问:
How to advance Physics?
I am currently in high school and I am an avid lover of Maths and Physics, I am currently doing my GCSE's and (without sounding arrogant) I don't really enjoy how basic our Physics is, I used to have this problem with Maths too until I picked up a (text)book on Linear Algebra (Linear Algebra: A Modern Introduction) which I'm enjoying very much. The issue is, I don't really know where to start: I am reading books on Physics like "QED by Richard Feynman" which I am really enjoying but I don't feel it's enough. I want to advance my Physics and I want to start applying Maths to Physics (why I posted on Maths.Exchange), I want to be able to pick up more challenging books on this topic and fully understand them. Yet I also really enjoy the "Physics-Physics side of Physics", I find it very interesting and I am wondering: how can I learn more about Physics (both Mathematically and Theoretically). Through experience I have seemed to grow a higher appreciation for learning stuff through books rather than online, so are there any recommendations of books I can pick up? (I mean as in textbooks, or something which I can study from rather than normal books- those which I am already reading). I know Physics is a very vague name for all the fascinating branches there are, but I would be content with anything - personally, I am really into Astrophysics, but I find everything interesting.

Nevertheless, thank you.

答:
I think the advice you will find here may be too situation independent for you as contributors here are from many different places(for those reading, GCSE is equivalent to precalc). Undergraduate textbooks are great, but you will have plenty of time to work through them at university. Shoot for the hoops first by doing well in the exams and getting into a good university.

I would recommend you start looking at A level maths/physics/further maths stuff. Not only will it be fun, you'll be ahead next year too. The Edexcel textbooks are great. Further mechanics modules FM1 and FM2 are more or less exactly what you are looking for.

OwnerDisplayName: anon

LastEditorDisplayName: anon

Title: How to advance Physics?


问:
Tensoring is thought as both restricting and extending?
I hope these questions are not too trivial.

Let  $I$ be an ideal in $R$. Write $I'\subseteq R[t]$. Then the notion of tensoring
$$
(R[t]/I')\otimes_{\,\mathbb{C}[t]} \mathbb{C}[t]/\langle t-c \rangle
$$
is thought to be restricting to the fiber over $t=c$.

On the other hand, considering $\mathbb{R}$,
$$
\mathbb{R}\otimes_{\,\mathbb{R}}\mathbb{C}
$$
is thought to be a base extension.

Question 1: So tensoring is not only thought of as a restriction, but it is also thought of as an extension?  Why do we need or when do we use base extensions?

Question 2: Geometrically, what are

$\operatorname{Spec}(\mathbb{C}[s]\otimes_{\,\mathbb{Z}}\mathbb{C}[u,v])$?

$\operatorname{Spec}(\mathbb{C}[s]\otimes_{\,\mathbb{R}}\mathbb{C}[u,v])$?

$\operatorname{Spec}(\mathbb{C}[s]\oplus\mathbb{C}[u,v])$?

答:
The tensor product of rings is in Hartshorne's book Thm. II.3.3 used to give a general construction of the fiber product of schemes. Given two schemes $X,Y$ over a scheme $S$ we define the fiber product $X\times_S Y$ (if it exists) using universal properties (see HH, Chapter II.3). It is not completely clear that the fiber product exists in general.

Example. If $S:=\mathrm{Spec}(R), X:=\mathrm{Spec}(A), Y:=\mathrm{Spec}(B)$ where $R$ is a commutative unital ring and $A,B$ are commutative unital $R$-algebras it follows

F1.  $X\times_S Y\cong \mathrm{Spec}(A\otimes_R B)$.

Using the "universal property" of $\otimes$ we may show that the fiber product in F1 is given by the tensor product of $A$ and $B$.  In HH Thm. II.3.3
one proves that the fiber product exists in general.

Question 1. So tensoring is not only thought of as a restriction, but it is also thought of as an extension? Why do we need or when do we use base extensions?

Question 2. Geometrically, what are $R_1:=\mathrm{Spec}(\mathbb{C}[s]⊗_{\mathbb{Z}}\mathbb{C}[u,v]), R_2:=\mathrm{Spec}(\mathbb{C}[s]⊗_{\mathbb{R}}\mathbb{C}[u,v])$ and $R_3:=\mathrm{Spec}(\mathbb{C}[s]⊕\mathbb{C}[u,v])$?

Example. In general if $\pi: X \rightarrow S$ is any morphism and if $i:U \rightarrow S$ is an open subscheme it follows the fiber product $U\times_S X\cong \pi^{-1}(U) $ is the definition of the inverse image scheme of $U$ via the morphism $\pi$. If $j: \mathrm{Spec}(\kappa(s))\rightarrow S$ is a point it follows $\mathrm{Spec}(\kappa(s))\times_S X :=\pi^{-1}(s)$ (by definition) is the fiber of $\pi$ at $s$. With this definition it follows the inverse image $\pi^{-1}(U)$ and the fiber $\pi^{-1}(s)$ have a canonical scheme structure and satifies a universal property.

Example. Base change. Let $k:=\mathbb{R}$ be the field of real numbers and let $K:=\mathbb{C}$ be the field of complex numbers. Let $A:=k[t]/(t^2+1)$
Let $C:=\mathrm{Spec}(A)$ and let $S:=\mathrm{Spec}(k)$ with $\pi: C \rightarrow S$ the canonical morphism induced by the inclusion $k \rightarrow A$. Let $T:=\mathrm{Spec}(K)$ and let $i:=T \rightarrow S$ be the canonical map induced by the field extension $k \subseteq K$. The fiber product $T\times_S C$ is canonically isomorphic to
$C':=\mathrm{Spec}(B)$ where $B:=K[t]/(t^2+1)$ and in $K[t]$ it follows the polynomial
$f(t):=t^2+1$ equals $f(t)=(t-i)(t+i)$ and since the two maximal ideals $(t-i)$ and $(t+i)$ are coprime, it follows there is an isomorphism of rings $B \cong B_1 \oplus B_2$ (here we use the Chinese remainder theorem) where $B_1\cong K[t]/(t-i)\cong K$ and $B_2\cong K[t]/(t+i)\cong K$. It follows

F1. $\mathrm{Spec}(B)\cong C_1 \cup C_2 \cong \mathrm{Spec}(B_1)\cup \mathrm{Spec}(B_2) =\mathrm{Spec}(K) \cup \mathrm{Spec}(K)$.

By $\cup$ I mean disjoint union. Here we have used the fact that the spectrum of a direct sum of rings is the disjoint union of their spectra. Hence the fiber product is a disjoint union of two copies of $\mathrm{Spec}(K)$. The scheme $\mathrm{Spec}(B)$ is reduced ($B$ is a product of fields) and reducible. The original scheme $C$ was integral since $A:=k[t]/(t^2+1)\cong K$ is an integral domain, but when we make a field extension from $k$ to $K$ it becomes reducible.

Example. Finite extensions of number fields. If $K \subseteq L$ is a finite extension of algebraic number fields with rings of integers $\mathcal{O}_K \subseteq \mathcal{O}_L$ and $\mathfrak{q} \subseteq \mathcal{O}_K$ is a maximal ideal, there is always an equality $\mathfrak{q}\mathcal{O}_L =\prod_{i=1}^m  \mathfrak{p}_i^{l_i}$ where $\mathfrak{p}_i \subseteq \mathcal{O}_L$ are maximal ideals. Here $\mathfrak{p_i}\neq \mathfrak{p}_j$ for $i\neq j$. It follows there is a direct sum decomposition of rings

$\mathcal{O}_L/\mathfrak{q}\mathcal{O}_L \cong D_1 \oplus \cdots \oplus D_m$

with $D_i \cong \mathcal{O}_L/\mathfrak{p}_i^{l_i}$. Let $C_K:=\mathrm{Spec}(\mathcal{O}_K)$ and $C_L:=\mathrm{Spec}(\mathcal{O}_L)$ with canonical morphism

$\pi: C_L \rightarrow C_K$.

The fiber of $\pi$ at a point $\mathfrak{q}\in C_K$ is by definition given by the ring

$\mathcal{O}_K/\mathfrak{q}\otimes_{\mathcal{O}_K} \mathcal{O}_L \cong D_1\oplus \cdots \oplus D_m$.

It follows the fiber of $\pi$ at $\mathfrak{q}$ is

$ \pi^{-1}(\mathfrak{q})\cong \mathrm{Spec}(D_1) \cup \cdots \cup \mathrm{Spec}(D_m)$.

The rings $D_i$ are Artinian local rings. The ring $D_i$ is non-reduced iff $l_i \geq 2$, hence non-reduced schemes appear in algebraic number theory when we calculate the fiber of the map $\pi$.

Question 1. We need tensor product when we calculate the fiber of a morphism. Sometimes the fiber has non-reduced structure and the tensor product gives a canonical scheme structure to the fiber of a map between affine schemes.

Question 2. The scheme in Q2.3 is the disjoint union $\mathbb{A}^1_{\mathbb{C}} \cup \mathbb{A}^2_{\mathbb{C}}$. The scheme in Q2.2 seems to be $\mathbb{A}^3_U$ where $U:=\mathbb{C}\otimes_{\mathbb{R}}\mathbb{C}$. The ring $U$ seems not to be a domain:

Eq1. $(i\otimes 1 -1\otimes i)(i\otimes 1+1\otimes i)=(i)^2\otimes 1 -1\otimes (i)^2=-1\otimes 1+1\otimes 1=0$.

It seems the following is true:

$\mathbb{C}\otimes_{\mathbb{R}}\mathbb{C} \cong \mathbb{C}[t]/(t^2+1)$ and there is an equality $(t^2+1)=(t-i)(t+i)$ in the ring $\mathbb{C}[t]$.
The ideals $I_1:=(t-i)$ and $I_1:=(t+i)$ are coprime hence there is an isomorphism of rings

$\mathbb{C}\otimes_{\mathbb{R}}\mathbb{C} \cong \mathbb{C} \oplus \mathbb{C}$.

Hence $\mathbb{A}^1_{\mathbb{C}} \times_{\mathbb{R}} \mathbb{A}^2_{\mathbb{C}} \cong \mathbb{A}^3_U \cong \mathbb{A}^3_{\mathbb{C}} \cup \mathbb{A}^3_{\mathbb{C}}$ is a disjoint union of two copies of affine 3-space over the complex numbers. Hence the fiber product of two irreducible schemes may no longer be irreducible.

Question 2.1: Let $R:=\mathbb{C}\otimes_{\mathbb{Z}}\mathbb{C}$. It seems there is an isomorphism

$\mathbb{C}[s]\otimes_{\mathbb{Z}} \mathbb{C}[u,v]\cong R[s,u,v]$. Hence

$\mathrm{Spec}(\mathbb{C}[s]\otimes_{\mathbb{Z}} \mathbb{C}[u,v]) \cong \mathbb{A}^3_T$ with $T:=\mathrm{Spec}(R)$. The ring $T$ is not an integral domain since Eq1 holds in $R$ as well.

Here is an other example: Let $S:=Spec(\mathbb{Q}[x])$ and let $k:=\mathbb{Z}$.
It follows $S\times_k S \cong Spec(R[x,y])$ where $R:=\mathbb{Q}\otimes_{\mathbb{Z}} \mathbb{Q}$. What is $R$?

There is an inclusion $\mathbb{Z} \subseteq \mathbb{Q}$ and $\mathbb{Q}$ is a torsion free $\mathbb{Z}$-module: For any element $s\in \mathbb{Q}$  it follows
the canonical map $\phi_s:\mathbb{Z}\rightarrow \mathbb{Q}$ defined by $\phi_s(m):=ms$ is an injection. But $\mathbb{Q}$ does not have a basis as left $\mathbb{Z}$-module and is not finitely generated.

It seems the following holds: For any element $z:=\frac{1}{p}\otimes \frac{1}{q}\in R$ there is the following equality:
$\frac{pq}{pq}z=\frac{1}{pq}(\frac{p}{p}\otimes \frac{q}{q})=\frac{1}{pq}(1\otimes 1)$.

Let $\rho: \mathbb{Q}\rightarrow R$ be defined by $\rho(r):=r\otimes 1$.
We may write any element in $w\in R$ as follows:

$w:=\sum_i \frac{p_i}{q_i}\otimes \frac{a_i}{b_i} =\sum_i \frac{p_ia_i}{q_ib_i}(1\otimes  1):=r(1\otimes 1)$ for a rational number $r$.  Hence it seems $\rho$ is an isomorphism giving an isomorphism

$\mathbb{A}^1_{\mathbb{Q}}\times_{\mathbb{Z}} \mathbb{A}^1_{\mathbb{Q}} \cong
\mathbb{A}^2_{\mathbb{Q}}$.

In general for any multiplicative subset $S\subseteq A$, there is an isomorphism $S^{-1}A\otimes_A S^{-1}A\cong S^{-1}A$ of rings.

ContentLicense: CC BY-SA 3.0

ContentLicense: CC BY-SA 4.0


问:
modulo question For a positive integer $k$, does $\{x^2 \pmod k \mid x \in\mathbb Z\} = \{x \pmod k \mid x \in\mathbb Z\} $?
For a positive integer $k$, does $\{x^2 \pmod k \mid x \in \mathbb{Z}\} = \{x \pmod k \mid x \in \mathbb{Z}\} $ ?

答:
The question is equivalent to ask does $\{x^2 \pmod k | x \in Z\} =\{0,...,k-1\}$. This is obviously false, take $k=3$, we have no $x$ s.t. $x^2 \equiv 2 \pmod 3$.

In fact, given a non-zero $n$ and an odd prime $p$, Euler's criterion tells us that ${\displaystyle n}$ has a square root (i.e., ${\displaystyle n}$ is a quadratic residue) if and only if:
$$
{\displaystyle n^{\frac {p-1}{2}}\equiv 1{\pmod {p}}}
$$


问:
Questions on atoms of a measure

In Kai Lai Chung's A course in
probability theory,

An atom of any probability measure $\mu$ on
  $(\mathbb{R}, \mathcal{B})$ is a
  singleton $\{x\}$ such that $\mu({x}) > 0$.

In Wikipedia:

Given a measurable space $(X,Σ)$ and a
  measure $\mu$ on that space, a set A
  in Σ is called an atom if $\mu (A) >0$
  and for any measurable subset B of A
  with $\mu(A) > \mu (B)$, one has
  $\mu(B) = 0$.

I was wondering if Chung's and Wikipedia's definitions agree with each other? If yes, does Chung's
definition mean that in the special probability measure space
$(\mathbb{R}, \mathcal{B}, \mu)$, there
is no  atom, as defined as in
Wikipedia, that is not a singleton?

Chung says in his book that the
number of atoms of any
$\sigma$-finite measure is
countable.

I was wondering if suppose there are    uncountably many atoms, how
will it    contradicts
$\sigma$-finite measure?

Is this conclusion also true for
general measure space $(X,Σ, \mu)$
instead of just for probability
measure space $(\mathbb{R},
        \mathcal{B}, \mu)$?

Thanks and regards!

答: Take a look at the exercise 22 in the page 32 of the Chung's book. In this exercise, he asks that be demonstrated the equivalence between these two definitions of atoms when the measure is defined on the euclidean Borel field.</p>

Title: Questions on atoms of a measure


问:
Proving a bounded set is Lebesgue Measurable
This is an exercise question from the book Measure,Integration and Real Analysis  by Sheldon Axler.

Suppose $ b < c$ and  $A\subset(b,c)$. Prove that A is Lebesgue measurable if and
only if $ | A | + |( b, c ) \setminus A | = c − b.$

Here $|A|$ represents outermeasure of A.

I'm trying to prove the converse part.

Definition of Lebesgue Measurable set :  A Set A is said to be Lebesgue Measurable if $ \forall  \epsilon>0 , \exists   F(closed set) \subset A$ such that $|A\setminus F|<\epsilon $
I'm trying to prove this by contradiction.

Suppose A isn't Lebesgue Measurable. $\exists \epsilon>0$ such that $|A\setminus F|>\epsilon $ for all closed  $F\subset A$.

Similarly  $(b,c)\setminus A$ isn't Lebesgue measurable $\exists \epsilon>0$ such that $|((b,c)\setminus A) \setminus F|>\epsilon $ for all closed  $F\subset (b,c)\setminus A$

Since (b,c) is Lebesgue measurable, $ \forall  \epsilon>0 , \exists   F(closed set) \subset (b,c)$ such that $|(b,c)\setminus F|<\epsilon $

If F  is closed subset of A and E is closed subset of (b,c)\A then $|(b,c)\setminus (F\cup E)| = |A\setminus F|+|((b,c)\setminus A)\setminus E|$

I dont know how to arrive at a contradiction from this.

答: Suppose that $ | A | + |( b, c ) \setminus A | = c − b$ and $\epsilon>0$  be any number. There exist countable covers $\mathcal C_1$ of $A$ and $\mathcal C_2$ of $( b, c ) \setminus A$ by open subintervals of $(b,c)$ such that $S_1+S_2\le c-b+\epsilon$, where $S_i=\sum_{C\in\mathcal C_i} |C| $ for each $i$. It is easy to check that if $|\bigcup\mathcal C_1\cap\bigcup \mathcal C_2|>\epsilon$ then $\mathcal C_1\cap\mathcal C_2$ cannot cover $(b,c)$, so $|\bigcup\mathcal C_1\cap\bigcup \mathcal C_2|\le\epsilon$. Put $F=(b,c)\setminus\bigcup \mathcal C_2$. Clearty $F$ is a closed set contained in $A$. Then $A\setminus F= A\setminus \bigcup \mathcal C_2\subset \bigcup \mathcal C_1\cap\bigcup \mathcal C_2$, so $| A\setminus F |\le\epsilon$.


问:
Show that F is a normal family
G is a domain in $C$, where $C$ is the complex plane and $G \neq C$ and $M>0$ we have that:
$F=\{f\in H(G): \int_{G}|f(z)|^2 dm_2(z)\leq M\}$

How do I show that F is a normal family? - Some of the other answers are missing some calculations and so I don't quite follow it.

答:
Fix $f \in F$, $z_0 \in G$. We can find $R > 0$ so that $\overline{D}(z_0, 2R) \subset G$. Let $z \in D(z_0, R)$, $0 \le r \le R$. By the mean value property:
$$
f(z) = \frac{1}{2\pi} \int_0^{2\pi} f(z + re^{i\theta}) \,d\theta
$$

Multiply both sides by $r$ and integrate with respect to $r$ from $0$ to $R$ to get:
$$
f(z) = \frac{1}{\pi R^2} \int_0^R \int_0^{2\pi} f(z + re^{i\theta})r \,d\theta dr
$$

By Cauchy-Schwarz:

\begin{align}
\left|f(z)\right| &\le \frac{1}{\pi R^2} \int_0^R \int_0^{2\pi} \left|f(z + re^{i\theta})r\right| \,d\theta dr \\
&\le \frac{1}{\pi R^2} \left\{\int_0^R \int_0^{2\pi} \left|f(z + re^{i\theta})\right|^2 r \,d\theta dr\right\}^{1/2} \left\{\int_0^R \int_0^{2\pi} r \,d\theta dr\right\}^{1/2}
\end{align}

The last expression is bounded by a bound that is independent of $f$. It only depends on $M$ and $R$. Hence it is valid for all $z \in D(z_0, R)$. It follows that all $f \in F$ are uniformly bounded in a neighborhood of $z_0$ for any $z_0 \in G$. Let $K$ be a compact subset of $G$. Cover $K$ with a finite set of such neighborhoods. It follows that $F$ is uniformly bounded on each compact subset of $G$. Thus $F$ is normal.

Title: Show that F is a normal family


问:
How to "fix" $\int_{-1}^1 \frac {dx}{x^2}$ with complex numbers?
Given the following definite integral,

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

I can see that it is an improper integral because of the asymptote at $x=0$ and I know from the graph of $\frac 1{x^2}$ that both parts on either side of the $y$-axis are identical.

Hence, computing only the part on the right of the $y$-axis I find that the integral $\int_{-1}^1 \frac {dx}{x^2}$ is divergent since $-$as expected from the graph of $\frac 1{x^2}-$ this part goes to $\infty$:

$$\lim_{\epsilon\to 0^+}\int_{0+\epsilon}^1 \frac {dx}{x^2} \;=\; \lim_{\epsilon\to 0^+}-\frac 1x \;\bigg|_{\,0+\epsilon}^1 \;=\; -1-\left(\lim_{\epsilon\to 0^+}-\frac 1{0+\epsilon}\right) \;=\; \infty\;.$$

If I intentionally overlook that and blindly compute the integral, I get the silly result

$$\int_{-1}^1 \frac {dx}{x^2} = -2\;.$$

That is, a negative value for the area under a function that is above the $x$-axis everywhere.

I was told that $\int_{-1}^1 \frac {dx}{x^2}$ can be "fixed" using imaginary numbers by a T.A. but neither of us had the time to expand on that point.

I understand complex numbers and how they work so I am left to wonder; how can I "fix" an improper integral using complex numbers?

答:
When we are speaking about the function $f(x)=\frac1{x^2}$ we can mean two different things:

$\,f_1(x)={\begin{cases}0,&{\mbox{if }}x=0\\\frac1{x^2},&{\mbox{otherwise}}\end{cases}}$

We consider $f_2(x)$ as a holomorphic function.

If we integrate the first case, we get the principal value and it will be infinite:

$$PV\int_{-1}^1\frac{dx}{x^2}=\infty$$

If we integrate the second case, we will get the Hadamard finite part and it will be $-2$:

$$\operatorname{f.p.}\int_{-1}^1\frac{dx}{x^2}=-2$$

Now, what happens? I will try to explain in simple language.

If we take the first case, the integral $\int_{-\infty}^\infty f_1(x)dx$ is equal to $\int_{-\infty}^\infty 1 dx = 2\pi\delta(0)$ formally (or germ of $\frac2x$ at $x\to 0^+$) which can be seen from the Laplace transform. This integral has the regularized value of $0$.

Since $\int_1^\infty \frac1{x^2} dx=1$, the integral $\int_{-1}^1 f_1(x)dx$ has the regularized value $-2$. We can write it formally as $2\pi\delta(0)-2$.

Lets now consider the second case.
Here, the function has singularity at zero... and that singularity behaves in such a way as if it was infinitely negative, so it balances out the other parts of the function:

$\int_{-\infty}^\infty f_2(x)dx=0$

This is analogous to the imaginary jump the logarithmic function has at zero, except in this case the jump is infinite (and not imaginary).

So, $\int_{-1}^1 f_2(x)dx=-2$.

This, second case is used in tables of Fourier transforms, for instance.


问:
Showing derivative operator is self-adjoint
Consider the Hilbert space $L^2(\mathbb{R})$, and unbounded operator $Au:=iu’$ with domain
$$D(A)= \{u \in L^2(\mathbb{R}) | u \text{ is absolutely continuous and } u’ \in L^2(\mathbb{R})\} $$
I’m showing that it’s self-adjoint. It’s not hard to show that it’s symmetric, and suffice to show that $D(A^*) \subseteq D(A)$, which I don’t know how to deal with.
Any help will be appreciated.

答:
Suppose $v\in\mathcal{D}(A^*)$. Then there exists $w\in L^2(\mathbb{R})$ such that, for all $u\in\mathcal{D}(A)$, one has
$$
           \int_{-\infty}^{\infty} (Au)vdx=\int_{-\infty}^{\infty}uwdx.
$$
Information about $w$ is obtained by choosing specific $u \in \mathcal{D}(A)$. For example, let
$$
         u_{a,b,\epsilon,\delta}=\int_{-\infty}^{x}\frac{1}{\epsilon}\chi_{[a-\epsilon,a]}(t)-\frac{1}{\epsilon}\chi_{[b,b+\delta]}(t)dt.
$$
This function is $0$ for $x < a-\epsilon$ and $0$ for $x > b+\delta$. It is $1$ in $[a,b]$, and it is linear on $[a-\epsilon,a]$ and on $[b,b+\delta]$. $u_{a,b,\epsilon,\delta}\in\mathcal{D}(A)$. So, the first equation applied with $u=u_{a,b,\epsilon}$ gives
$$
       \frac{1}{\epsilon}\int_{a-\epsilon}^{a}vdx-\frac{1}{\epsilon}\int_{b}^{b+\delta}vdx=\int_{-\infty}^{\infty}u_{a,b,\epsilon,\delta}wdx
$$
The limit as $\epsilon\downarrow 0$ exists, because it exists on the right. Similarly the limit as $\delta\downarrow 0$ exists. It follows that the following holds for almost all $a,b\in\mathbb{R}$:
$$
              v(a)-v(b)=\int_{a}^{b}w(x)dx
$$
Therefore, $v$ is equal a.e. to an absolutely continuous function that is in $L^2$ with a derivative $v'=w\in L^2$. So $v\in\mathcal{D}(A)$. In other words $\mathcal{D}(A^*)\subseteq\mathcal{D}(A)$. The opposite inclusion follows directly using the symmetry of $A$.


问:
If A is semisimple algebra then every A-module is completely reducible
A semisimple Algebra means the regular module is completely reducible. Let $A= \sum_{i}A_{i}$ where $A_{i}$ is irreducible sub modules of A. Suppose V is A-module. I tried to define a map between the $A_{i}$ and V.
$$\theta_{i}: A_{i} \longrightarrow V$$
$$ x_{i}\mapsto vx_{i} $$
the map is homomorphism so
$A_{i}\setminus ker(\theta_{i})$  is isomorphic to $\theta (A_{i}) \subseteq V$
the kernel is $\{0\}$ since $A_{i}$ is irreducible, so $A_{i}$  is isomorphic to $\theta_{i} (A_{i}) \subseteq V$.

$\theta_{i} (A_{i})$ is irreducible because of the isomorphic. Now, $\oplus \sum_{i} \theta_{i} (A_{i})$ is the direct sum of irreducible V submodules.

$\oplus \sum_{i}A_{i}$  is isomorphic to $\oplus \sum_{i} \theta_{i} (A_{i})$, i.e  $A$ is isomorphic to $\oplus \sum_{i} \theta_{i} (A_{i})$

If the steps are right, how can I show that $\oplus \sum_{i} \theta_{i} (A_{i})=  V$

答:
$A$ is semisimple as (right) $A$-module, $V$ is a (right) $A$-module and $v\in V$.

$\ker(A\to vA)$ is a submodule of $A$ thus it has a complementary module $A =\ker(A\to vA)\oplus B$ $$B\cong \ker(A\to vA)\backslash A\cong vA$$ so that $vA\cong B$ is a submodule of $A$, it is semisimple.

For another cyclic submodule $wA\subset V$ and $U\subset V$ semisimple we have that $$U= (wA\cap U)\oplus S \qquad \implies wA+U = wA\oplus S$$ is semisimple.

Thus every (finitely generated) $A$-module is semisimple.


问:
Statistics, Hypothesis and P-value - Check my answers
I am trying to solve a question but stuck with the steps. I can not find any similar questions. With help of some online resources to calculate some parts of the question but I can see that is not enough. I know my approach has lack of information but, this is the only thing I have reached, I was covid ill at the class hours and can not follow the class examples, I thought someone can help me to solve and learn the subject.

With help of the answers from here I try to give an answer. Still need some improvements but tried to do my best. I still do not have answer for question D and confused about CL(the part C) and Significance level(part B)

My answers:

$N\ =\ 9\ \ \ \ \ \ \ \ Sum\ of\ x\ =\ 3970\ \ \ \ \ \ \ \ Mean,\ µ = 441.1111  \      Variance,σ^2  = 161.1111$
$ \sigma\ =\ \sqrt{161.1111}  = 12.6929$

$t\ =\ \frac{m\ -\ \mu}{s\ /\ \sqrt n}$

$t\ =\ \frac{500\ -\ 441.1111}{12.6929\ /\ \sqrt9} = 13.918545$

We subtract 1 to get degrees free 9 - 1 = 8

Degrees of freedom = n – 1 = 8

$Probability: P( T ≤ 13.918545) = 0.00000069 $ So, this is the p-Value

$We\ will\ reject\ H_0\ at \ \alpha = 1% $ and also any > 1%

$$ (i) 0.10\
The\ information\ from\ the\ first\ question\, the\ critical\ t-value\ for\ α = 0.10\ and\ df = 8,\ t_c=\ 1.86
 \\ CI\ =\ (\bar{X}\ -\ \frac{t_c\ \times\ s}{\sqrt n},\ \bar{\ X}\ +\ \frac{t_c\ \times\ s}{\sqrt n})
 \\CI\ =\ (441.1111\ -\ \frac{1.86\ \times\ 12.6929}{\sqrt9},\ \ 441.1111+\frac{1.86\ \times\ 12.6929}{\sqrt9})
= (441.1111 – 7.868, 441.1111 + 7.868)
= (433.243, 448.979)
 $$

$For\ the\ other\ t_c\ values:$

(ii) 0.05  $t_c=\ 2.306$ CL = (431.354, 450.868) (iii) 0.01 $t_c=\
> 3.355$
= (426.915, 455.308)

Based on the answers in part 2 for (i) = 0.90, (ii) = 0.95, (iii) = 0.99, none of the confidence intervals contain 500.

The Question:

The worker says that the mean purchasing cost is 500 USD. We decide to test this.

For a random sample of 9 purchases drawn from a normally distributed population with unknown variance, the costs are:

```<code>430, 450, 450, 440, 460, 420, 430, 450, 440.
</code>```

A) Conduct a hypothesis test of whether the population mean purchasing equals 500 USD. Include all assumptions, the hypotheses, test statistic, and P-value and interpret the result in context.

B) For which significance levels can you reject $H_0?$ (i) 0.10, (ii) 0.05, or (iii)
0.01.

C) Based on the answers in part B), for which confidence levels would the confidence interval contain 500? (i) 0.90, (ii) 0.95, or (iii) 0.99.

D) Use part B) and part C) to illustrate the correspondence between results
of significance tests and results of confidence intervals.

答:
As population variance is unknown you have a t-score, not z-score.

I did not check your calculation but if the score is so high (in Absolute value) you will reject the null hypothesis for any significance level.

Thus you reject $H_0$ at $\alpha =1\%$ and also at any $\alpha>1\%$

The rest of the exercise follows as a consequence


问:
How do I find the equation of the sphere given a circle that passes through the sphere and a plane that is tangent to the sphere?
My problem is the following:
Find the equation of the sphere, that passes through the circle $$x^2+y^2=11$$, $$z=0$$ and is tangent to the plane $$x+y+z-5=0$$.
I have no idea how to solve this one. I know this looks easy, but I guess I'm stupid and I have an analytic geometry exam in a week. I would really appreciate any hint or solution. Thanks in advance!

P.S.Please don't bully me for asking stupid questions. I'm a sensitive person

答:
Assume the center of the sphere is $(0,0,t)$, then the equation of the sphere is
$$
x^2+y^2+(z-t)^2=t^2+11
$$
Using the formula of Distance from point to plane, we have
$$
d= \frac{1\cdot 0+1\cdot 0+1\cdot t-5}{\sqrt{1^2+1^2+1^2}} = \dfrac{t-5}{\sqrt{3}}
$$
Since it is tangent to the plane, the following relation holds:
$$
d = r = \sqrt{t^2+11}
$$
Simplify, and we get
$$
\left( \frac{t-5}{\sqrt{3}} \right) ^2=t^2+11\\
\dfrac{t^2-10t+25}{3}=t^2+11\\
t^2-10t+25=3t^2+33
\\
2t^2+8=-10t
\\
t^2+5t+4=0
$$
Thus
$$
t=-1 \text{ or } t = -4
$$
We are done.


问:
Vectors that form a triangle!
I have a problem here.

How can I prove that sum of vectors that form a triangle is equal to 0 $(\vec {AB}+\vec {BC}+\vec {CA}=\vec 0)$ ?

答:
Vector addition is defined as follows:
$$\vec{PQ}+\vec{QR}:=\vec{PR}$$
So that, $\vec{AB}+\vec{BC}+\vec{CA}=\vec{AC}+\vec{CA}=\vec{AA}={\bf 0}$.</p>

Title: Vectors that form a triangle!


问:
How can I prove this sequence $b_{n+1}=\frac{1}{3}(b_n^2+1)$ is contractive

A sequence $\{a_n\}$ is contractive if there exists $0<r<1$ such that $$\left|a_{n+1}-a_n \right| \le r\left|a_n-a_{n-1} \right|$$ for all $n\ge2$. Let $$b_{n+1}=\frac{1}{3}(b_n^2+1)$$, for $n\ge 1$ and $0<b_1<1$.
Please prove that $\{b_n\}$ is contractive.

I have thought about this question for a couple of days and I still can't figure out how to do it. I think how $b_n$ defined is meaningful but I can not find some useful inequality to solve this question. I thought arithmetic-geometric mean inequality would help but it fails. Hope someone can give me a hint how to do it.

答:
Let $f(x) = \frac{x^2+1}{3}$. Hence

$$|b_{n+1} - b_n| = |f(b_n) - f(b_{n-1})| = |f'(\xi_n)| |b_{n} - b_{n-1}|, $$
where $\xi_n$ lies between $b_{n}$ and $b_{n-1}$.

If $b_1 \in (0,1)$ then it's easy to prove by induction that $b_n \in (0,1)$, $n \ge 1$. Hence $|f'(\xi_n) | \le \sup_{x \in [0,1]} f'(x) = \frac23$.

Thus $$|b_{n+1} - b_n| \le \frac23 |b_{n} - b_{n-1}|$$
and the sequence is contractive with $r = \frac23$.


问:
Dot notation for group extensions
I came across the notation $A.B$ in many occassions while reading papers in Group Theory. However, I have not yet found any description of what this notation is supposed to mean. From context I suppose it means that if $G=A.B$, then $G$ is a solution to the extension problem $1\rightarrow A\rightarrow G\rightarrow B\rightarrow 1$, and the extension is not necessarily split. Is this the right way to interpret this symbol? In that case I suppose it would be useful in cases where the analysis does not depend on the particular extension.

答:
This is known as ATLAS notation, and it was introduced and widely used in the ATLAS of Finite Groups. It has become a standard notation among (particularly finite) group theorists, and it is very concise and useful, but it should be used with care.

In fact you have guessed the meaning correctly. If $A$ and $B$ are groups (or isomorphism types of groups), then $A.B$ means any group with normal subgroup isomorphic to $A$ with corresponding quotient group isomorphic to $B$.

There are refinements. For example $A:B$ denotes a split extension, and you would normally use $A \times B$ if you knew that it was a direct product, but just writing $A.B$ does not imply that it is not split or a direct product, it probably just means that you do not know or care.

Other conventions include using a number to denote a cyclic group of that order (as in $3.A_6$or $A_5.2 = S_5$), or a number like $[12]$ in square brackets to denote an unspecified group of order $12$.

Title: Dot notation for group extensions


问:
Finding angles of non-overlapping adjacent triangles whose vertices lie on the edges of an annulus

The problem

I have two concentric circles with radii $r$ and $R$, where $R \geq r$. These circles forms an annulus. On the outer circle $n$ equidistant points are placed, denote them by $(p_0 = p_n, p_1, p_2, \dots, p_n)$. I want to find $n$ equidistant points $\{q_1, \dots, q_n\}$ on the inner circle such that the triangles $\{\Delta(p_i, p_{i+1}, q_i) \mid 1 \leq i \leq n \}$ do not overlap and are adjacent to each other.

See the following picture for an illustration where $n = 5$ and $R = 2*r$: , and a larger example:
The problem comes down to essentially finding the angle $\alpha$ indicated in the first picture.

What I've tried

Using the fact that the angle of one corner of a regular $n$-gon is $\frac{(n-2)*\pi}{n}$ I could derive one angle of the coloured triangles. Similarly, the sides of the outer and inner polygon are found using $2R\sin{\frac{\pi}{n}}$ and $2r\sin{\frac{\pi}{n}}$. Also the area of one colored triangle can be calculated by $\frac{1}{2}\sin{(\frac{2\pi}{n})}(R^2 - r^2)$.

The visuals linked previously were made by manually finding the correct angle, for $n=5$ and $R=2*r$ the angle $\alpha \approx 0.527$ radians or around $30$ degrees.

Could someone give a hint on how to continue? This is not a homework question or anything; I want to make an animation and for that I need the angle $\alpha$.

答:
For outer radius $r$, inner radius $s$, and $\theta := \pi/n$ (ie, half the central angle of the $n$-gon), we have ...

$$s\cos\theta = r\cos(\theta+\alpha) \quad\to\quad \alpha = \arccos\left(\frac{s}{r}\,\cos\theta\right)-\theta$$

For $n=5$ and $r=2s$, this yields $\alpha = 0.52603\ldots$ (aka, $30.1396\ldots^\circ$).


答: Hint: consider the difference $b_{n+1}-b_n$, use the reccurence formula and get $r=2/3$.

Title: How can I prove this sequence $b_{n+1}=\frac{1}{3}(b_n^2+1)$ is contractive

Body:
A sequence $\{a_n\}$ is contractive if there exists $0<r<1$ such that $$\left|a_{n+1}-a_n \right| \le r\left|a_n-a_{n-1} \right|$$ for all $n\ge2$. Let $$b_{n+1}=\frac{1}{3}(b_n^2+1)$$, for $n\ge 1$ and $0<b_1<1$.
Please prove that $\{b_n\}$ is contractive.

I have thought about this question for a couple of days and I still can't figure out how to do it. I think how $b_n$ defined is meaningful but I can not find some useful inequality to solve this question. I thought arithmetic-geometric mean inequality would help but it fails. Hope someone can give me a hint how to do it.


问:
Why can $M$ be an abelian group?
In Rotman's, "An introduction to homological algebra", there is a proof of that the categories of modules has enough injectives.

In the proof, he proves that there is an injective map from $M$ as an abelian group to an injective module, why this proves that $M$ as a module can be embebed in such injective module?

答:
Every left $R$-module can be seen as an $R$-$\Bbb Z$-bimodule, and any Abelian group as a $\Bbb Z$-$\Bbb Z$ bimodule.

Now, if ${}_AM_B$ and ${}_AN_C$ are bimodules, we get a $B$-$C$-bimodule structure on $\hom_A(M,N)$ by defining
$$b\cdot f\cdot c\ :=\ m\mapsto  f(mb)\cdot c\,.$$

It follows that $\hom({}_{\Bbb Z}R,\,{}_{\Bbb Z}D)$ is a left $R$-module by the right $R$-action on $R$, and they verify that the injective map $M\to\hom_{\Bbb Z}(R,M)\to\hom_{\Bbb Z}(R,D)$  actually preserves the left $R$-action.

Now, if we know that $\hom_{\Bbb Z}(R,D)$ is an injective $R$-module whenever $D$ is a divisible Abelian group, then we're done.

Title: Why can $M$ be an abelian group?


答:
Welcome to MSE!

You mention the definition of measurability in terms of being "almost closed", however Axler actually gives a whole slew of equivalent definitions of measurable. These show up in the section 2D, and seeing as your exercise is 2D.12, it makes sense that you might want to use these definitions too! I've included a screenshot for reference:

A proof that these are equivalent can be found on page 53 of Axler's book. Again, this is all done before your exercise of interest, so I will freely use these equivalent definitions.

Let's show the easy direction first, you say you've done it, but I'll go ahead and do it too for completeness:

If $A$ is measurable, why is $|A| + |(b,c) \setminus A| = c-b$?

From chapter 2D again (Theorem 2.72), we know that $|\cdot|$ is a measure on the measurable sets (so at least the name makes sense :P). But from the definition of a measure (chapter 2C), we have additivity of the measures of disjoint sets.

In particular, since $A$ and $(b,c) \setminus A$ are disjoint, we have

$$|A| + |(b,c) \setminus A| = |A \cup ((b,c) \setminus A)| = |(b,c)| = c-b$$

What about the other direction? This is notably trickier, as I'm sure you're aware. To show it, we have to use a trick which will follow you throughout measure theory: go back to the basics. In measure theory even more than other areas of math, we define properties $P$ first for "simple" objects. Then we show how to approximate arbitrary objects by simple ones, and define $P$ for arbitrary objects to be the limit of $P$ of the simple objects. Oftentimes there's a follow-up step, where we restrict attention from all objects to only those objects on which $P$ is "nice".

We have already seen that for the definition of a measure.

First we defined $|\cdot|$ for open intervals.

Then we extended the definition to arbitrary sets (this is the "outer measure")

Then we restrict attention to those sets on which the outer measure behaves well (we call these sets "measurable")

You'll see soon enough that we do the same thing with integration!

Ok, but why bring this up? Because when we have these kinds of iterative definitions, it's often a good idea to go back to basics.

In particular, recall (definition 2.2)

$$|E| \triangleq \inf \left \{ \left . \sum_n |I_n| ~\right |~ E \subseteq \bigcup_n I_n \right \}$$

where the $I_n$ are open intervals, and we've already defined $|(a,b)| = \ell((a,b)) = b-a$.

But what does this buy us? For any set $E$ any $\epsilon$, we can find some intervals $I_n^\epsilon \subseteq (b,c)$ so that $E \subseteq \bigcup_n I_n^\epsilon$ and $\sum_n |I_n^\epsilon| \leq |E| + \epsilon$ (do you see why?).

Working with a sequence of intervals is kind of unwieldy, though. So we'll simplify with subadditivity. We'll suggestively write $G^\epsilon \triangleq \bigcup_n I_n^\epsilon$. Then $E \subseteq G^\epsilon$, and we know

$$|G^\epsilon| = \left | \bigcup_n I_n^\epsilon \right | \overset{\star}{\leq} \sum_n |I_n^\epsilon| \leq |E| + \epsilon$$

Here $\star$ follows from subadditivity of outer measure (Theorem 2.8).

Notice also that $G^\epsilon$ is a union of open intervals. In particular it's open.

Now we make one last step, and define $B = \bigcap_n G^{1/n}$. Then we have $E \subseteq B$ (do you see why?), with $B$ borel (do you see why?), and $|E| = |B|$. This last statement may be tricky as you're gaining experience with analysis, so I'll give a (do you see why?), but I'll also include a proof here:

 $E \subseteq B$, so by theorem 2.5 we have $|E| \leq |B|$.

 In the other direction, notice for every $n$, $E \subseteq G^{1/n}$, so

 $$|B| \leq |G^{1/n}| \leq |E| + \frac{1}{n}$$

 But that means $|B| \leq |E|$, as needed.

So, at the end of the day, what have we done? We've proven a lemma that says

For any set $E$, There is a borel set $B \supseteq E$ so that $|E| = |B|$.

We're in the home stretch now! Applying the lemma shows that we can find a set $B \supseteq A$ so that $|B| = |A|$. But by criterion (g) above, if we can show $|B \setminus A| = 0$, then $A$ is measurable and we're done! This doesn't seem that far off at all!

We're left with one last trick: We must also approximate $A^c$. That is, let's apply the lemma again to get a borel set $C \supseteq A^c$ with $|C| = |A^c|$.

Now we finally apply our assumption: $|A| + |A^c| = |(b,c)|$. This tells us that $|B| + |C| = |(b,c)|$ too.

But since $C$ is measurable (it's borel), we also know that $|C| + |C^c| = |(b,c)|$.

Thus $|C| + |C^c| = |(b,c)| = |B| + |C|$, and (since everything in sight is finite), we have $|C^c| = |B|$, and both of these sets are borel! Notice also that $C^c \subseteq A \subseteq B$.

Finally, then $|B \setminus A| \overset{\heartsuit}{\leq} |B \setminus C^c| \overset{\star}{=} |B| - |C^c| = 0$. And we're done by condition (g). And one last (do you see why $\heartsuit$ and $\star$ hold?) for the road.

Edit:

Also, this looks rather long and unwieldy, but that's primarily because I'm trying to provide motivation for every step. Once you become fluent with these concepts, the proof is quite short! Alex Ravsky's answer does a great job showing this.

I hope this helps ^_^

Title: Proving a bounded set is Lebesgue Measurable

Body:
This is an exercise question from the book Measure,Integration and Real Analysis  by Sheldon Axler.

Suppose $ b < c$ and  $A\subset(b,c)$. Prove that A is Lebesgue measurable if and
only if $ | A | + |( b, c ) \setminus A | = c − b.$

Here $|A|$ represents outermeasure of A.

I'm trying to prove the converse part.

Definition of Lebesgue Measurable set :  A Set A is said to be Lebesgue Measurable if $ \forall  \epsilon>0 , \exists   F(closed set) \subset A$ such that $|A\setminus F|<\epsilon $
I'm trying to prove this by contradiction.

Suppose A isn't Lebesgue Measurable. $\exists \epsilon>0$ such that $|A\setminus F|>\epsilon $ for all closed  $F\subset A$.

Similarly  $(b,c)\setminus A$ isn't Lebesgue measurable $\exists \epsilon>0$ such that $|((b,c)\setminus A) \setminus F|>\epsilon $ for all closed  $F\subset (b,c)\setminus A$

Since (b,c) is Lebesgue measurable, $ \forall  \epsilon>0 , \exists   F(closed set) \subset (b,c)$ such that $|(b,c)\setminus F|<\epsilon $

If F  is closed subset of A and E is closed subset of (b,c)\A then $|(b,c)\setminus (F\cup E)| = |A\setminus F|+|((b,c)\setminus A)\setminus E|$

I dont know how to arrive at a contradiction from this.


问:
For which $\alpha\in\mathbb{R}$ is $\frac{1}{1+\lvert\lvert x\rvert\rvert^{\alpha}}$ integrable on $\mathbb{R}^n\setminus\{0\}$?
Let $\lvert\lvert \,.\rvert\rvert$ be a norm on $\mathbb{R}^n$. For which $\alpha\in\mathbb{R}$ is $\frac{1}{1+\lvert\lvert x\rvert\rvert^{\alpha}}$ integrable on $\mathbb{R}^n\setminus\{0\}$?

My approach: I have tried to approximate $\frac{1}{1+\lvert\lvert x\rvert\rvert^{\alpha}}$ from below using simple functions but nothing useful has presented itself. Is there a better way for solving this?

答:
If you change to polar coordinates, you will see it all comes down to

$$\int_0^\infty {r^{n-1}\, dr\over 1 + r^\alpha}.$$

The integrand is continuous, so you only need to analyze what makes it integrate at $\infty$.


问:
The range of an inverse function
I study maths as a hobby. I have the following question:

Find the inverse function of $f(x) = \frac{1}{(px + q)}$, $x \ne \frac{-q}{p}$ where p and q are positive constants.

Now, I can find the inverse function easily enough. The book gives the answer as:

$f^{-1} = \frac{1 - qx}{px} (x \ne 0)$

It is this last part I cannot figure out. Why $(x \ne 0)$?

答: Observe that $0$ is not in the range of $f$ (even though the range and domain is not explicitly defined.), thus it is not in the domain of the inverse of the function $f$.

Title: The range of an inverse function


问:
A problem about pigeon hole principle.
Let $1 < a_1 < a_2 < a_3 < ... < a_{51} < 142.$ Prove that, among the 50 consecutive differences $(a_i − a_{i - 1})$ where $i = 1, 2, 3, ..., 51$, some value, must occur at least twelve times.

My attempt; let $d_i = a_{i+1}-a_i$, so like $d_1=a_2-a_1 , d_2 = a_3 - a_2 ......... d_{50} = a_{51} - a_{50}$ , adding all these, we get $d_1 + d_2 + ..... d_{50} = a_{51} - a_1 \le 139$, now I am not sure what to do, can someone please help?

Thanks!

答:
Hint:

Suppose you tried to make $d_1 + d_2 + ..... + d_{50}$ as small as possible without having any value appear more than eleven times

$11 \times 1 + 11 \times 2 + 11 \times 3 + 11 \times 4 + 6 \times 5=140$

Title: A problem about pigeon hole principle.


答:
Suppose each difference can only appear $11$ times. Then:

$$\sum_{i=1}^{50}d_i \ge (1+2+3+4)\times 11+5\times 6 = 140$$

Title: A problem about pigeon hole principle.

Body:
Let $1 < a_1 < a_2 < a_3 < ... < a_{51} < 142.$ Prove that, among the 50 consecutive differences $(a_i − a_{i - 1})$ where $i = 1, 2, 3, ..., 51$, some value, must occur at least twelve times.

My attempt; let $d_i = a_{i+1}-a_i$, so like $d_1=a_2-a_1 , d_2 = a_3 - a_2 ......... d_{50} = a_{51} - a_{50}$ , adding all these, we get $d_1 + d_2 + ..... d_{50} = a_{51} - a_1 \le 139$, now I am not sure what to do, can someone please help?

Thanks!


问:
RGB to HSV Color Conversion Algorithm
I'm a programmer looking to build an RGB to HSV color converter.  I found an algorithm, but I have very little mathematical background and I'm not quite sure what's going on with it.  A step-by-step breakdown of exactly what is happening would be tremendously helpful so that I could code it.  RGB and HSV are each sets of three values.  R, G, and B are each 0-255, while H is 0-360° and S and V are each 0%-100%.

Here's the algorithm:

The R,G,B values are divided by 255 to change the range from 0..255 to 0..1:

R' = R/255

G' = G/255

B' = B/255

Cmax = max(R', G', B')

Cmin = min(R', G', B')

Δ = Cmax - Cmin

Hue calculation:

Saturation calculation:

Value calculation:
V = Cmax

.

.

Here are my questions:

Is CMax the average of the three numbers R, G, and B?

Is CMin always equal to zero?  If so, wouldn't Delta just be CMax?

What does mod6 mean?

In the calculations for H, there are three lines each with two parts separated by commas.  What the heck is going on here?

With S, I have the same gap in understanding as H, and I also don't know what <> means.

.

As you can see, I basically have no clue what's going on here.  Like I said, your help in solving this problem would be extremely useful to me and would be immensely appreciated.

Thank you so much for your time, and hopefully your help as well!

答:
Well, I have been searching for the same algorithm for months! I actually did not get my algorithm from Wikipedia, I got it from GitHub$^1$. Anyways, here is my formula:

$$r,g,b = \frac{r'}{255},\frac{g'}{255},\frac{b'}{255}$$
$$M = \max(r,g,b)$$
$$m = \min(r,g,b)$$
$$c = M - m$$
$$s = (\frac{c}{M})100$$
$$R, G, B = \frac{M-r}{c},\frac{M-g}{c},\frac{M-b}{c}$$

$$h' = $$
$0$: if M = m

$0+B-G$: if M = r

$2+R-B$: if M = g

$4+G-R$: if M = b

$$h = (\frac{h'}{6}\mod{1})360$$
$$v = M100$$

$^1$: https://github.com/python/cpython/blob/3.9/Lib/colorsys.py

This is my code version (python):

```<code>def rgb_to_hsv(r, g, b):
  r /= 255
  g /= 255
  b /= 255
  maxc = max(r, g, b)
  minc = min(r, g, b)
  v = maxc
  if minc == maxc:
      return 0.0, 0.0, v
  s = (maxc-minc) / maxc
  rc = (maxc-r) / (maxc-minc)
  gc = (maxc-g) / (maxc-minc)
  bc = (maxc-b) / (maxc-minc)
  if r == maxc:
      h = 0.0+bc-gc
  elif g == maxc:
      h = 2.0+rc-bc
  else:
      h = 4.0+gc-rc
  h = (h/6.0) % 1.0
  return h * 360, s * 100, v * 100
</code>```

Title: RGB to HSV Color Conversion Algorithm

Body:
Well, I have been searching for the same algorithm for months! I actually did not get my algorithm from Wikipedia, I got it from GitHub$^1$. Anyways, here is my formula:

$$r,g,b = \frac{r'}{255},\frac{g'}{255},\frac{b'}{255}$$
$$M = \max(r,g,b)$$
$$m = \min(r,g,b)$$
$$c = M - m$$
$$s = (\frac{c}{M})100$$
$$R, G, B = \frac{M-r}{c},\frac{M-g}{c},\frac{M-b}{c}$$

$$h' = $$
$0$: if M = m

$0+B-G$: if M = r

$2+R-B$: if M = g

$4+G-R$: if M = b

$$h = (\frac{h'}{6}\mod{1})360$$
$$v = M100$$

$^1$: https://github.com/python/cpython/blob/3.9/Lib/colorsys.py

This is my code version (python):

def rgb_to_hsv(r, g, b):
  r /= 255
  g /= 255
  b /= 255
  maxc = max(r, g, b)
  minc = min(r, g, b)
  v = maxc
  if minc == maxc:
      return 0.0, 0.0, v
  s = (maxc-minc) / maxc
  rc = (maxc-r) / (maxc-minc)
  gc = (maxc-g) / (maxc-minc)
  bc = (maxc-b) / (maxc-minc)
  if r == maxc:
      h = 0.0+bc-gc
  elif g == maxc:
      h = 2.0+rc-bc
  else:
      h = 4.0+gc-rc
  h = (h/6.0) % 1.0
  return h * 360, s * 100, v * 100


问:
Bayes' theorem: how frequently is the medical test correct?
Suppose that 1 in 3 people showing symptoms of dry cough, fatigue (tiredness)

and fever have COVID19.There is a test for COVID-19 that gives a positive result 98.5%

of the time when given to someone with the disease. When given to someone without the disease.

80% of the time it gives a negative result.

What is the probability that a randomly selected person showing symptoms of dry cough, fatigue (tiredness) and fever has COVID-19 if the test returns a positive result?

What is the probability that a randomly selected person showing symptoms of dry cough, fatigue (tiredness) and fever does not have COVID-19 if the test returns a negative result?

I approached the problem following way but got stuck

Let +ve denote the people who test positive.

And covid denote people who actually have covid.

Now from the given statements we can say that

P(+ve | covid) = .985 -- Means Given a person has

covid the probability that the test result will be +ve is .985

P(-ve | non-covid) = .8

P(covid | symptoms) = 1/3

Now the Q1. asks to find out P(+ve | symptoms)

P(+ve | symptoms ) = P(symptoms | +ve) x P(+ve) / P(symptoms)

Now the test gives +ve 98.5% of time for one with covid and

when given to somone without covid it gives +ve 20% of the time.

P(+ve | non-covid)  =.2  Dont know how to proceed further

答:
I will answer the first questions, the second is very similiar.

$$p(\text{covid}|\text{positive}) = \frac{p(\text{positive}|\text{covid})p(\text{covid})}{p(\text{positive}|\text{covid})p(\text{covid})+ p(\text{positive}|\text{not covid})p(\text{not covid})} =\\
\frac{0.985\times0.33}{0.985\times0.33 + 0.2\times0.66} \sim 70\%
$$

This is why doctors make patients take more tests if a test is known to give false positives.
See pic for a geometric interpretation for this kind of problems: think about plugging a number of random event at top.

Bye


问:
Is "choosing" considered as some sort of "chance" in probability?
There are 3 baskets. Basket 1 contains 4 red balls and 2 green balls. Basket 2 contains 6 red balls, 3 green balls, and 1 yellow ball. Basket 3 contains 5 red balls and 5 green balls. If I got a green ball, what is the probability that the basket I chose was basket 1?

I am not sure if the term "chose" was some hint that would suggest that I should treat the problem differently?

Anyway, here's my approach:

$B_1$ = Basket 1

$B_2$ = Basket 2

$B_3$ = Basket 3

$p(R|B_1) = \left(\frac{1}{3}\right)\cdot\left(\frac{4}{6}\right) = 0.22$

$p(G|B_1) = \left(\frac{1}{3}\right)\cdot\left(\frac{2}{6}\right) = 0.11$

$p(Y|B_1) = \left(\frac{1}{3}\right)\cdot\left(\frac{0}{6}\right) = 0$

$p(R|B_2) = \left(\frac{1}{3}\right)\cdot\left(\frac{6}{10}\right) = 0.2$

$p(G|B_2) = \left(\frac{1}{3}\right)\cdot\left(\frac{3}{10}\right) = 0.1$

$p(Y|B_2) = \left(\frac{1}{3}\right)\cdot\left(\frac{1}{10}\right) = 0.03$

$p(R|B_3) = \left(\frac{1}{3}\right)\cdot\left(\frac{5}{10}\right) = 0.167$

$p(G|B_2) = \left(\frac{1}{3}\right)\cdot\left(\frac{5}{10}\right) = 0.167$

$p(Y|B_2) = \left(\frac{1}{3}\right)\cdot\left(\frac{0}{10}\right) = 0$

Using Bayes' Theorem:
\begin{align*}
p(B_1|G) &= \dfrac{p(G|B_1)}{p(G|B_1) + p(G|B_2) + p(G|B_3)}\\
&= \dfrac{0.11}{0.11 + 0.1 + 0.167}\\
&= \dfrac{0.11}{0.377}\\
&= \dfrac{0.11}{0.377}\\
&= 0.29 = 29\%
\end{align*}

Please tell me if my approach was right.

答:
No, it is not. Words such as "choose", "pick", "select", "draw", etc... are usually meant in their literal English sense and involve no hidden things. Sometimes the probability of getting something is higher than the rest but that Is usually mentioned (example: a coin where the probability of getting head is $0.70$), what you should pay attention to is if the choice is made with or without replacement as that leads to a totally different answer.
That being said, your solution is correct.


答:
Two ways to look at it: one is that given a (bijective) function $f\colon X\to Y$ we define the inverse to be a function $f^{-1}\colon Y \to X$. But your $f$ is a function of the form $1/k$ and this fraction is never zero, hence $0\notin Y$.

The second way is simply to note that setting $x=0$ would cause $f^{-1}$ to become undefined (due to division by zero), which would make the inverse invalid.

Title: The range of an inverse function

Body:
I study maths as a hobby. I have the following question:

Find the inverse function of $f(x) = \frac{1}{(px + q)}$, $x \ne \frac{-q}{p}$ where p and q are positive constants.

Now, I can find the inverse function easily enough. The book gives the answer as:

$f^{-1} = \frac{1 - qx}{px} (x \ne 0)$

It is this last part I cannot figure out. Why $(x \ne 0)$?


问:
Refine the finite open covering to compact included covering
Let K be compact set in $\Bbb{R}^n$,assume $U_1,U_2,...,U_n$ be the open covering for it.Prove there eixst refinement $V_i\subset \overline{V_i}\subset U_i$ such that $\overline{V_i}$ is compact and $V_1,V_2,...,V_n$ open covers $K$ also.

My attempt due to locally compact we can find $V$ such that $K\subset V\subset \overline{V}\subset \bigcup_{i=1}^n U_i$ now we take $V_i = V\bigcap U_i$ comes the problem since $\overline{V_i}$ may not lies $U_i$ although it's compact

答:
For each $x\in K$ then $x\in U_i$ for some $i$. Now find an open ball $B_r(x)$ ($r$ depends on $x$) such that $\overline{B_r(x)}\subset U_i$. The family of balls $\{ B_r(x)\}_{x\in K}$ covers $K$ and thus has a finite subcover

$$ K \subset \bigcup_{j=1}^N B_{r_j} (x_j).$$

For each $j$, let $i = i(j)$ so that $\overline{B_{r_j} (x_j)} \subset U_i$ (such an $i$ exists since this is how we choose the balls in the first place). Let

$$V_i = \bigcup_{j:\  i(j)=i} B_{r_j} (x_j).$$

Then $\overline{V_i} \subset U_i$, $\overline{V_i}$ is compact and $\{ V_i\}$ covers $K$.


问:
A property of polynomials with nonnegative coefficients and constant term 1.
Let $p(x)$ be a polynomial with non negative integer coefficients and constant term 1, having $n$ real roots. Prove that $p(2) ≥ 3^n$

I'm actually getting the opposite result, $p(2) < 3^n$

Since it has only positive (or zero) coefficients, it must only have negative roots. If the roots are $-a_i$, $1≤i≤n$, then $p(x) = (x+a_1)(x+a_2).....(x+a_n)$ where $a_i ≠ a_j$ for $i ≠ j$

Then $p(2) =  (a_1 + 2)....(a_n + 2)$. The only way for $p(2) > 3^n$ is if each of the $a_i$ are $≥ 1$, but this is impossible, since $a_1a_2..a_n = 1$. Of course we can still have $p(2) = 3^n$, but then the appropriate result to prove would be $3^n ≥ p(2)$. Am I wrong?

Edit: I realize now that the statement 'the only way for $p(2) > 3^n$ is if each of the $a_i$ are $≥1$ is wrong. So the question is, how would I prove this(the original problem)

答:
Let $P$ be a polynomial of degree $n$ with non negative real coefficients and constant term $1$, having $n$ real roots.

As you said, the roots have to be negative as $P(x) > 0$ for all $x \ge 0$.

As such, we can write $P(x) = \prod_{k=1}^n (x+r_k)$ such that $r_k \ge 0$ for all $k$ and $r_1 \cdot r_2 \cdot r_3 \ldots r_n = 1$

We can plug $2$ into that expression to get:
$$
P(2) = \prod_{k=1}^n (2+r_k)
$$

Here, we can use the AM-GM inequality to proceed.
\begin{align*}
\frac{r_k + 1 + 1}3 &\ge \sqrt[3]{r_k \cdot 1 \cdot 1} \\
\implies 2 + r_k &\ge 3 \sqrt[3]{r_k} \\
\implies  \prod_{k=1}^n (2+r_k) &\ge 3^n \sqrt[3]{\prod_{k=1}^n r_k} \\
\implies P(2) &\ge 3^n
\end{align*}

Here, I split the $2$ into $1+1$ in order to get $3$ in the denominator.

In the end, we didn't need to use the fact that the coefficients of $P$ were integers so I don't know what's up with that.


问:
modulo question For a positive integer $k$, does $\{x^2 \pmod k \mid x \in\mathbb Z\} = \{x \pmod k \mid x \in\mathbb Z\} $?
For a positive integer $k$, does $\{x^2 \pmod k \mid x \in \mathbb{Z}\} = \{x \pmod k \mid x \in \mathbb{Z}\} $ ?

答:
In order to have $\{0^2,1^2,2^2,\cdots,(k-1)^2\}=\{0,1,2,\cdots ,k-1\}$ we need that $x\to x^2$ be a bijection.

This is possible only when $k=2$ because $x^2=(-x)^2$. It is possible for $k=2$ because $1\equiv -1\pmod2$.


问:
What's the sign of $\det\left(\sqrt{i^2+j^2}\right)_{1\le i,j\le n}$?
Suppose $A=(a_{ij})$ is a $n×n$ matrix by $a_{ij}=\sqrt{i^2+j^2}$. I have tried to check its sign by matlab. l find that the determinant is positive when n is odd and negative when n is even. How to prove it？

答:
What follows is just an explication of fedja's beautiful ideas, with the minor gap of non-zero-ness filled in. There seemed like enough interest to warrant sharing it. Along the way I decided to make it very explicit to make it more accessible, since it's so pretty.

Let
$$A = \left(\sqrt{i^2+j^2}\right)_{1 \leq i, j \leq n}.$$

Theorem: $(-1)^{n-1} \det(A) > 0$.

Proof: Since $A$ is symmetric, it has an orthogonal basis of eigenvectors, and its determinant is the product of its eigenvalues. We'll show there are $n-1$ negative eigenvalues and $1$ positive eigenvalue, i.e. $A$ is non-degenerate with signature $(1, n-1)$.

Let $x_0 := (1, \ldots, 1)^\top$. Considering the quadratic form corresponding to $A$,
$$x_0^\top A x_0 = \sum_{1 \leq i, j \leq n} \sqrt{i^2 + j^2} > 0.$$
Thus there must be at least $1$ positive eigenvalue $\lambda_+>0$. On the other hand, we have the following.

Claim: If $x \cdot x_0 = 0$ and $x \neq 0$, then $x^\top A x < 0$.

Assume the claim for the moment. If $A$ had another eigenvalue $\lambda \geq 0$, then the $2$-dimensional subspace $\mathrm{span}\{\lambda_0, \lambda\}$ would intersect the $(n-1)$-dimensional subspace $\{x : x \cdot x_0 = 0\}$ non-trivially, but at that point $y$ we would have both $y^\top A y \geq 0$ and $y^\top A y < 0$, a contradiction. So, the theorem follows from the claim. For readability, we break the argument into two pieces.

Subclaim: If $x \cdot x_0 = 0$ and $x \neq 0$, then $x^\top A x \leq 0$.

Proof of subclaim: We first introduce
$$B(t) := \left(e^{-t\sqrt{i^2+j^2}}\right)_{1 \leq i,j \leq n}.$$
Intuitively, $A$ is the linear coefficient of the Taylor expansion of $B$ around $t=0$. More precisely, working coefficient-wise,
$$\lim_{t \to 0} \frac{B_0 - B(t)}{t} = A$$
where $B_0 := B(0) = (1)_{1 \leq i, j \leq n}$ is the matrix of all $1$'s.

Since $x \cdot x_0 = 0$, we have $B_0 x = 0$. Thus
$$\tag{1}\label{1}x^\top A x = \lim_{t \to 0} x^\top \frac{B_0 - B(t)}{t} x
= -\lim_{t \to 0} \frac{x^\top B(t)\,x}{t}.$$

The key insight is to express the quadratic form $x^\top B(t) x$ in a manifestly positive way. For that, it is a fact1 that for all $z \geq 0$,
$$e^{-\sqrt{z}} = \frac{1}{2\sqrt{\pi}} \int_0^\infty e^{-zs} s^{-3/2} \exp\left(-\frac{1}{4s}\right)\,ds.$$

Thus, for all $t \geq 0$, we have the following entry-wise equality:
$$B(t) = \left(e^{-t\sqrt{i^2+j^2}}\right)_{1 \leq i,j \leq n} = \int_0^\infty \left(e^{-t^2(i^2+j^2)s}\right)_{1 \leq i,j \leq n} \, g(s)\,ds$$
where
$$g(s) := \frac{1}{2\sqrt{\pi}} s^{-3/2} \exp\left(-\frac{1}{4s}\right) > 0\quad\text{for all }s > 0.$$
The integrand matrices can be decomposed as an outer product, namely
$$\left(e^{-t^2(i^2+j^2)s}\right)_{1 \leq i,j \leq n} = \left(e^{-t^2 i^2 s} e^{-t^2 j^2 s}\right)_{1 \leq i,j \leq n} = u(s, t)u(s, t)^\top$$
where
$$u(s, t) := (e^{-t^2 i^2 s})_{1 \leq i \leq n}$$
is a column vector. Thus,
$$\begin{align*}
x^\top B(t)\,x
&= x^\top\left(\int_0^\infty u(s, t) u(s, t)^\top \, g(s)\,ds\right)x \\
&= \int_0^\infty (u(s, t)^\top x)^\top (u(s, t)^\top x)\,g(s)\,ds \\
&= \int_0^\infty (u(s, t) \cdot x)^2\,g(s)\,ds.
\end{align*}$$
Hence \eqref{1} becomes
$$\tag{2}\label{2}x^\top A\,x = -\lim_{t \to 0^+} \int_0^\infty \frac{(u(s, t) \cdot x)^2}{t} g(s)\,ds.$$

It is now clear that $x^\top A\,x \leq 0$ since $g(s) \geq 0$, finishing the subclaim.

Proof of claim: It will take a little more work to show that $x^\top A\,x < 0$ is strict. Let $t = 1/\alpha$ for $\alpha > 0$. Apply the substitution $s = \alpha^2 r$ to the integral in \eqref{2} to get
$$\begin{align*}
\int_0^\infty \frac{(u(s, t) \cdot x)^2}{t} g(s)\,ds
&\geq \int_{\alpha^2/2}^{\alpha^2} \frac{(u(s, t) \cdot x)^2}{t} g(s)\,ds \\
&= \int_{1/2}^1 (u(\alpha^2 r, 1/\alpha) \cdot x)^2 \alpha g(\alpha^2 r)\,\alpha^2\,dr \\
&= \int_{1/2}^1 \alpha^3 (u(r, 1) \cdot x)^2 \frac{1}{2\sqrt{\pi}} \alpha^{-3} r^{-3/2} \exp\left(-\frac{1}{4\alpha^2 r}\right)\,dr \\
&= \frac{1}{2\sqrt{\pi}} \int_{1/2}^1 (u(r, 1) \cdot x)^2 r^{-3/2} \exp\left(-\frac{1}{4\alpha^2 r}\right)\,dr.
\end{align*}$$

Thus \eqref{2} becomes
$$\begin{align*}
x^\top A\,x
&\leq -\lim_{\alpha \to \infty} \frac{1}{2\sqrt{\pi}} \int_{1/2}^1 (u(r, 1) \cdot x)^2 r^{-3/2} \exp\left(-\frac{1}{4\alpha^2 r}\right)\,dr \\
&= -\frac{1}{2\sqrt{\pi}} \int_{1/2}^1 (u(r, 1) \cdot x)^2 r^{-3/2}\,dr.
\end{align*}$$

Hence it suffices to show that $u(r, 1) \cdot x \neq 0$ for some $1/2 \leq r \leq 1$. Indeed, $\{u(r_j, 1)\}$ is a basis for any $r_1 < \cdots < r_n$. One way to see this is to note that the matrix is $\left(e^{-i^2 r_j}\right)_{1 \leq i, j \leq n} = \left(q_i^{r_j}\right)_{1 \leq i, j \leq n}$ where $q_i := e^{-i^2}$. Since $0 < q_n < \cdots < q_1$, this matrix is an (unsigned) exponential Vandermonde matrix in the terminology of Robbin--Salamon, and therefore has positive determinant. Hence $u(r, 1) \cdot x = 0$ for all $1/2 \leq r \leq 1$ implies $x=0$. contrary to our assumption. This completes the claim and proof. $\Box$

1As fedja points out, existence of an appropriate expression follows easily from Bernstein's theorem on completely monotone functions. As you'd expect, this explicit formula can be done by residue calculus.


问:
Unexpected approximations which have led to important mathematical discoveries
On a regular basis, one sees at MSE approximate numerology questions like

Prove $\log_{{1}/{4}} \frac{8}{7}> \log_{{1}/{5}} \frac{5}{4}$,

Prove $\left(\dfrac{2}{5}\right)^{{2}/{5}}<\ln{2}$,

Comparing $2013!$ and $1007^{2013}$

or yet the classical $\pi^e$ vs $e^{\pi}$. In general I don't like this kind of problems since
 a determined person with calculator can always find two numbers accidentally close to each other - and then ask others
 to compare them without calculator. An illustration I've quickly found myself (presumably it is as difficult as stupid): show that $\sin 2013$ is between $\displaystyle \frac{e}{4}$ and $\ln 2$.

However, sometimes there are deep reasons for "almost coincidence". One famous example is
 the explanation of the fact that $e^{\pi\sqrt{163}}$ is an almost integer number (with more than $10$-digit accuracy) using the theory of elliptic curves with complex multiplication.

The question I want to ask is: which unexpected good approximations have led to important mathematical
 developments in the past?

To give an idea of what I have in mind, let me mention Monstrous Moonshine where the observation
 that $196\,884\approx 196\,883$ has revealed deep connections between modular functions,
 sporadic finite simple groups and vertex operator algebras.

Many thanks in advance for sharing your insights.

答: Not sure if this is unexpected or so but I think the fast inverse square root is kinda cool. Don't think it lead to any mathematical developments though it's been implemented more widely since. </p>


问:
Sum of integer coefficients sum of which equals $2020$ of a quadratic polynomial that has 2 integer roots differing by $1$
We have a quadratic polynomial $ax^2+bx+c$ for $a,b,c \in \mathbb{Z}, a+b+c=2020$ with roots $k, l \in \mathbb{Z}, |k-l| = 1$. The task of the entire exercise is to find how many such polynomials exist. I am stuck at this particular step:

Solution written in the book this problem comes from says

The polynomial can be rewritten as $a(x-k)(x-l)$. Considering that the roots differ by $1$, the sum of our coefficients can be simply expressed as $a(k-1)(l-1)$, which is equal to $2020$.

But my question is, why can the sum of coeffecients be expressed as $a(k-1)(l-1)$ and how can that be proven? The authors haven't really given much information about this step.

Knowing this "formula", the rest of the exercise just comes down to prime factorization of  $2020$ is not difficult to finish.

Body:
$$ax^2+bx+c=a(x-k)(x-l)$$ $$a\cdot 1^2+b\cdot 1+c=a(1-k)(1-l) \tag{put x=1}$$ $$a+b+c=a(k-1)(l-1)$$

Hope it helps


问:
Integral / Fourier Series
Why is the following integral written with Fourier series development like this?

$\displaystyle h_i\sum_{k,l=1}^{\infty}a_k a_l \int_0^1 \sin(k\pi t)\sin(l\pi t) \,dt = \frac{h_i}{2}\sum_{k,l=1}^{\infty} a_k a_l \delta_{kl}$

What is used to move from left limb to right limb?

Thanks! :)

答:
It is simply because, when $k\neq l$, we have
$$
\int_0^1 \sin (\pi  k t) \sin (\pi  l t) \, dt = \frac{l \sin (\pi  k) \cos (\pi  l)-k \cos (\pi  k) \sin (\pi  l)}{\pi  k^2-\pi  l^2} = 0
$$
And when $k=l$, we have
$$
\int_0^1 \sin (\pi  k t) \sin (\pi  k t) \, dt = \frac{1}{2}-\frac{\sin (2 \pi  k)}{4 \pi  k} = \dfrac{1}{2}
$$
Hence
$$
\int_0^1{\sin \left( k\pi t \right)}\sin \left( l\pi t \right) \,dt = \dfrac{1}{2} \delta_{kl}
$$
Thus your relation holds.

Title: Integral / Fourier Series


问:
How do I represent recursion in my filtration set?

Let's take a super simple stochastic process consisting of one coin toss.

Normally I would have $\mathcal{F}=\{\{\},\{T\},\{H\},\{T,H\}\}$

Now let's say that if I get heads, I toss again.

How would my new $\mathcal{F}'$ set look like?

Update: the process is meant to continue indefinitely if I keep getting $H$, so imagine the filtration symbol $\mathcal{F}'$ would have to be re-used in the set itself

答:
The smallest $\sigma -$algebra that describe your experiment is $$\mathcal F'=\sigma \{\{(H,H)\},\{(H,T)\}\},$$
but the problem with this $\sigma -$agebra it's that it's not complete. The best would be to take $$\mathcal F'=\mathcal F\otimes \mathcal F=2^{\Omega\times \Omega }.$$


问:
Calculate zeros of rational function directly from residues of partial fraction expansion without calculating polynomial coefficients
Imagine the following rational function in the Laplace-domain with $s = \mathrm{j}\omega$

$$G(s) = \sum_{i=1}^{m} \dfrac{c_i}{s-p_i}= h\,\dfrac{\prod\limits_{i=1}^{n}(s-z_i)}{\prod\limits_{i=1}^{m}(s-p_i)} = \dfrac{A_0 + A_1 s + \ldots A_n s^n}{B_0 + B_1 s + \ldots B_m s^m}$$

for which I know the poles $p_i$ and the residues $c_i$.

I would like to calculate the zeros $z_i$. This is actually trivial: one can calculate the coefficients $A_1 \ldots A_n$ of the numerator polynomial in various ways (most simple would be the Matlab function residue) and calculate the roots of this polynomial. However, in case floating-point-precision is an issue, the polynomial coefficients are the bottleneck and it would be advantageous to avoid that intermediate step.

Question

Do you know any approach, algortihm or source, which explains a possibility to directly calculate the zeros $z_i$ from the residues $c_i$ and poles $p_i$ without calculating the numerator polynomial as intermediate step?

At the moment I use Matlab's vpa to increase the precision, but it appears to be a waste of computational effort, if I am not interested in the polynomial but just the zeros and poles. So any hint would be appreciated.

Thank you very much!

答:
$\color{brown}{\textbf{Alternative approach.}}$

From the given equation
$$G(s)=\sum\limits_{i=0}^m\dfrac{c_i}{s-p_i} = h\;\dfrac{\prod\limits_{j=1}^n(s-z_j)}
{\prod\limits_{i=1}^m(s-p_i)},\quad n\le m-1,\tag1$$

immediately should

$$\text{If}\quad h_1= \lim\limits_{\large s\to \infty}sG(s) = \sum\limits_{i=1}^mc_i\not=0,\quad\text{then}\quad \dbinom nh = \dbinom {m-1}{h_1},\tag{2.1}$$

$$\text{Else if}\quad h_2= \lim\limits_{\large s\to \infty}s^2G(s) \not=0,\quad\text{then}\quad \dbinom nh = \dbinom {m-2}{h_2},\dots\tag{2.2}$$
Also, easily to check the expression for the residue $\;c_k\;$ of $k-$th pole $p_k$ in the form of
$$c_k =\lim\limits_{\large s\to p_k}(s-p_k)G(s) = h\;\dfrac{\prod\limits_{j=1}^n(p_k-z_j)}
{\prod\limits_{i=1}^m(p_k-p_i)^{\large1-\delta_{\large ki}}},\quad (k=1\dots m),\tag3$$
or
$$\prod\limits_{j=1}^n(p_k-z_j) = b_k,\quad (k=1\dots m),\tag4$$
where
$$b_k = \dfrac{c_k}{h}\prod\limits_{i=1}^m(p_k-p_i)^{\large1-\delta_{\large ki}}\tag5$$
are the constants,
and expression with the Kronekker symbol $\delta$ in $(3),(5)$ means the production of non-zero differences.

Solving of the system  $(4)$ does not contain  calculating the numerator polynomial and looks as the required alternative approach.

If some of the roots are known, then the model order decreases.

$\color{brown}{\textbf{Simple example.}}$

Let us consider the case $\quad\mathbf{ m=3,\; p_1=-1,\; p_2=0,\; p_3=1,\; c_i =1.}\quad$

$$G(s)=\dfrac1{s+1} + \dfrac1s+\dfrac1{s-1}\quad \left(=\dfrac{3s^2-1}{s^3-s}\right),$$
Then $\;h_1=\lim\limits_{s\to\infty} sG(s) =3\;\Rightarrow\;
\mathbf{h=3,\;n=2},$
and from $(1)$ should
$$G(s) = 3\dfrac{(s-z_1)(s-z_2)}{(s+1)s(s-1)},$$
$$\lim\limits_{s \to-1} (s+1)\left(\dfrac{c_1}{s+1} + \dfrac{c_2}s+\dfrac{c_3}{s-1}\right)
= \lim\limits_{s \to-1} 3\;\dfrac{(s-z_1)(s-z_2)}{s(s-1)},$$
$$(-1-z_1)(-1-z_2)  = b_1,$$
$$\mathbf{b_1 = \dfrac{(-1)(-1-1)}3 = \dfrac23},$$
$$b_2 = \dfrac{(0+1)(0-1)}3 = -\dfrac13,\quad b_3 = \dfrac{(1+1)(1+0)}3 = \dfrac23,$$
$$\begin{cases}
(-1-z_1)(-1-z_2) = \dfrac23\\
(-z_1)(-z_2) = -\dfrac13\\
(1-z_1)(1-z_2) = \dfrac23,
\end{cases}\;\Rightarrow\;
\begin{cases}
z_1z_2+(z_1+z_2) + 1 = \dfrac23\\
z_1z_2 = -\dfrac13\\
z_1z_2 -(z_1+z_2)+1 = \dfrac23,
\end{cases}\tag6
$$
$$z_1=\dfrac1{\sqrt3},\quad z_2 = -\dfrac1{\sqrt3}.$$

Therefore, proposed approach works.

$\color{brown}{\textbf{Solving.}}$

Solution of the system $(6)$ can be obtained analytically.

Also, this system can be solved as the optimization task in the form of
$$\vec z = \text{ argmin } \{((-1-z_1)(-1-z_2) - \,^2/_3)^2
+ ((-z_1)(-z_2) + \,^1\!/_3)^2
+((1-z_1)(1-z_2) - \,^2/_3)^2 \}.$$

If $n\ge3,$ then attempts of the analytical solving of $(4)$ possibly lead to the  same numerator polynomial. So the numeric solutions (such as the gradient descend method) can be recommended. Multivariable model allows to avoid the roots mixing via their resorting after each iteration step.

The alternative approach is the searching of the roots in the intervals between the poles via the dihotomy algorithm (without derivatives using). In this way, algorithm should use sufficient set of the initial points to obtain exactly $n$ roots.

At last, can be used combined algorithm, which uses

simple dihotomy to obtain some real roots between the poles,

the simplified model $(4)$ for the rest of roots in the complex presentation.

Good luck!


问:
If $f(x)\rightarrow 0$ as $x\rightarrow 0^{+}$ and $g(x)\geq 1$ then $\dfrac{g(x)}{f(x)}\rightarrow\infty$ as $x\rightarrow 0^{+}$

Prove if $f(x)\rightarrow 0$ as $x\rightarrow 0^{+}$ and $g(x)\geq 1$ for all $x\in\mathbb{R}$,then $\dfrac{g(x)}{f(x)}\rightarrow +\infty$ as $x\rightarrow 0^{+}$

$\underline{Attempt}$

Since $f(x) \rightarrow 0$ as $x   \rightarrow 0^{+},$

by definition $\forall  \varepsilon> 0$, $\exists
 $ $\delta>0$ such that $0<x<\delta$ imply $|f(x)-0|=|f(x)|<\varepsilon$ ,

and also  we know $g(x) \geq 1$
for all $x \in \mathbb{R}$

so for $|f(x)| \neq0$

$$\frac{1}{\varepsilon}<\frac{1}{|f(x)|}$$

therefore we can get

$$\frac{1}{\varepsilon} <\bigg| \dfrac{g(x)}{f(x)} \bigg| $$ so in either case $$\frac{1}{\varepsilon} < \dfrac{g(x)}{f(x)}  $$

this implies  $\forall M=\frac{1}{\varepsilon}\in \mathbb{R}$ , $\exists$ $ \delta>0$ such that $\dfrac{g(x)}{f(x)}>M $ whenever $0<x<\delta$ ,so
$\dfrac{g(x)}{f(x)}\rightarrow \infty$ as $x   \rightarrow 0^{+}$

I don't know whether my attempt correct or wrong if it is not give some help and if this is correct give me how to prove $f(x)=0$ this is crazy problem  to me what happen if $f(x)=0$,

答:
There's no need in my opinion to bother going straight into $\epsilon-\delta$. Instead, consider that:

$$g_2(x)>g_1(x)\geq 1 \overbrace{\implies}^{x<1, x\neq 0} \frac{g_2(x)}{x}>\frac{g_1(x)}{x}$$

You should prove this assertion. But, as a direct consequence of this, we note that setting $g(x)\equiv 1$ and evaluating the limit will give us the lower bound for the limit, as this is the smallest allowed function under the condition $g(x) \geq 1$. So now all you need to prove is that $$\lim_{x\to 0^+}\frac{1}{f(x)}=\infty$$
and then every other function will have a greater (that is, also infinite) limit.


问:
Direct proof for the irrationality of $\sqrt 2$.

Prove that $\sqrt 2$ is irrational using direct proof.

I have seen TONS indirect proofs (e.g. proof by contradiction) for it, and people say that it's difficult to proof this directly. So is this impossible?

Thank you.

答:
What do you exactly mean by a "direct proof"?

The most direct argument I can think of for showing that $\sqrt{2}$ is irrational uses continued fractions. $\sqrt{2}$ has an infinite continued fraction (namely: $[1,2,2,2,...,]$) and can as such not be rational.


问:
Logical Question about Exercise 7 on p.111 in “Analysis on Manifolds” by James R. Munkres.
I am reading "Analysis on Manifolds" by James R. Munkres.

There is the following exercise on p.111 in this book:

Munkres says "$B\subset E$ because the limit cannot be defined if $\mathbf{x_0}$ is not a limit point of $S$".

Logically no problem about this sentence?

When we discuss if $\lim_{\mathbf{x}\to\mathbf{x_0}} f(\mathbf{x})=0$ holds or not, I think $x_0$ must be a limit point of $S$.

So, I think we cannot discuss if $\lim_{\mathbf{x}\to\mathbf{x_0}} f(\mathbf{x})=0$ holds or not when $x_0$ is an isolated point of $S$.

I think "For an isolated point $x_0$, $\lim_{\mathbf{x}\to\mathbf{x_0}} f(\mathbf{x})=0$ fails to hold" is a nonsense sentence because "$\lim_{\mathbf{x}\to\mathbf{x_0}} f(\mathbf{x})$" cannot be defined.

答:
Munkres adopts the following convention:

This means that writing $\lim_{x\to x_0} f(x) = y_0$ is an abbreviation for the following two conditions:

$x_0$ is a limit point of $A$.

$f(x) \to y_0$ as $x \to x_0$.

If one of these two conditions is not satisfied, then $\lim_{x\to x_0} f(x) = y_0$ fails to hold. It is perhaps unusual to interpret $\lim_{x\to x_0} f(x) = y_0$ in that sense, and it may cause confusion, but it is consistent in itself.

By the way, we could of course define "$f$ approaches $y_0$ as $x$ approaches $x_0$" as above without any restriction on $x_0$, but if $x_0$ is an isolated point, then $f$ approaches each $y_0$ as $x$ approaches $x_0$ - simply because $x_0$ has an open neigborhood containing no other point than $x_0$ itself. This does not make much sense, and this is why Munkres says that he does not attempt to give a definition in that case.


问:
Find the probability that two persons are grouped in the same team
A group of 9 people was further divided into 3 teams of three persons. How do you find the probability that 2 persons (let's name them person A and person B) are in the same group?

I think this is similar to the word permutation problem wherein two letters must be right next to each other, so you just treat them as one.

What I would do is treat person A and person B as one person and continue in grouping them, but there is a special case that only in the group of this new person can there be only 2 persons and not 3 (since that new person represents two).

Using the combination formula: $\frac{n!}{r!(n-r)!}$

\begin{align*}
&= \dfrac{9!}{3!(9-3)!}\\
&= \dfrac{9!}{3!\cdot6!}\\
&= 84
\end{align*}

This is clearly wrong since it is not taking into account the fact that I am treating two persons as one.

答:
Instead of going with your route of grouping $A$ and $B$ as only one person that has the problem that you've found, try to start with calculating all possibilities of teams and all possibilities with $A$ and $B$ joined.

Always with this kind of problems it's useful to start with something small that you can manually write and check. Say that we want to divide 4 people into two groups. We have a total of $\frac{4!}{2!2!2!}$ possibilities. This is because we can rearrange people from the same team (so we need to divide by $2!2!$, one $2!$ for each team) but we also don't care about the order of the teams, so we can rearrange them too, that's why there are three $2!$ in the denominator. A total of $3$ possibilities. If you force $A$ and $B$ to be together, there is only one possibility for them to be together, so the total probability is $1/3$.

Try to repeat this process but with $9$ people and $3$ teams.


答:
If $A$ gets into any of the teams, there are still two places in the same team.

The probability of $B$ getting into one of those two places out of remaining $8$ people $ = \displaystyle \frac{2}{8} = \frac{1}{4}$.

Or you can think of it as below -

$A$ and $B$ can be together in any of the $3$ teams and whichever team they are on, there is just one slot in the team where $1$ from rest of the $7$ people can get in.

So, desired probability $ \displaystyle = \frac{{3 \choose 1} \times {2 \choose 2 } \times {7 \choose 1} }{9 \choose 3} = \frac{1}{4}$

Title: Find the probability that two persons are grouped in the same team

Body:
A group of 9 people was further divided into 3 teams of three persons. How do you find the probability that 2 persons (let's name them person A and person B) are in the same group?

I think this is similar to the word permutation problem wherein two letters must be right next to each other, so you just treat them as one.

What I would do is treat person A and person B as one person and continue in grouping them, but there is a special case that only in the group of this new person can there be only 2 persons and not 3 (since that new person represents two).

Using the combination formula: $\frac{n!}{r!(n-r)!}$

\begin{align*}
&= \dfrac{9!}{3!(9-3)!}\\
&= \dfrac{9!}{3!\cdot6!}\\
&= 84
\end{align*}

This is clearly wrong since it is not taking into account the fact that I am treating two persons as one.


问:
Solve Matrix Linear Least Squares with Frobenius Norm Regularization and Linear Equality Constraints
How to solve the following constrained minimization problem:
$$
\arg_S min_\; \frac{1}{2}\left \{ \left \| K_2SK_1^T-M \right \|_F^2 +\lambda \left \| S \right \|_F^2\right \} \\
s.t. \sum_{1}^{col}S=Sum1 \\
\sum_{1}^{row}S=Sum2 \\
$$
where $K_1$,$K_2$,$M$ and $S$ are 2d Matrix, and only $S$ is unknown. In the constraints, $Sum1$ is the sum along the column of $S$, which is a row vector. $Sum2$ is the sum along the row of $S$, which is a column vector.

Here is the data stored in mat format. How to solve this kind of problem?

```<code>    load('matlab.mat');
%　 min norm( K2*X*K1'-M,'fro')^2+lambda*norm(X,'fro')^2
%   s.t. sum(X,1) = Sum1 ;   sum(X,2) = Sum2;
</code>```

答:
You're already using MATLAB, so it's easy to use CVX.</p>
```<code>cvx_begin
variable S(size(K2,2),size(K1,2))
minimize(0.5*(square_pos(norm(K2*S*K1'-M,'fro'))+lambda*square_pos(norm(S,'fro'))))
sum(S,1) == Sum1
sum(S,2) == Sum2
cvx_end
</code>```

Body:
You're already using MATLAB, so it's easy to use CVX.

cvx_begin
variable S(size(K2,2),size(K1,2))
minimize(0.5*(square_pos(norm(K2*S*K1'-M,'fro'))+lambda*square_pos(norm(S,'fro'))))
sum(S,1) == Sum1
sum(S,2) == Sum2
cvx_end


问:
if $F'(ax) = G(x)$, then $F'(nax) = nG(nx)$?
A question I found stated that:

Given $$\frac{d}{dx}F(ax) = G(x)$$

then $$\frac{d}{dx}F(2ax) = \dots$$

with choices:

```<code>          1.) G(2ax)
          2.) G(ax)
          3.) 2G(2x)
          4.) 2G(x)
</code>```

I got the third option as the answer by assuming that

$F(ax) = \sin(ax)$ and therefore $F'(ax)= G(x) = a\cos(ax)$

and hence:

$F(2ax) = \sin(2ax)$ and therefore $F'(2ax)= 2a\cos(2ax) = 2G(2x)$

I then tested with $F(ax) = (ax)^2$ and $F(ax) = \log(ax)$ with same result as above.

Is there a general rule to proof the statement: if $F'(ax) = G(x)$, then $F'(nax) = nG(nx)$?

答:
I repost my answer. In the last one I made a silly mistake.

Using the chain rule:$$\frac{d}{dx}F(2ax) =(2ax)' F'(2ax) = 2a F'(2ax)$$

Since we have for hypothesis that $$G(x) := \frac{d}{dx}F(ax) = aF'(ax) \quad \Rightarrow \quad G(2x) = \frac{d}{d(2x)}F(2ax) \stackrel{\star}{=} aF'(2ax)$$

If you do not understand the derivation $\star$ define $y := 2x$, then
$$\frac{d}{d(2x)}F(2ax) = \frac{d}{d(y)}F(ay) = aF'(ay) = aF'(2ax)$$

You can now deduce from the equations above that we have
$$\frac{d}{dx}F(2ax) = 2a F'(2ax) = 2 G(2x)$$

As @reuns pointed out in the comments you have to undarstand that there is a difference between $(F(ax))'$ and $F'(ax)$


问:
Characterization of $\mathbb{Q}$ in $\mathbb{Q}_p$
Let’s assume $\mathbb{Z}_p$ is the ring of p-adic integers. Then, how can we determine the elements of $\mathbb{Q}$ as an element of $\mathbb{Z}_p$, when it is defined? More specifically, I want to show that any p-adic integer with periodic p-adic expansion belongs to $\mathbb{Q}$, but it is not clear for me what I should show.

答:
Let $$G = \{ \sum_{n\ge -N} a_n p^n \in \Bbb{Q}_p, a_n\in 0\ldots p-1,\exists m,k, \forall n\ge m, a_n= a_{n+k}\}$$

Show it is a group using finiteness of states during carry calculations in $\sum_{n\ge -N} a_n p^n-\sum_{n\ge -N} b_n p^n$

For $s=p^{-N} u/v\in \Bbb{Q}, p\nmid v$ let $r$ be the order of $p\bmod v$ then  $$s=p^{-N} \frac{w}{1-p^r}= w p^{-N}\sum_{n\ge 0} p^{rn}\in  wG\subset G$$

Conversely if $\sum_{n\ge -N} a_n p^n\in G$ then $$\sum_{n\ge -N} a_n p^n=\sum_{n=-N}^{km-1} a_n p^n + \sum_{l=0}^{k-1} a_l \sum_{n=0}^\infty p^{l+kn+km}=\sum_{n=-N}^{km-1} a_n p^n + \sum_{l=0}^{k-1} a_l\frac{p^{l+km}}{1-p^k}\in \Bbb{Q}$$


问:
Evaluating Complex Integral using Residues
Evaluate the integral $\int_C \frac{\cot z}{z^2}$ where $C: |z|  = \frac{\pi}{3}.$
I am planning to use residue theorem. The integrand has poles at $z  = 0, n\pi, n \in \mathbb{Z}.$ Only $z  = 0$ is inside $C$. However, I'm confused. Is $z  = 0$ a pole with multiplicity three? Thank you in advance!

答:
Yes. Since$$\frac{\cot z}{z^2}=\frac{\cos z}{z^2\sin z},$$since $\cos0\ne0$, and since $0$ is a zero of order $1$ of $\sin$, then $0$ is a zero of order $3$ of $\frac{\cos z}{z^2\sin z}$.

You can compute it as follows. Let $s(z)=1-\frac{z^2}{3!}+\frac{z^4}{5!}-\cdots$ and note that, if $z\ne0$, then $s(z)=\frac{\sin z}z$. Then$$\frac{\cos z}{z^2\sin z}=\frac1{z^3}\cdot\frac{\cos z}{s(z)}.$$It is easy to see that$$\frac{\cos z}{s(z)}=1-\frac{z^2}3+o(z^3)$$and therefore$$\frac{\cos z}{z^2\sin z}=\frac1{z^3}-\frac1{3z}+\cdots$$So, $\operatorname{res}_{z=0}\left(\frac{\cos z}{z^2\sin z}\right)=-\frac13.$


Title: Evaluating Complex Integral using Residues

Body:
Evaluate the integral $\int_C \frac{\cot z}{z^2}$ where $C: |z|  = \frac{\pi}{3}.$
I am planning to use residue theorem. The integrand has poles at $z  = 0, n\pi, n \in \mathbb{Z}.$ Only $z  = 0$ is inside $C$. However, I'm confused. Is $z  = 0$ a pole with multiplicity three? Thank you in advance!

Body:
Hint

$$\frac{\text{cot}(z)}{z^2}=\frac{1}{z^2\tan(z)}=\frac{1}{z^3(1+\frac{z^2}{3}+o(z^2))}=\frac{1}{z^3}\left(1-\frac{z^2}{3}+o(z^2)\right).$$


问:
Minimum number of Hamiltonian Cycle(s)
Given a graph of size n (n vertices). State the minimum number of Hamiltonian Cycle(s) it can have
In a document I was looking for, it is

```<code>2^[(n−3)/2]
</code>```

Is it true? How to prove it?

答:
In The minimum number of Hamilton cycles in a hamiltonian threshold graph of a prescribed order, Theorem 6 stated:

Theorem 6. The minimum number of Hamilton cycles in a hamiltonian threshold
graph of order $n$ is $2^{\lfloor(n-3) / 2\rfloor}$ and this minimum number is attained uniquely by the graph $G_n$.

You can take a look at the proof there.

Title: Minimum number of Hamiltonian Cycle(s)


问:
Show $2^{55} + 1$ divisible by 33 using only basic identities
I'm working my way through The Mathematical Olympiad Handbook by A. Gardiner.

In the section A little useful mathematics right at the start the following basic identities are laid out

$x^n -1 = (x-1)(x^{n-1} + ... + 1)$

$x^n - y^n = (x-y)(x^{n-1} + ... + y^{n-1})$

$x^n + 1 = (x+1)(x^{n-1} + ... + 1)$

$x^n + y^n = (x-y)(x^{n-1} + ... + y^{n-1})$

In this context the following two exercises are posed

Prove $2^{55} + 1$ is divisible by $33$

Prove $1900^{1990} -1$ is divisible by $1991$

So for (1) the strong implication seems to be to somehow use the fact that $33=2^5+1=(2+1)(2^4-2^3+2^2-2^1+1)$ and use that, but other than realising that I am slightly lost.

I can see using Fermat's Little Theorem allows an attack ($2^{55}=(2^5)^{11}$ etc.) but how to do it with just those identities I can't see it as Fermat's Little Theorem is used later as a second proof of this problem in the number theory section.

Can anyone see how to do it?

Body:
$$(2^{5})^{11} + 1 = 32^{11} + 1 \overset * \equiv (-1)^{11} + 1 = 0 \pmod{33}$$
$(*)$: $32 = 33 - 1 \equiv -1 \pmod {33}$


Title: Show $2^{55} + 1$ divisible by 33 using only basic identities

Body:
I'm working my way through The Mathematical Olympiad Handbook by A. Gardiner.

In the section A little useful mathematics right at the start the following basic identities are laid out

$x^n -1 = (x-1)(x^{n-1} + ... + 1)$

$x^n - y^n = (x-y)(x^{n-1} + ... + y^{n-1})$

$x^n + 1 = (x+1)(x^{n-1} + ... + 1)$

$x^n + y^n = (x-y)(x^{n-1} + ... + y^{n-1})$

In this context the following two exercises are posed

Prove $2^{55} + 1$ is divisible by $33$

Prove $1900^{1990} -1$ is divisible by $1991$

So for (1) the strong implication seems to be to somehow use the fact that $33=2^5+1=(2+1)(2^4-2^3+2^2-2^1+1)$ and use that, but other than realising that I am slightly lost.

I can see using Fermat's Little Theorem allows an attack ($2^{55}=(2^5)^{11}$ etc.) but how to do it with just those identities I can't see it as Fermat's Little Theorem is used later as a second proof of this problem in the number theory section.

Can anyone see how to do it?

Body:
Hint:

$$2^{55} +1 =(2^5)^{11} +1 = 32^{11} +1 = (32+1)(\dots) $$ using your third identity.


问:
How is "direct search" used, in general, to solve a system of nonlinear equations?
I'm interested in using direct search (see) to solve systems of nonlinear equations. Unfortunately, however, there isn't much detailed information out there on the direct search family of algorithms. So, I'm interested... how, precisely, are direct search algorithms used to solve nonlinear system in practice? I know that it can and has been done. I also know other methods are available (Newton-Raphson, etc). I'm simply interested in this method, which does not require differentiation.

From MathWorks:

Direct search is a method for solving optimization problems that does
not require any information about the gradient of the objective
function. Unlike more traditional methods... the direct search algorithm uses a set of points around the current point, looking for one where the value of the objective function is lower than the value at the current point. You can use direct search to solve problems for which the objective function is not differentiable, or is not even continuous.

答:
If you're looking for the worst possible way of solving the problem, you have found it. But if you insist:

For each nonlinear equation, form the left-hand side minus right hand side. Then square each of these. Then sum these over all the equations. This will be the objective function for an unconstrained nonlinear least squares (optimization) problem. Then apply MATLAB's Direct Search algorithm to this optimization problem.

If the algorithm finds a point having objective value equal zero, that point is a solution of the original system of nonlinear equations. If it finds a point having objective function value almost zero, that point is "almost" a solution of the original system of nonlinear equations, and you will have to judge whether it is close enough to being a solution to suit your purposes.


问:
Probability of hitting the “bullseye”
I'm trying to solve this task but I'm obviously stuck. Will anyone help me solve it?

John is throwing darts at a target. He is very good at the game, so
the chances of him hitting the “bullseye” (i.e. the center) are $25$%.
It is reasonable to assume that individual attempts are independent.
If he throws $5$ times in total, what is the probability that he will
hit the bullseye:

a) Exactly three times

b) At least three times

答:
This is a binomial distribution since these are independent Bernoulli trials so a trial outcome is either success (hitting the bullseye) or failure and you want the probability of getting a certain number of successes in a certain number of trials regardless of where they happen.

In Bionmial distribution $P(X=k)$ is of the form:

$\binom{n}{k}p^k(1-p)^{n-k}$

Where:

$p$ is the probability of success

$n$ is the total number of trials

$k$ is the number of successes

The general formula for your question would be:

$\binom{5}{k}\cdot (0.25)^k \cdot (0.75)^{5-k}$
For your first question we get:
$\binom{5}{3}\cdot (0.25)^3 \cdot (0.75)^{2}$

For your second question they are asking for the probability that AT LEAST $3$ successes so that is the probability of getting three successes+ the probability of getting four successes+ probability of getting four successes, we get the following sum:

$\sum_{k=3}^5 \binom{5}{k}\cdot (0.25)^k \cdot (0.75)^{5-k}$

Title: Probability of hitting the “bullseye”


答:
1) wikipedia has given a constructive proof, see http://en.wikipedia.org/wiki/Square_root_of_2

2) all rational numbers have a finite continued fraction expression, but $\sqrt2$ doesn't

Title: Direct proof for the irrationality of $\sqrt 2$.

Body:
Prove that $\sqrt 2$ is irrational using direct proof.

I have seen TONS indirect proofs (e.g. proof by contradiction) for it, and people say that it's difficult to proof this directly. So is this impossible?

Thank you.


问:
Problem based on Rigid Body Mechanics (Newtonian-Mechanics)
I got this problem in a tutorial sheet given to me by my Professor.
I am confused as to how to apply the torque about point C, which is hinged. Please be kind to ignore any wrong format of asking the question.
Without any further ado, Here's the question:

The rigid triangular element shown is hinged at C and supported at pin B (BC is horizontal). Assuming all contacts to be frictionless, given F=84.2 N in the direction shown, calculate

(a) the magnitude of reaction force at pin B, and

(b) the magnitude of the net reaction force at hinge C.

答:
$F_X=842 cos(15)^o=813, N$

$F_y=842 sin (15^o)=218, N$

$T_x=813 \times 75=61000 N-mm=61 N-m$

$T_y=218\times (75+150)=49033 N-mm=49 N-m$

Total moment(torque) on C:

$T_C=61+49=110 N-m$

This torque is resisted by reaction force on B, so total reaction force on B is:

$F_B=\frac{110}{0.15}=734 , N $

This force is $y^-$ direction and can be considered as the vector sum of two components because of slot.

The external force to system is $842, N$ which can be considered as total force on C.


问:
Let $A_1,A_2\in L(V,V)$, $A_i(V)=V$ and $\dim\ker A_i=n_i\in\mathbf{N}$. Is $\dim\ker (A_2A_1)=n_1+n_2$ even if $\dim V=\infty$?
Let $V$ be a vector space that is not necessarly finite dimensional. Consider two surjective linear operators $A_1,A_2\in L(V,V)$ with finite-dimensional kernel: $A_i(V)=V$ and $\dim\ker A_i=n_i\in\mathbf{N}$. Can it be proven that $\dim\ker (A_2A_1)=n_1+n_2$?

In words: Is the dimension of the kernel of the composition of surjective linear operators with finite dimensional kernels the sum of the dimensions of the kernels?

Looking at this formula, one might think that the formula $\dim\ker A_2A_1=n_1+_2$ is valid, but I have looked at the answers and they are based on the rank nullity theorem, which I can't use, since $V$ is not finite dimensional.

Motivation: I want to prove that the solution space of homogeneous linear differential equations of order $n$ is $n$-dimensional by writing the $n$-th order differential operator as composition of first-order differential order with one-dimensional kernel. Actually, the answer to this question completes my proof.

答:
Here's s solution without too many computations (Martin Argerami's is basically the same). We only need $A_1$ surjective. (or even less, that $\ker(A_2)\subseteq A_1(V)$). We don't actually need finite dimension, and the argument below can be adapted easily.

$$\ker(A_2A_1)=(A_2A_1)^{-1}(0)=A_1^{-1}(A_2^{-1}(0))=A_1^{-1}(\ker(A_2))$$
Choose a basis $\left\{b_1,\ldots,b_n\right\}$ of $\ker(A_2)$. Since $A_1$ is surjective, there are $a_i$ such that $A_1(a_i)=b_i$. Let $W=\operatorname{span}\left\{a_1,\ldots,a_n\right\}$

Clearly, $A_1$ restricts to an isomormorphism from $W$ to $\ker(A_2)$

The usual argument: indeed, the images of the $a_i$ are linearly independet, so the $a_i$ are linearly independent as well, generate $W$, so they form a basis, and $A_1$ takes this basis of $W$ to the basis $\left\{b_1,\ldots,b_n\right\}$ of $\ker(A_2)$, therefore it is an isomorphism. Phew!

Then $A_1(W)=\ker(A_2)$, by construction, and so $A_1^{-1}(\ker(A_2))=W+\ker(A_1)$. Since $A_1$ is injective on $W$ then the spaces $W$ and $\ker(A_1)$ are independent, i.e., $W\cap\ker(A_1)=\left\{0\right\}$, so the sum above is direct:
$$\ker(A_2A_1)=A_1^{-1}(\ker(A_2))=W\oplus\ker(A_1)$$
and therefore $$\dim(\ker(A_2A_1))=\dim(W)+\dim(\ker(A_1))=\dim(\ker(A_2))+\dim(\ker(A_1)).$$

You can also show that $W\cap\ker(A_1)=\left\{0\right\}$ directly: Suppose $x=\sum_i\lambda_i a_i\in W\cap\ker(A_1)$. Then
$$0=A_1(x)=\sum_i\lambda_i A_1(a_i)=\sum_i\lambda_i b_i$$
Since the $b_i$ are linearly independent, all $\lambda_i=0$ and so $x=\sum_i 0a_i=0$.


问:
Find the equation of the hyperbola, if it passes through $(1;0)$ and has the asymptotes $x=0$ and $y=1$.
I know I have to rotate it 45 degrees I just don't know how. I don't even know how this rotation thing works. I would really appreciate any help and if you can help me understand the rotation thing I promise to pray for you and your family every night. Thanks in advance!

P.S.Please do not bully me if I'm asking stupid stuff.I am a sensitive person.

答:
A rotated, translated equilateral hyperbola has equation
$$y=\frac{ax+b}{cx+d}$$
horizontal asymptote is $y=\frac{a}{c}$, vertical is $x=-\frac{d}{c}$.

Here we have
$$
\begin{cases}
\frac{a}{c}=1&a=c\\
-\frac{d}{c}=0&d=0\\
\frac{a+b}{c+d}=0&a+b=0\to b=-a\\
\end{cases}
$$
So the equation is
$$y=\frac{ax-a}{ax}\to y=\frac{x-1}{x}$$


问:
Domain and Continuity of Rational Functions
Question

From the equation

$x^{21} - 1 = 0 $

Define 20 complex roots as $\omega_{1},\omega_{2}, \omega_{3},...,\omega_{20}$

What is the value of
$(1-\omega_{1})(1-\omega_{2})(1-\omega_{3})...(1-\omega_{20})$?

We can rewrite the equation as

$x^{21}- 1 = (x-1)(x-\omega_{1})(x-\omega_{2})...(x-\omega_{20})$

Divide both sides by $x-1$

$ \dfrac{x^{21}- 1}{x- 1} =(x-\omega_{1})(x-\omega_{2})...(x-\omega_{20})$

On the other hand, we can also do the long division:

$x^{21}-1 = (x-1)(x^{20}+x^{19}+x^{18}+...+x+1)$

So substitute this:

$\dfrac{(x-1)(x^{20}+x^{19}+x^{18}+...+x+1)}{x- 1} =(x-\omega_{1})(x-\omega_{2})...(x-\omega_{20})$

$x^{20}+x^{19}+x^{18}+...+x+1 =(x-\omega_{1})(x-\omega_{2})...(x-\omega_{20})$

Plug in 1, we have

$(1-\omega_{1})(1-\omega_{2})(1-\omega_{3})...(1-\omega_{20}) = 21$.

But I'm not too sure if this process is entirely justified. Because ultimately the original equation is:

$ \dfrac{x^{21}- 1}{x- 1} =(x-\omega_{1})(x-\omega_{2})...(x-\omega_{20})$

And the LHS is NOT defined when $x = 1$, or rather take one more step backward we have:

$x^{21}- 1 = (x-1)(x-\omega_{1})(x-\omega_{2})...(x-\omega_{20})$

And if we were to divide both sides by $(x-1)$ we can only do this when $ x \neq 1$, because otherwise the process is undefined.

How can I resolve this?

答:
Hint:

You may consider $$\lim_{x\to1}\frac{x^{21}-1}{x-1},$$
which exists.


问:
Find the value of $\theta$ in parallelogram $ABCD$.
Question:

If $ABCD$ is a parallelogram then find the angle $\theta$ in degrees.

What I tried:

I assumed $BD$ a straight line as I don't think this question can be solved if $BD$ is not a straight line.(I will be happy if someone solves this without this assumption)

Construction: Drew $EF$ perpendicular to ($CD$ and $AB$) and $GH$ to ($BC$ and $AD$).

$$\angle DAO=\angle DCO=y$$ Alternate interior angles:$$\angle BDA=\theta\\\angle ABD=40^\circ$$

By Angle Sum Property:
$$\angle DOG=90^\circ-\theta\\\angle GOA=90^\circ-y\\\angle AOF=70^\circ\\\angle FOB=50^\circ\\\angle BOH=90^\circ-\theta\\\angle HOC=70^\circ\\\angle COE=90^\circ-y\\\angle EOD=50^\circ$$
Adding all these and equating to $360^\circ$:
$$240^\circ+360^\circ-2\theta-2y=360^\circ$$
$$\theta+y=120^\circ$$

After this, I drew parallel lines to ($AD$ and $BC$) and ($AB$ and $CD$) through $O$ but didn't get the value of $\theta$.

How to solve this question?
Can this be solved?

Thanks!

答:
The following solution assumes $BD$ is a straight line.

Extend $AO$ such that it intersects $BC$ at $X$, and extend $CO$ such that it intersects $AB$  at $Y$. Then $\angle AOY+\angle OAB=\angle OYB=\angle OCB+\angle CBA\implies \angle AOY=\angle COX=100^{\circ}-\theta$. Then $\angle AXB=120^{\circ}-\theta$, and angle sum in $\triangle BOX$ gives $\angle BOX=\angle AOD=60^{\circ}$. So $\angle COB=160^{\circ}-\theta$.

Let the height $EF$ be as in OP's attempt. Suppose that $OE:OF=m:n$. By similar triangles, $OE:OF=DO:BO=CO:OY=CD:BY=AO:OX=AD:BX=m:n$, and  $CO:AO=OX:OY=CX:AY$ since $\triangle COX \sim \triangle AOY$. Now, $CO:OY=AO:OX\implies CO:AO=OY:OX$, but $CO:AO=OX:OY$, so $OY:OX=OX:OY\implies OX^2=OY^2\implies OX=OY$. Hence $AO=CO$ and $\triangle BOC\cong\triangle BOA$.

Thus $\theta=40^{\circ}$.


答:
For divisibility of $2^{55}+1$ by $33,$ we can write
$$2^{55}+1=(2^{5})^{11}+1=(33-1)^{11}+1,$$ and apply the third identity.

Isn't there a typo in the second exercise? I suggest to prove divisibility of $1990^{1990} -1$ by $1991$ instead.  We have

$$1990^2 -1=1991\times 1989$$ and use the first identity.


问:
Showing characteristic and minimum polynomial coincide for this particular matrix
Given a matrix such as  $$\pmatrix{0 & 0 & 2 \\ 1 & 0 & -1 \\0 & 1 & 1   \\ },$$ whose characteristic polynomial is $-X^3+X^2-X+2.$
 $$$$How it could be deduced that it equals minus the minimum polynomial ?$$$$Thanks in advance. </p>

答:
Think of this as the matrix of an endomorphism $u$ expressed in a suitable basis $(e_1,e_2,e_3)$: the three vectors $(e_1,u(e_1),u^2(e_1))$ form a basis (actually, $u(e_1)=e_2$ and $u^2(e_1)=e_3$). Now suppose $\mu$ is $u$'s minimal polynomial. What does the preceding remark tell us about the degree of $\mu$?

Well, suppose $Q$ is a non-zero polynomial of degree at most $2$, say $Q(X)=aX^2+bX+c$. Then $Q(u)\neq 0$, for if we apply the endomorphism $Q(u)$ to the vector $e_1$, the result is non zero:
$$Q(u)\Big(e_1\Big)=au^2(e_1)+bu(e_1)+ce_1\neq 0.$$
That's because the family $(e_1,u(e_1),u^2(e_1))$ is linearly independent, and, by virtue of $Q\neq 0$, at least one of the three coefficients $a,b,c$ is non zero. Since $\mu(u)=0$, the preceding implies $\mathrm{deg}(\mu)\geq 3$. Also, by definition $\mu$ has leading coefficient $1$, and by the Cayley-Hamilton theorem it divides $P$ which has degree $3$ and leading coefficient $-1$. All of this together implies that $\mu=-P$.

The same reasoning will show that if $x,u(x),\dots,u^{p-1}(x)$ are linearly independent where $x$ is some vector in a finite dimensional vector space $V$, and $u$ is an endomorphism of $V$, then $u$'s minimal polynomial has degree $\geq p$.


答:
The roots of $f(z):=z^{21}-1$ form the cyclic $\Bbb C^*$ subgroup $\langle\omega\rangle$ of size $21$ where $\omega=\cos(\frac{2\pi}{21})+i\sin(\frac{2\pi}{21})$. $$f(z)=z^{21}-1=\prod_{j=1}^{21}(z-\omega^j)$$ $$f'(z)=21z^{20}=\cdots$$ $$(\text{Product Rule})\;\;\cdots=\prod_{k\neq 21}(z-\omega^k)\cdot\frac{d}{dz}(z-1)+(z-1)\frac{d}{dz}\prod_{k\neq 21}(z-\omega^k)$$
$$\therefore\;\; f'(1)=\prod_{k\neq 21}(1-\omega^k)=21$$

Title: Domain and Continuity of Rational Functions

Body:
Question

From the equation

$x^{21} - 1 = 0 $

Define 20 complex roots as $\omega_{1},\omega_{2}, \omega_{3},...,\omega_{20}$

What is the value of
$(1-\omega_{1})(1-\omega_{2})(1-\omega_{3})...(1-\omega_{20})$?

We can rewrite the equation as

$x^{21}- 1 = (x-1)(x-\omega_{1})(x-\omega_{2})...(x-\omega_{20})$

Divide both sides by $x-1$

$ \dfrac{x^{21}- 1}{x- 1} =(x-\omega_{1})(x-\omega_{2})...(x-\omega_{20})$

On the other hand, we can also do the long division:

$x^{21}-1 = (x-1)(x^{20}+x^{19}+x^{18}+...+x+1)$

So substitute this:

$\dfrac{(x-1)(x^{20}+x^{19}+x^{18}+...+x+1)}{x- 1} =(x-\omega_{1})(x-\omega_{2})...(x-\omega_{20})$

$x^{20}+x^{19}+x^{18}+...+x+1 =(x-\omega_{1})(x-\omega_{2})...(x-\omega_{20})$

Plug in 1, we have

$(1-\omega_{1})(1-\omega_{2})(1-\omega_{3})...(1-\omega_{20}) = 21$.

But I'm not too sure if this process is entirely justified. Because ultimately the original equation is:

$ \dfrac{x^{21}- 1}{x- 1} =(x-\omega_{1})(x-\omega_{2})...(x-\omega_{20})$

And the LHS is NOT defined when $x = 1$, or rather take one more step backward we have:

$x^{21}- 1 = (x-1)(x-\omega_{1})(x-\omega_{2})...(x-\omega_{20})$

And if we were to divide both sides by $(x-1)$ we can only do this when $ x \neq 1$, because otherwise the process is undefined.

How can I resolve this?


问:
Are the two sets equal?
Is the following statement True for any sets A and B?
$$(A\setminus B)^\complement= A^\complement\setminus B^\complement$$

Attempt:

Let $x\in (A\setminus B)^\complement$. Then, $x\notin A\setminus B$. So, $x\notin A$ and $x\in B$.

Let $x\in A^\complement\setminus B^\complement$. Then, $x\in A^\complement$ and $x\notin B^\complement$. So, $x\notin A$ and $x \in B$.

So, this is a true statement. But the Venn diagram I drew says different. So maybe the attempt was wrong.

Note: $A^\complement$ is complement of $A$ and $A\setminus B$ is the set difference between $A$ and $B$.

答:
First of all , you should know that $A \setminus B$ = $A \cap B^c$ and $ (A \cap B)^c$=$A^c \cup B^c$ i.e you can think it like de -morgans law.

so,$(A \setminus B)^c =A \cup B^c$ but ,$A^c \setminus B^c=A^c \cap B$

so, It is false statement

Title: Are the two sets equal?