问:
What does the inverse FFT matrix look like without using negative powers
Edit: I know screen readers can't read images but I think this image adds more context regarding my notation which @Ian clarifies in the comments.

This is from the Graduate Algorithms course on Udacity with Georgia Tech using FFT for polynomial multiplication.

First row is all 1's.
Second row is $1, W_n^{-1}$, ... $W_n^{-(n-1)}$ that can be rewritten not using negative powers as $1, W_n^{n-1}, W_n^{n-2},...,W_n^1$
The third row I can start but not finish:
$1, W_n^{n-2}, W_n^{n-4},...,$

The last term in the third row with negative exponents is $W_n^{-2(n-1)}$ so I tried to use my limited algebra skills in figuring out what multiplied by $W_n^{2(n-1)}$ would equal $W_n^n = W_n^0$?

I let $2(n-1) + x = n$ and got $x = -n +2$ so I have for that last term in the third row, $W_n^{-2n+2}$ which can be rewritten as $W_n^{-2(n+1)}$ but I cannot visualize where this Omega would be on the unit circle on the complex plane where as for the second row I can clearly see that it starts at the last Omega and goes around clock-wise.

答:
I followed @Ian's advice and tried for $n=8$

The last element of row 3 is $W_8^{2(8-1)}$ which is $W_8^{14}$ which on the unit circle on the complex plane is at $-i$ or in polar coordinates $(1, 6pi/4)$.  Multiply this by $(1, 8pi/4)$ since $W_8^8$ is at 1 or $(1, 2pi)$ which gives me $(1, 14pi/4)$ which is at positive $i$.  So the final term of the row 3 can be rewritten as $W_n^2$.


问:
$J$ be a $3\times 3$ matrix with all entries $1$ Then $J$ is
$J$ be a $3\times 3$ matrix with all entries $1$ Then $J$ is

Diagonalizable

Positive semidefinite

$0,3$ are only eigenvalues of $J$

Is positive definite

$J$ has minimal polynomial $x(x-3)=0$ so 1, 2,3 are true , am I right?

答:
You are correct. This is a multiple duplicate but I can't find any.

The matrix has rank $1$, which makes $0$ an eigenvalue of multiplicity $2$. Then the trace tells us what the other eigenvalue is, whence diagonalizability, since it adds the missing dimension. Since $0$ is an eigenvalue, it can't be positive definite. But it is indeed positive semidefinite by diagonalization in an orthonormal basis ($J$ is symmetric, eigenvalues are nonnegative). The minimal polynomial is obvious by diagonalization.

Alternative: we can observe that $J^2=3J$ so $J$ is annihilated by $X^2-3X=X(X-3)$, which is therefore divided by the minimal polynomial. Since $J$ is nether $I_3$ nor $3I_3$, the latter must be the minimal polynomial. It splits and has simple roots. Whence 1 and 3.

ContentLicense: CC BY-SA 3.0


问:
Proving $\sum_{k=0} \frac{(-1)^k}{1+2k} = \frac{\pi}{4} $ taking $4 \int_0^1 \sqrt{1 - x^2} \; \text{d}x$ as the definition of $\pi$.
I'm trying to prove the Madhava-Leibniz formula for $\pi$ using one quarter of the area of the unit circle as the definition of $\frac{\pi}{4}$.

$$ \int_{0}^{1} \sqrt{1-xx}\; \text{d}x = \sum_{k=0}^{\infty} \frac{(-1)^k}{2k+1} $$

I have a proof of this identity, but it uses a lot of trigonometric substitutions and makes direct use of $\pi$ as a constant.

This feels a bit circular since the left hand side is supposed to be the definition of $\frac{\pi}{4}$.

Is there a way to prove this identity directly, instead of proving that both sides are equal to $\frac{\pi}{4}$?

All integrals are over the interval $[0,1]$ in the variable $x$. Let $\theta$ refer to $\frac{\pi}{4} x$. $\theta$ is a constant multiple of $x$, I'm using it so I don't need to change the bound of integration.

$$ \sum_{k=0} \frac{(-1)^k}{1+2k} $$

Using Lemma 101.

$$ \sum_{k=0} \int_{0}^{1} (-1)^k x^{2k} \text{d}x $$
$$ \sum_{k=0} \int_{0}^{1} (-xx)^k \text{d}x $$

Exchanging the sum and integral.

$$ \int_{0}^{1} \sum_{k=0} (-xx)^k \text{d}x $$

Applying Lemma 102

$$ \int_{0}^{1} \frac{1}{1+xx} \text{d}x $$

Applying Lemma 103, with $g$ equal to $\tan(\theta)$ and $g'$ equal to $\frac{\pi}{4} \sec^2(\theta)$.

$$ \int_{0}^{1} \frac{\pi}{4} \cdot \sec^2(\theta) \cdot \frac{1}{1+\tan^2(\theta)} \text{d}x $$

$$  \frac{\pi}{4} \int_{0}^{1} \sec^2(\theta) \cdot \frac{1}{\sec^2(\theta)} \text{d}x $$

$$  \frac{\pi}{4} \int_{0}^{1} 1\; \text{d}x $$

$$ \frac{\pi}{4} $$

And for the other side.

$$ \int_0^1 \sqrt{1-xx} \; \text{d}x $$

Apply lemma 103, with $g = \sin(2\theta)$ and $g' = \frac{\pi}{2}\cos(2\theta)$.

$$ \int_0^1 \frac{\pi}{2} \cdot \cos(2 \theta) \sqrt{1-\sin^2(2 \theta)} \; \text{d}x $$

$$ \frac{\pi}{2} \int_0^1 \cos^2(2 \theta) \; \text{d}x $$

Apply lemma 104.

$$ \frac{\pi}{2} \cdot \frac{1}{2} $$

$$ \frac{\pi}{4} $$

Lemmas

Lemma 101: integral of monomials

$$ \int_0^1 x^n \text{d}x = \frac{1}{1+n} $$

Lemma 102: sum of geometric series

$$ \sum_{k=0} r^k = \frac{1}{1-r} $$

Lemma 103: Special case of change of variables for integration

Suppose $g(0)=0$ and $g(1)=1$, then the following holds.

$$ \int_0^1 f(x) \text{d}x = \int_0^1 g'(x) \cdot f'\circ g(x) \text{d}x $$

Lemma 104: Integral of $\cos^2 + \sin^2$

For all values of $x$, $\cos^2(x) + \sin^2(x) = 1$, therefore the following holds

$$ \int_0^1 \cos^2(2\theta) + \sin^2(2\theta) \text{d}x = 1 $$

However, $\cos\left(\frac{\pi}{2}-x\right) = \sin(x) $, therefore, over the interval $[0, 1]$, $\sin$ and $\cos$ take on the same values at the same offsets, but in the opposite order. Thus

$$ \int_0^1 \sin^2(2\theta) \text{d}x = \int_0^1 \cos^2(2\theta) \text{d}x $$

Thus

$$ \int_0^1 \cos^2(2\theta) \text{d}x = \frac{1}{2} $$

ContentLicense: CC BY-SA 4.0

Body: Substitute $x=t/\sqrt{1+t^2}$; then $dx=dt/(1+t^2)^{3/2}$ and $$\int_0^1\sqrt{1-x^2}\,dx=\int_0^\infty\frac{dt}{(1+t^2)^2}=\int_0^1\frac{dt}{(1+t^2)^2}+\int_0^1\frac{dt/t^2}{(1+1/t^2)^2}=\int_0^1\frac{dt}{1+t^2}.$$ Now you know what to do next.


问:
Computing $\log(2)$ as a period
I would like to prove
$$\int_0^1 \frac{xdx}{\log\frac{1}{1-x}} = \log 2$$
and for this I tried to make the change of variables $x \mapsto 2x-x^2$ in
$$\int_{2\varepsilon - \varepsilon^2}^1 \frac{dx}{\log (1-x)}$$
but I don't see the relation (and maybe I miss a classical integral representation of $\log 2$ too).

答:
Use $x\mapsto 1-x$ instead (this substitution preserves the interval of integration):

$$\int_0^1\frac{x\,\mathrm{d}x}{\log\frac1{1-x}}\overset{x\mapsto 1-x}=-\int_0^1\frac{1-x}{\log(x)}\ \mathrm{d}x=\int_0^1\frac{x-1}{\log x}\ \mathrm{d}x$$

The remaining integral is easily recognized as Frullani type after enforcing $x\mapsto e^{-x}$:

$$\int_0^1\frac{x-1}{\log x}\ \mathrm{d}x\overset{x\mapsto e^{-x}}=\int_0^\infty \frac{e^{-x}-e^{-2x}}{x}\ \mathrm{d} x=\log2$$

Title: Computing $\log(2)$ as a period


答:
Let $x \mapsto 1-x$ and use $\displaystyle \int_0^1 x^{\lambda} \, \mathrm{d}\lambda = \frac{x-1}{\log{x}}.$

$$\displaystyle \displaystyle \begin{aligned}I & = \int_{0}^{1}\frac{x-1}{\log{x}}\;{dx} \\&= \int_{0}^{1}\int_{0}^{1}x^{\lambda}\;{d\lambda}\;{dx} \\& =\int_{0}^{1}\int_{0}^{1}x^{\lambda}\;{dx}\;{d\lambda} \\&= \int_{0}^{1}\frac{1}{1+\lambda}\;{d\lambda} \\&= \log{2}.\end{aligned}$$

Title: Computing $\log(2)$ as a period

Body:
I would like to prove
$$\int_0^1 \frac{xdx}{\log\frac{1}{1-x}} = \log 2$$
and for this I tried to make the change of variables $x \mapsto 2x-x^2$ in
$$\int_{2\varepsilon - \varepsilon^2}^1 \frac{dx}{\log (1-x)}$$
but I don't see the relation (and maybe I miss a classical integral representation of $\log 2$ too).


问:
Forming a group from power set pairs
Question
I want to form a group out of disjoint pairs taken from the power set $\mathcal{P}(X)$ over a finite set $X$. An element of the group $G$ is a pair $u=(u_1,u_2) \in \mathcal{P} \times \mathcal{P}$ such that $u_1 \cap u_2 = \emptyset$. Are there any ways to form a group on these disjoint pairs of elements?

Attempt
I thought to use the symmetric difference $a \Delta b = (a \cup b) \backslash (a \cap b)$ as done for the regular power set group, but for both pairs. Take $u,v \in G$ and the group operation gives $w$ defined as:
\begin{eqnarray}
w = u \circ v &=& (u_1 \Delta v_1,\;u_2 \Delta v_2 \backslash [(u_1 \Delta v_1) \cap (u_2 \Delta v_2)])\\
&=&(u_1 \Delta v_1,\;(u_2 \Delta v_2) \backslash (u_1 \Delta v_1)).
\end{eqnarray}
The term in the square brackets ensures that the new pair is disjoint: $w_1 \cap w_2 = \emptyset$. This is closed in $G$ since $G$ contains every pair of disjoint sets. To form a group: (a) this has an identity $(\emptyset, \emptyset)$, (b) each element is its own inverse (an involution), (c) I'm unclear about associativity. I tried showing that it's associative by expanding both $a \circ (b \circ c)$ and $(a \circ b) \circ c$, but I'm running into some trouble. So far, I have:
\begin{eqnarray}
a \circ (b \circ c) &=& a \circ (b_1 \Delta c_1,\; b_2 \Delta c_2 \backslash [(b_1 \Delta c_1) \cap ( b_2 \Delta c_2)])\\
&=& (a_1 \Delta b_1 \Delta c_1,\; a_2 \Delta (b_2 \Delta c_2 \backslash  [(b_1 \Delta c_1) \cap (b_2 \Delta c_2)])\\
\end{eqnarray}
and
\begin{eqnarray}
(a \circ b) \circ c &=& (a_1 \Delta b_1,\; a_2 \Delta b_2 \backslash [(a_1 \Delta b_1) \cap ( a_2 \Delta b_2)]) \circ c\\
&=& (a_1 \Delta b_1 \Delta c_1,\; (a_2 \Delta b_2 \backslash [(a_1 \Delta b_1) \cap ( a_2 \Delta b_2)])\Delta c_2)\\
\end{eqnarray}
The left parts of the pairs match, but the right parts need some work. So we want to check whether the right parts are equal, i.e. that:
$$
a_2 \Delta (b_2 \Delta c_2 \backslash  [(b_1 \Delta c_1) \cap (b_2 \Delta c_2)] = (a_2 \Delta b_2 \backslash [(a_1 \Delta b_1) \cap ( a_2 \Delta b_2)])\Delta c_2
$$
I'm getting stuck here just because this expands (in $\Delta$) to a lot of terms. Perhaps there's a smart to check this?

答:
Your construction cannot give a group if your set has more than one element.

Consider $a=(\{1,2\},\varnothing)$, $b=(\varnothing,\{2\})$, and $c=(\varnothing,\varnothing)$. Then $b\neq c$, but
$$\begin{align*}
a\circ b&= (\{1,2\},\varnothing)\circ (\varnothing,\{2\})\\
&= (\{1,2\},\{2\}-\{1,2\}) = (\{1,2\},\varnothing).\\
a\circ c&= (\{1,2\},\varnothing)\circ (\varnothing,\varnothing) = (\{1,2\},\varnothing).
\end{align*}$$
So $b\neq c$, but $a\circ b= a\circ c$. This is impossible in a group.

In fact, one element suffices: set $X=\{\star\}$. Then
$(X,\varnothing)\circ(\varnothing,X)=(X,\varnothing)=(X,\varnothing)\circ(\varnothing,\varnothing)$, so again you cannot have a group.

(Of course, if $X$ is empty then you have the one-element group...)

You ask in comments whether you have a loop. The answer is no, unless $X$ is empty. Consider the elements $(X,\varnothing)$ and $(\varnothing,X)$. There is no element $A$ such that $(X,\varnothing)\circ A = (\varnothing, X)$. Indeed, such an element $A=(R,S)$ would necessarily have $X\Delta R=\varnothing$, hence $R=X$; but then the condition $R\cap S=\varnothing$ requires $S=\varnothing$, and $(X,\varnothing)\circ(X,\varnothing)=(\varnothing,\varnothing)\neq (\varnothing, X)$.

But in a loop, the equations $ax=b$ and $ya=b$ have solutions for all $a$ and $b$, so this is not a loop either.

Title: Forming a group from power set pairs


问:
Showing half plane is open.
Let $A=\{(x,y)\in \mathbb{R}^2|Ax+By<C\}$, if I want to show the half plane is open can I just choose $\epsilon=\frac{|Aa+Bb-C|}{\sqrt{A^2+B^2}}$, then would $B_{\epsilon}((a,b)) \subset A$, for an arbitrary $(a,b)$ in $A$?

答: Yes, but you dont need to be that explicit. Since the distance is greater than 0, such epsilon exists. I say this because it will prove helpful in future exercises, where there is not (or you just don't know) a distance formula suiting your problem.</p>

Title: Showing half plane is open.


答:
The upper half plane $U$ is open.  Therefore, with a rotation and translation, your half plane is open.

Let $z=(x,y)$ with $y>0.$  Then $B_{y/2}(z) \subset U.$

Title: Showing half plane is open.

Body: Let $A=\{(x,y)\in \mathbb{R}^2|Ax+By<C\}$, if I want to show the half plane is open can I just choose $\epsilon=\frac{|Aa+Bb-C|}{\sqrt{A^2+B^2}}$, then would $B_{\epsilon}((a,b)) \subset A$, for an arbitrary $(a,b)$ in $A$?


答:
Your operation need not be associative.

Let $X=\{0,1\}$, $a=\langle\{0\},\{1\}\rangle$, and $b=\langle\{1\},\{0\}\rangle$. Then $a\circ b=\langle X,\varnothing\rangle$, so

$$a\circ(a\circ b)=\langle\{0\},\{1\}\rangle\circ\langle X,\varnothing\rangle=\langle\{1\},\varnothing\rangle\,.$$

However, $(a\circ a)\circ b=b\ne\langle\{1\},\varnothing\rangle$.

Title: Forming a group from power set pairs

Body:
Question
I want to form a group out of disjoint pairs taken from the power set $\mathcal{P}(X)$ over a finite set $X$. An element of the group $G$ is a pair $u=(u_1,u_2) \in \mathcal{P} \times \mathcal{P}$ such that $u_1 \cap u_2 = \emptyset$. Are there any ways to form a group on these disjoint pairs of elements?

Attempt
I thought to use the symmetric difference $a \Delta b = (a \cup b) \backslash (a \cap b)$ as done for the regular power set group, but for both pairs. Take $u,v \in G$ and the group operation gives $w$ defined as:
\begin{eqnarray}
w = u \circ v &=& (u_1 \Delta v_1,\;u_2 \Delta v_2 \backslash [(u_1 \Delta v_1) \cap (u_2 \Delta v_2)])\\
&=&(u_1 \Delta v_1,\;(u_2 \Delta v_2) \backslash (u_1 \Delta v_1)).
\end{eqnarray}
The term in the square brackets ensures that the new pair is disjoint: $w_1 \cap w_2 = \emptyset$. This is closed in $G$ since $G$ contains every pair of disjoint sets. To form a group: (a) this has an identity $(\emptyset, \emptyset)$, (b) each element is its own inverse (an involution), (c) I'm unclear about associativity. I tried showing that it's associative by expanding both $a \circ (b \circ c)$ and $(a \circ b) \circ c$, but I'm running into some trouble. So far, I have:
\begin{eqnarray}
a \circ (b \circ c) &=& a \circ (b_1 \Delta c_1,\; b_2 \Delta c_2 \backslash [(b_1 \Delta c_1) \cap ( b_2 \Delta c_2)])\\
&=& (a_1 \Delta b_1 \Delta c_1,\; a_2 \Delta (b_2 \Delta c_2 \backslash  [(b_1 \Delta c_1) \cap (b_2 \Delta c_2)])\\
\end{eqnarray}
and
\begin{eqnarray}
(a \circ b) \circ c &=& (a_1 \Delta b_1,\; a_2 \Delta b_2 \backslash [(a_1 \Delta b_1) \cap ( a_2 \Delta b_2)]) \circ c\\
&=& (a_1 \Delta b_1 \Delta c_1,\; (a_2 \Delta b_2 \backslash [(a_1 \Delta b_1) \cap ( a_2 \Delta b_2)])\Delta c_2)\\
\end{eqnarray}
The left parts of the pairs match, but the right parts need some work. So we want to check whether the right parts are equal, i.e. that:
$$
a_2 \Delta (b_2 \Delta c_2 \backslash  [(b_1 \Delta c_1) \cap (b_2 \Delta c_2)] = (a_2 \Delta b_2 \backslash [(a_1 \Delta b_1) \cap ( a_2 \Delta b_2)])\Delta c_2
$$
I'm getting stuck here just because this expands (in $\Delta$) to a lot of terms. Perhaps there's a smart to check this?


问:
Polynomial Long Division Algorithm for Integers?
I would like to understand the mathematics behind the standard long division algorithm, specifically for the ring of $\mathbb{Z}$.  I'm looking for an algebraic or set theoretic explanation.   Can anyone explain or point me to an explanation?

I asked here in which I was told the Wiki algorithm works over any field except for $\mathbb{Z}$.  I was then referred to the non-monic algorithm proof in this post.  I tried using the theorem and followed the proof, but I did not get an expected result.  For instance using post:

$G = 1b^1 + 8b^0, F = 9b^0$, the leading coeff $a = 9$ then $i \ge max\{0, 1+deg G- degF\} = 2$.  Substitution:

$a^i G = QF + R = 9^2(1b^1 + 8b^0) = Q9b^0 + R$, where $Q = 1(b^1+8b^0)$ and at this point it appears something went wrong.

Appreciate your guidance

答:
First things first: $\Bbb Z$ is not a field (it's lacking multiplicative inverse for all non-zero elements distinct from $\pm1$). Polynomial long division works over any field $K$ as the polynomial ring $K[x]$ is a euclidean domain in this case. As I mentioned, $\Bbb Z$ is not a field and hence $\Bbb Z[x]$ not necessarily an euclidean domain (it's in fact even less; but that's not important right now).

We can, however, divide polynomials in arbitrary polynomial rings $R[x]$ whenever the leading coefficient of the divident is $1$ (or, more generally, invertible; this was mentioned in the post Bill Dubuque linked to you). In this case the usual algorithm works just fine. If the we have a field instead then every non-zero element is invertible which basically ensures that polynomial long division works out.

On to your specific case: you misread the precise statement of the divison algorithm on multiple occasions. As you have quoted

$$i\color{red}{\ge}\max\{1,1+\deg G-\deg F\}=2$$

which, in particular, includes the case $i>2$. You have only tested the case $i=2$. Moreover, the polynomials $Q,R$ are not of arbitrary degree; in fact $\deg R<\deg F$. But $\deg F=0$ and hence $\deg R<0$ which (by convention) implies $R\equiv0$ and we are only left with the possibility of $a^iG=QF$. This in turn does not leave that much choice for $Q$ other than $a^{i-1}G$ as $F=a$.

I am, honestly speaking, not sure what exactly you are asking.


问:
Equivalence classes of a kernel

Let $f$ be a function with domain $A$ and codomain $B$. Consider the relation $K \subseteq A \times A$ defined on the domain of $f$ by $(x, y) \in K$ if and only if $f(x) = f(y)$. The relation $K$ is called the kernel of $f$.

For the specific case of $A = \mathbb{Z}$, where $\mathbb{Z}$ is the set of integers, let $f: \mathbb{Z} \rightarrow \mathbb{Z}$ be defined by $f(x) = x^2$. Describe the equivalence classes of the kernel for this specific function.

As I understand, I can simplify this down to: find the set of functions where its table looks like:

$x$
$y$

...
...

$-2$
$4$

$-1$
$1$

$0$
$0$

$1$
$1$

$2$
$4$

...
...

I would say $\{f(y) = y^2, f(y) = (-y)^2\}$ is a subset of the equivalence classes. But I'm not sure if that is all. First of all, am I going in the right direction? "Describe the equivalence classes" is kind of vague. Second, how can I be sure that my answer is the complete set of equivalence classes of the kernel?

答:
First: you need to check that the relation is actually an equivalence relation. So define

$$x\sim y\ \iff \ f(x)=f(y)$$

and show reflexivity, symmetry and transitivity. Otherwise it does not make much sense to talk about equivalence classes and such (if this was already checked elsewhere: good; I just think it's an important thing to do).

Now, the equivalence class $[n]$ of $n\in\Bbb Z$ is the subset of $\Bbb Z$ consisting of all elements in $\Bbb Z$ which are equivalent to $n$ under the given equivalence relation. Formally:

$$[n]=\{m\in\Bbb Z\,\mid\,n\sim m\}=\{m\in\Bbb Z\,\mid\, f(n)=f(m)\}\subseteq\Bbb Z$$

You are expected to explictely describe $[n]$ given $n\in\Bbb Z$ arbitrarily. So, given $n\in \Bbb Z$ you want to determine for which $m\in\Bbb Z$ you have

$$n\sim m\ \iff\ f(n)=n^2=m^2=f(m)\,.$$

If not written as subset of $\Bbb Z$: you can write it as subset of $\Bbb Z\times\Bbb Z$ (in the end a relation on a set $S$ is nothing else than a subset of $S\times S$)

$$[n]=\{(n,m)\in\Bbb Z\times\Bbb Z\,\mid\,n\sim m\}=\{(n,m)\in\Bbb Z\times\Bbb Z \,\mid\, f(n)=f(m)\}\subseteq K\subseteq\Bbb Z\times\Bbb Z$$

but this depends on how you have defined equivalence relations and equivalence classes.

Can you do this? that is describe $[n]$ as an explicit subset of $\Bbb Z$?

(Hint: you have to make a minor case distinction)

Title: Equivalence classes of a kernel


问:
Divisibility by 7 Proof by Induction
Prove by Induction that

$$7|4^{2^{n}}+2^{2^{n}}+1,\; \forall n\in \mathbb{N}$$

Base case:

$$
\begin{aligned}
7&|4^{2^{1}}+2^{2^{1}}+1,\\
7&|7\cdot 3
\end{aligned}$$ Which is true.

Now, having $n=k$, we assume that:

$$7|4^{2^{k}}+2^{2^{k}}+1,\;\; \forall k\in \mathbb{N}$$

We have to prove that for $n=k+1$ that,

$$7|4^{2^{k+1}}+2^{2^{k+1}}+1,\;\; \forall k\in \mathbb{N}$$

We know that if $a$ is divisible by 7 then $b$ is divisible by 7 iff $b-a$ is divisible by 7.

Then,
$$
\begin{aligned}
b-a &= 4^{2^{k+1}}+2^{2^{k+1}}+1 - (4^{2^{k}}+2^{2^{k}}+1)\\
&= 4^{2^{k+1}}+2^{2^{k+1}} - 4^{2^{k}}-2^{2^{k}}\\
&= 4^{2\cdot  2^{k}}+2^{2\cdot  2^{k}} - 4^{2^{k}}-2^{2^{k}}
\end{aligned}
$$

I get stuck here, please help me.

答:
Modulo $7,$ $4^{2^n}+2^{2^n}+1 \equiv 2+4+1\equiv 0$ when $n$ is odd, and
to $4+2+1\equiv 0$ when $n$ is even.

(Start with $4^{2^n}\equiv 2$
and $2^{2^n}\equiv 4$ when $n=1.$)

Title: Divisibility by 7 Proof by Induction


问:
Determining the group homomorphism in semidirect product
We know that if $N$ is a normal subgroup, $H$ is a subgroup, and $\varphi$ is the group homomorphism such that $\varphi:H\to$Aut$(N)$. And this gives a unique group, called the outer semidirect product of $N$ and $H$.

But when a group and two of its subgroups are known, how to determine the homomorphisms? For instance, from here we know
$$D_3\times \mathbb{Z}_3=(\mathbb{Z}_3\times \mathbb{Z}_3) \rtimes \mathbb{Z}_2$$
And from here
$$\langle a,b,c| a^3=c^3=b^2=1, ac=ca,ab=ba^{-1},cb=bc^{-1}\rangle=(\mathbb{Z}_3\times \mathbb{Z}_3)\rtimes \mathbb{Z}_2$$
The group homomorphism $\varphi:\mathbb{Z}_2 \to$Aut$(\mathbb{Z}_3\times \mathbb{Z}_3)$ are surely different. But I have no idea on how to determine this homomorphism. Any hint or solutions are welcomed! Thanks!

答:
If $N\trianglelefteq G$ and $H$ is a subgroup such that $HN=G$ and $H\cap N=1$, then $G$ is an internal semi-direct product $N\rtimes H$. This means it can be constructed as an external semidrect product $N\rtimes H$ with the data of a homomorphism $H\to{\rm Aut}(N)$. If we want to characterize this homomorphism when we already know $G$ full well, we think of conjugation. That is, for $h\in H$ and $n\in N$ and homomorphism $\alpha:H\to{\rm Aut}(N)$, then $\alpha(h)$ applied to $n$ yields the element $hnh^{-1}\in N\subseteq G$.

I like to think of semidirect products as special quotients of free products. Recall that relations are equations of the form $\rm blah=blah$, and we can quotient by relations $\rm \color{Blue}{blah}=\color{Purple}{blah},\color{Teal}{blah}=\color{Magenta}{blah},$... by quotienting a group by the normal subgroup generated by $\rm \color{Blue}{blah}\color{Purple}{blah}^{-1}$, $\rm \color{Teal}{blah}\color{Magenta}{blah}^{-1}$, etc. (The normal subgroup generated by a set $S$ is the normal closure of $\langle S\rangle$, i.e. the group generated by the union of all conjugates of $S$.) Then $N\rtimes H$ is isomorphic to $N*H$ modulo $h(n)=hnh^{-1}$ for all elements $h\in H$ and $n\in N$ (the expression $h(n)$ is interpreted as the result of $h$'s action on $n$).


问:
Understanding Poincaré lemma in one variable Griffith and Harris
I would like to understand one little small detail in the proof. How come we end up with a factor of $\frac{-1}{\pi}$. I understand where the $-1$ came from, though I don't know what happened to the $2$. I understand everything else line by line.

$\bar\partial$-Poincaré Lemma in One Variable. Given $g(z)\in C^\infty(\bar\Delta)$, the function
$$f(z)=\frac{1}{2\pi\sqrt{-1}}\int_\Delta\frac{g(w)}{w-z}dw\wedge d\bar w$$
is defined and $C^\infty$ in $\Delta$ and satisfies
$$\frac{\partial f}{\partial\bar z}=g.$$

Proof. For $z_0\in\Delta$ choose $\varepsilon$ such that the disc $\Delta(z_0,2\varepsilon)\subset\Delta$ and write
$$g(z)=g_1(z)+g_2(z),$$
where $g_1(z)$ vanishes outside $\Delta(z_0,2\varepsilon)$ and $g_2(z)$ vanishes inside $\Delta(z_0,\varepsilon)$. The integral
$$f_2(z)=\frac{1}{2\pi\sqrt{-1}}\int_\Delta g_2(w)\frac{dw\wedge d\bar w}{w-z}$$
is well-defined and $C^\infty$ for $z\in\Delta(z_0,\varepsilon)$; there we have
$$\frac{\partial}{\partial\bar z}f_2(z)=\frac{1}{2\pi\sqrt{-1}}\int_\Delta\frac{\partial}{\partial\bar z}\left(\frac{g_2(w)}{w-z}\right)dw\wedge d\bar w=0.$$
Since $g_1(z)$ has compact support, we can write
\begin{align*}
\frac{1}{2\pi\sqrt{-1}}\int_\Delta g_1(w)\frac{dw\wedge d\bar w}{w-z} & =\frac{1}{2\pi\sqrt{-1}}\int_\mathbb{C} g_1(w)\frac{dw\wedge d\bar w}{w-z}\\
& =\frac{1}{2\pi\sqrt{-1}}\int_\mathbb{C} g_1(u+z)\frac{du\wedge d\bar u}{u},
\end{align*}
where $u=w-z$. Changing to polar coordinates $u=re^{i\theta}$ this integral becomes
$$f_1(z)=-\frac{1}{\pi}\int_\mathbb{C}f_1(z+re^{i\theta})e^{-i\theta}dr\wedge d\theta,$$
which is clearly defined and $C^\infty$ in $z$. Then
\begin{align*}
\frac{\partial f_1(z)}{\partial\bar z} & =-\frac{1}{\pi}\int_\mathbb{C}\frac{\partial g_1}{\partial\bar z}(z+re^{i\theta})e^{-i\theta}dr\wedge d\theta\\
& =\frac{1}{2\pi\sqrt{-1}}\int_\Delta\frac{\partial g_1}{\partial\bar w}(w)\frac{dw\wedge d\bar w}{w-z};
\end{align*}

答:
If $u = re^{i\theta}$, then $du = e^{i\theta}dr + ire^{i\theta}d\theta$, while $\bar{u} = re^{-i\theta}$ so $d\bar{u} = e^{-i\theta}dr - ire^{-i\theta}d\theta$. So

$$du\wedge d\bar{u} = (e^{i\theta}dr + ire^{i\theta}d\theta)\wedge (e^{-i\theta}dr - ire^{-i\theta}d\theta) = -irdr\wedge d\theta + ird\theta\wedge dr = -2irdr\wedge d\theta$$

and therefore $$\dfrac{1}{u}du\wedge d\bar{u} = \dfrac{-2ir}{re^{i\theta}}dr\wedge d\theta = -2ie^{-i\theta}dr\wedge d\theta.$$

This is why the coefficient $\dfrac{1}{2\pi i}$ in $g_1$ becomes $-\dfrac{1}{\pi}$ in $f_1$.


问:
Tic Tac Toe Combinations
I am creating a Tic-Tac-Toe game and was wondering how many combinations there are. I am an amateur coder and was going to code all the combinations by hand. Also if I do not do then what am I supposed to do? code in c# (using console)</p>

答:
It depends how you count symmetries, but there are at least $26830$ possible games and at most $255168$ possible games

I have a page doing the larger count in detail

Title: Tic Tac Toe Combinations


问:
Differentiating the maximum?
Consider a function $f(x,y) : \mathbb [0,1]^2 \to \mathbb R$, smooth in both variables (very strong assumption for the purposes). Define $f_{X}(x) = \max_{y \in [0,1]} f(x,y)$ and $f_Y{(y)} = \max_{x \in [0,1]} f(x,y)$ , which are well-defined. Oftentimes, in optimization, it is a useful statement to say that $f_X$ and $f_Y$ have unique maxima/minima. For this purpose, it is convenient to see if they are differentiable, and set the derivative to zero.

Under what sufficient conditions are $f_X$ and $f_Y$ differentiable (they are always continuous, as pointed out below)? Is the condition of "f smooth" acceptable?

How can I study the derivative of $f_X$ and $f_Y$, so that I can look at the maxima/minima of such functions?

EDIT : Since we have come up with a counterexample to non-differentiability, given by $(x,y) \to 1-xy$ on $[-1,1]^2$, I would like sufficient conditions under which this is true. One may use the language of sub-differentials etc. (from convex analysis) if required for this purpose.

答:
We may apply Danskin's theorem. See [1], [2], [3].

If $f(x, y)$ is convex in $x$ for each $y \in [0, 1]$, and at a given $x_0 \in (0, 1)$, $f(x_0, y)$
has a unique maximizer $y_0$ on $[0, 1]$, then $f_X(x)$ is differentiable at $x_0$ with
$$\frac{\mathrm{d} f_X(x)}{\mathrm{d} x}\Big\vert_{x=x_0}
= \frac{\partial f(x, y_0)}{\partial x}\Big\vert_{x=x_0}.$$

Reference

[1] https://en.wikipedia.org/wiki/Danskin%27s_theorem

[2] R. T. Rockafellar, “On a special class of convex functions”,
Journal of Optimization Theory and Applications volume 70, pages619–621(1991).

[3] Dimitri P. Bertsekas, “Nonlinear Programming”, 2nd Ed., Appendix B.5, page 717.

Title: Differentiating the maximum?


问:
Finding the slope of the steepest, shallowest line formed by two members of a set
Consider a non-empty set of cartesian coordinates in $2D$ space. We may assume that the set includes at least two unique members. It is possible that the set contains duplicated members. It may or may not be that all members are colinear.

I want to determine the slope of both the steepest possible line and the shallowest possible line that can be formed by two (or more) members of the set. Moreover, I want to do so as efficiently as possible.

The obvious, and perhaps the simplest answer, is to simply iterate over each member of the set, compute the slope of a line formed by that member and every other member, and keep track of that slope which is steepest/shallowest. As pointed out below, this probably has something like $O(\frac{1}{2}n^2)$ complexity. For a large number of points, however, this could be impractical.

Is there a better way?

答:
I'm not well-versed in algorithms, but perhaps you can approach the problem geometrically.

For the steepest line, start by sorting the points by their $x$-coordinate, from smallest to largest.  Then, for each $x$-coordinate, draw a vertical line and rotate it by small amounts until in intersects another point and compute the slope (start with clockwise rotations, and then check CCW rotations).  Store the slope and move to the next point.  Hopefully, you would only need to check a few neighboring points for each point as you move from left to right, so that it is close to order $n$.

The same idea, but sort according to $y$-coordinate and moving from bottom to top with a horizontal line and slowly rotating as needed.


问:
The weighted average absolute value of pairwise sum of any finite sequence is larger than the weighted average of absolute value.
How do we prove for any given real number $a_1,\ldots,a_n$, we have
$$\frac{1}{n^2}\sum_{i=1}^n\sum_{j=1}^n|a_i+a_j|\geq \frac{1}{n}\sum_{i=1}^n|a_i|$$

I have checked several easy cases for this:

When $n=1$, LHS $=2|a_1|$ and RHS $=|a_1|$, so it trivially holds.

When $n=2$, we are going to prove
$$2|a_1|+2|a_1+a_2|+2|a_2|\geq 2(|a_1|+|a_2|)\Longleftrightarrow 2|a_1+a_2|\geq 0$$
which is also trivial.

When $n=3$, we are going to prove
$$\sum_{i=1}^3\sum_{j=1}^3|a_i+a_j|\geq 3\sum_{i=1}^n|a_i|$$
which is equivalent to prove
$$2(|a_1+a_2|+|a_2+a_3|+|a_3+a_1|)\geq |a_1|+|a_2|+|a_3|$$
This can be proved simply by triangular inequality, and actually the coefficient $2$ is not sharp, $1.5$ is enough.

However, when $n$ gets larger, e.g., $n\geq 6$, simply applying triangular inequality is not enough because by in general it reduces to prove:
$$2\sum_{i<j}|a_i+a_j|\geq (n-2)\sum_{i=1}^n|a_i|$$

I don't know how to proceed, can anyone help me?

Edit: Since this problem is pointed out to be duplicate, I slightly generalize it. If $b_1,\ldots,b_n$ are convex combination coefficient, i.e., they are nonnegative and $b_1+\ldots+b_n=1$, then how to prove
$$\sum_{i=1}^n\sum_{j=1}^nb_ib_j|a_i+a_j|\geq \sum_{i=1}^nb_i|a_i|$$

答:
This is a discreet version of the problem which appeared at 64th Putnam in 2003.
The original was formulated in this format:
$$\int_{0}^{1}\int_{0}^{1}|f(x) + f(y)|dxdy >= \int_{0}^{1}|f(x)|$$
for continuous $f$.


问:
Solve trigonometric integral $\int \frac{\mathrm dx}{(a+b\cos x)^2}$
Source: Question bank on challenging integral problems for high school students. An advanced level problem

Problem: Evaluate the indefinite integral:
$$\int \frac{\mathrm dx}{(a+b\cos x)^2}$$ given that $a>b$

My try on it: It seems simple enough but upon  a deeper look I can't find a suitable substitution for it. Not even going to try integrating by parts! Tried constructing a function whose integral is the one above. Gotten close results but not quite it. And that is not the correct approach anyway! Also tried using Partial fractions by constructing a function

$${1\over(a+by)^2} = {Ay+B\over(a+by)^2} + {Cy+D\over a+by}$$

Pretty unsure about this one. Am I going right or can you please suggest a better, quicker alternative. I can upload the answer if someone needs to refer it.

答:
Utilize the simpler integral result
$$\int \frac{dx}{a+b\cos x}=\frac2{\sqrt{a^2-b^2}}\tan^{-1}
 \frac{\tan \frac x2 }{ \sqrt{\frac{a+b}{a-b}}}
$$
and integrate both sides of the equation below
$$\left( \frac{\sin x}{a+b\cos x}\right)’
= \frac{a}{b(a+b\cos x)}- \frac{a^2-b^2}{b(a+b\cos x)^2}
$$
to obtain

$$\int \frac{dx}{(a+b\cos x)^2}= \frac{b\sin x}{(b^2-a^2)(a+b\cos x)}+ \frac{2a}{(a^2-b^2)^{3/2}}\tan^{-1}
 \frac{\tan \frac x2 }{ \sqrt{\frac{a+b}{a-b}}}
$$


问:
Find a convex polygon $P$ s.t. $P\subset Q \subset (1+\epsilon) P $.
Problem : Assume that $Q$ is a compact convex plane figure containing $R$-ball of the center origin. Then prove that for any $\epsilon >0$, there is a convex polygon $P\subset Q$ s.t. $$ P\subset Q\subset
(1+\epsilon)\cdot P $$

Proof : Define $A = \bigg\{ \delta (x,y) \bigg| x,\ y\in \mathbb{Z}\bigg\}$ From this the book says that $\epsilon = \frac{10\ {\rm diam}\ Q}{R}\delta,\ \delta <
\frac{R}{10}$

Question : How can we obtain this ?

The following is my different approach :

(1) $x_i$ is point in $\partial Q$ s.t. $|x_i-x_{i+1} |
\leq R$

Then ${\rm conv}\ \{x_i\}$ contains $\frac{R}{2}$-ball.

(2) Redefine $x_i\in \partial Q$ s.t. the arc
$\widetilde{x_ix_{i+1}}$ in $\partial Q$ has a
length $\leq M=\min\ \{\ R\epsilon , R\ \} \ (\ast)$.

If
$$|o-x_i|,\ |o-x_{i+1}|\geq \frac{R}{2},$$
then consider a line containing $[x_ix_{i+1}]$ and a line containing $
(1+\epsilon )\cdot [x_i x_{i+1}]$.

Note that the distance between two lines is greater than
$\frac{R}{2} \epsilon$.

From $\ast$ the arc $\widetilde{x_ix_{i+1}}$ is between two triangles $ox_ix_{i+1},\ o(1+\epsilon )x_i(1+\epsilon)x_{i+1}$.

Hence we have the desired $P ={\rm conv}\ \{x_i\}$.

Here in $\ast$ we do not have diameter term.

Reference : From Euclid to Alexandrov; a guided tour - Petrunin and Yashinski.

答:
Fix $\varepsilon$. Define $A = \{ (m,n)\delta |m,\ n\in
\mathbb{Z}\}$ We let $\delta = \frac{R}{\sqrt{2}}\frac{1}{M}$ for
large positive integer $M$.

Then the convex polygon $C = {\rm conv} (A\cap Q)$ contains
$\frac{R}{2}$-ball whose center is origin $o$.

For $x\in \partial
Q$, let $y\in [xo] $ s.t. $$ |x-y|=d< \frac{R}{2} \varepsilon$$

Further there is $r>0$ s.t. a closed ball $B[r,y]$ is in $C$. Here
$r$ is independent of $x,\ y$.

If $M$ is large, then each $B[r,y]$ contains at least nine squares
generated from $A$.

Here we can assume that a square placed in
center contains a point $y$. Hence $C$ contains $y$.

If $C\cap
[ox]=[oz]$ where $z$ is a point in a line segment $[ox]$, then
\begin{align*}
|oz| (1+ \varepsilon )&\geq |oz| +
\frac{R}{2}\varepsilon
\\&\geq |oz| +d \\&\geq
|ox|
\end{align*}  which finishes the proof.


问:
Is there a notion of basis for Banach spaces?
Consider the Banach space $\ell^1(\mathbb N)$.

The sequence $(e_n)_{n\in\mathbb N}$ feels like a kind of basis because every element $a\in\ell^1(\mathbb N)$ can be written as an absolutely convergent infinite linear combination $\sum_{n\in\mathbb N}a(n)e_n$ in a unique way.

(Here $e_n$ denotes the vector whose $n$th entry is 1 and all of whose other entries vanish.)

The same is true for the Banach space $c_0(\mathbb N)$.

Is the above property of the sequence $(e_n)_{n\in\mathbb N}$ appropriate in order to abstractly define a basis of a Banach space? Has this been considered?

答: As Soarer points out: Yes, it is called a Schauder basis.</p>

ContentLicense: CC BY-SA 2.5


问:
If one of $n$ coins is fair, then find the probability that the total number of heads is even
A collection of $n$ coins is flipped. The outcomes are independent, and the $i$-th coin comes up heads with probability $\alpha_i, i=1, \dots, n$. Suppose that for some value of $j, 1 \leq j \leq n, \alpha_j=\frac{1}{2}$. Find the probability that the total number of heads to appear on the $n$ coins is an even number. </p>

<p>Now, I don't understand the importance of the statement "Suppose that for some value of $j, 1 \leq j \leq n, \alpha_j=\frac{1}{2}$" </p>

答:
That one coin guarantees that the probability is exactly $1/2$.

If the sum from the other coins is odd and that one turns up "heads", then the total is even; otherwise it's odd.

$$
\begin{align}
& {}\quad\Pr(\text{even}) = \mathbb E(\Pr(\text{even} \mid \text{other coins' outcomes})) \\[8pt]
& = \sum_{\text{other coins' outcomes}} \Pr(\text{even}\mid\text{particular other coins' outcome})\cdot\Pr(\text{particular other}\ldots) \\[8pt]
& = \sum_\text{outcomes} \frac12\cdot \Pr(\cdots) \\[8pt]
& = \frac12,\text{ since the sum of those probabilities is }1.
\end{align}
$$

Slightly later edit:  Just to be concrete: Suppose $\Pr(\text{1st coin H}) = 9/10$ and $\Pr(\text{2nd coin H})=1/2$.  Then
$$
\begin{align}
\Pr(\text{even}) & = \Pr(\text{1st H})\cdot\Pr(\text{even}\mid\text{1st H}) + \Pr(\text{1st
T})\cdot\Pr(\text{even}\mid\text{1st T}) \\[8pt]
& = \frac9{10}\cdot\frac12 + \frac1{10}\cdot\frac12 \quad = \frac12\left(\frac9{10}+\frac1{10}\right) \quad = \frac12\cdot1.
\end{align}
$$


问:
Let $f(x)$ be continuous everywhere and consider the function $g(x)$. Prove that $g(x)$ is continuous everywhere.

Let $f(x)$ be continuous everywhere and consider the function:
$$g(x)=\begin{cases}f(x) &\text{if}\ f(x)>0\\0 &\text{if}\ f(x)\le0\end{cases}$$
Prove that $g(x)$ is continuous everywhere.

Hello, can anyone please help me with this one? I have a feeling this has to do with composition but cannot figure out the correlation between $f(x)$ and $g(x)$.
Thank you!

答:
Hint.

The function $g$ can be written as
$$
g(x)=\frac12(f(x)+|f(x)|)
$$

and $x\mapsto |x|$ is a continuous function.

You may want to recall that (whenever compatible)

the sum of two continuous functions is continuous;

the composition of two continuous functions is continuous.


问:
Calculating the number of trials necessary for even odds
Abraham De Moivre gave a formula for finding how many trials are necessary to find even odds of a given even occurring at least once.

When $a$ is the odds of the event happening, and $b$ is the odds of it not happening, and $a:b=1:q$, then the equation can be written $(1+\frac{1}{q})^x=2$

For example - to determine when there are even odds of rolling 3 ones in a roll of three dice at least once you can solve for $x$ with $q=215$ (which is $6^3-1$).

How can I generalize this equation to find even odds of the event occuring at least twice, or three times etc?

答:
Your equation is based on the probability of it not happening in one attempt being  $\frac{q}{q+1}$ and so it not happening in $x$ attempts being $\left(\frac{b}{a+b}\right)^x=\left(\frac{q}{q+1}\right)^x$ which you set equal to $\frac12$ and solve for $x$ to give $x = \frac{\log(2)}{\log(a+b)-\log(b)} = \frac{\log(2)}{\log\left(1+\frac1q\right)}$.

Most $q$ will not give an integer result and your dice example has $x\approx 149.37$

For event occurring at least $n$ times you can do something similar with the cumulative distribution of a binomial distribution, though slightly more complicated, as you want to find the solution to $$\sum\limits_{j=0}^{n-1} {x \choose j} \frac{a^jb^{x-j}}{(a+b)^x}=\frac12\text{, i.e. }\sum\limits_{j=0}^{n-1} {x \choose j} \frac{q^{x-j}}{(q+1)^x}=\frac12$$

Again most $q$ will not give an integer result for $x$, and this time I would not expect a closed from for $x$.


问:
Most general exponential $\mathbb R\to\mathbb R$
I am familiar with the multivalued general definition of exponentiation, namely,
$$z^w = \exp(w\log z) = \exp(w(\log|z|+i\arg(z))).$$
My question is, what is the most general definition for $x^y$ as a map $\mathbb R\to\mathbb R$? I think it is this:
$$x^y=\begin{cases}
\exp(y\log x) & \text{if $x>0$}\\[5pt]
\hfil 0 \hfil & \text{if $x=0$ and $y\geqslant0$}\\[5pt]
\hfil\sqrt[b]{x^a}\hfil & \text{if $x<0$ and $y=\frac ab$ with $b$ odd}.
\end{cases}$$
In other words, for positive bases we define it as usual with the real logarithm (or as $\lim x^{y_n}$ where $y_n$ is a sequence of rational numbers converging to $y$, such as its decimal expansion). For negative bases, we can only define it for rational powers with odd denominators in reduced form, since these are the only powers which give a real root ($r^b = x^a$ only has real roots $r$ for odd values of $b$).

Is this the most general definition we can have? It seems strange that we can make sense of $x^{1.4}$ for all real $x$, but not of $x^{1.5}$.

答:
You need to be very careful when you are trying to define $x^{a/b}$ with $x<0$. You defined it as $\sqrt[b]{x^a}$ which is equivalent to $(x^a)^{1/b}$, but why?

Probably because you are thinking that $(x^a)^{1/b} = x ^{a \cdot (1/b)} = x^{a/b}$, but you are using the exponent rule $(x^r)^s = x ^{ rs}$ which is NOT valid for $x<0$.

That's why often $x^{a/b}$ is not defined in a general way, because it's not possible to define it in such a way that the exponent rules are valid.


问:
Question on basic theorems on differential manifold
I'm studying the definition of smooth manifolds and constantly thrown off by two atlases, $A=(\mathbb{R}, f:x\mapsto x^{1/3})$ and $B=(\mathbb{R}, f:x\mapsto x)$, each of which constutes of a single chart (I guess this is valid, correct?).

In Tu's An Introduction to Manifolds (second edition), he writes on pages 62 and page 63 the following two results which I cannot seem to digest with regard to $A$ and $B$.

Proposition 6.8. Let $N$ and $M$ be smooth manifolds, and $F: N\to M$ a continuous map.
The following are equivalent.

(i) The map $F:N\to M$ is $C^{\infty}$.

(ii) There are atlases $u$ for $N$ and $B$ for $N$ such that for every
chart $(U,\phi)$ in $u$ and $(V,\psi)$ in $B$, the map

$$\psi\circ F\circ \phi^{-1}:\phi(U\cap F^{-1}(V))\to \mathbb{R}^m$$
is $C^{\infty}$.

(iii) For every chart $(U,\phi)$ on $N$ and $(V,\psi)$ on $M$, the map
$$\psi\circ F\circ \phi^{-1}:\phi(U\cap F^{-1}(V))\to\mathbb{R}^m$$ is $C^{\infty}$.

Does this mean that the smoothness between two manifolds is independent of the choice of maximal atlases? Like, even if we impose some stupid atlas like $A$, it doesn't matter?

Proposition 6.10. If $(U,\phi)$ is a chart on a manifold $M$ of dimension $n$, then the coordinate map $\phi:U\to \phi(U)\subset\mathbb{R}^n$ is a diffeomorphism.

I'm really having trouble with this proposition. For example, $A$ is a chart on the manifold $\mathbb{R}$, but it is not a diffeomorphism at $x=0$. Could someone please explain? Thank you.

答:
As $M$ and $N$ are smooth manifolds, they are already equipped with maximal atlases. Therefore the charts in the statement refer to charts from those atlases; Tu makes this explicit in the last sentence of section $5.3$.

As for Proposition $6.10$, Tu is endowing $\phi(U)$ with its usual smooth structure. In the proof, $\phi(U)$ is equipped with the atlas $(\phi(U), \mathbb{1}_{\phi(U)})$.


问:
Is the following producing a list of all prime numbers without skipping in a consecutive order?
The following consecutive list is made as a table with the following columns :

column 1
column 2
column 3

$A$
$B=(A^2 -1) / 6$
prime factors of $B$

Where $A$ is odd and $A \mod 3 ≠ 0$

Example: $A = 7$ : $B=(7^2-1) / 6 = 8$ : prime factors of $8$ are $1,2$

You will notice that the results start building a sequence of all possible prime numbers, and never skipping a prime number in a consecutive way.

For example: If $19$ is a prime factor result of $A=37$, it means that $17$ had somewhere appeared before, and as if it would been impossible for $19$ to appear after a resulted $13$ while skipping a previous $17$. I have run the experiment until $A=1000$ and it seems to be true. Is there any proof of it being true? or is there a counter example?

Update: per @Ross Millikan comments 1 is not a prime and 5 and 7 appear in the list before 3 (making it the only exception so far). I didn't take into account 3 as a prime because the whole list is generated by the condition that  $$ is odd and $ \mod 3 ≠0$, and the reason $1$ was part of the results is just produced by my programming.

To better illustrate, here is a sample list where $A$ is growing consecutively (Where $A$ is odd and $A \mod 3 ≠ 0$):

$A$ : $B=(A^2 -1) / 6$ : prime factors of $B$

5 : 4 : prime factors of 4 are 2

7 : 8 : prime factors of 8 are 2

11 : 20 : prime factors of 20 are 2 , 5

13 : 28 : prime factors of 28 are 2 , 7

17 : 48 : prime factors of 48 are 2 , 3

19 : 60 : prime factors of 60 are 2 , 3 , 5

23 : 88 : prime factors of 88 are 2 , 11

25 : 104 : prime factors of 104 are 2 , 13

29 : 140 : prime factors of 140 are 2 , 5 , 7

31 : 160 : prime factors of 160 are 2 , 5

35 : 204 : prime factors of 204 are 2 , 3 , 17

37 : 228 : prime factors of 228 are 2 , 3 , 19

41 : 280 : prime factors of 280 are 2 , 5 , 7

43 : 308 : prime factors of 308 are 2 , 7 , 11

47 : 368 : prime factors of 368 are 2 , 23

49 : 400 : prime factors of 400 are 2 , 5

53 : 468 : prime factors of 468 are 2 , 3 , 13

55 : 504 : prime factors of 504 are 2 , 3 , 7

59 : 580 : prime factors of 580 are 2 , 5 , 29

61 : 620 : prime factors of 620 are 2 , 5 , 31

65 : 704 : prime factors of 704 are 2 , 11

67 : 748 : prime factors of 748 are 2 , 11 , 17

71 : 840 : prime factors of 840 are 2 , 3 , 5 , 7

73 : 888 : prime factors of 888 are 2 , 3 , 37

77 : 988 : prime factors of 988 are 2 , 13 , 19

79 : 1040 : prime factors of 1040 are 2 , 5 , 13

83 : 1148 : prime factors of 1148 are 2 , 7 , 41

85 : 1204 : prime factors of 1204 are 2 , 7 , 43

89 : 1320 : prime factors of 1320 are 2 , 3 , 5 , 11

91 : 1380 : prime factors of 1380 are 2 , 3 , 5 , 23

95 : 1504 : prime factors of 1504 are 2 , 47

97 : 1568 : prime factors of 1568 are 2 , 7

101 : 1700 : prime factors of 1700 are 2 , 5 , 17

103 : 1768 : prime factors of 1768 are 2 , 13 , 17

107 : 1908 : prime factors of 1908 are 2 , 3 , 53

109 : 1980 : prime factors of 1980 are 2 , 3 , 5 , 11

113 : 2128 : prime factors of 2128 are 2 , 7 , 19

115 : 2204 : prime factors of 2204 are 2 , 19 , 29

119 : 2360 : prime factors of 2360 are 2 , 5 , 59

121 : 2440 : prime factors of 2440 are 2 , 5 , 61

答:
In effect you are asking whether there exists a prime $p$ which is not a factor of $\frac{A^2-1}{6}$ for any $A$ not divisible by $2$ or $3$

Any $A$ not divisible by $2$ or $3$ will be of the form $A=6n+1$ or $6n-1$ for some $n$.

If $A=6n+1$ then $\frac{A^2-1}{6} = 2n(3n+1)$ which is divisible by $n$

If $A=6n-1$ then $\frac{A^2-1}{6} = 2n(3n-1)$ which is also divisible by $n$

So just let $n=p$ and $A=6p\pm1$ (your choice): you will then have $\frac{A^2-1}{6}$ divisible by $p$.

For example for $p=2$, $A=11$ or $13$ works; for $p=3$, $A=17$ or $19$ works;  for $p=5$, $A=29$ or $31$ works;   for $p=7$, $A=41$ or $43$ works;   for $p=11$, $A=65$ or $67$ works;  and so on.  There will always be such examples fitting this pattern on top of many other cases.  And you do not need to restrict $n$ to be prime: for example for $n=4$, $A=23$ or $25$ leads to $\frac{A^2-1}{6}$ being divisible by $4$.

So no, there are no such primes.


问:
What is $P(X=x)$ when $X\sim \text{Unif}(a,b)$?
I'm trying to solve this excursive A stick of length 1 is broken in a random place that is U(0, 1). Let X be the length of the longer piece and let Y be the length of the shorter piece. in another way. The question:

A stick of length 1 is broken in a random place that is $\text{Unif}(0,1)$. Let $X$ be the point chosen on the steak and $Y$ be the length of the longer piece. Find $E(Y)$.

So we get:
$$
Y=\begin{cases}
1-X & X<0.5\\
X & X\geq0.5
\end{cases}
$$
Then we can say:
$$
\begin{align*}
E\left(Y\right)&=\int_{-\infty}^{\infty}y\cdot P\left(Y=y\right)dy\\&=\int_{0}^{1}y\cdot P\left(Y=y\right)dy=\int_{0}^{0.5}y\cdot P\left(Y=y\right)dy+\int_{0.5}^{1}y\cdot P\left(Y=y\right)dy\\&=\int_{0}^{0.5}y\cdot P\left(Y=y\right)dy+\int_{0.5}^{1}y\cdot P\left(Y=y\right)dy\\&=\int_{0}^{0.5}(1-x)\cdot P\left(X=x\right)dx+\int_{0.5}^{1}x\cdot P\left(X=x\right)dx\\&=\int_{0}^{0.5}(1-x)\cdot P\left(X=x\right)dx+\int_{0.5}^{1}x\cdot P\left(X=x\right)dx
\end{align*}
$$
I want to say $P(X=x)=1$ when $x\in[0,1]$. But is it true to say that $P(X=x)=\frac{1}{b-a}$ when $x\in[a,b]$?

答:
We can rewrite this in terms of $f_X(x)=1:$ \begin{align*}
\mathbb E(Y)&=\int_{-\infty}^\infty y\cdot f_X(x)dx=\int_0^{0.5}(1-x)(1)dx+\int_{0.5}^1x(1)dx\\
&= 0.75
\end{align*}
You don't have to deduce the pdf of $Y$ to get its expectation, as $Y$ is a function of $X$.


问:
Bayes' Theorem probability testing for a virus
I am trying to translate the following question into input for the Bayes' theorem.

What i have currently is P(V|A) = 0.95 which is the probability of having the virus given that test A recognises it which is 95%, and P(V|~A) = 0.10 which is the prob of having the virus given that the test doesn't detect it, Also P(V) = 0.01 because 1% of people have the virus.

I am wondering if i have misinterpreted the "Test A" part of the question?

Consider two tests, A and B, for a virus.
Test A is 95% effective at recognizing the virus when it is present (that is 95% of the time that the virus is
present, the test detects it), but has a 10% false positive rate (that is, 10% of the time it indicates the virus
is present when it is not).
Test B is 90% effective at recognizing the virus, but has a 5% false positive rate.
The two test use different, independent, methods of indentifying the virus.
1% of all people have the virus.
Joe tests positive for the virus using test A. Bob tests positive using test B. Who is more likely to have the
virus?

答:
In the above response, everything looks good but there is a minor typo for the values for the equation for $P(V|B)$

In the bottom part, $P(B|~V)$ should have been $0.05$ (since it is only $5$% false positive), but it was written as $0.1$

The final answer for $P(V|B)$ is $0.1538461538461539$ accurate though.


问:
Adding locally compact Hausdorff space into product besides two spaces of a quotient mapping.
I am strugling to prove the next theorem. More precisely, I have no clue how to even start solving it.

The problem states:

Let $f:X\rightarrow Y$ be a quotient mapping and let $K$ be a locally compact Hausdorff space. Prove that $f^*:X\times K\rightarrow Y\times K:(x,k)\mapsto(f(x),k)$ is one quotient mapping.

I know that the title is terrible. I am sorry, I just couldn't think of any better.

答:
Let $U\subseteq Y\times K$ be such that $V:=(f^*)^{-1}(U)$ is open in $X\times K$.

For each $(y,k)\in U$, pick $x\in X$ with $f(x)=y$ and open sets $x\in V_1\subseteq X$ and $k\in W\subseteq K$ with $\overline{W}$ compact such that $V_1\times \overline{W}\subseteq V$. Our goal is to make $V_1$ both saturated and open.

We can enlarge $V_1$ to $V_2:=f^{-1}(f(V_1))$ to get a saturated set, but this may not be open. By the compactness of $\overline W$, we can enlarge $V_2$ to an open set $V_3$ (see below) with $V_3\times \overline W\subseteq V$, but this needs not be saturated. We can then generate a sequence of sets
$$V_1\subseteq V_2\subseteq V_3 \subseteq V_4\subseteq\cdots.$$
The trick would be to take the union of all these (call it $V'$), which will then be both saturated and open and satisfy $V'\times \overline W\subseteq V$ (why?). We then have $(y,k)\in f(V')\times  W\subseteq U$.

How to construct $V_3$:

For each point $p\in V_2$, $V\cap(X\times \overline W)$ is open in $X\times \overline W$. By the tube lemma, we can pick an open set $V_p$ such that $(p,k)\in V_p\times\overline W\subseteq V\cap(X\times \overline W)$. Then we can take $V=\bigcup_{p\in V_2} V_p$.


问:
Prove that $\sqrt{13}$ is irrational in 3 ways.
I am asked how to prove that the $\sqrt{13}$ is irrational in 3 ways. I only know of one, where we assume $\sqrt{13} = \frac{a}{b}$ where $a, b$ are coprime, and we prove with contradiction that $13 \mid a$ and $13 \mid b$.

However, I am not sure where to begin with the other two proofs. I thought that this was the only way to prove irrationality of a number. I would really appreciate some help on where to start/look for to start the two other proofs.

答: It is realtively easy to prove that if $q\in\Bbb Q$ is such that $q^2\in \Bbb Z$, then $q\in\Bbb Z$ as well. Once you have established that, you only need to prove that there is no integer with square $13$, and there must be three, four or perhaps ninety-seven trite ways to prove it.


问:
$X$ is a normal scheme iff $O_X (U)$ is normal integral domain.
This is a proposition in Liu's book.(proposition 4.1.5)
I can't understand the proof....

Proposition 1.5. Let $X$ be an irreducible scheme. The following properties are equivalent:
(i) The scheme $X$ is normal.
(ii) For every open subset $U$ of$X$, $O_X(U)$ is a normal integral domain.
Proof: Let us suppose $X$ is normal. Let $U$ be an open subset of $X$ and $α ∈ Frac(O_X(U))$ integral over $O_X(U)$. Let $V$ be an affine open subset of $U$. Then $α$ is integral over $A := O_X(V)$ and $α ∈ Frac(A)$. Let us set $I = Ann((αA+A)/A)$. If $I \neq A$, then $I ⊆ p$ for a prime ideal $p$ of $A$.As $A_p$ is normal, there exists an $s ∈ A \setminus p$ such that $sα ∈ A$. It follows that $s ∈ I$, whence a contradiction. Hence $1 ∈ I$ and $α ∈O_X(V)$. As this is true for every affine open subset $V$ of $U$,we have $α ∈O_X(U)$ (Proposition 2.4.18). This shows that (i) implies (ii).

In proof why $\alpha$ is integral over $A := O_X(V)$ and $\alpha ∈ Frac(A)$
And why is $I$ defined like that?
I really need help with this proof....

答:
Since $X$ is normal it is reduced and $X$ is irreducible so $X$ is integral. Therefore $\mathcal{O}_X(U)$ is a domain so we just need to prove it is integrally closed. Therefore we reduce to the following problem.

Proposition. Let $A$ be a domain such that $A_{\mathfrak{p}}$ is integrally closed for each prime then $A$ is integrally closed.

Proof. Take $x \in K$ where $K = \mathrm{Frac}(A)$ is integral. Notice that in $K$ we have,
$$ A = \bigcap_{\mathfrak{p} \subset A} A_{\mathfrak{p}} $$
Then $x$ is integral over each $A_{\mathfrak{p}}$ and thus $x \in A_{\mathfrak{p}}$ since each local ring is integrally closed. Thus $x \in A$.

Thus we have proven it for any affine $U$. Now for an arbitrary $U$ cover $U$ by affine opens $U_i$ and consider the map $\mathcal{O}_X(U) \to \mathcal{O}_{X}(U_i)$. If $f \in K$ is integral over $\mathcal{O}_X(U)$ then it is integral over each $\mathcal{O}_X(U_i)$ so $f \in \mathcal{O}_X(U_i)$ and therefore $f \in \mathcal{O}_X(U)$ by the sheaf property.


问:
Find sum of the digits of $1395t+2015$

Assume $S(x)$ be equal to sum of digits of the integer $x$. for
example $S(123)=1+2+3=6$. suppose $t$ be smallest natural number so
that $S(t)=1394^{1394}$. then what is the value of $S(1395t+2015)$?

$1)26
 \quad\quad2)35\quad\quad3)134\quad\quad4)143\quad\quad5)\text{None}$

To solve this problem first of all I should find the value of $t$. because $t$ is smallest natural number with digits sums to $1394^{1394}$, therefore we should use the digit $9$ as much as possible to reduce the number of digits.  $1394^{1394}\equiv(-1)^{1394}\equiv1\pmod{9}$so $$t=1\underbrace{999\cdots9}_{\tfrac{1394^{1394}-1}9}$$
But I don't know how to find sum of the digits of $1395t+2015$

答:
You have made a great start. To continue, let

$$n = \frac{1394^{1394}-1}{9} \tag{1}\label{eq1A}$$

This then gives

$$t = 2 \times 10^{n} - 1 \tag{2}\label{eq2A}$$

Thus,

$$\begin{equation}\begin{aligned}
1395t + 2015 & = 1395(2 \times 10^{n} - 1) + 2015 \\
& = 2790 \times 10^{n} - 1395 + 2015 \\
& = 2790 \times 10^{n} + 620
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$

Note $n$ is much larger than $3$, so there's no overlap between the the highest non-zero digits of $279$ and the final digits of $620$, with all of the digits in between being $0$. This means that

$$S(1395t + 2015) = 2 + 7 + 9 + 6 + 2 = 26 \tag{4}\label{eq4A}$$

i.e., the listed choice $1$.

Title: Find sum of the digits of $1395t+2015$


答:
Two other proof techniques you can use are :

The rational root theorem : Let $f(x) = a_0 + a_1x + \cdots + a_nx^n$ be a polynomial, and let $\frac pq$ be a rational root of $f$. Then $p$ divides $a_0$ and $q$ divides $a_1$. For using this to prove that $\sqrt {13}$ is irrational, we create the contrapositive : if $f(x)$ is a polynomial as above, and for every $p$ dividing $a_0$ and $q$ dividing $a_n$ we have $f(\frac{p}{q}) \neq 0$, then every root of $f$ is irrational. Now, $x_0 = \sqrt{13}$ satisfies $f(x)= x^2-13$. You can see that $f$ has no rational roots using the rational root theorem, so $x_0$ must be irrational.

Let $S = \{n \in \mathbb N : n\sqrt 13 \in \mathbb N\}$. If $\sqrt {13}$ is rational then $S$ is non-empty because $\sqrt{13} > 0$. Let $n \in S$ and let $q = n(\sqrt{13}-3)$. Then $q<n$, $q \in \mathbb N$ , and $q\sqrt{13} = 13n - 3n\sqrt{13} \in \mathbb N$ by $n \in S$. Therefore, $S$ has no minimal element despite being a subset of the natural numbers, a contradiction.

Title: Prove that $\sqrt{13}$ is irrational in 3 ways.

Body:
I am asked how to prove that the $\sqrt{13}$ is irrational in 3 ways. I only know of one, where we assume $\sqrt{13} = \frac{a}{b}$ where $a, b$ are coprime, and we prove with contradiction that $13 \mid a$ and $13 \mid b$.

However, I am not sure where to begin with the other two proofs. I thought that this was the only way to prove irrationality of a number. I would really appreciate some help on where to start/look for to start the two other proofs.


问:
Different quotient, what does 'a' represent?
My book says :

Different quotient represents the average rate of change of $f(x)$ between x = $a$ and $x = a + h$

the question ask for $f(a)$:

$f(x) = x^3$ therefore,

$f(a) =$

I have tried $f(a) = x^3$ and $f(a) = f(x)$ and this answers are incorrect. what confuses me is that if the definition says $x = a$ then why is $f(a) = f(x)$ not true.

Thanks.

答:
You are thinking too hard. If $f(x)=x^3$, then $f(a)=a^3$. The expression $f(a)$ means the value of the function $f$ evaluated at $x=a$.

To see more examples,
$$
f(1)=1^3,\quad f(2)=2^3,\quad f(\pi)=\pi^3,\cdots
$$


问:
Proof that every open set in $\mathbb{R}^n$ is the union of ...??
Can't figure this 2 part question out - I think the first part involves using open balls but I'm not sure. It's straightforward to prove this for $\mathbb{R}^1$ with disjoint segments but am a little lost as to what to do when it comes to $\mathbb{R}^n$.

(1) Prove that every open set in $\mathbb{R}^n$ is the union of an at most countable collection of disjoint segments.
Hint: You need to replace disjoint segments with the appropriate objects (which I'm thinking are open balls).

(2) Generalize the statement to a separable topological space.
Hint: you may need some extra assumption.

Any help would be much appreciated! The more detail, the better! Thanks!

答: Yes, open balls work here; just cover every point in the open set with an open ball; then because separable metric spaces are Lindelof, you can take a countable subcover.</p>


问:
What is the meaning of inner product of two circles?
I was reading this post about orthogonal circles (here) and in the question body something about inner product of circles was used.. it seemed quite mysterious but still interesting. I know of inner products of vector which is simply the dot product but what exactly is the inner product of circles? Does it having an associated geometric meaning?</p>

答:
We may stereographically project $\mathbb{R}^n$ to $S^n\subseteq\mathbb{R}^{1+n}$ given by $u\mapsto\big(\frac{1-|u|^2}{1+|u|^2},\frac{2u}{1+|u|^2}\big)$.

Let's say you're already slightly familiar with pseudo-Euclidean spaces $\mathbb{R}^{p,q}$, and know that the null cone consists of rays through a copy of $S^{p-1}\times S^{q-1}$; then you also know $S^n$ may be identified with $\{1\}\times S^n$ in the null cone and this forms a complete set of representatives for the projectivized null cone (i.e. null vectors mod rescaling). Then we may rescale by to write $\mathbb{R}^n\to S^n\to \mathbb{R}^{1,1+n}$ as

$$ u \mapsto \left(\frac{1-|u|^2}{1+|u|^2},\frac{2u}{1+|u|^2}\right) \mapsto \left(\frac{1+|u|^2}{2},\frac{1-|u|^2}{2},u\right). $$

Indeed, this map $\phi$ can be considered a form of stereographic projection. Now let $n=2$.

Note the pseudo- inner product $\xi\cdot \phi(u)$ is quadratic in $u$, and similarly the equation of a circle in the plane centered at $p$ of radius $r$, $|u-p|^2=r^2$, is similarly quadratic in $u$. Thus, for any choice of circle $C(p,r)$, there should be a corresponding $\xi$ for which $|u-p|^2=r^2 ~\Leftrightarrow~ \xi\cdot\phi(u)=0$. Explicitly, solving

$$ |u|^2-2p\cdot u + (|p|^2-r^2) = 0 $$

$$ \iff $$

$$ -\xi_1\frac{1+|u|^2}{2} +\xi_2\frac{1-|u|^2}{2}+\zeta\cdot u =0 $$

yields (after rescaling)

$$ \xi(p,r)=\left(\frac{1+|p|^2-r^2}{2},\frac{1-|p|^2+r^2}{2},p\right). $$

Of course, you're curious what the inner product of two circles is, so you compute

$$ \xi(p_1,r_1)\cdot\xi(p_2,r_2)=\frac{1}{2}(r_1^2+r_2^2-|p_1-p_2|^2). $$

If the two circles intersect, then the terms of this expression are the same as those that appear in the law of cosines for the triangle vertexed at the circle centers and one of their points of intersection. Indeed, the angle $\varphi$ between the legs of this triangle is the (acute) angle between the circles' tangent lines at their intersection, because the triangle legs are just the tangent lines rotated by a right angle. Therefore,

$$ \xi(p_1,r_1)\cdot\xi(p_2,r_2)=r_1r_2\cos\varphi $$

by the law of cosines.

It is also noteworthy that $\xi(p,r)\cdot\xi(p,r)=r^2$, so the above is reminiscent of the  norm-angle formula for the usual Euclidean inner product. This observation can also help generalize $\xi$ to a projection $\eta:\mathbb{R}^n\to\mathbb{R}^{2,1+n}$.

You may also look up "power of a point" or "Darboux product." The projection $\xi$ is the topic of "Mobius geometry" which leads to $\eta$ in "Lie sphere geometry." One source for this is Cecil.


问:
$\langle A,B \rangle$ is the unique Sylow $2$-subgoup of $SL_2(F_3)$
Prove that the subgroup of $SL_2(F_3)$ generated by $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ and $\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$ is the unique Sylow $2$-subgoup of $SL_2(F_3)$.

Proof: Let $G=SL_2(F_3)$. $|G|=24=2^3\cdot 3$. Let

$A =\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$
and
$B =\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$

By the Sylow Theorems, let $n_2$ be the number of Sylow $2$-subgroup. $n_2\equiv 1 \mod 2$ and $n_2 | 3 \Rightarrow n_2=1,3$, and Sylow $2$-subgroup(s) have order $8$. We need to check $|\langle A,B \rangle | = 8$. Note $A^4=B^4=I$. We find
$$
\langle A,B \rangle
= \left\{ \pm I, \pm A, \pm B, \pm \begin{pmatrix} -1 & 1 \\ 1 & 1 \end{pmatrix}\right\}
$$
What is left to show is $\langle A,B\rangle$ is the only one. We can do that by show $\langle A,B \rangle$ is normal.
i.e. $gng^{-1} \in \langle A,B \rangle$ $\forall g=\begin{pmatrix}
a & b \\ c & d \end{pmatrix}$ and $n\in N$.

So this is where I am stuck. I know I can do the computation for each element of $\langle A,B \rangle$ but that would suck. Is there a way to avoid that or not?

答:
Let $P = \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix}$.  Then $P^3 = I$ and $P^{-1} = P^2 = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$.  Since $| \langle P \rangle | = 3$ and $| \langle A, B \rangle | = 8$ we have $\langle P \rangle \cdot \langle A, B \rangle = \langle P, A, B \rangle = SL_2(\mathbb{F}_3)$.

Calculating $PAP^{-1} = \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} -1 & -1 \\ -1 & 1 \end{pmatrix} = -B \in \langle A, B \rangle$,

and

$PBP^{-1} = \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} -1 & 1 \\ 1 & 1 \end{pmatrix}= AB \in \langle A, B \rangle$.

Of course $P$ commutes with $\pm I$ and since these elements generate $\langle A, B \rangle$ we have that conjugation by $P$ fixes $\langle A, B \rangle$.  Since $\langle A, B \rangle$ is closed under conjugation by every element of itself, and by $P$, and since these elements generate $SL_2(\mathbb{F}_3)$, we have that $\langle A,B \rangle$ is closed under conjugation by any element, and thus is normal.  Since all Sylow 2-subgroups are conjugate, there is only one, namely $\langle A, B \rangle$.


问:
Find the least square solution x

Find the least square solution $x$ for $Ax = b$ if
$$A =\left(\begin{array}{rrr} 2 & 0\\
1 & 3 \\
0 & 2\end{array}\right) \quad b =\left(\begin{array}{rrr} 7\\
0\\-1\end{array}\right) $$

My Solution:

I found $A^T A =\left(\begin{array}{rrr} 7 & -1\\
-1 & 8 \\
\end{array}\right) $

$$A^T b =\left(\begin{array}{rrr} 2\\
2\\
\end{array}\right) $$

$$\left(\begin{array}{rrr} 5 & -1\\
-1 & 5 \\
\end{array}\right) x = \left(\begin{array}{rrr} 2\\
2\\
\end{array}\right)  $$

But I can't solve this equation to find $x$ as there is no solution.

My solution after finding $x$ was to then find this:

$$Ax = \left(\begin{array}{rrr} 2 & 0\\
1 & 1 \\
0 & 2\end{array}\right)  \left(\begin{array}{rrr} x\\
x\\ \end{array}\right) = \left(\begin{array}{rrr} a\\
b\\c\end{array}\right)  $$

and then use $b-Ax^ = \sqrt{1+0+1} - \sqrt{a^2 + b^2 +c^2} = \textrm{final  answer}$

Where am I going wrong with my approach? I am stuck as I can't solve for $x$?

答:
Your first error is in computing $A^\top b$. We have
$$A^\top b = \begin{pmatrix} 2 & -1 & 0 \\ 0 & 1 & 2 \end{pmatrix}\begin{pmatrix} 1 \\ 0 \\ -1\end{pmatrix} = \begin{pmatrix} 2 \\ -2 \end{pmatrix}$$

So, we need to solve
$$\begin{pmatrix} 5 & -1\\
-1 & 5
\end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2\\
-2
\end{pmatrix}.$$

I don't know where you went wrong specifically here, but there is a unique solution, as $A^\top A$ is invertible (compute its determinant to see this). We can use the old shortcut to compute the inverse of this matrix:
$$\begin{pmatrix} a & b \\ c & d\end{pmatrix}^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a\end{pmatrix}.$$
In particular,
$$\begin{pmatrix} 5 & -1 \\ -1 & 5\end{pmatrix}^{-1} = \frac{1}{24}\begin{pmatrix} 5 & 1 \\ 1 & 5\end{pmatrix},$$
so
$$\begin{pmatrix} x \\ y \end{pmatrix} = \frac{1}{24}\begin{pmatrix} 5 & 1\\1 & 5\end{pmatrix}\begin{pmatrix} 2\\-2\end{pmatrix} = \begin{pmatrix}\frac{1}{3} \\ -\frac{1}{3}\end{pmatrix}.$$
Therefore, the closest point to $(1, 0, -1)^\top$ in the columnspace of $A$ is
$$A\begin{pmatrix}\frac{1}{3} \\ -\frac{1}{3}\end{pmatrix} =\begin{pmatrix} 2 & 0\\-1 & 1 \\0 & 2\end{pmatrix}\begin{pmatrix}\frac{1}{3} \\ -\frac{1}{3}\end{pmatrix} = \begin{pmatrix} \frac{2}{3} \\ -\frac{2}{3} \\ -\frac{2}{3} \end{pmatrix}.$$

Title: Find the least square solution x


问:
$A$ be a $2\times 2$ real matrix with trace $2$ and determinant $-3$
$A$ be a $2\times 2$ real matrix  with trace $2$ and determinant $-3$, consider the linear map $T:M_2(\mathbb{R})\to M_2(\mathbb{R}):=B\to AB$ Then which of the following are true?

$T$ is diagonalizable

$T$ is invertible

$2$ is an eigen value of $T$

$T(B)=B$ for some $0\ne B\in M_2(\mathbb{R})$

if $ A=\begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{pmatrix}$ Then I have calculated that matrix of $T$ will be \begin{pmatrix}a_{11}&a_{12}&0&0\\a_{21}&a_{22}&0&0\\0&0&a_{11}&a_{12}\\0&0&a_{21}&a_{22}\end{pmatrix} so $T$ is invertible I can say as $\det T=(\det A)^2=9$, could any one help to find out others as true/false?

答:
Nicely done so far. Don't forget to precise that you took the matrix of $T$ with respect to the basis $\{E_{11},E_{12},E_{21},E_{22}\}$ if you are to write this down.

By assumptions, the characteristic polynomial of $A$ is $$X^2-(\mbox{tr} \,A)X+\det A=X^2-2X-3=(X-3)(X+1).$$
So the eigenvalues of $A$, $-1,3$, are distinct. Hence $A$ is diagonalizable. This should help you answer 1, 3, and 4. By block considerations. Note that 4 says that $1$ is an eigenvalue of $T$.

In case you are not used to matrix block computations, if $P$ is invertible in $M_2$, then
$$
\pmatrix{P&0\\0&P}\pmatrix{A&0\\0&A}\pmatrix{P^{-1}&0\\0&P^{-1}}=\pmatrix{PAP^{-1}&0\\0&PAP^{-1}}
$$
in $M_4$. So if $P$ diagonalizes $A$...


问:
How to write the negation of a biconditional?
I have two statements $p$ and $q$ such that:

Statement $p$: Ravi reads Mathematics

Statement $q$: Ravi reads Chemistry

Now, I am required to write the negation of $p \iff q$, i.e., I need to write $\lnot (p \iff q)$, which, if I understand correctly, is, by definition: $(p \land \lnot  q) \lor(\lnot p \land q )$

So, according to me, it should be as follows:

Ravi reads Mathematics and not Chemistry or Ravi doesn't read Mathematics and reads Chemistry.

But, the book states the answer to be as follows:

Ravi reads neither Mathematics nor Chemistry

Are the two statements equivalent? Or am I going wrong here?

Check question 3 (e) for the question asked here

The answer for question 3 (e)

答:
You are exactly right, and your book is wrong.

Some equivalent ways to write $\lnot (p \iff q)$ are

$$
\begin{split}
\lnot (p \iff q) &\equiv \lnot(p \implies q) \lor \lnot(q \implies p) \\
&\equiv (p \land \lnot q) \lor (\lnot p \land q) \\
&\equiv p \oplus q.
\end{split}
$$
(The last expression uses the XOR ("exclusive or") operator, $\oplus$.)

However you choose to write it, it is equivalent to what you said:

"Ravi reads Mathematics and not Chemistry or Ravi doesn't read Mathematics and reads Chemistry.
"

OP, don't be discouraged by the rude or confusing answers in this thread. Trust in your logic!


答: To add to Henry's answer:  your claim that the first appearances of primes (other than 3) come in increasing order is also correct.  More specifically, for all primes $p > 3$, the first occurrence of $p$ in the last column of the table is next to either $A = 2p-1$ or $A = 2p+1$.  Proof:  since $B = (A-1)(A+1)/6$ (and $p$ doesn't divide 6), $p$ divides $B$ if and only if it divides either $A-1$ or $A+1$.  The first such odd numbers $A > 1$ with this property are $A = 2p \pm 1$.  So $p$ will first appear next to either $A = 2p-1$ (if this isn't a multiple of 3) or $A = 2p+1$ (otherwise).

Title: Is the following producing a list of all prime numbers without skipping in a consecutive order?

Body:
The following consecutive list is made as a table with the following columns :

column 1
column 2
column 3

$A$
$B=(A^2 -1) / 6$
prime factors of $B$

Where $A$ is odd and $A \mod 3 ≠ 0$

Example: $A = 7$ : $B=(7^2-1) / 6 = 8$ : prime factors of $8$ are $1,2$

You will notice that the results start building a sequence of all possible prime numbers, and never skipping a prime number in a consecutive way.

For example: If $19$ is a prime factor result of $A=37$, it means that $17$ had somewhere appeared before, and as if it would been impossible for $19$ to appear after a resulted $13$ while skipping a previous $17$. I have run the experiment until $A=1000$ and it seems to be true. Is there any proof of it being true? or is there a counter example?

Update: per @Ross Millikan comments 1 is not a prime and 5 and 7 appear in the list before 3 (making it the only exception so far). I didn't take into account 3 as a prime because the whole list is generated by the condition that  $$ is odd and $ \mod 3 ≠0$, and the reason $1$ was part of the results is just produced by my programming.

To better illustrate, here is a sample list where $A$ is growing consecutively (Where $A$ is odd and $A \mod 3 ≠ 0$):

$A$ : $B=(A^2 -1) / 6$ : prime factors of $B$

5 : 4 : prime factors of 4 are 2

7 : 8 : prime factors of 8 are 2

11 : 20 : prime factors of 20 are 2 , 5

13 : 28 : prime factors of 28 are 2 , 7

17 : 48 : prime factors of 48 are 2 , 3

19 : 60 : prime factors of 60 are 2 , 3 , 5

23 : 88 : prime factors of 88 are 2 , 11

25 : 104 : prime factors of 104 are 2 , 13

29 : 140 : prime factors of 140 are 2 , 5 , 7

31 : 160 : prime factors of 160 are 2 , 5

35 : 204 : prime factors of 204 are 2 , 3 , 17

37 : 228 : prime factors of 228 are 2 , 3 , 19

41 : 280 : prime factors of 280 are 2 , 5 , 7

43 : 308 : prime factors of 308 are 2 , 7 , 11

47 : 368 : prime factors of 368 are 2 , 23

49 : 400 : prime factors of 400 are 2 , 5

53 : 468 : prime factors of 468 are 2 , 3 , 13

55 : 504 : prime factors of 504 are 2 , 3 , 7

59 : 580 : prime factors of 580 are 2 , 5 , 29

61 : 620 : prime factors of 620 are 2 , 5 , 31

65 : 704 : prime factors of 704 are 2 , 11

67 : 748 : prime factors of 748 are 2 , 11 , 17

71 : 840 : prime factors of 840 are 2 , 3 , 5 , 7

73 : 888 : prime factors of 888 are 2 , 3 , 37

77 : 988 : prime factors of 988 are 2 , 13 , 19

79 : 1040 : prime factors of 1040 are 2 , 5 , 13

83 : 1148 : prime factors of 1148 are 2 , 7 , 41

85 : 1204 : prime factors of 1204 are 2 , 7 , 43

89 : 1320 : prime factors of 1320 are 2 , 3 , 5 , 11

91 : 1380 : prime factors of 1380 are 2 , 3 , 5 , 23

95 : 1504 : prime factors of 1504 are 2 , 47

97 : 1568 : prime factors of 1568 are 2 , 7

101 : 1700 : prime factors of 1700 are 2 , 5 , 17

103 : 1768 : prime factors of 1768 are 2 , 13 , 17

107 : 1908 : prime factors of 1908 are 2 , 3 , 53

109 : 1980 : prime factors of 1980 are 2 , 3 , 5 , 11

113 : 2128 : prime factors of 2128 are 2 , 7 , 19

115 : 2204 : prime factors of 2204 are 2 , 19 , 29

119 : 2360 : prime factors of 2360 are 2 , 5 , 59

121 : 2440 : prime factors of 2440 are 2 , 5 , 61


问:
Two discrete time random processes taking the same value in the same time period
My question is kind of a basic one, but it confuses me. The answer will be probably very simple.

We have two discrete time random processes $\{X_i\}$ and $\{Y_i\}$, where $X_i \in \{1,2,\dots,K\}$ and $Y_i \in \{1,2,\dots,K,\dots,L\}$ ($L>K$), and $P(X_i=k)=\frac{1}{K}$ and $P(Y_i=k)=\frac{1}{L}, \ \forall k$ and they are i.i.d .

Consider the case that the time scale of the processes are different, i.e., for every time step of $\{X_i\}$, an integer number of steps pass for $\{Y_i\}$. Let $\gamma$ denote this scaling factor.

I will try to clarify the scaling with some illustration for $\gamma=2$.

$\{X_i\}$ : |1111|2222|1111|2222|----|

$\{Y_i\}$ : |11|22|11|22|11|22|11|22|--|--|

So the processes are taking values 1 and 2 alternatingly. A time step of the first process is of size 4 time unit, and a time step of the second process is of 2 time unit. (I know this is counterintuitive to discrete time processes, maybe this is the problem).

What I am trying to find is the probability that both processes have the same value in the same time period. I have for the general case, but I am not sure it is correct: $$1- P(X_i=k,Y_{\gamma i}\neq k,Y_{\gamma i -1 } \neq k,\dots ,Y_{\gamma (i-1)+1} \neq k) = 1- \frac{1}{K}(1-\frac{1}{L})^{\gamma}$$

In fact, I think when $\gamma=1$, this probability becomes $$P(X_i=k,Y_{\gamma i} = k) = \frac{1}{K}\frac{1}{L}$$

which is not the same as the general case.

Can you help me to write a correct formulation for this probability?

答:
I think I solved my problem. The answer is

$$P(\text{none of Y_i's equals to k in that period} |X_i=k)P(X_i=k)=(1-(1-\frac{1}{L})^{\gamma})\frac{1}{K}$$


问:
Infinite series and its upper and lower limit.
I am learning analysis on my own and I am puzzled with the following question.

Consider the series $$\frac{1}{2}+\frac{1}{3}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{2^3}+\frac{1}{3^3}+\cdots$$ Indicate whether this series converges or diverges.

The following is what I was able to get.

First I noticed that $$a_n = \frac{1}{2^n}+\frac{1}{3^n}$$ would be a simple and nice form of this sequence, so according to the ratio test, $$\lim_{n \to \infty} \frac{1}{6} \left( \frac{3+2(\frac{2}{3})^n}{1+(\frac{2}{3})^n}\right) = \frac{1}{2}$$ therefore the series must converge.

However, I later understood the concept of upper an lower limits a little, so taking $$ a_n =
\begin{cases}
\frac{1}{2^n},  & \text{if $n$ is odd} \\
\frac{1}{3^{n-1}}, & \text{if $n$ is even}  \\
\end{cases}$$ I figured out the following $$\lim_{n \to \infty} \frac{1}{2}\left(\frac{2}{3}\right)^n = 0$$ when $n$ is odd and $$\lim_{n \to \infty} \frac{1}{2}\left(\frac{3}{2}\right)^n = +\infty$$ when $n$ is even.

Through this experience I realized and really understood how a subsequence of a sequence may have multiple limits, and the supremum and the infimum of those limits give us the upper and lower limits, respectively. But the book does something different than I did, and I would like to have the following questions answered.

1),$\quad$ From my understanding, the ratio test tells us that $\Sigma a_n$ diverges when the upper limit is greater than 1. Clearly one of the limits goes to $+\infty$, so why cannot we conclude that the series diverge ?

2),$\quad$  I learned that by choice of the subsequence, the limit may or mayn't be different. In the case that it is different, how do we know what is the supremum and infimum? It would be clear that if they were $\pm \infty$ but in my case, why is it guaranteed that there is no other subsequence that has a negative limit ? Is there an algebraic way to guarantee them ? Or do we always have to conceptually understand all of the subsequential limits.

3),$\quad$  The book says the ratio test is inconclusive, which turns me back to question 1), but it used the ratio test to determine that the series converges. According to my even-odd wise definition of $a_n$, the "larger limit" (I am purposefully not saying supremum because I don't know how it's guaranteed that it is the supremum) using the ratio test can be found as $$\lim_{n \to \infty} \sqrt[n]{a_n} = \lim_{n \to \infty} \sqrt[n]{\frac{1}{2^n}} = {1 \over 2}$$ But the book says $$\lim_{n \to \infty} \sqrt[n]{a_n} = \lim_{n \to \infty} \sqrt[2n]{\frac{1}{2^n}} = {1 \over \sqrt{2}}.$$ Where did the $2n$ come from ?

Can someone help me out ?

Edit:Thanks for helping me out everyone! I really appreciate it.

答:
The even terms are a geometric series, with sum $\frac{1/3}{1-1/3}=\frac{1}{2}$.  The odd terms are a geometric series, with sum 1.  Because all the terms are positive, the series is absolutely convergent, so you may rearrange the odd terms to go first, then the even terms.  Hence the overall series is convergent, with sum $\frac{3}{2}$.

In answer to your specific questions, the ratio test is indeed inconclusive in this case, because the limit of the ratios does not exist.  That doesn't mean the series diverges, it means the test doesn't help.  The supremum is infinite, the infimum is finite, so even with that refinement of the ratio test it's inconclusive.

With regards to your third question, the n-th root test is about specifically the n-th term.  In this series, $\frac{1}{2^n}$ is the 2n-th term.  If you group the terms pairwise as you did, then you need to take the limit  of $\sqrt[n]{\frac{1}{2^n}+\frac{1}{3^n}}$.


问:
Evaluate $\lim_{n \rightarrow \infty} \int_0^1 \frac{nx^{n-1}}{2+x} dx$
Evaluate $\lim_{n \rightarrow \infty} \int_0^1 \frac{nx^{n-1}}{2+x} dx$.

Attempt:

Let $h(x)=x^n$. Then $nx^{n-1} = h'(x)$.

I'm thinking about using the Dominated Convergence Theorem.

So $f_n= \frac{nx^{n-1}}{2+x} $.

But I'm not sure what the limit of $f_n$. And I couldn't find any bounds for $f_n$.

So I don't know if DCT will work here.

If DCT doesn't apply here, what would be a good way to solve this?

Thanks.

答:
The integrand strongly suggest  a gaussian hypergeometric function with a last argument being $\pm \frac x 2$
$$I_n=\int\frac{nx^{n-1}}{x+2} dx=\frac{x^n}{2}-\frac{n\, x^{n+1} }{4 (n+1)}\, _2F_1\left(1,n+1;n+2;-\frac{x}{2}\right)$$ Assuming $n>1$
$$J_n=\int_0^1\frac{nx^{n-1}}{x+2} dx=\frac 12-\frac n{4(n+1)}\, _2F_1\left(1,n+1;n+2;-\frac{1}{2}\right)$$
$$\lim_{n\to \infty } \, \, _2F_1\left(1,n+1;n+2;-\frac{1}{2}\right)=\frac 23$$
$$J_n \sim \frac{1}{2}-\frac{n}{6 (n+1)}\quad \to\quad \frac 13$$


问:
Rotation about the y-axis and volume between two-curves.
I was doing this practice question that if given two functions of y=1/4x and y=x^2, find the volume of the solid around the y-axis.

I am just really confused on what to do - could someone please guide me through it?

Here are my steps so far

Doing 1/4x = x^2, we get x = 0, or x = 0.25.

Thus, continuing to an integral:

$$\int_0^4 pi  *((\frac{x}{4})^2 -(x^2)^2) = \int_0^4 pi  *((\frac{x^2}{16}) -(x^4)$$

But at this point I get a negative value, so I know I'm doing something wrong.

答:
First off, your integral is missing $dx$, and since you want to find the volume of the solid formed when rotated about the y-axis, you should be looking at the integral with respect to $y$, i.e. an integral $dy$.

This is because you want the cross sections to look about the same and be able to describe them (the cross sections) properly with a function of $y$. (as you move across your solid)

So we write the functions as $x = 4y, x = \sqrt{y}$.

You are correct in finding the points of intersection, but I will turn our focus to the y-coordinate of those points of intersection, and we see they are $y=0,y=\frac{1}{16}$.

So the integral should be phrased as $\displaystyle\int_{0}^{\frac{1}{16}} \pi ((\sqrt{y})^2 - (4y)^2)dy = \int_{0}^{\frac{1}{16}} \pi (y - 16y^2)dy$.


问:
Where does the function $f(x) = \frac{2x}{x - 7}$ have an increasing slope?
Where does the function $f(x) = \frac{2x}{x - 7}$ have an increasing slope?

$a. x \le 0, x > 7$

$b. x<7$

$c. x > 7$

$d. x \in \Bbb R, x \neq 7$

This question is from a test of mine in a pre-calculus course (so no calculus allowed in answering the question).I have no idea as to how to solve this problem. I can tell where the function is negative and where it is positive, but that's about it, and I'm fairly sure that's no use here.

Any ideas?

答:
$f(x)$ has an increasing slope when the second derivative of $f(x)$ with respect to $x$ is positive. The second derivative is $28/(x-7)^3$, which is $>0$ when $x>7$, $<0$ when $x<7$, and undefined when $x=7$. Therefore, the answer is $c (x>7)$.

Alternately, you could graph your function, draw some tangent lines to the curve at different points, and just notice where the slopes of those tangent lines are increasing. This way you won't have to do any differentiation. Hope this helps.


问:
Calculate Integral $\int_{0}^ady \int_{y}^a \sqrt{x^2-y^2}f'''(a-x)dx$
I need to show that

$\int_{0}^ady \int_{y}^a \sqrt{x^2-y^2}f'''(a-x)dx=\frac{\pi}2(f(a)-f(0)-af'(0)-\frac{a^2}2f''(0))$

I started first by doing integration by parts

$\int_{0}^ady \int_{y}^a \sqrt{x^2-y^2}f'''(a-x)dx=\int_{0}^a[-f''(a-x)\sqrt{x^2-y^2}]_{y}^ady+\int_{0}^ady \int_{y}^a f''(a-x)\frac{x}{\sqrt{x^2-y^2}}dx$

$=-f''(0)\frac{\pi a^2}4+\int_{0}^ady \int_{y}^a f''(a-x)\frac{x}{\sqrt{x^2-y^2}}dx$

I thought I was on the right track, but then when I do integration by parts once again

$\int_{0}^ady \int_{y}^a f''(a-x)\frac{x}{\sqrt{x^2-y^2}}dx=\int_{0}^a[-f'(a-x)\frac{x}{\sqrt{x^2-y^2}}]_{y}^a dy+\int_{0}^ady \int_{y}^a f'(a-x)\frac{y^2}{(x^2-y^2)^{3/2}}dx$

When I plug in the $y$, I get a diverging result

Am I missing something? or is there another way to show this besides the integration by parts?

Any help would very much be appreciated

答:
Switching the order of integration we get

$$\int_0^af'''(a-x)dx\int_0^x\sqrt{x^2-y^2}dy = \frac{\pi}{4}\int_0^ax^2f'''(a-x)dx$$

since it's just the area of a quarter circle with radius $x$. From here integration by parts should get the desired equality.


问:
Proving the product topology is discrete iff each component is discrete

Let $(X_i,T_i),i\in\Bbb N$, be a countably infinite (non-empty) family of topological spaces. Prove that $\prod(X_i,T_i)$ is a discrete space iff each $(X_i,T_i)$ is discrete and all but a finite number of the $X_i,i\in\Bbb N$ are singleton sets.

I think the backward direction is quite trivial so I will just do the forward direction here.

If $(X_i,T_i),i\in\Bbb N$ is discrete, then all singletons or points are open. Now all basic open sets are of the form $U=O_1\times O_2\times\cdots\times O_n×X_{n+1}\times X_{n+2}$ where all but a finite number of the open sets are of the form $X_i$.

How do I go from here to showing that each component is discrete and furthermore all but a finite number of the components are singleton sets?

答:
Let $=\prod_{n\in\Bbb Z^+}X_n$, and let $x=\langle x_n:n\in\Bbb Z^+\rangle\in X$. $X$ is discrete, so $\{x\}$ is open, and there is therefore a basic open set $B$ such that $x\in B\subseteq\{x\}$, i.e., such that $B=\{x\}$. Since $B$ is a basic open set, there are open sets $U_n\subseteq X_n$ for $n\in\Bbb Z^+$ such that

$$\{x\}=B=\prod_{n\in\Bbb Z^+}U_n\,,$$

and there is an $m\in\Bbb Z^+$ such that $U_n=X_n$ for all $n>m$. Clearly $x_n\in U_n$ for each $n\in\Bbb Z^+$. Suppose that there are an $n\in\Bbb Z^+$ and a $z\in U_n\setminus\{x_n\}$, and let $y=\langle y_k:k\in\Bbb Z^+\rangle$ be given by

$$y_k=\begin{cases}
z,&\text{if }k=n\\
x_k,&\text{otherwise.}
\end{cases}$$

Then on the one hand $y_k\in U_k$ for each $k\in\Bbb Z^+$, so $y\in B$, but on the other hand $y\ne x$, so $y\notin B$. This is impossible, so there is no such point $y$ in $X$, and therefore $U_n\setminus\{x_n\}=\varnothing$ for each $n\in\Bbb Z^+$. $U_n=X_n$ when $n>m$, so for $n>m$ we have $X_n=\{x_n\}$; and for $n\le m$ we have $U_n=\{x_n\}$, so $\{x_n\}$ is open in $X_n$. Thus, at most the first $m$ factor spaces have more than one point.

I leave it to you to finish the job of showing that each $X_n$ with $n\le m$ must have the discrete topology; just pick any $n\le m$ and any point $x_n'\in X_n$, apply the same argument to the point of $X$ obtained from $x$ by changing the $n$-th coordinate to $x_n'$, and conclude that $\{x_n'\}$ is open in $X_n$.


问:
show $(1+\frac{x}{n})^n \leq e^x$
I want to show that $(1+\frac{x}{n})^n \leq e^x$ for all $x \geq -n$, $n\in \mathbb{N}$.

I already showed that positive case ($x>0$) and zero case $(x=0)$.

Not sure how to approach the negative case, because in the case $x>0$ I used that $x^n$ is an increasing function on $(0, \infty)$. But I can't do the same when $x$ is negative.

答:
You first want to prove $1 + x \le e^x$ for all $x$, which holds because $e^x$ is convex and $1+x$ is the tangent line to $e^x$ at $(0,1)$, for example.

From $x \ge -n$ we get $\frac xn \ge -1$, or $1 + \frac xn \ge 0$. By the inequality, $1 + \frac xn \le e^{x/n}$. Because $t \mapsto t^n$ is an increasing function on $[0,\infty)$ and both sides are positive, $(1 + \frac xn)^n \le (e^{x/n})^n = e^x$.

Title: show $(1+\frac{x}{n})^n \leq e^x$


问:
χ 2 distribution with 50 degrees of freedom. Compute approximate value of P(49 < Y < 51). (whole question in body its long)
Let Y bar denote the mean of a random sample of size 100 from a χ 2 distribution with 50 degrees of freedom. Compute an approximate value of P(49 < Y < 51).

I've been away from statistics for two semesters so this stuff is really difficult for me now. I don't really remember how to approach these problems anymore and I feel like I have a lot of gaps in my knowledge. Ideally I'd like to see the step by step process on this so I can apply this to other problems that are relatively similar.

答:
The central limit theorem says that the average of $Y_1,...,Y_n$ where Y's are independent and identically distributed (iid) having common mean $\mu$ and variance $\sigma^2$ will be approximately normally distributed $\bar Y\sim N(\mu, \frac {\sigma^2} n)$ for n at least around 20 to 30. The question asks you to use the Central Limit Theorem to approximate $P(49<\bar Y<51)$. As you know $Y\sim ChiSq(k=50)$ with mean k=50 and variance 2k=100.

You get:

$$\begin{split}Z&\approx\frac{\bar Y-\mu}{\sigma/\sqrt n}\\&=\frac{\bar Y-k}{\sqrt{2k}/\sqrt {100}}\\
&=\frac{\bar Y-50}{1}\text {is approximately normally distributed}\end{split}$$

$$\begin{split}Z_1&=\frac{51-50}{1}\text{ find the upper bound}\\
&=1\\
Z_2&=\frac{49-50}{1}\text{ find the lower bound}\\
&=-1\end{split}$$

Now you must calculate the area under the normal distribution between -1 and 1. You can use a table or use R:

```<code>> pnorm(1)-pnorm(-1)
[1] 0.6826895
</code>```

Body:
The central limit theorem says that the average of $Y_1,...,Y_n$ where Y's are independent and identically distributed (iid) having common mean $\mu$ and variance $\sigma^2$ will be approximately normally distributed $\bar Y\sim N(\mu, \frac {\sigma^2} n)$ for n at least around 20 to 30. The question asks you to use the Central Limit Theorem to approximate $P(49<\bar Y<51)$. As you know $Y\sim ChiSq(k=50)$ with mean k=50 and variance 2k=100.

You get:

$$\begin{split}Z&\approx\frac{\bar Y-\mu}{\sigma/\sqrt n}\\&=\frac{\bar Y-k}{\sqrt{2k}/\sqrt {100}}\\
&=\frac{\bar Y-50}{1}\text {is approximately normally distributed}\end{split}$$

$$\begin{split}Z_1&=\frac{51-50}{1}\text{ find the upper bound}\\
&=1\\
Z_2&=\frac{49-50}{1}\text{ find the lower bound}\\
&=-1\end{split}$$

Now you must calculate the area under the normal distribution between -1 and 1. You can use a table or use R:

> pnorm(1)-pnorm(-1)
[1] 0.6826895


问:
Find $\lim_{n \rightarrow \infty} \int_0 ^{\infty} \left(1+ \frac{x}{n}\right)^n e^{-ex} dx$
Find $$\lim_{n \rightarrow \infty} \int_0 ^{\infty} \left(1+ \frac{x}{n}\right)^n e^{-ex} dx.$$

I know that $\lim_{n \rightarrow \infty}  \left(1+ \frac{x}{n}\right)^n = e^x.$

Also, I showed that $ \left(1+ \frac{x}{n}\right)^n \leq e^x$ for all $x \geq -n$.

So I'm thinking about the Lebesgue Dominated Convergence Theorem. But I'm not sure how to deal with the $e^{-ex}$ part.

答:
Let $f_n : [0, \infty) \to \Bbb R$ be defined by $$f_n(x) := \left(1 + \frac{x}{n}\right)^ne^{-ex}.$$

With a clever use of AM-GM, one can show that $f_n(x) \le f_{n+1}(x)$ for all $x  \ge 0$. (You only need to show the inequality for the part without the $e^{-ex}$.)

What can you conclude using the Monotone Convergence Theorem now?


答:
I think at least one of the suggested ways in metamorphy's comment should appear as answer:

Integration by parts gives:

\begin{eqnarray*}\int_0^1 \frac{nx^{n-1}}{2+x}dx
& = & \left.\frac{x^{n}}{2+x}\right|_0^1 + \underbrace{\int_0^1\frac{x^{n}}{(2+x)^2}dx}_{I_n} \\
& = & \frac 13 + I_n
\end{eqnarray*}

Now, without any dominated convergence you get

$$0\leq I_n \leq \frac 14\int_0^1x^n\;dx = \frac 1{4(n+1)}\stackrel{n\to\infty}{\longrightarrow}0$$

Body:
Evaluate $\lim_{n \rightarrow \infty} \int_0^1 \frac{nx^{n-1}}{2+x} dx$.

Attempt:

Let $h(x)=x^n$. Then $nx^{n-1} = h'(x)$.

I'm thinking about using the Dominated Convergence Theorem.

So $f_n= \frac{nx^{n-1}}{2+x} $.

But I'm not sure what the limit of $f_n$. And I couldn't find any bounds for $f_n$.

So I don't know if DCT will work here.

If DCT doesn't apply here, what would be a good way to solve this?

Thanks.


问:
Prove that $\sqrt{13}$ is irrational in 3 ways.
I am asked how to prove that the $\sqrt{13}$ is irrational in 3 ways. I only know of one, where we assume $\sqrt{13} = \frac{a}{b}$ where $a, b$ are coprime, and we prove with contradiction that $13 \mid a$ and $13 \mid b$.

However, I am not sure where to begin with the other two proofs. I thought that this was the only way to prove irrationality of a number. I would really appreciate some help on where to start/look for to start the two other proofs.

答:
\begin{align}
\sqrt{13} & = 3 + \text{a fractional part, since } 3^2<13<4^2 \\[8pt]
& = 3 + (\sqrt{13}-3) \\[8pt]
& = 3 + \frac{4}{\sqrt{13}+3} \text{ by rationalizing the numerator} \\[8pt]
& = 3 + \frac 1 {(\sqrt{13}+3)/4}
\end{align}
Since $3<\sqrt{13}<4,$ we have $1 < (\sqrt{13}+3)/4 < 2,$ so $(\sqrt{13}+3)/4$ is between $1$ and $2.$

So we have
\begin{align}
\sqrt{13} & = 3 + \frac 1 {1 + \left( \frac {\sqrt{13}-1} 4 \right)} \\[12pt]
& = 3 + \frac 1 {1 + \cfrac 3 {\sqrt{13} + 1}} \text{ by rationalizing the numerator as before}
\end{align}

Continuing in this way, we get
$$
\sqrt{13} = 3 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1{\sqrt{13}+3}}}}}.
$$
In the steps before this, the expression in the denominator was $\dfrac{\sqrt{13} + (\text{an integer})}{\text{an integer}}$ and the denominator in this last fraction was more than $1.$ But this time, the denominator is $1,$, i.e. we have $\dfrac{\sqrt{13}+3} 1.$ This means that now in place of $\sqrt{13}$ at the end of this long expression one can put this entire continued fraction, and we get

$$
\sqrt{13} = 3 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1{6 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1{\sqrt{13}+3}}}}}}}}}}.
$$

And then we can do that again. And again. We get this repeating sequence:

$$
3, \, \overbrace{1,1,1,1,\,6,}\, 1,1,1,1,\,6,\, 1,1,1,1,\,6,\, \ldots
$$

The part under the $\overbrace{\text{overbrace}}$ must keep repeating. This could not happen if this number were rational, as we will argue below. To see, why, suppose we start with a ration approximation to $\sqrt{13}$:

\begin{align}
\sqrt{13}\approx\frac{119}{33} & = 3 + \frac{20}{33} = 3 + \cfrac 1 {1 + \cfrac{13}{20}} \\[12pt]
& = 3 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 7 {13}}} = 3 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 6 7}}} \\[12pt]
& = 3 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 7 {13}}} = 3 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 6}}}}.
\end{align}

But we cannot keep getting smaller positive integers forever. So if we start with a rational number, this process cannot go on forever; it must end. But as shown above, if we start with $\sqrt{13},$ it does not end; it just keeps repeating.


问:
Maximum value of $k$ so that the inequality $\sqrt{y-5}+\sqrt{8-y} \ge k$ should have answer

Suppose $k$ be a real number so that the inequality
$\sqrt{y-5}+\sqrt{8-y} \ge k$ should have answer. what is the maximum
possible value of $k$?

$1)\sqrt{6}-1\quad\quad2)\frac12+\sqrt{3}\quad\quad3)\sqrt6\quad\quad4)2\sqrt2\quad\quad5)\sqrt6+1$

So I translated the question into finding the minimum value of $f(y)$ where $f(y)=\sqrt{y-5}+\sqrt{8-y}\quad$.(because for $k=\min(f(y))$ the inequality certainly has answer.) therefore we should solve  $f'(y)=0$:

$$f'(y)=\frac{1}{\sqrt{y-5}}-\frac{1}{\sqrt{8-y}}=0$$
$$y-5=8-y\rightarrow\quad y=6.5$$

And $f(6.5)=2\sqrt{\frac32}=\sqrt{6}$. so the answer is third choice.

Is my answer right? I know to find maximum or minimum of a function we take derivative of it and equate it to $0$. so here I don't know why $f(6.5)$ is minimum. can you please explain it to me?

EDIT:

I just noticed that $\sqrt6$ is not the minimum because $f(5)=f(8)=\sqrt3<\sqrt6$ maybe it is maximum! so I conclude the answer to this problem should be $\sqrt3$ but I don't have this option in the choices

答:
You found the correct derivative of $f(y)$, but you missed a couple cases.  Also, you are looking for the maximum of $f$, not the minimum, because you only need to have any solutions.  When $k$ is the maximum, only maximizers will solve the inequality.

When solving a problem like this, you should always graph it if you can, or try to visualize it if you are confident.  The graph has a domain of $[5,8]$ and is convex from above so the point $6.5$ is actually a maximum.

When testing for a maximum or minimum, you have to check not only the zeros of the derivative but points where it is undefined and boundaries of the search region as well.  In this case, $5$ and $8$ are both points where the derivative is undefined and boundaries of the search region (because the search region is the whole domain of $f(y)$).

Finally, you should always check if a zero of the derivative is a maximum or minimum.  This can be done in several ways: the second derivative test, comparison to another point which cannot be extremal, or you can simply calculate the function at all zeros/undefined points of the derivative and boundaries and find the largest/smallest value.

In this case, both $5$ and $8$ produce the smallest value of $\sqrt{3}$, so the value of $\sqrt{6}$ you found is indeed the answer, but because it is the maximum, not the minimum.

Edit: Also, you can solve this without calculus.  We want $$\sqrt{y-5}+\sqrt{8-y}\ge k$$ which implies (by squaring both sides) $$3+2\sqrt{-y^2+13y-40}\ge k^2.$$
The maximum of the lhs is $3+2\sqrt{\max(-y^2+13y-40)}$, but that polynomial is a parabola that is convex from above so its vertex is its maximum.  A parabola's vertex is at $\frac{-b}{2a}$, which here is $\frac{13}{2}$.  The value of the parabola at the vertex is thus $\frac{9}{4}$, which we can plug in to get the a maximum of 6 for $k^2$ and thus a maximum of $\sqrt{6}$ for $k$.


问:
Given a triangle with points in $\mathbb{R}^3$, find the coordinates of a point perpendicular to a side
Consider the triangle ABC in $\mathbb{R}^3$ formed by the point $A(3,2,1)$, $B(4,4,2)$, $C(6,1,0)$.

Find the coordinates of the point $D$ on $BC$ such that $AD$ is perpendicular to $BC$.

I believe this uses projections, but I can't seem to get started. I tried the projection of $AC$ onto $BD$ and $AB$ onto $BC$, but to no avail.

Any help is loved! Thanks.

答:
Edited:

There are two ideas used is this.

(1),$AD$ must be perpendicular to $BC$, so their dot product must equal 0.

(2),$D$ is on $BC$, so we must find an eqn. for line $BC$.

Let's start with (2).  We know that the direction vector of the line $BC$ is parallel to the vector $\overrightarrow BC$ which is $<2,-3,-2>$ and the line passes through $B(4,4,2)$. Therefore the parametric eqn. for line $BC$ must be $$x = 2t+4$$ $$y = -3t+4$$ $$z=-2t+2$$

This means that all points on the line $BC$, in particular, $D$, will have coordinates of the form $$D(2t+4,-3t+4,-2t+2)$$

Now let's move on to (1).

The vectors $\overrightarrow {AD}$ and $\overrightarrow{BC}$ are perpendicular to each other, so their dot product must be 0, i.e $$<3-(2t+4),2-(-3t+4),1-(-2t+2)> \cdot <2,-3,-2> = 0$$

All you have to do is to find $t$ and plug it back in.


答:
tl;dr: you can use the DCT after the change of variable.

If you want to use the DCT (it's overkill, but why not?): first, do the change of variable $u = x^{n}$ you were thinking of. Then,
$$
\int_0^1 \frac{n x^{n-1}}{2+x}dx = \int_0^1 \frac{1}{2+u^{1/n}}du
$$
and you can now apply the DCT to the new integral (the dominating function being, for instance, $g=1/2$). This gives you
$$
\int_0^1 \frac{n x^{n-1}}{2+x}dx = \int_0^1 \frac{1}{2+u^{1/n}}du \xrightarrow[n\to\infty]{} \int_0^1 \frac{1}{3}du = \frac{1}{3}
$$
since the pointwise limit of $f_n(u)=\frac{1}{2+u^{1/n}}$ on $(0,1]$ is $\frac{1}{3}$.

Title: Evaluate $\lim_{n \rightarrow \infty} \int_0^1 \frac{nx^{n-1}}{2+x} dx$


答:
There is already an answer that clears your question

We can find the best minimum without using derivatives $${(\sqrt{y-5}+\sqrt{8-y})}^2=3+2\sqrt{8-y}\sqrt{y-5}\ge 3$$  because  $\sqrt{8-y}\sqrt{5-y}\ge 0$ $$\implies \sqrt{y-5}+\sqrt{8-y}\ge \sqrt{3}$$

Title: Maximum value of $k$ so that the inequality $\sqrt{y-5}+\sqrt{8-y} \ge k$ should have answer

Body:
Suppose $k$ be a real number so that the inequality
$\sqrt{y-5}+\sqrt{8-y} \ge k$ should have answer. what is the maximum
possible value of $k$?

$1)\sqrt{6}-1\quad\quad2)\frac12+\sqrt{3}\quad\quad3)\sqrt6\quad\quad4)2\sqrt2\quad\quad5)\sqrt6+1$

So I translated the question into finding the minimum value of $f(y)$ where $f(y)=\sqrt{y-5}+\sqrt{8-y}\quad$.(because for $k=\min(f(y))$ the inequality certainly has answer.) therefore we should solve  $f'(y)=0$:

$$f'(y)=\frac{1}{\sqrt{y-5}}-\frac{1}{\sqrt{8-y}}=0$$
$$y-5=8-y\rightarrow\quad y=6.5$$

And $f(6.5)=2\sqrt{\frac32}=\sqrt{6}$. so the answer is third choice.

Is my answer right? I know to find maximum or minimum of a function we take derivative of it and equate it to $0$. so here I don't know why $f(6.5)$ is minimum. can you please explain it to me?

EDIT:

I just noticed that $\sqrt6$ is not the minimum because $f(5)=f(8)=\sqrt3<\sqrt6$ maybe it is maximum! so I conclude the answer to this problem should be $\sqrt3$ but I don't have this option in the choices