问:
Does the ratio of the $x-$values of two continuous functions $f(x), g(x) \in \mathbb{R^+}$ tell you anything about the ratio between their $y-$values?
Does the ratio of the $x-$values of two continuous functions $f(x), g(x) \in \mathbb{R^+}$ tell you anything about the ratio between their $y-$values?

In general, obviously not. But suppose the following is also true:

$\lim_{x \to \infty}f(x) = \lim_{x \to \infty}g(x) = 0$,

$\lim_{x \to \infty} \frac{f(x)}{g(x)}$ converges to $L_1$ and $\lim_{y \to 0+} \frac{f^{-1}(y)}{g^{-1}(y)}$ converges to $L_2$.

Then what can we say about $L_2$ in terms of $L_1$? e.g. If $L_1 = \infty$ then does $L_2 = \infty$? If $L_1$ is finite then is $L_2$ finite?

答:
To begin with, let us assume that $f,g$ are injective so that we have no problems with defining the inverses. Then by continuity, they are monotonic. To vahe nno problems with $x\to 0^+$, we better assume they are positive (and hence decreasing).
Even then, we cannot conclude a lot about $L_2$ from $L_1$:

By  swapping $f\leftrightarrow g$, we find that if $(L_1,L_2)$ is possible, then so is $(1/L_1,1/L_2)$.

Of course, if $1<L_1\le \infty$, then the graph of $f$ is eventually above that of $g$, hence $f^{-1}$ must also be above $g^{-1}$, i.e., $1\le L_2\le\infty$.

In other words, $L_1<1<L_2$ and  $L_1>1>L_2$ are impossible.

We can construct $f,g$ with limits to our liking as follows:

Let $\{a_n\}_{n\in\Bbb N}$ and $\{b_n\}_{n\in\Bbb N}$ be convergent (possibly to $\infty$) sequences of numbers $>1$. Let $x_n=\prod_{k=1}^n a_k$, $y_n=\prod_{k=1}^n b_k^{-1}$.
Assume $x_n\to \infty$ and $y_n\to 0$.
Let $f$ be the piecewise linear interpolation through the points $(x_n,y_n)$ and $g$ the piecewise linear interpolation through the points $(x_n,y_{n+1})$. Then one directly checks that $\frac{f(x)}{g(x)}=\frac{y_n}{y_{n+1}}=b_{n+1}$ at $x=x_n$.

One verifies that on the interval $[x_n,x_{n+1}]$, the quotient $\frac{f(x)}{g(x)}$ only varies between $b_{n+1}$ and $b_{n+2}$ (the quotient is essentially piecewise a hyperbola). We conclude that $$L_1=\lim_{x\to\infty}\frac{f(x)}{g(x)}=\lim_{n\to\infty} b_n.$$ By the same argument,
$$ L_2=\lim_{y\to0^+}\frac{f^{-1}(y)}{g^{-1}(y)}=\lim_{n\to\infty} a_n.$$

By picking $a_n=a^n$ with $1<a<\infty$ we can thus achieve $L_1=a$.
By picking $a_n=n!$, we achieve $L_1=\infty$, and by $a_n=1+\frac1n$ we achieve $L_1=1$. In summary, we can achieve any $L_1\in[1,\infty]$ by a suitable choice of  $\{a_n\}_{n\in\Bbb N}$.
Likewise, we can achieve any $L_2\in[1,\infty]$ by a suitable choice of  $\{b_n\}_{n\in\Bbb N}$.

As the choices are independent, any combination $(L_1,L_2)\in[1,\infty]\times [1,\infty]$ can be achieved (and likewise any $(L_1,L_2)\in [0,1]\times[0,1]$).


问:
I have found a system of 3 variables that lead to a quadratic equation, so how do I find the answer?/
$$p^2 + q^2 = 81$$
$$p^2 + (q-x)^2 = 225$$
$$(p-x)^2 + q^2 = 144$$
So these are the 3 equations.

ContentLicense: CC BY-SA 4.0

Body:
Let $(p,q)=9(\cos t,\sin  t)$, then

\begin{align}
  81+x^2-18x\sin t &= 225 \\
  81+x^2-18x\cos t &= 144 \\
  (18x)^2 &= (x^2-144)^2+(x^2-63)^2 \\
  0 &= 2x^4-738x^2+24705 \\
  x &= \pm 3\sqrt{\frac{41 \pm 3\sqrt{119}}{2}} \\
  p &= 9\cos t \\
  &= \frac{x^2-63}{2x} \\
  q &= 9\sin t \\
  &= \frac{x^2-144}{2x}
\end{align}


问:
Spherical balloon inflating
I've encountered this calculus problem:

1-A spherical balloon is being inflated in such a way that its radius is increasing at the constant rate of 3 cm/min. If the volume of the balloon is 0 at time 0, at what rate is the volume increasing after 5 minutes?

So I set the radius function as $r(t) = 3t$ and the volume function as $v(r) = 4\pi r^3/3$. I found the derivative to be $4\pi r^2$, used the chain rule and found the balloon to be increasing at a rate of $2700$ cm/min after $5$ minutes. Is that right?

Thanks!

答:
Just to summarize what's been said in the comments:

You've obviously got the right volume as a function of the radius. Differentiating that function with respect to $t$ yields $$\frac{dv}{dt}=4\pi r^2\frac{dr}{dt}$$ Now we need to evaluate this expression at $t=5\mathrm{\:min}$. We can tell from the problem statement that $$r(5\mathrm{\:min})=15\mathrm{\:cm}$$ We can now plug this into our expression for $dv/dt$ along with the given value of $dr/dt$. Keeping the units attached to all the quantities is helpful here as a sanity check: $$\frac{dv}{dt}(5\mathrm{\:min})=4\pi(15\mathrm{\:cm})^2(3\mathrm{\:cm/min})=2700\pi\mathrm{\:cm^3/min}$$

ContentLicense: CC BY-SA 3.0

Title: Spherical balloon inflating


问:
How can this come up from ¬∃x¬(Atom(x)∨EmptySpace(x)) to ∀x(Atom(x)∨EmptySpace(x))?
When I tried to answer this question

Atomism, the statement that everything is made up of tiny, indivisible, unchangeable particles, which are called “atoms”, was originally proposed by two Greek philosophers -- Democritus (460-371B.C.) and Leucippus (early 5th century B.C.). Its central argument, nothing exists except atoms and empty space, can be paraphrased as there exists no such thing that is neither an atom nor empty space and written as  ¬∃x(¬Atom(x)∧¬EmptySpace(x))  in FOL.

This FOL sentence, however, is somewhat cumbersome, and can be simplified using de Morgan's laws. Which of the following is equivalent to the sentence, and cannot be simplified further?

my answer was wrong. So, I checked the key answer and here is the correct:

This is a neat version of Atomism's original statement, which is achieved by first applying de Morgan's law to  ¬Atom(x)∧¬EmptySpace(x)  inside the parenthesis and get  ¬(Atom(x)∨EmptySpace(x)) , then applying de Morgan's law for quantifiers on  ¬∃x¬(Atom(x)∨EmptySpace(x))  and get  ∀x(Atom(x)∨EmptySpace(x)) , which says that everything (that exists) is either an atom or empty space.

In the correct answer, how can this come up from ¬∃x¬(Atom(x)∨EmptySpace(x)) to ∀x(Atom(x)∨EmptySpace(x)) ?

答:
To make things clearer, let us use $A$ and $E$ rather than $\text{Atom}$ and $\text{EmptySpace}$.

We start with: "There is no thing that is both not $A$ and not $E$". $$\lnot\exists x~(\lnot A(x)\land\lnot E(x))$$

Invoking deMorgan's Laws: "There is no thing that is not either $A$ or $E$." $$\lnot\exists x~\lnot(A(x)\lor E(x))$$

Invoking quantifier duality: "Every thing is either $A$ or $E$." $$\forall x~(A(x)\lor E(x))$$

That is all.

De Morgan's laws are that ${(\lnot p\land\lnot q)\iff\lnot(p\lor q)\\(\lnot p\lor\lnot q)\iff\lnot(p\land q)\\\text{et cetera}}$

Quantifier duality is that ${\lnot\exists x~\phi\iff\forall x~\lnot\phi\\\lnot\forall x~\phi\iff\exists x~\lnot\phi\\\text{et cetera}}$


问:
Puzzle identification and solving algorithm
I am trying to solve 8x8 puzzle (total 64 buttons). Similar to LightsOut, but in this rules are different. Goal is turn ON every button.

Example:

```<code>Button 1 is turned on/off by buttons 25, 36
Button 2 is turned on/off by buttons 25, 55
Button 3 is turned on/off by buttons 20, 58
...
Button 20 is turned on/off by buttons 4, 9
...
Button 25 is turned on/off by buttons 22, 59
...
Button 36 is turned on/off by buttons 42, 50
...
Button 55 is turned on/off by buttons 3, 24
...
Button 64 is turned on/off by buttons 29, 32
</code>```

Full list is available here: http://pastebin.com/9b0MKXCb

I see that every button can be turned ON/OFF by any of 2 buttons (it's always 2).

I succeeded solving it manually by trial & error method, but I would like to do it proper. Program starts with all lights (buttons) turned OFF. Target is to turn ON all of them.

Is this still LightOuts problem? How can I solve it?

Thank you!

UPDATE:

I've been asked why is this 8x8 puzzle. I called it that because there is total 64 buttons. Is this wrong?

For example. If I wish to turn ON button 1, I have to click on button 25 or 36. If I click two times on 25 or 36 state is restored - it's like nothing happened.

So if I wish to turn ON buttons 1 and 2 I have to click on button 25 OR button 36 AND button 55.

答:
Let $x_i$ be the number of times you push button $i$. You'll only ever push it $0$ times or $1$ time, so be can take $x_i$ to be modulo $2$.

You would like light 1 to go from OFF to ON. When we examine the chart in the link in your comments, we can only affect light 1 by switching buttons 25 or 36. This yields an equation: $$x_{25}+x_{36}\equiv1$$ so that light 1 will be left ON. Repeat this for every light, and you have 64 equations modulo $2$ in $64$ variables modulo $2$. This system of equations can be solved through row reduction of the corresponding matrix (or possibly row reduction will reveal there is no solution.) If there is a solution, it is either

a specific set of values of 0s and 1s for the $\{x_i\}$,

or it is a larger solution set that maybe parametrized, from which you can find an optimal specific solution my minimizing (the integer value of) $x_1+x_2+\cdots+x_{64}$.

Let's take a smaller, $2\times2$ example. Mimicking the link you provide:

1 = [2,4] (button 1 triggers state change on button 2 AND button 4)
2 = [1, 3] (button 2 triggers state change on button 1 AND button 3)
3 = [1, 2, 4] (button 3 triggers state change on button 1, button 2, AND button 3)
4 = [3]

$$\begin{align}
\text{To turn on light 1} &&x_2+x_3&=1\\
\text{To turn on light 2} &&x_1+x_3&=1\\
\text{To turn on light 3} &&x_2+x_4&=1\\
\text{To turn on light 4} &&x_1+x_3&=1\\
\end{align}
$$

This system of equations has matrix
$$\begin{align}\begin{bmatrix}
0&1&1&0&1\\
1&0&1&0&1\\
0&1&0&1&1\\
1&0&1&0&1
\end{bmatrix}\end{align}$$
Row reducing (if you do not know how to do this, please research it online - wikipedia should do) yields:
$$\begin{align}\begin{bmatrix}
1&0&0&1&1\\
0&1&0&1&1\\
0&0&1&1&0\\
0&0&0&0&0
\end{bmatrix}\end{align}$$

$x_4$ is a free variable, so we can use a parameter $t$ for $x_4$. $t$ can range between $0$ and $1$. For the rest, we read (using modulo 2 arithmetic) $x_1=1-t, x_2=1-t, x_3=t$. This yields two solutions. $t=0\implies(1,1,0,0)$ and $t=1\implies(0,0,1,1)$.

So we could either push buttons 1 and 2, or buttons 3 and 4. the order that we push them in is irrelevant.


问:
Why is this particular solution giving extraneous roots (solving $z^n= \bar z$)

I can't understand why this is giving extraneous roots. The real set of solutions is supposed to be fourth roots of unity.

答:
Squaring, cubing, or doing other such operations to both sides of an equation often introduces extraneous roots. Observe, for instance:

\begin{align*}
x+1 = -2
&\implies (x+1)^2 = (-2)^2 \\
&\implies (x+1)^2 = 4 \tag{$\ast$}\\
&\implies |x+1| = 2 \\
&\implies x \in \left\{ -1+2, -1-2 \right\} = \left\{ 1,-3 \right\}
\end{align*}

We start with the equation $x+1=-2$, and square both sides. After simplifying, note at $(\ast)$ in particular that what we have is a second-degree equation, a quadratic, which has two distinct solutions most of the time.

From there, we take the square root, and note that $\sqrt{z^2} = |z|$. From there, we can solve for $x$, and indeed have two solutions.

The reason this arises is that squaring both sides is not a bijective operation -- you cannot always "undo" it cleanly, so squaring both sides, in a sense, causes you to lose information.

The same is true for any integer power you raise both sides to, because you can be introducing a bunch of extraneous solutions. Say you have a degree $d$ polynomial set equal to something else, and then raise both sides to the $n$th power; you could have as many as $n \cdot d$ solutions to the resulting equation, because the polynomial is now one of degree $n \cdot d$.

This is also why I wrote my derivation above using forward implications and not $\iff$ ("if and only if") arrows, because it's not undoable.

Your situation - or at least the issue you have with your method of solution - is very analogous to that I mentioned with my much-simpler problem.


问:
What does $E_\theta(X) : R^D \rightarrow R$ mean?
I'm reading a paper. In this paper,

img1

I don't understand what "$E_\theta(X) : R^D \rightarrow R$" means.

I found that $R^D$ means D-dimensional vector on the web but can't still understand the rest.

答:
This notation just means it's a function, named $E_θ$, which takes in $D$-dimensional vectors $\vec x$ with real coordinates, which come from $\Bbb R^D$, and maps them to a scalar in $\Bbb R$. Such a function could be, say, if we let $\vec x = (x_1,\cdots,x_D)$,

$$f\Big((x_1,x_2,x_3,⋯,x_D)\Big)=x_1+x_2+x_3+⋯+x_D$$


答:
The quick answer is that you also get the solutions to $\bar{z}^3=-z$, because cubing and taking complex conjugate of this equation would also give
$$
z^9=\overline{(\bar{z}^3)^3}=\overline{(-z)^3}=-\bar{z}^3=-(-z)=z.
$$

Title: Why is this particular solution giving extraneous roots (solving $z^n= \bar z$)

Body: I can't understand why this is giving extraneous roots. The real set of solutions is supposed to be fourth roots of unity.


问:
Minimize $\{tr^Trt^T\}_{ii}$ subject to $r^Tt=1$
I want to solve the following optimization problem over $r,t\in\mathbb{R}^{n\times 1}$:

\begin{array}{ll} \text{minimize} & \{tr^Trt^T\}_{ii}\\\quad r,t\\ \text{subject to} & r^Tt=1\end{array}

My attempt: since $r^Tr$ is scalar and $\{tt^T\}_{ii}=t^2_{i}$ , we can rewrite above problem as:

\begin{array}{ll} \text{minimize} & r^Trt^2_{i}\\\quad r,t\\ \text{subject to} & r^Tt=1\end{array}

Can I replace $r^T$ by its pseudo-inverse $r^T=(t^Tt)^{-1}t^T$? Then

\begin{array}{ll} \text{minimize} & \frac{t^2_{i}}{t^2_{1}+t^2_{2}+\cdots+t^2_{n}}=0?\\\quad t\end{array}

What I want is to minimize each diagonal element of $\{tr^Trt^T\}_{ii}$ as much as possible., subject to $r^Tt=1$, we can assume that diagonal elements of $t$ are ordered in nondecreasing order.

答:
For any give $T$, we can rewrite the optimization problem as follows:
\begin{array}{cl}
\operatorname{minimize}_{R}& RR'\\
\operatorname{s.t.} & RT=1
\end{array}
The Lagrangian is
$$L=RR'+\lambda(RT-1)$$
The optimality conditions are
\begin{aligned}
\frac{\partial L}{\partial R}&=2R+\lambda T'=0\\
\frac{\partial L}{\partial \lambda}&=RT-1=0
\end{aligned}
So, $R=-\frac{\lambda}{2}T'\Rightarrow-\frac{\lambda}{2}T'T-1=0\Rightarrow\lambda=\frac{-2}{T'T}$

Therefore, the optimal $R$ is $R = (T'T)^{-1}T'$.

You are right!!!

If you want to minimize all elements, you should construct some proper problem, for example
\begin{array}{cl}
\operatorname{minimize}_{T,R}& \sum_{i}\{TRR'T'\}_{ii}\\
\operatorname{s.t.} & RT=1
\end{array}


问:
Calculate the angle between two curves $f(x)=x^2$ and $g(x)=\frac{1}{\sqrt{x}}$
I want to Calculate the angle between two curves on their intersect $f(x)=x^2$ and $g(x)=\frac{1}{\sqrt{x}}$, what I did so far is:

$$x^2=\frac{1}{\sqrt{x}} \rightarrow x=1$$then :
$$\tan(a)=\left |\frac{f'(a)-g'(a)}{1+f'(a)*g'(a)}\right|
$$ after set $x=1$ I get zero at the denominator.

Any suggestions? Thanks!

答:
Another approach that I've found useful in applications (and avoids the "undefined tangent value" problem) is to construct tangent vectors to the two curves, using the formulation $ \  \langle 1 \ ,  \frac{dy}{dx} |_{x = a}  \rangle  $ for each one.

In this problem, we have  $ \ \overrightarrow{T_1} \ = \ \langle 1 , f'(1) \rangle \ = \ \langle 1 , 2 \rangle $ for  $ \ y = x^2 \ $ and $ \ \overrightarrow{T_2} \ = \ \langle 1 , g'(1) \rangle \ = \ \langle 1 , -\frac{1}{2} \rangle \ $ for  $ \ y = \frac{1}{\sqrt{x}} \ . $  We find in this problem that

$$ \ \cos \theta \ = \ \frac{\overrightarrow{T_1} \cdot \overrightarrow{T_2} }{|| \overrightarrow{T_1}  || \ || \overrightarrow{T_2}  ||} \ = \ 0 \ , $$

so the curves are orthogonal at $ \ x = 1 \ . $  (This approach works more generally, however.)


答:
If $z \neq 0$ and $z^3= \bar{z}$ then $|z| = 1$ and so $z^3 = \bar{z} $iff $z^4 = 1$.
Hence you are looking for the fourth roots of one.

If $z^4 = 1$ then $z^{8} = 1$, but not the other way around, because the map $w \mapsto w^2$ is not injective.


问:
Newton's Method and Intervals
I'm trying to find the root of $x^4-2x^3+5x^2-6=0$ in the interval $[1,2]$.

For $x_1$, can I use $1$ or $2$ since they're in the interval, or do I have to choose a number between $1$ and $2$?

答: To summarize what's been said in the comments, any value in the interval $[1,2]$ will give you the right root in this example. In general, though, you have to be a little more careful. See this question and its answers for more detail on what can go wrong if you make a bad guess.

Title: Newton's Method and Intervals


问:
Why are sine and cosine identities always true (and thus sentences) even though they have free variables?
In mathematics, we often see statements such as $\sin ^2x + \cos^2 x = 1$.

Notice that this is not a sentence, as the variable $x$ is free. But we all agree that this statement is true, given the usual interpretations of the symbols. How can we square this with the claim that sentences are the formulas that can be either true or false?

答:
A formula that is always true does not have to be a sentence.

We have e.g. formula $x=x$. True in every model and looked at as logical axiom.

Dealing with sine and cosine we are busy in some model $\mathfrak A$.

There we can easily interpret function symbols in such a way that for distinct terms $t$ and $s$ we have:
$$\mathfrak A\vDash t=s[\alpha]$$
for every $\mathfrak A$-assignment $\alpha$.


问:
Prove $1^r+2^r+.....\equiv 0\pmod p$ , given that p is odd prime and $0<r<p-1$
An exercise reads as follows:
If $p$ is an odd prime and $0<r<p-1$ then prove that $s=1^r+2^r+...(p-1)^r\equiv 0\mod p$

I know one clever way to prove it is to use a primitive root $g$ and write the given sum as: $g^r+g^{2r}+g^{3r} +...+g^{(p-1)r}$ which is a geometric process easy to handle and deduce the result.

But I am looking for a more primitive way right now (suppose I know nothing about primitive roots). My thought is: the polynomial $$f=x^{p-1}-1$$ has $p-1$ roots distinct roots in the field $\Bbb{Z_p}$ so one can also write $$f= (x-1)(x-2)(x-3)...(x-(p-1))$$. By inspection of the coefficients of the 2 equal polynomials we get

$t_1=1+2+3+...+(p-1)\equiv0 \mod p$

$t_2=1\cdot 2+1\cdot 3+...+(p-2)\cdot( p-1)\equiv 0 \mod p$

$t_3=1\cdot 2\cdot3+1\cdot 2\cdot4+...+(p-3)\cdot (p-2)\cdot(p-1)\equiv 0\mod p$

and so on ....

now suppose $r=2$ then $s=t^2_1 -2t_2$, where we deduce that $p$ divides $s$

But how in general to compute $s$ by the $t_i$ mentioned above? Is there an algorithm for that? Thanks a lot.

答:
HINT:

The polynomial $1-t^{p-1}$ decomposes $\mod p$ as
$$1-t^{p-1}=(1-t)(1-2t)\cdots (1-(p-1)t)$$
so taking logarithmic derivative we get
$$-\frac{t^{p-2}}{1-t^{p-1}} = \sum_{a=1}^{p-1} \frac{a}{1-a t}$$
Now consider the series expansion of both sides in $\mathbb{F}_p[[t]]$.


问:
How to find the basis of the kernel of this linear transformation?
Say given the function:
$$ T: \mathbb R_{2}[t] \rightarrow \mathbb R_{1}[t]$$
$$ at^{2} + bt + c \rightarrow at + 2c - b $$

We were tasked to find the basis for the kernel of $T$ and $T(\mathbb R_2[t])$.

Where do I start? Could somebody point me in the right direction? I was told to do these steps:

But I don't even understand the first part. Could somebody explain how the matrix A is generated from writing the polynomials as column vectors? What is the map of T?

答:
Let's see. You have to understand that $\mathbb{R}_2[t]$ is a vector space. A basis you could choose is $\{t^2, t, 1\}$ (it is easy to check that really it is). Analogously for $\mathbb{R}_1[t]$, a basis is  $\{t, 1\}$.
Now, in order to obtain the matrix of the transformation $T$, you only have to check that the first vector of the first basis ($t^2=1\cdot t^2+0t+0$) get transformed to $1.t+0$, so the first column of the matrix will be $(1,0)$ (as a column!). And so on for the others two element of the basis to create the other columns.


答:
First of all ${(a^b)}^c=a^{b^c}$ is wrong .No such identity exists(try $a=2,b=3,c=4$).Hence you would have to take the 4th power on both sides and then adding which will give you result

Now that i have answered your query i present an easier solution.

By Cauchy schwarz inequality $$\frac{\sin^4 x}{2}+\frac{\cos^4x}{3}\ge \frac{{(\sin^2 x+\cos^2 x)}^2}{5}=\frac{1}{5}$$ For equality we must heve $$\frac{\sin^4 x}{4}=\frac{\cos^4 x}{9} \iff \tan^2 x=\frac{2}{3}$$ Ther rest can be done by your way

Title: Using $\frac{{{{\sin }^4}x}}{2} + \frac{{{{\cos }^4}x}}{3} = \frac{1}{5}$ finding the value of $\frac{{{{\sin }^8}x}}{8} + \frac{{{{\cos }^8}x}}{27}$

Body:
If $\frac{{{{\sin }^4}x}}{2} + \frac{{{{\cos }^4}x}}{3} = \frac{1}{5}$, then

(A)${\tan ^2}x = \frac{2}{3}$

(B)$\frac{{{{\sin }^8}x}}{8} + \frac{{{{\cos }^8}x}}{27} = \frac{1}{125}$

(C)${\tan ^2}x = \frac{1}{3}$

(D) $\frac{{{{\sin }^8}x}}{8} + \frac{{{{\cos }^8}x}}{27} = \frac{2}{125}$

The official answer is A and B.

My approach is as follow

$\frac{{{{\sin }^4}x}}{2} + \frac{{{{\cos }^4}x}}{3} = \frac{1}{5} \Rightarrow \frac{{{{\sin }^4}x}}{2} + \frac{{{{\left( {1 - {{\sin }^2}x} \right)}^2}}}{3} = \frac{1}{5}$

$ \Rightarrow \frac{{{{\sin }^4}x}}{2} + \frac{{1 + {{\sin }^4}x - 2{{\sin }^2}x}}{3} = \frac{1}{5} \Rightarrow 3{\sin ^4}x + 2 + 2{\sin ^4}x - 4{\sin ^2}x = \frac{6}{5}$

$ \Rightarrow 2 + 5{\sin ^4}x - 4{\sin ^2}x = \frac{6}{5} \Rightarrow 10 + 25{\sin ^4}x - 20{\sin ^2}x = 6 \Rightarrow 25{\sin ^4}x - 20{\sin ^2}x + 4 = 0 \Rightarrow {\left( {5{{\sin }^2}x - 2} \right)^2} = 0 \Rightarrow {\sin ^2}x = \frac{2}{5}$

${\cos ^2}x = \frac{3}{5}$, Hence ${\tan ^2}x = \frac{2}{3}$,

which is correct as per the official answer

$\frac{{{{\sin }^2}x}}{2} = \frac{1}{5}\& \frac{{{{\cos }^2}x}}{3} = \frac{1}{5}$

cubing and adding we get

$\frac{{{{\sin }^8}x}}{8} + \frac{{{{\cos }^8}x}}{{27}} = \frac{2}{{125}}$ but it is wrong as per the official answer key and the correcct answer is $\frac{{{{\sin }^8}x}}{8} + \frac{{{{\cos }^8}x}}{27} = \frac{1}{125}$

Where I am making mistake.


问:
Apparent Implication That $\pi_1(S^1, x_0)$ is Trivial
I know this isn't true, but I can't seem to figure out why. In my topology class, we recently went over the theorem in Munkres section 59 stating that if $X = U \cup V$, with $U, V$ open in $X$, and $U \cap V$ path-connected with $x_0 \in U \cap V$, then if $\iota \colon U \hookrightarrow X$ and $\jmath \colon V \hookrightarrow X$ are the respective inclusion maps, then the fundamental group $\pi_1(X, x_0)$ is generated by the images of $\pi_1(U, x_0)$ and $\pi_1(V, x_0)$ under the induced homomorphisms $\iota_*, \,\, \jmath_*$.

This obviously implies that when $\iota, \jmath$ are the identity map, that $\pi_1(X, x_0)$ is trivial. But this is where I run into a problem. $S^1$ is an open subset of itself, so let $X = U = V = S^1$. Then since $U \cap V = S^1$ is path-connected, and the inclusion maps are both just the identity map on $S^1$, this implies that $\pi_1(S^1, x_0)$ is trivial, a clear contradiction. But I can't find the hole in my reasoning. I'm sure it's something stupid and simple, but could someone point out what I did wrong?

答: The induced homomorphism of fundamental groups is not trivial for the identity map of $S^1$.  It corresponds to the element $1$ in $\pi_1(S^1)\cong\Bbb Z$.


问:
How would you divide $x$ number of things among $y$ people but each person can only have $z$ max things?
I've heard of using a strategy to solve problems like find the ways to divide $x$ things among $y$ people where you solve by saying you have $x$ stars and put $y-1$ bars in between to divide between them and so you'd do $\binom{x + y -1}{y-1}$. If there's a minimum amount, you need to to simply subtract the minimum of each person. So if each person has to have $1$ minimum, you'd subtract $x$ from $x+y-1$.

What if there was a maximum? Say each person could only have $5$ maximum? Is there a generalized way to solve this with variable $z$?

答:
Normally, you may have to apply inclusion-exclusion, but at times, there is a neat short cut. I shall illustrate with an example.

Suppose you have to distribute $18$ apples to $4$ people, with no one getting more than $6\;$ then applying inclusion-exclusion

The answer will work out to $\binom{21}3 -\binom4 1\binom{14}3 + \binom4 2\binom7 3 = 84$

[All ways - at least one person getting $7$ + at least two persons getting $7$]

The intuitive shortcut is to imagine $6$ apples each already given, ie $24$ apples in toto, and take out $24-18$ = $6$ apples in any which way, thus $\binom{6+4-1}{4-1} = 84$

Of course, you can only do this if there are enough apples to be thus taken out !


答:
Just to be a bit more explicit than the other answers.

The exact statement is that $\pi_1 (X) = \pi_1(U) * \pi_1(V) /N$ where $N$ basically describes how the loops of $U \cap V$ fits inside $U$ and inside $V$, then equates both of those.

So in your particular example we have that $\pi_1(U) =  $ as it's just the circle. $\pi_1(V) =  $ as well, and finally $\pi_1(U \cap V) = <c>$.

The inclusion maps are identity as you remarked so $c$ gets mapped to $a$ and $c$ gets mapped to $b$ respectively. Since these are the generators this fully describes everything.

So we finally get that $\pi_1(X) =  */<ab^{-1}> =  =  (or ) = \mathbb{Z}$

Title: Apparent Implication That $\pi_1(S^1, x_0)$ is Trivial

Body:
I know this isn't true, but I can't seem to figure out why. In my topology class, we recently went over the theorem in Munkres section 59 stating that if $X = U \cup V$, with $U, V$ open in $X$, and $U \cap V$ path-connected with $x_0 \in U \cap V$, then if $\iota \colon U \hookrightarrow X$ and $\jmath \colon V \hookrightarrow X$ are the respective inclusion maps, then the fundamental group $\pi_1(X, x_0)$ is generated by the images of $\pi_1(U, x_0)$ and $\pi_1(V, x_0)$ under the induced homomorphisms $\iota_*, \,\, \jmath_*$.

This obviously implies that when $\iota, \jmath$ are the identity map, that $\pi_1(X, x_0)$ is trivial. But this is where I run into a problem. $S^1$ is an open subset of itself, so let $X = U = V = S^1$. Then since $U \cap V = S^1$ is path-connected, and the inclusion maps are both just the identity map on $S^1$, this implies that $\pi_1(S^1, x_0)$ is trivial, a clear contradiction. But I can't find the hole in my reasoning. I'm sure it's something stupid and simple, but could someone point out what I did wrong?

Body:
Just to be a bit more explicit than the other answers.

The exact statement is that $\pi_1 (X) = \pi_1(U) * \pi_1(V) /N$ where $N$ basically describes how the loops of $U \cap V$ fits inside $U$ and inside $V$, then equates both of those.

So in your particular example we have that $\pi_1(U) = < a> $ as it's just the circle. $\pi_1(V) = < b> $ as well, and finally $\pi_1(U \cap V) = <c>$.

The inclusion maps are identity as you remarked so $c$ gets mapped to $a$ and $c$ gets mapped to $b$ respectively. Since these are the generators this fully describes everything.

So we finally get that $\pi_1(X) = <a> *<b>/<ab^{-1}> = <a, b: ab^{-1}> = <a> (or <b>) = \mathbb{Z}$


答:
\begin{align}
p^2+q^2 &= 81
\tag{1}\label{1}
,\\
p^2+q^2-2qx+x^2 &= 225
\tag{2}\label{2}
,\\
p^2-2px+x^2+q^2 &= 144
\tag{3}\label{3}
.
\end{align}

Subtract \eqref{1} from \eqref{2} and \eqref{3}
to get $p$ and $q$ in terms of $x$,
\begin{align}
p &= \tfrac12\,\frac{x^2-63}x
\tag{4}\label{4}
,\\
q &= \tfrac12\,\frac{x^2-144}x
\tag{5}\label{5}
,
\end{align}

then substitute \eqref{4} \eqref{5} into \eqref{1}
to get biquadratic equation in $x$

\begin{align}
2\,(x^2)^2-738\,x^2+24705
&=0
,
\end{align}

which gives
\begin{align}
x^2&=\tfrac{369}2\pm\tfrac{27}2\,\sqrt{119}
\end{align}
and hence, four solutions for $x$:

\begin{align}
x_1&=\tfrac32\,\sqrt{82+6\sqrt{119}}
,\\
x_2&=-\tfrac32\,\sqrt{82+6\sqrt{119}}
,\\
x_3&=\tfrac32\,\sqrt{82-6\sqrt{119}}
,\\
x_4&=-\tfrac32\,\sqrt{82-6\sqrt{119}}
.
\end{align}

Title: I have found a system of 3 variables that lead to a quadratic equation, so how do I find the answer?/

Body:
$$p^2 + q^2 = 81$$
$$p^2 + (q-x)^2 = 225$$
$$(p-x)^2 + q^2 = 144$$
So these are the 3 equations.


答:
The suggestion just notes that the term $at^2+bt+c $ can be identified with the triple $(a,b,c)$ (that is, the term has the coordinates $(c,b,a)$ with the basis $1,t,t^2$).

I think is is easier to compute directly.

Note that $at^2+bt+c = 0$ iff $a=b=c=0$.

We have
\begin{eqnarray}
\ker T &=& \{ at^2+bt+c | T(at^2+bt+c) = 0 \} \\
&=& \{ at^2+bt+c | at+2c-b = 0 \} \\
&=& \{ at^2+bt+c | a = 0, b=2c \} \\
&=&  \{ 2 ct +c | c \in \mathbb{R} \} \\
&=& \{ c (2t+1) | c \in \mathbb{R} \}
\end{eqnarray}
From this we see that $2t+1$ is a basis for $\ker T$.

The image is computed similarly.
\begin{eqnarray}
{\cal R} T &=& \{ T(at^2+bt+c) | at^2+bt+c \in \mathbb{R}_2[t]\} \\
&=& \{ at+2c-b | at^2+bt+c \in \mathbb{R}_2[t]\} \\
&=& \{ at+2c-b | a,b,c \in \mathbb{R}\} \\
&=& \{ at+d | a,d \in \mathbb{R}\} \\
\end{eqnarray}
And so we see that $t, 1$ form a basis for ${\cal R}T$.

Title: How to find the basis of the kernel of this linear transformation?

Body:
Say given the function:
$$ T: \mathbb R_{2}[t] \rightarrow \mathbb R_{1}[t]$$
$$ at^{2} + bt + c \rightarrow at + 2c - b $$

We were tasked to find the basis for the kernel of $T$ and $T(\mathbb R_2[t])$.

Where do I start? Could somebody point me in the right direction? I was told to do these steps:

But I don't even understand the first part. Could somebody explain how the matrix A is generated from writing the polynomials as column vectors? What is the map of T?


问:
Do $\limsup$ and $\liminf$ are the actual bounds of a sequence?
Can you say that $\limsup a_n$ is the upper bound of the sequence $a_n$ and $\liminf a_n$ is the lower bound of $a_n$?

答:
No.

If $a_n=(-1)^n + \frac{(-1)^n}{n}$, then

the upper bound of the sequence is $\frac{3}{2}$

the lower bound for the sequence is $-2$

$\limsup a_n = 1$

$\liminf a_n = -1$

so you have a case when the $\limsup$ is not the upper bound, and the $\liminf$ is not the lower bound.


问:
Trig Identities Problem
I am trying to find the trig identity of this problem:

$\sin(x)\cos(x)\tan(x)\sec(x)\csc(x)$

I already know the answer is $\tan(x)$, but everytime I've tried to solve it I always ended up changing it.

答:
It's very simple.
$$
\require{cancel}
\begin{align}
\sin x\cos x\tan x\sec x\csc x&=\cancel{\sin x}\cancel{\cos x}\frac{\sin x}{\cos x}\frac{1}{\cancel{\cos x}}\frac{1}{\cancel{\sin x}}\\
&=\frac{\sin x}{\cos x}\\
&=\tan x
\end{align}
$$

Title: Trig Identities Problem


问:
Solving $\frac{1}{x^2-1} \geqslant \frac{1}{x+1}$
How do I solve:

$$\frac{1}{x^2-1} \geqslant \frac{1}{x+1}$$

Here's my attempt to solve it:
I brought the fraction on the right side to the left side, then I've found the common denominator, $(x-1)(x+1)$, and after all the computations, this is my situation:
$$ \frac{2-x}{(x-1)(x+1)} \geqslant 0$$
Now, I've put the numerator and the denominator in a system of inequalities:
(I don't know how to type the system of inequalities in mathjax, sorry)

$$x<2$$
$$x < -1 \space \text{or} \space x>1 $$
My result is (I found the intersection): $$-1<x<1 \space \space \text{or} \space \space x>2$$

Let me know if the result is wrong.
Additionally, because in the test I'll have only 50 minutes (lots of questions), this method requires a lot of time; if you know a quicker method, let me know.

EDIT (solved in a quicker way):

when you are in this situation: $$\frac{2-x}{(x-1)(x+1)} \geqslant 0$$
change the sign of the inequality and the signs that appear in the numerator,

then, check each interval in the normal way, but don't forget you have changed the sign; therefore, if you had a greater sign, and you changed it, you have to take only negative intervals (not positive intervals, in order to express the correct solution).

答:
Use the difference of two squares to get $\frac{1}{(x+1)(x-1)} ≥ \frac{1}{x+1}$.

Case 1: If $x + 1 > 0 \Rightarrow x > -1$, we have:

$$\frac{1}{x-1} ≥ 1 \tag{$x \ne -1, 1$}$$

The $x \ne -1, 1$ comes from the denominator $(x+1)(x-1)$, which is undefined when $x = -1, 1$.

If $x - 1 > 0 \Rightarrow x > 1$, then multiplying both sides by $x-1$ gives $1 ≥ x - 1 \Rightarrow x ≤ 2$. If $x - 1 < 0$, then because $x > -1$, there are no solutions for this case. Hence the intersection of $x > -1, x > 1$, and $x ≤ 2$ imply $1 < x ≤ 2$.

Case 2: If $x + 1 < 0 \Rightarrow x < -1$, we have:

$$\frac{1}{x-1} ≤ 1 \tag{$x \ne -1, 1$}$$

$x - 1 > 0$ is not possible in this case because $x < -1$. Thus $x -1 < 0$, or just $x < -1$, and there are no more conditions.

So the solution to the inequality is $x < -1$, $1 < x ≤ 2$ for $x \in \mathbb R$.


问:
$\exists \mathfrak{M} \models T, T = \mathsf{Th}(\mathfrak{M})\implies\forall\mathfrak{M}, \mathfrak{N} \models T, \mathfrak{M} \equiv \mathfrak{N}$
$\newcommand{\Th}{\mathsf{Th}}$
$\newcommand{\M}{\mathfrak{M}}$
$\newcommand{\N}{\mathfrak{N}}$
$\newcommand{\L}{\mathscr{L}}$

I'm trying to prove equivalent statements of a complete theory by cyclic implication. However, I'm unable to prove $(4) \implies (5)$ or equivalently $(4) \implies (3)$ as $(3) \implies (5)$ is obvious.

Let $T$ be a consistent $\L$-theory. The following are equivalent:

$T$ is complete ;

The deductive closure of $T$ is a maximal consistent theory ;

$\forall \M \models T, T = \Th(\M)$ ;

$\exists \M \models T, T = \Th(\M)$ ;

$\forall \M, \N \models T, \M \equiv \N$

for any $\L$-sentence $\varphi$, $T \not\models \varphi \implies T \models \neg \varphi$.

($T$ is complete $\iff$ for all sentence $\varphi$, $T \models \varphi$ or $T \models \neg \varphi$.)

Draft for $(4) \implies (5)$:

Let $\M \models T$ s.t. $T = \Th(\M)$. Then for $\N_1, \N_2 \models T$, there is $\Th(\N_1) \supseteq \Th(\M)$ and $\Th(\N_2) \supseteq \Th(\M)$.

I have no idea how to prove the inclusion $\subseteq$.

答:
Let $M$ be as in 4.(I don't like the curly stuff)

The important point is that $T = Th(M)$ is complete. This is because for any $\varphi$ sentence, either $M \vDash \varphi$ or $M \vDash \neg \varphi$, so either $\varphi \in T$ or $\neg \varphi \in T$.

Well then given any $N \vDash T$. Let $\varphi$ be any sentence. We then have:

$M \vDash \varphi \iff \varphi \in T \iff N \vDash \varphi$

The first $\iff$ follows from $T  = Th(M)$.

The second $\implies$ follows from $N \vDash T$.

The second $\impliedby$ is by completeness of $T$. Either $\varphi \in T$ or $\neg \varphi \in T$. Since $N \vDash T$ and $N \vDash \varphi$, clearly $\neg \varphi \not \in T$ (else $N \vDash \varphi$ and $N \vDash \neg \varphi$, and so since $\neg \varphi  \not \in T$,we have $\varphi \in T$).

Thus for any $N \vDash T$, $N \equiv M$. So given $N, N' \vDash T$, $N \equiv M \equiv N' \rightarrow N \equiv N'$


问:
Showing that matrix $A$ and linear operator $T: M_{n\times n} \rightarrow M_{n\times n}$ ($T(B) = AB$) have the same eigenvalues
$T: M_{n\times n} \rightarrow M_{n\times n} \\ T(B) = AB$

Defining $S_A$ as the set of eigenvalues of $A$ and $S_T$ as the set of eigenvalues of $T$, I have to prove $1. \ S_T \subseteq S_A$ and $2. \ S_A \subseteq S_T$.

The eigenvalues of $A$ are derived from the operation of $A$ on $n\times 1$ matrices but $T$ operates on $n\times n$ matrices and I don't think I can just assume $A$ to be the matrice form of $T$. I attempted defining an isomorphism $M_{n\times n} \rightarrow \mathbb{F}^{n^2}$ but clearly that wouldn't work because we can no longer use the same matrix multiplication anymore.

答:
When $B$ has columns $b_1,b_2,\dots,b_n$ so that $B=(\ b_1\mid b_2 \mid \dots \mid b_n\ )$, then note that
$$
T(B) = AB = A(\ b_1\mid b_2 \mid \dots \mid b_n\ ) = (\ Ab_1\mid Ab_2 \mid \dots \mid Ab_n\ ).
$$

Hence $AB=\lambda B$ is equivalent to $Ab_i = \lambda b_i$ for all columns $b_i$ of $B$.

When $\lambda$ is an eigenvalue of $A$, let $b\neq 0$ an eigenvector so that $Ab=\lambda b$ and consider the matrix $B=(\ b \mid b \mid \dots \mid b\ )$. We have $B\neq 0$ and
$$
AB = (\ Ab \mid Ab \mid \dots \mid Ab\ ) = (\ \lambda b \mid \lambda b \mid \dots \mid \lambda b\ ) = \lambda B.
$$
Hence $B$ is an eigenvector of $T$ with eigenvalue $\lambda$.

When $\lambda$ is an eigenvalue of $T$, so that $AB=\lambda B$ for some non-zero matrix $B$, there is at least one non-zero column $b_i$ of $B$ and for that we have $Ab_i = \lambda b_i$. Hence $b_i$ is an eigenvector of $A$ with eigenvalue $\lambda$.

This shows that $A$ and $T$ have the same eigenvalues.


问:
Linearly independent set with dense span and Schauder basis.
Let $V$ be a Banach space. A Schauder basis of $V$ is a sequence $(v_n)$ in $V$ such that for each $v$ in $V$ there is a unique sequence $(a_n)$ of complex numbers such that $v=\sum_{n=1}^\infty a_nv_n$.

Now suppose that there is a linearly independent countable set $S\subseteq V$ such that $\text{span}(S)$ is dense in $V$.

My question is if one can necessarily find a Schauder basis of $V$ in $\text{span}(S)$.

答: Let $V=C[0,1]$ with sup norm and $S=\{1,x,x^{2},...\}$. Then your conditions are satisfied. If there is a sequence of polynomials $p_n$ such that every continuous function is a sum $\sum a_np_n$ (the series converging in sup norm) then every continuous function would be analytic, in particular a $C^{\infty}$ function,  which is false.


问:
Finding the limit as x approaches 1 of $\sum \frac {(-1)^{n+1}}{n} x^n$
I want to find $\displaystyle \lim_{x \rightarrow 1^-} \sum_{n \in \mathbb N} \frac {(-1)^{n+1}}{n} x^n$

What I'm guessing I'm supposed to do is use a certain property that states that if $\displaystyle \sum_{n \in \mathbb N} \frac {(-1)^{n+1}}{n} x^n$ converges for some $x$ in $[0,1]$ and the series of the derivatives $\displaystyle \sum_{n \in \mathbb N} (-1)^{n+1} x^{n-1}$ converges uniformly in $[0,1]$, then the original series converges uniformly and thus I can exchange the limit on $x$ and the limit on $n$, giving

$$\displaystyle \lim_{x \rightarrow 1^-} \lim_{N \rightarrow \infty} \sum_{n = 1}^N \frac {(-1)^{n+1}}{n} x^n = \lim_{N \rightarrow \infty} \lim_{x \rightarrow 1^-}  \sum_{n = 1}^N \frac {(-1)^{n+1}}{n} x^n$$

Which would result in the alternate Harmonic series (multiplied by $-1$).

The problem, however, is that while it is true that the derivatives converge uniformly on $[0,1)$, they don't on $[0,1]$ since when x = 1 the partial sums for the derivatives are $1,0,1,0 \dots$

However, the statement that the derivatives need to converge uniformly on $[0,1]$ is what's making my head spin. The original functions are defined on $[0,1]$ so shouldn't the derivatives be defined on $(0,1)$ ?. If not, what does it mean for the derivatives to converge uniformly on $[0,1]$ ?

答:
Here is one way forward.  We will write the series in terms of the even and odd parts of its summand.  Proceeding, we find that

$$\begin{align}
\sum_{n=1}^\infty \frac{(-1)^{n-1}x^n}{n}&=\sum_{n=1}^\infty \left(\frac{x^{2n-1}}{2n-1}-\frac{x^{2n}}{2n}\right)\\\\
&=\sum_{n=1}^\infty  x^{2n-1}\left(\frac{2n(1-x)+1}{2n(2n-1)}\right)\\\\
&=\sum_{n=1}^\infty \frac{(1-x)x^{2n-1}}{2n-1}+\sum_{n=1}^\infty \frac{x^{2n-1}}{2n(2n-1)}\tag1
\end{align}$$

Now, the second series on the right-hand side of $(1)$ is easily seen to be uniformly convergent for $0\le x\le 1$.

To show that the first series on the right-hand side of $(1)$ converges uniformly on $[0,1]$ we have for any $\varepsilon>0$ (and $N\ge1$)

$$\begin{align}
\sum_{n=N+1}^\infty \frac{(1-x)x^{2n-1}}{2n-1}&\le \sum_{n=N+1}^\infty \frac{(1-1/2n)^{2n-1}}{2n(2n-1)}\\\\
&\le \sum_{n=N+1}^\infty \frac{1}{4n(2n-1)}\\\\
&<\frac18\sum_{n=N+1}^\infty \left(\frac1{n-1}-\frac1n\right)\\\\
&=\frac1{8N}\\\\
&<\varepsilon
\end{align}$$

whenever $N>\frac1{8\varepsilon}$.   Hence, we can assure that

$$\begin{align}
\lim_{x\to 1^-}\sum_{n=1}^\infty \frac{(-1)^{n-1}x^n}{n}&=\sum_{n=1}^\infty \frac{1}{2n(2n-1)}\\\\
&=\sum_{n=1}^\infty \left(\frac{1}{2n-1}-\frac1{2n}\right)\\\\
&=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\\\\
&=\log(2)
\end{align}$$

as was to be shown!


问:
Set inclusion question
Suppose I want to show ${\{\begin{bmatrix} 1 \\ 2 \\ 0  \end{bmatrix}\}\ ^{\perp}=span\{{\begin{bmatrix} 2 \\ -1 \\ 0  \end{bmatrix},\begin{bmatrix} 0 \\ 0 \\ 1  \end{bmatrix}\\}}\}$

If I do this with set inclusions, for the forward direction I would start with $x\in \{\begin{bmatrix} 1 \\ 2 \\ 0  \end{bmatrix}\}\ ^{\perp}$, and by definition,only  $\begin{bmatrix} 0 \\ 0 \\ a \end{bmatrix}$ and $\begin{bmatrix} 2b \\ -b \\ 0 \end{bmatrix}$ would be in $\{\begin{bmatrix} 1 \\ 2 \\ 0  \end{bmatrix}\}\ ^{\perp}$ because their inner product with $\{\begin{bmatrix} 1 \\ 2 \\ 0  \end{bmatrix}\}$ is $0$, and so it is the span of of the right hand side. My question is, how do I do the inclusion formally? Like starting from $x\in \{\begin{bmatrix} 1 \\ 2 \\ 0  \end{bmatrix}\}\ ^{\perp}$ and ending at $x \in span\{{\begin{bmatrix} 2 \\ -1 \\ 0  \end{bmatrix},\begin{bmatrix} 0 \\ 0 \\ 1  \end{bmatrix}\\}\}$? I have no clue when dealing the above with an  "$x$"

答: If $$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \in \left\{ \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix} \right\}^\perp$$ this means $x_1+2x_2=0$, that is, $x_1=-2x_2$. Thus $$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -2x_2 \\ x_2 \\ x_3 \end{bmatrix} = -x_2 \begin{bmatrix} 2 \\ -1 \\ 0 \end{bmatrix} + x_3 \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \in \operatorname{span} \left\{ \begin{bmatrix} 2 \\ -1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \right\}.$$

Title: Set inclusion question


问:
Ramsey numbers Olympiad combo
Show that if l,s are positive integers, then
$r(l,s) \leq {l + s - 2 \choose l - 1} $

答:
In order to prove this claim, I will use the result $r(l, s) \leq r(l - 1, s) + r(l, s - 1)$. A proof of this inequality is shown here.

When $l = s = 2$, we have $r(2, 2) = 2$, and the claim holds. Now if the relation holds for $r(l - 1, s)$ and $r(l, s - 1)$, then one has

$$r(l, s) \leq r(l - 1, s) + r(l, s - 1) = {l + s - 2\choose l -1}, $$

where the last equality follows from using the induction hypothesis with Pascal's Identity. By induction, the conclusion follows.

Title: Ramsey numbers Olympiad combo


答:
Let $a= \begin{bmatrix} 1 \\ 2 \\ 0  \end{bmatrix}$, $b= \begin{bmatrix} 2 \\ -1 \\ 0  \end{bmatrix}$ and  $c=\begin{bmatrix} 0 \\ 0 \\ 1  \end{bmatrix}$

$V:=\{\begin{bmatrix} 1 \\ 2 \\ 0  \end{bmatrix}\} ^{\perp} $ and $ U:={span\{{\begin{bmatrix} 2 \\ -1 \\ 0  \end{bmatrix},\begin{bmatrix} 0 \\ 0 \\ 1  \end{bmatrix}\\}}\}$

Now,take any $x\in V$ then $x= m\cdot a$ so: $$\langle x,b\rangle  = m\langle a,b\rangle =0$$ and $$\langle x,c\rangle  = m\langle a,c\rangle =0$$ so $x\in U$ and thus $\boxed{V\subseteq U}$.

For converse, take any $x\in U$, then $x= m\cdot b+n\cdot c$ and we have $$\langle a,x\rangle  = m\langle a,b\rangle +n\langle a,c\rangle  = 0$$  so $x\in V$ and thus $\boxed {U\subseteq V}$ and so $U=V$.

Title: Set inclusion question

Body:
Suppose I want to show ${\{\begin{bmatrix} 1 \\ 2 \\ 0  \end{bmatrix}\}\ ^{\perp}=span\{{\begin{bmatrix} 2 \\ -1 \\ 0  \end{bmatrix},\begin{bmatrix} 0 \\ 0 \\ 1  \end{bmatrix}\\}}\}$

If I do this with set inclusions, for the forward direction I would start with $x\in \{\begin{bmatrix} 1 \\ 2 \\ 0  \end{bmatrix}\}\ ^{\perp}$, and by definition,only  $\begin{bmatrix} 0 \\ 0 \\ a \end{bmatrix}$ and $\begin{bmatrix} 2b \\ -b \\ 0 \end{bmatrix}$ would be in $\{\begin{bmatrix} 1 \\ 2 \\ 0  \end{bmatrix}\}\ ^{\perp}$ because their inner product with $\{\begin{bmatrix} 1 \\ 2 \\ 0  \end{bmatrix}\}$ is $0$, and so it is the span of of the right hand side. My question is, how do I do the inclusion formally? Like starting from $x\in \{\begin{bmatrix} 1 \\ 2 \\ 0  \end{bmatrix}\}\ ^{\perp}$ and ending at $x \in span\{{\begin{bmatrix} 2 \\ -1 \\ 0  \end{bmatrix},\begin{bmatrix} 0 \\ 0 \\ 1  \end{bmatrix}\\}\}$? I have no clue when dealing the above with an  "$x$"


问:
Matrix represents a bounded linear operator $T \in L (l^ 2 )$ with respect to the orthonormal canonical basis
Consider the following Jacobi matrix:
\begin{equation}
\begin{pmatrix}
a_1 & b_1 & 0 & 0 & 0 & \ldots\\
c_2 & a_2 & b_2 & 0 & 0 & \ldots\\
0 & c_3 & a_3 & b_3 & 0 & \ldots\\
0 & 0 & c_4 & a_3 & b_4 & \ldots\\
. & . & . & . & . & . \\
. & . & . & . & . & . \\
. & . & . & . & . & .
\end{pmatrix}
\end{equation}
Prove that if

$\beta=\sup\left\lbrace|c_n| + |a_n| + |b_n| : n = 1, 2, \ldots\right\rbrace <\infty$ and

$\alpha = \sup\left\lbrace|b_{n−1}|+|a_n|+|c_{n+1}| : n = 2, 3,\ldots\right\rbrace <\infty$

then the matrix represents a bounded linear operator $T\in L(l^2)$ with respect to the orthonormal canonical basis and that $||T|| ≤ (\alpha\beta)^{1/2}$.

Under what conditions is T compact?

Generalize the result for band matrices?

答:
This is known as Schur's test.

Let $(x_n)$ and $(y_n)$ be any sequences in $\ell^2$.

Then (interpreting the first row appropriately) \begin{align}|\langle (x_n),T(y_n)\rangle|&=|\sum_n\bar{x}_n(c_ny_{n-1}+a_ny_n+b_ny_{n+1})|\\
&\le\sum_{n,m}|x_n|(|c_n|\delta_{m,n-1}+|a_n|\delta_{m,n}+|b_n|\delta_{m,n+1})|y_m|\\
&=\sum_{n,mm}|x_n|d_{m,n}|y_m|\quad\textrm{where}\ d_{m,n}=|c_n|\delta_{m,n-1}+|a_n|\delta_{m,n}+|b_n|\delta_{m,n+1}\\
&\le\sqrt{\sum_{n,m}|x_n|^2d_{m,n}}\sqrt{\sum_{n,m}|y_m|^2d_{m,n}}\quad \textrm{(Cauchy-Schwarz)}\\
&=\sqrt{\sum_n|x_n|^2(|c_n|+|a_n|+|b_n|)}\sqrt{\sum_m(|c_{m+1}|+|a_m|+|b_{m-1}|)|y_m|^2}\\
&\le\sqrt{\alpha\beta}\|(x_n)\|\|(y_m)\|
\end{align}

Hence $\|T\|=\sup_{x,y\ne0}\frac{|\langle x,Ty\rangle|}{\|x\|\|y\|}\le\sqrt{\alpha\beta}$.

The result is easily generalized to banded matrices.


问:
$f+g \in \mathcal{L}^p$
Let $\Omega, \mathcal{M}, \mu$ be a measure space and $p\in \{1,2\}$. Let $f, g \in \mathcal{L}^p(\Omega,\mu).$

Show that $f+g \in \mathcal{L}^p$.

答: By Minkowski's inequality, $$\int_{\Omega}|f+g|^p d\mu \le \Big(\Big(\int_{\Omega}|f|^p d\mu\Big)^{\frac{1}{p}}+\Big(\int_{\Omega}|g|^p d\mu\Big)^{\frac{1}{p}}\Big)^p< +\infty$$ as both $f,g \in \mathcal{L}^p(\Omega)$ and hence $f+g \in \mathcal{L}^p(\Omega)$

Title: $f+g \in \mathcal{L}^p$


问:
Calculus Integral Question

Question: Given the point (a, b) in the first quadrant, find the downward-opening parabola that passed through the point (a, b) and the origin such that the area under the parabola is a minimum.

I've tried to solve it. But the answer can't be right, since it leads to a upward-opening parabola.

Here's what I've done so far.

Step One:

Let $$y = cx^2 + dx + e$$ be the equation for the parabola.

Because the parabola has a root at the origin, $$e = 0$$ The function for parabola can be written as $$y = cx^2 + dx$$

Step Two:

We are given that the parabola passes through the point (a, b) in the first quadrant. Plugging the numbers into the function above, $$b = c(a)^2 + d(a)$$ $$\frac{b-a^2c}{a}=d $$
Therefore, $$y=cx^2+\frac{b-a^2c}{a}x$$

Step Three:

Next, to find the area under the parabola, use the integral: $$\int_0^n(cx^2+\frac{b-a^2c}{a}x)dx$$where n is the second root of the parabola (first root is the origin).

For n, $$0 = cx^2+\frac{b-a^2c}{a}x$$ $$0 = cx+\frac{b-a^2c}{a}$$ $$x=-\frac{b-a^2c}{ac} = n$$

Now, substitute the expression for n, $$\int_0^{-\frac{b-a^2c}{ac}} (cx^2+\frac{b-a^2c}{a}x)dx$$ which equals $$\frac{c}{3}\bigl(-\frac{b-a^2c}{ac}\bigr)^3+\frac{b-a^2c}{2a}\bigl(-\frac{b-a^2c}{ac}\bigr)^2$$ which equals $$\frac{b-a^2c}{6a^3c^2}$$

It can be concluded that the area under the parabola has the function $$A=\frac{b-a^2c}{6a^3c^2}$$

Step Four:

To find the minimum area -- as requested, take the derivative of A. $$A'=\frac{6a^3c^2(-a)^2-(b-a^2c)(12a^3c)}{(6a^3c^2)^2}$$ $$=\frac{a^2c-2b}{6a^3c^3}$$ Find its root: $$0=\frac{a^2c-2b}{6a^3c^3}$$ $$0=a^2c-2b$$ $$c=\frac{2b}{a^2}$$

Since (a, b) is in the first quadrant, c is a positive number.

Can you tell me where have I gone wrong?

Thanks!

答:
Please note

$\displaystyle A = \frac{c}{3}\bigl(-\frac{b-a^2c}{ac}\bigr)^3+\frac{b-a^2c}{2a}\bigl(-\frac{b-a^2c}{ac}\bigr)^2 = \frac{(b - a^2 c)^3} {6 a^3 c^2}$

Taking derivatative wrt. $c$ and equating to zero,

$\displaystyle A' = 0 = \frac{2(b-a^2c)^3 + 3ca^2(b-a^2c)^2}{6a^3c^3}$

This gives us $\displaystyle c = \frac{b}{a^2}, c = - \frac{2b}{a^2}$

Title: Calculus Integral Question


问:
Failure of isomorphisms on stalks to arise from an isomorphism of sheaves
It is well-known (Hartshorne 2.1.1) that if $F$ and $G$ are sheaves on a space $X$, then $\phi:F\rightarrow G$ is an isomorphism if and only if the induced stalk map $\phi_p:F_p\rightarrow G_p$ is an isomorphism for all $p\in X$. However, if we have a collection of isomorphisms $\{\psi_p:F_p\rightarrow G_p\}_{p\in X}$, this does not guarantee that $F$ and $G$ are isomorphic, because the $\psi_p$ might not be related to each other, i.e. there might not be a sheaf map $\psi:F\rightarrow G$ such that $\psi_p$ is the induced stalk map for all $p\in X$.

However, I was recently making this point to someone and was unable to think of a good example of non-isomorphic $F$ and $G$ having isomorphisms $\psi_p:F_p\rightarrow G_p$. I'm sure I knew one at some point, but I'm blanking on it now. Can someone provide an illustrative example, e.g. an example that occurs in some natural or basic problem, or one that captures the essential pattern of any example where this issue arises, or one where it is clear that $F$ and $G$ could not be isomorphic?

答: A bit late to the party, but my favourite examples are the twisting sheaves. In the proof of Proposition 5.12(a) of Hartshorne, the isomorphism of restricted sheaves does not lift to an isomorphism of the entire sheaf.</p>


答:
We note on the side that $x^2\ne1$ and multiply by $x^2-1$, so need to distinguish two cases:

$$\begin{cases}x^2-1>0\to 1\ge x-1,\\x^2-1<0\to 1\le x-1.\end{cases}$$

The second case is impossible and we are left with

$$x<-1\lor1<x\le2.$$

You can also work in systematic way, using a table of sign variations,

$$\begin{array}{}&&-1&&1&&2\\\hline x+1&-&0&+&+&+&+&+\\x-1&-&-&-&0&+&+&+\\2-x&+&+&+&+&+&0&-\\\hline&+&|&-&|&+&0&-\end{array}$$

Title: Solving $\frac{1}{x^2-1} \geqslant \frac{1}{x+1}$

Body:
How do I solve:

$$\frac{1}{x^2-1} \geqslant \frac{1}{x+1}$$

Here's my attempt to solve it:
I brought the fraction on the right side to the left side, then I've found the common denominator, $(x-1)(x+1)$, and after all the computations, this is my situation:
$$ \frac{2-x}{(x-1)(x+1)} \geqslant 0$$
Now, I've put the numerator and the denominator in a system of inequalities:
(I don't know how to type the system of inequalities in mathjax, sorry)

$$x<2$$
$$x < -1 \space \text{or} \space x>1 $$
My result is (I found the intersection): $$-1<x<1 \space \space \text{or} \space \space x>2$$

Let me know if the result is wrong.
Additionally, because in the test I'll have only 50 minutes (lots of questions), this method requires a lot of time; if you know a quicker method, let me know.

EDIT (solved in a quicker way):

when you are in this situation: $$\frac{2-x}{(x-1)(x+1)} \geqslant 0$$
change the sign of the inequality and the signs that appear in the numerator,

then, check each interval in the normal way, but don't forget you have changed the sign; therefore, if you had a greater sign, and you changed it, you have to take only negative intervals (not positive intervals, in order to express the correct solution).


答: I plotted some of your solutions to see where the problem lies. $c>0$ is still a valid solution and also one that gives you a lower area:

Title: Calculus Integral Question

Body:
Question: Given the point (a, b) in the first quadrant, find the downward-opening parabola that passed through the point (a, b) and the origin such that the area under the parabola is a minimum.

I've tried to solve it. But the answer can't be right, since it leads to a upward-opening parabola.

Here's what I've done so far.

Step One:

Let $$y = cx^2 + dx + e$$ be the equation for the parabola.

Because the parabola has a root at the origin, $$e = 0$$ The function for parabola can be written as $$y = cx^2 + dx$$

Step Two:

We are given that the parabola passes through the point (a, b) in the first quadrant. Plugging the numbers into the function above, $$b = c(a)^2 + d(a)$$ $$\frac{b-a^2c}{a}=d $$
Therefore, $$y=cx^2+\frac{b-a^2c}{a}x$$

Step Three:

Next, to find the area under the parabola, use the integral: $$\int_0^n(cx^2+\frac{b-a^2c}{a}x)dx$$where n is the second root of the parabola (first root is the origin).

For n, $$0 = cx^2+\frac{b-a^2c}{a}x$$ $$0 = cx+\frac{b-a^2c}{a}$$ $$x=-\frac{b-a^2c}{ac} = n$$

Now, substitute the expression for n, $$\int_0^{-\frac{b-a^2c}{ac}} (cx^2+\frac{b-a^2c}{a}x)dx$$ which equals $$\frac{c}{3}\bigl(-\frac{b-a^2c}{ac}\bigr)^3+\frac{b-a^2c}{2a}\bigl(-\frac{b-a^2c}{ac}\bigr)^2$$ which equals $$\frac{b-a^2c}{6a^3c^2}$$

It can be concluded that the area under the parabola has the function $$A=\frac{b-a^2c}{6a^3c^2}$$

Step Four:

To find the minimum area -- as requested, take the derivative of A. $$A'=\frac{6a^3c^2(-a)^2-(b-a^2c)(12a^3c)}{(6a^3c^2)^2}$$ $$=\frac{a^2c-2b}{6a^3c^3}$$ Find its root: $$0=\frac{a^2c-2b}{6a^3c^3}$$ $$0=a^2c-2b$$ $$c=\frac{2b}{a^2}$$

Since (a, b) is in the first quadrant, c is a positive number.

Can you tell me where have I gone wrong?

Thanks!


问:
Finding probability based on mean and standard deviation.
Suppose the scores of an entrance test are normally distributed.
Also, suppose the test has a mean of m and a standard deviation of s.
You have set the cut off as 90 marks for the test. What is the probability of a student passing the test?

E.g. For a mean of 100 and standard deviation of 10, this probability is 0.84.

However, What is the formula or method to arrive at this number?

答:
This is a very easy, standard question related to the normal distribution. We have a normally distributed variable $X \sim N(100,10)$. Then
$$
P(X>90) = 1- P(X<90) = 1- \Phi \left( \frac{90-100}{10} \right) \approx 0.841~344
$$
It was necessary to normalize the value inside the cumulative density function $\Phi$ because it is calculated for the $N(0,1)$-case.


问:
Prove that If x and y are integers and x + y is odd , then x−y is odd.
I am facing a problems with this question : Prove that If x and y are integers and x + y is odd , then x−y is odd. , any suggestions ?</p>

答:
Your solution is wrong because you cannot just assume that $x$ is odd. Your proof says nothing about what happens if $x$ is even. You could separate the cases (there are four of them) and run the proof over each of those, but there is a much simpler way of proving the statement.

Hint:
Use the fact that
$$x-y = (x+y) - 2y.$$


问:
How to find the sample size for a level-$\alpha$-test?
Let $X_1,X_2,\ldots,X_n$ be normally distributed with known variance $\sigma^2 > 0$.

We test

$$H_0: \mu = \mu_0 \ \text{ and } \ H_A:\mu > \mu_0.$$

Let $\alpha, \beta \in (0,1)$ and $\delta$ be given such that $\alpha + \beta < 1$ and $\delta > 0$. How big has to be the sample size that under the common level-$\alpha$-test the power against $\mu = \mu_0 + \delta$ is at least $1-\beta$?

Remark: The "common level-$\alpha$-test" is given by the condition that for the critical area $K$ holds
$P_{\theta_0}(X \in K) \le \alpha \Longleftrightarrow P_{\theta_0}(X \in K^C) \ge 1-\alpha$.

I do not know how to do this. I know what the power function and the level-$\alpha$-test are, but I do not see how I should use this here to compute a sample size. Could you help me?

答:
Comment: This is not a general answer to your question, but a bit of a hint that may help you get started. Maybe take @heropup's advice and review
the definitions of significance level and power before you start.

A specific computation in Minitab statistical software for normal data with known $\sigma= 3$ is shown below. Suppose you want to test $H_0: \mu = 20$ vs. $H_a: \mu > 20$ at the 5% level of significance. Specifically, you want power $1-\beta = 0.8$ to detect a difference of $\delta = 1.$ That is, power 80% against the specific alternative
is $\mu_a = 21.$

```<code>  Power and Sample Size

1-Sample Z Test

Testing mean = null (versus > null)
Calculating power for mean = null + difference
α = 0.05  Assumed standard deviation = 3

            Sample  Target
Difference    Size   Power  Actual Power
         1      56     0.8      0.802222
</code>```

After you solve your general problem, you can plug in
my values for $\sigma,\delta,$ and power, to see if
you get $n = 56.$  The following graph shows power
for various values of $\delta.$

If you can't see how to solve the general problem, then try this specific
version first---then generalize.

[Start by finding the critical region (rejection region) for a test at level $\alpha = 0.05.$ Then find $n$ that gives the critical region probability $0.8$ when $\mu=21.]$

Perhaps consider the following figure;

```<code>hdr="n = 46: Dist'ns Under Null (blue) and Alternative"
curve(dnorm(x, 20, 3/sqrt(46)), 18,23, lwd=2, col="blue",
      ylab="Density", xlab="Sample Mean", main=hdr)
curve(dnorm(x, 21, 3/sqrt(46)), add=T, lwd=2, col="brown")
abline(h = 0, col="green2")
abline(v = qnorm(.95, 20, 3/sqrt(46)), lty="dotted")
</code>```

Body:
Comment: This is not a general answer to your question, but a bit of a hint that may help you get started. Maybe take @heropup's advice and review
the definitions of significance level and power before you start.

A specific computation in Minitab statistical software for normal data with known $\sigma= 3$ is shown below. Suppose you want to test $H_0: \mu = 20$ vs. $H_a: \mu > 20$ at the 5% level of significance. Specifically, you want power $1-\beta = 0.8$ to detect a difference of $\delta = 1.$ That is, power 80% against the specific alternative
is $\mu_a = 21.$

  Power and Sample Size

1-Sample Z Test

Testing mean = null (versus > null)
Calculating power for mean = null + difference
α = 0.05  Assumed standard deviation = 3

            Sample  Target
Difference    Size   Power  Actual Power
         1      56     0.8      0.802222

After you solve your general problem, you can plug in
my values for $\sigma,\delta,$ and power, to see if
you get $n = 56.$  The following graph shows power
for various values of $\delta.$

If you can't see how to solve the general problem, then try this specific
version first---then generalize.

[Start by finding the critical region (rejection region) for a test at level $\alpha = 0.05.$ Then find $n$ that gives the critical region probability $0.8$ when $\mu=21.]$

Perhaps consider the following figure;

hdr="n = 46: Dist'ns Under Null (blue) and Alternative"
curve(dnorm(x, 20, 3/sqrt(46)), 18,23, lwd=2, col="blue",
      ylab="Density", xlab="Sample Mean", main=hdr)
curve(dnorm(x, 21, 3/sqrt(46)), add=T, lwd=2, col="brown")
abline(h = 0, col="green2")
abline(v = qnorm(.95, 20, 3/sqrt(46)), lty="dotted")


问:
How to proof $a = \left\lfloor\frac{Log(N)}{Log(P)}\right\rfloor$ is the maximum exponent of prime P such that $P^a \le N$
I'm trying to solve Project Euler's Problem #5 which is:

What is the smallest positive number that is evenly divisible by all of the numbers from 1 to N?

I came across a solution here using prime factorization method. I had difficulty understanding how to calculate the maximum exponent a of a prime P such that $P^a \le N$. I could prove that $P^{\left\lfloor\frac{Log(N)}{Log(P)}\right\rfloor} \le N$, but it doesn't make sure that $a = \left\lfloor\frac{Log(N)}{Log(P)}\right\rfloor$ is the maximum exponent such that $P^a \le N$ and $P^{a + 1} > N$

I need a solid proof/explanation that $a = \left\lfloor\frac{Log(N)}{Log(P)}\right\rfloor$ is in fact, the largest natural number such that $P^a \le N$.

答:
Encouraged by @Gerry Myerson, I could answer my question in a much more straight forward way.

First, we have a prime P and a natural number N, the exponent a' such that $P^a = N$ is obviously $a' = \log_P N$

Second, we also know that $\log_P N = \frac{\log N}{\log P}$, according to logarithm change of base rule. So, $a' = \frac{\log N}{\log P}$

Since a' is a real number, depending whether or not N can be evenly divided by P. So the largest natural number a such that $P^a \le N$, by definition of floor (as mentioned by @Michal Adamaszek), is: $a = \left\lfloor\frac{\log N}{\log P}\right\rfloor$


答: The difference between $x+y$ and $x-y$ is an even number.

Body: I am facing a problems with this question : Prove that If x and y are integers and x + y is odd , then x−y is odd. , any suggestions ?</p>


答:
The reqyired parabola should be of the type $y=-kx^2+hx, k>0$ if it passes through $(a,b)$ then $h=(b+ka^2)/a$. So the eq. of the parabola is $$y=-kx^2+\frac{(b+ka^2)}{a}x$$
This cots $x$ axis at $x=0$, $x=x_0=a+b/(ak)$, so the area made by the parabola on x-axis is
$$A(k)=\int_{0}^{x_0}\left(-kx^2+\frac{(b+ka^2)x}{a}\right) dx, k>0$$
$$A(k)=\frac{ab}{2}+\frac{b^3}{6a^3k^2}+\frac{b^2}{2ak}+\frac{a^3k}{6}$$
$$\implies A'(k)=\frac{a^3}{6}-\frac{b^3}{2a^3k^3}-\frac{b^2}{2ak^2}=0$$
Check that have three roots: $$k=-\frac{b}{a^2},-\frac{b}{a^2},\frac{2b}{a^2}$$
Third one being positive is useful here. Also check that $A''(k)>0$. Thus,
$$A_{min}=A(2b/a^2)=\frac{9ab}{8}$$

Title: Calculus Integral Question


答:
They are $$\frac{|S_{m,2}|}{m!n}$$ where $S_{m,2}$ are the Stirling numbers of the first kind, $m$ is the column (starting from $m=2$), and $n$ is the row counting from the diagonal (i.e. row-col+1).

The Stirling numbers $|S_{m,k}|=\begin{bmatrix}m\\k\end{bmatrix}$ are defined here.

For $k=2$, they are $0, 1, 3, 11, 50, 274, 1764$, etc. So starting with $m=2$ and $n=1$, we get $$\begin{array}{l|lllll}
n=1&\frac{1}{2!}=\frac{1}{2}&\frac{3}{3!}=\frac{1}{2}& \frac{11}{4!}=\frac{11}{24}& \frac{50}{5!}=\frac{5}{12}& \frac{274}{6!}=\frac{137}{360}\\
n=2&\frac{1}{2!2}&\frac{3}{3!2}& \frac{11}{4!2}& \frac{50}{5!2}& \frac{274}{6!2}\\
n=3&\frac{1}{2!3}&\frac{3}{3!3}& \frac{11}{4!3}& \frac{50}{5!3}& \frac{274}{6!3}
\end{array}$$

Title: General expression of a (maybe 3 or 2 dim) sequence

Body:
$\frac{1}{2}$

$\frac{1}{4}$    $\frac{1}{2}$

$\frac{1}{6}$    $\frac{1}{4}$  $\frac{11}{24}$

$\frac{1}{8}$    $\frac{1}{6}$  $\frac{11}{48}$  $\frac{5}{12}$

$\frac{1}{10}$   $\frac{1}{8}$  $\frac{11}{72}$  $\frac{5}{24}$ $\frac{137}{360}$

$\frac{1}{12}$ $\frac{1}{10}$ $\frac{11}{96}$  $\frac{5}{36}$ $\frac{137}{720}$  $\frac{7}{20}$

$\frac{1}{14}$ $\frac{1}{12}$ $\frac{11}{120}$ $\frac{5}{48}$ $\frac{137}{1080}$ $\frac{7}{40}$ $\frac{363}{1120}$

...

Does anyone know what's the generate term expression for this number sequence


问:
$a + b + c, \ ab + bc + ca, \ abc$ Prove a, b and c are real.
I have found this problem in a book. If a, b and c are complex numbers, such that $|a| \neq |b| \neq |c| \neq |a|$ and $a + b + c, \ ab + bc + ca, \ abc$ are all real numbers, we must prove $a, \ b, \ c$ are also real numbers. We can get $(a + b + c) ^ 2 \in \ \mathbb {R}$, so $a ^ 2 + b ^ 2 + c ^ 2 \in \mathbb{R}$. Then, using $a ^ 3 + b ^ 3 + c ^ 3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab -bc-ac) =  3(a + b)(b + c)(a+c)$ I can get $a ^ 3 + b ^ 3 + c ^ 3 \in \mathbb{R}$ and $(a+b)(b+c)(a+c) \in \mathbb{R}$. But I cannot fit these to prove that a, b or is real.

答:
By Viète's relations, $a,b,c$ are the roots of
$$x^3-(a+b+c)x^2+(ab+bc+ca)x-abc$$
All the coefficients of this polynomial are real. If the polynomial has complex roots, there must be one such pair of them related by conjugation, but conjugate roots have the same magnitude, which is disallowed by the question. Thus the polynomial has only real roots, i.e. $a,b,c$ are real.


问:
Why isometric isomorphic between Banach spaces means we can identify them?
Is the "isometric" part really necessary? For what reason is that?

Eg. we prove that there is an isometric isomorphism between $(L^p)'$ and $L^q$ ($(p,q)$ conjugate) and then we identify them together as the same space. If they were isomorphic but not necessarily isometric, could we not identify them?

答:
We can identify two objects if there is a canonical isomorphism between them. The notion of isomorphism depends on context. In the theory of Banach spaces there are two common notions of an isomorphism:

Bijective linear map $T:X\to Y$ such that both $T$ and $T^{-1}$ are bounded operators.

Bijective linear map $T:X\to Y$ which is an isometry: $\|Tx\|=\|x\|$ for all $x$.

The second kind of isomorphism, called isometric isomorphism, preserves all the structure that a Banach space has. Hence, we lose nothing in identifying two spaces that are isometrically isomorphic to each other, provided that the isomorphism is canonical (e.g., does not depend on a choice of basis).

The first kind, called simply an isomorphism, throws out a bit of information. For example, one of two isomorphic spaces may be uniformly convex while the other one isn't. Generally, it is not practical to think of two isomorphic Banach spaces as being the same.


问:
If F is a monotonic function then $F(Z) \subseteq Z$ where Z is the intersection of power sets.
I have been presented the following problem:

Let $A$ be any set and $p(A)$ be the power set of A. Show that for any function $p(A)$ to $p(A)$, the set of all subsets X of A such that $F(X) \subseteq X$ is nonempty, and therefore the intersection Z of all such subsets exists. A function F from $p(A)$ to $p(A)$ is called monotone if for any subsets $X$ and $Y$of $A$, if $X \subseteq Y$, then $F(X) \subseteq F(Y)$.

Show that if F is monotone then $F(Z) \subseteq Z$.

Where I'm at so far:

In the question, X is defined as $p(A)$, thus the cardinal of X would be the cardinal of $p(A)$. Therefore, for any $x \in X$, $x \in p(A)$.

Monotonic functions are functions wherein ordered sets either have their order preserved or reversed. Thus, I understand the statement "A function F from $p(A)$ to $p(A)$ is called monotone if for any subsets $X$ and $Y$of $A$, if $X \subseteq Y$, then $F(X) \subseteq F(Y)$. "

What I don't understand is this portion of the question:

Show that for any function $p(A)$ to $p(A)$, the set of all subsets X of A such that $F(X) \subseteq X$ is nonempty, and therefore the intersection Z of all such subsets exists.

Can you help me visualise/understand the function $p(A)$ to $p(A)$ and the statements "$F(X) \subseteq X$ is non-empty therefore there exists an intersection Z of all such subsets"?

A simple example or something would help me a lot.

Thank you so much!

答:
To visualize a function $F$ just take a couple of examples.

First with $A = \{1,2\}$ you have $$\mathcal P(A)= \{\emptyset, \{1\},\{2\}, \{1,2\}\}.$$ And you can take $F_1(\emptyset) = \{1,2\}$, $F_1(\{1\}) = \{2\}$, $F_1(\{2\}) = \{1\}$ and $F_1(\{1,2\}) = \emptyset$. However $F_1$ is not increasing, I let you tell why.

Another example of increasing $F_2$ also defined on $\mathcal P(A)$ is $F_2(x) = \{1,2\}$ for all $x \in \mathcal P(A)$ which is increasing.

If you want an example of an increasing map with $A$ infinite, take $A = \mathbb N$ and $F (x) = x \cup \{1, 2\}$ for all $x \in \mathcal P(A) = \mathcal P(\mathbb N)$.

Now coming to the question you are asked. Proving that the set
$$S = \{X \in \mathcal P(A) \mid F(X) \subseteq X\}$$ is not empty is pretty easy... Just notice that $A$ belongs to $S$! And as $S \neq \emptyset$

$$\bigcap S$$ exists.


问:
infimum and supremum of $(sin(a))^n and (cos(a))^n$
infimum and supremum of  $(sin(a))^n$ and $(cos(a))^n$ with $a\in \mathbb{R}, n\in \mathbb{N}$

How do you show that $(sin(a))^n$ and $(cos(a))^n $ have supreme and infimum?

答: Since $\sin a,\,\cos a$ each have images $[-1,\,1]$ on $\Bbb R$, $\sin^na,\,\cos^na$ each have images $[0,\,1]$ for even $n$ and $[-1,\,1]$ for odd $n$ on $\Bbb R$, because these are the images of $x^n$ on $[-1,\,1]$. So the supremum is $1$, and the infimum is $0$ for even $n$ but $-1$ for odd $n$.


问:
Show $I_b \subseteq I_a$ when $b \leq a$
I am currently working on the following problem:

Let A be any nonempty set, $p(A)$ be the power set of A, $F$ be a function from $p(A)$ to $p(A)$ that is monotone, meaning that whenever $X \subseteq Y$ then $F(X) \subseteq F(Y)$.

By transfinite induction define a set $I_a \subseteq A$ for every ordinal $a$ as follows:

$I_a$ = empty set if $a$ = 0

$I_a$ = $F(I_b)$ if $a$ = $b+1$

$I_a$ = $\cup_(b\leq a)$

Show $I_b \subseteq I_a$ when $b \leq a$

My question is this:

Would it not be enough to cite the successor case in transfinite induction as enough to prove that statement?

Or is this question asking me to prove the successor case?

答:
In short, the answer is no. You also have to take care of the limit ordinal case to have a complete proof.

And by the way, there is probably a typo in your question. I suspect that
$$I_b = \bigcup_{b\leq a} I_a$$ in case of a limit ordinal.


问:
Probability of X or Y happening
I was trying to work this out last night and it was driving me mad, as I've forgotten how to calculate the probability of X or Y happening. I am on an on-call rota at work, and we get around 3 callouts in a week. So, that means the chances of getting a callout on any given night are 3/7. If I want to know the chances of getting a callout on either a Monday or a Tuesday night, how would I approach that please?
Thanks
Andrew</p>

答:
Let $M = \text{callout on Monday}$ and $T = \text{callout on Tuesday}$. We know that
$$
P(M) = P(T) = \frac{3}{7}
$$
How would we combine them with an "or" operator? Well, if you draw the Venn diagram, you will see that
$$
P(M~\text{or}~T) = P(M)+P(T)- P(M~\text{and}~T)
$$
Since the two events are independent, we get $P(M~\text{and}~T)=P(M)P(T)$. Can you finish the calculation now?

Note that there are several ways of mathematically modeling this. Here, we assumed that the events are independent, and it could be that this is not exactly true. The sentence "we get around 3 callouts in a week, so (...) the chances of getting a callout on any given night are 3/7" is not entirely true. Why? Consider that you got a callout on Monday and Tuesday. Is the probability of a callout on Wednesday still 3/7? How about Thursday? As you see, there are several ways of going about this ...

Title: Probability of X or Y happening


问:
Exercise 6.17 from Isaac's Character Theory of Finite Groups.
I am trying to solve the following problem in Isaac's Character Theory of Finite Groups and I am really sorry if this is a very simple question.

6.17: Let $N$ be a normal subgroup of $G$ with $G/N$ cyclic. Let $\eta\in \operatorname{Irr}(N)$ be invariant in $G$ and assume that $(\eta(1),|G:N|)=1.$ Show that $\eta$ is extendible to $G.$

Q1: The hint is to show that $G/\ker(\det\eta)$ is abelian. But I don't know how to prove the hint or how to solve the problem using it. I think proving that $\det \eta$ would be enough but I don't know how to do that.

Q2: There's a note right below the problem saying that this is true even without the assumption $(\eta(1),|G:N|)=1.$ Could someone please give me a hint for this? My only experience in Character Theory is reading up to chapter 6 in Isaac's book so I'm not sure if I know enough material to solve this without this assumption.

Thank you!!

答:
These are just hints.

To prove the hint in the book, note that $N/\ker(\det \eta) \le Z(G/\ker(\det \eta))$, since otherwise $\eta$ would not be invariant in $G$. Now $G/\ker(\det \eta)$ abelian follows from $G/N$ being cyclic.

To prove the result using the hint, use Theorem (6.25) in the book.

It is not part of the exercise to solve the problem without the assumption, so you should probably not worry about it! I have not thought about it myself, but it seems likely that the solution involves some results or techniques that have not been covered in the book at this point.


问:
how to determine the domain of this function: $\sqrt{(\sin(x))^2 + \sin x}$?
let $x$ be in this interval: $[0, 2\pi]$. the exercise's request is to find the domain of this function:$$\sqrt{(\sin(x))^2 + \sin x}$$

Here's my attempt to solve it:

I found the domain of the square root: $$(sin (x))^2 + \sin x \geq 0$$
In this case, I've found that $$\sin x \ge 0 $$ and that: $$\sin x \ge -1 $$

in the trigonometric circle, I have found that the set of solution is only the 4th quadrant, since the sin of -1 is $3/2\pi$ (270 degree).

Is that correct?

答:
What if $\sin x(\sin x+1)=0?$

Else  we need $\sin x(\sin x+1)>0$

If $\sin x>0,\sin x+1>0\iff\sin x>-1\implies \sin x>$ max$(0,-1)$

$\implies2n\pi\le x\le2n\pi+\pi$

What if $\sin x<0?$


问:
Why are sine and cosine identities always true (and thus sentences) even though they have free variables?
In mathematics, we often see statements such as $\sin ^2x + \cos^2 x = 1$.

Notice that this is not a sentence, as the variable $x$ is free. But we all agree that this statement is true, given the usual interpretations of the symbols. How can we square this with the claim that sentences are the formulas that can be either true or false?

答:
It follows from the definition of satisfaction in a structure:

$$\mathcal{M} \models \phi \\\Longleftrightarrow \text{ for all assignment functions } v: \mathcal{M}, v \models \phi$$

i.e., a formula is true in a structure if and only if it is true under all variable assignments in that structure.

The definition of satisfaction thus renders open formulas as implicitly universally quantified, and we have

$$\mathcal{M} \models \phi \text{ iff } \mathcal{M} \models \forall x_1 \ldots \forall x_n \phi \text{ (where } x_1, \ldots, x_n \text{ are the variables occurring free in } \phi \text{)}$$

that is, a formula is true in a structure exactly when its universal closure is.

So

$$\sin ^2x + \cos^2 x = 1$$

is a distinct formula from but has the same structure-level truth conditions as

$$\forall x (\sin ^2x + \cos^2 x = 1)$$


问:
$\exists M >0$ s.d. $ \vert R(z)\vert \leq \frac{M}{x^2}$, calculation of integral with the residue theorem
Given a polynomial of the form $R(z):=\frac{P(z)}{Q(z)}$ such that $R(z)$ has no real roots and $deg(Q) \geq deg(P) + 2$, then the integral can be expressed as

$$\int_{-\infty}^{+\infty} R(z)dz=2\pi i\sum_{a \in \Bbb H}{\mathrm{Res}\left ( f;a \right )}$$
where $\Bbb H:=\{z \in \Bbb C: Im(z)>0\}$.

Now for proving this statement, we have first to show that the following limit exists:
$$\int_{-\infty}^{+\infty} R(z)dz=\lim_{r \to -\infty}\int_{r}^{+\infty} R(z)dz+\lim_{r \to +\infty}\int_{-\infty}^{r} R(z)dz$$
The argument they used there, is that $\exists M >0$ s.d. $ \vert R(z)\vert \leq \frac{M}{x^2}$. But I don't see, why it exists such an $M$. Many thanks for some help!

答:
$z^{2} \frac {a_0+a_1z+...+a_nz^{n}} {b_0+b_1z+...+b_mz^{m}}$ tends to $0$ if $m >n+2$ and it tends to $\frac {a_n} {b_n}$ if $m=n+2$. Hence $z^{2} \frac {a_0+a_1z+...+a_nz^{n}} {b_0+b_1z+...+b_mz^{m}}$ is a bounded function.

[Since $x^{2}R(x)$ is continuous on the real line and has finite limits at $\pm \infty$ it is a bounded function].


问:
Find the value of $x^3-x^{-3}$ given that $x^2+x^{-2} = 83$

If $x>1$ and $x^2+\dfrac {1}{x^2}=83$, find the value of the expression$$x^3-\dfrac {1}{x^3}$$

a) $764$

b) $750$

c) $756$

d) $760$

In this question from given I tried to approximate the value of $x$ which should just above to 9 then I tried to calculate the value of cubic expression but all options are close enough to guess. Any idea to solve it?

答:
First, notice that

$$\left(x-\frac{1}{x}\right)^2=x^2-2+\frac{1}{x^2}=83-2=81$$

Then, use the difference of two cubes formula:

$$x^3-\frac{1}{x^3}=\left(x-\frac{1}{x}\right)\left(x^2+1+\frac{1}{x^2}\right)=9\cdot(83+1)=756$$

We take the positive root of $81$ because $x>1$.


答: The function $x\mapsto x^2R(x)=\frac{x^2P(x)}{Q(x)}$ is a rational function such that the degree of the numerator is smaller than or equal to the degree of the denominator, and therfore the limits $\lim_{x\to\pm\infty}\frac{x^2P(x)}{Q(x)}$ exist and are real numbers. So, that function is bounded and now you can take $M$ such that $(\forall x\in\Bbb R):\left|\frac{x^2P(x)}{Q(x)}\right|\leqslant M$, and then $(\forall x\in\Bbb R):\left|R(x)\right|\leqslant\frac M{x^2}$.

Title: $\exists M >0$ s.d. $ \vert R(z)\vert \leq \frac{M}{x^2}$, calculation of integral with the residue theorem

Body:
Given a polynomial of the form $R(z):=\frac{P(z)}{Q(z)}$ such that $R(z)$ has no real roots and $deg(Q) \geq deg(P) + 2$, then the integral can be expressed as

$$\int_{-\infty}^{+\infty} R(z)dz=2\pi i\sum_{a \in \Bbb H}{\mathrm{Res}\left ( f;a \right )}$$
where $\Bbb H:=\{z \in \Bbb C: Im(z)>0\}$.

Now for proving this statement, we have first to show that the following limit exists:
$$\int_{-\infty}^{+\infty} R(z)dz=\lim_{r \to -\infty}\int_{r}^{+\infty} R(z)dz+\lim_{r \to +\infty}\int_{-\infty}^{r} R(z)dz$$
The argument they used there, is that $\exists M >0$ s.d. $ \vert R(z)\vert \leq \frac{M}{x^2}$. But I don't see, why it exists such an $M$. Many thanks for some help!


答:
$X$ is not defined at all in the question: there is no specific set called $X$. Rather, the dummy variable $X$ is used to define a certain family of subsets of $X$. This might be clearer if we give that family a name:

$$z(A)=\{X\in p(A):F(X)\subseteq X\}\,.$$

Then $Z$ is defined as

$$Z=\bigcap z(A)=\bigcap\{X\in p(A):F(X)\subseteq X\}\,.$$

This makes $Z$ an intersection of subsets of $A$ and therefore itself just a subset of $A$.

For example, if $A=\{0,1,2\}$, and $F(X)=A\setminus(X\cup\{0\})$ for each $X\in p(A)$, we have:

$$\begin{align*}
&F(\varnothing)=\{1,2\},\\
&F(\{0\})=\{1,2\},F(\{1\})=\{2\},F(\{2\})=\{1\},\\
&F(\{0,1\})=\{2\},F(\{0,2\})=\{1\},F(\{1,2\})=\varnothing,\\
&F(A)=\varnothing
\end{align*}$$

For this function $F$ we can see that $F(X)\subseteq X$ if and only if $X=A$ (since $F(A)=\varnothing\subseteq A$) or $X=\{1,2\}$ (since $F(\{1,2\})=\varnothing\subseteq\{1,2\}$). Thus, $z(A)=\big\{A,\{1,2\}\big\}$, and

$$Z=\bigcap z(A)=A\cap\{1,2\}=\{1,2\}\,.$$

The first paragraph that you wrote describing where you are now is far off the mark. If $X\in z(A)$, then by definition $X$ is a subset of $A$, so $|X|\le|A|$; in particular, $|X|<|p(A)|$, since $|A|<|p(A)|$ no matter what $A$ is. However, cardinality is simply not relevant to this problem. Finally, if $X\in z(A)$ and $x\in X$, then $x\in A$; in general $x\notin p(A)$.

Your first task is to show that $z(A)\ne\varnothing$, i.e., that $A$ has at least one subset $X$ such that $F(X)\subseteq X$. By definition $F(X)\subseteq A$ for all $X\in p(A)$ (why?), and $A\in p(A)$, so $F(A)\subseteq A$. Thus, $A\in z(A)$ no matter what function $F$ is mapping subsets of $A$ to subsets of $A$. This means that we can form $Z=\bigcap z(A)$; it will be some subset of $A$.

Your second task is to show that if $F$ is monotone — i.e., if $F(X)\subseteq F(Y)$ whenever $X\subseteq Y\subseteq A$ — then $F(Z)\subseteq Z$. In other words, you’re to show that if $F$ is monotone, then $Z\in z(A)$: the intersection of all of the sets in $z(A)$ is itself in $z(A)$.

HINT: We know that $Z\subseteq X$ for each $X\in z(A)$. (Why?) Thus, if $F$ is monotone we know that $F(Z)\subseteq F(X)\subseteq X$ for each $X\in z(A)$. In particular, $F(Z)\subseteq X$ for each $X\in z(A)$. What does this tell you about the relationship between $F(Z)$ and $\bigcap z(A)$?

Title: If F is a monotonic function then $F(Z) \subseteq Z$ where Z is the intersection of power sets.

Body:
I have been presented the following problem:

Let $A$ be any set and $p(A)$ be the power set of A. Show that for any function $p(A)$ to $p(A)$, the set of all subsets X of A such that $F(X) \subseteq X$ is nonempty, and therefore the intersection Z of all such subsets exists. A function F from $p(A)$ to $p(A)$ is called monotone if for any subsets $X$ and $Y$of $A$, if $X \subseteq Y$, then $F(X) \subseteq F(Y)$.

Show that if F is monotone then $F(Z) \subseteq Z$.

Where I'm at so far:

In the question, X is defined as $p(A)$, thus the cardinal of X would be the cardinal of $p(A)$. Therefore, for any $x \in X$, $x \in p(A)$.

Monotonic functions are functions wherein ordered sets either have their order preserved or reversed. Thus, I understand the statement "A function F from $p(A)$ to $p(A)$ is called monotone if for any subsets $X$ and $Y$of $A$, if $X \subseteq Y$, then $F(X) \subseteq F(Y)$. "

What I don't understand is this portion of the question:

Show that for any function $p(A)$ to $p(A)$, the set of all subsets X of A such that $F(X) \subseteq X$ is nonempty, and therefore the intersection Z of all such subsets exists.

Can you help me visualise/understand the function $p(A)$ to $p(A)$ and the statements "$F(X) \subseteq X$ is non-empty therefore there exists an intersection Z of all such subsets"?

A simple example or something would help me a lot.

Thank you so much!


问:
What do these mean pdv . vdp . d(pv)?
I am a bit new to calculus and if we have two variables p and v
$$pdv \ vdp \ d(pv)$$
what will the above quantities representmean ?

答:
You can think of it as the relation of the changes between two quantities as we put some 'perturbation' into the system. It is more transparent for finite changes:

$$ \Delta (PV) = V \Delta P + P \Delta V + \Delta P \Delta V$$

Now, suppose we make the magnitude of changes very small $ \Delta P , \Delta V \to (0,0)$ then:

$$ d(PV) = V dP + P dV$$

This is useful for relating the 'change' in thermodynamic variables.

Illustration

Suppose we are dealing with an ideal gas and $PV= nRT$ , assuming constant moles :

$$ nR dT = V dP + PdV$$

Suppose we have a reversible isothermal process, then $ dT = 0$,

$$ V dP = - P dV$$

Or,

$$ \frac{-V}{P} = \frac{dV}{dP}$$

Since volume is always greater than zero and pressure is greater than zero, it is easy to see as we increase the volume $ \frac{-V}{P}$ becomes more negative, so the decrease in volume as we increase pressure more and more is much greater. Indeed, we have found the partial derivative of volume with pressure at constant temperature:

$$ \frac{-V}{P} = (\frac{\partial V}{\partial P})_T$$

tl;dr: the relation between 'small changes' in a system, helps us understand how the system behaves to be given 'small disturbances'. eg: I give a small disturbance in pressure, then how does my volume change if the process is reversible isothermal?

Edit: Oops! flipped the fraction by accident.

Title: What do these mean pdv . vdp . d(pv)?


问:
Can a certain monoid exist?
Is it possible to have an uncountable commutative monoid, where for every $a$ in the monoid, $a+a=a$?

I have a set which I am trying to define a group structure on (I am settling for a monoid structure though). the set is a bunch of equivalence classes of the real numbers, and I'm trying to see if it would be possible to define addition of classes by addition of representatives. The problem is adding an element of any class to itself, will always get you in the same class, so i'm not sure if this would mean I cant define the structure this way.

Also I should add, when I say i'm defining addition by addition of representatives, each class has to have a canonical representative which is used in the addition operation, as addition definitely is dependant on the choice of representative.

答:
Here is some background. An element satisfying $a + a = a$ is called idempotent. A commutative monoid in which every element is idempotent is called, naturally, an idempotent commutative monoid.

These things are plentiful, and in fact are more or less equivalent to (bounded) (join-)semilattices, which are partially ordered sets $(P, \le)$ with a bottom / least element $\bot$ (an element such that $\bot \le p$ for all $p \in P$) and binary joins $p \vee q$ (an element such that $p \le x$ and $q \le x$ iff $p \vee q \le x$). Joins are always commutative and idempotent, $\bot$ is the identity element, and the partial order can be recovered from the join operation since $p \le q$ iff $p \vee q = q$ (exercise).

Joins are a generalization of the $\text{max}$ operation; here are a few illustrative examples.

The nonnegative real numbers $\mathbb{R}_{\ge 0}$ are totally ordered (so in particular partially ordered). The bottom element is $0$ and the join is $\text{max}$.

If $X$ is any set, the set $P(X)$ of subsets of $X$ is partially ordered by inclusion. The bottom element is the empty subset, and the join is the union.

The set $\mathbb{N}$ of positive integers is partially ordered by divisibility. The bottom element is $1$, and the join is $\text{lcm}$.

If $V$ is any vector space, the set of subspaces of $V$ is partially ordered by inclusion. The bottom element is the zero subspace, and the join is the sum.

The first, second, and fourth examples all give examples of uncountable idempotent commutative monoids; the second and fourth examples in particular admit many variations. For example one can consider subgroups of a group, subrings of a ring, submodules of a module, etc.

Title: Can a certain monoid exist?


问:
Solving $\frac{1}{x^2-1} \geqslant \frac{1}{x+1}$
How do I solve:

$$\frac{1}{x^2-1} \geqslant \frac{1}{x+1}$$

Here's my attempt to solve it:
I brought the fraction on the right side to the left side, then I've found the common denominator, $(x-1)(x+1)$, and after all the computations, this is my situation:
$$ \frac{2-x}{(x-1)(x+1)} \geqslant 0$$
Now, I've put the numerator and the denominator in a system of inequalities:
(I don't know how to type the system of inequalities in mathjax, sorry)

$$x<2$$
$$x < -1 \space \text{or} \space x>1 $$
My result is (I found the intersection): $$-1<x<1 \space \space \text{or} \space \space x>2$$

Let me know if the result is wrong.
Additionally, because in the test I'll have only 50 minutes (lots of questions), this method requires a lot of time; if you know a quicker method, let me know.

EDIT (solved in a quicker way):

when you are in this situation: $$\frac{2-x}{(x-1)(x+1)} \geqslant 0$$
change the sign of the inequality and the signs that appear in the numerator,

then, check each interval in the normal way, but don't forget you have changed the sign; therefore, if you had a greater sign, and you changed it, you have to take only negative intervals (not positive intervals, in order to express the correct solution).

答:
First write the function such that both numerator an denominator have same sign.
$$ \frac{x-2}{(x-1)(x+1)} \leqslant 0$$
We use the sign scheme method for this. Now first we will find the critical points. Critical points are those points where the function changes its behaviour. Usually for this problem the zeroes and poles are those points.

Critical points are $x = 2,1,-1$. There are three critical points with each critical point having odd frequency i.e. all critical points occur odd number of times.

Now plot the points on a number line.

Find the maximum critical point. In this case it's $x =2$. The function is positive always when the value of x is greater than the maximum critical point. So the function is positive for $x \geq 2$.

The function between $x=2$ and $x = 1$ will be negative.

The function between $x=1$ and $x=-1$ will be positive.

The function beyond $x=-1$ will be negative.

Tricks to solve it quickly:

Always write the function in the above format. So that numerator and denominator have same sign.

Find the critical points. Any value of $x$ beyond the maximum critical point will make the function positive.

Now as we go backwards, see how many times the critical points occurred. Since $x=2$ occurred once, sign of the function will change once from positive to negative. And the function will behave such that its negative till the next critical point is encountered.

Now once you reach the critical point $x=1$ you will see it occurred odd number of times again the sign transition will happen once. So the function will behave such that its positive till it encounters the next critical point.

Once you hit the last critical point, see the frequency since its odd, one sign change. So beyond $x=-1$ the function will be negative.

Remember poles wont be included in the final solution as you will make the function infinitely large. Also zeroes will be included since the inequality also has an equal sign.

Now compile the solution from these regions: you want all the negative regions, the final solution being:
$$x\in (-\infty,-1) \cup (1,2]$$

I will give a completely different example for the same thing so that concepts are clear. Lets say your function was:
$$ \frac{(x-2)(x-3)^2(x+5)^3}{(x-1)^5x(x+1)^6} \lt 0$$

Format is correct, numerator denominator has same signs.

Find critical points with frequency (number of times each point occurred). So the critical points and frequency are : $x = 2(1),3(2),-5(3),0(1),-1(6),1(5)$.

The region greater than the maximum critical point is always positive. So for values $x \gt 3$ the function is always positive. Note I have not done $x \geq 3$ because thats not the requirement of the inequality. It will follow for every such critical point.

As we move backwards, observe the number of occurrences of the critical point (odd or even) and apply that many sign transitions. For example, as we reach $x=3$ we see it occurred two times so sign transition will be 2 times from + to - (1 transition) and from - to + (2nd transition). So overall when the values are just less than 3 the sign of function didnot change.

Go backwards, as you encounter critical points make sign transitions equal to the number of occurrences of the critical point. So if 2 occured once only one sign transition will happen as the function goes beyond it.

Even frequency of critical point: No sign change will happen

Odd frequency of critical point: Sign change will happen

Now compile the solution. In the inequality we need the function to be negative. So just compile the negative parts from the figure. The solution will be :
$$x \in (-5,-1) \cup (-1,0) \cup (1,2)$$
I had to break at -1 since the function will go to $-\infty$. If you want the solution such that the function can go to $-\infty$ you can include that.
Hope this helps. With practice you will solve these lenghty questions in seconds. Tested :D


问:
How to prove that one set of vectors forms the basis for another set of vectors?
Question: How do I prove that one given set of vectors forms the basis of another given set of vectors?

I know that if we could prove that both sets are identical to each other, then the first set would obviously span the second set. But are there any other simpler ways of proving that one set forms the basis of another set without having to go through every vector and see if they are present in both sets?

Question Two: If we combine the vectors in each set into a matrix and reduce the matrices to reduced row echelon form, could we simply say that one set spans the other if their row reduced echelon form are the same?

答:
I think you have not quite understood the difference between a vector space and its basis. You van certainly not prove that a set is a basis for another set by showing that they are the same. With the exception of the trivial vector space, no vector space equals its own basis. A basis for a vectorset $V$ is a subset $\mathcal{B}$ of $V$ such that $\mathcal{B}$ is linearly independent in $V$ and spans $V$. So to prove that a set forms a basis you need to check (1) linear independence and (2) span.

If you want to check if two bases are bases for the same vector space, which might be what you are really asking, you need to compare their spans. If they span they same set, the form bases for the same set.

As to your second question, you may check whether a set of vectors $v_1, \ldots v_n$ form a basis for $\mathbb{R}^n$ by considering the reduced row echelon form (rref) of the matrix with these vectors as column, i. e.

$$B = [ v_1 \quad v_2 \quad \cdots \quad v_n].$$

They form a basis for $\mathbb{R}^n$ if and only if the rref of $B$ is the identity matrix. If the vector space in question is not the entire $\mathbb{R}^n$, you need to check if the column spaces of the two matrices you compare are the same.


问:
Can a complex number ever be considered 'bigger' or 'smaller' than a real number, or vice versa?
I've always had this doubt.
It's perfectly reasonable to say that, for example, 9 is bigger than 2.

But does it ever make sense to compare a real number and a complex/imaginary one?

For example, could one say that $5+2i> 3$ because the real part of $5+2i
$ is bigger than the real part of $3$? Or is it just a senseless statement?

Can it be stated that, say, $20000i$ is bigger than $6$ or does the fact that one is imaginary and the other is natural make it impossible to compare their 'sizes'?

It would seem that the 'sizes' of numbers of any type (real, rational, integer, natural, irrational) can be compared, but once imaginary and complex numbers come into the picture, it becomes a bit counter-intuitive for me.

So, does it ever make sense to talk about a real number being 'more than' or 'less than' a complex/imaginary one?

答:
We can compare complex numbers just like we compare two-digit numbers,

For example, $53 > 42$ can be seen as $5 + 3i > 4 + 2i$

So, For $Z_{1} = a_{1} + ib_{1}$ and $Z_{2} = a_{2} + ib_{2}$ choose a $k$ such that $k>a_{1}$, $k>b_{1}$, $k>a_{2}$, $k>b_{2}$ and then, the complex number can be written as $ka_{1} + b_{1}$, $ka_{2} + b_{2}$, now, compare it just like you compare Real Numbers.


答: Hint: $P(a)-2020=0$, and notice that $P(x)-2020=325\Phi_7(x)$ where $\Phi_7(x):=x^6+x^5+x^4+x^3+x^2+x+1$ is the $7$-th Cyclotomic polynomial. Now use facts about the roots of $\Phi_7(x)$, in particular these are primitive $7$-th roots of unity.

Title: How to find $a^{2033}+\frac{1}{a^{2033}}$ from the equation?

Body:
I was Junior High School Student, please explain the answer like you teach a Junior High School Student

Given :
$$P(x) = 325^6+325^5+325^4+325^3+325^2+325+2345$$

Given a number $a$, $P(a) = 2020$,
Find $$a^{2033} + \frac{1}{a^{2033}}$$

I have
$P(a) = 2020$

$325a^6+325a^5+325a^4+325a^3+325a^2+325a+2345$ = $2020$

$325(a^6+a^5+a^4+a^3+a^2+a+1)=0$

divided by $a^3$, I get :

$a^3+a^2+a+1+\frac{1}{a}+\frac{1}{a^2}+\frac{1}{a^3}=0$

$(a+\frac{1}{a})^3-3(a+\frac{1}{a})+(a+\frac{1}{a})^2-2+(a+\frac{1}{a})+1=0$

Suppose that $y=a+\frac{1}{a}$

$y^3+y^2-2y-1=0$, i'm stuck:)


问:
When does a variety have a point over a finite field for sufficiently large primes p?
Let $X$ be an algebraic variety over the rational numbers. Suppose that $X$ has positive dimension. I would like to say that $X(\mathbb{F}_p)$ is non-empty for sufficiently large primes $p$. One idea is to use the Weil conjectures, but that seems like overkill. So my two questions are:

When can one deduce that a system of polynomial equations has a solution over $\mathbb{F}_p$ for sufficiently large primes $p$? I might have guessed that $\dim(X) > 0 $ is sufficient, but it seems like the affine variety $X^2+Y^2 = 0$ minus the point $(0,0)$ (which has no solutions for $p = -1 \mod 4$) suggests that extra conditions are necessary.

Is there an elementary proof of the (correct) version of part 1?

答:
Using the Lang-Weil bound and maybe also the Chebotarev density theorem or something like that, I believe the correct statement is that if $\dim X \ge 1$ then $X(\mathbb{F}_p)$ is non-empty for all sufficiently large primes $p$ if and only if the action of the absolute Galois group $\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ on the set of irreducible components of $X_{\overline{\mathbb{Q}}}$ fixes at least one component of positive dimension. This condition doesn't hold in your example; there are two irreducible components over $\mathbb{Q}(i)$, given by the lines $X + iY = 0$ and $X - iY = 0$ (minus the origin), and the action of complex conjugation switches them.

I don't see any way to use a weaker result than the Lang-Weil bound.

I also want to point out that strictly speaking if $X$ is a variety over the rational numbers then it doesn't make sense to talk about $X(\mathbb{F}_p)$ because $\text{Spec } \mathbb{F}_p$ is not a variety (or scheme or anything) over the rational numbers; when people write expressions like this they are implicitly choosing a lift of $X$ to a scheme defined over some localization of $\mathbb{Z}$ defined by the denominators of some collection of polynomials cutting out $X$, and the behavior of $X(\mathbb{F}_p)$ for any fixed prime $p$ could a priori depend on this choice. I think it's true that the question of whether $X(\mathbb{F}_p)$ is non-empty for sufficiently large $p$ does not depend on this choice but I can't quote a theorem off the top of my head that guarantees this. In the Lang-Weil argument above let's say we're fixing a particular lift to keep things safe.


问:
Does $x<y$ and $a\le b\implies x+a<y+b$? Similarly, does $x>y$ and $a\ge b\implies x+a>y+b$?
I know this is a very simple question, but I still have my doubts on this question:

Does $x<y$ and $a\le b\implies x+a<y+b$? Similarly, does $x>y$ and $a\ge b\implies x+a>y+b$? Why or why not?

Based on intuition, these seem to be true, but how would I "prove" this concretely? I'm not too sure if there are any rules that allow me to combine $<$ and $\le$(or $>$ and $\ge$) inequalities. It seems like there should be one, but I wasn't able to find any. Could someone help me on this?

答: If $x<y$, then $x+a<y+a$ and then, if $a\leqslant b$, $y+a\leqslant y+b$, and therefore $x+a<y+b$.


答:
There is perhaps a better way to approach,  consider:

$$ f= \sin^2 x + \sin x$$

Now, we can see there are zero set for this function: $ \sin x = \{0,-1 \}$, in the interval $[0, 2 \pi]$ the zero set in terms of 'x' : $\{0, \pi, \frac{ 3 \pi}{2} \}$. Hence, the domain of $x$ in which the function is greater than zero and also belongining the original domain constrained is given by:

$$ \left[ 0 , \pi \right] \cup \left[ \frac{3 \pi}{2} , 2 \pi\right]$$

Explanation:

The function 'sign flips' whenever it hits a root, so we can partition the given domain  based on root as as:

$$ \left[ 0, \pi \right] \cup \left[ \pi, \frac{ 3 \pi}{2} \right] \cup \left[ \frac{3 \pi}{2} , 2 \pi \right]$$

Now since in the first domain the function is positive, the second it will be negative and third it will be positive again. Hence, the final answer is given by:

$$ \left[ 0, \pi \right] \cup \left[ \frac{ 3\pi}{2} , 2\pi \right]$$

Edit

Oops! The function above is slightly more complicated than a regular quadratic in the sense that a zero doesn't correspond exactly as a sign flip!

As correctly pointed out by the @The 2nd, the function retain's it's negative sign even after the zero $ \frac{3 \pi}{2}$ , hence the whole domain of $x$ values greater than $ \frac{3 \pi}{2}$ is neglected. But, there is still one more root to consider, that is the one at $ x = 2 \pi$

Hence, the final answer is given by:

$$ \left[ 0 , \pi \right] \cup \left[ \frac{ 3 \pi}{2} \right] \cup \left[ 2 \pi \right]$$

Title: how to determine the domain of this function: $\sqrt{(\sin(x))^2 + \sin x}$?

Body:
let $x$ be in this interval: $[0, 2\pi]$. the exercise's request is to find the domain of this function:$$\sqrt{(\sin(x))^2 + \sin x}$$

Here's my attempt to solve it:

I found the domain of the square root: $$(sin (x))^2 + \sin x \geq 0$$
In this case, I've found that $$\sin x \ge 0 $$ and that: $$\sin x \ge -1 $$

in the trigonometric circle, I have found that the set of solution is only the 4th quadrant, since the sin of -1 is $3/2\pi$ (270 degree).

Is that correct?


答:
Let $t = \sin x \,\, (-1 \leq t \leq 1)$

The domain of the function will become:

$$t^2 + t \geq 0$$

$$\implies t \in (-\infty, -1] \cup [0, +\infty)$$

But $-1 \leq t \leq 1 \implies t = -1$ or $0 \leq t \leq 1$

$t = -1 \implies \sin x = -1 \implies x = -\dfrac{\pi}{2} + k2\pi \implies x = \dfrac{3\pi}{2}$ (since $x \in [0, 2\pi]$)

$0 \leq t \leq 1 \implies 0 \leq \sin x \leq 1 \implies x \in [0, \pi] \cup \{2\pi\}$

Therefore, the domain is:

$$D = [0, \pi] \cup \left\{\dfrac{3\pi}{2}, 2\pi \right\}$$