问:
How to expand x² ? Is that possible?
I have uploaded the steps in link provided.
I don't understand what just happened to x² in solution attached. Is there a Taylor series expansion for x²? I don't think so. Can you please explain the expansion of x² in attached solution.
Click here to see solution of above question

答:
As it says in the solution, $x^2$ is already a Taylor series: every polynomial is. If you like, though, you can expand it in the usual way about $0$: if $f(x)=x^2$, then

$$f(x)=\sum_{n\ge 0}\frac{f^{(n)}(0)}{n!}x^n=\frac2{2!}x^2=x^2\,,$$

since all of the derivatives except the second are $0$ or evaluate to $0$ at $x=0$. If you insist on writing out every term of the series, you have

$$x^2=\sum_{n\ge 0}b_nx^n\,,$$

where

$$b_n=\begin{cases}
1,&\text{if }n=2\\
0,&\text{otherwise.}
\end{cases}$$

The solution then simply adds this Maclaurin series to the one for $e^x$:

$$\begin{align*}
e^x+x^2&=\sum_{n\ge 0}\frac{x^n}{n!}+\sum_{n\ge 0}b_nx^n\\
&=\sum_{n\ge 0}\left(\frac1{n!}+b_n\right)x^n\\
&=1+x+\left(\frac1{2!}+1\right)x^2+\sum_{n\ge 3}\frac1{n!}x^n\\
&=\frac1{0!}+\frac1{1!}x+\frac3{2!}x^2+\sum_{n\ge 3}\frac1{n!}x^n\,,
\end{align*}$$

so it is the exponential generating function for the sequence $\langle a_n:n\ge 0\rangle$, where $a_2=3$, and $a_n=1$ if $n\ne 2$.

Title: How to expand x² ? Is that possible?


问:
Very confused about the definition of subbasis for a topology in Munkres's book
In Munkres's Topology textbook, it says,

Definition

A subbasis S for a topology on X is a collection of subsets of X whose union equals X.

The topology generated by the subbasis S is defined to be the collection T of all unions of finite intersections of elements of S.

I am having a very hard time understanding the first sentence. Let's say $X = \{1, 2, 3\}$. Then let $A = \{\{1\}, \{2\}, \{3\}\}$, or $A = \{\{1, 2\}, \{3\}\}$. Either way, $A$ is a collection of subsets of $X$, and the union of the elements in $A$ is $X$, but $A$ does not seem like a subbasis here...

What did I misunderstand here? What should be the correct interpretation of "a collection of subsets of X whose union equals X"?

Thank you in advance!

Edit:

So turned out $A$ is a subbasis in both cases, thanks @Surb for your comment.

But here is what I don't understand (which made me mistakenly think A may not even be subbasis in the first place). If $A = \{\{1\}, \{2\}, \{3\}\}$, then wouldn't "the collection T of all unions of finite intersections of elements of $A$" be {$\varnothing$}, since any intersection of any two elements in $A$ is empty? But since T is a topology, T must also contain $X$, but, in this case, it does not?

答:
Munkres has a slightly modified definition of subbase, from my perspective.

For any family $\mathcal{S}$ of subsets of $X$ there is a unique smallest topology $\mathcal{T}(\mathcal{S})$ that contains $\mathcal{S}$, which we can define abstractly as the intersection of all topologies on $X$ that contain $\mathcal{S}$; we at least have one such topology (the discrete one, which is just all subsets of $X$) and any intersection of topologies is again a topology.

But there is also a less abstract way to define $\mathcal{T}(\mathcal{S})$ which is quite natural once, you have seen bases of topologies:

Let $\mathcal{S}'$ be the set of all intersections of finite subfamilies of $\mathcal{S}$. (if $\mathcal{N} \subseteq \mathcal{S} \subseteq \mathscr{P}(X)$, we define $$\bigcap \mathcal{S} = \{x \in X\mid\forall O \in \mathcal{N}: x \in O\}\tag{1}$$

as usual in set theory and only take those intersections for finite subfamilies $\mathcal{N}$. Now topologies are always closed under finite intersections so if $\mathcal{T}$ is any topology that contains $\mathcal{S}$, it will also contain $\mathcal{S}'$, defined by all finite intersections from $\mathcal{S}$ in the above sense.

By the notion of "void truth" it holds that $\bigcap \emptyset = X$ as we have a universal quantifier over the $\mathcal{N} = \emptyset$ in $(1)$.
So we get "for free" that $X \in \mathcal{S}$, as the empty subfamily is surely finite. Also $\mathcal{S} \subseteq \mathcal{S}'$ as when $S \in \mathcal{S}$, $\{S\}$ is also a finite subfamily of $\mathcal{S}$ and $\bigcap \{S\} = S \in \mathcal{S}'$ by the definition. But of course also $S_1, S_2 \in \mathcal{S}$ implies $S_1 \cap S_2=  \bigcap \{S_1,S_2\} \in \mathcal{S}'$ and likewise for intersections of $3$,$4$ etc. members. All of these must be in any topology containing $\mathcal{S}$, so in $\mathcal{T}(\mathcal{S})$ in particular.

Anyway, we then recall the criteria that Munkres gives for a family $\mathcal{B}$ to be a base for some topology:

$\bigcup \mathcal{B} = X$

$\forall B_1, B_2 \in \mathcal{B}: \forall x \in B_1 \cap B_2 \exists B_3 \in \mathcal{B}: x \in B_3 \subseteq B_1 \cap B_2$

and if these conditions are satisfied, the set of all unions of subfamilies $\mathcal{B}' \subseteq \mathcal{B}$, where $$\bigcup \mathcal{B}' = \{x \in X\mid \exists B \in \mathcal{B}: x \in B\}\tag{2}$$
forms the unique minimal topology that has $\mathcal{B}$ as a base.

Now we easily check that our $\mathcal{S}'$ defined from $\mathcal{S}$ indeed satisfies 1. and 2. That 1. holds follows from $X \in \mathcal{S}'$ trivially. That 2. holds is because $\mathcal{S}'$ is closed under finite intersections: given $B_1 = \bigcap \mathcal{N}_1$ and $B_2 = \mathcal{N}_2$, with $\mathcal{N}_1, \mathcal{N}_2 \subseteq \mathcal{S}$ finite, we just pick for any $x$ in their intersection $B_3 = \bigcap (\mathcal{N}_1 \cup \mathcal{N}_2) \in \mathcal{S}'$ (as a union of two finite sets is finite).

So $\mathcal{S}'$ forms a base for $\mathcal{T}(\mathcal{S})$ by minimality and the topology is just the set of all unions of (finite intersections from $\mathcal{S}$).

This $\mathcal{S}$ is traditonally called the subbase for this minimal topology $\mathcal{T}(\mathcal{S})$ generated by $\mathcal{S}$, so just another word for generating set: we just specify (as minimally as possible, usually) a collection of sets that has to be open to define a topology. This happens quite a lot in topology. The order topology is one example : we just specify that all sets of the form $\{x\mid x < a\}$ and $\{x\mid x > a\}$ where $a \in X$ have to be open, so these form a subbase for the order topology. (But not yet a base).

Munkres almost does the same thing but seems not to want to rely on the void truth convention to ensure that $X \in \mathcal{S}'$, but just demands as a condition that $\bigcup \mathcal{S} = X$ so that 1. is also satisfied for the base $\mathcal{S}' \supseteq \mathcal{S}$. The rest of the arguments then stay the same.

In practice this doesn't make much difference as standard subbases automatically satisfy that condition (the order topology subbase on $X$ only does if $|X| \ge 2$) and the set $\{p_i^{-1}[O]\mid i \in I, O \subseteq X_i \text{ open }\}$ for the product topology on $\prod_{i \in I} X_i$ etc.

The main reason why subbases got their separate name is that the Alaxander subbase lemma (or theorem) can be used to prove compactness of a space: $X$ is compact iff every open cover with elements from $\mathcal{S}$ has a finite subcover. This allows for efficient proofs of Tychonoff's theorem and other applications.

SO for me a subbase can be any family, but Munkres wants to avoid logical discussions and demands the union be $X$. But the basic idea stays the same: specify a small set that generates the topology.

ContentLicense: CC BY-SA 4.0


问:
Showing that an open interval is an open set

Context:

I'm trying to algebraically prove that an open interval is an open set. If I sketch it, as suggested by @rschwieb in this answer, then it seems quite obvious that this is indeed true. But I would like to be able to show it algebraically and after having looked at various sources found on the internet, I have decided to ask it here.

Attempt:

For definiteness, let us consider:
\begin{equation}
(c,d) = \{ x \in \mathbb{R} \mid c < x < d \}
\end{equation}
Furthermore, let $a \in (c,d)$, and recall that the $\epsilon$-neighborhood of $a$ is the set:
\begin{equation}
V_\epsilon (a) = \{ x \in \mathbb{R} \mid |x-a| < \epsilon \}
\end{equation}
Now, if we take $\epsilon = \mathrm{min} \{ a-c,d-a \}$, then $a-\epsilon \geq c$ and $a+\epsilon \leq d$. Up until here I understand everything. But then I don't understand how we can conclude that $V_\epsilon (a) \subseteq (c,d)$?

答:
For any point $x$ in an open interval $(a,b)$,
let $p=|x-a|$ and $q = |b-x|,$
then $x \in (a+\frac{p}{2}, b-\frac{q}{2}).$
Since $(a+\frac{p}{2}, b-\frac{q}{2})$ is entirely contained in $(a, b)$, then $(a,b)$ is an open set


问:
How do I explain why $dA/dr = 2 \pi r$ geometrically?
There's this question in my calculus book that goes something like this:

The derivative of the area of a circle with respect to its radius is equal to the circle's circumference ($dA/dr = 2 \pi r$). Give a geometric explanation of why this is the case.

To me this is really obvious, but I find it hard to put into words.  If you increased the radius of the circle by putting your finger inside it and pushing the edge outward, then you would have to push it around the whole circumference of the circle.  Heh.  I don't know.

答: Draw a circle with area $A$ and radius $r$, then draw around it a larger circle. The distance from your first circle to the second is $dr$, and the difference of the areas is $dA$. As $dr \to 0$, $dA$ goes to the circumference of the interior circle, hence $dA \to 2\pi r\:dr$.</p>

ContentLicense: CC BY-SA 3.0


问:
Continuity of function $f(x)=2^{-\frac{1}{x^{2}}}$
I need help with this task, if anyone had a similar problem it would help me !

The task is: Determine the type of interruption at the point x = 0 for the function

$$f(x)=2^{-\frac{1}{x^{2}}}$$

I did:

$$L=\lim_{x\to 0^{-}} 2^{-\frac{1}{x^{2}}} = 0 $$
$$R=\lim_{x\to 0^{+}} 2^{-\frac{1}{x^{2}}} = 0 $$
$$L=R=x=0$$

And as I concluded the function is continuous at x = 0, but in the solution it says that the break is of the first kind.
So I don’t understand why a breakup is the first kind?

Thanks in advance !

答:
After OP's calculation of L and R which are equal to zero.

A prescription for $f(0)$ is required. if $f(0)=K=0$, the function is continuous at $x=0$. If $K\ne 0$,  the function $F(x)$ is discontinuous at $x=0$. So there is a removable discontinuity at $x=0$ which can be avoided by choosing $K=0$.


问:
Algorithm for dividing sums of square roots: $\tfrac{\sum_{i=1}^m \sqrt{x_i}}{\sum_{j=1}^n \sqrt{y_j}}$
As a part of a project for supporting symbolic computation of square roots (since using any sort of approximated calculation ends up accumulating errors after several iterations) I'm trying to implement an algorithm for division of sums of square roots, i.e. expression like
$$
\frac{\sum_{i=1}^m \sqrt{x_i}}{\sum_{j=1}^n \sqrt{y_j}}
$$
where $x_i, y_j \in \mathbb{Q}$.

We can assume that $n > 2$ since $n = 1$ is trivial and $n = 2$ can be solved by multiplying both numerator & denominator by $\sqrt{y_1} - \sqrt{y_2}$.

For example if we have expression
$$
\mathrm{expr} := \sqrt{2} + \sqrt{3} + \sqrt{5}
$$
when we square it
$$
\mathrm{expr}^2 = 2\sqrt{15} + 2\sqrt{10} + 2\sqrt{6} + 10
$$
and divide by the original one we should have it as a result of the division so
$$
\frac{2\sqrt{15} + 2\sqrt{10} + 2\sqrt{6} + 10}{\sqrt{2} + \sqrt{3} + \sqrt{5}} = \sqrt{2} + \sqrt{3} + \sqrt{5}
$$

I think there is an algorithm for this particular task (since WolframAlpha is capable of doing so), but I couldn't find it online, so any help would be appreciated.

Edit

If I understood correctly the approach described by @Empy2 we can iteratively multiply by $\sum_{j=1}^{n-1} \sqrt{y_j} - \sqrt{y_n}$, but this won't work for 4 terms, because after multiplication total number of terms will be $\tfrac{(n - 1)(n - 2)}{2} + 1$ which increases for $n > 3$.
For $n = 4$ we can multiply both numerator & denominator by $(\sqrt{y_1} + \sqrt{y_2}) - (\sqrt{y_3} + \sqrt{y_4})$ leading us to the case with $n = 3$, but that's it, for $n > 4$ I'm not aware of any working algorithm.

Body:
$$f=\sqrt2+\sqrt3+\sqrt5+\sqrt7+\sqrt{11}\\
g=(\sqrt2+\sqrt3+\sqrt5+\sqrt7)^2-(\sqrt{11})^2\\
=6+2(\sqrt6+\sqrt{10}+\sqrt{14}+\sqrt{15}+\sqrt{21}+\sqrt{35})\\
=2(3+\sqrt6+\sqrt{10}+\sqrt{15})+2\sqrt7(\sqrt2+\sqrt3+\sqrt5)\\
=g_1+\sqrt7g_2$$

$$h=(g_1+\sqrt7g_2)(g_1-\sqrt7g_2)=g_1^2-7g_2^2\\
=4(40+10\sqrt{15}+12\sqrt{10}+16\sqrt6)-28(10+2\sqrt6+2\sqrt{10}+2\sqrt{15})\\
=-120+8\sqrt6-8\sqrt{10}-16\sqrt{15}\\
=(-120+8\sqrt6)-\sqrt5(8\sqrt2+16\sqrt3)\\
=h_1+\sqrt5h_2$$

$$k=(h_1+\sqrt5h_2)(h_1-\sqrt5h_2)=h_1^2-5h_2^2\\
=14400+384-1920\sqrt6-5(128+768+256\sqrt6)\\
=10304-3200\sqrt6\\
N=10304^2-6(3200)^2$$

Given $f(\sqrt2,\sqrt3,\sqrt5)$ in the denominator, let

$$g(\sqrt2,\sqrt3)=f(\sqrt2,\sqrt3,\sqrt5)f(\sqrt2,\sqrt3,-\sqrt5)\\
h(\sqrt2)=g(\sqrt2,\sqrt3)g(\sqrt2,-\sqrt3)\\
K=h(\sqrt2)h(-\sqrt2)$$

EDIT

$$f(\sqrt2,\sqrt3,\sqrt5,\sqrt7,\sqrt{11})=f_1(\sqrt2,\sqrt3,\sqrt5,\sqrt7)+\sqrt{11}f_2(\sqrt2,\sqrt3,\sqrt5,\sqrt7)\\
g(\sqrt2,\sqrt3,\sqrt5,\sqrt7)=f(\sqrt{11})f(-\sqrt{11})=f_1^2-11f_2^2\\
=g_1(\sqrt2,\sqrt3,\sqrt5)+\sqrt7g_2(\sqrt2,\sqrt3,\sqrt5)\\
h(\sqrt2,\sqrt3,\sqrt5)=g(\sqrt7)g(-\sqrt7)=g_1^2-7g_2^2\\
k(\sqrt2,\sqrt3)=h_1^2-5h_2^2\\
m(\sqrt2)=k_1^2-3k_2^2\\
N=m(\sqrt2)m(-\sqrt2)$$


问:
How to apply and understand the pigeonhole principle in this question?
Now this question has been asked in this site and I am aware of that. But after some reading I still don't get the answer there.

This is a field with finitely many elements. I am asked to prove
that  a natural number $n$ is present such that:

based on other answers from quora: https://www.quora.com/Suppose-F-is-a-field-with-finitely-many-elements-How-do-I-prove-that-there-exists-a-natural-number-n-such-that-1-1-1-1-1-n-times-0

Almost everyone applies that pigeon hole principle. Many of the answers included said something like:  "there exists two elements in the sequence which equal each other."

I really don't see where (at least) two elements are equal to each other in the sequence? I am really confused on this part. I understand the lines after that (for example you can use the other element as an additive identity, if thats so, it is a field because $0$ - the unique element is present. But I don't get the two elements are equal part? How can I see this? Can someone provide another example in terms of math? Because I have read the wikipedia definition and that only make sense intuitively.

答:
So, if I understand the question correctly, you have a fiinite field $F$ and you want to prove that there exists an $n\neq0$ such that
$$
n\cdot1_F=\underbrace{1_F+\cdots+1_F}_n=0_F
$$
using the pigeonhole principle.

I would do as follows:since $|F|<\infty$ for $k>0$ sufficiently large you have to find some repetitions in the sequence
$$
1_F,\quad 2\cdot1_F,\quad 3\cdot1_F,\quad...,k\cdot1_F.
$$
This is because the "pigeons", i.e. the numbers $\{1,...,k\}$, are more than the "holes", i.e. the elements of $F$.

So, let $r<s$ such that $r\cdot1_F=s\cdot1_F$. Then if $n=s-r>0$ we have $n\cdot1_F=0$.

I must say that this is a unusual way to prove the claim. The usual strategy is to observe that there's a canonical ring homomorphism
$$
\phi:\mathbb{Z}\rightarrow F,\qquad\phi(k)=k\cdot1_F
$$
and the finiteness of $F$ implies that $\ker(\phi)$ is a non trivial ideal in $\mathbb{Z}$, so that $\ker(\phi)=n\mathbb{Z}$ for some $n\geq0$ that in fact can be assumed positive and shown to be prime.


问:
How can I prove that the Core of $H$ in $G$ is a normal subgroup?
Given $G$ a group and $H$ a subgroup of $G$, how can I prove that ${\rm Core}_H G:=\bigcap_{g\in G} gHg^{-1}$ is a normal subgroup of $G$?

I've proved that is a subgroup, but how can I prove that it is a normal subgroup?

If $x \in{\rm Core}_H G,\ y \in G$, how can I prove that $yxy^{-1}$ is in ${\rm Core}_H G$?

As $x \in{\rm Core}_H G,\ \forall g\in G,\ \exists a\in H:x=gag^{-1} $
Moreover, as ${\rm Core}_H G \subseteq H$, (since it is subgroup), $x\in H$

How can I continue and show that $yxy^{-1}\in {\rm Core}_H G$

答:
Another way to look at this: $core_G(H)$ is the largest normal subgroup of $G$ contained in $H$. For if $N \unlhd G$ and $N \subseteq H$, then for every $g \in G$, we have $g^{-1}Ng=N \subseteq g^{-1}Hg$, so $N \subseteq core_G(H)$.

Yet another observation: $core_G(H)$ is the subgroup of $H$ generated by all the $G$-conjugacy classes contained in $H$.


问:
What is the domain of k? $2|x-2|-3|x-3|=k$
What is the domain of k?

$2|x-2|-3|x-3|=k$

Using the definition of absolute function, I obtain:

$|x-2|= \begin{cases}
        x-2, & x\ge2 \\
        2-x, & x<2\\
        \end{cases}\\$

$|x-3|= \begin{cases}
        x-3, & x\ge3 \\
        3-x, & x<3 \\
        \end{cases} \\$

For $x<2$

$x-5=k$

For $2\le x<3$

$5x-13=k$

For $x \ge 3$

$-x+5=k$

But I need to find the domain of k.

答:
From the $3$ rules that you worked out, the graph of $2|x-2|-3|x-3|$ increases on $(-\infty, 3]$ and then decreases from $[3, \infty)$.

Hence the possible values of $k$ are $(-\infty, f(3)) = (-\infty, 2]$.


问:
Evaluate the following limit without using L'Hospital
Evaluate the following limit without using L'Hospital

$$\lim_{x\to \infty} \frac{5x-2x\sqrt{x^2+1}+\sqrt{4x^4+3}}{2-\sqrt{x^2+1}}$$

I have tried substituting $\sqrt{x^2+1}$ with a and tried to $\sqrt{4x^4+3}$ in terms of a but that failed miserably.

Then I wrote 5x as 4x + x and found out that I can get 2x + ... but that did not help either.

Thanks

Body:
$$\begin{align}
L=\lim_{x\to \infty} \frac{5x-2x\sqrt{x^2+1}+\sqrt{4x^4+3}}{2-\sqrt{x^2+1}}= &
\lim_{x\to \infty} \frac{x(5-2x\sqrt{1+\frac{1}{x^2}}+2x\sqrt{1+\frac{3}{4x^4}})}{x(\frac{2}x-\sqrt{1+\frac{1}{x^2}})}= &\\
\lim_{x\to \infty} \frac{5-2x\sqrt{1+\frac{1}{x^2}}+2x\sqrt{1+\frac{3}{4x^4}}}{\frac{2}x-\sqrt{1+\frac{1}{x^2}}}=&
\lim_{x\to \infty} \frac{5-2x\left(\sqrt{1+\frac{1}{x^2}}-\sqrt{1+\frac{3}{4x^4}}\right)}{\frac{2}x-\sqrt{1+\frac{1}{x^2}}}
\end{align}$$

Now we rewrite $2x\left(\sqrt{1+\frac{1}{x^2}}-\sqrt{1+\frac{3}{4x^4}}\right)$ as
$$2x\left(\sqrt{1+\frac{1}{x^2}}-\sqrt{1+\frac{3}{4x^4}}\right)\frac{\left(\sqrt{1+\frac{1}{x^2}}+\sqrt{1+\frac{3}{4x^4}}\right)}{\left(\sqrt{1+\frac{1}{x^2}}+\sqrt{1+\frac{3}{4x^4}}\right)}=
\frac{2x\left(1+\frac{1}{x^2}-1-\frac{3}{4x^4}\right)}{\left(\sqrt{1+\frac{1}{x^2}}+\sqrt{1+\frac{3}{4x^4}}\right)}=
\frac{\left(\frac{2}{x}-\frac{3}{2x^3}\right)}{\left(\sqrt{1+\frac{1}{x^2}}+\sqrt{1+\frac{3}{4x^4}}\right)}
$$
Hence using $\lim_{x\to \infty} \frac{1}{x}=0$ we get
$$
L= \lim_{x\to \infty} \frac{5-\frac{\left(\frac{2}{x}-\frac{3}{2x^3}\right)}{\left(\sqrt{1+\frac{1}{x^2}}+\sqrt{1+\frac{3}{4x^4}}\right)}}{\frac{2}x-\sqrt{1+\frac{1}{x^2}}}
= \frac {5}{-1}=-5$$


问:
circumcenter of the $n$-simplex
Given $m$ points $v_i\in\mathbb{R}^n$, $m<n$. How to find the circumcenter of the simplex formed by the points?

答:
In a simplex, the circumcircle goes through all vertices so the distance of the circumcenter $c$ is equal for all vertices $v_i$ of the simplex,
$$
r^2 = \|v_i - c\|^2 \quad\forall i.
$$
$r$ is the circumradius.
Writing the circumcenter in barycentric coordinates,
$$
c = \sum_i v_j \alpha_j, \quad \sum\alpha_j = 1,
$$
this can be expanded to
$$
2 V^T V \alpha + \left(r^2 - \alpha^T V^TV \alpha\right) = w
$$
with $V$ being the colum-matrix of the $v_i$ and $w_i=\|v_i\|^2$.

Calling $\lambda = r^2 - \alpha^T V^TV \alpha$, this can be written as a Lagrangian system
$$
\begin{pmatrix}
2V^T V & e\\
e^T & 0
\end{pmatrix}
\begin{pmatrix}
\alpha\\
\lambda
\end{pmatrix}
=
\begin{pmatrix}
w\\
1
\end{pmatrix}
$$
with $e = (1,\dots,1)^T$. Note that $V^T V$ is singular if the spatial dimension is smaller than the number of vertices, e.g., triangles in 2D space. (Quite the typical case really.)

I checked numerically and indeed, $\alpha=(0.5, 0.5)$ if two points are given. I would have hoped that for simple cases like this, this expression simplifies, too, but unfortunately I don't see it.

Title: circumcenter of the $n$-simplex


问:
Calculation of Confidence Interval
We randomly choose 1000 people and run a survey on whether they prefer Apples or Berries. The results are $71\%$ for Apples and $29\%$ for Berries.

We want to calculate the Margin of error of the poll, with confidence level $90\%$.

My attempt, as I have just started to study probability and statistics (out of personal interest - I am not a student):

For confidence level of $90\%$, we get the value of $z = 1.645$.

So the MOE is $z*\sqrt{\frac{p(1-p)}{n}} = 1.645*\sqrt{\frac{0.71(0.29)}{1000}} = 2.36\%$?

Thank you!

答:
Yes, it is correct. The Moe is $\pm 2.36\%$ in the sense that the confidence interval at $90\%$ is

$$(71 \pm 2.36) \%$$ for apples and

$$(29 \pm 2.36)\%$$ for berries

If you want to type % you have to type $\text{\%}$ otherwise the symbol will not appear.

Title: Calculation of Confidence Interval


答:
You need to say $$f(x<2)= x-5, f(2\le x<3)=5x-13, f(x\ge 3)= 5-x$$
It is a continuous function such that $f(\pm \infty)=-\infty$ its plot suggests that and $f_{max}=f(3)=2$, see below. So the domain of $k$, is the range of $f(x)$ and it is $(-\infty, 2]$.

Body:
What is the domain of k?

$2|x-2|-3|x-3|=k$

Using the definition of absolute function, I obtain:

$|x-2|= \begin{cases}
        x-2, & x\ge2 \\
        2-x, & x<2\\
        \end{cases}\\$

$|x-3|= \begin{cases}
        x-3, & x\ge3 \\
        3-x, & x<3 \\
        \end{cases} \\$

For $x<2$

$x-5=k$

For $2\le x<3$

$5x-13=k$

For $x \ge 3$

$-x+5=k$

But I need to find the domain of k.


问:
Runge-Kutta method ; order of convergence

Write the Butcher table of the Runge-Kutta method defined by $$y_{n+1} = y_n + \frac{h}{2} f(x_n, y_n) + \frac{h}{2} f(x_n + h, y_{n+1})$$ and determine its order of convergence.

The answer by Lutz is really helpful. Thanks. Now

$$b_1=1/2, b_2=1/2, c_1=0, c_2=1, a_{11}=0, a_{12}=0, a_{21}=1/2, a_{22}=1/2$$

Can I just use $\sum_{i=1}^{2}=1$ and $\sum_{i=1}{2}b_ic_i=1/2$ to prove the consistency order is $0$?

答:
Give the function values their traditional names
$$
k_1=f(x_n,y_n),~~k_2=f(x_{n+1},y_{n+1})
$$
then
$$
y_{n+1}=y_n+\frac{h}2(k_1+k_2)
$$
and consequently also
$$
k_2=f(x_n+h,y_n+h(\tfrac12k_1+\tfrac12k_2)).
$$


问:
The Galois group of $\bar{K}(t)/K(t)$ is isomorphic to the Galois group of $\bar{K}/K$?
Consider the Galois extensions $\bar{K}/K$ and $\bar{K}(t)/K(t)$, where $\bar{K}$ is the algebraic closure of $K$. Is it true that their Galois groups are isomorphic, and if so how do we prove it? Do we need to use Hilbert's irreducibility theorem?

答: The answer is yes, and the reason is nothing more than the observation that any $K$-automorphism $\overline{K} \to \overline{K}$ gives rise to a $K(t)$-automorphism $\overline{K}(t) \to \overline{K}(t)$, and any $K(t)$-automorphism $\overline{K}(t) \to \overline{K}(t)$ must send any $x \in \overline{K}$ to one of its conjugates, which must still be in $\overline{K}$, and thus gives rise to a $K$-automorphism $\overline{K} \to \overline{K}$, and this determines the $K(t)$-automorphism $\overline{K}(t) \to \overline{K}(t)$ simply because $t \in K(t)$ means that it must fix $t$.


问:
Is it possible to find function that contains every given point?
Let say we have a arbitrary number of given points and there is at least one function, for which every point lies on its graph. Is it possible to find that function using only X and Y coordinates of every given point?

Example:

We are given points $A(0,1)$; $B(0.27;0)$ and $C (3.73;0)$. All of these points lie on the graph of the function $f(x) = x^2 - 4x + 1$. Is it possible to find this function using only X and Y coordinates of the point A,B and C.

Note: The number of point is can be really big and also the exponent of the function.

答:
There are many such functions so in general it's not possible to recover the one you started with.

For example if you had a set of points $\{(n,n): n\in\mathbb N\}$ both the functions $f(x) = x$ and $g(x) = x+sin(\pi x)$ would pass through all the points.

If you have a finite number of points you can find a polynomial that passes through them all.

One way to find it would be the following algorithm.

Let the coordinates of the points be $(x_1,y_1), (x_2,y_2),\dots, (x_N,y_N)$

First we find a function $f_1$ that passes through $(x_1,y_1)$ that's easy. $F_1(x) = y_1$.

Now suppose we've constructed $F_n$ that passes through $(x_1,y_1), \dots,(x_n, y_n)$ (so far we've only done this for $n=1$).

Let $$g_n(x) = \prod_{i=1}^n (x-x_i)$$
then $g_n(x_i) = 0$ for every $i\leq n$. Therefore for every value of $\lambda$ the function
$F_n(x_i) + \lambda g_n(x_i) = y_i$ for every $i\leq n$.
So set $$\lambda_n = \frac{y_{n+1} - F_n(x_{n+1})}{g_n(x_{n+1})}$$ and $$F_{n+1}(x) = F_n(x) +  \lambda_n g_n(x).$$

Then $F_{n+1}$ passes through $(x_i,y_i)$ for every $i\leq n+1$.
So apply this step repeatedly until you get $F_N$, which passes through every point.

Let's try it with your equation I'm going to choose slightly different points so that we can use whole numbers. Set
$$\begin{array}{rlrl}
x_1 &=0 & y_1 &=1 \\
x_2 &=2 & y_2 &=-3 \\
x_3 &=5 & y_3 &=6
\end{array}$$

It's easy to check that $x^2-4x+1$ passes through all three points. So let's follow the recipe.

First we set $F_1(x) = y_1 = 1$ and $g_1(x) = (x-x_0) = x$.

Then we can calculate $$\begin{array}{rl}\lambda_1 &= \frac{y_2 - F_1(x_2)}{g_1(x_2)} \\&= -2\end{array}$$
Hence $$\begin{array}{rl}F_2(x) &= F_1(x) + \lambda_1 g_1(x) \\ &= 1 -2x \end{array}$$

We can check that $F_2(0)=1$ and $F_2(2)=-3$.

Next we have $g_2(x) = x(x-2) = x^2-2x$
and we can work out $$\begin{array}{rl}\lambda_2 &= \frac{y_3 - F_2(x_3)}{g_2(x_3)} \\ &= \frac{6 - (-3)}{40} \\&= 1\end{array}$$

So $\lambda_2 = -\frac 3{15}$ and we have

Then we have $$\begin{array}{rl}F_3(x) &= F_2(x) + \lambda_2 g_2(x) \\
&= (1-2x) + (x^2 - 2x) \\
&= x^2-4x+1
\end{array}$$

Then as we've run out of points we set $f(x) = F_3(x)$ and we are done.


问:
From four red, five green, and six yellow apples, how many selections of eight apples are possible if three of each color are to be selected?
How do you do this question? If there were 9 apples it would be easy, but not sure how to select 8, if we have 3 colours from each.</p>

答: Well it's simply not possible so the answer is zero?</p>


问:
How to address the numerical instability of a function involving erf and an exponential
We need to compute this function:

$f(x) = x + \sqrt {\frac 2 \pi} \cdot \sigma_x \cdot \frac {e^{- ( \frac {x} {\sqrt 2 \cdot \sigma_x} )^2}} {1 + erf(\frac {x} {\sqrt 2 \cdot \sigma_x})}$

We noticed that, for negative values of $\frac {x} {\sqrt 2 \cdot \sigma_x}$ lower than a certain value (and different depending on which software we use to compute it, say about $-5$ for instance), there is numerical instability (oscillations).

I saw this post suggesting how to achieve better stability, although on a rather different function:

Numerical Instability with this cumulate

Do you think a similar method would work here? And if so, should I express the $erf$ as an integral?

Or would you suggest some other method? Taylor expansion? Anything else...?

Thanks

EDIT after evaluation of the responses

Thank you all very much for your feedback!

I think there is a complication here that I had not explicitly foreseen.

First I will make the following substitution, so your alternative expressions of $erf$ can be used directly:

$x = y \cdot \sqrt 2 \cdot \sigma_x$

$f(y) = \sqrt 2 \cdot \sigma_x \cdot y + \sqrt {\frac 2 \pi} \cdot \sigma_x \cdot \frac {e^{- y^2}} {1 + erf(y)}$

Now, the theory we used to derive this equation requires that:

$f(y) > 0, \forall y$

$\lim_{y \to -\infty} f(y) = 0$

So, we expect $f(y)$ to be very slowly approaching $0$ as $y$ becomes more negative; but $f(y)$ should never be negative.

In fact, if I plot the second term of $f(y)$, it looks like, as $y$ becomes more and more negative:

$\sqrt {\frac 2 \pi} \cdot \sigma_x \cdot \frac {e^{- y^2}} {1 + erf(y)} \approx - \sqrt 2 \cdot \sigma_x \cdot y$

This is in substantial agreement with Claude's approximation for the $erf$ part.

I.e., using:

$\frac{e^{-y^2}}{1+\text{erf}(y)} \approx -\sqrt{\pi } \,y\Bigg[1+\frac{1}{2 y^2}-\frac{1}{2 y^4}+\frac{5}{4
   y^6}+O\left(\frac{1}{y^8}\right)\Bigg]$

I get:

$\sqrt {\frac 2 \pi} \cdot \sigma_x \cdot \frac {e^{- y^2}} {1 + erf(y)} \approx \sqrt {\frac 2 \pi} \cdot \sigma_x \cdot (-\sqrt{\pi }) \,y\Bigg[1+\frac{1}{2 y^2}-\frac{1}{2 y^4}+\frac{5}{4
   y^6}+O\left(\frac{1}{y^8}\right)\Bigg] =$

$= - \sqrt {2} \cdot \sigma_x \cdot \,y\Bigg[1+\frac{1}{2 y^2}-\frac{1}{2 y^4}+\frac{5}{4
   y^6}+O\left(\frac{1}{y^8}\right)\Bigg]$

which indeed approaches $- \sqrt 2 \cdot \sigma_x \cdot y$ when $y$ becomes very negative.

With the above approximation:

$f(y) \approx \sqrt 2 \cdot \sigma_x \cdot y - \sqrt {2} \cdot \sigma_x \cdot \,y\Bigg[1+\frac{1}{2 y^2}-\frac{1}{2 y^4}+\frac{5}{4
   y^6}+O\left(\frac{1}{y^8}\right)\Bigg] = $
$= - \sqrt {2} \cdot \sigma_x \cdot \,y\Bigg[\frac{1}{2 y^2}-\frac{1}{2 y^4}+\frac{5}{4
   y^6}+O\left(\frac{1}{y^8}\right)\Bigg] $

Numerically, this 'does the trick', i.e. for $y \le -4$ it seems really very close to the function I need to compute, as indicated by Claude. And in fact with the rational function version even up to $y \le -1$.

This seems to imply that only polynomial approximations are viable, not the $erfc$-based ones, or am I wrong?

Any thoughts?

答:
In the negatives, $$\text{erf(x)}$$ very quickly tends to $-1$ and it is no surprise that you soon reach the limits of the floating-point representation.

You will regain full accuracy by computing

$$\text{erf}(x)+1$$ directly. It is possible that you will obtain better results with $\text{erfc}(-x)$, but that depends on the numerical library you use.

Otherwise, you could use the fourth approximation in https://en.wikipedia.org/wiki/Error_function#Approximation_with_elementary_functions, and you will see a nice simplification with your numerator.


问:
Point lies inside a triangle ABC with $\measuredangle BAC=45^\circ$ and $\measuredangle ABC=30^\circ$
A triangle $ABC$ is given with $\measuredangle BAC=45^\circ$ and $\measuredangle ABC=30^\circ.$ Point $M$ lies inside the triangle and $\measuredangle MAB=\measuredangle MBA=15^\circ.$ Find $\measuredangle BMC$.

I am not sure how to approach the problem. We can find some angles:$$\measuredangle ACB=180^\circ-45^\circ-30^\circ=105^\circ\\ \measuredangle CAM=45^\circ-15^\circ=30^\circ\\ \measuredangle MBC=30^\circ-15^\circ=15^\circ\\ \measuredangle AMB=180^\circ-2\cdot15^\circ=150^\circ.$$

I will be very grateful if we can see a solution without using trigonometry.

答:
Please apply Trigonometric form of Ceva's theorem.

If $\angle BCM = x$,

$\sin \angle BAM \ \sin \angle ACM \sin CBM = \sin \angle MAC \ \sin \angle MCB \ \sin \angle MBA$

$\sin 15^0 \sin (105^0-x) \sin 15^0 = \sin 30^0 \sin x \sin 15^0$

$\cos (15^0-x) \sin 15^0 = 2 \sin 15^0 \cos 15^0 \sin x$

$\sin 30^0 \cos (15^0-x)  = \cos 15^0 \sin x$

$\sin (45^0-x) + \sin (15^0+x)  = \sin (15^0+x) + \sin(x-15^0)$

$45^0-x = x-15^0 \implies x = 30^0$

So $\angle BMC = 180^0 - 30^0 - 15^0 = 135^0$.

EDIT: For geometric solution, see the below construction that you can use. $AN$ is angle bisector of $\angle CAM$. You can prove $\angle ANM = \angle ANC = 60^0$ and $CN = NM$. So, $\angle MCB = 30^0$. $\therefore \angle BMC = 135^0$.


问:
Expected number of different colors selected in k picks from balls of n colors
Problem statement reads like this - There are n different colored balls available in a basket. What is value of expected number of distinct colors selected from k random picks with replacement. I approached the problem in following manner

Considering n = 4, k = 3.

Number of possibilities of selecting 3 different colors is $\frac{4\times3\times2}{4^3}$.

Number of possibilities of selecting 2 different colors is $\frac{4\times3\times3}{4^3}$ (selecting first color, then putting it any of three places, then choosing next color)

Number of possibilities of selecting 1 different color is $\frac{4}{4^3}$

This leads me to correct answer however, I'm not sure how to approach the problem when n and k is large like $10^5$.

答:
Let $X_i$ be the event: the $i$-th color was already drawn. The expected value of this event after $k$ picks is just the probability that this event happens:
$$
\mathbb E(X_i)=1-\left(\frac{n-1}n\right)^k.\tag1
$$
By the symmetry and linearity of expectation:
$$
\mathbb E\left(\sum_iX_i\right)=\sum_i\mathbb E(X_i)=
n\left[1-\left(\frac{n-1}n\right)^k\right].\tag2
$$


问:
Tangential Space of a differentiable manifold is always $\mathbb R^n$?
Let $\mathcal M$ be a differential manifold with a point $p$. Let U be an open set, $p\in U$, on $\mathcal M$ and let $\phi,\psi:U\to \mathbb R^n$ be a charts on $\mathcal M$. I'm having diffculties arranging all the concepts of a differential manifold. The "manifold" of required spaces gives me a headache. At the moment im utterly confused with the construction of the tangent space at $p$.
I'll start describing my thoughts from the start: $\require{AMScd}$

\begin{CD}
 U   @>\phi>>  \mathbb R_\phi^n \\
@V \psi VV @.\\
\mathbb R_\psi^n
\end{CD}

That's the open set U with charts $\psi,\phi$. I denoted $\mathbb R^n$ with the chart pointing to it, so we can make sure to differentiate between the different $\mathbb R^n$ -spaces we will encounter.

The tangent space of $\mathcal M$ or $U$ in $p$ is a $n$-dimensional $\mathbb R$-vector space, which has isomorpisms pointing to $T_{\phi(p)} \mathbb R^n$ and $T_{\psi(p)} \mathbb R^n$.
Where $T_{\phi(p)} \mathbb R^n$ and $T_{\psi(p)} \mathbb R^n$ are vector spaces we're inducing by the choice of our bases $B_\psi , B_\phi$ depending on our charts as follows after the diagram:

\begin{CD}
T_p \mathcal M   @> \cong >>  T_{\phi(p)} \mathbb R^n \\
@V \cong  VV @.\\
T_ {\psi(p)} \mathbb R^n
\end{CD}

$T_{\phi(p)} \mathbb R^n$ and $T_{\phi(p)} \mathbb R^n$ are induced by the vector space bases $B_\psi , B_\phi$ using isomorphisms from the images of our charts ($\phi(U)\subset \mathbb R_{\phi}^n, \psi(U)\subset\mathbb R_{\psi}^n)$ to the tangent space $T_{\phi(p)} \mathbb R^n$.
The first base vector of the base $B_\phi$ is given as $\frac{\partial}{\partial x_i}|_p := \phi^-1 (\phi(p)+ t e_i)$, the other base vectors of $B_\phi$ follow with $i \in {2,....,n}$.

Let's call the function that takes $p$ and gives us $\frac{\partial}{\partial x_i}|_p$ and $\frac{\partial}{\partial y_i}|_p$ by names  $\bar{\phi}_p$ and $\bar{\psi}_p$. In a diagram we now got two choices for the base of our ismorphic tangent spaces depending on $\phi,\psi$:

\begin{CD}
   T_{\phi(p)} \mathbb R^n   @. T_{\psi(p)} \mathbb R^n \\
@A \bar{\phi}_p AA @A \bar{\psi}_p AA\\
\mathbb R_\phi^n @. \mathbb R_\psi^n
\end{CD}

First question: Why is $T_p \mathcal M$ an $\mathbb R$-vector space, i don't know why the underlying field has to be $\mathbb R$. Or is it, that we just identify $T_p M$ with $T_{\phi(p)}\mathbb R^n$, if so, why can we do that?
I have the notion that $\mathcal M$ is of a abstract nature, so how comes it's tangential space is so $\mathbb R^n$ related?

Second question:
Assuming i understood why $T_p \mathcal M$ is a $\mathbb R$-vector space, i can now ask myself how i can change bases.
So we get to linear algebra and the pushforward as linear function between vector spaces.
The image of the coordinate vector $(1,0,...,0)^t$,of the first base vector $\frac{\partial}{\partial x_i}|_p$, is the first row of the matrix $D((\psi)\circ\psi^-1)$ giving us the coordinate vector expressed in the base $B_\psi$.

\begin{CD}
T_{\phi(p)} \mathbb R^n   @>D((\psi)\circ\phi^-1)>> T_{\psi(p)} \mathbb R^n
\end{CD}

Third question: The "last" notion that gives me a headache right now is that of the function:

$$\phi_*: T_p \mathcal M \to T_{\phi(p)} \mathbb R^n\text{ defined as }
\phi_*(\gamma) := \frac{d}{dt}(\phi \circ \gamma )|_{t=0}$$

where $\gamma$ is a tangent vector, for example a curve(geometric defintion of a tangent vector).

Is that definition base independend? If so taking the first base vector of $B_\phi$
namely $\frac{\partial}{\partial x_i}|_p := \phi^-1 (\phi(p)+ t e_i)$ and using $\phi_*$ on it, gives us another tangent vector in the tangent space $T_{\phi_*} \mathbb R^n$, or is it that same as $\mathbb R_\phi^n$ or $T_{\phi(p)}R^n$?

Thanks for any answers!

答:
First question:

Congratulations, you have discovered algebraic varieties and their Zariski tangent space. For classical differential geometry, the reason that the tangent space is an $\mathbb{R}$ vector space is by definition: the local model of a differentiable manifold is $\mathbb{R}^n$. The whole motivation of classical differential geometry is to study stuff that, at sufficiently small neighborhoods, look like $\mathbb{R}^n$. It would be strange if its tangent space does not look like that of $\mathbb{R}^n$, but be over some other field, no? If you model the local geometry by an affine variety associated to a different base field, you will get appropriately different vector spaces as the tangent space.

Second question:

Looks like you more or less got it, except you probably meant $\psi\circ \phi^{-1}$ rather than $\psi \circ \psi^{-1}$. Again, locally you are just working with functions defined over $\mathbb{R}^n$ (which may take $\mathbb{R}^k$ values).

Third question:

$\phi_*$ pushes forward to $T_{\phi(p)}\mathbb{R}^n$, not $T_p\mathbb{R}^n$. Since you didn't explain what $\gamma$ is, I have no idea what your second equation means. But the derivative with respect to $t$ makes me wonder if you intended $\gamma:\mathbb{R}\to M$ and you are actually thinking about the expression

$$ \phi_*(\frac{\mathrm{d}}{\mathrm{d}t}{\gamma}(t)) = \frac{\mathrm{d}}{\mathrm{d}t}( \phi\circ \gamma)(t)~? $$

In that case, this definition is base independent. You can explicitly verify it by composing with the smooth maps associated to changes of charts.

Your expression $\partial/\partial x_i |_p = \phi^{-1}(\phi(p) + t e_i)$ is wrong. The left hand side is an object living in $T_p M$ and the right hand side is an object living in $M$. You appear to have a type error.


答:
It can easily be solved using indicator variables.

Let $X_i$ be an indicator variable that equals $1$ if the $i_{th}$ color appears in $k$ trials, and $0$ otherwise

Then $\Bbb{P}(X_i) = 1 - (\frac{n-1}{n})^k$

Now the expectation of an indicator variable is just the probability of the event it indicates,

so $\Bbb{E}(X_i) = \Bbb{P}(X_i) = 1 - (\frac{n-1}{n})^k$

and by linearity of expectation, which applies even if the variables are not independent,

$\Bbb{E}(X) = \Bbb{E}(X_1) + \Bbb{E}(X_2) + .... \Bbb{E}(X_n)$

$= n\left(1 - (\frac{n-1}{n})^k\right)$

Title: Expected number of different colors selected in k picks from balls of n colors

Body:
Problem statement reads like this - There are n different colored balls available in a basket. What is value of expected number of distinct colors selected from k random picks with replacement. I approached the problem in following manner

Considering n = 4, k = 3.

Number of possibilities of selecting 3 different colors is $\frac{4\times3\times2}{4^3}$.

Number of possibilities of selecting 2 different colors is $\frac{4\times3\times3}{4^3}$ (selecting first color, then putting it any of three places, then choosing next color)

Number of possibilities of selecting 1 different color is $\frac{4}{4^3}$

This leads me to correct answer however, I'm not sure how to approach the problem when n and k is large like $10^5$.


问:
One-point compactification of a linear space
Is it possible to take the one-point compactification of a linear space and get again a linear space? Like if $X$ is a normed space add $*$ to it and define $|x-*| = 1$?</p>

答:
Suppose $\hat{\Bbb R}$ is a copmpatification of $\Bbb R$ and the normal norm on $\Bbb R$ can be extended to $\hat{\Bbb R}$. Then $\Bbb R$ is a topological subgroup of the topological group $\hat{\Bbb R}$. This means that $\Bbb R$ is precompact. Precompactness of $\Bbb R$ means any sequence in $\Bbb R$ must have a Cauchy subsequence. $\{n^2\}$ has not a Cauchy subsequence. A contradiction.

See also: this post and Alex Ravsky's comments in this thread.

If anything is wrong let me know. I'll fix or delete this post.


问:
What is the prob. that you find at least 2 typos?
Assume that the prob. of at least one typo in the slides is 0.2.

What is the prob. that you find at least 2 typos?

I realize that typos are a rare event, so I need to use a Poisson Approximation to find this probability. Let $X\sim\text{Poisson}(\lambda)$. The relevant formula is
$$\mathbb{P}(X=k)=e^{-\lambda}\frac{\lambda^k}{k!}$$
From the problem, I am given $\mathbb{P}(X\geq1)=0.2$. and the probability I am trying to find is $\mathbb{P}(X\geq 2)$. I think that it is easier to find $\mathbb{P}(X<2)$ so I will take its complement $\mathbb{P}(X\geq2)=1-\mathbb{P}(X<2)$. Then, my desired probability is found as follows:
$$\mathbb{P}(X\geq2)=1-\mathbb{P}(X=1)$$
My problem is that I do not know how to find $\lambda$ from the given information.

答:
$$\mathbb{P}(X\geq 1)=1-\mathbb{P}(X=0)=1-e^{-\lambda}=0.2$$

thus easy find $\lambda=-\log 0.8\approx 0.223$


问:
Using the power rule to find the derivative of $\frac{d}{dx} (\sqrt x)$
Using the power rule, $\frac{d}{dx} (x^n) = n x^{n-1}$, where $n = \frac{1}{2}$ to find the derivative of:

$$
\frac{d}{dx} (\sqrt x)
$$

Steps:

(1) $= \frac{d}{dx}(x^{1/2})$

(2) $= \frac{x^{-1/2}}{2}$

(3) $= \frac{1} {2\sqrt x}$

I get confused at the last two steps. How does the numerator become $x^{-1/2}$ and the denominator become ${2}$ in step 2? And then the numerator $1$ in step 3?

Body: $$\frac{d(\sqrt x)}{dx}=\frac{d(x^{1/2})}{dx}=\frac12x^{1/2-1}=\frac12x^{-1/2}=\frac{x^{-1/2}}2=\frac12\frac1{x^{1/2}}=\frac1{2x^{1/2}}=\frac12\frac1{\sqrt x}=\frac1{2\sqrt x}.$$


问:
Help understanding change of basis problem

So I'm struggeling with (ii). To be honest, I don't even understand what it means when the say that I'm supposed to assume that $\vec{u}$ in E-coordinates. And does the apostrophe mean that we are supposed to find the derivative? I'm really lost as one can see and I don't find anything in my book explaining this. Does someone have some tips or links a similar problem?

Edit: So I found a theorem in my book that might help after all:

$[x]_c=P_{c\leftarrow b}[x]_b$

So is this right then when it comes to my problem?

$\pmatrix{x'\\y'}$=$\pmatrix{cos\theta & sin\theta\\-sin\theta&cos\theta}\pmatrix{x\\y}$ where the $2\times 2$ matrix is the $P_{(\beta\leftarrow\epsilon)}$

$x'=xcos\theta+ysin\theta$

$y'=-xsin\theta+ycos\theta$

答:
So coordinate is coefficient of your basis.  I suppose x and y are coordinate. To get a concrete understanding , i recommand you to assume epsilon coordinate system to be e1=(1,0) e2=(0,1) and beta to be e'1=(1,1) e'2=(-1,1)  ex) u= x'e'1+ y'e'2 = (x'-y',x'+y') = (x,y) = xe1+ye2
now get abstract using your text book. coordinate change matrix</p>


问:
finding groups $G$ such that $G \simeq \operatorname{Aut}(G)$
I proved that the mapping $\phi: G \to \operatorname{Aut}(G)$ that sends $g \longmapsto c_g (x) = gxg^{-1}$ is a homomorphism, but I need to find a group $G$ for which this map establishes an isomorphism. I know that $S_3$ is isomorphic to $\operatorname{Aut}(G)$, but I don't know off-hand whether this bijection is the correct one. I referenced a similar answer here Proving that ${\rm Aut}(S_3)$ is isomorphic to $S_3$, but it doesn't give an explicit formula for the bijection, and though I understand that $S_3$ is generated by its transpositions and so any automorphism is determined completely by how it acts on $(12)$, $(13)$, and $(23)$, I don't understand how a bijection follows.

Updated Attempt:

I claim that $\phi: S_3 \to \mathrm{Aut}(S_3)$ sending $g \longmapsto c_g (x) = gxg^{-1}$ is an isomorphism. In general, $\phi: G \to \mathrm{Aut}(G)$ sending $g \longmapsto c_g$ is a homomorphism, so it suffices to show that $\phi$ is a bijection. We have:
\begin{align*}
g \in \mathrm{ker}(\phi) & \iff \phi(g) = c_g (x) = x, \; \forall x \in G \\
& \iff gxg^{-1} = x, \; \forall x \in G \\
& \iff gx = xg, \; \forall x \in G \\
& \iff g \in Z(G) = \{e\},
\end{align*}
so the kernel of $\phi$ is trivial, so $\phi$ is injective. I claim that $\phi$ is also surjective. We have $S_3 = \langle a = (12), b = (13), c = (23) \rangle$, each of which has order $2$, so given $g \in S_3$, we have $g = a^i b^j c^k$ for $i,j,k \in \{0,1,2\}$. Since an automorphism $f \in Aut(S_3)$ must preserve the order of elements, $f$ must send transpositions to transpositions. Furthermore, upon fixing where $f$ send these transpositions, the rest of the map is determined. Since there are $3!$ possibilities for where to send the permutations, there are exactly $6$ automorphisms, so $|S_3| = |\mathrm{Aut}(S_3)|$. But $\phi(S_3) \leq \text{Aut}(S_3)$, so that they have the same order immediately implies that $\phi(S_3) = \text{Aut}(S_3)$, so $\phi$ is surjective, hence bijective.

答:
Your map $\phi:G\to Aut(G)$ has image $im(\phi)=Inn(G)=\{c_g| \ g\in G\}$ and $Aut(G)/Inn(G)=Out(G)$ so as mentioned in the comments, $\phi$ is onto iff $Out(G)=1$. Also $ker(\phi)=Z(G)$ hence $\phi$ is $1-1$ iff $Z(G)=1$. To sum up, $\phi$ is an isomorphism iff $Out(G)=Z(G)=1$.

It could be $Aut(G)\cong G$ with the isomorphism between $G$ and $Aut(G)$ being othen than $\phi$.


答:
I would suggest a series expansion for $x \leq- 4$
$$f(x)=\frac{e^{-x^2}}{1+\text{erf}(x)}=-\sqrt{\pi } \,x\Bigg[1+\frac{1}{2 x^2}-\frac{1}{2 x^4}+\frac{5}{4
   x^6}+O\left(\frac{1}{x^8}\right)\Bigg]$$ For $x=-5$, the "exact" value is $9.03318$ while this truncated expansion gives $9.03313$.

Much better would be the Padé approximants
$$f(x)=-\sqrt{\pi } \,x \,\frac{8 x^6+84 x^4+210 x^2+105 } {8 x^6+80 x^4+174 x^2+48 }$$ which, for $x=5$, leads to an error of  $2.2\times 10^{-8}$.

Title: How to address the numerical instability of a function involving erf and an exponential

Body:
We need to compute this function:

$f(x) = x + \sqrt {\frac 2 \pi} \cdot \sigma_x \cdot \frac {e^{- ( \frac {x} {\sqrt 2 \cdot \sigma_x} )^2}} {1 + erf(\frac {x} {\sqrt 2 \cdot \sigma_x})}$

We noticed that, for negative values of $\frac {x} {\sqrt 2 \cdot \sigma_x}$ lower than a certain value (and different depending on which software we use to compute it, say about $-5$ for instance), there is numerical instability (oscillations).

I saw this post suggesting how to achieve better stability, although on a rather different function:

Numerical Instability with this cumulate

Do you think a similar method would work here? And if so, should I express the $erf$ as an integral?

Or would you suggest some other method? Taylor expansion? Anything else...?

Thanks

EDIT after evaluation of the responses

Thank you all very much for your feedback!

I think there is a complication here that I had not explicitly foreseen.

First I will make the following substitution, so your alternative expressions of $erf$ can be used directly:

$x = y \cdot \sqrt 2 \cdot \sigma_x$

$f(y) = \sqrt 2 \cdot \sigma_x \cdot y + \sqrt {\frac 2 \pi} \cdot \sigma_x \cdot \frac {e^{- y^2}} {1 + erf(y)}$

Now, the theory we used to derive this equation requires that:

$f(y) > 0, \forall y$

$\lim_{y \to -\infty} f(y) = 0$

So, we expect $f(y)$ to be very slowly approaching $0$ as $y$ becomes more negative; but $f(y)$ should never be negative.

In fact, if I plot the second term of $f(y)$, it looks like, as $y$ becomes more and more negative:

$\sqrt {\frac 2 \pi} \cdot \sigma_x \cdot \frac {e^{- y^2}} {1 + erf(y)} \approx - \sqrt 2 \cdot \sigma_x \cdot y$

This is in substantial agreement with Claude's approximation for the $erf$ part.

I.e., using:

$\frac{e^{-y^2}}{1+\text{erf}(y)} \approx -\sqrt{\pi } \,y\Bigg[1+\frac{1}{2 y^2}-\frac{1}{2 y^4}+\frac{5}{4
   y^6}+O\left(\frac{1}{y^8}\right)\Bigg]$

I get:

$\sqrt {\frac 2 \pi} \cdot \sigma_x \cdot \frac {e^{- y^2}} {1 + erf(y)} \approx \sqrt {\frac 2 \pi} \cdot \sigma_x \cdot (-\sqrt{\pi }) \,y\Bigg[1+\frac{1}{2 y^2}-\frac{1}{2 y^4}+\frac{5}{4
   y^6}+O\left(\frac{1}{y^8}\right)\Bigg] =$

$= - \sqrt {2} \cdot \sigma_x \cdot \,y\Bigg[1+\frac{1}{2 y^2}-\frac{1}{2 y^4}+\frac{5}{4
   y^6}+O\left(\frac{1}{y^8}\right)\Bigg]$

which indeed approaches $- \sqrt 2 \cdot \sigma_x \cdot y$ when $y$ becomes very negative.

With the above approximation:

$f(y) \approx \sqrt 2 \cdot \sigma_x \cdot y - \sqrt {2} \cdot \sigma_x \cdot \,y\Bigg[1+\frac{1}{2 y^2}-\frac{1}{2 y^4}+\frac{5}{4
   y^6}+O\left(\frac{1}{y^8}\right)\Bigg] = $
$= - \sqrt {2} \cdot \sigma_x \cdot \,y\Bigg[\frac{1}{2 y^2}-\frac{1}{2 y^4}+\frac{5}{4
   y^6}+O\left(\frac{1}{y^8}\right)\Bigg] $

Numerically, this 'does the trick', i.e. for $y \le -4$ it seems really very close to the function I need to compute, as indicated by Claude. And in fact with the rational function version even up to $y \le -1$.

This seems to imply that only polynomial approximations are viable, not the $erfc$-based ones, or am I wrong?

Any thoughts?


问:
Proving $x\left(1+e^x\sin^2x\right)^{-1}$ is Lebesgue integrable
How does one prove that $x\left(1+e^x\sin^2x\right)^{-1}$ is Lebesgue integrable on $(0,\infty)$?

The source of this problem includes a hint to compare this function with one of the form $\left(1+x^2\right)^{-1}+\text{a simple function}$, but I don't see how to do that either.

答:
This is a continuous function, so we just need to show that it is integrable on $[N, \infty)$ for some (large) $N>0$. In the following we assume $x\geq N$, we will decide what is $N$.

For any $k\pi\geq N-2\pi$, choose $r_k>0$ so that $e^{k\pi} r_k^2=(k\pi)^4$, that is,
$$
r_k=e^{-k\pi/2}(k\pi)^2.
$$
We can pick $N$ big so that $k\pi\geq N-2\pi$ implies
$e^{-k\pi/2}(k\pi)^2<k^{-3}$.
Thus
$$
r_k<k^{-3}.
$$
Now for any $x\geq N$ so that $x\notin (k\pi-r_k, k\pi+r_k)$ for any $k\pi\geq N-2\pi$, we find $n$ so that $n\pi-\frac{\pi}{2}\leq  x\leq n\pi+\frac{\pi}{2}$; so $|x-n\pi|\geq r_n$, so
$$
1+e^{x}\sin^2x\geq 1+e^{n\pi-\frac{\pi}{2}}\sin^2r_n\geq 1+c^2e^{n\pi}r_n^2=1+c^2(n\pi)^4.
$$
Here we used the well known fact that $\frac{\sin r}{r}> c=\frac{2}{\pi}$ when $|r|<\frac{\pi}{2}$. Thus if $N$ is sufficiently big (note $n\pi\geq N-2\pi$), we have
$$
1+e^{x}\sin^2x\geq 1+(n\pi+\pi/2)^3.
$$
In particular, since $x\leq n\pi+\frac{\pi}{2}$, we see
$$
1+e^{x}\sin^2x\geq 1+x^3.
$$
In conclusion, for $x\geq N$ outside every $(k\pi-r_k, k\pi+r_k)$, we have
$$
f(x)=\frac{x}{1+e^x\sin^2x}\leq \frac{x}{1+x^3}.
$$
Thus
$$
\int_{[N, \infty)-\cup_k (k\pi-r_k, k\pi+r_k)} f(x)dx<\int_N^\infty \frac{x}{1+x^3}dx<\infty.
$$

On $(k\pi-r_k, k\pi+r_k)$, we estimate simply $f(x)\leq \frac x1 \leq k\pi+r_k$, so
$$
\sum_k \int_{k\pi-r_k}^{k\pi+r_k}f(x)dx< \sum_k (k\pi+r_k)\cdot 2r_k\leq \sum_k 2(k+1)\pi k^{-3}<\infty.
$$
The summation is over those $k$ with $k\pi\geq N-2\pi$. Add these two estimates we get
$$
\int_N^\infty f(x)dx<\infty.
$$


问:
Integrability as prerequisite in the Monotone Convergence Theorem
I know the Monotone Convergence Theorem formulated as follow:

Let $(f)_{n \geq 1}$ be an sequence of measurable functions isotonely converging to a function $f$ that is numerically measurable. If either $f_n \geq 0 \,\, \forall n\geq1 \,\, \mu-a.e.$ $\textbf{or}$ $f_1 \in \mathcal{L^1}(\mu)$ with $\mu$ being a measure, are fulfilled, the follwing equation holds:

$\\\\ \int f d\mu = \int \lim_{n \to \infty} f_n d\mu =\lim_{n \to \infty}\int f_n d\mu $

I am wondering why it's possible to dispense the prerequisite of non-negativity if we have integrability of the function $f_1$. Can anyone help out?

One of my friends guessed that both conditions enable monotonicity of integrals which is needed to proof the theorem and are therefore valid.

Thanks in advance.

答: Substract f1 on all sequence. Since f1 is L1 we have finite integral of f1</p>


问:
Pointwise limit of seq.
$(f_n)_(n\geq2)$

$$f_n (x) =
 \begin{cases}
  1-n^2x &\quad\text{for}\quad x \in [0,\frac{1}{n^2}]\\
  0 &\quad\text{for}\quad x \in (\frac{1}{n^2},\frac{2}{n^2}) \\
  x - \frac{2}{n^2} &\quad\text{for}\quad x \in [\frac{2}{n^2},1].
 \end{cases} $$

We just need to find the pointwise limit and i'm a little stuck.

I know that for $x=0$ we can do $f_n(0) = 1$ and hence $lim$ here is 1

But then I want to fix x between $0,1$ would i do this in separate cases. One between $(0,\frac{1}{n^2})$  and $[\frac{1}{n^2},1]$

I'm kind of stuck if anyone could show a working to this, it'd be really helpful

答: If $x>0$, then $\frac2{n^2}<x$ if $n\gg1$ and then you have $f_n(x)=x-\frac2{n^2}$. Since $\lim_{n\to\infty} x-\frac2{n^2}=x$, you have $\lim_{n\to\infty}f_n(x)=x$.

Title: Pointwise limit of seq.


问:
What is the domain of k? $2|x-2|-3|x-3|=k$
What is the domain of k?

$2|x-2|-3|x-3|=k$

Using the definition of absolute function, I obtain:

$|x-2|= \begin{cases}
        x-2, & x\ge2 \\
        2-x, & x<2\\
        \end{cases}\\$

$|x-3|= \begin{cases}
        x-3, & x\ge3 \\
        3-x, & x<3 \\
        \end{cases} \\$

For $x<2$

$x-5=k$

For $2\le x<3$

$5x-13=k$

For $x \ge 3$

$-x+5=k$

But I need to find the domain of k.

答:
If $ -\infty <x≤2$, then you have

$$k=2(2-x)-3(3-x)=x-5 \Longrightarrow - \infty <k=x-5≤-3$$

If $2<x≤3$, then you have

$$k=2(x-2)-3(3-x)=5x-13 \Longrightarrow -3<k=5x-13≤2$$

If $3<x<\infty$, then you have

$$k=2(x-2)-3(x-3)=-x+5 \Longrightarrow 2>k=-x+5> - \infty \Longrightarrow -\infty <k<2$$

Finally, you get

$$k \in (- \infty, -3] ∪ (-3, 2] ∪ (-\infty, 2)= (-\infty,2].$$


问:
Probability of a sum being even
Say for every natural number z $\ge$ 1, we choose a number $a_z$ from the set {${1,...,z}$} randomly with equal probability for each number 1 to z. For example, $a_3$ we can choose 1, 2, or 3 all with probability 1/3. If b $\ge$ 2 what is the probability that the sum of $a_1$ + ... + $a_b$ is even for any b.

I know that whenever b is an even number the probability that the number chosen is odd is 1/2. I tried to use the fact that $a_1$ has to be 1 to try to see when I can have the sum of two odds since that is even. However, I am stuck on figuring out the probability that the entire sum is even.

答:
Let's call the sum $S_b = a_1+a_2+\cdots+a_{b-1}+a_b$.

If $b$ is even and $b\ge 2$ then the probability is $\frac12$ that $a_b$ is odd and so changes the parity of the sum, and $\frac12$ that $a_b$ is even and so does not change the parity of the sum, making the probability that $S_b$ is odd or even $\frac 12$ each.

If $b$ is odd then $b-1$ is even. So as long as $b-1\ge 2$, we can now say $S_{b-1}$ is equally likely to be odd or even. Meanwhile $a_b$ is odd with probability $\frac{b+1}{2b}$ and even with probability $\frac{b-1}{2b}$.  So the probability $S_b$ is even given $b$ is odd and $b\ge 3$ is $\frac12 \frac{b+1}{2b}+\frac12 \frac{b-1}{2b}= \frac12$ too.

That leaves the cases $b=0$ and $b=1$, which are easy to handle.

When $b=0$ then $S_0=0$ and even (the empty sum) so $\mathbb P(S_0 \text{ is even})=1$.

When $b=1$ then $S_0=1$ and odd so $\mathbb P(S_1 \text{ is even})=0$.

When $b\ge 2$ then $\mathbb P(S_b \text{ is even})=\frac12$.

Title: Probability of a sum being even


问:
Solve system of equations involving cos and sin
I have come up with the following system, I want to solve it for $a$ and $c$:

$
a \sin (x_0) - c \sin(x_0 - L) = 0\\
c \cos(x_0 - L) - a \cos(x_0) = 1
$

In this system $x_0$ and $L$ are arbitrary.

p.s. why can't $L$ be $n\pi$?

Body:
$
a \sin (x_0) - c \sin(x_0 - L) = 0\\
c \cos(x_0 - L) - a \cos(x_0) = 1
$

from eqn (1)

$$a \sin (x_0) =c \sin(x_0 - L)\implies a=\dfrac {c \sin(x_0 - L)}{\sin (x_0)}$$

in eqn (2)

$$c \cos(x_0 - L) - \cos(x_0)\cdot\dfrac{c \sin(x_0 - L)}{\sin (x_0)} = 1$$
$$c[\cos(x_0 - L)\cdot\sin (x_0)-\sin(x_0 - L)\cdot\cos(x_0)]=\sin {(x_0)}$$
$$c[\sin (x_0-(x_0 - L))]=\sin {(x_0)}$$
$$c=\dfrac{\sin {(x_0)}}{\sin (L)}$$

Now $$a=\dfrac {c \sin(x_0 - L)}{\sin (x_0)}$$
$$a=\dfrac {\dfrac{\sin {(x_0)}}{\sin (L)}\cdot\sin(x_0 - L)}{\sin (x_0)}$$
$$a=\dfrac{\sin(x_0-L)}{\sin L}$$


问:
What does "#" mean here? "[these structures] are diffeomorphic to $\#^kS^1\times B^3$"
I am reading The topology of 4-manifolds by Kirby.

At page 7 the author uses the symbol # what does it mean?

The sentence for context is this:

However, the 3-handles and 4-handle of a closed $M^4$ together are diffeomorphic to $\#^kS^1\times B^3$...

答:
It means connected sum, e.g. $S^1 \# S^1$. For details see, for example,

connected sum of surfaces is well defined proof attempt

For the "iterated" notation see here:

notation for connected sum $\#^n S^2 \times S^2$


问:
Split 3D representation of S3 in irreducible components
I saw from this post that you can prove that the 3d representation of S3 is reducible.

What if I want to split this representation in a sum of irreducible representation?

答: I'll assume that by "the 3d representation" you mean the representation afforded by $S_3$ acting on $\mathbb R^3$ by permuting components. Counting fixpoints shows that the character of this representation is $3$ at the identity, $1$ at transpositions and $0$ at $2$-cycles. Then forming scalar products with the rows of the character table of $S_3$ shows that this representation is the sum of the trivial representation and the two-dimensional irreducible representation.</p>


问:
Can I get a polynomial out of a polynomial division?
I have a to divide a thrid degree polynomial by another one,
is it possible for me to get another polynomial out of it?

By the way, I am working with decimal numbers (floats in 32bits).

If it is possible, this problem needs to be solved relatively quickly as I plan on using the solution in a videogame, basically every frame.
Let's say the division looks something like this:
$$\frac{ax^3+bx^2+cx+d}{ex^3+fx^2+gx+h}$$
How would you get a polynomial out of it?
Any suggestion is appreciated, thanks in advance.

EDIT: So it seems like my problem wasn't properly
explained what i wanted is to get h(x) where
$$h(x)  =\frac{ax^3+bx^2+cx+d}{ex^3+fx^2+gx+h}$$
written as
$$h(x) = jx^3+kx^2+lx+m$$
(or get a higher degree polynomial).

However, as @no lemon no melon said, it turns out that the result of the division of polynomials is not a polynomial but rather a rational function.
So I guess that my question was
"is the result of a polynomial division a polynomial?".

This is really new to me so hopefully this didn't seemed too stupid, again thanks a lot.

答:
Suppose $\frac{f}{g}$ gives you a polynomial $h$ of degree $k$.

If degree of $f$ is $3$ and degree of $g$ is $3$.

Then we have $f=gh$. Degree of $f$ is equal to the degree of $g$ plus the degree of $h$.

Hence degree of $h$ must be $0$, the constant polynomial.


问:
Integral of gamma - exponential product
The following integral is part of a calculation I have been working on:

$$
\int_{-\infty}^{+\infty} e^{ibx}x^{-a}\Gamma(a,ix)dx
$$

where a is a strictly natural number and $b=1/4$, $i$ is the imaginary unit and $\Gamma$ is the incomplete gamma function. Integration by parts is not helpful, since the argument will appear again in the process, no matter which combination between the three terms we choose. I have been stuck on this for quite a while, any advice is welcome!

Title: Integral of gamma - exponential product

Body:
For $a<1$, Mathematica gives for
$$I=\int_{-\infty}^{+\infty} e^{ibx}\,x^{-a}\,\Gamma(a,ix)\,dx$$
$$I=e^{-\frac{1}{2} i \pi  a}\, b^{a-1} \,\sin (\pi  a)
   \left(B_{b^2}\left(\frac{1}{2}-\frac{a}{2},0\right)+B_{b^2}\left(1-\frac{a}{2
   },0\right)\right)$$


问:
Radius of circle tangent to six tangent parabolas

Six congruent copies of the parabola $y = x^2$ are arranged in the plane so that each vertex is tangent to a circle, and each parabola is tangent to its two neighbors. Find the radius of the circle.

I'm honestly not sure how to start this problem. The vertices of the parabolas are spaced $60^\circ$ apart on the circumference of the circle, and the points where two parabolas are tangent are also spaced $60^\circ$ apart, $30^\circ$ from a vertex. But I'm not sure how this is useful.

I also know that the focus and directrix of each parabola are the same distance from the vertex, and the directrixes (is that the plural?) intersect $60^\circ$ apart, $30^\circ$ from a vertex. But again, I don't know how useful that is.

Normally on problems like these I'd try to draw a radius and form some right triangles but I really don't see how I can do that here.

TIA!

答:
This issue can be treated using complex numbers geometry.

Indeed, if we take the initial parabola with a horizontal axis of symmetry, it can be written as the locus of points

$$z=t^2+it+r, \ \  t \in \mathbb{R}\tag{1}$$

(with unknown $r$). The symmetry with respect to abscissas axis being obtained by changing $t$ into $-t$.

Therefore, as rotation by $\pi/3$ is obtained by

$$z \to \omega z, \ \ \text{with} \ \ \omega:=e^{i\pi/3}=\dfrac{1+\sqrt{3}}{2}\tag{2}$$

we can express the tangency of the initial parabola with its rotated version by saying that the following equation has a double root:

$$\omega(t^2+it+r)=(t^2\color{red}{-}it+r)\tag{3}$$

This equation can be written :

$$(\omega-1)t^2+i(\omega+1)t+(\omega-1)r=0\tag{4}$$

The double root condition is equivalent to nullify the discriminant of (4):

$$\Delta=-(\omega+1)^2-4(\omega-1)^2r=0$$

giving:

$$r=-\dfrac14\left(\dfrac{\omega+1}{\omega-1}\right)^2=-\dfrac14\left(\dfrac{e^{i\pi/3}+1}{e^{i\pi/3}-1}\right)^2=-\dfrac14\left(\dfrac{e^{i\pi/6}+e^{-i\pi/6}}{e^{i\pi/6}-e^{-i\pi/6}}\right)^2$$
$$r=-\dfrac14\left(\dfrac{\cos(\pi/6)}{i\sin(\pi/6)}\right)^2=\dfrac14\operatorname{cot}^2(\pi/6).\tag{5}$$

Extension: it happens that what has been done for $n=6$ parabola can be done for any value of $n \ge 3$ with the following immediate extension of formula (5):

$$r=\dfrac14\operatorname{cot}^2(\pi/n).\tag{5}$$

The figures represent resp. cases $n=4$ and $n=24$.

Matlab program for the figures:

```<code>   clear all;close all;hold on;axis equal off
   set(gcf,'color','w');
   n=24;
   om=exp(2*i*pi/n);
   a=0.25*cot(pi/n)^2;
   v=1+n/5;t=-v:0.01:v;
   s=t.^2+i*t+a; % initial parabola
   t=-1:0.01:1;plot(a*exp(i*pi*t),'color','r');
   for k=1:n;
       plot(s);
       s=s*om;% 2*pi/n rotation of previous parabola
   end;
</code>```

Body:
This issue can be treated using complex numbers geometry.

Indeed, if we take the initial parabola with a horizontal axis of symmetry, it can be written as the locus of points

$$z=t^2+it+r, \ \  t \in \mathbb{R}\tag{1}$$

(with unknown $r$). The symmetry with respect to abscissas axis being obtained by changing $t$ into $-t$.

Therefore, as rotation by $\pi/3$ is obtained by

$$z \to \omega z, \ \ \text{with} \ \ \omega:=e^{i\pi/3}=\dfrac{1+\sqrt{3}}{2}\tag{2}$$

we can express the tangency of the initial parabola with its rotated version by saying that the following equation has a double root:

$$\omega(t^2+it+r)=(t^2\color{red}{-}it+r)\tag{3}$$

This equation can be written :

$$(\omega-1)t^2+i(\omega+1)t+(\omega-1)r=0\tag{4}$$

The double root condition is equivalent to nullify the discriminant of (4):

$$\Delta=-(\omega+1)^2-4(\omega-1)^2r=0$$

giving:

$$r=-\dfrac14\left(\dfrac{\omega+1}{\omega-1}\right)^2=-\dfrac14\left(\dfrac{e^{i\pi/3}+1}{e^{i\pi/3}-1}\right)^2=-\dfrac14\left(\dfrac{e^{i\pi/6}+e^{-i\pi/6}}{e^{i\pi/6}-e^{-i\pi/6}}\right)^2$$
$$r=-\dfrac14\left(\dfrac{\cos(\pi/6)}{i\sin(\pi/6)}\right)^2=\dfrac14\operatorname{cot}^2(\pi/6).\tag{5}$$

Extension: it happens that what has been done for $n=6$ parabola can be done for any value of $n \ge 3$ with the following immediate extension of formula (5):

$$r=\dfrac14\operatorname{cot}^2(\pi/n).\tag{5}$$

The figures represent resp. cases $n=4$ and $n=24$.

Matlab program for the figures:

   clear all;close all;hold on;axis equal off
   set(gcf,'color','w');
   n=24;
   om=exp(2*i*pi/n);
   a=0.25*cot(pi/n)^2;
   v=1+n/5;t=-v:0.01:v;
   s=t.^2+i*t+a; % initial parabola
   t=-1:0.01:1;plot(a*exp(i*pi*t),'color','r');
   for k=1:n;
       plot(s);
       s=s*om;% 2*pi/n rotation of previous parabola
   end;


问:
On leap years and conditional probability
Edit

(Copying this from the end of my post and pasting it at the start because people are still voting to close this.)

Before anyone flags this as a duplicate, I understand that the following exact problem has been posted before, but my post is about why my logic (for the second question below) is wrong (and not about what the correct answer should be).

Problem

A year in the 2020s (i.e. from 2020 to 2029 inclusive) is selected uniformly at random, a month is selected uniformly at random from that year and a day is selected uniformly at random from that month.

What is the probability that the day is the 29th of February?

Given that the day is the 29th, what is the probability that the month is February?

My idea for Question 1

The first question is very clear-cut.

First, we observe that only leap years have a "29th of February" and out of the entire decade, only 3 are leap years (i.e. 2020, 2024 and 2028), so we have a $\frac 3 {10}$ chance of picking a leap year.

By the same logic, we have a $\frac 1 {12}$ chance of picking February out of the leap year we have chosen and we have a further $\frac 1 {29}$ chance of picking the 29th of February from the February of the leap year we have chosen.

Thus, $\mathbb{P}(\mathrm{29th\ of\ February}) = (\frac 3 {10})(\frac 1 {12})(\frac 1 {29}) = \frac 1 {1160}$.

Okay. Easy.

My idea for Question 2

Now, I understand that the second question is on conditional probability and Bayes' Theorem should be invoked. However, smart me thought I saw a short-cut (obviously incorrectly) and I have been wondering why my logic (as explained below) is wrong.

Since we are "given that the day is the 29th", I thought that we need only pick a year and a month. In other words, once we have a year and a month set, the probability that the day is the 29th should be $1$ right?

Thus, $\mathbb{P}(\mathrm{February\ |\ 29th}) = (\frac 3 {10})(\frac 1 {12}) = \frac 1 {40}$.

Again. Easy. Or so I thought.

Suggested solution for Question 2

From the solution given (as explained below), it seems I have oversimplified Question 2. According to my professor, we need to break the question down much more than I had.

First, we must observe that, out of the entire decade (i.e. 120 months), we have 3 months with 29 days, 40 months with 30 days (since 4 months in a year have 30 days) and 70 months with 31 days (since 7 months in a year have 31 days). We do not count the "normal Februarys (i.e. the Februarys with 28 days)" as it is obvious that we cannot possibly have a 29th day from those months.

Thus, $\mathbb{P}(\mathrm{February\ |\ 29th}) = \frac {(\frac 3 {120})(\frac 1 {29})} {(\frac 3 {120})(\frac 1 {29}) + (\frac {40} {120})(\frac 1 {30}) + (\frac {70} {120})(\frac 1 {31})} = \frac {279} {9965}$.

While I perfectly understand my professor's logic for the second question, I cannot help but wonder why my idea is flawed. Any intuitive explanations will be greatly appreciated!

答:
Correct!

Given that the date is 29, the probability that the month is Feburary results to me

$$\frac{\frac{3}{10}\cdot\frac{1}{12}\cdot\frac{1}{29}}{\frac{3}{10}\left[\frac{1}{12}\cdot\frac{1}{29}+\frac{4}{12}\cdot\frac{1}{30}+\frac{7}{12}\cdot\frac{1}{31}   \right]+\frac{7}{10}\left[\frac{4}{12}\cdot\frac{1}{30}+\frac{7}{12}\cdot\frac{1}{31}   \right]}=\frac{279}{9965}$$


问:
Is my solution of group theory question correct?
The following question appeared in my examination.

Let $G$ be the symmetric group $S_5$ of permutations of five symbols. Consider the set $\mathscr{S}$ of subgroups of $G$ that are isomorphic to the non-cyclic group of order $4$. Let us call these two subgroups $H$ and $K$ belonging to $\mathscr{S}$ as equivalent if they are conjugate (that is, there exists $g \in G$ such that $gHg^{-1}=K$). How many equivalence classes are there in $\mathscr{S}$?

I thought for constructing non cyclic subgroup we would need three elements of order $2$ and order two elements in $S_5$ either would be transposition or it would be product of two transpositions. Because two permutations are conjugate iff they have same cycle representations and thus we would get $2$ equivalence classes. Is this correct? What are other ways to solve this question?

答:
Your reasoning is not correct/complete, but you're on the right track.

You say an order 2 element in $S_5$ "would be transposition or it would be product of two permutations". Presumably you meant "two transpositions" in the sense that their cycle decomposition would be the product of two transpositions.

In any case, this just says that any subgroup of order 4 must consist of elements from two conjugacy classes (together with the identity). It says nothing about the conjugacy classes of such subgroups themselves. You may as well replace the number 4 with the number 2 up to this point--clearly you're missing something.

Edit: Some more discussion was requested. Let's consider $(12)(34)$. Which copies of the non-cyclic group of order 4 can it participate in, if any? Well, $\{e, (12), (34), (12)(34)\}$ comes to mind immediately. Try playing around and seeing if there are any others (hint: there are). Then consider which of them are conjugate (if any). My suggestion is essentially to classify the subgroups first and then ask about their conjugacy afterwards.


问:
Let $0<b<1$ and $f_n(x)=x_n$ defined on $[0,b]$. Show that $f_n\to0$ uniformly on $[0,b]$
Let $0<b<1$ and $f_n(x)=x_n$ defined on $[0,b]$. Show that $f_n\to0$ uniformly on $[0,b]$.

I tried almost everything to prove the statement. I tried to construct a sequence $x_{n_k}$ which is convergent (so Cauchy) as $x_n$ is bounded on $[0,b]$ and to apply uniform continuity of $f_n$ but I can't conclude the proof. If someone could give a hint, I would really appreciate it. Thank you in advance.

答:
If $f_n(x)=x^n$, then $$\lim_{n\to \infty} \sup_{x\in [0,b]} |f_n(x)-0|=\lim_{n\to \infty}b^n=0,$$ since $0<b<1$.

EDIT: If $b\ge 1$, then $\lim_{n\to \infty}b^n\neq 0$ and the convergence is not uniform.


问:
$x-\frac{1}{x} - \frac{1}{x-\frac{1}{x}} - \dots $ can we represent this infinite expression by an integral or a summation?
$$x-\frac{1}{x} - \frac{1}{x-\frac{1}{x}} - \frac{1} {x-\frac{1}{x} - \frac{1}{x-\frac{1}{x}}} - \dots$$
this expression is the result of recursively subtracting $1/x$ from $x$,

i.e.

$$x_{n+1} = x_n - 1/x_n$$

when executing the previous expression multiple times it does not converge, as the number of executions increase the value of $x$ tends to change slower and slower.

I wonder if this expression can be represented by some integral or summation

答:
As user @jjagmath has already noticed, there is no (convergence to a) fixpoint, or one might say, the only fixpoint is infinity.

But there can be said more: there are periodic points of any period length, and they can be numerically approximated to any precision. Since the iteration-process in the question produces divergence, any small error in these approximations shall escalate after a certain finite number of iterations. However, such periodic points can be found as roots of polynomials, and are thus algebraic numbers.

It came out to be helpful to decode the variable $x$ as fractional representation, for instance $x = \frac ab$. I'll use that below.

The iterative approach to find periodic points

The iteration $\frac ab \to { a^2-b^2\over ab}$ can be inverted; however the results in the reverse iteration are ambiguous having either $ \frac ab $ or $ \frac {-a}{-b}$ . So with the route of reverse iterations we must supply a vector containing the signs     for each iteration. If the vector is finite of length, say, $h$ and the provided signs are taken periodically, then the reverse iteration converges to a set of $h$-periodic points. I've done the calculation of iterations in $(a,b)$ in terms of a 2x2-matrix $A = \begin{bmatrix}a&-b\\b&a \end{bmatrix} $ and $D = \small \begin{bmatrix}1&2\\2&1 \end{bmatrix} $ and $D_r = \small \begin{bmatrix}1&1/2\\1/2&1 \end{bmatrix} $  to iterate $ A \to A^2 \star D_r $ where the "$\star$" means Hadamard-multiplication (=elementwise multiplication), and the reverse iteration $ A \to_\pm \sqrt{D  \star A} $ where -if we iterate- the signs are taken from a predefined vector or list.

So for instance I got with the Pari/GP-routine (containing a simple matrix-root-routine which also includes the Hadamard-multiplication with constant matrix $D$)

```<code>{M2=[a0,-b0;b0,a0];
 il=0;list=vectorv(1000);  \\ takes the protocol
 for(k=0,199,
     M2=matroot(M2);
     M2*=[1,-1][k % 2+1];   \\ vector of signs is [1,-1] and taken as infinitely repeating
     il++;list[il]=M2[1,];  \\ document current values of [a,b]
     );
 list=Mat(VE(list,il)); }  \\ reshape list to get it as matrix
</code>```

This got me the protocol:

```<code>    a                   b
-----------------------------------
   ...              ...             (a couple of hundred lines of
   ...              ...                     not yet well approximates)
 -1.34150376263   -1.02674252883    (approximated to more than 170 dec digits)
  0.745432124647   1.37737896568
 -1.34150376263    1.02674252883
  0.745432124647  -1.37737896568

 -1.34150376263   -1.02674252883   (repeating previous values precisely)
    ...                ...
</code>```

where, after $800$ iterations the difference from $ |a_{k+4}-a_k| \lt 1e-170$.
The last entry cutted to $40$ visible dec digits protocol gives

a =-1.341503762630577719675693514485165986352 and b=-1.026742528828304264759792215760033619885 .

You might try iteration starting at this values (with high enough internal precision) to see the periodicity.

Periodic points via solving polynomials

Periodic points in the reals can be found solving polynomials as they occur from 2- or more steps iteration.

The assumed fixpoint iteration at $L$ leads to $L=L-\frac 1L$ as mentioned correctly by user @jjagmath: if $x$ is expressed as $\frac ab$ then we find the fixpoint
$$  \frac ab = \frac ab - \frac ba = {a^2-b^2\over ab } \\
 \text{solving} \\
 \begin{array}{}
 (1)&  a &= a^2 - b^2 \\
 (2)&  b &= ab  &\qquad \to & a=1 \\
 \to & 1 &= 1^2 - b^2 &\qquad \to & b=0\\
 \end{array}
$$
which represents of course the singularity for the expression $x = \frac ab$.

However, this does not exhaust all interesting properties of the iteration in the OP.

Using the polynomials in $(a,b)$ which occur if we iterate 2 or more times symbolically (with vector of signs provided) allows to get infinitely many solutions over algebraic numbers defining periodic iterations.

For instance the above numerical approximation, initially been found by reverse iteration, gives the ratio
$$ {a \over b} = \sqrt{1+\sqrt{1/2}} = x $$
as value, which defines a 4-periodic set of values.
We find this value if we write $ (a_0,b_0) \to (a_1,b_1)=(a_0^2-b_0^2,a_0 b_0) \to (a_2,b_2)=(a_1^2-b_1^2,  a_1 b_1)$ and solve $(a_0,b_0)=(a_2,-b_2)$

The above heuristical protocol more detailed:

```<code>   a_k              b_k          x_k               x_k symbolic
-1.34150376263 ,-1.02674252883, 1.30656296488,  sqrt(1+sqrt(1/2))
 0.745432124647, 1.37737896568, 0.541196100146, sqrt(1-sqrt(1/2))
-1.34150376263 , 1.02674252883,-1.30656296488, -sqrt(1+sqrt(1/2))
 0.745432124647,-1.37737896568,-0.541196100146,-sqrt(1-sqrt(1/2))
</code>```

P.S. Wolfram Alpha decomposes the following values (from second row in the protocol)

```<code> [a_1 ≈] 0.74543212464725619 ≈ 1/(1 + sqrt(2))^(1/3) `
 [b_1 ≈] 1.377378965676004   ≈ 2^(1/4) (1 + sqrt(2))^(1/6) `
</code>```

Final remark: since we have periodic points for those points the infinite iteration-series (as intended in the OP) might be regularizable by the $\zeta(0)$ or $\eta(0)$ application.

Example (hopefully:simple)  and possibly regularization

If we do not assume a fixpoint $L$ to which the partial sums converge in the sense that in the limit $L=L - 1/L$ (which can only hold for infinite $L$) but a 2- periodic point, then we get for the limit:
$$L = M - 1/M  \text{ and } M= L - 1/L \tag 1 $$
Let's see, whether we can find some $L$:
$$ \begin{array} {}
L &= L-1/L - 1/(L-1/L) \\
  &=(L^2-1)/L - L/(L^2-1) \\
  &=((L^2-1)^2-L^2)/(L(L^2-1)) \\
L^2(L^2-1)&=(L^2-1)^2-L^2 \\
L^4&=(L^2-1)^2 \\
L^4&=L^4-2L^2+1 \\
2L^2&=1 \\
L&=\sqrt{1/2} \\
M &= \sqrt{1/2}-\sqrt2 \\
&=\sqrt{1/2}(1-2)=-\sqrt{1/2}=-L
\end{array} \tag 2
$$
With this we find the partial sums of the OP's expression, when $x=L$ as
$$ \begin{array}{r|r|l}
\text{ps} & \text{ps} & \text{partial expression in OP ; ps:partial sum} \\
 \hline
L & L & x \\
M & -L & x - 1/x \\
L & L & x - 1/x - 1/(x-1/x)\\
M & -L & x - 1/x - 1/(x-1/x) - 1/(...)\\
\vdots & \vdots & \vdots
\end{array}\tag 3 $$

One might consider to assign a regularized value in the sense of the alternating $\zeta(0)$-series; if I didn't mess it up we should get the (Euler-) regularized value for $f(x)$ at $x=L=1/\sqrt 2$ as
$$f(L) \underset{\mathfrak E}=0 \tag 4$$

Final remarks

In 2020 I've done a (formally) similar analyis (without the final attempt to regularization) on the question of periodic points in the iteration of the (complex-valued) function $z_{k+1} =\exp(z_k)$ and have discussed this to some extent here in MSE because I had some open questions with this. I find it quite funny to find here such a familarity with the derivations...

Form my earlier comment at the OP one may take the external references discussing interesting aspects of the OP's problem:

See oeis.org/A147985 for more information. Clark Kimberling has done an article on this in JIS see: cs.uwaterloo.ca/journals/JIS/VOL12/Kimberling/kimberling56.pdf

Body:
As user @jjagmath has already noticed, there is no (convergence to a) fixpoint, or one might say, the only fixpoint is infinity.

But there can be said more: there are periodic points of any period length, and they can be numerically approximated to any precision. Since the iteration-process in the question produces divergence, any small error in these approximations shall escalate after a certain finite number of iterations. However, such periodic points can be found as roots of polynomials, and are thus algebraic numbers.

It came out to be helpful to decode the variable $x$ as fractional representation, for instance $x = \frac ab$. I'll use that below.

The iterative approach to find periodic points

The iteration $\frac ab \to { a^2-b^2\over ab}$ can be inverted; however the results in the reverse iteration are ambiguous having either $ \frac ab $ or $ \frac {-a}{-b}$ . So with the route of reverse iterations we must supply a vector containing the signs     for each iteration. If the vector is finite of length, say, $h$ and the provided signs are taken periodically, then the reverse iteration converges to a set of $h$-periodic points. I've done the calculation of iterations in $(a,b)$ in terms of a 2x2-matrix $A = \begin{bmatrix}a&-b\\b&a \end{bmatrix} $ and $D = \small \begin{bmatrix}1&2\\2&1 \end{bmatrix} $ and $D_r = \small \begin{bmatrix}1&1/2\\1/2&1 \end{bmatrix} $  to iterate $ A \to A^2 \star D_r $ where the "$\star$" means Hadamard-multiplication (=elementwise multiplication), and the reverse iteration $ A \to_\pm \sqrt{D  \star A} $ where -if we iterate- the signs are taken from a predefined vector or list.

So for instance I got with the Pari/GP-routine (containing a simple matrix-root-routine which also includes the Hadamard-multiplication with constant matrix $D$)

{M2=[a0,-b0;b0,a0];
 il=0;list=vectorv(1000);  \\ takes the protocol
 for(k=0,199,
     M2=matroot(M2);
     M2*=[1,-1][k % 2+1];   \\ vector of signs is [1,-1] and taken as infinitely repeating
     il++;list[il]=M2[1,];  \\ document current values of [a,b]
     );
 list=Mat(VE(list,il)); }  \\ reshape list to get it as matrix

This got me the protocol:

    a                   b
-----------------------------------
   ...              ...             (a couple of hundred lines of
   ...              ...                     not yet well approximates)
 -1.34150376263   -1.02674252883    (approximated to more than 170 dec digits)
  0.745432124647   1.37737896568
 -1.34150376263    1.02674252883
  0.745432124647  -1.37737896568

 -1.34150376263   -1.02674252883   (repeating previous values precisely)
    ...                ...

where, after $800$ iterations the difference from $ |a_{k+4}-a_k| \lt 1e-170$.
The last entry cutted to $40$ visible dec digits protocol gives

a =-1.341503762630577719675693514485165986352 and b=-1.026742528828304264759792215760033619885 .

You might try iteration starting at this values (with high enough internal precision) to see the periodicity.

Periodic points via solving polynomials

Periodic points in the reals can be found solving polynomials as they occur from 2- or more steps iteration.

The assumed fixpoint iteration at $L$ leads to $L=L-\frac 1L$ as mentioned correctly by user @jjagmath: if $x$ is expressed as $\frac ab$ then we find the fixpoint
$$  \frac ab = \frac ab - \frac ba = {a^2-b^2\over ab } \\
 \text{solving} \\
 \begin{array}{}
 (1)&  a &= a^2 - b^2 \\
 (2)&  b &= ab  &\qquad \to & a=1 \\
 \to & 1 &= 1^2 - b^2 &\qquad \to & b=0\\
 \end{array}
$$
which represents of course the singularity for the expression $x = \frac ab$.

However, this does not exhaust all interesting properties of the iteration in the OP.

Using the polynomials in $(a,b)$ which occur if we iterate 2 or more times symbolically (with vector of signs provided) allows to get infinitely many solutions over algebraic numbers defining periodic iterations.

For instance the above numerical approximation, initially been found by reverse iteration, gives the ratio
$$ {a \over b} = \sqrt{1+\sqrt{1/2}} = x $$
as value, which defines a 4-periodic set of values.
We find this value if we write $ (a_0,b_0) \to (a_1,b_1)=(a_0^2-b_0^2,a_0 b_0) \to (a_2,b_2)=(a_1^2-b_1^2,  a_1 b_1)$ and solve $(a_0,b_0)=(a_2,-b_2)$

The above heuristical protocol more detailed:

   a_k              b_k          x_k               x_k symbolic
-1.34150376263 ,-1.02674252883, 1.30656296488,  sqrt(1+sqrt(1/2))
 0.745432124647, 1.37737896568, 0.541196100146, sqrt(1-sqrt(1/2))
-1.34150376263 , 1.02674252883,-1.30656296488, -sqrt(1+sqrt(1/2))
 0.745432124647,-1.37737896568,-0.541196100146,-sqrt(1-sqrt(1/2))

P.S. Wolfram Alpha decomposes the following values (from second row in the protocol)

 [a_1 ≈] 0.74543212464725619 ≈ 1/(1 + sqrt(2))^(1/3) `
 [b_1 ≈] 1.377378965676004   ≈ 2^(1/4) (1 + sqrt(2))^(1/6) `

Final remark: since we have periodic points for those points the infinite iteration-series (as intended in the OP) might be regularizable by the $\zeta(0)$ or $\eta(0)$ application.

Example (hopefully:simple)  and possibly regularization

If we do not assume a fixpoint $L$ to which the partial sums converge in the sense that in the limit $L=L - 1/L$ (which can only hold for infinite $L$) but a 2- periodic point, then we get for the limit:
$$L = M - 1/M  \text{ and } M= L - 1/L \tag 1 $$
Let's see, whether we can find some $L$:
$$ \begin{array} {}
L &= L-1/L - 1/(L-1/L) \\
  &=(L^2-1)/L - L/(L^2-1) \\
  &=((L^2-1)^2-L^2)/(L(L^2-1)) \\
L^2(L^2-1)&=(L^2-1)^2-L^2 \\
L^4&=(L^2-1)^2 \\
L^4&=L^4-2L^2+1 \\
2L^2&=1 \\
L&=\sqrt{1/2} \\
M &= \sqrt{1/2}-\sqrt2 \\
&=\sqrt{1/2}(1-2)=-\sqrt{1/2}=-L
\end{array} \tag 2
$$
With this we find the partial sums of the OP's expression, when $x=L$ as
$$ \begin{array}{r|r|l}
\text{ps} & \text{ps} & \text{partial expression in OP ; ps:partial sum} \\
 \hline
L & L & x \\
M & -L & x - 1/x \\
L & L & x - 1/x - 1/(x-1/x)\\
M & -L & x - 1/x - 1/(x-1/x) - 1/(...)\\
\vdots & \vdots & \vdots
\end{array}\tag 3 $$

One might consider to assign a regularized value in the sense of the alternating $\zeta(0)$-series; if I didn't mess it up we should get the (Euler-) regularized value for $f(x)$ at $x=L=1/\sqrt 2$ as
$$f(L) \underset{\mathfrak E}=0 \tag 4$$

Final remarks

In 2020 I've done a (formally) similar analyis (without the final attempt to regularization) on the question of periodic points in the iteration of the (complex-valued) function $z_{k+1} =\exp(z_k)$ and have discussed this to some extent here in MSE because I had some open questions with this. I find it quite funny to find here such a familarity with the derivations...

Form my earlier comment at the OP one may take the external references discussing interesting aspects of the OP's problem:

See oeis.org/A147985 for more information. Clark Kimberling has done an article on this in JIS see: cs.uwaterloo.ca/journals/JIS/VOL12/Kimberling/kimberling56.pdf


问:
Computing $\det(a_1 M_{ij} + a_2 b_i b_j + a_3 c_i c_j)$
I found interesting formula which seems hard to prove.

Let $M=(M_{ij})$ be an invertible $d\times d$ symmetric matrix. Then
\begin{align}
\det(a_1 M_{ij} + a_2 b_i b_j + a_3 c_i c_j)
= a_1^{d-2} \det(M_{ij}) \left( (a_1 + a_2 b^2)(a_1 + a_3 c^2) - a_2 a_3 (b\cdot c)^2 \right)
\end{align}
where
\begin{align}
b^2 = b_i b_j (M^{-1})^{ij}, \quad c^2 = c_i c_j (M^{-1})^{ij}, \quad b\cdot c = b_i c_j (M^{-1})^{ij}
\end{align}
and $a_1, a_2, a_3 \in \mathbb{R}$ with $a_1 \neq 0$, $b_i, c_i$ are arbitrary $d$-column vectors.

For $a_2=a_3=0$ it reduce $\det(a M_{ij}) = a^d \det(M_{ij})$ so it seems fine. and for $a_2=0$ or $a_3=0$ seems fine.

\begin{align}
\det(a_1 M_{ij} + a_2 b_i b_j) &=  \det\left( a_1 M_{ij} (Id + a_1^{-1}a_2 M^{ij}  b_i b_j) \right)  \\
&
=\det(a_1 M_{ij}) (1+\operatorname{tr}(a_1^{-1} a_2 (M^{-1})^{ij} b_i b_j)  \\
&= a_1^{d-2} \det(M_{ij}) \left( a_1^{2} + a_1 a_2 b^2\right)
\end{align}
In the process I used  "for the rank one matrix $A$, $\det(Id+A) = 1+\operatorname{tr}(A)$"

So for $a_2=0$ or $a_3=0$ is fine, but I am having trouble for $a_2 \neq 0, a_3 \neq 0$. Is there any nice approach?

答:
I think that what you are looking at is a double bordered determinant, the determinant of a matrix of this sort
$$
H:=
\begin{pmatrix}
A & K\\
L^{T} & B
\end{pmatrix}
$$
where $A$ is $d\times d$, $K,L$ are $d\times 2$, and $B$ is $2\times 2$. We assume that $A,B$ are invertible.

We can find two expressions for $\det H$ by factorising $H$ in the following ways:
$$
H=
\begin{pmatrix}
A& O\\O&I
\end{pmatrix}
\begin{pmatrix}
I& O\\L^{T} & I
\end{pmatrix}
\begin{pmatrix}
I & A^{-1}K\\O &B-L^{T}A^{-1}K
\end{pmatrix}
$$
and
$$
H=
\begin{pmatrix}
I& O\\O&B
\end{pmatrix}
\begin{pmatrix}
I& K\\O & I
\end{pmatrix}
\begin{pmatrix}
A-KB^{-1}L^{T} & O\\B^{-1}L^{T} & I
\end{pmatrix}.
$$

We will then have
$$
\det H = \det A\ \det(B-L^{T}A^{-1}K)=\det B\ \det(A-KB^{-1}L^{T}).
$$

So now let $A=a_1 M$, $B=\begin{pmatrix}a_2 & 0\\0& a_3\end{pmatrix}$, $K=\begin{pmatrix} b & c \end{pmatrix}$, and $L=\begin{pmatrix}  -b & -c \end{pmatrix}$. The two expressions for the determinant are the two expressions you wish to equate.


问:
Formal statement that an analytic function injective on the boundary is injective on the closure
I have been told that such a result guaranteeing injectivity on the closure of the domain exists, however I have not been able to find a formal statement of this result. If someone could provide I would be very grateful.</p>

答:
The correct version of this claim, sometimes known as Darboux's theorem (sometimes also Picard-Darboux theorem), requires the boundary to be a jordan curve, that is:

Let $f$ be a holomorphic function on a Jordan region such that it has
a continuous extension to the boundary. If $f$ is injective on the
boundary, it is injective on the region.

The proof, roughly sketched, is like this:

Let $\gamma$ be the jordan curve and $\Omega$ be the region bounded by the curve. Since $f$ is continuous and injective, $f(\gamma)$ is a jordan curve. First note that $f(\Omega)$ must be bounded by $f(\gamma)$, since if it is not there must be a $w_0:f(z)-w_0$ has at least a solution in $\Omega$ and not on $\gamma$. This is impossible by the argument principle. Similarly, $f(\Omega)\cap f(\gamma)=\emptyset$ for otherwise the open map theorem implies that $f(\Omega)$ is not bounded by $f(\gamma)$. Now let $w$ be such that $f(z)=w$ has at least one solution on $\Omega$. It is easy to see, by the argument principle and the fact that a positively oriented jordan curve has winding numer $1$, that it must have only one solution. Thus $f$ is one-one on $\Omega\cup \gamma$.

It is easy to see that the following weaker statement has a quite straightforward proof:

Let $\gamma$ be a closed jordan curve bounding a region $\Omega$, and
let $f$ be holomorphic on a neighbourhood of $\overline\Omega$. If it
is injective on $\gamma$, its injective on $\gamma\cup \Omega$.

As the example by @HagenVonEitzen shows, the hypotesis that $\partial \Omega$ is a jordan curve cannot be omitted.


问:
How to address the numerical instability of a function involving erf and an exponential
We need to compute this function:

$f(x) = x + \sqrt {\frac 2 \pi} \cdot \sigma_x \cdot \frac {e^{- ( \frac {x} {\sqrt 2 \cdot \sigma_x} )^2}} {1 + erf(\frac {x} {\sqrt 2 \cdot \sigma_x})}$

We noticed that, for negative values of $\frac {x} {\sqrt 2 \cdot \sigma_x}$ lower than a certain value (and different depending on which software we use to compute it, say about $-5$ for instance), there is numerical instability (oscillations).

I saw this post suggesting how to achieve better stability, although on a rather different function:

Numerical Instability with this cumulate

Do you think a similar method would work here? And if so, should I express the $erf$ as an integral?

Or would you suggest some other method? Taylor expansion? Anything else...?

Thanks

EDIT after evaluation of the responses

Thank you all very much for your feedback!

I think there is a complication here that I had not explicitly foreseen.

First I will make the following substitution, so your alternative expressions of $erf$ can be used directly:

$x = y \cdot \sqrt 2 \cdot \sigma_x$

$f(y) = \sqrt 2 \cdot \sigma_x \cdot y + \sqrt {\frac 2 \pi} \cdot \sigma_x \cdot \frac {e^{- y^2}} {1 + erf(y)}$

Now, the theory we used to derive this equation requires that:

$f(y) > 0, \forall y$

$\lim_{y \to -\infty} f(y) = 0$

So, we expect $f(y)$ to be very slowly approaching $0$ as $y$ becomes more negative; but $f(y)$ should never be negative.

In fact, if I plot the second term of $f(y)$, it looks like, as $y$ becomes more and more negative:

$\sqrt {\frac 2 \pi} \cdot \sigma_x \cdot \frac {e^{- y^2}} {1 + erf(y)} \approx - \sqrt 2 \cdot \sigma_x \cdot y$

This is in substantial agreement with Claude's approximation for the $erf$ part.

I.e., using:

$\frac{e^{-y^2}}{1+\text{erf}(y)} \approx -\sqrt{\pi } \,y\Bigg[1+\frac{1}{2 y^2}-\frac{1}{2 y^4}+\frac{5}{4
   y^6}+O\left(\frac{1}{y^8}\right)\Bigg]$

I get:

$\sqrt {\frac 2 \pi} \cdot \sigma_x \cdot \frac {e^{- y^2}} {1 + erf(y)} \approx \sqrt {\frac 2 \pi} \cdot \sigma_x \cdot (-\sqrt{\pi }) \,y\Bigg[1+\frac{1}{2 y^2}-\frac{1}{2 y^4}+\frac{5}{4
   y^6}+O\left(\frac{1}{y^8}\right)\Bigg] =$

$= - \sqrt {2} \cdot \sigma_x \cdot \,y\Bigg[1+\frac{1}{2 y^2}-\frac{1}{2 y^4}+\frac{5}{4
   y^6}+O\left(\frac{1}{y^8}\right)\Bigg]$

which indeed approaches $- \sqrt 2 \cdot \sigma_x \cdot y$ when $y$ becomes very negative.

With the above approximation:

$f(y) \approx \sqrt 2 \cdot \sigma_x \cdot y - \sqrt {2} \cdot \sigma_x \cdot \,y\Bigg[1+\frac{1}{2 y^2}-\frac{1}{2 y^4}+\frac{5}{4
   y^6}+O\left(\frac{1}{y^8}\right)\Bigg] = $
$= - \sqrt {2} \cdot \sigma_x \cdot \,y\Bigg[\frac{1}{2 y^2}-\frac{1}{2 y^4}+\frac{5}{4
   y^6}+O\left(\frac{1}{y^8}\right)\Bigg] $

Numerically, this 'does the trick', i.e. for $y \le -4$ it seems really very close to the function I need to compute, as indicated by Claude. And in fact with the rational function version even up to $y \le -1$.

This seems to imply that only polynomial approximations are viable, not the $erfc$-based ones, or am I wrong?

Any thoughts?

答:
$1 + erf(x)$ suffers from catastrophic cancellation in the negative half plane. This is avoided by replacing it with $erfc(-x)$, that is, use of the complementary error function. Most standard math libraries, for example the C++ standard math library, include this as a function erfc.

Many special function libraries also offer the exponentially scaled complementary error function, $e^{x^{2}} erfc(x)$, often as a function called erfcx. With that, one can compute $\frac{e^{-x^{2}}}{1+erf(x)}$ accurately as 1.0 / erfcx (-x). I also provided a double-precision implementation of erfcx on Stackoverflow.

As I became aware belatedly, there is a second instance of catastrophic cancellation in $f(x)$ for negative $x$ large in magnitude: $x$ and the rest of the expression are similar in magnitude but of opposite sign. Around $-10^{7}$ complete loss of accuracy occurs.


问:
Proving that a plane triangulation is 3-colorable iff it's even, using faces?
My textbook gave the theorem that a plane triangulation is 3-vertex-colorable if and only if all its vertices were of even degree. It offered a proof using that the graph is 2-face-colorable, which I do not understand.

The book basically states that all the faces that were colored red would be labelled a, b, c clockwise, and all the faces colored blue would be labelled a, b, c counterclockwise, and that "this vertex coloring can be extended to the whole graph", thus proving the theorem.

I feel like I'm missing something because this doesn't really seem to be a proof so much as an example. Is there some theorem regarding "oriented" labels, or some name on this concept I can read up more on?

答:
"that a plane triangulation is 3-vertex-colorable if and only if all its vertices were of even degree".
If ONE vertex has an odd degree you need 3 colours to colour the adjacent ones and a 4th colour for that vertex.

Colour the vertices of one triangle and give all opposite vertices the same colour. As all degrees are even this works always, for any (near-)triangulation.

Giving orientations (+1 or -1) to the triangles of any triangulation is the dual of a Heawoods vertex character for cubic graphs*. If the sum around each vertex is 0Mod3 then the graph is four-colourable. Giving alternate orientations around each even degree-vertex gives then always 0Mod3.

(*) See T.L.Saaty & P.C.Kainen,"The four-color problem" Dover 1986, page 120.


答:
For (1) you have a probability of 29th February chosen of $(\frac 3 {10})(\frac 1 {12})(\frac 1 {29}) = \frac 1 {1160}$

The equivalent probability of 29th March is $(\frac {10} {10})(\frac 1 {12})(\frac 1 {31})$ and there are $7$ months with $31$ days

The equivalent probability of 29th April is $(\frac {10} {10})(\frac 1 {12})(\frac 1 {30})$ and there are $4$ months with $30$ days

So that makes the probability of February given a 29th was chosen:

$$\dfrac{(\frac 3 {10})(\frac 1 {12})(\frac 1 {29}) }{1(\frac 3 {10})(\frac 1 {12})(\frac 1 {29}) +7(\frac {10} {10})(\frac 1 {12})(\frac 1 {31})+4(\frac {10} {10})(\frac 1 {12})(\frac 1 {30}) }= \dfrac{\frac 3 {29} }{\frac 3 {29}  +\frac{70}{31}+\frac {40} {30}} =\dfrac{279}{9965}$$ which is the suggested solution.

It is about $0.028$, slightly more than your $\frac1{40}=0.025$.  The reason is that because leap-year Februaries are shorter than Marches or Aprils, choosing a 29th is slightly more likely in such Februaries, so the reverse conditionality takes this into account.

Title: On leap years and conditional probability

Body:
Edit

(Copying this from the end of my post and pasting it at the start because people are still voting to close this.)

Before anyone flags this as a duplicate, I understand that the following exact problem has been posted before, but my post is about why my logic (for the second question below) is wrong (and not about what the correct answer should be).

Problem

A year in the 2020s (i.e. from 2020 to 2029 inclusive) is selected uniformly at random, a month is selected uniformly at random from that year and a day is selected uniformly at random from that month.

What is the probability that the day is the 29th of February?

Given that the day is the 29th, what is the probability that the month is February?

My idea for Question 1

The first question is very clear-cut.

First, we observe that only leap years have a "29th of February" and out of the entire decade, only 3 are leap years (i.e. 2020, 2024 and 2028), so we have a $\frac 3 {10}$ chance of picking a leap year.

By the same logic, we have a $\frac 1 {12}$ chance of picking February out of the leap year we have chosen and we have a further $\frac 1 {29}$ chance of picking the 29th of February from the February of the leap year we have chosen.

Thus, $\mathbb{P}(\mathrm{29th\ of\ February}) = (\frac 3 {10})(\frac 1 {12})(\frac 1 {29}) = \frac 1 {1160}$.

Okay. Easy.

My idea for Question 2

Now, I understand that the second question is on conditional probability and Bayes' Theorem should be invoked. However, smart me thought I saw a short-cut (obviously incorrectly) and I have been wondering why my logic (as explained below) is wrong.

Since we are "given that the day is the 29th", I thought that we need only pick a year and a month. In other words, once we have a year and a month set, the probability that the day is the 29th should be $1$ right?

Thus, $\mathbb{P}(\mathrm{February\ |\ 29th}) = (\frac 3 {10})(\frac 1 {12}) = \frac 1 {40}$.

Again. Easy. Or so I thought.

Suggested solution for Question 2

From the solution given (as explained below), it seems I have oversimplified Question 2. According to my professor, we need to break the question down much more than I had.

First, we must observe that, out of the entire decade (i.e. 120 months), we have 3 months with 29 days, 40 months with 30 days (since 4 months in a year have 30 days) and 70 months with 31 days (since 7 months in a year have 31 days). We do not count the "normal Februarys (i.e. the Februarys with 28 days)" as it is obvious that we cannot possibly have a 29th day from those months.

Thus, $\mathbb{P}(\mathrm{February\ |\ 29th}) = \frac {(\frac 3 {120})(\frac 1 {29})} {(\frac 3 {120})(\frac 1 {29}) + (\frac {40} {120})(\frac 1 {30}) + (\frac {70} {120})(\frac 1 {31})} = \frac {279} {9965}$.

While I perfectly understand my professor's logic for the second question, I cannot help but wonder why my idea is flawed. Any intuitive explanations will be greatly appreciated!


问:
A nontrivial everywhere continuous function with uncountably many roots?
This is my first post on SE, forgive any blunders.

I am looking for an example of a function $f:\mathbb{R} \to \mathbb{R}$ which is continuous everywhere but has uncountably many roots ($x$ such that $f(x) = 0$). I am not looking for trivial examples such as $f = 0$ for all $x$. This is not a homework problem. I'd prefer a nudge in the right direction rather than an explicit example.

Thanks!

Edit: Thanks all! I've constructed my example with your help.

答: This could be interesting for you.</p>

Body: This could be interesting for you.