问:
Finding Mode from Histogram
In our High School Mathematics, we often find mode of a grouped data from a histogram. I am not talking about that process here.My question is why the process works to find the mode using the graph?</p>

答:
The answer depends on what you mean by the mode of grouped data. Sometimes
this is based on a histogram, The modal interval is defined as the one with
the tallest histogram bar. Then there are various rules for picking one point
within the modal interval to designate as the mode of the grouped data.

Often the ultimate goal of such a procedure is to estimate the mode of the
population, which is at the highest point of the density function.
For example, the mode of a normal distribution is always at the mean. Suppose
you are trying to use a histogram to find the mode of the normal distribution
from which the the data in the histogram were sampled. If you have the $n$
individual observations that were grouped to make the histogram, then
the best estimate of the mode of the normal distribution from which the
data were sampled is just the sample mean of the observations,
$\bar X = \frac 1n \sum_{i=1}^n X_i.$

Example 1: Suppose I have $n = 500$ observations from the normal distribution
$\mathsf{Norm}(\mu = 50, \sigma=7).$ Below I use R statistical software to
make such a sample, rounding the observations to 2 decimal places:

```<code>set.seed(2021)
x = round(rnorm(500, 50, 7), 2)
summary(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  31.09   44.87   50.10   50.03   54.88   71.15
</code>```

We know that the mode of the normal distribution is at $\mu=50.$
From the data summary we would estimate this by $\bar X = 50.03.$

Below is one possible histogram of the data.  (I have shown the density
function of the population in the background.)  The modal interval of
the histogram is $(50, 60).$ So any formula for finding 'the mode' from
the histogram will pick some point in that interval.

```<code>hist(x, prob=T, ylim = c(0,.06), col="skyblue2", main="n=500: NORM(50, 7)")
 curve(dnorm(x, 50, 7), add=T, lwd=2, col="orange")
</code>```

If I just have the grouped data, then the frequencies $f_j$ of the intervals
and the midpoints $m_j$ of the intervals are as shown below.

```<code>f = hist(x,plot=F)<span class="math-container">$counts;  f
[1]   4  41  82 118 131  74  42   7   1
m = hist(x,plot=F)$</span>mids;  m
[1] 32.5 37.5 42.5 47.5 52.5 57.5 62.5 67.5 72.5
</code>```

Then I can use the formula $\bar X \approx \frac 1 n \sum_{j=1}^9 f_jm_j = 50.12$
to estimate the sample mean from the interval frequencies and midpoints.

```<code>sum(f*m)/500
[1] 50.12
</code>```

I don't know what formula you are using to find the mode from the histogram, but
whatever it is, for normal data it won't estimate the mode of the population
more reliably than $\bar X.$

I have used the default histogram in R. I could have use R along with additional
instructions to make a different histogram. Then estimates using the histogram
would be similar to the above, but a little different. One possibility is shown below. Now the modal bar is in the interval $(50,52).$

```<code>hist(x, prob=T, br = 20, col="skyblue2", main="n=500: NORM(50, 7)")
 curve(dnorm(x, 50, 7), add=T, lwd=2, col="orange")
</code>```

Here, the parameter br = 20 is only a rough suggestion; R chooses 21 instead in order
to make a histogram with 'round' numbers as interval boundaries. [I could pick
my own exact interval endpoints, and R would obey, if possible.]

Example 2. Now let's look briefly at a sample of $n = 1000$ observations from a non-normal population (gamma distribution), for which the population mode $40$ is not the same as the population mean $\mu = 50.$

```<code>set.seed(2011)
y = round(rgamma(1000, 5, 1/10), 2)
summary(y)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
   7.34   32.76   46.02   49.65   63.59  199.30
hist(y, prob=T, br=15, col="skyblue2", main="n=500, GAMMA(5, .1)")
 curve(dgamma(x, 5, .1), add=T, lwd=2, col="orange")
 lines(density(y), lwd=2, lty="dotted")
</code>```

For this histogram, the modal interval is $(30,40)$ and your method
of finding the sample mode would choose a point in that interval. Knowing
the formula for this density function (orange curve), I could use calculus to find
its maximum, which is the population mode. If I have only the sample
I can estimate the mode from the histogram.

Or, ignoring the histogram, I can use more advanced methods to estimate the shape and rate parameters of the distribution and estimate the mode from them. Alternatively, for moderately large samples, a kernel density estimator (KDE) can estimate the density from the sample and we can use that to estimate
the mode. The KDE for the sample above is shown as a dotted curve.

Title: Finding Mode from Histogram

Body:
The answer depends on what you mean by the mode of grouped data. Sometimes
this is based on a histogram, The modal interval is defined as the one with
the tallest histogram bar. Then there are various rules for picking one point
within the modal interval to designate as the mode of the grouped data.

Often the ultimate goal of such a procedure is to estimate the mode of the
population, which is at the highest point of the density function.
For example, the mode of a normal distribution is always at the mean. Suppose
you are trying to use a histogram to find the mode of the normal distribution
from which the the data in the histogram were sampled. If you have the $n$
individual observations that were grouped to make the histogram, then
the best estimate of the mode of the normal distribution from which the
data were sampled is just the sample mean of the observations,
$\bar X = \frac 1n \sum_{i=1}^n X_i.$

Example 1: Suppose I have $n = 500$ observations from the normal distribution
$\mathsf{Norm}(\mu = 50, \sigma=7).$ Below I use R statistical software to
make such a sample, rounding the observations to 2 decimal places:

set.seed(2021)
x = round(rnorm(500, 50, 7), 2)
summary(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  31.09   44.87   50.10   50.03   54.88   71.15

We know that the mode of the normal distribution is at $\mu=50.$
From the data summary we would estimate this by $\bar X = 50.03.$

Below is one possible histogram of the data.  (I have shown the density
function of the population in the background.)  The modal interval of
the histogram is $(50, 60).$ So any formula for finding 'the mode' from
the histogram will pick some point in that interval.

hist(x, prob=T, ylim = c(0,.06), col="skyblue2", main="n=500: NORM(50, 7)")
 curve(dnorm(x, 50, 7), add=T, lwd=2, col="orange")

If I just have the grouped data, then the frequencies $f_j$ of the intervals
and the midpoints $m_j$ of the intervals are as shown below.

f = hist(x,plot=F)$counts;  f
[1]   4  41  82 118 131  74  42   7   1
m = hist(x,plot=F)$mids;  m
[1] 32.5 37.5 42.5 47.5 52.5 57.5 62.5 67.5 72.5

Then I can use the formula $\bar X \approx \frac 1 n \sum_{j=1}^9 f_jm_j = 50.12$
to estimate the sample mean from the interval frequencies and midpoints.

sum(f*m)/500
[1] 50.12

I don't know what formula you are using to find the mode from the histogram, but
whatever it is, for normal data it won't estimate the mode of the population
more reliably than $\bar X.$

I have used the default histogram in R. I could have use R along with additional
instructions to make a different histogram. Then estimates using the histogram
would be similar to the above, but a little different. One possibility is shown below. Now the modal bar is in the interval $(50,52).$

hist(x, prob=T, br = 20, col="skyblue2", main="n=500: NORM(50, 7)")
 curve(dnorm(x, 50, 7), add=T, lwd=2, col="orange")

Here, the parameter br = 20 is only a rough suggestion; R chooses 21 instead in order
to make a histogram with 'round' numbers as interval boundaries. [I could pick
my own exact interval endpoints, and R would obey, if possible.]

Example 2. Now let's look briefly at a sample of $n = 1000$ observations from a non-normal population (gamma distribution), for which the population mode $40$ is not the same as the population mean $\mu = 50.$

set.seed(2011)
y = round(rgamma(1000, 5, 1/10), 2)
summary(y)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
   7.34   32.76   46.02   49.65   63.59  199.30
hist(y, prob=T, br=15, col="skyblue2", main="n=500, GAMMA(5, .1)")
 curve(dgamma(x, 5, .1), add=T, lwd=2, col="orange")
 lines(density(y), lwd=2, lty="dotted")

For this histogram, the modal interval is $(30,40)$ and your method
of finding the sample mode would choose a point in that interval. Knowing
the formula for this density function (orange curve), I could use calculus to find
its maximum, which is the population mode. If I have only the sample
I can estimate the mode from the histogram.

Or, ignoring the histogram, I can use more advanced methods to estimate the shape and rate parameters of the distribution and estimate the mode from them. Alternatively, for moderately large samples, a kernel density estimator (KDE) can estimate the density from the sample and we can use that to estimate
the mode. The KDE for the sample above is shown as a dotted curve.


问:
Rotation number of non-invertible continuous map of $S^1$
Recently I have been studying rotation numbers and all sources I came across define them for invertible circle maps $A$. But what breaks, when $A$ is not bijective? I.e...

Let $A: \mathbb{R} \to \mathbb{R}$ be a continuous map such that $A(y + 1) = A(y) + 1$. For $y \in \mathbb{R}$ let its rotation number be given by
$$
    \rho_y(A) := \lim\limits_{k \to \infty} \frac{A^k(y) - y}{k}.
$$

Suppose now $A$ is non-invertible.

Then can $\rho_y$ exist?

If so, is $\rho_y$ independent of $y$?

I would be very grateful about a hint or two where to start.

So I came up with the following example:
For $s \in \mathbb{R}$ let
$$
    A(y) = \begin{cases}
        \lfloor y \rfloor + s &\text{ if } & y \mod 1 \leq \frac{1}{2} \\
        2y - \lfloor y \rfloor - 1 + s &\text{ else } &
    \end{cases}.
$$
$A$ fulfills the requirements stated above and is not injective.

And empirically we have $\rho_y(A) = s$:

So $\rho$ does exists and is independent of $y$.

Proof (Idea):

Bound $A(y)$ from below by $L(y) = \lfloor y \rfloor + s$ and from above by $U(y) = y + s.$ Then sandwich $\rho_y(A)$.

答:
Here's a counterexample to $\rho_y$ being constant.

Suppose $A(0)=0, A(0.5) = 1.5, A(1)=1$, e.g. use linear interpolation, and extend this by periodicity. Note that $A$ cannot be bijective by the intermediate value theorem.

Clearly $\rho_0 = 0$. However, $A(0.5) = 1.5$, $A(1.5) = A(0.5) + 1 = 2.5$, and in general $A^n(0.5) = n + 0.5$. Thus $\frac{A^n(0.5) - 0.5}{n} = 1$, so $\rho_{0.5} = 1 \neq 0 = \rho_0$.

Edit: I think I have a counterexample in principle for existence of $\rho_y$ as well, but writing out the details precisely would be very messy.

The idea is to have two locations, $0$, and $y^*$, where for $y$ near $0$, $A(y) \approx y$, but for $y$ near $y^*$, $A(y^*) \approx y^* + 1$. Now you want to arrange it so that there is some orbit which alternates periods of time near 0 and $y^*$ in such a way that it spends exponentially more time near each limit point after each visit. In that case, the limit won't exist for roughly these reasons.

How do we get the orbit to behave this way? Well, if we consider $A(y) = 0.5 y$, then for $y>0$ small we'll get "attraction" towards $0$, and if we consider $A(y) = 2y$, then for $y<0$ small we'll get "repulsion" away from $0$. So, start with the $A(y) = 0.5 y$ behavior headed towards $0$, but then after some number of iterations make $A(y)$ slightly negative, letting the $A(y) = 2y$ repulsion kick in until we're far enough from $0$ that we can just head to $y^* + \epsilon$. Make a similar attractor/repulser setup near $y^*$. When you head back to $0+\epsilon'$, make sure you pick $\epsilon'$ small enough that it's "closer in" than the point where you made $A(y)$ slightly negative the first time, and then repeat the whole thing infinitely many times.

You do need to make sure the "slightly negative" values you pick will, after the repulsion step, get you to exactly where you want to be near the other limit point. That's possible using an $A(y) = 2y$ repulser directly since its iterates are simply $A^n(y) = 2^n y$ and you can reverse-engineer the initial $y$ easily from its target value.

ContentLicense: CC BY-SA 4.0


问:
How to find derivative of $\int_0^{\sin x} (1+t)^{\frac{1}{t}} dt$
I am asked to solve this question:
'Which one is the higher order infinitesimals when $\mathbf{x \rightarrow 0 \, \int_0^{\sin x} (1+t)^{\frac{1}{t}} dt}$' or $\mathbf{x^2}$.

I know that:I have to solve $\lim_{x \rightarrow 0}\frac{\int_0^{\sin x} (1+t)^{\frac{1}{t}} dt}{x}$ by using L 'Hospital's rule and compare the result with $x^2$
But I cant find the derivative.

I have recited that the derivative of $\int_{φ(x)}^{ψ(x)}f(t)dt$ (if x is not in f(t)) is $f[φ(x)]φ'(x)-f[(ψ)]ψ'(x)$, without actually knowing how it comes.

But for now, this equation seems impossible since I cannot get $(1+0)^{\frac{1}{0}}$, I don't think $\lim (1+0)^{\frac{1}{0}}$ should be right, since the hint said the $\int_0^{\sin x} (1+t)^{\frac{1}{t}} dt$ ~ $ex$

So what should I do? Is the equation I have recited succeed all the time?

答:
First note that, as $t\to 0$,
$$
(1 + t)^{\frac{1}{t}}  = \exp \left( {\frac{1}{t}\log (1 + t)} \right) = \exp \left( {1 + \mathcal{O}(t)} \right) = e\exp \left( {\mathcal{O}(t)} \right) = e + \mathcal{O}(t).
$$
Hence
$$
\int_0^{\sin x} {(1 + t)^{\frac{1}{t}} dt}  = e\sin x + \mathcal{O}(\sin ^2 x) = ex + \mathcal{O}(x^2 )
$$
as $x\to 0$.


问:
Whether the subspaces are homeomorphic or not
From topology without tears,

Let $X$ be a unit circle in $\mathbb{R}^2$,that is , $X=\{(x,y):x^2+y^2=1\}$ and has subspace topology.  $Y$ be a subspace in  $\mathbb{R}^2$ given by  $Y=\{(x,y):x^2+y^2=1\} \cup \{(x,y):(x-2)^2+y^2=1\} $
and $Z$ be a subspace in  $\mathbb{R}^2$ given by  $Z=\{(x,y):x^2+y^2=1\} \cup \left\{(x,y):\left(x-\frac{3}{2}\right)^2+y^2=1\right\} $

I have two questions:

1)How to show whether $Y$ is homeomorphic to a interval?

I have theorem:Let $f:(X, \tau_1) \rightarrow (Y, \tau_2)$ be a homeomorphism. Let $a \in X$, so that $X \setminus \lbrace a \rbrace$ is a subspace of $X$ and has induced topology $\tau_3$. Also, $Y\setminus\lbrace f(a)\rbrace$ is subspace of $Y$ and has induced topology $\tau_4$. Then $(X \setminus \lbrace a \rbrace, \tau_3)$ is homeomorphic to $(Y\setminus\lbrace f(a)\rbrace, \tau_4)$.

Using the above theorem  I can show $X$ is not homeomorphic to any interval say $(a,b)$.Because if it is,then $X\setminus\{a\}$ must be homeomorphic to $(a,b)\setminus\{f(a)\}$ for all $a \in X$ by above theorem. But $X\setminus\{a\}$ is always connected but $(a,b)\setminus\{f(a)\}$ is not connected. Therefore $X$ cannot be homeomorphic  to any interval.

But how should I show whether $Y$ is homeomorphic to any interval or not.Because here $Y\setminus\{(1,0)\}$ is not connected.So I cannot use the same argument as above and show that $Y$ is not homeomorphic to any interval.Is it homeomorphic?How should I show that?

2)To show that $Z$ is not homeomorphic to $X$ and $Y$

Using the above theorem I can show that $Z$ is not homeomorphic to $Y$,Since $Z-\{a\}$ is always connected but $Y\setminus\{(1,0)\}$ is not connected so I can define $f(a)=(1,0)$ and show $Z\setminus\{a\}$ cannot be homeomorphic to  $Y\setminus\{f(a)\}$ for  some $a$. And hence $Z$ cannot be homeomorphic to $X$

But how should I show $Z$ is not homeomorphic to $X$,because $Z\setminus\{a\}$ and $X\setminus\{b\}$ is always connected for all $a\in X$ and $b\in Z$. Connectedness cannot be used here.So how should I prove this?

Any hint will be of big help.

Thanks in advance!

答:
Since $Y$ is compact, if it is homeomorphic to some interval, then it has to be an interval of the form $[a,b]$. Suppose that there is such a homeomorphism $f\colon Y\longrightarrow[a,b]$. Note that $Y\setminus\{(-1,0),(3,0)\}$ is connected. But the only two points $c$ and $d$ from $[a,b]$ such that $[a,b]\setminus\{c,d\}$ is connected are $a$ and $b$. So, $f(-1,0)=a$ and $f(3,0)=b$ or $f(-1,0)=b$ and $f(3,0)=a$. Now, let $p=f(1,0)$. Then, since $Y\setminus\{(-1,0),(3,0),(1,0)\}$ has four connected components, $(a,b)\setminus\{p\}$ should also have four connected components. But it has two instead.

And there are two points $x$ and $y$ in $Z$ such that $Z\setminus\{x,y\}$ has four connected components. No such points exist in $X$ or in $Y$.


问:
Maximum distance from a point in hyperbola $ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ to its asymptotes
The canonical hyperbola is
$$ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$

I have already derive the equation for the distance from any point $(x_0,y_0)$ in hyperbola to the asymptote $ y=\frac{b}{a}x$ as
$$ d =\frac{|bx_0-ay_0|}{c}$$
where
$a^2+b^2=c^2$ and $x_0 \geq a$.

I know that the maximum distance is when $x_0=a$ but I don't know how to proof it. Can you help me? Thanks.

答:
WLOG $a=c\cos u,b=c\sin u$

and any point $P$ on the hyperbola can be $(a\sec v, b\tan v)=(c\cos u\sec v,c\sin u\tan v)$

So, the distance of $P$ from $bx-ay=0$ will be  $$\dfrac{|b(c\cos u\sec v)-a(c\sin u\tan v)|}{\sqrt{a^2+b^2}}$$

$$=\dfrac{|c\sin u|(c\cos u\sec v)-|c\cos u|(c\sin u\tan v)}{|c|}$$

$$=|c\sin u\cos u(\sec v-\tan v)|$$

$$=\dfrac{|c\sin2u\tan\left(\dfrac v2-\dfrac\pi4\right)|}2$$

which will be maximum

if $c\sin2u\ne0$  and

if $\tan\left(\dfrac v2-\dfrac\pi4\right)\to\infty$

which occurs if $\dfrac v2-\dfrac\pi4\to\pm\dfrac\pi2$


问:
function for second quadrant in unit circle
I am no "math-guy" and would really appreciate some help in writing a function for the second quadrant of the unit circle.

Conditions I want to be met:

Centre of the circle is moved so second quadrant is between 0 – 1 on the x-axis

Introduce parameters so I can bend the outer edges on the curve (black dashed lines in figure). What I thought of so far is writing two smaller circles if x < value 1 or x > value 2. Value 1 and 2 are values for when the curve should be bent.

Control the function so that x = 0.5 always returns wanted y-value. (blue dashed line in figure)

the derivative is equal to 0 at x = 1

答:
You can try the family

$$x^\alpha+y^\alpha=c$$ with $\alpha>2$ and $c\le1$. (Below, $\alpha=3,c=0.6,0.8,1$; the curves should continue to the axis.)

Mirror and translate as required.


问:
Arrangement of objects in 2 intersecting circles
In how many ways can 8 tennis balls and 8 baseballs be arranged in 2 intersecting circles (much like on the perimeter of a Venn diagram), if there is a ball placed at each of the 2 intersections of the circle perimeters, 3 on each segment of circle perimeter inside the other circle, and 5 on the segment of perimeter outside the other circle? The points are equally spaced and rotations are not counted.

Now approaching this question is difficult. Should the balls be sorted into the circles first, or sorted to fit the 3 types of points described in the question?

答:
Hint: Split into cases.

Case 1: Balls on the intersection are both tennis balls (or baseballs).

Case 2: Balls on intersection are 1 tennis, 1 baseball.

In each case, it's easy to count the number of possibilities directly.

Case 1: Both balls are tennis balls. Then, the rest of the 14 balls must be arranged in some order. Ignoring the restriction about rotation, there are ${14 \choose 8} $ ways to do so. Now, we account for rotation. Generally, the rotated version will be different (and hence we would have double counted), except when it is symmetrical. There are ${ 7 \choose 4} $ ways which are symmetrical (basically once you decide which 4 balls remaining in the 'left' circle are baseballs).

Hence, there are $ \frac{ {14 \choose 8} + { 7 \choose 4} } { 2}  $ ways.

Case 1b: Both balls are baseballs. Then, there are $\frac {{14 \choose 8} + { 7 \choose 4} } {2} $ ways.

Case 2: One ball is a tennis ball, one is a baseball. WLOG, let the 'top' ball be a tennis ball. Then, there are ${14 \choose 7} $ ways to arrange the rest of the balls. Note that each of these ways are not rotations of each other, since the 'top' ball is a tennis ball.

Hence, there are $ { 14 \choose 8} + {7 \choose 4} + { 14 \choose 7} $ ways to arrange this.

ContentLicense: CC BY-SA 3.0


问:
Nos Systems Ten
How do I find $4$ digit $10$'s complement of $-456$ (withbase $10$)?

I am able to do $4$ digit $10$'s complement of a positive number (base 10) using the formula

$base^n - number$

where $n$ is the number of digits, in this case $4$.

i.e. $4$ digit $10$'s complement of $456$ is

$10^4  - 456 = 9544$

But I can't seem to find the negative. I've tried googling and looking in StackExchange Math but none seems to fit what I need.

I would appreciate any help!

答:
Taking base($=$radix) as $r$ and $n$ number of digits, so called, Radix-Complement   Representation for the complement of $n$-digit number is obtained by subtracting it from $r^n$. If  $D$  is  between  $1$ and $r^n  −  1$, then  subtraction  produces again result between  $1$ and $r^n  −  1$. For example for number $1849$ it's $10$'s complement is $8151$, for $100$ it's $10$'s complement is $9900$. So, you computation is correct, you obtained $10$'s complement in $4$ digits of $456$ and it is $9544$.

"negative" $(-456)$ i.e. complement of $456$ is exactly $9544$ in $10$'s complement representation. Now if you take $10$'s complement of $9544$ i.e. $(-(-456))$, then you, obviously, obtain $456$.

Addition.

Let's look in more detail. In $n=4$ digits we have $10^n=10000$ possible decimal combinations from $0$ up to $9999$. Now, when we have only this combinations, but we want to have also negative numbers, then in $10$'s complement representation we make following: we leave $0$ in its place and keep first $\frac{10^n}{2}-1$ combinations as positive numbers i.e. from $1$ up to $\frac{10^n}{2}-1=4999$. And we call negative numbers all numbers from $\frac{10^n}{2}=5000$ up to $9999$. Finally we have positive $4999$ numbers,  negative $5000$ numbers and zero.

So, $9544$ is $10$'s complement representation of $(-456)$ and it is $10$'s complement of $456$.

Some books use circle for radix complement representation: $0$ is north pole. Increasing direction is clockwise and south pole is first negative number. BTW south pole is most bad place in this representation.

Title: Nos Systems Ten


问:
Question about a determinant property I encountered in a question.
Here's the question:

let $A,B\in \mathbb{R}^{3\times 3}$ such that $AB=|A|I$ and $|(2A)^2(A^t)^{-1}B|=16$ then $|A|=?$

I wanted to ask about how to deal with the information that $AB=|A|I$, can I just $|AB|=||A|I|=|A|$? am I allowed to do that?

From the other information I got that $4^3|A||B|=16$ so $|AB|=\frac {1}{4}$.

So what I said was $|A|=\frac {1}{4}$, but the answer was $|A|=(\frac{1}{4})^{\frac {1}{3}}$, so I think I have a miss understanding about how to calculate $||A|I|$.

Thanks in advance for all the help.

Body:
Hint:

$|kA|=k^3|A|$, if $A$ is $3 \times 3$.similarly $||A|I|=|A|^3$.


问:
How many winning hands are possible in Mahjong?
A set of Mahjong tiles consists of four copies of 34 unique tiles (disregard "Bonus" tiles for the purpose of this question):

There are three ways of creating a meld:

Eye: two copies of the same tile

Pong: three copies of the same tile

Kong: four copies of the same tile

Chow: three consecutive tiles (only possible for each of the suits)

A winning hand consists of exactly one eye and four non-eye melds. A tile cannot be shared across multiple melds.

My attempt at calculating an upper limit for the number of different winning hands is below:

34 unique tiles gives 34 unique eyes, 34 unique pongs and 34 unique kongs. Within a single suit, 7 $[(1,2,3), (2,3,4), (3,4,5), (4,5,6), (5,6,7), (6,7,8), (7,8,9)]$ chows are possible, giving 21 chows in total.

Choosing one pair and four kongs gives $34\times nCr(34+34+21, 4)= 83015284$

However, I realised an error in my calculations is that I assumed that a meld cannot be used twice, which is false as any chow can occur a maximum of four times. Therefore, the number above may not be the upper limit.

Could someone help with calculating how many winning hands are possible in Mahjong?

答:
An upperbound

The winning hand must contain $14$ tiles and so we have the following possibilities. I've explained the reasoning in the most difficult case of the four Chows.

Eye and three Kongs $$\begin{pmatrix}34\\4\\\end{pmatrix} \times 4=185504.$$
Eye and four Pongs $$\begin{pmatrix}34\\5\\\end{pmatrix} \times 5=1391280.$$
Eye, Chow and three Pongs $$21\times\begin{pmatrix}34\\4\\\end{pmatrix} \times 4=3895584.$$
Eye, two Chows and two Pongs $$\left (\begin{pmatrix}21\\2\\\end{pmatrix}+21\right)\times\begin{pmatrix}34\\3\\\end{pmatrix} \times 3=4146912.$$
Eye, three Chows and one Pong $$\left (\begin{pmatrix}21\\3\\\end{pmatrix}+\begin{pmatrix}21\\2\\\end{pmatrix}\times 2+21\right) \times\begin{pmatrix}34\\2\\\end{pmatrix} \times 2=1751442.$$
Eye and four Chows

The Chows can be all different $\begin{pmatrix}21\\4\\\end{pmatrix}$.

There can be three different Chows $\begin{pmatrix}21\\3\\\end{pmatrix}\times3$.

There can be two different Chows $\begin{pmatrix}21\\2\\\end{pmatrix}\times3$.

The Chows can be all the same $\begin{pmatrix}21\\1\\\end{pmatrix}$.

This adds up to $10626$ possibilities.

There is then a maximum of $34$ possibilities for the Eye.

$$10626\times34=361284.$$

Total

$$11,732,006$$

N.B.

I hope I've understood the rules correctly!

This answer is an upper bound. There will be cases where choosing overlapping Chows will limit the possibilities for the other tiles but these will be relatively few in number. For example, in the case of four identical Chows there will only be $31$ possibilities for the Eye. Thus our count has been $3\times 21=63$ too large for these possibilities.


问:
Is a set of prime numbers open in Furstenberg's topology?
I was reading Furstenberg's Proof of Infinitude of Primes and I wonder if a set of prime numbers is open in this topology.

Thanks!

答:
Every non-empty open set $U$ of $\Bbb{Z}$ contains a basis $a + b \Bbb{Z}$ for some $b > 0$. In particular,

$$ \liminf_{n\to\infty} \frac{\# (U \cap [-n, n])}{2n+1} \geq \frac{1}{b} > 0. $$

But if $\Bbb{P}$ denotes the set of primes, it is not hard to prove that

$$ \liminf_{n\to\infty} \frac{\# (\Bbb{P} \cap [-n, n])}{2n+1} = \prod_{p \in \Bbb{P}} \left( 1 - \frac{1}{p} \right) = 0. $$

Therefore $\Bbb{P}$ is not open in $\Bbb{Z}$.


答: From $AB=|A|\operatorname{Id}$ and from the fact that $A$ is a $3\times3$ matrix, you get that$$|A||B|=\bigl||A|\operatorname{Id}\bigr|=|A|^3.$$And from the fact that $\left|(2A)^2\left(A^T\right)^{-1}B\right|=16$, you get that$$64|A|^2\frac1{|A|}|B|=16\left(\iff|A||B|=\frac14\right).$$So, $|A|^3=\frac14$, and therefore $|A|=\frac1{\sqrt[3]4}$.

Title: Question about a determinant property I encountered in a question.

Body:
Here's the question:

let $A,B\in \mathbb{R}^{3\times 3}$ such that $AB=|A|I$ and $|(2A)^2(A^t)^{-1}B|=16$ then $|A|=?$

I wanted to ask about how to deal with the information that $AB=|A|I$, can I just $|AB|=||A|I|=|A|$? am I allowed to do that?

From the other information I got that $4^3|A||B|=16$ so $|AB|=\frac {1}{4}$.

So what I said was $|A|=\frac {1}{4}$, but the answer was $|A|=(\frac{1}{4})^{\frac {1}{3}}$, so I think I have a miss understanding about how to calculate $||A|I|$.

Thanks in advance for all the help.


问:
Show it's a metric
Let $X$ be set and $d:X \times X\rightarrow\mathbb{R}$ and I have to two 2 properties
$$\tag{A1}d(x,y)=0 \leftrightarrow x=y$$
$$\tag{A2}d(x,y)\leq d(x,z)+ d(y,z)$$ for all x,y,z in X

If we show the symmetry we can also change A2 such that the triangle inequaltiy is also given.

Now Let's use A2 for symmetry: $d(x,y) \leq d(x,z)+d(y,z)$
and $d(y,x)\leq d(y,z)+ d(x,z)$ so we see that $d(x,y)$ and $d(y,x) $ have the same estimates. But what enables me NOW to say that $d(x,y) = d(y,x)$?, it has to be axiomatic when $a \leq c; b\leq c $ then $a=b$ I could not show this.

I have made small mistakes, now its corrected

Thanks for the answer, I will post the longer version here as well.

$d(x,y) \leq d(x,z)+d(y,z) $ now use $z=x$ and you have $d(x,y)\leq 0+d(y,x)$
Second part:
$d(y,x) \leq d(y,z)+d(x,z) $ now use $z=y$ and you have $d(x,y)\leq 0+d(x,y)$

so you have: $d(x,y)\leq d(y,x)\&d(y,x)\leq d(x,y)$ at the end this guives us: $d(x,y)=d(y,x)$

答: Put $z=x$ to get $d(x,y) \leq 0+d(y,x)$. Now interchange $x$ and $y$ for the reverse inequlity.

Title: Show it's a metric


问:
For different integers there exists an integer so that exactly one of the sums is prime.
Let $i,j$ be distinct natural numbers. In a comuter science class the lecturer said

Since the prime numbers are not regularly distributed, there is some natural number $k$ such that $i+k$ is prime and $j+k$ is not.

I found this reasoning a bit unsatisfying, so I tried to show the claim, however it turned out to be not that trivial. If lets say $i$ has a prime factor $p$ that doesn't divide $j$ then by Dirichlet there is a prime $q\equiv j\pmod p$, $q>j$. Then choose $k=q-j$, so $j+k$ is prime but $i+k$ has the prime factor $p$, hence is not prime. But this reasoning doesn't work if $i,j$ have the same prime factors. I also feel like this can be proven more elementary but I don't see how.

Edit: Thanks everyone for the help, I learned something from every answer. Unfortunately I can only accept one of them, so I hope I didn't offend anyone for not accepting their answer.

答:
Suppose $i > j$. Further suppose that the statement is not true. Then there are two cases:

Case 1: $i + k$ is not prime for all $k\in \mathbb N$. This contradicts the infinitude of primes.

Case 2: $i+k$ is prime for some $k\in\mathbb N$ if and only if $j+k$ is also a prime. In that case:

Since $j + (k + i - j)$ is prime, so is $i + (k+i-j) = 2i -j + k$.

Since $j + (k + 2i - 2j)$ is prime, so is $i + (k + 2i - 2j) = 3i - 2j + k$.

Since $j + (k + 3i - 3j)$ is prime, so is $i + (k + 3i - 3j) = 4i - 3j + k$, and so on...

We have now created an infinite arithmetic progression of primes: $i + k + n(i-j)$ is prime for all $n \in \mathbb N$. To see why this is a contradiction, consider the remainders of this progression when dividing by $i-j+1$.


答:
First compute 9's complement, i.e.,

$= 9999 - 0456$
$= 9543$

Then, add 1 to the LSB, that will be 10's complement, i.e.,

$= 9543 + 1$
$= 9544$

Note that this is the magnitude of given negative number.

Title: Nos Systems Ten

Body:
How do I find $4$ digit $10$'s complement of $-456$ (withbase $10$)?

I am able to do $4$ digit $10$'s complement of a positive number (base 10) using the formula

$base^n - number$

where $n$ is the number of digits, in this case $4$.

i.e. $4$ digit $10$'s complement of $456$ is

$10^4  - 456 = 9544$

But I can't seem to find the negative. I've tried googling and looking in StackExchange Math but none seems to fit what I need.

I would appreciate any help!


答:
Suppose $i<j$. Add a number to each to make $j+k=p$, a prime.

Now $i+k$ is coprime to $p$ and so by Dirichlet there is a number $l$ such that $i+k+lp$ is prime, whereas $j+k+lp$ is a proper multiple of $p$.

Title: For different integers there exists an integer so that exactly one of the sums is prime.

Body:
Let $i,j$ be distinct natural numbers. In a comuter science class the lecturer said

Since the prime numbers are not regularly distributed, there is some natural number $k$ such that $i+k$ is prime and $j+k$ is not.

I found this reasoning a bit unsatisfying, so I tried to show the claim, however it turned out to be not that trivial. If lets say $i$ has a prime factor $p$ that doesn't divide $j$ then by Dirichlet there is a prime $q\equiv j\pmod p$, $q>j$. Then choose $k=q-j$, so $j+k$ is prime but $i+k$ has the prime factor $p$, hence is not prime. But this reasoning doesn't work if $i,j$ have the same prime factors. I also feel like this can be proven more elementary but I don't see how.

Edit: Thanks everyone for the help, I learned something from every answer. Unfortunately I can only accept one of them, so I hope I didn't offend anyone for not accepting their answer.


答: As each sequence $S(a,b) = a \mathbb{Z} + b$ contains non-primes, i.e. $b | ab + b$, this cannot be true.</p>

Title: Is a set of prime numbers open in Furstenberg's topology?

Body:
I was reading Furstenberg's Proof of Infinitude of Primes and I wonder if a set of prime numbers is open in this topology.

Thanks!


问:
Connecting the rank with the determinant of a skew symmetric matrix.
We need to find the rank of the matrix:

\begin{bmatrix}
0&c_{1}&-b_{1}&a_{2}\\
-c_{1}&0&a_{1}&b_{2}\\
b_{1}&-a_{1}&0&c_{2}\\
-a_{2}&-b_{2}&-c_{2}&0
\end{bmatrix}

Where all $a_{i},b_{i},c_{i}$ are non zero.

The only thing I could observe was the given matrix is skew-symmetric, and the determinant is $(a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2})^2.$However I could not proceed any further than that. The rank is ofcourse = the number of non zero rows in the row echelon form of the matrix, but taking that route seems to be too tedious , and I believe the purpose of giving a skew symmetric matrix was to exploit the (skewed)symmetry somehow.

The solution given in my book is :

Rank=$4$,if $det(A) \neq 0$. (This is straightforward)

Rank=$2$ otherwise, since "all determinants of the third order have the value zero".

Questions:

What does the statement in quotes mean, and how is it true?

What connection does it have to finding the rank?

答:
An argument which avoids to consider all the $3\times 3$ submatrices is the following.

Consider the "twin" matrix
$$A'=\begin{pmatrix}
0&c_{2}&-b_{2}&a_{1}\\
-c_{2}&0&a_{2}&b_{1}\\
b_{2}&-a_{2}&0&c_{1}\\
-a_{1}&-b_{1}&-c_{1}&0
\end{pmatrix}$$
If your matrix $A$ has zero determinant, then all the columns of $A'$ are in the kernel $\ker A$. By the Rank–nullity theorem it holds $rg(A)=4-\ker A\leq 4-rg(A')$, but $rg(A')\geq 2$, since $A'$ is at least a 2×2 submatrix with non zero determinant, hence $rg(A)\leq 2$. By the same reason $rg(A)\geq 2$.


问:
The equation of a sphere tangent to two planes
The center of a sphere belongs to the line
$$d : {{x-1}\over3} = {{y}\over 2} = {{z+2}\over-2} $$
Find the equation of the sphere if the planes
$$ (P):z+3=0$$
$$ (Q) :3x-4z-33=0 $$
are tangent to the sphere.

My solution: from the parametric equations of the line, I found that the center of the sphere is of the form $C(1+3t,2t,-2-2t)$. Now, if the planes are tangent to the sphere then the distance from the center of the sphere to each plane is equal with the radius. The thing is, I've computed both of these but I'm stuck at a point where I have the absolute value of two expressions. Would it also be viable to find the line as the intersection of the two planes and work with that? Thank you, in advance!

答:
Perpendicular to the first  plane from center is
$d_1=-2-2t+3=0=1-2t$
Similarly $d_2=(17t-22)/5$
Equate $d_1=d_2$ to get $t=1$, So now you know the center $(4,2,-4)$ and the radius $r=|d_1|=|d_2|=1.$ to write the Eq. of sphereas
$$(x-4)^2+(y-2)^2+(z+4)^2=1$$


问:
Car insurance - Probability
A car insurance policy insures as many men as women.

The probability that a woman will claim the insurance in any given year is independent of the other years and is equal to $\beta$.

The same holds to the male persons with the probability $\alpha$.

Now let's assume that the insurance company pulls a person at random.

(a) What is the probability that this random person will claim insurance this year?

(b) What is the probability that the person will claim the insurance in two consecutive years?

$$$$

For question (a) : The probability to take a woman is equal to the probability to take a man, which is $\frac{1}{2}$.

So is the probability that this random person will claim insurance this year equal to $p=\frac{1}{2}\cdot \beta+\frac{1}{2}\cdot \alpha$ ?

For question (b) : Since each year is independent is the probability equal to $p^2$, where $p$ is the probability from (a) ?

答:
The solution for a) is OK. So it is

$p = p_1 = \frac{1}{2}\cdot \beta+\frac{1}{2}\cdot \alpha$

The solution for b) is not this. It is

$p_2=\frac{1}{2}\cdot \beta^2+\frac{1}{2}\cdot \alpha^2$

because the problem says: first they pull a person at random

(either a man or woman - probability $1/2$),

and then this (particular) person makes claims in two consecutive years

(probability $\alpha^2$ or $\beta^2$ depending on if they picked
a man or woman).

I think what you have calculated as $p^2$ is the probability that:

a randomly chosen person A claims this year, say in 2021

AND

a randomly chosen person B (it may happen that $A=B$ or $A \ne B$) claims in the year 2022

But this is not the event whose probability they want in b).

Title: Car insurance - Probability


问:
Probability density function of Z where Z=X/Y, X,Y are exp random variables with parameter is 1?
What I thought is, for any a that Z = a, if Y = y, then X should be ay. The probability X = ay and Y = y is $e^{-ay}e^{-y}$. Then take intergal $\int_{0}^{\infty } e^{-y}e^{-ay}dy=\frac{1}{a+1}$.

But it's not the right answer, right answer is $\frac{1}{(1+a)^{2}}$.

I use the other method that calculate $F_{Z}(z)$ first then derivative it and gets the right p.d.f. But I am confused where my thinking above is wrong...

答:
You missed the Jacobian determinant.   The transformation $\langle Z,Y\rangle\mapsto\langle ZY,Y\rangle$ is not colinear.   You have to apply the chain rule.

$$\begin{align}f_{\small Z}(z) &=\int_\Bbb R f_{\small Z,Y}(z,y)\,\mathrm d y\\[1ex]&= \int_\Bbb R \left\lVert\dfrac{\partial \langle zy, y\rangle}{\partial\langle z, y\rangle}\right\rVert f_{\small X,Y}(zy,y)\,\mathrm d y\\[1ex]&=\int_0^\infty y\,\mathrm e^{-(z+1)y}\mathbf 1_{0\leqslant z}\,\mathrm d y\\[1ex]&=\dfrac 1{(z+1)^2}\mathbf 1_{0\leqslant z}\int_0^\infty u\,\mathrm e^{-u}\;\mathrm d u&&u:=(z+1)y, \mathrm d u:=(z+1)\mathrm d y\\[1ex]&=\dfrac{1}{(z+1)^2}\mathbf 1_{0\leqslant z}\end{align}$$

Alternatively

$\begin{align}\mathsf P(Z\leqslant z)&=\int_{\Bbb R^+}\mathsf P(Z\leqslant z\mid Y=y)\,f_{\small Y}(y)\,\mathrm d y\\[1ex] &=\int_{\Bbb R^+}\mathsf P(X\leqslant Yz\mid Y=y)~f_{\small Y}(y)\,\mathrm d y\\[1ex] &=\int_{\Bbb R^+}\mathsf P(X\leqslant yz)~f_{\small Y}(y)\,\mathrm d y\\[1ex]&=\int_{\Bbb R^+}\int_{0}^{yz} f_{\small X}(x)\,f_{\small Y}(y)\,\mathrm d x\,\mathrm d y\\[2ex]f_{\small Z}(z)&=\dfrac{\mathrm d \mathsf P(Z\leqslant z)}{\mathrm d z} \\[1ex]&=\int_{\Bbb R^+}\, y f_{\small X}(yz)\,f_{\small Y}(y)\,\mathrm d y\end{align}$


问:
Is local compactness a topological property? Moreover if any metric space say $Y$ is locally compact then is the space $Y^n$ also locally compact?
Is local compactness a topological property?

Moreover if any metric space say $Y$ is locally compact then is the space $Y^n$ also locally compact?

In particular, do homeomorphisms preserve localness of any topological property?

答:
Yes, it is a topological property: it's defined purely in terms of set theory and open sets.

And for any space $Y$ (not only metric) we have that $Y$ locally compact implies $Y^n$ is for finite $n$. (for all usual non-equivalent definitions of local compactness).

That homeomorphisms preserve this property is the definition of a topological property. The last question might depend on how you define "locally P" for a property $P$. If you say $X$ is locally $P$ iff every point has a neighbourhood (in the general sense) that has $P$, and $P$ is itself a topological property, then yes, so is locally $P$, obviously. You need that a homeomorphisms preserves open sets and hence neighourhoods and is surjective, so that all points in the image are reached.


答:
A set in any topological space is open iff for every point in it there exists an open set containing it and contained in the set.

Now, can you find an open set contained in "a set" of primes? In fact, can you find a basic  open set $\,a\Bbb Z+b\subset\Bbb Z\,$ contained in a set (any set) of primes?


答:
Well, so many questions. Local compactness is a topological property (i.e a space is locally compact if every point has a base of compact neighborhoods). When talking about the product, we need to ask a bit more of the space. There is a theorem (check this url for more info https://en.wikipedia.org/wiki/Locally_compact_space) that states the following: Let ${(X_j,T_j)_{j\in J}}$ be a non-empty family of non-empty spaces. Then the product is locally compact iff every space is locally compact and every space is compact minus a finite number of spaces.
Finally, local compacity is a topologically invariant property.

Body:
Is local compactness a topological property?

Moreover if any metric space say $Y$ is locally compact then is the space $Y^n$ also locally compact?

In particular, do homeomorphisms preserve localness of any topological property?


问:
Any good upper bounds for $\sum_{i = 0}^{m} 2^i {k+i \choose i}$?
I am trying to solve a recursion (I faced it when I wanted to analyze the running time of an algorithm precisely). I reduced it to compute a "good" upper bound for $\sum_{i = 0}^{m}2^i {k + i \choose i}$. We have these assumptions that $1 < m \leq Ck$, for some $C\in \mathbb{N}$.
I have tried many approximations but the bounds are not what I want. They are very loose. On the other hand, it doesn't look like a very sophisticated summation but I have so much difficulty solving it!
Does anyone have any ideas?

答:
Not a full answer and not rigorous, but somebody more knowledgeable may be able to justify the asymptotics below.

If we set a specific value for $k$, we can compute easily enough a formula of the form:

$$\sum_{i=0}^{m}{2^i{k+i \choose i}} = P_k(m)2^m+(-1)^k$$

where $P_k(m)=a_{k}m^k+a_{k-1}m^{k-1}+\ldots+a_0$, and experimentally (to be proved) it looks like, for $k \ge 2$:

$$a_k = \frac{2}{k!}$$

I have retrieved some of the $P_k(m)$ with the help of OEIS:

$$P_1(m) = m - 1$$
$$P_2(m) = m^2+m+2$$
$$P_3(m) = \frac{1}{3}(m^3+3m^2+8m)$$
$$P_4(m) = \frac{1}{12}(m^4+6m^3+23m^2+18m+24)$$
$$P_5(m) = \frac{1}{60}(m^5+10m^4+55m^3+110m^2+184m)$$

and, in general, a conjecture, for $m$ big and $k \ge 2$:

$$\sum_{i=0}^{m}{2^i{k+i \choose i}} \approx \frac{m^{k}2^{m+1}}{k!}$$


问:
Is this Perfect Binary Tree Countably Infinite or Uncountably Infinite?
Given a perfect binary tree which at a certain fixed level has infinitely many leaves, hence each leaf at that fixed level has infinitely many ancestors. From that fixed level the tree also extends to infinite height.

Are the nodes of this tree countably infinite or uncountably infinite?

I know that a rooted binary tree, even if extended to a height of infinity is countably infinite; just start at the root and enumerate level by level. However, I cannot imagine a way to count the nodes of this tree or to apply the idea that the union of countably infinite subtrees is countable. Thus, I’m leaning towards it being uncountably infinite but have serious doubts.

How can a tree look like that?

Imagine that instead of the tree simply growing in height it is also expanding sideways as shown below, so it is infinitely high, infinitely wide and infinitely deep. Getting infinitely wide is illustrated below.

答:
The number of nodes in your infinite binary tree is countable.

In particular, each node at the $k$-the level below the root can be uniquely labelled with a binary $k$-tuple (i.e. an element of the set $\{0,1\}^k$) corresponding to its "address", i.e. the path needed to reach it from the root node (identifying $0$ with a step from a node to its left child, and $1$ with a step from a node to its right child).  Thus, every node in the tree can be uniquely labelled with an element of the set of finite-length binary tuples $$\{0,1\}^* = \bigcup_{k \in \mathbb Z^+} \{0,1\}^k.$$

As a countable union of finite sets, $\{0,1\}^*$ is countable.  (The definition above also suggests a fairly simple way to construct an explicit enumeration, by starting with the elements of $\{0,1\}^0$, then those of $\{0,1\}^1$, then those of $\{0,1\}^2$, and so on.)  Thus, the set of nodes in your tree is also countable.

What might be confusing you, however, is that it's also possible to consider (countably) infinite descending paths through your tree, corresponding to infinite binary sequences.  The set $\{0,1\}^{\mathbb N}$ of such sequences is uncountable — as shown by Cantor's diagonal argument — and each such sequence corresponds to a distinct path through the tree (in the sense that, for any two distinct sequences in $\{0,1\}^{\mathbb N}$, there is a node in your tree where the corresponding paths diverge and never again meet).
However, these infinite paths never reach a final node, and thus do not correspond to the address of any node in the tree.

So if you're only interested in the number of nodes in your tree, it is countable.  But there are uncountably many different descending paths through it.  This is in sharp contrast to the corresponding situation with finite binary trees, where every descending path must eventually end at a specific node.

Ps. Above, I've assumed that your tree in fact remains rooted throughout the construction — i.e. that there is a unique root node that is not a child of any other node, and remains so even at the infinite limit.  Probably the conceptually simplest (though not the only) way to construct such an infinite rooted tree is to have it grow only at the branch tips — i.e. at every stage of the construction, take every node that does not yet have children and give it two new child nodes.

However, the picture in your question shows the tree also growing "to infinity" at the root.  It's not 100% clear how to interpret that, but one reasonable interpretation would be that, at every stage, the current tree is (also) duplicated, and the root nodes of the two copies are then made children of a new root node.

The tricky part with this "doubly infinite growth" construction is that the tree may or may not actually remain rooted — in particular, depending on how you relabel the (old and new) nodes at each stage, the root might end up escaping to infinity so that, in the final infinite tree, there are no longer any parentless nodes left(!).

However, this will not actually change the number of nodes in (or paths through) the resulting infinite tree.  In particular, even if the infinite tree has no root node distinguished by its lack of a parent, we can still pick any arbitrary node in the tree as a starting point and label all other nodes by the shortest path needed to reach them from the start node — which will involve first climbing $k$ steps up (i.e. from each node to its parent) to the closest common parent of the start node and the target node, and then descending $n$ steps towards the target from this common parent.  The resulting node addresses can thus be identified with elements of $\mathbb Z^+ \times \{0,1\}^*$, which is still a countable set.

Indeed, more generally, it's not hard to show that any construction that only has countably many stages, and only adds a finite number of new nodes at each stage, cannot produce more than a countable number of nodes in the final tree.


问:
What is a Manifold?
Now we always encounter definition of a manifold from a mathematical point of view where it is a topological space along with a family of open sets that covers it and the same old symphony. My question is from your own expertise and from what you have learned and taught to day, how can we get a deeper understanding to what's a Manifold? Maybe from a more physics-point of view?</p>

答:
As Bronstein and others have put it in Geometric deep learning: going beyond Euclidean data (Read the article here)

Roughly, a manifold is a space that is locally Euclidean. One of the
simplest examples is a spherical surface modeling our planet: around a
point, it seems to be planar, which has led generations of people to
believe in the flatness of the Earth. Formally speaking, a
(differentiable) d-dimensional manifold X is a topological space where
each point x has a neighborhood that is topologically equivalent
(homeomorphic) to a d-dimensional Euclidean space, called the tangent
space.

OwnerDisplayName: PhilosophicalPhysics

Title: What is a Manifold?


问:
Prove that an uncountable subset of a second countable space contains at least one limit point
This is a 2 part problem about second countable space.

i)Prove that the set of all isolated points of a second countable space is countable

Definition: $x$ is an isolated point if ${x}$ is open in the topology.

Since the topology is second countable, the set of basis is countable and open sets are defined to be unions of sets in the basis. Thus if x is an isolated point, then x is part of the basis set. Now if there are an uncountable number of isolated points, they are all basis elements, so we would have an uncountable number of basis elements.

(ii) Hence, show that any uncountable subset A of a second countable space contains at least one point which is a limit point of A.

Suppose A does not contain any point that is a limit point. Since the space is second countable, we have a countable basis set $B_i$, $i \in N$ such that their union is the entire space X. Since A is a subset of X, each point of A must be contained in some $B_i$ and each $B_i$ is open. Now I'm not sure if this is correct, but since A is uncountable and we only have countable number of $B_i$, is it the case that one of the $B_i$ must contain an uncountable number of points of A. I think that part is accurate. But how do I go from here to showing that every basis elements that contains points of A all contain another point of A that they all have in common, which would be the limit point of A?

答:
As to (i), you are right: if $X$ is second countable, as witnessed by a countable base $\{B_n\mid n \in \Bbb N\}$ of $X$, than for any isolated point $x$, by the definition of a base there should be a basic set $B_n$ such that $x \in B_n \subseteq \{x\}$ and it follows that $\{x\}=B_n$ is in the base to start with. All singletons of isolated points are members of any base for the space. As there is a countable base, there cannot be more than countably many isolated points.

If then $A$ is an uncountable subset, $A$ itself is a second countable space when given the subspace topology, so $\textrm{Isol}(A)$ (the (relatively) isolated points of $A$) is at most countable by $(i)$ and so $A\setminus \textrm{Isol}(A)$ is uncountable (or $A$ would have been countable, which it is not) and all points in it are by definition limit points of $A$ that (bonus!) are even members of $A$ themselves (there can be many limit points outside as well).


答:
In order for $\Bbb P$, the set of prime numbers, to be open in the Furstenberg topology on $\bf N$, it must be the union of the basis sets, namely a union of infinite arithmetic sequences. Since $\Bbb P$ is nonempty, this means it would have to contain at least one such sequence. However this is not possible, since any infinite arithmetic sequence contains a composite number.

Let $\{an+d:n\in{\bf Z}\}\cap{\bf N}$ be any such arithmetic sequence. Then $(an+d)(am+1)$ is composite for sufficiently large choices of $n,m$, and it can be written as $a(anm+n+dm)+d$ which is in the arithmetic sequence. Thus, every such arithmetic sequence contains a composite number.

If you work over $\bf Z$ instead, then a much simpler argument works: every infinite arithmetic sequence contains negative integers, but the set of primes contains no negative integers.

These same arguments apply to any subset of $\Bbb P$, i.e. any nonempty set of prime numbers.


问:
Let $G$ be a group and $g \in G$. Show that $o(g)=o(xgx^{-1})$, where $o(x)$ is the order.

Let $G$ be a group and $g \in G$. Show that $o(g)=o(xgx^{-1})$, where $o(x)$ is the order.

I've managed to show that the order of $xgx^{-1}$ is at most $o(g)$ by the following way.

Denote $o(g) = n$ and $o(xgx^{-1}) = m.$ Now $(xgx^{-1})^n=xgx^{-1} \cdots xgx^{-1} = xg^nx^{-1} = xx^{-1} = e.$

But I'm not sure how to show this the other way around? Any hints would be appreciated. This doesn't seem to yet imply that the orders would equal one another.

答:
Using your argument you obtain that $m|n$.

Now observe that $g=x^{-1}(x g x^{-1}) x$ and using the same argument you obtain $n|m$.


问:
Generator of direct product of modules is direct product of each generator?
Let assume I have an R-module M over a ring $R$ with unit and I want to consider the direct product (or direct sum, since I understand that they are both the same when the product has finitely many factors) $M \oplus M$. Furthermore, let $N_1 \oplus N_2 \subseteq M \oplus M$ be a submodule in $M \oplus M$ which is generated by
\begin{align*} \left\langle \begin{pmatrix} a_1 \\ b_1 \end{pmatrix}, \begin{pmatrix} a_2 \\ b_2\end{pmatrix} \right\rangle , a_i, b_i \in M \end{align*}
and lets assume I want to understand the structure of this submodule. My question is: Can I instead study the generator
\begin{align*} \left\langle  a_1 , a_2 \right\rangle\oplus \left\langle  b_1 , b_2 \right\rangle , \end{align*}
and is it true that
\begin{align*} N_1 \oplus N_2 = \left\langle  a_1 , a_2 \right\rangle\oplus \left\langle  b_1 , b_2 \right\rangle. \end{align*}

Sorry if this is a stupid question but I am not very familiar with these concepts. Thank you in advance!

EDIT: My intuition tells me that this is not possible. And if you know helpful literature regarding this topic and tell me some of them, it will be appreciated.

答:
It is not true take :

$M=\mathbb{Z}$ , $N_1=0, N_2=M$ . then $0 \oplus \mathbb{Z} \subseteq \mathbb{Z} \oplus \mathbb{Z}$ now take  : \begin{align*} S= \left\langle \begin{pmatrix} 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 \\ 1\end{pmatrix} \right\rangle \end{align*} $S$ is isomorphic to $0 \oplus \mathbb{Z}$  . in fact $0 \oplus \mathbb{Z}$ is isomorphic to  $\mathbb{Z}$ and $\mathbb{Z}$ is isomorphic to $S=\left\langle  (1 , 1) \right\rangle$ via the map : $x\rightarrow (x,x)$ .

Now do we have : \begin{align*} 0 \oplus \mathbb{Z} = \left\langle  0 , 1 \right\rangle\oplus \left\langle  0 , 1 \right\rangle. \end{align*}
The answer is obviously a no, since if we had this, then $\mathbb{Z}=\mathbb{Z} \oplus \mathbb{Z}$ which is not true.


问:
Prove $4(x^2+1)^3 \leq 5(x^3+2)^2$
I need to prove that $4(x^2+1)^3 \leq 5(x^3+2)^2$ ,  for all $x\geq 0$, in the context of an introductory analysis course. I thought I could somehow use convex functions to prove this, but I simply cannot get it to work. Hints anyone?

Title: Prove $4(x^2+1)^3 \leq 5(x^3+2)^2$

Body:
Hint :

$$ 5(x^3+2)^2 - 4(x^2+1)^3 = (x-2)^2(x^4+4x^3+4x+4)$$


问:
Understanding integral remainder for Taylor polynomials
I understand the Taylor polynomial. But when I get to the integral's remainder, I can't understand what it represents. I'm asking if someone could give me a clear explanation on that, as on the internet all I encountered was Lagrange's form.

The remainder Rn for the MacLaurin's polynomial is given by:
$$\frac{1}{n!}\int_{0}^{x}(x-t)^{n}f^{(n+1)}(t)dt$$
There's three points that are intriguing me

Is it supposed to solve the integral?

Does the $x$ in the $(x-t)^n$ represents the same $x$ of the integral? I'm a bit confused about this.

In general I don't understand what am I supposed to do with this.

I know my doubts are a bit childish (I started calculus a few weeks ago) but this is really messing with me. Any help is appreciated.

答:
No, you are not supposed to compute that integral and, yes, it is the same $x$.

Suppose that you want to compute $\frac1e$ with great accuracy. You can do it using Taylor polynomials with that formula for the remainder. First of all, note that $\frac1e=\exp(-1)$. So, for any $n\in\Bbb Z^+$, you have\begin{align}\frac1e&=\exp(-1)\\&=1-1+\frac1{2!}-\frac1{3!}+\cdots+\frac{(-1)^n}{n!}+R_n(-1).\end{align}So, you want to have a very small $|R_n(-1)|$. But\begin{align}|R_n(-1)|&=\left|\frac1{n!}\int_0^{-1}(-1-t)^n\exp^{(n+1)}(t)\,\mathrm dt\right|\\&=\frac1{n!}\left|\int_{-1}^0(-1-t)^ne^t\,\mathrm dt\right|\\&\leqslant\frac1{n!}\int_{-1}^0|t+1|^ne^t\,\mathrm dt\\&\leqslant\frac1{n!}\end{align}since, for each $t\in[-1,0]$, both $|t+1\$ and $e^t$ are smaller than or equal to $1$.

Therefore, if you want to have a numerical approximation of $\frac1e$ with an error smaller than, say, $\frac1{1000}$, just take $n=7$. That will do, since $n!>1000$. So,$$\frac1e\approx\frac1{2!}-\frac1{3!}+\cdots-\frac1{7!}=\frac{103}{280}$$and the error is smaller that $\frac1{1000}$.


问:
When can I use the formula $f_{X,Y}(x,y)=f_{X}(x)\cdot f_Y(y)$?
I have the pdf of $X$ and $Y$ to be:
$$
f_{X}(a)=f_{Y}(a)=ae^{-0.5a^{2}}I_{(0,\infty)}(a)
$$
It's also known that $X$ and $Y$ have the same distribution and are independent. I was asked to calculate the $f_{X,Y}(x,y)$. In the solution they say that:
$$
f_{X,Y}(x,y)=f_{X}(x)\cdot f_Y(y)
$$
Which I understand why is true. But what conditions should be true so I could use this formula?

Should $X$ and $Y$ have the same distribution?

Should $X$ and $Y$ be independent?

In any case that the condition is not true, how should I calculate $f_{X,Y}(x,y)$ then?

Also, can I say the same thing for $F_{X,Y}=F_X(x)\cdot F_Y(y)$?

答:
First of all, amend an obvious typo in your post as the two densities are

$$a e^{-a^2/2}$$

and not as you wrote:

$$a e^{-e^2/2}$$

just FYK, this is a known density named Rayleigh

Second, they explicitly said that $X\perp\!\!\!\!\!\!\perp Y$ and have the same distribution thus, using the definition of independence the joint density is simply

$$f_{XY}(x,y)=xye^{-(x^2+y^2)/2}\cdot\mathbb{1}_{(0;+\infty)}(x)\cdot\mathbb{1}_{(0;+\infty)}(y)$$

Also, can I say the same thing for $F_{X,Y}=F_X(x)\cdot F_Y(y)$?

Yes, you can!


Title: Prove $4(x^2+1)^3 \leq 5(x^3+2)^2$

Body: I need to prove that $4(x^2+1)^3 \leq 5(x^3+2)^2$ ,  for all $x\geq 0$, in the context of an introductory analysis course. I thought I could somehow use convex functions to prove this, but I simply cannot get it to work. Hints anyone?

Body:
we must prove :
$$4x^6+12x^4+12x^2+4 \leq 5x^6+20x^3+20$$
which rewrites to :
$$x^6-12x^4+20x^3-12x^2+16 \ge0$$
or :
$$(x-2)(x^5+2x^4-8x^3+4x^2-4x-8) \geq 0$$
or :
$$(x-2)^2(x^4+4x^3+4x+4)\geq 0$$
which is clearly true, with equality at $x=2$


Title: Prove $4(x^2+1)^3 \leq 5(x^3+2)^2$

Body:
By Holder's inequality  $$(x^3+2)(x^3+2)(4+1)\ge {\left (\sqrt[3]{x^3\cdot x^3 \cdot 4}+\sqrt[3]{2\cdot 2\cdot 1}\right)}^3=4{(x^2+1)}^3$$

Done!


答:
It's much simpler than what you think: conjugation in a graoup is an automorphism, hence
$$xg^nx^{-1}=e\iff g^n =e,$$
whence the orders of $xgx^{-1}$ and $g\:$ (least positive exponent such that $g^n=e$) are equal.

Body:
Let $G$ be a group and $g \in G$. Show that $o(g)=o(xgx^{-1})$, where $o(x)$ is the order.

I've managed to show that the order of $xgx^{-1}$ is at most $o(g)$ by the following way.

Denote $o(g) = n$ and $o(xgx^{-1}) = m.$ Now $(xgx^{-1})^n=xgx^{-1} \cdots xgx^{-1} = xg^nx^{-1} = xx^{-1} = e.$

But I'm not sure how to show this the other way around? Any hints would be appreciated. This doesn't seem to yet imply that the orders would equal one another.


答:
An elementary proof

Suppose $i<j$ and let $p$ be any prime greater than $i$. Let $a=j-i$.

Consider the sequence $p,p+a,p+2a,...,p+pa$. The initial number is prime and the final number is composite and so there is a successive pair of these numbers $u<v$ where precisely one is prime.

Then $k=u-i$ solves the problem.


问:
Maximum number of members in a club
Some people form $n=36$ clubs. There are no two clubs with same set of club members. No two people are in the same set of clubs.

If a person is not in a club, he/she must be friend with one person in that club.

However, nobody is friend with another person in the same club (yes, a weird bunch of people).

What's the maximum possible size of a club in this setup?

This is similar to the odd town/even town problem and the Fisher inequality. I suspect we could approach in a similar linear algebra setup: https://www.cs.utexas.edu/~panni/lec20.pdf

答:
My general advice is to try and solve this for small $n$ first. For me, it also helped to assume that friendship has to be an equivalence relation. Then the friendship rule becomes "every club must contain exactly one person from every equivalence class of friends", which might be easier to grasp.

Hint:

We can regard every person as a subset of the set of clubs. Since "no two people are in the same set of clubs", every person is a different subset.
So what we have to do is, find as large as possible a set of subsets of an $n$-set such that we can assign friendship in a way satisfying the rules and maximizing the largest club.

Solution:

Fortunately, we can just take all the subsets and say every set is friends with their complement set, so this setting should also give you the correct maximum size. Also note that we did not even have to assume that friendship is symmetrical nor does it matter whether a person must be friends with at least or exactly one person in every club they are not in. Of course, we also dropped the assumption that friendship has to be an equivalence relation. It just happens to be one in the solution.

Title: Maximum number of members in a club


答:
Let $i$ and $j$ be distinct positive integers and assume that there is some positive integer $k$ such that $j+k$ is prime whenever $i+k$ is prime.

Assume $i<j$

Then, for every prime number $p$ larger than $i$ which can be written as $i+(p-i)$ , the number $$j+(p-i)=p+(j-i)$$ is prime as well. Since $j-i$ is fixed, we can establish an upper bound for all prime gaps which contradicts the fact the prime gaps can become arbitarily large.

The case $i>j$ gives the same result : From some point on, the prime gaps are bounded by $i-j$ , since for every sufficient large prime $p$ , $p-(i-j)$ is prime as well.


问:
Basis for field extension $\mathbb{Q}(\sqrt{d})$ with composite d
Every time I see an example of a basis for $\mathbb{Q}(\sqrt{d})$ in my book, $d$ is prime. Is there a reason for this?

Let $d=p_1\times p_2$ where $p_1$ and $p_2$ are different primes, and $\sqrt{d} \not \in \mathbb{Q}$. Is the basis for $\mathbb{Q}(\sqrt{d})=\{a+ b\sqrt{d} \mid a,b \in \mathbb{Q}\}$?

Thanks in advance

Edit:
I meant: the basis for $\mathbb{Q}(\sqrt{d})=\{1, \sqrt{d}\}$ as pointed out in the answers. Thanks!

答:
$\{a + b\sqrt{d} \mid a, b \in \Bbb Q\}$ is not a basis for $\Bbb Q(\sqrt{d})$. It is not linearly independent. It contains $0$, for example.

One example for a $\Bbb Q$-basis for $\Bbb Q(\sqrt{d})$ is $\{1, \sqrt{d}\}$. This works not only for $d = p_1p_2$ (with $p_1 \neq p_2$) but also for any $d \in \Bbb Q$ which is not the square of a rational number. (How?)


问:
Maximum possible quantifier depth in FO$^k$ i.e First Order Logic with $k$ variables?
I have just always assumed that the quantifier depth of a function free Fist Order Logic language has to be bounded above by the number of variables in that language. Is this assumption correct ? If not can anyone give a counter example ?</p>

答:
Perhaps surprisingly, restricting the number of variables does not set a bound on quantifier depth.

In the context of finite-variable logics, variable reuse is generally permitted. (Or rather: variable reuse is generally permitted but unimportant in first-order logic in general, but when restricting the number of variables it becomes important that we're permitting variable reuse.) Permitting such reuse obviously allows formulas and sentences of arbitrary quantifier depth.

For example, the expression

$$\forall x\exists y(P(x,y)\wedge \forall x(Q(y,x))$$

is a well-formed sentence equivalent to the perhaps-more-pleasant

$$\forall x\exists y(P(x,y)\wedge\forall z(Q(y,z)).$$

See the beginning of Chapter $11$ of Libkin's Elements of finite model theory for an example of how permitting this sort of reuse leads to a rich theory. Specifically, Libkin shows how graph connectivity can be expressed as a three-variable infinitary first-order sentence, the key point being that for each $n\in\mathbb{N}$ we can express $\varphi_n(x,y)\equiv$ "the distance between $x$ and $y$ is at most $n$" as a three-variable first-order formula via judicious variable reuse. The quantifier depth of $\varphi_n$ is $2n-2$.


问:
The best way to calculate a risk score based on driving conditions
If I were to have a list of current weather conditions, including measurements like temperature, humidity, etc. What would the mathematical best way to calculate a risk score for bad driving conditions based on those measurements?</p>

答:
Get a dataset with different observations of those weather conditions and a risk score for bad driving conditions, perhaps from a human assessor or from the weather service.

Then estimate a statistical model that predicts those risk scores based on weather conditions. It could be a Logit/Probit model if your risk assessment is just binary (bad driving conditions vs good driving conditions), or ordered Logit if you have more than two categories. OLS might also be used. OLS would also have a precise meaning to "mathematical best way" - the statistical model would fit the data best in a least squares sense.


问:
Generating normal distribution by random generation of each binary bit
I'm making a circuit that draws a random amount of current by switching parallel resistors on or off. The resistances increase by factors of two from resistor to resistor, so this can be modeled as generating a word by randomly setting each bit. With a 50/50 chance of each one being on or off, the generated words follows a uniform distribution.

I want the current to be normally distributed. I can set the probability of each resistor being on or off arbitrarily. Is there any way to use this capability to get a normal distribution?

I'm leaning towards there not being one, or else there would be no need for the Box-Muller transform. But if I'm wrong, this method would be a big help.

答:
This can be done with random bits using an algorithm presented by C.F.F. Karney, "Sampling exactly from a normal distribution".

This algorithm involves building up uniform random numbers one bit at a time, from left to right, using unbiased random bits, and doing comparisons with these numbers. If these comparisons succeed, this means the initial bits of a normal random number have thus been generated, so the algorithm can fill up additional digits to the right with unbiased random bits. For details on the algorithm, see the paper.

In case the "Bernoulli trial generator" outputs biased random bits, the algorithm can produce unbiased random bits from those biased bits using a randomness extraction technique, such as von Neumann's algorithm, Peres's iterated von Neumann method, or Zhou and Bruck's "extractor tree" (see my note on randomness extraction).


问:
Number ways to distribute $k$ identical objects between $n$ people so that a special group of $a$ people receive odd number of objects among them
Suppose there is a group of $n$ people, and we need to choose exactly $k$ among them to give $k$ identical prizes. Also, exactly $a$ of these $n$ people are special. The problem is to calculate, how many ways are there to choose the $k$ people in such a way that, an odd number of special people are included in the choice?

For example, if $n = 5,\ k = 3,\ a = 3$ then there are $4$ possible ways. WLOG, we assume that the first $3$ people $p_1,\ p_2,\ p_3$ are special. Then the possible ways are:
$$
(p_1, p_2, p_3)\\
(p_1, p_4, p_5)\\
(p_2, p_4, p_5)\\
(p_3, p_4, p_5)
$$

I know the solution to a similar problem where the constraint of odd special people are not present, so we can pick any number of people from the $a$ special ones and the remaining from the $n - a$ non-special ones. There is bijection argument which shows the answer is simply ${n \choose k}$. But I can't seem to find any way to adapt this solution for the above problem with the odd constraint.

答:
The expression:
$$
\sum_{i=0}^{\left\lfloor\frac{a-1}2\right\rfloor}\binom{a}{2i+1}\binom{n-a}{k-2i-1}
$$
hardly can be generally simplified.


问:
f(n) for rows with $2^x$ bits on
This is similar to a question I asked here but now I need to alter it so I have brought across the relevant parts.

Given a binary table with n bits as follows:

$$\begin{array}{cccc|l}
2^{n-1}...&2^2&2^1&2^0&row\\ \hline \\ &0&0&0&1 \\ &0&0&1&2 \\ &0&1&0&3 \\ &0&1&1&4 \\ &1&0&0&5 \\ &1&0&1&6 \\ &1&1&0&7 \\ &1&1&1&8
\end{array}
$$

If I replace each $0$ with a $1$ and each $1$ with $g(n)$ as follows

$$\begin{array}{cccc|l}
2^{n-1}...&2^2&2^1&2^0&row\\ \hline \\ &1&1&1&1 \\ &1&1&g(0)&2 \\ &1&g(1)&1&3 \\ &1&g(1)&g(0)&4 \\ &g(2)&1&1&5 \\ &g(2)&1&g(0)&6 \\ &g(2)&g(1)&1&7 \\ &g(2)&g(1)&g(0)&8
\end{array}
$$

If I multiply the values in a row together and add up each row, thanks to Thomas Andrews, I now know that I can represent it as follow: $$f(n) = \prod_{k=0}^{n-1} (1+g(k))$$

But now I want to be more selective about the rows that I am adding, instead of adding all of them I only want to add rows that have a $2^x$ number of $g(k)$'s. i.e add all of the rows that have $2,4,8...$ g(k)'s in them.

In my example table above I would add rows 4,6,7 together as these each have 2 $g(k)$'s in them. If n=4 in my above example them we would also need to add row 16 as it would have 4 g(k)'s in its row. If n=5 there would be 5 such rows that have 4 g(k)s in them. I hope this makes sense.

If I were only concerned with rows that have 2 $g(k)$'s in them I could have something as follows:$$f(n) = \sum_{j=1}^{n-1} \sum_{k=j+1}^n g(j)g(k)$$

I can't seem to extend this to all $2^x$ rows and not only $2^1$ rows. I was considering some recursive function but can't seem to figure it out. Any help would be much appreciated.

Thanks in advance!

答:
Following the interpretation that you have, consider the expansion as $f(n, x) = \prod ( 1 + g(n) x )$, and then find the coefficients of the terms of degree $2^k$, and then add them up.

This is equivalent to adding it up manually though.

If you want to find the sum of those with an even number of $g(k)$, then that is much more approachable.

Title: f(n) for rows with $2^x$ bits on


问:
Is being locally compact a topological property?
I want to make sure that locally compact is a topological property. If it is right, then the rational number are not homeomorphic to positive integers numbers.</p>

答:
Yes, local compactness is a topological property, for it is defined purely in terms of set theory and open sets.

That homeomorphisms preserve this property is the definition of a topological property. The last question might depend on how you define "locally P" for a property P. If you say X is locally P iff every point has a neighbourhood (in the general sense) that has P, and P is itself a topological property, then yes, so is locally P, obviously. You need that a homeomorphisms preserves open sets and hence neighourhoods and is surjective, so that all points in the image are reached.


问:
Atiyah-MacDonald Ex.3.15: Every set of generators of $A^n$ has at least $n$ elements
There is the following question in Atiyah-Macdonald:

Let $A$ be a ring, and let $F$ be the $A$-module $A^n$. Show that every set of $n$-generators of $F$ is a basis of $F$.

Deduce that every set of generators of $F$ has at least $n$ elements

There is a detailed set of hints that give the first part of the question. Although I am not sure why they assume there are $n$ generators. They use the fact that we have $n$ generators $x_1,...,x_n$ to define $\phi:F\rightarrow F$ by $\phi(x_i)=e_i$. But if we have more than $n$ generators its not clear to me what to do. For example, if $x_1,...,x_n$ are generators, let's take the set of $n+1$ generators $x_1,x_1...,x_n$. Their image is no basis since we have two of the same elements.

I am not sure how to do the second part of the question.

The idea is (I imagine) to show that every basis of $F$ has at least $n$ elements. Then the result follows from the first part. Assume $x_1,...,x_d$ with $d<n$ is a set of generators for $F$. We can define a morphism $\phi: F\rightarrow A^d$ with $\phi(x_i)=e_i$. This is clearly subjective. I think the following works to show that it is injective but I'm not sure if this is  valid property for a basis of a free-module. Since $e_j$ is a basis for the free module $A^d$, $\sum a_i e_i=0$ implies each $a_i=0$. This then gives that if $0=\phi(f)=\phi(\sum a_ix_i)=\sum a_i\phi(x_i)=\sum a_ie_i$. Hence each $a_i=0$ and therefore $f=0$. So $\phi$ is an isomorphism. This gives that $F\cong A^d$ as $A$-modules. But the first part also gives that $F\cong A^n$ as $A$-modules. If we know that $A^d\cong A^n$ implies that $d=n$ we are done.

But this also confused me as this seems to prove something stronger. It proves that every set of generators has exactly $n$ elements. The question only asked to show that each set of generators has at least $n$ elements.

Why do they make the assumption that we have $n$ generators and what do we do if we have more than $n$ generators?

Is my solution for the second part correct?

Is it true that $A^d\cong A^n$ implies $d=n$ and if so how do we prove this? Or perhaps is there some easier result along the lines of: if $A^d\rightarrow A^n$ is a surjection then $d\geq n$, this seems like it would work better with the wording of the question being "at least $n$ elements.

答:
For the first part, suppose $x_1,...,x_n$ are generators of $A^n$. Since $\{e_1,...,e_n\}$ is a basis we know that the map which sends $e_i\to x_i$ for $1\leq i\leq n$ can be extended to a homomorphism of modules $\phi:A^n\to A^n$. It is obviously surjective because $x_1,x_2,...,x_n$ are all in the image of $\phi$. Now, there is a standard result that a surjective endomorphism of a finitely generated module is an isomorphism. Hence $\phi$ is an isomorphism, and so $\{x_1,...,x_n\}$ is a basis. Of course this solution wouldn't work if the set of generators had more than $n$ elements.

Now, you ask if $A^d\cong A^n$ implies $d=n$. The answer is yes. Let $M=A^n$ and suppose $\{e_1,...,e_r\}$ is some basis of $M$. Using Zorn's lemma we know that there exists some maximal ideal $I\subseteq A$. Then the quotient $M/IM$ is a vector space over the field $A/I$, where the scalar multiplication is $(a+I).(m+IM)=a.m+IM$. Try to show that $\{e_1+IM,...,e_r+IM\}$ is a basis of this vector space.

So, if $M$ has a basis of $r$ elements then $M/IM$ is an $r$-dimensional vector space over $A/I$. Since the dimension of a vector space is well defined, it follows that all bases of $M$ have the same number of elements.

Anyway, the second part of the question doesn't require that. If $\{x_1,...,x_d\}$ is a basis with $d<n$ then we again can define $\phi:A^n\to A^n$ which sends $e_i\to x_i$ for $1\leq i\leq d$ and, say, $e_i\to 0$ for $i>d$. It is a surjective endomorphism of a finitely generated module, and so has to be an isomorphism. But this is absurd, as it is clearly not injective.


问:
Discrete Fourier Transform real and imaginary parts
I am trying to find the discrete Fourier transform of a waveform with known x, y coordinates. I know that the Fourier transform produces transformed vectors in the form of sin and cosine statements. I use a Matlab plugin to generate the first 3 coefficients of the waveform, for example:

```<code>For the first coefficient: Cos:0.64, Sin:1.16
</code>```

However, I use a different implementation of the Fourier transform that generates sets of real and imaginary parts for this type of transform:

```<code>For the first coefficient: Real part:3.07, Imaginary part:-45.18
</code>```

I don't understand the relationship between sin and cosine statements and the real and imaginary parts of the first coefficient (How can I map output 1 to output 2). I need to compare the result of these two methods and see which one is more similar to the original form. Is there an online service that I can use to display each output in terms of sins and cosins and real and imaginary parts?

答:
Take a look at the complex plane:

Clearly:

$$z=a+ib=r\cos (\alpha)+i\cdot r\sin(\alpha)$$

Where:

$a$ is $z$'s real part.

$b$ is $z$'s imaginary part.

$r=\sqrt{a^2+b^2}$ (since $r$ is the hypotenuse in the triangle that is formed)

$\alpha=\arctan{(\frac ba)}$.

Thus, you can rewrite the relationship between the two parts of the equation as:

$$
\left\{
\begin{array}{c}
a=r\cos(\alpha)\\
b=r\sin(\alpha)
\end{array}
\right.
$$


答:
Lemma. For every $n\in\Bbb N$, $(xgx^{-1})^n=xg^nx^{-1}$.

Proof. Induction on $n$; trivially true for $n=1$ and true for $n$ (inductive hyp.); then:
$$(xgx^{-1})^{n+1}=(xgx^{-1})^{n}xgx^{-1}\stackrel{(i.h.)}{=}xg^nx^{-1}xgx^{-1}=xg^{n+1}x^{-1}$$

$\Box$

Claim 1. $o(g)\mid o(xgx^{-1})$.

Proof. $e=(xgx^{-1})^{o(xgx^{-1})}\stackrel{(Lemma)}{=}xg^{o(xgx^{-1})}x^{-1}\Rightarrow g^{o(xgx^{-1})}=e\Rightarrow o(g)\mid o(xgx^{-1})$.

$\Box$

Claim 2. $o(xgx^{-1})\mid o(g)$.

Proof. $(xgx^{-1})^{o(g)}\stackrel{(Lemma)}{=}xg^{o(g)}x^{-1}=xx^{-1}=e\Rightarrow o(xgx^{-1})\mid o(g)$.

$\Box$

By the two claims, $o(g)=o(xgx^{-1})$.

Title: Let $G$ be a group and $g \in G$. Show that $o(g)=o(xgx^{-1})$, where $o(x)$ is the order.


答:
The answer given would be correct in an ideal world.  Unfortunately there are (1) bugs in the implementation, and also, sadly (2) bugs in the description of the algorithm in Henri COhen's book, because he takes some formulas from Smart's book, which Smart gets from the monograph of Sinou David, but he (Smart) does so incorrectly.  (It's a question of which order you put the generating periods: is is w1/w2 or w2/w1 which is in the fundamental region.  HC knows about this (I informed him) but I don't think that there is yet a correction on his web page (which has 20 pages of errata for the two volumes).

I am preparing a better implementation -- have been doing so for a few years now -- but currently you can not assume that the list of integral points output by Sage for elliptic curves over Q is complete, even if you have the full Mordell-Weil group.  When I compute integral pints for the curve in the LMFDB I do so using both Sage and Magma and take the union.  It is some time since I saw any integral points found by Sage not found by Magma, but in fairness I should say that while I was testing the first implementation of integral points in Sage (in 2008) that happened a lot, and as a result the Magma code was completely rewritten by Steve Donnelly, using several tricks he thought up but which have never been published.

Title: Are results found of an Elliptic Curve by SageMathCell proven (does there exists no more solutions)?

Body:
Well, I have for example the following  SageMathCell-code:

```<code>sage: E = EllipticCurve(QQ, [0,63,-2205,-12348,0])
sage: E
sage: for P in E.integral_points():
....:     Q = -P
....:     print( "P = %8s and -P = %8s" % (P.xy(), Q.xy()) )
</code>```

This code computes the integral points of the Elliptic Curve that is defined by:

$$[0,63,-2205,-12348,0]\space\space\space\to\space\space\space y^2 - 2205y = x^3 + 63x^2 - 12348x\tag1$$

Are these results I get, proven to be the only ones out there? Or can there be more solutions that SageMathCell did not find?

Bytheway, the code gives the following output:

```<code>P = (-174, 1161) and -P = (-174, 1044)
P = (-147, 2205) and -P = (-147, 0)
P = (-98, 2548) and -P = (-98, -343)
P = (-68, 2528) and -P = (-68, -323)
P = (-54, 2484) and -P = (-54, -279)
P = (0, 2205) and -P =   (0, 0)
P = (57, 2052) and -P = (57, 153)
P = (84, 2205) and -P =  (84, 0)
P = (147, 3087) and -P = (147, -882)
P = (231, 4851) and -P = (231, -2646)
P = (309, 6840) and -P = (309, -4635)
P = (375, 8730) and -P = (375, -6525)
P = (378, 8820) and -P = (378, -6615)
P = (711, 20691) and -P = (711, -18486)
P = (1176, 42336) and -P = (1176, -40131)
P = (2107, 99127) and -P = (2107, -96922)
P = (2886, 157716) and -P = (2886, -155511)
P = (5412, 401472) and -P = (5412, -399267)
P = (5572, 419293) and -P = (5572, -417088)
P = (37275, 7203735) and -P = (37275, -7201530)
P = (26162409, 133818797385) and -P = (26162409, -133818795180)
</code>```


问:
How to find derivative of $\int_0^{\sin x} (1+t)^{\frac{1}{t}} dt$
I am asked to solve this question:
'Which one is the higher order infinitesimals when $\mathbf{x \rightarrow 0 \, \int_0^{\sin x} (1+t)^{\frac{1}{t}} dt}$' or $\mathbf{x^2}$.

I know that:I have to solve $\lim_{x \rightarrow 0}\frac{\int_0^{\sin x} (1+t)^{\frac{1}{t}} dt}{x}$ by using L 'Hospital's rule and compare the result with $x^2$
But I cant find the derivative.

I have recited that the derivative of $\int_{φ(x)}^{ψ(x)}f(t)dt$ (if x is not in f(t)) is $f[φ(x)]φ'(x)-f[(ψ)]ψ'(x)$, without actually knowing how it comes.

But for now, this equation seems impossible since I cannot get $(1+0)^{\frac{1}{0}}$, I don't think $\lim (1+0)^{\frac{1}{0}}$ should be right, since the hint said the $\int_0^{\sin x} (1+t)^{\frac{1}{t}} dt$ ~ $ex$

So what should I do? Is the equation I have recited succeed all the time?

答:
I expand on my comment for your benefit.

Let's first analyze the integral $\int_0^{\sin x} (1+t)^{1/t}\,dt$. You should note that the integrand is not defined at $t=0$. However, by definition the symbol $\int_a^b f(x) \, dx$ requires $f$ to be defined in whole interval $[a, b] $.

In such a situation when the integrand is not defined at a finite number of points in the interval of integration, the  convention is to define the function at those exceptional points in any manner whatsoever. This does not create any problem because the value of an integral or its existence is not affected by behavior of function at a finite number of points.

Next we define the integrand in question at $t=0$ in such a manner that the function becomes continuous there so that we can take advantage of the Fundamental Theorem of Calculus.

Let $f(0)=e$ and $f(t)=(1+t)^{1/t}$ for $t\neq 0$. Then $f$ is defined in some neighborhood of $0$ and moreover is continuous in that neighborhood.

Now $$\lim _{x\to 0}\frac{1}{x}\int_0^{\sin x} (1+t)^{1/t}\,dt=\lim_{x\to 0}\frac{\sin x}{x}\cdot\frac{1}{\sin x} \int_{0}^{\sin x} f(t) \, dt$$ The first fraction tends to $1$ and next we replace $\sin x$ with $u$ and note that $u$ tends to $0$ with $x$. Hence the desired limit is $$\lim_{u\to 0}\frac{1}{u}\int_0^u f(t) \, dt$$ By FTC this equals $f(0)=e$ as $f$ is continuous at $0$.


问:
$a \geq \frac{1}{4}$ prove that $ab(1-a^2b) \leq 1$
for all a,b $\in \mathbb{R}$ and $a \geq \frac{1}{4}$
prove that
$ab(1-a^2b) \leq 1$

答:
rewrite it as
$$(\frac{ab}{2}-1)^2+(a-\frac{1}{4})(ab)^2 \ge 0$$
which is trivial


答:
This is equivalent to

$$a^3b^2 - ab + 1 \ge 0$$

If you view this as a quadratic function of $b$, and compute the discriminant $D$, you will see that it's non-positive when $a \ge 1/4$

$D = a^2(1-4a) \le 0$

This proves the statement because the coefficient before $b^2$ in the quadratic function is positive.

Body:
for all a,b $\in \mathbb{R}$ and $a \geq \frac{1}{4}$
prove that
$ab(1-a^2b) \leq 1$


答:
Let $ab=x$ then we have to prove that $ax^2-x+1\ge0.$

The minimum of this quadratic occurs at $x=\frac{1}{2a}$ and is $1-\frac{1}{4a}\ge 0.$

Title: $a \geq \frac{1}{4}$ prove that $ab(1-a^2b) \leq 1$


问:
What does distance of a point from line being negative signify?
When we take distance from the line, we take

$$ d = \frac{ Ax_o + By_o + C}{ \sqrt{A^2 +B^2}}$$

usually with a modulus on top, now my question is if I evaluate this distance as negative what does it mean? Can I decide on which half-plane a point using this?

答:
To add to the answers, if we have a function $$ f(x,y) = Ax+By +C$$ Then we can understand that the family of parallel lines of the form $ Ax+By + C' = 0$ is level sets to the above. Taking the gradient, we find the normal of the line as:

$$ \nabla F = $$

This is what @enoztib was talking about

Body:
To add to the answers, if we have a function $$ f(x,y) = Ax+By +C$$ Then we can understand that the family of parallel lines of the form $ Ax+By + C' = 0$ is level sets to the above. Taking the gradient, we find the normal of the line as:

$$ \nabla F = <A,B>$$

This is what @enoztib was talking about


问:
Question about integral on the complex plane
That is the infernal integral:

$\int_{|z|=2}{\frac{e^{z+\frac{1}{z}}}{1-z^2}}dz$

I tried to solve it with residuals, but I can't find the expression of the function in its Laurent form.
Sorry if the question is already posted, i don't speak the language well yet.

答:
Okay, let me try to find the residue of $e^{z + \frac{1}{z}}$ in $z=0$ (outside of $0$, it is holomorphic). The other poles were treated in the comment already.

First comes the Taylor expansion of $e^x$:
$$
e^{z + \frac{1}{z}} = \sum_{n=0}^\infty \frac{1}{n!} \Big( z + \frac{1}{z} \Big)^n
$$
Using the binomial theorem, one gets
$$
 \Big( z + \frac{1}{z} \Big)^n = \sum_{k=0}^n \frac{n!}{k!(n-k)!} z^{n-k} (\frac{1}{z})^k = \sum_{k=0}^n \frac{n!}{k!(n-k)!} z^{n-2k}
$$
and the series becomes
$$
e^{z + \frac{1}{z}} = \sum_{n=0}^\infty  \sum_{k=0}^n \frac{1}{k!(n-k)!} z^{n-2k}
$$
To write this as a proper Laurent series in $z^k, k=-\infty...\infty$ may be difficult. For the residue, however, we only need to sum all coefficients with  $z^{-1}$, so all terms where $n - 2k = -1$ or $n = 2k-1$.

For each odd $n \geq 0$, there is exactly one $k$ such that $n= 2k-1$. We can turn this around and say for each $k\geq 1$ there is one $n = 2k-1$ such that $z^{n-2k} = z^{-1}$.

For the residue, we sum over all of the coefficients:
$$
\text{Res}[e^{z + \frac{1}{z}} ,0] = \sum_{\substack{k=1 \\ n=2k-1}}^\infty \frac{1}{k!(n-k)!}  = \sum_{k=1}^\infty \frac{1}{k!(k-1)!} = I_1(2) = 1.5906...
$$
where $I_1$ is the modified Bessel function.

I hope this is correct, let me know if I made any mistakes!

EDIT:

Seeing that the comment does not seem to properly adress the question, I add the treatment of the remaining poles:
$$
\frac{e^{z + \frac{1}{z}}}{1-z^2} = \frac{e^{z + \frac{1}{z}}}{(1+z)(1-z)}
$$
has three poles at $z=0, \pm 1$. The pole at $z= 0$ gives residue $\text{Res}\Big[\frac{e^{z + \frac{1}{z}}}{1-z^2}, 0\Big] = \text{Res}\Big[e^{z + \frac{1}{z}}, 0\Big] = I_1(2)$, as $\frac{1}{1-z^2}|_{z=0} = 1$.

The poles at $z= \pm 1$ yield residues:
$$
\text{Res}\Big[\frac{e^{z + \frac{1}{z}}}{1-z^2}, \pm 1\Big] = \mp \frac{e^{\pm 2}}{2} ~~,~~
$$
The integral then $\int_{|z|=2} f(z) = 2\pi i \sum_{z_0 \in \{\pm 1,0\}} \text{Res}[f(z), z_0]$ as we know from the residue theorem.


问:
What is the difference between a surface and the graph of a function?
When I was studying a book, Elementary classical analysis (Jerrold.E.Marsden), there was a confusing sentence.

"The unit sphere $x^2+y^2+z^2=1$ in $\mathbb R^3$ is a surface of the form $F(x,y,z)=c$ which is not the graph of a function."
What does this sentence mean?

Is there any relevance between above sentence and following problems?

1). find the tangent plane to the graph of $z=x^2 + y^2$ at $(0,0)$

2). Find the equation of the tangent plane to the surface $x^2 - y^2 + xyz = 1$ at $(1,0,1)$

I think I have some confusion between a surface and a graph. right?

please help me~

答:
The equation $x^2 + y^2 + z^2 = 1$ has two solutions for every $(x,y)$ satisfying $x^2 + y^2 < 1$. This means that we can't find a function $f$, such that the sphere is the graph $z = f(x,y)$. (A function has only one value for every choice of variables.)

A similar argument shows that the sphere is not the graph of a function of any other pair of the variables either.

This is similar to the one-variable setting: the circle $x^2 + y^2 = 1$ is not the graph of a function of $x$ or a function of $y$. Go through this simpler example first.


问:
How to find an expected value $E[X]$ if you're given the probabilities $P(X < 3) = \frac{1}{3}$ and $P(X \geq 6) = \frac{1}{6}$?
If I know the probabilities of a non-negative random variable constraints $P(X < 3) = \frac{1}{3}$ and $P(X \geq 6) = \frac{1}{6}$, how then I find all possible expected values E[X]?
I tried to use Markov Inequality for the second probability to then find that $E[X] \geq 1$, $X \geq 6$. I struggle to find its values if $X < 6$.

答:
Consider the three classes of value with the corresponding probabilities

$[0;3)$ with probability $1/3$

$[3;6)$ with probability $1/2$

$[6;+\infty)$ with probability $1/6$

looking at the classes, we can easy substitute $+\infty=9$, taking the center value of each class, and calcualate the expectation

$$\mathbb{E}[X]=1.5\cdot\frac{1}{3}+4.5\cdot\frac{1}{2}+7.5\cdot\frac{1}{6}=4$$

the possible values of the expectation are in the second class:

$$\mathbb{E}[X] \in [3;6)$$

Of course, in a very theoretically scenario, if you get that the third class is $[6; >33)$ the expectation can be in the third class... but considering that $83.3\%$ of the distribution is less than 6, the more consistent answer is that the expectation is in the second class of values...

Always theoretically speaking you could have an extreme situation with Expectation =2.5 when the first class is totally concentered in 0, the second in 3 and the third in 6, so a general answer is

$$\mathbb{E}[X]\geq 2.5$$


答:
Let $ab=x$, then
$$ab-a^3b^2-1\le 0 \implies ax^2-x+1 \ge 0, \forall x \in R$$
$$\implies (B^2-4AC)\le 0 \implies a\ge 1/4.$$


问:
Weakened version of Fano's inequality
How does the weakened version of the Fano's inequality $1+P_{e}\log(|X|-1) \geq H(X|Y)$ make sense because trivially $H(X|Y) \leq 1$. What am I missing?</p>

答:
The standard formulation is
$$
H(X | Y) \le H({1}\{X \ne Y\}) + P(X \ne Y) \log( |\mathcal{X}| - 1)
$$
but $1\{X \ne Y\}$ is just a Bernoulli($p$) random variable for some $p$. The entropy of this RV is maximised by $p= 1/2$ and at this value the entropy is $H(\text{Bernoulli}(1/2)= 1$.

Title: Weakened version of Fano's inequality


问:
Coefficients of $\frac{\frac{d^n}{dx^n} \sin(\ln(x))}{n!}$ are less then 1
The coefficients of $\frac{\frac{d^n}{dx^n} \sin(\ln(x))}{n!}$ are less then 1 for n >> 0?
how i prove it?

I asked a similar question some time ago, but I still can't solve it, the question can be found here Expression for $\frac{d^n}{dx^n} \sin(\ln(x))$

I am new in mathematics so sorry if this is a silly question

答:
I suppose that you are speaking about sequence $A231530$ in $OEIS$.

Unfortunately, they do not give any asymptotics for this sequence and the table gives values up to $n=22$.

I generated the values for $100\leq n \leq 1000$ and performed a quick and dirty nonlinear regression. The following seems to work
$$\log(|a_n|) \sim a n^b \big[\log(n)\big]^{c}$$  The results are reported below
$$\begin{array}{clclclclc}
 \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\
 a & 0.462783 & 0.000591 & \{0.461623,0.463943\} \\
 b & 0.954337 & 0.000240 & \{0.953866,0.954808\} \\
 c & 1.480720 & 0.001515 & \{1.477750,1.483700\} \\
\end{array}$$

For example for $n=500$, this would give for the $a_{500}$ a value of  $4.5106\times 10^{1131}$ while the exact value is $4.5488\times 10^{1131}$ while $500!=1.2201\times 10^{1134}$.

Now, considering in detail the sequence
$$b_n=\frac {|a_n|}{n!}$$ the first terms (starting at $n=5$) are
$$\left\{\frac{1}{12},\frac{1}{8},\frac{73}{504},\frac{11}{72},\frac{2795}{18144},\frac
   {611}{4032},\frac{58643}{399168},\frac{2561}{18144},\frac{322921}{2395008},\frac{8
   7805}{684288}\right\}$$ they go through a maximum value of $0.154$ for $n=9$, then decrease to almost $0$ for $n=83$, increase again up to $0.0075$ for $n=182$ and continue decreasing tending to $0$ (for $n=10^3$, the ratio is $0.00115$ and for $n=10^4$ it is $0.00019$)


问:
Evaluating an integral involving hyperbolic functions and exponentials
I'm interested in evaluating the following integral, which I encountered while working on a special partition function $\int_{0}^{+\infty} a^{2}\sinh(x)^{2} e^{-ba\cosh(x)}dx$. Even an aproximate solution is accepted.

答:
This is basically the same as

$$I(c)=\int_0^\infty \sinh^2(x) e^{-c \cosh(x)}dx$$

Use

$$\sinh^2(x) =\cosh^2(x)-1   $$

Modified Bessel function of the second kind:

$$K_0(c) =\int_0^\infty e^{-c \cosh(x)}dx $$

Using Feynman's trick of differentiation under the integral sign, we have:

$$I(c)=-K_0(c)+\frac{d^2}{d c^2} K_0(c)=K_1(c)/c$$

Here we used properties of Bessel functions from Wikipedia


问:
Solutions of a differential and its related integral equation
We are looking for solutions to the following differential equation
$$
tf'(t)-\mu f(\frac{t}{\mu})+\mu f(0)=0\;\; ;\;\;t\in \mu I,
$$
where $I$ is an interval containing $0$, $\mu\in (0,1)$ is a constant,
$\mu I=\{ \mu x: x\in I\}$, and $f:I\rightarrow \mathbb{R}$ is a differentiable function.

Note. The functions $f(t)=kt$ satisfy the equation, and the related integral equation is
$$
\int^x\frac{f(\frac{1}{\mu}t)}{t}dt=\frac{1}{\mu}f(x)+f(0)\log|x|\; ; \; x\neq 0.
$$

Now,

(1) Can one obtain its general solution?, if no,

(2) Are there some other infinite classes for its solutions?

(3) What about uniqueness conditions for a special solution?

Thanks in advance

答:
Let $u(x)=f(e^{-x})$. Then
$$
u'(x)=-e^{-x}f'(e^{-x})=-tf'(t)=−μf(t/μ)+μf(0)=−μ(u(x-\delta)-A)
$$δ
where $δ=-\lnμ$ and $A=\lim_{x\to\infty}u(x)$.

The solution formula is
$$
u(x)=u(0)−μ\int_0^{x-δ}(u(s)-A)\,ds,~~x>δ.
$$
The solution is completely determined by the values of $u$ on $[0,δ]$, or of $f$ on $[μ,1]$.

Using an exponential trial one finds a solution of $u(x)=A+Be^{-x}\implies f(t)=A+Bt$. This is of course only one special solution class among many more.


问:
Prove there exists a $y \in \mathbb{R}$ s.t. $P(X-y\in A) > \delta$
Let $X,Y$ be independent real random variables. Let $\delta \geq0 $ and $A\in \mathcal{B}(\mathbb{R})$, and suppose $P(X-Y \in A) > \delta$.

Prove there exists a $y \in \mathbb{R}$ s.t. $P(X-y\in A) > \delta$.

I have tried writing it a bit out:
$$\delta < P(X-Y \in A) = \mathbb{E}[1_{A}(X-Y)] = \int_AX-y \ P_Y(dy),$$
but I can't really see where to go from here (and if this is even the best approach). The book I am reading just says it follows from Fubini's Theorem. Can someone help me see that?

答: The joint distribution of $(X,Y)$ is the product of the marginals. Hence, Fubini/Tonelli's Theorem is applicable and we get $\delta <P(X-Y \in A)=\int \int I_{\{(x,y): x-y \in A\}} d(F_X\times F_Y) (x,y)=\int P(X-y \in A)dF_Y(y)$. Now suppose $P(X-y \in A) \leq \delta$ for all $y$. Then we get the contradiction $\delta <\int P(X-y \in A)dF_Y(y) \leq \int \delta dF_Y(y)=\delta$.


问:
Taylor expansion of $\frac{ \cos( \ln (n+1))}{(n+1)^{1/2}}$
Taylor Series Expansion of$$\frac{ \cos( \ln (n+1))}{(n+1)^{1/2}}$$

My try-

$$\cos(\ln (n+1))= \cos\left (\ln \ n+ \ln\left(1+\frac{1}{n}\right)\right)$$$$=\cos (\ln\  n) \cos\left(\ln\left(1+\frac{1}{n}\right)\right)-\sin(\ln \ n)\sin\left(\ln \left(1+\frac{1}{n}\right)\right) $$
How to proceed?

Body:
Near $x=0$, we have
$$
\log (x + 1) = x - \frac{1}{2}x^2  + \frac{1}{3}x^3  -  \cdots ,
$$
$$
\cos (x) = 1 - \frac{1}{2}x^2  +  \cdots
$$
and
$$
\frac{1}{{\sqrt {x + 1} }} = 1 - \frac{1}{2}x + \frac{3}{8}x^2  -\frac{5}{16}x^3 + \cdots .
$$
Thus
\begin{align*}
& \frac{{\cos (\log (x + 1))}}{{\sqrt {x + 1} }} = \\ &= \left( {1 - \frac{1}{2}\left( {x - \frac{1}{2}x^2  + \frac{1}{3}x^3  -  \cdots } \right)^2  +  \cdots } \right)\left( {1 - \frac{1}{2}x + \frac{3}{8}x^2  - \frac{5}{{16}}x^3  +  \cdots } \right)
\\ &
 = 1 - \frac{1}{2}x - \frac{1}{8}x^2  + \frac{7}{{16}}x^3  +  \cdots .
\end{align*}


问:
Factoring a determinant
I want to prove that $x+3$ and $x+4$ are both factors of the following determinant:

$\begin{vmatrix}x+4&2&3\\-1&x+1&-3\\0&2&x+7\end{vmatrix}$

I have done this by expanding the determinant, but I want to prove this without expansion. What I did so far is to replace row 1 by row 1+row 2, and this proves that$x+3$ is a factor, but I cannot seem to prove $x+4$ is a factor. Any help?

答:
After what you did, multiply the 1st row by $(-1)$ and add it to the 3rd row. And you get the $(x+4)$ factor too (in the 3rd row).

In other words, subtract the 1st row from the 3rd one.

Title: Factoring a determinant


答:
Note that from the expression of the determinant, you can conclude that it will be a polynomial in $x$. Let us call it $D(x)$.

Now, $x + 3$ being a factor of $D(x)$ is equivalent to $D(-3) = 0$. What do you get upon substituting $x = 3$ in the determinant? Do you see that the first row is a multiple of the second?

Similarly, what is $D(-4)$? Can you conclude now?

Title: Factoring a determinant

Body:
I want to prove that $x+3$ and $x+4$ are both factors of the following determinant:

$\begin{vmatrix}x+4&2&3\\-1&x+1&-3\\0&2&x+7\end{vmatrix}$

I have done this by expanding the determinant, but I want to prove this without expansion. What I did so far is to replace row 1 by row 1+row 2, and this proves that$x+3$ is a factor, but I cannot seem to prove $x+4$ is a factor. Any help?


问:
What is the minimum level of certainty so that the witness statement was correct with $50\%$ probability?
At night there was a serious traffic accident with a hit-and-run offence in front of the main train station.

There is only one witness who claims it was a green taxi.

There are only green and blue taxis in the city, $20\%$ green and $80\%$ blue taxis.

There is no doubt that the accident was caused by a taxi.

(a) What is the minimum level of certainty that the witness must be able to distinguish between green and blue taxis at night so that the witness statement "It was a green taxi." was correct with $50\%$ probability?

(b) How high is the probability of having made a correct testimony if the witness can distinguish between blue and green taxis at night with a $75\%$ probability?

$$$$

For question (b) I have done the following :

We consider the following events:

$A$: the taxi in the Accident was green

$B$: the witness identifies the taxi as green

From Bayes' theorem we get: $$P(A\mid B) = \frac{P(B\mid A)\cdot P(A)}{P(B)}$$

$P(B\mid A)=0.75$ (if it really was green, the witness identifies it as green with a probability of $75\%$)

$P(A) =0.20$ (since $20\%$ of the taxis in the city are green)

$P(B)=P(B\mid A)P(A)+P(B\mid \neg A)P(\neg A)=0.75\cdot 0.20+(1-0.75)\cdot 0.80=0.35$

So we get $$P(A\mid B) = \frac{0.75\cdot 0.20}{0.35}$$ Is that correct?

Could you give me a hint for question (a) ? Do we maybe do the same but with $P(A\mid B)=0.5$ and $P(B\mid A)$ is unknown?

答:
Q(b). Yes this is the correct answer which  could be given as $\frac{3}{7}.$

Q(a). If $p$ is the probability of correctly distinguishing , then
$$0.5=\frac{0.2\times p}{0.2\times p+0.8\times (1-p)}$$

and $p=\frac{4}{5}$.


问:
sum of two dedekind cuts
according to Dedekind cut definition in Pugh analysis book, if $ x=A|B $ and $y=C|D$ definition of two cuts addition is as following $x+y=E|F$ which E={e=a+c:$a\in A$ and $c\in C$} and F=Q\E I can understand that why E and F are nonempty and E doesn't have largest element but I can't understand why every element of E is less than F. can someone please help me with it.

答: If $x \in E$ and $y \le x$, then $x = a + c$ for some $a \in A$ and $c \in C$, but then $y = (a + y - x) + c$ and $a + y - x \le a$, so $a + y - x \in A$, because $A\mid B$ is a cut, giving us that $y \in E$. Hence, if $F = \Bbb{Q} \setminus E$ and $f \in F$ we cannot have $f \le x$ for any $x\in E$, so we must have that $f > x$ for every $x\in E$.

Title: sum of two dedekind cuts