问:
Distribution of sum of i.i.d random variables
Assume that we have a sequence of $n$ realizations $x_1, x_2, ..., x_n$ of an i.i.d random variable $X$ with cdf $F_X(x)$ and pdf $f_X(x)$. Now define $Y$ as the sum of $k$ consecutive realizations of $X$

$Y_i = \sum_{j=i}^{k} x_j$

which generates the sequence $Y_1, ..., Y_{n-k+1}$ where $Y_i$ contains $k-1$ elements of $Y_{i-1}$ (and $Y_{i+1}$).

What is the disribution $F_Y(x)$ of $Y$?

答:
I will denote $Y^k = \sum_{j=i}^k X_j$, as the dependence on $k$ is what matters.

As you said, the pdf $f_{Y^k}(x)$ is given by the $k$-fold convolution of $f_X(x)$. You can prove this using the characteristic function of $Y^k$, $\phi_{Y^k}(t) = \langle e^{it Y^k} \rangle_{Y^k} = \langle e^{it \sum_{j=i}^k X_j} \rangle_{X_i, \dots, X_{k+i}} $

You don't need to distinguish these two cases, as the pdf of $Y^k$ depends only on $k$ (not on $n$). In fact, you can prove that the distribution of $Y^k$ does not depend on $i$. For large $k$, you have a central limit theorem for $Y^k$.

That is true also for finite $n$ (see point $1$).


问:
Assume $f:[a,b] \to \mathbb R$ is continuous and for every $x \in \mathbb R :f(x) \ge 0$,if $M:=\sup\left\{f(x): x \in [a,b]\right\}$
Assume $f:[a,b] \to \mathbb R$  is continuous and for every $x \in \mathbb R :f(x) \ge 0$,if $M:=\sup\left\{f(x): x \in [a,b]\right\}$,show that the sequence $$\left\{\left(\int_{a}^{b}f^{n}\left(x\right)dx\right)^{\frac{1}{n}}\right\}_n$$ Is convergent to $M$.

If $M=\sup\left\{f(x): x \in [a,b]\right\}$ then exists $x \in [a,b]$ such that foe every $\epsilon>0:M - \epsilon < f(x) \le M$,so :

$$\left(M-\epsilon\right)^{n}\left(b-a\right)=\int_{a}^{b}\left(M-\epsilon\right)^{n}dx\le\int_{a}^{b}f^{n}\left(x\right)dx\le\int_{a}^{b}M^{n}dx=M^{n}\left(b-a\right)$$
Taking the $n$th root of the both sides yields:
$$\left(M-\epsilon\right)\left(b-a\right)^{\frac{1}{n}}\le\left(\int_{a}^{b}f^{n}\left(x\right)dx\right)^{\frac{1}{n}}\le M\left(b-a\right)^{\frac{1}{n}}$$

Taking the limit of both sides:

$$\left(M-\epsilon\right) \le \lim_{n \to \infty} \left(\int_{a}^{b}f^{n}\left(x\right)dx\right)^{\frac{1}{n}} \le M$$

I have some questions:

How do we know that $M-\epsilon$ is not negative?

We used that $(M - \epsilon )^n< f^n(x)\le M^n $ and from here $$\int_{a}^{b}\left(M-\epsilon\right)^{n}dx\le\int_{a}^{b}f^{n}\left(x\right)dx\le\int_{a}^{b}M^{n}dx$$

Why that's true??

答:
As long as $f$ is not null (which is a degenerate and simple case), one can choose $\epsilon$ positive as little as necessary to $M - \epsilon$ be positive. That's the common use of $\epsilon$.

Integrals are monotonic. Every good integral theory (Riemann, Lebesgue, Stieltjes, etc ...) should provide monotonicity.

ContentLicense: CC BY-SA 4.0


问:
Evaluate $\int_{|z-1|=3} \frac{z-2}{z(z-1)(z+2)}dz$ using residue theorem
I need to evaluate $$\int_{|z-1|=3} \frac{z-2}{z(z-1)(z+2)}dz$$
So the integrand has 3 poles of order 1: $z=0$, $z=1$ and $z=-2$. But $z=-2$ lies on the contour itself. I found out that I need to calculate Cauchy Principal Value, but I can't understand how.

答:
The principal value is normally assessed by cutting out a small portion, say length $2 \varepsilon$, symmetrically on either side of the singularity, evaluate the integral on the resulting contour and then allow $\varepsilon \to 0$.  If this has a limit then that is the principal value.

Applying it to your case, let $I$ be the integral taken round the circular contour but stopping at a distance of $\varepsilon$ each side of $z=-2$.  Then add each of two paths to $I$.  The first, $J_1$ joins the ends of $I$ in a nearly circular arc so that combined $I$ and $J_1$ enclose $z=-2$.  The second, $J_2$, also joins the ends of $I$ using a circular arc, but excluding $z=-2$.

If the residues at the poles are $r_1, r_2, r_3$, with that at $z=-2 = r_1$ then, irrespective of $\varepsilon$,
$$I+J_1 = 2\pi i (r_1+r_2+r_3), \quad I+J_2 = 2\pi i(r_2+r_3)$$
Now the "Indentation Lemma" states that if $f$ is holomorphic except at a simple pole at $z=a$ where it has residue $b$, taking a contour $\gamma=a + \varepsilon e^{i\theta}$ for $\theta_1 \leqslant \theta \leqslant \theta_2$, then as $\varepsilon \to 0$,
$$\int_{\gamma} f(z)dz \to ib(\theta_2-\theta_1).$$

Using it for $J_1$ and $J_2$, noting their respective orientations, as $\varepsilon \to 0$ both $J_1, J_2$ become semi-circular, and by this lemma, $J_1$ and $-J_2$ will have limit $\pi i r_1 $ and $J_1+J_2 \to 0$.  We now have,
$$2I = 2 \cdot 2\pi i (r_2+r_3) + 2\pi i r_1 - (J_1+J_2).$$
Let $\varepsilon \to 0$, to obtain,
$$I = 2\pi i \big(\tfrac{1}{2} r_1+r_2+r_3 \big)$$


问:
Simple Derivative $\frac{x}{\ln x}$
Okay,
I have find the derivative of $$\frac{x}{\ln x}$$ so I used the quotient rule and got: $$\frac{\ln x(1)-x(\frac{1}{x})}{(\ln x)^2}$$

I then simplified this to: $$\frac{-x}{\ln x}.$$

Now here come the problem, when I put $e$ into $f'(x)$ I am supposed to get $-1.5271$ and I do not and I cannot figure out why.  Could someone point out what I am doing wrong?

Thanks

ContentLicense: CC BY-SA 3.0

Title: Simple Derivative $\frac{x}{\ln x}$

Body:
It's
$$ {\ln(x) - 1\over (\ln(x))^2}.$$
The $\ln(x)\cdot 1$ parses to $\ln(x)$.


问:
Set notation for multiple indices
I have a set $A_{x}$ where $A_{x}$ is an indexed family of set and represent element $a$ from set $A_{x}$ as $a^{i}_{m}$ . Is this the right notation for elements with multiple indexing ?. All the resources that I have gone through represent multiple indices as subscripts in ordered pairs only (eg; $a_{i,m}$).

Also like set $A_{x}$, I have multiple sets say $A_{1}, A_{2},... $. Let the indexing set be $ I = \{i,j,k\}$ and $J = \{m,n,o\}$. But all combinations from $ I and J (ie.\{i,j,k\}\times \{m,n,o\} $ will not be considered as index for a particular set $A_{1}$. So if I want to define the indexing set for a particular set $A_{x}$, can I do it as follows: Let $I$,$J$, $S_{1}$ and $S_{2}$ be sets. Let $x_{1} : I \rightarrow S_{1}$ be one mapping and $x_{2}: J \rightarrow S_{2}$ be another mapping. (Two mapping functions are used to denote different indices set). Is it correct to write as follows: $$x_{1}(x_{2}(a)) = \langle a_{m}^{i} \rangle_{i \epsilon I , m \epsilon J}$$
What I understood from other resources is that, if I write $I \times J$, it means all possible combinations (Cartesian product) of elements from $I$ and $J$. How can I represent only a few combinations of elements only from $I$ and $J$. ie. for example if $A_{1}$ has indices ${(i,m),(j,m),(k,m}$ and $A_{2}$ has indices ${(i,n),(j,n)}$ only will occur? So set $A_{1}$ = $\{a_{i}^{m},a_{j}^{m},a_{k}^{m}\}$ and family of set $A$ is union of all $A_{x}$. Also what I understood is $A_{x}$ is a family of indexed sets and $A_{X}$ is also an subset of another family of sets $A$.

答:
Well, if you have a collection $(A_x)_x$ of sets, the elements of $A_x$ may be indexed by $a_{x,i}$ if the sets $A_x$ are denumerable.
I'd avoid superscripts if possible.

Title: Set notation for multiple indices


问:
How to find perimeter of right angle triangle given area and three angles

Hello everyone,

I’ve come across a question where I need to find the perimeter of a right angle triangle given its area and three sides (the only angle written in the picture is 40 degrees, but the other must be 50 degrees given that it is a right triangle). I honestly am at a complete loss at how to solve this, I’d appreciate any help.

Thanks!

答:
In $\triangle ABC,$ with $\angle ACB=90^o$ and $\angle BAC=40^o$: Let $a=BC,\,b=CA,\,c=AB.$ We have $$20 \text { cm}^2=ab/2=(c \cos 40^o)(c\sin 40^o)/2.$$ Therefore $c=\sqrt {\frac {40}{ (\cos 40^o)(\sin 40^o)}}\,$cm.

Knowing $c,$ we have $a=c\cos 40^o$ and $b=c\sin 40^0$.


问:
Do there exists $p$ and $q$ primes for which $p^3+1=2q^2$?
Do there exists $p$ and $q$ primes for which
$p^3+1=2q^2$?

答:
Suppose $p$ and $q$ are two prime numbers satisfying
$$p^3+1=2q^2.$$
Note that
$$p^3+1=(p+1)(p^2-p+1),$$
and let $d$ denote the greatest common divisor of these two factors. Then $d$ divides $3$ because
$$1\cdot(p^2-p+1)-(p-2)\cdot(p+1)=3.$$
If $d=1$ then either $p+1\leq2$ or $p^2-p+1\leq2$, both of which are clearly impossible.

If $d=3$ then $3^2$ divides $p^3+1=2q^2$ and so $q=3$. But then
$$p^3+1=2q^2=18,$$
which is again clearly impossible.


问:
unsure how to expand a LAX pair for KDV equation
This is my first post so im not sure how to make it all mathsy so im going to write it on here, I know that to find the lax equation you find [LM-ML]=0 but im struggling to follow the expansion,

for eg $$L=\left(\frac{d}{dx}\right)^2+u$$
$$M=a\left(\frac{d}{dx}\right)^3+bu\frac{d}{dx}+c\frac{du}{dx}$$

so LM would be $=a(d/dx)^5+$(on the book it says $b(d/dx)(u(d/dx)^2+(du/dx)(d/dx)$) i am unsure how this part comes about, ive tried to expand it and do it all but it doesn't simplify to that

im sorry if it doesnt look clear, again this is my first post.

答:
The commutator, like any product, is distributive. That is, you can compute the commutators of the terms and then add all up in the end. This might be simpler than herding the long total expression, also cancellations occur earlier.
Assuming again that $a,b,c$ are constants, one gets using $D=\frac{d}{dx}$
\begin{align}
[D^2,aD^3]&=0\\
[u,aD^3]&=-a(D^3u) -3a(D^2u)D-3a(Du)D^2\\
[D^2,buD]&=b(D^2u)D+2b(Du)D^2\\
[u,buD]&=-bu(Du)\\
[D^2,c(Du)]&=c(D^3u)+2c(D^2u)D\\
[u,c(Du)]&=0
\end{align}
To get $[L,M]=0$ would require, by comparing the coefficients of the free derivative powers,
\begin{align}
D^2:&&0&=(-3a+2b)Du\\
D: && 0&=(-3a+b+2c)D^2u\\
1:&& 0&=-aD^3u -bu(Du)
\end{align}
so that $b=\frac32a$, $c=\frac34a$, $u'''+\frac32uu'=0$.


问:
complex theory of Pell equation
In the paper entitled "Complex continued fractions: early work of the brothers Adolf and Julius Hurwitz" by Nicola M. R. Oswald and Jörn J. Steuding stated that "Moreover, Adolf Hurwitz mentioned that his approach could be used to build up a complex theory of the Pell equation $t^2−Du^2 = 1 $ where $D$ is a given number and solutions $t$ and $u$ are to be complex integers" (page 515). So, from this statement I understood that finding the complex integer solutions of the above equation is the complex theory of Pell equation. Whether I am correct or not?
What is the domain of $D$. Where can I find details about this complex theory of Pell equation?</p>

答: Check out the articles by R. Lakein on "Computation of the Ideal Class Group of Certain Complex Quartic Fields" in Math. Comp. for applications of Hurwitz continued fractions to the computation of units and class numbers. See also the thesis of B. Cijsouw.</p>

Title: complex theory of Pell equation


问:
determine P(X+Y < 3)
Let X and Y be two jointly continuous random variables with joint probability density function:

$f(x,y)$ = $\frac{1}{25}$ $(5x+y)$       for  $0<x<1$ , $0<y<5$

Hello can someone help me to solve this problem ? im stuck to find the integral limit for x because it is between 0 to 1

答:
are you able to integrate $f(x,y)$ in the purple area?

$$\mathbb{P}[X+Y<3]=\int_0^1\Bigg[\int_0^{3-x}f_{XY}(x,y)dy\Bigg]dx$$

Title: determine P(X+Y < 3)


问:
Need help with proof: $a \mathbb Z + b \mathbb Z = gcd(a,b) \mathbb Z$
As the title says I am trying to prove $a \mathbb Z + b \mathbb Z = gcd(a,b) \mathbb Z$.

My steps so far:

$$a \mathbb Z + b \mathbb Z = \{ a z_1 + bz_2 : z_1, z_2 \in \mathbb Z \} = \{ kp z_1 + lpz_2 : z_1, z_2 \in \mathbb Z \}$$

where $p=gcd(a,b)$, $kp=a$ and $lp=b$.

$$\ldots = \{ p \cdot (k z_1 + l z_2) : z_1, z_2 \in \mathbb Z \} = p \{(k z_1 + l z_2) : z_1, z_2 \in \mathbb Z \}$$

So far, so good. What remains to show is that

$$\{(k z_1 + l z_2) : z_1, z_2 \in \mathbb Z \} = \mathbb Z $$

since then it would follow that

$$p \{(k z_1 + l z_2) : z_1, z_2 \in \mathbb Z \} = p \mathbb Z $$

but here I am stuck. I suspect it has something to do with the fact that $k$ and $l$ are coprime? Maybe there is a theorem about linear combinations of coprime numbers that somehow generates all integers? In any case, help is much appreciated!

答:
It follows immediately from the (universal) definition of ideal sum and the fact that $\,\Bbb Z\,$ is a PID, i.e.

$\begin{align} \forall\ C\!:\,\ \ C\, \supseteq\ A,\,\ B\ \,&\iff\, C\,\supseteq \ A\,+\,B,\, \text{the def'n of ideal sum, becomes in a PID}\\[.3em]
\forall\,(c)\!:\ (c) \supseteq (a),(b)&\iff (c)\supseteq (a)+(b) =: (d),\ \text{becomes in divisibility language}\\[.3em]
\forall\, \ c\!:\,\ \ \ \  c\ \ \mid\,\ \ a,\,\ \ b\ \,&\color{#c00}\iff c\,\mid\, d,\,\ \text{by contains = divides for principal ideals}\\[.3em]
\!\!\!\iff \gcd(a,b) = |d|\ \ \ {\rm [\,or}\ &\,d\,\ {\rm is\ associate\ to}\ \,\gcd(a,b)\,\ \text{in a general domain } D\,]
\end{align}$

since $\ d\mid a,b\ $ by $\,c\!=\!d\;$ in $\,(\!\color{#c00}{\Longleftarrow}\!),\,$ so $\,|d|\,$ is a common divisor of $\,a,b,\,$ necessarily the greatest such since it is divisible by every common divisor $\,c\,$ by $\,(\!\color{#c00}{\Longrightarrow}\!)$

Remark $ $ The $\color{#c00}{\rm red}$ equivalence above is in fact the definition of a gcd in general domains, i.e. a gcd is a common divisor that is divisibly greatest, i.e. an infimum (greatest lower bound) in the poset induced by the divisibility partial order.


问:
Continuity of composition implies continuity of function
Let be $f:→$ function and $T$ topology in $$. If $∘:(,)→(,)$ is continuous, then is $f:(,)→(,)$ continuous?

I can't think if this is true (I don't think so), any help is appreciated!

答:
This is false in general.

Over $\Bbb R$ with the standard topology, let $f(x)$ be the characteristic function of rationals, that is, $$f(x)=\begin{cases}1&x\in\Bbb Q\\ 0&x\notin\Bbb Q.\end{cases}$$

It is well known that this function is nowhere continuous, but $(f\circ f)(x)=1$ for all $x\in\Bbb R$, so continuous over $\Bbb R$.


问:
inequality of integrals
Given that $f$ is continuous, decreasing and always positive on $[0,1]$, show that

$$ \frac{\operatorname{\Large\int}_0^1  \ xf^2(x)dx}{\operatorname{\Large\int}_0^1 xf (x)dx}  \leq \frac{\operatorname{\Large\int}_0^1  f^2(x)dx}{\operatorname{\Large\int}_0^1  f (x)dx}. $$

I am trying to prove this but I do not get any lead, any hint is appreciated

答:
\begin{align*}
  \int_0^1\int_0^1 & (x-y)f^2(x)f(y)\mathrm{d}x\mathrm{d}y   =
\int_0^1\int_0^y(x-y)f^2(x)f(y)\mathrm{d}x\mathrm{d}y\ + \\
& \int_0^1\int_0^x(x-y)f^2(x)f(y)\mathrm{d}y\mathrm{d}x =
\int_0^1\int_0^y(x-y)\big[f^2(x)f(y)-f(x)f^2(y)\big]\mathrm{d}x\mathrm{d}y
<0,
\end{align*}

since $f$ is decreasing.

Title: inequality of integrals


问:
Find $n$ such that ${n \choose 0}+{n \choose 1}+{n \choose 2}+{n \choose 3}$ is a factor of $2^{2018}$
Find all $n$ such that ${n \choose 0}+{n \choose 1}+{n \choose 2}+{n \choose 3}$ is a factor of $2^{2018}$.

I got this interesting question from a friend. I made an assumption and was able to get $n$ as $3$ and $7$ but am not able to get more solutions.

Any help to solve this question would be appreciated. Thanks in advance.

答:
Let

$$u(n)={n \choose 0}+{n \choose 1}+{n \choose 2}+{n \choose 3}=\frac{(n+1)(n^2-n+6)}6$$

First, we must have $n\ge0$, otherwise the left hand side is zero.

If $u(n)$ is a factor of $2^{2018}$, that means $u(n)$ must be a power of two. Since $3$ must divide either $n+1$ or $n^2-n+6$, this means that either:

$$\begin{eqnarray}
n+1&=&3\cdot 2^p\\
n^2-n+6&=&2^q
\end{eqnarray}
$$

or

$$\begin{eqnarray}
n+1&=&2^p\\
n^2-n+6&=&3\cdot2^q
\end{eqnarray}
$$

In the first case, since $n^2-n+6>\dfrac{n+1}{3}$, we must have $2^p|2^q$, but

$$n^2-n+6=(n-2)(n+1)+8=3(n-2)\frac{n+1}{3}+8$$

Hence $\frac{n+1}{3}$ divides $8$. The possible values of $n$ are then $\{2,5,11,23\}$, and it's easy to check that only $2$ and $23$ are valid solutions.

In the second case, we have either $2^p|2^q$ or $2^q|2^p$ (or both).

If $2^p|2^q$, then $n^2-n+6=(n-2)(n+1)+8$, hence $n+1$ divides $8$, and the possible values of $n$ are $\{0,1,3,7\}$, which are all valid solutions.

If $2^q|2^p$, then $\frac{3(n+1)}{n^2-n+6}$ is an integer, but it's $\ge1$ for only finitely many values of $n$, namely for $n\le3$, but we already have all values of $n\le3$ as valid solutions (but it's easy to check that $\frac{3(n+1)}{n^2-n+6}$ is an integer only for $n\in\{1,3\}$).

All in all, the set of solutions is

$$n\in\{0,1,2,3,7,23\}$$


问:
Calculate $x^3 + \frac{1}{x^3}$

Question

$x^2 + \frac{1}{x^2}=34$ and $x$ is a natural number. Find the value of $x^3 +  \frac{1}{x^3}$ and choose the correct answer from the following options:

198

216

200

186

What I have did yet

I tried to find the value of $x + \frac{1}{x}$. Here are my steps to do so:
$$x^2 + \frac{1}{x^2} = 34$$
$$\text{Since}, (x+\frac{1}{x})^2 = x^2 + 2 + \frac{1}{x^2}$$
$$\Rightarrow (x+\frac{1}{x})^2-2=34$$
$$\Rightarrow (x+\frac{1}{x})^2=34+2 = 36$$
$$\Rightarrow x+\frac{1}{x}=\sqrt{36}=6$$

I have calculated the value of $x+\frac{1}{x}$ is $6$. I do not know what to do next. Any help will be appreciated. Thank you in advanced!

答: Hint $(x+\frac{1}{x})^3=x^3+\frac{1}{x^3}+3(x+\frac{1}{x})$ also $(x+\frac{1}{x})=6$ not $36$ you seem to have typed an incorrect calculation.

Title: Calculate $x^3 + \frac{1}{x^3}$


问:
Compare $\mathbb{E}(X)$ and $\mathbb{E}(Y)$ if $X \leq Y$
Assume that non-neqative random variables $X$ and $Y$ are defined at the same probability space and $X(\omega) \leq Y(\omega)$ and $\mathbb{P}(X < Y) > 0$.

It is clear from in this case that

$$
\mathbb{E}(X) \leq\mathbb{E}(Y)
$$

The question: is the following stronger statement true

$$
\mathbb{E}(X) < \mathbb{E}(Y)
$$
?

答: Yes it is true, let $Z=Y-X$, then $Z\geqslant 0$ a.s and $\mathbf{P}(Z>0)>0$. Let us suppose that $\mathbf{E}(X)=\mathbf{E}(Y)$, then $\mathbf{E}(Z)=0$ and therefore $Z=0$ a.s (because $Z\geqslant 0$ a.s) which means that $X=Y$ a.s. This contradicts the fact that $\mathbf{P}(Z>0)>0$, therefore $\mathbf{E}(X)<\mathbf{E}(Y)$.


问:
A metric space whose length metric induces a different topolgy
In any metric space we can define the length of any continuous map $f:I\to X$ from the unit interval to $X$ as the supermum over all finite partitions $P$ of $I$ of the following sum $$\ell(f)=\sup_P\underset{x_i\in P}{\Sigma}d(f(x_i),f(x_{i+1}))$$

To any metric space $(X,d)$ we can associate a length (or intrinsic) metric $$d_\ell(x,y)=\inf\{\ell(f)|f:I\to X;\  x,y\in\operatorname{Im}(f)\}$$

The metrics $d$ and $d_\ell$ are generally different, and it is not difficult to check (by triangular inequality) that the topology induced by the latter is finer or equal. I can easily produce examples where the two metrics are different, for example any non-planar surface embedded in $\mathbb{R}^3$ with the subspace metric, but the two topologies are still the same.

Question: What are some nice examples where the topologies induced by $d$ and $d_\ell$ are different?

答:
As others have pointed out (namely, user DIdier_ ), if you have a non-rectifiable topological embedding $\gamma:I\to (X,d)$ (say, a portion of Koch curve) then the image of $\gamma$ is path-connected in the subspace topology, but it is disconnected in the intrinsic metric (say, $\gamma(0)$ and $\gamma(1)$ aren't in the same galaxy). In fact, a portion of Koch curve has discrete intrinsic topology.

A compact example where every two points are joined by a rectifiable path is this: call \begin{align}\Gamma_1&=\left\{(x,y)\in\Bbb R^2\,:\,  0<x\le\frac1\pi \land y=\sin\frac 1x\right\}\\\Gamma_2&=\{0\}\times [-1,2]\\ \Gamma_3&=\left(\left\{\pi^{-1}\right\}\times[0,2]\right)\cup\left(\left[0,\pi^{-1}\right]\times \{2\}\right)\end{align}

and consider $\Gamma=\Gamma_1\cup \Gamma_2\cup \Gamma_3$ with the Euclidean distance. Obviously, $(\Gamma,d_\ell)$ is homeomorphic (in point of fact, isometric) to $[0,\infty)$, so it isn't compact. But $\Gamma$ is.


Title: Calculate $x^3 + \frac{1}{x^3}$

Body:
Question

$x^2 + \frac{1}{x^2}=34$ and $x$ is a natural number. Find the value of $x^3 +  \frac{1}{x^3}$ and choose the correct answer from the following options:

198

216

200

186

What I have did yet

I tried to find the value of $x + \frac{1}{x}$. Here are my steps to do so:
$$x^2 + \frac{1}{x^2} = 34$$
$$\text{Since}, (x+\frac{1}{x})^2 = x^2 + 2 + \frac{1}{x^2}$$
$$\Rightarrow (x+\frac{1}{x})^2-2=34$$
$$\Rightarrow (x+\frac{1}{x})^2=34+2 = 36$$
$$\Rightarrow x+\frac{1}{x}=\sqrt{36}=6$$

I have calculated the value of $x+\frac{1}{x}$ is $6$. I do not know what to do next. Any help will be appreciated. Thank you in advanced!

Body:
You wrote

$$(x+\frac{1}{x}) = x^2 + 2 + \frac{1}{x^2}.$$

But it should read

$$(x+\frac{1}{x})^2 = x^2 + 2 + \frac{1}{x^2}.$$

Then we obtain $x+\frac{1}{x}=6.$

From $x^2 + \frac{1}{x^2}=34$ we obtain

$(1) \quad x^3+\frac{1}{x}=34 x$

and

$(2) \quad x+\frac{1}{x^3}=\frac{34}{x}.$

If we add (1) and (2) we get

$$x^3+\frac{1}{x^3}=33(x+\frac{1}{x}).$$

Ans so

$$x^3+\frac{1}{x^3}=6 \cdot 33=198.$$

Remark : if $x^2 + \frac{1}{x^2}=34$, then $x$ can not be a natural number !


问:
number of ways to make $2.00
How many different ways can you make $2.00 using only 1 cent, 5 cent, 10 cent, and 25 cent pieces, and 1 and 2 dollar bills (there are 100 cents in a dollar)? I have worked out an equation:

$$p + 5n + 10d + 25q + 100l + 200t = 200$$

Where all variables are non-negative integers. But I don't know what to do from here to figure out how many different sets of non-negative integers there are that satisfy this equation. What do I do next?

答: This is essentially the coin changing problem popularized by Pólya in his "On Picture-Writing", AMM 63(10), 689-697 (dec 1956). Graham, Knuth and Patashnik's "Concrete Mathematics" develops a closed form solution.</p>

Title: number of ways to make $2.00


问:
Self-adjoint and isometric then $A^2=Id$
If $E$ is a real vector space and $A:E\to E$ a linear operator self-adjoint and isometric (preserves inner product) then $A^2=Id$.

I think this problem is easy, but I think I'm missing something, could someone help me?

Body:
Let $(\cdot, \cdot )$ the inner product. For $x \in E$ we have

$(A^2x-x, A^2x-x)=(A^2x,A^2x)-2(A^2x,x)+(x,x)=(Ax,Ax)-2(Ax,Ax)+(x,x)=-(Ax,Ax)+(x,x)=-(x,x)+(x,x)=0.$


问:
Proof explanation of a simple sum
Can someone explain to me (step by step), how the following summation goes from the left to the right:

$\sum_{i=1}^{r}n(1-p)^{i-1} = n(1-(1-p)^r)/p$.

I'm having trouble with maths..

Title: Proof explanation of a simple sum

Body:
$$\sum_{i=1}^{r}n(1-p)^{i-1}+n(1-p)^r = n+\sum_{i=2}^{r+1}n(1-p)^{i-1}= n+(1-p)\sum_{i=1}^{r}n(1-p)^{i-1}.$$

You can draw $\displaystyle\sum_{i=1}^{r}n(1-p)^{i-1}$ from this.


问:
Distance of Limit Point of Subset of Metric Space from that Subset
During the course of working through exercises in various elementary topology works, we come to the question where we have to prove something along the lines that:

if $K$ is closed in a metric space $M = (A, d)$, then $d (K, x) = 0 \iff x$ is a limit point of $K$

or:

$K$ is closed iff ($d (K, x) = 0 \implies x$ is a limit point of $K$)

or however it is presented.

Both of these results is directly dependent upon what appears to be a much more fundamental result:

$d (K, x) = 0$ iff $x$ is a limit point of $K$

This is so straightforward to prove that even I can do it:

Necessary Condition

Let $x$ be a limit point of $K$.

Aiming for a contradiction, suppose $d (K, x) \ne 0$.

By definition of metric, that means:
$$d (K, x) > 0$$

Hence:
$$\forall y \in K: d (x, y) > 0$$

Then:
$$\exists \epsilon \in \mathbb R_{>0}: \forall y \in K: d (x, y) > \epsilon$$

That is:
$$\forall y \in H: y \notin B_\epsilon (x) \setminus \{x\}$$

where $B_\epsilon (x) \setminus \{x\}$ denotes the deleted $\epsilon$-neighborhood of $x$

That is:
$$(B_\epsilon (x) \setminus \{x\}) \cap H = \varnothing$$

Hence, by definition, $x$ is not a limit point of $H$.

From this contradiction it follows that it cannot be the case that $$d (K, x) \ne 0$$

That is:
$$d (K, x) = 0$$
$\Box$

Sufficient Condition

Let $d (K, x) = 0$.

Aiming for a contradiction, suppose $x$ is not a limit point of $H$.

Then, by definition, it is not the case that:
$$\forall \epsilon \in \mathbb R_{>0}: \{y \in H: 0 < d (y, x) < \epsilon\} \ne \varnothing$$

So:
$$\exists \epsilon \in \mathbb R_{>0}: \{y \in H: 0 < d (y, x) < \epsilon\} = \varnothing$$

which means:
$$\exists \epsilon \in \mathbb R_{>0}: \forall y \in H: d (y, x) \ge \epsilon$$

That is:
$$\inf_{y \mathop \in K} (d (y, x)) \ge \epsilon$$

By definition of the distance between $x$ and $K$:
$$d (x, K) \ge \epsilon > 0$$

From this contradiction it follows that it cannot be the case that $x$ is not a limit point of $K$.

That is:

$x$ is a limit point of $K$.

$\blacksquare$

I've taken great care to hammer this out carefully, and it may not be the most streamlined proof of this ever. But the point is, despite the fact that this is a pretty fundamental result, it is rare to find it explicitly presented (in other words, in the topology books that I've studied properly, it isn't).

Is it actually correct, and have I proved it properly?

If not, then where am I going wrong?

As I say, I intend to use it as a basic lemma in the result(s) stated at the start of this question.

答:
Suppose $K$ is a subset of a metric space and $x$ is a point in the space.

It is true that, if $x$ is a limit point of $K$, then $d(x,K)=0$.

It is not true, that, if $d(x,K)=0$, then $x$ is a limit point of $K$. For example, in the space $\mathbb R$, if $K=\mathbb Z$ and $x\in\mathbb Z$, then $d(x,K)=0$ but $x$ is not a limit point of the set $K$, which has no limit points.

A correct statement: $d(x,K)=0$ if and only if $x$ is a limit point or an element of $K$; in other words, $d(x,K)=0$ if and only if $x$ belongs to the closure of $K$.

The statement
$$\forall y\in K:d(x,y)\gt0$$
is equivalent to $x\notin K$, and certainly does not imply the existence of a positive number $\epsilon$ such that $d(x,y)\gt\epsilon$ for all $y\in K$.

However, if we assume that $d(x,K)\gt0$, then such an $\epsilon$ does exist, for instance $\epsilon=d(x,y)/2$.


问:
Finite ultraproduct
I stucked when trying to prove:

If $A_\xi$ are domains of models of first order language and $|A_\xi|\le n$ for $n \in \omega$ for all $\xi$ in index set $X$ and $\mathcal U$ is ultrafilter of $X$ then $|\prod_{\xi \in X} A_\xi / \mathcal U| \le n$.

My tries:

If $X$ is finite set then $\mathcal U$ is principal. Then singleton $\{x\}\in \mathcal U$ and $|\prod_{\xi \in X} A_\xi / \mathcal U| = |A_x|$. If $\mathcal U$ is not principal then for $x \in X$ there is $S_x \in \mathcal U$ with $x \notin S_x$. Then for every $k \in \omega$ there exists equivalence class corresponding to $S_{x_1} \cap \dots \cap S_{x_k}$ with size greater $|A_1|\cdot \dots \cdot |A_k|$.

Can there be said anything about a structure of the ultrafilter if $X$ is infinite? And how to prove it?

答:
For $k=1,2,3,\dots, n$, let $X_k$ be the set of all $\xi\in X$ such that $|A_{\xi}|=k$.

Since $X$ is the finite disjoint union $X_1\cup X_2\cup \cdots \cup X_n$, there is precisely one value of $k$ such that $X_k\in \mathcal{U}$. The ultrapower then has $k$ elements.

Why does it have $k$ elements? We can think in terms of sentences: the assertion that there are precisely $k$ elements  is true "almost everywhere" modulo $\mathcal{U}$.

The following is a perhaps somewhat more concrete reason, if the word concrete can be applied to ultrafilters. Work over the $X_k$ which is in $\mathcal{U}$. For $\xi\in X_k$, we can assume that $A_{\xi}$ consists of the numbers $1,2,\dots,k$. Then any element of the product $\prod A_{\xi}$ is almost everywhere (modulo $\mathcal{U}$) equal to one of the sequences of all $1$'s, all $2$'s, and so on up to all $k$'s.

Title: Finite ultraproduct


问:
Find the derivative of the function $f(x)=(P(x))^a$ where $a\in\Bbb{R}$ and $P(x)$ is a real polynomial with no real roots
I need help to clarify this statement. I found it in a friend's old homework notebook and I think that the answer is

$$f'(x)=aP'(x)\cdot(P(x))^{a-1}$$
However, I just don't get why the fact of $P(x)$ having no real roots is relevant. Is there a way to state the answer using only $P(x)$ instead of $P'(x)$? Am I missing something?

I think his teacher was trying to make this problem look more complex than it is, but I just wanted to check if maybe I am the one who is mistaken.

答: Unless you are allowed to extract the coefficients and modify them, there is no way to obtain $P'(x)$ from $P(x)$. Indeed, there is no functional relation $P'(x)=f(P(x))$ between them, as $P(x)$ and $P(x)+C$ have the same derivative.


问:
Let $T: \mathbb{R}^{3} \longrightarrow \mathbb{R}^{2\times 2}$ be a Linear Transformation such that…
$T(1,5,5)=\begin{bmatrix}5&0\\8&26\end{bmatrix}$

$T(0,1,5)=\begin{bmatrix}0&1\\5&25\end{bmatrix}$

$T(0,0,1)=\begin{bmatrix}5&5\\33&151\end{bmatrix}$

Find a matrix $A\in\operatorname{Im}T$ such that $\det(A)=-245$

What I tried to do: finding a general matrix for the image of $T$ (by using the $3$ given vectors and their images) , But I got a very complicated matrix, to the point that I didn't finish calculating it and came here to ask before.
 I would like to know if there's a trick that I didn't pay attention to or if there's an easier way to solve this question.
 Aprreciate any help.

Body: Since $\det\bigl(T(0,1,5)\bigr)=-5$, $\det\bigl((7T(0,1,5)\bigr)=-49\times5=-245$. So, take$$A=7T(0,1,5)=\begin{bmatrix}0&7\\35&175\end{bmatrix}.$$


问:
How many N-digit numbers have the sum of odd places digits equal the sum of even places digits?
We define a N-digit number as a sequence $d_{1}d_{2}d_3...d_N$, where $d_k\in\{0,1,2,3,4,5,6,7,8,9\}$. So the first digits can be equal to zero. N is a positive even integer.
The problem is to find, how many N-digit numbers are there, which have their sum of the digits on odd places equal to the sum of digits on the even places. Example number would be 2563 for 4-digit numbers as 2 + 6 = 5 + 3.

My thought was to choose the odd digits, we can do that in $10^{N/2}$ ways and then for each of that combinations make even digits a permutation of them ($\frac{N}{2}!$), but this, if correct at all, is only a small fraction of all the possibilities. Why? Because the digits do not have to be the same. Only the sums have to be equal.

Then my second though was that each of such numbers consists of digits from other (shorter) numbers that satisfy the requirement, but it is not true since only the sums have to be equal.

We can for each sum of odd/even digits calculate how many ways we can express it on N/2 digits and then sum the squares of such numbers of combinations. However that leads to very unfriendly calculations.

After you find a nice way, please provide some example calculations for N=4 and N=6.

答:
This is a separate answer to note that the cases $N=4$ and $6$ can be solved very easily by a geometrical method, although this method cannot be extended very easily.

The case N=4

If you consider the two dimensional possibility space with $0$ to $9$ on each axis, then the number of choices for each sum is given by counting vertices on a diagonal of the form $x+y=c$.

These numbers are $1,2,...,9,10,9,...,2,1$ and the sum of squares is $\frac{1}{3}.9.10.19+100=670$.

The case N=6

The  possibility space is three dimensional and we have to consider vertices on planes $x+y+z=c$. Most of these planes intersect the possibility space in a triangular shape and the numbers of vertices on these planes are two sets of the triangular numbers, $T_1,T_2,...,T_{10}$. For the remaining planes, the triangular shapes have truncated corners and the number of vertices on these planes are two sets of the numbers $$T_{11}-3T_1=63,T_{12}-3T_2=69,T_{13}-3T_3=73,T_{14}-3T_4=75.$$

Going across the cube, we then obtain the sum of squares by calculating $$2(\sum _{i=1} ^{10} T_i^2+63^2+69^2+73^2+75^2)=2(7942+19684)=55252.$$


问:
Proving that a field cannot be an ordered field
Suppose $F = \mathbb{R}^2$ is a field equipped with the following operations:

Addition: Defined entrywise i.e.

$$(x_1, x_2) + (y_1, y_2) = (x_1 + y_1, x_2 + y_2)$$

Multiplication:

$$(x_1, x_2) \cdot (y_1, y_2) = (x_1y_1 - x_2y_2, x_1y_2 + x_2y_1)$$

Provided with the operations above, and knowing that $F$ is a field, show that no ordered relation makes $F$ into an ordered field.

I'm attempting to solve the problem above. I started with "Assume that < is an ordered relation on $F$ and $(x_1, x_2), (y_1, y_2), (z_1, z_2) \in F$. We see that: " in an attempt to find a contradiction but I haven't been able to find any. I'm not exactly sure how I can proceed by using the axioms though I'm sure it has something to do with multiplication since it isn't defined entrywise. Also, I'm not sure if I can use this but I found the additive identity to be $(0, 0)$ and the multiplicative identity to be $(1, 0)$. Any hints or assistance are much appreciated.

答:
Hint: $(0,1)^2=(-1,0)=-(1,0)$. In an ordered field, every nonzero square is positive…

The exercise however should also tell you that the set is $F=K^2$, where $K$ is at least a commutative ring, so the operations are well defined.

When is $F$ a field? Precisely if and only if $K$ is a field and, for all $x,y\in K$, $x^2+y^2=0$ implies $x=y=0$.


答:
Let $B:=T(0,1,5)=\begin{bmatrix}0&1\\5&25\end{bmatrix}.$ Then $B \in Im(T)$ and $\det (B)=-5.$

Let $ \alpha \in \mathbb R$ and let $A:=\alpha B$. Then $A \in Im(T)$ and $\det(A)=-5 \alpha^2.$

Now choose $ \alpha $ such that $-5 \alpha^2= -245.$

Title: Let $T: \mathbb{R}^{3} \longrightarrow \mathbb{R}^{2\times 2}$ be a Linear Transformation such that…

Body:
$T(1,5,5)=\begin{bmatrix}5&0\\8&26\end{bmatrix}$

$T(0,1,5)=\begin{bmatrix}0&1\\5&25\end{bmatrix}$

$T(0,0,1)=\begin{bmatrix}5&5\\33&151\end{bmatrix}$

Find a matrix $A\in\operatorname{Im}T$ such that $\det(A)=-245$

What I tried to do: finding a general matrix for the image of $T$ (by using the $3$ given vectors and their images) , But I got a very complicated matrix, to the point that I didn't finish calculating it and came here to ask before.
 I would like to know if there's a trick that I didn't pay attention to or if there's an easier way to solve this question.
 Aprreciate any help.


问:
Are there numbers with the property that the sum of their squares equals the square of their sum?
Are there numbers a and b with the property that

$a^2 + b^2= (a+b)^2$

Note that the answer is yes. For instance, $a=0 , b= \mathbb{R}$ would work. Can you think of other numbers that would also work?

I have found that it will also work when $a=0 , b=0$

Can you think of others ?

答:
As has been said in the comments you can reduce the property in the following way
$$ a^2 + b^2 = (a+b)^2 \Leftrightarrow a^2+b^2 = a^2+2ab+b^2 \Leftrightarrow 0 = 2 a b$$
which just reduces to $a b = 0$. Therefore the solutions you found are all there are because for a product to be zero at least one of the factors has to be zero.


问:
Calculating flight distance of drone without losing line of sight due to obstacles
I hope everyone is off to a good star of the new year!

So I'm trying to figure out a way of calculate how far I'm able to fly my drone without losing visual sight of it when having an obstacle (a forest for example, but a house or hill or similar would also be applicable) between me and the drone. This would be helpful when planning the flight and choosing a take off site.

The problem:
Where I live we have lots of forests. There are clearings in the forests of different sizes however and I want to launch and control my drone from inside of these clearings. The problem is that I need to keep a clear visual line of sight (VLOS) to the drone, the reason for this is twofold: It's the law and the drone will lose its connection eventually if the obstacle is removing enough signal strength. Without knowing how far I'm able to fly the drone in VLOS while planning the mission I risk not being able to cover the planned area. So I need to be able to calculate the possible flight distance during planning and i figured simple trigonometry would be a logic way of doing that.

The solution:
Calculating the flight distance using simple trigonometry, using the Pythagorean theorem and Tangents, Sines, and Cosines.

How:
The illustration below shows the problem. The line i call VLOS need to be clear of any obstacle, in this case, trees. I want to know the maximum flight distance before the VLOS line hits a tree.

Assuming I know the height of the obstacle and the distance between it and where I will be standing with the remote, and also the altitude of the drone, I should be able to calculate the flying distance without losing VLOS.

As you see in this illustration I made (Flight distance illustration) if the drone goes any further the VLOS will hit the trees and thus making it impossible to see, and also weakening the signal. Assuming I'm already at maximum height (120 m) I'm unable to increase the altitude.

In this illustration (Flight distance illustration with values) I have changed the names of the lines to a,b,c etc. I have also made another triangle consisting of the distance between the RC and the bottom of the tree (b1), the height of the tree (a1) and the distance between the RC and the top of the tree. (c1).

Here are the values i will use:

a1= 36,7m

b1= 33m

c1= ?

a2= 120m

b2= ? (what I'm after)

c2= ?

v= ?

To start I would use the pythagorean theorem to get the length of c1 (distance between RC and treetop):

√(a1^2+b1^2)= c1

√36,7^2+33^2= 49,35m

The next step then would be to calculate the angle of v using arccos (degrees)

arccos(b1/c1)=v

arccos(33/49,35)=48,03°

Now the last step is to get the flight distance (b2) using Tangents

a2/Tan(v)=b2

120/Tan(48,03)=107m

So the maximum flight distance (while still being able to unhindered see the drone) is calculated to be 107m, however I have tried this exact scenario and the flight distance was instead 172m. Why is that?

The weaknesses that I have found to this method is:

My math. To put i simply I'm not very good at math and I could have easily made a mistake in my equations.

Using the wrong value for a1. This the easiest error to make here, since I don't have the exact height of the trees. Accuracy depending on method used for measuring the height. (in my case a tree height measuring application)

Using the wrong value for b1. Also easy to make. Depends on method used to measure. (in my case I measured in QGIS on satellite imagery)

Note: The drone doesn't change its altitude and it is relative to where the RC is, so the terrain under the drone doesn't affect the result of a2, it is constant)

If I change a1 and b1 to be -5m and +5m (a1=31,7 and b1=38) I still only get 148 m of flight distance. I find it rather unlikely that I would be that much off on both of these values.

While I was out doing this test I was also able to make a test where a1=26,7m and b1=94m. The result using the same calculation as above is 422 m  flight distance (or b2). The real result however was 843 m of flight distance!

So what is going on here? Why is the calculated flight distance much lower than the real flight distance?

Am I using the equations wrong?

I look forward to any input on this, thanks in advance!

UPDATE: I will do a third test using an obstacle with known height and distance since it all points to that there is something off with the values of a1 and b1. I will post the result as soon as this is done. Thanks for all the input so far!

Best regards
Patrik Forsberg

答:
There is a similarity between triangles and

$$b_2=b_1\frac{a_2}{a_1}=107.9\,m.$$

This can't be wrong.

How did you estimate the distance to the trees, their height, and the altitude of the drone ? How does the drone determine the travelled distance ?


问:
Computing homology groups of some quotient space (Hatcher 2.2.13)

Let $X$ be the $2$-complex obtained from $S^1$ with its usual cell structure by attaching
two $2$-cells by maps of degrees $2$ and $3$, respectively.

Hatcher ask us to compute the homology group of $X/A$ where $A$ is any subcomplex of $X$. The subcomplexes consist of $e^0, S^1, S^1 \cup_2 e^2_1, S^1 \cup_3 e^2_2$ and $X$. Using the cellular boundary formula, I could compute their homology groups, namely,
\begin{align*}
\tilde{H}_n(X) \cong \begin{cases} \mathbb{Z} & n=2 \\
0 & \text{otherwise},
\end{cases}
\end{align*}

\begin{align*}
\tilde{H}_n(S^1 \cup_2 e^2_1) \cong \begin{cases} \mathbb{Z}_2 & n=1 \\
0 & \text{otherwise},
\end{cases}
\end{align*}
and
\begin{align*}
\tilde{H}_n(S^1 \cup_3 e^2_2) \cong \begin{cases} \mathbb{Z}_3 & n=1 \\
0 & \text{otherwise}.
\end{cases}
\end{align*}
I am not able to compute all homology groups of $\tilde{H}_\ast(X/S^1)$ and $\tilde{H}_\ast(X/S^1 \cup_{2,3} e^2_{1,2})$ for those three subcomplexes. Here is what I have tried so far:

I used the long exact sequence in reduced homology for the pair $(X,A)$ with $A$ the subcomplex. For example, the pair $(X,S^1 \cup_2 e^2_1)$ gives
\begin{equation}
0 \longrightarrow \mathbb{Z} \longrightarrow H_2(X/(S^1 \cup_2 e^2_1)) \longrightarrow \mathbb{Z}_2 \longrightarrow 0 \longrightarrow H_1(X/(S^1 \cup_2 e^2_1)) \longrightarrow 0.
\end{equation}
Hence, we deduce that $H_1(X/(S^1 \cup_2 e^2_1))$ is trivial by exactness of the sequence. How can go I from there? I do not understand how to compute the induced inclusion $j_\ast : H_2(X) \rightarrow H_2(X/(S^1 \cup_2 e^2_1))$ and the connecting homomorphism $\partial_\ast : H_2(X/(S^1 \cup_2 e^2_1)) \rightarrow H_1(S^1 \cup_2 e^2_1)$.

答:
I will use $\cong$ to indicate homeomorphic. Note that $$\frac{X}{\Bbb S^1}=\frac{X^{(2)}}{X^{(1)}}\cong \bigvee_{2-\text{cells}}\Bbb S^2= \Bbb S^2\lor \Bbb S^2,$$ $$\frac{X}{\Bbb S^1\sqcup_2 e^2_1}\cong \frac{\Bbb D^2}{\Bbb S^1}\cong\Bbb S^2,\text{ and }\frac{X}{\Bbb S^1\sqcup_3 e^2_2}\cong \frac{\Bbb D^2}{\Bbb S^1}\cong\Bbb S^2.$$

Now, you can compute all $\widetilde H_*(X/A)$, keep in mind that $\widetilde H_*(\Bbb S^2\lor \Bbb S^2)=\widetilde H_*(\Bbb S^2)\oplus \widetilde H_*(\Bbb S^2).$

The quotient of $n$-th skeleton by $(n-1)$-th skeleton is a wedge of $\#n\text{-cells}$ many $n$-sphers.

Note that we are also using following general fact: Let $X=\sqcup_\alpha\Bbb D^n$ and $A=\sqcup_\alpha\Bbb S^{n-1}$ and $Y$ be arbitrary space with a map $f:A\to Y$. Now, $Y$ is a closed subspace of $Y\sqcup_f X$. Next, consider $Z=\text{pt}$ and the mapping $g:Y\to Z$. So, we have $$Z\sqcup_g \big(Y\sqcup_f X\big)\cong Z\sqcup_{g\circ f}X$$ by Law of Horizontal Composition. Next, note that both $g$ and $g\circ f$ is constants. Now, adjunction space of constant map is quotient space, so that $$\frac{Y\sqcup_f X}{Y}\cong Z\sqcup_g \big(Y\sqcup_f X\big)\text{ and }Z\sqcup_{g\circ f}X\cong\frac{X}{A}=\frac{\sqcup_\alpha\Bbb D^n}{\sqcup_\alpha\Bbb S^{n-1}}\cong\lor_\alpha\Bbb S^n$$$$\implies\frac{Y\bigsqcup_f \sqcup_\alpha\Bbb D^n}{Y}\cong\lor_\alpha\Bbb S^n.$$


问:
how to solve this into partial fractions
I'm having a bit of a hard time putting this into partial fractions:
$$\frac{10}{x^2+2x+1+\pi^2}.$$
I know that the roots of the denominator are $-1 \pm i\pi$, but I dont know how to proceed on puting this in partial fractions. Can anyone here help me, please?

Body:
\begin{gather*}
I=\frac{10}{x^{2} +2x+1+\pi ^{2}} =\frac{10}{( x+1+i\pi )( x+1-i\pi )}\\
=\frac{10}{2i\pi }\frac{2i\pi }{( x+1+i\pi )( x+1-i\pi )}\\
=\frac{10}{2i\pi }\frac{( x+1+i\pi ) -( x+1-i\pi )}{( x+1+i\pi )( x+1-i\pi )}\\
=\frac{10}{2i\pi }\left(\frac{1}{( x+1-i\pi )} -\frac{1}{( x+1+i\pi )}\right)
\end{gather*}


问:
Is the Erdős–Faber–Lovász conjecture open still?
Is the Erdős–Faber–Lovász   Conjecture open still? According to Wikipedia it is unsolved still, but I think this is not hard to solve this conjecture.

Conjecture: If $n$ complete graphs, each having exactly $n$ vertices, have the property that every pair of complete graphs has at most one shared vertex, then the union of the graphs can be colored with $n$ colors.

My attempt: Consider the, $n-1$ complete graphs $K_n$ which all has a common vertex, say $u$. then this graph can be colored with $n$ colors. by choosing $n-1$ vertices (one from each graph $K_n\backslash \{u\}$) by different color and construct a graph with this vertices and a new vertex, say $v$. we can assign to $v$ color of $u$. Why this is not correct or maybe I don't understand the problem ?

答: On January 2021 The problem was settled for sufficiently large values of $n$ https://arxiv.org/abs/2101.04698


问:
What does the order of a permutation say about it's cycle lengths?
Say we have a Permutation $a$ of $S_5$.

We know that a is of order 6. Since the order is the least common multiple of a permutation, $a$ probably looks something like this:

$(a_1, a_2, a_3)(a_4, a_5)$

My question is whether you can always determine the length of the disjunct cycle when you only have the order and the number of elements i.e. 5.

答: The lcm of the lengths of the cycles in its cycle decomposition is equal to the order.  That's about all you can say.</p>


问:
Separable extension iff $\Omega_{E/F}=0$?
I know if $E/F$ is a finitely generated algebraic extension of fields, then $E/F$ is separable iff the Kahler differential $\Omega_{E/F}=0$.

Question: I wonder if we only assume that $E/F$ is algebraic without finitely generated, and  $\Omega_{E/F}=0$, then is $E/F$ separable?

Edit: This question doesn't solve my question, since the answer there uses the assumption that $E/F$ is finitely generated.

答: Kahler differentials commute with colimits, and any algebraic extension is the colimit of its finite subextensions (more generally, any field extension is the colimit of its finitely generated subextensions). Applying the result in the finite case, we have your desired result as the colimit of a diagram of zero modules is zero.</p>


问:
Chain rule Differentiation help
Can someone help me differentiate the function $$\frac{\ln(x)+\sin(x)}{x^2}$$ With respect to $x$? I tryed applying the chain rule but I keep geting lost.</p>

答:
You can use the quotient rule to solve this. Given $f(x)=\frac{u}{v}$, $f'(x)=\frac{u'v-uv'}{v^2}$.

So we have:

$$f'(x)=\frac{(\frac{1}{x}+\cos(x))(x^2)-(\ln(x)+\sin(x))(2x)}{x^4}$$

Which simplifies to:

$$f'(x)=\frac{x+x^2\cos(x)-2x\ln(x)-2x\sin(x)}{x^4}$$

And then:

$$f'(x)=\frac{1+x\cos(x)-2\ln(x)-2\sin(x)}{x^3}$$

Alternatively, you can use the product rule ($f(x)=uv \implies f'(x)=uv'+u'v$) for this on the following function. The product rule is often more intuitive than the quotient rule, so this might work better for you.:

$$f(x)=(\ln(x)+\sin(x))(x^{-2})$$

Which gives:

$$f'(x)=(\ln(x)+\sin(x))(-2)(x^{-3})+x^{-2}(\frac{1}{x}+\cos(x))$$

This simplifies to:

$$f'(x)=(\ln(x)+\sin(x))(-2)(x^{-3})+x^{-3}(1+x\cos(x))$$

And then:

$$f'(x)=\frac{-2\ln(x)-2\sin(x)+1+x\cos(x)}{x^{3}}$$

We can see that the solutions from the applications of the quotient rule and the product rule are equal, so that's a great way to check the answer as well!

Title: Chain rule Differentiation help


答:
Let the fraction be $$\frac{A}{x+1-i\pi}+\frac{B}{x+1+i\pi}.$$ Then $$A(x+1+i\pi) + B(x+1-i\pi) \equiv 10.$$  Compare the coefficient of $x$ on the LHS to see that $A = -B$.  Then $2Ai\pi = 10$, so that $A = \dfrac{5}{i\pi}.$  Hence $$\frac{10}{x^2+2x+1+\pi^2} \equiv \frac{5}{i\pi}\left(\frac{1}{x+1-i\pi}-\frac{1}{x+1+i\pi}\right).$$

P.S. My answer simplified the fraction.

Title: how to solve this into partial fractions

Body:
I'm having a bit of a hard time putting this into partial fractions:
$$\frac{10}{x^2+2x+1+\pi^2}.$$
I know that the roots of the denominator are $-1 \pm i\pi$, but I dont know how to proceed on puting this in partial fractions. Can anyone here help me, please?


Body:
Welcome to MSE:$$\frac{10}{x^2+2x+1+\pi^2}=\frac{10}{(x+1)^2+\pi^2}=\\
\frac{a}{x+1+i\pi}+\frac{b}{x+1-i\pi}=\\\frac{a(x+1-i\pi)+b(x+1+i\pi)}{(x+1)^1+\pi^2}=\\
\frac{(x+1)(a+b)-i\pi(a-b)}{(x+1)^1+\pi^2}=\\\to a+b=0 \to a=-b\\
and \\(-i\pi)(a-b)=10\to a-b=\frac{10}{-i\pi}$$so $$-2b=\frac{10}{-i\pi}\to b=\frac{5}{i\pi}=\frac{5}{i\pi}\frac ii=-\frac{5i}{\pi}\\a=\frac{5i}{\pi}$$


答:
Remember than you can always process by identification of the coefficients as a fallback when you don't see some other shortcut.

$\dfrac{10}{x^2+2x+1+\pi^2}=\dfrac{a}{x-1+i\pi}+\dfrac{b}{x-1-i\pi}=\dfrac{x(a+b)-(a+b+ai\pi-bi\pi)}{x^2+2x+1+\pi^2}$

Therefore you get to solve

$\begin{cases}a+b=0\\(a+b)+(a-b)i\pi=-10\end{cases}\iff\begin{cases}b=-a\\2ai\pi=-10\end{cases}\iff\begin{cases}a=\frac {5i}{\pi}\\b=-\frac{5i}{\pi}\end{cases}$


答:
You get:

$\frac{10}{x^2-2x+1+\pi^2}=\frac{A}{x+(1-\pi i)}+\frac{B}{x+(1+\pi i)}=\frac{(A+B)x+A(1+\pi i)+B(1-\pi i)}{x^2-2x+1+\pi^2}$

So you have to solve the following system: $\left\{\begin{matrix}
A+B=0\\
A+B+(A-B)\pi i=10
\end{matrix}\right.$

Which is equivalent to: $\left\{\begin{matrix}
B=-A\\
(A-B)\pi i=10
\end{matrix}\right.$

The solution is: $\left\{\begin{matrix}
2A\pi i=10 \Rightarrow A=\frac{5}{\pi i}=-\frac{5 i}{\pi}\\
B=-A \Rightarrow B=\frac{5 i}{\pi}
\end{matrix}\right.$

So you end up with:  $\frac{10}{x^2-2x+1+\pi^2}=-\frac{\frac{5 i}{\pi}}{x+(1-\pi i)}+\frac{\frac{5 i}{\pi}}{x+(1+\pi i)}$


问:
Complex number question
For any complex numbers  $z_1, z_2$, is the quantity $S$: $$
S = 4 \left(| z_1|^6 + |z_2 |^6\right ) + 4 |z_1|^3 |z_2 |^3 + \left(2 |z_1|^2\times \overline{z_1}^2\times z_2^2\right) + \left(2 |z_2|^2\times \overline{z_2}^2\times z_1^2 \right)$$ always real and nonnegative?

Here overline means the complex conjugate.

答:
Hint for the question "Is $S$ always real":

What is $\overline{S}$?

Title: Complex number question


问:
Question 5 Hungerford algebra Pg 252
Consider the following problem :

If char K $\neq $2 and $f\in  K[x]$ is a cubic whose discriminant is a square in K, then f
is either irreducible or factors completely in K.

This question is from section "Galois Group of the Polynomial "  although I have read the section carefully but I am still not able to figure out which result should I use to solve the problem. So ,there is a result that says that Galois Group must be $S_3$ or $A_3$ and if it is $A_3$ then discriminant is equal to square of an element of K.

But Still I am not close to proving why it is either irreducible or factors completely.

So, Can you please help me with that?

答:
By using the assumption $\operatorname{char} K\ne2$ show the claim is true in the case where $f$ is inseparable. So we may assume that $f$ is separable. Let $F$ be a splitting field of $f$. Since the discriminant of $f$ is a square in $K$ the Galois group $G=G(F/K)$ is contained in $A_3$ (Here we again used $\operatorname{char}K\ne2$). In particular we have $\vert G\vert = 1$ or $\vert G\vert =3$. Now if $f$ is not irreducible argue why the case $\vert G\vert =3$ cannot happen. We then get $\vert G\vert=1$, i.e $F=K$.

Edit: Assume that $f$ is inseparable and reducible. So over some extension field $f$ factors as $(x-\alpha)(x-\beta)^2$ where at least one of $\alpha,\beta$ is already in $K$. If $\alpha=\beta$ we are done (since all linear factors are in $K$), so assume $\alpha\ne\beta$. If $\beta\in K$, then $(x-\beta)^2\in K[x]$, so we can divide $f$ byy $(x-\beta)^2$ over $K$ and get $(x-\alpha)\in K[x]$, so $\alpha\in K$. Assume $\alpha\in K$. Again by division we get $(x-\beta)^2\in K[x]$. In characteristic $p$ an irreducible inseparable polynomial always has degree $p^r$. Since we are not in characteristic $2$ this means that $(x-\beta)^2$ cannot be irreducible, hence $\beta\in K$.

In total we proved that if $f$ is inseparable and reducible then $f$ splis into linear factors over $K$.

Title: Question 5 Hungerford algebra Pg 252


问:
Pivot's test on Hessian matrix: why I can't, at the end, multiply the third row by $-1$ and say that it is negative definite?
Look at the matrix
$$A=\begin{pmatrix}
-1& 0& -1\\
0&-2&0\\
1&0&4
\end{pmatrix}.$$
It is the Hessian matrix of a certain function in 3-variables evaluated at a point $P$. In order to understand if $P$ is a relative minimum or a maximum point, I have to study the matrix $A$ by using the pivots. In particular, if all the pivots are strictly positive, the Hessian matrix is positive definite and then the point $P$ is a minimum.

Now, in order to have the matrix in echelon form, I proceed with the elementary operation of substituting the 3rd row with the sum between the first and the third one, obtaining the matrix
$$\begin{pmatrix}
-1& 0& -1\\
0&-2&0\\
0&0&3
\end{pmatrix}.$$
Since we have 2 positive pivots and one is negative, thus the matrix is neither positive definite nor negative definite, thus $P$ is a saddle point.

My question is: In these kind of exercise, in order to obtain the echelon form of the matrix, only linear substitutions of row with their linear combinations are allowed (e.g. permutation of rows are not allowed!). Why I can't, for example, multiply the third row by $-1$ obtaining
$$\begin{pmatrix}
-1& 0& -1\\
0&-2&0\\
0&0&-3
\end{pmatrix}$$
and say it is negative definite?

Could anyone please help me?

Thank you in advance!

答:
The determinant of a square matrix is unchanged by adding one row to another row, since

the determinant of a square matrix containing two identical rows is zero.

the determinant is multilinear (i.e. linear with respect to each row).

The solution consists of the row operation $R_3 \leftarrow R_3 + R_1$, so the determinant of the given matrix is unchanged by this row operation.

However, by multiplying a row of a square matrix by a constant $k$, the new determinant becomes $k$ times the determinant of the original square matrix, due to the multilinearity of the determinant.

In particular, when you multiply the third row by $-1$, you've changed the sign of the determinant, so that would give you the opposite of what you want.


问:
Ring of Endomorphism over $\mathbb{Z}_n$
I must have some definition problem which I can't put my finger on:

When defining homomorphism of rings $ f :R \mapsto S$, we require:

$f(1_R)=1_s$

$f(r_1+r_2)=f(r_1)+f(r_2)$

Now I have question asks to prove that End($\mathbb{Z}_n$) isomorphic to $\mathbb{Z}_n$. If every endomorphism $f$ must send 1 to itself, and preserve sum, how can I get 2 different enomorphisms?

答:
To elaborate on my comment:

The notation $\operatorname{End}(\mathbb Z_n)$ is used here to denote the endomorphisms of $\mathbb Z_n$ as a $\mathbb Z$-module (abelian group). Hence, $f\colon \mathbb Z_n\to\mathbb Z_n$ is in $\operatorname{End}(\mathbb Z_n)$ iff $f(x+y)=f(x)+f(y)$.

Those $\mathbb Z$-module endomorphisms form a ring with addition defined point-wise by $(f+g)(x)=f(x)+ g(x)$ and multiplication given by composition as $(f\circ g)(x)=f(g(x))$.

You can show that every such $f$ will come in the form $f_k(x)=kx$ for some $k\in\mathbb Z_n$ and then see that $f_k + f_l = f_{k+l}$ and $f_k \circ f_l = f_{k\cdot l}$, which turns $k \mapsto f_k$ into (bijective) a ring homomorphism $\mathbb Z_n\to \operatorname{End}(\mathbb Z_n)$, since also $1\mapsto \operatorname{id}_{\mathbb Z_n}$, which is the one-element in $\operatorname{End}(\mathbb Z_n)$.


问:
Finding joker numbers
Let $n$ be an integer, $q(n)$ be the smallest prime number which divides $n$ and $r(n)$ be the biggest prime number less than or equal to $\sqrt{n}$. We say that $n$ is joker if $q(n)=r(n)$.

Except $8$, joker numbers  are of the form $p_n\cdot p_{n+k}$, with $k\geq 0$.
The converse is obviously false.
If $(p_n)$ is the prime numbers sequence, it is easy to prove that a number of the form $p_n\cdot p_{n+k}$ is joker if and only if $p_{n+1}^2 > p_n\cdot p_{n+k}$. In particular, if a number is not joker for $k$, it is not joker for $k+1$ (with the same $n$).
So, all the numbers of the form $p_n^2$ and $p_n\cdot p_{n+1}$ are jokers.
Using a spreadsheet, we can see that for $n\leq 1000$, the percentage of numbers of the form $p_n\cdot p_{n+k}$ which are jokers, with $k=2, 3, 4, 5, 6$ and $7$ is respectively $52 \%, 19.2 \%, 7.8 \%, 2 \%, 0.5 \%$ and $0.1 \%$.

Question : for any $k$, does there exist $n$ such that the number $p_n\cdot p_{n+k}$ is joker ?

答:
With PARI/GP, I could find examples upto $k=22$ , listed in the following output :

```<code>gp > for(s=2,25,z=primes(s);[a,b,c]=[z[1],nextprime(z[1]+1),z[s]];q=1;while(b^2<=a*c,q=q+1;a=b;b=nextprime(b+1);c=nextprime(c+1));print(s-1,"   ",q))
1   1
2   2
3   4
4   30
5   180
6   462
7   890
8   1532
9   3385
10   19871
11   29040
12   31545
13   31545
14   31545
15   597311
16   1293698
17   1293698
18   1293698
19   2279181
20   2279181
21   118374763
22   118374763
</code>```

If the gap to the next prime is denoted with $a$ and the difference $p_{n+k}-p_n$ is denoted with $b$ , then $2a>b$ is sufficient for an example.

Considering that the merit of a prime gap defined as $$\frac{p_{n+1}-p_n}{\ln(p_n)}$$ can be arbitarily large, such a solution should exist for arbitarily large $k$ , but this is of course no proof.

Note that a solution immediately implies that there is a solution for all smaller $k$ as well. So, chances are very good that the answer to your question is "yes".

Title: Finding joker numbers

Body:
With PARI/GP, I could find examples upto $k=22$ , listed in the following output :

gp > for(s=2,25,z=primes(s);[a,b,c]=[z[1],nextprime(z[1]+1),z[s]];q=1;while(b^2<=a*c,q=q+1;a=b;b=nextprime(b+1);c=nextprime(c+1));print(s-1,"   ",q))
1   1
2   2
3   4
4   30
5   180
6   462
7   890
8   1532
9   3385
10   19871
11   29040
12   31545
13   31545
14   31545
15   597311
16   1293698
17   1293698
18   1293698
19   2279181
20   2279181
21   118374763
22   118374763

If the gap to the next prime is denoted with $a$ and the difference $p_{n+k}-p_n$ is denoted with $b$ , then $2a>b$ is sufficient for an example.

Considering that the merit of a prime gap defined as $$\frac{p_{n+1}-p_n}{\ln(p_n)}$$ can be arbitarily large, such a solution should exist for arbitarily large $k$ , but this is of course no proof.

Note that a solution immediately implies that there is a solution for all smaller $k$ as well. So, chances are very good that the answer to your question is "yes".


问:
when does a functor map products into products?
Motivation: wikipedia claims, that in algebraic topology, there holds: $\pi_1(X\times Y)\cong\pi_1(X)\times\pi_1(Y)$ and $\pi_1(X\vee Y)\cong\pi_1(X)\ast\pi_1(Y)$. A similar statement holds for arbitrary products and one-point unions, making the (covariant) fundamental group functor $\pi_1:\mathrm{TOP}^0 / h-\mathrm{TOP}^0\rightarrow GRPS$ preserve products and coproducts.

I'm guessing the same holds for the functors $\pi_k$ (homotopy groups), $H_k$ (homology groups)?

Definitions:

Examples:
In the category of sets, groups, rings, $R$-modules, vector spaces, topological spaces,etc, the product is the cartesian product. In the category of sets and topological spaces, the coproduct is the disjoint union/topological sum. In the category of groups, the coproduct is the free product $\ast$. In the abelian group / $R$-modules / vector spaces category, it is the direct sum $\oplus$. In the topological pointed spaces category, it is the one-point union $\vee$.

Question: I would very much like to prove this in a general way, so I'd like to know the following:
Theorem???: Suppose $F:\underline{A}\rightarrow\underline{B}$ is a covariant / contravariant functor. What are some (reasonably general) sufficient conditions on $F,\underline{A},\underline{B}$, that make $F$ send (products to products and coproducts to coproducts) / (products to coproducts and coproducts to products ), i.e.
$$F \text{ covariant }\Rightarrow F(\prod_{i\in I}A_i)=\prod_{i\in I}F(A_i)\text{ and }F(\coprod_{i\in I}A_i)=\coprod_{i\in I}F(A_i);$$
$$F \text{ contravariant }\Rightarrow F(\prod_{i\in I}A_i)=\coprod_{i\in I}F(A_i)\text{ and }F(\coprod_{i\in I}A_i)=\prod_{i\in I}F(A_i)?$$

The sufficient conditions that I'm looking for are primarily intended for important standard functors, such as $\pi_k$, $H_k$, $H^k$, tangent and cotangent bundle functor, ... so that I can prove the results in one sweep.

Counterexample: In the category of division rings / fields, we have $\mathbb{Z}_2\times\mathbb{Z}_2 = \mathbb{Z}_2$, thus the forgetful functor to the category of sets doesn't preserve products.

答:
A useful sufficient condition is that such a functor $F : C \to \text{Set}$ is representable; such functors preserve limits more or less by definition. For example:

The forgetful functor $\text{Grp} \to \text{Set}$ preserves limits because it's $\text{Hom}(\mathbb{Z}, -)$.

The forgetful functor $\text{Ring} \to \text{Set}$ preserves limits because it's $\text{Hom}(\mathbb{Z}[x], -)$.

For $R$ a commutative ring, the forgetful functor $R\text{-Mod} \to \text{Set}$ preserves limits because it's $\text{Hom}(R, -)$.

The forgetful functor $\text{Top} \to \text{Set}$ preserves limits because it's $\text{Hom}(\bullet, -)$ where $\bullet$ is the one-point space.

The homotopy group functors $\pi_k : \text{hTop}_{\ast} \to \text{Set}$, where $\text{hTop}_{\ast}$ is the homotopy category of pointed topological spaces, preserve limits because they are $\text{Hom}(S^k, -)$.

(This argument doesn't directly apply to functors which take values in categories other than $\text{Set}$, but there's a way to extend it that I'm not familiar with: see this MO question. The extended argument should handle cohomology by Brown representability.)

Another useful sufficient condition is that such a functor $F : C \to D$ is a right adjoint (equivalently, has a left adjoint). This is true of many forgetful functors (where the left adjoint is the corresponding free functor), including the ones above. In fact, the forgetful functor $\text{Top} \to \text{Set}$ has both a left and a right adjoint, which is why it preserves both limits and colimits. The left adjoint sends a set to the discrete topology on that set, and the right adjoint sends a set to the indiscrete topology on that set.

The two conditions are related. If $F : C \to \text{Set}$ has a left adjoint $G : \text{Set} \to C$, then

$$\text{Hom}_C(G(X), Y) \cong \text{Hom}_{\text{Set}}(X, F(Y))$$

implies that

$$\text{Hom}_C(G(1), Y) \cong \text{Hom}_{\text{Set}}(1, F(Y)) \cong F(Y)$$

hence $F$ is representable by $G(1)$. This general pattern explains the forgetful examples above.


问:
"Since $H^*(X)$ is a polynomial ring, $H^{q + 2^i} =0$ for $0 < 2^i <q$"?
I have some questions in the proof:

"Theorem: If $H^*(X,\mathbb{Z}_2)$ is a polynomial ring or a truncated polynomial ring in a generator $x$ of dimension $q$, and $x^2 \neq 0$, then $q=2^k$ for some $k$.

Proof:
Since $H^*(X)$ is a polynomial ring, $H^{q + 2^i} =0$ for $0 < 2^i <q$. Therefore $Sq^{2^i}x=0$ for $0<2^i<q$. But $Sq^{2^k} \neq 0$ for some $k$ such that $0<2^k \le q$. So $q=2^k$."

Could you help me to explain why "Since $H^*(X)$ is a polynomial ring,  $H^{q + 2^i} =0$ for $0 < 2^i <q$" ?

答: If the cohomology ring is a polynomial ring or a truncated polynial ring generated by an element of degree $q$, then the cohomology groups of positive degree less that $q$ are obviously zero.


问:
Calculate $x^3 + \frac{1}{x^3}$

Question

$x^2 + \frac{1}{x^2}=34$ and $x$ is a natural number. Find the value of $x^3 +  \frac{1}{x^3}$ and choose the correct answer from the following options:

198

216

200

186

What I have did yet

I tried to find the value of $x + \frac{1}{x}$. Here are my steps to do so:
$$x^2 + \frac{1}{x^2} = 34$$
$$\text{Since}, (x+\frac{1}{x})^2 = x^2 + 2 + \frac{1}{x^2}$$
$$\Rightarrow (x+\frac{1}{x})^2-2=34$$
$$\Rightarrow (x+\frac{1}{x})^2=34+2 = 36$$
$$\Rightarrow x+\frac{1}{x}=\sqrt{36}=6$$

I have calculated the value of $x+\frac{1}{x}$ is $6$. I do not know what to do next. Any help will be appreciated. Thank you in advanced!

答:
Let have a look at this: finding $x^6+y^6$ given $x+y$ and $xy$

$$U_n=x^n+\frac 1{x^n}$$

$s=x+\frac 1x=U_1$ and $p=x\times\frac 1x=1$

also we have $\begin{cases}U_2=34\\U_0=x^0+\frac 1{x^0}=1+1=2\end{cases}$

We have the relation $$U_{n+1}=U_1U_n-U_{n-1}$$

$U_2=U_1^2-U_0\implies U_1^2=34+2=36\implies U_1=\pm 6$

And $U_3=U_1U_2-U_1=\pm 6(34-1)=\pm 198\quad$ (among proposed answers the positive one fits).

Note: $x$ cannot be integer, since $x+\frac 1x=\pm 6\iff x=\pm(3\pm 2\sqrt{2})$

This method can be generalized to most $x^n+y^n$ types of questions.

Title: Calculate $x^3 + \frac{1}{x^3}$


问:
Sum of unbounded functions
It is a well-known fact that the sum of two unbounded functions may or may not be bounded.
One of the simplest examples to observe this is:

$f(x)=x$, $g(x)=-x$ are unbounded but $f+g$  is bounded and $f-g$ is unbounded.
We can easily construct examples of of two unbounded functions whose sum is bounded using the scheme:
Let $f(x)$ = (Arbitrary unbounded function-1) and $g(x)$ = (Arbitrary bounded function-2) - (Arbitrary unbounded function-1) then $f+g$ = (Arbitrary bounded function-2)

I am trying to construct an example of two unbounded functions whose sum is bounded but which is NOT constructed from the above scheme; that is, an example of two unbounded functions (preferably continuous functions) $f(x)$ and $g(x)$ whose sum cancels off the effect of unboundedness so that it becomes bounded.

答: It will always be of your form. Let $h(x)=f(x)+g(x)$. Then $g(x)=h(x)-(h(x)-g(x))$ where $h$ is bounded and $h-g=f$ is not.

Title: Sum of unbounded functions


Title: Chain rule Differentiation help

Body: Can someone help me differentiate the function $$\frac{\ln(x)+\sin(x)}{x^2}$$ With respect to $x$? I tryed applying the chain rule but I keep geting lost.</p>

Body:
Hint:$(\frac{u}{v}) '=\frac{u 'v-uv '}{v^2}$
$$u=\ln(x)+\sin(x) \to u ' =\frac1x+cos(x)$$

$$v=x^2\to v'=2x$$


问:
In how many ways can you place $12$ different toys in $3$ identical boxes?
I am a freshman studying Computer Science at a university, and currently struggling with some tasks from the Combinatorics part of the Discrete Math class. Any help would be appreciated.

In how many ways can you place $12$ different toys in $3$ identical boxes?

My vision of solving this problem:

For each toy there is $3$ boxes to put it into. $\implies 3^{12}$ possible combinations to place it.

Considering that the boxes are identical we have just counted doubles as well that should not be counted.

There is $3$ identical boxes $\implies$ they can be permuted $P_3 = 3!$ times for each combination of toys and these would actually correspond to the same combination.

Hence, Solution: $$\frac{3^{12}}{3!}$$ possible ways to place $12$ different toys in $3$ identical boxes.

I am not sure if it's really the $3!$ that I should be dividing them by and whether even I should do so. Does this solution make any sense?

答:
The "all $12$ in one box" arrangement corresponds to $3$ labeled-box arrangements.

All other arrangements correspond to $6$ labeled-box arrangements (since the $3$ boxes all hold different contents in all of these).

The number of labeled-box arrangements is 3^12 = 531441.

So the total you want is $1 + (531441 - 3)/6 = 88574$


问:
Why $C(S,F)$ is only a subspace for a number of finite elements in $ S$
The original question is:

Let $S$ be a nonempty set and F a field. Let $ C(S,F) $ denote the set of all functions $f \in F(S,F) $ such that $f(s) = 0$ for all but a finite number of elements of $S$.

Prove that $C(S,F)$ is a subspace of $F(S,F).$

Now it is fairly easy to verify that $ C(S,F)$ is a subspace, but I do not understand why $f(s) = 0$ can only hold for all but a finite number of elements of $S$. It seems that if we allow infinitely many elements, the zero vector still exists, and it is still closed under addition and scalar multiplication.

答:
A simple example will hopefully make it clearer:

$C(S,F)$, more commonly denoted $F^{(S)}$, as opposed to $\mathcal F(S,F)$, also denoted $F^S$, are the underlying vectors spaces of the ring of polynomials with coefficients in $F$, $F[X]$, and of the ring of formal power series, $F[\mkern-2mu[X]\mkern-2mu]$ respectively.


问:
Prove the following inequality for all real numbers a,b,c,d
In this question I think we take the cyclic summation.... I am not familiar with it and could not understand it.Could you please help me and explain it in a easy way. Also how to you go about solving this problem?

Thanks in advance

I am familiar with AM-GM-HM-RMS inequality and Cauchy Schwarz inequality.
$$\frac{a}{b+2c+3d}+\frac{b}{c+2d+3a}+\frac{c}{d+2a+3b}+\frac{d}{a+2b+3c}\geq\frac{2}{3}.$$

答:
Let
$$x = b+2c+3d, \quad y = c+2d+3a,  \quad z = d+2a+3b,  \quad t = a+2b+3c,$$
then
$$a = \frac{z+t-5x+7y}{24},  \quad b = \frac{t+x-5y+7z}{24},$$
$$c = \frac{x+y-5z+7t}{24},  \quad d = \frac{y+z-5t+7x}{24}.$$
The inequality become
$$\left(\frac x z+\frac z x\right)+\left(\frac t y+\frac y t\right)+\left(\frac x y+\frac y z+\frac  zt+\frac tx\right) + 7\left(\frac y x+\frac zy+\frac tz+\frac xt\right)\geqslant 36.$$
Which is true by the AM-GM inequality.


问:
$GL_{n+1}(\mathbb R)/GL_n(\mathbb R)$ is not compact
Let $GL_{n+1}(\mathbb R)\subset M_{n+1}(\mathbb R )$ be given the subspace topology and identify $GL_n(\mathbb R)$ with the subset of $GL_{n+1}(\mathbb R)$ consisting of the matrices of the form $\begin{bmatrix}1&0 \\ 0&A \end{bmatrix}$ with $A\in GL_n(\mathbb R)$ .Then I have to show that  $GL_{n+1}(\mathbb R)/GL_n(\mathbb R)$ is not compact in the quotient topology.

I understand  that $GL_{n+1}(\mathbb R)$ is not compact itself,because it is not bounded.But I have no clear idea about what happens when I mod out by the subset mentioned above.

Any help would be appreciated.Thanks in advance.

答: The function $$B\in GL_{n+1}(\mathbb R)\mapsto b_{11}\in\mathbb R$$ is continuous and constant on equivalent classes, so it factors thru the quotient giving a continuous unbounded function.  Therefore the quotient is not compact.


问:
Find the value of $k$ in $\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \ge \sqrt{a}+\sqrt{b}+\sqrt{c}+k$
It is also given that $abc = 1$.

I used AM-GM inequality to reach till
$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \ge \sqrt{\frac{bc}{a}} + \sqrt{\frac{ac}{b}} + \sqrt{\frac{ab}{c}} $

How to go further

答:
By AM-GM $$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \ge 2(\sqrt{\frac{bc}{a}} + \sqrt{\frac{ac}{b}} + \sqrt{\frac{ab}{c}} )=2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$$
Now using $x^2+y^2+z^2\ge xy+yz+zx$ $$\begin{align*}2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})  \ge 2(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ac}})=2(\sqrt{a}+\sqrt{b}+\sqrt{c})& \\ \ge\sqrt{a}+\sqrt{b}+\sqrt{c}+3\sqrt[3]{\sqrt{abc}}\end{align*}$$ as $abc=1$ we see that $k=3$

Equality when $a=b=c=1$


问:
Equivalence of events - CDF of distance of randomly chosen points.
Problem:

The points $X$ and $Y$ are independently and uniformly chosen on a
segment $[0,a]$. Find the CDF of the distance $D=|X-Y|$.

Solution:

The vector $(X,Y)$ is uniformly distributed in the square $Q=[0,a]$. The event $D<t$ is the same as the event that the point lies in the strip $S$ between the lines $y=x-t$ and $y=x+t$. So that $F_D(t)=\Pr(D\le t)=|S\cap Q|/|Q|=(2at-t^2)/a^2$ if $0\le t<a$ and $1$ if $t\ge a.$

Questions:

Isn't $Q=[0,a]$ a typo? Since this is not a square, it should be $[0,a]\times[0,a]$ right?

I need to intuitively see why those two events are the same, that is why $\{D<t\}\Leftrightarrow\{(X=x,Y=y)\in S\}.$

I've drawn a coordinate system with a $Y-$axis and an $X-$axis and marked $a$, $t$ and the chosen coordinates, but how can I myself show that $S$ is that stripe between those two lines?

答:
It is very easy.

first do a drawing representing you situation

The CDF of D is the white area divided by the total area $a^2$...

It is easier to calculate this ara doing the complement

$$F_D(d)=\frac{a^2-(a-d)^2}{a^2}$$

More precisely you CDF is the following

$$ F_D(d) =
\begin{cases}
0,  & \text{if $d<0$}  \\
\frac{a^2-(a-d)^2}{a^2},  & \text{if $0\leq d<a$ } \\
1, & \text{if $d \geq a$ }
\end{cases}$$


答:
To be completed

given equation

you missed a term $dx$ inside the bracket on the RHS

$x^2+y^2 = e^2d^2 + 2e^2dx + e^2x^2$

$x^2 - e^2x^2 - 2e^2dx = e^2d^2 -y^2$  (same logic with coefficient corrected)

$(1-e^2)x^2 - 2e^2dx= e^2d^2 - y^2$

$$x^2 - \frac{2e^2dx}{(1-e^2)}= \frac{e^2d^2 - y^2}{(1-e^2)}$$

$$x^2 - \frac{2e^2dx}{1-e^2} + (\frac{e^2d}{(1-e^2)})^2 = \frac{e^2d^2 - y^2}{(1-e^2)} + (\frac{e^2d}{(1-e^2)})^2$$ (same logic here)

The first term inside the bracket on the LHS should be $x$.  $$(x - \frac{e^2d}{(1-e^2)})^2 = \frac{e^2d^2 - y^2}{(1-e^2)} + (\frac{e^2d}{(1-e^2)})^2$$

Can you continue?

Title: How to get from $x^2 + y^2 = e^2(d + x)^2$ to $(\frac{(1-e^2)^2}{e^2d^2})(x-\frac{e^2d}{1-e^2})^2+(\frac{(1-e^2)}{e^2d^2})y^2=1$

Body:
I'm working through a problem in my pre-calc book and I don't know how to get the equation into the same form as the book.  We start with $x^2 + y^2 = e^2(d + x)^2$ and the book gets to the following form:

$(\frac{(1-e^2)^2}{e^2d^2})(x-\frac{e^2d}{1-e^2})^2+(\frac{(1-e^2)}{e^2d^2})y^2=1$

When I try I get stuck.  Here are my steps:

1.) $x^2 + y^2 = e^2(d + x)^2$

2.) $x^2+y^2 = e^2(d^2 + dx + x^2)$

3.) $x^2+y^2 = e^2d^2 + e^2dx + e^2x^2$

4.) $x^2 - e^2x^2 - e^2dx = e^2d^2 -y^2$ (Grouping like terms)

5.) $(1-e^2)x^2 - e^2dx= e^2d^2 - y^2$

6.) $x^2 - \frac{e^2dx}{(1-e^2)}= \frac{e^2d^2 - y^2}{(1-e^2)}$

7.) $x^2 - \frac{e^2dx}{1-e^2} + (\frac{e^2d}{2(1-e^2)})^2 = \frac{e^2d^2 - y^2}{(1-e^2)} + (\frac{e^2d}{2(1-e^2)})^2$ Completing the square.

8.) $(x^2 - \frac{e^2d}{2(1-e^2)})^2 = \frac{e^2d^2 - y^2}{(1-e^2)} + (\frac{e^2d}{2(1-e^2)})^2$

But from here I don't know where to go.  Any help is Much appreciated.


问:
Bounding $\|e^{2\pi iq}-1\|$ for $q$ an almost idempotent
Let $A$ be a unital $C^*$-algebra, and let $q\in A$ such that $\|q^2-q\|<\varepsilon$, and $\|q\|<K$.

Question:
How small does $\varepsilon$ need to be (possibly in terms of $K$) to guarantee that

$$\|e^{2\pi iq}-1_A\|<1?$$

答:
You can get a numerical condition by bounding the terms in the sum. Like the others I don't know what you mean with homotopic. Start with

$$\|q^n-q\| = \|q^{n-2}(q^2-q)+q^{n-1}-q \| < \epsilon K^{n-2} +\|q^{n-1}-q\| $$
so by induction
$$\|q^n-q\| < \epsilon \sum_{k=0}^{n-1} K^{n-2-k} =  \epsilon K^{-2} \dfrac{1-K^{n}}{1-K}$$
(for $K\neq 1$). Making use of $\sum_{k=1}^\infty \frac{(i2\pi)^k}{k!}=0$ we may evaluate
$$e^{i2\pi q}-1 =\sum_{k=1}^\infty \frac{(i 2\pi)^k q^k}{k!}= \sum_{k=1}^\infty \frac{(i2\pi)^k(q^k-q)}{k!}$$
Applying the bound we have derived for $q^k-q$ retrieves:
$$\|e^{i2\pi q}-1\| < \epsilon \frac{K^{-2}}{1-K}\sum_k \frac{(2\pi)^k(1-K^n)}{k!}= \epsilon\  \frac{e^{2\pi K}-e^{2\pi}}{K^3-K^2}$$
which provides you with a condition on $\epsilon$ depending on $K$. This bound is definitely not optimal.


问:
Prove difference between a set and the union of other two sets
This question asked which of the two following formulas was always right whilst the other one was sometimes wrong, and what was the necessary and sufficient condition for the formula which was sometimes incorrect to be always right:

(i) $\; A - (B - C) = (A - B) \cup C $

(ii) $\; A - (B \cup C) = (A - B) - C $

Using Venn diagrams, I guessed (ii) was always right, and (i) would be wrong if $A$ and $C$ are disjoint. Thus, the necessary and sufficient condition for (i) to be correct was $A$ and $C$ contain some elements in common.

My questions are:

Were my guesses correct?

How could I prove the two formulas above using words/expressions rather than using Venn diagrams?

Thanks a lot.

答:
Yes, your guesses are correct (except for “the necessary and sufficient condition for (i) to be correct” — I explain it in my other answer.)

Ad (i):

To prove, that it is not always true, it is sufficient to provide particular sets $A$, $B$, $C$ which don't satisfy the equation (i.e. so called contra example).

So let $A$, $B$, $C$ are as in this table, where are progressively calculated expressions from them:

Row
Expression
Value

1
$$A$$
$$\{1\}$$

2
$$B$$
$$\emptyset$$

3
$$C$$
$$\{2\}$$

4
$$(B-C)$$
$$\emptyset$$

5
$$A−(B−C)$$
$$\{1\}$$

6
$$(A−B)$$
$$\{1\}$$

7
$$(A−B)∪C$$
$$\{1, 2\}$$

By comparing rows 5 and 7 we see that the equation is not satisfied.

Ad (ii):

To prove the equality for every sets $A$, $B$, $C$, we need to prove that every element in the set on the left-hand side of the equality

$$A−(B∪C)=(A−B)−C$$

belongs to the right-hand set, and vice versa.

So let

$$x \in A−(B∪C).$$

Then by definition of the set subtraction

$$x \in A, \ \text{but}\ x \notin (B \cup C)$$

But if $x \notin (B \cup C)$, then by definition of the set union

$$\text{neither}\ x \in B, \ \text{nor}\ x \in C$$

which mean — again by definiton of the set subtraction — that

$$x \in (A - B), \ \text{ but }\ x \notin C$$

i.e. — once more by definiton of the set subtraction — that

$$x \in (A - B) - C$$

So we proved that the arbitrary element $x$ of the left-hand side belongs to the right-hand side, too.

The reversed statement we may prove by similar manner.

OwnerDisplayName: Nemo

OwnerDisplayName: MarianD


问:
Contest Problem linked to Fibonacci and Number Theory
Consider the Fibonacci sequence 1, 1, 2, 3, 5, 8, 13, ... What are the last three digits (from left to right) of the 2020th term?

I tried using formulae for nth term of Fibonacci series using generating function and then apply binomial expansion to eliminate irrational terms and then apply mod 1000. I couldn’t make much headway. Is there a smarter way to solve this problem given we have 7 to 8 minutes to solve in contest settings.

答:
By the CRT it’s enough to figure out $F_{2020}$ mod $8$ and $125$.

It’s easy to check that six applications of the linear recurrence relation turn $(0,1)=(F_0,F_1)$ into $(F_6,F_7)=(8,13)=5 \cdot (0,1)$ mod $8$. As the multiplicative order of $5$ mod $8$ is two, it follows that $F_n$ mod $8$ has a period of $12$. Now note that $2020$ is congruent to $4$ mod $12$ so $F_{2020}$ is congruent to $3$ mod $8$.

Now mod $125$. That one is trickier and I’ll need some matrix formalism. Note that five times the recurrence relation turn $(0,1)$ into $(5,8)=5 \cdot (1,1)+3 \cdot (0,1)$ and turn $(1,1)$ into $(F_6,F_7)=(8,13)=8 \cdot (1,1)+5 \cdot (0,1)$, so mod $5$, five times the recurrence relation means multiplication by three.

In other words, let $A=\begin{bmatrix}1 & 1\\1& 0\end{bmatrix}$ be the matrix representing the recurrence relation for the pair $(F_{n+1},F_n)$, then $A^5$ mod $5$ is three times the identity, ie $A^5=3I+5B_1$ for some integer matrix $B$. Thus $A^{20}=I+5B$ for some integer matrix $B$.

Now, mod $125=5^3$, $A^{20n}=I+5nB+25\frac{n(n-1)}{2}B^2$, so to get the identity matrix on the right-hand side we can choose $n=25$. It follows that $F_n$ mod $125$ is $500$-periodic and $F_{2020}$ mod $125$ is $F_{20}$ mod $25$.

Now, we can check (without a calculator) that (all mod $125$) $F_{12}=19,F_{13}=-17$, so $F_{14}=2,F_{15}=-15,F_{16}=-13,F_{17}=-28,F_{18}=-41,F_{19}=-69,F_{20}=15$.

Finally, it follows that $F_{2020}=515$ mod $1000$ (I hope).


问:
Simplifying Boolean Algebra
I am trying to prove that BC + !A!B + !A!C = ABC +!A

I have attempted using De Morgan laws, and substituting X for !A!B and Y for !A!C, however I made no headway in this.

I've also tried grouping the A's like so, !A(!B+!C), however again I couldn't get anywhere. If anyone could point me in the right direction, help me solve, show me a tool that can do it, etc. I'd be grateful.

答:
In standard logical notation, with which I'm much more familiar, you want to prove $$(B \land C) \lor (\lnot A \land \lnot B) \lor (\lnot A \land \lnot C) \;\equiv\; (A \land B \land C) \lor \lnot A$$
We can simplify the right hand side as follows:
$$
\begin{align}
& (A \land B \land C) \lor \lnot A \\
\equiv & \;\;\;\;\;\text{"use negation of $\lnot A$ in other side of $\lor$"} \\
& (\textrm{true} \land B \land C) \lor \lnot A \\
\equiv & \;\;\;\;\;\text{"simplify"} \\
& (B \land C) \lor \lnot A \\
\end{align}
$$
and the left hand side like this:
$$
\begin{align}
& (B \land C) \lor (\lnot A \land \lnot B) \lor (\lnot A \land \lnot C) \\
\equiv & \;\;\;\;\;\text{"factor out $\lnot A$"} \\
& (B \land C) \lor (\lnot A \land (\lnot B \lor \lnot C)) \\
\equiv & \;\;\;\;\;\text{"use De Morgan"} \\
& (B \land C) \lor \lnot(A \lor (B \land C)) \\
\equiv & \;\;\;\;\;\text{"use negation of $B \land C$ in other side of $\lor$"} \\
& (B \land C) \lor \lnot(A \lor \textrm{false}) \\
\equiv & \;\;\;\;\;\text{"simplify"} \\
& (B \land C) \lor \lnot A \\
\end{align}
$$
which shows that both sides are logically equivalent.</p>

Title: Simplifying Boolean Algebra


问:
A generalization of constant chord theorem
Can you provide a proof for the following claim:

Claim. Given two concentric circles $k_1$,$k_2$ and the third circle $k_3$ which intersects both $k_1$ and $k_2$ at the points $A$ and $B$ respectively. A point $P$ is selected on the circle $k_3$ on the outside arc. Let $C$ be projection of the point $P$ through point $A$ onto circle $k_1$ and $D$ projection of the point $P$ through point $B$ onto circle $k_2$. Then the length of the line segment $CD$ doesn't depend on the location of the point $P$ on its arc.

GeoGebra applet that demonstrates this claim can be found here.

In the case when radii of the circles $k_1$ and $k_2$ are equal we can prove the claim by observing that $\angle CPD$ doesn't depend on the position of the point $P$ and that its measure is defined by the difference of the arcs (on the circle $k_1$) $CD$ and $AB$. Since the arc $AB$ is fixed so is the arc $CD$ , hence the length of the line segment $CD$ is constant. But how to prove the claim in the case when the radii of the circles $k_1$ and $k_2$ are not equal?

答:
Let the centre of the concentric circles be $O_{1}$ and the centre of circle $k_{3}$ be $O_{2}$. Draw $O_{1}C$ and $O_{1}D$. Let another random point on circle $k_{3}$ be $P'$. Let points $C'$ and $D'$ be the equivalent points for this point $P'$. Draw $PC'$, $PD'$, $C'D'$, $O_{1}C'$ and $O_{2}D'$.

Proving $CD$ and $C'D'$ equal is equivalent to proving $\angle CO_{1}D=\angle C'O_{1}D'$ and thereafter proving $\triangle CO_{1}D$ and $\triangle C'O_{1}D'$ congruent.

Observe that, $\angle CO_{1}C'=2\angle CAC'=2\angle PAP'=2\angle PBP'=2\left(180-\angle DBD'\right)=\angle D'O_{1}D$.

$\Rightarrow  \angle CO_{1}C'=\angle D'O_{1}D$

Adding $\angle CO_{1}D'$ to both sides gives $\angle CO_{1}D=\angle C'O_{1}D'$.

In $\triangle C'O_{1}D'$ and $\triangle CO_{1}D$, $C'O_{1}=CO_{1}$, $D'O_{1}=DO_{1}$ and $\angle CO_{1}D=\angle C'O_{1}D'$. Hence, $\triangle C'O_{1}D'\cong \triangle CO_{1}D$ and therefore, $CD=C'D'=constant$.


问:
Interpretation of Value at Risk
Let $X$ be a Loss random variable (Positive values of X represents Losses) and let $p \in (0,1)$. I know that the Value at Risk at level $p$ of $X$ is defined as:

$$VaR_p(X) = inf{\{x \in \mathbb{R} : F(x) \ge p \}}= inf{\{x \in \mathbb{R} : P[X \gt x] \le 1- p \}}$$

(Also this infimum is equal to the minimum value because $F(VaR_p(X))\ge p$). My problem is the interepretation of this quantity:

In some books (for example: Loss Models) $Var_p(X)$ is interpreted as the minimum capital required such that the probability of being insolvent is at most $1-p$: that is : $P[X \gt VaR_p(X)] \le 1-p$. This interpretation is fine with me.

In some other references and books (for example Wikipedia) $VaR_p(X)$ is interpreted as the maximum possible loss (at the level $p$) such that the probability of loss being less than $VaR_p(X)$ is at least $p$: that is: $P[X \le VaR_p(X)] \ge p$

This second definition of maximum possible loss doesn't make sense to me because formally the definition of $VaR_p(X)$ is with an infimum (which coincides with the minimum)

I know that the value at risk is also equal to:

$$VaR_p(X) = sup{\{x \in \mathbb{R} : F(x) \lt p \}}= sup{\{x \in \mathbb{R} : P[X \gt x] \gt 1- p \}}$$

But if we try to intepret the $VaR_p(X)$ using this definition as a maximum possible loss it would be:
The maximum possible Loss such that the probability of having a loss $X$ less than $VaR_p(X)$ is less than $p$ but again it still doesn't make sense to me.

I would really appreciate if someone can help me understanding this concept.

答:
In the end VaR is just a quantile in the loss distribution. Then, if the given interpretation agrees with the formal definition of quantile it should be fine.

Start from the definition of the p-quantile of the loss distribution:

$$VaR_p(X) = inf{\{x \in \mathbb{R} : F(x) \ge p \}}$$

Or in English, the minimum (assuming it coincides with the infimum) value such that the probability of X being lower or equal than it is greater or equal than p. This definition is consistent with the first interpretation you give and the following:

It is the minimum loss such that lower losses will happen at least with probability p.

It is the minimum loss such that higher losses will happen at most with probability 1-p.

Now, you can show that

$$VaR_p(X) = inf{\{x \in \mathbb{R} : F(x) \ge p \}} = sup{\{x \in \mathbb{R} : F(x) \lt p \}}$$

Or in English, the maximum (assuming it coincides with the supremum) value such that the probability of X being lower or equal than it is lower than p. This allows for interpretations in terms of maximum losses:

It is the maximum loss such that lower losses will happen at most with probability p.

It is the maximum loss such that higher losses will happen at least with probability 1-p.

Since we are rearranging the formal definitions and writting them in words, all interpretations are equivalent. However, the ones based on the infimum (minimum) are also more intuitive for me.

Title: Interpretation of Value at Risk


问:
Calculator or another way to find the $a_n$ of a function
$$
a_{n+2} - 2a_{n+1} - 3a_n  = 0
$$

I have this  with $a_0=1$ and $a_1=2$.

$$z^{-2}\left[A(z)-a_0-a_1z\right]-2z^{-1}\left[A(z)-a_0\right]-3A(z)=0$$

Then I solve this $A(z)$ and I replace the values $a_0=1$ and $a_1=2$
and I find the result that it is $a_n=1/4(-1)^n+3/4\cdot3^n$ the
generator of the function.

What I want is another easier way to calculate this.It needs a lot of time to reach that. Is a calculator or another way easy way to do that? Thanks in advance

Body:
$$a_{n+2}-2a_{n+1}-3a_n=0;\;a_0=1,\;a_1=2$$
Characteristic equation is
$$x^2-2x-3=0\to x_1=-1;\;x_2=3$$
The general solution is $$a_n=A(-1)^n+B\cdot 3^n$$
Plug initial values
$$a_0=A+B=1;\;a_1=-A+3B=2\to A=\frac{1}{4},B=\frac{3}{4}$$
The actual solution is
$$a_n=\frac{1}{4}(-1)^n+\frac{3}{4} \cdot 3^n$$
simplified
$$a_n=\frac{1}{4} \left((-1)^n+3^{n+1}\right)$$


问:
What does "most" mean in "the most general solution of $\tan\theta=-1$ and $\cos\theta=\frac{1}{\sqrt2}$"?
What does "most" mean here?

The most general solution of $\tan\theta = -1$ and $\cos\theta = \dfrac{1}{\sqrt2}$ is:

(A) $n\pi +  \dfrac{7\pi}{4}$, $n\in I$

(B) $n\pi+{(-1)^n} \cdot \dfrac{7\pi}{4}$, $n\in I$

(C) $2n\pi+\dfrac{7\pi}{4}$, $n\in I$

(D) none of these

答:
The most general solutions means this: each of the $4$ suggested answers defines a set $A$ of some numbers. Most general solution means:

the set $A$ suggested in the answer is indeed a solution set

that set $A$ is super-set of any other solution set $B$


问:
"Since $H^*(X)$ is a polynomial ring, $H^{q + 2^i} =0$ for $0 < 2^i <q$"?
I have some questions in the proof:

"Theorem: If $H^*(X,\mathbb{Z}_2)$ is a polynomial ring or a truncated polynomial ring in a generator $x$ of dimension $q$, and $x^2 \neq 0$, then $q=2^k$ for some $k$.

Proof:
Since $H^*(X)$ is a polynomial ring, $H^{q + 2^i} =0$ for $0 < 2^i <q$. Therefore $Sq^{2^i}x=0$ for $0<2^i<q$. But $Sq^{2^k} \neq 0$ for some $k$ such that $0<2^k \le q$. So $q=2^k$."

Could you help me to explain why "Since $H^*(X)$ is a polynomial ring,  $H^{q + 2^i} =0$ for $0 < 2^i <q$" ?

答: If $x$ is your generator and $\deg(x)=q$, then you may only have nonzero cohomology groups $H^n(X)$ for $n$ a multiple of $q$, ($\deg(x^k)=kq$).  Since $q<q+2^i<2q$ for $0<2^i<q$, $H^{q+2^i}=0$.</p>


问:
Conditional probability with two conditions
I am trying to solve this conditional and joint probability quiz. I am not able to answer one question. Can anyone help? I am trying to understand how any of the answer can be correct, because none of my calculation is leading to these answers.

The question is:

Consider a set containing all Science students OR all male students. What is the Probability that a randomly selected person will belong to this set?

The answers provided are:

40/90

50/90

310/500

260/500

I am not sure which formula to use for finding out this answer. Any help is appreciated.
P.S - I also asked in stats group but I am not getting any answers so posting in math group as well hoping some math wiz can help me :)

Science
Others
Total

Male
80
180
260

Female
70
170
240

Total
150
350
500

答:
The correct answer results to me

$$\frac{150+260-80}{500}=\frac{330}{500}$$

Thus none of the provided answers


答: An alternative way: By the rearrangement inequality we have $$\frac{b}{\sqrt{a}}+\frac{c}{\sqrt{a}}+\frac{c}{\sqrt{b}}+\frac{a}{\sqrt{b}}+\frac{a}{\sqrt{c}}+\frac{b}{\sqrt{c}}\geq \frac{a}{\sqrt{a}}+\frac{a}{\sqrt{a}}+\frac{b}{\sqrt{b}}+\frac{b}{\sqrt{b}}+\frac{c}{\sqrt{c}}+\frac{c}{\sqrt{c}}=2(\sqrt{a}+\sqrt{b}+\sqrt{c})$$ Now proceed by AM-GM as in the answer by Albus Dumbledore.

Title: Find the value of $k$ in $\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \ge \sqrt{a}+\sqrt{b}+\sqrt{c}+k$

Body:
It is also given that $abc = 1$.

I used AM-GM inequality to reach till
$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \ge \sqrt{\frac{bc}{a}} + \sqrt{\frac{ac}{b}} + \sqrt{\frac{ab}{c}} $

How to go further


问:
Is it possible to define a topology on a set such that continuity of a map out of this space implies that map is an isometry?
I understand that the use of defining a topology on a set is to define what it means for a map out of the space to be continuous. It is then natural to ask for which types of maps out of a set  is it possible to define a topology on the set to allow only said maps to be continuous (e.g. any map continuity -> discrete topology, 'traditional' continuity -> euclidean topology).

This has not yet been addressed in the textbook I am currently reading so I decided to take it here.

My question is specifically about isometric maps, but any more general answer will suffice.

答:
In order to talk about "isometry", you need a "metric".

In a general topological space, this is no metric and the topology is not necessarily metrizable.

If you do have a metric space $(X,d)$, then the natural topology is the metric topology induced by the metric $d$. One can then ask whether a continuous function with respect to this topology is an isometry.

But there is no reason for an arbitrary continuous map to be an isometry: you can always consider the constant map.


问:
Theorem about non extendable solution of an ODE Cauchy problem
I' ve found in a book without a proof  the following theorem:
Let $A=I\times \mathbb{R}$, where  I=]a,b[ is an interval with   $-\infty\leq a<b\leq +\infty$, be a set in which  the  function f(x,y): $(x,y) \subset A  \rightarrow \mathbb{R} $ is defined, continous and locally
lipschitzian in y uniformly in x . Let
\begin{cases}
 y'(x)=f(x,y)  \\ y(x_0)=y_0
\end{cases}
be a Cauchy problem with $x_0\in I$.
If $\tilde{y}$  is a not exctendable unique solution of the problem defined in a maximal interval $]a_{m},b_m[$ then we have that:

(i)$b=b_m$ or $b_m<b $   and  $ \lim_{x \to b_m^{-}} |\tilde{y}(x)| = +\infty$

(ii)$a=a_m$ or $a_m>a $   and  $ \lim_{x \to a_m^{+}} |\tilde{y}(x)| = +\infty$

Now I ask where can I find the proof of this theorem and the possible usability in finding the maximal interval of existency of the solution of a Cauchy problem.

答:
A more general statement is that for any ODE IVP
$$y'(x)=f(x,y(x)),~~y(x_0)=y_0$$
where $f$ has an open domain
$$D\subset \Bbb R\times \Bbb R^n,~~ (x_0,y_0)\in D,$$
and $f$ is continuous and locally Lipschitz in $y$ on $D$, ... , then the (unique) maximal solution leaves any compact set inside of $D$.

The proof is essentially that any solution that ends in a compact set has a limit point, and can thus be continued by the local solution of the IVP starting at the limit point.

In your situation, take the compact sets $[a+\delta,b-\delta]\times[-N,N]\subset D$.


问:
Complex number question
For any complex numbers  $z_1, z_2$, is the quantity $S$: $$
S = 4 \left(| z_1|^6 + |z_2 |^6\right ) + 4 |z_1|^3 |z_2 |^3 + \left(2 |z_1|^2\times \overline{z_1}^2\times z_2^2\right) + \left(2 |z_2|^2\times \overline{z_2}^2\times z_1^2 \right)$$ always real and nonnegative?

Here overline means the complex conjugate.

答: No. The only part we need to check is $2\left|z_1\right|^2\,\bar{z}_1^2\,z_2^2 + 2\left|z_2\right|^2\,\bar{z}_2^2\,z_1^2$. Its two summands clearly have opposite argument (polar angles), but different moduli (radii) whenever $\left|z_1\right|\neq\left|z_2\right|$. Therefore the imaginary part of $S$ will be nonzero whenever $\left|z_1\right|\neq\left|z_2\right|$.</p>

Title: Complex number question


问:
Is the following matrix invertible or not? Provide condition if needed.

The matrix is $n\times n$

The nth row is $[n^1,n^2,n^3....n^n]$

I tried the first few examples:
$\begin{pmatrix}1\end{pmatrix}$ ,$\begin{pmatrix}1 & 1\\ 2 & 4\end{pmatrix}$,$\begin{pmatrix}1 & 1 & 1\\ 2 & 4 & 8\\3 & 9 & 27\end{pmatrix}$. They are obviously invertible.

But afterwards I got confused. Do I try to prove that the determinant keeps increasing, or should I try proving by induction?

答:
This matrix is related to the Vandermonde's matrix. It is always invertible because the determinant is always non-zero.

There's an easy procedure to compute the determinant of the Vandermonde matrix.

When calculating the determinant of your matrix, pull out a common multiplier $k$ from each row (where $k$ is the row number). Then you are left exactly with Vandermonde's determinant. The rest is standard theory.

https://en.wikipedia.org/wiki/Vandermonde_matrix


问:
Singularity or not
A point in which a function is not defined, for instance for $f(x)=\frac{1}{x-1}$ let consider $x=1$, it makes sense to know if it is a point of singularity? Or not since the study of derivability makes sense only in the domain?

答:
Well, you need to define what you mean by "singularity". People may call a point outside of the domain of a function "singular". Sometimes, a point where the function is not differentiable is also called "singular".

In complex analysis, there are several kinds of "singularities". For instance, near $z=0$, the functions $f(z)=\frac1z$ and  $g(z)=\frac{\sin z}{z}$ behave quite differently.

Title: Singularity or not


问:
Tangent parabolas
Given that $y^2=x$ and $x^2=y$ are both parabolas, what is the value of $c$ if $y^2=x$ and $x^2+c=y$ are both tangent?

What I have tried:

$$\begin{cases} y^2=x\\ x^2+c= y \end{cases}$$
Taking the tangent line equation I get
$$\begin{cases} y-y_1=\frac 1 {2y_1} (x-x_1) \\ y-y_2=2x_2(x-x_2) \end{cases}$$
Comparing the equations getting $8x_2c-8x_2^3=1$, finding the value of $c$ such that this equation has only one real solution, am I on the right track? What should I do?

答:
HINT

Essentially, the question is asking us to equate the derivatives of the $2$ functions.

If you need any more help, please don't hesitate to ask. I have a full solution ready to type up and post.

Title: Tangent parabolas


问:
Isogeny of algebraic groups
Let $f:G\to H$ be an isogeny between connected linear algebraic groups. What invariants (rank, semisimple rank, reductive rank, being semisimple...) do these groups share? Are there any properties that one might like these groups to share that they do not? I'm interested in cases when $G$ and $H$ are variously reductive and semisimple so if you need to assume something like this it's no problem.

The motivation is a proof I am reading in which the author seems to be asserting that in such a situation $G$ inherits various properties from $H$.

答:
For clarity, an isogeny is a surjective morphism of algebraic groups with finite kernel.

Since an isogeny $f$ induces an isomorphism of Lie algebras, any property of the group that is really a property of the Lie algebra is evidently invariant. So things like being semisimple and being reductive are invariant by isogeny, as are semisimple and reductive ranks. The fundamental example of a (necessary global) characteristic that is not invariant is the fundamental group. Evidently, if $G$ is simply connected the fundamental group of $H$ is $\mathrm{ker}(f)$.

Title: Isogeny of algebraic groups


答:
WLOG any point on $y^2=x$ will be $P(t^2,t)$

So, the derivative at $P,$  $$\dfrac{dy}{dx}=\dfrac1{2t}$$

The derivative of $y=x^2+c$ at $P$ is $$2t^2$$

So, we need $2t^2=\dfrac1{2t}\implies t=?$

$$\implies c=y-x^2_{(\text{at }P)}=t-t^4=?$$

Title: Tangent parabolas

Body:
Given that $y^2=x$ and $x^2=y$ are both parabolas, what is the value of $c$ if $y^2=x$ and $x^2+c=y$ are both tangent?

What I have tried:

$$\begin{cases} y^2=x\\ x^2+c= y \end{cases}$$
Taking the tangent line equation I get
$$\begin{cases} y-y_1=\frac 1 {2y_1} (x-x_1) \\ y-y_2=2x_2(x-x_2) \end{cases}$$
Comparing the equations getting $8x_2c-8x_2^3=1$, finding the value of $c$ such that this equation has only one real solution, am I on the right track? What should I do?


问:
Can I apply L'Hopital rule to this limit?
I have to compute the follow limit:
\begin{equation*}
\lim_{x\rightarrow 0} (1+\sin x)^{\frac{1}{2x}}
\end{equation*}
I have taken logarithm and then I have applied L'Hopital rule but in the book says that it's not valid. Any helps?

答: Since$$(1+\sin x)^{1/(2x)}=\exp\left(\frac{\log(1+\sin x)}{2x}\right),$$all you need to do is to compute the limit$$\lim_{x\to0}\frac{\log(1+\sin x)}{2x}.\tag1$$So, consider the function $f(x)=\log(1+\sin x)$. Then$$(1)=\frac12f'(0)=\frac12$$and therefore your limit is $e^{1/2}=\sqrt e$.


答:
Hint

Think about the relation with the limit $\lim_{x \to \infty} (1 + \frac{1}{x})^{x} = e$

Title: Can I apply L'Hopital rule to this limit?

Body:
I have to compute the follow limit:
\begin{equation*}
\lim_{x\rightarrow 0} (1+\sin x)^{\frac{1}{2x}}
\end{equation*}
I have taken logarithm and then I have applied L'Hopital rule but in the book says that it's not valid. Any helps?


答:
Let $f(x)=(1+\sin x)^{\frac{1}{2x}}$. Then

$$ g(x):=\ln (f(x))= \frac{\ln (1+\sin x)}{2x}.$$

L'Hospital gives

$$\lim_{x \to 0}g(x)=\lim_{x \to 0} \frac{ \cos x}{1+ \sin x} \cdot \frac{1}{2}=\frac{1}{2}.$$

Thus

$$\lim_{x \to 0}f(x)= \sqrt{e}.$$

Your book is wrong !


问:
When is a $p$-adic unit an $m$-th power
Given $x\in\mathbb{Q}_p^*$, we can write $x=p^nu$ where $n\in\mathbb{Z}$ and $u\in\mathbb{Z}_p^*$. Then we can decide whether or not $x$ is a square by looking at $n$ and $u$. If $p\neq 2$ then $x$ is a sqaure if and only if the "first component" of $x$ is a square as a member of $\mathbb{F}_p^*$ and $n$ is even. If $p=2$ then $x$ is a square if and only if $n$ is even and $u\in1+8\mathbb{Z}_p$.

Is there a similar characterization of when $x\in\mathbb{Q}_p^*$ is an $m$-th power?

答:
Incomplete Answer:
At least for the cube root case, this - like the square root case - follows from a theorem of Katok (2007). You can also get nice approximations in both the square root case and cube root case.

For more details on the above, see (link):

Ignacio, P. S., Addawe, J., Alangui, W., & Nable, J. (2013). Computation of Square and Cube Roots of $p$-Adic Numbers via Newton-Raphson Method. Journal of Mathematics Research, 5(2), p. 31.

Title: When is a $p$-adic unit an $m$-th power