问:
Check my proof of "overspill" in non-standard models of Peano arithmetic.
Proposition

Let $\mathcal{M}$ be a nonstandard model of Peano arithmetic, $\phi(v,\bar{w})$ a formula in the language of arithmetic, and $\bar{a} \in \mathbb{M}$. Show that if $\mathcal{M} \models \phi(n,\bar{a})$ for all $n < \omega$, then there is an infinite $c \in \mathbb{M}$ s.t. $\mathcal{M} \models \phi(c,\bar{a})$.

My Proof

Let $\mathcal{N}$ be the standard model of Peano arithmetic, and $\pmb{PA}$ the theory of Peano arithmetic so that $\mathcal{N} \models \pmb{PA}$; additionally, we know by definition $\mathcal{M} \models \pmb{PA}$. Suppose for the sake of contradiction that there exists a nonstandard model $\mathcal{M}$ so that $\mathcal{M} \models \phi(n,\bar{a})$ but $\mathcal{M} \not\models \phi(c,\bar{a})$. Since $\mathcal{M} \models \pmb{PA}$, $\mathcal{M} \models \phi(0,\bar{a})$ and by the assumption that $\mathcal{M} \models \phi(n,\bar{a})$ for all $n < \omega$, we know $\mathcal{M} \models \forall x (\phi(x,\bar{a}) \Rightarrow \phi(x + 1, \bar{a}))$. Hence by the induction axiom of $\pmb{PA}$:

$$Ind(\phi) := (\phi(0) \wedge \forall x(\phi(x) \Rightarrow (x+1))) \Rightarrow \forall x \phi(x),$$

we know $\mathcal{M} \models \forall x (\phi(x,\bar{v}))$, therefore $\mathcal{M} \models \phi(c,\bar{a})$. But this contradicts our hypothesis and we conclude:
$$\mathcal{M} \models \phi(c,\bar{a}).$$

My Problem

My primary concern is whether I can use the induction principle of Peano arithmetic to argue that I can "reach" this $c$ by induction. Since as I understand, this $c$ lies beyond $\omega$.

Additionally, is there another way to prove the proposition without invoking the details of Peano Arithmetic?

答:
To me, Rick's proof is correct, but does not give enough details. Here is a complete proof. Let's assume we have a formula $\psi$ such that : $$\forall n < \omega, \mathcal{M} \vDash \psi(n)$$
Let's show that there exists a non standard integer satisfying $\psi$. By contradiction, we assume that all non standard integers do not satisfy $\psi$.

We have $\mathcal{M} \vDash \psi (0)$. For $n < \omega$, we have $\mathcal{M} \vDash \psi(n+1)$, so we also have $\mathcal{M} \vDash \psi(n) \rightarrow \psi(n+1)$

By hypothesis, for a non standard integer $b$ we have $\mathcal{M}\nvDash \psi(b)$, so, exfalso, we have $\mathcal{M} \vDash \psi(b) \rightarrow \psi(b+1)$ (the left side of implication is false).

By induction hypothesis, that is true in $\mathcal{M}$ as it is a model of Peano's arithmetic, we have $\forall x, \mathcal{M} \vDash \psi(x)$, which contradicts our hypothesis. So there exists at least one non standard integer satisfying $\psi$.

One can easily extend this proof to show that once we have $\mathcal{M} \vDash \psi(a)$ for some-non standard integer $a$, then we have $\forall b < a, \mathcal{M} \vDash \psi(b)$.

Have a good day !


问:
Evaluate the I?
This  is question is inspired from this

find the value   of $$I=\int_{0}^{2\pi} \int_{0}^{\sqrt 2 a}\sqrt{ \frac{u^4}{a^2} +u^2} du dv$$

My attempt : $$I=\int_{0}^{2\pi} \int_{0}^{\sqrt 2 a}u\sqrt{ \frac{u^2}{a^2} +1} du dv$$

after that im not able to solved this

答:
Before I worked out a complete solution, @TitoEliatron has left a key step in his comment.

\begin{align}
I&=\int_{0}^{2\pi} \int_{0}^{\sqrt 2 a}u\sqrt{ \frac{u^2}{a^2} +1} \,du \, dv\\
&=\int_0^{2\pi}dv \; \frac12 \int_0^{\sqrt2 a} \sqrt{ \frac{u^2}{a^2} +1} \, d(u^2) \\
&= \pi \int_0^{2a^2} \sqrt{x/a^2 + 1} \, dx \tag{$x = u^2$} \\
&= \frac{\pi}{a} \int_0^{2a^2} \sqrt{x + a^2} \, dx \\
&= \frac{2\pi}{3a} \left[ (x+a^2)^{3/2} \right]_0^{2a^2} \\
&= \frac{2\pi (3\sqrt3 - 1) a^2}{3}
\end{align}

Edit : The $a^3$ in the numerator cancelled with $a$ in the denominator to give $a^2$.

ContentLicense: CC BY-SA 4.0

Title: Evaluate the I?


问:
What is the cardinality of $O(f(x))$?
I was wondering what the cardinality of

$$O(f(x))=\left\{g:\mathbb R\to\mathbb R\left|\lim_{x\to\infty}\sup\frac{|g(x)|}{f(x)}<\infty
\right.\right\}$$
is.

I know that if $g(x)=cf(x)$ for some $c\in\mathbb R$ then $g(x)\in O(f(x))$, hence $\{c\in\mathbb R|cf(x)\}\subset O(f(x))$. As $|\mathbb R|=\beth_1$, this implies $|\{c\in\mathbb R|cf(x)\}|=\beth_1$. At the same time, $O(f(x))\subset\{g:\mathbb R\to\mathbb R\}$ and $|\{g:\mathbb R\to\mathbb R\}|=\beth_2$.

The implication is that

$$\beth_1\leq|O(f(x))|\leq\beth_2$$

However, I'm not sure whether $|O(f(x))|=\beth_1$ or $|O(f(x))|=\beth_2$. Could anybody figure out what the cardinality of $O(f(x))$ is?

答:
Take any subset $A\subset [0,1]$ and define $f_A(x):=\chi_A(x)+f(x)\chi_{[0,1]^c}(x)$.

Since $A\subset[0,1]$, it is clear taht $f_A\in O(f(x))$ and $|\{f_A:\, A\subset [0,1]\}=|\cal P[0,1]|=\beth_2$

For the sake of completenes $$\chi_A(x)=\begin{cases}1&x\in A\\0&x\notin A\end{cases}$$

Title: What is the cardinality of $O(f(x))$?


问:
$x = -y^3/8$ and $ y= -2x^3$ Systems of equations
Algebra has been a while and I forgot how to do this.

Say we have the following 2 properties, how would we be able to use the system of equations?

$x = -y^3/8$ and $ y= -2x^3$

The result is supposed to be $(0,0), (-1, 2), (1, -2)$ but I'm not getting there when I try to substitute.

答:
Sub the value of $y$ from $ii$ in $i$ or the value of $x$ from $i$ in $ii$

I'm doing in the 1st way.

$\begin{align}&x = -\frac{y^3}{8} = -\frac{(-2x^3)^3}{8} = x^9 \\
\Rightarrow &x^9 - x = x(x^8-1) = x(x-1)(x+1)(x^2+1)(x^4+1) = 0\end{align}$

If you need only real values, then $x = 0,-1,1$ and correspondingly, $y = 0, 2,-2$


问:
Prove that there exists at most one vector $v$ orthogonal to $p-q$ and such that $x$ belongs to the range of $\alpha_{p,q,v}$
For any triple $p,q,v \in \mathbb{R^n}$ the path $$\alpha_{p,q,v}:[0,1] \to \mathbb{R^n},\alpha_{p,q,v}(t)=(1-t)p+tq+t(1-t)v$$

Is a parabolic arc with endpoints $p,q$.
Prove that if $p\neq q$, for any $x\in \mathbb{R^n}- \lbrace p,q \rbrace$ there exists at most one vector $v$ orthogonal to $p-q$ and such that $x$ belongs to the range of $\alpha_{p,q,v}$

I try to imagine it in the plane and fix $x$, and draw the points $p,q$ and the vector $-q$ and draw the vector $p-q$ and try to identify the way of build the vector $v$ orthogonal to $p-q$ but Simply any construction come to my thoughts, I have 3 hours trying it, but any idea comes.

If someone can give me a suggestion, hint, or idea that was very helpful,

答:
By a suitable translation we assume that $q=0$.  For any  $x$ (not necessarily belonging to $\alpha$) (and not parallel to $p$ as in this case clearly no solution exists) the usual suspect for $v$ would be a vector parallel to
$$\frac{\|p\|^2}{\langle x,p\rangle}x-p,$$
that is
$$v=k\cdot\left(\frac{\|p\|^2}{\langle x,p\rangle}x-p\right)$$
for some constant $k$.

Now we force $k$ such $x$ belongs to $\alpha$ setting
$$x=p(1-t)+t(1-t)k\cdot\left(\frac{\|p\|^2}{\langle x,p\rangle}x-p\right).$$

Multiply by $p$ to get
$$1-t=\frac{\langle x,p\rangle}{\|p\|^2}$$
and subtitute that $t$.

We want $0\leq t\leq1$, so we must have
$$0\leq\frac{\langle x,p\rangle}{\|p\|^2}\leq1.$$

Now multiply out and determine the coefficient of $p$.  As $p$ and $x$ are linearly independent the coefficient of $p$ must vanish, which yields to
$$k=\frac{\|p\|^2}{\|p\|^2-\langle x,p\rangle},$$
if $p$ and $p-x$ aren't orthogonal (in which case there is no solution).  Hence finally
$$v=\frac{\|p\|^2}{\|p\|^2-\langle x,p\rangle}\cdot\left(\frac{\|p\|^2}{\langle x,p\rangle}x-p\right).$$


问:
Probability that you win in rock scissors paper with 3 players?
Let's say person A, person B and person C are playing rock paper scissors.

Clearly $n(S)=3\times 3\times 3=27$. We want to find the probability that person A wins.

$n(\text{only A wins})=3$ (A wins with rock/scissors/paper).

$n(\text{A and one other person wins})=2\times 3=6$
($2$ ways to choose the other winner, $3$ ways to win (rock/scissors/paper))

$\therefore P(\text{A wins})=\frac{3+2\times3}{27}=\frac{1}{3}$, but this is flawed but this because it doesn't account for the probability where you draw. (i.e. if the probability of each person winning is $\frac{1}{3}$, then the total probability is already $1$, when we haven't accounted for the probability of drawing yet.)

What is the correct probability that you will win in a game of rock scissors paper with 3 players? Thanks.

答:
The sample space contains $3\times 3\times 3=27$ outcomes, as you say.

There are different situations. If all players pick the same symbol, there is not a player that stands out (3 outcomes). If the three players pick three different symbols, there is also not a player that stands out (6 outcomes).

You call these $3+6=9$ outcomes draws.

The remaining situations corresponds to two of the players having the same symbol, and the last player having a different symbol. If the symbol held by the two players beats the symbol of the remaining player, you use the terminology that the two players win together (9 outcomes). However, if the symbol shared by the two players is weaker than the symbol chosen by the last player, then the last player wins alone (9 outcomes).

You count the number of outcomes where player A wins alone as 3, corresponding to probability that A wins alone on $\frac{3}{27}=\frac19$.

You also correctly count the number of outcomes where player A wins together with another player as 6, corresponding to probability $\frac{6}{27}=\frac{2}{9}$. But in 3 of these 6 outcomes, player B also wins together with another player, and in the other 3 it is player C which also wins. So this overlaps.

This why you cannot say that the (correct!) probability $\frac13$ that A wins either alone or together with one other player does not leave room for the situation where we have a draw.

If we denote a draw with three different symbols $ABC$, the situation where A and B win together as $AB$, and the situation where A wins alone with $A$, and finally a draw with a single symbol repeated threes times $O$, then we have:
$$
P(ABC) = \frac{6}{27} \\
P(AB)=P(BC)=P(CA)=\frac{3}{27} \\
P(A)=P(B)=P(C)=\frac{3}{27} \\
P(O)=\frac{3}{27}
$$
and these eight events are disjoint and make up the entire sample space, and
$$
\frac{6}{27} + \left(\frac{3}{27}+\frac{3}{27}+\frac{3}{27}\right)
+ \left(\frac{3}{27}+\frac{3}{27}+\frac{3}{27}\right) + \frac{3}{27} = 1.
$$


问:
Function generation by input $y$ and $x$ values
I wonder if there are such tools, that can output function formulas that match input conditions. Lets say I will make input like that:

$y=0, x=0$

$y=1, x=1$

$y=2, x=4$

and tool should generate for me function formula y=x^2. I am aware its is not possible to find out exact function, but it would be great to get some possibilities. I'm game developer and i need to code some behaviours in game mechanics, that aren't simply linear, sometimes I need for example arcus tangens, when i want my value to increase slower and slower for higher arguments. Problem is that I finished my school long time ago and simply I don't remember how does many functions looks like and such tool would be great to quickly find out what i need.

答: One of my favorite curve-fitting resources is zunzun.  They have many, many possible types of curves that can fit the data you give it.</p>

ContentLicense: CC BY-SA 3.0


问:
Why is the subspace topology same as the Order topology?
Hi i am reading Topology by Munkres and there in the proof of theorem 27.1 it is written that

Given $a<b$, let $\mathcal A$ be a covering of $[a,b]$ by sets open in $[a,b]$ in the subspace topology(which is the same as the order topology).

I can't understand why is the subspace topology same as the order topology in this case? Is this always true? Btw i know how subspace topology and order topologies are defined. If $(X,\tau)$ is a topological space and $A$ be any subset of X then we define $\tau_A$ to be the set with subsets as : $\tau_A=\{U\cap A | U\in \tau\}$. And order topology is defined in a different way.So how can they be same here? For reference i am attaching the screenshot where i have highlighted the part where this is mentioned.

答:
Well, in general to check that two topologies are the same, you need to check that an open set in one topology is also open in the other, and viceversa.

Recall that in an ordered set, you define the order topology by declaring your open sets to have a basis of open rays, and open intervals. On the other hand, the subspace topology on the set $[a,b]$ consists precisely (has a basis) of open intervals. Thus, in this case both topologies agree.


问:
Binary String's set of elements
Example: How many 7-digit binary strings have three 1's?
Answer: $ { 7 \choose 3} = 35 $

definition of $ { n \choose k} $ : If n and k are integers, then $ { n \choose k} $ denotes the number of subsets that can be made by choosing k elements from an n-element set.

According to the above definition, $ { 7 \choose 3} = \frac{7!}{3!4!} = 35 $

My difficulty: in the example we're choosing 3 elements from from a 7-element set, however what is this 7-element set?

I thought about this because when we calculate 7! , it means we have some set of 7 elements, and in this example's case we have a binary string, which is composed of the set { 0 , 1 } , i.e. a 2-element set and not a 7-element set.

Title: Binary String's set of elements

Body:
No.

$${ 7 \choose 3} = \frac{7!}{3!4!} = \frac{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2}{(3 \cdot 2)(4 \cdot 3 \cdot 2)} = 7 \cdot 5 = 35.$$


问:
Span and Basis Linear Algebra
Let V = R4 and W = {( a-2b-c, a+b, b-2c, 2b+c):a, b, c € R}
a.) Find a set S that spans W.
b.) Find a subset S that forms a basis for W.
c.) What is the dimension of W?

I already got the set S where spans{v1,v2,v3,v4}
Where v1= [1,1,0,0], v2=[-2,1,1,2], v3= [-1,0,-2,1] v4=[0,0,0,0]

I tried to answer b and c but my answers were wrong

答: You only need to check whether the $v_1,v_2,v_3$ you gave are linearly independent.

Title: Span and Basis Linear Algebra


答:
The $7$ elements are the terms in the $7$-tuple that form the $7$-digit string.

Each of these elements is distinct.

The "choice" part of the exercise is: "Is this element a $1$ or a $0$?"

Hence this exercise is equivalent to the exercise of choosing $3$ elements from a $7$-element set.

It is "how many ways can I choose $3$ elements of this string of $7$ digits so that these $3$ elements are $1$ and all the others are $0$?"

And of course the answer is $\dbinom 7 3 = \dfrac {7!} {3!4!} = \dfrac {5040}{6 \times 24}$ which works out to be $35$.

Title: Binary String's set of elements

Body:
Example: How many 7-digit binary strings have three 1's?
Answer: $ { 7 \choose 3} = 35 $

definition of $ { n \choose k} $ : If n and k are integers, then $ { n \choose k} $ denotes the number of subsets that can be made by choosing k elements from an n-element set.

According to the above definition, $ { 7 \choose 3} = \frac{7!}{3!4!} = 35 $

My difficulty: in the example we're choosing 3 elements from from a 7-element set, however what is this 7-element set?

I thought about this because when we calculate 7! , it means we have some set of 7 elements, and in this example's case we have a binary string, which is composed of the set { 0 , 1 } , i.e. a 2-element set and not a 7-element set.


问:
Is game theory a part of math?
I'm going to write project paper for my course, "The History of Math". My field is 'game theory'.

However, I'm in doubt that game theory is really part of math, since it comes from economics (for me).

My questions are:

Is game theory a part of math?

What are some papers that heavily use game theory and involve some real math calculations?

答:
Question #1:
Well, I expect some people would say that math is a part of game theory :-).
But certainly some areas of game theory are highly mathematical.

The math subject classification system of the American Mathematical Society lists Game Theory as one of their categories: link here

Question #2:
Try this book . There are many other similar ones.

Title: Is game theory a part of math?


问:
When is the restriction of a quotient map $p : X \to Y$ to a retract of $X$ again a quotient map?
Quite a number of questions in this forum deal with the problem when the restriction $p' : A \stackrel{p}{\to} p(A)$ of a quotient map $p : X \to Y$ to a subspace $A \subset X$ is again a quotient map. An example is The restriction of a quotient map to a saturated closed set is a quotient map.

Here the question is

Given a quotient map $p : X \to Y$ and a subspace $A \subset X$. When is the restriction $p \mid_A : A \to Y$ a quotient map?

This question occurs in hidden form in the context of projective spaces. Let $\mathbb K = \mathbb R, \mathbb C$. Then
$$\mathbb{KP}^n = (\mathbb K^{n+1} \setminus \{0\})/ \sim$$
where the equivalence relation $\sim$ is defined by $x' \sim x$ if $x' = \lambda x$ for some $\lambda \in \mathbb K \setminus \{0\}$. It is well-known that an alternative representation of $\mathbb{KP}^n$ is
$$\mathbb{KP}^n = S(\mathbb K^{n+1}) / \sim$$
where $S(\mathbb K^{n+1})$ denotes the unit sphere in $\mathbb K^{n+1}$ (i.e $S(\mathbb R^{n+1}) = S^n$, $S(\mathbb C^{n+1}) = S^{2n+1}$). This shows in particular that $\mathbb{PK}^n$ is compact. Note that for $\mathbb K = \mathbb R$ the relation $\sim$ restricts to $S^n$ via $x' \sim x$ if $x' = \pm x$ and for $\mathbb K = \mathbb C$ restricts to $S^{2n+1}$ via $x' \sim x$ if $x' = \lambda x$ for some $\lambda \in S^1 \subset \mathbb C$.

The above alternative representation of $\mathbb{KP}^n$ is an obvious consequence of the following fact:

Given a map $p : X \to Y$ and a retraction $r : X \to A$ onto a subspace $A \subset X$ such that $p \mid_A \circ \phantom{.} r = p$. Then $p$ is a quotient map if and only if $p \mid_A : A \to Y$ is a quotient map.

The question is to give a proof of this fact to obtain a standard reference in this forum.

Note that $S(\mathbb K^{n+1})$ is a retract of $\mathbb K^{n+1} \setminus \{0\}$; simply define $r : \mathbb K^{n+1} \setminus \{0\} \to S(\mathbb K^{n+1}), r(x) = \dfrac{x}{\lVert x \rVert}$ and note that the quotient map $\pi : \mathbb K^{n+1} \setminus \{0\} \to \mathbb{KP}^n$ has the property $\pi \mid_{S(\mathbb K^{n+1})} \circ \phantom{.} r = \pi$.

答:
Let $p : X \to Y$ be map and $p = q \circ r$ be any factorization with maps $r : X \to Z, q : Z \to Y$.

If $p$ is a quotient map, then $q$ is a quotient map.

Proof. $q$ is surjective because $Y = p(X) = q(r(X)) \subset q(Z)$. Let $U \subset Y$ such that $q^{-1}(U)$ is open in $Z$. Then $p^{-1}(U) = (q \circ r)^{-1}(U) = r^{-1}(q^{-1}(U))$ is open in $X$, thus $U$ is open in $Y$ because $p$ is a quotient map.

Assume that $r$ has a right inverse map $i : Z \to X$ (i.e. $r \circ i = id_Z$). Note that this is slightly more general than $r$ being a retraction onto a subspace $Z \subset X$.  Then $r$ is a quotient map.

Proof. $r$ is surjective because $Z = (r \circ i)(Z) = r(i(Z)) \subset r(X)$. Let $V \subset Z$ be such that $r^{-1}(V)$ is open in $X$. Then $V = (r \circ i)^{-1}(V) = i^{-1}(r^{-1}(V))$ is open in $Z$.

If $q$ is a quotient map and $r$ has a right inverse map $i : Z \to X$, then $p$ is a quotient map.

Proof. By 2. $r$ is a quotient map and composition of quotient maps yields a quotient map.


答:
You can think of it this way: Your seven-element set is $\{1, 2, 3, 4, 5, 6, 7\}$, corresponding to the indices of the digits which form your string (i.e., $3$ corresponds to the third digit).

Now you choose three elements out of this set, corresponding to the spots of your string where a "one digit" should appear. (This is a binary decision, meaning that the other elements are "zero digits".)

Title: Binary String's set of elements


问:
How do we integrate the function $\frac{b\sqrt{[1+(\theta-1)2b]^{2} - 4\theta(\theta-1)b^{2}}}{4(\theta-1)}$?
I came across the problem of determining the following integral:

\begin{align*}
I(\theta) = \int_{0}^{1}\dfrac{b\sqrt{[1+(\theta-1)2b]^{2} - 4\theta(\theta-1)b^{2}}}{4(\theta-1)}\mathrm{d}b
\end{align*}

where $b\in[0,1]$, $\theta > 0$ and $\theta\neq 1$. Here, we are going to study the case where $0 < \theta < 1$.

I managed to rearrange the square root argument as $1 + 4b(1-b)(\theta-1)$.

One possible approach consists in trying the substitution $u = 2b(\theta -1)$, from whence we get
\begin{align*}
I(\theta) = \frac{1}{16(\theta - 1)^{3}}\int_{0}^{2(\theta-1)}u\sqrt{1 + 2u\left(1 - \dfrac{u}{2(\theta-1)}\right)}\mathrm{d}u
\end{align*}

Can someone help me to take it from here? Any help is appreciated.

答:
Rearranging
\begin{equation}
 [1+(\theta-1)2b]^{2} - 4\theta(\theta-1)b^2=(1-\theta)(2b-1)^2+\theta
\end{equation}
the integral can be written as
\begin{align}
 I(\theta) &= \frac{1}{4(\theta-1)}\int_{0}^{1}b\sqrt{[1+(\theta-1)2b]^{2} - 4\theta(\theta-1)b^{2}}\,db\\
 &=\frac{1}{4(\theta-1)}\int_{0}^{1}b\sqrt{(1-\theta)(2b-1)^2+\theta}\,db\\
 &=\frac{\sqrt \theta}{16(\theta-1)}\int_{-1}^{1}(x+1)\sqrt{1+\frac{(1-\theta)}{\theta}x^2}\,dx
\end{align}
(substitution $x=2b-1$ is used to obtain the latter expression). Due to parity reasons it can be simplified as
\begin{equation}
 I(\theta)=\frac{\sqrt \theta}{8(\theta-1)}\int_{0}^{1}\sqrt{1+\frac{(1-\theta)}{\theta}x^2}\,dx
\end{equation}
Now, with $x=\sqrt{\theta/(1-\theta)}\sinh t$,
\begin{equation}
 I(\theta)=\frac{ -\theta}{8(1-\theta)^{3/2}}\int_{0}^{\operatorname{arcsinh} \sqrt{\frac{1-\theta}{\theta}}}\cosh^2t\,dt
\end{equation}
After some simple manipulations, we find
\begin{equation}
 I(\theta)=\frac{-1}{16(1-\theta)^{3/2}}\left[\sqrt{1-\theta}+\theta\operatorname{arcsinh}\sqrt{\frac{1-\theta}{\theta}}\right]
\end{equation}


答: You are looking for something like a polynomial fitting or regression analysis. Take a look at http://en.wikipedia.org/wiki/Polynomial_regression .</p>

Title: Function generation by input $y$ and $x$ values

Body:
I wonder if there are such tools, that can output function formulas that match input conditions. Lets say I will make input like that:

$y=0, x=0$

$y=1, x=1$

$y=2, x=4$

and tool should generate for me function formula y=x^2. I am aware its is not possible to find out exact function, but it would be great to get some possibilities. I'm game developer and i need to code some behaviours in game mechanics, that aren't simply linear, sometimes I need for example arcus tangens, when i want my value to increase slower and slower for higher arguments. Problem is that I finished my school long time ago and simply I don't remember how does many functions looks like and such tool would be great to quickly find out what i need.


问:
2D system of hyperbolic equation (LeVeque)
Again I need help. This time it's exercise 18.1 from LeVeque Finite Volume Methods.

Consider the system $q_t+Aq_x+Bq_y=0$ with $A=\begin{pmatrix}3&1\\1&3\end{pmatrix}$ and $B=\begin{pmatrix}0&2\\2&0\end{pmatrix}$.

Show that these matrices are simultaneously diagonalizible and determine the general solution to this system with arbitrary initial data. In particular sketch how the solution evolves in the $x-y$ plane with data

$$\begin{align*}
&q^1(x,y,0)=\left\{\begin{array}{ll}1&\text{ if } x^2+y^2\leq 1\\0&\text{ otherwise }\end{array}\right.\\
&q^2(x,y,0)\equiv 0
\end{align*}$$

So far I've got $A=R\Lambda^xR^{-1}$ and $B=R\Lambda^yR^{-1}$ with

$$\begin{align*}
&R&&=\begin{pmatrix}1&1\\1&-1\end{pmatrix}\\
&R^{-1}&&=\frac{1}{2}\begin{pmatrix}1&1\\1&-1\end{pmatrix}\\
&\Lambda^x&&=\begin{pmatrix}4&0\\0&2\end{pmatrix}\\
&\Lambda^y&&=\begin{pmatrix}2&0\\0&-2\end{pmatrix}
\end{align*}$$

but now I don't know how to move on. Any tips are appreciated :)

答:
Setting $p = R^{-1}q$, we have
$$
p_t + \Lambda_x p_x + \Lambda_y p_y = 0 \, .
$$
Since the $\Lambda^\alpha$ matrices are diagonal, this system is decoupled.
Applying the method of characteristics componentwise yields
$$
p(x,y,t) = R^{-1} q(x,y,t) = \begin{bmatrix}
F_1(\lambda^x_1 x -t, \lambda^y_1 x - \lambda^x_1 y) \\
F_2(\lambda^x_2 x -t, \lambda^y_2 x - \lambda^x_2 y)
\end{bmatrix}
$$
where $F_1$, $F_2$ are two arbitrary functions, and $\lambda_i^\alpha$ denotes the diagonal entries of $\Lambda_i^\alpha$ (see this post, where the notations correspond to $\Phi = F_i$, $a=1$, $b=\lambda_i^x$, and $c=\lambda_i^y$). Now, it remains to apply the intial condition to determine the arbitrary functions:
$$
q(x,y,0) = R\, p(x,y,0) = \begin{bmatrix}
F_1(4 x, 2 x - 4 y) + F_2(2 x, -2 (x + y)) \\
F_1(4 x, 2 x - 4 y) - F_2(2 x, -2 (x + y))
\end{bmatrix} .
$$
From the initial condition $q_1(x,y,0) = \Bbb I_{x^2 + y^2 < 1}$ and $q_2(x,y,0) = 0$ in OP, we deduce that
\begin{aligned}
\tfrac12\Bbb I_{x^2 + y^2 < 1} &= F_2(2x,-2(x+y)) \\
& = F_2(X,Y) = \tfrac12\Bbb I_{X^2 + (Y+X)^2 < 4} \\
\tfrac12\Bbb I_{x^2 + y^2 < 1} &= F_1(4x,2 x - 4 y) \\
& = F_1(\xi,\eta) = \tfrac12\Bbb I_{\xi^2 + (\eta-\xi/2)^2 < 16}
\end{aligned}
where $\Bbb I$ is the indicator function. Finally,
$$
q(x,y,t) = \frac12 \begin{bmatrix}
\Bbb I_{(x-t/4)^2 + (y-t/8)^2 < 1} + \Bbb I_{(x-t/2)^2 + (y+t/2)^2 < 1} \\
\Bbb I_{(x-t/4)^2 + (y-t/8)^2 < 1} - \Bbb I_{(x-t/2)^2 + (y+t/2)^2 < 1}
\end{bmatrix}
$$
with the above expressions.


问:
Possible truth function in $n$ truth values and $m$ variables.
I am trying to do this problem but I am not that good with combinatorics . But I still tried and here is what I came up with so far:

for $1$ variable there are $n$ assignment.

for $2$ variable there are $n^2$ assignment.

for $3$ variable there are $n^3$ assignment.

.

.

.

for $m$ variable there are $n^m$ assignment.

On the other hand ,

for $1$ assignment there are $n$ possible truth values.

for $2$ assignments there are $n^2$ possible truth values.

for $3$ assignments there are $n^3$ possible truth values.

.

.

.

for $n^m$ assignments there are $n^{n^m}$ possible truth values.

So there are $n^{n^m}$ possible truth functions.

Is this correct?And if this is correct then is there a way to make this more rigorous?

答:
You are correct, and your way of doing things is also correct.

The barrier that naturally arises when we think of counting functions , is that we seem to count them using heuristics like : for each choice of xyz, there's these many choices of abc etc. which is what you have done. Probably  because it is still a bit handwavy and there isn't enough mathematical notation to back up words such as "choice" and "assignment", we must strive to make this more rigorous.

The point is, that functions themselves make up a set. So to rigorously express the argument made, we must create a set, assert that the given set represents what we want to count, and then count the number of elements of that set.

To see how we can do this, let's take a finite set $A = \{a_1,...,a_{|A|}\}$ and a finite set $B$. What is the set of functions from $A$ to $B$? We claim that it is in one-to-one bijection with the set $\underbrace{B \times B \times B \times ... \times B}_{|A| \mathrm{\ times}}$. Indeed, note that for each element from $A$, we have a choice of an element from $B$, so for example if $a_1$ goes to $b_1$, $a_2$ goes to $b_2$ and so on, then a function is nothing but the tuple $(b_1,b_2,...,b_{|A|})$, and similarly from a tuple you can get a function by mapping $a_1$ to the first element of the tuple, $a_2$ to the second element of the tuple and so on.

Therefore, the set of functions from $A$ to $B$, has cardinality equal to the cardinality of $\underbrace{B \times B \times B \times ... \times B}_{|A| \mathrm{\ times}}$, and by the definition of the cartesian product (and using a simple induction) the answer is $|B|^{|A|}$.

Now, we identity each function in $m$ variables with $n$ truth values as follows : basically, an assignment of variables is a function from $\{1,...,m\} \to \{1,...,n\}$ because we give variable $1$ the value the function assigns it, variable $2$ the value the function assigns it, and so on. Let us call this set $S$, then realize that $|S| = n^m$ from the previous argument.

Next, note that the set we want, is the set of functions from $S$ to $\{1,...,n\}$, because for each such assignment we want a truth value to be assigned. Therefore, the set we are looking for, has cardinality $n^{|S|} = n^{(n^m)}$, as desired.

The takeaway from this , is that the rigorous way of describing "free choice"  is using the cartesian product to express a set of functions in terms of another easy-to-describe set which is counted using the rules for Cartesian products.

Having said that, I would still be very surprised if you did not get partial credit for your explanation in an exam, because it pretty much contains the entire thinking behind the problem, if not the rigour.


问:
$\int_0^1 \sqrt {\tan^{-1}x}\space dx=$?
How to evaluate $$\int_0^1 \sqrt {\tan^{-1}x}\space dx\qquad ?$$ Is there an elementary expression or value for it? (though I know that there is no elementary expression for $\int \sqrt {\tan^{-1}x}\space dx$)  . </p>

答:
You can have a solution as an infinite sum

$$\int_0^1 \sqrt {\tan^{-1}x}\space dx = \frac{\sqrt {\pi }}{2}-\frac{1}{\sqrt {\pi }}\sum _{n=0}^{\infty }{\frac {\psi^{(2\,n+1
)} \left( \frac{1}{2} \right) }{ 16^n\left( 4\,n+3 \right)  \left( 2\,
n+1 \right) !}} \sim 0.6298233443,$$

where $\psi(x)$ is the digamma function.


问:
Is curl actually a cross product?
I tried deriving curl for some vector $\vec{v}$, everything in a cylindrical coordinate system.
I looked at cross products in this coordinates:
https://bridge.math.oregonstate.edu/workshops/mini04/13.pdf

So when I calculate the Curl, I get
$$\vec{\nabla} \times \vec{V} = \begin{vmatrix} \hat{r}\ & \hat{\theta} & \hat{z} \\ \partial/\partial r & (1/r)\partial/\partial \theta & \partial/\partial z \\ 0 & V_{\theta} & 0 \end{vmatrix}$$
Which are the Del-operator components, and the vector components. However, upon looking at curl formulas, that determinant is incorrect. If curl was a cross product between two vectors that exist in an orthonormal basis, then it should have this form. How come is the curl different than the way we would go about doing a cross product.

答:
Even in cartesian coordinates, the curl isn't really a cross product.

A cross product is a map with the following properties:

It takes two vectors from $\mathbb{R}^3$ and outputs a third vector in $\mathbb{R}^3$;

It's anticommutative;

It's rotationally invariant.

The curl has only property 3, not 1 or 2. It doesn't even make sense to discuss property 2, since it doesn't make sense to write the partial derivative operator on the right.

The cross-product notation for the curl is just a helpful mnemonic for certain properties of the curl, such as the fact that its output is a vector, it has a handedness, and that the divergence of a curl is zero.

OwnerDisplayName: RSM

Title: Is curl actually a cross product?


答:
As has been stated, the curl is not defined to be the cross product.  In what follows, I'll give an operational definition of what the curl is.  It won't be pretty and it won't be convenient, but hopefully it will motivate the equations with which you are already familiar.

Consider a 2D vector field $\mathbf V$. The circulation of $\mathbf V$ around a simple closed curve $C$ is defined to be $\oint_C \mathbf V \cdot d\mathbf r$ - that is, the line integral of $\mathbf V$ around $C$.

Example:  Let $\mathbf V = x^2y\hat x + x\hat y$, and let $C$ be a square of side length $a$.  The circulation of $\mathbf V$ around $C$ is not too hard to calculate, as we can break the integral up into four parts:
$$\oint_C\mathbf V \cdot d\mathbf r = \int_{-a/2}^{a/2} \frac{a}{2} dy + \int_{a/2}^{-a/2} x^2\left(\frac{a}{2}\right) dx + \int_{a/2}^{-a/2} \left(-\frac{a}{2}\right) dy+\int_{-a/2}^{a/2}x^2\left(-\frac{a}{2}\right) dx$$
$$=\frac{a^2}{2} -\frac{2}{3}\left(\frac{a}{2}\right)^4 + \frac{a^2}{2} - \frac{2}{3}\left(\frac{a}{2}\right)^4$$
$$ = a^2 - \frac{a^4}{12}$$

Next, we consider the circulation of $\mathbf V$ around $C$ divided by the area enclosed by $C$.  To my knowledge, this quantity doesn't have a good name, but it's essentially a measure of the average circulation.  In particular, we are interested in the limit of this average circulation as the loop shrinks down to a point.

Example: In the previous example, the circulation of $\mathbf V = x^2y\hat x + x \hat y$ around a square of side length $a$ centered at the origin was found to be $a^2 - \frac{a^4}{12}$.  The area of this square is simply $a^2$, and so the "average circulation" is immediately seen to be $1-\frac{a^2}{12}$.  Taking the limit as the square shrinks down to a point (the origin), we simply get $1$.

This is the definition of the curl.  In order to compute the curl of a vector field $\mathbf V$ at a point $p$, we choose a curve $C$ which encloses $p$ and evaluate the circulation of $\mathbf V$ around $C$, divided by the area enclosed.  We then take the limit of this quantity as $C$ shrinks down to $p$.

One might immediately ask if there is a more efficient means to calculate this quantity, and the answer is yes.  If we use Cartesian coordinates where $p=(x,y)$, then the most efficient thing is to encircle our point of interest with a square of side length $a$.  If $a$ is very small, we find

$$\oint_C \mathbf V \cdot d\mathbf r \approx a\bigg(V_y(x+a/2,y)-V_x(x,y+a/2)-V_y(x-a/2,y)+V_x(x,y-a/2)\bigg)$$
Dividing by $a^2$ and taking the limit as $a\rightarrow 0$ yields the quantity $$\frac{\partial V_y}{\partial x}-\frac{\partial V_x}{\partial y}$$.

If our coordinates are not cartesian, then a square is not a good shape.  For polar coordinates, we should use a wedge as shown e.g. here:

Evaluating this circulation yields
$$\oint \mathbf V \cdot d\mathbf r \approx V_r\left(r,\theta-\frac{\Delta\theta}{2}\right)\Delta r +V_\theta\left(r+\frac{\Delta r}{2},\theta\right)\cdot \left(r+\frac{\Delta r}{2}\right) \Delta \theta$$
$$ - V_r\left(r,\theta+\frac{\Delta\theta}{2}\right)\Delta r - V_\theta\left(r-\frac{\Delta r}{2},\theta\right)\left(r-\frac{\Delta r}{2}\right)\Delta \theta$$

The area of this wedge is approximately $r(\Delta r )(\Delta \theta)$.  Dividing and taking the limit as $\Delta r$ and $\Delta \theta$ go to zero yields the quantity

$$\frac{\partial V_\theta}{\partial r} + \frac{V_\theta}{r} - \frac{1}{r} \frac{\partial V_r}{\partial \theta} = \frac{1}{r}\frac{\partial (rV_\theta)}{\partial r} - \frac{1}{r}\frac{\partial V_r}{\partial \theta}$$

Lastly, of course curl is a 3D operation while we've been working in 2D.  In 3D, each unit vector defines a plane perpendicular to it, with an orientation given by the right hand rule; the corresponding component of the curl is this average infinitesimal circulation as computed in that plane.  For instance, what I calculated above is the $\hat z$ component of the curl in cartesian and cylindrical coordinates, because it was computed in the plane perpendicular to $\hat z$.

Title: Is curl actually a cross product?

Body:
I tried deriving curl for some vector $\vec{v}$, everything in a cylindrical coordinate system.
I looked at cross products in this coordinates:
https://bridge.math.oregonstate.edu/workshops/mini04/13.pdf

So when I calculate the Curl, I get
$$\vec{\nabla} \times \vec{V} = \begin{vmatrix} \hat{r}\ & \hat{\theta} & \hat{z} \\ \partial/\partial r & (1/r)\partial/\partial \theta & \partial/\partial z \\ 0 & V_{\theta} & 0 \end{vmatrix}$$
Which are the Del-operator components, and the vector components. However, upon looking at curl formulas, that determinant is incorrect. If curl was a cross product between two vectors that exist in an orthonormal basis, then it should have this form. How come is the curl different than the way we would go about doing a cross product.

OwnerDisplayName: J. Murray


问:
Is there any continuous function that is only differentiable on $\mathbb{Q}$?
I am looking for a continuous function $f: \mathbb R \rightarrow \mathbb R$ so that $f$ is differentiable in $x$, if and only if $x \in \mathbb Q$.

I already know there is no function that is continuous everywhere in $\mathbb Q$ but nowhere outside of it, but I wonder if something similar can be said about differentiability. If not, how could one prove that?

Thanks in advance!

答:
Here is a reference:

Mark Lynch. A Continuous Function That Is Differentiable Only at the Rationals. Mathematics Magazine 86, no. 2 (2013): 132-35. doi:10.4169/math.mag.86.2.132.

An explicit construction is given of a function that is continuous on an interval, and differentiable only at the rationals.

See also references within (Zahorski's Theorem) for a nonconstructive proof.


问:
How much edges you need to add to bipartite graph to make coloring = 8
Please I need to find answer to this question, how much edges we need to add to bipartite graph $K_{4,4}$ to make graph coloring = 8 ? And why?

答: Well, if there were two non-adjacent vertices, then we could color them using the same color, and color the rest of the vertices with six other (distinct) colors, so in this case the chromatic number would be at most seven. Thus we need to add an edge between every non-adjacent pair of vertices, which makes $8 \cdot 7 / 2 - 4 \cdot 4 = 12$ edges.


问:
Let $Z\sim N(0, 1)$ then $E(f(Z))\le E(f(X))$ for any convex $f$ and any $X$ with unit variance.
I thought about this question purely out of curiosity. I have a feeling that it is true, but I can't prove it.

Standard normal distribution arises if we maximize the entropy among all absolutely continuous probability measures with mean $0$ and variance $1.$ But I want to show that $Z\preceq_{cx} X.$ That is, Suppose $f$ is a convex function and $Z\sim N(0, 1)$ and $X$ is some (absolutely continuous) random variable with zero mean and variance $1.$ I want to show that
$$E(f(Z))\le E(f(X)),$$
provided both sides are finite.

答:
Take the random variables $X_k$ with densities
$$
f_k(x) = \mathbb{1}_{(\alpha,+\infty)}(|x|)c|x|^{-k}
$$
where
$$
k>3,\qquad
\alpha^2 = \frac{k-3}{k-1},  \qquad 2c = \frac{k-3}{\alpha^{3-k}}.
$$
They have mean 0 and variance 1 for every $k$, and their fourth moment (when $k>5$) is finite and it is
$$
\frac{(k-3)^2}{(k-3)^2-4}
$$
that is as large as you want when you take $k=5+\epsilon$.


问:
$a=\frac{1+i}{\sqrt2}$ $b=\frac{\sqrt3+i}{2}$ $a,b,z\in\mathbb{C}$ What is the real part of $z=\frac{a-b}{1+ab}$?
$a=\frac{1+i}{\sqrt2}$

$b=\frac{\sqrt3+i}{2}$

$a,b,z\in\mathbb{C}$

What is the real part of $z=\frac{a-b}{1+ab}$ ?

The answer is $0$ but i do not know why.

I tried simply substituting $a, b$ but i didn't get anything in a simple form.
Then i tried using trigonometric form because i had $\frac{1}{\sqrt2},\frac{\sqrt3}{2},\frac{1}{2}$ which are common cosine and sine values, but again i got nothing. I have also noted that $a^2=i$ but that didn't seem to help much.

Do you have any tips ? Thanks in advance !

Body:
If $|a|=|b|=1$ (and this is your case)
$$\frac{a-b}{1+ab}= \frac{(a-b)\overline{1+ab}}{(1+ab)\overline{1+ab}} = \frac{(a-b)(1+\bar a\bar b)}{|1+ab|^2} = \frac{a-b+a\bar a\bar b-\bar a b\bar b}{|1+ab|^2}= \frac{a-b+|a|^2\bar b-\bar a |b|^2}{|1+ab|^2}=$$ $$=\frac{(a-\bar a)-(b-\bar b)}{|1+ab|^2}=\frac{2i\Im(a)-2i\Im(b)}{|1+ab|^2}=i\cdot 2\frac{\Im(a)-\Im(b)}{|1+ab|^2}$$


问:
Prove $\frac{d}{dt} \Big|_{t=0}\mbox{tr}(e^{X+tY})=\mbox{tr}(e^XY)$
I’m asked to prove $\frac{d}{dt}\Big|_{t=0}\mbox{tr}(e^{X+tY})=\mbox{tr}(e^XY)$ for any $X,Y$ in $M_n(\mathbb{C})$. My attempt is to assume both $X$ and $Y$ are diagonalizable, and since the set of all diagonalizable matrices is dense in $M_n(\mathbb{C})$, if we can show this is true for diagonalizable matrices, then we are done. I expected this will somehow simplify the proof, but seems it does not work well unless I further assume $X,Y$ can be diagonalizable at the same time. Any suggestions on this?

答:
We may assume $X \neq 0$. Put $f_m(t)= \frac{1}{m!}\mbox{tr}(X+tY)^m$. Then $\mbox{tr}(e^{X+tY})=\sum_{m=0}^{\infty} f_m(t)$. We will interchange the differentiation and the infinite sum.

Let's find an upper bound for $|f_m'(t)|$ using the Frobenius norm $\lVert \cdot \rVert$. Recall that $|\mbox{tr}(A)| \leq \sqrt{n} \cdot \lVert A \rVert$ hold for any $A \in M_n(\mathbb C)$. Pick $a>0$ and suppose $t \in [-a, a]\setminus \{0\}$. Then

\begin{align*} |f_m'(t)| &\leq  \frac{\sqrt{n}}{m!}\sum_{r=1}^{m} \binom{m}{r}r|t|^{r-1} \lVert X\rVert^{m-r} \lVert Y\rVert^r  \\ &\leq \frac{\sqrt{n}}{m!}\sum_{r=1}^{m} \binom{m}{r}ra^{r-1} \lVert X\rVert^{m-r} \lVert Y\rVert^r \\ &=\sqrt{n}\cdot\frac{\lVert X \rVert^m}{m!}\sum_{r=1}^{m} \binom{m}{r}ra^{r-1} \lVert X\rVert^{-r} \lVert Y\rVert^r \\&= \sqrt{n}\cdot\frac{\lVert X \rVert^m}{m!} \frac{d}{da} (1+a\lVert X \rVert^{-1} \lVert Y \rVert )^m \\&= \frac{\sqrt{n} \cdot \lVert Y \rVert}{(\lVert X \rVert +a\lVert Y \rVert )}  \cdot\frac{1}{m!} (\lVert X \rVert+a \lVert Y \rVert )^{m}  \\ &:=M_m\end{align*}

Observe that \begin{align*}\sum_{m=0}^{\infty} M_m = \frac{\sqrt{n} \cdot \lVert Y \rVert}{(\lVert X \rVert +a\lVert Y \rVert )}  \cdot\exp(\lVert X \rVert+a \lVert Y \rVert) < \infty \end{align*}

By the Weierstrass M-test, $\sum_{m=0}^{\infty}f_m'$ converges uniformly in $[-a, a] \setminus \{ 0\}$. If $t=0$, then \begin{align} &f_m'(0)=\frac{\mbox{tr}(X^{m-1}Y)}{(m-1)!} \tag{if  $m \geq 1$} \\ &f_0'(0) = 0 \end{align}

whence $\sum_{m=0}^{\infty} f_m'(0)= \mbox{tr}\left(e^X Y \right)$. To sum up, $\sum_{m=0}^{\infty}f_m'$ converges uniformly on $[-a, a]$. It follows that $f(t):=\mbox{tr}(e^{X+tY})=\sum_{m=0}^{\infty}f_m(t)$ satisfies $f'(t) = \sum_{m=0}^{\infty} f_m'(t)$ for all $t \in [-a, a]$. In particular, $f'(0)=\sum_{m=0}^{\infty}f_m'(0) = \mbox{tr}(e^X Y)$.


问:
Show that $Cov(X, Y) = p(1-p)\left(E[Y|X=1]-E[Y|X=0]\right)$
Suppose that $X$ is a Bernouilli distributed random variable, i.e. $X = 1$ with probability $p \in [0, 1]$; and $X = 0$ otherwise. Let $Y$ denote another random variable. I am trying to show that
\begin{equation}
Cov(X, Y) = p(1-p)\left(E[Y|X=1]-E[Y|X=0]\right)
\end{equation}
but can't do it! Here is my attempt:
\begin{equation}
\begin{split}
Cov(X, Y) & = E[XY] - E[X]E[Y] \\
&  = pE[Y|X=1]-p(1-p)(pE[Y|X=1]+(1-p)E[Y|X = 0]) \\
& = E[Y|X=1](p - p^2(1-p))+ E[Y|X=0]p(1-p)^2
\end{split}
\end{equation}
but I can't tidy this up into the requisite form. What am I doing wrong?

Edit: As helpfully pointed out by @drhab, I had for some reason inserted $E[X] = p(1-p)$ instead of $E[X] = p$; and this explains why I couldn't obtain the desired expression.

答:
$$\mathbb{E}[X]=p$$

$$\mathbb{E}[Y]=p\mathbb{E}[Y|X=1]+(1-p)\mathbb{E}[Y|X=0]$$

$$\mathbb{E}[XY]=p\mathbb{E}[Y|X=1]$$

substitute in

$$cov(X,Y)=\mathbb{E}[XY]-\mathbb{E}[X]\mathbb{E}[Y]$$

and get the requested proof


问:
prove this is not true. $e^x-1 = O(x^2)$ as $x \rightarrow 0$
Hello I need to show that this assertion is not true and I ran into an issue. I was taught that I could prove this by doing $\lim_{x\to0} \frac{(e^x-1)}{x^2}=c\neq0$. I used l'hoptial rule once but then was stuck at $\frac{1}{0}$. When I plugged the limit into wolfram alpha online it used a laurent series to expand it but not quite sure how to do that. Is there an easier way to calculate this limit. Thanks

答:
You can easily compute the limit that you mentioned:
$$
\lim_{x\to 0}\dfrac{e^x-1}{x^2} = \underbrace{\lim_{x\to 0}\dfrac{e^x-1}{x}}_{=1}\cdot \lim_{x\to 0} \dfrac{1}{x}= \infty
$$

This means that $x^2$ tends to zero much faster than $e^x-1$ (as $x\to 0$). In fact, using Taylor's formula (as pointed out in comments), you see that $e^x-1 = O(x)$ as $x \to 0$.


问:
The Soup Problem: how to asymptotically fairly split a geometric series and a constant one using a single pattern?
Literally every time I'm serving some soup I'm thinking of this little mathematical problem I devised.

Imagine you have a very large (= infinite, for the purposes of the actual problem) bowl of soup and want to divide it into two halves using a (finite) ladle. The trouble is that some of the good stuff floated to the surface and every scoop takes a part $p$ of it, $p < \frac12$ (so its amount goes down geometrically with a quotient of $1-p$). The other good stuff is evenly dissolved in the volume. The protocol is that you always take a full scoop in this manner, so the only freedom is in what order do you empty the ladle into $A$ or $B$'s plate. What order would you take so that, asymptotically, both parties get fair shares in both respects?

Obviously the answer can't be $ABABAB\ldots$, because $A$ would always get $1/(1-p)$-times more than $B$. $\overline{ABBA}$ does not work either because $1+(1-p)^{-3}$ is always greater than $(1-p)+(1-p)^2$. In fact (except for a countable number of special values of $p$ (*)) it can't be any periodic infinite word of $A$ and $B$ for a slight variant of the same reason. One more thing I found is that the greedy algorithm (go $A$ until it has more than $B$, then swap) won't work either because while it would always satisfy splitting of the geometrical part, it would break the linear one (number of scoops). That's all I know at the moment.

Rephrased: for a given $1-p =: q \in (\frac12, 1)$, find a subset $M ∈ \mathbb{N}$ such that both

$\displaystyle\lim_{N→∞} \frac{\#\left( M \cap \{1,\ldots,N\} \right)}{N} = \frac12$,

$\displaystyle\sum_{n=1}^∞ q^n \chi_M(n) = \frac12 \sum_{n=1}^∞ q^n$.

I'm pretty sure the answer will lie somewhere in the theory of non-integer base systems or combinatorics on words or both, but I know little of those.

(*) For example, for $q = φ - 1$ the equality $q + q^2 + q^3 = 1 + q^4 + q^5$ holds, so for $p = 2 - φ \approx 0.381966$ the sequence $\overline{ABBBAA}$ would work.

答:
Let $\{a_i\}$ be a  G.P. with  ratio $r\ge \frac{1}{2}$and sum $S$. Then, for any number $T$ in $(0,S]$ there is a subsequence $\{b_i\}$ of $\{a_i\}$ with sum $T$.

Proof. Let $b_1=a_m$ where $m$ is minimal such that $a_m\le T$. Then
$$T<a_{m-1}=\frac{a_m}{r}\le \frac{a_m}{1-r}=a_m+\frac{a_{m+1}}{1-r}.$$
Therefore the initial conditions are reproduced for $T-a_m$ and $\{a_i\}_{i>m}$ and we can now successively choose $b_2, b_3,...$

Soup for $q\ge \frac{1}{\sqrt2}$

Let $r=q^2$. From the above result we can choose a set $I$ of positive integers such that $$\sum {{q^{2i}}_{i\in I}}= \frac{1}{2(1-q^2)}.$$

Then, for the sequence of As and Bs, place the As  in positions $2i-1$ if $i\in I$ and in position $2i$ otherwise.

The absolute differences between the amount of 'good' given to A and B in successive pairs are then proportional to $1,q^2,q^4, ...$ and, by the definition of $I$, the total signed difference is zero.


问:
Why is an open subset of a variety also called a variety?

Let $V$ be an algebraic set in $\Bbb{A}^n$. Any open subset $X$ of $V$ will be called a variety.

This confused me. How come an open set will be called a variety? Isn't a variety the zero set of a bunch of polynomials? This seems like the complement of that zero set (I'm assuming we're talking about the Zariski Topology here).

答:
I asked myself the very same question while reading W. Fulton's Book
'Algebraic Curves', the definitions of which I am referring to in the
following (Chapter 6, Section 2).

By definition, if $V$ is a nonempty irreducible algebraic set, then an open
subset of $V$ is called a variety, where 'open' refers to the Zariski
topology.

This definition includes $V$ itself, since $\mathbb{A}^n$ is an open subset of
itself (by definition of a topology), such that $V = V \cap \mathbb{A}^n$ is
open in the topology of $V$ induced by that of $\mathbb{A}^n$. Hence, $V$ is
an open subset of itself, and is thus a variety according to the topological
definition above.

In other words, the algebraic definition of a variety (zero locus of a prime
ideal) is more restrictive that the topological definition above. That is, the
topological definition does not contradict the algebraic definition, but it
extends it.


问:
Why this answer does not provide counter-example for $C([0,1/2],||\cdot||_\infty)$
Why this answer does not provide counter-example for $x^n\in C([0,1/2],||\cdot||_\infty)$

From:https://math.stackexchange.com/a/711371/342943

I have thought of a simpler answer, which I believe to be correct, it plays on the assumption of cauchy, and then letting $m=N$, and $n=N+1$.
Then: $\|x^m-x^n\|_\infty= ||x^{N}-x^{N+1}\|_\infty = \|x^{N}(1-x)\|_\infty   $
Now:
$ ||x^{N}(1-x)\|_\infty = \sup_{x\in [0,1]}|x^{N}| |(1-x)| \geq \frac{1}{2}^N\times\frac{1}{2} = \frac{1}{2}^{N+1}
$
Now let $$ \varepsilon =\frac{1}{4}^{N+1} $$, so $\|x^N-x^{N+1}\|_\infty\geq \frac{1}{2}^{N+1} \gt \frac{1}{4}^{N+1} = \varepsilon    $

I know that $x^n\in C([0,1/2],||\cdot||_\infty)$ is uniformly convergent but not in $C([0,1],||\cdot||_\infty)$
But I cannot understand why the above answer from Ellya, is not valid in the case $x^n\in C([0,1/2],||\cdot||_\infty)$ so with this counter-exampe doesnot Ellya, show that $x^n\in C([0,1/2],||\cdot||_\infty)$ is not uniform convergent, as well?

答: That answer makes no sense, because in that answer the number $\varepsilon$ is $4^{-N}$; it should be a fixed number greater than $0$.


问:
Evalute this limit ($x$ tends to $0$) $|x|^{\sin x}$
$$
\lim_{x\rightarrow 0}
|x|^{\sin x}
$$

my approach:-
Let $L = \lim_{x\rightarrow 0}  |x|^{\sin x}$
$$
\ln L = \lim_{x\to0} (\sin x)\ln|x|
$$
that's it...
please help me

thanks in advance

answer to this problem is 1
Also I'm new here and Idk how to write in mathematical notation sorry.
I'll learn it soon.

答:
Hint:

$$ \sin x \cdot\ln |x|= \frac{\sin x}{x} \cdot \frac{\ln |x|}{\frac{1}{x}}.$$

Can you proceed ?


问:
Why is a connected planar graph isomorphic to its double dual?
Let $G^*$ be the dual graph of a planar graph $G$ (see wikipedia article). How does one prove that if $G$ is connected then it is isomorphic to $G^{**}$?

答:
Notice that each face in $G^*$ is represented by one vertex in $G^{**}$. On the other hand in $G^*$ there's exactly one vertex in each face.

This is the case since one can represent $G^*$ by putting vertex $v^*_f$ of $G^*$ inside the corresponding face $f$ of $G$  and by drawing the edge $e^*$ of $G^*$ corresponding to an edge $e$ of $G$ in such a way that it intersects $e$ exactly once and does not intersect any other edge of $G^*$. Then one can use the Jordan curve theorem.

Now vertices $u,v$ are neighbours in $G$ if and only if the faces $f_u^*,f_v^*$ in $G^*$ that contain $v,u$ are adjoint.

Equivalently, vertices $u^{**},v^{**}$ in $G^{**}$ (that correspond to $f_u^*,f_v^*$) are neighbours.

This means any vertices $u,v$ from $G$ become $u^{**},v^{**}$ in $G^{**}$. Also, any $u,v$ are neighbours in $G$ if and only if $u^{**},v^{**}$ in $G^{**}$ are neighbours as well.

Therefore, $G$ is isomorphic to $G^{**}$.


问:
Function generation by input $y$ and $x$ values
I wonder if there are such tools, that can output function formulas that match input conditions. Lets say I will make input like that:

$y=0, x=0$

$y=1, x=1$

$y=2, x=4$

and tool should generate for me function formula y=x^2. I am aware its is not possible to find out exact function, but it would be great to get some possibilities. I'm game developer and i need to code some behaviours in game mechanics, that aren't simply linear, sometimes I need for example arcus tangens, when i want my value to increase slower and slower for higher arguments. Problem is that I finished my school long time ago and simply I don't remember how does many functions looks like and such tool would be great to quickly find out what i need.

答:
While it is possible to find e.g. a polynomial of degree $n-1$ that exactly passes through $n$ given points, this is probably not what you need: Such interpolating polynomials tend to oscillate in a nondesireable fashion.
Since I assume that speed of calculation is an issue, I suggest to simply use linear interpolation between consecutive points - this may be much faster than a convoluted expression involving $\arctan$ and others.
If you prefer a smoothe curve, try cubic spline interpolation - it's good enough to construct a roller coaster that doesn't get bumpy. Also it has the advantages that adjustment to one of the given points has no side-effects on points that are far away from it.


问:
How can I justify the substitution of a complex variable in the following integral?
$ \displaystyle I(k) = \int_{0}^{\infty} e^{(i-1)x}x^{k} \mathrm{d}x \tag*{} $

I want to know how can we justify the substitution $ (1-i)x=t $ while the upper and lower limits of the integral remains the same.

答:
This is done using Cauchy integral theorem. For any $0<r<R$, by the theorem, $$\left[\int_r^R+\int_R^{(1-i)R}+\int_{(1-i)R}^{(1-i)r}+\int_{(1-i)r}^r\right]z^k e^{-z}\,dz=0.$$ Here we assume that $\Re k>-1$, and the principal value of $z^k$ is taken.

Now let $r\to 0$ and $R\to\infty$. The 2nd and 4th integrals vanish. The 3rd one, after substituting $z=(1-i)x$, is $-(1-i)^{k+1}\int_r^R x^k e^{(i-1)x}\,dx$ and tends to $-(1-i)^{k+1}I(k)$. Finally, the 1st one tends to $\int_0^\infty z^k e^{-z}\,dz=\Gamma(k+1)$.


问:
f is a real valued continuous function and $\int_0^a f(x)dx=0$. Prove that $∃ c ∈ (0, a) $ such that $\int_0^c xf(x) dx = 0.$

Let $f : [0, a] → \mathbb{R}$ be a continuous function, where $a ∈ \mathbb{R^+}$ and $\int_0^a f(x)dx=0$. Prove that $∃ c ∈ (0, a) $ such that $\int_0^c xf(x) dx = 0.$

I feel I've to use the Fundamental Theorem of Calculus and then Lagrange's mean value theorem but I can't work out the details. From the condition we get, $F(a)=F(0)$ where $F'(x)=f(x)$. Then I tried considering the function $g(x)=xF(x)$ but I can't proceed.

Can a solution without using double integrals or flett's theorem be found? Using high school math.

答:
Let's have some names for the anti-derivatives of $f(x)$ and $xf(x)$.

$\displaystyle F(x)=\int_0^x\int_0^t f(u)\,du\,dt\quad$ thus $F''=f$ and by hypothesis $F'(a)=0$

$\displaystyle g(x)=\int_0^x tf(t)\,dt$

We also have trivially that $F(0)=F'(0)=g(0)=0$ since the lower bound of all the integrals involved is zero.

Applying integration by parts to $g$ we get:

$\displaystyle g(x)=\int_0^x tF''(t)\,dt=\bigg[tF'(t)\bigg]_0^x-\int_0^x F'(t)\,dt=xF'(x)-F(x)$

Now remark that $F'(0)=F'(a)=0$ so by application of Flett's theorem there exists $c\in(0,a)$ such that

$$F'(c)=\dfrac{F(c)-F(0)}{c-0}=\dfrac{F(c)}{c}$$

Therefore we have our result: $$g(c)=cF'(c)-F(c)=0$$


问:
Understanding of real numbers based on Dedekind cut?
I can not understand the construction of real numbers by Dedekind cuts. Can somebody help me with understanding? The problem which I have is that the cardinality of rationals is $ \aleph_0 $. Base on that, my assumption is that I can cut this only in $ \aleph_0 $ places in such a way that $ \forall a \in A, b \in B : a \lt b  $ and $ A \cup B = \mathbb{Q} $. If it is true then it could find only $ \aleph_0 $ real numbers from $ \mathbb{R} $. Please point me where is a logic mistake.

答: The mistake lies in your assertion that you can cut only in $\aleph_0$ points. Why? The set $\Bbb R$ is uncountable and, for each $x\in\Bbb R$, if you consider the cut$$\Bbb R=\bigl((-\infty,x)\cap\Bbb Q\bigr)\cup\bigl([x,\infty)\cap\Bbb Q\bigr),$$you get different cuts for distinct values of $x$.


问:
Differential equation where forcing function depends on variable
The force acting on a body depends on the displacement $x(t)$ of the body from the origin.
This dependence is given by

$$ F(x) =
\begin{cases}
F_0 - mg, &-L < &x < 0,\\
-mg,& &x > 0.
\end{cases}
$$

where $L$ is a positive number, $F_0 > 0$ is a constant resting force and $g$ is the gravitational acceleration of Earth.

Newton's second law gives $m\ddot{x} = F(x)$ and we impose the initial conditions $\dot{x}(0) = 0, x(0) = -L.$

How does one solve an equation like this? My main problem is that if we restrict to the branch $x > 0$ we get $x = -g\frac{t^2}{2} + At + B$. I don't know if I am allowed to plug in the boundary conditions here, but if I were to, I would get $x = -g\frac{t^2}{2} - L$. But this function is clearly always negative, so then how could this satisfy the branch $x>0$?

Any tips or clarifications?

答:
The motion is a sequence of two uniformly accelerated movements. The first is immediately found to be

$$x(t)=(f_0-g)\frac{t^2}2-L$$ and $$\dot x(t)=(f_0-g)t.$$

$x$ cancels after

$$t^*=\sqrt{\frac{2L}{f_0-g}}$$ seconds, with speed $$v^*=\sqrt{2L(f_0-g)}.$$

Then the motion continues as

$$x(t)=-g\frac{(t-t^*)^2}2+v^*(t-t^*).$$

$f_0:=\dfrac{F_0}m.$


问:
Explanation of solution of a question about polynomials
Taken from https://artofproblemsolving.com/wiki/index.php/1970_Canadian_MO_Problems/Problem_10

Problem

Given the polynomial $f(x)=x^n+a_{1}x^{n-1}+a_{2}x^{n-2}+\cdots+a_{n-1}x+a_n$ with integer coefficients $a_1,a_2,\ldots,a_n$, and given also that there exist four distinct integers $a, b, c$ and $d$ such that $f(a)=f(b)=f(c)=f(d)=5$, show that there is no integer $k$ such that $f(k)=8$.

Solution

Set $g(x) = f(x) − 5$. Since $a, b, c, d$ are all roots of $g(x)$, we must have

$g(x) = (x − a) (x − b) (x − c) (x − d) h(x)$

for some $h(x) ∈ Z[x].$
...

How do we know that h(x) has integral coefficients here? How is it obvious?

I tried simplifying the $g(x)$ expression and comparing coefficients, and it seems to be true, but cannot prove it. Moreover, how is it so obvious?

答:
Let $f(x)=\sum_{k=0}^n a_kx^k $ with $a_k\in\mathbb Z $ have a root $A $. Then by FTA:
$$
g (x)=\frac {f (x)}{x-A}=\sum_{k=0}^{n-1} b_kx^k
$$
is a polynomial of degree $n-1 $. By polynomial division one obtains:
$$b_{n-1}=a_n;\quad b_i=a_i+A\times b_{i+1}\ (n-2\ge i\ge0).$$

Since integer numbers are closed under operations of addition and multiplication one concludes: $$A  \in\mathbb Z\implies  b_k\in\mathbb Z .$$

The same conclusion is valid by induction in the case of several integer roots.


问:
Positive semidefinite matrix multiplied on both sides by a projector
Let $P$ be a projection matrix and $A$ be a positive semidefinite matrix. Is it true that $P A P$ is positive semidefinite?

答:
Let $(\cdot, \cdot)$ denote the inner product and let $B$ be a self-adjoint matrix, then

$$(BABx, x)=(ABx, B^*x)=(A(Bx), Bx) \ge 0.$$


问:
Transformation Jacobian for integration by substitution in single and multiple variables?
I wonder how to reconcile the transformation rules for integration by substitution in a single variable vs several variables.

In the single-variable case

the Jacobian factor $\varphi'(x)$ is just a derivative and may take on negative as well as positive values depending on the particular function and integration range.

However, in the multi-variable case

the Jacobian determinant factor Det$(D\varphi)(u)$ is taken in absolute value, which allows this transformation Jacobian to take on positive values only.

If we take the multi-variable expression and consider the single-variable special case, then the partial derivative determinant again reduces to simply $\varphi'(x)$. However, the additional absolute value of the determinant seems to introduce a discrepancy between the purely single-variable treatment and this special case.

Moreover, it is easy to find examples showing that the absolute value indeed should not be there in the single-variable case. How to reconcile the two? Is the absolute value truly needed and correct in the multi-variable case?

EDIT:

Since some doubt was expressed in the comments on whether $\varphi'(x)$ may or may not be negative in the single-variable case, consider the following example:

$$\int_0^1 du\,u^2=\frac{1}{3}$$

Now take $u\equiv\varphi(x)=-x$. If we consider the absolute value of the Jacobian to be true, we get:

$$\int_{0}^{-1}\left|\frac{d\varphi(x)}{dx}\right|dx\,(-x)^2=\int_0^{-1}dx\,x^2=-\frac{1}{3}$$

which gives a wrong result; while if we do not take the absolute value, we recover:

$$\int_{0}^{-1} \frac{d\varphi(x)}{dx}dx\,(-x)^2=-\int_0^{-1}dx\,x^2=\frac{1}{3}$$

which is correct.

答:
In the case $U=(a,b)$ and $n=1$ the first formula is actually the more general version, because the second statement requires $\varphi$ to be injective, whereas the first does not. If we restrict to injective functions, both statements agree. This is clear for $\varphi > 0$. So assume that $\varphi <0$. Then $\varphi((a,b)) = (\varphi(b),\varphi(a))$ and:
$$\begin{align}\int_{\varphi(U)} f(v) dv &= \int_{\varphi(b)}^{\varphi(a)} f(u) d(u) \\
&= -\int_{\varphi(a)}^{\varphi(b)} f(u) d(u) \\
&= -\int_{a}^{b} f(\varphi(x))\varphi'(x) dx \\
&= \int_{a}^{b} f(\varphi(x)) |\varphi'(x)| dx \end{align}$$


问:
$a=\frac{1+i}{\sqrt2}$ $b=\frac{\sqrt3+i}{2}$ $a,b,z\in\mathbb{C}$ What is the real part of $z=\frac{a-b}{1+ab}$?
$a=\frac{1+i}{\sqrt2}$

$b=\frac{\sqrt3+i}{2}$

$a,b,z\in\mathbb{C}$

What is the real part of $z=\frac{a-b}{1+ab}$ ?

The answer is $0$ but i do not know why.

I tried simply substituting $a, b$ but i didn't get anything in a simple form.
Then i tried using trigonometric form because i had $\frac{1}{\sqrt2},\frac{\sqrt3}{2},\frac{1}{2}$ which are common cosine and sine values, but again i got nothing. I have also noted that $a^2=i$ but that didn't seem to help much.

Do you have any tips ? Thanks in advance !

答: as $|a|=|b|=1\implies a=\dfrac{1}{ \bar a },b=\dfrac{1}{\bar b}$ thus $$Z=\frac{a-b}{1+ab}=\frac{\dfrac{1}{ \bar a }-\dfrac{1}{ \bar b }}{1+\dfrac{1}{ \bar a }\cdot \dfrac{1}{ \bar b }}=-\left(\frac{\bar a-\bar b}{1+\bar a\bar b}\right)=-\bar Z $$ so $\text{Re}( Z)=0$


问:
Determining the convexity of functions
I am trying to show the convexity of functions which include exponential variables with different powers.
My question is, for quadratic functions($x^TQx$) it is easier to show as showing Q is nonnegative is enough to prove. Or we know linear functions are convex. But for these problems it is different. And finding Hessian of each function can be very time consuming task. What I want to ask is, do I need to find all of the expressions'(the objective functions and the constraints) Hessians to prove they are all convex? Or is there any other shorter way to do that?

Sample problems:

Sample Problem 1

Minimize $|2x_1+3x_2+x_3|+x_1^2+x_2^2+x_3^2+\sqrt{2x_1^2+4x_1x_2+7x_2^2+10x_2+6}$

s.t. $\frac{x_1^2+1}{x_2}+2x_1^2+5x_2^2+10x_3^2+4x_1x_2+2x_1x_3+2x_2x_3 \leq7$

Sample Problem 2

Minimize $\sqrt{2x_1^2+3x_2^2+x_3^2+4x_1x_2+7}+(x_1^2+x_2^2+x_3^2)^2$

s.t. $\frac{(x_1+x_2)^2}{x_3+1}+x_1^8\leq7$

$x_1^2+x_2^2+4x_3^2+2x_1x_2+2x_1x_3+2x_2x_3\leq10$

$x_1,x_2,x_3\geq0$

I know that sum of two strictly convex is convex. Is there any similar properties to make these proofs easier?

答:
I found the answer. Using sum of convex functions are convex property is enough and very useful here. That way, I was able to split the problem into subproblems and solved it.

For instance, take the first sample. $|2x_1+3x_2+x_3|$ here, absolute value of a convex function is always convex. $x_1^2+x_2^2+x_3^2$ here, as this is a quadratic function, it can be written in the form $x^TQx$ where $Q$ matrix is the identity matrix which is positive definite. Therefore, it is convex, too. For the last expression of the objective function it is a bit trickier. $\sqrt{2x_1^2+4x_1x_2+7x_2^2+10x_2+6}$ it can be written in the following form as inside of the square root is quadratic: $\sqrt{x^TQx+k}$. Here Q is positive definite as well, I am not going to get into the proof of that one, but you can find it easily with a simple search. And, we are able to prove that the objective function is convex. I'll stop here but you should do that for the constraints to show that feasible region consists of a convex set.

Title: Determining the convexity of functions


问:
Seeking corresponding functional to show existence of weak solution to PDE
$$-\Delta  u = \lambda u \space, \space\space\space\ x∈Ω$$
$$u(x) = 0\space, \space \space\space\space\space x∈\partial Ω$$

What functional would form a good correspondence, and whose minimization could show that the above function, $u$, has a weak solution in $H^1_0(\Omega)$? Is this analogous solving the case of:

$$-\Delta  u = f(x) \space, \space\space\space\ x∈Ω$$
$$u(x) = 0\space, \space \space\space\space\space x∈\partial Ω$$
taking $f(x)=\lambda u$, or is the there a better way to solve this specific case?

答:
If $\lambda\leq 0$ and is known, then the equation is nothing but a regular Poisson equation with an $L^2$-term.

If $\lambda >0$ is is unknown, then this is an eigenvalue problem, the link toypajme's comment is pretty nice and readable. Here I just add a bit more about motivation. Normally for eigenvalue problem, we are minimizing the Rayleigh quotient:
$$
F(u) = \frac{\| \nabla u\|_{L^2(\Omega)}^2}{\| u\|_{L^2(\Omega)}^2},\tag{1}
$$
whereas
$$\int_\Omega u w_i = 0\quad \text{for eigenfunction } w_i, i=1,\ldots,n-1,$$
and the minimizer $u = w_n$ is another eigenfunction while $F(w_n)$ is the $n$-th eigenvalue. Notice problem (1) can be written as the same constraint minimization problem in the link in the toypajme's comment.

If $\lambda = \omega^2$ is known and is not one of the eigenvalues for the operator $-\Delta$, also we have a right hand side $f$, then this is a Helmholtz equation, and it weak formulation
$$
\int_{\Omega}\nabla u\cdot \nabla v - \omega^2 \int_{\Omega}uv = \int_\Omega fv
$$
has a unique solution in $H^1_0$ by Fredholm alternative.


答:
With  the complex exponential:

$a=\mathrm e^{\tfrac{i\pi}4},\quad b=\mathrm e^{\tfrac{i\pi}6}$.

For any  $z\in\mathbf C$, $\;\operatorname{Re}z=\frac12(z+\bar z)$.

Therefore
\begin{align}
2\operatorname{Re}\Bigl(\frac{a-b}{1+ab}\Bigr)&=\frac{\mathrm e^{\tfrac{i\pi}4}-\mathrm e^{\tfrac{i\pi}6}}{1+\mathrm e^{\tfrac{5i\pi}{12}}}+\frac{\mathrm e^{-\tfrac{i\pi}4}-\mathrm e^{-\tfrac{i\pi}6}}{1+\mathrm e^{-\tfrac{5i\pi}{12}}}\\
&=\frac{\Bigl(\mathrm e^{\tfrac{i\pi}4}-\mathrm e^{\tfrac{i\pi}6}\Bigr)\Bigl(1+\mathrm e^{-\tfrac{5i\pi}{12}}\Bigr)+\Bigl(\mathrm e^{-\tfrac{i\pi}4}-\mathrm e^{-\tfrac{i\pi}6}\Bigr)\Bigl(1+\mathrm e^{\tfrac{5i\pi}{12}}\Bigr)}{\Bigl(1+\mathrm e^{\tfrac{5i\pi}{12}}\Bigr)\Bigl(1+\mathrm e^{-\tfrac{5i\pi}{12}}\Bigr)},
\end{align}
and expanding the numerator, you can check one obtains $0$.

Title: $a=\frac{1+i}{\sqrt2}$ $b=\frac{\sqrt3+i}{2}$ $a,b,z\in\mathbb{C}$ What is the real part of $z=\frac{a-b}{1+ab}$?

Body:
$a=\frac{1+i}{\sqrt2}$

$b=\frac{\sqrt3+i}{2}$

$a,b,z\in\mathbb{C}$

What is the real part of $z=\frac{a-b}{1+ab}$ ?

The answer is $0$ but i do not know why.

I tried simply substituting $a, b$ but i didn't get anything in a simple form.
Then i tried using trigonometric form because i had $\frac{1}{\sqrt2},\frac{\sqrt3}{2},\frac{1}{2}$ which are common cosine and sine values, but again i got nothing. I have also noted that $a^2=i$ but that didn't seem to help much.

Do you have any tips ? Thanks in advance !


答:
$$L=\lim_{x\to 0} |x|^{\sin x}= \exp[\lim_{x \to 0} \sin x \log |x|] =\exp[\lim_{x \to 0} \frac{\ln |x|}{\csc x}]$$
Apply L'Hospital, to get
$$L=\exp[\lim_{x \to 0} \frac{\theta(x)/|x|}{-\cot x \csc x}]=\exp[\lim_{x \to 0} \frac{-\theta(x)\sin^2 x}{|x| \cos x}]=\exp[\lim_{x \to 0}\frac{-\theta(x)\sin x}{|x|} \tan x]=\exp[0]=1. $$
Here $\theta(x>0)=1, \theta(x<0)=-1, \lim_{x\to 0} \frac{\sin x}{x}=1$.

Title: Evalute this limit ($x$ tends to $0$) $|x|^{\sin x}$

Body:
$$
\lim_{x\rightarrow 0}
|x|^{\sin x}
$$

my approach:-
Let $L = \lim_{x\rightarrow 0}  |x|^{\sin x}$
$$
\ln L = \lim_{x\to0} (\sin x)\ln|x|
$$
that's it...
please help me

thanks in advance

answer to this problem is 1
Also I'm new here and Idk how to write in mathematical notation sorry.
I'll learn it soon.


问:
The curvature of a tangent to a curve considered as a curve in its own right
Suppose we have a curve $c:I \rightarrow \mathbb{R^3}$ that is parametrised with respect to arc length. We then consider $T,B,N$ (tangent, binormal and normal) as curve in there own right each defined at a time $t$. The task is to express the curvature of the curve $T$ in terms of the curvature and torsion of the orginal curve $c$. I am struggling to do this. So far I have:

$$
\begin{align*}
\kappa_T &= \frac{\|T' \times T''\|}{\|T'\|^3} \\
&=\frac{\|T' \times T''\|}{\kappa^3}\\
&=\frac{\|c'' \times c'''\|}{\kappa^3}.
\end{align*}
$$

I need a expression for that top product, I am trying to link it to the alternative formula for torsion given by:
$$
\tau = \frac{\det(c'(t),c''(t),c'''(t)}{\|c' \times c''\|^2}.
$$

Any help welcome :)

答:
I think I may have it now:

We can rewrite the numerator of the torsion formula in the following way
$
\tau = \frac{det(c'(t),c''(t),c'''(t)}{||c' \times c''||^2}
= \frac{c'\cdot (c'' \times c''')}{||c' \times c''||^2}
$
Now we use that the denominator is actually $\kappa^2$, multiply up and take norms of both sides:

$
||\tau \kappa^2|| = ||c'||||c'' \times c'''||\
= ||c'' \times c'''||
$

And substituting this into what I had earlier, I get:

$
\kappa_T = \frac{||\tau||}{\kappa}
$


问:
Questions of division formulas and their origin
Hi guys I came across a few neat formulas to calculate the number of intergers in an interval divisable by another integer!
I was wondering why does it work? what's the mathematics behind them?

If for example I want to find the number of integers between 99 and 993 that are divisable by 2
I can do the following, choose the highest number divisable by x and the lowest number divisable by x and then

$\frac{\left(992-100\right)}{2}+1=447$

and also I can do

$⌊\frac{993-99}{2}⌋=447$

in the case of the floor fucntion if 2 is to divide both the upper and lower bound I would add 1 to the result!

Thank you so in advance guys!

答:
You showed an example but not the original formula with variables but$\cdots$

If you mean how many numbers between $x$ and $y$ are divisible by $z$, the answer is that every $z^{th}$ number will be a multiple of $z$. for example, between $1$ and $21$ how many numbers will be divisible by $z$? Those numbers will be $3,6,9,12,15,18,21\quad$ seven.

In the "formula",

$$\bigg\lfloor\frac{21-1}{3}\bigg\rfloor=6\qquad\bigg\lceil\frac{21-1}{3}=7\bigg\rceil$$
however
$$\bigg\lfloor\frac{22-1}{3}\bigg\rfloor=7\qquad\bigg\lceil\frac{22-1}{3}=7\bigg\rceil$$
and
$$\bigg\lfloor\frac{23-1}{3}\bigg\rfloor=7\qquad\bigg\lceil\frac{23-1}{3}=8\bigg\rceil$$
I would say the formula is good for a estimate, $\pm1$.

If you choose  $0$ as the lowest number divisible by $z$ then you need no rounding but, for instance, between $10$ and $20$ where $12,15,18$ are divisible by $3$ we are "off" again
$$\frac{18-12}{3}=\frac{6}{3}=2$$


答:
This formula is based on arithmetic progression.

Consider the interval $[a,b]$ and suppose $x,y$ are respectively the smallest and largest numbers divisible by $k$ in this interval.

Note that the multiples of $k$ in this interval would be

$$x,x+k,x+2k,\cdots x+(n-2)k,y$$

where $y=x+(n-1)k$ is the $nth$ multiple of $k$ in this interval. We are required to find $n$

A simple rearrangement of $y=x+(n-1)k$ would yield $n=\frac{y-x}{k}+1$ which is the formula you have used

Title: Questions of division formulas and their origin

Body:
Hi guys I came across a few neat formulas to calculate the number of intergers in an interval divisable by another integer!
I was wondering why does it work? what's the mathematics behind them?

If for example I want to find the number of integers between 99 and 993 that are divisable by 2
I can do the following, choose the highest number divisable by x and the lowest number divisable by x and then

$\frac{\left(992-100\right)}{2}+1=447$

and also I can do

$⌊\frac{993-99}{2}⌋=447$

in the case of the floor fucntion if 2 is to divide both the upper and lower bound I would add 1 to the result!

Thank you so in advance guys!


问:
prove that $∃ c ∈ (0, 1)$ such that $c^2f(c) = 2\int_0^cxf(x) \,dx$ for a continuous function $f$
Let $f\colon [0, 1] → \mathbb{R}$ be a continuous function. Given, $f(0) = 0$ and $\int_0^1 f(x)\, dx = 0$, prove that $∃ c ∈ (0, 1)$ such that $c^2f(c) = 2\int_0^cxf(x) \,dx.$

I need some help on how to use the fundamental theorem to prove this result.

My attempt: Let $F'(x)=f(x)$, the $\int_0^1 f(x)\, dx = 0$ implies that $F(1)=F(0)$. Now we take $H(x)=xf(x)$ and $\int_0^c xf(x)dx=\frac{c^2f(c)}{2}-\int_0^c\frac{x^2f'(x)}{2}dx$. This is where I'm stuck.

答:
First (as here) we prove there exists $\xi \in (0,1)$ such that $\int_0^c tf(t)\,dt = 0$. Define $\phi : [0,1] \to \Bbb{R}$ as $$\phi(x) := \frac1x \int_0^x \int_0^t f(u)\,du\,dt, \quad x \in (0,1]$$
and $\phi(0)=0$. Then $\phi$ is continuous on $[0,1]$ and differentiable on $(0,1)$ with derivative
$$\phi'(x) = \frac{x\int_0^x f(t)\,dt - \int_0^x\int_0^t f(u)\,du\,dt}{x^2} = \frac1{x^2}\int_0^x tf(t)\,dt, \quad x \in (0,1)$$
by integration by parts. Mean value theorem gives a $\theta \in (0,1)$ such that
$$\phi(1) - \phi(0) = \phi'(\theta)$$
so Fubini gives $$\phi'(\theta) = \phi(1) = \int_0^1 \int_0^t f(u)\,du\,dt = \int_0^1 \int_u^1 f(u)\,dt\,du = \int_0^1(1-u)f(u)\,du = -\int_0^1 uf(u)\,du$$ On the other hand $\phi'(1) = \int_0^1 tf(t)\,dt = -\phi(1)$. The derivative $\phi'$ extended by $\phi'(0)=0$ (right derivative) is continuous on $[0,1]$ so from
$$\begin{cases} \phi'(\theta) = \phi(1), \\ \phi'(1)=-\phi(1)\end{cases}$$the intermediate value theorem gives that there exists $\xi \in [\theta,1]$ such that $\phi'(\xi) = 0$.

$\phi'$ is furthermore differentiable on $(0,1)$ with derivative
$$\phi''(x) = \frac{x^3f(x) - 2x\int_0^x tf(t)\,dt}{x^4} = \frac{x^2f(x) - 2\int_0^x tf(t)\,dt}{x^3}, \quad x\in (0,1).$$

Hence the mean value theorem implies there exists $c \in (0,\xi)$ such that
$$\phi'(\xi)-\phi'(0) = \phi''(c)(\xi - 0)$$
or $\phi''(c)=0$. This means
$$c^2f(c) = 2\int_0^c tf(t)\,dt.$$


问:
Riddle regarding mean value theorem

A man is
photographed at a tollbooth at 12:00, and then arrives another
tollbooth, 250 miles down the road, at 2:00. A cop pulls him over and
gives him a traffic ticket for driving 125 mph.

His defense lawyer claims in court, "You can't prove that there was
ever any particular moment when my client was actually driving 125
mph..."

Hi guys I came across this riddle and I was wondering how should I approach this while I know it might have something to do with the mean value theorem? :)

答:
By the mean value theorem the driver must have been driving at $\displaystyle \frac {250-0}{2:00-12:00}=125\text{mph}$ at some point. If you join the two points on a distance-time graph using a line or curve the slope will have to be at least 125 at some point between them.

This law was attempted to be enforced in the courts of Boston, MA, USA but didn’t go through.

Title: Riddle regarding mean value theorem


答:
The mistake may be that you are trying to extrapolate from simpler linearly ordered sets. If, instead of $\Bbb{Q}$ you had a finite linearly ordered set in which every element has an immediate successor (and all but the first element is a successor of a single element) then the number of cutting points would, indeed, be more or less equal to the number of elements ($\pm1$ depending on whether you include the ends). Every possible cut occurs between an element $x$ and its successor $s(x)$. The same holds for infinite sets that have a similar successor function. For example, you can cut $\Bbb{N}$ at only countably infinitely many points – the two "halves" being $[0,n]$ and $[n+1,\infty)$ for all $n$.

However, the ordering of $\Bbb{Q}$ is more complicated. For one, it is dense. Between any two rational numbers there are infinitely many others. This means that there cannot be a successor function. Implying that you can keep on adding more cuts between any two earlier cutting points. This is in sharp contrast to cutting $\Bbb{N}$. It is easy to see that this leads to uncountably many cuts, but the details of that argument may be better explained by other means (such as Cantor's diagonal argument).

Title: Understanding of real numbers based on Dedekind cut?

Body: I can not understand the construction of real numbers by Dedekind cuts. Can somebody help me with understanding? The problem which I have is that the cardinality of rationals is $ \aleph_0 $. Base on that, my assumption is that I can cut this only in $ \aleph_0 $ places in such a way that $ \forall a \in A, b \in B : a \lt b  $ and $ A \cup B = \mathbb{Q} $. If it is true then it could find only $ \aleph_0 $ real numbers from $ \mathbb{R} $. Please point me where is a logic mistake.


答: Assume integer speeds, because I’ve never heard of police giving rational tickets.  If the client had been driving at 124 mph for two hours, they could not have reached the second toll point 250 miles away 2 hours later.  Thus they had to exceed 124 mph at some point, hitting 125.</p>

Title: Riddle regarding mean value theorem

Body:
A man is
photographed at a tollbooth at 12:00, and then arrives another
tollbooth, 250 miles down the road, at 2:00. A cop pulls him over and
gives him a traffic ticket for driving 125 mph.

His defense lawyer claims in court, "You can't prove that there was
ever any particular moment when my client was actually driving 125
mph..."

Hi guys I came across this riddle and I was wondering how should I approach this while I know it might have something to do with the mean value theorem? :)


问:
Condition for a cubic equation to have a single root
If a cubic equation

$$ f(x) = ax^3+bx^2+cx+d$$

Is given, what is the condition for the equation to only have a single root (counting multiple roots as one)

答:
There is not a unique condition: as a cubic function always has at least one real root,

either $f'(x)\le 0$,

or $f(x)$ has a triple root, i.e. $f(x)=a(x-\alpha)^3$,

or $f'(x)>0$, so the function has a maximum and a minimum, and both extrema have the same sign,


答: $$f(x)=ax^3+bx^2+cx+d \implies f'(x)=3ax^2+2bx+c >0 \forall x \in R, ~if~ b^2 < 3 ac.$$ A monotonically increasing pr decreasing function has at most one root. Also $f(-\infty) f(\infty) <0$ $f(x)=0$ has at least  one real root by IVT. So when $b^2<3ac$, the given cubic has only one real root.

Title: Condition for a cubic equation to have a single root

Body:
If a cubic equation

$$ f(x) = ax^3+bx^2+cx+d$$

Is given, what is the condition for the equation to only have a single root (counting multiple roots as one)


问:
Two Lagrange multipliers with one equation

I have an equation as below,

$$Rw = \lambda_1R_aw +  \lambda_2R_bw $$

where, $R$, $R_a$, and $R_b$ are positive definite at least semi-positive definite and Hermitian matrix. $\lambda_1$ and $\lambda_2$ are getting from Langrage multipliers. We consider this equation in complex domain.
$R$, $R_a$, and $R_b$ are $M \times M$ matrix, respectively. $w$ is $M \times 1$ vector.

At this moment, can we get the general solution like a single Lagrange multiplier case?
In other words, only one Lagrange multiplier case considers as Eigenvalue problem i.e., $ Rw = \lambda w .$

If we need to get the closed form solution or numerical soultion from the above equation, do we need more information?

Although I found the articles concerning two parameter eigenvalue problem, it consists with two equations. However in this case, we get only one equation with two eigenvalues(or Lagrange multipliers).

Because the number of unknowns(two) is bigger than the number of equations(one), is there no general solution?

Thank you for reading this question.

答: You  actually have $M$ conditions. Even  if the equation is just one , it has to be sattisfied in all $M$ coordinates. Your uknowns are $M+2$ , one for each coordinate plus $\lambda_1$ $\lambda_2$. But remember that the solution of a system of non linear eqations is not necesserely unique ( I'm sorry but i don't know any theorems on the solvability of non linear equations).</p>


答:
You may look at its discriminant, which is zero if and only if the cubic equation has a multiple root.

For the equation $ax^3+ bx^2 + cx + d = 0$, the discriminant is
$$
18abcd – 4b^3d + b^2c^2 – 4ac^3 – 27a^2d^2.
$$
Therefore the condition you seek is
$$
18abcd – 4b^3d + b^2c^2 – 4ac^3 – 27a^2d^2=0.
$$


答:
Seeing that there are answers involving the derivative that miss the point:
It is about the Discriminant $ \Delta $. When $ \Delta \le 0 $ we have one real root.

See https://en.wikipedia.org/wiki/Cubic_equation#Nature_of_the_roots


问:
Volume of $n$-simplex in $\Bbb R^m$ without Cayley-Menger determinant
The $n$-volume of an irregular $n$-simplex in $\mathbb{R}^m$ with $m\gt n$ can be calculated by the Cayley-Menger determinant using squared edge lengths. Are there alternative approaches to calculate the $n$-volume if $m \gt n$ without using a Cayley-Menger determinant? Reshaped matrices (like here) that are in principal equal variants of the Cayley-Menger determinant  do not count.

答:
Here's one alternative method.

If $\ V_n\big(v_0,v_1,\dots,v_n\big)\ $ is the $n$-volume of the $n$-simplex $\ \mathcal{S}\ $with vertices $\ v_0,v_1,\dots,v_n\ $, then
$$
V_n\big(v_0,v_1,\dots,v_n\big)=\frac{V_{n-1}\big(v_0,v_1,\dots,v_{n-1}\big)\big\|\big(I-P_{n-1}\big)\big(v_n-v_0\big)\big\|}{n}\ ,
$$
where $\ P_n\ $ is the perpendicular projection onto the $\ n-1$-dimensional space spanned by $\ v_{n-1}-v_0,$$v_{n-2}-v_0,$$\dots, v_1-v_0\ $.  The factor $\ \big\|\big(I-P_{n-1}\big)\big(v_n-v_0\big)\big\|\ $ in this product is just the height of the vertex $\ v_n\ $ above the $n-1$-simplex with vertices $\ v_0,v_1,\dots,v_{n-1}\ $, taken as a base of $\ \mathcal{S}\ $. By induction, therefore,
$$
V_n\big(v_0,v_1,\dots,v_n\big)=\frac{1}{n!}\|v_1-v_0\|\prod_{j=1}^{n-1}\big\|\big(I-P_j\big)\big(v_{j+1}-v_0\big)\big\|\ .
$$

Derivation of the formula

If $\ h_n=\big\|\big(I-P_{n-1}\big)\big(v_n-v_0\big)\big\|\ $ is the height of the vertex $\ v_n\ $ of the simplex $\ \mathcal{S}\ $ above its base, $\ \text{Conv}\big(v_0,v_1,\dots,v_{n-1}\big)\ $, and $\ A_{n-1}(x)\ $ is the $\ n-1$-volume of the cross-section of  $\ \mathcal{S}\ $ at height $\ x\ $ above that base, then the volume of $\ \mathcal{S}\ $ is
$$
\int_0^{h_n}A_{n-1}(x)dx\ .
$$
But the cross-section of $\ S\ $ at height $\ x\ $ above the base is the $\ n-1$-simplex with vertices $\ v_0(x),v_1(x), \dots, v_{n-1}(x)\ $, where
$
v_i(x)=\left(1-\frac{x}{h_n}\right)v_i+\left(\frac{x}{h_n}\right)v_n
$. This simplex is similar to the base $\ \text{Conv}\big(v_0,v_1,\dots,v_{n-1}\big)\ $, scaled down by a linear factor $\ \frac{x}{h_n}\ $ $\big($since $\ v_i(x)-v_j(x)=\left(\frac{x}{h_n}\right)\big(v_i-v_j\big)\ $$\big)$. Its $n-1$-volume $\ A_{n-1}(x)\ $is therefore $\ \left(\frac{x}{h_n}\right)^{n-1}V_{n-1}\big(v_0,v_1,\dots,v_{n-1}\big)\ $. The volume of $\ \mathcal{S}\ $ is therefore
\begin{align}
\int_0^{h_n}A_{n-1}(x)dx&=\int_0^{h_n}\left(\frac{x}{h_n}\right)^{n-1}V_{n-1}\big(v_0,v_1,\dots,v_{n-1}\big)dx\\
&=\frac{h_n\,V_{n-1}\big(v_0,v_1,\dots,v_{n-1}\big)}{n}\ ,
\end{align}
as stated above.  The formula is a generalisation of the instances that the area of a triangle is half the length of its base multiplied by its height, and the volume of a tetrahedron is one third of the area of its base multiplied by its height.

Simplification

The above formula can be written more simply if the Gram-Schmidt procedure is used to orthonormalise the vectors $\ v_1-v_0,v_2-v_0,\dots, v_n-v_0\ $:
\begin{align}
u_1&=\frac{v_1-v_0}{\big\|v_1-v_0\big\|}\ ,\\
u_j&=\frac{v_j-v_0 -\sum_\limits{i=1}^{j-1}\left\langle v_j-v_0,u_i\right\rangle}{\big\|v_j-v_0 -\sum_\limits{i=1}^{j-1}\left\langle v_j-v_0,u_i\right\rangle\big\|}
\end{align}
for $\ j=2,3,\dots,n\ $. Then $\ \big(I-P_j\big)\big(v_{j+1}-v_0\big)=$$\left\langle\big(v_{j+1}-v_0\big),u_{j+1}\right\rangle u_{j+1}\ $, and $\ \big\|\big(I-P_j\big)\big(v_{j+1}-v_0\big)\big\|=$$\left\langle\big(v_{j+1}-v_0\big),u_{j+1}\right\rangle\ $. Therefore
$$
V_n\big(v_0,v_1,\dots,v_n\big)=\frac{\prod_\limits{j=0}^{n-1}\left\langle\big(v_{j+1}-v_0\big),u_{j+1}\right\rangle}{n!}\ .
$$

Postscript

It's occurred to me that the orthonormalisation, $\ u_1,u_2,\dots,u_n\ $, of $\ v_1-v_0,v_2-v_0,\dots, v_n-v_0\ $ allows one to define an isometry $\ v_0+\sum_\limits{i=1}^nx_iu_i\mapsto\big(x_1,x_2,\dots,x_n\big)\ $ from the affine hull of $\ \mathcal{S}\ $ onto $\ \mathbb{R}^n\ $.  The image of $\ \mathcal{S}\ $ under this isometry will be an $n$-simplex in $\ \mathbb{R}^n\ $  congruent to $\ \mathcal{S}\ $, and therefore having the same volume.  Therefore any of the well-known formulas for the volume of an $n$-simplex in $\ \mathbb{R}^n\ $ will give another method for calculating the volume of $\ \mathcal{S}\ $.


问:
Are the following function 1st derivatives continuous at (0,0)?
Apologies: I've had to rephrase this question quite a few times

Suppose the following function:
$$f(x,y)=\frac{2x^3y^2}{x^4+y^2}$$
$$f(0,0)=0$$
Show it is differentiable/not differentiable at $(0,0)$

I attempted this question first by applying the definition of differentiability which yielded the following expression:
$$\lim_{(x,y)\to(0,0)}\frac{f(x,y)}{\sqrt{x^2+y^2}} = \lim_{(x,y)\to(0,0)}\frac{2x^3y^2}{(x^4+y^2)\sqrt{x^2+y^2}}$$

If the above equals 0, then the function is infact differentiable at $(0,0)$. I can't seem to find a counterexample which proves that it is discontinuous so I attempted to prove its continuity by applying the $\epsilon-\delta$ definition of limits. Unfortunately, no matter whether I express the function in terms of polarcoordinates or not, I cannot bound the following
$$|\frac{2x^3y^2}{(x^4+y^2)\sqrt{x^2+y^2}}| < \epsilon$$
for all $\epsilon$.

I'd be greatly appreciative for a solution to this kind of problem!

答: Hints for proving your inequality: Write $|2x^{3}y^{2}|$ as $|2x^{2}y| \sqrt |xy| \sqrt |xy|$. Note that $|2x^{2}y| \leq x^{4}+y^{2}$ and $\sqrt |2xy| \leq \sqrt {x^{2}+y^{2}}$. Now you can easily make your expression less than $\epsilon$.


问:
If $f: \mathbb{R} \to \mathbb{R}$ such that $\displaystyle \lim_{x \to \pm \infty}\sqrt{|x|}(f(x)- cos(x))=1$ is continous, it is uniformly continous.
There is a continous function $f: \mathbb{R} \to \mathbb{R}$ such that: $\displaystyle \lim_{x \to \pm \infty}\sqrt{|x|}(f(x)- \cos(x))=1$. Prove that function $f$ is uniformly continous.

I know that there are proofs on this forum that if $g$ is bounded and continous, $g$ is uniformly continous, but I don't know how to apply those to my situation.

Maybe I need to show that if function if $\sqrt{|x|}(f(x)- \cos(x))$ is bounded then $f$ is bounded?

答:
This is a streamlined version of Teresa Lisbon's proof.

In fact,
$$\lim_{x\to\pm\infty}\sqrt{|x|}(f(x)-\cos(x))=1
\implies \lim_{x\to\pm\infty}(f(x)-\cos(x))=0.$$
So for any $\epsilon>0$, there exists $X_0>0$ such that
when $|x|\geq X_0$,
$$|f(x)-\cos(x)|<\frac{\epsilon}{3}.$$
The function $\cos x$ is uniformly continuous on $\mathbb{R}$,
there exists $\delta>0$ such that when $|x_2-x_1|<\delta$,
$$|\cos x_2-\cos x_1|<\frac\epsilon 3.$$
For any $x_1,x_2\in[X_0,+\infty)$, when $|x_2-x_1|<\delta$,
$$|f(x_2)-f(x_1)|\leq |f(x_2)-\cos x_2|+|\cos x_2-\cos x_1|+|f(x_1)-\cos x_1|
<\epsilon.$$
So $f$ is uniformly continuous on $[X_0,+\infty)$.

In the same way, we can prove that $f$ is uniformly continuous on $(-\infty,-X_0]$.

By Cantor's theorem, $f$ is uniformly continuous on $[-X_0,X_0]$.

So the function $f$ is uniformly continuous on $\mathbb R$.


问:
Let $f: A \to B$ and $g : B \to C$. Is it true that if $g$ is surjective, then $g\circ f$ is also surjective?

Let $f: A \to B$ and $g : B \to C$. Is it true that if $g$ is surjective, then $g\circ f$ is also surjective?

Since the surjectivity of $g$ only tells us that the cardinality of $B$ is greater than or equal to $C$ that doesn't necessarily imply that the same holds for $A$ and $C$ so it seems that the statement is not true? Any hints on how to show this formally?

答:
It is not true. Let fx. $A=B=C=\mathbb{N}$ with $g(x)=x$ and $f(x)=1$.

However, if $f$ is surjective aswell, then $g\circ f$ is surjective.


答:
The statement is false. Choose, as a counterexample, $$f: \mathbb R \to \mathbb R, \quad x\mapsto 0$$ and $$g: \mathbb R \to \mathbb R, \quad x\mapsto x.$$
(Obviously, $g$ is surjective, but $g\circ f: \mathbb R\to \mathbb R, x\mapsto 0$, is not.)

Body:
Let $f: A \to B$ and $g : B \to C$. Is it true that if $g$ is surjective, then $g\circ f$ is also surjective?

Since the surjectivity of $g$ only tells us that the cardinality of $B$ is greater than or equal to $C$ that doesn't necessarily imply that the same holds for $A$ and $C$ so it seems that the statement is not true? Any hints on how to show this formally?


问:
Equal number of marked cells
In an $n\times n$ board, some cells are marked, so that the number of marked cells is divisible by $n$. Is it always possible to divide the $n^2$ cells into $n$ connected subsets, each with $n$ cells, so that all subsets contain an equal number of marked cells? (Two cells are connected if they share a side.)

Two easy way to divide the $n^2$ cells into $n$ sets of $n$ cells is to consider the rows or consider the columns. But it is not always the case that one of these ways must work (the x cells are marked):

x o o

x x o

o o o

Doing the split either by row of by column does not work, but taking the first row and L-shapes in the remaining two rows works.

答:
This is not possible.

A counterexample would be $n = 5$ with $10 = n * 2$ marked cells. Place a grid of 3 by 3 markers in a corner, place the last marker anywhere.

Any connected set containing the marker in the corner would contain at least 3 marked values, while there should be only 2 in each set.

$$
\begin{matrix}
o & o & o & o & o\\
o & o & o & o & x\\
o & o & x & x & x\\
o & o & x & x & x\\
o & o & x & x & \mathbf{x}\\
\end{matrix}
$$

Title: Equal number of marked cells


答:
No, this is not always possible.

Consider a 4x4 square with a corner 2x2 block marked. You want to split this into 4 regions with 1 marked cell each.
The marked corner cell is not adjacent to any unmarked cell, so cannot be in a region with only one marked cell.

Title: Equal number of marked cells

Body:
In an $n\times n$ board, some cells are marked, so that the number of marked cells is divisible by $n$. Is it always possible to divide the $n^2$ cells into $n$ connected subsets, each with $n$ cells, so that all subsets contain an equal number of marked cells? (Two cells are connected if they share a side.)

Two easy way to divide the $n^2$ cells into $n$ sets of $n$ cells is to consider the rows or consider the columns. But it is not always the case that one of these ways must work (the x cells are marked):

x o o

x x o

o o o

Doing the split either by row of by column does not work, but taking the first row and L-shapes in the remaining two rows works.


问:
Given $y=e^{2x}(\cosh x + \sinh x).$ Show that the second derivative is $9y$
Given $y=e^{2x}(\cosh x + \sinh x)$. Show that the second derivative is $9y$. The answer I get is $e^x + 6e^{3x}$. I used the product rule to solve it. But how can the second derivative be $9y$? How can I get that answer? Can anyone explain to me ?

答:
$y(x)=\exp(2x) (\cosh(x)+\sinh(x))$, so by using the product rule and the fact that $\cosh'=\sinh$, $\sinh'=\cosh$, we get $$y'(x)=\exp(2x) (\cosh(x)+\sinh(x)) + 2 \exp(2x) (\cosh(x)+\sinh(x))=3y(x).$$

Because $y'(x)=3y(x)$, we get $y''(x)=(y'(x))'=3y'(x)=9y(x)$.


问:
$u_t+(u(1-u))_x=a(1-2u)$, transients to steady solution
We consider the non conserving equation
$$u_t+(f(u))_x=af'(u)$$
where $a$ is a constant, 0$\leq$ x $\leq$ 1 and $f(u)=u(1-u)$.

The steady solution of this equation with boundary condition $u(0)=u_0$ and $u(1)=u_1$ can be obtained by putting $u_t=0$ and the solution obtained are $u=1/2, u=ax+u_0$ and $u=ax+u_1$

I am trying to solve this equation by method of characteristics to obtain the transient solution with the some given initial condition u(x,0).

By method of characteristics, I have $\displaystyle \frac{dt}{1}=\frac{dx}{1-2u}=\frac{du}{a(1-2u)}$, this means that the characteristics equation is $$\displaystyle \frac{dx}{dt}=1-2u$$ along with $\displaystyle \frac{du}{dx}=a, \displaystyle \frac{du}{dt}=a (1-2u).$

Solving these equations, I reached upto $u(x,t)=\displaystyle\frac{1-(1-2u_0)e^{-2at}}{2}$.

As I am trying to reach to the steady solution, I must apply $t \rightarrow \infty$ which gives me $u=1/2$.

But I also have another solution $u=ax+u_0$ which is obtained at a steady state.

Does changing the initial condition in some way can help me reach the other solution or I am not solving this equation in a correct way. I have also tried to solve this equation using the Reimann condition this post still, I reach the solution $u=1/2$ on applying $t\rightarrow \infty$.
Thanks in advance for any help.

答:
Consider the initial-value problem $u(x,0) = U^0(x)$ with $U^0(x) = ax+C$. This problem amounts to setting $v=1-2 u$ initially equal to $f(x) = 1-2 U^0(x)$, using the notations of my previous answer. The solution deduced from the method of characteristics therein leads to
$$
v(x,t) = 1-2 U^0(x) , \qquad u(x,t) = U^0(x) ,
$$
showing that $u = U^0(x)$ is a steady state (i.e., the solution of the corresponding initial-value problem does not depend on time). Taking the limit $t\to \infty$ still provides $u \to U^0(x)$.

Now, consider the initial conditions $u(x,0) = U^\text{0}(x) + \epsilon U^{1}(x)$ where  $\epsilon$ is a small perturbation. The method of characteristics yields the implicit equation
$$
v^1 = -2U^1\!\left(x-(v^0 + \epsilon v^1)\frac{e^{2at}-1}{2a}\right) ,
$$
where we have inserted the Ansatz $v(x,t) = v^0(x) + \epsilon v^1(x,t)$ with $v^0(x) =1-2U^0(x)$. At leading order in $\epsilon$, we find
$$
v^1 = -2U^1\!\left(xe^{2at} + (2C-1)\frac{e^{2at}-1}{2a}\right) .
$$
Thus, if the perturbation $U^1(x)$ is vanishing fast enough at $x\to\pm\infty$, then the IVP's solution might still tend towards the steady state $U^0(x)$ as $t\to \infty$.


答:
Hint:

$\cosh x+\sinh x=\cdots=e^x$

Now  if $y=e^{mx},$

$ y_1=me^{mx},$

$y_2=?$

More generally if $$y=e^{mx}(a\cosh x+b\sinh x)$$

using General Leibniz rule or by repeated differentiation

$$y_2=m^2e^{mx}(a\cosh x+b\sinh x)+\binom21me^{mx}(a\sinh x+b\cosh x)+e^{mx}(a\cosh x+b\sinh x)$$

If $a=b$

$$y_2$$
$$=am^2e^{mx}(\cosh x+\sinh x)+\binom21me^{mx}(\sinh x+\cosh x)+e^{mx}(\cosh x+\sinh x)$$

$$=(m+1)^2y$$

Title: Given $y=e^{2x}(\cosh x + \sinh x).$ Show that the second derivative is $9y$

Body: Given $y=e^{2x}(\cosh x + \sinh x)$. Show that the second derivative is $9y$. The answer I get is $e^x + 6e^{3x}$. I used the product rule to solve it. But how can the second derivative be $9y$? How can I get that answer? Can anyone explain to me ?


问:
A separable, regular space which has cardinality of the continuum but is not first countable?
Actually the title says it all. Is there such a topological space which is separable, regular, has cardinality of the continuum but is not first countable?

If so, is there also an example of a topological vector space having this property?
Thank you!

答:
Let $D=\{p_\xi:\xi<2^\omega\}$ be a family of $2^\omega$ free ultrafilters on $\omega$. The space $X=\omega\cup D$; clearly $|X|=2^\omega$. Points of $\omega$ are isolated, and for each $\xi<2^\omega$ the family $$\mathscr{B}_\xi=\big\{\{p_\xi\}\cup U:U\in p_\xi\big\}$$ is a local base at $p_\xi$. Then

$$\big\{\{n\}:n\in\omega\big\}\cup\bigcup_{\xi<2^\omega}\mathscr{B}_\xi$$

is a clopen base for $X$, which is clearly $T_1$, so $X$ is Tikhonov (and hence certainly $T_3$). $X$ is separable, since $\omega$ is dense in $X$. Finally, $X$ is not first countable at any point of $D$.


Body: $$y=e^{2x}(\text{cosh} x+ \text{sinh} x)=e^{3x} \implies y'=3 e^{3x}, y''= 9 e^{3x}=9y.$$


问:
Euclid Division Lemma Proof for J < 0
I am reading George E. Andrews' Number Theory and am at the portion where he proves Euclid's Division Lemma.

The lemma states that

For any integers $k (k>0)$ and $j$, there exist unique integers $q$ and $r$ such that $0≤r<k$ and $j=qk+r.$

I am having difficulty understanding the final part of the proof, where the uniqueness of $q$ and $r$ when $j < 0$ is proven.

By this point it has been established that the lemma is true for $j>1$ because by the basis representation theorem, $j$ has a unique representation to the base $k$

$$j=a_sk^s+a_{s-1}k^{s-1}+...+a_1k+a_0$$
$$=k(a_sk^{s-1}+a_{s-1}k^{s-2}+...+a_1)+a_0$$
$$=kq+r$$

Where $0≤r=a_0<k$.

Supposing a second pair $q'$ and $r'$ exists, there is a representation for $q'$ to the base $k$:

$$q'=b_tk^t+b_{t-1}k^{t-1}+...+b_1k+b_0.$$

Thus,

$$j=kq'+r'$$
$$=b_tk^{t+1}+b_{t-1}k^t+...+b_1k^2+b_0k+r'.$$

But,

$$j=a_sk^s+a_{s-1}k^{s-1}+...+a_1k+a_0.$$

By the uniqueness of the representation of $j$ to base $k$, the following may be gleaned: $r'=r=a_0, \ a_{i+1}=b_i, \ \text{and} \ t=s-1$. Therefore,

$$q'=b_tk^t+...+b_1k+b_0$$
$$=a_sk^{a-s}+...+a_2k+a_1$$
$$=q.$$

Therefore the lemma is true when $j>0$. And if $j=0, \ q=r=0$ is the only possible solution where $0≤r<k$. It is left to prove that the lemma is true when $j<0$.

Here we run into my issue. He writes,

If $j<0$, then $-j>0$, and there exist unique integers $q'' \ \text{and} \ r''$ such that

$$-j=kq''+r''.$$

If $r''=0$, then $j=k(-q'')$; thus we may take $q=-q'' \ \text{and} \ r=0$.

If $r''\neq 0$, then

$$j=-kq''-r''$$
$$=k(-q''-1)+(k-r''),$$

and we may take $q=-q''-1, \ \text{and} \ r=k-r''$.

In either case, $q$ and $r$ satisfy equation $j=kq+r$. Uniqueness for negative $j$ follows from uniqueness for $-j$, which is then positive.

I understand the reason for using the equation $-j=kq''+r''$, since it has already been established that when the equation is greater than $0$, there exists unique integers $q''$ and $r''$. This is useful because it is equivalent to $j=-kq''-r''$, so if there exists unique integers $q''$ and $r''$ for $-j$, they must also exist for $j$, and so we may conclude that the lemma is true when $j<0$, correct? But this raises the question, why doesn't the proof end there? I am missing something.

There is a reason why we treat cases $r'' \neq 0$ and $r''=0$ separately. Why is that?

Furthermore, how is the fact that in the former case $q=-q''$ and $r=r''=0$ relevant to the proof? And in the latter case, how do $q=-q''-1$ and $r=k-r''$ satisfy the equation $j=kq+r$?

Finally, the phrasing of the last sentence is confusing:

Uniqueness for negative $j$ follows from uniqueness for $-j$, which is then positive.

It seems as if "negative $j$" and "$-j$" are different things, otherwise the statement would be begging the question. But what then is "negative $j$?" The "$j$" from the equation $j=kq+r$ multiplied by $-1$? I assume "$-j$" is from the equation $-j=kq''+r''.$

I realize I have asked a great many questions and probably have not been clear enough in stating my confusions. I am very new to number theory and am learning it independently, so I am very prone to misunderstanding the content. This is the only place I can reliably go for clarification. My apologies for the length of this question, I have tried to be as specific as possible to avoid miscommunication.

答:
The approach using base representation is really ridiculous! The mathematics needed to prove the base-representation theorem is much more involved than to prove the division-remainder theorem you asked about, and in fact any proof of the base-representation theorem would essentially need to use the division-remainder theorem! (Not only that, the use of "..." is a sure sign that the cited argument is not rigorous.)

Here is the proof that you ought to learn of the theorem you want. First deal with the non-negative case, namely $j ≥ 0$. Since $j-j$ is a multiple of $k$, by the well-ordering principle for $ℕ$ let $r∈ℕ$ be the minimum such that $j-r$ is a multiple of $k$. Then $r < k$ otherwise $j-(r-k)$ is a multiple of $k$ contradicting minimality of $r$. So we are done. Next deal with the negative case, namely $j < 0$. Since $-j ≥ 0$, using the first case let $q,r∈ℤ$ such that $-j = q·k+r$ and $0 ≤ r < k$. If $r = 0$ then $j = (-q)·k+0$ so we are done. If $r > 0$ then $j = (-q)·k-r = (-q-1)·k+(k-r)$ and $0 ≤ k-r < k$ so we are also done.

Title: Euclid Division Lemma Proof for J < 0