问:
$f_n$ is a sequence of bounded real functions converging uniformly to a function $f$ on every finite interval. Is $\limsup f_n(x) = \sup f(x)$?
I'm trying to answer whether $\lim_{n \to \infty}\sup f_n(x)$ equals the sup of $f(x)$.

Let $(f_n)_n$ be is a sequence of real valued functions defined and bounded on the real numbers such that $(f_n)_n$ converge uniformly to a function $f$ on every finite interval. Is it true that

$$\lim_{n \to \infty}\sup f_n(x)  = \sup_n f_n(x)~?$$

I'm not sure how to deal with the sup's in this situation. In fact, I'm not really quite sure what it's asking. Any hints would be appreciated.

答: No. Let $f_n$ equal $0$ on $[-n, n]$ and equal to $1$ otherwise.


问:
Convexity of sublevel set $\sin (x) \leq 1$
I am having trouble determining if the following set is convex

\begin{equation}
\left\{x \in \mathbb{R} : \sin (x) \leq 1 \right\}
\end{equation}

I know that the function itself is not convex function but on the other hand, a $\sin x$ is less than or equal $1$ on all periods so it creates a continuous line as a set which is a convex set. What is the convexity of this constraint and what is the proof ?

答: The set $A = \{x: x \in \mathbb{R}, \sin x \le 1\}= (-\infty,\infty)$ which is clearly convex.

ContentLicense: CC BY-SA 4.0


问:
Prove a subset is not a submanifold using dimension of the Tangent space
Let's consider the set :
$$\mathcal{A}=\left\{(x,y,z)\in \mathbb{R^3},x^2+y^2-z^2=0\right\}$$
I'm struggling to demonstrate it's not a submanifold of $\mathbb{R^3}$ for the following reasons :

A common way to demonstrate a subset is not a subvariety of global space would be to demonstrate it's tangent space has not the same dimension as $\mathcal{A}$.

Defining the function $f:(x,y,z)\rightarrow x^2+y^2-z^2$ , i would like to say the dimension of the manifold , would be 2, but f does not define a submersion of $\mathbb{R}^3$ in $\mathbb{R}$ as it's differential is equal to the null function in $(0,0,0)$.

But even if i knew the potential dimension of the manifold would be $2$, i don't really understand what property i should use to determine it . To what i could understand , i should be able to show the Tangent space in $0$ of $\mathcal{A}$ is a cone , but i have no idea how to demonstrate it properly.

The reader can see another question on the subject has been asked

Why cone of revolution is not a submanifold?

However , it seems it doesn't use the same criterion i'd like to use for the contradiction.

答:
Lets supposed for the moment that $\mathcal{A}$ is a submanifold of $\mathbb{R}^3$. Lets consider the tangent space at the point $(1,0,1)\in \mathcal{A}$.  At this point, the curves $\alpha(t)=(1+t,0,1+t)$ and $\beta(t)=(\cos t,\sin t,1)$ for $t\in (-\epsilon,\epsilon)$ are tangent to $\mathcal{A}$.  The tangent vector associated to $\alpha$ and $\beta$ at the point $p:=(1,0,1)$ is
$$
\dot{\alpha}(0)=\frac{\partial}{\partial x}|_p+\frac{\partial}{\partial z}|_p
$$
and
$$
\dot{\beta}(0)=\frac{\partial}{\partial y}|_p
$$
So if $\mathcal{A}$ is a submanifold, its tangent space must be of dimension at least 2.  Since $\dim \mathcal{A}\ge 2$, locally $\mathcal{A}$ looks like $\mathbb{R}^k$ for some $k\ge 2$.  Hence, if we remove a point from $\mathcal{A}$, $\mathcal{A}$ must remain connected since $\mathbb{R}^k-\{pt\}$ is connected for $k\ge 2$.  However, the set $\mathcal{A}-\{(0,0,0)\}$ consists of two components (upper and lower cones).  Therefore, $\mathcal{A}$ can't be a submanifold.


问:
Concise proof that $\lvert a-b\rvert < \frac{\lvert b \rvert}{2} \implies \lvert a\rvert > \frac{\lvert b \rvert}{2}$
I am only considering $\mathbb{R}$ in this context. I understand why the statement holds and can prove it geometrically, but I can’t think of a concise way to derive the consequent from the precedent in a purely algebraic way.

答: Hint: $|a-b| \ge |b| - |a|$. Can you take it from here?


问:
How to interpret "let" in mathematics?
Before explaining my issue, I wanted to first explain what things do make sense to me.

So, statements (used at the beginning of proofs) like "Suppose $x$ is an integer" or "Assume $x$ is an integer" make sense to me. The way I read them is like: "Let's just pretend that the symbol $x$ happens to represent an integer".

Statement (also used in proofs) like "Let $x$ be 3" or "Let $x$ equal 3" also make sense to me. The way I read them is like: "Let's just temporarily name 3 with the symbol $x$.

My issue comes with statements like "Let $x$ be an integer" or "Let $x$ $\in$ $\mathbb{Z}$". I really do not know how I should intuitively interpret such a statement. I can't interpret it in the same way as like "Let $x$ be 3" because there is not a specific object being assigned like 3. Should I interpret it like how I interpret "Suppose $x$ is an integer"?

EDIT:

So, from what I gathered from the responses, I think I understand now how I should think of "Let $x$ be an integer".

I could think of it as "Assume a newly created symbol x happens to represent an integer". However, this can cause issues as then "Let $x$ be an element of the empty set" is also completely valid.

Instead, I should not think of "Let $x$ be an integer" as an assumption, but an assignment/declaration, just like "Let $x$ be 3". One tangible way to think about this is to imagine myself assigning the newly created symbol $x$ to an integer chosen in secret by a friend. With this mindset, $x$ is not assumed to be an integer - it is an integer. It is just unknown to me what integer it is.

I hope this made sense to anybody with similar questions. If anyone thinks I have made an incorrect finding, please feel free to correct me.

答:
"Let" is usually used to introduce a new symbol along with an assumption about it ("Let $G$ be a group") or when defining notation ("Let $\mathcal P(S)$ denote the power set of $S$"). "Suppose" and "assume" are more conventional when introducing new assumptions using only existing symbols and notation ("Suppose $x^2<5$"). This isn't a rule, though - you can certainly introduce new symbols with an "assume" or "suppose".

I'd also say "suppose" is slightly more common when introducing a proposition that is "in doubt" and will shortly be refuted or discarded (e.g. with a proof by contradiction), whereas "assume" is more neutral in tone and is more common when introducing long-lasting assumptions or "background information".

The three words have the same technical meaning, in that we'd translate them into formal logic in the same way.

Title: How to interpret "let" in mathematics?


问:
$3\times3$ grid filled in randomly; probability of a $2\times2$ grid filled with one element
I have a $3 \times 3$ grid. For each of the $9$ spaces inside, I randomly populate it with either $A$ or $B$ ($\frac{1}{2}$ chance of either element). Each space is filled independently.

What's the probability that in a $2\times2$ grid within the $3 \times 3$ grid, it is filled with only $A$?

I know that there are $4$ possible $2\times2$ grids within the $3 \times 3$ grid. Any pointers on what to do from here?

答:
The $95$ you found is correct, by your use of inclusion-exclusion with $128−40+8−1$ and you have to divide this by the $2^9=512$ equally likely possibilities to get the probability

Doing inclusion-exclusion as probabilities ($4$ corners, less $4$ sides and $2$ diagonals, plus $4$ three-corners, less $1$ whole square) would have given $$\frac4{2^4}-\frac4{2^6}-\frac2{2^7}+\frac4{2^8}-\frac1{2^9} = \frac{95}{512}$$

Or counting would have given

$68=4 \times 17$ with just one $2\times2$ square $A$s

$20 = 4 \times 5$ with two $2\times2$ squares $A$s adjacent

$2 = 2 \times 1$ with two $2\times2$ squares $A$s diagonal

$4 = 4 \times 1$ with three $2\times2$ squares $A$s

$1 $ with all four $2\times2$ squares $A$s, in fact all $A$

adding up to $95$ again


问:
Find a formula for leaves in tree
Given a tree $T$ with $n$ nodes, every node that isn't a leaf has 2 children, can i drive  a formula for number of leaves in a such tree $T$?

My idea: i read this post:Full binary tree proof validity: Number of leaves $L$ and number of nodes $N$
, but it is about full binary tree, our problem maybe not full binary tree.

答:
Suppose that $T$ has $n$ nodes, $\ell$ of which are leaves; then the sum of the degrees of the nodes is

$$2+3(n-\ell-1)+\ell=3n-2\ell-1\,,$$

since the root has degree $2$, each of the other $n-\ell-1$ non-leaves has degree $3$, and each leaf has degree $1$. This is twice the number of edges, so $T$ has $\frac{3n-1}2-\ell$ edges.

On the other hand, $T$ is a tree, so it has $n-1$ edges, and we have the equation

$$\frac{3n-1}2-\ell=n-1\,,$$

so that

$$\ell=\frac{n+1}2\,.$$

The reason that $n$ has to be odd, as this formula clearly implies, is that the non-root nodes of $T$ come in sibling pairs, so there is an even number of them, and the root makes an odd total number of nodes.

Title: Find a formula for leaves in tree


问:
Simple Calculus question (Substitution)
My question is "To show
$ \int_{1}^\alpha {1\over1+x^2} dx = \int_{1\over\alpha}^1 {1\over 1+u^2} du$

such that
$u = \frac{1}{x}$ and $\alpha $ > 1,

I got to

$ \int_{1}^\alpha {1\over1+x^2} dx = -\int_{1}^{1\over\alpha} {1\over 1+{1\over u}^2} {du\over u^2}  $

but i don't know how to swap the upper bound and lower bound around, could someone please explain it clearly to me?

答:
Here is an alternative solution to the original problem that doesn't use substitution.

Define $\displaystyle f(\alpha) = \int_1^\alpha \dfrac{1}{1+x^2}dx$, and $\displaystyle g(\alpha)=\int_{1/\alpha}^1\dfrac{1}{1+x^2}dx$.

Note that $f(1)=g(1)=0$.  By the fundamental theorem of calculus, $$f'(\alpha)=\dfrac{1}{1+\alpha^2}.$$

By the fundamental theorem of calculus and the chain rule,  $$g'(\alpha)=\dfrac{-1}{1+(1/\alpha)^2}\cdot\dfrac{-1}{\alpha^2}=\dfrac{1}{1+\alpha^2}.$$

Because $f(1)=g(1)$ and $f'=g'$, it follows from the mean value theorem that $f=g$.

ContentLicense: CC BY-SA 3.0

Title: Simple Calculus question (Substitution)


问:
How many liters of a $25\%$ percent saline solution must be added to $3$ liters of a $10\%$ percent saline solution?

How many liters of a $25\%$ percent saline solution must be added to $3$ liters of a $10\%$ percent saline solution to obtain a $15\%$ percent saline solution?

Answer:

 1.5

But I don't know how to solve it. Help me, please.

答:
If 1 liter of 25% is added to 1 liter of 10% you get 2 liters of 17.5%...extra 2.5%/liter is enough to convert 2nd 10% liter to 15%. One 25% was enough to convert 2 liters, you only need .5 to convert the third liter. Total of 1.5 liters.

It’s the same as “how many tests do you have to score 25% in to bring up your average to 15% if you’ve already taken 3 tests at 10%”

Averages


问:
Is there any situation where the LDU decomposition is the same as the eigenvalue decomposition?
I was just wondering if there are any situation where the LDU decomposition is the same as eigenvalue decomposition (diagonalization)? The only way this can be possible if L and U are inverse so symmetric matrices seems like a good candidate, but do we need any more requirements.</p>

答: The inverse of an upper triangular matrix is always upper triangular. So, $L$ and $U$ must be diagonal. Hence, the original matrix must be diagonal.


答:
‘Let’ assigns meaning  (definitions or restrictions or values) to variables or phrases or symbols.

In “let $k$ be an integer and $x=2k$”, $x$ is an arbitrary even number (or: the representation or general specification of the even numbers).

In “let $x=7$”, $x$ has been instantiated with a specific value.

(or: “Put $x=7.$”)

The adjective ‘arbitrary’ is frequently tacit, and these sentences all mean the same:

Let $x$ be an arbitrary real number and suppose that $P(x).$

Let $x$ be real and suppose that $P(x).$

Let $x\in\mathbb{R}$ and suppose that $P(x).$

Let $x$ be an arbitrary real number such that $P(x).$

Let $x$ be real such that $P(x).$

Let $x\in\mathbb{R}$ such that $P(x).$

‘Any’ versus ‘arbitrary’.

Title: How to interpret "let" in mathematics?

Body:
Before explaining my issue, I wanted to first explain what things do make sense to me.

So, statements (used at the beginning of proofs) like "Suppose $x$ is an integer" or "Assume $x$ is an integer" make sense to me. The way I read them is like: "Let's just pretend that the symbol $x$ happens to represent an integer".

Statement (also used in proofs) like "Let $x$ be 3" or "Let $x$ equal 3" also make sense to me. The way I read them is like: "Let's just temporarily name 3 with the symbol $x$.

My issue comes with statements like "Let $x$ be an integer" or "Let $x$ $\in$ $\mathbb{Z}$". I really do not know how I should intuitively interpret such a statement. I can't interpret it in the same way as like "Let $x$ be 3" because there is not a specific object being assigned like 3. Should I interpret it like how I interpret "Suppose $x$ is an integer"?

EDIT:

So, from what I gathered from the responses, I think I understand now how I should think of "Let $x$ be an integer".

I could think of it as "Assume a newly created symbol x happens to represent an integer". However, this can cause issues as then "Let $x$ be an element of the empty set" is also completely valid.

Instead, I should not think of "Let $x$ be an integer" as an assumption, but an assignment/declaration, just like "Let $x$ be 3". One tangible way to think about this is to imagine myself assigning the newly created symbol $x$ to an integer chosen in secret by a friend. With this mindset, $x$ is not assumed to be an integer - it is an integer. It is just unknown to me what integer it is.

I hope this made sense to anybody with similar questions. If anyone thinks I have made an incorrect finding, please feel free to correct me.


问:
Why are there 48 symmetries of a cube?
I'm trying to prove that there are a total of 24 rotation and 24 reflection symmetries of a cube. I can show the first part, but I don't have a good proof for why there are also 24 reflections.

The argument I have so far is that we can pick a reflection $A$, and then $A$ acting on any element of the rotation group $SO(3)$ gives us a distinct reflection. But I don't know why this generates the full set of reflections.

In other words, why would another reflection $A'$ acting on the rotation group give us the same set of reflections? I guess I'm trying to show that for $B\in SO(3)$, there exists $B'\in SO(3)$ such that $AB=A'B' \Rightarrow (A')^{-1}AB=B'.$ Is it enough to say that this holds since $(A')^{-1}A$ is the composition of two reflections and thus a rotation, and composed with $B$, this means $(A')^{-1}AB$ is also a rotation?

答:
Here is a proof that there are at least 48 symmetries.  (Goal:  make it obvious and easy to understand.)

Go to a craps table and borrow one of the dice.

There are eight corners on your die, and so you can position the die in any of eight ways based on the corner you are looking at.

For each of those eight corners, there are three rotations.  For example, if you are looking at the corner where you can see sides with one dot, two dots, and three dots, you can rotate the die so that any of these three sides is on top.

So, that gets you 24 symmetries.

Now, the twist others have alluded to about a non-physical move.  If you got the die in Las Vegas, the one/two/three corner will go counter-clockwise.  If you got the die in Macau, the one/two/three corner will go clockwise.  This gives us an additional factor of two in our count of symmetries.  "Change" the die from left-handed to right-handed or vice versa.

This additional factor of 2 gives us 48 symmetries.

That 48 is an upper bound:  pick any of the 8 corners.  The three adjacent corners can be permuted in 3-factorial ways.  So, there can be no more than 48 permutations.

Title: Why are there 48 symmetries of a cube?


问:
Why does the fixed point method rely on the derivative of the root for convergence or divergence?
I'm currently studying the fixed point iterative method, and I am confused as to how the derivative of g(x) at a point between a guess and a root can tell us if the method will converge or diverge for this point. For example, for

$f(x) = e^x - 10x = 0$,

$x = g(x)$, where $g(x) = (1/10)e^x$.

When you rearrange it with the root solutions, you will get

$x_{i+1} − x_{r} = {\dfrac{dg}{dx}}(ξ) (x_{i} − x_{r})$,

where $x_{i+1}$ and $x_{i}$ are the guesses of the root, $x_{r}$ is the root, and the mean value theorem is used, with $ξ$ being a value inbetween $x_{i}$ and $x_{r}$.

I understand mathematically, the ${\dfrac{dg}{dx}}(ξ)$ term has to be below 1 for convergence to occur, but I do not understand the intuition behind it.

Does this mean the fixed point method only works for roots where the derivative near a point is relatively flat?

答:
A visual interpretation of the derivative is that we graph the function on the $xy$-plane, then "zoom in" near one point until the curve looks almost like a straight line. The derivative is the slope of this line which the curve looks like. So for simpler intuition, what happens if a piece of the function's graph actually is a straight line segment?

Suppose $g(x) = mx+b$, with $m \neq 1$. The unique point where $g(x)=x$ is $x_r = \frac{b}{1-m}$.

But if we ignore the simple algebraic solution and just try iteration from some guess, we'll have some sequence $(x_i)$ where $x_{i+1} = g(x_i)$. Is this sequence converging to, or getting farther from, the value $x_r$?

$$ \begin{align*}
 x_{i+1} - x_r &= g(x_i) - x_r \\
&= mx_i + b - x_r \\
&= m(x_i - x_r) + m x_r + b - x_r \\
&= m(x_i - x_r) + (m-1) x_r + b \\
x_{i+1} - x_r &= m(x_i - x_r)
\end{align*} $$

So the absolute value of the error in the sequence decreases if $|m|<1$, or increases if $|m|>1$.

Intuitively then, once the sequence is "close enough" to the correct point that the derivative is a "good enough" description of the overall behavior of the iterative algorithm, it will converge if $|g'(x_r)| < 1$, since the derivative approximates a linear slope in a neighborhood of the point. If $|g'(x_r)| > 1$, the sequence can't converge unless it happens to land on $x_N = x_r$ exactly, since for a point $x$ close to $x_r$, the iteration will send the next point farther from $x_r$. If the derivative doesn't exist, or if $g'(x_r) = \pm 1$, we need to look for other means of analyzing things.

Getting away from intuition to precise statements, using the mean value theorem's value $\xi$ is part of a way of proving something general about actual values which aren't "arbitrarily close to $x_r$" (since no such values really exist in ordinary real analysis).


问:
How do I prove that the cross product follows the right hand rule for vectors in the xz plane?
Thanks to this wonderful article I have been able to prove that the cross product follows the right hand rule when the cross product has a non-zero z component. In this case we show that the z component is the same for the vectors and their projections onto the xy plane, and that this component is positive for positivily oriented vectors and negative for negatively oriented ones.

The article (in less detail) describes how to deal with cases where the vectors $u$ and  $v$ have the same ratio of $y$ to $x$. In this case the z component of the cross product would be equal to zero, and so we project the vectors onto the zy-plane instead, and calculate the x-component of the cross product in the same manner as above.

There is however one more case where the cross product has a z-component equal to zero - when the vectors $u$ and $v$ lack an y-component, and span the $xz$ plane. In this case the cross product will be limited to only a y component.

This time we won't have to project anything onto the $xz$ plane (the vectors are already there).

I try to apply the same reasoning as the article. If $u=(a_1, b_1, c_1)$ and $v=(a_2, b_2, c_2)$ then this would mean that

$v \times u=\begin{bmatrix}0c_2-c_1 0\\-(a_1 c_2-c_1 a_2)\\a_1 0-0a_2 \end{bmatrix}$

If we measure our angles from the positive x axis, and if we let $\alpha$ denote the angle for $u$ and $\beta$ the angle for $v$, we should be able to write this as:

$$c_1a_2-a_1c_2=|u|sin\alpha |v|cos \beta-|u|cos \alpha |v|sin \beta$$

According to the sine subtraction identity, we can write this as:

$$|u|sin\alpha |v|cos \beta-|u|cos \alpha |v|sin \beta = |u||v|sin(\alpha-\beta)$$

This is where things get confusing. In the other two cases we had to make the component of the cross product equal to $sin (\beta-\alpha)$, the minus sign switches things around I guess.

If $(u,v)$ are positively oriented there are two options, either $\alpha$ is smaller than $\beta$ (figure to the left) or vice versa (figure to the right). In both cases the $v$ must be less than 180 degrees conterclockwise from $u$.

(picture from article, all credit to Eric Thurschwell)

In the first case, $\alpha-\beta$ would give us the negative difference between $\beta$ and $\alpha$, which by definition must be less than 180 degrees. The sine of something a value between 0 and -180 must be negative, which gives us a negative y component.

In the second case, $\alpha-\beta$ would give us the positive difference between $\alpha$ and $\beta$, but since the vectors are positively oriented, it means this value must be greater that 180 degrees (and smaller than 360 degrees) by definition, making the sine of that value negative.

...Which would mean the cross product doesn't follow the right hand rule at all, or at least not for these kinds of vectors.

Would someone mind helping me out? I've been working on this for two weeks. This is the last thing I need to straighten out before I can move on to other topics...

答:
To start with, you have transcribed the rule incorrectly.

In your source, the formula is given this way:
if $\mathbf a = \langle a_1,a_2,a_3\rangle$
and $\mathbf b = \langle b_1,b_2,b_3\rangle$,
their cross product is denoted $\mathbf a\times \mathbf b$
and deﬁned analytically as the vector
$\langle a_2b_3 − a_3b_2, a_3b_1 − a_1b_3, a_1b_2 − a_2b_1\rangle.$

Transcribing this into your notation, that is,
$\mathbf a \to v,$ $\mathbf b \to u,$
$a_1 \to a_2,$ $a_2 \to b_2,$ $a_3 \to c_2,$
$b_1 \to a_1,$ $b_2 \to b_1,$ and $b_3 \to c_1$:

If $v=(a_2, b_2, c_2)$ and $u=(a_1, b_1, c_1),$ then
$$
v \times u
 = \begin{bmatrix} b_2 c_1 - c_2 b_1 \\
                   c_2 a_1 - a_2 c_1 \\
                   a_2 b_1 - b_2 a_1 \end{bmatrix} .
$$

You set the second coordinate to $a_2 c_1 - c_2 a_1,$ opposite the correct sign.
On the other hand I'm not sure you actually meant to write $v \times u$;
if you change the left-hand side of your equation to $u \times v$ then the right-hand side you wrote will be correct as it is.

The next thing that is problematic in your analysis is that if you look "down" at the $xz$ plane from a place far along the positive $y$ axis, the positive $z$ axis should be on the right and the positive $x$ axis on the left, that is, what you see should be this:

The convention I would expect to follow is that angles are measured counterclockwise from the $z$ axis in this view, as the figure above illustrates for the vector $u.$

By saying that $c_1 = \lvert u\vert \sin \alpha,$
in effect you have decided to measure your angles clockwise from the positive $x$ axis. Reversing rotation direction from counterclockwise to clockwise is equivalent to flipping the direction of the normal vector.
Note that
$\langle 1,0,0\rangle \times \langle 0,0,1\rangle = \langle 0,-1,0\rangle,$
that is, $\hat x \times \hat z = -\hat y.$

There are so many inadvertent sign changes in your analysis that I find it hard to count them. But there are apparently an odd number of them, because they don't all cancel out to give you the expected answer.

Now for a personal opinion about the linked article,
https://www.researchgate.net/publication/305759205_An_Even_Simpler_Proof_of_the_Right-Hand_Rule.
I find this article unsatisfying.
Setting aside the failure to give any guidance about handling two vectors in the $x,z$ plane, where is the argument that projecting both vectors onto a coordinate plane will preserve the right-hand rule?
Such an argument possibly could be made, but it would make this "proof" not nearly as simple as it claims to be.

If you want a proof that is as simple as the one claimed in the linked paper,
I think you will have to accept as given that the cross-product formula is correct up to possible reversals of the signs of the three resulting coordinates.
That is, we assume that the magnitude of each coordinate of the cross-product is correct in the given formula, and we merely need to show that the sign is correct for the right-hand rule rather than for the left-hand rule.

In order to show this for the $y$ axis, it is sufficient to take any two vectors such that these two vectors and the $y$ axis do not all lie in one plane and show that the resulting $y$ coordinate has the correct sign.
In particular, we can take the vectors $\hat x = \langle 1,0,0\rangle$
and $\hat z = \langle 0,0,1\rangle.$
The right-hand rule requires that $\hat z \times \hat x = \hat y.$
We see that the formula satisfies this requirement, so it is correct for the $y$ component of the cross product.


问:
Derivatives of parametric curve
Given the curve $x = t^3 + 3t + 1 ,  y = t^2 + 4t - 4$

A. Find $\frac{dy}{dx}$ in terms of $t$.

B. Find $\frac{d^2y}{dx^2}$ in terms of t

I need help answering those please.
For A). the answer I got is $\frac{4}{3t}$ I'm not sure if it is correct.

Title: Derivatives of parametric curve

Body:
Need to use chain rule and quotient rule to get these derivatives.

$x = t^3 + 3t + 1 , y = t^2 + 4t$

$\frac{dx}{dt}$ = $3t^2$+3

$\frac{dy}{dt}$ = 2t+4

$\frac{dt}{dx}$ =$\frac{1}{3t^2+3}$

$\frac{dy}{dx}$ =$\frac{dy}{dt}$$\frac{dt}{dx}$=$\frac{2t+4}{3t^2+3}$

$\frac{d^2y}{d^2x}$ =$\frac{d\frac{2t+4}{3t^2+3}}{dt}$ $\frac{dt}{dx}$
=$\frac{2(3t^2+3)-(2t+4)(6t)}{(3t^2+3)^3}$ =  $\frac{-2t^2-8t+2}{9(t^2+1)^3}$


问:
Axiomatic proof in S4
I'm trying to prove the following formula in S4.

$\Box \Diamond \Box \Diamond P \rightarrow \Box \Diamond P$

I have mp, nec, K,D,T,S4 as well as the derived rule $\Diamond \Diamond \phi \rightarrow \Diamond \phi$. I also am able to make obvious moves from propositional logic. i.e. going from $\phi \rightarrow \psi$ to $(\phi \land \sigma) \rightarrow \psi$ etc.

But I can't seem to figure this deduction out. Each time I end up with an extra $\Diamond$ or $\Box$.

答:
Let's first note that in system K, we have derived rules: if $\varphi \rightarrow \psi$ is a theorem, then (a) $\square \varphi\rightarrow \square \psi$ is a theorem, and (b) $\lozenge \varphi\rightarrow \lozenge \psi$ is a theorem. For (a), assume $\varphi\rightarrow \psi$ is a theorem. By necessitation, $\square(\varphi\rightarrow \psi)$. By $(K)$ and modus ponens, $\square \varphi\rightarrow \square \psi$. For (b), assume $\varphi\rightarrow \psi$ is a theorem. Then $\lnot \psi\rightarrow \lnot \varphi$ is a theorem. By (a), $\square \lnot \psi\rightarrow \square \lnot \varphi$. Taking the contrapositive again, $\lozenge \varphi\rightarrow \lozenge \psi$.

Ok, now for the proof. By $(T)$, $\square \lozenge P\rightarrow \lozenge P$. By (b), $\lozenge \square \lozenge P\rightarrow \lozenge\lozenge P$. By the derived rule $\lozenge\lozenge \varphi\rightarrow \lozenge \varphi$ you mentioned in your question, $\lozenge \square \lozenge P\rightarrow\lozenge P$. By (a), $\square\lozenge \square \lozenge P\rightarrow\square\lozenge P$, as desired.

Note that the proof used modus ponens, necessitation, $(K)$, $(T)$, and $(4)$ in the form of the derived rule from your question. The proof did not use $(D)$, and indeed $(D)$ is usually not taken as an axiom in S4 modal logic.

Title: Axiomatic proof in S4


问:
Unique solution in a quasi-convex optimization problem?
Let $f:\mathbb{R}^n\mapsto \mathbb{R}$ and $g:\mathbb{R}^m\mapsto \mathbb{R}$ be two differentiable functions and consider the optimization problem
$$
\min_{x\in \mathbb{R}^n} f(x)
$$
subject to
$$
g(x)\leq 0
$$
I am considering two cases here:

$f$ and $g$ are both strictly quasi-convex.

$f$ is linear (but not constant) and $g$ is strictly quasi-convex.

My questions are:

Can we show that under both scenarios there is a unique global minimizer $x$ that solves the optimization problem?

Are the Karush-Kuhn-Tucker conditions sufficient to characterize that global minimizer?

Also, is there a good reference for (strictly) quasi-convex optimization? Many thanks for any help!

答:
First, the problem may not have a global minimum. Think about the simple case of $\min x^3 \; \text{s.t.}\; x^3\le 0$.

Second, suppose the problem has a global minimum. Then, it's quite easy to prove that a strictly quasi-convex function $f(x)$ over a nonempty convex feasible region $C$ has a unique minimum: Suppose $x\ne y$ in $C$ are both global minima. Then, $z=\lambda x + (1-\lambda)y\in C$ for any $\lambda\in(0,1)$ and $f(z)<f(x)=f(y)$, which is impossible.
Note that differentiability is not needed. And any local minimum is a global minimum.

Third, for a linear objective function over a convex set $C$ defined by a strictly convex function, if the minimum exists, then it must be unique -- which follows from a similar argument as above, except noting that if $x,y\in C$ then $\lambda x + (1-\lambda)y$ is in the interior of $C$.

Fourth, since a strictly quasi-convex function still may have stationary points that are not local/global minima, then KKT conditions probably cannot characterize its global minimizer, if exists.


问:
Prove that a sequence converges iff that same sequence, minus the first m elements, also converges
I am very unsure of where to start this proof, and how it deviates from the simple proof that $x_{n}$ converges? Would somebody be able to set me on the right path?

Additionally, is this statement equivalent to saying "a sequence converges iff it has a convergent subsequence"?

答:
This statement is saying that a sequence converges iff the sequence obtained by removing the first $m$ terms also converges.

$(\Rightarrow)$ Use the fact that a subsequence of a convergent sequence is also convergent and converges to the same limit.

$(\Leftarrow)$ Say $(x_{n+m})$ converges to $x$. Let $\epsilon > 0$. There exists $N \in \mathbb{N}$ such that $|x_{n+m}-x| < \epsilon$ for all $n+m \geq N$. Use this to find an $M \in \mathbb{N}$ such that $|x_n - x| < \epsilon$ for all $n \geq M$.

Note, the iff statement you wrote at the end is not true. The sequence $(1,-1,1,-1,...)$ has a convergent subsequence, but is not convergent itself.


问:
Draw a graph of the following sinusoidal situation. Then write an equation in the form
Draw a graph of the following sinusoidal situation. Then write an equation in the form
$\displaystyle y\,\, = \,\,a\cos {{2\pi \left( {t\,\, - \,\,c} \right)} \over p}\,\, + \,\,b$

```<code>    The high tide in a harbour occurs at 2:30 am and the depth at high tide is 16 metres.
    8.4 hours later the low tide occurs. The depth of low tide is 4 metres.
</code>```

答:
Since high tide is $16$ and low tide is $4$, the amplitude is half the difference, which gives us $a=\dfrac{16-4}{2}=6$. By taking the average, the "middle" tide corresponds to the vertical displacement of the graph, giving us $b=\dfrac{4+16}{2}=10$. Note that 1 period occurs after we go from high tide, to low tide, then back to high tide. Thus, by symmetry, we have $p=2(8.4)=16.8$.

Now assume that $t$ represent the number of hours that have passed after $12$am (midnight). Then since high tide occurs at $2$:$30$am, we know that the graph contains the point $(2.5, 16)$. Hence, since the graph of $y=\cos x$ contains the point $(0,1)$, the phase shift must be $c=2.5$.

This gives us the equation:

$$
y=6\cos{\left(\dfrac{2\pi}{16.8}(x-2.5) \right)}+10
$$

You can see how Wolfram|Alpha plots this here. The important points you need to plot are:

$(2.5, 16)$

$(6.7, 10)$

$(10.9, 4)$

$(15.1, 10)$

$(19.3, 16)$

where the $x$-coordinates increase by a quarter of the period each time and the $y$-coordinates oscillate from its max to its min then back to its max.


问:
Why do these two equivalent equations have different implicit differentiations?
Using implicit differentiation, we have that the derivative of $4x^2y-3/y=0$ is    $$\frac{dy}{dx} = -\frac{8xy^3}{4x^2y^2+3}.$$ But, if we multiply both sides of the original equation by y, we have the following equation $4x^2y^2-3=0$ which is seemingly equivalent for $y≠0$. Taking the derivative of this new function yields $-y/x$ What is causing the difference here. Is it a rule of implicit differentiation or because the $0$ in the original equation that allows us to rewrite it in such a form? I am new to implicit differentiation and any help is appreciated.

答:
Notice that if you take your equation $4x^2y^2 - 3 = 0$ then you know that

\begin{align}
4x^2y^2 &= 3\tag{1}\\
y^2 &= \frac{3}{4x^2}.\tag{2}
\end{align}

Now, taking the equation $$\frac{dy}{dx} = -\frac{8xy^3}{4x^2y^2+3}$$ and using $(1)$ and $(2)$ we get $$\frac{dy}{dx} = \frac{-8xy\left(\frac{3}{4x^2}\right)}{6} = -\frac{24xy}{24x^2} = -\frac{y}{x}.$$  So, your two forms of the derivative are equivalent, given the original equation.  It is frequently the case that when doing implicit differentiation you may be able to obtain several seemingly different formulations of the derivative, which are equivalent given the original equation.  As we see here, sometime one is much simpler than the other.


问:
counting numbers with specific Quality
What is the count of 10 digit numbers with the exact sequences 72 and 081 in them with no repetition .
Is there any fast way to calculate this because of the 081 the places of 72 are limited and not coherent. For example placing 081 nearly at the end with 1 digit next to it potentially removes the ability to place 72 there</p>

答:
Stick $081$ together, ditto for $72$, then $5$ other single digits remain.

These "blocks" can be permuted in $7!$ ways but a number can't start with a $0$, so multiply by the Pr that the $081$ block is not at the start to get $\left(\dfrac6 7 \cdot 7!\right)$

Title: counting numbers with specific Quality


问:
Representations of a semigroup over a Hilbert space
A representations of a discrete semigroup $S$ with an involution $\star$ over a Hilbert space $H$ is a semigroup homomorphism $\varphi : S \to B(H)$ that preserve the involution. That is, for any $a, b\in S$

$\varphi(ab)=\varphi(a)\varphi(b)$

$\varphi(a^{\star})=\varphi(a)^{\star}$

There is a construction that represent any inverse semigroup over a Hilbert space which I found from this paper, but do not understand some parts of it. Let me rephrase it here as I understood:

It start with the $\star$-group-algebra $\mathbb{C}[S]=\left\{\sum_{\text{finite}}az_a : a\in S, z_a\in\mathbb{C}\right\}$ whose objects are formal finite linear combinations and addition, scalar multiplication, product, $\star$ operations defined in most obvious ways. Then completion of $\mathbb{C}[S]$ under $l_2$ norm $\left\Vert \sum_{}az_a\right\Vert_2=\sqrt{\sum_{} |z_a|^2}$ makes it into a Hilbert space denoted by $l^2(S).$ So we have a chain of inclusions $S\hookrightarrow \mathbb{C}[S]\hookrightarrow l^2(S)$ and we have a representation $\varphi : S\to B(l^2(S))$ given by $$\varphi(a)(b) =
\begin{cases}
 ba,  & \text{if } aa^{\star}\ge b^{\star}b\\
 0, & \text{otherwise}\end{cases}$$ and extended linearly and continuously. Now my questions are:

Is this the construction mentioned in the third page of this paper?

Is $l^2(S)$ same as the $C^{\ast}$-algebra mentioned in the first page as $C^{\ast}(S)$?

Where can I read about this construction with more details?

答:
Regarding your description of Duncan and Paterson's regular representation (which they call $\lambda $ and you are
calling $\varphi $), I think you should define $\varphi (a)(b)=ab$, as opposed to $ba$, and
require $a^*a\geq  bb^*$, rather than $aa^{*}\ge b^{*}b$.  Notice that with your definition you get an
anti-representation,  that is, $\varphi $ reverses multiplication in the sense that  $\varphi (a_1a_2) = \varphi (a_2) \circ \varphi (a_1)$.

$l^2(S)$ is not a C*-algebra.  It is a Hilbert space.  $C^*(S)$ is the envelopping C$^*$-algebra of $l^1(S)$.

Paterson's book [1] has a lot more on C*-algebras of inverse semigroups.  You might also want to to check [2], where a
different construction is made, and which is more suitable to certain important applications, such as the Cuntz-Krieger algebras.

[1] Paterson, Alan L. T., Groupoids, inverse semigroups, and their operator algebras, Progress in Mathematics (Boston, Mass.). 170. Boston, MA: Birkhäuser. xvi, 274 p. (1999). ZBL0913.22001.

[2] Exel, Ruy, Inverse semigroups and combinatorial (C^*)-algebras, Bull. Braz. Math. Soc. (N.S.) 39, No. 2, 191-313 (2008). ZBL1173.46035.


问:
In a formal axiomatic system with both the law of the excluded middle and the axiom of infinity does an uncountable set have to exist?
This was something I have been thinking around when examining writings of some “philosophers” who seemingly argue against the existence of a continuation but at the same time accept the law of the excluded middle and also the axiom of infinity.

I know from cantor’s proof of the unaccountability of the reals if you have something that obeys the real number axions then the set of all of them will be uncountable (if you have LEM).  However what I am interested in is if you just have the natural numbers and LEM does that entail the existence of some sort of uncountable set.

Clearly introducing the power set axiom would also entail an uncountable set so that isn’t the type of FAS I am interested in.

答:
Certainly not: consider (in the setting of classical logic) any structure in which there is an infinite set but no countable set, such as $L_{\omega+1}$, and then just consider the theory of that structure.

(Note that in the case of $L_{\omega+1}$ we have something even better - a definable bijection with $\omega$. By contrast, something like $L_{\omega_1}$ is "locally countable" but not "internally globally countable.")

I suspect that you want some additional properties of the theory in question, but I'm not sure what.


问:
Simplifying a radical equation?
I'm lost on how to simplify this.

Simplify:
  $$\frac{\sqrt[\large 4]{144x^9y^8}}{\sqrt[\large 4]{9x^5y^{-3}}}$$

OK So I think I got it.

$$\frac{\sqrt[\large 4]{144x^9y^8}}{\sqrt[\large 4]{9x^5y^{-3}}} = \left(\frac{{144x^9y^8}}{{9x^5y^{-3}}}\right )^{1/4}$$

You can break that down like this -I think-:

$$\sqrt[\large 4]{16x^4y^{11}}$$

And then finally:

$$2xy^2\sqrt[\large 4]{y^3}$$

EDIT - You can go even further (Thanks amWhy):

$$2xy^{\large \frac{11}{4}}$$

答:
First, recall that

$$ \frac{\sqrt[\large4]{144x^9y^8}}{\sqrt[\large4]{9x^5y^{-3}}} = \sqrt[\large 4]{\frac{144x^9y^8}{9x^5y^{-3}}} = \left(\frac{144x^9y^8}{9x^5y^{-3}}\right)^{1/4}$$

Now, use the "laws" of exponents to simplify the rational expression, (the fraction inside parentheses), and then take distribute the exponent $1/4$ over the factors remaining after simplifying.

ADDED: You are almost there: you can also write the exponent of $y$ as a fraction. Recall that $\;\;(y^a)^b = y^{ab}.\;$ In your case, $\,a = 3$, and $\;b=1/4$:

$$2xy^2\sqrt[4]{y^3} = 2xy^2\left((y^3)^{1/4}\right) = 2x y^2 \left(y^{3/4}\right) = 2x\left(\color{blue}{\bf y^{\left(2 + \frac 34\right)}}\right)\quad ?$$

$\color{blue}{\bf \left(y^cy^d = y^{(c + d)}\right)}$

Title: Simplifying a radical equation?


问:
Uniform Problem - Expected Value
A fraternity is throwing a party for its members. The cost of booking a band to play at the party, the amount that the supplier is going to charge, the cost of renting some place and some other costs are random uniform variables over the intervals (1300,1800),(1800,2000),(800,1200) and (400,700) respectively. If the number of people invited is a random integer in the interval (150,200). What is the minimum amount, in average, the fraternity will have to charge each person to not lose money?

What I did was: The sum of all the costs and then the average, that gave me 5000, then the average of the attendees, 175. I divided them. But I don't know how to continue.

答:
Here is a simulation in R of total costs for the party and cost per ticket.

```<code>set.seed(2021)
m=10^5
x1 = runif(m, 1300, 1800)
x2 = runif(m, 1800, 2000)
x3 = runif(m, 800, 1200)
x4 = runif(m, 400, 700)
t = x1 + x2 + x3 + x4
n = sample(150:200, m, rep=T)

summary(t);  sd(t)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
   4333    4850    5000    5000    5150    5673
[1] 212.0116
</code>```

You can sum the means of the four uniform distributions
to get the mean of the total cost (about $\$5000).$ Also, you can sum
the variances of these four distributions (provided they're independent) to get the
variance of the total cost; the take the square root to
get the standard deviation (about $\$212.)$ The distribution is not exactly normal but nearly normal.

```<code>hist(t, prob=T, col="skyblue2",
     main="Simulated Dist'n of Cost")
</code>```

```<code>summary(t/n)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  21.88   26.62   28.55   28.77   30.81   37.40
q = quantile(t/n, .95);  q
     95%
33.45204
hist(t/n, prob=T, col="skyblue2",
     main="Approx. Dist'n per Ticket")
 abline(v = q, col="red", lwd=2)
</code>```

If the probability models for costs and attendance are correct, the fraternity is unlikely to lose
money charging $\$33.50$ per ticket. [If the ticket price were set at $\$28.55$ (median of distribution) there would be a 50:50 chance of losing money.]

Title: Uniform Problem - Expected Value

Body:
Here is a simulation in R of total costs for the party and cost per ticket.

set.seed(2021)
m=10^5
x1 = runif(m, 1300, 1800)
x2 = runif(m, 1800, 2000)
x3 = runif(m, 800, 1200)
x4 = runif(m, 400, 700)
t = x1 + x2 + x3 + x4
n = sample(150:200, m, rep=T)

summary(t);  sd(t)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
   4333    4850    5000    5000    5150    5673
[1] 212.0116

You can sum the means of the four uniform distributions
to get the mean of the total cost (about $\$5000).$ Also, you can sum
the variances of these four distributions (provided they're independent) to get the
variance of the total cost; the take the square root to
get the standard deviation (about $\$212.)$ The distribution is not exactly normal but nearly normal.

hist(t, prob=T, col="skyblue2",
     main="Simulated Dist'n of Cost")

summary(t/n)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  21.88   26.62   28.55   28.77   30.81   37.40
q = quantile(t/n, .95);  q
     95%
33.45204
hist(t/n, prob=T, col="skyblue2",
     main="Approx. Dist'n per Ticket")
 abline(v = q, col="red", lwd=2)

If the probability models for costs and attendance are correct, the fraternity is unlikely to lose
money charging $\$33.50$ per ticket. [If the ticket price were set at $\$28.55$ (median of distribution) there would be a 50:50 chance of losing money.]


问:
Colouring of lines in a polygon to find relations between people at a party
There are $7$ people at a party. Each of them has 2 friends, 2 enemies and 2 strangers. If all relations are mutual (i.e. if A is friend of B, then B is friend of A), find a way for such an arrangement.
 Note: this is not the original question, so it's ok if you can't find a solution (I'm not sure there is one myself).

 My attempt was to take a heptagon and colour all of its diagonals and sides red (enemy), blue (stranger) and green (friend). I'm not sure how to proceed, though as the number of cases gets really big.

 I'm sharing a Geogebra link to experiment with this... https://www.geogebra.org/m/n7pmwgvd

答:
This is one such coloring, which I believe answers your question

To achieve that, notice that each type of relation forms a cycle on the vertices, e.g., as everyone has exactly two enemies, you can start with one person, go to their enemy, then to the next enemy and so on without repeating anyone. The same applies to the other two relations as well.


答:
Assume $x_n \to L$, we show $x_{n+m} \to L,\forall m \ge 1$. Let $\epsilon > 0$ be arbitrary, $\exists N \ge 1$ such that: $|x_{n+m} - x_n| < \dfrac{\epsilon}{2}, |x_n - L| < \dfrac{\epsilon}{2}, n \ge N\implies |x_{n+m} - L| < |x_{n+m} - x_n|+|x_n - L| < \dfrac{\epsilon}{2}+\dfrac{\epsilon}{2} = \epsilon\implies x_{n+m} \to L$.
For the other direction, if $x_{n+m} \to L$, we show $x_n \to L$. Let $\epsilon > 0$ be given, $\exists N \ge 1$ such that if $n+m \ge N$ then $|x_{n+m} - L| < \epsilon$. Now let $N_1 = N + m$, then if $n \ge N_1\implies n - m \ge N\implies (n-m)+m \ge N+m > N \implies |x_{(n-m)+m} - L|< \epsilon\implies |x_n - L| <\epsilon\implies x_n \to L$.

Title: Prove that a sequence converges iff that same sequence, minus the first m elements, also converges

Body:
I am very unsure of where to start this proof, and how it deviates from the simple proof that $x_{n}$ converges? Would somebody be able to set me on the right path?

Additionally, is this statement equivalent to saying "a sequence converges iff it has a convergent subsequence"?


答:
No.  For instance, if you omit the power set axiom from ZFC, you get a theory that is consistent with the non-existence of uncountable sets.  In particular, working in ZFC, consider the set of hereditarily countable sets.  This is a model of every axiom of ZFC except for the power set axiom, and in this model every set is countable.

(This theory is also quite a bit more powerful than you might think and is strong enough for most of mathematics.  For instance, you don't have a set of real numbers, but you can still do basically all of real analysis by treating the real numbers as a class, and describing sufficiently nice functions $\mathbb{R}\to\mathbb{R}$ using a countable amount of data.)

Title: In a formal axiomatic system with both the law of the excluded middle and the axiom of infinity does an uncountable set have to exist?

Body:
This was something I have been thinking around when examining writings of some “philosophers” who seemingly argue against the existence of a continuation but at the same time accept the law of the excluded middle and also the axiom of infinity.

I know from cantor’s proof of the unaccountability of the reals if you have something that obeys the real number axions then the set of all of them will be uncountable (if you have LEM).  However what I am interested in is if you just have the natural numbers and LEM does that entail the existence of some sort of uncountable set.

Clearly introducing the power set axiom would also entail an uncountable set so that isn’t the type of FAS I am interested in.


问:
Use Schur to prove that for any norm $\lim_{n \rightarrow \infty} ||A^n|| = 0 \text{ iff } \rho(A) < 1$ where $\rho$ denotes spectral radius.
I know I need to prove both directions of the biconditional, but am having some trouble in both ways.

First, looking left to right, If we assume the limit equals zero, it means that the $A$ matrix keeps getting smaller each time we apply $A$. Thus, since eigenvalue represent the effect of the matrix the maximum would have to be less than one but this doesn't seem very rigorous.

Then, I'm not sure how to proceed by assuming the spectral radius is less than 1 except for doing the backwards reasoning of what I talked about above. Any help is appreciated.

答:
Let $A=UTU^\ast$ be a unitary triangulation. Then $\rho(T)=\rho(A)$. Since all norms are equivalent on a finite-dimensional vector space, we may assume that the norm in question is the induced $2$-norm (i.e. the largest singular value, a.k.a. operator norm or spectral norm) of a matrix. Thus $\|A^n\|=\|T^n\|$ and the problem reduces to showing that $\lim_{n\to\infty}\|T^n\|=0$ if and only if $\rho(T)<1$ for any upper triangular matrix $T$.

Suppose $\lim_{n\to\infty}\|T^n\|=0$. Since $0\le\rho(T)^n=\rho(T^n)\le\|T^n\|$, we have $\lim_{n\to\infty}\rho(T)^n=0$ and hence $\rho(T)<1$.

Conversely, suppose $\rho(T)<1$. Let $T$ be $m\times m$ and write $T=D+F$, where $D$ and $F$ are respectively the diagonal part and strictly upper triangular part of $T$. Since the product of a diagonal matrix and a strictly upper triangular matrix is strictly upper triangular, and the product of $m$ or more $m\times m$ strictly upper triangular matrices is zero, if we expand $T^n=(D+F)^n$ as a sum of monomials in $D$ and $F$, all terms containing at least $m$ copies of $F$ are zero. It follows that when $n\ge m-1$,
$$
\|T^n\|
\le\sum_{k=0}^{m-1}\binom{n}{k}\|F\|^k\|D\|^{n-k}
=\sum_{k=0}^{m-1}\binom{n}{k}\|F\|^k\rho(T)^{n-k}.
$$
Since $\rho(T)<1$, we have $\lim_{n\to\infty}\binom{n}{k}\rho(T)^{n-k}=0$ for each $k$. Consequently, $\lim_{n\to\infty}\|T^n\|=0$.


问:
How many outcomes are possible?
An experiment consists of rolling three dice, and then choosing 2 cards at random from a standard deck of 52 cards. How many outcomes are possible?</p>

答:
Since the results of dices and cards are independent.

The number of possible results of dices is

$$
N_{d}=\binom{6}{3}+2\times\binom{6}{2}+\binom{6}{1}=56
$$

And the number of possible results of cards is

$$
N_c=\binom{52}{2}=1326
$$

So the number of possible outcomes is

$$
N_d\times N_c=74256
$$

Title: How many outcomes are possible?


问:
Prove GCD statement with multiple variables
For positive integers $a_{1}, a_{2}\cdots, a_{k},$ define $\gcd(a_{1}, a_{2}\cdots, a_{k})$ to be the largest positive integer $d$ such that $d$ divides every $a_{i}$ and any positive integer $c$ that divides every $a_{i}$ also has to divide $d.$ Is it true that there are integers $m_{i},$ not necessarily positive, such that $d= \sum_{i= 1}^{k}m_{i}a_{i},$ right ?

How can i prove this ? I was thinking maybe one could use mathematical induction but i'm not quite sure.

答:
First Prove that if $\gcd(a_1, ...., a_k) = d_k$ then $\gcd(a_1,a_2,.....,a_k, a_{k+1}) = \gcd(d_k, a_{k+1})=d_{k+1}$.

...

Then we can prove our induction step: that if there exist $m_i$ so that $\sum_{i=1}^k m_i a_i = d_k$, and by Bezout, there are integers $N, P$ so that $Nd_k + Pa_{k+1} = d_{k+1}$--- so if we define $b_i = a_i*N$ for $i = 1...k$ and $b_{k+1} = P$ then we have $\sum_{i=1}^{k+1} b_i a_i = d_{k+1}$.


问:
Show Lipschitz condition for $f(x) = x+|x|$
I am struggling to prove $$x+|x|$$ to be locally Lipschitz. Since it is not continuously differentiable - not differentiable at $x=0$ - it is hence not globally Lipschitz. But how do I proceed for the local case?

$$|f(x) - f(y)| \leq L|x-y|$$
$$\Rightarrow |x+|x|-y-|y||=...$$

答: $|x+|x|-y-|y||=|(x-y)+(|x|-|y|)|\le |x-y|+||x|-|y|| \le |x-y|+|x-y| = 2|x-y|$. Thus $k = 2$. The key inequality is $xy \le |xy| = |x||y|$ which leads to $||x| - |y|| \le |x-y|$ which can be shown by simply squaring both sides.


答:
Hint:  Start by using the fact that $$\frac{\sqrt[4]{a}}{\sqrt[4]{b}}=\sqrt[4]{\frac ab},$$ and see what you can do on the inside of the radical. Then use the fact that $\sqrt[4]c=c^{1/4}$ with exponent rules. Don't forget that $16=2^4.$

Edit: Bingo! Nicely done.

Title: Simplifying a radical equation?

Body:
I'm lost on how to simplify this.

Simplify:
  $$\frac{\sqrt[\large 4]{144x^9y^8}}{\sqrt[\large 4]{9x^5y^{-3}}}$$

OK So I think I got it.

$$\frac{\sqrt[\large 4]{144x^9y^8}}{\sqrt[\large 4]{9x^5y^{-3}}} = \left(\frac{{144x^9y^8}}{{9x^5y^{-3}}}\right )^{1/4}$$

You can break that down like this -I think-:

$$\sqrt[\large 4]{16x^4y^{11}}$$

And then finally:

$$2xy^2\sqrt[\large 4]{y^3}$$

EDIT - You can go even further (Thanks amWhy):

$$2xy^{\large \frac{11}{4}}$$


答: $f(x)=x+|x|=\max\{0,2x\}$, so it's globally Lipschitz with constant 2.

Title: Show Lipschitz condition for $f(x) = x+|x|$

Body:
I am struggling to prove $$x+|x|$$ to be locally Lipschitz. Since it is not continuously differentiable - not differentiable at $x=0$ - it is hence not globally Lipschitz. But how do I proceed for the local case?

$$|f(x) - f(y)| \leq L|x-y|$$
$$\Rightarrow |x+|x|-y-|y||=...$$


问:
Cube root of $-2+i$
Edit:

My question comes from finding the solutions of this equation using Cardano's method(because our teacher said :D ):

$$x^3-6x+4=0$$

And finally I got:

$$x=(\sqrt[3]{2})\sqrt[3]{-2+\sqrt{-1}}⠀+(\sqrt[3]2)\sqrt[3]{-2-\sqrt{-1}}$$

I want to denest and find the cube roots of this two radicals.

According to this links answers:

First link

Second link

Third link

I tried to find this using norm(actually for dinest):

$$\sqrt[3]{-2+i}$$

So the problem's norm is $5$ and I must find a root with norm = $\sqrt[3]{5}$  .

And I found this :

$$(\sqrt{\sqrt[3]{5}⠀-1}⠀+i)$$

But:

$$(\sqrt{\sqrt[3]{5}⠀-1}⠀+i)^3\neq {-2+i}$$

Anybody can tell me whats my wrong?

And how to correct that?

答:
The given number can be presented in the forms of
$$-2+i=\sqrt5e^{i\varphi}=\sqrt5\left(\cos\varphi+i\sin\varphi\right),$$
where
$$\cos\varphi = -\dfrac2{\sqrt5},\quad \sin\varphi=\dfrac1{\sqrt5},\quad \tan\dfrac\varphi2 = \dfrac{\sin\varphi}{1+\cos\varphi}=\dfrac1{\sqrt5-2}=\sqrt5+2,$$
$$\varphi=2\arctan(\sqrt5+2)\approx2.678.\tag1$$
Taking in account the periodicity
$$\sqrt5e^{i\varphi} = \sqrt5e^{i\varphi+2i\pi} = \sqrt5e^{i\varphi-2i\pi},$$
one can get three different values of the required cubic root in the form of
$$\sqrt[3]{-2+i}\,=\sqrt[6]5e^{i\large\frac{\varphi+2\pi k}3}, k=0,1,-1.\tag2$$
Then, for example,
$$\sqrt[6]5e^{i\large\frac{2\arctan(5+\sqrt2)}3}\approx0.82036+1.01832i.$$

Using of the half-argument in the tangent function allows to avoid problems with the quadrant accounting.

Title: Cube root of $-2+i$


问:
Rudimentary Infinite Series question
A ball is dropped from a height of 10m. Every time it bounces, it returns to $\frac{1}{3}$ of its previous height. When is the first time that the bounce is below $10^{-2}$m, and when is it the first time below $10^{-7}$m? To do this I would obviously need an infinite series. It would be modeled by $10 + \frac{10}{3} + \frac{10}{3} + \frac{10}{9} + \frac{10}{9} + \frac{10}{27} + \frac{10}{27}...$. I had the idea of factoring one of the fractions and turning it into a series, solve for the total distance, etc. How would I figure out what the question asks though? How do I tell when the ball first reaches any distance?

答:
Welcome to MSE!

I think the infinite series is not necessary in this case, more so, an infinite sequence would be needed.

Let $i$ be the number of bounces, $h_i$ is the height of the ball after $i$ bounces

$h_0 = 10, h_1 = \frac{10}{3}, h_2 = \frac{10}{3^2},..., h_i =\frac{10}{3^i},...$

thus you would need to find $i$ approximately for $h_i \leq 10^{-2},10^{-7}$.

$h_i = \frac{10^1}{3^i} \leq \frac{1}{10^2}, \frac{1}{10^7}$

$3^i \geq 10^3 , 10^8 \:\:\:\:\text{and then}\:\:\:\: i\cdot\ln(3) \geq 3\cdot\ln(10), 8\cdot\ln(10)$

for $10^{-2}$, $i \geq 3\cdot\frac{\ln(10)}{\ln(3)} \approx 6.2877$, so after the $7th$ bounce

for $10^{-7}$, $i \geq 8\cdot \frac{\ln(10)}{\ln(3)} \approx 16.767226$, so after the $17th$ bounce

I think your series represents the total distance travelled by the ball.

To model the total distance travelled by the ball is not hard either

Let $S_n = \sum\limits_{i=0}^n \left(\frac{1}{3}\right)^i$ then $S_n -\frac{1}{3}S_n = \frac{2}{3}S_n = 1-\left(\frac{1}{3}\right)^{n+1}$

thus $S_n = \frac{3}{2}\cdot\left(1-\left(\frac{1}{3}\right)^{n+1}\right)$

and you can model the total distance by $2\cdot 10 \cdot S_n -10 = 10\cdot \left(2-\left(\frac{1}{3}\right)^{n}\right)$.

Title: Rudimentary Infinite Series question


问:
Existence of specific non-symplectic manifolds
I am fairly new to symplectic geometry but I started wondering about this :

From the definition of a symplectic manifold we can extract that $M$ has to be even dimensional , orientable , and in the case that it's compact $H^2_{dR}(M)\neq 0$.  Now is there extra conditions that we can impose on even-dimensional compact manifold with  $H^2_{dR}(M)\neq 0$ such that $M$ will be symplectic , other than having a symplectic form of a $2-$form $\omega$ such that $\omega^n$ is a volume form ? And does anybody know if there exists an example of a compact manifold that satisfies all those 3 conditions such that it is not symplectic ? Or if it will automatically be a symplectic manifold?

Thanks in advance.

答: In general there are no purely topological conditions one can impose to ensure the existence of a symplectic structure; it necessarily involves the smooth structure. There are examples of smooth 4-dimensional manifolds that are homeomorphic to symplectic (even K\”ahler) manifolds but which admit no symplectic structure themselves. Many such examples were constructed by Fintushel and Stern in their paper “Knots, links and 4-manifolds,” but I’m not sure if these were the first.</p>


问:
Is the image of closed unit ball under the linear operator closed?
I have the linear operator $A:l_1 \rightarrow L_3(0,+\infty)$ which is defined with the formula $(Ax)(t)=\sum\limits_{k=1}^{+\infty} \frac{x(k)}{\sqrt{k}+\sqrt{t}} $ for all $t\in (0,+\infty)$ and for all $x\in l_1$. The full task sounds like this: find the adjoint operator $A^*$ for the operator $A$, then check if the set $A(B_1(0))$ is totally bounded and closed in $L_3(0,+\infty)$. I have found the adjoint operator, then using the criterion of Kolmogorov and Frechet I proved total boundedness of $A(B_1(0))$. But I do not know how to check this set for closedness. It seems difficult to check it with the definition. Also, I cannot see what theorems I may apply. This task is given in the topic about adjoint operators, so maybe I should use some results about adjoint operators, but I am not sure about it.

答:
Consider the sequence $x_n\in B_1(0)\subset l_1$ and let $Ax_n\to y$ in $L_3(0,\infty)$. Then $\exists Ax_{n_k}\to y$ almost everywhere, i.e. $\displaystyle\sum\limits_{s=1}^{\infty} \frac{x_{n_k}(s)}{\sqrt{s}+\sqrt{t}}\to y(t)$ a.e. As you know, a closed unit ball in $l_1$ lies in a closed unit ball of space $l_2$. Since $l_2$ is reflexive, $\exists x_{n_{k_r}}\rightharpoonup z\in l_2$ and $\|z\|_2\leq1$ (since the closed unit ball in $l_2$ is weakly sequentially compact). Since $\forall N\in\mathbb{N}$ $\displaystyle\sum\limits_{s=1}^{N}|x_{n_{k_r}}(s)|\leq1$ and weak convergence implies the coordinatewise, then $\forall N\in\mathbb{N}$ $\displaystyle\sum\limits_{s=1}^{N}|z(s)|\leq1$, so $\|z\|_1\leq1$. Next, for $\varepsilon>0$ choose $M$ so, that $\dfrac{2}{\sqrt{M+1}}<\dfrac{\varepsilon}{2}$ and consider $u=\left(\dfrac{1}{\sqrt1+\sqrt t},\dfrac{1}{\sqrt2+\sqrt t},...,\dfrac{1}{\sqrt M+\sqrt t},0,0,...\right)\in l_2$, then $$\left|\sum\limits_{s=1}^{\infty} \frac{x_{n_{k_r}}(s)}{\sqrt{s}+\sqrt{t}}-\sum\limits_{s=1}^{\infty} \frac{z(s)} {\sqrt{s}+\sqrt{t}}\right|\leq\left|\sum\limits_{s=1}^{M} \frac{x_{n_{k_r}}(s)-z(s)}{\sqrt{s}+\sqrt{t}}\right|+\left|\sum\limits_{s=M+1}^{\infty} \frac{x_{n_{k_r}}(s)-z(s)}{\sqrt{s}+\sqrt{t}}\right|\leq$$$$\leq\left|\sum\limits_{s=1}^{\infty} (x_{n_{k_r}}(s)-z(s))u(s)\right|+\dfrac{2}{\sqrt{M+1}}<\varepsilon,$$
for large enough $r$, because first sum tends to $0$.

Thus, we have $Ax_{n_{k_r}}=\displaystyle\sum\limits_{s=1}^{\infty} \frac{x_{n_{k_r}}(s)}{\sqrt{s}+\sqrt{t}}\to\sum\limits_{s=1}^{\infty} \frac{z(s)} {\sqrt{s}+\sqrt{t}}=Az=y$ a.e., where $z\in B_1(0)$.


问:
Zeroes of $x^{p^n} - x$ are distinct in splitting field of $\mathbb{F}_p$
I am attempting to show that the zeroes of $x^{p^n} - x$ are distinct (as a part of showing the splitting field over $\mathbb{F}_p$ has $p^n$ elements). I understand that it is sufficient to show that $g(x)=\frac{x^{p^n} - x}{x-\alpha}$ is non-zero for all $\alpha$ that are zeroes of $x^{p^n} - x$ on the splitting field. Clearly, $0$ is one such $\alpha$, and $g(0)$ is undefined. But for the purpose of my proof, being undefined still implies that $g(0) \not = 0$, is that correct? Just asking for a sanity check.

答:
It is really bad idea to call concepts that are "undefined" an instance of not being $0$. You're supposed to be using legitimate calculations the whole time.

Method 1: Use derivatives of polynomials and the fact that $f(x)$ is separable if and only if $(f(x),f'(x))$ is a nonzero constant.  In characteristic $p$, $x^{p^n}-x$ has an especially simple derivative.

Method 2: If $\alpha$ is a root of $x^{p^n} - x$ in some field $K$ containing $\mathbf F_p$ then
$$
x^{p^n} - x = x^{p^n} - x - (\alpha^{p^n} - \alpha) = (x-\alpha)^{p^n} - (x-\alpha),
$$
where I rewrote $x^{p^n} - \alpha^{p^n}$ as $(x-\alpha)^{p^n}$ by taking advantage of the fact that the $p$th power mapping (as well every iteration of it) is additive in rings of characteristic $p$ like $K[x]$. Now go ahead and take out a factor of $x-\alpha$ from each terms in $(x-\alpha)^{p^n} - (x-\alpha)$ and see what the value at $\alpha$ is of the remaining part.  You should get a nonzero value, so $x-\alpha$ is not a repeated factor of $x^{p^n} - x$ in $K[x]$.


问:
Does there exist a differentiable function from $\mathbb R^3$ to $\mathbb R$ whose zeros are a line?
Suppose we are given a line in $\mathbb R^3$.  Is there a differentiable function, $f$, from $\mathbb R^3$ to $\mathbb R$ such that the solutions to $f(\vec{x})=0$ are precisely the points on the line?

I would like to be able to describe a line using the zeros of a single function rather than using a parametrization.

EDIT: I deleted a false claim about not being able to use a polynomial.

答:
If we assume your line is given by

$$\vec L(t) = \vec p + t\vec v$$

where $||\vec v || = 1$, then for any point $\vec x $, the point on $\vec L$ that is closest to $\vec x$  (let's call it $proj(\vec x) $) is given by

$$proj( \vec x ) = \vec p + ((\vec x - \vec p)\cdot\vec v) v.$$

Then you can let

$$f(\vec x) = ||\vec x - proj(\vec x)||^2.$$

Incidentally, you can write this out explicitly and see that it is differentiable. In fact, it should be a quadratic in the components of $\vec x$. Which means either I made a mistake, or your statement that "I know there are no polynomials with this property" has a problem. I'm not yet 100% sure which -- can you say why you believe that to be so?


问:
Rolling a die. Math problem
Consider the experiment of rolling a die until "6" appears. How many outcomes are in the event that exactly four rolls are required to get a "6"?</p>

答:
For the game to stop at the 4th turn $6$ must come at the fourth position and no 6 should come before it so the first 3 outcomes can conatain any number except 6 thus the answer is $5 \times 5 \times 5=125
$

Title: Rolling a die. Math problem


问:
Simple Graph Transformation Question $\rightarrow$ $1/f(x)$
for the graph:

such that the function is : $ y = \frac{a+x}{b+cx} $ where a = -2, b = 1 and c = 1/2

how do you sketch the graph of $ y = |\frac{b+cx}{a+x}| $  ??

i got that the VA of the new graph is "$+2$" and the root of the new graph is "$-2$", but then the answer says that the horizontal asymptote is "$\frac{1}{2}$" which i don't really understand..

Could anyone explain to me what to happen, if i am to transform a graph from $f(x)$ to $\frac{1}{f(x)}$.

i only know that the roots of $f(x)$ will become VA value of $1/f(x)$ and VA values of $f(x)$ becomes the zeros of $\frac{1}{f(x)}$

and that if $f(x) \gt 0$ , then $\frac{1}{f(x)}$ $\gt 0$ such that $f(x) \to 0 \implies \frac{1}{f(x)} \to \infty$ and $f(x) \to \infty \implies \frac{1}{f(x)} \to  0$.

答:
You have a function
$$f(x) = \frac{x-2}{\frac{x}{2} + 1}$$
and you want to sketch $y = \frac{1}{f(x)}$.

The $y$-intercept of the graph $y = f(x)$ is given by $f(0)$ (let this be $y_0$). The $y$-intercept of the graph $y = \frac{1}{f(x)}$ is given by $\frac{1}{f(0)}$, which is the reciprocal of $y_0$.

Similarly, the old horizontal asymptote, $y = \displaystyle\lim_{x\to\infty}f(x) = 2$, will become $y = \displaystyle\lim_{x\to\infty} \frac{1}{f(x)} = \frac{1}{\displaystyle\lim_{x\to\infty}f(x)} = \frac{1}{2}$, since $f(x)$ is being replaced by $\frac{1}{f(x)}$.

In general, the $y$ values of $y = f(x)$ will be the reciprocals of $y = \frac{1}{f(x)}$, i.e.,

$$y_{\text{new}} = \frac{1}{y_{\text{old}}}.$$

To then sketch the absolute value of this function, simply reflect any part below the $x$-axis above it, so there are no negative $y$ values for any $x$.

Pictures speak a thousand words...

Note that for each $x$ value, the $y$ values are reciprocals:

Reflecting the part below the $x$-axis:

Also, technically $f$ is not defined at $x = -2$, so neither should $\frac{1}{f}$ or $\left | \frac{1}{f} \right |$, i.e., there are holes at $x=-2$ on the blue graphs.


问:
What is an example of a discontinuous function from $\mathbb{R}^{\omega}$ such that sequential limit is preserved?
In point-set topology class we proved that if a topological space $X$ is first countable, then for any function $f:X\rightarrow Y$, where $Y$ is some topological space, the condition $x_n\rightarrow x \Rightarrow f(x_n)\rightarrow f(x)$ implies $f$ is continuous.

It is known that $\mathbb{R}^{\omega}$ with box topology is not first countable. An exercise is to find an example $f:\mathbb{R}^{\omega}\rightarrow Y$, such that $x_n\rightarrow x \Rightarrow f(x_n)\rightarrow f(x)$ but $f$ is not continuous. I have been struggling to find an example. Any ideas?

答:
I will sketch the construction and verification of one example. There may be others that are simpler, but this is the first that occurs to me, and filling in the many missing details is a good exercise.

Let $z$ be the zero sequence in $\Bbb R^\omega$, and let $\langle x_n:n\in\Bbb N\rangle$ be a sequence in $\Bbb R^\omega$. Suppose that for each $n\in\Bbb N$ there are $m(n),k_n\ge n$ such that $x_{m(n)}(k_n)\ne 0$. Then it is not hard to construct an open nbhd $U$ of $z$ such that $x_{m(n)}\notin U$ for each $n\in\Bbb N$, thereby showing that $\langle x_n:n\in\Bbb N\rangle$ does not converge to $z$. Thus, if $\langle x_n:n\in\Bbb N\rangle\to z$, there must be an $n_0\in\Bbb N$ such that $x_m(k)=0$ for all $m,k\ge n_0$. This extends easily to show that if $\langle x_n:n\in\Bbb N\rangle\to x$, then there is an $n_0\in\Bbb N$ such that $x_m(k)=x(k)$ for all $m,k\ge n_0$. This isn’t quite as straightforward as saying that the only convergent sequences are the ones that are eventually constant, but it’s similar, and for a suitable choice of $Y$ we can use it in much the same way.

Define an equivalence relation $\sim$ on $\Bbb R^\omega$ as follows:

$$x\sim y\text{ iff }\{n\in\Bbb N:x(n)\ne y(n)\}\text{ is finite.}$$ (Equivalently, $x\sim y$ iff there is an $n_0\in\Bbb N$ such that $x(k)=y(k)$ for all $k\ge n_0$.) For each $x\in\Bbb R^\omega$ let $[x]$ be the $\sim$-equivalence class of $x$. Let $Y=\Bbb R^\omega/\!\sim$ with the discrete topology, so that the points of $Y$ are the $\sim$-classes $[x]$ for $x\in\Bbb R^\omega$. Let

$$f:\Bbb R^\omega\to Y:x\mapsto[x]\,;$$

then $f$ is not continuous, but $\langle f(x_n):n\in\Bbb N\rangle\to f(x)$ in $Y$ whenever $\langle x_n:n\in\Bbb N\rangle\to x$ in $\Bbb R^\omega$.


问:
Matrix normal, then is the same true for the fundamental system?
Let $A : [0,T]\to\mathbb R^{n\times n}$ be a matrix function such that each $A(t)$ is normal, i.e., $A(t)^TA(t) = A(t)A(t)^T$ for all $t\in [0,T]$. Now, consider the ODE system $\dot x(t) = A(t)x(t)$. We know that there exists a $n\times n$ fundamental matrix function $\Phi(t)$ such that the columns of $\Phi$ are linearly independent solutions of the ODE, that is, $\dot\Phi(t) = A(t)\Phi(t)$ for $t\in [0,T]$ and fixing the value $\Phi(0)$ also fixes the fundamental system.

My question:

If $A$ is normal as above, does the same hold for the solution of
$\dot\Phi(t) = A(t)\Phi(t)$ with $\Phi(0) = I_n$?

For constant $A$ this is true as the solution is given by $\Phi(t) = e^{tA}$, which is easily seen to be normal for each $t$. It's also true when $A(t)$ is diagonal for each $t$.

答:
No, this is not true.  Let $$A(t)=\begin{cases}\begin{bmatrix}0&\log{(2)}\\\log{(2)}&0\end{bmatrix} & 0\leq t<1 \\
\begin{bmatrix}0&\frac{\pi}{4}\\-\frac{\pi}{4}&0\end{bmatrix} & t\geq1\end{cases}$$ Then $\Phi(0)=I_2$, $\Phi(1)=\frac{1}{4}\begin{bmatrix}5&3\\3&5\end{bmatrix}$ (so far so good) but $\Phi(2)=2\sqrt{2}\begin{bmatrix}4&4\\-1&1\end{bmatrix}$, which is not normal.

To see how I got this, note that this sort of phenomenon is generic: assuming $A$ constant, on a finite interval, the solution $\Phi$ is continuous in $A$.  So if we can approximate $A$ in the uniform norm, then $\Phi$ will depend continuous on the approximation (in, e.g., the uniform norm on $(\Phi,\dot{\Phi})$).   In particular, we can approximate $A$ by a piecewise-constant function.

So suppose $A$ attains value $M_1$ on $[0,t_1)$ and $M_2$ on $[t_1,t_2)$, etc.  Then $$\Phi(t_2)=e^{M_2t_2}e^{M_1t_1}\Phi(0)$$  Now, if $A$ is very nice, and $M_1$ and $M_2$ commute, then we can reduce this to $e^{M_2t_2+M_1t_1}\Phi(0)$.  If they don't commute, this value is off...but not by much; c.f. the Baker-Campbell-Hausdorff formula.  In any case, this is a real danger: a sum of normal matrices need not be normal!  A classic example is $$\begin{bmatrix}0&a+b\\a-b&0\end{bmatrix}=\begin{bmatrix}0&a\\a&0\end{bmatrix}+\begin{bmatrix}0&b\\-b&0\end{bmatrix}$$  The remainder is just messing around with the constants.

Edit: the OP has asked for a $C^1$ example.  The key idea is that the solution varies continuously in $A$, as measured in the $L^1$ norm of $A$.  More precisely: let $B$ match $A$ on $[0,1-\epsilon]$ and $[1+\epsilon,2]$, and quadratically interpolate between.  Say $\Phi_A$ and $\Phi_B$ solve the obvious systems; clearly $\|A-B\|$, $\|\Phi_A\|$, and $\|\Phi_B\|$ are all bounded (in norm) on $[0,1+\epsilon]$ by, say, $M$.  Then $$\|\Phi_B(1+\epsilon)-\Phi_A(1+\epsilon)\|\leq M((1+\epsilon)-(1-\epsilon))\sup_{s\in(1-\epsilon,1+\epsilon)}{\|A(s)-B(s)\|}\leq2M\epsilon$$  On $[1+\epsilon,2]$, we have again $A=B$, so that $$\frac{d}{dt}(\Phi_A-\Phi_B)=A(\Phi_A-\Phi_B)$$  Taking norms and solving, on $[1+\epsilon,2]$, that $2M\epsilon$-sized difference grows by a constant factor of $e^{\|A\|(1-\epsilon)}$.  But $$\lim_{\epsilon\to0^+}{2M\epsilon e^{\|A\|(1-\epsilon)}}\to0$$ and there are no normal matrices near $\left[\begin{smallmatrix}4&4\\-1&1\end{smallmatrix}\right]$.  (To see that last claim, note that $XX^{\mathsf{T}}-X^{\mathsf{T}}X$ is continuous in $X$.)


问:
There are 10 speakers in a conference $S_1, S_2..S_{10}$ and $S_1$ speaks only after $S_2$ has spoken...

There are 10 speakers in a conference $S_1, S_2..S_{10}$ and $S_1$ speaks only after $S_2$ has spoken, then find number of ways in which speakers speak.

Using string method, let $S_1$ and $S_2$ be one block. Also there is only one way to arrange $S_1$ and $S_2$ inside the block.

So total permutations are $9!$

But given answer is $\frac{10!}{2}$. Where am I going wrong?

答:
If you consider $S_1$ and $S_2$ are one block, you are counting the ways in which $S_1$ speaks immediately after $S_2$.

Hint There are $10!$ arrangements of 10 speakers, show that $S_1$ speaks after $S_2$ in half of them.


问:
When can we embed field extensions into one another?
I was recently informed of the following theorem:

Let $K$ be a field of characterisic $0$, and let $L$, $M$ be field extensions of $K$ such that

i) $L$ is algebraic over $K$,

ii) every irreducible polynomial $p(x) \in K[x]$ with a root in $L$ has a root in $M$,

then $L$ embeds over $K$ into $M$.

I was also told that this can also be shown by model theoretic compactness.

If $L/K$ is a finite extension, this immediately follows from the fact that finite extensions over fields of characteristic $0$ are simple. Could anyone outline an algebraic/ model-theoretic proof of this fact for the case L is not a finite extension of K?

答:
Let me give you a few more proofs, to complement those given in Eric Wofsey's answer, and to show you a variety of ways that compactness arguments can be phrased.

Let's start with a proof that only works under the assumption that $[L:K]$ is countable.

Proof 1 (Kőnig's Lemma): Assume $[L:K]$ is countable. Write $L = \bigcup_{i\in  \mathbb{N}}L_i$, where $L_0 = K$ and each $L_{i+1}$ is a finite extension of $L_i$. Let $T$ be the set of all embeddings $L_i\to M$ over $K$ with $i\in \mathbb{N}$, ordered by function extension. Then $T$ is an infinite finitely branching tree. It is infinite because each $L_i$ is a finite extension of $K$, and hence embeds in $M$ over $K$, so $T$ contains at least one embedding $L_i\to M$ for each $i$. It is finitely branching because each $L_{i+1} = L_i(\alpha)$ is a finite extension of $L_i$ and hence has only finitely many embeddings into $M$ extending a given embedding $L_i\to M$. By Kőnig's Lemma, $T$ has an infinite branch. This is a cohering system of embeddings $L_i\to M$ for all $i$, so their union is an embedding $L\to M$ over $K$. $\square$

Here Kőnig's Lemma captures the following idea: if we have a sequence of choices to make, any particular choice has only finitely many options, and for any $n$, there is a way to make the first $n$ choices consistently, then there is a way to make all the choices consistently.

If we want to handle the general case, we need the following adjustment of that idea: if we have an arbitrary family of choices to make, any particular choice has only finitely many options, and for any finitely many choices, there is a way to make these choices consistently, then there is a way to make all the choices consistently.

This is what's going on in the proof in the last paragraph of Eric's answer, using the compactness principle "Any inverse limit of finite nonempty sets is nonempty." Here's another version of essentially the same argument, but using Tychonoff's Theorem.

Proof 2 (Tychonoff's Theorem): For each $\alpha\in L$, let $f_\alpha$ be the minimal polynomial of $\alpha$ over $K$, and let $X_\alpha$ be the finite (but non-empty) set of roots of $f_\alpha$ in $M$. Note that for $\alpha\in K$, $X_\alpha = \{\alpha\}$. Think of each $X_\alpha$ as a topological space with the discrete topology. Let $X = \prod_{\alpha\in L}X_\alpha$. By Tychonoff's Theorem, $X$ is compact. For all $\alpha,\beta\in L$, let $P_{\alpha,\beta} = \{(x_\gamma)\in X\mid x_\alpha + x_\beta = x_{\alpha+\beta}\}$, and let $T_{\alpha,\beta} = \{(x_\gamma)\in X\mid x_\alpha x_\beta = x_{\alpha\beta}\}$. Then $P_{\alpha,\beta}$ and $T_{\alpha,\beta}$ are closed sets in $X$, and an element of $\bigcap_{\alpha,\beta\in L} (P_{\alpha,\beta}\cap T_{\alpha,\beta})$ encodes a homomorphism $L\to M$ over $K$ by $\alpha\mapsto x_\alpha$. An intersection of finitely many of these sets is non-empty, since it enforces the homomorphism conditions only on a finitely generated subfield $L'\subseteq L$ which is a finite extension of $K$, and $L'$ can always be embedded in $M$ over $K$ (such an embedding always maps $\alpha\in L'$ to an element of $X_\alpha$, and values of $x_\gamma$ for $\gamma\notin L'$ can be chosen arbitrarily). By compactness, $\bigcap_{\alpha,\beta\in L} (P_{\alpha,\beta}\cap T_{\alpha,\beta})$ is non-empty, so we are done. $\square$

The previous proof can be rephrased to use the compactness theorem for propositional logic instead of topological compactness.

Proof 3 (Compactness of propositional logic): For each $\alpha\in L$, let $f_\alpha$ be the minimal polynomial of $\alpha$ over $K$, and let $X_\alpha$ be the finite (but non-empty) set of roots of $f_\alpha$ in $M$. Note that for $\alpha\in K$, $X_\alpha = \{\alpha\}$. For each $\alpha\in L$ and $m\in X_\alpha$, let $Q_{\alpha,m}$ be a propositional variable. For all $\alpha\in L$, let $\varphi_{\alpha}$ be the sentence asserting that exactly one of the finitely many variables $\{Q_{\alpha,m}\mid m\in X_\alpha\}$ is true. For all $\alpha,\beta\in L$, let $\pi_{\alpha,\beta}$ be the sentence $Q_{\alpha,m}\land Q_{\beta,n}\rightarrow Q_{\alpha+\beta,m+n}$ and let $\tau_{\alpha,\beta}$ be the sentence $Q_{\alpha,m}\land Q_{\beta,n}\rightarrow Q_{\alpha\beta,mn}$. Let $T = \{\varphi_\alpha\mid \alpha\in L\}\cup \{\pi_{\alpha,\beta},\tau_{\alpha,\beta}\mid \alpha,\beta\in L\}$. A truth assignment satisfying $T$ encodes a homomorphism $L\to M$ over $K$ mapping each $\alpha$ to the unique $m$ such that $Q_{\alpha,m}$ is true. Any finitely many sentences in $T$ is consistent, since it enforces the homomorphism conditions only on a finitely generated subfield $L'\subseteq L$ which is a finite extension of $K$, and $L'$ can always be embedded in $M$ over $K$ (such an embedding always maps $\alpha\in L'$ to an element of $X_\alpha$, and the $Q_{\gamma,m}$ for $\gamma\notin L'$ can be chosen arbitrarily). By compactness, $T$ is consistent, and we are done. $\square$

Finally, here's an alternative proof using the compactness theorem for first-order logic. First-order compactness in the language of fields is actually bit less efficient as a tool for the job, because in general it forces us to pass to an elementary extension of the field we're interested in. But as we'll see, this is not a problem in this case.

Proof 4 (Compactness of first-order logic): Consider the language of fields augmented with constant symbols $(c_l)_{l\in L}$ and $(d_m)_{m\in M}$. Let $T = \mathrm{EDiag}(M)\cup \mathrm{Diag}(L)\cup \{c_k = d_k\mid k\in K\}$. Here $\mathrm{EDiag}(M)$ is the set of all sentences $\varphi(d_{m_1},\dots,d_{m_n})$ such that $M\models \varphi(m_1,\dots,m_n)$, and $\mathrm{Diag}(L)$ is the set of all atomic and negated atomic sentences $\psi(c_{l_1},\dots,c_{l_n})$ such that $L\models \psi(l_1,\dots,l_n)$. A finite subset $T'\subseteq T$ only mentions constant symbols from $M$ and a subfield $L'\subseteq L$ such that $K\subseteq L'$ is a finite extension. We can embed $L'$ in $M$ over $K$, by $f\colon L'\to M$, so interpreting each constant symbol $c_{l}$ as $f(l)$ for $l\in L'$ (and the rest of the $c_{l}$ for $l\notin L'$ arbitrarily), $M\models T'$. By compactness, $T$ is consistent. Let $M'\models T$. Then $M'$ is an elementary extension of $M$ (identifying each $m\in M$ with the interpretation of $d_m$ in $M'$), and $L$ embeds in $M'$ over $K$ (by mapping each $l\in L$ to the interpretation of $c_l$ in $M'$). For each $l\in L$, let $f_l(x)$ be the minimal polynomial of $l$ over $K$. Since $M'$ is an elementary extension of $M$, it does not add any new roots of $f_l$ (if $m_1,\dots,m_{k_l}$ are the roots of $f_l$ in $M$, then $M\models \forall x\, (f_l(x) = 0\rightarrow \bigvee_{i=1}^{k_l} x = m_i)$, so this formula is also true in $M'$). Since the image of $l$ in $M'$ is a root of $f_l$, this image is in $M$. Thus the embedding $L\to M'$ actually lands in $M$. $\square$


答:
If you make $S1$ and $S2$ one block, you make their relative speaking positions also fixed. As per question, they can speak in any position as long as $S1$ speaks after $S2$. It is easy to see that $S1$ will speak after $S2$ in half of all permutations and before $S2$ in rest half.

So the answer is $\displaystyle \frac{10!}{2}$.

Edit: on your point about not being convinced on the same. Let's see this way. If $S2$ is in the first position, there are $9!$ ways for arranging rest of the speakers and all are permissible. If $S2$ is in second position, $8$ permissible positions for $S1$ and $8!$ ways for rest of the speakers. Continue with same logic and total number of permutations come to

$9! + 8! \cdot (8 + 7 + 6 + 5 + 4 + 3 + 2 + 1) = 1814400$.

Title: There are 10 speakers in a conference $S_1, S_2..S_{10}$ and $S_1$ speaks only after $S_2$ has spoken...

Body:
There are 10 speakers in a conference $S_1, S_2..S_{10}$ and $S_1$ speaks only after $S_2$ has spoken, then find number of ways in which speakers speak.

Using string method, let $S_1$ and $S_2$ be one block. Also there is only one way to arrange $S_1$ and $S_2$ inside the block.

So total permutations are $9!$

But given answer is $\frac{10!}{2}$. Where am I going wrong?


问:
Uniform Problem - Expected Value
A fraternity is throwing a party for its members. The cost of booking a band to play at the party, the amount that the supplier is going to charge, the cost of renting some place and some other costs are random uniform variables over the intervals (1300,1800),(1800,2000),(800,1200) and (400,700) respectively. If the number of people invited is a random integer in the interval (150,200). What is the minimum amount, in average, the fraternity will have to charge each person to not lose money?

What I did was: The sum of all the costs and then the average, that gave me 5000, then the average of the attendees, 175. I divided them. But I don't know how to continue.

答:
The question in the exercise points IMO to an exercise where a so called "fair price" is to be calculated, which means that the expected costs should be equal to the expected revenue from the price.

You almost did the job already:

Expected costs: $5000$

Expected number of attendees: $175$

Fair price $p$: $5000 = 175p \Rightarrow p = \frac{200}7 \approx 28.57143$.

If you want to be picky, you may round it up to $28.58$ in order not to loose money "in average" (as the exercise text requires).

Title: Uniform Problem - Expected Value


问:
Shuffling items between two boxes
I have two initially identical boxes. They each contain 1 red ball and 3 green balls.

From the first box, I randomly take out a ball and put it into the second box.

From that second box, I randomly take out a ball and put it into the first box.

Finally, I randomly take out a ball from the first box.

What's the probability that the said ball is red?

I would appreciate any tips on getting started.

答:
The simplest solution is to note that in expectation, the two boxes are again identical after you've placed a ball from the second box back into the first, so the answer is $\frac{1}{4}$. Without any information, your best guess prior to drawing a ball a third time is that the boxes are the same as they were before you started the process. This is a consequence of the law of iterated expectations.

Here's a hint on how to work through it the long way and prove it to yourself:

There are three ways you can pull the red ball:

The first ball was red and the second red.

The first ball was not red and the second ball was red.

The first ball was not red and the second ball was not red.

If the first ball was red and the second not red, there will be $0$ red balls in the first box, so it would be impossible to pull a red ball.

For the first option, there is a $\frac{1}{4}$ chance of pulling the red ball first. After moving it, there are now $2$ red balls and $3$ green balls in the second box. So, there's a $\frac{2}{5}$ chance of pulling a red ball from the second box. There are now $1$ red and $3$ green balls in the first box, so we have a $\frac{1}{4}$ chance of pulling the red ball from it. Overall, this sequence has probability $\frac{1}{4}*\frac{2}{5}*\frac{1}{4} = \frac{1}{40}$.

You should get that the second sequence of events has probability $\frac{3}{40}$, and the third has probability $\frac{3}{20}$, giving you a grand total of $\frac{1}{4}$.

Title: Shuffling items between two boxes


问:
Possible Error in the 1990 B1 Putnam Answer Key
Find all real differentiable functions $f$ that satisfy the following.

$$f(a)^2 = \int_0^a (f(x)^2 + f'(x)^2)dx + 1990 \forall a \in \mathbb{R}$$

The solution manual is found here.

Putting $y = f(x)$ and differentiating the relationship gives $2y y' = y^2 + y'^2$ or $(y - y')^2 = 0$. So $y = y'$ or $y = -y'$. Integrating, $y = Ae^x$ or $y = - Ae^x$. But $y(0) = \pm \sqrt{1990}$, so $f(x) = \pm \sqrt{1990}e^x$.

Wouldn't the differential linear differential equation $y'=-y$ yield a solution of $y=Ae^{-x}$ as opposed to $y=-Ae^x$? Additionally, I am unsure why the solution $y=-y'$ is even necessary.

答: The procedure is correct, but you only get $y'=y$ (which follows from $(y'-y)^2=0$), hence $y=Ce^x$ and, setting $a=0$, $y=\pm\sqrt{1990}e^x$


问:
Dividing 10 students into three groups

There are 10 students. Find the number of ways in which they can be divided into 3 groups such that each group has at least 1 student and third group has at most 3

We can make 3 separate cases where $G_3$ has $1$, $2$, $3$ students respectively.

As for the remaining two groups, there are no restrictions other than that no group can have $0$ students.

But I am not sure about how to divide the remaining students into $2$ groups. Clearly the stars and bars method won’t work because students are not identical objects.

答:
Let's take the case where $G_3$ has one student. Pick the student there and you have nine left. Now, $G_1$ (say) can have $1$ to $8$ students. Pick the students for $G_1$ and then all of the others are on $G_2$.

So for this case you'll get

$$10\cdot \sum_{j=1}^8 {9 \choose j}$$

groups. And similarly for the other cases for $G_3$.

Title: Dividing 10 students into three groups


答:
You can imagine this as labelling the students with the group numbers $\{1,2,3\}$.

So, basically you are looking for all $10$-digit numbers with digits $\{1,2,3\}$ where there is at least one $1$, one $2$ and from one to three $3$'s.

Now, splitting up into the cases where exactly $1,2,3$ get the label $3$, you get

exactly $1$ with label $3$: $\color{blue}{\binom{10}1\cdot (2^9-2)}$

exactly $2$ with label $3$: $\color{blue}{\binom{10}2\cdot (2^8-2)}$

exactly $3$ with label $3$: $\color{blue}{\binom{10}3\cdot (2^7-2)}$

Adding this together gives:

$$\color{blue}{31650}$$

Title: Dividing 10 students into three groups

Body:
There are 10 students. Find the number of ways in which they can be divided into 3 groups such that each group has at least 1 student and third group has at most 3

We can make 3 separate cases where $G_3$ has $1$, $2$, $3$ students respectively.

As for the remaining two groups, there are no restrictions other than that no group can have $0$ students.

But I am not sure about how to divide the remaining students into $2$ groups. Clearly the stars and bars method won’t work because students are not identical objects.


问:
Showing that the theory of an infinite descending $\subseteq$ chain has Morley rank 2
Consider the language $\{P_i\mid i\in\omega\}$ and the theory asserting

$P_0\supseteq P_1\supseteq P_2\supseteq...$

$\neg P_0$ is infinite

Each $P_n\smallsetminus P_{n+1}$ is infinite

I want to show that, relative to this theory, the Morley rank of "$x=x$" is 2.

I think I am not having the right picture of what saturated models of this theory look like. The model that I use for guiding my intuition is the following: the domain is the rational numbers $\mathbb{Q}$, each $P_n$ is interpreted as $\{x\in\mathbb{Q}\mid x>n\}$. This picture allows one to see relatively easily that $\text{MR}(x=x)\geq 1$. This is because $\langle P_i(x)\wedge\neg P_{i+1}(x)\mid i\in \omega \rangle$ is a list of consistent formulas that are pointwise disjoint. However, I am having troubles finding the formulas to use to witness $\text{MR}(x=x)\geq 2$ and also why it is exactly 2.

答:
I think I am not having the right picture of what saturated models of this theory look like. The model that I use for guiding my intuition is the following: the domain is the rational numbers $\mathbb{Q}$, each $P_n$ is interpreted as $\{x\in\mathbb{Q}\mid x>n\}$.

That's a great picture to have in mind, but it's not saturated, because it doesn't have any elements satisfying all the $P_n$. A countable model of this theory is determined up to isomorphism by the number of elements which satisfy all the $P_n$. This number can be $0$ (as in your model) or any positive natural number, or $\aleph_0$ (this is the saturated model).

This picture allows one to see relatively easily that $\text{MR}(x=x)\geq 1$. This is because $\langle P_i(x)\wedge\neg P_{i+1}(x)\mid i\in \omega \rangle$ is a list of consistent formulas that are pointwise disjoint.

These formulas actually witness $\mathrm{MR}(x=x)\geq 2$, because they each have Morley rank $\geq 1$. It's good to remember the following interpretations of small Morley ranks:

$\mathrm{MR}(\varphi(x))\geq 0$ iff $\varphi(x)$ defines a non-empty set.

$\mathrm{MR}(\varphi(x))\geq 1$ iff $\varphi(x)$ defines an infinite set.

$\mathrm{MR}(\varphi(x))\geq 2$ iff $\varphi(x)$ can be partitioned into infinitely many infinite definable sets.

Explicitly, to see that $\mathrm{MR}(P_i(x)\wedge\neg P_{i+1}(x))\geq 1$, pick infinitely many distinct elements $(a_n)_{n\in \omega} \in M$ such that $M\models P_i(a_n)\land \lnot P_{i+1}(a_n)$ for all $n$. Then we have $\mathrm{MR}(x = a_n)\geq 0$ for all $n$, the formulas $\{x = a_n\mid n\in \omega\}$ are pairwise disjoint, and each formula $x = a_n$ implies $P_i(x)\wedge\neg P_{i+1}(x)$.

As for showing that $\mathrm{MR}(x=x)$ is exactly $2$, this is a bit trickier. Obtaining an upper bound on the Morley rank requires understanding all the definable sets (or, equivalently, all the complete types). As a strategy, you could prove (by quantifier elimination or an automorphism argument) that any infinite definable set is either

A finite union of sets of the form $\lnot P_0(x)$ or $P_i(x)\land \lnot P_{i+1}(x)$, plus or minus finitely many points, or

Some $P_n$, plus a set of the first form.

Then observe that (a) sets of the first form have Morley rank $1$, since they do not contain infinitely many disjoint infinite definable sets, and (b) no two sets of the second form are disjoint. So we cannot find an infinite family of disjoint sets of Morley rank $2$, and thus $x = x$ has Morley rank at most $2$.

A more efficient strategy is to use the equivalence between Morley rank and the Cantor-Bendixson rank on the space of types over the countable saturated model. This space has the realized types (Morley rank $0$), one non-realized type containing $\lnot P_0(x)$ and one containing $P_i(x)\land \lnot P_{i+1}(x)$ for each $i\in \omega$ (Morley rank $1$), and one non-realized type containing all the $P_i(x)$ (Morley rank $2$).


问:
Aproximating function by polynomials of degree at least $2$.
I was working in the next exercise but I don't know what to do. The exercise is the next:

Let $f:[0,1]\to\mathbb{R}$ be a continuous function and suposse that for some $\alpha>1$ the inequality $$\left|\dfrac{f(x)}{x^{\alpha}} \right|\leq 1$$holds for all $x\in (0,1]$. Prove that $f$ can be aproximated uniformly by polynomials of the form $a_2x^{2}+\cdots+a_nx^{n}$ with $n\geq 2$.

I really don't know how to proced. My first idea was to think in Stone-Weierstrass theorem because then I can find a sequence of polynomials that converges uniformly to $f$ but the problem is the degree. How can I "delete" the first terms of the polynomials and don't lose the convergence? Is it possible? Anothe idea was to think in Taylor series but I'm not sure that $f$ is an infinitely many differentiable function. Any hint? I really aprecciate any help you can provide me. Thanks.

答:
The Bernstein polynomials

Use the Bernstein proof of the Weierstrass theorem for your purpose.

Indeed, let $f \in C[0,1]$ and $f_n(x) = \sum_{k=0}^n \binom nk x^k(1-x)^{n-k}f(\frac kn)$. Then $f_n$ are polynomials. It is well known that $f_n(x) \to f(x)$ uniformly on $[0,1]$, a proof of this can be found here. If you do not know this fact, then you have just been enlightened! $f_n$ is called the $n$th Bernstein polynomial corresponding to $f$, and the attached proof is called Bernstein's proof of the Weierstrass theorem.

The Basic Idea

We will try to obtain our polynomial by removing constants and terms having just $x$ (and not higher powers) from this summation. Now, $f_n$ is a polynomial of degree $n$. Note, however, that if $k \geq 2$ then the term corresponding to $k$ has powers of $x$ only above $x^2$, because we are multiplying by $x^k$ with $k \geq 2$. So we can leave these terms as they are.

The term for $k=0$ is $(1-x)^n f(0)$. However, $|f(x)| \leq |x|^{\alpha}$ so letting $x\to 0$, by continuity of $f$ we get that $f(0)=0$. Thus, the term for $k=0$ is zero. This means that $f_n$ doesn't have any constant terms, since from $k>0$ onwards there are no constant terms, everything is multiplied by at least $x$.

The term for $k=1$ is $nx(1-x)^{n-1} f(\frac 1n)$. The only degree $1$ term in here, from the binomial expansion, is $nxf(\frac 1n)$. But this doesn't contribute very much. What is the reason? Well, $\left|f(\frac 1n)\right| \leq n^{-\alpha}$ so $|nxf(\frac 1n)| \leq n^{1 - \alpha}$ for all $x$.

Finishing the proof

So the final proof is as follows : Let $f$ be a continuous function with $|f(x)| \leq |x|^{\alpha}$ for all $x \in (0,1]$. Let $f_n$ be the $n$th Bernstein polynomial of $f$ and let for $n>1$, $g_n(x) = f_n(x) - nxf(\frac 1n)$. Then $g_n$ has degree at least $2$, and $g_n \to f$ uniformly as $f_n \to f$ uniformly and $nxf(\frac 1n) \to 0$ uniformly in $x$.

Expected extensions

One can probably extend this to the following :

Let $m$ be a natural number and $\alpha>m$ be a real number. Then any $f$ continuous on $[0,1]$ which  satisfies $|f(x)| \leq |x|^\alpha$ for all $x \in (0,1]$ can be uniformly approximated by polynomials of the form $a_{m+1}x^{m+1}+...+a_nx^n$ with $n \geq m+1$.

You can try to prove this using the Bernstein polynomial.


问:
How to calculate: $\sum_{n=1}^{\infty} n a^n$
I've tried to calculate this sum:

$$\sum_{n=1}^{\infty} n a^n$$

The point of this is to try to work out the "mean" term in an exponentially decaying average.

I've done the following:

$$\text{let }x = \sum_{n=1}^{\infty} n a^n$$
$$x = a + a \sum_{n=1}^{\infty} (n+1) a^n$$
$$x = a + a (\sum_{n=1}^{\infty} n a^n + \sum_{n=1}^{\infty} a^n)$$
$$x = a + a (x + \sum_{n=1}^{\infty} a^n)$$
$$x = a + ax + a\sum_{n=1}^{\infty} a^n$$
$$(1-a)x = a + a\sum_{n=1}^{\infty} a^n$$

Lets try to work out the $\sum_{n=1}^{\infty} a^n$ part:

$$let y = \sum_{n=1}^{\infty} a^n$$
$$y = a + a \sum_{n=1}^{\infty} a^n$$
$$y = a + ay$$
$$y - ay = a$$
$$y(1-a) = a$$
$$y = a/(1-a)$$

Substitute y back in:

$$(1-a)x = a + a*(a/(1-a))$$
$$(1-a)^2 x = a(1-a) + a^2$$
$$(1-a)^2 x = a - a^2 + a^2$$
$$(1-a)^2 x = a$$
$$x = a/(1-a)^2$$

Is this right, and if so is there a shorter way?

Edit:

To actually calculate the "mean" term of a exponential moving average we need to keep in mind that terms are weighted at the level of $(1-a)$. i.e. for $a=1$ there is no decay, for $a=0$ only the most recent term counts.

So the above result we need to multiply by $(1-a)$ to get the result:

Exponential moving average "mean term" = $a/(1-a)$

This gives the results, for $a=0$, the mean term is the "0th term" (none other are used) whereas for $a=0.5$ the mean term is the "1st term" (i.e. after the current term).

答:
Yes, you're right.

Shorter way: note your series is $$a\frac{d}{da}\frac{1}{1-a}=\frac{a}{(1-a)^2}$$

That is $$\frac{1}{1-a}=\sum_{n=0}^\infty a^n\implies a\frac{d}{da}\frac{1}{1-a}=a\sum_{n=1}^\infty na^{n-1}=\sum_{n=1}^\infty na^{n}$$

ADD. In general one can find $$\sum_{n=1}^\infty n^kx^n=\left(x\frac{d}{dx}\right)^k\frac{1}{1-x}=\frac{\varphi_k(x)}{k!}\left(\frac d{dx}\right)^k\frac 1 {1-x}= \frac{\varphi_k(x)}{k!}\frac{1}{(1-x)^{k+1}}$$

where $\varphi_k(x)$ are the Eulerian polynomials $$\varphi_1(x)=x$$ $$\varphi_{k+1}(x)=x(1-x)\varphi_{k}^\prime(x)+(k+1)x\varphi_k(x)$$

or $$\varphi_k(x)=\sum_{m=1}^k E(m,k)x^m$$ where $E(m,k)=0$ if $m\leq 0$ or $m>k$ and $$E(m,k+1)=mE(m,k)+(k-m+2)E(m-1,k)$$


问:
Find the spectrum of $Ax(t)=x(t+s)$
Let be $\mathcal{C}_\mathbb{R}$ the set of the continuous and bounded function in $\mathbb{R}$ with norm $ \|x\|=\underset{t\in\mathbb{R}}{sup}|x(t)|$ Consider the operator $A$ such that $Ax(t)=x(t+s)$ for some $s\in\mathbb{R}$. Show that the spectrum of A is the unit sphere $=\{\lambda\in\mathbb{C}: |\lambda|=1\} $.

Please, I don't know how to aboard this exersice. I find the norm of $A$, $\|A\|=1$ so $\sigma(A)\subseteq \bar{D}(0,1)$. Then I was trying to show that if $|\lambda|<1$ then $A-\lambda Id$ is invertible, so I began from show that it's injective but I only find that $|x(t+s)|<|x(t)|$ for all $ t\in\mathbb{R}$ and I don't know why this implies that $x=0$.
Or maybe this is not te correct way to make this exercise please could you give me some hints?

答:
This is definitely not enough to finish the exercise, but a couple things you could notice:

$A$ is an isometry, i.e. $\|Ax\|=\|x\|$.

if $|\lambda|<1$, then $$ \|(A-\lambda I)x\|=\|Ax-\lambda x\|\geq\|Ax\|-|\lambda|\,\|x\|=(1-|\lambda|)\|x\|,$$ so $A-\lambda I$ is bounded below which shows that it is injective and has closed range.

when $|\lambda|=1$, you can show that $\lambda$ is an eigenvalue by constructing an eigenvalue explicitly. Not needed, by the multiplicity of each eigenvalue is infinite.

Title: Find the spectrum of $Ax(t)=x(t+s)$


答:
First, you need to assume that the sum exists, which it might not. For example, if $a\geq 1$, then the sum is infinite.

But, assuming that the sum exists, a better way to rewrite what you are saying is

$$\begin{array} {l l l l l l l }
x & = 1a^1 & + 2a^2 &+ 3 a^3 & + \;\cdots \\
ax & =    & + 1a^2 & + 2 a^3 & + \;\cdots \\
\hline
(1-a) x &= 1a^1 & + 1a^2 & + 1 a^3 & + \;\cdots \\
\end{array}$$

[Note: This is exactly the same as your first paragraph of math, but presented in a slightly different way that makes it more obvious.]

At this stage, you can similarly show that $a + a^2 + \cdots = \dfrac{a}{1-a}$, to reach your conclusion that $x = \dfrac{a}{(1-a)^2}$.

Title: How to calculate: $\sum_{n=1}^{\infty} n a^n$

Body:
I've tried to calculate this sum:

$$\sum_{n=1}^{\infty} n a^n$$

The point of this is to try to work out the "mean" term in an exponentially decaying average.

I've done the following:

$$\text{let }x = \sum_{n=1}^{\infty} n a^n$$
$$x = a + a \sum_{n=1}^{\infty} (n+1) a^n$$
$$x = a + a (\sum_{n=1}^{\infty} n a^n + \sum_{n=1}^{\infty} a^n)$$
$$x = a + a (x + \sum_{n=1}^{\infty} a^n)$$
$$x = a + ax + a\sum_{n=1}^{\infty} a^n$$
$$(1-a)x = a + a\sum_{n=1}^{\infty} a^n$$

Lets try to work out the $\sum_{n=1}^{\infty} a^n$ part:

$$let y = \sum_{n=1}^{\infty} a^n$$
$$y = a + a \sum_{n=1}^{\infty} a^n$$
$$y = a + ay$$
$$y - ay = a$$
$$y(1-a) = a$$
$$y = a/(1-a)$$

Substitute y back in:

$$(1-a)x = a + a*(a/(1-a))$$
$$(1-a)^2 x = a(1-a) + a^2$$
$$(1-a)^2 x = a - a^2 + a^2$$
$$(1-a)^2 x = a$$
$$x = a/(1-a)^2$$

Is this right, and if so is there a shorter way?

Edit:

To actually calculate the "mean" term of a exponential moving average we need to keep in mind that terms are weighted at the level of $(1-a)$. i.e. for $a=1$ there is no decay, for $a=0$ only the most recent term counts.

So the above result we need to multiply by $(1-a)$ to get the result:

Exponential moving average "mean term" = $a/(1-a)$

This gives the results, for $a=0$, the mean term is the "0th term" (none other are used) whereas for $a=0.5$ the mean term is the "1st term" (i.e. after the current term).


问:
If $f$ is a function such that $f(f(x))=x^{2}-1$ determine the function $f(f(f(f(x))))$
I have tried $f(f(f(f(x))))=f\left(f\left(x^{2}-1\right)\right)$ . Since we know that $f(f(x))=x^{2}-1$, we have
$$
\begin{aligned}
f\left(f\left(x^{2}-1\right)\right) &=\left(x^{2}-1\right)^{2}-1 \\
&= x^{4}-2 x^{2}+1-1 \\
&=x^{4}-2 x^{2}.
\end{aligned}
$$But something doesn't feel right, is there any other way to solve this?

答:
The approach given is correct. Perhaps it would feel more natural to be  explicit in our substitution, and write $g(x) = f(f(x))$, so we have $g(x) = x^2-1$, and we are interested in $g(g(x))$.

Then it is easy to see that $g(g(x)) = (x^2-1)^2-1 = x^4-2x^2$.