问:
Question about example 7.1.7 Alaca/Williams.
So they are trying to prove that given $K=\mathbb{Q(\sqrt{2}+i)}$:

$O_K= \mathbb{Z}+\mathbb{Z}\sqrt{-1} + \mathbb{Z}\sqrt{2} + \mathbb{Z}(\frac{1}{2}(\sqrt{2}+\sqrt{-2}))$

You have that $K$'s subfields are $\mathbb{Q},\mathbb{Q}(\sqrt{-1}),\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{-2})$

Now,given $\alpha \in O_K$, if $\alpha$ belongs to any of the subfields, you can easily show that $\alpha \in \mathbb{Z}+\mathbb{Z}\sqrt{-1} + \mathbb{Z}\sqrt{2} + \mathbb{Z}(\frac{1}{2}(\sqrt{2}+\sqrt{-2}))$

Suppose that $\alpha$ does not belong to any of the subfields:

Now here is what I do not get, they state that since $\alpha \in K$, you ca express it as:

$\alpha = a + b i+ c\sqrt{2}+ d\sqrt{2}i$ with $a,b,c,d \in \mathbb{Q}$

(Why? Is this a general thing? Is it because the extension is Galois?)

And then they proceed with my second doubt, they state that the conjugates of $\alpha$ are:

$\alpha' = a - b i+ c\sqrt{2}- d\sqrt{2}i$

$\alpha'' = a + b i-c\sqrt{2}- d\sqrt{2}i$

$\alpha''' = a - b i- c\sqrt{2}+ d\sqrt{2}i$

I don't get why this is? I mean in this particular case the conjugate are directly related to the subfields, but I don't get how you get to those equations for the conjugates. This could just be a lot of operating though, this book has a habit of not showing any of the work (Although I get why, some of the work ommited is not at all trivial, for example, finding the subfields was a hassle, they just invoke them. And I am trying to do the work for this examples so when I get to do excercises I will have a better idea on how to solve them)

答:
Okay so the first step is to convince yourself that $K = \Bbb Q(\sqrt{2},i)$, if that hasn't been done in the mentioned argument. This isn't hard, but it's far from obvious if you haven't done it before, so let's delve into it.

Note that since $\sqrt{2}+i \in \Bbb Q(\sqrt{2},i)$, we already have $K \subset \Bbb Q (\sqrt{2},i)$. To achieve the equality, we can show that $\sqrt{2},i \in K$. But moreover, if we show that one of these is in $K$, since their sum is already an element of $K$, we get the other for free. So, by direct computation:

$$
K \ni (\sqrt{2}+i)^3 = (1+\sqrt{2}i)(\sqrt{2}+i) = 3i
$$

hence $i \in K$ and $K = \Bbb Q(\sqrt{2},i)$.

Thus we have a tower of extensions $K = \Bbb Q(i)(\sqrt{2}) - \Bbb Q(i) - \Bbb Q$, and there's a general way to compute a $k$-basis of $E$ if it lies in a tower $E - F -k$. Namely: consider a $k$-basis $\{x_i\}$ of $F$, an $F$-basis $\{y_j\}$ of $E$, and then do all possible products $\{x_iy_j\}_{i,j}$.

Moreover, when the extension is already in the form $\Bbb Q(\alpha)$, it's not hard to show that a $\Bbb Q$-basis is $\{1,\alpha,\ldots, \alpha^{d-1}\}$ with $d = \deg m(\alpha, \Bbb Q)$.

Thus as $\Bbb Q$-vector spaces we have $\Bbb Q(i) = \Bbb Q 1 + \Bbb Q i$ and $\Bbb Q(\sqrt{2}) = \Bbb Q 1+ \Bbb Q\sqrt{2}$, hence a $\Bbb Q$ basis for $K$ consists of all possible products of basic elements

$$
B = \{1,\sqrt{2},i,\sqrt{2}i\}.
$$

That solves the first question. As for the second: via our new characterization of $K$, we know that automorphisms are determined by the images of $i$ and $\sqrt{2}$, these have quadratic polynomials, and so their images must be $\pm i$ and $\pm \sqrt{2}$ respectively.

Hence we have $4$ different admissible morphisms, each corresponds to a different conjugate.

Edit: we did not need any Galois theory but the extension is indeed Galois, as it is the splitting field of $\{X^2-2,X^2+1\}$ and its Galois group then must be the Klein four group. Explicit generators are $$\sigma(a+ib+c\sqrt{2}+di\sqrt{2}) = a+ib-c\sqrt{2}-di\sqrt{2}$$ and $$c(a+ib+c\sqrt{2}+di\sqrt{2}) = a-ib+c\sqrt{2}-di\sqrt{2},$$ i.e. the map that sends $i\mapsto i, \sqrt{2} \mapsto - \sqrt{2}$ and complex conjugation.


问:
Confusion about the definition of maximal ideal
In my book, the definition of a maximal ideal is as follows:

Let $R$ be a commutative ring. A maximal ideal of $R$ is an ideal $I$ such that:

$I \neq R$.

There exists no ideal $J$ of $R$ such that $I \subsetneq J \subsetneq R$.

When trying to prove the following statement:

$I$ is a maximal ideal of $\mathbb{Z} \implies I = (p)$ with $p$ a prime number.

I came up with an error in my reasoning, but do not understand at all why it is an error. Because all ideals of $\mathbb{Z}$ are of the form $(n)$ with $n \in \mathbb{Z}$, I know that $I = (n)$. So I tried formulating the definition of maximal ideal as follows:

$\forall n' \in \mathbb{Z}: (n') = \mathbb{Z} \ $ or $ \ (n') \subset (n)$.

But this must obviously be false because then $n$ would be a divisor of all $n' \in \mathbb{Z}$, so that $n = 1$ and thus $(n) = \mathbb{Z}$. This cannot be the case since $I$ is a maximal ideal. Where did my reasoning go wrong ?

答:
If you follow the definition of maximal ideals, then for $\mathbb{Z}$, you have
that $(n)$ is a maximal ideal if and only if

$(n)\ne\mathbb{Z}$;

there is no ideal $(m)$ of $\mathbb{Z}$ such that $(n)\subsetneq (m)\subsetneq\mathbb{R}$.

The second condition does not imply that for every $m\in\mathbb{Z}$ (and $m\ne 1$), $(m)\subset (n)$.

For two integers $m$ and $n$, exactly one of the following is true:
$$
m<n,\quad m=n,\quad m>n
$$
Hence if $m\ge n$ is not true, you must have $m<n$.

However, for two ideals $(m)$ $(n)$, it is possible that none of the following is true
$$
(m)\subset(n),\quad (m)=(n),\quad (m)\supset(n)
$$

Consider for instance the example $m=3$ and $n=5$.

ContentLicense: CC BY-SA 4.0


问:
Combinations sum : $\sum (k+1)^2 \binom{n}{k}$
$$1^2 + 2^2\binom{n}{1} +\ldots+(n+1)^2\binom{n}{n}=?$$
I tried this splitting
$$\sum_{k=0}^n (k+1)^2 \binom{n}{k}=\sum_{k=0}^n (k^2+2k+1) \binom{n}{k}= n^2 \cdot 2^n + 2n \cdot 2^n + 2^n $$

It didn't work for most cases , I then just tried
$$2^n \cdot (n+1)^2  $$
but it also didn't work . So I'm most likely doing it completely wrong or I'm approaching it in the wrong way. Thank you in advance.

答:
Splitting it up is a good way to start.  Then use the binomial theorem $$(1+x)^n=\sum_{k=0}^n\binom nk x^k$$

Differentiate both sides to get $$n(1+x)^{n-1}=\sum_{k=0}^n\binom nk kx^{k-1}$$ and set $x=1$.  Then multiply both sides of the last equation by $x$ and differentiate again.


问:
Prove that semi-simple rings are Dedekind-finite
Just to be consistent with the terminology, let me define the words I'm using.

A module $M$ is simple if it has no proper non-zero submodules, and semi-simple if it can be written as a direct sum $M = \bigoplus_i A_i$ where the $A_i$'s are simple submodules.

A ring $R$ is semi-simple if the left $R$-module $_RR$ is semi-simple. This means that $R$ is a direct sum of "minimal" ideals. I want to prove that such rings are also Dedekind-finite, i.e. if $a,b\in R$ and $ab = 1$, then $ba = 1$. I want to do this without invoking Artin-Wedderburn. (EDIT: We have "semi-simple $\implies$ Noetherian $\implies$ Dedekind-finite", but I don't know how to prove the first implication without using composition series. It seems to me that my problem can be solved without using the fact that semi-simple rings are Noetherian.)

So far, I can show that a semi-simple ring $R$ is a direct sum of finitely many minimal ideals: if $R = \bigoplus_\lambda A_\lambda$ where each $A_\lambda$ is a minimal ideal of $R$, then write $1 = e_1 + \cdots + e_n$ where $e_i \in A_{\lambda_i}$. It follows that $r = re_1 + \cdots + re_n$ for any $r \in R$, so $R = \bigoplus_{i=1}^n A_{\lambda_i}$. Let's abbreviate the notation by setting $A_i := A_{\lambda_i}$.

Alright, now suppose $a,b\in R$ are such that $ab = 1$. We want to show that $ba = 1$. I have attempted to solve the problem in the case that $a \in A_i$ for some $i$. First, we get $ab = 1 \implies bab = b \implies (ba - 1)b = 0$. Since $b\neq 0$, we'd be done if $R$ is a domain. Even if $R$ is not a domain, I think we can show that we must have $ba - 1 = 0$, although my approach doesn't work.

Define an $R$-module homomorphism $F: R \rightarrow R$ by $F(r):= rb$. Suppose that $a\in A_i$ for some $i$. Then $F$ is non-zero on $A_i$, since $F(a) = ab = 1 \neq 0$. (Here's where we use that $a \in A_i$.) If we restrict $F$ to $A_i$, the kernel of this restriction is proper (since $F$ is non-zero) and therefore the kernel must be $\{0\}$ (because $A_i$ is minimal). Therefore the restriction of $F$ to $A_i$ is one-to-one.

Note that $F(ba - 1) = (ba - 1)b = 0$ as mentioned above. Since $F$ is one-to-one, we must have $ba - 1 = 0$, as required.

Problem with this last argument: we only know that $F$ is one-to-one on $A_i$, but we don't know that $ba - 1 \in A_i$. In fact, if $ba - 1 \in A_i$, then $1 = ba + -(ba - 1) \in A_i$ since $a \in A_i$. Thus $A_i$ is not proper, so $_RR = {_R}A_i$ is a simple $R$-module, which means that $R$ is a division ring (it's easy to prove that "$_RR$ simple $\implies$ $R$ is a division ring"). Since division rings are clearly Dedekind-finite, the problem is trivial in this case.

Anyway, so my attempt using a homomorphism $F$ doesn't really work. But it might be possible to salvage the approach by modifying it slightly. Even then, this would only give the result when $a \in A_i$ for some $i$, but hopefully this special case can be used to prove the more general claim.

Reminder that I want to prove that semi-simple rings are Dedekind-finite without using the Artin-Wedderburn theorem.

Cheers!

答:
Prepended: OK, here is an elementary argument based only on the simple submodules. We begin by assuming $R$ is a direct sum of $n$ minimal right ideals, and we know that $n$ is unique.

If $ab=1$, then the homomorphism $x\mapsto bx$ from $R\to R$ is injective. Thus $bR\cong R$ as $R$ modules. Find a complement $C$ to $bR$ so that $R=bR\oplus C$. If $ba\neq 1$, then $bR\neq R$ and $C\neq \{0\}$.

But look: $R\cong R\oplus C$, and since $C$ is nonzero it contributes at least one simple submodule to the sum. But that would mean that the right hand side of the equation is a sum of more than $n$ simple modules, whereas the left is the sum of $n$ simple modules. This isn't possible, so $bR$ must be $R$, showing that $ba=1$.

As soon as you have convinced yourself $R$ is a direct sum of finintely many minimal right ideals, then you immediately have that $R$ has a finite composition series, hence it is right Artinian (and right Noetherian.)

I think it is even easier to use the Artinian property than the Noetherian property.

Suppose $ab=1$. Clearly $b$ is left invertible. Consider the descending chain $bR\supseteq b^2R\supseteq\dots$. Since semisimple rings are right (and left) Artinian, this chain must stabilize, say at $b^nR=b^{n+1}R$. This implies $b^n=b^{n+1}r$ for some $r\in R$. Multiplying on the left by $a$ $n$ times, you get $1=br$, showing that $b$ is right invertible.

Thus $a=abr=r$, and so $ba=1$.

Note: There is a way to prove a semisimple ring is Dedekind finite independently of the chain conditions. Since semisimple rings are self-injective on both sides, they are Dedekind finite. Unfortunately, this route seems even more complicated than the methods above, but still, it is interesting to see that you can get Dedekind finiteness "two different ways" for semisimple rings.

ContentLicense: CC BY-SA 3.0


答:
Someone has proposed that you thought that the ideals of $\mathbb Z$ are linearly ordered, but I have a feeling that the mistake was actually simpler than that, and "upstream" so to speak. I think it is just an error in quantification.

It should read

“for all $n’$, if $(n’)\supseteq (n)$ it follows that A or B.”

and not

“for all $n’$, it follows that A or B.”

A stands for $(n')=(n)$ and B stands for $(n')=R$, but I wanted to disguise that the conditions themselves are not important. What was important was the missing "if."

The second one implies that all ideals of $R$ are comparable to $(n)$ (which can be exaggerated to "all ideals are in a chain" as the other solution thought.)

To connect this with your original definition, which was correct:

There exists no ideal $J$ of $R$ such that $I\subsetneq J \subsetneq R$

This would be logically equivalent to

"If $I\subseteq J \subseteq R$, then $J=I$ or $J=R$."

The misstep you wrote looked like this:

"If $J \subseteq R$, then $J=I$ or $J=R$."

Title: Confusion about the definition of maximal ideal

Body:
In my book, the definition of a maximal ideal is as follows:

Let $R$ be a commutative ring. A maximal ideal of $R$ is an ideal $I$ such that:

$I \neq R$.

There exists no ideal $J$ of $R$ such that $I \subsetneq J \subsetneq R$.

When trying to prove the following statement:

$I$ is a maximal ideal of $\mathbb{Z} \implies I = (p)$ with $p$ a prime number.

I came up with an error in my reasoning, but do not understand at all why it is an error. Because all ideals of $\mathbb{Z}$ are of the form $(n)$ with $n \in \mathbb{Z}$, I know that $I = (n)$. So I tried formulating the definition of maximal ideal as follows:

$\forall n' \in \mathbb{Z}: (n') = \mathbb{Z} \ $ or $ \ (n') \subset (n)$.

But this must obviously be false because then $n$ would be a divisor of all $n' \in \mathbb{Z}$, so that $n = 1$ and thus $(n) = \mathbb{Z}$. This cannot be the case since $I$ is a maximal ideal. Where did my reasoning go wrong ?


问:
Prob. 6 (d), Chap. 1, in Baby Rudin, 3rd ed: How to complete this proof?
Here is Prob. 6, Chap. 1, in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Fix $b>1$.

Problem 6(a):

Let $m$, $n$, $p$, $q$ be integers such that $n>0$, $q>0$, and $r = m/n = p/q$. Then [I've managed to] prove that
$$ b^{m/n} = b^{p/q}.$$ So we can reasonably define $b^r$ as $$b^r \colon = \sqrt[n]{b^m}.$$

Problem 6(b)

[From this definition we can] prove that $$ b^{r+s} = b^r \cdot b^s, $$ where $r$ and $s$ are any rational numbers.

Problem 6(c):

Now for a rational number $r$, let the set $B(r)$ be defined as follows:
$$B(r) \colon= \{ \, b^t \colon \, t\in \mathbb{Q}, \, t \leq r \, \}. $$
Then it is clear that $$b^r = \sup B(r).$$ So for every real $x$, we can define $b^x$ as follows: $$b^x \colon= \sup B(x) = \sup \big\{ \, b^t \, \colon \, t \in \mathbb{Q}, \, t \leq x \, \big\}. $$

Problem 6(d):

Using this definition, (we are required to) prove that, for every pair of real $x$ and $y$, the equation $$b^{x+y} = b^x \cdot b^y$$ holds.

My Attempt:

If $r$, $s \in \mathbb{Q}$ such that $r \leq x$ and $s \leq y$, then $r+s \in \mathbb{Q}$ also and $r+s \leq x+y$ so that we can write
$$ B(x) \cdot B(y)  = \{ \, b^r \cdot b^s \, \colon \, r \in \mathbb{Q}, s\in \mathbb{Q}, \, r \leq x, s\leq y \, \} \subseteq \{ \, b^t \, \colon \, t \in \mathbb{Q}, \, t \leq x+y \, \} = B(x+y),$$ and hence
$$b^x \cdot b^y = \sup B(x) \cdot \sup B(y) = \sup \{ \, b^r \, \colon \, r \in \mathbb{Q}, \, r \leq x \, \} \cdot \sup \{ \, b^s \, \colon \, s \in \mathbb{Q}, \, s \leq y \, \} = sup \{ \, b^{r+s} \, \colon \, r \in \mathbb{Q}, \, s\in \mathbb{Q}, \,  r \leq x, \, s \leq y \, \} \leq \sup \{ \, b^t \, \colon \, t \in \mathbb{Q}, \, t \leq x+y \, \} = \sup B(x+y) = b^{x+y}.$$
Here we have used the following definition: Given two non-empty sets $U$ and $V$, say of real numbers, we define the set $U \cdot V$ as follows:
$$U \cdot V \colon= \{ \, u \cdot v \, \colon \, u \in U, \, v \in V \, \}.$$

And we have also used the fact that if $W$ and $Z$ are two non-empty bounded above subsets of the set of positive real numbers such that $W \subseteq Z$, then we must have
$$ \sup W \leq \sup Z.$$

So far, we have shown that $$b^x \cdot b^y \leq b^{x+y}.$$
Now how to prove the reverse inequality using the machinery developed above?
That is, how to prove that $$b^{x+y} \leq b^x \cdot b^y?$$

答:
Right now, there are five answers: one that isn't actually an answer, two that use concepts like limits and continuity the Rudin book hasn't tackled by this point, and two that are correct but rely on redefining the set $B(x)$ from Rudin's definition $B(x) = \{b^t: t \in \mathbb{Q}, t \leq x\}$ to $\{b^t: t \in \mathbb{Q}, t < x\}$.

This redefinition is okay because $sup\ \{b^t: t \in \mathbb{Q}, t \leq x\} = sup\ \{b^t: t \in \mathbb{Q}, t < x\}$.  I thought it might be useful to show why this is true using the tools we have so far.

Let's define $B(x)$ as Rudin does, with $t\leq x$.  Let's define $C(x) = \{b^t: t \in \mathbb{Q}, t < x\}$.  I will show $sup\ B(x) = sup\ C(x)$.

Note that $C(x) \subset B(x)$. Thus, since $sup\ B(x) > y$ for all $y \in B(x)$, $sup\ B(x) > y$ for all $y \in C(x)$, and $sup\ B(x)$ is an upper bound of $C(x)$.

Pick $z < sup\ B(x)$.  We know $sup\ B(x) >0$ since $b^t>0$ for all $t\in \mathbb{Q}$.  Thus we can pick some positive real $w$ with $z<w<sup\ B(x)$.  Then $w$ is not an upper bound of $B(x)$ so there exists $p\in\mathbb{Q}$ with $p\leq x$ and $b^p>w$.

By the work in Problems 7a-c, we know there exists a positive integer $b$ such that $b^n < \frac{b^p}{w}$.  (Note that this is not a circular reference, since the work in 7a-c does not reference anything in Problem 6.)  Then $\frac{1}{b^n} > \frac{w}{b^p}$.  Then $b^p\frac{1}{b^n} > b^p\frac{w}{b^p}=w$.  Since $w>z$, we have $b^p\frac{1}{b^n} > z$.

Note $b^p\frac{1}{b^n} = b^{p-1/n}$, $p-\frac{1}{n}\in\mathbb{Q}$, and $p-\frac{1}{n}<p\leq x$.  Then $b^{p-1/n} \in C(x)$, and $z$  not an upper bound of $C(x)$.

Thus $sup\ B(x) = sup\ C(x)$, and it is equivalent to show that $sup\ C(x+y) = sup\ C(x)sup\ C(y)$ as two others have done above.


问:
Continuous Linear Mapping $C[0,1]\rightarrow C[0,1]$
Show that $L(f)(x)= \int_0^x f(t) dt $ is a continuous linear mapping from $C[0,1]$ into itself.

Do I only have to show that the operator is bounded? How to do I explicitly choose my $M$ such that $\|L \ f\|<M\|f\|$?

答:
Yes, continuous is equivalent to bounded, among linear mappings between normed spaces.

Choose $M$ to be the length of the given interval, now $M=1$ will do it.

 $\left| \displaystyle\int_0^x f(t)\,dt\right| \ \le \ \int_0^x|f(t)|\,dt\ \le \ \int_0^1|f(t)|\,dt\  \le\ \int_0^1\,\|f\|_\max = 1\cdot \|f\|_\max$


问:
If A and B are 6x6 matrices with det(A) = -10 and det(B) =5, how do i find det(-A), det(3A) and especially det(A^t B^-1)?
So i have these two matrices where all i know is that they are 6 by 6 and I have their determinants, how would I work out the above? I know I need to use, or can use, the properties of determinants but I feel like there might be more to it than it being as simple as multiplying by -1 or 3. Thanks</p>

答: Use some of these properties of the determinant to help you out.</p>


问:
Calculate coordinates of triangle corners when area is known
The three corners of triangle have following coordinates:

$A = (1,0,0)$

$B = (0,1,0)$

$C = (0,0,z)$

What is the positive value of $z$ when area of triangle is $4$?

I have tried:

$AB = [-1,1,0]$

$AC = [-1,0,z]$

$\begin{bmatrix}i & j & k\\-1 & 1 & 0\\-1 & 0 & z\end{bmatrix} = k + jz + iz$

$(k + jz + iz) / 2 = 4$

$k + jz + iz = 8$

But I cannot solve $z$. Could I have some advice?

答:
Hint.

You can't solve the equation since you incorrectly equating a vector with a scalar.

Instead, you should use the "length" of the vector on the left.


答:
The area $\mathcal{A}$ of a triangle $ABC$ can be obtained with
$$\mathcal{A}=\frac{1}{2}\left|\overrightarrow{AB}\times\overrightarrow{AC}\right|.$$
Then you should write
$$\frac{1}{2}\sqrt{z^2+z^2+1^2}=4.$$
EDIT:
\begin{align*}
\frac{1}{2}\sqrt{z^2+z^2+1^2}=4 & \Longrightarrow\sqrt{2z^2+1}=8\Longrightarrow\\
& \Longrightarrow2z^2+1=8^2\Longrightarrow\\
& \Longrightarrow2z^2=63\Longrightarrow\\
& \Longrightarrow z=+\sqrt{\frac{63}{2}}.
\end{align*}

Title: Calculate coordinates of triangle corners when area is known

Body:
The three corners of triangle have following coordinates:

$A = (1,0,0)$

$B = (0,1,0)$

$C = (0,0,z)$

What is the positive value of $z$ when area of triangle is $4$?

I have tried:

$AB = [-1,1,0]$

$AC = [-1,0,z]$

$\begin{bmatrix}i & j & k\\-1 & 1 & 0\\-1 & 0 & z\end{bmatrix} = k + jz + iz$

$(k + jz + iz) / 2 = 4$

$k + jz + iz = 8$

But I cannot solve $z$. Could I have some advice?


问:
Lattice points on $x+y=n$ always visible from the origin.
Let $m \in \mathbb{Z_{\gt 0}}$ and $\{n \in \mathbb{Z_{\gt 0}}:n\mod 10\}$  = $\{1, 3, 7, 9\}$.

Define

$\Large a_{m} = \frac {5^m-3}2$

Claim

Any lattice points $(x,y)$ of the form

$\Large(\left \lfloor{\frac{n}2}\right \rfloor + a_{m} + 2,\left \lfloor{\frac{n}2}\right \rfloor - a_{m} - 1)$

where $x+y=n$ are always visible from the origin.

Proof

Let

\begin{align*}
x=\frac {n}2 + a_{m} + 2\\
y=\frac {n}2 - a_{m} - 1
\end{align*}

Point $(x,y)$ is coprime when

\begin{align*}
gcd(x,y)=1 \iff gcd(x+y,x-y)=1
\end{align*}

Calculating

\begin{align*}
gcd(x+y,x-y) &= gcd(n,\frac {n}2 + a_{m} + 2 - (\frac {n}2 - a_{m} - 1))\\
&=gcd(n,2a_{m} + 3)\\
&=gcd(n,5^m)\\
&=1\ or \ 5^m\\
\end{align*}

However, the only prime divisor of $5^m$ is $5$ and that is not a factor of $n$, so

\begin{align*}
gcd(x+y,x-y) &= 1\\
\end{align*}

Calculating

\begin{align*}
gcd(x, y) &= gcd(x−y, y),\ if\ (x > y)\\
&= gcd(\frac {n}2 + a_{m} + 2 - (\frac {n}2 - a_{m} - 1),\frac {n}2 - a_{m} - 1)\\
&=gcd(2a_{m} + 3,\frac {n}2 - a_{m} - 1)\\
&=gcd(5^m,\frac {n}2 - a_{m} - 1)\\
&=5^m\ or \ 1\\
\end{align*}

However, for any integers

\begin{align*}
\frac {gcd(x+y, x-y)}{gcd(x,y)}
\end{align*}

because anything that divides both $x$ and $y$ divides their sum and difference.

Substituting yields

\begin{align*}
\frac {1}{gcd(x,y)}
\end{align*}

Therefore

\begin{align*}
gcd(x,y) = 1
\end{align*}

$\blacksquare$

Examples

$\xrightarrow{\text{(n = 31, m = 1)}} (18, 13)$

$\xrightarrow{\text{(n = 31, m = 2)}} (28, 3)$

$\xrightarrow{\text{(n = 77777, m = 1)}} (38891, 38886)$

$\xrightarrow{\text{(n = 77777, m = 2)}} (38901, 38876)$

$\xrightarrow{\text{(n = 123456789, m = 10)}} (66611207, 56845582)$

$\xrightarrow{\text{(n = 123456789, m = 11)}} (86142457,37314332)$

$\xrightarrow{\text{(n = 9876543213, m = 12)}} (5060341919, 4816201294)$

$\xrightarrow{\text{(n = 9876543213, m = 13)}} (5548623169, 4327920044)$

Note: I only know basic high-school level maths.

答:
You are overthinking this.

First, for any integers
$$
\gcd(x,y) \text{ divides } \gcd(x+y, x-y)
$$
because anything that divides both $x$ and $y$ divides their sum and their difference.

In this question, $x -y$ = $5^m$. It's a little harder to see that $x+y = n$. (Since $n$ is odd, rounding down $n/2$ and doubling yields $n-1$).

The only prime divisor of $5^m$ is $5$ and $5$ is not a factor of $n$ so $\gcd(n, 5^m) = 1$.

Thus
$$
\gcd(x,y) \text{ divides } \gcd(n, 5^m) = 1
$$
so $x$ and $y$ are relatively prime.

(Note: be a little careful with odds and evens if you want to use this trick again. $\gcd(5,3) = 1$ but $\gcd(5-3,5+3) = 2$.)


问:
CAS with a standard language
I hope this question is suitable for the site.

I recently had to work with Mathematica, and the experience was, to put it kindly, unpleasing. I do not have much experience with similar programs, but I remember not liking much Matlab or Maple either. The result is that I am a mathematician who likes programming, but I never managed to learn how to work with a computer algebra system.

Does there exist a CAS which can be programmed using a standard language? I guess the best thing would be just an enormous library of mathematical algorithms implemented for C or Python or whatever.

I know SAGE is based on Python, but as far as I understand (which is not much) it just collects preexisting open source software, so (I assume) one has to learn how to use a new tool for every different problem.

答:
If you don't want to install all of Sage (which is quite big), you could perhaps try one of its components, SymPy, on its own to be begin with.

If you use C++ maybe GiNaC is useful (I haven't tried it).

ContentLicense: CC BY-SA 2.5

Title: CAS with a standard language


问:
A naive inquiry of Godel's incompleteness--or why does mathematics need proofs of unprovability?
My question stems from reading Swetz, 1994 (mostly excerpts from the journal Mathematics Teacher) and Berlinski, 2005 (a popular book on 10 most important mathematical breakthroughs in history).

1) I'm having difficulty understanding why Godel's theorem (if I understand what I have read) requires both formulas provable and unprovable. My naive concept of mathematics is a system containing only "positive" theorems and laws and I imagine the same for science. That the scientific method culls anything proven wrong, so as not to carry failures as baggage together with its laws. Where failure, mistakes, and problems are separated as historical study or challenge. Why do unprovable statements need be admitted to a proper system?

In anticipation of my enlightenment, I have a few follow-up Qs,

2) Its my understanding the Hilbert project meant to take the point of view that mathematics and formalism needed to take a meta-point of view for the purpose of separating formulas (as symbols) from the discussion (in natural language) about mathematics. If these unprovable formulas are necessary to prove other provable proofs, then why not segregate them to yet another meta-level?

3) What I gather about Godel's argument from these sources includes a point about mathematics' use of symbols (representing numbers, variables for formulas, variables for sets and sets of sets, etc.). Then is it his argument to suggest that where a proof can be admitted to the system which can be neither proved or disproved, the substitution of variables will propagate this error?

Answers need not be technical, only sufficient to sort my "Godel baggage".

Chris

PS. I found this January post very useful for its references and look forward to finding a volume for my naive appetite for the history of formalism:Understanding Gödel's Incompleteness Theorem---

答:
When people set about formalizing Mathematics, they did it this way:

List the symbols you're allowed to use to make mathematical statements,

Give some rules to distinguish "well-formed formulas" (like $3+4=6$) from other strings of symbols (like $3\div\gt4=$),

List the axioms (a finite list),

List the "rules of inference" (the methods you are allowed to use to get from one well-formed formula to another).

The theorems are then the well-formed formulas you can get to, starting from the axioms and using the rules of inference; they are the things you can prove.

The naive hope was that for every well-formed formula $P$, there would either be a proof of $P$, or a proof of the negation of $P$ (but not both).

What Godel proved was that if your axioms and rules of inference are strong enough to do ordinary arithmetic, then the naive hope cannot be realized. Either the system is inconsistent (meaning that you can prove every statement, including the negation of statements you can prove), or it is incomplete (meaning there are well-formed formulas $P$ such that you can't prove $P$ and you can't prove the negation of $P$).

I don't understand what you mean by "admitting unprovable statements to a proper system." We don't accept them as theorems; we note that neither they nor their negations are theorems. What would you have us do with unprovable statements?


问:
why $f(z)= \frac{(-e_1 + 2e_2)}{\sqrt 5}?$
I have some confusion in Kreyszig book

My confusion is given below in red box

My doubts : Im not getting  how $f(z)= \frac{(-e_1 + 2e_2)}{\sqrt 5}?$

My attempt :$ \delta = \inf_{y \in Y}\|y-x_0\|$ . Here $y= (e_1,e_1/2,0)$ and $x_0= (1,3,0)$
from $ \delta = \inf_{y \in Y}\|y-x_0\|$  we have $\| 1-1 ,3-1/2,0-0\|$  by taking $e_1=1$

Therefore $\delta = \sqrt 5$

so $f(z)= \alpha \sqrt 5$

But here im not getting  why  $f(z)= \frac{(-e_1 + 2e_2)}{\sqrt 5}\text{?}$

答:
$$z = y + \alpha x_0$$ For any $z \in Z$, there is only one value of $\alpha$ such that $z - \alpha x_0 \in Y$, and $y$ is defined as $y = z - \alpha x_0$ for that value of $\alpha$. So each $z$ has its own value of $y$, and its own value of $\alpha$.

Let $z = (\xi_1, \xi_2, 0)$. $z - \alpha x_0 = (\xi_1 - \alpha, \xi_2 - 3\alpha, 0)$ and we want the value of $\alpha$ that puts this point on $Y$. That is $$\xi_2 - 3\alpha = \frac {\xi_1 - \alpha}2$$
which solves to $$\alpha = \frac {-\xi_1+2\xi_2}5$$


答:
You may also try the following approach.

Consider the function $f(a)=\sum_{k=0}^n\binom nk \exp(ak)=\sum_{k=0}^n\binom nk (exp(a))^k=\left(1+\exp(a)\right)^n$.

Then $\sum_{k=0}^n \binom{n}{k}=f(a)|_{a=0}=2^n$

$\sum_{k=0}^nk \binom{n}{k}=\frac{d}{da}f(a)|_{a=0}=n2^{n-1}$

$\sum_{k=0}^nk^2 \binom{n}{k}=\frac{d^2}{da^2}f(a)|_{a=0}=n(n+1)2^{n-2}$, etc.

Title: Combinations sum : $\sum (k+1)^2 \binom{n}{k}$

Body:
$$1^2 + 2^2\binom{n}{1} +\ldots+(n+1)^2\binom{n}{n}=?$$
I tried this splitting
$$\sum_{k=0}^n (k+1)^2 \binom{n}{k}=\sum_{k=0}^n (k^2+2k+1) \binom{n}{k}= n^2 \cdot 2^n + 2n \cdot 2^n + 2^n $$

It didn't work for most cases , I then just tried
$$2^n \cdot (n+1)^2  $$
but it also didn't work . So I'm most likely doing it completely wrong or I'm approaching it in the wrong way. Thank you in advance.


问:
Prove: minimal origin-to-ellipse $|z+a|+|z-a|=2r$ lies on minor axis, using complex numbers
Let $|z+a|+|z-a|=2r$ where $a,z\in\mathbb{C}$ and $r>|a|$.

Prove the minimum of $|z|$ is $\sqrt{r^2-|a|^2}$

The equation describes an ellipse, and it's clear from drawing a picture that the minimal $|z|$ are the two vertices of the ellipse on the minor axis. How can we prove this claim analytically using complex numbers?

Thank you in advance!

答:
$|z+a|+|z-a|=2r$ is equation of an ellipse with foci at $-a$ and $a.$

Focal distance is $|2a|$ and the length of one axis is $2r.$ Thus the other axis has length $$2|d|=2\sqrt{r^2-|a|^2}.$$
Since the axes of ellipse are perpendicular, extremal points (vertices) lying on this axis are $$\pm d=\pm i\times |d|\times {a\over |a|}.$$

The nearest points to the origin are vertices of the ellipse that lie on the minor axis.

If $|a|<\sqrt{r^2-|a|^2},$ then $z_{1,2}=\pm a.$

If $|a|>\sqrt{r^2-|a|^2},$ then $z_{1,2}=\pm d.$


答: The orthogonal projection of $C$ on the segment $AB$ is a point $P$ of the form $(t,1-t,0)$. The value of $t$ must be chosen so that $PC$ and $AB$ are orthogonal, that is, so that$$(-t,t-1,z).(-1,1,0)=0.$$Therefore, $t=\frac12$. So, the area of the triangle is$$\frac12\times\text{length of }AB\times\text{length}PC=\frac12\sqrt2\sqrt{\frac12+z^2}=\sqrt{\frac14+\frac{z^2}2}.$$So, take $z=3\sqrt{\frac72}$.


问:
Conditional probability compliment, AIMA Excercise
In AIMA 2nd edition, there is this exercise

13.8 After your yearly checkup, the doctor has bad news and good news. The bad news is that you tested positive for a serious disease and that the test is 99% accurate (i.e., the probability of testing positive when you do have the disease is 0.99, as is the probability of testing negative when you don't have the disease). The good news is that this is a rare disease, striking only 1 in 10,000 people of your age. Why is it good news that the disease is rare? What are the chances that you actually have the disease?

I solved it no problem, but the problem solution manual did this

We got the same result at the end, but p(test|~disease) is not the compliment of p(test|disease), how this unfolded the same result?

Edit 1:

Why p(test|~disease) = 0.01 ? I know that p(test|disease)=0.99 but those two probabilities do not complement each other, yet p(test|~disease) was replaced by 0.01 without any explanation?

答:
Let ${}^c$ denote the complement. Recall that
\begin{align*}
\mathbb{P}(A) &= \mathbb{P}(A \cap B) + \mathbb{P}(A \cap B^c) \\
&= \mathbb{P}(A \, | \, B)\mathbb{P}(B) + \mathbb{P}(A \, | \, B^c)\mathbb{P}(B^c)
\end{align*}
for any set $B$ with $\mathbb{P}(B) \in (0,1)$.


问:
Five digit re-write game
In the habit of factoring numbers, a notebook I bought had a five digit item number $77076$, which factors as $2^2 3^2 2141$, which may also be $9 \cdot 8564$, and in this form the count of digits is again five. (Repeated digits count separately). [Note I thank Calvin Lin for pointing out I initially had the wrong product for $77076$.]

So I wondered how often this can work for five digit numbers, and started at $10001=73 \cdot 167$, $10002=2 \cdot 3 \cdot 1667=6\cdot 1667,$ and so on. The rules I decided to stick to were only that the five digit number has to be re-written as a product of two or more factors, where the total number (with repetitions) of digits occurring in the factors is again five.

Of course one runs into problems if the five digit number is itself a prime, for example at $10007$ and $10009$. There may be other general restrictions, statable in terms of the form of the prime factorization of the given five digit number; if so I'd be interested in that.

Sometimes the initial factorization into primes has to be juggled with. For example $10010=2 \cdot 5 \cdot 7 \cdot 11 \cdot 13$, which as it stands is two digits over the goal of five. We can lower the digit count by $1$ if we can for example multiply a one digit by a two digit prime in the factorization, and get only a two digit result. For this case we can use $2 \cdot 11=22,\ 5 \cdot 7=35,$ to get
$$10010=7 \cdot 22 \cdot 65,$$
so the five digit requirement is met. In this same example we could instead use $2 \cdot 13=26,\ 5 \cdot 11=55$ and get another re-write:
$$10010=7 \cdot 26 \cdot 55.$$

I know this is not serious math, hence the recreational tag; maybe someone might find it amusing to look at these rewrites.

Just to make a few specific questions: Is there a nice characterization, say in terms of the numbers of digits in the primes occurring in the factorization of $n$, which would say which composite $n$ with five digits did not have five digit rewrites as above? What happens in case we increase the number of digits to say 6 or 7?

答:
This is an interesting idea. A few things that immediately come to mind:

For five digit numbers $n$, $10000 \leq n \leq 99999$. Notice the number of digits required to represent a positive integer $n$ in base 10 is equal to $\lfloor {\log_{10}{n}}\rfloor+1$, which in this case you want to be 5, so $\lfloor {\log_{10}{n}}\rfloor=4$. Also remember that $\log{abc}=\log{a}+\log{b}+\log{c}$.

What happens if $n$ can be factored into, say $abc$? (to take a specific example). Then the factorization requires $\lfloor {\log_{10}{a}}\rfloor+1+\lfloor {\log_{10}{b}}\rfloor+1+\lfloor {\log_{10}{c}}\rfloor+1=\lfloor {\log_{10}{a}}\rfloor+\lfloor {\log_{10}{b}}\rfloor+\lfloor {\log_{10}{c}}\rfloor+3$ digits to represent, and you want this number to also be 5. So you want $\lfloor {\log_{10}{a}}\rfloor+\lfloor {\log_{10}{b}}\rfloor+\lfloor {\log_{10}{c}}\rfloor$=2 which means that either one of a, b, c is between 100 and 999 and the others are between 2 and 9 or two are between 10 and 99 and the other is between 2 and 9.

Can you extend this to more/less than three factors? What if $n$ has 6, 7 digits?

Title: Five digit re-write game


问:
Formula of intersection of 2 points with the x axis
I'm trying to figure out how to get the point x = 3 :
What's given here are the points S and G .
(Assuming the 2 angles are equal)

Apparently, we can assume that the ball does not bounce off the x-axis. Instead, we can change the target with respect to the symmetrical point on the x-axis so that we can now look at the intersection of the x-axis and the line connecting the current and target point.

The Formula to get x=3 is below :

$$\frac{S_x.G_y + G_x.S_y}{S_y+G_y}$$

Is there any explanation for this formula ?

答:
If $G = (G_x, G_y)$, then $G' = (G_x, ??)$.

Next, can you use two-point form to write the equation of SG'?

$ P(P_x, 0)$ is a point on SG'.

Putting the above together with S = (1, 1) and G=(7,2) , you will get $P_x = 3$.


答:
Hint:

This is an isosceles triangle, with basis $AB$, hence, if $I$ is the midpoint of $[AB]$, its area is equal to $\frac12 AB\cdot CI$. Note that $I$ has coordinates  $\bigl(\frac12,\frac12,0\bigr)$.


问:
How do I find a matrix representation of a linear transformation with two bases?
I've seen other examples of how to do this, but I still can't seem to figure this question out:

I believe I am supposed to plug the elements of the basis alpha into the transformation to see which coefficients of beta it is in terms of, using these as the columns for the vector.
My attempts at this so far have been unsuccessful, any help would be greatly appreciated.

答:
Yes, for some weird reason people struggle greatly with this. And yet, the algorithm is extremely simple: if $T:V \to W$ is a linear map and $V$ and $W$ have (ordered) bases $B$ and $C$, you

(1) evaluate $T$ at the $j$th element of $B$;

(2) write the result as a linear combination of the elements in $C$;

(3) place the coefficients of this combination in the $j$th column of a matrix.

This matrix is $[T]_{B,C}$.


答:
You're pulling some unreal magic trick by writing
$$\sum k \binom{n}{k}=n\cdot 2^n , \quad \sum k^2 \binom{n}{k}=n^2\cdot 2^n \tag{✗✗}$$
These of course aren't true.

Instead, by using the identity repeatedly
$$k \binom{n}{k}=n \binom{n-1}{k-1}$$
you can simplify as
$$\sum k \binom{n}{k}=n\sum \binom{n-1}{k-1}=n\cdot 2^{n-1}$$
and
$$\sum k^2 \binom{n}{k}=n\sum k \binom{n-1}{k-1}=$$
$$n\sum (k-1)\binom{n-1}{k-1}+n\sum \binom{n-1}{k-1}=n(n-1) 2^{n-2}+n\cdot 2^{n-1}$$
to yield the final result as
$$\boxed{(n+1)(n+4)\cdot 2^{n-2}}$$

Another approach would be combinatorial method.

Consider a class of $n$ students numbered $1,2,\ldots,n$ along with their teacher. The teacher decides to form a club of any strength, which has a captain and a vice-captain. The teacher must be present in the club and same people are allowed to occupy both positions.

Then we have three cases :

Teacher is both  captain and vice-captain. The students can join or not join the club freely. $1\cdot 2^n$ ways.

Teacher occupies exactly one position. Then choose one of the students for second position. Other $n-1$ can freely join. $2\cdot n\cdot2^{n-1}$ ways.

Teacher doesn't occupy any position. Choose the captain and vice-captain. Remaining students can join freely. $n\cdot 2^{n-1}+ n(n-1)\cdot 2^{n-2}$ ways.

Summing, we obtain
$$\boxed{(n+1)(n+4)\cdot 2^{n-2}}$$


问:
For a continuous function $f :\mathbb{R}\rightarrow \mathbb{R}$ satisfying $\int_{\mathbb{R}}|f(x)|dx<\infty$, which of the following is true?
Problem: For a continuous function $f :\mathbb{R}\rightarrow \mathbb{R}$ satisfying $$\int_{\mathbb{R}}|f(x)|dx<\infty$$ and for some $\alpha >0$ let $d_f(\alpha)$ be the Lebesgue measure of the set $$\{x\in \mathbb{R} : |f(x)|>\alpha\}$$
Then, for all $\alpha \geq 0$ which of the following is true?

$\alpha d_f(\alpha)\leq \int_{\mathbb{R}}|f(x)|dx$

$\alpha^2 d_f(\alpha)\leq \int_{\mathbb{R}}|f(x)|dx$

$d_f(\alpha)\leq \alpha \int_{\mathbb{R}}|f(x)|dx$

$d_f(\alpha)\leq \alpha^2 \int_{\mathbb{R}}|f(x)|dx$

My solution: As $\alpha<|f(x)|$ on $\{x\in \mathbb{R} : |f(x)|>\alpha\}$ we have
$$\int_{\{x: \alpha<|f(x)|\}}\alpha<\int_{\{x: \alpha<|f(x)|\}}|f(x)|<\int_{\mathbb{R}}|f(x)|dx$$
i.e., $$\alpha.d_f(\alpha)<\int_{\mathbb{R}}|f(x)|dx$$

So, I can see that first option is true.

I belive all the other three options are not necessarily true but i could not come up with an example.

Please help me to see if my argument for first bullet is sufficient/clear and help me to see other in detail.

Thank you.

答:
Question: I believe all the other three options are not necessarily true but i could not come up with an example.

Answer: Yes, you are right. Examples are as follows:

For option $\bf{(2)}$, Take $$f(x)=\begin{cases}
      x &\text{if}~~~~ x\in [0,20] \\
      40-x &\text{if}~~~~ x\in [20,40]\\
      0 &\text{elsewhere}
         \end{cases}$$
Take $~\alpha=10~$ and $~d_f(\alpha)=20~$ as $~d_f(\alpha)=m\left(\{x\in\mathbb R:|f(x)|>10\}\right)=m(10,30)~$
$\therefore~\alpha^2d_f(\alpha)=100\times20=2000~$ and $~\int_{\mathbb R}|f(x)|dx=\frac{x^2}{2}\big|_0^{20}+\left(40x-\frac{x^2}{2}\right)\big|_{20}^{40}=400~$
Thus $~\alpha^2d_f(\alpha)\not\le\int_{\mathbb R}|f(x)|dx~.$
Hence option $(2)$ is incorrect.

For option $\bf{(3)}$, Take $$f(x)=\begin{cases}
      x &\text{if}~~~~ x\in [0,3] \\
      6-x &\text{if}~~~~ x\in [3,6]\\
      0 &\text{elsewhere}
         \end{cases}$$
Take $~\alpha=0.1~$ and $~d_f(\alpha)=5.8~$ as $~d_f(\alpha)=m\left(\{x\in\mathbb R:|f(x)|>0.1\}\right)=m(0.1,5.9)~$
 and $~\int_{\mathbb R}|f(x)|dx=9\implies \alpha\int_{\mathbb R}|f(x)|dx=0.1\times 9=0.9~$
Clearly $~d_f(\alpha)\not\le\alpha\int_{\mathbb R}|f(x)|dx~$ and hence option $(3)$ is incorrect.

For option $\bf{(4)}$, From the previous example as on option $(3)$, $~\alpha^2=0.01~$ and
 $~\alpha^2\int_{\mathbb R}|f(x)|dx=0.01\times 9=0.09~$
Thus $~d_f(\alpha)\not\le\alpha^2\int_{\mathbb R}|f(x)|dx~.$
Hence option $(4)$ is incorrect.

Note: I know that the answer given here is an accepted answer. But I think my answer will also useful for solving the problem to the future reader. So if you downvote my answer, please tell me, where I am wrong. Thank you.


问:
Picard approximations
I'm homelearning calculus and trying to solve the following problem:

Find the first two Picard approximations for the solution of the
equation:

$x''=(x')^2x$

so that conditions are met: $x(0)=-\frac{1}{2}$ and $x'(0)=0$

However, I'm not sure how to calculate it for a differential equation of second degree. I only know it for the first degree. Could you please help me or give me a hint?

答:
As always in the general theory or in the application of general solvers, you transform into a first-order system
$$
x'=y,\\
y'=xy^2.
$$
Then the initial conditions transfer to the initial constant approximations
$$
x_0(t)=x_0=-\frac12,\\
y_0(t)=y_0=0.
$$
Now you can iterate the Picard-iteration
$$
x_1(t)=x_0+\int_0^ty_0(s)\,ds,\\
y_1(t)=y_0+\int_0^tx_0(s)y_0(s)^2\,ds,
$$
then
$$
x_2(t)=x_0+\int_0^ty_1(s)\,ds,\\
y_2(t)=y_0+\int_0^tx_1(s)y_1(s)^2\,ds,
$$

etc.

Title: Picard approximations


问:
High school pigeonhole principle question
Let p(x) be a polynomial with real coefficients and such that the product of all the roots is negative.
Show that if the degree of p(x) is 6, then p(x) has at least one positive root.</p>

答:
If a polynomial has real coefficients then for any complex root $z$, the conjugate $\bar{z}$, is also a root .

And $z\cdot \bar{z}=|z|^2>0$.

Therefore if the degree is $6$ it can have $0,2,4,6$ real roots, including multiple roots.

If the product of all roots is negative, then real roots can't be all negative: there must be a positive one, at least.


问:
Three line equations form a triangle. How do I find the equations for the heights?
The line equations:
$$x+2y-1=0$$
$$5x+4y-17=0$$
$$x-4y+11=0$$
form a triangle,

How can I find the line equations for the heights of that triangle, without searching for the coordinates of the vertices?
I know it has to do with the intersection of the lines, but I don't know how to find the symmetrical line of that angle.

Thanks in advance

答:
without searching for the coordinates of the vertices?

There isn't any need to "search" for the coordinates of the vertices, because the coordinates of a vertex are nothing but the solution of the equations of the two lines that intersect to form that vertex. I will therefore solve the three equations to get the vertex coordinates since I have been provided with them.

Suppose the vertices are $A, B$ and $C$, and the equations correspond to lines $AB, BC$ and $AC$. Thus we have three equations for three lines:
$$AB: x + 2y - 1 = 0$$
$$BC: 5x + 4y -17 = 0$$
$$AC: x - 4y + 11 = 0$$

By solving the equations for AB & AC, AB & BC, and BC & AC respectively, we get the following coordinates: $A(5, -2)$, $B(1, 3)$, $C(-3, 2)$. Also, by expressing the equations in slope-intercept form, we get $m_{AB} = \frac{-1}2$, $m_{BC} = \frac{-5}4$ and $m_{AC} = \frac{1}4$.

Again suppose, the three heights are $AD, BE$ and $CF$. That means, line $AD$ is perpendicular to $BC$, $BE$ is perpendicular to $AC$ and $CF$ is perpendicular to $AB$. We now recall that the slope of a perpendicular line is the negative inverse of the slope of the base (the proof is fairly easy so I will leave that out). Using this knowledge, we get the following slopes: $m_{AD} = -\frac{1}{m_{BC}} = \frac{4}5$, $m_{BE} = -\frac{1}{m_{AC}} = -4$ and $m_{CF} = -\frac{1}{m_{AB}} = 2$.

Since we have the coordinates of the vertices and the slopes of the perpendiculars from the vertices (which are of course lines passing through the vertices), we can easily construct equations for the three perpendiculars using $m = \frac{y-y_1}{x-x_1}$ which we get from the definition of slope. Therefore, the three required equations for three perpendiculars are:
$$AD: 4x - 5y + 30 = 0$$
$$BE: -4x - y + 7 = 0$$
$$CF: 2x - y + 8 = 0$$


问:
Working out the analysis of a circuit with non-linear resistor
I am working on a problem from the Strogatz book, specifically problem 2.2.12. I am having trouble evaluating the fixed point that I found. I have summarized the problem statement here, and show my work below:

Suppose a series circuit with a voltage source $V_0$, a non linear resistor and a capacitor. Assume that the the non linear nature of the resistor is such that $I = g(V)$, which is graphed in the book and I have tried to reproduce:

So, to model the equation:

$$V_0 = V_r + Q/C \Leftrightarrow V_r = V_0 - Q/C$$

Where $V_r$ is the voltage drop across the non-linear resistor. Coupling this with the fact (given in the text) that $\dot{Q} = I$, the function $g(V)$ can be rewritten as

$$\dot{Q} = g(V_0 - \frac{Q}{C})$$

and we see that $\dot{Q} = 0$ When $Q = \frac{V_0}{C}$

I thought to use the linearization technique mentioned in this chapter, namely to find out the sign of $f'(Q, t) = \ddot{Q} = -\frac{1}{C}g'(V_0 - \frac{Q}{C})$ at the fixed point.

So here is where I get a little stumped. I see that $g'(V_0 - \frac{Q}{C})$ is positive and non-zero at (0,0) and therefore the overall value is negative (stable) due to the factor of $-\frac{1}{C}$, but I don't know how to determine end behavior. The task was to determine the qualitative differences between this model and the one where voltage drop is simply given by $V_r = I_rR$.

That means, in my mind, that $Q(t) \rightarrow Q^* = V_0 / C$ as $t \rightarrow \infty$, which is different than the fixed point in the example of a linear resistor. Is that all that the question is asking? But the stability remains the same... except that I recall stability analysis for non-linear equations is supposed to be local only, so maybe there is something else I am missing?

答:
I got bugged by this exercise too. I must say that you put me in the right direction for a better understanding of it, and therefore I will put my thoughts here for the sake of completeness.

Once you have the flow equation:
$$
\dot{Q} = g(V_0 - \frac{Q}{C})
$$

The rest is to be solved and explained qualitatively (the linear stability analysis comes later in the chapter so I don't think it should be necessary to answer here).

A small mistake in your post though, from the equation above, and given that $g(0) = 0$, we find $Q^* = CV_0$, exactly like in the linear resistor case (when $I = V_r/R$). That fixed point is indeed stable, but we can draw the phase portrait from the sketch of $g(V)$: you just need to flip around the graph w.r.t the origin and translate of $V_0$ along the x-axis (this is a qualitative sketch, there is also a scaling by $1/C$ to be applied to be exact, and we don't even know $g$ despite the sketch of its graph). Also in the plot below, I probably did not reproduce exactly the intended $g$, I don't think there should be an infinite derivative at the fixed point.

The only qualitative difference that I conclude from that is that for the non-linear case the approach to the stable fixed point is much faster in its vicinity, while slower when starting far away from it...


问:
About the existence of the diagonal set of Cantor
The classic proof of the Cantor set start with the assumption that the set

$$B=\{x\in A:x\notin f(x)\}$$

exists, where $f: A\to\mathcal P(A)$ is a bijective function. I understand the proof but I dont understand the assumptions where you start to make this proof.

To be clear, why a set $B$ can be constructed? How you can justify this assumption? To me the proof of the Cantor theorem is far to be clear or complete if there are not an explanation about why $B$ must be possible.

Then, can someone explain to me or justify, via other theorems if possible, why $B$ must exist? Thank you in advance.

P.S.: can someone explain to me the downvotes in the question?

答:
Another example, from the infinite set of natural numbers:
\begin{eqnarray}
S=\mathbb{N}\\
P\left(S\right)=\{\phi, S, \forall U\neq\phi;U\subset S\}\\
f\left(x\right)=\{x\}
\end{eqnarray}
Then f is 1-1.

$B=\{x\in S;x\notin f\left(x\right)\}$

Then
$\forall x\notin B$

$B=\phi \in P\left(S\right)$

Which exists and again is not in the image of $f$.


Body:
\begin{align}
   A &= (1,0,0) \\
   B &= (0,1,0) \\
   C &= (0,0,z) \\
\end{align}

\begin{align}
   a &= BC &=\sqrt{1+z^2} \\
   b &= AC &=\sqrt{1+z^2} \\
   c &= AB &=\sqrt 2 \\
\end{align}

$$s = \frac 12(a+b+c) = \sqrt{1+z^2} + \frac{1}{\sqrt 2}$$

\begin{align}
   s-a &= \frac{1}{\sqrt 2} \\
   s-b &= \frac{1}{\sqrt 2} \\
   s-c &= \sqrt{1+z^2} - \frac{1}{\sqrt 2}
\end{align}

$$s(s-c) = (1+z^2) - \frac 12 = z^2 + \frac 12$$
$$(s-a)(s-b) = \frac 12$$

\begin{align}
   |ABC| &= \sqrt{s(s-a)(s-b)(s-c)} &\text{(Heron's formula)}\\
       4 &= \sqrt{\frac 12 \left(z^2 + \frac 12 \right)} \\
      32 &= z^2 + \frac 12 \\
      63 &= 2z^2 \\
     z^2 &= \dfrac{63}{2} \\
       z &= \pm \sqrt{\frac{63}{2}}
\end{align}


问:
Category Theory: Equality if All Projections are Equal?
Let $\{C_\alpha\}_{\alpha \in A}$ be objects in a category: in response to comments, a concrete category.

Let the product object $C = \Pi C_{\alpha \in A}$ exist with corresponding projections {$\pi_\alpha$}

If $x, y \in C$ and $\pi_\alpha (x) = \pi_\alpha (y) $ for all $\alpha$ does it follow that $x = y$ ?

I think I can prove this in the category of vector spaces by assuming the form of $x = (x_\alpha)$ and linearity of the projections, but is it readily provable (even true ?) in general, say from the universal property ?

答:
You category is concrete, i.e. you have a fully faithful functor $U \colon \mathsf{C} \to \mathbf{Set}$. Suppose this has some right adjoint $F \colon \mathbf{Set} \to \mathsf{C}$ i.e. a "free construction".

Elements of $C$ as a set are maps $x,y \colon \ast \to  UC$ in the category of sets. This is the first step - but very much in the spirit - in the direction of generalized elements. You want to show $x = y$, as elements of the set $UC$, which is the same as saying that they are the same arrow in $\mathbf{Set}$.

By the adjunction, this is to say that the maps $x^\flat,y^\flat \colon R(\ast) \to C$ induced by the adjunction coincide.

Since $R$ is a right-adjoint, it preserves limits and sends $\ast$ to the terminal object $1$ in $\mathsf{C}$, so we ought to decide wether $x^\flat,y^\flat \colon 1 \to C$ are equal, and since $C$ is a product, it suffices to check wether composing with each projection they do, that is wether $\pi_\alpha x^\flat = \pi_\alpha y^\flat$ for all $\alpha$.

But these are maps $1 \to C_\alpha$ which correspond via $U$ with $\ast \to UC_\alpha$ associated with the elements $\pi_\alpha(x)$ and $\pi_\alpha(y)$, assumed to be equal, completing the proof. Here we use that $U 1 = \ast$, as left adjoints preserve finite limits.

Just to be clear: the hypothesis I introduced were

the elements $x,y$ are maps $x,y \colon \ast \to UC$.

the forgetful functor $U$ has some right adjoint.

for each projection we have $U\pi_\alpha x^\flat = U \pi_\alpha y^\flat$.

This coincides with the intuitive notion for classical concrete categories such as the category of vector spaces, topological spaces, abelian groups, etc.

Edit: the adjunction hypothesis is only to have a well defined notion of starting with an element of a set (i.e a map $x : \ast \to UC$) and lifting it to a map $x^\flat \colon 1 \to C$ in $\mathsf{C}$.

In general, what is true is that maps $f,g \colon B \to C$ are equal iff $\pi_\alpha f = \pi_\alpha g$ for all $\alpha$, but since $U$ is fully faithful this occurs precisely if $U \pi_\alpha f = U \pi_\alpha g$ for all $\alpha$. Hence we can check at the level of functions.

So if "elements" for you are maps $1 \to C$ to begin with, you only need the concreteness hypothesis.


问:
Looking for analytical proof that this function given as a power series is constant.
Answering a recent question I came across the following function ($t$ is a positive integer), defined for $0\le x\le 1$ as:
$$
P_t(x)=x^t\sum_{n=0}^\infty \frac1{n+1}\binom{t(n+1)}n\left[(1-x)x^{t-1}\right]^n.
$$

Is there an analytical way to prove that $P_t(x)$ is a continuous function on $x\in(0,1)$ and:

$P_t(x)=1$ for all $x:\ 0\le 1-x\le\frac1t$,

$P_t(x)<1$ for all $x:\ \frac1t< 1-x\le1$.

I proved this for trivial cases $t=1,2$ but did not find a way to deal with general $t$.

The behavior of the function for $t=1,2,3,4$ is demonstrated below:

Body:
comment

Interesting.  I tried $n=7$ in Maple, and the result was
$$
P_7(x) = {x}^{7}
{\mbox{$_7$F$_6$}\left(1,{\frac{8}{7}},{\frac{9}{7}},{\frac{10}{7}},{\frac{11}{7}},{\frac{12}{7}},{\frac{13}{7}};\,\frac43,\frac32,\frac53,{\frac{11}{6}},2,{\frac{13}{6}};\,-{\frac { \left( -823543+823543\,x \right) {x}^{6}}{46656}}\right)}
$$
That graph does, indeed, look like yours.  Why?


问:
Why are $\text{versin}(x)=1-\cos(x)$, $\text{coversin}(x)=1-\sin(x)$, $\text{vercosin}(x)=1+\cos(x)$ no longer used in maths?
I was preparing for a calculus exam and I came across the Wikipedia article for all the trigonometric identities. There I came across some terms that I had never seen before. They were:
$\text{versin}(x)=1-\cos(x)$, $\text{coversin}(x)=1-\sin(x)$, $\text{vercosin}(x)=1+\cos(x)$ and other similar ones.

In the article it states that these were historically used but nowadays they have no real use. Why is that?  Why are these not used anymore?

答:
The versine function is well documented in this Wikipedia article: https://en.wikipedia.org/wiki/Versine

The versine or versed sine is a trigonometric function found in some of the earliest (Vedic Aryabhatia I) trigonometric tables. The versine of an angle is 1 minus its cosine.

It is very likely that people need all sorts of trig functions to write trigonometric tables in the old days. But now we have computers.


答:
As θ goes to zero, versin(θ) is the difference between two nearly equal quantities, so a user of a trigonometric table for the cosine alone would need a very high accuracy to obtain the versine in order to avoid catastrophic cancellation, making separate tables for the latter convenient.

These days we walk around with devices capable of approximating the versine to a degree of accuracy that would in the past be considered an extraordinary degree of accuracy and capable of looking up tables the size of books for exact results.

That is the practical aspect, the other aspect is that it has fallen out of taste, and there is no accounting for taste so take it as you will.

Title: Why are $\text{versin}(x)=1-\cos(x)$, $\text{coversin}(x)=1-\sin(x)$, $\text{vercosin}(x)=1+\cos(x)$ no longer used in maths?

Body:
I was preparing for a calculus exam and I came across the Wikipedia article for all the trigonometric identities. There I came across some terms that I had never seen before. They were:
$\text{versin}(x)=1-\cos(x)$, $\text{coversin}(x)=1-\sin(x)$, $\text{vercosin}(x)=1+\cos(x)$ and other similar ones.

In the article it states that these were historically used but nowadays they have no real use. Why is that?  Why are these not used anymore?


问:
Given a board of size $1 \times n$,in how many this board can be colored with the colors white,yellow,blue assuming the number of blue cells is even
Given a board of size $1 \times n$,in how many this board can be colored with the colors white,yellow and blue assuming the number of blue cells is an even number.

My try:

Given a board of size $1 \times n$,then there are two possibilities for the color of the first cell:

If the first cell is blue then the it's needed to find another cell to color it with the blue color ,this can be done in $n-1$ ways and for each one of these choices there are $b_{n-2}$ ways to color the remaining $1 \times (n-2)$ board with the colors yellow,white and blue assuming the number of blue cells is an even number.

If the first cell is yellow or white then the remaining $1 \times (n-1)$ board can be colored in $b_{n-1}$ ways assuming the restriction on the number of blue cells is satisfied.

Gathering these cases follows:

$$b_n=(n-1)b_{n-2}+2b_{n-1}$$

With $b_0=1$ and $b_1=2$ as initial conditions.

Calling $G(x)=\sum_{k\ge 0}^{} b_k x^k$,omitting some elementary calculations follows:

\begin{align*}
G(x)&=\frac{-1}{\left(n+1\right)^{2}x^{2}+2x-1}\\
&=\frac{-1}{\left(n+1\right)^{2}}\frac{1}{(x-a)(x-b)}\\
&=\frac{1}{\left(n+1\right)^{2}(a-b)}\left(\frac{1}{a-x}-\frac{1}{b-x}\right)\\
&=\frac{1}{\left(n+1\right)^{2}(a-b)}\left(\frac{1}{a}\frac{1}{1-\frac{x}{a}}-\frac{1}{b}\frac{1}{1-\frac{x}{b}}\right)\\
&
\end{align*}

From which we conclude:
\begin{align*}
b_n=\left[x^{n}\right]G\left(x\right)
&=\frac{1}{\left(n+1\right)^{2}(a-b)}\left(\left(\frac{1}{a}\right)^{n+1}-\left(\frac{1}{b}\right)^{n+1}\right)\\
\end{align*}

Where $$a=\frac{-1+\sqrt{n+2}}{n+1},b=\frac{-1-\sqrt{n+2}}{n+1}$$

But the closed form does not coincide with the recurrence that I've found.(I only want the question to be solved using recurrence relations.)

答:
Let $a_n$ be the number of colorings with an odd number of blues, and $b_n$ be the number of colorings with an even number of blues.

Then $$a_n=b_{n-1}+2a_{n-1}\tag1$$ because to get an odd number of blues, we can either color the first square blue and color the remaining $n-1$ squares with an even number of blues, or color the first square white or yellow, and then color the remaining squares with an odd number of blues.  Similarly, we have $$b_n=a_{n-1}+2b_{n-1}\tag2$$  From $(2)$, $a_{n-1}=b_n-2b_{n-1}$, and substituting this in $(1)$ gives $$b_{n+1}-4b_n+3b_{n-1}=0,$$ which can be solved in the usual manner.


问:
Define $g(x)=xf(x)$. If $f$ is continuous at $0$, then prove that $g$ is differentiable at $0$.

Given $f\colon \mathbb{R} \rightarrow \mathbb{R}$ and define $g\colon\mathbb{R} \rightarrow \mathbb{R}$ by $g(x)=xf(x)$. If $f$ is continuous at $0$, then prove that $g$ is differentiable at $0$ with $g'(0)=f(0)$.

Here's my attempt:

We know that if $f(x)$ is continuous at $0$, then $g(x)$ is continuous at $0$, because multiplication of $x$ and $f(x)$ again a continuous function. It follows that:
\begin{align}
g(0)=0
\end{align}
Also, because we know that $f$ is continuous at $x=0$:
\begin{align}
\lim_{x\to\ 0} f(x)=f(0)
\\ \lim_{x\to\ 0} \frac{f(x)-f(0)}{x-0}(x-0) \end{align}
That means $f$ also differentiable, so by knowing $x$ and $f(x)$ are differentiable we can say $g(x)$ is also differentiable at $x=0$. If we take the derivative at the definition of the function we get
\begin{align}
g'(0)=f(0)
\end{align}
Is it true logic? Or does it make sense? Thanks for helping.

答:
No, in the middle of your argument you conclude that $f$ is differentiable at $x=0$, but $f$ is only assumed to be continuous at $x=0$.

Hint.

Study the quotient
$$
\frac{g(x)-g(0)}{x-0}=\frac{xf(x)}{x}
$$
and use the continuity of $f$ at $x=0$.


答:
First, you don't need a complicated argument regarding continuity of $g$ to conclude that $g(0)=0$, you can simply substitute
$$g(0) = 0 \cdot f(0) = 0
$$
But the main problem with your proof is that in order to determine whether $g'(0)$ exists, you must write out the difference quotient $\frac{g(x)-g(0)}{x-0}$ and simplify it and try to find its limit as $x \to 0$. I do not see anywhere in your solution that you have written out the different quotient.

Title: Define $g(x)=xf(x)$. If $f$ is continuous at $0$, then prove that $g$ is differentiable at $0$.

Body:
Given $f\colon \mathbb{R} \rightarrow \mathbb{R}$ and define $g\colon\mathbb{R} \rightarrow \mathbb{R}$ by $g(x)=xf(x)$. If $f$ is continuous at $0$, then prove that $g$ is differentiable at $0$ with $g'(0)=f(0)$.

Here's my attempt:

We know that if $f(x)$ is continuous at $0$, then $g(x)$ is continuous at $0$, because multiplication of $x$ and $f(x)$ again a continuous function. It follows that:
\begin{align}
g(0)=0
\end{align}
Also, because we know that $f$ is continuous at $x=0$:
\begin{align}
\lim_{x\to\ 0} f(x)=f(0)
\\ \lim_{x\to\ 0} \frac{f(x)-f(0)}{x-0}(x-0) \end{align}
That means $f$ also differentiable, so by knowing $x$ and $f(x)$ are differentiable we can say $g(x)$ is also differentiable at $x=0$. If we take the derivative at the definition of the function we get
\begin{align}
g'(0)=f(0)
\end{align}
Is it true logic? Or does it make sense? Thanks for helping.


问:
Continuous Linear Mapping $C[0,1]\rightarrow C[0,1]$
Show that $L(f)(x)= \int_0^x f(t) dt $ is a continuous linear mapping from $C[0,1]$ into itself.

Do I only have to show that the operator is bounded? How to do I explicitly choose my $M$ such that $\|L \ f\|<M\|f\|$?

答:
If $f$ is continuous, then $F=\displaystyle\int_0^x f$ is differentiable, thus clearly continuous.

On the other hand, suppose that $$\lVert f-g\rVert_\infty<\frac{\epsilon}2$$

Then for $0\leq x \leq 1$,$$\left|\int_0^x(f-g)\right|\leq \int_0^x|f-g|\leq\int_0^1 \lVert f-g\rVert_\infty<1\cdot\frac{\epsilon}2=\frac{\epsilon}2$$

which means $$\lVert F-G\rVert_\infty\leq \frac{\epsilon}{2}<\epsilon$$


问:
Group of symmetries of a parallelepiped in three dimension.
Describe the group of symmetries of a parallelepiped $-a\leq x\leq a$, $-b\leq y\leq b$, $-c\leq z\leq c$ in $\mathbb{R}^3$ with $a\neq b\neq c.$

I know that a dihedral group is the group of symmetries of a regular polygon. But in our problem,  the parallelepiped is not a regular polygon. How to think about this group?

答:
It is going to depend very heavily on what you mean by symmetries.

It is probably meaning either actions induced by GL(n,F) on the space in which the parallelepiped is embedded which fix the parallelepiped in some way, or restricting the actions further by looking at actions induced by some subgroup of GL(n,F) which fix the parallelepiped.

You are probably looking at linear isometries in 3d which act invariantly on the parallelepiped. So moving along as if that is what you mean by symmetries, your group will have the 3 generators

$f(x,y,z) = (-x,y,z)$

$f(x,y,z) = (x,-y,z)$

$f(x,y,z) = (x,y,-z)$

There would be more maps in the case $a=b$, $b=c$, or $a=c$.

The reason I chose isometries as the maps you were probably interested in is that you took the time to specify $a\neq b \neq c$.

This rules out rotations along with changes of coordinate.

So the group is $Z_{2} + Z_{2} + Z_{2}$. Sorry for the notation I'm still getting used to math jax.


问:
Stabilising chain of submodules
Take $S$ to be a commutative ring with unit, and $N \subset M$ to be $S$-modules. I am interested in the necessary and sufficient condition for it to be the case that for every collection of submodules $\{M_i\}_{i=1,\dots,n}$, such that $M_i \cap N = 0$, they form a stabilising chain,

$$M_1 \subset M_2 \subset \cdots \subset M_{n}.$$

I believe a necessary and sufficient condition is that every $M_i$ such that $M_i\cap N=0$ is finitely generated. I know that if every submodule was finitely generated, then $M$ would be Noetherian, and thus I am guaranteed this stabilising chain.

However, if I am only requiring that $M_i$'s such that $M_i \cap N = 0$ be finitely generated, this does not exclude the possibility of submodules which are not finitely generated to exist in $M$ so $M$ is not being required to be Noetherian, and so I am not guaranteed this stabilising chain. How can I approach this?

答:
EDIT: This answer is completely wrong, so don't bother reading it. I'll leave it up for the record; details are in the comments, but I amateurishly confused the set $M \setminus N \cup \{ 0 \}$ with the submodule that it generates.

Let $M_N$ be the smallest submodule of $M$ containing $\left( M \setminus N \right) \cup \{ 0 \}$. The submodules of $M_N$ are exactly the submodules $M_i \leq M$ s.t. $M_i \cap N = \{ 0 \}$. The property in question is thus equivalent to whether $M_N$ is Noetherian or not. There are two cases.

Trivially, $M_N$ could simply be $M$ itself, i.e. $M$ has a generating set in $M \setminus N$.

$M_N$ could be a proper submodule of $M$. Since $M \setminus M_N = N \setminus \{ 0 \}$, this works out to say that at least one non-identity of element $N$ cannot be written as a linear combination of elements of $M_N$ over $S$.

Title: Stabilising chain of submodules


问:
Proving that transitive $R-$action implies $M$ is a simple module.
Here is the question I want to answer:

A module is simple if it is not the zero module and it has no proper nonzero submodule.

$(a)$ Let $M$ be an $R-$module. Show that the following conditions are equivalent.

$M$ is a simple $R-$module.

for any nonzero $x,y \in M$ there exists $r \in R$ such that $rx = y.$(i.e. the action of $R$ on $M$ is transitive)

I managed to prove $1 \implies 2.$ For $2 \implies 1,$ here is my trial (after getting some help from kind volunteers here):

Assume that $M$ is an $R-$module and assume that for any nonzero $x,y \in M$ there exists $r \in R$ such that $rx = y$ (i.e. the action of $R$ on $M$ is transitive). We want to show that $M$ is a simple $R-$module i.e. it is non-zero and it has no non-zero proper submodules.

First, $M \neq 0$ because the $R-$action on $M$ is a non-zero action. Is this reasoning correct?

Second, Suppose that $S \subset M$ is a non-zero submodule. We want to show that $S = M.$ If $S \neq M,$ then $\exists m \in M \setminus S$ But then I do not know how to complete this argument, could anyone help me in completing it please?

答:
Hint: to show that $M$ is simple, you can equivalently prove that $M \neq 0$ but $M = Rx$ for all nonzero $x \in M$.

 Indeed if $M$ is simple and $x$ nonzero, then $(x)$ is a nonzero submodule, and it must coincide with $M$. Conversely, let's show that $M$ is simple. If $S$ is a nonzero submodule, it contains some nonzero $x$. But then $M = (x) \subset S \subset M$, hence $M = S$.

Can you take it from there?


问:
Find $\min(\max(x+y,2x-y+2))$ under $x+2y\ge 4, x\ge 0, y\ge 0$
Find $\min(\max(x+y,2x-y+2))$ under $x+2y\ge 4, x\ge 0, y\ge 0$.

My attempt: In the region defined by the constraints, we have that $f(x,y)=\max(x+y,2x-y+2)=2x-y+2$. So, I believe that it is enough to minimize $2x-y+2$ under the given constraints. There is no solution, because there is no level set that shares a unique point or side with the defined region.

However Wolfram Alpha tells that there is a minimum, which is the point $(0,2)$. How can I come to this conclusion? This point is intersection of $2x-y+2=0$ with $x+2y=4$.

答: You can get to the solution graphically. If you draw the restrictions and the contours of $f(x,y)= \max\{x+y,2x-y+2\}$, you'll easily realise that the minimum is actually attained at $(0,2)$.


问:
Argument Principle: roots of $z^6+9z^4+z^3+2z+4$
I'm having trouble understanding the argument principle as seen in Gamelin complex analysis.

I want to go over the example in the text, so I will link the text of the example and go through it.

We want to find the roots of polynomial $z^6+9z^4+z^3+2z+4$ in the first quadrant given by $\{0 < \text{arg} z < \pi/2 \}$. So we consider the quarter disk $D_{R}$ in the first quadrant, consiting of poitns in the first quadrant satisfying $|z| < R$.

The boundary of $D_R$ is given by going through the real axis, then the quarter circle, and then going down the imaginary axis.

We can see that going through the real axis, the change in argument is $0$.

In the quarter circle, the text says that the term $z^6$ dominates, so we just consider $\text{arg} p(z) \approx 6 \text{arg} z$, so the change in argument is $3 \pi$.

My question is, how do we see that $z^6$ "dominates"?

And in the part where we go down the imaginary axis, $z^6$ does not dominate. Why is this so?

Thanks for your help!

答:
"$z^6$ dominates" means that when $|z|$ is "large" you have
$$
p(z)\approx z^6
$$
(All other terms $9z^4+z^3+2z+4$ are relatively small compared with $z^6$.)
In particular, when $R$ is large, on $\Gamma_R$, you have (1).

To be precise, you have
$$
\lim_{|z|\to\infty}\frac{|p(z)|}{|z^6|}=1
$$


问:
How to express $2 \cos X = \sin X$ in terms of $\sin X$?
The Question was:

Express $2\cos{X} = \sin{X}$ in terms of $\sin{X}$ only.

I have had dealings with similar problems but for some reason, due to I believe a minor oversight, I am terribly vexed.

答: Try using $\cos{x}^2 + \sin{x}^2 =1$ or some other known identity, but solving for $\cos{x}$ in terms of $\sin{x}$. Then just substitute.</p>


答:
You can convert this question (and any similar question) into a standard problem as follows.

If $x+y\ge 2x-y+2$ then you must minimise $x+y$. So make $2y\ge x+2$ an extra constraint and find the  minimum of $x+y$ in a standard way.

Similarly, minimise $2x-y+2$ when $2y<x+2$.

Finally compare the two answers.

Body:
Find $\min(\max(x+y,2x-y+2))$ under $x+2y\ge 4, x\ge 0, y\ge 0$.

My attempt: In the region defined by the constraints, we have that $f(x,y)=\max(x+y,2x-y+2)=2x-y+2$. So, I believe that it is enough to minimize $2x-y+2$ under the given constraints. There is no solution, because there is no level set that shares a unique point or side with the defined region.

However Wolfram Alpha tells that there is a minimum, which is the point $(0,2)$. How can I come to this conclusion? This point is intersection of $2x-y+2=0$ with $x+2y=4$.


问:
For the given function $f(x)=\frac{1}{\sqrt {1+x}} +\frac{1}{\sqrt {1+a}} + \sqrt{\frac{ax}{ax+8}}$, prove that $1<f(x)<2$ for positive a and $x\ge 0$

For the given function $f(x)=\frac{1}{\sqrt {1+x}} +\frac{1}{\sqrt {1+a}} + \sqrt{\frac{ax}{ax+8}}$, prove that $1<f(x)<2$ for +ve $a$ and $x\ge 0$

I tried the most rudimentary method ie. differentiating wrt x and hoped to get at least 2 solutions for $f’(x)$, unfortunately I wasn’t able to find a solution for implying function is monotonous . Now whether it’s sis increasing or decreasing still remains a mystery, given that we don’t know what $a$ is.

What method can be used to solve this?

答:
This is actually a very devious inequality problem disguised as a calculus problem, and it took me very long to realise this. But eventually what gave it away was that calculating the second derivative of $f(x)$ would be a monstrous task, and that the $\sqrt{\dfrac{ax}{ax+8}}$ term seemed very suspicious (As we shall see, dividing both numerator and denominator by $ax$ would yield something very interesting and useful.)

First let us rename $x$ as $b, b \geq 0$. The case where $b=0$ is trivial, since we have $\dfrac{1}{\sqrt{1+a}} + \dfrac{1}{\sqrt{1+b}} + \sqrt{\dfrac{ab}{ab+8}} = \dfrac{1}{\sqrt{1+a}} +1 $. Thus, we may assume $a,b >0$. Let $a=2x,b=2y$, where $x, y >0$.

Thus, $$\dfrac{1}{\sqrt{1+a}} + \dfrac{1}{\sqrt{1+b}} + \sqrt{\dfrac{ab}{ab+8}} =\dfrac{1}{\sqrt{1+2x}} + \dfrac{1}{\sqrt{1+2y}} + \dfrac{1}{\sqrt{1+\dfrac{2}{xy}}}.$$
Let $2z=\dfrac{2}{xy} \Rightarrow xyz=1$. So really, this problem is equivalent to us proving the following inequality:

Inequality:

Let $x,y,z \in \mathbb{R^+}$, $xyz=1$. Prove that $1<\dfrac{1}{\sqrt{1+2x}} + \dfrac{1}{\sqrt{1+2y}} + \dfrac{1}{\sqrt{1+2z}} <2. $

Now, the lower bound is relatively easy, but the upper bound is a real killer.

Lower Bound:

Since $1+2x >1 , \sqrt{1+2x} < 1+2x \Rightarrow \dfrac{1}{\sqrt{1+2x}} > \dfrac{1}{1+2x}$.  Thus it suffices for us to prove that:

\begin{align}
\sum_{\text{cyc}} \dfrac{1}{1+2x} \geq 1 & \iff \dfrac{(1+2y)(1+2z) + (1+2x)(1+2z) + (1+2x)(1+2y)}{(1+2x)(1+2y)(1+2z)} \geq 1 \\
& \iff 3+4(x+y+z) + 4(yz+xz+xy) \geq 9+2(x+y+z) +  4(yz+xz+xy)  \\
& \iff 2(x+y+z) \geq 6 \\
& \iff x+y+z \geq 3 \\
\end{align}
Which follows immediately from AM-GM and the given condition.

Upper Bound:

The following proof borrowed an important idea introduced in Yufei Zhao's inequalities notes: Inequalities. Sometimes switching the roles of the constraint and the inequality can yield wonders!

We begin by letting $p=\dfrac{1}{\sqrt{1+2x}}, q=\dfrac{1}{\sqrt{1+2y}}, r=\dfrac{1}{\sqrt{1+2z}} $, and $p,q,r<1$. We have to prove that $xyz=1 \Rightarrow p+q+r <2$. Instead of proving this directly, we will prove the contrapositive: $p+q+r \geq 2 \Rightarrow xyz <1 \Rightarrow xyz \neq 1$. Next we carry out the following algebraic manipulations:

\begin{align}
p=\dfrac{1}{\sqrt{1+2x}} & \Rightarrow p^2=\dfrac{1}{1+2x} \\
& \Rightarrow \dfrac{1}{p^2} = 1+2x \\
& \Rightarrow x=\dfrac{1-p^2}{2p^2} = \dfrac{(1+p)(1-p)}{2p^2} < \dfrac{1-p}{p^2}\\
\end{align}

But
\begin{align}
p+q+r \geq 2 & \Rightarrow p \geq 2-q-r \\
& \Rightarrow 1-p \leq -1+q+r = qr-(1-q)(1-r) < qr \\
& \Rightarrow x < \dfrac{qr}{p^2}.
\end{align}
Similarly, we obtain that $y < \dfrac{pr}{q^2}$ and $z < \dfrac{pq}{r^2}$. Thus, $xyz <\dfrac{qr}{p^2} \cdot \dfrac{pr}{q^2} \cdot  \dfrac{pq}{r^2} =1$, and we are done.

Final note: In fact, I believe that we have the following stronger upper bound:
$$\dfrac{1}{\sqrt{1+2x}} + \dfrac{1}{\sqrt{1+2y}} + \dfrac{1}{\sqrt{1+2z}} \leq  \sqrt{3}.$$
However, I have absolutely no clue on how to proceed. Perhaps someone else can come up with a nice idea?


问:
(Algebraically) prove $a^x$ is an increasing function
How to prove $a^x$ (for $a>1$) is an increasing function, independently of $\int_1 ^x \frac{dx}{x}$ or $\operatorname{Ln}(x)$. Is there a way to algebraically prove $(x_1>x_2 \implies a^{x_1}>a^{x_2})$ with as little reference possible to calculus?

I understand we technically define $a^x$ through $\operatorname{Ln}(x)$, but is there a different algebraic perspective that we can give to $a^x$? At least for the rational numbers (since we need the $\operatorname{Ln}(x)$-based definition for defining irrational powers.)

Body: HINT: $a^{x_1}=a^{x_2}a^{x_1-x_2}$


答:
No, we don't need to use logarithms to define $a^x$ for $a>0$. You can define it first when $x$ is rational: if $x=\frac mn$, with $m\in\Bbb Z$, and $n\in\Bbb N$, then $a^x=\sqrt[n]{a^m}$. It is easy to see that this makes sense, that is, that if $\frac mn =\frac pq$, then $\sqrt[n]{a^m}=\sqrt[q]{a^p}$. Then, for an arbitrary $x$, you can define $a^x$ as follows:  take a sequence $(q_n)_{n\in\Bbb N}$ of rational numbers such that $\lim_{n\to\infty}q^n=x$ and define $a^x=\lim_{n\to\infty}a^{q_n}$. It can be proved that this definition doesn't depend upon the choice of the sequence $(q_n)_{n\in\Bbb N}$ and that the limit $\lim_{n\to\infty}a^{q_n}$ always exists.

Now, if $a>1$, if $m_1,m_2\in\Bbb Z$ and if $n\in\Bbb N$, it is easy to prove that$$\frac{m_1}n<\frac{m_2}n(\iff m_1<m_2)\implies a^{m_1/n}<a^{m_2/n}.$$So, the restriction of $x\mapsto a^x$ is strictly increasing. And now it's not hard to prove that if $x,y\in\Bbb R$ are such that $x<y$ and that if $q\in(x,y)\cap\Bbb Q$, then$$a^x<a^q<a^y.$$

Title: (Algebraically) prove $a^x$ is an increasing function

Body:
How to prove $a^x$ (for $a>1$) is an increasing function, independently of $\int_1 ^x \frac{dx}{x}$ or $\operatorname{Ln}(x)$. Is there a way to algebraically prove $(x_1>x_2 \implies a^{x_1}>a^{x_2})$ with as little reference possible to calculus?

I understand we technically define $a^x$ through $\operatorname{Ln}(x)$, but is there a different algebraic perspective that we can give to $a^x$? At least for the rational numbers (since we need the $\operatorname{Ln}(x)$-based definition for defining irrational powers.)


答:
So you asked to show that $a^{q} > a^{p}$ for rationals $q>p$.

Divide both sides by $a^{p}$. You want to show, $a^{q-p} > 1$. You can get a common denominator d for p and q and use the fact that $x^{n}$ is increasing in $[1,\infty)$ for natural number n to get that $a^{q-p} > 1$ if and only if $a^{qd-pd} > 1$. Now it suffices to show that $a^{n}> 1$ for a natural number n if $a>1$. You will use the fact that $x^{n}$ is increasing in $[1,\infty)$ for natural number n here again.


问:
Fundamental group of some quotient of the annulus
This problem comes from the book Topology and Geometry of Bredon :

Consider the annulus. Identify antipodal points on the outer circle. Also identify antipodal points on the inner circle. Calculate the fundamental group of this surface.

Let us denote $X$ the annulus with identifications. Consider an open cover $\{U,V\}$ of the surface with $U$ and $V$ being annulus inside the big annulus where the former has its outer circle being the outer circle of the annulus with identifications, and where the latter has its inner circle being the inner circle of the annulus with identifications.

Now, the way I understand it, $U$ and $V$ strongly deformation retract to the outer and inner circle of the annulus with identifications respectively, which is $\mathbb{RP}^1$ and thus
\begin{equation*}
\pi_1(U) \approx \pi_1(V) \approx \pi_1(U \cap V) \approx \mathbb{Z}.
\end{equation*}

By the Seifert-van Kampen's theorem, the canonical maps of the fundamental group of $U$, $V$, and $U \cap V$ into that of $X$ induce an isomorphism
\begin{equation*}
\pi_1(X) \approx \pi_1(U) \ast_{\pi_1(U \cap V)} \pi_1(V).
\end{equation*}

I am not sure of how one could compute the free product with amalgamation? If I am not mistaken, a generator of $\pi_1(U \cap V)$ is sent to both the generator of $\pi_1(U) = \langle a \rangle$ represented by the image of the word $a^2$ and the generator of $\pi_1(V) = \langle b \rangle$ represented by the image of the word $b^2$. Therefore, the fundamental group should look like
\begin{equation*}
\pi_1(X) \approx \big \langle a, b \big | a^2 = b^2\big \rangle.
\end{equation*}

Is this correct? If not, could someone explain me where are the flaws in the argument?

答:
Your calculation of the group is correct.

Just a couple of comments though. First, instead of $\approx$, it is much more common to use $\cong$ (\cong) to denote isomorphic groups. Secondly, you said the following:

... a generator of $\pi_1(U \cap V)$ is sent to both the generator of $\pi_1(U) = \langle a \rangle$ represented by the image of the word $a^2$ and the generator of $\pi_1(V) = \langle b \rangle$ represented by the image of the word $b^2$.

The words $a^2$ and $b^2$ are not generators of $\pi_1(U)$ and $\pi_1(V)$ respectively. Instead you could say

... a generator of $\pi_1(U \cap V)$ is sent to both the word $a^2$ in $\pi_1(U) = \langle a \rangle$ and the word $b^2$ in $\pi_1(V) = \langle b \rangle$.


问:
Is $e^x=\sum_{n=0}^{\infty}\frac{e^{x-1}}{n!}$ a known thing?
I don't know much about math though I want to learn everything about it.
I was just in the wikipedia page of the Euler's number which says

$e=\displaystyle\sum_{n=0}^{\infty}\frac{1}{n!}$

I also know something about the Exponential function $f(x)=e^x$

So what I just done is multiplying $\displaystyle\sum_{n=0}^{\infty}\frac{1}{n!}$ by itself to get $e^2$

and I got something like this:
$e+\frac{e}{2}+\frac{e}{6}+\frac{e}{24}+\frac{e}{120}...$

and then i figured out that $e^x$ equals $\displaystyle\sum_{n=0}^{\infty}\frac{e^{x-1}}{n!}$

Is this a known thing? Sorry for wasting your time if it is, I just want to know more about these things.

答: If this all started because you were squaring the series $\sum_n\tfrac{1}{n!}$, you probably want to see how we show the exponential series satisfies $\exp(x+y)=\exp x\exp y$. It follows by the binomial theorem:$$\begin{align}\exp(x)\exp(y)&=\sum_{k,\,l\ge0}\frac{x^ky^l}{k!l!}\\&=\sum_{k,\,l\ge0}\frac{\binom{k+l}{k}x^ky^l}{(k+l)!}\\&\stackrel{\star}{=}\sum_{n\ge0}\frac{1}{n!}\sum_{k=0}^n\binom{n}{k}x^ky^{n-k}\\&=\sum_{n\ge0}\frac{(x+y)^n}{n!}\\&=\exp(x+y).\end{align}$$In particular, $\star$ defines $n:=k+l$ and rearranges a double sum by the Fubini principle, exploiting the fact that $\tfrac{1}{n!}$ decays superexponentially, and therefore there are no convergence issues. (The above principle's equivalent on integrals is what we would normally intend by "Fubini's theorem".)


问:
Finding a rate of sphere area increase given its volume increase
Volume of sphere, $V = \dfrac43 \pi r^3$

Surface area of sphere $S = 4 \pi r^2$

If we know, $\dfrac{dV}{dt} = R$

Let us consider both volume and area as composite functions, thus
$$\dfrac{dV}{dt} = \dfrac{dV}{dr} \times \dfrac{dr}{dt} = 4 \pi r^2 \times \dfrac{dr}{dt} = R$$
whence $\dfrac{dr}{dt} = \dfrac {R}{4 \pi r^2}$
since $\dfrac{dS}{dt} = 6 \pi r  \dfrac{dr}{dt}$, let the value of the $\dfrac{dr}{dt}$ into the second equation, to get the answer.
Is this approach logically correct? Thank you.

答:
Everything is fine until you did

whence $\dfrac{dr}{dt} = \dfrac {R}{4 \pi r^2}$
since $\dfrac{dS}{dt} = 6 \pi r  \dfrac{dr}{dt}$

$$\frac{dS}{dt}=8 \pi r  \frac{dr}{dt}$$
therefore
$$\frac{dS}{dt}=8 \pi r \frac {R}{4 \pi r^2}$$
and finally
$$\frac{dS}{dt}=\frac{2R}{r}$$


问:
From universal measurability to measurability

Let $(\Omega,\Sigma)$ be a measurable space and $K$ be a compact
  metrizable space endowed with its Borel $\sigma$-algebra
  $\mathcal{B}(K)$. Let $A\subseteq\Omega\times K$ be universally
  measurable and such that $$C_\omega=\{x\in K:(\omega,x)\in A\}$$ is
  closed for all $\omega\in\Omega$. Let $\Sigma_u$ the universal
  completion of $\Sigma$.  Is it then the case that $A$ is in
  $\Sigma_u\otimes \mathcal{B}(K)$?

I face the problem when I want to obtain a certain set-valued function that satisfies a strong measurability condition. It is enough for my purposes to get a closed-valued set-valued function with a graph measurable with respect to the completion on my underlying probability space. I can obtain the desired set-valued function as a projection, but this only gives me universal measurability of the graph.

As a first step, it might be intersting to know whether the conjecture holds in the case that $A(\omega)$ contains a single element for all $\omega$, so that it really is a function.

If $f:\Omega\to K$ has a universally measurable graph, is $f$ then
  $\Sigma_u$-$\mathcal{B}(K)$-measurable?

I have asked this now also on Mathoverflow.

答:
I found a solution that suffices for what I do. It is based on strengthening the assumption that the graph $A$ is universally measurable to it being analytic. The notion of analyticity being used is that a subset $S$ of a measurable space $(M,\mathcal{M})$ is analytic if there is a compact metric space $T$ with Borel $\sigma$-algebra $\mathcal{B}(T)$ and a product measurable set $X\in\mathcal{M}\otimes\mathcal{B}(T)$ such that $S=\pi_M(X)$, where $\pi_M$ is the projection onto $M$. Analytic sets are always universally measurable. This notion of analyticity is extensively developed in the book Probability and Potential by Dellacherie and Meyer. A nice guide to the essentials can be found in this paper (JSTOR required).

The countable intersection or union of analytic sets is again analytic. If an analytic set is analytic in the product of some measurable space and a compact metric space, then the projection on the first coordinate is again analytic. An important fact from the theory of set-valued functions on a measurable space is that if the values are closed subsets of separable metric space, then the following condition is sufficient for the graph to be product measurable: The set $$\{\omega:C_\omega\cap O\neq\emptyset\}$$ is measurable for each open set $O$.

So assume that $A$ is analytic and $O$ is open. Then $$\{\omega:C_\omega\cap O\neq\emptyset\}=\pi_\Omega \big(A\cap(\Omega\times O)\big)$$ and hence analytic by the facts above and therefore universally measurable.


问:
Question about tangent circles related to complex analysis
Consider two circles $C_1$ and $C_2$, tangent at a point $p$, with
$C_2$ contained in the disk bounded by $C_1$. In the region between $C_1$ and $C_2$,
construct an infinite chain of circles $S_1$, $S_2$, $S_3$,..., each of them tangent to both
$C_1$ and $C_2$, and with $S_n$ tangent to $S_{n+1}$ at a point $p_n$, as seen in the figure

Now the strange thing thing about this question is that it is an exercise coming from an example sheet about conformal maps in the complex plane and Möbius transformations. Using geometry I am unable to solve this question, but since it is asked in a complex analysis course it suggests that the problem should be solved using complex analysis techniques. Unfortunately, I don't quite see the link between this question and complex analysis. Any help would be welcome!

答:
Möbius transformations preserve angles and map circles to circles or lines. Here it is convenient to map the entire configuration with $f(z) = 1/(z-p)$. $f(p) = \infty$ (on the extended complex plane) so that the circles are mapped to parallel lines, and the region between them is mapped to a parallel strip.

If the circles are symmetric with respect to the real axis then their images are lines parallel to the y-axis.

Finding a chain of circles tangent to both lines and tangent to each other is not difficult. Then map the configuration back with $f^{-1}$.

Remark: Circles and lines in the complex plane are sometimes called “generalized circles” because they correspond exactly to circles on the Riemann sphere via the stereographic projection. With that notation one can simply say that Möbius transformations map generalized circles to generalized circles.


问:
Understanding why this linear operator is diagonalizable.
let me show you the context of my question:

Let $T:V \to V$ a linear operator, where $V$ is a finite dimensional vector space and $m_t(x)=(x - \lambda_1)^{m_1} \cdots(x-\lambda_k)^{m_k}$.
In the proof of Primary decomposition Theorem we take $q_j=\displaystyle\prod_{{i=1}}^{k}(x- \lambda_i)^{m_i}$, with $j \neq i$ then since $gdc(q_1,q_2,\dots,q_k)=1$ there exists $f_1,f_2,\dots,f_k$ s.t $f_1q_1+f_2q_2+\dots +f_kq_k = 1$.

So, if we call $g_j= f_jq_j$ we have that the operator $P_j(T) = g_j(T)$ is a projection s.t

$P_jP_i = 0$ if $j \neq i$

$I = P_1+\cdots+P_k$

$Im Pj = \ker (T - \lambda_jId)^{m_j}= W_j$.

Using that we prove that $V = W_1 \oplus\cdots \oplus W_k$.

Ok, I have a simple question, in the condiction above, why $D = \lambda_1P_1 + \dots+ \lambda_kP_K$ is diagonalizable??

I know that for each $v \in V$, by 2. we have $v = P_1(v)+ \cdots + P_k(v)$ and by 1. we have $$D(v) = D(P_1(v)+ \cdots+ P_k(v))= (\lambda_1P_1(v)+ \cdots+ \lambda_kP_k(v))(P_1(v)+ \cdots+ P_k(v)) = \lambda_1P_1(v)+ \cdots+\lambda_kP_k(v).$$

I have no Idea what can I do with that, so can you help me?

答: Since $P_j$ is a projection (so $P_j^2=P_j$) with image$~W_j$, it acts as the scalar $1$ (i.e., as identity) on vectors in $W_j$, and by condition 1 it acts as the scalar$~0$ on any vector in $W_i$ with $i\neq j$. Then the linear operator $D=\lambda_1P_1+\cdots+\lambda_kP_k$ acts as the scalar$~\lambda_i$ on vectors in $W_i$, in other words such (nonzero) vectors are eigenvectors of$~D$ for the eigenvalue $\lambda_i$. Since the (direct) sum of the subspaces $W_i$ equals the whole space, this makes $D$ diagonalisable, and you can find a basis of eigenvectors by combining bases of the subspaces$~W_i$.


问:
What is the opposite of a regular curve?
The definition in Wikipedia is:

The curve $\gamma: I \to \mathbb R^n$ $r$-times continuously differentiable with $I$ a non-empty interval of real numbers is regular of order $m$ (where $m\leq r)$ if, for every $t\in I,$

$$\{\gamma'(t), \gamma''(t),\cdots,\gamma^{(m)}(t)\}$$

is a linearly independent subset of $\mathbb R^n.$ In particular, a parametric $C^1$-curve $\gamma$ is regular iff $\gamma'(t)\neq \mathbf 0$ for any $t\in I.$

Does this imply that a particle moving through the curve constantly changes velocity? What is the motivation of this definition? What is the opposite of a regular curve?

答:
If the $C^1$ curve $\gamma\colon I\to\mathbb R^n$ is regular, then by definition, $\gamma'\left(x\right)\ne\mathbf 0$ for all $x\in I$. This means that curve $\gamma$ never "slows down" to stop.

For example, constant curve $\gamma_1\left(t\right)=\mathrm{x}_0$ is not regular because $\gamma'_1=\mathbf 0$. On the other hand, straight line $\gamma_2\left(t\right)=\mathrm x_0 + t\cdot \mathrm x_1$ where $\mathrm x_1\ne \mathbf 0$ is regular because $\gamma_2'=\mathrm x_1\ne\mathbf 0$. This curve is not regular of order 2 (2-regular) because $\gamma_2''=\mathbf 0$. Circle, $\gamma_3\left(t\right)=\left(\cos t, \sin t\right)$, is regular of order 2. Notice that all three curves are $C^{\infty}$.


问:
How to use method of moments to estimate mean?
I have given the density function
$$f(\epsilon,\mu) = (1-\epsilon)\frac{1}{3}e^{-x/3}+\epsilon\frac{1}{\mu}e^{-x/\mu}.$$
And would like to estimate $\mu$ using the method of moments and then numerically
maximize the likelihood as a function of $\epsilon$. I know that mgf is
$$M_n(x) = E[(X-EX)^n],$$
but I don't know how to calculate this to use method of moemnts in thus task. Will appreciate a hint, thank you!

答: Since $f$'s support being $[0,\,\infty)$ implies$$E[X^n]=n!((1-\epsilon)3^n+\epsilon\mu^n),$$we get $(1-\epsilon)3+\epsilon\mu=M_1$ and $(1-\epsilon)9+\epsilon\mu^2=M_1^2+M_2$. In particular,$$\epsilon\mu(3-\mu)=3M_1-M_1^2-M_2$$is a quadratic in $\mu$ we can solve to express it in terms of $M_1,\,M_2,\,\epsilon$ (just make sure you take $\mu>0$). MLE is then equivalent to maximizing the sample mean of $\ln f$ as a function of $\epsilon$, after we've removed its explicit $\mu$-dependence by substitution.


问:
An integral that can be solved by trigonometric substitution must be solved by algebraic substitution
EDIT:THANK YOU FOR THE HELP, I am sorry I did not include my progress so far I will learn from all of your feedback

I am new to calculus 2 and I would appreciate some help, our teacher gave us this question:

$$\int \frac{\sqrt{z^2-49}}{z}\,dz$$

He showed us the answer which is

$$-7\sec^{-1}\frac{z}{7}+\sqrt{z^2+49}+C$$

But the catch is we have to solve it using algebraic substitution, not trigonometric substitution. I am really stumped I can't seem to get that answer,  I would really be grateful for the help, I apologize if this question seems easy to you

答:
Let
\begin{align}
u &= \sqrt{z^2-49} \implies z^2 = u^2 + 49\\
du &= \frac{z}{\sqrt{z^{2} - 49}}\,dz \implies dz = \frac{u}{z}\,du
\end{align}
which gives us
\begin{align}
\int \frac{\sqrt{z^2-49}}{z}\,dz &= \int \frac{u^2}{z^2}\,du\\
&= \int\frac{u^2}{u^2 + 49}\,du\\
&= \int \frac{u^2 + 49 - 49}{u^2 + 49}\,du\\
&= \int \left(1 + \frac{1}{u^2+49}\right)\,du\\
\end{align}

Can you take it from there?


问:
Failures of Spherical Coordinates: What do we do with points on the $z$-axis?
For a set of spherical coordinates ($\rho, \theta, \phi$), we'll define $\rho= \sqrt{x^2 + y^2 + z^2}$, $\theta \in [0, 2\pi]$ as the rotation away from the $x$-axis (forward-back horizontal axis), and $\phi \in [0, \pi]$ as the rotation down from the $z$-axis (vertical axis). So, routine stuff, but just to dispel any ambiguities.

Consider the point with Cartesian/ rectangular coordinates (0,0,10); really we could generalize this to some sort of (0,0,$z$) for $z \in \mathbb{R}$. Suppose we want to convert this into spherical coordinates.

What are we supposed to do here?

From some self-teaching, I've essentially just thought of going from rectangular to spherical coordinates as solving this "system":

$\begin{cases}
x = \rho \sin\phi \cos \theta \\
y = \rho \sin \phi \sin \theta \\
z = \rho \cos \phi
 \end{cases}$

It's quickly clear that we run into an issue, and I haven't yet found a definitive answer anywhere; only tangentially related answers and wiki's. $\phi = 0 \implies \sin \phi = 0$, so we cannot determine what $\theta$ is.

Thinking geometrically, wouldn't we be able to give any theta considering we're going straight up the $z$-axis? Would that not indicate the same point? Testing my hypotheses with some online calculators yielded nothing: the ones I tried just didn't spit out an answer, so the other possibility I can imagine is just that there is... well, no spherical representation for these such points. I assume this system is rigorous, well-defined, etc. (and that, therefore, there is an answer to this), but this seems to be quite an oversight.

TLDR; how do we handle points on the $z$-axis with the spherical coordinate system?

答:
It is hard to answer this because it is hard to know what you are asking precisely. You haven't yet run into any difficulties but you are suspicious of the idea that the change of coordinates map is not a bijection.

This can be a problem when you are trying to push around jacobians near the z axis or something.

The answer is that in 3d minus a neighborhood around the z axis everything will work well. And in the case where you have to do something fancy around the z axis you will want to do a change of coordinates so that you are working with a different set of spherical coordinates near that problem area.

This is why we have the machinery of atlases and tangent bundles, etc ..in differential geometry. To deal with these problems that might arise in a general way by looking locally, however you may be best to not worry too much about it while you go through your earlier classes.


问:
is first hitting time of a general non-cadlag process a stopping time?
Suppose I have an increasing random process $A_t$ with non-right-continuous filtration ($A_t$ can have double jumps). $A_t$ is predictable and its right limit $A_{t+}$ is optional. Consider
$$\tau=\inf \{t>0 : A_{t+}>c\} ,$$
$c$ is some positive constant. Is the random time $\tau$ a stopping time?
I guess, we can write $\{\tau>t\}= \cap_{s\le t}\{A_{s+}\le c\} \in \mathcal{F}_t$, so it is a stopping time, am I correct?

答:
Your description of the event $\{\tau>t\}$ is not quite right. Consider, for example, deterministic $A_t(\omega) = t$ for all $t$, $\omega$. Then $\tau(\omega) = c$ so $\{\tau>t\}$ is either empty (if $t\ge c$)  or the entire sample space (if $t<c$), but $\cap_{s\le t}\{A_{s+}\le c\}$ is either empty (if $t>c$) or the entire sample space (if $t\le c$).

The example hints at the truth: In general, $\tau$ is a stopping time of $(\mathcal F_{t+})$, but not necessarily of $(\mathcal F_t)$. This because $\{\tau<t\}=\cup_{s<t, s\in\Bbb Q}\{A_{s+}>c\}\in \mathcal F_t$.


问:
about isomorphic between polynomial ring
let $k$ be algebraic close field , why following rings are isomorphic ?

$k[x, y] /\left(y-x^{2}\right) \cong k\left[x, x^{2}\right] \cong k[x]$

i need theorem for general case for example when $k[x, y] /\left(f(x,y)\right)  \cong k[x]$ ? for what $f(x,y)$ this is true ?

答:
There is an isomorphism $f:k[x]\to k[x,y]/(y-x^2)$ sending $x\mapsto x$. The inverse homomorphism is $g:k[x,y]/(y-x^2)\to k[x]$ sending $x\mapsto x$ and $y\mapsto x^2$.

Note that this does not require $k$ to be algebraically closed.


问:
Proving Bijections in $\mathbb{R}^n$
I have a question that may seem trivial or silly, however I'll try to make my point clear. I'm a student of Mathematical Physics and unfortunately my college doesn't offer a set theory course, so that we are not taught to prove bijections between arbitrary sets. The single variable calculus course we have teaches us to prove bijections by looking for one inverse function in a very clear way: we write $y = f(x)$, try to solve for $x$ and then create the function desired (it's of course more formal than that, first we prove injectivity showing that $f(x)=f(y)$ implies $x=y$ and then we look for the inverse to prove surjectivity).

The multivariable calculus course on the other hand doesn't talk about proving bijections from subsets of $\mathbb{R}^n$ to subsets of $\mathbb{R}^m$. On the other hand we have the analysis course, the topology course and the differential geometry course that all presumes we know how to prove those bijections and so on.

Now, I'm confused, because none of our courses teaches to prove that kind of thing. If we have some function $f: \mathbb{R}^n \to \mathbb{R}^m$, proving injectivity is the easiest part, but for surjectivity, seeking for some inverse seems kind of painful. By now I'm studying differential geometry of curves and surfaces (it uses Do Carmo's book), and the only thing I'm needing is this basic notion on how to prove bijections in these cases (for instance, when we come to show that something is a chart, proving continuity is rather easy, we have been taught on how to do it, but proving the bijection is complicated).

Is there any general guidelines as there is in the case of functions from $\mathbb{R}$ to $\mathbb{R}$? Is there any reference that can be given that treats this in a good way? Again, I know this is basic and that we all should know before coming to things like differential geometry, but since the course doesn't teach this, I came to ask here to fill this gap.

Thanks very much in advance, and sorry if this question is silly.

答:
Let $X$ and $Y$ be sets, and let $f: X \to Y$ be a function.   We say that $f$ is injective if for all $x, y \in X$,  $x \ne y \implies f(x) \ne f(y)$.

We say that $f$ is surjective if for every $y \in Y$, there exists $x \in X$ such that $f(x) = y$.  A function is bijective if it is both injective and surjective.

One does not need to find an inverse to prove that a function is surjective, and in general, there will be no inverse unless the function is bijective.

Does that clear things up?

Title: Proving Bijections in $\mathbb{R}^n$


问:
Limit of $L^p$ norm when $p\to0$
Let ($\Omega$, $\cal{F}$, $\mu$) be a probability space and $f\in L^1(\Omega)$. Prove that

$$\displaystyle\lim_{p\to 0} \left[ \int_{\Omega}|f|^pd\mu \right]^{\frac{1}{p}}=\exp \left[ \int_{\Omega}\log|f| d\mu  \right],$$

where $\exp[-\infty]=0$. To simplify the problem, we may assume $\log|f|\in L^1(\Omega).$

答:
@DavideGiraudo: This is a long comment an so I laid it as an answer.

I did not find a direct way to fix the problem for the case where $\log|f|\notin L_1(\mu)$ in the spirit of our solution.

I did however obtained a slightly different solution  that applies to whether $\log|f|$ is integrable or $\int(\log|f|)_-=\infty$. Here is a sketch:

Basically one notices that for any $a>0$, the map $\phi_a(p)=\frac{a^p-1}{p}$ is monotone nondecreasing on $(0,\infty)$ (due to convexity of $p\mapsto a^p$) and so, $g_p:=(|f|-1)-\frac{|f|^p-1}{p}$ is positive and nondecreasing as $p\searrow0$.  Monotone convergence implies that
$$\lim_{p\rightarrow0+}\int g_p=\int \lim_{p\rightarrow0+}g_p=\int(|f|-1-\log|f|)$$
Thus, $\lim_{p\rightarrow0+}\int\frac{|f|^p-1}{p}=\int\log|f|$ regardless of integrability of $\log|f|$. The conclusion follows now from the inequality $\log(a)\leq a-1$ for all $a>0$ and  Jensen's inequality:
\begin{align}
\int_\Omega\log|f|\,d\mu&= \frac{1}{p}\int_\Omega\log(|f|^p)\,d\mu\leq \frac{1}{p}\log\Big(\int_\Omega|f|^p\,d\,\mu\Big)=\log\|f\|_p\\
&\leq \frac{\|f\|^p_p-1}{p}=\int_\Omega\frac{|f|^p -1}{p}\,d\mu\xrightarrow{p\rightarrow0+}\int_\Omega\log|f|\,d\mu
\end{align}

Title: Limit of $L^p$ norm when $p\to0$