问:
Need help understanding proof to their being a prime gap $G(p_k, p_{k+1})$ with $p_k \leq n$ and $p_{k+1} - p_k > \frac{1}{2}\ln(n)$

Let $p_k ( k \geq 1)$ be an enumeration of all the positive primes with $p_1 = 2$ and $p_k < p_{k+1}$ for all $k \geq 1$. Prove that if $n$ is sufficiently large, then there is a prime gap $G(p_k, p_{k+1})$ with $p_k \leq n$ and $p_{k+1} - p_k > \frac{1}{2}\ln(n)$.

HINT: Let $m$ be the largest integer with $p_m \leq n$. Consider $\sum_{k=1}^m (p_{k+1} - p_k)$ and apply the Prime Number Theorem to $\pi(x)$.

$\pi(x)$ is the prime counting function.

I didn't have a clue how to do this so I looked in the answers and this is what it says:

If $n$ is sufficently large given $n$, we have $\pi(n) \leq \frac{5n}{4 \ln(n)}$. If $m$ is the largest integer with $p_m \leq n$, then $\pi(n) = m$. If $p_{k+1} - p_k \leq \frac{1}{2} \ln(n)$ for all $k \leq m$ then

$$n \leq p_{m+1} - 1 \leq 1 + \sum_{i = 1}^m(p_{k+1} - p_k) \leq 1 + \frac{1}{2} \ln(n) \cdot m \leq 1 + \frac{1}{2} \cdot \ln(n) \cdot \frac{5}{4} \cdot \frac{n}{\ln(n)} = 1 + \frac{5n}{8}.$$

This gives a contradiction if $\frac{3n}{8} > 1$, in paarticular, for $n \geq 3$.

Now, I kind of understand most of everything that's going on, but one this that's making it very hard is the very first line. Where does the $\frac{5}{4}$ come from? I know they've re-aranged the prime number theorem to get it in this form, but I don't get where that number comes from?

Can someone tell me?

答: If you substitute $\alpha$ for your $5/4$ and run your proof you will end up with the inequality $n\le1+\frac{\alpha}{2}n.$ Now to get a contradiction, you may take anything to satisfy $\alpha/2<1.$ So there is nothing special about $5/4$ here, except that it is $> 1$ (to satisfy the convergence in the Prime number theorem) and $\frac{5}{4}\cdot\frac{1}{2}<1.$ </p>


问:
Question about John Lee's proof of Proposition 3.13 in ISM on the identification of the tangent space of a vector space $V$ to $V$
I have a question about the proof of Proposition 3.13 from John Lee's Introduction to Smooth Manifolds.

We give $V$ the natural smooth structure given by the following Example.

Now the proposition is as follows:

It says we can use the same argument as in the proof of Proposition 3.2, to show that $D_v|_a$ is an isomorphism. I can see that the exact same argument applies to show injectivity. However, I do not understand how we can show surjectivity, as in the proof below, this requires using Taylor's theorem. However, Taylor's theorem does not apply here as $f$ is a smooth function from a vector space $V$ that may not be the Euclidean space. So how can we adapt the proof for a general finite-dimensional vector space $V$ here?

答:
Maybe this question is a bit old but I've managed to find a way.

I'll use the same notations as in the OP.
Let $(V,E^{-1})$ be the chart and $E^{-1} = (x^1, \ldots ,x^n)$ the coordinate maps. Your choice of the vector $v \in V$ is the right one, indeed take $v = w(x^i) \, E_i$ where $(E_1, \ldots ,E_n)$ is the basis associated with $E$ and $w \in T_aV$ is given.

Now, for any $f \in C^\infty(V)$ holds $ w(f) = (d(E^{-1})_a (w))(f \circ E)$ because $E \in C^\infty$ and we applied the definition of differential. You can now use Taylor's theorem, as in the proposition 3.2 of Lee's book, on $f \circ E \in  C^\infty(\mathbb{R}^n)$ at the point $E^{-1}(a)$ to get

$$ (f \circ E)(x) = f(a) + \frac{\partial (f \circ E)}{\partial e^i}|_{E^{-1}(a)} \, (e^i(x)-x^i(a)) \, + \sum_{i,j}... $$

where $e^i$ are the coordinate maps of the standard basis of $\mathbb{R}^n$ (to not get confused with $x^i$).

So we have
\begin{align}
w(f) &= \frac{\partial (f \circ E)}{\partial e^i}|_{E^{-1}(a)} \, (d(E^{-1})_a (w))(e^i) = \frac{\partial (f \circ E)}{\partial e^i}|_{E^{-1}(a)} \, w(x^i) =\\
&= \frac{\partial (f \circ E)}{\partial e^i}|_{E^{-1}(a)} \, (E^{-1}(v))^i = \frac{d}{dt}|_{t=0} (f \circ E)(E^{-1}(a)+tE^{-1}(v)) = \\
&= \frac{d}{dt}|_{t=0} f(a + tv) = D_v|_a f
\end{align}

Notice that we exploited the linearity of $E^{-1}$ to use the chain rule for maps $\mathbb{R} \rightarrow \mathbb{R}^n \rightarrow \mathbb{R}$, after we expressed $f = (f \circ E) \circ E^{-1}$.

ContentLicense: CC BY-SA 4.0


问:
finding the coordinate that lies in the vector and through it
I'm trying to find the coordinates of A, where the line $l_1$ is given by this equation:

$$\begin{pmatrix}8 \\ 1 \\ -3 \end{pmatrix} +\mathbf{\mu} \begin{pmatrix}-5 \\ 4 \\ 3 \end{pmatrix} $$

Where $\mu$ is a scalar parameter = 1. The point A lies on the equation provided above.

Find the coordinates of A.

My approach:

Given $\mu$ = 1; Then is A = $8  -5\mu=3?$

Furthermore, The point P has position vector:

$$\begin{pmatrix}  1  \\ 5 \\ 2 \end{pmatrix} $$

The line $l_2$ passes through the point P and is parallel to the line $l_1$, whats the vector equation for line $l_2$?

Do I have to do this?

$$\begin{pmatrix}8 \\ 1 \\ -3 \end{pmatrix} +\mathbf{\mu} \begin{pmatrix}1 \\ 5 \\ 2 \end{pmatrix} $$

Given that $\mu$ = 1; then

$$\begin{pmatrix}  10 \\ 11 \\ 1 \end{pmatrix} \implies \begin{pmatrix}10 \\ 11 \\ 1 \end{pmatrix} +\mathbf{\mu} \begin{pmatrix}1 \\ 5 \\ 2 \end{pmatrix} $$

Then If I'm to find the exact distance AP, giving an answer in the form $k\sqrt{2}$, where is to be determined how to I proceed?

And if the acute angle between AP and $l_2$ is $\theta$, how to find $\cos\theta?$

答:
It looks like you're struggling a bit with the concept of a line expressed through vectors. I think understanding the equation for the line will help answer your questions.

Take a line in two dimensions, say $ \begin{pmatrix} 0 \\ 0 \end{pmatrix} + \mu *\begin{pmatrix} -1\\ 1 \end{pmatrix}$. Then this tells you that starting from the point $\begin{pmatrix} 0 \\ 0 \end{pmatrix}$, all points in the space that can be reached through the vector $\mu *\begin{pmatrix} -1\\ 1 \end{pmatrix}$ are a part of the line.

The rule for vector addition is very simple, you just add the coordinates. I.e. the points on this line are $\begin{pmatrix} 0 \\ 0 \end{pmatrix} + 1 *\begin{pmatrix} -1\\ 1 \end{pmatrix} =\begin{pmatrix} 0+\mu *(-1) \\ 0 +\mu *1\end{pmatrix}=\begin{pmatrix} -\mu \\ \mu \end{pmatrix}$ for all values of $\mu$.

For instance, let point A be that point on the line where $\mu =1$. Then $A=\begin{pmatrix} 0 \\ 0 \end{pmatrix} + 1 *\begin{pmatrix} -1\\ 1 \end{pmatrix} =\begin{pmatrix} -1 \\ 1\end{pmatrix}$

Now, the same holds for your problem. Your A is a point in a three-dimensional space, and what you've done is calculate just the first coordinate of A. You want to do the same for the other two coordinates as well, so $A=\begin{pmatrix} 3\\ x_2 \\ x_3 \end{pmatrix}$.

If you understand how the line works, then you also know how to create a line that is parallel to l1 and goes through a different point. What makes the line parallel is the direction vector, which is $\begin{pmatrix} -5\\ 4 \\ 3 \end{pmatrix}$ in your case. So, the simplest solution for l2 is $P+\mu *\begin{pmatrix} -5\\ 4 \\ 3 \end{pmatrix}$.

Last part of this answer: Think about how you calculate a "distance" in one dimension. Say, the distance between the numbers 3 and 1. That's 3-1=2. Same goes in your vector space: the distance between $\begin{pmatrix} x_1\\ x_2 \\ x_3 \end{pmatrix}$ and $\begin{pmatrix} y_1\\ y_2 \\ y_3 \end{pmatrix}$ is the length of the vector $\begin{pmatrix} x_1\\ x_2 \\ x_3 \end{pmatrix} - \begin{pmatrix} y_1\\ y_2 \\ y_3 \end{pmatrix} = \begin{pmatrix} x_1 - y_1\\ x_2 - y_2 \\ x_3 - y_3 \end{pmatrix} $. You can look up easily how to calculate the length of a vector.

Hope this helps!


问:
When is $a^2 \equiv b^2 \pmod n \implies a\equiv b \pmod n$
I know that: $$a\equiv b \pmod n \implies a^2 \equiv b^2 \pmod n $$ But this is not an equivalence relation so we cannot say that if:
$$a^2 \equiv b^2 \pmod n \implies a\equiv b \pmod n $$

But are there some restrictions that makes this proposition true?

答:
We can say
$$
a^2 \equiv b^2 \mod n \implies (a - b)(a + b) \equiv 0 \mod n
$$
Bezout says $\exists s, t \in \mathbb{Z}$
$$
s(a + b) + tn = \gcd (a + b, n)
$$
Which means
$$
s(a + b) \equiv \gcd(a + b, n) \mod n
$$
So that
$$
(a - b) \gcd(a + b, n) \equiv 0\mod n
$$
And I don't know if we can say anything more than this.

Of course if $\gcd(a + b, n) \equiv 1 \mod n$, then we can say
$$
a \equiv b \mod n
$$
Otherwise, I'm not sure we can say much.


问:
Notation for the set of symmetric matrices and symmetric positive definite matrices
I would like to know if there exists a notation for the set of symmetric matrices and symmetric positive definite matrices. For instance, the set of $N \times N$ matrices with real entries is denoted as $\mathbb{R}^{N \times N}$.</p>

答: From Stephen Boyd & Lieven Vandenberghe's Convex Optimization:

ContentLicense: CC BY-SA 3.0


问:
General approach to construct a basis for a vector space
I know there are already threads, which discuss the construction of a basis for a specific vector space. But in general, if I am given a vector space and am asked to construct a basis for that vector space, how do I proceed?

For example, I am asked to construct a basis for \begin{pmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4} \end{pmatrix}

with $x_{1}, x_{2}, x_{3}, x_{4} \in \Bbb R$ . The problem I am working on already gives me two vectors of a basis. These are:
$$\begin{pmatrix}2\\3\\-1\\5\end{pmatrix}  and  \begin{pmatrix}1\\3\\-1\\5\end{pmatrix}$$ So how do I approach finding two other vectors ( I know the basis consists of 4 elements since I already have seen the solution) that would complement the two vectors that are already given, so that all these vectors form a basis? I know how to check for linear independency. But how do I find vectors so that the span of these vectors equals $\Bbb R_{4}$ (all vectors with 4 rows and 1 column with components $\in \Bbb R$.

答:
Look  the following points:

a set containing $ n $ independent  vectors in a $ n$ dimensional vector space is a basis of that vector space.

Now for your problem  $dim( R^4)=4$   so you are to construct a linearly independent set of vectors containing 4 vectors  ( two vectors  are given , two are to be determined), then use my result 1.

Now consider the third vector $(0,0,1,0)$

Look at the  combination

$ (0,0,1,0)=a ( 2,3,-1,5) + b (1, 3,-1,5) $

and equating the coefficient you would get  $ a=0, b=0 $

Which shows that  {(2,3,-1,5),(1,3,-1,5),(0,0,1,0)}
is linear independent.

Now consider the fourth vector  $(0,0,0,1)$
Look at the combination

$(0,0,01)= a( 2,3,-1,5) +b (1,3,-1,5)+c(0,0,1,0)$

Equating both sides and solving you will get $ a=0,b=0,c=0$

Which implies   {(2,3,-1,5),(1,3,-1,5),(0,0,1,0),(0,0,0,1)} is linearly independent.

So we get the desired  independent set containing 4 vectors and  using result we get the desired basis.
Since  the chosing such vectors is not unique, so such vector vary from person to person.


问:
Inequality containing logarithms
I've been reading an algorithm book lately.
There is this one question that I've been struggling with and can't solve it:

Consider a computer that can do 10^10 operations per second.
With an algorithm that has a time complexity of $n *log(n)$.
What is the largest input size $n$ that this computer can complete within an hour?

Perhaps this question is pretty simple, but none of the online calculators that I know of (things like symbolab.com) have been able to solve it.

Any help would be greatly appreciated.

答: You need to solve $n \log(n) = x$ which you can do either iteratively or calling some library functions like ProductLog (see https://mathworld.wolfram.com/LambertW-Function.html). It is not possible to solve it analytically

Title: Inequality containing logarithms


Title: Inequality containing logarithms

Body:
I've been reading an algorithm book lately.
There is this one question that I've been struggling with and can't solve it:

Consider a computer that can do 10^10 operations per second.
With an algorithm that has a time complexity of $n *log(n)$.
What is the largest input size $n$ that this computer can complete within an hour?

Perhaps this question is pretty simple, but none of the online calculators that I know of (things like symbolab.com) have been able to solve it.

Any help would be greatly appreciated.

Body:
Hint

$\log (n) = [\log (10) \times \log_{10} (n)] \approx
[(2.30) \times \log_{10} (n)].$

Thus, for $n = 10^{k}$,

$n \log(n) \approx [(2.30) \times k \times 10^k].$


答:
Two things:

If $(-a)^2 = a^2$ so $(-a)^2 \equiv a^2 \pmod n$ so unless $a\equiv -a \pmod n$ is always true this can't ever be true.

And if $n$ isn't prime and $n = jk$ then $a^2 \equiv b^2 \mod n$ will imply $a^2 -b^2=(a+b)(a-b)\equiv 0 \pmod n$.  We could have $a+b \equiv j\pmod n$ and $a-b\equiv k\pmod n$ and that would be true.

......

To tie this all together.

1a)  If $n = 1$ then $a \equiv 0 \pmod 1$. Always.  And $a^2 \equiv b^2\pmod 1 \implies a\equiv b \equiv 0 \pmod 1$.  But this is trivial and silly.

1b)  If $n=2$ then $a \equiv -a\pmod 2$. ($1 \equiv -1\pmod 2$ and $0\equiv 0$) and $a^2 \equiv b^2 \pmod 2 \implies a \equiv b\pmod 2$.  But this is the last case this is true.

If $n > 2$ then there will be some $a$ where $a \not \equiv -a \pmod n$ so $a^2 \equiv (-a)^2 \pmod n$ but $a \not \equiv -a \pmod n$ so this will never be true for $n > 2$

BUT  $a^2 \equiv b^2 \pmod n \implies a \equiv \pm b \pmod n$ if $n$ is prime.

If $n$ is prime then

$a^2 \equiv b^2 \pmod n \implies (a-b)(a+b)\equiv 0 \pmod p$.

So that means $p|(a-b)(a+b)$.  But if $p$ is prime then by Euclid's lemma we either have $p|a-b$ so $a\equiv b \pmod n$.  Or we have $p|a+b$ and $a \equiv -b\pmod n$.

If $n$ is not prime.... well it could still be true that $a^2 \equiv b^2 \implies a\equiv \pm b$.  For example $n=6$ we have $0^2,1^2, 2^2, 3^2, 4^2, 5^2\equiv 0,1,4,3,4,5$ and the result is true.

But it doesn't have to be.  If $a^2 \equiv b^2 \pmod n$ we well have $n|(a+b)(a-b)$.

If $n$ is not prime and it is possible that $d|n$ and $d|a+b$ where $\frac nd|(a-b)$ we can have a contradiction.

One issue is that $(a+b),(a-b)$ are either both even or both odd so $n = 2p$ where $p$ is prime will be impossible.  But otherwise it will be.

So .....

If $n\le 2$ then $a^2 \equiv b^2 \mod n \implies a\equiv b \pmod n$.

If $n > 2$ and $n$ is prime or $n = 2p; p$ prime then $a^2 \equiv b^2 \mod n\implies a \equiv \pm b \pmod n$.

Otherwise there are exceptions.

Title: When is $a^2 \equiv b^2 \pmod n \implies a\equiv b \pmod n$

Body:
I know that: $$a\equiv b \pmod n \implies a^2 \equiv b^2 \pmod n $$ But this is not an equivalence relation so we cannot say that if:
$$a^2 \equiv b^2 \pmod n \implies a\equiv b \pmod n $$

But are there some restrictions that makes this proposition true?


问:
A power series equality involving binomial coefficients
I believe the following identity holds for any non-negative integers $m,k$:

$$\binom{m+k}{k}^2 = \sum_{n=0}^m \binom{k}{m-n}^2\binom{2k+n}{n}.$$

It seems like there should be a slick double counting way of proving it, but I am not able to see it. I got this identity because I believe the following power series equality holds for $|x| < 1$,

$$(1-x)^{2k-1} \sum_{m=0}^\infty \binom{m+k-1}{k-1}^2 x^m = \sum_{j=0}^{k-1} \binom{k-1}{j}^2 x^j. $$

The binomial identity above comes from dividing by $(1-x)^{2k-1}$, applying the binomial theorem replacing $k$ with $k+1$ and comparing the coefficients of both sides.

But this power series equality doesn't seem any easier to prove than the binomial coefficient identity, since I don't really have a handle on the sums on either side.

Does anyone know a reference for an identity of this nature or have an idea for a double counting argument?

答:
Permit me to contribute an algebraic proof. We seek

$${m+k\choose k}^2 =
\sum_{q=0}^m {k\choose m-q}^2 {2k+q\choose q}$$

Starting with the RHS we find

$$\sum_{q=0}^m {k\choose q}^2 {2k+m-q\choose m-q}
\\ = \sum_{q=0}^m {k\choose q}
[z^k] z^q (1+z)^k [w^{m}] w^q (1+w)^{2k+m-q}.$$

Now we may extend $q$ to infinity because the coefficient extractor
$[w^m]$ enforces the upper limit. We get

$$[z^k] (1+z)^k [w^m] (1+w)^{2k+m}
\sum_{q\ge 0} {k\choose q} z^q
w^q (1+w)^{-q}
\\ = [z^k] (1+z)^k [w^m] (1+w)^{2k+m}
(1 + zw/(1+w))^k
\\ = [z^k] (1+z)^k [w^m] (1+w)^{k+m}
(1 + w + zw)^k$$

Re-expanding we find

$$[z^k] (1+z)^k [w^m] (1+w)^{k+m}
\sum_{q=0}^k {k\choose q} w^q (1+z)^q.$$

We may set the upper limit of the sum to $m.$ (If $k\lt m$ the values
$k\lt q\le m$ produce zero from the binomial coefficient and we may
raise $q$ to $m.$ If $k\gt m$  the values $m\lt q\le k$ produce zero by
the coefficient extractor  $[w^m]$ and we may lower $q$ to $m.$) We get

$$[z^k] (1+z)^k [w^m] (1+w)^{k+m}
\sum_{q=0}^m {k\choose q} w^q (1+z)^q
\\ = \sum_{q=0}^m {k\choose q} {k+q\choose k} {k+m\choose m-q}.$$

Now observe that

$${k+q\choose k} {k+m\choose m-q}
= \frac{(k+m)!}{k! \times q! \times (m-q)!}
= {m+k\choose k} {m\choose m-q}.$$

This yields for our sum

$${m+k\choose k} \sum_{q=0}^m {k\choose q} {m\choose m-q}.$$

Using Vandermonde we obtain at last

$$\bbox[5px,border:2px solid #00A000]{
{m+k\choose k}^2.}$$


问:
Flawed proof by induction that $a^{n-1}=1$
Donald Knuth "The Art of Computer Programming" Section 1.2.1 has an exercise No.2, the solution to which I don't understand. Exercise:

There must be something wrong with this proof. What is it? Theorem. Let $a$ be any positive number. For all positive integers $n$ we have $a^{n-1}=1$. Proof. If  $n = 1, a^{n-1}=a^{1-1}=a^0=1$. And by induction, assuming that the theorem is true for $1, 2, ..., n$, we have
$a^{(n+1)-1}=a^n=\frac{a^{n-1}\times a^{n-1}}{a^{(n-1)-1}}=\frac{1\times1}{1}=1$; so the theorem is true for $n+1$ as well.

As far as I understand the proof by induction, the solution should be approximately following:
$a^{1-1}=1$
Assume that $a^{n-1}=1$;
Therefore $a^{(n+1)-1}=1\times a$;
So $a^{(n+1)-1}$ does not equal $a^{n-1}$ and does not equal $1$, unless $a=1$

The solution to exercise reads:

The theorem has not been proved for $n=2$; in the second part of the proof, take $n=1$; we assume there that $a^{-1}=1$. If this condition is true (so that $a=1$), the theorem is indeed valid.

I want to compare it to another, correct, proof by induction in the example from here:

Adding up Odd Numbers
$1+3+5+...+(2n−1)=n^2$

Show it is true for $n=1$

$1=1^2$ is True

Assume it is true for $n=k$

$1+3+5+...+(2k−1) = k^2$ is True
(An assumption!)

Now, prove it is true for $(k+1)$

$1 + 3 + 5 + ... + (2k−1) + (2(k+1)−1) = (k+1)^2$
We know that $1 + 3 + 5 + ... + (2k−1) = k^2$ (the assumption above),
so we can do a replacement for all but the last term:
$k^2 + (2(k+1)−1) = (k+1)^2$
Now expand all terms:
$k^2 + 2k + 2 − 1 = k^2 + 2k+1$
And simplify:
$k^2 + 2k + 1 = k^2 + 2k + 1$
They are the same! So it is true.
So: $1 + 3 + 5 + ... + (2(k+1)−1) = (k+1)^2$ is True

Returning to the first example, I didn't quite understand the meaning of the phrase

The theorem has not been proved for $n=2$;

In the second (correct) proof there is no proving that it holds for $n=2$, despite covering the same topic. And Knuth himself has the proof of the theorem $1+3+5+...+(2k−1) = k^2$ similar to the one above, where he too doesn't prove it for $k=2$. So what did he mean by this phrase? Does he mean that there already is an exception in the rule stated when we take arbitrary $n$, for example $n=2$, so therefore it wouldn't hold for all examples anyway? But taking arbitrary numbers and putting it into equation instead of $n$ doesn't seems to be an effective and systematic way to debunk such proofs. How would you explain what is wrong with the exercise?

答:
To understand what's happening in the first example, it suffices to substitute actual values for $n$.  We clearly have no argument with $a^0 = 1$, which corresponds to the case $n = 1$.  But when $n = 2$, the inductive reasoning step looks like this:

$$a^{2-1} = \frac{a^{1-1} a^{1-1}}{a^{0-1}} = \frac{a^0 a^0}{a^{-1}}.$$  So the next step fails immediately because the base case begins at $n = 1$, not $n = 0$; that is to say, $a^{-1} \ne 1$ as required for the inductive step to remain true.

The reason why the second example does not suffer from this defect is because the inductive step does not rely on an incorrect assumption about an earlier case being true.  You don't have this problem in your example because all of the algebraic steps are valid and supported by the induction hypothesis.  In the first example, this is not so, because the very next case is assuming that $a^{-1} = 1$.

Indeed, if $a = 1$, thereby satisfying this assumption, then it is true that $a^{n-1} = 1$ for all $n$.  The way in which the inductive step fails actually illustrates precisely when the theorem is true.


问:
Equivalences of bounded sets
$\newcommand{\diam}{\operatorname{Diam}}$
$\newcommand{\R}{\mathbb{R}}$

Definition: Let $X$ be a metric space, $M \subseteq X$ is bounded if $\diam(M) < \infty$

Let $(X,d)$, $X \not=  \emptyset$, be a metric space and $M \subseteq X$. Show that the following are equivalent:

(i) $M$ is bounded.

(ii) $\exists_{x \in X}\left\{\exists_{r > 0}\left[ M \subseteq B_{r}(x)\right]
\right\}$

(iii) $\forall_{x \in X}\left\{\exists_{r > 0}\left[ M \subseteq B_{r}(x)\right]\right\}$

Attempt (following suggestions from you):

Suppose $M$ is bounded. We want to see that $\exists_{x \in X}\left\{\exists_{r > 0}\left[ M \subseteq B_{r}(x)\right]\right\}$.

(iii) $\rightarrow$ (ii)

Suppose that $\forall_{x \in X}\left\{\exists_{r > 0}\left[ M \subseteq B_{r}(x)\right]\right\}$. We want to see that $\exists_{x \in X}\left\{\exists_{r > 0}\left[ M \subseteq B_{r}(x)\right]\right\}$.

By hypothesis it is true for any $x \in X$. Therefore, in particular, it is true for some $x \in X$. Hence, $\exists_{x \in X}\left\{\exists_{r > 0}\left[ M \subseteq B_{r}(x)\right]\right\}$.

(ii) $\rightarrow$ (i)

Suppose that $\exists_{x \in X}\left\{\exists_{r > 0}\left[ M \subseteq B_{r}(x)\right]\right\}$. We want to see that $M$ is bounded. Suppose by contradiction that $M$ is not bounded.

By hypothesis above, we have that $\diam(M) = \infty$. So there exists $m_{1},m_{2} \in M$ such that they are arbitrary far. It is a contradiction! Because the ball $B_{r}(x)$, which has a finite radius $r$, can not contain its points $m_{1}$ and $m_{2}$ but they are inside $M$.

How can I deduce (i) $\rightarrow$ (iii)?

答:
Your treatment of the case $M=\varnothing$ is unclear: at one point you say ‘Notice that $M\ne\varnothing$’, which is simply false. And while the last sentence of that paragraph is correct, you really should justify it better. The whole paragraph could be reduced to this:

If $M=\varnothing$, let $x\in X$; then $M\subseteq B_1(x)$, so (ii) is satisfied.

In your treatment of the case $M\ne\varnothing$ you define $\theta=\inf(M)$. This makes no sense: $X$ need not be any kind of ordered set, so the notion of infimum need not even apply to subsets of $X$. You can accomplish what you were trying to do simply by letting $x\in M$ be any point. Then $d(x,y)\le c$ for all $y\in M$, so $M\subseteq B_{c+1}(x)$.

Let me say something about the rest of the theorem. Notice that (iii) immediately implies (ii), so an efficient way to prove that (i), (ii), and (iii) are equivalent is to prove that (i) implies (iii) and that (ii) implies (i). Proving that (i) implies (iii) is almost as easy as proving that it implies (ii). There is no change in the argument if $M=\varnothing$. If $M\ne\varnothing$, let $x\in X$, and let $r=\inf\{d(x,y):y\in M\}$, the distance from $x$ to $M$.

Show that $d(x,y)\le r+c$ for each $y\in M$, and conclude that $M\subseteq B_{r+c+1}(x)$.

After that it remains only to show that (ii) implies (i).

Show that if $M\subseteq B_r(x)$, then $\operatorname{diam}(M)\le 2r$.

Title: Equivalences of bounded sets


问:
Prove that the zeros of the polynomial $P(z)=z^7+7z^4+4z+1$ lie inside the disk of radius $2$ centered at the origin.

Prove that the zeros of the polynomial $P(z)=z^7+7z^4+4z+1$ lie inside the disk of radius $2$ centered at the origin.

Assuming the contrary that there exists an $|z|\geqslant2$ such that $P(z)=0$ I have that $$z^7+7z^4+4z+1 = 0 \implies 1+\frac{7}{z^3}+\frac{4}{z^6} +\frac{1}{z^7}=0 \implies 1+\frac{7}{8}+\frac{4}{64}+\frac{1}{128} = \frac{249}{128}\ne0$$

which is a contradiction so the zeros would lie inside the disk of radius $2$?

Body:
If there is a solution, $|z^7|=|7z^4+4Z+1|$ Suppose that $|z|\geq 2$,

$2^7=128>7 \times 2^4 +4 \times  2+1=127$


问:
Conditional Mutual Information - Cover exercise
Following is the exercise from T. Cover's Elements of Information Theory

While I understand that $$I(X_{n-1}; X_n \mid X_1, \ldots, X_{n-2}) = H(X_{n-1} \mid X_1, \ldots, X_{n-2}) - H(X_{n} \mid X_1, \ldots, X_{n-2}, X_{n-1}), $$
$$I(X_{n-1}; X_n \mid X_1, \ldots, X_{n-2}) = H(X_{n-1}).$$
But I don't understand how $H(X_{n-1})$ = 1 bit (meaning how do we know that $X \sim \text{Ber}(1/2)$).

答:
Your first equality is wrong, it should be

$$I(X_{n-1}; X_n \mid X_1, \ldots, X_{n-2}) = H(X_{n} \mid X_1, \ldots, X_{n-2}) - H(X_{n} \mid X_1, \ldots, X_{n-2}, X_{n-1})$$

The last term is zero because, given $X_1, \cdots X_{n-1}$ is known.

For the other term:

Given $X_1,…,X_{n−2}$ there are two alternatives for the last two bits.

If $X_1,…,X_{n−2}$ has even number of $1$'s, then either $(X_{n-1},X_n)=(0,0)$ or $(X_{n-1},X_n)=(1,1)$ with equal probability.

If $X_1,…,X_{n−2}$ has odd number of $1$'s, then either $(X_{n-1},X_n)=(1,0)$ or $(X_{n-1},X_n)=(0,1)$ with equal probability.

Hence $H(X_{n} \mid X_1, \ldots, X_{n-2}) =1$ bit


问:
Are the partial derivatives of the real and imaginary part of a holomorphic function always continuous?
I'm quite confused by my course notes for Complex Analysis. On the one hand, it says that a complex function $f:\mathbb{C} \to \mathbb{C}$ is differentiable at a point $z_0$ iff the corresponding vector-function $F:\mathbb{R^2} \to \mathbb{R^2}$ is differentiable and the Cauchy-Riemann equations are satisfied. Now differentiability of a multivariate real function is equivalent to the continuity of partial derivatives, so this would imply for $f(z) = u(x,y)+iv(x,y)$ that $u(x,y)$ and $v(x,y)$ have continuous partial derivatives.

On the other hand, the proof that is given for the Cauchy-Goursat theorem is an extremely long, convoluted one, and at the end of it, it mentions that a much more elegant and short proof is possible if one assumes that $u(x,y)$ and $v(x,y)$ have continuous partial derivatives (it makes use of Green's theorem). Why is this an 'assumption'? Doesn't it necessarily follow from the assumption of $f$ differentiable in the theorem statement?

答:
No, differentiability of a multivariate real function is not equivalent to the continuity of partial derivatives. For instance, if$$f(x,y)=\begin{cases}x^2\sin\left(\frac1x\right)&\text{ if }x\ne 0\\0&\text{ if }x=0,\end{cases}$$then $f$ is differentiable at $(0,0)$, but $\frac{\partial f}{\partial x}$ is dicontinuous there.

However, it is true that the continuity of partial derivatives implies differentiability.


答:
For $\vert z \vert =2$ you have

$$\vert 7z^4 +4z+1\vert \le 121 \lt 128 = \vert z^7 \vert$$
Hence according to Rouché’s theorem $P$ has seven zeros in the disk of radius equal to $2$.

Body:
Prove that the zeros of the polynomial $P(z)=z^7+7z^4+4z+1$ lie inside the disk of radius $2$ centered at the origin.

Assuming the contrary that there exists an $|z|\geqslant2$ such that $P(z)=0$ I have that $$z^7+7z^4+4z+1 = 0 \implies 1+\frac{7}{z^3}+\frac{4}{z^6} +\frac{1}{z^7}=0 \implies 1+\frac{7}{8}+\frac{4}{64}+\frac{1}{128} = \frac{249}{128}\ne0$$

which is a contradiction so the zeros would lie inside the disk of radius $2$?


问:
Sketch the phase line of ODE: $\frac{dx}{dt}=f(t,x)$
I was asked to sketch the phase line of scalar ODE, and using the phase lines, determine whether all equilibrium point(s) are sinks, sources, or nodes. Here is the scalar ODE:
$$\frac{dx}{dt}=f(t,x)$$ where

$f(t,x):=\frac{2-π}{π}$, if $x\ge \frac{π}{2}$

$f(t,x):=\frac{sinx}{x}-1$, if $0\lt x\lt \frac{π}{2}$

$f(t,x):=xe^x$, if $x\le 0$

For $x\le 0$, I calculated $f(t,x)=0$ and got $x=0$ as the equilibrium point.

For $0\lt x\lt \frac{π}{2}$, I have no idea how to calculate and proceed on $f(t,x)=0$ as I would result in $sinx=x$.

For $x\ge \frac{π}{2}$, I don't think there exists any equilibrium points.

I do know how to distinguish between sinks, sources, and nodes, but I do not have a clear picture on way to determine all the equilibrium points. Thank you.

答:
Here is the stream plot of your function:

In Mathematica:

```<code>f[t_, x_] :=
  Which[x <= 0, x Exp[x], 0 < x <= \[Pi]/2, Sin[x]/x - 1,
   x > \[Pi]/2, (2 - \[Pi])/\[Pi]];
StreamPlot[
 Evaluate[D[
   f[t, x], {{t, x}}]], {x, -\[Pi], \[Pi]}, {t, -\[Pi], \[Pi]},
 StreamScale -> Medium]
</code>```

Body:
Here is the stream plot of your function:

In Mathematica:

f[t_, x_] :=
  Which[x <= 0, x Exp[x], 0 < x <= \[Pi]/2, Sin[x]/x - 1,
   x > \[Pi]/2, (2 - \[Pi])/\[Pi]];
StreamPlot[
 Evaluate[D[
   f[t, x], {{t, x}}]], {x, -\[Pi], \[Pi]}, {t, -\[Pi], \[Pi]},
 StreamScale -> Medium]


问:
Closure of infinite set in finite complement topology
I just asked a question to know if a proof using "or" to make a justification was valid logically.

Here is an example of my confusion.

Show that if $A$ is an infinite subset of $X$, then $\bar A=X$ in the  finite complement topology.

Here was my proof. Let $x \in X, U$ be an open set containing $x$.If $U=X$, then $U \cap A \neq \varnothing$. Suppose $U \neq X$. Then $U=X-S$ for some finite set $S$. Consider $A \cap (X-S)$. Then this intersection is nonempty (here is where I am unsure if this is correct) otherwise $A$ or $X-S$ would have to be finite.(A would be finite since $A \subset S$ if the intersection is empty, however I noticed the second statement is not necessarily true). So $U \cap A \neq \varnothing$ for each open set $U$, and since $x$ was arbitrary, $\bar A=X$.

So the statement "otherwise,..." is my justification for the contradiction that will arise if $(X-S) \cap A = \varnothing$. So I want to know if including the not necessarily true statement $X-S$ would have to be finite, ruins the validity of my proof.

答:
Yes, the assertion that

if $A\cap(X\setminus S)=\varnothing$, then either $A$ is finite, or $X\setminus S$ is finite

is technically true, because in fact one can conclude that $A$ must be finite, but in ordinary mathematical writing it will be understood to mean that there are two ways in which $A\cap(X\setminus S)=\varnothing$ can be true, $A$ finite or $X\setminus S$ finite, that both can occur, and that they exhaust the possibilities. But the fact that $A$ is infinite does not imply that its complement is finite, so the fact that $A$ and $X\setminus S$ are disjoint doesn’t mean that at least one must be finite.

Fortunately, the argument can be fixed quite easily. Let $x\in X$, and let $U$ be an open nbhd of $x$. Then $U=X\setminus S$ for some finite $S\subseteq X$, so

$$A\cap U=A\cap(X\setminus S)=(A\cap X)\setminus S=A\setminus S\ne\varnothing\,,$$

since $A$ is infinite.

Note that there is no need to consider the case $U=X$ separately: that’s simply $U=X\setminus S$ for $S=\varnothing$.


问:
Is there a possibility to choose fairly from three items when every choice can only have 2 options
Me and my wife are often not knowing which DVD to watch. If we have two options we have a simple solution, I put one DVD in one hand behind my back and the other DVD in the other hand. She will randomly choose a hand and the DVD I have in that hand will be the one to watch.

This procedure is easy to expand to any power of 2. If we have 4 DVD's I hold 2 in one hand, 2 in the other. When a pair of DVD's is chosen, I split them out to two hands and she choses again.

The question is, what can we do when we have  3 DVD's. The assumptions we make are:

I am not impartial. If I can influence the result somehow I will try to do that

My wife really choses randomly a side every-time, independent of what she chose earlier.

I don't have any other place to hide the DVD's, so every DVD is either visible or in one of the two hands.

As requirement we have that it must be a procedure with a predetermined number of steps. Not more, not less. If this is not possible, a solution that guarantees to finish with an upperbound, this is a good second choice. Off course the DVD we choose must be really random!

Body:
Here's a table of n 2^n m 3^m and p(fail) where 3^m is the largest value < 2^m. The first row represents the standard algorithm.

If you care about the near-worst-case performance, the n=65 row suggests an interesting possibility: take a random 65-bit number, reject if if it's too large (1.14% of cases), then use repeated div/mod to get 41 1-in-3 choices. This algorithm is 99% efficient with the input entropy vs. 59% for the standard algorithm.

On 64-bit hardware you can have two cases depending on the 65'th bit being 0 or 1.

2 4 1 3 25.00
3 8 1 3 62.50
4 16 2 9 43.75
5 32 3 27 15.62
6 64 3 27 57.81
7 128 4 81 36.72
8 256 5 243 5.08
9 512 5 243 52.54
10 1024 6 729 28.81
11 2048 6 729 64.40
12 4096 7 2187 46.61
13 8192 8 6561 19.91
14 16384 8 6561 59.95
15 32768 9 19683 39.93
16 65536 10 59049 9.90
17 131072 10 59049 54.95
18 262144 11 177147 32.42
19 524288 11 177147 66.21
20 1048576 12 531441 49.32
21 2097152 13 1594323 23.98
22 4194304 13 1594323 61.99
23 8388608 14 4782969 42.98
24 16777216 15 14348907 14.47
25 33554432 15 14348907 57.24
26 67108864 16 43046721 35.86
27 134217728 17 129140163 3.78
28 268435456 17 129140163 51.89
29 536870912 18 387420489 27.84
30 1073741824 18 387420489 63.92
31 2147483648 19 1162261467 45.88
32 4294967296 20 3486784401 18.82
33 8589934592 20 3486784401 59.41
34 17179869184 21 10460353203 39.11
35 34359738368 22 31381059609 8.67
36 68719476736 22 31381059609 54.33
37 137438953472 23 94143178827 31.50
38 274877906944 23 94143178827 65.75
39 549755813888 24 282429536481 48.63
40 1099511627776 25 847288609443 22.94
41 2199023255552 25 847288609443 61.47
42 4398046511104 26 2541865828329 42.20
43 8796093022208 27 7625597484987 13.31
44 17592186044416 27 7625597484987 56.65
45 35184372088832 28 22876792454961 34.98
46 70368744177664 29 68630377364883 2.47
47 140737488355328 29 68630377364883 51.24
48 281474976710656 30 205891132094649 26.85
49 562949953421312 30 205891132094649 63.43
50 1125899906842624 31 617673396283947 45.14
51 2251799813685248 32 1853020188851841 17.71
52 4503599627370496 32 1853020188851841 58.85
53 9007199254740992 33 5559060566555523 38.28
54 18014398509481984 34 16677181699666569 7.42
55 36028797018963968 34 16677181699666569 53.71
56 72057594037927936 35 50031545098999707 30.57
57 144115188075855872 35 50031545098999707 65.28
58 288230376151711744 36 150094635296999121 47.93
59 576460752303423488 37 450283905890997363 21.89
60 1152921504606846976 37 450283905890997363 60.94
61 2305843009213693952 38 1350851717672992089 41.42
62 4611686018427387904 39 4052555153018976267 12.12
63 9223372036854775808 39 4052555153018976267 56.06
64 18446744073709551616 40 12157665459056928801 34.09
65 36893488147419103232 41 36472996377170786403 1.14 ***
66 73786976294838206464 41 36472996377170786403 50.57
67 147573952589676412928 42 109418989131512359209 25.85
68 295147905179352825856 42 109418989131512359209 62.93
69 590295810358705651712 43 328256967394537077627 44.39
70 1180591620717411303424 44 984770902183611232881 16.59
71 2361183241434822606848 44 984770902183611232881 58.29
72 4722366482869645213696 45 2954312706550833698643 37.44
73 9444732965739290427392 46 8862938119652501095929 6.16
74 18889465931478580854784 46 8862938119652501095929 53.08
75 37778931862957161709568 47 26588814358957503287787 29.62
76 75557863725914323419136 47 26588814358957503287787 64.81
77 151115727451828646838272 48 79766443076872509863361 47.21
78 302231454903657293676544 49 239299329230617529590083 20.82
79 604462909807314587353088 49 239299329230617529590083 60.41
80 1208925819614629174706176 50 717897987691852588770249 40.62
81 2417851639229258349412352 51 2153693963075557766310747 10.93
82 4835703278458516698824704 51 2153693963075557766310747 55.46
83 9671406556917033397649408 52 6461081889226673298932241 33.19
84 19342813113834066795298816 52 6461081889226673298932241 66.60
85 38685626227668133590597632 53 19383245667680019896796723 49.90
86 77371252455336267181195264 54 58149737003040059690390169 24.84
87 154742504910672534362390528 54 58149737003040059690390169 62.42
88 309485009821345068724781056 55 174449211009120179071170507 43.63
89 618970019642690137449562112 56 523347633027360537213511521 15.45
90 1237940039285380274899124224 56 523347633027360537213511521 57.72
91 2475880078570760549798248448 57 1570042899082081611640534563 36.59
92 4951760157141521099596496896 58 4710128697246244834921603689 4.88
93 9903520314283042199192993792 58 4710128697246244834921603689 52.44
94 19807040628566084398385987584 59 14130386091738734504764811067 28.66
95 39614081257132168796771975168 59 14130386091738734504764811067 64.33
96 79228162514264337593543950336 60 42391158275216203514294433201 46.49
97 158456325028528675187087900672 61 127173474825648610542883299603 19.74
98 316912650057057350374175801344 61 127173474825648610542883299603 59.87
99 633825300114114700748351602688 62 381520424476945831628649898809 39.81
100 1267650600228229401496703205376 63 1144561273430837494885949696427 9.71
101 2535301200456458802993406410752 63 1144561273430837494885949696427 54.86
102 5070602400912917605986812821504 64 3433683820292512484657849089281 32.28
103 10141204801825835211973625643008 64 3433683820292512484657849089281 66.14
104 20282409603651670423947251286016 65 10301051460877537453973547267843 49.21
105 40564819207303340847894502572032 66 30903154382632612361920641803529 23.82
106 81129638414606681695789005144064 66 30903154382632612361920641803529 61.91
107 162259276829213363391578010288128 67 92709463147897837085761925410587 42.86
108 324518553658426726783156020576256 68 278128389443693511257285776231761 14.30
109 649037107316853453566312041152512 68 278128389443693511257285776231761 57.15
110 1298074214633706907132624082305024 69 834385168331080533771857328695283 35.72
111 2596148429267413814265248164610048 70 2503155504993241601315571986085849 3.58
112 5192296858534827628530496329220096 70 2503155504993241601315571986085849 51.79
113 10384593717069655257060992658440192 71 7509466514979724803946715958257547 27.69
114 20769187434139310514121985316880384 71 7509466514979724803946715958257547 63.84
115 41538374868278621028243970633760768 72 22528399544939174411840147874772641 45.76
116 83076749736557242056487941267521536 73 67585198634817523235520443624317923 18.65
117 166153499473114484112975882535043072 73 67585198634817523235520443624317923 59.32
118 332306998946228968225951765070086144 74 202755595904452569706561330872953769 38.99
119 664613997892457936451903530140172288 75 608266787713357709119683992618861307 8.48
120 1329227995784915872903807060280344576 75 608266787713357709119683992618861307 54.24
121 2658455991569831745807614120560689152 76 1824800363140073127359051977856583921 31.36
122 5316911983139663491615228241121378304 76 1824800363140073127359051977856583921 65.68
123 10633823966279326983230456482242756608 77 5474401089420219382077155933569751763 48.52
124 21267647932558653966460912964485513216 78 16423203268260658146231467800709255289 22.78
125 42535295865117307932921825928971026432 78 16423203268260658146231467800709255289 61.39
126 85070591730234615865843651857942052864 79 49269609804781974438694403402127765867 42.08
127 170141183460469231731687303715884105728 80 147808829414345923316083210206383297601 13.13


问:
Distributional derivative of $\left|\cos|\pi x|\right|$
I am new to distributions and I have a problem finding the distributional derivative for the following function: $$\left|\cos|\pi x|\right|$$

I would be grateful if you could help to find the distributional derivative for this case. Please find the attached figure with this link.
Problem

I have tried up to some extent and I have attached a figure of that one here too. However, I could not find a way to extend that solution. Please find that figure with this link.attempt_one

答: If $f$ is continuous and piecewise $C^1$ then its distributional derivative is just $g(a)=f'(a)$ at the $a$ where $f$ is differentiable.


问:
Real analysis: Prove that a sequence in $(0,1)$ converging to $1$ has an increasing subsequence
The question is  "Suppose $\{a_n\}$ is a sequence of real numbers, all of which lie in the open interval $(0, 1)$, and suppose $\lim a_n = 1.$ Prove that $\{a_n\}$ has an increasing subsequence".

I understand that for a sequence to be increasing its elements must be such that $ a_1 < a_2 < a_3 $ and so on, but how do I prove that using the fact that it is bounded and has a limit rather than with using the elements?

Thanks

答: Let $n_1=1$. Since $\lim_{n\to\infty}a_n=1>a_{n_1}$, there is some $n_2\in\Bbb N$ such that $a_{n_2}>a_{n_1}$. And, since $\lim_{n\to\infty}a_n=1>a_{n_2}$, there is some $n_2\in\Bbb N$ such that $a_{n_3}>a_{n_2}$. And so on…


问:
2021 AMC 10 A Problem #22

Question: Hiram's algebra notes are $50$ pages long and are printed on $25$ sheets of paper; the first sheet contains pages $1$ and $2$, the second sheet contains pages $3$ and $4$, and so on. One day he leaves his notes on the table before leaving for lunch, and his roommate decides to borrow some pages from the middle of the notes. When Hiram comes back, he discovers that his roommate has taken a consecutive set of sheets from the notes and that the average (mean) of the page numbers on all remaining sheets is exactly $19$. How many sheets were borrowed?

$\textbf{(A)} ~10\qquad\textbf{(B)} ~13\qquad\textbf{(C)} ~15\qquad\textbf{(D)} ~17\qquad\textbf{(E)} ~20$

Does anyone have a solution that's easy to understand?

So here's what I've tried so far: I set $x$ as the first page his roommate took and $y$ as the total number of sheets his roommate took. I tried setting up an equation to find the average (19):
$$
\frac{1275-\text{sum of the  pages his roomate took}}{50-2y}=19.
$$
1275 is the sum of all the pages and 50-2y is the number of pages that are left. But I can't figure out how to express sum of the  pages his roommate took in terms of x and y

答:
Let $n$ be the number of sheets borrowed with average page number $k+25.5$. Since the sheets start with an odd number and end with an even number, we have $k\in \mathbb N$, and the sum of all page numbers are

$$\sum_{i=1}^{50} i = 50 \times 25.5 = 2n\times (k+25.5) + 2(25-n)\times 19 \tag 1$$

$$ \implies 0 = 2nk + 2(25-n)(19-25.5) = 2nk - 13(25-n)\tag 2$$
$$ \implies 13(25-n)=2nk \implies 13 | 2nk \implies 13 |n \text{ or }13|k$$

Case 1: If $13|n$, since $n<25$ we have $n=13, k=6$.

Case 2: If $13|k$, notice that $k \le (49+50)/2-25.5=24$ so $k=13 \implies 25-n=2n$, impossible because $n\in \mathbb N$.

Remark 1: We could obtain $(2)$ directly without $(1)$ by arguing the weighted decrease of average page number (from $25.5$ to $19$) is compensated by weighted increase of average page number (from $25.5$ to $k+25.5$), with the weights being the page (or sheet) number in each group (left vs. borrowed). In fact the first thing I wrote down when I solved this problem was $$(25-n)(25.5-19)=nk$$

Remark 2: if I were in the contest I'd plug in $n=13$, get $k=6$ and move on to the next problem.)

Title: 2021 AMC 10 A Problem #22


Title: Real analysis: Prove that a sequence in $(0,1)$ converging to $1$ has an increasing subsequence

Body:
The question is  "Suppose $\{a_n\}$ is a sequence of real numbers, all of which lie in the open interval $(0, 1)$, and suppose $\lim a_n = 1.$ Prove that $\{a_n\}$ has an increasing subsequence".

I understand that for a sequence to be increasing its elements must be such that $ a_1 < a_2 < a_3 $ and so on, but how do I prove that using the fact that it is bounded and has a limit rather than with using the elements?

Thanks

Body:
What about $b_1= a_1$ and

$$b_{n+1}=\inf \{a_k \mid a_k \gt \{b_1, \dots b_n,1-1/n\}\}$$


问:
Show that $f(n, k) = \binom{n}{k}$ is an onto function
I have to rigorously prove the statement:

Let $f(n, k) = \binom{n}{k}$, where $n, k \in \mathbb{N}$. Then $f: \mathbb{N} \times \mathbb{N}$ is an onto function.

First I translated the above statement into predicate logic:

$\forall n, k \in \mathbb{N}, f(n,k) = \binom{n}{k} \implies (\forall m \in \mathbb{Z}^{+}, \exists x, y \in \mathbb{N}, f(x, y) = m)$.

How should I approach this proof?

答: Because $$\binom{n}{1} = n$$ for all $\mathbb N$, it follows that $f$ is onto:  for every $n \in \mathbb N$, the pair $(n, 1)$ is a preimage of $n$ under $f$.


问:
Prove that $ \sum \frac{2^k}{k}$ is divisible by $2^M$
For each integer $M > 0$, there ${\bf exists}$ an $n$ such that

$$ \sum_{k=1}^n \dfrac{ 2^k}{k} $$

is divisible by $2^M$

${\bf try}$ Im struggling a bit to visualize this exercise. So, I tried to see for concrete number, for instance take $M=1$, then $n=2$ works: as

$$ 2 + \dfrac{2^1}{2} = 2^1 (1 + 2 )$$

Now, take $M=2$ and factor

$$ 2^{2} \underbrace{ \left( \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{2}{3} + \dfrac{2^4}{4} ... + \dfrac{2^{n-2} }{n} \right) }_{(*)}$$

now, we need to choose $n$ so that $(*)$ is an integer. Im unable to do so. Any help?

答:
First, an important definition

Definition the $2$-adic valuation $v_2$ is given by
$$v_2\left( 2^n\frac{p}{q} \right) = n$$
where $p, q$ are odd (not divisible by $2$).

For instance $v_2(12/5) = 2$ since we can write $12/5$ as $2^2 \frac{3}{5}$.

In this language, what we want to show is that

$$\lim_{n \to \infty} v_2\left( \sum_{k = 1}^n \frac{2^k}{k} \right) = \infty. \tag{$*$}$$

In my opinion the best/most conceptual/requires the least tricks way to prove this is to identify the infinite series

$$ \sum_{k = 1}^\infty \frac{2^k}{k} $$

as the $2$-adic logarithm of $-1$. That is, it is $-\log(1 - 2) = -\log(-1)$. Recall that the logarithm is given by the series

$$-\log(1 - x) = \sum_{k = 1}^\infty \frac{x^k}{k}. $$

Given this, and assuming that basic properties of logarithms still hold in the $2$-adics, we have

$$\log(-1) + \log(-1) = 2 \log(-1) = \log((-1)^2) = \log(1) = 0. $$

Therefore $\log(-1) = 0$ as a $2$-adic number.

Well the series converges to $0$ in the $2$-adics if and only if the partial sums grow to $\infty$ in $2$-adic valuation. I.e. The series converges to $0$ if and only if $(*)$ holds.

This requires some work to understand:

what does it mean to converge in the $2$-adic valuation

do these series converge?

is $\log(xy) = \log x + \log y$ still true and when?

For all these questions, I must refer to Keith Conrad's notes. The identity $(*)$ is discussed at the top of page 29 (Example 8.10) and I'm not going to fit 29 pages of exposition in here.

However, I will say that this approach is much nicer than the non-logarithm approach. Because that approach requires several very hard-to-find tricks. If I were locked in a room with no internet or any other references and I had to prove this fact, I would pick the p-adic logarithm approach because at least then I have some idea what needs to be done.

I'll try to find a "more elementary" approach in a separate answer (this one's getting sort of long). But I'll warn you now: don't expect it to be easy to understand or enlightening.


问:
Prove whether or not $f(x,y) = \frac{x^2y}{x^2+y^2}$ is differentiable at $(0,0)$.
I have the following function:
$f: \mathbb{R}^2 \to \mathbb{R}$ with $f(x,y) = \frac{x^2y}{x^2+y^2}$ for $(x,y) \neq (0,0)$ and $f(0,0) = 0$. Now I want to proof that $f$ is differentiable in $(0,0)$, but I get stuck with an $\epsilon-\delta$ proof for the limit
\begin{equation}
\lim_{(k,h) \to (0,0)} \frac{f(0+k,0+h) - f(0,0) - kf_1(0,0) - hf_2(0,0)}{\sqrt{k^2 + h^2}} = 0
\end{equation}
I think that $f_1(0,0) = f_2(0,0) = 0$, because $\lim_{h \to 0} \frac{f(h,0) - f(0,0)}{h} = \lim_{h \to 0} \frac{\frac{h^2 \cdot 0}{h^2 + 0}-0}{h} = 0$. And the same for $y+h$.

So the actual limit I want to solve is
\begin{equation}
\lim_{(k,h) \to (0,0)} \frac{\frac{k^2h}{k^2+h^2}}{\sqrt{k^2+h^2}} = \lim_{(k,h) \to (0,0)} \frac{k^2h}{(k^2 + h^2)^{\frac{3}{2}}} = 0
\end{equation}
Perhaps I made a mistake earlier on, but I can't seem to give a right proof of the existence of the limit...

答: The function $f$ is not differentiable at $(0,0)$. If it was, then, since $f_x(0,0)=f_y(0,0)=0$, then $f'(0,0)$ would be the null function. In other words,$$\lim_{(x,y)\to(0,0)}\frac{x^2y}{(x^2+y^2)^{3/2}}=0.$$But if $x=y>0$, $\frac{x^2y}{(x^2+y^2)^{3/2}}=\frac1{2\sqrt2}$.


答:
You are right, $f:\mathbb{C} \to \mathbb{C}$ is differentiable at a point $z_{0}$ if and only if the corresponding function $F = Re(f)+Im(f):\mathbb{R^2} \to \mathbb{R^2}$ is differentiable at $(x_{0}, y_{0})$ where $z_{0} = x_{0} + i y_{0}$ and the Cauchy-Riemann equations are satisfied. However, the sufficient condition for differentiabllity theorem only says that if the partial derivatives of $F$ are continuous at $c$ then it is differentiable at $c$, it's not a bidirectional implication as you can find a counterexample in José Carlos answer.

In relation to the Cauchy-Goursat theorem, take into account that the function is holomorphic at $U \setminus \,\{p\,\}$ where p is a point in $U$.

Title: Are the partial derivatives of the real and imaginary part of a holomorphic function always continuous?

Body:
I'm quite confused by my course notes for Complex Analysis. On the one hand, it says that a complex function $f:\mathbb{C} \to \mathbb{C}$ is differentiable at a point $z_0$ iff the corresponding vector-function $F:\mathbb{R^2} \to \mathbb{R^2}$ is differentiable and the Cauchy-Riemann equations are satisfied. Now differentiability of a multivariate real function is equivalent to the continuity of partial derivatives, so this would imply for $f(z) = u(x,y)+iv(x,y)$ that $u(x,y)$ and $v(x,y)$ have continuous partial derivatives.

On the other hand, the proof that is given for the Cauchy-Goursat theorem is an extremely long, convoluted one, and at the end of it, it mentions that a much more elegant and short proof is possible if one assumes that $u(x,y)$ and $v(x,y)$ have continuous partial derivatives (it makes use of Green's theorem). Why is this an 'assumption'? Doesn't it necessarily follow from the assumption of $f$ differentiable in the theorem statement?


问:
Prove $\sin \left(1/n\right)$ tends to $0$ as $n$ tends to infinity.
I'm sure there is an easy solution to this but my mind has gone blank! Any help on proving that $\sin( \frac{1}{n})\longrightarrow0$ as $n\longrightarrow\infty$ would be much appreciated.

This question was set on a course before continuity was introduced, just using basic sequence facts. I should have phrased it as:

Find an $N$ such that for all $n > N$ $|\sin(1/n)| < \varepsilon$

答:
Consider the Taylor series expansion of $\sin(1/n)$, then</p>

<p>$$ \frac{1}{n}- \frac{1}{n^3} \leq\sin \left(\frac{1}{n}\right) \leq \frac{1}{n}. $$ </p>

Body:
Consider the Taylor series expansion of $\sin(1/n)$, then

$$ \frac{1}{n}- \frac{1}{n^3} \leq\sin \left(\frac{1}{n}\right) \leq \frac{1}{n}. $$


问:
$(a_n)_{n=1}^\infty$ & $(b_n)_{n=1}^\infty$ are seq st $(a_n)_{n=1}^\infty$ & $[{(a_n)_{n=1}^\infty + (b_n)_{n=1}^\infty}]$ con. Prove $(b_n)$ con
Suppose $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ are sequences such that $(a_n)_{n=1}^\infty$ and $[{(a_n)_{n=1}^\infty + (b_n)_{n=1}^\infty}]$ converge. Prove that $(b_n)_{n=1}^\infty$ converges

I can say $b_n=(a_n + b_n)-a_n$. Since both $(a_n)_{n=1}^\infty$ and $[{(a_n)_{n=1}^\infty + (b_n)_{n=1}^\infty}]$ converge, isnt' there a subtraction rule that says that because those both converge, that $b_n$ would also converge?

答:
hint

Given $ \epsilon>0 $ ,Assume
$$\lim_{n\to+\infty}a_n=A$$
and
$$\lim_{n\to+\infty}(a_n+b_n)=C$$

then
$$|b_n-(C-A)|=$$
$$|(a_n+b_n)-C-(a_n-A)|$$
$$\le |(a_n+b_n)-C|+|a_n-A|$$
$$<\frac{\epsilon}{2}+\frac{\epsilon}{2}$$
for $ n $ large enough.


问:
Rudimentary question about differential equations.
One thing I have never understood about differential equations is how we can just consider the homogenous case in order to solve the equation.

for example for any second-order linear ODE:
$$ \alpha\frac{d^2y}{dx^2} + \beta\frac{dy}{dx} + \gamma y = P(x) $$

why do we solve this for $P(x) = 0$ as its own seperate solutions, in other words why is the paticular integral not the only solution?

答:
Don’t forget that when we speak of a solution, we also speak of the initial conditions $y(x_0)=y_0$ and $y^\prime(x_0)=y_1$.

The homogeneous equation has a unique solution given the initial conditions. This can be used by addition  to find a solution of the equation with the RHS equal to $P$ and the initial conditions.

This works as a sum of solutions of a linear equation is also a solution of the equation.


问:
Probability of time waiting for the bus
The time (in min.) to wait for my bus can be modeled as an exponential r.v.

I know that the average time I need to wait for the bus is 10 min.

What is the prob. that I would wait more than 15min?

My Attempt:

$X\text{~}Exp(\lambda)$: $f(x)=10e^{-10x}$

I know that $\mathbb{X}[E]=\int^\infty_{-\infty}xf(x)dx$, so I want to solve: (lower bound is 15 because we want to calculate the probability that we wait at least 15 min).

$$\int^\infty_{15}x(10e^{-10x})dx=\left[-\frac{(10x+1)e^{-10x}}{10}\right]^\infty_{15}=\frac{151e^{-150}}{10}$$

My concern is that this number is so incredibly small so I don't think I did it correctly. Can I get pointers in the right direction? Thank you.

Title: Probability of time waiting for the bus

Body: $$P(X \geq 15)=1-\int_{0}^{15}\frac{1}{10}e^{-\frac{1}{10}x}dx$$


问:
On uniquely determining a bounded surface in 3D space when the planar projections are given
Let $f(x,y,z)=0$ is a bounded closed surface defined in $D^3$ Where $D$ is a finite closed interval included in real number set. The planar projections of the surface are given:

On $z=0$ plane the Projection is given.

On $y=0$ the Projection is given.

On $x=0$ the  Projection is given.

From the above mentioned planar projections  can we uniquely determine the original surface?

答:
Let $D=[-1,1]$ and consider the surface $f(x,y,z)=0$ where
$$
f(x,y,z)=(x^2+y^2+z^2-1)(x^2+y^2+z^2-r^2)
$$
for some fixed $r\in (0,1)$.

Then the projections are independent of the choice of $r$.

To address your comment, to show an example where the surface is connected and bounds a connected region, we can modify the above example as follows . . .

Let $D=[-1,1]$ and consider the surface $f(x,y,z)=0$ where
$$
f(x,y,z)
=
\Bigl(z-\sqrt{1-x^2-y^2}\Bigr)
\Bigl(z-a\sqrt{1-x^2-y^2}\Bigr)
$$
for some fixed $a\in (0,1)$.

Then the projections are independent of the choice of $a$, and the region bounded by the surface is connected.


问:
Norm turning $C^\infty(\overline{B(0,1)})$ into a Banach space?
I've been trying to find a proof that for $K=\overline{B(0,1)} \subseteq \mathbb R^n$, we can find some norm or notion of convergence (I am indifferent as to how we specify the topology) on $\mathcal C^\infty (K, \mathbb R) = \mathcal C^\infty (K)$ such that it is a Banach space. To my surprise, I couldn't find anything of the sort. This fact is stated in the answer here, so I have reason to believe it's true. Also, I saw here how this works but only for $n=1$, and there the proof of the limit function also having a continuous derivative depends on being able to express the limit function with respect to the integral of the derivative. Can this approach be generalised to $\mathcal C^\infty(K)$ (there would be an obvious problem  of convergence with taking the sum of the supremum norms of all partial derivatives of all orders), or how would one endow $\mathcal C^\infty (K)$ with a topology such that it is complete?

答:
Part 1

If you can show that the algebraic dimension of the space $C^\infty(K)$ is equal to $\frak{c}$, then since $\frak{c}^{\aleph_0}=\frak{c}$, this post shows that there does exist a norm that makes it a Banach space:

Can every vector space (over $\mathbb{R}$ or $\mathbb{C}$) can be a Banach space (or Hilbert space)?

Part 2

Of course, the above is not practical at all and not constructive at all. Let's see a construction for $n=1$. The answer I am going to give yields a metric and not a norm, but it makes your space complete.

Let $f,g\in C^\infty([-1,1])$.
We define
$$d(f,g)=\sum_{n=0}^\infty\frac{\min\{1,\|f^{(n)}-g^{(n)}\|_\infty\}}{2^n}$$
where $f^{(n)}$ denotes the $n$-th derivative of $f$. It is immediately verified that $d$ defines a metric on our space. We show that the space is also complete.

Let $(f_n)\subset C^\infty([-1,1])$ be a Cauchy sequence with respect to $d$. by the inequality $\|f_n^{(k)}-f_m^{(k)}\|_\infty\leq2^k\cdot d(f_n,f_m)$ which holds for all $n,m,k\geq0$ we can easily deduce that for any $k\geq0$ the sequence $(f_n^{(k)})$ is Cauchy with respect to the supremum norm. Since uniformly Cauchy sequences of continuous functions converge to some continuous function, we have that each sequence $(f_n^{(k)})$ converges uniformly to some continuous function.

So, for $k=0$, let's say that $f_n\to f$ and for $k=1$, we have that $f_n'\to g$ uniformly.
Note that
$$\sup_{x\in[-1,1]}\bigg|\int_{-1}^xf_n'(t)dt-\int_{-1}^xg(t)dt\bigg|\leq\int_{-1}^1|f_n'(t)-g(t)|dt\leq2\|f_n'-g\|_\infty\to0 $$
but $\int_{-1}^xf_n'(t)dt=f_n(x)-f_n(-1)\longrightarrow f(x)-f(-1)$. We conclude that $f(x)=\int_{-1}^xg(t)dt+f(-1)$, so $f$ is differentiable and moreover $f'(x)=g(x)$.

Following a similar argument for higher derivatives, one can easily deduce that $f\in C^\infty([-1,1])$ and that $f_n^{(k)}\to f^{(k)}$ for all orders $k$. We will now show that $d(f_n,f)\to0$: Let $\varepsilon>0$. We find $k_0\in\mathbb{N}$ large enough so that $\sum_{k=k_0+1}^\infty\frac{1}{2^k}<\varepsilon/2$. Then
$$d(f_n,f)=\sum_{k=0}^\infty\frac{\min\{1,\|f_n^{(k)}-f^{(k)}\|_\infty\}}{2^n}\leq\sum_{k=0}^{k_0}\frac{\|f_n^{(k)}-f^{(k)}\|_\infty}{2^k}+\sum_{k=k_0+1}^\infty\frac{1}{2^k}\leq$$
$$\leq\sum_{k=0}^{k_0}\frac{\|f_n^{(k)}-f^{(k)}\|_\infty}{2^k}+\varepsilon/2 $$

But now we can take $n_0$ large enough so that $\sum_{k=0}^{k_0}\frac{\|f_n^{(k)}-f^{(k)}\|_\infty}{2^k}<\varepsilon/2$ for all $n\geq n_0$, which will in turn yield $d(f_n,f)<\varepsilon$ for all $n\geq n_0$. As $\varepsilon>0$ was arbitrary, we conclude that $f_n\to f$ with respect to $d$. Thus $C^\infty([-1,1])$ is complete when equipped with the metric $d$.

Comment: The reason that $d$ fails to give a norm is because we apply these minima $\min\{1,-\}$. Those are necessary: if for example we tried to define a norm as $\|f\|=\sum_{n}\frac{\|f^{(n)}\|_\infty}{2^n}$, then the function $f(x)=e^{2x}$ will give $f^{(n)}(x)=2^ne^{2x}$, so the series will not converge. In order to avoid such pathologies we have to be more flexible and settle for a metric, using this $\min\{1,-\}$ trick that resolves such problems.

Comment: I am not completely sure but I guess that the analogue for greater dimensions $d\geq1$ would have to be something with multi-indexes, like
$$d(f,g)=\sum_{\alpha}\frac{\min\{1,\|\partial^\alpha f-\partial^\alpha g\|_\infty}{2^{|\alpha|}}$$
where $\alpha\in\mathbb{N}_0^d$, $\alpha=(\alpha_1,\dots,\alpha_d)$, $|\alpha|=\alpha_1+\dots\alpha_d$ and
$$\partial^\alpha f=\prod_{j=1}^d\frac{\partial^{\alpha_j}f}{\partial x_j^{\alpha_j}}.$$

I believe that the same arguments can be modified to get a complete metric on this space, but this is only my opinion, I am not sure.


Title: Prove $\sin \left(1/n\right)$ tends to $0$ as $n$ tends to infinity.

Body:
I'm sure there is an easy solution to this but my mind has gone blank! Any help on proving that $\sin( \frac{1}{n})\longrightarrow0$ as $n\longrightarrow\infty$ would be much appreciated.

This question was set on a course before continuity was introduced, just using basic sequence facts. I should have phrased it as:

Find an $N$ such that for all $n > N$ $|\sin(1/n)| < \varepsilon$


答:
You're approaching the problem incorrectly:  The exponential should be $-t/10$.

$$\int\limits_{t=15}^\infty \frac{1}{10} e^{-t/10}\ dt = \frac{1}{e^{3/2}} \approx 0.22313.$$

Title: Probability of time waiting for the bus

Body:
The time (in min.) to wait for my bus can be modeled as an exponential r.v.

I know that the average time I need to wait for the bus is 10 min.

What is the prob. that I would wait more than 15min?

My Attempt:

$X\text{~}Exp(\lambda)$: $f(x)=10e^{-10x}$

I know that $\mathbb{X}[E]=\int^\infty_{-\infty}xf(x)dx$, so I want to solve: (lower bound is 15 because we want to calculate the probability that we wait at least 15 min).

$$\int^\infty_{15}x(10e^{-10x})dx=\left[-\frac{(10x+1)e^{-10x}}{10}\right]^\infty_{15}=\frac{151e^{-150}}{10}$$

My concern is that this number is so incredibly small so I don't think I did it correctly. Can I get pointers in the right direction? Thank you.


问:
Point within the interior of a given angle
The point $M$ is within the interior of given angle $\alpha$. Find the distance between $M$ and the vertex of the angle ($OM=?$) if $a$ and $b$ are the distances from $M$ to the sides of the angle.

We can see that $$OM^2=OP^2+PM^2=OK^2+KM^2$$
Let $OP=x;PK=y$. Then $$2OM^2=OP^2+PM^2+OK^2+KM^2\\=x^2+b^2+y^2+a^2.$$ I can't approach the problem further. How can we use the given angle $\alpha$? Thank you in advance!

答:
Hint

Denote by $A$ the angle opposite to side with length $a$, similarly $B$.

You have
$$\sin A=\frac{a}{OM}, \, \sin B =\frac{b}{OM}$$ and

$$A+B= \alpha.$$


答:
Hint

$$\quad0 \leq\sin\left(\frac{1}{n}\right)\leq \frac{1}{n}$$


Title: Point within the interior of a given angle

Body:
The point $M$ is within the interior of given angle $\alpha$. Find the distance between $M$ and the vertex of the angle ($OM=?$) if $a$ and $b$ are the distances from $M$ to the sides of the angle.

We can see that $$OM^2=OP^2+PM^2=OK^2+KM^2$$
Let $OP=x;PK=y$. Then $$2OM^2=OP^2+PM^2+OK^2+KM^2\\=x^2+b^2+y^2+a^2.$$ I can't approach the problem further. How can we use the given angle $\alpha$? Thank you in advance!

Body:
Let $m=OM$, then $${a\over m} =  \sin \alpha _1$$ and $${b\over m} =  \sin \alpha _2$$

so $$\cos \alpha _1 = \sqrt{1-{a^2\over m^2}}\;\;\;\;\;\;\;\;\;\; \wedge \;\;\;\;\; \;\;\;\;\; \cos \alpha _2 = \sqrt{1-{b^2\over m^2}}$$

so $$\sin \alpha = {a\over m}\sqrt{1-{b^2\over m^2}}  + {b\over m} \sqrt{1-{a^2\over m^2}}$$

Now you have to find out $m$. :)


答:
Denote the mid-point of $OM$ be $N$. Then $N$ is the center of the circle passing through the points $O,P,M,K$.

We have:

$$\angle PNK = 2\alpha,\quad \angle PMK = \pi-\alpha$$

By Cosine Theorem,

$$PK^2 = a^2+b^2-2ab\cos \angle PMK = PN^2 + KN^2 - 2PN\cdot KN \cos 2\alpha$$

Writing $PN = KN = \frac12 OM = r$, we have:

$$a^2 + b^2 +2ab\cos \alpha = 2r^2(1-\cos 2\alpha)$$

or

$$r = \sqrt\frac {a^2+b^2+2ab\cos \alpha}{2(1-\cos2\alpha)}$$

To find $OM$ we just double this value:

$$OM = 2r = 2\sqrt\frac {a^2+b^2+2ab\cos \alpha}{2(1-\cos2\alpha)} = 2\sqrt\frac {a^2+b^2+2ab\cos \alpha}{2(2 \sin^2 \alpha)} = \frac {\sqrt{a^2+b^2+2ab\cos \alpha}}{\sin \alpha}$$


问:
On what interval is $f(x)=x-\tan(x)$ increasing or decreasing?
My attempt at the solution is:
$$
f'(x) = 1- \frac{1}{\cos^2(x)} $$
We check when $f'(x)=0$ to determine when the derivative is zero.
$$
1- \frac{1}{\cos^2(x)} = 0 \iff 1 = \frac{1}{\cos^2(x)} \iff \cos^2(x)=1
$$
From that we can see that
$$
x=n\pi, n \in \mathbb Z.
$$
I don't understand how to continue. Normally we use a "signs table" in this situation when the given function is a regular one and not trigonometric.

答: Maybe it's better to work with the inequality directly.  $f$ is increasing if $f' > 0$.  So we require $$1 - \frac{1}{\cos^2 x} > 0,$$ or $$\cos^2 x - 1 > 0, \quad \cos x \ne 0.$$  But because $-1 \le \cos x \le 1$, we have $\cos^2 x \le 1$, hence this inequality is never satisfied, and $f$ is never strictly increasing.


答: Let the derivative of your function: $f'(x)=1-\frac{1}{\cos^2{x}}=\frac{\cos^2{x}-1}{\cos^2{x}}\leq 0$ $\forall x\in\mathbb{R}$, since $\cos^2{x}\leq 1$. So the function is always decreasing.

Title: On what interval is $f(x)=x-\tan(x)$ increasing or decreasing?

Body:
My attempt at the solution is:
$$
f'(x) = 1- \frac{1}{\cos^2(x)} $$
We check when $f'(x)=0$ to determine when the derivative is zero.
$$
1- \frac{1}{\cos^2(x)} = 0 \iff 1 = \frac{1}{\cos^2(x)} \iff \cos^2(x)=1
$$
From that we can see that
$$
x=n\pi, n \in \mathbb Z.
$$
I don't understand how to continue. Normally we use a "signs table" in this situation when the given function is a regular one and not trigonometric.


问:
Is any subset of $\mathbb{R}$ a Borel set?
Consider $S \subset \mathbb{R}$. Define for all $n \in \{1,2,3,\ldots\}$

\begin{align*}
S_n := \bigcup_{a \in S} \left (a - \frac{1}{n}, a + \frac{1}{n}\right)
\end{align*}

Since arbitrary union of open sets is open, $S_n$ is open for all $n$.

Now
\begin{align*}
S = \bigcap_{n=1}^{\infty} S_n
\end{align*}

Hence $S$ is the intersection of countably many open sets, therefore, it is a Borel set.

Are my steps correct or is there something wrong? If there is something wrong, could you point out which step exactly?

答:
Your claim that $\bigcap_{n=1}^{\infty} S_n =S$ is false.

Take $S=\mathbb Q.$ Then $S_n=\mathbb R$ for all $n.$

More generally, the closure of $S$ is contained in $\bigcap S_n.$ You can prove that $\bigcap S_n=\overline S,$ but all closed sets are Borel already.


答:
Reflect $M$ across $OA$ and $OB$ and you get $A'$ resp. $B'$.

Let $t=AB$, then $A'B' = 2t$ and $OA'= OB' = OM = m$. Also $\angle A'OB' = 2\alpha$.

By the Law of cosine in triangle $ABM$ we have $$ t = \sqrt{a^2+b^2+2ab\cos \alpha}$$

and finally in isosceles triangle $A'B'O$ we have $$\boxed{m = {t\over \sin \alpha}}$$


答:
Right so the "elementary answer." There are actually a couple that I've seen. One is cited by Gouvêa in his p-adic numbers book (Gouvêa, like me, also said the problem was too difficult to work out the answer from scratch).

The solution Gouvêa cites is Exercise 10.10 of D. P. Parent Exercises in Number Theory (1984).

The rough idea there (assuming I'm summarizing correctly) is

$$ 0 = \frac{1 - (1 - 2)^{2^n}}{2^n} = \sum_{k = 1}^{2^n} (-1)^{k+1}2^k \frac{1}{2^n}\binom{2^n}{k} \equiv \sum_{k=1}^{2^n} \frac{2^k}{k} \pmod{2^h} $$

for $n \ge h$. Plus there's some other work to compare this with the partial sums that are not powers of $2$.

The next two solutions I found cited by the OEIS (https://oeis.org/A087910). Namely the two solutions given for the 2002 Sydney University Mathematical Society Problems Competitions Problem 9.

Solution 1 Summary

If $n$ is even then

$$ 1 = (-1)^n = (1 - 2^n) = \sum_{k = 0}^n \binom{n}{k}(-2)^k. \tag{1} $$

Subtract 1 and divide by $n$ to get

$$ 0 = \sum_{k = 1}^n \frac1n \binom{n}{k}(-2)^k. $$

Then

$$ \frac1n \binom{n}{k} = \frac{(n-1)(n-2)\cdots(n-k+1)}{k!} = \frac{(-1)^{k-1}}{k} + n\frac{m_{n,k}}{k!} \tag{2}$$

for some integer $m_{n,k}$ (separate the term $(-1)(-2)\cdots(-k+1)$ from all the terms divisible by $n$).

By $(1)$ and $(2)$,

$$ \sum_{k = 1}^n \frac{2^k}{k} = n \sum_{k = 1}^n \frac{(-2)^km_{n,k}}{k!}$$

Then you use a well known fact that $v_2(k)! \le O(k)$. Thus

$$ v_2\left( \sum_{k = 1}^n \frac{2^k}{k} \right) = v_2(n) + v_2\left( \sum_{k = 1}^n \frac{(-2)^km_{n,k}}{k!} \right) \ge v_2(n). $$

Now we can see that when $n = 2^k$ this tends to infinity.

Solution 2 Summary

First show that

$$ \sum_{k = 1}^n \frac{2^k}{k} = \frac{2^n}{n} \sum_{k = 0}^{n-1} \frac{1}{\binom{n-1}{k}}. $$

Next, use the well-known formula $$v_2(n!) = \sum_{i = 0}^\infty \left\lfloor \frac n{2^i} \right\rfloor$$ to get

$$ v_2\left( \binom{n}{k} \right) = v_2(n!) - v_2(k!) - v_2((n - k)!) = \sum_{i = 0}^\infty \left\lfloor \frac n{2^i} \right\rfloor - \left\lfloor \frac k{2^i} \right\rfloor - \left\lfloor \frac {n-k}{2^i} \right\rfloor. $$

Then by some analysis,

$$v_2\left( \binom{n-1}{k} \right) \le r \text{ if } 2^r + 1 \le n \le 2^{r + 1}. $$

So if $n$ is even and $2^r + 1 \le n \le 2^{r + 1}$ then

$$ v_2\left( \sum_{k = 1}^n \frac{2^k}{k} \right) \ge n - r. $$

And the result follows.

You'll have to see the solutions I linked to if you want all the details, they wouldn't fit in one answer. I hope this gives you some appreciation for the $2$-adic logarithm approach.

Title: Prove that $ \sum \frac{2^k}{k}$ is divisible by $2^M$

Body:
For each integer $M > 0$, there ${\bf exists}$ an $n$ such that

$$ \sum_{k=1}^n \dfrac{ 2^k}{k} $$

is divisible by $2^M$

${\bf try}$ Im struggling a bit to visualize this exercise. So, I tried to see for concrete number, for instance take $M=1$, then $n=2$ works: as

$$ 2 + \dfrac{2^1}{2} = 2^1 (1 + 2 )$$

Now, take $M=2$ and factor

$$ 2^{2} \underbrace{ \left( \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{2}{3} + \dfrac{2^4}{4} ... + \dfrac{2^{n-2} }{n} \right) }_{(*)}$$

now, we need to choose $n$ so that $(*)$ is an integer. Im unable to do so. Any help?


问:
Is correct this way to prove that $\dim(V)<\infty$?

Let be $F$ a field, $V$ and $W$ are vector spaces over $F$. Prove that
if it exist a linear tranformation $T:V\rightarrow W$ with
$\dim(\operatorname{Im}(T))$ and $\dim(\operatorname{Ker}(T))$ both finite, then $\dim(V)$ is finite.

My attempt:

Since we know that $\dim(\operatorname{Ker}(T))<\infty$, there exist a finite basis $B_0 \subset V$ of $\operatorname{Ker}(T)$
Also, there exists $B$ a basis of $V$.

On  another side, let be $x \in \operatorname{Im}(T)$, then $x=T(y)$ for some $y \in V$. We have to note that $y$ is not a linear combination of elements from $B_{0}$, since $T(B_{0})=0$.

Then, $y$ is a linear combination of elements of $B \setminus B_{0}$.

\begin{align}
&\Rightarrow x \text{ is a linear combination of elements from } T(B \setminus B_{0}) \text{ for every } x \in Im(T) \\
&\Rightarrow T(B \setminus B_{0}) \text{ is a basis of } Im(T)\\
&\Rightarrow  T(B \setminus B_{0}) \text{ is finite }\\
&\Rightarrow B \setminus B_{0} \text{ is finite } \\
&\Rightarrow  B \text{ is finite}\\
&\Rightarrow  \dim(V)<\infty
\end{align}

Is my proof correct? In case that it is not, please help me to write it fine. I really appreciate your help.

答: Hint Take a finite basis $B_0$ of $\text{Ker} T$ and a finite basis $Y$ of $\text{Im} T$. Use this  to create a basis $B_0 \cup X$ of $V$.


答:
In the intervalle $ I_k= (\frac{-\pi}{2}+2k\pi,\frac{\pi}{2}+2k\pi) $,
$ f $ is continuous,
$$f'(x)=1-1-\tan^2(x)\le 0$$
and
$$f'(x)=0\iff x=2k\pi$$
So, $ f $ is Strictly decreasing at $ I_k$.

The same at $ J_k=(\frac{\pi}{2}+2k\pi,\frac{3\pi}{2}+2k\pi)$.


问:
Prove that : $ \int_{0}^{2\ln{\varphi}}{\theta\ln{\left(2\sinh{\frac{\theta}{2}}\right)}\,\mathrm{d}\theta}=-\frac{1}{5}\zeta\left(3\right) $
Denoting $ \varphi=\frac{1+\sqrt{5}}{2}=\mathrm{Golden\ Ratio} $. How would you prove that : $$  \int_{0}^{2\ln{\varphi}}{\theta\ln{\left(2\sinh{\frac{\theta}{2}}\right)}\,\mathrm{d}\theta}=-\frac{1}{5}\zeta\left(3\right) $$

I came accross this strange identity while reading the following article : VALUES OF THE RIEMANN ZETA FUNCTION AND
INTEGRALS INVOLVING $\log{\left(2\sinh{\frac{\theta}{2}}\right)}$ AND $\log{\left(2\sin{\frac{\theta}{2}}\right)}$. (No complete proof was given in the article). I'm still trying to prove it. For now I did try to set $ x=2\ln{\left(\sinh{\frac{\theta}{2}}\right)} $ to get rid of the golden ratio, but it doesn't seem to get any better. I know I'll have to expand the integrand to get to $ \zeta $ using, for example, the fact that : $ \ln{\left(\sinh{x}\right)}=x-\ln{2}-\sum\limits_{n=1}^{+\infty}{\frac{\mathrm{e}^{-2nx}}{n}} $, but I'm sure I cannot do it now, because of that golden ratio in the upper bound. Any ideas, hints will be appreciated.

答:
$$I=\int_0^{2\ln\varphi}\theta\ln\left(2\sinh\frac\theta2\right)\,d\theta$$
if we let $\alpha=\frac\theta2\Rightarrow d\theta=2d\alpha$ soL
$$I=\int_0^{\ln\varphi}2\alpha\ln\left(2\sinh\alpha\right).2\,d\alpha=4\int_0^{\ln\varphi}\alpha[\ln(2)+\ln(\sinh\alpha)]d\alpha$$
now you can break it up: $I=4(I_1+I_2)$ so:
$$I_1=\int_0^{\ln\varphi}\alpha\ln(2)\,d\alpha=\frac{\ln2}{2}\left[\alpha^2\right]_0^{\ln\varphi}=\frac12\ln(2)\ln^2(\varphi)$$
now:
$$I_2=\int_0^{\ln\varphi}\alpha\ln(\sinh(\alpha))\,d\alpha=\int_0^{\ln\varphi}\left(\alpha^2-\alpha\ln(2)-\alpha\sum_{n=1}^\infty\frac{e^{-2n\alpha}}{n}\right)d\alpha$$
now notice the middle term is just $I_1$ so that will cancel nicely so we may as well redefine: $I=I_3-I_4$
$$I_3=\int_0^{\ln\varphi}\alpha^2\,d\alpha$$
$$I_4=\sum_{n=1}^\infty\frac1n\int_0^{\ln\varphi}\alpha e^{-2n\alpha}\,d\alpha$$
now the first integral is easy, and with the second integral you can just use IBP, you should get a series involving powers of $\varphi$ for which relations do exist, hope this helps :)

The bottom integral would evaluate to:
$$\frac{1}{4n^2}-\frac{\varphi^{-2n}\left(2n\ln\varphi+1\right)}{4n^2}$$
which does look tricky to link to $1/n^3$

Remember that the $n$th term of the Fibonacci sequence is given by:
$$F(n)=\frac{\varphi^n-(-\bar{\varphi})^n}{\sqrt{5}}$$
so maybe there is a weird link here somewhere


答:
This certainly doesn't seem true

If any $n > 2$ there is a prime $p$ so that $2<p \le n < 2p$ (Bertands postulate) and so

And so if $2^m| \sum_{k=1}^n \frac {2^k}k$ then $\sum_{k=1}^n \frac {2^k}k=W$ is as an integer.  And so $W-\frac {2^p}p$ is not an integer.

But $W - \frac {2^p}p= \frac {pW - 2^p}p$ is a fraction in lowest terms.

Which means $\sum_{k=1;k\ne p}^n \frac {2^k}k = \frac {pW -2^p}p$.  But be figuring out the least common denominator of $\frac {2^k}{k}; k\ne p$ we get the lets common denominater is $\operatorname{lcm}(k; k\ne p)$.  But that will not reduce so $\frac{pW-2^p}p$ unless $p$ divides into $\operatorname{lcm}(k; k\ne p)$.

Which it clearly can't as $\operatorname{lcm}(k; k\ne p)$ does not have $p$ as a prime divisor of any of the $k$ terms.

SO.... unless I'm making a bonehead error, this is not possible.

Not only is this not true  $\sum_{k=1}^n \frac {2^k}k$ is never an integer if $n > 2$.

..... unless I'm making a boneheaded error.


问:
If $I\pi ^{-k}\subseteq D$ for all $k\leq m$ but $I\pi ^{-m-1}\nsubseteq D$, then $I\pi ^{-m}\nsubseteq (\pi )$
I'm working through a proof and I've got stuck on one detail, it seems like it is supposed to be totally obvious, but need help to figure out why.

Let $D$ be a Noetherian integrally closed domain with unique prime ideal $(\pi)$. For an ideal $I$ of $D$ we have that $I\pi ^{-k} \subseteq D$ for all $k\leq m$ but $I\pi ^{-m-1}\nsubseteq D$. I wonder how to make the conclusion that $I\pi ^{-m}\nsubseteq (\pi )$.

This is used in proving that every ideal in the ring $D$ above is a power of $(\pi )$. I hope I didn't leave out any crucial information...

答: Suppose, for the sake of contradiction, that $I\pi^{-m}\subseteq(\pi)$. Then every element of $I\pi^{-m}$ is of the form $\pi a$ for some $a\in D$. Thus, every element of $I\pi^{-m-1}$ is an element of $D$, i.e. $I\pi^{-m-1}\subseteq D$; but we chose $m$ so that this is not the case.</p>


问:
Hypergeometric differential equation for $c=1$ and $a+b+1=0$
I would like to find the base of solutions for the following differential equation,
$$z(1-z)f''(z)+f'(z)+ \alpha \cdot f(z)=0$$
where $\alpha$ is a parameter and the prime indicates derivative w.r.t z.  The above equation is the hypergeometric one say, $$z(1-z)f''(z)+(c-z(a+b+1))f'(z) -ab\cdot f(z)=0$$ in the case with $c=1$ and $a+b+1=0$. BUT it is not clear at all what are the two independent solutions. One has to be the hypergeometric but  the other one I do not know beacuse of the degeneracy of the parameters [ $c=1$ and $a+b+1=0$].

Many thanks!!!

答:
WolframAlpha gives two solutions (link). One of them is the hypergeometric function $f(z)={_2}F_1(a,b;c;z)$ that you expected, with

$$a,b=-\frac12 \pm \frac12\sqrt{4\alpha+1},\quad c=1$$
(The order of $a,b$ is irrelevant.) The second is expressed in terms of the MeijerG function. If I use Mathematica and force it to expand the MeijerG function (via the FunctionExpand command) I obtain

$$f(z)=\frac12 (z-1)^2 {_2}F_1(1+a,1+b;3;1-z)$$
where $a,b$ are chosen as above. However, Mathematica indicates that this representation is only valid for $|z|<1$. For $|z|>1$, it instead concludes that the relevant MeijerG function is identically zero. I take this to mean that a different second solution is valid outside the unit circle. As such, the utility of this basis will depend on where you're trying to solve $f(z)$.


问:
Constructing a Multivariate Probability Distribution Formula
Suppose there are three individuals playing a simple game, Player $1$, Player $2$, and Player $3$.

In each round of the game, one of the players is uniformly randomly selected to receive the point for that round.  The game ends when any player achieves a score of $10$ points.

How can a multivariate probability distribution formula be constructed that would express the probability of each possible combination of the total number of rounds a game will contain, and an arbitrarily chosen player's (i.e., Player $1$'s) score at the end of the game?  How does this formula change if we introduce the assumption that the arbitrarily chosen player did not win the game?

I started by noting that the chosen player has a $\frac{1}{3}$ probability of winning each round.  I think that for a fixed number of rounds $n$, the formula would be the binomial distribution ${n \choose a} (\frac{1}{3})^a (\frac{2}{3})^{1 - a}$, where by plugging a possible point score into $a$ one obtains the probability of the chosen player attaining that exact point score, but I'm not sure if this is right or where to go from here.

答:
That's the correct formula for the probability that a specified play scores exactly $a$ points in $n$ rounds, but I see that it helps you much.

If a specified player wins, then he scores $10$ points and each of the others scores at most $9$ (so $10\leq n\leq 28)$.  So the probability that he wins is
$$\frac13\sum_{k=0}^9\sum_{j=0}^9\binom{k+j+9}{k,j,9}\left(\frac13\right)^{k+j+9}$$ because the winning toss, when the specified player gets his tenth point, must be the last one.

Now you just have to modify this formula to account for the fact that the game lasts $n$ rounds, that is $k+j+10=n$.


Body:
Let $d=OM$

$$ \arcsin(a/d)+ \arcsin(b/d) = \alpha$$

$$ a \sqrt{d^2-b^2} + b \sqrt{d^2-a^2} = d^2\sin \alpha$$

Let Mathematica solve fourth order equation for $d$

$$sal= \sin \alpha$$

{d -> 0.5` Sqrt[(4.` (1.` + 1.` b^2))/sal^2 +
    2.` Sqrt[(4.` (1.` + 1.` b^2)^2)/sal^4 - (
      16.` (0.25` - 0.5` b^2 + 0.25` b^4 + 1.` b^2 sal^2))/sal^4]]}


问:
Is there a symbol for set operation $A \cup B^\complement$?
Many of the 16 binary set operations (corresponding to the 16 logical connectives) don’t seem to have common dedicated symbols. Instead, they are often simply expressed in terms of $\cup, \cap, \setminus$, etc.

My question is whether there is any known precedent of a symbol for the operation $A \cup B^\complement$. This would complete the following (where ? is a placeholder for the operation in question):

$A \setminus B = A \cap B^\complement$

$A \operatorname{?} B = A \cup B^\complement$

For more context (based on discussion in the comments below):

In everyday work, of course it makes sense to use just a few symbols (e.g. complement, $\cup, \cap, \setminus$) and express other operations in terms of these. This approach represents a practical tradeoff between two extremes:

At one extreme, only one symbol is needed (just NAND or NOR, cf. functional completeness). The drawback is that most expressions will be quite large when written this way.

At the other extreme, we would have a unique symbol for each of the 16 operations. Expressions may be written more compactly with these symbols, but then there is more to memorize.

(Why are the complement, $\cup, \cap, \setminus$ operations in particular given special status? My only guess is that they correspond with concepts in spoken language: NOT, OR, AND, WITHOUT, which, in turn, may correspond to which operations are most naturally computed by our brains, but I digress...)

A comment below suggests simply using $(B \setminus A)^\complement$ to represent $A \cup B^\complement$, but the same could be said of any operation, e.g. why do we need a symbol for set difference $A \setminus B$ when we can just write $A \cap B^\complement$? Or why do we need a special logical connective for implication (which actually corresponds to the set operation in question here)?

My point is that sometimes things can be written more concisely with dedicated symbols for the uncommon cases. It may be rare that these symbols are needed, but in those instances where one of the uncommon set operations is heavily utilized ($A \cup B^\complement$ in my case here), it would be nice to have some agreed-upon symbol. And here I simply want to know if there is any precedent in the literature for those weird cases.

The diagram below (originally found on Wikipedia here or here) shows that all 16 logical connectives have dedicated symbols. Why not extend this luxury to the corresponding set operations?

答:
Update: This answer lists sources which use various symbols for this set operation. (No strong consensus, but at least there is some precedent in the literature.)

I have not been able to find any literature containing a symbol for $A \cup B^\complement$, so I'm tentatively saying the answer here is: no, there is no precedent for such a symbol.

Here are some ideas for possible symbols (some inspired by comments on the question above):

Some set operations are visually similar to their corresponding logical connectives ($\cup$ and $\vee$, $\cap$ and $\wedge$). Logical implication corresponds to the set operation in question here, so we could just use the arrow: $A \leftarrow B = A \cup B^\complement$.

We could use a more "rounded" version of an arrow to distinguish it from the implication connective, e.g. the "lollipop" symbol (used for linear implication in linear logic): $B \multimap A = A \cup B^\complement$. (That's \multimap in LaTeX. \multimapinv is the opposite direction, but that appears not to work on stackexchange.)

We could just use a forward slash, mirroring the set difference backslash, so: $A \big{/} B = A \cup B^\complement$. This feels like the obvious solution when you look at the visual symmetries here ($\cup, \cap, \setminus, /$). My only hesitation is that having both forward and back slashes might be a bit hard to parse visually (in the same way that left/right are generally harder to learn and process than up/down).

As the operation in question here is just the complement of the set difference operation, maybe just add an overbar to the set difference symbol: $B \overline{\setminus} A = (B \setminus A)^\complement = A \cup B^\complement$. (This might not look great in LaTeX without a lot of tweaking. Alternatively, imagine a single symbol, like a number 7 mirrored left-to-right; but then one big drawback is that this is not a standard symbol in LaTeX or Unicode.)


问:
Complex Integration Using Residue Theorem with branch cut
How to calculate $$\int_{0}^{\infty }\frac{ln(1+x^{2})}{(1+x^{2})} dx $$ ? How to choose the branch cut and proceed?

答:
First, as $\frac{ln(1+x^{2})}{1+x^{2}}$ is an even function, $\int_{0}^{\infty }\frac{ln(1+x^{2})}{(1+x^{2})} dx = \frac{1}{2} \int_{- \infty}^{\infty }\frac{ln(1+x^{2})}{(1+x^{2})} dx$.

Now, there's a general theorem which says that if $lim_{z \to \infty}zf(z)=0$ then
\begin{equation}
\int_{- \infty}^{\infty }f(x) dx = 2 \pi i \sum_{\substack{a \in Sing(f) \\ a \in \mathbb{H}}}Res(f,a) \ + \ \pi i \sum_{\substack{a \in Sing(f) \\ Im(a) = 0}}Res(f,a)
\end{equation}
where $\mathbb{H}= \{\,z \in \mathbb{C}: Im(z) > 0\,\} $.

You know that $i$ and $-i$ are the only singularities of your function. As $-i \not\in \mathbb{H}$ your problem is reduced to find $Res(f,i)$.


问:
When using a bounded existential quantifier, why can't I use an implication instead of a conjunction when expanding it to an unbounded quantifier?

  ∃
  x
  ∈
  A
  P
  (
  x
  )
  ∨
  ∃
  x
  ∈
  B
  P
  (
  x
  )

=
∃
x
(
x
∈
A
∧
P
(
x
)
)
∨
∃
x
(
x
∈
B
∧
P
(
x
)
)

  (
  ∀
  x
  ∈
  A
  P
  (
  x
  )
  )
  ∧
  (
  ∀
  x
  ∈
  B
  P
  (
  x
  )
  )

  =
  (
  ∀
  x
  (
  x
  ∈
  A
  →
  P
  (
  x
  )
  )
  )
  ∧
  (
  ∀
  x
  (
  x
  ∈
  B
  →
  P
  (
  x
  )
  )
  )

Why does the first expansion turn the expression inside the parentheses to a conjunction, while the second turns the expression into an implication statement?

When should I turn a bounded quantifier to a conjunction of the bound or an implication of the bound?
Does it have to do with the Universal vs existential bound?

I understand why the universal bound expansion is the implication, but I don't understand why the existential qualifier expansion cannot have the implication statement instead of the conjunction.

答:
Does

There is some $A$ which has property $P$

mean

There is something which is in $A$ and has property $P$

or

There is something which, if it's in $A$, has property $P$?

These statements correspond to "$\exists x\in A(P(x))$," "$\exists x(x\in A\wedge P(x)),$" and "$\exists x(x\in A\rightarrow P(x))$," respectively.

Intuitively it should be clear that the third candidate has a "silly" way of being true: as long as there is a single object which is not in $A$ in the first place, the statement "$\exists x(x\in A\rightarrow P(x))$" is true by virtue of the behavior of the material conditional. In particular, "$\exists x(x\in A\rightarrow P(x))$" is automatically true if $A$ is empty (and the universe isn't). This is pretty obviously not the situation described by the first statement! Note that this is clearer if we think explicitly in terms of "truth conditions" (or "falsity conditions") rather than trying to just "guess the meaning" of the statement in question. Complicated logical expressions are a bit unintuitive, and it's often helpful to think of them a bit abstractly.

This leaves the question of why we don't see a "simple duality" between the universal and existential versions. In fact there is a duality, but it's obscured a bit by the notation.

The key duality between $\forall$ and $\exists$ is given via negation: we have $$\forall x\varphi(x)\iff\neg\exists x\neg\varphi(x)\quad\mbox{and}\quad\exists x\varphi(x)\iff\neg\forall x\neg\varphi(x).$$ A duality of exactly the same form is at play with the bounded quantifiers: "$\forall x\in A$" is the same as "$\neg \exists x\in A\neg$," and similarly "$\exists x\in A$" is the same as "$\neg\forall x\in A\neg$." This should make sense when thinking in terms of natural language ("Something in $A$ has property $P$" is the same as "It is not the case that everything in $A$ has property $\neg P$"), and granting this duality we can justify one symbolization of the bounded quantifier in terms of the other.

For example, you're happy with the rephrasing of "$\forall x\in A(P(x))$" as "$\forall x(x\in A\rightarrow P(x))$," and less happy with the rephrasing of "$\exists x\in A(P(x))$" as "$\exists x(x\in A\wedge P(x))$." So let's convince ourselves of this latter equivalence, using the former better-understood equivalence and the above-mentioned duality between the bounded quantifiers. I've broken things into separate hints since I think that's ultimately more useful; it starts, as all good things do, with a (double) double negation:

 By applying double negation elimination twice, once to the "outside" and once to the "inside," the original statement $\exists x\in A(P(x))$ is equivalent to $$\neg(\color{red}{\neg\exists x\in A(\neg}\neg P(x))).$$ This looks messier, but that highlighted bit is quite nice since we can ....

  ... apply the bounded quantifier duality described above! We know that the red "$\neg\exists x\in A\neg$" can be replaced with "$\forall x\in A$," so the above is in fact equivalent to $$\neg\color{red}{(\forall x\in A(\neg P(x)))}.$$ The highlighted bit above can be simplified because it ...

 ... is in terms of a bounded quantifier we already understand! So we know how to unpack it, and we get $$ \color{red}{\neg\forall x}(x\in A\rightarrow\neg P(x)).$$ That bit in red reminds us of ...

 ... the way negation plays with ordinary quantifiers! We now use the original duality between the universal and existential quantifiers to bring that outermost negation inside a quantifier: "$\neg\forall$" is the same as "$\exists\neg$," and so we get $$\exists x(\color{red}{\neg(x\in A\rightarrow\neg P(x))}).$$

After negating the conditional and killing off one last double negation, this gives us, as desired, $$\exists x(x\in A\wedge P(x)).$$

I think the key point, in terms of convincing yourself that there really is a tight parallel between the universal and existential bounded quantifiers, is the result of the fourth step above, that $$\exists x\in A(P(x))\mbox{ means }\exists x(\color{red}{\neg}(x\in A\rightarrow\color{red}{\neg} P(x))).$$ This nicely parallels a fact you already understand, namely that $$\forall x\in A(P(x))\mbox{ means }\forall x(x\in A\rightarrow P(x)),$$ the difference being a careful insertion of two negation signs. It's just that in rewriting the bounded existential quantifier we keep going ... and, while we wind up with something easier to use in the end, this obscures the simple parallel between the two types of bounded quantifier.


问:
Subspace of $M_2(\mathbb{R})$

Let $V = M_2(\mathbb{R})$ and $W = \lbrace  A \in M_2(\mathbb{R}) : A = A^T\rbrace$.

Determine if $W$ is a subspace of $V$.

I am a bit confused about the vector space of $2\times 2$ matrices and how to start over with the transpose function.

答:
Take $A,B\in W$ and arbitrary scalars $a,b\in\mathbb{R}$, then : \begin{align}(aA+bB)^T&=(aA)^T+(bB)^T\\
 &= a^TA^T+b^TB^T\\& = aA+bB\end{align}

so $aA+bB\in W$ and thus it is a vector subspace.

Title: Subspace of $M_2(\mathbb{R})$


问:
Surjective etale morphism from an open subscheme.
I'm looking for an example of a surjective etale morphism from an open subscheme of $\mathbb{A}^d$ (which is not the whole $\mathbb{A}^d$) to $\mathbb{A}^d$. The base field is a finite field or its algebraic closure. I am just wondering whether something like this is possible.

答:
There is an infinitude of examples. Namely, let us assume that $f:\mathbb{A}^d_k\to \mathbb{A}^d_k$ is a non-trivial finite etale cover (e.g. an Artin-Schreier cover) where $k$ is some finite extension of $\mathbb{F}_p$. Choose any closed point $p$ of $\mathbb{A}^d_k$. After possibly taking a finite extension of $k$, we can assume that $f^{-1}(p)$ consists of $n:=\deg(f)$ $k$-points. Let us write $f^{-1}(p)=\{p_1,\ldots,p_n\}$. Then, evidently

$$f:U\to \mathbb{A}^d_k,\qquad U:=\mathbb{A}^d_k-\{p_1\}$$

is an example of what you're after.


问:
a convex closed set of a Banach space contains no closest point
I need to prove that:

For the Banach space $E=C([0,1])$ with $||.||_{\infty}$ and the closed convex set $A=\{f \in E : f(1)=0 \}$, can we find an element $f \in E$ such that $A$ contains no closest point to $f$.

答:
I think this is not true.

Consider the functional $\delta_1 \colon E \to \mathbb R $ defined via $\delta_1(f)=f(1)$ (the dirac functional). Then $ ||δ_1||=1$. Notice that $A=\ker \delta_1$. Thus for all  $ f \in E$
$$d(f,A) = d(f, \ker \delta_1) =\frac{|\delta_1(f)|}{||\delta_1|| } = |f(1)|$$
(see here). Let $ f \in E$ be given. If we pick $g \in E$ such that $g(x)=f(x)-f(1)$ we have that $g \in A $. Then
$$||f-g||= \sup | f(x)-f(x)+f(1)| =|f(1)| = d(f,A)$$
and so the distance is always attained.

Edit: If $X$ is a nonreflexive Banach space (like $C([0,1]))$, then by James's theorem, there exists $x^* \in X^*$ that is not norm attaining (that is, there is no $x \in B_x$ such that $ ||x^*||=|x^*(x)|$. It is not hard to see that the subspace $ V=\ker x^*$ doesn't have the best approximation property. To see this, note that for all $x \in X$
$$d(x,V)= \frac{|x^*(x)|}{||x^*||} .$$
Arguing by contradiction, suppose that there exists $x \in X$ such that $ d(x,V)=||x-v||$ for some $v \in V$. Then, $ |x^*(x-v)| = |x^*(x)|= ||x^*|| \ || x-v||$. In other words,
$$ ||x^*|| = |x^* \left ( \frac{ x-v}{||x-v||}\right )|$$
which is impossible since $x^*$ is not norm
attaining (note that $\frac{ x-v}{||x-v||} \in B_X$)


问:
Closure of infinite set in finite complement topology
I just asked a question to know if a proof using "or" to make a justification was valid logically.

Here is an example of my confusion.

Show that if $A$ is an infinite subset of $X$, then $\bar A=X$ in the  finite complement topology.

Here was my proof. Let $x \in X, U$ be an open set containing $x$.If $U=X$, then $U \cap A \neq \varnothing$. Suppose $U \neq X$. Then $U=X-S$ for some finite set $S$. Consider $A \cap (X-S)$. Then this intersection is nonempty (here is where I am unsure if this is correct) otherwise $A$ or $X-S$ would have to be finite.(A would be finite since $A \subset S$ if the intersection is empty, however I noticed the second statement is not necessarily true). So $U \cap A \neq \varnothing$ for each open set $U$, and since $x$ was arbitrary, $\bar A=X$.

So the statement "otherwise,..." is my justification for the contradiction that will arise if $(X-S) \cap A = \varnothing$. So I want to know if including the not necessarily true statement $X-S$ would have to be finite, ruins the validity of my proof.

答: Why not say: If $A$ is infinite, it cannot be contained in a finite subset and the only infinite closed subset is $X$ itself. So $\overline{A}=X$, and $A$ is dense.


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

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

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

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

答:
Rather basic, but it was surprising for me:

For any matrix, column rank = row rank.

LastEditorDisplayName: anon


问:
Integral over a ball
Let $a=(1,2)\in\mathbb{R}^{2}$ and $B(a,3)$ denote a ball in $\mathbb{R}^{2}$ centered at $a$ and of radius equal to $3$.

Evaluate the following integral:
$$\int_{B(a,3)}y^{3}-3x^{2}y \ dx dy$$

Should I use polar coordinates? Or is there any tricky solution to this?

答:
Here is the brute force approach.
Let $f(x,y) = y^3-3x^2y$.
Shift so the center of the ball is at the origin and go to polar coordinates.
The integral is then<br>
$$\int_0^{2\pi}\int_0^3 f(r\cos\theta+1,r\sin\theta+2)
r dr d\theta.$$
The integral is straightforward but tedious, with value $18\pi$ as found by @robjohn in his much slicker answer. </p>

Title: Integral over a ball

Body:
Here is the brute force approach.
Let $f(x,y) = y^3-3x^2y$.
Shift so the center of the ball is at the origin and go to polar coordinates.
The integral is then

$$\int_0^{2\pi}\int_0^3 f(r\cos\theta+1,r\sin\theta+2)
r dr d\theta.$$
The integral is straightforward but tedious, with value $18\pi$ as found by @robjohn in his much slicker answer.


问:
Notation for a proof containing group actions
I am writing a proof to Sylow's 2nd Theorem. I have done it using group actions. My tutor told me it would be wise to introduce notation to make it more clear what group action I am talking about. I am unsure on how to do this. Do I just assign letters to each group action?</p>

答:
There is not perfect choice of notation. Look into the lecture notes or ask your tutor/other students.

I like this notation for group actions:
For a left action of $g \in G$ acting on an element $x$ you could write
$$
g \rhd x.
$$

To differentiate between different group actions, you could introduce for example $\rhd_A$ and $\rhd_B$ to make clear which group action you refer to.


问:
calculate this integral $\int^{8}_{0} \pi y(y^{\frac{4}{3}}+1)dy$
I'm trying to calculate the following integral, however, I'm stuck on how to proceed after this point:

$\int^{8}_{0} \pi y(y^{\frac{4}{3}}+1)dy$

$\pi\int^{8}_{0}  y(y^{\frac{4}{3}}+1)dy$

Multiply by conjugate
2. $\pi\int^{8}_{0}  y(y^{\frac{4}{3}}+1)dy$ = $\pi\int^{8}_{0} y(\frac{y^{\frac{8}{3}}-1}{y^{\frac{4}{3}}-1})dy$

Where to go from here

Body:
Hint:$$\int^{8}_{0} \pi y(y^{4/3}+1)\,dy$$
$$\int (y^{7/3}+y)\,dy=\frac{3 y^{10/3}}{10}+\frac{y^2}{2}+C$$


问:
Complex Integration and trigonometric functions
I am just begining with complex integration. Please help with this.
Evaluate $\int_{-i\pi}^{i\pi}\cos z dz$ where $z$ is a complex number.

This is my effort,
I know that $\int_{C}{f(z)dz} = \int_a^b f(z(t))z'(t)dt$ where C:[a,b] is a complex curve and $a \leq t \leq b$

Step one would be to determine z(t) from the path of integration and this is the toughest part for me.

Once I have obtaine z(t) I hope the rest would be easy

答:
$f(z) = \cos(z)$ is the derivative of $F(z) = \sin(z)$. Therefore, for any path $C$ joining $a=-i\pi$ and $b= i\pi$ is
$$
\int_{C}{f(z)dz} = \int_a^b f(z(t))z'(t)dt = \int_a^b \frac{d}{dt}(F(z(t)) \,  dt = F(b) - F(a) \\
= \sin(i\pi) - \sin(-i \pi) = 2 \sin(i\pi)\, .
$$

Generally, $\int_{C}{f(z)dz}$ depends only on the endpoints of $C$ if $C$ is a path in a simply-connected domain $D$ where $f$ is holomorphic, that is the contents of Cauchy's integral theorem.


Title: calculate this integral $\int^{8}_{0} \pi y(y^{\frac{4}{3}}+1)dy$

Body:
I'm trying to calculate the following integral, however, I'm stuck on how to proceed after this point:

$\int^{8}_{0} \pi y(y^{\frac{4}{3}}+1)dy$

$\pi\int^{8}_{0}  y(y^{\frac{4}{3}}+1)dy$

Multiply by conjugate
2. $\pi\int^{8}_{0}  y(y^{\frac{4}{3}}+1)dy$ = $\pi\int^{8}_{0} y(\frac{y^{\frac{8}{3}}-1}{y^{\frac{4}{3}}-1})dy$

Where to go from here

Body: Apply power rule : \begin{aligned}\int_{0}^{8}{\pi y\left(y^{\frac{4}{3}}+1\right)\mathrm{d}y}&=\pi\int_{0}^{8}{y^{\frac{7}{3}}\,\mathrm{d}y}+\pi\int_{0}^{8}{y^{1}\,\mathrm{d}y}\\ &=\pi\left[\frac{y^{\color{red}{1+}\frac{7}{3}}}{\color{red}{1+}\frac{7}{3}}\right]_{0}^{8}+\pi\left[\frac{y^{\color{red}{1+}1}}{\color{red}{1+}1}\right]_{0}^{8}\\ &=\frac{3\pi}{10}\times 2^{10}+\frac{\pi}{2}\times 8^{2}\\ \int_{0}^{8}{\pi y\left(y^{\frac{4}{3}}+1\right)\mathrm{d}y}&=\frac{1696\pi}{5}\end{aligned}


问:
Intersection of open dense subsets in a complete metric space is nonempty?
The proof for Baire category theorem in my book assumes that the intersection of dense subsets of a complete metric space is nonempty. Why can this be assumed, why is it true?</p>

答:
It's not true the way you stated it (think $\Bbb Q$ and the irrationals $\Bbb P$ in $\Bbb R$, e.g.)

But two (or finitely many) open dense sets always intersect, because dense sets intersect all (non-empty) open sets... In any space the finite intersection of finitely many dense open sets is dense and open, Baire strengthens this to countably infinite intersections in some spaces (we lose openness of the intersection, but keep dense-ness).