答:
I'm afraid, this approach won't work, as $A\cap B$ alone is possibly greater than $A\cap B \cap C$. I like to always get rid of minus using $A-B=A\cap B^c$. Now it is much easier as you can use distributivity and De Morgan's law:
$$(A-B)\cup (B-C)\cup (C-A)=(A\cap B^c)\cup (B\cap C^c)\cup (C\cap A^c)=$$
$$(A\cup B\cup C)\cap (A\cup B\cup A^c)\cap (A\cup C^c\cup C)\cap (A\cup C^c\cup A^c)\cap (B^c\cup B\cup C)\cap (B^c\cup B\cup A^c)\cap (B^c\cup C^c\cup C)\cap (B^c\cup C^c\cup A^c)$$
While it looks overwhelming, it actually simplifies to:
$$(A\cup B\cup C)\cap (B^c\cup C^c\cup A^c)=(A\cup B\cup C)\cap (B\cap C\cap A)^c$$
And now we can go back to set-minus representation:
$$(A\cup B\cup C)\cap (B\cap C\cap A)^c=(A\cup B\cup C)\setminus (B\cap C\cap A)$$
So finally:
$$(A-B)\cup (B-C)\cup (C-A)\cup (A\cap B \cap C)=A\cup B\cup C$$
While such approach is always possible, it isn't the shortest one.

Title: $A \cup B \cup C = (A - B) \cup (B - C) \cup (C - A) \cup (A \cap B \cap C)$

Body:
I am using the fact that $X = (X - Y) \cup (X \cap Y)$

So far I've got:

$$A \cup B \cup C = (A - B) \cup (B - C) \cup (C - A) \cup (A\cap B)\cup(B\cap C)\cup(C\cap A)$$

This is almost done, except now I have to deal with $(A\cap B)\cup(B\cap C)\cup(C\cap A)$. How can I use this to get to $A\cap B \cap C$?


问:
What's the best way to think about the Hessian?
I've always thought about the Hessian like this:

Let $f:\mathbb R^n \to \mathbb R$ be smooth.  Let $g:\mathbb R^n \to \mathbb R^n$ such that $g(x) = \nabla f(x)$.  (I am using the convention that $\nabla f(x)$ is a column vector.)  Then the Hessian of $f$ at $x$ is, by my definition, the $n \times n$ matrix $H(x) = g'(x)$.

However, with this way of looking at the Hessian, I'm not thinking of $H(x)$ as being a quadratic form.  I'm worried that there is a "quadratic form" viewpoint of the Hessian that I am missing.  There is a rule of thumb I've heard that when a matrix (such as the Hessian) is automatically symmetric, then it's often most natural to think of it as defining a quadratic form.  I realize that you can define a quadratic form at $x_0$ by $x \mapsto \langle x, H(x_0) x \rangle$, and that this quadratic form appears in Taylor's formula.  But I still think I'm missing something, because I don't see why it is fundamentally most natural to think of the Hessian as being a quadratic form.

Is it true that there is a quadratic form viewpoint that I'm missing out on?  If so, what is it?  More generally, what do you think is the best way to think about the Hessian?

答:
One way to understand the Hessian better is to think of it in the language of curves.

Fix $p \in \mathbb{R}^n$. Let $q:T_p\mathbb{R}^n \to \mathbb{R}$  be the quadratic form defined by $q(v) = v^T H(p) v$. $T_p\mathbb{R}^n$ is the tangent space of $\mathbb{R}^n$ at $p$, which is just the set of all vectors starting at $p$. It's basically a copy of $\mathbb{R}^n$ but with $p$ as the origin, and is the space of all possible directions I can go starting from $p$.

$q(v)$ in some sense is the acceleration of the function $f$ in the direction of $v$. Let's see how.

Fix a vector $v\in T_p\mathbb{R}^n$. Let $\gamma:(-1,1) \to \mathbb{R}^n$ be any curve passing through $p$ with velocity $v$. So $\gamma(0) = p, \gamma'(0) = v$.

We can look at the function $f$ restricted to the curve $\gamma$ and study how fast it changes as we pass through $p$ on the curve. So we are now looking at the single variable function $g := f \circ \gamma :(-1,1) \to \mathbb{R}$.

Notice that the speed at which the function changes as we pass through $p$ on the curve is

$$g'(0) = \nabla f \cdot v$$

and the acceleration of the function as we pass through $p$ on the curve is

$$g''(0) = q(v)$$

Now observe the remarkable fact that the right side of both equations does not depend on the curve $\gamma$, but only depends on $p$ and $v$. That is why we can call $g'(0)$ the speed at which $f$ changes as we pass through $p$ in the direction of $v$ (without any mention of $\gamma$). And similarly, we can call $g''(0)$ the acceleration at which $f$ changes as we pass through $p$ in the direction of $v$ (also without any mention of $\gamma$).

ContentLicense: CC BY-SA 3.0

ContentLicense: CC BY-SA 4.0


问:
$A \cup B \cup C = (A - B) \cup (B - C) \cup (C - A) \cup (A \cap B \cap C)$
I am using the fact that $X = (X - Y) \cup (X \cap Y)$

So far I've got:

$$A \cup B \cup C = (A - B) \cup (B - C) \cup (C - A) \cup (A\cap B)\cup(B\cap C)\cup(C\cap A)$$

This is almost done, except now I have to deal with $(A\cap B)\cup(B\cap C)\cup(C\cap A)$. How can I use this to get to $A\cap B \cap C$?

答:
Venn Diagrams are great! Here is a visual "proof" of your statement:

Note that all regions are mutually exclusive (i,e, an element in $A - B$ cannot be in $B - C$, and an element in $C - A$ cannot be in $A\cap B\cap C$). Also note that an element of $A\cup B\cup C$ must be in at least one of these regions.


问:
proof of Invertible matrix
Let $U$ be a n by n matrix, satisfy $U^T=-U$

Prove that $U-I$ is invertible.

So, I have some hints, $\vec{v}^T\vec{v}$ for $\vec{v} \in Null(U-I)$

答:
I assume $U$ is a real matrix.

Note that

$(I + U)(I - U) = I - U^2; \tag 1$

given

$U^T = -U, \tag 2$

we have

$U^2 = -U^T U = -UU^T, \tag 3$

whence

$(I + U)(I - U) = I + U^TU; \tag 4$

now if $I - U$ were not invertible, there would be a vector $v \ne 0$ such that

$(I - U)v = 0, \tag 5$

whence

$(I + U^TU) v = (I + U)(I - U)v = 0; \tag 6$

but

$\langle v, (I + U^TU) v \rangle = \langle v, Iv \rangle + \langle v, U^TUv \rangle$
$= \langle v, v \rangle + \langle Uv, Uv \rangle = \Vert v \Vert^2 + \Vert Uv \Vert^2  \ge \Vert v \Vert^2 > 0, \tag 7$

since $v \ne 0$, in contradiction to (6); thus there is no $v$ such that (5) holds, and thus $I - U$ is invertible, as is

$U - I = -(I - U). \tag 8$

Title: proof of Invertible matrix


问:
Turning points of parametric curve

Find the slope of the curve at $t=\frac{1}{4}\pi$.

$$\begin{cases}x=\sin t+\cos t \\ y=\frac{1}{2}\sin(2t)\end{cases}$$

$$\frac{dy}{dx}=\frac{\cos(2t)}{-\sin t+\cos t}$$

$$\left.\frac{dy}{dx}\right|_{\frac{1}{4}\pi}=\lim_{t\to \frac{1}{4}\pi}\frac{\cos(2t)}{-\sin t+\cos t}$$

Multiplying by the conjugate of the denominator won't help, so I applied L'Hôpitals Rule

$$\lim_{t\to \frac{1}{4}\pi}\frac{\cos(2t)}{-\sin t+\cos t}=\lim_{t\to \frac{1}{4}\pi}\frac{-2\sin(2t)}{-\cos t-\sin t}=\sqrt{2}$$

Is there another way to evaluate this limit?

答:
$$L=\lim_{t\to \frac{1}{4}\pi}\frac{\cos(2t)}{-\sin t+\cos t}=\lim_{t\to \frac{1}{4}\pi}\frac{2\cos^2 t -1}{\cos t-\sin t}$$
$$L=\lim_{t\to \frac{1}{4}\pi}\frac{\cos^2 t -\sin ^2 t}{\cos t-\sin t}$$
$$L=\lim_{t\to \frac{1}{4}\pi}\frac{\cos^2 t -\sin ^2 t}{\cos^2 t-\sin^2 t}(\cos t + \sin t)$$
$$\implies L= \lim_{t\to \frac{1}{4}\pi}{(\cos t+\sin t)}=\sqrt 2$$

Title: Turning points of parametric curve


问:
Proof If A is open then $\ A \bigcap \delta A=\emptyset$
So this is my proof, if someone can help me is right or wrong

We know $A$ is open and $\delta A =\overline{A} \bigcap A^C $ so $A \bigcap (\overline{A}\bigcap A^C)=(A\bigcap \overline{A})\bigcap A^C=\emptyset$ so $A\bigcap\delta A=\emptyset$

Thanks.

答:
It looks fine! Maybe you can add these steps.
Since $A$ is open, we have:

$$\delta A=\overline{A}\setminus\mathring{A}=\overline{A}\setminus A=\overline{A}\cap A^C$$
Then...


问:
Compact set in the order topology
Let $(X, \leq)$ be a linearly ordered set and let $\mathcal{T}$ denote the order topology on $X$. Prove that $(X, \mathcal{T})$ is compact if and only if every nonempty set of $X$ has a greatest lower bound and a least upper bound.

I don't know how to apply the definition of compactness in this topology. Can anyone give an idea?

答:
For one direction, suppose that $A\subseteq X$ is a non-empty set with no least upper bound. Let $U$ be the set of upper bounds for $A$; show that $U$ is open. ($U$ may of course be empty.) Then show that

$$\{U\}\cup\{(\leftarrow,a):a\in A\}$$

is an open cover of $X$ with no finite subcover. You can use much the same idea if you have a non-empty set with no greatest lower bound.

For the other direction suppose that every non-empty subset of $X$ has a greatest lower bound and a least upper bound. In particular, $X$ has a least element $a$ and a greatest element $b$. Now imitate the proof that $[0,1]$ is compact to show that $X$ is compact.

Title: Compact set in the order topology


问:
How to derive: $\left(\forall x\right)\left[Lx\rightarrow\left(Mx\land N x\right)\right]\vdash\left(\forall x\right)\left(Lx\rightarrow M x\right)$
I'm trying to derive:
$\left(\forall x\right)\left[Lx\rightarrow\left(Mx\land N x\right)\right]\vdash\left(\forall x\right)\left(Lx\rightarrow M x\right)$

I thought I was on the right track with:

$1 \hspace{5 mm}(1) \left(\forall x\right)\left[Lx\rightarrow\left(Mx\land N x\right)\right]\hspace{10 mm}A\\
1\hspace{5 mm}(2)\hspace{10 mm}   La\rightarrow\left(Ma\land N a\right) \ \hspace{8 mm}1,\forall E\\
2\hspace{5 mm}(3)\hspace{10 mm}   La \hspace{42 mm}A\\
2\hspace{5 mm}(4)\hspace{10 mm}   Ma\ \land Na \ \hspace{20 mm}2,3\rightarrow E\\
2\hspace{5 mm}(5)\hspace{10 mm}   Ma \hspace{40 mm}4, \land E4\\
1,2\hspace{1 mm}(6)\hspace{10 mm}   La\rightarrow Ma \hspace{23 mm}2,4\rightarrow I\\
1,2\hspace{5 mm}(7)\hspace{10 mm}   \left(\forall x\right)\left(Lx\rightarrow M x\right) \hspace{10 mm}6, \forall I\\
$

However, I think by assuming the antecedent in step (3) that I've done something wrong.
Is that step allowed or is there a more efficient way to derive the conclusion?
Or can anyone provide any insights that I may have missed?

答:
Your derivation is all correct; well done!

The assumption in (3) is unproblematic because the $\to I$ rule in step 6 allows to discharges this assumption: When using $\to I$ to turn the assumption from which the succedent was derived into the antecedent of a conditional, that assumption can be dropped because the conclusion no longer depends on the truth of that assumption.


问:
My proof about $\text{sup}(AB) = \text{inf}(A) \text{inf}(B)$ for real negative nonempty sets $A,B$
Given two nonempty bounded below sets $A,B\subset \mathbb R$ of negative numbers.

Define $AB=\left\{ab:a \in A,b \in B\right\}$

I concluded that $$\text{sup}(AB) =  \text{inf}(A)  \text{inf}(B)$$

$$ab \le \text{inf}(A)b \le\text{inf}(A)\text{inf}(B)$$

So $AB$ is a nonempty subset of $\mathbb R$ bounded above and so its supremum does exist, by the minimality of supremum :

$$  \text{sup}(AB) \le \text{inf}(A)\text{inf}(B)\tag{I}$$

On the other hand:

$$\text{sup}(AB)\ge ab \iff \frac{1}{a}\text{sup}(AB)\le b$$
Which is true for all $a \in A,b \in B$,from the definition of infimum:
$$\frac{1}{a}\text{sup}(AB)\le \text{inf}(B) \iff \frac{1}{\text{inf}(B)}\text{sup}(AB)\le a$$

Again from the definition of infimum:

$$\frac{1}{\text{inf}(B)}\text{sup}(AB)\le \text{inf}(A) \iff \text{sup}(AB) \ge  \text{inf}(A)  \text{inf}(B)\tag{II}$$

From $\text{(I)}$ and $\text{(II)}$ the result follows.

The result looks like the relation $$\text{sup}(AB) =  \text{sup}(A)  \text{sup}(B)$$ for two positive nonempty real subsets $A,B$.

Can someone check the validity of my proof?

答:
Your proof is fine - but, I think that this is the sort of problem where a reader will be more easily convinced of the correctness of the proof if you reduce the problem to one involving positive numbers first. It's a bit jarring if the first statement in the proof is
$$ab\leq \inf(A)b\leq \inf(A)\inf(B)$$
because then the reader needs to explicitly think that one must have $\inf(A)\leq a$ but that this get reversed when we multiply by a negative $b$. This doesn't affect the correctness of your proof, of course, but multiplying by negative numbers everywhere makes things much more difficult for the reader, since one has to worry about the directions of inequalities getting reversed.

Instead, I propose you would most cleanly start this proof by presenting a lemma:

Lemma: If $A$ and $B$ are sets of positive real numbers, then $\sup(AB)=\sup(A)\sup(B).$

Then, prior to proving this lemma, you can mention its application to your given problem:

This lemma implies the given problem since, if $A$ and $B$ were sets of negative real numbers, then $(-A)$ and $(-B)$ would be sets of positive real numbers. Assuming the lemma, this yields $\sup(-A)\sup(-B)=\sup((-A)(-B))$. However, note that $(-A)(-B)=AB$ and that $\sup(-A)=-\inf(A)$, so we can derive $\inf(A)\inf(B)=\sup(AB)$ for sets of negative numbers from the lemma.

You might consider breaking out the fact that $-\inf(A)=\sup(-A)$ as a lemma of its own if this is not something you want to use without proof. Doing it this way breaks out all the difficulties of negative numbers into one small section and leaves us with only positive numbers (where multiplication by a constant is an increasing function) - and it nicely takes advantage of the idea that the negative numbers sort of look like the positive numbers, just reversed.

With this prelude out of the way, let's take a look at what you've written, and how it might be better formatted and fit into such a proof. First of all, your first equation is correct, but it should be given context. Your ultimate claim is now that $\sup(A)\sup(B)$ is the least upper bound for $\sup(AB)$, so it's reasonable to have two steps: first, that it is an upper bound, and second, that it is the least upper bound.

Lemma: $\sup(A)\sup(B)$ is an upper bound for $AB$.

Proof: Let $ab\in AB$ where $a\in A$ and $b\in B$. By definition $a\leq \sup(A)$ and $b\leq \sup(B)$. As multiplication by a constant preserves the order, we can derive that $$ab\leq \sup(A)b\leq \sup(A)\sup(B).$$

This is essentially what you wrote, except that it is explicit about both why we care about the equation and how the variables are quantified. You don't have to write in the rigid lemma and proof format, but it is important to let the reader know why an equation is desirable.

The other part of your proof shows that this is the least upper bound. You might notice that you don't actually need to assume $\sup(AB)$ exists - everything you do in that section is true of any upper bound, and ultimately, that's what you want to do. You might continue your proof as:

Lemma: $\sup(A)\sup(B)$ is less than or equal to any other upper bound $L$ for $AB$.

Proof: Suppose that $L$ is an upper bound for $AB$. Then, for every $a\in A$ and $b\in B$ we have
$$L\geq ab$$
which implies
$$\frac{L}a \geq b.$$
This means that $\frac{L}a$ is an upper bound for $B$ and therefore, by the definition of the supremum, we must have
$$\frac{L}a \geq \sup(B).$$
Rearranging further gives
$$\frac{L}{\sup(B)}\geq a$$
where we may then invoke that $\sup(A)$ is the least upper bound for $A$ to receive that
$$\frac{L}{\sup(B)}\geq \sup(A)$$
$$L\geq \sup(A)\sup(B).$$

Then, you can simply conclude:

These two lemmas together imply that $\sup(A)\sup(B)$ is the least upper bound of $AB$, as desired.

Note that I have not touched your algebra at all, except to convert things to positive numbers - what you wrote is perfectly solid, but could be improved by telling the reader why some algebra is important before just jumping to a result and by moving to an easier context for the algebra.
Also observe that, in splitting the proof up according to our requirements, we also eliminated any usage of the completeness of the real numbers - indeed, I might note that your proof method is quite a good one, as it avoids using epsilons or convergence and sticks close to the algebraic and order theoretic definitions.

A slight aside on the second half of the proof: I often like to think about the step from
$$L\geq ab$$
to
$$L\geq \sup(A)b$$
as a single operation of "taking the supremum of both sides over $A$" - since then the whole second half becomes that we take the supremum over $a\in A$ then over $b\in B$ - and can generalize this slightly to allow non-negative reals instead of just positive reals. You need some additional lemma to do this formally - something like:

If $f$ is a continuous non-decreasing function then $\sup(f[X])=f(\sup(X))$

Although, in this case, it suffices to just prove this where $f(x)=\alpha x$ for $\alpha\geq 0$. I suppose, most conservatively, you might just prove an extra lemma:

If $c$ is a positive real, then $\sup(cS)=c\sup(S)$

and applying this twice in the second half to see that $L\geq ab$ implies $L\geq \sup(A)b$ and $L\geq \sup(A)\sup(B)$ in two steps.

Either way, this breaks out some of the algebra into something a bit more conceptual and the idea of turning an inequality quantified over a set into an inequality of extrema has often proved to be useful when I've wanted to do these harder problems in this sort of style.

It's definitely overkill for this problem, but if you're really so inclined, you can also do this sort of manipulation in a more set theoretic way by additionally noting that if $\mathcal F$ is a set of subsets of $\mathbb R$ and $\bigcup \mathcal F$ is the union of all elements of $\mathcal F$, then $$\sup\left(\bigcup \mathcal F\right) = \sup\{\sup(S) : S\in \mathcal F\}.$$
Then you just observe that $AB=\bigcup_{b\in B}bA$ so \begin{align*}\sup(AB)&=\sup\{\sup(bA) : b\in B\}\\&= \sup\{b\sup(A) : b\in B\}\\&=\sup(\sup(A)B)\\&=\sup(A)\sup(B).\end{align*}
where you use the given lemma for the first step, then use the prior lemma that $\sup(cS)=c\sup(S)$ several times - although a lot of the algebra you did ends up being packed into the proof of these lemmas so this ends up being essentially the same argument you put forth, except that it works using sets instead of equations with quantifiers. Sometimes this is a useful way to think about things, although I typically think that the algebraic methods you're using are clearer.


问:
Why are these two numbers relatively prime
I am studying Linear Congruential Generators (LCG). In the proof of the main theorem of these generators:

Theorem: The sequence defined by $X_{n+1}=(aX_n+c) \;mod\;m$, has full period $m$, provided that:

$(i)$ $c$ is relatively prime to $m$

$(ii)$ $a\equiv 1 \pmod{p}$ if $p$ is a prime factor of $m$.

$(iii)$ $a\equiv 1 \pmod {4} $, if 4 is a factor of $m$

Thet sequence can be written as
$$
\frac{(a^n-1)(x_0(a-1)+c)}{a-1}\equiv 0\pmod{m}
$$
To reduce that expression to the following:
$$
\frac{(a^n-1)}{a-1}\equiv 0\pmod{m}
$$
The author says that that can be done because due to "the conditions of the theorem (I think $i$ and $ii$), $x_0(a-1)+c$ is relatively prime to $m$.

Why that's true? And why with that can the congruence be simplified?

答:
A $1$-line proof shows $\,b+c\,$ is coprime to $m$ if every prime factor $p$ of $m$ divides $b$ or $c$ but not both. OP is special case: all $\,p\mid b\! =\! x_0(a\!-\!1)\,$ by $\,p\mid  a\!-\!1;\,$  no $\,p\mid c\,$ (else $\,p\mid c,m\,$ contra $c,m$ coprime).

Remark $ $ As explained in the linked post. this simple idea (going back to Stieltjes) allows us to generalize Euclid's proof of infinitely many primes to a proof of infinitely many $\rm\color{#c00}{co}primes$ in an arithmetic progression (a much simpler $\rm\color{#c00}{co}prime$ analog of Dirichlet's famous theorem). This very simple result is often overlooked — leading to more complex and less conceptual proofs.


问:
Finding out all questions in test bank
Let’s Say that a school has a test bank for an online multiple choice math exam which contains 100 questions. When a student starts his exam the system randomly selects 15 questions for him.
If my friends and I decide to find out all the 100-questions, how many trials do we need if that is possible??

Edit:(as suggested in comments)

How many trials are needed, on average, to see all the questions??

答: Let's instead assume that we're sampling from the bank of 100 multiple choice questions one-by-one and with replacement. For each $i\in \mathbb{N}$ let $X_i$ denote the number of distinct questions we observed after sampling the the $i^{th}$ multiple choice question. Then $X_1=1$ and $\{X_i\}_{i\in \mathbb{N}}$ is an absorbing Markov Chain defined on the state space $ \{1,\ldots,100\} $ having transition probabilities $$P(X_{i+1}=r|X_i=r)=\frac{r}{100}$$ $$P(X_{i+1}=r+1|X_i=r)=\frac{100-r}{100}$$ Here $r=1,\ldots ,99$ while the final state of observing all 100 multiple choice questions is absorbing; techniques for computing the expected number of steps until we reach this absorbing state can be found using the fundamental matrix. If $E$ is this expected value, we can evaluate the ceiling of $(E+1)/15$ to get the answer you're looking for.

Title: Finding out all questions in test bank


问:
Interesting theoretic proof with polynomials
Four polynomials with real coefficients $p_1, p_2, p_3, p_4$ are given. The sum of any two of them has no real roots. Prove that if the polynomial $p_1+p_2+p_3+p_4$ has a real root, then at least one of the polynomials $p_1, p_2, p_3, p_4$ doesn't have a real root. $$$$
Unfortunately I'm quite new to polynomials and don't have the best understanding of them, so I would greatly appreciate if someone could guide me through this with explanation.

答:
Let us use the fact that a polynomial with no real roots can be represented by one of the two forms:

$\ \ \ P(x)^2+Q(x)^2$ [case (+)] ($P$ and $Q$ having no common roots).

$−(P(x)^2+Q(x)^2)$ [case (-)]

(see there), corresponding to the cases where the dominant coefficient is positive (resp. negative).

The first hypothesis means that

$$p_1+p_2, \ \ \ \ p_1+p_3, \ \ \ \ p_1+p_4, \ \ \ \  p_2+p_3,  \ \ \ \ p_2+p_4,  \ \ \ \ p_3+p_4$$

are each one either in case (+) or in case (-).

Let us assume that

$$p_1+p_2+p_3+p_4 \ \ \ \text{has a real root.}\tag{1}$$

Therefore, we have as many (+) and (-) cases (see Remark 1 below), like this:

$$\begin{cases}(a)&p_1+p_2&=&A^2+B^2\\(b)&p_3+p_4&=&-(C^2+D^2)\end{cases},
\ \ \begin{cases}(c)&p_1+p_3&=&E^2+F^2\\(d)&p_2+p_4&=&-(G^2+H^2)\end{cases}, \ \ \begin{cases}(e)&p_1+p_4&=&I^2+J^2\\(f)&p_2+p_3&=&-(K^2+L^2)\end{cases}\tag{2}$$

for certain polynomials $A,B,C,\cdots K,L$.

Combining (a)+(c)-(f), one gets

$$2p_1=A^2+B^2+E^2+F^2+K^2>0, \ \ \text{therefore with no real roots} \tag{3}$$

as was desired.

Remark 1: In (2), if we have had for example $$\begin{cases}(a)&p_1+p_2&=&A^2+B^2\\(b)&p_3+p_4&=&(C^2+D^2)\end{cases},$$ the sum $p_1+p_2+p_3+p_4$ would have always been $>0$, contradicting (1).

Remark 2: One could object that the cases considered in (2) are one among many. In fact, we can WLOG attribute indices to the $p_k$s in such a way that equations (a),(b),(c),(d) hold. It remains the last two equations (e) and (f) for which we could have a (+) sign for (f) and (-) sign for (e). The reader will not have difficulty to see that in this case as well, one can conclude as in (3).


问:
Prove that the recursively defined sequence is Cauchy.
Define a sequence recursively by $x_1=1$, $x_{n+1}=\frac{5+5x_n}{5+x_n}$. Prove that this sequence is Cauchy. Then find the limit.

We call a sequence $\{a_n\}$ Cauchy, if for any $\epsilon>0$, there exists $N\in\mathbb{N}$ such that for all $n,m\geq M$, $|a_n-a_m|<\epsilon$.

答:
If the sequence converges, then $\lim_{n \to \infty} x_n = L$ for some finite real number $L$. Then:

$$L(5+L) = 5+5L$$

$$5L+L^2 = 5+5L$$

$$L = \sqrt{5}$$

where we have discard $-\sqrt{5}$ as a solution because it is clear that the sequence $(x_n)$ consists only of elements greater than or equal to $0$. Now, we prove two things;

$(x_n)$ is nondecreasing

$(x_n)$ is bounded above by $\sqrt{5}$.

We prove the second statement by induction. So, clearly, the result holds when $n=1$. Suppose that it holds for arbitrary $n$. Then:

$$x_{n+1} = \frac{5+5x_n}{5+x_n} = 5 \cdot \frac{x_n+1}{x_n+5} = 5 \cdot \left(1 - \frac{4}{x_n+5} \right)$$

$$x_{n+1} = 5-\frac{20}{x_n+5}$$

Then, $x_n <  \sqrt{5} \implies x_n+5 < \sqrt{5}+5 \implies -\frac{1}{x_n+5} < -\frac{1}{5+\sqrt{5}}$. So:

$$x_{n+1} = 5-\frac{20}{x_n+5} < 5-\frac{20}{5+\sqrt{5}} = 5-\frac{20}{5^2-5} \cdot (5-\sqrt{5}) = 5-(5-\sqrt{5}) = \sqrt{5}$$

So, of course, it follows that $(x_n)$ is bounded above by $\sqrt{5}$.

We will now prove that it is nondecreasing. We have the following;

$$x_{n+1}-x_n = \frac{5+5x_n}{5+x_n}-x_n = \frac{5-x^2_n}{5+x_n}$$

However, we just proved that $(x_n)$ is bounded above by $\sqrt{5}$. So:

$$\forall n \in \mathbb{N}: 0 < x_n \leq \sqrt{5}$$

$$\forall n \in \mathbb{N}: x^2_n \leq 5$$

$$\forall n \in \mathbb{N}: 5-x^2_n \geq 0$$

which implies that $x_{n+1} \geq x_n$. By the Monotone Sequence Theorem, $(x_n)$ is convergent and, hence, is a Cauchy Sequence. $\Box$

I'm not entirely sure if there is a fast way to prove that this is Cauchy using only the definition. I suspect that you'd have to do quite a bit of work for it but I certainly haven't tried thinking about it too hard.


问:
calculate $\lim _{n\to \infty}\sum _{ k=1 }^{ n }{ \left| { e }^{ \frac { 2\pi ik }{ n } }-{ e }^{ \frac{ 2 \pi i(k-1) }{ n }} \right|}$
$\lim _{ n\to \infty  } \sum _{ k=1 }^{ n }{ \left| { e }^{ \frac { 2\pi ik }{ n }  }-{ e }^{ \frac { 2\pi i(k-1) }{ n }  } \right|  } $

We know that $e^{i\pi}=-1$ $\rightarrow$ $e^{2i\pi}=1$ $\rightarrow$ $e^{2i\pi k}=1$, $e^{2i\pi(k-1)}$= $\frac{e^{2i\pi k}}{e^{2i\pi}}=\frac{1}{e^{2i\pi }}=1$. Then

\begin{align}
\lim_{n\to \infty} \sum _{k=1}^{n} \left| e^{\frac{ 2\pi ik }{ n }}  -e^{ \frac { 2\pi i(k-1) }{ n }  } \, \, \, \right| &= \lim _{n\to \infty} \sum _{ k=1 }^n \left| (e^{ 2\pi ik })^{ 1/n }-(e^{ 2\pi i(k-1) })^{ 1/n } \right| \\
&=\lim _{n\to \infty} \sum _{ k=1 }^n \left| (1)^{1/n}-(1)^{1/n} \right| =0
\end{align}

I am not convinced by the result. Any suggestion?

答: Consider these as points in the complex plane. Each value $e^{2\pi ik/n}$ is a vertex of a regular $n$-gon centered at $0$ with all vertices lying on the unit circle, and your sum is just the perimeter of this $n$-gon. As $n\to\infty$, the sides approximate the circumference of the unit circle, giving an answer of $2\pi$.


问:
Implicit differentiation $e^{(x^2y)}=\cos y$
I took the natural log of both sides, giving me $x^2y=\ln(\cos y)$.

Then I differentiated: $2xydx + (x^2)dy = (-\tan y)dy$

That turned into $\frac{dy}{dx} = \frac{2xy}{-\tan y-x^2}$

This was marked incorrect. I differentiated with another method and got the right answer. What did I do in this process that was illegal?

答:
We have:

$e^{x^{2}y} = \cos(y)$

$x^{2}y = \ln(\cos(y))$

Implicitly differentiating:

$2xy + x^{2}y' = \frac{d}{dx}(\cos(y))\frac{1}{\cos(y)}$

$2xy + x^{2}y' = -\frac{y'\sin(y)}{\cos(y)}$

$2xy = y'(-\tan(y)-x^{2})$

$y' = -\frac{2xy}{\tan(y)+x^{2}}$

To be safe, you should use derivative notation instead of differentials because differentials aren't exactly "formal".


Title: Prove that the recursively defined sequence is Cauchy.

Body:
Define a sequence recursively by $x_1=1$, $x_{n+1}=\frac{5+5x_n}{5+x_n}$. Prove that this sequence is Cauchy. Then find the limit.

We call a sequence $\{a_n\}$ Cauchy, if for any $\epsilon>0$, there exists $N\in\mathbb{N}$ such that for all $n,m\geq M$, $|a_n-a_m|<\epsilon$.

Body:
Clearly $x_n>0$. Noting
$$ x_{n+1}-\sqrt5=\frac{5+5x_n}{5+x_n}-\sqrt5=\frac{5+5x_n-5\sqrt5-\sqrt5x_n}{5+x_n} =\frac{5-\sqrt5}{5+x_n}(x_n-\sqrt5)$$
one has
$$ |x_{n+1}-\sqrt5| =\frac{5-\sqrt5}{5+x_n}|x_n-\sqrt5|\le\frac{5-\sqrt5}{5}|x_n-\sqrt5|.$$
This gives
$$ |x_n-\sqrt5|\le\bigg(\frac{5-\sqrt5}{5}\bigg)^{n-1}|x_1-\sqrt5|.$$
Thus
$$ \lim_{n\to\infty}|x_n-\sqrt5|=0 $$
or
$$ \lim_{n\to\infty}x_n=\sqrt5. $$


答:
Your answer is correct. I would write it as $-\frac{2xy}{x^2+\tan y}$, but that's just a matter of personal preference.

I assume that the "correct answer" that was given was $\dfrac{dy}{dx}=-\displaystyle\frac{2xye^{x^2y}}{x^2e^{x^2y}+\sin y}$, because that's what Wolfram Alpha gives. However, this answer is exactly equivalent to yours. To see this, remember that $e^{x^2y}=\cos y$, so
$$\frac{dy}{dx}=-\frac{2xy\cos y}{x^2\cos y+\sin y}=\frac{2xy\cos y}{x^2\cos y+\sin y}\cdot\frac{\sec y}{\sec y}=-\frac{2xy}{x^2+\frac{\sin y}{\cos y}}=\frac{2xy}{x^2+\tan y}$$
I would go talk to whoever marked it incorrect to see if the grade was a mistake. It could also be that your teacher/professor wanted you to do the problem a certain way, like not using $\ln$ and instead implicitly differentiating in the exponential. Either way, the result is correct.

Title: Implicit differentiation $e^{(x^2y)}=\cos y$

Body:
I took the natural log of both sides, giving me $x^2y=\ln(\cos y)$.

Then I differentiated: $2xydx + (x^2)dy = (-\tan y)dy$

That turned into $\frac{dy}{dx} = \frac{2xy}{-\tan y-x^2}$

This was marked incorrect. I differentiated with another method and got the right answer. What did I do in this process that was illegal?


问:
Nonexistence of a simple group of order 420
From Dummit & Foote, Abstract Algebra, $\S6.2$, Exercise 17(a).

Prove that there is no simple group of order 420.

Suppose not; label such group $G$. the number of Sylow 7-subgroups of $G$ is 15. Let $G$ act on the set of Sylow 7-subgroups (denoted hereon by "letters") by conjugation. There is only one orbit of size 15 (by Sylow 2nd), thus each stabilizer on one letter has size 420/15 = 28.

The action induces an injective homomorphism of $G$ into $A_{15}$. Each stabilizer on a single letter should have an element of order 7 (by Cauchy), permuting the remaining 14 letters. Naturally, this element then generates the unique Sylow 7 within the stabilizer.

I now assume that the element of order 7 is a product of 2 disjoint 7-cycles. Is this valid, and why? In particular, am I able to eliminate the possibility of the element being a single 7-cycle?

If the above assumption is valid, then I am now able to eliminate the possibility of order 14 and 28 elements, since order 14 implies single 14-cycle (odd) or product of single 7-cycle and some 2-cycles (2nd power is single 7-cycle), likewise for order 28.

Now use the fact that the Sylow 7 is normal within the stabilizer: the permutations that sends a 7-cycle to its power by conjugation is either a product of 3 2-cycles (sends to inverse), 2 3-cycles (sends to 2nd/4th power), or a 6-cycle (sends to 3rd/5th power). By similar calculations, permutations that switch between the two 7-cycles are either 7 2-cycles, a 2-cycle and 3 4-cycles, a 2-cycle and 2 6-cycles, a 2-cycle and a 12-cycle, or a 14-cycle. Since the remaining elements are either order 2 and 4, our choices are either 6 2-cycles, 7 2-cycles (odd), or a 2-cycle and 3 4-cycles (third power breaks the pairing of the two 7-cycles).

Are my calculations and reasoning correct here?

The pair of fixed letters in the 2 7-cycles then determines the whole of the permutation, but noting that the two different fixed letter 2-cycles (per 1 7-cycle) result in a product of a 7-cycle, we conclude that the pairings of fixed letters must be disjoint for two different elements. Then, the number of possible remaining elements is 7 out of a required 21; contradiction.

In general, is there a cleaner way to go about this exercise, or proving nonexistence of groups of some highly composite order? I only know the method of embedding the group in an alternating group and trying to derive a contradiction from there (outside the usual repertoire).

答:
A15 does have subgroups of order 14 or 28. Actually an odd cycle is an even permutation, so 7-2-2 and 7-cycle are in A15. Without futher deduction that Sylow 2 group is isomophic to V4, the group of 28 could be <(1,2,3,4,5,6,7),(8,9,10,11)(12,13,14,15)>, i.e. A group generated by a 7-cycle and a disjoint 4-cycles.
By contradiction assume that G is simple, |G|=4 x 3 x 5 x 7, so no subgroup of index less than 7, making n5=21, n7=15. Denote N7 be the normalizer of a sylow-7 group, so |N7|=28 and |N5|=20.

We first prove that n3=70 and n2=35

show that N2=A4 (N2 isomorphic to A4) and P2=V4 (Klein 4 group)

There is an element of order 14 in N7 and an element of order 10 in N5.

G acts transitively on all subgroups of order 2 by conjugation.

There is a subgroup U of order 2, s.t. <P2, P5, P7> <= NG(U), whose index is less than 7.

By sylow's theorem, n3=10, 28 or 70, n2=7, 15, 21, 35 or 105.
If n3=7 (or |N3|=60), since N3 is not simple group of order 60, n5=1. That means a P5 normalizes a P3, contradicting to |N5|=20. By the same method, we rule out |N2|=60.

If n3=10 (or |N3|=42), then there is a P7 commute with P3 in N3, contradict to |N7|=28.

Leaving only n3=70.

Next we rule out n2=15 (or N2=28). If it were, we first prove that N7=N2=C28, i.e. N7 and N2 are in the same conjugation class, then contradict by counting elements. If N7 is not isomorphic to C28, we can conjugate any N2 so that it contains the same P7 subgroup. The the unqiue sylow-2 subgroup Q2 of N2 normalizes P7 but distincts from P2 in N7, since N7 is not isomorphic to C28. So P7 is normal in <P7, P2, Q2>, contradicts to |N7|=28. So N7=N2=C28. The total elements of order 3 and 5 is 2x70+4x21=224, and the total unique elements of all conjugations of N2 (or N7) is greater than 196, overflowing |G|.

Now we rule out n2=21 (or N2=20). By the same way, we argue that N2 and N5 are in the same conjugation class and isomorphic to C20. We first find out U of order 2 commute with a P7, then find a P5 and P2 commute with U, so U=<P7, P5, P2>, whose order is greater than 140.

Next we rule out n2=105, leaving only n2=35 (or N2=12).

Now we show that N2=A4. If not, then it is isomorphic to an abelian group, contradicts to N3=6.

Since A4 act transitively on subgroups of order 2 in V4 (the order 4 subgroup in A4), and G act transitively on all P2, by conjugation, completing the proof that G acts transitively on all order 2 subgroups of G.

Since P2=V4, a homomorphism from V4 to Aut(P7) cannot be injective, so there is an element of order 2 compute with element of order 7 in N7. The same reason, There is an element of order 2 compute with element of order 5 in N5.

Now that G acts transitively on any subgroup U of order 2, we can choose a P5, P7, P2 normalizing U, making order of NG(U) bigger or equal to 70. QED.


问:
Solve $\sin(z) = z$ in complex numbers
Show that $\sin(z) = z$ has infinitely many solutions in complex numbers.
Little Picard theorem should help, but using big Picard theorem is undesirable.
Thanks a lot!</p>

答:
Let $k \in \mathbb{Z}$, $y \in \mathbb{R}$, $y>0$ and consider the solid rectangle $$S = \{z \in \mathbb{C} \mid 0 \leq \operatorname{Im}(z) \leq y \textrm{ and } (2k - \tfrac{1}{2})\pi \leq \operatorname{Re}(z) \leq (2k+\tfrac{1}{2})\pi \}.
$$

The function $\sin(z)$ is single valued on $S$ and maps onto to the upper half of a solid ellipse with foci $\pm 1$.  The lower, left and right sides of $S$ map onto the real interval and the top side onto the elliptic arc. (The following image may help to visualize this.)

<img src="https://i.stack.imgur.com/LKmbq.png" alt="image of $\sin(z)$" />

On the top side (where $\operatorname{Im}(z) = y$) the inequality $|\sin(z)| \geq \sinh(y)$ holds.  Choose $y>0$ such that $$\sinh(y) \geq \max_{z\in S}|z|$$ (note that $S$ itself depends on $y$ but $\sinh$ grows fast enough).  Then it follows that $$S \subset \sin(S).$$  If $\sin^{-1}\!: \sin(S) \to S$ is the inverse this can be written as $$S = \sin^{-1}(\sin(S)) \subset \sin(S).$$

By Brouwer's fixed point theorem $\sin^{-1}$ must have a fixed point in $S$, which will also be a fixed point of $\sin$. By symmetry of $\sin$ there is also a fixed point in $\overline{S}$, $-S$ and $-\overline{S}$.  Varying $k$ shows that $\sin$ has infinitely many fixed points.

There are also relevant complex analytic fixed point results (see here and here for example).  To name some consequences: there is at most one fixed point in the interior of $S$ and if there is one then iterating $\sin^{-1}$ on any starting point will converge to the fixed point.  (There is the technical difficulty that $\sin^{-1}$ does not map $\sin(S)$ into the interior of $S$.  Note for example that for $k=0$ there is no fixed point in the interior of $S$!)

Title: Solve $\sin(z) = z$ in complex numbers


问:
Rewriting basis of vector in terms of another another basis vector
So I'm currently studying linear algebra and came across an example from the textbook which I don't understand. The example is: Let $V=\mathbf{R}^2$ and let $\beta$  be the basis consisting of the vectors $u_1$ = $(1, 1)$ and $u_2 = (1, — 1)$  and $\beta \prime$ be the basis consisting of the vectors $u_1 \prime = (1,2)$ and $u_2 \prime = ( — 2, 1)$. If $v$ is any vector in $V$, then $v$ can be expressed uniquely as $v = x_1 u_1 + x_2u_2$, and also uniquely as $v = x_1\prime u_1\prime +  x_2\prime u_2\prime$

I don't have any problem understanding the part above but what I don't get is the following:

Notice that $u_1$  and $u_2 $ in particular can be expressed in terms of $u_1 \prime$ and $u_2 \prime$
:

$u_1 = (3/5)u_1\prime - ( 1 / 5 )u_2\prime$ and $u_2 = ( - 1 / 5 ) u_1\prime - (3 /5 )u_2\prime$

The textbook skips the steps to show how $u_1$ and $u_2$ are rewritten in terms of $u_1 \prime$ and $u_2 \prime$ and I'm confused to how it's possible. Could someone help me with the hidden step and break down how  $u_1$ and $u_2$ are expressed in terms of $u_1 \prime$ and $u_2 \prime$

答:
Since $\mathcal{B}' = \{u'_{1},u'_{2}\}$ is a basis, we can particularly express $u_{1}$ and $u_{2}$ in terms of it.

More precisely, one has that
\begin{align*}
\begin{cases}
u_{1} = (1,1) = a_{11}u'_{1} + a_{12}u'_{2} = (a_{11} - 2a_{12},2a_{11} + a_{12})\\\\
u_{2} = (1,-1) = a_{21}u'_{1} + a_{22}u'_{2} = (a_{21} - 2a_{22},2a_{21} + a_{22})
\end{cases}
\end{align*}

When you solve such systems of equations, you get the desired relation.

Hopefully this helps.


答:
Your proof is correct, but the last step is perhaps unclear: I would've wrote:
$$
(A \cap \bar{A}) \cap A^c = (\bar{A} \cap A) \cap A^c = \bar{A} \cap (A \cap A^c) = \emptyset,
$$
so it's evident from the fact that $A \cap A^c = \emptyset$.

Here is another way to prove the same result: suppose $A$ is open yet $A \cap \partial A \neq \emptyset$. Let $x \in A \cap \partial A$. By definition of boundary, for every neighborhood $N$ of $x$, we have $N \cap A \neq \emptyset$ and $N \cap A^c \neq \emptyset$. But since $x \in A$ and $A$ is open, there exists a neighborhood $N'$ of $x$ such that $N' \subseteq A$. But this would imply $N' \cap A^c = \emptyset$, contradicting the fact $x$ is in the boundary of $A$. So it must be that $A \cap \partial A = \emptyset$.

Title: Proof If A is open then $\ A \bigcap \delta A=\emptyset$

Body:
So this is my proof, if someone can help me is right or wrong

We know $A$ is open and $\delta A =\overline{A} \bigcap A^C $ so $A \bigcap (\overline{A}\bigcap A^C)=(A\bigcap \overline{A})\bigcap A^C=\emptyset$ so $A\bigcap\delta A=\emptyset$

Thanks.


问:
Marking the prime points on a circle
If you travel around a circle and mark all the points on the circle where the distance you travelled is a prime number, where you would go through many rotations*, do you end up marking the entire circumference of the circle as the number of points you pick goes to infinity?

Or do you leave  "holes" in the circle by virtue of only marking the prime points of the circumference?

*I understand that you would have to define the circumference of your circle. Define it as a finite number, prime or not. How does the circumference you pick affect the answer/whether the circumference is prime or not?

Please let me know how I can explain this question further or whether this question does not make sense.

答:
Let's look at some examples, calling the starting point $0$

If you have a circle of circumference $2$, then you mark just two points. Point $0$ gets marked just once for the prime $2$, and all the odd primes fall at point $1$.

If the circumference is $6$ you miss points $0$ and $4$ and mark $2$ and $3$ just once, while $1$ and $5$ are each marked infinitely often.

This shows a link with Dirichelet's theorem on primes in arithmetic progression.

A different kind of question would be to take a circle of radius $1$ (circumference $2\pi$, so close to the $6$ of the previous example) and to ask whether the prime points form a dense set on the circumference of the circle.

Title: Marking the prime points on a circle


问:
Inequalities with entropy with bracketing
Show that for all $1\leq p < \infty$:

$$H_p(\delta, \mathcal G, Q) \leq H_{p, B}(\delta, \mathcal G, Q) \leq H_\infty (\delta / 2, \mathcal G)$$

where $H_p$ is the $\delta$-entropy of the class of functions $\mathcal G$ under the $L_p(Q)$ norm and $H_{p, B}$ is the entropy with bracketing and $H_\infty$ is the $\delta$ entropy under the sup norm.

What I've tried

For the first inequality we can show that a set of brackets of size $\delta$ for $\mathcal G$ yields a $\delta$ covering of $\mathcal G$. Notice that for a function $g \in \mathcal G$ and it's bracket $[g_j^L, g_j^U]$ we have:

$$\|g_j^L - g\|_{p, Q} + \|g_j^U - g\|_{p, Q}= \|g_j^L - g_j^U\|_{p, Q} \leq \delta$$

So either the $g_j^L$ or $g_j^U$ suffice as a covering set.

For the second inequality I'm having more trouble. My original thought was to try the same method as for the first inequality or try to construct a bracket from a sup norm $\delta$ covering but I have not found a way to do that in a way that implies the inequality. I've also tried rephrasing the requirement as:

If $H_\infty(\delta/2, \mathcal G) \leq N$ then $H_{p, B}(\delta, \mathcal G, Q) \leq N$

答: Given a sup-norm covering of radius $\delta/2$, call it $\{g_j\}$, consider the brackets $[g_j^L, g_j^U]$ where $g_j^L(x) = g_j(x) - \delta/2$ and $g_j^U(x) = g_j (x) + \delta/2$. If $Q$ is a probability measure, then $\|g^L_j - g^U_j\|_{p, Q} = \delta$, and by construction these brackets cover $\mathcal{G}$.


问:
What is the geometric meaning of this vector equality? $\vec{BC}\cdot\vec{AD}+\vec{CA}\cdot\vec{BD}+\vec{AB}\cdot\vec{CD}=0$
I was doing some exercises for linear algebra. One of them was to prove that for any four points $A, B, C, D \in \mathbb{R}^3$ the following equality holds:
$$\overrightarrow{BC} \cdot \overrightarrow{AD}\ +\ \overrightarrow{CA} \cdot \overrightarrow{BD}\ +\ \overrightarrow{AB} \cdot \overrightarrow{CD}\ = 0$$
The proof is easy; you just make three vectors starting in $A$ and then see that all the terms cancel out.

My question is: what is the geometric interpretation of this equality? How can I visualize it or understand its deeper meaning? Does this equality have a name or where can I read more about it?

I'm asking this because it turns out that it is not just a random equality and is rather useful. For example, if we want to prove the existence of orthocenter, we can do it surprisingly easily and quickly using this equality.

答:
Here is another proof, maybe it will be of use: change $D$ by adding any vector $v$ to it. The sum changes by $\left(\overrightarrow{AB}\ +\ \overrightarrow{BC}\ +\ \overrightarrow{CA}\right)\cdot v=0$. So this is an expression independent of $D$. Similarly it is independent of $A$, $B$ and $C$, so is constant. Clearly this constant is $0$.

(In fact one could just move $D$ to $A$ and get zero right away. One of the proposed solutions moves $D$ to orthocenter $O$,  but that is not really necessary.)

EDIT: To see independence from $A$ massage the formula by swapping direction of arrows so that $A$ is last:

$$\overrightarrow{BC} \cdot \overrightarrow{AD}\ +\ \overrightarrow{CA} \cdot \overrightarrow{BD}\ +\ \overrightarrow{AB} \cdot \overrightarrow{CD}\ = \overrightarrow{CB} \cdot \overrightarrow{DA}\ +\  \overrightarrow{BD} \cdot \overrightarrow{CA} \ +\ \overrightarrow{DC} \cdot \overrightarrow{BA}  $$

Now adding $v$ to $A$ changes the sum by $ (\overrightarrow{CB}  +  \overrightarrow{BD}  + \overrightarrow{DC} )\cdot v=0$.

Same works for $B$ and $C$.


问:
How many ways to distribute these balls?
There is a box of balls containing six different colors of balls. More precisely there are
nine, eight, five, seven, two, and four of the various colors for a total of 35 balls. If eight people
take balls from the box, in how many ways can the balls be distributed? (There are no restrictions how many balls one person can get, so one person can get all 35 and the others get nothing).

Is it correct that there are $8^{35}$ ways to distribute the balls? The balls are distinct, but there are different numbers of each color, so that's what is getting me. I think the answer might involve n choose k but I'm not sure how to apply it.

答:
It would be $8^{35}$ if the $35$ balls were numbered or otherwise distinguishable. As it is, the $9$ balls of the first color are presumably indistinguishable, so there’s only one way for the first person to get $5$ of them and the second person $4$, not the $\binom95$ ways that there would be if the $9$ balls were numbered or otherwise made individually identifiable.

Suppose that you knew that there were $n_1$ ways to distribute the $9$ balls of the first color amongst the $8$ people, $n_2$ ways to distribute the $8$ balls of the second color, and so on. You could then calculate the number of possible distributions of all $35$ balls amongst the $8$ people; how?

Calculating each these numbers $n_1,\ldots,n_6$ is a stars and bars problem. And your problem as a whole is similar to this one.


问:
How to check that given polynomials form a Groebner basis
I am wondering if some polynomials are given, how do we know whether they form Groebner basis or not.
Note that it is not necessary that given poly's form a reduced Groebner basis.

I know how to find a Grobner basis of an ideal but Grobner basis is not unique. So I don't know how to determine if they are Groebner basis.

答:
You can use a variant of Buchberger's algorithm to check if a given set of generators is a Gröbner basis.  Specifically:

For each pair of generators, compute the $S$-polynomial.

Use the multivariate division algorithm to reduce the $S$-polynomial with respect to the generators as much as possible.

If the remainder is zero for every possible $S$-polynomial, then the generating set is a Gröbner basis.


答:
Yes, your counterexample to $\mathcal{P}(A \,\triangle\, B) \setminus \{\emptyset\}= \mathcal{P}(A) \,\triangle\, \mathcal{P}(B)$ is correct. And your argument is  correct as well.

By the way, you can find an easier counterexample to  $\mathcal{P}(A \,\triangle\, B) \setminus \{\emptyset\}= \mathcal{P}(A) \,\triangle\, \mathcal{P}(B)$.
Take
\begin{align} A &= \{1\} & B &= \{2\}.
\end{align}
Then, $A \cap B = \emptyset$ and $A \cup B = \{1,2\}$ and so $A \,\triangle\, B = (A \cup B) \setminus (A \cap B) = \{1,2\}$. Therefore, $\mathcal{P}(A \,\triangle\, B) = \{\{1\},\{2\},\{1,2\},\emptyset\}$.

On the other hand, $\mathcal{P}(A) = \{\{1\},\emptyset\}$ and $\mathcal{P}(B) = \{\{2\},\emptyset\}$, hence $\mathcal{P}(A) \cup \mathcal{P}(B) = \{\{1\}, \{2\}, \emptyset\}$ and $\mathcal{P}(A) \cap \mathcal{P}(B) = \{\emptyset\}$.
Thus, $\mathcal{P}(A) \,\triangle\, \mathcal{P}(B) = \big(\mathcal{P}(A) \cup \mathcal{P}(B)\big) \setminus \big(\mathcal{P}(A) \cap \mathcal{P}(B)\big) = \{\{1\}, \{2\}\}$

Summing up,
\begin{align}
\mathcal{P}(A \,\triangle\, B) \setminus \{ \emptyset\} = \{\{1\},\{2\},\{1,2\}\} \neq \{\{1\},\{2\}\} = \mathcal{P}(A) \,\triangle\, \mathcal{P}(B)
\end{align}

Title: Disprove $\mathcal{P}(A\Delta B) \setminus \{\emptyset\} = \mathcal{P}(A) \Delta\mathcal{P}(B) $. Is my proof good?

Body:
Proof that the following equality is wrong:
$$\mathcal{P}(A\Delta B) \setminus \{\emptyset\} = \mathcal{P}(A) \Delta\mathcal{P}(B) $$

I choose elements of sets $A$ and $B$ so that they are separate.

$A = \{1, 2\}$

$ \mathcal{P}(A) = \{\{1\}, \{2\}, \{1, 2\}, \emptyset\}$

$B = \{3, 4\}$

$\mathcal{P}(B) = \{\{3\}, \{4\}, \{3, 4\}, \emptyset\}$

$A\cup B = \{1, 2, 3, 4\}$

$A\cap B = \emptyset$

Now I see that:

On the left side:

$\mathcal{P}(A\Delta B) = \mathcal{P}((A\cup B) \setminus (A\cap B)) = \mathcal{P}((A\cup B) $
$ \mathcal{P}((A\cup B) =\{\{1\}, \{2\}, \{3\}, \{4\}, \{1, 2\}, \{1, 3\}, \{1, 4\}, \{2, 3\}, \{2, 4\}, \{3, 4\}, \{1, 2, 3\}, \{1, 2, 4\}, \{1, 3, 4\}, \{2, 3, 4\}, \{1, 2, 3, 4\},\emptyset \} $

$ \mathcal{P}((A\cup B) \setminus \{\emptyset\} =\{\{1\}, \{2\}, \{3\}, \{4\}, \{1, 2\}, \{1, 3\}, \{1, 4\}, \{2, 3\}, \{2, 4\}, \{3, 4\}, \{1, 2, 3\}, \{1, 2, 4\}, \{1, 3, 4\}, \{2, 3, 4\}, \{1, 2, 3, 4\}\} $

On the right side:

$\mathcal{P}(A) \Delta\mathcal{P}(B) = (\mathcal{P}(A) \cup \mathcal{P}(B)) \setminus (\mathcal{P}(A) \cap \mathcal{P}(B)) = \{ \{1\}, \{2\}, \{3\}, \{4\}, \{1, 2\}, \{3, 4\} \} $

Therefore:

$\{\{1\}, \{2\}, \{3\}, \{4\}, \{1, 2\}, \{1, 3\}, \{1, 4\}, \{2, 3\}, \{2, 4\}, \{3, 4\}, \{1, 2, 3\}, \{1, 2, 4\}, \{1, 3, 4\}, \{2, 3, 4\}, \{1, 2, 3, 4\}\} \neq \{ \{1\}, \{2\}, \{3\}, \{4\}, \{1, 2\}, \{3, 4\} \} $

i.e. $\mathcal{P}(A \,\Delta\,B) \setminus \{\emptyset\} \neq \mathcal{P}(A) \Delta\mathcal{P}(B)$.

Is my proof valid? Are there any errors? I could take smaller $A$ and $B$ sets but I was scared that it wouldn't be so obvious that way.


答:
We use this property: any $v\in\mathbb{R}^{2}$ could be expressed as a linear combination of $u_1’$ and $u_2’$.
So $$
u_1=λ_1u_1’+μ_1u_2’$$
$$u_2=λ_2u_1’+μ_2u_2’
$$
And this leads to
$$1=λ_1-2μ_1$$
$$1=2λ_1+μ_1$$
$$1=λ_2-2μ_2$$
$$-1=2λ_2+μ_2$$
By solving these equations, we get
$$λ_1=\frac{3}{5},λ_2=-\frac{1}{5}$$
$$μ_1=-\frac{1}{5},μ_2=-\frac{3}{5}$$
That’s exactly what we get finally.

Title: Rewriting basis of vector in terms of another another basis vector

Body:
So I'm currently studying linear algebra and came across an example from the textbook which I don't understand. The example is: Let $V=\mathbf{R}^2$ and let $\beta$  be the basis consisting of the vectors $u_1$ = $(1, 1)$ and $u_2 = (1, — 1)$  and $\beta \prime$ be the basis consisting of the vectors $u_1 \prime = (1,2)$ and $u_2 \prime = ( — 2, 1)$. If $v$ is any vector in $V$, then $v$ can be expressed uniquely as $v = x_1 u_1 + x_2u_2$, and also uniquely as $v = x_1\prime u_1\prime +  x_2\prime u_2\prime$

I don't have any problem understanding the part above but what I don't get is the following:

Notice that $u_1$  and $u_2 $ in particular can be expressed in terms of $u_1 \prime$ and $u_2 \prime$
:

$u_1 = (3/5)u_1\prime - ( 1 / 5 )u_2\prime$ and $u_2 = ( - 1 / 5 ) u_1\prime - (3 /5 )u_2\prime$

The textbook skips the steps to show how $u_1$ and $u_2$ are rewritten in terms of $u_1 \prime$ and $u_2 \prime$ and I'm confused to how it's possible. Could someone help me with the hidden step and break down how  $u_1$ and $u_2$ are expressed in terms of $u_1 \prime$ and $u_2 \prime$


问:
Which of these statements about vector fields is false?
A: $\iint\limits_{S}\mathbf{f}\cdot\mathbf{g} dS=(\iint\limits_{S}\mathbf{f}\cdot\mathbf{n} dS)(\iint\limits_{S}\mathbf{g}\cdot\mathbf{n} dS)$

B: $\iint\limits_{S}(f+g) dS=\iint\limits_{S}{f} dS +\iint\limits_{S}gdS$.

C: $\iint\limits_{S}(c{f}) dS=c\iint\limits_{S}f dS$ where c is any constant.

D: $\iint\limits_{S\cup S'}{f} dS=\iint\limits_{S}f dS+\iint\limits_{S'}f dS$

What I have deduced so far:

I think B and C are $true$ due to the same addition property of double integrals.

答:
Your intuition for $B$ and $C$ is correct. This is not difficult to prove. Let me use a basic example to convince you that $D$ is true. Consider two disjoint subsets of $\mathbb{R}^3$.  For example, take $S$ to be the unit sphere, and $S'$ to be another unit sphere in a different location in $\mathbb{R}^3$. (Notice these are very informal terms, but this is to give you a general idea of why.) It should make sense to you that if you take the surface integral over both of them ($S \cup S'$), you just do the surface integrals separately. How else would you do it? Just think of it like doing two different surfaces separately.

On the other hand, think about why $A$ cannot be true? Is this true for single integrals? Namely:

$$\int f(x)g(x)dx \neq \int f(x) dx \int g(x)dx$$

It's not even true for derivatives. Therefore, it is not true for double integrals, and thus not true for surface integrals. Try coming up with a simple counterexample. For single integrals, I would disprove the above equation by letting $f(x)=g(x)=x$.

$$\int x \times x dx \neq \int x dx \times \int x dx$$

Try to come up with another very simple example like this and disprove the statement.

Comment with questions.

Cheers.


问:
If $AB = I$ then $BA = I$

If $A$ and $B$ are square matrices such that $AB = I$, where $I$ is the identity matrix, show that $BA = I$.

I do not understand anything more than the following.

Elementary row operations.

Linear dependence.

Row reduced forms and their relations with the original matrix.

If the entries of the matrix are not from a mathematical structure which supports commutativity, what can we say about this problem?

P.S.: Please avoid using the transpose and/or inverse of a matrix.

答:
For a more elementary treatment ...

Fact. If the rows of $A$ are linearly dependent, then the rows of $A B$ are linearly dependent.

Proof of fact. Consider the 3x3 case, where the linearly-dependent rows of $A$ are $\mathbf{a}_1$, $\mathbf{a}_2$, $\mathbf{a}_3 = h \mathbf{a}_1 + k \mathbf{a}_2$ (for some scalars $h$ and $k$):

$$A = \begin{bmatrix}\mathbf{a}_1 \\ \mathbf{a}_2 \\ h\mathbf{a}_1 + k\mathbf{a}_2\end{bmatrix} = \begin{bmatrix}p & q & r \\ s & t & u \\ hp + ks & hq + kt & hr + ku \end{bmatrix}$$

Writing $\mathbf{b}_1$, $\mathbf{b}_2$, and $\mathbf{b}_3$ for the rows of $B$, we have

$$A B = \begin{bmatrix}p & q & r \\ s & t & u \\ hp + ks & hq + kt & hr + ku \end{bmatrix} \begin{bmatrix}\mathbf{b}_1 \\ \mathbf{b}_2 \\ \mathbf{b}_3\end{bmatrix} = \begin{bmatrix}p
\mathbf{b}_1+ q\mathbf{b}_2 + r\mathbf{b}_3 \\ s\mathbf{b}_1 + t\mathbf{b}_2 + u\mathbf{b}_3 \\ (hp + ks)\mathbf{b}_1 + (hq + kt)\mathbf{b}_2 + (hr + ku)\mathbf{b}_3 \end{bmatrix}$$

$$= \begin{bmatrix}p
\mathbf{b}_1+ q\mathbf{b}_2 + r\mathbf{b}_3 \\ s\mathbf{b}_1 + t\mathbf{b}_2 + u\mathbf{b}_3 \\ h(p\mathbf{b_1}+q\mathbf{b}_2+r\mathbf{b}_3) + k(s\mathbf{b}_1 + t\mathbf{b}_2 + u\mathbf{b}_3) \end{bmatrix}$$

Generally, the linear dependence of the rows of $A$ carries over to the rows of the product, proving our Fact. (This reasoning actually shows the more-precise Fact that $rank(AB)\le rank(A)$.)

We can restate the Fact this way:

Re-Fact. If the rows of $AB$ are linearly independent, then the rows of $A$ are linearly independent.

To your question: If $A B = I$, then (by the Re-Fact) the rows of $A$ must be linearly independent. This implies that $A$ can be row-reduced to a diagonal matrix with no zero entries on that diagonal: the row-reduced form of $A$ must be the Identity matrix.

Note that row-reduction is actually an application of matrix multiplication. (You can see this in the equations above, where (left-)multiplying $B$ by $A$ combined the rows of $B$ according to the entries in the rows of $A$.) This means that, if $R$ is the result of some row combinations of $A$, then there exists a matrix $C$ that "performed" the combinations:

$$C A = R$$

If (as in the case of your problem) we have determined that $A$ can be row-reduced all the way down to the Identity matrix, then the corresponding $C$ matrix must be a (the) left-inverse of $A$:

$$C A = I$$

It's then straightforward to show that left and right inverses of $A$ must match. This has been shown in other answers, but for completeness ...

$$A B = I \;\; \to \;\; C (A B) = C \;\; \to \;\; (C A) B = C \;\; \to \;\; I B = C \;\; \to \;\; B = C$$

Once you start thinking (ahem) "outside the box (of numbers)" to interpret matrices as linear transformations of vectors and such, you can interpret this result in terms of mapping kernels and injectivity-vs-surjectivity and all the kinds of sophisticated things other answers are suggesting. Nevertheless, it's worth noting that this problem is solvable within the realm of matrix multiplication, plain and simple.

Title: If $AB = I$ then $BA = I$

ContentLicense: CC BY-SA 2.5


问:
Numbers which are their own inverse are exactly $1$ and $p − 1$?
We begin by listing all the positive integers from $1$ to $p − 1$:
$$1, 2, 3, . . . , p − 1. $$

$p$ is a prime. The numbers on the list that are their own inverse are exactly $1$ and $p − 1$.

For $x, 1\leq x \leq p − 1, x$ is its own inverse if and only if $x^2 ≡ 1 \pmod p$. Because $p$ is prime, this can happen only if $p$ divides either $x − 1$ or $x + 1$.

Then the author writes -

Thus, the numbers on the list that are their own inverse are exactly $1$ and $p − 1$.

I just couldn't see through the last line, how from  $p$ divides either $x − 1$ or $x + 1$ we deduce  the numbers which are their own inverse are exactly $1$ and $p − 1$?

答:
Conceptually this is an immediate consequence of the uniqueness of the remainder, i.e.

$$ x \equiv r\!\!\!\pmod{\! n} \ \ {\rm and} \ \ 0\le x,r < n\,\Longrightarrow\, x = r$$

Your special case is $\,r = 1\,$ and $\,r = n\!-\!1\ [\:\!\equiv -1\pmod{\!n}\:\!]$

Remark $ $ Any complete system of residues comprises a set of normal forms ("remainders") for $\,\Bbb  Z/n = \Bbb Z\bmod n = $ integers $\!\bmod n,\,$ i.e. every integer is congruent to exactly one of them.

The above displayed inference implies congruence is equivalent to equality on normal forms, i.e.

$$ r'\equiv r\!\!\!\pmod{n}\iff    r'\bmod n\,=\, r\bmod n$$

where here $\,k\bmod n\,$ denotes the normal form from the chosen complete set of representatives. This is the key property of normal forms.

It is important to understand this conceptual viewpoint since else it will prove very cumbersome to rediscover or reprove these results every time they are (frequently) used in number theory and algebra. While here the proof is not too difficult, it may be much more complex in other cases (e.g. proving that two fractions are equivalent iff they have the same reduced form).


问:
$t^2 + t$ is a polynomial, then what is $t^{-2} + t^{-1}$ called?
$t^2 + t$ is a polynomial.

What is

$t^{-2} + t^{-1}$ called?

More generally,

$t^{-n} + t^{-(n-1)} + \cdots + t^{-2} + t^{-1} + c$.

答: Let $k$ be a field. The linear map $\overline{\cdot}: k[t, t^{-1}] \to k[t, t^{-1}]$ which sends $t$ to $t^{-1}$ is often called the "bar involution'' in certain subjects in algebra (for example, Hecke algebras and quantum groups). So $t^{-2} + t^{-1}$ could be called "the image of $t^2 + t$ under the bar involution", i.e., $\overline{t^2 + t} = t^{-2} + t^{-1}$. Of course, as one of the commenters has noted, the image under the bar involution is not the multiplicative inverse of the polynomial.


问:
the cosine between two vectors represented by the XYZ component vectors.
the original text is:

Let $n$ be any direction in the space, The projection of vector $A$ on
the $n$ direction is :  $A_n = |A| cos (n,A)$.\ or, notice the
equation we have already know in the analytical geometry that the
cosine of the angle between two vectors is :\ $A_n =| A|
[cos(n,X)cos(A,X)+ cos(n,Y)cos(A,Y)+cos(n,Z)cos(A,Z)] = A_x cos(n,X) +A_ycos(n,Y)+A_z cos(n,Z)$

OK,I know I am very stupid, but I just can't understand how he get

$cos(n,A)=cos(n,X)cos(A,X)+ cos(n,Y)cos(A,Y)+cos(n,Z)cos(A,Z)$

I have tried to search this on wikipedia related to the vector topic, but gain nothing basically.

Thank you!

答: Your term is an awkward way of writing out $n_x \cdot A_x + n_y \cdot A_y + n_z \cdot A_z$, that is, a simple dot product ${\bf n} \cdot {\bf A}$.


问:
How to calculate probability of this random variable if all is known is the mean and variance?
Suppose a light bulb turns on with a certain probability. The number of light bulbs that turn on per hour is a random variable with mean 1000 and variance 1000. What is the probability that more than 1200 light bulbs will turn on in an hour?

Kind of stuck with this question as all is known is the mean and variance. Nothing about the distribution is known, but $E(X) = \operatorname{Var}(X)$ suggests that $X\sim \operatorname{Poisson}(1000)$. Am I allowed to jump to this assumption though?

答:
Poisson distribution? I agree with @user1, that it's reasonable to assume the number of lightbulbs turning on in an hour is Poisson. It is typical to talk of Poisson events as taking place with a certain average number within a particular period of time.

Also, Poisson distributions are among those that have mean and variance equal. It would add to the strength of this supposition if you have recently covered Poisson distributions. So I'd say to go ahead and "jump."

Desired probability. So consider $X \sim \mathsf{Pois}(\lambda = 1000).$ Then you seek $P(X > 1200) = 1 - P(X \le 1200) \approx 0.$

Normal approximation: With such a large mean as $\lambda = 1000,$ one can use the normal approximation to Poisson distributions to find this
probability in terms of a normal distribution with $\mu = 1000, \sigma = \sqrt{1000}= 31.62278.$

So you can get the answer by standardizing and using printed normal tables.
Because $(1200 - 1000)/31,62278 \approx 6.32$ standard deviations above the mean, you can guess that the probability is essentially $0.$

Software: You can also use statistical computer software or a statistical calculator to find the exact Poisson probability. In R, where ppois is a Poisson CDF, the
computation is as shown below:

```<code>1 - ppois(1200, 1000)
[1] 3.884939e-10
</code>```

In terms of the normal approximation, R gives nearly $0$ again:

```<code>1 - pnorm(1200, 1000, sqrt(1000))
[1] 1.269814e-10
</code>```

If you do this by standardizing and using normal tables,
you will see that z-scores above about 3.5 are off the table, so you would have to understand that a z-score above $6$ corresponds to an answer near $0$ (which might be one point of this exercise).

Below is a plot of the Poisson probabilities (with values between 860 and 1200) along with the density function of he approximating normal distribution.

```<code>x = 860:1200; pdf = dpois(x, 1000)
plot(x, pdf, type="h", col="blue")
 abline(h=0, col="green2")
 curve(dnorm(x,1000,sqrt(1000)), add=T, col="brown", lwd=2, lty="dashed")
 abline(v = 1200)
</code>```

Body:
Poisson distribution? I agree with @user1, that it's reasonable to assume the number of lightbulbs turning on in an hour is Poisson. It is typical to talk of Poisson events as taking place with a certain average number within a particular period of time.

Also, Poisson distributions are among those that have mean and variance equal. It would add to the strength of this supposition if you have recently covered Poisson distributions. So I'd say to go ahead and "jump."

Desired probability. So consider $X \sim \mathsf{Pois}(\lambda = 1000).$ Then you seek $P(X > 1200) = 1 - P(X \le 1200) \approx 0.$

Normal approximation: With such a large mean as $\lambda = 1000,$ one can use the normal approximation to Poisson distributions to find this
probability in terms of a normal distribution with $\mu = 1000, \sigma = \sqrt{1000}= 31.62278.$

So you can get the answer by standardizing and using printed normal tables.
Because $(1200 - 1000)/31,62278 \approx 6.32$ standard deviations above the mean, you can guess that the probability is essentially $0.$

Software: You can also use statistical computer software or a statistical calculator to find the exact Poisson probability. In R, where ppois is a Poisson CDF, the
computation is as shown below:

1 - ppois(1200, 1000)
[1] 3.884939e-10

In terms of the normal approximation, R gives nearly $0$ again:

1 - pnorm(1200, 1000, sqrt(1000))
[1] 1.269814e-10

If you do this by standardizing and using normal tables,
you will see that z-scores above about 3.5 are off the table, so you would have to understand that a z-score above $6$ corresponds to an answer near $0$ (which might be one point of this exercise).

Below is a plot of the Poisson probabilities (with values between 860 and 1200) along with the density function of he approximating normal distribution.

x = 860:1200; pdf = dpois(x, 1000)
plot(x, pdf, type="h", col="blue")
 abline(h=0, col="green2")
 curve(dnorm(x,1000,sqrt(1000)), add=T, col="brown", lwd=2, lty="dashed")
 abline(v = 1200)


问:
Uniform probability density function
If I have uniform probability density function:
$\pi(r)=1 \space \forall r\in [0,1] \space and \space 0 \space \forall r \notin [0,1]$
I just don't get it. If I integrate function f(x) from 0 to 0,5 and then from 0 to 0,7 how I get from both integrates 1 like $\pi(r)$ shows? That can't be true unless f(x) is zero between 0,5 and 1.

edit:
I'm thinking that $\pi (r)$ is a new (density)function which comes from integrated f(r). And that's probably wrong thinking because: $\pi (r)$ is always 1 on every X from 0 to 1 and if I integrate f(r) i.e from 0 to 0,5 and from 0 to 0,7 I won't get same result which should be 1..
I know this is a confusing question..

答:
I believe here there is a misunderstanding.

[From Wikipedia ...] "probability distribution function" may be used
  when the probability distribution is defined as a function over
  general sets of values, or it may refer to the cumulative distribution
  function [...]

So here $\pi(r)$ is the probability distribution and not the cumulative distribution function. Therefore it is not the integration from $0$ to $r$ of a $f(r)$.

Now, if you'd like to calculate the cumulative distribution function $\Pi(r)$, by integrating from $-\infty$ and $r$ your $\pi(r)$, you get:

$$
\Pi(r) = \int_{-\infty}^r\!\pi(x)\,\mathrm{d}x=\int_0^r\!\pi(x)\,\mathrm{d}x=\left\{
  \begin{array}{l l}
    0 & \quad r<0\\
    r & \quad 0\le r<1\\
    1 & \quad r\ge 1
  \end{array} \right.
$$

Title: Uniform probability density function


答:
For the sake of completeness, I summarize the comments above:

As pointer out by Torsten and manthanomen, there are specific names (Laurent polynomial) for such functions. However, these are not widely known, so I agree with Brian and Deepak and suggest using a more general concept called rational function:

A function $f(x)$ is called a rational function if and only if it can be written in the form $f(x) = \frac{P(x)}{Q(x)}$ where $P$ and $Q$ are polynomial functions of $x$ and $Q$ is not the zero function.

If you do not want to call it rational function and want to define these functions merely through the regular polynomials then saying $f(t) = P\left(\tfrac{1}{t}\right)$ as suggested by N. S. is the way to go.

Finally, as pointed out by David, $$(t+t^2)^{-1} = \frac{1}{t + t^2} \ne \frac{1 + t}{t^2} = t^{-1} + t^{-2},$$ hence not an inverse number in the usual meaning, and as noticed by symplectomorphic, $f(t) = t + t^2$ as a function does not have an inverse function in the usual meaning because it is not injective: $f(0) = f(-1) = 0$.

Title: $t^2 + t$ is a polynomial, then what is $t^{-2} + t^{-1}$ called?

Body:
$t^2 + t$ is a polynomial.

What is

$t^{-2} + t^{-1}$ called?

More generally,

$t^{-n} + t^{-(n-1)} + \cdots + t^{-2} + t^{-1} + c$.


问:
Prove that if $-2 \leq x_0 \leq 2$, then $-2 \leq 3x_0 - x_0^3 \leq 2$.
Given $-2 \leq x_0 \leq 2$, I have to prove $-2 \leq x_n \leq 2$, where $x_n = 3x_{n - 1} - x_{n - 1}^3$ and $n \in \mathbb{Z}_+$. I thought this problem can be solve by mathematical induction, but I can't prove the initial step, i.e., $n = 1$.

I try this two ways:

1st. If $-2 \leq x_0 \leq 2$, then $-6 \leq 3x_0 \leq 6$ and $-8 \leq x_0^3 \leq 8$. But, this provied the inequality

\begin{equation}
   -14 \leq 3x_0 - x_0^3 \leq 14,
\end{equation}

that its true.

2nd. I try prove it separately, with $0 \leq x_0 \leq 2$ first, but its wrong too.

How can I prove it?

Body:
$$f(x)=3x-x^3 \implies f'(x)=3-3x^2, f'(x)=0 \implies x=0,\pm 1$$
So
$$f_{max}=f(1)=2, f(2)=-2, f(0)=0 \implies -2\le f(x) \le 2$$


问:
vector space without addition operation
Sorry for the strange question.

If you take a vector space, which has addition and multiplication by scalar (e.g. real or complex) and a certain finite dimension, and you remove the addition operation. What do you end up with? Does it have a name?

I know that this is probably a very useless space but the question is mainly about nomenclature.
Perhaps "scalable space".

It looks to me that you end up with set sum of (infinite) one dimensional spaces (rays?) and the elements can be in one of them at time and it has a null element in each, which is the same in all rays.

$1\cdot \mathbf{v} = \mathbf{v} \quad \forall \mathbf{v}$

$0\cdot \mathbf{v} = 0\cdot \mathbf{w} \quad \forall \mathbf{v}, \mathbf{w}$

Not many other interesting properties it seems.

(The reason I ask is because, in a certain computer language I have a "interface" that multiplies values by a scalar-like thing, but there is no summation in general.)

答:
Have a look at the axioms that are satisfied if you were to remove $+: V \times V \to V$, where $V$ is the original vector space in question. We would still have:

$\forall x \in V: 1 \cdot x = x$

$\forall \alpha, \beta \in \mathbb{F}: \forall x \in V:(\alpha \beta)x = \alpha(\beta x)$

That is all you would be left with if you only had scalar multiplication defined on the set $V$. As far as I can see, this sort of structure doesn't have a specific name (probably because it doesn't seem to be extremely interesting). It's really just a field $\mathbb{F}$ that's providing scalars so that the two statements above hold.

In such a space, it wouldn't even make sense to talk about a basis. Remember, every vector in a vector space can be written as some linear combination of its basis. Since addition is crucial to forming linear combinations, there's no sense in which we can talk about such things in the set  above.

In that way, talking about dimensionality would also be pointless. I guess "scalable space" would be a good name for such a structure.

Title: vector space without addition operation


问:
Different approach to finding projection of a line on a plane
We are given a  plane x+y+z=7 (3D rectangular coordinates) and two points namely A(5,-1,4) and B(4,-1,3).we have to find projection of AB on the given plane.

My problem is that while first time i approached this problem, i found out the feet of perpendicular on the plane from both the points and then used distance formula between those feet. But im getting the wrong answer. Is there any lapse in my understanding?

actual answer is sqrt(2/3) but i am getting the answer as 1

the solution given in the book finds the projection by dotting the vector AB with the normal of the plane and ive understood that but still i dont find any problem in my method. Please help me pointing out my mistake

答:
I believe the method you describe should give you the correct answer. Are you sure you've made no mistakes in your calculations?  For the foot of the perpendicular to the plane from point A I get $$P_A=\frac{1}{3}\left(14,-4,11\right)\ ,$$
for the foot of the perpendicular to the plane from point B I get
$$
P_B=\frac{1}{3}(13,-2,10)\ ,
$$
and the distance between $\ P_A\ $ and $\ P_B\ $ is
$$
\frac{\sqrt{(14-13)^2+(-4+2)^2+(11-10)^2}}{3}=\frac{\sqrt{6}}{3}=\sqrt{\frac{2}{3}}\ .
$$


答: Prove that $f(x)+2$ is positive,  and $f(x)-2$ is negative.  Hint: factors

Title: Prove that if $-2 \leq x_0 \leq 2$, then $-2 \leq 3x_0 - x_0^3 \leq 2$.

Body:
Given $-2 \leq x_0 \leq 2$, I have to prove $-2 \leq x_n \leq 2$, where $x_n = 3x_{n - 1} - x_{n - 1}^3$ and $n \in \mathbb{Z}_+$. I thought this problem can be solve by mathematical induction, but I can't prove the initial step, i.e., $n = 1$.

I try this two ways:

1st. If $-2 \leq x_0 \leq 2$, then $-6 \leq 3x_0 \leq 6$ and $-8 \leq x_0^3 \leq 8$. But, this provied the inequality

\begin{equation}
   -14 \leq 3x_0 - x_0^3 \leq 14,
\end{equation}

that its true.

2nd. I try prove it separately, with $0 \leq x_0 \leq 2$ first, but its wrong too.

How can I prove it?


答: In some areas of algebraic combinatorics, the related operation $f(q) \mapsto q^{\mathrm{deg}(f)} f(q^{-1})$ is called  "reversal". Here one usually has $f(0) \neq 0$, making it an involution.


问:
Is there any flaw in applying Maximum Modulus and Minimum modulus theorem?
What additional information can be concluded about the function?

(a)A function analytic in the closed disk:$|z|\leq 4$ with $\inf|f(z)|=5$ on the circle $|z|=4$ and with $f(1)=i$

(b)A function analytic in the closed disk:$|z|\leq 1$ with $\sup|f(z)|=2$ on the circle $|z|=1$ and with $f(0)=-2i$

My Attempt.

(a) $f(z)=i, \forall z:|z|\leq 4$, By Minimum modulus principle.

(b) $f(z)=-2i, \forall z:|z|\leq 1$, By maximum modulus principle.
Am I correct?
Is there any flaw in applying Maximum Modulus and Minimum modulus theorem?

答:
I'm going to put our conversation into an answer that I hope will help for (a).

If you have a statement of the form $P$ implies $Q$.  This is logically equivalent to not $Q$ implies not $P$.

Minimum modulus principle:: If $f$ is analytic on and inside $D$ and $f(z)\neq 0$ for all $z$, then $f$ attains its minimum modulus on the boundary.

P: $f$ is analytic on and inside $D$ and $f(z)\neq 0$ for all $z$.

Q: $f$ attains its minimum modulus on the boundary.

In your problem, $f$ does not attain its minimum modulus on the boundary. We know this since $\inf |f(z)|=5$ on the boundary, but for a point inside, $|f(1)|=|i|=1$.

Hence, you have not Q, which implies not P.

Not P: $f$ is not analytic on and inside $D$ or $f(z)=0$ for some $z$. What can you conclude?


问:
How to understand $f$ is one-to-one in Jech, 'Set theory' Lemma 3.10.?
I saw T. Jech, 'Set theory' Lemma 3.10:

An infinite cardinal $\kappa$ is singular iff there exists a cardinal $\lambda<\kappa$ and a family $\{S_\xi :\xi<\lambda\}$ of subsets of $\kappa$ s.t. $|S_\xi|<\kappa$ for each $\xi<\lambda$, and $\kappa=\bigcup_{\xi<\lambda}S_\xi$. The least cardinal $\lambda$ that satisfies the condition is $\operatorname{cf}\kappa$.

and this book gives proof of this lemma:

Proof:
  (I omitted the front of this proof)

If the conditions holds, let $\lambda<\kappa$ be the least cardinal for which there is a family $\{S_\xi :\xi<\lambda\}$ s.t. $\kappa=\bigcup_{\xi<\lambda}S_\xi$ and $|S_\xi|<\kappa$ for each $\xi<\lambda$. For every $|S_\xi|<\kappa$, let $\beta_\xi$ be the order-type of $\bigcup_{\nu<\xi}S_\nu$. The sequence $\langle \beta_\xi :\xi<\lambda\rangle$ is nondecreasing, and by the minimality of $\lambda$, $\beta_\xi<\kappa$ for all $\xi<\lambda$. We shall show that $\lim_\xi\beta_\xi=\kappa$, thus proving that $\operatorname{cf}\kappa\le\lambda$.

Let $\beta=\lim_\xi\beta_\xi$. There is one-to-one mapping $f$ of $\kappa=\bigcup_{\xi<\lambda}S_\xi$ into $\lambda \times \beta$: If $\alpha\in\kappa$, let $f(\alpha)=(\xi,\gamma)$, where $\xi$ is the least $\xi$ s.t. $\alpha\in S_\xi$ and $\gamma$ is order-type of $S_\xi\cap\alpha$. Since $\lambda<\kappa$ and $|\lambda\times\beta|=|=\lambda\cdot|\beta|$, it follows that $\beta=\kappa$.

But I don't understand why $f$ is one-to-one function. How to understand it? How to prove that $f$ is one-to-one? Thanks in advance.

答:
Assume that $f(\alpha)=(\beta,\xi)=f(\alpha')$. Then $\alpha\in S_\xi$, and $\gamma$ is the order type of $S_\xi\cap \alpha$.  But the same is true for $\alpha'$. </p>

<p>So $S_\xi \cap \alpha$ and
$S_\xi \cap \alpha'$ have the same order type.  But if $\alpha<\alpha'$, the
 order type of $S_\xi \cap \alpha$ would be a proper initial segment of the order type
of $S_\xi \cap \alpha'$, which is impossible (as a proper initial segment of a well-order
is never isomorphic to the whole well-order. </p>

Body:
Assume that $f(\alpha)=(\beta,\xi)=f(\alpha')$. Then $\alpha\in S_\xi$, and $\gamma$ is the order type of $S_\xi\cap \alpha$.  But the same is true for $\alpha'$.

So $S_\xi \cap \alpha$ and
$S_\xi \cap \alpha'$ have the same order type.  But if $\alpha<\alpha'$, the
 order type of $S_\xi \cap \alpha$ would be a proper initial segment of the order type
of $S_\xi \cap \alpha'$, which is impossible (as a proper initial segment of a well-order
is never isomorphic to the whole well-order.


问:
Proving all exponential properties from $b^{x+k}=b^x\cdot b^k$
Let's say that there is a function $f(x)$ which satisfies the following property:
$$f(x+k)=f(x)\space{}\cdot\space{}f(k)$$
In addition, $f(1)=b\gt1$. I am trying to prove the following property without explicitly relying on the fact that $f(x)=b^x$ (circular reasoning) or requiring that $n\in\mathbb{Z}^+$:
$$[f(x)]^n=f(nx)$$
The only way I've been able to prove it is thinking about $nx$ as being $x+x+x+x+\space{}...\space{}$ n times, and then using the assumed property of $f(x)$ to break this up as a product $n$ times which can then be written as an exponent by definition. However, this argument only really makes sense if $n$ is a natural number, but this property should hold for all $n\in\mathbb{R}$. Is there a way of extending or altering this argument so that it is still sensible for any real number value of $n$?

答:
First let examine $f(0)$:

$f(0)=f(0+0)=f(0)f(0)\iff f(0)(f(0)-1)=0$ therefore $f(0)\in\{0,1\}$

If $f(0)=0$ then $f(x)=f(x+0)=f(x)f(0)=0$ and $f$ is the null function, which is not very interesting.

From now on let set $f(0)=1$ and $f(1)=b$.

Let examine $f(n)$:

$f(n+1)=f(n)f(1)=bf(n)=b^2f(n-1)=\cdots=b^{n+1}f(0)=b^{n+1}$

So by induction (base case $f(0)$ and $f(1)$ verified, then $f(n)=b^n,\ \forall n\in\mathbb N$

Let examine $f(-x)$:

$1=f(0)=f(x-x)=f(x)f(-x)\implies f(-x)=\dfrac 1{f(x)}$

In particular $f(-n)=\dfrac 1{f(n)}=\dfrac 1{b^n}=b^{-n}$ and we have extended to all $\mathbb Z$.

Let examine $f(\frac pq)$:

By the same induction used for $f(n)$ we have $f(nx)=f(x)^n$ for $n$ natural, and use $f(-nx)=\frac 1{f(nx)}$ to extend to all integers.

In particular $b=f(1)=f(\frac qq)=f(\frac 1q)^q$ therefore $f(\frac 1q)=b^{\frac 1q}$

And $f(\frac pq)=f(\frac 1q)^p=b^{\frac pq}$.

Note in the same way we have $f(\frac pqx)=f(x)^{\frac pq}$

Now we use continuity of $f$:

If we don't assume continuity we are stuck with the rationals. So since $\mathbb Q$ is dense in $\mathbb R$, we can extend $f$ to the reals and we have $f(x)=b^x$ and the formula $f(xy)=f(x)^y$ by continuity too from the last note.


问:
Understanding why GeoGebra fails to show the correct range of a function over certain domains
I need help. Below are several images from this GeoGebra lesson, and I should understand why GeoGebra has problems showing the correct range of the function for certain domains $[a,b]$ (controlled by "$a$" and "$b$" sliders).

For the domain $[-5,-1.9]$, the plot shows not actual range, but for $[-5, -1.8]$ all seems well. The same issue for $[-5,-0.9]$ and $[-5,-0.8]$ and many times after. I understood that it's somehow connected with the second derivative of the function and "interval arithmetic", but how???

If anybody have any idea or way which can help me to find best answer please help ;-)

Here's a related Desmos project: https://www.desmos.com/calculator/aappxay2yc

答:
The surest way to discover why this example plots the wrong range in some cases is first to ask the person who made this page to explain the method they used in order to compute the range. Then, if you still cannot spot the error, explain it in the question so that someone else might figure it out.

But I will make a guess.
By trying different functions, you can see that the upper and lower ends of the plotted "range" are always a local maximum and local minimum of the function.
I think these values are found by applying some optimization algorithm that finds a local extremum, not necessarily a global extremum, and that is sensitive to the choice of a starting point.
As you change the domain, the starting point changes, and (if the starting point changes enough) so does the local extremum that the algorithm finds.

When the chosen local minimum and maximum are coincidentally the global minimum and maximum, the algorithm appears to be correct, but in other cases it gives wrong answers.

If you want to know exactly why the algorithm chooses wrong answers when it does, I suggest you try to find out what algorithm was used.


问:
Proof for the existence of a limit using the $(\varepsilon, \delta)$ definition
Question: Suppose that $\lim_{z \to z_0}g(z)$ exists and is finite. Assume that $f$ is another function defined on the same domain as $g$. Show that $\lim_{z \to z_0}f(z)$ exists iff $\lim_{z \to z_0}f(z) +g(z)$ exists.

My Proof: I believe it's sufficient to state that if $\lim_{z \to z_0}g(z) = L$ and $\lim_{z \to z_0}f(z) +g(z) = K$, then

\begin{align}
\lim_{z \to z_0}f(z) +g(z) &= K\\
\left[\lim_{z \to z_0}f(z) \right] + \left[\lim_{z \to z_0}g(z) \right] &= K\\
\left[\lim_{z \to z_0}f(z) \right] + L &= K\\
\lim_{z \to z_0}f(z) &= K - L.
\end{align}

How can I verify this result with the epsilon-delta definition of a limit? My attempt so far is as follows:

Since we are given that $\lim_{z \to z_0}g(z) = L$, we know that $\forall \varepsilon, \exists\delta $ such that $|g(z) - L| < \varepsilon$ and $0<|z-z_0|<\delta$. Similarly, we know that $\lim_{z \to z_0}f(z) +g(z) = K$, and so $\forall \varepsilon', \exists\delta'$ such that $|f(z) + g(z) - K| < \varepsilon'$ and $0<|z-z_0|<\delta'$. If the limit does exist, then $\forall \varepsilon'', \exists\delta''$, we must show that $|f(z) - (K-L)| < \varepsilon''$ and $0<|z-z_0|<\delta''$.

I believe this setup is correct, however I'm unsure where to go from here. Any help or guidance would be greatly appreciated!

答:
Suppose that $\lim_{z \to z_0} (f(z)+g(z)) = K$ and $\lim_{z \to z_0} g(z) = L$. Now, we want to show that:

$$\lim_{z \to z_0} f(z) = K-L$$

Let $\epsilon > 0$ be given. Then, we want a $\delta  > 0$ such that:

$$0 < |z-z_0| < \delta \implies |f(z)-(K-L)| < \epsilon$$

Then, we can see that:

$$|f(z)-(K-L)| = |(f(z)+g(z)-K)+(L-g(z))| \leq |(f(z)+g(z))-K| + |L-g(z)|$$

We know that:

$$\exists \delta_1 > 0: 0 < |z-z_0| < \delta_1 \implies |(f(z)+g(z))-K| < \frac{1}{2}\epsilon$$

$$\exists \delta_2 > 0: 0 < |z-z_0| < \delta_2 \implies |g(z)-L| < \frac{1}{2} \epsilon$$

Define $\delta = \max\{\delta_1,\delta_2\}$. Then:

$$0 < |z-z_0| < \delta \implies |f(z)-(K-L)| \leq  |(f(z)+g(z))-K| + |g(z)-L| < \epsilon$$

which proves the desired result. I will leave you to try and prove the forward conditional on your own. $\Box$

I hope that makes sense.


问:
evaluation of chi-squared density
I am trying to work out the cdf of the chi-squared distribution.
My query is whether it is possible to evaluate this integral:

$$\int_0^y t^{n/2-1} e^{-t/2} \mathrm{d}t$$

答:
Make $t=2x$
$$\int_0^y t^{\frac n2-1} e^{-\frac t2} \, dt=2^{\frac n2}\int_0^{\frac y 2} x^{\frac{n}{2}-1}e^{-x}\,dx$$  and use the gamma function.

$$\int_0^y t^{\frac n2-1} e^{-\frac t2} \, dt=2^{\frac n2} \left(\Gamma \left(\frac{n}{2},0\right)-\Gamma
   \left(\frac{n}{2},\frac{y}{2}\right)\right)$$

Title: evaluation of chi-squared density


问:
partial fraction expansion with As+B and irreducible quadratic
My question is at the bottom.

In an electrical engineering text on Laplace methods for solving electrical circuits a section on discussing p.f.e. the author says that when the denominator has an irreducible quadratic (does not factorize without imaginary terms), the partial fraction expansion should have a term with As+B for numerator as below,

$$ \frac{n(s)}{q(s)(s^2+cs+d)} = \frac{As+B}{s^2+cs+d} + ...$$

I solved a circuit problem (see this question) that had an irreducible quadratic in the denominator of the resulting current, i(s),
$$ i(s) = \frac{264.15}{s^2+37.74s+37735.85}$$

However, I worked it with simple, complex roots as,
$$i(s)=\frac{264.15}{s^2+37.74s+37735.85}=\frac{A}{s+α+jβ}+\frac{B}{s+α-jβ}    $$

With result of,
$$i(s)=\frac{j0.683}{s+18.87+j193.24}+\frac{-j0.683}{s+18.87-j193.24}    $$
My solution after inverse Laplace being,
$$ i(t)=1.366e^{-18.87t} sin(193.24t)$$

After verifying my analytic solution was correct (verified with numerical simulation), I wondered how it would have turned out if i had used the book recommended method with As+B approach.

$$i(s)=\frac{264.15}{s^2+37.74s+37735.85}=\frac{As+B}{s^2+37.74s+37735.85}    $$

But, when i eliminate the fraction and equate coefficients i get,
$$A = 0$$
$$B = 264.15$$

Which gets me nowhere.  I've read several good pages on this method and i think i'm doing it properly.

My question:  Can this "As+B" method be used to solve this problem?  If so, where am i going off track.

答: When $q(s)$ is an irreducible quadratic, the term $\frac{As+B}{q(s)}$ is considered in final partial-fraction form and is meant to be integrated directly. One can do so using a combination of $\int q'(s)/q(s)\,ds = \log|q(s)|$ (which handles the linear term in the numerator, perhaps at the cost of changing the constant term) and then using a change of variables to transform $\int B/q(s)\,ds$ into $\int B'/(u^2+1)\,du$, which can be evaluated in terms of $\arctan u$.


问:
Show $\|\cdot\|$ is a norm of the space of $C^1[0,1]$
Let $C^1([0,1])$ be the space of all functions having continuous derivatives. For each $f \in C^1([0,1])$, set

$$\|f\| = \left(\int_{0}^{1} (|f|^2 + |f'|^2) dx \right)^{1/2}$$

Show that $\|\cdot\|$ is a norm of the space of $C^1([0,1])$.

I mainly want to know to answer this question do I simply need to show the following properties: positivity, non-degenerate, homogeneity & the triangle inequality?

答:
It suffices to remark that
$$<f,g>=\int_0^1(fg+f'g')$$
is an inner product on $C^1([0,1])$.</p>


答:
There is an elementary way to prove this using AM-GM. If $x^2<3$, then

$$
\left( x(3-x^2) \right)^2
=\frac{1}{2}  2x^2 \cdot (3-x^2) \cdot (3-x^2)\\
\leqslant \frac{1}{2} \left(\frac{2x^2+3-x^2+3-x^2}{3} \right)^3 =4.
$$

If $x^2\ge 3$, then
$$
\left( x(3-x^2) \right)^2
=x^2 \cdot (x^2-3)^2  \leqslant 4\cdot 1^2=4.
$$


问:
A puzzling point in proof of Eisenstein Criterion for irreducible polynomials on Integral Domain
Theorem (Eisenstein Criterion).
Let $P$ be a prime ideal of the integral domain
$R$ and let $f(x) = x^n +a_{n-1}x^n+\cdots+a_1 x +a_0$ be a polynomial in $R [x]$ (here $n \geq 1$).
Suppose $a_{n-1},\dots, a_1, a_0$ are all elements of $P$ and suppose $a_0$ is not an element of $P^2$. Then $f(x)$ is irreducible in $R[x]$.

Proof. Suppose $f(x)$ were reducible, say $f(x) = a(x)b(x)$ in $R[x]$, where $a(x)$
and $b(x)$ are nonconstant polynomials. Reducing this equation modulo $P$ and using
the assumptions on the coefficients of $f(x)$ we obtain the equation $x^n = \overline{a(x)b(x)} $ in $(R/ P)[x]$, where the bar denotes the polynomials with coefficients reduced mod $P$.
Since $P$ is a prime ideal, $R/P$ is an integral domain, and it follows that both $\overline{a(x)}$ and
$\overline{b(x)}$ have $0$ constant term, i.e., the constant terms of both $a (x)$ and $b(x)$ are elements
of $P$ . But then the constant term $a_0$ of $f(x)$ as the product of these two would be an
element of $P^2$, a contradiction. $\Box$

This proof is from Abstract Algebra by Dummit & Foote. The puzzling point is bold-italic in the proof. Why it is not possible that only one constant of $\overline{a(x)}, \overline{b(x)}$ is $0$ and the other is not? I don't think it violates the rule of integral domain in this case.

Could any one give me some idea ? Thank you in advance!

答:
Match coefficients. You obtain a system of equations where if $\deg a = i$ and $\deg b = j$ then $i+j = n$, and, denoting their coefficients by $\alpha_s$ and $\beta_t$, you have $\alpha_i \beta_j = 1$ and then intermediary sums of cross terms, all of which must be $0$. Then the proof as in Dummit and Foote resumes - if these are zero, they are elements of $P$. Ideal arithmetic says that such products are elements of $P^2$, obtaining the contradiction.

Edit: I did not see your more refined question at the end. Suppose that $a(x)$ has $0$ constant term, and it's $b(x)$ that doesn't. Then $a \mod P$ would have to be identically $0$. If not, you could look at its lowest nonzero coefficient and this would eventually be multiplied by the constant term in $b \mod P$. The result is a nonzero monomial with $\deg < n$ in the product $ab \mod P$, since $P$ is a prime ideal. But that means that the whole polynomial $a(x)$ is in $P$, and therefore the whole polynomial is in $P$ since ideals are closed under multiplication. But $f$ is monic, so this is impossible.


问:
expected value of permutation
Suppose you have a set X with 2 or greater distinct values and x1,x2...xn are permutations of set X. For i=2,3,...n the position i in the permutation is a step if xi−1<xi. Position 1 is also considered a step. What is the expected value of steps in random permutation of X?

Please explain if possible.

So far I solved by brute force so when n=2 I got 3 steps. So probability 3/2

For n=3 I got 12/6 and so on.

I noticed the denominator is changing by n! but I cannot find the correlation in the numerator.

答:
For $n=1,2,3$, and $4$ one gets $\frac11,\frac32,\frac{12}6$, and $\frac{60}{24}$, respectively. The ratios between consecutive values are $\frac32,\frac43$, and $\frac54$. We have pretty minimal data, but there is a pattern there: it suggests that there are altogether $\frac{(n+1)!}2$ steps in the $n!$ permutations of $[n]$. One might reasonably try to prove this by induction on $n$.

SKETCH: Suppose that $n>1$, and there are $\frac{n!}2$ steps in the $(n-1)!$ permutations of $[n-1]$; then there are

$$\begin{align*}
(n-1)(n-1)!-\frac{n!}2&=\left((n-1)-\frac{n}2\right)(n-1)!\\
&=\left(\frac{n}2-1\right)(n-1)!
\end{align*}$$

non-steps in those permutations, and the same number of non-steps in the permutations of $\{2,3,\ldots,n\}$.

Each permutation of $[n]$ can be obtained uniquely by inserting a $1$ into a permutation of $\{2,\ldots,n\}$. Show that inserting the $1$ adds a step if and only if we insert it at the front or before a non-step, and use this to complete the induction step by showing that there are $\frac{(n+1)!}2$ steps in the $n!$ permutations of $[n]$.

Title: expected value of permutation


问:
Show that $2x^2 + 4$ is reducible in $\Bbb C$
Show that $2x^2 + 4$ is reducible in $\Bbb C$.

By the definition of irreducible, $2x^2 + 4$ is reducible if I can find $f(x)$ and $g(x)$ such that $2x^2 + 4 = f(x)g(x)$ and $f(x)$ and $g(x)$ are both not units.  But I can't seem to find a non-unit product of $2x^2 + 4$ in $\Bbb C$.

答:
polynomial over C[x] does not have a unit when x is put to be roots...
As 2(x-i√2)(x+i√2) for x to be +-i√2 factors are not unit over C[x]


问:
Check the continuity and differentiability of $f(x)= \sin^{-1}(\cos x)$ at $x=0$

Check the continuity and differentiability of  $f(x)= \sin^{-1}(\cos x)$  at $x=0$

This is how I tried to solve the problem:

$$f(x)= \sin^{-1}(\cos x)=???$$

$$\lim_{x \to 0^+}f(x)=\lim_{x \to 0^-}f(x)=f(0)=???$$
Therefore the function is continuous at $x=0$
$$\lim_{x \to 0^+}\frac{f(x)-f(0)}{x-0}=???$$and
$$\lim_{x \to 0^-}\frac{f(x)-f(0)}{-x}=???$$

答:
EDIT: with some precisions following Erick's comment :)

I suppose $x\in[0,\pi]$ so $\cos$ is bijective to $[-1,1]$.
If you use $\cos x = \sin\left(\frac{\pi}{2}-x\right)$, then $\left(\frac{\pi}{2}-x\right)\in[-\frac{\pi}{2},\frac{\pi}{2}]$ and you can safely apply $\sin^{-1}$:
$$
f(x) = \frac{\pi}{2}-x
$$
which gives us $f(0^+)=\frac{\pi}{2}$ and
$
\frac{f(x)-f(0^+)}{x} = -1
$.

Now, to get the same with $0^-$, let's assume $x\in[-\pi,0]$. This time $\left(\frac{\pi}{2}-x\right)\in[\frac{\pi}{2},\frac{3\pi}{2}]$ and given that $f(x)=f(-x)$ as a result of being composed with $\cos$, we have:
$$
f(x) = \frac{\pi}{2}+x
$$
which gives us $f(0^-)=\frac{\pi}{2}$ and
$
\frac{f(x)-f(0^-)}{x} = 1
$.

Therefore $f$ is continuous, but not differentiable at $0$ :)


问:
Prove that the recursively defined sequence is Cauchy.
Define a sequence recursively by $x_1=1$, $x_{n+1}=\frac{5+5x_n}{5+x_n}$. Prove that this sequence is Cauchy. Then find the limit.

We call a sequence $\{a_n\}$ Cauchy, if for any $\epsilon>0$, there exists $N\in\mathbb{N}$ such that for all $n,m\geq M$, $|a_n-a_m|<\epsilon$.

答:
$$
x_{n+1} - \sqrt 5 = \frac{5+5x_n}{5+x_n} - \sqrt{5} = \frac{(5-\sqrt{5})(x_n-\sqrt 5)}{5+x_n}
$$

$$
x_{n+1} + \sqrt 5 = \frac{5+5x_n}{5+x_n} + \sqrt{5} = \frac{(5+\sqrt{5})(x_n+\sqrt 5)}{5+x_n}
$$

Therefore
$$
\frac{x_{n+1}-\sqrt 5}{x_{n+1}+\sqrt 5}=\frac{5-\sqrt 5}{5+\sqrt 5} \cdot \frac{x_n-\sqrt 5}{x_n+\sqrt 5}\\
\Rightarrow \frac{x_n-\sqrt 5}{x_n+\sqrt 5}=\left(\frac{5-\sqrt 5}{5+\sqrt 5}\right)^{n-1} \cdot \frac{1-\sqrt{5}}{1+\sqrt{5}}
$$

then you can get a closed form solution for $x_n$ and the rest should be straightforward.

Please check this post: Recursive sequence depending on the parameter

I just learned about Möbius transformation the other day.


问:
How do you find the number of contestants in a tv show based on their eye colors?
The problem is as follows:

To participate in an edition of The Price is Right $60$ people
showed up. The producer of the show collected the following
information:

$10$ women have black eyes $16$ women don't have black eyes $14$ women
do not have brown eyes $10$ men do not have brown or black eyes

Based on information from the show's producer. How many men have brown
or black eyes?

The alternatives in my book are as follows:

$\begin{array}{ll}
1.&\textrm{24 men}\\
2.&\textrm{22 men}\\
3.&\textrm{18 men}\\
4.&\textrm{28 men}\\
5.&\textrm{20 men}\\
\end{array}$

What I attempted to do is summarized in the table from below which showcases my findings. I must indicate that I arranged the characters of the story in two groups. One for women and the other for men. One column indicates $F$ for false of those who do not have a certain characteristic, i.e. A man who does not have black eyes would be in the F column below black eyes, a woman who has black eyes will lie below T in the black eyes column.

Hence this is represented as:

Where $x+y=10$

Adding all of these yields:

$14+10+16 +x+y+a+b=60$

Since it mentions $x+y=10$ and they want to know what is a+b then this would mean:

$a+b=60-50=10$

However this is not within the alternatives.

In my effort to better represent this situation and avoid counting twice I made the diagram from below:

This different representation using an Euler diagram shows what I thought, the set of those who do not have brown eyes must account for those who have black eyes.

There is no interesection in any of these sets as there cannot be someone who has brown and black eyes at the same time. Thus I concluded what they want is the number of men who have brown eyes and those who have black eyes.

This part I require clarification, what is the meaning of OR translated in set theory language?. Does it mean is it a Union?. Is it $\cup$?. Can someone help me with this part?.

Returning to my attempt I then thought that the number $14$ of those who do not have brown eyes will include those who have black eyes therefore.

Adding those who have black eyes with those who have brown eyes with those who dont have brown neither black in the women side. With those on the men side who dont have neither black or brown with those who have brown and those who have black will equate to $60$. This translated into an equation would be as this:

14+16+10+u=60

Assuming $u=\textrm{unknown males with brown with those with black eyes}$

This would become into:

$u=20\,men$

But this is not the right answer. Why?.

Did I made a mistake or something?. Because of this reason I need a very detailed help with this.

Please an answer must include a Venn or Euler diagram strategy or using a table. To better aid the understanding of this problem. I have tried all sorts, but I don't seem to get the right answer.

Can someone include also the representation of set notation using algebraic notation accompanied with an Euler or Table diagram?.

答:
You are given:

10 women have black eyes

16 women don't have black eyes

14 women do not have brown eyes

10 men do not have brown or black eyes

Assuming no one have heterochromia, each person only has one single eye color.

Hence either they have black eyes, or they do not.

From the first two conditions we see that there are $26$ women. Thus there are $34$ men.

Now $10$ men do not have brown or black eyes, and we are to find the number of men with brown or black eyes, which is the complement of those $10$ men, so there are $24$ of them.

The major mistake in your answer is the equation

$$14+10+16+x+y+a+b=60$$

Since the $14$ women without brown eyes and those $16$ women without black eyes are not mutually exclusive, you can't just add them up; you will overcount.

To answer your side question, OR translates to the union $\cup$ and AND translates to the intersection $\cap$. However these do not come into play in this specific question.


问:
Discrete Mathematics > The Logic of Quantified Statements > Predicates and Quantified Statements > Negation of A Universal Conditional Statement
There are various questions in this topic, but none were covering my particular question.
Can you please help me with the following: I have a Universal Conditional Statement (Universal Implication). Please note that it is NOT a simple straight forward Conditional (Implication) Statement. It has a UNIVERSAL QUANTIFIER:

$\forall , ,  \in \Bbb Z$,  $(−)$    $(−)$   ℎ $(−)$  .

Now I am looking to find the negation of the "Universal Conditional Statement.
If I rewrite the question:

Originally, we have:

$P(x) = (−)$    $(−)$   = $R(x) \wedge S(x)$

$R(x) = (−)$

$S(x) = (−)$

$Q(x) = (−)$

which is

$\forall , ,  \in \Bbb Z$,  $P(x)$ ℎ $Q(x)$

or in another word:

$\forall , ,  \in \Bbb Z$,  $R(x) \wedge S(x)$ ℎ $Q(x)$

Will the negation of the above 'Universal Conditional Statement' be:

$\exists , ,  \in \Bbb Z$,  $P(x)$ ℎ $Q(x)$

$\exists , ,  \in \Bbb Z$,  $(−)$    $(−)$   ℎ $(−)$  not e.

This is as per page $111$ of Discrete Mathematics from Susan Epp $4$th edition (or page $125$ of the same book $5$th edition)

or will it be:

$\exists , ,  \in \Bbb Z, P(x)$ and $\neg Q(x)$?

$\exists , ,  \in \Bbb Z, (−)$    $(−)$   and $(−)$  not e.

答: The negation of $\forall m(T(m))$ is $\exists m(\neg T(m))$. Here $m$ represents the ordered triplet $(a,b,c)\in\Bbb Z^3$ and $T(m)$ is the statement $P(a,b)\implies Q(a,b)$. The negation of $T(m)$ is $P(a,b)\wedge\neg Q(a,b)$ (read this post!) so the required negation is$$\exists m(\neg T(m))\iff \exists a,b,c\in\Bbb Z,(a-b)\text{ is even and }(b-c)\text{ is even and }(a-c)\text{ is not even.}$$i.e. your second statement.