问:
Fitting circle in $f(x)=x^3-3x^2$ what is the radius?

The whole problem is the above.

I tried to solve it analytically but I couldnot find enough equation to solve it.

Assume the center is $(a,-r)$ where $r$ is the radius of the circle.

The points on the sphere $(x-a)^2+(y+r)^2=r^2$

and $(x,f(x))$ must satisfy it so if we denote the intersection points $(x_1,y_1),(x_2,y_2)$:

$$(x_1-a)^2+(x_1^3-3x_1^2+r)^2=r^2$$
$$(x_2-a)^2+(x_2^3-3x_2^2+r)^2=r^2$$

Are two equation with 4 unknown.

Can you give me elegant relations between these two unknowns?

答:
Here is a different path that can be considered maybe as "elegant" by building a kind of (generalized) angle bissector.

Take a look at the following figure: the red curve is the locus of the center of circles that are tangent to the (blue) curve and the (green) $x$ axis. The final answer to your question is that the center of the circle has coordinates

$$a=1.80155263360, \ \ -r=-1.06021721502$$

The abscissas of the tangency points with the curve being:

$$x_1=0.79999363787 \ \ \text{and} \ \ \ x_2=2.85193070410.
$$

How have these results been found ?

Let

$$p:=\sqrt{1+f'(m)^2}\tag{1}$$

This swallow-tailed curve has parametric equations:

$$\begin{cases} \ \ a_m&=&m-\dfrac{f'(m)r}{p}&=&\ \ \ \color{red}{m}&+&(p-1)\dfrac{f(m)}{f'(m)}\\
-r_m&=&\dfrac{f(m)p(p-1)}{f'(m)^2}&=&\color{red}{f(m)}&-&(p-1)\dfrac{f(m)}{f'(m)^2}\end{cases}\tag{2}$$

(where $m$ has the following meaning: $(m,f(m))$ is the point of tangency of a circle tangent (at least once) to the blue curve with center $(a_m,-r_m)$ ).

All boils down to the obtention of a double point in a parametric curve. I have used numerical methods for this purpose (an analytical solution looks out of reach, due in particular to the square root in (1)).

Explanation for (2): Let us express that the center of the current circle is at the same distance $r$ from the curve and from the $x$ axis by finding a unit vector on the normal to the blue curve at point $\binom{m}{f(m)}$, a property that can be expressed in the following way:

$$\binom{m}{f(m)}+r \underbrace{\dfrac{1}{p}\binom{-f'(m)}{1}}_{\text{unit normal vector}}=\binom{a}{-r},  \ \ \ p \ \text{has been defined in (1)}$$

giving two equalities from which one can extract $r$ then $a$ (see (2)).


问:
Best book for preparation to read The Art of the Computer Programming

I decided to read The Art of the Computer Programming by Donald Knuth. The problem is, that i'm in high school now, so i need a help with mathematics oriented book which help me to be prepared for it.

Thank you!

答:
The book is challenging, no doubt about it. The key is repetition. The book isn't a one time read. You have to keep going back to it after referencing several other books like "Concrete Mathematics" or the MIT OCW course on "Mathematics for Computer Science". It is a multi-faceted book and you might miss a few things the first time but you will start figuring stuff out by reading and talking to people as you go. So get started without a doubt.

Helpful Tip:

The book has a lot of proofs, and the foundation of any mathematical endeavour and in turn one that of computer science is to prove that a specific algorithm works. You can watch a few videos about proofs in the link I've pasted below: [link]

ContentLicense: CC BY-SA 3.0

ContentLicense: CC BY-SA 4.0


问:
If a smooth function defined on $\mathbb{R}^n$ is convex and bounded below, can we assert that its minima must be a connected set?
If a function defined on $\mathbb{R}^n$ is

infinitely differentiable,

convex, i.e., $f\left( t x_1 + (1-t) x_2 \right) \le t f(x_1) + (1-t) f(x_2)$, and

bounded below,

can we assert that its minima, if exist, form a connected set?

Background

I encounter this question when dealing with the maximum likelihood estimation of probabilistic graphical models.
The negative log-likelihood is bounded below by zero (since probabilities are bounded above), and its Hessian is positive semi-definite (but not positive definite).
By intuition, I guess its minima form a connected set, thus optimization of the loss function will give a good estimation of model parameters in principle, if the true model is not pathological.

Here is a specific example.

All variables are real.
The probability of an observed data point $s$ is given by
$$
P(s | h, J)
= \frac{1}{Z(h,J)} \exp\left( h^T s + \frac{1}{2} s^T J s \right)
= \frac{1}{Z(h,J)} \exp\left( \sum_{i=1}^{N} h_i s_i + \sum_{i=1}^N \sum_{j>i}^N J_{ij} s_i s_j \right).
$$
Here $s = (s_1, \cdots, s_N)^T$ and $s_i = \pm 1$,
$N$ is a given integer,
$h=(h_1, \cdots, h_N)^T$ is real,
and $J$ is a $N \times N$ real symmetric matrix whose diagonal element are zero, i.e.,
$$
J=
\begin{pmatrix}
  0      & J_{12} & \cdots & J_{1N} \\
  J_{12} & \ddots &        & J_{2N} \\
  \vdots &        & \ddots & \vdots \\
  J_{1N} & J_{2N} & \cdots & 0      \\
\end{pmatrix}.
$$
The normalization factor (partition function) is therefore
$$
Z(h,J) = \sum_{s_1=\pm 1} \cdots \sum_{s_N=\pm 1} \exp\left( h^T s + \frac{1}{2} s^T J s \right) .
$$
(FYI, this is just the generalized Ising model in physics or the Markov random field in statistics.)

The negative log-likelihood (loss function) for the above probability is
$$
L(h,J)
=
- h^T s - \frac{1}{2} s^T J s + \log Z(h,J) ,
$$
which is infinitely differentiable, convex with respect to $h$ and $J$, bounded below by zero.

答:
Firstly, $f$ may not have any minimum. For example, if $f(x) = e^x$, $f$ is smooth, bounded from below but does not have any minimum.

Moreover, suppose $f$ has a global minimum. By translation, suppose $f \geqslant 0$ and $\min f = 0$. Suppose there exists two point $x_1\neq x_2$ such that $f(x_1)=f(x_2) = 0$. Thus by convexity, for all $t \in [0,1]$, $f((1-t)x_1+tx_2) \leqslant (1-t)f(x_1)+tf(x_2) = 0$. As $f\geqslant 0$, $f((1-t)x_1+tx_2)=0$. Consequently, the entire interval $[x_1,x_2]$ minimize $f$, and thus, the set of $x$ such that $f(x) = 0$ is a convex subset of $\mathbb{R}^n$.

Notice that the smoothness of $f$ is never used here.


问:
Monic Generator and Ideals of Polynomials

Is the set of all polynomials $P \in \mathbb{Q}[x]$ s.t. $P(1)=P(3)=0$ an ideal? If so find the monic generator.

I do not think it is because if we have $P(x)=x^2$, then $x*x^2=x^3$ which means $P(3)$ is not $0$, am i right?

I also do not know what monic generator is and have a hard time finding a good explanation so I was wondering if one could offer a simple explanation on what monic generator is and give an example on it, e.g. all polynomials of degree $\ge n$ (since i think that is ideal.)

答: Hint: Consider the map $\mathbb Q[x] \to \mathbb Q \times \mathbb Q$ given by $P \mapsto (P(1),P(3))$. Prove that this map is a ring homomorphism whose kernel is the given set.


答:
The set of all minimizers of a convex function is a convex set. (See proof below.) And of course, every convex set is connected.

Proof:
Suppose $x$ and $y$ are minimizers of a convex function $f$. So $f(x) = f(y) = f^*$, where $f^*$ is the minimum value of $f$.If $0 \leq \theta \leq 1$, and $z = \theta x + (1-\theta) y$, then
$$
f(z) \leq \theta f(x) + (1-\theta) f(y) = f^*.
$$
So $z$ is also a minimizer for $f$.

Title: If a smooth function defined on $\mathbb{R}^n$ is convex and bounded below, can we assert that its minima must be a connected set?

Body:
If a function defined on $\mathbb{R}^n$ is

infinitely differentiable,

convex, i.e., $f\left( t x_1 + (1-t) x_2 \right) \le t f(x_1) + (1-t) f(x_2)$, and

bounded below,

can we assert that its minima, if exist, form a connected set?

Background

I encounter this question when dealing with the maximum likelihood estimation of probabilistic graphical models.
The negative log-likelihood is bounded below by zero (since probabilities are bounded above), and its Hessian is positive semi-definite (but not positive definite).
By intuition, I guess its minima form a connected set, thus optimization of the loss function will give a good estimation of model parameters in principle, if the true model is not pathological.

Here is a specific example.

All variables are real.
The probability of an observed data point $s$ is given by
$$
P(s | h, J)
= \frac{1}{Z(h,J)} \exp\left( h^T s + \frac{1}{2} s^T J s \right)
= \frac{1}{Z(h,J)} \exp\left( \sum_{i=1}^{N} h_i s_i + \sum_{i=1}^N \sum_{j>i}^N J_{ij} s_i s_j \right).
$$
Here $s = (s_1, \cdots, s_N)^T$ and $s_i = \pm 1$,
$N$ is a given integer,
$h=(h_1, \cdots, h_N)^T$ is real,
and $J$ is a $N \times N$ real symmetric matrix whose diagonal element are zero, i.e.,
$$
J=
\begin{pmatrix}
  0      & J_{12} & \cdots & J_{1N} \\
  J_{12} & \ddots &        & J_{2N} \\
  \vdots &        & \ddots & \vdots \\
  J_{1N} & J_{2N} & \cdots & 0      \\
\end{pmatrix}.
$$
The normalization factor (partition function) is therefore
$$
Z(h,J) = \sum_{s_1=\pm 1} \cdots \sum_{s_N=\pm 1} \exp\left( h^T s + \frac{1}{2} s^T J s \right) .
$$
(FYI, this is just the generalized Ising model in physics or the Markov random field in statistics.)

The negative log-likelihood (loss function) for the above probability is
$$
L(h,J)
=
- h^T s - \frac{1}{2} s^T J s + \log Z(h,J) ,
$$
which is infinitely differentiable, convex with respect to $h$ and $J$, bounded below by zero.


问:
Trying to define an abstract notion of a function that turns sets of sentences into the set of models satisfying those sentences.
Let $\mathsf{Sen}$ denote a Boolean algebra, thought of as a collection of sentences, and let $\mathsf{Mod}$ denote a set without any additional structure, thought of as a collection of models. I'm trying to define (abstractly) the notion that a function $f : \mathcal{P}(\mathsf{Sen}) \rightarrow \mathcal{P}(\mathsf{Mod})$ turns a set of sentences $A \in \mathcal{P}(\mathsf{Sen})$ into the set of all models satisfying those sentences $f(A) \in \mathcal{P}(\mathsf{Mod})$.

Now its clear that $f$ ought to turn arbitrary unions into intersections. That is, for all $\mathcal{A} \subseteq \mathcal{P}(\mathsf{Sen})$ we require that $$f(\bigcup_{A \in \mathcal{A}} A) = \bigcap_{A \in \mathcal{A}}f(A).$$

So my first question is: going on our intuition involving first/second-order logic and first/second-order models, ought we require that $f$ turn arbitrary intersections into unions? And should we require that $f$ interact nicely with complements?

Now furthermore, $f$ ought to be compatible with the Boolean Algebra structure on $\mathsf{Sen}$. For instance, if $\phi,\psi \in \mathsf{Sen},$ then we want $$f(\{\phi \wedge \psi\}) = f(\{\phi\}) \cap f(\{\psi\}),\quad f(\{\phi \vee \psi\}) = f(\{\phi\}) \cup f(\{\psi\}),\quad f(\{\phi^c\}) = \mathsf{Mod} \setminus f(\{\phi\})$$

My second question is: Are the requirements of the above paragraph sufficient, given that they're only written in terms of singleton sets, when after all $f$ can potentially operate on infinite collections of sentences? Again, let us go with the intuition that comes from thinking about first/second-order logic and first/second-order models.

答:
$f$ is a Galois connection between the poset of sets of sentences and the poset of sets of models. Here is a general way to induce Galois connections between such posets: let $X, Y$ be sets and let $R \subseteq X \times Y$ be a relation, which we will write $xRy$. Then we get a Galois connection between the poset of subsets of $X$ and the poset of subsets of $Y$ given by the functions

$$f(S) = \{ y : xRy \forall x \in S \}$$

where $S \subseteq X$ and $f(S) \subseteq Y$, and

$$g(T) = \{ x : xRy \forall y \in T \}$$

where $T \subseteq Y$ and $g(T) \subseteq X$. We recover several classical examples of Galois connections this way.

The motivating example behind the name: let $L/K$ be a Galois extension, let $X = K$, and let $Y = \text{Gal}(L/K)$. The relation is $xRy$ iff the group element $y$ fixes the element $x$. This induces the Galois connection between subextensions of $L/K$ and subgroups of $Y$ which is the subject of the fundamental theorem of Galois theory

Let $X = k[x_1, ... x_n]$ ($k$ an algebraically closed field) and let $Y = k^n$. The relation is $xRy$ iff the function $x$ vanishes on the point $y$. This induces the Galois connection between ideals of $k[x_1, ... x_n]$ and subvarieties of $k^n$ which is the subject of the Nullstellensatz.

In this case, let $X$ be the set of sentences and let $Y$ be the set of models. The relation is $xRy$ iff the sentence $x$ is true in the model $y$. This induces the Galois connection between sets of sentences and sets of models which is the subject of the completeness theorem (in first-order logic).

I think it was Lawvere who first noticed this. The slogan is "syntax is adjoint to semantics." Connecting the last two examples together, you can think of sentences as functions on the space of models and taking the set of models satisfying a collection of sentences as taking the subspace on which a collection of functions vanishes.


问:
Find a linear map with conditions for kernel and image
I need to find a linear map T: M_2(R) -> R[t]_3 such that the kernel is the subspace of symmetric 2 x 2 matrices (a, b, b, c), a, b, c real numbers and the Image need to contain a polynomial of grade 3.

I tried to write (a, b,b,c ) = a + bt + btˆ2 + btˆ3 and write the kernel basis in function of a, but I think that wrong. I'm really lost in this exercise. How should I start?

答:
Hint: Try the linear transformation of the form
$$
T \pmatrix{a & b\\c & d} = (b - c) t^3.
$$
Verify that this transformation has a kernel consisting of the symmetric matrices and that its image contains a polynomial of degree $3$.


问:
Finding points of Continuity of functions
There are three functions where I have to show the points of continuity of. While I think it makes a lot of sense, I'm struggling with the notation. I apologize for the lengthy post.

$(1)$ $f: \mathbb{R} \rightarrow \mathbb{R}$ with $f(x) = 1 \space for\space \lvert x\rvert \lt \sqrt{3}$ and $f(x) = -1 \space for\space \lvert x\rvert \geq \sqrt{3}$

$(2)$ $f: [0,\infty) \rightarrow \mathbb{R}$ with $f(x) = x^2\lfloor\frac{1}{x}\rfloor \space$ for $x \ne 0$ and $f(0) = 0$

$(3)$ $f:\mathbb{Q} \rightarrow \mathbb{R}$ with $f(x) = 1 \space for\space \lvert x\rvert \lt \sqrt{3}$ and $f(x) = -1 \space for\space \lvert x\rvert \geq \sqrt{3}$

$(1)$ I've tried solving the following: The possible points of discontinuity could be $\sqrt{3}$ and $-\sqrt{3}$. I'm really not sure how to put this formally. If we evaluate for $x = \sqrt{3}$:$\space$for values less than $\sqrt{3}$ we get 1. For values $\geq$$\sqrt{3}$ we get $-1$.

Comparing the left and right limits we get: $\lim_{x\to \sqrt{3},\space x \lt \sqrt{3}}=1 \ne -1 =\lim_{x\to -\sqrt{3},\space x \geq \sqrt{3}}$
So there is a point of discontinuity there. The same thing can be done for $-\sqrt{3}$. Therefore, since I'm asked for the points of continuity, we get $\mathbb{R}\setminus\{\sqrt{3},-\sqrt{3}\}$

For $(2)$ I'm not really sure. I've tried plotting it and realised, that after $x=1, \space \lfloor\frac{1}{x}\rfloor$ will always be $0$, therefore $f(x) = 0$ for $x>1$. However, this obviously is not a proof and for values between $0$ and $1$, this function seems to have many points of discontinuity.
How can I try to show this or even calculate all the points of discontinuity?

For $(3)$ I have to admit I can't really make much sense of. Since the inputs are rationals the outcome will be different from $(1)$. Is there an easy way to calculate or prove this? Does $\sqrt{3}$ being irrational have something to do with it?

Thanks in advance for your patience and apologies for my lack of knowledge and the mistakes I probably made while calculating $(1)$

答:
1)) You already showed that at points $\pm \sqrt{3}$ the function $f$ is discontinuous. It remains to show
that $f$ is continuous at $x\ne \pm \sqrt{3}$. This holds because for each such $x$ there exists $\varepsilon>0$ such that
$f$ is constant on an interval $(x-\varepsilon, x+\varepsilon)$. Remark that since $\sqrt{3}$ is irrational, such $\varepsilon$
exists for each rational $x$, so the function $f$ in (3) is continuous too.

2)) If $x\ne 0$ and $1/x\not\in\Bbb N$ then there exists $\varepsilon>0$ such that
$\lfloor 1/y\rfloor=\lfloor 1/x\rfloor=\operatorname{const}$ when $y\in (x-\varepsilon, x+\varepsilon)$. Then
$f(y)= y^2\lfloor 1/x\rfloor $ for such $y$, so $f$ is continuous at $x$.

If $1/x\in\Bbb N$ then it is easy to check that $\lim_{y\to x+} f(y)=x^2\lfloor 1/x\rfloor=f(x)$,
but $\lim_{y\to x-} f(y)= x^2(\lfloor 1/x\rfloor+1) \ne f(x)$, so $f$ is discontinuos at $x$.

Finally, if $x=0$ then $$0=\lim_{y\to x+} f(y)\le \lim_{y\to x+} y^2\frac 1{y}=\lim_{y\to x+} y=0,$$
so $f$ is continuous at $x$.


问:
If $X,Y$ are two independent random variables how can you calculate $\mathbb{E}[e^{XY}]$
Let $X,Y$ be two independent random variables where $X$ is standard normal distributed and $Y$ is distributed as $P(Y=1)=P(Y=-1)=\frac{1}{2}$  how can one calculate $\mathbb{E}[e^{XY}]$? I don't see how to use the joinst density of $X,Y$ to find the expectation (since one is continious and the other is discrete). I also don't see how I can factor the density since we dont have $\mathbb{E}[f(X)g(Y)]=\mathbb{E}[f(X)]\mathbb{E}[g(Y)]$ but rather $\mathbb{E}[h(X,Y)]$.

答:
$Z=XY$ is still a standard normal, thus $\mathbb{E}[e^Z]$ is a lognormal...

Thus

$$\mathbb{E}[e^{XY}]=\sqrt{e}$$

EDIT: to derive that $Z=XY\sim \Phi$ just observe that if $Z=-X$ you get

$$f_Z(z)=f_X(-z)=\frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}$$


问:
Point $M$ lies inside $\triangle ABC$, $\angle MAC = 10^\circ$ and $\angle MCA = 30^\circ$. Find $(180^\circ - \angle BMC)$

In $\Delta ABC, \angle CAB = 30^\circ$ and $\angle ABC = 80^\circ$. Point $M$ lies inside the triangle such that $\angle MAC = 10^\circ$ and $\angle MCA = 30^\circ$. Find $(180^\circ - \angle BMC)$.

What I Tried: Here is a picture to keep track of the angle-chasing .

The red angle is $30^\circ$ , the green one is $10^\circ$ , the purple one is $20^\circ$ . Now I extend $MC$ to $AB$ such that it meets at $K$. First I noticed that $AK = KC$ but that didn't seem to help. Also the $2$ yellow ones are equal to $40^\circ$ each, the brown one is equal to $80^\circ$ . The light green angle is $60^\circ$ . We have to find the blue angle.

After all these information, it seems like I am still missing something, because from here you actually cannot deduce the value of the blue angle. Can anyone help?

Thank You.

答:
Let $\angle MBA = \theta$, $\angle MBC = 80 - \theta$

Using trig form of Ceva's theorem,

$$ \sin MAC \sin MCB \sin MBA = \sin MCA \sin MBC \sin MAB$$

$$ \sin 10 \sin 40 \sin \theta = \sin 30 \sin (80-\theta) \sin 20 $$

$$ 4\sin 10 \cos 20 =\dfrac{\sin (80-\theta)}{\sin \theta} $$

$$ 2(\sin 30 - \sin 10) =\sin 80 \cot \theta - \cos 80 $$

$$ 1 - \sin 10 =\sin 80 \cot \theta $$

$$ \cot \theta = \dfrac{1 - \sin 10}{\cos 10} $$

$$ \cot \theta = \dfrac{(\cos 5 - \sin 5)^2}{\cos^2 5 - \sin^2 5} $$

$$ \cot \theta = \dfrac{\cos 5 - \sin 5}{\cos 5 + \sin 5} $$

$$ \cot \theta = \dfrac{1 - \tan 5}{1 + \tan 5} = \tan 40$$

$$\therefore \angle MBA = 50$$

$$\Rightarrow \boxed{180 - \angle BMC = 70}$$


问:
Equivalence class and how to find it
I need to find equivalence classes of relation $\sim$
\begin{array}{l}
x\sim y \Leftrightarrow 4 \mid\left(x^{3}-y^{3}\right) \\
\text { for } x, y \in \mathbb{Z}
\end{array}

I have shown that $\sim$ is equivalence  relation. I blind guessed that all even numbers will be one equivalence  class, so my question is is there some  systematic  approach to this problem?
Any help is appreciated :)

答:
We have $x\sim y \iff $ $x^{3}$ and $y^{3}$ leave the same remainder when divided by $4$.

Write $x=4q+r$ with $r\in\{0,1,2,3\}$. Then $x \sim r$ (check!) and so it suffices to consider the equivalence class of $r$. In particular, there are at most four equivalence classes. In fact, it turns out that there are three equivalence classes: $[0]$, $[1]$, $[3]$, because $[2]=[0]$.

In other words, all even numbers are in the same class, but the odd numbers fall into two classes.

Title: Equivalence class and how to find it


问:
An example of a non-rectifiable curve
How can I solve this problem?

Let the curve in $\mathbb{R}^2$ of the equation $\alpha(t)=(t,g(t)), t\in [0,1]$, where $$g(t)=\left\{ \begin{aligned} t \cos\left( \frac{\pi}{2t}\right), \quad t\not=0 \\ 0, \quad t=0\end{aligned}\right.$$
For $n \in \mathbb{N}$, let the partition $P$ of $[0,1]$ given by $$P=\left\{ 0, \frac{1}{2n},\frac{1}{2n-1},\ldots, \frac{1}{3},\frac{1}{2},1\right\},$$ prove the length $\ell_n$ of the polygonal inscribed satisfies $$\ell_n>\sum_{k=1}^{2n}\frac{1}{k}$$and and deduce that the curve $\alpha$ is not rectifiable.

My approach:  I know for example that if I prove that arc length of $\alpha$ is infinite, so I can say that $\alpha$ is not rectifiable, so I need to calculate $I_{n}$.

I know that the arc-length of a curve $\alpha$ is $$s(t)=\int_{t_0}^t \|\dot{\alpha}(t)\| \, dt$$
so I need to prove that $s(t)\to \infty \implies \alpha$ is not rectifiable.

I know how calculate $\dot{\alpha}(t)$, but how can I choose $t_0$ and $t$?

Questions:
Now, How can I prove the first part? It's to say, how can I prove that $\ell_n>\sum_{k=1}^{2n}\frac{1}{k}$? and how can I relate that result to the fact that alpha is not rectifiable?

答:
Here is the idea. In the euclidean plane, the shortest path between two point is the straight line. Let $P = \{ 0,\frac{1}{2n}, \frac{1}{2n-1},\ldots, \frac{1}{2},1\} = \{ x_0,x_1,\ldots x_{2n}\}$. Consider the points $M_k$ on the graph of $g$ defined by $M_k = (x_k,g(x_k))$. Because the straight line is the shortest path in the plane, one has $\mathrm{length}(g|_{[x_k,x_{k+1}]}) \geqslant \mathrm{length}([M_k,M_{k+1}])$. As $P$ is a partition of $[0,1]$, $\mathrm{length}(g|_{[0,1]})=\sum_{k=0}^{2n-1}\mathrm{length}(g|_{[M_k,M_{k+1}]})$.

So now you have the inequality:
\begin{align}
\forall n \geqslant 1,~ \mathrm{length}(g) \geqslant A_n
\end{align}
where $A_n = \sum_{k=0}^{2n-1} \mathrm{length}\left([M_k,M_{k+1}]\right)$. Try to bound from below these last terms and you are done.

Title: An example of a non-rectifiable curve


问:
Analytic function on D restriction to U is one-to-one,then it is one-to-one on D?

problem: let $f(z)$ be analytic function on $D=\{|z|＜1\}$. there is an annulus $U=\{r＜|z|＜1\}$ such that the restriction of $f(z)$ to $U$ is one-to-one. Show that $f(z)$ is one-to-one on $D$.

I don't know how to use the condition ''the restriction of $f(z)$ to $U$ is one-to-one''.

答:
Fix some $r <\rho <1$ and let $C$ the circle $|z|=\rho$ and $J=f(C)$ an analytic Jordan curve; if $w \notin J$ then (by the argument principle) the number of zeroes of $f(z)-w, |z| < \rho$ is given by the winding number $\chi(J,w)=\frac{1}{2 \pi i}\oint_J\frac {d\xi}{\xi-w}=\frac{1}{2 \pi i}\oint_C\frac {f'(\zeta)d \zeta}{f(\zeta)-w}$ and said number is $0$ if $w$ is exterior to $J$ and $\pm 1$ if $w$ is interior.

In particular we get that $f$ takes all values inside $J$ precisely once when $|z| < \rho$ and no values outside $J$; now $f$ is an open map so the previous fact implies that $f$ cannot take any value on $J$ when $|z| <\rho$ so indeed $f$ is $1-1$ on $|z| \le \rho$. Letting $\rho \to 1$ shows that $f$ is $1-1$ on $D$ so done!

(edit later to clarify as asked)

by definition, if $J$ is a (nice enough - eg rectifiable) closed curve, the winding number of $w \notin J$ is $\chi(J,w)=\frac{1}{2 \pi i}\oint_J\frac {d\xi}{\xi-w}$. This is an integer and is continuous in $w \notin J$ so, in particular, is constant on the connected components of the complement of $J$ and by letting $|w| \to \infty$ it is clearly zero on the unbounded component (which is unique regardless whether $J$ is Jordan or has self-intersections); if now $J$ is Jordan (simple) there is only one bounded component called the inside of $J$ and it is easy to see that the result there is $\pm 1$ since the winding number is invariant to homotopy (as long as we keep the crucial condition that $w \notin J_t, J_t, 0 \le t \le 1$ being a homotopy) and one can deform $J$ to a circle still having $w$ inside and for the circle, the computation is straightforward (in our case as $J$ is analytic one can show it is actually $1$ not $-1$ but that is less important)


答:
For $n \in \mathbb N$, you have

$$\begin{aligned}\left\Vert \alpha(\frac{1}{n+1}) - \alpha(\frac{1}{n}) \right\Vert &=\sqrt{\left(\frac{1}{n+1}-\frac{1}{n}\right)^2+\left(\frac{\cos\left(\frac{\pi(n+1)}{2}\right)}{n+1}-\frac{\cos\left(\frac{\pi n}{2}\right)}{n}\right)^2}\\
&\ge \frac{1}{n+1}
\end{aligned}$$

as one of $\cos\left(\frac{\pi(n+1)}{2}\right) , \cos\left(\frac{\pi n}{2}\right)$ vanishes while the absolute value of the other one is equal to one. This enables to get the required inequality

$$l_{n}>\sum_{k=1}^{2n}\frac{1}{k}.$$

As the arc length of  a curve is greater than a polygonal path approximating it and that the harmonic series $\sum \frac{1}{n}$diverges, you get that $\alpha$ is not rectifiable.

Title: An example of a non-rectifiable curve

Body:
How can I solve this problem?

Let the curve in $\mathbb{R}^2$ of the equation $\alpha(t)=(t,g(t)), t\in [0,1]$, where $$g(t)=\left\{ \begin{aligned} t \cos\left( \frac{\pi}{2t}\right), \quad t\not=0 \\ 0, \quad t=0\end{aligned}\right.$$
For $n \in \mathbb{N}$, let the partition $P$ of $[0,1]$ given by $$P=\left\{ 0, \frac{1}{2n},\frac{1}{2n-1},\ldots, \frac{1}{3},\frac{1}{2},1\right\},$$ prove the length $\ell_n$ of the polygonal inscribed satisfies $$\ell_n>\sum_{k=1}^{2n}\frac{1}{k}$$and and deduce that the curve $\alpha$ is not rectifiable.

My approach:  I know for example that if I prove that arc length of $\alpha$ is infinite, so I can say that $\alpha$ is not rectifiable, so I need to calculate $I_{n}$.

I know that the arc-length of a curve $\alpha$ is $$s(t)=\int_{t_0}^t \|\dot{\alpha}(t)\| \, dt$$
so I need to prove that $s(t)\to \infty \implies \alpha$ is not rectifiable.

I know how calculate $\dot{\alpha}(t)$, but how can I choose $t_0$ and $t$?

Questions:
Now, How can I prove the first part? It's to say, how can I prove that $\ell_n>\sum_{k=1}^{2n}\frac{1}{k}$? and how can I relate that result to the fact that alpha is not rectifiable?


问:
Mean number of particle present in the system: birth-death process, $E(X_t|X_0=i)$, $b_i=\frac{b}{i+1}$, $d_i=d$
Let $\{X_t\}$ be a birth–and–death process with birth rate
$$
b_i = \frac{b}{i+1},
$$
when $i$ particle are in the system, and a constant death rate
$$
d_i=d.
$$
Find the expected number of particle in the system at time $t$, given that $X_0=i$.

Define
$$
f(t)=E(X_t),
$$
and
$$
p_n=P\left(X_t=n | X_0=i \right).
$$
Using the foward equation,
$$
f'(t)=\sum_{n=1}^\infty n \left( p_{n-1} \frac{b}{n} + p_{n+1}d + p_n\left( 1 - \frac{b}{n+1}-d\right)\right).
$$
After simplification, I have
$$
f'(t)=p_0 b - \sum_{n=1}^\infty\frac{b}{n+1}p_n + d + f(t),
$$
and I don't see how to solve this differential equation.

答:
Not sure one can get explicit formulas for $E[X_t]$ but anyway, your function $f$ is not rich enough to capture the dynamics of the process.

The canonical way to go is to consider $u(t,s)=E[s^{X_t}]$ for every $t\geqslant0$ and, say, every $s$ in $(0,1)$. Then, pending some errors in computations done too quickly, the function $u$ solves an integro-differential equation similar to
$$
\frac{s}{1-s}\cdot\frac{\partial u}{\partial t}(t,s)=d\cdot(u(t,s)-u(t,0))-b\int_0^su(t,r)\mathrm dr,
$$
with initial condition $u(0,s)=s^i$. Assuming one can solve this (which does not seem obvious at first sight), your answer is
$$
E[X_t]=\frac{\partial u}{\partial s}(t,1).
$$


问:
$\{1,\dots,n\}$ - Without further definitions, does that imply $n \geq 1$?
I have another extremely basic notation question:

In many books, wikipedia and other places they write $\{1,\dots,n\}$ without ever defining n.

Does that imply that $ n \in \mathbb{N} $ or does that also allow $ n \in \mathbb{Z} $, which would mean that the set could also be empty?

(Sorry if this has already been answered somewhere, I find it very hard to find the right search terms for these kind of questions)

答: Well, if you consider the set $\{1,\ldots,n\}$ where $n\geq 0$ is an integer, then the case $n=0$ is defined as the empty set.


答:
Usually, when we have a set of the form

$$\{1, \dots, n\}$$

we are referring to the set of natural numbers from $1$ to $n,$ where $n$ is some natural number ($n \in \mathbb{N}$).

Also, note that $\emptyset$ is not “included” in the above set, for some object $x$ be included in a set $X$ it has to be the case that $x \in X,$ i.e., $x$ is an element of $X.$

In this case, for $n \in \mathbb{N}$ it is clear that $\emptyset \notin \{1,\dots, n\}.$

Although, the empty set is a subset of $\{1,\dots,n\},$ i.e., $\emptyset \subseteq \{1,\dots,n\}$ (it is the case that $\emptyset$ is a subset of any set).

Title: $\{1,\dots,n\}$ - Without further definitions, does that imply $n \geq 1$?

Body:
I have another extremely basic notation question:

In many books, wikipedia and other places they write $\{1,\dots,n\}$ without ever defining n.

Does that imply that $ n \in \mathbb{N} $ or does that also allow $ n \in \mathbb{Z} $, which would mean that the set could also be empty?

(Sorry if this has already been answered somewhere, I find it very hard to find the right search terms for these kind of questions)


答:
It depends on the context and the intentions of the author. In many
instances we define $\,[n] := \{1,2,\dots,n\}\,$ and intend it to be
used when $\,n>0.\,$ However, in other instances, we allow $\,n=0\,$
in which case $\,[0] = \emptyset := \{\} \,$ is the set with zero elements. It
all depends on the situation and the author. Sometimes $\,n=0\,$ does
makes sense, but at other times it does not. You can think of
this as merely a matter of convention exactly similar to the
convention that an empty summation is equal to zero.

Note that $\,[m]\subseteq [n]\,$ iff $\,m\le n\,$ which includes
the special case $\,\emptyset = [0]\subseteq [n]\,$ for all
$\,n\ge 0.\,$


问:
Find the residue of $1/(z^2+1)^{n+1}$ at $z=i$.
I'm trying to find the residue of $\frac{1}{(z^2+1)^{n+1}}$ at $z=i$.  I believe this should be approached combinatorially but I am getting stuck in the combinatorics.  Here's what I have so far.

$$ \frac{1}{(z^2+1)^{n+1}} = \frac{1}{(z-i)^{n+1}} \cdot \frac{1}{(z-i+2i)^{n+1}} $$

$$ = \frac{1}{(z-i)^{n+1}} \cdot \frac{1}{(2i)^{n+1}(1-[-z+i]/(2i))^{n+1}} $$

$$ = \frac{1}{(z-i)^{n+1}} \cdot \frac{1}{(2i)^{n+1}}\cdot \left[\sum_{k=1}^\infty \left((-z+i)/(2i)\right)^k\right]^{n+1} $$

$$ = \frac{1}{(z-i)^{n+1}} \cdot \frac{1}{(2i)^{n+1}}\cdot \left[\sum_{k=1}^\infty \left(\frac{-1}{2i}\right)^k(z-i)^k\right]^{n+1} $$

To get the residues we want the $(z-i)^{-1}$ term, which we can determine once we know the $(z-i)^n$ term of the factor which is expressed above as a series.  This will come from making $n+1$ choices of a term from each of the $n+1$ copies of the series, such that the exponents sum to $n$.  However, this now gets into a pretty famous combinatorial problem of finding the ways to partition an integer.  That sounds like a lot of tricky combinatorial stuff when you're learning about residues.  Am I doing this the hard way--is there an easier one?

Body:
There is agreement between the residue and the integral of $f(x)$ along the real axis.  We show that $$\int_{-\infty}^\infty \frac{dx}{(x^2+a^2)^{n+1}} = \frac{\pi}{2^{2n}a^{2n+1}} \frac{(2n)!}{n!n!}.$$

This is true for $n=0$:

$$\int_{-\infty}^\infty \frac{dx}{x^2+a^2} = \frac{1}{a}\int_{-\infty}^\infty \frac{\frac{dx}{a} }{\left(\frac{x^2}{a^2}+1\right)}=\frac{1}{a} \int_{-\infty}^\infty \frac{dq}{q^2+1}=\frac{1}{a} \left.\tan^{-1} q\right|_{-\infty}^\infty=\frac{\pi}{a}.$$

Assume true for $n-1$ and differentiate with respect to $a$.

$$\frac{d}{da}\int_{-\infty}^\infty \frac{dx}{(x^2+a^2)^{n}} = \frac{d}{da}\left[\frac{\pi}{2^{2n-2}a^{2n-1}} \frac{(2n-2)!}{(n-1)!(n-1)!}\right].$$

$$-2 n a\int_{-\infty}^\infty \frac{dx}{(x^2+a^2)^{n+1}} =-(2n-1)\left[\frac{\pi}{2^{2n-2}a^{2n}} \frac{(2n-2)!}{(n-1)!(n-1)!}\right].$$

$$\int_{-\infty}^\infty \frac{dx}{(x^2+a^2)^{n+1}} =\frac{2n}{2n}\left[\frac{\pi}{2^{2n-1}a^{2n+1}} \frac{(2n-1)!}{n!(n-1)!}\right].$$

$$\int_{-\infty}^\infty \frac{dx}{(x^2+a^2)^{n+1}} =\frac{\pi}{2^{2n}a^{2n+1}} \frac{(2n)!}{n!n!}.$$

With $a=1$,

$$\int_{-\infty}^\infty \frac{dx}{(x^2+1)^{n+1}} =\frac{\pi}{2^{2n}} \frac{(2n)!}{n!n!}=2\pi i \cdot \text{Res}_{z=i}\, f(z).$$


问:
Fixed point theorem for $f :\mathbb{R}\to \mathbb{R}$, alternative proof
I would like to poof the fixed point theorem that sais that if $f:\mathbb{R}\to \mathbb{R}$ is Lipschitz continious map with lipschitz constant $L<1$ then the map has a fixed point, i.e. $\exists x \in \mathbb{R}$ such that $f(x)=x$. I don't want to consider the sequence $x_{n+1}=f(x_n)$ (I already understood this proof). I would like to do the following : take $x_0\in \mathbb{R}$. Either $f(x_0)=x_0$ and we would be done. Or without loss of generality, $f(x_0)>x_0$. Then since the slope of the curve is always lower than $1$, I would like to say that at some point $x>x_0 f(x)$ is goint to cross the slope y=x. But I can't make this formal.

答:
We have $f(x) \leq L \cdot (x - x_0) + f(x_0)$ for $x > x_0$. Taking $x_1 = \max\left(\frac{f(x_0) - L x_0}{1 - L}, x_0\right) + 1$ we get $f(x_1) < x_1$.

Now, consider $g(x) = f(x) - x$. We have $g(x_0) > 0$ and $g(x_1) < 0$. As $g$ is continous, $g(x_2) = 0$ for some $x_2 \in (x_0, x_1)$.


答:
\begin{align}
\mathbb{E}\big[e^{XY}\big]&= \mathbb{E}\left[\mathbb{E}\left[e^{XY}\big|Y\right]\right]\\
&= \frac{1}{2}\mathbb{E}\left[e^{XY}\big|Y=-1\right]+ \frac{1}{2}\mathbb{E}\left[e^{XY}\big|Y=1\right]\\
&= \frac{1}{2}\mathbb{E}\left[e^{-X}\right]+ \frac{1}{2}\mathbb{E}\left[e^{X}\right]\\
&=\frac{1}{2\sqrt{2\pi}}\int_{-\infty}^\infty\big(e^{-x}+e^x\big)e^{-\frac{x^2}{2}}dx\\
&= \frac{1}{2\sqrt{2\pi}}\int_{-\infty}^\infty e^\frac{1}{2}\big(e^{-\frac{(x+1)^2}{2}}+e^{-\frac{(x-1)^2}{2}}\big)dx\\
&=e^\frac{1}{2}
\end{align}

Title: If $X,Y$ are two independent random variables how can you calculate $\mathbb{E}[e^{XY}]$

Body: Let $X,Y$ be two independent random variables where $X$ is standard normal distributed and $Y$ is distributed as $P(Y=1)=P(Y=-1)=\frac{1}{2}$  how can one calculate $\mathbb{E}[e^{XY}]$? I don't see how to use the joinst density of $X,Y$ to find the expectation (since one is continious and the other is discrete). I also don't see how I can factor the density since we dont have $\mathbb{E}[f(X)g(Y)]=\mathbb{E}[f(X)]\mathbb{E}[g(Y)]$ but rather $\mathbb{E}[h(X,Y)]$.


问:
Proving that $\mathbb{Z}[x]/\langle x^2 + 1\rangle$  is isomorphic to $\mathbb{Z}[i]$
I have some general idea of going about this. I'm trying to come up with a suitable mapping. I've come up with $\phi\colon a + \langle x^2 + 1\rangle\longmapsto a$ where $a$ is an irreducible element in $\mathbb{Z}[i]$ but this mapping has some issues. For example take
$5x + x^2 + 1$, which is $5x \pmod{x^2+1}$. $5x$ does not belong to $\mathbb{Z}[i]$. So my question is, is there a better mapping?

I know for a fact that if you add any integer call it $b$ to $\langle x^2 + 1\rangle$, you will get $b$ back which belongs to $\mathbb{Z}[i]$. I'm stuck when $b$ is not an integer.

答:
When you want to show that a quotient is isomorphic to a given object (ring, group, semigroup, etc), it is usually simplest to get the Isomorphism Theorem to do your work for you. Instead of trying to define an isomorphism from $\mathbb{Z}[x]/\langle x^2+1\rangle$ to $\mathbb{Z}[i]$, which will usually require you to show the map is well-defined, try to find a surjective map from $\mathbb{Z}[x]$ to $\mathbb{Z}[i]$ whose kernel is exactly $\langle x^2+1\rangle$. Then the Isomorphism Theorem guarantees the existence of an isomorphism
$$\frac{\mathbb{Z}[x]}{\langle x^2+1\rangle} = \frac{\mathbb{Z}[x]}{\mathrm{ker}(\phi)} \cong \mathbb{Z}[i].$$

Now, to define a map from $\mathbb{Z}[x]$ to $\mathbb{Z}[i]$, it is enough to specify what happens to the coefficients and what happens to $x$. There is an obvious thing to do with the coefficients (just let them be). Is there an obvious thing you can map $x$ to? I think there is. Then check what the kernel of this map, and let the Isomorphism Theorem do all the hard work. After all, that's why we proved it in the first place.

Well, apparently that's not the case here, as the Isomorphism Theorems have not been proven yet in your class.

Still: it's usually annoying to define a map from a quotient to a ring; because the elements of the quotient are equivalence classes, and that usually means that you need to show that your definition of map does not depend on the representative. It's much easier to map into a quotient. So instead of trying to find an isomorphism from $\mathbb{Z}[x]/\langle x^2+1\rangle$ to $\mathbb{Z}[i]$, it might be easier to find a map $\varphi$ from $\mathbb{Z}[i]$ to $\mathbb{Z}[x]/\langle x^2+1\rangle$.

Given an element $a+bi\in\mathbb{Z}[i]$, with $a,b\in \mathbb{Z}$, where should we map it? We certainly want to map $1$ to $1 + \langle x^2+1\rangle$ (identity to identity). Which means we want to map $a$ to $a+\langle x^2+1\rangle$. Which means that the only thing we need to do is figure out what to do with $i$. If we map $i$ to $p(x)+\langle x^2+1\rangle$, then this will completely determine our homomorphism, since
$$\begin{align*}
\varphi(a+bi) &= \varphi(a) + \varphi(b)\varphi(i)\\
 &= a+\langle x^2+1\rangle + \left(b+\langle x^2+1\rangle\right)\left(p(x)+\langle x^2+1\rangle\right)\\
 &= a + bp(x)\langle x^2+1\rangle.\end{align*}$$

Okay, what should $\varphi(i)$ be? Since $\varphi(i)^2 = \varphi(i^2) = \varphi(-1) = -1+\langle x^2+1\rangle$, we need $\varphi(i)$ to be the congruence class of a polynomial $p(x)\in\mathbb{Z}[x]$ such that $p^2(x)\equiv -1\pmod{\langle x^2+1\rangle}$. Is there some simple choice that might work? Again, I think there is; check to see that it actually works.


问:
Holomorphic function such that $Ref + Imf \geq 1$
Find all holomorphic functions $f:\mathbb{C} \rightarrow \mathbb{C}$ such that $Ref + Imf \geq 1$. I tried to use Liouville theorem, but it was no effective. Do you have any hints?

答: Notice that $\frac1f:\Bbb C\to\Bbb C$ is entire and bounded, and therefore constant.


问:
Find the Laurent Series expansion for $\displaystyle{f(z) = ze^{1/(z-1)}}$ valid for $\left|z-1\right|> 0$
Can you help me and discuss me on the question?

Expand $\displaystyle{f(z) = ze^{1/(z-1)}}$ in a Laurent series valid for $\displaystyle{\left|z-1\right|> 0}$.

I have no idea anything about the exponential complex form to fnd the Laurent series for that.  I know that the laurent series of $e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!}$ for every complex $z$.  Can I just substitiute it?

答:
Since $$e^{w}=\sum_{k\geq0}\frac{w^{k}}{k!},\,\left|w\right|<+\infty$$ we have, taking $w=\frac{1}{z-1}$, that for $0<\left|z-1\right|<+\infty$ (these bounds follow from $\left|w\right|=\left|\frac{1}{z-1}\right|<+\infty$) we obtain

\begin{align}
ze^{\frac{1}{z-1}}=\sum_{k\geq0}\frac{z}{k!\left(z-1\right)^{k}} & =\sum_{k\geq0}\frac{1}{k!\left(z-1\right)^{k-1}}+\sum_{k\geq0}\frac{1}{k!\left(z-1\right)^{k}}
\\ & =z-1+\sum_{k\geq1}\frac{1}{k!\left(z-1\right)^{k-1}}+\sum_{k\geq0}\frac{1}{k!\left(z-1\right)^{k}}
\\ & =z-1+\sum_{k\geq0}\frac{1}{\left(k+1\right)!\left(z-1\right)^{k}}+\sum_{k\geq0}\frac{1}{k!\left(z-1\right)^{k}}
\\ & =z-1+\sum_{k\geq0}\left(\frac{1}{\left(k+1\right)!}+\frac{1}{k!}\right)\frac{1}{\left(z-1\right)^{k}}
\\ &= \color{red}{z-1+\sum_{k\geq0}\frac{k+2}{\left(k+1\right)!}\frac{1}{\left(z-1\right)^{k}}}.
\end{align}


问:
Why is the linear space V = $\mathbb{C}$ a real linear space when scalars are from $\mathbb{R}$
This started with a question in the book Linear Algebra by S. Friedberg, A. Insel and L. Spence in which they define a vector space $\text{V} = \{(a_1, a_2, ...,a_n):a_i \in \mathbb{C}\}$, so a vector space over $\mathbb{C}$, and ask if it is a vector space over the field of real numbers.

I'm inclined to say this is false because $a_i$ could be any complex number, and multiplying by a scalar from $\mathbb{R}$ could still keep it in $\mathbb{C}$ right? In other words, if $\text{V}$ is defined over $\mathbb{C}$ and an element of $\text{V}$ with complex components is multiplied by a real scalar, its components could still be complex and non-real. In which case it is not an element in the field of $\mathbb{R}$.

The problem is that I've checked community solutions of the book online which say that it is in fact a vector field over the field of $\mathbb{R}$ and also consulted Apostol's Calculus: Vol 2, which says "Let $\text{V}= \mathbb{C}$, the set of all complex numbers, and define $ax$ to be multiplication of the complex number $x$ by the real number $a$. Even though the elements of $\text{V}$ are complex numbers, this is a real linear space because the scalars are real."

So I'm clearly wrong, but I fail to see why. How can a space be over a subset of a certain field by only limiting its scalar multiplication to that subset?

答:
If we have a vector space $V = \mathbb{C}^n$ over the field $\mathbb{R}$ you have indeed a vector space. Note that the operation that you have are the usual addition of vectors and the multiplication by aclarar is defined to be

$$\lambda \cdot (x_1,\dots,x_n) = (\lambda \cdot x_1,\dots, \lambda \cdot x_n) \quad \quad \forall (x_1,\dots,x_n) \in \mathbb{C}^n, \quad\forall \lambda \in \mathbb{R}.$$

Note that his operation if well-defined, since when you multiply $\lambda \in \mathbb{R}$ by each $x_i \in \mathbb{C}$ you get $\lambda \cdot x_i \in \mathbb{C}$ for all $i \in \{1,\dots,n\}.$ So the RHS of the equation above is in $\mathbb{C}^n.$

After realising this it is easy to prove that $V$ in addition with the usual sum of vectors in $\mathbb{C}^n$ and the multiplication by a scalar $\cdot$ as defined above is a vector space over $\mathbb{R}.$


问:
Comparison test for the series
I want to test that the given series is convergent or divergent.

$\Sigma_{n=2}\frac {(-i)^n}{\ln n} $, ($z_n = \frac {(-i)^n}{\ln n}$)

I know that for $n >1 $, $\ln n< n$ holds. So by using the comparison test, $\Sigma_{n=2} \frac{1}{n} $ diverges so $\Sigma_{n=2}\frac {1}{\ln n} $ diverges too. But what about $\Sigma_{n=2}\frac {(-i)^n}{\ln n} $? I must check the divergence of series $z_n$, not for the series $|z_n|$ and I got confused...

p.s. When checking the convergence of series, it's quite obvious to only test for  the series $|z_n|$ case because absolute convergence implies convergence. But what about the divergence case?

答:
Your p.s. is wrong. There are series like $\sum (-1/n)^n$ which converge while the series of the absolute values diverges.

Your series is one of this. Indeed we can write
$$\sum _{n=2}^{\infty } \frac{(-i)^n}{\log (n)}=\sum _{n=1}^{\infty } \frac{(-i)^{2 n}}{\log (2 n)}+\sum _{n=1}^{\infty } \frac{(-i)^{2 n+1}}{\log (2 n+1)}$$
The first one can be rewritten as
$$\sum _{n=1}^{\infty } \frac{(-1)^{ n}}{\log (2 n)}$$
which converges for the Leibniz criterion because general term tends to zero

The other series can be written as
$$\sum _{n=1}^{\infty } \frac{(-i)^{2 n+1}}{\log (2 n+1)}=i\sum _{n=1}^{\infty } \frac{(-1)^{n+1}}{\log (2 n+1)}$$
which converges for the same test as above.

Title: Comparison test for the series


问:
If $f$ is symmetrically continuous, then $f$ is not necessarily continuous
I've already proved that continuous functions are symmetrically continuous.

Now I want to show that the converse does not hold.

Define $f:X\rightarrow \mathbb{R}, f$ is symmetrically continuous at $x$

So, $\lim_{h\to 0} [f(x+h)-f(x-h)]=0$

I can't come up with a function that's symmetrically continuous but not continuous. Any ideas or hints? The Cantor set comes to mind, but is the Cantor set symmetrically continuous?

答: Choose your favourite continuous function $g:\Bbb R\to\Bbb R$ and call $f(t)=\begin{cases}g(x)+1&\text{if }t=x\\ g(t)&\text{if }t\ne x\end{cases}$


问:
Fitting circle in $f(x)=x^3-3x^2$ what is the radius?

The whole problem is the above.

I tried to solve it analytically but I couldnot find enough equation to solve it.

Assume the center is $(a,-r)$ where $r$ is the radius of the circle.

The points on the sphere $(x-a)^2+(y+r)^2=r^2$

and $(x,f(x))$ must satisfy it so if we denote the intersection points $(x_1,y_1),(x_2,y_2)$:

$$(x_1-a)^2+(x_1^3-3x_1^2+r)^2=r^2$$
$$(x_2-a)^2+(x_2^3-3x_2^2+r)^2=r^2$$

Are two equation with 4 unknown.

Can you give me elegant relations between these two unknowns?

答:
In figure suppose AC is straight line.We first find the angle BAC the curve makes with x axis:

$y=x^3-3x^2$ ⇒ $y'=3x^2-6x$

⇒ $tan (\widehat{BAC})=y'$ at point A(3, 0):

$tan (\widehat{BAC})=3^3-6\times 3=9$

⇒ $ \widehat{BAC} ≈83^o$

⇒$ \widehat{BAC}=\widehat BAD ≈41.5^o$

So the center of circle is on line $y=(m=tan 41.5)(x-3)$

Now we use plain geometry. we have:

$AB^2+(BD=r)^2=AD^2$

$BD^2+r^2=OD^2$

$AB+BO=3$

$(AB+BO)^2=AB^2+OB^2+2\cdot AB\cdot BO$

⇒ $AB^2+OB^2-2r^2=9-(DA^2+OD^2)$

$tan(41.5)=\frac r{AB}$

$OD^2=OB^2+r^2=(3-AB)^2+r^2$

These two relations give $OD^2$ in terms of r.

$sin(41.5)=\frac r{AD}$

This relation gives AD in term of r.

Now in relation:

$AB^2+OB^2-2r^2=9-(DA^2+OD^2)$

$OB=3-AB$

So OB can also be found in term of r. Plugging, DA, OA, AB and OB in above relation give an equation in terms of r

Note, r ≈1.1 and coordinates of its center is D( ≈1.8,  ≈-1.1)


问:
Kernel of direct sum of module homomorphisms
Let $M, N, P, Q$ be $R$-modules and $\varphi:M\to P$, $\psi:M\to Q$, $\phi:N\to P$ and $\lambda:N\to Q$ module homomorphisms. Consider

$$ M\oplus N \xrightarrow{f:=\left(\begin{matrix}
    \varphi & \phi \\
    \psi & \lambda  \\
    \end{matrix}\right)}  P\oplus Q $$
that is for $(a,b)\in M\oplus N$ we have

$$f(a, b) = \left(\begin{matrix}
    \varphi & \phi \\
    \psi & \lambda  \\
    \end{matrix}\right)\left(\begin{matrix}a \\ b \end{matrix}\right) = \left(\begin{matrix}\varphi(a)+\phi(b) \\ \psi(a) +\lambda(b) \end{matrix}\right)$$

(Please excuse the slight abuse of notation here, just trying to make the matrix notation look neat) I'm trying to describe $\ker (f)$ in terms of the kernels of $\varphi$, $\psi$, $\phi$ and $\lambda$. I know that the kernel of

$$\left(\begin{matrix}\varphi & 0 \\0 & \lambda  \\\end{matrix}\right)$$

is $\ker(\varphi)\oplus\ker(\lambda)$ since this is just the generic direct sum of maps. But is it possible to have a similar representation with this more general case? I can see that $$(\ker(\varphi)\cap\ker(\psi))\oplus(\ker(\phi)\cap\ker(\lambda))\subsetneq \ker(f)$$ but past that I'm unsure. Anyone have any ideas?

答: There's not much else you can say in general, as the following example illustrates: Let $R = k$ be a field, and let $M = N = P = Q = k$, so that $\varphi, \psi, \phi, \lambda$ are just scalars and the matrix is literally a $2\times 2$ matrix with entries in $k$. If we know that all of the kernels of $\varphi, \psi, \phi, \lambda$ are zero, that means the scalars are nonzero—but what does a matrix having all nonzero entries tell us about its kernel? Really not much. We know the kernel of $f$ isn't all of $k^2$, and more generally we can say that $\ker(f)$ doesn't contain the standard basis vectors $(1, 0)$ or $(0, 1)$, but I can't see what more there is to say in general. The kernel could be trivial, or it could be any one-dimensional subspace other than the coordinate factors.


答:
It is an operation known as ‘restriction  of scalars’. Any $\mathbf C$-vector space $V$ is also an $\mathbf R$-vector space just by asking the scalar  multiplication to be restricted to the real numbers, and we have a relation between the dimensions as a $\bf C$- and $\bf R$-vector space if $V$ is finite-dimensional:
$$\dim_{\bf  R}V=\dim_{\bf C}V\dim_{\bf R}\mathbf C=2\dim_{\bf C}V.$$
Likewise, the same vector space cab be seem as a $\bf Q$-vectorspace, and in this case, it has infinite dimension as a vector space over $\bf Q$, because $\bf C$ has infinite dimension over $\bf Q$.

Title: Why is the linear space V = $\mathbb{C}$ a real linear space when scalars are from $\mathbb{R}$

Body:
This started with a question in the book Linear Algebra by S. Friedberg, A. Insel and L. Spence in which they define a vector space $\text{V} = \{(a_1, a_2, ...,a_n):a_i \in \mathbb{C}\}$, so a vector space over $\mathbb{C}$, and ask if it is a vector space over the field of real numbers.

I'm inclined to say this is false because $a_i$ could be any complex number, and multiplying by a scalar from $\mathbb{R}$ could still keep it in $\mathbb{C}$ right? In other words, if $\text{V}$ is defined over $\mathbb{C}$ and an element of $\text{V}$ with complex components is multiplied by a real scalar, its components could still be complex and non-real. In which case it is not an element in the field of $\mathbb{R}$.

The problem is that I've checked community solutions of the book online which say that it is in fact a vector field over the field of $\mathbb{R}$ and also consulted Apostol's Calculus: Vol 2, which says "Let $\text{V}= \mathbb{C}$, the set of all complex numbers, and define $ax$ to be multiplication of the complex number $x$ by the real number $a$. Even though the elements of $\text{V}$ are complex numbers, this is a real linear space because the scalars are real."

So I'm clearly wrong, but I fail to see why. How can a space be over a subset of a certain field by only limiting its scalar multiplication to that subset?


问:
Growth rate of $f(n) = \frac{1-\cos(\frac{n\pi}{n+1})}{1-\cos(\frac{\pi}{n+1})}$
I encountered this expression when studying the growth of condition numbers on a linear system. I was trying to verify that the order of the growth is $O(n^2)$. However I'm stuck upon proving this. Any hints would be appreciated. Thanks.

答:
Hint

$$1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right)$$

Then $$\begin{aligned}
f(n) &= \frac{1-\cos(\frac{n\pi}{n+1})}{1-\cos(\frac{\pi}{n+1})}\\
&= \frac{\sin^2\left(\frac{n \pi }{2(n+1)}\right)}{\sin^2\left(\frac{\pi}{2(n+1)}\right)}\\
& \le \frac{1}{\sin^2\left(\frac{\pi}{2(n+1)}\right)}\\
&\simeq \frac{\pi^2}{4}n^2
\end{aligned}$$


答:
Well if you have an ambient vector space $V= K^n = \{(a_1,\ldots,a_n)\mid a\in K\}$, where $K$ is a field, then $V$ is also a vector space over any subfield $L$ of $K$.

For this, consider the scalar multiplication

$\kappa (a_1,\ldots a_n) = (\kappa a_1,\ldots,\kappa a_n)$

where $\kappa\in K$. This multiplication is also defined if $\kappa$ is from a subfield $L$. The addition of vectors doesn't change.


问:
Derivative of integral with varying domain? Fundamental theorem of calculus?
Say I want to find the $n^{th}$ derivative of
$$\int_{V_x} F{(x_1,...,x_k)}dV$$
where $V_x$ is some $k$-dimensional volume dependent on $x$. In the $1$-dimensional case this is the fundamental theorem of calculus for $n=1$ and we can take higher derivatives after applying the fundamental theorem. To be concrete, say $V_x$ is the cube $[0,x]^k$. What can I do in general? Is this Stokes theorem?

答:
The general problem would be to compute the derivative of
$$ F(\mathbf{x},\mathbf{u}) = \int_{\Omega(\mathbf{u})} f(\mathbf{x})d\mathbf{x} $$
with respect to $\mathbf{x}$ with $\mathbf{u} = \mathbf{T}(\mathbf{x})$ (in this case $T = I$ is the identity map). The generalized Leibniz rule gives:
$$ \frac{\partial F}{\partial \mathbf{u}} = \int_{\partial \Omega(\mathbf{u})} f(\mathbf{x})\frac{\partial \mathbf{x}}{\partial \mathbf{u}}^{\top} \mathbf{n}(\mathbf{x})d\Gamma $$ Combining this result with Leibniz rule again for differentiation under the integral sign and using the chain rule  the gradient with respect to $\mathbf{x}$ is:
$$ \frac{\partial F}{\partial \mathbf{x}} =  \int_{\Omega(\mathbf{u}))} \nabla_{\mathbf{x}}f(\mathbf{x})d\mathbf{x} + \frac{\partial T}{\partial \mathbf{x}}^{\top}\int_{\partial \Omega(\mathbf{u})} f(\mathbf{x})\frac{\partial \mathbf{x}}{\partial \mathbf{u}}^{\top} \mathbf{n}(\mathbf{x})d\Gamma   $$
The hidden complexity here is how to compute the "velocity" of the boundary points as the domain change. As an example consider the two-dimensional domain $\Omega = [0,x_1] \times [0,x_2]$ and the function $f=1$ then $F = \int_{\Omega} f d\mathbf{x} = x_1 x_2$ and obviously $\nabla F = [x_2 \, \,  x_1]^{\top}$. Applying the previous machinery we get:
$$  \frac{\partial F}{\partial \mathbf{x}} =  \int_{\Omega(\mathbf{u}))} \nabla_{\mathbf{x}}f(\mathbf{x})d\mathbf{x} + \frac{\partial T}{\partial \mathbf{x}}^{\top}\int_{\partial \Omega(\mathbf{u})} f(\mathbf{x})\frac{\partial \mathbf{x}}{\partial \mathbf{u}}^{\top} \mathbf{n}(\mathbf{x})d\Gamma =  \int_{\partial \Omega(\mathbf{u})} f(\mathbf{x})\frac{\partial \mathbf{x}}{\partial \mathbf{u}}^{\top} \mathbf{n}(\mathbf{x})d\Gamma = \int_{\Gamma_2(\mathbf{u})} f(\mathbf{x})\frac{\partial \mathbf{x}}{\partial \mathbf{u}}^{\top} \mathbf{n}(\mathbf{x})d\Gamma + \int_{\Gamma_3(\mathbf{u})} f(\mathbf{x})\frac{\partial \mathbf{x}}{\partial \mathbf{u}}^{\top} \mathbf{n}(\mathbf{x})d\Gamma $$
Being $\Gamma_2$ and $\Gamma_3$ the right and upper sides of the rectangle that are the only ones that have non-zero velocity of their boundaries. Since $\Gamma_2$ is moving in the horizontal direction we have $\frac{\partial \mathbf{x}}{\partial\mathbf{u}}^{\top} = \begin{bmatrix} 1 & 0 \\ 0 & 0\end{bmatrix}$ on $\Gamma_2$ while $\frac{\partial \mathbf{x}}{\partial\mathbf{u}}^{\top} = \begin{bmatrix} 0 & 0 \\ 0 & 1\end{bmatrix}$ on $\Gamma_2$ (i.e. $\Gamma_2$ is moving vertically) then we have:
$$ \frac{\partial F}{\partial \mathbf{x}} = \int_{\Gamma_2(\mathbf{u})} f(\mathbf{x})\frac{\partial \mathbf{x}}{\partial \mathbf{u}}^{\top} \mathbf{n}(\mathbf{x})d\Gamma + \int_{\Gamma_3(\mathbf{u})} f(\mathbf{x})\frac{\partial \mathbf{x}}{\partial \mathbf{u}}^{\top} \mathbf{n}(\mathbf{x})d\Gamma = \int_{\Gamma_2(\mathbf{u})} \begin{bmatrix} 1 \\ 0 \end{bmatrix} d\Gamma + \int_{\Gamma_3(\mathbf{u})} \begin{bmatrix} 0 \\ 1 \end{bmatrix} d\Gamma  = \begin{bmatrix} x_2 \\ x_1 \end{bmatrix}$$


问:
Probability of breast cancer
I'm having trouble with a probability problem I've been trying to solve for a while. It's about the accuracy of breast cancer testing. Relevant probabilities are listed below, where:

"$\text{cancer}$" is the event "has breast cancer".

"$+$" is the event "tests positive for breast cancer".

$P(\text{cancer}) = \frac{12}{1000}$

$P(+|\text{cancer}) = \frac{11}{12}$

$P(+) = \frac{31}{1000}$

$P(\text{cancer}|+) = 0.355$

This last line is a result from a previous problem. The next part involves updating the probability of having cancer, but I'm having trouble figuring out what the answer is.

In the next part of the question, there's a woman who has tested positive and her doctor says she's part of a population for which there's a 40% chance of breast cancer.

I need to find the probability that the woman has cancer.

I am confused by this update to the cancer probability, but i will assume that this means $P(\text{cancer})$ has changed.

I also assume this means I need to find a new value for $P(\text{cancer}|+)$, but I'm not getting this right.

$P(+ | \text{cancer}) = \frac{11}{12} = \frac{P(\text{cancer} | +)\cdot P(+)}{P(\text{cancer})} = \frac{P(\text{cancer} | +) \cdot \frac{31}{1000}}{0.40}$

$P(\text{cancer} | +) = \frac{11}{12} \cdot 0.40 \cdot \frac{1000}{31} = 11.828$

The result can't be correct because it's way over 1.

How can i fix this? Thank you in advance for any insight.

答:
Just to check the figures given early in your question:

\begin{align*}
P(\text{cancer}|+) &= \frac{P(+|\text{cancer}) \cdot P(\text{cancer})}{P(+)} \\
&= \frac{\frac{11}{12} \cdot \frac{12}{1000}}{\frac{31}{1000}} \\
&= \frac{11}{31} \\
&\approx 0.355
\end{align*}

So at least that part is correct.

Now, what happens when you change $P(\text{cancer})$ to $0.40$, but keep $P(+|\text{cancer}) = \frac{11}{12}$ and $P(+) = \frac{31}{1000}$? You have already calculated this, but there is a more direct way of writing your calculation:

\begin{align*}
P(\text{cancer}|+) &= \frac{P(+|\text{cancer}) \cdot P(\text{cancer})}{P(+)} \\
&= \frac{\frac{11}{12} \cdot 0.40}{\frac{31}{1000}} \\
&= \frac{11}{12} \cdot 0.40 \cdot \frac{1000}{31} \\
&\approx 11.828
\end{align*}

This calculation shows that this combination of probabilities does not work. This conclusion is correct, but why?

Here is a simple explanation: Increasing $P(\text{cancer})$ but keeping $P(+|\text{cancer})$ the same increases $P(\text{cancer} \cap +)$. In fact, in this case,

\begin{align*}
P(\text{cancer} \cap +) &= P(\text{cancer}) \cdot P(+|\text{cancer}) \\
&= 0.40 \cdot \frac{11}{12} \\
&> \frac{1}{3} \\
&\gg \frac{31}{1000} \\
&= P(+)
\end{align*}

So what should you do? It’s hard to tell from such a vague question. My guess is to keep $P(+|\text{cancer})$ and $P(+|\neg \text{cancer})$ the same as they were in the original situation (because these should depend only the test and therefore be independent of the cancer distribution):

\begin{align*}
P(+|\text{cancer}) &= \frac{11}{12} \\
P(\neg \text{cancer}) &= 1 - P(\text{cancer}) \\
&= \frac{988}{1000} \\
P(\text{cancer} \cap +) &= P(\text{cancer}) \cdot P(+|\text{cancer}) \\
&= \frac{12}{1000} \cdot \frac{11}{12} \\
&= \frac{11}{1000} \\
P(\neg \text{cancer} \cap +) &= P(+) - P(\text{cancer} \cap +) \\
&= \frac{31}{1000} - \frac{11}{1000} \\
&= \frac{20}{1000} \\
P(+|\neg \text{cancer}) &= \frac{P(\neg \text{cancer} \cap +)}{P(\neg \text{cancer})} \\
&= \frac{\left( \frac{20}{1000} \right)}{\left( \frac{988}{1000} \right)} \\
&= \frac{5}{247}
\end{align*}

(There’s a big red flag here: apparently, this test has performed the miracle of minimising both the false positive and false negative rates. But I see nothing better, so I will continue on my original path.)

Applying these values of $P(+|\text{cancer})$ and $P(+|\neg \text{cancer})$ to the new value of $P(\text{cancer}) = 0.40$ gives a new value of $P(+)$:

\begin{align*}
P(\neg \text{cancer}) &= 1 - P(\text{cancer}) \\
&= 0.60 \\
P(\text{cancer} \cap +) &= P(\text{cancer}) \cdot P(+|\text{cancer}) \\
&= 0.40 \cdot \frac{11}{12} \\
&= \frac{11}{30} \\
P(\neg \text{cancer} \cap +) &= P(\neg \text{cancer}) \cdot P(+|\neg \text{cancer}) \\
&= 0.60 \cdot \frac{5}{247} \\
&= \frac{3}{247} \\
P(+) &= P(\text{cancer} \cap +) + P(\neg \text{cancer} \cap +) \\
&= \frac{11}{30} + \frac{3}{247} \\
&= \frac{2807}{7410}
\end{align*}

Now we can apply Bayes’ theorem with this new value of $P(+)$:

\begin{align*}
P(\text{cancer}|+) &= \frac{P(+|\text{cancer}) \cdot P(\text{cancer})}{P(+)} \\
&= \frac{\frac{11}{12} \cdot 0.40}{\frac{2807}{7410}} \\
&\approx 0.968
\end{align*}

Finally, let’s think about whether this answer is reasonable. Since our miracle test has a low rate of false results, we would expect:

$$P(\text{cancer}|+) \approx 1 \tag{1}$$

With the original value of $P(\text{cancer}) = \frac{12}{1000}$, the approximation $(1)$ was false, because $P(\neg \text{cancer})$ was so high that even a low false positive rate results in a high number of false positives (compared to the number of people tested) – in this case, higher than the number of false negatives.

But with the new value of $P(\text{cancer}) = 0.40$, the approximation $(1)$ is true. Or, at least, it’s a much better approximation than it was before.

So it looks like this answer is reasonable.

Title: Probability of breast cancer


问:
Is my prove that $A \setminus B \subseteq \overline{B \setminus A}$ acceptable?
We should prove that $A \setminus B \subseteq \overline{B \setminus A}.$

Let $x \in A \setminus B.$ Then

$$\begin{align}
x \in A \setminus B & \implies x \in A \wedge x \notin B\\
& \implies x \in A\\
& \implies x \notin B \setminus A\\
& \implies x \in \overline{B \setminus A} \quad \quad \square
\end{align}$$

答:
I’m gonna just present a simple explanation of why these implications work, but the proof is fine!

Let $x \in A \setminus B.$ Then

$$\begin{align}
x \in A \setminus B & \implies x \in A \wedge x \notin B && \text{(by definition of “$\setminus$”)}\\
& \implies x \in A && \text{(Simplification)}\\
& \implies x \notin B \setminus A && \text{(by definition of “$\setminus$“)}\\
& \implies x \in \overline{B \setminus A} && \text{(definition of complement)}
\end{align}$$


问:
Poisson limit process with divergence condition
Please help me with problem 3.26 of 'Probability Theory' by Varadhan:

For each $n\in \mathbb{N}$, $\ X_{n,j}:(\Omega,P) \to \mathbb{R}$ with $1\leq j\leq k_n$ are $k_n$ independent random variables with $$P[X_{n,j}=1]=p_{n,j} \text{ and } P[X_{n,j}=0]=1-p_{n,j}.$$

Let $S_n:=\sum_{j} X_{n,j}$. Let $\lambda_n:=\sum_{j} p_{n,j}$ be the mean of the $S_n$.

Question: If $\lambda_n \to \infty$, show that the distribution of $\frac{S_n-\lambda_n}{\sqrt{\lambda_n}}$ converges to the standard normal distirbution.

Though not explicitly mentioned, I think the assumption that the $X_{n,j}$ be uniformly infinitesimal is required. That is, we assume
$$\lim\limits_{n\to \infty} \sup\limits_{1\leq j\leq k_n} p_{n,j} = 0.
$$

Attempt: This looks similar to the central limit theorem without the variables being identically distributed. So anyhow, I try to compute characteristic function -

\begin{equation}
\begin{split}
\widehat{\frac{S_n-\lambda_n}{\sqrt{\lambda_n}}}(t) &= \int\exp \left[it\frac{S_n(\omega)-\lambda_n}{\sqrt{\lambda_n}}\right]dP(\omega) \\
&= \int\exp\left[ \frac{itS_n(\omega)}{\sqrt{\lambda_n}}\right] \exp\left[\frac{-it\lambda_n}{\sqrt{\lambda_n}}\right]dP(\omega)\\
&= \exp\left[\frac{-it\lambda_n}{\sqrt{\lambda_n}}\right]\int\exp\left[\frac{it}{\sqrt{\lambda_n}}\sum x_j\right]d((X_{n,1},\dots,X_{n,k_n})_*P)(x_1,\dots,x_{k_n})\\
&= \exp\left[\frac{-it\lambda_n}{\sqrt{\lambda_n}}\right]\int\dots\int\exp\left[\frac{it}{\sqrt{\lambda_n}}\sum x_j\right]d((X_{n,1})_*P)(x_1)\dots d((X_{n,k_n})_*P)(x_{k_n})\\
&= \exp\left[\frac{-it\lambda_n}{\sqrt{\lambda_n}}\right] \prod \int \exp\left[\frac{itx}{\sqrt{\lambda_n}}\right]d((X_{n,j})_*P)(x)\\
&= e^{\frac{-it}{\sqrt{\lambda_n}}\sum p_{n,j}} \prod \left[(1-p_{n,j}) + p_{n,j}e^{\frac{it}{\sqrt{\lambda_n}}}\right]\\
&= \prod \left[(1-p_{n,j})e^{\frac{-itp_{n,j}}{\sqrt{\lambda_n}}} + p_{n,j}e^{\frac{it(1-p_{n,j})}{\sqrt{\lambda_n}}}\right].
\end{split}
\end{equation}

Not sure what to do now. Maybe take log, but that doesn't really look too promising. Or should I be looking at the Levy-Khintchine representation?

Question: Why are people interested in such a sequence of $S_n$? I understand that the iid case with $p_{n,j}=1/2$ is talking about someone tossing a bunch of coins.

Edit: I see that this question is very similar although I'm not sure if it's exactly the same since there is some different assumption involving some variables following a Poisson distribution.

答:
This exercise is more elementary than that. I'm not sure if it was created to use Levy-Khintchine. Everything comes from the following observation:

Lemma 1:
Let $X$ be a bounded random variable.

Denote $\phi_X(t) = \log\left[ \mathbb{E}\left( e^{tX} \right) \right]$, then we have following assertions:

$\phi_X(0)=1$

$\phi_X'(0)= \mathbb{E}(X)$

$\phi_X''(0) = Var(X)$

$\phi_X^{(3)}(t) = \mathbb{E}^t(X^3)-3\mathbb{E}^t(X^2)\mathbb{E}^t(X)+2\left[\mathbb{E}^t(X)\right]^3$ for all $t$

where $\mathbb{E}^t$ is the expectation taking in the probability space $\mathbb{P}^t$ which is definied as:
$$ \dfrac{d\mathbb{P}^t }{d\mathbb{P}  } = \dfrac{e^{tX}}{\mathbb{E}(e^{tX})}$$
Proof:( Just derivation)

Then now, I'll add another lemma for the sake of approximation.

Lemma 2:

If $B$ is a $\mathcal{B}(p)$, then let $X:= B-p$. We have that forall $t \in \mathbb{R}$:

$|X| \le 1$

$ |X(1-X^2) | \le 2p(1-p)$ almost surely.

$ \vert \mathbb{E}^t(X)\vert \le 2p(1-p)e^{|t|}$

and as a direct consequence of the three above:
$$|\phi_X^{(3)}(t)| \le 7p(1-p)e^{3|t|}$$

Back to our intial problem, let $Z_n$ denote $\dfrac{S_n-\lambda_n}{\sqrt{\lambda_n}} $ , that is:
$$ Z_n:= \dfrac{S_n-\lambda_n}{\sqrt{\lambda_n}} $$
For the sake of calculation, we also change $\lambda_n$ in the denumerator to $ \sum_{j \ge 1} p_{nj}(1-p_{nj})$. (the $\lambda_n$ in the numerator stays put, and let call $\tilde{\lambda_n}$ for that old lambda)

( the new and old lamdas are equivalent under your additional condition)

By using our lemmas and Taylor's development, see that:

$$\phi_{Z_n}(t) = \sum_{j \ge 1} \phi_{ X_{nj}-p_{nj}}\left( \dfrac{t}{\sqrt{\lambda_n}} \right) =  \sum_{j \ge 1} \left[ \frac{1}{2}p_{nj}(1-p_{nj})+O(1) \frac{|t|}{\sqrt{\lambda_n}}p_{nj}(1-p_{nj})e^{ \frac{3|t|}{\sqrt{\lambda_n}}} \right]\dfrac{t^2}{\lambda_n}$$

where the grand O notation is uniform for all $n,j$

So,
$$ \phi_{Z_n}(t)= \frac{1}{2}t^2+O(1)\frac{|t|}{\sqrt{\lambda_n}}e^{ \frac{3|t|}{\sqrt{\lambda_n}}}t^2$$
Hence, when if $\lambda_n \xrightarrow{n \rightarrow \infty}$, we have the limit of moment generating function for $Z_n$, that is:
$$ \phi_{Z_n}(t) \longrightarrow \frac{1}{2}t^2$$

Hence conclusion.


问:
Find $\lim_{n \to \infty}\frac{\sin 1+2\sin \frac{1}{2}+\cdots+n\sin \frac{1}{n}}{n}$

Find $$\lim_{n \to \infty}\frac{\sin 1+2\sin \frac{1}{2}+\cdots+n\sin \frac{1}{n}}{n}.$$

This is a recent exam question, which I couldn't figure out in the exam. My guess is it doesn't exist.

答:
Or we can use Stolz–Cesàro theorem to find that $$\lim_{n \rightarrow \infty}\frac{\sin 1+2\sin \frac{1}{2}+\cdots+n\sin \frac{1}{n}}{n}=\lim_{n\to \infty}\frac{(n+1)\sin {\frac{1}{n+1}}}{1}=1.$$

This is similar with @DonAntonio's solution.


问:
Maxima of expression on given interval
How do I find the maximum value of $x(1-x)^n$ on the interval $[0,1]$?

答:
Let $f(x) = x(1-x)^{n}$. Then, $f'(x) = (1-x)^{n} - nx(1-x)^{n-1}$. We wish to find critical points of this function. These are the endpoints $0$ and $1$, as well as the points where $f'(x) = 0$. We have:

$$f'(x) = 0$$

$$(1-x)^{n} - nx(1-x)^{n-1} = 0$$

$$(1 - x) - nx = 0\ \ (x\neq 1)$$

$$(n+1)x = 1$$

$$x = \frac{1}{n+1}$$

Then, note that $f'(x) > 0$ if $0 < x < \frac{1}{n+1}$, and $f'(x) < 0$ if $\frac{1}{n+1}< x < 1$. Thus, $x = \frac{1}{n+1}$ is at a relative maximum (global in that interval), and the maximum value is thus $f(\frac{1}{n+1}) = \boxed{\frac{n^{n}}{(n + 1)^{n + 1}}}$

Title: Maxima of expression on given interval


问:
If A is a n*n matrix with only 1 eigenvalue a, how can we show that A is an upper triangular matrix with all diagonal entries as a.
I have been trying for a long time now but not reaching anywhere. Can you please give me a hint on how to appraoch this question?
Will be very grateful.</p>

答:
As stated, and working with real matrices, the statement may be wrong. If that would be the case, the characteristic polynomial of $A$ would always split which may not be the case. For example if

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

$a$ is the only real eigenvalue of $A$ though.


问:
Group action of $\mathbb{Z}_3$ on the complex plane by rotations.
Consider the action of $\mathbb{Z}_3$ on the complex plane by rotations. Which of these pairs of complex numbers are in the same orbit?

A: $1$ and $e^{\frac{i\pi}{2}}$

B: $1$ and $e^{\frac{i\pi}{3}}$

C: $e^{\frac{i\pi}{3}}$ and $e^{\frac{-2i\pi}{3}}$

D: $e^{\frac{i\pi}{3}}$ and $e^{\frac{-i\pi}{3}}$

What I know so far:

For $\bar{k}\in\mathbb{Z}_3$ and $z\in\mathbb{C}$  $$\bar{k} \cdot z= e^{\frac{2ik\pi}{n}} \times z $$

Hence for $\bar{0}$ we get $1$ which rules out C and D.
For $\bar{1}$ we get $e^{\frac{2i\pi}{3}}$ but that is not an option.

Any help is greatly appreciated!

答: Hint/solution: Each orbit has three elements, except for the orbit of the origin, which is a singleton. Two complex numbers $z$ and $w$ are in the same orbit if $|z|=|w|$ and the angle between them is $2\pi/3$. This immediately rules out everything except for option D, whose elements form an angle of $(\pi/3)-(-\pi/3)=2\pi/3$.


问:
Best resource for concentration of measure?
What are the best resources to learn concentration of measure phenomenon and concentration inequalities? I have heard that Talagrand's papers are good, but they are not particularly good read.

Any help is appreciated. Thanks a lot!

答:
In Marseille, France, Sudeep Kamath delivered an excellent four-part video lecture on concentration of measure, and writes mathematical proofs for entropic and transport-related concentration inequalities:

https://www.youtube.com/results?search_query=Sudeep+Kamath+%3A+Concentration+of+Measure

Variance bounds

Information inequalities

Entropy method

Transportation method

Based on this outline alone, his lectures seem to resemble the table of contents of:

Concentration Inequalities: A Nonasymptotic Theory of Independence by  Stéphane Boucheron, Gábor Lugosi, and Pascal Massart, 2013.

For those interested, the transportation-information inequalities that appear in concentration of measure can be discussed here.


问:
Point $M$ lies inside $\triangle ABC$, $\angle MAC = 10^\circ$ and $\angle MCA = 30^\circ$. Find $(180^\circ - \angle BMC)$

In $\Delta ABC, \angle CAB = 30^\circ$ and $\angle ABC = 80^\circ$. Point $M$ lies inside the triangle such that $\angle MAC = 10^\circ$ and $\angle MCA = 30^\circ$. Find $(180^\circ - \angle BMC)$.

What I Tried: Here is a picture to keep track of the angle-chasing .

The red angle is $30^\circ$ , the green one is $10^\circ$ , the purple one is $20^\circ$ . Now I extend $MC$ to $AB$ such that it meets at $K$. First I noticed that $AK = KC$ but that didn't seem to help. Also the $2$ yellow ones are equal to $40^\circ$ each, the brown one is equal to $80^\circ$ . The light green angle is $60^\circ$ . We have to find the blue angle.

After all these information, it seems like I am still missing something, because from here you actually cannot deduce the value of the blue angle. Can anyone help?

Thank You.

答:
Let $O$ be the circumcenter of $\triangle ABC$. Consider $\triangle OAC$,

$\angle AOC = 2\angle ABC = 160°$

$\angle OAC = \angle OCA = \frac{180° - 160°}{2} = 10°$

Hence, $O$ lies on $AM$.

$\angle MOC = 10° + 10° = \angle MCO \Rightarrow MO = MC$.

$\angle OCB = \angle OBC = \frac{180° - 60°}{2} = \angle COB \Rightarrow BC = OB$.

Hence, $\triangle BMO \cong \triangle BMC$.

$\therefore 180° - \angle BMC = \frac{\angle OBC}{2} + \angle BCO - 20° = 70°$


问:
Probability on a given sample mean given standard deviation.
The standard deviation for a given jobs' wage is $ \sigma = 4200 $.

What is the probability that the mean $ \overline{X}_{40}$ of a random sample $ X_1,X_2,X_3 \dots X_{40}$ is greater than the population's mean.

I haven't encountered a similar question to this one , so I have no attempts.

答:
as per $n=40$ is greater enough, you can use CLT to say that

$$\overline{X}_{40}\dot{\sim}N\Bigg(\mu;\frac{4200^2}{40}\Bigg)$$

Thus

$$\mathbb{P}[\overline{X}_{40}>\mu]=\frac{1}{2}$$

EDIT: answering to  your comment,

$$\mathbb{P}[\overline{X}_{40}>\mu+50]=\mathbb{P}[Z>\frac{50}{4200}\sqrt{40}]=1-\Phi(0.075)\approx 0.47$$


问:
Integration of hypergeometric function involving power and exponential function
I would be grateful if someone can explain how to evaluate the following
$$
\int_{0}^{\infty} y^a exp(-y/b)\,_2F_1(1,1;2;-c y^2) dy
$$
where $a>0$, $b>0$, and $c>2$. I searched on special functions (using Gradestien book and others) but I could not find an answer.

Body:
I do not know if a closed form exist.

Let $y=\frac{x}{\sqrt{c}}$ and $k=\frac{1}{b \sqrt{c}}$ to make the integral to be
$$ c^{-\frac{a+1}{2} }\int_{0}^{\infty}x^{a-2} e^{-k x} \log \left(x^2+1\right)\,dx$$ Let
$$J_a=\int_{0}^{\infty}x^{a-2} e^{-k x} \log \left(x^2+1\right)\,dx$$
For $J_2$ it is simple. Now appear  the Meijer G function in all the cases I tried
$$J_1=\frac{G_{2,4}^{4,1}\left(\frac{k^2}{4}|
\begin{array}{c}
 0,1 \\
 0,0,0,\frac{1}{2}
\end{array}
\right)}{2 \sqrt{\pi }}\qquad \qquad  J_3=\frac{2 G_{1,3}^{3,1}\left(\frac{k^2}{4}|
\begin{array}{c}
 0 \\
 0,0,\frac{3}{2}
\end{array}
\right)}{\sqrt{\pi } k^2}$$

$$J_4=\frac{4 G_{2,4}^{4,1}\left(\frac{k^2}{4}|
\begin{array}{c}
 0,1 \\
 0,0,\frac{3}{2},2
\end{array}
\right)}{\sqrt{\pi } k^3}\qquad \qquad  J_5=\frac{8 G_{2,4}^{4,1}\left(\frac{k^2}{4}|
\begin{array}{c}
 0,1 \\
 0,0,2,\frac{5}{2}
\end{array}
\right)}{\sqrt{\pi } k^4}$$

$$J_6=\frac{16 G_{2,4}^{4,1}\left(\frac{k^2}{4}|
\begin{array}{c}
 0,1 \\
 0,0,\frac{5}{2},3
\end{array}
\right)}{\sqrt{\pi } k^5}\qquad \qquad J_7=\frac{32 G_{2,4}^{4,1}\left(\frac{k^2}{4}|
\begin{array}{c}
 0,1 \\
 0,0,3,\frac{7}{2}
\end{array}
\right)}{\sqrt{\pi } k^6}$$

$$J_8=\frac{64 G_{2,4}^{4,1}\left(\frac{k^2}{4}|
\begin{array}{c}
 0,1 \\
 0,0,\frac{7}{2},4
\end{array}
\right)}{\sqrt{\pi } k^7}\qquad \qquad J_9=\frac{128 G_{2,4}^{4,1}\left(\frac{k^2}{4}|
\begin{array}{c}
 0,1 \\
 0,0,4,\frac{9}{2}
\end{array}
\right)}{\sqrt{\pi } k^8}$$ and, I hope, we could conjecture that for $a >3$
$$\color{blue}{J_a=\frac {2^{a-2}}{k^{a-1}\,\sqrt \pi}G_{2,4}^{4,1}\left(\frac{k^2}{4}|
\begin{array}{c}
 0,1 \\
 0,0,\frac{k-1}2,\frac{k}{2}
\end{array}
\right)}$$


问:
Recursive sequence is increasing
I have to show, that the sequence $g(n+1) = \frac{2ag(n)}{a+g(n)^2}$
is increasing for $a>0$ and $g(0)>0$. I know that it holds for $g(n)^2<a$
since $2a/(a+g(n)^2)>1$. But for $g(n)^2>a$ the opposite seems to be true.

I assume my professor forgot some initial restrictions but maybe I am overseeing something. I’d appreciate your thoughts on the matter.

答:
Rewrite the recursion as

$${1\over g(n+1)}={a+g(n)^2\over2ag(n)}={1\over2}\left({1\over g(n)}+{g(n)\over a} \right)$$

and now let $g(n)=1/h(n)$ and $a=1/A$, so that the recursion is

$$h(n+1)={1\over2}\left(h(n)+{A\over h(n)}\right)$$

and

$$h(n)-h(n+1)={1\over2}\left(h(n)-{A\over h(n)}\right)$$

To show that $g(1)\le g(2)\le g(3)\le\cdots$, we need to show $h(1)\ge h(2)\ge h(3)\ge\cdots$, i.e., $h(n)-h(n+1)\ge0$. But, by AGM, we have, for $n\ge0$,

$$h(n+1)={1\over2}\left(h(n)+{A\over h(n)}\right)\ge\sqrt{h(n)\cdot{A\over h(n)}}=\sqrt A$$

and therefore

$$h(n)-h(n+1)={1\over2}\left(h(n)-{A\over h(n)}\right)\ge{1\over2}\left(\sqrt A-{A\over\sqrt A}\right)=0$$

Remark: As has been noted, the increasing nature of $g(n)$ need not kick in at $n=0$. For example, if $a=1$ and $g(0)=2$, then $g(1)=4/5\lt g(0)$.

Title: Recursive sequence is increasing


问:
Finding the tangent vector to a surface in 3D space
Consider the surface $:$ $()=−ℎ(,)=0$, where $ℎ(,)$ is an arbitrary single-valued continuous and differentiable function of  and

For the specific function $h(x,y)=\frac{1}{1+x^2+y^2}$ we make a sketch in the plane $y=0$

where i have drawn one of the two distinct vectors which are locally tangent to $$ at each point $(,)=++ℎ(,)$ lying on

I was wondering where the other tangent vector be $t_y$

Would it coming out (or in) of the page or does it exists since y=0

答:
The gradient at any point of the surface is

$$(-h_x,-h_y,1)$$

(partial derivatives) and any vector orthogonal to it is tangent to the surface.

E.g. $(1,0,h_x)$ and $(h_y,-h_x,0)$. Notice that for $y=0$, $h_y=0$.


问:
Prove that a Lipschitz function is continuous
Define $f:[a.b] \rightarrow \mathbb{R}$ as Lipschitz

By definition, then, $\exists M>0, M \in \mathbb{R}:|f(x)-f(y)\leq M|x-y|$  $ $ $\forall x,y\in[a,b]$

Rearranging this, we have $\frac{|f(x)-f(y)|}{x-y} \leq M$, positing that the slope must be finite on the interval $[a,b]$

I understand conceptually that Lipschitz functions must be continuous but I'm having trouble showing it.

Should I go about this by showing assuming the negation of Lipschitz and showing that for some $\epsilon,  \exists \delta>0: |f(x)-L|>\epsilon$ ?

答:
We first show that Lipshitz continuity implies uniform continuity, from which it follows that Lipshitz continuity implies continuity, because uniform continuity implies continuity.

Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces and suppose $f:X\to Y$ is Lipshitz continuous. Then exists a $K\geq 0$ such that for all $x,y\in X$, we have $d_Y(f(x),f(y))\leq K\,d_X(x,y)$. Fix $\varepsilon>0$. Choose $\delta=\frac{\varepsilon}{K}$. Then if $v,u\in X$ satisfying $d_X(v,u)<\delta$, we have $$d_Y(f(v),f(u))\leq K\,d_X(v,u)<K\delta=K\frac{\varepsilon}{K}=\varepsilon$$

If $K=0$, then choose $\delta=\varepsilon$.


问:
Orthogonal Linear Transformation of Subspace
given that U is an orthogonal n by n matrix, and linear transformation T map from $R^n$ to $R^n$ be $T(x) = Ux$. Let W be a subspace of $R^n$ such that $T(W)  \subseteq   W$. Prove $T(W) = W$ and $T(W^\perp) = W^\perp$.
Here is my attempt:

For the first question, I constructed an orthogonal basis U for W, and showed that $T(U)$ is orthogonal and linearly independent with dimension dim(W).

Any hint will be appreciated. Thanks in advance.

答:
Just curious- For the first part is it sufficient to prove T is injective to say T(W)=W. Since T being injective implies kernel is a nullity, which implies T is surjective and hence a bijection?

Thanks!


问:
The second Friedrichs' inequalities?
In paper On the Validity of Friedrichs' Inequalities，$\Omega$ is a bounded convex domain of $\mathbb{R}^d$, $d=2,3$. Then
$$
\tag{1}\qquad \|\mathbf{u}\|_{1,\Omega} \le C\Big(\|\nabla\cdot\mathbf{u}\|_{0,\Omega} +\|\nabla\times\mathbf{u}\|_{0,\Omega}\Big)
$$
for all $\mathbf{u}\in\mathbf{H}_0(\operatorname{div};\Omega) \cap \mathbf{H}(\operatorname{curl};\Omega)$ or $\mathbf{u}\in\mathbf{H}(\operatorname{div};\Omega) \cap \mathbf{H}_0(\operatorname{curl};\Omega)$.

If $\mathbf{u}\in\mathbf{H}(\operatorname{div};\Omega) \cap \mathbf{H}(\operatorname{curl};\Omega)$ with mixed boundary conditions:
$$\mathbf{n}\cdot\mathbf{u} = 0 \text{ on }\Gamma_1, \quad \text{ and }\quad \mathbf{n} \times \mathbf{u} = \mathbf{0} \text{ on }\Gamma_2, $$
where $\Gamma_1,\Gamma_2\neq\emptyset$, $\Gamma_1\cap \Gamma_2 = \emptyset$, and $\overline{\Gamma}_1\cup \overline{\Gamma}_2 = \partial \Omega$. Does the inequality $(1)$ hold?

答:
I know this is an old question, but I think it's worth to improve the previous answer with some remarks for the benefit of future readers.

The space $\{ \mathbf u \in \mathbf{H}(\operatorname{\textbf{curl}}; \Omega) \cap H(\operatorname{\textbf{div}};\Omega) : \mathbf
u \cdot \mathbf n |_{\Gamma_1} = 0, \ \mathbf u \times \mathbf n |_{\Gamma_2} = \mathbf 0 \}$ is in fact compactly embedded into $\mathbf L^2(\Omega)$ if $\Omega$ is merely weak Lipschitz (see reference $[1]$). On the other hand, it is not embedded into $\mathbf{H}^1(\Omega)$.

One can thus obtain a Poincaré-Friedrichs inequality similar to $(1)$ even with mixed boundary conditions, but with the $\| \cdot \|_{0,\Omega}$ norm of the LHS and imposing an orthogonality condition with respect to a space of so-called "Dirichlet-Neumann" harmonic fields - vectors with zero curl and divergence and suitable null normal/tangential traces -. You may look at [$1$, Thm. 5.1].  Interestingly enough, the additional requirement prevents the above counterexample.

Reference:

$[1]$ Sebastian Bauer, Dirk Pauly and Michael Schomburg. "The Maxwell compactness property in bounded weak Lipschitz domains with mixed boundary conditions." SIAM Journal on Mathematical Analysis 48.4 (2016): 2912-2943.

Title: The second Friedrichs' inequalities?


问:
Generalization of Radon-Nikodym Theorem
I'm wondering if the following statement holds:

Let $(X, \mathfrak{B})$ be a measurable space, and $\mu, \nu$ be complex measures. Assume $\nu << \mu$. Is there a complex valued measurable function $f \colon X\to \mathbb{C}$, which satisfies $$\forall B\in \mathfrak{B}; \int_B f d\mu = \nu(B)$$

If so, I want to call $f = \frac{d\nu}{d\mu}$ the Radon-Nikodym derivative. My textbook says it holds (only) when $\mu$ is $\sigma$-finite positive measure. Any help will be appriciated.

P.S. Here complex measure means the function $\mu \colon \mathfrak{A}\to \mathbb{C}$ which satisfies $\mu(\varnothing)=0$ and $\sigma$-additivity in  $\mathbb{C}$ (with "Euclidean topological convergence").

答:
Recall that $|\mu|$, the total variation of $\mu$ is a finite measure, $\mu << |\mu|$ and $$|\frac{d\mu}{d|\mu|}| = 1 \text{(i.e. invertible in $\mathbb{C}$, $|\mu|$-a.e.)}$$ $$\int_A f d\mu = \int_A f \frac{d\mu}{d|\mu|}d|\mu|$$
holds when $f$ is $|\mu|$-integrable.

Now let $$g=\frac{d\nu}{d|\mu|}(\frac{d\mu}{d|\mu|})^{-1}$$ and calculate the integration of $g$. We see that $$\int_A gd\mu = \int_A \frac{d\nu}{d|\mu|}(\frac{d\mu}{d|\mu|})^{-1}\frac{d\mu}{d|\mu|}d|\mu| = \nu(A)$$ so $g$ is the Radon-Nikodym derivative.

Note that $g$ is unique up to $|\mu|$-a.e., since if $g$ and $h$ are both Radon-Nikodym derivative, $\int_A (g-f)\frac{d\mu}{d|\mu|}d|\mu| = 0$ for all $A$, then $(g-f)\frac{d\mu}{d|\mu|} = 0$, so $g-f=0$, $|\mu|$-a.e. .

Title: Generalization of Radon-Nikodym Theorem


问:
Is $P(G,x)= (x-1)^n+(-1)^n(x-1)$ a sufficient condition to G is an n-cycle?
The way to prove the chromatic polynomial of a n-cycle graph G is $P(G,x)= (x-1)^n+(-1)^n(x-1)$ is well known. While, I find it not so easy to prove another direction.

The connectivity of G is the first problem I met. I try to solve it by factoring. $P(G,x)=(x-1)\left[(x-1)^n+(-1)^n\right]$. Intuitively, It might work. However, I have no idea to explain it in plain words.

Then, I don't know what to do to solve the rest problem as well. The only clue I can get is when $x=3, P(G,x)>0 \Rightarrow 3=\chi(G)\ge\omega(G)$.

Any hint or idea would be helpful. Thanks a lot!

答:
QAQ I figure out a possible solution.

Above theorems can be easily proved by Birkhoff-Lewis reduction thereom $c_G(k)=c_{G-e}(k) - c_{G/e}(k)$.

Based on above thereoms, this question would be much easier! o(≧v≦)o

Firstly, $G$ is connected. This can be induct from #3, since if $G$ is disconnected $c_G(k)$ would equal to the product of its component and the coefficient of x would be $0$.

Then, $G$ can not be tree because $P(T,x)=x(x-1)^{n-1}$ for a tree T. Therefore, there should be at least one cycle in G. By #5 of above theorem, there is $n$ edges in $G$, so there can only be one cycle.

At last, to prove $G$ is the cycle itself, instead of a cycle attached with some edges. Supposing G contain the cycle $C$ and $|C|=k\le n$. Then $P(G,x)$ should be $[(k-1)^k+(-1)^k(x-1)](x-1)^{n-k}$ which equals to $(x-1)^n+(-1)^n(x-1)$ only when $k=n$.

Q.E.D.  (o゜▽゜)o☆[BINGO!]

update: fix tons of grammatical error(I'm not a native English speaker...)


答:
Ah, I think I am late. May this gives you more geometric insight.

You can take $\angle BAF$ right-angle and $AF$ meet $BE$ at $F$ then $ABCF$ share same circle since $\angle CAF=\angle CBF$, that makes $\angle BCF$ also a right-angle, and $\angle I=60$, which indicates similar proportional

$$\frac{AC}{BF}=\frac{AI}{FI}=\frac1{2}$$

thus we spot $H$ the middle point of $BF$ for $\angle BAF$ and $\angle BCF$ both right-angle, $H$ is the circumcenter of four point system $ABCF$, moreover $\triangle AHC$ is equilateral, in which $CD$ is one of its perpendicular line. And $AE$ is actually the perpendicular line of $\triangle ABC$.

And now you can solve any angle in the graph with almost no calculation.

Title: Point $M$ lies inside $\triangle ABC$, $\angle MAC = 10^\circ$ and $\angle MCA = 30^\circ$. Find $(180^\circ - \angle BMC)$

Body:
In $\Delta ABC, \angle CAB = 30^\circ$ and $\angle ABC = 80^\circ$. Point $M$ lies inside the triangle such that $\angle MAC = 10^\circ$ and $\angle MCA = 30^\circ$. Find $(180^\circ - \angle BMC)$.

What I Tried: Here is a picture to keep track of the angle-chasing .

The red angle is $30^\circ$ , the green one is $10^\circ$ , the purple one is $20^\circ$ . Now I extend $MC$ to $AB$ such that it meets at $K$. First I noticed that $AK = KC$ but that didn't seem to help. Also the $2$ yellow ones are equal to $40^\circ$ each, the brown one is equal to $80^\circ$ . The light green angle is $60^\circ$ . We have to find the blue angle.

After all these information, it seems like I am still missing something, because from here you actually cannot deduce the value of the blue angle. Can anyone help?

Thank You.


问:
How many sentences are there (up to tautologies) in a finite langugage $L = \left \{ P_1,...,P_n \right \}$
I know that there are $2^n$ possible truth evaluations for a langauge $L$ of cardinality $n$

However, how can I count how many sentences are there constructed by the languge up to tautologies?

Is the answer $2^{(2^n)}$?

答:
If you mean by up to tautologies the fact that you identify two sentences that have the same truth table, then the number you're looking for is indeed $2^{(2^n)}$.

This is because a truth table over $L = \left \{ P_1,…,P_n \right \}$ is just a map from $\{0,1\}^n$ to $\{0,1\}$. As $\{0,1\}^n$ has $2^n$ elements, the desired result is effectively $2^{(2^n)}$.


问:
question about sets
I have this as a beggining to a question:

$A\subseteq Z^2$

$$
A = \left \langle \left ( 1,7 \right );\left ( 7,2 \right );(2,3) \right \rangle = \left \{  \right.n_{1}*(1,7)+n_{2}*(7,2)+n_{3}*(2,3)\left.  \right \},\left \{  \right.n_{i}\in Z\left.  \right \}
$$

I understand the first part, $A$ is a subset of $Z^2$, which means $A$ is a set of ordered pairs in $Z^2$.

I don't understand the second part. how did the $n1,n2,n3$ get there and how is $A$ suddenly a sum of numbers?

Thanks in advance,

Yaron.

答:
The second part is the definition of $A$. $A$ is defined as the subgroup of $\mathbb{Z}^2$ generated by $(1,7),(7,2),(2,3)$, meaning that $A$ is the set of ordered pairs of integers that can be expressed as a linear combination of those three ordered couples with integers coefficients.

You can prove that $A$ is the smallest subgroup of $\mathbb{Z}^2$ that contains $(1,7),(7,2)$ and $(2,3)$, i.e. if $B$ is a subgroup of $\mathbb{Z}^2$ and $B$ contains $(1,7),(7,2)$ and $(2,3)$, then $B$ also contains $A$.

Title: question about sets


问:
Using induction to prove $(ab)^n=a^n*b^n, a,b \neq 0 a,b \in \mathbb R, n \in \mathbb Z$
I am trying to prove that $(ab)^n=a^n*b^n, a,b \neq 0, a,b \in \mathbb R, n \in \mathbb Z$

I am not sure if we can use induction here.

If yes, we can see three options:

1) $n=0$, so $(ab)^0=1=1*1=a^0*b^0$

2) $n>0$, so for $n=1$ we have $(ab)^1=ab=a^1*b^1$

for $n+1$: $(ab)^{n+1}=(ab)^n*ab=a^n*a*b^n*b=a^{n+1} * b^{n+1}$

3) $n<0$, so here is the question. Am I allowed to use induction here?

If it is true for $-(n)$ and $-(n+1)$, does it mean that we prove whole thing using induction?

$(ab)^{-(n+1)}=(ab)^{-n}*(ab)^{-1}$=$(a^{-n}*b^{-n})/ab$=$a^{-n}*a^{-1}*b^{-n}*b^{-1}=a^{-(n+1)}*b^{-(n+1)}$

答:
Yes, you can use Mathematical Induction here. Take substitution like $a_1 = a^{-1}$ and $b_1 = b^{-1}$ if you wish. Then everything takes the form of your previous work. Note that $a,b \in \mathbb R\setminus \{0\} \implies a_1,b_1 \in \mathbb R\setminus \{0\}$.

For the Commented Question:

We need to prove $(a\cdot b)^p=a^p\cdot b^p$, where $p\in\mathbb Q$. Let $p = n/m$.

Then, we have $a^{n/m} = (\sqrt[m]{a})^n$ and $b^{n/m} = (\sqrt[m]{b})^n$ by definition.
$$(a\cdot b)^{n/m}=(\sqrt[m]a \cdot\sqrt[m]b)^n$$
By taking $c = \sqrt[m]a$ and $d=\sqrt[m]b$, we see that, again, it comes down to your previous work.


问:
Find all positive integers $m$ with the property that every unit $a$ in $\mathbb{Z}_m$ satisfies $a^2 = 1$.
I am having troubles solving this exercise. I've tried using Euler's Theorem but I'm not too experienced with it. It also that, until $m=8$, the only integers that satisfy this are powers of $2$. Any help?

答:
Hint If $d|m$ then every unit in $\mathbb Z_d$ also satisfies $a^2=1$.

Hint 2 If $p$ is prime and $a^2 =1 \pmod{p}$ then $a = \pm 1 \pmod{p}$


问:
Show that $\int_{1}^{x} \frac{\sin t}{t}dt -x +1 <0,\: x>1$?
I made the following attempt but it did not feel convincing. I also learned afterward that differentiation is not allowed across inequalities.

The original expression on the LHS is equivalent to

$\frac{d}{dx}\int_{1}^{x}\left(\frac{\sin(t)}{t}-1\right)\:dt = \frac{\sin(x)}{x}-1$
So the original inequality now says $\sin(x)<x, \: x>1$

But this is a trivial inequality as $\sin(x)$ is bounded by $[-1,1]$.
Therefore, the inequality holds for $x>1$.

Body:
From $$\frac{\sin t}{t}< 1;\;t\ge1 $$
as $x>1$ follows
$$\int_{1}^{x} \frac{\sin t}{t}dt < \int_1^x 1\,dx=x-1$$
thus
$$\int_{1}^{x} \frac{\sin t}{t}dt < x-1\to \int_{1}^{x} \frac{\sin t}{t}dt-x+1< 0$$


答:
You already made an important, though incomplete  observation.

For non-powers of $2$, let $p$ be an odd prime dividing $m$, say $k\ge1$ and $m=p^kr$ with $p\nmid r$.
Note that $2$ is a unit $\bmod p^k$ and $1$ is a unit $\bmod r$. By the Chinese Remainder Theorem, there exists $a$ with  $a\equiv 2\pmod{p^k}$ and $a\equiv 1\pmod r$. Then $a$ is a unit $\bmod m$ and $a^2\equiv 4\not\equiv 1\pmod {p^k}$, so $a^2\not\equiv 1\pmod m$ -- except perhaps when $p^k=3$.

We conclude that the only possible $m$ with the desired poperty are of the form $m=2^k$ or $m=3\cdot 2^k$ with $k\ge 0$.
But we also cannot have $m=2^k$ with  $k\ge 4$ because in that case $3$ is a unit $\bmod {2^k}$ but its square is $9\not\equiv 1\pmod m$.
By another invocation of the Chinese Remainder Theoram or by explicitly looking at the unit(!) $a=5\bmod 3\cdot 2^4$ we see that $m=3\cdot 2^4$ also requires $m\le 3$. Now check the remaining cases $m=1,2,3,4,8,12,24$ carefully again: for which of these does the property hold?

Title: Find all positive integers $m$ with the property that every unit $a$ in $\mathbb{Z}_m$ satisfies $a^2 = 1$.

Body: I am having troubles solving this exercise. I've tried using Euler's Theorem but I'm not too experienced with it. It also that, until $m=8$, the only integers that satisfy this are powers of $2$. Any help?


问:
Probability of girls and boys around a table
Marco and a group of seven friends, four girls and four boys in all, meet for a dinner. The dinner table is round and the seats are numbered. The group decides to assign the seats at random, each drawing their own seat number. Calculate the following probabilities:
a) PA: that Marco has two boys on either side;
b) PB: that Marco has a boy and a girl on either side;
c) PC: that Marco has two girls on either side.

For the first probability I obtained $ P_A = \frac{ \binom{3}{1} \frac{1}{3} \binom{2}{1} }{ \binom {4}{3} } = 0,5 $
For the second probability I obtained $ P_B = \frac{ \binom{4}{1} \frac{1}{4} \binom{3}{1} }{ \binom {15}{3} } = 0,00659 $
For the third probability I obtained $ P_c = 0,00659 $

But I don’t know if my result are correct.. thank you !

答:
There are
$$\binom{7}{2}$$
ways to select two people to sit next to Marco (without considering who sits to his left and who sits to his right).  Since Marco is one of the four boys, three of the seven friends who could sit next to him are boys.  Thus, there are
$$\binom{3}{2}$$
ways to select two boys to sit next to Marco (again, without considering who sits to his left and who sits to his right).  Hence, the probability that Marco is seated between two boys is
$$P_A = \frac{\dbinom{3}{2}}{\dbinom{7}{2}}$$
If you prefer to think about order, note that there are seven ways to choose the person who sits to Marco's left and six ways to choose the person who sits to Marco's right.  For the favorable cases, there are three ways to choose the boy who sits to Marco's left and two ways to choose the boy who sits to his right.  Thus,
$$P_A = \frac{3 \cdot 2}{7 \cdot 6}$$
which agrees with the answer obtained above.

For part (b), choose a boy and a girl who will sit next to Marco.

For part (c), choose two girls who will sit next to Marco.


问:
Insurance Policy Probability Question

Hi, I was wondering if anyone could help me consolidate my answer/workings for this question.

So I have initially just stated that there are 100 people in each policy group just to make the percentages a nice number of people.

I know from the figures that 40 women have policy A and 26 women have policy B making a total of 66 women having a policy in general.

Then because we are choosing a policy from the women selection we know that Policy B will have a 26/66=0.394 chance of being selected and Policy A will have a 40/66= 0.606.

Is my method correct or am I going wrong here. Thanks in advance.

答:
I know from the figures that 40 women have policy A and 26 women have policy B

That is wrong! You know from the figure that 40% of the policyholders are female, thus if you have 100 guys total only 20 are female and only $20\cdot0.40=8$ are female AND policy A

With similar reasoning, starting with 1000 policyholders total, you have that 288 are females and 80 are female with policy A. Thus the requested probability is

$$\frac{80}{288}\approx 27.78\%$$

In these situation a table like the following is very useful to represent the problem

(I used 1000 guys to avoid "non-integer" guys)

Title: Insurance Policy Probability Question


问:
A question on the Substitution for predicate letters theorem as in Kleene Introduction to Metamathematics (1952)
I'm self studying Kleene's IM (1952). On page 149 Kleene defines a name form thus:

"Let $x_1, ... x_n$ be distinct variables, and $A(x_1, ... x_n)$ a formula. When we are interpreting $A(x_1, ... x_n)$ by a predicate, or performing formal operations with it which are in keeping with an interpretation by a predicate (even though the interpretation is not involved in the formal operations), we call $A(x_1, ... x_n)$ a name form in $x_1, ... x_n$ as the name form variables ..."

My first question here is which are the formal operations which are in keeping with an interpretation by a predicate and which are not in keeping with an interpretation by a predicate? If, as he says, the interpretation is not involved in the formal operations, why should there even be formal operations in or out of keeping with with the interpretation?

On page 157/8 is goes on with the following definitions in preparation for the SUBSTITUTION FOR PREDICATE LETTERS Theorem (Theorem 15 on page 159):

"By an occurrence of a predicate letter $P(a_1, ... a_n)$ with attached variables in a predicate letter formula $E$ we shall mean a (consecutive) part of E of the form $P(t_1, ... t_n)$ where $t_1, ... t_n$ are terms. A predicate letter formula $E$ is said to be a predicate letter formula in the distinct predicate letters

(1) $P_1(a_1, ... a_{n_1})$, ..., $P_m(a_1, ... a_{n_m})$, where   $n_1, ... n_m \geq 0; m \geq 1$

if no predicate letter other than (1) occur in $E$.

The substitution of formulas

(2) $A_1(a_1, ... a_{n_1})$, ..., $A_m(a_1, ... a_{n_m})$

considered as name forms in the respective variables shown, for the predicate letters (1) in $E$ (with result $E^*$) shall consist in replacing, simultaneously for each $j(j=1, ..., m)$, each occurrence $P_j(t_1, ... a_{t_j})$ of $P_j(a_1, ... a_{n_j})$ in $E$ by $A_j(t_1, ... t_{n_j})$."

My question here is, why the expressions in (2) have to be name forms and not just formulas?

Thanks for your help.

答:
See Stephen Cole Kleene, Mathematical logic (1967, Dover reprint), page 77:

The issue is about the proper way to use symbols to represent a predicate.

The predicate (also called propositional function) named by "$x < y$" can also be named simply "$<$".

The issue is similar to that concerning e.g. the $\sin$ function, that usually we write as $\sin(x)$.

A notation for predicate using variables we may call a name form for the predicate.

See page 78: the "official" syntax is made of ions $P(-,-)$ for e.g. binary predicates (with two argument-places) and (individual) variables: $x_1,x_2,x_3$.

Thus, an atom (or prime formula) $P(x_1,x_2)$ is a ion "applied" to variable.

IMO, with a name form $P(x,y)$ Kleene means a "generic" name for the ion $P(-,-)$, understanding that it is different from the atom.

See page 78:

We call $P(x,y,z), P(y,z,x)$, etc. different name forms for the same 3-place prime predicate expression or ion $P(-,-,-)$.

And see page 98 for the Substitution theorem:

Say $E$ is a formula which contains only the distinct ions having the respective name forms

$$\text {(1)  } P_1(w_1,\ldots,w_{P_1}), \ldots P_N(w_1,\ldots,w_{P_1})$$

[...] The substitution of formulas

$$\text {(2)  } A_1(w_1,\ldots,w_{P_1}), \ldots A_N(w_1,\ldots,w_{P_1})$$

for (1) in $E$ with result $E^*$ [...]


问:
About parallel time computation
I am studying a paper where it is mentioned that Newton iteration may be used to compute the inverse of $n \times n$, well- conditioned matrix in parallel time $o(\log^2n)$ and that this computation is processor efficient. I want to understand meaning of this term. What is parallel time computation and how can this be useful for any iteration method?

Here is the link

Thank you very much.

答:
What is meant here is more in the realm of computer science than pure mathematics, so if you don't have a strong computer background it may help to read the Wiki entry on Parallel Computing. Basically, a parallel algorithm is used to let multiple processor cores act at the same time rather than a single core acting alone. The efficiency of a parallel algorithm is based on a measurement of parallel time, which is defined as the time it takes for the parallel algorithm to run. To be considered efficient, the algorithm must satisfy: $$T_{sequential} \approx T_{parallel} * N_{cores}$$
Where $T_{sequential}$ is the run time of best known serial algorithm.

The efficiency of the parallel algorithm you referenced is then given in terms of poly-logarithmic time: $O (\log^c(n))$ where c is a constant and n is the size of the data set. See the Wiki article on time complexity for a more detailed explanation of poly-logarithmic (and other) time scales.

It is also interesting to note that iterative algorithms such as Newton's method are usually considered inherently serial, i.e. quite difficult to (efficiently) parallelize.

Title: About parallel time computation


问:
About the cardinality of a set
I'm trying to understand the following exercise about set theory:

We have:     $\Omega= \{a,b,c,d\}$

and we are supposed to figure out the cardinality of the following set:

$\textbf{{Y$\subseteq$ $\Omega$ | Y $\cap$ {a,b} = Y \ {c}}}$

According to my tutor the cardinality is $8$, because this set is a power set using the formula of the cardinality for a power set, $2^k$, $2^3$ we get $8$.

I think that I more or less understand what a power set is, in general, and how to compute the cardinality of a given power set.

But I still do not know how I can come to that conclusion in this exercise, or the answer for the cardinality of the set above. Or why the set above has anything to do with a power set.

答:
The set in question is the set of subsets of $\Omega$ whose intersection with $\{a,b\}$ is the same as removing $c$ from it.

Notice that all subsets of $\{a,b,c\}$ are elements of this set and no other subset can be since that would imply $d\in Y$ but $d\notin Y\cap\{a,b\}$ and $d\in Y\setminus\{c\}$ and these two are equal, so...

Title: About the cardinality of a set