问:
Associativity of addition in integers modulo $n$.
My proof that addition in $G =\mathbb{Z}/n\mathbb{Z}$ is associative is as follows:

Proof：Before taking (mod $n$), we add members of $G$ in the same way as members of $H=(\mathbb{Z}, +)$ where we know that addition in $H$ is associative. Since taking (mod n) does not affect this addition, the addition in $G$ is also associative.

Does this proof seem like enough? I think I'm right but I often risk falling short in my proofs at times.

答:
Before taking (mod n), we add members of G in the same way as members of H=(Z,+)

This definitely needs to be proven. It's a breezy statement and as stated isn't clear what you even mean.

We DO need to state and DEFINE what if $[a],[b] \in \mathbb Z/n\mathbb Z$ what $[a]+[b]$ means.

And defining it is not as easy as it looks.  $[a] = \{k\in \mathbb N: n|a-k\}$ so we can define $[a] + [b]:= [a+b]$ OR we could define it as $[a]+[b]=\{m+n|m\in [a];n\in [b]\}$ BUT 1) we have no idea whether $[a]+[b]:=[a+b]$ is well defined; we don't know that if $\alpha\in [a]$ and $\beta \in [b]$ that $\alpha + \beta \in [a+b]$ and 2) we don't have any reason to believe that $\{m+n|m\in [a];n\in [b]\}$ as a set is an equivalence class in $\mathbb Z/n\mathbb Z$.

However if we can prove that if $\alpha \in [a]=\{k\in \mathbb N: n|a-k\}$ and $\beta\in [b] = \{k\in \mathbb N: n|a-k\}$ then $\alpha+\beta\in [a]+[b]$, then the definition $[a]+[b] :=[a+b]$ will be well defined.

Pf:  $\alpha\in [a];\beta\in [b]\implies n|a-\alpha;n|b-\beta\implies n|(a-\alpha)+(b-\beta)=(a+b) - (\alpha + \beta)\implies \alpha + \beta\in [a+b]$.

....

NOW you can use your one line breezy argument.

$([a] + [b]) +[c] = ([a+b])+c = [(a+b) + c]= [a+(b+c)] = [a] + [b+c]=[a]+([b]+[c])$.


问:
How can any sentence be valid?
I've just started working through C. Chang and H. Keisler's Model Theory independently right now, and I'm reading through the first chapter, but I'm a little confused by the nature of a valid sentence. So suppose we have language, $\mathscr{S}$, and a sentence built from these, $\varphi$. According to the text, we say that $\models\varphi$ if and only if $$\forall A\subseteq \mathscr{S},A\models \varphi$$
However, earlier the text also says,

Two distinguished models are the empty set $\emptyset$ and the set $\mathscr{S}$ itself.

It seems to me that this implies that no $\varphi$ may be valid then, as $$\nexists \varphi,\emptyset\models\varphi$$
due to the fact that there are no sentence symbols from which to construct any $\varphi$. Is it the case then that no sentence ever valid in our model theory of sentential logic? In which case why do we even care about validity? Or is it to be inferred that in defining validity we must ignore the empty model?

答: Let $S$ be a sentence symbol, and let $\varphi$ be $\neg S$; then $\varnothing\not\vDash S$, since $S\notin\varnothing$, so $\varnothing\vDash\varphi$.

ContentLicense: CC BY-SA 4.0

Title: How can any sentence be valid?


问:
Why is the sum of two functions expressed as simple functions is the sum of the weighted indicator variable of the intersection?
If

$$g=\sum_{i=1}^n a_i \mathbb I_{A_i}$$

and

$$h=\sum_{i=1}^m b_j \mathbb I_{B_j}$$

why is

$$g + h = \sum_{i=1}^n \sum_{j=1}^m (a_i + b_j) \mathbb I_{A_i\cap B_j}$$

I would have guessed that the sum would use the indicator variable $\mathbb I_{A_i \cup B_j},$ which is to say the union of both measurable sets.

答:
Suppose $g = 3 \mathbb I_{A_1}$ and $h = 4 \mathbb I_{B_1}$.
The definition is multiplying the coefficients.  The combination has the value $12$ only in the intersection, not the union.


问:
Rationals with denominator $1, 2$ or $3$ under addition
Do the rationals with denominator $1, 2$ or $3$ form a group under addition if we include $\dfrac{0}{1}$? What if we don't? They clearly have the additive identity if we include $\dfrac{0}{1}$. Associativity and additive inverse is also satisfied. But I'm having trouble proving closure. Any pointers are welcome.

答: No.  Closure is not satisfied.  For instance, $1/2+1/3=5/6$ is not in the set.


问:
Proving a set X is dense in [0,1] equivalence relation
Let the relation in $\mathbb{R}: x \equiv y \ \mbox{mod} \ \mathbb{Z}$, when $x-y \in \mathbb{Z}$. For each $n \in \mathbb{N}$, let $x_n \in [0,1)$ such that $x_n \equiv \sqrt{n} \ \mbox{mod} \ \mathbb{Z}$. Prove $X = \{x_n : n \in \mathbb{N} \}$ is dense in $[0,1]$. I don't know where to start, but I have doubts about using the greatest integer.

答:
Hint: For any $k \in \Bbb N$: if $k^2 \leq n < (k+1)^2$, then we find that
$$
\sqrt{(k+1)^2} - \sqrt{(k+1)^2 - 1} \leq \sqrt{n+1} - \sqrt{n} \leq \sqrt{k^2 + 1} - \sqrt{k}.
$$
Note that the upper bound
$$
\sqrt{k^2 + 1} - \sqrt{k} = \frac{1}{\sqrt{k^2 + 1} + \sqrt{k}}
$$
decreases towards $0$ as $k \to \infty$.


问:
$\int_0^{\pi} \frac{x \sin x}{2 + \cos x}dx$
Problem : $$\int_0^{\pi} \frac{x \sin x}{2 + \cos x}dx$$

I tried $x=\pi - t$ but this just made integrand more messy
because its denominator isn't $2 + \cos x$.

I want to try this without $\tan(x/2)=t$ (like symmetric or etc.)

Thanks for help.

答:
The present answer is quite applicable and straightforward but I want to interpret my way of solving this kind of question.

Let us start with the substitution: $$ \bbox[yellow] { 2+\cos (x)=\frac{4-\cos^2 (x)}{2-\cos (x)} }$$ This is followed by: $$\int_{0}^{\pi} \frac{(2-\cos (x))\space x \sin (x)}{4-\cos^2 (x)} dx = \int_{0}^{\pi} \frac{2x \sin (x)}{4-\cos^2 (x)} dx-\int_{0}^{\pi} \frac{x \cos (x) \sin (x)}{4-\cos^2 (x)} dx$$ The evaluation of the first part of the integral comes from the lemma: $$\int_{0}^{\pi} x f(\sin (x)) dx=\frac{\pi}{2} \int_{0}^{\pi} f(\sin (x))dx$$ If we set $f(\sin (x))$ to be $\frac{2\sin(x)}{3+\sin^2(x)}$: $$\int_{0}^{\pi} x f(\sin(x))dx=\frac{\pi}{2} \int_{0}^{\pi} \frac{2 \sin(x)}{3+\sin^2(x)} dx=\frac{\pi}{2} \int_{0}^{\pi} \frac{2\sin(x)}{4-\cos^2 (x)} dx$$ We can integrate it using the method of substitution: $$\begin{align} y=\frac{\cos(x)}{2}\rightarrow -\frac{\pi}{2} \int_{\frac{1}{2}} ^{-\frac{1}{2}} \frac{dy}{1-{y^2}}
&=\Biggl[-\frac{\pi}{2}\tanh^{-1}y\Biggl]^{-\frac{1}{2}}_{\frac{1}{2}} \\
&=-\frac{\pi}{2} \tanh^{-1}\Biggl(-\frac{1}{2}\Biggl)+\frac{\pi}{2}\tanh^{-1}\Biggl(\frac{1}{2}\Biggl) \\
&=\pi \tanh^{-1} \Biggl(\frac{1}{2}\Biggl) \\
&= \frac{\pi}{2} \ln(3) \end{align}$$
If we apply integration by parts for the second integral: $$\begin{align} \int_{0}^{\pi}\frac{x \sin(x) \cos(x)}{4-\cos^2(x)}dx
&=\Biggl[x\frac{\ln(8-2\cos^2(x))}{2}\Biggl]_{0}^{\pi}-\frac{1}{2}\int_{0}^{\pi} \ln(8-2\cos^2(x))dx \\
&= \frac{\pi}{2} \ln(6)-\frac{1}{2} \int_{0}^{\pi}\ln(8-2\cos^2(x))dx \\
&= \frac{\pi}{2} \ln(6)-\frac{1}{2} \int_{0}^{\pi} \ln(2)+\ln(2-\cos(x))+\ln(2+\cos(x)) dx \end{align}$$ The final result is: $$\begin{align}
& \frac{\pi}{2} \ln(3)-\frac{\pi}{2}\ln(6)+\frac{\pi}{2} \ln(2)+\frac{1}{2}\int_{0}^{\pi} \ln(2-\cos(x))dx+\frac{1}{2} \int_{0}^{\pi} \ln(2+\cos(x))dx  \\
&= \pi \ln\Biggl(\frac{1+\sqrt{3}}{2}\Biggl)\space+\space \pi \ln\Biggl(\frac{1+\sqrt{3}}{2}\Biggl)  \\
&= 2\pi \ln\Biggl(\frac{1+\sqrt{3}}{2}\Biggl)  \end{align}$$


问:
Set of piecewise continuous real-valued functions on $[0,1] $ is convex
I've to proof that the set $E=\{f:[0,1]\rightarrow\mathbb{R}\}$ with $f$ a piecewise continuous function is a convex set, for that I have the next tip:

Proof that for every $\lambda\in[0,1]$ exists $a\in[0,1]$ such that
$$\lambda \int_0^1 f(t)dt=\int_0^a f(t)dt$$

答:
from my perspective, because f is piecewise continuous function on [0,1].
(1) let F(x) be the integral of f(x) on [0,x], x≤1，proof that the value of F(x) is intervel [c，d]，c丶d∈R.
（2）for any λ∈[0,1],and F(1)∈[c，d]，let m=λF(1). proof that m∈[c,d].  This is not hard, just find out the possible value range of λF(1) , we can always find out"m" ∈[a,b],m=λF(1).</p>


问:
How to evaluate the following limit?
How to evaluate this limit?

$$\underset{n\to \infty }{\mathop{\lim }}\,\left( {{n}^{-2}}\sum\limits_{i=1}^{n}{\sum\limits_{j=1}^{{{n}^{2}}}{\frac{1}{\sqrt{{{n}^{2}}+ni+j}}}} \right)$$

Thanks!

答:
The idea here is to rearrange in terms of a Riemann sum that leads to a double integral.  Here, a little manipulation produces

$$\frac{1}{n} \sum_{i=1}^n \frac{1}{n^2} \sum_{j=1}^{n^2} \frac{1}{\sqrt{1+\frac{i}{n}+\frac{j}{n^2}}}$$

Then as $n \to \infty$, the double sum becomes

$$\begin{align}\int_0^1 dx \, \int_0^1 dy \frac{1}{\sqrt{1+x+y}} &= 2 \int_0^1 dx \left [ \sqrt{2+x}-\sqrt{1+x}\right ]\\ &= \frac{4}{3} \left [\left (3^{3/2}-2^{3/2} \right ) - \left (2^{3/2}-1 \right )\right ]\\&=4 \sqrt{3} - \frac{16}{3} \sqrt{2}+1\end{align}$$

ContentLicense: CC BY-SA 3.0

Title: How to evaluate the following limit?


问:
Checking differentiability at $(0,0)$
Check
$$f(x,y) =
\begin{cases}
\frac{x^3y^3}{x^2+y^2} & x^2+y^2 \neq 0 \\
0 & x = y = 0
\end{cases}$$

is differentiable at $(0,0)$  or not
.

I have a very vague understanding of differentiability in case of multi-variable calculus . I know that if a function $f(x,y)$ is differentiable at a point $(a,b)$ , then it can be uniquely represented in the form
$$f(a+h,b+k)-f(a,b)=Ah +Bk+h\phi(h,k)+k\psi(h,k)$$ where A and B are dependent on the point and the function that we are choosing , while $\phi(h,k)$  , $\psi(h,k)$ tend to $0$ as $(h,k)$ approaches $0$. This doesn't seem to fit here.

Hints please.

答:
The derivative at $(0,0)$ exists if there is a linear operator $Df(0,0) = L$ from $\mathbb{R}^2$ to $\mathbb{R}$ such that

$$\lim_{(h,k) \to (0,0)}\frac{|f(h,k) - f(0,0) - L \cdot (h,k)|}{\sqrt{h^2 + k^2}}= 0$$

Note that

$$\frac{|f(h,k) - f(0,0)|}{\sqrt{h^2+k^2}} = \frac{|h|^3|k|^3}{(h^2+k^2)^{3/2}} = \frac{1}{2^{3/2}}\underbrace{\left(\frac{2|h||k|}{h^2 + k^2} \right)^{3/2}}_{\leqslant 1}|h|^{3/2}|k|^{3/2} \\\leqslant |h|^{3/2}|k|^{3/2} \underset{(h,k) \to (0,0)}\longrightarrow 0$$

Hence the derivative at $(0,0)$ exists and is the zero operator,

$$Df(0,0): (h,k) \mapsto 0$$

You could have obtained a clue that this was a suitable candidate for the derivative by noting that the partial derivatives are both $0$, but this alone is not sufficient for existence.  A sufficient condition is that the partial derivatives exist in a neighborhood of a point and are continuous at the point.

Title: Checking differentiability at $(0,0)$


问:
Show that Maximum and Minimum are Global
So I have the following question:

Find the extrema of the function
$$
f(x,y)=4x-6y
$$
Given the constraint
$$
4x^2-4x+9y^2-6y-2=0
$$
And determine whether these extrema are local/global on the constraint.

I found a max and min respectively at
$$
(\frac{1+\sqrt2}{2},\frac{1-\sqrt2}{3}) ,(\frac{1-\sqrt2}{2},\frac{1+\sqrt2}{3})
$$
with the value of f(x,y) at those points being
$$
4\sqrt2 ,-4\sqrt2
$$

I know these points a global maxima on the restriction/constraint, but I am having trouble proving that they are global.

答:
weimiao gives the essential answer to your question:  when the function is linear, changing the value of $ \ ax \ + \ by \ = \ c \ $ shifts the line graph of the function parallel to itself.  For a closed bounded constraint curve, there will be tangent lines to that curve which have largest and smallest values of $ \ c \ $ .  There is something of further interest to remark about this situation.

When we have a constraint curve which has four-fold symmetry about the origin, a linear function will have level curves  $ \ ax \ + \ by \  =  \ \pm c \ $ at tangent points that are symmetric about the origin $ \ ( x_m \ , \ y_m ) \ $ and $ \ ( -x_m \ , \ -y_m ) \ $ .  This symmetry can be often exploited to find the absolute maximum $ \ +c \ $ and minimum $  - c \ $ for the function fairly easily; this can also be the case for other sorts of functions.

What is interesting about this extremization problem is that the constraint curve $$ 4x^2 \ - \ 4x \ + \ 9y^2 \ - \ 6y \ \ - \ \ 2 \ \ = \ 0 \ \ \rightarrow \ \ \frac{(x \ - \ \frac{1}{2})^2}{1} \ + \ \frac{(y \ - \ \frac{1}{3})^2}{4/9} \ \ \ = \ \ 1 \ \ , $$

while an ellipse with its major and minor axes parallel to the coordinate axes, is not centered on the origin, and thus has no symmetry about the origin.  We expect  that the tangent points to this ellipse for parallel lines are not symmetrically placed in this way, so the maxima and minima of linear functions will not generally be the negatives of one another.

The exception happens to be the linear function $ \ 4x \ - \ 6y \ $  .  For the value $ \ c \ = \ 0 \ $ , the line $ \ 4x \ - \ 6y \ = \ 0 \ $  passes through the origin and the center of the constraint ellipse, so the ellipse is symmetrical about that line.  We could then choose coordinate variables relative to the center of the ellipse, $ \ X \ = \ x \ - \ \frac{1}{2} \ \ , \ \ Y \ = \ y \ - \ \frac{1}{3} \ $ to write the constraint curve as $ \ \ 4X^2 \ + \ 9Y^2 \ = \ 4 \ \ $ and the linear function as

$$ \ 4 · \left(x \ - \ \frac{1}{2} \right) \ - \ 6 · \left(y \ - \ \frac{1}{3} \right) \ = \ c \ - \ 2 \ + \ 2 \ \ \rightarrow \ \ 4X \ - \ 6Y \ = \ c \ \   .  $$

We can now solve for the absolute extrema of the linear function without a sophisticated technique such as Lagrange multipliers.  Implicit differentiation of the constraint curve yields

$$ \ 8X \ + \ 18YY' \ = \ 0 \ \ \Rightarrow \ \ Y' \ = \ \frac{-8X}{18Y} \ =  \  -\frac{4X}{9Y} \ \ . $$

Since the linear function produces parallel lines $ \ Y \ = \ \frac{2X - c}{3} \ ,  \ $ we find

$$ Y' \ = \   -\frac{4X}{9Y} \ \ = \ \ \frac{2}{3} \ \ \Rightarrow \ \ -12X \  = \  18Y \ \ \Rightarrow \ \ Y \  =  \ -\frac{2}{3}X  $$

(for instance)
$$ \Rightarrow \ \ 4X^2 \ + \ 9 · \left(-\frac{2}{3}X \right)^2 \ \ = \ \ 4 \ \ \Rightarrow \ \ 4X^2 \ + \ 4  X^2 \ \ = \ \ 4  \ \ \Rightarrow \ \ X^2 \ = \ \frac{1}{2} $$
$$ \Rightarrow \ \ X \ = \ \pm \frac{\sqrt{2}}{2} \ \ , \ \ Y \ = \ -\frac{2}{3} \ · \ \pm \frac{\sqrt{2}}{2} \ \ = \ \  \mp \frac{\sqrt{2}}{3} \ \ . $$

The absolute extrema for our linear function are

$$ f(X,Y) \ \ = \ \ 4X \ - \ 6Y \ \ = \ \ 4· \left(\pm \frac{\sqrt{2}}{2} \right)  - \ 6· \left(\mp \frac{\sqrt{2}}{3} \right) \  = \ \ \pm 2 \sqrt{2} \ \pm \ 2 \sqrt{2} \ \ = \ \ \pm 4 \sqrt{2} \ \ . $$

So this symmetry occurs even though the associated tangent points, which you show, are not symmetrical about the origin:

$$ X \ = \ \frac{\sqrt{2}}{2} \  ,  \ Y \ = \ -\frac{\sqrt{2}}{3} \ \ \rightarrow \ \ \left( \frac{1 + \sqrt{2}}{2} \ , \ \frac{1 - \sqrt{2}}{3} \right) \ \ \approx \ \ (1.207 \ , \ -0.138) \ \ \text{for} \ \ f \ = \ 4\sqrt{2} \ \ ,  $$
$$ X \ = \ -\frac{\sqrt{2}}{2} \  ,  \ Y \ = \ \frac{\sqrt{2}}{3} \ \ \rightarrow \ \ \left( \frac{1 - \sqrt{2}}{2} \ , \ \frac{1 + \sqrt{2}}{3} \right) \ \ \approx \ \ (-0.207 \ , \ 0.805) \ \ \text{for} \ \ f \ = \ -4\sqrt{2} \ \ ,  $$


答: If $(D_uf)(0,0)$ is the directional derivative of $f$ at $(x,y)=(0,0)$ in the direction of the unit vector $u=(u_1,u_2)$ you can compute $(D_uf)(0,0)$ directly as the following limit: $$(D_uf)(0,0)=\lim_{t \rightarrow 0}\frac{f(u_1t,u_2t)-f(0,0)}{t}=\lim_{t\rightarrow 0}u_1^3u_2^3t^3=0$$ Note that $(D_uf)(0,0)=\vec{0}\cdot u$ is a linear function of $u$, so $f'(0,0)$ exists. If the above limit were not a linear function of $u,$ then $f$ wouldn't be differentiable at the origin. Check here. As pointed out in the comments, a standard way to prove that a function is differentiable at $(0,0)$ is to show that $f_x$ and $f_y$ exist and are continuous at $(0,0)$.

Title: Checking differentiability at $(0,0)$

Body:
Check
$$f(x,y) =
\begin{cases}
\frac{x^3y^3}{x^2+y^2} & x^2+y^2 \neq 0 \\
0 & x = y = 0
\end{cases}$$

is differentiable at $(0,0)$  or not
.

I have a very vague understanding of differentiability in case of multi-variable calculus . I know that if a function $f(x,y)$ is differentiable at a point $(a,b)$ , then it can be uniquely represented in the form
$$f(a+h,b+k)-f(a,b)=Ah +Bk+h\phi(h,k)+k\psi(h,k)$$ where A and B are dependent on the point and the function that we are choosing , while $\phi(h,k)$  , $\psi(h,k)$ tend to $0$ as $(h,k)$ approaches $0$. This doesn't seem to fit here.

Hints please.


问:
Significant digits/rounding
I have a bunch of criteria to evaluate for a product, and each is scored on a scale from 0 to 5. Each criterion has a weight associated with it.

If I find the weighted score of a criterion, it is

weighted score = score * weight

Does the significant digits apply here where I should only take one significant digit for my weighted score (since there is only one significant digit in my scoring scale (0-5))? Is it necessary?

答:
mean and standard deviation: round to one more decimal place than your original data. [...]
  This suggestion follows Sullivan, Michael, Fundamentals of Statistics 3/e (Pearson Prentice Hall, 2011) page 118.

Source

So, I suggest keeping one digit after the decimal dot when reporting the weighted average.

Title: Significant digits/rounding


问:
Is there an elegant exact formula for the zeta zero counting function?
I've noticed the zeta zero counting function defined in formula (1) below isn't exactly correct as was pointed out in this related question and answer on MathOverflow. The function $\vartheta(T)$ is the  Riemann–Siegel theta function.

(1) $\quad N_0(T)=1+\frac{\vartheta(T)}{\pi}+\frac{\arg\left(\zeta\left(\frac12+i\,T\right)\right)}{\pi}$

$\qquad\qquad\quad\ =1+\frac{\vartheta(T)}{\pi}+\frac{\Im\left(\log\zeta\left(\frac12+i\,T\right)\right)}{\pi}$

Some of the comments on the question and answer linked above seem to argue that formula (1) above is exactly correct so I'll attempt to clarify with a few illustrations.

Figure (1) below illustrates $\frac{\vartheta(T)}{\pi}$ in orange, $\frac{\Im(\log\zeta(1/2+i\,T))}{\pi}$ in green, and $N_0(T)$ in blue. The red discrete dots represent the evaluation points $(\Im(\rho_k),k-1/2)$. Note $N_0(T)$ seems to evaluate exactly as desired in the range of $T$ values illustrated in Figure (1) below.

Figure (1): Illustration of $\frac{\vartheta(T)}{\pi}$ (orange), $\frac{\Im(\log\zeta(1/2+i\,T))}{\pi}$ (green), and $N_0(T)$ (blue)

Now consider Figures (2) and (3) below which illustrate the evaluations of $\frac{\Im(\log\zeta(1/2+i\,T))}{\pi}$ and $N_0(T)$ in the neighborhood of Gram point $g_{126}$ and zeta zero $\rho_{127}$. The green dots in Figures (2) and (3) below represent the evaluations of $\frac{\Im(\log\zeta(1/2+i\,T))}{\pi}$ and $N_0(T)$ at $T=g_{126}+\epsilon$, and the red dots in Figures (2) and (3) below represent the evaluations of of $\frac{\Im(\log\zeta(1/2+i\,T))}{\pi}$ and $N_0(T)$ at $T=\Im\left(\rho_{127}\right)+\epsilon$. Note $g_{126}$ is the first Gram point where Gram's law is violated (see Wikipedia: Gram points), and the evaluation of formula (1) for $N_0(T)$ illustrated in Figure (3) below corrects itself at $T=\Im\left(\rho_{127}\right)$.

Figure (2): Illustration of $\frac{\Im(\log\zeta(1/2+i\,T))}{\pi}$

Figure (3): Illustration of $N_0(T)$

Question: Is there an elegant exact formula for the zeta zero counting function that is practical to evaluate?

By "elegant" and "practical to evaluate" I mean a formula that has been completely derived and preferably a closed form formula that can be evaluated similar to formula (1) above. For the purposes of this question I'm not interested in complex formulas with complex integrals that have been left unevaluated such as the formula for $N^*(T)$ illustrated in this Math StackExchange question.

In an attempt to further clarify the source of the problem Figures (4) and (5) below illustrate the real and imaginary parts of $\zeta\left(\frac12+i\,T\right)$ in the neighborhood of Gram point $g_{126}$ and zeta zero $\rho_{127}$. The green dots in the two figures below represent the evaluation of $\zeta\left(\frac12+i\,T\right)$ at $T=g_{126}$ and the red dots in the two figures below represent the evaluation of $\zeta\left(\frac12+i\,T\right)$ at $T=\Im\left(\rho_{127}\right)$. The source of the problem is the brief positive transition of $\Im\left(\zeta\left(\frac12+i\,T\right)\right)$ at $T=g_{126}$ illustrated in Figure (5) below. This transition violates Grams's law and causes undesirable branches in $\arg\left(\zeta\left(\frac12+i\,T\right)\right)$ and $\Im(\log\zeta(1/2+i\,T))$. I'll note I believe that the built-in Mathematica functions all support analytic continuation.

Figure (4): Illustration of $\Re\left(\zeta\left(\frac12+i\,T\right)\right)$

Figure (5): Illustration of $\Im\left(\zeta\left(\frac12+i\,T\right)\right)$

The following four figures contain 3D plots intended to clarify the Mathematica analytic continuation of $\log\zeta(s)$ in the range $\Re(s)\in (-1,3)$ and $\Im(s)\in (-50,50)$.

Figure (5): 3D Plot of $|\log\zeta(s)|$

Figure (6): 3D Plot of $\Re(\log\zeta(s))$

Figure (7): 3D Plot of $\Im(\log\zeta(s))$

Figure (8): 3D Plot of $\arg(\log\zeta(s))$

答:
Your edit shows that Mathematica is just taking $\arg \zeta(s)$ in $(-\pi,\pi)$. The obtained function doesn't have good analytic properties, it is isn't what we need for $N(T)$ (the non-trivial zeros counting function).

What we need, the standard definition of $\arg \zeta(s)$. (ie. the one you'll find in every books and articles, except the cranks ones)

We start with $\log \zeta(s)$ analytic for $\Re(s)>1$ and $\log \zeta(2)=\ln \pi^2/6>0$ (the Dirichlet series $\sum_{p^k} \frac{p^{-sk}}{k}$), because $\zeta(s)$ has some poles and zeros, the analytic continuations to larger domains depend on which path we are doing it (rotating around $s=1$ will subtract $2i\pi$). To get a direct connexion to $N(T)$ we choose to make $\log \zeta(s)$ analytic on horizontal lines (thus on horizontal strips) not containing a zero/pole. And mathematica is doing something different, that you have to decipher.

Is it clear to you?

We choose this definition for $\log \zeta(s)$ because with $C$ a contour enclosing the zeros up to height $T$ we have by the residue theorem (the argument principle) and the symmetry around $\Re(s)=1/2,\Im(s)=0$: $$N(T)=\frac1{4i\pi}\int_C \frac{\xi'(s)}{\xi(s)}ds=\Re(\frac1{i\pi}\int_2^{2+iT}+ \int_{2+iT}^{1/2+iT} \frac{\xi'(s)}{\xi(s)}ds)$$
the RHS is $=\frac1{\pi}\Im \log \xi(1/2+iT)$ under the above analytic continuation scheme. Then $\frac1\pi\Im \log \xi(1/2+iT)$ decomposes as $1+\vartheta(T)/\pi+\frac1\pi\Im \log \zeta(1/2+iT)$

The asymptotic of $\vartheta(T)/\pi$ is found from Stirling (it is $\sim \frac{T}{2\pi} \log T$), and from Jensen's formula we get that $\frac1\pi\Im \log \zeta(1/2+iT)=O(\log T)$.


问:
How to solve for the function $f(x)$ such that the area underneath is always equal to the arclength?
First, I set up my equation. The area is obviously $\int f(x) \space dx$. The arclength is $\int \sqrt{1+(\frac{dy}{dx})^2}  \space dx$. Thus, I have the equation $\int f(x) \space dx = \int \sqrt{1+(\frac{dy}{dx})^2}  \space dx$. Now I differentiatie both sides to get $f(x) = \sqrt{1+(\frac{dy}{dx})^2} $. I will now make some substitutions to make it look simpler like this. $y = \sqrt{1 +y'^2}$. This becomes $y^2 = 1+y'^2$. Since we have a function that is similar to its own derivative, I will guess the solution looks like $y = Ae^{kx}$. $y' = Ake^{kx}$. Pugging these values into our equation, I get $(Ae^{kx})^2 = (Ake^{kx})^2 + 1$ or  $A^2e^{2kx} = A^2k^2e^{2kx} + 1$. I am now stuck here, please help.

答:
The equation you get is going to be
$$y' = \sqrt{1+y^2}$$
which is nonlinear, but separable. Integrating, you get
$$x = \int \frac{dy}{\sqrt{1+y^2}}$$
and you can solve this with the substitution $y = \sinh (x + C)$.


问:
What is the expected number of ingredients used to make $4$ pizzas if each pizza must contain $4$ of $100$ possible ingredients?
I'm trying to solve an optimization problem of the Google Hashcode contest and by analyzing the dataset I reduced a part of it down to a statistics problem, which I'm not able to solve. Any help would be appreciated.

Assume we have $100$ types of ingredients (from each type infinitely many) and we want to make $4$ pizzas. Each pizza must contain exactly $4$ distinct ingredients. All of the $100$ ingredients are equally likely for a candidate position of an ingredient for a pizza. What is the expected value of the total distinct ingredients used in all $4$ pizzas?

Here is my initial thought which is unlikely to be true:

$$E[N] = \sum_{i = 4}^{16}{i \times \frac{\binom{100}{i} {\binom{i}{4}}^4 - \binom{100}{i - 1} {\binom{i - 1}{4}}^4}{\binom{100}{16}{\binom{16}{4}}^4}} =15.9. $$
Numerator is essentially difference of total number of states with $i$ ingredients at most and with $i-1$ ingredients at most which gives the total number of states with exactly $i$ ingredients, while denominator is the total number of states with $16$ ingredients.

I think the answer for the above method ($15.9$) is not very intuitive. It seems more logical for the answer to be much more lower.

ps. I previously tried to put ${\binom{100}{4}}^4$ in the denominator but after doing a simple summation I ended up with a very unreasonable quantity.

ps2. Order of pizzas are not important.

答:
Consider a particular ingredient:

The probability it is in the first pizza is $\frac4{100}=0.04$

The probability it is not in the first pizza is $1-0.04=0.96$

The probability it is not in any of the four pizzas is $0.96^4=0.84934656$

The probability it is in any of the four pizzas is $1-0.84934656=0.15065344$

But there are $100$ possibilities so the expected number of distinct ingredients in the four pizzas is $100 \times 0.15065344$ which is $$15.065344$$

To confirm the answer, here is a simulation in R:

```<code>set.seed(2021)
distinctingredients <- function(numpizzas, totingr, ingrperpizza){
length(unique(as.vector(replicate(numpizzas, sample(totingr, ingrperpizza)))))
}
mean(replicate(10^5, distinctingredients(4,100,4)))
# 15.06479
</code>```

which seems close to my answer.

Body:
Consider a particular ingredient:

The probability it is in the first pizza is $\frac4{100}=0.04$

The probability it is not in the first pizza is $1-0.04=0.96$

The probability it is not in any of the four pizzas is $0.96^4=0.84934656$

The probability it is in any of the four pizzas is $1-0.84934656=0.15065344$

But there are $100$ possibilities so the expected number of distinct ingredients in the four pizzas is $100 \times 0.15065344$ which is $$15.065344$$

To confirm the answer, here is a simulation in R:

set.seed(2021)
distinctingredients <- function(numpizzas, totingr, ingrperpizza){
length(unique(as.vector(replicate(numpizzas, sample(totingr, ingrperpizza)))))
}
mean(replicate(10^5, distinctingredients(4,100,4)))
# 15.06479

which seems close to my answer.


问:
Closed form of $\sum_{n=-\infty}^\infty \frac{(-1)^n}{\sinh (z+n)}$?
I considered the following function:
$$f:\, \mathbb{C}\mapsto\mathbb{C}_{\infty},\, z\mapsto \left(\sum_{n=-\infty}^\infty \frac{(-1)^n}{\sinh (z+n)}\right)^{-1}.$$

It can be seen that $f(z)=0$ at every integer (division of a non-zero complex number by zero is assumed to be complex infinity, so its reciprocal is $0$). It also seems that the function is periodic on the real axis with period $2$, i.e.
$$\forall x\in\mathbb{R}:\, f(x+2)=f(x),$$
though I was not able to prove that rigorously. I tried to use the $\sinh$ addition formula, but that's probably not useful.

On the whole real axis, $f$ is very close to
$$g:\, \mathbb{C}\mapsto \mathbb{C},\, z\mapsto 0.31837572\sin \pi z,$$
but on the imaginary axis, there's a big difference between $f$ and $g$, for instance:
$$f(2i)\approx 5.3796i,\,\text{but}\, g(2i)\approx 85.2499i.$$

Why is that? Can a closed form of $f$ in terms of Weierstrass/Jacobi elliptic functions be found?

Note:
Curiously, any
$$h:\,\mathbb{R}\mapsto\mathbb{R},\, x\mapsto \left(\sum_{n=-\infty}^\infty \frac{(-1)^n}{a^{x+n}-a^{-x-n}}\right)^{-1}$$
where $a\gt 0$ and $a\ne 1$ seems to be $2$-periodic...

Edit:
The function
$$f:\, \mathbb{C}\mapsto\mathbb{C}_{\infty},\, z\mapsto \left(\sum_{n=-\infty}^\infty \frac{(-1)^n}{\sinh (z+n)}\right)^{-1}$$
is probably a doubly-periodic elliptic function with periods $2$ and $2\pi i$, so I added an appropriate tag.

答:
Let $$g(z) = \sum_{n\in \mathbb{Z}} \frac{(-1)^n}{\sinh(z+n)}$$
$g$ has $2$ and $2\pi i$ as period, so an elliptic function. Poles of $g$, which are all simple, are exactly $0,1,\pi i,1+\pi i$ modulo periods, with residues $1,-1,-1,1$.

$\DeclareMathOperator{\sn}{sn}\DeclareMathOperator{\ns}{ns}$
Since $g$ has only simple poles, it is easier to express it as Jacobi elliptic function. Recall that $\ns(z,k)$ has periods $4K,2iK'$ and residues $1$ at $0$. Hence $g(z)$ must be of form
$$g(z) =a \ns(az,k) - a \ns(a(z-\pi i),k) + C$$
then it is easy to determine $K'/K = 2\pi, a = 2K, C=0$. This puts $g(z)$ in terms of Jacobi-elliptic function:
$$g(z) = 2K[\ns(2Kz, k) - k \sn(2Kz,k)] \qquad K' = 2\pi K$$

Mathematica code to calculate $g(z)$ numerically:

```<code>g[z_] := Module[{m = InverseEllipticNomeQ[Exp[-2 Pi^2]], K},
   K = EllipticK[m];
   2 K (JacobiNS[2 K*z, m] - Sqrt[m]*JacobiSN[2 K*z, m])];
</code>```

This can give, within an eyeblink, value of any $g(z)$ to great precision. For example, try

```<code>N[g[1/3], 1000]
</code>```

Body:
Let $$g(z) = \sum_{n\in \mathbb{Z}} \frac{(-1)^n}{\sinh(z+n)}$$
$g$ has $2$ and $2\pi i$ as period, so an elliptic function. Poles of $g$, which are all simple, are exactly $0,1,\pi i,1+\pi i$ modulo periods, with residues $1,-1,-1,1$.

$\DeclareMathOperator{\sn}{sn}\DeclareMathOperator{\ns}{ns}$
Since $g$ has only simple poles, it is easier to express it as Jacobi elliptic function. Recall that $\ns(z,k)$ has periods $4K,2iK'$ and residues $1$ at $0$. Hence $g(z)$ must be of form
$$g(z) =a \ns(az,k) - a \ns(a(z-\pi i),k) + C$$
then it is easy to determine $K'/K = 2\pi, a = 2K, C=0$. This puts $g(z)$ in terms of Jacobi-elliptic function:
$$g(z) = 2K[\ns(2Kz, k) - k \sn(2Kz,k)] \qquad K' = 2\pi K$$

Mathematica code to calculate $g(z)$ numerically:

g[z_] := Module[{m = InverseEllipticNomeQ[Exp[-2 Pi^2]], K},
   K = EllipticK[m];
   2 K (JacobiNS[2 K*z, m] - Sqrt[m]*JacobiSN[2 K*z, m])];

This can give, within an eyeblink, value of any $g(z)$ to great precision. For example, try

N[g[1/3], 1000]


问:
Why do these complements of closed sets induced by annihilators form an open cover of the spectrum?
For an important characterization of morphisms to projective spaces, one step in "The Geometry of Schemes" by Eisenbud and Harris is to understand morphisms from affine schemes to projective spaces (Proposition III-40). The beginning is:

Let $T$ be a ring and $K$ a free summand of $T^{n+1}$ of rank $n$. Write $P$ for the invertible (i.e. locally free of rank 1) module $T^{n+1}/K$ which is generated by the $n+1$ images of the canonical generators $e_i$ of $T^{n+1}$.

Now, my problem is that I cannot show the following:

Let $I_j$ be the annihilator of $(P/T e_j)$ and $U_j$ the complement of $V(I_j)$, which is the set of all prime ideals $q$ of $T$ with $I_j\subset q$, in $\mathrm{Spec}(T)$. Then the $U_j$ form an open cover of $\mathrm{Spec}(T)$.

I think I miss something important about annihilators and the set $V(-)$ as I seem to not be able to show that the $U_j$ cover $\mathrm{Spec}(T)$ with the properties I know about locally free, invertible and/ or projective modules.

答: Let $q\in\operatorname{Spec}(T)$.  Since $P$ is invertible, the localization $P_q$ is free of rank $1$ over $T_q$, and so it is generated by some single $e_j$ (namely, any $e_j$ that is not in $qP_q$).  That is, every $e_i$ becomes a multiple of this particular $e_j$ after localizing at $q$.  This means that for each $i$, there is some element $a_i\in T\setminus q$ such that $a_ie_i$ is a multiple of $e_j$.  Letting $a=\prod_i a_i$, we see that $ae_i$ is a multiple of $e_j$ for all $i$.  Since the $e_i$ generate $P$, this means $aP\subseteq Te_j$, so $a\in I_j$.  Since $a\in T\setminus q$, this means $q\in U_j$.


问:
Functions in Group Theory
I am reading a math book and in it, it says, "Let $V$ be the set of all functions $f: \mathbb{Z^n_2} \rightarrow \mathbb{R}.$ I know that $\mathbb{Z^n_2}$ is just the cyclic group of order $2$ taken to the $n$th power, but I don't get what the actual function means.

How would such a function, $f$, work? Could someone please give an example?

答:
[...] I don't get what the actual function means.

$f:\bf{Z}_2^n \to R$ simply assigns to each element of $\bf{Z}_2^n$ an element of $\bf{R}$.

How would such a function, f , work?

Each element of $\bf{Z}_2^n$ is assigned an element of $\bf{R}$.

Could someone please give an example?

Say $n=2$. Define $f:\bf{Z}_2^n \to R$ such that

$$f((0, 0)) = 1$$
$$f((0, 1)) = 2$$
$$f((1, 0)) = 3.14159$$
$$f((1, 1)) = 1738$$
That's all a function does - it assigns elements of one set to elements of another. That's it.

Title: Functions in Group Theory


问:
Probability across an interval using Poisson
Working through some Poisson probability questions and am wondering whether the knowledge of how many events have occurred changes a calculation.

The question is thus:

The number of prank calls received by an bartender is modeled using a Poisson process with rate 3 per hour. If we consider a time interval and we know the exact number of calls received in that time interval, then the calls are distributed independently and uniformly within that time interval. We know that exactly one call was received from 10pm to 10.15pm: what is the probability that this call was received after 10.10pm?

Normally my approach would have been the following:
$ λ = \frac {60}{20} $ across whole interval

Poisson PMF: $\frac {(e^{-λ})*λ^x)}{x!}$

$ λ = \frac {10}{20} $ for the fraction of the interval we are interested in

PMF = $1-\frac {(e^{-.5})*.5^1)}{1}$

PMF = $1-.303$

PMF = $.697$

However the questions states that 'we know that one call was received from 10pm to 10.15pm'. Does this change the λ from 3 to 4? Or should I be looking at an exponential distribution instead - eg. $ fx(x) = e^{(−λx)}$?.

Thanks in advance

Regards,

Greg

答:
You're told that the conditional distribution of the event time, given occurrence in a fixed interval, is uniform:

If we consider a time interval and we know the exact number of calls received in that time interval, then the calls are distributed independently and uniformly within that time interval.

That means, if exactly one call was received between 10:00 and 10:15 PM, then the probability it occurred after 10:10 PM is simply $\frac{15-10}{15} = \frac{1}{3}.$


问:
Smoothness of a map from an open subset of $\mathbb{H}^n$ to $\mathbb{R}^k$
Please take a look at the following picture and find the statement underlined with red. I'm curious about the meaning of "the usual sense". For your information, $\mathbb{H}^n$ denotes the closed $n$-dimensional upper half-space, that is,
$$\mathbb{H}^n=\{(x_1,\ldots,x_n)\in\mathbb{R}^n:x_n\geq 0 \}.$$
To me, in order to talk about the smoothness of a function in the usual sense,  the domain of the function has to be an open subset of a Euclidean space. In view of this, I doubt that $U\cap\textrm{Int}\mathbb{H}^n$ is an open subset of $\mathbb{R}^n$. Does anyone have an idea? Thank you.

答: I'm sorry about being silly. If $U$ is relatively open in $\mathbb{H}^n$, there exists an open subset $A$ of $\mathbb{R}^n$ s.t. $U=A\cap\mathbb{H}^n$. By using the fact that the upper half-space is a disjoint union of its boundary and interior, one can see that $U\cap\textrm{Int}\mathbb{H}^n=A\cap\textrm{Int}\mathbb{H}^n$, which is obviously an open set in the Euclidean space.


问:
How to prove that $\int_{[-\pi,\pi]}\log(\vert 1- \exp(it)\vert)\mathrm{d}\lambda(t)=0$?
Let $r>1$ and $\displaystyle\int_{[-\pi,\pi]}\log(\vert 1- r\exp(it)\vert)\mathrm{d}\lambda(t)= 2\pi\log(r)$.

We want to prove that :

$\displaystyle \lim \limits_{r\to 1} \int_{[-\pi,\pi]}\log(\vert 1- r\exp(it)\vert)\mathrm{d}\lambda(t)=\int_{[-\pi,\pi]}\log(\vert 1- \exp(it)\vert)\mathrm{d}\lambda(t)$.

We want to apply the dominated convergence theorem. Let $(r_n)_{n\ge 0}$ a sequence of real numbers such that $\lim \limits_{n\to +\infty}r_n = \lim \limits_{r \to 1} r$.

For all $t \in [-\pi,\pi]$, the sequence of functions $t \mapsto \log(\vert 1- r_n\exp(it)\vert)$ is continuous. So the sequence of functions is in $L^1([-\pi,\pi])$.

$\lim \limits_{n\to +\infty} \log(\vert 1- r_n\exp(it)\vert)=\lim \limits_{r\to 1} \log(\vert 1- r\exp(it)\vert)=\log(\vert 1- \exp(it)\vert)$.

Note that the function $t\mapsto \log(\vert 1- \exp(it)\vert)$ is continuous on $[-\pi,\pi]- \{0\}$ and in a neighborhood of $t=0$ we have $\log(\sqrt{(2-2\cos(t)}))=o \ (\dfrac{1}{\sqrt t})$. So the function $t\mapsto \log(\vert 1- \exp(it)\vert)$ is in $L^1([-\pi,\pi])$.

For the hypothesis of domination I am stuck.

Thanks in advance !

答:
Let $r_n \to 1$

You can assume that $r_n<2,\forall n \in \Bbb{N}$

$|1-r_ne^{it}|\leq 1+r_n<3$ almost everywhere

Thus $\log|...| \leq \log 3 \in L^1[-\pi,\pi]$


问:
Does $\log z$ have a laurent series about $0$?
I think not because $\log z$ isn't analytic on any neighborhood of 0. Is this correct? thanks

答:
I would use Tristan Needham's approach found in his book "Visual Complex Analysis" to define the formal Laurent series of the complex logarithm as
$$\forall n\in\mathbb{Z}, \ c_n = \frac{1}{2\pi i}\oint_0 \frac{\log(z)}{z^n}\frac{dz}{z}$$
and the formula would read
$$\log(x)=\sum_{n\in\mathbb{Z}} c_n x^n.$$

Whether this contour integral works or not for any $n\in\mathbb{Z}$ seems to be solved nowadays, regarthless it exists for $n=-1$. We would need to find these explicit values $c_n.$

Nevertheless, everyone says there does not exists a Laurent series expansion for logarithm, for example:

Now what about log(z)? That one has singular points called "logarithmic branch points" of infinite order since circuits around the origin or infinity around the function never return to the starting value. These too are not poles since they also are not issolated and so therefore do not have Laurent series expansions about their singular points.

Source https://www.physicsforums.com/threads/log-z-singularity-series-and-pole.601750/


问:
Show that the area vectors for a general $n$-sided closed shape sum to zero
It is possible to show that the sum of the area vectors for a general, closed, $n$-sided figure in $\mathbb{R}^3$ (3-space) is zero.

Hint: it may be easiest to consider orientable and non-orientable surfaces separately.

答:
A three-dimensional "body" $B\subset{\mathbb R}^3$, e.g., a ball, an octahedron, or a cylinder, has a boundary $\partial B$ which is an oriented $2$-chain, i.e., a union of smooth oriented pieces of surfaces, such that the "positive normal" ${\bf n}$ points to the outside at all points ${\bf p}\in\partial B$, where it is defined. For this situation one has Gauss' divergence theorem which says that
$$\int_{\partial B}{\bf v}\cdot{\bf n}\ {\rm d}\omega=\int_B{\rm div}({\bf v})\ {\rm d}{\bf x}$$
for all $C^1$ vector fields ${\bf v}$ on $B$. When ${\bf v}$ is a constant vector field then ${\rm div}({\bf v})\equiv0$, and we can write
$${\bf v}\cdot\int_{\partial B}{\bf n}\ {\rm d}\omega=\int_{\partial B}{\bf v}\cdot{\bf n}\ {\rm d}\omega=\int_B{\rm div}({\bf v})\ {\rm d}{\bf x}=0\ .$$
Since this is true for arbitrary ${\bf v}\in{\mathbb R}^3$ it follows that
$$\int_{\partial B}{\bf n}\ {\rm d}\omega=0\ .$$


问:
How to find critical points of definite integral
Say I have a function

$$g(x) = \int_a^b (f(t)-x)^3dt$$

how would I go about finding the critical points of this function?

I know that FTC gives you that if

$$h(x) = \int_0^x f(t)dt$$ then $$h'(x) = f(x)$$

but I just keep thinking that

$$g(x) = (f(b)-x) - (f(a)-x)$$

so wouldn't the derivative be 0? And since you find critical points where g'(x)=0 wouldn't that mean everywhere is a critical point?

I know this can't be true but having hard time figuring out what to do.

答:
If you want to figure out where $g'$ is zero, compute it as
\begin{align*}
g'(x) &= -3\int_a^b\, (f(t) - x)^2\, dt \\
&= -3\int_a^b\, f^2(t)\, dt + 6x \int_a^b f(t)\, dt -3x^2(b - a).
\end{align*}
It's a quadratic in $x$. Set it to zero and solve for the $x$s that are the critical points.

If you set $I_0 = b - a$, $I_1 = \int_a^b\, f(t)\, dt$, $I_2 = \int_a^b\, f^2(t)\, dt$ your critical points occur at
\begin{equation*}
\hat{x} = \frac{I_1}{I_0} \pm \sqrt{\biggl( \frac{I_1}{I_0}\biggr)^2 - \frac{I_2}{I_0}}
\end{equation*}


问:
Finding population median and IQR(x)

I have done the following for the median

$2(x-1) = .50$

results in $x = 1.25$

For the quartiles, I have $q_{1}$--> $2(x-1)=.25$ which results in $q_{1} = 1.125$

for $q_{3}$--> $2(x-1)=.75$ which results in $q_{3}=1.375$ and then subtract them to get $.25$
Please correct me if I am wrong! thank you in advance

答:
The median is the number $m$ for which $$\int_{x=-\infty}^m f(x) \, dx = \frac{1}{2}.$$  So you need to solve $$\frac{1}{2} = \int_{x=1}^m 2(x-1) \, dx = \left[x^2 - 2x\right]_{x=1}^m = m^2 - 2m - (1-2) = (m-1)^2.$$

For the IQR, you need to solve for the first and third quartiles in a similar fashion:  $q_1$ satisfies $$\frac{1}{4} = \int_{x=1}^{q_1} f(x) \, dx,$$ and $q_3$ satisfies $$\frac{3}{4} = \int_{x=1}^{q_3} f(x) \, dx.$$  Then the IQR is simply $q_3 - q_1$.

Title: Finding population median and IQR(x)


问:
Prove that the set $a+b\sqrt{2}$ where a and b are rational without zero is a group under multiplication
We are given $G=\{a+b\sqrt{2}: a,b\in\mathbb{Q}\}$ and asked to show that $G-\{0+0\sqrt{2}\}$ is a group under multiplication.

Let us first specify that multiplication takes place as expected:

$$(a+b\sqrt{2})(c+d\sqrt{2})=a(c+d\sqrt{2})+b\sqrt{2}(c+d\sqrt{2})=ac+2bd+(ad+bc)\sqrt{2}$$

It is clear that $1+0\sqrt{2}$ is the multiplicative identity.

But I have hit a blocker on finding the inverse. Dummit and Foote's hint is to 'rationalize the denominator' but I'm not sure what this means in this case. If I try straight up equating coefficients in

$$(a+b\sqrt{2})(c+d\sqrt{2})=1+0\sqrt{2}$$ then that's just $$ac+2bd=1$$ and $$ad+bc=0$$. That's not a very nice set up at all. Any hints?

答:
Hint:

Multiply top and bottom of $\;\dfrac1{a+b\sqrt2}\;$ by $\;{a-b\sqrt2}$


答:
Well, for any $a+b\sqrt{2}$ you have to find an inverse $c+d\sqrt{2}$, so your setup is actually pretty nice: it consists of a linear system

$$
\begin{pmatrix}
a & 2b\\
b & a
\end{pmatrix} \cdot
\begin{pmatrix}
c\\
d\\
\end{pmatrix} =
\begin{pmatrix}
1\\
0
\end{pmatrix}.
$$

You can compute the inverse of the matrix explicitly in terms of the determinant, multiply on the left and you will get a formula for $c$ and $d$ depending on $a$ and $b$.

Edit: also, note that if $a+b \sqrt{2} \neq 0$, then we cannot have $a^2 = 2b^2$ by a factorization argument. Hence the determinant is always nonzero for nonzero elements.

Title: Prove that the set $a+b\sqrt{2}$ where a and b are rational without zero is a group under multiplication

Body:
We are given $G=\{a+b\sqrt{2}: a,b\in\mathbb{Q}\}$ and asked to show that $G-\{0+0\sqrt{2}\}$ is a group under multiplication.

Let us first specify that multiplication takes place as expected:

$$(a+b\sqrt{2})(c+d\sqrt{2})=a(c+d\sqrt{2})+b\sqrt{2}(c+d\sqrt{2})=ac+2bd+(ad+bc)\sqrt{2}$$

It is clear that $1+0\sqrt{2}$ is the multiplicative identity.

But I have hit a blocker on finding the inverse. Dummit and Foote's hint is to 'rationalize the denominator' but I'm not sure what this means in this case. If I try straight up equating coefficients in

$$(a+b\sqrt{2})(c+d\sqrt{2})=1+0\sqrt{2}$$ then that's just $$ac+2bd=1$$ and $$ad+bc=0$$. That's not a very nice set up at all. Any hints?


问:
How to imagine (or visualize) the convergence of $x,\; f(x),\; f(f(x)),\; f(f(f(x))), \dots\; $?
I found a problem (Calculus 3rd edition by Michael Spivak, Chapter 22 question 21) that goes as follows:

21. (a) Suppose that $f$ is continuous on $[0,1]$ and that $0\leq f(x) \leq 1$ for all $x$ in $[0,1]$. Problem 7-11 shows that $f$ has a fixed point. If $f$ is increasing, a much stronger statement can be made:

For any $x$ in $[0,1]$, the sequence
$$x,\; f(x),\; f(f(x)),\dots$$
has a limit (which is necessarily a fixed point by problem 20). Prove this assertion, by examining the behavior of the sequence for $f(x)>x$ and $f(x)<x$ or by looking at Figure 10. [...]

I managed to prove the assertion but can't seem to visualize the concept. Could someone explain Figure 10 and its implications? Is $\varphi_n(x)=f^n(x)$ (where $n$ is the number of compositions) getting as close as desired to $id_{[0,1]}(x)=x$ given a sufficiently large $n$ ($\forall x\in [0,1]$)? What is happening to $\varphi_n(x)$ as $n$ gets larger? What about the case $f=id_{[0,1]}$?

Proof of the assertion

Let $x\in\{\alpha\in [0,1]\mid 1\geq f(\alpha)>\alpha\}=A$, then, since $f$ is increasing, $f(f(x))>f(x)$ and $f(f(f(x)))>f(f(x))>f(x)>x$ so the sequence is increasing $\forall x\in A$ but $f(x)\leq 1$ so it's also bounded and thus a limit exists. For $x$ such that $f(x)<x$ the proof is similar $\square$.

答:
Basically, if you go along a vertical line until you hit the curve, then you get to the point $(x, f(x))$.

Now you go along the horizontal line until you hit the line $y = x$, then you get to the point $(f(x), f(x))$.

Going again along the vertical line, you get to $(f(x), ff(x))$.

Again along the horizontal line, you get to $(ff(x), ff(x))$.

The procedure continues forever. Repeating these two steps is thus equivalent to iterations of $f$.

The convergence should then be obvious from tracing the picture.


问:
Prove that the tetragon ABCD also has an inscribed circle
There is given convex tetragon ABCD and some points J,K,L,M on the line segments AB,BC,CD,DA respectively. The intersection of  JL and KM is T.  Each of the tetragons AJTM,BKTJ,CLTK, and DMTL have an inscribed circle.

Prove that the tetragon ABCD also has an inscribed circle.

答: there is a similar problem with a geometry Theorem。 you may proof this geometry theorem first if you haven't see it before. then use it to proof your question.</p>


答:
It suffices to show that $G\le (\Bbb R^*, \times)$, since any subgroup of a group is itself a group.

I will use the one-step subgroup test.

Clearly $1\in G$, so $G\neq \varnothing$.

Let $x=a+b\sqrt{2}, y=c+d\sqrt{2}\in G$. Then $a,b,c,d\in \Bbb Q$, so, in particular, $x\in \Bbb R^*$ as $x\neq 0$. Hence $G\subseteq \Bbb R^*$. Furthermore, we have

$$\begin{align}
xy^{-1}&=\frac{a+b\sqrt{2}}{c+d\sqrt{2}}\\
&=\frac{a+b\sqrt{2}}{c+d\sqrt{2}}\times \frac{c-d\sqrt{2}}{c-d\sqrt{2}}\tag{1}\\
&=\frac{(a+b\sqrt{2})(c-d\sqrt{2})}{c^2-2d^2}\\
&=\frac{ac-2bd}{c^2-2d^2}+\frac{bc-ad}{c^2-2d^2}\sqrt{2},
\end{align}$$

which is in $G$ as neither $x$ nor $y$ is zero. Hence $xy^{-1}\in G$.

Hence $G\le (\Bbb R^*, \times)$.

Hence $G$ is a group.

$(1)$: Here is what is meant by "rationalise the denominator".


问:
Is it possible for a limit point not to be a sequential limit point?
I'm studying elementary topology, and I'm trying to understand the difference between limit points and sequential limit points. So far, I have only studied several basic topologies on subsets of the natural numbers, such as the discrete topology and the cofinite topology. In order to develop some intuition, I'm attempting to come up with an example of a space composed strictly of natural numbers, with a subset such that some point x is a limit point of the subset, but not a sequential limit point of it. Is this possible on the natural numbers, or will every limit point also be a sequential limit point?</p>

答:
It is possible to construct such a space, but it’s not trivial. I’ll give you two examples. One is easily described directly in terms of $\Bbb N$, but you have to know something about ultrafilters; the other is a space most easily defined on the set $\Bbb N\times\Bbb N$, not on $\Bbb N$, but you can use a bijection between the two sets to transfer the topology to $\Bbb N$ itself, if you really want to do so.

Note: My $\Bbb N$ includes $0$.

For the first example let $\mathscr{U}$ be a free (i.e., non-principal) ultrafilter on $\Bbb N$. Points of $\Bbb Z^+$ are isolated, and the open nbhds of $0$ are precisely the members of $\mathscr{U}$ that contain $0$. Every member of $\mathscr{U}$ is infinite, so $0$ is clearly a limit point of $\Bbb Z^+$.

Suppose, now, that $\sigma=\langle a_n:n\in\Bbb N\rangle$ is any sequence of positive integers, and let $A=\{a_n:n\in\Bbb N\}$. If $A\notin\mathscr{U}$, then $\Bbb N\setminus A\in\mathscr{U}$ by a basic property of ultrafilters, so $\Bbb N\setminus A$ is an open nbhd of $0$ that contains no point of $\sigma$, and hence $\sigma$ certainly does not converge to $0$. If $A\in\mathscr{U}$, then $A$ is infinite, so we can partition it into two infinite subsets $A_0$ and $A_1$. It is a basic fact about ultrafilters that exactly one of $A_0$ and $A_1$ belongs to $\mathscr{U}$; assume without loss of generality that $A_0\in\mathscr{U}$. Let $U=\{0\}\cup A_0$; then $U\in\mathscr{U}$, so $U$ is an open nbhd of $0$. And $U\cap A_1=\varnothing$, so $U$ does not contain a tail of $\sigma$: there is no $n_0\in\Bbb N$ such that $a_n\in U$ for all $n\ge n_0$. Thus, $\sigma$ does not converge to $0$ in this case, either, and it follows that $0$ is not the limit of any sequence in $\Bbb Z^+$.

My second example is the Arens-Fort space; it’s similar in spirit but more elementary. Let $X=\Bbb N\times\Bbb N$, and let $p=\langle 0,0\rangle\in X$. Points of $X\setminus\{p\}$ are isolated. For $n\in\Bbb N$ let $C_n=\{n\}\times\Bbb N$. A set $U\subseteq X$ is an open nbhd of $p$ if and only if $p\in U$, and the set $\{n\in\Bbb N:C_n\setminus U\text{ is infinite}\}$ is finite.

If you sketch $\Bbb N\times\Bbb N$, you’ll see that the set $C_n$ is the column of points with first coordinate $n$. Say that a set $S$ almost contains $C_n$ if it contains all but at most finitely many points of $C_n$, i.e., if $C_n\setminus S$ is finite. Then a set containing $p$ is open if and only if it almost contains all but finitely many of the columns $C_n$, i.e., if and only if there are only finitely many columns that it does not almost contain.

Clearly $p$ is a limit point of the set $X\setminus\{p\}$. Let $\sigma=\big\langle\langle n_k,m_k\rangle:k\in\Bbb N\big\rangle$ be a sequence in $X\setminus\{p\}$. Suppose that for each $n\in\Bbb N$ the set $\{k\in\Bbb N:n_k=n\}$ is finite. Then for each $n\in\Bbb N$ there is an $\ell_n\in\Bbb N$ such that $m_k\le\ell_n$ whenever $n_k=n$. Let $$U=\{p\}\cup\{\langle n,m\rangle\in X:m>\ell_n\}\,;$$ then $U$ is an open nbhd of $p$ that contains no point of $\sigma$.

Otherwise there is an $n\in\Bbb N$ such that $N=\{k\in\Bbb N:n_k=n\}$ is infinite. Let $$U=\{p\}\cup\{\langle\ell,m\rangle\in X:\ell\ne n\}\,;$$ then $U$ is an open nbhd of $p$ that contains no point of the infinite subsequence $\big\langle\langle n_k,m_k\rangle:k\in N\big\rangle$ of $\sigma$. In either case $\sigma$ does not converge to $p$, so $p$ is not the limit of a sequence in $X\setminus\{p\}$.


问:
Four conditions imply invertibility of a filter - or not?
I have an image (which is just three matrices $R,G,B$ in the usual red/green/blue configuration) and four transformations, that is four matrices $A,B,C,D$ which act on the image by means of multiplication. So $A$ for example is both a matrix and a transformation acting on $R$ and returning $AR$.

I've managed to determine the following about the filters:

$AC-BD$ leaves the image unchanged so $AC-BD=I$ the identity matrix.

$AD+BC$ makes the image flat so $AD+BC=0$ the zero matrix.

$CA-DB=I$

$DA+CB=0$

Based on way $AC$ works on the image, I suspect it is almost always invertible with $det(AC)\geq 0$ in every case. Is there a way to prove this using 1) to 4)?

答:
A special case . . .

The claim holds if $A,B$ commute.

Proof:

Assume $A,B,C,D$ are square matrices of the same size with real entries such that the given equations are satisfied, and suppose $A,B$ commute.

We want to show that $\det(AC)\ge 0$.

If $A$ is singular, then $\det(AC)=\det(A)\det(C)=0$, and we're done.

So assume $A$ is nonsingular.

From the equation $AD+BC=0$, we get $D=-A^{-1}BC$.
\begin{align*}
\text{Then}\;\;&
AC-BD=I
\\[4pt]
\implies\;&
AC-B\bigl(-A^{-1}BC\bigr)=I
\\[4pt]
\implies\;&
AC+A^{-1}B^2C=I
\\[4pt]
\implies\;&
A^2C+B^2C=A
\\[4pt]
\implies\;&
(A^2+B^2)C=A
\\[4pt]
\end{align*}
hence $A^2+B^2$ and $C$ are both nonsingular.

Since $C$ is nonsingular, we have $\det(C^2) > 0$.

Since $A,B$ commute, we have $A^2+B^2=(A+iB)(A-iB)$.

Since $A^2+B^2$ is nonsingular, we get that $\det(A+iB),\det(A-iB)$ are nonzero.

Since $A,B$ are real matrices, we get that $\det(A+iB),\det(A-iB)$ are complex conjugates.

It follows that $\det(A^2+B^2) > 0$.
\begin{align*}
\text{Then}\;\;&
(A^2+B^2)C=A
\\[4pt]
\implies\;&
(A^2+B^2)C^2=AC
\\[4pt]
\implies\;&
\det(AC) > 0
\\[4pt]
\end{align*}
which completes the proof.


答:
Bear in mind the multiplication in $G$ is the same as the multiplication of the reals so whatever is true about multiplication of $\mathbb R$ will still be true of the multiplication of $G$.

Multiplication is associative.  We don't have to prove that.

We do have to prove that mulitplicaition is closed on $G$.  That is: if $w \in G$ and $u \in G$ then $wu \in G$. (which you did).

The identity in $\mathbb R$ is $1$ so the identity for $G$ will be $1$.  We don't have to prove that $1*w = w$ for all $w\in G$ and we don't have to solve that if $e*w = w$ then $e=1$.  We know that must be true.  The only thing we have to prove is that $1$ is actually a member of $G$.  (Which as $1 = 1 + 0\sqrt 2$ it obviously is.

And as the invers of $x$ is $\frac 1x$, we do know that the inverse of $w \in G$ will  HAVE to be $\frac 1w$.  The issue though is we have to prove that if $w \in G$ then $\frac 1w \in G$ also.

SO let's prove that:

If $w = a+ b\sqrt 2\ne 0$ then $\frac 1w = \frac 1{a+b\sqrt 2}$.  And that is in $G$????  We can write $\frac 1{a+b\sqrt 2}$ as $c +d \sqrt 2$ where $c$ and $d$ is rational?  Is that actually true?  How would we show that? How would we figure out what $c,d $ are?

Well, the hint "rationalize the denominator" is very apt.

$\frac 1{a+b\sqrt 2} = \frac 1{a+b\sqrt 2}\cdot \frac {a-b\sqrt 2}{a-b\sqrt 2}=$

$\frac {a-b\sqrt 2}{a^2 -2b^2 } =\frac a{a^2 -2 b^2}- \frac b{a^2 -2b^2}\sqrt 2$

And as $a,b \in \mathbb Q$ we have $\frac a{a^2 -2b^2},-\frac b{a^2 -2b^2}\in \mathbb Q$.  So $\frac 1{a+b\sqrt 2} = \frac a{a^2 -2 b^2}- \frac b{a^2 -2b^2}\sqrt 2\in G$.

And that's that.

=====

Actually $ac + 2bd = 1$ and $ad + bc = 0$ is a perfectly fine set up.

$d = -\frac {bc}a$ (assuming $a \ne 0$)

$ac + 2b(-\frac {bc}a) = ac- \frac {2b^2c}a =2$

$c (a-\frac {2b^2}a) = 1$

$c = \frac 1{a-\frac{2b^2}a}=\frac {a}{a^2 -2b^2}$ (assuming $a-\frac{2b^2}a\ne 0$.... which it cant because $(a-\frac{2b^2}a)c = 1 \ne 0$.)

So $d =\frac {b \frac {1a}{a^2 -2b^2}}a= \frac {b}{a^2 - 2b^2}$

....

And if $a$ does equal $0$ we have

$ac + 2bd = 2bd =1$ and so $b\ne 0$  and $ad +bc = bc = 0$.

Now $b \ne 0$ as 1) $2bd =1 \ne 0$ and also because we are told $a + b\sqrt 2 \ne 0$ and if $a =0$ then....

SO $b\ne 0$ and $bc = 0$ so $c = 0$.

And $2bd= 1$ so $b = \frac 1{2b}$.

... in other words  $b\sqrt 2\cdot x = 1 \implies x = \frac 1{2b}\sqrt 2$


答:
We have $$ac+2bd=1,ad+bc=0$$ and that is nice because it is a system of linear equations in two variables. Multiplying first equation by $b$ and second one by $a$ and subtracting the two we get $$(2b^2-a^2)d=b$$ ie $$d=\frac{b} {2b^2-a^2}$$ and $$c=-ad/b=\frac{a} {a^2-2b^2}$$ thus $$c+d\sqrt {2}=\frac{a}{a^2-2b^2}+\frac{b}{2b^2-a^2}\sqrt{2}$$ is the inverse of $(a+b\sqrt{2})$.

I wonder why you stopped at those equations. Not every question is difficult and try to have more confidence in yourself.


问:
Expectation of e^xbar
I trying to find the expectation of $(1-\bar X)e^{\bar X}$, where $X_1,...X_n$ are i.i.d poisson random variables with mean $\lambda$. I am stuck because I have no idea how to find the mean of $e^{\bar x}$. Any tips or hints to get going would be appreciated.

答:
If $X\sim Po(\lambda)$ then $Y=\overline{X}_n\sim Po\left(n\lambda\right)$, even if it has a modified support.

$Y=0,\frac{1}{n},\frac{2}{n},...$

Remember also that $E(e^Y)$ is the definition of MGF with $t=1$

Title: Expectation of e^xbar


问:
Prove that $(f \text{ continuous on } \mathbb{R}) \land (\forall x<c,\; f(x)<0)\land (\forall x>c,\; f(x)>0)\implies f(c)=0$
Problem

Let $f:\mathbb{R}\to \mathbb{R}$ and $c\in\mathbb{R}$. Prove that
$$(f \text{ continuous on } \mathbb{R}) \land (\forall x<c,\; f(x)<0) \land (\forall x>c,\; f(x)>0) \implies f(c)=0$$

My proof

Since $f$ is continuous, we have that, for all $\varepsilon$, there is a $\delta$ such that for all $x\in [c-\delta,c+\delta]$
$$-\varepsilon \leq f(c)-f(x)\leq \varepsilon$$
This implies that $f(c)-\varepsilon\leq f(x)$

By contradiction, assume $f(c)>0$ (the case $f(c)<0$ is similar) and lets use $f(c)$ as our $\varepsilon$, then we have a $\delta$ such that $\forall x\in [c-\delta,c+\delta]$
$$0=f(c)-f(c)\leq f(x)$$
This says that $\forall x\in [c-\delta,c+\delta],\; f(x)\geq0$ but there are values on $[c-\delta,c+\delta]$ that are $>c$ and some that are $<c$ thus $f(x)$ cannot have the same sign for all of them. Contradicting  $\forall x\in [c-\delta,c+\delta],\; f(x)\geq 0\;\square$.

Is this a valid proof? I find it particularly simple given that I didn´t use the intermediate value theorem. (I'm not supposed to know it for this problem)

答: The proof is correct. It could also be proven by the use of the IVT.</p>


问:
Geometric Maximization Exercise
What is the height of a circular cone with surface line ("Mantellinie") s, which has the maximal volumina?
My problem here is, that I am not really sure what surface line is..is this the same thing as just to say surface?</p>

答:
The surface line should be a line on the surface which, when rotated, will form the surface in question. Usually I'd expect such  surface line to lie in the same plane as the axis of rotation. So in the case of the cone, this would be a line segment starting at the origin, assuming your axis of rotation passes through the origin as well.

I'd think about this as an angle problem. Let's use $\alpha$ to denote the angle between surface line and axis of rotation. Then the height of the cone will be $s\cos\alpha$, the radius of its circular base will be $s\sin\alpha$, the area of the base will be $\pi s^2\sin^2\alpha$ and the volume will be $\frac13s^3\pi\cos\alpha\sin^2\alpha$. As you can see, the actual length of $s$ is irrelevant, you can choose your unit of length arbitrarily. All that matters is $\alpha$. You're looking for the value which maximizes $\cos\alpha\cdot\sin^2\alpha$. Simply typing this into Wolfram Alpha will, among others, tell you:

$$\max\left(\cos(\alpha) \sin^2(\alpha)\right) =
\frac2{3\sqrt3}
\quad\text{at}\quad
\alpha = 2 n \pi\pm2 \tan^{-1}{\sqrt{2-\sqrt3}}
\quad\text{for }\pi\in\mathbb Z$$

So your best solution is something close to $\alpha\approx54.73561°$. And since you can construct square roots using compass and ruler, you could even construct this angle using the classic set of construction tools. You can even Google for the above decimal representation. If you do, you will find various situations where that angle plays a certain role, and some documents will even give a nicer formula for the optimal angle:

$$\alpha=\arccos\frac{1}{\sqrt3}$$

All the above computation assumes that the base cap of the cone is not counted as part of the surface, since I did not count its radius towards the length of the surface line. If you do count that as well, things will become a lot more difficult. In particular, you'd have

\begin{align*}
r &= h\tan\alpha \\
s &= h+r
\end{align*}

From these two you could compute $h$ and $r$ given $s$ and $\alpha$, and then again maximize $V=\frac13\pi h r^2$. But since the equations are very non-linear, the result will likely look even uglier than the previous one. But the term “Mantellinie” in particular does usually not include the cap, so you probably won't have to worry about this.

Title: Geometric Maximization Exercise


答:
The moment generating function for the underlying Poisson distributions is

$$\mathbb E\left[e^{X_i t}\right] = \exp\left(\lambda (e^t-1)\right)$$ so for the sum of $n$ i.i.d. copies is $$\mathbb E\left[e^{\sum X_i t}\right] = \exp\left(n\lambda (e^t-1)\right)$$ and letting $t=\frac1n$ gives $$\mathbb E\left[e^{\bar{X}}\right] = \mathbb E\left[e^{\frac1n \sum X_i }\right] = \exp\left(n\lambda (e^{\frac1n}-1)\right)$$

Title: Expectation of e^xbar

Body: I trying to find the expectation of $(1-\bar X)e^{\bar X}$, where $X_1,...X_n$ are i.i.d poisson random variables with mean $\lambda$. I am stuck because I have no idea how to find the mean of $e^{\bar x}$. Any tips or hints to get going would be appreciated.


问:
Show that $A$ can be inverted when $AB$ = Identity matrix
If A and B are both n x n matrices and AB = I, how would you prove that A can be inverted?

答: $AB = I \implies det(AB) = det(I) \implies det(A)det(B) = 1 \implies det(A) \neq 0$. Therefore, $A$ is invertible.


答:
Let $x_0\in[0,1]$. You have already proved that $\varphi(x_0)$ given by $\varphi_n(x_0) = f^n(x_0)$ converges to, let's say, $l$. Therefore, you may write
$$l=\lim_{n\to\infty}\varphi_n(x_0) = \lim_{n\to\infty}\varphi_{n+1}(x_0) = \lim_{n\to\infty} f(\varphi_n(x_0)) = f\left(\lim_{n\to\infty}\varphi_n(x_0)\right) = f(l),$$
where we have used the continuity of $f$. Hence, $l$ is a fixed point of $f$.

What Figure 10 is displaying can be seen as the following iteration:

You begin with $q_0 = \left(x_0,0\right)$.

At step $n$, apply $f$ to move vertically from $q_n$ to $p_n = \left(\varphi_n(x_0),f\left(\varphi_{n}(x_0)\right) = \left(\varphi_n(x_0),\varphi_{n+1}(x_0)\right)\right)$.

Move horizontally from $p_n$ to $q_{n+1} = \left(\varphi_{n+1}(x_0),\varphi_{n+1}(x_0)\right)$.

Update $n = n + 1$ and go to step 2.

Notice that a fixed point will be reached in the limit as
$$p_n =  \left(\varphi_{n}(x_0), f\left(\varphi_{n}(x_0)\right)\right) = \left(\varphi_{n}(x_0), \varphi_{n+1}(x_0)\right) \to \left(l,l\right).$$

Title: How to imagine (or visualize) the convergence of $x,\; f(x),\; f(f(x)),\; f(f(f(x))), \dots\; $?

Body:
I found a problem (Calculus 3rd edition by Michael Spivak, Chapter 22 question 21) that goes as follows:

21. (a) Suppose that $f$ is continuous on $[0,1]$ and that $0\leq f(x) \leq 1$ for all $x$ in $[0,1]$. Problem 7-11 shows that $f$ has a fixed point. If $f$ is increasing, a much stronger statement can be made:

For any $x$ in $[0,1]$, the sequence
$$x,\; f(x),\; f(f(x)),\dots$$
has a limit (which is necessarily a fixed point by problem 20). Prove this assertion, by examining the behavior of the sequence for $f(x)>x$ and $f(x)<x$ or by looking at Figure 10. [...]

I managed to prove the assertion but can't seem to visualize the concept. Could someone explain Figure 10 and its implications? Is $\varphi_n(x)=f^n(x)$ (where $n$ is the number of compositions) getting as close as desired to $id_{[0,1]}(x)=x$ given a sufficiently large $n$ ($\forall x\in [0,1]$)? What is happening to $\varphi_n(x)$ as $n$ gets larger? What about the case $f=id_{[0,1]}$?

Proof of the assertion

Let $x\in\{\alpha\in [0,1]\mid 1\geq f(\alpha)>\alpha\}=A$, then, since $f$ is increasing, $f(f(x))>f(x)$ and $f(f(f(x)))>f(f(x))>f(x)>x$ so the sequence is increasing $\forall x\in A$ but $f(x)\leq 1$ so it's also bounded and thus a limit exists. For $x$ such that $f(x)<x$ the proof is similar $\square$.


问:
Which one result in mathematics has surprised you the most?
A large part of my fascination in mathematics is because of some very surprising results that I have seen there.

I remember one I found very hard to swallow when I first encountered it, was what is known as the Banach Tarski Paradox. It states that you can separate a ball $x^2+y^2+z^2 \le 1$ into finitely many disjoint parts, rotate and translate them and rejoin (by taking disjoint union), and you end up with exactly two complete balls of the same radius!

So I ask you which are your most surprising moments in maths?

Chances are you will have more than one. May I request post multiple answers in that case, so the voting system will bring the ones most people think as surprising up. Thanks!

答:
One surprise for me -- What is the optimal way to cover an equilateral triangle with two squares?

It wasn't solved correctly until 2009.  https://erich-friedman.github.io/packing/squcotri/

LastEditorDisplayName: anon


问:
unexplained inequalities from Niven's 1971 proof $\sum 1/p$ diverges
Ivan Niven published a much shorter and easier proof that the prime harmonic series $\sum 1/p$ diverges. The proof is now available as open access here in this link.

In the short proof he makes use of two inequalities:

$$ \left( \sum_{k<n}'1/k \right) \left( \sum_{j<n}1/j^2 \right) \geq \sum_{m<n} 1/m$$

$k$ are the square-free integers, $j$ and $m$ are ordinary integers, and

$$\prod_{p<n}(1+1/p) \geq \sum_{k<n}'1/k$$

Question: Is there a simple logical rationale for why these inequalities hold?

I can only verify these inequalities by hand, multiplying out series for small $n$. Obviously this isn't proof, but doing this can sometimes reveal the more general logic. In this case I can't see it.

I would appreciate solutions and comments that avoid technical terminology as I, and my students, are not university trained mathematicians.

Update: Since Niven's article is now open access, here is a screenshot and a transcription:

$$\text{A PROOF OF THE DIVERGENCE OF $\Sigma 1 / p$}$$
$$\text{Ivan Niven, University of Oregon}$$
First we prove that $\Sigma^{\prime} 1 / k$ diverges, where $\Sigma^{\prime}$ denotes the sum over the squarefree positive integers. Each positive integer is uniquely expressible as a product of a squarefree positive integer and a square, so for any positive integer $n$,
$$
\left({\sum_{k<n}}^{\prime} 1 / k\right)\left(\sum_{j<n} 1 / j^{2}\right) \color{red}\geqq \sum_{m<n} 1 / m
$$
Here the second sum is bounded but the third sum is unbounded as $n$ increases, so the first sum must be unbounded. Next suppose that $\Sigma 1 / p$ converges to $\beta$, the sum taken over all primes $p$. By dropping all terms beyond $x$ in the series expansion of $e^{x}$ or $\exp (x),$ we see that $\exp (x)>1+x$ for $x>0 .$ Hence for each positive integer $n$
$$
\exp (\beta)>\exp \left(\sum_{p<n} 1 / p\right)=\prod_{p<n} \exp (1 / p)>\prod_{p<n}(1+1 / p) \color{red}\geqq {\sum_{k<n}}^{\prime} 1 / k
$$
But this contradicts the unboundedness of the last sum, so $\Sigma 1 / p$ diverges.

答:
He explains his reasoning with respect to the first inequality in the paragraph above the inequality.  Since each integer can be uniquely expressed as the product of a square-free integer and a square, each term on the right-hand sum occurs as a product of a single term from the square-free sum and a single term from the square sum.  The inequality isn't strict because as $k$ and $j$ approach $n$, their product may exceed $n$, yielding some additional terms in the sum.

Similarly, in the second inequality, if you simply expand the product on the left, you will get all of the terms in the right-hand sum, and possibly some additional terms that you can throw away.


答:
Each $m<n$ can be uniquely expressed as $m=kj^2$ where $k$ is a square-free integer, therefore
$$ \sum_{m<n}\frac{1}{m}=\sum_{\substack{k,j \\ kj<n}}\frac{1}{kj^2}\leqslant\sum_{k,j<n}\frac{1}{kj^2}=\left(\sum_{k<n}'\frac{1}{k}\right)\left(\sum_{j<n}\frac{1}{j^2}\right) $$
where $k$ is a square free integer in the second and the third sum. As for the last inequality, let $p_1,\ldots,p_r$ be the prime numbers $<n$, then developping the product gives
$$ \prod_{p<n}\left(1+\frac{1}{p}\right)=\sum_{I\subset[\![1,r]\!]}\prod_{i\in I}\frac{1}{p_i}$$
Each square-free $k<n$ has their prime divisors among $p_1,\ldots,p_r$ and thus is of the form $\prod_{i\in I}p_i$ where $I\subset[\![1,r]\!]$, therefore
$$\prod_{p<n}\left(1+\frac{1}{p}\right)\geqslant\sum_{k<n}'\frac{1}{k} $$

Title: unexplained inequalities from Niven's 1971 proof $\sum 1/p$ diverges

Body:
Ivan Niven published a much shorter and easier proof that the prime harmonic series $\sum 1/p$ diverges. The proof is now available as open access here in this link.

In the short proof he makes use of two inequalities:

$$ \left( \sum_{k<n}'1/k \right) \left( \sum_{j<n}1/j^2 \right) \geq \sum_{m<n} 1/m$$

$k$ are the square-free integers, $j$ and $m$ are ordinary integers, and

$$\prod_{p<n}(1+1/p) \geq \sum_{k<n}'1/k$$

Question: Is there a simple logical rationale for why these inequalities hold?

I can only verify these inequalities by hand, multiplying out series for small $n$. Obviously this isn't proof, but doing this can sometimes reveal the more general logic. In this case I can't see it.

I would appreciate solutions and comments that avoid technical terminology as I, and my students, are not university trained mathematicians.

Update: Since Niven's article is now open access, here is a screenshot and a transcription:

$$\text{A PROOF OF THE DIVERGENCE OF $\Sigma 1 / p$}$$
$$\text{Ivan Niven, University of Oregon}$$
First we prove that $\Sigma^{\prime} 1 / k$ diverges, where $\Sigma^{\prime}$ denotes the sum over the squarefree positive integers. Each positive integer is uniquely expressible as a product of a squarefree positive integer and a square, so for any positive integer $n$,
$$
\left({\sum_{k<n}}^{\prime} 1 / k\right)\left(\sum_{j<n} 1 / j^{2}\right) \color{red}\geqq \sum_{m<n} 1 / m
$$
Here the second sum is bounded but the third sum is unbounded as $n$ increases, so the first sum must be unbounded. Next suppose that $\Sigma 1 / p$ converges to $\beta$, the sum taken over all primes $p$. By dropping all terms beyond $x$ in the series expansion of $e^{x}$ or $\exp (x),$ we see that $\exp (x)>1+x$ for $x>0 .$ Hence for each positive integer $n$
$$
\exp (\beta)>\exp \left(\sum_{p<n} 1 / p\right)=\prod_{p<n} \exp (1 / p)>\prod_{p<n}(1+1 / p) \color{red}\geqq {\sum_{k<n}}^{\prime} 1 / k
$$
But this contradicts the unboundedness of the last sum, so $\Sigma 1 / p$ diverges.


答: For the first one, as Niven says every positive integer $m$ can be expressed uniquely in the form $m = kj^2$ where $k$ is squarefree. Obviously if $m<n$, then $k, j < n$ too. By distributivity, the left-hand side is $\sum_{k,j} 1/(kj^2)$ where $k, j < n$ and $k$ is squarefree, so every term on the right is included (along with a few more).


问:
When does $A^{1/2}$ exists? for $A$ a square matrix.
I know that if $A$ is a square hermitian matrix positive definite (positive semi-definite) then $A^{1/2}$ exists and is also a square hermitian matrix positive definite (positive semi-definite respectively). However I want to know how weaker can I make the hypothesis for the existance of $A^{1/2}$. Does it work for any square matrix? Do I need $A$ to be at least positive semi-definite?

答:
If $A$ is diagonalizable, $QAQ^{-1} = D$ with $D = \mathrm{diag}(d_1,\ldots,d_n)$; and each $d_i$ has a square root in the field you are considering (for example, this is always the case in $\mathbb C$), then yes: take $D' = \mathrm{diag}(s_1,\ldots,s_n)$ with $s_i^2 = d_i$ and then

$$
(Q^{-1}D'Q)^2 = Q^{-1}D'^2Q = Q^{-1}DQ = A.
$$

Note that, however, a square root in general is not unique. Moreover, without any hypothesis they may fail to exist.


问:
Question about solution to Putnam 1995 A2.
To provide context Putnam 1995 A2 states:

For what pairs of integers $(a,b)$ of positive real numbers does the improper integral

$$ \int_{b}^{\infty} \left( \sqrt{ \sqrt{x+a \vphantom{b}} -\sqrt{x}} - \sqrt{\sqrt{x} - \sqrt{x-b}} \right) $$
  converge?

The solution then proceeds to make the following claims:

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

So:

$$\sqrt{x-a} - \sqrt{x} = x^{1/2}(\sqrt{1 + a/x} - 1)$$

Subbing in the expansion we have:

$$x^{1/2}(1 + a/(2x) + O(x^{-2}))$$

Here I didn't agree, I thought it would be:

$$x^{1/2}(1 + a/(2x) + O(x^{-2}) - 1)$$

which becomes:

$$x^{1/2}(a/(2x) + O(x^{-2}))$$

Furthermore, after taking another square root they get:

$$x^{1/4}\left(a/(4x) + O(x^{-2})\right)$$

I would imagine it should be:

$$x^{1/4}\sqrt{a/(2x) + O(x^{-2})}$$

Can someone explain what I'm doing wrong or not understanding?

答:
As you point out, the solution there is incorrect but it can be salvaged.  It's correct that

$$\sqrt{x+a}-\sqrt{x} = x^{1/2}(a/2x + O(x^{-2}))$$

and so if you factor out the $a/2x$ term you get

$$\sqrt{x+a}-\sqrt{x} = a/2\cdot x^{-1/2}(1 + O(x^{-1})).$$
This means that

$$\sqrt{\sqrt{x+a}-\sqrt{x}} = \sqrt{a/2}\cdot x^{-1/4}(1+O(x^{-1}))$$
where we again used binomial expansion.  You can do the same thing to obtain

$$\sqrt{\sqrt{x}-\sqrt{x-b}} = \sqrt{b/2}\cdot x^{-1/4}(1+O(x^{-1})).
$$

So ultimately the integral becomes

$$\int_b^\infty (\sqrt{a/2}-\sqrt{b/2})x^{-1/4}\,dx + \int_b^\infty O(x^{-5/4})\,dx
.$$
The second integral converges because $5/4 > 1$, and since $0 < 1/4 \le 1$ the first integral will converge precisely when $\sqrt{a/2}-\sqrt{b/2}=0$ or equivalently when $a=b$.


问:
Is there a power series that converges to the function $f(x)= \lvert x\rvert$ for all $x$?
Is there a power series that converges to the function $f(x)= \lvert x\rvert$ for all $x$?

I am pretty lost on how to even start this.

答: A more algebraic way to see that there is no such power series is to consider its square $f(x)^2$, which must be equal to $x^2$. Even in the ring of formal power series (which is an integral domain) there are only two sqaure roots of the power series $x^2$, namely $x$ and $-x$; these obviously don't qualify.</p>


问:
Diagonalizing a matrix with 4 circulant blocks
I have the following matrix:

$$\mathbf{M} = \begin{pmatrix}
G_{1}^{(N)} & G_{2}^{(N)} \\
G_{2}^{(N)} & G_{3}^{(N)}
\end{pmatrix}$$,

where $G^{(N)}_{j}$ are symmetric circulant matrices of size $N-\text{by}-N$ with complex-valued entries.
Can this be diagonalized (not block-diagonalized!)?

答:
This is not a full diagonalization, but we can end up with a block-diagonalization where each block has size only $2 \times 2$.

Let $F$ denote the DFT matrix of size $N$. Denote
$$
P_1 = I_2 \otimes F = \pmatrix{F & 0\\0 & F}.
$$
Here, $\otimes$ denotes the Kronecker product. Note that $P$ is unitary, which is to say that $P^{-1} = P^*$ (the conjugate-transpose of $P$). We find that
$$
P_1^{-1}MP_1 = \pmatrix{F^{-1}G_1^{(N)}F & F^{-1}G_2^{(N)}F\\ F^{-1}G_2^{(N)}F & F^{-1}G_3^{(N)}F} = \pmatrix{D_1 & D_2\\D_2 & D_3},
$$
and each of the matrices $D_1,D_2,D_3$ is diagonal. Now, let $P_2$ denote the commutation matrix $P_2 = K^{(2,N)}$. $P_2$ is a symmetric permutation matrix, so that $P^{-1} = P^T$. We have
$$
P_2^{-1}(P_1^{-1}MP_1)P_2 = \pmatrix{A_1\\ & \ddots \\ && A_N}.
$$
Each matrix $A_j$ has size $2 \times 2$. If $D_{jk}$ denotes the $k$th diagonal entry of $D_j$, then we have
$$
A_k = \pmatrix{D_{1k} & D_{2k}\\ D_{2k} & D_{3k}}.
$$
$M$ is diagonalizable if and only if this last matrix $P_2^{-1}P_1^{-1}MP_1P_2$ is diagonalizable, which holds if and only if each matrix $A_k$ is diagonalizable. We can see that the matrices $A_k$ are symmetric with complex entries, but this does not guarantee that they are diagonalizable.


问:
Non-associative: set with a binary operation, but has inverses and identity
I've been thinking about an example of some set with a binary operation which would satisfy all axioms of groups except for associativity. I'm new to Group Theory, so I would appreciate your knowledgeable insight.

My example:

$S={\mathbb{C} - \{0\}}$; $H = (S, \text^)$, where ^ is exponentiation. Can we consider this as a correct example? The identity under ^ appears to be 1. The inverse appears to be $\frac{2i\pi}{\ln(x)}$.

答: Consider the set $\{0,1,2\}$ with the binary operation $a*b = |a-b|$. It has an identity element (namely $0$) and every element has an inverse (namely itself) and it is not associative: $(1*1)*2 = 0*2 = 2$ but $1*(1*2) = 1*1 = 0$.


问:
How do I evaluate $\int_{0}^{\sqrt{2}}\int_{x}^{\sqrt{4-x^2}}{\sqrt{x^2+y^2}} \, dy \, dx$?
I'm having troubles evaluating this double integral. Can somebody help me? I've gone to the part that I need to use trigonometric substitution, but performing the said sub, I think I'm kind of unsure of the next steps.

$$\int \limits_{0}^{\sqrt{2}}\int \limits_{x}^{\sqrt{4-x^2}}{\sqrt{x^2+y^2}} \, dy \, dx $$

答:
Note that the integration region is in the first quadrant, between $y=x $ and $x^2+y^2=4$, which, in polar coordinates, leads to
$$\int \limits_{0}^{\sqrt{2}}\int \limits_{x}^{\sqrt{4-x^2}}{\sqrt{x^2+y^2}} \ dy \ dx
=\int_{\pi/4}^{\pi/2} \int_0^2 r \>rdrd\theta=\frac{2\pi}3
$$


问:
Does there exist a continuous partition of the sphere into sets of cardinality 4?
Define $X^{\{n\}}:=\{A\subseteq X:|A|=n\}$, the set of subsets of cardinality $n$. If $X$ is a topological space, $X^{\{n\}}$ can be given a topology by considering it to be a quotient of $X^n$ minus the extended diagonal.

Define a continuous $n$-partition of a space $X$ to be a partition of $X$ into sets of cardinality $n$ such that the associated function $X\to X^{\{n\}}$, sending $x$ to the element of the partition that contains it, is continuous.

(The notation and terminology is my own. I don't know if there is standard notation or terminology for these.)

Does there exist a continuous $4$-partition of the sphere?

My assumption is no, because most obvious $4$-partitions fail. If we could find a $2$-partition (also known as a fixed-point-free involution) of the projective plane ${\rm\Bbb RP}^2$, we'd be golden, since each element of ${\rm\Bbb RP}^2$ corresponds to an antipodal pair of the sphere $S^2$, but this is impossible by the Lefschetz fixed point theorem.

There's probably some simple algebraic topology solution for this, but unfortunately, I am unskilled in algebraic topology.

For a further conjecture, I believe $S^2$ has size $2$ mod $4$, in the following sense: if $k$ points are deleted from $S^2$, then the result has a continuous $4$-partition iff $k\equiv2 \pmod{4}$.

答:
Let $X$ be a manifold and suppose $f:X\to X^{\{n\}}$ is a continuous $n$-partition.  Let $p:X\to Y$ be the quotient map associated to this partition of $X$.  I claim that in fact $p$ is a covering map.  (Conversely, it is easy to see that any partition that comes from an $n$-sheeted covering map is continuous.)

To prove this, let $y\in Y$, so $y=\{x_1,\dots,x_n\}$ is an $n$-element subset of $X$.  Let $U_1,\dots,U_n$ be pairwise disjoint neighborhoods of $x_1,\dots,x_n$.  Then there is an open subset $U$ of $X^{\{n\}}$, consisting of sets which contain exactly one element from each $U_i$, and the topology of $U$ is just the product topology when you identify it with $\prod U_i$.  By continuity of $f$, there is an open neighborhood $V_1\subseteq U_1$ of $x_1$ such that $f(V_1)\subseteq U$.  Restricting $f$ to a map $V_1\to U\cong \prod U_i$, we can think of it as a collection of continuous maps $f_i:V_1\to U_i$ for each $i$.  Since $f$ is a partition, each $f_i$ is injective.  By invariance of domain (this is the one place we are using the assumption that $X$ is a manifold), this means $f_i(V_1)$ is open for each $i$ and $f_i$ is a homeomorphism $V_1\to f_i(V_1)$.  Let us write $V_i=f_i(V_1)$.

So, we have open neighborhoods $V_1,\dots,V_n$ of $x_1,\dots,x_n$ and homeomorphisms $f_i:V_1\to V_i$ such that our partition just splits $\bigcup V_i$ into the sets of the form $\{f_1(x),f_2(x),\dots,f_n(x)\}$ for each $x\in V_1$.  So, the set $W=p(V_1)=p(V_2)=\dots=p(V_n)$ is an open neighborhood of $p$ in $Y$, and it is evenly covered by $p$, with $p^{-1}(W)=\bigcup V_i$ and $p$ mapping each $V_i$ to $W$ homeomorphically.

In particular, in the case of $S^2$, a continuous $n$-partition would give a surface that has $S^2$ as an $n$-sheeted covering space.  By the classification of surfaces, this is possible for only $n=1,2$.

Furthermore, if you have a continuous $n$-partition of an $k$-times punctured sphere $X$, this gives an $n$-sheeted covering map $p:X\to Y$.  The space $Y$ must then be a connected surface with finitely generated fundamental group, and thus is a surface of finite type (you can show it has finitely many ends and its end compactification is a closed surface).  It follows that $\chi(X)=n\chi(Y)$.  Since $\chi(X)=2-k$, this means $k$ must be $2$ mod $n$.


答:
When you multiply out the product

$$\left(\sum_{k<n}'\frac1k\right)\left(\sum_{j<n}\frac1{j^2}\right)\,,\tag{1}$$

you get the sum of all possible terms of the form $\frac1{kj^2}$, where $k$ is a square-free integer less than $n$, and $j$ is any integer less than $n$. Every positive integer $m<n$ can be expressed as the product of such a $k$ and $j^2$, so these product terms include every fraction $\frac1m$ for $m<n$. In general, however, they will also include some fractions $\frac1m$ with $m\ge n$: if $n-1$ is square-free, for instance, we can take $k=j=n-1$ and get $\frac1{(n-1)^3}$. Thus, every term of $\sum_{m<n}\frac1m$ appears in $(1)$ after the latter is multiplied out, and $(1)$ generally contains extra terms as well, so

$$\left(\sum_{k<n}'\frac1k\right)\left(\sum_{j<n}\frac1{j^2}\right)\ge\sum_{m<n}\frac1m\,.$$

For the second inequality let $P=\{p_1,\ldots,p_\ell\}$ be the set of primes less than $n$. When the product

$$\prod_{p<n}\left(1+\frac1p\right)=\prod_{i=1}^\ell\left(1+\frac1{p_i}\right)\tag{2}$$

is multiplied out, there is one term for each possible choice of one factor, $1$ or $\frac1{p_i}$, from each of the $\ell$ factors. Specifically, if $S\subseteq P$, one of the terms of the product is

$$\frac1{\prod_{i\in S}p_i}\,,\tag{3}$$

obtained by choosing $1$ when $i\notin S$ and $\frac1{p_i}$ when $i\in S$.

Every square-free integer less than $n$ is a product of distinct primes less than $n$, so every term of $\sum\limits_{k<n}'\frac1k$ is one of the terms $(3)$. However, there may also be some $S\subseteq P$ such that $\prod_{i\in S}p_i\ge n$, so the terms of $(2)$ after it’s multiplied out include all of the terms of $\sum\limits_{k<n}'\frac1k$ and possible some others as well. Thus,

$$\prod_{p<n}\left(1+\frac1p\right)\ge\sum_{k<n}'\frac1k\,.$$


问:
Why is $\infty \cdot 0$ not clearly equal to $0$?
I did a bit of math at school and it seems like an easy one - what am I missing?

$$n\times m = \underbrace{n+n+\cdots +n}_{m\text{ times}}$$

$$\quad n\times 0 = \underbrace{0 + 0 + \cdots+ 0}_{n\text{ times}} = 0$$

(i.e add $0$ to $0$ as many times as you like, result is $0$)

So I thought an infinite number of $0$'s cannot be anything but $0$? But someone claims different but couldn't offer a reasonable explanation why. Google results seemed a bit iffy on the subject - hopefully this question will change that.

答:
The symbol $\infty$ was originally introduced by Wallis in the 17th century.  He used it to denote a specific infinite number, and went on to consider partitions of intervals into $\infty$ parts of width $\frac{1}{\infty}$ in applications such as calculation of areas of plane figures.

Such giants as Leibniz, Euler, and Cauchy exploited infinite quantities to obtain results in analysis.  More details can be found for instance in the recent article here.

In an enriched number system containing such infinite numbers, it is correct that an infinite number times $0$ is indeed "an easy Zero answer". One could take such a number system to be the hyperreals, but there are many such number systems.  So long as the number system is a field, anything multiplied by $0$ will necessarily give $0$, even if the "anything" is an infinite number.

The customary interpretation of the expression "infinity times zero" is in terms of the so-called "indeterminate forms" (see also Whats infinity divided by infinity?), and then the answer is far from easy and in fact is not zero in general.

However, when interpreted literally, the OP's hunch that $\infty \times 0=0$ is "easy" can be fully justified as above.

For a related question on "infinity times infinitesimal", see infinity times infinitesimal - what happens?


问:
How can you calculate the probability of failure of a product after a given time, given the expected failures per unit time?
For example, if a product has an expected failure rate of 1 failure every 10 years of operation, what's the probability that the product fails after 1 year of use? How does this probability change if we ask the same question, but now we are using 2 of the same product in parallel? Thanks!</p>

答:
You can use an exponential distribution to model this type of problem. Let $t$ be a random variable with distribution $t \sim \text{Exp}(\lambda)$. Here, $t$ represents the amount of time to failure, and $\lambda$ is the rate parameter, where $E[t] = \frac{1}{\lambda}$. Then, we have the following cumulative distribution function:

$$ P[t < T] = 1 - e^{-\lambda T} . $$

We're given that the expected failure rate is $1$ failure in $10$ years of operation. Interpreting that as the mean time to failure, we have $\lambda = \frac{1}{10}$. Plugging in $\lambda$, we have

$$ P[t < T] = 1 - e^{-\frac{T}{10}} . $$

The probability that the product fails within $T = 1$ year is

$$ P[t < 1] = 1 - e^{-\frac{1}{10}} \approx 9.5 \% . $$

Now, let's say we have $n$ of these products running in parallel, and we can assume that the failure of any one product is independent of the failure of any other products. Let $t_i$ represent the time to failure of the $i$th product, where each $t_i$ is independently and identically distributed the same way as $t$ above.

Probability of at Least One Failure

If we want the probability that at least one of the products fails within time T, we're looking for

$$ P[t_1 < T \cup t_2 < T \cup \ ... \ \cup t_n < T] . $$

This isn't a particularly easy way to frame this calculation - we'd have to calculate the probability of exactly one of them failing, or exactly two of them failing, etc., and we'd have to account for all possible combinations of products. But, there's a way to reframe it that's a lot easier.

The complement of the event that at least one product fails is that none of the products fail. In that case, we have

$$ P[t_1 < T \cup t_2 < T \cup \ ... \ \cup t_n < T] = 1 - P[t_1 > T \cap t_2 > T \cap \ ... \ \cap t_n > T] . $$

Since we assumed these failures were independent, we have

$$ 1 - P[t_1 > T \cap t_2 > T \cap \ ... \ \cap t_n > T] $$
$$ = 1 - P[t_1 > T]* P[t_2 > T]* \ ... \ *P[t_n > T] $$
$$ = 1 - (P[t > T])^n $$
$$ = 1 - (1 - P[t < T])^n $$
$$ = 1 - \left(1 - \left(1 - e^{-\frac{T}{10}} \right) \right)^n $$
$$ = 1 - \left(e^{-\frac{T}{10}} \right)^n $$
$$ = 1 - e^{-\frac{nT}{10}} . $$

When $n = 2$ and $T = 1$, we have

$$ P[t_1 < 1 \cup t_2 < 1] = 1 - e^{-\frac{2*1}{10}} \approx 18.1 \%. $$

Probability of $n$ Failures

Now, if we want to find the probability that all $n$ products fail within time $T$, we want

$$ P[t_1 < T \cap t_2 < T \cap \ ... \ \cap t_n < T] . $$

Again, through independence, we can split this up into individual probabilities and calculate:

$$ P[t_1 < T \cap t_2 < T \cap \ ... \ \cap t_n < T] $$
$$ = P[t_1 < T]*P[t_2 < T]* \ ... \ *P[t_n < T] $$
$$ = (P[t < T])^n $$
$$ = \left(1-e^{-\frac{T}{10}}\right)^n . $$

For $n = 2$ and $T = 1$, we have

$$ P[t_1 < 1 \cap t_2 < 1] = \left(1-e^{-\frac{1}{10}}\right)^2 $$
$$ \approx 0.9 \% . $$


问:
Trigonometry - distance and bearing

A yacht and a catamaran leave the harbour at $09:00$ hours. The yacht sails at $24$ knots on course $037$ and the catamaran at $15$ knots on $139$. Find the bearing and the distance in nautical miles of the yacht from the catamaran at $1400$ hours ($1$ knot is a speed of $1$ nautical mile per hour).

The final answer I am getting is $135$m using the cosine rule. However this doesn’t match with the solutions. Please could someone clarify both the length and the bearing?

答:
Put the harbor at $(0,0)$ on a Cartesian plane.

Let's measure angles clockwise from the positive $y$ axis, because the $y$ axis is in the "north" direction of a conventional Mercator projection and angles are measured clockwise from north in navigation.

Then after $5$ hours, the coordinates of the yacht are

$$ (120 \sin 37^\circ, 120 \cos 37^\circ) $$

and the coordinates of the catamaran are

$$ (75 \sin 139^\circ, 75\cos 139^\circ). $$

Use the usual formulas to get the distance and direction from the catamaran to the yacht, being careful to measure the angle clockwise from the positive $y$ axis.

Title: Trigonometry - distance and bearing


问:
Double coin toss
Double coin toss. We wanna know what's the probability of throwing at least one head.

First we model it as $\Omega$ = {{H $\cap$ H}, {H $\cap$ T}, {T $\cap$ H}, {T $\cap$ T}}. I often see here that people solve: P({H $\cap$ H} $\cup$ {H $\cap$ T} $\cup$ {T $\cap$ H}) = P(H $\cap$ H) + P(H $\cap$ T) + P(T $\cap$ H). But this is not possible because Kolmogorov's axiom #3 just allows addition of disjunct events/sets, but (H $\cap$ T) and  (T $\cap$ H) are not disjunct. So in this model we'd need a different way to answer the question, e.g. thru using the probability of the complement.

Now we model it as $\Omega$ = {{H1 $\cap$ H2}, {H1 $\cap$ T2}, {T1 $\cap$ H2}, {T1 $\cap$ T2}}. Now we can apply the addition rule: P({H1 $\cap$ H2} $\cup$ {H1 $\cap$ T2} $\cup$ {T1 $\cap$ H2}) = P(H1 $\cap$ H2) + P(H1 $\cap$ T2) + P(T1 $\cap$ H2) because all events/sets are disjunctive.

In model 1, Bayes' theorem holds, because (H $\cap$ T) = (T $\cap$ H) by commutativity and therefore its probabilites are the same which is the condition to derive Bayes' theorem. But in model 2, Bayes' theorem does not hold because you do not have a relation of commutativity like in 1, but you need it, else it's not a theorem there.

Can you tell me if I am right in all three points, and if not why?

答:
If a student wrote something like (1), I'd assume they meant (2) but weren't being careful with notation. I'd encourage them to define the events in words before doing any symbolic reasoning. ("Let $H_1$ be the event that the first toss comes up heads," etc.)

For (2), assuming the definitions of $H_1,H_2,T_1,T_2$ as above, you're right that the union is disjoint. We can treat this as obvious or prove it from the fact that $H_1,T_1$ are disjoint and $H_2,T_2$ are disjoint.

As for (3), Bayes' theorem holds for any pair of events. Of course, $H_1$ and $H_2$ are not the same event, so it would be a mistake to treat them as such (by accidentally calling them both $H$, for example) when applying the theorem.

Title: Double coin toss


问:
Is $\mathcal{O} \otimes_{\mathbb{Z}_p} F$ a finite direct product of fields?
Consider the $p$-adic field $\mathbb{Q}_p$ with ring of integers $\mathbb{Z}_p$ and residue field $\mathbb{F}_p$. Let us take a finite  extension $K$ of $\mathbb{Q}_p$, with ring of integers $\mathcal{O}_K$ and $k$ be the residue field of $K$. Let $\pi \in \mathcal{O}_K$ be a uniformizer. Consider the ring of power series $\mathfrak{S}\mathrel{:=}W(k)[[u]]$, where $W(k)$ is the ring of Witt vectors on the residue field $k$. Let $\mathcal{O}:=\widehat{\mathfrak{S}[\frac{1}{u}]}$ be the $p$-adic completion of $\mathfrak{S}[\frac{1}{u}]$. Also let $F$ be finite field.

Then $\mathcal{O} \otimes_{\mathbb{Z}_p} F$ decomposes as a product of fields.

Question:
Is $\mathcal{O} \otimes_{\mathbb{Z}_p} F$ a finite product of fields ? Or an arbitrary product of fields ?

For example, $W(k) \otimes_{\mathbb{Z}_p} W(k)$ is a finite ($[k: \mathbb{F}_p]$ copies) direct product of $W(k)$.

But I couldn't conclude  the question.

Any hints please

答:
Correct me if I'm wrong but $W(k)= O_L=\Bbb{Z}_p[\zeta_{q-1}]$ where $L=\Bbb{Q}_p(\zeta_{q-1}),q=|k|$ is the largest unramified extension $\subset K$ $$\mathfrak{S}[\frac{1}{u}]=O_L[[u]][u^{-1}]=\{ \sum_{n\ge -N} a_n u^n, a_n\in O_L\}$$ whose $p$-adic completion is $$\mathcal{O} = \{ \sum_{n=-\infty}^\infty a_n u^n, a_n\in O_L, \lim_{n\to -\infty} v_p(a_n)=\infty\}$$
Note that $\mathcal{O}/(p)=\Bbb{F}_q((u))$.

For a finite field $F$, when $char(F)\ne p$ then $\mathcal{O}\otimes_{\Bbb{Z}_p}F=\mathcal{O}/(char(F))\otimes_{\Bbb{Z}_p}F=0$,

when $char(F)=p,F=\Bbb{F}_p[x]/(f(x))$ then $$\mathcal{O}\otimes_{\Bbb{Z}_p}F=\mathcal{O}/(p)\otimes_{\Bbb{F}_p}\Bbb{F}_p[x]/(f(x))$$ $$=\Bbb{F}_q((u))\otimes_{\Bbb{F}_p} \Bbb{F}_p[x]/(f(x))=\Bbb{F}_q((u))[x]/(f(x))$$ $$=\Bbb{F}_q((u))[x]/(\prod_j f_j(x))=\prod_j\Bbb{F}_q((u))[x]/(f_j(x))=\prod_j \Bbb{F}_{q^{\deg(f_j)}}((u))$$


问:
Probability and Straight line
Two points are taken at random on the given straight line segment $AB$ of length $L$. Determine the probability that the length of none of the $3$ parts exceeds the given value $c$, where $L/3 \le c < L$.

I have tried dividing the line into segments for example, $AX = x$, the distance between $A$ and $X$.
and $XY = y$, the distance between the first and second arbitrary point and $YB = z$ the distance between $Y$ and the tip of the straight line.

Intuitively, I think the distance should be $1/3$ between them. But I don't know how to solve my problem.

答:
let $L=1$

Let $x$ and $y$ be the locations of the 2 cuts

Let $S$ be the region defined by $0<y<x<1$

The three pieces will have lengths of $y, x-y$, and $1-x$

the condition that all three pieces have length less than $c$ is that the point $(x,y)$ is in the region $T$ of points satisfying$$(y<c) \cap(y>x-c)\cap ( x>1-c)$$
Then the probability that all three pieces have length less than $c$ is given by the ratio of the areas...
$$ P = \frac {A_{S\cap T}}{A_S}$$

When $\frac 13 < c < \frac 12$ a plot of the regions looks as in the image below where the region $T$ is shown in red...

$T$ is a single isosceles right triangle with two sides of length $3c-1$
so $$P=(3c-1)^2$$
When When $\frac 12 < c < 1$ a plot of the regions looks as in the image below ...

The 3 blue isosceles right triangles each have two sides of length $1-c$, so the probability in this regime is $$P = 1 - 3(1-c)^2 $$

Putting it all together ...
$$ P(c) = \begin{cases}
(3c-1)^2  & \frac 13 \le c < \frac 12
\\ 1-3(1-c)^2  &  \frac 12 \le c \le 1 \end{cases}  $$

Title: Probability and Straight line


问:
Find a non-constant $f_n : \mathbb{R} \rightarrow \mathbb{R}$ so $f: \mathbb{R} \rightarrow \mathbb{R}^{\omega}$ is continuous.
Find, for every $n \in \omega$, a non-constant function $f_n : \mathbb{R} \rightarrow \mathbb{R}$ so that $f: \mathbb{R} \rightarrow \mathbb{R}^{\omega}$, defined as $f(x) = (f_n(x))_{n \in \omega}$, is continuous.

In the question, $\mathbb{R}^{\omega}$ comes with the box topology. To be honest, I am quite confused in trying to solve this problem. Up until now, i've studied the definition of the box topology and tried to see what it takes the functions $f_n$ to conform a continuous function $f$, but i've struggled with the fact that it can't be constant. I don't see how can be possible to give an example of a function like this.

答: Hint: So, (and I am completely using @EricWofsey's answer linked above), you could let $f_n(x)=\begin{cases}x,x\in[n,n+1]\\n,x\lt n\\n+1,x\gt n+1\end{cases}$


问:
Expectation of e^xbar
I trying to find the expectation of $(1-\bar X)e^{\bar X}$, where $X_1,...X_n$ are i.i.d poisson random variables with mean $\lambda$. I am stuck because I have no idea how to find the mean of $e^{\bar x}$. Any tips or hints to get going would be appreciated.

答:
In order to calculate $E(\overline{X} e^\overline{X})$ I think it is better to do it in an analytical way

Set $Y=\Sigma_i X_i\sim Po(n\lambda)$ you get

$$\mathbb{E}(\overline{X} e^\overline{X})=\frac{1}{n}\mathbb{E}(Ye^{Y/n})=\frac{e^{-n\lambda}}{n}\sum_{y=0}^{\infty}ye^{y/n}\frac{(n\lambda)^y}{y!}=$$

$$=\frac{e^{-n\lambda}}{n}\cdot e^{1/n}n\lambda\sum_{y-1=0}^{\infty}\frac{(e^{1/n}n\lambda)^{y-1}}{(y-1)!}=e^{1/n}\lambda e^{-n\lambda}e^{e^{1/n}n\lambda}=\lambda e^{1/n+n\lambda(e^{1/n}-1)}$$

With the same easy procedure, you can calculate also $E(e^{\overline{X}})$ in a couple of passages and without invoking MGF

$$\mathbb{E}\left[e^{\overline{X}}\right]=\sum_{y=0}^{\infty}e^{y/n}\cdot\frac{e^{-n\lambda}(n\lambda)^y}{y!}=e^{-n\lambda}\sum_{y=0}^{\infty}\frac{(e^{1/n}n\lambda)^y}{y!}=e^{-n\lambda}\cdot e^{e^{1/n}n\lambda}=e^{n\lambda(e^{1/n}-1)}$$

as previously showed by @Henry using MGF

Title: Expectation of e^xbar


问:
Convergence of integral implies integrand is asymptotically bounded above by a power law that has a convergent integral?
Let $f$ be a continuous, real-valued function on $[1,\infty)$ such that
$\int_{1}^{\infty} f(x) \, dx$ is finite.

Does this necessarily imply that there exists $\kappa > 0$ such that $f(x) = O\!\left( \frac{1}{x^{1+\kappa}} \right)$ ?

That is (expanding the big-O notation), there exists $\kappa > 0$, $M > 0$, $x_0 > a$ such that $|f(x)| \leq \frac{M}{x^{1+\kappa}}$ for all $x \geq x_0$?

Remark: When $\kappa > 0$ we have $\int_{1}^{\infty} \frac{1}{x^{1+\kappa}} \, dx = \left[ \frac{-1}{\kappa x^{\kappa}}\right]_{1}^{\infty} = \frac{1}{\kappa}$ which is finite.

Why I am asking: I have a situation where I have a convergent integral, and want to use feature of the integrand to establish whether another integral is convergent.  Many thanks in advance.

Edit: See this question & answer for a revised question that got to an answer that I applied.

答:
You can't.

Example: Let $f$ be mostly zero, except centered at every half-integer $n+\frac12$, $n\geq 1$, we have a triangle function of height $1$ and base $1/n^2$.  In other words,
$$
f(x)=\sum_{n\geq 1}\max(0,1-2n^2\lvert n+\tfrac12-x\rvert)
$$
Then $\sup_{x\geq N}f(x)=1$ for all $N$ so it doesn't decay.  But the total area of triangles, $\frac12\zeta(2)$, is finite.

In fact, you can tweak this example so that $\sup_{x\geq N}f(x)=\infty$, and if you know about bump functions, get $f$ infinitely differentiable too.


问:
Unitary group is a properly embedded Lie subgroup of $GL(n, \mathbb C)$
I would like to show that $U(n)$, the unitary group of degree $n$, is a properly embedded Lie subgroup of $GL(n, \mathbb C)$ of dimension $n^2$, and find the matrices that are in the tangent space at the identity.

To show the first part, I think I can argue it is a kernel $\Phi^{-1}(I_n)$, where $\Phi(A) = \overline {A^T}A$. However, I am not sure how to show its dimension, and how to identify matrices that are in the tangent space at the identity.

答:
To figure out what the tangent space to $U(n)$ at the identity, use the trajectories of curves definition of the tangent space. Let the tangent space to the identity of $U(n)$ be denoted $\mathfrak{u}(n)$. Let $A\in \mathfrak{u}(n)$ and let $\gamma:(-\varepsilon,\varepsilon)\to U(n)$ so that $\gamma(0)=I_n$ and $\gamma'(0)=A$. By definition, as $\gamma(t)\gamma(t)^\dagger=I_n$, and so differentiating at $t=0$ we get
$$
0=\frac{d}{dt}\bigg|_{t=0}\gamma(t)\gamma(t)^\dagger=\gamma'(0)\gamma(0)^\dagger+\gamma(0)\gamma'(0)^\dagger=A+A^\dagger.
$$
You can use this idea to argue that $\mathfrak{u}(n)$ consists of the set of $n\times n$ skew-Hermitian matrices. I.e.
$$
\mathfrak{u}(n)=\{A\in \mathfrak{gl}(n,\Bbb{C}):A=-A^{\dagger}\}.
$$
Next, you need to come up with a basis for this space and hence count its dimension. For $n=2$, a basis over $\Bbb{R}$ is given by
$$
\begin{bmatrix}
i&0\\
0&0
\end{bmatrix},
\:\:
\begin{bmatrix}
0&0\\
0&i
\end{bmatrix},\:\:
\begin{bmatrix}
0&1\\
-1&0
\end{bmatrix},\:\:
\begin{bmatrix}
0&i\\
i&0
\end{bmatrix}.
$$
I leave the generalization of this to you.