问:
In the notation $T\colon\mathbb{R}^{n} \to \mathbb{R}^m$, does $\mathbb{R}^m$ refer to the codomain or the image of $T$?

Question:

It is my understanding that the term codomain refers to a set, of which the image is a subset (although, not necessarily a proper subset).

However, when speaking of vector transformations using notation like "$T\colon\mathbb{R}^{n} \to \mathbb{R}^m$",

am I correct in understanding that $\mathbb{R}^m$ refers to the codomain, not the image?

in the case of vector transformations, is it appropriate to speak of the image as a subspace of the codomain (Vs. a subset)?

Apologies in advance for asking two questions, but given that the first is basically yes/no, I figured it was okay.

My Work (Thoughts)

I am fairly certain that the image of a linear transformations is a subspace of its codomain, because, in that case, the transformation has a matrix representation

$$\boldsymbol{M} \, \in \, \mathbb{M}_{{\color{red}m}\times n}(\mathbb{R}),$$

and $\textrm{Im}\{\boldsymbol{M}\}$ is, in general, a subspace of $\mathbb{R}^{\color{red}m}$; however, I'm always wary of assuming the language can cross over (i.e. from the matrix to the transformation, in this case).

Moreover, based on the little I know about non-linear transformations, I would not expect their images to define a subspace. As such, I would assume that the term subset should always be used when describing the general relationship between the image and codomain.

Background / Level

Aside from real-valued functions and the usual fare seen in vector-calculus, I have no real experience with non-linear vector transformations. I have also never properly understood the importance of the distinction between image and codomain.

答:
You are correct in that $\mathbb{R}^m$ is the codomain of this map $T$. However, lets explore the roots of your confusion as noted at the end of your question.

The image of $T$ is defined as $T(\mathbb{R}^n) := \{T(x): x \in \mathbb{R}^n \}$. As you can see, this is a subset (and if $T$ is linear, also a subspace ${}^\dagger$) of $\mathbb{R}^m$ and is read as "the set of all those $T(x)$ such that $x$ lies in $\mathbb{R}^n$". This set certainly can be equal to the whole codomain if $T$ is a surjective map, i.e $T(\mathbb{R}^n) = \mathbb{R}^m$. If $T$ is not surjective, then there is a vector in $\mathbb{R}^m$ who is not the image of some vector in $\mathbb{R}^n$.

Note, we can adapt this notion of the image of a set to subsets of the domain: e.g., let $A \subset \mathbb{R}^n$ then the image of $A$ is $T(A) := \{T(x): x \in A\}$.

Note: It is common to see other notations to denote the image of a map such as $im \ T$ or rather uncommonly, the name of the map will appear as subscript to aide in specifying the image of a particular set (with respect to $T$): $im_{T}(A)$.

${}^\dagger$ Proof that if $A$ is a subspace of $\mathbb{R}^n$, and $T$ is a linear transformation, $T(A)$ is a subspace of $\mathbb{R}^m$.

Suppose that $w_1, w_2 \in T(A)$. Then, by definition of $T(A)$ there exists $u_1, u_2 \in A$ such that $w_{1} := T(u_{1})$ and $w_{2} := T(u_{2})$. Now, we want to show that $T(A)$ is closed under linear combinations; consider the linear combination $\alpha w_{1} + \beta w_{2}$ where  $\alpha, \beta$ are scalers. Therefore we can compute: $$\begin{align*} \alpha w_{1} + \beta w_{2} &= \alpha T(u_1) + \beta T(u_2) \\ &= T(\alpha u_1) + T(\beta u_2) \ \ \ by \ the \ linearity \ of \ T \\ &= T(\alpha u_1 + \beta u_2) \ \ \ again \ by \ the \ linearity \ of \ T.    \end{align*}$$ So since $A$ is a subspace, $u_{1}, u_2 \in A \implies \alpha u_1 + \beta u_2 \in A.$ Thus,  $\alpha w_{1} + \beta w_{2} \in T(A)$, since it is the image of the vector $\alpha u_1 + \beta u_2 \in A$. So $T(A)$ is closed under linear combinations and hence is a subspace of $\mathbb{R}^m$ $\blacksquare$


问:
Benford's law – formula
May I ask what does the "$k$" represent in Benford's law?

The formula I'm struggling with is in here –
$$\mathbb P(d) =\sum\limits_{k=10^{n-2}}^{10^{n-1}-1} \log_{10}\left(1+\frac1{10k+d}\right)$$

Thanks in advance.

答:
$$\mathbb P(d) =\sum\limits_{k=10^{n-2}}^{10^{n-1}-1} \log_{10}\left(1+\frac1{10k+d}\right)$$ is an expression for the probability under Benford's law that the $n$th digit being considered is $d$, with $d$ in $\{1,2,3,4,5,6,7,8,9\}$.

For this sum to to be meaningful, you need $n\ge 2$; if instead you are considering the first digit then you have $\mathbb P(d) = \log_{10}\left(1+\frac1{d}\right)$.

As an example, suppose $d=7$ and $n=3$ so you are considering the probability that the third digit is $7$ with Benford's law.  You want the probability that numbers start $107\ldots$ or  $117\ldots$ or $127\ldots$ and so on up to $997\ldots$.

That will be $\log_{10}\left(1+\frac1{107}\right)+\log_{10}\left(1+\frac1{117}\right)+\log_{10}\left(1+\frac1{127}\right) +\cdots+\log_{10}\left(1+\frac1{997}\right)$ which could be written as $\sum\limits_{k=10}^{99} \log_{10}\left(1+\frac1{10k+7}\right)$, i.e. the original expression with $d=7$ and $n=3$.

ContentLicense: CC BY-SA 4.0

Title: Benford's law – formula


问:
show that : $\int \sqrt{a^2-x^2}dx = \frac{1}{2}x\sqrt{a^2-x^2} + \frac{1}{2}a^2 \arcsin(\frac{x}{a}) + C \ (a > 0)$
How to show that : $\int \sqrt{a^2-x^2}dx = \frac{1}{2}x\sqrt{a^2-x^2} + \frac{1}{2}a^2 \arcsin(\frac{x}{a}) + C \ (a > 0)$

My first idea is to take the derivative of both sides. The right side will yield $\sqrt{a^2-x^2}$, the left side is a bit harder to deal with. If $a > 0$ and $a > x$ in order to avoid a complex result, then I can safely say that $x > 0$. Hence we can create a new function for the integral:

$$F(x) = \int_{0}^{x} \sqrt{a^2-x^2}dx$$

By the fundamental theorem of calculus, the derivative of this function is $\sqrt{a^2-x^2}$. Hence showing that the two results are equal on both sides.

Is this solution correct?

答:
Integrate by parts as follows
\begin{align*}
\int\sqrt{a^{2} - x^{2}}\mathrm{d}x & = \int\frac{\sqrt{a^{2} - x^2}}{2x}d(x^2) \\
 & = \frac{1}{2}x\sqrt{a^2-x^2} +\frac12  a^2\int\frac{1}{\sqrt{a^{2} - x^2}}dx\\
& =  \frac{1}{2}x\sqrt{a^2-x^2} + \frac{1}{2}a^2 \arcsin\frac{x}{a}+C\\
\end{align*}


问:
How do you prove by mathematical induction that $n(n+2)$ is divisible by $8$, if n is even and greater than 1
Everything is ok except for when I try to substitute $n = k + 1$. I get $8J + 4k + 8$ (where J is an integer) and cannot further factorize out an 8 to prove it is divisible. Is this even a correct substitution?

答:
If $n$ is even, then both $n$ and $n + 2$ are even (i.e, divisible by 2). If $n$ is also divisible by $4$, then $n(n + 2)$ is divisible by $4\cdot 2 = 8$. However, if $n$ is not divisible by $4$, then $n$ must have a remainder of $2$ when divided by $4$, which means $n+2$ is divisible by $4$. Then, $n(n+2)$ is divisible by $2\cdot 4 = 8$.

Thus, $\boxed{n(n+2)\text{ is always divisible by }8.}$


问:
sum of binomial coefficients expansion to prove equation
I want to prove that
$$
\sum_{i=1}^{n}{\binom{i}{2}} = \binom{n+1}{3}
$$

I already expanded

$$
\binom{n+1}{3}
$$
to
$$
\binom{n+1}{3} = \frac{1}{6} * (n+1) *n*(n-1)
$$
and I know that the following equation must be right
$$
\sum_{i=1}^{n}{\binom{i}{2}} = \frac{1}{6} * (n+1) *n*(n-1)
$$
but I do not get the expansion right, I tried starting with writing the sum explicit
$$
\binom{1}{2}+\binom{2}{2}+\ldots+\binom{n-1}{2}+\binom{n}{2}
$$
since 1 over 2 is zero it can be shortened to
$$
\sum_{i=2}^{n}{\binom{i}{2}} = \binom{2}{2}+\ldots+\binom{n-1}{2}+\binom{n}{2}
$$
then I expanded the binomial coefficients to the corresponding factorial form
$$
\binom{n}{k} = \frac{n!}{k! * (n-k)!}
$$
but I do not get it right, could someone please help me?

Body:
$\begin{array}\\
\binom{n+1}{3}-\binom{n}{3}
&=\dfrac{(n+1)n(n-1)}{6}-\dfrac{n(n-1)(n-2)}{6}\\
&=\dfrac{n(n-1)((n+1)-(n-2))}{6}\\
&=\dfrac{3n(n-1)}{6}\\
&=\dfrac{n(n-1)}{2}\\
&=\binom{n}{2}\\
\end{array}
$

Therefore
$\sum_{i=1}^{n}\binom{i}{2}
=\sum_{i=1}^{n}(\binom{i+1}{3}-\binom{i}{3})
=\binom{n+1}{3}
$.

Note that
$\binom{n}{m} = 0$
for $m > n$.

In general,
since
$\binom{n}{m}
=\dfrac{\prod_{k=0}^{m-1}(n-k)}{m!}
$,

$\begin{array}\\
\binom{n+1}{m}-\binom{n}{m}
&=\dfrac{\prod_{k=0}^{m-1}(n+1-k)}{m!}-\dfrac{\prod_{k=0}^{m-1}(n-k)}{m!}\\
&=\dfrac{\prod_{k=-1}^{m-2}(n-k)}{m!}-\dfrac{\prod_{k=0}^{m-1}(n-k)}{m!}\\
&=\dfrac{(n+1)\prod_{k=0}^{m-2}(n-k)-(n-m+1)\prod_{k=0}^{m-2}(n-k)}{m!}\\
&=\dfrac{((n+1)-(n-m+1))\prod_{k=0}^{m-2}(n-k)}{m!}\\
&=\dfrac{m\prod_{k=0}^{m-2}(n-k)}{m!}\\
&=\dfrac{\prod_{k=0}^{m-2}(n-k)}{(m-1)!}\\
&=\binom{n}{m-1}\\
\end{array}
$

so that

$\sum_{i=1}^{n}\binom{i}{m-1}
=\sum_{i=1}^{n}(\binom{i+1}{m}-\binom{i}{m})
=\binom{n+1}{m}
$.


问:
Find the volume of solid of revolution ($y$-axis)
$R$ is a region that is defined by the curve $y=1/(4+x^2)$ and the lines $x=0$, $y=0$ and $x=2$. I need to find the volume of solid of revolution about the $y$-axis.

We know that $x^2 = (1-4y)/y$ and the boundaries are $y=0$ and $y=4$.

Then we know that the $$V = \pi \int_0^4(1-4y)/y \text{ }dy.$$

The answer should be $\pi \ln{2}$ which is not equal to my volume. What went wrong?

答:
Notes:

First of all, at $x=2, ~y = (1/8)~~$
and at $~~x = 0, ~y = (1/4).$

Next, for $y$ between $0$ and $(1/8)$, the radius will be 2.

For $y$ between $(1/8)$ and $(1/4)$, the radius will be

$$ \sqrt{\left[\frac{1}{y} - 4\right]}.$$

Next, as you integrate between $y = 0$ and $y = (1/4),$ the cross sectional area is

$$\pi \times (\text{radius})^2.$$

Therefore the volume is

$$\pi \times \int_0^{(1/8)} (2)^2 dy $$

$$ + ~~\pi \times \int_{(1/8)}^{(1/4)} \left[\frac{1}{y} - 4\right] dy.$$

The $\pi \ln 2$ answer given is correct.

Can you take it from here?


问:
Finding the conditional expectation of a max/min random variable

Problem

Suppose that $ Z_1 $ and $ Z_2 $ are independent with common density
$$
f_Z(z) =
\begin{cases}
e^{-z} & \text{ if } z > 0 \\
0 & \text{ otherwise }.
\end{cases}
$$
Let $ X_1 = \min\left\{ Z_1, Z_2 \right\} $ and $ X_2 = \max\left\{ Z_1, Z_2 \right\} $.
Compute $ \mathrm{E}\left[ X_2 - X_1 | X_1 = x_1 \right] $.

Attempted Solution

Proceed with using the law of iterated expecation with the following finite
partition of the sample space
\begin{gather*}
D_1: \left\{ Z_1 > Z_2 \right\} \\
D_2: \left\{ Z_2 > Z_1 \right\} \\
D_3: \left\{ Z_1 = Z_2 \right\}
\end{gather*}
We should have
\begin{align*}
\mathrm{E}\left[ X_2 - X_1 | X_1 = x_1 \right]
&= \mathrm{E}\left[ X_2 - X_1| D_1, X_1 = x_1 \right] \Pr(D_1) \\
&\quad+ \mathrm{E}\left[ X_2 - X_1| D_2, X_1 = x_1 \right] \Pr(D_2) \\
&\quad+ \mathrm{E}\left[ X_2 - X_1| D_3, X_1 = x_1 \right] \Pr(D_3) \\
&= \mathrm{E}\left[ X_2 - X_1| Z_1 > Z_2, X_1 = x_1 \right] \Pr(Z_1 > Z_2) \\
&\quad+ \mathrm{E}\left[ X_2 - X_1| Z_2 > Z_1, X_1 = x_1 \right] \Pr(Z_2 > Z_1) \\
&= \mathrm{E}\left[ Z_1 - Z_2 | Z_2 = x_1 \right]\Pr(Z_1 > Z_2) \\
&\quad+ \mathrm{E}\left[ Z_2 - Z_1 | Z_1 = x_1 \right](1 - \Pr(Z_1 > Z_2)) \\
&= \left( \mathrm{E}\left[ Z_1 | Z_2 = x_1 \right] - x_1 \right)\Pr(Z_1 > Z_2) \\
&\quad+ \left( \mathrm{E}\left[ Z_2 | Z_1 = x_1 \right] - x_1 \right)(1 - \Pr(Z_1 > Z_2)).
\end{align*}
Where we have employed that $P(D_3|X_1 = x_1) = 0$ since $ Z_1 = Z_2 $ occurs
with measure $0$ probability and that $\Pr(Z_2 > Z_1) = 1 - \Pr(Z_1 \ge Z_2) = 1 - \Pr(Z_1 > Z_2)$.
Notice that $ Z_1, Z_2 $ are identical and independent with mean $ 1 $ (since
they are exponential distribution with $ \lambda = 1 $, hence $ \mathrm{E}[Z_i]
= \frac{1}{\lambda} = 1 $). Therefore we have
\begin{gather*}
\mathrm{E}\left[ Z_1 | Z_2 = x_1 \right] = \mathrm{E}[Z_1] = 1 \\
\mathrm{E}\left[ Z_2 | Z_1 = x_1 \right] = \mathrm{E}[Z_2] = 1
\end{gather*}

We can conclude that
\begin{align*}
\mathrm{E}\left[ X_2 - X_1 | X_1 = x_1 \right]
&= \left( \mathrm{E}\left[ Z_1 | Z_2 = x_1 \right] - x_1 \right)\Pr(Z_1 > Z_2)
+ \left( \mathrm{E}\left[ Z_2 | Z_1 = x_1 \right] - x_1 \right)(1 - \Pr(Z_1 > Z_2)) \\
&= \left( 1 - x_1 \right)\Pr(Z_1 > Z_2)
+ \left( 1 - x_1 \right)(1 - \Pr(Z_1 > Z_2)) \\
&= 1 - x_1
\end{align*}

Question

Now here is the issue, the answer is supposed to be $ \mathrm{E}\left[ X_2 - X_1 | X_1
= x_1 \right] = 1 $. Both a friend and I have gone through the
attempted solution above but could not find the error with it, yet it gives the
wrong answer. Can anyone help out with identifying where the error is and how you might
modify the approach to get the right answer?

I'd also be open to hear if Math StackExchange could find a different solution method (preferably one that is as succinct).

答:
Your evaluation of $\mathsf E(X_2{-}X_1\mid X_1{=}x, Z_2{>}Z_1)$ is awry.

The event of $\{X_1{=}x, Z_2{>}Z_1\}$ is the event of $\{Z_1{=}x,Z_2{>}x\}$. Also under that condition, then $X_2-X_1$ will equal $Z_2-Z_1$

$$\begin{align}
\mathsf E(X_2{-}X_1\mid X_1{=}x, Z_2{>}Z_1) &= \mathsf E(Z_2{-}Z_1\mid Z_1{=}x,Z_2{>}x) \\ &=\mathsf E(Z_2\mid Z_2{>}x,Z_1{=}x)-\mathsf E(Z_1\mid Z_2{>}x,Z_1{=}x)
\end{align}$$

Next use that the $Z_\star$ are independent, and recall that exponential distributed random variables have the memoryless property.

$$\begin{align}
\mathsf E(X_2{-}X_1\mid X_1{=}x, Z_2{>}Z_1)  &=\mathsf E(Z_2\mid Z_2{>}x)-\mathsf E(Z_1\mid Z_1{=}x) \\ &= \mathsf E(Z_2\mid Z_2{>}x)-x\\ &= \mathsf E(Z_2)\\&=1
\end{align}$$

Likewise $\hspace{12ex}\mathsf E(X_2{-}X_1\mid X_1{=}x, Z_1{>}Z_2)=1$

Therefore:...

$$\begin{align}\mathsf E(X_2{-}X_1\mid X_1{=}x)&= {\mathsf E(X_2{-}X_1\mid X_1{=}x, Z_2{>}Z_1)\,\mathsf P(Z_2{>}Z_1)\\+\mathsf E(X_1{-}X_2\mid X_2{=}x, Z_1{>}Z_2)\,\mathsf P(Z_1{>}Z_2)}\\&=1\end{align}$$


答:
If $x\in\mathbb R$, then $x^2+1\ge1$.

On the other hand, $(x-1)^2\ge0$, so $x^2+1\ge 2x$, so $1\ge\dfrac{2x}{x^2+1}$.

Therefore, $x^2+1\ge\dfrac{2x}{x^2+1}$.

Title: If $x$ $\in \mathbb{R}$, then $(x^{2} + 1)^{2}\geq 2x$

Body:
If  $x \in \mathbb{R}$, then:

$x^2+1 \geq \dfrac{2x}{x^2+1}$?


问:
Automorphism of $A[t]/(t^m)$
Let $A$ be a commutative ring and $t$ an indeterminate over $A$. If $f$ is an automorphism of the ring $A[t]/(t^m)$ satisfying $f(x)\equiv x\pmod{(t)}$ for each $x\in A[t]/(t^m)$ with $m$ a positive integer, then can we deduce that $f(t)=t$?</p>

答:
The answer to your question is no. The condition on $f$ (that $f(x) \equiv x \pmod{t}$) is equivalent to $f$ acting as the identity on $A$, but it can do "anything" within the maximal ideal (recall that $A[t]/(t^m)$ has a decomposition $A \oplus \bigoplus_{i=1}^{m-1} t^i\cdot A$).

So, for example, if $2$ is a unit in $A$, the function defined by $f(t)=2t$ satisfies your condition, and is an isomorphism of $A[t]/(t^m)$.

ContentLicense: CC BY-SA 3.0

Title: Automorphism of $A[t]/(t^m)$


答:
HINT

To begin with, notice that $x^{2} + 1 \geq 1$. Thus we have

\begin{align*}
(x^{2} + 1)^{2} \geq x^{2} + 1 \geq 2|x| \geq 2x
\end{align*}

Hopefully this helps!


问:
$f: X \rightarrow \mathbb{R}$ is measurable iff $\{(x, t): t=f(x)\} \in \mathcal{F} \otimes \mathcal{B}(\mathbb{R})$.
A function $f: X \rightarrow \mathbb{R}$ is measurable if and only if its graph is measurable, $\{(x, t): t=f(x)\} \in \mathcal{F} \otimes \mathcal{B}(\mathbb{R})$.

This is easy when we consider the points under the graph, but not sure how to do the graph itself

答:
It is also easy to prove that if $f: X \rightarrow \mathbb{R}$ is measurable, then $\{(x, t): t=f(x)\} \in \mathcal{F} \otimes \mathcal{B}(\mathbb{R})$.

Since  $f: X \rightarrow \mathbb{R}$ is measurable, we have that $g: X \times  \mathbb{R} \rightarrow  \mathbb{R} $  defined by $g(x,y)= y-f(x)$ is $\mathcal{F} \otimes \mathcal{B}(\mathbb{R})$-measurable.

So $g^{-1}(\{0\}) \in  \mathcal{F} \otimes \mathcal{B}(\mathbb{R})$. But $g^{-1}(\{0\}) = \{(x, t): t=f(x)\}$.

So $\{(x, t): t=f(x)\} \in \mathcal{F} \otimes \mathcal{B}(\mathbb{R})$.


问:
Proving that a complex function is real valued on an interval
Let $V = \{ z \in \mathbb{C}: |z| > 1\}$ and $f \in H(V)$. Assume that $f$ is real-valued on $(1,\infty)$. Show that it is real valued on $(-\infty, -1)$.

This is a problem from a complex analysis course. I'm not sure what technique I should use here to prove this. At first, I thought the Cauchy-Riemann equations might be useful here but I don't think that is the right path anymore. Any help would be appreciated.

答: Since $V$ is conjugate invariant the function $g(z)=\overline {f(\bar z)}$ is defined and holomorphic on $V$. By hypothesis $g=f$ on $(1,\infty)$ hence everywhere on $V$, so in particular $f$ is real on all reals in $V$ hence on $(-\infty,-1)$ too


答: The case for $n=2$ is clear. Then we we must show that if the expression is true for some $n$ then it is true for $n+2$. For some $n$ we know that $8|n(n+2) $and we must show that from this information $8|(n+2)(n+4)$. Notice that $(n+2)(n+4) = n(n+2) + 4(n+2)$. Since $n $is even, $n = 2k$ so that $4(n+2) = 4n + 8= 8k + 8$ which $8$ divides. Then $8|n(n+2)$ and $8|4(n+2)$ so that $8|(n+2)(n+4).$

Title: How do you prove by mathematical induction that $n(n+2)$ is divisible by $8$, if n is even and greater than 1

Body: Everything is ok except for when I try to substitute $n = k + 1$. I get $8J + 4k + 8$ (where J is an integer) and cannot further factorize out an 8 to prove it is divisible. Is this even a correct substitution?


问:
How to negate if and only if
How can I get $(A ↔ ¬B)$ or $(¬A ↔ B)$ from $¬(A ↔ B)$ using propositional laws?

I tried expanding $¬(A ↔ B)$ and I got to $(A ∧ ¬B) ∨ (B ∧ ¬A)$

I also tried expanding $(A ↔ ¬B)$ and I got to $(¬A ∨ ¬B) ∧ (B ∨ A)$

I don't know where in my expansion I went off the wrong path

答:
Your expansion of $\lnot(A\leftrightarrow B)$ into $(A\land\lnot B)\lor(B\land\lnot A)$ was correct.

Your expansion of $(A\leftrightarrow\lnot B)$ into $(\lnot A\lor\lnot B)\land(B\lor A)$ was also correct.

Well done. As @zkutch pointed out, the negation of $A\leftrightarrow B$ can also be though of as exclusive or (XOR), meaning one or the other but not both.

To address your comment, we first expand $\lnot(A\leftrightarrow B)$ into $\lnot((A\land B)\lor(\lnot A\land\lnot B))$. Then we use De Morgan's laws (which flip $\land$'s, $\lor$'s, and negate everything which isn't one) to get $\lnot(A\land B)\land\lnot(\lnot A\land\lnot B)$, which turns into $(\lnot A\lor \lnot B)\land(A\lor B)$.

We then distribute over the $\land$ to arrive at $(\lnot A\land A)\lor (\lnot A\land B)\lor(\lnot B\land A)\lor(\lnot B\land B)$, which simplifies to $(\lnot A\land B)\lor(\lnot B\land A)$.

Since $(\lnot A\land B)\lor(\lnot B\land A)$ reads as $A$ is true and not $B$, or $B$ is true and not $A$, it is equivalent to $A\leftrightarrow\lnot B$.

Title: How to negate if and only if


问:
Polar Coordinates and Double Integrals
Problem 1:

Find the area enclosed by the ellipse $\displaystyle \frac {1} {r} = 1 – 0.6 \cos(\theta)$.

We know $0\leq \theta\leq 2\pi$.

We know $0\leq r\leq 1/(1-0.6\cos(\theta))$.

Questions:

What I’m not sure about it how to set up the double integral.
What am i integrating?
And when I sketched the picture the ellipse went through $(2.5, 0)$ and $(-0.5, 0)$ and $(0, 1)$ and $(0, -1)$. Is that useful?

Similar problem:

Given $x = ar\cos(\theta)$ and $y = br\sin(\theta)$, find the area enclosed by the curve:

$(x^2/a^2 + y^2/b^2)^2 = xy/c^2$

Again, I can find the limits but I’m not sure how to set this up correctly.

答:
If you do want to do this with a double integral, the answer to your question "what am I integrating" is $1$. But unless you explicitly have to do that, you can save yourself some work. An ellipse is a stretched circle -- a unit circle stretched by a factor of the two semi-axes in two perpendicular directions. Thus its area is just the area of the unit circle, $\pi$, times the stretching factors, which are the semi-axes. So all you have to do is to find the semi-axes.

For the first problem, you can see in your sketch (or from the fact that $r$ is an even function of $\theta$) that the semi-axes are parallel to the coordinate axes. The horizontal one is immediate from Matthew's comment: $a= (2.5+0.625)/2=\frac{5}{4}+\frac{5}{16}=\frac{25}{16}=\left(\frac{5}{4}\right)^2$. For the vertical one, you have to maximize the vertical coordinate:

$$r\sin\theta = \frac{\sin\theta}{1-0.6\cos\theta}\to\max$$
$$\left(\frac{\sin\theta}{1-0.6\cos\theta}\right)'=\frac{\cos\theta(1-0.6\cos\theta)-0.6\sin^2\theta}{(1-0.6\cos\theta)^2}=\frac{\cos\theta-0.6}{(1-0.6\cos\theta)^2}=0$$
$$\cos\theta=0.6$$
$$\frac{\sin\theta}{1-0.6\cos\theta}=\frac{\sqrt{1-0.6^2}}{1-0.6^2}=\frac{1}{\sqrt{1-0.6^2}}=\frac{1}{0.8}=\frac{5}{4}$$

Thus, the area of the ellipse is $\left(\frac{5}{4}\right)^2\frac{5}{4}\pi=\left(\frac{5}{4}\right)^3\pi$.

Title: Polar Coordinates and Double Integrals


问:
$X$ topological space and $D$ dense in $X$. If $Y$ is $T_2$ and $f,g$ continuous s.t $f(d)=g(d)$ for all $d\in D$ then $f(x)=g(x)$ for all $x\in X$
This is my attempt of proving this:

Let $f(x),g(x)\in Y$. Because $Y$ is Hausdorff we can find open disjoint subsets of $Y$, namely $U,V$ of $f(x)$ and $g(x)$ respectively such that $f(x)\in U$ and $g(x)\in V$. Applying the continuity of $f$ and $g$ we have that $f^{-1}(U)$ and $g^{-1}(V)$ are open sets of $X$ both containing $x$.

Because $D$ is dense in $X$, $x\in \bar{D}$, this means that for all open subsets $U$ of $X$ we have $U\cap D\neq\emptyset$. Of course, this works for the open sets $f^{-1}(U)$ and $g^{-1}(V)$. Then we have that $f^{-1}(U)\cap D\neq\emptyset$ and $g^{-1}(V)\cap D\neq\emptyset$.

We take $z\in f^{-1}(U)\cap D$ and $z'\in g^{-1}(V)\cap D $. This means that $f(z)\in U\cap f(D)$ and $g(z')\in V\cap g(D)$. But because $z,z' \in D$, by our hypothesis $f(z)=g(z)$ and $f(z')=g(z')$. Because $f(x),g(x)$ were arbitrary, $f(x)=g(x)$ for all $x\in X$

答:
You go a bit astray right at the start: unless you explicitly assume that $f(x)\ne g(x)$, which you did not do, you cannot say that $f(x)$ and $g(x)$ have disjoint open nbhds. For the approach that you’re taking, you should begin something like this:

Suppose that there is an $x\in X$ such that $f(x)\ne g(x)$.

Then you can argue that $f(x)$ and $g(x)$ have disjoint open nbhds $U$ and $V$, respectively, and that $f^{-1}[U]$ and $g^{-1}[V]$ are open nbhds of $x$. It’s true that there are $z\in D\cap f^{-1}[U]$ and $z'\in D\cap g^{-1}[V]$, but nothing in the hypotheses lets you conclude that $f(z)=g(z')$. The hypothesis says only that $f(z)=g(z)$ for each $z\in D$; it says nothing about the values of $f$ and $g$ at two different points of $D$. You need a single point $z$ that is in both $D\cap f^{-1}[U]$ and $D\cap g^{-1}[V]$.

Fortunately, this is not a problem: $f^{-1}[U]$ and $g^{-1}[V]$ are open nbhds of $x$, so their intersection is also an open nbhd of $x$ and therefore contains a point of $D$. That is, there is a $z\in D\cap f^{-1}[U]\cap g^{-1}[V]$, and now you can conclude the argument along the lines that you had in mind: $f(z)\in U$, $g(z)\in V$, and $U\cap V=\varnothing$, so $f(z)\ne g(z)$, contradicting the hypothesis that $f$ and $g$ agree on $D$. This contradiction shows that no such point $x$ exists and hence that $f(x)=g(x)$ for all $x\in X$.


问:
Benford's law and voting in Georgia
With Benford's law "the leading digits of the number found in real-world data sets" should make sense.  But this video

https://www.youtube.com/watch?v=DoF3WS42w3M&ab_channel=RobertA.Bonavito%2CCPA

claims that by applying Benford's law they can prove electoral fraud.

He claims that since Benford's law is "broken" in these graphs, it appears that there has been fraud committed, but I wonder if the artificiality of breaking things into counties and only having 156 of them violates the "naturally occuring" idea that is the foundation of Benford's Law.

I am curious why Benford's law does not seem to apply here.

答:
As I said in an answer on Skeptics SE, you can get a "Benford plot" by starting with a simple histogram of the values on a $\log_{10}$ scale, then "wrapping" the histogram by merging buckets that have the same fractional part, then redistributing the values into nine buckets of unequal size. Here's an example of doing this on about 2000 random log-normally distributed values with a standard deviation of 0.5:

The buckets on the right follow Benford's law to exactly the same extent that the buckets in the middle are filled uniformly. The buckets in the middle will generally be filled uniformly if the original distribution on the left is "wide enough"; of course they could also end up uniformly filled for other reasons, but that's the most common reason.

Here's what happens if you use the same number of random values with the same mean, but a standard deviation of 0.2:

Benford's law is violated not because the distribution is any less realistic, but just because it's narrower. The fact that the mean is 10.0 and not, say, 10.5 makes a big difference here; the Benford plot would look very different if the mean was shifted, though the original histogram wouldn't.

In my other answer you can see similar charts using real data from Chicago precincts.

This video used data scraped from here. Here are my histograms of that data for Trump (top) and Biden (bottom):

The charts on the right don't quite match the ones in the question, presumably because the numbers have been updated, but they're close.

I didn't do any goodness-of-fit analysis, but what I see here is a whole lot of nothing. The distributions for both Trump and Biden look like what I'd expect from a random simulation, keeping in mind that there are only 159 counties in Georgia, to the 2069 precincts in Chicago. The Benford diagrams on the right are not so obviously random, but that's simply because they use uneven buckets; they are harder to eyeball. There is no advantage to bucketing the data that way. It only obfuscates it. The rationale for checking Benford's law in base 10 specifically is that numbers made up by human beings don't have the correct distribution of leading digits, but that would still show up in the histograms on the left as a pattern with a period of 1. I'm not convinced that there's any advantage to doing a Benford's law analysis over just analyzing the original histogram.

Title: Benford's law and voting in Georgia


问:
Showing that $\lim_{x\to\infty}c=c$ using the definition of limit
$$\lim_{x\to\infty} c =c$$

From the definition of limit, we want to prove that $\forall\varepsilon>0$, $\exists N>0$ such that $\forall x>N \implies |f(x)−L|<\varepsilon$; in this particular case, $|c−c|<\varepsilon$.

Proof. Let $\varepsilon>0$. There exists $N=\varepsilon$ such that...

I'm stuck, as I don't know how to link $c$ with $x>N$ in the subsequent steps.

Does someone have a clue?

答: Let $\epsilon>0$.  Then for any $N>0$, for all $x>N$, $|f(x)-L|=|c-c|=0<\epsilon.$


问:
example of a particular type of graph
I am searching for those classes of graphs where distance between any two vertices of the graph is $d$. Are there such type of graphs, after leaving the $Complete Graphs, K_n$.
Thanks for your help. </p>

答:
There are no such class of graphs except for d=1.

Say a graph for d=2 exists. $d(a,c)=2$. There must be a path $a-b-c$. So $d(a,b)=1$.

Title: example of a particular type of graph


答:
I'd say no. If the graph is not complete, there are going to be two nodes $a$,$b$ without an edge $(a,b)$. Pick a path from $a$ to $b$, and let $n > 1$ be its length. Then the distance from $a$ to $b$ is $nd \neq d$. Unless of course, $d=0$, i.e. unless you assign the weight $0$ to every edge.

This assumes that the graph is connected. If you require distance $d$ only for vertices which are reachable from one another, possible solutions are graphs whose connected components are complete graphs.

Title: example of a particular type of graph

Body:
I am searching for those classes of graphs where distance between any two vertices of the graph is $d$. Are there such type of graphs, after leaving the $Complete Graphs, K_n$.
Thanks for your help. </p>


问:
Solve $ y’’ - 3 \frac{cosx}{sinx}y’ + (\frac{3}{sin^2x}-2)y = 0 $
At the interval $(0, \pi)$ solve:

$$
y’’ - 3 \frac{cosx}{sinx}y’ + (\frac{3}{sin^2x}-2)y = 0
$$

If it is known that $\sin x$ Is a solution to the ODE.

What i tried:

I wanted to use the known solution in order to reduce the order of the ODE.

Setting:

$$
y = \sin x z
$$

Getting:

$$
y’ = \cos x z + \sin xz’
$$
$$
y’’ = -\sin xz +2cosxz’ + sinxz’’
$$

Implementing those derivatives into the ODE we get:

$$
-sinxz + 2cosxz’ + sinxz’’ - 3\frac{cosx}{sinx}(cosx + sinxz’) + (\frac{3}{sin^2x} - 2)sinxz = 0
$$

Finally getting to:

$$
sin^2xz’’ - sin2xz’ - 3z + 3 = 0
$$

The problem is that the order of the ODE has not changed, namely, it is still 2nd order ODE.

I tried to look for technical errors, but failed in finding.

So, what did i do wrong?

Thanks.

答:
The coefficient of the $z$ term should be

$$ -\sin(x)-3\frac{\cos^2(x)}{\sin(x)}+\frac{3}{\sin(x)}-2\sin(x) $$

which reduces to

$$ -3\sin(x)+3\sin(x)=0 $$

The coefficient of the $z^\prime$ term should be

$$ 2\cos(x)-3\cos(x)=-\cos(x) $$

So you get

$$ \sin(x)z^{\prime\prime}-\cos(x)z^\prime=0 $$

Letting $u=z^\prime$ reduces to a first order equation and you proceed accordingly.


问:
Variants of Bernoulli factory problems
I am interested in the following two questions related to the famous Bernoulli factory problems.

Bernoulli factory for certain functions: suppose we are given a coin with arbitrary (unknown) head probability $p$, I am wondering if there is an easy-to-implement algorithm for generating a $\min\{p, 0.5\}$ coin for any $p\in [0,1]$. Function $f(p) = \min\{p, 0.5\} \geq \min\{p, 1-p\}$ so the famous paper by  Keane and O’Brien guarantees such an algorithm exists. Meanwhile, the function looks slightly simpler than the function discussed in Nacu and Peres, so I am wondering if there is any existing simple algorithm for simulating this. Similarly, can we simulate $f(p) = \min\{p,a\}$ for fixed $a\in (0,1)$ efficiently?

High dimensional Bernoulli factory: How to simulate a coin with head probability $f(p_1, p_2)$ (for example $f(p_1, p_2) = \min\{p_1, p_2, 0.5\}$), or more generally, simulating a coin with head probability $f(p_1, \cdots, p_n)$. The setting here is that we are able so simulate a $p_i$ coin, but without further knowledge on the value of $p_i$, which is consistent with the traditional Bernoulli factory's setting.  When $f$ is a single variable function then the problem is reduced to the Bernoulli factory problem.

Any reference/idea will be much appreciated for the multivariate case, thanks!

答:
I reproduce my answers from Cross Validated:

On number 1:

Since the function min(λ, c) meets the Keane–O'Brien theorem, this implies that there are polynomials that converge from above and below to the function. This is discussed, for example, in Thomas and Blanchet (2012) and in Łatuszyński et al. (2009/2011). However, neither paper shows how to automate the task of finding polynomials required for the method to work, so that finding such a sequence for an arbitrary function (that satisfies the Keane–O'Brien theorem) remains far from trivial (and the question interests me to some extent, too).

But fortunately, there is an alternative way to simulate min(λ, 1/2) without having to build a sequence of polynomials explicitly. This algorithm I found is given below, and I have a page describing the derivation of this algorithm.

With probability 1/2, flip the input coin and return the result.

(Random walk.) Generate unbiased random bits until more zeros than ones are generated this way for the first time. Then set m to (n −1)/2+1, where n is the number of bits generated this way.

(Build a degree-m*2 polynomial equivalent to (4*λ*(1− λ ))m/2.) Let z be (4m/2)/choose(m*2, m). Define a polynomial of degree m*2 whose (m*2)+1 Bernstein coefficients are all zero except the mth coefficient (starting at 0), whose value is z. Elevate the degree of this polynomial enough times so that all its coefficients are 1 or less. Let d be the new polynomial's degree.

(Simulate the polynomial, whose degree is d (Goyal and Sigman 2012).) Flip the input coin d times and set h to the number of ones generated this way. Let a be the hth Bernstein coefficient (starting at 0) of the new polynomial. With probability a, return 1. Otherwise, return 0.

I suspected that the required degree d would be floor(m*2/3)+1. With help from the MathOverflow community, steps 3 and 4 of the algorithm can be described more efficiently as follows:

(3.) Let r be floor(m*2/3)+1, and let d be m*2+r.

(4.) (Simulate the polynomial, whose degree is d.) Flip the input coin d times and set h to the number of ones generated this way. Let a be (1/2) * 2m*2*choose(r, h − m)/choose(d, h) (the polynomial's hth Bernstein coefficient starting at 0; the first term is 1/2 because the polynomial being simulated has the value 1/2 at the point 1/2). With probability a, return 1. Otherwise, return 0. (Here, choose(n, k) is a binomial coefficient.)

In addition, there is an approximate way to sample min(λ, c) and most other continuous functions f that map (0, 1) to (0, 1). Specifically, it's trivial to simulate an individual polynomial with Bernstein coefficients in [0, 1], even if the polynomial has high degree and follows the desired function closely (Goyal and Sigman 2012):

Flip the input coin n times (where n is the polynomial's degree), and let j be the number of ones.

With probability a[j], that is, the j-th control point, starting at 0, for the polynomial's corresponding Bézier curve, return 1. Otherwise, return 0.

To use this algorithm, simply calculate a[j] = f(j/n), where n is the desired degree of the polynomial (such as 100). Each a[j] is one of the Bernstein coefficients of a polynomial that closely approximates the function; the higher n is, the better the approximation.

On number 2:

A related problem is the Dice Enterprise problem first given by Morina et al. 2019/2020, which involves simulating a m-faced die with a n-faced die, where the faces have separate probabilities of occurring. This is not the same as the problem in your question, though, as these probabilities are interrelated rather than independent.

REFERENCES:

Goyal, V. And Sigman, K., 2012. On simulating a class of Bernstein polynomials. ACM Transactions on Modeling and Computer Simulation (TOMACS), 22(2), pp.1-5.

Łatuszyński, K., Kosmidis, I., Papaspiliopoulos, O., Roberts, G.O., "Simulating events of unknown probabilities via reverse time martingales", arXiv:0907.4018v2 [stat.CO], 2009/2011.

Morina, G., Łatuszyński, K., et al., "From the Bernoulli Factory to a Dice Enterprise via Perfect Sampling of Markov Chains", arXiv:1912.09229 [math.PR], 2019/2020.

EDIT (Nov. 13):

A very relevant paper was just made available and came to my attention:

Niazadeh, R., Leme, R.P., Schneider, J., "Combinatorial Bernoulli Factories: Matchings, Flows, and Polytopes", arXiv:2011.03865v1 [cs.DS], Nov. 7, 2020.

(Edited Apr. 4):

Let $f:\mathcal{P}\to [0, 1]$ be a function.

In the traditional Bernoulli factory problem, we have a coin with unknown probability of heads $\lambda$ and we seek to sample the probability $f(\lambda)$. In this case, the domain $\mathcal{P}$ is either $[0, 1]$ or a subset thereof.

The paper cited above, however, studies Bernoulli factories when $f$ has a different domain: namely when $\mathcal{P}$ is a "polytope contained in the unit hypercube", that is, either $[0,1]^n$ or a subset thereof. Now the Bernoulli factory problem is as follows:

Define a polytope $\mathcal{P}$ as above.

We have $n$ coins with unknown probabilities of heads of $\lambda_1, ..., \lambda_n$. These probabilities form a vector $x = (\lambda_1, ..., \lambda_n)$ lying on inside the polytope.

Assign a specially-designed coin to each vertex of $\mathcal{P}$.

Sample a vertex (coin) of the polytope such that the expected value equals $x$.

For example, given two coins the vector is $x = (\lambda_1, \lambda_2)$, and one possible choice of $\mathcal{P}$ is $[0, 1]\times[0, 1]$. In that case, there are four vertices, namely $(0, 0)$, $(0, 1)$, $(1, 0)$, and $(1, 1)$. When this procedure is done enough times, the sampled vertices average to $x$.

The main question studied in this paper is whether and how a polytope of the kind just given admits a Bernoulli factory.

Unfortunately, this is a different problem from the Bernoulli factory problem you give in your question, where the goal is to sample the probability $f((\lambda_1, \lambda_2, ..., \lambda_n))$ given $n$ coins each with unknown probability of heads of $\lambda_1, ..., \lambda_n$ as the case may be. Moreover, the use of the term "Bernoulli factory" as studied in the paper may be misleading, because the output of the algorithm is not necessarily a Bernoulli random variable. (This is so even though the paper defines "one-bit Bernoulli factories" similarly to traditional Bernoulli factories.) As a result, the paper didn't study any conditions on $f$ that are necessary for a Bernoulli factory of the kind you give in your question, such as whether $f$ has to be continuous or bounded away from its domain.

Nacu & Peres found algorithms to solve the Bernoulli factory problem using approximation theory. Currently, polynomials and rational functions are the main kinds of multivariate functions I am aware of that have Bernoulli factory algorithms (Morina et al.) Many multivariate analytic functions are also taken care of by the composition rule (e.g., proposition 14(ii) of Nacu and Peres).

However, although the function $g = \min(\lambda_0, \lambda_1)$ is continuous, it's not differentiable, which presents an apparent difficulty.

But by the Stone--Weierstrass theorem, any continuous function on $[0, 1]^d$ can be approximated arbitrarily well by polynomials, so I suspect the following:

The necessary conditions of Keane and O'Brien extend to multivariate functions: the function must be continuous on $[0, 1]^d$ and either constant or polynomially bounded, which seems to be the case for $g$. I also suspect that similarly to the proof of Keane and O'Brien, any such function $f$ can be simulated by finding multivariate, continuous and polynomially bounded functions $f_k$ that approximate $f$ from below, although the algorithm in that proof is far from practical as it requires finding the degree of approximation for each function $f_k$.

Since an approximate polynomial exists for any such continuous function, the algorithm for simulating polynomials given by Goyal and Sigman can be easily extended to the multivariate case: flip each coin n times, count the number of heads for each coin, then return 0 with probability equal to the chosen monomial's coefficient.

There may be an algorithm that works similarly in the multivariate case to the general Bernoulli factory algorithms in the univariate case, including the one by Nacu & Peres.

This will require looking at the research on multivariate polynomial approximation (especially research that gives bounds on the approximation error with polynomials, especially multivariate polynomials in Bernstein form).

It's also possible that for the particular function $g$, a series expansion exists whose terms can be "tucked" under a discrete probability mass function, which enables an algorithm to simulate $g$ via convex combinations (Wästlund 1999, Theorem 2.7), as is the case for $\min(\lambda, 1/2)$, for example.

EDIT (Sep. 28, 2021): Also, see chapter 3 of G. Morina's doctoral dissertation (2021), which shows that multivariate Bernoulli factories require the function to be continuous and polynomially bounded.

EDIT (Feb. 18, 2022):  A new paper by Leme and Schneider (2022), "Multiparameter Bernoulli Factories", deals with the problem you're asking about.  Among other things, they show that a function $f(p_1, ..., p_n)$ admits a Bernoulli factory if and only if $f$ is continuous and meets a polynomial boundedness condition that reduces in the 1-dimensional case to that found in Keane and O'Brien.

REFERENCES:

Goyal, V. And Sigman, K., 2012. On simulating a class of Bernstein polynomials. ACM Transactions on Modeling and Computer Simulation (TOMACS), 22(2), pp.1-5.

Wästlund, J., "Functions arising by coin flipping", 1999.

Morina, Giulio (2021) Extending the Bernoulli Factory to a dice enterprise. PhD thesis, University of Warwick.

Title: Variants of Bernoulli factory problems


问:
Prove that every tangent line to the function $f(x)$ only intersects the function once.
Prove that every tangent line to the function $f(x)$ only intersects the function once.

$$f(x) = \int_{0}^{x^2}e^{t^3}dt, x \in(0,\infty)$$

My Attempt:

Lemma 1: $f'(x)$ is an increasing function.

By the fundamental theorem of calculus we get $f'(x) = 2xe^{x^6}$. Assume $x_1, x_2 \in (0, \infty)$ such that $x_1 < x_2$. Then:

$$2x_1 < 2x_2$$
$$2x_1e^{x_1^6} < 2x_2e^{x_2^6} \rightarrow f'(x_1) < f'(x_2)$$

This lemma allows us to conclude that every slope of $f(x)$ is unique. Assume that there is an intersection point for some slope. Let the slope be at $(x_1, y_1)$ and the intersection point be $(a,b)$. By the MVT there is a $f'(c) = f'(x_1)$ where $c \in (x_1, a)$. This contradicts lemma 1. QED

Is this strategy correct?

答:
The strategy is good. There is a small, somewhat inconsequential error though: the derivative of $f$ is $2xe^{x^6}$ not $2xe^{x^6} + C$. Remember, you're not integrating here, but differentiating a function (defined by a definite integral, which wouldn't need a constant of integration anyway).

What you're essentially doing here is showing that $f$ is a strictly convex function; if a differentiable function $f$ has a strictly increasing derivative, then $f$ is strictly convex. Tangents to convex functions lie on or below the graph of the function. In the case of strictly convex functions, the tangent only touches the function once.


问:
Construct the splitting field for the polynomial of $x^4-3$ over $\mathbb{Q}$?
I've been given this problem. I think I must proceed in the following way:

Given $a=\sqrt[4]{3}$ and $b=\cos\frac{2\pi}{4}+ \sin\frac{2\pi}{4}i$, the roots of $x^4-3$ are $a$, $ab$, $ab^2$, $ab^3$.

We have $\mathbb{Q}\subset \mathbb{Q}[a]\subset\mathbb{Q}[ab]$ so, is $\mathbb{Q}[ab]$ the desired spliting field?

Given a previous example in the book, it seems it is this way it should be done. However, we extend a previous field in various steps, it's not clear to me when this process "ends", for example: I could extend further and find $\mathbb{Q}[ab^2]$ but in a previous example, it stops before adjoining all the roots, I think the same applies here.

答:
Just addressing your concerns about finiteness and stopping.

$p(x) = x^4 - 3$ has at most four roots in any field.  Call them $r_1$, $r_2$, $r_3$, and $r_4$.  Then, at the very least $\Bbb{Q}(r_1, r_2, r_3, r_4)$, whatever field that is, is a field that splits $p$.  More generally, for a polynomial of degree $d$, extending $\Bbb{Q}$ by its (up to (see "separable")) $d$ roots yields a splitting field of the polynomial.  Proceeding in this way, the process is evidently finite.

It may turn out that $\{r_1, r_2, r_3, r_4\}$ are not independent.  For instance, it may occur that $r_4 \in \Bbb{Q}(r_1, r_2, r_3)$.  So the field we get may be more directly described using a shorter sequence of extensions.  To detect this, we ask "Is $r_2 \in \Bbb{Q}(r_1)$?  To really ask this, we have to be explicit about which roots are which.  That can be difficult, so another way to go is to see if you can partially factor the polynomial in some (smaller) extension field.

We see that $x^4 - 3 = (x^2 + \sqrt{3})(x^2 - \sqrt{3})$.  This is not a factorization in $\Bbb{Q}$, but is a factorization in $\Bbb{Q}(-\sqrt{3}, \sqrt{3})$.We get to decide which of these we label as the first root and which the second.  In order to potentially reduce the numbers of minus signs, let's try to determine whether $-\sqrt{3} \in \Bbb{Q}(\sqrt{3})$.  To do so, we look at the $\Bbb{Q}(\sqrt{3})$ vector space over $\Bbb{Q}$.  Generally, this is
$$  \{q_0 + q_1 \sqrt{3} + q_2 (\sqrt{3})^2 + \cdots q_n (\sqrt{3})^n \mid n \in \Bbb{Z}_{\geq 0}, q_i \in \Bbb{Q} \}  \text{.}  $$
This is a $\Bbb{Q}$-linear vector space with spanning set $\{1, \sqrt{3}, \sqrt{3}^2, \dots \}$.  However, this spanning set is redundant.  We know this because $\sqrt{3}$ is an algebraic number, of some degree, $m$.  So $\sqrt{3}$ is a root of a polynomial of degree $m$.  So there is a $\Bbb{Q}$-linear combination of the first $m$ powers of $\sqrt{3}$ that is zero -- that is, those $m$ powers are not $\Bbb{Q}$-linearly independent.  For $\sqrt{3}$, its minimal polynomial over $\Bbb{Q}$ is $x^2 - 3 = 0$, so $\sqrt{3}^2 \in \Bbb{Q}$.  Therefore, a basis for $\Bbb{Q}(\sqrt{3})$ is $\{1, \sqrt{3}\}$ and (as a set)
$$  \Bbb{Q}(\sqrt{3}) = \{q_0 + q_1\sqrt{3} \mid q_0, q_1 \in \Bbb{Q} \}  \text{.}  $$

All of that setup lets us observe $-\sqrt{3} = 0 + (-1)\sqrt{3} \in \Bbb{Q}(\sqrt{3})$, so $\Bbb{Q}(-\sqrt{3}, \sqrt(3)) = \Bbb{Q}(\sqrt{3})$.

Now what?  We need a further extension.  The factor $x^2 - \sqrt{3} = (x - \sqrt[4]{3})(x + \sqrt[4]{3})$  As above, any field containing $\sqrt[4]{3}$ contains $-\sqrt[4]{3}$, so we will extend to $\Bbb{Q}(\sqrt{3}, \sqrt[4]{3})$.  But notice that $\sqrt{3} = (\sqrt[4]{3})^2$ is in the $\Bbb{Q}$-linear span of the powers of $\sqrt[4]{3}$, so $\sqrt{3} \in \Bbb{Q}(\sqrt[4]{3})$.  So our new field extension is still principal.  (We first chopped $3$ into finer factors to get the first field extension.  If we chop it even finer, there is a second extension that contains all the slices.  For instance, $\Bbb{Q}(\sqrt{3}, \sqrt[3]{3})$ is contained in $\Bbb{Q}(\sqrt[6]{3})$ -- all the generators of the first extension are powers of the generator of the second.)

That leaves the other factor: $x^2 + \sqrt{3} = (x - \mathrm{i}\sqrt[4]{3})(x + \mathrm{i}\sqrt[4]{3})$.  Since the field $\Bbb{Q}(\sqrt[4]{3})$ is a bunch of real numbers, $\Bbb{Q}$, extended by a real number, $\sqrt[4]{3}$, it contains neither of these complex roots.  We showed this slowly, above, but more quickly here: any field containing $\mathrm{i}\sqrt[4]{3}$ contains $(-1)\mathrm{i}\sqrt[4]{3}$.  So we study the extension $\Bbb{Q}(\sqrt[4]{3})(\mathrm{i}\sqrt[4]{3})$.  Just as we can use field operations to go from the element $\mathrm{i}\sqrt[4]{3}$ to $(-1)\mathrm{i}\sqrt[4]{3}$, we can use field elements to replace $\mathrm{i}\sqrt[4]{3}$ with $\frac{\mathrm{i}\sqrt[4]{3}}{\sqrt[4]{3}} = \mathrm{i}$.  (That is, $\sqrt[4]{3}$ is a (nonzero) coefficient in the base field $\Bbb{Q}(\sqrt[4]{3})$ of $\Bbb{Q}(\sqrt[4]{3})(\mathrm{i}\sqrt[4]{3})$, so we are free to add, subtract, multiply or divide any generator of the further extension by it.)  So a more compact way to write our extension is $\Bbb{Q}(\sqrt[4]{3}, \mathrm{i})$.  We know the second generator is not in the span of the first by comparing real to nonreal numbers.  Do we know the first is not in the span of the second?

So we study the question: $\sqrt[4]{3} \in \Bbb{Q}(\mathrm{i})$?  The basis here is $\{1, \mathrm{i}\}$.  We write
$$  \sqrt[4]{3} \in q_0 + q_1 \mathrm{i}  $$
and solve for $q_0$ and $q_1$.  From our long experience with complex numbers, we know the only way for this to happen is $q_0 = \sqrt[4]{3}$ and $q_1 = 0$, which is not a pair of rational numbers.  Therefore, $\sqrt[4]{3} \not\in \Bbb{Q}(\mathrm{i})$ and our splitting field is $\Bbb{Q}(\sqrt[4]{3}, \mathrm{i})$.

How does that respond to your question?  First, we extend by the (degree of polynomial) many roots.  Everything after that is reduction -- at each step we determine whether each generator is in the field extended by the other generators, so in at most (degree of polynomial) steps we have a subset of the roots that are independent extension generators.  So this is a finite process.

We have used the fact that the extension generators are algebraic to get finite lists of powers to use in writing generic elements.

What does this have to do with your $a$ and $b$?  In either $\Bbb{Q}(\mathrm{i})(\sqrt[4]{3})$ or $\Bbb{Q}(\sqrt[4]{3})(\mathrm{i})$ (which is two ways to describe the same field extension of $\Bbb{Q}$), we have $a = \mathrm{i}$ and $b = \sqrt[4]{3}$ and a complete set of generators for the field extension is $\{1,b, b^2, b^3, a, ab, ab^2, ab^3\}$.  We are free to see this as "powers of $b$ with coefficients from powers of $a$" or "powers of $a$ with coefficients from powers of $b$", precisely corresponding to the two ways we wrote the extension.  However, this is also true for the other versions of the field extension that we wrote down or passed by without explicitly writing down.  For instance, it also holds for $\Bbb{Q}(-\sqrt[4]{3})(-\mathrm{i}\sqrt[4]{3})$.  We do get a different basis, but it spans the same extension field.

Again, this is finite because we start with a finite number of roots, $r_1, \dots, r_n$, each of finite degree, $d_1, \dots d_n$, and the resulting set of $\prod_{i=1}^n d_i$ generators is finite.  We then eliminate redundant generators (which induces a partial order on the field extensions) to obtain a minimal set of generators, a basis.  There are only finitely many generators to eliminate and there are only finitely many partial orders of intermediate extensions, so the entire process is finite.


问:
Give an expression that describe the language generated by this grammar
the grammar:

I tried to breakdown the problem but i am not sure what the end result expression could be.

I have to give an expression and try to describe this grammar.

Here's what i did for now :

I see that there's two possible starting words ( S non-terminal means start)

For $S \Rightarrow WTba$ i get :

$W^*\$(ba)^*ba$

W can be Wb or ab so we know the word will start by ab for sure : $ab(b^*(ab)^*)^*\$(ba)^*ba$

For $S \Rightarrow abUV$ i get :

$ab(ab)^*\$V$
since V can be Va or ba the word will end by ba for sure : $ab(ab)^*\$(a^*(ba)^*)^*ba $

From these two end result i can see that  they are ALMOST the reversed language of each other but one has the letter b instead of a in the big loop $(a^*(ba)^*)^*$ vs $(b^*(ab)^*)^*$

But i can't manage to figure out a general expression and correctly describe this grammar, so if anyone could help out it would be greatly appreciated. Thanks.

答:
You have the right idea, but you’ve lost the relationship between the part to the left of the $\$$ and the part to the right of it: you cannot express it in terms of Kleene stars. For instance, when you apply the production $T\to WTba$ $n$ times, you get the same number of $W$s as you do $ba$s, so you don’t get $W^*T(ba)^*$.

First, it’s clear that the language produced by $W$ is $abb^*$: the only possible derivations starting with $W$ have the form

$$W\Rightarrow^nWb^n\Rightarrow abb^n$$

for $n\ge 0$. Now let’s look at a derivation that starts by using the production $S\to WTba$. Since we know what $W$ does, we should concentrate on $T$: we can apply the production $T\to WTba$ $n$ times for any $n\ge 0$, but eventually we must apply $T\to\$$ in order to get rid of the $T$. Thus, we get derivations of the form

$$S\Rightarrow WTba\Rightarrow^n WW^nT(ba)^nba\Rightarrow WW^n\$(ba)^nba$$

for each $n\ge 0$. This will evidently generate the language described by the expression

$$(abb^*)^n\$(ba)^n\,,\tag{1}$$

where $n$ can be any positive integer.

The other half of the language can be analyzed similarly. $V$ generates everything of the form $baa^*$, so we end up with everything of the form

$$(ab)^n\$(baa^*)^n\tag{2}$$

for $n\ge 1$. Let $L_1$ be the language described by $(1)$ and $L_2$ the language described by $(2)$, so that we want the language $L=L_1\cup L_2$. $L$ can be described in just this way, but the description can perhaps be made a little nicer if we notice that $L_2$ can be obtained from $L_1$ (and vice versa) in a fairly simple way. Let

$$f:\{a,b\}\to\{a,b\}:x\mapsto\begin{cases}
a,&\text{if }x=b\\
b,&\text{if }x=a\,,
\end{cases}$$

and let $\hat f$ be the natural extension of $f$ to $\{a,b\}^*$. (E.g., $\hat f:(abbbab)=baaaba$.) Then

$$L_2=\{\hat f(v)\$\hat f(u):u\$v\in L_1\}\,,$$

and

$$L_1=\{\hat f(v)\$\hat f(u):u\$v\in L_2\}\,.$$

Thus, $L$ can be described simply in terms of $(1)$ and $\hat f$.


答:
Distribute:

$$(A\land\lnot B)\lor(B\land\lnot A)=(A\lor B)\land(\lnot B\lor B)\land(A\lor\lnot A)\land(\lnot B\lor\lnot A)$$

and then recognize that $(P\lor\lnot P)$ is always true, so the middle two pairs drop out and the right hand side reduces to

$$(A\lor B)\land(\lnot B\lor\lnot A)$$

Title: How to negate if and only if

Body:
How can I get $(A ↔ ¬B)$ or $(¬A ↔ B)$ from $¬(A ↔ B)$ using propositional laws?

I tried expanding $¬(A ↔ B)$ and I got to $(A ∧ ¬B) ∨ (B ∧ ¬A)$

I also tried expanding $(A ↔ ¬B)$ and I got to $(¬A ∨ ¬B) ∧ (B ∨ A)$

I don't know where in my expansion I went off the wrong path


问:
What is the remainder when $1^n + 2^n + 3^n + \ldots + 99^n$ is divided by $1 + 2 + 3 + \ldots + 99$?
My first idea on how to approach was to create a polynomial expansion for the first few terms and then try to find a pattern for the rest but this became cumbersome and I don't think that this is a right approach as this would quickly become a problem that involves factorials and I have not covered modular arithmetic, is there a way to approach this problem in such a way that does not involve using factorials with mods.</p>

答:
An ad-hoc solution:

$n$ odd one can group $1^n+99^n, 2^n+98^n,...50^n$ to get $S_n$ divisible by $50$ and similarly $1^n+98^n,..49^n+50^n$ to get $S_n$ divisible by $99$ so as noted above the remainder is zero.

For $n$ even we notice that modulo $3$ each group $3k-2,3k-1, k=1,..33$ gives a sum that is $2$ modulo $3$ and since there are $33$ such sums, $S_n$ is then $3$ modulo $9$ as the multiples of $3$ give zero modulo $9$ for $n \ge 2$ but there is an odd number of them $15$ that are odd so $S_n=9\times (2m+1)+q$

Modulo $5$ we have to split into $n=4k+2$ when we have $20$ sums that are $0$ modulo $5$ so $S_n$ is divisible by $25$ and then $n=4k$ when similarly we get $S_n=80=5$ modulo $25$ and since $S_n$ is always even, $S_n$ is $30$ modulo $50$

Modulo $11$ we have two cases - $n$ multiple of $10$ we get then $S_n=90=2$ modulo $11$ ($9$ sums of $10$ each) and $n$ not multiple of $10$ when $S_n$ is divisble by $11$ (easily seen for $n \le 8$ since $S_n$ has a factor of $99$ and a denominator that doesn't contain $11$ and then by periodicity since $S_n=S_{n+10}$ modulo $11$)

Putting it together we get:

$n=2k+1$ we get $0$

$n=4k+2, n\ne 10m$ the remainders modulo $50,9,11$ are $0,3,0$ so we get $1650$

$n=4k, n \ne 10m$ the remainders modulo $50,9,11$ are $30,3,0$ so we get $3630$

$n=20k+10$ the remainders modulo $50,9,11$ are $0,3,2$ so we get $750$

$n=20k$ the remainders modulo $50,9,11$ are $30,3,2$ so we get $2730$ and that's all!


问:
Contour integration with pole
I wanted to integrate

$$\oint \frac{\sqrt{z+5}}{z^5}$$

around a circular contour radius $1$ center $0$.

So the function has pole at 0. How can I proceed from here?

Title: Contour integration with pole

Body:
Cauchy's integral formula is

$$f^{(n)} (z_0) = \frac{n!}{2\pi i} \oint \frac{f(z)}{(z-z_0)^{n+1}} dz$$
so
$$ \oint \frac{\sqrt{z+5}}{z^5} dz = \frac{2\pi i}{4!} \,\left.\frac{d^4}{dz^4}(z+5)^{1/2}\right|_{z=0}$$

$$ \oint \frac{\sqrt{z+5}}{z^5} dz = -\frac{2\pi i}{4!} \, \left. \frac{15}{16(5+z)^{7/2}}\right|_{z=0}$$

$$ \oint \frac{\sqrt{z+5}}{z^5} dz = -\frac{\pi i}{64\cdot 5^{5/2}} $$


问:
Embeddings of $\mathrm{GL}(n-1,q)$ into $\mathrm{GL}(n,q)$
Let $n>2$ and $q$ be a prime power. I want to find all the embeddings of $\mathrm{GL}(n-1,q)$ into $\mathrm{GL}(n,q)$. An embedding I first thought of is as following. Let $V$ be an $n$-dimensional vector space over $\mathrm{GF}(q)$ and $H$ is a subgroup of $\mathrm{GL}(V)$ fixing a nonzero vector in $V$. Then $H$ is isomorphic to $\mathrm{GL}(n-1,q)$.

答:
If your last sentence is you definition of "natural", then the answer is that other embeddings are possible. Since the ones below follow a simple pattern, you will probably want to extend "natural" to include them as well.

You can always embed $\mathbf{GL}(n-1,F)$ into $\mathbf{GL}(n,F)$ by sending $A$ to the block matrix
$$
\begin{pmatrix} A&0\\0&\det A\end{pmatrix}.
$$
Unless $F=\mathbf{GF}(2)$, the image of this embedding does not have any fixed vectors.


问:
Evaluate integral using Cauchy's integral for derivatives.

Integrate counterclockwise around the unit circle.

$$\int_c \frac{e^{z}}{z^{n}}dz$$ where n = 1,2,...

Where do I even begin this?

I know the integral formula that I probably want to use is:

$$\int \frac{f(z)}{(z-a)^{n+1}}dz = \frac{2\pi \cdot i}{n!} \cdot f^{n}(a)$$

Is the answer just going to be in general form?

So $e^{z}$ derivatives are  just $e^{z}$. What should I do with z? Do I have to convert it?

答:
That's it!

$$\oint_C \frac{e^z}{z^n} dz = \frac{2\pi i}{(n-1)!} \left. \frac{d^{n-1}}{dz^{n-1} }e^z \right |_{z=0}$$

$$\oint_C \frac{e^z}{z^n} dz = \frac{2\pi i}{(n-1)!}$$

Another way $\cdots$

$$\oint_C \frac{e^z}{z^n} dz = \oint \frac{dz}{z^n}\left[1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots \right]=2\pi i\oint \frac{dz}{z^n}\frac{z^{n-1}}{(n-1)!}=\frac{2\pi i} {(n-1)!},$$  since all other terms evaluate to zero.


问:
Do we really need polynomials (In contrast to polynomial functions)?
In the following I'm going to call

a polynomial expression an element of a suitable algebraic structure (for example a ring, since it has an addition and a multiplication)
that has the form $a_{n}x^{n}+\cdots+a_{1}x+a_{0}$, where $x$ is
some fixed element of said structure,

a polynomial function a function of the form $x\mapsto a_{n}x^{n}+\cdots+a_{1}x+a_{0}$ (where the same algebraic considerations as above apply)

a polynomial an element of the form $a_{n}X^{n}+\cdots+a_{1}X+a_{0}$ with indeterminates $X$ (these
can be formalized, if we are in a ring, with strings of $0$s and
a $1$).

Note that when we are very rigorous/formal, polynomial expressions
and polynomials are something different (although in daily life we
often use them synonymously). Polynomial functions and expressions
are also different from each other although in this case the relationship
is a closer one, since every polynomials expression can be interpreted
as a polynomial function evaluated at a certain point (thus "polynomial
functions" are something more general than "polynomial expressions").

My question is: Why do we use polynomials ? It seems to me that every
piece of mathematics I have encountered so far, one could replace
every occurrence of polynomials with polynomial expressions/functions
without any major change in the rest of the proof/theorem/definition
etc.

My thoughts:

The only reasons that I can see to use polynomials are the following
two:

After one makes the idea precise that one can plug "suitable"
elements into polynomials (which may lie a ring containing the ring
in which the coefficients live in), one can save time in certain setting,
by handling the "plugging of suitable elements into polynomials"
more elegantly: For example in case of the theorem of Cayley-Hamilton,
which in its "polynomial function" version would look like:

Let $A$ be an $n\times n$ matrix over $K$, whose characteristic
polynomial (function) is $x\mapsto a_{n}x^{n}+\cdots+a_{1}x+a_{0}$.
Then
$$
a_{n}A^{n}+\cdots+a_{1}A+a_{0}I=0.
$$

whereas the "polynomial" version looks more elegant:

Let $A$ be an $n\times n$ matrix over $K$, whose characteristic
polynomial is $p_{A}\in K\left[X\right]$. Then
$$
p_{A}\left(A\right)=0.
$$

The only thing that polynomials can "do", but algebraic expressions/functions can't, is to be different, when the algebraic expressions/functions are the same (i.e. there's a theorem that tells us that the mapping of polynomials to polynomials expressions/functions
isn't injective, if the field is finite). Maybe this small difference makes a big enough difference
to consider polynomials after all, but as I said, I haven't encountered
any situation in which this difference could manifest itself.

(I'm guessing that maybe cryptography or higher number theory really needs
polynomials and not just polynomial expressions/functions. Since I don't know
anything about these subjects, I would be very happy with an example
of a theorem (whose content isn't merely technical as it is the case with the theorem above) involving polynomials, where these absolutely cannot
be replaced by polynomial expressions/functions. Conversely I would also be happy with an authoritative statement from someone knowledgeable that indeed we could dispense of polynomials, if we wanted to.)

答: Over some fields, the map from polynomials to polynomial functions is not injective; for example, over the finite field $\mathbb{F}_2$, the polynomial function $x^2+x$ is the same as the $0$ function - try plugging in $0$ and $1$. So in this case the (infinite) set of polynomials over $\mathbb{F}_2$ is much larger than the (finite) set of polynomial functions $\mathbb{F}_2\to\mathbb{F}_2$.</p>


问:
Commutative local Artinian ring without a section in characteristic $0$
Let $R$ be a commutative local Artinian ring, with unique maximal ideal $\mathfrak m$ and residue field $k = R/\mathfrak m$.

If $R$ is a finitely generated algebra over a perfect field $F$, then the Jordan-Chevalley decomposition applied to the regular representation of $R$ seems to imply that that the surjection $R \twoheadrightarrow k$ necessarily has a (unique!) section $k \hookrightarrow R$ of rings. I'm pretty sure my reasoning there is solid (you have to apply the JC decomposition to multiple commuting operators, but I'm pretty sure you can do that).

An example of $R$ where the residue field has no section is $R = \mathbb Z/p^n \mathbb Z$, where $p$ is a prime integer and $n > 1$.

It seems like this failure of the existence of a section occurs naturally in an "arithmetic setting", whereas it never seems to happen in a "geometric setting", if these phrases can be given any precise meaning.

For the sake of a precise question to address my intuition here: Is it possible to find an example for which the section $k \hookrightarrow R$ does not exist in characteristic $0$?

I'm not sure whether it's fair to categorize non-perfect fields as an "arithmetic not geometric" setting; feel free to pitch in your opinion on that point as well if you think such a discussion would be productive.

答:
The Cohen structure theorem answers your question. First, recall the following definition.

Definition: Let $(R,\mathfrak{m})$ be a complete local ring. A subring $\Lambda\subseteq R$ is called a coefficient ring if the following conditions hold:

$\Lambda$ is a complete local ring with maximal ideal $\Lambda\cap\mathfrak{m},$

the residue field of $\Lambda$ maps isomorphically to the residue field of $R,$ and

$\Lambda\cap\mathfrak{m} = p\Lambda,$ where $p$ is the characteristic of the residue field of $R.$

In particular, if $(R,\mathfrak{m})$ is a complete local ring with residue field $k$ of characteristic $0$ and $\Lambda$ is a coefficient ring, then $\Lambda$ is the residue field of $R$, as it has unique maximal ideal $\Lambda\cap\mathfrak{m} = (0).$ Thus $\Lambda$ is a field, and the second condition above forces $\Lambda = k.$

Notice also that any Artinian local ring $(R,\mathfrak{m})$ is a complete local ring, as $\mathfrak{m}^n = 0$ for some $n.$

Theorem: Let $(R,\mathfrak{m})$ be a complete local ring. Then

$R$ has a coefficient ring, and

if $\mathfrak{m}$ is a finitely generated ideal, then $R$ is isomorphic to a quotient $\Lambda[\![x_1,x_2,\dots, x_r]\!]/I$ where $\Lambda$ is either a field or a Cohen ring.

You of course do not need the full power of this theorem for your result. Let us prove directly that there is a section in your case.

Claim: If $(R,\mathfrak{m})$ is a complete local ring with residue field $k$ of characteristic $0,$ then there is a section $k\to R$ of the quotient map $R\to R/\mathfrak{m}.$ (In particular, a section exists if $R$ is Artinian.)

Proof: We will show inductively that a section exists to the canonical quotient map $R/\mathfrak{m}^n\to k.$ The base case is obvious. Then suppose that a section of $R/\mathfrak{m}^r\to k$ exists for $r = n-1.$ Since $\operatorname{char}k = 0,$ $k/\Bbb{Q}$ is separable, hence formally smooth (see here), and thus a morphism $k\to R/\mathfrak{m}^n$ exists making the diagram
$$
\require{AMScd}
\begin{CD}
R/\mathfrak{m}^n @>>> R/\mathfrak{m}^{n-1}\\
@AAA @AAA\\
\Bbb{Q} @>>> k\\
\end{CD}
$$
commutative (with the arrow $k\to R/\mathfrak{m}^n$ included, of course). This completes the proof, as $R\cong\varprojlim_n R/\mathfrak{m}^n.$ $\square$

Final remark. The map $\Bbb{Q}\to R/\mathfrak{m}^n$ exists because we have a map $\Bbb{Q}\to R.$ As $\operatorname{char}k = 0$ it follows that for any prime $p\in\Bbb{Z},$ $p\not\in\mathfrak{m}.$ Thus, $p\in R^\times,$ so the canonical map $\Bbb{Z}\to R$ extends to a map $\Bbb{Q}\to R$ by the universal property of localization.


问:
One Dimension Wave Equation (Calculus of several variables)
I can't find any way to get to that expression. Honestly I don't how to start this. Any recommendations?

Consider the One Dimension Wave Equation
$$
\frac{\partial^{2} u}{\partial t^{2}}=c^{2} \frac{\partial^{2} u}{\partial x^{2}}
$$
where $u:=u(x, t) .$ If now we make a change of variable $\xi=x+c t, \eta=x-c t,$ show that the wave equation can be written as $$\frac{\partial^{2} u}{\partial \eta \partial \xi}=0$$

答:
The solution is quite simple: Try to substitude $\xi$ and $\eta$ in

$\frac{\partial^2 u}{\partial \eta \partial \xi}=0$,

with the given relations and try to recover your first equation. The rest is up to you. It is done in one second.

Regards


问:
The Basic Fraenkel Model
I was working through Lemma 9.3 in Jech's Axiom of Choice (on page 136).

Here $V_2$ is the basic Fraenkel model determined by the group of all permutation of the set of atoms and the finite-subset ideal. I didn't understand the reasoing starting from "Since $x \subseteq \pi x$". Any help will be appreciated!

答:
Start with $x\subseteq\pi x$ and apply $\pi$ to both sides: $\pi x\subseteq\pi^2x$, so $x\subseteq\pi x\subseteq\pi^2x$. By an easy induction $x\subseteq\pi^ix$ for each $i\in\omega$. In particular, if we can find $k$ and $\rho$ as described, we’ll have

$$x\subseteq\pi^kx=\rho x\,.$$

Applying $\rho^{-1}$ to both ends, we see that $\rho^{-1}x\subseteq x$. But $\rho^{-1}x=\rho x$, so

$$x\subseteq\pi^kx=\rho x=\rho^{-1}x\subseteq x\,,$$

and it follows that $x=\pi^kx$. If $x$ were a proper subset of $\pi x$, we’d also have $x\subsetneqq\pi^ix$ for each $i>0$, so we must have $x=\pi x$.

I have no idea, however, how he goes on to find $k$ and $\rho$.

Title: The Basic Fraenkel Model


问:
What is the difference between P(W1,W2) vs P(W2|W1) in this context
What is the conceptual difference between the probability of winning two successive games in a playoff series vs winning the second game given that you won the first game?

I understand P(W2|W1) as the probability of winning the second game given you've won the first game, but I can't see how to interpret P(W2, W1), or if it even applies in this context; I think to get to W2 you have to have W1 as a given.

I can find similar examples here and here but they lack the successive nature of a series of games.

I've mapped out my cases as tree of possibilities to try to see how P(W2, W1) would fit in. Is there a better way to think about this?

答:
$\mathsf P(W_1,W_2)$ is the joint probability of the events; that is the probability of their intersection, also written as $\mathsf P(W_1\cap W_2)$. The probability that both events have occurred (win game 1 and game 2).

$\mathsf P(W_2\mid W_1)$ is the conditional probability that $W_2$ occurs when it is given that $W_1$ has.

By the definition for conditional probability, we have that:

$$\mathsf P(W_2\mid W_1)=\dfrac{\mathsf P(W_1, W_2)}{\mathsf P(W_1)}$$

Or $\mathsf P(W_1,W_2)=\mathsf P(W_1)\cdot \mathsf P(W_2\mid W_1)$


问:
Compute the desired area
I solved it on my own but I need some feedback on whether I solved it correctly or not.

Consider the function $$f(x)=x^3-2x^2-x+2$$

Let $h(x)$ be the function that results when $f(x)$ is shifted by $−1$ in the $x$-direction.

a) Sketch the graph of the function $f(x)$.

I sketched it.

b) Sketch the graph of the function $h(x)$ and give an algebraic definition
for $h(x)$.

I sketched it and I formulated the formula $$h(x)=f(x+1)=(x+1)^3-2(x+1)^2-(x+1)+2= x^3-2x^2-2x$$

Is this correct?

c) Find the intersection points of $f(x)$ and $h(x)$.

I solved this simultaneously and got $x=-2$
Then I plugged this value in at $f(x)$ and got $y=-12$

d) Compute the desired area.

this is the part i am confused about. At what interval do I need to do this? -2 to 2? I used -2 to 2 and got the Area= 16.

Title: Compute the desired area

Body:
I think you are supposed to find this area ($f(x)$ is in red, $h(x)$ is in blue):

First, we wish to find the $2$ intersection points. Note that $h(x) = f(x + 1) = x^{3} -x^{2} + 2x$. Then, we want to find:

$$f(x) = h(x)$$

$$x^{3}-2x^{2} - x + 2 = x^{3} + x^{2} -2x$$

$$3x^{2} - x - 2 = 0$$

$$(3x + 2)(x - 1) = 0$$

$$x = -\frac{2}{3}, 1$$

Note that since $f(x)>h(x)$ for $x\in(-\frac{2}{3}, 1)$, we wish to find:

$$\int_{-\frac{2}{3}}^{1}f(x)-h(x)\ dx$$

$$\int_{-\frac{2}{3}}^{1}-3x^{2} + x + 2\ dx$$

$$(-x^{3} + \frac{x^{2}}{2} + 2x)\bigg\vert_{-\frac{2}{3}}^{1}$$

$$\frac{3}{2} + \frac{22}{27}$$

$$\boxed{\frac{125}{54}}$$


问:
Improper integration and boundedness of function
This question came to my mind when I was following the book of Goldberg on Real Analysis for introduction of improper integrals.

Suppose $f:[a,\infty)$ is a function such that (i) $f$ is integrable on $[a,b]$ for every $b>a$. (ii) $\lim_{b\rightarrow\infty} \int_a^b f(x)\,dx$ exist.

Does it follow from these two conditions that $f$ is bounded on whole domain $[a,\infty)$?

(I am studying integrals of this kind seriously first time, so this question come to my mind. I do not know example; if there is trivial example, one may put in comment, and then vote for close, that is also fine to me.)

答:
Not necessarily. Let $f : [1, \infty) \to \mathbb{R}$ be defined as:
$$
f(x) = \begin{cases} x &\text{ if } x \in \mathbb{N} \\
e^{-x} &\text{ otherwise}
\end{cases}
$$ Then, for any $b \geq 1$, $f$ only has a finite number of discontinuities over $[1, b]$; so, of course, $\int_{1}^b f(x)\ dx = \int_{1}^b e^{-x}\ dx$ . Hence, $\int_{1}^\infty f(x)\ dx$ exists and is just $\int_{1}^\infty e^{-x}\ dx$. However, obviously $f$ is not bounded on $\mathbb{N}$.


问:
Permutations with a String
How many permutations of the string “ABCDEFGHIJKLMNOPQRSTUVWXYZ” contain
the substrings “HAVE” and “FUN”? Substrings must appear exactly as written.
e.g., LCQOPGRKFUNTYDIWSHAVEZXMJB is a valid substring.

My initial thought on this problem is maybe we could say that each work is three and four letters long, and by doing $\frac{26!}{4!3!}$, but I am wrong. I was wondering if someone could explain the solution to me.

The solution is: $21!$

答:
Assume that FUN and HAVE as a single letter in that order.
The different number of arrangements of B,C,D,G,I,J,K,L,M,O,P,Q,R,S,T,W,T,Y,Z,FUN,HAVE is $21!$

Title: Permutations with a String


问:
How to show that $\sum_{i=1}^n 1-e^{-n p^i/i} = \Omega(\log n)$ for $0 < p < 1/2$?
For a fixed $0 < p < 1/2$, how can I show the following lower bound?

$$ \sum_{i=1}^n 1-\exp\left(- \frac{np^i}{i}\right) = \Omega(\log n) $$

答:
Note that $x \mapsto 1 - e^{-x}$ is increasing for $x \geq 0$. Now let $a_n = \tfrac{1}{2}\log_p(1/n)$. Then for any $n$ large enough so that $a_n \leq \sqrt{n}$ holds, we get

\begin{align*}
\sum_{i=1}^{n} \bigl( 1 - e^{-np^i/i} \bigr)
&\geq \sum_{i \leq a_n} \bigl( 1 - e^{-np^i/i} \bigr) \\
&\geq \sum_{i \leq a_n} \bigl( 1 - e^{-np^{a_n}/\sqrt{n}} \bigr) \\
&= \lfloor a_n \rfloor \bigl( 1 - e^{-1} \bigr).
\end{align*}

Therefore the desired lower bound is obtained.

Addendum. In fact, we can prove more. Write $p = e^{-\beta}$ for the convenience, i.e., $\beta = \log(1/p)$. Then it turns out that the following holds:

$$ \lim_{n\to\infty} \frac{1}{\log n} \sum_{k=1}^{n} \bigl( 1 - \exp\bigl\{ -n/ke^{k\beta} \bigr\} \bigr) = \frac{1}{\beta} $$

Lower bound. For any $0 < r < 1$, let $a_n = \frac{r}{\beta} \log n$. Then for large $n$,

\begin{align*}
\sum_{k=1}^{n} \bigl( 1 - \exp\bigl\{ -n/ke^{k\beta} \bigr\} \bigr)
&\geq \sum_{k \leq a_n} \bigl( 1 - \exp\bigl\{ -n/ke^{k\beta} \bigr\} \bigr) \\
&\geq \sum_{k \leq a_n} \bigl( 1 - \exp\bigl\{ -n/ a_n e^{\beta a_n} \bigr\} \bigr) \\
&= \lfloor a_n \rfloor \bigl( 1 - e\bigl\{ -n^{1-r}/a_n \bigr\} \bigr).
\end{align*}

This prove that

$$ \liminf_{n\to\infty} \frac{1}{\log n} \sum_{k=1}^{n} \bigl( 1 - \exp\bigl\{ -n/ke^{k\beta} \bigr\} \bigr) \geq \frac{r}{\beta} $$

holds for all $r \in (0, 1)$, and so, letting $r \uparrow 1$ produces the lower bound.

Upper bound. Write

\begin{align*}
&\sum_{k=1}^{n} \bigl( 1 - \exp\bigl\{ -n/ke^{k\beta} \bigr\} \bigr) \\
&\quad \leq \int_{0}^{\infty} \bigl( 1 - \exp\bigl\{ -n/xe^{\beta x} \bigr\} \bigr) \, \mathrm{d}
= - \frac{1}{\beta} \int_{0}^{\infty} \bigl( 1 - e^{-\beta y} \bigr) \, \mathrm{d}W(n/y) \\
&\qquad = \int_{0}^{\infty} e^{-\beta y} W(n/y)  \, \mathrm{d}y.
\end{align*}

In the second step, we utilized the substitution $x=\frac{1}{\beta}W\bigl(\frac{n}{y}\bigr)$, where $W(\cdot)$ is the Lambert W-function. Then by using that $W_0(x) = \mathcal{O}(\log x)$ and $W_0(x) \sim \log x$ as $x \to \infty$, the dominated convergence theorem shows:

\begin{align*}
&\limsup_{n\to\infty} \frac{1}{\log n} \sum_{k=1}^{n} \bigl( 1 - \exp\bigl\{ -n/ke^{k\beta} \bigr\} \bigr) \\
&\quad \leq \limsup_{n\to\infty}\int_{0}^{\infty} e^{-\beta y} \frac{W(n/y)}{\log n}  \, \mathrm{d}y
= \int_{0}^{\infty} e^{-\beta y}  \, \mathrm{d}y
= \frac{1}{\beta}.
\end{align*}


答:
You have a mistake in your $h(x)$ and intersection points. Curves intersect at two points.

$h(x) = (x+1)^3 - 2 (x+1)^2 - (x+1) + 2 = x^3 + x^2 - 2x$

Please see the graph for area bound by them. You can use single integral of function $(f(x)-h(x))$ between bounds of $x$ to find area.

Title: Compute the desired area

Body:
I solved it on my own but I need some feedback on whether I solved it correctly or not.

Consider the function $$f(x)=x^3-2x^2-x+2$$

Let $h(x)$ be the function that results when $f(x)$ is shifted by $−1$ in the $x$-direction.

a) Sketch the graph of the function $f(x)$.

I sketched it.

b) Sketch the graph of the function $h(x)$ and give an algebraic definition
for $h(x)$.

I sketched it and I formulated the formula $$h(x)=f(x+1)=(x+1)^3-2(x+1)^2-(x+1)+2= x^3-2x^2-2x$$

Is this correct?

c) Find the intersection points of $f(x)$ and $h(x)$.

I solved this simultaneously and got $x=-2$
Then I plugged this value in at $f(x)$ and got $y=-12$

d) Compute the desired area.

this is the part i am confused about. At what interval do I need to do this? -2 to 2? I used -2 to 2 and got the Area= 16.


问:
Second Course on ODEs - Self Study - Book Recommendation
I'm looking for a book that would serve as a a second course on Ordinary Differential Equations. The book should be theoretically-oriented and for self study. I just finished the book by George Simmons - Differential Equations with Historical notes and Applications and in all honestly didn't really like it, finished it for the sake of completeness but it isn't what I was looking for.
I would like the book to contain a decent number of exercises after each section/chapter and, I stress, to be a theoretical approach to the topic. I originally started with Birkhoff and Rota's ODEs but one of the requierements was basic knowledge with complex functions, which I haven't studied yet.
I have knowledge of Calculus I (Spivak and Apostol), Calculus II (Apostol), Linear Algebra (Friedberg), Abstract Algebra (Herstein). Also, I will be finishing Introduction to Analysis (Mattuck) this week and will be starting with Rudin's Principles of Mathematical Analysis afterwards.
I have added the authors of the books, all of which I've liked, I used to study the previous topics for reference.
I'm looking for a book that I will be able to complete with the current knowledge that I have. I don't know if learning complex functions first, which I beleive I can take after finishing Rudin's Analysis, and then going back to Birkhoff's would be an option, but if it is I'm open to it too.

Thanks in advance

答:
I will give you a recommendation of three texts on ordinary differential equations that I took from the "theorem-proof-proposed problems" approach.

However, I am sure that here at MathSE there are many experts who will be able to give you a better recommendation.

Barbu, V. (2016). Differential Equations. (First edition). Cham,
Switzerland: Springer.

Hirsch, M. & Smale, S. & Devaney, R. (2013). Differential
Equations, Dynamical Systems and an Introduction to Chaos.
(3rd Edition). New York, USA: Academic Press.

Arnold, V. (2006). Ordinary Differential Equations. (First
edition). Berlin, Germany: Springer.

I highly recommend the first text, since it is a book that deals with the study of differential equations in a very formal way. However, it requires the knowledge of solid concepts in real and complex analysis in addition to a previous course in differential equations with a solid background.
It seems to me that if you have studied from Simmons' book of differential equations, you will have no trouble adding solid analytical knowledge to that.


答:
If you want to approach this from the other direction and take the brute force approach, note that by the chain rule,

$\frac{\partial u}{\partial t} = \frac{\partial u}{\partial \eta}\frac{\partial \eta}{\partial t}+\frac{\partial u}{\partial \xi}\frac{\partial \xi}{\partial t}$,

$\frac{\partial u}{\partial x} = \frac{\partial u}{\partial \eta}\frac{\partial \eta}{\partial x}+\frac{\partial u}{\partial \xi}\frac{\partial \xi}{\partial x}$.

Since $\xi=x+c t, \eta=x-c t$,

$
\frac{\partial \eta}{\partial t}=-c, \frac{\partial \eta}{\partial x}=1,\\
\frac{\partial \xi}{\partial t}=c, \frac{\partial \xi}{\partial x}=1.
$

Replacing these in the first two equations:

$\frac{\partial u}{\partial t} = -c\frac{\partial u}{\partial \eta}+c\frac{\partial u}{\partial \xi}$,

$\frac{\partial u}{\partial x} = \frac{\partial u}{\partial \eta}+\frac{\partial u}{\partial \xi}$.

Now, take the derivatives again,

$\frac{\partial^2 u}{\partial t^2} = -c\frac{\partial^2 u}{\partial \eta \partial t}+c\frac{\partial^2 u}{\partial \xi \partial t}$,

$\frac{\partial^2 u}{\partial x^2} = \frac{\partial^2 u}{\partial \eta \partial x}+\frac{\partial^2 u}{\partial \xi \partial x},$

and use the chain rule again:

$\frac{\partial^2 u}{\partial t^2} = c^2\frac{\partial^2 u}{\partial^2 \eta}-c^2\frac{\partial^2 u}{\partial \eta \partial \xi}-c^2\frac{\partial^2 u}{\partial \xi \partial \eta}+c^2\frac{\partial^2 u}{\partial^2 \xi }$,

$\frac{\partial^2 u}{\partial x^2} = \frac{\partial^2 u}{\partial^2 \eta} +\frac{\partial^2 u}{\partial \eta \partial \xi} +\frac{\partial^2 u}{\partial \xi \partial \eta} +\frac{\partial^2 u}{\partial^2 \xi}$.

Finally, replace these in
$$\frac{\partial^{2} u}{\partial t^{2}}=c^{2} \frac{\partial^{2} u}{\partial x^{2}}$$
to get
$$ c^2\frac{\partial^2 u}{\partial^2 \eta}-c^2\frac{\partial^2 u}{\partial \eta \partial \xi}-c^2\frac{\partial^2 u}{\partial \xi \partial \eta}+c^2\frac{\partial^2 u}{\partial^2 \xi } = \frac{\partial^2 u}{\partial x^2} = c^2\frac{\partial^2 u}{\partial^2 \eta} +c^2\frac{\partial^2 u}{\partial \eta \partial \xi} +c^2\frac{\partial^2 u}{\partial \xi \partial \eta} +c^2\frac{\partial^2 u}{\partial^2 \xi},$$ which simplifies to
$$\frac{\partial^2 u}{\partial \xi \partial \eta}=0$$
after cancellations.

Title: One Dimension Wave Equation (Calculus of several variables)

Body:
I can't find any way to get to that expression. Honestly I don't how to start this. Any recommendations?

Consider the One Dimension Wave Equation
$$
\frac{\partial^{2} u}{\partial t^{2}}=c^{2} \frac{\partial^{2} u}{\partial x^{2}}
$$
where $u:=u(x, t) .$ If now we make a change of variable $\xi=x+c t, \eta=x-c t,$ show that the wave equation can be written as $$\frac{\partial^{2} u}{\partial \eta \partial \xi}=0$$


问:
A curve has the property that the normal line through any point on the curve passes through $(2,0)$. Find its equation.
A curve has the property that the normal line through any point on the curve passes through $(2,0)$.  If the curve contains the point $(2,3)$ find its equation.

My attempt:

Assume $(a,b)$ is a point on the function. The normal line has a slope $m=\frac{b}{a-2}$ and hence the tangent line has a slope $m_T = \frac{2-a}{b}$. Hence, this is the slope we can integrate to get the equation:

$$F(x) = \int \frac{2-a}{b}x dx = \frac{2-a}{2b}x^2+C$$

Since we know that $(2,3)$ is a point we can isolate for $C$ to get:

$$F(x) = \frac{2-a}{2b}x^2+\frac{3b-4+2a}{b}$$

I am getting stuck here because I am not sure how to get values for a $a$ and $b$...

答:
A somewhat more general proof, but one which is nevertheless in the spirit of our colleague Andrew Chin's answer, may be had using elementary differential geometry:

A Little Lemma:  Let $\alpha(s)$ be an arc-length parametrized curve in $\Bbb R^2$ such that every normal line to $\alpha(s)$ passes through the point $C \in \Bbb R^2$.  Then $\alpha(s)$ lies in a circle centered at $C$.

Proof of Little Lemma: Let $T(s)$ be the unit tangent vector field to $\alpha(s)$; that is

$T(s) = \dot \alpha(s); \tag 1$

by the first Frenet-Serret equation, $T$ is related to the curvature $\kappa(s)$ of $\alpha(s)$ via the formula

$\dot T(s) = \kappa(s) N(s), \tag 2$

where

$\kappa(s) > 0 \tag 3$

and $N(s)$ is the unit normal vector field to $\alpha(s)$.  Then the line through
$\alpha(s)$ in the direction $N(s)$ passes through $C$; if $l(s)$ denotes the signed length of the segment joining $\alpha(s)$ and $C$, we have

$\alpha(s) + l(s)N(s) = C, \tag 4$

holding for all $s$ in the domain of $\alpha(s)$.  (Note that $l(s)$ is differentiable in $s$ since we may write (4) in the form

$l(s)N(s) = C - \alpha(s), \tag{4.1}$

from which

$l(s) = l(s)N(s) \cdot N(s) = (C - \alpha(s)) \cdot N(s), \tag{4.2}$

with the right-hand side differentiable in $s$.)

We differentiate equation (4) with respect to $s$, and obtain

$\dot \alpha(s) + \dot l(s) N(s) + l(s) \dot N(s) = 0, \tag 5$

since $C$ does not depend on $s$.

The companion equation to (2),

$\dot N(s) = -\kappa(s)T(s), \tag 6$

which may be derived starting from the orthogonality of $T$ and $N$,

$T(s) \cdot N(s) = 0, \tag 7$

and then differentiating:

$\dot T(s) \cdot N(s) + T(s) \cdot \dot N(s) = 0, \tag 8$

and substituting $\dot T(s)$ from (2):

$\kappa(s) N(s) \cdot N(s) + T(s) \cdot \dot N(s) = 0; \tag 9$

since

$N(s) \cdot N(s) = 1, \tag{10}$

(9) may be written

$T(s) \cdot \dot N(s) = -\kappa(s); \tag{11}$

now, differentiating (10) yields

$\dot N(s) \cdot N(s) = 0, \tag{12}$

which shows that $\dot N(s)$ is parallel to $T(s)$; then in light of (11) we infer

$\dot N(s) = -\kappa(s)T(s), \tag{13}$

and substituting this and (1) into (5) we write

$T(s) + \dot l(s) N(s) - l(s) \kappa(s) T(s) = 0, \tag{14}$

i.e.,

$(1 - l(s) \kappa(s))T(s) + \dot l(s) N(s) = 0; \tag{15}$

the orthnormality of $T(s)$ and $N(s)$ now allows us to conclude that

$1 - l(s) \kappa(s) = 0, \tag{16}$

$\dot l(s) = 0, \tag{17}$

which show that both $l(s) = l$ and $\kappa(s) = \kappa$ are constant with

$\kappa = \dfrac{1}{l}, \tag{18}$

and hence in light of (3) that

$l > 0; \tag{18.1}$

finally, from (4)

$\vert \alpha(s) - C \vert = \vert l(s)N(s) \vert = l(s) \vert N(s) \vert = l, \tag{19}$

which shows that $\alpha(s)$ lies in the circle of radius $l$ centered at $C$.  End:  Proof of Litte Lemma.

Now taking

$C = (2, 0) \tag{20}$

and

$\alpha(s) = (2, 3) \tag{21}$

for some $s$, we find that

$l = 3, \tag{22}$

and thus see that $\alpha(s)$ lies in the circle of radius $3$ about $(2, 0)$, the same result as obtained by Andrew Chin in his answer.  The equation of the circle in which $\alpha(s)$ lies is of course

$(x - 2)^2 + y^2 = 9. \tag{23}$


问:
For what values of $z$ does the series converge?
Consider the following series:

$$\sum^{\infty}_{n=0}\frac{1}{1+z^{n}}.$$

For what values of $z$ ($z$ complex) does this converge?

Thoughts:

The series $\frac{1}{z^{n}}$ converges for all $|z|>1$. Since $1+z^{n}>z^{n}$, the above series must converge for all $|z|>1$ as well. I wanted to know how one proves rigorously that $\frac{1}{z^{n}}$ converges for all $|z|>1$, and whether this is indeed the radius of convergence for the above series as well.

答: Inequalities like $1+z^{n}>z^{n}$ have no meaning for complex numbers. The series converges absolutely for $|z|>1$ because $|\frac  1 {1+z^{n}}| \leq \frac 1 {|z|^{n}-1}$ and $\sum \frac 1 {|z|^{n}-1}$ is convergent. For $|z| \leq 1$ the series is not convergent because the general term does not tend to $0$. [It is not even defined for $z=-1$].


问:
Making deductions: Are there any trolls scared of goats
Question You come across four trolls playing bridge. They declare:

Troll 1: All trolls here see at least one knave.

Troll 2: I see at least one troll that sees only knaves.

Troll 3: Some trolls are scared of goats.

Troll 4: All trolls are scared of goats.

Are there any trolls that are not scared of goats? Recall, of course,
that all trolls are either knights (who always tell the truth) or knaves
(who always lie).

What I have done so far

I have represented all the four statements using propositional logic.

If troll $ p $ sees a knave lets say $ K(p) $

If troll $ p $ sees a troll lets say $ T(p) $

If a troll $ p $ is scared of goats lets say $ G(p) $

Thus our four statements can be written like so:

$ \forall p K(p) $

$ \exists p \exists q  T(p) \land K(q) $

$ \exists p G(p) $

$ \forall p G(p) $

I see that perhaps to solve the problem, first I need to find out whether we are dealing with lying knaves or we are dealing with truth-telling knights. To do that we use statements 1 and 2. Once we know if they are knights or knaves we use that knowledge to find out whether any trolls are not scared of goats, that is, $\exists p \lnot G(p) $, from statements 3 and 4.

However I do not know how to deal with the first part or where to start from. Any help will be appreciated.

答:
I’d not bother with the formal notation.

If Troll 1 is a knave, the other three must all be knights in order for his statement to be false. However, Troll 2’s statement would then be false, so Troll 1 must be a knight. That means that there must be at least two knaves amongst the other three trolls in order to make has statement true.

The only person who could possibly see only knaves is Troll 1, since he’s a knight. This means that Troll 2’s statement is true if and only if Trolls 1, 2, and 3 are all knaves, in which case Troll 2’s statement cannot be true. Thus, Troll 2’s statement must be false, and every troll sees at least one knight. Of course Trolls 2, 3, and 4 see Troll 1, so what that really tells us is that Troll 1 sees a knight: one of Trolls 2, 3, and 4 is a knight, the other two are knaves, and Troll 2 is one of the knaves. We conclude that one of Trolls 3 and 4 is a knight.

Note that if Troll 4 is telling the truth, so is Troll 3, so if Troll 4 is a knight, so is Troll 3. This is impossible, so Troll 4 must be a knave, and we conclude that not all trolls are scared of goats.


问:
Finding base for subspace of cubic polynomials with vanishing integrals
$U \leq P_3(\mathbb{R}), \\ U = \{P_3(\mathbb{R}) : \int_{-1}^{1}p = 0 \} $

How do I find a base for this subspace? I know how to do that for when the condition for $U$ is having a specific root for $p(x)$. For example $U = \{P_3(\mathbb{R}) : p(6) = 0\}$. We can write the constant in the polynomial using the other coefficients and substitute then write the polynomials as a linear transformation of the coefficients.

My intuition is that $p(x)$ is odd so it has to have the following form: $p(x) = a_1x^3 + a_2x$.

In this example, I write down the integrated function and substitute the upper and lower bounds (-1 and 1) and the only thing I get is $c = \frac{-a}{2}$ for $p(x) = ax^3 + bx^2 + cx + d$. I don't know where to go from there. All I know is that I should probably somehow write $p(x)$ as a linear transformation that includes $x^3$ and $x$.

答:
Let $f \in U$ and $f(x) = ax^3 + bx^2 + cx + d$

$\int_{-1}^{1}f(x)\mathrm{d}x = \frac{a}{4}x^4 + \frac{b}{3}x^3 + \frac{c}{2}x^2 + dx \ \Big|_{-1}^{1}$

$ = (\frac{a}{4} + \frac{b}{3} + \frac{c}{2} + d) - (\frac{a}{4} - \frac{b}{3} + \frac{c}{2} - d)$

$\rightarrow b = -3d$

Hence:

$f(x) = ax^3 + (-3d)x^2 + cx + d\\
= a(x^3) + d(1 - 3x^2) + c(x)$

So we can write every member of $U$ as a linear combination of $B = \{x^3, 1 - 3x^2, x \}$ therefore, $B$ generates $U$. $B$ is also clearly linearly independent so $B$ is a basis for $U$.


问:
Geometric Series and its infinite sum
when we say $1+r+r^2+r^3+\cdots=S$ do we mean $S$ is the partial sum of first $n$ terms of this geometric series where $n$ goes to infinity (which is not boundless due to the existence of $n$th term) or do we say $S$ is sth that has literally infinite, boundless terms so that $S-1=rS$?

In other words, are we saying when we talk about $S$, are we talking about the partial sum of the first n terms where n is getting really really large or do we literally mean the bound does not exist at all, so that $S-1=rS$?

This matters a lot to me because it seems like there can be some values of things like that even if the partial sum is divergent!

If the meaning of $S$ is the 'limit' of partial sum then it's obvious that $S-1=rS$ if and only if the absolute value of $r$ is less than one. but when we treat $S$ as sth that does not have the 'nth term (the bound)' no matter how far away then I can't think of any reason of saying $S-1=rS$ is wrong. So, to sum up, is $S$ limit of the partial sum or is it sth that is endless? I hope I can get satisfying answer of this question.

答:
The official meaning of the assertion that
$$1+r+r^2+r^3+\cdots=S$$
is that
$$\lim_{n \to \infty}(1+r+r^2+\cdots +r^n) =S$$.

By the usual formula for the sum of a finite geometric series,
$$1+r+r^2+\cdots +r^n=\frac{1-r^{n+1}}{1-r}$$
(unless, of course, $r=1$).
It is reasonably clear that if $|r|<1$,
$$\lim_{n\to\infty}\frac{1-r^{n+1}}{1-r}=\frac{1}{1-r}$$
and that if $|r|\ge 1$, then
$$\lim_{n \to \infty}(1+r+r^2+\cdots +r^n)$$
does not exist.

But that is not really your question.  What I think you are saying is roughly this: let
$$S=1+r+r^2+r^3+\cdots$$
Then at the level of formal "algebraic" manipulation, we have (??)
$$S-1=rS$$
and "therefore" (??)
$S=1/(1-r)$
(at least if $r \ne 1$).

But do "formal" manipulations always yield correct results?  Not necessarily.  For example, you have probably already seen plausible-looking algebraic manipulations that appear to yield the absurd "result" that $0=1$ (usually here the flaw is a carefully hidden division by $0$.)

The formal manipulation yields, if we take $r=1/2$, the perfectly correct result that
$$1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots =2$$
This makes physical sense.  If you look at the real number line, and put a dot at $1$, then at $1+1/2$, then at $1+1/2+1/4$, and so on, your dots are clearly approaching $2$, and are after a while indistinguishably close to $2$.

Now look at the formal sum
$$S=1+2+2^2+2^3+\cdots$$
Purely formal manipulation then yields that $S=-1$.  Does this make any kind of physical sense?  Certainly if you add more and more terms of the above series, you are not getting in any sense close to $-1$.  And at the crude informal level, it is clear that the sum is "infinite," whatever that may mean.  Formal manipulations on non-existent objects can yield absurd results.

So in this case formal manipulation has produced a result that makes no physical sense.  There are a number of instances where blind manipulation yields wrong results, so early in university, students are trained to apply
algebraic "rules" only in situations where these rules are valid.

However, the situation is more complicated than that!  Once upon a time complex numbers, when they came up in a formal calculation, were dismissed as "absurd."  Now they are part of the essential toolkit of the electrical engineer!  There are similar phenomena with series.

In combinatorics, for example, formal power series are successfully used, at the purely manipulational level, even when the series technically do not converge.  And in various areas of mathematics, divergent series of a suitable type can be very useful, even in computations!

But at this stage, you should be disciplined, and apply formal manipulations only in situations where you know, or have been assured that, they are "safe."  Later, perhaps, you can explore situations where venturing beyond safe confines yields interesting and useful results.

Title: Geometric Series and its infinite sum


问:
Prove that if a number can be written in the form $n(n+1) , n+2$ can never be a factor of the number .
I actually did not find this in any book , what actually happened is i was proving an equation and I came across the question I posted .If anyone can help me that would be great.</p>

答: Begin by noting that $ n+2 $ shares no common factors with $ n+1 $.  So we would need $ n + 2 $ to be a factor of $ n $, which is absurd (for positive $n$).


答:
Suppose $n+2$ is a factor of $n(n+1)$. Since $n+2$ and $n+1$ are coprime, $n+2$ divides $n$ by Euclid's Lemma. For positive $n$ this is impossible since the absolute value of the divisor cannot exceed the original number (in this case, $|n+2|\le |n|$).

However, if one considers the nonpositive integers (which I am clearly not accustomed to), a few more solutions crop up. We can still use Euclid's Lemma to conclude $n+2$ divides $n$.

We write:

$$k(n+2) = n \leadsto n=-\frac {2k}{k-1}=-\left(2+\frac2{k-1}\right)$$

so $k-1$ must be a divisor of $2$. This gives $k=3,2,0,-1$, which corresponds to the solutions $n = -3,-4,0,-1$. Checking that our original expression is zero at $0,-1$, no more solutions can be found.

Title: Prove that if a number can be written in the form $n(n+1) , n+2$ can never be a factor of the number .

Body: I actually did not find this in any book , what actually happened is i was proving an equation and I came across the question I posted .If anyone can help me that would be great.</p>


问:
Find the Taylor series of $f(x) =\sqrt{1 + \sin{x}}$ around $x_0= 0$
Find the Taylor series of $f(x) =\sqrt{1 + \sin{x}}$  around $x_0= 0$

I am aware of the taylor series expansion formula, but that's where I get stuck. Any help/a "model" solution on how to solve such problems would be helpful!

答:
Hint

Use the  fact that
$$\sqrt{1+\sin(x)}=\sin \left(\frac{x}{2}\right)+\cos \left(\frac{x}{2}\right)$$


问:
Are there other solutions to the functional equation $f(x,y)\circ f(y,x)=e$?
Let $(G,\circ)$ be a group with identity $e$. We want to solve for $f$ in the functional equation $$f(x,y)\circ f(y,x)=e\qquad \forall x,y \in G$$

A trivial solution is $f(x,y) := e$. Working backwards from known examples for $(\mathbb R, +)$ and  $(\mathbb R_+, \cdot)$, I see that two possible solutions are:
$$ f(x,y) := x \circ y^{-1}$$  and $$ f(x,y) := y \circ x^{-1}$$

Are there any other solutions, or can it be shown that these are the only two non-trivial solutions?

答: Let $f\colon X\times X\to G$ where $X$ is an arbitrary non-empty set. Define an equivalence relation $\sim$ on $X\times X$ by $(x,y)\sim (u,v)$ iff $\{ x,y\}=\{u,v\}$, Then the equivalence classes are of the form $\{(x,x)\}$ or of the form $\{(x,y),(y,x)\}$ with $x\not=y$. Then all functions $f$ such that $f(x,x)\circ f(x,x)=e$ for all $x$ and $f(x,y)\circ f(y,x)=e$ for all $x\not=y$ have the desired property. Thus you may define $f$ according to these rules on the equivalence classes. But there is no need to have relations between different classes.


答:
If Troll 2's statement is true, then the troll referred to in his statement sees Troll 2, so that Troll 2 is a knave.  This is impossible, since a knave wouldn't make a true statement, so Troll 2 is a knave.  Suppose that no troll is afraid of goats.  Then Trolls 3 and 4 are also knaves, and Troll 1 sees only knaves, so that Troll 2's statement was true after all.  We have seen that this is impossible, so some trolls fear goats, and Troll 3 is a knight.  If Troll 1 is a knave, then his statement is true, since everyone see Troll 1 or Troll 2.  This is absurd, so Troll 1 is a knight.  Now Troll 2 must see a knave and that can only be Troll 4.

Not all trolls are afraid of goats.

Title: Making deductions: Are there any trolls scared of goats

Body:
Question You come across four trolls playing bridge. They declare:

Troll 1: All trolls here see at least one knave.

Troll 2: I see at least one troll that sees only knaves.

Troll 3: Some trolls are scared of goats.

Troll 4: All trolls are scared of goats.

Are there any trolls that are not scared of goats? Recall, of course,
that all trolls are either knights (who always tell the truth) or knaves
(who always lie).

What I have done so far

I have represented all the four statements using propositional logic.

If troll $ p $ sees a knave lets say $ K(p) $

If troll $ p $ sees a troll lets say $ T(p) $

If a troll $ p $ is scared of goats lets say $ G(p) $

Thus our four statements can be written like so:

$ \forall p K(p) $

$ \exists p \exists q  T(p) \land K(q) $

$ \exists p G(p) $

$ \forall p G(p) $

I see that perhaps to solve the problem, first I need to find out whether we are dealing with lying knaves or we are dealing with truth-telling knights. To do that we use statements 1 and 2. Once we know if they are knights or knaves we use that knowledge to find out whether any trolls are not scared of goats, that is, $\exists p \lnot G(p) $, from statements 3 and 4.

However I do not know how to deal with the first part or where to start from. Any help will be appreciated.