac{\pi t}{360} &= \frac{\pi}{11}(2n+1) \\
 t &= \frac{360}{11}(2n+1)
\end{align*}

Since we're dealing with time, we want:
\begin{align*}
 t &> 0 \\
 ... &> ... \\
 n &\geq 0
\end{align*}
There are 1440 minutes in a single day:
\begin{align*}
 t &\leq 1440 \\
 ... &\leq ... \\
 n &\leq 21
\end{align*}

There are 22 integers between 0 and 22 (including 0 and 22). So, the hour and minute hands of a clock face opposite positions to each other 22 times a day.


问:
Show that $H(x) := |x|^{-1} u(x/|x|^2) $ is harmonic if $u$ is harmonic
I have a "simple" question but I'm not able to solve it.

Suppose that $u$ is a harmonic function on $\mathbb{R}^3$. Prove that the function $H(x) := |x|^{-1} u(x/|x|^2) $ is harmonic on $\mathbb{R}^3 \setminus\{0\}$

I tried brute force evaluating $\Delta H$ but even there I had some problems.. Any hint or partial solution is welcomed.

答:
My attempt:

$$D_iH
= \frac{x_i}{|x|^3}u+\frac 1{|x|}\sum_{j=1}^3D_ju\cdot\left(\frac
{\delta_{ij}}{|x|^2}-\frac{2x_ix_j}{|x|^4}\right)
$$
and
\begin{align*}
D_{ii}H
& = \left(\frac{1}{|x|^3}-\frac{3x_i^2}{|x|^5}\right)u +\frac{x_i}{|x|^3}\sum_{j=1}^3D_ju\cdot\left(\frac
{\delta_{ij}}{|x|^2}-\frac{2x_ix_j}{|x|^4}\right)\\
& -\frac{x_i}{|x|^3}\sum_{j=1}^3D_ju\cdot\left(\frac
{\delta_{ij}}{|x|^2}-\frac{2x_ix_j}{|x|^4}\right)\\
& + \frac 1{|x|}\cdot\frac{D_{ii} u}{|x|^2}+\frac 1{|x|}\frac{2D_iu\cdot x_i}{|x|^4}\\
&-\frac{1}{|x|}\sum_{j=1}^3\left(D_{ij}u\cdot \frac{2x_ix_j\delta_{ij}}{|x|^4} + D_j u\cdot \left(\frac{2x_j}{|x|^4} + \frac{2x_i\delta_{ij}}{|x|^4} + \frac{4x_i^2x_j}{|x|^6} \right)\right)\\
& = \frac{1}{|x|^3}-\frac{3x_i^2}{|x|^5} + \frac{D_{ii}}{|x|^3}.
\end{align*}
Using the harmonicity of $u,$ we get $\Delta H=0$ if I don't miscalculate anything (when cancelling some terms).


问:
handshaking lemma and Erdos-Gallai theorem
The conditions for a sequence to be the degree sequence of a simple graph are given by the Erdos-Gallai theorem in addition to the handshaking lemma. Is there an example of a degree sequence where the handshaking lemma is satisfied, but the Erdos-Gallai theorem is not satisfied and thus, the sequence is not graphic (or vice versa).

Or, does satisfying one of these conditions ensure the other is always satisfied?

答:
The simplest example would be
$$
(2,0,0)
$$
The handshaking lemma is satisfied, but the Erdős–Gallai inequalities are not, as $2\not\le 1\cdot (1-1)+0+0$.


问:
Dimension of the set of self-adjoint operators
I'm trying to figure out what the dimension of the set of self-adjoint operators on V would be, or in more concrete terms:

Let $dim V =n$. Let $S(V)$ denote the set of self-adjoint linear operators on V. What is its dimension?

The only thing I know that somewhat resembles this might be that $dim L(V) = (dim V)^2$ but I'm not sure how I might be able to apply that to this problem.

Any tips or assistance would be greatly appreciated. Thanks!

EDIT: Also, V is a finite dimensional inner product (or Hermitian) space.

答:
Use matrix representations to figure out the answer. Let $\mathcal B$ be an orthonormal basis of $V$, and let $u\in\mathrm{End}_{\Bbb C}(V)$ be an endomorphism of $V$. Then $u$ is selfadjoint iff its matrix in the basis $\mathcal B$ is hermitian i.e. $$u\text{ is selfadjoint iff }^t\overline{\mathrm{Mat}(u,\mathcal B)}=\mathrm{Mat}(u,\mathcal B).$$
What is the (real) dimension of the subspace of hermitian matrices $H_n\subset M_n(\Bbb C)$? Consider a matrix $(u_{ij})\in  M_n(\Bbb C)$: it is hermitian iff for all $i,j\in\lbrace 1,\dots,n\rbrace,~\overline{u_{ji}}=u_{ij}$. You can thus choose the upper coefficients ($i<j$) freely in $\Bbb C$ , the diagonal coefficients $(i=j$) have to be real but can be chosen arbitrarily, and the coefficients below the diagonal ($i>j$) are completely determined by those above the diagonal. Therefore the real dimension of $H_n$ is $2\frac{n(n-1)}{2}+n=n^2$.</p>


问:
Mapping vs function
I am reading an engineering mathematics book whose author says that the term mapping and function have the same meaning and whether one calls $f:X\rightarrow Z$ a function or mapping is a matter of choice. Nevertheless, I do remember from calculus course that the two terms cannot be used interchangeably and I think that mapping is a more general term than function. So, what am I missing here? Is the author right?

答: The author is correct. What you're thinking of is a relation, which is defined as any subset of a Cartesian product of two sets $A \times B = \{(a,b) : a \in A, b \in B\}$. The relation is then true for a pair $a,b$ if $(a,b)$ is an element of the subset. The difference between this and a function/mapping is that with a function $f$, for every $a \in A$ there must be exactly one corresponding element $(a,b) \in f$. This in turn means that a function maps exactly one $b\in B$ to every $a \in A$ and we write $f(a)=b$.

Title: Mapping vs function


问:
Dantzig's algorithm for longest path
What I am looking for is a algorithm from Dantzig which should find a longest way in digraph. Couple of science sources mention this algorithm so afterwhile I have been able to track it down to this book:

https://books.google.cz/books?id=YfdgAQAAQBAJ&pg=PA72&lpg=PA72&dq=Gondran,+Michel+and+Minoux,+Michel.+Graphes+et+algorithmes.+dantzig&source=bl&ots=T4cjgcB38K&sig=ACfU3U0HSGy4bjfpXckDf9DjBIihagX02g&hl=cs&sa=X&ved=2ahUKEwiLus709fztAhXN8qQKHUceB4IQ6AEwEXoECBYQAg#v=onepage&q=du%20plus%20long&f=false

since I don't speak french it is matter of time until I translate all of that. I am unable to find any english-written resources. If you have knowledge of this algorithm, I would be much obliged if you pointed me to alternative resource.

答:
Okay I think I found it. The simple trick is to:

multiply all arcs by -1

find the shortest path in graph using Dantzig's algorithm (desribed in the book as Algorithm 10)

multiply result by -1

Good thing is this algorithm should also work for multigraphs.

UPDATE: I ended up using Bellman-Ford algorithm with the modification described above. Worked for directed multigraph with negative edges.

Title: Dantzig's algorithm for longest path


问:
Prove or disprove this statement: For any $y \in \mathbb{Z}$ and $y \neq \pm 1$, $(y+1)^{2}$ is not divisible by $y$.
I want to prove or disprove this statement:

For any $y \in \mathbb{Z}$ and $y \neq \pm 1$, $(y+1)^{2}$ is not divisible by $y$.

The case where $y$ is even can be easily proved. However, I am stuck at the case where $y$ is odd. I try to start with the fact that "$y+1$ is not divisible by $y$" and expect to use something like the Euclid's Lemma. Any idea on how to proceed?

答: Just observe that $y$ does not divide any number in the form of $ky+ 1$ since it divides $ky$. Now note that $(y+1)^2  = 1 + ky$ where $k = y + 2$.


问:
Determinant of a matrix defined by $a_{i+i,j+1}=a_{i,j}+a_{i+1,j}+a_{i,j+1}$
Let $A_n$ be a $n\times n$ matrix with entries defined by $a_{i+i,j+1}=a_{i,j}+a_{i+1,j}+a_{i,j+1}$ for all $1\leq i,j \leq n-1$ and $a_{1,j}=a_{i,1}=1$ for all $1\leq i,j\leq n$.

What is $\det(A_n)$?

Example:

$$A_3 = \begin{bmatrix}
1 & 1 & 1\\
1 & 3 & 5 \\
1 & 5 & 13
\end{bmatrix}$$

Initial values: $\det(A_1)=1$, $\det(A_2)=2$ and $\det(A_3)=8$.

Maybe it is $$\det(A_n)=2^{n(n-1)/2}$$ but it is just a guess idk.

答:
Hint: As is suggested by the setup, perform row/column reductions by subtracting each row from the row above it (starting at the bottom).

Hint: Read the previous hint $n$ times (with a suitable modification to make things easy).

What do you end up with?

 You end up with a upper triangular matrix whose diagonal entries are $ 2^{i-1}$.

 Hence, the determinant is $ 2^ { n(n-1) / 2}$ as you hypothesized.

Note: I currently don't have a simple explanation for the diagonal entries. The best that I can come up with is

 The generating function for the $i$th row is $ \frac{ (1+x)^{i-1} } { (1 - x)^i }$, and so when you take the successive difference, you end up with $2^{i-1}$.


问:
Solve the system of equations for $\sqrt{xy}$
$$x + y\sqrt{x} = \frac{95}{8}$$
$$y + x\sqrt{y} = \frac{93}{8}$$
$$x, y \in \mathbb{R}$$

I can't solve this system of equations I got asked in a group. I added and substracted them to find the following equations but I don't really know what to do with them.

$$\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}-\sqrt{xy}\right) = \frac{1}{4}$$

$$\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}+\sqrt{xy}\right) - 2\sqrt{xy} = \frac{47}{2}$$

This is all I have. I checked the answer using Wolframalpha and it was $\frac{15}{4}$ for $\sqrt{xy}$.

I have been looking for this question's solution but didn't find any related question/solution on Stack Exchange or Web. Sorry if it is duplicate, thank you in advance.

答:
This is an algebraic proof by a method which can be useful when there is a known solution and the task is to prove there are no further ones. Since finding the intersection of the two original curves is messy, use a suitable third curve through the same point.

Define a third, continuous curve to be $\sqrt{x}+\sqrt{y} = \frac{143}{36}$ for $x\le \frac{20}{9}$ and to be $18\sqrt{x}+20\sqrt{y} = 75$ for $x\ge \frac{20}{9}.$
Finding the intersection of this new curve with either of the original curves gives a cubic (in $\sqrt{x}$ and $\sqrt{y}$, respectively).

In each case, the only real solution gives the intersection point $(\frac{25}{4},\frac{9}{4})$. Thus the new curve separates the two original curves except at the given point and this is therefore the unique solution to the original problem.