问:
What would be an "or otherwise" method to solve this question?
A game is played where a blue die (with faces $4,6,8,9,10$ and $12$)and a pink die (with faces $2,3,5,7,11$ and $13$) are thrown once. The player wins if the number on the pink die is higher than the blue die.

The question guides me to draw a table of possible outcomes or otherwise. The table was easy enough (and answer of $\frac{7}{18}$) but what would the otherwise method refer to? I can't think of any combinatoric argument to solve the question. Probably a method that avoids cases would be the best but it seems as if that is unavoidable in this case.

So I know for sure that the two numbers must be distinct and if the pink dies lands on $2$ or $3$ it is impossible to win. How do I get an expression in $\frac{n!}{r!(n-r)!}$ from here?

答:
Approximation by simulation. In R, I simulated 10 million rolls of the blue and pink die.
The pink die has the larger number in very nearly $7/18$
of the 10 million. With $10^7$ iterations, one can expect
three-place accuracy. The simulation illustrates that
your answer is correct.

```<code>set.seed(306)
blu = c(4,6,8,9,10,12)
pnk = c(2,3,5,7,11,13)
x = sample(blu, 10^7, rep=T)
y = sample(pnk, 10^7, rep=T)
mean(y > x)
[1] 0.3890144  # approx P(Y > X) = 7/18
7/18
[1] 0.3888889

hdr = "ECDF Plots: Blue (Mostly) Dominates Pink"
plot(ecdf(x), col="blue", main=hdr)
 lines(ecdf(y), col="pink", lwd=3, lty="dotted")
</code>```

An alternative probabilistic solution. For each possible pink
number (probability $1/6)$ determine what fraction of blue numbers
it can beat. Each term is a probability $(1/6)$ multiplied by a conditional probability. [Essentially, you already mentioned two of the six terms.]

```<code>(1/6)*(0 + 0 + (1/6) + (2/6) + (5/6) + 1)
[1] 0.3888889
</code>```

Body:
Approximation by simulation. In R, I simulated 10 million rolls of the blue and pink die.
The pink die has the larger number in very nearly $7/18$
of the 10 million. With $10^7$ iterations, one can expect
three-place accuracy. The simulation illustrates that
your answer is correct.

set.seed(306)
blu = c(4,6,8,9,10,12)
pnk = c(2,3,5,7,11,13)
x = sample(blu, 10^7, rep=T)
y = sample(pnk, 10^7, rep=T)
mean(y > x)
[1] 0.3890144  # approx P(Y > X) = 7/18
7/18
[1] 0.3888889

hdr = "ECDF Plots: Blue (Mostly) Dominates Pink"
plot(ecdf(x), col="blue", main=hdr)
 lines(ecdf(y), col="pink", lwd=3, lty="dotted")

An alternative probabilistic solution. For each possible pink
number (probability $1/6)$ determine what fraction of blue numbers
it can beat. Each term is a probability $(1/6)$ multiplied by a conditional probability. [Essentially, you already mentioned two of the six terms.]

(1/6)*(0 + 0 + (1/6) + (2/6) + (5/6) + 1)
[1] 0.3888889


问:
A strange characterization of the minimum singular value of a matrix
Let $A = \begin{bmatrix}a_1 | \cdots|a_n \end{bmatrix} \in\mathbb{R}^{n\times n}$. Is it true that

$$\sigma_n(A) = \min_{1\leq j\leq n} \min_{\alpha\in\mathbb{R}^{n}}\|a_j - \sum_{i=1,i\neq j}^{n}\alpha_i a_i\|_2?$$

If it is not true, then what would be a counterexample?

答:
It is true in some cases, e.g. when $A$ is singular or when $A$ is a scalar multiple of an orthogonal matrix, but in general it is false. The quantity $\min\limits_{\mathbf c\in\mathbb R^n}\left\|a_j-\sum_{i\ne j}c_ia_i\right\|_2$ is the norm of the component of $a_j$ orthogonal to other columns of $A$. Now consider
$$
A=\pmatrix{2&1\\ 1&1}.
$$
The component of $a_1$ in $\{a_2\}^\perp$ is $a_1-\frac{\langle a_1,a_2\rangle}{\|a_2\|_2^2}a_2=\pmatrix{2\\ 1}-\frac32\pmatrix{1\\ 1}=\pmatrix{\frac12\\ -\frac12}$. Its norm is $\frac{1}{\sqrt{2}}$.

The component of $a_2$ in $\{a_1\}^\perp$ is $a_2-\frac{\langle a_1,a_2\rangle}{\|a_1\|_2^2}a_1=\pmatrix{1\\ 1}-\frac35\pmatrix{2\\ 1}=\pmatrix{-\frac15\\ \frac25}$. Its norm is $\frac{1}{\sqrt{5}}$.

However, $\sigma_2(A)=\frac{3-\sqrt{5}}{2}\ne\min(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{5}})$.

From another perspective, consider a generic square matrix $A$ whose singular values are all distinct and positive. Let $A=USV^T$ be a singular value decomposition. Then
$$
\min_{1\le j\le n}\min\limits_{\mathbf c\in\mathbb R^n}\left\|a_j-\sum_{i\ne j}c_ia_i\right\|_2
=\min\limits_{\substack{\mathbf x\in\mathbb R^n\\ x_j=1\text{ for some }j}}\left\|Ax\right\|_2
=\min\limits_{\substack{\mathbf x\in\mathbb R^n\\ x_j=1\text{ for some }j}}\left\|SV^Tx\right\|_2
\ge\sigma_n(A).
$$
Since $S$ or $A$ have distinct positive singular values, in order that equality holds in the above, the minimiser $x$ must satisfy $V^Tx=\pm e_n$. Hence $x=\pm Ve_n$ is, up to a sign, a column of $V$. But then the constraint that $x_j=1$ for some $j$ implies that this column of $V$ is equal to $\pm e_j$. That is, some column of $V$ is (up to a sign) a vector in the standard basis. Obviously this condition is not always satisfied.

ContentLicense: CC BY-SA 4.0


问:
Quotient group of normal subgroups is cyclic if quotient group of intersection is cyclic
Let $M$, $N$ be normal subgroups of a group $G$.

Prove that if $G/M\cap N$ is cyclic, then $G/M$ and $G/N$ are cyclic. Give a counter example to show that the converse is not always true.

I proved until now that if $G/M\cap N$ is abelian, then $G/M$ and $G/N$ are abelian, I don't know if that helps a lot but it's the only thing I could come up with.

答:
Hint for $\implies$:

Any quotient of a cyclic group is a cyclic group.

The third isomorphism theorem.

Hint for $\;\;\,\not\!\!\!\!\impliedby$:

There is a counterexample to the converse where $G$ has four elements (there are not very many such groups, so it should be easy to work out).

ContentLicense: CC BY-SA 3.0


问:
Relation between $\lim a_n$ and $\lim_{x\to \infty} f(x)$
Suppose that we are given a sequence $(a_n)$, where $a_n=\frac n{n+1}$. Then, $a_n\to 0$.

It may be observed that if we define $f:\mathbb R\to\mathbb R:f(x)=\frac x{x+1}$, that is $f(n)=a_n$ for all $n\in \mathbb N$. It follows that $\lim_{x\to \infty}f(x)=0=\lim a_n$.

My question is: For any given sequence $(a_n)$ and function $f:\mathbb R\to \mathbb R$ defined as $f(x)=a_x$, where $a_x=$ expression obtained by replacing $n$ by $x$ in the expression of $a_n$, is it true that $(a_n)$ converges if and only if $\lim_{x\to \infty}f(x)$ exists? $\tag 1$

If I were to find: $\lim_{x\to \infty}(1+\frac 1{x})^x$, then if $(1)$ were true, I would simply find this limit by $ \lim_{x\to \infty}(1+\frac 1{x})^x=\lim(1+\frac 1{n})^n=e$

If $(1)$ is not true then this is how I would calculate the aforementioned limit: For sufficiently large $x$, there exists $n$ such that $n\le x\le n+1\implies \frac 1{n+1}\le \frac 1{x}\le \frac 1{n}\implies \frac {n+2}{n+1}\le 1+\frac 1{x}\le \frac {n+1}{n}\implies (\frac {n+2}{n+1})^n\le (1+\frac 1{x})^x\le (\frac {n+1}{n})^{n+1}$. The result  follows by squeeze theorem.

If $(1)$ is not true then what are the conditions(continuity, differentiability of $f$) under which $(1)$ is true?

答:
If $\lim_{x \to \infty} f(x)$ exists, then $\lim_{x \to \infty} f(x) = \lim_{n \to \infty} a_n$. This includes when the limit is infinite. This direction is quite trivial with $\delta-\epsilon$.

On the other hand, it's possible for $\lim_{n \to \infty} a_n$ to exist without $\lim_{x \to \infty} f(x)$ existing. Consider the very simple case of $f(x) = \cos (2 \pi x)$. In this case, $a_n = \cos (2 \pi n) = 1$ for all $n$, so $\lim_{n \to \infty} a_n = 1$. On the other hand, $\lim_{x \to \infty} f(x)$ does not exist.


问:
When throwing a die $3$ times, what is the probability of the product of the three results being a multiple of $9$?
When throwing a die $3$ times, find the probability of the product of the three results being a multiple of $9$.

I tried drawing a table for the possible outcomes of the first TWO dice. Then I tried considering the cases for each when the third die is rolled - as you can imagine not very efficient. Perhaps I could use $x+y+z = 9k$ as a divisibility test, where $k$ is an integer. I'm pretty sure this would involve stars and bars, but I'm not sure how to implement it in this case. Would I have to solve for $k$?

I also worked out the boundaries but I'm not sure if that will help either: $9≤x+y+z≤216$.

Do I need to use stars and bars for this question?

答:
I agree with tommik's solution. Here is a simulation in R to verify.

```<code># Number of simulations
nsims <- 1e7

roll <- function() sample(1:6, size=1)

# Generate products
products <- purrr::map_dbl(1:nsims, ~roll()*roll()*roll())

# Get proportion divisible by 9
sum((products %% 9) == 0) / nsims
</code>```
```<code>0.2592488
</code>```

which is about $\frac{7}{27} = 0.259259259...$

Body:
I agree with tommik's solution. Here is a simulation in R to verify.

# Number of simulations
nsims <- 1e7

roll <- function() sample(1:6, size=1)

# Generate products
products <- purrr::map_dbl(1:nsims, ~roll()*roll()*roll())

# Get proportion divisible by 9
sum((products %% 9) == 0) / nsims

0.2592488

which is about $\frac{7}{27} = 0.259259259...$


问:
A very small disc winds around a larger circular disc of radius $R$ connected to it by a string. How long is the spiral it travels?
A very small disc winds around a larger circular disc of radius $R$. It is connected to it by a string of length l that remains tight. What distance does it travel before it hits the larger disc of radius $R$ ? I've estimated that it is $2\pi(l-R)$ based on assuming a spiral is a little like a series of concentric circles.

答:
Another way to add up your diminishing circle circumferences
is to use the formulas $2\pi R = \frac LN$ and  $N\pi = \frac{L}{2R},$
both of which are derived from $N = \frac{L}{2\pi R}.$

\begin{align}
\left(L - \frac LN\right)2\pi &+ \left(L - 2\frac LN\right)2\pi
 + \left(L - 3\frac LN\right)2\pi +  \cdots \\
&= \left(1 - \frac 1N\right)2\pi L + \left(1 - \frac 2N\right)2\pi L
 + \left(1 - \frac 3N\right)2\pi L +  \cdots \\
&= \left(N - 1\right)\frac{2\pi L}{N}
 + \left(N - 2\right)\frac{2\pi L}{N}
 + \left(N - 3\right)\frac{2\pi L}{N} +  \cdots \\
&= \left((N - 1) + (N - 2) + (N - 3) + \cdots + 1\right)\frac{2\pi L}{N} \\
&= \left(\frac{N(N - 1)}{2}\right)\frac{2\pi L}{N} \\
&= (N - 1)\pi L \\
&= N\pi L - \pi L \\
&= \frac{L^2}{2R} - \pi L. \\
\end{align}

But this is an underestimate, because at the beginning the end of the string is tracing a path of length approximately $L \theta$, where $\theta$ is the change in the angle of the string, although you have assumed it is only
$\left(1 - \frac 1N\right)L \theta.$
You could get a better approximation by wrapping the string around a regular polygon of $k$ sides and then letting $k$ increase to infinity
(effectively computing the integral of the path length as the string wraps around the circle);
the $\pi L$ term then drops out.

Your answer would also come up with the same sum except for a couple of errors.
The first error is that $2\pi \times 2\pi = 4\pi^2$ but you wrote $4\pi.$
The second error is on the very last step. Note that $2\pi N = \frac LR$
and $2\pi RN = L$ but $\pi(N + 1) = \frac{L}{2R} + \pi,$
so the result actually should be
\begin{align}
2\pi NL - 2\pi^2 RN(N+1)
 &= \frac{L^2}{R} - L\left(\frac{L}{2R} + \pi\right) \\
 &= \frac{L^2}{R} - \frac{L^2}{2R} - \pi L \\
 &= \frac{L^2}{2R} - \pi L.
\end{align}


问:
are there infinitely many primes $p,q$ such that $pq=a^2+b^4$
Are there infinitely many primes $p<q$, $p,q\neq 2,3$ such that $pq=a^2+b^4$ where $a,b\in \mathbb{Z}$ ? I've no idea if this is a very easy or very hard question. Any known result about this ?
Thank you for your comments !

EDIT : using mod 4 arguments you can derive very easily conditions on a and b. Also using Gaussian integers, this product pq boils down to the product of 4 gaussian irreducible elements. Note that p,q must be congruent to 1 mod 4.

答:
The answer should certainly be yes.
For example, with $b=1$ and $a = 2 + 5 x$, we'll have $a^2 + b^4 = 5 (1+4x+5x^2)$, and Bunyakovsky's conjecture implies there are infinitely many integers $x$ for which $1+4x+5x^2$ is prime.

However, no proof of this is known.


问:
Why is the arccos of a negative ratio the obtuse angle?
This is a repost of this question since it was not answered and here it was not answered clearly.

if we can no longer think about triangles when in any quadrant other than the first, how else can I visualize or think about the angle and it's sin,cos,tan ?

Thank you in advance

答:
Thinking in terms of the sides of a right triangle with angle $\theta$ so $\cos \theta = \frac {adjacent}{hypotenuse}$ and $\sin \theta = \frac {opposite}{hypotenuse}$ only works for $0 < \theta < 90$.

To extend it to any angle it is better to think of it this way:

Imagine a unit circle centered at $(0,0)$ and with radius $1$.  Imagine a point $(x,y)$ on the circle (that would mean $x^2 + y^2 = 1$) so that the angle made by the point, the center of the circle, and the $x$-axis in the positive direction, forms the angle $\theta$.  Then we call $x = \cos \theta$ and $y = \sin \theta$.

If $0 < \theta < 90$ then the point is in the first quadrant and $0 < x=\cos \theta < 1; 0< y=\sin x < 1$.  And $(x,y), (x,0), (0,0)$ form a right triangle with hypotenuse$=1$ and base angle $\theta$.  And $x = \cos \theta = adjecent = \frac {adjacent }1=\frac {adjacent}{hypotenuse}$ and $y = \sin \theta = \frac {opposite}{hypotenuse}$.

If $90 < \theta < 180$ then the point is in the second quadrant and $-1 < x =\cos \theta < 0; 0 < y= \sin  theta< 1$.  And $(x,y), (x,0), (0,0)$ form a right triangle with hypotenuse $1$ but it's extending the adjacent leg in the opposite (negative direction) and the base angle is $180 - \theta$.  And we have $y=\sin \theta = \frac {opposite}{hypotenuse = 1}$ and $x =\cos \theta =  -\frac {adjacent}{hypotenuse = 1}$.

And if $180 < \theta < 270$ then the point is in the third quadrant and $-1 < x =\cos \theta < 0; -1 < y=\sin \theta <0$.  And $(x,y), (x,0),(0,0)$ forms a right triangle but it's extended in the negative direction.  The base angle is $\theta - 180$ and $y = \sin \theta = -\frac {opposite}{hypotenuse = 1}$ and $x = \cos \theta = -\frac {adjacent}{hypotenuse = 1}$.

And if $270 < \theta < 360$ then the point is in the fourth quadrant and $x$ is positive and $y$ is negative.  We get a right triangle with base angle $360 - \theta$ and $x = \cos \theta = \frac {adjacent}{hypotenuse}$ and $y = \sin \theta =-\frac {opposite}{hypotenuse}$

=======

Okay, full disclosure.... I stole those images from other sites without permission.  But credit where due they are from https://www.expii.com/t/using-unit-circle-to-find-sine-or-cosine-5219 and http://flipping-a-classroom.weebly.com/unit-circle-and-the-six-trig-functions.html

I haven't read those sites but... they look like they'll have the information you are interested in if you want more information.


问:
A mathematical expression that gives 0 if variable is different from constant and 1 if variable same as constant
Example,
∆ = any mathematical expression;
c = constant;
x = variable;

c∆x=1 (if c = x)
c∆x=0 (if c ≠ x)

am just asking?

答:
What you are describing is the Dirac delta function $\delta(x)$, which is often used in theoretical physics.

In your case, the function $\delta(x-c)$ has the property that it has value $1$ if $x=c$, and value $0$ if $x \neq c$.

For more details, you may wish to refer to this Wikipedia article:
https://en.wikipedia.org/wiki/Dirac_delta_function


答: There is a natural surjective map $G/(M \cap N) \to G/M$.  This means $G/M$ is a quotient of the group $G/(M \cap N)$.  Did you know that quotients of cyclic groups still cyclic?  If not, prove it yourself.  It's not too hard :)</p>

Title: Quotient group of normal subgroups is cyclic if quotient group of intersection is cyclic

Body:
Let $M$, $N$ be normal subgroups of a group $G$.

Prove that if $G/M\cap N$ is cyclic, then $G/M$ and $G/N$ are cyclic. Give a counter example to show that the converse is not always true.

I proved until now that if $G/M\cap N$ is abelian, then $G/M$ and $G/N$ are abelian, I don't know if that helps a lot but it's the only thing I could come up with.


问:
Center of Mass for a Lamina. Why is the value inside the integral $x$ and not $x/2$?
TLDR

Why is the center of mass formula for a lamina equal to $A_1$ but not $A_2$? (Specifically the $0.5$ in the $x$-coordinate integral).

$$ A_1=\left(\frac{\int_a^bxf(x)dx}{\int_a^bf(x)dx}, \frac{\int_a^b 0.5 f(x)^2dx}{\int_a^bf(x)dx}\right)$$

$$ A_2=\left(\frac{\int_a^b0.5xf(x)dx}{\int_a^bf(x)dx}, \frac{\int_a^b 0.5 f(x)^2dx}{\int_a^bf(x)dx}\right)$$

Longer Explanation and Question

The center of mass of $n$ point masses in the $xy$-plane is point $P$ where $P$ is defined as:

$$ P = \left(\frac{\sum_{i=1}^n m_ix_i}{\sum_{i=1}^n m_i}, \frac{\sum_{i=1}^n m_iy_i}{\sum_{i=1}^n m_i} \right) .$$

When applying this to functions we could find the center of mass for an area under a curve bounded by two endpoints. For a given function $f(x)$ bounded by points $a$ and $b$,  we can chop up the function into smaller and smaller rectangles and average out their center of masses. Point $Q$ represents the center of mass within this interval and it is defined as:

$$ Q = \left(\frac{\int_a^bxf(x)dx}{\int_a^bf(x)dx}, \frac{\int_a^b 0.5 f(x)^2dx}{\int_a^bf(x)dx} \right) . $$

When taking the center of mass for the $y$-coordinate, it makes sense that we would be summing up the halfway height of every rectangle. However, we do not sum up the halfway width of every rectangle. The center of mass of a rectangle lies directly in the middle of the $dx$ slice we use to integrate the region. However, the formula doesn't include this, and it is wrong.

I cannot find an explanation as to why it is wrong. As the widths of our rectangles approach zero, I guess the center of mass can be approximated as being $x$ instead of $\frac{x}{2}$. We could just as easily factor out the $\frac{1}{2}$ value.

答:
In the $x$-dimension, the midpoint of your infinitesimal is not $x / 2$. Given a slice between $x_1$ and $x_2$ with width $\Delta x = x_2 - x_1$, the midpoint is

$$ \frac{x_1 + x_2}{2} $$
$$ = \frac{x_1 + x_1 + \Delta x}{2} $$
$$ = x_1 + \frac{\Delta x}{2} . $$

In the limit as $x_2 \to x_1$, we have $\Delta x \to 0$, and

$$ x_1 + \frac{\Delta x}{2} \to x_1 .$$


问:
Require help to solve an Inequality
$|0.022\cdot e^{-2t}\cdot sin(11t)|<0.01$

Values of $t$ for which the inequality holds.

Thanks in advance

Title: Require help to solve an Inequality

Body:
Since $|\sin(x)| \le 1$,
a sufficient condition is
$|0.022 e^{-2t}| < 0.01$
or
$e^{-2t} < 0.01/0.022=1/2.2$
or
$e^{2t} > 2.2$
or
$t > \ln(2.2)/2$.


问:
Trouble understanding the quotient morphism of subgroups.
In my lecture notes that I'm studying right now for a course on Abstract Algebra, I was given the following proposition:

Suppose $N\trianglelefteq G$ and let $\pi:G\to G/N$ be the quotient morphism. The $\pi$ induces a bijection, $$ \begin{align*} \Phi:\{\text{subgroups }H\leq G \text{ that contain }N\} &\longrightarrow \{\text{subgroups }\bar{H}\leq G/N\}  \end{align*} $$ such that $$ H\longmapsto \pi(H) = H/N. $$
The inverse of $\Phi$ is given by $\Psi(\bar{H}):=\pi^{-1}(\bar{H})$.

My problem with this statement is the way that $\Psi$ is defined. It seems to assume that $\pi^{-1}$ exists, and although $\pi$ is certainly surjective, I have a hard time believing it's always injective. So why can we define $\Psi$ such that it's contingent on $\pi^{-1}$ if it's not necessarily a bijection? Any help would be appreciated, thank you!

答:
Here, $\pi^{-1}$ denotes not an inverse function for $\pi$, but the preimage (or inverse image), which is defined for any function and takes subsets of the codomain to subsets of the domain (rather than acting on elements). Explicitly, given a function $\pi\colon A \to B$ and a subset $C \subseteq B$, we define
$$\pi^{-1}(C) = \{a \in A : \pi(a) \in C\}.$$
In your case, we have $A = G$, $B = G/N$, and $C = \bar{H}$, so $\pi^{-1}(\bar{H})$ is the subgroup of $G$ consisting of all elements of $G$ whose image under the quotient map $\pi$ is contained in the subgroup $\bar{H}$ of $G/N$.

(Also, indeed, the quotient map $\pi\colon G \to G/N$ is injective only in the trivial case where $N = \{1\}$.)


问:
Feedback on my proof of showing if $|a - b| \leq 1$, then $|a| \leq |b| + 1$
i'm taking a non-credit real analysis course online and this is a proof i wrote for one of my answers in my first problem set.

my solution is different from the instructor's, but he said in the video that there's more than one valid way to solve this problem, so i'd like to know if i did that. i'm not sure if what i did makes any sense because instead of proving if A then B, i proved if B then A, and it's not always the case that converses of true propositions are also true.

Any feedback? thanks in advance.

prove: if $|a - b| \leq 1$, then $|a| \leq |b| + 1$

proof:

$|a| \leq |b| + 1$

$|a| - |b| \leq 1$

$(|a| - |b|)^2 \leq 1$

$(|a| - |b|)(|a| - |b|) \leq 1$

$|a|^2 - 2|a||b| + |b|^2 \leq 1$

note: $|a| \geq a, |b| \geq b,$ so

$a^2 - 2ab + b^2 \leq |a|^2 - 2|a||b| + |b|^2 \leq 1$

$(a - b)^2 = |a - b|^2 \leq 1$

答:
No, this is wrong. The main problem is that you're trying to prove the converse of the statement, which as you point out yourself, isn't the same thing as proving the statement.

In this case, the converse is false. Try $a=0, b=2$, for example, and see where the logic falls apart.


答:
First, there's the major mistake you mentioned, that you're attempting to prove the converse instead of the statement in the question, so even if your proof were correct, it wouldn't be proving the right implication.

And note I said "even if your proof were correct", from which you might correctly guess it's not correct.  The easiest way to see this is that $|a| \leq |b| + 1$ is true for $a=-1, b=3$, but then $|a-b| \leq 1$ is false, so the converse isn't even true.

There are two steps you made an error in:

$$|a|-|b| \leq 1 \implies (|a|-|b|)^2 \leq 1 \\ |a|\geq a, |b|\geq b \implies a^2 - 2ab+b^2 \leq |a|^2 - 2|a||b| + |b|^2$$

Both of these are invalid steps.  Can you see why?

Title: Feedback on my proof of showing if $|a - b| \leq 1$, then $|a| \leq |b| + 1$

Body:
i'm taking a non-credit real analysis course online and this is a proof i wrote for one of my answers in my first problem set.

my solution is different from the instructor's, but he said in the video that there's more than one valid way to solve this problem, so i'd like to know if i did that. i'm not sure if what i did makes any sense because instead of proving if A then B, i proved if B then A, and it's not always the case that converses of true propositions are also true.

Any feedback? thanks in advance.

prove: if $|a - b| \leq 1$, then $|a| \leq |b| + 1$

proof:

$|a| \leq |b| + 1$

$|a| - |b| \leq 1$

$(|a| - |b|)^2 \leq 1$

$(|a| - |b|)(|a| - |b|) \leq 1$

$|a|^2 - 2|a||b| + |b|^2 \leq 1$

note: $|a| \geq a, |b| \geq b,$ so

$a^2 - 2ab + b^2 \leq |a|^2 - 2|a||b| + |b|^2 \leq 1$

$(a - b)^2 = |a - b|^2 \leq 1$


答:
You're right in that you've proved the converse of the statement (incorrectly, as other responses have pointed out), not the statement itself. Generally, one way you can check whether your proof still holds is to reverse the order of your proof, and see if it still works. There may be a step that no longer works, in which case the proof is invalid. But, if it still works in the reverse order, then you're good to go!

So, for this proof, you would start with:

$$ |a-b|^2 \le 1 $$
$$ \Rightarrow (a-b)^2 \le 1 ,$$

and continue back up your proof until you find a mistake or get back to the other end.

Another common way to approach something like this is to prove the contrapositive. In this case, you could do the following.

We want to prove that $|a-b| \le 1 \Rightarrow |a| \le |b| - 1 $. The contrapositive of this statement is:

$$ \neg (|a-b| \le 1) \Leftarrow \neg (|a| \le |b| - 1) $$
$$ \Leftrightarrow (|a| > |b| - 1 \Rightarrow |a-b| > 1) . $$

So, if you can prove that $|a| > |b| - 1$ implies $|a-b| > 1$, then you've also proved that $|a-b| \le 1$ implies $|a| \le |b| - 1$.


问:
automorhpism of a vector space leaving direct sum invariant
I was reading Sadri Hassani book in which the author proves that if an automorphism of a vector space leaves one summand of a direct sum invariant, then the other one is invariant as well. The proof is as follows:-

I cannot understand why you have $T(V_1\oplus V_2) = T(V_1)\oplus T(V_2)$. The statement also seems wrong to me as if you suppose a 2 dimensional space with the automorphism operator $T$ given by(in matrix notation):-

\begin{bmatrix}
    1 & 1 \\
    0 & 1
  \end{bmatrix}
The vector $V_1$ could be your x axis and $V_2$ be the y axis. Now, T leaves the x axis invariant but doesn't leave the y axis invariant. Where is my understanding wrong?

答:
As a side note to my comment, the step you had issues with is correct, in fact:
$T(V_1 \oplus V_2) = T(V) = V$
For every $v \in V$ we have that $v = T(w)$ and therefore $v = T(v_1+v_2) = T(v_1)+T(v_2)$ so $T(V_1) + T(V_2) = V$.
To see that the sum is direct, if $v \in T(V_1) \cap T(V_2)$ then $v = T(v_1) = T(v_2)$ but then $T(v_1-v_2) = 0$ which would imply $v_1 = v_2$ and therefore $v_1=v_2=0 \Rightarrow v=0$.

The problem lies in the assumption that $V_1 \oplus V_2 = V_1 \oplus V_2' \Rightarrow V_2 = V_2'$, as a counterexample you can take the spaces you just mentioned, $V_1 = \{(x,0)|x \in \mathbb{R}\}$, $V_2 = \{(0,y)| y \in\mathbb{R}\}$ and $V_2'=\{(x,x)|x \in \mathbb{R}\}$.


问:
What should $\mu$ be here in order for the logistic map to be stable?
I think this might be an error in A Survey of Computational Physics: Introductory Computational Science by Landau. On page 292, he mentions that in order for the logistic map to be stable, we must have

$$\left | \frac{df}{dx} \right |_{x_*}<1$$

The logistic map is given by

$$x_{i+1}=\mu x_i(1-x_i)$$

For the fixed point $x_*=\frac{\mu-1}{\mu}$, he claims that the map is stable if $\mu<3$. But shouldn't he have written that the map is stable if $1<\mu < 3$? My reasoning was that

$$\left | 2-\mu \right | <1$$
$$\to 2-\mu <1 \quad ; \quad -2+\mu < 1$$
$$\to \mu>1 \quad ; \quad \mu <3$$

So why didn't he write $\mu >1$ as a second condition for the map's stability?

答: You're right, but.... It's usual to consider the logistic map as a map on $[0,1]$. I don't know whether the author makes this restriction explicit in the text. With that restriction, $x=(\mu-1)/\mu$ is not in the domain for $\mu<1$, so it doesn't count as a fixed point. So as far as fixed points given by that formula go, we're only considering $\mu\ge1$ anyway, and in that range we have stability for $\mu<3$.


问:
Differentiability at endpoints of curves
This might sound like a silly question, but, why is it that for many theorems requiring a function be continuous on a curve on $[a,b]$, that whether the endpoints are differentiable don't matter?
For example, the Mean Value Theorem requires that for a continuous function $f$ on $[a,b]$, it must be differentiable on $(a,b)$. Why is the interval open instead of closed now?

答:
Generally, those types of assumptions are based on what's minimally required for the theorem to be true, and just those bare minimum requirements.

In the example you give, the MVT doesn't hold generally if $f$ is continuous on $(a,b)$ - think about what happens if there's a jump discontinuity at one of the endpoints. So, we need continuity on the closed interval or the theorem can be violated.

Similarly, the MVT is true when $f$ is differentiable on $(a,b)$. Requiring differentiability on $[a,b]$ instead would be a stricter requirement, and would needlessly limit the scope of the theorem. It would no longer apply to any functions that are not differentiable at either endpoint, even though the conclusion of the theorem would still be true for them.


问:
The proof of inequality about diameter.

Let $A(\neq \phi)$ be a set , $\epsilon >0$ and $G=\{ x \in \mathbb{R^n} | d(x, A) < \dfrac{\epsilon}{2} \delta (A) \}$. $\delta (A)$ means diameter of $A$, i.e $\delta (A)=$sup {$ d(x,y) | x,y\in A\} .$

I want to prove $\delta (G)\leqq (1+\epsilon) \delta (A)$.

I could understand by drawing a picture but I couldn't prove it with a mathematical formula.

I would like you to give me some ideas.

答: Take $x, y\in G$, and let $a, a' \in A$ such that $d(x,a) < \frac{\epsilon}{2}\delta(A)$ and $d(y,a) < \frac{\epsilon}{2}\delta(A)$. The triangle inequality gives $d(x,y) \leqq d(x,a) + d(a,a') + d(a',y)$. Show that this implies $d(x,y) \leqq (1 + \epsilon)\delta(A)$. Since $x$ and $y$ are arbitrary, $\delta(G) \leqq (1 + \epsilon)\delta(A)$.

Title: The proof of inequality about diameter.


问:
Regarding limits and $1^\infty$

Possible Duplicate:

  Why is $1^{\infty}$ considered to be an indeterminate form

I have some questions about limits and the undefinability of $1^\infty$.

For example, is $\lim_{x\to\infty}1^x$ indefinite? Why is it not $1$? Or do mathematicians, when saying that $1^\infty$ is indefinite, actually refer to cases such as $lim_{x\to\infty} \left(1 + \frac{a}{x}\right)^x$ where even though at a first glace the result is $1$, this is actually a special case and it is equal to $e^a$?

答:
Look up indeterminate forms. It depends on how you interpret $1^{\infty}$

The indeterminate form $1^{\infty} = \lim_{n \rightarrow \infty} f(n)^{g(n)}$ where $\lim_{n \rightarrow \infty} f(n) = 1$ and $\lim_{n \rightarrow \infty} g(n) = \infty$

For instance, all of the following limits are of the form $1^{\infty}$, yet they all evaluate to different numbers

$$\lim_{n\rightarrow \infty} 1^n = 1$$

$$\lim_{n \rightarrow \infty} \left(1 + \frac{1}{n} \right)^n = e$$

$$\lim_{n \rightarrow \infty} \left(1 + \frac{1}{\log(n)} \right)^n = \infty$$

$$\lim_{n \rightarrow \infty} \left(1 - \frac{1}{\log(n)} \right)^n = 0$$

It is a question of which one tends faster whether $f(n) \rightarrow 1$ or $g(n) \rightarrow \infty$.

When you get a finite number which is non-zero and not one as the limit, there is in some sense a balance between the rate at which $f(n) \rightarrow 1$ and $g(n) \rightarrow \infty$.

When you get infinity or zero as the limit, $g(n) \rightarrow \infty$ faster than $f(n) \rightarrow 1$.

When you get $1$ as the limit, $f(n) \rightarrow 1$ faster than $g(n) \rightarrow \infty$.

Title: Regarding limits and $1^\infty$


问:
Max-Flow Min-Cut Theorem
Max-Flow Min-Cut theorem uses three assertions about flow graphs and relates them proving that one implies other.

f is the maximum flow in Graph G

The residual graph contains no augmenting paths

The flow value of f (|f|) is c(S,T) (capacity of the cut) for some cut (S,T) of graph G.

Though the theorem carries the title min-cut, the third assertion only loosely mentions about some-cut but not about the specific cut namely the minimum cut. Is there any reason for this ?

答:
Welcome to MSE!

The max-flow/min-cut theorem is called that because it asserts the following:

The value of the maximum flow in a graph equals the weight of a minimum cut

Said in symbols:

MaxFlow = MinCut

whence the name.

Recall a "cut" is just a way of disconnecting the source and the sink. So the flow cannot be larger than the weight of some minimal cut. After all, if we remove all the cut edges, then the flow must be $0$ (since there's no longer a way to get from source to sink).

The max-flow/min-cut theorem tells you that this "obvious" obstruction is the only one. We can actually achieve this upper bound for any graph. As a remark, this notion that "the obvious obstruction is the only one" is a very common trope in combinatorics. You might think of other theorems in this vein.

I hope this helps ^_^

Title: Max-Flow Min-Cut Theorem


问:
Prove inequality $\tan \left( \frac\pi2 \frac{(1+x)^2}{3+x^2}\right) \tan \left( \frac\pi2 \frac{(1-x)^2}{3+x^2}\right)\le\frac13 $
How to determine the range of the function

$$f(x)=\tan \left( \frac\pi2 \frac{(1+x)^2}{3+x^2}\right)
\tan \left( \frac\pi2 \frac{(1-x)^2}{3+x^2}\right)
$$

It it is straightforward to verify that $f(x)$ is even and

$$f(0)= \frac13, \>\>\>\>\> \lim_{x\to\pm \infty} f(x) \to -\infty$$

which implies $f(x) \in (-\infty,\frac13]$, i.e.

$$\tan \left( \frac\pi2 \frac{(1+x)^2}{3+x^2}\right)
\tan \left( \frac\pi2 \frac{(1-x)^2}{3+x^2}\right)\le \frac13
$$

and is visually confirmed below

However, it is not obvious algebraically that $f(x)$ monotonically decreases away from $x=0$. The standard derivative tests are not viable due to their rather complicated functional forms.

So, the question is how to prove the inequality  $f(x) \le \frac13$ with rigor.

Note that it is equivalent to proving

$$\cot \left(\frac{\pi(1+x)}{3+x^2}\right)
\cot \left(\frac{\pi(1-x)}{3+x^2}\right)\le \frac13
$$

答:
The problem is not too hard if you consider
$$f(x)=\tan (A(x))\,\tan (B(x))$$ Using logarithmic differentiation
$$\frac{f'(x)}{f(x)}=A'(x) \csc (A(x)) \sec (A(x))+B'(x) \csc (B(x)) \sec (B(x))$$

This gives $f'(0)=0$.

Repeating the process knowing that $f(0)=\frac 13$ and $f'(0)=0$ (this simplifies a lot the calculations), we have
$$f''(0)=\frac{16}{81} \left(\sqrt{3}-\pi \right) \pi <0$$ So, by the second derivative test $x=0$ corresponds to the maximum of the function.

We could also have composed Taylor series around $x=0$ to obtain
$$f(x)=\frac{1}{3}+\frac{8}{81} \left(\sqrt{3}-\pi \right) \pi  x^2+\frac{8 \pi
   \left(\pi  \left(27+2 \sqrt{3} \pi -3 \pi ^2\right)-9 \sqrt{3}\right)
   }{2187}x^4+O\left(x^6\right)$$ which a perfect approximation of the function for $-\frac 12 \leq x \leq \frac 12$ (maximum error  $< 0.00012$ at the bounds).


问:
probablity function of $XY+Z$ if $X,Y,Z\sim_{i.i.d} Bern(p)$
The question asks why would $T = XY+Z$ not be a sufficient statistic for $X,Y,Z \sim_{i.i.d} Bern(p)$ As I try to check $\frac{\mathbb{P}(X,Y,Z,T=t)}{\mathbb{P}(T=t)}$ I cannot figure out what $\mathbb{P}(T)$ is.

答:
$T$ is not sufficient. To show this it is not necessary to derive the whole conditional distribution. A counterexample is enough

$$f_{X,Y,Z|T=0}(0,1,0|0)=\frac{\mathbb{P}[X=0;Y=1;Z=0;T=0]}{\mathbb{P}[T=0]}=\frac{(1-p)^2p}{(1-p)^3+2(1-p)^2p}=\frac{p}{1+p}$$

As you can see the conditional distribution (for this example) still depends on $p$ thus $T$ is not sufficient.

You can find other counterexamples or, as an exercise, all the distribution.

EDIT, answering to you comments

To calculate $\mathbb{P}[T=0]$ observe that it is attained only in the following cases

$$(X,Y,Z)=\{(0,0,0);(0,1,0);(1,0,0)\}$$

The conditional distribution is the discrete one defined on the above 3 realizations wiht the following probability, respectively:

$$\left\{ \frac{1-p}{1+p} ; \frac{p}{1+p} ; \frac{p}{1+p}\right\}$$


问:
Summing Concentric Circles.
This should be a simple problem. I want to sum lengths of concentric circles with the outer circle at radius L. Then each smaller circle is at radius $2\pi R$ less than the previous until the inner most circle hits $R$ and has circumference $2\pi R$. (This is intended to approximate a string wrapping around a disc R)

Assuming $L>>R$.

Assume $N=\frac{L}{2\pi R}$ is an integer in suitable units for the particular problem.

$\text{sum of diminishing circle circumferences}=$

$(L-2\pi R)2\pi + (L-2\pi R -2\pi R)2\pi +...$

$=2\pi NL - 4\pi^2 R (1+2 +3+...N)=2\pi NL - 4\pi^2 R(N(N+1))/2$

$=\frac{L^2}{2R} +\pi L$

Note: This problem was also answered here (where was considered as an approximation to a more difficult problem):

A very small disc winds around a larger circular disc of radius $R$ connected to it by a string. How long is the spiral it travels?

答:
Maximum/minimum radii are $(L,R)$. A finite thickness of circles in a continuous tape  packing is assumed and taken into account :

Let the thickness be $t$ and let each circle contact the next without a gap. Then
$$ n= \dfrac{L-R}{t}, \text {total length from each average circle  }= 2n \pi \dfrac{R+L}{2}$$

This can be also seen as total concentric circles lateral edge area divided by $t$ to be its total unwound length:

$$= \dfrac{\pi (L^2-R^2)}{t}.$$

Title: Summing Concentric Circles.


问:
Help me solve this equation: $x^\frac23 - 9x^\frac13+8=0$
Not sure what to call this type of equation so please let me know. I'm having trouble solving it though.

Solve the equation:
$$x^\frac23 - 9x^\frac13+8=0$$

答: Hint: try the substitution $y = x^{\frac{1}{3}}$. This should give you a quadratic. </p>


问:
Several ways to prove that $\sum\limits^\infty_{n=1}\left(1-\frac1{\sqrt{n}}\right)^n$ converges
I believe there are several ways to prove that $\sum\limits^{\infty}_{n=1}\left(1-\frac{1}{\sqrt{n}}\right)^n$ converges. Can you, please, post yours so that we can learn from you?

HERE IS ONE

Let $n\in\Bbb{N}$ be fixed such that $a_n=\left(1-\frac{1}{\sqrt{n}}\right)^n.$ Then,
\begin{align} a_n&=\left(1-\frac{1}{\sqrt{n}}\right)^n \\&=\exp\ln\left(1-\frac{1}{\sqrt{n}}\right)^n\\&=\exp \left[n\ln\left(1-\frac{1}{\sqrt{n}}\right)\right] \\&=\exp\left[ -n\sum^{\infty}_{k=1}\frac{1}{k}\left(\frac{1}{\sqrt{n}}\right)^k\right]\\&=\exp\left[ -n\left(\frac{1}{\sqrt{n}}+\frac{1}{2n}+\sum^{\infty}_{k=3}\frac{1}{k}\left(\frac{1}{\sqrt{n}}\right)^k\right)\right]\\&=\exp \left[-\sqrt{n}-\frac{1}{2}-\sum^{\infty}_{k=3}\frac{n}{k}\left(\frac{1}{\sqrt{n}}\right)^k\right]\\&\equiv\exp \left(-\sqrt{n}\right)\exp \left(-\frac{1}{2}\right)\end{align}
Choose $b_n=\exp \left(-\sqrt{n}\right)$, so that
\begin{align} \dfrac{a_n}{b_n}\to\exp \left(-\frac{1}{2}\right).\end{align}
Since $b_n \to 0$, there exists $N$ such that for all $n\geq N,$
\begin{align} \exp \left(-\sqrt{n}\right)<\dfrac{1}{n^2}.\end{align}
Hence, \begin{align}\sum^{\infty}_{n=N}b_n= \sum^{\infty}_{n=N}\exp \left(-\sqrt{n}\right)\leq \sum^{\infty}_{n=N}\dfrac{1}{n^2}<\infty,\end{align}
and so, $\sum^{\infty}_{n=1}b_n<\infty\implies \sum^{\infty}_{n=1}a_n<\infty$ by Limit comparison test.

答:
For $n\ge2$,
$$
\begin{align}
\left(1-\frac1{\sqrt{n}}\right)^n
&=\left(\left(1+\frac1{\sqrt{n}-1}\right)^{\sqrt{n}}\right)^{-\sqrt{n}}\tag{1a}\\
&\le\left(1+\frac{\sqrt{n}}{\sqrt{n}-1}\right)^{-\sqrt{n}}\tag{1b}\\[9pt]
&\le2^{-\sqrt{n}}\tag{1c}
\end{align}
$$
Explanation:

$\text{(1a)}$: $1-\frac1{\sqrt{n}}=\left(1+\frac1{\sqrt{n}-1}\right)^{-1}$

$\text{(1b)}$: Bernoulli's Inequality

$\text{(1c)}$: $1+\frac{\sqrt{n}}{\sqrt{n}-1}\ge2$

Since $(1)$ holds for $n=1$, it is true for $n\ge1$.
$$
\begin{align}
\sum_{n=1}^\infty\left(1-\frac1{\sqrt{n}}\right)^n
&\le\sum_{n=1}^\infty2^{-\sqrt{n}}\tag{2a}\\
&=\sum_{k=1}^\infty\sum_{n=(k-1)^2+1}^{k^2}2^{-\sqrt{n}}\tag{2b}\\
&\le\sum_{k=1}^\infty(2k-1)\,2^{-k+1}\tag{2c}\\[9pt]
&=6\tag{2d}
\end{align}
$$
Explanation:

$\text{(2a)}$: apply $(1)$

$\text{(2b)}$: group the terms

$\text{(2c)}$: $2k-1$ terms of size less than $2^{-k+1}$

$\text{(2d)}$: compute sum using

$\phantom{\text{(2d):}}$ $\sum\limits_{k=1}^\infty x^k=\frac{x}{1-x}\quad$ and $\quad\sum\limits_{k=1}^\infty kx^k=\frac{x}{(1-x)^2}$


问:
$\sum_{n=2}^{\infty}(2+c)^{-n}=b$ infinite series
Consider $\sum_{n=2}^{\infty}(2+c)^{-n}=b$. $b$ is a positive constant, and $c \in \mathbb{R}$. For which values of b does the infinite series permit a single solution of c? First, evaluating this series by the divergence test it has to be 0 for it to certainly converge. There's also a possibility of multiple solutions, but this is not very likely.

答:
The infinite series is a geometric series. It has a nice closed form, and has radius of convergence $1$. That is:

The sum converges only when $\left|\frac1 {2+c}\right| <1$, or $|2+c| > 1$, or ($c<-3$ or $c>-1$)

When it does converge, $\sum_{n=2}^\infty (2+c)^{-n} = \frac {(2+c)^{-2}}{1-(2+c)^{-1}} = \frac1{(2+c)(1+c)}$

The equation $\frac1{(2+c)(1+c)} = b$, or $(c+1)(c+2) = \frac1b$ is a quadratic equation in $c$. Any positive $b$ would yield two solutions for $c$, unless one of the solutions violates our convergence rule above.

Therefore we need to find when the quadratic equation $(c+1)(c+2)= \frac1b$ gives a solution $c\in[-3,-1]$. As one of the solutions is always greater than $-1$ and the other always less than $-2$, this invalid solution must be within $[-3, -2)$, which corresponds to $\frac1b \in (0,2]$, or $b\in[1/2,\infty)$.


问:
A very small disc winds around a larger circular disc of radius $R$ connected to it by a string. How long is the spiral it travels?
A very small disc winds around a larger circular disc of radius $R$. It is connected to it by a string of length l that remains tight. What distance does it travel before it hits the larger disc of radius $R$ ? I've estimated that it is $2\pi(l-R)$ based on assuming a spiral is a little like a series of concentric circles.

答:
As stated on that physics page, it is easiest to consider the length and angle of the tight straight piece of rope.  If the angle $\theta$ starts at $0$ then winding the cord around gives a maximal angle $L/R$ (in radians).  The length of the straight piece of rope decreases linearly in $\theta$, so it is $$L(\theta) = L - R \theta.$$ Then the distance that the end of the rope travels is given by $$ \int_0^{L/R}L(\theta) \mathrm d \theta = \int_0^{L/R}(L - R \theta) \mathrm d \theta = \frac{L^2}{2R}.$$

This formula can also be obtained from an explicit parameterization of the path and calculating its length.  Again expressed in the angle $\theta$ the path can be given by $$\begin{eqnarray}
x(\theta) & = & R \sin(\theta) + (L - R \theta) \cos(\theta)\\
y(\theta) & = & -R \cos(\theta) + (L - R \theta) \sin(\theta)
\end{eqnarray}$$

The integrand of the path length integral simplifies quite a bit since $$\dot x(\theta)^2 + \dot y(\theta)^2 = (L - R \theta)^2.$$


答:
Hint: put $\,t:=x^{1/3}\,$ and get the easy quadratic

$$0=t^2-9t+8=(t-8)(t-1)\ldots\ldots$$

Title: Help me solve this equation: $x^\frac23 - 9x^\frac13+8=0$

Body:
Not sure what to call this type of equation so please let me know. I'm having trouble solving it though.

Solve the equation:
$$x^\frac23 - 9x^\frac13+8=0$$


问:
Could Someone Please Verify My Proof? (fun with two primes)
Am I wrong? Is my proof clear?
Is this already a known result? If so, are there any simple implications or corollaries of this proof? (if that makes sense) I know some about groups, rings and fields, but not enough to argue with them.

Suppose $n,k\in\mathbb{N}$ such that $n=2k$, and $k>2$.

$\exists x,y\in\mathbb{N}$ such that $x+y=k-1$.

I claim that if $2x+1$ and $2y+1$ are prime then $n$ is the sum of two primes.

$$\begin{align*}
n&=2k,\\
&=2(x+y+1),\\
&=4xy+2(x+y+1)-4xy,\\
&=4xy+2x+2y+2-4xy,\\
&=(4xy+2x+2y+1)+1-4xy,\\
&=(4xy+2x+2y+1)+1-((2x+1)-1)((2y+1)-1),\\
&=(2x+1)(2y+1)+1-((2x+1)-1)((2y+1)-1).
\end{align*}$$

Let $m,l\in\mathbb{N}$ such that $m=2x+1$ and $l=2y+1$.

Then $$n=ml+1-(m-1)(l-1)$$ and

$
\forall p_{1},p_{2}\in\mathbb{N}$ such that $p_{1}$ and   $p_{2}$ are prime, if
$m=p_{1}$ and $l=p_{2}$

$$\begin{align*}
n&=2k,\\
&=ml+1-(m-1)(l-1),\\
&=p_{1}p_{2}+1-(p_{1}-1)(p_{2}-1),\\
&\therefore n=p_{1}+p_{2}
\end{align*}$$

答:
I'm afraid the result is quite trivial:
$$x+y=k−1$$
$$2x+2y=2k-2=n-2$$
$$(2x+1)+(2y+1)=n$$
So if $2x+1$ and $2y+1$ are prime then $n$ is the sum of two primes.


问:
Calculation of finding way to make a password
A computer password requires you to use exactly 1 uppercase, exactly 3 lowercases, 3 digits and 2 special charecters(given that there are 33 special charecters that can be used)(every thing can be repeated) . In how many ways you can create password?

My attempt: 1 uppercase can be choosen by ${26 \choose 1}$ ways, 3 lowercases can be choosen by ${26\choose 1}$$\times {26\choose 1}$$\times {26\choose 1}$ ways, 3 digits can be choosen by ${10\choose 1}$$\times {10\choose 1}$$\times {10\choose 1}$ways and 2 special charecters can be choosen by ${33\choose 1}$$\times {33\choose 1}$ ways. But I am stucking for further progress.
Please help me. Thanks in advance

答:
You want a configuration like $\,U\,L\,L\,L\,D\,D\,D\,S\,S\,$ and you already know how many ways there are to choose for this configuration.

Now configurations can be permuted in $\dfrac{9!}{1!3!3!2!}$ ways, just multiply


问:
Functions and Mapping question?

Determine all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying the
  condition

$$2f(x) = f(x + y) + f(x + 2y)$$

for all real numbers $x$ and all non-negative real numbers $y$.

I just don't know where to start.

答:
See my answer is x=y and simultaneously x=-y both are true but for a real number they cant be true simultaneously.

In case of mod x= mod y for x less than 0 its not possible hence we cant go for the above problem as they are fixed for a single run.
So the above is false for the given mapping.

Title: Functions and Mapping question?


问:
What should be proved in the binomial theorem?
I'm following Cambrige mathematics syllabus, from the list of contents of what should be learned:

Induction as a method of proof, including a proof of the binomial theorem with non-negative integral coefficients.

I know what it is, but I'm not sure of what should be proved here. The first thing that comes to mind is the idea of proving it it for $n+1$, but I thought about writing:

$(a+b)^n$

$(a+b)^{n+1}$

But I am missing what premise I should prove. I guess that the proof involves the nature of the coefficients of the expansion of $(a+b)^n$ but from here, I have no idea on how to proceed.
Can you help me?

Edit: I guess I've made some progress. First I evaluated

$$(x+y)^0=1$$

Then I've evaluated it with the summation form

$$\sum_{j=0}^{0}{n \choose j}x^{n-j}y^{j}$$

And confirmed that it's equal to $1$ (I guess this is the base step).

The I did the same for $n+1$:

$$(x+y)^1=x+y$$

Then I've evaluated again for the summation form and checked that the results are the same. Is this all that should be done?

答: If you know the expansion of $(a+b)^n$, it follows that the expansion of $(a+b)^{n+1}=(a+b)(a+b)^n=a(a+b)^n+b(a+b)^n$ can be found by distributing term by term and collecting coefficients. This is the type of reasoning you should use when doing inductive proofs.</p>


问:
Center of Mass for a Lamina. Why is the value inside the integral $x$ and not $x/2$?
TLDR

Why is the center of mass formula for a lamina equal to $A_1$ but not $A_2$? (Specifically the $0.5$ in the $x$-coordinate integral).

$$ A_1=\left(\frac{\int_a^bxf(x)dx}{\int_a^bf(x)dx}, \frac{\int_a^b 0.5 f(x)^2dx}{\int_a^bf(x)dx}\right)$$

$$ A_2=\left(\frac{\int_a^b0.5xf(x)dx}{\int_a^bf(x)dx}, \frac{\int_a^b 0.5 f(x)^2dx}{\int_a^bf(x)dx}\right)$$

Longer Explanation and Question

The center of mass of $n$ point masses in the $xy$-plane is point $P$ where $P$ is defined as:

$$ P = \left(\frac{\sum_{i=1}^n m_ix_i}{\sum_{i=1}^n m_i}, \frac{\sum_{i=1}^n m_iy_i}{\sum_{i=1}^n m_i} \right) .$$

When applying this to functions we could find the center of mass for an area under a curve bounded by two endpoints. For a given function $f(x)$ bounded by points $a$ and $b$,  we can chop up the function into smaller and smaller rectangles and average out their center of masses. Point $Q$ represents the center of mass within this interval and it is defined as:

$$ Q = \left(\frac{\int_a^bxf(x)dx}{\int_a^bf(x)dx}, \frac{\int_a^b 0.5 f(x)^2dx}{\int_a^bf(x)dx} \right) . $$

When taking the center of mass for the $y$-coordinate, it makes sense that we would be summing up the halfway height of every rectangle. However, we do not sum up the halfway width of every rectangle. The center of mass of a rectangle lies directly in the middle of the $dx$ slice we use to integrate the region. However, the formula doesn't include this, and it is wrong.

I cannot find an explanation as to why it is wrong. As the widths of our rectangles approach zero, I guess the center of mass can be approximated as being $x$ instead of $\frac{x}{2}$. We could just as easily factor out the $\frac{1}{2}$ value.

答:
The actual mass of the lamina is not distributed along a line but is distributed over a region of the plane. To work out the equivalent of
$$\sum_{i=1}^n m_i x_i $$
for the area between the graph of $y = f(x)$ and the line $y = 0$
between the lines $x = a$ and $x = b$,
where $f(x)$ is a positive function over $[a,b]$,
you can integrate a uniform mass density (density $1$ is fine)
over the entire region.
You can do this with a double integral like this:

$$
\int_a^b \int_0^{f(x)} 1x\,dy\,dx.
$$

The integrals $\int_a^b \int_0^{f(x)}$ correspond to $\sum_{i=1}^n,$
the density $1$ corresponds to $m_i,$
and the variable $x$ in the integrand corresponds to $x_i.$

The equivalent integral for
$$ \sum_{i=1}^n m_iy_i $$
is quite similar, except instead of $x_i \sim x$ we have $y_i \sim y$:
$$
\int_a^b \int_0^{f(x)} 1y\,dy\,dx.
$$

The denominator is similar, but without multiplication by either $x$ or $y$;
$$ \sum_{i=1}^n m_i $$
becomes
$$
\int_a^b \int_0^{f(x)} 1\,dy\,dx.
$$

Where things start to look less symmetric is when we start to do the integration.
Note that
$$ \int_0^{f(x)} 1x\,dy = x \int_0^{f(x)} 1\,dy = x f(x), $$
but
$$ \int_0^{f(x)} 1y\,dy = \int_0^{f(x)} y\,dy = \frac12 (f(x))^2 $$
and
$$ \int_0^{f(x)} 1\,dy = f(x). $$

These first steps of integration work out this way for any positive $f(x)$,
so it is possible to write the formula for $Q$ that you have shown.
But you could also have written the same solution as
$$
Q = \left(\frac{\int_a^b\int_0^{f(x)} x\,dy\,dx}{\int_a^b\int_0^{f(x)} 1\,dy\,dx},
 \frac{\int_a^b\int_0^{f(x)} y\,dy\,dx}{\int_a^b\int_0^{f(x)} 1\,dy\,dx} \right).
$$


问:
Change of coordinate matrix proof
Let $V$ be a finite deminsional vector space over a field $\mathbb F$, and let $\beta=${$x_1,x_2,....x_n$} be an ordered basis for $V$. Let $Q$ be an $n \times n$ invertible matric with entries from $\mathbb F$. Define $$x'_j=\sum_{i=1}^n Q_{i,j}x_i , 1\leq j \leq n$$

and set $\beta'=${$x'_1,x'_2,....,x'_n$}. Prove that $\beta'$ is a basis for $V$ and hence that $Q$ is the change of coordinate matrix changing $\beta'$-coordinates into $\beta$-coordinates.

Since $|\beta'|=n=|\beta|$, then I only have to show linear independence to prove $\beta'$ is a basis.

I'm a little unsure about my indexing, but here is what I think is the next step:

If $$\sum_{j=1}^n \alpha_j x'_j=0$$
($\alpha_j \in \mathbb F$) then
$$ \alpha_1 x'_1+\alpha_2x'_2+....+\alpha_nx'_n=0$$
$$=\alpha_1\sum_{i=1}^nQ_{i,1}x_i+\alpha_2\sum_{i=1}^nQ_{i,2}x_i+...+\alpha_n\sum_{i=1}^nQ_{i,n}x_n=0$$

Then, since $Q$ is not the zero matrix since it's invertible. Also, $x_is$ are not equal to zero since they are the basis elements for $\beta$. This means $\alpha_1=\alpha_2=....=\alpha_n=0$. So $\beta'$ is linearly independent and therefore a basis for $V$.

Did I do this correctly? I feel that this is not a correct approach since I'm supposed to show that $Q$ is the change of coordinate matrix.

Any help/feedback is appreciated.
Thanks.

答:
If we rewrite the last line $$0 = \alpha_1\sum_{i=1}^nQ_{i,1}x_i+\alpha_2\sum_{i=1}^nQ_{i,2}x_i+\cdots+\alpha_n\sum_{i=1}^nQ_{i,n}x_n,$$
as follows
$$(\sum_{j=1}^{n}Q_{1,j}a_{j})x_{1}+(\sum_{j=1}^{n}Q_{2,j}a_{j})x_{2}+\cdots+(\sum_{j=1}^{n}Q_{n,j}a_{j})x_{n} = \sum_{i=1}^{n} (\sum_{j=1}^{n}Q_{i,j}a_{j})x_{i}$$
then we observe a system of linear equations. Hence, we have $Qa=0,$ where $a$ is a column vector with entries $a_{1}, a_{2}, \dots, a_{n},$ and we obtain $a = 0$ by multiplying both sides by the inverse $Q^{-1}.$ Then, $a_{1}=a_{2}\cdots=a_{n}=0$ implies $\beta^{'}$ is a linearly independent subset of $V,$ and from here, one can easily show that $\beta^{'}$ is a basis for $V.$

Subsequently, we write the coordinate vector of $x_{j}^{'}$ relative to $\beta$
$$[x_{j}^{'}]_{\beta} = [I_{V}(x_{j}^{'})]_{\beta} = [I_{V}]_{\beta^{'}}^{\beta} [x_{j}^{'}]_{\beta^{'}} = Q [x_{j}^{'}]_{\beta^{'}},$$
where $I_{V}$ is the identity transformation on $V$ and $[x_{j}^{'}]_{\beta^{'}} = e_{j},$ i.e., a $1$ in the $j$th component and 0 elsewhere. So, $Q$ changes $\beta^{'}$-coordinates into $\beta$-coordinates.

Title: Change of coordinate matrix proof


问:
Understanding the Cauchy Integral Formula
This is the statement of the Cauchy Integral Formual.  I have seen its proof, and I have used this formula many times to compute integrals. However, I feel like I do not understand this formula completely.

Please note that Wikipedia says, "it expresses the fact that a holomorphic function defined on a disk is completely determined by its values on the boundary of the disk." This statement seems very strong and insightful. My question is: What does this exactly mean? I have some sort of vague intuitive understanding but I am not satisfied with my understanding at all. Since the integral of
${\frac {1}{z-a}}$ is  $2 \pi i$ along the boundary of the disc, "roughly" we are taking "the average" of the function along the boundary of the disc to compute the function value  $f(a)$ for every $a$ in the interior of $D.$ Could you help me understand the following sentence?

"It (the Cauchy Integral Formula) expresses the fact that a holomorphic function defined on a disk is completely determined by its values on the boundary of the disk."

Thanks so much.

答: Note that, when $|a-z_0|<r$, we have$$f(a)=\frac1{2\pi i}\int_{|z-z_0|=r}\frac{f(z)}{z-a}\,\mathrm dz.\tag1$$Since$$\int_{|z-z_0|=r}\frac{f(z)}{z-a}\,\mathrm dz=\int_0^{2\pi}\frac{f(re^{it})ire^{it}}{re^{it}-a}\,\mathrm dt,$$which depends only on the values that $f$ takes on the circle centered at $z_0$ with radius $r$, $(1)$ tells us that values that $f$ takes at $a$ (which does not belong to that circle) using only the values that $f$ takes on the circle. In other words, knowing only how $f$ behaves on the boundary of the disk centered at $z_0$ with radius $r$ will allow you to compute the values that $f$ takes inside that disk.


问:
Show non-standard interpretation on language of arithmetic s.t. for any formula, if $\mathcal{N} \vDash F(n)$, then $\mathcal{M} \vDash F[m]$

Let $\mathcal{L}$ be the language of arithmetic. Let $\Gamma$ be the set of sentences true in the standard interpretation $\mathcal{N}$ [natural numbers]. Let $F(x)$ be any formula. Show that if $\mathcal{N} \vDash F(n)$, for every numeral $n$, then there is a non-standard model $\mathcal{M}$ of $\Gamma$ and a non-standard number $m$ such that $\mathcal{M} \vDash F[m]$

(Hint: use compactness)

The compactness theorem says that if every finite subset of a set of sentences has a model, then the whole set has a model. I'm not sure how that helps.

One common nonstandard interpretation is one with rational numbers $\mathcal{Q}$ instead of natural numbers. However, that is not a model for $\Gamma$. Consider the formula $\forall x \exists y (x < y \, \& \, \lnot \exists z (x < z \, \& \, z < y))$. That is true in $\mathcal{N}$ and therefore is in the set $\Gamma$ but it's not true in $\mathcal{Q}$.

What other kind of non-standard models for $\Gamma$ exist?

Can I get a hint on this?

EDIT: Revised with help from @Noah Schweber.

答:
Mark has provided what I suspect is the intended solution, namely a compactness argument. However, this is an odd problem - a much stronger statement is true, and compactness plays no role at all.

(I'm assuming here that the existence of nonstandard models at all has already been established; if not, then of course compactness is needed but now there's no obvious reason for talking about a formula $F$.)

Suppose $F$ is a formula such that, for every numeral $m$, we have $\mathcal{N}\models F(m)$. There is a single sentence $\varphi$ which captures this situation: $$\forall x(F(x)).$$ Since every natural number has a corresponding numeral, the assumption "$\mathcal{N}\models F(m)$ for every numeral $m$" is equivalent to "$\mathcal{N}\models\forall x(F(x))$."

So the sentence $\forall x(F(x))$ is in $\Gamma$. But then if $\mathcal{M}$ is any nonstandard model of $\Gamma$ we must have $\mathcal{M}\models\forall x(F(x))$, or put another way $\mathcal{M}\models F(a)$ for every $a\in\mathcal{M}$ (nonstandard or not). Consequently we have:

Suppose $F$ is a formula such that, for every numeral $m$, we have $\mathcal{N}\models F(m)$. Then for every nonstandard model $\mathcal{M}$ of $\Gamma$ and every element $a$ of $\mathcal{M}$ we have $\mathcal{M}\models F(a)$.

This is a stronger result than what is claimed in the OP - the latter merely says that some nonstandard model has some nonstandard element satisfying $F$.

That said, there is a version of this problem which does use compactness:

Suppose $\Theta$ is a first-order theory in the language of arithmetic which proves $m\not=n$ for all distinct numerals $m,n$, and $F$ is a formula in the language of arithmetic such that $\Theta\models F(m)$ for every numeral $m$. Show that there is an $\mathcal{M}\models\Theta$ with some $a\in\mathcal{M}$ such that $\mathcal{M}\models a\not=m$ for every numeral $m$ but $\mathcal{M}\models F(a)$.

Basically, there is no version of arithmetic which "pins down" the numerals exactly. Here Mark's answer (with $\Theta$ replacing $\Gamma$) is exactly what does the job. I suspect your professor actually intended to ask this, or something like it, since here compactness really is necessary.


问:
Naturally ordered semigroups unique up to isomorphism theorem
I'm having some trouble understanding the theorem stating that naturally ordered semigroups are unique up to isomorphism. I'm reading the version from Proofwiki:

http://www.proofwiki.org/wiki/Naturally_Ordered_Semigroup_Unique_up_to_Isomorphism

Let $\left({S, \circ, \preceq}\right)$ and $\left({S\,', \circ',
\preceq'}\right)$ be naturally ordered semigroups.

Let:

  $\bullet$ $0\,'$ be the smallest element of $S\,'$

  $\bullet$ $1'$ be the smallest element of $S\,' \setminus \left\{{0\,'}\right\}
= S\,'^*$.

  Then the mapping $g: S \to S\,'$ defined as:   $$\forall a \in S: g \left({a}\right) = \circ'^a 1'$$ is an isomorphism from
  $\left({S, \circ, \preceq}\right)$ to $\left({S\,', \circ',
 \preceq'}\right)$.

  This isomorphism is unique.

  Thus, up to
  isomorphism, there is only one naturally ordered semigroup.

I'm confused by the mapping definition, $\circ'^a1'$. I'm not sure what it's saying: the operator-superscript notation's not explained anwhere I can find, and it seems to be combining operators and elements from two separately defined semigroups. I'd have thought that $\circ'$ would be a mapping with its domain within $S\,'$, so you couldn't just feed in an element from $S$ unless you knew or assumed $S$ was a subset of $S\,'$. (I may well have the wrong end of a few sticks: I'm just a nosy layperson reading online maths texts out of curiosity, not a real student.)

答: The map you refer to is defined in http://www.proofwiki.org/wiki/Recursive_Mapping_to_Semigroup/Powers_of_Commutative_Elements. You'll have to refer to http://www.proofwiki.org/wiki/Recursive_Mapping_with_Identity to make sense of the definition. As far as I can see the notation $\circ'^a1'$ is basically what would usually be denoted $1'^a$ if $a$ were a natural number, though this notation doesn't explicitly state the operator.</p>


问:
Why those integral limits are wrong?
Let $$\displaystyle{ f_{X,Y}(x,y)=\left\{ \begin{array}{cc} \frac18 & |x|+|y|\leq 2 \\ 0 & |x|+|y|>2 \end{array}\right. }$$ be density function of continuous random variable $X,Y$. for every $|y|<2$ find $f_{X|Y}(x|y)$

So my idea here was to find $f_y$ and than use the formula of $$f_{X|Y}(x|y)=\frac{f_{X,Y}(x,y)}{f_Y(y)}$$

so the  because $|x| \leq 2-|y|$ I can say that $-2 \leq x \leq 2$ and than $$f_Y = \int_{-2}^{2}f_{X,Y}dx$$.

Why is this wrong?

答:
Why is this wrong?

Because $(X,Y)$ support is a square with vertices $(0,2);(2,0);(0,-2);(-2,0)$

thus integrating you get

$$f_Y(y)=\frac{1}{8}\int_{-2-y}^{y+2}dx=\frac{y+2}{4}$$

for $-2<y\leq 0$

and

$$f_Y(y)=\frac{1}{8}\int_{y-2}^{2-y}dx=\frac{2-y}{4}$$

for $0<y<2$

Concluding:

$$f_Y(y)=\frac{y+2}{4}\cdot\mathbb{1}_{(-2;0)}(y)+\frac{2-y}{4}\cdot\mathbb{1}_{(0;2)}(y)$$

... a triangle distribution

Title: Why those integral limits are wrong?


问:
$x^{x}$ as a composite function
I understand that this question has been asked a lot of times but I haven't yet got a clear explanation
$$x^{x}$$
is not called a composite function.

The reason they give is that it cant be written as $$f(g(x))$$
and then they start talking about logarithms

What I want to know is that why it cant be written as a composite function with some examples without involving logarithms in the explanation

Just a clear reason with example

答: We can write $x^x = e^{x\ln(x)}$ for $x>0$. In this way, it is a composite function.

Title: $x^{x}$ as a composite function


问:
Proof of Frullani's theorem
How can I prove the Theorem of Frullani? I did not even know all the hypothesis that $f$ must satisfy, but I think that this are

Let $\,f:\left[ {0,\infty } \right) \to \mathbb R$ be a a continuously differentiable function such that $$
\mathop {\lim }\limits_{x \to \infty } f\left( x \right) = 0,
$$
and let $
a,b \in \left( {0,\infty } \right)$.
Prove that $$
\int\limits_0^{\infty}  {\frac{{f\left( {ax} \right) - f\left( {bx} \right)}}
{x}}dx  = f\left( 0 \right)\left[ {\ln \frac{b}
{a}} \right]
$$
 If you know a more general version please give it to me )= I can´t prove it.

答:
$$F(a) = \int_0^\infty \frac{f(ax)}x dx, $$
$$F'(a) = \int_0^\infty f'(ax) dx = \frac{f(\infty) - f(0)}{a}.$$

$$\Rightarrow F(a) = (f(\infty) - f(0)) \ln a + C_1.$$

Similarly

$$F(b) = (f(\infty) - f(0)) \ln b + C_2$$

Here $C_1 = C_2$ since basically function are the same. Subtracting gives

$$F(a) - F(b) = (f(\infty) - f(0)) \ln (a/b).$$

I think we can prove it by splitting the integral into two, treat each part as functions of a and b respectively, differentiate wrt a and b respectively (after this step finding the integral is trivial). This will be followed by the integration of the obtained functions wrt to a and b. The constants of integration will be the same as the function essentially remains the same, only the variables are changed. So on subtraction, the constants get cancelled, and hence the result! At least for an undergraduate, this should be fine.

Title: Proof of Frullani's theorem


问:
Differentiate the integral $\frac{{\rm d}}{{\rm d}x} \int_{x_0}^{x_1} \rho u_{tt} {\rm d}x$
In Partial Differential Equations by Walter Strauss, Ch 1.3 Example 2, the author differentiates the integral $$\int_{x_0}^{x_1} \rho u_{tt} {\rm d}x$$ with respect to $x$ and finds that

$$\frac{{\rm d}}{{\rm d}x} \int_{x_0}^{x_1} \rho u_{tt} {\rm d}x=\rho u_{tt}$$

Why is this the case ? Did they use the Leibniz integral rule or the fundamental theorem of calculus?

答: I think its just the fundamental theorem of calculus. $x_1$ is a variable and you treat $x_0$ as fixed so when you differentiate the integral you get $\rho u_{tt}|^{x=x_1}$ and differentiating $T{u_x}|_{x_0}^{x_1}$ wrt $x_1$ is the same as differentiating $T{u_x}$ wrt x and then evaluating at $x_1$ so you get $(T{u_x})_x|^{x_1}$. Therefore $Tu_x=\rho u_{tt}$


问:
Prove the equivalence of sets
Suppose that $A \subseteq$ B $\Leftrightarrow$ $A \cap B^c = \emptyset$

I tried Venn diagrams, but I didn't help

答:
It's simpler to prove the contrapositive
$$A \nsubseteq B \Longleftrightarrow A \cap \overline{B} \neq \emptyset$$
First suppose that $A \nsubseteq B$. This means that NOT all elements of $A$ belong to $B$.
This is equivalent to say that there is some element of $A$ not belonging to $B$.
But the set of elements of $A$ not belonging to $B$ is exactly $A \cap \overline{B}$.  Hence this is equivalent to say that $A \cap \overline{B}$ is not empty.

Title: Prove the equivalence of sets


问:
Find the minimum of $\sqrt{4y^2-12y+10}+\sqrt{18x^2-18x+5}+\sqrt{18x^2+4y^2-12xy+6x-4y+1}$
Find the minimum of $$f(x,y)=\sqrt{4y^2-12y+10}+\sqrt{18x^2-18x+5}+\sqrt{18x^2+4y^2-12xy+6x-4y+1}$$ It seems that $f_x=f_y=0$ is very hard to compute. Is there any easier idea?

Body:
We have
\begin{align}
f(x, y) &= \sqrt{4y^{2}- 12y+ 10}+ \sqrt{18x^{2}- 18x+ 5} \\
&\qquad + \sqrt{18x^{2}+ 4y^{2}- 12xy+ 6x- 4y+ 1} \\[6pt]
&= \sqrt{\frac{ ( 2y- 6 )^{2} + ( 6y- 8 )^{2}}{10}} + \sqrt{\frac{( 6x- 5 )^{2}+ ( 12x- 5 )^{2}}{10}} \\[6pt]
&\qquad + \sqrt{\frac{ ( 6x+ 2y- 1  )^{2}+  ( 12x- 6y+ 3  )^{2}}{10}}\\
&\ge \frac{6- 2y}{\sqrt{10}}+ \frac{5- 6x}{\sqrt{10}}+ \frac{6x+ 2y- 1}{\sqrt{10}} \\[10pt]
&= \sqrt{10}.
\end{align}


问:
Division algebra over a an algebraically closed field
I am reading my notes and I stumbled upon a proof that I dont fully understand, and I was hoping maybe someone could clear the details.

The main goal was to show that if $k$ is algebraically closed, then the group ring $kG\cong M_{n_1}(k)\times....\times M_{n_r}(k)$ (where the characteristic of the field does not divide $|G|$).

First we applied Maschkes theorem and obtained $kG\cong M_{n_1}(D_1)\times...\times M_{n_r}(D_r)$, second we wanted to prove that in this case each $D_i$ equals $k$.

This are my poorly taken notes:

Let $x\in D$ then $k[x]\in D$ is a finite dimensional $k$-algebra which is a domain (because $D$ is division): Either $k[x]/k$ is integral or $\forall y\in k[x], y\neq 0$ the map induced by left multiplication by $y$ is an injection, hence a surjection, hence an isomorphism. So $k[x]/k$ is finite with $k$ algebraically closed then $k=k[x]$ so $x\in k$.

So on the above proof I see if $k[x]/k$ was integral then its finite and then we are done, but I dont understand the "or" part.

Thanks

答: I don't really understand what you wrote either. Here is the statement and then the proof: we claim that if $D$ is a finite-dimensional division algebra over an algebraically closed field $k$, then in fact $D \cong k$. As proof, if $x \in D$, then consider the inverse closed subring of $D$ generated by $k$ and $x$. This is a finite, hence algebraic, extension of $k$, hence must be equal to $k$. </p>


问:
When throwing a die $3$ times, what is the probability of the product of the three results being a multiple of $9$?
When throwing a die $3$ times, find the probability of the product of the three results being a multiple of $9$.

I tried drawing a table for the possible outcomes of the first TWO dice. Then I tried considering the cases for each when the third die is rolled - as you can imagine not very efficient. Perhaps I could use $x+y+z = 9k$ as a divisibility test, where $k$ is an integer. I'm pretty sure this would involve stars and bars, but I'm not sure how to implement it in this case. Would I have to solve for $k$?

I also worked out the boundaries but I'm not sure if that will help either: $9≤x+y+z≤216$.

Do I need to use stars and bars for this question?

答:
You can solve it quickly by counting the cases when the product is not divisible by $9$.

Number of all throws: $\color{blue}{6^3}$

Number of throws containing neither $3$ nor $6$: $\color{blue}{4^3}$

Number of throws containing exactly one $3$: $\color{blue}{3\cdot 4^2}$

Number of throws containing exactly one $6$: $\color{blue}{3\cdot 4^2}$

Hence, the probability $P$ of a throw with a product divisible by $9$ is

$$P = \color{blue}{\frac{6^3-4^3 -3\cdot 4^2 - 3\cdot 4^2}{6^3}}= \frac{7}{27}$$


问:
Can we define a reverse mapping for the $\phi : Y \rightarrow C(X, Y)$, as $\phi(y) = y' $, where $y' $ is a constant map in $C(X, Y)$.
For a continuous map $\phi : Y \rightarrow C(X, Y)$, as $\phi(y) = y' $, where $y' $ is a constant map in $C(X, Y)$.
Can we define a reverse mapping as $\phi' : C(X, Y) \rightarrow Y$ for a metric space $Y$. If yes how can it be defined? And suppose $C(X, Y)$ is endowed with uniform topology.

答:
Ideas: if $X$ is a connected space and $Y$ is totally disconnected (like $Y=\Bbb Q$ or $Y=C$ the Cantor set etc.) then $C(X,Y) = \phi[Y]$, i.e. all continuous maps from $X$ to $Y$ are constant, and $\phi$ has an obious inverse that is also continuous (in most natural topologies that we could put on $C(X,Y)$).

Otherwise I don't really see a natural map $C(X,Y) \to Y$, unless in specific situations like $X$ pseudocompact and $Y$ an ordered space, where $f \to \max f[X]$ or $f \to \min f[X]$ seems natural. I think it will be continuous in the uniform topology, though I haven't checked the details.

Even more specific, if $Y=\Bbb R^n$ and $C(X,Y)$ has such a topology that the set of constant maps $\phi[Y]$ is closed in it and $C(X,Y)$ is normal, then Tietze applies to show there is a continuous extension of $\phi^{-1}: \phi[Y] \to Y$ to $C(X,Y)$.


问:
Expectation of a distribution function of a random variable
$\mathbf {The \ Problem \ is}:$ If $X$ is a random variable with distribution function $F,$ then show that $E[F[X]]= \frac{1}{2} + \frac{1}{2}\sum_{x} P\{X=x\}.$

$\mathbf {My \ Approach }:$ Actually, if $P_X$ is the probability measure corresponding to $X$, then by Lebesgue Decomposition Theorem, $P_X=P_1+P_2$ where $P_1$ is absolutely continuous and $P_2$ is discrete .

Then $\frac{1}{2}$ is appearing after integrating $F_1$(distribution function corresponding to $P_1$) but how to bring $\frac{1}{2}\sum_{x} P\{X=x\}$ ?

A hint is very much required at this moment .....Thanks in advance .

答:
This is false. If $x$ takes the values $-1$ and $+1$ with probability $\frac 1 2 $ each then $EF(X)=\frac {F(-1)+F(1)} {2}=\frac {\frac 1 2 +1} 2=\frac 3  4$ but $\frac  1 2+\frac 1  2 \sum P(X=x)=1$ for any discetre random variable.

Actually, $\int F(x)dF(x)+ \int F(x-)dF(x)=1$ for any distribution function $F$. And since $F(x)=F(x-) +P(X=x)$, this gives $2\int F(x)dF(x) =1+\int P(X=x)dF(x)$ which implies that $EF(X)=\frac 1  2 +\frac 1  2\sum_x (P(X=x))^{2}$.

Ref: Integration by parts for Lebesgue Steiltjes Integrals in Real and Abstract Analysis by Hewitt and Stromberg.


问:
Prove that the convolution of $f$ and $g$ in $C^1[-\pi, \pi]$ if either $f$ or $g$ in $C^1[-\pi, \pi]$
Prove that the convolution of $f$ and $g$ in $C^1[-\pi, \pi]$ if either $f$ or $g$ in $C^1[-\pi, \pi]$. Here, $C^1$ is the space of all continuously differentiable functions. Here we obtain the same for $C^0$

But, I am not sure how to proceed for $C^1$ space. Kindly help me out.

Thanks in advance!

答:
I'll give my suggestion:

As stated by @Davide Giraudo in the post you linked to, you know that $f\star g= g\star f$, so we can assume without loss of generality that $f\in C^1$.

Using  this version of Leibniz intergral rule, we obtain that

$$\frac{d }{d x}\big[ f\star g \big] = \frac{1}{2\pi} \int_{-\pi}^\pi \frac{\partial }{\partial x} \big[ g(y) \cdot f(x-y) \big]dy= $$

$$=\frac{1}{2\pi} \int_{-\pi}^\pi g(y)\cdot f'(x-y)dy $$

which is continuous using the related post since $f'\in C^0$. You do have to note that $f,f'\in L^\infty$ since they are continuous on a compact set.


问:
Calculation of finding way to make a password
A computer password requires you to use exactly 1 uppercase, exactly 3 lowercases, 3 digits and 2 special charecters(given that there are 33 special charecters that can be used)(every thing can be repeated) . In how many ways you can create password?

My attempt: 1 uppercase can be choosen by ${26 \choose 1}$ ways, 3 lowercases can be choosen by ${26\choose 1}$$\times {26\choose 1}$$\times {26\choose 1}$ ways, 3 digits can be choosen by ${10\choose 1}$$\times {10\choose 1}$$\times {10\choose 1}$ways and 2 special charecters can be choosen by ${33\choose 1}$$\times {33\choose 1}$ ways. But I am stucking for further progress.
Please help me. Thanks in advance

答:
As the repetition is allowed, first find number of ways to choose positions of $1U, 3L, 3D$ and $2S$ ($U$ - uppercase, $L$ - lowercase, $D$ - digit, $S$ - special character).

That can be done in $\displaystyle 9 \cdot {8 \choose 3} \cdot {5 \choose 3}$ ways.

Now uppercase can be chosen in $26$ ways, lowercase can be chosen in $26^3$ ways, digits can be chosen in $10^3$ ways and special characters can be chosen in $33^2$ ways.

So total number of passwords of length $9$ that can be made

$ = \displaystyle 9 \cdot {8 \choose 3} \cdot {5 \choose 3} \cdot 26^4 \cdot 10^3 \cdot 33^2$


问:
Does $(0, \infty)$ inherit its usual topology as a subspace of $\mathbb{R}_K$?
Recall that $\mathbb{R}_K$ denotes the real line with the $K$ topology. In other words, its basis elements are of the form $(a, b) - K$, where $K = \{1/n: n \in \mathbb{Z}_+\}$. On page 178 of Munkres's 'Topology' book, the author states that $(0, \infty)$ inherits its usual topology as a subspace of $\mathbb{R}_K$. But I disagree because $(0,1) - K$ is open in $\mathbb{R}_K$ and so its intersection with $(0,\infty)$ is open in $(0, \infty)$. In other words, $((0,1) - K) \cap (0, \infty) = (0,1) - K$ is open in $(0, \infty)$, but the standard topology on $(0, \infty)$ does NOT have $(0,1) - K$ as one of its open sets. So where am I wrong in my argument here? I am assuming that Munkres is correct and I am not. Thanks!

答: $K$ is not closed in $\Bbb R$ in the usual topology, but it is closed in $(0,+\infty)$ in its subspace topology (as $K = (K \cup \{0\}) \cap (0, +\infty)$, so the intersection of a closed subset of $\Bbb R$ with the subspace). So the $K$-topology adds no new sets to the subspace topology on $(0,+\infty)$, hence Munkres' remark.


问:
a topological question on semialgebraic sets
Let $g_1,\cdots,g_m \in \mathbb{R}[x]=\mathbb{R}[x_1,\cdots,x_d]$ be polynomials with real coefficients and define the semi-algebraic sets $K=\left\{x \in \mathbb{R}^d: g_i(x) \ge 0, \forall i\right\}, K'=\left\{x \in \mathbb{R}^d: g_i(x) >0, \forall i\right\}$.

Proposition: The interior of $K$ is nonempty if and only if $K'$ is nonempty.

Proof: If $K' \neq \emptyset \Rightarrow Int(K) \neq \emptyset$, since $K' \subset Int(K)$. Conversely, suppose $Int(K) \neq \emptyset$ but $K'= \emptyset$. Then there exists an $\epsilon$-ball $\mathcal{B}$ inside $K$. Consider the polynomial $g=g_1 \cdots g_m$. Since $K'=\emptyset$, $g$ vanishes on $K$ and thus $g$ vanishes on the ball $\mathcal{B}$.

Question: Why does the vanishing of $g$ on $\mathcal{B}$ imply that $g=0$?

答: You can prove that if $g \in \mathbb R[x_1, \ldots, x_n]$ is zero on any epsilon ball then $g = 0$.  You do this by induction on $n$.  For the base case $n = 1$ use that non-zero polynomials only have finitely many roots.  For the inductive step write $g = \sum_ix_n^ig_i$ where each $g_i$ is a polynomial in $x_1, \ldots, x_{n-1}$.  If $g \neq 0$ then some $g_i \neq 0$ so there is a point $a = (a_1, \ldots, a_{n-1}) \in \mathbb R^{n-1}$ such that $g_i(a) \neq 0$ (you can choose $a$ such that for some $a_n$ the point $(a_1, \ldots, a_n) \in \mathbb R^n$ is in the epsilon ball).  Then $g(a_1, \ldots, a_{n-1}, x_n)$ is a nonzero polynomial in one variable that vanishes on some epsilon ball, which is a contradiction.</p>


问:
Constructing element which does not belong to a set
If $A$ is an arbitrary set, is it possible to construct an element $a\not\in A$ without the use of the Axiom of Regularity?

It is clear that such $a$ exists (else $A$ would be the set of all
sets), but this is not an explicit construction.

Under assumption of the Axiom of Regularity, one could choose $a=A$.

To give more context, I'm starting with a family $\{X_i\mid i\in I\}$ of non-empty sets, and would like to find a sequence $\langle y_i \rangle_{i\in I}$ such that $y_i\not\in X_i$ for each $i\in I$, without using the Axiom of Choice and the Axiom of Regularity.

答: This might not be constructive enough, but there exists an element in the Hartogs ordinal of $A$ that is not a member of $A$. Otherwise it would be a subset of $A$, a contradiction (there can be no injection from the Hartogs ordinal into $A$).


答:
Here's one observation (which is essentially the same as Tesla Daybreak's answer):

$\mathsf{ZF-Reg}$ is strong enough to prove that no set contains every ordinal. Now given a set $x$, let $y$ be the least ordinal not in $x$; regardless of whether regularity holds in the universe, the ordinals themselves are well-ordered as usual. This is totally explicit and pins down $y$ uniquely in terms of $x$.

Title: Constructing element which does not belong to a set

Body:
If $A$ is an arbitrary set, is it possible to construct an element $a\not\in A$ without the use of the Axiom of Regularity?

It is clear that such $a$ exists (else $A$ would be the set of all
sets), but this is not an explicit construction.

Under assumption of the Axiom of Regularity, one could choose $a=A$.

To give more context, I'm starting with a family $\{X_i\mid i\in I\}$ of non-empty sets, and would like to find a sequence $\langle y_i \rangle_{i\in I}$ such that $y_i\not\in X_i$ for each $i\in I$, without using the Axiom of Choice and the Axiom of Regularity.


问:
Uniformly distributed transformation
We let X be uniformly distributed  on the interval $(0,1)$. We must find the distribution of $Y=-log(X)$. I'm not sure how to do that but maybe I think maybe I can use that if X be uniformly distributed  on the interval $(a,b)$ then will the distribution be:
$$P(Y \leq y)=\frac{y-a}{b-a}$$, but how to use this and what if $P(Y > y)$?

答:
$Y=-\log X=g(X)$ is a monotonic transformation thus $f_Y(y)$ can be derived immediately by derivating $g^{-1}(y)$

$$f_Y(y)=e^{-y}$$

$y>0$

that is a Negative Exponential distribution with mean 1

This because you can apply the fundamental transformation theorem

$$f_Y(y)=\underbrace{f_X[g^{-1}(y)]}_{=1}\cdot|\frac{d}{dy}g^{-1}(y)|$$

Title: Uniformly distributed transformation


问:
Lemma 8.2.7 Tao Analysis 1
I am trying to prove the following Lemma from Tao's Analysis 1

Let $\sum_{n=0}^{\infty}a_{n}$ be a series of real numbers which is conditionally convergent, but not absolutely convergent. Define the sets $A_{+}\colon\!=\{n\in\mathbb{N}:a_{n}\geq 0\}$ and $A_{-}\colon\!=\{n\in\mathbb{N}:a_{n} < 0\}$. Thus $A_{+}\cup A_{-}=\mathbb{N}$ and $A_{+}\cap A_{-} = \phi$. Then both the series $\sum_{n\in A_{+}}a_{n}$ and $\sum_{n\in A_{-}}a_{n}$ are not absolutely convergent.

I am able to prove that both $A_{+}$ and $A_{-}$ must be countably infinite. Then, I proceed with proof by contradiction. The case when both $\sum_{n\in A_{+}}a_{n}$ and $\sum_{n\in A_{-}}a_{n}$ are absolutely convergent is easy. However, I am not sure about the case when (WLOG) $\sum_{n\in A_{+}}a_{n}$ is absolutely convergent but $\sum_{n\in A_{-}}a_{n}$ is not. I am able to show that tail sum of the series $\sum_{n\in A_{-}}a_{n}$ will eventually be small (which, if correct, would imply that $\sum_{n\in A_{-}}a_{n}$ is absolutely convergent, a contradiction). However, I am not sure if this is the correct approach. It seems to me that one should obtain a contradiction by showing that $\sup\big\{\sum_{n\in A}a_{n}: A\subseteq A_{-}, A\text{ finite}\big\}$ is finite which is the definition of absolute convergence for sums over arbitrary sets (Definition $8.2.4$ in the book). Could someone let me know the correct approach?

答:
First note  that both A+ and A− are at most countable, because they are subsets of ℕ. I will restrict myself to both being countable. There is a bijeciton f: ℕ → A+. Likewise for A− there is g: ℕ → A−.

In the second step we prove, that if one of the subseries is conditionally convergent it is also absolutely convergent. For every an ∈ {n∈A+: an} we know that an = |an|. If Σn∈A+ an is convergent and equal to L, Then L =  Σn∈A+ an = Σn∈A+ |an|. Thus Σn∈A+ an is also absolutely convergent.
If A− is conditionally convergent with Σn∈A− an = K, then Σn∈A− |an| = 0=iΣ∞ |ag(i)| = 0Σ∞ −ag(i) =  − 0Σ∞ ag(i) = − K. So this series is also absolutely convergent.

Now we try a proof of the statement by contradiction. Assume that not both subseries need to be conditionally convergent. We have 2 main cases:
both are conditionally convergent and only one of them is conditionally convergent.

case 1:
Both are conditionally convergent. Then both are also absolutely convergent.  By 8.2.6c  also Σn∈A+∪A− an = Σn∈ℕ an  is absolutely convergent, a contradiction .

case 2:
Assume to the contrary Σn∈A+ an  is conditionally convergent, again it is also absolute convergent. Sadly the prerequisites for P8.2.6c are not there, we know only of one subset that it is absolutely convergent.

Let Σn∈A+∪A− an = M and  Σn∈A+ an = L. We make use of the bijections. I will only cover the case that both sets are countable, the cases with one of them finite are similar. Set up the bijection h : ℕ → A+ ∪ A−
h(i)   =  f(i/2)             if i is even
              g(⌊(i/2)⌋)     if i is odd

We have 0=iΣ2N+1 ah(i) = 0ΣN af(i) + 0ΣN ag(i)  and when taking the limit of N to ∞ we get
 0Σ∞ ah(i) = 0Σ∞ af(i) + 0Σ∞ ag(i).
So M = Σn∈A+∪A− an = 0Σ∞ ah(i) = 0Σ∞ af(i) + 0Σ∞ ag(i) = L + Σn∈A− an.
Reordered this gives Σn∈A− an = M − L, a finite value. So the other subseries is also conditionally convergent and we can refer back to case 1.

The case that only A− is conditionally convergent is symmetrical.

Title: Lemma 8.2.7 Tao Analysis 1


问:
Finding sheet number of torus using universal cover
I have a question from my lecture notes that I need clearing up:
Given a covering $p: S^1\times S^1 \rightarrow S^1\times S^1$ by $p(z,w)=(z^a w^b,z^c w^d), a,b,c,d\in\mathbb{Z}$, we want to find the sheet number of the covering. So we consider universal coverings of $S^1\times S^1$ to obtain a lift of $p$ from $\mathbb{R}^2$ to $\mathbb{R}^2$. We call the lift $L$ say, which always exist because $\pi(\mathbb{R}^2)=1$.
\begin{matrix}
  \mathbb{R}^2 & \rightarrow & \mathbb{R}^2 \\
  \downarrow &  & \downarrow \\
  S^1\times S^1 & \rightarrow & S^1\times S^1
 \end{matrix}
Now, after some work, we get that $L(x,y)=(ax+by+Q,cx+dy+R)$ for some integers $Q,R$. How do we find the sheet number from here? I think that it maybe the number of lattice points in the parallelogram obtained from $L$, but why?
Thanks for the help in advance.</p>

答: Think about the linear map $f\colon \mathbb R^2\to\mathbb R^2$ given by $f(x,y)=(ax+by,cx+dy)=A(x,y)$. When $a,b,c,d\in\mathbb Z$, why does $ \det A$ tell us the sheet number (counting signs) of the induced map $\mathbb R^2/\mathbb Z^2\to\mathbb R^2/\mathbb Z^2$?</p>


问:
Solve $x^{15} \equiv 7$ (mod $19$)
Using the brute force approach, I found that $5$, $16$ and $17$ satisfied the congruence. How to find all possible solutions?

答:
First, let us use Fermat's little theorem.  Since $x^{18}\equiv 1 \pmod{19}$ for any $x\not \equiv 0 \pmod{19}$, we can rewrite the equation as $x^{-3}\equiv 7\pmod{19}$, and since $7^{-1}\equiv 11\pmod{19}$, the problem is equivalent to solving

$$ x^{3}\equiv 11 \pmod{19}$$

Since this is a 3rd degree polynomial equation an $19$ is prime, there are at most $3$ solutions, all of which have been found.

However, if you only had one solution, here is a way you could generate the others.  Suppose that $\omega^{3}\equiv 1 \pmod{19}$.  If $x^{3}\equiv 11$, then $(\omega x)^{3}\equiv \omega^{3}x^{3}\equiv x^{3}\equiv 11$, and similarly with $\omega^2 x$.  Therefore, if you have one solution, finding such an $\omega$ will give you the rest.

But how do you find such an $\omega$? If you have a generator $g$, then set $\omega=g^{18/3}=g^6$.  If you don't have a generator, but you do have $h$ such that $h\neq 0, h^2\neq 1$, then either $h^6=1$, in which case we can take $\omega = h^2$, or else we can take $\omega=h^6$.

Title: Solve $x^{15} \equiv 7$ (mod $19$)


问:
prove that Petersen graph has no cycles less than or equal to 4
I studying the proof that the Petersen graph is not Hamiltonian, and in the proof, they used an observation that seems intuitively correct but I want to provide rigorous proof for it, given that I'm taking an introductory course to discrete mathematics.

Claim: Petersen graph has no cycles less than or equal to 4

I want to prove it by contradiction using the basics ideas, can you help me with that?

答:
We can use a simple argument by coloring the edges of the Petersen's graph as shown below:

First note that there are 3 types of edges:

5 blue edges forming outer cycle of length 5

5 pink edges forming inner cycle of length 5

5 green edges connecting outer nodes with inner nodes

Now, since there is a blue cycle of length 5 with 5 outer blue nodes, it makes it impossible to have a blue monochrome cycle of length $\leq 4$, because

for creating a length-4 or length-3 cycle we must connect two non-adjacent blue nodes by an additional blue edge.

in order to accommodate that additional blue edge, we must remove an exisiting blue edge from the blue cycle.

but, minimum number of blue edges needed to construct a length-5 blue cycle with 5 blue nodes is 5, with each node having one incoming and one outgoing blue edge.

hence, we can't remove an edge from the existing length-5 blue cycle, which contains 5 blue edges only.

Using same argument we can show that we can't have an inner monochrome pink cycle of length $\leq 4$.

Also, a cycle which is not monochromatic must have a green edge (connecting an inner and an outer node). Now, in order to have a non-monochrome cycle of length $\leq 4$, we must have

either a couple of green edges from a pink node, which is impossible (since there is exactly one green edge incident on each pink node),

or a couple of green edges incident on a blue node (again impossible),

or it would need at least one additional blue edge (connecting two non-adjacent blue nodes) or one additional pink edge (connecting two non-adjacent pink nodes), making it impossible to have blue or pink cycles of length 5.


问:
Can a finite group act transitively on an infinite set?
I had a question on my algebra exam, asking me to show that in case of a transitive action of a finite group $G$ on a set $X$, $|X|$ divides $|G|$. In case of a finite set, this is trivial, so I tried to see if I could come up with a possible counterexample with $X$ infinite (I failed to do so). Intuitively I feel like an infinite $X$ is "too big" for a finite group to act transitively on, but how would one show this formally?

答:
If $G$ acts transitively on $X$, then, given $x\in X$, for every $y\in X$ there is some $g\in G$ such that $y=g\cdot x$. This is precisely the condition for a surjection $f_x\colon G\longrightarrow X$ to exist. Therefore the cardinality of $G$ is at least equal to the one of $X$.

(Incidentally, since by group action's axioms distinct $x$'s induce distinct $f_x$'s, there are $|X|$ of such surjections.)


问:
Can we have a factorial of a factorial?
Can we have a factorial of a factorial?I tried searching Google but I was unable to find the answer.Please help me.</p>

答:
Yes, for example $(3!)!$ means $(3\cdot 2\cdot1)!$ which is $6!$, quite large, $6\cdot5\cdot4\cdot3\cdot2\cdot1 =720$

They get large very quickly, $(5!)!$ gives an error on a typical calculator and so does $[(3!)!]!$

(Teresa and bof are mentioning that the above should not be confused with !!, which is different and explained here https://en.wikipedia.org/wiki/Double_factorial )

Title: Can we have a factorial of a factorial?


问:
$({\overline{E}})^C=\mathring{(E^C)}$
I wanted to show that $({\overline{E}})^C=\mathring{(E^C)}$ using only definitions and well-known set operations (distributivity of intersection for example, etc) . I did it in a way that I think is very very long.
First I show three smalls properties :

$\partial E = \partial(E^C)$ (easy)

$(\partial E)^C = \mathring{E} \cup (\mathring{E^C})$ (easy with double inclusion)

$\mathring {(E^C)} \subset (\mathring{E})^C$. I struggle to show this point : I have the following : Let $x \in \mathring {(E^C)}$,$ \exists \delta_x > 0 \text{ such that } B(x,\delta_x) \subset E^C$ $\implies x \notin \partial (E^C) \iff x \notin \partial E$. Then I don't know what to do.

For the final step I show the equality using only set algebraic properties.

Hence, $({\overline{E}})^C = \mathring {(E^C)}$.

答:
An alternative view of the interior and closure: let $\mathcal{N}_x$ be the set of neighbourhoods of $x \in X$ (if in a metric space, you an just take all balls $B(x,r), r>0$ around $x$, as these form a base, which is good enough).
Then for any $E \subseteq X$:

$$x \in \mathring{E} \iff \exists N \in \mathcal{N}_x: N \subseteq E$$

$$ x \in \overline{E} \iff \forall N \in \mathcal{N}_x: N \cap E \neq \emptyset$$

$$x \in \partial E \iff \forall N \in \mathcal{N}_x: (N \cap E \neq \emptyset) \land (N \cap E^\complement \neq \emptyset)$$

Then $x \in (\overline{E})^\complement \iff \exists N \in \mathcal{N}_x: N\cap E = \emptyset$ by simple logic

and the latter is equivalent to

$\exists N \in \mathcal{N}_x: N \subseteq E^\complement \iff x \in \mathring{(E^\complement)}$ by definition.

So your statement is then quite easy to prove.

In my first course in topology the above were the definitions of interior and closure etc. (it was a course in metric topology in the first year at uni, so we used the set of balls as $\mathcal{N}_x$). In the second year (general topology I) we started with the open set axioms for a topology $\mathcal{T}$ and defined

$$\mathring{E} = \bigcup\{O \in \mathcal{T}\mid O \subseteq E\}$$

(so the maximal open subset of $E$), defined closed as being the complement of an open set and then the dual definition

$$\overline{E} = \bigcap \{C\mid X\setminus C\in \mathcal{T}, E \subseteq C\}$$

(the minimal closed superset of $E$), and $\partial E = \overline{E}\setminus \mathring{E} = \overline{E} \cap \overline{E^\complement}$, where the equivalence of these expressions follows from the statement from the question.

We then proceeded to define neighbourhoods and neighbourhood systems (alternative axioms for topology) and prove the equivalence of our first and second year definitions etc. All quite helpful IMO. It really helped us to solidify the concepts.

Title: $({\overline{E}})^C=\mathring{(E^C)}$


问:
Integrate $\frac{\log(x^2+4)}{(x^2+1)^2}$.

Using residue calculus show that
$$\int_0^{\infty}\frac{\log(x^2+4)}{(x^2+1)^2}dx=\frac{\pi}2\log 3-\frac{\pi}6.$$

I was thinking of using some keyhole or semi-circular contour here. But the problem is apart from poles at $x=-i$ and $x=i$, the logarithm has singularities when $x=\pm 4i$.

I consider $C_R$, a semicircle contour oriented clockwise with radius $R$ centered at origin. The semicircle resides in the lower half plane, so that it encloses $x=-i$ as a pole.  I set
$$\int_{C_R} \frac{\log(2-ix)}{(x^2+1)^2}dx$$
But it seems like it leads to wrong answer.

答:
One thing you could do to avoid a keyhole is to denote

$$I[a] = \int_0^\infty \frac{\log\left(a(x^2+1)+3\right)}{(x^2+1)^2}\:dx \implies I'[a] = \frac{1}{2}\int_{-\infty}^\infty \frac{1}{x^2+1}\cdot \frac{1}{ax^2+a+3}\:dx$$

then apply a semicircular contour and calculate the residues as normal.


问:
Let $f(x)$ be an invertible function and $x\in [1,5]$. Let g be the inverse of f such that $g(3)=1$ and $g(6)=5$, then find..

Let $f(x)$ be an invertible function and $x\in [1,5]$. Let g be the inverse of f such that $g(3)=1$ and $g(6)=5$, then find $\int_1^5 f(x).dx +\int_5^6 g(x).dx$

The only thing I managed to find out was $f(1)=3$ and $f(5)=6$

I don’t see how it’s possible to integrate without even know what the function is. What am I missing?

答:
Assuming that the second integral starts at $3$ one can use the following rule (A very nice proof of this rule without words is given on that page):

$$\int_{{\color{green}{1}}}^{\color{blue}{5}} f(x)\,dx +\int_{{\color{green}{3}}}^{\color{blue}{6}} g(x)\,dx =\color{blue}{5\cdot 6} - \color{green}{1\cdot 3}=27$$


问:
What is $\lim_{z \to 0} \frac{1}{z}$ in $\mathbb{C}$?
I have the limit $$\lim_{z \to 0} \frac{1}{z}$$

that I want to find for $z \in \mathbb{C}$. If for $z = x+iy$ with $x = \operatorname{Re}{z}$ and $y = \operatorname{Im}{z}$, I restrict $z$ to the positive real axis $z = x$, I obtain the limit $$\lim_{x \to 0} \frac{1}{x} = \pm \infty.$$ Since the limit approaches different values, the limit does not exist. Is this correct? This thread
Limit of $f(z)=\frac 1z$ as $z$ approaches $0$? mentions the limit to be $\infty_{\mathbb{C}}$ so I am confused.

答:
In $\mathbb{C}$, there is no distinction between $\infty$ and $- \infty$. This is because, in $\mathbb{C}$, one can approach $\infty$ in any direction. Therefore, use the convention that

$$\lim_{z \to a} f(z) = \infty \iff \lim_{z \to a} |f(z)| = \infty,$$

where the last limit is a real limit and is defined in the real sense. The concept of a point at infinity in $\mathbb{C}$ has a very elegant interpretation through Stereographic projection and the Riemann-sphere. To put it very shortly, we can define what is sometimes called the extended complex plane $\mathbb{C}_{\infty} = \mathbb{C} \cup \{ \infty \}$, which is isometric to the Riemann-sphere (read about it here: https://en.wikipedia.org/wiki/Riemann_sphere). Topoligists will sometimes refer to $\mathbb{C}_{\infty}$ as the one-point compactification of $\mathbb{C}$, meaning that it is a compact metric space differing from $\mathbb{C}$ by just one point.

Edit: Now to the limit in question. You have already observed that the limit does not exist in the sense that it does not approach a complex number. With that in mind, try to evaluate the limit using the equivalence above. We consider

$$\lim_{z \to 0} \left| \frac{1}{z} \right| = \lim_{z \to 0} \frac{1}{|z|} =  \lim_{r \to 0^+} \frac{1}{r}=\infty.$$


问:
Lagrange multipliers - missing critical points?

Given $S = \left \{ (x,y,z) : z^2-xy=1\right \}$

a. Prove that there is at least one point which is closest to $(0, 0, 0)$ and find the distance.

b. In section a you should have found several "critical points". show that those which are not a global minimum are also not a local extremum.

My problem is that I only get global minimum points in section a:

Defining $f = || \ ||^2=x^2+y^2+z^2$ and using Lagrange multipliers theorem we know that for any local extremum $p$ s.t $\nabla F(p) \neq 0$, we have $\nabla F(p) = \lambda \nabla f(p)$.

So, $\lambda (2x_0, 2y_0, 2z_0) = (-y_0, -x_0, 2z_0)$, thus $\lambda = 1$, and $x_0=y_0=0$

So we get 2 solutions $(0, 0, 1), (0, 0, -1)$.

Which are both a global minimum.

Can anyone explain what am I missing?

答:
For part (a), paxtibimarce and Math Lover have identified the pairs of critical points of the "distance-squared" function $ \ x^2  +  y^2  +  z^2 \ $ on the surface $ \ z^2 \ = \ 1  +  xy \  $ as $ \ (0 \ , \ 0 \ , \ \pm 1) \ $ and
$ \ (\pm 1 \ , \ \mp 1 \ , \ 0 ) \ . $  Here, we'll briefly discuss part (b) as to why there are two sets of critical points, only one of which is extremal.

The surface is an elliptic "hyperboloid of one sheet" oriented with the symmetry axis of its "throat" along the line $ \ y  =  x \ $ .  All of the critical points are located in the narrowest part of the throat [the first pair marked in yellow, the second in light blue].

By direct calculation, we find $ \ f(0, 0, \pm 1) \ = \ 1 \ , $ while $ \ f(\pm 1, \mp 1 , 0) \ = \ 2 \ . $  The second partial derivatives all of limited help here,
as the Hessian is simply a diagonal matrix with all three entries being $ \ 2 \ . $  So while this confirms that the points at distance $ \ 1 \ $ from the origin are at the global minimum distance, it doesn't tell us much about the pair at distance $ \ \sqrt{2} \ . $  (It is clear that there is no global maximum for distance from the origin.)

Instead, we need to examine the surface itself.  If we add the plane $ \ y \ = \ -x \ $ , the intersection with the surface is an ellipse centered on the origin, with its minor axis along the $ \ z-$ axis and its major axis along the line $ \ y \ = \ -x \ $ .  The points $ \ (\pm 1 \ , \ \mp 1 \ , \ 0 ) \   $ are the endpoints of this major axis, so their distance from the origin on this curve is maximal.  However, if we intersect our surface with the plane $ \ z \ = \ 0 \ , \ $ we obtain the hyperbola $ \ xy \ = \ -1 \ , $ for which it is a familiar fact (or easily calculated) that the points under discussion lie on the curve at the minimal distance from the origin.  Thus, the points $ \ (\pm 1 \ , \ \mp 1 \ , \ 0 ) \ $ serve as saddle points for the function on this surface.


问:
Proving prime $p$ divides $\binom{p}{k}$ for $k\in\{1,\ldots,p-1\}$
Prove if $p$ is a prime then $p \mid \binom pk$ for $k\in\{1,\ldots,p-1\}$

I don't really know where to begin with this one.

I can see that I have to use the fact that $p$ is prime somewhere - the same is not true for composite numbers, for example $4\nmid 6=\binom42$.

I have checked that this is true for the first few primes:

$3$ divides $\binom 31=\binom32=3$

$5$ divides both $\binom 51=\binom 54=5$ and $\binom 52=\binom 53=10$

$7$ divides $\binom71=\binom76=7$, $\binom72=\binom75=21$ and $\binom73=\binom74=35$.

答:
By definition, $\binom{p}{k}$ can be expanded as follows:

$\binom{p}{k}$ = $\frac{p!}{k!(p - k)!}$

Since $\binom{p}{k}$ is a positive integer number by definition, let's denote it as $n$, $n \in \mathbb N$.

$n$ = $\frac{p!}{k!(p - k)!}$

Now, move the denominator to the left side:

$n$ $\cdot$ $k!(p - k)!$ = $p!$

$p!$ is obviously divisible by $p$.

Hence, $n$ $\cdot$ $k!(p - k)!$ is also divisible by $p$.
So, at least one factor must be divisible by $p$:

$k!$ is not divisible by $p$ because $k < p$ and $p$ is prime.

$(p - k)!$ is not divisible by $p$ because $(p - k) < p$ and $p$ is prime.

Thereof, $p \mid n \Rightarrow p \mid \binom{p}{k}$