问:
Smallest topology on $\mathbb{C}$ such that all singletons are closed
This question was part of an exam for which I am preparing.

Question was related to finding smallest topology T on $\mathbb{C}$ in which all singletons are closed.
and then findind which options are true amog these:

A ($\mathbb{C}$,T) is haursdorff.

B ($\mathbb{C}$,T) is compact.

C ($\mathbb{C}$,T) is connected.

D  $\mathbb{Z}$ is dense in ($\mathbb{C}$,T) .

But  I am just a beginner in topology and self studying from Wayne patty . I was unable to find what topology should be . So, can you please tell what topology should be . rest I would prefer to work by mysely.

Thank you.

答:
I think the topology you are looking for is the co-finite topology, i.e.

$$ \tau= \{ A\subseteq X: \vert A^c\vert<\infty \} \cup \{ \emptyset\}.  $$

You just have to verify that every topology in which the singletons are closed, contains the co-finite topology, and that the co-finite topology is one of them.


问:
Determine convergence and limit
I want to determine whether each of the sequences converges, and if so, to what limit.

$$\begin{align} 1. \quad \quad & a_n=\frac{n^{2}}{n+1}-\frac{n^{2}+1}{n}\\
2. \quad \quad & a_n=\sqrt{n+1}-\sqrt{n}
\end{align}$$

For the first one, I tried to use ratio test, but it's not working. For the second, I tried to multiply it by $\sqrt{n+1}+\sqrt{n}$ and then devided it. But I'm stuck again and don't know which test should I use to determine its convergence.

答:
For the first sequence, note that

$$\begin{align}
a_n & = \frac{n^2}{n+1} - \frac{n^2+1}{n}\\
& = \frac{n^3-n^3-n-n^2-1}{n^2+n}\\
& = \frac{-n^2-n-1}{n^2+n}.
\end{align}$$

And taking the limit you obtain

$$\lim_{n \to \infty} a_n = -1 \quad (\in \mathbb{R})$$

so it converges.

For the second sequence, as you said, multiply the by $\frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}}$ and you will obtain

$$\begin{align}
a_n & = (\sqrt{n+1}-\sqrt{n}) \cdot \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}}\\
& = \frac{(\sqrt{n+1})^2-(\sqrt{n})^2}{\sqrt{n+1} + \sqrt{n}}\\
& = \frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}\\
& = \frac{1}{\sqrt{n+1}+\sqrt{n}}.
\end{align}$$

Taking the limit you get

$$\lim_{n \to \infty} a_n = 0 \quad (\in \mathbb{R})$$

so it converges too.

ContentLicense: CC BY-SA 4.0

Title: Determine convergence and limit


问:
Immersion in both ways yet not an isomorphism
i was wondering if it's possible for two total orders A and B that there exist two injective functions f and g such that f is an immersion of A into B and g is an immersion of B into A, yet $A\ncong B$.

I came up with $A=\mathbb{R},B=\mathbb{Q^{\geqslant0}}$ so $f:A\Rightarrow(0,1)\cap\mathbb{Q^{\geqslant0}}$ and $g:B\Rightarrow[0,1)\cap\mathbb{R}$ but i am not sure it is correct at all.

答:
Let $A=[0,\infty)$ and let $B=(0,\infty)$.$\;$Then

$A$ can be order-embedded in $B$ via the map $a\mapsto a+1$.$\\[4pt]$

$B$ can be order-embedded in $A$ via the map $b\mapsto b$.

but $A,B$ are not order-isomorphic since $A$ has a least element but $B$ does not have a least element.

As regards your proposed example, it doesn't work; there is no injection of $A=\mathbb{R}$ into $B=\mathbb{Q^{\geqslant0}}\;$since $A$ is uncountable whereas $B$ is countable.


问:
Why do we study prime ideals?
I hope this isn't an inappropriate question here!

I'd like to ask the following (perhaps slightly ill-posed) question: why do we study prime ideals in general (commutative or non-commutative) rings? The situation is somewhat clear to me in some nice cases such as Dedekind rings, but even in slightly more general settings (e.g. general commutative Noetherian rings) the theory becomes rather horrible. What benefit do we gain from an understanding of the prime ideals of a ring?

Thank you!

答:
One of Qiaochu Yuan's blogposts addresses exactly this question.

The central fact is more or less obvious (that the prime ideals are exactly the kernels of homomorphisms of $R$ into domains) but it does a great job of explaining why that is useful and natural to study.

ContentLicense: CC BY-SA 3.0

Title: Why do we study prime ideals?


问:
Poisson kernel for half disc
I want to derive the Poisson kernel for half disc $\Omega = D(0,1) \cap \{ z \in \mathbb{C}: \operatorname{Im} z > 0 \}$. Suppose we have the BVP
$$ \left\{ \begin{array}{ll}
\Delta u =0 \qquad & \text{in } \Omega \\
u= \phi \qquad & \text{on } \partial \Omega
\end{array} \right. $$
where $\phi$ is continuous and real-valued.
I know that the composition of
$$ z \mapsto \frac{2z}{z^2+1}, \qquad \text{and} \qquad z \mapsto \frac{z-i}{z+i}$$ is a conformal map from $\Omega$ onto $D(0,1)$. If you compute this, $f: \Omega \rightarrow D(0,1)$ given by
$$ f(z) = \frac{2z - i (z^2+1)}{2z + i (z^2+1)}$$
is a biholomorphism.
I tried to apply Poisson integral formula for $D(0,1)$ on the following BVP:

$$ \left\{ \begin{array}{ll}
\Delta (u \circ f^{-1}) =0 \qquad & \text{in } D(0,1) \\
u \circ f^{-1}= \phi \circ f^{-1} \qquad & \text{on }  \partial D(0,1)
\end{array} \right. $$

Then using change of variables,
\begin{eqnarray*}
 &&u (\tilde{\omega}) \\
&=&u \circ f^{-1}(z) \\
&=& \frac{1}{2\pi} \int_{0}^{2\pi} \phi(f^{-1}(e^{i\theta})) \frac{1-|z|^2}{|z-e^{i \theta}|^2} d \theta \\
&=& \cdots \\
&=& \frac{1}{2\pi} \int_{\Omega} \phi(\eta) \frac{1 - |f(\tilde{\omega})|^2}{|f(\tilde{\omega}) - f(\eta)|^2} \frac{4 (1- \eta)^2}{4 \eta^2 + (\eta^2 + 1)^2} d \eta
\end{eqnarray*}
where $\tilde{\omega} = f^{-1}(z)$.

But this integral is too complicated and I have no idea to proceed from here. Any suggestions about my argument? Thank you.

答:
You argument should work but the calculations might be simpler with
the following lemma.

Lemma. Let $z \in \Omega$. There exists only one biholomorphism from $\Omega$ onto $\mathbb{D}$ that sends $z$ on the origin given by
$$ \forall w \in \Omega,\ f_z(w) = \frac{w-z}{1-w \bar{z}} \frac{1-w z}{w-\bar{z}}. $$
Furthermore, the function $f_z$ admits a smooth extension to some neighbourhood of $\Omega$.

Now, we proceed as you suggested.
Let $u \in \mathrm{H}(\Omega)$ continuous on the boundary. Then
$$ \forall z \in \Omega,\ u(z) = (u \circ f_z^{-1})(0) = \int_0^{2\pi} (u \circ f_z^{-1})(e^{i\theta}) \frac{d\theta}{2\pi}.$$
Let us do the change of variable $\xi = f_z^{-1}(e^{i\theta})$ in the above integral. But
$$ d\xi = ie^{i \theta} (f_z^{-1})'(e^{i\theta}) d\theta = i \frac{f_z(\xi)}{f_z'(\xi)} d\theta.$$
Hence
$$ \forall z \in \Omega,\ u(z) = \int_{\partial \Omega} u(\xi) P(z, \xi) d\xi\ \mbox{ with }\ P_{\Omega}(z, \xi) = -\frac{i}{2\pi} \cdot \frac{f_z'(\xi)}{f_z(\xi)}. $$
Using the additive property of the logarithmic derivative, we got
$$ \frac{f'_z(\xi)}{f_z(\xi)} = \frac{1-\xi\bar{z}}{\xi-z} \frac{1-|z|^2}{(1-\xi\bar{z})^2} - \frac{1-\xi z}{\xi-\bar{z}} \frac{1-|z|^2}{(1-\xi z)^2} = \frac{1-|z|^2}{(\xi-z)(1-\xi \bar{z})} - \frac{1-|z|^2}{(\xi-\bar{z})(1-\xi z)}. $$
Thus, the Poisson kernel of the half-disk is given by
$$ P_{\Omega}(z, \xi) = \frac{1}{2\pi i}\cdot \left\{\frac{1-|z|^2}{(\xi-z)(1-\xi \bar{z})} - \frac{1-|z|^2}{(\xi-\bar{z})(1-\xi z)}\right\}. $$

If you split the boundary of $\Omega$ in two parts, namely the semi-circle and the interval $(-1, 1)$, you can produce more explicit formulae. See
Value Distribution of Meromorphic Functions of Goldberg and Ostrovskii, page 3. For instance, the Poisson kernel along the semi-circle is given by
\begin{align} P_{\Omega}(z, \xi) d\xi &= P_{\Omega}(z, e^{i\theta}) ie^{i\theta} d\theta = \frac{1}{2\pi}\frac{1-|z|^2}{|z-e^{i\theta}|^2}d\theta - \frac{1}{2\pi} \frac{1-|z|^2}{|\bar{z}-e^{i\theta}|^2}d\theta \\ &= P_{\mathbb{D}}(z, e^{i\theta}) d\theta- P_{\mathbb{D}}(\bar{z}, e^{i\theta})d\theta. \end{align}
This relation between the Poisson kernels of the disk and the half-disk is not uncanny as soon as you are familiar with the Schwarz reflection principle.

Title: Poisson kernel for half disc


问:
Expression of sum of squares as a sum of a specific form
This question is a follow-up of this one.

Let $x,y \in \mathbb{Z}$, and suppose that $x^2+y^2 \ge 4$, and that $x,y$ are not both odd.

Do there exist $a,b,c,d \in \mathbb{Z}$ such that
$
(a+d)^2+(b-c)^2=x^2+y^2
$
and $ad-bc=1$?

This question is motivated by an attempt to characterise the norms of matrices in $SL_2(\mathbb{Z})$: If $A =\begin{pmatrix} a & b \\\ c & d \end{pmatrix} \in SL_2(\mathbb{Z})$, then
$$
(a+d)^2+(b-c)^2=\|A\|^2+2,
$$
so I wonder whether $\|A\|^2+2$ can be any sum of squares $x^2+y^2$ where $x,y$ are not both odd.

This parity restriction on $x,y$ is certainly necessary for the existence of such a representation:

If $x,y$ are both odd, then $a+d,b-c$ must be odd (by looking at equality $\mod 4$). Thus one of $a$ and $d$, and one of $b$ and $c$ must be even, hence $ad-bc$ is even which is incompatible with $ad-bc=1$.

答:
Let $x=0$ and $y=3$ so that $x^2+y^2=9$.  Then if $\tbinom{a\ b}{c\ d}\in\operatorname{SL}_2(\Bbb{Z})$ is such that
$$(a+d)^2+(b-c)^2=x^2+y^2=9,\tag{1}$$
then $\{a+d,b-c\}=\{0,\pm 3\}$ for some for some choices of the signs. Then of course the matrices
$$\begin{pmatrix}-a&\hphantom{-}b\\\hphantom{-}c&-d\end{pmatrix},
\begin{pmatrix}\hphantom{-}a&-b\\-c&\hphantom{-}d\end{pmatrix},
\begin{pmatrix}\hphantom{-}b&\hphantom{-}a\\-d&-c\end{pmatrix}
\in\operatorname{SL}_2(\Bbb{Z}),$$
also satisfy $(1)$, so without loss of generality $a+d=0$ and $b-c=3$. Then
$$1=ad-bc=a(-a)-b(b-3),$$
which shows that $b$ is an integral root of a quadratic with discriminant
$$\Delta=-4a^2+77.$$
But this is never a perfect square; a contadiction! Hence no such matrix exists.


问:
The U.S. Senate consists of $100$ senators, with $2$ from each of the $50$ states
The U.S. Senate consists of $100$ senators, with $2$ from each of the $50$ states. There are $50$ Democrats in the Senate. A committee of size $10$ is formed, by picking a random set of senators such that all sets of size $10$ are equally likely.

a) Find the expected number of Democrats on the committee.

b) Find the expected number of states represented on the committee (by at least one senator).

c) Find the expected number of states such that both of the state’s senators are on the committee.

For part a), I defined $D$ as a random variable for democrats, got the Hypergeometric Distribution to be $D$ ~ $(100, c, d),$ and the expectation to be $E(D) = c(\frac{d}{100}) = \frac{cd}{100}$. Then I plugged in $10$ for $c$ and $50$ for $d$ and got $E(D) = 5.$

For part b), I defined $I_j$ as the random variable for the j-th state represented/ not represented in the committee, got $P(I_j = 1) = 1 - P(I_j = 0) = 1 - \frac{\binom{98}{c}}{\binom{100}{c}} = 1 - \frac{(100 - c)(99 - c)}{(100)(99)}$, and finally $E(\sum_{j} I_j) = \sum_{j}E(I_j) = \sum_{j}P(I_j = 1) = 50(1 - \frac{(100 - c)(99 - c)}{(100)(99)})$.

Then I plugged in $10$ for $c$ and got $P(I_j = 1) = 1 - \frac{8010}{9900}$ = $\frac{21}{110}$ and $E(\sum_{j} I_j) = 50(1 - \frac{(100 - 10)(99 - 10)}{(100)(99)})$ = $50(\frac{21}{110})$ = $\frac{105}{11} $

For part c), I defined $K_j$ as the random variable for both being from/ not from j-th state in the committee, got $P(K_j = 1) = \frac{\binom{98}{c}}{\binom{100}{c}} = \frac{(100 - c)(99 - c)}{(100)(99)}$ and finally $E(\sum_{j} K_j) = \sum_{j}E(K_j) = \sum_{j}P(K_j = 1) = 50( \frac{(100 - c)(99 - c)}{(100)(99)})$

Then I plugged in $10$ for $c$ and got $P(K_j = 1) = \frac{8010}{9900}$ = $\frac{89}{110}$ and $E(\sum_{j} K_j) = 50(\frac{(100 - 10)(99 - 10)}{(100)(99)})$ = $50(\frac{89}{110})$ = $\frac{445}{11} $.

However, I am not sure if my answers are correct. Any help would be much appreciated!

答:
A) and B) looks fine, for C) I think the right definition for $K_j$ would be let $K_j$ be the indicator variable denoting if there is representation for state $j$ in the committee from both the senators.

Then $P[K_j = 1 ]$ = $\frac{{2 \choose {2}}*{{98}\choose{8}}}{{100}\choose{10}}$

Therefore the expected number of such states in the committee is $50*P[K_j = 1] = \frac{45}{99}$


问:
representing an open set in countable union of closed sets
This question was asked in my analysis quiz and I marked every option wrong. I admit I didn't had much clue about this question.

Let $(X,d)$ be a metric space. Then:

A An arbitrary open set $G$ in $X$ is a countable union of closed sets.

B An arbitrary open set $G$ in $X$ cannot be countable union of closed sets if X is connected.

C  An arbitrary open set $G$ in $X$ is a countable union of closed sets only if $X$ is countable.

D An arbitrary open set $G$ in $X$ is a countable union of closed sets only if $X$ is locally compact.

Answer

 A

I marked B and was clueless about D . But now even if I have answer key I am unable to reason open sets with union of Closed sets and also with locally compact and connected sets .

Can you please give hints for A,B,D.

I shall be really thankful

答:
If we prove that  (A) is true it follows immediately that the other options are all false.

Any closed set $C $ in a metric space is the intersection of the  sets $\{x: d(x,C) <\frac 1 n\}$ and these sets are open. Taking complements we see that (A) is always true.


问:
For what exactly we need that the submartingale is a finite sequence in theorem 6.7.3 of Ash's book?
I want to check if my understanding of the following proof is correct. The following is a slight rephrasing of a theorem of Ash's probability book:

6.7.3 Theorem. Let $\{X_n:n=1,\ldots ,m\}$ a submartingale and let $T_1,T_2,\ldots $ an increasing sequence of stopping times for the $\mathscr{F}_n$ (that is, $\{\omega :T_n(\omega )\leqslant j\}\in \mathscr{F}_j$ for each $n$, and each $T_n$ take values in $\{1,\ldots ,m\}$). Now let $$\mathscr{F}_{T_i}:=\{A\in \mathscr{F}: A\cap \{\omega :T_i(\omega )\leqslant k\}\in \mathscr{F}_k\}$$ Then the $X_{T_n}$ form a submartingale respect to the $\mathscr{F}_{T_n}$.

There we defines $X_{T}(\omega ):=X_n(\omega )$ whenever $T(\omega )=n$. Now the proof:

Set $Y_n:=X_{T_n}$ and note that $$\int_{\Omega }|X_{T_n}|\mathop{}\!d P=\sum_{k=1}^m\int_{\{\omega : T_n(\omega )=k\}}|X_k|\mathop{}\!d P\leqslant \sum_{k=1}^m\mathrm{E}[|X_k|]<\infty\tag1 $$ As the $T_n$ increases with $n$ so does $\mathscr{F}_{T_n}$. Now if $A\in \mathscr{F}_{T_n}$ then we must show that $\int_{A}Y_{n+1}\mathop{}\!d P\geqslant \int_{A}Y_n\mathop{}\!d P$. Since $A=\bigcup_{j}A\cap \{T_n=j\}$ it suffices to replace $A$ in the integral by $D_j:=A\cap \{T_n=j\}$, wich belong to $\mathscr{F}$.

Now note that if $k>j$ then $T_n=j$ implies $T_{n+1}\geqslant j$, hence $$\int_{D_j}Y_{n+1}\mathop{}\!d P=\sum_{i=j}^k\int_{D_j\cap \{T_{n+1}=i\}}Y_{n+1}\mathop{}\!d P+\int_{D_j\cap \{T_{n+1}>k\}}Y_{n+1}\mathop{}\!d P\tag2$$ so
$$\int_{D_j}Y_{n+1}\mathop{}\!d P=\sum_{i=j}^k\int_{D_j\cap \{T_{n+1}=i\}}X_i\mathop{}\!d P+\int_{D_j\cap \{T_{n+1}>k\}}X_k\mathop{}\!d P-\int_{D_j\cap \{T_{n+1}>k\}}(X_k-Y_{n+1})\mathop{}\!d P\tag3$$ and from the definitions its easy to see that $$\int_{D_j\cap \{T_{n+1}\geqslant k\}}X_k\mathop{}\!d P\geqslant \int_{D_j\cap \{T_{n+1}\geqslant k\}}X_{k-1}\mathop{}\!d P\tag4$$ Then by a finite recursion we find that $$\int_{D_j}Y_{n+1}\mathop{}\!d P\geqslant \int_{D_j\cap \{T_{n+1}\geqslant j\}}X_j\mathop{}\!d P-\int_{D_j\cap \{T_{n+1}>k\}}(X_k-Y_{n+1})\mathop{}\!d P\tag5$$

Now $\{T_{n+1}>k\}$ is empty for $k\geqslant m$. Finally, $D_j\cap \{T_{n+1}\geqslant j\}=D_j$ since $D_j\subset \{T_{n}=j\}$, and $X_j=Y_n$ in $D_j$. Thus $$\int_{D_j}Y_{n+1}\mathop{}\!d P\geqslant \int_{D_j}Y_n\mathop{}\!d P\tag6$$

This means that $-\int_{D_j\cap \{T_{n+1}>k\}}(X_k-Y_{n+1})\mathop{}\!d P>0$, what is a consequence of $\{X_n\}$ being a submartingale, ok. But this is my question: for what exactly was used that $\{X_n\}$ was a finite sequence? Just to prove that the $Y_j\in L^1(P)$? It seems that it is used for a different step, as Ash says just after the end of this proof

Theorem 6.7.3 extends immediately to the case of an infinite sequence if
each $T_n$ is bounded, that is, for each $n$ there is a positive constant $K_n$ such that $T_n \le K_n$ a.e. The same proof may be used; the key point is that $\{T_{n+1} > k\}$ is still empty for sufficiently large $k$.

This seems to imply that the finiteness of the sequence $\{X_k\}$ is used in the conclusion (6) from (5), that is, that the RHS of (5) is finite or so. It is not so clear. Can someone clarify it?

答:
Lots of typos in the question---just to point out a couple:

"...$\{\omega :T_n(\omega )\leqslant j\}\in \mathscr{F}_n...$" should be "...$\{\omega :T_n(\omega )\leqslant j\}\in \mathscr{F}_j$...".

"...$X_{T}(\omega ):=T(\omega )$..." should be "...$X_{T}(\omega ):=X_{T(\omega )}(\omega)$...".

Now to answer your question:

...where exactly was the assumption that $m < \infty$ (in $\{X_n:n=1,\ldots ,m\}$) used in the
argument?

The answer is nowhere, really. Note that the argument only involves two adjacent elements $Y_{n+1} = X_{T_{n+1}}$ and $Y_{n} = X_{T_{n}}$. This would remain the same regardless whether $m$ is finite or infinite.

Rather, the key to the argument is the assumption that the sequence of stopping times $\{ T_n \}$ is bounded. For example, the assumption you mention that $T_n < K_n\;\; a.s.$ would suffice (where the sequence $K_n$ may diverge to $\infty$).

Under the assumption that $T_n < K_n\;\; a.s.$, the same argument goes through verbatim for a submartingale $\{  X_n \}_{n \geq 1}$:

Proof:

Integrability of $Y_n$ holds exactly the same as before:
$$
\int |Y_n| dP = \sum_{j = 1}^{K_n} \int_{\{T_n = j\}} |X_{j}| dP < \infty.
$$

Let $A \in \mathscr{F}_{T_n}$, then $A\stackrel{(1)}{=}\bigsqcup\limits_{j=1}^{K_n } A\cap \{T_n=j\}$. Therefore it suffices to consider
$$
D_j = A \cap \{T_n=j\}
$$
for some $1 \leq j \leq K_n$. So $Y_n \cdot 1_{D_j} = X_{T_j} \cdot 1_{D_j}$.

On the other hand,
$$
\int_{ D_j } Y_{n+1} dP \stackrel{(2)}{=}
\sum_{i = j }^{K_{n+1}} \int_{ D_j  \cap \{ T_{n+1} = i \} } Y_{n+1} dP
= \sum_{i = j }^{K_{n+1}} \int_{ D_j  \cap \{ T_{n+1} = i \} } X_i dP
\geq  \int_{ D_j } X_j dP = \int_{ D_j } Y_{n} dP.
$$
End of Proof.

With the additional assumption that $\{  X_n \}_{n \geq 1}$ is, say, dominated by an $Z \in L^1_+$, i.e. $|X_n| \leq Z \; a.s.$, each element of the stopping time sequence $\{ T_n \}$ can be allowed to be unbounded:

$\int |Y_n| dP = \sum\limits_{j = 1}^{\infty} \int_{\{T_n = j\}} |X_j| dP \leq \int Z dP < \infty$.

"$\stackrel{(1)}{=}$" becomes $A = \bigsqcup\limits_{j=1}^{\infty } A\cap \{T_n=j\}$. Therefore it still suffices to consider
$$
D_j = A \cap \{T_n=j\}.
$$

"$\stackrel{(2)}{=} $" becomes $\int\limits_{ D_j } Y_{n+1} dP =
\sum\limits_{i = j }^{\infty} \int\limits_{ D_j  \cap \{ T_{n+1} = i \} } Y_{n+1} dP$. Now, for each finite $m$,
$$
\sum\limits_{i = j }^{m} \int\limits_{ D_j  \cap \{ T_{n+1} = i \} } Y_{n+1} dP
\geq \sum\limits_{i = j }^{m} \int\limits_{ D_j  \cap \{ T_{n+1} = i \} } Y_{n} dP.
$$
So the submartingale property follows from the Dominated Convergence Theorem.

One can probably construct counterexamples where the $L^1$-dominated condition does not hold, each $T_n$ is unbounded, and $\{ X_{T_n} \}$ is no longer a submartingale.


问:
Wanted : for more formulas to find the area of a triangle?
I know some formulas to find a triangle's area, like the ones below.

Is there any reference containing most triangle area formulas?

If you know more, please add them as an answer

$$s=\sqrt{p(p-a)(p-b)(p-c)} ,p=\frac{a+b+c}{2}\\s=\frac{h_a*a}{2}\\s=\frac{1}{2}bc\sin(A)\\s=2R^2\sin A \sin B \sin C$$
Another symmetrical form is given by :$$(4s)^2=\begin{bmatrix}
a^2 &  b^2 & c^2
\end{bmatrix}\begin{bmatrix}
-1 & 1  & 1\\
1 &  -1 & 1\\
1 & 1 & -1
\end{bmatrix} \begin{bmatrix}
a^2\\
b^2\\
c^2
\end{bmatrix}$$

Expressing the side lengths $a$, $b$ & $c$ in terms of the radii $a'$, $b'$ & $c'$ of the mutually tangent circles centered on the triangle's vertices (which define the Soddy circles)
$$a=b'+c'\\b=a'+c'\\c=a'+b'$$gives the paticularly pretty form $$s=\sqrt{a'b'c'(a'+b'+c')}$$
If the triangle is embedded in three dimensional space with the coordinates of the vertices given by $(x_i,y_i,z_i)$ then $$s=\frac{1}{2}\sqrt{\begin{vmatrix}
y_1 &z_1  &1 \\
 y_2&z_2  &1 \\
y_3 &z_3  &1
\end{vmatrix}^2+\begin{vmatrix}
z_1 &x_1  &1 \\
 z_2&x_2  &1 \\
z_3 &x_3  &1
\end{vmatrix}^2+\begin{vmatrix}
x_1 &y_1  &1 \\
 x_2&y_2  &1 \\
x_3 &y_3  &1
\end{vmatrix}^2}$$
When we have 2-d coordinate $$ s=\frac{1}{2}\begin{vmatrix}
x_a &y_a  &1 \\
x_b &y_b  &1 \\
x_c &y_c  & 1
\end{vmatrix}$$

In the above figure, let the circumcircle passing through a triangle's vertices have radius $R$, and denote the central angles from the first point to the second $q$, and to the third point by $p$ then the area of the triangle is given by:
$$ s=2R^2|\sin(\frac{p}{2})\sin(\frac{q}{2})\sin(\frac{p-q}{2})|$$

答:
Consider $\triangle ABC$ and inscribed circle with radius $r$.
Three smaller similar triangles
$\triangle AcBcC,\ \triangle AbBCb,\ \triangle ABaCa$
are cut by
tangent lines to the incircle parallel to
the corresponding sides of $\triangle ABC$.

Given the radii $r_a,\ r_b,\ r_c$ of the incircles of
$\triangle ABaCa,\ \triangle AbBCb,\ \triangle AcBcC$,
respectively,
the area of $\triangle ABC$ is found as

\begin{align}
\bbox[5px,border:2px solid #C0A000]{
S=
\sqrt{\frac{(r_a+r_b+r_c)^7}{r_a\,r_b\,r_c}}
}
\tag{1}
.
\end{align}

Also, inradius $r$, semiperimeter $\rho$, and circumradius $R$
of $\triangle ABC$ can be found as

\begin{align}
r&=r_a+r_b+r_c
\tag{2}
,\\
\rho&=
\sqrt{\frac{(r_a+r_b+r_c)^5}{r_a\,r_b\,r_c}}
\tag{3}
,\\
R&=
\tfrac14\,\frac{r(r-r_a)(r-r_b)(r-r_c)}{r_a r_b r_c}
\\
&=
\tfrac14\,\frac{(r_a+r_b+r_c)(r_a+r_b)(r_b+r_c)(r_c+r_a)}{r_a r_b r_c}
\tag{4}
.
\end{align}

And the side lengths of $\triangle ABC$ can be found explicitly as

\begin{align}
a&=r\,(r_b+r_c)\,\sqrt{\frac{r}{r_a\,r_b\,r_c}}
,\\
b&=r\,(r_c+r_a)\,\sqrt{\frac{r}{r_a\,r_b\,r_c}}
,\\
c&=r\,(r_a+r_b)\,\sqrt{\frac{r}{r_a\,r_b\,r_c}}
.
\end{align}

Related answer.


问:
Discrete Mathematics - Modular arithmethics
I've been trying to solve a task for some while now but I'm currently very stuck.

For the integers $a$ and $b$ it applies that $b \equiv a \pmod{91}$ and $\gcd (a, 91) = 1$. Determine a positive number $k> 1$ such that $b^k \equiv a \pmod{91}.$

I know that  $b^{\phi(91)} = 1\pmod{91}$ and  $b^{\phi(91)} = 72$.

After this I'm pretty empty. Would anyone be so kind to guide me or help me in the right direction?

Body:
$k$ calculate as discrete logarithm by modulo $91$ and have form $k=r\pmod{m}$, where $r$=2..11 and $m$=3,4,6,12.

gp-code:

k91()=
{
 K= [];
 for(a=0, 90, if(gcd(a,91)==1, for(b=0, 90, if(gcd(b,91)==1,
  m= Mod(b, 91);
  h= znorder(m);
  k= znlog(a, m, h);
  if(k, if(k>1,
   K= concat(K, [[k,h]]);
  ))
 ))));
 print(vecsort(K,,8))
};

Output [r,m]:

[[2, 3], [2, 4], [2, 6], [2, 12], [3, 4], [3, 6], [3, 12], [4, 6], [4, 12], [5, 6], [5, 12], [6, 12], [7, 12], [8, 12], [9, 12], [10, 12], [11, 12]]


问:
Is there a simple explanation why degree 5 polynomials (and up) are unsolvable?
We can solve (get some kind of answer) equations like:

$$ ax^2 + bx + c=0$$

$$ax^3 + bx^2 + cx + d=0$$

$$ax^4 + bx^3 + cx^2 + dx + e=0$$

But why is there no formula for an equation like $$ax^5 + bx^4 + cx^3 + dx^2 + ex + f=0$$

I'm not sure if this has anything to do with the Galois theory, but is there a dumbed-down simple explanation as to why degree 5 polynomials (and up) are unsolvable?

答:
There is already one "dumbed" answer above, so forgive me for adding an even "dumber".

What it means by solvable is to be able to write down a formula for a solution of a polynomial equation like $ax^2 + bx + c = 0$ such this one:

${x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}}$

In other words, we need the solution (whatever it would be in $x$) could be expressed in some algebraic expressions. In the example above: $-, b, \pm, \sqrt, \: ^2, 4, a, c, /,$ and $2$ are all algebraic expressions.

For most of all numbers we know, they could be expressed in algebraic expressions. For example: $2, 6/3, \sqrt{2}, \sqrt{-1},$ etc and even for a non-repeating irrational number like $\pi$ could be expressed in $\frac{C}{D}$ where $C$ is the circumference of a circle and $D$ is the diameter.

Amazingly there are numbers that could not be expressed in any algebraic expressions we know.

Historically, it was discovered when observing that polynomial equations of degree higher than 4 may not necessarily have a solution that could be expressed in algebraic expressions. It was first observed by Joseph-Louis Lagrange in 1770, partly proven by Paolo Ruffini in 1799, and then completed by Niels_Henrik_Abel in 1824, establishing Abel–Ruffini theorem. But not until 1830 Évariste Galois at the very young age of 18 showed the necessary condition of a polynomial equation to have a solution hence called solvable.

The theory is called Galois theory, but unfortunately it is too complex to be explained satisfactorily here. If we have enough background in fundamental algebra, it would still need at least 200 hundreds pages of hard math to understand it. Nevertheless the necessary condition showed by Galois theory also shows that there are polynomial equations that their solutions could not be expressed in any algebraic expressions.

An example of a such unsolvable polynomial equation is $x^5 - x - 1 = 0$. Note to not be confused with the algebraic expression mentioned above. In the equation $x^5 - x - 1 = 0$, we cannot transform it into a solution expression where a single $x$ is put on the left side of the equation such $x = \: ...$ because, well, we don't have the algebraic expression to be written down in the right side of the equation.

Of course we could approximate the solution with some number close enough, but it is still NOT the solution to the equation because simply it is just close enough BUT not the exact solution. The exact solution is not expressible yet. Amazing right!


问:
Divide polynomials and conclude:
I have to divide $x^3-a^3$ by $x-a$ and conclude that $x^3 - a^3 = (x-a)*(x^2+ax+a^2)$, but, I'm not sure about how to divide this polynomials, basically doing what I think I should do I will get just "$x^2+a^2$", from where in the division came the "$ax$"?</p>

答:
here some prove i think you want to see this
$$(x-a)^3=x^3-y^3-3xa(x-a)$$
$$(x-a)^3+3xa(x-a)=x^3-y^3$$
$$(x-a)((x-a)^2+3xa)=x^3-y^3$$
$$(x-a)((x^2+a^2-2xa+3xa)=x^3-y^3$$
$$(x-a)((x^2+a^2+xa)=x^3-y^3$$

now we have value of   $x^3-y^3$ and we want to divide it (x-a).
so
$$\frac{(x-a)((x^2+a^2+xa)}{(x-a)}$$
clearly (x-a) is cancelling out above (x-a)
so answer will be
$$x^2+a^2+xa$$

Title: Divide polynomials and conclude:


问:
Propositional Logic "Riddle/Puzzle"
I have this kind of 'riddle' as a question that i need to complete, however I'm not sure what to do of it.

This is the question:

Determine who out of the following is guilty of doping. The suspects are: Sam, Michael, Bill, Richard, Matt.

1) Sam said: Michael or Bill took drugs, but not both.

2) Michael said: Richard or Sam took drugs, but not both.

3) Bill said: Matt or Michael took drugs, but not both.

4) Richard said: Bill or Matt took drugs, but not both.

5) Matt said: Bill or Richard took drugs, but not both.

^ Of these 5 statements, 4 are true, one is false.

6) Tom said: If Richard took drugs, then Bill took drugs.

^  This statement is guaranteed to be true.

From this information, I deduced:

p : Michael took drugs

q : Bill took drugs

r : Richard took drugs

s : Sam took drugs

t : Matt took drugs

So given this I came up with this:

1) (p ^ ~q) v (~p ^ q)

2) (r ^ ~s) v (~r ^ s)

3) (t ^ ~p) v (~t ^ p)

4) (q ^ ~t) v (~q ^ t)

5) (q ^ ~r) v (~q ^ r)

6) (~r v q)

However, I'm not sure where to go from here. I suppose, I could connect each statement with an ^ as that seems like the next step to do. Then that entire equation would essentially tell me who was guilty? The next step, would obviously be to simplify, and come up with a name. However, I'm not sure how to do this.

Could anyone please shed some light and give me some tips on how to do this?

Thanks.

答:
Based on what @Tim said I have framed the following answer:

Out of statements from 1 to 5, exactly one is false. While statement 6 is true. On perusal of statements 1 to 5, we see that statements, 1, 3 and 4 are mutually contradictory i.e. one of these three statements is false.
This implies statements 2, 5 and 6 are true.

2: Richard or Sam took drugs, but not both.

5 : Bill or Richard took drugs, but not both.

6: If Richard took drugs then Bill also took drugs.

We deduce that Sam and Bill took drugs, because if Richard is to take drugs then as per statement 6, Bill also took drugs however that will violate statement 5, hence Richard didn’t take drugs. Thus going by statement 2, Sam has taken drugs for sure.

1: Michael or Bill took drugs, but not both.

3: Matt or Michael took drugs, but not both.

4: Bill or Matt took drugs, but not both.

As Bill took drugs, then going by statements 1 and 4 Matt and Michael didn’t take drugs.

Thus, statement 3 is false, in the sense that neither of the two took drugs (this is where my logic is contradicting that of Tim's).
Therefore, Sam and Bill are the ones who took drugs.

Title: Propositional Logic "Riddle/Puzzle"


问:
To prove that a curve $\gamma_0$ is homotopic to another curve $\gamma_1$ as closed curves
Let U be an open subset of C. I need to prove the following

I am struggling with this, since I am new to topology and the idea of homotopy. I would appreciate some way to make headway.

I am adding here a definition of homotopy with fixed end points and closed curves,as requested by @paulFrost

The concatenation($\gamma_1+\gamma_2$) of two curves is defined as follows:

答:
Let $X$ be a topological space (if you want, you can restrict to $X = U \subset \mathbb C$). By $\mathcal P(X)$ let us denote the set of all curves $\gamma : [a,b] \to X$ which are defined on any closed interval $[a,b]$. Let $\mathcal L(X) \subset \mathcal P(X)$ denote the subset of all closed curves. We say that a closed curve $\gamma : [a,b] \to X$ is based at $\xi \in X$ if $\gamma(a) = \gamma(b) = \xi$.

On $\mathcal P(X)$ we have two equivalence relations:

Homotopic with fixed endpoints. Here the curves have to be defined on the same interval.

Equivalent with respect to order preserving reparameterization. Here the curves may be defined on different intervals.

On $\mathcal L(X)$ we have one more equivalence relation:

Homotopic as closed curves. Here the curves again have to be defined on the same interval.

Taking both 1. and 2. generates a more general equivalence relation on $\mathcal P(X)$. More precisely, given curves $\gamma_i : [a_i,b_i] \to X$, $i = 0,1$, we write   $\gamma_0 \equiv \gamma_1$ if there exists an order preserving reparameterization $u : [a_0,b_0] \to [a_1,b_1]$ such that $\gamma_0$ and $\gamma_1 \circ u$ are homotopic with fixed endpoints. It is easy to verify that this is in fact an equivalence relation. Moreover, the choice of $u$ is irrelevant, i.e. we may take any reparameterization $u$ to check whether $\gamma_0$ and $\gamma_1 \circ u$ are homotopic with fixed endpoints. In fact, if $u ,v  : [a_0,b_0] \to [a_1,b_1]$ are order preserving reparameterizations, then they are homotopic keeping the endpoints fixed (a homotopy is given by $H(s,t) = su(t) + (1-s)v(t)$; note that for $s \in (0,1)$ both $su$ and $(1-s)v$ are strictly increasing, thus also their sum is). This means that $\gamma_1 \circ u$ and $\gamma_1 \circ v$ are homotopic with fixed endpoints.

What you want to prove is this:

Let $\gamma_0, \gamma_1 :[a,b] \to X$ be two closed curves based at $\xi$. Then $\gamma_0, \gamma_1$ are homotopic as closed curves if and only there exists a closed curve $\gamma_2$ based at $\xi$ such that $\gamma_0 \equiv \gamma_2 + \gamma_1 + (-\gamma_2)$.

(1) Let  $\gamma_0, \gamma_1$ be homotopic as closed curves. Thus there exists a map $\gamma : [0,1] \times [a,b] \to X$ such that $\gamma(0,t) = \gamma_0(t)$ and $\gamma(1,t) = \gamma_1(t)$ for all $t \in [a,b]$ and $\gamma(s,a) = \gamma(s,b)$  for all $s \in [0,1]$. Define $\gamma_2 : [0,1] \to X, \gamma_2(t) = \gamma(t,a)$. Then $\bar \gamma = \gamma_2 + \gamma_1 + (-\gamma_2)$ lives on $[0,2+b-a]$. Note that
$$\bar \gamma(t) = \begin{cases}  \gamma_2(t) & 0 \leq t \leq 1 \\ \gamma_1(t - 1 +a ) & 1 \leq t \leq 1 + b - a \\ -\gamma_2(t - (2 + b -a))) = \gamma_2(2 + b -a - t) &  1 + b - a \leq t \leq 2 + b - a \end{cases}$$
Reparameterizing $\bar \gamma$ via $u : [a,b] \to [0,2+b-a], u(t) = p(t-a)$ with $p = \frac{2+b-a}{b-a}$, gives us a curve $\gamma^* = \bar \gamma \circ u$ and now it suffices to show that $\gamma_0, \gamma^*$ are homotopic with fixed endpoints. Note that
$$\gamma^*(t) = \begin{cases} \gamma_2(p(t-a)) & a \leq t \leq a + \frac{1}{p} \\ \gamma_1(a + \frac{b - a}{b - a - \frac{2}{p}}(t - a - \frac{1}{p})) & a + \frac{1}{p} \leq t \leq b - \frac{1}{p} \\ \gamma_2(p(b-t)) &  b - \frac{1}{p} \leq t \leq b \end{cases}$$
Define
$$\delta : [0,1] \times [a,b] \to X, \delta(s,t) = \begin{cases} \gamma_2(p(t-a)) & a \leq t \leq a + \frac{s}{p} \\ \gamma(s,a + \frac{b - a}{b - a - \frac{2s}{p}}(t - a - \frac{s}{p})) & a + \frac{s}{p} \leq t \leq b - \frac{s}{p} \\ \gamma_2(p(b-t)) &  b - \frac{s}{p} \leq t \leq b \end{cases}$$
This is a homotopy with with fixed endpoints such that $\delta(0,t) = \gamma_0(t)$ and $\delta(1,t) = \gamma^*(t)$.

(2) Let $\gamma_2 : [c,d] \to X$ be a closed curve based at $\xi$ such that $\gamma_0 \equiv \bar \gamma = \gamma_2 + \gamma_1 + (-\gamma_2)$. The latter lives on $[c,  2d - c + b-a]$. Reparameterize $\bar \gamma$ via $u : [a,b] \to [c,2d -c + b - a], u(t) = c + q(t-a)$ with $q = \frac{2d+b-a}{b-a}$ to obtain $\gamma^* : [a,b] \to X$. We know that $\gamma_0$ and $\gamma^*$ are homotopic with fixed endpoints. Thus there exists a map $\gamma : [0,1] \times [a,b] \to X$ such that $\gamma(0,t) = \gamma_0(t)$ and $\gamma(1,t) = \gamma^*(t)$ for all $t \in [a,b]$ and $\gamma(s,a) = \gamma(s,b) = \xi$  for all $s \in [0,1]$. Let $\partial R = [0,1] \times \{a,b\} \cup  \{0,1\} \times [a,b]$ be the boundary of $R = [0,1] \times[a,b]$. Consider the points $A = (0,a), B = (0,b), A' = (1/2,a), B' = (1/2,b), A'' = (1,a), B'' = (1,b)$ and $C = (1,u^{-1}(d)), D = (1,u^{-1}(d + b - a))$ on $\partial R$. There is a homeomorphism $h : \partial R \to \partial R$ which is the identity on the oriented line segment $\overline{AB}$ from $A$ to $B$ and maps linearly $\overline{AA'}$ to $\overline{AA''}$, $\overline{A'A''}$ to $\overline{A''C}$, $\overline{A''B''}$ to $\overline{CD}$, $\overline{B''B'}$ to $\overline{D'B''}$ and $\overline{B'B}$ to $\overline{B''B}$. Drawing a picture is helpful. There is an extension of $h$ to a homeomorphism $H : R \to R$. Define
$$\delta = \gamma \circ H : R \to X .$$
It is then easy to check that $\delta$ is a homotopy of closed curves from $\gamma_0$ to $\gamma_1$.


问:
Calculating Point on a Sphere
Given a sphere, I am attempting to calculate the coordinates of a point on that sphere given some initial point, the radius of the sphere, and some arbitrary distance d the point will travel (along with the vector it is headed). This sphere is a planet and the problem is essentially getting the coordinate of point #2 after traveling some distance.

I've been reading up on Spherical Trigonometry but I'm having a hard time applying it here. I'm thinking applying Rodrigues' formula might be a solution..?

I do have the North Pole position if that's any help at all.

CONTEXT

Planet is a perfect sphere with radius r. The coordinates are in a Cartesian plane with the center of the planet being (0, 0, 0). The "height" of the object relative to the surface of the sphere should remain the same.

答:
That's how I understand the problem in mathematical language: We have a sphere with radius $r$ and a Cartesian coordinate system with origin in the center of the sphere. Given is a point with coordinates $x_1,y_1,z_1$ on the sphere, that means $$x^2_1+y^2_1+z^2_1=r^2.$$ This point is to move for a distance $d$ on the sphere, starting in a direction given by a vector $\overrightarrow{v}=(v_x,v_y,v_z)$. Of course, it's crucial that this direction be tangential to the sphere (otherwise the point would take off into space, or bury into the ground), i.e. $$x_1v_x+y_1v_y+z_1v_z=0.$$ We'll assume that the length of this vector is one, i.e. $$v^2_x+v^2_y+v^2_z=1.$$ Then, the movement of the point is a rotation in the plane given by the origin, the point and the direction $\overrightarrow{v}$, the rotation angle defined by the arc, $$\alpha=\frac{d}r.$$ Since the normal vector $\overrightarrow{n}=(x_1/r,y_1/r,z_1/r)$ and $\overrightarrow{v}$ are orthogonal, this will transform $\overrightarrow{n}$ into $\cos\alpha\,\overrightarrow{n}+\sin\alpha\,\overrightarrow{v}$, so the new position is
$$r\,(\cos\alpha\,\overrightarrow{n}+\sin\alpha\,\overrightarrow{v})=(x_1\,\cos\alpha+r\,v_x\,\sin\alpha,y_1\,\cos\alpha+r\,v_y\,\sin\alpha,z_1\,\cos\alpha+r\,v_z\,\sin\alpha).$$ For a very small distance $d$, we have $\sin\alpha\approx\alpha=d/r$ and $\cos\alpha\approx1$, so this becomes just $(x_1+d\,v_x,y_1+d\,v_y,z_1+d\,v_z)$, as one would expect. If, instead, $d=\pi\,r$, i.e. we're travelling half the circumference, we have $\cos\pi=-1$ and $\sin\pi=0$, and the new position will be $(-x_1,-y_1,-z_1)$ irrespective of the direction $\overrightarrow{v}$.

Title: Calculating Point on a Sphere


问:
Is it possible to prove $a^{5n}b^{3m}c^{n}d^{m}$ is irregular knowing $a^nb^n$ is irregular?
The proof using the pumping lemma is super easy but is it possible to solve it without pumping lemma and knowing $a^nb^n$ is irregular ?

答:
we can use homomorphism to solve problem.
Define h(L) :
h(a) = a
h(b) = a
h(c) = bbbbb
h(d) = bbb

h(L) = a^(5n)a^(3m) (bbbbb)^n(bbb)^m =
a^(5n)a^(3m) b^(5n)b^(3m) = a^(5n+3m) b^(5n+3m)

Define C = 5n+3m  ---->  h(L) = a^C b^C
Then we can show that a^C b^C is irregular with pumping lemma. Or show a^C b^C is irregular with pigeonhole principle easily.
If h(L) is irregular, L is irregular too.
So we solve the problem :)


问:
What's the actual volume of a Great Stellated Dodecahedron? (Im getting seemingly different formulas)
So I was trying to derive the formula for the Great Stellated Dodecahedron starting from a unitary side length Dodecahedron. I managed only to get so far.

So when I went to check the actual formula, I found that Mathworld and WolframAlpha were giving me seemingly different formulas, but both assuming unit edge length.

The WolframAlpha formula:

$V = \frac{1}{4}(7 \sqrt5 - 15) $

The Mathworld formula:

$V = \frac{5}{4}(3 + \sqrt5 ) $

I could be misreading the entry on Mathworld (I don't speak english natively)
Now im curious about this difference, both are clearly formulas for something but they can't be both the volume of the Great Stellated Dodecahedron. What am I missing?

答:
Suppose the pentagram faces have unit edge length. Then the $20$ spikes of the great stellated dodecahedron have edge lengths $\varphi^{-3}$ (base) and $\varphi^{-2}$ (from tip) where $\varphi$ is the golden ratio. The base circumradius is $\frac1{\sqrt3}\varphi^{-3}$, yielding a height of $\varphi^{-3}\sqrt{\varphi^2-\frac13}$ and a volume for each spike of
$$\frac13×\frac{\sqrt3}4\varphi^{-6}×\varphi^{-3}\sqrt{\varphi^2-\frac13}=\frac1{12}\varphi^{-9}\sqrt{3\varphi^2-1}=\frac1{12}\varphi^{-7}$$
The whole solid has $20$ spikes, so we multiply this by $20$ to get the spike volume as $\frac53\varphi^{-7}$. To this we add the hidden icosahedron's volume of
$$\frac5{12}(3+\sqrt5)\varphi^{-9}=\frac56\varphi^{-7}$$
and obtain the volume of the great stellated dodecahedron as
$$\frac52\varphi^{-7}=0.086104\dots$$

We can rewrite the MathWorld $V$ as $\frac52\varphi^2$. This means the MathWorld formula is correct if the edge length of the hidden icosahedron (length of each side of the spike base) is $1$. The Wolfram Alpha formula seems completely wrong.

MathWorld says

Cumulating a unit dodecahedron to produce a great stellated dodecahedron…

"Cumulation" is another word for augmentation, adding pyramids to each face of the original polyhedron.

This is rubbish. You need to augment an icosahedron to produce a great stellated dodecahedron.


问:
Technical explanation for an obvious discontinuity
Say $f(x)= \frac{1}{x^2-17x+66}$, and you're supposed to find where $f(\frac{2}{x-2})$ is discontinuous. It's pretty clear that, apart from the points you get from the roots of the binomial in the denominator, the other required point would be $x=2$, since $f(\frac{2}{x-2})$ becomes $f(\frac{2}{0})$.

Is there a technical explanation for this, though? Because when you substitute $x$ for $\frac{2}{x-2}$ in the original expression for $f(x)$, the result is $\frac{(x-2)^2}{4-34(x-2)+66(x-2)^2}$. Evaluating this with $2$ for $x$ gives the final answer $0$, not an undefined value.

Why is this way of looking at it incorrect?

答:
Substituting $\frac 2 {x-2}$ for $x$ into the expression for $f(x)$ does not yield $\frac{(x-2)^2}{4-34(x-2)+66(x-2)^2}$, it yields
$$
\frac{1}{\left(\frac 2 {x-2}\right)^2-17\cdot\frac 2 {x-2}+66},
$$
which is still undefined for $x=2$. The simplification you did yields a term that can then be evaluated at $x=2$, but the defining term can't. While this leaves $x=2$ a value for which $f\left(\frac 2 {x-2}\right)$ is not defined, it does show that this gap in the domain can be closed: there is a unique continuous extension of $f\left(\frac 2 {x-2}\right)$ at $x=2$. This is an example of a removable singularity.

An easier example would be $g(x) = \frac{x}{x}$, which has maximal domain $\mathbb R\setminus\{0\}$, since $\frac 0 0$ is not defined. However, both one sided limits $\lim_{x\to 0^-} \frac x x$ and $\lim_{x\to 0^+} \frac x x$ exist and are equal to $1$. Hence $\hat g(x)=1$ is the unique continuous extension of $g$ to all of $\mathbb R$.


问:
Can a disconnected set disconnect a connected set?
Suppose $V$ is a connected open set in $\mathbb{R}^n$ and $F$ is a closed set in $\mathbb{R}^n$. If $V\setminus F$ is disconnected, can we conclude that $F\cap V$ is connected? The answer for $n=1$ is obviously no and we can take as a counter example $V=\mathbb{R}$ and $F=[-1,0]\cup[1,2]$, for example. But what if $n>1$? Is the answer still no?

答: Take $V=\mathbb{R}^n$ and $F = \{x\in \mathbb{R}^n:x_n=0 \text{ or } x_n=1\}$. Then $V\setminus F$ and $V\cap F$ are disconnected.


问:
How do I reason with Skolem functions?
Given the first-order sentence.

$\forall x \forall y : (Overlap(x,y) \iff  (\exists z Part(z,x) \land Part(z,y)))~~~~~~~~~~~~~~~~$  (1)

Here is an outline (partial) proof that $Overlap$ is symmetric:

Proof 1
\begin{align}
& \text{1}~~\forall x \forall y : (Overlap(x,y) \iff  (\exists z Part(z,x) \land Part(z,y))) & \tag{Axiom}\\
&  \text{2}~~~~~~~Overlap(a,b) \tag{Assume} \\
&  \text{3}~~~~~~~\exists z (Part(z,a) \land Part(z,b)) \tag{1,2}\\
&  \text{4}~~~~~~~\exists z (Part(z,b) \land Part(z,a)) \tag{3}\\
&  \text{5}~~~~~~~Overlap(b,a)  \tag{1,2,4}\\
&  \text{6}~~\forall x \forall y : (Overlap(x,y) \implies Overlap(y,x)) \tag{$\forall$-Intro 2-5} \\
\end{align}
Note in line 4 the same existential $z$ is used.

Now I wish write the  axiom (1) in an equisatisfiable form that can be proved using equational logic and term rewriting. This usually requires Skolemization.  Sentence (1) can be Skolemized as follows:

$\forall x \forall y : (Overlap(x,y) \iff   (Part(Sk(x,y),x) \land Part(Sk(x,y),y))~~~~~~~~~~~$ (2)

Now I try to prove symmetry using (2) with a similar approach as in Proof 1:

Proof Attempt 2
\begin{align}
& \text{1}~~\forall x \forall y : (Overlap(x,y) \iff   (Part(Sk(x,y),x) \land Part(Sk(x,y),y))& \tag{Axiom}\\
&  \text{2}~~~~~~~~Overlap(a,b) \tag{Assume} \\
&  \text{3}~~~~~~~~Part(Sk(a,b),a) \land Part(Sk(a,b),b)  \tag{1,2}\\
&  \text{4}~~~~~~~~Part(Sk(a,b),b) \land Part(Sk(a,b),a)  \tag{3}\\
&  \text{5}~~~~~~~~Overlap(b,a)\tag{XX}\\
&  \text{6}~~\forall x \forall y : (Overlap(x,y) \implies Overlap(y,x)) \tag{XX} \\
\end{align}

I am stuck at lines 4-5 in Proof Attempt 2.By following the approach of Proof 1 there is an issue with the order of arguments to Skolem functions.

答:
We can translate the original bi-conditional axiom into three implications. Using the quantifier equivalence, $\exists x (Q(x) \implies P) \equiv \forall x Q(x) \implies P $, the existential in the original  (1) can be replaced by a universal $z$, so line 3 below does not need a Skolem function. I believe these 3 sentences and the FOL sentence (1) are equisatisfiable. I believe breaking  (1) into three parts makes the proof fairly straightforward.

Proof
\begin{align}
& \text{1}~~\forall x \forall y : (Overlap(x,y) \implies  Part(Sk(x,y),x)) & \tag{Axiom1}\\
& \text{2}~~\forall x \forall y : (Overlap(x,y) \implies  Part(Sk(x,y),y))& \tag{Axiom2}\\
& \text{3}~~\forall x \forall y \forall z : ((Part(z,x) \land Part(z,y)) \implies Overlap(x,y))& \tag{Axiom3}\\
&  \text{4 (a,b)}~~Overlap(a,b) \tag{Assume} \\
&  \text{5}~~~~~~~~~~~~~(Part(Sk(a,b) ,a) \land Part(Sk(a,b) ,b))  \implies Overlap(a,b)   \tag{$\forall$-Elim 3}\\
&  \text{6}~~~~~~~~~~~~~Overlap(a,b)  \implies  Part(Sk(a,b),a)   \tag{$\forall$-Elim 1}\\
&  \text{7}~~~~~~~~~~~~~Overlap(a,b)  \implies  Part(Sk(a,b),b)   \tag{$\forall$-Elim 2}\\
&  \text{8}~~~~~~~~~~~~~Part(Sk(a,b),a)  \tag{$\implies$-Elim 4,6}\\
&  \text{9}~~~~~~~~~~~~~Part(Sk(a,b),b)  \tag{$\implies$-Elim 4,7}\\
&  \text{10}~~~~~~~~~~~~~Part(Sk(a,b),b)  \land  Part(Sk(a,b),a)  \tag{$\land$-Intro 8,9}\\
&  \text{11}~~~~~~~~~~~~~(Part(Sk(a,b) ,b) \land Part(Sk(a,b) ,a))  \implies Overlap(b,a)   \tag{$\forall$-Elim 3}\\
&  \text{12}~~~~~~~~~~~~~Overlap(b,a) \tag{$\implies$-Elim 10,11}\\
&  \text{13}~~\forall x \forall y : (Overlap(x,y) \implies Overlap(y,x)) \tag{$\forall$-Intro 4-12} \\
\end{align}

Title: How do I reason with Skolem functions?


问:
Does this patter occur somewhere? Does it have a name?
Consider the following pattern:

```<code>n = 1 -> (1)
n = 2 -> (2,0), (1,1)
n = 3 -> (3,0), (2,1), (1,1,1)
n = 4 -> (4,0), (3,1), (2,2), (2,1,1), (1,1,1,1)
</code>```

Does this occur somewhere in mathematics (or Computer Science)? Does it have a name?

答:
Unordered number theoretic partitions:

$P(n,k)$ is the number of words $x_1\ldots x_k$ of length $k$ over the alphabet $\Bbb N$ such that $x_1\geq x_2\geq \ldots \geq x_k$ and $x_1+\ldots+x_k=n$.

For instance, $n=5$:

$k=1$: $5$

$k=2$: $41,32$

$k=3$: $311,221$

$k=4$: $2111$

$k=5$: $11111$.

There is an iterative formula for $P(n,k)$, but no closed form.


问:
How can I describe the pattern of Fibonacci numbers modulo 5 non-recursively?
In my work, I've proved that the period of the pattern is 20, and so you can state that $[F_{[n]_{20}}]_5$ is equal to $[F_n]_5$. But this is still a recursive description of the pattern right? Is there any way that I can describe the pattern without using recursion?

答:
One could use the methods explained in https://en.m.wikipedia.org/wiki/Linear_difference_equation , but work in the field of residues modulo $5$, which I will call $\mathbb Z_5$. In this field, the characteristic polynomial of the Fibonacci sequence:

$$x^2-x-1$$

has a double zero $x=3$, so the general solution of the Fibonacci recurrence relation in $\mathbb Z_5$ is:

$$F_n=(a+bn)\cdot 3^n$$

Tailoring $a,b$ to the initial conditions $F_0=0, F_1=1$ yields $a=0,b=2$, so: $F_n=2n\cdot 3^n$, or, back in $\mathbb Z$ we have $F_n\equiv 2n\cdot 3^n\pmod 5$.


问:
Solve the differential equation with variable separable method
I am not able to figure out how to integrate after separating

Equation is:
$$y\sqrt{1-x^2}dy+x\sqrt{1-y^2}dx=0$$

In my book there is no solution but have answer:
$$\sqrt{1-x^2}+\sqrt{1-y^2}=c$$

答:
Note that $\int \frac{v}{\sqrt{1-v^2}}dv=-\sqrt{1-v^2}+c_1$ which can be obtained by the substitution $u=1-v^2$ for example.

Then separating the differential equation we have $$y\sqrt{1-x^2}dy+x\sqrt{1-y^2}dx=0$$
$$y\sqrt{1-x^2}dy=-x\sqrt{1-y^2}dx$$
$$\frac{y}{\sqrt{1-y^2}}dy=-\frac{x}{\sqrt{1-x^2}}dx$$

and integrating gives $$-\sqrt{1-y^2}+c=\sqrt{1-x^2}$$ or $$\sqrt{1-x^2}+\sqrt{1-y^2}=c$$


问:
Find all natural $n$ numbers such that
Find all natural $n$ numbers such that
$15(n!)^2+1$ is divisible by $2n-3$.
My try: First I assumed $2n-3$ is not prime number. Let $a$ be divisor of $2n-3$. It's clear that $a<n-1$, so $15(n!)^2$ is divisible by $a$. Which means $15(n!)^2+1$ is not divisble by $a$. But it is given that $15(n!)^2+1$is divisble by $n$ which means it is divisible by $a$ too. But we have already proved it is not. Contradiction!
So $2n-3$ must be prime number. Now if we change $2n-3$ as $p$. We can say $15*((p+3)/2)!*((p+3)/2)!$ is congruent to $-1$ by p module. By Wilson's theorem $15*((p+3)/2)!*((p+3)/2)!$ is congruent to $(p-1)!$ by p module. From here I don't know how to continue.

答:
You do need Wilson's Theorem, but observe:

\begin{align}n &\equiv -(n-3) \pmod {2n-3}\\
n-1 &\equiv -(n-2) \pmod {2n-3}\\
&\ \vdots\\
1 &\equiv -(2n-4)\pmod {2n-3}\end{align}

This gives:

\begin{align}(n!)^2&\equiv n!(-1)^n(2n-4)(2n-5)\dots(n-2)(n-3) \pmod {2n-3}\\&\equiv(-1)^n(2n-4)!(n)(n-1)(n-2)(n-3) \pmod {2n-3}\\\text{(Wilson)} &\equiv(-1)^{n+1}(n)(n-1)(n-2)(n-3)\pmod {2n-3}\\
&\equiv(-1)^{n+1}n^2(n-1)^2 \pmod {2n-3}\end{align}

Now we seperate into cases where $n$ is odd or even. For $n$ odd satisfying the division condition,

$$15(n!)^2+1\equiv 15n^2(n-1)^2+1 \equiv 0 \pmod {2n-3}$$

$z = \dfrac {15n^2(n-1)^2+1}{2n-3}$ is an integer iff $16z$ is. Simplifying we have

$$16z=120 n^3 - 60 n^2 + 30 n + 45 + \frac {151} {2 n - 3}$$

So the above is an integer only if $2n-3$ divides $151$, which is a prime, giving $2n-3 = 151$, $n = 77$.

The case for $n$ even should be similar. Technically, you also need to consider the cases $2n-3 = \pm 1$ separately, since $\pm1$ are neither prime nor composite.

Title: Find all natural $n$ numbers such that


问:
Closed image of $T_0$-space
Let $X$ and $Y$ be topological spaces, $X$ is $T_0$ and suppose $f\colon X\to Y$ is a continuous closed surjection. Must $Y$ be $T_0$?</p>

答: HINT: For each $n\in\Bbb N$ let $V_n=\{k\in\Bbb N:k<n\}$, and let $\tau=\{\Bbb N\}\cup\{V_n:n\in\Bbb N\}$; then $\langle\Bbb N,\tau\rangle$ is a $T_0$-space. Take it as your $X$, and let $Y$ be the two-point space with the indiscrete topology. Can you find a surjection $f:\Bbb N\to Y$ that is continuous and closed, something a bit similar to Arthur Fischer’s example in his answer to your previous question?</p>

Title: Closed image of $T_0$-space


问:
Remainder for a GP
What is the remainder of  : $\sum_{i=0}^{2019} 3^i$  divided by $3^4$ ?

I know that $$\sum_{i=0}^{2019} 3^i = \frac{3^{2020}}{3-1} = \frac{3^{2020}}{2} $$

so  $$\frac{\sum_{i=0}^{2019} 3^i}{3^4} = \frac{3^{2016}}{2}$$

how do I proceed from here? I tried running a program in python on it but I think the number is too big

Title: Remainder for a GP

Body: Hint: Your sum is $3^0+3^1+3^2+3^3+3^4(3^0+3^1+...+3^{2015})$.


问:
how to calculate $\lim_{x\to 0} \frac{x^{3}-2\sin(x)+\sin(2x)}{\arctan(x^{3})-\arctan^{3}(x)}$
is there any way to solve this limit I tried many tricks but all of them leads me to an Indeterminate form.
$$\lim_{x\to 0} \dfrac{x^{3}-2\sin(x)+\sin(2x)}{\arctan(x^{3})-\arctan^{3}(x)}$$

sorry, it was a typo I rectified.

答:
You can use the first few terms of the power series expansions of $\sin $ and $\arctan $ to find that for small $x$ the numerator is $2x + \cdots$ and the denominator is $x^5+\cdots$, so that the limit will be infinite.  A calculator will illustrate the point.

A more interesting limit is
$$
\lim_{x\to0} \frac{x^3-2 \sin x + \sin 2x}{\arctan x - \arctan^3 x}
$$
in which the same approach tells you the limit is $1/4$.

Seeing the question has been corrected, I can expand on the 'more interesting' limit.  Using power series expansions, the numerator becomes $x^5/4 + \cdots$ and the denominator is still $x^5 + \cdots$ where the dots indicate higher powers of $x$.  Thus the limit can be seen to be $1/4$


问:
Show that Riemann sum does not depend on rectangle endpoint choice
To calculate a Riemann sum by assuming right endpoints, we have $$ S= \lim_{n\to\infty} { \sum_{i=1}^{n} f(x_i^*)\Delta x } =  \lim_{n\to\infty} { \sum_{i=1}^{n} f\left( i\Delta x+ a\right)\Delta x .}$$

Then, to calculate $S$ using any endpoint, I introduce a parameter $\alpha, \quad 0\leq\alpha\leq1$ such that $$ S =  \lim_{n\to\infty} { \sum_{i=1}^{n} f\left( (i-\alpha)\Delta x+ a\right)\Delta x }.$$ For $\alpha = 0$, we have right endpoints, for $\alpha = 1$ we use left endpoints, for $\alpha = \tfrac{1}{2}$ we use mid point, and so on.

How could I prove that $S$ does not depend on $\alpha$ for any integrable $f$ analytically? I have done so numerically, but am struggling with the analytical method. Since I am not working with a particular function, I can't use the classic textbook formulas for the series for $\{i^n\}$. I would be inclined to try to show that partial derivative wrt $\alpha$ is 0, but I'm not sure under which conditions I can pass a derivative operator through an infinite sum.

Are there any helpful documents/references/answers you could give me?

EDIT: mistake.

答:
A good reference is Wikipedia Riemann integral article.

The article mentions two definitions of Riemann integral and proves that those are equivalent.

What you ask is related to that. The proof goes through Darboux integral.

It only supposes that the function $f$ is Riemann integrable. Easier proves can be done with further assumptions of $f$ like you devise, i.e. if $f$ is supposed to be continuously differentiable.


问:
Use skills to solve this integral
How to use technique to integrate this integral? $\int_{0}^{\frac{\pi}{4}} \cos t\cdot\sin t e^{\sin t}-\sin^2t \cdot \tan te^{\cos t} +\tan te^{\tan t} dt$

I use integration by parts for many many times.

答:
Let us consider the three antiderivatives
$$I_1=\int   \cos (t)\sin (t)\,e^{\sin (t)} dt$$ Let $\sin(t)=x$ to arrive at
$$I_1=\int x e^x \,dx= (x-1)e^x= (\sin (t)-1)\,e^{\sin (t)}$$
$$I_2=\int \sin ^2(t)  \tan (t)\,e^{\cos (t)}\,dt$$ Let $\cos(t)=x$ to arrive at
$$I_2=\int \frac{e^x \left(x^2-1\right)}{x}\,dx=\int x e^x \,dx-\int\frac{e^x} x \,dx=(x-1)e^x-\int\frac{e^x} x \,dx$$ that is to say
$$I_2=(\cos (t)-1)\,e^{\cos (t)}-\int \frac {e^{\cos(t)}}{\cos(t)} dt$$
$$I_3=\int  \tan (t)\, e^{\tan (t)}\,dt$$ Let $\tan(t)=x$ to arrive at
$$I_3=\int\frac{x\,e^x }{x^2+1}\,dx=\frac{1-i}2\left(\int \frac{e^x}{x+i}\,dx-\int \frac{e^x}{x-i}\,dx \right)$$ All the unsolved integrals are very simple if you are familiar with the exponential integral function.

If you are not, compute
$$I_3=\int  \tan (t)\, e^{\tan (t)}\,dt \qquad \text{and} \qquad I_4=\int \frac {e^{\cos(t)}}{\cos(t)} dt $$ using series expansions and integrate termwise.

Finsih using the bounds.

Title: Use skills to solve this integral


问:
Calculating the Variance of $a^x, a \in R$
Please give me a hint/direction and not a full answer, as those are my homeworks

I have a question where $X \sim Po(3)$ and i need to calculate: $Var[2^X]$.

What i tried:

$$
2^x = e^{\ln 2^x} = e^{x\ln 2} \Rightarrow V[2^x] = V[e^{x\ln 2}]
$$

On the other hand, i know that $X \sim Po$ and i know that the $Var$ of a poisson distribution equalls $\lambda$ so i think i should use it somehow but I'm not sure how.

Here im stuck, i know its not much but i dont have an idea.

Thank you.

答: You can use the MGF of poisson distribution: $M_X(t)=\Bbb E[(e^t)^X]\implies V[e^{X\ln 2}]=\Bbb E[e^{X\ln4}]-\Bbb E[e^{X\ln2}]^2=M_X(\ln 4)-M_X(\ln2)^2$


问:
How to solve $x^2 \equiv 12 \pmod {13}$?

Solve $x^2 \equiv 12 \pmod {13}$

By guessing I can say that the solutions are $5$ and $8$, but is there another way to find the solution besides guessing?

答:
In the field ${\displaystyle \mathbb {Z} /13\mathbb {Z}}$, $\,[1] + [1] \ne [0]$, and therefore there are either zero or two distinct $\text{modulo-}13$ solutions for,

$\tag 1 x^2 \equiv 12 \pmod{13}$

When one solution $[u]$ has been found the other solution is $-[u]$.

We have

$\; \large x^2 \equiv 12 \pmod{13} \; \text{ iff }$

$\; \large x^2 \equiv 2^{-2} \cdot (2^2 \cdot 12) \pmod{13} \; \text{ iff }$

$\; \large x^2 \equiv 2^{-2} \cdot (3^2) \pmod{13} \; \text{ iff }$

$\; \large x^2 \equiv \bigr(2^{-1} \cdot 3^1\bigr)^2 \pmod{13}$

Now the inverse of $[2]$ is easily calculated,

$\quad [2]^{-1} = [\frac{13 + 1}{2}] = [7]$

and so one solution to $\text{(1)}$ is given by

$\quad x \equiv 7 \cdot 3 \equiv 8 \pmod{13}$

The other solution is given by

$\quad x \equiv -8 \equiv 5 \pmod{13}$

Note: Examining this question resulted in a conjecture,

$\quad$ A new method for finding a solution (when they exist) to $x^2 = a \pmod p$?


答: As seen in this other question, the expectation is easy to compute.  The expectation of the square is also easy, since $$\operatorname{E}[(2^X)^2] = \operatorname{E}[4^X].$$  Therefore, $$\operatorname{Var}[2^X] = \operatorname{E}[4^X] - \operatorname{E}[2^X]$$ and use the answer provided in the link.

Title: Calculating the Variance of $a^x, a \in R$

Body:
Please give me a hint/direction and not a full answer, as those are my homeworks

I have a question where $X \sim Po(3)$ and i need to calculate: $Var[2^X]$.

What i tried:

$$
2^x = e^{\ln 2^x} = e^{x\ln 2} \Rightarrow V[2^x] = V[e^{x\ln 2}]
$$

On the other hand, i know that $X \sim Po$ and i know that the $Var$ of a poisson distribution equalls $\lambda$ so i think i should use it somehow but I'm not sure how.

Here im stuck, i know its not much but i dont have an idea.

Thank you.


问:
Do continuous injections preserve open sets?
Do continuous injections preserve open sets?

I'm pretty sure that's true in euclidean space.

If we let the singleton sets of integers generate the topology of the domain, and then identity map it to the real with standard topology, is that a counterexample?

If they don't, what combination of injective, surjective, continuous, and inverse continuous is the minimum to be an open map?

Edit:  Either I was really tired and distracted on the bus when I typed this into my phone and somehow forgot to say continuous (entirely possible), or whoever put "from $\mathbb{R}^m$ to $\mathbb{R}^n$" in my title deleted it.  Should I start a new one?

答:
It is not true over multiple Euclidean spaces.

Just consider the natural projection $f: \mathbb R \to \mathbb R^2$ where $f(x) = \left[\begin{matrix}x \\ 0\end{matrix}\right]$. $f(S)$ is never open for $\emptyset \ne S \subset \mathbb R$, because for any $x \in S$, $f(x) + \left[\begin{matrix}0 \\ r\end{matrix}\right]$ is not in the open ball $B(f(x), 2r)$. Illustratively, $f((0, 1))=(0,1) \times \{0\}$ which is obviously not open (nor closed).


问:
When $f(x+1)-f(x)=f'(x)$, what are the solutions for $f(x)$?
The question is:

When $f(x+1)-f(x)=f'(x)$, what are the solutions for $f(x)$?

The most obvious solution is a linear function of the form $f(x)=ax+b$. Is this the only solution?

Edit

I should add that $f:\mathbb R\to\mathbb R$ to the question.

答:
This is an example of a delay differential equation, and it has a very large number of solutions.  I will begin by discussing solutions $f\colon[0,\infty)\to\mathbb{R}$, and then extend to solutions on the real line.

Suppose first that we have defined $f$ on the interval $[0,1]$, say $f(x) = f_0(x)$ for some differentiable function $f_0\colon[0,1]\to\mathbb{R}$.  For extending solutions forwards, the best way to write the equation is
$$
f(x+1) \;=\; f(x) + f'(x).
$$
Using this equation, it is not hard to define $f$ on the interval $[1,2]$. Specifically, if $x\in[1,2]$, then
$$
f(x) \;=\; f_0(x-1) + f_0'(x-1).
$$
Note that this function will only be differentiable if $f_0$ was twice differentiable on $[0,1]$.  Moreover, the derivatives at $1$ will only match up if $f_0'(1) = f_0'(0) + f_0''(0)$.  Assuming this is the case, it is easy to extend again to $[2,3]$.  Specifically, if $x\in[2,3]$, then
$$
f(x) \;=\; f(x-1) + f'(x-1) \;=\; f_0(x-2) + 2 f_0'(x-2) + f_0''(x-2).
$$
In general, assuming that:

The function $f_0$ is $C^\infty$ on $[0,1]$ and,

$f_0^{(k)}(1) = f_0^{(k)}(0) + f_0^{(k+1)}(0)$ for all $k\geq 0$,

we can extend $f$ to all of $[0,\infty)$ by the formula
$$
f(x) \;=\; \sum_{k=0}^{\lfloor x\rfloor} \,\binom{\lfloor x\rfloor}{k}f_0^{(k)}(x-\lfloor x\rfloor)
$$

Now let's discuss how to extend backwards.  Assuming we have defined $f$ on $[0,\infty)$, the formula for $f$ on $[-1,0]$ is determined by the initial-value problem
$$
f'(x) + f(x) \;=\; f_0(x+1),\qquad f(0) = f_0(0).
$$
This is a first-order differential equation, so it must have a solution.  Moreover, assuming $f_0$ is $C^\infty$ on $[0,1]$, it follows from the Picard-Lindelöf Theorem that there is a unique $C^\infty$ solution for $f$ on $[-1,0]$.  Note that the derivatives at $0$ will automatically match, since $f_0(1) - f(0) = f_0'(0)$.  Continuing in this fashion, we can extend any solution backwards all the way to $-\infty$.

We conclude that any function $f_0\colon[0,1]\to\mathbb{R}$ satisfying the two conditions listed above extends to a solution on the entire real line.

Edit: Incidentally, there is a simple integral formula for $f$ on $(-\infty,0]$, which can be used to compute values for $f$ on a computer:
$$
f(x) \;=\; f(0) + e^{-x-1}\int_{x+1}^1 e^{t}\,f(t)\,dt
$$
This can also be used to provide an alternative proof that the function extends backwards, without appealing to existence and uniqueness of ODE's.  I found this formula by solving the initial value problem $f(x) + f'(x) = f(x+1)$ using integrating factors.


问:
Is it appropriate to use minus-plus with inequalities?
Is it appropriate to use $\mp$ with inequalities instead of using it with a $\pm$ , like in the following examples?

$$\mp\ 2\leq 2\implies -2<2\ ,\ 2=2\\ \text{or}\\
x+4\ge\mp\ (x+4)\implies x+4>-(x+4)\ ,\ x+4=x+4$$

答:
Generally speaking, $\mp$ signs are used sparingly in mathematics. Here is an example of when it would be appropriate to use this symbol:
$$
\cos (\alpha \pm \beta) = \cos \alpha \cos \beta \mp \sin \alpha \sin \beta \, .
$$
The rationale behind using it here is that when you are adding $\alpha$ and $\beta$ on the LHS, you are  subtracting $\cos \alpha \cos \beta$ and $\sin \alpha \sin \beta$ on the RHS. But this 'add-subtract' guideline is, well, just a guideline. If $x=\pm 5$, then you would be forgiven for writing $-x = \pm 5$ rather than $-x = \mp 5$. I would especially warn about using this symbol when solving inequalities, as the mental effort of keeping track of the 'flips' between $\pm$ and $\mp$ might make you more prone to error.

Overall, mathematical notation is not set in stone—it's about clear communication. Sometimes being pedantic about whether you should use $\pm$ or $\mp$ can do more harm than good.


问:
Showing that this process is a Markov Chain
I have a Markov chain $\{X_n\}$, with the states $S=\{0,1\}$ and the probability matrix

$$p_{11}=1-a \quad p_{12}=a \quad p_{21}=b \quad p_{22}=1-b \quad 0<a, b<1$$

I want to show that the stochastic process $Z_n=(X_{n-1},X_n) \quad n \geq 1,$ is the Markov chain and write its probability matrix.

I don't know where to start and what does it really mean that $Z_n=(X_{n-1},X_n)$. Thank you for your help!

答:
There are $4$ states for $Z_n$, namely
$$
(1,1),\;\;(1,2),\;\;(2,1),\;\;(2,2)
$$
For example, $Z_n$ is in state $(1,2)$ when $X_{n-1}=1$ and $X_n=2$.

Note that the value of $X_{n-1}$ plays no role in the transition from $Z_n=(X_{n-1},X_n)$ to $Z_{n+1}=(X_n,X_{n+1})$.

Based on that observation, I've filled in the second row of the transition probability matrix for $Z_n$.
\begin{array}{|c|c|c|c|c|}
\hline
&(1,1)&(1,2)&(2,1)&(2,2)\\
\hline
(1,1)&&&&\\
\hline
(1,2)&0&0&b&1-b\\
\hline
(2,1)&&&&\\
\hline
(2,2)&&&&\\
\hline
\end{array}
In the above example, from state $(1,2)$, the next state must be either $(2,1)$ or $(2,2)$, and the probabilities for those two outcomes are $p_{21}$ and $p_{22}$ respectively.

See if you can fill in the rest.


问:
Discrete change of multiple variables?
Suppose I have a sum of the form:

$$ \sum_{j=0}^n \sum_{i=0}^m c_{i,j}$$

And I sub :

$$ j+ i \to x$$

And,

$$ i-j \to y$$

Then:

$$ \frac{i+j}{2} = x$$

And,

$$ \frac{i-j}{2} = y$$

If I sub:

$$ \sum_{j=0}^n \sum_{i=0}^m c_{ \frac{x+y}{2} , \frac{x-y}{2} }$$

How do I figure out the new bounds of summation..?

答:
I will give an answer for the infinite case:

$$\sum_{j=0}^{\infty} \sum_{i=0}^{\infty}  c_{i,j}=\sum_{x=0}^{\infty} \sum_{\begin{matrix}y=-x\\ \color{red}{\text{step} \ 2} \end{matrix}}^{y=x} c_{x,y} $$

The "$\color{red}{\text{step} \ 2}$" meaning that the values taken by $y$ are:

$$y=-x, \ -x+2, \ -x+4 \ \cdots x. \tag{1}$$

Why that ? let us write the $i,j$ tables under this form:

$$\text{Abscissas i : } \ \ \begin{matrix}
..&..&..&..&..&\\
0&1&2&3&4&\cdots\\
0&1&2&3&4&\cdots\\
0&1&2&3&4&\cdots\\
0&1&2&3&4&\cdots\\
0&1&2&3&4&\cdots\\
\end{matrix} \ \ \ \ \
\text{Ordinates j : } \ \ \begin{matrix}
..&..&..&..&..&\\
4&4&4&4&4&\cdots\\
3&3&3&3&3&\cdots\\
2&2&2&2&2&\cdots\\
1&1&1&1&1&\cdots\\
0&0&0&0&0&\cdots\\
\end{matrix}$$

From which we deduce the tables of $x$ and $y$:

$$\text{Values of  x : } \ \ \begin{matrix}
..&..&..&..&..&\\
4&5&6&7&8&\cdots\\
\color{blue}{3}&4&5&6&7&\cdots\\
2&\color{blue}{3}&4&5&6&\cdots\\
1&2&\color{blue}{3}&4&5&\cdots\\
0&1&2&\color{blue}{3}&4&\cdots\\
\end{matrix} \ \ \ \ \ \ \
\text{Values of y : } \ \ \begin{array}{rrrrrc}
..&..&..&..&..&\\
-4&-3&-2&-1&0&\cdots\\
\color{red}{-3}&-2&-1&0&1&\cdots\\
-2&\color{red}{-1}&0&1&2&\cdots\\
-1&0&\color{red}{1}&2&3&\cdots\\
0&1&2&\color{red}{3}&4&\cdots\\
\end{array}$$

(check that $i+j=x$ and $i-j=y$).

In the table of $x$ values, one sees that $x$ can take (along diagonals with "equations" $i+j=x$, one of them has been highlighted in $\color{blue}{\text{blue}}$) all positive integer values, indicating that it has to be taken as the "driving" variable. $x$ being fixed, the corresponding values taken by $y$ (in $\color{red}{\text{red}}$) are indeed described by (1).

Title: Discrete change of multiple variables?


问:
Self Inverse in Integral Domains
I need to show that if R is an integral domain and it has unity then the only elements of R which are inverse to itself are 1 and -1 (with respect to multiplication). But I don't know where to start. Can anyone help? Thanks..</p>

Title: Self Inverse in Integral Domains

Body:
Here's a hint:
$$(x+1)(x-1)=x^2-1.$$


问:
What are the equations for the modeling?
We have a system of spring-mass-spring-mass-spring-mass and required to write the motion equation of the described system.

If I apply hooke's law to the system I got the following :

Are these the correct equation form?

答:
The movement equations can be derived from the Lagrangian $L = T - U$ with

$$
\cases{
U = m g \sum_{k=1}^n x_k\\
T =\frac 12 m\sum_{k=1}^n \dot x_k^2+\frac 12 k\sum_{k=1}^n(x_{k-1}-x_k)^2
}
$$

with $x_0 = 0$


问:
Solve the quadratic equation with complex number $b$
Solve the quadratic equation:
$$x^2 +xi+2=0$$

I was thinking to set $x^2=-2-xi$ but what $i$ have to do when $x+yi$ in this case $x=-2$ and $y=x$...

答:
Exactly the same as in the real case...taking into account that we are working in $\;\Bbb C\;$ , of course:

$$x^2+ix+2=0\implies \Delta=i^2-4\cdot2=-9=(3i)^2$$

so the solutions are

$$x_{1,2}=\frac{-i\pm\sqrt\Delta}2=\frac{-i\pm3i}2=\begin{cases}\cfrac{-4i}2=-2i\\{}\\\cfrac{2i}2=i\end{cases}$$

Another way: complete the square:

$$0=x^2=ix+2=\left(x+\frac i2\right)^2+\frac94\implies\left(x+\frac i2\right)^2=-\frac94=\left(\frac32i\right)^2\implies$$

$$\implies x+\frac i2=\pm\frac32i\implies x=\pm\frac32i-\frac i2=\begin{cases}-2i\\{}\\i\end{cases}$$

Third way: as you began to do, with $\;x=a+bi\;$ :

$$a^2-b^2+2abi=x^2=-xi-2=b-ai-2=(b-2)-ai$$

Now compare real and imaginary parts in both sides:

$$\begin{cases}I\;\;\;a^2-b^2=b-2\\{}\\II\;\;\;2ab=-a\end{cases}$$

Now we have two cases: if $\;a\neq0\;$, then from $II$ we get $\;b=-\cfrac12\;$, and then in $\;I\;$ we get:

$$a^2-\frac14=-\frac52\implies a^2=-\frac94$$

and tyhis can't be as both $\;I,\,II\;$ are real equations, thus it must be that $\;a=0\;$ and then:

$$-b^2\stackrel{II}=-b-2\implies =b^2-b-2=(b-2)(b+1)\implies b=-2,1$$

and again we get the same solutions, as it should be.


问:
Among all shapes with the same area, a circle has the shortest perimeter
Is it true that "Among all shapes with the same area, a circle has the shortest perimeter"  ? and how to prove it ?</p>

答:
The question can be asked such as "Find the plane curve which encloses given area with shortest perimeter". Then by parametric formulation of coordinates $x(t),y(t)$ and assuming non-selfintersecting curve the area can be written as
$$A=\frac 12 \int_{t_1}^{t_2}\big(xy'-x'y\big)dt$$
and the perimeter
$$I=\int_{t_1}^{t_2}\sqrt{x'^2+y'^2}dt$$
Then the Euler-Lagrange becomes
$$-\frac{d}{dt}\bigg(\frac{x'}{\sqrt{x'^2+y'^2}}\bigg)-\frac{\lambda}2\bigg(y'+\frac{d}{dt}\bigg(y\bigg)\bigg)=0$$
$$-\frac{d}{dt}\bigg(\frac{y'}{\sqrt{x'^2+y'^2}}\bigg)-\frac{\lambda}2\bigg(-x'-\frac{d}{dt}\bigg(x\bigg)\bigg)=0$$
If $ds$ is the element of arc the equations can be reduced to
$$\frac{d^2x}{ds^2}+\lambda \frac{dy}{ds}=0$$
$$\frac{d^2y}{ds^2}-\lambda \frac{dx}{ds}=0$$
which has the solution
$$x(t)=K_1+C_1\cos(\lambda s)+C_2\sin(\lambda s)$$
$$y(t)=K_2-C_2\cos(\lambda s)+C_2\sin(\lambda s)$$
It can be further simplified by "Angle sum and difference identities"
$$x(t)=K_1+D\sin(\lambda s+\alpha)$$
$$y(t)=K_2-D\cos(\lambda s+\alpha)$$
where $D\sin(\alpha)=C_1$ and $D\cos(\alpha)=C_2$; and this solution represents a circle of radius $D$ and centre $(K_1,K_2)$</p>


问:
How to calculate the integer coefficients for infinitely many integers $n$, $P(P(n)+1)$ is prime?
The question is: "Calculate all polynomials P with integer coefficients s. t. for infinitely many integers $n$, $P(P(n)+n)$ is a prime."

After using google, I have found proof of contradiction for $P(n)$: Link

My questions are:

Do you think that, is it sufficient to find a polynomial which does not fulfill the statement. Is there a famous polynomial for that, what I don't know? Sorry, I am not a mathematician.

I posted a line to a contradiction. Can I also adapt that to solve my problem?

答: If $P(n)=an+b$, then $P(P(n)+1)=P(an+b+1)=a(an+b+1)+b=a^2n+ab+a+b$, so it will represent infinitely many primes if and only if $\gcd(a^2,ab+a+b)=1$, which is the same as $\gcd(a,b)=1$. This settles the case where $P$ is linear.


答:
Hint:

By inspection (thinking of Vieta's formulas) we factor$$(ix)^2-(ix)-2=(ix+1)(ix-2).$$

Title: Solve the quadratic equation with complex number $b$

Body:
Solve the quadratic equation:
$$x^2 +xi+2=0$$

I was thinking to set $x^2=-2-xi$ but what $i$ have to do when $x+yi$ in this case $x=-2$ and $y=x$...


问:
Euler substitutions - indefinite integral
See please https://en.wikipedia.org/wiki/Euler_substitution

Let us look at the most general indefinite integral

$$\int f(x) dx \tag{1} $$ where $x \in D$ and $D$ is some interval.

It is my understanding that to do any substitution $$x = \phi(t)$$

in $(1)$, it must be the case that $\phi(t)$ is either strictly increasing or strictly decreasing i.e. that $\phi$ maps in a bijective manner some interval $D_1$ into the interval $D$ and then of course $x$ takes values in $D$.

In other words to have a valid substitution, the derivative $\phi'$

must be either strictly positive or strictly negative in the whole interval $D_1$

But in the first Euler substitution, it is not always the case that

$$\phi(t) = \frac{t^2-c}{b-2t\sqrt{a}}$$

has either a positive or a negative derivative (for all values of $t$).

So... why for this substitution there is no additional requirement which would ensure that?

I mean... in this sense is it a valid substitution in general?

I assume it is after being checked by so many people. But why?

I guess the same question applies to the other two Euler substitutions, but I haven't yet calculated $\phi'$ for their cases. I wonder if for the other two substitutions $\phi'$ is strictly positive/negative, or it is but only under certain conditions, and I wonder what these conditions are. I will check that.

But let's say my question is ONLY about the first substitution in particular.

答:
According to [Fich, Ch. VIII, §1, 268], a substitution of a variable in an indefinite integral is based on a local rule of a differentiability of a composite function at a point. Namely,  if $\int g(t)dt=G(t)+C$  then $\int g(\omega(x))\omega’(x)dx =G(\omega(x))+C$, provided functions $g(t)$, $\omega(x)$, $\omega’(x)$ are continuous. This straightforwardly follows from an equality

$$\frac {\partial d}{dx} G(\omega(x))=G’(\omega(x))\omega’(x)= g(\omega(x))\omega’(x),$$
taking into account that $G’(t)=g(t)$.

I copied a more detailed explanation and an example from [Fich] here and here.

References

[Fich] (Grigorii Fichtenholz, Differential and Integral Calculus, v. II, 7-th edition, M.: Nauka, 1970 (in Russian).


问:
Why do we study prime ideals?
I hope this isn't an inappropriate question here!

I'd like to ask the following (perhaps slightly ill-posed) question: why do we study prime ideals in general (commutative or non-commutative) rings? The situation is somewhat clear to me in some nice cases such as Dedekind rings, but even in slightly more general settings (e.g. general commutative Noetherian rings) the theory becomes rather horrible. What benefit do we gain from an understanding of the prime ideals of a ring?

Thank you!

答:
There is a strong conceptual reason to study the prime ideals of a ring: in algebraic geometry, we can study / think about a ring $R$ using an associated geometric object called $\mathrm{Spec}(R)$, the points of which are precisely the prime ideals of $R$. This idea is one of the many Grothendieck introduced, ushering in "modern" algebraic geometry.

(on the left: the frontispiece of Reid's Undergraduate Commutative Algebra)

(on the right: Mumford's famous sketch of $\mathrm{Spec}(\mathbb{Z}[x])$; note that $\mathbb{Z}[x]$ is not Dedekind.)

However, I don't have enough time at the moment to give a detailed explanation of the above; I'm sure someone else will be able to. But just to offer a perhaps more mundane answer: there are many properties $\mathsf{X}$ that an arbitrary commutative ring might have that are local, in the sense that
$$R\text{ has property }\mathsf{X}\iff R_P\text{ has property }\mathsf{X}\text{ for all prime ideals }P\subset R,$$
where $R_P$ is the localization of the ring $R$ at the prime ideal $P$. The same is true for arbitrary modules over arbitrary commutative rings; that is, if $M$ is a module over $R$, then there are many module properties $\mathsf{Y}$ for which
$$M\text{ has property }\mathsf{Y}\iff M_P\text{ has property }\mathsf{Y}\text{ for all prime ideals }P\subset R.$$

Title: Why do we study prime ideals?


问:
the limit of complex conjugate of characteristic function
In Durrett's probability theory, we want to show that if $\lim_{t \downarrow 0} (\varphi(t) - 1)/t^2 = c > -\infty$, then $EX = 0$ and $E|X|^2 = -2c < \infty$.

I know I have to use this theorem to get $E|X|^2 < \infty$:

Theorem 3.3.21 If $\limsup_{t \downarrow 0} \{ \varphi(t) - 2 \varphi(0) + \varphi(-t) \} / t^2 > -\infty$, then $E|X|^2 < \infty$

I'm having trouble satisfying the condition of the above theorem for $\lim_{t \downarrow 0} (\varphi(t) - 1)/t^2 = c > -\infty$. I can rewrite the limit as
$$\limsup_{t \downarrow 0} { \varphi(t) - 2 \varphi(0) + \varphi(-t)  \over t^2}= \underbrace{\limsup_{t \downarrow 0} {\varphi(t) - 1 \over t^2}}_{=c} + \underbrace{\limsup_{t \downarrow 0}{\varphi(-t) -1 \over t^2}}_{?}$$
I believe we need to use the fact that $\varphi(-t) = \overline{\varphi(t)}$. Is
$$\limsup_{t \downarrow 0} {\varphi(t) - 1 \over t^2} = \limsup_{t \downarrow 0}{\overline{\varphi(t)} -1 \over t^2}$$

答: $${\overline{\varphi(t)} -1 \over t^2}$$  is the complex conugate of $$ {\varphi(t) - 1 \over t^2} $$ But here it is assumed that $c$ is a real number so $c$ is its own complex conjugate.


问:
Finding $\lim_{\varepsilon\to 0}\int_{\varepsilon}^{\frac{\varepsilon}{1+\varepsilon}}\frac{e^{-x^2}}{x^2}dx$
$$\lim_{\varepsilon\to 0}\int_{\varepsilon}^{\frac{\varepsilon}{1+\varepsilon}}\frac{e^{-x^2}}{x^2}dx$$

Okay this seems weird to me-- couldn't i just write down the answer 0 because $\int_{0}^{0} f(x)dx=0$? (The answer is $-1$ btw)
There was even a hint for the question: $\displaystyle \lim_{x\to0}\frac{e^{-x^2}-1}{x^2}$ may be useful.

Am I supposed to go  $$\displaystyle \int_{\varepsilon}^{0}\frac{e^{x^2}}{x^2}dx+\int_{0}^{\frac{\varepsilon}{1+\varepsilon}}\frac{e^{x^2}}{x^2}dx\,\,\,?$$ And I couldn't even go L'Hospital with that.....
Any help would be appreciated ! Thanks!

Of course, it would be nice if the answer was simply 0!

答:
Consider

$I:=\displaystyle{\int_{\epsilon/(\epsilon+1)}^{e}} \dfrac{e^{-x^2}}{x^2}dx, \epsilon >0$;

$e^{-(\epsilon/(1+\epsilon))^2} \displaystyle{\int_{\epsilon/(\epsilon+1)}^{\epsilon}}\dfrac{1}{x^{2}}dx> I >$

$e^{-\epsilon^{2}}\displaystyle{\int_{\epsilon/(1+\epsilon)}^{\epsilon}}\dfrac{1}{x^2}dx;$

Calculate the above integral and take the limit $\epsilon \rightarrow 0^{+}$ (Squeeze Theorem).

Similarly for $\epsilon \rightarrow 0^{-}.$


问:
Probability normal distribution question
I had this question on a quiz, since I don't have the solution for it I solved it. can anyone tell me if I did it right?

The weight of a coffee capsule is normally distributed with a mean of 13.6 grams and a standard deviation of 0.8 gram. Suppose that 25 coffee capsules are placed in a package. Also, suppose that the weights of coffee capsules are independent. What is the probability that the weight of a package of 25 coffee capsules is less than 335 grams?

25 coffees capsule

$Y=X_1+X_2+X_3+\dots+X_{25}$

$E(X)=13.6$

$E(Y)=13.6\cdot25$

$V(X)=0.8^2$

$V(Y)=(0.8^2)\cdot25=16$

standard deviation of $Y=4$

$P(X<335)=P\Big(Z\leq \frac{335-340}{4}\Big)=P(Z\leq -1.25)=0.1056$

Is this correct?

Just one more question, can we add standard deviations instead?
because the first time I attempted this problem  I multiplied standard deviation by 25. But I heard that its not a linear or something. Just want to make sure.

答:
Is this correct?

Yes, absolutely.

Just one more question, can we add standard deviations instead?

No, you cannot.


问:
In $\mathbb{R}$ find the solutions of the equation $x^4-x^3-18x^2+3x+9=0$
I couldn't solve this question and hence looked at the solution which goes as follows:

$0$ is no a root of the equation. Hence:

$$x^2-x-18+\frac{3}{x}+\frac{9}{x^2}=0. \text{ So,  }
x^2+\frac{9}{x^2}-(x-\frac{3}{x})-18=0$$

I state that $y=x-\frac{3}{x}$, so we have that $y^2=x^2+\frac{9}{x^2}-6$, in other words $x^2+\frac{9}{x^2}=y^2+6$. Hence the equation is written as $y^2+6-y-18=0$ so $y^2-y-12=0$, hence $y=4$ or $y=-3$ so $x=2\pm\sqrt{7}$, or $x=\frac{-3\pm\sqrt{21}}{2}$

Could you please explain to me why the solution author of the solution thought of originally dividing by the equation by x and after that substituting $x-\frac{3}{x}$ with $y$? Also if you can think of a more intuitive approach could you please show it?

答:
For the factorization, using the long, long way ,consider
$$(x^2+ax+b)(x^2+cx+d)-(x^4-x^3-18x^2+3x+9)=0$$ Expand and group terms to gat
$$(b d-9)+x (a d+b c-3)+x^2 (a c+b+d+18)+x^3 (a+c+1)=0$$ So, $c=-a-1$; replace
$$(b d-9)+x ((-a-1) b+a d-3)+x^2 ((-a-1) a+b+d+18))=0$$ So, $b=a^2+a-d-18$; replace
$$\left(d \left(a^2+a-d-18\right)-9\right)+x (-a (a (a+2)-2
   d-17)+d+15)=0$$ So, $d=\frac{a^3+2 a^2-17 a-15}{2 a+1}$; replace
$$\frac {(a-3)(a+4)(a^4+2 a^3-23 a^2-24 a-3 )} {(1+2a)^2 }=0$$ The quartic does not show any real root : so $a=3\implies c=-4$ or $a=-4\implies c=3$. Choose the one you prefer and run back for $b$ and $d$.

Edit

I made a mistake when I wrote "the quartic does not show any real root". What I was supposed to write is that "the quartic does not show any rational root". This correction followed  @Macavity's comment.


问:
How do I solve $\overline{(A \cap B) \cup (\overline{A} \cap C)} = (A \cap \overline{B}) \cup (\overline{A} \cap \overline{C})$?
The question is to prove  $\overline{(A \cap B) \cup (\overline{A} \cap C)} = (A \cap \overline{B}) \cup (\overline{A} \cap \overline{C})$ using laws of set theory.

I got stuck after the following steps:
$$
\overline{(A \cap B) \cup (\overline{A} \cap C)} = \overline{(A \cap B)} \cap \overline{(\overline{A} \cap C)}
\\ = (\overline{A} \cup \overline{B}) \cap (A \cup \overline{C})
\\ = ((\overline{A} \cup \overline{B}) \cap A) \cup ((\overline{A} \cup \overline{B}) \cap \overline{C})
\\ = (\overline{A} \cap A) \cup (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}) \cup (\overline{B} \cap \overline{C})
\\ = (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}) \cup (\overline{B} \cap \overline{C})
$$

Not sure what to do next. I somehow feel that the absorption law could be used here, but don't know how.

答: Hint: To show that $$ (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}) \cup (\overline{B} \cap \overline{C})$$ is same as $$ (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}) $$ it is enough to show that $$(\overline{B} \cap \overline{C})$$ is  a subset of $$ (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C})$$ Prove this by writing $E$ as $(E \cap A) \cup (E \cap \overline A)$ where $E$ stands for $$(\overline{B} \cap \overline{C})$$


问:
Prove that $\int_{\frac{-f(0)}{M}}^{\frac{f(0)}{M}}f(x)\, dx \geq f(0)^{2}/M$
Let $f:\mathbb{R}\to\mathbb{R}$ be a differentiable function with bounded derivative. Assume that $|f'(x)|<M$ for all $x\in\mathbb{R}$ and $f(0)>0$. Then prove that $$\int_{\frac{-f(0)}{M}}^{\frac{f(0)}{M}}f(x)\, dx \geq \frac{f(0)^2}{M}$$

The only result i know is the mean value theorem which asserts $f'(c)=\frac{f(b)-f(a)}{b-a}$. But I don't think it can be used here. I am completely out of ideas. A detailed solution would prove helpful.

答:
According to Mean Value Theorem

$$\vert f(x) - f(0) \vert \le M \vert x \vert$$ for all $x \in \mathbb R$.

Therefore

$$-M \vert x \vert + f(0) \le f(x) \le M \vert x \vert + f(0)$$ and by integration over the interval $[- \frac{f(0)}{M},\frac{f(0)}{M}]$

$$3\frac{f(0)^2}{M} \ge\int_{- \frac{f(0)}{M}}^{\frac{f(0)}{M}} f(x) \ dx \ge 2 \frac{f(0)^2}{M} - \frac{f(0)^2}{M} = \frac{f(0)^2}{M}$$ as desired.


问:
Representation of a closed ideal in $L^1(T)$.
Let $T$ be the unit circle. Given that $I$ is a closed ideal in $L^1(T)$. I conjecture that there exists a set $S \subseteq \mathbb{Z}$ such that
\begin{equation}
I=\{f \in L^1(T) : \widehat{f}(n)=0, n \in S\}.
\end{equation} What I have shown so far is that if $I$ has this form, then it is indeed a closed ideal in $L^1(T)$. What should I consider next?

答:
TRUE.

Assume that for every $k\in\mathbb Z$, there exists an $f_k\in I$, with $\hat f(k)\ne 0$. Then
$$
\mathrm{e}^{ikx}=\left(\frac{1}{\hat f(k)}\mathrm{e}^{ikx}\right)*f(x)\in I,
$$
and hence all trigonometric polynomials belongs to $I$, and as $I$ is closed, then $I=L^1(T)$.

Note. I have assume the the multiplication is the ring $L^1(T)$ is the standard convolution.


问:
A machine can compare at most three integers in one run. Show that at least $k(n) := \lceil log_6(n!) \rceil$ runs are needed to sort $n$ integers.
My idea is to use a proof by induction on $n$.

For $n = 1,2,3$, $1$ run is needed so we are done.

Assume the statement is true for sets of $n$ or less integers, $n > 3$. Let $S$ be a set of $n+1$ integers, and $p \in S$. By the induction hypothesis, it takes at least $k(n) = \lceil \operatorname{log}_6(n!) \rceil$ runs of the machine to sort $S \setminus \{p\}$.

At this point we’d have a sorted list $S \setminus \{p\}$ and want to know where $p$ should be inserted into this list. Let $k’$ be the number of runs required to do this task. A ternary search on the sorted list $S \setminus \{p\}$ could result in $k’ = \operatorname{log}_3(n)$ runs in the worst-case scenario (say, if $p$ is in the middle of the sorted list $S \setminus \{p\}$). Now we’d have:

\begin{align}
k(n+1)
&= k’ + k(n) \\
&= \operatorname{log}_3(n) + \lceil \operatorname{log}_6(n!) \rceil \\
&\geq \operatorname{log}_6(n+1) + \lceil \operatorname{log}_6(n!) \rceil \\
&\geq \operatorname{log}_6(n+1) + \operatorname{log}_6(n!) \\
&= \operatorname{log}_6((n+1)!),
\end{align}

where the first inequality is true since $\operatorname{log}_3(n) \geq \operatorname{log}_6(n+1)$ holds for $n \geq 2$. So $k(n+1) \geq \lceil \operatorname{log}_6((n+1)!) \rceil$.

What I’m not sure of is whether I’m assuming too much what the algorithm would do in obtaining $k’$, or if it’s clear ternary search is the fastest way to do this task. Please give some comment.

答: This is an exercise in decision trees. Each run has $3!=6$ possible results, but we need to distinguish between $n!$ results (corresponding to all permutations of $n$ distinct integers). The least depth of the decision tree that will allow for accommodating $n$ leaves is $\lceil\log_6n!\rceil$.


答:
Since $\varepsilon\to 0$, the domain of integration is very small ($\frac{\varepsilon}{1+\varepsilon}\sim \varepsilon-\varepsilon^2$). Using series
$$\frac{e^{-x^2}}{x^2}=\frac{1}{x^2}-1+\frac{x^2}{2}+O\left(x^4\right)$$
$$\int\frac{e^{-x^2}}{x^2}dx\sim -\frac{1}{x}-x+\frac{x^3}{6}+O\left(x^5\right)$$

Using the bounds,
$$I=\left(\frac{\varepsilon ^3}{6 (\varepsilon +1)^3}-\frac{\varepsilon }{\varepsilon
   +1}-\frac{\varepsilon +1}{\varepsilon } \right) -\left(\frac{\varepsilon ^3}{6}-\varepsilon -\frac{1}{\varepsilon } \right)$$

Using Taylor
$$I=-1+\varepsilon ^2-\varepsilon ^3+O\left(\varepsilon ^4\right)$$

Title: Finding $\lim_{\varepsilon\to 0}\int_{\varepsilon}^{\frac{\varepsilon}{1+\varepsilon}}\frac{e^{-x^2}}{x^2}dx$

Body:
$$\lim_{\varepsilon\to 0}\int_{\varepsilon}^{\frac{\varepsilon}{1+\varepsilon}}\frac{e^{-x^2}}{x^2}dx$$

Okay this seems weird to me-- couldn't i just write down the answer 0 because $\int_{0}^{0} f(x)dx=0$? (The answer is $-1$ btw)
There was even a hint for the question: $\displaystyle \lim_{x\to0}\frac{e^{-x^2}-1}{x^2}$ may be useful.

Am I supposed to go  $$\displaystyle \int_{\varepsilon}^{0}\frac{e^{x^2}}{x^2}dx+\int_{0}^{\frac{\varepsilon}{1+\varepsilon}}\frac{e^{x^2}}{x^2}dx\,\,\,?$$ And I couldn't even go L'Hospital with that.....
Any help would be appreciated ! Thanks!

Of course, it would be nice if the answer was simply 0!


答:
There are $n!$ different possible ways to order the $n$ elements. After the first comparison, and subsequent ordering of the three compared elements, there are at still at least $\frac{n!}{6}$ possible orders the elements could have. After the second comparison, and subsequent rearrangement, there are at least $\frac{n!}{6^2}$ possible orders left.

At the very least, we need to keep going until that number reaches $1$. This will take $\lceil\log_6(n!)\rceil$ comparisons.

Title: A machine can compare at most three integers in one run. Show that at least $k(n) := \lceil log_6(n!) \rceil$ runs are needed to sort $n$ integers.

Body:
My idea is to use a proof by induction on $n$.

For $n = 1,2,3$, $1$ run is needed so we are done.

Assume the statement is true for sets of $n$ or less integers, $n > 3$. Let $S$ be a set of $n+1$ integers, and $p \in S$. By the induction hypothesis, it takes at least $k(n) = \lceil \operatorname{log}_6(n!) \rceil$ runs of the machine to sort $S \setminus \{p\}$.

At this point we’d have a sorted list $S \setminus \{p\}$ and want to know where $p$ should be inserted into this list. Let $k’$ be the number of runs required to do this task. A ternary search on the sorted list $S \setminus \{p\}$ could result in $k’ = \operatorname{log}_3(n)$ runs in the worst-case scenario (say, if $p$ is in the middle of the sorted list $S \setminus \{p\}$). Now we’d have:

\begin{align}
k(n+1)
&= k’ + k(n) \\
&= \operatorname{log}_3(n) + \lceil \operatorname{log}_6(n!) \rceil \\
&\geq \operatorname{log}_6(n+1) + \lceil \operatorname{log}_6(n!) \rceil \\
&\geq \operatorname{log}_6(n+1) + \operatorname{log}_6(n!) \\
&= \operatorname{log}_6((n+1)!),
\end{align}

where the first inequality is true since $\operatorname{log}_3(n) \geq \operatorname{log}_6(n+1)$ holds for $n \geq 2$. So $k(n+1) \geq \lceil \operatorname{log}_6((n+1)!) \rceil$.

What I’m not sure of is whether I’m assuming too much what the algorithm would do in obtaining $k’$, or if it’s clear ternary search is the fastest way to do this task. Please give some comment.


问:
Polynomials maps on $\phi \mathbb{A}^1_k \rightarrow \mathbb{A}^2_k$
I have the following question which seems to come up every year in exam papers, but just different numbers...Im really stuck on some parts...

Define $\phi: \mathbb{A}^1_k\rightarrow \mathbb{A}^2_k$ by $\phi(t)=(t^2-1,t^3+2)$

(i) Describe the induced map $\phi$*:k[x,y] $\rightarrow$k[t] and compute $\phi$*$(x^2y-3xy)$

(ii) Describe the ideal $J_\phi$ and explain how to use $J_\phi$to compute the kernel of $\phi$*

(iii) Give the gernerating set for the kernel of $\phi$*

(iv)Is the point (2,-1) in the Zariski closure of the image $\phi$?

Working

I have computed

(i) $\phi$*$(f(x,y))=f(t^2-1,t^3+2)$

$\phi$*$(x^2y-3xy)=(t^2-1)^2(t^3+2)-3(t^2-1)(t^3+2)$

$=t^7-5t^5+2t^4+4t^3-10t^2+8$

.

(ii)

(iii)

(iv)

Please help me!!!

答:
Write $k[x,y]$ as  $k[a,b]$ with $a = x+1$ and $b=y-2$.

In $ab$-coordinates the map $\phi$ is $(t^2, t^3)$ which implicitizes to $a^3=b^2$.

This answers some parts of the question, and for the rest the definition of $J_\phi$ is missing.


问:
What does it mean to some logical statement to be provable and decidable?
I am not a mathematician and trying to get deeper insight into modern logic.

It happens all the time, that statements like statement P is unprovable arise, or, more formally, $\lnot \text{Provable}(P)$. That could mean:

There exists no path from assumed propositions (axioms) to the $P$ that obeys rules of inference. As such, we are not able to prove $P$ and yet we might be able to prove $\lnot P$.

There exists no path from assumed propositions (axioms) to $P$ nor to $\lnot P$, thus we are not able to prove neither one about $P$: it is something we don't know how to reason about.

Both $P$ and $\lnot P$ can be proved from assumed propositions (axioms) with respect to rules of inference. That would mean that our system is not consistent and has hidden contradiction inside. That's a broken dangerous system and does not deserve our trust.

I am not that good in math to be able to illustrate all of the above mentioned options, but it feels natural to me to state unprovability in such a way.

How does modern logic answers my question?
And do "provable" and "decidable" convey exactly the same meaning and are interchangeable in all contexts?

答:
Provability and decidability are two distinct concepts and they are not interchangeable at all. The difference is subtle.

Saying that a statement $P$ is unprovable (in a given system) means exactly what you said in Point 1:

There exists no path from assumed prepositions (axioms) to $P$ that obeys rules of inference. As such, we are not able to prove $P$ and yet we might be able to prove $¬P$.

Saying that a statement $P$ is undecidable (in a given system) means exactly what you said in Point 2:

There exists no path from assumed prepositions (axioms) to $P$ nor to $\lnot P$, thus we are not able to prove neither one about $P$: it is something we don't know how to reason about.

However, I wouldn't say that if $P$ is undecidable then $P$ "is something we don't know how to reason about". First, saying "how to reason about it" is a bit ambiguous. Moreover, it would be better to say that the given system doesn't know how to reason about it. Indeed, in other systems $P$ can be decidable: for instance, take the system where $P$ is undecidable and add $P$ (resp. $\lnot P$) as an axiom; in the new system, $P$ is decidable and more precisely $P$ (resp. $\lnot P$) is provable.

Clearly, the fact that $P$ is provable in a system implies that $P$ is decidable in such a system (which amounts to say that undecidability of a statement implies unprovability of that statement), but the converse fails: in a system, it is possible that $\lnot P$ is provable (and hence $ P$ is decidable) but $P$ is not provable.

By system, I mean a set of axioms and of inference rules. Note that it is possible that in a system both $P$ and $\lnot P$ are provable (your Point 3). In that case, we say that the system is incoherent, and as a consequence (called principle of explosion) everything is provable in an incoherent system. Said differently, incoherent systems are not informative at all.

A system is said to be coherent if there is no statement $P$ such that both $P$ and $\lnot P$ are provable. In a coherent system, the situation you described in Point 3 is impossible.
In a coherent system it is possible (but not necessary) to have unprovable and/or undecidable statements.

Concerning decidability, what I mentioned above is the meaning of "decidable" when referred to a statement. Unfortunately, in logic there is another meaning of "decidable", but referred to a system, not a statement.
A system is decidable if there is an effective method (an algorithm) for determining whether arbitrary statements are provable or not is that system.


问:
Show that there exist infinitely many matrices $\gamma \in \Gamma_1$ with $\gamma x_1 = x_2$
I am trying to solve the following exercise:

For $x \in \bar{\mathbb{Q}} := \mathbb{Q} \, \cup \, \{ \infty \}$ and $\gamma =
  \left[ {\begin{array}{cc}
   a & b \\
   c & d \\
  \end{array} } \right] \in \Gamma_1 \ $where $\Gamma_1 = SL_2(\mathbb{Z}),$ we define $\gamma x \in \bar{\mathbb{Q}}$ by

$$ \gamma x = \left\{
\begin{array}{ll}
\frac{a}{c}, & \textrm{if} \, x = \infty \\
\infty, & \textrm{if} \, x = -\frac{d}{c} \\
\frac{ax+b}{cx+d}, & \textrm{else}
\end{array}
\right. $$

Let $x_1,x_2 \in \bar{\mathbb{Q}}.$ Show that there exists infinitely many matrices $\gamma \in \Gamma_1$ with $\gamma x_1 = x_2$

My approach would be to look at all the cases and see what comes out of it. I just don’t know how to draw backs to the matrix and why there are so many of them. Could someone give me an approach? Or how to solve such tasks in general? Thank you in advance!

答:
Hint Remember that you must ensure that $a d - b c = 1$. Start by finding infinitely many such matrices $A$ such that $\gamma_A x_1 = 0$. Then find infinitely many such matrices $B$ such that $\gamma_B 0 = x_2$. Then use the matrices $B A$.

Note that $\gamma_B 0 = x_2$ is easy to solve because it reduces to $\gamma_{B}^{-1} x_2 = 0$, so that there is only one problem.

First problem
Let us assume that $x = p/q \in \mathbb{Q}$ with $p$ and $q$ coprimes. Let $\gamma=\pmatrix{a& b\cr c& d}$ be a matrix such that $\gamma x=0$. Note that $a$ and $b$ are coprime because $a d - b c= 1$, it follows easily that $p/q = -b/a$ and finally $(a, b) = \pm (q, -p)$. Let $(u, v)$ be a Bezout pair such that $u p + v q = 1$. The condition on the determinant implies that for some $k\in \mathbb{Z}$ one has $c = u - k q$ and $d = v + k p$, finally
\begin{equation}
\gamma = \pm \begin{pmatrix}q & -p\cr u- k q& v + kp\end{pmatrix}
\qquad k\in\mathbb{Z}
\end{equation}

Second problem With the same assumptions, let us look for a matrix $\gamma$ such that $\gamma 0 = x$. Then one has $\gamma^{-1}x = \gamma^{-1}\gamma 0 = 0$ and $\gamma^{-1}$ is a matrix of the previous form. It follows that here
\begin{equation}
\gamma = \pm \begin{pmatrix}v+k p & p\cr -u+ k q& q\end{pmatrix}
\qquad k\in\mathbb{Z}
\end{equation}
Initial problem
Assuming that $x_1,x_2\in \mathbb{Q}$ take two matrices $\gamma_1$
and $\gamma_2$ having the previous forms (for $x_1=p_1/q_1$ and $x_2=p_2/q_2$). The matrices $\gamma = \gamma_2\gamma_1$ solve the initial problem. There are infinitely many of them by letting $k_1$ or $k_2$ vary.

It remains to handle the special cases where one or both $x_i$s can be equal $\infty$.


问:
Reference Request for Optimal Transport
I am reading through Villani's Optimal Transport: Old and New (2009) and I have a solid understanding of all the details presented in the first six chapters (essentially just Kantorovich duality, Wasserstein distances).  However, when I am attempting to read any modern papers in the subject, I find myself having a nearly complete lack of ability to understand most of what is going on.  I am looking for some suggestions for papers to read that might be accessible to someone with this fairly minimal background in OT.

In terms of my math background, it's not perfect by any stretch of the imagination.  I know probability/measure theory to the level of Billingsley and real analysis to the level of the first 10 chapters in Rudin.  Hopefully this is a good enough set of tools with which to understand deeper ideas in OT.  Thanks for any help you can provide!

答:
Well, I mean OT is a vast research area. I guess you will need to specify a little in which direction you wanna go. Besides, I personally wouldn't recommand Villani's opus magum as introductory literature. His other book 'topics in optimal transportation' or the one of Santambrogio are better for getting an overall idea of OT (personal opinion). But since you were asking for papers, I'd say the following four are amongst the 10 most influencial papers in OT of the last decades (which at the same time are not too hard to understand, I think. You may need however some background in PDE, stochastic analysis and/or Riemannian geometry (at least for the first 3), but that you will need in most of modern OT papers, anyway.):

Benamou, Brenier; A computational fluid mechanics solution to the Monge-Kantorovich mass transfer problem

Jordan, Kinderlehrer, Otto; The Variational Formulation of the Fokker-Planck Equation

Otto; The geometry of dissipative evolution equations: the porous medium equation

Cuturi; Lightspeed Computation of Optimal Transport

Title: Reference Request for Optimal Transport