问:
upper bounds and ordered fields
the problem I am dealing with is N contained in R is not bounded above. the proof I have says assume the contrary and prove but I don't understand completely.

the work is as follows

there exists A z := supN in R
n<=z every n in N
n+1<=z every n in N
n<=z-1 every n in N

and somehow z-1 proves the contradiction. I don't know, help, please and thank you.

答:
So your proof assumes $\Bbb N$ is bounded above. So $z:=\sup \Bbb N$ exists (every non-empty subset of $\Bbb R$ that is bounded above has a least upper bound, the sup, a basic fact about $\Bbb R$).

Now let $n \in \Bbb N$, then $n+1 \le z$ because $n+1 \in \Bbb N$ as well and $z$ is an upperbound for $\Bbb N$ by definition.

So it follows that $n \le z-1$ by basic arithmetic facts and as $n$ was arbitrary, it turns out $z-1$ is in fact also an upperbound for $\Bbb N$! But $z$, being the supremum is the smallest upper bound for $\Bbb N$ while $z-1 < z$, a contradiction.

So the assumption that $\Bbb N$ is bounded above must be false.

Title: upper bounds and ordered fields


问:
solving nonlinear equation with logarithmic terms
Let $0<a,b<1.$ How should one solve (for $x$)
$$
x+a \log x= \log y -b \log \log y.
$$
Please note the two opposite signs on $a,b.$ We can assume $y$ is very large and positive. Maybe the $W$ function is useful, but I am not an expert.

Even locating the solution in a manageable interval, whose two ends are in closed form would be good.

答:
As @Semiclassical wrote,

$$x+a\log(x)=k \implies x=a W\left(\frac{1}{a}e^{\frac{k}{a}}\right)$$ If you cannot use Lambert function, you will need a numerical method and take into account the fact that $\forall t \geq e$
$$W(t) \geq\log (t)-\log (\log (t))+\frac 12\frac{\log (\log (t))}{ \log (t)}=f(t)$$
$$W(t) \leq \log (t)-\log (\log (t))+\frac e{e-1}\frac{\log (\log (t))}{ \log (t)}=g(t)$$

In the past, I minimized
$$\Phi(a)=\int_e^{1000} \Big[a\, f(t)+(1-a) \,g(t)-W(t) \Big]^2\,dt$$ and obtained $a_{opt}=0.57958$ for which $\Phi(a_{opt})=0.134235$; for comparison, $\Phi(0)=35.6714$, $\Phi(0.5)=0.80423$ and $\Phi(1)=18.8333$.

Making numbers rational, a good approximation is
$$W(t) \sim \log (t)-\log (\log (t))+\frac{11133}{11659}\frac{\log (\log (t))}{ \log (t)}$$ Over the considered range, the maximum error is $\sim 0.007$.

With such an estimate, Newton method would converge to high accuracy within a couple of iterations.

Edit

For illustration, let use use $y=10^{99}$, $b=\frac 12$ and $a=\pi$. This would give $t\sim 4.36783\times 10^{30}$, $W(t)\sim 66.3531$ and then $x\sim 208.454$ while the "exact" solution is $x \sim 208.466$.

ContentLicense: CC BY-SA 4.0


答:
And since noone mentioned this yet, here is a direct proof.

If $n=2k$ is even we need to show that $10^n \cong 1 \mod 11$. But $10^{2k}-1=999....999$ where there are $2k$ 9's. Thus

$$10^{2k}-1= 90909090..909 *11 $$

If $n=2k+1$ is even we need to show that $10^n \cong -1 \mod 11$.

But $10^{2k+1}+1=999....9990+11$ where there are $2k$ 9's. Thus

$$10^{2k}+1= 90909090..9090 *11 +11$$

Second direct solution

We will prove instead that

$$(-10)^n \equiv 1 \pmod{11}

using the well known geometric formula:

$$1-10+10^2-10^3+...+(-10)^{n-1}= \frac{1-(-10)^n}{1-(-10)} \,.$$

Since the left side is an integer, the right side is an integer, which proves the claim. THis is overkill though....

ContentLicense: CC BY-SA 3.0

Title: Complete induction of $10^n \equiv (-1)^n \pmod{11}$

Body:
To prove $10^n \equiv (-1)^n\pmod{11}$, $n\geq 0$, I started an induction.

It's $$11|((-1)^n - 10^n) \Longrightarrow (-1)^n -10^n = k*11,\quad k \in \mathbb{Z}. $$

For $n = 0$:
$$ (-1)^0 - (10)^0 = 0*11 $$

$n\Rightarrow n+1$
$$\begin{align*}
 (-1) ^{n+1} - (10) ^{n+1} &= k*11\\
(-1)*(-1)^n - 10*(10)^n &= k*11
\end{align*}$$
But I don't get the next step.


问:
How to verify if a span of vectors is a subset of another span of vectors

$\displaystyle b_1=\left\lbrace\frac{12}{9},\frac{12}{9},2\right\rbrace^T,b_2=\{-18,-18,21\}^T$ and $\displaystyle v_1=\{-1,-1,2\}^T,v_2=\{3,3,-3\}^T$. $b_1 \in \operatorname{Span}\{v_1,v_2\} \text{ and } b_2 \in \operatorname{Span}\{v_1,v_2\}$. Can we conclude that (Without performing further verification) $\operatorname{Span}\{b_1,b_2\} \subseteq \operatorname{Span}\{v_1,v_2\}$? What about $ \operatorname{Span}\{v_1,v_2\} \subseteq  \operatorname{Span}\{b_1,b_2\}$?

As the question has provided, both $b_1,b_2$ belongs to the span of $v_1,v_2$. However, the answer key only pointed out that we can only conclude $\operatorname{Span}\{b_1,b_2\} \subset \operatorname{Span}\{v_1,v_2\}$ but not $ \operatorname{Span}\{v_1,v_2\} \subset  \operatorname{Span}\{b_1,b_2\}$ without any further explanation.

Is the fact that both $b_1 \in \operatorname{Span}\{v_1,v_2\} \text{ and } b_2 \in \operatorname{Span}\{v_1,v_2\}$ implies that $\operatorname{Span}\{b_1,b_2\} \subseteq \operatorname{Span}\{v_1,v_2\}$ ?

More generally, as my question's title suggests, how do I verify if a Span of vectors is a subset of another span of vectors? (I am capable of verifying if a single vector belongs to a span of vectors) but I can't make any connections between the two

答:
If $b_1 \in \text{Span}(v_1, v_2) $ and $b_2 \in \text{Span}(v_1, v_2)$, then $\text{Span}(b_1, b_2) \subset \text{Span}(v_1, v_2)$.

Proof:

$b_1 \in \text{Span}(v_1, v_2) $ $\implies $ $b_1 = k_1 v_1 + k_2 v_2$. Similarly, $b_2 \in \text{Span}(v_1, v_2)$ $\implies $ $b_2 = c_1 v_1 + c_2 v_2$.

Hence, an arbitrary vector $w \in \text{Span}(b_1, b_2)$ satisfies

\begin{align}
  w &= m_1 b_1 + m_2 b_2\\
    &= (m_1 k_1 + m_2 c_1) v_1 + (m_1 k_2 + m_2 c_2 ) v_2
\end{align}

This implies $w \in \text{Span}(v_1, v_2)$. Hence, $\text{Span}(b_1, b_2) \subset \text{Span}(v_1, v_2)$.

End Proof

If $b_1 \in \text{Span}(v_1, v_2) $ and $b_2 \in \text{Span}(v_1, v_2)$, then $\text{Span}(v_1, v_2) \subset \text{Span}(b_1, b_2)$ is in general False.

Counterexample

Take $b_1 = b_2 = v_1$. Then, Span$(b_1, b_2)$ = Span$(v_1)$. Hence,  $\text{Span}(v_1, v_2) \nsubseteq \text{Span}(b_1, b_2)$

Response to comments:

If $b_1 \in \text{Span}(v_1, v_2) $ and $b_2 \in \text{Span}(v_1, v_2)$, then $\text{Span}(b_1, b_2) \subset \text{Span}(v_1, v_2)$.

Now if $\text{Span}(b_1, b_2) \subset \text{Span}(v_1, v_2)$  and dim(Span($b_1, b_2$)) $=$ dim(Span($v_1, v_2$)), then indeed Span($b_1, b_2$) $=$ Span($v_1, v_2$). This follows from a general rule that any subspace $A$ of a vector space $V$ satisfying dim$(A) = $ dim($V$) implies that $A = V$.


问:
Find a joint distribution of two random variables $X$ and $Y$ that are not independent and that $Y$ has a marginal distribution $ Y\thicksim Ber(p)$
Find a joint distribution of two random variables $X$ and $Y$ that are not independent and that $Y$ has a marginal distribution $ Y\thicksim Ber(p)$

The problem I have is that for every X with a certain distribution I propose for this question, it ends up fulfilling that X and Y are independent

答:
Find a joint distribution

This is one of the $\infty$ example you can do

Y=0
Y=1
Total

X=0
1-p

1-p

X=1

p
p

Total
1-p
p
1

This is another one

Y=0
Y=1
Total

X=0
1-p
p/2
1-p/2

X=1

p/2
p/2

Total
1-p
p
1

... et cetera et cetera

The fact that $X$ and $Y$ are not independent is immediately shown by the fact that there are some emply cells in the table (the joint distribution)


问:
Is this additive subgroup of $\Bbb R$ a Bernstein set?
This Q Construct an additive group by transfinite induction asked for proof-verification and also asked for other ways to construct a Bernstein subset of $\Bbb R$ that is also an additive group.

Consider $\Bbb R$ as a vector space over the field $\Bbb Q$ and let $B$ be a Hamel basis for $\Bbb R$ with $1\in B.$ For each $x\in \Bbb R$ there is a unique $\{x_b:b\in B\}\subset \Bbb Q$ such that $x=\sum_{b\in B}bx_b.$ ( Only finitely many $x_b$ are non-$0$.).

Q: Is the additive group  $S=\{x\in \Bbb R: x_1\in \Bbb Z\}$ a Bernstein set?

The set $S\cap [0,1)$ is an old textbook example of a non-Lebesgue-measurable set and it has inner Lebesgue measure $0.$ But does $S$ have an uncountable closed Lebesgue-null subset? I have not got anywhere on this.

答:
It is at least possible for $S$ to have a perfect subset (which is by definition uncountable and closed, and must be null since $S$ has inner measure $0$). I suspect some $B$s will yield Bernstein $S$s, though I haven't verified that.

Consider the following result (where $\{q_w: w\in\mathbb{N}\}$ is some fixed enumeration of the rationals):

Fix a natural number $k$ and finitely many disjoint nondegenerate closed intervals $A_1,...,A_n$ with $1\in A_1$. There are disjoint nondegenerate closed intervals $B_1,...,B_{2n}$ such that $1\in B_1$ and for each $i\in\{1,...,2n\}$ we have:

$B_{2i}\cup B_{2i+1}\subseteq A_i$, and

for each $f\in B_i$ and each $g_1,...,g_k\in \bigcup_{j\in\{1,...,2n\}\setminus\{i\}}B_j$, there are no $c_1,...,c_k\in \{q_1,...,q_k\}\cup\{0\}$ such that $$f=\sum_{u=1}^kc_ug_u.$$

The proof is straightforward but tedious. The point is that by iterating this - increasing $k$ as we go along - we get a Cantor-set-like construction of a perfect $\mathbb{Q}$-linearly-independent set $Y$ with $1\in Y$; by cutting out a small open neighborhood of $1$, we get a $\mathbb{Q}$-linearly-independent perfect set $X$ whose span doesn't contain $1$. But now if we take a Hamel basis $B\supseteq Y$ (by Zorn), we get that $X$ is a perfect subset of the corresponding $S$.


答:
Indeed you cannot say that $Span\{v_1,v_2\} \subseteq Span\{b_1,b_2\}$ in general. To see that consider the trivial case where:
$v_1=\begin{pmatrix} 1 \\ 0 \end{pmatrix}$, $v_2=\begin{pmatrix} 0 \\ 1 \end{pmatrix}$

and $b_1=b_2=\begin{pmatrix} 1 \\ 0 \end{pmatrix}$.

It is clear that $Span\{b_1,b_2\}=Span\{b_1\} \subseteq Span\{v_1,v_2\}$ but not the contrary (you cannot generate $v_2$ using $b_1$).

Now, about your question "Is the fact that both $b_1 \in Span\{v_1,v_2\} \text{ and } b_2 \in Span\{v_1,v_2\}$ implies that $Span\{b_1,b_2\} \subseteq Span\{v_1,v_2\}$ ?" The answer is yes.

To see that, go back to the definition of $Span\{v_1,...,v_n\}$ : it is the set of vectors that can be genereated by a linear combination of $(v_1,...,v_n)$.

In your case, as $b_1$ and $b_2$ can be computed as linear combinations of $v_1$ and $v_2$, thus any linear combination of $b_1$ and $b_2$ can be computed as linear combinations of $v_1$ and $v_2$. Hence the result.

Now a general method to see if a Span of vectors $b_1,...,b_k$ is a subset of another span of vectors $v_1,...,v_k$ is to check if for all i in 1,...,k $b_i$ is a linear combination of $v_1,...,v_k$.

Hope it helps you!

Title: How to verify if a span of vectors is a subset of another span of vectors

Body:
$\displaystyle b_1=\left\lbrace\frac{12}{9},\frac{12}{9},2\right\rbrace^T,b_2=\{-18,-18,21\}^T$ and $\displaystyle v_1=\{-1,-1,2\}^T,v_2=\{3,3,-3\}^T$. $b_1 \in \operatorname{Span}\{v_1,v_2\} \text{ and } b_2 \in \operatorname{Span}\{v_1,v_2\}$. Can we conclude that (Without performing further verification) $\operatorname{Span}\{b_1,b_2\} \subseteq \operatorname{Span}\{v_1,v_2\}$? What about $ \operatorname{Span}\{v_1,v_2\} \subseteq  \operatorname{Span}\{b_1,b_2\}$?

As the question has provided, both $b_1,b_2$ belongs to the span of $v_1,v_2$. However, the answer key only pointed out that we can only conclude $\operatorname{Span}\{b_1,b_2\} \subset \operatorname{Span}\{v_1,v_2\}$ but not $ \operatorname{Span}\{v_1,v_2\} \subset  \operatorname{Span}\{b_1,b_2\}$ without any further explanation.

Is the fact that both $b_1 \in \operatorname{Span}\{v_1,v_2\} \text{ and } b_2 \in \operatorname{Span}\{v_1,v_2\}$ implies that $\operatorname{Span}\{b_1,b_2\} \subseteq \operatorname{Span}\{v_1,v_2\}$ ?

More generally, as my question's title suggests, how do I verify if a Span of vectors is a subset of another span of vectors? (I am capable of verifying if a single vector belongs to a span of vectors) but I can't make any connections between the two


问:
Prove that $f\left ( x \right )- x^{2021}$ always has at least one root $x_{0}\in\left ( 0, 1 \right )$

Given positive continuous function $f\left ( x \right )$ on the interval $\left [ 0, 1 \right ]$ so that $\int_{0}^{1}f\left ( x \right ){\rm d}x< \frac{1}{2022}.$ Prove that $f\left ( x \right )- x^{2021}$ always has at least one root $x_{0}\in\left ( 0, 1 \right ).$

Suppose that there no exists $x_{0}\in \left ( 0, 1 \right )$ as the statement, let $g\left ( x \right )= f\left ( x \right )- x^{2021}.$ Because $f\left ( x \right )$ is a continuous function on $\left [ 0, 1 \right ]$ so $g\left ( x \right )$ too, then $g\left ( x \right )= kF\left ( x \right )$ with $F\left ( x \right )> 0, k= constant.$ On the other hand $g\left ( 0 \right )= f\left ( 0 \right )> 0\Rightarrow k> 0\Rightarrow g\left ( x \right )> 0,$

let $G\left ( x \right )= \int_{0}^{1}g\left ( x \right ){\rm d}x.$ We obtain ${G}'\left ( x \right )> 0,$ it reminds me the following inequality with strictly decreasing $f\left ( x \right )$ for all $x\in \left [ 0, 1 \right ]$ by Cauchy-Schwarz $\int\limits_{0}^{1}f\left ( x \right ){\rm d}x+ (\int\limits_{0}^{1}xf\left ( x \right ))^{2}> -1.$ How should I try next ? I need to the help, thank you.

答:
$f(0)>0$ is crucial for this. (Otherwise $\epsilon x^{2021}$ would be a counter-example for small enough $\epsilon$).

If $f(x) -x^{2021}$ is never zero then it is always positive or always  negative (by IVP). If it is always negative we get a  contradiction by letting $x \to 0$. If it is always positive we get a contradiction to $\int_0^{1} f(x) dx <\frac 1 {2022}$.


答: Since $\int_0^1 g(x)dx=\int_0^1 f(x)-x^{2021}dx<0$ there has to be some $x_1\in (0,1)$ with $g(x_1)<0$. Since $g(0)>0$ by the intermediate value theorem there has to be a $x_0\in (0,1)$ with $g(x_0)=0$.

Title: Prove that $f\left ( x \right )- x^{2021}$ always has at least one root $x_{0}\in\left ( 0, 1 \right )$

Body:
Given positive continuous function $f\left ( x \right )$ on the interval $\left [ 0, 1 \right ]$ so that $\int_{0}^{1}f\left ( x \right ){\rm d}x< \frac{1}{2022}.$ Prove that $f\left ( x \right )- x^{2021}$ always has at least one root $x_{0}\in\left ( 0, 1 \right ).$

Suppose that there no exists $x_{0}\in \left ( 0, 1 \right )$ as the statement, let $g\left ( x \right )= f\left ( x \right )- x^{2021}.$ Because $f\left ( x \right )$ is a continuous function on $\left [ 0, 1 \right ]$ so $g\left ( x \right )$ too, then $g\left ( x \right )= kF\left ( x \right )$ with $F\left ( x \right )> 0, k= constant.$ On the other hand $g\left ( 0 \right )= f\left ( 0 \right )> 0\Rightarrow k> 0\Rightarrow g\left ( x \right )> 0,$

let $G\left ( x \right )= \int_{0}^{1}g\left ( x \right ){\rm d}x.$ We obtain ${G}'\left ( x \right )> 0,$ it reminds me the following inequality with strictly decreasing $f\left ( x \right )$ for all $x\in \left [ 0, 1 \right ]$ by Cauchy-Schwarz $\int\limits_{0}^{1}f\left ( x \right ){\rm d}x+ (\int\limits_{0}^{1}xf\left ( x \right ))^{2}> -1.$ How should I try next ? I need to the help, thank you.


问:
The $n-1$ order principle minor of the adjoint matrix of $A$ has determinant $|A|^{n-2}a_{nn}$.
Let $A$ be a $n\times n$ matrix, $A^*$ is its adjoint, that is, $A=\begin{pmatrix}A_{11}&\cdots&A_{n1}\\
\vdots&&\vdots\\
A_{1n}&\cdots&A_{nn}\end{pmatrix}$, where $A_{ij}$ is the cofactor of $a_{ij}$ in $A$. Show that $det\begin{pmatrix}A_{11}&\cdots&A_{n1}\\
\vdots&&\vdots\\
A_{1,n-1}&\cdots&A_{n-1,n-1}\end{pmatrix}=|A|^{n-2}a_{nn}$. How to show? We do not know whether or not $A$ or the $n-1$ matrix is invertible?

答:
To prove an identity of this sort, the usual technique is to prove it for a matrix $A$ containing $n^2$ indeterminates, so that $A$ is always invertible. Then, by "substituting" some concrete numbers into the indeterminates, we automatically get an identity for this new $A$ with number entries, even if it is singular.

(There is also a similar but subtly different perspective from Zariski topology, which loosely says that the set of all invertible matrices are "dense" in the set of all matrices, so that in most cases, if a matrix identity holds for all invertible matrices, we can pass the identity to the limit to make it valid for all square matrices. If the field in question is $\mathbb F=\mathbb R$ or $\mathbb C$, you can also ignore Zariski topology and consider the usual topology of $M_n(\mathbb F)$.)

Anyway, suppose $A$ is invertible. The identity then follows from a simple block-matrix argument that was portrayed in Prasolov, Problems and Theorems in Linear Algebra, theorem 2.5.1. Partition $A$ as $\pmatrix{X&y\\ z&a_{nn}}$ and partition $\operatorname{adj}(A)$ in a conforming manner as $\pmatrix{U&v\\ w&m_{nn}}$, where $a_{nn}$ is the bottom right entry of $A$ and $m_{nn}$ is the corresponding $(n-1)$-rowed minor. Since $A\operatorname{adj}(A)=\det(A)I_n$, we have
$$
\pmatrix{X&y\\ z&a_{nn}} \pmatrix{U&v\\ w&m_{nn}}=\det(A)I_n.
$$
Hence
$$
\pmatrix{z&a_{nn}} \pmatrix{U&v\\ w&m_{nn}}=\det(A)\pmatrix{0&1}=\pmatrix{0&\det(A)}
$$
and
$$
\pmatrix{I&0\\ z&a_{nn}} \pmatrix{U&v\\ w&m_{nn}}=\pmatrix{U&v\\ 0&\det(A)}.
$$
Take determinants on both sides, we obtain $a_{nn}\det(\operatorname{adj}(A))=\det(A)\det(U)$. However, from $A\operatorname{adj}(A)=\det(A)I_n$ we also obtain $\det(\operatorname{adj}(A))=\det(A)^{n-1}$. Therefore $a_{nn}\det(A)^{n-2}=\det(U)$.


问:
Is $\{0\}/I$ an ideal in quotient ring?
Let $R$ be a ring with ideal $I \neq \{0\}$. Is $\{0\}/I$ an ideal of $R/I$? I know that $I$ is supposed to be an ideal (and hence a subset) of $R$ according to the definition of a quotient ring. But $\{0\}/I = \{0+I\} = \{I\} = \{i+I\ \mid i \in I \} = I/I$. So if these two objects are the same, what's going on? Any help understanding why this does not make sense would be appreciated.

答:
Lets fix notation. $R$ is a ring, $I\subseteq R$ an ideal. So far so good. Now we form the quotient ring $R/I=\{x+I\ |\ x\in R\}$. It is typical at this point to assume that $I\subseteq R$, ergo $\{0\}/I$ doesn't really make sense unless $I=\{0\}$.

However there is a way to make this sensible. Consider the quotient homomorphism $\pi:R\to R/I$, $\pi(x)=x+I$. Then given any subset $A\subseteq R$ we can write $A/I$ meaning $\pi(A)$. However I wouldn't use "$A/I$" notation in this case, since typically quotients are reserved for the "$I\subseteq A$" case.

I think it is a good place to mention the correspondence theorem: let $R$ be a ring and $I\subseteq R$ an ideal. Write

$$Ideals(R, I)=\{J\subseteq R\ |\ J\text{ is an ideal and }I\subseteq J\}$$

Now let $I\subseteq R$ be a fixed ideal. Define

$$\Theta:Ideals(R, I)\to Ideals(R/I, 0)$$
$$\Theta(J)=J/I=\pi(J)$$

Here "$0$" on the right side means the zero ideal in $R/I$, i.e. $\{0+I\}$.

The theorem states that $\Theta$ is a well defined bijection. In particular ideals of $R/I$ are fully described by ideals of $R$ containing $I$ and vice versa.

So if these two objects are the same, what's going on?

What exactly surprises you? That a function maps two subsets into the same subset? Here are few examples:

$R=\mathbb{Z}$ and $I=(p)$ for prime $p$, then any ideal $J=(n)$ with $p$ not dividing $n$ will give $\pi(J)=R/I$.

$R$ any ring, $I\subseteq R$ any ideal, $A\subseteq I$ any subset, including $A=\{0\}$. Then $\pi(A)=\{0+I\}$.

Title: Is $\{0\}/I$ an ideal in quotient ring?


问:
Range of $\frac{\cos\theta_1+\cdots+\cos\theta_{10}}{\sin\theta_1+\cdots+\sin\theta_{10}}$ given $\sin^2\theta_1+\cdots+\sin^2\theta_{10}=1$

Given that for $ \theta_i \in \left[0, \dfrac{\pi}{2}\right]$, where $1 \le i \le 10$ , $\sin^2\theta_1+\sin^2\theta_2+\cdots+\sin^2\theta_{10}=1$, find the minimum and maximum value of $$\dfrac{\cos \theta_1+\cos\theta_2+\cdots+\cos\theta_{10}}{\sin\theta_1+\sin\theta_2+\cdots+\sin\theta_{10}}$$

I tried using Cauchy Schwarz inequality which just gives $\displaystyle \sum_{i=1}^{10} \sin\theta_i \le \sqrt{10}$ and $\displaystyle \sum_{i=1}^{10} \cos \theta_i \le \sqrt{90}$ because $\displaystyle \sum_{i=1}^{10}\cos^2\theta_i =9$ but I don't think it would be helpful here.

Maybe some inequality can be applied by writing it as $$\dfrac{\displaystyle \sum_{i=1}^{10} \sqrt{1-\sin^2\theta_i}}{\displaystyle \sum_{i=1}^{10} \sin \theta_i}$$

Any hints would be appreciated!

答:
lower bound:

Notice that $$\cos ^2 \theta_1=\sin^2 \theta_2+\sin^2 \theta_3+\cdots+\sin^2 \theta_{10}$$ similarly $$\cos^2 \theta_i=\sum_{j=1 \rightarrow 10 , j\neq i }\sin^2 \theta_j$$

$$S_i=\sum_{j=1\rightarrow 10,j\neq i} \sin \theta_i \tag {say}$$
thus by power mean inequality

$$\cos \theta_i\ge \sqrt{\frac{S_i^2}{9}}=\frac{S_i}{3}$$ Hence  $$\dfrac{\cos \theta_1+\cos\theta_2+\cdots+\cos\theta_{10}}{\sin\theta_1+\sin\theta_2+\cdots+\sin\theta_{10}}\ge \frac{S_1+S_2+S_3+\cdots+S_{10}}{3(\sin \theta_i+\sin \theta_2+\cdots+\sin \theta_{10})}=\boxed 3$$

Upper bound: see See Hai's beautiful proof!


问:
Differentiate with respect to w
$\dfrac{-\pi*A^2(3aA^2+8k-8mw^2)}{32w}$

This is really embarrassing but the important point to make is that I keep getting the same thing.
I'm using the product rule with

$u = -\pi A^2(3aA^2+8k - 8mw^2)$
I get as the derivative of u
as $\pi A^2 16mw$ using the chain rule

$n = \dfrac{1}{32w}$ the derivative is $-\dfrac{1}{32w^2}$

$\dfrac{\pi A^2(3aA^2+8k -8mw)}{32w^2} + \dfrac{\pi A^2m}{2}$

Ive been staring at this for an hour and a half

I know that there should be a four as that last two (on the denominator) Somebody help. This is really embarrassing I'm a uni student

答:
To expand on Willie's comment (which to be honest should say it all):

$$\begin{align}\dfrac{\pi A^2(3aA^2+8k -\boxed{8mw^2})}{32w^2} + \dfrac{\pi A^2m}{2} &= \dfrac{\pi A^2(3aA^2+8k)}{32w^2} \boxed{-\dfrac{\pi A^2\cdot 8mw^2}{32w^2} + \dfrac{\pi A^2m}{2}}
\\ &= \cdots + \boxed{\pi A^2m\left(-\dfrac1 4 + \dfrac1 2\right)}
\\ &= \cdots + \boxed{\pi A^2m\times \dfrac1 4} \end{align}$$

Title: Differentiate with respect to w


答:
As @ironX & @pietro said, and also by definition of vector spaces, they contain the $Span$ of any subset of themselves. But for inverse, also as they said truly, you can't conclude it. Of course, if you are curious to find such condition:

If you have $\dim(Span\{ v_1, v_2 \}) \leqslant \dim(Span\{ b_1, b_2 \})$, beside $Span\{ b_1, b_2 \} \subseteq Span\{ v_1, v_2 \}$, then you can conclude the inverse: $Span\{ v_1, v_2 \} \subseteq Span\{ b_1, b_2 \}$ and so they become equal with equal dimension.

In some cases investigating the independency of vectors can help much.


问:
Computing principal curvatures in local coordinates
I was wondering how to compute the principal curvatures when working with a local parametrization. (Context: Surfaces, i.e. 2-dim Manifolds). Since I have formulas for the Gaussian and the mean curvature, I used them to derive:

$$k_i = H \pm \sqrt{H^2 - K} \, ,$$

where the $k_i$ denote the two principal curvatures, K the Gaussian curvature and H the mean curvature. Now when looking through the Wikipedia page about Differential geometry of surfaces I saw a different formula (in the bottom row of the table I have linked):

$$k_i = \frac{H \pm \sqrt{H^2 - 4K}}{2}.$$

Now I have two questions:

Why is the formula different? Is it a different context? As far as I could tell everything else matches what I am working with...

Is there a better way to compute the principal curvatures when working with local coordinates? Using the formulas for the Gaussian and mean curvature is pretty straight forward, which I like, but it does involve a lot of computation. Maybe there is a more effective way to do it?

Thanks!

答:
It seems that in your formulas the definition of the mean curvature differs. There is a choice to be made and both options are equally valid and are used in the literature. One way would be be
$$\frac{\kappa_1+\kappa_2}{2}=H,$$
i.e. $H$ is the mean of the principal curvatures. The other option is that $H$ is the sum of the principal curvatures, i.e.
$$\kappa_1+\kappa_2=H.$$
Wikipedia uses this version.
Since the underlying math and geometry is not really changed, both version are valid. Only the calculations may differ here and there.

You could calculate the principal curvatures as the eigenvalues of the second fundamental form. It may be a bit shorter, see https://en.wikipedia.org/wiki/Principal_curvature


问:
If $X_1$ and $X_2$ are individually independent to $\{Z_k\}$ does it follow that their sum is too.
If $X_1$ and $X_2$ and $\{Z_k\}_{k}$ denote Random variables on a probability space and it holds that

$\{X_1\}\cup\{Z_k\}$ is a collection of independent r.v.s

$\{X_2\}\cup\{Z_k\}$ is a collection of independent r.v.s

does it then hold that $\{X_1+X_2\}\cup\{Z_k\}_k$ is a collection of independent random variables? We do not assume that $X_1$ and $X_2$ are independent.

We know that $\sigma(X_1+X_2) = \sigma(\sigma(X_1)\cup \sigma(X_2))$, however this doesn't say much?

答:
No. Let $A,B,C$ be pair-wise independent events with $P(A\cap B\cap C) \neq P(A)P(B)P(C)$. Let $X=I_A, Y=I_B$ and $Z=I_C$. Then the hypothesis is satisfied. If $X+Y$ is indepdent of $Z$ then we would have $E((X+Y)^{2}Z)=E(X+Y)^{2}EZ$. A little simplification of this equation gives $P(A\cap B\cap C) =P(A)P(B)P(C)$ which is not true.

For an eample of such events see https://en.wikipedia.org/wiki/Pairwise_independence


答:
It is just a matter of convention on what you call the mean curvature $H$.

Suppose $\Sigma^2\subset \mathbb{R}^3$ is an embedded surface and $S$ is its shape operator. Then the principal curvatures are the eigenvalues of $S$ (and the principal curvature directions are the corresponding eigen directions).

Some author refer to the mean curvature $H$ as the trace of $S$, that is, the sum of the principal curvatures, while some others refer to $H$ as the mean of the principal curvatures, that is, in dimension $2$, $\frac{1}{2}\mathrm{trace}S$.

Changing $H$ by $H/2$ in your first formula gives the second.

Concerning the computations of the principal curvatures: it is in general a hard and impossible problem because it leads to computing eigenvalues of an endomorphism and roots of a polynomial, which we know is impossible in degree greater than 5. But in dimension $2$, the formula you gave is just the usual formula for computing the roots of a degree 2 polynomial!

Title: Computing principal curvatures in local coordinates

Body:
I was wondering how to compute the principal curvatures when working with a local parametrization. (Context: Surfaces, i.e. 2-dim Manifolds). Since I have formulas for the Gaussian and the mean curvature, I used them to derive:

$$k_i = H \pm \sqrt{H^2 - K} \, ,$$

where the $k_i$ denote the two principal curvatures, K the Gaussian curvature and H the mean curvature. Now when looking through the Wikipedia page about Differential geometry of surfaces I saw a different formula (in the bottom row of the table I have linked):

$$k_i = \frac{H \pm \sqrt{H^2 - 4K}}{2}.$$

Now I have two questions:

Why is the formula different? Is it a different context? As far as I could tell everything else matches what I am working with...

Is there a better way to compute the principal curvatures when working with local coordinates? Using the formulas for the Gaussian and mean curvature is pretty straight forward, which I like, but it does involve a lot of computation. Maybe there is a more effective way to do it?

Thanks!


问:
Pulling out Conditional Expectation from a Rational Function
Let $X,Y$ be two (possibly dependent) random variables with $X,Y>-1$. I would like to evaluate the expected value of a ratio
$$\mathbb{E}\left[\frac{1+Y}{1+X}\right]$$ and stuck on some steps; see detail below.

My thought: Using conditonal expectation. For $y>-1$,
$$
\begin{align*}
\mathbb{E}\left[\frac{1+Y}{1+X}\right] &= \mathbb{E}\left[ \mathbb{E}\left[\frac{1+Y}{1+X} | \ Y=y\right]\right]\\
&= \mathbb{E}\left[ (1+y) \mathbb{E}\left[\frac{1}{1+X} | \ Y=y\right]\right]\\
&\geq \mathbb{E}\left[ (1+y) \frac{1}{\mathbb{E}[1+X| \ Y=y]} \right] \;\;\; \rm{By \ Jensen's \ inequality.}
\end{align*}
$$
However, I am not sure if the conditional expection (I know if standard expectation, I can, but not sure for conditional one.) can be pulled out? That is, I wonder
$$
\mathbb{E}\left[ (1+y) \frac{1}{\mathbb{E}[1+X| \ Y=y]} \right] = ? = \underbrace{\mathbb{E}\left[ 1+y \right]\frac{1}{\mathbb{E}[1+X| \ Y=y]}}_{=(1+y)\frac{1}{\mathbb{E}[1+X| \ Y=y]}}.
$$

Any thought/comment is appreciated.

答:
It is not very clear to me the symbol $y$ that you introduce. If the question is if this holds as an equality :

$$
\mathbb{E}\left[ (1+Y) \frac{1}{\mathbb{E}[1+X| \ Y]} \right] = ? = \mathbb{E}\left[ 1+Y \right]\frac{1}{\mathbb{E}[1+X| \ Y]}
$$

Than I would argue that it does not make much sense since on the right we have a Y measurable random variable, and on the left a (deterministic) scalar.

This instead would work if $\frac{1}{\mathbb{E}[1+X| \ Y]}$ and $1+Y$ were independent:

$$
\mathbb{E}\left[ (1+Y) \frac{1}{\mathbb{E}[1+X| \ Y]} \right] = ? = \mathbb{E}\left[ 1+Y \right]\mathbb{E} \left[\frac{1}{\mathbb{E}[1+X| \ Y]}\right]
$$

but not in a general case.


问:
Number of pairs with sum $m^2$

Let $a$ and $b$ are chosen at random from $S = \{1,2,\dots,n\}$. Then the number of pairs $(a, b)$ with sum $m^2$ is $m^2-1$ for $m^2\leq n$, and $2n-m^2+1$ for $n<m^2\leq 2n$.

I can easily understand the first part of the second statement but not the 2nd part. Can anyone explain how are we getting this?

For $n<m^2\leq 2n$ there are $m^2-1$ pairs which give sum $m^2$ but out of these pairs there are some of them whose entries do not belong to the set $S$. So, we have to remove those pairs. But I don't know how many of them are there in general. I tried with some examples but in vain.

Thanks!

答:
For each $s\in S$ let $s'=n+1-s$; the map $S\to S:s\mapsto s'$ is an involution. For $a,b\in S$ we have $a+b=m^2$ if and only if $a'+b'=2n-m^2+2$.

Now assume that $n<m^2\le 2n$; then $2\le 2n-m^2+2\le n+1$, so that if $c+d=2n-m^2+2$ for positive integers $c$ and $d$, then $c,d\in S$. There are $2n-m^2+1$ ordered pairs $\langle a,b\rangle$ of positive integers whose sum is $2n-m^2+2$. Let $\langle a,b\rangle$ be such a pair; then $\langle a,b\rangle\in S\times S$, so $\langle a',b'\rangle\in S\times S$, and $a'+b'=m^2$. Thus, there are $2n-m^2+1$ pairs $\langle a,b\rangle\in S\times S$ such that $a+b=m^2$.

Title: Number of pairs with sum $m^2$


问:
Which definition of Haar measure is correct?
I have encountered several different definitions of left Haar measure that don't all seem to agree.

The setting I care about is Locally Compact Groups.

The first seems to completely disagree with the other two with the regularity dispute.

$\bf{\text{Which definition is correct? or are they all correct in different contexts?}}$

Let $G$ be a locally compact group, and by a measure I mean a Borel measure.

$\bf{\text{Definition 1:}}$

On Wikipedia: http://en.wikipedia.org/wiki/Haar_measure

A left Haar measure is a non-zero measure $\mu$ on $G$ such that

(i) $\mu(K) < \infty$ for compact $K$

(ii) $\mu$ is left $G$-invariant

(iii) $\mu$ is outer regular on Borel sets

(iv) $\mu$ is inner regular on open Borel sets

(And then an example is given to show that $\mu$ need not be inner regular on all Borel sets!)

$\bf{\text{Definition 2:}}$

Supplied in an answer: Why are Haar measures finite on compact sets?

A left Haar measure is a non-zero measure $\mu$ on $G$ such that

(i) $\mu$ is regular

(ii) $\mu > 0$ on open sets.

(iii) $\mu < \infty$ on compact sets.

(iv) $\mu$ is left $G$-invariant

$\bf{\text{Definition 3:}}$

Given in an appendix to a book.

A left Haar measure is a non-zero measure $\mu$ on $G$ such that

(i) $\mu$ is regular

(ii) $\mu$ is left $G$-invariant.

答: The first two are equivalent and both correct.  The third one requires finite-ness on compact sets.  Positivity on open sets follows from local compactness of $G$ and non-triviality of $\mu$.</p>


问:
Represent a system of nonlinear equations in compact form
If a system of linear equations can be represented by the compact form, $$Ax=b$$ how will a system of nonlinear equations (system of quadratic equations and system of exponential equations) be represented in compact form?

答:
This is way too general. Perhaps something can be said about quadratic equations: you can write them as $x^*Ax + b^*x +c=0$, where $x$ is the vector of unknowns, b is a vector and c is a scalar. But a system of such things... That would require 3-dimensional matrices, and it would not be nice. If you think about the degree of freedom: you need two dimensions for all variations of the form $x_ix_j$, and a third one for the different equations.

As for exponential equations, I am not even sure what you mean, and what the degree of freedom would be. I do not think there is a compact form.


答:
If $1\leq a,b\leq n$ and $m^2\leq 2n$ then
$$n\geq \max(a,b)\geq \frac{a+b}{2}\geq \frac{m^2}{2}$$
is always satisfied. Therefore for any $a\in \{m^2-n,\dots,n\}$  (note that $m^2-n\geq 1$ because $m^2>n$), we have a proper couple $(a,m^2-a)$. Finally observe that the cardinality of the set of values for $a$, that is $\{m^2-n,\dots,n\}$ is just
$$n-(m^2-n)+1=2n-m^2+1.$$

Title: Number of pairs with sum $m^2$

Body:
Let $a$ and $b$ are chosen at random from $S = \{1,2,\dots,n\}$. Then the number of pairs $(a, b)$ with sum $m^2$ is $m^2-1$ for $m^2\leq n$, and $2n-m^2+1$ for $n<m^2\leq 2n$.

I can easily understand the first part of the second statement but not the 2nd part. Can anyone explain how are we getting this?

For $n<m^2\leq 2n$ there are $m^2-1$ pairs which give sum $m^2$ but out of these pairs there are some of them whose entries do not belong to the set $S$. So, we have to remove those pairs. But I don't know how many of them are there in general. I tried with some examples but in vain.

Thanks!


问:
Is it true that $M$ has $0$ as eigenvalue with multiplicity exactly $1$.
I am stuck on the following question: Its a true/false statement.

It says : Let $M$ be an $n\times n$ matrix. If none of the entries of $M$ are 0, but the row sum is zero for each row and no two rows are scalar multiples of each other, is it true that $M$ has $0$ as eigenvalue with multiplicity $1$.

My try:  Since $M$ is an $n\times n$ matrix with row sum  $0$ for each row, so $0$ is an eigenvalue of $M$ with multiplicity atleast $1$.
But is it true that it is an eigenvalue with multiplicity exactly $1$. I am stuck. I am getting it false. Is it correct?

What shall i do ? Can someone help.

Body:
A counterexample is given by
$$\begin{pmatrix} -3&1&1&1 \\ 1&-3&1&1\\-4&2&1&1\\2&-4&1&1 \end{pmatrix} $$


问:
Clarifying the definition of a logical system on wikipedia

Logical system is a deductive system together with additional (non-logical) axioms and a semantics. An example of a logical system is Peano arithmetic.  Wikipedia

Can you please explain what is non-logical axioms and a semantics in this context?

答:
As per discussions on previous similar posts we have many terms with similar meaning.

Maybe the first one is Formal language: an alphabet made of an initial (usually finite) stock of basic symbols and a set (usually finite) of rules to produce expressions (usually with finite length) that are called formulas.

With it we build a Formal system (also called a logical calculus, or a logical system) consisting of a formal language together with a deductive apparatus (also called a deductive system).

The deductive apparatus is a set (usually finite) of rules, called rules of inference, that we use to produce formulas from an initial stock of formulas: the axioms.

This is the syntactical part.

The logical system is used to derive logical truths (like e.g. $\vdash p \lor \lnot p$) and to derive consequences from assumptions: $\Gamma \vdash \varphi$.

We may call it the "pure" logical theory.

When the set of assumptions $\Gamma$ is the formalized version of the collection of axioms of some mathematical theory, like e.g. the firts-order arithmetic that formalizes the theory based on Peano axioms, we have an "applied" logical theory.

A formalized mathematical theory can be interpreted: for first-order arithmetic, the "natural" intepretation is the mathematical structure base on the set $\mathbb N$ of natural numbers.

The intuitive notion of interpretation, i.e. the way to assign meaning to symbols and formulas of a formal language, is made rigorous through the mathematical definition of Interpretation of a formal language.

This is the semantics.


问:
Integral of Bessel function $K_0$
I find myself dealing with the following integral $$
\int_{-1} ^1 dt \,e^{it} K_0(b t),
$$
and I was thinking if this can be somehow expressed in terms of something nice, like other Bessel functions for instance. $b $ is a constant and $K_0$ is the modified Bessel function of the second kind.

Thanks for your help!

Title: Integral of Bessel function $K_0$

Body:
I am not sure (at all) that there is a solution to this problem and I am afraid that only numerical integration could be the only way to go.

What I would do is to generate tables for a series of values of $b$ and use any method of interpolation (independently for the real and imaginary parts of the result); the functions are really smooth.

I give below a few values
$$\left(
\begin{array}{ccc}
b & \text{real part} & \text{imaginary part} \\
 0.25 & +2.98678 & -2.65532 \\
 0.50 & +1.83156 & -2.69092 \\
 0.75 & +1.14941 & -2.75129 \\
 1.00 & +0.64764 & -2.83803 \\
 1.25 & +0.23109 & -2.95345 \\
 1.50 & -0.14555 & -3.10066 \\
 1.75 & -0.50939 & -3.28370 \\
 2.00 & -0.87976 & -3.50766 \\
 2.25 & -1.27265 & -3.77889 \\
 2.50 & -1.70300 & -4.10521 \\
 2.75 & -2.18611 & -4.49626 \\
 3.00 & -2.73868 & -4.96380 \\
 3.25 & -3.37976 & -5.52223 \\
 3.50 & -4.13165 & -6.18909 \\
 3.75 & -5.02096 & -6.98583 \\
 4.00 & -6.07974 & -7.93861 \\
 4.25 & -7.34693 & -9.07937 \\
 4.50 & -8.87001 & -10.4472 \\
 4.75 & -10.7072 & -12.0898 \\
 5.00 & -12.9298 & -14.0658
\end{array}
\right)$$


问:
Does existential import rejects subcontraries?
In Categorical logic (AEIO) I and O are subcontraries. They both can not be False at the same time.

What happens after integrating existential import?

My book says:

Two propositions are subcontraries if they cannot be false together.
Yet I- and O-propositions are both false when there is no S. If there
is no S, then there is no S that is P and there is no S that is not P.
It just is nothing at all. So I- and O-propositions are no more
subcontraries.

Lee, Siu-Fan. Logic: A Complete Introduction: Teach Yourself (Complete
Introductions) (p. 158). John Murray Press. Kindle Edition.

But I and O propositions always have existential import. And this post confirms that subcontraries retain.

答:
There is a difference between the Traditional Square of Oppositions and the modern standard translation of so-called Categorical propositions.

According to Aristotle's Logic, the Universal Affirmative ($\text {Aab}$): "a  belongs to all b" implies the corresponding Particular ($\text {Iab}$): "a belongs to some b".

With modern symbolization we have:

$\text {(Aab)} \ \ \forall x (b(x) \to a(x))$

and

$\text {(Iab)} \ \ \exists x (b(x) \land a(x))$.

The issue with so-called Existential Import is that $\text {Aab}$ implies $\text {Iab}$ under the condition that there are $\text A$'s.

The plausible explanation is that, according to Aristotle theory of science, terms are universals and they cannot be empty.

The origin of the Square is Aristotle's De Interpretatione (17b.17–26):

A and O are contradictories [they cannot both be true and they cannot both be false]; E and I are contradictories, and A and E are contraries [they cannot both be true but can both be false].

From this, the rest of the Square follows:  I and O are subcontraries [they cannot both be false but can both be true], i.e. they cannot both be false, as well as subalternation [a proposition is a subaltern of another iff it must be true if its superaltern is true, and the superaltern must be false if the subaltern is false].

Subcontraries: suppose that I is false. Then its contradictory, E, is true. So E’s contrary, A, is false. An thus A’s contradictory, O, is true. The conclusion is that I and O cannot be both false.

Subalternation: suppose that A is true. Then its contrary E must be false. But then the E’s contradictory, I, must be true. Thus if the A form is true, so must be the corresponding I.

According to modern translation, $\text {Iab}$ is $\exists x (a(x) \land b(x))$ and $\text {Oab}$ is $\exists x (a(x) \land \lnot b(x))$.

If there are no $a$, both are false. Thus, according to modern point of view, two subcontraries can both be false.


问:
Looking for a theorem concerning similar triangles and proportionality
While trying to provide a proof for this question, I stumbled upon a theorem that I have probably seen before:

Theorem. If line segments joining corresponding vertices of two similar triangles in the same orientation (not reflected) are split into equal proportions, the resulting points form a triangle similar to the original triangles.

In the diagram above, $RZ: ZU = TA_1:A_1W = SB_1:B_1V$, and the purple triangle is similar to the orange triangles. Note that the lines does not need to be parallel or concurrent, so the triangles are not necessarily in perspective or share any parallel sides. This theorem can also be extended easily to polygons.

I am confident that this is indeed a theorem, but I am having a hard time recalling its name, and searching for geometric theorems is practically impossible.

For those who are interested, a solution to the linked question is immediate using this theorem and a simple reflection:

since $CH:HE = B'G:GB = AI:ID = 2:1$ where $B'$ is the reflection of $B$ across the line $AC$.

答:
Let us use complex numbers geometry.

Two triangles $T=(z_1,z_2,z_3)$ and $T'=(z'_1,z'_2,z'_3)$ are similar iff there exists a similitude $Z=az+b$ (with $a,b \in \mathbb{C}$ ) sending the first one onto the second one.

Therefore if a third triangle $T''$ is for example at the two-thirds from $T$ and $T'$ with the relationship described above, we can write :

$$\begin{cases}z"_1&=&\frac13 z_1+\frac23 z'_1&=&\frac13 z_1+\frac23 (az_1+b)&=& \frac{2a+1}{3}z_1+\frac23 b\\
z"_2&=&\frac13 z_2+\frac23 z'_2&=&\frac13 z_2+\frac23 (az_2+b)&=& \frac{2a+1}{3}z_2+\frac23 b\\
z"_3&=&\frac13  z_3+\frac23 z'_3&=&\frac13 z_3+\frac23 (az_3+b)&=& \frac{2a+1}{3}z_3+\frac23 b\end{cases}$$

exhibiting a similitude

$$Z=\frac{2a+1}{3}z+\frac23 b$$

sending $T$ onto $T''$, therefore establishing that $T$ and $T''$ are similar.

Remark: Of course, what we have done with coefficients $\frac13$ and $\frac23$ can be done  with any other "weighting" on triangles $T$ and $T'$.


问:
people liars or trustworthy
I have a visitor that meets A and B.We have trustworth and liars.
A declares that he is like B.
B declares that one of them is only trustworth.

So I have solved this exercise with truth table

A: trustworthy so p
B: trustworthy  so q (cause he is saying that one of them is trustworth)
I have the table

```<code>p        q       p  ^ q
true    true      lie
true     lie      lie
lie      true     lie
lie      lie      true
</code>```

from their  statement we have true and true  so both of them are liars.What the visitor assume ?

1)A and B trustworth 2)A trustworth and B liar 3)A liar and B
trustworth 4)both liars

I find 4.both liars
is it?

答:
As @user619755 said, you can see:

If A is trustworthy, so it said the truth, therefore B become trustworthy like it and say the truth only, but it contradict with its statement, since there is 2 trustworthy, hence A can't be trustworthy.

If B is liar, its statement become wrong, so A can't be trustworthy, therefore it also its statement become wrong, but it contradict with both of them being like each other (liar), hence B can't be liar.

Now the only condition remains:

A is liar and B is trustworthy, So they aren't like each other, as we expect from liarness of A, also there is just one trustworthy, as we expect from trustworthiness of B. Conditions are compatible. So it's the answer.

Title: people liars or trustworthy


问:
Homeomorphism examples
What's a homeomorphism?

a. any countable $f\colon \mathbb {N} \to \mathbb {Q}$

b. $f\colon \mathbb {R} \to \mathbb {R}$, $x \mapsto x^3$

c. $f\colon [0,2\pi ) \to S^1 = \{ z \in \mathbb {C}\,|\,|z|=1\}$, $\varphi \mapsto e^{i \varphi }$

I'm very sure that b doesn't work because it's not bijective and I think that everything is fine with a. But I'm really unsure about c, can someone explain it for me?

答:
a. There exist bijections $f : \mathbb N \to \mathbb Q$, but none of them can be a homeomomorhism. In fact, let $x \in \mathbb N$. Then $\{x\}$ is open in $\mathbb N$, but $f(\{x\}) = \{f(x)\}$ is never open in $\mathbb Q$.

b. It is a homeomorphism. Its inverse is $f^{-1}(x) = \sqrt[3]{x}$ which is continuous.

c. $[0,2\pi)$ is non-compact, $S^1$ is compact. Thus they cannot be homeomorphic.

Title: Homeomorphism examples


问:
A matrix problem :)

If $l_i,m_i,n_i$  ; $i=1,2,3$ denote the direction cosines of three mutually perpendicular vectors in space, provided that $AA^T=I$ ,where
  $$A=\begin{bmatrix}
        l_1 & m_1 & n_1 \\
        l_2 & m_2 & n_2 \\
        l_3 & m_3 & n_3 \\
        \end{bmatrix} $$

I couldn't quite understand the question. Sorry for not posting 'how I tried to solve the answer'.

Please provide assistance.

Thank you. :)

答: Hint: Consider first $$A^TA=  \begin{bmatrix}l_1 & l_2 & l_3 \\m_1 & m_2 & m_3 \\n_1 & n_2 & n_3 \\ \end{bmatrix} \begin{bmatrix}l_1 & m_1 & n_1 \\l_2 & m_2 & n_2 \\l_3 & m_3 & n_3 \\ \end{bmatrix} = \cdots $$ by working out a few of the elements in terms of the components of $\mathbf {l,m,n}$. Then ask yourself: if $AB=I$, then what is $A^{-1}$? Therefore, what is $BA$?

Title: A matrix problem :)


问:
Non-unique solutions for ODE with boundary conditions at infinity
Ello,

I am looking for solutions to equations such as

$$ U ^ \prime (y) - y^{\prime \prime} + \beta y^{\textit{IV}} = 0$$
where $\beta >0$ is a constant and $U(x)$ is a function whose Taylor expansion at the origin equals $ \alpha x^2 + \dots$ for $\alpha >0$ and such that $\lim_{x \to \infty} = -\infty$.
A prototypical example would be the quartic double-well, $$ U(x) = x^2 -x^4$$.

There is a trivial solution $y(x) = 0$. On intuitive grounds (please see physical background below), as well as noting that the origin is a hyperbolic point for a dynamical system governed by the related system of equations, another solution is expected as well, with boundary conditions $$ \begin{cases} \lim_{x \pm \infty} y(x) = 0  \\
\lim_{x \pm \infty} y^ \prime (x) = 0 \end{cases}$$

This fact does not contradict the unicity theorem, as far as I understand, as the latter deals with boundary conditions on the real line, I believe.

Now to my problem, what numerical scheme could be used to get the non-trivial solution? Anything I could think of will pick the trivial solution branch only.
Further, I would be grateful for theory references on the possibility of multiple solutions for boundary conditions at infinity.

Things would be different if $\beta$ were equal to $0$.
In that case it is easy to determine two boundary conditions (also clarified in the physical background section below) for $x = 0$, and any numerical method will handle it easily.

Physical Background

The ODE is the Euler-Lagrange equation for the functional

$$ \int _{- \infty} ^{\infty } \frac{1}{2} y^{\prime 2} + U(y) + \frac{\beta}{2} y^{\prime \prime 2} \mathrm{d}x $$
representing the energy of a stretchable beam in the potential $U$. The first term represents the stretching energy, the second the potential energy, the third the bending energy.
Given this physical picture, the trivial solution represents a beam lying in the potential energy "valley" at the origin, un-stretched and straight. The non-trivial solution represents a beam lying on the valley as before, then deforming over the potential energy "bump", kept in the equilibrium by the pull from the descending potential energy on the other side, and going back to the valley at the origin.

If $\beta = 0$ the non-trivial solution is amenable to a simple solution.
The energy functional reduces in such case to

$$ \int _{- \infty} ^{\infty } \frac{1}{2} y^{\prime 2} + U(y)  \mathrm{d}x $$ which could be interpreted as the action of a particle rolling in a potential energy $-U(x)$ (with $x$ now representing time). The turning point is then easily found by solving the equation $ U(x_1) = 0$: that is, the particle starts from $x = 0$, rolls down and up until $x_1$, and rolls back in a homoclinic orbit.
This gives two boundary conditions

$$ \begin{cases}  y(0) = x_1  \\
 y^ \prime (0) = 0 \end{cases}$$

and these allow any numerical scheme to pick the non-trivial branch.

If $\beta \neq 0$, this simple reasoning fails for the full fourth-order ODE (I also thought about converting the fourth-order ODE to a system of equations, to attempt a similar interpretation for a system of four particles, but did not get anywhere yet, apart confirming that a second non-trivial solution exists). I am hence left with the boundary conditions a infinity, and cannot get any solution other than the trivial one.

To sum up, I would like to be pointed towards references for the non-unicity of solutions when boundary conditions at infinity are applied, and to understand how to numerically solve similar problems.

The ODE has a physical interpretation and I was unsure whether the question were more apt for Physics StackExchange, or even SciComp StackExchange. I would certainly re-route if advised, I thought to start here as ultimately the question is primarily on the maths, thanks a lot.

答:
You can use a linearization procedure to solve this nonlinear boundary value. For instance using a linear approximation scheme consisting in iterations over the linear ODE

$$
y''_{k+1}-V(y_k)+V'(y_k)(y_{k+1}-y_k)-\beta y''''_{k+1}=0
$$

given $y_0$ and the boundary conditions

$$
\cases{
y_{k+1}(-x_{max})=0\\
y'_{k+1}(x_{max})=0\\
y''_{k+1}(-x_{max})=0\\
y''_{k+1}(x_{max})=0
}
$$

Let us proceed with the calculations for $\beta = 0$ which is known to have as solution

$$
y=\sqrt{1-\tanh ^2\left(\sqrt{2} x\right)}
$$

We will approach this solution using negative $\beta$'s as follows in the attached MATHEMATICA script.

```<code>Clear[V, dV, y1, y0]
V[x_] := -2 x + 4 x^3;
dV[x_] := -2 + 12 x^2

y0 = Exp[-x];
beta = -0.00001;
xmax = 5;
nmax = 20;
bcs = {y1[-xmax] == 0, y1[xmax] == 0, y1'[-xmax] == 0, y1'[xmax] == 0};
SOLS = {Plot[y0, {x, -xmax, xmax}]};
thickness = Thin;
color = Blue;

For[k = 1, k <= nmax, k++,
   ode = y1''[x] + V[y0] + dV[y0] (y1[x] - y0) - beta y1''''[x];
   sol = NDSolve[Join[{ode == 0}, bcs], y1, {x, -xmax, xmax}][[1]];
   y0 = y1[x] /. sol;
   If[k == nmax, thickness = Thick; color = Black];
   AppendTo[SOLS, Plot[y0, {x, -xmax, xmax}, PlotStyle -> {thickness, color}]]
]

grfin = SOLS[[Length[SOLS]]];
Show[Plot[Sqrt[1 - Tanh[Sqrt[2] x]^2], {x, -xmax, xmax}, PlotStyle -> {Thick, Red}], grfin]
Show[SOLS, grfin, PlotRange -> All]
</code>```

so with $\beta = -0.00001$ we obtain

Body:
You can use a linearization procedure to solve this nonlinear boundary value. For instance using a linear approximation scheme consisting in iterations over the linear ODE

$$
y''_{k+1}-V(y_k)+V'(y_k)(y_{k+1}-y_k)-\beta y''''_{k+1}=0
$$

given $y_0$ and the boundary conditions

$$
\cases{
y_{k+1}(-x_{max})=0\\
y'_{k+1}(x_{max})=0\\
y''_{k+1}(-x_{max})=0\\
y''_{k+1}(x_{max})=0
}
$$

Let us proceed with the calculations for $\beta = 0$ which is known to have as solution

$$
y=\sqrt{1-\tanh ^2\left(\sqrt{2} x\right)}
$$

We will approach this solution using negative $\beta$'s as follows in the attached MATHEMATICA script.

Clear[V, dV, y1, y0]
V[x_] := -2 x + 4 x^3;
dV[x_] := -2 + 12 x^2

y0 = Exp[-x];
beta = -0.00001;
xmax = 5;
nmax = 20;
bcs = {y1[-xmax] == 0, y1[xmax] == 0, y1'[-xmax] == 0, y1'[xmax] == 0};
SOLS = {Plot[y0, {x, -xmax, xmax}]};
thickness = Thin;
color = Blue;

For[k = 1, k <= nmax, k++,
   ode = y1''[x] + V[y0] + dV[y0] (y1[x] - y0) - beta y1''''[x];
   sol = NDSolve[Join[{ode == 0}, bcs], y1, {x, -xmax, xmax}][[1]];
   y0 = y1[x] /. sol;
   If[k == nmax, thickness = Thick; color = Black];
   AppendTo[SOLS, Plot[y0, {x, -xmax, xmax}, PlotStyle -> {thickness, color}]]
]

grfin = SOLS[[Length[SOLS]]];
Show[Plot[Sqrt[1 - Tanh[Sqrt[2] x]^2], {x, -xmax, xmax}, PlotStyle -> {Thick, Red}], grfin]
Show[SOLS, grfin, PlotRange -> All]

so with $\beta = -0.00001$ we obtain


问:
Simplify a complicated trigonometric expression

My question is to simplify the following expression:$$\frac{1}{2\sin x}\left(\sin 2x + \cos \frac{\pi}{4} + \cos \left( \frac{3\pi}{4} - 2x\right) + \cos 0 + \cos \left(\pi - 2x\right) + \cos\left(-\frac{\pi}{4}\right) + \cos\left(\frac{5\pi}{4} - 2x\right) + \cos\left(-\frac{\pi}{2}\right) + \cos\left(\frac{3\pi}{2} - 2x\right)\right)$$

I have this complicated trigonometric expression as above, it is obtained in a geometry problem but I don't know what's the best way to simplify it. The answer for this is $(1 + \sqrt{2})\sin x$ according to WolframAlpha.

My attempt

\begin{align}
&\frac{1}{2\sin x}\left(\sin 2x + \cos \frac{\pi}{4} + \cos \left( \frac{3\pi}{4} - 2x\right) + \cos 0 + \cos \left(\pi - 2x\right) + \cos\left(-\frac{\pi}{4}\right) + \cos\left(\frac{5\pi}{4} - 2x\right) + \cos\left(-\frac{\pi}{2}\right) + \cos\left(\frac{3\pi}{2} - 2x\right)\right)\\
&=\frac{1}{2\sin x}\left(\cos \frac{\pi}{4} + \cos \left( \frac{3\pi}{4} - 2x\right) + \cos 0 + \cos \left(\pi - 2x\right) + \cos\left(-\frac{\pi}{4}\right) + \cos\left(\frac{5\pi}{4} - 2x\right) + \cos\left(-\frac{\pi}{2}\right)\right) \\
&=\frac{1}{2\sin x}\left(1 + \sqrt{2} + \cos \left( \frac{3\pi}{4} - 2x\right) + \cos \left(\pi - 2x\right) + \cos\left(\frac{5\pi}{4} - 2x\right)\right)
\end{align}
$$$$
How to compute after this? Should I use the compound angle formula to expand the $\cos(a + b)$ terms? It seems a bit heavy to me cuz I don't see how it will cancel nicely with $\sin x$. Or should I use the sum to product formula for cos? Thanks in advance!

答:
You are almost there.

$\frac{1}{2\sin x}\left(\sin 2x + \cos \frac{\pi}{4} + \cos \left( \frac{3\pi}{4} - 2x\right) + \cos 0 + \cos \left(\pi - 2x\right) + \cos\left(-\frac{\pi}{4}\right) + \cos\left(\frac{5\pi}{4} - 2x\right) + \cos\left(-\frac{\pi}{2}\right) + \cos\left(\frac{3\pi}{2} - 2x\right)\right) = \frac{1}{2\sin x}\left(\sin 2x + 1 + \sqrt2 + \cos \left(2\pi - (\frac{3\pi}{4} - 2x\right)\big) - \cos \left(2x\right) + \cos\left(\frac{5\pi}{4} - 2x\right) - \sin2x\right)$

$ = \frac{1}{2\sin x} \ (1 + \sqrt2 + \cos (\frac{5\pi}{4} + 2x) + \cos(\frac{5\pi}{4} - 2x) - \cos(2x) )$

Using $\cos(A+B) + \cos(A-B) = 2 \cos A \cos B$,

$ = \frac{1}{2\sin x} \ (1 + \sqrt2 - \sqrt2 \cos (2x) - \cos(2x) )$

$ = \frac{1+\sqrt2}{2\sin x} \ (1 - \cos(2x)) = = \frac{1+\sqrt2}{2\sin x} \ (2 \sin^2x) = (1+\sqrt2) \sin x$


答:
At least, simplify as (for example)

$$\cos \left( \frac{3\pi}{4} - 2x\right)=-\cos \left(2 x+\frac{\pi }{4}\right)$$
$$\cos \left(\pi - 2x\right)=-\cos (2 x)$$
$$ \cos\left(\frac{5\pi}{4} - 2x\right)=-\cos \left(2x-\frac{\pi }{4}\right)$$
$$\cos \left(2 x+\frac{\pi }{4}\right)+\cos \left(2 x-\frac{\pi }{4}\right)=\sqrt{2} \cos (2 x)$$

Title: Simplify a complicated trigonometric expression

Body:
My question is to simplify the following expression:$$\frac{1}{2\sin x}\left(\sin 2x + \cos \frac{\pi}{4} + \cos \left( \frac{3\pi}{4} - 2x\right) + \cos 0 + \cos \left(\pi - 2x\right) + \cos\left(-\frac{\pi}{4}\right) + \cos\left(\frac{5\pi}{4} - 2x\right) + \cos\left(-\frac{\pi}{2}\right) + \cos\left(\frac{3\pi}{2} - 2x\right)\right)$$

I have this complicated trigonometric expression as above, it is obtained in a geometry problem but I don't know what's the best way to simplify it. The answer for this is $(1 + \sqrt{2})\sin x$ according to WolframAlpha.

My attempt

\begin{align}
&\frac{1}{2\sin x}\left(\sin 2x + \cos \frac{\pi}{4} + \cos \left( \frac{3\pi}{4} - 2x\right) + \cos 0 + \cos \left(\pi - 2x\right) + \cos\left(-\frac{\pi}{4}\right) + \cos\left(\frac{5\pi}{4} - 2x\right) + \cos\left(-\frac{\pi}{2}\right) + \cos\left(\frac{3\pi}{2} - 2x\right)\right)\\
&=\frac{1}{2\sin x}\left(\cos \frac{\pi}{4} + \cos \left( \frac{3\pi}{4} - 2x\right) + \cos 0 + \cos \left(\pi - 2x\right) + \cos\left(-\frac{\pi}{4}\right) + \cos\left(\frac{5\pi}{4} - 2x\right) + \cos\left(-\frac{\pi}{2}\right)\right) \\
&=\frac{1}{2\sin x}\left(1 + \sqrt{2} + \cos \left( \frac{3\pi}{4} - 2x\right) + \cos \left(\pi - 2x\right) + \cos\left(\frac{5\pi}{4} - 2x\right)\right)
\end{align}
$$$$
How to compute after this? Should I use the compound angle formula to expand the $\cos(a + b)$ terms? It seems a bit heavy to me cuz I don't see how it will cancel nicely with $\sin x$. Or should I use the sum to product formula for cos? Thanks in advance!


问:
Why does skolemization just happen under the scope of universal?
I understand you can skolemize an existential under the scope of a universal and get :

$\forall$x $\exists$y.P(x,y) $\Leftrightarrow$ $\forall$x. P(x, f(x))

what if the universal quantifier is under the scope of negation? Can you get the following?

$\neg\forall$x $\exists$y.P(x,y)$\Leftrightarrow$ $\neg\forall$x. P(x, f(x))

As I understand skolemization usually is used in NNF, but why? what is wrong with the above?

my other question is how would you read this? $\neg\forall$x. P(x, f(x))

答:
Welcome to MSE!

We skolemize so that we can find explicit witnesses to existential quantifiers. That is, if $\forall x . \exists y . P(x,y)$ we should be able to choose such a $y$ for each $x$. But a consistent way of making such a choice is exactly the function $f_P$ so that $\forall x . P(x, f_P x)$.

Let's think about what $\lnot \forall x . \exists y . P(x,y)$ says. It says that it isn't the case that every $x$ has such a $y$. Pushing to NNF, we can rewrite this as $\exists x . \forall y . \lnot P(x,y)$. That is, for some distinguished value of $x$, there is no choice of $y$ making $P(x,y)$ true.

Now I ask you, if we wanted to skolemize, what should $f_P(x)$ be? There's no choice of $y$ which makes $P(x,y)$ true. You ask in the question about the formula $\lnot \forall x . P(x,fx)$, and you certainly could consider a function which satisfies this equation. But it (probably) isn't a useful thing to study. Why?

The equation $\forall x. P(x,f_P x)$ determines the behavior of $f_P$. You can imagine there are very few $y$ for which $P(x,y)$, and so there aren't many choice of $f_P(x)$. In fact, this condition is strict enough to make skolemization a very useful procedure.

In contrast, what does the equation $\lnot \forall x. P(x,f_P x)$ tell us? Again, moving to NNF we see $\exists x . \lnot P(x, f_P x)$. But this tells us almost nothing! Indeed, to satisfy this formula, we only need to determine $f_P(x)$ for one choice of $x$. Moreover, we have almost free reign! If we again interpret $P(x,y)$ as being something which doesn't happen often, then almost any choice of $y$ will satisfy $\lnot P(x,y)$. So knowing that $f_P$ satisfies this condition gives very little control over $f_P$.

As for your last question, I've already answered it above. But explicitly,
$\lnot \forall x . P(x,fx)$ should be read as

There is some $x$ which is not $P$ related to $fx$.

I hope this helps ^_^


问:
Find $\int_{\left(C\right)}xy{\rm d}x+y^{2}{\rm d}y$ with $\left(C\right)$ bound by $y\geq 0,x^{2}+y^{2}=4\left({\rm clockwise}\right).$
Prob. Find $\int_{\left ( C \right )}xy{\rm d}x+ y^{2}{\rm d}y$ with $\left ( C \right )$ closed by the path $y\geq 0, x^{2}+ y^{2}= 4\left ( {\rm clockwise} \right ).$

My attempt: $\int_{\left ( C \right )}xy{\rm d}x+ y^{2}{\rm d}y= \int_{\left ( C \right )}xy{\rm d}x+ \frac{x^{2}-4}{2}{\rm d}y+ \int_{0\leq y\leq 2}\frac{3y^{2}}{2}{\rm d}y,$ let $P\left ( x, y \right )= xy, Q\left ( x, y \right )= \frac{x^{2}- 4}{2},$ so we have $\frac{\partial}{\partial y}P\left ( x, y \right )= \frac{\partial}{\partial x}Q\left ( x, y \right )= x$ on whole plane $\left ( Oxy \right )$ then the integral isn't up to the route. Replace $x= 2\cos t, y= 2\sin t\geq 0, 0\leq t\leq \pi,$ solve $\int_{\left ( C \right )}xy{\rm d}x+ y^{2}{\rm d}y= 0.$

This is my 1st time using Green's theorem, and I'm not sure of my opinion, I replaced $\int_{\left ( C \right )}\frac{3y^{2}}{2}{\rm d}y$ by $\int_{0\leq y\leq 2}\frac{3y^{2}}{2}{\rm d}y,$ it's no use. What is your thought on that ? Thanks a real lot.

答:
If it is a closed path,

$\int_{\left(C\right)} \ xy \ {\rm d}x+y^{2} \ {\rm d}y$

$Q_x = 0, P_y = x$

Given it is clockwise, the line integral is equivalent to

$\int_C (P_y - Q_x) \ dx \ dy = \int_C x \ dx \ dy$

As $x$ is an odd function and we have semicircular region symmetric to $y-$axis, the  integral is zero.


问:
The law of excluded middle and the abstraction of "actual infinity".
The encyclopedia of mathematics states:

"Logically, acceptance of the abstraction of actual infinity leads to the acceptance of the law of the excluded middle as a logical principle."

This is mentioned in justifying constructive approaches.

What is the intuition for (and presumably proof of?) this "logical" connection?

答:
See Brouwer’s Development of Intuitionism:

on Brouwer’s view, there is no determinant of mathematical truth outside the activity of thinking, a proposition only becomes true when the subject has experienced its truth (by having carried out an appropriate mental construction); similarly, a proposition only becomes false when the subject has experienced its falsehood (by realizing that an appropriate mental construction is not possible). Hence Brouwer can claim that “there are no non-experienced truths”.

in his dissertation On the Foundations of Mathematics, defended in 1907, [...] Brouwer sets out to reconstruct Cantorian set theory. When an attempt at making constructive sense out of Cantor’s second number class (the class of all denumerably infinite ordinals) and higher classes of even greater ordinals fails, he realises that this cannot be done and rejects the higher number classes, leaving only all finite ordinals and an unfinished or open-ended collection of denumerably infinite ordinals [emphasis added]. Thus, as a consequence of his philosophical views, he consciously puts aside part of generally accepted mathematics. Soon he would do the same with a principle of logic, the principle of the excluded middle (PEM), but in the dissertation he still thinks of it as correct but useless, interpreting $p \lor \lnot p$ as $\lnot p \to \lnot p$.

In “The Unreliability of the Logical Principles” of 1908, Brouwer formulates, in general terms, his criticism of PEM: although in the simple form of $p \lor \lnot p$, the principle will never lead to a contradiction, there are instances of it for which one has, constructively speaking, no positive grounds. Brouwer names some. Because they do not in the strict sense refute PEM, they are known as “weak counterexamples”.

Later Brouwer [showed] a refutation of PEM in the form $\forall x \in \mathbb R (Px \lor \lnot Px)$, by showing that it is false that every real number is either rational or irrational (the first of "strong counterexamples").

The interplay between PEM (or LEM) and actual infinity is one of the basic ground for Intuitionistic Logic:

the constructive validity [of PEM: $A ∨ ¬A$] would mean that we have a method that, for any $A$, either gives us a construction for $A$, or shows that such a construction is impossible. But we do not have such a general decision method, and there are many open problems in mathematics. Brouwer states “Every number is finite or infinite” as an example of a general proposition for which so far no constructive proof has been found. As a consequence, he says, it is at present uncertain whether problems such as the following are solvable:

Is there in the decimal expansion of π a digit which occurs more often than any other one?

Do there occur in the decimal expansion of π infinitely many pairs of equal consecutive digits?

In effect, Brouwer is saying that we can assert the weak negations of the propositions expressed in these questions; hence, these propositions are so-called “Brouwerian counterexamples” or “weak counterexamples” to PEM. On the constructive reading of PEM, of course any as yet unsolved problem is a weak counterexample to PEM.

In a nutshell, the basic Brouwerian "rejections" are: LEM applied to infinite collections (for finite sets, there is no issue with it) and the indirect proof of existence, i.e. deriving $\exists x Px$ from a derivation of a contradiction from assumption $\forall x \lnot Px$.


问:
Solving 3 simultaneous cubic equations
I have three equations of the form:

$$i_1^3L_1+i_1K+V_1+(i_2+i_3+C)Z_n=0$$
$$i_2^3L_2+i_2K+V_2+(i_1+i_3+C)Z_n=0$$
$$i_3^3L_3+i_3K+V_3+(i_1+i_2+C)Z_n=0$$

where $L_1,L_2,L_3,K,V_1,V_2,V_3,C$ and $Z_n$ are all known constants.

What methods can I use to obtain the values of $i_1,i_2$ and $i_3$ ?

答:
A numerical way to solve this would be to use the Newton-Raphson method. This method can be extended to 3 dimensions as follows:

$$\vec{i_{n+1}}=\vec{i_n}-J^{-1}(i_n)\vec{f}(i_n)$$

Where $J$ is the Jacobian matrix of the system:

$$J=
        \begin{bmatrix}
        3i_1^2L_1+K & Z_n & Z_n \\
        Z_n & 3i_2^2L_2+K & Z_n \\
        Z_n & Z_n & 3i_3^2L_3+K \\
        \end{bmatrix}
$$

Choose an initial "guess" $\vec{i_0}$, and repeat this process. Since it's an iterative process, the more times you evaluated it, the closer you get to the solution.

Title: Solving 3 simultaneous cubic equations


答:
The $y^2dy$ piece is conservative so it will equal

$$\frac{1}{3}y^3\Bigr|_{(-2,0)}^{(2,0 )} = 0$$

by the fundamental theorem of line integrals. The first integral is

$$\int_{-2}^2x\sqrt{4-x^2}dx = 0$$

because the function is odd. Hence the whole thing is $0$.

Title: Find $\int_{\left(C\right)}xy{\rm d}x+y^{2}{\rm d}y$ with $\left(C\right)$ bound by $y\geq 0,x^{2}+y^{2}=4\left({\rm clockwise}\right).$

Body:
Prob. Find $\int_{\left ( C \right )}xy{\rm d}x+ y^{2}{\rm d}y$ with $\left ( C \right )$ closed by the path $y\geq 0, x^{2}+ y^{2}= 4\left ( {\rm clockwise} \right ).$

My attempt: $\int_{\left ( C \right )}xy{\rm d}x+ y^{2}{\rm d}y= \int_{\left ( C \right )}xy{\rm d}x+ \frac{x^{2}-4}{2}{\rm d}y+ \int_{0\leq y\leq 2}\frac{3y^{2}}{2}{\rm d}y,$ let $P\left ( x, y \right )= xy, Q\left ( x, y \right )= \frac{x^{2}- 4}{2},$ so we have $\frac{\partial}{\partial y}P\left ( x, y \right )= \frac{\partial}{\partial x}Q\left ( x, y \right )= x$ on whole plane $\left ( Oxy \right )$ then the integral isn't up to the route. Replace $x= 2\cos t, y= 2\sin t\geq 0, 0\leq t\leq \pi,$ solve $\int_{\left ( C \right )}xy{\rm d}x+ y^{2}{\rm d}y= 0.$

This is my 1st time using Green's theorem, and I'm not sure of my opinion, I replaced $\int_{\left ( C \right )}\frac{3y^{2}}{2}{\rm d}y$ by $\int_{0\leq y\leq 2}\frac{3y^{2}}{2}{\rm d}y,$ it's no use. What is your thought on that ? Thanks a real lot.


问:
$\left|\frac{\sin(nx/2)}{\sin(x/2)}\right| \leq n$
I found a solution but I'm not sure how they got the last few steps

Clarification would be awesome!
Thanks

答:
We proceed with mathematical induction in the following formf1:

we prove that $\left|\frac{\sin\left(\frac{1}{2}n\theta\right)}{\sin\left(\frac{1}{2}\theta\right)}\right|=\left|1+e^{i\theta}+e^{i2\theta}+...+e^{i\left(n-1\right)\theta}\right|$,

we prove that $n=1+\left|e^{i\theta}\right|+\left|e^{i2\theta}\right|+...+\left|e^{i\left(n-1\right)\theta}\right|$,

from $\left|1+e^{i\theta}+e^{i2\theta}+...+e^{i\left(n-1\right)\theta}\right|\le1+\left|e^{i\theta}\right|+\left|e^{i2\theta}\right|+...+\left|e^{i\left(n-1\right)\theta}\right|$, we deduce $\left|\frac{\sin\left(\frac{1}{2}n\theta\right)}{\sin\left(\frac{1}{2}\theta\right)}\right|\le n$.

We know that

$~~~~~1+r+r^{2}+...+r^{n}=\dfrac{1-r^{n+1}}{1-r}.$

For $r=e^{i\theta}$,

$
\begin{align}
~~~~~\left|1+e^{i\theta}+e^{i2\theta}+...+e^{i\left(n-1\right)\theta}\right|&=\left|\frac{1-e^{in\theta}}{1-e^{i\theta}}\right|\\
&=\left|\frac{-e^{in\theta/2}\left(e^{in\theta/2}-e^{-in\theta/2}\right)}{-e^{i\theta/2}\left(e^{i\theta/2}-e^{-i\theta/2}\right)}\right|\\
&=\left|\frac{e^{in\theta/2}}{e^{i\theta/2}}\right|\left|\frac{\frac{1}{2i}\left(e^{in\theta/2}-e^{-in\theta/2}\right)}{\frac{1}{2i}\left(e^{i\theta/2}-e^{-i\theta/2}\right)}\right|\\
&=\left|e^{i\left(n-1\right)\theta/2}\right|\left|\frac{\sin\left(\frac{1}{2}n\theta\right)}{\sin\left(\frac{1}{2}\theta\right)}\right|
&\text{by }\sin\left(\theta\right)=\frac{1}{2i}\left(e^{i\theta}-e^{-i\theta}\right)
\end{align}
$

is equivalent to $\left|\frac{\sin\left(\frac{1}{2}n\theta\right)}{\sin\left(\frac{1}{2}\theta\right)}\right|~\left(\#\right)$, because $\left|e^{i\left(n-1\right)\theta/2}\right|=1$. Generally we know that $\left|k.e^{ix}\right|=k~~$f2.

For $\color{#592720}{k=1}$ and $x=\theta,\ 2\theta,\ ...,\ \left(n-1\right)\theta$ in $\left|k.e^{ix}\right|=k$ we have

$~~~~~1+\left|e^{i\theta}\right|+\left|e^{i2\theta}\right|+...+\left|e^{i\left(n-1\right)\theta}\right| = \overbrace{1+1+...+1}^{\text{n times}} = n~\left(\#\#\right).$

From Triangle Inequality we have

$~~~~~\left|1+e^{i\theta}+e^{i2\theta}+...+e^{i\left(n-1\right)\theta}\right|\le1+\left|e^{i\theta}\right|+\left|e^{i2\theta}\right|+...+\left|e^{i\left(n-1\right)\theta}\right|,$

which is equivalent to $\left|\dfrac{\sin\left(\frac{1}{2}n\theta\right)}{\sin\left(\frac{1}{2}\theta\right)}\right|\le n$ after plugging $\left(\#\right)$ and $\left(\#\#\right)$.

1.This answer is just the clarification of the image linked in the question.

2.See this answer for proof.


问:
Moment generating function of standard normal
As far as I know, given a random variable $X$, we define its moment generating function as
$$M_X(t) = \mathbb{E} \left[ e^{ tX} \right] \ \ , \ t \in \mathbb{R}$$
I read that MGF for a general random variable may not be defined for negative values of $t$. What about if $X \sim N(0,1)$? Can we be sure its MGF is well defined for negative t?

答: For $X \sim N(0,1)$ we have $Ee^{tX}=e^{t^{2}/2}$ for all real numbers $t$. One way to prove this is to use the characteristic function and use  a basic result from Complex  Analysis [The Identity Theorem].


问:
Prove that every convex function is continuous

A function $f : (a,b) \to \Bbb R$ is said to be convex if
$$f(\lambda x+(1-\lambda)y)\le \lambda f(x)+(1-\lambda)f(y)$$
whenever $a < x, y < b$ and $0 < \lambda <1$. Prove that every convex function is continuous.

Usually it uses the fact:

If $a < s < t < u < b$ then $$\frac{f(t)-f(s)}{t-s}\le \frac{f(u)-f(s)}{u-s}\le\frac{f(u)-f(t)}{u-t}.$$

I wonder whether any other version of this proof exists or not?

答:
Here, I just like to continue @Jonathan Honório's proof with the comment by @Ben Ciotti.

Let $x\in(a,b)$ and take $y\in(a,x)$. Suppose in the case that
\begin{align}
f(\lambda_n x+(1-\lambda_n)y)\le f(x)-\epsilon
\end{align}
for some fixed $\epsilon>0$ and for some sequence $(\lambda_n)$ that satisfies
\begin{align}
0<\lambda_1<\lambda_2<\cdots<1;\quad \lambda_n\to 1.
\end{align}
Take $u\in(x,b)$, and denote $x_n:=\lambda_nx+(1-\lambda_n)y$, and define the sequence $(\theta_n)$ with
\begin{align}
x=\theta_nx_n+(1-\theta_n)u.
\end{align}
Clearly, the definition of $x_n$ implies $0<\theta_n<1$. Since $x_n\to x$, we obtain $\theta_n\to 1$. Now, the convexity of $f$ and the assumption gives
\begin{align}
f(x)\le\theta_nf(x_n)+(1-\theta_n)f(u)\le\theta_n(f(x)-\epsilon)+(1-\theta_n)f(u).
\end{align}
Subtracting $f(x)$ on both sides gives
\begin{align}
0\le(\theta_n-1)f(x)-\theta_n\epsilon+(1-\theta_n)f(u)
\end{align}
As $\theta_n\to 1$, we obtain $0\le-\epsilon$, which is a contradiction.


答:
Edit: Looks like you updated the link to a new page (was abstraction of potential realizability, is now the abstraction of actual infinity). I will take a look and see how that changes my answer below.

Edit 2: I don't think the link update really changes the general ideas in my response below. The first link was about disregarding that reaching infinity is not doable in practice when assessing the validity of conclusions, whereas the second is saying, pretend it is actually reachable and treat it as if it has been reached. The ideas are similar. I'm not planning to modify my response further.

In the same encyclopedia website, searching for the law of the excluded middle, there is a section that says:

Since there is no general method for establishing in a finite number of steps the truth of an arbitrary statement, or of that of its negation, the law of the excluded middle was subjected to criticism by representatives of the intuitionistic and constructive directions in the foundations of mathematics (cf. Intuitionism; Constructive mathematics).

I think the connection is this finite steps idea. By embracing the abstraction of potential realizability, one accepts unbounded constructions as valid, even if it would be impossible to physically carry out an infinite number of logical steps or calculations. Applied to the law of the excluded middle, a given proposition may require an infinite number of steps to establish it's truth or that of its negation. The abstraction of potential realizability essentially says that you can treat things that require an infinite number of steps as valid.

This reminds me of something I read in the book Gödel, Escher, Bach, which I highly recommend. It explains how Gödel's incompleteness theorem involves the construction of statements that essentially say "This statement is not provable within this logical system". It is analogous to the "this statement is a lie" paradox. At the end of the book, there is discussion about adding an axiom to state a proof to the paradoxical statement does exist. By doing so, one finds that such a proof is essentially infinitely long (contains an infinite number of deduction steps). Incidentally this then implies the existence of a whole new class of numbers called surreal numbers, which is pretty interesting. The relevance here is: there was a paradoxical proposition that seemed neither true nor false, apparently falling within the excluded middle zone. By asserting there is an answer, one can resolve the paradox (placing the proposition squarely in either the true or false realm), but doing so requires acceptance of logical arguments which are infinitely long, embracing the abstraction of potential realizability.

Title: The law of excluded middle and the abstraction of "actual infinity".

Body:
The encyclopedia of mathematics states:

"Logically, acceptance of the abstraction of actual infinity leads to the acceptance of the law of the excluded middle as a logical principle."

This is mentioned in justifying constructive approaches.

What is the intuition for (and presumably proof of?) this "logical" connection?


问:
If two finite groups on the same set have the same homomorphisms in and out, must they be the same group?
Let $X$ be a finite set and let $*,*'$ be binary operators on $X$ such that $(X,*)$ and $(X,*')$ are both groups.

First, assume that for each finite group $(H,\cdot)$ and for each map $f:H\rightarrow X$, $f$ is a homomorphism from the group $(H,\cdot)$ into the group $(X,*)$ if and only if $f$ is a homomorphism from the group $(H,\cdot)$ into the group $(X,*')$.

Second, assume that for each finite group $(H,\cdot)$ and for each map $f:X\rightarrow H$, $f$ is a homomorphism from the group $(X,*)$ into the group $(H,\cdot)$ if and only if $f$ is a homomorphism from the group $(X,*')$ into the group $(H,\cdot)$.

Must $*$ and $*'$ be the same binary operator? That is, must it hold that $\forall a,b\in X, a*b=a*'b$.

答:
As Joshua has noted in the comments, this is true. You're also onto a useful insight for the rest of your mathematical career: we can understand an object itself (up to isomorphism) by understanding the morphisms to/from it.

Let's see why this particular result is true:

Say $(H,\cdot) = (X,\ast)$ and $f : H \to X$ is the identity map. Then by your condition, $f : (X,\ast) \to (X, \ast')$ is also a homomorphism. Similarly, taking $(H,\cdot) = (X, \ast')$ we see the identity map is also a homomorphism in the other direction. That is:

$$(X, \ast) \overset{\text{id}}{\longrightarrow} (X, \ast') \overset{\text{id}}{\longrightarrow} (X, \ast)$$

Since $\text{id} \circ \text{id} = \text{id}$ we see these maps are inverse, and so $(X,\ast) \cong (X, \ast')$. In fact, since the identity map is the isomorphism, we see

$$a \ast b = \text{id}(a \ast b) = \text{id}(a) \ast' \text{id}(b) = a \ast' b$$

so the operations must agree exactly.

As a (fun?) exercise can you use a similar argument if we work with the condition that maps from $(X, \ast) \to H$ are the same as maps $(X, \ast') \to H$? The same idea works, but it will be good for you to go through it.

As an aside, this theorem is a special case of the famous (infamous?) Yoneda Lemma. Roughly, the Yoneda Lemma (one version of it at least) says that in very general settings, this idea always works.

That is, if you have two objects $A$ and $B$, and you know that they have "the same" morphisms to them (or from them) then $A \cong B$. Here the notion of "the same" morphisms is slightly more general than the strict equality that you've asked for (it's the notion of a natural isomorphism) but morally the same idea is at play.

I hope this helps ^_^


问:
I need help understanding factoring better for this problem
Here is the equation:
$$\frac{-3(x+h)^2-6(x+h)+4+3x^2+6x-4}{h}$$

I am getting the wrong answer every time.
\begin{align*}
f & = \frac{-3(x^2+3xh+h^2)-6x+6h+4+3x^2+6x-4}{h}\\
f & = \frac{-3x^2-6xh-3h^2-6x+6h+4+3x^2+6x-4}{h}\\
f & = \frac{-6xh-3h^2}{h}\\
f & = 3(-2x-h^2)
\end{align*}

Two questions, what am I doing wrong when factoring down this problem and where is the best place to really learn factoring super well?

答:
it looks like you are differentiating $-(3x^2 + 6x-4)$

$$ -3(x+h)^2 -6(x+h)+4 = -3x^2 -6xh-3h^2 -6x -6h+4 $$

from which $-3x^2 -6x +4 $ cancel out with $3x^2 +6x -4$

$$ -6xh-3h^2-6h$$

you are missing the $-6h$ term,

and $\frac{-6xh-3h^2-6h}{h}=-6x-3h-6= -3(2x+h+2)$

Furthermore, taking the limit as $h \rightarrow 0$, $-3(2x+2)=-6x-6$.

Check:

$\frac{d}{dx}[-3x^2-6x+4] = (-3)\cdot 2 x -6 +0= -6x-6$.

Included the limit and derivative comments as a bonus, you tagged the question as pre-calculus, I’m not sure what level you’re at, if they confuse you for the moment, you can disregard them.


问:
Proof that $\frac{9}{8} < \sum\limits_{n=1}^{\infty} \frac{1}{n^3} < \frac{5}{4}$.
I'm trying to prove that $\frac{9}{8} < \sum\limits_{n=1}^{\infty} \frac{1}{n^3} < \frac{5}{4}$. I've seen similar proofs to this that tend to approach the proofs geometrically, using the upper and lower bounds of the remainder, the sum of the series less its $k$th partial sum, generated from the integral test. It usually involves some trick like "excluding $k$ terms." My professor tends to start at the second term to find a "lower bound," from which the proof follows with algebraic manipulation, but seemed to suggest to me that this followed from trial and error.

So, with that said, I really can't say that I understand the intuition behind finding or proving this. The first step seems to be plotting $y = \frac{1}{x^3}$ and then considering upper and lower Riemann sums (geometrically, "boxes" above the curve that will generate a sum greater than the area below it and boxes below the curve that will generate a sum less than the area below it). Though I believe I can draw the graph, I'm struggling with how to approach the problem, even if I have the abstract idea right, which I  quite doubt.

I'd very much appreciate if someone could shed some light on this. Thanks in advance.

Body:
Here's another approach:
\begin{align}\frac{9}{8} =\frac{1}{1^3} + \frac{1}{2^3} <\sum_{n=1}^\infty\frac{1}{n^3} = 1+\sum_{n=2}^\infty \frac{1}{n^3}<1+\sum_{n=2}^\infty\frac{1}{n^3-n} &= 1+\sum_{n=2}^\infty\frac{1}{n(n-1)(n+1)} \\ &= 1+\frac{1}{2}\sum_{n=2}^\infty\left(\frac{1}{n(n-1)}-\frac{1}{n(n+1)}\right) \\ &= 1 + \frac{1}{2}\lim_{k\to\infty}\left(\frac{1}{2}-\frac{1}{k(k+1)}\right) \\ &= \frac{5}{4}.\end{align}


问:
How to factor quadratic $ax^2+bx+c$?
How do I shorten this? How do I have to think?
$$ x^2 + x - 2$$

The answer is
 $$(x+2)(x-1)$$

I don't know how to get to the answer systematically. Could someone explain?
Does anyone have a link to a site that teaches basic stuff like this? My book does not explain anything and I have no teacher; this is self-studies.

Please help me out; thanks!

答:
You might appreciate the series of Khan Academy lectures (+ practice problems, etc).

Each lesson is given via a short 12 minute video, which you can replay if needed: the link I'm including below should take you to "Algebra" lessons; but you might also want to look at "Developmental math I and II".

This is a great way to learn via video demonstrations, practice, repetition, etc., and if you stumble, say, with negative exponents, e.g., you can go directly to a lesson addressing the issue at hand.  I believe there is diagnostic testing available (this is all free, no fees for watching videos, doing practice sets, tests, etc.), and such testing can really be helpful in learning where you need to direct your energies to move to the next level.

Edit: I've also come across this site, devoted to tutoring/demonstrating how to factor polynomials: it's developmental in approach, in that it starts slowly, with very common patterns used when factoring, and progresses in difficulty, building on what you've already learned, so best to approach the tutorial in the order given.

Title: How to factor quadratic $ax^2+bx+c$?


问:
If the norm of a path is a constant, is the derivative always zero?
Suppose $\gamma(t):[a,b]\to\Bbb{R^n}$ is a continuously differentiable path and $\vert \gamma(t)\vert=\int_a^b\vert\gamma'(t)\vert \enspace dt=C$, where $C$ is a constant, and this relation is true for all $t\in[a,b].$ Now, if we differentiate both sides of the relation with respect to $t$, we have $\vert \gamma'(t) \vert =0$ and therefore $\gamma'(t)=0$ for all $t\in[a,b]$ correct?

答:
No, it is not correct. Consider $\gamma(t)=( \cos t, \sin t)$ for $t \in [a,b]=[0,1]$ and $C=1$

Then $| \gamma(t)|= |\gamma'(t)|=1.$


问:
Sequence/series with triangular(?) pattern emerging
What i thought was a geometric series of the following form:

$$N = \sum_{i=1}^n[N_{i-1} + (T-N_{i-1})P]$$ where $$ lim_{i \to \infty} N = T$$ and $$N_{i=0}=0$$
I find the series to do the following:
$$i=1 : N = TP$$
$$i=2 : N = 2TP-TP^2$$
$$i=3 : N = 3TP-3TP^2 + TP^3$$
$$i=4 : N = 4TP-6TP^2+4TP^3-TP^4$$
$$i=5 : N = 5TP-10TP^2+10TP^3-5TP^4+TP^5$$
$$etc$$

I have been unable to find a solution to reduce this series, and am wondering in general about the pattern emerging.

added:
I see that this series mimics a converging inverse exponential, approximately:
$$N_n=T(1-e^{-Pn})$$

答:
Your sequence $\{N_i\}$ appears to be defined by $$N_{i+1} = N_{i} + (T-N_{i})P.$$
Therefore $$N_{i+1}-T = (N_{i}-T)(1-P) $$

$$N_k-T = (N_{0}-T)(1-P)^k=-T(1-P)^k.$$

We thus have  $$N_k =T(1-(1-P)^k)$$ which, of course, produces the binomial pattern of the terms you have found.


问:
What are some good ways of visualizing the type of substitution of this integral.
I came across the given integral while solving a problem:

$$\int_u^{1} v^{\beta-1} (1-v)^{\gamma-1}\left(\frac{v-u}{v}\right)^{\beta-1}dv$$

I couldn't find a way of simplifying it at all using substitution. After giving up I checked some proposed hints and they showed up this substitution:

$$
y = \frac{v-u}{1-u}, dy= \frac{dv}{1-u}
$$

Then the integral simplifies to a beta:

$$\int_0^{1} y^{\beta-1} (1-y)^{\gamma-1}dy$$

which is known.

I known the beta integral very well and I am aware that it was probably the best solution to try to get to a known integral from the original one. However, I still don't know a systematic way of coming up with this substitution. Seems out of the blue.

Can someone help developing techniques to be able to quickly visualize this kind of substitution?

Edit: I understand that substitution technique comes from the "reversed chain rule" in integration. But in this case I couldn't see the pattern. The real question is: how to develop some technique to be able to come up to this substitution? Or also, in this case, what are the main patterns in the integral that would suggest this substitution?

Thanks a lot.

答:
Mh, your example is somewhat rigged, because

$$ v^{\beta-1} (1-v)^{\gamma-1}\left(\frac{v-u}{v}\right)^{\beta-1}$$ immediately simplifies as

$$ (1-v)^{\gamma-1}\left(v-u\right)^{\beta-1}.$$

Then $t:=1-v$ brings you very close to a Beta:

$$ t^{\gamma-1}\left((1-u)-t\right)^{\beta-1}.$$ Just rescale.


问:
how to calculate $E(|Z^\frac{4}{3}|)$
As we all know, $E(|Z|)$ is $\sqrt(\frac{\pi}{2})$, where Z~N(0,1)

How to calculate $E(|Z^\frac{4}{3}|)$

Or someone can just tell me the value without the process, because typing Latex is so annoying.

答:
As we all know, $E(|Z|)$ is $\sqrt{\frac{\pi}{2}}$, where $Z\sim N(0,1)$

I do not know this fact....but calculating it I get

$$\mathbb{E}[|Z|]=\int_{-\infty}^{+\infty}|z|\phi(z)dz=\frac{1}{\sqrt{2\pi}}\Bigg\{e^{-z^2/2}\Big]_{-\infty}^0-\{e^{-z^2/2}\Big]_{0}^{+\infty}\Bigg\}=\sqrt{\frac{2}{\pi}}$$

Title: how to calculate $E(|Z^\frac{4}{3}|)$


问:
Finding the limit of a 2 variables function using epsilon and delta
I have been reviewing the notion of limit. With function of one variable, it is perfectly clear and the demonstration make perfect sense to me. I was curious how this would work in the world of two variables.
I have constructed a small example but I got stuck in the middle while trying to find a value for $\delta$.

Would you mind to finish the demonstration so that I could understand how this is working ?

$f : \mathbb{R^2} \mapsto \mathbb{R} $

$(x,y) \mapsto x^2 + y^2$

Find the limit : $$\lim_{x,y \to (1,1)} f(x,y) = 2$$

I started this way : $$ f(x,y) - 2 = x^2 + y^2 - 2 = (x^2 - 1) + (y^2 - 1)$$

$$ \lvert f(x,y) - 2 \rvert = \lvert (x+1)(x-1) + (y-1)(y+1) \rvert \le \epsilon$$

This is where I am stucked..

答:
The definition of a limit is usually extended to $\mathbb{R}^2$ as follows: for $f: \mathbb{R}^2 \to \mathbb{R}$, we say that $\lim_{(x, y) \to (x_0, y_0)} f(x,y) = L$ if for all $\epsilon > 0$, there exists a $\delta > 0$ such that $|f(x,y) - L| < \epsilon$ whenever $\lVert (x,y) - (x_0, y_0) \rVert < \delta$. Here $\lVert (x,y) \rVert = \sqrt{x^2 + y^2}$ is the usual Euclidean norm, which generalises the absolute value.

If $f(x,y) = x^2 + y^2$, we can continue from what you found. Fix $ \epsilon > 0$, and we want
$$
|f(x,y) - 2| = (x^2 - 1) + (y^2 - 1) < \epsilon .
$$
Now, if
$$
\lVert (x,y) - (1,1) \rVert = \lVert (x-1,y-1) \rVert = \sqrt{(x-1)^2 + (y-1)^2} < \delta ,
$$
can you see what value of $\delta = \delta(\epsilon)$ will complete the argument?


答:
We can calculate more generally

$$\mathbb{E}(|Z|^r) = \sqrt{ \frac{2}{\pi} } \int_0^{\infty} x^r e^{- \frac{x^2}{2} } \, dx$$

as follows. Substituting $y = \frac{x^2}{2}, dx = \frac{dy}{x} = \frac{dy}{\sqrt{2y}}$ gives

$$\sqrt{ \frac{2}{\pi} }  \int_0^{\infty} (2y)^{\frac{r-1}{2}} e^{-y} \, dy = \boxed{ \frac{2^{\frac{r}{2}}}{\sqrt{\pi}} \Gamma \left( \frac{r+1}{2} \right) }.$$

Substituting $r = \frac{4}{3}$ gives

$$\mathbb{E} \left( |Z|^{\frac{4}{3}} \right) = \boxed{ \frac{2^{\frac{2}{3}}}{\sqrt{\pi}} \Gamma \left( \frac{7}{6} \right) \approx 0.831 \dots }$$

where $\Gamma \left( \frac{7}{6} \right) \approx 0.928 \dots$ does not appear to have a closed form.

Title: how to calculate $E(|Z^\frac{4}{3}|)$

Body:
As we all know, $E(|Z|)$ is $\sqrt(\frac{\pi}{2})$, where Z~N(0,1)

How to calculate $E(|Z^\frac{4}{3}|)$

Or someone can just tell me the value without the process, because typing Latex is so annoying.


问:
System of Delay Differential Equation
In my differential equations course we are looking at systems of DDEs and I was wondering if there is a particular method that is used to solve these types of systems? One example that was given in class is the following;
$$\begin{align}v_1^\prime(t)&=\phi(t)v_1(t)\\  v_2^\prime(t)&=\alpha(v_1(t-1)-v_2(t-1)) \end{align}$$
Where $\alpha\in\mathbb{R}$ and $\phi(t)$ is some quasi-periodic function. Does an analytical solution exist? Or should I relay on numerical solutions as I have got here

Where x1 and x2 are v1 and v2 respectively, solutions to the above equation with $\phi(t)=1+2(t+\omega) (mod\ 1)$ for $w$ random.

答:
From the first ODE we have

$$
v_1(t) = C_0 e^{\int_0^t \phi(\tau)d\tau}
$$

and the second after substitution

$$
v_2'(t) + \alpha v_2(t-1) = C_0 e^{\int_0^{t-1}\phi(\tau)d\tau}
$$

to solve this ODE you can use many numerical methods and also when $\phi(t)$ is a linear function like $b t$, the Laplace transform can provide algebraically suitable methods to make parametric considerations, as follows

$$
\alpha  \left(e^{-s} \left(\mathcal{L}_t[v_2(t)](s)\right)+e^{-s} \int_{-1}^0 e^{-s t} v_2(t) \, dt\right)+s \left(\mathcal{L}_t[v_2(t)](s)\right)-\frac{\sqrt{\pi
   } C_0 e^{\frac{(c-s)^2}{4 b}} \left(\text{erf}\left(\frac{c-s}{2 \sqrt{b}}\right)+1\right)}{2 \sqrt{b}}-v_2(0)=0
$$

or calling $V_2(s) = \mathcal{L}_t[v_2(t)](s)$

$$
V_2(s) = \frac{\sqrt{\pi } C_0 e^{\frac{(c-s)^2}{4 b}+s} \left(\text{erf}\left(\frac{c-s}{2 \sqrt{b}}\right)+1\right)-2 \sqrt{b} \left(\alpha  \int_{-1}^0
   e^{-s t} v_2(t) \, dt-e^s v_2(0)\right)}{2 \sqrt{b} \left(\alpha +e^s s\right)}
$$

now making some assumptions like $v_2(t) = 0$ for $-1\le t\le 0$ and using for $e^s$ the Padé expansion we can proceed making stability considerations involving $\{\alpha, b, c\}$ as

$$
V_2(s) = \frac{\sqrt{\pi } C_0e^{\frac{(c-s)^2}{4 b}+s} \left(\text{erf}\left(\frac{c-s}{2 \sqrt{b}}\right)+1\right)}{2 \sqrt{b} \left(\alpha +e^s s\right)}
$$

Title: System of Delay Differential Equation


问:
Question on groups of order $pq$

Let $G$ be a group of order $pq$, $p>q$  and $p$, $q$ are primes. Then prove that

If $q\mid p-1$ then there exists a non abelian group of order $pq$.

Any two non-abelian groups of order $pq$ are isomorphic.

I have proved that if $q\not\mid p-1$ then $G$ is cylic . But how to prove this one I have no idea.
Any kind of hint is very much welcome. This problem is in Herstein book, page 75.

答:
If $q\mid p-1$ then $\rm{Aut}(C_p)$ has a unique subgroup of order $q$, and the map embedding $C_q$ in $\rm{Aut}(C_p)$ gives a semidirect product, which is not abelian (easy to check).

On the other hand, if $G$ is some other non-abelian group of order $pq$ then it is an easy exercise that $G$ has a normal subgroup of order $p$ and since $G$ also has a subgroup of order $q$, it must be a semidirect product. But since the subgroup of order $q$ in $\rm{Aut}(C_p)$  was unique, the only possibility is the one we already accounted for.

Title: Question on groups of order $pq$


问:
Proving $abcd+3\geq a+b+c+d$

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

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

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

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

Its from here

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

答:
This is a remark on Dr. Mathva's answer, which may seem very mysterious at first but becomes more transparent when you consider a simpler problem:

If $a, b \geq 0$ with $a^2 + b^2 = 1$, prove that
$$ ab+1 \geq a+b \,.$$

(It is the problem from the question if you impose $c = d = 1$.)

Here it is clear that you can factorize it as $(a-1)(b-1) \geq 0$, so you are naturally led to distinguish cases according to whether $a, b \geq 1$ or $\leq 1$. If both are $\geq 1$ or both are $\leq 1$, we are done. So suppose wlog $a \geq 1 \geq b$. Then plug in the constraint to get
$$ab + a^2+b^2 \geq a+b \,.$$
Now play a bit to recover every term in the RHS from the LHS. Because $a \geq 1$, we can just remove factors $a$ from the LHS. That is, $ab \geq b$ and $a^2 \geq a$, and we are done.

By considering this simpler problem, you can more easily get the idea of assuming wlog $a \geq b \geq c \geq d$, and of distinguishing cases according to the position of $1$.

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


问:
coordinate form of an affine transformation
I am having trouble with this seemingly simple question: Write the standard coordinate form of an affine transformation in $\mathbb{A}^{2}(\mathbb{R})$ that maps
the point (1, −2) to the point (0, 10), and the lines $10x_{1} − 4x_{2} = 1$ and $3x_{1} − 3x_{2} = −7$ to the lines $x_{1} − 2x_{2} = −3$ and $x_{1} − x_{2} = 6$, respectively.

Now as I understand for the point transformation
$\begin{pmatrix}0\\10\\0\end{pmatrix}=\begin{pmatrix}a_{11} & a_{12} & b_{1}\\a_{21} & a_{22} & b_{2}\\0 & 0 & 1\end{pmatrix}\begin{pmatrix}1\\-2\\0\end{pmatrix}$ and I need to find this matrix of transformation, this matrix equation gives me two equations with 6 unknowns. So I guess 4 more equations will be coming from the transformations of the two lines. Now here is where I am having trouble, I don't know how to get those equations. Any kind of hints would be really helpful. Thanks in advance.

答:
I think the added third coordinates should be $1$ instead of $0$ to obtain the translation effect of the matrix multiplication.

Write $x=(x_1,x_2,1)^T$.

If an expression $e(x)$ defines a given set $E$, in the sense that $e(x)$ is true iff $x\in E$, and $A$ is an invertible transformation, then the image of $E$ can be described by the expression
$$f(y):=e(A^{-1}(y))$$
because $y\in A(E)\iff A^{-1}(y)\in E\iff e(A^{-1}(y))$.

Since we don't want to calculate the inverse of the unknown matrix, we can rewrite the above formula by the $x=A^{-1}(y)$ substitution as
$$f(A(x))=e(x)\,.$$
So, we have the following connection between the first equations:
$$(a_{11}x_1+a_{12}x_2+b_1)\ -\ 2(a_{21}x_1+a_{22}x_2+b_2)\ =\ -3\ \iff\ 10x_1-4x_2\,=\,1\,,$$
which means that the two equations are a scalar multiple of each other. This actually brings in one more variable and 3 equations.

And the same to be done with the other two lines.


问:
What are some of the Websites/web-tools which must be a part of every mathematician's arsenal?
What are some of the web-tools which must be a part of every mathematician's arsenal? Tools which help mathematicians research, solve problems or even collaborate together!

Here is a list I gathered. Many among you may already know these, But still, in case.

MathQuill--
This is a great website for easy typing in LaTeX.

Integral calculator and the Derivative calculator-- Especially useful when one wants to crack a tough indefinite question. Steps included for those who need them!

OEIS--A list of all mathematically interesting sequences that contains over 3 million of such. Many things are discovered thanks to OEIS.

Desmos--One of the best online graphing tool with loads of computing power.

WolframAlpha and Mathematica-- Who doesn't know them! A unique engine for computing answers and providing knowledge.

MATLAB--Math. Graphics. Programming.
Whether you’re analyzing data, developing algorithms, or creating models, MATLAB is designed for way you think and the work you do.

Software such as Maple, MATLAB, Mathcad, Pari/GP, R, and the WolframAlpha website, etc are famous, and this question is not asking for similar software alternatives.

I haven't added math stack exchange to this list  for obvious reasons :~)

Thanks in advance to all for the over-enthusiastic help :-)

You might want to check out
Online tools for doing symbolic mathematics

PS-No intention of advertisement or promotion of any website or software.

答:
Here is another list that I managed to collect, and was too long to be posted in the comments--

Latex to word converter

Matrix Calculator--Really powerful computing

number empire--Useful for number theory

3d math tool

Pi search page-for pi lovers

Prime number finder--by numbereducation

Meta numbers--MetaNumbers is a free math tool providing information about any positive integer (up to 9223372036854775807), such as its factorized form, its divisors, its classification, or its arithmetic properties (widely used in the field of number theory).

Crazy numbers--Mathematical operations, numerology, symbolic numbers, representations, mathematical games, find everything you've always wanted to know about numbers.

other tools and calculators  by math is fun

calculators (for those who cannot code python, these prove very useful sometimes)--
a) Big number calculator--can calculate even 1,00,000 factorial !!
b) Google calculator-Calculate on the go
c) Desmos calculator--Quite powerfull
d) Google sheets-I had used them to generate lots of Random numbers I needed for a complex algorithm and tabular calculation

truth-table generator;

another one;

this one generates a URL to the required truth table, and also a LaTeX truth table

tree-proof generator--even provides counterexample


问:
exponential functions and calculus
Assume the rate of change of a quantity T is proportional to the difference between the quantity T and some fixed constant A.
That is
dT/dt = K(T-A)
show that T=A+Ce^(kt) (where C is a constant) is a solution to this differential equation.
I dont really know what the question is asking me to even try and begin to attempt it, if anyone can help me out that would be great thanks :) </p>

答: You are given $T=A+Ce^{kt}$ as a possible solution for the differential equation.  Take $\frac {dT}{dt}$ and plug it into the equation.  See that it is satisfied once you make the two $K$'s the same.</p>

Title: exponential functions and calculus


问:
Does convergence theorem holds for decreasing function?
I was wondering there is no convergence monotone theorem for decreasing function, i.e. if $\varphi _n\searrow f$ and $f\geq 0$, why it may not hold that $$\lim_{n\to \infty }\int_X \varphi _n=\int_Xf\ \ ?$$

So, I saw this link, but they don't give an example of such sequence.

答:
Take for example $$\varphi _n(x)=\infty \boldsymbol 1_{(0,\frac{1}{n})}.$$

Then $\varphi _n\searrow f$, but $\int \varphi _n=\infty $ for all $n$. Notice that this theorem holds if $\int \varphi _n<\infty $ for some $n$ (by DCT).