问:
How to deal with Statistical outliers in Rstudio
I have some outliers in a data set that I would like to deal with.

After doing some reading into the topic I came to the conclusion that the trimmed mean function and the boxplot (in rstudio) function might come in handy when trying to deal with the outliers.

I like to work with concrete baby examples, so let me just give a concrete case and show you what I have done so far, perhaps you could guide me in the correct direction.

$a$ <- $c(1,2,3,100,5,3,200,5,2,3) \implies mean(a) = 32,4.$

Obviously the two outliers are 100 and 200. Thus, if we want to find a more "realistic" mean value we simply remove 100 and 200 from the data set.

Since n = 10 in the data set a, and since there are two outliers (100,200), the "correct" way to trim a from the outliers would be the following command:

$mean(a, trim = 0.2) = mean(3+5+3+5+2+3) = 3.5$

In this baby example it was easy to find the "correct" trim value, which in this case was $0.2 = 20%$.

What I am looking for is a more general method on, how I can (given a large data set) find the correct trim value such that I can remove the outliers that are artificially skyrocketing the mean value.

Boxplot seems to be a good way to visualize the outliers, but in a very large dataset it wouldn't give me any concrete numbers regarding how many outliers there might.

While writing this question I just came up with an idea. Perhaps I could:

Loop through the columns in the row that I want to analyze for outliers.

If the number in the column > the real mean then count.

Now I have a variable which contain the number of columns in the row > the real mean.

If say, (number of columns in the row > the real mean) = 10 and the total amount of columns in the row is 100. Then to remove the 10 outliers from the 100 columns, we say that the trim value should be $\frac{10}{100} = 0.1$

So the command dealing with the 10 outliers would be
$mean(vector, trim = 0.1)$

Any comments or help is appreciated.

答:
If you have a samples of size $n = 100$ from a normal
distribution, you will typically see outliers in half or more of the samples.
Here are boxplots of 20 samples, each of size 100.

```<code>x = rnorm(2000, 100, 15)
g = rep(1:20, each=100)
boxplot(x ~ g, col="skyblue2", pch=20)
</code>```

More than half have at least one outlier.

For exponential data, almost all sample of size 100 have outliers; outliers are a typical characteristic of
this distribution.

```<code>x = rexp(2000, .01)
g = rep(1:20, each=100)
boxplot(x ~ g, col="skyblue2", pch=20)
</code>```

You really want to delete all of those outliers?

Body:
If you have a samples of size $n = 100$ from a normal
distribution, you will typically see outliers in half or more of the samples.
Here are boxplots of 20 samples, each of size 100.

x = rnorm(2000, 100, 15)
g = rep(1:20, each=100)
boxplot(x ~ g, col="skyblue2", pch=20)

More than half have at least one outlier.

For exponential data, almost all sample of size 100 have outliers; outliers are a typical characteristic of
this distribution.

x = rexp(2000, .01)
g = rep(1:20, each=100)
boxplot(x ~ g, col="skyblue2", pch=20)

You really want to delete all of those outliers?


问:
locus of points with sum of distance

Find the quadratic equation for the locus of points whose sum of its distance from $(1, 0)$ and $(−1, 0)$ is $6$.

I know to start with $\sqrt{(x-1)^2 +y^2} + \sqrt{(x+1)^2+y^2}=6$, then square both sides. I'm not sure about the algebra after squaring both sides as I got that part wrong.

答:
$$\sqrt{(x-1)^2 +y^2} + \sqrt{(x+1)^2+y^2}=6$$
$$\sqrt{(x+1)^2+y^2}=6-\sqrt{(x-1)^2+y^2}$$
square both sides ($LHS>0;\;RHS>0$)
$$(x+1)^2+y^2=\left(6-\sqrt{(x-1)^2+y^2}\right)^2$$
$$x^2+2 x+y^2+1=36+x^2-2 x+1+y^2-12 \sqrt{(x-1)^2+y^2}$$
bring every term but the square root in the RHS
$$12 \sqrt{(x-1)^2+y^2}=36-4 x$$
simplify dividing both sides by $4$
$$3 \sqrt{(x-1)^2+y^2}=9-x$$
Notice that both sides are positive, so square again
$$9 \left((x-1)^2+y^2\right)=(9-x)^2$$
expand and bring variables into LHS and constants in the RHS
$$8 x^2+9 y^2=72$$
divide both sides by $72$
$$\frac{x^2}{9}+\frac{y^2}{8}=1$$
Which is the equation of an ellipse center $(0,0)$ and semiaxes $a=3;\;b=2\sqrt{2}$

ContentLicense: CC BY-SA 4.0

Title: locus of points with sum of distance


问:
Prove if $\lim_{k \rightarrow \infty} A_k = A$, then $\lim_{k \rightarrow \infty} A_k^{-1} = A^{-1}$
Let $\{A_k\}_{k=1}^{\infty}$ be a sequence of real $n \times n$ matrices. Suppose $A_k$ is invertible for all $k \in \mathbb{N}$, that $\lim_{k \rightarrow \infty} A_k = A$, and that $A$ is invertible. Is $\lim_{k \rightarrow \infty} A_k^{-1} = A^{-1}$?

This feels like it should be true but I'm having trouble coming up with a proof.

Also, I think defining $\lim_{k \rightarrow \infty} A_k = A$ as componentwise convergence makes the most sense for my problem but any other insights are welcome!

答: Sine $A_k$ converge, so do their determinants (which are polynomial functions of the entries), and the limit of the determinant is NOT zero, so do their cofactors (which are polynomial functions of the entries) and so do the cofactors over the determinant, which are the entries of the inverses.


问:
Find the value of $\lfloor (\sqrt{21}+\sqrt{29})^{1984}\rfloor \mod 100$
My question is to find the value of $$\lfloor (\sqrt{21}+\sqrt{29})^{1984}\rfloor \mod 100$$
Mathematica told me the result is 71 and $$\lfloor (\sqrt{21}+\sqrt{29})^{20k+4}\rfloor \mod 100$$ is always 71 for $k\in \mathbb N$, can anyone explain the result?

Body:
Let $\alpha = (\sqrt{21}+\sqrt{29})^{2} = 50 + 2 \sqrt{609}$. Then $\alpha^n = x_n + y_n \sqrt{609}$ with $x_n, y_n \in \mathbb Z$.

Let $\beta = (\sqrt{21}-\sqrt{29})^{2} = 50 - 2 \sqrt{609}$. Then $\beta^n = x_n + y_n \sqrt{609}$ and so $\alpha^n+\beta^n = 2x_n \in \mathbb Z$.

Since $0 < \beta < 1$, we get $\lfloor \alpha^n \rfloor = 2x_n-1$.

Since $\alpha^2 = 100 \alpha -64$, we get $x_{n+2} = 100 x_{n+1} - 64 x_n$, with $x_0=1$ and $x_1=50$.

Let $z_n = x_n \bmod 100$. Then $z_{n+2} \equiv 36 z_n \bmod 100$ and so $z_{2n} \equiv 36^n \bmod 100$.

Now, $(\sqrt{21}+\sqrt{29})^{1984} = \alpha^{992}$ and so $z_{992} \equiv 36^{496} \bmod 100$.

Finally, $36^n \bmod 100$ is a pre-periodic sequence of period $5$ starting at $n=1$:
$$
1,36,96,56,16,76,36,96,56,16,76,\dots
$$
that is, $36^{n+5} \equiv 36^n \bmod 100$ for $n\ge 1$. In particular, $36^{5n+1} \equiv 36 \bmod 100$ for all $n$.

Since $496 =5 \cdot 99 + 1$, we get $36^{496} \equiv 36 \bmod 100$.

Therefore, $\lfloor (\sqrt{21}+\sqrt{29})^{1984}\rfloor = 2x_{992}-1 \equiv 2z_{992}-1 \equiv 2 \cdot 36 -1 = 71\bmod 100$.


问:
Group Theory - Prove that power set expression is a subset of another
I've been asked to prove that: $P(A\setminus B)\subseteq P(A)\setminus P(B) \cup \{\emptyset\}$.

By the laws of group subtraction, I've concluded that:

$\alpha =P(A\setminus B) = P(A \cap B^c) $

$\beta = P(A)\setminus P(B) \cup \{\emptyset\} = P(A)\cap P(B)^c \cup \{\emptyset \}$

But I'm not sure where to go on from here.

It can be said that both sets have a member in the form of the empty set.
But how can I prove that each member of set $\alpha$ is also a member in $\beta$?

答: If $X\in\mathcal P(A\setminus B)$, and if $X\ne\emptyset$, then $X\subset A$; therefore, $X\in\mathcal P(A)$. But $X$ contains no element of $B$; in particular, and since $X\ne\emptyset$, $X\notin\mathcal P(B)$. So, $X\in\mathcal P(A)\setminus\mathcal P(B)$.


问:
Does there exist a sequence of polynomials converging uniformly to $f(x) = |x| $ on $ℝ$?
Does there exist a sequence of polynomials converging uniformly to $f(x) = |x|$ on $ℝ$?

I understand this is possible on a closed interval due to Weierstrass Approximation Theorem.

Intuitively, I think there's no such sequence on $ℝ$ but I'm not sure how to prove this rigorously.

Any hints would be so useful!

答: There are not any $p_1 \neq p_2$ such that $|p_1 -p_2|$ is bounded, unless $p_1 - p_2$ is a constant. This means that for a sequence of polynomials to converge uniformly, all the coefficients but the constant terms must be (eventually) fixed, so the limit of such a sequence must be a polynomial. $|x|$ is not a polynomial.


问:
Write permutation as disjoint cycles $(5 1 2)(3 1 2)(4 1 3)$
I'm really confused with permutations overall, I know the "algorithm" yet every time I get a mistake...

Write this permutation in $S_6$ as disjoint cycles and find its order: $(5 1 2)(3 1 2)(4 1 3)$

My try:

$$\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6\\
2 & 3 & 4 & 5 & 1 & 6\\
\end{pmatrix}$$
so $(5 1 2)(3 1 2)(4 1 3)=(1 2 3 4 5)$.

Yet the solution in WolframAlpha states: $(1 4)(2 5)$ (order $2$ yet in my case the order is $5$). Am I mistaken?

Is there any good website for checking this sort of calculation (or is WolframAlpha actually correct)?

Thank you in advance.

答: WolframAlpha evaluates permutations from left to right. Your calculation is correct when evaluating from right to left.</p>


问:
How can I show that a set is dense in $\mathbb{R}?$
Excuse me, the problem is:

If $u>0$ is any number and $x<y$, show that exists $r\in\mathbb{Q}$ such that $$x<ru<y.$$

I know that the problem is show that $\{ru:r\in\mathbb{Q},u>0\}$ is dense in $\mathbb{R}$. I had tried with the archimedean property, and other thing but I can't do it. I feel mentally blocket.

Thanks.

答: You want to find a rational number between $\frac{x}{u}$ and $\frac{y}{u}.$


问:
Approximating the roots of $\epsilon^{2}x^{3}+x+1$
I saw the following in my lecture notes, and I am having difficulties
verifying the steps taken.

The question is:

Assuming $0<\epsilon\ll1$ find all the roots of the polynomial
  $$\epsilon^{2}x^{3}+x+1$$ which are $O(1)$ up to a precision of
  $O(\epsilon^{2})$

and the solution given was

Assume that $x=O(1)$ and that $$x(\epsilon)=x_{0}+\epsilon
 x_{1}+O(\epsilon^{2})$$ Then by setting it in the equation and letting
  $\epsilon\to0$ we get $$x_{0}=-1,x_{1}=0$$

Hence $x(\epsilon)=-1+O(\epsilon^{2})$

I have two questions:

Where did we use the assumption that $x=O(1)$

How did they get $$x_{0}=-1,x_{1}=0 ?$$

When I did the step setting
it in the equation and letting $\epsilon\to0$ I got $$x_{0}+1+O(\epsilon^{2})=0$$
and so I don't know anything about $x_{1}$.

Should I ignore $O(\epsilon^{2})$
and from that I should get $x_{0}=-1$

答:
Much simpler: As $x(\epsilon)\in O(1)$, we have immediately from rewriting the cubic that
$$x(\epsilon)=-1-\epsilon^2x(\epsilon)^3\in -1+O(\epsilon^2).$$</p>

ContentLicense: CC BY-SA 3.0


问:
floor function inequality $n+ \left \lfloor \tfrac{n}{3} \right \rfloor \le x$
I have the following problem which has two parts:
Given any $x>0$, I define a function $A$ by
\begin{equation}
A(x)=\max \{ n \in \mathbb{N} : n+ \left \lfloor \tfrac{n}{3} \right \rfloor  \le x\}
\end{equation}
I want to compute $A(x)$ and
\begin{equation}
\lim_{x \to \infty} \dfrac{A(x)}{x}.
\end{equation}

I tried to use the fact that $ x \le \lfloor x \rfloor \le x+1 $ but I got stuck.
Do I have to do a case by case analysis of if $n$ is divisible by $3$?

答: If you drop the floors, then $n=\frac{3 x}{4}.$ Putting the floors back in, check that $\lfloor \frac{3 x}{4}\rfloor$ is your $n.$ The limit is then easy to compute.


问:
Prove that the inverse image of an open set is open
Let $ X \subset \mathbb{R}$ be a non-empty, open set and let $f: X \rightarrow \mathbb{R}$ be a continuous function. Show that the inverse image of an open set is open under f, i.e. show:

If $M \subset \mathbb{R}$ is open, then $f^{-1}(M)$ is open as well.

I think that this should basically follow from the definitions, but I'm still having some troubles with the proof.

Suppose $M \subset \mathbb{R}$ is open, then it follows that $\forall z \in M, \exists r > 0: K_r(z) \subseteq M$

Since f is continuous it follows for each $x_0 \in X$ that for each $\epsilon > 0 ,\exists \delta > 0 : |f(x)-f(x_0)|< \epsilon, \forall x \in X: |x-x_0|<\delta$

Now I need to show that $f^{-1}(M)$ is open, so that for each $z \in M, \exists r>0: K_r(f^{-1}(z_0)) \subseteq f^{-1}(M)$

I'm not sure how to prove this. I guess since $z = f(x)$ and since M is open we have that $\exists r>0:\{z:|z-z_0|<r\}\subseteq M$. And since f is continuous: $|f(x)-f(x_0)| = |z-z_0|< \epsilon$ for all $|x-x_0|<\delta$, but since $|x-x_0|=|f^{-1}(z)-f^{-1}(z_0)|<\delta$ it somehow follows from that that $f^{-1}(M)$ is open?!

Which is just horrible and probably completely wrong. But I'm really confused from all the definitions right now and need some help please.

答:
Consider a point $f(z)$ in $U$. For clarity, the theorem assumes $U$ is open.

Since all the points in U are mapped to the set $f^{-1}(U)$, now consider the equivalent point $z$ in $f^{-1}(U)$.

Remember, we know that $f(z)$ is in $U$. All points in $U$ have a ball around them of at least $\epsilon$. Since $f$ is continuous, there must exist a $\delta>0$ such that $z$ is in a ball of radius $\delta$, and that all points in that ball are mapped by $f$ to $U$, in fact, they end up within $\epsilon$ of $f(z)$. This is true for all $z$ such that $f(z)$ is in $U$. The fact that a ball exists of non-zero radius around each point $z$ in some set $X \subset \mathbb{R}$ is equivalent to that set being open. So it is done.


问:
Open maps of spaces vs open maps of locales
A continuous function between topological spaces $f:X\to Y$ is called open, if $f[U]\in\mathcal{O}(Y)$ for all $U\in\mathcal{O}(X)$.

A morphism of locales $f:X\to Y$ is called open, if the associated morphism of frames $f^*:\mathcal{O}(Y)\to\mathcal{O}(X)$ has a monotone left adjoint $f_!$ which satisfies the Frobenius condition $f_!(U)\wedge V\leq f_!(U\wedge f^*(V))$ for all $U\in\mathcal{O}(X)$ and $V\in\mathcal{O}(Y)$.

It is easy to see that every continuous $f$ that is open in the topological sense is also open in the localic sense, but the converse is false in general. For example,  the maps $1\to\nabla(2)$ and $\mathsf{cofin}(\mathbb{N})\to\mathsf{sobrify}(\mathsf{cofin}(\mathbb{N}))$ are open in the localic, but not in the topological sense, since they're not surjective, but the induced locale-morphisms are iso.

These two examples involve non-sober spaces, so the natural thing to ask is: Do the two notions coincide for sober spaces? Or more generally, does the canonical functor from locales to spaces preserve open maps?

Or are there other criteria for when the two notions coincide? For example, I think I have convinced myself that continuous maps that are open in the localic sense are open in the topological sense whenever the codomain is $T_1$.

答:
Sketches of an Elephant notes (before Lemma C1.5.3)

Of course, we say a [locale morphism] $f: X \to Y$ is open if, for every open sublocale $U$ of $X$, its image $f_!(U)$ in $\mathrm{Sub}(Y)$ is an open sublocale of $Y$....this is not quite equivalent to the usual notion of openness for continuous maps of spaces; it is implied by the latter, but in the opposite direction it only implies that the set-theoretic image of each open subset of $X$ is 'almost open' in $Y$, in the sense that its subclosure is open. Even when X and Y are both sober, this does not suffice to make all such images open..., but it obviously does so if $Y$ is a $T_D$-space.

The distinction here seems to correlate with the distinction between subspaces and sublocales. The latter extends to a distinction between image of a subspace and image of a sublocale, which are invoked in the definition of open map of spaces and locales, respectively.


问:
Problem with central limit theory
I understand a proof of CLT and I'm using it frequantly, but just realised, that I dont understand one thing - how is it possibile that mean from uniform distribution can have Normal distribution? My problem od that mean from uniform is strictly limited, and cant exced given value, where in normal distribution you can have values form (-inf to inf). I hope I formulated the problem clearly :)</p>

答: The reason that the theorem is called the "central limit theorem" is that it describes the functional form of the center (middle) of the distribution. It really has nothing to say about the tails (something that is frequently misunderstood).

Title: Problem with central limit theory


问:
How to calculate an inverse Fourier transform of product $\operatorname{rect}(f)$ with $\frac{1}{f+K}$?
I need to calculate an inverse Fourier transform $\int_{-\infty}^{\infty} \bigl( \frac{1}{f+K} \bigr) \, \operatorname{rect}(Bf) \, e^{j 2\pi f x} \, df$,

where $K$ is a constant, and $B$ is the width of the rectangular box:
$\operatorname{rect}(Bf) = \begin{cases}1 &, -B/2 \leqslant f \leqslant B/2\\ 0 &, \text{otherwise.} \end{cases}$.

I ended up at a stage where I need to calculate the convolution $ (e^{-j2\pi Kx} j\pi\operatorname{sgn}(x)) * \operatorname{sinc}(Bx)$.

Any hints? Thanks in advance.

答:
Using the notation:

$$\Pi(t) = \begin{cases} 1 & |t| < \frac{1}{2} \\
0 & |t| > \frac{1}{2}
\end{cases}$$
Hint:

$$\begin{align*} I &= \int_{-\infty}^\infty \dfrac{1}{f+K}\Pi\left(\dfrac{f}{B}\right)e^{j2\pi xf}df\\
\\
&= \int_{-\frac{B}{2}}^{\frac{B}{2}} \dfrac{1}{f+K}e^{j2\pi xf}df\\
\\
&= e^{-j2\pi x K}\int_{-\frac{B}{2}}^{\frac{B}{2}} \dfrac{1}{f+K}e^{j2\pi x(f+K)}df\\
\\
u &= f+K\\
\\
\dots
\end{align*}$$

Finish the $u$ substitution and do the integration by parts.


问:
Calculation of determinant for differential matrix equations
Here is the differential equation $$N'_x(x)=G(x)N(x)$$ where $N, G$ are $2\times2$ matrices depending on $x$, and $G$ satisfies $\mathrm{trace}  (G)=0$. My question is:

How can one then calculate the determinant of $N$, since for $G$ not a constant, one cannot use an exponential of a matrix and the formula $\det e^{A}=e^{\mathrm{trace} (A)}$?

答: See http://en.wikipedia.org/wiki/Jacobi's_formula - using this together with the cyclic nature of the trace and properties of the adjugate should get you where you need.</p>


问:
Struggle with obtaining same answer for both Disk-Washer Method and Shell Method on same Problem
The problem is straightforward with the Disk-Washer method.  To help me fully understand these rotation problems I like to work the problems using both Disk-Washer and Shell (some problems become too difficult to solve for the x or y variable, but most work at the Calculus 1 level).

The problem is:  Find the volume of the solid that results when the region is rotated about the y-axis.

$y=3-2x, y=2, y=0, x=0$

Using $V=\pi \int_{0}^2 (\frac{3-y}{2})^2 dy$ I get the correct answer of $13\pi/6$

When evaluating how to do this via the Shell method I notice that I will need to break this region into 2 pieces - a rectangle and a triangle.

Since the line $y=2$ intersects the line $y=3-2x$ at $(\frac{1}{2},2)$ I chose the two shell integrals as follows:

$V=2 \pi\int_{0}^\frac{1}{2} (x)(2) dx + 2\pi\int_{\frac{1}{2}}^{\frac{3}{2}} (x-\frac{1}{2})(3-2x)dx $

This does NOT produce the same answer. I think the correct answer is produced if I use:

$2\pi\int_{\frac{1}{2}}^{\frac{3}{2}} (x)(3-2x)dx $ for the second integral.  But I don't understand this.  For my second integral, it would seem to me that my radius is NOT $x$ but $(x-\frac{1}{2})$.

Can you assist with doing this problem via the Shell Method?

答: In the second integral, the cylinder radius should be $x$, not $x-\tfrac{1}{2}$, as it is the distance of the position of the cylindrical shell from the rotation axis: the line $x=0$, not the line $x=\tfrac{1}{2}$.  With integrand $x(3-2x)$, you obtain the result $\frac{13\pi}{6}$ as with disks.


问:
Is $f(n)=n!$, multiplicative, completely multiplicative, or both?
I do not understand how $f(n)=n!$ is multiplicative or completely multiplicative. I know that if gcd$(m,n)=1$ and $f(mn)=f(m)f(n)$, then it is multiplicative. If $f(mn)=f(m)f(n)$ is true for all integers, then it is completely multiplicative.

Let's take $2$, and $3$, for example. $f((2)(3)=6)=6!$ which is not equal to $2!3!=f(2)f(3). So it is not completely multiplicative.

Now let's take $2$, and $4$, for example. $f((2)(4)=8)=8!$ which is not equal to $2!4!=f(2)f(4). So it is not multiplicative.

The directions given said it has to be at least one but I think it is neither. Any hint would be helpful. Thank you!

答: It is neither, as you can tell by observing that $2! 3! \neq 6!.$


问:
Expected value of 3 to the power of sum of 2 dice
Two dice are rolled and then 3 gets raised to the power of the sum of these two dice's values. What is the expected value of the expression.

Solution
The expected value of one die is $3.5$ and the expected value of their sum should be $7$, because the dice are independent. Thus, the expected value is $3^7$.

Question: Is this a correct solution?

答:
Easiest approach is to take off your shoes.

$\sum_{k=2}^{12} ~p(k)3^k$, where $p(k) =$ probability that $k$ was rolled.

If you get to the point where you accept that the above expression is accurate, then any alternative expression that does not result in the same final computation can't be correct.


答:
Following @user2661923 (and yes... my shoes are off), here's the distribution of dice sums:

and here is the derived distribution:

...and the mean of the final distribution is: $33124$.

For those familiar with Mathematica:

```<code>q = TransformedDistribution[
   x + y,
   {x \[Distributed] DiscreteUniformDistribution[{1, 6}],
    y \[Distributed] DiscreteUniformDistribution[{1, 6}]}];

DiscretePlot[PDF[q, x], {x, 2, 12}]

z = TransformedDistribution[3^x, x \[Distributed] q];

Mean[z]
</code>```

(* 33124 *)

```<code>DiscretePlot[{3^y, PDF[z, x^y]}, {y, 2, 12},
 Ticks -> {Table[{y, 3^y}, {y, 2, 12}], Automatic},
 PlotRange -> {{1, 12}, Automatic}]
</code>```

Title: Expected value of 3 to the power of sum of 2 dice

Body:
Two dice are rolled and then 3 gets raised to the power of the sum of these two dice's values. What is the expected value of the expression.

Solution
The expected value of one die is $3.5$ and the expected value of their sum should be $7$, because the dice are independent. Thus, the expected value is $3^7$.

Question: Is this a correct solution?

Body:
Following @user2661923 (and yes... my shoes are off), here's the distribution of dice sums:

and here is the derived distribution:

...and the mean of the final distribution is: $33124$.

For those familiar with Mathematica:

q = TransformedDistribution[
   x + y,
   {x \[Distributed] DiscreteUniformDistribution[{1, 6}],
    y \[Distributed] DiscreteUniformDistribution[{1, 6}]}];

DiscretePlot[PDF[q, x], {x, 2, 12}]

z = TransformedDistribution[3^x, x \[Distributed] q];

Mean[z]

(* 33124 *)

DiscretePlot[{3^y, PDF[z, x^y]}, {y, 2, 12},
 Ticks -> {Table[{y, 3^y}, {y, 2, 12}], Automatic},
 PlotRange -> {{1, 12}, Automatic}]


问:
Proving $\mathbb{Q\times Z}$ is countable
I want to prove that  $\mathbb{Q\times Z}$ is countable. I found bijection from $\Bbb N\times \Bbb N$ to $\Bbb N$. Can I use it?

答:
As someone noticed, you only need an injection: indeed, an infinite subset of $\mathbb{N}$ is itself countable! It turns out that your ingredient $\psi: \mathbb{N} \times \mathbb{N} \overset{\simeq}{\to} \mathbb{N}$ is enough to cook up the desired injection.

Let's see the $\psi$ power at work: a fraction can be written in a unique way as $\pm \frac{p}{q}$, where $p,q$ are coprime. We can define an injection $r: \mathbb{Q} \to \mathbb{Z} $ as

$$r(\pm p/q) = \pm \psi(p,q) \in \mathbb{Z}$$

Note that $\mathbb{Z}$ is also countable: just send $+m $ to $2m$ and $-m$ to $2m+1$ for $m\ge 0$. Denote this bijection by $\varphi: \mathbb{Z} \to \mathbb{N}$.

We are left with the final acrobatics:

$$ \mathbb{Q} \times \mathbb{Z} \overset{(r,1)}{\hookrightarrow} \mathbb{Z} \times \mathbb{Z} \overset{(\varphi, \varphi)}{\hookrightarrow} \mathbb{N} \times \mathbb{N} \overset{\psi}{\hookrightarrow} \mathbb{N} $$

Yey! A little remark: we only used that $\psi$ is injective. A simple example of such $\psi$ is $(n,m) \mapsto 2^n 3^m$.


问:
Integral with arctan and e: $\int_{0}^{\infty}\frac{\arctan(x^{3})}{e^{2\pi x}-1}\,\mathrm dx$
Here is an integral I ran across that appears to be tough.

It seems to me I have seen integrals like this before.  I have looked around the site, but saw nothing like this.

$$\displaystyle \int_{0}^{\infty}\frac{\arctan(x^{3})}{e^{2\pi x}-1}\,\mathrm dx=\frac{1}{4}\ln(2\pi)-\frac{\pi}{4\sqrt{3}}-\frac{1}{2}\ln(1+e^{-\pi\sqrt{3}})$$

I was wondering if it is possible to do this with residues/branch cuts?.

I have not seen many integrals with inverse trig functions done using residues.  Maybe there is a reason for that :)

I realize that $\arctan(z^{3})=\frac{i}{2}\ln\left(\left|\frac{z^{3}-i}{z^{3}+i}\right|\right)-\frac{1}{2}\operatorname{Arg}\left(\frac{z^{3}-i}{z^{3}+i}\right)$.

Then, the zeros of $z^{3}+i$ and $z^{3}-i$ would have to be dealt with, as well as those of $e^{2\pi z}-1=0$.

But, I am not sure how to proceed...if possible.

Besides using contours, any clever method would be nice to see and appreciated.

Thanks all.

Body: you can seehttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=67&t=493626


问:
Finding the number of positive integers that are less than 1000000
How  many  positive  integers  less  than  $1,000,000$  have  exactly  one  digit  that  is  $8$  and  the product of this digit $(8)$ with the sum of other digits is between $49$ and $69$?

Been stuck on this problem and trying to come up with a pretty solution because I can't to seem to find one.

We figured out that this has six spots because it cannot have $7$ or else it would be greater than a million: $$\_ \space \_ \space\_ \space\_ \space\_ \space\_ \space$$

And there is always an 8 and not counted toward the sum: $$\_ \space \_ \space\_ \space\_ \space\_ \space 8 \space$$

This sum has to add to $7$ or $8$, becuase $49 \le (7\cdot8) \lor (8\cdot8) \le 69$.

An example of a valid way to find a sum: $$348000 \rightarrow (3+4)\cdot8 = 56$$

Can anyone help me to create a simple counting method to find all the integers?

答:
The punchline is going to be that you first choose the location of the $8$ in the six slots in $6$ ways.

You then distribute $7$ units among the remaining five spots in $\binom{7+5-1}{5-1}$ ways for the case of the overall product being $8\times 7=56$ or distribute $8$ units among the remaining five spots in $\binom{8+5-1}{5-1}$ ways in the case of the product being $8\times 8 = 64$, as per Stars-and-bars.

We recognize that among the outcomes counted in that second case were some we did not want to, namely those where we had two $8$'s and the rest of the digits are $0$'s.  There are $\binom{6}{2}$ of these.  We had accidentally counted them twice when we wanted to count them none.

Our total is then:

$$6\times \left(\binom{7+5-1}{5-1}+\binom{8+5-1}{5-1}\right)-2\times \binom{6}{2}$$


问:
Change of basis matrix represented by linear maps
Consider the basis $\alpha=\{v_{1},...,v_{n}\}$.

Suppose now that we have a vector $a$ which is expressed as coordinate
vectors in respect to $\alpha=[a]_{\alpha}.$

The vector $[a]_{e}=\lambda_{1}v_{1}+,...,+\lambda_{n}v_{n}=\begin{pmatrix}\\
v_{1} & ... & v_{n}\\
\\
\end{pmatrix}\begin{pmatrix}\lambda_{1}\\
...\\
\lambda_{n}
\end{pmatrix}$

Where $\begin{pmatrix}\lambda_{1}\\
...\\
\lambda_{n}
\end{pmatrix}=[a]_{\alpha}$ = the coordinate vector of $a$ wrt $\alpha$.

$\begin{pmatrix}\\
v_{1} & ... & v_{n}\\
\\
\end{pmatrix}=C_{\alpha}^{E}$ is simply the basis that we would like to convert from in columns
(to $E$)

So the change of basis matrix to the canonical basis is simply the
old basis in columns.

So $[a]_{E}=C_{\alpha}^{E}[a]_{\alpha}\implies[a]_{\alpha}=(C_{\alpha}^{E})^{-1}[a]_{E}=C_{E}^{\alpha}[a]_{E}$,
so to go from the canonical basis to our desired basis, we multiply
by the inverse of the change of basis matrix.

Suppose then we want to transform $[a]_{\alpha}\to[a]_{\beta}$

Sequentially it is $\underbrace{[a]_{\alpha}\to[a]_{E}}_{C_{\alpha}^{E}[a]_{\alpha}}$.
$\underbrace{[a]_{E}\to[a]_{\beta}}_{(D_{\beta}^{E})^{-1}C_{\alpha}^{E}[a]_{\alpha}=D_{E}^{\beta}C_{\alpha}^{E}[a]_{\alpha}}$

So $[a]_{\beta}=(D_{\beta}^{E})^{-1}C_{\alpha}^{E}[a]_{\alpha}$

In terms of maps:

$[a]_{\beta}=((\phi_{\beta}^{E})^{-1}\circ\phi_{\alpha}^{E})([a]_{\alpha})$,
which reading from right to left, makes sense.

So the change of basis map from $\alpha\to\beta$ is $(\phi_{\beta}^{E})^{-1}\circ\phi_{\alpha}^{E}$

Is this correct?

答:
That is correct.  A notation often found in introductory linear algebra texts is to write $P_{\beta \leftarrow \alpha}$ for the change-of-basis matrix that has two properties:

The columns of $P_{\beta \leftarrow \alpha}$ are the $\alpha$ basis vectors expressed in the $\beta$ basis, and

$P_{\beta \leftarrow \alpha}$ converts from $\alpha$ coordinates to $\beta$ coordinates in the sense that for every vector $v$, we have $[v]_\beta = P_{\beta \leftarrow \alpha} [v]_{\alpha}$

Your last equation then follows from the facts that $P_{\gamma \leftarrow \beta} P_{\beta \leftarrow \alpha} = P_{\gamma \leftarrow \alpha}$ and $P_{\alpha \leftarrow \beta} = \left( P_{\beta \leftarrow \alpha} \right)^{-1}$.


问:
Describing a section of a solid cone using spherical coordinates
I have to convert the following set to spherical coordinates.
$$S = \{(x, y, z)| 0 \le x \le 3, 0 \le y \le \sqrt{9 - x^2}, \sqrt{x^2 + y^2} \le z \le 9\}.$$
By visualizing the set, I have been able to figure out the bounds on $\theta$ and $\phi$. They are
$$0 \le \theta \le \pi/2$$
$$0 \le \phi \le \pi/4$$
Clearly $0$ is a lower bound for $\rho$, but I have been unable to get an upper bound. It think the upper bound must depend on just $\phi$, but I'm not sure how to get it.

答:
$S = \{(x, y, z)| 0 \le x \le 3, 0 \le y \le \sqrt{9 - x^2}, \sqrt{x^2 + y^2} \le z \le 9\}$

In spherical coordinates, we have two subsets, $S = S_1 \cup S_2$.

$x = \rho \cos\theta \sin\phi, y = \rho \sin\theta\sin\phi, z = \rho \cos\phi$

At the intersection of cone and cylinder,

$z = \rho \cos \phi = 3, \rho \sin \phi = 3 \implies \phi = \frac{\pi}{4} \ $ (equation of the cylinder in spherical coordinates is $\rho \sin \phi = 3$).

At the intersection of plane $z=9$ and the cylinder,

$z = \rho \cos\phi = 9, \rho \sin \phi = 3 \implies \phi = \arctan({\frac{1}{3}})$

(i) between $0 \leq \phi \leq \arctan({\frac{1}{3}}), 0 \leq \rho \leq 9 \sec \phi$
(as we are bound by the plane $z=9$)

(ii) Between $\arctan({\frac{1}{3}}) \leq \phi \leq \frac{\pi}{4}, 0 \leq \rho \leq 3 \csc \phi$ (as we are bound by the cylinder $\rho \sin \phi = 3)$

Lastly for both (i) and (ii), $0 \leq \theta \leq \frac{\pi}{2}$.


问:
Limit of sequences of $C^{\infty}$ function
I have a continuous bounded function $f\in C^{\infty}(\mathbb{R}^k, \mathbb{R})$ with non-zero, bounded, derivative in some neighbourhood of a point $c\in\mathbb{R}^k$. I also have two sequences $(a_n)_{n \in \mathbb{N}}$ and $(b_n)_{n \in \mathbb{N}}$ which both tend to $c$.

For all $n$, $d(a_n,c)>d(b_n,c)$. Furthermore, for all $n$, $f(a_n)>0$ and $f(b_n)>0$ and $f(c)=0$.

I want to show that there exists some $N$ such that $f(a_n)>f(b_n)$ for all $n>N$. Formally, I want to show:

$$\exists N \in \mathbb{N}\text{ such that } \forall n > N, f(a_n)>f(b_n)$$

Is this possible? At first I thought it would be easy and could be done using the fact a $C^{\infty}$ function is locally lipschitz continuous and considering a Lipshitz continuous region about $c$ but I can't get this to work... does anyone know how to show this?

答:
Edit: In fact the result is true as stated if $k=1$. We do that after the counterexample for $k=2$:

In $\Bbb R^2$ a counterexample is utterly trivial:

Take $c=(0,0)$, $a_n=(2/n,0)$, $b_n=(0,1/n)$, $f(x,y)=x+3y$.

(Seems to me $f$ being linear puts constraints on what sort of proof of a positive result might possibly work under what sort of additional hypotheses...)

News: The result is true for $k=1$ (i.e. for $f:\Bbb R\to\Bbb R$).

Proof. Wlog $c=0$. Since $f'(0)\ne0$, wlog $f'(0)>0$ and then wlog $f'(0)=1$.

Now $f<0$ on $(-\delta,0)$, so wlog $a_n,b_n>0$; hence $$a_n-b_n>0.$$

And $f\in C^1$ implies $$\frac{f(a_n)-f(b_n)}{a_n-b_n}\to1;$$

hence $f(a_n)-f(b_n)>0$ for large $n$.


问:
Ergodicity of surjective continuous endomorphism of compact abelian group (confused about a step)
Let $G$ be a compact abelian group with Haar measure and $A$ surjective continuous endomorphism. Then $A$ is ergodic $\iff$ $\phi(A^n)=\phi$ ($\phi$ are characters) for some $n>0$ implies $\phi$ is the trivial character.

In the proof here: Ergodicity of surjective continuous endomorphism of compact abelian group

I dont understand the following: $$\sum_{\chi \in \hat{X}} c_\chi \chi = f = f\circ T = \sum_{\chi \in \hat{X}} c_\chi (\chi\circ T).$$ seems to imply $c_\chi=c_{\chi \circ T}$. I do not see why. On the LHS we have the Fourier series. However, on the RHS we have an rearrangement of the Fourier series. Thus unless we have absolute convergence, we cannot rearrange our sums and conclude those coefficents are equal. What am I missing here?

答:
The argument in question is not a rearrangement but a selection of Fourier coefficients.

Let's first recall some relevant preliminaries. A reference is Einsiedler & Ward's Ergodic Theory, with a view towards Number Theory, Appendix C.3: Pontryagin Duality (pp. 433-440).

Let $G$ be an LCH abelian group. Then the group $\widehat{G}$ of topological group homomorphisms $\chi: G\to S^1 = \{z\in \mathbb{C}||z|=1\}$ endowed with the compact-open topology (= topology of uniform convergence on compact subsets) itself is an LCH abelian group and members of $\widehat{G}$ are called characters of $G$. Let us denote by $\langle\cdot|\cdot\rangle$ the natural bilinear pairing $\widehat{G}\times G\to S^1$.

If $G$ is compact abelian, then $\widehat{G}$ is countable discrete. Further, any Haar measure $dg$ on $G$ is unimodular and finite, so we may as well take it to be a probability measure. Let us consider $\widehat{G}$ with counting measure.

The Fourier transform on $L^2(G,\mathbb{C})$ acts like so:

$$f \mapsto \left[\hat{f}: \widehat{G}\to \mathbb{C},\; \chi \mapsto \int_G f(\chi) \overline{\langle \chi | g\rangle} dg \right].$$

The crucial theorem is this:

Theorem: For any compact abelian group $G$, $\widehat{G}$ is a complete orthonormal basis of $L^2(G,\mathbb{C})$.

This means that for computational purposes:

$\forall f\in L^2(G,\mathbb{C}): f(g) = \sum_{\chi\in\widehat{G}}\hat{f}(\chi)\langle\chi|g\rangle$ for almost every $g\in G$.

$\forall f_1,f_2\in L^2(G,\mathbb{C}), \forall \chi\in \widehat{G}: \hat{f_1}(\chi)=\hat{f_2}(\chi) \implies f_1(g)= f_2(g)$ for almost every $g\in G$ (and vice versa, of course).

Now let $G$ be compact abelian, $T: G\to G$ be a topological group homomorphism. Pullback by $T$ (the Koopman operator in ergodic theory cycles)

$$U_T: f \mapsto f\circ T$$

acts unitarily on $L^2(G,\mathbb{C})$ and $\widehat{G}$.

If $T$ is onto (= right-cancellable), then it preserves $dg$ (see Endomorphisms preserve Haar measure or Surjective endomorphism preserves Haar measure). Further, $U_T$ is injective (= left-cancellable). In particular we have a bijection

$$U_T : \widehat{G} \to U_T\left(\widehat{G}\right).$$

(This is what I meant by "selection" above.)

Let $f\in L^2(G,\mathbb{C})$ be $T$-invariant (i.e. $U_T$-invariant), so that $f=U_T(f)=f\circ T$ (almost everywhere). The Fourier expansion of $f$ at a generic point $g$ gives:

$$f(g) = \sum_{\chi\in\widehat{G}}\hat{f}(\chi)\langle\chi|g\rangle = \sum_{\chi\in U_T\left(\widehat{G}\right)}\hat{f}(\chi)\langle\chi|g\rangle + \sum_{\chi\in \widehat{G}\setminus U_T\left( \widehat{G}\right)}\hat{f}(\chi)\langle\chi|g\rangle.$$

Since $T$ is measure preserving $T(g)$ is also a generic point (of course one could be more rigorous here), and the Fourier expansion of $f$ at $T(g)$ gives:

$$f(T(g)) = \sum_{\rho\in\widehat{G}}\hat{f}(\rho)\langle\rho|T(g)\rangle = \sum_{\rho\in\widehat{G}}\hat{f}(\rho)\langle U_T(\rho)|g\rangle.$$

Here in the orthonormal basis $\widehat{G}$ the only characters we are using are the ones included in $U_T\left(\widehat{G}\right)$. Doing a change of variables $\chi = U_T(\rho)$, which we can for $\rho\in U_T\left(\widehat{G}\right)$ unambiguously, we get

$$f(T(g)) = \sum_{\chi\in U_T\left(\widehat{G}\right)}\hat{f}\left(U_T^{-1}(\chi)\right)\langle \chi|g\rangle = \sum_{\chi\in U_T\left(\widehat{G}\right)}\hat{f}\left(U_T^{-1}(\chi)\right)\langle \chi|g\rangle + \sum_{\chi\in \widehat{G}\setminus U_T\left( \widehat{G}\right)}0\;\langle\chi|g\rangle.$$

Comparing this with the Fourier expansion of $f$ at $g$ (which we can by the theorem), we have that

$$\forall \chi \in \widehat{G}: \hat{f}(\chi) = \begin{cases}  0,& \text{ if } \chi \in \widehat{G}\setminus U_T\left( \widehat{G}\right)\\
\hat{f}\left(U_T^{-1}(\chi)\right),& \text{ if } \chi \in U_T\left( \widehat{G}\right)\end{cases}
$$

Reusing the change of variables $\chi = U_T(\rho)$ in tandem with the second row, we have:

$$\forall \rho\in \widehat{G}: \hat{f}(U_T(\rho)) = \hat{f}(\rho).$$

Thus $\hat{f}\circ U_T = \hat{f}$.

Finally let me remark that doing these calculations for $G=\mathbb{T}$ and $T$ an irrational rotation or multiplication by some positive integer is very instructive.


问:
non-negative s-measurable functions on a finite measure space
Assume that $(X,s,\mu)$ is a finite measure space and $f,g$ are non-negative $s$-measurable functions such that $fg \ge 1$ a.e. Show that $(\int f \,d\mu) (\int g \, d\mu) \ge (\mu(X) )^2$

This inequality is clearly true if either integral is infinite, so I am looking for a solution for the finite case. I have an answer, but it does not depend on anything that we have recently covered, so I am pretty sure I am missing something big. Any help would be appreciated.

答:
Simply use Hölder's inequality:
$$\mu(X)=\int_X \sqrt{1} \mathrm d \mu\leq\int_X \sqrt{f}\sqrt{g} \mathrm d \mu\leq\left(\int_X f \mathrm d \mu\right)^{1/2}\left(\int_X g \mathrm d \mu\right)^{1/2}.$$


问:
Write permutation as disjoint cycles $(5 1 2)(3 1 2)(4 1 3)$
I'm really confused with permutations overall, I know the "algorithm" yet every time I get a mistake...

Write this permutation in $S_6$ as disjoint cycles and find its order: $(5 1 2)(3 1 2)(4 1 3)$

My try:

$$\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6\\
2 & 3 & 4 & 5 & 1 & 6\\
\end{pmatrix}$$
so $(5 1 2)(3 1 2)(4 1 3)=(1 2 3 4 5)$.

Yet the solution in WolframAlpha states: $(1 4)(2 5)$ (order $2$ yet in my case the order is $5$). Am I mistaken?

Is there any good website for checking this sort of calculation (or is WolframAlpha actually correct)?

Thank you in advance.

答:
I've been there. It takes a while to get used to working with permutations, and cycle notation isn't the easiest to work with. I would write out each cycle in 2-line notation:
$$ (512) = \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 5 & 3 & 4 & 1 & 6\end{pmatrix},
(312) = \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 3 & 1 & 4 & 5 & 6\end{pmatrix},
(413) = \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6\\ 3 & 2 & 4 & 1 & 5& 6\end{pmatrix}
$$
To "multiply" these (what we are really doing is composing them), you can put them "on top of each other"

$$ \begin{matrix}
\begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 5 & 3 & 4 & 1 & 6\end{pmatrix} \\
\begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 3 & 1 & 4 & 5 & 6\end{pmatrix} \\
\begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 \\ 3 & 2 & 4 & 1 & 5 & 6\end{pmatrix}
\end{matrix}
$$

and then align the top row of the second permutation with the second row of the first one.

$$
\begin{matrix}
\begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6\\ 2 & 5 & 3 & 4 & 1 & 6\end{pmatrix} \\
\begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6\\ 2 & 3 & 1 & 4 & 5 & 6\end{pmatrix} \\
\begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6\\ 3 & 2 & 4 & 1 & 5 & 6\end{pmatrix}
\end{matrix}
=
\begin{matrix}
\begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6\\ 2 & 5 & 3 & 4 & 1 & 6\end{pmatrix} \\
\begin{pmatrix}2 & 5 & 3 & 4 & 1 & 6\\ 3 & 5 & 1 & 4 & 2 & 6\end{pmatrix} \\
\begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6\\ 3 & 2 & 4 & 1 & 5 & 6\end{pmatrix}
\end{matrix}
=
\begin{matrix}
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6\\ 2 & 5 & 3 & 4 & 1 & 6\\
2 & 5 & 3 & 4 & 1 & 6\\ 3 & 5 & 1 & 4 & 2 & 6\end{pmatrix} \\
\begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6\\ 3 & 2 & 4 & 1 & 5 & 6\end{pmatrix}
\end{matrix}
=
\begin{matrix}
\begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6\\ 3 & 5 & 1 & 4 & 2 & 6\end{pmatrix} \\
\begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6\\ 3 & 2 & 4 & 1 & 5 & 6\end{pmatrix}
\end{matrix}
$$

and now do the same thing again:

$$
\begin{matrix}
\begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6\\ 3 & 5 & 1 & 4 & 2 & 6\end{pmatrix} \\
\begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6\\ 3 & 2 & 4 & 1 & 5 & 6\end{pmatrix}
\end{matrix}
=
\begin{matrix}
\begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6\\ 3 & 5 & 1 & 4 & 2 & 6\end{pmatrix} \\
\begin{pmatrix}3 & 5 & 1 & 4 & 2 & 6 \\ 4 & 5 & 3 & 1 & 2 & 6\end{pmatrix}
\end{matrix}
=
\begin{matrix}
\begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6\\ 3 & 5 & 1 & 4 & 2 & 6 \\
3 & 5 & 1 & 4 & 2 & 6 \\ 4 & 5 & 3 & 1 & 2 & 6 \end{pmatrix}
\end{matrix}
$$
which equals: $$
\begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6\\ 4 & 5 & 3 & 1 & 2 & 6\end{pmatrix}
$$

Now that we have our permutation we can write it in cycle notation:[ 1->4, 4-> 1 ][ 2->5,5->2 ][ 3->3 ] gives (14)(25)(3). People normally drop the (3), so it can be written as (14)(25).

The order is 2 because if you compose the permutation (14)(25) with itself you get the identity.


问:
What about the roots of $P+Q$ can be gleaned from $P,Q$?
Let $P,Q$ be polynomials (with whatever coefficients you desire.) In general, the roots of $P+Q$ are not immediately calculable (except by direct computation) even if we have full information of $P$ and $Q$, including their roots.

My question: what are some non-trivial* theorems that give information on the roots of $P+Q$ given information on $P$ and $Q$, probably with certain restrictions? Ie. their location, sum, exact value(s)...anything!

Example: One theorem which fits the bill is Rouché's theorem, which can be applied to polynomials.

*By non-trivial, I mean $P \neq-Q$, or things of that nature.

答:
If $P$ and $Q$ have some identic roots, then $P+Q$ has these roots too.

Since the Euclidean algorithm allows to obtain the polynomial with the common roots of $P$ and $Q,$ then the claim above looks non-trivial.


问:
Condition numbers of invertible 2x2 matrices
After learning about condition numbers, I worked through some MATLAB examples to compute condition numbers of several 2x2 matrices to gain some intuition. I noticed that for invertible 2x2 matrices, the 2-norm condition number seems to always be equal to the infinity-norm condition number. My question is twofold, does this result hold in general, and how can this be proven?</p>

答:
If $A \in \mathbb{R}^{m \times n}$ then you have the following bound

$$ \frac{1}{\sqrt{n}}\| A\|_{\infty} \leq \| A\|_{2} \leq  \| A \|_{\infty}$$

In this case $n=2$ so

$$ \frac{1}{\sqrt{2}}\| A\|_{\infty} \leq \| A\|_{2} \leq  \| A \|_{\infty}$$


问:
Number of subsets of $\{1,2, \ldots, n\}$ containing no three consecutive integers: recurrence equation?
I'm thinking about the problem below. I know that I have to find a polynomial formula for that first, and then from that polynomial formula I can find the recurrence relation. I actually attempted finding a polynomial formula, but I think I'm leaving some options while thinking about three consecutive integers. Anyway, without further ado, here is the problem and your suggestions are appreciated:

Let $f_n$ be the number of subsets of $\{1,2,3,\ldots, n\}$ that contain no three
  consecutive integers. Find a recurrence for $f_n$.

答: My Hint: $f_n$ can be any set from $f_i$ such that $i<n-1$ plus the element $\{n\}$. it can also be any set from $f_i$ such that $i<n-2$ plus the elements $\{n-1,n\}$. </p>


问:
Find the derivative of the solution to the I.V.P $ \binom{x}{y}’ = \binom{x(a-x)}{x} $ with respect to $a$
Find the derivative of the solution to the following I.V.P with respect to $a$, at $a = 1$.

$$
\binom{x}{y}(0) = \binom{1}{0}
$$
$$
\binom{x}{y}’ = \binom{x(a-x)}{x}
$$

What i tried:

Let $a = 1$

Substitute:

$$
z = \frac{\partial x}{\partial a}, f = \frac{\partial y}{\partial a}
$$

Therefore i get:

$$
\binom{z}{f}’ = \binom{z-2xz}{z}
$$

Now, i have the term “$-2xz$” which is a bit problematic.

Did i start right or there is another way? How do i proceed? Maybe i can substitute $x$ using the initial conditions somehow?

Thank you.

答: I'd go there brutally. As you see, only the first equation $x'=x(a-x)$ is really important. Once you have it, you can calculate y. But $x' = a x - x^2$ is a Bernoulli equation, which suggests that you replace $z = 1/x$. Then $z' = -x'/x^2 = -a/x + 1 = 1 - a z$ which is easy to solve by hand. The replace back (attention to signs!) Once you know x(a, t), take derivatives!


问:
prove f is separable if and only if f is relatively prime to its derivative.
Given the definition; a polynomial f in the complex numbers is separable if it has only simple roots.
My question is how do I prove f is separable if and only if f is relatively prime to its derivative.

Can someone help me go about doing this proof?

答: Hint: The discriminant of $f$ is (up to a constant) $\displaystyle \prod_{i \ne j} (\alpha_i - \alpha_j)$ where $\alpha_1, \cdots, \alpha_n$ are the roots of $f$.


问:
Does there exist a sequence of polynomials converging uniformly to $f(x) = |x| $ on $ℝ$?
Does there exist a sequence of polynomials converging uniformly to $f(x) = |x|$ on $ℝ$?

I understand this is possible on a closed interval due to Weierstrass Approximation Theorem.

Intuitively, I think there's no such sequence on $ℝ$ but I'm not sure how to prove this rigorously.

Any hints would be so useful!

答: That is not possible uniformly on $\mathbb R$. Clearly, it is not possible with a sequence of polynomials of degree less than $N$, whatever $N$ you chose. Therefore degrees of the approximating sequence MUST explode. But polynomial function grow so much faster at infinity than $|x|$ that uniform convergence is impossible.


问:
Solution verification relating determinants and planes alongside solution to a system of equations
I have this multipart question here. I did all of it except the last part and I was wondering if I could get a bit of help with that.

Let $$\vec{u}=(u_1,u_2,u_3),\vec{v}=(v_1,v_2,v_3),\vec{w}=(w_1,w_2,w_3)$$
be $3$ vectors in $\mathbb{R}^3$. Define a matrix whose rows are these three vectors:
$$\begin{bmatrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3\\ w_1 & w_2 & w_3\\
\end{bmatrix}$$
a) Show that $$\det(A)=\vec{u} \cdot (\vec{v} \times \vec{w})$$

This proof was trivial. Simply compute $\vec{v} \times \vec{w} = \begin{vmatrix} i & j & k \\ v_1 & v_2 & v_3\\ w_1 & w_2 & w_3\\
\end{vmatrix}$ and dot the result with $\vec{u}=(u_1,u_2,u_3)$. Compute the determinant (I used cofactor expansion) and you'll get both sides to match.

b) Suppose that $\vec{v},\vec{w}$ are not colinear. Show that $\vec{v} \times \vec{w}$ is normal to the plane given by $\text{Span}\{\vec{v},\vec{w}\}$.

Since $\vec{v}$ and $\vec{w}$ are not collinear, this implies any vecotor can be written as the span of the two vectors. Let $\vec{u}=\alpha\vec{v}+\beta\vec{w}$ where $\alpha, \beta$ are real numbers. If $\vec{u}$ is normal to $\vec{v} \times \vec{w}$ then $\vec{u}\cdot(\vec{v} \times \vec{w})$ implies:

$\vec{u}\cdot(\vec{v} \times \vec{w})$

$(\alpha\vec{v}+\beta\vec{w})\cdot(\vec{v} \times \vec{w})$

$(\alpha\vec{v})\cdot(\vec{v} \times \vec{w})+(\beta\vec{w})\cdot(\vec{v} \times \vec{w})=0$

It's $0$ since two of the rows are the same so the determinant (Since the is the same as part a) evaluates to $0$ as well.

c) Use parts (a) and (b) to argue that $\vec{u},\vec{v},\vec{w}$ all lie on the same plane if and only if $\det(A)=0$.

They have to. Part a and b directly implies that $\vec{u}$ is orthogonal to the span of $\vec{v}$ and $\vec{w}$, then they must lie on the same plane. $\vec{u}$ is orthogonal to $\vec{v}$ and $\vec{w}$ but that's fine. They're still in the same plane. I think this was obvious...

d) Suppose that the matrix $A$ row reduces to
$$\begin{bmatrix} 1 & 2 & 0 \\ 0 & 0 & 1\\ 0 & 0 & 0\\
\end{bmatrix}$$
Now consider the system of equations:
\begin{cases}
          u_1x_1+u_2x_2+u_3x_3=0 \\
          v_1x_1+v_2x_2+v_3x_3=0 \\
          w_1x_1+w_2x_2+w_3x_3=0 \\
       \end{cases}
Each equation defines a plane in $\mathbb{R}^3$. Describe how the three planes intersect.

After row reduction we get the same matrix as given. They intersect along a line given by $\vec{x}=t\{-2,1,0\}$.

e) Consider the system of equations
\begin{cases}
          u_1x_1+u_2x_2+u_3x_3=0 \\
          v_1x_1+v_2x_2+v_3x_3=0 \\
          w_1x_1+w_2x_2+w_3x_3=0 \\
       \end{cases}
Assume only that the vectors $\vec{u}$,$\vec{v}$,$\vec{w}$ are non-zero and $\det(A)=0$ but that you don't know how the matrix row reduces. Each equation defines a plane in $\mathbb{R}^3$. Describe the possible ways that the $3$ planes can intersect and explain why.

This is the only part I am not sure about. I know when the determinant is $0$, this means the matrix isn't invertible so there is no unique solution to the system (I think?). But besides that, I'm not sure what other kind of intersections should happen.

答:
You're right, the system cannot have a unique solution.  It is homogeneous (all RHSs zero) so it is consistent (the planes all pass through the origin so the origin is in the intersection), so there are infinitely many solutions.  But what is the geometry of the solution space?

Think about the rank of $A$.  Since $\det A = 0$, the rank cannot be 3.

If the rank is 2, then you more or less get the scenario of part (d): a line of intersection.  It doesn't matter whether the leading 1s are in columns 1 and 3 as in the matrix in part (d), or in columns 1 and 2, or even in columns 2 and 3.

What happens if the rank is 1?  Suppose for example that $A$ reduces to $\begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$: What is the solution space in that case?

Can the rank be zero?  Note that the given vectors are nonzero.


问:
Showing a convolution of a $f\in C_c(\mathbb{R}^n)$ with $g\in L^p_{loc}(\mathbb{R}^n)$s continuous
I am trying to use Lebesgue's dominated convergence theorem to show that if $f\in C_c(\mathbb{R}^k)$ and $g\in L_{loc}^{1}(\mathbb{R}^k)$ then we will have that $f* g\in C(\mathbb{R}^k)$. Now my idea was to take a sequence $x_n\rightarrow x$ and show that $(f*g)(x_n)\rightarrow (f*g)(x)$.

We have that $(f*g)(x_n)=\int_{\mathbb{R}^n}f(x_n-y)g(y)\,dy$ and $(f*g)(x)=\int_{\mathbb{R}^n}f(x-y)g(y)\,dy$, and so we want to see that $\lim_{n\rightarrow \infty}\int_{\mathbb{R}^n}f(x_n-y)g(y)\,dy=\int_{\mathbb{R}^n}f(x-y)g(y)\,dy$, and so the idea would be to apply the dominated convergence theorem to $h_n(y):=f(x_n-y)g(y)$ and $h(y):=f(x-y)g(y)$

Now Using the fact $f$ has compact support and is continuous, I was able to prove that $f(x_n-y)\rightarrow f(x-y)$ uniformly.

First we check that we have that $f(x_n-y)\rightarrow f(x-y)$ uniformly. Since the function $f$ is continuous we have that for every fixed $y$, $f(x_n-y)\rightarrow f(x-y)$, and around each point we can find a neighborhood $V_y$ such that if $y'\in V_y$, $|f(x_n-y')-f(x_n-y)|<\epsilon_y/3$ and $|f(x-y')-f(x-y)|<\epsilon_y/3$. Now since $f(x_n-y)\rightarrow f(x-y)$ there exists $N_y\in \mathbb{N}$ such that for every $n>N_y$ we will have that $|f(x_n-y)-f(x-y)|<\epsilon_y/3$. Using the triangle inequality we see that for every $y'\in V_y$ we have that $|f(x_n-y)-f(x-y)|<\epsilon_y$ for every $n>N_y$. Now since the support of $f$ is compact , we can take a finite subcovering of this covering by open sets $\{V_y\}$ and choose $N:=\max N_y$ and $\epsilon:=\min \epsilon_y$, so that for every $z\in K:=supp f$ and $n>N$ we have that $|f(x_n-z)-f(x-z)|<\epsilon$. Since the choice of $\epsilon$ was arbitrary we get that $f(x_n-y)\rightarrow f(x-y)$ uniformly.

Now I am not sure this can give me that $h_n\rightarrow h$ almost everywhere.And also we would need the fact that there exists an $h'\in L^1(\mathbb{R}^n)$ such that $|h_n(x)|\leq |h'(x)|$ almost everywhere in $\mathbb{R}^n$ for every $n$, but this seems to be a bit hard since I only know that $g\in L_\text{loc}^1(\mathbb{R}^n)$. Does anyone have any suggestions ? Thanks in advance.

Attempt at solving this difficulties :

Since $x_n\rightarrow x$ we can find an $N$ such that $x_n\in K$ for every $n\geq N$, where $K$ is a compact set and so since $f$ has compact support we will have that $h_n$ will have it's support inside a compact set $K'$ for every $n$ greater than $N$. And so know we can use the hypothesis that $g\in L_{loc}^1(\mathbb{R}^n)$ to get that there exists $M>0$ such that $|g(y)|\leq M$ for every $y\in K'$, and so we can get that $h_n\rightarrow h$ since $|f(x_n-y)g(y)-f(x-y)g(y)|\leq |f(x_n-y)-f(x-y)||M|$ for every $x\in \mathbb{R}^n$, and that $|h_n|\leq M'$ for every $n$ since $f$ has compact support.

What do you guys think ? I would appreciate some input just to see if I made any mistake or not.

答:
Once you proved all this, it's easy to finish:

Since $f(x_n-y) \to f(x-y)$ pointwise (even uniformly), then also $f(x_n-y)g(y) \to f(x-y)g(y)$ pointwise. Possibly not uniformly, but this is not necessary here.

Since you already observed that $h_n$ are supported in some compact set $K'$ (for large enough $n$), the distinction between $L^1_{loc}$ and $L^1$ disappears. To be precise, $|f|$ is bounded by some constant $M$, so $|h_n(y)| \le M|g(y)|$ everywhere. Knowing that $|h_n(y)| = 0$ outside $K'$, we have
$$
|h_N(y)| \le M |g(y)| \cdot \chi_{K'} =: h'(y),
$$
and $h'$ is the integrable function we seek. ($\chi_{K'}$ denotes the characteristic function of $K'$)


问:
An Ehrenfeucht-Fraïssé game

Let $\sigma = \{E \}$ be a signature with a binary relation symbol $E$. The $\sigma-$ structures $\mathcal{A}$ and $\mathcal{B}$ are given as follows:

$(i)$ Show that the duplicator wins the game $\mathfrak{G}_2(\mathcal{A}, \mathcal{B})$.

$(ii)$ Give a formula $\varphi \in FO[\sigma]$ with minimal quantifier rank, such that $\mathcal{A} \models \varphi$ and $\mathcal{B} \not\models \varphi$.

For $(i)$, I don't see how the duplicator can win in round two. For example, if the spoiler picks the vertex $c$, then the duplicator has to pick a vertex in $\mathcal{A}$ which also has two outgoing edges, but no such vertex exists. What am I missing here?

$(ii) \forall x \forall y \forall z (E(x, y) \land y \neq z \land x \neq y \rightarrow \lnot E(x, z))$

Here I'm saying that every vertex has at most one outgoing edge.

What do you think of my answers?

答:
Re: $(i)$, you write:

if the spoiler picks the vertex $c$, then the duplicator has to pick a vertex in $\mathcal{A}$ which also has two outgoing edges.

But that's not true. Remember that Duplicator wins if, at the end of the game, the atomic types of the tuples built are the same. For example, if Spoiler picks $c$, Duplicator picks $4$, Spoiler picks $a$, and Duplicator picks $3$, Duplicator will win the play. We don't check at the end of a play of the Ehrenfeucht-Fraisse game that the tuples built have matching out- (or in-)degrees.

In fact, there is very little Spoiler can do in two moves.

Title: An Ehrenfeucht-Fraïssé game


问:
Is there a more efficient equation for the Mandelbrot lemniscates?
A common coloring algorithm for the Mandelbrot Set is the escape time algorithm, which produces solid bands of color around the boundary of the set. These bands are bounded by a group of polynomial lemniscates of the form

$$x_n^2 + y_n^2 = r^2$$

Where
$$x_n = x_{n-1}^2 − y_{n-1}^2 + x,$$
$$y_n = 2 x_{n-1} y_{n-1} + y,$$
$$x_0 = x,$$
$$y_0 = y,$$
And $r$ is the escape threshold (usually two, but other values can be used).

This family of curves starts with the circle $x^2 + y^2 = 4$, then the Cassini oval $(x^2 − y^2 + x)^2 + (2 x y + y)^2 = 4$, then the curve $((((x^2 − y^2 + x))^2 − ((2 x y + y))^2 + x))^2 + ((2(x^2 − y^2 + x) (2 x y + y) + y))^2 = 4$, and they get exponentially more complicated as you go on. As $n$ increases, the curves converge to the boundary of the Mandelbrot Set.

Working with this implicit form is extremely unwieldy, and plotting them is inefficient. Is there a better way to represent these curves? Say, a polar equation, or a complex function, or a set of parametric equations? I'd like a way to graph the curves that doesn't require such cumbersome expressions.

答:
It's much more natural to use complex numbers.  Regarding efficiency, complex iteration of $z \to z^2 + c$ gives a degree $2^{n-1}$ polynomial in $c$ in $O(n)$ work.  I don't know if it's possible to do better than that: starting from a pre-expanded polynomial will be $O(2^n)$ work which is exponentially worse.

Suppose $c = x + i y$, $f_c(z) = z^2 + c$, $f_c^{\circ (n+1)}(z) = f_c^{\circ n}(f_c(z))$. Then $\frac{d}{dc} f_c^{\circ(n+1)}(z) = 2 f_c^{\circ n}(z) \frac{d}{dc} f_c^{\circ n}(z) + 1$.  These can be calculated together in one inner loop (being careful not to overwrite old values that are still needed).

Now you can use Newton's root finding method to solve the implicit form $f_c^{\circ n}(0) = r e^{2 \pi i \theta} = t$ by $c_{m+1} = c_{m} - \frac{f_{c_m}^{\circ n}(0) - t}{\frac{d}{dc}f_{c_m}^{\circ n}(0)}$.  Use the previous $c$ as initial guess for next $\theta$.  The increment in $\theta$ needs to be smaller than about $\frac{1}{4}$ for the algorithm to be stable.  Note that $\theta$ (measured in turns) wraps around the unit circle $2^{n-1}$ times before $c$ gets back to its starting point.

This approach is inspired by An algorithm to draw external rays of the Mandelbrot set by Tomoki Kawahira.

Compare with binary decomposition colouring (here with $r = 25$) (this particular image is rendered implicitly as a function from pixel coordinates to pixel colour, with edge detection of iteration bands $n$ as well as regions where $\Im f_c^{\circ n}(0) > 0$, where $n$ is the first iteration where $|f_c^{\circ n}(0)| > r$):


问:
How to get the weight of the orange sphere?
The problem is as follows:

The figure from below represents two scales, one is not in equilibrium while the other is at equilibrium. Assume both identical objects have the same weight and those with different shapes have different weights and each object weights an integer amount of pounds. With all this information, what would it be the least amount of the orange sphere?.

The choices given are:

$\begin{array}{ll}
1.&\textrm{3 pounds}\\
2.&\textrm{2 pounds}\\
3.&\textrm{1 pounds}\\
4.&\textrm{4 pounds}\\
\end{array}$

From looking at the figure what I've attempted to do was as follows:

I'm using the notation:

$\textrm{Circle = C}$

$\textrm{Triangle = T}$

$\textrm{Square = S}$

The first figure shows:

$2c+t>s+2t$

The second figure shows:

$s+c=2t$

Hence it can be inferred that:

$2c>s+t$

$c=2t-s$

$2(2t-s)>s+t$

$4t-2s>s+t$

$3t>3s$

$t>s$

Thus the triangle weighs more than the square.

But from only these it is not possible to know an exact weight for each of these figures. How can I assume which weight goes to which?

If I assume the square weighs $1$ pound, then triangle can be $2$, $3,$4$, etc.

Because the weights of the square and circle must account for two triangles and the weight of all objects is an integer this means that either the square and circle are both odds or both evens.

We don't know which is more heavier than the other so I'm setting that the sphere might be heavier.

Since I set the square to be $1$ pound, then the circle must be $3$, $5$, $7$, etc.

In order to reassure the least possible weight for that sphere, I shall assume that it is ligther in weight than the square.

Thus I rearrange my initial analysis and get to:

The square is $3$ pounds, the circle is $1$ pound, but triangle would be $2$ pounds, this is a contradiction as the triangle weight is more than the square.

$s=2$, $c=2$, $t=2$, also yields a contradiction.

I can't go up for the square as it will remain in contradiction. Thus the only logical choice would be what I originally set. $s=1$, $c=3$, $t=2$.

Therefore the minimum weight that it can have the sphere is three pounds or choice 1.

Is there any way to confirm this?. Is my logic correct?.

答:
Here is a concrete way to think. $T$ can be identified as arithmetic mean of $C,S$. As they are distinct, we have two possibilities : $C<T<S$ or $C>T>S$.

From first diagram, $2C > T+S$.

First case : $C<T, C<S \Rightarrow 2C < T+S$. Second case : $C>T, C>S \Rightarrow 2C > T+S$. Thus $C>T>S$ is true.

For minimum, let $S=1, C=3$ so that $T=(S+C)/2=2$ is an integer.


问:
Probability using a coin and a six-sided die
Assume a coin and a six-sided die. At the first move, you flip a coin. If you get a heads, you get to roll the die. If you get a tails, you have to flip the coin again. When you roll the die, if you get a $6$, the game is over. If you don't get a $6$, you have to flip the coin again. What is the probability of getting a $6$ on the die within $6$ moves? Each coin flip or die roll is a move.

I tried it with $4$ moves using brute force method with a tree diagram, and found an answer of $4$ in $60$, or $1$ in $15$. What I want to know is the general solution of this game within $n$ moves.

答:
We can use Markov transition matrix to solve this problem in general. Repeating  the following procedure  n  times to get the probability to roll 6 to exit (for example, when n = 4, the probability is  shown in the last row in $S_4$ below to be $\frac{13}{72}$).
$\to$

Set up the transition matrix T and initial state $S_0$, then calculate the next state with thie formula

$S_{i+1}$=T.$S_i$
(where i =0,1,2,3...).

$T= \begin{bmatrix} 0.5&\frac{5}{6}& 0 \\ 0.5&0 & 0\\0&\frac{5}{6}&1 \end{bmatrix}$
$S_0= \begin{bmatrix} 1 \\0\\0 \end{bmatrix}$

$S_1=T.S_0 = \begin{bmatrix} 0.5\\0.5\\0 \end{bmatrix}$

$S_2=T.S_1 = \begin{bmatrix} \frac{2}{3}\\\frac{1}{4}\\\frac{1}{12} \end{bmatrix}$

.
$S_3=T.S_2 = \begin{bmatrix} \frac{13}{24}\\\frac{1}{3}\\\frac{1}{8} \end{bmatrix}$

$S_4=T.S_3 = \begin{bmatrix} \frac{79}{144}\\\frac{13}{48}\\\frac{13}{72} \end{bmatrix}$


问:
prove the following: using a similar proof as Bolzano's Theorem
Let $f : [0, 1] → R$ be continuous with $f (0) = f (1) = 0$. Further suppose that whenever $f(a) = f(b) = 0$ for some $0 \leq a < b \leq 1$ there exists at least one $c \in (a, b)$ such that $f (c) = 0$. Show that $f (x) = 0$ for all $x \in [0, 1]$.

I have this to prove but I'm not sure with what idea to start this proof.

答:
We know that $f(1)=f(0)=0$ and by the property of the function given in the question, then $\exists^{\geq 1} c \in [0,1]$ such that $f(c)=0$.

Define the sets as follow:

$$A:=\{x \in [0,c) | f(x)=0\} \;\; B:=\{x\in (c,1] | f(x)=0\} $$

Since we have $f(0)=0$, then we know $A \neq \emptyset$. Similarly we know $B \neq \emptyset$.

Note that the intervals $[0,c)$ and $(c,1]$ are bounded (above by $c$ and below by $c$ respectively.
Therefore we know that $A$ is bounded above, and $B$ is bounded below. Thus $\exists \sup{A}, \inf{B}$.

Particularly, we have $0 \leq \sup{A} \leq c \leq \inf{B} \leq 1$.

Since $c \in (0,1)$, we have $\sup{A}, \inf{B} \in [0.1]$ and so $f$ has a value for these inputs.

Now, by the approximation property of the supremum, that is: $\exists a \in A \;\text{such that}\;\sup{A}+\frac{1}{n} < a \leq \sup{A}$

Thus, we have:

$$\exists(x_n)_{n \in \mathbb{N}} \subset A \;\;\text{such that}\;\; x_n\rightarrow \sup{A}$$

By continuity, $f(\sup{A}) \rightarrow f(x_n)$. However, by definition of $A$, we have $f(x_n)=0 \;\forall n \in \mathbb{N}$. Thus, $f(\sup{A})=0$.

We do the same procedure for finding $f(\inf{B})$ and then conclude that $f(x)=0 \; \forall x \in [0,1]$ via contradiction of the supremum/infimum properties.


问:
$f$ is continuous and $f(0)\ge f(x)$, for all $x\in [0,\delta]$
I am trying to prove (or disprove) the following statement:

$f:[0,\delta]\to [0,\infty[$, is a continuous function and $f(0)\ge f(x)$ for all $x$. Is there $\overline{\delta}<\delta$ such that $f$ is decreasing on $[0,\overline{\delta}]$?

It seems to be true but I am struggling to prove it.

I am trying to use contradiction but not successfully so far.

答: Doesn't look like it's true.  Consider for example $f(x) = -2x + x \sin \left( \tfrac{1}{x} \right)$ for $x \ne 0$, and $f(0) = 0$.


问:
floor function inequality $n+ \left \lfloor \tfrac{n}{3} \right \rfloor \le x$
I have the following problem which has two parts:
Given any $x>0$, I define a function $A$ by
\begin{equation}
A(x)=\max \{ n \in \mathbb{N} : n+ \left \lfloor \tfrac{n}{3} \right \rfloor  \le x\}
\end{equation}
I want to compute $A(x)$ and
\begin{equation}
\lim_{x \to \infty} \dfrac{A(x)}{x}.
\end{equation}

I tried to use the fact that $ x \le \lfloor x \rfloor \le x+1 $ but I got stuck.
Do I have to do a case by case analysis of if $n$ is divisible by $3$?

答:
Since $n\in\mathbb{N}$, it follows by contradiction that

$$n+\left\lfloor{\frac{n}{3}}\right\rfloor\leq x \iff n+\left\lfloor{\frac{n}{3}}\right\rfloor\leq \lfloor x\rfloor$$

Suppose that it is possible to make the above equation an equality. Writing $n=3k+r,0\leq r\leq 2$ we note that

$$\lfloor x\rfloor=4k+r$$

Thus the only cases where such an equality is allowed is when $\lfloor x\rfloor\equiv 0,1,2 \mod 4$.  Explicitly, in this case we can write the solution $n=\lfloor x\rfloor-\lfloor x/4\rfloor$.

Now when $x=3\mod 4$ we see that, since we cannot achieve equality with $\lfloor x\rfloor$ we can move to the closest integer to that, $\lfloor x \rfloor-1.$ We can certainly achieve equality here, with $n=\frac{3(\lfloor x \rfloor-1)+2}{4}=\lfloor x\rfloor-\lfloor x/4\rfloor-1$. Thus

$$A(x)=\begin{Bmatrix}\lfloor x\rfloor-\lfloor x/4\rfloor&x\neq 3\mod 4\\ \lfloor x\rfloor-\lfloor x/4\rfloor-1&x= 3\mod 4 \end{Bmatrix}$$

It should be clear that

$$\lim_{x\to\infty}\frac{A(x)}{x}=\frac{3}{4}$$

EDIT: It turns out a more elegant form exists:

$$A(x)=\lfloor x\rfloor-\left\lfloor \frac{x+1}{4}\right\rfloor$$

One can see this by noticing that the following Iverson bracket, the difference between the two branches above,  is succintly given by

$$[x=3\mod 4]=\left\lfloor \frac{x}{4}+\frac{1}{4}\right\rfloor-\left\lfloor \frac{x}{4}\right\rfloor$$


问:
How to get $2700$ grams of corn from a bag of $6.8$ kilograms?
The problem is as follows:

A merchant has a kiosk and has one bag with $6.8$ kilograms of corn and a two pan scale and three weights. One measures $5$ kilograms, the other $2$ kilograms, and the other $1$ kilogram. In how many weighing trials at minimum could he serve an order of $2700$ grams of corn?

The choices given are:

$4$ trials

$3$ trials

$1$ trial

$2$ trials

I'm not sure exactly how to get to the right combination for this problem. It would be very nice if the answer could include the steps in a logical and very detailed way so I can understand.

To my best of ability I thought that I could arrange the weights as follows:

$5$, $2$, $1$, $6.8$

To get $2.7$ from these numbers is somewhat convoluted to me. Thus I need some help. What can I do?

The thing is either method which I've tried from subtracting this amount from $6.8$ does give me more than what I need, thus I am going in circles.

答:
Personally, I see no algorithmic alternative to trial and error.

Hint:

If you can weigh out 1.4 kg, then the remaining corn, when cut in half, will suffice.

If you cut the original corn in half, you get to 3.4.....

Not sure if there is alternative approach that involves fewer weighings.


答:
This is not true.

A counterexample is

$$f(x)= -x\left(1+\frac{1}{2} \sin \left(\frac{1}{x}\right)\right)$$ for $x\neq 0$ and $f(0)=0$.

The derivative is taking positive and negative values on intervals included in any neighborhood of zero.

You can even have a counterexample when $f$ is supposed to be differentiable. See https://www.mathcounterexamples.net/a-function-whose-derivative-at-0-is-one-but-which-is-not-increasing-near-0/.

Title: $f$ is continuous and $f(0)\ge f(x)$, for all $x\in [0,\delta]$

Body:
I am trying to prove (or disprove) the following statement:

$f:[0,\delta]\to [0,\infty[$, is a continuous function and $f(0)\ge f(x)$ for all $x$. Is there $\overline{\delta}<\delta$ such that $f$ is decreasing on $[0,\overline{\delta}]$?

It seems to be true but I am struggling to prove it.

I am trying to use contradiction but not successfully so far.


问:
imaginary numbers - how can I understand them - especially as they occur in 'roots' of polynomials?
In another question here, about roots of equations being imaginary, the accepted answer
said something interesting about "imaginary" (as a technical word in math) not meaning "not real".  I understand that imaginary numbers are "manipulable and usable to do all kinds of things".   And thus, "they exist" at least, in the realm of mathematics.

So my question is this. I was playing around in wxMaxima, and solving polynomials, to get their roots.  In maxima, $\sqrt{-1}$ is represented as %i.

Editor's note: all "%i"s have been replaced with $i$.

Here's an example of a polynomial, say $3x^6 + 4x^4$:

solve($3x^6 + 4x^4$,$x$);

Maxima's results:
[$x=\displaystyle -\frac{2i}{\sqrt{3}}$,$x=\displaystyle \frac{2i}{\sqrt{3}}$,$x=0$]

My exact question:

Is it common to have the same imaginary result twice with only the sign different (both q and -q are imaginary roots)?  (At first I thought it had reported the same root twice as I overlooked the negative sign.)

One can easily plot a real root and see how it intercepts the x axis.  How does one "visualize" or think of an imaginary root -- does it correspond to a local minimum that does not occur at the x axis?

What uses do imaginary roots of polynomials have in math?

答:
I will answer the first question, and perhaps the others if no one else does.  It is not a double root, there is a minus sign in front of one of the answers.  The fact that the exponents are all even here forces any root to be accompanied by its negative.

The root $x=0$ is a quadruple root, but Maxima seems to mention it only once.


答:
In don't think there is a logical step-by-step solution. Some trial and error seems inevitable to figure out which numbers can be constructed, and in how many steps. So I give you my almost unfiltered thought process:

Weigh off $5$kg. Then $1.8$kg remains. Weigh off another $1.8$kg against this. Now you have two $1.8$kg piles. Split a $1.8$kg into two $0.9$kg piles by weighing it against itself. Then $2.7=1.8+0.9$.

Is this minimal?

You can also put the $5$kg weight on the scale, and distribute all the corn such that the scale is balanced. Then the two sides have $0.9$kg and $5.9$kg. Use the $0.9$kg to weigh off $0.9$kg twice more, to get $2.7=0.9+0.9+0.9$.

Is this minimal?

You can also put $1$kg on one side of the scale, and $5$kg on the other side. Then distribute all the corn such that the scale is balanced. Then the two sides have $5.4$kg and $1.4$kg. Split the $5.4$kg in half by weighing it against itself to get $2.7$kg.

Is this minimal?

Title: How to get $2700$ grams of corn from a bag of $6.8$ kilograms?

Body:
The problem is as follows:

A merchant has a kiosk and has one bag with $6.8$ kilograms of corn and a two pan scale and three weights. One measures $5$ kilograms, the other $2$ kilograms, and the other $1$ kilogram. In how many weighing trials at minimum could he serve an order of $2700$ grams of corn?

The choices given are:

$4$ trials

$3$ trials

$1$ trial

$2$ trials

I'm not sure exactly how to get to the right combination for this problem. It would be very nice if the answer could include the steps in a logical and very detailed way so I can understand.

To my best of ability I thought that I could arrange the weights as follows:

$5$, $2$, $1$, $6.8$

To get $2.7$ from these numbers is somewhat convoluted to me. Thus I need some help. What can I do?

The thing is either method which I've tried from subtracting this amount from $6.8$ does give me more than what I need, thus I am going in circles.


问:
Number of N step Random walks on 2D infinite grid or lattice
What is the number of N step random walks starting from a point (x0,y0) to a point (x1,y1) assuming each direction (right,left,up,down) has equal probability. I know the expression for the one dimension case is:

$$ \binom{N}{\frac{N+(y1-y0)}{2}} $$

is there a similar expression for two dimensions?
I saw this post (Probability Distribution of a 2D lattice random walk) but the answer given is clearly wrong.

答:
Here's another approach, less slick than that of @MikeEarnest (+1):

Wlog we are taking $N$ steps from $(0,0)$ to $(x,y)$. Suppose $k$ of these steps are horizontal (either left or right). Use the one-dimensional formula to count the number of ways to make $k$ (horizontal) steps starting at $0$ and ending at $x$. Next, multiply by the number of ways to embed a given sequence of $k$ horizontal steps into your total $N$ steps. Then given that embedding, count the ways to make $N-k$ (vertical) steps starting at $0$ and ending at $y$. This yields the sum
$$\sum_k{k\choose \frac12(k+x)}{N\choose k}{N-k\choose\frac12(N-k+y)}.$$
Change index of summation to $a:=\frac12(k-x)$ and define $m:=\frac12(N-x-y)$. This gives
$$
\sum_a\frac{N!}{(x+a)!a!(m+y-a)!(m-a)!}={N\choose m}\sum_a{m\choose a}{N-m\choose m+y-a}$$ after some rearranging along with the identity $N-m=(x+a)+(m+y-a)$. The rightmost sum evaluates to ${N\choose m+y}$, yielding
$${N\choose m}{N\choose m+y}={N\choose \frac12(N-x-y)}{N\choose\frac12(N-x+y)}.$$


问:
2-Connected Graph from a Connected Graph

Let $G = (V, E)$ be a connected graph with at least three vertices. Let $G'$ be the graph on V with an edge (v, w) if there is a path in G of length at most two between v and w. Show that G' is 2-connected.

My proof: For any two connected vertices v and w in G', either v or w has to be connected to another vertex, or else v and w would be an isolated component of G' (which contradicts that G' is connected since G is connected.) WLOG, let w be connected to x, then we have (v,w) and (w,x), but then we have to connect (w,x) for G'. So v, w, and x all have degree $\ge 2$. Thus G' is at least 2 connected.

Is my proof sufficient? I feel like something might be missing.

答:
Degree at least $2$ does not mean $G'$ is two-connected, and the graph

v-w-x seems to give your proof problems.


问:
How to show $\lim\frac{n^2+1}{n(n+1)}= 1$ by using the epsilon-delta definition.
How to show $$\lim_{n \to \infty}\frac{n^2+1}{n(n+1)}= 1$$ by using the epsilon-delta definition.

Step:

Fix $\varepsilon > 0$. There exists an $N \in \mathbb{N}$, such that
$\left|\frac{n^2+1}{n(n+1)}-1 \right|< \varepsilon$ ?

答:
Maybe you start with $\frac{n^2+1}{n^2+n} = 1 - \frac{n -1 }{n^2+n}$. Now
$\left| \frac{n -1 }{n^2+n} \right| \le \left| \frac{n + 1 }{n^2+n} \right| = 1/n$ should finish it.


问:
Free finite index submodule of $\mathcal{O}_L$
Let $L/K$ be a finite Galois extension of $p$-adic fields, with Galois group $G$. I've read somewhere that it is well-known that $\mathcal{O}_L$ contains a free finite index $\mathcal{O}_K[G]$-module, where $\mathcal{O}_L$ and $\mathcal{O}_K$ are the valuation rings. I'm triyng to prove this result, but I'm stuck.

My attempt: by the normal basis theorem, $L=K[G]\cdot \alpha$, where $\alpha=\beta/\gamma\in L^\times$. It is not hard to prove that we can assume $\beta\in\mathcal{O}_L$ and $\gamma\in\mathcal{O}_K$.  I believe that the module we are looking for is $\mathcal{O}_K[G]\cdot \beta\subseteq \mathcal{O}_L$, but I don't know how to conclude...

答:
Let me try to answer my question.

FACT 1: $O_L$ is the integral closure of $O_K$ in $L$.
This is clear in the number field case, but it is also true for $p$-adic fields. See, for example, Neukirch, 'Algebraic number theory' (first step of the proof of Theorem 4.8 of Chapter II).

FACT 2: If $\alpha\in L$, then there exists an element $0\ne\gamma\in O_K$ such that $\gamma\alpha =\beta\in O_L$. In particular, $\alpha=\beta/\gamma$. This is Milne, 'Algebraic number theory', Proposition 2.6.

FACT 3: If $\alpha$ is a generator of a normal basis, then since $1/\gamma$ is fixed by every $\sigma\in G$, also $\beta$ is a generator of a normal basis.

FACT 4: $O_L$ is free of finite rank over $O_K$. This is Milne, 'Algebraic number theory', Proposition 2.29.

Now we show that $|O_L:O_K[G]\cdot \beta|<\infty$. Since $K\otimes_{O_K} O_L\cong L$ vía multiplication map, we have $$K\otimes_{O_K}O_L/(O_K[G]\cdot \beta)\cong L/(K[G]\cdot \beta)=0.$$ Nơw $O_L/(O_K[G]\cdot \beta)$ is a finitely generated over the PID $O_K$, so is has a free part isomorphic to some copies of $O_K$, and a finite torsion part. If it is infinite, then the tensor product with $K$ would contain some copies of $K$, and this would be a contradiction.


问:
Sum of residue systems is complete iff they are both complete
Let $p$ and $q$ be distinct prime numbers. The integers $a_1, \ldots, a_p$ and $b_1,\ldots,b_q$ are such that the sums $a_i + b_j$ form a complete residue system modulo $pq$ (that is, there is precisely one sum which is $0$ mod $pq$, precisely one which is $1$ mod $pq$, etc.). Prove that $a_1,\ldots,a_p$ form a complete residue system mod $p$ and $b_1,\ldots,b_q$ form a complete residue system mod $q$.

Any idea on how to approach this? Any help appreciated!

答:
Here's an idea.

Suppose $\{a_i+b_j\}_{i,j}$ forms a complete residue system mod $pq$. Then $$0 = \sum_{k=0}^{pq-1} k = \sum_{i,j} (a_i+b_j) = q(a_1+\dots+a_p)+p(b_1+\dots+b_q) \pmod{pq}.$$ This implies $$a_1+\dots+a_p \equiv 0 \pmod{p}$$ and $$b_1+\dots+b_q \equiv 0 \pmod{q}.$$ Using these, and that $$\sum_{i,j} (a_i+b_j)^2 = q(a_1^2+\dots+a_p^2)+p(b_1^2+\dots+b_q^2)+2(a_1+\dots+a_p)(b_1+\dots+b_q),$$ we obtain $$a_1^2+\dots+a_p^2 \equiv 0 \pmod{p}$$ and $$b_1^2+\dots+b_q^2 \equiv 0 \pmod{q}.$$ We can continue like this to get higher and higher moments (it stops at $p-1$ for $a_i$'s and $q-1$ for $b_i$'s, I think). I wonder if this implies $\{a_i\}_i$ is a complete residue system or is all $0$s. Same with $\{b_j\}_j$.


答:
$6.8$ divided by two gives $3.4$. Divided by two again and you get $1.7$.

Add $1$kg and you are done.


问:
Question about absolute continuous functions (preservation of null sets)
I'm trying to prove that a function $ f:[a,b] \to \mathbb{R} $ is absolutely continuous iff $ \mu(A) = 0 \implies \mu( f(A)) = 0$ for all such $A \subseteq [a,b]$. I'm quite stuck. I'm trying to work from the definition to prove that if f is a.c. then the condition holds, but I am stuck since I can't seem to come up with an appropriate cover for $ f(A)$. I saw on another site that it follows from the fact that f being a.c. is equivalent to $f(x) = f(a) + \int_0^x \phi(t)dt$ where $\phi = f'$. but I can't see how it works. Could anyone provide help? Many thanks.</p>

答: One direct way of dealing with this is using the fact that sets of measure 0 in the line have outer measure 0, which means that given any $\delta$, we can cover the set with open intervals whose total length is less than $\delta$. The definition of absolute continuity then shows that we can make the image have outer measure less than any $\epsilon>0$, so it must have measure 0.</p>


问:
Is a local continuous martingale square integrable.
As the headline suggests, I'm wondering whether a local continuous martingale is actually necessarily square integrable.

In the text I'm reading this is only mentioned without explanation (the answer may be trivial but I don't see it). I know that the quadratic variation of a continuous local martingale is continuous, does this somehow with Doob-Meyer's decomposition yield the answer?

答:
If you are asking whether a continuous local martingale is square integrable process the answer is NO, as the counterexample of @Jochen demonstrated.

But if you are asking whether a continuous local martingale is locally a square integrable martingale (meaning  both properties should hold by the same localizing sequence), the answer is YES. This can be seen by the property that every continuous process locally bounded. By taking the minimum of both sequences (the one for the martingale-property and the one for the boundedness), both properties remain. So you have a localizing sequence which makes your process a bounded (and thus square integrable) martingale.


答:
No. Not all minimum-degree 2 graphs are 2 connected. Consider a graph consisting of 2 cycles that intersect at precisely one vertex.

HINT to show the result: Assume $G$ is not 2-connected otherwise we'd be done already. Also assume $G$ has 3 or more vertices. In fact, if $G \setminus \{v\}$ is connected so is $G' \setminus \{v\}$ So it suffices to show the following: If $v$ is such that $G \setminus \{v\}$ is disconnected then $G' \setminus \{v\}$ is connected. To this note, if $v$ has degree at least 2 in $G$, there is an edge between every pair of vertices in $N_G(v)$ [why]. [What if $v$ has degree-1 in $G$?] So this implies $G' \setminus \{v\}$ is connected [why is that].

Title: 2-Connected Graph from a Connected Graph

Body:
Let $G = (V, E)$ be a connected graph with at least three vertices. Let $G'$ be the graph on V with an edge (v, w) if there is a path in G of length at most two between v and w. Show that G' is 2-connected.

My proof: For any two connected vertices v and w in G', either v or w has to be connected to another vertex, or else v and w would be an isolated component of G' (which contradicts that G' is connected since G is connected.) WLOG, let w be connected to x, then we have (v,w) and (w,x), but then we have to connect (w,x) for G'. So v, w, and x all have degree $\ge 2$. Thus G' is at least 2 connected.

Is my proof sufficient? I feel like something might be missing.


问:
Find the derivative of the solution to the I.V.P $ \binom{x}{y}’ = \binom{x(a-x)}{x} $ with respect to $a$
Find the derivative of the solution to the following I.V.P with respect to $a$, at $a = 1$.

$$
\binom{x}{y}(0) = \binom{1}{0}
$$
$$
\binom{x}{y}’ = \binom{x(a-x)}{x}
$$

What i tried:

Let $a = 1$

Substitute:

$$
z = \frac{\partial x}{\partial a}, f = \frac{\partial y}{\partial a}
$$

Therefore i get:

$$
\binom{z}{f}’ = \binom{z-2xz}{z}
$$

Now, i have the term “$-2xz$” which is a bit problematic.

Did i start right or there is another way? How do i proceed? Maybe i can substitute $x$ using the initial conditions somehow?

Thank you.

答:
In the case where the parameter is inside the equation, you need also to include its derivative in the derived equation. Here
$$
\dot z=\dot x_a=(ax-x^2)_a=x+ax_a-2xx_a=x+az-2xz.
$$


问:
Period of the $p$-adic expansion of a rational number
I have a problem I have been very stuck on.

Show that if $p \nmid n$, then the length of the period [of the expansion of $\frac{a}{n} \in \mathbb Q$] divides $e$, where $e$ is the least positive integer such that $p^e \equiv 1 \pmod n$.

I went to my professor for a hint, and his hint was to write $\frac{a}{n}$ as
$$\frac{a k}{p^e - 1},$$
for some $k \in \mathbb Z$, then multiply both sides of the $p$-adic expansion
$$\frac{a}{n} = \sum_{i = 0}^\infty q_i p^i$$
by $p^e - 1$ and group common powers of $p$. Thus, if $a k < 0$ (I imagine there is a similar argument for $a k \geq 0$ if I can make this one work),
$$-a k = \sum_{i = 0}^{e - 1}q_i p^i + \sum_{i = e}^\infty (q_i - q_{i - e}) p^i,$$
and so if $q_i \geq q_{i - e}$ for all $i \geq e$, then this is the $p$-adic expansion of the non-negative integer $-a k$, which we know to be finite. Thus, $q_i = q_{i - e}$ for sufficiently large $i$, and I'm pretty sure I can show that the period divides $e$ from there. However, I can't figure out the details if there are any $i \geq e$ such that $q_i < q_{i - e}$.

答:
Let $$S_e=\{ \sum_{m\ge 0} b_m p^m, b_m\in 0\ldots p-1,b_{m+e}=b_m \text{ for m large enough}\}$$

Show that for $B\in S_e$ then $B+\frac1{1-p^e}\in S_e$.

Whence $\frac{k}{1-p^e}+l\in S_e$ for all $k\in \Bbb{Z}_{\ge 0},l\in \Bbb{Z}$

Show that for $B\in S_e$, $\frac1{1-p^e}-B\in S_e$.

Whence $\frac{a}{n}=\frac{a k}{p^e - 1}\in S_e$, your $q_i$ sequence is $e$-periodic for $i$ large enough, which implies that its least period divides $e$.


问:
About the existence of homotopy operator
What shown below is a reference from the text Analysis on Manifolds by James Munkres. I point out that the sybols $\Omega$ and $^*$ are used by Munkres to denote a set of differential forms and the pull back map respectively.

Definition

Let be $A$ and $B$ open sets in $\Bbb R^n$ and $\Bbb R^m$ respectively and let be $g$ and $h$ two $C^r$ map defined from $A$ into $B$. So we say taht $g$ and $h$ are differentiably homotpic of class $C^r$ if there exists a $C^r$ mapd $H$ from $A\times[0,1]$ into $B$ such that
$$
H(x,0):=g(x)\,\,\,\text{and}\,\,\,H(x,1)=h(x)
$$
for any $x\in A$. The map $H$ is called a differentiable homotopy between $g$ and $h$.

Theorem

Let be $A$ and $B$ open sets in $\Bbb R^n$ and $\Bbb R^m$, respectively. Let $g,h:A\rightarrow B$ be $C^\infty$ maps that are differentiably homotopic. Then there is a linear transformation
$$
\mathcal P:\Omega^{k+1}(B)\rightarrow\Omega^k(A)
$$
defined for $k>0$ such that for any form $\eta$ of order $k>0$
$$
d(\mathcal P\eta)+\mathcal P(d\eta)=h^*\eta-g^*\eta
$$
while for a form of $f$ of order $0$,
$$
\mathcal P(df)=h^*f-g^*f
$$

So to follow I summarise the proof of the previous theorem but if you like you can read it
here.

Well first of all Munkres consider a very special case. So given a neighborhood $U$ of $A\times [0,1]$ in $\Bbb R^{n+1}$ he defines the map $\alpha,\beta:A\rightarrow U$ given by the equations
$$
\alpha(x):=(x,0)\,\,\,\text{and}\,\,\,\beta(x):=(x,1)
$$
and thus observing that these two maps are differentiably homotopic he proves the existence of a linear operator $P$ from $\Omega^{k+1}(U)$ to $\Omega^k(A)$ havig the desidred properties. Then the general case follows by this. Inded if $g,h:A\rightarrow B$ are differetiably homotopic with respect $H:A\times [0,1]\rightarrow B$ then he defines the operator $\mathcal P$ through the equation
$$
\mathcal P\eta:=P(H^*\eta)
$$
for any $\eta\in\Omega^{k+1}(B)$ because (Munkres himself states what to follow) $H^*\eta$ is a $(k+1)$-form defined in a neighborhood of $A\times[0,1]$ and thus $P(H^*\eta)$ is a $k$-form defined in $A$.

So I do not really understand why the pull back of $H$ with respect $H$ is a $(k+1)$-form defined in $A\times[0,1]$. Indeed if $H$ goes from $A\times[0,1]$ to $B$ then the pull back $H^*$ goes from $\Omega^{k+1}(B)$ to $\Omega^{k+1}(A\times[0,1])$ and thus this is a very big problem: in particular it seems to me that if we consider the restriction $\tilde H$ of $H$ to $A\times[0,1)$ then the pull back $\tilde H^*$ define a $(k+1)$-form on $A\times[0,1)$ of class $C^r$ that can be extended to a $(k+1)$-form of class $C^r$ defined in an open neighborhood of $A\times[0,1)$ but unfortunately this open neighborhood is not generally a open neighborhood of $A\times[0,1]$. So why $H^*\eta$ is a form defined in an OPEN neighborhood $A\times[0,1]$? Could someone explain this to me, please?

答:
I don't understand the relevance of restricting $H$ to $A\times [0,1)$, but I think I see what's bothering you.

According to Munkres's definition, to say that $H\colon A\times [0,1]\to B$ is differentiable means that there is a neighborhood $U$ of $A\times [0,1]$ in $\mathbb R^{n+1}$ and a differentiable map $\widetilde H\colon U\to \mathbb R^m$ whose restiction to $A\times [0,1]$ is equal to $H$. The problem is that the image of $\widetilde H$ might not be contained in $B$, and therefore $\widetilde H^*\eta$ might not be defined on all of $U$.

But this is easily remedied. Because $B$ is open in $\mathbb R^m$ and $\widetilde H$ is continuous, the preimage $\widetilde H^{-1}(B)$ is an open subset of $\mathbb R^{n+1}$ containing $A\times [0,1]$. Let's call this preimage $U_0$, and let $H_0$ denote the restriction of $\widetilde H$ to $U_0$. Then $H_0$ is a differentiable map from a neighborhood of $A\times [0,1]$ into $B$, and $H_0^*\eta$ is a well-defined $(k+1)$-form defined on a neighborhood (namely $U_0$) of $A\times [0,1]$.


问:
closed form for arbitrary derivatives of gaussian function
What is the closed form expression for the $n$th order derivative of the Gaussian function?
$$ \frac{d^n}{dx^n}\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{1}{2}\frac{(x-\mu)^2}{\sigma^2}}$$
Wikipedia says that it can be expressed in terms of the Gaussian times Hermite polynomials, but it does not illuminate any further.

答:
The relation

Indeed, via Rodrigues' formula, the Hermite polynomials are
$$H_n(x) = (-1)^n \, e^{x^2} \, \frac{d^n}{dx^n} e^{-x^2} \; , $$
for $n \in \mathbb{N}$ including zero. And by changing variables $x \to  \frac{x-\mu}{\sigma\sqrt{2}}$ ones get
$$ H_n\left(\frac{x-\mu}{\sigma\sqrt{2}}\right) = (-1)^n \, e^{\frac{1}{2}\frac{(x-\mu)^2}{\sigma^2}} \, 2^{\frac{n}{2}} \, \sigma^n \,  \frac{d^n}{dx^n} e^{-\frac{1}{2}\frac{(x-\mu)^2}{\sigma^2}}  \; , $$
and the answer for your question is
$$ \frac{d^n}{dx^n} e^{-\frac{1}{2}\frac{(x-\mu)^2}{\sigma^2}} = (-1)^n \left( \frac{1}{2 \sigma^2} \right)^{\frac{n}{2}} \, e^{-\frac{1}{2}\frac{(x-\mu)^2}{\sigma^2}} \, H_n\left(\frac{x-\mu}{\sigma\sqrt{2}}\right) \; . $$

The Hermite polynomials

The above expression is useful only if you know the polynomial that you are interested in. For n = 0, ... , 10  there is a table here and some polynomials are
\begin{align}
H_0(x) &= 1 \; , \\
H_1(x) &= 2x \; , \\
H_2(x) &= 4x^2 - 2 \; ,\\
& \;  \; \vdots
\end{align}
and the polynomial $H_n(x)$ is of order $n$ and the polynomial has the same parity (odd or even) of $n$. The general expression for this polynomials is not trivial and normally one use its property instead of its explicitly expression. If you desire the explicitly expression I copy it from the wiki and it is
\begin{align}
H_{2n}(x) \; & = \; (2n)! \, \sum_{l=0}^{n} \frac{(-1)^{n-l}}{(2l)!(n-l)!} \, (2x)^{2l} \; ,\\
H_{2n+1}(x) \; & = \; (2n+1)! \, \sum_{l=0}^{n} \frac{(-1)^{n-l}}{(2l+1)!(n-l)!} \, (2x)^{2l+1} \; ,\\
\end{align}
where I separate the case for odd and even polynomials for ease notation.


问:
$p$ is irreducible if and only if the only divisors of $p$ are the associates of $p$ and the unit elements of $R$
Let $R$ be an integral domain and $p ∈ R$ be such that $p$ is nonzero and a nonunit. Then $p$ is irreducible if and only if the only divisors of $p$ are the associates of $p$ and the unit elements of $R$.

Proof. Suppose the only divisors of p are the associates of $p$ and the unit elements of $R$. Let $p = ab$ for some $a, b ∈ R$. Suppose $a$ is not a unit. Then $a$ is an associate of $p$. Therefore, $a = pu$ for some unit $u ∈ R$. Now
$p = pub$. Since $R$ is an integral domain, it follows that $ub = 1$. Hence, $b$ is a unit and so $p$ is irreducible. We leave the converse as an exercise.

this is a theorem from a book and I have tried to proof the converse in the following way:-

suppose $p$ is irreducible and $a|p$. Then there exist $b\in R$ such that $p=ab$ . Since $p$ is irreducible so one of $a$ & $b$ is unit.Let $a$ is unit.Then we need to show that $b$ is associate of $p$. Since $a$ is unit there exist $u \in R$ such that $au=1$. hence $aup=p \implies aup=ab \implies up=b $.So the result follows.

Am I correct?

答:
Yes. The key idea is this: $ $ a factor $\,b\,$ of $\,\color{#c00}{p = ab}\,$ is a unit $\iff$ its cofactor $\,a\,$ is associate to $\,p,\,$ by

$$\,b\ \ {\rm unit} \iff {\overbrace{b\mid 1 \iff\!\!\ p\mid a}^{\Large\color{#c00}{ 1/b\ \ \  \ =\ \ \ \ a/p}}}\!\overset{\color{#c00}{\large a\,\mid\, p}}\iff a\ \ {\rm associate\ to}\ \ p\qquad\ $$


问:
Simple asymptotic function
(I have seen this question but it is too complicated for my needs, and my math skills are not good enough to convert the answer.)

I am writing a game and I need a way to increase the armor of the character in a meaningful way: I want the increase in armor to mean a decrease in damage taken (meaning an increase in damage absorbed), approaching but never reaching 100% absorption. In other words, I need a simple, polynomial function f(x) so that $$\lim\limits_{x \to \infty} f(x) = 1$$ (x will never be less than zero.)

Of course, the question has trivial answers, like f(x) = 0.9; I need a non-trivial one, preferably one where f(x) grows faster at first and then "slows down".

[Edit] Removed the polynomial requirement... duh!

[Edit] I found something stupidly simple... $$x / (x + 5)$$ This is close enough to my requirements. Unfortunately, I have absolutely no idea how to allocate the "solution" checkmark, so I'll pick the responder with less points, and add a +1 to the other. I hope it won't create problems.

答:
$$f(x) = x/(x + 1)$$
you can replace 1 with any larger value to slow down the change in values.

 for example $f(x)=x/(x+2)$ will give you slower change in values.
So you could make it dynamic, and values could change slower or faster based on some condition.

ContentLicense: CC BY-SA 2.5

Title: Simple asymptotic function


问:
Is the formula $\exists x [f(x) = a]$ logically valid?
Since the variable $x$ belongs to the universal set, then we could say that $f^{-1}(a)$ is one of the values we could assign to $x$. Therefore, $\exists x [(f(x) = a]$ is equivalent to $f(x_0)=a \lor f(x_1)=a \lor \ldots f[f^{-1}(a)]=a \ldots f(x_n)=a$. Since $f[f^{-1}(a)]=a$ is true, then $\exists x [(f(x) = a]$ is always valid. Is it correct?

答:
Of course not and you know it.

Let the universal set be $\mathbb R$ and $f(x) = x^2$ and $a = -1$.  So is there an $x \in \mathbb R$ so that $x^2 = -1$?  No.

Your error is when you say $f(f^{-1}(a)) = a$ always.  That just is not true.

$f(f^{-1}(-1)) = (f^{-1}(a))^2 = \sqrt {-1}^2 \ne -1$ because $\sqrt{-1}$ does not exist so you can not square it.

More generally  $f^{-1}(a) = \{x| f(x) = a\}$ could very well be empty.

And if so.... $f(f^{-1}(a)) = \{f(x)| x \in f^{-1}(a)\}= \{f(x)|x\in \emptyset\} =\emptyset$.

......

Okay.  Let $U$ be your "universal set of unrestricted domain of discourse".

Now let $f(x) = \begin{cases}1& \text{if }x\in U\\x&\text{if }x \not \in U\end{cases}$

Then what is the solution to $f(x)= 2$?

As $f(x) \ne 1$ we must have $x \not \in U$.  So that means $f(x) =x$  So $f(x) = x = 2$.  So $2 \not \in U$. But $2\in U$ because surely a universal set of unrestricted domain can not be restricted to exclude $2$.  A contradiction.

....

Or more bluntly.  Let $f(x) = 1$.  Let $x\in f^{-1}(2)$.  Then $f(x) = 2$. But $f(x) =1$.  A contradiction.  Doesn't matter which domain we choose.


问:
How does Ito's formula work at a point where the function/derivative is undefined? E.g. $x=(0,...,0)$ and $f(x)=\lvert x\rvert^{p}$
How does Ito's formula work at a point where the function is undefined? Consider $(0,...,0)=x\in \mathbb R^{n}$ and $f(x)=\lvert x\rvert^{p},\; p\geq 1$ where $\lvert x\rvert^{p}:=(\sum\limits_{i=1}^{n}x_{i}^{2})^{\frac{p}{2}}$

Now for the Ito vector process $X_{t}:=(X_{t}^{1},...,X_{t}^{n})$ we obtain, where $X_{t}\neq 0$,

$d\lvert X_{t}\rvert^{p}=\lvert X_{t} \rvert^{p-2}\sum\limits_{j=1}^{n}X_{t}^{j}+\frac{1}{2}\sum\limits_{k,l=1}^{n} (p-2)\lvert X_{t}\rvert ^{p-4}X_{t}^{k}X_{t}^{l}+1_{\{k=l\}}\lvert X_{t}\rvert^{p-2}$

I hope that is is correct, but anyhow do I just "ignore" the case when $X_{t}=(0,...,0)$ or is there a special way I can treat it?

答:
Both comments you got are correct, but I'll just expand a bit. Probably you need to handle these on a case-by-case basis. As Michal says, you can add an $\epsilon$ to make the function smooth, and then try to let $\epsilon$ go to zero and show that everything works properly. In many cases in higher dimensions (e.g. Brownian motion in three dimensions or more, as user6... says) the process will avoid the trouble set with probability 1 anyway, since it is transient, and so you would hopefully be ok.

There's a special case worth mentioning, that user6... alludes to. That is when you are working in one dimension, let's just say with Brownian motion, and $f(x) = |x|$. Then, formally, the first derivative is $sgn(x)$, which is 1 for positive $x$ and $-1$ for negative $x$ (its definition at $x=0$ doesn't matter here), and the second derivative is $\delta(x)$, the Dirac delta function. Ito's formula should therefore give

$$
|B_t|-|B_0| = \int_0^t sgn(B_s)dB_s + \int_0^t \delta(B_s)ds.
$$

Remarkably, although this is just a formal identity it turns out to be correct, with the last term equal to a new process, the local time of the Brownian motion. This is not a simple matter, and many researchers have worked on developing the properties of this and related processes. A good book if you're interested in this topic is "Markov Processes, Gaussian Processes, and Local Times", by Marcus and Rosen.

Hope that helps.


答:
For this, I will find what Wikipedia https://en.wikipedia.org/wiki/Hermite_polynomials calls the "probabilist's Hermite polynomials", which differ from the "physicist's" version by the factor of $\frac{1}{2}$ in the exponent.

There are actually multiple equivalent definitions. The first one given by the article actually solves your problem by definition!  There are other equivalent versions: defined by orthogonality conditions, or through a generating function, or through the Hermite differential equation.

First of all, change variables so that $u = \frac{x - \mu}{\sigma}$. That way $\frac{d}{dx} = \frac{1}{\sigma} \cdot \frac{d}{du}$. Also let $k = \frac{1}{2\pi \sigma^2}$.

Let $$G = k\cdot e^{\frac{-u^2}{2}}$$ denote that Gaussian function $G(u)$, and $G^{(n)}$ denote its $n$th derivative with respect to $u$.

$G^{(0)} = (-1)^0 \cdot 1 \cdot G$

$G^{(1)} = -uG = (-1)^1 \cdot u \cdot G$

$G^{(2)} = -G + u^2G = (-1)^2 \cdot (u^2 - 1)\cdot G $

$G^{(3)} = (2u)G + (u^2-1)(-u)G = (-1)^3 \cdot (u^3 - 3 u)\cdot G$

The first few Hermite polynomials are $1, u, (u^2 -1), (u^3 - 3u),...$
To get the physicist version, multiply by $2^n$.

Title: closed form for arbitrary derivatives of gaussian function

Body:
What is the closed form expression for the $n$th order derivative of the Gaussian function?
$$ \frac{d^n}{dx^n}\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{1}{2}\frac{(x-\mu)^2}{\sigma^2}}$$
Wikipedia says that it can be expressed in terms of the Gaussian times Hermite polynomials, but it does not illuminate any further.


问:
Computation of fundamental group of the Möbius strip
How would one go about computing the fundamental group of the Möbius strip using its polygon representation?</p>

答: It seems like the fastest way to solve this is to note that the Mobius strip is homotopy-equivalent to the circle, which has fundamental group $\mathbb{Z}$. Intuitively, draw a line in the middle of the Mobius strip that makes a complete circle around it. Then you map the strip onto the central circle - this is a homotopy equivalence.