问:
$\begin{align}\color{#c00}a&\mid b\\\color{#c00}c&\mid d\end{align}\Rightarrow\ \color{#c00}{ac}\mid bd$ $\ \ \bf\small [Divisibility\ Product\ Rule]$
I just need to check the reasoning in my proof is correct, I think it is valid although I'm not totally convinced because I can't follow the logic; does proving that $x$ is an integer prove that $ac|bd$?

Theorem: Let $a$, $b$, $c$, $d$ be integers. If $a|b$ and $c|d$, then $ac|bd$.

Proof: $aj=b$ and $ck=d$ for integers $j$, $k$. Then, $ac|bd$ implies $acx=ajck$ and thus $x=jk$ for some $x$. Since the product $jk$ is an integer, $x$ is an integer and thus $acx=ajck$ and thus $ac|bd$.

答:
If you're having trouble with these kinds of proofs, it's often best to start by writing down what you're given on the left side of the paper and where you're trying to get on the right side.

```

a|b                                                ac | bd
c|d

```

Next fill in what you can from definitions:

```

b = aj                                              bd = ack
d = ci
```

Now use inferences to play with the things on the left and aim for what you have on the right.  Each time, you need to ensure that what you have implies what you write next.

You can also work from right to left, but you have to be careful.  Here you need to make sure the steps are "if and only if" so that you can go backwards from left to right if you eventually meet in the middle of the paper.

Knowing when it's safe to make a particular step either left to right or right to left comes from being extremely careful and from knowing exactly what's going on.  There are puzzles from false proofs where they test you on this.  A common example is

```
a=b
(a-b)c = (a-b)d                   so        c = d
```

Looks fine, right? But given that a=b we should have noticed that (a-b)=0 and you can't cancel zero from both sides.

Anyway, getting good at this stuff is called "mathematical maturity" and it, like most kinds of maturity, takes time.

Body:
If you're having trouble with these kinds of proofs, it's often best to start by writing down what you're given on the left side of the paper and where you're trying to get on the right side.

a|b                                                ac | bd
c|d

Next fill in what you can from definitions:

b = aj                                              bd = ack
d = ci

Now use inferences to play with the things on the left and aim for what you have on the right.  Each time, you need to ensure that what you have implies what you write next.

You can also work from right to left, but you have to be careful.  Here you need to make sure the steps are "if and only if" so that you can go backwards from left to right if you eventually meet in the middle of the paper.

Knowing when it's safe to make a particular step either left to right or right to left comes from being extremely careful and from knowing exactly what's going on.  There are puzzles from false proofs where they test you on this.  A common example is

a=b
(a-b)c = (a-b)d                   so        c = d

Looks fine, right? But given that a=b we should have noticed that (a-b)=0 and you can't cancel zero from both sides.

Anyway, getting good at this stuff is called "mathematical maturity" and it, like most kinds of maturity, takes time.


问:
Find RV distribution and prove its convergence to zero
Let $X_n$ be a sequence of independent and uniformly distributed over $[0; 1]$ random variables. The task is to find distribution of the following random variable: $m_n=\min\{X_1, X_2, \ldots, X_n\}$ and also proof that it converges almost surely to zero, in other words: $$\Pr(\lim_{n \rightarrow \infty}m_n=0)=1$$

答:
$$
\min\{X_1,\ldots,X_n\} \ge x \text{ if and only if } \big[X_1\ge x\ \& \cdots \ \&\ X_n\ge x\big].
$$
If you can find the probability of that, then that gives you the distribution of the minimum.

ContentLicense: CC BY-SA 4.0


问:
Gradient´s direction of constraints (Lagrange multiplier)
I am trying to understand Lagrange´s multiplier method. I have one question. I´ ve seen sketches where the gradient of function f and constraint g point to the same direction at a possible extreme value. However, there are also sketches where g´s gradient points to the opposite direction of f´s gradient at the possible extreme value. Is there any meaningful interpretation of the direction of the constraint´s gradient?</p>

答: The basic idea of this method is that in the point $p$ of possible constrained extremum a derivative of $f$ along any curve passing through $p$ is zero at $p$. Hence the gradient of $f$ belongs to the orthogonal complement of a tangent space to $g$ in $p$. Hence it can be represented as a linear combination of gradients of functions defining the constraints.


问:
Linear independence and pairwise angles
Given $p+1$ vectors with pairwise angles $> \pi/2$ in $\mathbb{R}^n$ then every $p$ of them are linearly independent. How to prove this fact?

The motivation for this question is the idea of extending the notion of "linear independence" for points on tangent cones on some metric spaces (Alexandrov spaces), on which we do not have a vector space structure, but have a good notion of angle.

答:
Suppose this is true for p. Let's prove this for p+1

$\sum_\alpha\alpha v_\alpha=0$. Separate nonnegative $\alpha$'s from negative ones
$\sum_{\alpha^+} \alpha^+ v_{\alpha^+}=\sum_{\alpha^-} -\alpha^- v_{\alpha^-}$

$(\sum \alpha^+ v_{\alpha^+})^2=\sum_{\alpha^+, \alpha^-}-\alpha^+\alpha^-\langle v_{\alpha^-}v_{\alpha^+}\rangle\le 0$ and $\ge 0$ hence

$(\sum \alpha^+ v_{\alpha^+})^2=0$ and all $\alpha^+=0$ since otherwise we would have $n\le p$ linearly dependent vectors, contradicting the induction assumption.
The scalar square of $(\sum \alpha^- v_{\alpha^-})$ gives the same for $\alpha_-$'s

ContentLicense: CC BY-SA 3.0

Title: Linear independence and pairwise angles


问:
If $f$ is surjective from $A$ to $B$ and $A$ is well-orderable, then $B$ is well-orderable.
I think the logic is correct, but I would greatly appreciate any pointers. In particular, the way in which I defined my function relies on a cumbersome FOL statement, which I would like to avoid (as I have seen recommendations from more experienced individuals to reduce the formalisms).

Given:

$f$ is a surjective function from $A$ to $B$. $\quad f:A\xrightarrow[\text{onto}]{} B$

$A$ is well-orderable. That means there is a relation $R$ that well-orders $A$.

Claim to Prove:

Show that $B$ is well-orderable (i.e. show that there exists a relation that well-orders $B$)

Definition in my book: $R$ well-orders $A$ iff $R$ totally orders $A$ strictly and $R$ is well-founded on $A$.

Importantly, one can demonstrate that all subsets of a well-ordered set are also well-ordered.

Because $f$ is surjective and $A$ is well-ordered by $R$: $\forall b \in B\ \    \exists! a' \in A \Big (f(a')=b \ \ \land \ \ \forall a \in A\  [f(a)=b \rightarrow a \not R a'] \Big)$.

That is to say: for a given $b$, of the element(s) from $A$ that map to $b$, there is always one minimum element $a'$ that is taken to $b$ through $f$ .

Let $g(b)$ map to this unique $a'$.

So we have:

$g: B \to A$ where $b \mapsto a'$

$g$ can be shown to be injective. Let $b_1 \neq b_2$. By definition, $g(b_1)=a_1'$ and $g(b_2)=a_2'$, where $f(a_1')=b_1$ and $f(a_2')=b_2.$ If $g(b_1)=g(b_2)$, then we see that $f$ maps the same element in $A$ to two different (non-equal) elements in $B$. $f$ is a function so this is a contradiction.

To ensure that $g$ is surjective, the codomain can be modified so that:

$g: B \xrightarrow[\text{onto}]{1-1}g[B]$, where $g[B] \subseteq A$. Thus, $g$ is a bijection.

Define a new relation $R'$ as follows:

$\forall b_1, b_2 \in B\ [b_1 \ R' \ b_2 \leftrightarrow g(b_1)\ R \ g(b_2)]$

$R'$ therefore inherits all of the "well-orderedness" of $g[B]$, which is a subset of the well-ordered set $A$.

$R'$ well-orders $B$ $\quad \square$

答:
I think your proof is quite complicated, much more so than necessary.

Defn: A set $A$ is well-ordered (under $<$) if every nonempty subset of $A$ has a least element with respect to $<$.

Thm: If $A$ is well-ordered and $f:A\rightarrow B$ is surjective, then $B$ has a well-ordering.

Proof: We can demonstrate such a well-ordering directly. For each $b \in B$, $f^{-1}(b)$ is nonempty by the surjectivity of $f$. Define $a_b$ to be the smallest element of $f^{-1}(b)$, which is guaranteed to exist by the well-ordering of $A$.

For $b_1$ and $b_2$ in $B$, define $b_1 \prec b_2$ iff $a_{b_1} < a_{b_2}$. We claim that $\prec$ is a well-ordering of $B$:

Suppose $X$ is any nonempty subset of $B$, and let $a_X = \{ a_b \mid b \in X \}$. Since $A$ is well-ordered, $a_X$ has a least element, call it $a_\ell$. Then by definition of $\prec$, we see that $b_\ell = f(a_\ell)$ is the least element of $X$. $\Box$


答: No, there is not. Suppose that you have two constraints $g_1$ and $g_2$, with $g_2=-g_1$. You want to find the local extremes of the restriction of $f$ to the set $\{x\in\Bbb R^n\mid g_1(x)=c\}$, for some $c\in\Bbb R$, which is equal to the set $\{x\in\Bbb R^n\mid g_2(x)=-c\}$. If $p$ is such that $\nabla f(p)=\lambda\nabla g_1(p)$ for some $\lambda\in\Bbb R$, then $\nabla f(p)=-\lambda\nabla g_2(p)$. So, either $\nabla f(p)$ and $\nabla g_1(p)$ have the same direction or $\nabla f(p)$ and $\nabla g_2(p)$ have the same direction. But, again, its the same set. So, having the same direction or opposite directions means nothing.

Title: Gradient´s direction of constraints (Lagrange multiplier)

Body: I am trying to understand Lagrange´s multiplier method. I have one question. I´ ve seen sketches where the gradient of function f and constraint g point to the same direction at a possible extreme value. However, there are also sketches where g´s gradient points to the opposite direction of f´s gradient at the possible extreme value. Is there any meaningful interpretation of the direction of the constraint´s gradient?</p>


问:
2D Poisson's Equation with constant source function and Dirichlet boundary conditions on rectangular boundary
I'm trying to solve the torsion problem for a rectangular cross-section of shape $2a$ by $2b$ in the x-y plane. The coordinate system is set up so that the origin is the centroid of the cross-section and so that the boundary of the rectangle is given by $-a≤x≤a$ and $-b≤y≤b$. The problem setup is illustrated here (ignore the asked question though, I understand that answer):

https://en.wikiversity.org/wiki/Introduction_to_Elasticity/Torsion_of_rectangular_sections

The governing partial differential equation is given by:

$$\Delta^2\phi(x,y)=-2\mu\theta$$
Or:
$$\frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} = -2\mu\theta$$

And is subject to the following:
$$ϕ(±a,y) = 0$$

$$ϕ(x,±b) = 0$$

$$-a≤x≤a$$

$$-b≤y≤b$$

$$a≥0$$

$$b≥0$$

i.e. $ϕ = 0$ on the boundary of the rectangular cross-section.

Unfortunately, I have not taken a class on partial differential equations and clearly need that to solve this problem. If someone could walk me through the process of solving this it would be greatly appreciated. Honestly, if anyone could even tell me anything about it it would be greatly appreciated.

What I do know is that this is a Poisson's equation in two dimensions with a constant source function and Dirichlet boundary conditions. I think I can use separation of variables, but I do not know how to choose my functions because it isn't homogeneous.

The solutions I have been able to find for this problem jump directly into Fourier series and I don't understand that jump. My general understanding of the Fourier series is that if your answer contains a trigonometric function with a constant multiplied by a variable inside it, then you can express the answer as an infinite sum of that function changing only multiples of the angle that satisfy the boundary conditions.

So, for a function $T(x,t)$:

$$T(x,t)=Csin(\sqrt \lambda x)$$

With the boundary condition:

$$T(a,t) = 0$$

We get, for the non-trivial solution $C\neq 0$:

$$sin(\sqrt \lambda a) = 0$$

$$\sqrt \lambda a= 0, \pi, 2\pi, ..., n\pi$$

$$\lambda = \frac{n^2\pi^2}{a^2}$$

Meaning:

$$T(x,t) = Csin(\frac{n\pi}{a}x)$$

Or, as a Fourier Series Representation:

$$T(x,t) = \sum_{n=1}^\infty C_n sin(\frac{n\pi}{a}x)$$

And we can find $C$ and $C_n$ using other boundary conditions. If this is wrong at all please let me know as well, but my issue is determining the actual function itself, $T(x,y)$ in the above example, $\phi(x,y)$ in my question. Thank you for all of your help.

答:
One can change variables such that without loss of generality the domain is $[0,\pi] \times [0,\pi]$ and the forcing is $1$. This is a fairly straightforward exercise.

The appropriate Fourier series for $f(x,y)=1$ with these boundary conditions is $\frac{4}{\pi^2} \sum_{m,n \text{ both odd }} \frac{4}{mn} \sin(mx) \sin(ny)$. The $\frac{4}{mn}$ and the restriction to odd $m,n$ is obtained by evaluating the integrals $\int_0^\pi \int_0^\pi \sin(mx) \sin(ny) \cdot 1 dx dy$; the $\frac{4}{\pi^2}$ is obtained by evaluating the integrals $\int_0^\pi \int_0^\pi \sin(mx)^2 \cdot \sin(ny)^2 dx dy$.

The corresponding eigenvalues of the Laplacian are $-(m^2+n^2)$, so you obtain the series solution $\frac{4}{\pi^2} \sum_{m,n \text{ both odd}} -\frac{4}{mn(m^2+n^2)} \sin(mx) \sin(ny)$.

Note that there is a serious difficulty implementing this formula in practice, because of the fact that the forcing does not actually satisfy the boundary conditions.


问:
Asymptotic analysis - if f(n) = Ω(g(n)), how to prove ln(f(n)) = Ω(ln(g(n)))?
Is the following statement true, if so, how can I prove it?

```<code>if f(n) = Ω(g(n)), is also true that ln(f(n)) = Ω(ln(g(n)))?
</code>```

Since f(n) >= c*g(n), I divided the problem into two branches:

If f(n) = g(n), it's also true that ln(x) = ln(x) and the statement it's true.

If f(n) > g(n), then the value for ln is greater, if the argument is greater:

```<code>x > y; ln(x) > ln(y)
</code>```

Is there something wrong with my reasoning?

答:
This is false in general without some further conditions.

For example take $f(n) = n$ and $g(n) = 2n$.  Then $f(n) \geq \frac{1}{2} g(n)$ on the domain $n \geq 1$ but $\log f(n) \not\geq c \log g(n)$ on the domain $n \geq 1$ for any $c > 0$.  In this case it is only true that $\log f(n) = \Omega(\log g(n))$ as $n \to \infty$.

For another example take $f(n) = \sin(n\pi/6) + 2$ and $g(n) = \sin(n\pi/6) + 3$.  Then $f(n) \geq \frac{1}{2} g(n)$ for all $n$ but $\log f(n) \not\geq c \log g(n)$ for any $c > 0$ on any domain that includes the points $n=9,21,33,45,\ldots$

The statement is true if $g(n) \to \infty$ as $n \to \infty$ and $f(n) = \Omega(g(n))$ as $n \to \infty$.  For then we can find a $c > 0$ such that

$$
f(n) \geq c\,g(n)
$$

for all $n$ sufficiently large, from which

$$
\log f(n) \geq \log c + \log g(n) \geq \frac{1}{2}\log g(n)
$$

for $n$ large enough.  Thus $\log f(n) = \Omega(\log g(n))$ as $n \to \infty$.


问:
Updating success probability of coin tosses with uniform prior
I have a coin with unknown success probability $\theta$. I regard $\theta$ to be an independent random variable having uniform distribution $U(0,1)$. I am trying to understand how I can get a good estimate of $\theta$ using Bayes rule after each observation.

Let us say, in the first trial I sampled for $\theta$ from $U(0,1)$ to get $\theta = 0.4$ and observed that I got a success (head). For the second trial, I got $\theta = 0.2$ and observed that I got a fail. For the third trial, I got $\theta = 0.4$ and observed that I got a fail. How can I use this info to sequentially get an estimate of $\theta$.

I think general update rule is as follows.

$\pi_{n+1}(\theta) \propto \pi(\theta) f(x_n|\theta)$

I believe that $\pi$ is the $U(0,1)$, $\pi_{n+1}$ is the posterior distribution and $f$ is the Bernoulli.
But I am confused at how to use the above equation to get the correct distribution estimate of $\theta$ from the above observations.  Any one can help me please ?

The one part which confuses me is that since $\theta$ is assumed to be from $U(a,b)$, can't we simply take $\theta$ as the mean of $U(a,b)$ rather than assuming that $\theta$ is unknown but $U(a,b)$ is known. (BTW, this is not a made up question. I am following a work where they assume such a setting)

答: If $\theta$ changes before each flip, then on the next coin flip the previous flips give you no informaiton about the next $\theta$. It will still be Unif(0,1).  But, if the $\theta$ is fixed (but unknown and assumed that anything between 0 and 1 are all equally likely) then after $n$ flips with $s$ successes (i.e. heads), this Beta distribution with mean $\frac{s+1}{n+2}$ is the posterior distribution of $\theta$.


问:
Proving $abcd+3\geq a+b+c+d$

If $a,b,c,d$ are non negatives and $a^2+b^2+c^2+d^2=3$ prove that  $$abcd+3\ge a+b+c+d$$

The inequality is not as simple as it looks.The interesting part is that the equality occurs when $a=0,b=c=d=1$ upto permutation .(I don't know if there are further equality cases.)

I tried rewriting the inequality as $$a^2+a(bcd-1)+b^2+c^2+d^2-b-c-d\ge 0$$ As its a quadratic in $a$ it suffices to show  the discriminant$$\Delta_a={(bcd-1)}^2-4(b^2+c^2+d^2-b-c-d)\le 0$$ which unfortunately is wrong (when $a=\sqrt{3},b=c=d=0$)

P.S ;I am not aware of using Lagranges multipliers

Its from here

Update :An answer has been posted in the link above (AOPS) similar to Dr Mathva's answer

答:
See, in the link below, our extension to n real variables (GMA Problem 494, page 57, with the our solution):
https://ssmr.ro/gazeta/gma/2020/gma3-4-2020-continut.pdf

Screenshots from the link provided:

Title: Proving $abcd+3\geq a+b+c+d$


问:
Is pointwise maximum of affine functions convex and continuous on a compact set $K$
Actually, I've already asked a similar question but I specified my question here.

Consider a max-affine function $f:K \rightarrow \mathbb{R}$ such that $f(x) = \max_{i \in I}f_i(x)$ where $K \subset \mathbb{R}^n$ is a compact set and $f_i(x) = a_i^T x + b_i, \forall i \in I$ are affine functions.

(Assumption) Suppose that $a_i$ and $b_i$ are bounded by $M > 0$ for all $i \in I$.

(Statement) The max-affine function $f$ is continuous on $K$.

(Proof) Since the max-affine function $f$ is convex (see this) and (Assumption) is assumed,
$f$ is a proper convex function on $K$.
Furthermore, $dom(f) = \mathbb{R}^n$ in this case. By Theorem 10.4 in R.T. Rockafella, "Convex Analysis", a proper convex function in $ri(dom(f))$ is continuous, it concludes the statement.

I wonder if 1) the proof is correct, 2) the assumption is necessary to make sure that $f$ is proper, and 3) this holds even if $I$ is uncountable.

答:
The proof is correct, although for the sake of argument you should clarify what $ri(dom(f))$ is in this case.

Yes, you need this assumption. It is automatically satisfied when $I$ is finite. When $I$ is not finite and you do not make this assumption, you could take $a_i=i$ resulting in $\sup_{i \in \mathbb{N}} f_i(x) = \infty$. Bounded should be in absolute sense, so $|a_i|\leq M$ and $|b_i|\leq M$.

When $I$ is uncountable you may need $\sup$ instead of $\max$, but the proof is still valid.


问:
Fixed point theorem for discontinuous functions
I'm trying to solve the below exercise from Pugh's analysis book.

Let $f: [0,1] \to [0,1]$ be a continuous function. Prove that $f$ has at least one fixed point: an $a \in A$ such that $f(a) = a$. Is the same true for discontinuous functions?

Here is my attempt.

Notice that if $f(0) = 0$ or $f(1) = 1$, the theorem is proved. Suppose not. Define $g(x) = f(x) - x$, which is a continuous function from $[0,1]$ to $[0,1]$. Then since $f(0) \neq 0$, $g(0) \neq 0$, so $g(0) > 0$. Since $f(1) \neq 1$, $g(1) < 0$. By the intermediate value theorem, there exists $a \in [0,1]$ such that $g(a) = 0$. So $f(a) - a = 0$, so $f(a) = a$.

I am fairly sure that the result is not true for discontinuous functions, largely because I required continuity of $g$ to invoke the intermediate value theorem. I am having trouble finding a counterexample, however. Do I define a function piece-wise, with jumps at $x = 0$ or $x = 1$, to try to break the intermediate value theorem?

Any help would be appreciated.

答:
Counterexamples abound:  draw some pictures to find some.  For one, try
$$
f(x) = \begin{cases} 1, & x \in [0,1) \\ 0, & x =1. \end{cases}
$$


问:
why we let $d \neq 0$ be orthogonal to $x_1-x_0?$
If $A\subset\mathbb{R^2}$ is countable, is $\mathbb{R^2}\setminus A$ path connected?

My attempt : I found  the answer here

Here is the outline of the given answer

Here is a marginal variation:

Choose $x_0,x_1 \in A^c$. Let $d \neq 0$ be orthogonal to $x_1-x_0$. Define the collection of paths $\gamma_\alpha(t) = x_0 + t(x_1-x_0) + \alpha t(1-t) d$. Since $\{\gamma_\alpha\}_{\alpha \in [0,1]}$ is uncountable, there exists $\alpha_0 \in [0,1]$ such that $\gamma_{\alpha_0}[0,1] \cap A = \emptyset$. Hence $A^c$ is path connected.

My doubt : why we let $d \neq 0$ be orthogonal to $x_1-x_0?$

My thinking : I draw the diagram in my mind by taking $x_0=(0,0) $ ,$x_1=(1,0)$ and $d=(0,1)$

I think here $d$ will be meaningless  and orthogonal mean  $x_1-x_0 \perp d =0$ i,e $(1,0)-(0,0) \perp (0,1)= (1,0)(0,1)=(0,0)$

Actually im not getting what is the motive  behind   $x_1-x_0 \perp d$ ?

Lastly i need some book /reference where i can find this kind of concept

答:
The motive behind $x_1- x_0 \perp d$ is $d$ is the perpendicular bisector of the segment from $x_0$ to $x_1$

construct  a line from $x_0 $and $x_1$ with different slope  i,e two line should not be parallel
and not intersect with $ A$

Choose  a line $L_1$ through $x_0$ and $L_2 $ line through $x_1$

Both $L_1$ and $L_2$ touch at some point of perpendicular bisector $d$

Now we have pair of lines intersecting at some point of $d$ i,e choose $r \in d$ and $r=(r_1,r_2)$

$\implies$  there is path  $f$ from $x_0$ to $r$ and again path $g$ from $r$ to $x_1$

By using pasting lemma  we get  a path $h$  which is connecteding  $x_0$ to $x_1$  via $r$

therefore $\mathbb{R}^2 \setminus A$ is path connected


问:
My ODE system's state consists only of the differences in the location and velocities of two particles. How can I solve for the absolute values?
I have a linear ODE describing the relative motion of two particles ($T$ and $M$) in 3 dimensions. I have the individual accelerations for $T$ and $M$, but the location and velocity coordinates are only given as the difference between the two particles (So e.g. in the $x$ direction the location difference is $D_x=T_x-M_x$ and the velocity difference is $\dot{D_x}=\dot{T_x}-\dot{M_x}$)

All in all the state vector is

$$ \vec{x} = (D_x,D_y, D_z,  \dot{D_x}, \dot{D_y}, \dot{D_z}, \ddot{T_x}, \ddot{M_x}, \ddot{T_y} \ddot{M_y}, \ddot{T_z}, \ddot{M_z}) $$

and I have a matrix $A$ which gives the dynamics

$$\dot{\vec{x}} = A\vec{x}$$

Question

I'd like to (given some specified initial conditions), solve for the absolute locations and velocities of the two particles $M$ and $T$. Is this possible?

答:
Maybe this will help a bit. I'm changing notation on you, just to make it easier to type, but I think you'll get the idea.

You have the state vector
\begin{equation}
 X =
 \begin{bmatrix}
  M - T \\
  M' - T' \\
  M'' \\
  T''
 \end{bmatrix}
\end{equation}
and the system of differential equations $X' = AX$ it satisfies.
You don't say much about $A$ but I'll assume the system is linear and
consistent and of the form
\begin{equation}
 A =
 \begin{bmatrix}
  0 & I & 0 & 0 \\
  0 & 0 & I & -I \\
  a_1 & a_2 & a_3 & a_4 \\
  b_1 & b_2 & b_3 & b_4
 \end{bmatrix}
\end{equation}
where $I$ is the $3 \times 3$ identity matrix and the $a$s and $b$s also
are $3 \times 3$ matrices. Augment $X$ with $M$ and $M'$ to get a new state
vector
\begin{equation}
 S =
 \begin{bmatrix}
  M \\
  M' \\
  M - T \\
  M' - T' \\
  M'' \\
  T''
 \end{bmatrix}
\end{equation}
Then you have $S' = B S$ where
\begin{equation}
 B =
 \begin{bmatrix}
  0 & I & 0 & 0 & 0 & 0 \\
  0 & 0 & 0 & 0 & I & 0 \\
  0 & 0 & 0 & I & 0 & 0 \\
  0 & 0 & 0 & 0 & I & -I \\
  0 & 0 & a_1 & a_2 & a_3 & a_4 \\
  0 & 0 & b_1 & b_2 & b_3 & b_4
 \end{bmatrix}.
\end{equation}
If you also know initial conditions $M(0)$ and $M'(0)$ you can
solve it to get all the information you need.


问:
GCD of a prime number and and increasing number
I saw from How to calculate "gcd product" $\operatorname{gcdp}(n,m)=\gcd(n,1)\gcd(n,2)\cdots\gcd(n,m)$
that for  any prime number $n$, $$\gcd(n,1)\gcd(n,2)\cdots\gcd(n,m) =n^{\lfloor m/n\rfloor}.$$

I can see that the product must be some sort of power of $n$.

But I don't see why it's true.

PS: I'm not a number theory expert and this is not for a number theory class.

Thanks.

答:
Let $p$ be a prime. Given say $m > 0$, we are asked to calculate
$\gcd(p, 1)\gcd(p, 2) \dots  \gcd(p, m)$. Since $p$ is a prime and thus has only two divisors, for each factor $\gcd(p, k)$ there are only two cases:

if $p$ does not divide $k$, then $\gcd(p, k) = 1$.

if $p$ does divide $k$, then $\gcd(p, k) = p$.

With this observation, we see that
$$\gcd(p, 1)\gcd(p, 2) \dots  \gcd(p, m) = p^{\#\{k \in [m]: p\ \text{divides}\ k  \}}. $$

It remains to show that $\#\{k \in [m]: p\ \text{divides}\ k  \} = \lfloor m / p \rfloor$. In short, the left set is counting multiples of $p$ that are less than $m$. You should convince yourself that this floor counts the same thing.


问:
Interpretation of proof by contradiction
If I derive ‘not-A’ from ‘A’, have I ‘proven “not-A” by contradiction’? Thank you.</p>

答: The implication $A \Rightarrow \overline{A}$ is false when $A$ is true. So if you derive $\overline{A}$ from $A$ correctly, then you have proven that $A$ is false. Then $\overline{A}$ is necessary true.


答: No. You've assumed A and arrived at a contradiction (namely that A and "not A" are both true), so you've disproved A.

Title: Interpretation of proof by contradiction

Body: If I derive ‘not-A’ from ‘A’, have I ‘proven “not-A” by contradiction’? Thank you.</p>


问:
Proof of $\varliminf a_n + \varlimsup b_n \leq \varlimsup (a_n + b_n) $ using the subsequential definition
$a_n$ and $b_n$ are assumed to bounded sequences. The definition I'm strictly adhering to is

$\varliminf a_n = \inf\{\text{subsequential limits of } a_ n \} $

$\varlimsup a_n = \sup\{\text{subsequential limits of } a_ n \} $

Below is an attempt I made; would highly appreciate any feedback on it. I have both seen and done other proofs of this statement, using alternative definitions of limit superior/inferior, so I wanted to write up a new one on my own.

Let $L_2$ be any subsequential limit of the sequence $b_n$. Then some $b_{n_k} \to L_2$.

Study the sequence $a_{n_k}$; this is bounded and thus has a convergent subsequence $a_{l_k}$. Denote the limit $L_1$. We know $\varliminf a_n \leq L_1$.

Looking at exactly the $l_k$-indices of $b_{n_k}$, we know $b_{l_k}$ is a subsequence of $b_{n_k}$ and hence also converges to $L_2$.

But then $a_{l_k} + b_{l_k}$ must converge to $L_1 + L_2$. But then of course $L_1 + L_2 \leq \varlimsup (a_n + b_n)$.

The statement we have is then: $\varliminf a_n + L_2 \leq L_1 + L_2 \leq \varlimsup (a_n +b_n) $, that is

$L_2 \leq \varlimsup (a_n +b_n) - \varliminf a_n$

So we have found an upper bound to the set of subsequential limits of $b_n$, which finally leaves us with:

$\varlimsup b_n \leq \varlimsup (a_n + b_n) - \varliminf a_n$. We are done $\blacksquare$.

答:
Your proof is fine. I only would use a different notation for the “iterated subsequence” (that is what caused my initial confusion): If $(b_{n_k})$ is a convergent subsequence of $(b_n)$ with limit $L_1$ then we can choose a convergent subsequence  $(a_{n_{k_j}})$ of $(a_{n_k})$. Then
$$
L_2 = \lim_{j \to \infty} a_{n_{k_j}} \ge \varliminf a_n
$$
and
$$
L_1 + L_2 = \lim_{j \to \infty} (a_{n_{k_j}} +  b_{n_{k_j}})  \le \varlimsup (a_n + b_n)
$$
so that
$$
  L_1 =
(L_1 + L_2) - L_2 \le \varlimsup (a_n + b_n) - \varliminf a_n
$$
This holds for all convergent subsequences of $(b_n)$, which implies
$$
 \varlimsup b_n \le \varlimsup (a_n + b_n) - \varliminf a_n \, .
$$


问:
Tensor product of a number field $K$ and the $p$-adic integers
In the paper A database of local fields, J. Jones and D. Roberts  introduced an isomorphism $K \otimes \mathbb{Q}_p \cong \prod\limits_{i=1}^g K_{p,i}$, where $K$ is some finite dimensional extension of $\mathbb{Q}$ and $\{K_{p,i} \}$ is some collection of finite dimensional extensions of $\mathbb{Q}_p$.

The isomorphism is not mentioned again, and it is not clear why this is even an isomorphism of modules (over what ring?), algebras (over what ring?), abelian groups etc.

$K \otimes \mathbb{Q}_p$ is called an "associated $p$-adic algebra," perhaps suggesting that this tensor product may be interpreted not as a module over $\mathbb{Q}$ but as an algebra over $\mathbb{Q}_p$.  But this doesn't make sense, since $K$ doesn't contain $\mathbb{Q}_p$ as a subfield.

I wanted to ask whether anyone can shed some light on this isomorphism-what it is an isomorphism of, how the isomorphism is given, how the fields $K_{p,i}$ are chosen etc., as well as the significance of these notions.  From the introduction in the paper it only reads "for investigating some problems about number fields, it suffices to know just basic invariants of the $K_{p,i}$, such as ramification index and residual degree."

Any explanation of the above, or link to a reference in which the details are given, would be greatly appreciated.

答:
$K \otimes \mathbb{Q}_p$ is a $\mathbb{Q}_p$ algebra, where $\mathbb{Q}_p$ is embedded in $K \otimes \mathbb{Q}_p$ via $x \mapsto 1 \otimes x$.  The fact that $\mathbb{Q}_p$ isn't contained in $K$ doesn't matter.  The tensor product is over $\mathbb{Q}$, their common base field.

In general, for any 2 fields $K$ and $F$ which are extensions of a common field $E$, we may regard $K \otimes_E F$ as an $F$-algebra in the same way.


问:
Solutions to $\sqrt{3 - \tan^2(\frac{3x}{2})} \cdot \sin x - \cos x = 2$.
$$\sqrt{3 - \tan^2(\frac{3x}{2})} \cdot \sin x - \cos x = 2.$$

How can such an equation be solved? I don't know where to start?

My attempt: $$(3 - \frac{\sin^2 3 y}{\cos^2 3 y}) \cdot 2 \cos y \sin y = (3 - 2 \sin^2 y)^2$$, where $y = \frac{x}{2}$. This does not simplify anything at all, I do not understand what to do next

答:
By transferring $\cos x$ to the RHS and squaring, we have
$$(3 - \tan^2\tfrac32x)\sin ^2x=(2+\cos x)^2.$$
Using the fact that $\cos^2x+\sin^2x=1$, some rearrangement gives
$$(\tan\tfrac32x\sin x)^2+(1+2\cos x)^2=0.$$
A real solution to this equation would simultaneously require, modulo $2\pi$, that $x=0$ and $x=\pm\frac23\pi$. Since these requirements are inconsistent, the original equation has no solution.


答: The clue is in your proof. You defined g(x) = f(x) - x, and found an x-intercept of g. Graphically speaking, a fixed point of the function f is any point where the graph of f crosses the line y = x. So all you have to do is draw a graph that does not cross this line.</p>

Title: Fixed point theorem for discontinuous functions

Body:
I'm trying to solve the below exercise from Pugh's analysis book.

Let $f: [0,1] \to [0,1]$ be a continuous function. Prove that $f$ has at least one fixed point: an $a \in A$ such that $f(a) = a$. Is the same true for discontinuous functions?

Here is my attempt.

Notice that if $f(0) = 0$ or $f(1) = 1$, the theorem is proved. Suppose not. Define $g(x) = f(x) - x$, which is a continuous function from $[0,1]$ to $[0,1]$. Then since $f(0) \neq 0$, $g(0) \neq 0$, so $g(0) > 0$. Since $f(1) \neq 1$, $g(1) < 0$. By the intermediate value theorem, there exists $a \in [0,1]$ such that $g(a) = 0$. So $f(a) - a = 0$, so $f(a) = a$.

I am fairly sure that the result is not true for discontinuous functions, largely because I required continuity of $g$ to invoke the intermediate value theorem. I am having trouble finding a counterexample, however. Do I define a function piece-wise, with jumps at $x = 0$ or $x = 1$, to try to break the intermediate value theorem?

Any help would be appreciated.


问:
Number of points chosen form a polygon to have no isosceles and equilateral triangles.

Let $\Omega$ be a regular polygon with $n$ sides. Let's choose $\Gamma$ a set of vertices, for which any triangle with the vertices in $\Gamma$ is neither isosceles, nor equilateral. Find $\max |\Gamma|$. (Bulgarian Team Selection Test)

Attempt:

I have tried to prove for small cases: $3$, $4$, $5$, $6$, $7$ etc., but no rule confirms, to be proven by induction. And, technically, induction is not possible in this case, because, polygon $\Omega \setminus A$, where $A$ is a vertex, won't be regular.

答:
The condition that $\Gamma$ be a set of points such that no three of them forms an isosceles or equalateral triangle is the same as saying that the distances between the pairs of points in $\Gamma$ are all distinct. In this case, if $\Gamma$ contains $k$ points then there are $\binom k2 = k(k-1)/2$ distinct distances.

Since $\Omega$ is a regular $n$-gon, it does not contain many distinct distances between its vertices. By symmetry we can count the number of distinct distances starting from a fixed vertex. Observe that there are $\lfloor n/2 \rfloor$ distinct distances.

If we number the vertices $0,1,2, \dots n-1$ going around $\Omega$ in order, we can pick a candidate set for $\Gamma$. Let's choose vertices $0,1,3,6,10,15,\dots, d$ where $d$ is the largest value less than $\lfloor n/2 \rfloor$. It's easy to check that all the distances in $\Gamma$ are distinct. Using the above we can see that if there were any more points in $\Gamma$ there would be a repeated distance in $\Gamma$.

EDIT: As pointed out in the comments this answer is wrong. A similar idea, using the fact that: for each point $p \in \Gamma$, the set of distances from $p$ to the other points of $\Gamma$ must all be distinct, which shows that $|\Gamma| \le \lfloor n/2 \rfloor + 1$.


答: If you assume $A$ is true, and you show that this implies that not $A$ is true, then yes, you have arrived at a contradiction. Therefore, your assumption that $A$ is true is false, and it must be the case that not $A$ is true.


问:
Question about existence of solution for an inequality
Question

Let $a_1,a_2,b_1,b_2$ and $x_1,x_2, y_1,y_2$ be non-negative numbers. Assume that:

$(2 x_1+a_2)(2 y_1+b_2) > (2 a_1+a_2)(2 b_1+b_2)$

$(2 a_1+x_2)(2 b_1+y_2) > (2 a_1+a_2)(2 b_1+b_2)$

Does there exist non-negative numbers $z_1,z_2,w_1,w_2$ such that:

$z_1+z_2=x_1+x_2$,

$w_1+w_2=y_1+y_2$,

$(2 z_1+z_2)(2 w_1+w_2) > (2 a_1+a_2)(2 b_1+b_2)$

If exists, then the statement still holds true if we replace $2$ by any $\lambda >0$?

COMMENTS

I feel that the answer is yes. For example, we can choose $z_1=x_1,z_2=x_2,w_1+y_1,w_2=y_2$ in the simple case when $x_1\geq a_1,x_2 \geq a_2,y_1 \geq b_1,y_2 \geq b_2$.

I try to make a simple calculation by finding $z_1,z_2,w_1,w_2$ as follows:

$$\begin{align}
2z_1+z_2=(2x_1+a_2)^\theta(2a_1+x_2)^{1-\theta}\\
2w_1+w_2=(2y_1+b_2)^\theta (2b_1+y_2)^{1-\theta}
\end{align}$$
for some $\theta \in (0,1)$. Combining with equations 1 and 2, I'm able to compute $z_1,z_2,w_1,w_2$. But I cannot show that they are non-negative.

EDIT

Thanks a lot for the answer by S.Dolan.
Now I change a bit on the question as follow

Let $a_1,a_2,b_1,b_2$ and $x_1,x_2, y_1,y_2$ be non-negative numbers. Assume that:

$e^{-x_1+\frac{a_2}{2}}+e^{-y_1+\frac{b_2}{2}} < e^{-a_1+\frac{a_2}{2}}+e^{-b_1+\frac{b_2}{2}}$

$e^{-x_2+\frac{a_1}{2}}+e^{-y_2+\frac{b_1}{2}} <e^{-a_2+\frac{a_1}{2}}+e^{-b_2+\frac{b_1}{2}}$

Does there exist non-negative numbers $z_1,z_2,w_1,w_2$ such that:

$z_1+z_2=x_1+x_2$,

$w_1+w_2=y_1+y_2$,

$e^{-z_1+\frac{z_2}{2}}+e^{-w_1+\frac{w_2}{2}} < e^{-a_1+\frac{a_2}{2}}+e^{-b_1+\frac{b_2}{2}}$

$e^{-z_2+\frac{z_1}{2}}+e^{-w_2+\frac{w_1}{2}} <e^{-a_2+\frac{a_1}{2}}+e^{-b_2+\frac{b_1}{2}}$

If exists, then the statement still holds true if we replace $\dfrac{1}{2}$ by any $1>\lambda >0$?

Btw any comments and discussions are very welcome.

Thanks in advance.

答:
This is not generally true.

Let $y_1=y_2=0,a_i=b_i=1$  and let the $x_i$ be any sufficiently large numbers.


问:
Closed subsets for the quotient topology
Let $X$ be a topological space and let $\pi :X\rightarrow Y$ be a surjective map. Let Y be equipped with the quotient topology determined by $\pi$, obtained by declaring a subset $U\subseteq Y$ to be open if and only if $\pi^{-1}(U)$ is open in $X$.

If $K\subseteq Y$ is such that $\pi^{-1}(K)$ is closed in $X$, then there exists an open subset $V\subseteq X$ such that $\pi^{-1}(K)=X\setminus V$.

How can it be deduced that $K$ is closed for the quotient topology on $Y$?
My problem is that from the definition of the quotient topology on $Y$ we cannot say that there exists some open subset $U\subseteq X$ such that $V=\pi^{-1}(U)$ (but if this holds then we could conclude from the surjectivity of $\pi$).

答: If $\pi^{-1}(K)$ is closed in $X$ then $X\setminus\pi^{-1}(K)=\pi^{-1}(Y\setminus K)$ is open, and so $Y\setminus K$ is open in $Y$. Thus $K$ is closed.


问:
Proving the concurrence of three lines.
Let $p_{1}, p_{2} \text{ and } p_{3}$ be three planes which intersect in a straight line (and not a point, which is generally the case).

Let a fourth plane $p_{4}$ cut these planes (not at the line of intersection of the three planes). Evidently, the intersections of the planes with $p_{4}$ would form three distinct straight lines.

Introduction to Higher Geometry by Graustein says that the three lines are concurrent. Any help regarding the proof would be great.

Thanks in advance.

答:
Let $l$ be the line where $p_1,p_2,p_3$ intersect.
The following cases cases are possible:

$p_4$ intersects $l$ in a point. Then this point is on all three intersection lines $p_i\cap p_4$.

$p_4$ contains $l$. This case was explicitly disallowed.

$p_4$ is parallel to $l$. Then the intersection linse $p_i\cap p_4$ are all parallel to $l$, hence are parallel. (In case of projective geometry, this case does not occur; or to put it differently: If $p_4$ and $l$ intersect at a point at infinity, then so do the $p_i\cap p_4$, i.e. we have just case 1 with a point at infinity (which of course doesn't matter as points at infinity are not distinguished))

Title: Proving the concurrence of three lines.


问:
How to construct and prove the construction of the DFA D' from D?
Let D = (Q, Σ, δ, q0, F) be a DFA. How do we construct a new DFA D'
from D such that
L(D'
) = {
w ∈ Σ*
: wβ ∈ L(D) for some β ∈ Σ*}

that is L(D'
) contains every string w that is a prefix of some string in L(D).

答:
Welcome to MSE!

Hint:

A word is in $L(D)$ if it ends in an accepting state.

So if you want $w$ to be a prefix of some word in $L(D)$, what can we say about the end state of $w$?

Well, if $w\beta \in L(D)$ that means if we start where $w$ ended and then follow $\beta$, we find ourselves in an accepting state.

Can you show the converse? That is, if there is a path to an accepting state from wherever $w$ ends up, then $w$ is a prefix of a word in $L(D)$?

Once you've shown this, you'll know that $D'$ should accept a word if and only if that word can see an accepting state in $D$. Do you see a way to modify $D$ in order to make it have this behavior?

I hope this helps ^_^


问:
Prove $\operatorname{Supp}(M / IM) = V(I) \cap \operatorname{Supp} M$
Let $M$ be a finitely generated $A$-module, and let $I$ be an ideal of $A$.
Claim: $\operatorname{Supp}(M / IM) = V(I) \cap \operatorname{Supp} M$

One implication is trivial: if $p \in \operatorname{Supp}(M / IM)$, then the definition of support tells me that the
localized in $p$ module $(M / IM)_p \neq 0$. So $M_p \neq 0$ as well and
$I \subset p$ (otherwise $I$ would countain $a \in I$ which is
invertible in $M_p$ and therefore $M_p = (aM)_p \subset IM_p$ and we
obtain $(M / IM)_p = 0$, a contradiction).

The other implication I not manage to show. Can any body help?
I tried to do it by contraposition. Assume
$p \notin \operatorname{Supp}(M / IM)$, then $(M / IM)_p = 0$.
If $M_p=0$ ($= p \notin \operatorname{Supp} M$ we are done), so we can assume
$M_p \neq 0$. And I have to show that $p \notin V(I)$. Assume
that $I \subset p$. And now?

答: Hint: If $(M/IM)_p=0$ then $M_p=I_pM_p$. Now assume $I\subset p$ and apply Nakayama's lemma (this is where we need the finiteness assumption)


问:
Non-Linear Additive function Existance from $R$ to $R$ Proof Explaination
This question is regarding the answer provided to exercise 39, section 2.2 of Friedbergs Linear Algebra Book (ISBN: 0130084514).

The exercise reads: "Prove that there is an additive function, $T:R \rightarrow R$ that is not a linear transformation."

He defines a function $f:β \rightarrow R$ where $β$ is a basis for $R \ over \ Q$ such that for some $x,y \in β, f(x) = y \ and \ f(y) = x \ and \ f(z)=z , \ otherwise$

Then he applies the theorem which states that there exists a unique linear transformation, $T:V \rightarrow V$ such that $T(u) = f(u), for \ any \ u \in β$

Finally he states that $T$ is additive but $T(c*x) \neq c*T(x), \ for \ c=y/x$

My problem is that if $c \in Q$ and $T(c*x) \neq c*T(x)$ then $T$ is not a linear transformation but we have defined it as one by the theorem.

If $c \notin Q$ and $T(c*x) \neq c*T(x)$ then this does not prove that T is not a linear transformation since $R$ was defined over $Q$, $R$ being the domain of $T$

So what am I missing?

Thanks in advance!!

答:
What you're missing is that $\Bbb R$ is a vector space in two ways:

It's a 1-dimensional vector space $V$ over the field $\Bbb R$.

It's an infinite-dimensional vector space $W$ over the field $\Bbb Q$.

Your function is linear as a mapping from the second vector space ($W$) to itself, but NOT as a mapping from the first vector space ($V$) to itself.


问:
Prove that if $x,$ $y,$ and $z$ are integers such that $x^2 + y^2 = z^2$ , then $3$ divides $xy.$
Prove that if $x,$ $y,$ and $z$ are integers such that $x^2 + y^2 = z^2$ , then $3$ divides $xy.$

答:
For any $a \in I$, $a^2 \equiv 1 \pmod 3$ or  $a^2 \equiv 0 \pmod 3$

If $x^2$ or $y^2 \equiv 0 \pmod 3$ then we are done.

Therefore, $x^2 \equiv 1 \pmod 3$ and $y^2 \equiv 1\pmod 3$ but then $z^2 \equiv 2 \pmod 3$ which is impossible


答:
Imagine that $K=\mathbb{Q}[\alpha]$ for some algebraic number $\alpha$. Let $m(x)$ be the minimal polynomial of $\alpha$. Then we know that $K\cong \mathbb{Q}[x]/\langle m(x)\rangle$. As $\mathbb{Q}_p$ is an extension field of the rationals, the polynomial $m(x)$ (irreducible in $\mathbb{Q}[x]$) may factor into a product of irreducible factors
$$
m(x)=\prod_i m_i(x),
$$
where $m_i(x)$ are irreducible polynomials in $\mathbb{Q}_p[x]$. Because $m(x)$ is separable, the polynomials $m_i(x)$ are distinct. Therefore the Chinese Remainder Theorem says that
$$
\mathbb{Q}_p[x]/\langle m(x)\rangle\cong\bigoplus_i \mathbb{Q}_p[x]/\langle m_i(x)\rangle.
$$
Here on the right hand side the summands $\mathbb{Q}_p[x]/\langle m_i(x)\rangle$ are all extension fields of $\mathbb{Q}_p$, because the polynomials $m_i(x)$ are irreducible. OTOH
$$
\mathbb{Q}_p[x]/\langle m(x)\rangle\cong\mathbb{Q}[x]/\langle m(x)\rangle\otimes\mathbb{Q}_p\cong K\otimes\mathbb{Q}_p.
$$
Undoubtedly you can guess the rest - the fields $\mathbb{Q}_p[x]/\langle m_i(x)\rangle$ are the fields $K_{p,i}$.</p>

Title: Tensor product of a number field $K$ and the $p$-adic integers

Body:
In the paper A database of local fields, J. Jones and D. Roberts  introduced an isomorphism $K \otimes \mathbb{Q}_p \cong \prod\limits_{i=1}^g K_{p,i}$, where $K$ is some finite dimensional extension of $\mathbb{Q}$ and $\{K_{p,i} \}$ is some collection of finite dimensional extensions of $\mathbb{Q}_p$.

The isomorphism is not mentioned again, and it is not clear why this is even an isomorphism of modules (over what ring?), algebras (over what ring?), abelian groups etc.

$K \otimes \mathbb{Q}_p$ is called an "associated $p$-adic algebra," perhaps suggesting that this tensor product may be interpreted not as a module over $\mathbb{Q}$ but as an algebra over $\mathbb{Q}_p$.  But this doesn't make sense, since $K$ doesn't contain $\mathbb{Q}_p$ as a subfield.

I wanted to ask whether anyone can shed some light on this isomorphism-what it is an isomorphism of, how the isomorphism is given, how the fields $K_{p,i}$ are chosen etc., as well as the significance of these notions.  From the introduction in the paper it only reads "for investigating some problems about number fields, it suffices to know just basic invariants of the $K_{p,i}$, such as ramification index and residual degree."

Any explanation of the above, or link to a reference in which the details are given, would be greatly appreciated.


问:
Showing that the mapping det either homogenous nor additive
I'm currently preparing myself for an exam in Mathematics. I got stuck on this task:

Show in the space of real 2 × 2 matrices that the mapping det is neither homogeneous
nor additive, so in particular it is not linear.

Could anybody please explain to me, how to handle it? I would be very grateful for any advices!

答:
I have to make an educated guess with the term "homogeneous," but you're being asked to show that it is not the case that
$$\det(cA) = c \det(A)\quad \text{for all}\ A \in \mathbb R^{2 \times 2}\ \text{and}\ c \in \mathbb R,$$
and
$$\det(A + B) = \det(A) + \det(B) \quad \text{for all}\ A, B \in \mathbb R^{2 \times 2}.$$

Were these two equations to hold, then $\det : \mathbb R^{2 \times 2} \to \mathbb R$ would be an example of a linear map. So my suggestion is to play around with some examples to see why the equations don't always hold.


问:
Approximation of an integral of logarithmic function
How can I approximate $\int_{0}^{n} dx \,\ln(x)$ for large values of $n$?

I am able to get indefinite integration is:
$$\int \ln(x) \ dx = x  \ \ln(x) - x + C, \quad C\in \Bbb R$$

答: $$\int^{n}_{0} dx \hspace{0.1cm}\ln(x) = \lim_{a \to 0^{+}} \int^{n}_{a} dx \hspace{0.1cm}\ln(x)\\ \hspace{6cm}= \lim_{a \to 0^{+}} n\hspace{0.1cm} \ln(n) - b\hspace{0.1cm}\ln(a) - n + \ln(a)\\\hspace{6.3cm}= \lim_{a \to 0^{+}}\bigg[ n\hspace{0.1cm}\ln\bigg(\frac{n}{e}\bigg) + a\hspace{0.1cm}[1 - \ln(a)]\bigg]\\ \hspace{5.8cm}=n\hspace{0.1cm}\ln\bigg(\frac{n}{e}\bigg) + \lim_{a \to 0^{+}}\frac{1 - \ln(a)}{\frac{1}{a}}\\\hspace{4.2cm}= n\hspace{0.1cm}\ln\bigg(\frac{n}{e}\bigg) + \lim_{a \to 0^{+}}a\\\hspace{2.cm} = n\hspace{0.2cm}\ln\bigg(\frac{n}{e}\bigg)$$


问:
What is the probability of ending up at a corner of a 3×3 square after 4 moves starting at the center?
In my programming class, we've been practicing trying to calculate the worst-case scenario. Recently, I was assigned a problem as homework that was interesting, but I wasn't able to find a solution to it. The problem goes a little like this:

Suppose your robot is playing a game on a 3x3 board of unit squares. The robot starts at the center square, and has four options. It can either move to the square above, to the left, to the right, or below. Because your robot has not been optimized yet, it will move randomly to either of the 4 options. The robot will have the option to move up to 4 times, but will stop if it ever reaches a corner square. what is the probability that the robot will end up at a corner square after the end of the 4 moves? Additionally, if a robot ever moves off the edge of the board, it will be warped to the opposite square. For example, if it were at the  leftmost square that was in the middle row and chose to move to the left for the next move, it would be warped to the rightmost square in the middle row.

I'm still pretty new to probability and all of this, but I'm guessing the first step is to find the total amount of sequences, and then finding the sequences that end up on a corner square. I was able to calculate the first 3 scenarios fine. The minimum moves needed to reach a corner square is at least 3 moves. I've found $4^4$ possibilities for the 4 moves. For the 2 moves and 3 moves, I've figured out there's 8 possibilities of getting to the corner for each. Is this correct? After this, I've got pretty stumped, as I'm not sure how to calculate the number of times it can get to the corner squares with $4$ moves.

EDIT: Ok so I've done a bit more work. For doing 2 moves, there are only 8 possibilities. The robot has to move out of the center, and then to one of the corner squares around it. For doing 3 moves, there were also 8 possibilities. The robot has to move out of the center, then moves the same direction again to warp to the opposite side, and then can go to one of the 2 corners. And for 4 moves, it turns out there's no way to utilize the warp to go to a corner. So the only way is either going to a side square, back to center, and then repeating the scenario with 2 squares. There's 4 ways to do the first part and 8 for the second part, so another 32. This adds up to 48 ways. However, I'm still having difficulty calculating the entire total. Does anyone have any ideas on how to get the amount of any possible moves?

答:
A direct count of possibilities isn’t too hard if you organize it properly. There are $4^4=256$ possible four-move sequences. I’ll use $L,R,U$, and $D$ to denote moves to the left, right, up, and down, respectively.

Any sequence that starts $LU,LD,RU,RD,UL,UR,DL$, or $DR$ reaches a corner. The robot actually stops at that point, but each of these two-move sequences can be extended to a four-move sequence in $4^4=16$ ways, so these $8$ paths account for $8\cdot16=128$ of the $256$ four-move sequences: we just ignore the last two moves.

Any sequence that starts $LLU,LLD,RRU,RRD,UUL,UUR,DDL$, or $DDR$ also reaches a corner; each of these $8$ sequences can be extended to a four-move sequence in $4$ ways, so they account for another $8\cdot4=32$ four-move sequences altogether.

Any sequence that starts $LR,RL,UD$, or $DU$ returns to the centre, and we’ve already seen that there are $8$ two-move sequences from the centre that reach a corner; that accounts for another $4\cdot8=32$ four-move sequences that reach a corner. There are altogether $4\cdot4=16$ possible two-move sequences, so the other $8$ of them do not reach a corner from centre; this accounts for $4\cdot8=32$ four-move sequences that do not reach a corner.

Any sequence that starts $LLR,RRL,UUD$, or $DDU$ reaches an edge square, as if just one move had been made, and has $2$ one-move continuations that reach a corner; that accounts for another $4\cdot2=8$ four-move sequences that reach a corner. From an edge square there are also $2$ moves that do not reach a corner, so this category accounts for $4\cdot2=8$ four-move sequences that do not reach a corner.

The first and third categories above cover all sequences that change direction on the second move, and the second and fourth categories account for all sequences that first change direction on the third move. If the first three moves are the same, i.e., $RRR,LLL,UUU$, or $DDD$, the robot is back at the centre and cannot reach a corner. From that point $4$ moves are possible, so this final category covers $4\cdot4=16$ four-move sequences, none of which reaches a corner.

We’ve now accounted for

$$128+32+32+32+8+8+16=256$$

four-move sequences, which is all of them, and found that $200$ of them reach a corner. The probability that the robot reaches a corner within four moves is therefore

$$\frac{200}{256}=\frac{25}{32}\,.$$

And we have the information to say more: the first category above shows that the probability that the robot reaches a corner in two moves, for instance, is

$$\frac{128}{256}=\frac12\,.$$

Similarly, the second category shows that the probability that the robot reaches a corner in three moves is

$$\frac{32}{256}=\frac18\,.$$

And of course this means that the probability that it takes the robot all four moves to reach a corner is

$$\frac{25}{32}-\frac12-\frac18=\frac5{32}\,.$$


答:
A formula for generating any Pythagorean triple is
$$x = m^2 - n^2$$
$$y = 2mn$$
$$z = m^2 + n^2$$
Suppose that $y$ is not a multiple of 3. Then $x$ must be a multiple of 3. Assume that $m > n$. Then if $y$ is not a multiple of 3, then neither is $m$ nor $n$. Assume that $m = 3a + 1$ and $n = 3b + 1$. Then $$m^2 - n^2 = 9(a^2 - b^2) + 6(a - b)$$ which is a multiple of 3. The other three cases can be proven similarly (i.e. $m = 3a + 1$ and $n = 3b + 2$, or $m = 3a + 2$ and $n = 3b + 1$ or $m = 3a + 2$ and $n = 3b + 2$).

Title: Prove that if $x,$ $y,$ and $z$ are integers such that $x^2 + y^2 = z^2$ , then $3$ divides $xy.$


答:
We can use Euclids Formula: https://en.wikipedia.org/wiki/Pythagorean_triple#Proof_of_Euclid's_formula

To say $x=mn$ and $y= \frac{(m-n)(m+n)}{2}$ for some positive integers $m,n$. Therefore $xy=\frac{mn(m-n)(m+n)}{2}$. From this we can deduce that:

$$xy \equiv 0 \text{ (mod 3)} \leftrightarrow mn(m-n)(m+n) \equiv 0 \text{ (mod 6)}$$

The right hand side can be shown trivially case by case.

Body: Prove that if $x,$ $y,$ and $z$ are integers such that $x^2 + y^2 = z^2$ , then $3$ divides $xy.$


问:
Counting integer solutions to $x^2 + y ^2 < n$
How can I count the number of integer solutions, $\mu(n)$, of $x^2+y^2 < n$, and then hopefully look at behavior as $n \to +\infty$?</p>

答:
The behaviour as $n\to\infty$ is simpler than counting exactly in the finite case:
You count exactly the integer points in a circle of radius $\sqrt n$. Since the circle is not too "chaotic", the number of points is essentially the area of the circle: $\mu(n)\sim \pi n$.</p>


问:
Analycity of Gamma function
I am stuck with something that may be simple, but I just cannot find the answer.

We define, for $z$ a complex number of real part strictly positive, $\Gamma(z) = \int_{0}^{\infty} t^{z - 1}exp(-t) dt$. How do we show that $\Gamma$ is an analytic function on $\{ z \in \mathbb{C}, \Re(z) > 0 \}$ ?

I started by rewritting the expression of $\Gamma$ this way:

$\Gamma(z) = \int_{- \infty}^{\infty} exp(zu)exp(-e^{u}) du$

Then, if $a$ is a complex number with $\Re(a) > 0$, we have:

$\Gamma(z) = \int_{- \infty}^{\infty} exp((z - a)u)exp(au)exp(-e^{u}) du = \int_{- \infty}^{\infty} \sum \limits_{n = 0}^{+\infty} \frac{(z-a)^{n}}{n!}u^{n}exp(-e^{u}) du$.

However after that, I do not see how I could exchange $\sum$ and $\int$ ...

Thank you for your help.

答: It is analytical, because it is differentiable as a function of a complex variable $z$. The integral corresponding to the derivative converges uniformly, so you can differentiate with respect to the parameter.

Title: Analycity of Gamma function


问:
How do you call an edge that returns to the original vertex?
This should have been asked before but I can't seem to find an answer to this question:

How do you call an edge that returns to the original vertex?

So basically given some graph:

$G = (E, V)$

$E \subseteq \{\{x,y\} | x,y \in V\}$

I'm looking for a name for these guys:

$E_? \subseteq = \{\{x,y\} | x,y \in V\, x=y\}$

In university in Frankfurt we called them Eigenkanten (roughly own-edge) but it seems that this term is not used very widespread anyhow in German.

答: This is called a loop in English.</p>

Body: This is called a loop in English.


问:
Prove that $\{Y:Y \subseteq X\}$ is a set.
I am a beginner and self-learning Real Analysis. Problem 3.4.6 from Tao's Analysis-I poses the following question. I'd like to someone to verify my proof, and help me write a technically correct proof.

Let $X$ be a set. Then the set $\{Y : Y \text{ is a subset of } X\}$ is a set.

(Hint given with the question: start with the set $\{0, 1\}^X$ and apply the replacement axiom, replacing each function $f$ with the object $f^{-1}(\{1\})$.)

My Attempt.

Using the Power set Axiom, if $X$ and $Y$ are sets, then $Y^X$ consists of all functions for
$X \to Y$

I assume for simplicity that set $X$ has two elements $\{x_1,x_2\}$.

By definition of the power set axiom, the set $Y^X$ consists of 4 functions as follows :

$f_1 \text{ maps } x_1\to 0 \text{ and } x_2 \to 0$

$f_2 \text{ maps } x_1\to 0 \text{ and } x_2 \to 1$

$f_3 \text{ maps } x_1\to 1 \text{ and } x_2 \to 0$

$f_4 \text{ maps } x_1\to 1 \text{ and } x_2 \to 1$

I want to apply the replacement axiom, replacing each function $f_i$ with the object $f_i^{-1}(\{1\})$.

$\{ f_1^{-1}(\{1\}),f_2^{-1}(\{1\}),f_3^{-1}(\{1\}),f_4^{-1}(\{1\})\}$

$\implies$ $\{ \emptyset,\{x_2\},\{x_1\},\{x_1,x_2\} \} $  which consists of all subsets of $X$.

*** My Question***

I have simply replaced the functions$(f_1,f_2,f_3,f_4)$ with the object $f^{-1}(\{1\})$. In order to use the replacement axiom, I must show that for each $x$, the statement $P(x,y)$ is true for atmost one $y$. What should be the proposition $P(x,y)$?

$\{x_1,x_2\}$ consists of all the elements of $X$, but from the definition of a function we know that bijective functions are invertible, but in the last function both  $\{x_1,x_2\}$ map to $1$, in that case, I don't think I can write
$f_4^{-1}(\{1\})$=$\{x_1,x_2\}$.

答: If you are working within the ZF set theory, this is an axiom.</p>


问:
partition set into two subsets minimize sum of some value function of both subsets
Given a set of finite numbers $X \subset \mathbb{R}$, we can partition $X$ into the upper half $U(X) := \{x|x \in X, x\gt median(X)\}$ and lower half $L(X) = \{ x| x \in X, x \lt median(X)\}$. As you can see, $L(X) \subset X, U(X) \subset X,L(X) \cup U(X) \subseteq X  $

And we define our value function $f(X) := \sum_{i \in U(X)} i - \sum_{j \in L(X)} j$. (I'm also looking for names for this value function)

Our interest is to find a two-way partition $A, B$ of $X$, $A\cup B =X, A\cap B = \emptyset$.

Such that sum of the value from both $A$ $B$ is minimized.

Namely, optimize:
$A^*, B^* = \text{argmin}_{A, B} f(A) + f(B), s.t. A,B \subset X, A\cup B =X, A\cap B = \emptyset$.

And So far I've tried the brute force solution by enumerating the power set of $X$ and evaluate all the configurations. Hoping for some faster solutions.

答:
As the brute force solution would have at least $O(2^{n-1})$ complexity. With some effort, I'm able to make a $O(n^6)$ complexity algorithm.

The trick is to find an anchor point to both $A, B$, which is the median.

Since there can maximum be $O(n^2)$ different medians one can get from any subset of $X$.

The unique pairs $(m_A, m_B)$ of medians for $A, B$ are bounded by $O(n^4)$.

Now that the rest of the elements in $X$ not used in the median can be grouped into pairs $(i, j)$, which is $O(n^2)$.

Now we just need two determine, which $(i, j)$ merges with $m_A$ or $m_B$ that maximize the value function.

So in total, there should exist a $O(n^6)$ complexity algorithm.


答: This is the Gauss circle problem; you’ll find considerable information there and at MathWorld. The latter has numerous references, and there are more references at OEIS A000328.</p>

Title: Counting integer solutions to $x^2 + y ^2 < n$

Body: How can I count the number of integer solutions, $\mu(n)$, of $x^2+y^2 < n$, and then hopefully look at behavior as $n \to +\infty$?</p>


问:
Unable to solve this differentiation
$$\mathbf{v}=d\mathbf{r}/dt=R( -\sin(\theta)d\theta/dt, \cos(\theta)d\theta/dt)=R\omega(-\sin(\theta), \cos(\theta))$$

Now, it’s differentiation equals in textbook

$$\mathbf{a}=d\mathbf{v}/dt= R (-d\omega/dt \sin(\theta)-\omega^2 cos(\theta), d\omega/dt\cos(\theta)-\omega^2 \sin(\theta))$$

I am not getting is that from where did we get $- \omega^2 \sin \theta - \omega ^2 \cos \theta$.

What I got is

$$- R w^2 \cos ^2( \theta) \quad , \quad R w^2 \sin^2(\theta)$$

答:
$$\frac{d(-R\omega \sin(\theta))}{dt}=-R\frac{d(\omega sin(\theta))}{dt}$$ according to product rule and chain rule,

$$\frac{d(\omega \sin(\theta))}{dt}=\frac{d\omega}{dt}\sin(\theta)+\omega (\frac{d\theta}{dt}\cos(\theta))=\frac{d\omega}{dt}\sin(\theta)+\omega^2\cos(\theta)$$

so that $$\frac{d(-R\omega \sin(\theta))}{dt}=-R\frac{d\omega}{dt}\sin(\theta)-R\omega^2\cos(\theta)$$

so the problem with your solution is that you forgot to apply chain rule to $\sin(\theta)$.

Title: Unable to solve this differentiation


问:
Variance of sine and cosine of a random variable
Suppose $X$ is a random variable drawn from a normal distribution with mean $E$ and variance $V$. How could I calculate variance of $\sin(X)$ and $\cos(X)$?

(I thought the question was simple and tried to do a search, but did not find any good answer.)

What if there is no assumption about the distribution of $X$, and only sample mean and variance are provided?

答:
Here I will call the expected value $\mu$ and the variance $\sigma^2.$
\begin{align}
\operatorname E(\sin X) = {} & \operatorname E\left( \frac{e^{iX} - e^{-iX}}{2i} \right) \\[8pt]
= {} & \operatorname E \left( \frac{e^{i(\mu+\sigma Z)}-e^{-i(\mu+\sigma Z)}}{2i} \right) \\[8pt]
= {} & \frac 1 {2i} \left( e^{i\mu} \operatorname E( e^{i\sigma Z}) - e^{-i\mu} \operatorname E(e^{-i\sigma Z})  \right).
\end{align}

$$
\text{And } \operatorname E(e^{i\sigma Z}) = \int_{-\infty}^{+\infty} e^{i\sigma z} \frac 1 {\sqrt{2\pi}} e^{-z^2/2} \, dz. \tag 1
$$

The exponent is
\begin{align}
-\tfrac 1 2 z^2 + i\sigma z = {} & -\tfrac 1 2 \left( z^2 - 2i \sigma z \right) \\[8pt]
= {} & -\tfrac 1 2 \left( z^2 - 2i\sigma z - \sigma^2 \right) - \tfrac 1 2 \sigma^2 \\[8pt]
= {} & -\tfrac 1 2 \left( z-i\sigma \right)^2 - \tfrac 1 2 \sigma^2
\end{align}
The integral on line $(1)$ above becomes
$$
e^{-i\sigma^2/2} \int_{-\infty}^{+\infty} \frac 1 {\sqrt{2\pi}} e^{-(z-i\sigma)^2/2} \, dz.
$$
The integral is equal to $1,$ and so the question is: how do we know that?

The integral is equal to $\displaystyle \lim_{A\to\infty} \int_{-A}^{+A} \frac 1 {\sqrt{2\pi}} e^{-(z-i\sigma)^2/2} \, dz.$

So we consider
$$
\left( \int_{+A}^{-A} +\int_{-A}^{-A-\sigma i} + \int_{-A-\sigma i}^{+A-\sigma i} + \int_{+A-\sigma i}^{+A} \right) \frac 1 {\sqrt{2\pi}} e^{-z^2/2} \, dz
$$
and observe that

the first integral above approaches $1$ as $A\to+\infty,$ and

the second and fourth integrals are easily seen to approach $0,$ and

the third integral is the one whose limit we seek, and

the sum of the four integrals is $0$ because we are integrating along a path that returns to its starting point and the function being integrated has no sigularities in the interior of the surrounded region.

We conclude that $\operatorname E(e^{i\sigma Z}) = e^{-i\sigma^2/2}.$ Similarly $\operatorname E(e^{-i\sigma Z}) = e^{+i\sigma^2/2}.$


问:
$\lim_{b \to \infty} \int_{f(0)}^{f(b)} \sqrt{|u|} \, \mathrm{d}u \quad \text{exists} \implies \lim_{b\to\infty } f(b) \quad \text{exists}$
I'm doing an exercise in which I think the below statement is true, but I can not prove it rigorously. Could you please elaborate if it's true or not. Thank you so much for your help!

Consider $f:[0, \infty) \to \mathbb R$ is continuously differentiable. $$\text{If} \quad\lim_{b \to \infty} \int_{f(0)}^{f(b)} \sqrt{|u|} \, \mathrm{d}u \quad \text{exists, then} \quad\lim_{b\to\infty } f(b) \quad \text{exists} $$

Update: Assume $F(u)$ is the antiderivative of $\sqrt{|u|}$. If $f(0) \le f(b)$, then $$\int_{f(0)}^{f(b)} \sqrt{|u|} \, \mathrm{d}u = F(f(b)) - F(f(0)).$$

If $f(0) > f(b)$, then $$\int_{f(0)}^{f(b)} \sqrt{|u|} \, \mathrm{d}u = - F(f(b)) + F(f(0)).$$

Hence the sign of $f(b) - f(0)$ is important. And I can not tackle this sign.

答:
Hint:

1.) We have
$$\int_{f(0)}^{f(b)} \sqrt{\vert u\vert } du =F(f(b))-F(f(0)).$$
See my comment above why the sign doesn't matter.

2.) Convergence of $\int_{f(0)}^{f(b)} \sqrt{\vert u\vert } du =F(f(b))-F(f(0))$ is equivalent to convergence of $F(f(b))$.

3.) We have $F'(x)=\sqrt{\vert x\vert}$, thus $F$ is strictly monotonically increasing and as $F$ is continuous, we also have that $F$ is a bijection on its image. Furthermore, the inverse of $F$ (defined on the image of $F$) is continuous.

4.) We have $f(b)= F^{-1} (F(f(b)))$.

5.) Conclude.


问:
Plot Line that gets exponentially larger around a point
I'm building a program that calculates the cost of an item based on it's size (let's say a bamboo pole). As the customer requests a longer pole, it gets hard to find a bamboo, plus requires more resources to grow, therefore, I would want to charge more per inch for the piece of bamboo based on it's length approaching a particular length. Then after that point the cost per inch would really escalate.

I believe that log would be the function that I need to use, but I just can't figure out how to make my formula. I've tried log(-x), log(x-1), log(y-x), I can't figure out how to get the log to shoot up to infinity, nor target a specific point.

Referencing the example above, I would want the cost/inch of the bamboo to stay reasonable up to 72", but after that, the cost/inch should rapidly increase, until it gets to 100", where it would become ridiculously expensive. Before 72", it should rise in cost/inch, but at a slow rate (it costs a little bit more per inch to grow a 72" stick than a 6" stick). I'm looking for a uniform growth, not a split formula. No f(x) where x<72, g(x) where x>72.

I'm not necessarily looking for the formula to solve the above question. I am looking for the HOW to research and solve the above question.

Many Thanks,
Matt

答: Try looking into statistics distributions. By messing with the degrees of freedom you can get something that could do what you want. If you take the inverse of the probability of something happening, then you can get your arbitrarily large number. Bonus: you can probably set 75" as the average parameter for the distro and not have to worry about it. </p>


问:
Jacobson Radical and Finite Dimensional Algebra
In general, it is usually not the case that for a ring $R$, the Jacobson radical of $R$ has to be equal to the intersection of the maximal ideals of $R$.

However, what I do like to know is, if we are given a finite-dimensional algebra $A$ over the field $F$, is it true that the Jacobson radical of $A$ is precisely the intersection of the maximal ideals of $A$?

答:
Yes, it is true. See

J.A. Drozd, V.V. Kirichenko,
Finite dimensional algebras,
Springer, 1994


问:
Combinatorics - Painting the Unit Squares of a 3x3 Square
I need some help with the exercise I was assigned in math class. We were given a 3x3 square made of unit squares, and we wanted to paint all the squares. We had 3 sets of paint, red, blue, and green. Each paint was distinct from each other, but two unit squares with the same color were the same. Additionally, there's only enough paint of each color to color 3 unit squares, and all the paint has to be used. The goal of the exercise was to find the total amount of distinct possible ways such that no two unit squares that share an edge (that is to say they are next to each other vertically or horizontally, having same color across a diagonal is fine).

I've started by trying to take a look at each possible arrangement of the bottom 3 unit squares, but I can't seem to find a pattern between it. Is there a better way than continuing by brute force? Any ideas on how I should proceed from here?

答:
** NOT A SOLUTION **

Hint: Consider diagonals.  There are 3 classes

3 different values

2 different values

1 value

Easier to enumerate them all in each class and by definition they are mutually exclusive and collectively exhaustive.

For example for $\text{diag}(1,2,3)$ this is the only solution $\{132,321,213\}$

For $\text{diag}(1,1,1)$ there are two.


问:
Confusion in chain rule of differentiation
Let sin(wt ) be the one.

Let z = wt

Then , we find dz / dt = w .

Then , let y = sin z

I am having confusion to why dy/dz will be - cos z and not only - cos .

Since , in dz /dt , t gets cut right . Why not in dy/dz also ?

答:
So, there are several confusions.  To begin with, there is no such thing as just $\cos$.  It alway requires an argument.  Also, I am unclear of whether $w$ is supposed to be a variable or a constant.  Let's start with $w$ as a variable.

So, with $z = wt$, and $y = \sin(z)$, then

$$ \frac{dy}{dz} = \cos(z)\frac{dz}{dz} \\
\frac{dy}{dz} = \cos(z)$$

Note that we take the derivative of the interior and multiply the result by that derivative.  It turns out to be $1$ ($\frac{dz}{dz}$), but I put in the step to be clear.

Now, let's find $\frac{dz}{dt}$ for $z = wt$. If $w$ is a variable, then, because of the product rule:

$$ \frac{dz}{dt} = w\frac{dt}{dt} + t\frac{dw}{dt}$$

But the $\frac{dt}{dt}$ simplifies to $1$, giving:

$$ \frac{dz}{dt} = w + t\frac{dw}{dt}$$

Now, if $w$ is a constant, then $w$ is not changing, and therefore $dw$ (the change in $w$) is zero.  So, if $w$ is a constant, this simplifies to:

$$ \frac{dz}{dt} = w + 0 = w $$

So, since $\frac{dy}{dt} = \frac{dy}{dz}\frac{dz}{dt}$, then:

$$ \frac{dy}{dt} = \frac{dy}{dz}\frac{dz}{dt} = \cos(z)\,w \\
\frac{dy}{dt} = \cos(wt)\,w $$

Or, as they probably want you to write it,
$$\frac{dy}{dt} = w\,\cos(wt)$$


问:
Contour Integrals and Residues
I'm trying to figure out what it is all about, but my mind is blowing up. First of all, I have turned back and looked at the general definitions of integrals. Then I have looked to line integrals. Basically, the integral is helping us to find the area between two points along the curve. For example when force $F$ is constant but height of the object varies point to point, we can use the integral to find the total work done by $F$.

The line integral is helping us to find the total work done by any vector or scalar fields along any path. Arch length formula gives us the length of curve. So if all these definitions are Ok, please help me to understand about Contour and Residues? Why do we need to use them? Is there any physical meaning or are we just expanding our number set to use The Euler formula and obtain the result easily?

答:
An analytic function $f:\ \Omega\to{\mathbb C}$ can be viewed as a kind of "complex force field" in $\Omega$, so that it makes sense to integrate $f$ along curves $\gamma\subset\Omega$.

A physical force field $F$ is conservative (meaning: it conserves energy) when the integral of $F$ along any closed curve $\gamma\subset\Omega$ is zero. Now it is a basic property of  analytic functions $f$ that they are locally conservative, meaning that integrals along "short" closed curves are zero. The reason for this is the following: The "complex curl" $${\partial f\over\partial\bar z}$$ of such a function is identically zero  – this is the content of the CR equations.

But "in the large" such a complex force field $f$ need not be conservative, as is exemplified by the function $$f(z):={e^z\over z^3}$$ in the punctured complex plane. When $\partial D$ is the counterclockwise oriented unit circle then $$\int_{\partial D} f(z)\ dz=\pi i\ne0\ .$$ In this example the non-conservativeness of $f$ can be precisely localized: It is the pole, an isolated singularity, of $f$ at $0$. But that's not all: The value of the integral need not be computed the hard way; it can be extracted from  the Laurent expansion of $f$ at $0$, which is given by
$$f(z)={1\over z^3}+{1\over z^2}+{1\over 2z}+{1\over 6}+{z\over 24}+\ldots\quad.$$
The residue of $f$ at $0$ is on the one hand the coefficient ${1\over2}$ of the ${1\over z}$-term in this expansion, and on the other hand it is  the value of the above integral, divided by $2\pi i$. That these two numbers coincide is the content of the residue theorem (which, of course, covers much more complicated cases).

Title: Contour Integrals and Residues


问:
Q about mathematical induction
Could somebody help me with this induction?

$1 \cdot 2 \cdot 3 \cdot 4 ... n>2^n , n \ge 4$  , n is an integer n.

I tried to solve it.
Sorry if my handwriting is not clear.

Sorry, I meant 4>1

答:
The image of your solution suggests that you were in the right track, but as you reached the 3rd step, you made a mistake.

For proof by induction, we need to prove the following things:

Check if $P(n_1)$ is true.

Suppose that $P(k)$ is true.

Prove that $P(k+1)$ is true.

Now, $$P(4) = 1*2*3*4 = 24 > 2^4$$
Thus, $P(4)$ is true.

Let,$$P(k) = 1*2*3...*k > 2^k$$

Now,$$P(k+1) = 1*2*3...*k*(k+1)$$
Since, $k≥4$,

therefore, $k+1>4$.

Now, multiply $k+1$ in $P(k)$.

Thus, $$1*2*3...*k*(k+1) > 2^k *(k+1)$$

Now, using the fact that if,

$$a>b$$

Then, $$K*a>K*b [for K>0]$$

We see that,

$$1*2*3...*k*(k+1) > 2^k*4$$

$$P(k+1) > 2^{k+2}$$
Thus,$$P(k+1) > 2^{k+1}$$
Hence, Proved!

Title: Q about mathematical induction


问:
If $ a,b,c \gt 0 $ and $ \frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} = 1 $ Prove $ abc \geq 8 $
If $ a,b,c \gt 0 $  and  $ \frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} = 1 $

Prove $ abc \geq 8 $

I have tried AM-GM and substituting the 1, without results.

答:
This simplifies to $$a+b+c+2=abc$$
$$⇒\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}+\frac{2}{abc}=1$$

Applying AM-GM inequality we get
$$\frac{1}{4}=\frac{\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}+\frac{2}{abc}}{4}≥(\frac{2}{a^3b^3c^3})^\frac{1}{4}$$

Riasing both sides to the power of 4 we get
$$\frac{1}{256}≥(\frac{2}{a^3b^3c^3})$$

$$⇒abc≥8$$


问:
Find subsequence with even and odd indices $(a_{2k})_{k}$ and $(a_{2k-1})_{k}$
I have the sequence $(a_{n})_{n}$ with first term $a_{1}=2$ and recursive relation $$a_{n+1}=\frac{2}{1+a_{n}}$$
So I have to find subsequence with even and odd indices $(a_{2k})_{k}$ and $(a_{2k-1})_{k}$

Any help?

答:
$$a_{n+2}=\frac{2}{1+a_{n+1}}=\frac{2+2a_n}{3+a_{n}}$$

Therefore the 'odd' terms satisfy

$$b_1=a_1=2, b_{n+1}=\frac{2+2b_n}{3+b_{n}}$$

and the 'even' terms satisfy the same relation with $$b_1=a_2=\frac{2}{3}.$$


问:
Confusion on pairwise disjoint and disjoint
I've just started trying to learn some set theory and topology and I've come across the definition of disjoint sets quite a lot I've seen lots of Definitions such as

A set (of sets) $\mathcal{A}$ is disjoint if $\bigcap \mathcal{A} = \emptyset$.

The set $\mathcal{A}$ is pairwise disjoint when $\forall x \in A: \forall y \in A: x \neq y \implies x \cap y = \varnothing$

I can't quite understand what the difference actually is as I saw on s.e that a pairwise disjoint set is related to a $k$-wise disjoint.

I also saw this definition

An indexed collection $\{A_i\}_{i\in I}$ of subsets of $X$ is said to be pairwise disjoint if $A_i\cap A_j=\varnothing$ whenever $i\neq j$.

This confused me even further as an indexed collection can be a surjective function so when $i=j$ it could well be that $A_i$ = $A_j$ in that case how is their intersection empty?.

What I'm really trying to understand is what is pairwise and k wise disjoint sets?
Thanks in advance

答: $A$ is the set of even numbers, $B$ is the set of odd numbers, and $C$ is the set of multiples of $3$.  The 3 sets are collectively disjoint (i.e. no element is in all 3 sets), but not pairwise disjoint, because (for example) the number $3$ is in both sets $B$ and $C$.


问:
Proof of the equality of the difference of two sets iff sets are equal (direct vs. indirect)
I have a problem with the following (really) basic result:

$$A\backslash B=B\backslash A \Longleftrightarrow A=B$$

More specifically, I am able to prove it only by contradiction (in particular in the necessary condition - for the sufficient condition I get a contradiction following the steps of a direct proof).

So my problem is, can we actually prove it without using contradiction?

Moreover (and this is probably a very dumb question, so I apologize), is there a general way to know if a given result that we know is true can be proven only by contradiction or there exist a direct proof of it?

Looking forward to any feedback!

答:
Here is the beginning of a proof, using the definition $$x \in A \setminus B \;\equiv\; x \in A \land \lnot(x \in B)$$  Let's just start at the most complex side, the left hand side, and see if we can simplify it at the logic/element level:
$$
\begin{align}
& A \setminus B \;=\; B \setminus A \\
\equiv & \;\;\;\;\;\text{"extensionality, i.e., two sets are equal iff they have the same elements"} \\
& \langle \forall x :: x \in A \setminus B \;\equiv\; x \in B \setminus A \rangle \\
\equiv & \;\;\;\;\;\text{"definition of $\setminus$, twice"} \\
& \langle \forall x :: x \in A \land \lnot(x \in B) \;\equiv\; x \in B \land \lnot(x \in A) \rangle \\
\equiv & \;\;\;\;\;\text{"negate one side using De Morgan -- to use the Golden Rule (see below)"} \\
& \langle \forall x :: \lnot(x \in A \land \lnot(x \in B) \;\equiv\; \lnot(x \in B) \lor x \in A)) \rangle \\
\end{align}
$$
Now apply (what Dijkstra/Gries/etc. call) the Golden Rule, viz. $$P \;\equiv\; Q \;\equiv\; P \land Q \;\equiv\; P \lor Q$$ for every boolean expression $P$ and $Q$.</p>


问:
Functions manipulation of variables question (inverse function)

I am trying to find the inverse function of the following function:
$$f(x)=\frac{x^2+5x+5}{x+2}$$

However, I always end up with $y^2$ and $y$ on one side and am unsure how to find $y$ by itself. There is a hint that says to use derivatives, but I do not know in what way they are used here.

答:
Hint:

Outlawing $(x = -2),$ you have that

$$x^2 + 5x + 5 = (x+2)y.$$

This can be construed to be a quadratic equation in $x$, where

$Ax^2 + Bx + C = 0$ will have solutions

$x = \frac{1}{2A} \left[-B \pm\sqrt{B^2 - 4AC}\right].$

It is presumed that you will have the constraint on $y$ that $(B^2 - 4AC) \geq 0.$

Note that (for example) the functional equation $x^2 = y$ means that for each positive value of $y$, there are two values of $x$ such that $x^2 = y$.  Compare this with the situation where in the above quadratic, $(B^2 - 4AC) > 0.$


问:
$\frac 21 \times \frac 43 \times \frac 65 \times \frac 87 \times \cdots \times \frac{9998}{9997} \times \frac {10000}{9999} > 115$
Prove that
$$x = \frac 21 \times \frac 43 \times \frac 65 \times \frac 87 \times \cdots \times \frac{9998}{9997} \times \frac {10000}{9999} > 115$$

saw some similar problems like
show $\frac{1}{15}< \frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}<\frac{1}{10}$ is true

but didn't manage to get 115. I could get a weaker conclusion of $x>100$ though.

\begin{align}
x^2 &= \left(\frac 21 \times \frac 21\right) \times \left(\frac 43 \times \frac 43\right) \times \cdots \times \left(\frac{10000}{9999} \times \frac {10000}{9999}\right) \\
&\ge \left(\frac 21 \times \frac 32\right) \times \left(\frac 43 \times \frac 54\right) \times \cdots \times \left(\frac{10000}{9999} \times \frac {10001}{10000}\right) \\
&= 10001
\end{align}

so $x > 100$

Body:
Show this product is equal to:

$$\frac{4^{5000}}{\binom{10000}{5000}}$$

Then use the inequality about the central binomial coefficient:

$$\frac{4^n}{\sqrt{4n}}\leq\binom{2n}{n}\leq\frac{4^n}{\sqrt{3n+1}}$$

Setting $n=5000$ this gives:

$$\frac{4^{5000}}{\sqrt{20000}}<\binom{10000}{5000}<\frac{4^{5000}}{\sqrt{15001}}$$

Or

$$122<\sqrt{15001}<\frac{4^{5000}}{\binom{10000}{5000}}<\sqrt{20000}<142$$


问:
Prove that $\{Y:Y \subseteq X\}$ is a set.
I am a beginner and self-learning Real Analysis. Problem 3.4.6 from Tao's Analysis-I poses the following question. I'd like to someone to verify my proof, and help me write a technically correct proof.

Let $X$ be a set. Then the set $\{Y : Y \text{ is a subset of } X\}$ is a set.

(Hint given with the question: start with the set $\{0, 1\}^X$ and apply the replacement axiom, replacing each function $f$ with the object $f^{-1}(\{1\})$.)

My Attempt.

Using the Power set Axiom, if $X$ and $Y$ are sets, then $Y^X$ consists of all functions for
$X \to Y$

I assume for simplicity that set $X$ has two elements $\{x_1,x_2\}$.

By definition of the power set axiom, the set $Y^X$ consists of 4 functions as follows :

$f_1 \text{ maps } x_1\to 0 \text{ and } x_2 \to 0$

$f_2 \text{ maps } x_1\to 0 \text{ and } x_2 \to 1$

$f_3 \text{ maps } x_1\to 1 \text{ and } x_2 \to 0$

$f_4 \text{ maps } x_1\to 1 \text{ and } x_2 \to 1$

I want to apply the replacement axiom, replacing each function $f_i$ with the object $f_i^{-1}(\{1\})$.

$\{ f_1^{-1}(\{1\}),f_2^{-1}(\{1\}),f_3^{-1}(\{1\}),f_4^{-1}(\{1\})\}$

$\implies$ $\{ \emptyset,\{x_2\},\{x_1\},\{x_1,x_2\} \} $  which consists of all subsets of $X$.

*** My Question***

I have simply replaced the functions$(f_1,f_2,f_3,f_4)$ with the object $f^{-1}(\{1\})$. In order to use the replacement axiom, I must show that for each $x$, the statement $P(x,y)$ is true for atmost one $y$. What should be the proposition $P(x,y)$?

$\{x_1,x_2\}$ consists of all the elements of $X$, but from the definition of a function we know that bijective functions are invertible, but in the last function both  $\{x_1,x_2\}$ map to $1$, in that case, I don't think I can write
$f_4^{-1}(\{1\})$=$\{x_1,x_2\}$.

答:
To answer your last question first, if $f\in Y^X$, the notation $f^{-1}$ has two possible meanings. If $f$ is a bijection, it can denote the inverse function to $f$. In all cases, however, it can be applied to subsets of $Y$: for any $A\subseteq Y$,

$$f^{-1}[A]=\{x\in X:f(x)\in A\}\,.$$

In particular, $f_4^{-1}[\{1\}]$ is $\{x_1,x_2\}$ in your example. (See Tao’s Definition $\mathbf{3.4.4}$.)

It should be intuitively clear that

$$\{Y:Y\subseteq X\}=\left\{f^{-1}[\{1\}]:f\in\{0,1\}^X\right\}\,,\tag{1}$$

so we have two tasks: we must show that $(1)$ is actually true, and we must show that the object on the righthand side of $(1)$ is a set.

It’s for the second task that we want replacement. For each $f\in\{0,1\}^X$ let $P(f,y)$ be the statement

$$y=\{x\in X:f(x)=1\}\,.$$

It is straightforward to prove that for each $f$ in $X^{\{0,1\}}$ there is a unique $y$ such that $P(f,y)$ is true, and that $y$ is evidently $f^{-1}[\{1\}]$, so we can apply replacement to conclude that

$$\left\{f^{-1}[\{1\}]:f\in\{0,1\}^X\right\}$$

is a set.

I’ll leave it to you to prove $(1)$.


问:
Tensor product of a number field $K$ and the $p$-adic integers
In the paper A database of local fields, J. Jones and D. Roberts  introduced an isomorphism $K \otimes \mathbb{Q}_p \cong \prod\limits_{i=1}^g K_{p,i}$, where $K$ is some finite dimensional extension of $\mathbb{Q}$ and $\{K_{p,i} \}$ is some collection of finite dimensional extensions of $\mathbb{Q}_p$.

The isomorphism is not mentioned again, and it is not clear why this is even an isomorphism of modules (over what ring?), algebras (over what ring?), abelian groups etc.

$K \otimes \mathbb{Q}_p$ is called an "associated $p$-adic algebra," perhaps suggesting that this tensor product may be interpreted not as a module over $\mathbb{Q}$ but as an algebra over $\mathbb{Q}_p$.  But this doesn't make sense, since $K$ doesn't contain $\mathbb{Q}_p$ as a subfield.

I wanted to ask whether anyone can shed some light on this isomorphism-what it is an isomorphism of, how the isomorphism is given, how the fields $K_{p,i}$ are chosen etc., as well as the significance of these notions.  From the introduction in the paper it only reads "for investigating some problems about number fields, it suffices to know just basic invariants of the $K_{p,i}$, such as ramification index and residual degree."

Any explanation of the above, or link to a reference in which the details are given, would be greatly appreciated.

答:
Others will have fuller, more accurate, and more efficient explanations than mine, but let me give it a try.

Let $\mathscr O$ be the ring of algebraic integers of $K$. Then $p\mathscr O=\prod_i\mathfrak p_i^{e_i}$, where $\mathfrak p_1,\ldots,\mathfrak p_g$ are the prime ideals of $\mathscr O$ dividing $p$. The $\mathfrak p_i$-adic completion of $K$ is a field $K_{p,i}$ that’s of degree $n_i$ over $\mathbb Q_p$, and we have $n_i=e_if_i$ where $e_i$ is as above and $f_i$ is the residue-class field degree at $p_i$, and furthermore $\sum_in_i=[K\colon\mathbb Q]$.

Then the map $K\to\bigoplus_iK_{p,i}$ by sending an element to the $g$-tuple of all its images in the various completions extends to $K\otimes\mathbb Q_p$. All the remaining details and verifications are left to the reader.


问:
**trajectories** of the permutation of $S_6$ , $\sigma = (1 2 356) (345)$
I have been asked to find the trajectories of the permutation of $S_6$  , $\sigma = (1 2 356) (345)$ and I have to write $\sigma$ as the product of  discrete cycles.

Can anyone please tell me what are the meanings of the bold words ?

答:
The decomposition of a permutation $\sigma\in S_n$ into disjoint ("discrete"(?)) cycles pops up when you consider the natural action of $\langle\sigma\rangle$ on $X=\{1,\dots,n\}$ (i.e. the action as a group of permutations). In fact, this action partitions $X$ into orbits ("trajectories"(?)), say $X=\bigsqcup_{i=1}^NO_i$, and the disjoint cycles are the permutations $\alpha_i\in S_n$ defined, for each $i=1,\dots,N$, as the extension to $X\setminus O_i$ by the map $Id_{X\setminus O_i}$, of the restriction $\sigma_{|O_i}$:

\begin{alignat}{1}
&\alpha_i(k)=\sigma(k), \space\text{for} \space k\in O_i \\
&\alpha_i(k)=k, \space\text{for} \space k\in O_{j\ne i} \\
\end{alignat}

The "disjoint" refers to the fact that $\operatorname{supp}(\alpha_i)\cap \operatorname{supp}(\alpha_j)=\emptyset$, for every $i,j\in \{1,\dots,N\}, i\ne j$, where $\operatorname{supp}(\alpha_i):=\{k\in X\mid \alpha_i(k)\ne k\}$.

The following deploys this general framework into your particular case of $\sigma=(12356)(345)\in S_6$.

The permutation you are given is, in matrix notation, $\sigma=\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 \\
2 & 3 & 4 & 6 & 5 & 1
\end{pmatrix}$. Then:

$$\sigma^2=\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 \\
3 & 4 & 6 & 1 & 5 & 2
\end{pmatrix}$$

$$\sigma^3=\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 \\
4 & 6 & 1 & 2 & 5 & 3
\end{pmatrix}$$

$$\sigma^4=\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 \\
6 & 1 & 2 & 3 & 5 & 4
\end{pmatrix}$$

$$\sigma^5=\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 \\
1 & 2 & 3 & 4 & 5 & 6
\end{pmatrix}=()$$

Now, consider the natural action of $\langle\sigma\rangle$ on $X=\{1,2,3,4,5,6\}$. What you call the "trajectories of $\sigma$" are most probably the orbits under such a $\langle\sigma\rangle$-action on $X$, namely:

\begin{alignat}{1}
O(1) &=\{\sigma^k(1), k=1,2,3,4,5\} = \{2,3,4,6,1\} \\
O(2) &=\{\sigma^k(2), k=1,2,3,4,5\} = \{3,4,6,1,2\}=O(1) \\
O(3) &=\{\sigma^k(3), k=1,2,3,4,5\} = \{4,6,1,2,3\}=O(2)=O(1) \\
O(4) &=\{\sigma^k(4), k=1,2,3,4,5\} = \{6,1,2,3,4\}=O(3)=O(2)=O(1) \\
O(5) &=\{\sigma^k(5), k=1,2,3,4,5\} = \{5\} \\
O(6) &=\{\sigma^k(6), k=1,2,3,4,5\} = \{1,2,3,4,6\}=O(4)=O(3)=O(2)=O(1) \\
\end{alignat}

So, there are:

one "trajectory" of size $5$, $O_1=\{1,2,3,4,6\}\space(=O(6)=O(4)=O(3)=O(2)=O(1))$;

one "trajectory" of size $1$, $O_2=\{5\}\space(=O(5))$.

Now, consider the permutations $\alpha_1,\alpha_2\in S_6$ defined by: $\alpha_i(k)=\sigma(k)$ for $k\in O_i$, and $\alpha_i(k)=k$ for $k\notin O_i$, $i=1,2$. You can verify that $\sigma=\alpha_1\alpha_2$ and $\operatorname{supp}(\alpha_1)\cap \operatorname{supp}(\alpha_2)=\emptyset$, where, for any $\tau\in S_6$, $\operatorname{supp}(\tau):=\{k\in X\mid \tau(k)\ne k\}$. Such $\alpha_1,\alpha_2$ are the disjoint ("discrete"(?)) cycles which $\sigma$ can be split into. According to this definition, for your $\sigma$ we get $\sigma=(12346)(5)=(12346)$.


问:
Linear congruence mod 50
I am revising for my exam but could not figure a way to solve the linear congruence

$$ 57x \equiv -3 \pmod{50}.$$

I have done $-3 = 50 × (-1) + 47$ but don't think i am on right track.

May someone give me a hint please.

Thanks

答:
$$ 57x \equiv 7x \pmod {50}$$

so we just need to find inverse of $7$ modulo $50$, and then multiply it by $-3$.

Observe that $7 \cdot 7 = 49 \equiv -1 \pmod{50}$, so $7^{-1} \equiv -7 \pmod{50}$. Therefore

$$x \equiv (-3)\cdot 7^{-1} \equiv (-3)\cdot(-7) =21 \pmod{50}$$

Title: Linear congruence mod 50


问:
Prove that this dot product property does not hold
I'm asked to disprove the following statement.

For vectors $\vec{x},\vec{y}\in \mathbb{R}^3:$

$(\vec{x}\cdot2\vec{y})+\vec{x}-\vec{y}=2(\vec{y}\cdot\vec{x})+\vec{x}-\vec{y}=(2\vec{x} \cdot \vec{y})+\vec{x}-\vec{y}$.

I don't see anything wrong with this though. Why is any of this wrong? I can pull the $2$ out and dot product is commutive so I am not sure why this is wrong.

Can someone explain?

答:
Usually one considers adding vectors and scalars to be an invalid operation -- in programming terms, it doesn't "typecheck".

However, there actually are contexts, where they can be combined in "formal expressions" as "multivectors", and the whole thing does make sense.  See the entire field of geometric algebra for example.

As applied to this particular statement, you can either take the first stance that "this isn't a valid expression", thus it is meaningless to ask whether it's true or false, or you can take the second stance, in which case it appears to be true, but only in a context where combining scalars and vectors this way is meaningful.


问:
Evaluating $\int_{0}^{\infty} \frac{x^8 -1}{x^{10} + 1} dx$

How to evaluate $$\int_{0}^{\infty} \frac{x^8 -1}{x^{10}  + 1} dx$$

Attempt:

$$f(x) = \frac{x^8-1}{x^{10}+1}$$

$$I = \int_{0}^{\infty} \frac{x^8 -1}{x^{10}  + 1} dx$$

Substituting $t = \frac{1}{x}$,

$$I = \int_{0}^{\infty} \dfrac{\frac{1}{x^8} -1}{\frac{1}{x^{10}}  + 1} x^2dx$$

$$I = \int_{0}^{\infty} \frac{x^{4} - x^{12}}{x^{10}  + 1} dx$$

Adding the two versions of $I$,

$$2I =  \int_{0}^{\infty} \frac{x^8 - x^{12} -1 + x^4}{x^{10}  + 1} dx = \int_{0}^{\infty} \frac{(x^8-1)(1 - x^{4}) }{x^{10}  + 1} dx$$

However, that does not help much.

Putting it in WolframAlpha gives $I = 0$

I am pretty sure there is an elegant solution which I am not able to find it.

答: As @DinosaurEgg and @StefanLafon note, in fact $I=\int_0^\infty\frac{1-x^8}{x^{10}+1}dx=-I$ after the substitution, so $I$ is $0$ if it converges. Indeed, the integrand is $O(1)$ for small $x$ and $O(1/x^2)$ for large $x$.


问:
Conditional probability for an exponential variable
Let $X$ be an exponential random variable with parameter $\lambda$. I have to find $\mathbb{E}[X\mid X<\alpha]$ where $\alpha >0$.

I must find $\mathbb{P}(X\mid X<\alpha)$ first.

What I did is as follows:

$\mathbb{P}(X\mid X<\alpha)=\dfrac{\mathbb{P}(X<t \;\text{and}\; X<\alpha)}{\mathbb{P}(X<\alpha)}$, where I assumed $t> \alpha$ then I get the conditional probability to be equal to $1$. I'm not sure if my assumption is right or wrong.

答:
The following is an approach that is more conceptual.

Let the  conditional expectation of $X$ given that $X\lt \alpha$ be $A$.

Note that the expectation of $X$ given that $X\ge \alpha$ is $\alpha+\frac{1}{\lambda}$. This is a consequence of the fact that by the memorylessness property of the exponential, the additional "waiting time" given that we have waited an amount $\alpha$, is exponentially distributed with parameter $\lambda$.

It follows by the Law of Total Expectation that
$$E(X)=\frac{1}{\lambda}=(1-e^{-\lambda\alpha})A+e^{-\lambda\alpha}\left(\alpha+\frac{1}{\lambda}\right).$$
Now we can solve for $A$.

Remark: Your approach from basics also works nicely. Define random variable $Y$ by $\Pr(Y\le y)=0$ if $y\le 0$, $\Pr(Y\le y)=1$ if $y\ge \alpha$, and $\Pr(Y\le y)=\Pr(X\le y|X\lt \alpha)$ otherwise. Then
$$F_Y(y)=\frac{1-e^{-\lambda y}}{1-e^{-\lambda\alpha}}$$
for $0\lt y\lt \alpha$. Now we can compute $E(Y)$ in several ways. The most basic is to note that $f_Y(y)=\frac{\lambda e^{-\lambda y}}{1-e^{\lambda\alpha}}$ on $(0,\alpha)$ and $0$ elsewhere.  Calculate $\int_0^\alpha y\frac{\lambda e^{-\lambda y}}{1-e^{\lambda\alpha}}\,dy$.


答:
Looking at your method, you've actually proven $\frac{3}{4}x^2 > 10000$ as well – just, instead of writing $\frac{2}{1} \geqslant \frac{3}{2}$, you can incorporate that $\frac{3}{4}$ here, and you'll have exactly $\frac{3}{2} = \frac{2}{1} \cdot \frac{3}{4}$. Now we can deduce that

$$ x > \sqrt{\frac{40000}{3}} \approx 115.47 > 115$$

Title: $\frac 21 \times \frac 43 \times \frac 65 \times \frac 87 \times \cdots \times \frac{9998}{9997} \times \frac {10000}{9999} > 115$

Body:
Prove that
$$x = \frac 21 \times \frac 43 \times \frac 65 \times \frac 87 \times \cdots \times \frac{9998}{9997} \times \frac {10000}{9999} > 115$$

saw some similar problems like
show $\frac{1}{15}< \frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}<\frac{1}{10}$ is true

but didn't manage to get 115. I could get a weaker conclusion of $x>100$ though.

\begin{align}
x^2 &= \left(\frac 21 \times \frac 21\right) \times \left(\frac 43 \times \frac 43\right) \times \cdots \times \left(\frac{10000}{9999} \times \frac {10000}{9999}\right) \\
&\ge \left(\frac 21 \times \frac 32\right) \times \left(\frac 43 \times \frac 54\right) \times \cdots \times \left(\frac{10000}{9999} \times \frac {10001}{10000}\right) \\
&= 10001
\end{align}

so $x > 100$


问:
Kernel-trick in the primal support vector machine problem
I have recently found the following statement of the support vector machine problem in a paper while doing research for a seminar. Let $P$ be the matrix with rows of positive samples, $N$ of negatives, and $w, \gamma$ define the classifier.
$$\begin{aligned}
\min (1-\lambda)(1^Ty+1^Tz)+\frac\lambda 2\lVert w\rVert_2^2\\
s.t.\quad\quad-Pw+1\gamma+1\le y,\\
Nw-1\gamma+1\le z,\\
y,z\ge 0,
\end{aligned}$$
The way I understand the kernel trick is that if we want to pull our data into a higher dimension using some transformation $\varphi$ to then solve the SVM problem in that higher dimension, we only need a function $k(x_1, x_2)=\varphi(x_1)\cdot \varphi(x_1)$ called a kernel function, and never need to apply $\varphi$ directly.

Now usually the kernel trick is introduced as a specific advantage of the dual problem, not the primal. However, in the above statement of the primal SVM problem, only dot products of vectors appear! The dot products of $w$ with itself for $\lVert w\rVert_2^2$ and of samples with $w$ in the constraints.

EDIT: I understand now that one key advantage is being able to precalculate dot products of the training data before optimizing. But still isn't it possible (and useful) to apply a nonlinear transformation into a higher space in the primal?

Is there a reason why we can't replace the dot products in the problem with the kernel function, thereby applying the kernel trick in the primal?

答:
The Kernel trick has two advantages:

It allows you to summarize the relevant aspects of the data in the sense that you do not need the original data anymore, as long as you have the kernel.

You can apply a nonlinear transformation to the data without explicitly stating what the transformation is, e.g., by using the Gaussian kernel.

You cannot do either in the primal problem.


答:
Here is some another proof. Basically, it uses the fact that $(p \iff q) \iff [(p \wedge q)\vee(\neg p \wedge \neg q)]$.
\begin{align}
A\setminus B = B \setminus A &\iff [x \in A \setminus B \iff x \in B \setminus A]\\
&\iff [(x \in A \setminus B \wedge x \in B \setminus A) \vee (x \not\in A \setminus B \wedge x \not\in B \setminus A)]\\
&\iff [0 \vee (x \in A \wedge x \in B) \vee (x \not\in A \wedge x \not\in B)]\\
&\iff [x \in A \iff x \in B]\\
&\iff A = B.
\end{align}</p>

Title: Proof of the equality of the difference of two sets iff sets are equal (direct vs. indirect)

Body:
I have a problem with the following (really) basic result:

$$A\backslash B=B\backslash A \Longleftrightarrow A=B$$

More specifically, I am able to prove it only by contradiction (in particular in the necessary condition - for the sufficient condition I get a contradiction following the steps of a direct proof).

So my problem is, can we actually prove it without using contradiction?

Moreover (and this is probably a very dumb question, so I apologize), is there a general way to know if a given result that we know is true can be proven only by contradiction or there exist a direct proof of it?

Looking forward to any feedback!


问:
If $|G|>1$ is not prime, there exists a subgroup of $G$ which is not $G$ or $e$.
My question is in the title: Prove this statement.

Immediately before this question, I proved that $\langle g \rangle$ is a subgroup of $G$. The question specifies $G$ is a finite group.

My attempt:

By the previous question, provided that g is in G, then $\langle g \rangle$ is a subgroup of G. $\langle g\rangle$ is not the identity nor is it G itself unless G = $\langle g \rangle$ (in which case G is cyclic). So we can write $G = ${$e,g,g^2,...,g^{o(g)-1}$}.  If G is cyclic, it's order may be prime or non prime. For the purposes of the question, say |G| is nonprime. Then we can write a subgroup as {$g^n$}, where $0<n<o(g)$ so as to exclude the trivial case. Done (I think).

If |G| ≠ |$\langle g \rangle$|, and in particular, |G| < |$\langle g \rangle$| by Lagrange, then |G| is not prime (since it is not equal to it's cyclic group) and so $\langle g \rangle$ qualifies as a subgroup.$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\blacksquare$

Is it okay?

MASSIVE EDIT: Question should say |G|$> 1$ is NOT prime. Apologies.

答:
I would start with stating that for $g \in G$ we know that $\langle g \rangle \leq G$. At this stage, we either have $\langle g \rangle < G$ is a proper subgroup as you desire, or $G$ is cyclic and $\langle g \rangle = G$.

Let us denote $|G|=mn$ for $m,n >1$. In the case where $G$ is cyclic, then consider the element $g^m \in G$. We can now generate a new cyclic subgroup with this element, $\langle g^m \rangle$.

It is known that $g^{|G|} = e \;\;\forall g \in G$. Thus,

$$g^{mn}=e \implies (g^m)^n=e \implies o(g^m)=n$$

We also know that the order of a cyclic subgroup is equal to the order of the generating element, thus:

$$|\langle g^m \rangle|=n<|G|$$

Now, we have a proper cyclic subgroup strictly not equal to $G$, thus you have proved what was asked of you.


问:
What is the algebraic way to solve inequalities like this: $|x+4|<2x$?
What is the algebraic way to solve inequalities like this: $|x+4|<2x$?

Can I rewrite this as $-2x<x+4<2x$ or [$x+4<2x$ and $-(x+4)<2x$]?

I found out that the answer was $x>4$, but how can I do it algebraically instead of by inspection? I also did some algebra that lead me to $x>{-4\over 3}$ when I considered $-(x+4)<2x$, this can't be right.

答:
Since $|x+4|\geq 0$ we get $\color{red}{2x> 0}$ so we can square the inequality:

$$ |x+4|^2< 4x^2\implies x^2+8x+16< 4x^2$$

or $$3x^2-8x-16> 0$$

Since $x_1 = 4$ and $x_2 = -{4\over 3}$ so $x\in \color{red}{\mathbb{R}^+\setminus[ -{4\over 3},4]}$.


答:
Your latter rewriting is correct and (in my opinion) the best one. Once you have that

$$\text{both} \; \begin{cases}
x+4 < 2x \\
-(x+4) < 2x
\end{cases}$$

solve each inequality for $x$. You get

$$\text{both} \; \begin{cases}
4 < x \\
-4/3 < x
\end{cases}$$

However, both need to be true, and since $-4/3 < 4$, only the first inequality is necessary to state.

Body:
What is the algebraic way to solve inequalities like this: $|x+4|<2x$?

Can I rewrite this as $-2x<x+4<2x$ or [$x+4<2x$ and $-(x+4)<2x$]?

I found out that the answer was $x>4$, but how can I do it algebraically instead of by inspection? I also did some algebra that lead me to $x>{-4\over 3}$ when I considered $-(x+4)<2x$, this can't be right.


答:
$$|x+4|<2x \implies 2x\ge 0$$
$$\implies x+4>0 \implies x+4<2x$$
$$\implies x>4$$

conversely

$$x>4\implies 0<x+4<2x $$
$$\implies |x+4|<2x$$

Title: What is the algebraic way to solve inequalities like this: $|x+4|<2x$?