问:
An interesting problem about functions in set theory
I am working on the following problem:

First prove the following proposition: let $\mathscr{C}$ be a set of functions which has the following property:
\begin{equation*}
\left(\forall f, g \in \mathscr{C}\right)\left(f \subseteq g \vee g \subseteq f\right),
\end{equation*}
then $\bigcup{\mathscr{C}}$ is a function. Then describe the domain of $\bigcup{\mathscr{C}}$ with respect to functions in $\mathscr{C}$.

I have proved that $\bigcup\mathscr{C}$ is indeed a function. The proof goes as follows.

To prove that $\bigcup{\mathscr{C}}$ is a function, we have to show that for arbitrary $a, b, c $, if $\langle a,b \rangle \in \bigcup{\mathscr{C}}$ and $\langle a,c \rangle \in \bigcup{\mathscr{C}}$, then $b = c$. Assume that $z \in \bigcup{\mathscr{C}}$, then there has to exist a $y \in \mathscr{C}$ such that $z \in y$. As it is known that $\mathscr{C}$ is a set of functions, we can see that $y$ is a function and $z = \langle a,b\rangle \in y$. Next we assume that $\langle a,b \rangle \in \bigcup{\mathscr{C}}$ and $\langle a,c \rangle \in \bigcup{\mathscr{C}}$, and $b \neq c$. As a result, there have to exist two different functions $f,g \in \mathscr{C}$ such that $\langle a,b \rangle \in f$ and $\langle a,c \rangle \in g$ while $\langle a,b \rangle \not\in g$ and $\langle a,c \rangle \not\in f$. On the other hand, we know $f \subseteq g$ or $g \subseteq f$. If $f \subseteq g$, then we require $\langle a,b \rangle \in g$, which leads to contradiction. Similarly, if $g \subseteq f$, then we require $\langle a,c \rangle \in f$, which also leads to a contradiction. As a result, the assumption that $\langle a,b \rangle \in \bigcup{\mathscr{C}}$ and $\langle a,c \rangle \in \bigcup{\mathscr{C}}$, and $b \neq c$ has to be false. We may conclude that $\langle a,b \rangle \in \bigcup{\mathscr{C}}$ and $\langle a,c \rangle \in \bigcup{\mathscr{C}}$ lead to $b = c$, and $\bigcup{\mathscr{C}}$ has to be a function.

However, I am stuck on describing the domain of $\bigcup{\mathscr{C}}$. It seems that the domain of $\bigcup{\mathscr{C}}$ is related to the "largest" element in $\mathscr{C}$ in terms of $\subseteq$. I cannot find a precise description. Could anyone help me verify my proof and try to propose a description?

答:
Your proof is correct but more complicated than necessary. Suppose that $\langle a,b\rangle,\langle a,c\rangle\in\bigcup\mathscr{C}$. Then there are $f,g\in\mathscr{C}$ such that $\langle a,b\rangle\in f$ and $\langle a,c\rangle\in g$. Without loss of generality we may assume that $f\subseteq g$, so $\langle a,b\rangle\in g$, and since $g$ is a function, it follows that $b=c$ and hence that $\bigcup\mathscr{C}$ is a function.

The domain of $\bigcup\mathscr{C}$ is just what you’d expect:

$$\operatorname{dom}\bigcup\mathscr{C}=\bigcup\left\{\operatorname{dom}f:f\in\mathscr{C}\right\}\,.$$

(The proof is very straightforward.) If $\mathscr{C}$ has a largest element, $\bigcup\mathscr{C}$ will be equal to that element and will of course have the same domain, but there is no reason to expect that $\mathscr{C}$ will have a largest element. The union, however, is the least upper bound of $\mathscr{C}$, and it does the trick.


问:
Covariance between coordinate x and distance in a circle.
Let $\Omega$ be a circle of radius $r$ and let $Z$ be the random variable that measures the distance from the center of the circle to a point. Let $X$ be the random variable that given $(x,y)$ associates the value of $x$. Calculate $cov(X,Z)$.

I'm having trouble thinking in a way to relate $X$ and $Z$ in order to find $E[XY]$. What limits should I consider when integrating? What I have done so far is define $p_{X,Y}(x,y)$ as:
$$
p_{X,Y}(x,y) = \frac{1}{\pi r^2},\;\text{for } x^2+y^2 ≤ r^2
$$
and
$$
p_{X,Y}(x,y) = 0\;\text{otherwise}
$$

Because I need to find $E[X]$, $E[Z]$ and $E[XZ]$ I defined first $p_{Z}$ as:
$$
p_{Z}(t) = \frac{2x}{r^2},\;\text{for } 0<t≤r
$$
Which resulted in: $E[Z]=\frac{2}{3}r$.

Then with $p_{X,Y}$ I integrated with the limits where $x$ is inside the circle to find $p_{X}$ as follows:
$$
p_X(x) = \int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}p_{X,Y}dy = \int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}\frac{1}{\pi r^2}dy = \frac{1}{\pi r^2}\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}dy = \frac{1}{\pi r^2}\left(\sqrt{r^2-x^2} + \sqrt{r^2-x^2}\right)= \frac{2}{\pi r^2}\sqrt{r^2-x^2}
$$

This left me with:
$$
E[X] = \int_{-r}^{r}xp_{X}(x)dx = \int_{-r}^{r}x\cdot\frac{2}{\pi r^2}\sqrt{r^2-x^2}dx = 0
$$

My main problem comes when finding $E[XZ]$. How should I define $p_{X,Z}$? I think I should be doing the double integral of $p_{X,Z}$, but how? I guess it is with establishing it as follows:
$$
\int_{-r}^{r}\int_{\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}xtp_{X,Z}dxdt
$$
I don't know if that's the way to go, because I've seen the limits of $p_{Z}$ defined from $0$ to $r$ instead of $-r$ to $r$. Any help in defining $p_{X,Z}$, the limits of these integrals or in telling me what should I do is appreciated.

答:
Here is a nicely written solution for your own reference.

Let $\Sigma$ be a disc of radius $R$ centered about the origin, and suppose $(X,Y)\sim \mathcal{U}(\Sigma)$. The density of $(X,Y)$, namely $f_{XY}$, has formula $f_{XY}(x,y)=\frac{1}{\pi R^2}$ for $(x,y)\in \Sigma$ and $f_{XY}(x,y)=0$ elsewhere. You're trying to compute $$\text{cov}(X,Z)=\mathbb{E}\Big(X\sqrt{X^2+Y^2}\Big)-\mathbb{E}(X)\mathbb{E}\Big(\sqrt{X^2+Y^2}\Big)$$ From the law of the unconscious statistician, $$\mathbb{E}\Big(X\sqrt{X^2+Y^2}\Big)=\int_{-R}^R \int _{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}}x\sqrt{x^2+y^2}f_{XY}(x,y)dydx=0$$ $$\mathbb{E}\big(X\big)=\int_{-R}^R \int _{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}}xf_{XY}(x,y)dydx=0$$ $$\mathbb{E}\Big(\sqrt{X^2+Y^2}\Big)=\int_{-R}^R \int _{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}}\sqrt{x^2+y^2}f_{XY}(x,y)dydx=\frac{2R}{3}$$ Putting this all together we get $$\text{cov}(X,Z)=0$$

ContentLicense: CC BY-SA 4.0


问:
Is this enough info to solve this time dilation problem
There are two clocks.  One is a regular clock measuring regular time $\tau$.  The other is a clock measuring time $t$ which also advances clockwise, but does not advance uniformly--it accelerates w.r.t. $\tau$ from 12 to 6 o'clock, and decelerates from 6 to 12, such that the two clocks always tell the same time at 12 and 6.  It is given that the alternating acceleration and deceleration of $t$ w.r.t. $\tau$ is uniform--$\frac{d^2t}{d\tau^2}=K$ where $K$ is a constant.  Is this enough info to determine the "dilation" $\frac{dt}{d\tau}$?  If not, what other info is needed?</p>

答:
This seems to be a straightforward application of integration... Not sure what this has to do with "time dilation" since there's nothing about traveling at relativistic speeds.

You have: $\frac{d^2t}{d\tau^2}=K$ and you wish to find $\frac{dt}{d\tau}$. Consider the situation piecewise and integrate: from 12 to 6, $\int{\frac{d^2t}{d\tau^2}d\tau}=\frac{dt}{d\tau}=K\tau+C$. You know that at t=12, $\tau=12$. For convenience, set 12 o'clock = 0. So $\frac{dt}{d\tau}=K\tau$. Be careful: there is a discontinuity at 6 o'clock (as measured by $\tau$), and K changes to -K so your $\frac{dt}{d\tau}=-K\tau$ from 6 to 12.

It is not possible for both clocks to read the same time at 6 under the given conditions since $\frac{dt}{d\tau}=K\tau$ implies t will be ahead of $\tau$ for the entire interval from 12 to 6.

ContentLicense: CC BY-SA 3.0


问:
Find $z$ such that $|z-w| - |z-w^2|$ is maximal where $z$ lies on a given ellipse

Find $z$ such that $|z-w| - |z-w^2|$ is maximal where $z$ lies on the curve $|z-4| + |z+4| = 10$, where $w$ is a non real cube root of unity.

I found the locus of the $z$ from the given equation, which is an ellipse but am unable to proceed after it.

The locus that I found was $ \frac{x^2}{25} + \frac{y^2}{9} = 1$ where I assumed $z = x + iy$ .

I am a class 12 student who came across this problem in one of the online exams that I was supposed to give for my coaching.

答:
The points that maximize the difference of the distance will lie, by triangle inequality, on the line joining the two points i.e on $x =-\dfrac{1}{2}$. This line intersects the ellipse at $\left(-\dfrac{1}{2}, \dfrac{9\sqrt{11}}{10} \right)$ and at  $\left(-\dfrac{1}{2}, -\dfrac{9\sqrt{11}}{10} \right)$

From this we readily obtain the difference of distances as $\sqrt 3$


问:
Show that $f$ is measurable with respect to the product measure $c \times c$.
I am really stuck on this question.  I am new in real analysis but interested. yet, I am not familiar with this.

Let $(X, \mathcal{A},\mu) = (Y, \mathcal{B}, v) = (\mathbb{N}, \mathcal{M}, c)$.  Here, $\mathbb{N}$ is the set of natural numbers, $\mathcal{M} = 2^{\mathbb{N}}$, and $c$ is the counting measures defined by setting $c(E)$ equal to the number of points in $E$ if $E$ s finite and $\infty$ if $E$ is infinite set.  Define $\displaystyle{f: \mathbb{N} \times \mathbb{N} \implies \mathbb{R}}$ by setting

$f(x,y) = 2-2^{-x}$ if $x = y$, or $f(x,y) = -2 + 2^{-x}$ if $x = y + 1$ or $f(x,y) = 0$ otherwise

Show that $f$ is measurable with respect to the product measure $c \times c$.

How to do this? I don't know how but can you help me through this?

答: Every function from $\mathbb N \times \mathbb N$ to $\mathbb R$ is measurable w.r.t. the product sigma algebra. Hence, it is measurable w.r.t. any measure on the product space. [ Measurable w.r.t. to a measure means it is measurable w.r.t the completion of the sigma algebra on $\mathbb N \times \mathbb N$ w.r.t. the measure, so if a function is already measurable w.r.t the product sigma algebra it is also measurble w.r.t  any measure].


问:
completeness of metric space $C_b(X,Y)$

Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces. Given $\emptyset \neq B\subseteq Y,$ let $\mathrm{diam}_Y(B) = \sup_{y_1, y_2\in B} d_Y(y_1, y_2).$ Let $C_b(X,Y) = \{f\in Y^X : \text{ f is continuous and } \mathrm{diam}_Y (f(X)) < \infty\}$ where $f(X) $ is the range of $f.$

Show that $d_C : C_b(X,Y) \times C_b(X,Y) \to [0,\infty)$ given by $d_C(f,g) = \sup_{x\in X}d_Y (f(x), g(x))$ defines a metric on $C_b(X,Y).$

Show that if $(Y,d_Y)$ is complete, then $(C_b(X,Y) , d_C)$ is also complete.

$1$ seems fairly straightforward; most of the properties of a metric (e.g. positive definiteness, symmetry, etc.) are fairly easy to verify. Also, the triangle inequality follows from properties of the supremum.

But I'm not sure how to justify why $d_C(f,g) < \infty$ for all $f,g$. I tried to come up with a contradiction, but it seems that that method isn't working too well.

For $2,$ I think the proof is similar to the one that $C([0,1])$ is a Banach space. So I need to show that every Cauchy sequence $(f_n)\subseteq C_b(X,Y)$ converges to some $f\in C_b(X,Y)$. Let $\epsilon>0.$ It'll be useful to use the fact that $(f_n(x))$ is Cauchy for each $x.$ So since $Y$ is complete, we can define $f(x):= \lim_{n} f_n(x)$ for each $x$. By the Cauchyness of $(f_n),\exists N\in\mathbb{N}$ so that $n,m\geq N$ implies $d_C(f_n, f_m) < \frac{\epsilon}3$. Then for $n\geq N, d_Y(f(x), f_n(x)) = \lim_{m\to \infty}d_Y(f_m(x), f_n(x)) \leq \lim_{m\to\infty} d_C(f_m, f_n) \leq \frac{\epsilon}3 <\frac{\epsilon}2 $.

Hence $d_C(f, f_n) \leq \frac{\epsilon}2 < \epsilon$ for $n\geq N$ so $f_n \to f.$ Also, $\mathrm{diam}_Y(f(X)) < \infty$ as $\sup_{x,y\in X}d_Y(f(x), f(y)) \leq \sup_{x\in X} d_Y(f_N(x), f(x)) + \sup_{x,y \in X}d_Y(f_N(x), f_N(y)) + \sup_{y\in X} d_Y(f_N(y), f(y)) = 2d_C(f_N, f) + \mathrm{diam}_Y(f_N(X)) < \infty$ and $f$ can be shown to be continuous as $f_n$ converges uniformly to $f$. Indeed if $(a_n)\subseteq Y, a_n\to x \in Y, $ since $f_n\to f,$ there exists $N_2\in \mathbb{N}$ so that $n\geq N_2$ implies $d_C(f_n, f) < \epsilon/3.$ Also, $\exists N_3\in\mathbb{N}$ so that $n\geq N_3\Rightarrow d_Y(a_n, x) < \epsilon/3.$ Then for $n\geq N, d_Y(f(a_n), f(x)) \leq d_Y(f(a_n), f_{N_2}(a_n)) + d_Y(f_{N_2}(a_n), f_{N_2}(a)) + d_Y(f_{N_2}(a), f_n(a)) < 2d_C(f,f_{N_2}) + d_Y(f_{N_2}(a_n), f_{N_2}(a)) < \epsilon.$ So $f$ is continuous.

答:
For 1) fix any point $x_0 \in X$.We have $$d_Y(f(x),g(x))$$ $$ \leq d_Y(f(x),f(x_0))+d_Y(f(x_0),g(x_0))+d_Y(g(x_0), g(x))$$ $$ \leq diam (f(X))+d_Y(f(x_0,g(x_0))+diam (g(X))<\infty$$.

Your argument for 2) looks OK.

Title: completeness of metric space $C_b(X,Y)$


问:
If $X$ is locally path connected then the components and the path components are the same
Hi i am reading Topology by Munkres and there in theorem 25.5 i have two doubts.In the proof it says that

Since $P$ is connected, $P\subset C$.

Then at another line it says

each of them necessarily lies in $C$.

My doubts are as following:

How does P is connected imply $P\subset C$?

How does each of them lies necessarily in $C$?

Why are we taking the union of all path components of $X$ that are different from $P$ and intersect $C$?

For reference i am attaching the screenshot of the theorem's proof where i have highlighted the part where the above quoted statements are given.

答:
Every path-connected set is connected. $P$ is the path component containing $x$, so $P$ is a connected set containing $x$. $C$, the component containing $x$, is the union of all connected sets containing $x$, so $P\subseteq C$.

Suppose that $P_0$ is a path component of $X$ that intersects $C$. Then $C\cup P_0$ is connected, so it is a connected set containing $x$ and is therefore contained in $C$, the largest connected set containing $x$. That is, $P_0\subseteq C$ for each path component of $X$ that intersects $C$, and $Q$ is the union of some of those path components, so $Q\subseteq C$.

The answer to your third question is implicit in the final paragraph of the proof. We’ve supposed that $P\subsetneqq C$ in order to get a contradiction. Showing that $C$ is not connected would give us a contradiction, since by definition the component $C$ is connected. And it turns out that $P$ and $Q$ give us a separation of $C$ and therefore give us the desired contradiction.

It may help if I expand that last paragraph a bit. Every point of $C$ lies in a unique path component of $X$ that intersects $C$. Either that path component is $P$, or it is a path component of $X$ that intersects $C$ but is not $P$. Thus, every point of $C$ is either in $P$ or in $Q$, and we’ve just seen that $P\cup Q\subseteq C$, so in fact $P\cup Q=C$, and of course $P\cap Q=\varnothing$. The final step is to use the hypothesis that $X$ is locally path-connected, which implies that each path component is open. In particular, $P$ is open, and $Q$, which is a union of path components, is also open. Thus, $C$ is the union of the disjoint, non-empty open sets $P$ and $Q$ and is therefore not connected — and there’s the desired contradiction showing that $P$ cannot be a proper subset of $C$ after all: it must be all of $C$.


问:
Prove the locus of $M$ is a conic passing though $I$ and $J$.
Problem: In the perspective plane $\mathbb{P}^2$, given a conic $(S)$ and a line $d$ having no intersection with $(S)$. Fixed two points $I$ and $J$ on $d$. Take a variety point $M$ such that the tangents from $M$ to $(S)$ intersect $d$ at two points $R$ and $S$ such that $(IJRS)=-1$. Prove the locus of $M$ is a conic passing through $I$ and $J$.

Here is what I've discovered so far.

Let the tangents from $I$ and $J$ touch $(S)$ at $I_1,I_2$ and $J_1,J_2$ respectively. Then the locus of $M$ is the conic passing through $I,I_1,I_2,J,J_1,J_2$. I also proved those 6 points lie on a conic (let's call it $(S')$).

Take any point on $(S')$, call $A$. Then the intersection $A_1, A_2$ of two tangents from $A$ to $(S)$ and $d$ satisfy $(IJ,A_1A_2)=-1$. Now I'm stucking at proving this claim.

2'. Something that might be useful for proving 2: Let $P$ the polar of $d$ wrt $(S)$. Then $PI$ and $PJ$ tangent to $(S')$.

Can I have some hints for the next steps? Any ideas would be appreciated. Thanks for reading.

答:
This locus is discussed in Milne, Cross Ratio Geometry, Article 185.

There, $P,A,B$ correspond to OP's $M,I,J$ and the condition that the lines $PA,PB$ are conjugate is equivalent to OP's $R,S$ being harmonic with $I,J$.

The proof relies on the locus conic being defined by the intersections of two homographic pencils centered at $A$ and $B$.

The excerpt is below.  But the full reference is given above if background and definitions need to be consulted.

I've added a discussion of the claim "range ($c$) = range ($d$)" in a separate answer.


问:
Book Recommendation for mathematical finance
Does anyone know a book which covers topics on:

Brownian Motion

Martingales

Stochastic Calculus

Stochastic Differential Equations

Options pricing. Black-Scholes model

Fundamental Theorems. Interest Rates

Random Walk

Applications in Insurance

Simulations. Convergence

Simulation methods

I would like something in-depth, but at an undergraduate level.

答:
This is answered on the Quantitative Finance S.E.

https://quant.stackexchange.com/q/38862

https://quant.stackexchange.com/q/2019

https://quant.stackexchange.com/q/15013

https://quant.stackexchange.com/q/1

https://quant.stackexchange.com/q/2391


问:
Compute $\int \frac{x+3}{\sqrt{x^3+1}} dx$
Compute  $\int \frac{x+3}{\sqrt{x^3+1}} dx$

Was thinking about the taylor expansion of $1/\sqrt{x^3+1}$ but is also rather hairy.  Any other tricks?

答:
Using series is not bad since
$$\frac{1}{\sqrt{x^3+1}}=\frac 1{\sqrt{\pi } }\sum_{n=0}^\infty (-1)^n\frac{ \left(n-\frac{1}{2}\right)! }{ n!} x^{3n}$$
So
$$\frac{x+3}{\sqrt{x^3+1}}=\frac 1{\sqrt{\pi } }\sum_{n=0}^\infty (-1)^n\frac{ \left(n-\frac{1}{2}\right)! }{ n!} x^{3n+1}+\frac 3{\sqrt{\pi } }\sum_{n=0}^\infty (-1)^n\frac{ \left(n-\frac{1}{2}\right)! }{ n!} x^{3n}$$
Just integrate termwise and this gives you the expansion of some nasty elliptic integral.


问:
Will $y=x$ and $y=\sin^n \left(\frac{\pi x}{2} \right)$ meet in the interval $(0,1)$ for all $n \ge 2$?
I am currently learning the basics of real analysis, and still haven't learned the concepts and properties of pointwise convergence and uniform convergence of functions.

Still, I was thinking of a function $f_n : (0,1) \to \mathbb{R}$,
$$ f_n(x) = \sin^n \left(\frac{\pi x}{2} \right) $$
and if we took the $n$ to infinity, it would give
$$ \lim_{n \to \infty} f_n(x) = 0 $$

From that, I thought that there might exist some $f_n$ that $y=f_n(x)$ does not meet $y=x$.

I thought it this way since for every $x \in (0,1)$, $\lim_{n \to \infty}f_n (x) = 0$, I could use $\epsilon - N$. To be specific, for given $x \in (0,1)$, I have taken $\epsilon = x$, then exists a natural number $N$ such that the following holds:
$$ n \ge N \Rightarrow |f_n(x)| < x $$

I thought that this means that for that point $x$, $(x,f_n(x))$ is below $(x,x)$, and hence will not meet $y=x$.

Since I can take such $N$ for every point in $(0,1)$, I thought that $y=f_n(x)$ and $y=x$ will not meet for some large $N$.

But my instructor told me it will always meet since the function is pointwise convergent, but not uniformly convergent. I do not have enough knowledge of these concepts, so I am asking help for them.

My own guess to the failure of my conjecture(?) is that taking the large $N$ is impossible.
To take $N$, it has to be bigger or equal to all the other $N$ taken from each $x \in (0,1)$. In other words,
$$ N := \sup\{N_x : x \in (0,1)\}$$
where $N_x$ is the $N$ for each $x$.
I thought that the set is not upper-bounded hence unable to take $N$.

Thank you in advance for your considerate help.

Body:
If $x_n=1-\frac1{2n}$, then

$$f_n(x_n)=\cos^n\left(\frac\pi{4n}\right)\\
=\left(1-\frac{\pi^2}{32n^2}+O(n^{-4})\right)^n\\
=1-\frac{\pi^2}{32n}+O(n^{-2})\gt x_n$$


问:
Greatest Integer Function and graphing
Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ and let $n^3 + 3n = 6 (\lfloor n \rfloor ^2) + \frac {478}{27}$, then solve for $n$?
Using plot function in wolfarm gives me different solution when I try to solve analytically. Also when I try to impose condition that fractional part is less than 1 but greater than equal to zero, the equation turns out to be unwieldy for me. What is the right way to tackle this problem?

答:
Let
$$f(x) = x^3 - 6 \lfloor x \rfloor^2 + 3x - 478/27 \tag{A}$$

Then we are trying to solve $f(x) = 0$. Suppose that $x = n + \delta$, where $n \in \mathbb Z$ and $\delta \in [0,1)$.

Note that, for $n \le x < n+1$, $\frac{d}{dx} \lfloor x \rfloor = 0$.

For $x \in [n, n+1)$, $f'(x) = 3x^2+3 > 0$.

On each interval, $x \in [n, n+1)$, f(x) is increasing.

$f(6) = \dfrac{8}{27}$.

For all $x \ge 6$, $f(x) > 0$.

$f(3) = -\frac{964}{27}$ and $f(3.99)=3.7875$

For all $x \le 3$, $f(x) < 0$.

It follows that $f(x) = 0$ implies that $3 < x < 6$.

$x=3+\delta$

$\qquad f(x) = \delta^3 + 9\delta^2 + 30\delta - 964/27$

$\qquad f(x) = 0 \implies x \approx 3.91403$

$x=4+\delta$

$\qquad f(x)= \delta^3 + 12\delta^2 + 51\delta - \dfrac{101}{27}$

$\qquad f(x) = 0 \implies x \approx 4.072117 $

$x=5+\delta$

$\qquad f(x) = \delta^3 + 15\delta^2 + 78\delta - 748/27$

$\qquad f(x) = 0 \implies x = \dfrac{16}{3}$

Title: Greatest Integer Function and graphing


问:
Right-angled triangle on a cylinder
I have a question about trigonometry on the surface of a cylinder, and whether it's different from regular trigonometry on a plane.

Suppose I have a straight line of length $L$, and I tilt it by an angle $\alpha$. Then the horizontal and vertical projections of this tilted line are given by $L \cos \alpha$ and $L \sin \alpha$ resp.

Now, consider the equivalent problem on the surface of a right circular cylinder of radius $R$. Then the straight line is an arc of radius $R$. If I now tilt this arc by an angle $\alpha$, so that the arc still lies on the cylinder, what are the horizontal and vertical projections in the cylinder?

I tried doing this by unwrapping the cylinder onto a plane. But I got the same values as before: $L \cos \alpha$ and $L \sin \alpha$. Does this make sense? I would expect it to be different.

答:
I realised in hindsight that I should not expect the two answers to be different. The 'rolling' map that takes the plane to the cylinder, will also roll up the flat line and the tilted line. Moreover, the map is isometric and conformal, thus the angle between the two lines will remain the same.

I hope this question is useful to someone else in the future.

Title: Right-angled triangle on a cylinder


问:
Is provability predicate a truth predicate?
Let T be a consistent, axiomatizable theory containing Q. Because the set of provable sentences of T is semi-recursive, we have a provability predicate of the form $\exists yPrf(x,y)$ with $Prf(x,y)$  rudimentary. This predicate is correct under the standard interpretation of arithmetic if and only if the sentence represented by $x$ is provable. Because this predicate is of the $\exists$-rudimentary form, it is correct if and only if it is provable from T, which is to say $T\vdash A$ if and only if  $T\vdash\exists yPrf(x,y)$, or equivalently $T\vdash A\leftrightarrow\exists yPrf(x,y)$, with x is the Godel number of the sentence A, for all A. This property of the provability predicate satisfies the condition of being a truth predicate. However, by the diagonal lemma, we can prove that there is no truth predicate. Could you help me explain why provability predicate is not a truth predicate? Thank you a lot!

答:
Your "or equivalently" claim is false: in general "$(T\vdash\varphi)\leftrightarrow(T\vdash\psi)$" does not imply "$T\vdash (\varphi\leftrightarrow\psi)$."

For example: under the usual assumption that $\mathsf{PA}$ is $\Sigma_1$-sound (this is stronger than mere consistency) we have $\mathsf{PA}\not\vdash 0=1$ and $\mathsf{PA}\not\vdash \neg Con(\mathsf{PA})$, so the statement "$(\mathsf{PA}\vdash 0=1)\leftrightarrow (\mathsf{PA}\vdash \neg Con(\mathsf{PA})$" is correct, but it is not the case that  $\mathsf{PA}\vdash (0=1\leftrightarrow \neg Con(\mathsf{PA}))$ since that would contradict the second incompleteness theorem.


答:
First, it is not necessarily the case that $T\vdash \exists yPrf(\ulcorner A\urcorner, y)$ if and only if $T\vdash A.$ For that we need another assumption, that $T$ does not prove any false $\exists$-rudimentary sentences. That is a mild correctness assumption in the grand scheme of things, but doesn't follow from $T$ being a consistent axiomatizable extension of Q. Consider for example the extension like PA + $\lnot$Con(PA). (The other direction, that $T$ proves every correct $\exists$-rudimentary sentence, does follow, though.)

But the main issue is that it's not true that $T\vdash \varphi$ if and only if $T\vdash \psi$ means the same thing as that  $T\vdash \varphi\leftrightarrow \psi.$ (Where $(\varphi, \psi)$ ranges over some family of pairs of sentences.)

Title: Is provability predicate a truth predicate?

Body: Let T be a consistent, axiomatizable theory containing Q. Because the set of provable sentences of T is semi-recursive, we have a provability predicate of the form $\exists yPrf(x,y)$ with $Prf(x,y)$  rudimentary. This predicate is correct under the standard interpretation of arithmetic if and only if the sentence represented by $x$ is provable. Because this predicate is of the $\exists$-rudimentary form, it is correct if and only if it is provable from T, which is to say $T\vdash A$ if and only if  $T\vdash\exists yPrf(x,y)$, or equivalently $T\vdash A\leftrightarrow\exists yPrf(x,y)$, with x is the Godel number of the sentence A, for all A. This property of the provability predicate satisfies the condition of being a truth predicate. However, by the diagonal lemma, we can prove that there is no truth predicate. Could you help me explain why provability predicate is not a truth predicate? Thank you a lot!


问:
$f(x)=x-x^2$ if $x$ is rational, $x+x^2$ if $x$ is irrational

$$f(x)= \left\{\begin{array}{ll}
  x-x^2 &\mbox{if $x$ is rational,}\\
  x+x^2 &\mbox{if $x$ is irrational.}
  \end{array}\right.$$
  Show that $f'(0)=1$ and yet there is no neighborhood $I$ of the point $0$ on which this function is monotonically increasing.

答:
$f'(x) := \lim_{\Delta x \to 0} \frac{f(x+\Delta x) - f(x)}{\Delta x}$

since f(0) = 0,

$f'(0) = \lim_{\Delta x \to 0} \frac{f(\Delta x)}{\Delta x}$

Using $\epsilon - \delta$ method : above equation is equal to (substitute k to $\Delta x$) "for every $\epsilon > 0$, there exists $\delta > 0$ that

$k \in (-\delta,\delta), k \neq 0 \implies \frac{f(k)}k \in (1-\epsilon,1+\epsilon)$."

If k is rational, then $f(k) = k - k^2$, and $\frac{f(k)}k = 1-k$. Therefore, if $\delta = \epsilon$, then $\frac{f(k)}k$ is between $1+\delta$ and $1-\delta$, which is $(1-\epsilon,1+\epsilon)$.

Similar method works if k is irrational ($\delta = \epsilon$)

$\therefore$ whether k is rational or not, for every $\epsilon>0$, there exists $\delta$.

$\therefore$, $f'(0) = 1$.

PS : some error can exists in $\epsilon - \delta$ method


问:
Area of circles: represent $x$ in terms of $r_1$ and $r_2$
See the image. Area of green and red regions are equal. Can you represent $x=|O_2D|$ in terms of $r_1$ and $r_2$ for $r_1> r_2$ ?

Edit: The point $O_1$ does not enter in the region of small circle.

答:
Consider this pink rhombus:

If we split the green area into two sections (P and Q) along line M, we can find each of their areas in terms of the radius and angle of one of the circles, by subtracting the area of the triangle formed by the two radii and M from the area of the whole sector. (I'm going to use big $R$ for $r_1$ and little $r$ for $r_2$.)

$$ Area\space of\space P = \frac{\theta}{2\pi}{\pi{r^2}} - r^2\cos{\frac{\theta}{2}}\sin{\frac{\theta}{2}}   $$
$$ Area\space of\space Q = \frac{\phi}{2\pi}{\pi{R^2}} - R^2\cos{\frac{\phi}{2}}\sin{\frac{\phi}{2}}   $$

$$Area \space of\space Q + \space Area \space of \space P = \frac{\phi}{2\pi}{\pi{R^2}} - R^2\cos{\frac{\phi}{2}}\sin{\frac{\phi}{2}} + \frac{\theta}{2\pi}{\pi{r^2}} - r^2\cos{\frac{\theta}{2}}\sin{\frac{\theta}{2}} = \frac{1}{2}\pi {r^2}\tag{1}$$

That's four variables. We can relate $\theta$ to $\phi$ by expressing the length of $M$ in terms of each.
$$ 2r\sin{\frac{\theta}{2}} = M = 2R\sin{\frac{\phi}{2}}$$
$$ \phi = 2\arcsin{(\frac{r}{R}\cdot\sin{\frac{\theta}{2}})}\tag{2}$$
Pluging $(2)$ into $(1)$ should be sufficient to express $\theta$ in terms of $r$ and $R$.

Now, $x$ is $R$ less the length of the rhombus.
$$ x = R - r\cos{\frac{\theta}{2}} - R\cos{\frac{\phi}{2}} $$
and the rest is algebra.


问:
How can I proove this using integrals
I want to know how to prove this: Assume a function, $f:[0,\pi] \rightarrow \mathbb{R}$ is continuous. Then prove $$\int_{0}^{\pi}xf(\sin(x))dx=\frac{\pi}{2}\int_{0}^{\pi}f(\sin(x))dx$$ First time, I tried using the fundamental theorem and substitute $u= \pi - x$. How can I continue?

I can still use this to evalute $$\int_{0}^{\pi}\frac{x\sin(x)}{1+\cos^{2}x}dx$$Can I do that using the proof here?

答:
You were on the right track let the integral be denoted by $\mathcal{I}$ then after substituiting $u = \pi - x$ we get
$$ \begin{align*} 2\mathcal{I} &= \int_{0}^{ \displaystyle \pi} xf(\sin(x)) dx \ + \ \int_{0}^{\displaystyle \pi} (\pi -x)f(\sin(x))dx \\ \\
&= \int_{0}^{\displaystyle \pi} \pi f(\sin(x))dx   \end{align*} $$
Which gives the result

Title: How can I proove this using integrals


问:
Can I do this to retrieve the Euler numbers from the power series of $\sec(x)$
I know that we can derive the general formula series for functions like $\tan(x)$ or $\sec(x)$ elegantly using the compact sigma notation. I am still very clumsy with manipulating with these compacts so instead I just divide 1 for $\cos(x)$ (since $\sec(x)=\frac{1}{\cos(x)}$) and I obtain the following power series:

$\sec(x)=\color{green}{1}+\dfrac{\color{green}{1}x^2}{2!}+\dfrac{\color{green}{5}x^4}{24}+\dfrac{\color{green}{61}x^6}{720}+\dfrac{277x^8}{8064}+\dfrac{50521x^{10}}{3628800}...$

We know that the compact and useful general formula for $\sec(x)$ is

$$\sum_{n=0}^{\infty}\dfrac{(-1)^{n}E_{2n}x^{2n}}{(2n)!}$$

The even Euler numbers are $E_0=1$, $E_2=1$, $E_4=5$, $E_6=61$, $E_8=1385$, $E_{10}=50521$, $E_{12}=2702765...$

So the colored number matches the Euler numbers

But as you can see, due to the long division, and using symbolab to sum up terms in between during the process of long division, the fifth term $\dfrac{277}{8064}x^8$ is obscured, and it doesn't show the eighth Euler number, which is $1385$.

So, in order to improve the visibility, I reason as followed:

The Taylor formula tells us that the 8th term of this power series of $\sec(x)$ must be $\dfrac{f^{8}(0)}{(8!)}x^8$, the denominator of $\dfrac{277}{8064}x^8$ must be a number $x$ that is multiplied with $8!$ to produce 8064, in other words, $8!x=8064$. I then divide $\frac{8064}{8!}$ and obtain $\frac{1}{5}$. This means multiply the numerator with $5$, you get $277\cdot{5}=1385$, the  $E_8$.

To confirm this, I simply use the compact formula to calculate the 8th power and I get $\dfrac{1385}{40320}$, simplify both for $5$, I get the simplified version made by symbolab.

To further confirm this, I repeat the process for the $12th$ term of this series, which is $\dfrac{540553}{95800320}$, take $\frac{95800320}{12!}$, I get $5$, multiply this to the numerator, and I obtain the 12th Euler number $2702765$

My question is, is this the right way to retrieve the original numbers after simplification made by symbolab? Why does this work?

答:
You were right, to see why the method works you can compare each term with the series given by symbolab i.e.
$$ \dfrac{E_{2n}}{(2n)!} = \dfrac{a}{b} $$ now you are multiplying a number $c$ such that $bc = (2n)!$ and if you put this in the equation you can easily see that $E_{2n} = ac$


问:
The languages $A, B \in \Sigma^*$ are two semi-decidable but not decidable languages. Show that $A \cup B$ is semi-decidable.

The languages $A, B \in \Sigma^*$ are two semi-decidable but not decidable languages. Show that $A \cup B$ is semi-decidable.

If a language is semi-decidable then it is also recursively enumerable. And vice-verse. Therefore, if I can show that $A \cup B$ is recursively enumerable, then it is also semi-decidable. A recursively enumerable language must be total and computable. Here is my attempt:

We define the following sets:
$$f_A(\mathbb{N}) = \{f_A(0), f_A(1),f_A(2),...\}$$
$$f_B(\mathbb{N}) = \{f_B(0), f_B(1),f_B(2),...\}$$
$$f_{A \cup B}(\mathbb{N}) = A \cup B = ...$$

The problem here could be that the number of words in $A$ might be infinite, therefore you would never actually start enumerating $B$. The solution would thus be to alternatively enumerate elements from both sets, like the following:

$$f_{A \cup B}(\mathbb{N}) = A \cup B = \{f_A(0), f_B(0),f_A(1),f_B(1),...\}$$

A more formal definition:
$$ f_{A \cup B}(n) = \begin{cases}
          f_A\left({\frac{n}{2}}\right) & n \ \text{is even,} \\
          f_B\left({\frac{n-1}{2}}\right) & n \ \text{is uneven}
       \end{cases}
    $$

Here, $n$ is the position of the words in the set $A \cup B$, starting from $n=0$. Now we must show that $f_{A \cup B}$ is a total and computable function. The function is computable because:

We know that determining whether a number is even or odd is computable.

This means that $\frac{n}{2}$ and $\frac{n-1}{2}$ are also computable.

$A$ and $B$ are semi-decidable and therefore, $f_A$ and $f_B$ are also computable because there exists $f_A(\mathbb{N})$ and $f_B(\mathbb{N})$ that recursively enumerate $A$ and $B$.

Therefore, $f_{A \cup B} $ is computable.  The function is total because:

There exists  a lower and upper  bound.

$f_A$ and $f_B$ are total.

Therefore, we have shown that $f_{A \cup B}$ is computable and total thus is recursively enumerable. It follows from the equivalence that $f_{A \cup B}$ is semi-decidable.

I'm a little unsure if my explanation for totality is adequate enough for this. How can I improve it?

答:
I am first going to provide some definitions in the context of Turing machines, which I am going to use later.

Let us first say that the collection of strings that a Turing machine $M$ accepts is called the language of $M$, or the language recognized by $M$, and let us denote this lanugage by $L(M)$.

Def. 1: A language is called Turing-recognizable (or recursively enumerable) if some Turing machine recognizes it. We also call such a language semi-decidable.

A Turing machine has three potential outcomes: accept, reject, or loop. A loop means that our Turing machine does not halt. A Turing machine that halts on all inputs is called a decider. We say that a Turing machine, which is a decider and recognizes a given language, decides that language. Consequently, every decidable language is also semi-decidable.

Why is a semi-decidable language also called recursively enumerable in the first place? There is a special type of Turing machine called enumerator. It starts with a blank input on its tape and outputs all the strings for its language (possibly with repition). It may print an infinite list of strings, if it does not halt.

Theorem 1: A language is semi-decidable (Turing-recognizable) if and only if some enumerator enumerates it.

Def. 3: A (total) function $f: \Sigma^* \to \Sigma^*$ is computable if there is a Turing machine $M$ that halts on every input $\omega$ with $f(\omega)$ as its output on the tape.

And we say that a set enumerated by some enumerator is said to be recursively enumerable. Hence, building a computable function which recursively enumerates the given set is sufficient.

Now, the nice thing about your proof is, that your function alternates the elements of the sets $A$ and $B$. This is essentially the same as constructing a $2$-tape Turing machine simulating a Turing machine $M_1$ recognizing $L_1$ on the first tape and a Turing machine $M_2$ recognizing $L_2$ on the second tape. And since $M_1$ or $M_2$ could run forever, we must alternate this process!

To see this more clearly, I am going to provide a proof using Turing machines.

We know that a language is semi-decidable if there exists a Turing machine that recognizes it (by definition $1$). Since our theorem states that the union of two such languages is also semi-decidable, it is sufficient if we explicitly construct a Turing machine which recognizes such a union.

Theorem 2: Given two semi-decidable languages $L_1$ and $L_2$. Then $L_1 \cup L_2$ is semi-decidable.

Proof: Let $M_1, M_2$ be Turing machines recognizing $L_1$ and $L_2$, respectively. We will construct $M_{L_1 \cup L_2}$ which recognizes $L_1 \cup L_2$ using $2$-tapes as follows: copy each input $\omega$ from tape one to tape two and position both heads at the start of the tapes. Simulate $M_1$ on the first tape and $M_2$ on the second tape. Simulate the first step of $M_1$, then simulate the first step on $M_2$, and so on. If either $M_1$ or $M_2$ accepts $\omega$, we know that $\omega \in L_1 \lor \omega \in L_2$, hence we let $M_{L_1 \cup L_2}$ accept; otherwise we run forever since $M_1$ and $M_2$ do not halt, hence $\omega \notin L_1 \land \omega \notin L_2$. Hence, $M_{L_1 \cup L_2}$ recognizes $L_1 \cup L_2$, and therefore $L_1 \cup L_2$ is semi-decidable.

(Notice that you can apply nearly the same to the intersection of two semi-decidable languages.)

I like your proof, however, I think that constructing a Turing machine directly is more satisfying, since we do not need any assumptions other than the definition of "semi-decidable" and the ones given in the theorem itself. Just show that there is a Turing machine that accepts the given language, and we are done.

Your proof on the other hand needs some more (of course somewhat trivial) knowledge in the given context. The things I have also stumbeled upon are mentioned in the comments by Kolmin and Manlio. $f_A$ and $f_B$ are total and computable of course, because they recursively enumerate $A$ and $B$ by assumption, but I would state that much earlier in your proof.

You say

$A$ and $B$ are semi-decidable and therefore, $f_A$ and $f_B$ are also computable because there exists $f_A(\mathbb{N})$ and $f_B(\mathbb{N})$ that recursively enumerate $A$ and $B$.

This information is already given by assumption, so I would suggest to state these fact right at the beginning by saying something like "$A$ and $B$ are semi-decidable by assumption, hence there exist computable (total) functions $f_A$ and $f_B$ that recursively enumerate $A$ and $B$, respectively."

Saying that these enumerations could be infinite is good (that is the key idea of the proof)! Now, just state that we construct a third function using $f_A$ and $f_B$ and provide a way of dealing with this infinite case. And that is precisely what you did. As Manlio already stated, the boundedness is irrelevant. Then concluding that $f_{A \cup B}$ is total is an imediate consequence of the definition of $f_A$ and $f_B$.

Please notice that what I have said might be nitpicky since your proof is already correct, but I think that it can be improved by the small details that I've provided.


问:
Sequences $(f_{n})^{\infty}_{n=1}$ that converge in $d_{l^{\infty}}$ but not in $d_{l^{1}}$
We are working in $ X = \lbrace f : \mathbb{N} \rightarrow \mathbb{R} : \sum^{\infty}_{j = 1} |f(j)| < + \infty \rbrace $. We define $d_{l^1}(x_n, y_n) = \sum_{i \in \mathbb{N}} |x_i - y_i|$ and $d_{l^{\infty}}(x_n, y_n) = \sup_{i \in \mathbb{N}}|x_i - y_i|$.

My goal is to identify a sequence $ (f_{n})^{\infty}_{n=1} \in X$ which converges with respect to $d_{l^{\infty}}$ but doesn't converge with respect to  $d_{l^{1}}$. I also want to show that a sequence in $X$ which converges with respect to $d_{l^1}$ must also converge with respect to $d_{l^{\infty}}$.

I'm a little confused about this because my understanding is that $X$ is just the $L^p$ space with $p = 1$ and I'm struggling to understand what it means for a sequence to converge in this space under a metric other than $d_{l^1}$.

Questions: what does it mean for a sequence to converge with respect to the $d_{l^{\infty}}$ metric in the $L^1$ space? When would a sequence in the $L^1$ space converge with respect to $d_{l^{\infty}}$ but fail to converge with respect to $d_{l^1}$?

EDIT: I think I actually found a sequence that works and proved it thanks to a friendly comment. See my answer and let me know if I made any mistakes in my proof.

答:
Consider the sequence in $X$ defined by

This sequence converges to the sequence $f(j) = 0$ with respect to $d_{l^{\infty}}$. This is because all non-zero terms of $f_{n}(j)$ are equal to $\frac{1}{n} > 0$. Equivalently, $\sup_{j \in \mathbb{N}}|f_{n}(j)| = \frac{1}{n}$, and so
$$\lim_{n \rightarrow \infty} \sup_{j \in \mathbb{N}} |f_{n}(j)| = \lim_{n \rightarrow \infty} \frac{1}{n} = 0$$
Now suppose there exists a function $g \in X$ to which $f_{n}(j)$ converges with respect to $d_{l^{1}}$. Then for every $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that when $n \geq N$, $\sum_{j = 1}^{\infty}|f_{n}(j) - g(j)| < \epsilon$.

For contradiction, let $\epsilon < \sum_{j=1}^{\infty} |g(j)|$ and suppose $N$ satisfies the definition above. In the specific case where $n = N$, we have that

$$\sum_{j=1}^{\infty} |f_{n}(j) - g(j)| = |f_{N}(1) - g(1)| + \cdots + |f_{N}(N) - g(N)| + \cdots$$

$$= \Bigg|\frac{1}{N} - g(1)\Bigg| + \cdots + \Bigg|\frac{1}{N} - g(N)\Bigg| + |g(N+1)| + \cdots$$

$$= \sum_{j=1}^{N}\Bigg|\frac{1}{N} - g(j) \Bigg| + \sum_{j = N+1}^{\infty}|g(j)| < \epsilon$$

Taking the limit of both sides of the inequality as $N \rightarrow \infty$, we have that
$$\lim_{N \rightarrow \infty} \Bigg[\sum_{j=1}^{N}\Bigg|\frac{1}{N} - g(j) \Bigg| + \sum_{j = N+1}^{\infty}|g(j)|\Bigg] < \lim_{N \rightarrow \infty} \epsilon$$
$$\sum_{j=1}^{\infty}|0 - g(j)| + 0 < \epsilon$$
$$\sum_{j=1}^{\infty}|g(j)| < \epsilon$$
This is a contradiction because we chose $\epsilon < \sum_{j=1}^{\infty}|g(j)|$, and we conclude that $f_{n}(j)$ does not converge with respect to $d_{l^{1}}$.


问:
Questions about $\mathrm{Aut}(\mathbb{Q} (\sqrt{2}, \sqrt{3})/\mathbb{Q})$

Consider the extension $\mathbb{Q} \subset\mathbb{Q} (\sqrt{2}, \sqrt{3})$. How many elements are there in $\text{Aut}(\mathbb{Q} (\sqrt{2}, \sqrt{3})/\mathbb{Q})?$ Describe all elements in $\text{Aut}(\mathbb{Q} (\sqrt{2}, \sqrt{3})/\mathbb{Q})$. Find all subgroups of $\text{Aut}(\mathbb{Q} (\sqrt{2}, \sqrt{3})/\mathbb{Q})$ and their fixed fields.

I'm using Dummit & Foote Chapter 14 (Galois Theory) if it needs to be referenced. Chapter 13 was a breeze for me so I just feel like I need someone to explain how easy of a connection this is and it'll just click (or maybe it's not and I need to be told that, too). I really just need direction.

答:
Hint:

$\text{Aut}(\mathbb{Q} (\sqrt{2}, \sqrt{3}):\mathbb{Q})$ $\cong$ $\mathbb{Z_2} \times\mathbb{Z_2}$

This together with the other hints should give you the answer. You will still need to find the subgroups and their corresponding fixed fields, which is not too difficult.


问:
Prove that $\lim_{n\to\infty} \left(\frac{\sqrt{n}(1 + 8\sqrt{n})}{4n - 1}\right) =2$

Prove that $$\left(\displaystyle\frac{\sqrt{n}(1 + 8\sqrt{n})}{4n - 1}\right)_{n \in \mathbb{N}}\to 2$$

My idea was to find some $N$ to do the definition job of convergence.

\begin{align}\left |\dfrac{\sqrt{n}(1 + 8\sqrt{n})}{4n - 1}-2 \right |&=\left |\dfrac{\sqrt{n}(1 + 8\sqrt{n})-8n+2}{4n - 1} \right |\\
&=\left |\dfrac{\sqrt{n}+8n-8n+2}{4n - 1} \right |\\
&=\left |\dfrac{\sqrt{n}+2}{4n - 1} \right |\\
&\leq \left |\dfrac{\sqrt{n}+2}{n}\right |\\
&\leq \left |\dfrac{\sqrt{n}}{n}\right |+ \left |\dfrac{2}{n}\right |\\
&\leq \dfrac{1}{\sqrt{n}}+2\\
&\leq 3
\end{align}

so I concluded that the sequence is bounded, so my idea was trying to show that this sequence is monotonous, so I could conclude that it converges. Taking $n = 4k$ I get a subsequence that converges to $2$, so if I knew that the sequence is convergent, the problem would be over.

答:
What you could have also done is
$$\left(\frac{\sqrt{x}(1 + 8\sqrt{x})}{4x - 1}\right)'=-\frac{4 \left(x+4 \sqrt{x}\right)+1}{2 (4 x-1)^2 \sqrt{x}} <0$$ So, the squence is decreasing.

Now, for large values of $n$
$$\frac{\sqrt{n}(1 + 8\sqrt{n})}{4n - 1}=2+\frac 1 {4\sqrt n}+O\left(\frac{1}{n}\right)$$


问:
Understanding Fourier transform example in Matlab
I'm studying about Fourier series and transform and I get confused with the following Matlab example of Fourier transformation:

```<code>Fs = 1000;                    % Sampling frequency
T = 1/Fs;                     % Sample time
L = 1000;                     % Length of signal
t = (0:L-1)*T;                % Time vector
% Sum of a 50 Hz sinusoid and a 120 Hz sinusoid
x = 0.7*sin(2*pi*50*t) + sin(2*pi*120*t);
y = x + 2*randn(size(t));     % Sinusoids plus noise
plot(Fs*t(1:50),y(1:50))
title('Signal Corrupted with Zero-Mean Random Noise')
xlabel('time (milliseconds)')
</code>```

This gives:

I have no problem with this part, but the Fourier transform is where I get lost:

```<code>NFFT = 2^nextpow2(L); % Next power of 2 from length of y
Y = fft(y,NFFT)/L;
f = Fs/2*linspace(0,1,NFFT/2+1);

% Plot single-sided amplitude spectrum.
plot(f,2*abs(Y(1:NFFT/2+1)))
title('Single-Sided Amplitude Spectrum of y(t)')
xlabel('Frequency (Hz)')
ylabel('|Y(f)|')
</code>```

, which gives:

This is the code which confuses me:

```<code>NFFT = 2^nextpow2(L); % Next power of 2 from length of y
Y = fft(y,NFFT)/L;
f = Fs/2*linspace(0,1,NFFT/2+1);

% Plot single-sided amplitude spectrum.
plot(f,2*abs(Y(1:NFFT/2+1)))
</code>```

I don't understand all the confusing multiplications and divisions done here....can someone explain why we are for example here: plot(f,2*abs(Y(1:NFFT/2+1))) multiplying by two etc. etc. Why not just do:  plot(f, abs(Y(1:NFFT/2+1))). Why are we dividing by L in this line: Y = fft(y,NFFT)/L;. What are we doing in this line: f = Fs/2*linspace(0,1,NFFT/2+1);??

The definitions for the Matlab functions are the following:

```<code>p = nextpow2(A) returns the smallest power of two that is greater than or equal to the absolute value of A. (That is, p that satisfies 2^p >= abs(A)).

abs(X) returns an array Y such that each element of Y is the absolute value of the corresponding element of X.

Y = fft(X,n) returns the n-point DFT. fft(X) is equivalent to fft(X, n) where n is the size of X in the first nonsingleton dimension. If the length of X is less than n, X is padded with trailing zeros to length n. If the length of X is greater than n, the sequence X is truncated. When X is a matrix, the length of the columns are adjusted in the same manner.

y = linspace(a,b,n) generates a row vector y of n points linearly spaced between and including a and b. For n < 2, linspace returns b.
</code>```

In other words I would just want someone to better explain the four given code lines to me. What are we doing an why :)

Thank you for any help!

P.S.

here is the reference:

http://www.mathworks.se/help/matlab/ref/fft.html

答:
Based on Parseval's theorem

$${\displaystyle \sum _{n=0}^{N-1}|x[n]|^{2}={\frac {1}{N}}\sum _{k=0}^{N-1}|X[k]|^{2}}$$

This expresses that the energies in the time- and frequency-domain are the same. It means the magnitude $|X[k]|$ of each frequency bin $k$ is contributed by $N$ samples. In order to find out the average contribution by each sample, the magnitude is normalized as $|X[k]|/N$, which leads to

$${\displaystyle \frac{1}{N}\sum _{n=0}^{N-1}|x[n]|^{2}=\sum _{k=0}^{N-1}\left|\frac{X[k]}{N}\right|^{2}},$$

where the LHS is the power of the signal.

The multiplication by two is due to the conversion from the two-sided to single-side spectrum. However, it was wrong in the old version as it should exclude Y(1) which corresponds to the zero frequency and is fixed in the newer version as Y(2:end-1) = 2*Y(2:end-1);

The answer by OmidS was misleading as the MATLAB example is actually NOT wrong, at least for the normalization part Y = fft(y,NFFT)/L; In the newer version of Matlab (I am using 2020a), the example has been modified so that L is not equal to Fs and Y=fft(y)/L.

However, he explained the conversion from the CTFT and DTFT.


问:
Completeness of function space with uniform metric
Let $A$ be a non empty set and let $(X , d)$ be a metric space. I know that the set of all functions from $A$ into $X$, endowed with the uniform metric is complete if $(X , d)$ is complete. But what can we say about the converse, that is if $(X , d)$ is not complete then is it true that the mentioned function space is not complete. If it is true then how to prove and if not can we have a counter example. Please help me. Thanks in advance.

答:
If $\langle X,d\rangle$ is not complete, let $\langle x_n:n\in\Bbb N\rangle$ be a Cauchy sequence in $X$ that does not converge. For each $n\in\Bbb N$ let $f_n:A\to X:a\mapsto x_n$, so that $f_n$ is the constant function at $x_n$. Then $\langle f_n:n\in\Bbb N\rangle$ is a Cauchy sequence in ${^A}X$ that does not converge.

(Note that I am assuming that you’re restricting yourself to bounded functions, so that the uniform metric is defined.)


答: You cannot define the uniform metric on this space if you consider all functions from $A$ to $X$. You have to restrict yourself to bounded functions. If you do this then converse is true. Given any Cauchy sequence $(x_n)$ in $X$ which is not convergent define $f_n: A \to X$ by $f_n(a)=x_n$ and $f_n (b)= y$ for  $b \neq a$ where $a \in A$ and $y \in X$ are fixed elements. Then $(f_n)$ is Cauchy but not convergent in the uniform metric.

Title: Completeness of function space with uniform metric

Body: Let $A$ be a non empty set and let $(X , d)$ be a metric space. I know that the set of all functions from $A$ into $X$, endowed with the uniform metric is complete if $(X , d)$ is complete. But what can we say about the converse, that is if $(X , d)$ is not complete then is it true that the mentioned function space is not complete. If it is true then how to prove and if not can we have a counter example. Please help me. Thanks in advance.


问:
Equivalence of definition of definite integral with sequences and with nets
Related but not quite what I'm looking for.

I just started learning about nets and I found that the main motivation for defining nets was the need for a unified definition of the definite integral with Riemann Sums. This answer explains it very well. So, using nets we can define the definite integral of a function $f$ on $[a, b]$ as follows:

Let $D$ be the set of all tagged partitions of the interval $[a, b]$, represented as pairs $(\Delta, \xi)$ where $\Delta$ is a finite system of points $a = x_0 < x_1 < \dots < x_{n-1} < x_n = b$ and $\xi$ is a system of intermediary points $(\xi)_{i=1}^n$ such that $\xi_i \in [x_{i-1}, x_i]$.

Define a preorder relation $\geq$ on the set $D$ for $P_1 = (\Delta_1, \xi_1)$ and $P_2 = (\Delta_2, \xi_2)$ as $P_2 \geq P_1 \iff \Delta_1 \subseteq \Delta_2$. We say that $P_2$ is a refinement of $P_1$. Then, $(D, \geq)$ is a directed set (because $\Delta_1 \cup \Delta_2$ with any suitable $\xi$ is a refinement of both $P_1$ and $P_2$)

Denote the Riemann Sums as $\sigma(f, P) = \sum_{i=1}^n f(\xi_i)(x_i - x_{i-1})$

Now, we can define integration with and without nets:

With nets: The integral $\displaystyle{{\int\limits_a^b} f(x) dx = \lim_{P\in D} \sigma(f, P)}$

Without nets: The norm of a partition $(\Delta, \xi)$ is $\displaystyle{\|\Delta\| = \max_{1\leq i \leq n} (x_i - x_{i-1})}$.

The function $f$ is integrable on $[a, b] \iff \exists L =: \displaystyle{{\int\limits_a^b} f(x)dx}$ s.t. $\forall[ (P_n)_{n\in\mathbb{N}}$ sequences of partitions from $D$, with $\displaystyle{\lim_{n\to\infty }\|\Delta_n\| = 0}]$, $\displaystyle{\lim_{n\to\infty} \sigma(f, P_n) = L}$

My question is: How can we prove that the two definitions above are equivalent (/are they?), without involving a third definition, the one with upper and lower Riemann Sums, as in the related question?

The way I tried to prove it (at least the direction "with nets" $\implies$ "without nets" which I'm more interested in) was using subnets. The definition of subnet I'm using is:

Let $(x_\alpha)_{\alpha\in A}, (y_\beta)_{\beta\in B}$ be nets. We say that $y$ is a subnet of $x$ if there exists a monotone function $h : B \to A$ such that $x_{h(\beta)} = y_\beta$ and $h(B)$ is cofinal in $A$.

Using the fact that, if all subnets of a net have a subsubnet that converges to $L$, then the net converges to $L$, it follows that if all subnets of a net converge to $L$ then the net converges to $L$ (see this question for the issue with the different definition of subnet). It's also not hard to prove the other direction.

So, since the monotone sequences (with respect to the direction on $D$) in the definition "without nets" are subnets of the net in the definition "with nets", they converge. I don't know how to go about the ones that are not monotone. I also have no idea how to go about the other implication.

Any help would be much appreciated.

答:
"with sequences" ⟹ "with nets"

Prove the contrapositive by contradiction. Suppose that there is a function $f$ that is not integrable with the nets definition, but is integrable with the sequences definition with integral $L$.

So there exist an $\varepsilon>0$ such that $\forall P∈D \ \exists Q∈D$ such that $P≥Q$ and $\vertσ(f,Q)-L\vert>\varepsilon$. Let $Δ_n$ a sequence of partitions such that $\displaystyle\lim_{n\to∞}∥Δ_n∥=0$, so there exist a $E_n$, such that $Δ_n≥E_n$ and $\vertσ(f,E_n)-L\vert>\varepsilon$, for every $n$. We have that $\displaystyle\lim_{n\to∞}∥E_n∥=0$, but $\displaystyle\lim_{n\to∞}σ(f,E_n)\neq L$ a contradiction.

"with nets" ⟹ "with sequences"

Suppose that $f $ is integrable with the net definition with integral $L$.
Let  $Δ_n$ a sequence of partitons such that $\displaystyle\lim_{n\to∞}∥Δ_n∥=0$, and let $P$ a partition such that $\vertσ(f,P)-L\vert<\frac{\varepsilon}{2}$. Define $U_n=P\cup Δ_n$
.We have that
$$
\vertσ(f,Δ_n)-L\vert<\vertσ(f,U_n)-L\vert+\vertσ(f,Δ_n)-σ(f,U_n)\vert<\frac{\epsilon}{2}+\vertσ(f,Δ_n)-σ(f,U_n)\vert
$$
Because $U_n$ is a refinement of $P$.

Suppose that $P$ has $m$ points.
Note that if a partition $P$ has $m$  points more that a partition $Q$, then
$$
\vertσ(f,P)-σ(f,Q)\vert\leq2mK∥Q∥
$$
$K$ is a upper bound of $f$.

So $\vertσ(f,Δ_n)-σ(f,U_n)\vert\leq2mK∥Δ_n∥$.

Finally if we take a $p$ such that if $n>p$ implies that $∥Δ_n∥<\frac{\varepsilon }{4mK}$, then
$$
\vertσ(f,Δ_n)-L\vert\leq\vertσ(f,U_n)-L\vert+\vertσ(f,Δ_n)-σ(f,U_n)\vert<\varepsilon\\
\forall n>p
$$


问:
if a curve is asked and you are asked to find a specific value in that curve
If a curve $y = f(x)$ passes through the point $(1, –1)$ and satisfies the differential equation, $y(1 + xy) dx = x dy$, then $f(-1/2)$ is equal to:

$$
\begin{align}
(1) & &4/5\\
(2) & &-2/5\\
(3) & &-4/5\\
(4) & &2/5
\end{align}
$$

$(1)$, $(2)$, $(3)$ and $(4)$ are the options. This is a multiple choice question.

My working:

$$ -d(x/y) = xdx $$

Till here I have found out

答:
$y=ux$ and $y(1 + xy) dx = x dy \Rightarrow uxdx+ux^3dx=uxdx+x^2du$

$uxdx+u^2x^3dx=uxdx+x^2du \Rightarrow u^2x^3dx=x^2du$

$u^2x^3dx=x^2du \Rightarrow x dx=\frac{du}{u^2}$

$x dx=\frac{du}{u^2} \Rightarrow \frac{x^2}{2}+c=\frac{-1}{u}$

$\frac{x^2}{2}+c=\frac{-1}{u} \Rightarrow y=\frac{-2x}{x^2+2c}$ and according to "$f(x)$ passes through the point $(1,-1)$"

$c=\frac{1}{2}$ so

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

so $f(-1/2)=4/5$


问:
Find a complete sufficient statistic T for the interior of a triangle
This is a textbook problem from a book by Keener.

For $\theta\gt0$, let $\Delta_\theta=\{(x,y)\in\mathbb R^2:x\gt0,y\gt0,x+y\lt\theta\}$, which is the interior of a triangle. Let $(X_1,Y_1),...,(X_n,Y_n)$ be iid from the uniform distribution on $\Delta_\theta$, so their common density is $\frac 2 {\theta^2}1_{\Delta_\theta}$ (indicator divided by area of triangle). Find a complete sufficient statistic T and the UMVU estimators of $\theta$ and $\cos\theta$.

Work:

$f((X_1,Y_1),...,(X_n,Y_n)|\theta)=\displaystyle \prod\frac 2{\theta^2}\textbf 1\{X_i>0\}\textbf 1\{Y_i>0\}\textbf 1\{X_i+Y_i<\theta\}=\frac {2^n}{\theta^{2n}}\textbf 1\{X_{(1)}>0\}\textbf 1\{Y_{(1)}>0\}\textbf 1\{\max\{X_i+Y_i\}<\theta\}=\underbrace{\textbf 1\{X_{(1)}>0\}\textbf 1\{Y_{(1)}>0\}}\underbrace{\frac {2^n}{\theta^{2n}}\textbf 1\{\max\{X_i+Y_i\}<\theta\}}$ so by the factorization theorem $T=\max\{X_i+Y_i\}$ is a sufficient statistic (right?).

Now to show if it's complete, I need to show that $E(f(T))=c$ implies that $f(T)=c$ i.e. is the constant function. And I'm not sure how to do it. $E(f(T))=\displaystyle \int_0^\theta f(x)f_T(x)dx=c$ where $f(x)$ is any function and $f_T(x)$ is the pdf of $T$. What is the pdf of $\max\{X_i+Y_i\}$?

答:
So far so good.

To find the density of $\max(X+Y)$ first derive the distribution of $Z=X+Y$

$$F_Z(z)=\frac{z^2}{2}\cdot\frac{2}{\theta^2}=\frac{z^2}{\theta^2}$$

Thus

$$f_Z(z)=\frac{2 z}{\theta^2}\cdot\mathbb{1}_{[0;\theta]}(z)$$

$Z_i$ are independent thus you surely will be able to find the law of $\max(Z_i)$


问:
Give a sequence of random variables that converges to a non-constant limit in $L^2$
Intuitively, I feel like there shouldn't exist such a sequence.

We need $\mathbb{E}((X_n - X)^2) \rightarrow 0$, and if $X_n$ indeed did converge to $X$, we would probably expect $\mathbb{E}(X_n) = \mathbb{E} (X)$ as $n \rightarrow \infty$.

But then, roughly speaking, as $n \rightarrow \infty$, $\mathbb{E}((X_n - X)^2) = \mathbb{E}(X_n^2 - 2X_n \cdot X + X^2) = \mathbb{E}(X^2) - 2\mathbb{E}(X)^2 + \mathbb{E}(X^2) = 2(\mathbb{E}(X^2) - \mathbb{E}(X)^2) = 2 \ Var(X)$.

So we need X to be a random variable whose variance goes to $0$ as $n \rightarrow \infty$. But the variance of X wouldn't depend on n, so we need Var$(X) = 0$, i.e. a constant random variable.

Where did my intuition go wrong, because the question implies that such a sequence of random variables does exist, and what is an example of such a random variable?

Thanks!

答:
Let $X\sim N(0,1)$ and $X_n =X$ for all $n$. Then $X_n \to X$ in $L^{2}$ (in fact, in every sense!) and $X$ is not  a constant.

Of course, we can take $X$ to be any non-constant random variable instead of a standard normal random variable.


问:
Does every automorphism of a field of fractions restrict to the original ring?
Let $R$ be an integral domain, and let $F$ be the field of fractions of $R$.  Then my question is, does every field automorphism of $F$ restrict to a ring automorphism of $R$?  By $R$ I mean the natural embedding of $R$ in $F$.

I’m pretty sure that it’s true, but I’m not sure how to prove it.

答:
EDIT: Berci's comment beat me to it!

No, this needn't be true in general. Consider an "ambiguous" $F$ - a field such that there are two different $R_0,R_1\subseteq F$ each of which has $F$ itself as its field of fractions. If $R_0$ and $R_1$ can be "swapped" by an automorphism of $F$, this gives a counterexample.

The simplest example I can think of is $\mathbb{R}[t]$ and $\mathbb{R}[t^{-1}]$ as subdomains of $\mathbb{R}(t)$: there is an automorphism of $\mathbb{R}(t)$ which swaps $t$ and $t^{-1}$, and this clearly doesn't restrict to an automorphism of $\mathbb{R}[t]$ (or $\mathbb{R}[t^{-1}]$).


问:
Definition of convergence in $C^\infty(\Omega)$
I am not convinced or may be don't understand, the way they define convergence and then topology as a consequence of convergence.

$\Omega$ is open subset of $\Bbb R^n.$Define standard topology on $\mathcal C^\infty(\Omega)$ through the following notion of convergence:

Which says, that a sequence $h^k $, $h^k \in \mathcal C^\infty(\Omega)$ converges to $h$ in $\mathcal C^\infty(\Omega)$ if $h^k$ converges to h uniformly with all partial derivatives of any order on any compact subset of $\Omega$.

I am wondering, why did they define convergence in such a way and how does it give birth to a topology?

[In above definition they are talking about the convergence of the sequence of derivatives of a function $h$ i.e. ($h^1, h^2....\to h)$. Now, if we want to discuss the convergence of any random sequence of functions (I mean, a sequence which is not made of derivatives of a particular function), how will we consider that sequence and the convergence....?]

Then writer says:

If $K$ is a compact subset of $\Omega$, we use the standard norm i.e $||\mathcal u||_{C^k(K)}= \sum _{|\alpha|\le k} sup_{x\in K}|D^\alpha\mathcal u(x)|$

Here, I've a basic question - why does he call it a standard norm and why does he consider only this specific norm...??

I will appreciate, if you can give me few minutes out of your precious time. Thank you so much.

答:
Note: I am pretty sure this is done very well in Rudin's Functional Analysis. But I don't have it with me so I can't point to a precise page.

In a metric space $(X,d)$, the topology is characterized by convergent sequences and their limits. Indeed, a set $S$ is closed in $X$ if whenever $(s_n)$ in $S$ converges to $x\in X$, we have $x\in S$. One says that an $X$ is a sequential space. Every first countable space is sequential. Metric spaces in particular.

So you need a metric on $C^\infty(\Omega)$ for which convergent sequences obey the description above.

One way to do that is to use the family of seminorms $\|\cdot \|_{C^k(K)}$ where $K$ is compact in $\Omega$ (they have positive homogeneity and triangular inequality, but not the separation).

Actually, you need only countably many of them, and that's why the topology is metrizable. Write $\Omega=\bigcup_{n\geq 1}K_n$ with $K_n$ a nondecreasing sequence of compact sets (this is called a compact exhaustion). You can define, for instance, the metric
$$
d(u,v):=\sum_{n\geq 1}\frac{\min \left(1,\|u-v\|_{C^n(K_n)}\right)}{2^n}.
$$
Then you have to check that this is indeed a distance (that's easy) on $C^\infty(\Omega)$, and that (more tedious) $d(u_n,u)\longrightarrow 0$ if and only if $u_n$ converges to $u$ uniformly on every compact set, together with its derivatives.


问:
Is orthogonal subspace to Eigen subspace another eigensubspace?

Is Orthogonal subspace to eigen subspace another eigen subspace?

Say A is matrix. $\lambda$ is an eigen value and $U=E_{\lambda} $ is its corresponding eigen subspace. $U^{\perp}$ is another eigen subspace?

$Au = \lambda u , u \in E_{\lambda}\\ $

Now say $\vec v$ is another vector in orthogonal subspace to U. $v^Tu=0 \\ u^TAv = v^TAu = v^T(\lambda u) = 0$

It means $A \vec v$ is also perpendicular to $\vec u$.

It means $A \vec v$ lies in V(orthogonal subspace of $E_{\lambda}$). Would that mean that $A \vec v = \lambda' v$ ???? making V another eigen subspace for $\lambda'$. If V happened to be spanned by single vector then V becomes eigen subspace right? It can happen that V may be spanned by more than 1 vector. In that case Av though lies in V (orthogonal subspace), Av may not be equal to scaled version of $\vec v$(Only in scaled version of v vector right, we call that eigen subspace!?)

答: This is not true in general. Just consider a nontrivial rotation $\mathbb{R}^3$. The axis of rotation is an eigenspace with eigenvalue $1$, but the orthogonal complement is not a (real) eigenspace (there are no other real eigenvalues).


问:
If $x,y∉ℚ$ and $x^2-y^2∈ℚ_{≠0}$ then $x+y,x-y∉ℚ$?
Show that if $x$ and $y$ are irrational numbers such that $x^2-y^2$ is a non-zero rational, then $x+y$ and $x-y$ are both irrational numbers.

I know that $\mathbb{Q}$ is closed under addition and product, however I have no clue how to solve this problem. I would really appreciate help. I tried assuming that the sums are rational to reach a contradiction, but I failed.

答:
A rational number times a rational number gives a rational number and a rational number plus a rational number gives a rational number.   $\mathbb Q$ is a field.

When we multiply two irrational numbers it is possible that we get a rational number. e.g. $(\sqrt 2)(\sqrt 8) = 4.$  But, a non-zero rational number times an irrational number will always be irrational. The similar rules apply for addition.

$x^2 - y^2 = (x+y)(x-y)$

If $x^2-y^2$ is rational then either $(x+y)$ and $(x-y)$ are both rational or both irrational.

If $(x+y)$ and $(x-y)$ are both rational, then $(x+y) + (x-y) = 2x$ must be rational.  But the proposition states that $x,y$ are both irrational.  So, it can't be the case that $(x+y)$ and $(x-y)$ are both rational, thus the must be both irrational.


答:
With

$x, y \notin \Bbb Q \tag 1$

but

$0 \ne x^2 - y^2 \in \Bbb Q, \tag 2$

first note that (2) implies that

$x - y, x + y \ne 0, \tag 3$

lest

$x^2 - y^2 = (x - y)(x + y) = 0; \tag 4$

then in light of (3), we have either

$x - y, x + y \in \Bbb Q \tag 5$

or

$x - y, x + y \notin \Bbb Q, \tag 6$

for

$x + y = \dfrac{x^2 - y^2}{x - y} \tag 7$

and

$x - y = \dfrac{x^2 - y^2}{x + y}. \tag 8$

Now if (5) binds, then

$x = \dfrac{(x + y) + (x - y)}{2} \in \Bbb Q \tag 9$

and

$y = \dfrac{(x + y) - (x - y)}{2} \in \Bbb Q \tag{10}$

in contradiction to (1); therefore (6) must hold.

$OE\Delta.$

Title: If $x,y∉ℚ$ and $x^2-y^2∈ℚ_{≠0}$ then $x+y,x-y∉ℚ$?

Body:
Show that if $x$ and $y$ are irrational numbers such that $x^2-y^2$ is a non-zero rational, then $x+y$ and $x-y$ are both irrational numbers.

I know that $\mathbb{Q}$ is closed under addition and product, however I have no clue how to solve this problem. I would really appreciate help. I tried assuming that the sums are rational to reach a contradiction, but I failed.


问:
Find UMVU estimators for an exponential distribution
Let $X_1, \ldots,X_n$ be iid from an exponential distribution with density

$$
f(x)=\begin{cases}\theta e^{-\theta x},&x>0, \\ 0,&\text{otherwise}.\end{cases}$$ Find UMVU estimators for $\theta$ and $\theta^2$.

$$f(\textbf x\mid\theta)=\displaystyle \prod\theta e^{-\theta x_i}1 \{x_i>0\} = \underbrace{\theta^ne^{-\theta\sum x_i}} \underbrace{1\{x_{(1)}>0\}}$$

so $\sum x_i$ is sufficient. But I do not know how to show it is complete. If this is the right way to find UMVU estimators, could someone teach me how to show if $\sum x_i$ is complete? And if it isn't how do you find an UMVU estimator?

答:
As you surely know, Negative Exponential distribution belongs to the Exponential Family thus $S=\Sigma_i X_i$ is CSS (Complete, minimal and Sufficient Statistic).

The UMVUE of $\theta$ has been discussed many times, one of them clearly linked in the above comment, so let's see how to find the UMVUE of $g(\theta)=\theta^2$

To find the UMVUE of $g(\theta)$ you can use Lehmann Scheffé Lemma by which, to find it it is enough to find an unbiased estimator, function of a CSS.

Let's set

$$T=\widehat{\theta^2}=\frac{1}{(\sum_i X_i)^2}$$

and set $Y=\sum_i X_i$. As known,

$$Y\sim \operatorname{Gamma}(n;\theta)$$

Thus

$$\mathbb{E}[T]=\int_0^\infty \frac{1}{y^2}\frac{\theta^n}{\Gamma(n)} y^{n-1}e^{-\theta y} \, dy = \frac{\theta^2}{(n-1)(n-2)} \underbrace{ \int_0^\infty \frac{\theta^{n-2}}{\Gamma(n-2)}y^{(n-2)-1}e^{-\theta y} \, dy}_{=1}=\frac{\theta^2}{(n-1)(n-2)}$$

The estimator, function of $S,$ is BIASED but with a correctable bias...

I think you can proceed and easy conclude...


答:
No, in addition to the example GReyes gave of a $3\times 3$ matrix with only one eigenvalue, if you have a matrix with more than one eigenvalue then $U^{\perp}$ will contain all of the other eigenspaces and so will not usually itself be an eigenspace.

In fact for $U^{\perp}$ to be an eigenspace, the matrix must have precisely two eigenvalues, and be diagonalizable (ie there must exist a basis for $\mathbb{R}^{n}$ consisting entirely of eigenvectors).

Title: Is orthogonal subspace to Eigen subspace another eigensubspace?

Body:
Is Orthogonal subspace to eigen subspace another eigen subspace?

Say A is matrix. $\lambda$ is an eigen value and $U=E_{\lambda} $ is its corresponding eigen subspace. $U^{\perp}$ is another eigen subspace?

$Au = \lambda u , u \in E_{\lambda}\\ $

Now say $\vec v$ is another vector in orthogonal subspace to U. $v^Tu=0 \\ u^TAv = v^TAu = v^T(\lambda u) = 0$

It means $A \vec v$ is also perpendicular to $\vec u$.

It means $A \vec v$ lies in V(orthogonal subspace of $E_{\lambda}$). Would that mean that $A \vec v = \lambda' v$ ???? making V another eigen subspace for $\lambda'$. If V happened to be spanned by single vector then V becomes eigen subspace right? It can happen that V may be spanned by more than 1 vector. In that case Av though lies in V (orthogonal subspace), Av may not be equal to scaled version of $\vec v$(Only in scaled version of v vector right, we call that eigen subspace!?)


答:
The sequence in a previous answer is sufficient, but doesn't answer the other direction and I think the reasoning can be slightly more slick in this regard.

Note straight from the definitions that $\Vert \cdot \Vert_\infty \leq \Vert \cdot \Vert_1$, the former takes a $\sup$ over the same set of values where the latter takes a sum. This implies any convergent sequence in $\ell_1$ converges in $\ell_\infty$.

This additionally means that if $f_n$ converges to $f$ in $\ell_\infty$ and converges to $g$ in $\ell_1$ then $f=g$. That's because $\ell_\infty$ is a metric space, so has unique limits, and $\Vert f_n - g \Vert_\infty \leq \Vert f_n - g \Vert_1 \to 0$ i.e. $f_n$ also converges to $g$ in $\ell_\infty$.

Let $f_n = \chi_{\{m \leq n\}}\frac{1}{n}$ as in previously suggested answers. It converges to $0$ in $\ell_\infty$.

Assume that $f_n$ converges to $g$ in $\ell_1$. We must have $g=0$, but it is easy to check that $f_n$ does not converge to $0$ in $\ell_1$.

Title: Sequences $(f_{n})^{\infty}_{n=1}$ that converge in $d_{l^{\infty}}$ but not in $d_{l^{1}}$

Body:
We are working in $ X = \lbrace f : \mathbb{N} \rightarrow \mathbb{R} : \sum^{\infty}_{j = 1} |f(j)| < + \infty \rbrace $. We define $d_{l^1}(x_n, y_n) = \sum_{i \in \mathbb{N}} |x_i - y_i|$ and $d_{l^{\infty}}(x_n, y_n) = \sup_{i \in \mathbb{N}}|x_i - y_i|$.

My goal is to identify a sequence $ (f_{n})^{\infty}_{n=1} \in X$ which converges with respect to $d_{l^{\infty}}$ but doesn't converge with respect to  $d_{l^{1}}$. I also want to show that a sequence in $X$ which converges with respect to $d_{l^1}$ must also converge with respect to $d_{l^{\infty}}$.

I'm a little confused about this because my understanding is that $X$ is just the $L^p$ space with $p = 1$ and I'm struggling to understand what it means for a sequence to converge in this space under a metric other than $d_{l^1}$.

Questions: what does it mean for a sequence to converge with respect to the $d_{l^{\infty}}$ metric in the $L^1$ space? When would a sequence in the $L^1$ space converge with respect to $d_{l^{\infty}}$ but fail to converge with respect to $d_{l^1}$?

EDIT: I think I actually found a sequence that works and proved it thanks to a friendly comment. See my answer and let me know if I made any mistakes in my proof.


问:
Compute $\int \frac{x+3}{\sqrt{x^3+1}} dx$
Compute  $\int \frac{x+3}{\sqrt{x^3+1}} dx$

Was thinking about the taylor expansion of $1/\sqrt{x^3+1}$ but is also rather hairy.  Any other tricks?

答:
By Byrd & Friedman 239.07 and 239.00
$$I=\int_{-1}^y\frac{x+3}{\sqrt{x^3+1}}\,dx=g\left((-1-\sqrt3)F(\varphi,m)+2\sqrt3\int_0^{F(\varphi,m)}\frac1{1+\operatorname{cn}u}\,du+3F(\varphi,m)\right)$$
where the parameter $m$ is $\frac12+\frac{\sqrt3}4$, $\cos\varphi=\operatorname{cn}u=\frac{\sqrt3-1-y}{\sqrt3+1+y}$ and $g=3^{-1/4}$. B&F 341.53 solves the second integral, yielding after some simplification
$$I=g((2+\sqrt3)F(\varphi,m)-2\sqrt3E(\varphi,m))+\frac{2\sqrt{y^3+1}}{\sqrt3+1+y}$$


问:
Covering dimensions of a locally compact Hausdorff space and its compactifications
Let $X$ be a topological space. Its Lebesgue dimension / topological covering dimension $dim(X)$ is defined as the minimal number $n$ for which, for every open cover $\{U\}$ of $X$ there is an open cover $\{V\}$ that refines $\{U\}$ such that every point $x\in X$ is contained in no more than $n+1$ sets of $\{V\}$. If no such minimal $n$ exists, $X$ is said to have infinite covering dimension.

Now, let $X$ be a locally compact Hausdorff space that is not compact, $X^\star$ be its one-point compactification, and $\beta X$ be its Stone-Cech compactification. It is direct to see that when $X=\mathbb{R}^n$, $dim(X)=dim(X^\star)$.

Question: Can one comment on $dim(X)$, $dim(X^\star)$, and $dim(\beta X)$ in the general case (as in, is there any ordering between them in general?)

I think there is another question and an answer that deals with somewhat related issues (for normal spaces), but that does not answer this question directly, I think.

$X$ is not a closed subspace of $X^\star$ or of $\beta X$, as it is not compact, so most of the usual theorems about closed subspaces having lower covering dimension, do not apply here (I think).

答:
For any $T_4$ space $X$ $\dim(X)=\dim(\beta X)$ (e.g. see Engelking's dimension theory book, chapter 4; but I believe it's also in his "general topology" book), and it's clear also that $\dim(X^\ast)=\dim(X)$ for all locally compact Hausdorff spaces (addition theorems will show this e.g.)

Engelking later in chapter 4 shows also that a $T_4$ space has a compactfication of the same covering dimension such that the compactification has the same weight as well (which $\beta(X)$ will not have, in general).


问:
Prove $\frac{d}{dt} \Big|_{t=0}\mbox{tr}(e^{X+tY})=\mbox{tr}(e^XY)$
I’m asked to prove $\frac{d}{dt}\Big|_{t=0}\mbox{tr}(e^{X+tY})=\mbox{tr}(e^XY)$ for any $X,Y$ in $M_n(\mathbb{C})$. My attempt is to assume both $X$ and $Y$ are diagonalizable, and since the set of all diagonalizable matrices is dense in $M_n(\mathbb{C})$, if we can show this is true for diagonalizable matrices, then we are done. I expected this will somehow simplify the proof, but seems it does not work well unless I further assume $X,Y$ can be diagonalizable at the same time. Any suggestions on this?

答:
The trace is just the sum of diagonal values, so you can put the derivative inside the trace:
$$\frac{d}{dt}|_{t=0}tr(e^{X+tY})=tr(\frac{d}{dt}|_{t=0}e^{X+tY})=tr(e^{X+tY}Y)|_{t=0}=tr(e^XY)$$


问:
How does inclusion exclusion relate to the complements
Let $A_1,A_2,A_3,A_4$  be events such that for every $i,j,k=1,2,3,4$,

$P(A_i) = \frac{1}{2}$,

$P(A_i \cap A_j)= \frac{1}{3},\quad i\ne j$,

$P(A_i\cap A_j\cap A_k) = \frac{1}{4},\quad i\ne j, j\ne k, k\ne i$,

$P(A_1\cap A_2\cap A_3\cap A_4) = \frac{1}{5}$.

Find $P(A^c_1\cap A^c_2\cap A^c_3)$.

Progress: I have obtained $$P( A_1 \cup A_2 \cup A_3 \cup A_4) = \frac{4}{5}$$

Any clues on how I can find the intersection of the complements? Thanks!

答:
Use De Morgan's law :

$$\bigcap \overline{A_i} =  \overline{\left(\bigcup A_i \right)}$$

So
$$P(A_1^c \cap A_2^c \cap A_3^c)=P(\overline{A_1 \cup A_2 \cup A_3})$$
$$=1-P(A_1 \cup A_2 \cup A_3)$$


问:
What does $m(v\cdot d/dr)v$ means?
I dont understand the meaning of $$m\left(\vec v \frac \partial {\partial \vec r}\right)\vec v$$ What does $\dfrac \partial {\partial \vec r}$ operate on? Why is it in brackets?

答:
It seems to me like you need to take the scalar product first. Say $\vec r=(x,y,z)$ and $\vec v=(v_x,v_y,v_z)$, then
$$m(\vec v\frac{\partial}{\partial\vec r})\vec v=m(v_x\frac{\partial}{\partial x}+v_y\frac{\partial}{\partial y}+v_z\frac{\partial}{\partial z})\vec v\\=mv_x\frac{\partial\vec v}{\partial x}+mv_y\frac{\partial\vec v}{\partial y}+mv_z\frac{\partial\vec v}{\partial z}$$
Note that the final answer is a vector:
$$m(v_x\frac{\partial v_x}{\partial x}+v_y\frac{\partial v_x}{\partial y}+v_z\frac{\partial v_x}{\partial z})\hat x\\+m(v_x\frac{\partial v_y}{\partial x}+v_y\frac{\partial v_y}{\partial y}+v_z\frac{\partial v_y}{\partial z})\hat y\\+m(v_x\frac{\partial v_z}{\partial x}+v_y\frac{\partial v_z}{\partial y}+v_z\frac{\partial v_z}{\partial z})\hat z$$

Title: What does $m(v\cdot d/dr)v$ means?


问:
The Soup Problem: how to asymptotically fairly split a geometric series and a constant one using a single pattern?
Literally every time I'm serving some soup I'm thinking of this little mathematical problem I devised.

Imagine you have a very large (= infinite, for the purposes of the actual problem) bowl of soup and want to divide it into two halves using a (finite) ladle. The trouble is that some of the good stuff floated to the surface and every scoop takes a part $p$ of it, $p < \frac12$ (so its amount goes down geometrically with a quotient of $1-p$). The other good stuff is evenly dissolved in the volume. The protocol is that you always take a full scoop in this manner, so the only freedom is in what order do you empty the ladle into $A$ or $B$'s plate. What order would you take so that, asymptotically, both parties get fair shares in both respects?

Obviously the answer can't be $ABABAB\ldots$, because $A$ would always get $1/(1-p)$-times more than $B$. $\overline{ABBA}$ does not work either because $1+(1-p)^{-3}$ is always greater than $(1-p)+(1-p)^2$. In fact (except for a countable number of special values of $p$ (*)) it can't be any periodic infinite word of $A$ and $B$ for a slight variant of the same reason. One more thing I found is that the greedy algorithm (go $A$ until it has more than $B$, then swap) won't work either because while it would always satisfy splitting of the geometrical part, it would break the linear one (number of scoops). That's all I know at the moment.

Rephrased: for a given $1-p =: q \in (\frac12, 1)$, find a subset $M ∈ \mathbb{N}$ such that both

$\displaystyle\lim_{N→∞} \frac{\#\left( M \cap \{1,\ldots,N\} \right)}{N} = \frac12$,

$\displaystyle\sum_{n=1}^∞ q^n \chi_M(n) = \frac12 \sum_{n=1}^∞ q^n$.

I'm pretty sure the answer will lie somewhere in the theory of non-integer base systems or combinatorics on words or both, but I know little of those.

(*) For example, for $q = φ - 1$ the equality $q + q^2 + q^3 = 1 + q^4 + q^5$ holds, so for $p = 2 - φ \approx 0.381966$ the sequence $\overline{ABBBAA}$ would work.

答:
Proposition. Let $(a_n)$ be an absolutely convergent series of real numbers such that for any natural $k$ we have
$$|a_{2k-1}-a_{2k}|\le \sum_{n=k+1}^\infty |a_{2n-1}-a_{2n}|.\label{1}\tag{1}$$
Then there exists a subset $M$ of natural numbers such that $|M\cap \{2k-1,2k\}|=1$ for each natural $k$ and $\sum_{n\in M} a_n=\tfrac 12\sum_{n=1}^\infty a_n$.

Proof. For each natural $k$ put $b_k=|a_{2k-1}-a_{2k}|$. It suffices to show that we can consecutively choose signs for $b_k$, providing $\sum_{k=1}^\infty \pm b_k=0$. At the beginning choose a sign “$+$” for $b_1$ and for each $k>1$ choose for $b_k$ a sign “$+$”, if $\sum_{i=1}^{k-1} \pm b_i\ge 0$ and “$-$”, otherwise. It is easy to check that Condition \eqref{1} provides $\sum_{k=1}^\infty \pm b_k=0$. $\square$

Corollary. For each $q\in (1/\sqrt{2},1)$ there exists a subset $M$ of natural numbers such that $|M\cap \{2k-1,2k\}|=1$ for each natural $k$ and $\sum_{n\in M} q^n=\tfrac 12\sum_{n=1}^\infty q^n$.

Proof. It suffices to check that for any any natural $k$ we have $$q^{2k-1}-q^{2k}\le \sum_{n=k+1}^\infty q^{2n-1}-q^{2n}=\frac{q^{2k+1}}{1+q}. \square$$


问:
Eliminating Space using Mathematical or Probability Equation
Let us say that there are one transmitter and two receivers. The receivers R1 is located at $(-c,0)$ and R2 is located at $(c,0)$ while the transmitter is located at $(x,y)$ (All in Cartesian Plane). I want to eliminate some quadrants from Cartesian plane when I have the time difference of signal arrival. For example if the transmitter is located at Q1, I should be able to eliminate Q2 and Q3 as the possible location of transmitter because the time arrival of R2 is lower than R1.

I just don't know how to write it in mathematical formula or probabilistic equation. Any help please? Thank you in advance.

答: The set of points in the plane that are closer to $R_2$ than to $R_1$ is given by the condition $x > 0$; similarly, the points that are closer to $R_1$ than $R_2$ is given by $x < 0$, and the $y$-axis, or $x = 0$, is the locus of points that are equidistant from the two receivers.  Since the converse is true, based on the observed signal arrival times, you can determine whether $x$ is positive, negative, or zero; but you cannot determine the value of $y$, because there are in general two points on either side of the $x$-axis that will have the same signal arrival times.  In order to pinpoint the transmitter's location, you need a third receiver that is not collinear with the other two.


问:
Show that the solution of this (nonlinear) ODE cannot remain bounded as $t\to\infty$
Preliminary properties: Let the state vector $x(t)=[x_1(t),\dots,x_n(t)]^T\in\mathbb{R}^n$ be constrained to the dynamical system
$$
\dot{x} = Ax +
\begin{bmatrix}
\phi_1(x_1) \\
\vdots \\
\phi_n(x_1) \\
\end{bmatrix}, \ \ \ \ x(0) = x_0
$$
where $A$ is defined by:
$$
A =
\begin{bmatrix}
\lambda_1 & 1 & 0 &\cdots& 0\\
0 & \lambda_2 & 1 &\ddots&\vdots\\
\vdots&\ddots&\ddots&\ddots&0\\
0&\cdots&0&\lambda_{n-1}& 1\\
0&\cdots&0&0&\lambda_n
\end{bmatrix}
$$
with $\lambda_i>0$, and $\phi_i(x_1) = \beta_i |x_1|^{\alpha_i}\text{sign}(x_1), \beta_i>0$, $0<\alpha_i<1$.

Question: Is it possible to show that for any initial condition $x_0\neq 0$, the solution $x(t)$ either converge to the origin, or $
\lim_{t\to\infty}\|x(t)\| = +\infty
$, but cannot remain in a bounded trajectory different from staying at the origin?

Concretelly, what additional structure or conditions on the system or the initial condition do we require to show this?

In case you find this useful, here are my attempts to understand/solve the problem.

Attempt 1: I was trying to use results such as the ones from here which can conclude what I want, but require to find a Lyapunov-like function (not necesarilly positive definite) for which $\ddot{V}\neq 0, x\neq 0$. However, I haven't been able to come up with a suitable such function.

Attempt 2: The differential equation have "explicit" solution (not precisely explicit but can be expressed as)
$$
x(t) = e^{At}x_0 + e^{At}\int_0^se^{-As}\Phi(x_1(s))ds
$$
where $\Phi(x_1) = [\phi_1(x_1),\dots,\phi_n(x_1)]^T$. So I wanted to proceed by contradiction: assume that there exists $b,B>0$ and $T>0$ such that $b\leq \|x(t)\|\leq B$ for all $t\geq T$. Hence,
$$
b\leq \left\|e^{At}x_0 + e^{At}\int_0^se^{-As}\Phi(x_1(s))ds\right\|\leq B
$$
And noticing that in this case there should be $c,C>0$ such that $0<c\leq\|\Phi(x_1(t))\|\leq C $, for all $t\geq T$. Thus, try to obtain a contradiction, for example by using $C\geq\|\Phi(x_1(t))\|$ to show that $B\leq\|x(t)\|$. But unfortunately I haven't obtained anything positive in this direction neither.

Attempt 3: Can Bendixon's/Dulac criterion (see Theorem 11  here) be used to conclude something for this system? It is easy to verify that if we write this system as $\dot{x} = f(x)$, we obtain $\nabla\cdot f(x)>0$.

I know that neither my attempts nor my exposition here are perfect. However, I'm looking for suggestions/references or any idea which might help me understand more this problem.

答:
Instead of the general case let us focus on the case where $\alpha_{i} = 1$ for all i. In this special case $sign(x_{1})|x_{1}|=x_{1}$ the $\phi(x_{1})$ term can then be absorbed into $A$, yielding $A' = A+\Phi(x_{1})$. So in this special case the equation is linear homogeneous.

This may seem like an oversimplification, however there are several salient points. First, it is a good first place to look for counterexamples. Second, when $\alpha > 1$ you can see that the local structure near the zero equilibrium is dominated by $A'$ for small |x(t)|. So the behavior of a subset of solutions will always be reliant on $A'$ in that way it is always worth looking at the structure of $A'$.

Lets look at the system of two equations version just to keep the notation to a reasonable level.

$$\frac{dx_{1}}{dt} = (\lambda_{1} + \beta_{1})x_{1}(t) +x_{2}(t)$$

$$\frac{dx_{2}}{dt} = \beta_{2}x_{1}(t) + \lambda_{2}x_{2}(t)$$

Solutions of which are given by:

$x(t) = e^{A't}x_{0}$

The same reasoning works in the full system case where all $\alpha_{i}=1$

Now in the general case but with system of two equations

$$|\frac{dx_{1}}{dt}| = |(\lambda_{1}x_{1}(t) + sign(x_{1}(t))\beta_{1})|x_{1}(t)|^{\alpha_{1}} +x_{2}(t)| = \lambda_{1}|x_{1}(t)| + \beta_{1}|x_{1}(t)|^{\alpha_{1}} + |x_{2}(t)|$$

$$|\frac{dx_{2}}{dt}| = |sign(x_{1}(t))\beta_{2})|x_{1}(t)|^{\alpha_{1}} + \lambda_{2}x_{2}(t) |= \beta_{2}|x_{1}(t)|^{\alpha_{1}} + \lambda_{2}|x_{2}(t)|$$

So we get the inequality:

$$|\frac{dx_{1}}{dt}| \geq \lambda_{1}|x_{1}(t)|+ |x_{2}(t)|$$

$$|\frac{dx_{2}}{dt}| \geq \lambda_{2}|x_{2}(t)|$$

Integrating against the inequality we get that

$|x_{1}(t)| > |y_{1}(t)|$ and $|x_{2}(t)| > |y_{2}(t)|$

Where $y(t) = e^{At}x_{0}$.

Note we make heavy use of the fact that the $\lambda_{i}$ and $\beta_{i}$ are all positive in the above inequality. If you want to generalize the problem to include negative $\lambda_{i}$ and $\beta_{i}$ the structure of $A'$ will be the place to start looking for counterexamples.

I will add the assumption that $x_{0}$ is either in the positive or negative orthant for now, until I update the argument.

Ok. So let's try this approach so that we can use the comparison theorem in your textbook. Let's let $U(t) = x_{1}(t) + x_{2}(t)$

$$|\frac{d(x_{1}(t) + x_{2}(t))}{dt}| = |\lambda_{1}x_{1}(t)| + |sign(x_{1}(t))\beta_{1}|x_{1}(t)|^{\alpha_{1}}| + |x_{2}(t)| + |sign(x_{1}(t))\beta_{2})|x_{1}(t)|^{\alpha_{1}}| + |\lambda_{2}x_{2}(t)|$$

$$\geq |\lambda_{1}x_{1}(t)| + |x_{2}(t)| + |\lambda_{2}x_{2}(t)| \geq min(\lambda_{1},1+\lambda_{2}) |x_{1}(t)+x_{2}(t)|$$

So $|\frac{dU}{dt}| > min(\lambda_{1},1+\lambda_{2}) |U(t)|$.

Then you can use the comparison theorem from your book to show that:

$|U(t)| > e^{min(\lambda_{1},1+\lambda_{2})t}|x_{0}|$.

You can do something similar in the kth order case.

Sorry folks, when I read the question I just assumed we were in the positive orthant. Here is a quick counterexample.

Choose so that $\beta_{1}+\lambda_{1}=1$ and $\beta_{2} = 1$ and $\lambda_{2}=1$

The alphas can be free.

$$\frac{dx_{1}}{dt} = \lambda_{1}x_{1}(t) + sign(x_{1}(t))\beta_{1}|x_{1}(t)|^{\alpha_{1}} +x_{2}(t)$$

$$\frac{dx_{2}}{dt} = sign(x_{1}(t))|x_{1}(t)|^{\alpha_{1}} + x_{2}(t)$$

$x_{0}=(1,-1)$ is a non-zero equilibrium point.


问:
Awkward notation in Bert Mendelson's "Intro to Topology" - supremum of union of difference between functions
I'm looking at Bert Mendelson's "Introduction to Topology" (3rd edition, 1975: Dover pubs) and I am having difficulty understanding the reason for the complexity of his (horribly cumbersome) notation in exercise 5 of section 2 of chapter 2 (Metric Spaces) as follows:

"Let $X'$ be the set of all bounded functions $f:[a,b] \to \mathbb R$. For $f, g \in X'$ define:
$$d'(f,g) = \operatorname {l.u.b} \cup_{x \in [a,b]} \{|f(x) - g(x)|\}$$

Prove that $(X',d')$ is a metric space."

In this context "bounded" just means that there is a real $K$ such that $|f(x) \le K$ for all $x \in [a,b]$.

Now I glanced at this first time into this book and casually brushed it aside thinking: oh yes I've seen this before in the context of W.A. Sutherland's "Introduction to Metric and Topological Spaces" (Oxford, 1975) where we have the following definition:

"Example $2.2.8$: Let $A$ be the set of all bounded functions $f:[a,b]\to \mathbb R$. Given two points $f$ and $g$ in $A$, let
$$d(f,g) = \sup_{x \in [a,b]} |f(x) - g(x)|$$"

which is, basically, the greatest difference between $f$ and $g$ over the domain of interest $[a,b]$.

Now I'm looking back at Mendelson's definition and wondering: is there actually any logical difference between these two definitions? It seems that Mendelson is also just trying to set up a distance function which is no more complicated than "the greatest difference between $f$ and $g$ on $[a,b]$".

Is Mendelson just clumsy in his notation here, or is there a subtle reason why he needs to include that $\cup$ in his notation? I can't see what it's there for.

答:
His definition is harder to read, but his notation is correct. He has an instance of

$$\bigcup_{x\in A}C_x\,,$$

where $A=[a,b]$, and $C_x=\{|f(x)-g(x)|\}$ for each $x\in[a,b]$. By definition

$$\begin{align*}
\bigcup_{x\in A}C_x&=\bigcup\big\{C_x:x\in A\big\}\\
&=\{y:\exists x\in A\,(y\in C_x)\}\,.\tag{1}
\end{align*}$$

In his context $(1)$ becomes

$$\big\{y:\exists x\in[a,b]\,\left(y\in\{|f(x)-g(x)|\}\right)\big\}\,,$$

which is equal to

$$\{|f(x)-g(x)|:x\in[a,b]\}\,.$$


问:
Is there a procedure to solve Diophantine Equations?
How would you go about solving a multivariable, non-linear Diophantine Equation?</p>

答: No, there is no general procedure. This was Hilbert's 10th problem.</p>


问:
Bipartite graph with even degrees
Given a bipartite graph with an equal number of vertices in its parts. Each vertex in the first part is connected with an even number of vertices in the second part. Prove that we can remove some (maybe none, but NOT all) vertices from the first part so that every vertex in the second part will have an even degree?

I tried to build an eulerian subgraph of such graph using some traverses like in Kuhns algorithm but got nothing good.

答:
First of all let's say that the vertices are in two different groups $A,B$, size of each of them equal to $n$. Let's call the vertices in $B$ that have odd degree as bad vertices.  Now since the sum of degrees of the vertices in the $A$ is even, (since the degree of each of them is even) the sum of degrees of the vertices in $B$ is even too. So clearly the number of bad vertices in $B$ is always even. Now let's evaluate the number of cases in which some of the vertices in $B$ can be bad vertices. (Through this solution we'll assume that there is at least one bad vertex in $B$ in the beginning of the problem. Otherwise we don't have to remove any of the vertices.) Since the number of bad vertices is even, the number of cases is equal to the number of subsets of $\{1,2,\dotsb,n\}$ with even number of elements and chosing zero vertices is not an option since it means that there are no bad vertices which we have discussed before. So the number of cases is equal to $$\binom{n}{2}+\binom{n}{4}+\dotsb=2^{n-1}-1$$

On the other hand the number of possibilities for removing a set of vertices in $A$ is equal to $2^n-2$. (not counting the cases in which none or all of the vertices are removed) Now for the sake of the contradiction, assume that none of these $2^n-2$ choices cause $B$ to have no bad vertices. Finally we conclude that according to the pigeonhole principle one of the following cases happen:

There are three of the $2^n-2$ cases that result in the same set of bad vertices in $B$.

For each set of vertices of $B$ like $S$ with even number of vertices, there are exactly two different choices of vertices of $A$ removing which the set of bad vertices is equal to the set $S$.

Before proving the cases note that if for two different sets $S_1,S_2$ of vertices of $A$, if $\{1,2,\dotsb,n\}-S_1\not =S_2$, then by removing the set of vertices in $A$ achieved by the binary sum of $S_1,S_2$ (which is neither containing all vertices nor an empty set), all of the vertices in $B$ will have even degrees.

Now the first case is trivial: Since there are three sets $S_1,S_2,S_3$ removing which will cause in the same set of bad vertices in $B$, by what we just said and because clearly $A+B=\{1,2,\dotsb,n\}$ and $A+C=\{1,2,\dotsb,n\}$ can't hold simultaneously, the case is proven.

But in the second case, note that if $S_1,S_2$ are the two sets removing which we'll have the same set of bad vertices in $B$, then $S_1+S_2\not =\{1,2,\dotsb,n\}$. (otherwise the case is proven) And because these two sets both have even number of elements $n$ is even, too. So we only have to prove the problem for even $n$.On the other hand if $n$ is even the set $\{1,2,\dotsb,n\}$ isn't counted in the number of sets with even number of elements. That's because if removing each of the sets $S_1,S_2$ such that $S_1+S_2=\{1,2,\dotsb,n\}$ results in all of the vertices of $B$ to have odd degrees, then it means that each vertex in $B$ is connected to odd number of vertices in $S_1$ and odd number of vertices in $S_2$ and hence each of the vertices in $B$ have even degrees and therefore this is equivalent to the case we ommited in the beginning. So once again, we assume that there are $2^n-2$ choices for removing some vertices of $A$ and $2^{n-1}-2$ choices for choosing an even number of vertices from the vertices of $B$ and hence by the pigeonhole principle, there are at least $\lceil \frac{2^n-2}{2^{n-1}-2}\rceil=3$sets of vertices in $A$ by removing which the same set of bad vertices in $B$ will be made and hence the problem will be solved by what we proved in the first case and therefore the proof is complete!

Title: Bipartite graph with even degrees


问:
Fibonacci variant with small exponent
The standard Fibonacci sequence defined by $F(n) := F(n-1) + F(n-2)$ has exponential asymptotic growth as stated e.g. here (with the approximate base being the golden ratio). What happens if the recurrence equation is altered by adding an exponent to $F(n-2)$? Intuitively, instead of 'add the previous number at every step', we get 'add a constant power of the previous number at every step'. I.e.,

$$ F(n) := F(n-1) + F(n-2)^c  \;\;\;\;\;\text{ where }\;\;\;\; c \in (0,1) $$

What is the asymptotic growth of such a sequence, assuming starting values are still 0 and 1? (Or if this problem is too difficult, what is the asymptotic growth for  $c = \frac12$ specifically?)

答: The Ansatz $F(n)\sim An^p$ gives$$1\approx\tfrac{A(n-1)^p+A^c(n-2)^{cp}}{An^p}\approx1-\tfrac{p}{n}+A^{c-1}n^{-(1-c)p}-2cpA^{c-1}n^{-(1-c)p-1}.$$We want the penultimate (last) term to scale as $n^{-1}$ ($n^{-2}$), so $p=\tfrac{1}{1-c},\,A=p^{1/(c-1)}$, i.e.$$F(n)\approx(n(1-c))^{1/(1-c)}.$$The $c\to0^+$ behaviour is consistent with @Joffan's answer. A comparison with the exponential $c\to1^-$ result is achievable by noting$$\exp(\alpha n)=\lim_{c\to1}[1+(1-c)\alpha n]^{1/(1-c)}.$$If we want a more precise approximation of $F(n)$, we can define$$\begin{align}\epsilon(n)&:=F(n)(n(1-c))^{-1/(1-c)}-1\\&=[F(n-1)+F(n-2)^c](n(1-c))^{-1/(1-c)}-1\\&=(1-1/n)^{1/(1-c)}(1+\epsilon(n-1))-1\\&+(1+\epsilon(n-2)(n-2)^{1/(1-c)}(1-c)^{1/(1-c)})^c(n(1-c))^{-1/(1-c)}\\&\approx\epsilon(n-1)-\tfrac{n^{-1}}{1-c}+n^{-1/(1-c)}(1-c)^{-1/(1-c)}+c\epsilon(n-2)(1-\tfrac{2n^{-1}}{1-c})\end{align}$$through repeated use of $(1+u)^v\approx1+uv$. In particular, $0\approx\epsilon(n)\approx\epsilon(n-1)+c\epsilon(n-2)$, so $\epsilon(n)\in O((-c)^n)$. In other words, $F$ is a power law plus an exponentially decaying error whose sign alternates.

Title: Fibonacci variant with small exponent


答:
By a famous result of Matiyasevich, there is no universal algorithm which, when fed any Diophantine equation, will determine whether or not that equation has a solution in integers.

Interestingly, it is still unknown whether there is an algorithm that will always determine whether or not a Diophantine equation has a solution in rationals.

For quadratic Diophantine equations, in any number of variables, there is an algorithm. For degree $4$ equations, it is known that there is not. The question for cubic equations is unresolved.

Title: Is there a procedure to solve Diophantine Equations?

Body: How would you go about solving a multivariable, non-linear Diophantine Equation?</p>


问:
How can I show that the "binary digit maps" $b_i : [0,1) \to \{0,1\}$ are i.i.d. Bernoulli random variables?
In this post What is the Lebesgue measure of the set of numbers in $[0,1]$ that has two thirds of ones in their infinite base-2 expansion? we needed the fact that if we let $b_i (x) \in \{0,1\}$ for $x \in [0,1)$ denote the $i$-th digit in the infinite binary expansion of $x$, i.e.

$$
x = \sum_{n=1}^\infty b_n (x) \cdot 2^{-n},
$$

where $b_n (x) = 1$ for all but finitely many $n$ (except for $x=0$), then the $(b_n)_n$ form a family of independent Bernoulli random variables, if they are considered as maps defined on the probability space $([0,1), \mathcal{B}, \lambda)$, where $\mathcal{B}$ are the Borel sets and $\lambda$ denotes Lebesgue measure.

A quick search did not turn up any useful results, so my question is for a nice, more or less elementary, proof of this result, which can also be used for future reference.

I will "answer my own question", i.e. provide one possible proof. But I would also be interested in other proofs!

答:
The infinite binary expansion means the non-terminating binary expansion, i.e., when there exist two expansions for $x$, one a non-terminating expansion $0.\ldots 01111\cdots$ with $1$ trailing forever, and the equivalent expansion $0.\ldots 10000\cdots$ which terminates, choose the former. (You don't want the condition "$b_i(x)=1$ for all but finitely many $n$", which is satisfied for only a countable number of $x$.) With this in mind, the interval $(0,1]$ is a more natural domain for the $b_i$'s.

Let $e_1,e_2,\ldots,e_n\in\{0,1\}$.  The set $$\{b_1=e_1,b_2=e_2,\ldots,b_n=e_n\}:=\bigcap_{k=1}^n\{x\in(0,1]:b_k(x)=e_k\}\tag1$$ consists of all $x$ whose binary expansion begins $0.e_1e_2\cdots e_n$. This set has measure $2^{-n}$, since it coincides with the interval $(d, d+2^{-n}]$ whose left endpoint is the dyadic rational $d:=\sum_i^n e_i/2^i$. (Or prove by symmetry: all $2^n$ possible initial expansions $0.e_1\cdots e_n$ of $n$ elements from $\{0,1\}$ must have equal measure.) Hence
$$\lambda(b_1=e_1,b_2=e_2,\ldots,b_n=e_n)=\frac1{2^n}.\tag2$$ By representing the set $\{b_k=e_k\}$ as a disjoint union of sets of the form (1), conclude for each $k$
$$\lambda(b_k=e_k) = \frac12.\tag3$$ Combining (2) and (3) gives, for each $e_1,\ldots,e_n$,
$$\lambda(b_1=e_1,b_2=e_2,\ldots,b_n=e_n)=\prod_{k=1}^n \lambda(b_k=e_k).\tag4$$ So (4) is saying that $b_1,\ldots, b_n$ are independent, while (3) is saying each $b_k$ has Bernoulli($1/2$) distribution.