问:
Pumping Lemma proof $L= \{a^nb^m2^{n*m} \mid m,n \ge 0 \} $ not regular
I'm trying to proof that the following language is not a regular language using the pumping lemma:

$L= \{a^nb^m2^{n*m} \mid m,n \ge 0 \} $

I tried to build the string by assuming that m,n are the same number p, $w = 0^p1^p2^{p^2}$.
but I'm stuck with what should I do next?

I know how to solve for example the problem $L= \{0^n1^n \mid n \ge 0 \} $.
I read many times the theorem:

If $L$ is regular, we are able to divide $w$ up into three parts
$s=xyz$ where:

$|xy|\leq p$,

$|y| > 0$,

$xyz \in L \Leftrightarrow xy^{i}z\in L $ for all $i\in\mathbb{N}$

I would be very grateful if somebody will help me

答:
Your choice of $w$ works. If $w=xyz$, where $|xy|\le p$ and $|y|>0$, then $xy$ is contained in the $0^p$ part of $w$. In fact, there are $r\ge 0$ and $s\ge 1$ such that $x=0^r$, $y=0^s$, and $r+s\le p$. Then $z=0^{p-r-s}1^p2^{p^2}$,

$$\begin{align*}
xy^iz&=0^r0^{si}0^{p-r-s}1^p2^{p^2}\\
&=0^{r+si+p-r-s}1^p2^{p^2}\\
&=0^{p+(i-1)s}1^p2^{p^2}\,.
\end{align*}$$

This is in $L$ if and only if $p\big(p+(i-1)s\big)=p^2$, which is true if and only if $p(i-1)s=0$, and since neither $p$ nor $s$ is $0$, that is true if and only if $i=1$. Thus, any change in $i$ away from $1$ gives you a word not in $L$, which therefore isn’t regular.

Note that you could have started with the simpler word $w=0^p12^p$ and made essentially the same argument.


问:
bijection map from set of all positive real numbers to the interval $(0,1)$.
Let $S = \{x \in \Bbb R \mid 0 < x < 1 \}$.
Show that $\Bbb R^+ \sim S$.

Attempt:
Let $f:(1,\infty) \to (0,1)$ defined by $f(x)=\frac{1}{x}$ for all $x \in (1,\infty)$. Let $x,y \in (1,\infty)$ and assume that $f(x) = f(y)$. Then, by definition of $f$, $\frac{1}{x} = \frac{1}{y}$ i.e. $x=y$. Hence, $f$ is injective.
Now, for all $y \in (0,1)$, there exists
$x=\frac{1}{y} \in (1,\infty)$ such that $f(x)=f\left(\frac{1}{y} \right) = \frac{1}{\frac{1}{y}} = y$. Hence, $f$ is surjective. Thus, $f$ is bijective i.e. $(1,\infty) \sim (0,1)$.

Next, define $g : \Bbb R^+ \to (1,\infty)$ by $g(x) = x+1$ for all $x \in \Bbb R^+$.
Let $x,y \in \Bbb R^+$ and assume that $g(x) = g(y)$. Then, by definition of $g$, $x+1=y+1$ i.e. $x=y$ and hence $g$ is injective. Now, for all $y \in (1,\infty)$, there exists $x= y-1 \in \Bbb R^+$ such that $f(x) = f( y-1 ) = (y-1)+1 = y$ and thus $g$ is surjective.

Hence, $g$ is bijective i.e. $\Bbb R^+ \sim (1,\infty)$.

Finally, define a mapping from $\Bbb R^+ \to (0,1)$ by compositioning $g$ on $f$ i.e. $f \circ g$. Then, $(f \circ g)(x) = \frac{1}{x+1}$ for all $x \in \Bbb R^+$ is a bijection map from $\Bbb R^+$ to $(0,1)$ i.e. $\Bbb R^+ \sim (0,1)$.

Am I true?

答: Correct. (I can not write comment directly, not enough rep)</p>

ContentLicense: CC BY-SA 4.0


问:
Prove $H$ is a normal subgroup of $K$
$G$ is a set.
$$G=\left\{\begin{vmatrix}1 & 0 \\ 0  & 1\end{vmatrix}, \begin{vmatrix}0 & 1 \\-1 & 0\end{vmatrix}, \begin{vmatrix}-1 & 0 \\ 0  & -1\end{vmatrix}, \begin{vmatrix}0 & -1 \\ 1  & 0\end{vmatrix}, \begin{vmatrix}i & 0 \\ 0  & i\end{vmatrix}, \begin{vmatrix}0 & i \\ -i  & 0\end{vmatrix}, \begin{vmatrix}-i & 0 \\ 0  & i\end{vmatrix}, \begin{vmatrix}0 & -i \\ i  & 0\end{vmatrix}\right\}$$ and $(G, \times)$ is a group.

If
$$H=\left\{\begin{vmatrix}1 & 0 \\ 0  & 1\end{vmatrix}, \begin{vmatrix}-1 & 0 \\ 0  & -1\end{vmatrix}\right\}$$ and
$$K=\left\{\begin{vmatrix}1 & 0 \\ 0  & 1\end{vmatrix}, \begin{vmatrix}-1 & 0 \\ 0  & -1\end{vmatrix}, \begin{vmatrix}i & 0 \\ 0  & i\end{vmatrix}, \begin{vmatrix}-i & 0 \\ 0  & -i\end{vmatrix}\right\},$$
then $H$ and $K$ is a subgroup of $G$, $H$ is a normal subgroup of $K$, and $K$ is a normal subgroup of $G$.

My questions are :

a. How do I prove that $H$ is a normal subgroup of $K$?
I knew that to prove if $H$ is a normal subgroup I have to prove that each left coset of $H$ in K is a right coset of $H$ in $K.$ But I don't know how apply that to the matrices.

b. What is every element of factor group $G/H?$

答:
a.  $H=\{I,-I\}$, so, for all $k\in K$, $kH=\{kI,-kI\}=\{k,-k\}$,

and also $Hk=\{Ik,-Ik\}=\{k,-k\}$,

so indeed each left coset of $H$ in $K$ is a right coset of $H$ in $K$, so $H\lhd K$.

b.  One element of $G/H$, for example,

is  $\left\{I\begin{bmatrix}0 & i \\ -i  & 0\end{bmatrix},   -I\begin{bmatrix}0 & i \\ -i  & 0\end{bmatrix}\right\}$ =$ \left\{\begin{bmatrix}0 & i \\ -i  & 0\end{bmatrix},   \begin{bmatrix}0 & -i \\ i  & 0\end{bmatrix}\right\}.$

Now can you find the others?

Title: Prove $H$ is a normal subgroup of $K$


问:
Some horrid integrals I am confused with
$$(1) ~~~~ \int_{1/e}^{\infty} \sqrt{\frac{\ln{x} + 1}{x^3 + 1}}~~dx$$

Speaking as an example on (1).

The only thing I could do from here is only to do a u-sub:

$$ u = \ln(x) + 1 \\ x = e^{u-1} \\ du = \frac{1}{x} dx \\ dx = x (du) \Rightarrow dx = e^{u-1} (du) $$

And so this becomes:

$$ \int \sqrt{ \frac{u}{e^{3u-3} + 1} } ~~ du \\ $$

And from here it's basically a deadend.. I would appreciate if you could give some insights..

Interestingly, these similar integrals have interesting solutions:

$$(2) ~~~~ \int_{1/e}^{\infty} \sqrt{\frac{\ln{x} +1}{x^3}}~~dx =  \sqrt{2 e \pi}$$

$$(3) ~~~~ \int_{1/e}^{\infty} \sqrt{\frac{\ln{x}}{x^3 + 1}}~~dx = \text{a complex number ?!}$$

(2) - where and why does the pi comes into play? this looks like an interesting solution..

(3) - why is the solution a complex number if the area is right infront my eyes, and it is real just like every other areas? What is happening here?

Mystery solved about the complex number - the lower bound should be $1$ and not $1/e$ my bad! But this is not the main mystery ;)

Thank you :-)

答:
First, notice that $\log(x)+1=\log(ex)$. With this, first substitute $u=ex$, then $k=\log u$. Your integral is now
$$e^{1/2}\int_0^\infty k^{1/2} \,e^{-k/2}\,dk.$$
This is a well known integral that can be evaluated in terms of the Gamma function:
$$e^{1/2}\,\Gamma(3/2)\,2^{3/2}=\sqrt{2\pi e}.$$
I used $\Gamma(3/2)=\sqrt{\pi}/2$. To learn more about the gamma function, see here. You can think about it as an analytic continution of the factorial $n!$ for $n\in\mathbb{C}$ instead of just $n\in\mathbb{N}$.


问:
Let $a,b\in\mathbb{Z}$ and $p$ prime such that $ord_p(a)=j$, $ord_p(b)=k$ and $ord_p(ab)=h$, then $a^{hk}\equiv1$ (mod $p$)
Let $a,b\in\mathbb{Z}$ and $p$ prime, such that:

order of $a$ modulo $p$ is $j$.

order of $b$ modulo $p$ is $k$.

order of $ab$ modulo $p$ is $h$.

Then $a^{hk}\equiv1$ (mod $p$).

I am trying this problem, but still not got the result

答:
We have $b^k\equiv1$ and $(ab)^h\equiv1$, so $(ab)^{hk}\equiv1$, so $a^{hk}b^{hk}\equiv1$, so $a^{hk}(b^k)^h\equiv1$,

so $a^{hk}\equiv1\pmod p$.


问:
Is there a way of defining infinity so that it both meets the criterion of being a number as well as being greater than any real number?
This number would not be equivalent to the concept of infinity, but would be a number analogous to that concept. I'd also be interested to how the different ways of doing this (I'm assuming there are many), how they differ, and what their different applications are.

So this version of infinity would behave like this:

$\infty$ > X for all X $\in$ $\mathbb{R}$

$\infty$ + 2 > $\infty$ + 1

2 > 1

Rather than:

$\infty$ is the largest number

$\infty$ + 2 = $\infty$ + 1

2 = 1

答:
As Noah says there are a bunch of inequivalent ways to do this depending on what properties of ordinary arithmetic you want to preserve. Here's what happens when you try to add a single infinite element.

Starting with the real numbers $\mathbb{R}$ (or the integers or whatever else) you can add a new element $\infty$ and declare that $x + \infty = \infty$ for any $x$ (including $\infty$). This works fine (considering only addition for now, no multiplication) except that you lose the ability to subtract; this addition is no longer cancellative, meaning that $x + y = x + z$ no longer implies $y = z$. Formally (ignore this if you don't care), addition only makes $\mathbb{R} \cup \{ \infty \}$ a monoid rather than a group, and $\infty$ becomes an absorbing element. But we can still define an order in which $x \le \infty$ for all $x \in \mathbb{R}$ and this order still has the property that if $x \le y$ then $x + z \le y + z$; this means we still have an ordered monoid.

Now suppose we want to get multiplication in the game. Probably we should ask that $x \cdot \infty = \infty$ for $x > 0$ and in particular for $x = \infty$. But we need to decide what $0 \cdot \infty$ is. If we insist that multiplication continue to distribute over addition then we need

$$0 \cdot \infty = 0 \cdot (\infty + \infty) = 0 \cdot \infty + 0 \cdot \infty$$

which means the only possible options are that $0 \cdot \infty = 0$ or $\infty$. If we pick $\infty$ then we lose the familiar fact that $0$ times anything is zero. I don't think $0$ is an entirely satisfactory option but let's stick with it for now.

Now we need to decide what $(-1) \cdot \infty$ is. Again, if we insist that multiplication continue to distribute over addition, this requires

$$0 = 0 \cdot \infty = (1 - 1) \cdot \infty = \infty + (-1) \cdot \infty.$$

But with the way we've defined addition this is impossible; there is no element which, when added to $\infty$, produces $0$. The general problem here is that distributivity forces $-1$ to act as the additive inverse, and we already gave that up. So we have a couple different choices here:

We could give up distributivity. That would be terrible.

We could add a new element $-\infty$ and declare that $(-1) \cdot \infty = - \infty$ and that $\infty + (-\infty) = 0$. But this breaks associativity: for example, we would have $0 = \infty + (-\infty) = (1 + \infty) + (-\infty) = 1 + (\infty + (-\infty)) = 1$. The lesson is that when we gave up subtraction we really gave up subtraction and we shouldn't try to get it back. Subtraction is gone forever.

We could give up negative numbers. This is actually a perfectly fine thing to do: the non-negative reals $\mathbb{R}_{\ge 0} \cup \{ \infty \}$ together with infinity, addition, and multiplication as defined here is a perfectly well-behaved ordered semiring (which basically means we can add, multiply, and compare $\le$, but we can't subtract), and the subring consisting of the non-negative integers together with $\infty$ is exactly the ordered semiring given by the at-most-countable cardinal numbers.

We have to give up something, though.


答:
Alternatively, one may use Myhill-Nerode (which I am aware is not being asked in the question, but I usually find much cleaner!):

Consider the set of strings $s_i = ab^i$, where $i \geq 0$.

Then if $i \neq j$, $s_i \cdot 2^i = ab^i 2^i$ is in the language, but $s_j \cdot 2^i = ab^j 2^i$ is not. Therefore, $2^i$ distinguishes $s_i$ and $s_j$, so there are infinitely many equivalence classes.

Title: Pumping Lemma proof $L= \{a^nb^m2^{n*m} \mid m,n \ge 0 \} $ not regular

Body:
I'm trying to proof that the following language is not a regular language using the pumping lemma:

$L= \{a^nb^m2^{n*m} \mid m,n \ge 0 \} $

I tried to build the string by assuming that m,n are the same number p, $w = 0^p1^p2^{p^2}$.
but I'm stuck with what should I do next?

I know how to solve for example the problem $L= \{0^n1^n \mid n \ge 0 \} $.
I read many times the theorem:

If $L$ is regular, we are able to divide $w$ up into three parts
$s=xyz$ where:

$|xy|\leq p$,

$|y| > 0$,

$xyz \in L \Leftrightarrow xy^{i}z\in L $ for all $i\in\mathbb{N}$

I would be very grateful if somebody will help me


答: In $\Bbb Z_p$, which is cyclic, hence abelian, we have that $ab$ and $b^{-1}$ commute.  It follows that the order of their product divides the product of their orders.

Title: Let $a,b\in\mathbb{Z}$ and $p$ prime such that $ord_p(a)=j$, $ord_p(b)=k$ and $ord_p(ab)=h$, then $a^{hk}\equiv1$ (mod $p$)

Body:
Let $a,b\in\mathbb{Z}$ and $p$ prime, such that:

order of $a$ modulo $p$ is $j$.

order of $b$ modulo $p$ is $k$.

order of $ab$ modulo $p$ is $h$.

Then $a^{hk}\equiv1$ (mod $p$).

I am trying this problem, but still not got the result


问:
How to find the angle formed below a triangle joined by another?
The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&30^{\circ}\\
2.&40^{\circ}\\
3.&20^{\circ}\\
4.&45^{\circ}\\
\end{array}$

I'm stuck with this problem. What exactly should it be done here?. The thing is that I was only able to spot these relationships:

$\angle ABD = \angle ADB$

I'm assuming the intended approach is to draw a line between $DC$. But I can't really find a case of congruence to find the requested angle. Can someone help me here?.

How to solve this using only euclidean geometry?. The thing here is that. I'm stuck. I guess that the approach maybe is to spot an isosceles?. But other than the ones stated there I can't find others.

I've attempted to look if this figure could be inscribed in a circle but it doesn't seem the case. So I don't know what else to do?.

Can someone help me here please?. Please include a drawing in the answer so I can understand what to do.

答:
The inspiration here is to join $BC$. Note that $\triangle ABC$ is equilateral (why?), hence we have $BD=BC=AB$. That means we can draw a circle with $B$ as the center and $BA, BC, BD$ as the radii of the circle.

Can you work from here?


问:
The bartender's problem: grouping a list of orders into easy to handle small sets
Suppose that a bartender gets the orders just before the party gets started. There are many people attending this party and every person can order only one time and up to three different beverages from a list of, say, 5 different beverages: 1, 2, 3, 4, and 5.

The bartender is looking at the list of all orders, which is a simple table with the customer's name and his order. Let us look at a possible fragment from the list:

Jack  1,2,3

Beth  1,3,5

Ben   5       (meaning that Ben orders just one drink for the whole evening)

Liz   1,2,4

John  2,3,4

Mary  2,5

Tom   1,2,3

Our bartender wishes to optimize his work and go home as fast as he can since when he finishes with the preparations of the drinks, there is no longer a need for him. The total number of orders is, say, 20. The maximum number of orders per tray is 3. So he thinks to himself the following: I am willing to group the orders in such a way that each tray will contain the most intersected items(beverages). For example here is a possible list of trays:

Orders 1,7 and 5,

orders 2,3 and 6,

order 4.

Aiming to group the list into sets of three elements each, containing up to nine maximally common beverages possible. How should he proceed?

Remark 1: The above scenario was developed with a friend seeking an optimal strategy or algorithm which resolves the bartender's objective.

Remark 2: One can consider an order as a 3-tuple. Ben's order number 5, will look like $(0,0,5)$. Then a perfect candidate for $(a_1,a_2,a_3)$  sharing a tray with $(b_1,b_2,b_3)$ will be in case where $a_i=b_i$ for $i=1,2,3$.

Remark 3: One can think of the orders as vertices in a complete weighted graph, where each wight describes a measure of numbers of matches, that is: (1,2,3) and (1,2,4) this measure takes the value 2 since only the first two coordinates are same. Then finding triangular with maximum weight on the edges.

Remark 4: I still have the feeling that this problem is not yet enough well-defined. So any relevant resource to this or similar problem would be very appreciated.

Remark 5: The numbers chosen in the above problem description are almost arbitrary, and serve to ease the presentation of the problem.

Thank you.

答: It seems a programming task . The concept starts with all-vs-all intersection sets n-by-n. The result must be ordered by set cardinality. It presents factorial complexity, so beware. The decrescent picking must avoid orders already considered.</p>


问:
Fundamental Theorem of Homomorphism
$G$ is a group dan $K$ is a normal subgroup of $G$. If $H$ is a subgroup of $G$ then there is a one on one correspondence between every elemen in $H/K$ and ${H}/{H\cap K}$, so that $HK/K\approx H/H\cap K$.

With Fundamental Theorem of Homomorphism, I want to prove that

a. $K$ is a normal subgroup of $HK$

b. $H\cap K$ is a normal subgroup of H

c. There is $\psi : H \to HK/K$ with $\psi(a)=Ka$, $a \in H$ is a homomorphism from group $H$ onto $HK/K$

d. $Ker(\psi)=H\cap K$

For a. I proved them by proving $HK$ is normal in $G$ first then proving that since $K$ is normal in $G$ and $K \subseteq HK \subseteq G$, we have that $K$ is normal in $HK$.

For b. Since $H$ and $K$ are normal in $G$,  given any $g\in G$,
$gHg^{-1} \subseteq H$ and $gKg^{-1} \subseteq K$. So we have
$g(H\cap K)g^{-1} = gHg^{-1} \cap gKg^{-1} \subseteq (H\cap K)$.
So $(H\cap K)$ is normal in $G$. Since $(H\cap K) \subseteq H \subseteq G$, we have that $(H\cap K)$ is normal in $H$.

I have no idea for how to prove c and d.

答: Well, every element of $HK/K$ is a coset of the form $(hk)K$ $(=K(hk)$ since $K$ is normal in $HK)$ where $h \in H$ and $k \in K$. So, the function $\psi : H \to HK/K$ such that $$(\forall a \in H) \quad \psi(a) = aK \,(=(a1)K \in HK/K)$$ at least makes sense. Now, if $h \in H$ and $k \in K$, note that $hkK = hK$ because $h^{-1}(hk) = k \in K$. Thus, every element of $HK/K$ is in fact of the form $hK = \psi(h)$ for some $h \in H$, that is, $\psi$ is surjective. Now, $(ab)K = (aK)(bK)$ for every $a,b \in H$ tells us that $\psi$ is a group homomorphism, and, since $K$ is the identity element of $HK/K$, we also have that $$\ker \psi = \{a \in H : aK = K\} = \{a \in H : a \in K\} = H \cap K.$$

Title: Fundamental Theorem of Homomorphism


问:
What is the derivative of the inverse of the unit quaternion?
the unit quaternion is a good way to represent the rotation. but does anyone know the following formula is right or not?I am trying to prove it by myself but I am stuck on it.
$\dot{(q^{-1})}=-q^{-1}*\dot{q}*q^{-1}$ where q is a unit quaternion,* is the quaternion multiplication.

答:
I assume you are assuming $q(t)$ is a quaternion-valued function of a real variable $t$.

One thing you can check is that the product rule works for quaternion-valued functions. Indeed, the usual proof works, you just have to be careful not to use the commutative proprety.

Then, defining $p(t):=q(t)^{-1}$, we may differentiate $p(t)q(t)=1$ to obtain

$$ \frac{\mathrm{d}p}{\mathrm{d}t}q+p\frac{\mathrm{d}q}{\mathrm{d}t} = 0. $$

Can you finish?


问:
Calculating population size at any future time
If current population size is $P$ and average life-time of any member is $L$. Any pair of member of the population is allowed to have $C$ children on an average. They can have children between the age of $A$ to $B$.Then how can I tell population size at any  future time?</p>

答:
Ross has given you a formula for the first $A$ years.  Beyond that you could model this, or as an approximation assume that the shape of the population distribution does not change, in which case you will get exponential population change.

For simplification, let's assume everybody lives exactly $L$ years and each person gives birth to $C/2$ children (or that the half of them that are female each give birth to $C$)   after exactly $A$ years.  We want to find $k$ where the population after $N$ years is $Pk^N$.  The shape of the population distribution will be a truncated exponential or geometric distribution with each year group having $k$ times as many people as the one a year older, so the number dying each year will be $P\dfrac{k^N-k^{N-1}}{k^L-1}$ and the number being born  $Pk^{L}\dfrac{k^N-k^{N-1}}{k^L-1}$.  But the number being born is also $\dfrac{C}{2}Pk^{L-A}\dfrac{k^N-k^{N-1}}{k^L-1}$ so $k=\left(\dfrac{C}{2}\right)^{1/A}$ and that means that the population after $N$ years is

$$P\left(\frac{C}{2}\right)^{N/A}.$$

This seems fairly intuitive: note that it suggests that earlier parental age at childbirth means faster population growth (or decline) even if lifetime fertility stays the same, and that the population will grow iff $C>2$.

To make this more sophisticated, you can introduce $B$, so childbirth takes place over an interval (this will affect the denominator of the exponent, putting it between $A$ and $B$), and you can also widen the distribution of death with substantial effects if it is possible to die before childbearing age $A$ and the average number of children per couple does not take this into account. You could also start with a different population shape, or allow random effects, or allow fertility to vary over time, in a not particularly complicated model.

ContentLicense: CC BY-SA 3.0


问:
Proof by induction that $\left(\sum_{k=1}^nk\right)^2\ge\sum_{k=1}^nk^2$
I am a bit new to logical induction, so I apologize if this question is a bit basic.

I tried proving this by induction:

$$\left(\sum_{k=1}^nk\right)^2\ge\sum_{k=1}^nk^2$$

Starting with the base case $n=1$:

$$1^2\ge1^2$$

Then to prove that $P(n)\to P(n+1)$:

$$\left(\sum_{k=1}^{n+1}k\right)^2\ge\sum_{k=1}^{n+1}k^2$$
$$\left((n+1)\sum_{k=1}^nk\right)^2\ge(n+1)^2\sum_{k=1}^nk^2$$
$$(n+1)^2\left(\sum_{k=1}^nk\right)^2\ge(n+1)^2\sum_{k=1}^nk^2$$

Since they're both multiplied by $(n+1)^2$, it's easy to see that if $\left(\sum_{k=1}^nk\right)^2\ge\sum_{k=1}^nk^2$, then $(n+1)^2\left(\sum_{k=1}^nk\right)^2\ge(n+1)^2\sum_{k=1}^nk^2$. But if the $\ge$ were replaced with a $\le$, the proof would still be valid, even though it's demonstrably not true.

I can see that it would take strong induction would fix this problem, but if I didn't know by observation that $\left(\sum_{k=1}^nk\right)^2\le\sum_{k=1}^nk^2$ isn't true by observation, how would I know to use strong induction?

Body:
Hints:

$$\sum_{k=1}^n k=\frac{n(n+1)}2$$

$$\sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}6$$


问:
Intersection-exponent for one-dimensional Brownian motion
We let $B^1,B^2$ be independent, one-dimensional Brownian Motions with $B^1(0)=-1$ and $B^2(0)=1$ and $T_n^i=\inf\{t\geq0:|B^i(t)|=n\}$.

In Gregory Lawler's: Hausdorff Dimension of Cut-Points for Brownian Motion, it is claimed, that

$\textbf{P}\{B^1[0,T_n^1]\cap B^2[0,T_n^2]=\emptyset\} \approx n^{-2}$

where for functions $f$ and $g$, $f\approx g $ means $\lim_{n\to \infty}\frac{\ln f(n)}{\ln g(n)}=1$.

Now I think, that in this case even

$\lim_{n\to\infty}\frac{\textbf{P}\{B^1[0,T_n^1]\cap B^2[0,T_n^2]=\emptyset\}}{n^{-2}}=1\quad$ without the logarithms holds true.

Note that

$\textbf{P}\{B^1[0,T_n^1]\cap B^2[0,T_n^2]=\emptyset\}=\textbf{P}\{\max\limits_{t\in[0,T_n^1]}{B^1(t)}\leq \min\limits_{t\in[0,T_n^2]}{B^2(t)}\}$,

and we get an upper bound by the application of the Optional Stopping Theorem:

$\textbf{P}\{\max\limits_{t\in[0,T_n^1]}{B^1(t)}\leq \min\limits_{t\in[0,T_n^2]}{B^2(t)}\}\leq \textbf{P}\{\max\limits_{t\in[0,T_n^1]}{B^1(t)}\leq 1; \min\limits_{t\in[0,T_n^2]}{B^2(t)}\leq -1\}=(\frac{1}{n+1})^2$.

However I am struggling with finding a lower bound. I was thinking to use the distribution of the maximum-process $M_t=\min\limits_{0\leq s\leq t}{B(s)}$, but I was not successful so far.

Any help would be greatly appreciated!

答:
For the lower bound, the following pretty naive estimate works:

$$\mathbb{P}(B^{1}[0,T_{n}^{1}]\cap{B^{2}[0,T_{n}^{2}]}=\emptyset)\geq{\mathbb{P}(\{B^{1} \hspace{3pt}\text{hits $-n$ before hitting $0$}\}\cap\{B^{2} \hspace{3pt}\text{hits $n$ before hitting $0$}\})}= \\ =\mathbb{P}(B^{1} \hspace{3pt}\text{hits $-n$ before hitting $0$})\cdot\mathbb{P}(B^{2} \hspace{3pt}\text{hits $n$ before hitting $0$}\})=\frac{1}{n^{2}}$$

Also, $\frac{1}{n^{2}}$ is larger than your upper bound of $\frac{1}{(n+1)^{2}}$ which suggests that you probably made a calculation error somewhere. For what it's worth, you can get an upper bound by the same reasoning as the one I used for the lower bound above:

$$\mathbb{P}(B^{1}[0,T_{n}^{1}]\cap{B^{2}[0,T_{n}^{2}]}=\emptyset)\leq{\mathbb{P}(\{B^{1} \hspace{3pt}\text{hits $-n$ before hitting $1$}\}\cap\{B^{2} \hspace{3pt}\text{hits $n$ before hitting $-1$}\})}= \\ =\mathbb{P}(B^{1} \hspace{3pt}\text{hits $-n$ before hitting $1$})\cdot\mathbb{P}(B^{2} \hspace{3pt}\text{hits $n$ before hitting $-1$}\})=\frac{4}{(n+1)^{2}}$$


问:
Maximizing compound interest with fee per compound
Given the following:

initial principle $P$, interest rate $r$, flat fee whenever interest is applied $x$, period between applying interest $p$, total time period $t$

How do I calculate the final amount $A$?

the objective is to maximize $A$ by tuning $p$.

The context of this problem is staking cryptocurrencies and compounding their rewards

答:
Presuming that both $p$ and $t$ are measured in years, the number of compoundings per year is $\frac 1p$, and the interest rate for each period is the yearly interest rate divided by the number of compounding per year, or $r_p = \dfrac r{\frac 1p} = pr$.

At the end of the first period, an amount equal to $Ppr - x$ is added to the account (where $r$ is expressed as a fraction out of $1$, not as a percent), totalling

$$P_1 = P + Ppr - x = P(1+pr) - x$$

To make this a little easier, let $a = (1+pr)$, so $P_1 = Pa - x$.

At the end of the second period, the calculation is the same, except the beginning principle $P$ is replaced by the amount $P_1$ in the account at the beginning of the period.
$$\begin{align}P_2 &= P_1a - x\\&= (Pa - x)a - x\\&= Pa^2 - xa - x\\&=Pa^2 - x(a +1)\end{align}$$
Continuing we see that
$$P_3 = Pa^3 - x(a^2 + a + 1)\\P_4 = Pa^4 - x(a^3 + a^2 + a + 1)\\\vdots$$

Now this sum of powers is well known. If you multiply it by $a-1$ you get
$$\begin{align}(a-1)(a^{n-1} + a^{n-2} + \cdots + a + 1) &= a(a^{n-1} + a^{n-2} + \cdots + a + 1)\\&\qquad-(a^{n-1} + a^{n-2} + \cdots + a + 1)\\&=(a^n + a^{n-1} + \cdots + a^2 + a)\\&\qquad - (a^{n-1} + a^{n-2} + \cdots + a + 1)\\&=a^n - 1\end{align}$$
so
$$a^{n-1} + a^{n-2} + \cdots + a + 1 = \dfrac{a^n - 1}{a - 1}$$
And we can write
$$P_n = Pa^n - x\dfrac{a^n - 1}{a - 1}$$

The number $n$ of periods is going to be $n = \frac tp$. And noting that $a - 1 = pr$, this becomes
$$\begin{align}A &= Pa^{t/p} - x \frac{a^{t/p} - 1}{pr}\\&=Pa^{t/p}-\frac x{pr}a^{t/p}+\frac x{pr}\\&=\left(P-\frac x{pr}\right)a^{t/p} + \frac x{pr}\end{align}$$
$$\bbox[5px,border:2px solid] {A =\left(P-\frac x{pr}\right)\left(1+pr\right)^{t/p} + \frac x{pr}}$$


问:
number of ways to make $2.00
How many different ways can you make $2.00 using only 1 cent, 5 cent, 10 cent, and 25 cent pieces, and 1 and 2 dollar bills (there are 100 cents in a dollar)? I have worked out an equation:

$$p + 5n + 10d + 25q + 100l + 200t = 200$$

Where all variables are non-negative integers. But I don't know what to do from here to figure out how many different sets of non-negative integers there are that satisfy this equation. What do I do next?

Title: number of ways to make $2.00

Body:
Recursive algorithm for general monetary systems:

Let $f(x;a_0,a_1...a_k)$ be the number of ways to make $x$c with coins/notes of value $a_i$.

$f(200;1,5,10,25,100,200)=1+f(200;1,5,10,25,100)$

$f(200;1,5,10,25,100)=1+f(100;1,5,10,25)+f(200;1,5,10,25)$

$f(200;1,5,10,25)=1+f(25;1,5,10)+...+f(200;1,5,10)$

Carry on like this, using $f(0;...)=1$ and

$f(x;a_0..a_k)=1+f(a_k;...a_{k-1})+...+ f(na_k;...a_{k-1})$

where $n$ is the greatest integer such that $na_k \leq x$.


问:
Dual Killing form is a $\mathbb{Q}$-bilinear form
Given a semisimple Lie algebra $\mathfrak{g}$, with Cartan subalgebra $\mathfrak{h}$, s.t. $ \mathfrak{g}=\mathfrak{h}\oplus \bigoplus_{\alpha \in R}\mathfrak{g}_\alpha $, I'm trying to show that the dual killing form on $\sum_{\alpha \in R} \mathbb{Q}\alpha$ the roots is a positive definite bilinear form. Since we have that $$n_{\alpha \beta}=2\frac{(\alpha,\beta)}{(\alpha,\alpha)} \in \mathbb{Z} \ \ \ \forall \ \alpha, \ \beta \ \in  R$$
If I could show that $(\alpha,\alpha) \in \mathbb{Q} \ \forall \ \alpha \in R$, I would be able to conclude the proof quite easily, but I'm stuck at this passage, does anyone know how to prove this?

答:
By definition, $K(h_\alpha, h_\alpha)$ is the trace of $ad(h_\alpha) \circ ad(h_\alpha) \in End(\mathfrak g)$.

Now you already have the root space decomposition which gives a very convenient basis w.r.t. which the endomorphism $ad(h_\alpha)$ (and hence its composition with itself) is diagonal. In particular, note that if $v \in \mathfrak g_\beta$ for any root $\beta$, we have $ad(h_\alpha) (v) = [h_\alpha, v] = \beta(h_\alpha) \cdot v = \check{\alpha}(\beta) \cdot v$.

See how this implies that the trace of $ad(h_\alpha) \circ ad(h_\alpha)$ is the sum of squares of certain rational numbers, which are not all zero?


问:
Complete Statistic: Uniform distribution
Take a random sample $X_1, X_2,\ldots X_n$ from the distribution
$f(x;\theta)=1/\theta$ for $0\le x\le
\theta$.

I need to show that $Y=\max(X_1,X_2,...,X_n)$ is complete.

Now, I know I should multiply the sample distribution of $Y$ and multiply it with a function of $Y$, then integrate over the range of $\theta$ and equate them to zero. But how do I get the sampling distribution of $Y$?

答:
The p.d.f of $Y=X_{(n)}$ is $$\delta_{\theta}(y)=\frac{n}{\theta^n}y^{n-1},y\in[0,\theta]$$
For any function $f$ satisfying $E_{\theta}\left[f(y)\right]=0,\forall\theta\in\mathbb{R}^+$, we have $$\int_0^\theta f(y)y^{n-1}dy=0,\forall\theta\in\mathbb{R}^+$$
Since $[0,\theta]$ generate the Borel $\sigma$-algebra of $\mathbb{R}^+$, we have $$\int_A f(y)y^{n-1}dy=0, \text{ for all Borel set }A\subset\mathbb{R}^+$$
Therefore $$f(y)y^{n-1}= 0, a.e.\iff f(y)\equiv0, a.e.$$
This means $Y=X_{(n)}$ is complete.


问:
Matrix inverse of $ T = \begin{pmatrix} z & A^* \\ A & z \end{pmatrix}$
Consider the block matrix

$$ T = \begin{pmatrix} z & A^* \\ A & z \end{pmatrix}$$
where $z$ is a scalar times the identity on the Hilbert space and $A$ is a bounded operator.

I am wondering if there is a simpler block matrix inversion formula in this special case?

答:
Note that
$$
\underbrace{\pmatrix{z & -A^*\\ -A& z}}_S \ \underbrace{\pmatrix{z & A^*\\ A & z}}_T = TS = \pmatrix{z - A^*A & 0\\0 & z - AA^*}.
$$
With this, we can conclude that $T$ has inverse
$$
T^{-1} = \pmatrix{z - A^*A & 0\\0 & z - AA^*}^{-1} \pmatrix{z & -A^*\\-A & z}
\\= \pmatrix{z(z - A^*A)^{-1} & -(z-A^*A)^{-1}A^*\\
-(z-AA^*)^{-1}A & z(z - AA^*)^{-1}}.
$$
In order to verify that $T^{-1}$ is indeed self-adjoint (for real $z$), note that

$$
(z - AA^*)A = zA - AA^*A = A(z - A^*A) \implies\\
A(z - A^*A)^{-1} = (z - AA^*)^{-1} A.
$$
So, the adjoint of the bottom-left entry satisfies
$$
[-(z - AA^*)^{-1}A]^* = -[A(z - A^*A)^{-1}]^* = -(z - A^*A)^{-1}A^*,
$$
which coincides with the upper-right entry as required.


问:
How to go from general form to standard form of quadratic equation? (with a deeper question about forms as well)
I'm learning how to convert quadratic equations from general form to standard form, in order to make them easier to graph. We know the general form is ax^2+bx^2+c, and the standard form is a(x-h)^2+k. To help with the conversion, we can expand the standard form, and see that it turns into the general form. I totally get how to go from standard to general. I can easily memorize what h and k are, and use them to consistently derive standard forms.

What I'm curious about is how to, a priori, go from the general form to the standard form? Is there a way to see that ax^2+bx+c can turn into a(x-h)^2+k without knowing that form ahead of time? How was the form a(x-h)^2+k discovered in the first place? How are alternate forms of equations discovered in general? I honestly wouldn't know where to begin.

I ask out of curiosity, and because I believe knowing how to go in the other direction will help really solidify this concept for me. Even if that knowledge is above my skillset at the moment, at least an overview of what kind of math is involved may supplement this concept for me.

答:
To get the standard form from the general form, first factor out $a$, then complete the square, and finally adjust the constant term:

$$\begin{align*}
ax^2+bx+c&=a\left(x^2+\frac{b}ax+\frac{c}a\right)\\
&=a\left(\left(x+\frac{b}{2a}\right)^2+u\right)\,,\tag{1}
\end{align*}$$

where I don’t yet know what $u$ is. To find out, I multiply out $(1)$ to get

$$a\left(x^2+\frac{b}ax+\frac{b^2}{4a^2}\right)+au\,,$$

which is

$$ax^2+bx+\frac{b^2}{4a}+au\,.$$

If this is to be the same as the original quadratic, we must have

$$\frac{b^2}{4a}+au=c\,,$$

meaning that

$$u=\frac{c-\frac{b^2}{4a}}a=\frac{c}a-\frac{b^2}{4a^2}\,,$$

and our quadratic can be written as

$$a\left(x+\frac{b}{2a}\right)^2+\left(c-\frac{b^2}{4a}\right)\,.$$

Thus, the $h$ and $k$ of the standard form must be $$h=-\frac{b}{2a}$$ and $$k=c-\frac{b^2}{4a}\,.$$

Take the quadratic $2x^2+3x+4$ as an example. Factoring out the leading coefficient of $2$ results in

$$2\left(x^2+\frac32x+2\right)\,.$$

Then

$$x^2+\frac32x+2=\left(x+\frac34\right)^2+u$$

for some $u$, and since $\left(x+\frac34\right)^2=x^2+\frac32x+\frac{9}{16}$, $u$ must be $2-\frac9{16}=\frac{23}{16}$. Thus,

$$\begin{align*}
2x^2+3x+4&=2\left(\left(x+\frac34\right)^2+\frac{23}{16}\right)\\
&=2\left(x-\left(-\frac34\right)\right)^2+\frac{23}4\,,
\end{align*}$$

with $h=-\frac34$ and $k=\frac{23}8$.


问:
For x < 5 what is the greatest value of x
It can't be $5$. And it can't be $4.\overline{9}$ because that equals $5$. It looks like there is no solution... but surely there must be?</p>

答:
There isn't one.  Suppose there were; let's call it $y$, where $y<5$.

Let $\epsilon = 5 -y$, the difference between $y$ and 5.  $\epsilon$ is positive, and so $0 < \frac\epsilon2 < \epsilon$, and then $y < y+\frac\epsilon2 < y+\epsilon = 5$, which shows that $y+\frac\epsilon2$ is even closer to 5 than $y$ was.

So there is no number that is closest to 5. Whatever $y$ you pick, however close it is, there is another number that is even closer.

Consider the analogous question: “$x < \infty$; what is the greatest value of $x$?”  There is no such $x$.


答:
Brian M. Scott's answer gives a good algebraic explanation; let me try to give a geometric summary of what's going on.

Remember that there are four basic "geometric transformations" of the graph of an equation $y=f(x)$. Specifically, for a constant $u$, we have:

The graph of $y=f(x-u)$ is the graph of $y=f(x)$ shifted to the right by $u$.

The graph of $y=f(ux)$ is the graph of $y=f(x)$ "compressed horizontally" by a factor of $u$ (and possibly reflected about the $y$-axis, depending on the sign of $u$).

The graph of $y=f(x)+u$ is the graph of $y=f(x)$ shifted up by $u$.

The graph of $y=uf(x)$ is the graph of $y=f(x)$ "stretched vertically" by a factor of $u$ (and possibly reflected about the $x$-axis, depending on the sign of $u$).

Now - granting that a parabola is a transformation of the graph of $y=x^2$ via stretching, shifting, and flipping - this tells us that every parabola has the form $$y=u_0(u_1x-u_2)^2+u_3$$ for some $u_0,u_1,u_2,u_3$. Moreover, at least for parabolas (this doesn't work in general) we can combine two of these variables into a single one: the expression above is equivalent to $$y=(u_0u_1^2)(x-{u_2\over u_1})^2+u_3.$$ So we really only need three "transformation constants:"

Every parabola is the graph of an equation of the form $$y=w_0(x-w_1)^2+w_2$$ for some constants $w_0,w_1,w_2$.

And at this point there's just one last step remaining before the transformation into standard form is something guessable: realizing that the graph of $y=ax^2+bx+c$ is a parabola. This is something you can notice just by graphing a few examples (always graph a few examples!).

Title: How to go from general form to standard form of quadratic equation? (with a deeper question about forms as well)

Body:
I'm learning how to convert quadratic equations from general form to standard form, in order to make them easier to graph. We know the general form is ax^2+bx^2+c, and the standard form is a(x-h)^2+k. To help with the conversion, we can expand the standard form, and see that it turns into the general form. I totally get how to go from standard to general. I can easily memorize what h and k are, and use them to consistently derive standard forms.

What I'm curious about is how to, a priori, go from the general form to the standard form? Is there a way to see that ax^2+bx+c can turn into a(x-h)^2+k without knowing that form ahead of time? How was the form a(x-h)^2+k discovered in the first place? How are alternate forms of equations discovered in general? I honestly wouldn't know where to begin.

I ask out of curiosity, and because I believe knowing how to go in the other direction will help really solidify this concept for me. Even if that knowledge is above my skillset at the moment, at least an overview of what kind of math is involved may supplement this concept for me.


问:
A step in deriving the solution to heat equation
I'm trying to understand how to solve the heat equation. I'm stuck on this particular step:

$$Q(x,t) = \dfrac{1}{2} + \dfrac{1}{\sqrt\pi}\int_{0}^{\dfrac{x}{\sqrt{4kt}}}e^{-s^2}ds, t > 0.$$

I want to compute $\frac{\partial Q}{\partial x}.$ This should turn out to be
$$ \dfrac{1}{\sqrt {4k\pi}t}e^{-\dfrac{x^2}{4kt}}.$$ I think we have to use Leibniz integral rule but not sure how to do it. Any help appreciated.

答:
Let $$H(x,t) = Q(x\sqrt{4kt},t) = \frac{1}{2}+\frac{1}{\sqrt{\pi}}\int\limits_0^x e^{-s^2}ds $$

So $$ \frac{\partial H}{\partial x} = \frac{1}{\sqrt{\pi}}e^{-x^2} $$ by Leibnitz or just plain differentiation under the integral sign.

Can you finish from this point on?


问:
Let $(A,+, \cdot)$ be a ring with the given tables, determine which is the zero and inverse of each element
Let $(A,+, \cdot)$ be a ring with the given tables for operations $+$ and $\cdot$

\begin{array}{c|cccc}
 + & s & t & x & y \\ \hline
 s & y & x & s & t \\
 t & x & y & t & s \\
 x & s & t & x & y\\
 y & t & s & y & x\end{array}

\begin{array}{c|cccc}
 \cdot & s & t & x & y \\ \hline
 s & y & y & x & x \\
 t & y & y & x & x \\
 x & x & x & x & x\\
 y & x & x & x & x\end{array}

determine:
a) which is the zero of the ring
b) the inverse of each element;

for a) what I can see is that $x$ act as element $0$ for $+$

also $x$ act as element $0$ for $\cdot$ but  for $y$ the result equals $x$ too

for b) $s+t=x$ so if $x=0$ then $s$ is inverse of $t$

in $\cdot$ I can't see how is $s\cdot s= s \cdot t =y$, because as said before $s$ and $t$ is inverse of each other...

I would appreciate if someone can give some insight on how to solve this problem

答:
$x$ is the additive identity.

The additive inverses are $-s=t,-t=s,-x=x,-y=y$.

This isn't a ring with unit.


问:
Proving that the cohomology ring of $\mathbb{R}P^n$ is isomorphic to $\mathbb{Z}_{2}[x]/(x)^{n+1}$
We have to prove that $H^{*}(\mathbb{R}P^n, \mathbb{Z}_2) \simeq \mathbb{Z}_{2}[x]/(x)^{n+1}$ as ring.
So we have to find an isomorphism $$ \phi: \mathbb{Z}_{2}[x]/(x)^{n+1} \rightarrow H^{*}(\mathbb{R}P^n, \mathbb{Z}_2)   $$
We can send $x$ in a sub-manifold of $\mathbb{R}P^n$ of codimention $1$ in $H_{*}(\mathbb{R}P^n, \mathbb{Z}_2)$ (? $\mathbb{R}P^{n-1}$, why isn't it a $0$-class in homology?).
How can we build this isomorphism? And why can we use duality among $H^{*}$ and $H_{*}$ (if $n$ is odd $\mathbb{R}P^n$ isn't oriented)?</p>

答:
Here is a quick way to get what you want:

Note that $\mathbb{R}P^{n-1}\hookrightarrow\mathbb{R}P^n$ induces isomorphisms on $H_i(-;\mathbb{Z}_2)$ for $i\le n-1$.

Now $x$ generates $H_1(\mathbb{R}P^n;\mathbb{Z}_2)$ and we want to show that $H^*(\mathbb{R}P^n;\mathbb{Z}_2)\cong\mathbb{Z}_2[x]/(x^{n+1})$, so by induction we only need to show that $x\cup x^{n-1}\ne 0$.

Now just plug-n-chug through Poincare-duality with cup/cap products: $[\mathbb{R}P^n]\cap(x\cup x^{n-1})=([\mathbb{R}P^n]\cap x)\cap x^{n-1}=1$, where this last equality is because $[\mathbb{R}P^n]\cap x$ is a generator of $H_{n-1}(\mathbb{R}P^n;\mathbb{Z}_2)$! Thus $x^n\ne 0$.


问:
Calculate radius of an equilateral triangle such that its edges intersect with points of a smaller internal equilateral triangle.
If I have an equilateral triangle (the black one), with a known circumscribed circle radius R = h1, and a known angle θ, how can I find the value h2, which is the R for the outer equilateral triangle with edges that intersect with the points of the inner triangle?

I'm using this for a generative art project where I want a sequence of nested triangles.

答:
Use the sine rule to set up

$$\frac{x_1}{h_1}=\frac{\sin\theta}{\sin 30},\>\>\>\>\>
\frac{x_2}{h_1}=\frac{\sin(120-\theta)}{\sin 30}
$$

$$\implies x_1+x_2 = \sqrt3h_2 = 2h_1\sin\theta+ 2h_1\sin(120-\theta)$$

which yields

$$h_2 = h_1(\sqrt3\sin\theta+\cos\theta)$$


问:
How to find the angle formed below a triangle joined by another?
The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&30^{\circ}\\
2.&40^{\circ}\\
3.&20^{\circ}\\
4.&45^{\circ}\\
\end{array}$

I'm stuck with this problem. What exactly should it be done here?. The thing is that I was only able to spot these relationships:

$\angle ABD = \angle ADB$

I'm assuming the intended approach is to draw a line between $DC$. But I can't really find a case of congruence to find the requested angle. Can someone help me here?.

How to solve this using only euclidean geometry?. The thing here is that. I'm stuck. I guess that the approach maybe is to spot an isosceles?. But other than the ones stated there I can't find others.

I've attempted to look if this figure could be inscribed in a circle but it doesn't seem the case. So I don't know what else to do?.

Can someone help me here please?. Please include a drawing in the answer so I can understand what to do.

答:
$$AB = AC = BD$$

$$m\angle BAD = 50^\circ$$

$$m\angle DAC = 10^\circ$$

Since $m\angle BAC = 60^\circ$ and $AB = AC$, then
$\triangle ABC$ is an equilateral triangle.

So $AB = BC = CA$.

So $m\angle ABC = m\angle ACB = 60^\circ$.

So $m\angle AFB = m\angle ACF + m\angle FAC = 70^\circ$.
(External angle theorem.)

Since $AB = BD$, $\triangle ABD$ is isosceles.
So $m\angle ADB = 50^\circ$ and $m\angle ABD = 80^\circ$.

Since $BC = BD \dots$


问:
Maclaurin of $f(x)=\ln\left(\sum_{k=0}^{1000}\frac{x^k}{k!}\right)$
The 1001st order is required.

Here is how I went about the question:

We know that near $x=0$:

$$e^x=\sum_{k=0}^{1000}\frac{x^k}{k!}+\frac{x^{1001}}{1001!}+o(x^{1001})$$
So,
$$e^x-\frac{x^{1001}}{1001!}+o(x^{1001})=\sum_{k=0}^{1000}\frac{x^k}{k!}$$
And after taking the logarithm of both sides,
$$f(x)=\ln\left(e^x-\frac{x^{1001}}{1001!}+o(x^{1001})\right)$$

But what do I do next? I intuitively understand that the answer should be:
$$f(x)=x-\frac{x^{1001}}{1001!}+o(x^{1001})$$
But how do I prove it rigorously?

Thank you!

答:
Re-posting my comment as an answer since it resolved the question:

Factor out $e^x$ in the argument of $\ln$, then apply the fact that the log of a product is the sum of the logs. After this you can apply the Taylor series for the logarithm.


问:
Some horrid integrals I am confused with
$$(1) ~~~~ \int_{1/e}^{\infty} \sqrt{\frac{\ln{x} + 1}{x^3 + 1}}~~dx$$

Speaking as an example on (1).

The only thing I could do from here is only to do a u-sub:

$$ u = \ln(x) + 1 \\ x = e^{u-1} \\ du = \frac{1}{x} dx \\ dx = x (du) \Rightarrow dx = e^{u-1} (du) $$

And so this becomes:

$$ \int \sqrt{ \frac{u}{e^{3u-3} + 1} } ~~ du \\ $$

And from here it's basically a deadend.. I would appreciate if you could give some insights..

Interestingly, these similar integrals have interesting solutions:

$$(2) ~~~~ \int_{1/e}^{\infty} \sqrt{\frac{\ln{x} +1}{x^3}}~~dx =  \sqrt{2 e \pi}$$

$$(3) ~~~~ \int_{1/e}^{\infty} \sqrt{\frac{\ln{x}}{x^3 + 1}}~~dx = \text{a complex number ?!}$$

(2) - where and why does the pi comes into play? this looks like an interesting solution..

(3) - why is the solution a complex number if the area is right infront my eyes, and it is real just like every other areas? What is happening here?

Mystery solved about the complex number - the lower bound should be $1$ and not $1/e$ my bad! But this is not the main mystery ;)

Thank you :-)

答:
I don't think integral (1) has a closed form.  It can be represented by a series, though:

$$ \sum_{k=0}^\infty \frac{2 \sqrt{\pi}\, {\mathrm e}^{3 k +\frac{1}{2}} \left(-1\right)^{k} \left(2 k \right)! 4^{-k}}{\left(6 k +1\right) \sqrt{12 k +2}\, k !^{2}}$$

This comes from replacing $x^3+1$ by $x^3+a$, expanding the integrand in a series in powers of $a$, integrating term-by-term, and then substituting $a=1$.


问:
Prove that there is a two-sheeted covering of the Klein bottle by the torus
Prove that there is a two-sheeted covering of the Klein bottle by the torus.

OK, so we take the the polygonal representation of the torus and draw a line in the middle as follows:

Then there are two Klein bottles in there, but how do I write down the actual covering map $q:S^1 \times S^1 \to K$?

答: I would like to give my formula. Let us think of $T^2$ as $S^1\times S^1$ and $S^1$ as the unit circle in $\mathbb{C}$. Let $f$ act on $T^2$ by $f(z,w)=(-z, \bar{w})$, then $f^2$ is the identity. $f(z,w)\sim(z,w)$ gives an equivalence relation on $T^2$ and $T^2/\sim$ provides the desired two-fold covering map from $T^2$ to the Klein bottle, as illustraded in the figure by OP.


问:
How to understand $\frac{d}{dt}\{(\exp(tX))_*(Y)\}|_{t=0}=[X,Y]$?
Let $G$ be a Lie group on which $X$ and $Y$ are two vector fields. Let $G\xrightarrow{\exp(tX)} G$ be the (Lie theory) exponential map corresponding to $X$. Then of fundamental importance is \begin{equation}
\frac{d}{dt}\{(\exp(tX))_*(Y)\}|_{t=0}=[X,Y]
\end{equation}
where $[\cdot,\cdot]$ is the Lie bracket, and $\exp(tX)_*$ is the differential map induced from $\exp(tX)$.

Staring at this equation for a while, the only way I can prove it is by relating this exponential map to the (Remannian) exponential map. This proof involves a lot differential geometry, which seems rather unnecessary, and does not seem very instructive.

I am wondering whether someone can share a more elegant proof (more direct, less differential geometry) or just some intuitive idea about how one should understand this equation.

Thanks very much!

答:
This is a very general fact, and does not need fancy extra structure like that of a Lie group (or even a Riemannian structure):

If you have two smooth vector fields $X,Y$ on a smooth manifold $M$ (for simplicity, assume complete vector fields), and if $\phi_t: M \to M$ denotes the flow of $X$, then
$$\frac{d}{dt}\Big|_{t=0} (\phi_t)_* Y = [X,Y].$$
(Important: Note that the $(\phi_t)_*$ on the left hand side denotes the push-forward of a diffeomorphism acting on vector fields, and not mereley the fiber-wise map!)

Add.: There are two maps, which are sometimes denoted with a $*$: Suppose you have a smooth map $\psi: M \to N$ between manifolds (we could have $M=N$, but the general version might be clearer).

Given a tangent vector $v\in T_pM$ at a point, there is the fiber-wise map $\psi_*:T_pM \to T_{\psi(p)}N$ that maps $ v\mapsto \psi_* v$. It takes a vector at $p$ and maps it to a vector at the image point $\psi(p)$. This works for any smooth map $\psi:M \to N$.

If $\psi$ is a diffeomorphism, then we can also do the following. Given a vector field $X\in \Gamma(TM)$, we can map it via $\psi$ to vector field $\psi_*X\in \Gamma(TN)$. This is the push-forward map. We obtain $\psi_*X$ as follows: Given $q\in N$, let $p = \psi^{-1}(q)\in M$ be its preimage. Note that $X_p\in T_pM$ is a tangent vector. We can map $X_p$ to a tangent vector $\psi_* (X_p) \in T_{\psi(p)}N$ with the map from (1). The value of $\psi_*X$ at $q = \psi(p)$ is obtained by setting $(\psi_*X)_q := \psi_* (X_p)$. Doing this for every $q\in N$ gives us a vector field $[q\mapsto (\psi_*X)_q]\in \Gamma(TN)$.
By this process, we obtain a map $\psi_*: \Gamma(TM)\to \Gamma(TN)$ which is explicitly given by $$\psi_*: \Gamma(TM)\to \Gamma(TN), \quad [p\mapsto X_p] \mapsto [q\mapsto \psi_*(X_{\psi^{-1}(q)})].$$ The push-forward acting on vector fields. It only makes sense for diffeomorphisms in general and is a way of transporting vector fields from one place to another.

Now let $f\in C^\infty(M)$ be an arbitrary smooth function. To prove the above identity, we need to show that $\left(\frac{d}{dt}|_{t=0} (\phi_t)_\ast Y \right) f = [X,Y]f$. Remember that the push-forward $(\phi_t)_\ast X$ is defined by $$((\phi_t)_* Y )_pf = (\phi_{t})_\ast (Y_{\phi_{-t}(p)})f = Y_{\phi_{-t}(p)}(f\circ \phi_{t})$$ for all $p\in M$. Thus
\begin{align}
 \left(\frac{d}{dt}|_{t=0} (\phi_t)_\ast Y \right) {}_p f &= \lim_{t\to 0} \frac{((\phi_t)_* Y )_pf - Y_pf}{t} \\
&= \lim_{t\to 0} \frac{Y_{\phi_{-t}(p)}(f\circ \phi_{t}) - Y_pf}{t} \\
&= \lim_{t\to 0} \frac{Y_{\phi_{-t}(p)}(f\circ \phi_{t}) - Y_{\phi_{-t}}f}t +\lim_{t\to 0} \frac{Y_{\phi_{-t}(p)}f -Y_pf}{t} \\
&= \lim_{t\to 0} \frac{Y_{\phi_{-t}(p)}(f\circ \phi_{t}-f)}t +\lim_{t\to 0} \frac{(Yf)(\phi_{-t}(p)) -(Yf)(p)}{t} \\
&= Y_p\left(\lim_{t\to 0}\frac{f\circ \phi_{t} - f}{t}\right) +\lim_{t\to 0} \frac{(Yf)(\phi_{-t}(p)) -(Yf)(p)}{t}  \\
&= Y_p(Xf) - X_p(Yf) \\
&= [X,Y]_p f.
\end{align}

(the sign convention for the Lie brackets is $[X,Y]= YX-XY$ or $[X,Y] = XY-YX$, depending on preference.)

In your case $\phi_t = \exp(tX)$ is the flow of $X$.


问:
Is $\emptyset \subseteq P(A)$ or $\{\emptyset\}\subseteq P(A)$, where $A$ is a set and $P(A)$ is the power set of $A$
I think my professor made an error on his answer key and I'm trying to confirm it before I bring it to his attention. He asserts only 1.) is false. I believe both 1 and 4 are false. This class is only using naïve set theory.

$A = \{1,2,3,4\}$
Select the statement that is false

1.) $\{2,3\} \subseteq P(A)$
2.) $\{2,3\} \in P(A)$
3.) $ \emptyset \in P(A)$
4.) $\emptyset \subseteq P(A)$

$P(A)$ is the set of all the subsets of $A$

$P(A) = P(A) = \{\emptyset,\{1\},\{2\},\{3\},\{4\},\{1,2\},\{1,3\},\{1,4\},\{2,3\},\{2,4\},\{3,4\},\{1,2,3\},\{1,2,4\},\{1,3,4\},\{2,3,4\},\{1,2,3,4\}\}$

FALSE

$2$ is not an element of $P(A)$

$3$ is not an element of $P(A)$

∴ $\{2,3\}$ cannot be a subset of $P(A)$

$\{{2,3}\}$ would be a subset of $P(A)$

TRUE

The element $\{2,3\}$ can be found in the set $P(A)$

TRUE

The element $\emptyset$ can be found in the set $P(A)$

FALSE

Both operands of the subset operator requires a set. $\emptyset$ is the empty set were as $\{\emptyset\}$ is an element that is the empty set. Therefore $\emptyset$ is not a subset of $P(A)$.

答: The empty set is a subset of every set, so it is also true that $\emptyset \subset P(A)$.


答:
#4 is true. The empty set is a subset of every set.

It happens that the empty set us also an element of this set.

So both $\varnothing \subseteq \mathscr{P}(A)$ and $\varnothing \in \mathscr{P}(A)$ are true.

Title: Is $\emptyset \subseteq P(A)$ or $\{\emptyset\}\subseteq P(A)$, where $A$ is a set and $P(A)$ is the power set of $A$

Body:
I think my professor made an error on his answer key and I'm trying to confirm it before I bring it to his attention. He asserts only 1.) is false. I believe both 1 and 4 are false. This class is only using naïve set theory.

$A = \{1,2,3,4\}$
Select the statement that is false

1.) $\{2,3\} \subseteq P(A)$
2.) $\{2,3\} \in P(A)$
3.) $ \emptyset \in P(A)$
4.) $\emptyset \subseteq P(A)$

$P(A)$ is the set of all the subsets of $A$

$P(A) = P(A) = \{\emptyset,\{1\},\{2\},\{3\},\{4\},\{1,2\},\{1,3\},\{1,4\},\{2,3\},\{2,4\},\{3,4\},\{1,2,3\},\{1,2,4\},\{1,3,4\},\{2,3,4\},\{1,2,3,4\}\}$

FALSE

$2$ is not an element of $P(A)$

$3$ is not an element of $P(A)$

∴ $\{2,3\}$ cannot be a subset of $P(A)$

$\{{2,3}\}$ would be a subset of $P(A)$

TRUE

The element $\{2,3\}$ can be found in the set $P(A)$

TRUE

The element $\emptyset$ can be found in the set $P(A)$

FALSE

Both operands of the subset operator requires a set. $\emptyset$ is the empty set were as $\{\emptyset\}$ is an element that is the empty set. Therefore $\emptyset$ is not a subset of $P(A)$.


答:
You are wrong: both $\varnothing$ and $\{\varnothing\}$ are subsets of $\mathcal P(A)$. $\varnothing$ is a subset of any set $X$, because the condition of all its elements being elements of $X$ is vacuously true; $\{\varnothing\}$ is a subset of $\mathcal P(A)$ because $\varnothing\in\mathcal P(A)$, seeing as $\varnothing\subseteq A$.

$\{\varnothing\}$ is a set, specifically the set containing exactly the element $\varnothing$. Its existence is implied by the axiom of pair and the axiom of empty set together.


问:
The bartender's problem: grouping a list of orders into easy to handle small sets
Suppose that a bartender gets the orders just before the party gets started. There are many people attending this party and every person can order only one time and up to three different beverages from a list of, say, 5 different beverages: 1, 2, 3, 4, and 5.

The bartender is looking at the list of all orders, which is a simple table with the customer's name and his order. Let us look at a possible fragment from the list:

Jack  1,2,3

Beth  1,3,5

Ben   5       (meaning that Ben orders just one drink for the whole evening)

Liz   1,2,4

John  2,3,4

Mary  2,5

Tom   1,2,3

Our bartender wishes to optimize his work and go home as fast as he can since when he finishes with the preparations of the drinks, there is no longer a need for him. The total number of orders is, say, 20. The maximum number of orders per tray is 3. So he thinks to himself the following: I am willing to group the orders in such a way that each tray will contain the most intersected items(beverages). For example here is a possible list of trays:

Orders 1,7 and 5,

orders 2,3 and 6,

order 4.

Aiming to group the list into sets of three elements each, containing up to nine maximally common beverages possible. How should he proceed?

Remark 1: The above scenario was developed with a friend seeking an optimal strategy or algorithm which resolves the bartender's objective.

Remark 2: One can consider an order as a 3-tuple. Ben's order number 5, will look like $(0,0,5)$. Then a perfect candidate for $(a_1,a_2,a_3)$  sharing a tray with $(b_1,b_2,b_3)$ will be in case where $a_i=b_i$ for $i=1,2,3$.

Remark 3: One can think of the orders as vertices in a complete weighted graph, where each wight describes a measure of numbers of matches, that is: (1,2,3) and (1,2,4) this measure takes the value 2 since only the first two coordinates are same. Then finding triangular with maximum weight on the edges.

Remark 4: I still have the feeling that this problem is not yet enough well-defined. So any relevant resource to this or similar problem would be very appreciated.

Remark 5: The numbers chosen in the above problem description are almost arbitrary, and serve to ease the presentation of the problem.

Thank you.

答:
You can solve this set partitioning problem via integer linear programming as follows.  Let $O$ be the set of orders.  For each subset $S \subset O$ of up to $3$ orders ($|S| \le 3$), let binary decision variable $x_S$ indicate whether that subset appears as a tray, and let $w_S$ be the weight of that subset.  The problem is to maximize the total weight $\sum_S w_S x_S$ subject to
$$\sum_{S:\ o \in S} x_S = 1 \quad \text{for $o\in O$} \tag1$$
Constraint $(1)$ forces each order to appear on exactly one tray.


问:
Which trigonometric identities involve trigonometric functions?
Once upon a time, when Wikipedia was only three-and-a-half years old and most people didn't know what it was, the article titled functional equation gave the identity
$$
\sin^2\theta+\cos^2\theta = 1
$$
as an example of a functional equation.  In this edit in July 2004, my summary said "I think the example recently put here is really lousy, because it's essentially just an algebraic equation in two variables."  (Then some subsequent edits I did the same day brought the article to this state, and much further development of the article has happened since then.)

The fact that it's really only an algebraic equation in two variables, $x^2+y^2=1$, makes it a lousy example of a functional equation.  It doesn't really involve $x$ and $y$ as functions of $\theta$, since any other parametrization of the circle would have satisfied the same equation.  In a sense, that explains why someone like Norman Wildberger can do all sorts of elaborate things with trigonometry without ever using trigonometric functions.

But some trigonometric identities do involve trigonometric functions, e.g.
$$
\sin(\theta_1+\theta_2)=\sin\theta_1\cos\theta_2+\cos\theta_1\sin\theta_2
$$
$$
\sec(\theta_1+\cdots+\theta_n) = \frac{\sec\theta_1\cdots\sec\theta_n}{e_0-e_2+e_4-e_6+\cdots}
$$
where $e_k$ is the $k$th-degree elementary symmetric polynomial in $\tan\theta_1,\ldots,\tan\theta_n$.  These are good examples of satisfaction of functional equations.

So at this point I wonder whether all trigonometric identities that do seem to depend on which parametrization of the circle is chosen involve adding or subtracting the arguments and no other operations.  In some cases the addition or subtraction is written as a condition on which the identity depends, e.g.
$$
\text{If }x+y+z=\pi\text{ then }\tan x+\tan y+\tan z = \tan x\tan y\tan z.
$$

QUESTION: Do all trigonometric identities that do involve trigonometric functions, in the sense that they are good examples of satisfaction of functional equations by trigonometric functions, get their non-triviality as such examples only from the addition or subtraction of arguments?  Or is there some other kind?  And if there is no other kind, can that be proved?

Postscript: Wikipedia's list of trigonometic identities is more interesting reading than you might think.  It has not only the routine stuff that you learned in 10th grade, but also some exotic things that probably most mathematicians don't know about.  It was initially created in September 2001 by Axel Boldt, who was for more than a year the principal author of nearly all of Wikipedia's mathematics articles.

答:
If I understood correctly, the question is how uniquely we can determine trigonometric functions using functional equations. First thing is together with a strong enough functional equation we always need a some sort of "smoothness" condition. For example consider the system $$\begin{cases}
f(x+y)=f(x)g(y)+g(x)f(y) \\
g(x+y)=g(x)g(y)-f(x)f(y)
\end{cases}$$

Under any assumption sufficient for Cauchy functional equation we can deduce that it has a solution involving sine-and-cosine, but there are wildly discontinuous solutions one can construct using a Hamel bases of rationals over reals.

Next, since the set of all interesting functional equations is too big to consider, we can focus on somewhat special class called algebraic Addition theorems. For example $$h(x+y)=h(x)\sqrt{1-(h(y))^2}+\sqrt{1-(h(x))^2}h(y).$$ It is a theorem by Weierstrass that characterize all such meromorphic functions as either rational or periodic depending on the singularity type of $h$ at $\infty.$ You can see here for more details. Since the Jacobi's impossibility theorem of triply periodic meromorphic functions, we can find all of them explicitly. Roughly these are:

trigonometric functions (including hyperbolic, and exponential functions)

elliptic functions (I think of theses as some sort of generalized trigonometric functions, and I am not completely wrong as the theory of Jacobi elliptic functions can develop entirely analogous to trigonometry)

rational functions (in particular polynomials).

So yes!, under some reasonable assumptions, algebraic addition theorems recover a generalized class of trigonometric functions. Here being meromorphic is important otherwise we should include algebraic functions (and possibly more!). Alternatively, I will roughly sketch a proof to convince you that any complex differentiable function  with an algebraic addition theorem is the inverse of a certain kind of integral. Let $\varphi$ be an analytic function (on some domain of $\mathbb{C}$) with an addition theorem, then there is a polynomial $G$ with constant coefficients satisfying $$G(\zeta, \eta, \theta) =0,\tag1$$ where $\zeta=\varphi(z), \eta=\varphi(w)=$ and $\theta=\varphi(w+z).$ Since $\zeta$ and $\eta$ are independent variables, by considering total derivatives of $G$ with respect to them gives: $$\dfrac{\partial G}{\partial \zeta}\dfrac{d \zeta}{d z}+\dfrac{\partial G}{\partial \theta}\dfrac{d \theta}{d z}=\dfrac{\partial G}{\partial \eta}\dfrac{d \eta}{d w}+\dfrac{\partial G}{\partial \theta}\dfrac{d \theta}{d w}=0.$$ Also we know that $\dfrac{d \theta}{d z}=\dfrac{d \theta}{d w},$ and consequently $$\dfrac{\partial G}{\partial \zeta}\dfrac{d \zeta}{d z}-\dfrac{\partial G}{\partial \eta}\dfrac{d \eta}{d w}=0.\tag2$$ If $\theta$ does not appear in $\dfrac{\partial G}{\partial \zeta}, \dfrac{\partial G}{\partial \eta},$ this lase equation is an expression that contains only  $\zeta, \dfrac{d \zeta}{d z}, \eta$ and $\dfrac{d \eta}{d w}.$ On the other hand, even if they contain $\theta$ we can eliminate it using both $(1), (2)$ to obtain a polynomial relation with constant coefficients $$H\left(\zeta, \dfrac{d \zeta}{d z}, \eta, \dfrac{d \eta}{d w}\right)=0.$$ Since this is true for all $z, w$ in the domain, we can evaluate $v$ at some specific point to obtain a first order differential equation $\tilde{H}\left(\zeta, \dfrac{d \zeta}{d z}\right)=0.$ If this evaluation of $H$ vanish in all its coefficient, we can try different values of $w$ until we find a non-zero equation $\tilde{H}.$ We can solve this equation in terms of an algebraic function $K$ such that $\dfrac{d \zeta}{d z}=K(\zeta),$ and hence after separating variables $$z-z_0=\displaystyle\int_{\zeta_0}^{\zeta}\dfrac{d\zeta}{K(\zeta)}.$$
where $\varphi(z_0)=\zeta_0$ are appropriate constants. Then the addition thereom of $\varphi$ can be rewritten as $$\displaystyle\int_{\varphi(z_0)}^{\zeta}\dfrac{d\zeta}{K(\zeta)}+\displaystyle\int_{\varphi(z_0)}^{\eta}\dfrac{d\zeta}{K(\zeta)}=\displaystyle\int_{\varphi(z_0)}^{\theta}\dfrac{d\zeta}{K(\zeta)}.$$

You can find the original argument of this algorithm in this book with all the details and much more. For example to what I described here:
\begin{array}{|l|c|r|}
\zeta=\varphi(z) & \theta & \tilde{H} \\
\hline
\sin z & \zeta\sqrt{1-\eta^2}+\eta\sqrt{1-\zeta^2} & \left(\dfrac{d \zeta}{d z}\right)^2+\zeta^2-1 \\
\tan z & \dfrac{\zeta+\eta}{1-\zeta\eta} & \dfrac{d \zeta}{d z}-\zeta^2-1\\
\sec z & \dfrac{\zeta\eta}{1-\sqrt{1-\zeta^2}\sqrt{1-\eta^2}} & \left(\dfrac{d \zeta}{d z}\right)^2-\zeta^2(\zeta^2-1)
\end{array}

I am too lazy to continue this for other trigonometric and hyperbolic functions. In the next simplest case $K(\zeta)=K(\zeta, \sqrt{P(\zeta)})$ is a rational function, where $P$ is a polynomial of degree $3$ or $4$ with no repeated roots, we recover the  Jacobian elliptic functions. Similar computations can carry on for some other elliptic functions, in particular for lemniscate functions mentioned in this solution to one of your question.


答:
If $x<5$ then $2x<x+5$ so $x<\frac{x+5}{2}$. Similarly, $x<5$ means $x+5<5+5=10$ or $\frac{x+5}{2}<5$.  So if $x<5$ we have $x<\frac{x+5}{2}<5$, and therefore there is a larger number, $\frac{x+5}{2}$ less than  $5$.

Basically, the average of two different numbers must be strictly between those two numbers.

Title: For x < 5 what is the greatest value of x

Body: It can't be $5$. And it can't be $4.\overline{9}$ because that equals $5$. It looks like there is no solution... but surely there must be?</p>


问:
Prove that/Explain how for independent random variables $X_i$, we have $f_i(X_i)$ are independent (in particular without measure theory)
I have seen a lot of posts that describe the case for just 2 random variables.

Independent random variables and function of them

Are functions of independent variables also independent?

If $X$ and $Y$ are independent then $f(X)$ and $g(Y)$ are also independent.

If $X$ and $Y$ are independent. How about $X^2$ and $Y$? And how about $f(X)$ and $g(Y)$?

Are squares of independent random variables independent?

Prove that if $X$ and $Y$ are independent, then $h(X)$ and $g(Y)$ are independent in BASIC probability -- can we use double integration? (oh I actually asked the 2 variable elementary case here, but there's no answer)

I have yet to see a post that describes the case for at least 3.

Please answer in 2 situations

1 - for advanced probability theory:

Let $X_i: \Omega \to \mathbb R$ be independent random variables in $(\Omega, \mathscr F, \mathbb P)$. Let $i \in I$ for any index set I think (or maybe has to be countable). Of course, assume $card(I) \ge 3$. Then show $f_i(X_i)$ are independent. Give conditions on $f_i$ such that $f_i(X_i)$ is independent. I read in above posts that the condition is 'measurable', which I guess means $\mathscr F$- measurable, but I could've sworn that I read before that the condition is supposed to be 'bounded and Borel-measurable', as in bounded and $\mathscr B(\mathbb R)$-measurable for $(\mathbb R, \mathscr B(\mathbb R), Lebesgue)$

2 - for elementary probability theory

Let $X_i: \Omega \to \mathbb R$ be independent random variables that have pdf's. Use the elementary probability definition of independence that is 'independent if the joint pdf splits up', or something. I guess the index set $I$ need not be finite, in which case I think the definition is that the joint pdf of any finite subset of is independent. Give conditions on $f_i$ such that $f_i(X_i)$ is independent. Of course we can't exactly say that $f_i$ is 'measurable'.

Context for the elementary case: I'm trying to justify the computation for the formula for the  moment-generating function for linear combination of independent random variables. See here: Proving inequality of probabilty to derive upper bound for moment-generating functions

Based on the application of Riemann–Stieltjes integral (or Lebesgue–Stieltjes integral) to probability, I think the condition is any $f_i$ such that $E[f_i(X_i)]$ exists (i.e. $E[|f_i(X_i)|]$ is finite).

This is the same condition in Larsen and Marx - Introduction to Mathematical Statistics and Its Applications.

I think $f$ bounded implies this but not conversely.

Update: Also related through another question If $g$ is a continuous and increasing function of $x$, prove that $g(X)$ is a random variable. --> More generally for what functions $g$ is $g(X)$ is a random variable? Of course in advanced probability just say $g$ is Borel-measurable or $\mathscr F$-measurable or whatever, but I think in elementary probability we say $g$ such that $E[g(X)]$ exists i.e. $E[|g(X)|] < \infty$, EVEN THOUGH this is, I believe, a stronger condition than that $g$ is 'measurable', whatever this means in elementary probability. But then again this is kind of weird since we don't even necessarily expect $E[X]$ to exist (i.e. $E[|X|] < \infty$) or well any higher moment $E[X^n]$ I guess.

答:
measure-theoretic:

The measure-theoretic answer is extremely general. It requires nothing special about the real line or Borel sets, just pure measurability. Suppose $(X)_{i \in I}$ is a family (countable is not needed) of random elements, where $X_i: (\Omega, \mathscr{F}) \to (A_i, \mathscr{A}_i)$, i.e. each $X_i$ takes values in some space $A_i$ and $X_i$ is measurable, but all $X_i$ live on the same input space $\Omega$. No assumptions are made about the spaces $\Omega, A_i$ or $\sigma$-algebras $\mathscr{F},
 \mathscr{A}_i$.

Let a corresponding family of functions $(f_i)_{i \in I}$ be given such that for each $i$, $f_i: (A_i, \mathscr{A}_i) \to (B_i, \mathscr{B}_i)$ is measurable. That is, each $f_i$ accepts inputs from $A_i$ (the codomain of $X_i$) and takes values in some space $B_i$ such that $f_i$ is measurable. (This ensures that for each $i$, $f_i(X_i): (\Omega, \mathscr{F}) \to (B_i, \mathscr{B}_i)$ makes sense and is measureable.) Again, no assumptions are made about the spaces $B_i$ or $\sigma$-algebras $\mathscr{B}_i$.

Now suppose $(X_i)_i$ is an independent family under some probability measure $P$ on $(\Omega, \mathscr{F})$, i.e. that for any finite subset $J \subseteq I$ of indices and any measurable subsets $U_i \in \mathscr{A}_i$ one has
$$P(X_i \in U_i \text{ for all } i \in J) = \prod_{i \in J} P(X_i \in U_i).$$

Then we claim that $(f_i(X_i))_{i \in I}$ is also an independent family under $P$.
Indeed, let $J \subseteq I$ be some finite subset of indices and let measurable subsets $V_i \in \mathscr{B}_i$ be given. For each $i \in J$, by the measurability of $f_i$ and $V_i$, one has that $f_i^{-1}(V_i) \in \mathscr{A}_i$ and thus
$$
P(f_i(X_i) \in V_i \text{ for all } i \in J)
= P(X_i \in f^{-1}_i(V_i) \text{ for all } i \in J)
$$
$$
= \prod_{i \in J} P(X_i \in f^{-1}_i(V_i))
$$
$$
= \prod_{i \in J} P(f_i(X_i) \in V_i).
$$
Thus, $f_i(X_i))_{i \in I}$ is an independent family.

elementary probability:

As for the elementary probability solution, it really depends on what your definition of independence is. In all cases, the definition only involves finite subsets of the random variables. I would say that without the definition of a $\sigma$-algebra, the proof is out of grasp unless you make extra (unnecessary) assumptions. If your definition is that densities split as a product, then you must assume some conditions to ensure that $f_i(X_i)$ has a density and that you can apply the usual density transformation rules. If your functions take values in a countable space, the above proof can be repeated essentially verbatim replacing arbitrary $U_i, V_i$ with singletons, i.e. look at $P(f_i(X_i) = y_i, \forall i)$.

Alternatively, since you are avoiding a measure-theoretic answer to a question whose very definition is measure-theoretic, perhaps correctness of the argument is not a requirement? Just tell your students the independence condition must hold for "all sets (verbal asteristk)" and then give the above proof without mentioning the measurability. Or if your students are perhaps more comfortable with topology, you could use only continuous functions and look at preimages of open sets.


问:
Hahn Decomposition $(P,P^c)$
Let $\mu$ be a signed measure with Hahn decomposition $(P,P^c)$. Show that $(-\mu)^+=\mu^-$ and $(-\mu)^-=\mu^+$ and that $(P^c,P)$ is a Hahn decomposition of $-\mu$

I would like to know the following would help me solve this exercise, and if so, how could I apply it:

$\mu^+(E):=\mu(E\cap P)\ge0$, and $\mu^-(E):=-\mu(E\cap P^c)\ge0$

答:
Suppose $E \subseteq P$. By assumption $\mu(E) \geq 0$, so $ -\mu(E) \leq 0$.

Similarly, if $E \subseteq P^c$, then $-\mu(E) \geq 0$. So, $(P^c,P)$ decomposes $-\mu$.

Now, by the above, $(-\mu)^+(E) = -\mu(E \cap P^c) = \mu^-(E)$. And similarly $(-\mu)^- = \mu^+$.

Title: Hahn Decomposition $(P,P^c)$


问:
Combining two identical uniform distributions
Say two random variables, $X$ and $Y$, are such that  $X$ ~ $U(0,a)$ and $Y$ ~ $U(0,a)$.

What will the pdf be for $Z$, where $Z=X-Y$?

答:
We need to assume something about the relationship between $X$ and $Y$. For example, if $X=Y$, then $X-Y$ is identically $0$. We will assume what we are expected to assume but should have been mentioned, that $X$ and $Y$ are independent.

We will first find the cumulative distribution function $F(z)$ of $Z$, that is, the probability that $Z\le z$. First deal with the easy cases. If $z\lt -a$, then $F(z)=0$. If $z\gt a$, then $F(z)=1$. It remains to deal with the harder part, when $-a\le z\le a$.

Draw the square with corners $(0,0)$, $(0,a)$, $(a,a)$, and $(0,a)$. For a fixed value of $z$, draw the line $x-y=z$. We want the probability that we end up in the part of the square that has $x-y\le z$, that is, $y\ge x-z$. So we want to be above the line $y=x-z$. The answer looks different depending on whether $z$ is negative or positive. We deal with negative $z$, and leave it to you to deal with positive $z$.

When $z$ is negative, the part of the square where $y\ge z$ is an isosceles right triangle with  side $a+z$. This triangle has area $\frac{(a+z)^2}{2}$. The probability we land in that triangle is the area divided by the total area of the square. Thus if $-a\le z\le 0$ then
$$F(z)=\frac{(a+z)^2}{2a^2}.$$
For the density function, we will differentiate.

Now it's your turn: deal with $0\lt z\le a$.

Alternately, if you want to integrate, for $-a\le z\le 0$, we have
$$F(z)=\int_{x=0}^{a+z}\int_{y=x-z}^a\frac{1}{a^2}\,dy\,dx.$$
To find the limits of integration, it is useful to draw the same picture as the one that gives us the geometric solution.


问:
Ring of polynomials in $\mathbb F_2$ with roots at a (specific) root of unity
WLOG I'm going to use $\zeta = \sqrt[12]1$ in this example. I'm trying to find some way to work with sums of distinct powers of $\zeta$, specifically so that I can analyze sums that are equal to zero (such as $1 + \zeta + \zeta^2 + \zeta^5 + \zeta^6 + \zeta^8 + \zeta^9$) (EDIT: I want the coefficients in the sum to be either $0$ or $1$, so each distinct power is only included once at most). I initially tried to work in the quotient ring $\mathbb F_2[x] / ( x^{12} + 1)$, but it quickly became clear that while I expected $P=1 + x+ \cdots + x^{11}$ to be equal to $0$ (since $Px = P$), it was not. I'm not sure what the flaw in my logic here was (though I suspect it has something to do with $1$ being a root of $x^{12}+1$), but moving to $\mathbb F_2[x]/P$ obviously fixed that issue.  My problem is that I can't find a sense in which the polynomial (for example) $1 + x +  x^2 + x^5 + x^6 + x^8 + x^9 = 0$, even though that's what I should expect to see if $x = \zeta$. Where am I going wrong?

答:
Over $\mathbb{F}_2[x]$ we have $x^{12} - 1 = (x^3 - 1)^4$ so the quotient ring $\mathbb{F}_2[x]/(x^{12} - 1)$ is less well-behaved than you might have initially guessed and in particular is not a field or even an integral domain ($x^3 - 1$ is nilpotent); this means it is not the subfield of $\overline{\mathbb{F}_2}$ obtained by adjoining a $12^{th}$ root of unity, which may have been what you were actually going for. The problem persists for $\frac{x^{12} - 1}{x - 1} = (x - 1)^3 (x^2 + x + 1)^4$.

Linear combinations of powers of $\zeta_{12}$ equal to zero with integer or rational coefficients can be completely understood as follows. The minimal polynomial of $\zeta_{12}$ over $\mathbb{Z}$ is the cyclotomic polynomial

$$\Phi_{12}(x) = x^4 - x^2 + 1$$

and every linear relation between the powers of $\zeta_{12}$ is a consequence of the relation $\Phi_{12}(\zeta_{12}) = 0$ in the sense that it corresponds to a polynomial relation $p(\zeta_{12}) = 0$ and $\Phi_{12}(x) \mid p(x)$. In your example, WolframAlpha gives

$$p(x) = (x^4 - x^2 + 1)(x^5 + x^4 + x^3 + 2x^2 + x + 1).$$


问:
Help simplyfing a sum
Hi I was working on a Probability problem  and got as result this expression

$$P(Z = k) = \sum\limits_n \left({n \choose k} p^k (1-p)^{n-k}\right)\left(e^{- θ} * \frac{θ^n}{n!}\right)$$

And now I'm trying to simplify the expresion, but so far I only achieved this:

$$\sum\limits_n \left({n \choose k} p^k (1-p)^{n-k}\right)\left(e^{- θ} * \frac{θ^n}{n!}\right) $$

$$=\sum\limits_n \frac{1}{k!(n-k)!} p^k (1-p)^{n-k}e^{- θ}θ^n  $$

$$=\frac{p^{k}e^{-θ}}{k!}\sum\limits_n  \frac{(1-p)^{n-k}θ^n}{(n-k)!} $$

But now I'm at this point and I don't what else to do to simplify. Can someone help me?

答:
Implicitly your sum is over $n\geq k$, so
$$\frac{p^ke^{-\theta}}{k!}\sum_{n\geq k}\frac{(1-p)^{n-k}}{(n-k)!}\theta^n=\frac{p^ke^{-\theta}}{k!}\sum_{m\geq0}\frac{(1-p)^m}{m!}\theta^{m+k}=\frac{p^ke^{-\theta}}{k!}\cdot\theta^k\cdot e^{(1-p)\theta}$$
i.e. $Z\sim\operatorname{Po}(p\theta)$.

Title: Help simplyfing a sum


问:
Supremum and infimum of $S = \left\{\frac{n-1}{n+1} \cdot\cos\left(\frac{2n \pi}{3}\right) : n \in \Bbb N\right\}$
Let us consider the set $$S = \left\{\frac{n-1}{n+1} \cdot\cos\left(\frac{2n \pi}{3}\right) : n \in \Bbb N\right\}$$
I am trying to find the supremum and infimum of the set.

We know that $\cos\frac{2n \pi}{3}$ is either $-\frac12$ or $1$. Also we can write $$\frac{n-1}{n+1} = 1 - \frac{2}{n+1}$$

I am not able to proceed further.

答:
Hint: Let
$$f(n) = \frac{n-1}{n+1} \cos\left( \frac{2n\pi}{3}\right).$$
Then,
$$f(3n) = \frac{3n-1}{3n+1} \ \ \ \ \text{and} \ \ \ \ f(3n+1) = -\frac{1}{2} \frac{(3n-1) -1}{(3n+1)+1}.$$
Hence, as $n\to\infty$, the sequences $f(3n)$ and $f(3n+1)$ converge to limits (which are...?). Then note that
$$ -\frac{1}{2} \frac{n-1}{n+1} \le f(n) \le \frac{n-1}{n+1}$$
because, as you observed, $\cos(2n\pi/3)$ is either $-1/2$ or $1$.


问:
Non-cyclic module over the first Weyl algebra
Let $U$ be an open subset of the complex plane $\mathbb{C}$. Denote by $H(U)$ the $\mathbb{C}$-algebra of holomorphic functions $U\to\mathbb{C}$. The first Weyl algebra $A_1 = A_1(\mathbb{C})$ acts on $H(U)$ in the obvious way: the polynomials acts by multiplication and the derivation $\partial$ acts as the differentiation operator: $\partial f = f'$. Thus $H(U)$ is a left $A_1$-module.

I need to show that $H(U)$ is not a cyclic $A_1$-module. For this, assume the opposite, and let $h\in H(U)$ be such that $H(U)=A_1 h$. It is easy to see that $h$ cannot be a constant function (otherwise $H(U) = \mathbb{C}[x]$, which is absurd since $\exp\in H(U)$). I will be done if I can show that $\exp\circ h\not\in A_1 h$, that is, if $h$ does not satisfy a differential equation of the form
$$
e^{h(z)} = \sum_{j=0}^n f_j(z)\frac{d^j h}{dz^j}(z), \quad z\in U,
$$
where $f_j\in \mathbb{C}[x]$. But I'm stuck here. I have really no idea about how to prove this last assertion.

This proof is suggested by Coutinho in Exercise 4.7 on Chapter 5 in his book A primer of Algebraic $D$-modules.

Any help will be appreciated!

答: A cyclic module over the Weyl algebra is in particular countable-dimensional, and $H(U)$ is uncountable-dimensional, e.g. the functions $\exp(rz)$ for $r \in \mathbb{C}$ are linearly independent (exercise; there are several nice ways to do this, and the same result holds for the functions $\exp(rx), r \in \mathbb{R}$ on any open interval in $\mathbb{R}$ also, so the continuous, smooth, or analytic functions on an open interval are also uncountable-dimensional, and the functions $\exp(-rx^2), r > 0$ show the same for the Schwartz functions).


问:
Trigonometric substitution integral
Trying to work around this with trig substitution, but end up with messy powers on sines and cosines... It should be simple use of trigonometric properties, but I seem to be tripping somewhere.

$$\int x^5\sqrt{x^2+4}dx $$

Thanks.

答:
It comes out pretty well if you put $x=2\tan\theta$. Doing it carefully, remove a $\tan\theta\sec\theta$ and everything else can be expressed as a polynomial in $\sec\theta$, hence it's easily done by substitution.

But a more "efficient" substitution is $u=\sqrt{x^2+4}$, or $u^2=x^2+4$. Then $2u\,du = 2x\,dx$ andthe integral becomes
$$\int (u^2-4)^2 u^2\,du\,.$$

Title: Trigonometric substitution integral


问:
Given functions $h,k:\Bbb R\to \Bbb R$, is it possible to determine whether $f,g:\Bbb R\to\Bbb R$ exist so that $g\circ f=h$ and $f\circ g=k$?
Let's say I have two functions $h,k:\Bbb R\to \Bbb R$.  I want to find $f,g:\Bbb R\to \Bbb R$ such that $g\circ f=h$ and $f\circ g=k$.  I know that $f,g$ might not exist (for example, Functional equation involving composition and exponents).  Do we know at least a condition for $h,k$ such that $f,g$ exist?

Which condition guarantees the uniqueness of $f,g$ (provided that they exist)?  Note that there are $h,k$ such that $f,g$ are not unique. For example, $h=k=0$, where $f=0$ works and $g$ is any function s.t. $g(0)=0$.  Or when $h=k$ is the identity function, and we take $f$ to be any bijection and $g=f^{-1}$.

At the very least, what do we know about this problem when $h,k$ are polynomial functions?  Is there a simple test that tells us there are polynomials $f,g$ that satisfy the conditions for a given pair of polynomials $h,k$?  Again, what about the uniqueness of polynomial solutions?

If the general problem is too hard, I am most interested in this specific problem. I want to find $f,g:\Bbb R\to\Bbb R$ such that
$$g\circ f(x)=x^3+1$$
and
$$f\circ g(x)=x^3+3x^2+3x+2.$$
Clearly $f,g$ are bijective functions if they exist.  So, can we determine the value of $g\circ f^{-1}(-7)$?

I found $f,g$ that almost work.  When $f(x)=x^3$ and $g(x)=x+1$, we have $g\circ f(x)=x^3+1$ but $f\circ g(x)=x^3+3x^2+3x+1$.  Unfortunately they don't quite work.  I know also that there are no polynomial functions $f,g$ that work.

Note that
$$f(x^3+1)=f(x)^3+3f(x)^2+3f(x)+2$$
and
$$g(x^3+3x^2+3x+2)=g(x)^3+1.$$
$\therefore$ if $a,b$ are the unique real numbers such that $a^3+1=a$ and $b^3+3b^2+3b+2=b$, we see that $f(a)=b$ and $g(b)=a$.  These are the only values of $f$ and $g$ that I know.  But I can also see that
$$ f^{-1}(-7)=g(-3)$$
if that helps.

Let $h(x)=x^3+1$ and $k(x)=x^3+3x^2+3x+2$.  Due to $f\circ g(x)$ and $g\circ f(x)$ are given; find $f$ and $g$, if $f=f_0$ and $g=g_0$ satisfy the conditions, then $f=f_0\circ \phi$ and $g=\phi^{-1}\circ g_0$ form a solution for any bijection $\phi:\Bbb R\to\Bbb R$ such that $h\circ \phi=\phi\circ h$.  Because any iteration of $h$ commutes with $h$, we can see that there are infinitely many $f$ and $g$, if $f_0,g_0$ exist.  How do I see whether $f_0,g_0$ exist?

答:
This is an addendum to the very brilliant analysis already given by orangeskid. In light of their analysis, I'll provide some easy facts about topological conjugation over the reals.

Claim 1: If $f:\mathbf{R}\to\mathbf{R}$ is strictly increasing, continuous, unbounded above and below, and such that $f(0)>0$, then there is a strictly increasing and continuous $\varphi:\mathbf{R}\to\mathbf{R}$ such that $\varphi(0)=0$ and $f\circ\varphi(x)=\varphi(x+1)$. Moreover if $f(x)>x$ for all $x\in\mathbf{R}$, then $\varphi$ is also unbounded above and below.

Proof: Since we know $f(0)>0$, let $\varphi(a)=af(0)$ for all $a\in[0,1)$. We will define the rest of $\varphi$ by extending in the obvious fasion: $\varphi(x)=f^{(\lfloor x\rfloor)}\circ\varphi\left(x-\lfloor x\rfloor\right)$, where $f^{(-)}$ denotes functional iteration, as $f$ is bijective. Clearly the next thing to do is to check that this fits the requirements:

We forced $f\circ\varphi(x)=\varphi(x+1)$ by contsruction, so that is done.

To check continuity, note that $f^{(\lfloor x\rfloor)}$ is always continuous, so by functional composition $\varphi$ is continuous over $\mathbf{R}\smallsetminus\mathbf{Z}$. To check continuity on $\mathbf{Z}$, it suffices to check continuity as $x\to 1^-$. For this note that
$$\varphi(1)=f\circ\varphi(0)=f(0)=\lim_{x\to 1^-}\varphi(x)$$

To see $\varphi$ is strictly increasing, note that $f^{(\lfloor x\rfloor)}$ is strictly increasing by assumption and that $\varphi$ is strictly increasing over $[0,1)$, so we get $\varphi$ is strictly increasing over all intervals $[z,z+1)$ where $z\in\mathbf{Z}$. However $\varphi$ is continuous, and so it is strictly increasing over $\mathbf{R}$.

Now to check the "moreover" part.

If $\varphi$ is not unbounded, then by monotone convergence, there is a bound $M=\lim_{x\to A}\varphi(x)$ where $A\in\pm\infty$. However, as $f$ is continuous,
$$f(M)=f\left(\lim_{x\to A}\varphi(x)\right)=\lim_{x\to A}f(\varphi(x))=\lim_{x\to A}\varphi(x+1)=M$$
This contradicts that $f(x)>x$ for all $x\in\mathbf{R}$.

Claim 2: If $f:[0,\infty)\to[0,\infty)$ is strictly increasing and continuous, such that $f(0)=0$ and $f(x)>x$ for all $x>0$, then there is a strictly increasing, continuous, and unbounded $\varphi:[0,\infty)\to[0,\infty)$ such that $\varphi(0)=0$ and $f\circ\varphi(x)=\varphi(2x)$.

Proof: Let $g:\mathbf{R}\to\mathbf{R}$ be given by $g(x)=\log_2 f(2^x)$. By Claim 1, there is some $\psi:\mathbf{R}\to\mathbf{R}$ that is strictly increasing, continuous, unbounded above and below, and such that $g\circ\psi(x)=\psi(x+1)$. Then let $\varphi(x)=2^{\psi(\log_2 x)}$, so we see that
$$\varphi(2x)=2^{\psi(1+\log_2 x)}=2^{g\circ\psi(\log_2 x)}=f(2^{\psi(\log_2 x)})=f\circ\varphi(x)$$

Claim 3: If $f:\mathbf{R}\to\mathbf{R}$ is strictly increasing, continuous, and has exactly one unstable fixed point $c$, that is, $f(x)>x$ for all $x>c$ and $f(x)<x$ for all $x<c$, then there is an increasing homeomorphism $\varphi:\mathbf{R}\to\mathbf{R}$ such that $\varphi^{-1}\circ f\circ \varphi(x)=2x$.

Proof: Let $g:\mathbf{R}\to\mathbf{R}$ be given by $g(x)=f(x+c)-c$, thus $g$ shares all properties with $f$ except $0$ is the fixed point of $g$. By Claim 2, there are increasing homeomorphisms $\varphi_{\pm}:[0,\infty)\to[0,\infty)$ such that $\varphi_{\pm}(0)=0$, and moreover both $\varphi_+^{-1}\circ g\circ\varphi_+(x)=2x$ and $\varphi_-^{-1}(-g(-\varphi_-(x)))=2x$. Let $\psi:\mathbf{R}\to\mathbf{R}$ be given by
$$\psi(x)=\begin{cases}
\varphi_+(x)&\text{if }x\ge 0\\
-\varphi_-(-x)&\text{if }x<0
\end{cases}$$
Then it is not hard to see that $\psi$ is an increasing homeomorphism such that $\psi^{-1}\circ g\circ\psi(x)=2x$. Finally let $\varphi:\mathbf{R}\to\mathbf{R}$ be given by $\varphi(x)=\psi(x)+c$, so then
$$2x=\varphi^{-1}(\psi(2x)+c)=\varphi^{-1}(g\circ\psi(x)+c)=\varphi^{-1}\circ f\circ\varphi(x)$$

As a corollary, note that both $x^3+1$ and $x^3+2$ satisfies Claim 3, so both are conjugate to $2x$.

Also note that it is completely possible to modify the proof such that both $x^3+1$ and $x^3+2$ are conjugate to $2x$ via a homeomorphism that is smooth on all of $\mathbf{R}$ except at the fixed point.

This is unavoidable:

Added Claim 4: Consider the two linear functions $f(x)=2x$ and $g(x)=4x$. Let $\varphi:\mathbf{R}\to\mathbf{R}$ be any homeomorphism such that $\varphi\circ f=g\circ\varphi$. Then $\varphi$ cannot be continuously twice differentiable at $0$.

Proof: Assuming not, then by Taylor's theorem we have
$$\varphi(x)=ax+bx^2+h(x)\cdot x^2$$
where $h$ is continuous at $h(0)=0$. Then by expanding on $\varphi\circ f=g\circ\varphi$, we eventually get
$$h(2x)-h(x)=\frac{a}{2x}$$
Taking the limit $x\to 0$ on both sides, we see that $a=0$, and $h(2x)=h(x)$. However the continuity of $h$ at $0$ implies that $h$ is identically $0$, meaning that $\varphi(x)=bx^2$, and $\varphi$ cannot be a homeomorphism.


问:
Finding the angle of a line in 3D space projected onto a 2D plane
I'm actually a programmer here. I have a math problem, but I don't major in math so I may not explain this clearly.

I have a line in a 3D space with points A and B (e.g. (3,4,8) to (6,9,2)). I (or my eyes) am at point C looking towards a direction that allows me to see point A, as well as part of the line or the whole line.

If whatever I'm seeing is being projected onto a 2D plane, how do I calculate the angle between the line and the top of the 2D plane?

答:
You do not have sufficient information.

What you need to know is

The 3D direction vector $\vec{w}$ perpendicular to the viewing plane: from $C$ into the 3D direction that passes through the center of the 2D viewing plane.  This is the direction you are viewing at.

The 3D direction vectors $\vec{u}$ and $\vec{v}$ corresponding to "right" and "up", both perpendicular to $\vec{w}$.  Often only one of them is specified, because the other can be obtained from the other two via vector cross product: $\vec{v} = \vec{w} \times \vec{u}$ and $\vec{u} = \vec{v} \times \vec{w}$.

Since their lengths are irrelevant, they are usually unit vectors.  The following assumes so, that $\lVert\vec{u}\rVert = \lVert\vec{v}\rVert = \lVert\vec{w}\rVert = 1$.

You have a direction vector in 3D, $\vec{d} = B - A = (3, 5, -6)$.  Project this to the visual plane using
$$\begin{aligned}
\vec{p} &= \vec{d} - \frac{\vec{d} \cdot \vec{w}}{\left\lVert \vec{w} \right\rVert} \vec{w} \\
~ &= \vec{d} - (\vec{d} \cdot \vec{w}) \vec{w} \\
\end{aligned}$$
Then, the dot products between the projected direction vector with the "right" and "up" vectors give you the direction cosines,
$$\begin{aligned}
x_\Delta &= \vec{p} \cdot \vec{u} \\
y_\Delta &= \vec{p} \cdot \vec{v} \\
\theta &= \operatorname{atan2}\left(y_\Delta, x_\Delta\right) \\
\end{aligned}$$
where $\theta$ is the direction in 2D: $0°$ and $\pm 360°$ right, $+90°$ and $-270°$ up, $\pm 180°$ left, and $+270°$ and $-90°$ down, and $\operatorname{atan2}(y, x)$ is the two-parameter form of $\arctan(y/x)$, except covering all angles by taking both signs into account.  Most programming languages provide this two-parameter form.

Because $\vec{u}$ and $\vec{v}$ are perpendicular to $\vec{w}$, $x_\Delta$ and $y_\Delta$ are unchanged by projecting the direction vector to the viewing plane.  Thus, we can simplify the calculation to the following:

Let $\vec{u} = ( x_u , y_u , z_u )$ be the 3D vector that is directly right in the 2D view, and $\vec{v} = ( x_v , y_v , z_v )$ be the 3D vector that is directly up in the 2D view, and $\vec{d} = ( x_d , y_d , z_d ) = B - A$ the 3D direction vector.  Then,
$$\begin{aligned}
x_\Delta &= \vec{d} \cdot \vec{u} \\
y_\Delta &= \vec{d} \cdot \vec{v} \\
\theta &= \operatorname{atan2}\left( y_\Delta , x_\Delta \right) \\
\end{aligned}$$
or, in Cartesian coordinates,
$$\theta = \operatorname{atan2}\left( x_d x_v + y_d y_v + z_d z_v , \; x_d x_u + y_d y_u + z_d z_u \right)$$


问:
What is the order of convergence of Newtons root finding method? And when does it converge?
Given a function $f(x)$, we can approximate $x_r$ where $f(x_r)=0$ , by using Newton's method:

$$x_{n+1}=x_n -\frac{f(x_n)}{f'(x_n)} $$

The method only works when 'you choose an $x_0$ near enough to $x_r$'.

My questions are:

When does this series converge?

Given that the series converges, what is the order of convergence?

If it depends on $f(x)$, then let me define $f(x)=e^x-x-1.9\cos{x}$

答: When $|f'(x)|<1 \ \forall x \in (a,b), \ f(x)$ will converges to a root in $(a,b)$ given a start value in $(a,b)$. The rate of convergence is quadratic for roots of multiplicity $m=1$. for $m>1$ convergence is linear but in that case $x_{n+1}=x_n-m\frac{f(x_n)}{f'(x_n)}$ will have a quadratic convergence rate.    </p>


问:
Integral part of an infinite simple continued fraction
Let $[a_0,a_1,a_2,\ldots]$ be an infinite simple continued fraction.

How can one show that $a_0$ is the integral part of $[a_0,a_1,a_2,\ldots]$?

There is a similar theorem for finite continued fractions with length $n\geq 1$ and $a_n \geq 2$ that I know, however I don't know how to extend this to the infinite case. The infinite continued fraction is the limit of the sequence of its convergents but as each convergent will not always have $2$ as its last partial coefficient, I don't know how to proceed.

答:
For $n \geq 2$, every finite continued fraction $[a_2,a_3,\ldots,a_n]$ is at least $a_2$.  Therefore the infinite continued fraction $\beta := [a_2,a_3,\ldots]$, which is defined to the the limit of the finite continued fractions $[a_2,a_3,\ldots,a_n]$ is at least $a_2$: $\beta \geq a_2 > 0$.

Setting $\alpha = [a_0,a_1,a_2,\ldots]$, we have
$$
\alpha = a_0 + \frac{1}{a_1 + 1/\beta}.
$$
Since $\beta > 0$ we have $0 < 1/\beta$, so $0 < a_1 < a_1 + 1/\beta$, so
$$
0 < \frac{1}{a_1 + 1/\beta} < \frac{1}{a_1} \leq 1 \Longrightarrow 0 < \frac{1}{a_1 + 1/\beta} < 1.
$$
Add $a_0$ to each term in the last inequality and you get $a_0 < \alpha < a_0 + 1$, so $a_0 = \lfloor{\alpha}\rfloor$.


问:
Why is a "list" of values associated with a variable (e.g. X1, X2, ... , Xn) often called variables (plural)?
One of the most common beginnings of any statistics or probability proofs goes something like:

```<code>let X = (X1, X2, ... ,Xn) be a random vector with the variables assumed to be drawn from a multivariate Gaussian.
</code>```

...or something along these lines. The list of values (X1, X2, ... ,Xn) are being called variable(s), plural. But isn't this in fact ONE variable only, and the subscript versions are the various values that have been drawn for that ONE variable?

Why is the list of values being called variables?

答:
A variable is something which has (potentially temporarily, if you solve for it) indeterminate value, nothing more, nothing less. That said, it is important to know what your variable belongs to.

That is to say, when we have the variable $x$ in the number:

$$3x+12$$

we know from context that in fact, $x$ is a number as well. Let $nX+b = \{nx+b : x \in X\}$. Then we can also have the expression:

$$3X + 12$$

where our variable is a set. Now, we do have a few things to say about this. One, we don't really refer to it as a variable in common parlance. Generally, the term 'variable' is reserved for things made of numbers, with some strict definition. That said, that's mostly a convention of convenience and notation, as the essence of an indeterminate identity is what makes variables what they are. Reinforcing this is my second point: we can treat our second formula like the first.

For example, given $y = 3x+12$, we can see that the value of $x$ is determined by the value of $y$. Similarly, given:

$$Y = 3X + 12$$

then $X$ is is totally determined by $Y$; in fact:

$$\frac{1}{3}Y - 4 = X$$

So given a set $Y$ of numbers, we can uniquely determine $X$.

In general, this idea of an indeterminate form is what gives us generality. For example, every time a theorem starts with "Let $A$ be a [insert mathematical object here]," we are essentially declaring $A$ to be a "variable" of the form [mathematical object]. Indeed, just as we solve for $x$ given some parameter $y$, many theorems are of the form "Let $A$ be [xyz], and fix some constraints upon it. This implies that $A$ is actually specifically [abc], and only [abc]." That, structurally, is the same as fixing $x$ a random number, saying that some known number $y$ is $3x+12$, and concluding that $x$ must be known as well.

More practically, anything that depends on a variable is a variable. So let $x$ be a variable, and let:

$$\mathbf{a} = \begin{bmatrix} 1 \\ x \\ 123 \end{bmatrix}$$

This is a vector whose value uniquely depends on a variable, so in some sense we may write $\mathbf{a}(x)$ as in function notation, and see that $\mathbf{a}$ is a variable exactly in the way that $x$ is: as long as $x$ is a variable, so is $\mathbf{a}$.


问:
Prove equality of composition
I have to prove or disprove $f∘f^{-1}=f^{-1}∘f\ (f:A\to B$ and $f$ is a bijective function$)$.

I get $f∘f^{-1}$ and $f^{-1}∘f$ are both identity functions. But is this fact make them equal?

Here is what I think.
$a\in A$ and $b\in B$ with  $f(a)=b$
$$f∘f^{-1}(b) = f(a)=b$$
$$f^{-1}∘f(a) = f^{-1}(b)=a$$

So if $a \not= b$ we can not say $f∘f^{-1}=f^{-1}∘f$.

I am really confused with this question. Please give me a help.

答:
Your proof (in your original post) has the right idea. And if you distill that idea from your work, then you don't even need all that work… Let me explain.

One thing you need to understand here is that for any set $X$ there's a unique identity function $\operatorname{Id}_X:X\to X$ on this set. But such identity functions for two different sets — say, $\operatorname{Id}_X:X\to X$ and $\operatorname{Id}_Y:Y\to Y$, where $X\neq Y$ — are by definition different functions, since their domains and codomains are different.

That's all you need here. You're right in your proof: generally speaking, $a\neq b$, and so in general $f\circ f^{-1}\neq f^{-1}\circ f$. More specifically, this conclusion is only valid if $A\neq B$ are two distinct sets. In this case, given a bijective function $f:A\to B$, the two compositions are two different identity functions. To see that, note that $f^{-1}:B\to A$. And now answer the following questions:

What are the domain and codomain of $f^{-1}\circ f$? And therefore, which identity function is it?

What are the domain and codomain of $f\circ f^{-1}$? And therefore, which identity function is it?

On the other hand, if $f$ is a bijection from a set $A$ to itself, $f:A\to A$, then the two compositions are equal: $f\circ f^{-1}=f^{-1}\circ f=\operatorname{Id}_A$.

Title: Prove equality of composition


问:
Quirky Primes Test
Wrote a program to list all primes $n$ below $1,000,000$ and noticed they never satisfied (except $n=5$)

$\frac {n-1}2 \equiv 2 \pmod{10}$

or

$\frac {n-1}2 \equiv 7 \pmod{10}$

Is that true for all primes? Thanks.

```<code>#include <stdio.h>
#include <stdint.h>

int isPrimeNumber(int number)
{
    int iLoop = 0;
    int iPrimeFlag = 1;

    if (number <= 1)
    {
        iPrimeFlag = 0;
    }
    else
    {
        for (iLoop = 2; iLoop < number; iLoop++)
        {
            if ((number % iLoop) == 0)
            {
                iPrimeFlag = 0;
                break;
            }
        }
    }
    return iPrimeFlag;
}

int main(int argc, char* argv)
{
    uint64_t n, i, m;

    for (n = 2; n < 1000000; n++)
    {
        if (isPrimeNumber(n) == 1)
        {
            i = (n - 1) / 2;
            m = i % 10;

            if (m == 2 || m == 7)
            {
                printf("n=%llu i=%llu\n", n, i);
            }
        }
    }

    return 0;
}
</code>```

答: Your two congruences simplify to $n\equiv5\bmod10$, which is obviously not true for any prime except $5$.

Title: Quirky Primes Test


问:
Finding the number of non-decreasing traversals of a grid
I'm trying to find the number of non-decreasing traversals of an $n$ by $m$ grid ($n$ rows, $m$ columns) with $n<m$ from the bottom-left corner square to the top-right corner square. Here is an example for $m=8, n=4$:

Equivalently, I'm trying to find the number of non-decreasing (i.e. $k_1 = 1 \leq k_2 \leq ... \leq k_m = n$) sequences where $k_i \leq n$ and every number in $\{1, ..., n\}$ appears at least once.

This initially looked like a stars-and-bars problem with $\sum k_i = M$ but this construction doesn't quite work (I couldn't find a way to encode the "every number in $\{1, ..., n\}$ appears at least once" criterion).

答:
Replace each diagonal step by an up-step, and you get a path by rook moves from $\langle 1,1\rangle$ to $\langle n-m+1,m\rangle$. This path consists of $m-1$ up-steps and $n-m$ right-steps, for a total of $n-1$ steps. The steps can be taken in any order, and the path is completely determined when we know which steps are up-steps, so there are $\binom{n-1}{m-1}$ possible paths. On your $8\times 4$ board, for instance, there are $\binom73=35$ such paths.

Note that the transformation from one of your paths to a rook path is reversible simply by replacing each up-step by a diagonal step, so it really does define a bijection between your paths on an $n\times m$ board on and rook paths on an $(n-m+1)\times m$ board.