问:
Finding minimum and maximum points with Lagrange Multipliers
I need to use Lagrange Multipliers to find the maximum and minimum values of the function:

$f(x,y)=2e^{xy}$

subject to the given constraints:

$2x^2+y^2=32$

So I went through some examples, and I got:

$x=\pm2\sqrt{2}$ and $y=\pm4$ (Wolfram confirms).

Now I'm having trouble finding the maximum and the minimum.
I understood that If I want to find out if  $(2\sqrt2,4)$ is maxima or minima, then I'll take for example $x=3$, and by the constraint, this point will be $(3,\sqrt{14})$, and check if this value is greater or smaller than (for example) $(2\sqrt2,4)$, and if it is bigger - then my $(2\sqrt2,4)$ is a minimum.

But in my example, $3>2\sqrt2$ and $4>\sqrt{14}$, isn't it problematic?

This seems a bit messy. First, am I right? Second, any other way to do so, that is not much complicated?

答:
You don't really need to "plug" any numbers into the function in order to answer your question.  The function  $ \ f(x,y) \ = \ 2e^{xy} \ $ has  symmetry about the origin [ $ \ f(x,y) \ = \ f(-x,-y) \ $ ] .  The constraint curve $ \ 2x^2 \ + \ y^2 \ = \ 32 \ $ is an ellipse centered on the origin, so we expect a critical point in any quadrant to be "mirrored" by a critical point of the same kind in the diagonally-opposite quadrant.

You found two critical values for $ \ x \ $ and $ \ y \ $ each, producing four critical points in all, one in each quadrant as anticipated.  The product $ \ xy \ $ is positive in the first and third quadrants and negative in the second and fourth.  Thus, the exponent in $  \ 2e^{xy} \ $ is positive at $ \ (2\sqrt2 \ , \ 4) \ $  and $ \ (-2\sqrt2 \ , \ -4) \ $  , producing the absolute maxima for our function, with the exponent being negative at $ \ (2\sqrt2 \ , \ -4) \ $ and $ \ (-2\sqrt2 \ , \ 4) \ $  , where the absolute minima are located.  (The values are as shown by user.)


问:
Compute $\mathbb{E}[X]$ and $Var[X]$ for $\lambda=1$
Let X be a discrete random variable such that ($\lambda\geq0)$
$$P(X=k)=\frac{\lambda^k}{k!}e^{-\lambda}\quad\text{for}\;k\in\{0,1,...\} $$

Check that $\displaystyle{\sum^\infty_{k=0}P(X=k)=1}$

Compute $\mathbb{E}[X]$ for $\lambda=1$

Compute Var$[X]$ for $\lambda=1$

My attempt:

For the first part, we can use the fact that the taylor expansion of $e^x$ is $e^x=\displaystyle{\sum_{k=1}^\infty\frac{x^k}{k!}}\;\forall\;x\in\mathbb{R}$.
Then, $$\displaystyle{\sum^\infty_{k=0}P(X=k)=\sum^\infty_{k=0}\frac{\lambda^k}{k!}e^{-\lambda}=e^\lambda\cdot e^{-\lambda}=e^0=1}$$

For the second part, since we know that this is a valid p.d.f (from the first part), we can use the formula $\mathbb{E}[X]=\displaystyle{\sum_{k\in X}kp(k)}$. Then,
$$\mathbb{E}[X]=\sum k\cdot\frac{1^k}{k!}e^{-1}=\sum \frac{1^k}{(k-1)!} e^{-1}=1$$

This is where I am stuck. I don't know if I am finding the expectation correctly because the result seems strange.

Once I am able to find this, I can find the third part by using the formula $\text{Var}[X]=\mathbb{E}[X^2]-\mathbb{E}[X]^2$ to find the variance.

I also have a question about finding $\mathbb{E}[X^2]$. Is it $\displaystyle{\sum_{k\in X}(kp(k))^2}$ or $\displaystyle{\sum_{k\in X} k^2p(k^2)}$?

答:
The key is to always be aware of the supports for your series.   Then you may notice that the first term vanishes, and a change of variables will do the rest.

$$\begin{align}
\mathsf E(X) &= \sum_{k=0}^\infty k\cdot\dfrac{1^k\mathrm e^{-1}}{k!}
\\[1ex]&=\mathrm e^{-1}\left(0\cdot\dfrac{1}{0!}+\sum_{k=1}^\infty\dfrac{1}{(k-1)!}\right)\\[1ex]&=\mathrm e^{-1}\sum_{j=0}^\infty\dfrac{1}{j!}\\[1ex]&= \mathrm e^{-1}\,\mathrm e^1\\[1ex]&=1
\end{align}$$

ContentLicense: CC BY-SA 4.0


问:
Algebraic simplification of $(a-b)(2b-a+2)(2b+1) - (b+1)(2a-2b-1)(2b-a)$
When trying to replicate a proof about Catalan numbers, I came across the following simplification $$(a-b)(2b-a+2)(2b+1) - (b+1)(2a-2b-1)(2b-a)= a^2+a$$
This is confirmed by writing out all the terms (and by WolframAlpha). My question - is there a clever substitution or factorisation that makes this obvious / less tedious? (Or can anyone provide a more succinct explanation of the answer linked above?)

Thanks

Body:
Original problem : $$\color{green}{(a-b)}\color{red}{(2b-a+2)}\color{green}{(2b+1)} - (b+1)\color{blue}{(2a-2b-1)}(2b-a)$$

) Notice how

$\color{green}{(a-b)(2b+1)}=-2b^2+2ab-b+a=b(-2b+2a-1)+a=b\color{blue}{(2a-2b-1)}+a$

$$\Rightarrow\color{blue}{(2a-2b-1)}=\dfrac{\color{green}{(a-b)(2b+1)}-a}{b}$$

) We find the same for the second part of the equation:

$(b+1)(2b-a)=2b^2-ab+2b-a=b\color{red}{(2b-a+2)}-a$

$$\Rightarrow\color{red}{(2b-a+2)}=\dfrac{(b+1)(2b-a)+a}{b}$$

Plugging in we get:

$$(a-b)\color{red}{\dfrac{(b+1)(2b-a)+a}{b}}(2b+1) - (b+1)(2a-2b-1)(2b-a)=$$

$$\dfrac{(a-b)}{b}\cdot[(b+1)(2b-a)+a]\cdot(2b+1) - (b+1)(2a-2b-1)(2b-a)=$$

$$\dfrac{(a-b)}{b}\cdot(2b+1)\cdot[(b+1)(2b-a)+a] - (b+1)(2a-2b-1)(2b-a)=$$

$$\dfrac{(a-b)}{b}\cdot(2b+1)\cdot a + \dfrac{(a-b)}{b}(2b+1)(b+1)(2b-a) - (b+1)(2a-2b-1)(2b-a)=$$

$$\dfrac{(a-b)}{b}(2b+1)a + (b+1)(2b-a)\left(\dfrac{(a-b)}{b}(2b+1)+(2a-2b-1)\right)$$

$$\dfrac{b\color{blue}{(2a-2b-1)}+a}{b}a + (b+1)(2b-a)\left(\dfrac{(a-b)}{b}(2b+1)+(2a-2b-1)\right)$$

$$\dfrac{b\color{blue}{(2a-2b-1)}+a}{b}a + (b+1)(2b-a)\left(\dfrac{b\color{blue}{(2a-2b-1)}+a}{b}+\dfrac{\color{green}{(a-b)(2b+1)}-a}{b}\right)$$

$$\dfrac{b\color{blue}{(2a-2b-1)}+a}{b}a + (b+1)(2b-a)\left(\dfrac{b\color{blue}{(2a-2b-1)}+\color{green}{(a-b)(2b+1)}}{b}\right)$$

$$\dfrac{b\color{blue}{(2a-2b-1)}+a}{b}a + (b+1)(2b-a)\left(\dfrac{(2ab-2b^2-b)+\color{green}{-2b^2+2ab-b+a}}{b}\right)$$

$$\dfrac{b\color{blue}{(2a-2b-1)}+a}{b}a + (b+1)(2b-a)\left(\dfrac{4ab-4b^2-2b+a}{b}\right)$$

$$\dfrac{b\color{blue}{(2a-2b-1)}+a}{b}a + \dfrac{b\color{red}{(2b-a+2)}-a}{b}\left({4ab-4b^2-2b+a}\right)$$

$$\color{blue}{(2a-2b-1)}a+\dfrac{a^2}{b} + \left(\color{red}{(2b-a+2)}-\dfrac{a}{b}\right)\left({4ab-4b^2-2b+a}\right)$$

There is definitely something in this 'equation', but it definitely won't save you either time or paper.


问:
Show that two finite two generator groups are isomorphic
I was faced by a question that I can't solve. Any help would be great! Let $A$ and $B$ be groups with the following properties:

\begin{cases}
    |A| = 9 \cdot 3=27\\
A = \left \\
a^{9} = e \\
b^{3} = e \\
ba = a^{4}b
\end{cases}

\begin{cases}
    |B| = 9 \cdot3=27\\
B = \left<\alpha,\beta\right> \\
\alpha^{9} = e \\
\beta^{3} = e \\
\beta\alpha = \alpha^{7}\beta
\end{cases}
I need to prove that $A \cong B$.

I really can't think of any isomorphism (or how to use some theorem) to solve this. Thanks in advance!

答:
Let’s drop the assumption that $A$ and $B$ have order $27$ (it is not needed). Consider the presentations:
$$\begin{align*}
A &= \langle a,b\mid a^9 = b^3=e,\ ba=a^4b\rangle\\
B &= \langle \alpha,\beta\mid \alpha^9=\beta^3=e,\ \beta\alpha=\alpha^7\beta\rangle.
\end{align*}$$
By von Dyck’s Theorem, we get a morphism from $A$ to $B$ by identifying two elements of $B$ that satisfy the same relations as $a$ and $b$ do, in $B$; and we get a morphism from $B$ to $A$ by identifying two elements of $A$ that satisfy in $A$ the same relations that $\alpha$ and $\beta$ do in $B$.

Consider the pair $(\alpha,\beta^2)$ in $B$. I claim that they satisfy the same relations in $B$ that $a$ and $b$ do in $A$. Indeed, $\alpha^9 =e$, $(\beta^2)^3= (\beta^3)^2 = e^2 = e$. And
$$\beta^2\alpha = \beta(\alpha^7\beta) = \alpha^{49}\beta^2 = \alpha\beta^2.$$
Thus, the map $a\mapsto \alpha$, $b\mapsto \beta^2$ extends uniquely to a morphism $A\to B$.

Now consider the elements $(a,b^2)$ in $A$. We claim they satisfyt the same relations as $\alpha$ and $\beta$ do in $B$. Indeed, we have $a^9 = e$, $(b^2)^3 = (b^3)^2 = e^2 = e$, and
$$b^2a = b(a^4b) = a^{16}b^2 = a^7b^2.$$
Therefore, the map $\alpha\mapsto a$ and $\beta\mapsto b^2$ extends uniquely to a morphism $B\to A$.

Now note that the two morphisms are inverses of each other: we have
$$\begin{align*}
a&\mapsto \alpha\mapsto a&\quad b&\mapsto \beta^2\mapsto (b^2)^2 = b\\
\alpha&\mapsto a\mapsto \alpha &\quad \beta&\mapsto b^2\mapsto (\beta^2)^2 = \beta.
\end{align*}$$
Thus, the two maps are isomorphisms, so $A\cong B$.

Now, we can show that $A$ and $B$ are in fact of order $27$. It’s easy to show $A$ has cardinality at most $27$ by showing every element can be written in the form $a^ib^j$, $0\leq i\lt 9$, $0 \leq j \lt 3$. And then we can consider the semidirect product $C_7\rtimes_{\phi} C_3$, with $C_7$ generated by $x$ and $C_3$ generated by $y$, and where $\phi\colon C_3\to \mathrm{Aut}(C_7)\cong C_6$ sends $y$ to the automorphism $x\mapsto x^4$. Then $x$ and $y$ in $C_7\rtimes_{\phi}C_3$ satisfy the same realtions as $a$ and $b$ to in $A$, so $a\mapsto x$ and $b\mapsto y$ extends to a homomorphism $A\to C_7\rtimes_{\phi} C_3$. Since $x$ and $y$ generate the latter, the morphism is surjective, so $|A|\geq 27$, proving the desired equality.


问:
Questions on Fraenkel models
Halbeisen on page 172 contains a section entitled "The Second Fraenkel Model". The original paper by Fraenkel containing this model can be found here. I have several questions regarding this model and $\mathsf{FM}$-models in general.

If this 1922 paper by Fraenkel is not the first describing $\mathsf{FM}$-models, can someone please point me to his very fist paper in which he invents the technique? (given that Halbeisen calls it the second model I suppose there must be an earlier paper containing his first model although looking at this list of publications it appears to be his first.)

In this 1922 paper he writes that for this model construction we can keep the axioms of Zermelo I,II,IV-VII as they are but in Halbeisen on page 168 it is stated that we need to use modified versions of Empty Set and Extensionality. Hence my question 2.a) is: Is this a mistake in Fraenkel's paper? (Unlikely, I would think) and my question 2.b) is: Empty Set is a consequence of all the other axioms hence wouldn't it be enough to just use a modified version of Extensionality?

Last but not least the question that is most important to me: What exactly did Fraenkel's 1922 paper prove? The aim of course was to show that $\mathsf{AC}$ is independent of $\mathsf{ZF}$ by constructing a model of $\mathsf{ZF}$ in which $\mathsf{AC}$ fails. But the model he constructed is a model of $\mathsf{ZFA}$ and not of $\mathsf{ZF}$ and it seems to be the case that those transfer methods used to embed these Fraenkel models into symmetric models weren't known for another 30 years or so. Hence: what is the relation of $\mathsf{ZFA}$ and $\mathsf{ZF}$? (without using transfer methods)

答:
This is just a footnote to the other responses. But it is interesting that there seems to be a bit of a resurgence of interest in Fraenkel-Mostowski permutation models going on (for a small local example, this year 'Permutation Models for Set Theory' is on the list of essay topics for Part III Maths Tripos).

One source of renewed interest is from computer scientists interested in "capture-avoiding substitution": for some general thoughts, see e.g. this paper.

Another recent source is that Randall Holmes's announced proof of the consistency of NF set theory uses FM techniques. Interesting times!

ContentLicense: CC BY-SA 3.0

Title: Questions on Fraenkel models


问:
Single variable Integration word problem
In 2000 the yearly world petroleum consumption was about 77 billion barrels and the yearly exponential rate of increase in use was 2%. How many years after 2000 are the world's total estimated oil reserves of 1020 billion barrels likely to last?

What I did was I set up an Integral bounded from 0 to n with the integrand as 77(1.02)^t dt, and I got roughly 11.76, however my textbook has the integrand set as 77e^(0.02t) which I don't understand why they set it as. Sorry for the bad formatting, somewhat new to mathstackexchange.

I don't understand why the textbook would use e^(0.02t) since exponential increase is a(1+r)^t if I am not mistaken

答:
It is the same formula. $77(1.02)^t=77e^{.02t}$.  You are correct that exponential growth is of the form $a(1+r)^t$. You can convert between the forms $ae^{kt}=a\left(1+r\right)^{t}$ via $k=\ln(1+r)$. For this problem, $k=\ln1.02\approx0.0198$ or about .02. Then $77(1.02)^{t}=77e^{.02t}$.

However, I am unsure about integrating. Are you sure you're supposed to integrate? Just set $77e^{.02t}=1020\Rightarrow t=129.19$ years.


答:
This is not really a general method but just an insight. I would look at the terms and see if they can be written as sums of common terms. In your example:
$$2b+1 = b + (b+1)$$
$$2b-a = b - (a-b)$$
$$2b-a+2 = (b+1) + (-a+b+1)$$
$$2a-2b-1 = (a-b) - (-a+b+1)$$
Using the above, the expression can be re-written as:
$$(a-b)[(b+1)+(-a+b+1)][b+(b+1)] \; - \; (b+1)[(a-b)-(-a+b+1)][b-(a-b)]$$
At this point it seems that the expression can be simplified, because it involves some products of same terms with opposite signs that cancel each other out. If you like something tidier, you can take $X=(a-b)$ , $Y=(b+1)$ , $Z=(-a+b+1)$ , and $T=b$ . Then the above equation becomes:
$$X(Y+Z)(Y+T) + Y(X-Z)(X-T) $$
which makes it more intuitive to continue:
$$X(Y^2 + (Z+T)Y + ZT) \; + \; Y(X^2 -(Z+T)X +ZT)$$
$$= \; (X+Y)(XY + ZT)$$
Now getting back to the definitions of our auxiliary variables we see that
$$X+Y=a+1$$
$$XY + ZT = (a-b)(b+1) + (-(a-b)+1)(b) = (a-b)b + a-b -(a-b)b + b = a$$
So the original expression simplifies to $(a+1)a$ .

Title: Algebraic simplification of $(a-b)(2b-a+2)(2b+1) - (b+1)(2a-2b-1)(2b-a)$

Body:
When trying to replicate a proof about Catalan numbers, I came across the following simplification $$(a-b)(2b-a+2)(2b+1) - (b+1)(2a-2b-1)(2b-a)= a^2+a$$
This is confirmed by writing out all the terms (and by WolframAlpha). My question - is there a clever substitution or factorisation that makes this obvious / less tedious? (Or can anyone provide a more succinct explanation of the answer linked above?)

Thanks


问:
How to simplify $\frac{\sqrt{2}+2(\cos 20^\circ+\cos 25^\circ)}{\sin \left(90-\frac{45}{2}\right)\sin 55^\circ \sin 57.5^\circ}$?
The problem is as follows:

Simplify the following expression:

$B=\frac{\sqrt{2}+2(\cos 20^\circ+\cos 25^\circ)}{\sin \left(90-\frac{45}{2}\right)\sin 55^\circ \sin 57.5^\circ}$

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&7.5\\
2.&6\\
3.&8\\
4.&5\\
\end{array}$

I'm not sure how to proceed here because the division seems kind of complicated to simplify.

But I could spot that suspicioulsy $\sqrt{2}= \csc 45^\circ$

and also $2= \csc 30^\circ$

But I don't know if these would come into play in solving this problem as it is challenging. Perhaps does it exist a way to solve this without much fuss?.

I could also spot that:

$\sin \left(90-\frac{45}{2}\right)= \cos \frac{45}{2}$

$\sin 55^\circ = \cos 35^\circ$

and

$\sin 57.5^\circ= \cos 32.5^\circ = \sin \frac{65}{2}^\circ$

The rest I presume that involves the simplifcation of the expresion using sum to product formulas. But I got stuck with those. Can someone help me here?.

答:
Hint

$$\sqrt2=2\sin45^\circ=2\sin2\cdot\dfrac{45}2^\circ=?$$

Use https://mathworld.wolfram.com/ProsthaphaeresisFormulas.html for $$\cos20^\circ+\cos25^\circ$$

Finally $\cos\dfrac52^\circ+\cos\left(90-\dfrac{45}2\right)^\circ=?$


问:
Is the composition of a strictly concave function with a concave function strictly concave?
Let $h:\mathbb{R}\mapsto \mathbb{R}$ be a strictly concave and nondecreasing function and $g:\mathbb{R^n}\mapsto \mathbb{R}$ be a concave nondecreasing function. Is $f=h(g(x))$ strictly concave?

I thought that the answer would be "yes" since, for all $0\leq a\leq1$,
\begin{align}
h\left(g(ax+(1-a)y)\right)&\geq h\left(ag(x)+(1-a)g(y)\right)\\
& > ah(g(x))+(1-a)h(g(y))
\end{align}
where the first inequality follows since $h$ is nondecreasing and $g$ is concave, and the second inequality follows because $h$ is strictly concave.

But then if I set $h(y)=\sqrt{y}$ and $g(x)=x_1+x_2$ I get $h(x)=\sqrt{x_1+x_2}$ which is not strictly concave. What is wrong with my argument?

答:
A simpler counterexample, which helps shed some light: $g$ can be a constant function.

The issue in your argument is that you implicitly assume $g(x)\neq g(y)$ to obtain the strict inequality.

In you take $g(x) = x_1+x_2$, then you can have $g(x)=g(y)$ even if $x\neq y$: take for instance $x=(x_1,x_2)$ and $y=(x_2,x_1)$. You need injectivity somewhere for your argument to go through.


问:
Lagrange Multipliers to find the maximum and minimum values
Using Lagrange Multipliers I need to find the maximum and minimum values of the function $f(x,y,z)=x^2+y^2+z^2$ subject
to the given constraints: $g(x)= x^2/\alpha^2 +y^2/\beta^2 +z^2/\gamma^2=1$ and also $\alpha>\beta>\gamma>0$

I come to the conclusion that $x=\pm\alpha$ using the derivatives in the picture and also applying $\alpha^2=\lambda$,              but I'm stuck from this point, can anyone help me continue and find the min max vales?

答:
This extremization problem has a geometrical interpretation.  The function  $ \ f(x,y,z) \ = \ x^2+y^2+z^2 \ $ represents the distance-squared from the origin of a given point The constraint equation $ \ x^2/\alpha^2 \ + \ y^2/\beta^2 \ + \ z^2/\gamma^2 \ = \ 1 \ $ represents a triaxial ellipsoid centered on the origin (with $ \ \alpha  >  \beta  >  \gamma \ >  0 \ $ ) , with its axes along the coordinate axes.

Level surfaces of the function are spheres of varying radii centered on the origin.  The smallest one tangent to the ellipsoid from within (the inscribed sphere) contacts the ellipsoid at the endpoints of its shortest axis, where $ \ f(0 \ , \ 0 \ , \ \pm \gamma) \ = \ \gamma^2 \ $  [Andrei's first solution] .
The largest tangent sphere (the circumscribed sphere) meets the ellipsoid at the endpoints of its longest axis, with $ \ f(\pm \alpha \ , \ 0 \ , \ 0) \ = \ \alpha^2 \ $ [his third solution] .  Hence, these are the absolute minima and maxima, respectively.  The other solution represents a sphere tangent to the endpoints of the "midling" axis and does not produce an extremum for the function with the specified conditions of the problem.


问:
How to find $K$ in $2\csc 30 ^\circ(\cos 8x+\cos 6x)(\cos 6x + \cos 2x)-1=\sin (Kx)\cdot \csc x$?
The problem is as follows:

Which value should have $K$ in the expression from below so it makes an identity?

$2\csc 30 ^\circ(\cos 8x+\cos 6x)(\cos 6x + \cos 2x)-1=\sin (Kx)\cdot \csc x$

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&11\\
2.&12\\
3.&14\\
4.&15\\
\end{array}$

I'm confused exactly how to get that particular value for $K$.

The only thing which I was able to spot is that: (Using manipulations on the left side of the equation)

$2\csc 30 ^\circ(\cos 8x+\cos 6x)(\cos 6x + \cos 2x)-1=\sin (Kx)\cdot \csc x$

$2\csc 30 ^\circ (2\cos 7x \cos x)(2\cos 4x \cos 2x)-1$

$4 (2\cos 7x \cdot \cos 2x)(2 \cos x \cdot 4x)-1$

$4(\cos 9x + \cos 5x)(\cos 5x+\cos 3x)-1$

$4(\cos 9x \cos 5x +\cos 9x \cos 3x + \cos 5x \cos 5x + \cos 5x \cos 3x)-1$

$2(\cos 14x + \cos 4x + \cos 11x + \cos 6x + 1+\cos 10x + \cos 8x + \cos 2 x) - 1$

However at the point I reached here then I got stuck it seems that this might not be the intended strategy. What could be done here to solve this problem?. Is there any trick or what?.

Could someone help me with an orderly method to solve this without much fuss?.

答:
I show what $K$ must be. (I do not prove the identity.)
$\lim_{x\to 0} RHS=K.$
$\lim_{x\to 0} LHS=2\cdot 2(1+1)((1+1)-1=15.$

Therefore, $K$ is $15.$


答:
I found a solution from Lord Commander's working. As they note:
$$(a-b)(2b+1) = b(2a-2b-1) + a \tag{1}\label{eq1}$$

And:
$$(b+1)(2b-a) = b(2b-a+2) -a \tag{2}\label{eq2}$$

For simplicity, let $x = 2a-2b-1$ and $y = 2b-a+2$.

Thus $\eqref{eq1}$ becomes $(a-b)(2b+1) = bx + a$ and $\eqref{eq2}$ becomes $(b+1)(2b-a) = by - a$

Applying the equations:

\begin{align} (a-b)(2b+1)(2b-a+2) - (b+1)(2b-a)(2a-2b-1) &= y(bx + a) - x(by-a) \\
&= bxy + ay -bxy + ax \\
&= a(x+y) \\
&= a(2a-2b-1+2b-a+2) \\
&= a(a+1) \\
&= a^2 + a
\end{align}

Thanks Lord Commander!


问:
Prove by induction on $n$ that $(n – r)!r!$ divides $n!$ for all $0 ≤ r ≤ n$
Define $n!$ for $n ∈ \mathbb{N}_{0}$ by $0! = 1, (n + 1)! = n! (n + 1)$.
Prove by induction on $n$ that $(n – r)!r!$ divides $n!$ for all $0 ≤ r ≤ n$.

Here's my attempt at the proof:

Let $S = \{n ∈ \mathbb{N}_{0};  (n-r)!r! \mid n! \}$.

Base case, $n = 0$:

$n! = q((n-r)!r!)$

$0! = q((0-0)!0!)$ ($r$ being 0 is the only possibility when $n = 0$)

$0! = q\cdot1$

$ q = 1 $ (which shows that $0 ∈ S$)

Induction step:
Let's assume the proposition holds for all $k ∈ \mathbb{N}_{0}$ up to and including $n$. We have to show that for all $r ∈ \mathbb{N}_{0}; 0 ≤ r ≤ n + 1$ the following holds: $(n + 1)! = p(((n+1)-r)!r!)$ , where $p ∈ \mathbb{N}_{0}$.

Hence,

$n!\cdot(n+1) = q((n-r)!r!) \cdot (n+1)$

$(n+1)! = q((n-r)!r!) \cdot (n+1)$

$(n+1)! = q(n-r)!r! \cdot ((n-r+1)+r)= r!(n-r+1)!\cdot (q+qr(n-r)!\frac{1}{n-r+1})$

$(n+1)! =  r!(n-r+1)!\cdot \frac{q\cdot(n+1)}{(n+1)-r}=r!(n-r+1)!\cdot \underbrace{\frac{q}{1-\frac{r}{n+1}}}_{p}$

I have a feeling that something went wrong somewhere, since I don't know how to prove that $\frac{q}{1-\frac{r}{n+1}}$ is indeed an element of $\mathbb{N}_{0}$ for all $0 ≤ r ≤ n + 1$, because there is a a problem with $r = n + 1$ due to division by $0$. Also it seems that there is a simpler way to a proof.

答:
You can prove Pascal's identity completely algebraically or combinatorically if you please to avoid using two inductive proofs in a row.

To finish your prove this statement, it suffices to show that $\binom{n}{r}$ is always an integer.

We use Pascal's identity, given by
$$\binom{n}{r}=\binom{n-1}{r}+\binom{n-1}{r-1}$$
with no restrictions on the size of $n$ or $r$, since if $r>n$, we have $\binom{n}{r}=0$ and the identity holds.

So let $n=0$. Then $\binom{0}{0}=1\in\mathbb{N}$ so we are done with the base case.

Now suppose that $\binom{n}{r}$ is an integer for some $n\in\mathbb{N}$ and for every $r$ with $0\leq r\leq n$.

Let $1\leq r\leq n+1$. The case when $r=0$ holds. Then,

$$\binom{n+1}{r}=\binom{n}{r}+\binom{n}{r-1}$$ By hypothesis, $\binom{n}{r}$ is an integer if $r<n+1$ and is $0$ if $r=n+1$. Also by hypothesis, $\binom{n}{r-1}$ is an integer, so their sum is an integer. This completes the proof by induction.

You could also just give a combinatorial argument for this as well, where $\binom{n}{r}$ is the number of $r$ element subsets of an $n$ element set.

The argument of this goes as follows:

Suppose we want to select a $k$ element string from an $n$ element set. There are $n$ ways to choose the first element, $n-1$ ways to choose the second element, $n-2$ ways to choose the third element, and continuing in this way, there are $n-(k-1)$ ways to choose the $k$th element of the string.

Then $$n(n-1)(n-2)\cdots(n-(k-1))=\frac{n!}{(n-k)!}$$ is the number of strings of length $k$. Since the ordering of the elements in a subset does not matter, then we must account for that. We notice that each $k$ element subset occurs $k!$ times among these strings, as the elements in the strings can be permuted in $k!$ different ways. Therefore the number of $k$ element subsets of an $n$ element set is given by $$\binom{n}{k}=\frac{n!}{(n-k)!\cdot k!}$$


问:
Euclid's Algorithm - Finding positive solutions (x,y)
Ok, so maybe I'm missing some crucial steps here, but I just can't see it.

So, I have got $\gcd(144,252) = 36$

$144x+252y=36$

I need to find ALL positive solutions. I can find $x=2$ and $y=-1$ by reversing the algorithm. The answer is given as follows.

$36=144-108
  =144-(252-144)
  =2\times144-252     $

But then says

$\times 117$

$4212=118 \times2 \times144 -117 \times252$

solution of $4212 = 144x + 252y$
are $x= 234- 252t/36 = 243 - 7t$
   $ y= -117 + 144t/36 = -117 + 4t$

$x,y>0$ if $234/7 > t > 117/4$

$t=30,31,32,33$

$(x,y)=(24,3),(17,7),(10,11),(3,15)$

FIRSTLY - where on earth does the multiplier $117$ come from? this seems to have been plucked from thin air,

and secondly, please, how to convert the $t$ values into the $(x,y)$ values. ie, how does $30$ relate to $(24,3)$

Many thanks

答:
If $ax+by=c$,
then $a(x+nb)+b(y-na) = c$
for all $n$,
since
$a(x+nb)+b(y-na)
= ax+nab+by-nab
= c
$</p>


问:
3D Vectors linear algebra question
I'm stuck and would appreciate help this question:

u = (2, -1, 1)
v = (-3, 6, 3)
find components of w so that u , v and w are linearly dependent,
(w - u)⊥ u and (w - v) ⊥ v

visual of the problem

答:
You are given $u=\begin{pmatrix}2\\-1\\1\end{pmatrix},v=\begin{pmatrix}-3\\6\\3\end{pmatrix}$, and let $w=\begin{pmatrix}x\\y\\z\end{pmatrix}$. You have several conditions. The first is that they are linearly dependent, which can be expressed as for some a and b, $au+bv=w$ or $a\begin{pmatrix}2\\-1\\1\end{pmatrix}+b\begin{pmatrix}-3\\6\\3\end{pmatrix}=\begin{pmatrix}x\\y\\z\end{pmatrix}$. You're also given that $w-u\perp u$ and $w-v\perp v$. Two vectors are perpendicular if their dot product is 0. You get $\begin{pmatrix}x-2\\y+1\\z-1\end{pmatrix}\begin{pmatrix}2\\-1\\1\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}$ and $\begin{pmatrix}x+3\\y-6\\z-3\end{pmatrix}\begin{pmatrix}-3\\6\\3\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}$. This gives us the equations

```<code>2(x-2)-(y+1)+(z-1)=0
-3(x+3)+6(y-6)+3(z-3)=0

Solve in terms of x.

y=x+4
z=10-x

From the linear dependent condition previously,
2a-3b=x
-a+6b=x+4
a+3b=10-x

Solve for a, b, and x.
a=10/3, b=14/9, x=2

Get y and z.
y=6, z=8
</code>```

So the solution is $w=(2,6,8)^T$.

Title: 3D Vectors linear algebra question

Body:
You are given $u=\begin{pmatrix}2\\-1\\1\end{pmatrix},v=\begin{pmatrix}-3\\6\\3\end{pmatrix}$, and let $w=\begin{pmatrix}x\\y\\z\end{pmatrix}$. You have several conditions. The first is that they are linearly dependent, which can be expressed as for some a and b, $au+bv=w$ or $a\begin{pmatrix}2\\-1\\1\end{pmatrix}+b\begin{pmatrix}-3\\6\\3\end{pmatrix}=\begin{pmatrix}x\\y\\z\end{pmatrix}$. You're also given that $w-u\perp u$ and $w-v\perp v$. Two vectors are perpendicular if their dot product is 0. You get $\begin{pmatrix}x-2\\y+1\\z-1\end{pmatrix}\begin{pmatrix}2\\-1\\1\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}$ and $\begin{pmatrix}x+3\\y-6\\z-3\end{pmatrix}\begin{pmatrix}-3\\6\\3\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}$. This gives us the equations

2(x-2)-(y+1)+(z-1)=0
-3(x+3)+6(y-6)+3(z-3)=0

Solve in terms of x.

y=x+4
z=10-x

From the linear dependent condition previously,
2a-3b=x
-a+6b=x+4
a+3b=10-x

Solve for a, b, and x.
a=10/3, b=14/9, x=2

Get y and z.
y=6, z=8

So the solution is $w=(2,6,8)^T$.


问:
Show that if all points in space lie in a collection of subbasic open sets, all points lie in two of the subbasic open sets
Let $\{0,1\}^A$ be the collection of functions from an index set $A$ to $\{0,1\}$. Having subbasis $\{U(a,\delta)|a \in A,\delta \in \{0,1\}\},U(a,0)=\pi_{a}^{-1}(0),U(a,1)=\pi_{a}^{-1}(1)$.Show if a collection of subbasic open sets of $\{0,1\}^A$ has the property that every point of $\{0,1\}^A$ lies in at least one of those subbasic open sets, then there are two subbasic open sets in that collection such that every point of $\{0,1\}^A$ lies in one of those subbasic sets.

Attempt:Let $S$ be such a collection of subbasic sets. If $\{0,1\}^A \in S$. There is nothing to show. Suppose $\{0,1\}^A \notin S$.Then there are two subbasic sets $\pi_{a_1}^{-1}(b_1),\pi_{a_2}^{-1}(b_2),b_i \in \{0,1\},a_1\neq a_2$ in $S$. Then $\pi_{a_1}^{-1}(b_1)\cup\pi_{a_2}^{-1}(b_2)=\{0,1\}^A$, since the only factor of $\pi_{a_1}^{-1}(b_1)$ with points restricted to either $0$ or $1$ is the $a_1$'st factor. And the only factor of $ \pi_{a_2}^{-1}(b_2)$ with points restricted to $0$ or $1$ is the $a_2$'nd factor. So the union must contain all points.

答:
The first case in your proof is impossible: $\{0,1\}^A$ is not one of the sets $U(a,\delta)$ with $a\in A$ and $\delta\in\{0,1\}$, so by definition $\{0,1\}^A\notin S$. The argument in your second case is simply incomprehensible. If it were correct, the union of any two members of $S$ would be $\{0,1\}^A$, which certainly need not be the case: $S$ might contain $U(a_1,0)$, $U(a_2,0)$, and $U(a_2,1)$ for distinct $a_1,a_2\in A$, and

$$U(a_1,0)\cup U(a_2,0)\ne\{0,1\}^A\,:$$

for instance, $U(a_1,0)\cup U(a_2,0)$ does not contain the function that is $1$ on every element of $A$. Of course in this case $U(a_2,0)\cup U(a_2,1)$ is $\{0,1\}^A$, but your argument applies to any two members of $S$, not to some carefully chosen pair.

In fact $U(a_1,\delta_1)\cup U(a_2,\delta_2)=\{0,1\}^A$ if and only if $a_1=a_2$ and $\delta_1\ne\delta_2$.

Prove this.

Thus, if $S$ does not contain two members whose union is $\{0,1\}^A$, then for each $a\in A$ it must contain at most one of the sets $U(a,0)$ and $U(a,1)$. This means that for each $a\in A$ there is a $\delta_a\in\{0,1\}$ such that $U(a,\delta_a)\notin S$. Define $x\in\{0,1\}^A$ by letting $x(a)=\delta_a$ for each $a\in A$.

Prove that $x\notin\bigcup S$.

Conclude that if $S$ does not contain two members whose union is $\{0,1\}^A$, then $\bigcup S\ne\{0,1\}^A$, and hence that if $\bigcup S=\{0,1\}^A$, then $S$ contains both $U(a,0)$ and $U(a,1)$ for some $a\in A$ and hence contains two subbasic open sets whose union is $\{0,1\}^A$.


问:
Question on primes.
Let $p(n)$ denote the number of primes less than $n$. Show that there are infinitely many $n$ for which $n$ is divisible by $p(n)$.

Source : here

答:
Hint: Let $f(n): \mathbb{N} \rightarrow \mathbb{N}$ be a function such that $f(n+1) - f(n) = 0 $ or $1$ and $\lim_{n\rightarrow \infty} \frac{ n} { f(n) } \rightarrow \infty$.

Show that  the image of $\frac{ n}{f(n) } $ includes all integers greater than $\frac{ 1}{f(1)} $. It will help for you to consider how this function behaves going from $\frac{n}{f(n)} $ to $\frac{n+1} { f(n+1)}$.

Proof: Consider the integer valued function $g(n) = n - k f(n)$, where $k$ is an integer greater than $\frac{1}{f(1)}$.  Observe that $g(n+1)-g(n)$ is equal to 1 (if $f(n+1)=f(n)$) or is negative (if $f(n+1) > f(n)$).  From the conditions, we get that $g(1)$ is negative, and $g(n)$ is eventually positive for a large enough value of $n$. Hence, there is a value $n^*$ such that $g(n^*) = 0$, which gives us $\frac{n^*}{f(n^*)} = k $.

The reason for presenting the answer this way, is that it has nothing to do with the prime counting function, apart from the properties listed in the hint.

Title: Question on primes.


问:
Which elements in $\mathbb{Z}_{26}$ do not have multiplicative inverses?
How would I find which elements in $\mathbb{Z}_{26}$ do not have multiplicative inverses? I assume I would find them by finding which elements in $\mathbb{Z}_{26}$ have a $\gcd(a, m) \neq 1$. Is this the correct approach? I have gotten that 2, 4, 6, 8, 10, 12, 13, 14, 16, 18, 20, 22, and 24 don't have multiplicative inverses in $\mathbb{Z}_{26}$.

答:
The invertible elements form the group of units $\Bbb Z_{26}^×$, which has order $\varphi(26)=12$.

It's complement, $\Bbb Z_{26}\setminus\Bbb Z_{26}^×$ then consists of $14$ elements; namely, $0,2,4,6,8,10,12,13,14,16,18,20,22,24$.  Those are the elements which are not coprime with $26$.

In addition they are all divisors of zero.


问:
Continuous water-filling
Disclaimer: I x-posted this question on OR-stackexchange as suggested in the comment below. I reposted it there since I did not receive any satisfactory answer on math-stackexchange up to now.

Let $x\in\mathbb{R}^n$ be an optimization variable and $\alpha\in\mathbb{R}^n$ be a given $n$-dimensional vector. The standard water-filling problem is formulated as

\begin{equation}
\begin{array}{ll}
\underset{x}{\operatorname{minimize}} & -\displaystyle\sum_{i=1}^{n} \log \left(\alpha_{i}+x_{i}\right) \\
\text { subject to } & x \succeq 0, \quad \mathbf{1}^{T} x=1,
\end{array}
\end{equation}

and it has a well-known solution (see Boyd & Vandenberghe page 245).

I was thinking about the case in which we have continuous communication channel slots. Intuitively this may be thought of as the case when the sizes of the communication channels approach zero. How is it possible to generalize this optimization problem to this case?

I believe that $\alpha$ and $x$, which from now on we will denote by $\alpha(x)$ and $f(x)$ respectively, would in this case be continuous real function and the problem would be:

\begin{equation}
\begin{array}{ll}
\underset{f \in \mathcal{F}}{\operatorname{minimize}} & -\displaystyle\int_{x} \log \left(\alpha(x)+f(x)\right)dx \\
\text { subject to } & f(x) \geq 0\; \forall x, \quad \int_x f(x)dx=1
\end{array}
\end{equation}

where $\mathcal{F}$ is a given class of functions (e.g. Hölder continuous, Lipschitz, etc). We can also assume that the domain of the functions $f(x)$ and $\alpha(x)$ is compact, i.e. $x\in\mathcal{X}\subseteq \mathbb{R}$, with $\mathcal{X}$ a compact subspace of $\mathbb{R}$.

Is this the right path to follow? Do you have any other on how to solve these types of problems? It looks it is related to the calculus of variations, but I have never seen these types of problems and I have no idea how to solve them. Thanks!

答:
You're correct that the problem falls into the class treated by the calculus of variations. A nice modern approach to this subject is to treat the objective as a function on a topological vector space, which contains the space of functions over which you want to optimise the objective.  The optimality conditions are then often quite analogous to the Karush-Kuhn-Tucker conditions for optimisation over a finite-dimensional vector space, although there can be some messy technical details which depend on the specific class of functions over which you want to optimise. David Luenberger's Optimization by Vector Space Methods contains a nice exposition of this approach.

For the specific problem you've given, the optimality conditions are similar to those of the finite-dimensional version.  You introduce Lagrange multipliers $\ \lambda\in\mathbb{R}\ $ for the constraint
$$
\int_{\mathcal{X}}f(x)=1\ ,
$$
and $\ \mu\in\mathcal{F}^*\ $ for the constraint
$$
f(x)\ge0\ ,
$$
where $\ \mathcal{F}^*\ $ is the function-space dual of $\ \mathcal{F}\ $.  This is where some of the messy technical details mentioned above come in.  The duals of spaces of continuous functions are typically spaces of finitely additive measures, which can be difficult to deal with.  A way out, in this case, is to relax the conditions on $\ f\ $ by allowing it to be any function that's square-integrable with respect to Lebesgue measure over $\ \mathcal{X}\ $.  The space $\ \mathcal{H}\ $ of functions square-integrable over $\ \mathcal{X}\ $ is a real Hilbert space, which is self-dual.  The Lagrange multiplier $\ \mu\in\mathcal{H}^*\ $ corresponding to the constraint $\ f(x)\ge0\ $ is now itself just a square-integrable function $\ \mu:\mathcal{X}\rightarrow\mathbb{R}\ $.  The Lagrangian for the problem becomes
$$
\int_{\mathcal{X}}-\log(\alpha(x)+f(x))dx+\lambda\left(\int_{\mathcal{X}}f(x)dx-1\right)-\int_{\mathcal{X}}\mu(x)f(x)dx\ ,
$$
and the optimality conditions are
$$
\frac{-1}{\alpha(x)+f(x)}+\lambda-\mu(x)=0\\
\mu(x)\ge0,\ \ f(x)\ge0,\ \  \mu(x)f(x)=0,\ \ \int_{\mathcal{X}}f(x)=1\ ,
$$
which are very closely analogous to those of the finite dimensional case.

Now let $\ \mathcal{X}_1=\{ x\in\mathcal{X}\,|\,f(x)>0\,\}\ $. The complementary slackness conditions, $\ \mu(x)f(x)=0\ $, tell us that $\ \mu(x)=0\ $ for $\ x\in\mathcal{X}_1\ $, and the first optimality condition then gives
\begin{align}
\frac{1}{\alpha(x)+f(x)}&=\lambda\ \text{, or}\\
f(x)&=\frac{1}{\lambda}-\alpha(x)>0
\end{align}
for $\ x\in\mathcal{X}_1\ $.  For $\ x\in\mathcal{X}\setminus\mathcal{X}_1\ $, where $\ f(x)=0\ $, we must have
\begin{align}
\lambda-\frac{1}{\alpha(x)}&=\mu(x)\ge0\ \text{, or}\\
\frac{1}{\lambda}&\le\alpha(x)\ ,
\end{align}
since $\ \lambda\ $ must be positive.  It follows from this that $\ f(x)=$$\max\left(0,\frac{1}{\lambda}-\alpha(x)\right)\ $.  The constraint
$$
1=\int_\mathcal{X}f(x)dx=\int_\mathcal{X}\max\left(0,\frac{1}{\lambda}-\alpha(x)\right)dx
$$
now allows us to determine the value of $\ \lambda\ $. The function $\ \phi(\nu)=\int_\limits{\mathcal{X}}\max\left(0,\nu-\alpha(x)\right)dx\ $ is a strictly increasing continuous function of $\ \nu\ $, with $\ \phi(0)=0\ $ and $$\ \phi\left(\frac{1}{\int_\limits{\mathcal{X}}1dx}+\max_\limits{ x\in\mathcal{X}}\alpha(x)\right)\ge1\ ,$$ so there is a unique $\ \nu^*>0\ $ such that $\ \phi\big(\nu^*\big)=1\ $, and then $\ \lambda=\frac{1}{\nu^*}\ $.

Because the objective function is strictly concave in this case, the optimality conditions are both necessary and sufficient, so $\ f(x)=\max\left(0, \frac{1}{\nu^*}-\alpha(x)\right)\ $ achieves the minimum of the objective function over the class of square-integrable functions over $\ \mathcal{X}\ $.  Since $\ f\ $ happens to be continuous, and all continuous functions over the compact set $\ \mathcal{X}\ $ are square-integrable, it must also achieve the minimum of the objective over the set of continuous functions on $\ \mathcal{X}\ $.

Title: Continuous water-filling


问:
Calculate the third partial derivative
If $u = vt^{\frac{1}{3}}$ and $v=h(y), y=it^{\frac{1}{3}}\sin(iwx)$ I need to determine $ u_{xxx}$ and $u_t$

I have :

\begin{align}
u_x &= u_v \cdot v_y \cdot y_x\\
&= t^{\frac{1}{3}}v_y(-wt^{\frac{1}{3}}\cos(iwx))\\
&=-wt^{\frac{2}{3}}\cos(iwx)v_y
\end{align}

答:
You have $y=it^{1/2}\sin iwx$. So $\frac{dy}{dx}=-wt^{1/2}\cos iwx$. Then

$$\begin{split}u_x&=-wt^{2/3}h'(y)\cos iwx\end{split}$$

To get from $u_x$ to $u_{xx}$ you need to use the product rule. $u_{xx}=-wt^{2/3}\left[\left(\frac{d}{dx}h'(y)\right)\cos iwx+h'(y)\frac{d}{dx}\cos iwx\right]$. Now you can apply the chain rule now that you've applied the product rule.

$$\begin{split}u_{xx}&=-wt^{2/3}\left[h''(y)(-wt^{1/2}\cos^2 iwx)+h'(y)(-iw\sin iwx)\right]\\
&=w^2t^{2/3}\left[h''(y)t^{1/2}\cos^2 iwx+h'(y)i\sin iwx\right]\\
u_{xxx}&=w^2t^{2/3}\left[h'''(y)t^{1/2}(-wt^{1/2}\cos^3 iwx)+h''(y)t^{1/2}2\cos iwx (-iw\sin iwx)+h''(y)i\sin iwx (-wt^{1/2}\cos iwx)+h'(y)i^2w\cos iwx\right]\\
&=w^2t^{2/3}\left[-wt^{2/3}h'''(y)\cos^3 iwx-2iwt^{1/2}h''(y)\cos iwx \sin iwx-it^{1/2}wh''(y)\sin iwx \cos iwx-h'(y)w\cos iwx\right]\\
&=w^2t^{2/3}\left[-wt^{2/3}h'''(y)\cos^3 iwx-3it^{1/2}wh''(y)\sin iwx \cos iwx-h'(y)w\cos iwx\right]\\
&=-w^3t^{2/3}\cos iwx\left[t^{2/3}h'''(y)\cos^2 iwx+3it^{1/2}h''(y)\sin iwx +h'(y)\right]\end{split}$$

Title: Calculate the third partial derivative


问:
Elements of the ring $F[x,y]/(x^2-y^2)$
Let $F$ be a field, and $R=F[x,y]$. Does $R/(x^2-y^2)$ have any nilpotent elements? Idempotent elements?

I know that $R/(x^2-y^2)$ isn't an integral domain, as $x-y$ and $x+y$ are non-trivial zero divisors. I feel like there are no nilpotent/idempotent elements, as I can't think of any more zero divisors other than $x-y$ and $x+y$, neither of which are nilpotent/idempotent.

答:
Try writing out the definitions to get started:

For a polynomial $p(x,y) \in F[x,y]$, let $\overline{p}(x,y) := p(x,y) \pmod{(x^2 - y^2)}$. If $\overline{p}(x,y)$ is nilpotent, that means that $p(x,y)^N \equiv 0 \pmod{(x^2 - y^2)}$ for some $N > 1$, so $x^2 - y^2 \mid p(x,y)^N$. What can you conclude from here?

Similarly $\overline{p}(x,y)$ is idempotent iff $x^2 - y^2 \mid p(x,y)^2 - p(x,y)$. What can you conclude from here?

Lastly, keep in mind that $x - y = x + y$ in $F[x,y]$ if $2 = 0$ in $F$.

Title: Elements of the ring $F[x,y]/(x^2-y^2)$


问:
The set of the limit points and being dense

Let $A$ be a subset of the metric space $(X,d)$. Is the following statement true?
$$A' \ \text{is dense in}\  X \iff A \ \text{is dense in}\  X$$Where $A'$ denotes the set of the limit points.

My try: I tried to work with some examples. Let $A=\mathbb{Q}$, then $A' = \mathbb{R}$. So in this case the statement is true. If we know $A'$ is dense in $X$, according to the definition, $\text{cl}(A') = X$. Also we have $\text{cl}(A') = A' \cup (A')'$. In addition, if $A$ is dense in $X$, then $\text{cl}(A) = A \cup A' = X$. So it seems that if we find a set $A$ such that $(A')' \not= A$ we can disprove the statement but I couldn't find such set.

答:
$\Rightarrow)$ Suppose $A'$ is dense in $X$. Fix  $x\in X$ and $\varepsilon>0$. By assumption, there is some  $b\in A'$ such that $0<d(x,b)<\frac{\varepsilon}{2}$. Since $b\in A'$, there is some $a\in A$ with $0<d(a,b)<\frac{\varepsilon}{2}$. Note $d(x,a)<\varepsilon$ by the triangle inequality. This shows  that $A$ is dense.

The other implication is false. Let $X=\left\{\frac{1}{n+1}:n\in\mathbb{N}\right\}\cup \{0\}$, with the metric it inherits from $\mathbb R$. Put $A=\left\{\frac{1}{n+1}:n\in\mathbb{N}\right\}$. Then $A$ is dense in $X$, but $A'=\{0\}$ is not.


问:
Arithmetic operations on numbers represented by prime factors (optimizing calculations)
Shortly, I'm asking for ideas on how to do 2 things:

Given the prime factors of two numbers $x,y$, find the prime factors of $x+y$ without finding $x+y$ and factoring it.

Given the prime factors of two numbers, determine which number is bigger, or at least find a partial check that will determine the answer for a significant amount of the possible cases.

If you're interested in the long story and the bigger picture of what I'm playing with, read on. This is just for fun, I am not doing this with any particular practical applications in mind, but who knows maybe it will turn out to be useful. To speed up calculations with very big numbers and many operations, I was thinking about representing rational numbers by prime factors. I defined some arithmetic operations that are indeed faster with this representation, but I failed defining addition and also failed finding a way to compare numbers without converting them back into standard representation.

First, the representation:

Let $x$ be an infinite ordered tuple of whole numbers, and $p(n)$ be the $n$th prime number. $x$ represents the following number:

$$\prod_{n=1}^{\infty} p(n)^{x_n}$$

For example:

$$(1,1,-2,0,0,0,...) = 2^1 \cdot 3^1 \cdot 5^{-2}$$

But this represents only positive numbers, so let's add another number (a "sign number") to the beginning of the tuple ($0$th position - $x_0$) that must be in $\{-1,0,1\}$. Now the tuple represents the number:

$$x_0\prod_{n=1}^{\infty} p(n)^{x_n}$$

This is the final definition. So now that we have a way to represent all rational numbers, let's look at the arithmetic with this representation.

Let $x,y$ be rational numbers, and $x_i,y_i$ the $i$th digits in the infinite ordered tuples representing these numbers.

Multiplication is easy:

$$x\cdot y = (x_0 \cdot y_0, x_1 + y_1, x_2 + y_2, ..., x_i + y_i, ...)$$

The first numbers are multiplied together, and the rest are summed.

Division is the same but with subtraction instead of addition:

$$\frac{x}{y} = (x_0 \cdot y_0, x_1 - y_1, x_2 - y_2, ..., x_i - y_i, ...)$$

Powers are also easy:

$$x^n = (x_0^{|n|}, nx_1, nx_2, ..., nx_i, ...)\\
n\in \Bbb{Z}$$

Addition is where I'm stuck (I don't want to calculate the actual numbers, add them and then factor, this would defeat the purpose of representing numbers by their prime factors). I guess this is an unsolved problem because solving it would provide an easy way to find all the prime numbers, but I would appreciate any pointers to potentially useful directions.

I did find a way to at least somewhat optimize the calculation when I convert them to standard representation.

First I'll define something that will be useful here and also later on. Let's say I want to get a more convenient representation of $x$ and $y$ (still with tuples). The first thing I'll do is factor out the greatest common divisor, and then I'll also factor out the smallest number that is required to turn both numbers into integers. This leaves me with two integers that have no common factors, and the number I factored out to get these integers.

For example if:

$$x=(1,2,5,4,0,4,0,...)\\ y=(1,3,3,4,-2,4,0,...)$$

I can divide both of them by $(1,2,3,4,-2,4,0,...)$ and get:

$$\frac{(1,2,5,4,0,4,0,...)}{(1,2,3,4,-2,4,0,...)}=(1,0,2,0,2,0,0,...)\\
\frac{(1,3,3,4,-2,4,0,...)}{(1,2,3,4,-2,4,0,...)}=(1,1,0,0,0,0,0,...)$$

It's easy to find this convenient common factor, for every $i>0$, the $i$th number in the tuple representing the common factor is $\min(x_i,y_i)$. I have no better name for it, so let's just call it Convenient Common Factor (CCF for short).

Back to the topic of addition when using the tuple representation. I said I found a way to optimize the calculation despite the need to convert the numbers into standard representation. The idea is to factor out the CCF before the conversion from tuples to standard representation, so I can perform the conversion and addition on only the convenient integers, and once I convert their sum back into tuples I can multiply it by the CCF. This process makes both the addition and the conversion easier, including the factoring to convert back to tuples (simply because the number is smaller), but it's still going to be a ridiculously slow process because big numbers take a long time to factor (at least without quantum computers).

I'm also interested in comparing the size of two numbers. If $a>b$ then dividing or multiplying both of them by the same positive number doesn't change the inequality. This means that I can factor out the CCF and then compare the more convenient numbers. Using the numbers from the CCF example:

$$(1,2,5,4,0,4,0,...) - (1,3,3,4,-2,4,0,...) = (1,2,3,4,-2,4,0,...) \cdot \big( (1,0,2,0,2,0,0,...) - (1,1,0,0,0,0,0,...) \big)$$

While in this example it's obvious that the first number is bigger, I didn't find an algorithm that will give the correct answer in every case without multiplying some numbers in the tuple by their corresponding primes. Intuitively it seems like this problem is easier than the addition problem and might have an easier solution or at least a decent partial solution to narrow down the cases that require conversion into standard representation (by performing a relatively quick check to see if the answer can be found without conversion, like "are $x$'s factors bigger than all of $y$'s factors and at least as numerous?" but hopefully something that covers more cases).

Another way to look at the problem is to divide the numbers by each other and compare the result to 1, in the case of this example we get $(1,-1,2,0,2,0,0,...) > 1$. Not sure if it's helpful, just throwing ideas.

答:
Answer to question 2 -  a check that always works which doesn't involve calculating the actual numbers (nor multiplying any primes together).

Instead of comparing the numbers, compare their logarithms.

With very close numbers, their logarithms are also very close, so using calculations of limited precision is doomed to fail. The solution is to calculate the logarithms iteratively with ever increasing precision, and at the same time also calculate the error range, and as soon as the sum of logarithms exits the error range I can be certain about the answer.

I chose to use base 2 logarithm because it's convenient for a computer to work with base 2. This Wikipedia article explains how to calculate $log_2$ iteratively.

I wrote a program that uses this method and in each iteration chooses the logarithm with (approximately) worst precision and increases it's precision in the way the Wikipedia article describes, until the overall precision is enough. It works, and it works much faster than calculating the actual numbers. In fact, it worked even with such big numbers that my computer couldn't calculate their exact value within a reasonable amount of time at all! But of course I can go even bigger until this algorithm will become slow too.

Also, the answer for most numbers (at least for the random numbers my computer program generated) can be found using the check I described in my previous answer, so I get great speed gains by trying it first and using the logarithms as a plan B.


问:
Does weak*-convergence imply convergence of the operator norms?
Let $\mathcal A$ be a unital C*-algebra with topological dual $\mathcal A^*$ and denote the unit ball as $B_1^*:=\{\phi \in \mathcal A^* : \vert\vert \phi \vert\vert_{sup}\leq 1\}$.

If $\phi_n \rightarrow \phi$ is a weak*-convergent sequence with $\vert\vert \phi_n \vert\vert_{sup}= 1, \forall n \in \mathbb N$, does this imply that $\vert\vert \phi \vert\vert_{sup} = 1$?

答:
No.  Consider the Fourier coefficient functionals on $C[0,2\pi]$.

Or if $H$ is a separable Hilbert space with orthonormal basis $(e_n)$, consider the functionals $T\mapsto\langle Te_1,e_n\rangle$ on $B(H)$.

In general you have $\|\phi\|\leq \liminf_n\|\phi_n\|$, but the inequality can be strict as these examples show.


问:
A standard deck of 52 playing cards is shuffled and the cards are dealt face up one at a time until an ace appears.
Determine how many cards are needed to make the probability that an ace appears greater than 1/2.</p>

答:
Let $X$ equal the count of draws until the first ace appears.

Evaluate $\mathsf P(X>x)$ for any $x\in\{1,\ldots,49\}$

 $\mathsf P(X{>}x) $ is the probability that the four aces are among the later $52-x$ positions, when placing the four aces among the 52 positions without bias.

 $$\mathsf P(X{>}x) = \dfrac{\binom{52-x}{4}}{\binom {52}4} = \dfrac{(52-x)(51-x)(50-x)(49-x)}{52\cdot 51\cdot 50\cdot 49}$$

Find $\min\{m:\mathsf P(X{>}m)\lt 1/2\}$


问:
If $2+\log_2a=3+\log_3b=\log_6(a+b)$, find $\frac1a+\frac1b$ and $\log_{ab}\left(\frac1a+\frac1b\right)$
I have a kind of system of equations with logarithms.
How do I get to the expression that the problem asks for?
(Taken from the net.)

If $a>0$ and $b>0$ and it is given that
$$2+\log_{2}{a} =3+\log_{3}{b}=\log_{6}{(a+b)}$$

find the value of

a) $\quad\dfrac{1}{a}+\dfrac{1}{b}$

b) $\quad\log_{ab}{\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}$

Edit 1
I have related the first with the third and the second with the third to form a system of equations, but the values that come out are not very comfortable to deal with, I do not know if I am wrong or if the problem is like that.

$$2+\log_{2}{a}=\log_{6}{(a+b)}$$
$$3+\log_{3}{b=}\log_{6}{(a+b)}$$

Edit 2 :I have reformulated the problem by splitting it in two

答:
Per OP's request

This is not an answer.

However, the work is too long-winded to be placed as a series of comments.

$a>0$ and $b>0$

$$2+\log_{2}{a} =3+\log_{3}{b}=\log_{6}{(a+b)}$$

Compute
$$ \log_{ab}{\left(\frac{1}{a}+\frac{1}{b}\right)}$$

Let $x = \log_2 a, ~y = \log_3 b, ~z = \log_6(a+b) \implies$

$2^x = a, 3^y = b,~$ and $~6^z = (a+b).$

Also implied is that $2 + x = 3 + y = z.$

To Compute

Find the value $w$ such that $(ab)^w = \frac{a+b}{ab}
= \left(\frac{1}{a}+\frac{1}{b}\right).$

Clearly, $(ab)^{(-1)} = \frac{1}{ab}.$

Therefore, the problem reduces to finding $v$ such that $(ab)^v = (a+b) \implies w = (v-1)$.

From the premises, you know that

$(ab) = (2^x)(3^y).$

$(6^z) = (a + b)$.

$z = x+2$ and $y = x-1$.

Therefore, $(ab) = (2^x)[3^{(x-1)}] = \frac{6^x}{3}.$

Also, $36 \times 6^x = 6^{(x+2)} = 6^z = (a+b).$

Thus, you are looking for $v$ such that

$\left[\frac{6^x}{3}\right]^v = 36 \times 6^x.$

Therefore, $6^{xv} = (3^v) \times 36 \times 6^x \implies
6^{xv-x} = 36 \times (3^v) \implies$

$$6^{xw} = 36 \times 3 \times 3^w.\tag1$$

Therefore

$[(2^x)]^w \times [3^{(x-1)}]^w = 108 \implies$

$$a^w \times b^w = 108 = \frac{a+b}{ab} \implies
108(ab) = (a+b). \tag2 $$

The above analysis assumes (perhaps wrongly) that I haven't made any algebra mistakes.

Note For what it's worth, equation (2) above establishes that $\frac{1}{a} + \frac{1}{b} = 108.$

After I got to this point, I spent some time trying to solve for the exact value of $w$.  I got nowhere.  This is ugly.  Personally, I outlaw using computer software here, because it (in general) renders analytical attempts somewhat meaningless.

I experimented with setting $r = \log_2(3)$ to see if this yielded some elegant simplification.  It didn't, for me.  I then considered brute force, trying to specifically solve for $a,b,x,y,z,w.$  For me, that approach got ugly, fast.

I emphasize: this response is not an answer, but is posted as such per OP's request.


问:
Rational parametric equation of a circle from a line
I found out that we can define a circle equation as follows:
$$\begin{cases}x(t)=\dfrac{t}{t^2+(kt+b)^2},\\y(t)=\dfrac{kt+b}{t^2+(kt+b)^2},\end{cases}$$
where $k, b$ are real numbers.
For example, if we set $b=1, k=-1$, the parametric equation seems to correspond — except for the $t$ such that denominator is zero — to a circle $$\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2=\frac{1}{2}$$.

I want to understand why is this parametrization valid, and how can we get circle center and radius using the corresponding line equation $y=kx+b$.

In fact, I've been doing a textbook exercise involving complex numbers, and the first part of the exercise asked to prove that $f(z)=\overline{z}^{-1}$ is a self-inverse, that is $f(f(z))=z$, and the second part asked what the curve of the image of this function with the domain restricted to $z=t+(k\cdot t+b)i$, for some fixed constants $k,b$ and a parameter $t$, was. Such complex numbers $z$ can be visualized with the line equation $y=k\cdot x + b$. The parametric equation of the circle corresponds to a real and imaginary parts of an image of $f$. Thus, as it was a part of an exercise, I am especially interested in explaining this relation between a circle and a line using the property that $f$ is an involution. Other explanations are welcome, too.

答:
Squaring parametric equations and adding them gives
$$x^2+y^2=\frac {t^2+(kt+b)^2}{(t^2+(kt+b)^2)^2}=\frac 1{t^2+(kt+b)^2}.$$
Hence parametric equations can be rewritten as
$$x=t{(x^2+y^2)}, \quad y=(kt+b)(x^2+y^2).$$
Eliminating $t$ and rearranging gives
$$x^2+y^2+\frac kb x - \frac1b y=0\\\left(x+\frac k{2b}\right)^2+\left(y-\frac 1{2b}\right)^2=\left(\frac{\sqrt{k^2+1}}{2b}\right)^2$$
So centre is
$$\left(-\frac k{2b}, \frac 1{2b}\right)$$
and radius is
$$\frac {\sqrt{k^2+1}}{2b}.$$


问:
Show: (A ∪ B) × C = (A × C) ∪ (B × C).
We are on splitting cases and this problem has me stumped. I know it's not that hard but I've tried so many different ways to solve it and nothing seems correct to me.</p>

答:
$\underline{(A\cup B)\times C \subseteq (A\times C)\cup (B\times C)}$: Let $\mathbf{x}\in (A\cup B)\times C$, so $\mathbf{x} = (x_1, x_2)$ where $x_1\in (A\cup B)$ and $x_2\in C$. It follows that $x_1\in A$ or $x_1\in B$ (by definition of set union), and $x_2\in C$. But then we have $(x_1, x_2)\in A\times C$ or $(x_1, x_2)\in B\times C$, so $(x_1, x_2)\in (A\times C)\cup (B\times C)$. Therefore $\mathbf{x}\in (A\times C)\cup (B\times C)$, and we can conclude that
$$
(A\cup B)\times C \subseteq (A\times C)\cup (B\times C).
$$

Try proving $(A\times C)\cup (B\times C) \subseteq (A\cup B)\times C$ by following what I wrote above as a guide.

Title: Show: (A ∪ B) × C = (A × C) ∪ (B × C).


答: The probability that an ace does not appear for the first card is $\frac{48}{52}$ after two cards it is $\frac{48}{52}\times \frac{47}{51}$ so that after $k$ cards it is $P_k=\frac{48\times 47\times ...(49-k)}{52\times 51\times ...(53-k)}=\frac{(52-k)\times (51-k)\times(50-k)\times(49-k)}{52\times 51\times 50\times 49}$ - the last for $k\ge 4$.  The solution is find smallest $k$ for which $P_k\lt 1/2$.

Title: A standard deck of 52 playing cards is shuffled and the cards are dealt face up one at a time until an ace appears.

Body: Determine how many cards are needed to make the probability that an ace appears greater than 1/2.</p>


问:
Is there any inverse operator for mod?
It seems to me that operators usually come in pairs: Plus and minus, Multiplication and Division, Exponentiation and Logarith, Derivative and Integral.

Is there such inverse operation of modulus?

答: I don't think there's an inverse operation to modulus, however, there are additive inverses. Consider that $x = a$ mod $m$. Then the additive inverse is $(m - a)$ mod $m$. In particular, $x + m - a = a + m - a = m = 0$ mod $m$. Notice though additive inverses are not unique. For instance, an additive inverse to $x$ is $k(m-a)$ for $k \in \mathbb{Z}$

Title: Is there any inverse operator for mod?


问:
Fixed field with non-normal subgroup and Galois property
Let $L/K$ be a finite Galois extension with Galois group $G$. Furthermore, let $g \in G$ and $M = L^{\langle g \rangle}$.

I know that $\langle g \rangle$ does not always have to be a normal subgroup of $G$ in general, and therefore $M/K$ does not have to be Galois. But is it still correct that $L/M$ is Galois even if $\langle g \rangle$ is not a normal subgroup of $G$? I expect $\operatorname{Gal}(L/M) = \langle g \rangle$ but I am not so sure about it.

If someone could confirm or object, that would be nice!

答:
$L/M$ is separable and normal (the $K$-minimal polynomials split completely in $L$ thus so do the $M$-minimal polynomials), therefore $L/M$ is Galois.

If you know the primitive element theorem then $L=K(a)$,

$Gal(L/M)$ is a subgroup of $Gal(L/K)$,

It is generated by $g$ because $a$ is a root of $f=\prod_{j=1}^{|\langle g\rangle|} (x-g^j(a))\in M[x]$,

An element of $Gal(L/K)$ is determined by where it sends $a$.

Any $h \not \in \langle g \rangle$ is not permuting the root of $f$ thus $h$ is not acting trivially on its coefficients, ie.  $h$ is not leaving $M$ fixed.


问:
Equivalence of definitions of simplicial manifolds, and which ones imply "no branching"
I've found a couple of different definitions of simplicial manifolds with boundary:

A pure abstract simplicial $n$-complex such that the (geometric realization of the) link of every simplex $\sigma$ of dimension $k$ is homeomorphic to a sphere or ball of dimension $n - 1 - k$. (e.g., these notes or these notes.)

A pure abstract simplicial $n$-complex such that the (geometric realization of the) link of every vertex $v$ is homeomorphic to a sphere or ball of dimension $n - 1$. (i.e., #1 but just for $k=0$). (e.g., these notes.)

#1 obviously implies #2, but I'm wondering if #2 implies #1.

In particular, I'm interested in showing that a simplicial manifold obeys the "no branching" condition (part of the definition of a pseudomanifold), which is that every $(n-1)$-simplex is the proper face of 1 or 2 $n$-simplices. #1 implies no branching immediately, and if #2 implied #1, then #2 would also imply no branching. But failing that, does #2 imply no branching also?

答:
So I'm not aware of "simplicial manifold" as a thing. On the other hand the category of "PL manifolds" is a thing, and its definition is slightly more intricate than what you are saying, namely a PL version of 2: A simpicial $n$-complex is a PL-manifold if the link of every vertex is a PL-manifold of dimension $n-1$ that is PL-homeomorphic to the standard PL manifold structure on the sphere of dimension $n-1$ (e.g. the boundary of the $n$-simplex). And then this definition does indeed imply the correponding version of 1. Here's a quick proof, by induction on $n$.

Consider a PL manifold $M$ and a $k$-simplex $\sigma \subset M$, with $k \ge 1$. Pick a $0$-simplex $v \in \sigma$ and therefore $S = \text{Link}_M(v)$ is an $n-1$-sphere with the standard PL structure. The intersection $\tau = \sigma \cap S$ is a $k-1$ simplex in $S$. The key observation is that $\text{Link}_M(\sigma)$ is simplicially isomorphic to $\text{Link}_S(\tau)$, and so it follows by induction on $n$ that this link is a standard PL sphere of dimension $n-k-1$.

But in general, 2 does not imply 1. A counterexample is given by the double suspension theorem of Cannon and Edwards, which produces a simplicial structure on $S^5$ that satisfies 2, and a 1-simplex in that simplicial structure whose link is a 3-dimensional manifold that is not even homeomorphic to $S^3$.


问:
For any $n\times n$ matrix, the set of $\{x | x^T Mx \le c\}$ is a closed set.
I'm trying to prove that for any $n \times n$ matrix $M,$ for any $c>0$, the set of $S=\{x\in \mathbb{R}^n:x^TMx\le c\}$ is a closed set.

I have tried to expand the matrix into the function, however, it doesn't help at all. And I don't remember there is any properties to apply to all square matrix.

Can someone give a hint?

答: HINT: The inverse image of a closed set under continuous function is closed. The function $f:\mathbb R^n\rightarrow \mathbb R$ defined by $f(x)=x^TMx \ \ \forall x\in\mathbb R^n $ is continuous.


问:
Quadratic cost function solution
Why the solution to the following cost function:

$$\frac{1}{2}\|Lm-d\|^2 + \frac{1}{2} \mu \|W_m m\|^2_2$$

the below equation:

$$(L^Td+\mu W^T_m W_m)^{-1} L^Td  $$

答:
Suppose the problem was discretized. I.e. $L$ and $W$ are matrices, $\vec{m}$ and $\vec{d}$ are vectors. Thus the problem is to find the minimum of the following cost function
\begin{equation}
f(\vec{m}) = \frac{1}{2}(L_{ij}m_j - d_i)(L_{ik}m_k - d_i) + \frac{1}{2}\mu W_{ij}m_jW_{ik}m_k
\end{equation}
by arguments $m_p$. To find the extremum of the function, we have to solve the following equation
\begin{equation}
\frac{\partial f}{\partial m_p} = 0,
\end{equation}
which brings us to the following system of equations
\begin{equation}
(L_{ip}L_{ik} + \mu W_{ip}W_{ik})m_k = L_{ip}d_i.
\end{equation}
The last equation can be rewritten in the matrix form as follows
\begin{equation}
(L^{\intercal}L+\mu W^{\intercal}W)\vec{m}=L^{\intercal}\vec{d}.
\end{equation}
The solution of this equation is
\begin{equation}
\vec{m} = (L^{\intercal}L + \mu W^{\intercal}W)^{-1}L^{\intercal}\vec{d}.
\end{equation}

Title: Quadratic cost function solution


问:
How to use combinatorial methods to solve multivariable equation (stars and bars)?
I have the following question:

Let $A,B,C,D \in \mathbb{Z}$, such that $-21 \leq A,B,C,D \leq 21$. How many integer solutions are there for $A+B+C+D > 0$. I'm trying to use combinatorial methods, namely stars and bars, but I'm not sure how I'd be able to use stars and bars here with an inequality and a particular set of integer values for $A,B,C,D$. I'm thinking of using a change of variables but still trying to troubleshoot that.

Would someone know how to use combinatorics to solve the question?
Any help would be much appreciated!

答:
One way to solve it is to combine Karl’s hint in the comments with a change of variables. Let $n$ be the number of integer quadruples $\langle A,B,C,D\rangle$ such that $-21\le A,B,C,D\le 21$.

Among those $n$ quadruples let $n^-$ be the number such that $A+B+C+D<0$, $n^+$ the number such that $A+B+C+D>0$, and $n_0$ the number such that $A+B+C+D=0$.

What is $n$?

We want $n^+$.

Use the hint: explain why $n^+=n^-$, and conclude that $n^+=\frac12(n-n_0)$.

Now you just need to find $n_0$. Let $a=A+21$, $b=B+21$, $c=C+21$, and $d=D+21$; then $A+B+C+D=0$ if and only if $a+b+c+d=84$, and $-21\le A,B,C,D\le 21$ if and only if $0\le a,b,c,d\le 42$. Thus, you want the number of solutions to

$$x_1+x_2+x_3+x_4=84\tag{1}$$

in non-negative integers that do not exceed $42$. You can find $n_0$ by combining stars and bars with an inclusion-exclusion argument; my answer to this question illustrates the technique.


问:
Does $y = x/(1/x)$ contain $x = 0$ in its domain?
Is $y = x/(1/x)$ equivalent to $y = x^2$, or is there a hole at $x=0$?

答: When you say that $x/(1/x)$ and $x^2$ are equivalent, you have already assumed that $x\neq 0$. The two expressions are not equivalent at $x=0.$ The first does not have $x=0$ in its domain.


问:
Laplace transform vanishing at all integers
I am looking to find a function $f$ which is  not zero almost everywhere ($f\neq 0 ~a.e.$) and such that $$ \int_{\mathbb{R}} f(t) e^{-ts} dt = 0~\forall s\in\mathbb{Z}$$

I was thinking of taking an inverse Laplace transform of sine to help me with this, but apparently it does not exist. I am not thinking it should be a more elaborate construction, perhaps some sort of piecewise function.

答:
The OP made it more or less clear that we are dealing with the bilateral Laplace transform.

It follows from the Cauchy integral theorem that $$\int_{-\infty}^\infty e^{-(t+2i\pi)^2} e^{-st}dt=\int_{-\infty}^\infty e^{-t^2} e^{-s(t-2i\pi)}dt$$

whence take $$f(t)=e^{-t^2}-e^{-(t+2i\pi)^2}$$


问:
The stock of a warehouse consists of boxes of high, medium and low quality light bulbs in respective proportions 1:2:2.
The stock of a warehouse consists of boxes of high, medium and low
quality light bulbs in respective proportions $1:2:2$. The probabilities of bulbs of three types being unsatisfactory are $0.3, 0.1$ and $0.2$ respectively. If a box is chosen at random and two bulbs in it are tested and found to be satisfactory, what is the probability that it contains bulbs of high quality.

I was thinking of solving through Bayes theorem. I calculated $P($high quality $|$ satisfactory$)$ from Bayes theorem but couldn't proceed further.

答:
If you calculated $P(high|2satisf)$, then that is your answer. If you calculated $P(high|1satisf)$, then it's not quite right.

```<code>            / 2 satisfactory bulbs .3^2    0.018
/ high  1/5 -

            / 2 satisfactory bulbs .1^2   .004
- med   2/5 -

            / 2 satisfactory bulbs .2^2   .016
\ low   2/5 -

</code>```

Then $$\begin{split}P(high|2satis)&=\frac{P(high,2satisf)}{P(2satis)}\text { Bayes theorem}\\
&=\frac{P(high,2satisf)}{P(2satis,high)+P(2satis,med)+P(2satis,low)}\\
&=\frac{.018}{.018+.004+.016}\\
&=\frac 9 {19}\end{split}$$

So given that you have gotten 2 satisfactory bulbs out of the selected box, the probability that it is a high-quality box is $9/19$. This is what P(high | 2 satisfactory) means, e.g. given that 2 satisfactory bulbs are observed from the box, the probability that it was a high quality box is P(high | 2 satisfactory).

Body:
If you calculated $P(high|2satisf)$, then that is your answer. If you calculated $P(high|1satisf)$, then it's not quite right.

            / 2 satisfactory bulbs .3^2    0.018
/ high  1/5 -

            / 2 satisfactory bulbs .1^2   .004
- med   2/5 -

            / 2 satisfactory bulbs .2^2   .016
\ low   2/5 -

Then $$\begin{split}P(high|2satis)&=\frac{P(high,2satisf)}{P(2satis)}\text { Bayes theorem}\\
&=\frac{P(high,2satisf)}{P(2satis,high)+P(2satis,med)+P(2satis,low)}\\
&=\frac{.018}{.018+.004+.016}\\
&=\frac 9 {19}\end{split}$$

So given that you have gotten 2 satisfactory bulbs out of the selected box, the probability that it is a high-quality box is $9/19$. This is what P(high | 2 satisfactory) means, e.g. given that 2 satisfactory bulbs are observed from the box, the probability that it was a high quality box is P(high | 2 satisfactory).


问:
How can I construct $H$ and $G$ such that $T \circ G=0$ and $H \circ T=0$
I need your help to complete my proof please. This is what I've to prove:

Let be $F$ a fied, $V$ a finite dimensional vector space over $F$, and
a lineal operator $T:V \rightarrow V$ distinct to zero and not
invertible. Prove that there exists $G$ and $H$ lineal operations in
$V$ distinct to zero sush that: $T \circ G = 0$ and $H \circ T = 0$

My attempt:

As we know that $T$ is not injective, then $Ker(T) \neq \left \{ 0 \right \}$. Let be $\mathcal{B}=\left \{ e_1,e_2...,e_k,...e_n \right \}$ a basis of $V$, such that $\mathcal{B}_{W_T}=\left \{ e_1,...,e_k \right \}$ is a basis of the subspace $W_T=Ker(T)$. So if we define $G:V \rightarrow W_T$, it satisfies $\forall \alpha \in V$ that $(T \circ G) \alpha = 0$.

By the other side, we have also that $T$ is not invertible so that implies that $T$ is not surjective. So, $T:V \rightarrow W_H$, with $W_H \subset V$. We can construct $H$ a linear operator such that $W_H=Ker(H)$. So, it would satisfies $(H \circ T)\alpha=0$ for any $\alpha \in V$.

My doubt:

Is it necessary to construct $H$ and $G$ with that characteristics to complete my proof? Or is it enough to name those properties. I think that it is neccesary to construct them explicitly but I don't know how, can you help me please to construct them?

答:
Choose $k$ distinct integers $i_1,\dots,i_k \in \{1,\dots,n\}$ and define the linear map $G : V \to V$ by sending $e_{i_j}$ to $e_j$ (for $j=1,\dots,k$) and the others $n-k$ basis vectors to $0$. For example, $$G(e_1)=e_1,\dots,G(e_k)=e_k,G(e_{k+1})=0,\dots,G(e_n)=0.$$

Since $T \neq 0$ and $T$ is not surjective, $\operatorname{im} T$ is a proper nonzero subspace of $V$. Thus, we can find a basis $v_1,\dots,v_n$ for $V$ such that $v_1,\dots,v_l$ $(l< n)$ is a basis for $\operatorname{im} T$. Now define the linear map $H : V \to V$ by sending $v_1,\dots,v_l$ to $0$ and the others $n-l$ basis vectors not all to $0$.


答:
You can write this as

$\min \frac{1}{2} \| Am-b \|_{2}^{2}$

where

$A=\left[
\begin{array}{c}
L \\
\sqrt{\mu} W_{m}
\end{array}
\right]$

and

$b=\left[
\begin{array}{c}
d \\
0
\end{array}
\right]$.

Applying the normal equations to this linear least squares problem gives (assuming that the inverse exists)

$m=(A^{T}A)^{-1}A^{T}b.$

This simplifies to

$m=\left( \left[ L^{T} \sqrt{\mu}W_{m}^{T} \right] \left[ \begin{array}{c}
L \\
\sqrt{\mu}W_{m}
\end{array}\right]\right)^{-1} \left[ L^{T} \sqrt{\mu} W_{m}^{T} \right] \left[ \begin{array}{c}
d \\
0
\end{array}
\right]$

and finally

$m=\left(L^{T}L+\mu W_{m}^{T}W_{m} \right)^{-1}L^{T}d$.

Title: Quadratic cost function solution

Body:
Why the solution to the following cost function:

$$\frac{1}{2}\|Lm-d\|^2 + \frac{1}{2} \mu \|W_m m\|^2_2$$

the below equation:

$$(L^Td+\mu W^T_m W_m)^{-1} L^Td  $$


问:
How to simplify $\frac{\sqrt{2}+2(\cos 20^\circ+\cos 25^\circ)}{\sin \left(90-\frac{45}{2}\right)\sin 55^\circ \sin 57.5^\circ}$?
The problem is as follows:

Simplify the following expression:

$B=\frac{\sqrt{2}+2(\cos 20^\circ+\cos 25^\circ)}{\sin \left(90-\frac{45}{2}\right)\sin 55^\circ \sin 57.5^\circ}$

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&7.5\\
2.&6\\
3.&8\\
4.&5\\
\end{array}$

I'm not sure how to proceed here because the division seems kind of complicated to simplify.

But I could spot that suspicioulsy $\sqrt{2}= \csc 45^\circ$

and also $2= \csc 30^\circ$

But I don't know if these would come into play in solving this problem as it is challenging. Perhaps does it exist a way to solve this without much fuss?.

I could also spot that:

$\sin \left(90-\frac{45}{2}\right)= \cos \frac{45}{2}$

$\sin 55^\circ = \cos 35^\circ$

and

$\sin 57.5^\circ= \cos 32.5^\circ = \sin \frac{65}{2}^\circ$

The rest I presume that involves the simplifcation of the expresion using sum to product formulas. But I got stuck with those. Can someone help me here?.

答:
I think you may find some help here at Wikipedia. There are general trig identities and there are some... surprising identities. I've linked to a section with some identities that only hold when $\alpha + \beta + \gamma = \pi = \frac{\tau}{2} = 180°$, which if you look closely might (might!) help in the denominator.

In particular, $\sin \alpha \cdot \sin \beta \cdot \sin \gamma = \frac14(\sin 2\alpha + \sin 2 \beta + \sin 2\gamma)$

Earlier on the same page is a section of identities without variables, which might also help, though I haven't tried to fully solve based on either of these--just looking at resources.

Edit: Is there the possibility you're supposed to use small-angle identities to [cough] "ignore" some of the tiny leftover bits. $\cos 2.5° = 0.999$. It's not rigorous but it might nonetheless be intended behavior on the part of the instructor.


问:
How to solve this equation from a FOE
How do I solve this

$$5000 =  21t - 111.37e^\frac{-12t}{6.5625} + 1207.12$$

I am using the image below as a guide for my own question (above) but they mention to use math software to solve the equation but I have used Symbolab, Mathway and Photomath to solve and they wont work.

Have I done something wrong?

答:
Working with whole numbers and making the problem more general, you need to fing the zero of function
$$f(t)=15 t-\frac{379}{5} e^{-\frac {32}{15} t}+\frac{11503}{10}-k$$ and, as @vitamin d showed using Wolfram Alpha, the only explicit solution is given by
$$t=\frac {10k-11503}{150}+\frac{15}{32}W\left(\frac{12128}{1125}\exp\Big[ -\frac{16 (10 k-11503)}{1125}\Big]\right)$$ where $W(t)$ is Lambert function.

For $k=5000$, the argument of Lambert function is extremely small  $(\sim 1.78 \times 10^{-237})$ and can be neglected. So, avery good approximation is
$$t \sim \frac{38497}{150} \sim 256.6466667$$ while the exact solution would be  the same !

Title: How to solve this equation from a FOE


问:
Find, with calculus shapes of function $y=f(x)$, $y = g(x)$
There's a question of integrals. The region between some unknown function, $y=f(x)$,$y = g(x)$, and $f(x) \geq g(x) \geq 0$ in intervals $[2,5]$ is rotated about the lines, $x=9$ to make a 3D shape. So how do I find the volume of the shape? I personlly use disk and sphere method, and found $ \int_{2}^{5} \pi (f(x)+9)^2-(g(x)+9)^{2}dx$ is correct answer. So I'm here to ask questions to check if this is correct, or this is wrong. But if it's wrong answer, then please tell me if which one is right.

答:
Your formula is not correct and since you didn't show how you found it, we cannot check where you made mistakes.

Note, that for $x\in [2,5]$, the distance to the axis of rotation $x=9$ is
$r(x) = 9-x$. Rotating about the given axis gives rise to a cylinder with infinitesimal thickness $dx$ and a height of $h(x) = f(x)-g(x)$.

The Volume of such an infinitesimal cylinder is

$$dV(x) = 2\pi r(x)h(x)dx\Rightarrow V =2\pi\int_2^5(9-x)(f(x)-g(x))\;dx$$


问:
midline and amplitude of the curve $y = 3\csc (\frac{\theta}{2} - \frac{\pi}{3})+2$
Can anyone please tell me what is midline and amplitude of the curve $y = 3\csc \left(\frac{\theta}{2} - \frac{\pi}{3}\right)+2.$

My Attempt:  As maximum and minimum of $y = 3\csc (\frac{\theta}{2} - \frac{\pi}{3})+2$ do not exist. The amplitude of it will also not exist. $3\csc (\frac{\theta}{2} - \frac{\pi}{3})$ does not exist when $(\frac{\theta}{2} - \frac{\pi}{3}) = 0$. Then how we will find the midline ?

can anyone please help me ?

答: If you plot the cosecant function, you notice that it consists of repeating intervals of length $\pi$, alternating one above $y=1$, the next below $y=-1$. scaling and shifting $\theta$ will just scale/shift the intervals, but not the amplitude. When you multiply by $3$ you change the amplitude, and then you shift it up by $2$. So the midline will be $y=2$.


问:
To need a way of thinking about Ji_Chen's nice result $(a- c)^{2}+ (b- d)^{2}\geq\frac{7}{9}ab- \frac{7}{20}(c^{2}+ 4d^{2})$
given four real numbers $a, b, c, d$

Ji Chen gave a nice result on.AoPS
$$\left ( a- c \right )^{2}+ \left ( b- d \right )^{2}\geq\frac{7}{9}ab- \frac{7}{20}\left ( c^{2}+ 4d^{2} \right )$$

The original problem consists a condition $ab= c^{2}+ 4d^{2}= 4$ and hides the rightside of the earlier inequality, I wondered that how he could find this.. of course I searched in AoPS because of him but I also found another interesting work, which made me sense.

He (AoPS/@ye109) claimed that
$$\left ( a- c \right )^{2}+ \left ( b- d \right )^{2}\geq\frac{2\left ( k^{4}- 4 \right )}{3k^{2}}ab+ \frac{k^{4}- 4}{k^{4}- 16}\left ( c^{2}+ 4d^{2} \right )$$
with the best $k\cong 1.672955253$ is a real root of $k^{12}- 28k^{8}- 36k^{6}- 128k^{4}+ 1024$

That guy is very good at making second solution, we just need explain his solution from Ji Chen's 1st one, if Ji Chen used discriminant to find $7/9, -7/20$ then how could he use discriminant to find $\frac{2\left ( k^{4}- 4 \right )}{3k^{2}}, \frac{k^{4}- 4}{k^{4}- 16},$ maybe there could exist discriminant for polynomials not only for constants. What's your idea ? Who can give me a way of thinking ? Thanks a real lot !

Body:
it's about a symmetric matrix, the Hessian of your quadratic form, being positive semidefinite.

$$ P^T H P = D  $$
$$\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
 \frac{ 7 }{ 18 }  & 1 & 0 & 0 \\
 \frac{ 324 }{ 275 }  &  \frac{ 126 }{ 275 }  & 1 & 0 \\
 \frac{ 18 }{ 5 }  &  \frac{ 12 }{ 5 }  &  \frac{ 8 }{ 3 }  & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
180 &  - 70 &  - 180 & 0 \\
 - 70 & 180 & 0 &  - 180 \\
 - 180 & 0 & 243 & 0 \\
0 &  - 180 & 0 & 432 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
1 &  \frac{ 7 }{ 18 }  &  \frac{ 324 }{ 275 }  &  \frac{ 18 }{ 5 }  \\
0 & 1 &  \frac{ 126 }{ 275 }  &  \frac{ 12 }{ 5 }  \\
0 & 0 & 1 &  \frac{ 8 }{ 3 }  \\
0 & 0 & 0 & 1 \\
\end{array}
\right)
 = \left(
\begin{array}{rrrr}
180 & 0 & 0 & 0 \\
0 &  \frac{ 1375 }{ 9 }  & 0 & 0 \\
0 & 0 &  \frac{ 1701 }{ 55 }  & 0 \\
0 & 0 & 0 & 0 \\
\end{array}
\right)
$$
$$ Q^T D Q = H  $$
$$\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
 -  \frac{ 7 }{ 18 }  & 1 & 0 & 0 \\
 - 1 &  -  \frac{ 126 }{ 275 }  & 1 & 0 \\
0 &  -  \frac{ 324 }{ 275 }  &  -  \frac{ 8 }{ 3 }  & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
180 & 0 & 0 & 0 \\
0 &  \frac{ 1375 }{ 9 }  & 0 & 0 \\
0 & 0 &  \frac{ 1701 }{ 55 }  & 0 \\
0 & 0 & 0 & 0 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
1 &  -  \frac{ 7 }{ 18 }  &  - 1 & 0 \\
0 & 1 &  -  \frac{ 126 }{ 275 }  &  -  \frac{ 324 }{ 275 }  \\
0 & 0 & 1 &  -  \frac{ 8 }{ 3 }  \\
0 & 0 & 0 & 1 \\
\end{array}
\right)
 = \left(
\begin{array}{rrrr}
180 &  - 70 &  - 180 & 0 \\
 - 70 & 180 & 0 &  - 180 \\
 - 180 & 0 & 243 & 0 \\
0 &  - 180 & 0 & 432 \\
\end{array}
\right)
$$

See https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia


问:
How to solve these set of ODEs?
I am in need of directions to obtain the result (shown below) for the following problem:

For some constants, $\alpha$ and $Q$, I need to solve the following set of ODEs to find $b(x)$ and $u(x)$:

$$\frac{d}{dx} (b^2(x)u(x)) = 2\alpha bu$$

$$\frac{d}{dx} (b^2(x)u^2(x)) = \frac{Q}{u}$$

The result is presented as $b = \frac{6 \alpha}{5}x$ and $u = \frac{5}{6 \alpha} (\frac{9}{10} \alpha Q)^\frac{1}{3}x^{-\frac{1}{3}}$ in the paper by Morton et al.

答: If you set $(\mathbb{U},\mathbb{B})=(b^2u,b^2u^2)$ you can rewrite your system as $$\frac{d\mathbb{U}}{dx}=2\alpha\sqrt{\mathbb{B}}$$ $$\frac{d\mathbb{B}}{dx}=\frac{Q\mathbb{U}}{\mathbb{B}}$$ Dividing these equations yield $$\frac{d\mathbb{B}}{d\mathbb{U}}=\frac{Q\mathbb{U}}{2\alpha \mathbb{B}\sqrt{\mathbb{B}}}$$ which is separable. Can you finish?

Title: How to solve these set of ODEs?


答:
First of all $\frac{\log\ |1-z|}{z}$ is integrable on $|z|=1$ (it is continuous away from a logarithmic singularity at $z=1$)

$$\int _{|z|=1} \frac{\log\ |1-z|}{z}dz=i \Im(\int _{|z|=1} \frac{\log(1-z)}{z}dz)=\lim_{r\to 1} i\Im(\int _{|z|=r} \frac{\log(1-z)}{z}dz)=0$$
by the Cauchy integral theorem (as $\frac{\log(1-z)}{z}$ is analytic for $|z|<1$)

Title: Evaluate$ \oint _{|z|=1} \frac{\log\ |1-z|}{z}dz $

Body:
Evaluate

$$ \oint _{|z|=1} \frac{\log\ |1-z|}{z}dz $$
My Attempt
$$ I=\oint _{|z|=1} \frac{\log\ |1-z|}{z}dz $$
$$z=e^{i\theta} \Rightarrow dz =i e^{i\theta}d\theta$$
$$I=i \int_{0}^{2\pi} \log\  |1-e^{i\theta}|d\theta$$
$$\text{ define }\, f(z)= |1-z|$$
$f(0)=1$

Now, $f(z)=0$ gives $z=1$.

By Jensen's Formula,
$$\frac{1}{2\pi} \int_{0}^{2\pi} \log\  |1-e^{i\theta}|d\theta = \log |f(0)|+\log(1)$$
$f(0)=1$ gives,
$$\frac{1}{2\pi} \int_{0}^{2\pi} \log\  |1-e^{i\theta}|d\theta =0$$
$$I=0$$


问:
Show $\frac{x_1}{x_n} + \frac{x_2}{x_{n-1}} + \frac{x_3}{x_{n-2}} + \dots + \frac{x_n}{x_1} \geq n$
I was recently asked this question which stumped me.

How can you show $\dfrac{x_1}{x_n} + \dfrac{x_2}{x_{n-1}} + \dfrac{x_3}{x_{n-2}} + \dots + \dfrac{x_n}{x_1} \geq n$ for any positive reals $x_1, x_2, \dots, x_n$?

答: Hint: $\frac a b + \frac b a\ge2$</p>

Body: Hint: $\frac a b + \frac b a\ge2$


问:
How to formally prove that $\gcd(a,1) = 1$ when $a \in Z$
I was trying to prove that $$\gcd(2a+1,9a+4) = 1 , a\in Z$$

My proof went like this:

$$ (9a+4) = 4(2a+1) + a$$
$$ (2a+1) = 2(a) + 1 $$
$$ \Rightarrow \gcd(9a+4,2a+1) = \gcd(2a+1,a)=\gcd(a,1) = 1$$
The last line of reasoning seems intuitive for me; The greatest common divisor of any integer with 1 is definitely one because the only divisor of 1 is itself. However, the question of how to prove it formally arose.

Thanks for any suggestions!

答:
Let $D_a$ be the set of divisors of $a$. Note that $-1, 1 \in D_a$. On the other hand, $D_1 = \{\pm 1\}$.

Thus, $D_a \cap D_1 = \{\pm 1\}$. By definition, the gcd is the largest element in this (finite) set. Since $-1 < 1$, we get $\gcd(a, 1) = 1$.

If your definition works with positive integers, then just remove $-1$ everywhere from the above argument.

Some also define $\gcd(a, b)$ (where at least one of $a$ or $b$ is non-zero) to be the smallest positive element in $$S_{a, b} = \{am + bn : m, n \in \Bbb Z\} \subset \Bbb Z.$$
With this definition, it is clear that $\gcd(a, 1) \ge 1$ (since $1$ is the smallest positive integer).

On the other, $1 \in S_{a, 1}$ since $1 = a\cdot0 + 1\cdot1$ and thus, $1 \ge \gcd(a, 1)$.

Again, we get $\gcd(a, 1) = 1$.

Note that all the above arguments are also valid if $a = 0$.


答:
The formal definition for the greatest common divisor function is:
$$\gcd(n,m)=\max\{d\in\mathbb{N}:d\mid n \,\wedge\,d\mid m\}.$$
Set $m=1$. The only number $d$ that satisfies $d\mid n$ and $d\mid1$ is $1$. So we have
$$\gcd(n,1)=\max\{1\}=1.$$

Title: How to formally prove that $\gcd(a,1) = 1$ when $a \in Z$

Body:
I was trying to prove that $$\gcd(2a+1,9a+4) = 1 , a\in Z$$

My proof went like this:

$$ (9a+4) = 4(2a+1) + a$$
$$ (2a+1) = 2(a) + 1 $$
$$ \Rightarrow \gcd(9a+4,2a+1) = \gcd(2a+1,a)=\gcd(a,1) = 1$$
The last line of reasoning seems intuitive for me; The greatest common divisor of any integer with 1 is definitely one because the only divisor of 1 is itself. However, the question of how to prove it formally arose.

Thanks for any suggestions!


问:
$F^m \otimes F^n \cong F^{mn}$, so what is the bilinear map?
Let $F$ be a field and $m, n \in \mathbb{N}$. By definition of the tensor product, there is a bilinear map $B : F^m \times F^n \to F^m \otimes F^n$ such that if $V$ is an $F$-vector space and $C : F^m \times F^n \to V$ is bilinear, then there exists a unique linear map $\tilde{C} : F^m \otimes F^n \to V$ such that $C = \tilde{C} \circ B$.

Since $\dim F^m \otimes F^n = \dim F^{mn}$, it follows that $F^m \otimes F^n \cong F^{mn}$. Is there a bilinear map $B' : F^m \times F^n \to F^{mn}$ that behaves just like $B$?

答:
The universal bilinear map $F^m \times F^n\to F^m\otimes F^n$ is just $$(\sum_j a_j e_j,\sum_i b_i e_i)\to (\sum_j a_j  e_j)\otimes (\sum_i b_i  e_i)=\sum_j\sum_i a_j b_i e_j\otimes e_i$$ Sending $e_j\otimes e_i\to e_{ji}$ you get your universal bilinear map $F^m \times F^n\to F^{mn}$.

It is the same as sending $(u,v)$ to the matrix $u v^\top$.


问:
Mapping Between Sequences: Example
Take $0\leq r < m$, and let all values be nonnegative and integer.  Consider the function on a sequence ${x(n)}$, $\Phi_mx(mn+r)=mx(n)+\frac{r}{m}(x(n+1)-x(n))$, where we consider $x(0)=0$.

As an example, $\Phi_3$ maps $1,1,2,3$ to $1,2,3,3,3,3,4,5,6,7,8,9$.  Would anyone be able to walk me through why this is?  I end up with different values.

答:
It appears that the definition of $\Phi_m$ was written incorrectly in the original article, and that the correct definition is
$$\begin{aligned}(\Phi_mx)(mq+r)&=m\left(x(q)+\frac{r}{m}\left(x(q+1)-x(q)\right)\right)\\
&=mx(q)+r(x(q+1)-x(q)).\end{aligned}$$
Here $x$ is a sequence with initial term $x(0)=0,$ $m$ is a fixed positive integer, and $r$ is an integer satisfying $0\le r<m.$

If $y=\Phi_mx,$ then the $(mq)^\text{th}$ term of $y$ is $m$ times the $q^\text{th}$ term of $x.$  The $(mq)^\text{th}$ through $(mq+m)^\text{th}$ terms of $y$ form an arithmetic progression.  So $y$ can be thought of as a piecewise linear function on the nonnegative integers.

To perform some explicit computations, let $x$ be the sequence $0,1,1,2,3.$  Then $(\Phi_3x)(0)=0,$ $(\Phi_3x)(3)=3x(1)=3,$ $(\Phi_3x)(6)=3x(2)=3,$ $(\Phi_3x)(9)=3x(3)=6,$ and $(\Phi_x)(12)=3x(4)=9.$  The full sequence $\Phi_3x$ contains two terms between each two successive terms in the sequence $0, 3, 3, 6, 9,$ and these two terms interpolate linearly between successive terms.  Hence, one gets
$$\mathbf{0}, 1, 2, \mathbf{3}, 3, 3, \mathbf{3}, 4, 5, \mathbf{6}, 7, 8, \mathbf{9}.$$
Alternatively, we may just apply the formula.  For example, $(\Phi_3x)(7)=(\Phi_3x)(3\cdot2+1)=3x(2)+1\cdot(x(3)-x(2))=3\cdot1+1\cdot(2-1)=4,$
which agrees with the result above.

To take another example, $(\Phi_2x)$ is the sequence
$$\mathbf{0}, 1, \mathbf{2}, 2, \mathbf{2}, 3, \mathbf{4}, 5, \mathbf{6}.$$
The even indexed terms (in boldface) are simply $2$ times the sequence $x,$ while the remaining terms interpolate linearly between the terms in boldface.  An explicit computation of $(\Phi_2x)(3)$ is
$$(\Phi_2x)(3)=(\Phi_2x)(2\cdot1+1)=2\cdot x(1)+1\cdot(x(2)-x(1)=2\cdot1+1\cdot(1-1)=2.$$

The functions $\Phi_m$ have the sequence $z(0)=0,$ $z(1)=1,$ $\ldots,$ $z(n)=n,$ $\ldots,$ that is, the sequence of nonnegative integers, as a fixed point.  In other words, $(\Phi_mz)(n)=z(n)=n$ for all integers $m>0$ and $n\ge 0$.  To see this, simply compute
$$(\Phi_mz)(mq+r)=mz(q)+r(z(q+1)-z(q))=mq+r((q+1)-q)=mq+r.$$

Title: Mapping Between Sequences: Example


答:
using Euclid's Lemma, GCD (a,b) = GCD(a,a-b), , You can continue like this ,by subtracting smaller number from larger number.

${(2a+1 , (9a+4-2a-1))=(2a+1,7a+3)...  (a,a-1)=1}$

-|  but there is a faster way to do this.

${ GCD(2a+1,9a+4)= (2a+1, 4(2a+1)+a)= (2a+1,a)= (a,2a+1)=(1,a)=1 }$


问:
$\int_\gamma\textbf{u}\ d\textbf{r}$, wtih $\gamma$ a curve along an ellipse.
Let $$\textbf{u}(x,y)=(y(8x+1),2y^2)\\ \gamma: 4x^2+y^2=1, \ \text{from}\ (-\frac{1}{4},\frac{\sqrt{3}}{2})\ \text{to} \ (\frac{\sqrt2}{4},\frac{\sqrt2}{2})$$
and calculate the integral $\int_\gamma\textbf{u}\ d\textbf{r}$.

So my take is to use Green's formula to get: $$\int_\gamma y(8x+1) dx+2y^2dy=\iint_D \frac{\partial}{\partial x}(2y^2)-\frac{\partial}{\partial y}(y(8x+1))\ dxdy=\iint_D -8x-1\ dxdy $$

I can substitute to polar coordinates; $0\leq r\leq 1, t_1=(-\frac{\pi}{2}-\arctan{(\frac{1}{2\sqrt3})})\leq \theta \leq \arctan{(2)}=t_2$) and get $$x=\frac{1}{2}r\cos(\theta)\\y=r\sin(\theta)\\J=\frac{1}{2}r$$

When I integrate $$\int_{t_1}^{t_2}\int_1^0 2r^2cos(\theta)+\frac{r}{2} drd\theta$$

This gives me the wrong evaluation, but I am not sure why. Is it because I have not defined the the path of the curve back to its starting point and thus not getting a well defined shape or something else?

答:
You cannot use Green's Theorem as the path is not a closed curve. So you will have to do line integral directly.

$\vec F = (8xy+y,2y^2)$

$\gamma: 4x^2+y^2=1, \ \text{from}\ \text{point A} (-\frac{1}{4},\frac{\sqrt{3}}{2})\ \text{to} \ \text{B} (\frac{\sqrt2}{4},\frac{\sqrt2}{2})$.

Parametrize the ellipse as $ \ \gamma(t) = (\frac{1}{2} \cos t, \sin t)$. Based on values of $x, y$ coordinates, point A is represented by $t = \frac{2\pi}{3}$ and point B by $t = \frac{\pi}{4}$.

$\gamma'(t) = (-\frac{1}{2} \sin t, \cos t)$

$\vec F(\gamma(t)) = (4 \sin t \cos t + \sin t, 2 \sin^2t)$

$\vec F(\gamma(t)) \cdot \gamma'(t) = - \frac{1}{2} \sin^2t$.

Now integrate going from point A to point B.


问:
Complex logic binary and M
Here's the problem:

a and b are continuous between -15 and 15. Explain how the following conditions can be represented as linear constraints using binary variables: At least two of the conditions must be met.

Condition 1:
$(a-b)^2 \le 16$

Condition 2:
$ b \le 3$ and $a \ge -1$

Condition 3:
$a = -1 $

Note that big-M parameters must be the best possible big Ms.

I think there should be a binary variable for each condition, and that condition 2 must be split out into two binary variables, but I feel pretty lost on how I can apply big M logic.

答:
Yes, you want three binary variables $x_i$, with linear constraint $\sum_i x_i \ge 2$.  Note that the first condition is equivalent to $|a-b|\le 4$. The big-M constraints are
\begin{align}
a-b-4&\le M_1(1-x_1)\\
-a+b-4&\le M_2(1-x_1)\\
b-3&\le M_3(1-x_2)\\
-a-1 &\le M_4(1-x_2)\\
-a-1 &\le M_5(1-x_3)\\
a+1 &\le M_6(1-x_3)\\
\end{align}
I’ll leave it you to supply the big-M values.

Title: Complex logic binary and M


问:
How can I construct a specific sigmoid function?
The simple sigmoid function $$f(x)=1/(1+e^{−x})$$ approaches zero as x tends to negative infinity, and approaches $1$ as x tends to positive infinity. But I want to set $1$ and $20$ instead of $0$ and $1$. Simply my sigmoid function should behave like this:
$$\lim_{x \to +\infty} f(x) = 20,$$ and $$ \lim_{x \to -\infty} f(x) = 1.$$

答:
For a more general case where,

$$\lim_{x \to +\infty} f(x) = u,$$ and $$ \lim_{x \to -\infty} f(x) = v.$$

Try,

$$
\sigma(x) = \frac{v+ue^x}{1+e^x}
$$


问:
Prove that: $\vdash \forall x(\forall y\alpha)\to \forall y(\forall x \alpha)$
How does one prove that

$$\vdash \forall x(\forall y\alpha)\to \forall y(\forall x \alpha)$$

in first order logic?

I have tried using the specialization and generalization rules on various wffs but they don't lead me to anything concrete.

答:
I have tried using the specialization and generalization rules on various wffs but they don't lead me to anything concrete.

They should.

Just assume $\forall x~(\forall y~\alpha)$, instantiate using fresh arbitrary variables, and generalise in reverse order to derive $\forall y~(\forall x~\alpha)~$, under that assumption.

Now you may deduce $(\forall x~(\forall y~\alpha))\to(\forall y~(\forall x~\alpha))$.

$\blacksquare$


问:
Choquet Theorem and Rudin RCA problem 5.12
Let $K$ be a triangle (two-dimensional figure) in the plane, let $H$ be the set consisting of the vertices of $K$, and let $A$ be the set of all real functions $f$ on $K$, of the form
$$
f(x, y) = \alpha x + \beta y + \gamma, \quad (\alpha, \beta,\gamma \in\mathbb{R}\ ).
$$
Show that to each $(x_0, y_0)  \in K $ there corresponds a unique measure $\mu$ on $H$ such that
$$
f(x_0, y_0) = \int_H fd\mu.
$$
(Compare Sec. 5.22.)

Replace $K$ by a square, let $H$ again be the set of its vertices, and let $A$ be as above. Show that to
each point of $K$ there still corresponds a measure on $H$, with the above property, but that uniqueness
is now lost.
Can you extrapolate to a more general theorem? (Think of other figures, higher dimensional
spaces.)

I can prove the exist but Cannot figure out the uniqueness and the general theorem.

答:
Thanks for Kavi Rama Murthy comments.
The triangle and square just a special case of The Choquet_theory.
The form of function in $A$ just are liner. The interior value can be represented by its vertices. As in triangle there is only uniqueness but for square there two.


问:
Why is $2$-adic roots of unity $\{1,-1\}$
I am reading a proof of the fact that the only root of unity in $\mathbb Q_2$ is $\pm1$. But I am stuck at one point: The proof says that all of the root of unity in  $\mathbb Q_2$ has an order of a power of $2$. And since $-1$ is not a square in $\mathbb Q_2$, the root of unity in $\mathbb Q_2$ has to have order $2$ and thus they are just $\pm1$.

I understand the second sentence here but have no idea whether root of unity in  $\mathbb Q_2$ has order of a power of $2$. Any help is appreciated!

答: Clearly any root of unity in $\mathbb{Q}_2$ is actually in $\mathbb{Z}_2$. But now note that the group of units of $\mathbb{Z}/(2^n)$ has order $2^{n-1}$, so its only element of odd order is $1$.  This means that any root of unity in $\mathbb{Z}_2$ of odd order must be $1$ mod $2^n$ for all $n$, and thus must be equal to $1$.


问:
Constructing injections between sets proof
$\DeclareMathOperator{\card}{card}$
Show that $\card(\mathbb{R}) = \card(\mathbb{R} \setminus \mathbb{Q})$.

My attempt:

We have to find two injections $\mathbb{R}\setminus\mathbb{Q} \rightarrow \mathbb{R}$ and $\mathbb{R} \rightarrow \mathbb{R}\setminus\mathbb{Q}$.

For the first injection, $\mathbb{R} \setminus \mathbb{Q} \subset \mathbb{R}$ so a natural injection $\phi: \mathbb{R} \setminus \mathbb{Q} \rightarrow \mathbb{R}$ exists.

As for the second injection, define $\psi: \mathbb{R} \rightarrow \mathbb{R}\setminus\mathbb{Q}$ by:

$$\psi(x)= \left\{
  \begin{array}{lr}
      x & \mbox{if } x \notin \mathbb{Q} \\
      \sqrt2 + x & \mbox{if } x \in \mathbb{Q}
      \end{array}
\right.$$

Let $a, b \notin \mathbb{Q}$. Assume $\psi(a) = \psi(b)$ then automatically $a = b$.

Let $a, b \in \mathbb{Q}$. Assume $\psi(a) = \psi(b)$, then $\sqrt2 + a = \sqrt2 + b \implies a = b$.

The Cantor-Berstein theorem holds that $\mathbb{R}$ has the same cardinal as $\mathbb{R} \setminus \mathbb{Q}$.

QED.

Is this proof correct?

In spite of writing this, I still don't understand what a natural injection is. Is it just $\phi(x) = 1$? Or how is it defined? Rather is there proof that for two sets $A, B$, if $A \subset B$ then there is an injection $\rho: A \rightarrow B$?

Any assistance is appreciated.

答:
To answer your question about injections: Given sets $A \subset B$, there's a natural injection $i : A \to B$ given by $i(a) = a$ for all $a \in A$.

Since $A \subset B$, the above function is well-defined. It is easy to see that it is indeed injective.

As pointed out in the comments, your $\psi$ does not exactly work. However, a small modification does work:

Let $A = \{a + b\sqrt 2 \mid a \in \Bbb Q, b \in \Bbb Z_{\ge 0}\} \subset \Bbb R.$ (Note $\Bbb Q \subset A$.)

$$\psi(x) = \begin{cases}x + \sqrt 2 & x \in A, \\x & x \notin  A \end{cases}.$$

You'd have to check that the above is well-defined. (Just need to show that output is always irrational.)

After that, simply check that it's one-one by taking three cases. (Which three cases?) The key point here is that $x \in A \iff f(x) \in A$. This is what was missing from your example.

In fact, you can actually show that the above is a bijection, not just an injection.


问:
Solving recursive definition while factorials are involved
I've recently started studying recursive definitions of sequences and have encountered a recurrence formula which seems to include a factorial. Until now I have not seen any like this.

$$t_n = 5  \,\frac{(n + 1)!}{4^n} ,\quad n ≥ 1.$$

If I was to take a look at this segment (n + 1)! and replace n with position 1 would the factorial simply be 2! which I can then input back into the recurrence formula like :

$$t_n = 5\, \frac{2!}{4^n} ,\quad n ≥ 1.$$

Apologies for any incorrect terminology.

答:
$$T_n=5 \frac{(n+1)!}{4^n}~~~~(1)$$ will follow the recurrence relation given by
$$T_{n}=\frac{1}{4}(n+1)T_{n-1}~~~~(2)$$

EDIT for OP:

Now in order to get (1) from (2) one can do multiplicative telescoping or
divide (2) by $(n+1)!$ to get
$$\frac{T_n}{(n+1)!}=\frac{1}{4}\frac{(n+1)T_{n-1}}{(n+1)!}=\frac{1}{4}\frac{T_{n-1}}{n}~~~(3)$$
Next, define $\frac{T_{n-1}}{n!}=V_{n-1}$
to get $$V_{n}=\frac{1}{4} V_{n-1} \implies V_n=(\frac{1}{4})^n C \implies \frac{T_n}{(n+1)!}=\frac{C}{4^n},$$
giving back (1), where the constant $C=5$.


问:
Is there exist countable union of half-open $(a,b]$ in Interval $(0,1]$
We know that $\mathbb{X} = (0,1]$,and $\mathcal{M} = \{\emptyset,\text{finite unions of disjoint intervals that are open on the left and closed on the right} \}$ is not a $\sigma$-algebra.

However,if I let $\mathcal{M_1} = \{\emptyset,\text{countable unions of disjoint intervals that are open on the left and closed on the right} \}$, is $M_1$  a $\sigma$-algebra?

答: No. $\bigcup_{n=2}^{\infty} (\frac 1 2-\frac  1 n, \frac 1 2-\frac  1 {n+1}] \cup (\frac 1  2, 1]=(0,1]\setminus \{\frac 1 2\}$. The left side is in your class. If it is  a sigma algebra then $\{\frac 1 2\}$ would be in the class, but this is not true.


问:
Show that a topology is a first countable space

Let $\tau$ be the topology on the real line $\mathbb{R}$ generated by the closed open intervals $[a,b)$. Show that it is a first countable space.

So, what I think I have to do is to show that for any point $p \in \mathbb{R}$ there is a countable local base, for example $$B_{p}= \left\{\left[p,n^{-1} \right)\; n \in \mathbb{N}\right\}$$

My reasoning is that the open spheres in this topology are the same close open intervals and that the open sphere is getting smaller as $n$ increases and all of those balls include $p$. Am i correct?

答:
That’s not going to work unless $p=0$. If $p\ge 1$, the sets $\left[p,\frac1n\right)$ are all empty. However, you could let

$$\mathscr{B}_p=\left\{\left[p,p+\frac1n\right):n\in\Bbb Z^+\right\}\,.$$

To prove that this is a local base at $p$ you must show that if $U$ is any open nbhd of $p$ in this topology, there is an $n\in\Bbb Z^+$ such that $p\in\left[p,p+\frac1n\right)\subseteq U$. You can start by observing that the sets of the form $[a,b)$ with $a<b$ are a base for the topology, so if $U$ is an open nbhd of $p$, there must be $a,b\in\Bbb R$ such that $p\in[a,b)\subseteq U$. Now you just have to explain why there is an $n\in\Bbb Z^+$ such that

$$\left[p,p+\frac1n\right)\subseteq[a,b)\,.$$


答:
No it isn't.

The intervals $(1-\frac1n,1-\frac1{n+1}]$, $n=1,2,\dots$ are disjoint, the countable union is $(0,1)$, whose complement in $(0,1]$ is the singleton $\{1\}$ not in $\mathcal{M}_1$.

Title: Is there exist countable union of half-open $(a,b]$ in Interval $(0,1]$

Body:
We know that $\mathbb{X} = (0,1]$,and $\mathcal{M} = \{\emptyset,\text{finite unions of disjoint intervals that are open on the left and closed on the right} \}$ is not a $\sigma$-algebra.

However,if I let $\mathcal{M_1} = \{\emptyset,\text{countable unions of disjoint intervals that are open on the left and closed on the right} \}$, is $M_1$  a $\sigma$-algebra?


问:
Proof that $2^n-1$ is divisible by $3$ for all $n\in\mathbb{N}$ even
Basis: This is true for $ n = 2 $
$$ 2^2 - 1 = 3 $$

Hypothesis: There exists a number $ k $ that satisfies the expression $ 2^k -1 = 3a $.

Induction step: If $ k $ is even so the next valid number is $ k + 2 $:
$$ 2^k - 1 = 2^{k+2} - 4 = 2^{k+2}-1 = 3\cdot a \cdot 4 + 3 = 3b $$

I think that is is the proof but I don't know if is right or fully complete.

答:
Induction step: Suppose given statement holds for $k$, an even number. To show: the given statement holds for $k+2$.

$$\begin{align}2^{k+2} - 1 &= 4\cdot2^k - 1\\&=3\cdot2^k + (2^k - 1)\\ &=3\cdot2^k + 3a \textrm{, since } 3|(2^k-1) \\&=3\cdot(2^k +a) \end{align}$$


答: Your induction step is somehow unclear.$$2^{k+2}-1=2^2{2^k}-1=\\4\times 2^k-1\\=3\times 2^k+2^k-1\\=\underbrace{3\times 2^k}_{3a}+\underbrace{2^k-1}_{3b}$$

Title: Proof that $2^n-1$ is divisible by $3$ for all $n\in\mathbb{N}$ even

Body:
Basis: This is true for $ n = 2 $
$$ 2^2 - 1 = 3 $$

Hypothesis: There exists a number $ k $ that satisfies the expression $ 2^k -1 = 3a $.

Induction step: If $ k $ is even so the next valid number is $ k + 2 $:
$$ 2^k - 1 = 2^{k+2} - 4 = 2^{k+2}-1 = 3\cdot a \cdot 4 + 3 = 3b $$

I think that is is the proof but I don't know if is right or fully complete.


问:
Supremum of $x^2y^2(x^2+y^2)$
Let $S$ be the set of all tuples $(x, y) $ with $x, y$ non-negative real numbers satisfying $x+y=2n$,for a fixed $n\in \Bbb{N} $. Then the supremum value of $x^2y^2(x^2+y^2)$ on the set $S$ is

$3n^2$

$2n^6$

$4n^6$

$n^6$

My thinking : take $f(x, y) =x^2y^2(x^2+y^2)$ then express $f(x, y) $ as a function of $x$ by putting $y=2n-x$. Then applying calculus we can find the maximum for some value of $x$.

But calculation will be cumbersome. Is there any smart way?

Thanks in advance.

答:
Hint: In a situation like this, it seems intuitively plausible that the maximum is at the symmetric point ($x=y=n$) (since the other special point ($x=0$) clearly is not the maximum).

To test this, rewrite the variables as $x=n+d$, $y=n-d$. Can you take it from here?

Title: Supremum of $x^2y^2(x^2+y^2)$