答:
Lemma: if you repeat an experiment with probability $P$ for event $A$ and probability $Q$ for event $B$ (constant in each repetition) then the probability that you will see an $A$ before a $B$ is $\tfrac{P}{P+Q}$.

Based on the above lemma, let $A$ be the event of getting 7 or 11, and $B$ the event of getting 9. The desired probability is

$$\tfrac{P}{P+Q}+\tfrac{Q}{P+Q}\tfrac{P}{P+Q}$$

The first term is getting 7 or 11 before the first occurrence of 9. The second term is that the first 9 came before the first 7 or 11, but then the second 9 came after. The idea is that you can after the first 9 restart the rolls, waiting for only one 9.

Title: Probability Q on rolling two 8-sided dice

Body:
Let's say I roll two 8-sided dice. I win if the sums '7' and '11' show up before we see the sum '9' TWICE. What is my probability of winning?

So this is my answer and please correct me if I am wrong:

Prob of getting either 7 or 9 —> 6/64 + 6/64 = 12/64 —> 18.75%

Prob of getting 9 twice —> 8/64 x 8/64 = 1/64 —> 1.6%

But now to find the probability of winning, would it be 18.75 - 1.6 = 17.15% ? Am I right or wrong?

Thanks!

EDIT

From the answers provided below, we have three different results:

84%
49%
60%

Which one would be the correct answer?


问:
When calculating a summation will there be a difference in answer if the index of summation is 0 or 1?
When calculating a summation will there be a difference in answer if the index of summation is 0 or 1?

There was a question, have forgotten but when i = 1, the whole sum was equal to 15, and when i = 0, the sum was equal to 16.

Who closed this question? I promise this question will help many peoples.

答:
Note that $$\sum_{i=0}^n a_i=a_0+\sum_{i=1}^n a_i=a_0+a_1+\sum_{i=2}^n a_i =\ldots$$
hence $\sum_{i=0}^n a_i=\sum_{i=1}^n a_i$ if and only if $a_0=0$.

(The same holds with convergent series, i.e., $\sum_{i=0}^\infty a_i=\sum_{i=1}^\infty a_i$ if and only if either $a_0=0$)

ContentLicense: CC BY-SA 4.0


问:
Using trigonometric substitution to integrate $\int\frac{x^3dx}{\sqrt{25-x^2}}$

I tried solving this calculus problem so many times but I never got an answer that at least looked similar to the choices in the image. Can someone help?
I don't understand where the Sine came from. This is the answer I get

$$\frac13(25-x^2)^{3/2} - 25\sqrt{25-x^2} + C $$

Edits are appreciated

答:
You want to evaluate the $$\int \frac{x^3\mathrm d x}{\sqrt{25-x^2}}.$$ If you use the substitution $x=5\sin y,$ the integral is transformed into $$125\int \sin^3 y\mathrm dy=-125\int \sin^2y \mathrm d(\cos y)=-125\left(\cos y - \frac{\cos^3y}{3}\right)+\text{constant}.$$

To answer you directly from the solution you obtained, recall that $x=5\sin x,$ so that your result is equivalent to $$\frac{125\cos^3y}{3}-125\cos y+C.$$


问:
If a 3D-cake is cut by $n$ planes yielding the maximum number of pieces, then what is the number of pieces with the cake crust?
It is known that a 3D-cake can be cut by $n$ plane cuts at most into $N$ pieces, defined by Cake Number $N=\frac {1}{6}(n^3+5n+6)$. However, some of the pieces would have a crust of the cake as one of their boundaries. How many pieces are with crust? (Entire surface of the cake is assumed to be crust).</p>

答:
To get the maximum number, the cutting planes are in general position.

Consider the 2-D surface (crust) of the cake. Let the intersection of the cutting planes and the crust be denoted as cutting circles. Remove 1 point that is not on any of the cutting lines. Project what remains onto the plane. We now have $n$ circles in general position cutting the plane. We know that the number of regions it divides it into is

$$n^2-n+2.$$

ContentLicense: CC BY-SA 3.0


问:
Intuition behind f-related vector fields
Let $f: M \mapsto N $ be a smooth map between differentiable manifolds. Then a vector field $X$ on $M$ is f-related to a vector field $Y$ in N if $$(f_*)_p (X_p) = Y_{f(p)} \ \ \forall p \in M$$

That is the derivative of the map $f$ at $p$ acting on a tangent vector at $p$ of $X$ is equal to a tangent vector at $Y$ at $f(p)$.

I'm struggling to understand the intuition behind this definition. How should one think about this? Is there a geometric interpretation?

答:
Concretely, if $X$ is a vector field on $M$ and if $f : M \to N$ is an injective smooth map, then $\mathrm{d}f(X) = f_*X$ is a "vector field" on $f(M) \subset N$. $f(M)$ may not be a submanifold or anything, so it is not a "real" vector field. But if you can extend it in $N$, that is, if there exists a vector field $Y$ such that $f*X = Y$ at each $f(p) \in N$, then you say that $Y$ is $f-$related to $X$.

This definition is here to work with images of vector field by smooth maps as if they were "true" vector fields on $N$. Sometimes, one can not extend it (think of the tangent vector field of a self-crossing cuve in the plane).

Note that I asked $f$ to be injective, but one can just ask that if $f(p) = f(q)$, then $\mathrm{d}f(p)X_p = \mathrm{d}f(q)X_q$ so that the value of $f_*X$ at a point $z \in f(M)$ would not depend on the point in $f^{-1}(z)$.


问:
Rewrite Lebesgue integral over a centered interval
Given a function $f\in L^1(\mathbb{R})$, define the integral over an interval centered at $x$: $$f_c(x)=\frac{1}{2c}\int_{x-c}^{x+c} f(t)\,dt$$

There is a stated result that we can rewrite $f_c(x)$ into $$f_c(x)=\int_{\mathbb{R}}f(x-t)\frac{1}{2c}\mathbf{1}_{[-c,c]}\,dt$$where $\mathbb{1}_{[-c,c]}$ is the indicator function of $[-c,c]$.

But it seems to be counter-intuitive and I don't know how to proceed. Any idea how to show this is true?

Thanks in advance.

答:
Using the definition of the indicator function,
$$
\int_{\mathbb{R}} f(x-t) \frac{1}{2c} 1_{[-c,c]} \,dt = \frac{1}{2c} \int_{-c}^{c} f(x-t) \,dt.
$$
Now just make a change of variables $s=x-t$.


问:
Show that $f(x)= e^{-1/x} \sin e^{1/x}$ for $x>0$ and $f(0)=0$ is continuous at $x=0$
$$f(x) =  \begin{cases}
      e^{-1/x} \sin e^{1/x}  & x\in (0,\infty) \\
      0 & x=0
   \end{cases}
$$

How do I show this function is continuous at $0$? So far, I have determined that
$$\lim_{x \to 0} f(x) = 0$$ but  $$\lim_{x \to 0} e^{-\frac{1}{x}} \sin e^{\frac{1}{x}}$$ diverges. I don't know how to proceed.

答:
$\sin(g(x))$ is always bounded, and $\lim_{x \to 0} e^{-\frac{1}{x}} = e^{-\infty}=0$. So, multiplying these yields

$$ \lim_{x \to 0} e^{-\frac{1}{x}} \sin e^{\frac{1}{x}} = 0 \times \mathrm{bounded} = 0 = f(0)$$

So, $f$ is continuous at $x=0$


问:
Prove that $\log X < X$ for all $X > 0$
I'm working through Data Structures and Algorithm Analysis in C++, 2nd Ed, and problem 1.7 asks us to prove that $\log X < X$ for all $X > 0$.

However, unless I'm missing something, this can't actually be proven. The spirit of the problem only holds true if you define several extra qualifiers, because it's relatively easy to provide counter examples.

First, it says that $\log_{a} X < X$ for all $X > 0$, in essence.

But if $a = -1$, then $(-1)^{2} = 1$. Therefore $\log_{-1} 1 = 2$. Thus, we must assume
$a$ is positive.

if $a$ is $< 1$, then $a^2 < 1$.  Therefore we must assume that $a \geq 1$.

Now, the book says that unless stated otherwise, it's generally speaking about base 2 for logarithms, which are vital in computer science.

However, even then - if $a$ is two and $X$ is $\frac{1}{16}$, then $\log_{a} X$ is $-4$. (Similarly for base 10, try taking the log of $\frac{1}{10}$ on your calculator: It's $-1$.) Thus we must assume that $X \geq 1$.

...Unless I'm horribly missing something here. The problem seems quite different if we have to prove it for $X \geq 1$.

But even then, I need some help solving the problem. I've tried manipulating the equation as many ways as I could think of but I'm not cracking it.

答:
Suppose that $x>0$ and $\log x > x$.  Because the exponential map is monotonic increasing, it follows that $x > e^x$.  This contradicts the well-known series representation for $e^x$:

$$x\not>\sum_{k=0}^\infty \frac{x^k}{k!}=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots$$

We also have the following, stronger result: since $e^x>1+x$ for all $x>0$, we have $x > \log(1+x)$ for $x>0$.

Edit: It appears that the question concerns $\log_2$ and not $\log$, which requires a different proof.  Thus, for which $b >1$ do we have $\log_b x <x$ for all $x>0$?

The proof above shows that the set of admissible $b$ is non-empty.  As $b \to 1$ from above, the curves $\log_b x$ sweep down towards the line $f(x)=x$.  It's not hard to see that these functions sweep past $f(x)$, hence there exists a maximal $b$ such that $\log_{b} x \not < x$ for all $x>0$.  It follows that $\log_{b} x$ share a tangent line for some unique $b_0$.

Let $x_0$ denote such a point of tangency.  Comparing slopes, we must have $1/(x_0\log b_0)=1$, i.e $x_0=1/\log b_0$.  As the functions $\log_{b_0} x$ and $x$ must moreover agree at $x_0$, we compute $\log_{b_0}(1/\log b_0)=1/\log b_0$, hence $\log b_0=1/e$.  This gives a value of
$$b_0 =e^{1/e} \approx 1.44467,$$
such that $b>b_0$ implies that $\log_b x < x$ for all $x >0$.  In particular, $b_0<2$, so that $\log_2 x < x$ for all $x>0$.

There may be some details swept under the rug, but the idea (and resulting constant) is certainly correct.

Title: Prove that $\log X < X$ for all $X > 0$


Title: Rewrite Lebesgue integral over a centered interval

Body:
Given a function $f\in L^1(\mathbb{R})$, define the integral over an interval centered at $x$: $$f_c(x)=\frac{1}{2c}\int_{x-c}^{x+c} f(t)\,dt$$

There is a stated result that we can rewrite $f_c(x)$ into $$f_c(x)=\int_{\mathbb{R}}f(x-t)\frac{1}{2c}\mathbf{1}_{[-c,c]}\,dt$$where $\mathbb{1}_{[-c,c]}$ is the indicator function of $[-c,c]$.

But it seems to be counter-intuitive and I don't know how to proceed. Any idea how to show this is true?

Thanks in advance.

Body:
\begin{eqnarray}
f_c(x) &=& \int {1 \over 2c} 1_{[x-c,x+c]}(t) f(t) dt \\
&=& \int {1 \over 2c} 1_{[-c,+c]}(t-x) f(t) dt \\
&=& \int {1 \over 2c} 1_{[-c,+c]}(x-t) f(t) dt \\
&=& \int {1 \over 2c} 1_{[-c,+c]}(\tau) f(x-\tau) dt \\
\end{eqnarray}


问:
Game theory/probability puzzle
Part 1:

You play a game where there is a box with 100$ and there are two players, each of you should write a number 0-100 on paper, then you show your numbers, if the sum is higher than 100 then each of you get 0 dollars, else you get what you wrote. What is your strategy?

Part 2:

You play the same game but your opponent told you that he is putting 80 (he might change his mind) - what is your plan?

For part 1, I think the Nash equilibria are (my number, opponent's number) = (x, 100 - x) for x between 0 and 100? Is this right, and if so, how to choose between them to pick your move? I was thinking about saying 50 as my number for symmetry/aesthetic reasons but would be interested in how one is supposed to tackle this problem. I drew out a reward matrix and tried iterated removal of weakly dominated strategies to get (50, 50) as the only cell remaining - would be grateful for any thoughts on the validity of this.

For part 2, I thought, if you're allowed to say something to the opponent, you could say that you were going to put 50, to try to force your opponent down to 50? Again I have no idea how to approach this. Any help would be much appreciated.

答:
The question of choosing a NE among several options is an open question - there is no general mechanism to choose among them (think about Battle of the Sexes, for a very simple example). In many papers, the "punch line" is adding some restriction or addition to the model that ruins other equilibria or somehow helps players choose one.
Here, the only symmetric equilibrium is indeed $(50,50)$, but all the $(x,100-x)$ equilibria are also good.

The same goes for the second part. Without ability to credibly commit to $80$, his $80$ is more or less worthless.

Title: Game theory/probability puzzle


答:
Well, I would note that $x\to 0^+\implies e^{\frac1x}\to+\infty$. Then you have $$\lim_{x\to 0^+}\frac{\sin(e^{\frac 1x})}{e^{\frac 1x}}=\lim_{y\to +\infty}\frac{\sin y}{y}=0$$

Using the substitution $y=e^\frac 1x$. Note the second equality is seen because $|\sin y|\leq 1$, while $y$ is unbounded.

Title: Show that $f(x)= e^{-1/x} \sin e^{1/x}$ for $x>0$ and $f(0)=0$ is continuous at $x=0$

Body:
$$f(x) =  \begin{cases}
      e^{-1/x} \sin e^{1/x}  & x\in (0,\infty) \\
      0 & x=0
   \end{cases}
$$

How do I show this function is continuous at $0$? So far, I have determined that
$$\lim_{x \to 0} f(x) = 0$$ but  $$\lim_{x \to 0} e^{-\frac{1}{x}} \sin e^{\frac{1}{x}}$$ diverges. I don't know how to proceed.


问:
Square Root Trick
$\sqrt{\mathbf{X}}
\approx \frac{\mathbf{X}+\mathbf{Y}}{2 \sqrt{\mathbf{Y}}}$

And $|\sqrt{\mathbf{X}} -\frac{\mathbf{X}+\mathbf{Y}}{2 \sqrt{\mathbf{Y}}}| \le 10^{-1}$

$Y=$ The Nearest number have root

For example :

$\sqrt{23}=\frac{23+25}{2 \sqrt{25}}=4.8$ ,
$\sqrt{50}=\frac{50+49}{2 \sqrt{49}}=7.071$

Is this  true above approximate?

答:
Yes: this is the Taylor approximation of the square root, in the following sense. I will only cover the case $Y \le X$; the other is completely symmetric.

Let $X= n^2+k$ with $k\le n$. Then $Y=n^2$ so that

$$ \frac{X+Y}{2\sqrt{Y}} = \frac{2 n^2+k}{2n} = n+ \frac{k}{2n}$$

On the other side, note that for small $x$ you have $\sqrt{1+x} \approx 1+\frac{x}{2}$, by the Taylor expansion around zero.

Now we have that

$$\sqrt{n^2+k}= n\sqrt{1+(k/n^2) }\approx n(1+ (k/2n^2)) = n+ \frac{k}{2n}$$

As above. About the error, we can use the Lagrange reminder and get that there exist a $0\le \xi \le k/n^2$ so that

$$ | X - \frac{X+Y}{2\sqrt{Y}} | = n \frac{| f''(\xi) |(k/n^2)^2 }{2! }$$

Where $f(x) =\sqrt{1+x}$. You can see that the second derivative is $ -(1/4) (1+x) ^{-3/2}$ , which is maximum in absolute value where $x=0$ and it is $1/4$. In this way you get that the above remainder is at most
$$ \frac{k^2}{8n^3}\le \frac{1}{8n}$$

That is, it gets better and better as $X$ grows, surely $\le 1/10$ for $X\ge 4$.

Title: Square Root Trick


问:
Weyl's inequality for largest eigenvalues of two real symmetric matrices
I am trying to show that

$| \lambda_n(A) - \lambda_n(B)| \leq \lVert{\mathbf{A} - \mathbf{B}} \rVert_F $

where $A, B \in \mathbb{R}_{sym}^{nxn}$ and $\lambda_n(A)$ denotes the largest eigenvalue of $A$.

The largest eigenvalue for $A$ is defined as $\max_{v \in \mathbb{R}^n,   \lVert v \rVert_2 = 1}  v^TAv$, and for B, is similar.

I know this is related to Weyl's Inequality in some sense but cannot construct proof to show this.

答:
If we use the property you have it says

"The largest eigenavalue for A is defined as $ \max_{v \in \mathbb{R} ,  \|v\|=1} v^{T}Av$ "

Now if $A,B$ are symmetric then $A-B$ is also symmetric and using that we have

$$ \lambda_{\text{max}}(A-B) = \max_{v \in \mathbb{R} ,  \|v\|=1} v^{T}(A-B)v $$

Now you have to show the rest using properties of $\max$ note that

$$ \max\{ f(x) + g(x) \} \leq \max \{ f(x)\} + \max \{ g(x) \}$$

and if $\theta < 0$
$$ \max \theta f(x) = \theta \min f(x)  $$

I think this may help.


问:
How many ordered pairs $(x,y)$ satisfy the property $|x|+|y|\lt2019$?
How many ordered pairs $(x,y)$ satisfy the property $|x|+|y|\lt2019$? There must be a shorter way to get an answer here. $x$ and $y$ are natural numbers.

答:
For $k=1,\ldots,2018$ there are $k-1$ pairs $\langle x,y\rangle$ of positive integers such that $x+y=k$, one for each $x\in\{1,\ldots,k\}$. Thus, there are $$\sum_{k=1}^{2018}(k-1)=\sum_{k=0}^{2017}k=\frac{2017\cdot2018}2=2\,035\,153$$ solutions in positive integers.

For many of us, however, the natural numbers include $0$. That adds $2018$ solutions with $y=0$ and $x>0$, another $2018$ solutions with $x=0$ and $y>0$, and the solution $x=y=0$, for a total of $$2\,035\,153+4037=2\,039\,190$$ solutions.

Finally, if we allow $x$ and $y$ to range over all integers, as is suggested by the absolute values, there are $2\,035\,153$ solutions in the interior of each of the four quadrants, $2018$ on the positive $x$-axis, $2018$ on the positive $y$-axis, $2018$ on the negative $x$-axis, $2018$ on the negative $y$-axis, and one at the origin for a grand total of

$$4(2\,035\,153+2018)+1=8\,148\,685$$

solutions.


问:
Number of Tuesdays in five consecutive calendar years taken together is exactly 260. Which day of the week was 1st Jan. of the first of five years?
Number of Tuesdays in five consecutive calendar years taken together is exactly 260. Which day of the week was 1st January of the first of these five years?

I don't know at all how to do this question. I know that in leap year there are 2 odd days and in ordinary year there is 1 odd day.

答:
Since $5*52=260$, the question is equivalent to asking what day of the week the first year must begin on if none of five consecutive years has 53 Tuesdays.

A year will have 53 Tuesdays if the dominical letter is F, FE, or GF.

Examining the possibilities for the dominical letters (including the cases involving non-leap century years) gives the following:

AG, F, E, D, CB (second year has 53 Tuesdays)

GF, E, D, C, BA (first year has 53 Tuesdays)

FE, D, C, B, AG (first year has 53 Tuesdays)

ED, C, B, A, GF (fifth year has 53 Tuesdays)

DC, B, A, G, FE (fifth year has 53 Tuesdays)

CB, A, G, F, ED (fourth year has 53 Tuesdays)

BA, G, F, E, DC (third year has 53 Tuesdays)

A, GF, E, D, C (second year has 53 Tuesdays)

G, FE, D, C, B (second year has 53 Tuesdays)

F, ED, C, B, A (first year has 53 Tuesdays)

E, DC, B, A, G

D, CB, A, G, F (fifth year has 53 Tuesdays)

C, BA, G, F, E (fourth year has 53 Tuesdays)

B, AG, F, E, D (third year has 53 Tuesdays)

A, G, FE, D, C (third year has 53 Tuesdays)

G, F, ED, C, B (second year has 53 Tuesdays)

F, E, DC, B, A (first year has 53 Tuesdays)

E, D, CB, A, G

D, C, BA, G, F (fifth year has 53 Tuesdays)

C, B, AG, F, E (fourth year has 53 Tuesdays)

B, A, GF, E, D (third year has 53 Tuesdays)

A, G, F, ED, C (third year has 53 Tuesdays)

G, F, E, DC, B (second year has 53 Tuesdays)

F, E, D, CB, A (first year has 53 Tuesdays)

E, D, C, BA, G

D, C, B, AG, F (fifth year has 53 Tuesdays)

C, B, A, GF, E (fourth year has 53 Tuesdays)

B, A, G, FE, D (fourth year has 53 Tuesdays)

$2096, 2097, 2098, 2099, 2100$ ($2097$ has 53 Tuesdays)

$2097, 2098, 2099, 2100, 2101$ ($2097$ has 53 Tuesdays)

$2098, 2099, 2100, 2101, 2102$

$2099, 2100, 2101, 2102, 2103$

$2100, 2101, 2102, 2103, 2104$ ($2104$ has 53 Tuesdays)

$2196, 2197, 2198, 2199, 2200$ ($2199$ has 53 Tuesdays)

$2197, 2198, 2199, 2200, 2201$ ($2199$ has 53 Tuesdays)

$2198, 2199, 2200, 2201, 2202$ ($2199$ has 53 Tuesdays)

$2199, 2200, 2201, 2202, 2203$ ($2199$ has 53 Tuesdays)

$2200, 2201, 2202, 2203, 2204$

$2296, 2297, 2298, 2299, 2300$

$2297, 2298, 2299, 2300, 2301$ ($2301$ has 53 Tuesdays)

$2298, 2299, 2300, 2301, 2302$ ($2301$ has 53 Tuesdays)

$2299, 2300, 2301, 2302, 2303$ ($2301$ has 53 Tuesdays)

$2300, 2301, 2302, 2303, 2304$ ($2301$ has 53 Tuesdays)

For the cases that do not involve non-leap century years, one can see that the first year must be a non-leap year beginning on a Wednesday.

For the cases that involve non-leap century years, the above $15$ cases are the only ones that one need to check for because the Gregorian calendar repeats every $400$ years. Of those $15$ cases, $4$ of them do not contain a year with $53$ Tuesdays. The years $2098, 2200,$ and $2296$ begin on Wednesdays, while the year $2099$ begins on a Thursday.

So, the answer is that the first year must usually begin on a Wednesday. Once every $400$ years, there is a solution beginning with a year that begins on a Thursday (namely, the years $2099$ to $2103$).


问:
Inclusions between $H^1_0(\Omega) \cap H^2(\Omega)$ and $H^2_0(\Omega)$
Let $\Omega$ be a bounded domain in $\mathbb{R}^n$. I would like to understand something about the relationships between these two spaces: $$H^1_0(\Omega) \cap H^2(\Omega)\quad \text{and}\quad H^2_0(\Omega).$$

Is there any inclusion? If there is not, could you exhibit a counterexample?
Thank you!

答:
Just to continue Chris Wong's argument.

Let $\Omega = \{x\in \mathbb{R}^n: \|x\|_{\infty}= \max|x_i|<1 \}$ be an $n$-cube with side length $2$, then it can be verified that:
$$
u = \prod_{i=1}^n \sin(2\pi x_i)
$$
is in $H^1_0(\Omega)\cap H^2(\Omega)$, since on boundary of the $n$-cube, there is some $x_i = \pm 1$. However for $\partial u/\partial x_i$ will give you a cosine which does not vanish on the boundary when dot product with normal to the boundary, hence $\partial u/\partial \nu$ is not zero (though the tangentially it does vanish).

To visualize just consider $n=2$, the square $(-1,1)^2$, and $u = \sin(2\pi x)\sin(2\pi y)$.


问:
Probability Q on rolling two 8-sided dice
Let's say I roll two 8-sided dice. I win if the sums '7' and '11' show up before we see the sum '9' TWICE. What is my probability of winning?

So this is my answer and please correct me if I am wrong:

Prob of getting either 7 or 9 —> 6/64 + 6/64 = 12/64 —> 18.75%

Prob of getting 9 twice —> 8/64 x 8/64 = 1/64 —> 1.6%

But now to find the probability of winning, would it be 18.75 - 1.6 = 17.15% ? Am I right or wrong?

Thanks!

EDIT

From the answers provided below, we have three different results:

84%
49%
60%

Which one would be the correct answer?

答:
This answer assumes the person wins if there is a sum of $\, 7\,$ or $\,11\,$ in any toss before a sum of $\,9\,$ appears twice.

Probability of winning in first toss $ = \frac{12}{64}$

a) If the person gets a sum of $9$ once, the chance of winning from there

$\displaystyle P(W1) = \frac{12}{64} + \frac{44}{64} \times (\frac{12}{64} + \frac{44}{64} \times ...)... = \frac{12}{64} (1 + \frac{44}{64} + (\frac{44}{64})^2 + ...)$

$\displaystyle = \frac{3}{5}$

[Using sum of infinite geometric series $a + ar + ar^2 + ... = \frac{a}{1-r}$]

b) If the first toss is any sum other than $7, 9, 11, \,$starting 2nd toss

$P(W2) = \frac{12}{64} + \frac{8}{64} \times P(W1) + \frac{44}{64} \times (\frac{12}{64} + \frac{8}{64} \times P(W1) + ...)... $

$$= (\frac{12}{64} + \frac{8}{64} \times P(W1))(1 + \frac{44}{64} + (\frac{44}{64})^2 + ...) = \frac{21}{25}$$

$\,$

So person's chance of winning $ \displaystyle = \frac{12}{64} + \frac{8}{64} \times P(W1) + \frac{44}{64} \times P(W2) = \frac{21}{25}$


问:
Evaluation of $\int_{0}^{\infty}\frac{1}{\sqrt{x^4+x^3+x^2+x+1}}dx$
Evaluation of Integral $\displaystyle \int_{0}^{\infty}\frac{1}{\sqrt{x^4+x^3+x^2+x+1}}dx$

$\bf{My\; Try::}$ First we will convert $x^4+x^3+x^2+x+1$ into closed form, which is $\displaystyle \left(\frac{x^5-1}{x-1}\right)$

So Integral is $\displaystyle \int_{0}^{\infty}\frac{\sqrt{x-1}}{\sqrt{x^5-1}}dx$

now i did not understand how can i solve it

Help me

Thanks

Body:
$$
I=\int^{\infty}_{0}\frac{dx}{\sqrt{x^4+x^3+x^2+x+1}}
$$

Making the change of variable $x\rightarrow y/(1+y)$ we get

$$
I=\int^{1}_{0}\frac{dy}{\sqrt{y^4-2y^3+4y^2-3y+1}}=
$$

$$
=\int^{1}_{0}\frac{dy}{\sqrt{1/16(2y-1)^4+5/8(2y-1)^2+5/16}}.
$$

Also setting $(2y-1)\rightarrow u$, then

$$
I=\frac{1}{2}\int_{-1}^{1}\frac{du}{\sqrt{1/16 (u^2+5)^2-5/4}}=\int_{0}^{1}\frac{du}{\sqrt{1/16 (u^2+5)^2-5/4}}
$$

Then setting $(u^2+5)/4\rightarrow w$, we get

$$
I=\int_{5/4}^{6/4}\frac{dw}{\sqrt{w^2-5/4}}\frac{1}{\sqrt{w-5/4}}=\int_{0}^{1/4}\frac{w^{-1/2}dw}{\sqrt{(w+5/4-\sqrt{5/4})(w+5/4+\sqrt{5/4})}}=
$$

$$
=-2i\frac{5-\sqrt{20}}{\sqrt{5}\sqrt{5-\sqrt{20}}}\int^{-1/(5-\sqrt{20})}_{0}\frac{t^{-1/2}}{\sqrt{(1-t)(1-mt)}}dt,
$$

where $m=\frac{5-\sqrt{20}}{5+\sqrt{20}}$. From the deffinition of the elliptic integrals (in Mathematica's notation):

$$
F(\sin\phi,m)=\int_{0}^{\phi}\frac{1}{\sqrt{(1-t^2)(1-mt^2)}}dt=\frac{1}{2}\int_{0}^{ \phi^2}\frac{1}{\sqrt{t(1-t)(1-mt)}}dt\Leftrightarrow
$$

$$
2F\left(\sin\sqrt{\phi},m\right)=\int^{ \phi}_{0}\frac{1}{\sqrt{t(1-t)(1-mt)}}dt.
$$

Hence from (1) and (2) we have

$$
I=-4i\frac{5-\sqrt{20}}{\sqrt{5}\sqrt{5-\sqrt{20}}}\cdot F\left(\arcsin\left(\frac{i}{\sqrt{5-\sqrt{20}}}\right),\frac{5-\sqrt{20}}{5+\sqrt{20}}\right).
$$


答:
I'm assuming that you don't count $0$ as a natural number, though Brian's answer also addresses the case including $0$. This answer doesn't involve a large summation.

We know that $x+y<2019$ and that $x<x+y$ (since we assume that $y$ is not $0$). Hence, $\{x, x+y\}$ is a two element subset of $[2018]=\{1,2,3,\ldots, 2018\}$. Furthermore, every two element subset of $[2018]$ corresponds to an $(x,y)$ pair. Just pick the smaller element in your subset to be $x$ and the larger element to be $x+y$. Then the number of $(x,y)$ pairs is the number of two element subsets of $[2018]$, which is ${2018\choose 2}=\frac{2018\cdot 2017}{2}=2035153$.

Body: How many ordered pairs $(x,y)$ satisfy the property $|x|+|y|\lt2019$? There must be a shorter way to get an answer here. $x$ and $y$ are natural numbers.


问:
Beads on a necklace but they’re all the same colour
Okay, so I know that if 15 beads, 5 red, 5 yellow, 5 blue, the number of possible combinations is:
14! / 2*5!*5!*5!

But say all 15 were of the same colour. (the answer is obviously 1).
But wouldn’t that be 14! / 2 * 15! ? Because of the (n-1)! For a circle, divide by 2 for flipping, then the 15! for beads of the same colour. I know it’s super trivial, but I don’t understand what’s wrong with it. Thanks

答:
The actual calculation would be 15!/15! = 1 (which you know to be the correct final result).

Where your error comes from is by just applying formulaic rules without thinking through what those are saying, specifically about distinguishable and indistinguishable objects.  So lets expand on this.

To start:
15 beads can be arranged in 15! permutations.

Because this is a circle, when you have multiple colors you can create 15 distinguishable permutations that differ only by rotation.   This gives you 15!/15 permutations which is the same as 14! or (n-1)!  But with only one color, you don't have distinguishable rotations.
You also can flip the circle, with different colors then having distinguishable permutations differing only by flipping, thus divide by 2.  With only one color, the flip is indistinguishable, so no division.

The key is "distinguishable".

Finally is within each distinct color set, the ways those indistinguishable beads can be arranged is (color quantity)!

Thus with only one color, you have 15! permutations (no distinguishable rotations or flipping), divided by 15! permutations of that color, e.g. 15!/15!


问:
Expressing $\sin(2x)$ as a polynomial of $\sin{x}$
Using trignometric identities ( double angle forumlas) one can see that $\cos{2x} = 2 \: \cos^{2}{x} - 1$ can be expressed as a polynomial of $\cos{x}$, where $p(\cos{x})=2 \: \cos^{2}{x}-1$. Then its natural to ask the same question for the sine function.

Can we express $\sin{2x}$ as a polynomial of $\sin{x}$?

答: No, you can't get any simpler than $2\sin\;x\;\cos\;x$. In general, only sines of odd angle multiples ($\sin((2k+1)x)$) are expressible as polynomials in $\sin(x)$, because any powers of cosines that may appear are raised to even exponents, and thus the identity $\sin^2(x)=1-\cos^2(x)$ applies, to turn it into an expression entirely made up of powers of sines.</p>

ContentLicense: CC BY-SA 2.5


问:
Riemann--Stieltjes integral wrt. uniformly vanishing integrator?
I'm interested in uniform limits of Riemann--Stieltjes integrals of the form:

$\lim_{N \rightarrow \infty} \sup_{t \in [0,T]} \int_0^t f(s)\,\text{d}G^N(s)$

where $f(t)$ is non-increasing (but not necessarily continuous) and $|f(t)| \leq C$ for all $t \in [0,T]$. Also $\sup_{t \in [0,T]} |G^N(t)| \rightarrow 0$ as $N \rightarrow \infty$. The $G^N(t)$ may not be continuous or monotone but it is known that the integrals exist and are finite for all $N$.

My intuition is that the above limit of the integrals should be zero but I seem unable to prove this... perhaps I am wrong. Does anyone have any proof ideas or indeed a counterexample?

Thanks!

答:
The limit is indeed zero. This follows after  integration by parts
$$\int_0^t f\,dG^N = f(t)G^N(t)-f(0)G^N(0)-\int_0^t G^N\,df \tag1$$
because the uniform smallness of $G^N$ makes all three terms in (1)
 converge to $0$.

(You may have to work to justify the integration by parts, depending on how much of the Riemann-Stieltjes theory you have at your disposal.)


问:
Calculate the following limit:
Calculate $\;\lim\limits_{x\to\infty} \left[x^2\left(1+\dfrac1x\right)^x-ex^3\ln\left(1+\dfrac1x\right)\right]$.

It is a $\frac00$ case of indetermination if we rewrite as $\lim_{x\to\infty} \frac{((1+\frac1x)^x-e\ln(1+\frac1x)^x)}{\frac{1}{x^2}}$, since $\lim_{x\to\infty} \frac{1}{x^2} = 0$, $\lim_{x\to\infty} (1+\frac1x)^x = e$ and $\lim_{x\to\infty} \ln(1+\frac1x)^x = 1$.

I think that it is the type that has a solution without l'Hospital's rule, but it's quite difficult to find, so l'Hospital still remains the best try to me. I tried using it with different rewrites, but it seems that it needs to be used multiple times, and the expression gets harder and harder to calculate, so I assume that some other limit must be applied first to make the expression nicer.

Also, I futilely tried to use the following known limits by changing $x$ into $y = \frac1x$ if needed (and adding and substracting $ex^2$ in the main parenthesis and trying to use the last 2 limits), but maybe it can help you: $\lim_{x\to0} \frac{a^x-1}{x} = \ln a$, $\lim_{x\to0} \frac{\ln(1+x)}{x} = 1$, $\lim_{x\to0} \frac{(1+x)^r-1}{x} = r$, $\lim_{x\to0} \frac{(1+x)^\frac1x-e}{x} = -\frac{e}{2}$, $\lim_{x\to\infty} (x-x^2ln(1+\frac1x)) = \frac12$.

Can you help me with this problem?

答:
I am going to calculate your limit without using Taylor series but only the two following notable limits:

$\lim\limits_{x\to0}\dfrac{\ln(1+x)}{x}=1\;,\quad\lim\limits_{x\to0}\dfrac{x-\ln(1+x)}{x^2}=\dfrac12\;.$

$\lim\limits_{x\to\infty}\left[x^2\left(1+\dfrac1x\right)^x-ex^3\ln\left(1+\dfrac1x\right)\right]=$

$=\lim\limits_{x\to\infty}\left[x^2\left(1+\dfrac1x\right)^x-ex^2\ln\left(1+\dfrac1x\right)^x\right]=$

$=\lim\limits_{x\to\infty}\dfrac{\left(1+\frac1x\right)^x-e\ln\left(1+\frac1x\right)^x}{\left[1-\ln\left(1+\frac1x\right)^x\right]^2}\cdot\lim\limits_{x\to\infty}\left[x-x\ln\left(1+\frac1x\right)^x\right]^2$

Now I am going to calculate the first limit by using the following substitution:

$t=\dfrac1e\left(1+\dfrac1x\right)^x-1\;.$

$\lim\limits_{x\to\infty}\dfrac{\left(1+\dfrac1x\right)^x-e\ln\left(1+\dfrac1x\right)^x}{\left[1-\ln\left(1+\dfrac1x\right)^x\right]^2}=$

$=\lim\limits_{t\to0}\dfrac{e(1+t)-e\ln\big[e(1+t)\big] }{\left\{1-\ln\big[e(1+t)\big]\right\}^2}=$

$=e\cdot\lim\limits_{t\to0}\dfrac{1+t-1-\ln(1+t)}{\big[1-1-\ln(1+t)\big]^2}=$

$=e\cdot\lim\limits_{t\to0}\dfrac{t-\ln(1+t)}{\ln^2(1+t)}=$

$=e\cdot\lim\limits_{t\to0}\dfrac{t-\ln(1+t)}{t^2}\cdot\lim\limits_{t\to0}\left[\dfrac{t}{\ln(1+t)}\right]^2=$

$=e\cdot\dfrac12\cdot1^2=\dfrac e2\;.$

Now I am going to calculate the second limit by using the following substitution:

$u=\dfrac1x\;.$

$\lim\limits_{x\to\infty}\left[x-x\ln\left(1+\dfrac1x\right)^x\right]^2=$

$=\lim\limits_{x\to\infty}\left[x-x^2\ln\left(1+\dfrac1x\right)\right]^2=$

$=\lim\limits_{u\to0}\left[\dfrac1u-\dfrac1{u^2}\ln\big(1+u\big)\right]^2=$

$=\lim\limits_{u\to0}\left[\dfrac{u-\ln(1+u)}{u^2}\right]^2=$

$=\left(\dfrac12\right)^2=\dfrac14\;.$

Consequently,

$\lim\limits_{x\to\infty}\left[x^2\left(1+\dfrac1x\right)^x-ex^3\ln\left(1+\dfrac1x\right)\right]=$

$=\dfrac e2\cdot\dfrac14=\dfrac e8\;.$

Title: Calculate the following limit:


问:
Search a word in a matrix runtime comlexity
Trying to analyze the runtime complexity of the following algorithm:

Problem: We have an $m \cdot n$ array $A$ consisting of lower case letters and a target string $s$. The goal is to examine whether the target string appearing in $A$ or not.

algorithm:

```<code>for(int i = 0; i < m; i++){
    for(int j = 0; j < n; j++){
        if(A[i][j] is equal to the starting character in s) search(i, j, s)
    }
}

boolean search(int i, int j, target s){
    if(the current position relative to s is the length of s) then we find the target
    looping through the four possible directions starting from i, j: {p,q} = {i+1, j} or {i-1, j} or {i, j+1} or {i, j-1}, if the coordinate is never visited before
    search(p, q, target s)
}

</code>```

One runtime complexity analysis that I read is the following:

At each position in the array $A$, we are first presented with $4$ possible directions to explore. After the first round, we are only given 3 possible choices because we can never go back. So the worst runtime complexity is $O(m \cdot n \cdot 3^{len(s)})$

However, I disagree with this analysis, because even though we are only presented with 3 possible choices each round, we do need to spend one operation to check whether that direction has been visited before or not. For instance, in java you probably just use a boolean array to track whether one spot has been visited before, so in order to know whether a spot has been visited or not, one needs a conditional check, and that costs one operation. The analysis I mentioned does not seem to take into account this.

What should be the runtime complexity?

update:

Let us suppose that the length of the target string is l and the runtime complexity at a given position in the matrix is T(l). Then we have:

$T(l) = 4 T(l- 1) + 4 = 4(3T(l - 2) + 4) + 4 = 4(3( 3T(l -3) + 4) + 4)) + 4 = 4 \cdot 3^{(l - 1)} + 4 + 4 \cdot 4 + 4 \cdot 3 \cdot 4 + ...$

the $+4$ is coming from the fact that we are looping through four directions in each round besides recursively calling itself three times.

答:
Let $T(l)$ denote the time it takes to search for a string of length $l$, and let $T'(l)$ be the time it takes to search for a string of length $l$ when one direction has been eliminated. Then
$$
T(l)=4T'(l-1)+c,\\
T'(l)=3T'(l-1)+c
$$
where $c$ is $O(1)$, and includes the time it takes to check if the square is visited, mark the square as visited, and compare the current letter against $s$.

We solve the second recurrence for $T'$, and plug that solution in to the first equation to get $T$.

$$
T'(l)=c(1+3+3^2+\dots+3^{(l-1)})+3^{l}\cdot T(0)=c\cdot \frac{3^l-1}{2}+3^l\cdot T(0)
$$
so you still get $T'(l)=O(3^l)$, and the same for $T(l)$. Note that the $+c$ only adds another factor of $3^l$, so it does not change the asymptotic runtime.


问:
Concerns formalizing a "Minimal Element" proof by contradiction
I came across an exercise where my proposed solution strikes me as a little informal/hand-wavy at the end. I was hoping that someone could lend a hand!

Consider the set $A$ and two subsets of $A$,  $a\downarrow$ and $b\downarrow$, defined as follows:

$a\downarrow := \{x \in A | xRa\}$

$b\downarrow := \{x \in A | x R b\}$

Of importance, $R$ is well-ordered on $A$ (and any subset of $A$). Specifically, that means trichotomy, transitivity, irreflexivity, and well-foundedness are properties of $R$.

The aim of this proof is to demonstrate that if $a \neq b$ then no isomorphism $k$ exists such that $(b \downarrow;R) \cong (a \downarrow; R)$.

I proceed by contradiction and assume that such an isomorphism exists: call it $k: b \downarrow \to a \downarrow$.

Without Loss of Generality, let $a \downarrow \subset b \downarrow$. This is a consequence of trichotomy: either $aRb, bRa,$ or $a=b$. Choose $aRb$...thus $a \in b\downarrow$ but $a \notin a\downarrow$.

To demonstrate that $a\downarrow$ is a proper subset, pick an element $x \in a\downarrow$. By definition, $xRa$. But we know that $aRb$. Therefore, by transitivity of $R$, $xRb \implies x \in b \downarrow$.

Additionally, the definition of these sets prohibits $b \downarrow$ from containing elements that are "beneath" the minimal element of $a \downarrow$ (quick proof by contradiction).

Because $k$ is an isomorphism, it is surjective. Therefore, looking at the minimal element in $a \downarrow$, call it $m$, there are two possible categories of mappings:

Case 1) An element $x \in b\downarrow$ such that $mRx$, and $x$ gets mapped to $m \in a \downarrow$ through $k$.

$$k(x)=m$$

Case 2) $m \in b \downarrow$ gets mapped to $m \in a \downarrow$ through $k$.

$$k(m)=m$$

Case 1 results in a contradiction: order will not be preserved

If $k(x)=m$, where $mRx$, then there is a $y \in b \downarrow$ such that $yRx$. This $y$ must be mapped to some element in $a \downarrow$, call it $s$. So we have:

$$k(x)=m \land k(y)=s$$

$yRx$ but $[k(x)=m] \ R \  [s=k(y)]$. The order was flipped! Contradiction.

$\color{red}{\text{Case 2 is where I feel like my argument is hand-wavy}}$

We have $k(m)=m$. That is to say, the $m \in b\downarrow$ is being mapped to the $m \in a \downarrow$. Now consider the following algorithm:

Choose the element $n$ in $a \downarrow$ such that the following is true: $\forall z( z\in a\downarrow \setminus \{m\} \rightarrow nRz \lor n=z)$. You can view this $n$ as the "the next minimal element".

Notice that there are two subcases of possible mappings from elements of $b \downarrow$ to $n \in a\downarrow$ :

Subcase 1) An element $x \in b\downarrow$ such that $nRx$ gets mapped to $n \in a \downarrow$ through $k$.

$$k(x)=n$$

Subcase 2) $n \in b \downarrow$ gets mapped to $n \in a \downarrow$ through $k$.

$$k(n)=n$$

But this is just a "slightly different" version of the argument we encountered earlier with $m$.

This continues to repeat until you've run out of elements in $a \downarrow$ but are still left with $a \in b \downarrow$, which will have no where to map to without violating injectivity and ordering. Therefore, no such isomorphism $k$ exists.

That final piece of the argument, where I say," This continues to repeat...," strikes me as unsatisfactorily informal. Any suggestions would be greatly appreciated! Cheers~

答:
You’re right that it’s hand-wavy at that point: what you have is an and so on argument. You can convert it to a genuine minimum counterexample argument like this.

Since $a\!\downarrow\subsetneqq b\!\downarrow$, it’s clear that $\{x\in b\!\downarrow:k(x)\ne x\}\ne\varnothing$. Let

$$m=\min\{x\in b\!\downarrow:k(x)\ne x\}\,.$$

Then $k(x)=x$ for each $x\in m\!\downarrow$, so $k(m)\notin m\!\downarrow$. And by construction $k(m)\ne m$, so $m\mathrel{R}k(m)$. Now show that $m\notin\operatorname{ran}k$.


问:
Choosing a number from infinite numbers
Imagine infinite set of, let's say, natural numbers. I choose one of the infinite numbers randomly. Let's call that number n. If I choose another number too, can it be the same number (n), theoretically?

答:
I'd loved to put this in a comment, but I can't because I am new here.

I just wanted to say that since you picked that particular number the first time, it has a non-zero probability of being picked. Thus it definitely can be picked again.

Title: Choosing a number from infinite numbers


问:
If $u,v$ are orthogonal vectors in $\mathbb{R}^3$ and $||u|| = 3||v||$, then the set $\{u+3v,u-3v\}$ is also orthogonal
I can come up with examples where this is true, but I am having trouble generating a formal proof for this. Tried to prove this false with a counter-example but soon realized that wasn't gonna work.</p>

答:
By properties of the dot product, $(u+3v)\cdot(u-3v)=u\cdot u-9v\cdot v - 3u\cdot v+3v\cdot u$.

Now $u\cdot u - 9v\cdot v=|u|^2-|3v|^2=0$ because $|u|=|3v|$,

and $u\cdot v=v\cdot u=0$ because $u$ and $v$ are orthogonal.

Therefore $(u+3v)\cdot(u-3v)=0$; i.e., $u+3v$ and $u-3v$ are orthogonal.


问:
$\sum (B_{t_{i}^{n}}+B_{t_{i-1}^{n}})^{2} \left( \left(B_{t_{i}^{n}}-B_{t_{i-1}^{n}}\right)^2-[t^n_{i}-t^n_{i-1}]\right)$ converges in probability
Let $T > 0$, $B$ be a Brownian motion and consider a sequence of partitions $\{0=t_{0}^{n}<...<t_{l(n)}^{n}=T\}, \; n \in \mathbb N$ such that the mesh tends to $0$ as $n \to \infty$.

How can I show that $\sum\limits_{i = 1}^{l(n)}(B_{t_{i}^{n}}+B_{t_{i-1}^{n}})^{2} \left(\left(B_{t_{i}^{n}}-B_{t_{i-1}^{n}}\right)^2-(t^n_{i}-t^n_{i-1})\right)$ converges in probability as $n \to \infty$?

Of course I know that $E[\left(B_{t_{i}^{n}}-B_{t_{i-1}^{n}}\right)^2-(t^n_{i}-t^n_{i-1})]=0$ and $B_{t}^2-t$ is a martingale but there is no independence or other notions that will help me prove $L^{2}$ convergence and by extension convergence in probability.

Any tips/hints appreciated.

答:
Write $\Delta B_i = B_{t^n_i} - B_{t^n_{i-1}} $ and $\Delta t_i = t^n_i - t^n_{i-1}$. Then we may decompose the sum into three parts:

\begin{align*}
\sum_{i=1}^{l(n)} (2B_{t^n_{i-1}} + \Delta B_i)^2 \bigl( (\Delta B_i)^2 - \Delta t_i \bigr)
= S_{1,n} + S_{2,n} + S_{3,n},
\end{align*}

where

\begin{align*}
S_{1,n} &= 4 \sum_{i=1}^{l(n)} B_{t^n_{i-1}}^2 \bigl( (\Delta B_i)^2 - \Delta t_i \bigr) \\
S_{2,n} &= \sum_{i=1}^{l(n)} 4B_{t^n_{i-1}}(\Delta B_i) \bigl( (\Delta B_i)^2 - \Delta t_i \bigr) \\
S_{3,n} &= \sum_{i=1}^{l(n)} (\Delta B_i)^2 \bigl( (\Delta B_i)^2 - \Delta t_i \bigr).
\end{align*}

1. For the first sum, note that for $i < j$,

\begin{align*}
&\mathbb{E}\bigl[ B_{t^n_{i-1}}^2 \bigl( (\Delta B_i)^2 - \Delta t_i \bigr) B_{t^n_{j-1}}^2 \bigl( (\Delta B_j)^2 - \Delta t_j \bigr) \bigr] \\
&= \mathbb{E}\bigl[ B_{t^n_{i-1}}^2 \bigl( (\Delta B_i)^2 - \Delta t_i \bigr) B_{t^n_{j-1}}^2 \mathbb{E}\bigl[ (\Delta B_j)^2 - \Delta t_j \bigm| \mathcal{F}_{t^n_{j-1}} \bigr] \bigr] \\
&= 0.
\end{align*}

Also, by the independence,

\begin{align*}
&\mathbb{E}\bigl[ \bigl( B_{t^n_{i-1}}^2 \bigl( (\Delta B_i)^2 - \Delta t_i \bigr) \bigr)^2 \bigr]
= \mathbb{E}[ B_{t^n_{i-1}}^4 ] \mathbb{E}\bigl[ \bigl( (\Delta B_i)^2 - \Delta t_i \bigr)^2 \bigr]
= C (t^n_{i-1})^2 \Delta t_i^2
\end{align*}

for some absolute constant $C>0$. (We may choose $C = \mathbb{E}[B_1^4]\mathbb{E}[(B_1^2-1)^2]$, although its value is not important here.) Combining altogether, it follows that

$$ \mathbb{E}\bigl[ S_{1,n}^2 \bigr]
= C \sum_{i=1}^{l(n)} (t^n_{i-1})^2 \Delta t_i^2
\leq CT^3 \Bigl( \max_{1\leq i \leq l(n)} \Delta t_i \Bigr). $$

This tells that $S_{n,1} \to 0$ in $L^2$.

2. For the second sum, note that

$$ \mathbb{E} \bigl[ \bigl\lvert B_{t^n_{i-1}}(\Delta B_i) \bigl( (\Delta B_i)^2 - \Delta t_i \bigr) \bigr\rvert \bigr] \leq C_2 (t^n_{i-1})^{1/2} (\Delta t_i)^{3/2} $$

for another absolute constant $C_2 = \mathbb{E}[\lvert B_1\rvert]\mathbb{E}[\lvert B_1(B_1^2 - 1)\rvert] > 0$, and so,

$$ \mathbb{E} \bigl[ \bigl\lvert S_{2,n} \bigr\rvert \bigr] \leq C_2 T^{3/2} \Bigl( \max_{1\leq i \leq l(n)} \Delta t_i \Bigr). $$

This shows that $S_{2,n} \to 0$ in $L^1$.

3. A similar reasoning shows that $S_{3,n} \to 0$ in $L^1$ as well.

Therefore the original sum converges to $0$ in probability.


问:
Find the value of $\tan^2\alpha+\cot^2\beta$
A circle with centre o have two chords AC and BD,which are intersecting each other at P.If $\angle AOB=15^\circ$ and $\angle APB=30^\circ$,then find out value of
$$\tan^2\angle APB+\cot^2\angle COD$$

I want to do it as quick as possible this is an objective question so there is no need to  describe steps.

答:
Hint: $\angle COD = 45^\circ$ (or $315^\circ$, depending on how you measure it). As John mentioned, we know $\angle APB = 30^\circ$.

This is the quickest way, and I've not described any steps like you requested.


答: I'll spell out what @Mindlack meant. In general $\int_0^{\pi/2}\cos^{2a-1}\theta d\theta=\tfrac12\operatorname{B}(a,\,\tfrac12)=\frac{\sqrt{\pi}\Gamma(a)}{2\Gamma(a+\tfrac12)}$ in terms of Beta and Gamma functions. The case $a=\tfrac34$ obtains your integral as $\frac{\sqrt{\pi}\Gamma(\tfrac34)}{2\Gamma(\tfrac54)}$. The reflection formula $\Gamma(s)\Gamma(1-s)=\pi\csc\pi s$ gives $\Gamma(\tfrac14)\Gamma(\tfrac34)=\pi\sqrt{2}$ at $s=\tfrac14$, so the integral is$$\frac{\sqrt{2\pi^3}}{2\Gamma(\tfrac14)\Gamma(\tfrac54)}=\frac{\sqrt{2\pi^3}}{8\Gamma(\tfrac54)^2}=\frac{(2\pi)^{3/2}}{16\Gamma(\tfrac54)^2},$$since $\Gamma(s)=\frac1s\Gamma(s+1)$, again at $s=\tfrac14$.

Title: Prove that $\int_0^{\pi/2}\sqrt{\cos(\theta)}d\theta=\frac{(2.\pi)^{3/2}}{16.[\Gamma(5/4)]^2}$

Body:
To prove that $\int_0^{\pi/2}\sqrt{\cos(\theta)}d\theta=\frac{(2.\pi)^{3/2}}{16.[\Gamma(5/4)]^2}$, I've already done:

$\int_0^{\pi/2}\sqrt{\cos(\theta)}d\theta=\int_0^{\pi/2}{\cos(\theta)}^{2.(3/4)-1}.{\sin (\theta)}^{2.(1/2)-1}d\theta=\frac{1}{2}.B(3/4,1/2)=\frac{1}{2}.\Gamma(3/4).\Gamma(1/2).[\Gamma(5/4)]^{-1}$

but I don't know how to continue from this to reach $\frac{(2.\pi)^{3/2}}{16.[\Gamma(5/4)]^2}$.

By the first comment, I get it now. Here is the solution:

$\int_0^{\pi/2}\sqrt{\cos(\theta)}d\theta=\int_0^{\pi/2}{\cos(\theta)}^{2.(3/4)-1}.{\sin (\theta)}^{2.(1/2)-1}d\theta=\frac{1}{2}.B(3/4,1/2)=\frac{1}{2}.\Gamma(3/4).\Gamma(1/2).[\Gamma(5/4)]^{-1} = \frac{1}{2}.[\Gamma(3/4).\Gamma(1/4)].[\Gamma(1/4)]^{-1}.\sqrt{\pi}.[\Gamma(5/4)]^{-1}=\frac{1}{2}.\frac{\pi}{\sin(\pi/4)}.[4.\Gamma(5/4)]^{-1}.\sqrt{\pi}.[\Gamma(5/4)]^{-1}= \frac{1}{8}.\sqrt{2}.\pi^{3/2}.[\Gamma(5/4)]^{-2}=\frac{1}{16}.2^{3/2}.\pi^{3/2}.[\Gamma(5/4)]^{-2}$.


问:
Can there be a p-Sylow subgroup of a p-group
Have two particular questions about the definition of p-Sylow subgroups:

In all the definitions I've seen, p-Sylow subgroups of a group $G$ are defined as maximal subgroups of $G$ such that their order is a power of p. After a definition makes that point, something like "i.e. a p-subgroup that is not contained in any other p-subgroup". So does "maximal" mean that a given p-Sylow subgroup is not contained in any proper subgroup of $G$, or that that a given p-Sylow subgroup may or may not be contained in a subgroup of $G$ if that subgroup is not a p-group (so maximal only with respect to the $p-$ part.

The main question: Suppose $G$ is a p-group. Then if $H$ is a subgroup of $G$, by one of the definitions above (the latter one), it cannot be a p-Sylow subgroup of $G$ since then it would be contained in a (in this case, not proper, but nowhere is the exclusion made, as far as I know) p-subgroup of $G$ ($G$ itself).

This info has direct impact on a problem I'm working on. So at some point in that problem, we essentially consider a group $G$ and two subgroups $H$ and $K$ with $K$ defined as a p-Sylow subgroup. We eventually prove that $K$ is a subgroup of $H$ and that $H$ is a p-group. If the first definition in my first point is true and $K$ is maximal in the sense that it isn't contained in any proper subgroup of $G$, we need to consider both the case $H$ = $K$ and $H$ = $G$. But if we define it by the later definition and that definition doesn't restrict the 'maximality' of $K$ to proper subgroups of $G$, we need only consider the case $H = K$, because if $G = H$ but $H \neq K$ then that case immediately fails because $H$ not being a proper subgroup of $G$ would not free it from the requirement not to be a p-group.

答:
Maximal means that a $p$-Sylow is not (properly) contained in any other $p$-subgroup of $G$; in other words, it means maximal in the poset of $p$-subgroups. Other subgroups are irrelevant.

Every $p$-group is its own $p$-Sylow. The keyword "other" is important here; it actually implies proper containment if you squint at it, although this is a bit confusing and it would be better to say "properly contained" to be clearer.


问:
why is $(dX)_{q}: \mathbb{R}^2 \rightarrow \mathbb{R}^3 $ injective equivalent to saying that $X_{u}(q) $ and $X_{v}(q) $ are linearly independent?
I'm studying surfaces and in the definition I found a note that refers to an equivalence for one of the conditions in the definition, but I don't understand why those two things are equivalent. That is, why is $(dX)_{q}: \mathbb{R}^2 \rightarrow \mathbb{R}^3 $ injective equivalent to saying that $X_{u}(q) = \frac{\partial X}{\partial u}(q)$ and $X_{v}(q) = \frac{\partial X}{\partial v}(q)$ are linearly independent ? I'm referring to definition 2.2 and remark 2.5. I have added images:

答:
Just try to write down the matrix associated to the linear mapping $(dX)_q$.
Then remember that an injective linear mapping has trival kernel.


问:
Is there 2 ways to write a parabolic equation, without facing your teacher's fury?
This is me being extremely stupid, but a parabola with a vertex $(2,6)$ and $x$-intercepts of $(-4,0),(8,0)$ can be described with the equation $y=\frac{-(x+4)(x-8)}{6}$(my answer) which would also be the same as $\frac{-(x-2)^2}{6}+6$(the goddamn textbook's answer), yes? So if I were to write that on a test, let's say, would I be wrong since the textbook answers are the exact opposite of mine or are my answer(s) acceptable. I Am Confusion, even though this is extremely basic, any help welcome.

答:
Both ways you wrote are correct, and both are ways to present a parabola. Will your teacher mark off points? I don't know.

Mathematically, they are both correct. Personally, I would write it as

$$ y = -\frac{1}{6}x^2 + \frac{2}{3}x + \frac{16}{3} $$

but again, these are all correct!


问:
Prove that $\log X < X$ for all $X > 0$
I'm working through Data Structures and Algorithm Analysis in C++, 2nd Ed, and problem 1.7 asks us to prove that $\log X < X$ for all $X > 0$.

However, unless I'm missing something, this can't actually be proven. The spirit of the problem only holds true if you define several extra qualifiers, because it's relatively easy to provide counter examples.

First, it says that $\log_{a} X < X$ for all $X > 0$, in essence.

But if $a = -1$, then $(-1)^{2} = 1$. Therefore $\log_{-1} 1 = 2$. Thus, we must assume
$a$ is positive.

if $a$ is $< 1$, then $a^2 < 1$.  Therefore we must assume that $a \geq 1$.

Now, the book says that unless stated otherwise, it's generally speaking about base 2 for logarithms, which are vital in computer science.

However, even then - if $a$ is two and $X$ is $\frac{1}{16}$, then $\log_{a} X$ is $-4$. (Similarly for base 10, try taking the log of $\frac{1}{10}$ on your calculator: It's $-1$.) Thus we must assume that $X \geq 1$.

...Unless I'm horribly missing something here. The problem seems quite different if we have to prove it for $X \geq 1$.

But even then, I need some help solving the problem. I've tried manipulating the equation as many ways as I could think of but I'm not cracking it.

答:
One way to approach this question is to consider the minimum of $x - \log_a x$ on the interval $(0,\infty)$. For this we can compute the derivative, which is
$1 - 1/(\log_e a )\cdot x$.  Thus the derivative is zero at a single point, namely $x = 1/\log_e a,$ and is negative to the left of that point and positive to the right. Thus $x - \log_a x$ decreases as $x$ approaches $1/\log_e a$ from the left, and then increases as we move away from this point to the right.  Thus the minimum
value is achieved at $x = 1/\log_e a$.  (Here I'm assuming that $a > 1$, so that $\log_e a > 0$; the analysis of the problem is a little different if $a < 1$, since then for $x < a < 1$, we have $log_a x > 1 > x,$ and the statement is
not true.)

Now this value is equal to $1/\log_e a + (\log_e \log_e a)/\log_e a,$  and you want this to be $> 0 $.  This will be true provided $a > e^{1/e}$ (as noted in the comments).

Title: Prove that $\log X < X$ for all $X > 0$


问:
Sequences in $\mathbb{R}$ notation
Suppose $(x_n)$ is a sequence in $\mathbb{R}$. I have seen it written before that $(x_n)\subseteq \mathbb{R}$. Is this correct or is it abuse of notation since sequences aren’t simply just sets.

答:
Let $ (x_n)_{n\in\Bbb N} $ be a real sequence.

Here is what we are allowed to write :

$$(\forall n\in \Bbb N)\;\; x_n\in \Bbb R$$

$$(x_n)\in \Bbb R^{\Bbb N}$$

$$\{x_n : n\in \Bbb N \}\subset \Bbb R$$

Title: Sequences in $\mathbb{R}$ notation


答:
$(dX)_q$ is a linear map represented by a matrix having 3 rows and 2 columns. the first column is $\frac{\partial X}{\partial u}(q)$ and the second one $\frac{\partial X}{\partial v}(q)$.

Saying that $(dX)_q$ is injective is equivalent to say that its kernel is reduced to the zero vector which is equivalent to say that the two column vectors are linearly independent.

Title: why is $(dX)_{q}: \mathbb{R}^2 \rightarrow \mathbb{R}^3 $ injective equivalent to saying that $X_{u}(q) $ and $X_{v}(q) $ are linearly independent?

Body: I'm studying surfaces and in the definition I found a note that refers to an equivalence for one of the conditions in the definition, but I don't understand why those two things are equivalent. That is, why is $(dX)_{q}: \mathbb{R}^2 \rightarrow \mathbb{R}^3 $ injective equivalent to saying that $X_{u}(q) = \frac{\partial X}{\partial u}(q)$ and $X_{v}(q) = \frac{\partial X}{\partial v}(q)$ are linearly independent ? I'm referring to definition 2.2 and remark 2.5. I have added images:


问:
area of triangle in terms of squared edge lengths
It is well-known that the area of a triangle can be computed from the squared edge lengths $a^2$, $b^2$, $c^2$ as
$$
A
= \frac{1}{4}\sqrt{2a^2b^2 + 2b^2c^2 + 2c^2a^2 - a^4 - b^4 - c^4}\\
= \frac{1}{4}\sqrt{(a^2 + b^2 + c^2)^2 - 2 (a^4 + b^4 + c^4)}
$$
(See, e.g., here).

When computing the above expression numerically, the expression under the root can sometimes become negative (in the order of machine precision). One can of course ignore that and set those values to zero, but nevertheless the my question:

Is there an alternative expression for $A$ in terms of $a^2$, $b^2$, $c^2$ that is numerically positive, e.g., the sum of squares?

答:
As shown in the link you posted about Heron's formula, the expression that you have under the root comes from:
$(a+b+c)  (b+c-a)  (a+c-b)  (a+b-c)$

You can't have a triangle where one of the edges is bigger than the sum of the other two, so your expression is always positive if a, b and c are the edges of  a triangle.

You get negative values only when considering three segments that can't be put together to form  a triangle.


问:
Equation involving logarithms
So I've been stuck with this equation -  $\log_{\frac{x}{2}}(x^2) - 14\log_{16x}(x^3) + 40\log_{4x}(\sqrt{x}) = 0$. I was thinking of using identities such as $\log_{a}(x) = \frac{\log_{b}(x)}{\log_{b}(a)}$ but that didn't simplify much. Do you have an idea on how to proceed?

答:
$$\log_{\frac{x}{2}}(x^2) - 14\log_{16x}(x^3) + 40\log_{4x}(\sqrt{x}) = 0$$
$$\frac{\ln(x^2)}{\ln(x/2)}-14\frac{\ln(x^3)}{\ln(16x)}+40\frac{\ln \sqrt x}{\ln(4x)}=0$$
$$\frac{2\ln x}{\ln x-\ln 2}-14\frac{3\ln x}{\ln 16+\ln x}+40\frac{\frac{1}{2}\ln x}{\ln 4+\ln x}=0$$
Now set $\ln x=u$ and solve the algebraic equation.

Remember that $x>0;\;x\ne 2;\;x\ne \frac{1}{16};\;x\ne\frac{1}{4}$ otherwise the bases of the logarithm in the given equation become $1$ and this is not allowed

Title: Equation involving logarithms


问:
How to prove that this stochastic process converges in mean square as $t \to \infty$
I have that $\{ X_t ; t=0,1...\}$ is a martingale with finite second moments (meaning that $E(X_t^2) < \infty$). Assume that

$E(X_t^2) \to C \quad $ as $t \to \infty$

where $C < \infty$ is a constant. Prove that $\{ X_t;t=0,1,..\}$ converges in mean square as $t \to \infty$. I dont know where to start. Can somebody help me?

答:
Since $\mathbb{E}[X_t^2] \rightarrow C < \infty$, we have that $\sup_{t} \mathbb{E}[X_t^2] < \infty$ so $(X_t)$ is uniformly integrable (this is discrete time, but I'm not certain that this exact argument goes through in continuous time without some additional assumptions).  Doob's martingale convergence theorem then guarantees that there exists $X_\infty$ such that $(X_t) \rightarrow X_\infty$ almost surely, and since $(X_t)$ is uniformly integrable this implies $(X_t) \rightarrow X_\infty$ in $L^2$.  I'll also prove that for completeness: Fix $\varepsilon > 0$.  Then

\begin{align*}
\mathbb{E}[|X_t - X_\infty|^2] &= \mathbb{E}[|X_t - X_\infty|^21_{|X_t - X_\infty| > \varepsilon}] + \mathbb{E}[|X_t - X_\infty|^21_{|X_t - X_\infty| \le \varepsilon}] \\
&\le 2(\mathbb{E}[|X_t|^21_{|X_t - X_\infty| > \varepsilon}] + \mathbb{E}[ |X_\infty|^21_{|X_t - X_\infty| > \varepsilon}]) + \varepsilon^2.
\end{align*}

Since $(X_t) \rightarrow X_\infty$ almost surely, we can choose $T > 0$ to make $\sup_{t \ge T} \mathbb{P}(|X_t-X_\infty| > \varepsilon)$ arbitrarily small, so by uniform integrability we can choose $T> 0$ such that  $\mathbb{E}[|X_t|^21_{|X_t - X_\infty| > \varepsilon}] + \mathbb{E}[ |X_\infty|^21_{|X_t - X_\infty| > \varepsilon}] \le \frac{\varepsilon}{2}$ for all $t \ge T$.  Thus we conclude

\begin{align*}
\lim_{t \rightarrow \infty}\mathbb{E}[|X_t - X_\infty|^2] &\le \lim_{t \rightarrow \infty}2(\mathbb{E}[|X_t|^21_{|X_t - X_\infty| > \varepsilon}] + \mathbb{E}[ |X_\infty|^21_{|X_t - X_\infty| > \varepsilon}]) + \varepsilon^2 \\
&\le \varepsilon + \varepsilon^2
\end{align*}

and since $\varepsilon$ was arbitrary, we conclude $\lim_{t \rightarrow \infty}\mathbb{E}[|X_t - X_\infty|^2] = 0.$


问:
Diophantine equation in two variables involving floor function
The problem is about finding all $a,b\in\mathbb{N}$ such that $$\left \lfloor{\frac{a^2}{b}}\right \rfloor + \left \lfloor{\frac{b^2}{a}}\right \rfloor +1=\left \lfloor{\frac{a^2+b^2}{ab}}\right \rfloor + ab$$ On the book, there is a hint saying that, based on that we would have $\frac{a^2}{b}+\frac{b^2}{a}+1 > \frac{a^2+b^2}{ab}-1+ab$, and then, that $1 + (a + b − 1)(a^2 + b^2-ab)>a^2b^2$ and then, that $a\geq b\Rightarrow b=1$ but I don't know how to get to the first inequation.

答:
Note that the definition of the floor function gives that

$$x \ge \lfloor x \rfloor \gt x - 1 \tag{1}\label{eq1A}$$

Using the $x \ge \lfloor x \rfloor$ part with the left hand side of your equation, this gives

$$\frac{a^2}{b} + \frac{b^2}{a} + 1 \ge \left\lfloor{\frac{a^2}{b}}\right\rfloor + \left\lfloor{\frac{b^2}{a}}\right\rfloor + 1 \tag{2}\label{eq2A}$$

Using the $\lfloor x \rfloor \gt x - 1$ part with the right hand side of your equation, this gives

$$\left\lfloor{\frac{a^2 + b^2}{ab}}\right\rfloor + ab \gt \frac{a^2 + b^2}{ab} - 1 + ab \tag{3}\label{eq3A}$$

Putting \eqref{eq2A} and \eqref{eq3A} together, and using your equation, gives

$$\frac{a^2}{b} + \frac{b^2}{a} + 1 \gt \frac{a^2+b^2}{ab} - 1 + ab \tag{4}\label{eq4A}$$

Since you only ask about the first inequation, I assume you can deal with the rest of the given hint.


问:
Deriving the "Thermal wind" equation from the geostropic balance equation
The geostrophic balance equation is given by:

$$2 \mathbf{\Omega} \times \mathbf{v} = \mathbf{g}-\frac{1}{\rho}\nabla p$$

with $\mathbf{\Omega}=\Omega \mathbf{\hat{z}}$ and $\mathbf{g}=-g \mathbf{\hat{z}}$

I want to show that

$$\frac{\partial\mathbf{v}_h}{\partial z} \approx \frac{g}{2\Omega} \mathbf{\hat{z}} \times \nabla \ln(T)$$

using the approximation $\nabla p \approx \frac{\partial p}{\partial z}\mathbf{\hat{z}}$ and where $\mathbf{v}_h=(v_x,v_y,0)$

So far starting from the geostrophic balance equation and using the relation $p=R\rho T$ where $R$ is a constant and by taking the curl of the geostropic balance equation I get

$$2\mathbf{\Omega}(\nabla\cdot\mathbf{v})-2(\mathbf{\Omega}\cdot\nabla)\mathbf{v}=-R\nabla T \times \nabla \ln(p)$$

then I can plug in what I know I can arrive at

$$\frac{\partial\mathbf{v_h}}{\partial z}-(\nabla\cdot\mathbf{v}_h)\mathbf{\hat{z}}=-R\nabla T \times \nabla \ln(p)$$

I feel like this is very close but for the life of me I can't see how to get this in the right form any help is much appreciated.

答:
Taking the curl of the geostrophic balance, we get

$$2\tag{1}\nabla \times (\mathbf{\Omega}\times \mathbf{v}) = \nabla \times \mathbf{g} - \nabla \times \left(\frac{1}{\rho} \nabla p \right)$$

Since $\mathbf{g}$ is a constant vector, we have $\nabla \times \mathbf{g} = 0$. Applying product rules for curl  on both sides of (1), we get

$$\tag{2}-2\mathbf{\Omega}\cdot \nabla \mathbf{v} + 2 (\nabla \cdot \mathbf{v})\mathbf{\Omega} = - \nabla\left(\frac{1}{\rho} \right)\times \nabla  p - \frac{1}{\rho} \nabla \times \nabla p = - \nabla\left(\frac{1}{\rho} \right)\times \nabla  p,$$

where the last equality follows because the curl of a gradient ($\nabla \times\nabla p$)  is zero. Substituting with $\mathbf{\Omega} = \Omega \mathbf{\hat{z}}$ and $\mathbf{v} = \mathbf{v}_h$ into (2), we get

$$\tag{3} -2\Omega\,\left ( \frac{\partial \mathbf{v}_h}{\partial z}-  (\nabla \cdot \mathbf{v}_h)\mathbf{\hat{z}}\right) = - \nabla\left(\frac{1}{\rho} \right)\times \nabla  p$$

With the equation of state $p = R\rho T$, it follows that

$$-\nabla \left( \frac{1}{\rho}\right) = \frac{1}{\rho^2} \nabla \rho =  \frac{1}{\rho}\nabla \ln \rho = - \frac{1}{\rho}\nabla \ln T+ \frac{1}{\rho}\nabla \ln p$$

Hence, substituting into (3) along with the approximation $\nabla p = \rho g \mathbf{\hat{z}}, $ justified below, we get

$$\tag{4} \frac{\partial \mathbf{v}_h}{\partial z}  = -\frac{g}{2\Omega}\mathbf{\hat{z}} \times  \nabla \ln T + \frac{g}{2\Omega}\mathbf{\hat{z}} \times  \nabla \ln p +  (\nabla \cdot\mathbf{v}_h)\mathbf{\hat{z}}$$

Note that $\mathbf{\Omega}\times \mathbf{v_h} = (0,0,\Omega) \times (v_x,v_y,0) = (-\Omega v_y, \Omega v_x, 0)$ and the components of the geostrophic balance are

$$-2\Omega v_y = - \frac{1}{\rho}\frac{\partial p}{\partial x}, \quad 2\Omega v_x = - \frac{1}{\rho}\frac{\partial p}{\partial y} ,\quad 0 = g - \frac{1}{\rho}\frac{\partial p}{\partial z}$$

Thus,

$$\frac{\partial v_x}{\partial x} = -\frac{1}{2 \Omega}\frac{\partial^2 p}{\partial x \partial y} + \frac{1}{2\Omega \rho^2} \frac{\partial \rho}{\partial x}\frac{\partial p}{\partial y}, \\\frac{\partial v_y}{\partial y} = \frac{1}{2 \Omega}\frac{\partial^2 p}{\partial x \partial y} - \frac{1}{2\Omega \rho^2} \frac{\partial \rho}{\partial y}\frac{\partial p}{\partial x}$$

$$\tag{5}\nabla \cdot \mathbf{v}_h = \frac{\partial v_x}{\partial x} + \frac{\partial v_y}{\partial y} = \frac{1}{2\Omega \rho^2} \left(\frac{\partial \rho}{\partial x}\frac{\partial p}{\partial y}-\frac{\partial \rho}{\partial y}\frac{\partial p}{\partial x}\right)$$

Making the approximation $\nabla p \approx \frac{\partial p}{\partial z} \mathbf{\hat{z}}$, we have $\frac{\partial p}{\partial x}  = \frac{\partial p}{\partial y} \approx 0$ and  substituting into (5), we get $\nabla \cdot \mathbf{v}_h \approx 0$.

With the approximation we can rewrite (4) as

$$\tag{6} \frac{\partial \mathbf{v}_h}{\partial z}  \approx -\frac{g}{2\Omega}\mathbf{\hat{z}} \times  \nabla \ln T + \frac{g}{2\Omega}\mathbf{\hat{z}} \times  \nabla \ln p  \\= -\frac{g}{2\Omega}\mathbf{\hat{z}} \times  \nabla \ln T + \frac{g}{2\Omega p}\mathbf{\hat{z}} \times  \nabla  p  $$

Using  $\frac{\partial p}{\partial z} = \rho g$ from the third component equation along with  the approximation $\nabla p = \frac{\partial p}{\partial z} \mathbf{\hat{z}} $ , we get $\nabla p = \rho g \mathbf{\hat{z}}, $ and, hence,

$$\mathbf{\hat{z}} \times  \nabla  p = \mathbf{\hat{z}} \times (\rho g\mathbf{\hat{z}} ) = 0, $$

and (6) reduces to

$$ \frac{\partial \mathbf{v}_h}{\partial z}  \approx -\frac{g}{2\Omega}\mathbf{\hat{z}} \times  \nabla \ln T$$


问:
Existence of solution for a linear system mod 2
Let $A$ be a (skew-) symmetric matrix over $\mathbb{Z}/2$.
(In fact, I would take $A$ as the linking matrix of an oriented framed link in $S^3$ or the matrix representing the intersection form on a closed smooth 4-manifold. The following statement however does seem to hold in general.)
I am interested in the following linear system over $\mathbb{Z}/2$,
$$a_{i1}x_1+a_{i2}x_2\cdots+a_{in}x_n=a_{ii},\quad i=1,\cdots,n.$$

This system is known to always have a solution. (c.f. Saveliev's Lectures on the Topology of 3-Manifolds.)
But I cannot see why is this true unless $A$ is nonsingular over $\mathbb{Z}/2$.
Is there a general method to deal with these kinds of linear systems?

答:
This is true, but it is a bit tricky. The idea is simply to write the matrix in the form
$$
A=BB^T
$$
in such a way that the column space of $B$ is equal to that of $A$. All the columns of $A$ are linear combinations of columns of $B$, but it is not clear to me how to achieve the reverse inclusion (it is clearly not true for all choices of $B$).

So at this time I cannot write a completely self-contained proof, I need to refer to two articles:

A. Lempel, Matrix factorization over $GF(2)$ and trace-orthogonal bases of $GF(2^m)$, SIAM J. Comput., vol. 4, pp. 175-186, June 1975.

G. Seroussi, A. Lempel, Maximum Likelihood Decoding of Certain Reed-Muller Codes, IEEE Transactions on information theory, Vol. IT-29, NO. 3, May 1983.

IIRC only the first is needed. I include the latter, because I found the former by reading it.

The problem Lempel (of Lempel-Ziv fame) solves in the first article is the following. He wants to write a given symmetric $n\times n$ matrix $A$ over $\Bbb{Z}_2$ in the form $A=BB^T$
as efficiently as possible. That is, he wants to minimize the number of columns $m$ of $B$.
His answer is that

Normally $m$ is equal to the rank $r(A)$ of $A$. The exception comes when the diagonal of $A$ is all zeros, when $m=1+r(A)$ is the best we can do.

We can apply Lempel's result to settle this question as follows.

If the diagonal of $A$ is all zeros, the claim is trivial. We can use $x_i=0$ for all $i$.

When that is not the case, the number of columns of $B$ is equal to the rank of $A$.
As $A=BB^T$ the column space of $A$ is then equal to that of $B$.

So it suffices to show that the diagonal of $A$ is contained in the column space of $B$.

The equation $A=BB^T$ means that $a_{ii}$ is equal to the inner product $(B_i,B_i)$ of the $i$th row $B_i$ of $B$ with itself.

But $B_i$ is binary, so $(B_i,B_i)$ is simply the sum of the entries of that $i$th row as $x^2=x$ for all $x\in\Bbb{Z}_2$.

Therefore the diagonal of $A$ is the sum of the columns of $B$.

Therefore the diagonal of $A$ is also in the column space of $A$ and we are done.

This feels unnecessarily kludgy. The idea of using $A=BB^T$ came to me intuitively. I calculated several examples and noticed that the columns of $B$ sum up to the diagonal. Light bulb time!


问:
Biased outcomes from unbiased coin
I know that by combining heads and tails, we can generate unbiased outcomes using a biased coin. I am trying to understand if we can generate biased outcomes using an unbiased coin. My guess is that it has to do with binary representation of probability. Say the desired probability of winning is denoted by $p\in(0,1)$. We can write this in binary form
$$p=0.p_1p_2p_3.....p_n $$ where $p_i \in \{0,1 \}$. Let us toss the fair coin and take heads as $1$ and tails as $0$.

Is there any hope ahead through this approach?

Many thanks.

答:
Yes there is. Suppose you flip a coin randomly, generating a infinite sequence of heads and tails given by $f_1,f_2,...$, representing tails as $0$ and heads as $1$. Then if you use those flips for binary digits and let
$$
f=0.f_1f_2f_3...
$$
then there is exactly a probability $p$ that $f<p.$

In principle this could take infinitely many flips to determine, but if you were satisfied with finding an event of probability $p$ when $p=\frac{k}{2^n}$, then finding out whether $f<p$ would take at most $n$ flips.

Title: Biased outcomes from unbiased coin


问:
Differentiablity of a function $f(z)$ in the complex vs reals
I am quite confused about the differentiability of a function in the complex plane vs the real numbers.
Consider the function $f(z) = |z|^2$ for $z \in \mathbb{C}$. Using the definition of a derivative, and taking the limit as $\Delta z \rightarrow 0$, we can see that it is only differentiable at $z = 0$. Elsewhere, the limits are not unique, so do not exist. However, if we consider a similar function $F(x) = |x|^2$ for $x \in \mathbb{R}$, then $F'(x) = 2x$ and exists everywhere in $\mathbb{R}$.

We know that $\mathbb{C}$ is a field extension of $\mathbb{R}$, so $\mathbb{R} \subset \mathbb{C}$. The domain of $f(x)$ is then a subset of the domain of $f(z)$. Why isn't $f(z)$ instead differentiable for all $z = x + i0$?

Likewise for a similar function, $g(z) = \operatorname{Re}(z)$. $g(z)$ is nowhere differentiable in $\mathbb{C}$, yet $G(x) = x$ is differentiable everywhere in $\mathbb{R}$. Why is this the case? Is it due to the limit being used in the definition of a derivative? That the limit requires that $|f(z) - f(z_0)| < \epsilon$, which means that the limit of $f(z)$ must approach $f(z_0)$, as $z \rightarrow z_0$, independent of the direction for the limit to exist. If we were only working with the reals, the limit as $x \rightarrow x_0^+$ must equal that of $x \rightarrow x_0^-$ for the limit to exist, but in the complex plane, the limit must be equal as $z \rightarrow z_0$ from all sides of $z_0$, which is why the derivative for a function in $\mathbb{R}$ can exist, but not a similar one in $\mathbb{C}$?

Does this mean that given two similar functions $f(z)$ and $F(x)$ acting on both $\mathbb{C}$ and $\mathbb{R}$, respectively. Let $f(z=x+iy) = u(x,y) + iv(x,y)$,  $u(x,y) = F(x)$, and $z_0 = x_0 + iy_0$, then
$$f(z) \text{ differentiable at point } z_0 \in \mathbb{C} \text{ and } u'(z) \text{ exists at } z_0 \implies F(x) \text{ differentiable at }
x_0 \in \mathbb{R}?$$

答:
Why isn't $f(z)$ instead differentiable for all $z = x + i0$?

Because the existence of $\lim\limits_{z \to z_0}\frac{f(z)-f(z_0)}{z-z_0}$ is a much stronger requirement than the existence of $\lim\limits_{x \to x_0}\frac{f(x+i0)-f(x_0+i0)}{x-x_0}$. Intuitively, the first limit involves points in "two dimensional neighborhood" of $z_0$ while the second one only involves points lying on a line.

$$f(z) \text{ differentiable at point } z_0 \in \mathbb{C} \text{ and } u'(z) \text{ exists at } z_0 \implies F(x) \text{ differentiable at }
x_0 \in \mathbb{R}?$$

The answer is yes.


问:
A problem in calculus(an inequality regarding limits) to show f(x) < (1 +1/e).
f : R → R is a function such that f(1) = 1 and f'(x) = $e^{−xf(x)}$ holds for all x∈R.
P.T. $lim_{x→∞}$ f(x) < (1 +1/e).

I encountered this problem in a textbook of calculus. Now what I immediately thought of having such a function which satisfies the condition f'(x)=$e^{−xf(x)}$ .I need to identify the function and then prove the inequality of the limit.
Kindly shower some help....thank you.

And yes... This is not a homework problem and I have mentioned my thought process.

答:
Notice that $f$ is strictly increasing over $\mathbb{R}$ because the inequality

$$f'(x)=e^{-xf(x)}>0$$

holds for all $x\in\mathbb{R}$.

It follows that for $x\geq 1$,

\begin{align*}
f(x) &\geq f(1)=1\\
xf(x) &\geq x\\
-xf(x) &\leq-x\\
e^{-xf(x)}&\leq e^{-x}
\end{align*}

so $f'(x)\leq e^{-x}$ for $x\geq 1$.

Basic integral inequalities and two applications of $\text{FTC}2$ then establish that

\begin{align*}
f(x)-1 &= f(x)-f(1)\\
&= \int_{1}^{x}f'(t)\text{ }dt\\
&\leq \int_{1}^{x}e^{-t}\text{ }dt\\
&=e^{-1}-e^{-x}
\end{align*}

so $f(x)\leq 1+\frac{1}{e}-e^{-x}\leq 1+\frac{1}{e}$. This means that $f$ is eventually monotonically increasing and is bounded above by $1+\frac{1}{e}$, so it must have a limit. Denoting the limit by $L$, we deduce that

$$L=\lim_{x\to\infty}f(x)\leq \lim_{x\to\infty}\left(1+\frac{1}{e}-e^{-x}\right)=1+\frac{1}{e}$$

I'm not sure how to prove that $L\neq 1+\frac{1}{e}$. Hope this still helps!


问:
Foruier transform of spherical measure is radial
We define the surface measure of the sphere $\sigma$ as the unique measure on $\mathbb{S}^2$ satisfying
$$\int_\mathbb{S^2} f \operatorname{d}\hspace{-0.5mm}\sigma:=\int_0^\pi\int_0^{2\pi} f(\sin\phi \cos\theta,\sin\phi \sin\theta,\cos\phi) \sin\phi \operatorname{d}\hspace{-0.5mm}\theta \operatorname{d}\hspace{-0.5mm}\phi$$
for every $f \in L^\infty(\mathbb{S^2})$

Then, we define its Fourier transform as
$$\hat\sigma(\xi)=\int_\mathbb{S^2} e^{-2\pi i \langle x, \xi\rangle} \operatorname{d}\hspace{-0.5mm}\sigma(x)$$

So, I'm trying to see that $\hat\sigma$ is a radial function, and I know it must be by a change of variable with a rotation (similar to what is done when you have a radial function and you show its Fourier transform is radial) but I'm not able to do it rigurously since in the definition we go from working on $\mathbb{R}^3$ to $\mathbb{R}^2$.

Could someone prove this fact in detail?

答:
It's easier to prove this result using a different definition of $\sigma$, so let's first prove the two definitions are equivalent.  I claim that the definition of $\sigma$ given above satisfies
$$
\sigma(E) = 3 m(\overline{E})
$$
for all measurable $E \subseteq S^2$, where $m$ is $3$-dimensional Lebesgue measure and $\overline{E} := \{ tx : 0 \leq t \leq 1, x \in E \}$.  To prove this, evaluate the right hand side using spherical coordinates:
$$
m(\overline{E}) = \int_{\mathbb{R}^3} 1_E = \int_{0}^{\infty} \int_0^\pi \int_0^{2\pi} 1_\overline{E}(r\sin\phi\cos\theta, r\sin\phi\sin\theta, r\cos\phi) r^2 \sin\phi \,d\theta \,d\phi \,dr.
$$
By definition, $(r\sin\phi\cos\theta, r\sin\phi\sin\theta, r\cos\phi) \in \overline{E}$ if and only if $(\sin\phi\cos\theta, \sin\phi\sin\theta, \cos\phi) \in E$ and $0 \leq r \leq 1$.  Therefore the above is equal to
$$
\int_0^1 r^2 \int_0^\pi \int_0^{2\pi} 1_E(\sin\phi\cos\theta, \sin\phi\sin\theta, \cos\phi) \sin\phi \,d\theta \,d\phi \,dr = \sigma(E) \cdot \int_0^1 r^2\,dr = \frac13 \sigma(E)
$$
as claimed.

Now, in order to prove that $\widehat{\sigma}$ is radial, we need to prove that $\widehat{\sigma}(A\xi) = \widehat{\sigma}(\xi)$ for any $3\times3$ orthogonal matrix $A$.  First we have
$$
\widehat{\sigma}(A\xi) = \int_{S^2} e^{-2\pi i \langle x, A\xi \rangle} \,d\sigma(x) = \int_{S^2} e^{-2\pi i \langle Ax, \xi \rangle} \,d\sigma(x)
$$
because orthogonal matrices move across inner products.  Another property of orthogonal matrices is that they preserve distances between points, and therefore they also preserve the $3$-dimensional Lebesgue measure (in the sense that $m(A^{-1} E) = m(E)$ for any measurable $E$).  Therefore since we've shown that $\sigma(E)$ is just a constant multiple of $m(\overline{E})$, $\sigma$ must also be preserved by the action of orthogonal matrices, which allows us to conclude
$$
\int_{S^2} e^{-2\pi i \langle Ax, \xi \rangle} \,d\sigma(x) = \int_{S^2} e^{-2\pi i \langle x, \xi \rangle} \,d\sigma(x) = \widehat{\sigma}(\xi)
$$
as desired.


问:
Power series, interval of convergence
Let a power series be $$\sum_{n=0}^{\infty}a_nx^n$$ and if $$\lim_{n \to \infty}\frac{a_{n+1}}{a_n}=0$$, then is it true that the power series converges for all $x \in \mathbb{R}$?

If that limit has the absolute value, then using the Ratio Test, this is indeed true, but does it work without the absolute value?

答:
If the limit is zero without absolute value, then it is also zero with absolute value.... as the absolute value map is continuous at zero.

So you get that the radius of convergence is infinite and that your power series converges everywhere.

Title: Power series, interval of convergence


问:
Find the value of $\int_0^{\infty}\frac{x^3}{(x^4+1)(e^x-1)}\mathrm dx$
I need to find a closed-form for the following integral. Please give me some ideas how to approach it:
$$\int_0^{\infty}\frac{x^3}{(x^4+1)(e^x-1)}\mathrm dx$$</p>

答:
$$\int_0^{\infty}\frac{x^3}{(x^4+1)(e^x-1)}\mathrm dx=-\frac{\pi}{\sqrt8}-\log\sqrt{2\pi}-\frac{1}{2}\Re\ \psi\left(\frac{\sqrt[4]{-1}}{2\pi}\right),$$
where $\Re\ \psi(z)$ denotes the real part of the digamma function, $\psi(z)=\frac{\Gamma'(z)}{\Gamma(z)}$.

Solution: Use the approach from sos440's answer.
To calculate the infinite sum and simplify the result we need the following:
$$\frac{1}{n\left((2\,\pi\,n)^4+1\right)}=\frac{1}{n}-\frac{1}{4}\left(\frac{1}{n+\frac{(-1)^{1/4}}{2 \pi }}+\frac{1}{n+\frac{(-1)^{-1/4}}{2\pi}}+\frac{1}{n+\frac{(-1)^{3/4}}{2 \pi }}+\frac{1}{n+\frac{(-1)^{-3/4}}{2\pi}}\right),$$
$$\sum_{n=1}^\infty\left(\frac{1}{n}-\frac{1}{n+x}\right)=\gamma+\psi(1+x),$$
and the formulas (8), (9) from here.


答: In general absolute convergence is stronger than convergence. So knowing that $\sum_{n=0}^\infty |a_n|x^n$ converges for all n, then this tells us that $\sum_{n=0}^\infty a_nx^n$ must also converge. for example by the squeeze theorem.

Title: Power series, interval of convergence

Body:
Let a power series be $$\sum_{n=0}^{\infty}a_nx^n$$ and if $$\lim_{n \to \infty}\frac{a_{n+1}}{a_n}=0$$, then is it true that the power series converges for all $x \in \mathbb{R}$?

If that limit has the absolute value, then using the Ratio Test, this is indeed true, but does it work without the absolute value?


问:
Perfect sets are uncountable. Proof via $G_\delta$-sets
To prove the statement, we need the two theorems:

T1. Let $X$ be a complete metric space and $A\subset X$. Then $A$, with the metric of $X$, is a complete metric space if and only if $A$ is closed in $X$.

T2. If $X$ is a (nonvoid) complete metric space having no isolated points and $E$ is a countable dense subset of $X$, then $E$ is not a $G_\delta$ set.

The statement:

If A is a perfect nonvoid subset of a complete metric space, then $A$ is uncountable. [Hint: Use T1 and T2].

I think I'm stuck because I cannot see the usefulness of $G_\delta$ sets.

If $A$ is perfect, then it is equal to its limit points (has no isolated points) and thus $A$ is closed. By T1, $A$ is complete. In particular, if $A$ is a (dense) countable subset, then by T2 $A$ is not a $G_\delta$ set.

This means that $A$ cannot be expressed as the countable intersection of open subsets, fine, but this should rise a contradiction that I cannot see.

答: $A$ is closed and therefore complete by the first theorem. Now work in the space $A$. If $A$ were countable, it would be a countable, dense, open subset of itself.


问:
Discount Problem
A shopkeeper allows a discount of 10% on the marked price. How much above the cost price must he mark his goods to gain 8%?

I assume cost price to be P. So the shopkeeper must get 1.08P to make a profit of 8%. He allows a discount of 10%. I am told that 90% of marked price is 1.08P because the shopkeeper offers a discount of 10%. Can somebody please explain why 90% of marked price is 1.08P?

答:
To flesh out Math Lover's brief description:

Let $M$ be the marked price of the prpoduct and $P$ be the cost price.
As you've noted, the shopkeeper wants to get at least $1.08\cdot P$ from the sale.

Because of the $10\%$ discount, the amount the shopkeeper will get from the sale is $0.9\cdot M$.

So for the shopkeeper to get the amount they want, these two need to be equal:
$$0.9\cdot M = 1.08 \cdot P$$
Solving for $M$,
$$M = \dfrac{1.08}{0.9}P$$

Title: Discount Problem


问:
Probability Q on rolling two 8-sided dice
Let's say I roll two 8-sided dice. I win if the sums '7' and '11' show up before we see the sum '9' TWICE. What is my probability of winning?

So this is my answer and please correct me if I am wrong:

Prob of getting either 7 or 9 —> 6/64 + 6/64 = 12/64 —> 18.75%

Prob of getting 9 twice —> 8/64 x 8/64 = 1/64 —> 1.6%

But now to find the probability of winning, would it be 18.75 - 1.6 = 17.15% ? Am I right or wrong?

Thanks!

EDIT

From the answers provided below, we have three different results:

84%
49%
60%

Which one would be the correct answer?

答:
We can ignore all the other possible rolls and consider that we are given that we roll a $7$, with a $\frac6{64}$ chance, an $11$, also with a $\frac6{64}$ chance, or a $9$, with an $\frac8{64}$ chance.

Once we've rolled a $7$ or an $11$, we take it out of the denominator because it is now a race between the other and $9$.

Below, we compute the probability of the first of two equal outcomes, then double it.

$\overbrace{\ \ 7{\to}11\ \ \vphantom{\frac{\frac11}{\frac11}}}^{\rightarrow}\text{or }11{\to}7:2\cdot\overbrace{\frac{\frac6{64}}{\frac6{64}+\frac6{64}+\frac8{64}}}^{7\text{ before }9,11}\cdot\overbrace{\ \ \frac{\frac6{64}}{\frac6{64}+\frac8{64}}\ \ }^{11\text{ before }9}=2\cdot\frac3{10}\cdot\frac37=\frac9{35}$

$\overbrace{7{\to}9{\to}11\vphantom{\frac{\frac11}{\frac11}}}^{\rightarrow}\text{ or }11{\to}9{\to}7:2\cdot\overbrace{\frac{\frac6{64}}{\frac6{64}+\frac6{64}+\frac8{64}}}^{7\text{ before }9,11}\cdot\overbrace{\ \ \frac{\frac8{64}}{\frac6{64}+\frac8{64}}\ \ }^{9\text{ before }11}\cdot\overbrace{\ \ \frac{\frac6{64}}{\frac6{64}+\frac8{64}}\ \ }^{11\text{ before }9}=2\cdot\frac3{10}\cdot\frac47\cdot\frac37=\frac{36}{245}$

$\overbrace{9{\to}7{\to}11\vphantom{\frac{\frac11}{\frac11}}}^{\rightarrow}\text{ or }9{\to}11{\to}7:2\cdot\overbrace{\frac{\frac8{64}}{\frac6{64}+\frac6{64}+\frac8{64}}}^{9\text{ before }7,11}\cdot\overbrace{\frac{\frac6{64}}{\frac6{64}+\frac6{64}+\frac8{64}}}^{7\text{ before }9,11}\cdot\overbrace{\ \ \frac{\frac6{64}}{\frac6{64}+\frac8{64}}\ \ }^{11\text{ before }9}=2\cdot\frac25\cdot\frac3{10}\cdot\frac37=\frac{18}{175}$

Thus, the probability of winning is $\frac9{35}+\frac{36}{245}+\frac{18}{175}=\frac{621}{1225}$


问:
Simplify $\Gamma(n + x)/\Gamma(x)$
Is there a way to simplify $\Gamma(n + x)/\Gamma(x)$ for $n \in \mathbb{N}$ ? It's easy for the first few values of $n$ but I was wondering whether there is a known general formula (expressed as a polynomial of $n$ and $x$?).

答:
Hint: start with $\,n=1\,$ and use this property.

Observe further that $\displaystyle\dfrac{\Gamma(2 + x)}{\Gamma(x)}=\dfrac{\Gamma(1+(1 + x))}{\Gamma(1+x)}\dfrac{\Gamma(1 + x)}{\Gamma(x)}$ and so on...

Title: Simplify $\Gamma(n + x)/\Gamma(x)$


问:
Suppose that $\{f_n\}$ is a sequence of real valued functions on $\mathbb{R}$.
Suppose that $\{f_n\}$ is a sequence of real valued functions on $\mathbb{R}$. Suppose it converges to a continuous function $f$ uniformly on each closed and bounded subset of $\mathbb{R}$. Which of the following statements are true?

The sequence $\{f_n\}$ converges to $f$ uniformly on $\mathbb{R}$.

The sequence $\{f_n\}$ converges to $f$ pointwise on $\mathbb{R}$.

For all sufficiently large $n$, the function $f_n$ is bounded.

For all sufficiently large $n$, the function $f_n$ is continuous.

I have discarded options 1, 4 by choosing $f_n(x)=0$ for all $|x|\leq n$ and 1 otherwise. Further option 2 is true, by the given condition. please give a hint for  option number 3.

答: You may want to try $f_n(x)=f(x)=x$.

Body: You may want to try $f_n(x)=f(x)=x$.


答: What about $f_n(x) = 0$ for $\vert x \vert \le n$ and $f_n(x) = n$ otherwise?

Body:
Suppose that $\{f_n\}$ is a sequence of real valued functions on $\mathbb{R}$. Suppose it converges to a continuous function $f$ uniformly on each closed and bounded subset of $\mathbb{R}$. Which of the following statements are true?

The sequence $\{f_n\}$ converges to $f$ uniformly on $\mathbb{R}$.

The sequence $\{f_n\}$ converges to $f$ pointwise on $\mathbb{R}$.

For all sufficiently large $n$, the function $f_n$ is bounded.

For all sufficiently large $n$, the function $f_n$ is continuous.

I have discarded options 1, 4 by choosing $f_n(x)=0$ for all $|x|\leq n$ and 1 otherwise. Further option 2 is true, by the given condition. please give a hint for  option number 3.


答:
Well, if by "simplifying" you mean get rid of the Gamma function, then, yes, you can simply start by using your definition of Gamma.

If $ n $ is an arbitrary integer, we have for any $ k\in\mathbb{N} $, $ x>0 $ : \begin{aligned} \Gamma\left(x+k+1\right)&=\left(x+k\right)\Gamma\left(x+k\right)\\ \Longrightarrow \prod_{k=0}^{n-1}{\frac{\Gamma\left(x+k+1\right)}{\Gamma\left(x+k\right)}}&=\prod_{k=0}^{n-1}{\left(x+k\right)}\\ \iff \ \ \ \ \ \ \ \ \ \ \ \frac{\Gamma\left(x+n\right)}{\Gamma\left(x\right)}&=\prod_{k=0}^{n-1}{\left(x+k\right)} \end{aligned}

Title: Simplify $\Gamma(n + x)/\Gamma(x)$

Body: Is there a way to simplify $\Gamma(n + x)/\Gamma(x)$ for $n \in \mathbb{N}$ ? It's easy for the first few values of $n$ but I was wondering whether there is a known general formula (expressed as a polynomial of $n$ and $x$?).


问:
Prove that each prime divisor of $(2^p)-1$, where $p$ is a prime, is greater than $p$.
How would I prove that if $(2^p)-1$ is not prime, there is not a prime number less than p that divides it? (any hints for a proof by contradiction?) I have assumed that there is a prime divisor less than p, let's say q. But I am unsure of where to go from there.

答: By Fermat, we know $q\mid 2^{q-1}-1$. Than note that $q\mid 2^a-1$ and $q\mid 2^b-1$ implies $q\mid 2^{\gcd(a,b)}-1$.


问:
Can you compute the volume of a 4-Ball by using solids of revolution on a 3-Ball?
So in first year calculus most people at some point will be asked to compute the volume of the unit sphere in 3 dimensions, by rotating $y=\sqrt{1-x^2}$ about the x axis in $\mathbb{R}^3$. and integrating over it.

Im wondering if something similar can be used to compute the volume of the 4-ball in $\mathbb{R}^4$ by taking $z=\sqrt{1-x^2-y^2}$ and rotating it about the x,y plane in $\mathbb{R}^4$. Its been a while since ive done this kind of calculus so im very rusty, but it seems like it should be the same thing were you just multiply the function by $2\pi$ and do the double integral normally to get your answer, however I feel like there must be a mistake here. Also is there a way to get the 4 dimensional case by doing a revolution on the 2 dimensional case twice, so that you can solve it while doing a single variable integral?

答:
Well I feel a bit silly for asking this now, but I figured it out, so I'll answer my own question incase someone else has this question in the future.

The answer is yes, all you need to do is remember exactly how we found the volume of the sphere in the 3 dimensional case. What we did was found a formula for the area of a disk of a fixed radius and integrated over that function. In particular, if out sphere is $x^2+y^2+z^2\leq r^2$ then for each value of $x$ we consider the set of all $(x,y,z)$ in the circle with that particular x value. that set will be a disk of radius $\sqrt{r^2-x^2}$ and then we integrate over all such disks. the area of a disk of radius $r$ is $\pi r^2$ so we get the volume by solving $\int_{-r}^r\pi(\sqrt{r^2-x^2})^2dx$ which indeed gives us $\frac{4}{3}\pi r^3$

In the 4 dimensional case, we do the same thing, except instead of having an area of a disk at each x value, we have a volume of a sphere at each x value. so we just substitute the area formula for the volume formula and find get the integral $\int_{-r}^r\frac{4}{3}\pi(\sqrt{r^2-x^2})^3dx$ which is a harder integral to solve, but it does give us the right answer.


问:
Evaluate the Sum $\sum_{i=0}^\infty \frac {i^N} {4^i}$

$$\displaystyle\sum_{i=0}^\infty \frac {i^N} {4^i}$$

I'm supposed to evaluate this as I'm working through Data Structures and Algorithm Analysis in C++.  I've solved similar problems, and after trying all of the algebra-fu manipulations I could think of, I'm stumped.

My first efforts were, similar to how I solved the three previous problems in the series, to try doing things like setting it equal to S, dividing both sides by four, and subtracting the equations, in various ways in an attempt to cancel terms.  However, I'm missing some step to make it happen.

答:
Related techniques:(I), (II), (III). Recalling the identity for the operator $(xD)^n$, where $D=\frac{d}{dx}$,

$$ (xD)^n = \sum_{k=0}^{n}\begin{Bmatrix} n\\k \end{Bmatrix}x^k D^k , $$

where $\begin{Bmatrix} n\\k \end{Bmatrix}$ are the Stirling numbers of the second kind. One can get a closed form for the sum in terms of a finite sum. Applying this operator to the function $\frac{1}{1-x}$

$$ \sum_{k=0}^{\infty}{k^n}{x^k}= (xD)^n\frac{1}{1-x} = \sum_{k=0}^{n}\begin{Bmatrix} n\\k \end{Bmatrix}x^k D^k \frac{1}{1-x} $$

$$\implies  \sum_{k=0}^{\infty}{k^n}{x^k}= \sum_{k=0}^{n}\begin{Bmatrix} n\\k \end{Bmatrix} k!\frac{x^k}{(1-x)^{k+1}}. $$

Subs $x=\frac{1}{4}$ in the above identity gives

$$ \sum_{k=0}^{\infty}\frac{k^n}{4^k} = 4\sum_{k=0}^{n}\begin{Bmatrix} n\\k \end{Bmatrix} \frac{k!}{3^{k+1}}. $$

Here is some values of the last formula for $1\leq n \leq 7$

$$ \left\{ \frac{4}{9},{\frac {20}{27}},{\frac {44}{27}},{\frac {380}{81}},{
\frac {4108}{243}},{\frac {17780}{243}},{\frac {269348}{729}}
 \right\}
. $$


问:
Concerns formalizing a "Minimal Element" proof by contradiction
I came across an exercise where my proposed solution strikes me as a little informal/hand-wavy at the end. I was hoping that someone could lend a hand!

Consider the set $A$ and two subsets of $A$,  $a\downarrow$ and $b\downarrow$, defined as follows:

$a\downarrow := \{x \in A | xRa\}$

$b\downarrow := \{x \in A | x R b\}$

Of importance, $R$ is well-ordered on $A$ (and any subset of $A$). Specifically, that means trichotomy, transitivity, irreflexivity, and well-foundedness are properties of $R$.

The aim of this proof is to demonstrate that if $a \neq b$ then no isomorphism $k$ exists such that $(b \downarrow;R) \cong (a \downarrow; R)$.

I proceed by contradiction and assume that such an isomorphism exists: call it $k: b \downarrow \to a \downarrow$.

Without Loss of Generality, let $a \downarrow \subset b \downarrow$. This is a consequence of trichotomy: either $aRb, bRa,$ or $a=b$. Choose $aRb$...thus $a \in b\downarrow$ but $a \notin a\downarrow$.

To demonstrate that $a\downarrow$ is a proper subset, pick an element $x \in a\downarrow$. By definition, $xRa$. But we know that $aRb$. Therefore, by transitivity of $R$, $xRb \implies x \in b \downarrow$.

Additionally, the definition of these sets prohibits $b \downarrow$ from containing elements that are "beneath" the minimal element of $a \downarrow$ (quick proof by contradiction).

Because $k$ is an isomorphism, it is surjective. Therefore, looking at the minimal element in $a \downarrow$, call it $m$, there are two possible categories of mappings:

Case 1) An element $x \in b\downarrow$ such that $mRx$, and $x$ gets mapped to $m \in a \downarrow$ through $k$.

$$k(x)=m$$

Case 2) $m \in b \downarrow$ gets mapped to $m \in a \downarrow$ through $k$.

$$k(m)=m$$

Case 1 results in a contradiction: order will not be preserved

If $k(x)=m$, where $mRx$, then there is a $y \in b \downarrow$ such that $yRx$. This $y$ must be mapped to some element in $a \downarrow$, call it $s$. So we have:

$$k(x)=m \land k(y)=s$$

$yRx$ but $[k(x)=m] \ R \  [s=k(y)]$. The order was flipped! Contradiction.

$\color{red}{\text{Case 2 is where I feel like my argument is hand-wavy}}$

We have $k(m)=m$. That is to say, the $m \in b\downarrow$ is being mapped to the $m \in a \downarrow$. Now consider the following algorithm:

Choose the element $n$ in $a \downarrow$ such that the following is true: $\forall z( z\in a\downarrow \setminus \{m\} \rightarrow nRz \lor n=z)$. You can view this $n$ as the "the next minimal element".

Notice that there are two subcases of possible mappings from elements of $b \downarrow$ to $n \in a\downarrow$ :

Subcase 1) An element $x \in b\downarrow$ such that $nRx$ gets mapped to $n \in a \downarrow$ through $k$.

$$k(x)=n$$

Subcase 2) $n \in b \downarrow$ gets mapped to $n \in a \downarrow$ through $k$.

$$k(n)=n$$

But this is just a "slightly different" version of the argument we encountered earlier with $m$.

This continues to repeat until you've run out of elements in $a \downarrow$ but are still left with $a \in b \downarrow$, which will have no where to map to without violating injectivity and ordering. Therefore, no such isomorphism $k$ exists.

That final piece of the argument, where I say," This continues to repeat...," strikes me as unsatisfactorily informal. Any suggestions would be greatly appreciated! Cheers~

答:
Some comments:

What you call a "well order" is actually a strict well order (in my experience, most people take the non-strict version of (partial) orders as the primitive notion, hence the one without qualifier).

Your argument that $a\downarrow$ is a proper subset of $b\downarrow$ is not well written. The first part, that $a\downarrow\neq b\downarrow$ is not motivated and just comes out of nowhere. It should be after the statement that $a\downarrow$ is a proper subset.

There's a typo in your first enunciation of case 1: $mRx$ is not mapped anywhere, as it is a statement of relation (that is, that $(m,x)\in R$). You mean $x$ is such that $mRx$ (which under your definition implies $m\neq x$), and that $x$ is mapped to $m$ via $k$.

Your contradiction argument for case 1... why not use $m$ directly? You already know $mRx$. So you would get $k(m)R m$, which is impossible since $m$ is the least element of $a\downarrow$.

There is a way of doing your algorithm without doing your algorithm by instead describing the sets that are results of your algorithm. This is the same idea that George Bergman uses in his handout with a proof of Zorn's Lemma (the link is to a PostScript file). Though Brian Scott has given a way to get around it, I thought it might be instructive to show this for your argument.

You could describe the sets you get "along the way" as follows:

Let $\mathscr{C}$ be the collection of all subsets $T$ of $A$ with the following properties:

$m\in T$ (that is, $T\neq\varnothing$).

$T$ is an initial segment of $A$ (that is, if $y\in T$ and $xRy$, then $x\in T$).

$T\subseteq a\downarrow$ and $T\subseteq b\downarrow$.

The restriction of $k$ to $T$ is the identity map.

Note that $\mathscr{C}$ is nonempty, since $\{m\}$ itself is an element of $\mathscr{C}$, as we are in case 2.

These are, essentially, the sets you get at the end of each "step" of your algorithm.

Now, if $T_1$ and $T_2$ are two elements of $\mathscr{C}$, then either $T_1\subseteq T_2$ or $T_2\subseteq T_1$. Indeed, if $T_1$ is not contained in $T_2$, let $x\in T_1$, $x\notin T_2$. If $y\in T_2$, then we must have $yRx$ (if $xRy$, then $x\in T_2$ as $T_2$ is an initial segment). But then $y\in T_1$, as $T_1$ is an initial segment; hence $T_2\subseteq T_1$.

Let $U=\cup\mathscr{C}$. Then $m\in U$; $U$ is an initial segment of $A$; $U\subseteq a\downarrow$ and $U\subseteq b\downarrow$; and $k$ restricted to $U$ is the identity map. (I'll leave you to check the details).

That is, $U\in\mathscr{C}$. I claim that $U=a\downarrow$. Indeed, if $U$ is a proper initial segment of $a\downarrow$, then let $y$ be the least element of $(a\downarrow)\setminus U$; this is also the least element of $(b\downarrow) \setminus U$, and so using the argument like you did, we must have $k(y)=y$. Then $U\cup\{y\}\in\mathscr{C}$, but $y\notin U$, a contradiction. Thus, $U=a\downarrow$.

That means that $k$, restricted to the proper subset $a\downarrow$ of $b\downarrow$, is the identity map; hence it is already surjective. But then $k(a)=k(x)$ for some $x\in a\downarrow$, which contradicts the assumption that $k$ is one-to-one.


问:
Convergence of $\big ( 1 -\frac{1}{\sqrt{n}} \big )^n$.
I am trying to investigate the convergence of the series:

\begin{equation}
\sum_{n=1}^{\infty} \bigg ( 1 -\frac{1}{\sqrt{n}} \bigg )^n.
\end{equation}

According to the Mathematica, the series converges, but I am unable to come up with a series with which I can compare the series above etc. Any leads would be appreciated.

答:
Since $ \left(1-\frac{1}{\sqrt{n}}\right)^{\sqrt{n}}\underset{n\to +\infty}{\longrightarrow}\mathrm{e}^{-1} $, we have that $ \left(1-\frac{1}{\sqrt{n}}\right)^{n}\underset{n\to +\infty}{\sim}\mathrm{e}^{-\sqrt{n}} $

And since $ \mathrm{e}^{-\sqrt{n}}=\underset{\overset{n\to +\infty}{}}{o}\left(\frac{1}{n^{2}}\right) $, we get that $ \sum\limits_{n\geq 1}{\mathrm{e}^{-\sqrt{n}}} $ converges, which means $ \sum\limits_{n\geq 1}{\left(1-\frac{1}{\sqrt{n}}\right)^{n}} $ also converges.


问:
Self Adjoint Operators
I was wondering if someone could give me a few examples of some self adjoint operators, I understand this topic in terms of matrices with inner product spaces. However I have seen integrals that are inner products and I don't know any examples for the inner product space on an integral?

Could someone give me an example of a self adjoint operator and non self adjoint operator on the integral inner product space please! Thank you :)

答: Sure. Let $K\colon [0,1]\times [0,1] \to \Bbb C$ be a continuous function and define $T_K\colon \mathcal{C}([0,1], \Bbb C) \to \mathcal{C}([0,1], \Bbb C)$ by $$T_K(f)(s) = \int_0^1 K(t,s) f(t)\,{\rm d}t.$$Assume that the inner product in $\mathcal{C}([0,1], \Bbb C)$ is given by $$\langle f,g\rangle = \int_0^1 f(s) \overline{g(s)}\,{\rm d}s.$$If $K$ satisfies the Hermitian symmetry $K(s,t) = \overline{K(t,s)}$, then $T_K$ is self-adjoint. Because $$\langle T_K(f),g\rangle = \int_0^1 T_K(f)(s)\overline{g(s)}\,{\rm d}s = \int_0^1\int_0^1 K(t,s)f(t)\overline{g(s)}\,{\rm d}t\,{\rm d}s,$$as well as $$\langle f, T_K(g)\rangle = \int_0^1 f(s) \overline{T_K(g)(s)}\,{\rm d}s = \int_0^1\int_0^1 f(s) \overline{K(t,s) g(t)}\,{\rm d}t\,{\rm d}s. $$Using the Hermitian symmetry of $K$ and renaming $t \leftrightarrow s$ if needed, we have $\langle T_K(f),g\rangle = \langle f,T_K(g)\rangle$. For a non self-adjoint operator, take $T_K$ where $K$ does not satisfy that Hermitian symmetry.

Title: Self Adjoint Operators


问:
Prove that the integers modulo 8 is not an internal direct product of two proper subgroups.
Prove that $\mathbb{Z}_8$ is not an internal direct product of two proper subgroups.

I'm really struggling to get the wheels turning on this one. I know what it means for group to be an internal product of two subgroups, I'm just not sure how to get started in showing that $\mathbb{Z}_8$ is not.

答: The orders of the factors would have to be $2$ and $4$. Then the product has no elements of order $>4$, but $\Bbb Z_8$ has.


答: I am a bit rusty with group theory, so there may be another way of doing this, but $\mathbb{Z}_4$ only has so many subgroups, in fact it only has 2 proper subgroups so there are only 4 possibilities for you to check. I dont remember if there is an easy way to find the size of a product, but if there is then try to show that non of those things could have size 8. alternatively you can show none of them will have an element of order 8, without needing to compute the entire group in any of the 4 cases.

Title: Prove that the integers modulo 8 is not an internal direct product of two proper subgroups.

Body:
Prove that $\mathbb{Z}_8$ is not an internal direct product of two proper subgroups.

I'm really struggling to get the wheels turning on this one. I know what it means for group to be an internal product of two subgroups, I'm just not sure how to get started in showing that $\mathbb{Z}_8$ is not.